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Let $R$ be a commutative ring. Let $G\subset \mathrm{GL}_m$ be a linear algebraic subgroup. Has the group cohomology $H^i(G(R),R^m)$ been studied in the literature?

For example, do we know

(1) $H^i(\mathrm{Sp}_{2n}(\mathbb{Z}),\mathbb{Z}^{2n})$ for $i=1,2$?

(2) $H^i(\mathrm{Sp}_{2n}(\mathbb{C}),\mathbb{C}^{2n})$ for $i=1,2$?

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  • $\begingroup$ Isn't that defined by the cohomology of the Lie algebras? I think Chevally studied it during 1950s... $\endgroup$
    – Bombyx mori
    Commented Sep 17, 2019 at 15:53
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    $\begingroup$ @Bombyxmori If it's isomorphic to a Lie algebra cohomology group, this is a theorem, not a definition. In the $\mathbf{Z}$ case this sounds quite doubtful, by the way. $\endgroup$
    – YCor
    Commented Sep 17, 2019 at 15:54
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    $\begingroup$ As regards (2), do you mean group cohomology or continuous group cohomology? For instance $H^2(\mathrm{Sp}_{2n}(\mathbf{C}),\mathbf{C})$ is huge but the continuous cohomology is trivial. Maybe the same phenomenon occurs with the module $\mathbf{C}^{2n}$, I'm not sure. In general guess that the continuous cohomology in degree $\ge 1$ of any finite-dimensional $G$-module is zero, for $G$ semisimple. $\endgroup$
    – YCor
    Commented Sep 17, 2019 at 16:00
  • $\begingroup$ @YCor Thanks for pointing out, let's focus on continuous cohomology. And would you explain the calculation of the two versions of $H^2(\mathrm{Sp}_{2n}(\mathbb{C}),\mathbb{C})$? Thanks! $\endgroup$
    – user39380
    Commented Sep 17, 2019 at 16:32
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    $\begingroup$ For the continuous one, it's a standard fact about central extensions of semisimple Lie groups. For the non-continuous one, it's related to (symplectic) K-theory and more complicated (oh, well I'm sure that $H_2$ is huge; that $H^2$ valued in $\mathbf{C}$ is still huge, should be checked); see mathoverflow.net/questions/63529 for a related discussion. $\endgroup$
    – YCor
    Commented Sep 17, 2019 at 17:29

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Let me focus on (1). It is answered in Lemma A.3 of this paper by Krannich. However, let me explain why his answers are 2-torsion. This argument also works for (2) in either interpretation.

That $H^i(Sp_{2n}(\mathbb{Z});\mathbb{Z}^{2n})$ is 2-torsion is equivalent to it being zero after inverting 2 in the coefficients. This can be proven using the "centre kills" trick. The group $Sp_{2n}(\mathbb{Z})$ has $C_2 = \{\pm id\}$ in its center, so we can take the quotient $Sp_{2n}(\mathbb{Z})/C_2$ and attempt to compute the desired cohomology groups using the Serre spectral sequence with coefficients in $\mathbb{Z}[1/2]^{2n}$ for the fibration sequence

$$BC_2 \to BSp_{2n}(\mathbb{Z}) \to B(Sp_{2n}(\mathbb{Z})/C_2),$$

But $H^*(BC_2;\mathbb{Z}[1/2]^{2n})$ vanishes; by a transfer argument it vanishes in positive degrees and is given by the invariants in degree 0. Thus the $E^2$-page of the spectral sequence vanishes and hence so does its abutment $H^*(Sp_{2n}(\mathbb{Z});\mathbb{Z}[1/2]^{2n})$.

If you care about the 2-torsion in higher degrees, it becomes more difficult. I suggest combining homological stability with polynomial coefficients with a variation of work of Djament and Vespa expressing the stable homology with twisted coefficients in terms of functor homology (e.g. this paper).

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    $\begingroup$ If you are only interested with the stable value (that is, for $n$ big enough with respect to $i$), you can get very general computations (meaning: for any cohomological degree and more general coefficients, and more general fields/rings for taking the points of the symplectic groups) with the technology of polynomial functors, assuming the (co)homology with constant coefficients to be known --- see my article Sur l'homologie des groupes unitaires à coefficients polynomiaux, Journal of K-theory / Volume 10 / Issue 01 / August 2012, pp 87-1394 (on the arXiv: arxiv.org/abs/1104.1507). $\endgroup$
    – Aurélien Djament
    Commented Sep 19, 2019 at 6:25

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