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Let $G$ be a free group of finite rank. Consider any commutative ring $R$ containing $\mathbb{Z}$. Consider the group ring $RG$.

Q) What can we say about the Hochschild cohomology groups of $RG$ with coefficients in $R$?

I can find the $HH^1(RG)$ in terms of outer derivations but again I don't know the complete description of derivations on a free group. Also, I could not find a single element of $HH^n(RG)$ for $n\geq 2$.

Any comment, reference, or suggestion will be extremely helpful. Thanks in advance.

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Hochschild cohomology of group rings is reducible to ordinary group cohomology. (Not really if you're interested in multiplicative structure, but that's another story). Everyting works over any commutative base ring, you can assume it's $\Bbb Z$ if you'd like.

Namely, there's an adjunction between $kG-bimod$ and $kG-mod$ (I mean left modules): forgetful right adjoint functor $\rho$ sends a bimodule $A$ to the module $\rho A$ where action is conjugation, and left adjoint $\lambda$ is $M \mapsto \lambda M := M \otimes_k kG$. Right action is multiplication in $kG$, and left action is $h \cdot \sum_G m \otimes g = \sum_G hm \otimes hg$.

So, now we can get a string of identifications

$$HH^*(kG, A) = Ext_{bimod}(kG, A) = Ext_{bimod}(\lambda k, A) = Ext_{mod}(k, \rho A) = H^*(G, \rho A)$$

In particular, Hochschild cohomological dimension of a group ring is equal to comohomological dimension of a group itself, and $HH^*(kG, k) = H^*(G, k)$. (so if your group is free, there's no second cohomology at all)

Everything above is pretty much copied verbatim from chapter X of Homological algebra by Cartan and Eilenberg.

I'd suggest looking at Dan Burghelea and Sarah Witherspoons papers for further reading, but cannot provide explicit references for a while (because searching old papers from a phone is somewhat difficult).

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    $\begingroup$ Maybe it would be helpful to explicitly say a free group has cohomoligical dimension one since the rest of your answer is about a generic group.but that part is specific to free groups. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 24, 2022 at 13:20
  • $\begingroup$ Thanks for improving mediocre clarity of my writing! $\endgroup$
    – Denis T
    Commented Jan 24, 2022 at 13:23
  • $\begingroup$ Your answer was fine. $\endgroup$
    – Benjamin Steinberg
    Commented Jan 24, 2022 at 13:39
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    $\begingroup$ In order to get the adjunction to also provide an isomorphism for Ext, you probably want to say something about $\rho$ being flat (or that $\lambda$ preserves projectives). This guarantees that a projective resolution of $k$ turns into one for $\lambda k$ and hence can be used to compute Ext. It also shows directly that the projective dimension of $kG$ as a bimodule is bounded by the projective dimension of $k$ as a $kG$-module $\endgroup$
    – Benjamin Steinberg
    Commented Jan 24, 2022 at 13:56

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