K_0-theory of n-potents in rings and algebras

TL;DR

This paper introduces a new K-theory group based on n-potents in rings and algebras, exploring its properties, isomorphisms with classical K-theory, and functorial behavior under generalized homomorphisms.

Contribution

It defines and studies the K_0^n group for n-potents, establishing isomorphisms with classical K-theory for complex algebras and analyzing functorial properties.

Findings

K_0^n(A) is isomorphic to (K_0(A))^{n-1} for complex algebras

The isomorphism does not hold in general over cyclotomic fields

K_0^n is functorial for n-homomorphisms

Abstract

Let $n \geq 2$ be an integer. An \emph{$n$-potent} is an element $e$ of a ring $R$ such that $e^n = e$. In this paper, we study $n$-potents in matrices over $R$ and use them to construct an abelian group $K_0^n(R)$. If $A$ is a complex algebra, there is a group isomorphism $K_0^n(A) \cong \bigl(K_0(A)\bigr)^{n-1}$ for all $n \geq 2$. However, for algebras over cyclotomic fields, this is not true in general. We consider $K_0^n$ as a covariant functor, and show that it is also functorial for a generalization of homomorphism called an \emph{$n$-homomorphism}.

Equations

$$a=e^{n-1}a=a{f}^{n-1}=e^{n-1}a{f}^{n-1}$$

$$b=f^{n-1}b=b{e}^{n-1}=f^{n-1}b{e}^{n-1}.$$

$$\begin{gathered} ab=(e^{n-1}\tilde{a}{f}^{n-1})(f^{n-1}\tilde{b}{e}^{n-1})=e^{n-1}\tilde{a}{f}^{n-1}\tilde{b}{e}^{n-1} \\ =e^{n-1}(\tilde{a}\tilde{b}{)}^n{e}^{n-1}=e^{n-1}{e}^n{e}^{n-1}=(e^{n-1}{)}^2{e}^n=e^{n-1}e=e^n=e. \end{gathered}$$

$$st=a{f}^{n-2}cdb=a{f}^{n-1}b=a(ba{)}^{n-1}b=(ab{)}^n=e^n=e$$

$$ts=dba{f}^{n-2}c=d{f}^{n-1}c=d(cd{)}^{n-1}c=(dc{)}^n=g^n=g.$$

$$\left(\begin{array}{cc}e& 0\\ 0& 0\end{array}\right){\sim }_s\left(\begin{array}{cc}f& 0\\ 0& 0\end{array}\right)$$

$$u=\left(\begin{array}{cc}1-f^{n-1}& b\\ a{f}^{n-2}& 1-e^{n-1}\end{array}\right)$$

$$v=\left(\begin{array}{cc}1-e^{n-1}& e^{n-1}\\ e^{n-1}& 1-e^{n-1}\end{array}\right).$$

$$z\left(\begin{array}{cc}e& 0\\ 0& 0\end{array}\right)z^{-1}$$

$$=\left(\begin{array}{cc}bea{f}^{n-2}& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}f& 0\\ 0& 0\end{array}\right)$$

$$a_1{b}_2=a_1{f}_1^{n-1}{f}_2^{n-1}{b}_2=0.$$

$$(a_1+a_2)(b_1+b_2)=a_1{b}_1+a_2{b}_2=e_1+e_2$$

$$(b_1+b_2)(a_1+a_2)=b_1{a}_1+b_2{a}_2=f_1+f_2,$$

$$a=\left(\begin{array}{cc}0& e\\ f& 0\end{array}\right)\text{and}b=\left(\begin{array}{cc}0& f^{n-1}\\ e^{n-1}& 0\end{array}\right)$$

$$ab=\left(\begin{array}{cc}0& e\\ f& 0\end{array}\right)\left(\begin{array}{cc}0& f^{n-1}\\ e^{n-1}& 0\end{array}\right)=\left(\begin{array}{cc}{e}^n& 0\\ 0& f^n\end{array}\right)=\left(\begin{array}{cc}e& 0\\ 0& f\end{array}\right)$$

$$ba=\left(\begin{array}{cc}0& f^{n-1}\\ e^{n-1}& 0\end{array}\right)\left(\begin{array}{cc}0& e\\ f& 0\end{array}\right)=\left(\begin{array}{cc}{f}^n& 0\\ 0& e^n\end{array}\right)=\left(\begin{array}{cc}f& 0\\ 0& e\end{array}\right),$$

$$a=\left(\begin{array}{cc}e& 0\\ f& 0\end{array}\right)\text{and}b=\left(\begin{array}{cc}{e}^{n-1}& f^{n-1}\\ 0& 0\end{array}\right).$$

$$ab=\left(\begin{array}{cc}e& 0\\ f& 0\end{array}\right)\left(\begin{array}{cc}{e}^{n-1}& f^{n-1}\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}{e}^n& e{f}^{n-1}\\ f{e}^{n-1}& f^n\end{array}\right)=\left(\begin{array}{cc}e& 0\\ 0& f\end{array}\right)$$

$$ba=\left(\begin{array}{cc}{e}^{n-1}& f^{n-1}\\ 0& 0\end{array}\right)\left(\begin{array}{cc}e& 0\\ f& 0\end{array}\right)=\left(\begin{array}{cc}{e}^n+f^n& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}e+f& 0\\ 0& 0\end{array}\right),$$

