The dominant dimension of cohomological Mackey functors

TL;DR

This paper demonstrates that the dominant dimension of cohomological Mackey functors for a p-block of a finite group with a nontrivial defect group is always 2, using properties of symmetric algebras and separable equivalences.

Contribution

It establishes the invariant nature of dominant dimensions of certain endomorphism algebras under separable equivalences and applies this to cohomological Mackey functors.

Findings

Dominant dimension of coMack(B) is 2 for nontrivial defect groups.

Separable equivalences preserve dominant dimensions of specific endomorphism algebras.

The result applies to p-blocks of finite groups with nontrivial defect groups.

Abstract

We show that a separable equivalence between symmetric algebras preserves the dominant dimensions of certain endomorphism algebras of modules. We apply this to show that the dominant dimension of the category coMack(B) of cohomological Mackey functors of a p-block B of a finite group with a nontrivial defect group is 2.

Equations

$$0\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}\text{ignorespaces}$$

Full text

1 Introduction

Let $k$ be a field. Following Tachikawa [7], the dominant dimension of a finite-dimensional $k$-algebra $A$, which we will denote by $\mathrm{ddim}(A)$, is the largest nonnegative integer $d$ such that there exists an injective resolution

[TABLE]

of $A$ as a right $A$-module with the property that $I^{n}$ is projective for $0\leq n\leq d-1$, provided there is such an integer. If for any injective resolution $I$ of $A$ the term $I^{0}$ is not projective, then $\mathrm{ddim}(A)=$ [math], and if there exists an injective resolution $I$ of $A$ such that $I^{n}$ is projective for all $n\geq$ [math], then we adopt the convention $\mathrm{ddim}(A)=$ $\infty$. By a result of Müller [6, Theorem 4] the dominant dimension is equal to the obvious left module analogue. In order to calculate $\mathrm{ddim}(A)$ it suffices to consider a minimal injective resolution of $A$ as a right $A$-module.

If $k$ has prime characteristic $p$ and if $A$ is a source algebra of a block $B$ of a finite group algebra $kG$ with a defect group $P$, then, by the source algebra version [3, Theorem 1.1] of a result of Yoshida in [8], the category $\mathrm{coMack}(B)$ of cohomological Mackey functors of $G$ associated with $B$ is equivalent to the right module category of the endomorphism algebra $E=$ $\mathrm{End}_{A}(\oplus_{Q\leq P}\ A\otimes_{kQ}k)$. The dominant dimension of $\mathrm{coMack}(B)$ is defined as the dominant dimension of $E$.

Theorem 1.1**.**

Suppose that $k$ has prime characteristic $p$. Let $G$ be a finite group and $B$ a block of $kG$ with a nontrivial defect group $P$. The dominant dimension of $\mathrm{coMack}(B)$ is equal to $2$.

A finite-dimensional $k$-algebra $A$ is called symmetric if $A$ is isomorphic to its $k$-dual $A^{*}$ as an $A$-$A$-bimodule. Given two symmetric $k$-algebras $A$ and $B$, we say that an $A$-$B$-bimodule $M$ induces a separable equivalence between $A$ and $B$ if $M$ is finitely generated projective as a left $A$-module and as a right $B$-module such that $A$ is isomorphic to a direct summand of $M\otimes_{B}M^{*}$ as an $A$-$A$-bimodule and $B$ is isomorphic to a direct summand of $M^{*}\otimes_{A}M$ as a $B$-$B$-bimodule. In that case we say that $A$ and $B$ are separably equivalent. See [2] for details and more general versions of this notion. We will show that Theorem 1.1 is an immediate consequence of results from [3], [4], [6], and the following.

