On the classification of group actions on C*-algebras up to equivariant KK-equivalence

TL;DR

This paper investigates the classification of group actions on C*-algebras using equivariant KK-theory, establishing key equivalences and classifications for actions of various groups, including cyclic groups of prime order.

Contribution

It demonstrates that any group action is equivariantly KK-equivalent to an action on a simple, purely infinite C*-algebra and relates a conjecture of Izumi to cocycle conjugacy and KK-equivalence.

Findings

Any group action is KK-equivalent to an action on a simple, purely infinite C*-algebra.

Classifies actions of cyclic groups of prime order on stabilized Cuntz algebras.

Establishes the equivalence of Izumi's conjecture with cocycle conjugacy and KK-equivalence.

Abstract

We study the classification of group actions on C*-algebras up to equivariant KK-equivalence. We show that any group action is equivariantly KK-equivalent to an action on a simple, purely infinite C*-algebra. We show that a conjecture of Izumi is equivalent to an equivalence between cocycle conjugacy and equivariant KK-equivalence for actions of torsion-free amenable groups on Kirchberg algebras. Let G be a cyclic group of prime order. We describe its actions up to equivariant KK-equivalence, based on previous work by Manuel K\"ohler. In particular, we classify actions of G on stabilised Cuntz algebras in the equivariant bootstrap class up to equivariant KK-equivalence.

Equations

$$F(A):={\mathrm{K}}_{\ast }(A⋊G)\oplus {\mathrm{K}}_{\ast }(A)\oplus {\mathrm{KK}}_{\ast }^G(D,A),$$

$$\mathfrak{S}_{2}\mathrel{\vcentcolon=}\mathbb{Z}[s,t]\bigm{/}\bigl{(}(1-s)\cdot(1-t),N(s)+N(t)-p\bigr{)}.$$

$$M_1$$

$${\gamma }_g(a\cdot \xi )={\alpha }_g(a)\cdot {\gamma }_g(\xi ),{\gamma }_g(\xi \cdot a)={\gamma }_g(\xi )\cdot {\alpha }_g(a),{\alpha }_g(⟨\xi |\eta ⟩)=⟨{\gamma }_g(\xi )|{\gamma }_g(\eta )⟩$$

$${\mathrm{Ad}}_{{\gamma }_z(U)}={\gamma }_z{\beta }_g{\gamma }_z^{-1}={\beta }_g={\mathrm{Ad}}_U.$$

$$W\cdot \phi (a)\notin \phi (A)+\mathbb{K}(E)$$

$$T_{\lambda_{g}h}^{*}W\varphi(a)T_{h}(\xi)=\int_{G}\overline{h(x)}W_{1}V(\pi(\alpha_{x}^{-1}(a))\otimes 1_{E})h(x)\xi\,\mathrm{d}x\\ =W_{1}V\pi\biggl{(}\int_{G}\lvert h(x)\rvert^{2}\alpha_{x}^{-1}(a)\,\mathrm{d}x\biggr{)}\otimes 1.$$

$${\int }_G|h(x){|}^2{\alpha }_x^{-1}(a)\mathrm{d}x=0.$$

$$f^{\ast }:{\mathrm{KK}}^{G⋉X}\to {\mathrm{KK}}^{G⋉Y}$$

$${\mathrm{KK}}^{G⋉\mathcal{E}G}\to {\mathrm{KK}}^G,A\mapsto A{\otimes }_{\mathcal{E}G}\mathcal{P}.$$

$$p_1^{\ast }({\mathrm{C}}_0(\mathcal{E}G,A))\cong {\mathrm{C}}_0(\mathcal{E}G\times \mathcal{E}G,A)\cong p_2^{\ast }({\mathrm{C}}_0(\mathcal{E}G,A)),$$

$$\begin{gathered} {\mathrm{C}}_0(\mathcal{E}G)\otimes (\mathcal{B}{\otimes }_{\mathcal{E}G}\mathcal{P})\cong {\mathrm{C}}_0(\mathcal{E}G,\mathcal{P}){\otimes }_{\mathcal{E}G\times \mathcal{E}G}({\mathrm{C}}_0(\mathcal{E}G)\otimes \mathcal{B}) \\ {\sim }_{{\mathrm{KK}}^{G⋉\mathcal{E}G}}{\mathrm{C}}_0(\mathcal{E}G,\mathcal{P}){\otimes }_{\mathcal{E}G\times \mathcal{E}G}(\mathcal{B}\otimes {\mathrm{C}}_0(\mathcal{E}G))\cong \mathcal{B}\otimes \mathcal{P}{\sim }_{{\mathrm{KK}}^{G⋉\mathcal{E}G}}\mathcal{B}\otimes \mathbb{C}\cong \mathcal{B}. \end{gathered}$$

$$M:{\mathrm{KK}}^{BG}\to {\mathrm{KK}}^{G⋉\mathcal{E}G},A\mapsto {\mathrm{C}}_0(\mathcal{E}G){\otimes }_{BG}A,$$

$$\mathcal{M}(\mathcal{B}{)}^G:=\{b\in \mathcal{M}(\mathcal{B}):{\beta }_g(b)=b\text{ for all }g\in G\text{ and }{\mathrm{C}}_0(\mathcal{E}G)\cdot b\subseteq \mathcal{B}\}.$$

$${\mathcal{B}}^G:={\mathrm{C}}_0(BG)\cdot \mathcal{M}(\mathcal{B}{)}^G$$

$$\mathrm{KK}^{G}\bigl{(}(A,\alpha),(B,\beta)\bigr{)}\cong\mathrm{KK}^{BG}(\mathcal{A},\mathcal{B}).$$

$$L_{\ast }(A):={\mathrm{KK}}_{\ast }^G(\mathbb{C},A)\oplus {\mathrm{KK}}_{\ast }^G(\mathrm{C}(G),A)\cong {\mathrm{KK}}_{\ast }^G(\mathbb{C}\oplus \mathrm{C}(G),A),$$

$$\Sigma B\to C\to A\stackrel{f}{\to }B.$$

$${\mathrm{KK}}_{\ast }^G(\mathbb{C},A)\cong {\mathrm{KK}}_{\ast }(\mathbb{C},A⋊G)$$

$${\mathrm{KK}}_{\ast }^G(\mathrm{C}(G),A)\cong {\mathrm{KK}}_{\ast }(\mathbb{C},A)$$

$${\mathrm{KK}}_{\ast }(\mathbb{C},A⋊G)\cong {\mathrm{K}}_{\ast }(A⋊G)$$

$$L_{\ast }(A)\cong {\mathrm{K}}_{\ast }(A⋊G)\oplus {\mathrm{K}}_{\ast }(A).$$

$$\mathfrak{T}:={\mathrm{KK}}_{\ast }^G(\mathbb{C}\oplus \mathrm{C}(G),\mathbb{C}\oplus \mathrm{C}(G){)}^{\mathrm{op}}$$

$$\begin{gathered} \mathfrak{T}=L_{\ast }(\mathbb{C}\oplus \mathrm{C}(G))\cong L_{\ast }(\mathbb{C})\oplus L_{\ast }(\mathrm{C}(G)) \\ \cong {\mathrm{K}}_{\ast }(\mathbb{C}⋊G)\oplus {\mathrm{K}}_{\ast }(\mathbb{C})\oplus {\mathrm{K}}_{\ast }(\mathrm{C}(G)⋊G)\oplus {\mathrm{K}}_{\ast }(\mathrm{C}(G)) \\ \cong \mathbb{Z}[G]\oplus \mathbb{Z}\oplus \mathbb{Z}\oplus \mathbb{Z}[G]\cong {\mathbb{Z}}^{2p+2}. \end{gathered}$$

$$\left(\begin{array}{cc}{x}_{00}& x_{01}\\ x_{10}& x_{11}\end{array}\right),\begin{array}{rlrl}{x}_{00}& \in {\mathrm{KK}}_0^G(\mathbb{C},\mathbb{C})\cong \mathbb{Z}[G],& x_{01}& \in {\mathrm{KK}}_0^G(\mathbb{C},\mathrm{C}(G))\cong \mathbb{Z},\\ x_{10}& \in {\mathrm{KK}}_0^G(\mathrm{C}(G),\mathbb{C})\cong \mathbb{Z},& x_{11}& \in {\mathrm{KK}}_0^G(\mathrm{C}(G),\mathrm{C}(G))\cong \mathbb{Z}[G].\end{array}$$

$$\mathfrak{S}:=\mathbb{Z}[t]/(t^p-1).$$

$$\mathrm{BS}:{\mathrm{KK}}^G\stackrel{\sim }{\to }{\mathrm{KK}}^{\hat{G}},A\mapsto A⋊G,{\mathrm{KK}}_{\ast }^G(A,B)\stackrel{\sim }{\to }{\mathrm{KK}}_{\ast }^{\hat{G}}(A⋊G,B⋊G).$$

$${\mathrm{KK}}_0^G(\mathbb{C},\mathbb{C})\cong {\mathrm{KK}}_0^G(\mathrm{C}(G),\mathrm{C}(G)).$$

$$\mathbb{Z}/p\times \mathbb{Z}/p\to \mathrm{U}(1),(g,h)\mapsto \exp (2\pi \mathrm{i}\cdot g\cdot h/p).$$

$${\mathrm{KK}}_0^G(\mathbb{C},\mathrm{C}(G))\cong {\mathrm{K}}_0(\mathrm{C}(G)⋊G)\cong {\mathrm{K}}_0(\mathbb{K}({\ell }^{2}G))\cong {\mathrm{K}}_0(\mathbb{C})\cong \mathbb{Z}.$$

Full text

1. Introduction

The Kirchberg–Phillips Classification Theorem says that two purely infinite, simple, separable, nuclear $\mathrm{C}^{*}$-algebras – briefly called Kirchberg algebras – are isomorphic if and only if they are KK-equivalent (see [Phillips:Classification]). Kirchberg [Kirchberg:Michael] has generalised this theorem to non-simple $\mathrm{C}^{*}$-algebras that are separable, nuclear and “strongly” purely infinite. To be isomorphic, they must have homeomorphic primitive ideal spaces. Assume this and fix such a homeomorphism. This lifts to an isomorphism of $\mathrm{C}^{*}$-algebras if and only if the $\mathrm{C}^{*}$-algebras are equivalent in a suitable equivariant KK-theory, which takes the underlying primitive ideal space into account. To apply these theorems, we still need a criterion when two $\mathrm{C}^{*}$-algebras (with the same primitive ideal space) are equivariantly KK-equivalent.

The best criterion in the simple case follows from the Universal Coefficient Theorem: if both $\mathrm{C}^{*}$-algebras belong to the bootstrap class – the class of $\mathrm{C}^{*}$-algebras for which the Universal Coefficient Theorem holds – then any isomorphism of $\mathbb{Z}/2$-graded Abelian groups between their $\mathrm{K}$-theories lifts to a $\mathrm{KK}$-equivalence. It can also be shown that any $\mathbb{Z}/2$-graded countable Abelian groups may be realised by some Kirchberg algebra. So isomorphism classes of stable Kirchberg algebras in the bootstrap class are in bijection with isomorphism classes of $\mathbb{Z}/2$-graded countable Abelian groups. There are generalisations of this result to the non-simple case as well (see [Meyer-Nest:Filtrated_K, Bentmann:Thesis, Bentmann-Koehler:UCT, Bentmann-Meyer:More_general, Meyer:Compute_obstruction]). But the results are not as complete, and there is no evidence for a general classification result for non-simple Kirchberg algebras.

There has been a lot of work aiming at a classification of suitable group actions on $\mathrm{C}^{*}$-algebras. This theory differs from the corresponding classification of group actions on von Neumann factors because of several $\mathrm{K}$-theoretic obstructions. These are particularly complex for finite groups. In this article, we shall focus on these $\mathrm{K}$-theoretic issues by aiming only at a classification up to equivariant $\mathrm{KK}$-equivalence. Already in the classification of non-simple Kirchberg algebras, the problem may be decomposed into two aspects. On the one hand, there is the problem of lifting an equivalence in a suitable KK-theory to an isomorphism. On the other hand, there is the problem of detecting a KK-equivalence by simpler invariants. Whereas the first problem is analytic in nature and depends on assumptions such as pure infiniteness and nuclearity, the second problem is algebraic topology and may be addressed using homological algebra in triangulated categories. Separating these two aspects of the classification problem seems particularly helpful in the more complicated case of group actions.

Let $G$ be a locally compact group. The analytic question here is under which conditions a $\mathrm{KK}^{G}$-equivalence between two $G$-actions $(A,\alpha)$ and $(B,\beta)$ on separable $\mathrm{C}^{*}$-algebras lifts to an equivariant Morita–Rieffel equivalence; this becomes a cocycle conjugacy between the group actions if $A$ and $B$ are stable. Such a result is only plausible if $A$ and $B$ are Kirchberg algebras and the actions are pointwise outer. It also seems necessary to assume the group $G$ to be amenable. A recent result by Szabó [Szabo:Equivariant_Absorption] shows that two pointwise outer actions of an amenable group on unital Kirchberg algebras are cocycle conjugate once they are $G$-equivariantly homotopy equivalent. Equivariant homotopy is a much more flexible notion than cocycle conjugacy. So it seems conceivable that $\mathrm{KK}^{G}$-equivalent outer actions on Kirchberg algebras are cocycle conjugate if $G$ is amenable.

The existing classification theorems for group actions on Kirchberg algebras use direct methods and mix the analytic and topological aspects of the problem. The starting point here is Nakamura’s result for single automorphisms in [Nakamura:Aperiodic]. Similar results have been shown for certain actions of $\mathbb{Z}^{2}$ and $\mathbb{Z}^{N}$ (see [Izumi-Matui:Z2-actions, Matui:Outer_ZN_O2]). Izumi [Izumi:Group_actions] has conjectured a classification result for actions of torsion-free groups, and recent preprints by Izumi and Matui prove this conjecture for poly-$\mathbb{Z}$-groups (see [Izumi-Matui:Poly-Z, Izumi-Matui:Poly-Z_II]). Several results in this direction are still limited by assumptions that reduce the complexity of the $\mathrm{K}$-theoretic invariants. One of our results here shows that Izumi’s conjecture in [Izumi:Group_actions] is equivalent to an analogue of Kirchberg’s Classification Theorem for non-simple $\mathrm{C}^{*}$-algebras. More precisely, let $A$ be a Kirchberg algebra and $G$ a torsion-free, amenable, second countable, locally compact group. We prove a bijection between $\mathrm{KK}^{G}$-equivalence classes of actions of $G$ on $A$ and isomorphism classes of principal $\operatorname{Aut}(A)$-bundles over the classifying space $BG$. If $G$ is discrete, Izumi conjectures that the latter classify outer $G$-actions on $A$ up to cocycle conjugacy, not just up to $\mathrm{KK}^{G}$-equivalence. So Izumi’s conjecture is equivalent to the conjecture that two $\mathrm{KK}^{G}$-equivalent outer $G$-actions on $A$ are already cocycle conjugate.

For finite groups or, more generally, groups with torsion, the existing classification results need extra assumptions on the group actions such as the Rokhlin property (see [Izumi:Finite_group_I, Izumi:Finite_group]). The Rokhlin property is a severe restriction that drastically simplifies the $\mathrm{K}$-theory groups related to the group action. Here we study actions without the Rokhlin property, but only for the group $G=\mathbb{Z}/p$ for a prime number $p$. Manuel Köhler [Koehler:Thesis] has proven a Universal Coefficient Theorem for $\mathrm{KK}^{G}$ for these groups $G$. It implies a classification of objects in the equivariant bootstrap class in $\mathrm{KK}^{G}$ up to $\mathrm{KK}^{G}$-equivalence, as in the classical Kirchberg–Phillips Theorem.

The equivariant bootstrap class $\mathfrak{B}^{G}$ in $\mathrm{KK}^{G}$ is defined as the localising subcategory generated by $\mathrm{C}(G)$ with the free translation action and by $\mathbb{C}$. That is, it is the smallest class of separable $G$-$\mathrm{C}^{*}$-algebras that contains $\mathbb{C}$ and $\mathrm{C}(G)$ and is closed under $\mathrm{KK}^{G}$-equivalence, direct sum and suspensions and has the two-out-of-three property for extensions of $G$-$\mathrm{C}^{*}$-algebras with a $G$-equivariant completely positive contractive section. The equivariant bootstrap class $\mathfrak{B}^{G}$ is equal to the class of all $G$-actions that are $\mathrm{KK}^{G}$-equivalent to an action on a Type I $\mathrm{C}^{*}$-algebra (see [dellAmbrogio-Emerson-Meyer:Equivariant_Lefschetz]*§3.1). It is unclear whether we can replace “Type I” by “commutative” here. For the group $\mathbb{T}=\mathbb{R}/\mathbb{Z}$, however, Emerson has found an action in $\mathfrak{B}^{\mathbb{T}}$ that cannot be $\mathrm{KK}^{\mathbb{T}}$-equivalent to an action on a commutative $\mathrm{C}^{*}$-algebra. If $A\in\mathfrak{B}^{G}$, then $A$ and $A\rtimes G$ belong to the usual bootstrap class in $\mathrm{KK}$. We do not know whether the converse of this is true.

For an action of $G=\mathbb{Z}/p$ on a general separable $\mathrm{C}^{*}$-algebra $A$, Köhler’s invariant is the direct sum

[TABLE]

where $D$ is the mapping cone of the unit map $\mathbb{C}\hookrightarrow\mathrm{C}(G)$. The natural transformations $F\Rightarrow F$ form a $\mathbb{Z}/2$-graded ring $\mathfrak{K}$, and $F(A)$ is a countable $\mathbb{Z}/2$-graded $\mathfrak{K}$-module. This module structure is part of Köhler’s invariant. Let $A$ and $B$ be objects of $\mathfrak{B}^{G}$. Köhler shows that $A$ and $B$ are $\mathrm{KK}^{G}$-equivalent if and only if $F(A)\cong F(B)$ as $\mathbb{Z}/2$-graded $\mathfrak{K}$-modules. His proof also shows which countable $\mathbb{Z}/2$-graded $\mathfrak{K}$-modules belong to the range of $F$. Namely, the condition is that two chain complexes must be exact. One is the Puppe six-term exact sequence for the unit map $\mathbb{C}\hookrightarrow\mathrm{C}(G)$. The other is the Puppe sequence for the induced map $\mathrm{C}(G)\cong\mathbb{C}\rtimes G\hookrightarrow\mathrm{C}(G)\rtimes G\cong\mathbb{K}(\ell^{2}(G))$. An important point here is that the latter cone is $\mathrm{KK}^{G}$-equivalent to the suspension of $D$. A $\mathbb{Z}/2$-graded $\mathfrak{K}$-module is called exact if these two Puppe sequences are exact.

So isomorphism classes of objects in the bootstrap class in $\mathrm{KK}^{G}$ are in bijection with isomorphism classes of countable, exact $\mathbb{Z}/2$-graded $\mathfrak{K}$-modules. These are pretty complicated objects. Let $M$ be a $\mathbb{Z}/2$-graded $\mathfrak{K}$-module. Then $M=M_{0}\oplus M_{1}\oplus M_{2}$, where $M_{0}$, $M_{1}$ and $M_{2}$ are the images of the idempotent natural transformations that project $F(A)$ to the three summands $\mathrm{K}_{*}(A\rtimes G)$, $\mathrm{K}_{*}(A)$, and $\mathrm{KK}^{G}_{*}(D,A)$. The summands $M_{0}$ and $M_{1}$ are modules over $\mathbb{Z}[G]\cong\operatorname{Rep}(G)\cong\mathbb{Z}[x]/(x^{p}-1)$, where the actions are induced by the given action of $G$ on $A$ and the dual action of $G\cong\widehat{G}$ on $A\rtimes G$. The summand $M_{2}$ is a module over the more complicated ring

[TABLE]

Then there are further natural transformations that link $M_{0}$, $M_{1}$ and $M_{2}$.

Since Köhler’s invariant is complicated, we are going to consider several classes of exact $\mathfrak{K}$-modules that may be understood through simpler objects. My goal has been to classify those exact $\mathfrak{K}$-modules where $M_{1}$ is a cyclic group. Given Köhler’s classification theorem, these exact $\mathfrak{K}$-modules correspond to $G$-actions in the bootstrap class $A\in\mathfrak{B}^{G}$ where $\mathrm{K}_{*}(A)$ is cyclic. In fact, any such $A$ is $\mathrm{KK}^{G}$-equivalent to an action on a stabilised Cuntz algebra. The main issue here is to make the underlying $\mathrm{C}^{*}$-algebra of the action simple. That this is possible is proven in complete generality for actions of locally compact groups and is also used in the torsion-free case to reformulate Izumi’s conjecture. Since $A\in\mathfrak{B}^{G}$ is assumed, the results in this article only cover actions in $\mathfrak{B}^{G}$, so that $A\rtimes G$ must be in the non-equivariant bootstrap class. Thus we have nothing to say about the important question whether $A\rtimes G$ belongs to the bootstrap class for any $G$-action, say, on $\mathcal{O}_{2}$ (see, for instance, [Barlak-Li:Cartan_UCT]).

The end result of our classification of the actions of $G$ on Cuntz algebras is remarkably complicated. In Section 9, we list nine examples of such exact modules; some of these involve parameters, so that they rather form a family of examples. There is a tenth family of examples only for the case $p=2$, which is about a general method to modify a given example to produce another one. Our main theorem says that any exact module where $M_{1}$ is cyclic contains one of these ten examples as an exact submodule $M^{\prime}$, such that the quotient $M^{\prime\prime}\mathrel{\vcentcolon=}M/M^{\prime}$ has $M^{\prime\prime}_{1}=0$. We also show that any such $M^{\prime\prime}$ is uniquely $p$-divisible and may be described through a $\mathbb{Z}/2$-graded module over $\mathbb{Z}[\vartheta_{p},1/p]$. The examples in Section 9 are all quite different, and some of them are not obvious. I have found these examples by a case-by-case study of the possible exact modules over Köhler’s ring with cyclic $M_{1}$. The results of Köhler also imply that the module extension $M^{\prime}\rightarrowtail M\twoheadrightarrow M^{\prime\prime}$ in our main theorem lifts to an exact triangle in the bootstrap class in $\mathrm{KK}^{G}$. This explains why the piece $M^{\prime\prime}$ has to be there. It corresponds to adding $\mathcal{O}_{2}$ with a suitable action to a given $G$-action and then passing to a $\mathrm{KK}^{G}$-equivalent action on a Kirchberg algebra.

In order to classify the exact $\mathfrak{K}$-modules with cyclic $M_{1}$, we prove several more general results. First, we describe Köhler’s ring through generators and relations. This simplifies writing down a $\mathfrak{K}$-module. As it turns out, a module over Köhler’s ring consists of three $\mathbb{Z}/2$-graded Abelian groups $M_{j}$, $j=0,1,2$, with homomorphisms $\alpha_{jk}^{M}\colon M_{k}\to M_{j}$ for $j\neq k$ such that the two triangle-shaped diagrams

[TABLE]

are chain complexes. Exactness means that these two chain complexes are exact. There are also some additional relations involving the products $\alpha_{jk}^{M}\circ\alpha_{kj}^{M}$ (see Theorem 5.11). Köhler’s invariant for a $G$-$\mathrm{C}^{*}$-algebra $A$ has $M_{0}=\mathrm{K}_{*}(A\rtimes G)$ and $M_{1}=\mathrm{K}_{*}(A)$, and the maps $\alpha_{01}^{M}$ and $\alpha_{10}^{M}$ are induced by the inclusion maps $A\subseteq A\rtimes G$ and $A\rtimes G\subseteq A\otimes\mathbb{K}(\ell^{2}G)$. We show by an example that $M_{0}$ and $M_{1}$ alone with their natural extra structure do not yet classify up to $\mathrm{KK}^{G}$-equivalence, by exhibiting two non-isomorphic exact modules over Köhler’s ring for which the $M_{0}$- and $M_{1}$-parts are isomorphic (see Example 9.9).

We describe several classes of exact modules over Köhler’s ring using simpler ingredients. First, we describe modules where one of the groups $M_{j}$ is uniquely $p$-divisible (see Example 7.1); in particular, this covers the case when one of them vanishes, say, $M_{1}=0$. In the latter case, $M_{0}\cong M_{2}$ is a $\mathbb{Z}/2$-graded module over $\mathbb{Z}[\zeta_{p},1/p]$ with the primitive $p$th root of unity $\zeta_{p}\mathrel{\vcentcolon=}\exp(2\pi\mathrm{i}/p)$, and any such $\mathbb{Z}/2$-graded module corresponds to a unique exact $\mathfrak{K}$-module with $M_{1}=0$. Then we describe modules where one of the maps $\alpha_{jk}^{M}$ is zero. By exactness, this covers all cases where $\alpha_{01}^{M}$ or $\alpha_{10}^{M}$ is zero, injective, or surjective. Third, we describe modules with $M_{0}=\operatorname{im}\alpha_{01}^{M}\oplus\operatorname{im}\alpha_{02}^{M}$ or similar. As it turns out, these different cases suffice to describe all exact $\mathfrak{K}$-modules where $M_{1}$ is a cyclic group.

The starting point for the study of exact modules over Köhler’s ring here is the Bachelor’s thesis of Vincent Grande [Grande:Bachelor], which contains several examples of such modules and partial results on the different cases of submodules considered here.

It is a great pleasure to thank the referee for a very careful reading of the manuscript and a number of helpful suggestions.

2. Realising actions on Kirchberg algebras

The goal of this section is the following theorem:

Theorem 2.1**.**

Let $G$ be a second countable, locally compact group. Let $A$ be a separable $\mathrm{C}^{*}$-algebra and let $\alpha$ be a continuous action of $G$ on $A$. Then there is a non-zero, simple, purely infinite, stable $\mathrm{C}^{*}$-algebra $B$ with a pointwise outer, continuous action $\beta$ of $G$ such that $(A,\alpha)$ is $\mathrm{KK}^{G}$-equivalent to $(B,\beta)$. If $A$ is nuclear or exact, then so is $B$.

Proof.

The proof of [Benson-Kumjian-Phillips:Symmetries_Kirchberg]*Proposition 4.2 shows this, but the theorem makes only a weaker statement. So we go through the main construction again. The main tool is the Toeplitz algebra for $\mathrm{C}^{*}$-correspondences defined by Pimsner [Pimsner:Generalizing_Cuntz-Krieger], in the special case studied by Kumjian [Kumjian:Cuntz-Pimsner]. A similar construction appeared recently in [Deaconu:Pimsner_representation].

Before we really start, we need an extra step for the trivial case $A=0$. Then we replace $A$ by $\mathcal{O}_{2}$ with the trivial action of $G$. This is non-zero, nuclear and $\mathrm{KK}^{G}$-equivalent to [math]. So we may assume without loss of generality that $A\neq 0$.

First we build a regular covariant representation of $(A,\alpha,G)$ as in [Pedersen:Cstar_automorphisms]*§7.7.1. There is a faithful representation $\pi\colon A\to\mathbb{B}(\mathcal{H})$ of $A$ on a separable Hilbert space $\mathcal{H}$. Taking an infinite direct sum of copies of $\pi$, we arrange that $\pi(A)$ contains no compact operators. Now represent $G$ and the $\mathrm{C}^{*}$-algebra $A$ on $L^{2}(G,\mathcal{H})$ by the left regular representation $(\lambda_{g}f)(x)\mathrel{\vcentcolon=}f(g^{-1}x)$ and by $(\tilde{\pi}(a)f)(x)\mathrel{\vcentcolon=}\pi(\alpha_{x^{-1}}(a))f(x)$ for all $g,x\in G$, $a\in A$, $f\in L^{2}(G,\mathcal{H})$. This is a covariant representation, that is, $\lambda_{g}\tilde{\pi}(a)\lambda_{g^{-1}}=\tilde{\pi}(\alpha_{g}(a))$ for all $g\in G$, $a\in A$.

Let $E$ be the Hilbert $A$-module $L^{2}(G,\mathcal{H})\otimes A$, equipped with the left action $\varphi\colon A\to\mathbb{B}(E)$, $a\mapsto\tilde{\pi}(a)\otimes 1$, and the diagonal $G$-action $\gamma_{g}(f\otimes a)\mathrel{\vcentcolon=}\lambda_{g}(f)\otimes\alpha_{g}(a)$. This makes $E$ a $G$-equivariant $\mathrm{C}^{*}$-correspondence from $A$ to itself, that is, the left action is by adjointable operators and the $G$-action satisfies

[TABLE]

for all $g\in G$, $a\in A$, $\xi,\eta\in E$. If $\varphi(a)$ is compact, then $\pi(a)$ must be compact and then $a=0$ by construction. In particular, the left action of $A$ on $E$ is injective.

Let $\mathcal{T}_{E}$ be the Toeplitz $\mathrm{C}^{*}$-algebra of $E$ as in [Pimsner:Generalizing_Cuntz-Krieger]. The $G$-actions on $A$ and $E$ satisfy the right compatibility assumptions to induce a continuous action of $G$ on $\mathcal{T}_{E}$ (see [Pimsner:Generalizing_Cuntz-Krieger]*Remark 4.10.(2)). The natural inclusion of $A$ into $\mathcal{T}_{E}$ is a KK-equivalence by [Pimsner:Generalizing_Cuntz-Krieger]*Theorem 4.4, and this remains true in $\mathrm{KK}^{G}$ because all Kasparov cycles and homotopies between them used in the proof of that theorem are defined naturally and hence are exactly $G$-equivariant. This is already observed by Pimsner in [Pimsner:Generalizing_Cuntz-Krieger]*Remark 4.10.(2).

Since no non-zero element of $A$ acts on $E$ by a compact operator, [Kumjian:Cuntz-Pimsner]*Theorem 2.8 shows that $\mathcal{T}_{E}=\mathcal{O}_{E}$ is purely infinite and simple. To make it stable, we replace $\mathcal{T}_{E}$ by its $\mathrm{C}^{*}$-stabilisation $\mathcal{T}_{E}\otimes\mathbb{K}(\ell^{2}\mathbb{N})$, which remains purely infinite and simple and $\mathrm{KK}^{G}$-equivalent to $A$. The structural analysis of the Toeplitz algebra in [Pimsner:Generalizing_Cuntz-Krieger] or [Kumjian:Cuntz-Pimsner] shows that it inherits both nuclearity and exactness from $A$.

Finally, we show that the induced action $\beta$ of $G$ on $\mathcal{T}_{E}$ is pointwise outer (the $\mathrm{C}^{*}$-stabilisation does not affect this). Assume the contrary. So there are $g\in G\setminus\{1\}$ and a unitary multiplier $U$ of $B$ with $\beta_{g}=\operatorname{Ad}_{U}$. Let $\gamma\colon\mathbb{T}\to\operatorname{Aut}(\mathcal{T}_{E})$ be the gauge action on the Toeplitz algebra $\mathcal{T}_{E}$. It commutes with the action of $G$. Therefore,

[TABLE]

This implies that conjugation by the unitary $\gamma_{z}(U)U^{*}$ is the identity automorphism. That is, this unitary is central. Since $\mathcal{T}_{E}$ is simple, $\gamma_{z}(U)=c_{z}\cdot U$ for some scalar $c_{z}\in\mathbb{C}$ with $\lvert c_{z}\rvert=1$. The map $z\mapsto c_{z}$ is a character on $\mathbb{T}$. Since the dual group of $\mathbb{T}$ is $\mathbb{Z}$, there is $n\in\mathbb{Z}$ with $\gamma_{z}(U)=z^{n}\cdot U$ for all $z\in\mathbb{T}$. Recall that the Fock representation of $\mathcal{T}_{E}$ is a faithful and non-degenerate representation on the Hilbert $A$-module $\mathcal{F}\mathrel{\vcentcolon=}\bigoplus_{k=0}^{\infty}E^{\otimes_{A}k}$. The unitary $U$ gives a unitary operator on $\mathcal{F}$. We describe it by a block matrix $(U_{k,l})_{k,l\in\mathbb{N}}$ in the direct sum decomposition of $\mathcal{F}$. The gauge action $\gamma_{z}$ on $\mathcal{T}_{E}$ acts on the matrix coefficients by $\gamma_{z}(U_{k,l})=z^{k-l}U_{k,l}$ for all $k,l\in\mathbb{N}$. So the homogeneity condition $\gamma_{z}(U)=z^{n}\cdot U$ says that $U_{k,l}=0$ for $k-l\neq n$. If $n>0$, then this implies that $k>0$, so that the range of $U$ is orthogonal to the first summand $A$ in $\mathcal{F}$. This is impossible for a unitary operator. Similarly, $U^{*}$ cannot be unitary if $n<0$. So we must have $n=0$.

So $U$ is a multiplier of the fixed-point subalgebra $\mathcal{T}_{E}^{\mathbb{T}}\subseteq\mathcal{T}_{E}$ of the gauge action. This subalgebra acts on $\mathcal{F}$ by diagonal operators. Projecting to the first summand $A$ in $\mathcal{F}$ is a non-degenerate representation of $\mathcal{T}_{E}^{\mathbb{T}}$. It maps $U$ to a unitary multiplier $V$ of $A$ with $\operatorname{Ad}_{V}=\alpha_{g}$. Projecting to the second summand $E$ in $\mathcal{F}$ is another non-degenerate representation of $\mathcal{T}_{E}^{\mathbb{T}}$, which maps $U$ to a unitary operator $W$ on $E$. Conjugation by $W$ must give the restriction of $\beta_{g}$ on $\mathbb{K}(E)\cong\mathbb{K}(L^{2}(G,\mathcal{H}))\otimes A$. Conjugation by $\lambda_{g}\otimes V$ has the same effect. So $W=W_{0}\cdot(\lambda_{g}\otimes V)$ for some unitary $W_{0}$ in the centre of the multiplier algebra of $\mathbb{K}(E)$. Such operators must be of the form $W_{0}=1_{L^{2}(G,\mathcal{H})}\otimes W_{1}$ for some central unitary multiplier $W_{1}$ of $A$. So $W=\lambda_{g}\otimes(W_{1}\cdot V)$. The image of $\mathcal{T}_{E}^{\mathbb{T}}$ in $\mathbb{B}(E)$ is the linear span of $\mathbb{K}(E)$ and $\varphi(A)$, where $\varphi\colon A\to\mathbb{B}(E)$ is the left action chosen above. We are going to prove that

[TABLE]

for $a\in A$, $a\neq 0$. Then $W$ is not a multiplier of $\varphi(A)+\mathbb{K}(E)$. But it must be one, coming from a multiplier of $\mathcal{T}_{E}^{\mathbb{T}}$. This contradiction shows that the automorphism $\beta_{g}$ on $\mathcal{T}_{E}$ cannot be inner.

It remains to prove (2.2). Let $a\in A$ be such that $W\cdot\varphi(a)\in\varphi(A)+\mathbb{K}(E)$. We are going to prove that $a=0$. Since $g\neq 1$, there is $h\in\mathrm{C}_{\mathrm{c}}(G)$ so that $h$ and $\lambda_{g}h$ have disjoint support. We use the creation operator $T_{h}\colon\mathcal{H}\otimes A\to E$, $\xi\mapsto h\otimes\xi$. Then $T_{\lambda_{g}h}^{*}\varphi(A)T_{h}=0$ because $\varphi(A)$ does not change the support of functions on $G$. So $T_{\lambda_{g}h}^{*}W\varphi(a)T_{h}\in\mathbb{K}(\mathcal{H}\otimes A)$. Let $\xi\in\mathcal{H}\otimes A$. Then

[TABLE]

Since $W_{1}V$ is unitary and $\pi(A)\cap\mathbb{K}(\mathcal{H})=0$, this is only compact if

[TABLE]

Then $a=0$ because the action on $A$ is continuous. This proves the claim. ∎

There are not yet any definite classification results for general group actions on $\mathrm{C}^{*}$-algebras. So it is unclear which conditions are sufficient for classification. If a generalisation of Kirchberg’s classification theorem exists, it should probably assume a $G$-action on a simple, nuclear, separable $\mathrm{C}^{*}$-algebra that tensorially absorbs a suitable model $G$-action on $\mathcal{O}_{\infty}$, which is $\mathrm{KK}^{G}$-equivalent to $\mathbb{C}$ with the trivial action. This is analogous to the strong pure infiniteness requirement in Kirchberg’s classification theorem for non-simple $\mathrm{C}^{*}$-algebras, which says that the $\mathrm{C}^{*}$-algebra tensorially absorbs $\mathcal{O}_{\infty}$. The model action on $\mathcal{O}_{\infty}$ should tensorially absorb itself, so that tensoring with it produces a $\mathrm{KK}^{G}$-equivalent object that tensorially absorbs the model action. Such model actions are constructed in [Goldstein-Izumi:Quasi-free]*Theorem 5.1, [Izumi-Matui:Poly-Z]*Theorem 4.13 and [Szabo:Equivariant_Absorption]*Corollary 3.7. I thank the referee for pointing out these references. If a classification theorem for group actions will be proven, then one may use Theorem 2.1 to show that any object of $\mathrm{KK}^{G}$ whose underlying $\mathrm{C}^{*}$-algebra is nuclear is $\mathrm{KK}^{G}$-equivalent to a classifiable object, by tensoring the $G$-action from Theorem 2.1 with the model action. It may well be that the actions produced by Theorem 2.1 already absorb the model action. We do not discuss this further here because, anyway, our classification results are only up to $\mathrm{KK}^{G}$-equivalence.

3. Actions of torsion-free, amenable groups and a conjecture of Izumi

Throughout this section, we assume that $G$ is an amenable, second countable, Hausdorff, locally compact group. Let $\mathcal{E}G$ denote a second countable, locally compact, Hausdorff space with a proper, continuous $G$-action that is the universal proper action in the following sense:

1. (1)

for any other locally compact, proper $G$-space $Z$ there is a continuous $G$-equivariant map $Z\to\mathcal{E}G$, and 2. (2)

any two continuous $G$-equivariant maps $Z\rightrightarrows\mathcal{E}G$ are homotopic through a continuous $G$-equivariant homotopy $Z\times[0,1]\to\mathcal{E}G$.

Such a space exists by [Kasparov-Skandalis:Bolic]*Lemma 4.1 (take $X=G$). But smaller models are preferable for actual computations.