$$e=\sum _{k=1}^{n-1}{\omega }_k{e}_k.$$

$$p_k(x)=\frac{{\prod }_{j\ne k}(x-{\omega }_j)}{{\prod }_{j\ne k}({\omega }_k-{\omega }_j)}.$$

$$\sum _{k=0}^{n-1}{p}_k(\alpha )=p_0(\alpha )+\cdots +p_{n-1}(\alpha )=1$$

$$\alpha =\sum _{k=0}^{n-1}{\omega }_k{p}_k(\alpha ).$$

$$\sum _{k=0}^{n-1}{e}_k=\sum _{k=0}^{n-1}{p}_k(e)=1.$$

$$e_j{e}_k=p_j(e)p_k(e)=0$$

$$e=\sum _{k=1}^{n-1}{\omega }_k{p}_k(e)=\sum _{k=1}^{n-1}{\omega }_k{e}_k$$

$$i_k(a)=\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)$$

$$M_{\infty }(R)=\bigcup _{k=1}^{\infty }{M}_k(R),{\mathcal{P}}_{\infty }^n(R)=\bigcup _{k=1}^{\infty }{\mathcal{P}}_k^n(R)={\mathcal{P}}^n(M_{\infty }(R)).$$

$$e\oplus f=\mathrm{diag}(e,f)=\left(\begin{array}{cc}e& 0\\ 0& f\end{array}\right)\in {\mathcal{P}}_{k+l}^n(R)\subset {\mathcal{P}}_{\infty }^n(R).$$

$$\left(\begin{array}{cc}e& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right)\left(\begin{array}{cc}b& 0\\ 0& 0\end{array}\right)$$

$$\left(\begin{array}{cc}f& 0\\ 0& 0\end{array}\right)=\left(\begin{array}{cc}b& 0\\ 0& 0\end{array}\right)\left(\begin{array}{cc}a& 0\\ 0& 0\end{array}\right),$$

Full text

$2010$ Mathematics Subject Classification: 18F30, 19A99, 19K99

1. Introduction

For more than forty years, $K$-theory has been an essential tool in studying rings and algebras [1, 7]. Given a ring $R$, a simple functorial object associated to $R$ is the abelian group $K_{0}(R)$. There are multiple ways of defining $K_{0}(R)$, but the most useful characterization when working with operator algebras is to define $K_{0}(R)$ in terms of idempotents (or projections, if an involution is present) in matrix algebras over $R$; *i.e., *elements $e$ in $M_{k}(R)$ for some $k$ with the feature that $e^{2}=e$ ($p=p^{*}=p^{2}$ in the involutive case). In this paper, we define, for each natural number $n\geq 2$, a group which we denote $K_{0}^{n}(R)$. This group is constructed from matrices $e$ over $R$ with the property that $e^{n}=e$; we call such matrices $n$-potents. We define $K_{0}^{n}(R)$ for all rings, unital or not, and show that $K_{0}^{n}$ determines a covariant functor from rings to abelian groups.

Let $\mathbb{Q}(n-1)$ be the cyclotomic field obtained from the rationals by adjoining the $(n-1)$-th roots of unity. We show that $K_{0}^{n}$ is half-exact on the subcategory of $\mathbb{Q}(n-1)$-algebras, and given any such algebra $A$, we show that $K_{0}^{n}(A)$ is isomorphic to a direct sum of $n-1$ copies of $K_{0}(A)$. Since a $\mathbb{C}$-algebra $A$ is a $\mathbb{Q}(n-1)$-algebra for all $n$, whatever invariants are contained in $K_{0}^{n}(A)$ are already contained in $K_{0}(A)$. However, $K_{0}^{p}$ for $p\neq n$ may generate new groups for cyclotomic algebras, e.g., $K_{0}^{4}(\mathbb{Q}(4))\cong\mathbb{Z}\oplus 2\mathbb{Z}$ (Theorem 3.15) which is not isomorphic to $K_{0}^{4}(\mathbb{Q}(3))\cong\mathbb{Z}^{3}$. Thus, $K_{0}^{4}$ distinguishes between the fields $\mathbb{Q}(3)$ and $\mathbb{Q}(4)$, but idempotent, and also tripotent ($n=3$), $K$-theory does not.

The paper is organized as follows. In Section 2, we define various notions of equivalence on the set of $n$-potents, and explore the relationships between these equivalence relations. Most of our results in this section mirror analogous facts about idempotents, but in many cases the proofs differ or are more delicate for $n$-potents. In Section 3, we define $n$-potent $K$-theory and study its properties and compute some examples. Finally, in Section 4, we consider $n$-homomorphisms on rings and algebras [2, 3, 4], and show that $n$-potent $K$-theory is functorial for such maps; this is a phenomenon that does not appear in ordinary idempotent $K$-theory.

The authors thank Dana Williams and Tom Shemanske for their helpful comments and suggestions.

Note: Unless stated otherwise, all rings and algebras have a unit; *i.e., *a multiplicative identity, and all ring and algebra homomorphisms are unital.

2. Equivalence of $n$-potents

Fix a natural number $n\geq 2$. In this section, we develop the basic theory of $n$-potents, including various equivalence relations among them. We begin by looking at $n$-potents over general rings, but eventually we will specialize to get a well-behaved theory.