Theorem 1.2**.**

Let $A$, $B$ be symmetric $k$-algebras. Let $M$ be an $A$-$B$-bimodule inducing a separable equivalence between $A$ and $B$. Let $U$ be a finitely generated $A$-module such that $A$ is isomorphic to a direct summand of $U$, and let $V$ a finitely generated $B$-module such that $B$ is isomorphic to a direct summand of $V$. Suppose that $M^{*}\otimes_{A}U\in$ $\mathrm{add}(V)$ and that $M\otimes_{B}V\in$ $\mathrm{add}(U)$. Then the dominant dimensions of $\mathrm{End}_{A}(U)$ and of $\mathrm{End}_{B}(V)$ are equal.

2 Proof of Theorem 1.2

Let $A$ be a finite-dimensional $k$-algebra, and let $U$, $V$ be finitely generated $A$-modules. We use without further reference the following standard facts. If $V$ belongs to $\mathrm{add}(U)$, then $\mathrm{Hom}_{A}(U,V)$ is a projective right $\mathrm{End}_{A}(U)$-module, and any finitely generated projective right $\mathrm{End}_{A}(U)$-module is of this form, up to isomorphism. Given two idempotents $i$, $j$ in $A$, every homomorphism of right $A$-modules $iA\to$ $jA$ is induced by left multiplication with an element in $jAi$. Multiplication by $i$ is exact; in particular, if $Z$ is a complex of $A$-modules which is exact or which has homology concentrated in a single degree, the same is true for finite direct sums of complexes of the form $iZ$. Translated to endomorphism algebras this implies that for any two $A$-modules $V$, $W$ in $\mathrm{add}(U)$, any homomorphism of right $\mathrm{End}_{A}(U)$-modules $\mathrm{Hom}_{A}(U,V)\to$ $\mathrm{Hom}_{A}(U,W)$ is induced by composition with an $A$-homomorphism $V\to$ $W$. Thus any complex of finitely generated projective right $\mathrm{End}_{A}(U)$-modules is isomorphic to a complex obtained from applying the functor $\mathrm{Hom}_{A}(U,-)$ to a complex of $A$-modules $Z$ whose terms belong to $\mathrm{add}(U)$. Moreover, if $\mathrm{Hom}_{A}(U,Z)$ is exact, then so is any complex of the form $\mathrm{Hom}_{A}(U^{\prime},Z)$, where $U^{\prime}\in$ $\mathrm{add}(U)$. We use further the well-known fact that if $A$, $B$ are symmetric algebras and if $M$ is an $A$-$B$-bimodule which is finitely generated projective as a left $A$-module and as a right $B$-module, then the functors $M\otimes_{B}-$ and $M^{*}\otimes_{A}-$ between $\mathrm{mod}(A)$ and $\mathrm{mod}(B)$ are biadjoint.

Proof of Theorem 1.2.

The following argument from the proof of [3, Theorem 3.1] shows that $\mathrm{add}(M\otimes_{B}V)=$ $\mathrm{add}(U)$ and $\mathrm{add}(M^{*}\otimes_{A}U)=$ $\mathrm{add}(V)$. By the assumptions, we have $\mathrm{add}(M\otimes_{B}V)\subseteq$ $\mathrm{add}(U)$. Thus $\mathrm{add}(M^{*}\otimes_{A}M\otimes_{B}V)\subseteq$ $\mathrm{add}(M^{*}\otimes_{A}U)\subseteq$ $\mathrm{add}(V)$. Since $B$ is isomorphic to a direct summand of the $B$-$B$-bimodule $M^{*}\otimes_{A}M$, it follows that $V$ is isomorphic to a direct summand of $M^{*}\otimes_{A}M\otimes_{B}V$. Thus the previous inclusions of additive categories are equalities. The same argument with reversed roles shows the second equality.