We will soon also assume $G$ to be torsion-free, but the first step in our argument does not yet need this. For an amenable, torsion-free group, we will show that $\mathrm{KK}^{G}$-equivalence classes of actions of $G$ on a stable Kirchberg algebra $A$ are in bijection with isomorphism classes of locally trivial $\operatorname{Aut}(A)$-bundles over the classifying space $BG$. This relates a conjecture by Izumi [Izumi:Group_actions] to Kirchberg’s Classification Theorem. First we recall a key step in the proof of the Baum–Connes conjecture for amenable groups by Higson and Kasparov [Higson-Kasparov:E_and_KK].

Definition 3.1**.**

A $G$-$\mathrm{C}^{*}$-algebra $\mathcal{P}$ is proper if there are a locally compact, proper $G$-space $X$ and a $G$-equivariant, nondegenerate ∗-homomorphism from $\mathrm{C}_{0}(X)$ to the centre of the multiplier algebra of $\mathcal{P}$. In other words, $\mathcal{P}$ carries an action of the transformation groupoid $G\ltimes X$. Since the latter groupoid is proper, proper group actions on $\mathrm{C}^{*}$-algebras behave much better than general actions.

Theorem 3.2** ([Higson-Kasparov:E_and_KK]).**

Let $G$ be an amenable group or, more generally, a group with the Haagerup approximation property. Then there is a proper $G$-$\mathrm{C}^{*}$-algebra $\mathcal{P}$ that is $\mathrm{KK}^{G}$-equivalent to $\mathbb{C}$. In addition, $\mathcal{P}$ is nuclear.

The $\mathrm{KK}^{G}$-equivalence between $\mathcal{P}$ and $\mathbb{C}$ consists of two elements $D\in\mathrm{KK}^{G}_{0}(\mathcal{P},\mathbb{C})$, $\eta\in\mathrm{KK}^{G}_{0}(\mathbb{C},\mathcal{P})$, called Dirac and dual Dirac elements. Their composites in both direction are $1$. This data is built in [Higson-Kasparov:E_and_KK] and then used to prove the Baum–Connes conjecture for $G$. If $X$ is as in Definition 3.1, then there is a map $X\to\mathcal{E}G$, which induces a $G$-equivariant, nondegenerate ∗-homomorphism $\mathrm{C}_{0}(\mathcal{E}G)\to\mathcal{M}(\mathrm{C}_{0}(X))$. So we may replace $X$ by $\mathcal{E}G$ in Definition 3.1. Then $\mathcal{P}$ carries an action of the transformation groupoid $G\ltimes\mathcal{E}G$ as in [LeGall:KK_groupoid]. Given a $G$-$\mathrm{C}^{*}$-algebra $A$, the $\mathrm{C}^{*}$-algebra $\mathrm{C}_{0}(\mathcal{E}G)\otimes A\cong\mathrm{C}_{0}(\mathcal{E}G,A)$ carries a canonical action of $G\ltimes\mathcal{E}G$, using the diagonal $G$-action and the equivariant nondegenerate ∗-homomorphism $\mathrm{C}_{0}(\mathcal{E}G)\to Z\mathcal{M}(\mathrm{C}_{0}(\mathcal{E}G)\otimes A)$, $f\mapsto f\otimes 1$.

The construction $A\mapsto\mathrm{C}_{0}(\mathcal{E}G,A)$ above is part of a functor $p_{\mathcal{E}G}^{*}\colon\mathrm{KK}^{G}\to\mathrm{KK}^{G\ltimes\mathcal{E}G}$. This may be proven directly by writing down what $p_{\mathcal{E}G}^{*}$ does with Kasparov cycles. An elegant proof using the universal property of $\mathrm{KK}^{G}$ is explained in [Meyer-Nest:BC]*Section 3.2. More generally, the same proof gives pull-back functors

[TABLE]

for two locally compact $G$-spaces $X$ and $Y$ and a $G$-equivariant continuous map $f\colon Y\to X$; the functor $p_{\mathcal{E}G}^{*}$ is the special case where $f$ is the constant map from $\mathcal{E}G$ to the one-point space.

Definition 3.3**.**

Let $p_{1},p_{2}\colon\mathcal{E}G\times\mathcal{E}G\rightrightarrows\mathcal{E}G$ be the two coordinate projections. An object of $\mathrm{KK}^{G\ltimes\mathcal{E}G}$ is called symmetric if $p_{1}^{*}(\mathcal{B})$ and $p_{2}^{*}(\mathcal{B})$ are $\mathrm{KK}^{G\ltimes(\mathcal{E}G\times\mathcal{E}G)}$-equivalent.

Proposition 3.4**.**

Let $G$ be a locally compact group and $\mathcal{E}G$ a universal proper $G$-space. Assume that there is a nuclear, proper $G$-$\mathrm{C}^{*}$-algebra $\mathcal{P}$ that is $\mathrm{KK}^{G}$-equivalent to $\mathbb{C}$. The functor $\mathrm{KK}^{G}\to\mathrm{KK}^{G\ltimes\mathcal{E}G}$ defined above is an equivalence from $\mathrm{KK}^{G}$ onto the full subcategory of symmetric objects in $\mathrm{KK}^{G\ltimes\mathcal{E}G}$. This functor restricts to an equivalence from the subcategory of $G$-actions on nuclear $\mathrm{C}^{*}$-algebras to the full subcategory of symmetric objects in $\mathrm{KK}^{G\ltimes\mathcal{E}G}$ whose underlying $\mathrm{C}^{*}$-algebra is nuclear.

Proof.

The proper $G$-$\mathrm{C}^{*}$-algebra $\mathcal{P}$ gives rise to functor

[TABLE]

On the object level this functor first forms the tensor product $A\otimes\mathcal{P}$, which is a $\mathrm{C}^{*}$-algebra over $\mathcal{E}G\times\mathcal{E}G$, and then restricts this to the diagonal $\{(x,x)\,{:}\,\mathopen{}x\in\mathcal{E}G\}$. Then the $\mathrm{C}_{0}(\mathcal{E}G)$-$\mathrm{C}^{*}$-algebra structure is forgotten. The group action of $G$ goes along in these constructions. The universal property of $\mathrm{KK}^{G\ltimes X}$ shows that it defines a functor on Kasparov theory (see [Meyer-Nest:BC]*Section 3.2). Both functors $\mathrm{KK}^{G}\leftrightarrow\mathrm{KK}^{G\ltimes\mathcal{E}G}$ preserve nuclearity of $\mathrm{C}^{*}$-algebras. The composite functor $\mathrm{KK}^{G}\to\mathrm{KK}^{G}$ maps a $G$-$\mathrm{C}^{*}$-algebra $A$ to $\mathrm{C}_{0}(\mathcal{E}G,A)\otimes_{\mathcal{E}G}\mathcal{P}$, which is canonically isomorphic to $A\otimes\mathcal{P}$. And this is $\mathrm{KK}^{G}$-equivalent to $A$ because of the $\mathrm{KK}^{G}$-equivalence between $\mathcal{P}$ and $\mathbb{C}$. So the composite $\mathrm{KK}^{G}\to\mathrm{KK}^{G}$ is equivalent to the identity functor. Hence the other composite functor $S\colon\mathrm{KK}^{G\ltimes\mathcal{E}G}\to\mathrm{KK}^{G\ltimes\mathcal{E}G}$ is a projection onto a subcategory of $\mathrm{KK}^{G\ltimes\mathcal{E}G}$. Since

[TABLE]

the objects of this subcategory must be symmetric. Conversely, assume that $\mathcal{B}$ is a symmetric object of $\mathrm{KK}^{G\ltimes\mathcal{E}G}$. The composite functor $S$ above maps $\mathcal{B}$ to

[TABLE]

All objects in (3.5) carry the diagonal action of $G$ and the $\mathrm{C}_{0}(\mathcal{E}G)$-$\mathrm{C}^{*}$-algebra structure that acts only on the first leg, coming from the projection $p_{1}\colon\mathcal{E}G\times\mathcal{E}G\to\mathcal{E}G$. The first $\mathrm{KK}^{G\ltimes\mathcal{E}G}$-equivalence in (3.5) uses that $\mathcal{B}$ is symmetric. The second $\mathrm{KK}^{G\ltimes\mathcal{E}G}$-equivalence uses once again that $\mathcal{P}$ is $\mathrm{KK}^{G}$-equivalent to $\mathbb{C}$. So all objects in the range of $S$ are symmetric, and $S$ is equivalent to the identity functor on the full subcategory of symmetric objects. This finishes the proof. ∎

Theorem 3.2 shows that Proposition 3.4 applies if $G$ is amenable or has the Haagerup approximation property.

Now assume also that $G$ is torsion-free, that is, the only compact subgroup in $G$ is the trivial subgroup $\{1\}$. Then any proper $G$-action is free. So $\mathcal{E}G$ is also a universal free and proper action. Since $G$ has only the trivial compact subgroup, Abels’ Slice Theorem for proper actions in [Abels:Universal]*Theorem 3.3 says here that any $x\in\mathcal{E}G$ has a $G$-invariant open neighbourhood $U\subseteq\mathcal{E}G$ that is $G$-equivariantly homeomorphic to $G\times Y$ with the translation action on $G\times Y$. In other words, the projection $\mathcal{E}G\to\mathcal{E}G/G$ is locally trivial. So $\mathcal{E}G/G$ is a model for the classifying space of $G$, and it is legitimate to define $BG\mathrel{\vcentcolon=}\mathcal{E}G/G$. We are going to build a functor

[TABLE]

and show that it is an equivalence of categories. An action of $BG$ on a $\mathrm{C}^{*}$-algebra is simply a $\mathrm{C}_{0}(BG)$-$\mathrm{C}^{*}$-algebra structure. The bundle projection $q\colon\mathcal{E}G\to BG$ is $G$-equivariant if we let $G$ act trivially on $BG$. Equipping a $\mathrm{C}_{0}(BG)$-$\mathrm{C}^{*}$-algebra with the trivial action of $G$ gives a functor $\mathrm{KK}^{BG}\to KK^{BG\rtimes G}$, which we may compose with the pull-back functor $q^{*}$ to get the functor $M$ above. The explanation why $M$ is an equivalence of categories is that the $G$-action on $\mathcal{E}G$ is free and proper, so that the groupoid $G\ltimes\mathcal{E}G$ is equivalent to the space $BG$.

Proposition 3.6**.**

Let $G$ be a torsion-free, locally compact group. The functor $M$ above is an equivalence of categories $\mathrm{KK}^{G\ltimes\mathcal{E}G}\simeq\mathrm{KK}^{BG}$. The equivalence restricts to the subcategories of objects whose underlying $\mathrm{C}^{*}$-algebra is nuclear.

Proof.

Kasparov [Kasparov:Novikov] builds a functor from $\mathrm{KK}^{G\ltimes\mathcal{E}G}$ to $\mathrm{KK}^{BG}$ by taking a fixed-point algebra. Let $\mathcal{B}$ be a $G\ltimes\mathcal{E}G$-$\mathrm{C}^{*}$-algebra. Let

[TABLE]

We map $\mathrm{C}_{0}(BG)$ to $\mathrm{C}_{\mathrm{b}}(\mathcal{E}G)=\mathcal{M}(\mathrm{C}_{0}(\mathcal{E}G))$ using $q^{*}$ and further to $\mathcal{M}(\mathcal{B})$. The image of $\mathrm{C}_{0}(BG)$ belongs to the centre of $\mathcal{M}(\mathcal{B})^{G}$. So

[TABLE]

is a well defined ideal in $\mathcal{M}(\mathcal{B})^{G}$. This is the generalised fixed-point algebra for the $G$-action on $\mathcal{B}$. It is naturally a $\mathrm{C}_{0}(BG)$-$\mathrm{C}^{*}$-algebra. This construction is equivalent to the definition in [Kasparov:Novikov] and to the construction of generalised fixed-point algebras for generalised proper actions by Rieffel [Rieffel:Proper]. The fibre of the $\mathrm{C}_{0}(BG)$-$\mathrm{C}^{*}$-algebra $\mathcal{B}^{G}$ at $x\in BG$ is isomorphic to the fibre of $\mathcal{B}$ at any $y\in\mathcal{E}G$ with $q(y)=x$; these fibres in $\mathcal{B}$ are canonically isomorphic using the $G$-action. So in terms of upper semicontinuous fields of $\mathrm{C}^{*}$-algebras, our construction is simply viewing a $G$-equivariant upper semicontinuous field of $\mathrm{C}^{*}$-algebras over $\mathcal{E}G$ as an upper semicontinuous field of $\mathrm{C}^{*}$-algebras over $BG$. The construction of generalised fixed-point algebras above is clearly functorial for nondegenerate $G$-equivariant, $\mathrm{C}_{0}(\mathcal{E}G)$-linear ∗-homomorphisms. And a $G$-invariant ideal in $\mathcal{B}$ gives an ideal in $\mathcal{B}^{G}$ in an obvious way. This allows to extend the functoriality of the generalised fixed-point algebra construction to degenerate $G$-equivariant, $\mathrm{C}_{0}(\mathcal{E}G)$-linear ∗-homomorphisms. The functor so obtained is split-exact and $\mathrm{C}^{*}$-stable in the appropriate sense that it defines a functor $\mathrm{KK}^{G\ltimes\mathcal{E}G}\to\mathrm{KK}^{BG}$ by the universal property of equivariant $\mathrm{KK}$-theory. This functor is built by Kasparov [Kasparov:Novikov] in terms of Kasparov cycles, and shown in [Kasparov:Novikov]*Theorem 3.4 to be fully faithful.

We have built two functors $\mathrm{KK}^{G\ltimes\mathcal{E}G}\leftrightarrow\mathrm{KK}^{BG}$ in opposite directions. We claim that they are inverse to each other up to natural equivalence. Indeed, when we start with a $\mathrm{C}_{0}(BG)$-$\mathrm{C}^{*}$-algebra $A$, we first map it to $\mathrm{C}_{0}(\mathcal{E}G)\otimes_{BG}A$ and then take the generalised fixed-point algebra. In terms of upper semicontinuous fields, the field over $\mathcal{E}G$ has the same fibres as $A$ over $BG$, and so we get back $A$. Conversely, let $\mathcal{B}$ be a $G\ltimes\mathcal{E}G$-action. Then the structural homomorphism $\mathrm{C}_{0}(\mathcal{E}G)\otimes_{\mathrm{C}_{0}(BG)}\mathcal{B}^{G}\to\mathcal{B}$, $f\otimes b\mapsto f\cdot b$, is an isomorphism of $\mathrm{C}_{0}(\mathcal{E}G)$-algebras because it is an isomorphism fibrewise. Therefore, $\mathcal{B}$ is in the essential range of the functor $M$ above. This equivalence of categories is checked in different notation in [Huef-Raeburn-Williams:Brauer_semigroup]. ∎

Theorem 3.7**.**

Let $G$ be a torsion-free, amenable, second countable group. Let $BG$ be a second countable, locally compact model for the classifying space of $G$. The category $\mathrm{KK}^{G}$ is equivalent to the subcategory of $\mathrm{KK}^{BG}$ whose objects are the locally trivial bundles of $\mathrm{C}^{*}$-algebras over $BG$. This equivalence restricts to an equivalence between the subcategories of objects whose underlying $\mathrm{C}^{*}$-algebra is nuclear.

Proof.

Any model of $BG$ carries a universal principal $G$-bundle, whose total space is a model for $\mathcal{E}G$. Therefore, the choice of $BG$ does not matter.

Proposition 3.4 applies here because of Theorem 3.2. Proposition 3.6 applies because $G$ is torsion-free. Both propositions together give an equivalence of categories from $\mathrm{KK}^{G}$ onto a full subcategory of $\mathrm{KK}^{BG}$. Let $BG^{(2)}\mathrel{\vcentcolon=}(\mathcal{E}G\times\mathcal{E}G)/G$. The coordinate projections $p_{1},p_{2}\colon\mathcal{E}G\times\mathcal{E}G\rightrightarrows\mathcal{E}G$ induce continuous maps $\check{p}_{1},\check{p}_{2}\colon BG^{(2)}\rightrightarrows BG$. The description of the range of the equivalence from $\mathrm{KK}^{G}$ to a full subcategory of $\mathrm{KK}^{G\ltimes\mathcal{E}G}$ implies that a $\mathrm{C}_{0}(BG)$-$\mathrm{C}^{*}$-algebra $\mathcal{B}$ belongs to the essential image of $\mathrm{KK}^{G}$ if and only if $\check{p}_{1}^{*}(\mathcal{B})$ and $\check{p}_{2}^{*}(\mathcal{B})$ are $\mathrm{KK}^{BG^{(2)}}$-equivalent. We must show that this holds if and only if $\mathcal{B}$ is $\mathrm{KK}^{BG}$-equivalent to a locally trivial bundle of $\mathrm{C}^{*}$-algebras.

In one direction, we may directly compute the functor $\mathrm{KK}^{G}\to\mathrm{KK}^{BG}$ on objects. Let $A$ be a separable $G$-$\mathrm{C}^{*}$-algebra, viewed as an object of $\mathrm{KK}^{G}$. The functor to $\mathrm{KK}^{BG}$ maps $A$ to the generalised fixed-point algebra $\mathrm{C}_{0}(\mathcal{E}G,A)^{G}$. This bundle of $\mathrm{C}^{*}$-algebras over $BG$ is locally trivial because the projection $\mathcal{E}G\to BG$ is locally trivial. So any object in the essential range of our functor is $\mathrm{KK}^{BG}$-equivalent to a locally trivial bundle of $\mathrm{C}^{*}$-algebras. Conversely, assume $\mathcal{B}$ to be a locally trivial bundle of $\mathrm{C}^{*}$-algebras over $BG$. The maps $\check{p}_{1},\check{p}_{2}\colon BG^{(2)}\rightrightarrows BG$ are homotopic to each other because $p_{1},p_{2}$ are $G$-equivariantly homotopic. The quotient spaces $BG$ and $BG^{(2)}$ are second countable and locally compact, Hausdorff because $\mathcal{E}G$ and $\mathcal{E}G\times\mathcal{E}G$ are so and $G$ acts properly on them. So $BG$ and $BG^{(2)}$ are paracompact. Hence the pull-backs of $\mathcal{B}$ along the maps $\check{p}_{1},\check{p}_{2}\colon BG^{(2)}\rightrightarrows BG$ are isomorphic as locally trivial bundles over $BG^{(2)}$ (see [Husemoller:Fibre_bundles]*Theorem 2.9.9 on p. 51). Hence they are $\mathrm{KK}^{BG^{(2)}}$-equivalent as needed.

Finally, the constructions above all preserve nuclearity. So the subcategory of all $G$-actions on separable nuclear $\mathrm{C}^{*}$-algebras is equivalent to the subcategory of all locally trivial bundles of nuclear $\mathrm{C}^{*}$-algebras over $BG$. ∎

Remark 3.8*.*

The equivalence from $\mathrm{KK}^{G}$ to a full subcategory of $\mathrm{KK}^{BG}$ maps a $\mathrm{C}^{*}$-algebra $A$ with $G$-action $\alpha$ to a locally trivial bundle of $\mathrm{C}^{*}$-algebras with fibre $A$. This is equivalent to a locally trivial $\operatorname{Aut}(A)$-principal bundle. The latter is equal to the principal bundle $\mathcal{P}_{\alpha}$ that appears in the conjectures of Izumi [Izumi:Group_actions].

Definition 3.9**.**

An isomorphism or $\mathrm{KK}^{G}$-equivalence $A\to A$ is called $\mathrm{KK}$-trivial if the forgetful functor maps it to the identity in $\mathrm{KK}(A,A)$. An isomorphism of $\operatorname{Aut}(A)$-principal bundles is $\mathrm{KK}$-trivial if the induced isomorphism of locally trivial $\mathrm{C}^{*}$-algebra bundles with fibre $A$ restricts to the identity on $A$ in each fibre.

The notion of a $\mathrm{KK}$-trivial isomorphism of $\operatorname{Aut}(A)$-principal bundles is well defined, that is, it does not depend on the trivialisation of the locally trivial bundle. This is because the identity in $\mathrm{KK}(A,A)$ is invariant under conjugation by automorphisms.

Theorem 3.10**.**

Let $A$ be a purely infinite, simple, nuclear, $\mathrm{C}^{*}$-stable, separable $\mathrm{C}^{*}$-algebra and let $G$ be a torsion-free, amenable, second countable, locally compact group. Let $BG$ be a second countable, locally compact classifying space for $G$. Then ($\mathrm{KK}$-trivial)* $\mathrm{KK}^{G}$-equivalence classes of $G$-actions on $A$ are in bijection with ($\mathrm{KK}$-trivial*)* isomorphism classes of $\operatorname{Aut}(A)$-principal bundles over $BG$.*

Proof.

Let $\alpha,\beta$ be two actions of $G$ on $A$. Let $\mathcal{A}$ and $\mathcal{B}$ be the corresponding locally trivial bundles of $\mathrm{C}^{*}$-algebras over $BG$. Theorem 3.7 shows that

[TABLE]

In particular, our functor maps $\mathrm{KK}^{G}$-equivalences between $(A,\alpha)$ and $(B,\beta)$ bijectively to $\mathrm{KK}^{BG}$-equivalences between $\mathcal{A}$ and $\mathcal{B}$. Since the fibres of the bundles $\mathcal{A}$ and $\mathcal{B}$ are isomorphic to the Kirchberg algebra $A$, these bundles are purely infinite. We claim that they are even strongly purely infinite. If $BG$ is finite-dimensional, this follows from [Blanchard-Kirchberg:Non-simple_infinite]*Theorem 5.8. If $BG$ is infinite-dimensional, the result is still true for a different reason. It is easy to see that $\mathcal{A}$ is (“weakly”) purely infinite. Since $\mathrm{C}_{0}(BG)$ is in the centre of the multiplier algebra of $\mathcal{A}$ and $A$ has an approximate unit of projections, it follows that $\mathcal{A}$ has the “locally central decomposition property” (see [Kirchberg-Rordam:Infinite_absorbing]*Definition 6.1). And a purely infinite $\mathrm{C}^{*}$-algebra with the the locally central decomposition property is strongly purely infinite by [Kirchberg-Rordam:Infinite_absorbing]*Theorem 6.8. Then $\mathcal{A}\cong\mathcal{A}\otimes\mathcal{O}_{\infty}$ by [Kirchberg-Rordam:Infinite_absorbing]*Theorem 8.6. In addition, $\mathcal{A}$ is $\mathrm{C}^{*}$-stable by [Hirshberg-Rordam-Winter:Stability]*Proposition 3.9; I thank the referee for clarifying the need for this reference.

Hence $\mathcal{A}$ and $\mathcal{B}$ are covered by Kirchberg’s Classification Theorem (see [Kirchberg:Michael]*Folgerung 4.3). Namely, any $\mathrm{KK}^{BG}$-equivalence $\mathcal{A}\cong\mathcal{B}$ is induced by an isomorphism $\mathcal{A}\cong\mathcal{B}$ as $\mathrm{C}_{0}(BG)$-$\mathrm{C}^{*}$-algebras. We are also interested in $\mathrm{KK}^{BG}$-equivalences between two actions $\alpha,\beta$ on the same stable Kirchberg algebra $A$ that restrict to the identity in $\mathrm{KK}(A,A)$ in each fibre. This happens if and only if the corresponding equivalence in $\mathrm{KK}^{G}((A,\alpha),A,\beta))$ is $\mathrm{KK}$-trivial. We see that any $\mathrm{KK}$-trivial equivalence in $\mathrm{KK}^{G}((A,\alpha),A,\beta))$ comes from a $\mathrm{KK}$-trivial isomorphism $\mathcal{A}\cong\mathcal{B}$ as $\mathrm{C}_{0}(BG)$-$\mathrm{C}^{*}$-algebras. And the latter are the same as $\mathrm{KK}$-trivial isomorphisms of $\operatorname{Aut}(A)$-principal bundles.

To prove the desired bijection, we must show that any locally trivial bundle of $\mathrm{C}^{*}$-algebras $\mathcal{A}$ over $BG$ with fibre $A$ comes from some action $\alpha$ on $A$. To begin with, Theorem 3.7 shows that there is some object $(B,\beta)$ of $\mathrm{KK}^{G}$ so that $B$ is nuclear and the locally trivial $\mathrm{C}^{*}$-algebra bundle over $BG$ associated to $(B,\beta)$ is $\mathrm{KK}^{BG}$-equivalent to $\mathcal{A}$. Theorem 2.1 gives a $\mathrm{KK}^{G}$-equivalence from $(B,\beta)$ to some action $(\tilde{B},\tilde{\delta})$ where $\tilde{B}$ is purely infinite, simple, nuclear, $\mathrm{C}^{*}$-stable and separable. By the Kirchberg–Phillips Theorem, $\tilde{B}\cong A$. So the action $\tilde{\delta}$ transfers to an action $\alpha$ on $A$ so that the associated locally trivial $\mathrm{C}^{*}$-algebra bundle over $BG$ is $\mathrm{KK}^{BG}$-equivalent to our given $\mathcal{A}$. As above, this $\mathrm{KK}^{BG}$-equivalence may be improved to an isomorphism. And if the isomorphism is not yet $\mathrm{KK}$-trivial, we simply transfer the action along this isomorphism to get another action on $A$ such that the resulting bundle over $BG$ is even $\mathrm{KK}$-trivially isomorphic to the given bundle. ∎

Example 3.11*.*

Let $G=\mathbb{R}^{n}$. This group is torsion-free and the translation action on $\mathbb{R}^{n}$ is a model for $\mathcal{E}G$ with $BG=\mathrm{pt}$. So Theorem 3.10 implies that all $\mathbb{R}^{n}$-actions on $A$ are $\mathrm{KK}$-trivially $\mathrm{KK}^{G}$-equivalent to the trivial action.

Example 3.12*.*

Let $G=\mathbb{Z}$. The translation action of $\mathbb{Z}$ on $\mathbb{R}$ is a universal proper action. So $BG=\mathbb{R}/\mathbb{Z}$ is a circle. Isomorphism classes of $\operatorname{Aut}(A)$-principal bundles over the circle are naturally in bijection with the path connected components of $\operatorname{Aut}(A)$. Any automorphism of $A$ determines an action of $\mathbb{Z}$. Thus any $\operatorname{Aut}(A)$-principal bundle over the circle comes from an action of $\mathbb{Z}$ on $A$. Theorem 3.10 says, in addition, that two $\mathbb{Z}$-actions are $\mathrm{KK}$-trivially $\mathrm{KK}^{G}$-equivalent if and only if the generating automorphisms of $A$ are homotopic.

Example 3.13*.*

Let $G=\mathbb{Z}^{2}$. Then the translation action of $\mathbb{Z}^{2}$ on $\mathbb{R}^{2}$ is a universal proper action. So $BG=\mathbb{R}^{2}/\mathbb{Z}^{2}$ is a $2$-torus. Theorem 3.10 says that any $\operatorname{Aut}(A)$-principal bundle over the $2$-torus is associated to an action of $\mathbb{Z}^{2}$ on $A$. Using the standard CW-complex decomposition of the $2$-torus and some obstruction theory, we may describe an $\operatorname{Aut}(A)$-principal bundle by two elements $\alpha_{1},\alpha_{2}$ of $\operatorname{Aut}(A)$ together with a homotopy $H$ between $\alpha_{1}\alpha_{2}$ and $\alpha_{2}\alpha_{1}$ in $\operatorname{Aut}(A)$. The existence part of Theorem 3.10 says that there are automorphisms $\beta_{1},\beta_{2}$ of $A$ with $\beta_{1}\beta_{2}=\beta_{2}\beta_{1}$ and homotopies $K_{j}$ from $\alpha_{j}$ to $\beta_{j}$ for $j=1,2$, such that the concatenation of the homotopies $K_{2}\cdot K_{1}$ reversed, $H$ and $K_{1}\cdot K_{2}$ is homotopic to the constant homotopy on $\beta_{1}\beta_{2}=\beta_{2}\beta_{1}$. The pair $\beta_{1},\beta_{2}$ with $\beta_{1}\beta_{2}=\beta_{2}\beta_{1}$ is equivalent to an action of $\mathbb{Z}^{2}$.

Take two actions of $\mathbb{Z}^{2}$ on $A$ and describe the corresponding principal $\operatorname{Aut}(A)$-bundles through triples $(\alpha_{1},\alpha_{2},H)$ and $(\alpha_{1}^{\prime},\alpha_{2}^{\prime},H^{\prime})$ as above. Since we start with $\mathbb{Z}^{2}$-actions, we may arrange that $\alpha_{1}\alpha_{2}=\alpha_{2}\alpha_{1}$, $\alpha^{\prime}_{1}\alpha^{\prime}_{2}=\alpha^{\prime}_{2}\alpha^{\prime}_{1}$, and $H$ and $H^{\prime}$ are the constant homotopies. The principal bundles are isomorphic if and only if $\alpha_{j}$ is homotopic to $\alpha_{j}^{\prime}$ for $j=1,2$ in such a way that composing the resulting homotopy from $\alpha_{1}\alpha_{2}\alpha_{1}^{-1}\alpha_{2}^{-1}=\mathrm{id}_{A}$ to $\alpha_{1}^{\prime}\alpha_{2}^{\prime}(\alpha_{1}^{\prime})^{-1}(\alpha_{2}^{\prime})^{-1}=\mathrm{id}_{A}$ is homotopic to the constant homotopy. The uniqueness part of Theorem 3.10 says that this happens if and only if the two $\mathbb{Z}^{2}$-actions on $A$ are $\mathrm{KK}$-trivially $\mathrm{KK}^{\mathbb{Z}^{2}}$-equivalent. Both the existence and uniqueness part of Theorem 3.10 are non-trivial in this case and depend on $A$ being a Kirchberg algebra.

These examples end the discussion of actions of torsion-free groups. We now turn to the opposite end: actions of cyclic groups of prime order.

4. The equivariant bootstrap class

Definition 4.1** ([dellAmbrogio-Emerson-Meyer:Equivariant_Lefschetz]*§3.1).**

Let $G$ be a compact group. A $G$-action $(A,\alpha)$ on a separable $\mathrm{C}^{*}$-algebra $A$ belongs to the $G$-equivariant bootstrap class $\mathfrak{B}^{G}$ if it is $\mathrm{KK}^{G}$-equivalent to an action $(B,\beta)$ where $B$ is a $\mathrm{C}^{*}$-algebra of Type I.

The equivariant bootstrap class is also described in [dellAmbrogio-Emerson-Meyer:Equivariant_Lefschetz] as the localising subcategory generated by certain “elementary” building blocks. These building blocks are actions of $G$ on $\mathrm{C}^{*}$-algebras that are contained in $\mathbb{K}(\mathcal{H})$ for a separable Hilbert space $\mathcal{H}$. Up to taking direct sums, we may get them all by induction applied to an action of a closed subgroup of $G$ on a matrix algebra. Actions on matrix algebras, in turn, correspond to projective representations. Two projective representations with the same cocycle give equivariantly Morita equivalent actions on matrix algebras. So it suffices to take one projective representation from each cohomology class.

The equivariant bootstrap class $\mathfrak{B}^{G}$ is closed under $\mathrm{KK}^{G}$-equivalence by definition. So $(B,\beta)$ in Theorem 2.1 belongs to the equivariant bootstrap class if $(A,\alpha)$ does.

From now on, we specialise to a finite cyclic group of prime order, that is, $G=\mathbb{Z}/p$ for a prime $p$. The only proper subgroup of $G$ is the trivial subgroup, and neither $G$ nor the trivial group have non-trivial projective representations. Thus the equivariant bootstrap class $\mathfrak{B}^{G}$ for $G$ is generated by two objects, namely, $\mathbb{C}=\mathrm{C}(G/G)$ with the trivial action of $G$ and $\mathrm{C}(G)$ with the translation action of $G$. Any $G$-action on a finite-dimensional $\mathrm{C}^{*}$-algebra is $G$-equivariantly Morita–Rieffel equivalent to a direct sum of copies of these two actions. Given a separable $G$-$\mathrm{C}^{*}$-algebra $A$, its little invariant is defined by

[TABLE]

where the isomorphism uses the well-known additivity of $\mathrm{KK}^{G}$.

The little invariant does not classify actions up to $\mathrm{KK}^{G}$-equivalence; that is, there are isomorphisms $L_{*}(A)\cong L_{*}(B)$ that do not lift to isomorphisms $A\cong B$ (see Example 9.9). It has enough information, however, to detect whether an object in the equivariant bootstrap class is non-zero:

Lemma 4.2**.**

Let $A,B\in\mathfrak{B}^{G}$. Then $L_{*}(A)\cong 0$ if and only if $A$ is $\mathrm{KK}^{G}$-equivalent to [math]. And $f\in\mathrm{KK}^{G}_{0}(A,B)$ is invertible if and only if $L_{*}(f)\colon L_{*}(A)\to L_{*}(B)$ is invertible.

Proof.

This uses basic triangulated category theory. Any arrow $f$ is part of an exact triangle

[TABLE]

Here the object $C$ of $\mathfrak{B}^{G}$ is unique up to isomorphism. It is called the cone of $f$. Versions of the Five Lemma show that $f$ is invertible if and only if $C\cong 0$ and that $L_{*}(f)$ is invertible if and only if $L_{*}(C)\cong 0$. So the second statement follows from the first statement for $C$. It is clear that $C\cong 0$ implies $L_{*}(C)\cong 0$. Conversely, assume $L_{*}(C)\cong 0$. Let $Z\subseteq\mathfrak{B}^{G}$ be the class of all objects $D$ with $\mathrm{KK}^{G}_{*}(D,C)=0$. This is localising and contains $\mathbb{C}$ and $\mathrm{C}(G)$ by assumption. Hence it contains all of $\mathfrak{B}^{G}$. So $\mathrm{KK}^{G}_{*}(C,C)=0$. This is equivalent to $C\cong 0$. ∎

Next we relate $L_{*}(A)$ to $\mathrm{K}$-theory. We use the Green–Julg isomorphism

[TABLE]

(see [Julg:K_equivariante] or [Blackadar:K-theory]*Theorem 20.2.7.(a)), the induction–restriction adjunction

[TABLE]

for the inclusion of the trivial subgroup in $G$ (see [Meyer-Nest:BC]*Section 3.2) and the isomorphism

[TABLE]

(see [Kasparov:Operator_K] or [Blackadar:K-theory]*Proposition 17.5.5). These give a natural isomorphism

[TABLE]

In particular, $L_{*}(A)$ is countable for any separable $G$-$\mathrm{C}^{*}$-algebra $A$.

The invariant $L_{*}(A)$ is a $\mathbb{Z}/2$-graded right module over the $\mathbb{Z}/2$-graded ring $\mathrm{KK}^{G}_{*}(\mathbb{C}\oplus\mathrm{C}(G),\mathbb{C}\oplus\mathrm{C}(G))$ by the Kasparov composition product. We will later describe modules as representations of the underlying ring. Then it becomes confusing to use right modules. Therefore, we use the opposite $\mathbb{Z}/2$-graded ring

[TABLE]

and treat $L_{*}(A)\cong\mathrm{KK}^{G}_{*}(\mathbb{C}\oplus\mathrm{C}(G),A)$ as a left $\mathfrak{T}$-module. We denote reverse order composition products in $\mathrm{KK}^{G}$ by the symbol $\otimes$; this symbol is also used by Kasparov, but with a subscript denoting the $\mathrm{C}^{*}$-algebra over which tensor products of $\mathrm{KK}$-cycles are balanced. We will never use exterior products here, so $\otimes$ for $\mathrm{KK}^{G}$-cycles will always mean the reverse-order composition product.

We may compute the underlying $\mathbb{Z}/2$-graded Abelian group of $\mathfrak{T}$ using the additivity of $\mathrm{KK}^{G}$ and the isomorphisms that relate $L_{*}(A)$ to $\mathrm{K}$-theory:

[TABLE]

In particular, the odd part of $\mathfrak{T}$ vanishes. The ring structure on $\mathfrak{T}$ is also transparent from this computation. First, the direct sum decomposition of $\mathfrak{T}$ used in (4.3) gives $\mathfrak{T}$ a $2\times 2$-matrix structure with elements written as

[TABLE]

These are multiplied like $2\times 2$-matrices. The diagonal entry $\mathrm{KK}^{G}_{0}(\mathbb{C},\mathbb{C})$ is isomorphic to the representation ring of $G$. It acts on $\mathrm{K}_{*}(A\rtimes G)\cong\mathrm{K}_{*}^{G}(A)$ by exterior tensor product of $G$-equivariant projective modules with representations because the Kasparov composition product with elements of $\mathrm{KK}^{G}_{0}(\mathbb{C},\mathbb{C})$ is equal to the exterior product. The diagonal entry $\mathrm{KK}^{G}_{0}(\mathrm{C}(G),\mathrm{C}(G))$ is isomorphic to the group ring of $G$, realised by the $\mathrm{KK}^{G}$-classes of translations $G\to G$, $x\mapsto g\cdot x$, for $g\in G$. Since $G$ is Abelian, the map on $\mathrm{KK}^{G}_{*}(\mathrm{C}(G),A)$ induced by a translation in $\mathrm{C}(G)$ is the same as the map induced by applying the corresponding automorphism $\alpha_{g}$ of $A$. Therefore, the induced action of $G$ on $\mathrm{KK}^{G}_{*}(\mathrm{C}(G),A)\cong\mathrm{K}_{*}(A)$ is the canonical one induced by the $G$-action on $A$.

Since $G$ is the cyclic group of order $p$, its representation ring and its group ring are canonically isomorphic to

[TABLE]

This ring will play a crucial role in our study. Next we explain the isomorphism between the representation ring and the group ring of $G$ using Baaj–Skandalis duality (see [Baaj-Skandalis:Hopf_KK]). This duality will generate an important involutive automorphism of Manuel Köhler’s ring as well.

