# Functoriality of associating to a group algebra $kG$ its center $Z(kG)$, where $G$ is finite

It's well known that associating a group $$G$$ to its center $$Z(G)$$ is not functorial (read: doesn't extend to give us a functor $$\mathsf{Grp} \to \mathsf{Ab}$$). A simple counterexample is given by considering the inclusion of $$C_2 \hookrightarrow S_3$$, which has a retract given by $$S_3 \to S_3^{\text{ab}} = S_3/A_3 \cong C_2$$. You arrive at a contradiction for the functoriality of $$Z(-)$$ because $$Z(S_3)$$ is trivial. You can find more on this here https://math.stackexchange.com/a/3380460/395669.

Along these lines, one can construct a counterexample to show the failure of functoriality of $$Z(-)$$ for $$k$$-algebras (read: doesn't extend to give us a functor $${}_k\mathsf{Alg} \to {}_k\mathsf{CommAlg}$$), where $$k$$ is a unitcal commutative ring, by considering the algebras $$k[F_1]$$ and $$k[F_2]$$. Here, $$F_1$$ is the free group with one generator $$x$$ seen as a subgroup of $$F_2$$, the free group with two generators $$x$$ and $$y$$. Then, as above, the inclusion $$k[F_1] \hookrightarrow k[F_2]$$ has a retract given by $$k[F_2] \to k[F_2]/(y-x) \cong k[F_1]$$. You arrive at a contradiction for the functoriality of $$Z(-)$$ because $$Z(k[F_2])$$ is trivial, while $$k[F_1]$$ is a commutative $$k$$-algebra and hence its center is itself.

All this fine, what I've been unable to do is come up with a counterexample that shows the failure of $$Z(-)$$ to be a functor in the case of group algebras, when the group is finite. Here, the center of a group algebra has a different flavour, since it's a free $$k$$-module. Any ideas?

Addendum: A recent reading gave me a very partial answer to this question, which may or may not have a connection with Hochschild Cohomology. For a $$k$$-algebra $$A$$, we can talk about its commutator space $$[A,A]$$ defined as the $$k$$-submodule generated by $$[a,b] = ab - ba$$. Then for obvious reasons given an algebra map $$A \to B$$, we get an induced map $$A/[A,A] \to B/[B,B]$$. So we have a functorial construction here.

Now, in the case of a group algebra $$kG$$ for a finite group $$G$$, we define the $$k$$-dual of $$Z(kG)$$ obviously as $$Z(kG)^* = \mathrm{Hom}_k(Z(kG),k)$$. Then, we have a $$k$$-module isomorphism $$kG/[kG,kG] \cong Z(kG)^*$$. And so, given an algebra map $$kG \to kH$$, we do get a map $$Z(kH) \to Z(kG)$$, which is $$k$$-linear but not necessarily multiplicative. So, the center gives a contravariant functor but not quite straightforward.

• Have you already tried $k[C_2] \to k[S_3]$? Jul 14, 2021 at 7:37
• Sure, but have you tried to adapt the proof from the finite groups to these algebras? Jul 14, 2021 at 13:44
• This is not quite what you want, but there is a functor from the category of finite-dimensional $k$-algebras to $\mathsf{Vec}_k$ sending $\Lambda \mapsto HH^i(\Lambda, \Lambda ^*)$, where $\Lambda^*$ is the $k$-dual (you need the dual in the second position so that the variance is right). For symmetric algebras $\Lambda$, like $kG$, the image is the $i$th Hochschild cohomology group. For $i=0$ that's the centre. Jul 15, 2021 at 18:55
• This is definitely interesting! I have a feeling I've seen this in disguise of [will edit my question to include this.] Jul 15, 2021 at 18:57
• @NaweedG.Seldon Oh right, $Z(kG)$ has a basis corresponding to conjugacy classes of $G$ no matter what $k$ is, and you can use this to define an isomorphism $kG/[kG,kG] \cong Z(kG)$ as $k$-modules. (There are many possible choices for this. I believe I was using $[g]\mapsto \frac{1}{\lvert G\rvert} \sum_{h\in G} hgh^{-1}$, which doesn't work in all characteristics.) Jul 15, 2021 at 19:58

Let's suppose that $$kG$$ is semisimple (for instance when $$k$$ is algebraically closed and of characteristic $$0$$), since that lets us identify $$Z(kG)$$ with the subalgebra of central idempotents, a free subalgebra whose generators are in one-to-one correspondence with isomorphism classes of irreducible representations of $$G$$.

