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Jul 24, 2020 at 14:25 comment added Ian Agol Well, you need to know also that the cohomology is finite so that the Euler characteristic is well-defined. This follows from the compactification, or from a finite spine. But as I said, thinking about strata can be avoided by passing to a manifold cover (a level 3 principal congruence subgroup should suffice).
Jul 24, 2020 at 9:32 comment added sadok kallel This seems right. I think we can avoid the compactification by restricting the orbi-bundle over strata (locally closed). Over some strata of the base, the fiber is SO(n) and over other strata it is a quotient of SO(n) by a finite group I believe. In all cases, the fiber is compact with zero Euler characteristic. We then get a stratification of $SL_n(\mathbb R)/SL_n(\mathbb Z))$ by bundles whose $\chi_c=0$ since $\chi_c$ is multiplicative on bundles. Since $\chi_c$ is additive over strata, this gives the answer $0$ as well.
Jul 23, 2020 at 17:25 history edited Ian Agol CC BY-SA 4.0
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Jul 23, 2020 at 16:05 history edited Ian Agol CC BY-SA 4.0
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Jul 22, 2020 at 18:18 history answered Ian Agol CC BY-SA 4.0