# Euler characteristic with compact support of spaces of Euclidean lattices

Has the Euler characteristic with compact support of $$mathrm{SL}_n(mathbb R)/mathrm{SL}_n(mathbb Z)$$ been computed ? References? Thanks.

I think that the euler characteristic is 0 for the following reasons.

Firstly, the space $$SL_N(mathbb{R})$$ is a bundle over the symmetric space $$SO(N,mathbb{R})backslash SL_N(mathbb{R}) = SP(n,mathbb{R})=X$$, the space of symmetric positive-definite real matrices of determinant 1. For a discussion of this symmetric space, see e.g. Bridson-Haefliger II.10. Then $$SL_N(mathbb{R})/SL_N(mathbb{Z})$$ is a bundle over $$X/SL_N(mathbb{Z})$$ with fiber $$SO(N,mathbb{R})$$. Note that this is an orbifold bundle, but that by passing to a torsion-free subgroup, one can assume that it is a manifold (and since you're interested in euler characteristic, this just multiplies by the index).

Now the space $$X/SL_N(mathbb{Z})$$ admits a bordification by Borel-Serre. Hence $$SL_N(mathbb{R})/SL_N(mathbb{Z})$$ has a bordification by an $$SO(N,mathbb{R})$$-bundle over the Borel-Serre bordification. Hence it is the interior of a manifold with boundary $$M$$. In this case, $$H^*_c(SL_N(mathbb{R})/SL_N(mathbb{Z}))cong H^*(M,partial M)$$. Then by Lefschetz duality, $$chi(H^*_c(M,partial M))=chi(M)$$.

But since $$M$$ is a bundle with fiber $$SO(N,mathbb{R})$$, and $$chi(SO(N,mathbb{R}))=0$$ (any Lie group has a nowhere vanishing vector field), we have $$chi(M)=chi(SO(N,mathbb{R}))times chi(X/SL_N(mathbb{Z})) =0$$, since the euler characteristic of bundles is the product of the euler characteristic of the base and the fiber.

Answered by Ian Agol on November 9, 2021

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