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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.

Naweed G. Seldon
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    Have you already tried $k[C_2] \to k[S_3]$? – Martin Brandenburg Jul 14 '21 at 07:37
  • I know that if $\phi: A \to B$ is an algebra map, then the image of $\phi\vert_{Z(A)}$ is not in $Z(B)$. All that says is that $Z(\phi) = \phi\vert_{Z(A)}$ does not work. What's to say that there isn't any other association that does work. – Naweed G. Seldon Jul 14 '21 at 07:39
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    Sure, but have you tried to adapt the proof from the finite groups to these algebras? – Martin Brandenburg Jul 14 '21 at 13:44
  • Yes, and the result was the example I have above of group algebras for free groups. The one trouble is that for finite group algebras is that the center is never trivial, it's always at least of $k$-rank $2$ , since a group will have at least $2$ conjugacy classes. – Naweed G. Seldon Jul 14 '21 at 20:24
  • It's not clear to me what you are asking: $Z(\cdot)$ is only a mapping from algebras (or rings) to commutative algebras (or rings). So it doesn't quite make sense to ask if it "is" a functor (or rather, it's not, because you haven't said what should happen with morphisms). Do you mean to ask: "is there some definition of $Z(\phi)$ for a morphism $\phi$ of algebras for which $Z$ is a functor?" – Stephen Jul 14 '21 at 20:25
  • Indeed, you can say for sake of contradiction I'm denoting $Z(-)$ as the hypothetical functor. So, yes, we want to show there's no reasonable definition for $Z(\phi)$ that makes things work, or not. Specifically in the case of finite group algebras. – Naweed G. Seldon Jul 14 '21 at 20:26
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    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. – Matthew Towers Jul 15 '21 at 18:55
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    This is definitely interesting! I have a feeling I've seen this in disguise of [will edit my question to include this.] – Naweed G. Seldon Jul 15 '21 at 18:57
  • @MatthewTowers Thanks, I suspected that held (I'm not familiar enough with Hochschild (co)homology). This is known as the "derived center" of an algebra, right? And by "symmetric algebra," do you mean that it has a nondegenerate invariant symmetric form? Is this the only condition needed for $HH^0$ to be the center of the algebra? – Kyle Miller Jul 15 '21 at 19:37
  • @NaweedG.Seldon I'm not sure how clear it is, but in my answer $(kG)_0 := kG/[kG,kG]$, and the calculations were to show that since $kG$ is semisimple there's an isomorphism between $kG/[kG,kG]$ and $Z(kG)$, so at least as $k$-modules you can make $Z$ be functorial by passing through $kG/[kG,kG]$. – Kyle Miller Jul 15 '21 at 19:40
  • @KyleMiller, looked at it again and it made sense! But what I read seemed to indicate that this is true for any group-algebra, regardless of what $k$ is, when the group is finite. – Naweed G. Seldon Jul 15 '21 at 19:44
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    @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.) – Kyle Miller Jul 15 '21 at 19:58
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    @KyleMiller one definition of symmetric algebra is that there is a bimodule isomorphism between it and its dual, though there are several equivalents (details). Zeroth Hochschild cohomology, defined as $HH^0(\Lambda, \Lambda)$, is always the centre of $\Lambda$. OTOH $HH^0(\Lambda, \Lambda^*)$ is certainly the centre if $\Lambda$ is symmetric, but otherwise might not be. – Matthew Towers Jul 15 '21 at 21:01
  • @MatthewTowers Thanks -- this appears to be equivalent to what I had in mind. Symmetric algebras with a choice of symmetrizing form seem to be the same as symmetric Frobenius algebras. – Kyle Miller Jul 15 '21 at 21:14

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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?)

Kyle Miller
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