Baaj–Skandalis duality for a locally compact group $G$ says that taking the crossed product and giving it the dual coaction defines an equivalence of $\mathbb{Z}/2$-graded categories

[TABLE]

For our group $G$, a coaction of $G$ is equivalent to an action of the dual group $\widehat{G}$, and $\widehat{G}\cong G$. So we get an automorphism $\mathrm{KK}^{G}\to\mathrm{KK}^{G}$, mapping a separable $G$-$\mathrm{C}^{*}$-algebra $A$ to $A\rtimes G$ with the dual action of $G$. Taking this isomorphism twice gives $(A\rtimes G)\rtimes\widehat{G}\cong A\otimes\mathbb{K}(\ell^{2}G)$, which is equivariantly Morita–Rieffel equivalent to $A$. Now $\mathrm{C}(G)\rtimes G\cong\mathbb{K}(\ell^{2}G)$ is equivariantly Morita–Rieffel equivalent to $\mathbb{C}$ and $\mathbb{C}\rtimes G\cong\mathrm{C}^{*}(G)\cong\mathrm{C}(G)$ equivariantly. So $(\mathbb{C}\oplus\mathrm{C}(G))\rtimes G\sim_{\mathrm{KK}^{G}}\mathbb{C}\oplus\mathrm{C}(G)$, but this $\mathrm{KK}^{G}$-equivalence exchanges the two summands. And Baaj–Skandalis duality contains an isomorphism

[TABLE]

Actually, the Baaj–Skandalis duality isomorphism depends on the choice of the isomorphism $\widehat{G}\cong G$ that is used to turn $G$-coactions into $G$-actions. Different choices differ by an automorphism $\varphi$ of $G$, and the resulting automorphisms $\mathrm{BS}$ of $\mathrm{KK}^{G}$ differ by the automorphism of $\mathrm{KK}^{G}$ that replaces an action $\alpha\colon G\to\operatorname{Aut}(A)$ by $\alpha\circ\varphi$. We choose the isomorphism $\widehat{G}\cong G$ that is given by the symmetric pairing

[TABLE]

For this pairing, the Baaj–Skandalis duality automorphism is involutive, that is, $\mathrm{BS}\circ\mathrm{BS}$ is naturally equivalent to the identity functor on $\mathrm{KK}^{G}$.

Baaj–Skandalis duality induces a ring automorphism of $\mathfrak{T}$, which exchanges the two copies of $\mathfrak{S}$ on the diagonal and the two copies of $\mathbb{Z}$ off the diagonal. One copy of $\mathbb{Z}$ is

[TABLE]

Both the representation ring and the group ring of $\mathbb{Z}$ act on this by the augmentation character

[TABLE]

This is because the regular representation $\lambda$ of $G$ satisfies $\chi\otimes\lambda\cong\lambda$ for all characters $\chi$ of $G$ and because translations in $G$ become inner automorphisms of $\mathrm{C}(G)\rtimes G$ and thus act trivially on $\mathrm{K}$-theory. This gives the multiplication of the diagonal entries with the $01$-entries of $2\times 2$-matrices in $\mathfrak{T}$. By Baaj–Skandalis duality, the multiplication of the $10$-entries with diagonal entries is given by the same formula.

To compute the multiplication of two off-diagonal entries, we describe the generators of these entries more concretely. Let $\alpha_{01}\colon\mathbb{C}\to\mathrm{C}(G)$ be the unit map. We also denote its class in $\mathrm{KK}^{G}_{0}(\mathbb{C},\mathrm{C}(G))$ by $\alpha_{01}$. The first isomorphism in (4.5) maps $\alpha_{01}$ to the projection onto the trivial representation of $G$ in $\ell^{2}(G)$. Since this has dimension $1$, it follows that $\alpha_{01}$ is a generator of $\mathrm{KK}^{G}_{0}(\mathbb{C},\mathrm{C}(G))\cong\mathbb{Z}$.

Baaj–Skandalis duality maps $\alpha_{01}$ to a generator

[TABLE]

for the $10$-summand $\mathrm{KK}^{G}_{0}(\mathrm{C}(G),\mathbb{C})\cong\mathbb{Z}$ in $\mathfrak{T}$. Concretely, this is represented by the $G$-equivariant ∗-homomorphism

[TABLE]

combined with the canonical equivariant Morita–Rieffel equivalence $\mathbb{K}(\ell^{2}G)\sim\mathbb{C}$. A short computation identifies this $G$-equivariant ∗-homomorphism with the canonical inclusion $\mathrm{C}(G)\to\mathbb{K}(\ell^{2}G)$ by pointwise multiplication.

Lemma 4.8**.**

The natural transformations $\mathrm{K}_{*}(A\rtimes G)\leftrightarrows\mathrm{K}_{*}(A)$ induced by $\alpha_{10}$ and $\alpha_{01}$ are equal to the natural transformations induced by the inclusions $A\hookrightarrow A\rtimes G\hookrightarrow A\otimes\mathbb{K}(\ell^{2}G)$.

Proof.

The inclusions $A\hookrightarrow A\rtimes G$ are natural and hence form a natural transformation $\eta_{A}\colon\mathrm{K}_{*}(A)\to\mathrm{K}_{*}(A\rtimes G)$. By the Yoneda Lemma, this must come from a class in $\mathrm{KK}^{G}_{0}(\mathbb{C},\mathrm{C}(G))$. Since $\mathrm{KK}^{G}_{0}(\mathbb{C},\mathrm{C}(G))\cong\mathbb{Z}$ with generator $\alpha_{01}$, the natural transformation $\eta$ must be a multiple of the natural transformation induced by $\alpha_{01}$. Now $\eta_{\mathbb{C}}$ maps the class of the unit in $\mathrm{K}_{0}(\mathbb{C})$ onto the class of the unit in $\mathrm{K}_{0}(\mathrm{C}^{*}(G))\cong\mathbb{Z}[G]$. So does the natural transformation induced by $\alpha_{01}$. This proves the claim for the inclusion $A\hookrightarrow A\rtimes G$. The other inclusion is treated similarly. ∎

The computation of $\alpha_{01}$ and $\alpha_{10}$ implies that $\alpha_{01}\otimes\alpha_{10}\in\mathrm{KK}^{G}_{0}(\mathbb{C},\mathbb{C})$ is represented by the unit map $\mathbb{C}\to\mathbb{K}(\ell^{2}G)$. Its image in the representation ring is the class of the regular representation. The isomorphism $\mathrm{KK}^{G}_{0}(\mathbb{C},\mathbb{C})\cong\mathfrak{S}$ maps this to the norm element

[TABLE]

because each character appears in the regular representation with multiplicity one. We compute the other composition using that Baaj–Skandalis duality is an involutive ring automorphism:

[TABLE]

This completely describes the multiplication in the ring $\mathfrak{T}$. The following lemma summarises our results:

Lemma 4.10**.**

The elements in the ring $\mathfrak{T}$ are equivalent to $2\times 2$-matrices

[TABLE]

and these are multiplied by

[TABLE]

The augmentation character $\tau$ and the norm element $N(t)$ are defined in (4.6) and (4.9). Baaj–Skandalis duality acts on this ring by the involutive automorphism

[TABLE]

The ring $\mathfrak{T}$ described in Lemma 4.10 is generated by the following elements:

[TABLE]

These are subject to the following relations:

1. (1)

$1_{0}$ and $1_{1}$ are idempotent and $1_{0}+1_{1}$ is a unit element in $\mathfrak{T}$; 2. (2)

$1_{j}\otimes t_{j}=t_{j}=t_{j}\otimes 1_{j}$ for $j=0,1$ and $1_{0}\otimes\alpha_{01}=\alpha_{01}=\alpha_{01}\otimes 1_{1}$, $1_{1}\otimes\alpha_{10}=\alpha_{10}=\alpha_{10}\otimes 1_{0}$; 3. (3)

$\alpha_{01}\otimes t_{1}=\alpha_{01}=t_{0}\otimes\alpha_{01}$ and $\alpha_{10}\otimes t_{0}=\alpha_{10}=t_{1}\otimes\alpha_{10}$; 4. (4)

$\alpha_{01}\otimes\alpha_{10}=N(t_{0})$ and $\alpha_{10}\otimes\alpha_{01}=N(t_{1})$.

The relations $t_{j}^{p}=1_{j}$ for $j=0,1$ follow from this by writing

[TABLE]

with $k=1-j$. So the generators $1_{j}$ are redundant.

5. Manuel Köhler’s invariant

Now we extend the invariant $L_{*}(A)\cong\mathrm{K}_{*}(A\rtimes G)\oplus\mathrm{K}_{*}(A)$ studied above as in the thesis of Manuel Köhler [Koehler:Thesis]. Let

[TABLE]

be the mapping cone of the embedding $\mathbb{C}\to\mathrm{C}(G)$ as constant functions. This $\mathrm{C}^{*}$-algebra is commutative. We identify it with $\mathrm{C}_{0}(X)$ for

[TABLE]

for any $\varphi_{0}\in\mathbb{R}$, with $G$ acting by rotations around the origin, cyclically permuting the rays in $X$. This particular description will be used in the proof of Proposition 5.5.

Adding $D$ as an extra generator does not enlarge the equivariant bootstrap class $\mathfrak{B}^{G}$. So $\mathfrak{B}^{G}$ is also the localising subcategory of $\mathrm{KK}^{G}$ generated by $\mathbb{C}$, $\mathrm{C}(G)$ and $D$. And the same localising subcategory is generated by the direct sum

[TABLE]

Here we use that idempotents in localising subcategories of $\mathrm{KK}^{G}$ split because they are triangulated categories with countable direct sums (see [Neeman:Triangulated]*§1.3).

Let $\mathfrak{K}\mathrel{\vcentcolon=}\mathrm{KK}^{G}_{*}(B,B)^{\mathrm{op}}$ be the opposite of the $\mathbb{Z}/2$-graded endomorphism ring of $B$ in $\mathrm{KK}^{G}$, that is, $\mathfrak{K}_{0}\mathrel{\vcentcolon=}\mathrm{KK}^{G}_{0}(B,B)^{\mathrm{op}}$ and $\mathfrak{K}_{1}\mathrel{\vcentcolon=}\mathrm{KK}^{G}_{1}(B,B)^{\mathrm{op}}$, with the Kasparov composition product in reverse order as composition. Let $A$ be an object of $\mathrm{KK}^{G}$. Then

[TABLE]

is a $\mathbb{Z}/2$-graded left $\mathfrak{K}$-module by the Kasparov product in reverse order.

Let $1_{0},1_{1},1_{2}$ be the projections to the summands $\mathbb{C}$, $\mathrm{C}(G)$ and $D$ in $B$, respectively. These are even elements of $\mathfrak{K}$, and they are orthogonal idempotents whose sum is the unit element in $\mathfrak{K}$, that is,

[TABLE]

Using these idempotent elements, we may describe $\mathfrak{K}$ as a ring of $3\times 3$-matrices with entries

[TABLE]

The multiplication follows the matrix multiplication pattern as for the ring $\mathfrak{T}$ that acts on the little invariant. The idempotents $1_{j}$ also decompose any $\mathbb{Z}/2$-graded $\mathfrak{K}$-module $M$ as a direct sum $M=M_{0}\oplus M_{1}\oplus M_{2}$ with $\mathbb{Z}/2$-graded subgroups $M_{j}\mathrel{\vcentcolon=}1_{j}\cdot M$. The action of $\mathfrak{K}$ on $M$ is given by matrix-vector multiplication when we write elements of $M$ as column vectors with entries in $M_{j}$ for $j=0,1,2$. We shall tacitly identify $\mathfrak{K}$-modules with such triples $(M_{j})_{j=0,1,2}$ in the following. Let $A$ be an object of $\mathrm{KK}^{G}$. Then

[TABLE]

where $\mathrm{K}_{*}$ denotes ordinary K-theory, because

[TABLE]

There are canonical equivariant ∗-homomorphisms

[TABLE]

namely, $\alpha_{12}\colon\Sigma\mathrm{C}(G)=\mathrm{C}_{0}(G\times\mathbb{R})\to D=\mathrm{C}_{0}(X)$ is the ideal inclusion corresponding to the subset $X\setminus\{0\}\subseteq X$, and $\alpha_{20}\colon\mathrm{C}_{0}(X)=D\to\mathbb{C}$ is evaluation at [math]. The maps in (5.2) form an exact triangle in $\mathrm{KK}^{G}$ (see [Meyer-Nest:BC]). This exact triangle induces a long exact sequence of $\mathrm{KK}$-groups, the Puppe sequence for $\mathbb{C}\to\mathrm{C}(G)$:

[TABLE]

This long exact sequence implies that $\mathrm{KK}_{*}^{G}(D,A)$ is countable because $\mathrm{K}_{*}(A)$ and $\mathrm{K}_{*}(A\rtimes G)$ are countable for any separable $\mathrm{C}^{*}$-algebra $A$. Thus $F_{*}(A)$ is a countable $\mathbb{Z}/2$-graded $\mathfrak{K}$-module. The functor $F$ from $\mathrm{KK}^{G}$ to the category of countable $\mathbb{Z}/2$-graded $\mathfrak{K}$-modules is the invariant studied by Manuel Köhler. This functor is a stable homological functor as in [Meyer-Nest:Homology_in_KK, Meyer:Homology_in_KK_II], so that the homological algebra machinery developed in these articles applies to it. Even more, $F$ is the universal stable homological functor defined by its kernel on morphisms. Roughly speaking, this says that the homological algebra in $\mathrm{KK}^{G}$ that is produced by $F$ is equivalent to the homological algebra in the category of $\mathbb{Z}/2$-graded countable $\mathfrak{K}$-modules. While this is an important point for the proof of the main theorem of [Koehler:Thesis], there is no need to discuss this here. The following is the main result of [Koehler:Thesis]:

Theorem 5.4** ([Koehler:Thesis]).**

Let $A,C$ be separable $G$-$\mathrm{C}^{*}$-algebras. Assume $A\in\mathfrak{B}^{G}$. Then there is a natural short exact sequence

[TABLE]

If $A,C\in\mathfrak{B}^{G}$, then an isomorphism $\varphi\colon F_{*}(A)\to F_{*}(C)$ of $\mathbb{Z}/2$-graded $\mathfrak{K}$-modules lifts to a $\mathrm{KK}^{G}$-equivalence in $\mathrm{KK}^{G}_{0}(A,C)$. In particular, $A,C$ are $\mathrm{KK}^{G}$-equivalent if and only if $F_{*}(A)\cong F_{*}(C)$ as $\mathbb{Z}/2$-graded $\mathfrak{K}$-modules.

The main part in the proof of the Universal Coefficient Theorem exact sequence in Theorem 5.4 is to show that $F_{*}(A)$ for $A$ in $\mathrm{KK}^{G}$ has a projective resolution of length $1$ in the category of $\mathbb{Z}/2$-graded $\mathfrak{K}$-modules. The naturality of the UCT exact sequence means that composing the image of $\operatorname{Ext}^{1}_{\mathfrak{K}}$ with an element $\varphi\in\mathrm{KK}^{G}_{*}(A,C)$ is equivalent to composing with $F_{*}(\varphi)$ in the Ext-theory over $\mathfrak{K}$. In particular, the $\operatorname{Ext}^{1}_{\mathfrak{K}}$-part is nilpotent. Therefore, any lifting of an isomorphism in $\operatorname{Hom}_{\mathfrak{K}}(F_{*}(A),F_{*}(C))$ is invertible in $\mathrm{KK}^{G}$, provided the UCT applies to both $A$ and $C$. The general machinery of homological algebra in triangulated categories in [Meyer-Nest:Homology_in_KK] also shows that a separable $G$-$\mathrm{C}^{*}$-algebra $A$ belongs to $\mathfrak{B}^{G}$ if and only if $\mathrm{KK}^{G}_{*}(A,C)$ is computed by the UCT for all $C$, if and only if $\mathrm{KK}^{G}_{*}(A,C)=0$ for all $C$ with $F_{*}(C)=0$.

In addition to Theorem 5.4, Köhler also describes the range of the invariant $F_{*}$. This result is particularly important for us. Put in a nutshell, a countable $\mathfrak{K}$-module belongs to the essential range of $F_{*}$ if and only if two sequences built from the $\mathfrak{K}$-module structure are exact. One of these exact sequences is the Puppe sequence in (5.3). The most elegant way to get the second exact sequence uses that the object $D$ above is self-dual with respect to Baaj–Skandalis duality. This is a special feature of working with $\mathrm{C}^{*}$-algebras and equivariant $\mathrm{KK}^{G}$. It seems to have no analogue in the homological algebra of $\mathfrak{S}$-modules or in the algebraic topology of $G$-spaces.

Proposition 5.5**.**

$\mathrm{BS}(D)\cong\Sigma D$, where $\Sigma$ denotes suspension. Hence Baaj–Skandalis duality induces an involutive automorphism of the ring $\mathfrak{K}$ and a compatible isomorphism $F_{*}(A)\cong F_{*}(\mathrm{BS}(A))$ – but these do not preserve the $\mathbb{Z}/2$-gradings.

Proof.

This is shown in [Koehler:Thesis]. We sketch a proof that uses the geometric description of equivariant bivariant $\mathrm{K}$-theory developed in [Emerson-Meyer:Correspondences]. Define $\alpha_{10}\in\mathrm{KK}^{G}_{0}(\mathrm{C}(G),\mathbb{C})$ as in (4.7). Then $\mathrm{BS}(D)$ is a mapping cone for $\alpha_{10}$. We shall describe this cone in a different way to show that it is $\mathrm{KK}^{G}$-equivalent to $\Sigma D$. The groups $\mathrm{KK}^{G}_{*}(\mathrm{C}_{0}(X),\mathrm{C}_{0}(Y))$ for two $G$-manifolds $X,Y$ are described in [Emerson-Meyer:Correspondences] through equivariant correspondences, following Connes and Skandalis [Connes-Skandalis:Longitudinal]. In particular, some geometric $\mathrm{KK}^{G}$-cycle must give $\alpha_{10}$. Actually, the geometric cycles that we need are only wrong-way elements of equivariantly $\mathrm{K}$-oriented smooth maps, which are treated in [Emerson-Meyer:Normal_maps]. We claim that $\alpha_{10}$ is the wrong-way element $p!$ associated to the constant map $p\colon G\to\mathrm{pt}$ with the obvious $\mathrm{K}$-orientation, where $\mathrm{pt}$ denotes the one-point space. The Kasparov cycle associated to $p!$ for a surjective submersion is, in general, represented by the Kasparov cycle of the family of Dirac operators along the fibres of the submersion; this is a form of the analytic index map. In our [math]-dimensional case, this boils down to the class in $\mathrm{KK}^{G}_{0}(\mathrm{C}(G),\mathrm{C}(\mathrm{pt}))$ of $\ell^{2}(G)$ with $\mathrm{C}(G)$ acting by poinwise multiplication and with zero Fredholm operator. This is exactly our $\alpha_{10}$. The analytic index is always equal to the topological index, which is defined using only Thom isomorphisms and open embeddings. In general, the topological index map for a smooth $\mathrm{K}$-oriented map $X\to Y$ uses an equivariant embedding of $X$ into $\mathbb{R}^{n}$ with some linear action of $G$. In the case at hand, let $\mathbb{R}^{2}_{\varrho}$ denote $\mathbb{R}^{2}$ with the action of $G=\mathbb{Z}/p$ by rotations by angles $2\pi k/p$; this is $\mathrm{K}$-oriented in a canonical way. Then

[TABLE]

is a $G$-equivariant $\mathrm{K}$-oriented embedding. Its normal bundle is $G\times\mathbb{R}^{2}_{\varrho}\to G$. We choose a diffeomorphism from $G\times\mathbb{R}^{2}_{\varrho}$ onto an open neighbourhood of $G$ in $\mathbb{R}^{2}_{\varrho}$. The topological index element $p!_{\mathrm{top}}\in\mathrm{KK}^{G}_{0}(\mathrm{C}(G),\mathbb{C})$ is the composite of the Thom isomorphism class in $\mathrm{KK}^{G}_{0}(\mathrm{C}(G),\mathrm{C}_{0}(G\times\mathbb{R}^{2}_{\varrho}))$, the inclusion $\mathrm{C}_{0}(G\times\mathbb{R}^{2}_{\varrho})\hookrightarrow\mathrm{C}_{0}(\mathbb{R}^{2}_{\varrho})$ as an ideal from the open inclusion $G\times\mathbb{R}^{2}_{\varrho}\hookrightarrow\mathbb{R}^{2}_{\varrho}$, and the class of the Bott isomorphism in $\mathrm{KK}^{G}_{0}(\mathrm{C}_{0}(\mathbb{R}^{2}_{\varrho}),\mathbb{C})$. The general index theorem in [Emerson-Meyer:Normal_maps]*Theorem 6.1 contains the statement that $p!=p!_{\mathrm{top}}$ in $\mathrm{KK}^{G}_{0}(\mathrm{C}(G),\mathbb{C})$. So $\mathrm{BS}(D)$ is also a cone for $p!_{\mathrm{top}}$. Here the “cone” is defined as the third entry of an exact triangle

[TABLE]

using the triangulated category structure on $\mathrm{KK}^{G}$. This cone is determined uniquely up to $\mathrm{KK}^{G}$-equivalence. The $\mathrm{KK}^{G}$-cycle above does not yet come from an equivariant ∗-homomorphism. But since the Thom isomorphism and Bott isomorphism classes are invertible in $\mathrm{KK}^{G}$, a cone for the ideal inclusion $\mathrm{C}_{0}(G\times\mathbb{R}^{2}_{\varrho})\hookrightarrow\mathrm{C}_{0}(\mathbb{R}^{2}_{\varrho})$ is also a cone for $p!_{\mathrm{top}}$. The extension $\mathrm{C}_{0}(G\times\mathbb{R}^{2}_{\varrho})\rightarrowtail\mathrm{C}_{0}(\mathbb{R}^{2}_{\varrho})\twoheadrightarrow\mathrm{C}_{0}(\mathbb{R}^{2}_{\varrho}\setminus(G\times\mathbb{R}^{2}_{\varrho}))$ has long exact sequences in $\mathrm{KK}^{G}$. So there is an exact triangle

[TABLE]

in $\mathrm{KK}^{G}$. Hence the suspension of $\mathrm{C}_{0}(\mathbb{R}^{2}_{\varrho}\setminus(G\times\mathbb{R}^{2}_{\varrho}))$ is also a cone of $p!_{\mathrm{top}}$. Now we choose the tubular neighbourhood diffeomorphism $G\times\mathbb{R}^{2}_{\varrho}\to\mathbb{R}^{2}_{\varrho}$ so that each $\{k\}\times\mathbb{R}^{2}$ is mapped diffeomorphically onto an entire open sector

[TABLE]

Its complement in $\mathbb{R}^{2}$ is the union of the rays $(r\cos\varphi,r\sin\varphi)$ with $0\leq r<\infty$ and $\varphi\in(2k-1)\cdot\pi/p$, $k\in\mathbb{Z}/p$. This is diffeomorphic to the space $X$ in (5.1). So Baaj–Skandalis duality maps $D$ to something $\mathrm{KK}^{G}$-equivalent to $\Sigma D$.

Now we replace the generator $B$ by $B\oplus\Sigma B$. This is more precise because we are, anyway, taking the $\mathbb{Z}/2$-graded group $\mathrm{KK}^{G}_{*}(B,A)=\mathrm{KK}^{G}_{0}(B,A)\oplus\mathrm{KK}^{G}_{0}(\Sigma B,A)$. The isomorphisms $\mathrm{BS}(\mathbb{C})\cong\mathrm{C}(G)$, $\mathrm{BS}(\mathrm{C}(G))\cong\mathbb{C}$, $\mathrm{BS}(D)\cong\Sigma D$ in $\mathrm{KK}^{G}$ and their suspensions combine to an isomorphism $\mathrm{BS}(B\oplus\Sigma B)\cong B\oplus\Sigma B$. This induces an involutive automorphism $\mathrm{BS}$ of the ring $\mathfrak{K}=\mathrm{KK}^{G}_{0}(B\oplus\Sigma B,B\oplus\Sigma B)^{\mathrm{op}}$, and it induces an isomorphism

[TABLE]

which is compatible with the $\mathfrak{K}$-module structure in the sense that $\mathrm{BS}(x)\otimes\mathrm{BS}(y)=\mathrm{BS}(x\otimes y)$ for all $x\in\mathfrak{K}$, $y\in F_{*}(A)$. We must be aware, however, that $\mathrm{BS}$ does not preserve the $\mathbb{Z}/2$-gradings on $\mathfrak{K}$ and $F_{*}(A)$. When we decompose them according to the direct sum decomposition $B=\mathbb{C}\oplus\mathrm{C}(G)\oplus D$, then the grading is preserved on $\mathbb{C}$ and $\mathrm{C}(G)$ and reversed on $D$. ∎

Baaj–Skandalis duality applied to the exact triangle (5.2) gives another exact triangle in $\mathrm{KK}^{G}$:

[TABLE]

The $\mathrm{KK}^{G}$-equivalences $\mathrm{BS}(\mathrm{C}(G))\sim\mathbb{C}$, $\mathrm{BS}(\mathbb{C})\sim\mathrm{C}(G)$ and $\mathrm{BS}(D)\sim\Sigma D$ show that it is isomorphic to an exact triangle

[TABLE]

Implicitly, this defines $\alpha_{02}$, $\alpha_{21}$ and $\alpha_{10}$ as the images of $\alpha_{12}$, $\alpha_{20}$ and $\alpha_{01}$ under $\mathrm{BS}$, composed with the chosen $\mathrm{KK}^{G}$-equivalences. In other words, the automorphism $\mathrm{BS}\colon\mathfrak{K}\to\mathfrak{K}$ satisfies

[TABLE]

Here we have viewed the $\mathrm{KK}^{G}$-classes of $\alpha_{jk}$ as elements of $\mathfrak{K}$. Since $\mathrm{BS}$ is an involution, we also get

[TABLE]

The elements $\alpha_{jk}\in\mathfrak{K}$ are even or odd depending on the suspensions in (5.2) and (5.6). Namely, $\alpha_{12}$ and $\alpha_{21}$ are odd and $\alpha_{20}$, $\alpha_{01}$, $\alpha_{02}$, $\alpha_{10}$ are even. Since $F_{*}$ is a homological functor, the exact triangle (5.6) induces a second exact sequence, called the dual Puppe sequence,

[TABLE]

Definition 5.8**.**

Let $M$ be a $\mathfrak{K}$-module, $j,k\in\{0,1,2\}$ and $x\in\mathfrak{K}$ with $x=1_{j}\otimes x\otimes 1_{k}$. Write $M=M_{0}\oplus M_{1}\oplus M_{2}$ as above. We write $x^{M}$ for the map $M_{k}\to M_{j}$, $y\mapsto y\otimes x$. In particular, this defines $\alpha_{jk}^{M}$ for $j\neq k$. Later, we will also get elements $t_{j}^{M}$ for $j=0,2$ and $s_{j}^{M}$ for $j=1,2$.

We can now characterise the range of the invariant $F_{*}$.

Definition 5.9**.**

A $\mathfrak{K}$-module $M$ is exact if the two sequences of Abelian groups

[TABLE]

are exact.

Theorem 5.10** ([Koehler:Thesis]).**

Let $M$ be a countable $\mathbb{Z}/2$-graded $\mathfrak{K}$-module. The following are equivalent:

1. (1)

$M=F_{*}(A)$* for some $A$ in $\mathfrak{B}^{G}$;* 2. (2)

$M=F_{*}(A)$* for some $A$ in $\mathrm{KK}^{G}$;* 3. (3)

$M$* is exact;* 4. (4)

$M$* has a projective $\mathfrak{K}$-module resolution of length $1$.*

Straightforward parts of the proof.

It is trivial that (1) implies (2). Since the Puppe sequence (5.3) and the dual Puppe sequence (5.7) are exact, (2) implies (3). Condition (4) implies (1) by lifting projective resolutions in the Abelian approximation to projective resolutions in the triangulated category (see [Meyer-Nest:Homology_in_KK]*Theorem 59). The highly non-trivial point is that (3) implies (4), that is, exact modules have projective resolutions of length $1$. We refer to [Koehler:Thesis] for this. This together with the abstract UCT in [Meyer-Nest:Homology_in_KK]*Theorem 66 also implies Theorem 5.4. ∎

5.1. Generators and relations for Köhler’s ring

We are going to describe the ring $\mathfrak{K}$ by generators and relations. This makes it easier to apply Theorems 5.4 and 5.10 to classify objects of $\mathfrak{B}^{G}$ up to $\mathrm{KK}^{G}$-equivalence. Recall that

[TABLE]

Theorem 5.11**.**

The ring $\mathfrak{K}$ is the universal ring generated by the elements $1_{j}$ for $j=0,1,2$ and $\alpha_{jk}$ for $0\leq j,k\leq 2$ with $j\neq k$ with the following relations:

[TABLE]

The $\mathbb{Z}/2$-grading on $\mathfrak{K}$ is such that $\alpha_{12}$ and $\alpha_{21}$ are odd and all other generators are even.

This theorem follows from the computations in [Koehler:Thesis], but is not stated there as such. It will take a while to prove it. The relations (5.12) and (5.13) say that the elements $1_{j}$ are orthogonal idempotents that sum up to $1$. This is equivalent to the $3\times 3$-matrix decomposition of the ring $\mathfrak{K}$. Equation (5.14) says that $\alpha_{jk}$ belongs to the $j,k$-th entry in the matrix description of $\mathfrak{K}$. Condition (5.15) consists of six equations and says that the composite of any two consecutive maps in the two exact triangles (5.2) and (5.6) vanishes – as needed for an exact triangle. Our next goal is to understand the relations (5.16) and (5.17). First we add some convenient definitions to the relations above:

[TABLE]

Since $\alpha_{jk}$ belongs to the $j,k$-component of $\mathfrak{K}$, $t_{j}$ for $j=0,2$ and $s_{j}$ for $j=1,2$ belong to the $j,j$-component of $\mathfrak{K}$, respectively. The elements $t_{j}$ and $s_{j}$ are even. The relations (5.16), (5.17) and (5.18) in Theorem 5.11 say that

[TABLE]

The definitions of $t_{j}$ and $s_{j}$ say that

[TABLE]

The equations (5.20), (5.21) and (5.23)–(5.26) express the products $\alpha_{jk}\otimes\alpha_{kj}$ for all $j,k\in\{0,1,2\}$ with $j\neq k$ as a polynomial in $t_{j}$ or $s_{j}$. All other products of two $\alpha$-generators vanish by (5.15). Before we prove Theorem 5.11, we list further relations among the elements $t_{j}$, $s_{j}$ and $\alpha_{jk}$ that follow from the relations in Theorem 5.11.

Proposition 5.27**.**

The relations in Theorem 5.11 imply the following relations:

[TABLE]

Proof.

Equations (5.31), (5.33), (5.37) and (5.39) follow from the definitions of the elements $t_{j}$ and $s_{j}$. One of these computations is

[TABLE]

The others are the same with different indices. The definitions of the elements $t_{j}$ and $s_{j}$ and Equation (5.15) imply (5.30), (5.32), (5.34), (5.35), (5.36), (5.38), (5.40) and (5.41). One of these computations is

[TABLE]

The others are the same with different indices. A similar computation shows (5.42) and (5.43). One computation suffices:

[TABLE]

Equations (5.22) and (5.38) imply (5.44):

[TABLE]

Similar computations give (5.47), (5.48) and (5.51). Equation (5.45) holds because

[TABLE]

Similar computations give (5.46), (5.49), (5.50). We use $t^{p}-1=-N(t)(1-t)$ and (5.15) to check $t_{0}^{p}=1_{0}$:

[TABLE]

A similar computation gives $s_{1}^{p}-1_{1}=0$. Finally,

[TABLE]

This vanishes by (5.43) because the polynomial $p-N$ vanishes at $1$ and therefore may be written as $(1-s)f(s)$ for some polynomial $f$. A similar computations gives $1_{2}-s_{2}^{p}=0$. ∎

As a first step in the proof of Theorem 5.11, we justify the definitions of $t_{0}$ and $s_{1}$ above. Namely, we show that these elements are the canonical generators of $\mathfrak{S}\mathrel{\vcentcolon=}\mathbb{Z}[t]/(t^{p}-1)$ in the description of $\mathfrak{T}$ in Lemma 4.10.

Lemma 5.52**.**

Describe $\mathrm{KK}^{G}_{0}(\mathbb{C},\mathbb{C})$ and $\mathrm{KK}^{G}_{0}(\mathrm{C}(G),\mathrm{C}(G))$ as $\mathfrak{S}\mathrel{\vcentcolon=}\mathbb{Z}[t]/(t^{p}-1)$ as in Lemma 4.10. Then $\alpha_{02}\otimes\alpha_{20}$ and $\alpha_{12}\otimes\alpha_{21}$ become $1-t\in\mathfrak{S}$.

Proof.

The proof of Proposition 5.5 describes $\alpha_{02}$ as the composite of the Thom isomorphism in $\mathrm{KK}^{G}_{0}(\mathbb{C},\mathrm{C}_{0}(\mathbb{R}^{2}_{\varrho}))$ and the restriction map $\mathrm{C}_{0}(\mathbb{R}^{2}_{\varrho})\to\mathrm{C}_{0}(X)=D$. And $\alpha_{20}$ is represented by the ∗-homomorphism $\mathrm{C}_{0}(X)\to\mathbb{C}$, $f\mapsto f(0)$. Thus $\alpha_{02}\otimes\alpha_{20}$ is represented by the composite of the Thom isomorphism in $\mathrm{KK}^{G}_{0}(\mathbb{C},\mathrm{C}_{0}(\mathbb{R}^{2}_{\varrho}))$ and the evaluation map $\mathrm{C}_{0}(\mathbb{R}^{2}_{\varrho})\to\mathbb{C}$, $f\mapsto f(0)$. The Thom isomorphism uses the $\mathrm{K}$-orientation on $\mathbb{R}^{2}_{\varrho}$ that is given by the obvious complex structure on the tangent bundle. Namely, $T(\mathbb{R}^{2}_{\varrho})\cong\mathbb{C}_{\varrho}$, where the action of $G$ on $\mathbb{C}_{\varrho}$ is given by the character $k\bmod p\mapsto\exp(2\pi\mathrm{i}k/p)$. The Thom isomorphism involves a Fredholm operator on the $\mathbb{Z}/2$-graded vector bundle $\Lambda^{*}T(\mathbb{R}^{2}_{\varrho})=\mathbb{C}_{1}\ominus\mathbb{C}_{\varrho}$, where $\mathbb{C}_{1}$ denotes the trivial vector bundle with the trivial action of $G$ in even parity, and $\ominus$ signifies that $\mathbb{C}_{\varrho}$ has odd parity. Restricting to [math], the Fredholm operator becomes redundant and we get the class in the representation ring $\mathrm{KK}^{G}_{0}(\mathbb{C},\mathbb{C})$ of $1\ominus\varrho$, the trivial representation minus the character $\varrho$. This corresponds to the element $1-t$ in $\mathrm{KK}^{G}_{0}(\mathbb{C},\mathbb{C})\cong\mathfrak{S}$.

The ring automorphism $\mathrm{BS}$ maps $\alpha_{02}\otimes\alpha_{20}$ to $\alpha_{12}\otimes\alpha_{21}$ and it induces the identity map on $\mathfrak{S}$. So the claim about $\alpha_{12}\otimes\alpha_{21}$ follows. ∎

Lemma 5.52 shows that we do not need the generators $t_{0}$ and $s_{1}$ of $\mathfrak{T}$ as generators of $\mathfrak{K}$. Together with the relations of $\mathfrak{T}$ listed after Lemma 4.10, Lemma 5.52 implies the relations (5.16) and (5.17) in Theorem 5.11.

The ring $\mathfrak{T}$ described in Lemma 4.10 gives four of the nine matrix entries of the ring $\mathfrak{K}$. Now we describe the remaining five. First, we describe $\mathrm{KK}^{G}_{*}(\mathbb{C},D)$. The exact triangle (5.2) induces an exact sequence

[TABLE]

Lemma 4.10 shows that the map $(\alpha_{01})_{*}$ here is equivalent to the augmentation character $\tau\colon\mathfrak{S}\to\mathbb{Z}$, which is surjective. So $\mathrm{KK}^{G}_{1}(\mathbb{C},D)=0$ and $(\alpha_{20})_{*}$ is an isomorphism $\mathrm{KK}^{G}_{0}(\mathbb{C},D)\xrightarrow{\sim}\ker\tau$. Now $\ker\tau$ is the principal ideal generated by $t-1\in\mathfrak{S}$, and it is isomorphic to $\mathbb{Z}[t]/(N(t))$ because $f\in\mathbb{Z}[t]$ satisfies $t^{p}-1\mid(t-1)f$ if and only if $N(t)\mid f$. Lemma 5.52 shows that $1-t\in\mathfrak{S}$ is the composite $\alpha_{02}\otimes\alpha_{20}$. Hence there is an isomorphism

[TABLE]

Then Baaj–Skandalis duality implies $\mathrm{KK}^{G}_{0}(\mathrm{C}(G),D)=0$ and that the following map is an isomorphism:

[TABLE]

The exact triangle (5.2) also implies an exact sequence for $\mathrm{KK}^{G}_{*}({\ldots},\mathbb{C})$. This has the following form:

[TABLE]

Lemma 4.10 shows that the map $(\alpha_{01})^{*}$ here is equivalent to the map $\mathbb{Z}\to\mathfrak{S}=\mathbb{Z}[t]/(t^{p}-1)$, $k\mapsto k\cdot N(t)$. This map is injective. As above, it follows that $\mathrm{KK}^{G}_{1}(D,\mathbb{C})=0$ and that there is an isomorphism

[TABLE]

Baaj–Skandalis duality now gives an analogous isomorphism

[TABLE]

Since $\alpha_{21}$ is odd, this component of $\mathfrak{K}$ is only odd. Finally, it remains to describe the diagonal entry $\mathrm{KK}^{G}_{*}(D,D)$. Here we use the cyclic long exact sequence

[TABLE]

Here we have used that $\mathrm{KK}^{G}_{*}(D,\mathbb{C})$ lives in even and $\mathrm{KK}^{G}_{*}(D,\mathrm{C}(G))$ in odd parity. So $\mathrm{KK}_{1}^{G}(D,D)=0$, the map $(\alpha_{12})_{*}\colon\mathrm{KK}^{G}_{1}(D,\mathrm{C}(G))\to\mathrm{KK}^{G}_{0}(D,D)$ is injective, and its cokernel is identified by $(\alpha_{20})_{*}$ with $\mathrm{KK}^{G}_{0}(D,\mathbb{C})$. Define $t_{2}$ and $s_{2}$ by (5.19). Then (5.25) and (5.26) say that

[TABLE]

It follows from the definition of $s_{1}$ above that

[TABLE]

The computations above imply that the elements of the form $\alpha_{21}\otimes(1_{1}-s_{1})^{j}$ for $j=0,\dotsc,p-2$ form a $\mathbb{Z}$-module basis of $\mathrm{KK}^{G}_{1}(D,\mathrm{C}(G))$. Their images in $\mathrm{KK}^{G}_{0}(D,D)$ are $(1_{2}-s_{2})^{j}$ for $j=1,\dotsc,p-1$. These form a $\mathbb{Z}$-module basis of the image of $\mathrm{KK}^{G}_{1}(D,\mathrm{C}(G))$ in $\mathrm{KK}^{G}_{0}(D,D)$. The definitions of $t_{0}$ and $t_{2}$ in (5.19) imply

[TABLE]

So $(1_{2}-t_{2})^{j}\otimes\alpha_{20}=\alpha_{20}\otimes(1_{0}-t_{0})^{j}$. These elements for $j=0,\dotsc,p-2$ form a $\mathbb{Z}$-module basis of $\mathrm{KK}^{G}_{0}(D,\mathbb{C})$ by the computation above. Putting things together, we conclude that the elements $(1_{2}-t_{2})^{j}$ for $j=0,\dotsc,p-2$ and $(1_{2}-s_{2})^{j}$ for $j=1,\dotsc,p-1$ form a $\mathbb{Z}$-module basis of $\mathrm{KK}^{G}_{0}(D,D)$.