Set $E=$ $\mathrm{End}_{A}(U)$ and $F=$ $\mathrm{End}_{B}(V)$. By the assumptions, $\mathrm{add}(U)$ contains all finitely generated projective $A$-modules; similarly for $\mathrm{add}(V)$. In particular, if $U^{\prime}$ is a projective $A$-module, then $\mathrm{Hom}_{A}(U,U^{\prime})$ is a projective right $E$-module, and by [4, Proposition 3.2], $\mathrm{Hom}_{A}(U,U^{\prime})$ is also an injective right $E$-module. Let

[TABLE]

be an injective resolution of $E$ as a right $E$-module. Suppose that there is a positive integer $d$ such that $I^{n}$ is projective for $0\leq$ $n\leq$ $d-1$. We will show that there is an injective resolution

[TABLE]

of $F$ as a right $F$-module such that $J^{n}$ is projective for $0\leq$ $n\leq$ $d-1$. Since the right $E$-modules $I^{n}$ are projective and injective for $0\leq$ $n\leq$ $d-1$, it follows from [4, Proposition 3.2] that there are finitely generated projective $A$-modules $U_{n}$ for $0\leq$ $n\leq$ $d-1$ such that the sequence

[TABLE]

is isomorphic to a sequence of the form

[TABLE]

which is obtained from applying the functor $\mathrm{Hom}_{A}(U,-)$ to a sequence of $A$-modules

[TABLE]

It follows from the remarks at the beginning of this section that for any $A$-module $U^{\prime}$ in $\mathrm{add}(U)$, applying the functor $\mathrm{Hom}_{A}(U^{\prime},-)$ to the previous sequence of $A$-modules yields an exact sequence of right $\mathrm{End}_{A}(U^{\prime})$-modules of the form

[TABLE]

By the assumptions, the $A$-module $U^{\prime}=$ $M\otimes_{B}V$ belongs to $\mathrm{add}(U)$. Thus we obtain an exact sequence of the form

[TABLE]

Since $M\otimes_{B}-$ is left adjoint to $M^{*}\otimes_{A}-$, it follows that the previous exact sequence is isomorphic to an exact sequence of the form

[TABLE]

Since the $A$-modules $U_{n}$ are projective for $0\leq$ $n\leq$ $d-1$, it follows that the $B$-modules $M^{*}\otimes_{A}U_{n}$ are projective as well, hence in $\mathrm{add}(V)$. By [4, Proposition 3.2], the projective right $F$-modules $\mathrm{Hom}_{B}(V,M^{*}\otimes_{A}U_{n})$ are therefore also injective. Thus the preceding sequence is the beginning of an injective resolution of the right $F$-module $\mathrm{Hom}_{B}(V,M^{*}\otimes_{A}U)$ which has the property that its first $d$ terms are projective. Since $V$ belongs to $\mathrm{add}(M^{*}\otimes_{A}U)$, it follows that $F=$ $\mathrm{Hom}_{B}(V,V)$ is isomorphic, as a right $F$-module, to a direct summand of a direct sum of finitely many copies of the right $F$-module $\mathrm{Hom}_{B}(V,M^{*}\otimes_{A}U)$. This implies that $F$ has an injective resolution as a right $F$-module whose first $d$ terms are projective. This shows that $\mathrm{ddim}(F)\geq$ $\mathrm{ddim}(E)$ (including the case where both are $\infty$). Exchanging the roles of $A$, $E$ and $B$, $F$, respectively, yields the result. ∎

3 Proof of Theorem 1.1

We suppose in this section that $k$ is a field of prime characteristic $p$. Let $G$ be a finite group and $B$ a block of $kG$ with a nontrivial defect group $P$. Let $i\in$ $B^{P}$ be a source idempotent of $B$ and set $A=$ $iBi$; that is, $A$ is a source algebra of $B$. Note that $A$, $B$, $kP$ are symmetric algebras. It is well-known that $A$ and $kP$ are separably equivalent via the bimodule $A_{kP}$ and its dual, which is isomorphic to ${{}_{kP}A}$; see e. g. [1, Proposition 4.2] for a proof (the hypothesis on $k$ being algebraically closed in that paper is not needed for this result).

Proof of Theorem 1.1.