I'm only going to partially answer the question, unfortunately. I'm going to define a functor from $$k$$-algebras to $$k$$-modules that sends $$kG$$ to $$Z(kG)$$. When the algebra homomorphism $$f:kG\to kH$$ is induced by a surjective group homomorphism, then the induced map $$f_*$$ will be the algebra homomorphism $$f|_{Z(kG)}$$. If the homomorphism is not surjective, then it appears $$f_*$$ is not an algebra homomorphism in general, and if $$f$$ is an arbitrary algebra homomorphism, I don't expect $$f_*$$ to be an algebra homomorphism.

Maybe something in here will shed light on your problem. (But partly I'm answering because I want to write out this cyclic tensor product calculation somewhere.)

There are two ways to get the center of $$kG$$. The first is the definition: $$Z(kG) = \{a \in kG\mid \text{for all }b\in kG,ab=ba\}$$ The second is the horizontal trace, which unlike $$Z(kG)$$ is not obviously an algebra: consider $$kG$$ as a $$kG$$-bimodule, and let $$(kG)_0$$ denote in this answer $$kG$$ modulo the relation $$am\sim ma$$ for $$a\in kG$$ and $$m\in kG$$. In other words, this is $$kG$$ modulo cyclic shifts of words, and it's not hard to see that it's a vector space whose dimension is the number of conjugacy classes of $$G$$. (Note: this is not the abelianization of the algebra!) A priori, this is merely a $$k$$-algebra, but surprisingly it has a multiplication operation.

Before getting into that, let's talk about the idea behind the construction. An important part of the theory of representations of finite groups is that $$Z(kG)$$ is a free algebra whose generators are orthogonal idemponents $$\pi_1,\dots,\pi_n$$, one corresponding to each isomorphism class of irreducible representation $$V_1,\dots,V_n$$ of $$G$$. Let $$V=V_1\oplus \cdots\oplus V_n$$ be thought of as a left $$kG$$-module. It is also a right $$Z(kG)$$-module (and hence a $$(kG,Z(kG))$$-bimodule) by using this left action, which works out because elements of $$Z(kG)$$ commute with those of $$kG$$. Let $$V^{*}$$ be the dual representation $$V$$ but as a $$(Z(kG),kG)$$-bimodule (when you swap it like this, you don't need to insert inverses in the action). We can form two tensor products $$V\otimes_{Z(kG)} V^*$$ and $$V^*\otimes_{kG}V$$. Let's determine what they are isomorphic to before proceeding.

For the first tensor product: $$V \otimes_{Z(kG)} V^* = \bigoplus_{ij} V_i \otimes_{Z(kG)} V_j^* = \bigoplus_i V_i\otimes_k V_i^* = \bigoplus_i \operatorname{End}_k(V_i) \cong kG$$ where the first equality is from expanding direct sums, the second is from observing that when $$i\neq j$$ then there is a projector that kills the term and the remaining projector $$\pi_i$$ just acts by the identity on $$V_i$$, the third is since each $$V_i$$ is finite-dimensional, and the final isomorphism is the Artin-Wedderburn theorem.

For the second tensor product: $$V^* \otimes_{kG} V = \bigoplus_{ij} V^*_i \otimes_{kG} V_j = \bigoplus_i V^*_i \otimes_{kG} V_i \cong \bigoplus_i \operatorname{End}_{kG}(V_i) = \operatorname{End}_{kG}(V) \cong Z(kG)$$ The first two equalities are similar (and this time the second equality is from the fact that $$\pi_1,\dots,\pi_n\in kG$$). For the next step, we have isomorphisms $$V^*_i\otimes_{kG} V_i \cong k \cong \operatorname{End}_{kG}(V_i)$$ by the fact that $$V_i$$ is a cyclic module and Schur's lemma. Then, the direct sum commutes with $$\operatorname{End}_{kG}$$ since each irreducible representation appears exactly once, and the final isomorphism is that $$Z(kG)$$ is identified with the free algebra described earlier.

The idea with the horizontal trace $$(kG)_0$$ is that we form the tensor product of $$V$$ and $$V^*$$ cyclically to get a $$k$$-module. There isn't very good notation for it, so let's settle with $$V\otimes_{Z(kG)} V^* \otimes_{kG}{}$$ where that trailing tensor product means it's between $$V^*$$ as a right module and $$V$$ as a left module. The interesting this is that on one hand we have $$V\otimes_{Z(kG)} V^* \otimes_{kG}{} \cong kG\otimes_{kG}{} = (kG)_0$$ and on the other we have $$V\otimes_{Z(kG)} V^* \otimes_{kG}{} = V^* \otimes_{kG}V\otimes_{Z(kG)}{} \cong Z(kG)\otimes_{Z(kG)}{} = Z(kG)$$ Thus, we've identified $$(kG)_0$$ with $$Z(kG)$$, and with this we can give $$(kG)_0$$ an algebra structure!