Lemma 5.54**.**

$N(t_{2})+N(s_{2})=p$.

Proof.

The definitions of $t_{0}$ and $t_{2}$ in (5.19) imply $f(t_{2})\alpha_{20}=\alpha_{20}f(t_{0})$ for all polynomials $f$. Thus $N(t_{2})\alpha_{20}=\alpha_{20}N(t_{0})$. Using Lemmas 4.10 and 5.52, we rewrite this further as $\alpha_{20}N(t_{0})=\alpha_{20}\otimes\alpha_{01}\otimes\alpha_{10}=0$. Since the sequence (5.53) is exact, $N(t_{2})$ belongs to the image of $\mathrm{KK}^{G}_{1}(D,\mathrm{C}(G))$ in $\mathrm{KK}^{G}_{0}(D,\mathbb{C})$. And $(1-s_{2})^{j}$ for $j=1,\dotsc,p-1$ is a $\mathbb{Z}$-module basis for this image. This is also a basis for the principal ideal in $\mathbb{Z}[s_{2}]/(s_{2}^{p}-1)$ generated by $1-s_{2}$. So there is a unique element $f(s_{2})$ in this principal ideal with $N(t_{2})=f(s_{2})$. Since $N(1)=p$, $p-N$ belongs to this principal ideal, and we must show that $f=p-N$. The Baaj–Skandalis duality involution $\mathrm{BS}$ on the ring $\mathfrak{K}$ acts on the component $\mathrm{KK}^{G}_{0}(D,D)$ by exchanging $s_{2}\leftrightarrow t_{2}$. Therefore, the relation $N(t_{2})=f(s_{2})$ also implies $f(t_{2})=N(s_{2})$. We rewrite this as $(p-f)(t_{2})=(p-N)(s_{2})$ in order to get another $\mathbb{Z}$-relation among the elements $(1-s_{2})^{j}$ for $j=1,\dotsc,p-1$ and $(1-t_{2})^{j}$ for $j=0,\dotsc,p-1$. Our computation shows that $\mathrm{KK}^{G}_{0}(D,D)$ is a free Abelian group of rank $2(p-1)$. So any two $\mathbb{Z}$-linear relations among the elements above must be multiples of each other. Hence there is $c\in\mathbb{Z}$ with $p-f=c\cdot N$ and $p-N=c\cdot f$. So $f=p-c\cdot N$ and $p-N=c\cdot p-c^{2}\cdot N$. Plugging in a primitive $p$-th root of unity, this implies $p=c\cdot p$. So $f=p-N$ as desired. ∎

Our computations so far show that the ring $\mathfrak{K}$ is generated by the elements $1_{j}$ and $\alpha_{jk}$ and that all relations in Theorem 5.11 hold in it. It remains to show that these relations already imply all relations in $\mathfrak{K}$. To begin with, any element of $\mathfrak{K}$ is a linear combination of $1_{j}$ and of products in the generators $\alpha_{jk}$. Here we may restrict attention to products of the form $(\alpha_{jk}\otimes\alpha_{kj})^{\ell}$ or $(\alpha_{jk}\otimes\alpha_{kj})^{\ell}\otimes\alpha_{jk}$ for $j\neq k$ because all other products vanish by (5.15). Now we have to go through the nine matrix entries and check that the relations in Proposition 5.27, which follow from those in Theorem 5.11, allow to reduce any such expression to terms in a $\mathbb{Z}$-module basis of that entry. This is easier for the $j,k$-th entries with $j\neq k$ because only terms of the form $(\alpha_{jk}\otimes\alpha_{kj})^{\ell}\otimes\alpha_{jk}$ for $\ell\in\mathbb{N}$ may occur. And $(\alpha_{jk}\otimes\alpha_{kj})^{\ell}$ is a polynomial in $t_{j}$ or $s_{j}$, respectively. For the $0,1$-entry, we get

[TABLE]

because of (5.30). This implies by induction that $(\alpha_{01}\otimes\alpha_{10})^{\ell}\otimes\alpha_{01}=p^{\ell}\alpha_{01}$. So the $01$-component in the universal ring given by the relations in Theorem 5.11 is at most $\mathbb{Z}$, which is the corresponding component in $\mathfrak{K}$. A similar computation works for the $10$-component. The $02$-component in the universal ring is spanned by

[TABLE]

and $N(t_{0})\otimes\alpha_{02}=0$ implies that it has rank at most $p-1$. This is also the rank of the corresponding component in $\mathfrak{K}$. Similar computations work for the $12$-, $20$- and $21$-components, which all have rank $p-1$ and are identified with $\mathbb{Z}[t]/(N(t))$.

Now consider the $00$-component. Here $\alpha_{01}\otimes\alpha_{10}=N(t_{0})$ and $\alpha_{02}\otimes\alpha_{20}=1_{0}-t_{0}$ are polynomials in $t_{0}$, and the relation $t_{0}^{p}=1_{0}$ in (5.28) implies that the $\mathbb{Z}$-rank is at most $p$ as desired. A similar computation works for the $11$-component. Finally, consider the $22$-component. This is spanned by polynomials in $s_{2}$ and $t_{2}$; (5.42) and (5.43) allow us to get rid of products of these types of generators. The relations $t_{2}^{p}=1_{2}$ and $s_{2}^{p}=1_{2}$ from (5.28) and (5.29) reduce to polynomials in $s_{2}$ and $t_{2}$ of degree at most $p-1$. The constants in these two types of polynomials are the same, and (5.18) makes one more generator redundant, say, $s_{2}^{p-1}$. So we remain with a generating set with at most $2p-2$ elements. This is just enough to span the ring $\mathrm{KK}^{G}_{0}(D,D)$ computed above. So the canonical map from the universal ring defined by the generators and relations in Theorem 5.11 to $\mathfrak{K}$ is an isomorphism. This finishes the proof of Theorem 5.11.

Remark 5.55*.*

We have used Baaj–Skandalis duality several times to simplify the description of the ring $\mathfrak{K}$. It is visible from the generators and relations in Theorem 5.11 that $\mathfrak{K}$ has an involutive automorphism $\mathrm{BS}$. Namely, it is defined on generators by $\mathrm{BS}(1_{j})=1_{\sigma(j)}$ and $\mathrm{BS}(\alpha_{jk})=\alpha_{\sigma(j)\sigma(k)}$ for the transposition $\sigma=(01)$ acting on $j,k\in\{0,1,2\}$, $j\neq k$. This preserves the relations in Theorem 5.11. Hence it defines an automorphism $\mathrm{BS}$ of $\mathfrak{K}$. Since $\mathrm{BS}\circ\mathrm{BS}$ is the identity on generators, it is the identity on the whole ring $\mathfrak{K}$. We warn the reader once again that $\mathrm{BS}$ does not preserve the $\mathbb{Z}/2$-grading on $\mathfrak{K}$.

We also note a canonical anti-automorphism of $\mathfrak{K}$:

Proposition 5.56**.**

There is a unique ring anti-automorphism $x\mapsto x^{*}$ of $\mathfrak{K}$ with $1_{j}^{*}=1_{j}$ and $\alpha_{jk}^{*}=\alpha_{kj}$ for all $j,k\in\{0,1,2\}$ with $j\neq k$. It is grading-preserving and involutive.

Proof.

The formulas for $x^{*}$ on generators imply that the idempotent generators $1_{j}$ and the products $\alpha_{jk}\otimes\alpha_{kj}$ are self-adjoint. Hence $x\mapsto x^{*}$ leaves most of the relations in Theorem 5.11 unchanged, and it replaces the relations (5.14) and (5.15) for $j,k$ and $j,k,m$ by the same relations for $k,j$ and $m,k,j$, respectively. So there is indeed a unique anti-automorphism of $\mathfrak{K}$ with the given values on the generators. Since it preserves the grading on generators and satisfies $(x^{*})^{*}=x$ for all generators, it is grading-preserving and involutive. ∎

This anti-automorphism should be explicable by Poincaré duality in $\mathrm{KK}^{G}$ (see [Kasparov:Novikov, Emerson-Meyer:Dualities]). It can be shown that the three generators $\mathbb{C}$, $\mathrm{C}(G)$ and $D$ are Poincaré self-dual to themselves. This duality induces an anti-automorphism on the ring $\mathfrak{K}$. I guess that this automorphism is the one described in Proposition 5.56. The computations needed to verify this seem excessive, however, and therefore I decided not to discuss this any further in this article.

5.2. The prime 2

Let $p=2$. Then the ring $\mathfrak{K}$ is much simpler: the group ring of $\mathbb{Z}/2$ is quite small, namely, $\mathfrak{S}=\mathbb{Z}[t]/(t^{2}-1)$, and $N(t)=1+t$ and $1-t$ differ only by a sign. The relations (5.16), (5.17) and (5.18) are equivalent to

[TABLE]

In brief, we get the relation $\alpha_{ij}\otimes\alpha_{ji}+\alpha_{ik}\otimes\alpha_{ki}=2\cdot 1_{i}$ whenever $\{i,j,k\}=\{0,1,2\}$. These relations are symmetric under $\alpha_{jk}\mapsto\alpha_{\sigma(j)\sigma(k)}$ for any permutation $\sigma$ of $\{0,1,2\}$. The other relations in Theorem 5.11 are also symmetric under index permutations. So the ring $\mathfrak{K}$ carries an action of the symmetric group on three letters if $p=2$. In particular, the component $\mathrm{KK}^{G}_{*}(D,D)$ in the ring does not behave differently from the other components $\mathrm{KK}^{G}_{*}(\mathbb{C},\mathbb{C})$ and $\mathrm{KK}^{G}_{*}(\mathrm{C}(G),\mathrm{C}(G))$. This is a special feature of the prime $2$. In general, the $\mathbb{Z}$-ranks of these rings are $2p-2$, $p$ and $p$, respectively, and these are only equal for $p=2$.

The $\mathrm{C}^{*}$-algebra $D$ for $p=2$ in (5.1) is isomorphic to $\mathrm{C}_{0}(\mathbb{R})$ with the action of $\mathbb{Z}/2$ by reflection at the origin. The group $\mathrm{KK}^{\mathbb{Z}/2}_{*}(D,A)$ for a $\mathbb{Z}/2$-$\mathrm{C}^{*}$-algebra is canonically isomorphic to the $\mathrm{K}$-theory of $A$ as a $\mathbb{Z}/2$-graded $\mathrm{C}^{*}$-algebra (see also [Haag:Graded, Haag:Algebraic, Meyer:Equivariant]). Hence this piece in Köhler’s invariant is also a “standard” $\mathrm{K}$-theory group in this case.

Lemma 5.60**.**

Let $p=2$. Then the functor $\mathrm{KK}^{G}\to\mathrm{KK}^{G}$, $A\mapsto A\otimes D$, is an involutive automorphism up to natural equivalence, that is, $A\otimes D\otimes D$ is naturally $\mathrm{KK}^{G}$-equivalent to $A$ for all $A$ in $\mathrm{KK}^{G}$. And $\mathbb{C}\otimes D\sim D$, $D\otimes D\sim\mathbb{C}$, $\mathrm{C}(G)\otimes D\sim\Sigma\mathrm{C}(G)$. So tensoring with $D$ induces an automorphism of $\mathfrak{K}$. It corresponds to the automorphism that acts on generators by the permutation $(02)$.

Proof.

The tensor product $D\otimes D$ corresponds to $\mathrm{C}_{0}(\mathbb{R}^{2})$ with the action by point reflection at the origin. This is orientation-preserving, even $\mathrm{K}$-oriented – unlike the action on $\mathbb{R}$. Hence equivariant Bott periodicity gives a $\mathrm{KK}^{G}$-equivalence $D\otimes D\sim\mathbb{C}$. Thus tensoring with $D$ is an involutive automorphism of $\mathrm{KK}^{G}$. The tensor product $\mathrm{C}(G)\otimes D$ is isomorphic to the suspension of $\mathrm{C}(G)$ because the diagonal action on $\mathrm{C}(G)\otimes D$ is conjugate to the action only on the factor $\mathrm{C}(G)$. ∎

5.3. Exact modules versus group actions on C*-algebras

The following theorem combines Köhler’s results with Theorem 2.1.

Theorem 5.61**.**

There is a bijection between isomorphism classes of exact, countable $\mathfrak{K}$-modules and $\mathrm{KK}^{G}$-equivalence classes of pointwise outer actions of $G$ on stable, UCT Kirchberg algebras in the equivariant bootstrap class $\mathfrak{B}^{G}$.

Proof.

By Theorem 5.10, a $\mathfrak{K}$-module $M$ is of the form $F_{*}(A)$ for an object $A$ of the equivariant bootstrap class $\mathfrak{B}^{G}$ if and only if it is exact and countable. Then the underlying $\mathrm{C}^{*}$-algebra of $A$ may be chosen to be of Type I and hence nuclear, and it satisfies the UCT. Theorem 2.1 shows that $A$ may be chosen to be a stable UCT Kirchberg algebra with a pointwise outer action of $G$. Two actions of $G$ in $\mathfrak{B}^{G}$ that lift the same exact, countable $\mathfrak{K}$-module are $\mathrm{KK}^{G}$-equivalent by Theorem 5.4. ∎

For a pointwise outer action on a (purely infinite and) simple $\mathrm{C}^{*}$-algebra $A$, the crossed product $A\rtimes G$ is again (purely infinite and) simple (see [Kishimoto:Outer_crossed, Jeong-Kodaka-Osaka:Purely_infinite_crossed]). Hence $A$ viewed as a Hilbert bimodule over the fixed point algebra $A^{G}$ and $A\rtimes G$ is full, so that $A\rtimes G$ is Morita–Rieffel equivalent to the fixed point algebra $A^{G}$. We may, therefore, identify $\mathrm{K}_{*}(A\rtimes G)\cong\mathrm{K}_{*}(A^{G})$. Then the map $(\alpha_{10})^{*}\colon\mathrm{K}_{*}(A)^{G}\cong\mathrm{K}_{*}(A\rtimes G)\to\mathrm{K}_{*}(A)$ becomes the map induced by the inclusion $A^{G}\hookrightarrow A$; this follows from Lemma 4.8. So the pieces $M_{0}$ and $M_{1}$ and the maps $(\alpha_{01})^{*}$ and $(\alpha_{10})^{*}$ between them are standard ingredients in the study of $\mathbb{Z}/p$-actions on simple $\mathrm{C}^{*}$-algebras. The exactness of the two sequences in Definition 5.9 restricts $M_{2}$ considerably. In particular, if both $M_{0}$ and $M_{1}$ vanish, then $M=0$. However, $M_{0}$ and $M_{1}$ with the maps $t_{0}^{M}$, $s_{1}^{M}$, $\alpha_{01}^{M}$ and $\alpha_{10}^{M}$ do not yet determine $M$. A concrete counterexample in the case $p\neq 2$ is Example 9.9. This example describes exact $\mathfrak{K}$-modules that depend on a parameter $\tau\in\{1,\dotsc,p-1\}$, which enters only in the map $\alpha_{21}^{M}$. The maps $t_{0}^{M}$, $s_{1}^{M}$, $\alpha_{01}^{M}$ and $\alpha_{10}^{M}$ do not depend on $\tau$. Nevertheless, these examples for different choices of $\tau$ are not isomorphic as $\mathfrak{K}$-modules, and there are at least two choices for $\tau$ if $p\neq 2$.

5.4. Some consequences of exactness

This section collects some general properties of exact $\mathfrak{K}$-modules. Our starting point is the following elementary lemma, which we are going to apply to the composites $\alpha_{jk}\otimes\alpha_{kj}$:

Lemma 5.62**.**

Let $X$, $Y$ and $Z$ be Abelian groups. Let $f\colon X\to Y$ and $g\colon Y\to Z$ be composable homomorphisms and $h=g\circ f$. Then $\ker f\subseteq\ker h$ and $\operatorname{im}h\subseteq\operatorname{im}g$, and

[TABLE]

by applying $f$ or $g^{-1}$, respectively. So there is a short exact sequence

[TABLE]

Proof.

Let $x\in X$. Then $h(x)=0$ if and only if $f(x)\in\ker g$, that is, $x\in f^{-1}(\ker g)$. Since $f$ maps $f^{-1}(\ker g)$ onto $\operatorname{im}f\cap\ker g$ and $f^{-1}(\ker g)\supseteq\ker f$, the map $f$ induces an isomorphism \ker h\bigm{/}\ker f\cong\operatorname{im}f\cap\ker g. Now let $z\in Z$. If $z\in\operatorname{im}h$, then there is $x\in X$ with g\bigl{(}f(x)\bigr{)}=z. So $z\in\operatorname{im}g$. The map $g$ induces a bijection from $Y/\ker g$ onto $\operatorname{im}g$. The inverse of this bijection maps $\operatorname{im}h\subseteq\operatorname{im}g$ onto the set of all elements of the form $[f(x)]\in Y/\ker g$ for $x\in X$. This is the quotient $(\operatorname{im}f+\ker g)/\ker g$. This implies the claim about $\operatorname{im}g/\operatorname{im}h$. And then the claim about cokernels also follows. ∎

When we combine (5.20)–(5.26) with Lemma 5.62, we get twelve short exact sequences involving kernels and cokernels of the maps $\alpha_{jk}^{M}$. And these are grouped into six pairs because exactness implies that some of the third entries in the short exact sequences are equal. These pairs of exact sequences are collected in Table 1. The next lemma condenses these into isomorphisms between certain quotients of kernels or images.

Lemma 5.63**.**

Let $M$ be an exact $\mathfrak{K}$-module. There are isomorphisms

[TABLE]

For each quotient $X/Y$ above, it is also asserted implicitly that $Y\subseteq X$.

6. Integral representations of groups of prime order

The group ring of $G=\mathbb{Z}/p$ is isomorphic to $\mathfrak{S}\mathrel{\vcentcolon=}\mathbb{Z}[t]/(t^{p}-1)$. This section collects some facts about $\mathfrak{S}$-modules, which are certainly known to the experts. Recall that $N(t)\mathrel{\vcentcolon=}1+t+\dotsb+t^{p-1}$ and that $(1-t)\cdot N(t)=1-t^{p}\equiv 0$ in $\mathfrak{S}$. The ring $\mathbb{Z}[t]/(N(t))$ is isomorphic to $\mathbb{Z}[\vartheta]$, where $\vartheta$ denotes a primitive $p$-th root of unity. Let $M$ be an $\mathfrak{S}$-module. As in Definition 5.8, we define $t^{M}\colon M\to M$ by $t^{M}(x)\mathrel{\vcentcolon=}x\cdot t$ for all $x\in M$.

Lemma 6.1**.**

$N(t^{M})x=p\cdot x$* for all $x\in\ker(1-t^{M})$. And*

[TABLE]

Proof.

If $x=t^{M}x$, then $x=(t^{M})^{j}x$ for all $j\in\mathbb{N}$ and hence $N(t^{M})x=p\cdot x$. The inclusions $\operatorname{im}N(t^{M})\subseteq\ker(1-t^{M})$ and $\operatorname{im}(1-t^{M})\subseteq\ker N(t^{M})$ follow immediately from $(1-t^{M})N(t^{M})=N(t^{M})(1-t^{M})=(t^{M})^{p}-1=0$. Since $N(1)=p$, there is a polynomial $f$ with $f\cdot(1-t)=N(t)-p$. Therefore, if $x\in\ker N(t^{M})$, then

[TABLE]

That is, $p\cdot(\ker N(t^{M})\cap\ker(1-t^{M}))=0$. If $x\in\ker N(t^{M})\cap\ker(1-t^{M})$, then this computation shows $p\cdot x=0$. If $y\in M$, then

[TABLE]

Lemma 6.2**.**

There is a unit $u$ in the ring $\mathbb{Z}[\vartheta]$ with $p=u\cdot(1-\vartheta)^{p-1}$.

Proof.

Let $j\in\{2,\dotsc,p-1\}$. The fraction $u_{j}\mathrel{\vcentcolon=}\frac{1-\vartheta^{j}}{1-\vartheta}=\sum_{k=0}^{j-1}\vartheta^{k}$ lies in $\mathbb{Z}[\vartheta]$. The Galois group of the extension $\mathbb{Q}\subseteq\mathbb{Q}(\vartheta)$ acts simply transitively on the primitive $p$th roots of unity $\vartheta^{j}$, $j=1,\dotsc,p-1$. Since $N$ is the minimal polynomial of $\vartheta$, $N(t)=\prod_{j=1}^{p-1}(t-\vartheta^{j})$. So

[TABLE]

Since $1-\vartheta^{j}$ and $1-\vartheta$ are Galois conjugates, they have the same norm. Hence $u_{j}$ is a unit in $\mathbb{Z}[\vartheta]$. Now (6.3) implies

[TABLE]

So the unit $u\mathrel{\vcentcolon=}\prod_{j=1}^{p-1}u_{j}$ does the job. ∎

Definition 6.5**.**

An Abelian group $M$ (or a module over some ring) is called uniquely $p$-divisible if multiplication by $p$ on $M$ is invertible.

Example 6.6*.*

A finite group is uniquely $p$-divisible if and only if $p$ does not divide its order.

Definition 6.7**.**

An $\mathfrak{S}$-module $M$ is called cohomologically trivial if $\ker(1-t^{M})=\operatorname{im}N(t^{M})$ and $\operatorname{im}(1-t^{M})=\ker N(t^{M})$.

Proposition 6.8**.**

Let $M$ be an $\mathfrak{S}$-module. The following are equivalent:

1. (1)

$M$* is uniquely $p$-divisible;* 2. (2)

$M$* is cohomologically trivial and $M=\ker N(t^{M})\oplus\ker(1-t^{M})$;* 3. (3)

$1-t^{M}$* restricts to an invertible map $\ker N(t^{M})\to\ker N(t^{M})$ and $N(t^{M})$ restricts to an invertible map $\ker(1-t^{M})\to\ker(1-t^{M})$.*

Proof.

We prove that (1) implies (2). So assume $M$ to be uniquely $p$-divisible. If $x\in\ker(1-t^{M})$, then $p^{-1}\cdot x\in\ker(1-t^{M})$ as well because $M$ is uniquely $p$-divisible. So $\ker(1-t^{M})=p\cdot\ker(1-t^{M})$. Then Lemma 6.1 implies $\ker(1-t^{M})=\operatorname{im}N(t^{M})$. Similarly, $\ker N(t^{M})=p\cdot\ker N(t^{M})$ and hence $\ker N(t^{M})=\operatorname{im}(1-t^{M})$. Since $M$ is uniquely $p$-divisible, the last two equations in Lemma 6.1 imply $\ker N(t^{M})\cap\ker(1-t^{M})=0$ and $M=\operatorname{im}N(t^{M})+\operatorname{im}(1-t^{M})$. Now $M=\ker N(t^{M})\oplus\ker(1-t^{M})$ follows. We have shown that (1) implies (2).

Next we prove that (2) implies (3). So we assume $M=\ker N(t^{M})\oplus\ker(1-t^{M})$ and that $M$ be cohomologically trivial. Then $\ker N(t^{M})\cap\ker(1-t^{M})=0$. So the restrictions of $(1-t^{M})$ to $\ker N(t^{M})$ and of $N(t^{M})$ to $\ker(1-t^{M})$ are injective. Since $1-t^{M}$ vanishes on the complementary summand $\ker(1-t^{M})$, it follows that $(1-t^{M})(\ker N(t^{M}))=\operatorname{im}(1-t^{M})=\ker N(t^{M})$. Similarly, $N(t^{M})(\ker(1-t^{M}))=\operatorname{im}N(t^{M})=\ker(1-t^{M})$. So $1-t^{M}$ restricts to an invertible map $\ker N(t^{M})\to\ker N(t^{M})$ and $N(t^{M})$ restricts to an invertible map $\ker(1-t^{M})\to\ker(1-t^{M})$. We have shown that (2) implies (3).

We prove that (3) implies (1). The restriction of $N(t^{M})$ to $\ker(1-t^{M})$ is multiplication by $p$ by Lemma 6.1. Hence (3) implies that $\ker(1-t^{M})$ is uniquely $p$-divisible. Then $\ker(1-t^{M})=\operatorname{im}N(t^{M})$ by Lemma 6.1. The $\mathfrak{S}$-module structure on $\ker N(t^{M})$ descends to one over the ring $\mathbb{Z}[t]/(N(t))\cong\mathbb{Z}[\vartheta]$. Since $1-t^{M}$ is invertible on $\ker N(t^{M})$, Lemma 6.2 shows that multiplicationby $p$ is invertible on $\ker N(t^{M})$. That is, $\ker N(t^{M})$ is uniquely $p$-divisible. There is a short exact sequence $\ker N(t^{M})\rightarrowtail M\twoheadrightarrow\operatorname{im}N(t^{M})$. Since the kernel and quotient in it are uniquely $p$-divisible, the Snake Lemma implies that $M$ is uniquely $p$-divisible. ∎

Next we describe $\mathfrak{S}$-modules where $\ker(1-t^{M})=0$ or $\operatorname{coker}(1-t^{M})=0$. These will appear in the classification of exact $\mathfrak{K}$-modules.

Lemma 6.9**.**

Let $M$ be an $\mathfrak{S}$-module with $\ker(1-t^{M})=0$. Then $N(t^{M})=0$ and $\operatorname{coker}(1-t^{M})$ is a $\mathbb{Z}/p$-vector space. Let $(e_{i})_{i\in I}$ be elements of $M$ that represent a $\mathbb{Z}/p$-basis of $\operatorname{coker}(1-t^{M})$. Then the map

[TABLE]

is injective, and its cokernel is uniquely $p$-divisible.

Proof.

Lemma 6.1 implies $\operatorname{im}N(t^{M})=0$ – that is, $N(t^{M})=0$ – and $p\cdot M\subseteq\operatorname{im}N(t^{M})+\operatorname{im}(1-t^{M})=\operatorname{im}(1-t^{M})$. Thus $\operatorname{coker}(1-t^{M})=M/\operatorname{im}(1-t^{M})$ is a $\mathbb{Z}/p$-vector space. And the map $\psi$ is well defined. Since $1-t^{M}$ acts by an injective map on $M$, so does $(1-t^{M})^{k}$ for all $k\in\mathbb{N}$. We claim that the maps

[TABLE]

are isomorphisms for all $k\in\mathbb{N}$. They are well defined surjections by definition. If $\xi\in M$ is such that $(1-t^{M})^{k}\xi\in(1-t^{M})^{k+1}M$, then $(1-t^{M})^{k}\xi=(1-t^{M})^{k+1}\eta$ for some $\eta\in M$. Then $\xi=(1-t^{M})\eta$ because $(1-t^{M})^{k}$ is injective. So the map above is indeed bijective.

Now let $(f_{i})_{i\in I}$ belong to the kernel of $\psi$. That is, $\sum_{i\in I}f_{i}(t^{M})e_{i}=0$. In particular, $\sum_{i\in I}f_{i}(t^{M})e_{i}\in(1-t^{M})M$. Since $e_{i}$ is a basis for $M/(1-t^{M})M$, this implies $f_{i}(t^{M})e_{i}\in\operatorname{im}(1-t^{M})$ for all $i\in I$. Then $1-t\mid f_{i}$ for all $i\in I$ because the map $\mathbb{Z}[t]/(t-1,N(t))\to\mathbb{Z}/p$, $f\mapsto[f(1)\bmod p]$, is an isomorphism. Similarly, an induction over $k$ shows that $(1-t)^{k}\mid f_{i}$ for all $i\in I$, using that $(1-t^{M})^{k-1}e_{i}$ is a basis for $(1-t^{M})^{k-1}M/(1-t^{M})^{k}M$ as a $\mathbb{Z}/p$-vector space. But $(1-t)^{k}\mid f_{i}$ for all $k\in\mathbb{N}$ forces $f_{i}=0$ in the ring $\mathbb{Z}[t]/(N(t))$. So the map $\psi$ is injective.

The quotient $X\mathrel{\vcentcolon=}\operatorname{coker}\psi$ is a module over $\mathbb{Z}[\vartheta]\cong\mathbb{Z}[t]/(N(t))$. The cokernel of multiplication by $1-t$ on $\mathbb{Z}[t]/(N(t))$ is $\mathbb{Z}[t]/(1-t,N(t))\cong\mathbb{Z}/p$. Therefore, $\psi$ induces an isomorphism on the cokernels of multiplication by $1-t$. Now the Snake Lemma for the morphism of extensions

[TABLE]

shows $\ker(1-t^{X})=0$ and $\operatorname{coker}(1-t^{X})=0$. This is equivalent to $X$ being uniquely $p$-divisible by Proposition 6.8. ∎

Lemma 6.10**.**

Let $M$ be an $\mathfrak{S}$-module with $\operatorname{im}(1-t^{M})=M$. Then $N(t^{M})=0$ and $\ker(1-t^{M})$ is a $\mathbb{Z}/p$-vector space. Let $(e_{i})_{i\in I}$ be a basis of $\ker(1-t^{M})$. Then there is an injective $\mathfrak{S}$-module map

[TABLE]

with uniquely $p$-divisible cokernel.

Proof.

Lemma 6.1 implies $\ker N(t^{M})=M$ – that is, $N(t^{M})=0$ – and $p\cdot\ker(1-t^{M})\subseteq\operatorname{im}N(t^{M})=0$. That is, $\ker(1-t^{M})$ is a $\mathbb{Z}/p$-vector space. By Lemma 6.2, making $p$ or $1-\vartheta$ invertible has the same effect: $\mathbb{Z}[\vartheta,1/p]=\mathbb{Z}[\vartheta,(1-\vartheta)^{-1}]$. We may write any element of this ring as $f\cdot(1-\vartheta)^{-n}$ for some $f\in\mathbb{Z}[\vartheta]$, $n\in\mathbb{N}$.

Since $1-t^{M}$ is surjective, we may recursively choose $e_{i,n}\in M$ with $e_{i,0}=e_{i}$ and $(1-t^{M})e_{i,n+1}=e_{i,n}$ for all $n\in\mathbb{N}$. There is a well defined map

[TABLE]

because $(1-t^{M})e_{i,n+1}=e_{i,n}$ for all $i\in I$, $n\in\mathbb{N}$. This map vanishes on elements of $\bigoplus_{i\in I}(1-\vartheta)\mathbb{Z}[\vartheta]$ because $(1-t^{M})^{n+1}e_{i,n}=(1-t^{M})e_{i}=0$. We claim that the induced map $\psi$ on the quotient

[TABLE]

is injective. An element of $\bigoplus_{i\in I}\mathbb{Z}[\vartheta,(1-\vartheta)^{-1}]$ may be written as a finite linear combination $(1-\vartheta)^{-n}\sum_{i\in S}f_{i}\delta_{i}$ with $f_{i}\in\mathbb{Z}[\vartheta]$, $S\subseteq I$ finite, and $n\geq 0$. Assume that this is in the kernel of $\psi$, that is, $\sum_{i\in S}f_{i}(t^{M})e_{i,n}=0$. Applying $(1-t^{M})^{n-1}$ shows that $\sum_{i\in S}f_{i}(t^{M})e_{i}=0$. Since the elements $e_{i}$ form a basis for $\ker(1-t^{M})$, this implies $f_{i}\in\ker(\mathbb{Z}[\vartheta]\to\mathbb{Z}/p)$ for all $i\in S$. This kernel is the ideal generated by $1-\vartheta$ because $\mathbb{Z}[t]/(t-1,N(t))=\mathbb{Z}[t]/(t-1,p)\cong\mathbb{Z}/p$. So we may rewrite $f_{i}=(1-\vartheta)g_{i}$ for some $g_{i}\in\mathbb{Z}[\vartheta]$. Thus

[TABLE]

We continue like this until $n=0$. Then we reach a term in $\bigoplus_{i\in S}(1-\vartheta)\mathbb{Z}[\vartheta]$. This proves that $\psi$ is injective. Multiplication with $1-\vartheta$ is surjective on \bigoplus_{i\in I}\mathbb{Z}[\vartheta,1/p]\bigm{/}(1-\vartheta)\mathbb{Z}[\vartheta], and its kernel is the $\mathbb{Z}/p$-linear span of $\delta_{i}$ for $i\in I$. Hence $\psi$ induces an isomorphism between the kernels and cokernels of $1-t$ acting on the domain and codomain of $\psi$. Now the Snake Lemma implies as in the proof of Lemma 6.9 that $1-t$ acts by an invertible map on $\operatorname{coker}\psi$. Then $\operatorname{coker}\psi$ is uniquely $p$-divisible by Proposition 6.8. ∎

7. The uniquely divisible case

We are going to describe uniquely $p$-divisible exact $\mathfrak{K}$-modules in terms of more simple objects. This is used in [dellAmbrogio--Meyer:Spectrum_Kasparov_prime] to describe the Balmer spectrum of $\mathcal{B}^{G}$.

Example 7.1*.*

Let $X$ be a $\mathbb{Z}/2$-graded Abelian group and let $Y$ and $Z$ be two $\mathbb{Z}/2$-graded $\mathbb{Z}[\vartheta]$-modules. Assume that they are all uniquely $p$-divisble. Equivalently, they are $\mathbb{Z}/2$-graded modules over the rings $\mathbb{Z}[1/p]$ and $\mathbb{Z}[\vartheta,1/p]$, respectively. We define an exact $\mathfrak{K}$-module:

[TABLE]

Here $\Sigma Z$ means $Z$ with opposite parity. So $\alpha_{21}^{M}$ and $\alpha_{12}^{M}$ are odd. The relations (5.12)–(5.15) are easy to check. And we compute

[TABLE]

Hence

[TABLE]

The relations (5.16)–(5.18) follow. So we have defined a $\mathfrak{K}$-module by Theorem 5.11. The two sequences in Definition 5.9 are exact because multiplication by $p$ is invertible on $X$ and multiplication by $1-\vartheta$ is invertible on $Y$ and $Z$ by Proposition 6.8.

The following theorem implies that this already gives all uniquely $p$-divisible exact $\mathfrak{K}$-modules:

Theorem 7.2**.**

Let $M$ be an exact $\mathfrak{K}$-module. Assume that one of the pieces $M_{0}$, $M_{1}$ or $M_{2}$ is uniquely $p$-divisible. Then the other two pieces are uniquely $p$-divisible as well. And there are modules $X$ over $\mathbb{Z}[1/p]$ and $Y,Z$ over $\mathbb{Z}[\vartheta,1/p]$ such that $M$ is isomorphic to the exact $\mathfrak{K}$-module built out of $X,Y,Z$ in Example 7.1.

Some ideas used in the proof are also useful for more general examples. We develop these systematically in Section 7.1 and then use them in Section 7.2 to prove Theorem 7.2. Section 7.3 spells out what Theorem 7.2 says about $\mathbb{Z}/p$-actions on Cuntz algebras with uniquely $p$-divisible $\mathrm{K}$-theory. In Section 7.4, we generalise Theorem 7.2 to a statement that extends a uniquely $p$-divisible submodule in $M_{1}$ to a uniquely $p$-divisible submodule in $M$. This will be used to reduce the study of actions on Cuntz algebras with arbitrary finite $\mathrm{K}$-theory to the special case where the $\mathrm{K}$-theory is purely $p$-torsion.

7.1. Some general proof techniques for exact $\mathfrak{K}$-modules

Proposition 7.3**.**

Let $M$ be an exact $\mathfrak{K}$-module.

• •

If the $\mathfrak{S}$-module $(M_{0},t_{0})$ is cohomologically trivial, then

[TABLE]

• •

If the $\mathfrak{S}$-module $(M_{1},s_{1})$ is cohomologically trivial, then

[TABLE]

• •

If

[TABLE]

then

[TABLE]

Proof.

The middle equality in the six equations that we assert are the six conditions that say that a $\mathfrak{K}$-module is exact. When we replace the first and third equality in our six equations by “$\subseteq$”, they become the twelve inclusions that are implicitly asserted in Lemma 5.63. In each equation, the first and last item are equal by cohomological triviality or by (7.8), respectively. This forces the inclusions in between to be equalities. ∎

Corollary 7.11**.**

If $M_{0}$ is uniquely $p$-divisible, then (7.4) and (7.5) hold. If $M_{1}$ is uniquely $p$-divisible, then (7.6) and (7.7) hold.

Proof.

The pieces $M_{0}$ and $M_{1}$ in an exact $\mathfrak{K}$-module are $\mathfrak{S}$-modules. Uniquely $p$-divisible $\mathfrak{S}$-modules are cohomologically trivial by Proposition 6.8. Therefore, the statements follow from Proposition 7.3. ∎

We now carry this argument over to the case where $M_{2}$ is uniquely $p$-divisible. The group $M_{2}$ in a $\mathfrak{K}$-module is a module over the commutative ring

[TABLE]

The relations $t^{p}-1=0$ and $s^{p}-1=0$ hold in $\mathfrak{S}_{2}$ because

[TABLE]

Definition 7.13**.**

An $\mathfrak{S}_{2}$-module is called cohomologically trivial if it satisfies the two conditions $\operatorname{im}(1-t^{M})=\ker(1-s^{M})$ and $\operatorname{im}(1-s^{M})=\ker(1-t^{M})$ in (7.8).

Lemma 7.14**.**

Let $M$ be an $\mathfrak{S}_{2}$-module. Then

[TABLE]

Proof.