With the notation above, set $U=$ $\oplus_{Q}\ A\otimes_{kQ}k$, where $Q$ runs over the subgroups of $P$, and set $E=$ $\mathrm{End}_{A}(U)$. By [3, Theorem 1.1] we have $\mathrm{coMack}(B)\cong$ $\mathrm{mod}(E^{\mathrm{op}})$. Set $V=$ $\oplus_{Q}\ kP\otimes_{kQ}k$, where $Q$ runs as before over the subgroups of $P$, and set $F=$ $\mathrm{End}_{kP}(V)$. As in the proof of [3, Theorem 1.6], the functor $A\otimes_{kP}-$ sends $V$ to $\mathrm{add}(U)$ and the functor ${{}_{kP}A}\otimes_{A}-$ sends $U$ to $\mathrm{add}(V)$, because $A$ has a $P\times P$-stable $k$-basis, hence preserves the classes of $p$-permutation modules. By definition, the dominant dimension of $\mathrm{coMack}(B)$ is equal to $\mathrm{ddim}(E)$. It follows from Theorem 1.2 that $\mathrm{ddim}(E)=$ $\mathrm{ddim}(F)$. We have $\mathrm{ddim}(F)\geq$ $2$ by the general Morita-Tachikawa correspondence. Since the argument is very short, we sketch it (the dual argument for left modules is in the proof of [6, Theorem 2], for instance). Let $0\to$ $V\to$ $I^{0}\to$ $I^{1}\to\cdots$ be an injective resolution of $V$ as a $kP$-module, with $I^{n}$ finitely generated for all $n\geq$ [math]. The modules $I^{n}$ are also projective since $kP$ is symmetric. Applying $\mathrm{Hom}_{kP}(V,-)$ yields an exact sequence

[TABLE]

of right $F$-modules. The last two terms are projective as right $F$-modules because the finitely generated injective $kP$-modules $I^{0}$ and $I^{1}$ are in $\mathrm{add}(V)$, and the last two terms are also injective right $F$-modules by [4, Proposition 3.2] or by [5, (17.2)].

By Müller’s Lemma 3 in [6], in order to show that $\mathrm{ddim}(F)=$ $2$, it suffices to show that $\mathrm{Ext}^{1}_{kP}(V,V)$ is nonzero. The summand of $V$ indexed by $P$ is the trivial $kP$-module. Since $P$ is nontrivial, it follows that $\mathrm{Ext}^{1}_{kP}(k,k)\neq$ $\{0\}$, and hence $\mathrm{Ext}^{1}_{kP}(V,V)\neq$ $\{0\}$. Theorem 1.1 follows. ∎

Remark 3.1**.**

One can prove Theorem 1.1 also without using Theorem 1.2, by showing directly that $\mathrm{Ext}_{A}^{1}(U,U)$ is nonzero, and then applying [6, Lemma 3] as in the proof above. Indeed, the separable equivalence between $A$ and $kP$ implies that $kP$ is isomorphic to a direct summand of $A$ as a $kP$-$kP$-bimodule. Thus $A\otimes_{kP}k$ has a trivial summand as a left $kP$-module. Note that $U\cong$ $A\otimes_{kP}V$. A standard adjunction implies that we have an isomorphism $\mathrm{Ext}_{A}^{*}(U,U)\cong$ $\mathrm{Ext}^{*}_{kP}(V,\mathrm{Res}^{A}_{kP}(U))$. By the above, both $V$ and $\mathrm{Res}^{A}_{kP}(U)$ have a trivial summand, and hence $\mathrm{Ext}^{*}_{kP}(k,k)$ is a summand of $\mathrm{Ext}^{*}_{A}(U,U)$ as a graded $k$-vector space. In particular, $\mathrm{Ext}^{1}_{A}(U,U)$ is nonzero since $\mathrm{Ext}^{1}_{kP}(k,k)$ is nonzero.