On $$(kG)_0$$, this should define the multiplication operation: $$[g][g'] = \frac{1}{\lvert G\rvert}\sum_{h\in G} [ghg'h^{-1}]$$ where $$[g]\in (kG)_0$$ denotes the image of $$g\in G$$. (The scaling factor is chosen such that the composition $$Z(G)\hookrightarrow kG \twoheadrightarrow (kG)_0$$ is an algebra homomorphism.) In other words, by representing elements of $$(kG)_0$$ as linear combinations of conjugacy classes, to form the product of two conjugacy classes we take a representative of the first class, multiply it by each element of the second, and take the conjugacy class of each product.

Now that we've understood $$(kG)_0$$ better, let's talk about functoriality. Consider an algebra homomorphism $$f:kG\to kH$$. There is an induced map $$f_*:(kG)_0 \to (kH)_0$$ defined by $$[m]\mapsto [f(m)]$$, which is well-defined since $$f(am-ma)=f(a)f(m)-f(m)f(a)$$ for $$a\in kG$$ and $$m\in kG$$.

If $$f$$ is induced by a surjective group homomorphism $$\phi:G\to H$$, then we can calculate \begin{align*} f_*([g][g']) &= \frac{1}{\lvert G\rvert} \sum_{h\in G} [f(g)f(h)f(g')f(h)^{-1}] \\ &= \frac{\lvert \ker \phi\rvert}{\lvert G\rvert} \sum_{h\in H} [f(g)hf(g')h^{-1}] \\ &= \frac{1}{\lvert H\rvert} \sum_{h\in H} [f(g)hf(g')h^{-1}] \\ &= [f(g)][f(g')]. \end{align*} Thus $$f$$ induces an algebra homomorphism $$Z(kG)\to Z(kH)$$ functorially.

If $$\phi$$ is an arbitrary group homomorphism, then instead \begin{align*} f_*([g][g']) &= \frac{1}{\lvert G\rvert} \sum_{h\in G} [f(g)f(h)f(g')f(h)^{-1}] \\ &= \frac{1}{\lvert \operatorname{im} \phi\rvert} \sum_{h\in \operatorname{im}\phi} [f(g)hf(g')h^{-1}] \end{align*} and, with $$h_1,\dots,h_k\in H$$ being coset representatives for $$\operatorname{im}\phi$$, with $$h_1=1$$ and with $$k=\lvert H\rvert / \lvert \operatorname{im}\phi\rvert$$), \begin{align*} [f(g)][f(g')] &= \frac{1}{\lvert H\rvert} \sum_{h\in H} [f(g)hf(g')h^{-1}] \\ &= \frac{1}{\lvert H\rvert} \sum_{i=1}^k\sum_{h\in \operatorname{im}\phi} [f(g)h_ihf(g')h^{-1}h_i^{-1}] \\ &= \frac{1}{k\lvert G\rvert} \sum_{i=1}^k \sum_{h\in G} [f(g) h_i f(hg'h^{-1}) h_i^{-1}] \\ &= \frac{1}{k}\left( f_*([g][g']) + \frac{1}{\lvert G\rvert} \sum_{i=2}^k \sum_{h\in G} [f(g) h_i f(hg'h^{-1}) h_i^{-1}]\right). \end{align*} So we can see that it's not quite an algebra homomorphism. There's probably something more that could be said, but I'm not seeing it right now.

If I understand things correctly, the horizontal trace is related to the Hochschild homology of $$kG$$ with $$kG$$ coefficients. The only nontrivial homology group is $$HH_0(kG,kG)$$, I believe, and it is isomorphic to $$(kG)_0$$ (hence why I wrote the subscript-$$0$$). Hochschild homology is functorial, explaining, in some sense, why $$(kG)_0$$ is.

One thing I'm confused about is why I got Hochschild homology rather than cohomology. The complex $$HH^\bullet(kG,kG)$$ is, I believe, known as the derived center of an algebra. The correspondence between homology and cohomology here might be a coincidence for group algebras (or semisimple Hopf algebras?)