Let $x\in M$. Then $N(s^{M})(x)=(p-N(t^{M}))(x)$ because $N(s)+N(t)-p=0$ in $\mathfrak{S}_{2}$. Since $1-t\mid p-N(t)$, it follows that $N(s^{M})(x)\in\operatorname{im}(1-t^{M})$. The relation $(1-s)(1-t)=0$ in $\mathfrak{S}_{2}$ implies $\operatorname{im}(1-t^{M})\subseteq\ker(1-s^{M})$. If $(1-s^{M})(x)=0$, then $(N(s^{M})-p)(x)=0$. So the relation $N(s)+N(t)-p=0$ in $\mathfrak{S}_{2}$ implies $N(t^{M})(x)=0$. That is, $\ker(1-s^{M})\subseteq\ker N(t^{M})$. The proof of (7.16) is analogous. ∎

Proposition 7.17**.**

Let $M$ be an $\mathfrak{S}_{2}$-module. The following are equivalent:

1. (1)

$M$* is uniquely $p$-divisible;* 2. (2)

$M$* is cohomologically trivial and $M=\ker(1-s^{M})\oplus\ker(1-t^{M})$;* 3. (3)

$1-s^{M}$* restricts to an invertible map $\ker(1-t^{M})\to\ker(1-t^{M})$ and $1-t^{M}$ restricts to an invertible map $\ker(1-s^{M})\to\ker(1-s^{M})$.*

Proof.

If (1) holds, then Proposition 6.8 implies that $M$ is cohomologically trivial as a module over the two copies of $\mathfrak{S}$ inside $\mathfrak{S}_{2}$. Then $\operatorname{im}N(s^{M})=\operatorname{im}(1-t^{M})=\ker(1-s^{M})$ and $\operatorname{im}N(t^{M})=\operatorname{im}(1-s^{M})=\ker(1-t^{M})$ by Lemma 7.14. So $M$ is cohomologically trivial as an $\mathfrak{S}_{2}$-module and $M=\operatorname{im}N(s^{M})\oplus\operatorname{im}(1-s^{M})=\ker(1-s^{M})\oplus\ker(1-t^{M})$. So (1) implies (2).

Now assume (2). Then $\ker(1-s^{M})\cap\ker(1-t^{M})=0$ says that $1-s^{M}$ is injective on $\ker(1-t^{M})$. And $M=\ker(1-s^{M})+\ker(1-t^{M})$ implies $(1-s^{M})(M)=(1-s^{M})(\ker(1-t^{M}))$. Thus $1-s^{M}\colon\ker(1-t^{M})\to\operatorname{im}(1-s^{M})=\ker(1-t^{M})$ is also surjective. The same argument works with $t$ instead of $s$. So (2) implies (3).

Now assume (3). The subgroup $\ker(1-s^{M})$ is a module over $\mathbb{Z}[t]/(N(t))$ because both $1-s^{M}$ and $N(t^{M})$ vanish there by (7.15). Lemma 6.2 says that $p$ and $(1-\vartheta)^{p-1}$ differ only by a unit in the ring $\mathbb{Z}[t]/(N(t))$. Since $1-t^{M}$ is invertible on $\ker(1-s^{M})$, $\ker(1-s^{M})$ is uniquely $p$-divisible. Similarly, $\ker(1-t^{M})$ is uniquely $p$-divisible. Hence $\ker(1-t^{M})=\operatorname{im}N(t^{M})$ by Lemma 6.1. Then $\ker(1-t^{M})=\operatorname{im}(1-s^{M})$ by Lemma 7.14. Then the Snake Lemma applied to multiplication by $p$ in the short exact sequence $\ker(1-s^{M})\rightarrowtail M\twoheadrightarrow\operatorname{im}(1-s^{M})$ shows that $M$ is uniquely $p$-divisible. Hence (3) implies (1). ∎

Now the same argument as above gives us an analogue of Corollary 7.11 for $M_{2}$:

Corollary 7.18**.**

If $M_{2}$ is uniquely $p$-divisible, then (7.9) and (7.10) hold.

Our next goal are criteria for an exact $\mathfrak{K}$-module to satisfy $M_{0}\cong\operatorname{im}(\alpha_{01}^{M})\oplus\operatorname{im}(\alpha_{02}^{M})$, and similarly for $M_{1}$ or $M_{2}$. These criteria help to prove Theorem 7.2 because such decompositions clearly happen in the exact $\mathfrak{K}$-modules in Example 7.1.

Lemma 7.19**.**

Let $M$ be an exact $\mathfrak{K}$-module. The following are equivalent:

1. (1)

$M_{0}\cong\operatorname{im}(\alpha_{01}^{M})\oplus\operatorname{im}(\alpha_{02}^{M})$; 2. (2)

$\ker(\alpha_{01}^{M})=\ker N(s_{1}^{M})$* and $\operatorname{im}(\alpha_{10}^{M})=\operatorname{im}N(s_{1}^{M})$;* 3. (3)

$\ker(\alpha_{02}^{M})=\ker(1_{2}^{M}-t_{2}^{M})$* and $\operatorname{im}(\alpha_{20}^{M})=\operatorname{im}(1_{2}^{M}-t_{2}^{M})$.*

These equivalent conditions hold if the $\mathfrak{S}$-module $(M_{1},s_{1})$ is cohomologically trivial or if the $\mathfrak{S}_{2}$-module $(M_{2},s_{2},t_{2})$ is cohomologically trivial.

Proof.

The direct sum decomposition in (1) is equivalent to $\operatorname{im}(\alpha_{01}^{M})\cap\operatorname{im}(\alpha_{02}^{M})=0$ and M_{0}\bigm{/}(\operatorname{im}(\alpha_{01}^{M})+\operatorname{im}(\alpha_{02}^{M})\bigr{)}=0. By the exact sequences in Table 1, the first condition holds if and only if $\ker(\alpha_{02}^{M})=\ker(1_{2}^{M}-t_{2}^{M})$, if and only if $\ker N(s_{1}^{M})=\ker(\alpha_{01}^{M})$; and the second condition holds if and only if $\operatorname{im}(\alpha_{20}^{M})=\operatorname{im}(1_{2}^{M}-t_{2}^{M})$, if and only if $\operatorname{im}(\alpha_{10}^{M})=\operatorname{im}N(s_{1}^{M})$. Here we use that two maps with the same codomain have equal image if and only if they have equal cokernels.

If $M_{1}$ is cohomologically trivial, then (7.6) and (7.7) are valid and imply our statement (2). If $M_{2}$ is cohomologically trivial, then (7.9) and (7.10) are valid and imply (3). ∎

Assume that we are in the situation of Lemma 7.19. The direct sum decomposition in Lemma 7.19.(1) is one of $\mathfrak{S}$-modules because both $\operatorname{im}(\alpha_{01}^{M})$ and $\operatorname{im}(\alpha_{02}^{M})$ are $\mathfrak{S}$-submodules by (5.30) and (5.31). Using Lemma 7.19 and exactness, we identify

[TABLE]

Let $\varphi$ temporarily denote the first isomorphism. Then $\varphi\circ\alpha_{01}^{M}=N(s_{1}^{M})$ and $\alpha_{10}^{M}\circ\varphi^{-1}=\alpha_{10}^{M}\circ\alpha_{01}^{M}=N(s_{1}^{M})$. Therefore, when we use $\varphi$ to identify $\operatorname{im}(\alpha_{01}^{M})$ with $\operatorname{im}N(s_{1}^{M})$, then $\alpha_{01}^{M}$ becomes $N(s_{1}^{M})$ and $\alpha_{10}^{M}$ becomes the inclusion map on $\operatorname{im}N(s_{1}^{M})$. And $\alpha_{10}^{M}$ vanishes on the other summand $\operatorname{im}(\alpha_{02}^{M})$ by (5.15). Similarly, $\alpha_{02}^{M}$ becomes $1_{2}^{M}-t_{2}^{M}$, and $\alpha_{20}^{M}$ becomes the inclusion of the summand $\operatorname{im}(1_{2}^{M}-t_{2}^{M})$ and vanishes on the other summand. So the $\mathfrak{K}$-module is determined by $M_{1}$ and $M_{2}$ with the maps $t_{2}^{M}$, $\alpha_{12}^{M}$ and $\alpha_{21}^{M}$; these determine $s_{j}^{M}$ for $j=1,2$ by (5.24) and (5.26). This data gives an exact $\mathfrak{K}$-module if and only if

1. (1)

$N(t_{2}^{M})+N(s_{2}^{M})=p\cdot 1_{2}^{M}$; 2. (2)

$\ker(\alpha_{12}^{M})=\operatorname{im}(1_{2}^{M}-t_{2}^{M})$, $\operatorname{im}(\alpha_{21}^{M})=\ker(1_{2}^{M}-t_{2}^{M})$; 3. (3)

$\ker(\alpha_{21}^{M})=\operatorname{im}N(s_{1}^{M})$, $\operatorname{im}(\alpha_{12}^{M})=\ker N(s_{1}^{M})$.

We may arrange the exactness conditions in (2) and (3) in a periodic long exact sequence

[TABLE]

Lemmas 7.21 and 7.22 are the analogues of Lemma 7.19 for $M_{1}$ and $M_{2}$. Lemma 7.21 is formally equivalent to Lemma 7.19 because of the ring automorphism $\mathrm{BS}$ of $\mathfrak{K}$. The proof of Lemma 7.22 is similar to the proof of Lemma 7.19.

Lemma 7.21**.**

Let $M$ be an exact $\mathfrak{K}$-module. The following are equivalent:

1. (1)

$M_{1}\cong\operatorname{im}(\alpha_{10}^{M})\oplus\operatorname{im}(\alpha_{12}^{M})$; 2. (2)

$\ker(\alpha_{10}^{M})=\ker N(t_{0}^{M})$* and $\operatorname{im}(\alpha_{01}^{M})=\operatorname{im}N(t_{0}^{M})$;* 3. (3)

$\ker(\alpha_{12}^{M})=\ker(1_{2}^{M}-s_{2}^{M})$* and $\operatorname{im}(\alpha_{21}^{M})=\operatorname{im}(1_{2}^{M}-s_{2}^{M})$;*

These equivalent conditions hold if the $\mathfrak{S}$-module $(M_{0},t_{0})$ is cohomologically trivial or if the $\mathfrak{S}_{2}$-module $(M_{2},s_{2},t_{2})$ is cohomologically trivial.

Lemma 7.22**.**

Let $M$ be an exact $\mathfrak{K}$-module. The following are equivalent:

1. (1)

$M_{2}\cong\operatorname{im}(\alpha_{20}^{M})\oplus\operatorname{im}(\alpha_{21}^{M})$; 2. (2)

$\ker(\alpha_{20}^{M})=\ker(1_{0}^{M}-t_{0}^{M})$* and $\operatorname{im}(\alpha_{02}^{M})=\operatorname{im}(1_{0}^{M}-t_{0}^{M})$;* 3. (3)

$\ker(\alpha_{21}^{M})=\ker(1_{1}^{M}-s_{1}^{M})$* and $\operatorname{im}(\alpha_{12}^{M})=\operatorname{im}(1_{1}^{M}-s_{1}^{M})$.*

These equivalent conditions hold if one of the $\mathfrak{S}$-modules $(M_{0},t_{0})$ or $(M_{1},s_{1})$ is cohomologically trivial.

7.2. Proof of Theorem 7.2

Assume first that $M_{0}$ is uniquely $p$-divisible. Then $M_{0}$ is cohomologically trivial and $M_{0}\cong\ker N(t_{0}^{M})\oplus\ker(1_{0}^{M}-t_{0}^{M})$ by Proposition 6.8. And (7.4) and (7.5) hold. Hence the statements (1) in Lemma 7.19, (2) in Lemma 7.21 and (2) in Lemma 7.22 are true. So the other equivalent statements in these three lemmas are also true. In particular,

[TABLE]

Let $j,k\in\{0,1,2\}$ with $j\neq k$. The map $\alpha_{kj}^{M}$ maps the summand $\operatorname{im}(\alpha_{jk}^{M})$ to the summand $\operatorname{im}(\alpha_{kj}^{M})$ in $M_{j}$ and vanishes on the other summand in $M_{k}$. The exactness of $M$ shows that the restriction

[TABLE]

is invertible. As a consequence, $\alpha_{kj}^{M}\circ\alpha_{jk}^{M}$ is invertible on the direct summand $\operatorname{im}(\alpha_{kj}^{M})$. These statements for $k=1,2$ are one of the equivalent characterisations of unique $p$-divisibility in Propositions 6.8 and 7.17, respectively. So $M_{0}$ and $M_{2}$ are also uniquely $p$-divisible, and then so are all direct summands in them. Now it is routine to check that our exact $\mathfrak{K}$-module is isomorphic to the one built in Example 7.1 with $X=\operatorname{im}(\alpha_{01}^{M})\cong\operatorname{im}(\alpha_{10}^{M})$, $Y=\operatorname{im}(\alpha_{02}^{M})\cong\operatorname{im}(\alpha_{20}^{M})$, $Z=\operatorname{im}(\alpha_{12}^{M})\cong\operatorname{im}(\alpha_{21}^{M})$. This proves the theorem when $M_{0}$ is uniquely $p$-divisible. We may reduce the case where $M_{1}$ is assumed uniquely $p$-divisible to this case using the ring automorphism $\mathrm{BS}$, which exchanges the roles of $M_{0}$ and $M_{1}$.

Now let $M_{2}$ be uniquely $p$-divisible. Proposition 7.17 shows that $M_{2}$ is cohomologically trivial as an $\mathfrak{S}_{2}$-module and decomposes as $M_{2}=\operatorname{im}(1_{2}-s_{2}^{M})\oplus\operatorname{im}(1_{2}-t_{2}^{M})$. Thus the statements in (7.9) and (7.10) hold. Therefore, $M_{2}=\operatorname{im}(\alpha_{21}^{M})\oplus\operatorname{im}(\alpha_{20}^{M})$ as well. Then in each of the Lemmas 7.19, 7.21 and 7.22, one of the equivalent statements is true. So all the equivalent statements in these lemmas are true. And now the argument proceeds as above. This finishes the proof of Theorem 7.2.

7.3. Application to Cuntz algebras

Theorem 7.2 applies, in particular, if $M_{1}=0$ because the trivial group is uniquely $p$-divisible. So any exact $\mathfrak{K}$-module with $M_{1}=0$ is isomorphic to one as in Example 7.1. And $M_{1}=0$ forces $X=0$ and $Z=0$. So the only remaining ingredient is a uniquely $p$-divisible $\mathbb{Z}/2$-graded $\mathbb{Z}[\vartheta]$-module $Y$, which may be chosen at will. The resulting exact $\mathfrak{K}$-module with $M_{1}=0$ has $M_{0}=Y$, $M_{2}=Y$, $\alpha_{20}^{M}=1$ and $\alpha_{02}^{M}=1-\vartheta$. The following theorem summarises our results in this case:

Theorem 7.23**.**

Let $Y$ be a countable, uniquely $p$-divisible Abelian group with a representation of $\mathbb{Z}/p$, $k\mapsto t^{k}$, such that $N(t)=0$. Then there is a pointwise outer action of $\mathbb{Z}/p$ on $\mathcal{O}_{2}\otimes\mathbb{K}(\ell^{2}\mathbb{N})$ that belongs to the equivariant bootstrap class $\mathfrak{B}^{G}$, such that $\mathrm{K}_{*}(\mathcal{O}_{2}\otimes\mathbb{K}(\ell^{2}\mathbb{N})\rtimes\mathbb{Z}/p)\cong Y$ and the action on $K$-theory induced by the dual action becomes the one on $Y$ generated by $t$. The $G$-action on $\mathcal{O}_{2}\otimes\mathbb{K}(\ell^{2}\mathbb{N})$ above is unique up to $\mathrm{KK}^{G}$-equivalence.

Proof.

Combine Theorem 5.61 with the classification of exact $\mathfrak{K}$-modules with $M_{1}=0$. Since $F_{*}(A)_{1}=\mathrm{K}_{*}(A)$, the condition $M_{1}=0$ corresponds to an action on a $\mathrm{C}^{*}$-algebra with vanishing $\mathrm{K}$-theory such as $\mathcal{O}_{2}$. And then $M_{0}=\mathrm{K}_{*}(A\rtimes G)$. ∎

A representation of $\mathbb{Z}/p$ on a uniquely $p$-divisible group with $N(t)=0$ may also be considered as a module over the ring $\mathbb{Z}[\vartheta,1/p]$. The action of the group ring $\mathbb{Z}[\mathbb{Z}/p]$ descends to an action of $\mathbb{Z}[\vartheta]$ if and only if $N(t)=0$. And the underlying group is uniquely $p$-divisible if and only if the action of $\mathbb{Z}[\vartheta]$ extends to $\mathbb{Z}[\vartheta,1/p]$.

More generally, Theorem 7.2 classifies the actions in the equivariant bootstrap class on any UCT Kirchberg algebra $A$ whose $\mathrm{K}$-theory is uniquely $p$-divisible. Namely, $\mathrm{KK}^{G}$-equivalence classes of such actions are in bijection with isomorphism classes of countable modules over $\mathbb{Z}[\vartheta,1/p]$, exactly as for $\mathcal{O}_{2}$. The exact $\mathfrak{K}$-module corresponding to a given countable module $Y$ over $\mathbb{Z}[\vartheta,1/p]$ is as in Example 7.1, where $X$ and $Z$ are obtained by decomposing $\mathrm{K}_{*}(A)$ into $\ker(1-t)=\operatorname{im}N(t)$ and $\operatorname{im}(1-t)=\ker N(t)$, where $t\colon\mathrm{K}_{*}(A)\to\mathrm{K}_{*}(A)$ is induced by the generator of the $\mathbb{Z}/p$-action on $A$.

We consider the classification of $\mathbb{Z}/p$-actions on Cuntz algebras as a test case for our theory. The result in the uniquely $p$-divisible case covers $\mathbb{Z}/p$-actions on $\mathcal{O}_{n+1}$ for all $n\in\mathbb{N}_{\geq 1}$ that are coprime to $p$. In each case, there are as many such actions as there are countable modules over $\mathbb{Z}[\vartheta,1/p]$.

An exact $\mathfrak{K}$-module as in Example 7.2 is a direct sum of three pieces, which are exact $\mathfrak{K}$-modules where two of the three pieces $X,Y,Z$ vanish. Theorem 7.2 says that any object of $\mathfrak{B}^{G}$ where one of the pieces in $F_{*}(A)$ is uniquely $p$-divisible is such a direct sum. Hence the rather strong lack of uniqueness of $\mathbb{Z}/p$-actions in the situation above is caused by the many actions of $\mathbb{Z}/p$ on $\mathcal{O}_{2}$. Given any object $A$ of $\mathfrak{B}^{G}$ and any action on $\mathcal{O}_{2}$, we get an action on $A\oplus\mathcal{O}_{2}$. We may then use Theorem 2.1 to replace $A\oplus\mathcal{O}_{2}$ by a stable UCT Kirchberg algebra, which is again isomorphic to $A$ if $A$ is already a stable UCT Kirchberg algebra.

7.4. Removing a uniquely divisible submodule

We shall need a generalisation of Theorem 7.2 where we weaken the assumption that $M_{1}$ be uniquely $p$-divisible and only require a given uniquely $p$-divisible submodule of $M_{1}$. Similar statements hold for submodules in the other entries of $M$, but we shall only use this case.

Theorem 7.24**.**

Let $M$ be an exact $\mathfrak{K}$-module. Let $M_{1}^{\prime}\subseteq M_{1}$ be an $s_{1}^{M}$-invariant, uniquely $p$-divisible $\mathbb{Z}/2$-graded subgroup. Let $M^{\prime}_{0}\mathrel{\vcentcolon=}\alpha_{01}^{M}(M^{\prime}_{1})$ and $M^{\prime}_{2}\mathrel{\vcentcolon=}\alpha_{21}^{M}(M^{\prime}_{1})$. Then $M^{\prime}$ is an exact $\mathfrak{K}$-submodule of $M$. It is uniquely $p$-divisible and isomorphic to an exact $\mathfrak{K}$-module as in Example 7.1, with $X=M^{\prime}_{0}$ and $Z=M^{\prime}_{2}$. The quotient $M/M^{\prime}$ is also exact.

Proof.

Each generator $\alpha_{jk}$ maps $M^{\prime}$ into itself because $M^{\prime}_{1}$ is $s_{1}^{M}$-invariant. So $M^{\prime}$ is a $\mathfrak{K}$-submodule. By construction, $\alpha_{01}^{M^{\prime}}$ and $\alpha_{21}^{M^{\prime}}$ are surjective and hence $\alpha_{20}^{M^{\prime}}=0$ and $\alpha_{02}^{M^{\prime}}=0$. Next we claim that $\alpha_{10}^{M^{\prime}}$ and $\alpha_{12}^{M^{\prime}}$ are injective. To prove this, we apply Proposition 6.8 to $M^{\prime}$. This gives us equalities

[TABLE]

and shows that the restrictions of $1_{1}^{M^{\prime}}-s_{1}^{M^{\prime}}$ and $N(s_{1}^{M^{\prime}})$ to the appropriate direct summands are invertible. The map $\alpha_{01}^{M^{\prime}}$ vanishes on $\operatorname{im}(1_{1}^{M^{\prime}}-s_{1}^{M^{\prime}})$ and is injective on $\operatorname{im}N(s_{1}^{M^{\prime}})$ because $N(s_{1}^{M^{\prime}})=\alpha_{10}^{M^{\prime}}\circ\alpha_{01}^{M^{\prime}}$ is injective there. So $M^{\prime}_{0}=\alpha_{01}^{M^{\prime}}(M^{\prime}_{1})\cong\operatorname{im}N(s_{1}^{M^{\prime}})$. Similarly, $M^{\prime}_{2}=\alpha_{21}^{M^{\prime}}(M^{\prime}_{1})\cong\operatorname{im}(1_{1}^{M^{\prime}}-s_{1}^{M^{\prime}})$. And since $N(s_{1}^{M^{\prime}})$ and $1_{1}^{M^{\prime}}-s_{1}^{M^{\prime}}$ are invertible on these direct summands in $M_{1}^{\prime}$, the maps $\alpha_{10}^{M^{\prime}}$ and $\alpha_{12}^{M^{\prime}}$ are injective, and their images are $\operatorname{im}N(s_{1}^{M^{\prime}})$ and $\operatorname{im}(1_{1}^{M^{\prime}}-s_{1}^{M^{\prime}})$, respectively, which are also the kernels of the maps $\alpha_{21}^{M^{\prime}}$ and $\alpha_{01}^{M^{\prime}}$. So the $\mathfrak{K}$-module $M^{\prime}$ is exact. The proof also describes it so explicitly that its unique $p$-divisibility is manifest. And we have directly shown that it has the form of Example 7.1 with $X=\operatorname{im}N(s_{1}^{M^{\prime}})$, $Y=0$ and $Z=\operatorname{im}(1_{1}^{M^{\prime}}-s_{1}^{M^{\prime}})$. A quotient of an exact $\mathfrak{K}$-module by an exact $\mathfrak{K}$-submodule inherits exactness by the long exact sequence in homology. So $M/M^{\prime}$ is exact. ∎

8. Exact modules where one generator vanishes

Section 7 already describes exact $\mathfrak{K}$-modules where one of the pieces $M_{j}$ vanishes or, equivalently, where one of the idempotents $1_{j}$ vanishes. Now we study exact $\mathfrak{K}$-modules where one of the generators $\alpha_{jk}$ vanishes. Then one of the long exact sequences in Definition 5.9 reduces to a short exact sequence. This is particularly helpful if the object in the middle of the short exact sequence is not $M_{2}$ because the internal structure of $M_{0}$ and $M_{1}$ is simpler than that of $M_{2}$.

8.1. The case $\alpha_{21}^{M}=0$

Let $M$ be an exact $\mathfrak{K}$-module on which $\alpha_{21}^{M}$ vanishes. This puts a restriction on $M_{1}$: if $\alpha_{21}^{M}=0$, then $s_{1}^{M}=1_{1}^{M}$ and $s_{2}^{M}=1_{2}^{M}$ by (5.24) and (5.26). By exactness, $\alpha_{21}^{M}=0$ is equivalent to $\alpha_{10}^{M}$ being surjective and to $\alpha_{02}^{M}$ being injective. And there is a short exact sequence

[TABLE]

Theorem 8.4 will describe the category of exact $\mathfrak{K}$-modules with $\alpha_{21}^{M}=0$ through certain triples. To clarify this, we first simplify an exact $\mathfrak{K}$-module with $\alpha_{21}^{M}=0$. Its central piece is $M_{0}$, which is a module over $\mathfrak{S}\mathrel{\vcentcolon=}\mathbb{Z}[t]/(t^{p}-1)$. We briefly write $t$ instead of $t_{0}^{M}$ to reduce clutter. This operator will also determine the operator $t_{2}^{M}$. And $\alpha_{21}^{M}=0$ and (5.26) and (5.24) imply $s_{1}=1_{1}$ and $s_{2}=1_{2}$.

First, we use the exact sequence above to identify $M_{2}$ and $M_{1}$ with the image and the cokernel of the injective map $\alpha_{02}^{M}$. So $M_{2}$ becomes a $\mathbb{Z}/2$-graded subgroup of $M_{0}$ and $M_{1}$ becomes the quotient $M_{0}/M_{2}$, and $\alpha_{02}^{M}$ and $\alpha_{10}^{M}$ become the inclusion map $M_{2}\rightarrowtail M_{0}$ and the quotient map $M_{0}\twoheadrightarrow M_{1}$, respectively.

The relations (5.23) and (5.16) imply that $\alpha_{20}^{M}\colon M_{0}\to M_{2}\subseteq M_{0}$ is the corestriction of $1-t\colon M_{0}\to M_{0}$ and that $\alpha_{01}^{M}\colon M_{1}\to M_{0}$ is the map on the quotient induced by $N(t)$. For this to make sense, we need

[TABLE]

Let

[TABLE]

The $\mathbb{Z}/2$-graded $\mathfrak{S}$-module structure on $M_{0}$ makes this subquotient a $\mathbb{Z}/2$-graded module over $\mathbb{Z}[t]/(1-t,N(t))\cong\mathbb{Z}/p$. A $\mathbb{Z}/2$-graded subgroup $M_{2}$ as in (8.2) is equivalent to a $\mathbb{Z}/2$-graded $\mathbb{Z}/p$-vector subspace $\dot{M}_{2}\subseteq H_{0}(M_{0})$. For an exact $\mathfrak{K}$-module, the kernel and image of $\alpha_{12}^{M}\colon M_{2}\to M_{1}$ must be

[TABLE]

So $\alpha_{12}^{M}$ is induced by an isomorphism

[TABLE]

The map $\dot{\alpha}$ must reverse the grading because $\alpha_{12}^{M}$ is odd, and

[TABLE]

Theorem 8.4**.**

The category of exact $\mathfrak{K}$-modules with $\alpha_{21}^{M}=0$ is equivalent to the category of triples $(M_{0},\dot{M}_{2},\dot{\alpha})$ where

1. (1)

$M_{0}$* is a $\mathbb{Z}/2$-graded $\mathfrak{S}$-module with $\ker(1-t)=\operatorname{im}N(t)$;* 2. (2)

$\dot{M}_{2}\subseteq H_{0}(M_{0})$* is a $\mathbb{Z}/2$-graded $\mathbb{Z}/p$-vector subspace;* 3. (3)

$\dot{\alpha}$* is a grading-reversing isomorphism $\dot{\alpha}\colon\dot{M}_{2}\to H_{0}(M_{0})/\dot{M}_{2}$.*

Morphisms of triples $(M_{0},\dot{M}_{2},\dot{\alpha})\to(L_{0},\dot{L}_{2},\dot{\beta})$ are grading-preserving $\mathfrak{S}$-module homomorphisms $f_{0}\colon M_{0}\to L_{0}$ such that $H_{0}(f_{0})\colon H_{0}(M_{0})\to H_{0}(L_{0})$ maps $\dot{M}_{2}$ to $\dot{L}_{2}$ and intertwines the maps $\dot{\alpha}$ and $\dot{\beta}$. The homomorphism of exact $\mathfrak{K}$-modules induced by a morphism of triples $f_{0}\colon M_{0}\to L_{0}$ is injective if and only if $f_{0}$ is injective and $f_{0}^{-1}(L_{2})=M_{2}$; it is surjective if and only if $f_{0}$ is surjective and $f_{0}(M_{2})=L_{2}$.

Proof.

We have already associated $M_{0}$, $\dot{M}_{2}$ and $\dot{\alpha}$ to an exact $\mathfrak{K}$-module. The exactness of one of the two sequences in Definition 5.9 is built into our Ansatz. By construction, the sequence $\alpha_{01}^{M},\alpha_{20}^{M},\alpha_{12}^{M}$ is exact at $M_{2}$ and $M_{1}$ if and only if the map $\dot{\alpha}$ is invertible. In addition,

[TABLE]

because $\alpha_{10}^{M}$ is surjective, and

[TABLE]

because $\alpha_{02}^{M}$ is injective. Exactness at $M_{0}$ requires $\ker(\alpha_{20}^{M})=\operatorname{im}(\alpha_{01}^{M})$, which is equivalent to $\ker(1-t)=\operatorname{im}N(t)$. So the triple $(M_{0},\dot{M}_{2},\dot{\alpha})$ associated to an exact $\mathfrak{K}$-module satisfies the conditions in the theorem. Conversely, assume that such a triple is given. Reading our construction backwards, let $M_{2}\subseteq\operatorname{im}(1-t)$ be the pre-image of $\dot{M}_{2}$ and let $M_{1}\mathrel{\vcentcolon=}M_{0}/M_{2}$. Let $\alpha_{02}^{M}$ be the inclusion map. Let $\alpha_{10}^{M}$ be the quotient map. Then we may define $\alpha_{20}^{M}$ and $\alpha_{01}^{M}$ as above, induced by $1-t$ and $N(t)$. Let $\alpha_{12}^{M}$ be induced by $\dot{\alpha}$ and let $\alpha_{21}^{M}\mathrel{\vcentcolon=}0$. Let $M=M_{0}\oplus M_{1}\oplus M_{2}$ and let $1_{j}$ for $j=0,1,2$ be the projections onto these summands. The operators defined above clearly satisfy the relations (5.12)–(5.15). By construction, $t_{0}^{M}$ is the given operator $t$ on $M_{0}$, and (5.16) holds. Since $M_{1}$ is a quotient of $M_{0}$, this implies (5.17). And (5.18) is equivalent to $N(t)|_{M_{2}}=0$, which holds because $M_{2}\subseteq\ker N(t)$. So all conditions in Theorem 5.11 hold and we have built a $\mathfrak{K}$-module. The arguments above also show that it is exact.

Let $M$ and $L$ be exact $\mathfrak{K}$-modules with $\alpha_{21}^{M}=0$ and $\alpha_{21}^{L}=0$ and let $f\colon M\to L$ be a $\mathfrak{K}$-module homomorphism. This specialises to an $\mathfrak{S}$-module homomorphism $f_{0}\colon M_{0}\to L_{0}$. Any $\mathfrak{S}$-module homomorphism maps $\operatorname{im}N(t^{M})$, $\ker N(t^{M})$, $\operatorname{im}(1-t^{M})$ and $\ker(1-t^{M})$ to the corresponding subgroups in $L_{0}$. So the condition $f_{0}(\alpha_{02}^{M}(M_{2}))\subseteq\alpha_{02}^{L}(L_{2})$ is equivalent to $H_{0}(f_{0})(\dot{M}_{2})\subseteq\dot{L}_{2}$. If an $\mathfrak{S}$-module homomorphism $f_{0}$ satisfies this, then it induces maps $f_{j}\colon M_{j}\to L_{j}$ for $j=1,2$ that intertwine the actions of $\alpha_{01}$, $\alpha_{10}$, $\alpha_{20}$, $\alpha_{02}$. They also intertwine the vanishing actions of $\alpha_{21}$. And $f_{1}\circ\alpha_{12}^{M}=\alpha_{12}^{M}\circ f_{2}$ is equivalent to $H_{0}(f_{0})\circ\dot{\alpha}=\dot{\beta}\circ H_{0}(f_{0})$. This proves the asserted equivalence of categories.

The homomorphism $f$ is injective or surjective if and only if $f_{2},f_{0},f_{1}$ are injective, respectively. View these maps as a transformation between the short exact sequences (8.1) for our two exact $\mathfrak{K}$-modules. If $f_{0}$ is injective, then so is $f_{2}$, and $f_{1}$ is injective if and only if $f_{0}^{-1}(L_{2})=M_{2}$. If $f_{0}$ is surjective, so is $f_{1}$, and $f_{2}$ is surjective if and only if $f_{0}(M_{2})=L_{2}$. These are the criteria in the theorem for $f$ to be injective or surjective, respectively. ∎

The case where both $\alpha_{12}^{M}$ and $\alpha_{21}^{M}$ vanish is particularly simple:

Corollary 8.5**.**

The category of exact $\mathfrak{K}$-modules with $\alpha_{12}^{M}=0$ and $\alpha_{21}^{M}=0$ is equivalent to the category of $\mathbb{Z}/2$-graded cohomologically trivial $\mathfrak{S}$-modules. A homomorphism $f_{0}\colon M_{0}\to L_{0}$ between two $\mathbb{Z}/2$-graded cohomologically trivial $\mathfrak{S}$-modules induces an injective or surjective homomorphism of exact $\mathfrak{K}$-modules if and only if $f_{0}$ is injective or surjective, respectively.

Proof.

Under the equivalence in Theorem 8.4, the condition $\alpha_{12}^{M}=0$ corresponds to $\dot{\alpha}=0$. Then $\dot{M}_{2}=\{0\}$ and hence $\operatorname{im}(1-t)=\ker N(t)$. So the pieces $\dot{M}_{2}$ and $\dot{\alpha}$ in the triples in Theorem 8.4 become redundant in this case. Both $M_{1}$ and $M_{2}$ are naturally isomorphic to subobjects and quotients of $M_{0}$. This implies the criterion for injectivity and surjectivity of the induced $\mathfrak{K}$-module homomorphism. ∎

The exact $\mathfrak{K}$-module built from a cohomologically trivial $\mathfrak{S}$-module $M_{0}$ has

[TABLE]

The map $\alpha_{01}^{M}\colon\operatorname{im}N(t)\rightarrowtail M_{0}$ is the canonical inclusion map and $\alpha_{10}^{M}\colon M_{0}\twoheadrightarrow\operatorname{im}N(t)$ is $N(t)$. When we identify $M_{2}$ with $\operatorname{im}(1-t)$, then $\alpha_{02}^{M}\colon\operatorname{im}(1-t)\rightarrowtail M_{0}$ also becomes the canonical inclusion map and $\alpha_{20}^{M}\colon M_{0}\twoheadrightarrow\operatorname{im}(1-t)$ becomes $1-t$, respectively. This is because the isomorphism $M_{0}/\ker(1-t)\cong\operatorname{im}(1-t)$ applies the map $1-t$. If $M_{0}$ is $\mathbb{Z}/2$-graded, then the pieces $M_{1}$ and $M_{2}$ have the induced $\mathbb{Z}/2$-grading.

Izumi [Izumi:Finite_group] has met a stronger property than cohomological triviality in his classification of group actions with the Rokhlin property. Roughly speaking, his assumption of complete cohomological triviality says that an $\mathfrak{S}$-module and all the submodules of $n$-torsion elements for $n\in\mathbb{N}_{\geq 2}$ are cohomologically trivial. He shows that this condition is necessary and sufficient for an $\mathfrak{S}$-module to be isomorphic to $\mathrm{K}_{*}(A)$ for an action with the Rokhlin property on a simple unital $\mathrm{C}^{*}$-algebra. Whereas Izumi can study all finite groups, we must limit our study to cyclic groups of prime order. As a benefit, we can treat actions without the Rokhlin property. The class of examples treated in Corollary 8.5 seems $\mathrm{K}$-theoretically quite close to the examples studied by Izumi in [Izumi:Finite_group]. Our classification theorem for these modules only needs the $\mathfrak{S}$-module $\mathrm{K}_{*}(A)$ and no further data. There are cohomologically trivial $\mathfrak{S}$-modules that are not completely cohomologically trivial and hence cannot correspond to actions with the Rokhlin property. In fact, Example 9.16 is of this type. It corresponds to certain $\mathbb{Z}/p$-actions on the Cuntz algebras $\mathcal{O}_{p^{k}+1}$ for $k\geq 1$.

8.2. The case $\alpha_{12}^{M}=0$

Now let $M$ be an exact $\mathfrak{K}$-module on which $\alpha_{12}^{M}$ vanishes. By exactness, this is equivalent to $\alpha_{20}^{M}$ being surjective and to $\alpha_{01}^{M}$ being injective. In other words, there is a short exact sequence

[TABLE]

This case is very similar to the case $\alpha_{21}^{M}=0$ studied in Section 8.1. Hence we go through the construction more quickly. We write $t\mathrel{\vcentcolon=}t_{0}^{M}$ for the generator of the $\mathfrak{S}$-module structure on $M_{0}$. We identify $M_{1}$ with a subgroup of $M_{0}$ and $M_{2}$ with the quotient $M_{0}/M_{1}$, such that $\alpha_{01}^{M}$ and $\alpha_{20}^{M}$ become the inclusion map $M_{1}\rightarrowtail M_{0}$ and the quotient map $M_{0}\twoheadrightarrow M_{1}$, respectively. The map $\alpha_{10}^{M}\colon M_{0}\to M_{1}\subseteq M_{0}$ is the corestriction of $N(t)\colon M_{0}\to M_{0}$, and $\alpha_{02}^{M}\colon M_{2}\to M_{0}$ is the map on the quotient induced by $1-t$. This makes sense if and only if

[TABLE]

Let

[TABLE]

This is naturally a module over $\mathbb{Z}[t]/(1-t,N(t))\cong\mathbb{Z}/p$. A subgroup $M_{1}$ as in (8.7) is equivalent to a $\mathbb{Z}/p$-vector subspace $\dot{M}_{1}\subseteq H^{0}(M_{0})$. Equation (5.15) and exactness say that the remaining generator $\alpha_{21}^{M}\colon M_{1}\to M_{2}$ has kernel $\operatorname{im}N(t)=\operatorname{im}\alpha_{10}^{M}\subseteq M_{1}$ and image $\ker\alpha_{02}^{M}=\ker(1-t)/M_{2}\subseteq M_{0}/M_{1}=M_{2}$. So $\alpha_{21}^{M}$ must be induced by a grading-reversing isomorphism

[TABLE]

The construction also shows that

[TABLE]

Theorem 8.9**.**

The category of exact $\mathfrak{K}$-modules with $\alpha_{12}^{M}=0$ is equivalent to the category of triples $(M_{0},\dot{M}_{1},\dot{\alpha})$ where

1. (1)

$M_{0}$* is a $\mathbb{Z}/2$-graded $\mathfrak{S}$-module with $\ker N(t)=\operatorname{im}(1-t)$;* 2. (2)

$\dot{M}_{1}\subseteq H^{0}(M_{0})$* is a $\mathbb{Z}/2$-graded $\mathbb{Z}/p$-vector subspace;* 3. (3)

$\dot{\alpha}$* is a grading-reversing isomorphism $\dot{\alpha}\colon\dot{M}_{1}\to H^{0}(M_{0})/\dot{M}_{1}$.*

Morphisms of triples $(M_{0},\dot{M}_{1},\dot{\alpha})\to(L_{0},\dot{L}_{1},\dot{\beta})$ are grading-preserving $\mathfrak{S}$-module homomorphisms $f_{0}\colon M_{0}\to L_{0}$ such that $H^{0}(f_{0})\colon H^{0}(M_{0})\to H^{0}(L_{0})$ maps $\dot{M}_{1}$ to $\dot{L}_{1}$ and intertwines the maps $\dot{\alpha}$ and $\dot{\beta}$. The homomorphism of exact $\mathfrak{K}$-modules induced by a morphism of triples $f_{0}\colon M_{0}\to L_{0}$ is injective if and only if $f_{0}$ is injective and $f_{0}^{-1}(L_{1})=M_{1}$; it is surjective if and only if $f_{0}$ is surjective and $f_{0}(M_{1})=L_{1}$.

