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).

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