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.




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



Matsushima-Murakami Isomorphism for $L^2$-cohomology
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).


