Let $\mathbf{G}$ be a reductive connected linear algebraic group over a totally real global number field, say $\mathbb{Q}$. Let $\mathbb{A}=\mathbb{R}\times\mathbb{A}_f$ be the ring of rational adele. Fix a maximal compact subgroup $K_\infty$ of $\mathbf{G}(\mathbb{R})$ and an open compact subgroup $K_f$ of $\mathbf{G}(\mathbb{A}_f)$. Fix a finite dimensional complex representation $V$ of $\mathbf{G}(\mathbb{Q})$ with corresponding local system $\mathcal{V}$ on the Shimura variety $S_{K_f}=\mathbf{G}(\mathbb{Q})\backslash \mathbf{G}(\mathbb{A})/K_\infty K_f$.

Then there should be a Hecke-equivariant isomorphism $$\boxed{H^\bullet_{(2)}(S_{K_f},\mathcal{V})\cong H^\bullet(\mathfrak{g},K_\infty;L^2(\mathbf{G}(\mathbb{Q})\backslash \mathbf{G}(\mathbb{A}))^{K_f}\otimes V)}$$ between the $L^2$-cohomology and the relative Lie algebra cohomology, where $\mathfrak{g}=Lie(\mathbf{G})$.

I have found several references for this and this seems to be well-known. However, everybody refers the article of Borel and Casselman: "$L^2$-cohomology of locally symmetric manifolds of finite volume" in Duke vol 50, p. 625-647. Indeed, Proposition 5.6 in this article makes almost exactly this statement, but only for semisimple $\mathbf{G}$. It seems to me that the proof does not use this extra assumption.

My question: Does this isomorphism hold for arbitrary connected reductive $\mathbf{G}$?

up vote 4 down vote accepted

Morally yes, by

Franke, Jens Harmonic analysis in weighted $L_{2}$-spaces. Ann. Sci. École Norm. Sup. (4) 31 (1998), no. 2, 181--279

See especially theorem 4 (and note that if you want an equivariant isomorphism, you need to restrict to the twisted action by some character).

Your Answer

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.

Not the answer you're looking for? Browse other questions tagged or ask your own question.