Definition 2.1**.**

Let $R$ be a ring. An element $e$ in $R$ is called an $n$-potent if $e^{n}=e$. For $n=2,3,4$, we use the terms idempotent, tripotent, and quadripotent, respectively. The set of all $n$-potents in $R$ is denoted $\mathcal{P}^{n}(R)$.

We begin with a very simple but useful fact about $n$-potents:

Lemma 2.2**.**

Suppose $e$ is an $n$-potent. Then $e^{n-1}$ is an idempotent.

Proof.

$(e^{n-1})^{2}=e^{n-1}e^{n-1}=e^{n}e^{n-2}=ee^{n-2}=e^{n-1}.$ ∎

Definition 2.3**.**

Let $e$ and $f$ be $n$-potents in a ring $R$. We say that $e$ and $f$ are algebraically equivalent and write $e\sim_{a}f$ if there exist elements $a$ and $b$ in $R$ such that $e=ab$ and $f=ba$. We say that $e$ and $f$ are similar and write $e\sim_{s}f$ if there exists an invertible element $z$ in $R$ with the property that $f=zez^{-1}$.

Lemma 2.4**.**

Suppose that $e$ and $f$ are algebraically equivalent $n$-potents in a ring $R$. Then the elements $a$ and $b$ described in Definition 2.3 can be chosen so that

[TABLE]

Proof.

Choose elements $\tilde{a}$ and $\tilde{b}$ in $R$ so that $\tilde{a}\tilde{b}=e$ and $\tilde{b}\tilde{a}=f$. Set $a=e^{n-1}\tilde{a}f^{n-1}$ and $b=f^{n-1}\tilde{b}e^{n-1}$. Using Lemma 2.2, we have

[TABLE]

Similarly, $ba=f$. The two strings of equalities in the statement of the lemma then follow easily. ∎

Proposition 2.5**.**

The relations $\sim_{a}$ and $\sim_{s}$ are equivalence relations on $\mathcal{P}^{n}(R)$.

Proof.

The only nonobvious point to establish is that $\sim_{a}$ is transitive. Let $e$, $f$, and $g$ be elements of $\mathcal{P}^{n}(R)$, and suppose that $e\sim_{a}f\sim_{a}g$. Choose elements $a$, $b$, $c$ and $d$ in $R$ so that $e=ab$, $f=ba=cd$, and $g=dc$, and set $s=af^{n-2}c$ and $t=db$. Then

[TABLE]

and

[TABLE]

∎

Proposition 2.6**.**

If $e$ and $f$ are similar $n$-potents in a ring $R$, then they are algebraically equivalent.

Proof.

Choose an invertible element $z$ in $R$ such that $f=zez^{-1}$, and set $a=ez^{-1}$ and $b=ze^{n-1}$. Then $ab=e^{n}=e$ and $ba=ze^{n}z^{-1}=f$. ∎

As is the case with idempotents, algebraic equivalence does not imply similarity in general. However, we do have the following result, just as for idempotents:

Proposition 2.7**.**

Suppose that $e$ and $f$ are algebraically equivalent $n$-potents in a ring $R$. Then

[TABLE]

in the ring $M_{2}(R)$ of $2\times 2$ matrices over $R$.

Proof.

Choose elements $a$ and $b$ in $R$ so that $e=ab$ and $f=ba$; without loss of generality, we assume that $a$ and $b$ satisfy the conclusions of Lemma 2.4. Define

[TABLE]

and

[TABLE]

Straightforward computation yields that both $u^{2}$ and $v^{2}$ equal the identity matrix in $M_{2}(R)$, and thus each is its own inverse. Set $z=uv$. Then we compute that

[TABLE]

since $beaf^{n-2}=b(ab)a(ba)^{n-2}=(ba)^{n}=f^{n}=f.$ ∎

Definition 2.8**.**

We say $n$-potents $e$ and $f$ in a ring $R$ are orthogonal if $ef=fe=0$, in which case we write $e\perp f$.

The next result follows immediately by mathematical induction.

Proposition 2.9**.**

Let $e$ and $f$ be orthogonal $n$-potents in a ring $R$. Then $(e+f)^{k}=e^{k}+f^{k}$. In particular, $e+f$ is an $n$-potent.

Proposition 2.10**.**

For $i=1,2$, let $e_{i}$ and $f_{i}$ be algebraically equivalent $n$-potents in a ring $R$. Suppose that $e_{1}$ and $f_{1}$ are orthogonal to $e_{2}$ and $f_{2}$, respectively. Then $e_{1}+e_{2}$ and $f_{1}+f_{2}$ are algebraically equivalent.

Proof.

For $i=1,2$, choose $a_{i}$ and $b_{i}$ so that $e_{i}=a_{i}b_{i}$, $f_{i}=b_{i}a_{i}$, and so that $a_{i}$ and $b_{i}$ satisfy the conclusion of Lemma 2.4. Then

[TABLE]

Similarly, $b_{2}a_{1}$, $a_{2}b_{1}$, and $b_{1}a_{2}$ are also zero. Thus

[TABLE]

and

[TABLE]

whence $e_{1}+e_{2}$ is algebraically equivalent to $f_{1}+f_{2}$. ∎

Proposition 2.11**.**

Let $e$ and $f$ be $n$-potents in a ring $R$.

