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

MathOverflow Asked by sadok kallel on November 9, 2021

1 AnswersI 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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