# Cohomology of linear algebraic groups

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 $$i=1,2$$?

• Isn't that defined by the cohomology of the Lie algebras? I think Chevally studied it during 1950s... – Bombyx mori Sep 17 at 15:53
• @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. – YCor Sep 17 at 15:54
• 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. – YCor Sep 17 at 16:00
• @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! – Qixiao Sep 17 at 16:32
• 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. – YCor Sep 17 at 17:29

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 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). – Aurélien Djament Sep 19 at 6:25