1. (a)

$\begin{pmatrix}e&0\\ 0&f\end{pmatrix}\sim_{a}\begin{pmatrix}f&0\\ 0&e\end{pmatrix}$* and $\begin{pmatrix}e&0\\ 0&0\end{pmatrix}\sim_{a}\begin{pmatrix}0&0\\ 0&e\end{pmatrix}$.* 2. (b)

If $e\perp f$ then $\begin{pmatrix}e&0\\ 0&f\end{pmatrix}\sim_{a}\begin{pmatrix}e+f&0\\ 0&0\end{pmatrix}$.

Proof.

Define

[TABLE]

in $M_{2}(R)$. Then

[TABLE]

and

[TABLE]

which establishes the first part of (a); to obtain the second part, simply take $f$ to be zero.

To prove (b), first observe that if $e\perp f$, then $e+f$ is an $n$-potent by Proposition 2.9. Define

[TABLE]

Then

[TABLE]

and

[TABLE]

whence the result follows. ∎

Later in this paper we will restrict our attention to $n$-potent $K$-theory of cyclotomic algebras:

Definition 2.12**.**

For each integer $n\geq 2$, the cyclotomic field $\mathbb{Q}(n-1)$ is the field obtained by adjoining the $(n-1)$st primitive root of unity $\zeta_{n-1}=e^{2\pi i/(n-1)}$ to the field $\mathbb{Q}$ of rational numbers. A cyclotomic algebra is a $\mathbb{Q}(n-1)$-algebra for some $n\geq 2$.

Observe that $\mathbb{Q}(n-1)\subset\mathbb{C}$, and therefore every $\mathbb{C}$-algebra is canonically a $\mathbb{Q}(n-1)$-algebra for all $n$.

Definition 2.13**.**

Let $\mathbb{F}$ be a field and let $A$ be an $\mathbb{F}$-algebra with unit. An $n$-partition of unity is an ordered $n$-tuple $(e_{0},e_{1},\dots,e_{n-1})$ of idempotents in $A$ such that

1. (1)

$e_{0}+e_{1}+\cdots+e_{n-1}=1$; 2. (2)

$e_{0},e_{1},\dots,e_{n-1}$* are pairwise orthogonal; i.e., $e_{j}e_{k}=\delta_{jk}e_{k}$ for all $0\leq j,k\leq n-1$.*

Note that $e_{0}=1-(e_{1}+\cdots+e_{n-1})$ is completely determined by $e_{1},e_{2},\dots,e_{n-1}$ and is thus redundant in the notation for an $n$-partition of unity.

Cyclotomic algebras admit a distinguished $n$-partition of unity. Set $\omega_{0}=0$ and let $\omega_{k}=e^{2\pi i(k-1)/(n-1)}$ for $1\leq k\leq n-1$. Note that $\omega_{1},\dots,\omega_{n-1}$ are the $(n-1)$st roots of unity, and $\Omega_{n}=\{\omega_{0},\omega_{1},\dots,\omega_{n-1}\}$ is the set of roots of the polynomial equation $x_{n}-x=0$.

Theorem 2.14**.**

Let $A$ be a $\mathbb{Q}(n-1)$-algebra with unit, and suppose $e$ is an $n$-potent in $A$. Then there exists a unique $n$-partition of unity $(e_{0},e_{1},\dots,e_{n-1})$ in $A$ such that

[TABLE]

Proof.

Let $p_{0},p_{1},\dots,p_{n-1}\in\mathbb{Q}(n-1)[x]$ be the Lagrange polynomials

[TABLE]

In particular, $p_{0}(x)=1-x^{n-1}$. Each polynomial $p_{k}$ has degree $n-1$ and satisfies $p_{k}(\omega_{k})=1$ and $p_{k}(\omega_{j})=0$ for all $j\neq k$. We claim that for all numbers $\alpha\in\mathbb{Q}(n-1)\subseteq\mathbb{C}$,

[TABLE]

and that

[TABLE]

Indeed, these identities follow from the fact that these polynomial equations have degree $n-1$ but are satisfied by the $n$ distinct points in $\Omega_{n}$.

Now, given any $\omega_{i}^{n}=\omega_{i}$ in $\Omega_{n}$ it follows that $p_{k}(\omega_{i})^{2}=p_{k}(\omega_{i})$. Hence, for any $n$-potent $e\in A$, if we define $e_{k}=p_{k}(e)$, then each $e_{k}$ is an idempotent in $A$, and Equation (1) implies that

[TABLE]

These idempotents are pairwise orthogonal, because

[TABLE]

for $j\neq k$. Finally,

[TABLE]

by Equation (2). ∎

3. $K_{0}$-theory with $n$-potents

We can now proceed to construct our $n$-potent $K$-theory groups.