The proof of Theorem 8.9 is analogous to the proof of Theorem 8.4, and we omit further details. Theorem 8.9 also specialises to the classification for exact $\mathfrak{K}$-modules with $\alpha_{12}^{M}=0$ and $\alpha_{21}^{M}=0$ in Corollary 8.5.

8.3. Other cases of a vanishing generator

The cases where one of the maps $\alpha_{20}^{M}$ or $\alpha_{02}^{M}$ vanishes are analogous to the cases where $\alpha_{21}^{M}$ or $\alpha_{12}^{M}$ vanishes. We may, in fact, reduce these cases to the cases treated above using the automorphism $\mathrm{BS}$ of the ring $\mathfrak{K}$, which exchanges the role of the summands $M_{0}$ and $M_{1}$ in a $\mathfrak{K}$-module. The analogues of Theorems 8.4 and 8.9 for these situations are:

Theorem 8.10**.**

The category of exact $\mathfrak{K}$-modules with $\alpha_{20}^{M}=0$ is equivalent to the category of triples $(M_{1},\dot{M}_{2},\dot{\alpha})$ where $M_{1}$ is a $\mathbb{Z}/2$-graded $\mathfrak{S}$-module that satisfies

[TABLE]

$\dot{M}_{2}\subseteq H_{0}(M_{1})$* is a $\mathbb{Z}/2$-graded $\mathbb{Z}/p$-vector subspace and $\dot{\alpha}$ is a grading-preserving isomorphism $\dot{\alpha}\colon\dot{M}_{2}\to H_{0}(M_{1})/\dot{M}_{2}$.*

Here a morphism $(M_{1},\dot{M}_{2},\dot{\alpha})\to(L_{1},\dot{L}_{2},\dot{\beta})$ is a grading-preserving $\mathfrak{S}$-module homomorphism $f_{1}\colon M_{1}\to L_{1}$ such that $H_{0}(f_{1})\colon H_{0}(M_{1})\to H_{0}(L_{1})$ maps $\dot{M}_{2}$ to $\dot{L}_{2}$ and intertwines the maps $\dot{\alpha}$ and $\dot{\beta}$. The homomorphism of exact $\mathfrak{K}$-modules induced by a morphism of triples $f_{1}\colon M_{1}\to L_{1}$ is injective if and only if $f_{1}$ is injective and $M_{2}=f^{-1}(L_{2})$; it is surjective if and only if $f_{1}\colon M_{1}\to L_{1}$ is surjective and $L_{2}=f(M_{2})$.

Theorem 8.12**.**

The category of exact $\mathfrak{K}$-modules with $\alpha_{02}^{M}=0$ is equivalent to the category of triples $(M_{1},\dot{M}_{0},\dot{\alpha})$ where $M_{1}$ is a $\mathbb{Z}/2$-graded $\mathfrak{S}$-module that satisfies

[TABLE]

$\dot{M}_{0}\subseteq H^{0}(M_{1})$* is a $\mathbb{Z}/2$-graded $\mathbb{Z}/p$-vector subspace and $\dot{\alpha}$ is a grading-preserving isomorphism $\dot{\alpha}\colon\dot{M}_{0}\to H^{0}(M_{1})/\dot{M}_{0}$.*

Morphisms $(M_{1},\dot{M}_{0},\dot{\alpha})\to(L_{1},\dot{L}_{0},\dot{\beta})$ are grading-preserving $\mathfrak{S}$-module homomorphisms $f_{1}\colon M_{1}\to L_{1}$ such that $H^{0}(f_{1})\colon H^{0}(M_{1})\to H^{0}(L_{1})$ maps $\dot{M}_{0}$ to $\dot{L}_{0}$ and intertwines the maps $\dot{\alpha}$ and $\dot{\beta}$. The homomorphism of exact $\mathfrak{K}$-modules induced by a morphism of triples $f_{1}\colon M_{1}\to L_{1}$ is injective if and only if $f_{1}$ is injective and $f_{1}^{-1}(L_{0})=M_{0}$; and it is surjective if and only if $f_{1}$ is surjective and $L_{0}=f(M_{0})$.

Since $\alpha_{21}$ and $\alpha_{12}$ are odd,

[TABLE]

in the situation of Theorem 8.10 and $M_{2}\mathrel{\vcentcolon=}\Sigma(M_{1}/M_{0})$ in the situation of Theorem 8.12. That is, the grading on $M_{2}$ is reversed in both cases.

There is also an analogue of these results for the cases where $\alpha_{01}^{M}=0$ or $\alpha_{10}^{M}=0$. Since the automorphism $\mathrm{BS}$ maps these two cases onto each other, it suffices to discuss one of them. The condition $\alpha_{01}^{M}=0$ holds if and only if $\alpha_{12}^{M}$ is surjective, if and only if $\alpha_{20}^{M}$ is injective, if and only if these maps fit into an exact sequence

[TABLE]

As above, we may use this to identify $M_{0}$ with a subgroup of $M_{2}$ and $M_{1}$ with $\Sigma(M_{2}/M_{0})$, so that $\alpha_{20}^{M}$ becomes the inclusion map and $\alpha_{12}^{M}$ the quotient map, made grading-reversing. Now $M_{2}$ is a module over the ring $\mathfrak{S}_{2}$ defined in (7.12), and we denote its generators more briefly as $t^{M}=t_{2}^{M}$ and $s^{M}=s_{2}^{M}$.

The map $\alpha_{02}^{M}\colon M_{2}\to M_{0}\subseteq M_{2}$ must be the corestriction of $\alpha_{20}^{M}\circ\alpha_{02}^{M}=1-t^{M}$ by (5.25), whereas $\alpha_{21}^{M}\colon M_{1}\to M_{2}$ must be the map on the quotient induced by $\alpha_{21}^{M}\circ\alpha_{12}^{M}=1-s^{M}$ by (5.26). So we need $\operatorname{im}(1-t^{M})\subseteq M_{0}\subseteq\ker(1-s^{M})$. Let

[TABLE]

The maps on the subquotient $H(M_{2})$ induced by $1-s^{M}$ and $1-t^{M}$ vanish, and then so does multiplication by $p$ because $p=(p-N(t))+(p-N(s))$ holds in $\mathfrak{S}_{2}$. So $H(M_{2})$ is a vector space over $\mathbb{Z}/p$. A subgroup $M_{0}$ as above is the preimage of a unique $\mathbb{Z}/p$-linear subspace $\dot{M}_{0}$ in $H(M_{2})$. Exactness dictates that the missing map $\alpha_{10}^{M}$ is a composite map

[TABLE]

for an isomorphism $\dot{\alpha}\colon\dot{M}_{0}\xrightarrow{\sim}H(M_{2})/\dot{M}_{0}$ and that $\ker(1-t^{M})=\operatorname{im}(1-s^{M})$. Conversely, any triple $(M_{2},\dot{M}_{0},\dot{\alpha})$ as above comes from an exact $\mathfrak{K}$-module with $\alpha_{01}^{M}=0$, and the construction above is an equivalence of categories.

Theorem 8.14**.**

The category of exact $\mathfrak{K}$-modules with $\alpha_{01}^{M}=0$ is equivalent to the category of triples $(M_{2},\dot{M}_{0},\dot{\alpha})$ where $M_{2}$ is a $\mathbb{Z}/2$-graded $\mathfrak{S}_{2}$-module that satisfies

[TABLE]

$\dot{M}_{0}\subseteq H(M_{2})$* is a $\mathbb{Z}/2$-graded $\mathbb{Z}/p$-vector subspace and $\dot{\alpha}$ is a grading-preserving isomorphism $\dot{\alpha}\colon\dot{M}_{0}\to H(M_{2})/\dot{M}_{0}$.*

Here a morphism $(M_{2},\dot{M}_{0},\dot{\alpha})\to(L_{2},\dot{L}_{0},\dot{\beta})$ is a grading-preserving $\mathfrak{S}_{2}$-module homomorphism $f_{2}\colon M_{2}\to L_{2}$ such that $H(f_{2})\colon H(M_{2})\to H(L_{2})$ maps $\dot{M}_{0}$ to $\dot{L}_{0}$ and intertwines the maps $\dot{\alpha}$ and $\dot{\beta}$. The homomorphism of exact $\mathfrak{K}$-modules induced by a morphism of triples $f_{2}\colon M_{2}\to L_{2}$ is injective if and only if $f_{2}$ is injective and $M_{0}=f^{-1}(L_{0})$; it is surjective if and only if $f_{2}\colon M_{2}\to L_{2}$ is surjective and $L_{0}=f(M_{0})$.

As above, the case where both $\alpha_{01}^{M}=0$ and $\alpha_{10}^{M}=0$ corresponds to cohomologically trivial $\mathfrak{S}_{2}$-modules, and then $\dot{M}_{0}$ and $\dot{\alpha}$ are redundant:

Corollary 8.16**.**

The category of exact $\mathfrak{K}$-modules with $\alpha_{01}^{M}=0$ and $\alpha_{10}^{M}=0$ is equivalent to the category of $\mathbb{Z}/2$-graded cohomologically trivial $\mathfrak{S}_{2}$-modules.

A homomorphism $f_{2}\colon M_{2}\to L_{2}$ between two $\mathbb{Z}/2$-graded cohomologically trivial $\mathfrak{S}_{2}$-modules induces an injective or surjective homomorphism of exact $\mathfrak{K}$-modules if and only if $f_{2}$ is injective or surjective, respectively.

9. Examples of exact modules

We now describe some exact $\mathfrak{K}$-modules where $M_{1}$ is a cyclic group. These correspond to actions of $G\mathrel{\vcentcolon=}\mathbb{Z}/p$ on Cuntz algebras. The examples in this section are representative in the following sense:

Theorem 9.1**.**

Let $M$ be an exact $\mathfrak{K}$-module where $M_{1}$ is a cyclic group. Then there is an extension $M^{\prime}\rightarrowtail M\twoheadrightarrow M^{\prime\prime}$ of exact $\mathfrak{K}$-modules where $M^{\prime}$ is one of the exact $\mathfrak{K}$-modules described in Examples 9.2, 9.4, 9.7, 9.9, 9.11, 9.12, 9.14, 9.15, 9.16 and 9.17 and $M^{\prime\prime}$ is uniquely $p$-divisible. So $M^{\prime\prime}$ is as in Example 7.1. The extension above need not split.

Theorem 9.1 will be proven in Section 10 by a case-by-case study.

For each example, we first define $\mathbb{Z}/2$-graded Abelian groups $M_{j}$ for $j=0,1,2$. So $M=M_{0}\oplus M_{1}\oplus M_{2}$ and $1_{j}$ for $j=0,1,2$ is the projection onto the direct summand $M_{j}$. Then we define maps $\alpha_{jm}^{M}\colon M_{m}\to M_{j}$ for $j,m\in\{0,1,2\}$ with $j\neq m$. We compute the resulting maps $s_{j}^{M}\colon M_{j}\to M_{j}$ for $j=1,2$ and $t_{j}^{M}\colon M_{j}\to M_{j}$ for $j=0,2$ as defined in (5.19). Then we check that the relations in Theorem 5.11 hold for these maps – so our data defines a $\mathfrak{K}$-module – and that the exact sequences in Definition 5.9 are exact – so our $\mathfrak{K}$-module is exact. Here the relations (5.12)–(5.14) in Theorem 5.11 hold by construction, and (5.15) is part of exactness. The remaining relations are equivalent to (5.20)–(5.22), which is what we check.

We always define $M_{0}$ so that it has an obvious module structure over the ring $\mathfrak{S}\mathrel{\vcentcolon=}\mathbb{Z}[t]/(t^{p}-1)$ and write $f(t)$ for the map of multiplication by $f(t)$ in this module for $f\in\mathfrak{S}$. Our definitions are always arranged so that $t_{0}^{M}=t$ in this notation. In many examples, $(M_{2},t_{2}^{M})$ is also a module over $\mathfrak{S}$ and the same notation applies. Sometimes, we need the full $\mathfrak{S}_{2}$-module structure on $M_{2}$ with $\mathfrak{S}_{2}$ as in (7.12).

The notation $(f)$ or $(f_{1},f_{2})$ in a context such as $\mathfrak{S}/(f)$ or $\mathfrak{S}/(f_{1},f_{2})$ denotes the ideal in $\mathfrak{S}$ generated by $f$ or by $f_{1},f_{2}$, respectively. Recall that $\mathbb{Z}[\vartheta]\cong\mathfrak{S}/(N(t))$ is the ring generated by a primitive $p$th root of unity.

We begin with two examples of exact $\mathfrak{K}$-modules with $M_{1}=\mathbb{Z}$ and $s_{1}^{M}=1_{1}^{M}$.

Example 9.2*.*

Let

[TABLE]

All these groups are in even degree. Let $\alpha_{21}^{M}\mathrel{\vcentcolon=}0$ and $\alpha_{12}^{M}\mathrel{\vcentcolon=}0$. Let $\alpha_{01}^{M}\colon M_{1}\to M_{0}$ be multiplication with $N(t)$. Let $\alpha_{10}^{M}\mathrel{\vcentcolon=}\mathrm{ev}_{1}$ be evaluation at $1$. Let $\alpha_{20}^{M}$ be the quotient map, and let $\alpha_{02}^{M}$ be induced by multiplication with $(1-t)$; this is well defined and injective because $1-t^{p}\mid(1-t)\cdot f$ for $f\in\mathbb{Z}[t]$ is equivalent to $N(t)\mid f$. We compute $t_{0}^{M}=t$, $t_{2}^{M}=t$, $s_{1}^{M}=1$, $s_{2}^{M}=1$. So $N(t_{0}^{M})=N(t)$, $N(t_{2}^{M})=0$, $N(s_{1}^{M})=p$, $N(s_{2}^{M})=p$. Now the relations (5.20)–(5.22) become obvious. We have already checked that $\alpha_{02}^{M}$ is injective. And $\alpha_{01}^{M}$ is clearly injective and $\alpha_{10}^{M}$ and $\alpha_{20}^{M}$ are surjective. The kernel of $\alpha_{20}^{M}$ is the ideal in $\mathfrak{S}$ generated by $N(t)$, which is isomorphic to $\mathbb{Z}$ because $(1-t)\cdot N(t)=1-t^{p}=0$ in $\mathfrak{S}$. So the maps $\alpha_{01}^{M}$ and $\alpha_{20}^{M}$ form a short exact sequence. So do the maps $\alpha_{02}^{M}$ and $\alpha_{10}^{M}$; the image of $\alpha_{02}^{M}$ is the kernel of $\alpha_{10}^{M}$ because any $f\in\mathbb{Z}[t]$ decomposes uniquely as $f=f(1)+(t-1)f_{2}$ for some $f_{2}\in\mathbb{Z}[t]$. Together with $\alpha_{12}^{M}=0$ and $\alpha_{21}^{M}=0$, this says that the remaining relation (5.15) holds and that the sequences in Definition 5.9 are exact. So the data above defines an exact $\mathfrak{K}$-module.

Let $\mathrm{tv}$ denote the trivial action of $G$ on the Cuntz algebra $\mathcal{O}_{\infty}$. Then $F_{*}(\mathcal{O}_{\infty},\mathrm{tv})$ is easily seen to be isomorphic to the $\mathfrak{K}$-module in Example 9.2. This gives another proof that Example 9.2 defines an exact $\mathfrak{K}$-module.

Example 9.3*.*

We prove by an example that the sequence in Theorem 9.1 need not split. This example is a variant of Example 9.2. Let

[TABLE]

Let $M$ be the exact $\mathfrak{K}$-module in Example 9.2. Then $\mathbb{Z}_{(p)}\otimes M$ is another exact $\mathfrak{K}$-module. Define

[TABLE]

Inspection shows that this is a $\mathfrak{K}$-submodule of $\mathbb{Z}_{(p)}\otimes M$. It contains $M\cong\mathbb{Z}\otimes M$. The quotient $M^{\prime\prime}\mathrel{\vcentcolon=}M^{\prime}/M$ is the uniquely $p$-divisible exact $\mathfrak{K}$-module with $M_{1}^{\prime\prime}=0$ and $M_{0}^{\prime\prime}\cong M_{2}^{\prime\prime}\cong\mathbb{Z}_{(p)}[\vartheta]/\mathbb{Z}[\vartheta]$. Since $M$ and $M^{\prime\prime}$ are exact, so is $M^{\prime}$. The extension $M\rightarrowtail M^{\prime}\twoheadrightarrow M^{\prime\prime}$ does not split because $M_{2}\rightarrowtail M^{\prime}_{2}\twoheadrightarrow M^{\prime\prime}_{2}$ does not split.

Example 9.4*.*

Let $Q$ be the quotient $\mathbb{Z}[\vartheta]$-module

[TABLE]

Let

[TABLE]

Here $\Sigma$ means that $Q$ is in degree $1$, whereas $\mathbb{Z}$ is in degree [math]. Define

[TABLE]

Let $\alpha_{21}^{M}\colon\mathbb{Z}\to Q$ be the composite of the unital ring homomorphism $\mathbb{Z}\hookrightarrow\mathbb{Z}[\vartheta,1/p]$ and the quotient map to $Q$. Notice that $\alpha_{12}^{M}$ and $\alpha_{21}^{M}$ are grading-reversing, whereas the other $\alpha_{jk}$ are grading-preserving. This is why we put $Q$ into odd degree. We compute $t_{0}^{M}=1\oplus\vartheta$, $t_{2}^{M}=\vartheta$, $s_{1}^{M}=1$, $s_{2}^{M}=1$. So $N(t_{0}^{M})=p\oplus 0$, $N(t_{2}^{M})=0$, $N(s_{1}^{M})=p$, $N(s_{2}^{M})=p$. The relations (5.20)–(5.22) are now clear. The maps $\alpha_{01}^{M}$ and $\alpha_{20}^{M}$ clearly form a short exact sequence, which matches $\alpha_{12}^{M}=0$. The image of $\alpha_{10}^{M}$ is $p\cdot\mathbb{Z}\subseteq\mathbb{Z}$. This is equal to the kernel of $\alpha_{21}^{M}$ because

[TABLE]

The kernel of $\alpha_{10}^{M}$ is the direct summand $\Sigma Q$. This is also the image of $\alpha_{02}^{M}$ because multiplication by $1-\vartheta$ on $Q$ is surjective by Lemma 6.10. This lemma also shows that the kernel of multiplication by $1-\vartheta$ on $Q$ is the subgroup isomorphic to $\mathbb{Z}/p$ generated by the class of $1$. So $\ker(\alpha_{02}^{M})=\operatorname{im}(\alpha_{21}^{M})$. Hence the remaining relation (5.15) holds and the $\mathfrak{K}$-module defined above is exact.

The remaining examples are exact $\mathfrak{K}$-modules with $M_{1}=\mathbb{Z}/p^{k}$ for some $k\in\mathbb{N}_{\geq 1}$. The following lemma will be used in several of these examples:

Lemma 9.6**.**

Let $X$ be an $\mathfrak{S}$-module with $(1-t^{X})^{2}=0$. Then

[TABLE]

If $p\neq 2$ and $p\cdot(1-t^{X})=0$, then $N(t^{X})=p$.

Proof.

Any polynomial $f\in\mathbb{Z}[t]$ may be written as $f=f(1)+f^{\prime}(1)\cdot(t-1)+f_{2}(t)\cdot(t-1)^{2}$. Then $f(t^{X})=f(1)+f^{\prime}(1)(t^{X}-1)$ because $(1-t^{X})^{2}=0$. Since $N(1)=p$ and $N^{\prime}(1)=\sum_{j=1}^{p-1}j=p(p-1)/2$, this implies the first claim. The second claim follows from the first one because $2\mid p-1$ if $p\neq 2$. ∎

We begin with the only example where the action of $\mathbb{Z}/p$ on $M_{1}$ generated by $s_{1}^{M}$ is non-trivial:

Example 9.7*.*

Assume $k\geq 2$ if $p\neq 2$ and assume $k\geq 3$ if $p=2$. Let $\tau\in\{1,\dotsc,p-1\}$. Let $\nu\mathrel{\vcentcolon=}p$ if $p\neq 2$ and $\nu\mathrel{\vcentcolon=}2(1-2^{k-2}\tau)$ if $p=2$. Define

[TABLE]

where $\Sigma$ denotes that $M_{2}$ is in degree $1$. For $x\in\mathbb{Z}$, let $[x]$ denote its class in one of the quotients $M_{j}$, $j=0,1,2$. Define

[TABLE]

Notice that $\alpha_{12}^{M}$ and $\alpha_{21}^{M}$ are grading-reversing and the other generators grading-preserving. We compute $t_{0}^{M}=1$, $t_{2}^{M}=1$, $s_{2}^{M}=1$ because $k\geq 2$, $s_{1}^{M}=1-\tau p^{k-1}$. So $N(t_{0}^{M})=p$, $N(t_{2}^{M})=p\equiv 0$, $N(s_{2}^{M})=p\equiv 0$. We claim that $N(s_{1}^{M})=\nu$. If $p=2$, then $N(s_{1}^{M})=1_{1}^{M}+s_{1}^{M}$ and the claim is trivial. Let $p\neq 2$. We compute

[TABLE]

and $p\cdot(1-s_{1}^{M})=0$. Since $p\neq 2$, Lemma 9.6 implies $N(s_{1}^{M})=p=\nu$. Hence (5.21) holds. Equation (5.20) is trivial for $p\neq 2$ and holds for $p=2$ because $\nu\equiv 2\bmod 2^{k-1}$. And (5.22) is trivial. The map $\alpha_{10}^{M}$ is injective with image $p\mathbb{Z}/p^{k}$, and $\alpha_{21}^{M}$ is surjective with kernel $p\mathbb{Z}/p^{k}$. The map $\alpha_{01}^{M}$ is surjective with kernel $p^{k-1}\mathbb{Z}/p^{k}$, and $\alpha_{12}^{M}$ is injective with image $p^{k-1}\mathbb{Z}/p^{k}$. Hence (5.15) holds and the sequences in Definition 5.9 are exact. So we have defined an exact $\mathfrak{K}$-module.

Example 9.9*.*

Assume $p\neq 2$ or $k\geq 2$. Let $\ell\geq 2$ and $\tau\in\{1,\dotsc,p-1\}$. Define

[TABLE]

[TABLE]

Recall that the inclusion $\mathbb{Z}\to\mathbb{Z}[\vartheta]$ induces an isomorphism $\mathbb{Z}/p\cong\mathbb{Z}[\vartheta]/(1-\vartheta)$ because $(1-t,N(t))=(1-t,p)$. The map $\alpha_{12}^{M}$ is well defined as the composite of this isomorphism with the map $\mathbb{Z}/p\to\mathbb{Z}/p^{k}$, $x\mapsto p^{k-1}x$. The map $\alpha_{12}^{M}$ is well defined as the composite of the inverse of this isomorphism with the map $\mathbb{Z}[\vartheta]/(1-\vartheta)\to\mathbb{Z}[\vartheta]/(1-\vartheta)^{l}$, $f\mapsto f\cdot(1-\vartheta)^{l-1}$.

If $k=1$, then the summand $\mathbb{Z}/p^{k-1}$ in $M_{0}$ disappears; this simplifies the formulas for the maps $\alpha_{jm}^{M}$ in this case. We compute

[TABLE]

Here $s_{1}^{M}=1_{1}^{M}$ follows because $\operatorname{im}\alpha_{21}^{M}=(1-\vartheta)^{\ell-1}/(1-\vartheta)^{\ell}$ is contained in $\ker\alpha_{12}^{M}=(1-\vartheta)/(1-\vartheta)^{\ell}$ because $\ell\geq 2$. We claim that $N(s_{2}^{M})=p$. This is clear if $k\geq 2$. Let $k=1$. Since $s_{1}^{M}=1$, a computation analogous to (9.8) shows that $(1-s_{2}^{M})^{2}=0$. And $p\cdot(1-s_{2}^{M})=0$ because $(1-\vartheta)^{\ell}$ divides $p(1-\vartheta)^{\ell-1}$. Since $p\neq 2$ if $k=1$, Lemma 9.6 implies $N(s_{2}^{M})=p$. Since $N(t_{2}^{M})=N(\vartheta)=0$, the condition $N(s_{2}^{M})+N(s_{2}^{M})=p\cdot 1_{2}^{M}$ in (5.22) holds in all allowed cases. Equations (5.21) and (5.20) are easy to check.

To check exactness, we will use that multiplication by $(1-\vartheta)^{\ell-1}$ and $1-\vartheta$ induce isomorphisms

[TABLE]

respectively. Then we compute

[TABLE]

This finishes the proof that $M$ is an exact $\mathfrak{K}$-module.

The little invariant contained in $M$ does not depend on the parameter $\tau$. The following lemma shows, however, that the isomorphism class of $M$ depends on $\tau$. Since there are actions of $G$ that realise the exact $\mathfrak{K}$-modules in this example, we get counterexamples of objects in $\mathfrak{B}^{G}$ that are not $\mathrm{KK}^{G}$-equivalent, but have isomorphic little invariants.

Lemma 9.10**.**

The $\mathfrak{K}$-modules $M$ in Example 9.9 for different values of $\tau$ are not isomorphic.

Proof.

Let $(x,y)\in M_{1}\oplus M_{1}$. There is $b\in M_{2}$ with $(1_{2}^{M}-t_{2}^{M})^{\ell-1}(b)=\alpha_{21}^{M}(x)$ and $\alpha_{12}^{M}(b)=y$ if and only if $y=\tau p^{k-1}\cdot x$. Hence the map of multiplication by $\tau p^{k-1}$ on $\mathbb{Z}/p^{k}$ is an isomorphism invariant of $M$. The information in this map is exactly the class of $\tau$ in $\mathbb{Z}/p$. ∎

Roughly speaking, Example 9.7 is the case $\ell=1$ of Example 9.9, which we have not allowed there. If we took $\ell=1$ in Example 9.9, we could simplify $\mathbb{Z}[\vartheta]/(1-\vartheta)^{\ell-1}=0$ and $\mathbb{Z}[\vartheta]/(1-\vartheta)^{\ell}\cong\mathbb{Z}/p$. Then the formulas for $\alpha_{jm}^{M}$ for $(j,m)\neq(0,1)$ are exactly those in Example 9.7. The formulas for $\alpha_{10}^{M}$ are the same as well if $p\neq 2$, but not if $p=2$. The subtle choice of $\alpha_{10}^{M}$ in Example 9.7 ensures that (5.21) also holds if $p=2$.

Example 9.11*.*

Define Q\mathrel{\vcentcolon=}\mathbb{Z}[\vartheta,1/p]\bigm{/}(1-\vartheta)\mathbb{Z}[\vartheta] as in (9.5). Define

[TABLE]

We describe elements of $M_{0}$ as equivalence classes of triples $[x,y,z]$ with $x\in\mathbb{Z}/p^{k}$, $y,z\in Q$. Define

[TABLE]

We have implicitly used the isomorphism $\mathbb{Z}/p\cong\mathbb{Z}[\vartheta]/(1-\vartheta)$ induced by the inclusion $\mathbb{Z}\to\mathbb{Z}[\vartheta]$ and the inclusion $\mathbb{Z}[\vartheta]/(1-\vartheta)\subseteq Q$. Notice that the maps $\alpha_{10}^{M}$ and $\alpha_{20}^{M}$ are well defined. And the maps $\alpha_{12}^{M}$ and $\alpha_{21}^{M}$ reverse the grading, while the other $\alpha_{jm}^{M}$ preserve the grading. We compute

[TABLE]

Now the relations (5.21), (5.20) and (5.22) are clear. And

[TABLE]

Here we have used that multiplication by $1-\vartheta$ on $Q$ is surjective and has the kernel $\mathbb{Z}[\vartheta]/(1-\vartheta)$ (see Lemma 6.10). The equivalence class of $([x],[f_{1}],[f_{2}])$ belongs to the kernel of $\alpha_{10}^{M}$ if and only if there is $y\in\mathbb{Z}/p$ with $x=p^{k-1}y$; then

[TABLE]

And the equivalence class of $([x],[f_{1}],[f_{2}])$ belongs to the kernel of $\alpha_{20}^{M}$ if and only if $[f_{2}]=0$ and $[f_{1}]\in\mathbb{Z}[\vartheta]/(1-\vartheta)\cong\mathbb{Z}/p$; then

[TABLE]

Thus $M$ is an exact $\mathfrak{K}$-module.

Example 9.12*.*

Let $k\geq 1$, $\ell\in\mathbb{N}_{\geq 1}$ and $c\in\{1,\dotsc,p-1\}$. Define

[TABLE]

Let $I$ be the ideal \bigl{(}(1-t)^{\ell}-cp^{k-1}N(t),p^{k}N(t)\bigr{)} in $\mathfrak{S}$. Let $\alpha_{21}^{M}=0$ and $\alpha_{12}^{M}=0$. Let $\alpha_{20}^{M}\colon M_{0}\to M_{2}$ be the quotient map; this exists because I\subseteq\bigl{(}(1-t)^{\ell},N(t)\bigr{)}. Let $\alpha_{02}^{M}\colon M_{2}\to M_{0}$ be induced by multiplication with $1-t$; this is well defined because (1-t)\cdot\bigl{(}(1-t)^{\ell}-cp^{k-1}N(t)\bigr{)}\equiv(1-t)^{\ell+1}\bmod t^{p}-1, so that $(1-t)\cdot((1-t)^{\ell},N(t))\subseteq I$ as ideals in $\mathfrak{S}$. Let $\alpha_{10}^{M}\colon M_{0}\to M_{1}$ be induced by $\mathrm{ev}_{1}\colon\mathfrak{S}\to\mathbb{Z}$; this is well defined because $N(1)=p$ and hence $\mathrm{ev}_{1}(I)\subseteq(p^{k})$. Let $\alpha_{01}^{M}\colon M_{1}\to M_{0}$ be induced by $\mathbb{Z}\to\mathfrak{S}$, $j\mapsto N(t)\cdot j$; this is well defined because $p^{k}N(t)\in I$.

We compute that $t_{0}^{M}=t$ and $t_{2}^{M}=t$ are multiplication with $t$ on $M_{0}$ and $M_{2}$, respectively, whereas $s_{j}^{M}=1$ for $j=1,2$. The composite maps $\alpha_{01}^{M}\circ\alpha_{10}^{M}\colon M_{0}\to M_{0}$ and $\alpha_{10}^{M}\circ\alpha_{01}^{M}\colon M_{1}\to M_{1}$ are multiplication with $N(t)$ and with $p$, respectively. Hence the relations (5.20), (5.21) and (5.22) hold.

The maps $\alpha_{20}^{M}$ and $\alpha_{10}^{M}$ are surjective by construction. And $\operatorname{im}(\alpha_{01}^{M})=\ker(\alpha_{20}^{M})$ holds if and only if the map $\operatorname{coker}(\alpha_{01}^{M})\to M_{2}$ induced by $\alpha_{20}^{M}$ is injective. Now $M_{2}=\mathfrak{S}/((1-t)^{\ell},N(t))$ and $\operatorname{coker}(\alpha_{01}^{M})=\mathfrak{S}/(I+\mathbb{Z}\cdot N(t))=\mathfrak{S}/(N(t),(1-t)^{\ell})$ as needed. Similarly, we compute

[TABLE]

because $(c,p)=1$. Hence the map from $\operatorname{coker}(\alpha_{02}^{M})=\mathfrak{S}/((1-t)+I)$ to $M_{1}\cong\mathfrak{S}/(1-t,p^{k})$ induced by $\alpha_{10}^{M}$ is injective, so that $\operatorname{im}(\alpha_{02}^{M})=\ker(\alpha_{10}^{M})$. To check that $M$ is an exact $\mathfrak{K}$-module, it remains to show that $\alpha_{02}^{M}$ and $\alpha_{01}^{M}$ are injective. These two claims translate to the following more elementary statements. First, if $f\in\mathfrak{S}$ satisfies $(1-t)f\in I$, then $f\in((1-t)^{\ell},N(t))$. Secondly, if $j\in\mathbb{Z}$ satisfies $jN(t)\in I$, then $p^{k}\mid j$. We check these elementary claims. Let $f\in\mathbb{Z}[t]$. First, if $(1-t)f\in I$, then there are $f_{1},f_{2},f_{3}\in\mathbb{Z}[t]$ with

[TABLE]

It follows that $pf_{3}(1)-cf_{2}(1)=0$. Hence $p\mid f_{2}(1)$. Then $f_{2}\in(1-t,p)$ and $(1-t)f$ belongs to the ideal generated by $1-t^{p},(1-t)^{\ell+1},p(1-t)^{\ell},p^{k}N(t)$. We may drop the generator $p(1-t)^{\ell}$ because $p\equiv N(t)\bmod(1-t)$ and $1-t^{p}\mid N(t)\cdot(1-t)^{\ell}$; here we need $\ell\geq 1$. So $f\in((1-t)^{\ell},N(t))$ as asserted. For the second claim, assume $jN(t)\in I$. Then there are $f_{1},f_{2},f_{3}\in\mathbb{Z}[t]$ with

[TABLE]

Hence $N(t)\mid f_{2}\cdot(1-t)^{\ell}$. Since multiplication by $1-t$ is injective on $\mathfrak{S}/(N(t))\cong\mathbb{Z}[\vartheta]$, it follows that $N\mid f_{2}$. Hence $p\mid f_{2}(1)$. Then (9.13) implies $p^{k}\mid j$.

Example 9.14*.*

Assume $k\geq 2$ or $p\neq 2$. Define

[TABLE]

[TABLE]

The map $\alpha_{12}^{M}$ is well defined because $\mathbb{Z}[t]/(N(t),t-1)=\mathbb{Z}[t]/(t-1,p)\cong\mathbb{Z}/p$ or, equivalently, $\mathbb{Z}[\vartheta]/(1-\vartheta)\cong\mathbb{Z}/p$. (If $k=1$, then the summand $\mathbb{Z}/p^{k-1}=0$ disappears and the formulas for the maps $\alpha_{jm}^{M}$ simplify accordingly. In particular, $\alpha_{01}^{M}$ and $\alpha_{10}^{M}$ become [math] in this case.) We compute

[TABLE]

The relations (5.20) and (5.21) are clear, and (5.22) is equivalent to $N(s_{2}^{M})=p$ because $N(t_{2}^{M})=0$. If $k\geq 2$, then $\alpha_{21}^{M}\circ\alpha_{12}^{M}=0$. Hence $s_{2}^{M}=1\oplus 1$ and $N(s_{2}^{M})=p$. If $k=1$, then $1_{2}^{M}-s_{2}^{M}=\alpha_{21}^{M}\circ\alpha_{12}^{M}$ maps $(f_{1},[f_{2}])\mapsto(0,[f_{1}(1)])$, where we use the canonical inclusion $\mathbb{Z}/p\cong\mathbb{Z}[\vartheta]/(1-\vartheta)\subseteq Q$. Hence $(1_{2}^{M}-s_{2}^{M})^{2}=0$. Since $p\cdot(1_{2}^{M}-s_{2}^{M})=0$ as well and $p\neq 2$, Lemma 9.6 implies $N(s_{2}^{M})=p$ as needed. The definitions imply

[TABLE]

Multiplication by $1-\vartheta$ is injective on $\Sigma\mathbb{Z}[\vartheta]$ with cokernel $\mathbb{Z}/p$ and surjective on $\Sigma Q$ with kernel $\mathbb{Z}/p$ by Lemmas 6.9 and 6.10. This implies

[TABLE]

It follows that $M$ is an exact $\mathfrak{K}$-module.

Example 9.15*.*

Define

[TABLE]

[TABLE]

The map $\alpha_{10}^{M}$ is well defined because $p^{k-1}N(1)=p^{k}$, and $\alpha_{12}^{M}$ is well defined because $\mathfrak{S}/(N(t),1-t)\cong\mathbb{Z}[\vartheta]/(1-\vartheta)\cong\mathbb{Z}/p$. We compute

[TABLE]

So $N(t_{0}^{M})=N(t)\oplus N(t)=N(t)\oplus 0$ and $N(t_{2}^{M})=0$. Now $\alpha_{01}^{M}\circ\alpha_{10}^{M}=N(t_{0}^{M})$ follows from $N(t)\cdot f=N(t)\cdot f(1)$ for all $f\in\mathfrak{S}$, giving (5.20). The relations (5.21) and (5.22) are clear. The map $\alpha_{10}^{M}$ is clearly surjective and $\alpha_{02}^{M}$ is injective (see Lemma 6.9). Since the kernel of $\mathfrak{S}\to\mathbb{Z}$, $f\mapsto f(1)$, is the ideal generated by $1-t$, it follows that $\ker(\alpha_{10}^{M})=\operatorname{im}(\alpha_{02}^{M})$. Together with $\alpha_{21}^{M}=0$, this is one of the exact sequences in Definition 5.9. The isomorphism $\mathbb{Z}[\vartheta]/(1-\vartheta)\cong\mathbb{Z}/p$ implies that

[TABLE]

And $\ker(\alpha_{20}^{M})=(N(t))/(p^{k-1}N(t))\oplus 0$ because multiplication by $1-t$ on $\mathfrak{S}/(N(t))$ is injective by Lemma 6.9. So $\ker(\alpha_{20}^{M})=\operatorname{im}(\alpha_{01}^{M})$. And $\ker(\alpha_{01}^{M})=p^{k-1}\mathbb{Z}/p^{k}=\operatorname{im}(\alpha_{12}^{M})$. So $M$ is an exact $\mathfrak{K}$-module.

