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Let $K$ be a field and $G=\pi_1(\Sigma_g)$ the surface group of genus $\geq 2$. I want to know the Hochschild cohomology of the group algebra $A=K[G]$ with coefficients in $A$ and $A\otimes A$, namely, $ \mathrm{HH}^\bullet(A)$ and $\mathrm{H}^\bullet(A,A\otimes A)\,. $

This includes many important spaces such as the center, outer derivations and double derivations. The degree $\geq 3$-part is zero, since one can take a free resolution associated to a cellular decomposition of the upper half plane.

Any comments or references on this topic (including outer derivations of a general group rings) are appreciated.

Edit: What I really want to know is whether $\mathrm{H}^1(A,A\otimes A)=0$. If so, the rest should be (hopefully) easy.

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  • $\begingroup$ I don't know if this will help, but if I recall correctly, for $A=kG$ and $M$ an $A$-bimodule, the Hochschild cohomology groups $H^n(A,M)$ are isomorphic in a canonical way to $H^n(A,M^\circ)$ where $M^\circ$ is the new $A$-bimodule defined as follows: the left G-action is defined by conjugation, $g \bullet x := gxg^{-1}$, and the right $G$-action is defined by augmentation, $x\bullet g = x$ (and then one extends this G-bimodule structure to an $A$-bimodule structure by linearity). $\endgroup$
    – Yemon Choi
    Commented Oct 26, 2023 at 10:29
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    $\begingroup$ For the case $M=A$, one can then view $M^\circ$ as a direct sum $\bigoplus_C kC$ where the sum is over all conjugacy classes $C$ and $kC$ is just the free-vector space on $C$ with the natural $G$-action. What I don't recall is if there is a way to get at $H^n(kG, kC)$ via Ext machinery. Things would work more nicely if one takes $M=A^\ast = {\rm Hom}_k(A,k)$, the dual of $A$ as an $A$-bimodule, for then $H^n(kG, (kC)^\ast) = {\rm Ext}^n_{G} (kC, k) = {\rm Ext}^n_H(k,k)$ where $H$ is the centralizer of any choice of $y\in C$, but of course that's not the question you were asking about. $\endgroup$
    – Yemon Choi
    Commented Oct 26, 2023 at 10:52
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    $\begingroup$ @Yemon Choi: Since this $G$ is a Poincare duality group, these Ext groups can be rewritten as Tor groups, which behave well with respect to direct sums. $\endgroup$
    – Tom Goodwillie
    Commented Oct 28, 2023 at 22:22

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In general, if $A=K[G]$ for a group $G$ then an $A$-bimodule is the same as a vector space with action of $G\times G$.

$HH^0(A,M)$ is the vector space of $A$-bimodule maps $A\to M$, which (in the case $K[G]$) is the same as the subspace of $M$ consisting of elements $m$ such that $gmg^{-1}=m$ for all $g\in G$, in other words the group cohomology $H^0(G,M_{conj})$, where $M_{conj}$ means the vector space $M$ with $G$ acting by $m\mapsto gmg^{-1}$.

Therefore $HH^i(A,-)$, the $i$th derived functor of $HH^0(A,-)$, is given by $$ HH^i(A,M)=H^i(G,M_{conj}). $$

When $G$ is the fundamental group of a closed oriented aspherical $2$-manifold then there are duality isomorphisms $$ H^i(G,M_{conj})\cong H_{2-i}(G,M_{conj}). $$ In the case when $M=A\otimes A$, the vector space $M_{conj}$ has a $K$-basis consisting of the elements $g_1\otimes g_2$, and the action of $G$ is freely permuting these basis elements, with orbit representatives $g\otimes 1$. So the only nontrivial (co)homology group is $H^2=H_0\cong K[G]$.

In the case when $M=A$, the space $M_{conj}$ is a direct sum over conjugacy classes of $G$, and we find that $H^i=H_{2-i}$ is the direct sum, over conjugacy classes, of $H_{2-i}(Z_g,K)$, where $Z_g$ is the centralizer of a representative $g$ of the conjugacy class.

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