Definition 3.1**.**

Let $R$ be a ring. For all $k\geq 1$, let $\mathcal{P}^{n}_{k}(R)$ denote the set of $n$-potents in $M_{k}(R)$, and let $i_{k}$ denote the inclusion

[TABLE]

of $M_{k}(R)$ into $M_{k+1}(R)$, as well as its restriction as a map from $\mathcal{P}^{n}_{k}(R)$ to $\mathcal{P}^{n}_{k+1}(R)$. Define $M_{\infty}(R)$ and $\mathcal{P}^{n}_{\infty}(R)$ to be the (algebraic) direct limits

[TABLE]

We define a binary operation $\oplus$ on $\mathcal{P}^{n}_{\infty}(R)$ as follows: let $e$ and $f$ be elements of $\mathcal{P}^{n}_{\infty}(R)$, choose the smallest natural numbers $k$ and $\ell$ such that $e\in M_{k}(R)$ and $f\in M_{l}(R)$, and set

[TABLE]

Definition 3.2**.**

Let $R$ be a ring, and define an equivalence relation $\sim$ on $\mathcal{P}^{n}_{\infty}(R)$ as follows: take $e$ and $f$ in $\mathcal{P}^{n}_{\infty}(R)$, and choose a natural number $k$ sufficiently large that $e$ and $f$ are elements of $\mathcal{P}^{n}_{k}(R)$. Then $e\sim f$ if $e\sim_{a}f$ in $M_{k}(R)$. We let $\mathcal{V}^{n}(R)$ denote the set of equivalence classes of $\sim$.

Note that if $e=ab$ and $f=ba$ in $M_{k}(R)$, then

[TABLE]

and

[TABLE]

and therefore the equivalence relation described in Definition 3.2 is well-defined.

Note that for any $n$-potent $e,f$ in $M_{\infty}(R)$, we get

[TABLE]

Thus, the binary operation $\oplus$ induces a binary operation $+$ on $\mathcal{V}^{n}_{\infty}(R)$ as follows: take $e$ and $f$ in $\mathcal{P}^{n}_{\infty}(R)$, and define

[TABLE]

This operation is well-defined and commutative by Propositions 2.9 and 2.11.

The next proposition is straightforward and left to the reader.

Proposition 3.3**.**

For every ring $R$ and natural number $n\geq 2$, $\mathcal{V}^{n}(R)$ is an abelian monoid under the addition defined above, and whose identity element is the class of the zero $n$-potent. If $\alpha:R\longrightarrow S$ is a unital ring homomorphism, then the induced map $\mathcal{V}^{n}(\alpha):\mathcal{V}^{n}(R)\longrightarrow\mathcal{V}^{n}(S)$ given by

[TABLE]

is a well-defined homomorphism of abelian semigroups. The correspondences $R\mapsto\mathcal{V}^{n}(R)$ and $\alpha\mapsto\mathcal{V}^{n}(\alpha)$ induce a covariant functor from the category of rings and ring homomorphisms to the category of abelian monoids and monoid homomorphisms.

Definition 3.4**.**

Let $R$ be a ring and let $n\geq 2$ be a natural number. We define $K_{0}^{n}(R)$ to be the Grothendieck completion [6] of the abelian monoid $\mathcal{V}^{n}(R)$. Given an $n$-potent $e$ in $\mathcal{P}^{n}_{\infty}(R)$, we denote its class in $K_{0}^{n}(R)$ by $[e]$.

In light of Propositions 2.6 and 2.7, we could have alternatively used similarity to define $\mathcal{V}^{n}(R)$, and hence $K_{0}^{n}(R)$.

Proposition 3.5**.**

The assignments $R\mapsto K_{0}^{n}(R)$ determines a covariant functor from the category of rings and ring homomorphisms to the category of abelian groups and group homomorphisms.

Proof.

Proposition 3.3 states that $\mathcal{V}$ is a covariant functor from the category of rings to the category of abelian monoids, and Grothendieck completion determines a covariant functor from the category of abelian monoids to the category of abelian groups; we get the desired result by composing these two functors. ∎

The following result shows that for (unital) algebras over a field of characteristic $\neq 2$, the tripotent $K$-theory functor $K_{0}^{3}$ offers us no new invariants over ordinary idempotent $K$-theory. However, we will see later (Theorem 3.15) that the situation is subtly different for $K_{0}^{4}$.

Theorem 3.6**.**

Let $\mathbb{F}$ be a field with characteristic $\neq 2$. If $A$ is a unital algebra over $\mathbb{F}$ then there is a natural isomorphism

[TABLE]

of abelian groups.

Proof.

If $e=e^{3}\in M_{\infty}(A)$ is a tripotent, then one can easily check that

[TABLE]

are (unique) idempotents in $M_{\infty}(A)$ such that $e=e_{1}-e_{2}$. It follows that we have a natural bijection of abelain monoids

[TABLE]

with inverse map $[e_{1}]\oplus[e_{2}]\mapsto[e_{1}\oplus-e_{2}]$. Since these maps are additive, the result easily follows. ∎

While $K_{0}^{n}(R)$ is well-defined for any ring $R$, to obtain a well-behaved theory where the usual exact sequences exist, we must restrict our attention to a smaller class of rings. The problem is that unlike the situation for idempotents, it is not generally true that if $e$ is an $n$-potent, then so is $1-e$. However, given an $n$-potent in an algebra over the cyclotomic field $\mathbb{Q}(n-1)$, there is an adequate substitute:

Definition 3.7**.**

Let $e$ be an $n$-potent in a $\mathbb{Q}(n-1)$-algebra $A$, and write

[TABLE]

as in the conclusion of Theorem 2.14. We define an $n$-potent

[TABLE]

and call $e^{\perp}$ the complementary $n$-potent of $e$.