Example 9.16*.*

Let $k\geq 1$. Let $\mathbb{Z}[\vartheta]\mathrel{\vcentcolon=}\mathfrak{S}/(N(t))$ and Q\mathrel{\vcentcolon=}\mathbb{Z}[\vartheta,1/p]\bigm{/}(1-\vartheta)\mathbb{Z}[\vartheta] as in (9.5). Define

[TABLE]

Define $\alpha_{12}^{M}\mathrel{\vcentcolon=}0$, $\alpha_{21}^{M}\mathrel{\vcentcolon=}0$, and

[TABLE]

The maps $\alpha_{10}^{M}$ and $\alpha_{20}^{M}$ are well defined – that is, they vanish on $(p^{k}N(t),0)$ and $(p^{k-1}N(t)j,j)$ for $j\in\mathbb{Z}/p$ – because $p^{k-1}N(1)=p^{k}\equiv 0$ in $M_{1}$ and $(1-t)j=0$ in $Q$ for all $j\in\mathbb{Z}/p$. And the map $\alpha_{02}^{M}$ is well defined – that is, it vanishes on $(N(t),0)$ – because $(1-t)N(t)=0$ in $\mathfrak{S}$. We compute

[TABLE]

This implies $N(t_{2}^{M})=N(\vartheta)=0$ and $N(s_{2}^{M})=p$, giving (5.22). And $\alpha_{10}^{M}\circ\alpha_{01}^{M}[x]=[p\cdot x]$ and $\alpha_{01}^{M}\circ\alpha_{10}^{M}[f_{1},f_{2}]=[N(t)\cdot f_{1},0]$; so (5.21) and (5.20) also hold.

The map $\alpha_{01}^{M}$ is injective. Its cokernel is the quotient of $\mathfrak{S}\oplus Q$ by $(N(t))\oplus\mathbb{Z}/p$ because elements of the form $(x\cdot N(t),0)$ are killed in it, and $[0,j]=[-p^{k-1}jN(t),0]$ holds in $M_{0}$. The map $\alpha_{20}^{M}$ induces an isomorphism from this to $\mathbb{Z}[\vartheta]\oplus Q$ because of Lemmas 6.9 and 6.10. So the maps $\alpha_{01}^{M}$ and $\alpha_{20}^{M}$ form a short exact sequence. The map $\alpha_{02}^{M}$ is injective because $(1-t)\cdot\mathfrak{S}\cap N(t)\cdot\mathfrak{S}=0$. Its cokernel is the quotient of $\mathfrak{S}\oplus Q$ by the subgroup generated by $(1-t)\mathfrak{S}\oplus Q$, $(p^{k}N(t))\oplus 0$ and $\{(p^{k-1}N(t)j,j)\,{:}\,\mathopen{}j\in\mathbb{Z}/p\}$. All of $Q$ is killed in this cokernel, and evaluation at $1$ is an isomorphism $\mathfrak{S}/(1-t)\cong\mathbb{Z}$, which maps $N(t)\mapsto p$. So the cokernel of $\alpha_{02}^{M}$ is isomorphic to $\mathbb{Z}/p^{k}$ through the map $\alpha_{10}^{M}$. Thus both sequences in Definition 5.9 are exact. This verifies the relation (5.15) and shows that the data above defines an exact $\mathfrak{K}$-module.

Example 9.17*.*

Let $p=2$. Then the ring $\mathfrak{K}$ has an automorphism that maps $\alpha_{jm}\mapsto\alpha_{\sigma(j)\sigma(m)}$ for the permutation $\sigma=(02)$. It also switches the $\mathbb{Z}/2$-grading on $M_{2}$. And it maps $s_{1}$ to $-s_{1}$. Hence each example above produces another example where the sign of $s_{1}$ is switched, by permuting the summands $M_{j}$ and the maps $\alpha_{jm}^{M}$ according to the permutation $\sigma$. We denote it by $M^{\sigma}$.

We claim that $M^{\sigma}$ is truly a new example except in Examples 9.11 and 9.15 in the special case $k=1$. The only example above where $s_{1}^{M}\neq 1_{1}^{M}$ is Example 9.7, where $s_{1}^{M}=1-\tau p^{k-1}$. Since $k\geq 3$ is assumed if $p=2$, it follows that $1-\tau p^{k-1}\equiv 1\bmod 4$ and hence $-(1-\tau p^{k-1})\equiv 3\bmod 4$. So the automorphism applied to Example 9.7 gives new examples. If $s_{1}^{M}=1_{1}^{M}$, then $-s_{1}^{M}=s_{1}^{M}$ if and only if multiplication by $2$ is the identity map on $M_{1}$. This only happens if $M_{1}=\mathbb{Z}/2$. This is not possible in Examples 9.2, 9.4, 9.9, 9.14, so they always give new examples. The other examples Examples 9.11, 9.12, 9.15 and 9.16 also give new examples if $p=2$ and $k\geq 2$. We examine what happens in the latter examples if $p=2$ and $k=1$. It can be seen that $M\cong M^{\sigma}$ in the situation of Examples 9.11 and 9.15 if $k=1$ and $p=2$. In contrast, we claim that Examples 9.12 and 9.16 give new examples also for $k=1$ and $p=2$. Since $\alpha_{21}^{M}=0$ and $\alpha_{12}^{M}=0$ in these two examples, $\alpha_{01}^{M^{\sigma}}=0$ and $\alpha_{10}^{M^{\sigma}}=0$. No other example with $k=1$ and $p=2$ has this property. As a result, $M^{\sigma}$ is indeed a new example except for Examples 9.11 and 9.15 with $k=1$.

10. Proof of Theorem 9.1

Let $M$ be an arbitrary exact $\mathfrak{K}$-module where $M_{1}$ is a cyclic group. First, we study the possibilities for the action of $\mathbb{Z}/p$ on $M_{1}$.

We begin with the case $M_{1}=\mathbb{Z}$. The automorphism group of $M_{1}$ has two elements, and the non-trivial one acts by $x\mapsto-x$. This automorphism of order $2$ only occurs if $p=2$. In the case $p=2$, there is an automorphism of the ring $\mathfrak{K}$ that maps $s_{1}$ to $-s_{1}$. Since Example 9.17 takes this automorphism into account, Theorem 9.1 holds in the case $M_{1}=\mathbb{Z}$ once it holds in the case where both $M_{1}=\mathbb{Z}$ and $s_{1}^{M}=1_{1}^{M}$. Then $N(s_{1}^{M})=p$. So

[TABLE]

Next, let $M_{1}$ be a finite cyclic group. Then $M_{1}\cong\mathbb{Z}/p^{k}\oplus M_{1}^{\prime}$ with a cyclic group $M_{1}^{\prime}$ of order coprime to $p$. So $M_{1}^{\prime}$ is uniquely $p$-divisible. Theorem 7.24 extends $M^{\prime}_{1}$ to an exact $\mathfrak{K}$-submodule $M^{\prime}$ of $M$, which has the form of Example 7.1 with $Y=0$. The quotient $M^{\prime\prime}\mathrel{\vcentcolon=}M/M^{\prime}$ is an exact $\mathfrak{K}$-module with $M_{1}^{\prime\prime}=\mathbb{Z}/p^{k}$.

Lemma 10.2**.**

The submodule $M^{\prime}$ is a direct summand, so that $M=M^{\prime\prime}\oplus M^{\prime}$.

Proof.

The projection $e_{1}$ onto $M_{1}^{\prime}$ that kills $\mathbb{Z}/p^{k}$ commutes with $s_{1}^{M}$. Since $M^{\prime}_{1}$ is uniquely $p$-divisible, it splits as $M_{1}^{\prime}=N(s_{1}^{M})(M_{1}^{\prime})\oplus(1-s_{1}^{M})(M_{1}^{\prime})$, and $1-s_{1}^{M}$ is invertible on $(1-s_{1}^{M})(M_{1}^{\prime})$ (see Proposition 6.8). Let $e_{1N}$ and $e_{1-}$ be the projections onto $N(s_{1}^{M})(M_{1}^{\prime})$ and $(1-s_{1}^{M})(M_{1}^{\prime})$, respectively. Now the maps

[TABLE]

are well defined. Equations (5.21) and (5.24) imply $e_{0}^{2}=e_{0}$, $e_{2}^{2}=e_{2}$, and their images are easily seen to be $M^{\prime}_{0}$ and $M^{\prime}_{2}$, respectively. The idempotent maps $e_{j}$ for $j=0,1,2$ form a $\mathfrak{K}$-module map. So they give the asserted direct sum decomposition. ∎

As a result, if Theorem 9.1 holds in the case where $M_{1}=\mathbb{Z}/p^{k}$ for some $k\in\mathbb{N}_{\geq 1}$, then it holds whenever $M_{1}=\mathbb{Z}/n$ with $n\in\mathbb{N}_{\geq 1}$. So in the case of finite $M_{1}$, we may assume without loss of generality that $M_{1}=\mathbb{Z}/p^{k}$.

Now we consider the possible representations of $\mathbb{Z}/p$ on $M_{1}=\mathbb{Z}/p^{k}$. Any automorphism of $\mathbb{Z}/p^{k}$ is of the form $x\mapsto x\cdot m$ for a unit $m$ in the ring $\mathbb{Z}/p^{k}$. To get an action of $\mathbb{Z}/p$, we need $m^{p}\equiv 1\bmod p$. This forces $m\equiv 1\bmod p$. So only the trivial automorphism is possible if $k=1$. Assume now that $k\geq 2$. If $p\neq 2$, then the multiplicative group of elements in $1+p\mathbb{Z}\bmod p^{k}$ is isomorphic to the additive group $p\mathbb{Z}/p^{k}\cong\mathbb{Z}/p^{k-1}$ because the power series defining the exponential function and the logarithm converge. Hence $m^{p}=1\bmod p$ holds if and only if $m=1-p^{k-1}\tau$ for some $\tau\in\mathbb{Z}/p$. If $p=2$, then the argument above only works if $m\equiv 1\bmod 4$. It shows that the possible solutions of $m^{p}=1\bmod 2^{k}$ with $m\equiv 1\bmod 4$ are $m=1-2^{k-1}\tau$ for $\tau\in\mathbb{Z}/2$, and $\tau\neq 0$ only if $k\geq 3$. So all solutions in $\mathbb{Z}/2^{k}$ are $\pm(1-2^{k-1}\tau)$ for the two signs and $\tau\in\mathbb{Z}/2$, and $\tau\neq 0$ is only allowed if $k\geq 3$. An automorphism of $\mathfrak{K}$ that switches the sign of $s_{1}$ is described in Section 5.2. Hence the cases $m=-(1-2^{k-1}\tau)$ are covered by Example 9.17. Therefore, if $M_{1}=\mathbb{Z}/p^{k}$, then we may assume without loss of generality that $s_{1}^{M}$ is multiplication with $m=1-p^{k-1}\tau$ for some $\tau\in\mathbb{Z}/p$, where $\tau=0$ if $k=1$ or if $k=2$ and $p=2$. Then we compute $(s_{1}^{M})^{j}(x)=m^{j}\cdot x=(1-p^{k-1}j\tau)x$ for $j\in\mathbb{N}$. If $p\neq 2$, then $N(s_{1}^{M})(x)=p\cdot x$ because $p\mid\sum_{j=1}^{p-1}j$. If $p=2$, then $N(s_{1}^{M})(x)=2\cdot(1-2^{k-2}\tau)\cdot x$. Since $1-2^{k-2}\tau$ for $k\geq 3$ is a unit in $\mathbb{Z}/2^{k}$, this implies

[TABLE]

Hence $\operatorname{coker}N(s_{1}^{\mathbb{Z}/p^{k}})\cong\mathbb{Z}/p$. Lemma 5.63 implies

[TABLE]

the third claim in (10.4) is equivalent to the second. Since $M$ is exact, the map $\alpha_{21}^{M}$ induces an isomorphism $\operatorname{coker}\alpha_{10}^{M}\cong\operatorname{im}\alpha_{21}^{M}$. Equations (10.1) and (10.3) show that $\ker N(s_{1}^{M})$ and $\operatorname{coker}N(s_{1}^{M})$ are both $\mathbb{Z}/p$. Since the only subgroups of $\mathbb{Z}/p$ are $\{0\}$ and $\mathbb{Z}/p$, our classification problem splits into four cases:

1. (1)

$\ker\alpha_{01}^{M}=\{0\}$ and $\operatorname{coker}\alpha_{10}^{M}\cong\operatorname{im}\alpha_{21}^{M}=\{0\}$; 2. (2)

$\ker\alpha_{01}^{M}=\{0\}$ and $\operatorname{coker}\alpha_{10}^{M}\cong\operatorname{im}\alpha_{21}^{M}\cong\mathbb{Z}/p$; 3. (3)

$\ker\alpha_{01}^{M}\cong\mathbb{Z}/p$ and $\operatorname{coker}\alpha_{10}^{M}\cong\operatorname{im}\alpha_{21}^{M}=\{0\}$; 4. (4)

$\ker\alpha_{01}^{M}\cong\mathbb{Z}/p$ and $\operatorname{coker}\alpha_{10}^{M}\cong\operatorname{im}\alpha_{21}^{M}\cong\mathbb{Z}/p$;

We split the proof of Theorem 9.1 into several propositions. The first one handles exact $\mathfrak{K}$-modules with $M_{1}=\mathbb{Z}$ (Proposition 10.5). Then we treat exact $\mathfrak{K}$-modules with $M_{1}=\mathbb{Z}/p^{k}$ in the first three cases above in Propositions 10.8, 10.14 and 10.16, respectively. The fourth case is split further into Propositions 10.17 and 10.18. These propositions together will prove Theorem 9.1.

Proposition 10.5**.**

Let $M$ be an exact $\mathfrak{K}$-module with $M_{1}=\mathbb{Z}$ and $s_{1}^{M}=1_{1}^{M}$. Then $\operatorname{coker}\alpha_{10}^{M}=\{0\}$ or $\operatorname{coker}\alpha_{10}^{M}=\mathbb{Z}/p$. In the first case, let $M^{\prime}$ be the exact $\mathfrak{K}$-module in Example 9.2. In the second case, let $M^{\prime}$ be the exact $\mathfrak{K}$-module in Example 9.4. There is a $\mathfrak{K}$-module extension $M^{\prime}\rightarrowtail M\twoheadrightarrow M^{\prime\prime}$ where $M^{\prime\prime}$ is a uniquely $p$-divisible exact $\mathfrak{K}$-module as in Example 7.1 with $X=0$ and $Z=0$.

Proof.

The assumption $s_{1}^{M}=1_{1}^{M}$ implies $N(s_{1}^{M})=p$. So $\ker N(s_{1}^{M})=0$ and $\operatorname{coker}N(s_{1}^{M})\cong\mathbb{Z}/p$. Hence $\ker\alpha_{01}^{M}=\{0\}$ and either $\operatorname{coker}\alpha_{10}^{M}=\{0\}$ or $\operatorname{coker}\alpha_{10}^{M}=\mathbb{Z}/p$. Thus $\alpha_{01}^{M}$ is injective, $\alpha_{12}^{M}=0$ and $\alpha_{20}^{M}$ is surjective because $M$ is exact. And the image of $\alpha_{21}^{M}$ is isomorphic to either [math] or $\Sigma\mathbb{Z}/p$; the suspension comes in because $\alpha_{21}^{M}$ reverses the grading.

Exact $\mathfrak{K}$-modules with $\alpha_{12}^{M}=0$ are studied in Section 8.2. Theorem 8.9 shows that their isomorphism classes are in bijection with isomorphism classes of triples $(M_{0},\dot{M}_{1},\dot{\alpha})$, where $M_{0}$ is an $\mathfrak{S}$-module with $\ker N(t)=\operatorname{im}(1-t)$, $\dot{M}_{1}\subseteq H^{0}(M_{0})\mathrel{\vcentcolon=}\ker(1-t)/\operatorname{im}N(t)$ is a $\mathbb{Z}/p$-vector subspace and $\dot{\alpha}$ is a grading-preserving isomorphism $\dot{\alpha}\colon\dot{M}_{1}\xrightarrow{\sim}H^{0}(M_{0})/\dot{M}_{1}$. Equation (8.8) shows that $\dot{M}_{1}\cong\operatorname{im}(\alpha_{21}^{M})$. The two cases for $\operatorname{coker}\alpha_{10}^{M}$ correspond to triples with

1. (1)

$\dot{M}_{1}=0$; 2. (2)

$\dot{M}_{1}\cong\Sigma\mathbb{Z}/p$.

The existence of a grading-reversing isomorphism $\dot{\alpha}\colon\dot{M}_{1}\xrightarrow{\sim}H^{0}(M_{0})/\dot{M}_{1}$ implies $\ker(1-t)=\operatorname{im}N(t)$ in case (1) and $\ker(1-t)/\operatorname{im}N(t)\cong\mathbb{Z}/p\oplus\Sigma\mathbb{Z}/p$ in case (2). By construction, $M_{1}\subseteq M_{0}$ is an extension of $\operatorname{im}N(t)$ by $\dot{M}_{1}$. So $\operatorname{im}N(t)\cong\mathbb{Z}$ holds in both cases.

Let $M^{\prime}$ be the exact $\mathfrak{K}$-module in Example 9.2 if $\dot{M}_{1}=0$ and the exact $\mathfrak{K}$-module in Example 9.4 if $\dot{M}_{1}\cong\Sigma\mathbb{Z}/p$. Notice that $M^{\prime}_{1}=\mathbb{Z}$, $s_{1}^{M^{\prime}}=1_{1}^{M^{\prime}}$ in both examples, so that $M^{\prime}$ also corresponds to a triple $(M_{0}^{\prime},\dot{M}_{1}^{\prime},\dot{\alpha}^{\prime})$. And $\dot{M}_{1}^{\prime}\cong\dot{M}_{1}$ holds in both cases. We are going to define a morphism of triples $\psi\colon M_{0}^{\prime}\to M_{0}$ that restricts to an isomorphism $M_{1}^{\prime}\to M_{1}$ and is injective as a map $M_{0}^{\prime}\to M_{0}$. The criterion in Theorem 8.9 shows that this map $\psi$ induces an injective $\mathfrak{K}$-module homomorphism $\psi_{*}\colon M^{\prime}\hookrightarrow M$. Let $M^{\prime\prime}\mathrel{\vcentcolon=}M/M^{\prime}$. This is an exact $\mathfrak{K}$-module as well because both $M^{\prime}$ and $M$ are exact. And $M^{\prime\prime}_{1}=0$. Hence Theorem 7.2 shows that $M^{\prime\prime}$ is uniquely $p$-divisible and has the form of Example 7.1 with $X=0$ and $Z=0$. It remains to build the homomorphism of triples $\psi$ above.

We first do this in the case $\dot{M}_{1}=0$. Then $\ker(1-t)=\operatorname{im}N(t)$ and $\ker N(t)=\operatorname{im}(1-t)$. That is, $M_{0}$ is cohomologically trivial, and we are in the situation of Corollary 8.5. And

[TABLE]

where the isomorphism $M_{0}/\ker(1-t)\cong\operatorname{im}(1-t)$ is induced by $1-t$.

Lemma 10.6**.**

Let $a\in\operatorname{im}N(t)\cong\mathbb{Z}$ be a generator and choose $y\in M_{0}$ with $N(t)y=a$. Then $(1-t)y$ is a generator of $\operatorname{im}(1-t)/\operatorname{im}(1-t)^{2}$, which is isomorphic to $\mathbb{Z}/p$. And $\ker(1-t)\cap\operatorname{im}(1-t)=0$.

Proof.

The map $N(t)$ acts by multiplication by $p$ on $\operatorname{im}N(t)$ by Lemma 6.1. This is injective because $\operatorname{im}N(t)\cong\mathbb{Z}$. So $\ker N(t)\cap\operatorname{im}N(t)=0$. This is equivalent to $\ker(1-t)\cap\operatorname{im}(1-t)=0$ by cohomological triviality. Equivalently, $1-t$ restricts to an injective map on $\operatorname{im}(1-t)$. Let $a\in\operatorname{im}N(t)\cong\mathbb{Z}$ be a generator. Let $x\in\operatorname{im}(1-t)$. This means that there is $z\in M_{0}$ with $x=(1-t)z$. This element is unique up to adding an element of $\ker(1-t)=\operatorname{im}N(t)\cong\mathbb{Z}$. We have $x\in\operatorname{im}(1-t)^{2}$ if and only if $z$ may be chosen in $\operatorname{im}(1-t)=\ker N(t)$. Equivalently, there is $m\in\mathbb{Z}$ so that $N(t)(z+m\cdot a)=0$. Now $N(t)(m\cdot a)=p\cdot m\cdot a$ because $t\cdot a=a$. So there is $m\in\mathbb{Z}$ with $N(t)(z+m\cdot a)=0$ if and only if $N(t)(z)$ belongs to the subgroup generated by $p\cdot a$. Hence there is a well defined injective map from $\operatorname{im}(1-t)/\operatorname{im}(1-t)^{2}$ to $\mathbb{Z}/p$ that maps the class of $x$ to the class of $N(t)z$ in $a\mathbb{Z}/pa\mathbb{Z}$. Since $a\in\operatorname{im}N(t)$, there is $y\in M_{0}$ with $N(t)y=a$. Then the map $\operatorname{im}(1-t)/\operatorname{im}(1-t)^{2}\to\mathbb{Z}/p$ maps $(1-t)y\in\operatorname{im}(1-t)$ to a non-trivial element. Hence this map is also surjective. ∎

Lemma 10.7**.**

The map $\psi\colon\mathfrak{S}\to M_{0}$, $f\mapsto f(t)y$, is an injective $\mathfrak{S}$-module homomorphism. Its image contains $M_{1}$ and its cokernel is uniquely $p$-divisible.

Proof.

The image of $\psi$ contains $\mathbb{Z}\cdot a$, which is $M_{1}$ by the choice of $a$. So $\psi$ has the same cokernel as the map $\mathbb{Z}[\vartheta]\to\operatorname{im}(1-t)$, $f\mapsto f(t)(1-t)y$. This map is injective with a uniquely $p$-divisible cokernel by Lemmas 6.9 and 10.6. It remains to show that $\psi$ is injective as well. The principal ideal in $\mathfrak{S}\mathrel{\vcentcolon=}\mathbb{Z}[t]/(t^{p}-1)$ generated by $t-1$ is isomorphic to $\mathbb{Z}[\vartheta]$ because $(t^{p}-1)\mid f\cdot(t-1)$ if and only if $N(t)\mid f$. The map $\psi$ restricts to an injective map from $(t-1)\cdot\mathfrak{S}$ to $\operatorname{im}(1-t)\subseteq M_{0}$ by Lemma 6.9. Let $\psi(f)=0$. Then $\psi((t-1)\cdot f)=0$ as well. This implies $N(t)\mid f(t)$ because $\psi$ is injective on $(t-1)\cdot\mathfrak{S}\cong\mathfrak{S}/(N(t))$. Then $f\equiv m\cdot N(t)\bmod(t^{p}-1)$ for some $m\in\mathbb{Z}$. Now $\psi(f)=m\cdot N(t)y=m\cdot a$, and this is only [math] if $m=0$. Then $t^{p}-1\mid f$, that is, $f=0$ in $\mathfrak{S}$. So $\psi$ is injective. ∎

The exact $\mathfrak{K}$-module $M^{\prime}$ in Example 9.2 also satisfies $\alpha_{21}^{M^{\prime}}=0$ and $\alpha_{12}^{M^{\prime}}=0$. So Corollary 8.5 shows that it is associated to a cohomologically trivial $\mathfrak{S}$-module, namely, $M_{0}^{\prime}=\mathfrak{S}$. So the map $\psi$ above induces a $\mathfrak{K}$-module homomorphism $M^{\prime}\to M$. It is injective by the criterion in Corollary 8.5. This finishes the proof in the case of exact $\mathfrak{K}$-modules with $M_{1}=\mathbb{Z}$, $\alpha_{21}^{M^{\prime}}=0$ and $\alpha_{12}^{M^{\prime}}=0$.

Now we treat the case where $\dot{M}_{1}\cong\Sigma\mathbb{Z}/p$. Since $N(t)|_{M_{1}}$ is multiplication by $p$, the maps $\alpha_{01}^{M}$ and $N(s_{1}^{M})$ are both injective and have the same image, namely, the subgroup $p\cdot\mathbb{Z}\subseteq\mathbb{Z}$. Hence the statement in Lemma 7.19.(2) holds. So the equivalent statements (1) and (3) in Lemma 7.19 hold as well. Thus

[TABLE]

as an $\mathfrak{S}$-module; here we have used (5.23) and that $\alpha_{20}^{M}$ is surjective. Lemma 6.10 applies and gives an injective $\mathfrak{S}$-module homomorphism $M_{2}^{\prime}\hookrightarrow M_{2}$ with $M_{2}^{\prime}=\Sigma Q$ as in Example 9.4, such that $M_{2}/M_{2}^{\prime}$ is uniquely $p$-divisible. The map $\alpha_{02}^{M}\colon\Sigma Q\to M_{0}$ has kernel $\Sigma\mathbb{Z}/p$ and $Q/(\mathbb{Z}/p)\cong Q$ by multiplication with $1-t$. Hence $M_{0}\cong M_{1}\oplus\Sigma Q$ as an $\mathfrak{S}$-module, and $\dot{M}_{1}$ is the image of $M_{1}$ in $H^{0}(M_{0})$. We may choose the generator $e\in\mathbb{Z}/p$ in the proof of Lemma 6.10 to be $\dot{\alpha}([1])$. Then the injective homomorphism $M_{0}^{\prime}\mathrel{\vcentcolon=}\mathbb{Z}\oplus\Sigma Q\hookrightarrow M_{0}$ is a morphism of triples as needed for Theorem 8.9. It is injective on $M_{0}^{\prime}$ and induces an isomorphism $M_{1}^{\prime}\xrightarrow{\sim}M_{1}$. Hence the induced homomorphism of exact $\mathfrak{K}$-modules is injective by Theorem 8.9. Its cokernel $M^{\prime\prime}$ is an exact $\mathfrak{K}$-module with $M_{1}^{\prime\prime}=0$. This finishes the proof in the case of exact $\mathfrak{K}$-modules with $M_{1}=\mathbb{Z}$, $\alpha_{21}^{M^{\prime}}=0$ and $\operatorname{im}(\alpha_{21}^{M})\cong\mathbb{Z}/p$. So Proposition 10.5 is proven. ∎

We have proven Theorem 9.1 in the case $M_{1}=\mathbb{Z}$. So from now on, we may assume that $M_{1}=\mathbb{Z}/p^{k}$ for some $k\geq 1$. If $s_{1}^{M}\neq 1$, then $\alpha_{12}^{M}\neq 0$ and $\alpha_{21}^{M}\neq 0$ by (5.19). We shall first treat the three cases where at least one of the maps $\alpha_{12}^{M}$ and $\alpha_{21}^{M}$ is zero. So we shall only meet exact $\mathfrak{K}$-modules with $s_{1}^{M}=1$ for a while.

The first case in our list has $\ker\alpha_{01}^{M}=\{0\}$ and $\operatorname{coker}\alpha_{10}^{M}=\{0\}$. The study of this case is parallel to the case $\dot{M}_{1}=0$ in the proof of Proposition 10.5. Again, $\alpha_{12}^{M}=0$ and $\alpha_{21}^{M}=0$ hold, so that we are in the situation of Corollary 8.5. Exact $\mathfrak{K}$-modules in this case are equivalent to cohomologically trivial $\mathfrak{S}$-modules with

[TABLE]

Proposition 10.8**.**

Let $M$ be an exact $\mathfrak{K}$-module with $M_{1}=\mathbb{Z}/p^{k}$ and $\ker\alpha_{01}^{M}=\{0\}$ and $\operatorname{coker}\alpha_{10}^{M}=\{0\}$. Then there is a $\mathfrak{K}$-module extension $M^{\prime}\rightarrowtail M\twoheadrightarrow M^{\prime\prime}$, where $M^{\prime}$ is one of the exact $\mathfrak{K}$-modules in Example 9.12 or 9.16 and $M^{\prime\prime}$ is a uniquely $p$-divisible exact $\mathfrak{K}$-module as in Example 7.1 with $X=0$ and $Z=0$.

Proof.

We begin by studying $M_{2}$, which we identify with $\ker N(t)=\operatorname{im}(1-t)$ as above. We compute the kernel and cokernel of $1-\vartheta$ acting on $M_{2}$. We use the Snake Lemma for the endomorphism $1-t$ of the extension $M_{2}\rightarrowtail M_{0}\twoheadrightarrow\operatorname{im}N(t)$, where the quotient map is $N(t)$. Using that $M_{0}$ is cohomologically trivial, this gives a long exact sequence

[TABLE]

Since $N(t)\colon\operatorname{im}N(t)\to\operatorname{im}N(t)$ is multiplication by $p$ on $\mathbb{Z}/p^{k}$, this implies

[TABLE]

More precisely, $\ker(1-t)|_{M_{2}}\cong\mathbb{Z}/p$ is generated by the class of $p^{k-1}$ in $\mathbb{Z}/p^{k}=M_{1}\cong\operatorname{im}N(t)$. The boundary map in the Snake Lemma maps the generator $1$ of $\mathbb{Z}/p^{k}$ to a generator of $\operatorname{coker}(1-t)|_{M_{2}}$. Choose $b\in M_{0}$ with $N(t)b=[1]$ in $\mathbb{Z}/p^{k}$. Then $(1-t)b\in\operatorname{im}(1-t)=M_{2}$ represents the generator of $\operatorname{coker}(1-t)|_{M_{2}}$. Next we classify the resulting modules over $\mathbb{Z}[\vartheta]$:

Lemma 10.10**.**

Let $M_{2}$ be a $\mathbb{Z}[\vartheta]$-module with (10.9). Then there is a submodule $M_{2}^{\prime}\subseteq M_{2}$ such that $M_{2}/M_{2}^{\prime}$ is uniquely $p$-divisible, and which has the following form. Either $M_{2}^{\prime}=\mathbb{Z}[\vartheta]/(1-\vartheta)^{\ell}$ for some $\ell\in\mathbb{N}_{\geq 1}$ or $M_{2}^{\prime}=\mathbb{Z}[\vartheta]\oplus Q$.

Proof.

Let $a_{0}\in M_{2}$ generate $\ker(1-\vartheta)$. We recursively construct $a_{n}\in M_{2}$ with $(1-\vartheta)a_{n}=a_{n-1}$ as long as $a_{n-1}\in\operatorname{im}(1-\vartheta)$. Assume first that this process stops at some $\ell\in\mathbb{N}$, that is, $a_{0},\dotsc,a_{\ell-1}$ are defined and $a_{\ell-1}\notin\operatorname{im}(1-\vartheta)$. Define $\psi\colon\mathbb{Z}[\vartheta]\to M_{2}$, $f\mapsto f\cdot a_{\ell-1}$. So \psi\bigl{(}(1-\vartheta)^{j}\bigr{)}=a_{\ell-j-1} for $j=0,\dotsc,\ell-1$ and \psi\bigl{(}(1-\vartheta)^{\ell}\bigr{)}=0. An argument as in the proof of Lemma 6.10 shows that the map $\mathbb{Z}[\vartheta]/(1-\vartheta)^{\ell}\to M_{2}$ induced by $\psi$ is injective. And $\psi$ induces an isomorphism between the kernel and cokernel of multiplication by $1-\vartheta$ on $\mathbb{Z}[\vartheta]/(1-\vartheta)^{\ell}$ and $M_{2}$. By the Snake Lemma, multiplication by $1-\vartheta$ is invertible on the cokernel of $\psi$. Then the cokernel of $\psi$ is uniquely $p$-divisible by Proposition 6.8. So $M_{2}$ has the asserted form with $M_{2}^{\prime}=\mathbb{Z}[\vartheta]/(1-\vartheta)^{\ell}$.

Now assume that $a_{\ell}$ is defined for all $\ell\in\mathbb{N}$. Let $X_{0}\subseteq M_{2}$ be the subgroup generated by the $a_{\ell}$. This is a submodule because $(1-\vartheta)a_{\ell}=a_{\ell-1}$ for $\ell\geq 1$ and $(1-\vartheta)a_{0}=0$. By construction, multiplication by $1-\vartheta$ on $X_{0}$ is surjective and its kernel is $(\mathbb{Z}/p)\cdot a_{0}\cong\mathbb{Z}/p$. So the map $\psi_{0}\colon Q\to X_{0}$ defined by $(1-\vartheta)^{-n}f\mapsto f(\vartheta)a_{n}$ for all $n\in\mathbb{N}$, $f\in\mathbb{Z}[\vartheta]$ is an isomorphism by Lemma 6.10. The Snake Lemma implies that multiplication by $1-\vartheta$ is injective on $M_{2}/X_{0}$ and still has the same cokernel $\mathbb{Z}/p$. So Lemma 6.9 gives an injective $\mathbb{Z}[\vartheta]$-module homomorphism $\dot{\psi}_{1}\colon\mathbb{Z}[\vartheta]\to M_{2}/X_{0}$ with uniquely $p$-divisible cokernel. We may lift $\dot{\psi}_{1}$ to a $\mathbb{Z}[\vartheta]$-module homomorphism $\psi_{1}\colon\mathbb{Z}[\vartheta]\to M_{2}$ because $\mathbb{Z}[\vartheta]$ is free. Since $\dot{\psi}_{1}$ is injective, the resulting map

[TABLE]

is still injective. Its cokernel is isomorphic to the cokernel of $\dot{\psi}_{1}$, which is uniquely $p$-divisible. So $M_{2}^{\prime}$ has the asserted form. ∎

Let $M_{2}^{\prime}\subseteq M_{2}$ be the submodule given by Lemma 10.10. We study the two possibilities for $M_{2}^{\prime}$ separately.

Lemma 10.11**.**

Let $M_{2}^{\prime}=\mathbb{Z}[\vartheta]/(1-\vartheta)^{\ell}$ in Lemma 10.10. There are $c\in\{1,\dotsc,p-1\}$ and a submodule $M_{0}^{\prime}\subseteq M_{0}$ that contains $M_{1}=\operatorname{im}N(t)$, such that

[TABLE]

and $M_{0}/M_{0}^{\prime}\cong M_{2}/M_{2}^{\prime}$ is uniquely $p$-divisible.

Proof.

Above Lemma 10.10, we have chosen $b\in M_{0}$ with $N(t)b=[1]$ in $\mathbb{Z}/p^{k}$. We have seen that $(1-t)b$ generates $\operatorname{coker}(1-t)|_{M_{2}}$. We may add any element of $\ker N(t)=\operatorname{im}(1-t)=M_{2}$ to $b$ without changing $N(t)b=[1]$. This way, we may add an arbitrary element of $(1-t)M_{2}$ to $(1-t)b$ and arrange that $(1-t)b\in M_{2}^{\prime}$. The group $M_{2}^{\prime}$ is also a ring, and $(1-t)b$ must be invertible in its ring structure because it generates $\operatorname{coker}(1-t)|_{M_{2}^{\prime}}\cong\mathbb{Z}/p$. There is a group automorphism of $M_{2}^{\prime}$ that maps $(1-t)b$ to the unit element in $M_{2}^{\prime}$. Then we may assume without loss of generality that $(1-t)b=1$. Then $(1-t)^{\ell}b=(1-\vartheta)^{\ell-1}\in M_{2}^{\prime}$ is a generator of $\ker(1-t)|_{M_{2}}$. Above Lemma 10.10, we have seen that $[p^{k-1}]\in M_{1}\cap M_{2}$ is also a generator for $\ker(1-t)|_{M_{2}}$. So there is $c\in\{1,\dotsc,p-1\}$ such that

[TABLE]

That is, $(1-t)^{\ell}-cp^{k-1}N(t)$ belongs to the annihilator of $b$. So we get a well defined map

[TABLE]

We have built an exact $\mathfrak{K}$-module $M^{\prime}$ with M_{0}^{\prime}=\mathfrak{S}\bigm{/}\bigl{(}(1-t)^{\ell}-cp^{k-1}N(t),p^{k}N(t)\bigr{)} in Example 9.12. This example has $\alpha_{21}^{M}=0$ and $\alpha_{12}^{M}=0$, and so it falls under the classification in Corollary 8.5. So $M_{0}^{\prime}$ is cohomologically trivial and $M^{\prime}$ is the exact $\mathfrak{K}$-module associated to $M_{0}^{\prime}$ by the construction in Corollary 8.5. By Corollary 8.5, the $\mathfrak{S}$-module homomorphism $\psi$ above induces a homomorphism of exact $\mathfrak{K}$-modules $\psi_{*}\colon M^{\prime}\to M$, which is injective if and only if $\psi$ is injective. Actually, $\psi$ clearly induces an isomorphism $M_{1}^{\prime}\xrightarrow{\sim}M_{1}$. So injectivity on $M_{0}^{\prime}$ follows if the restriction of $\psi$ to $M_{2}^{\prime}=\ker N(t^{M^{\prime}})=\operatorname{im}N(t^{M^{\prime}})\subseteq M_{0}^{\prime}$ is injective. And this follows from Lemma 10.10 and the description of $M_{2}^{\prime}$ in Example 9.12. ∎

Next we consider the case with $M_{2}^{\prime}=\mathbb{Z}[\vartheta]\oplus Q$ in Lemma 10.10. As above, choose $b\in M_{0}$ with $N(t)b=[1]\in\mathbb{Z}/p^{k}$. Then $(1-t)b$ generates $\operatorname{coker}(1-t)|_{M_{2}}$. We use this element to generate the direct summand $\mathbb{Z}[\vartheta]$ in $M_{2}^{\prime}$. Recall that $M_{2}=\ker N(t)=\operatorname{im}(1-t)$ and that $\ker(1-t)\cap\operatorname{im}(1-t)=\ker N(t)\cap\operatorname{im}N(t)\cong\mathbb{Z}/p$ corresponds to the subgroup $p^{k-1}\mathbb{Z}/p^{k}$ in $M_{1}=\mathbb{Z}/p^{k}$. So $j\in\mathbb{Z}/p^{k}$ belongs to $M_{2}$ if and only if $p^{k-1}\mid j$. And $[p^{k-1}]=p^{k-1}N(t)b$ generates $\ker(1-t)|_{M_{2}}\cong\mathbb{Z}/p$. We use this generator to build the embedding $\psi_{Q}\colon Q\hookrightarrow M_{2}$ in the proof of Lemma 10.10. Then we define a map $\psi\colon\mathfrak{S}\oplus Q\to M_{0}$ by $\psi(f)\mathrel{\vcentcolon=}f(t)\cdot b$ for $f\in\mathfrak{S}$ and $\psi(x)\mathrel{\vcentcolon=}\psi_{Q}(x)$ for $x\in Q$.