Observe that if $n=2$, this definition agrees with the usual one for idempotents; i.e., $e^{\perp}=1-e$. Note also that $e\oplus e^{\perp}\sim_{s}\omega$, where

[TABLE]

Proposition 3.8** (Standard Picture of $K_{0}^{n}(A)$).**

Let $n\geq 2$ be a natural number and let $A$ be a $\mathbb{Q}(n-1)$-algebra. Then every element of $K_{0}^{n}(A)$ can be written in the form $[e]-[\omega]$, where $e$ in an $n$-potent in $M_{k}(A)$ for some natural number $k$ and $\omega$ is a diagonal $n$-potent in $M_{k}(\mathbb{Q}(n-1))$.

Proof.

Start with an element $[\tilde{e}]-[\tilde{f}]$ in $K_{0}^{n}(A)$, and take $\tilde{f}^{\perp}$ to be the complementary $n$-potent of $f$ as defined in Definition 3.7. Then

[TABLE]

The $n$-potents $\tilde{f}$ and $\tilde{f}^{\perp}$ are orthogonal, and therefore

[TABLE]

where $\omega$ has the desired form. Finally we take $e$ to be $\tilde{e}\oplus\tilde{f}^{\perp}$, and by enlarging the matrix $\omega$, we obtain the desired result. ∎

Proposition 3.9**.**

Let $n\geq 2$ and let $A$ be a $\mathbb{Q}(n-1)$-algebra. Suppose $e$ and $f$ are $n$-potents in $M_{\infty}(A)$. Then $[e]=[f]$ in $K_{0}^{n}(A)$ if and only if $e\oplus\omega$ is similar to $f\oplus\omega$ for some $n$-potent $\omega$ in $M_{\infty}(\mathbb{Q}(n-1))$.

Proof.

The “only if”  direction is obvious. To show the inference in the opposite direction, suppose that $[e]=[f]$ in $K_{0}^{n}(A)$. By the definition of the Grothendieck completion, $e\oplus\tilde{e}$ is similar to $f\oplus\tilde{e}$ for some $n$-potent $\tilde{e}$ in $M_{\infty}(A)$. Then $e\oplus\tilde{e}\oplus\tilde{e}^{\perp}$ is similar to $f\oplus\tilde{e}\oplus\tilde{e}^{\perp}$. But if we write $\tilde{e}=\sum_{k=1}^{n-1}\omega_{k}\tilde{e}_{k}$ as in Theorem 2.14, then Proposition 2.11(b) implies that

[TABLE]

Therefore $\tilde{e}\oplus\tilde{e}^{\perp}$ is similar to an $n$-potent in $M_{\infty}(\mathbb{Q}(n-1))$, and the proposition follows. ∎

We next turn our attention to $n$-potent $K$-theory for nonunital algebras. Given a nonunital $\mathbb{Q}(n-1)$-algebra $A$, we define its unitization $A^{+}$ as the unital $\mathbb{Q}(n-1)$-algebra $A^{+}=\{(a,\lambda):a\in A,\lambda\in\mathbb{Q}(n-1)\}$, where addition and scalar multiplication are defined componentwise, and multiplication is given by $(a,\lambda)(b,\tau)=(ab+a\tau+b\lambda,\lambda\tau)$.

Definition 3.10**.**

Let $A$ be a nonunital $\mathbb{Q}(n-1)$-algebra, and let $A^{+}$ be its unitization. Let $\pi:A^{+}\longrightarrow\mathbb{Q}(n-1)$ be the algebra homomorphism $\pi(a,\lambda)=\lambda$. Then we define $K_{0}^{n}(A)=\ker\pi_{*}$.

It is easy to see that $\pi_{*}$ is surjective, so by definition of $K_{0}^{n}(A)$ we have a short exact sequence

[TABLE]

with splitting induced by the map $\psi:\mathbb{Q}(n-1)\longrightarrow A^{+}$ defined by $\psi(\lambda)=(0,\lambda)$. In addition, it is easy to check that if $A$ already has a unit and we form $A^{+}$, then $\ker\pi_{*}$ is naturally isomorphic to our original definition of $K_{0}^{n}(A)$.

Proposition 3.11**.**

Let $A$ be a nonunital $\mathbb{Q}(n-1)$-algebra. Then every element of $K_{0}^{n}(A)$ can be written in the form $[e]-[s(e)]$, where $e$ is an $n$-potents in $M_{k}(A^{+})$ for some integer $k\geq 1$, and $s=\psi\circ\pi:A^{+}\to A^{+}$ is the scalar mapping [6, Sect. 4.2.1].

Proof.

Follows directly from Proposition 3.8 and Definition 3.10. ∎

Proposition 3.12** (Half-exactness).**

Every short exact sequence

[TABLE]

of $\mathbb{Q}(n-1)$-algebras, with $A$ unital, induces an exact sequence

[TABLE]

of abelian $n$-potent $K$-theory groups.

Proof.

Since $q\circ i=0$, we have by functoriality that $q_{*}\circ i_{*}=0$ and so the image of $K_{0}^{n}(I)$ under $i_{*}$ in $K_{0}^{n}(A)$ is contained in the kernel of $q_{*}$. To show the reverse inclusion, suppose we have $[e]-[\lambda]$ in $K_{0}^{n}(A)$ such that q_{*}\bigl{(}[e]-[\lambda]\bigr{)}=0. Then $[q(e)]=[q(\lambda)]=[\lambda]$ in $K_{0}^{n}(A/I)$. By Proposition 3.9, there exists an $n$-potent $\tau$ in $M_{\infty}(\mathbb{Q}(n-1))$ so that

[TABLE]

Choose $N$ sufficiently large so that we may view $e$, $\lambda$, and $\tau$ as $N$ by $N$ matrices, and choose $z$ in $GL_{2N}(A/I)$ so that

[TABLE]

By Proposition 3.4.2 and Corollary 3.4.4 in [1], we can lift ${\operatorname{diag}}(z,z^{-1})$ to an element $u$ in $GL_{4N}(A)$. Set $f=u(e\oplus\tau)u^{-1}$. Then

[TABLE]

and thus $f$ and $\lambda\oplus\tau$ are in $M_{4N}(I^{+})$. Therefore

[TABLE]

is in the image of $K_{0}^{n}(I)$ under $i_{*}$ as desired. ∎

Note that our proof of Proposition 3.12 relies critically on Proposition 3.9, which in turn is proved using the standard picture of $K_{0}^{n}(A)$. We do not have a standard picture for $K_{0}^{k}(A)$ when $k\neq n$, and it seems likely to the authors that $K_{0}^{k}$ is, in fact, not half-exact in this case. However, we do not have a counterexample where half-exactness fails to hold.

While it is not at all obvious from its definition, $K_{0}^{n}(A)$ can be identified with a more familiar object.

Theorem 3.13**.**

Let $n\geq 2$ be a natural number and let $A$ be a not necessarily unital $\mathbb{Q}(n-1)$-algebra. Then there is a natural isomorphism

[TABLE]

of abelian groups.

Proof.

First consider the case where $A$ is unital. We define a homomorphism \tilde{\psi}:\mathcal{V}^{n}(A)\longrightarrow\bigl{(}\mathcal{V}_{0}(A)\bigr{)}^{n-1} in the following way: for each $n$-potent $e=\sum_{k=1}^{n-1}\omega_{k}e_{k}$ in $M_{\infty}(A)$, set

[TABLE]

It is easy to check that $\tilde{\psi}$ is additive and well-defined. Next, define a homomorphism \tilde{\phi}:\bigl{(}\mathcal{V}(A)\bigr{)}^{n-1}\longrightarrow\mathcal{V}^{n}(A) by the formula

[TABLE]

Note that

[TABLE]

and

[TABLE]

where the last equality is a consequence of Proposition 2.11(b). The universal mapping property of the Grothendieck completion implies that $\tilde{\psi}$ extends uniquely to an abelian group isomorphism

[TABLE]

and thus the theorem is true for unital $\mathbb{Q}(n-1)$-algebras.

Now suppose that $A$ does not have a unit. Then we have the following commutative diagram with exact rows:

[TABLE]

An easy diagram chase shows that there is a unique group isomorphism from $K_{0}^{n}(A)$ to \bigl{(}K_{0}(A)\bigr{)}^{n-1} that makes the diagram commute. ∎

Since a complex algebra is a $\mathbb{Q}(n-1)$-algebra for all values of $n$, we have the following immediate corollary.

Corollary 3.14**.**

If $A$ is a $\mathbb{C}$-algebra, there are natural isomorphisms

[TABLE]

of abelian groups for all natural numbers $n\geq 2$.

We now arrive at the result that suggests why we should consider all $K_{0}^{n}$-functors for algebras over a cyclotomic field.

Theorem 3.15**.**

Let $\mathbb{Q}(4)=\mathbb{Q}[i]$ be the $4$th cyclotomic field. Then we have the following isomorphisms of abelian groups:

[TABLE]

Thus, $K_{0}^{4}(\mathbb{Q}(4))\not\cong\mathbb{Z}^{3}\cong K_{0}^{4}(\mathbb{Q}(3))$.

Proof.

Since $\mathbb{Q}(4)$ is a field [7], we have $K_{0}^{2}(\mathbb{Q}(4))=K_{0}(\mathbb{Q}(4))\cong\mathbb{Z}$. The field $\mathbb{Q}(4)$ has characteristic $0\neq 2$, so Theorem 3.6 implies that K_{0}^{3}(\mathbb{Q}(4))\cong\bigl{(}K_{0}(\mathbb{Q}(4)\bigr{)}^{2}\cong\mathbb{Z}^{2}. Theorem 3.13 implies that we have an isomorphism K_{0}^{5}(\mathbb{Q}(4))\cong\bigl{(}K_{0}(\mathbb{Q}(4)\bigr{)}^{4}\cong\mathbb{Z}^{4}.

However, the spectrum of $4$-potents is contained in

[TABLE]

which is not contained in $\mathbb{Q}(4)$ since the two primitive $3$rd roots of unity $\omega=\zeta_{3}=-\frac{1}{2}+\frac{\sqrt{3}}{2}i$ and $\bar{\omega}=\bar{\zeta}_{3}=-\frac{1}{2}-\frac{\sqrt{3}}{2}i$ are not in $\mathbb{Q}(4)=\mathbb{Q}[i]$.