Lemma 10.13**.**

The kernel of the map $\mathfrak{S}\oplus Q\to M_{0}$ above is the subgroup generated by $(p^{k-1}N(t),1)\in\mathfrak{S}\oplus Q$.

Proof.

We have built the embedding $\psi_{Q}$ so that $\psi(p^{k-1}N(t),1)=0$. Now let $(f_{1},f_{2})\in\mathfrak{S}\oplus Q$ satisfy $\psi(f_{1},f_{2})=0$. We must show that $(f_{1},f_{2})$ is a multiple of $(p^{k-1}N(t),1)$. Since the image of $\psi(f_{1},f_{2})$ in $M_{0}/M_{2}$ vanishes, $p^{k}$ divides $f_{1}(1)$. So we may write $f_{1}=j\cdot p^{k-1}N(t)+(1-t)f_{3}$ for some $j\in\mathbb{Z}$, $f_{3}\in\mathfrak{S}$. Then $(f_{1},f_{2})=j\cdot(p^{k-1}N(t),1)+((1-t)f_{3},f_{2}-j)$. Lemma 10.10 shows that $\psi$ restricts to an injective map on $(1-t)\mathfrak{S}\oplus Q\cong\mathbb{Z}[\vartheta]\oplus Q$. Hence $((1-t)f_{3},f_{2}-j)=0$. ∎

Let $M^{\prime}$ be the exact $\mathfrak{K}$-module in Example 9.16. Lemma 10.13 shows that the map $\psi\colon\mathfrak{S}\oplus Q\to M_{0}$ descends to an injective map $M_{0}^{\prime}\to M_{0}$. Now the same argument as above shows that the map $M^{\prime}\to M$ associated to $\psi$ is injective and has uniquely $p$-divisible cokernel. This finishes the proof of Proposition 10.8. ∎

The following proposition deals with the case $\ker\alpha_{01}^{M}=\{0\}$ and $\operatorname{coker}\alpha_{10}^{M}\cong\operatorname{im}\alpha_{21}^{M}\cong\mathbb{Z}/p$.

Proposition 10.14**.**

Let $M$ be an exact $\mathfrak{K}$-module. Assume that $M_{1}=\mathbb{Z}/p^{k}$ as an Abelian group, that $\ker\alpha_{01}^{M}=\{0\}$ and $\operatorname{im}\alpha_{21}^{M}\cong\mathbb{Z}/p$. Then there is a $\mathfrak{K}$-module extension $M^{\prime}\rightarrowtail M\twoheadrightarrow M^{\prime\prime}$, where $M^{\prime}$ is the exact $\mathfrak{K}$-module in Example 9.11 and $M^{\prime\prime}$ is a uniquely $p$-divisible exact $\mathfrak{K}$-module as in Example 7.1 with $X=0$ and $Z=0$.

Proof.

The assumption $\ker\alpha_{01}^{M}=\{0\}$ is equivalent to $\alpha_{12}^{M}=0$ by exactness. So we are in the situation of Theorem 8.9. Thus exact $\mathfrak{K}$-modules of this type are equivalent to certain triples $(M_{0},\dot{M}_{1},\dot{\alpha})$. Here $M_{0}$ is an $\mathfrak{S}$-module with $\ker N(t)=\operatorname{im}(1-t)$; $\dot{M}_{1}$ is a $\mathbb{Z}/p$-subvector space of H^{0}(M_{0})\mathrel{\vcentcolon=}\ker(1-t)\bigm{/}\operatorname{im}N(t) and $\dot{\alpha}$ is an isomorphism $\dot{\alpha}\colon\dot{M}_{1}\xrightarrow{\sim}H^{0}(M_{0})/\dot{M}_{1}$. The assumption $\operatorname{im}\alpha_{21}^{M}\cong\mathbb{Z}/p$ implies $\operatorname{im}\dot{\alpha}\cong\mathbb{Z}/p$ or, equivalently, $\dot{M}_{1}\cong\mathbb{Z}/p$. Hence $\ker(1-t)/M_{1}\cong H^{0}(M_{0})/\dot{M}_{1}\cong\Sigma\mathbb{Z}/p$. And $\operatorname{im}(1-t)\cong\mathbb{Z}/p^{k-1}$.

Lemma 10.15**.**

There is an injective $\mathfrak{S}$-module homomorphism $\psi^{\prime}\colon Q\oplus\Sigma Q\hookrightarrow\ker N(t)$ such that $\ker N(t)/\operatorname{im}(\psi^{\prime})$ is uniquely $p$-divisible.

Proof.

Since $pM_{1}=N(t)(M_{1})$ and $\operatorname{im}N(t)=N(t)(M_{0})$ are two subgroups in $M_{1}\cong\mathbb{Z}/p^{k}$ of index $p$, they are equal. Therefore, any $x\in M_{0}$ may be written as $x=x_{1}+x_{2}$ with $x_{1}\in M_{1}$ and $x_{2}\in\ker N(t)$: choose $x_{1}\in M_{1}$ with $N(t)x_{1}=N(t)x$ and put $x_{2}\mathrel{\vcentcolon=}x-x_{1}$. Since $(1-t)|_{M_{1}}=0$, it follows that $(1-t)(M_{0})=(1-t)(\ker N(t))$. Since $\ker N(t)=\operatorname{im}(1-t)$, the restriction $1-t\colon\ker N(t)\to\ker N(t)$ is surjective.

Next we claim that $\ker(1-t)\cap\ker N(t)\cong\mathbb{Z}/p\oplus\Sigma\mathbb{Z}/p$. Recall that $\ker(1-t)/M_{1}\cong\Sigma\mathbb{Z}/p$ in the case under study. Let $x\in\ker(1-t)$ represent the (odd) generator of $\ker(1-t)/M_{1}$. We have seen above that there is $x_{1}\in M_{1}$ with $N(t)x=N(t)x_{1}$. Then $x-x_{1}$ is another representative of the generator of $\ker(1-t)/M_{1}$ that also belongs to $\ker N(t)$. So any element of $\ker(1-t)/M_{1}\cong\Sigma\mathbb{Z}/p$ lifts to an element of $\ker(1-t)\cap\ker N(t)$. This lifting is unique modulo $\ker N(t)\cap M_{1}$. And $M_{1}=\mathbb{Z}/p^{k}$ implies that $\ker N(t)\cap M_{1}\cong\mathbb{Z}/p$. This finishes the proof that $\ker(1-t)\cap\ker N(t)\cong\mathbb{Z}/p\oplus\Sigma\mathbb{Z}/p$. Since the restriction of $1-t$ to $\ker(1-t)$ is surjective, Lemma 6.10 gives an embedding $\psi^{\prime}\colon Q\oplus\Sigma Q\to\ker(1-t)$ with uniquely $p$-divisible cokernel. ∎

Let $M^{\prime}$ be the exact $\mathfrak{K}$-module built in Example 9.11. Since $\alpha_{12}^{M^{\prime}}=0$, it is also associated to a suitable triple as in Theorem 8.9. Let $X\mathrel{\vcentcolon=}\operatorname{im}(\psi^{\prime})\subseteq\ker(1-t)\subseteq M_{0}$. Since $X$ contains $\ker(1-t)\cap\ker N(t)$, it follows from the proof of Lemma 10.15 that $M_{1}\cap X=M_{1}\cap\ker N(t)=\mathbb{Z}/p$. We may choose the map $\psi^{\prime}$ so that the canonical generator $[p^{k-1}]\in M_{1}=\mathbb{Z}/p^{k}$ of $M_{1}\cap X$ goes to $(1,0)$. Then $M_{1}+X\subseteq M_{0}$ is canonically isomorphic to $M_{0}^{\prime}$ as an $\mathfrak{S}$-module. We choose the generator of $\Sigma Q$ in $X$ to be $\alpha_{21}^{M}([1])$. Then the injective $\mathfrak{S}$-module homomorphism $\psi\colon M_{0}^{\prime}\to M_{0}$ is a homomorphism of triples as in Theorem 8.9. The criterion in Theorem 8.9 shows that the induced homomorphism of exact $\mathfrak{K}$-modules $M^{\prime}\to M$ is injective. As above, it follows that the quotient $M^{\prime\prime}\mathrel{\vcentcolon=}M/M^{\prime}$ is an exact $\mathfrak{K}$-module with $M^{\prime\prime}_{1}=0$. Then it is isomorphic to an exact $\mathfrak{K}$-module as in Example 7.1 with $X=0$ and $Z=0$. This finishes the proof of Proposition 10.14. ∎

Now we turn to the case $\ker\alpha_{01}^{M}=p^{k-1}\mathbb{Z}/p^{k}$ and $\operatorname{coker}\alpha_{10}^{M}=\{0\}$.

Proposition 10.16**.**

Let $M$ be an exact $\mathfrak{K}$-module. Assume that $M_{1}=\mathbb{Z}/p^{k}$ as an Abelian group, that $\ker\alpha_{01}^{M}=p^{k-1}\mathbb{Z}/p^{k}$ and $\operatorname{coker}\alpha_{10}^{M}=\{0\}$. Then there is a $\mathfrak{K}$-module extension $M^{\prime}\rightarrowtail M\twoheadrightarrow M^{\prime\prime}$, where $M^{\prime}$ is the exact $\mathfrak{K}$-module in Example 9.15 and $M^{\prime\prime}$ is a uniquely $p$-divisible exact $\mathfrak{K}$-module as in Example 7.1 with $X=0$ and $Z=0$.

Proof.

The assumptions $\ker\alpha_{01}^{M}=p^{k-1}\mathbb{Z}/p^{k}$ and $\operatorname{coker}\alpha_{10}^{M}=\{0\}$ are equivalent to $\operatorname{im}(\alpha_{12}^{M})\cong\mathbb{Z}/p$ and $\alpha_{21}^{M}=0$ by exactness. So we are in the situation of Theorem 8.4. Exact $\mathfrak{K}$-modules with $\alpha_{21}^{M}=0$ are equivalent to certain triples $(M_{0},\dot{M}_{2},\dot{\alpha})$, where $M_{0}$ is an $\mathfrak{S}$-module with $\ker(1-t)=\operatorname{im}N(t)$, $\dot{M}_{2}$ is a $\mathbb{Z}/p$-subvector space of $H_{0}(M_{0})\mathrel{\vcentcolon=}\frac{\ker N(t)}{\operatorname{im}(1-t)}$, and $\dot{\alpha}$ is a grading-reversing isomorphism $\dot{\alpha}\colon\dot{M}_{2}\xrightarrow{\sim}H_{0}(M_{0})/\dot{M}_{2}$. We have $\dot{M}_{2}\cong\operatorname{im}(\alpha_{12}^{M})\cong\mathbb{Z}/p$. We also need

[TABLE]

to be isomorphic to $\mathbb{Z}/p^{k}$. And $\alpha_{21}^{M}=0$ implies $1_{1}^{M}-s_{1}^{M}=0$. So Lemma 5.63 implies $\ker(1_{0}^{M}-t_{0}^{M})=\ker(\alpha_{20}^{M})$. This is equal to $\operatorname{im}(\alpha_{01}^{M})$ because $M$ is exact, and this is isomorphic to $\mathbb{Z}/p^{k-1}$ because $\ker(\alpha_{01}^{M})=\operatorname{im}(\alpha_{12}^{M})=p^{k-1}\mathbb{Z}/p^{k}$. In terms of our triples, this means that

[TABLE]

One short exact sequence in Table 1 implies $\ker(\alpha_{10}^{M})\cap\operatorname{im}(\alpha_{01}^{M})=0$ because $\alpha_{01}^{M}$ and $N(s_{1}^{M})$ have the same kernel, namely, $\mathbb{Z}/p$. Hence $\operatorname{im}(\alpha_{02}^{M})\cap\ker(1_{0}^{M}-t_{0}^{M})=0$. In terms of our triples above, this means that $\ker(1-t)\cap M_{2}=0$, that is, $1-t\colon M_{2}\to M_{2}$ is injective. And $\operatorname{coker}(\alpha_{10}^{M})=0$ and $\operatorname{coker}N(s_{1}^{M})\cong\mathbb{Z}/p$ imply

[TABLE]

The map $\alpha_{12}^{M}$ is a grading-reversing isomorphism from $\operatorname{coker}(\alpha_{20}^{M})$ onto $\operatorname{im}(\alpha_{12}^{M})\cong\mathbb{Z}/p$. So $\operatorname{coker}(\alpha_{20}^{M})\cong\Sigma\mathbb{Z}/p$. Now one of the short exact sequences in Table 1 gives an extension $\mathbb{Z}/p\rightarrowtail\operatorname{coker}(1_{2}^{M}-t_{2}^{M})\twoheadrightarrow\Sigma\mathbb{Z}/p$. So

[TABLE]

The proof also allows us to choose representatives $e_{0},e_{1}\in M_{2}$ for the generators of $M_{2}/(1-t)M_{2}$. First, let $a\in M_{0}$ be an even-degree element that lifts the generator of $M_{0}/M_{2}=M_{1}=\mathbb{Z}/p^{k}$. The image of $a$ in $\mathbb{Z}/p^{k}$ does not belong to $\operatorname{im}N(t)=\ker(1-t)$. Thus the element $e_{0}\mathrel{\vcentcolon=}(1-t)a$ of $\operatorname{im}(1-t)\subseteq M_{2}$ cannot be of the form $(1-t)x$ with $x\in M_{2}$ because $a-x_{2}$ is another representative of the generator of $\mathbb{Z}/p^{k}$ and so $N(t)(a-x_{2})\neq 0$. Secondly, there is an odd-degree element $e_{1}\in M_{2}$ with $\alpha_{12}^{M}(e_{1})=[p^{k-1}]\in\mathbb{Z}/p^{k}$. Then $e_{1}\notin(1-t)M_{2}$.

Now Lemma 6.9 shows that the grading-preserving map

[TABLE]

is injective with uniquely $p$-divisible cokernel. To include $M_{1}$, we use the grading-preserving map

[TABLE]

Its image surjects onto $M_{1}$ because $a$ lifts a generator of $M_{1}$. We claim that

[TABLE]

First, $\psi(f_{0},f_{1})\in M_{2}$ for $f_{0}\in\mathfrak{S}$, $f_{1}\in\Sigma\mathbb{Z}[\vartheta]$ is equivalent to $f_{0}\in(t-1,p^{k})=(t-1,p^{k-1}N(t))$ because $M_{1}\cong\mathbb{Z}/p^{k}\cong\mathfrak{S}/(t-1,p^{k})$ as an $\mathfrak{S}$-module. The element $p^{k-1}N(t)a$ becomes [math] in $M_{1}=M_{0}/M_{2}$, so that it belongs to $M_{2}$. And it also belongs to $\ker(1-t)$. Since $M_{2}\cap\ker(1-t)=0$, it follows that $p^{k-1}N(t)\cdot a=0$. That is, $(p^{k-1}N(t))\oplus 0$ is contained in the kernel of $\psi$. Conversely, let $\psi(f_{0},f_{1})=0$. Then $\psi(f_{0},f_{1})\in M_{2}$, so that $f_{0}\in(t-1,p^{k-1}N(t))$. That is, we may write $f_{0}=(1-t)f_{2}+p^{k-1}N(t)f_{3}$. The second summand is contained in the kernel of $\psi$. So $\psi((1-t)f_{2},f_{1})=f_{2}\cdot e_{0}+f_{1}\cdot e_{1}=0$. The injectivity of $\varphi$ above shows that $f_{2},f_{1}\in(N(t))$. Hence $(1-t)f_{2}=0$ in $\mathfrak{S}$ and $f_{1}=0$ in $\mathbb{Z}[\vartheta]$. So $p^{k-1}N(t)$ generates the kernel of $\psi$ as asserted. We have found an injective map from the $\mathfrak{S}$-module at [math] in Example 9.15 to $M_{0}$. This induces a homomorphism of exact $\mathfrak{K}$-modules by Theorem 8.4. Since it restricts to an isomorphism $M^{\prime}_{1}\xrightarrow{\sim}M_{1}$, the criterion in Theorem 8.4 shows that the map $M_{2}^{\prime}\to M_{2}$ is injective as well. The quotient $M^{\prime\prime}\mathrel{\vcentcolon=}M/M^{\prime}$ is an exact $\mathfrak{K}$-module with $M^{\prime\prime}_{1}=0$. Thus it is as in Example 7.1 with $X=0$ and $Z=0$. This finishes the proof of Proposition 10.16. ∎

Now we turn to the fourth case, $\ker(\alpha_{01}^{M})\cong\mathbb{Z}/p$ and $\operatorname{coker}(\alpha_{10}^{M})\cong\mathbb{Z}/p$. This is the only case where exact $\mathfrak{K}$-modules with a non-trivial $\mathbb{Z}/p$-action on $M_{1}$ may occur. We have already shown at the beginning of Section 10 that, without loss of generality, $s_{1}$ is multiplication with some element of $\mathbb{Z}/p^{k}$ of the form $1-p^{k-1}\tau$ for some $\tau\in\mathbb{Z}/p$, where $\tau=0$ if $k=1$ or if $k=2$ and $p=2$. This leaves out about half of the cases for $p=2$, which are covered by Example 9.17 because it allows to change the sign of $s_{1}^{M}$ if $p=2$ and $k>1$. The following proposition describes the exact $\mathfrak{K}$-modules with $M_{1}=\mathbb{Z}/p^{k}$ and $s_{1}^{M}\neq 1$. It gives Theorem 9.1 for all exact $\mathfrak{K}$-modules with $M_{1}=\mathbb{Z}/p^{k}$ and a non-trivial $\mathbb{Z}/p$-action on $M_{1}$.

Proposition 10.17**.**

Let $\tau\in\{1,\dotsc,p-1\}$. Assume $p\neq 2$ and $k\geq 2$, or $k\geq 3$. Let $M$ be an exact $\mathfrak{K}$-module with $M_{1}=\mathbb{Z}/p^{k}$ and $s_{1}^{M}(x)=(1-p^{k-1}\tau)\cdot x$. Then there is an injective homomorphism from the exact $\mathfrak{K}$-module $M^{\prime}$ in Example 9.7 into $M$ whose cokernel is a uniquely $p$-divisible exact $\mathfrak{K}$-module as in Example 7.1 with $X=0$ and $Z=0$.

The proof of Proposition 10.17 is parallel to the proof of the next proposition:

Proposition 10.18**.**

Let $M$ be an exact $\mathfrak{K}$-module with $M_{1}=\mathbb{Z}/p^{k}$, $s_{1}^{M}=1_{1}^{M}$, $\ker(\alpha_{01}^{M})\cong\mathbb{Z}/p$, and $\operatorname{coker}(\alpha_{10}^{M})\cong\mathbb{Z}/p$. Assume $k\geq 2$ or $p\neq 2$. Then there is an extension of exact $\mathfrak{K}$-modules $M^{\prime}\rightarrowtail M\twoheadrightarrow M^{\prime\prime}$, where $M^{\prime}$ is isomorphic to one of the exact $\mathfrak{K}$-modules described in Examples 9.9 and 9.14 and $M^{\prime\prime}$ is uniquely $p$-divisible and has the form described in Example 7.1 with $X=0$ and $Z=0$.

Proof of Propositions 10.17

and 10.18.

In the situation of Proposition 10.17, $s_{1}^{M}\neq 1_{1}^{M}$, forcing $\alpha_{12}^{M}\neq 0$ and $\alpha_{21}^{M}\neq 0$. Then $\alpha_{01}^{M}$ cannot be injective and $\alpha_{10}^{M}$ cannot be surjective. So we must be in the case where $\ker(\alpha_{01}^{M})\cong\mathbb{Z}/p$, and $\operatorname{coker}(\alpha_{10}^{M})\cong\mathbb{Z}/p$. This is already assumed in Proposition 10.18, so it holds in the situation of both propositions. It implies

[TABLE]

Equations (10.19) and (10.20) give statement (2) in Lemma 7.19. That lemma shows that

[TABLE]

The discussion after Lemma 7.19 culminates in a long exact sequence (7.20) involving the maps $N(s_{1}^{M})$, $\alpha_{21}^{M}$, $1_{2}^{M}-t_{2}^{M}$, $\alpha_{12}^{M}$. Our next goal is to understand $M_{2}$ with the maps $s_{2}^{M}$ and $t_{2}^{M}$. We claim that $N(s_{2}^{M})=p$ and hence $N(t_{2}^{M})=0$ by (5.22); so $(M_{2},t_{2}^{M})$ is a module over $\mathbb{Z}[\vartheta]=\mathbb{Z}[t]/(N(t))$. If $k\geq 2$, then (10.19) and (10.20) imply $\operatorname{im}(\alpha_{12}^{M})\subseteq\ker(\alpha_{21}^{M})$, so that

[TABLE]

This implies $N(s_{2}^{M})=p$. Now let $k=1$ and $p\neq 2$. Then $s_{1}^{M}=1$, and this implies $(1-s_{2}^{M})^{2}=0$ by a computation as in (9.8). The assumption $\operatorname{im}(\alpha_{21}^{M})\cong\operatorname{coker}(\alpha_{10}^{M})\cong\mathbb{Z}/p$ implies $p\cdot\alpha_{21}^{M}=0$ and hence $p\cdot(1-s_{2}^{M})=0$. Now Lemma 9.6 shows that $N(s_{2}^{M})=p$. So this holds in the situation of both propositions.

Equation (10.19) implies $\operatorname{im}(\alpha_{12}^{M})\cong\mathbb{Z}/p$. The map $\alpha_{21}^{M}$ induces a grading-reversing isomorphism from $M_{1}/\ker(\alpha_{21}^{M})$ to $\operatorname{im}(\alpha_{21}^{M})$. So (10.20) implies $\operatorname{im}(\alpha_{21}^{M})\cong\Sigma\mathbb{Z}/p$. Then the long exact sequence in (7.20) shows that

[TABLE]

We have classified $\mathbb{Z}[\vartheta]$-modules with this property in Lemma 10.10 and found that there are two basic examples, such that all others are extensions of them by uniquely $p$-divisible $\mathbb{Z}[\vartheta]$-modules. So the proof now splits into two subcases. Namely, $M_{2}$ contains either $M_{2}^{\prime}=\Sigma\mathbb{Z}[\vartheta]\oplus\Sigma Q$ or $M_{2}^{\prime}=\Sigma\mathbb{Z}[\vartheta]/(1-\vartheta)^{\ell}$ for some $\ell\in\mathbb{N}_{\geq 1}$, such that the quotient $M_{2}/M_{2}^{\prime}$ is uniquely $p$-divisible. Here the $\Sigma$ reminds us that these $\mathbb{Z}[\vartheta]$-submodules occur in odd parity.

Assume first that $M_{2}$ contains $M_{2}^{\prime}\mathrel{\vcentcolon=}\Sigma\mathbb{Z}[\vartheta]\oplus\Sigma Q$. By our description of the image and kernel of $\alpha_{21}^{M}$, it must factor through the quotient map $\mathbb{Z}/p^{k}\twoheadrightarrow\mathbb{Z}/p$, an isomorphism $\dot{\alpha}_{21}^{M^{\prime}}\colon\mathbb{Z}/p\xrightarrow{\sim}\ker(1_{2}^{M}-t_{2}^{M})$, and the inclusion $\ker(1_{2}^{M}-t_{2}^{M})\hookrightarrow M_{2}^{\prime}\subseteq M_{2}$. Similarly, $\alpha_{12}^{M^{\prime}}$ factors through the quotient map $M_{2}\twoheadrightarrow\operatorname{coker}(1_{2}^{M}-t_{2}^{M})=\operatorname{coker}(1_{2}^{M_{2}^{\prime}}-t_{2}^{M_{2}^{\prime}})$, an isomorphism $\dot{\alpha}_{12}^{M^{\prime}}\colon\operatorname{coker}(1_{2}^{M_{2}^{\prime}}-t_{2}^{M_{2}^{\prime}})\xrightarrow{\sim}\mathbb{Z}/p$, and the injective map $\mathbb{Z}/p\to\mathbb{Z}/p^{k}$, $c\mapsto c\cdot p^{k-1}$. The proof of Lemma 10.10 starts with arbitrary generators of $\ker(1_{2}^{M}-t_{2}^{M})$ and $\operatorname{coker}(1_{2}^{M}-t_{2}^{M})$. We may choose them so that $\dot{\alpha}_{21}^{M^{\prime}}[1]$ is the image of $1$ in \Sigma Q=\Sigma\mathbb{Z}[\vartheta,1/p]\bigm{/}(1-\vartheta)\mathbb{Z}[\vartheta] and $\dot{\alpha}_{12}^{M^{\prime}}$ maps the unit element in $\Sigma\mathbb{Z}[\vartheta]$ to the unit in $\mathbb{Z}/p$. Let $M_{1}^{\prime}=M_{1}$ and $M_{0}^{\prime}=\alpha_{01}^{M^{\prime}}(M_{1})\oplus\alpha_{02}^{M^{\prime}}(M_{2}^{\prime})\subseteq M_{0}$; the decomposition of $M_{0}^{\prime}$ is direct by (10.21). We may identify

[TABLE]

where the last isomorphism is multiplication by $1-\vartheta$. Now inspection shows that $M^{\prime}$ is isomorphic to the exact $\mathfrak{K}$-module in Example 9.14. The quotient $\mathfrak{K}$-module $M^{\prime\prime}\mathrel{\vcentcolon=}M/M^{\prime}$ is again exact and has $M_{1}^{\prime\prime}=0$. So Theorem 7.2 implies that $M^{\prime\prime}$ has the form of Example 7.1 with $X=0$ and $Z=0$. This deals with the subcase where $M_{2}$ contains $\Sigma\mathbb{Z}[\vartheta]\oplus\Sigma Q$.

Next assume that $M_{2}$ contains $M_{2}^{\prime}=\Sigma\mathbb{Z}[\vartheta]/(1-\vartheta)^{\ell}$ for some $\ell\geq 1$. As above, the map $\alpha_{21}^{M^{\prime}}$ must factor through the quotient map $\mathbb{Z}/p^{k}\twoheadrightarrow\mathbb{Z}/p$, an isomorphism $\dot{\alpha}_{21}^{M^{\prime}}\colon\mathbb{Z}/p\xrightarrow{\sim}\ker(1_{2}^{M}-t_{2}^{M})$, and the inclusion $\ker(1_{2}^{M}-t_{2}^{M})\hookrightarrow M_{2}^{\prime}\subseteq M_{2}$; and $\alpha_{12}^{M^{\prime}}$ factors through the quotient map $M_{2}\twoheadrightarrow\operatorname{coker}(1_{2}^{M}-t_{2}^{M})$, an isomorphism $\dot{\alpha}_{12}^{M^{\prime}}\colon\operatorname{coker}(1_{2}^{M}-t_{2}^{M})\xrightarrow{\sim}\mathbb{Z}/p$, and the injective map $\mathbb{Z}/p\to\mathbb{Z}/p^{k}$, $c\mapsto c\cdot p^{k-1}$. Here the unit element of $\mathbb{Z}[\vartheta]/(1-\vartheta)^{\ell}$ represents a generator of $\operatorname{coker}(1_{2}^{M}-t_{2}^{M})\cong\mathbb{Z}/p$, and $(1-\vartheta)^{\ell-1}$ generates $\ker(1_{2}^{M}-t_{2}^{M})$. By an automorphism of $\mathbb{Z}[\vartheta]/(1-\vartheta)^{\ell}$, we may arrange that $\dot{\alpha}_{12}^{M^{\prime}}$ is the canonical map. So $\alpha_{12}^{M^{\prime}}$ restricts to $p^{k-1}\mathrm{ev}_{1}$ as in Examples 9.7 and 9.9. The other map $\dot{\alpha}_{21}^{M^{\prime}}$ involves an isomorphism of $\mathbb{Z}/p$, that is, multiplication with some $\tau\in\{1,\dotsc,p-1\}$. As above, we put $M_{1}^{\prime}=M_{1}$ and $M_{0}^{\prime}=\alpha_{01}^{M^{\prime}}(M_{1})+\alpha_{02}^{M^{\prime}}(M_{2}^{\prime})$ and then show that $M_{0}^{\prime}\cong\mathbb{Z}/p^{k-1}\oplus\Sigma\mathbb{Z}[\vartheta]/(1-\vartheta)^{\ell-1}$. If $\ell=1$, then the second summand goes away and the generator of $\ker(1-\vartheta)$ in $M_{2}$ also represents a generator of $\operatorname{coker}(1-\vartheta)$. Therefore, $M^{\prime}$ is isomorphic to the exact $\mathfrak{K}$-module in Example 9.7 with the given $\tau$. If $\ell>1$, then we get the exact $\mathfrak{K}$-module in Example 9.9 with the given $\tau$. As in the case above, the quotient $M^{\prime\prime}\mathrel{\vcentcolon=}M/M^{\prime}$ is an exact $\mathfrak{K}$-module with $M_{1}^{\prime\prime}=0$. Hence it is described by Example 7.1. ∎

Finally, we are left with the case where $\ker(\alpha_{01}^{M})=\mathbb{Z}/p$, $\operatorname{coker}(\alpha_{10}^{M})=\mathbb{Z}/p$, $k=1$ and $p=2$. Here the equation $N(s_{2}^{M})=p$ breaks down, so that the structure of $M_{2}$ may be rather different. We may treat this case using the automorphisms of Köhler’s ring $\mathfrak{K}$ for $p=2$ in Section 5.2. Since $k=1$, our assumptions say that $\alpha_{01}^{M}=0$ and $\alpha_{10}^{M}=0$. The involutive automorphism of $\mathfrak{K}$ that exchanges [math] and $2$ replaces $M$ by an exact $\mathfrak{K}$-module $M^{\sigma}$ with $\alpha_{21}^{M}=0$ and $\alpha_{12}^{M}=0$ or, equivalently, $\ker\alpha_{01}^{M}=\{0\}$ and $\operatorname{coker}\alpha_{10}^{M}=\{0\}$. So we are in the situation of Proposition 10.8. It gives an extension $M^{\prime}\rightarrowtail M^{\sigma}\twoheadrightarrow M^{\prime\prime}$, where $M^{\prime}$ is as in Example 9.12 or 9.16 and $M^{\prime\prime}$ is a uniquely $p$-divisible exact $\mathfrak{K}$-module as in Example 7.1 with $X=0$ and $Z=0$. So $M$ itself contains $(M^{\prime})^{\sigma}$ with a uniquely $p$-divisible quotient. And this is covered by Example 9.17. We have treated all the possibilities for exact $\mathfrak{K}$-modules with cyclic $M_{1}$. So the proof of Theorem 9.1 is finished.

11. Equivalence at the prime $p$

The examples of exact $\mathfrak{K}$-modules in Section 9 only give all examples up to uniquely $p$-divisible quotients. We now study what this gives on the level of $\mathrm{KK}^{G}$. We introduce a weaker notion of equivalence in $\mathrm{KK}^{G}$ which, roughly speaking, looks only at the $p$-torsion information in an object. When we classify up to this notion of equivalence, then the uniquely $p$-divisible quotients become irrelevant, and also some of the examples in Section 9 are identified.

Let

[TABLE]

be an extension of countable, exact $\mathfrak{K}$-modules. By Theorems 5.4 and 5.10, there are objects $A^{\prime},A$ of $\mathfrak{B}^{G}$ with $F_{*}(A^{\prime})\cong M^{\prime}$ and $F_{*}(A)\cong M$, and there is $j\in\mathrm{KK}^{G}_{0}(A^{\prime},A)$ with $F_{*}(j)=i$ (up to the chosen isomorphisms $F_{*}(A^{\prime})\cong M^{\prime}$ and $F_{*}(A)\cong M$). Since $\mathfrak{B}^{G}$ is a triangulated category, the map $j$ is part of an exact triangle

[TABLE]

Here $\Sigma A^{\prime\prime}$ is a “cone” of $j$. Since $F_{*}$ is a homological functor, it maps this exact triangle to a long exact sequence of $\mathfrak{K}$-modules. Since $F_{*}(j)$ is injective, $F_{*}(\delta)=0$ and $F_{*}(r)$ is a cokernel for $F_{*}(j)=i$. So there is an isomorphism $F_{*}(A^{\prime\prime})\cong M^{\prime\prime}$ such that $F_{*}(r)=q$ (up to the chosen isomorphisms $F_{*}(A)\cong M$ and $F_{*}(A^{\prime\prime})\cong M^{\prime\prime}$). In other words, we have lifted the $\mathfrak{K}$-module extension (11.1) to an $F_{*}$-exact, exact triangle in $\mathfrak{B}^{G}$. Theorem 9.1 gives such extensions (11.1) where $M^{\prime\prime}_{1}=0$. So $M^{\prime\prime}$ describes an action on an object that is non-equivariantly $\mathrm{KK}$-equivalent to [math]. By Theorem 7.23, this corresponds to an action on $\mathcal{O}_{2}$. So the exact triangle above involves the objects of $\mathfrak{B}^{G}$ that realise the exact $\mathfrak{K}$-modules $M^{\prime}$ and $M$ and an action of $G$ on $\mathcal{O}_{2}$. In this sense, Theorem 9.1 must allow extensions as in (11.1) merely to accommodate exact triangles involving actions on $\mathcal{O}_{2}$.

In the situation of Theorem 9.1, we also know that $M^{\prime\prime}$ is uniquely $p$-divisible. So $j$ is a $p$-equivalence in the following sense:

Definition 11.2**.**

Let $A,B$ be objects of $\mathfrak{B}^{G}$. We call $f\in\mathrm{KK}^{G}_{0}(A,B)$ a $p$-equivalence if $F_{*}(\operatorname{cone}(f))$ is uniquely $p$-divisible.

We define $p$-equivalence for objects of $\mathfrak{B}^{G}$ to be the equivalence relation that is generated by $A\sim B$ if there is a $p$-equivalence in $\mathrm{KK}^{G}_{0}(A,B)$. We make this more transparent using localisation techniques. Let $\mathrm{UHF}(p^{\infty})$ be the UHF-algebra of type $p^{\infty}$. Its $\mathrm{K}$-theory is $\mathbb{Z}[1/p]$. Let $u\colon\mathbb{C}\to\mathrm{UHF}(p^{\infty})$ be the unit map. On $\mathrm{K}$-theory, this is the inclusion $\mathbb{Z}\hookrightarrow\mathbb{Z}[1/p]$. Let $L_{p}\mathrel{\vcentcolon=}\operatorname{cone}(u)$. Then $\mathrm{K}_{0}(L_{p})=0$ and $\mathrm{K}_{1}(L_{p})=\mathbb{Z}[1/p]/\mathbb{Z}$. We equip $\mathbb{C}$, $\mathrm{UHF}(p^{\infty})$ and $L_{p}$ with the trivial action of $G$. Given any $A$ in $\mathfrak{B}^{G}$, there is an exact triangle

[TABLE]

and we may identify $A\otimes\mathbb{C}\cong A$. Since $\mathrm{K}$-theory commutes with inductive limits, so does our invariant $F_{*}$. So

[TABLE]

This is uniquely $p$-divisible. So the boundary map $A\otimes L_{p}\to A$ in the exact triangle above is a $p$-equivalence.

Lemma 11.3**.**

Two objects $A,B$ of $\mathfrak{B}^{G}$ are $p$-equivalent if and only if $A\otimes L_{p}$ and $B\otimes L_{p}$ are $\mathrm{KK}^{G}$-equivalent.

Proof.

Since the boundary maps $A\otimes L_{p}\to A$ and $B\otimes L_{p}\to B$ are $p$-equivalences, a $\mathrm{KK}^{G}$-equivalence between $A\otimes L_{p}$ and $B\otimes L_{p}$ implies that $A$ and $B$ are $p$-equivalent. Let $f\in\mathrm{KK}^{G}_{0}(A,B)$ be a $p$-equivalence. We claim that $f\otimes 1\in\mathrm{KK}^{G}(A\otimes L_{p},B\otimes L_{p})$ is a $\mathrm{KK}^{G}$-equivalence. It follows from this claim that $A\otimes L_{p}$ and $B\otimes L_{p}$ are $\mathrm{KK}^{G}$-equivalent if $A$ and $B$ are $p$-equivalent. Let $C$ be the cone of $f$. Then $C\otimes L_{p}$ is a cone of $f\otimes\mathrm{id}_{L_{p}}$. By assumption, $F_{*}(C)$ is uniquely $p$-divisible. Hence the canonical map $F_{*}(C)\to F_{*}(C)\otimes_{\mathbb{Z}}\mathbb{Z}[1/p]$ is an isomorphism. This is the map induced by the inclusion map $C\to C\otimes\mathrm{UHF}(p^{\infty})$. So $F_{*}(C\otimes L_{p})=0$. Since $C\otimes L_{p}$ belongs to $\mathfrak{B}^{G}$, it follows that $C\otimes L_{p}$ is $\mathrm{KK}^{G}$-equivalent to [math]. And this is equivalent to $f\otimes\mathrm{id}_{L_{p}}$ being a $\mathrm{KK}^{G}$-equivalence. ∎

So the objects of $\mathfrak{B}^{G}$ of the form $A\otimes L_{p}$ are representatives for all $p$-equivalence classes in $\mathfrak{B}^{G}$. The special property of these representatives is that

[TABLE]

So $p$-equivalence classes of objects of $\mathfrak{B}^{G}$ are in bijection with isomorphism classes of countable exact $\mathfrak{K}$-modules $M$ with $M\otimes_{\mathbb{Z}}\mathbb{Z}[1/p]=0$ by Lemma 11.3. Inspecting our examples, we see that the $\mathfrak{K}$-modules in Examples 9.7, 9.9, 9.11 have the property that $M\otimes_{\mathbb{Z}}\mathbb{Z}[1/p]=0$. So have the examples for $p=2$ built as in Example 9.17 from these examples. It is also easy to see that the actions on $\mathcal{O}_{\infty}$ for Examples 9.2 and 9.4 are $p$-equivalent. The criterion above does not directly apply here because $M_{1}=\mathbb{Z}$ forces $M\otimes_{\mathbb{Z}}\mathbb{Z}[1/p]$ to be non-trivial.

So when we classify actions of $\mathbb{Z}/p$ only up to $p$-equivalence, then the number of cases to consider is reduced considerably.