Given any $4$-potent $e\in M_{n}(\mathbb{Q}(4))\subset M_{n}(\mathbb{C})$ we can uniquely write

[TABLE]

where $e_{1},e_{2},e_{3}$ are orthogonal idempotents in $M_{n}(\mathbb{C})$ that sum to an idempotent $e_{1}+e_{2}+e_{3}=e^{3}$ in $M_{n}(\mathbb{Q}(4))$ by Lemma 2.2. We thus have that

[TABLE]

because $\omega^{2}=\bar{\omega},\bar{\omega}^{2}=\omega$, and $\omega^{3}=\bar{\omega}^{3}=1$. Since $\omega+\bar{\omega}=-1$, this implies that the first idempotent

[TABLE]

and the sum of the last two idempotents

[TABLE]

are both in $M_{n}(\mathbb{Q}(4))$. Using a simple trace argument and the fact that $\omega,\bar{\omega}\not\in\mathbb{Q}(4)$, we conclude that

[TABLE]

and so ${\operatorname{rank}}(e_{2}+e_{3})={\operatorname{trace}}(e_{2}+e_{3})=2{\operatorname{trace}}(e_{2})$ is even. We then have a well-defined map

[TABLE]

this is because the classes of $e_{1}$ and $e_{1}+e_{2}$ are preserved by (stable) similarity, and the $K_{0}$-class of an idempotent in a matrix ring over a number field (or a PID) is the rank (= trace). It is easy to check that this map is injective (using $e_{1}\perp e_{2}+e_{3}$ in $M_{n}(\mathbb{Q}(4))$) and additive. The only question is surjectivity. It suffices to show that there is a $4$-potent $e$ over $\mathbb{Q}(4)$ whose stable similarity class is mapped to the generator $1\oplus 2$ of $\mathbb{N}\oplus 2\mathbb{N}$. Consider the block diagonal matrix

[TABLE]

which is easily checked to be quadripotent. The lower right quadripotent $2\times 2$ invertible block has the desired eigenvalues $\omega$ and $\bar{\omega}$, and so does not diagonalize over $\mathbb{Q}(4)$. The result now follows easily. ∎

4. $n$-Homomorphisms and $K_{0}^{n}$ Functorality

We know from Proposition 3.5 that $K_{0}^{n}$ is a covariant functor from the category of (unital) rings and ring homomorphisms to the category of abelian groups and group homomorphisms. However, $K_{0}^{n}$ is actually functorial for a more general class of ring mappings.

Definition 4.1**.**

Let $R$ and $S$ be rings. An additive map (not necessarily unital) $\phi:R\longrightarrow S$ is called an $n$-homomorphism if

[TABLE]

for all $a_{1},a_{2},\dots,a_{n}$ in $R$.

Obviously every (ring) homomorphism is an $n$-homomorphism, but the converse is false in general. For example, an $AE_{n}$-ring is a ring $R$ such that every additive map $\phi:R\to R$ is an $n$-homomorphism. Feigelstock [2, 3] classified all unital $AE_{n}$-rings. The algebraic version of $n$-homomorphism was introduced for complex algebras in [4] and has been carefully studied in the case of $C^{*}$-algebras in [5].

Proposition 4.2**.**

Let $\phi:R\to S$ be an $n$-homomorphism between unital rings. Then $\phi$ induces a group homomorphism

[TABLE]

Furthermore, the assignment $R\mapsto K_{0}^{n}(R)$ is a covariant functor from the category of unital rings and $n$-homomorphisms to the category of abelian groups and ordinary group homomorphisms.

Proof.

For each natural number $k$, we extend $\phi$ to a map from $M_{k}(R)$ to $M_{k}(S)$ by applying $\phi$ to each matrix entry; it is easy to check this also gives us an $n$-homomorphism. Moreover, $\phi$ is compatible with stabilization of matrices; the only nonobvious point to check is that $\phi$ respects algebraic equivalence.

Let $e$ and $f$ be algebraically equivalent $n$-potents in $M_{k}(R)$ for some $k$, and choose $a$ and $b$ in $M_{k}(R)$ so that $e=ab$ and $f=ba$. Define elements $a^{\prime}=\phi(ea)\phi(f)^{n-2}$ and $b^{\prime}=\phi(b)$ in $M_{k}(S)$. We compute:

[TABLE]

A similar argument shows that $b^{\prime}a^{\prime}=\phi(f)$. Therefore $\phi$ determines a monoid homomorphism from $\mathcal{V}^{n}(R)$ to $\mathcal{V}^{n}(S)$, and hence a group homomorphism $\phi_{*}:K_{0}^{n}(R)\longrightarrow K_{0}^{n}(S)$. We leave it to the reader to make the straightforward computations to show that we have a covariant functor. ∎

Note that while we have an isomorphism K_{0}^{n}(A)\cong\bigl{(}K_{0}(A)\bigr{)}^{n-1} for $\mathbb{Q}(n-1)$-algebras, it is not at all clear from the right hand side of this isomorphism that $K_{0}^{n}(A)$ is functorial for $n$-homomorphisms.