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What is the rational cohomological dimension of the lattices in $SL_n(\mathbb{Q}_p)$, where $n\geq 3$ ? A reference would be appreciated.

For the definition of "cohomological dimension of a group over ring" one can look at here: https://en.wikipedia.org/wiki/Cohomological_dimension, and the term `rational cohomological dimension' is used when the ring is taken as the field of rational numbers $\mathbb{Q}$.

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    $\begingroup$ It would help if you gave examples, since the term "lattice" has a different flavor when the field is real or global or $p$-adic. And the term "rational cohomological dimension" may not be familiar to everybody without a little explanation. $\endgroup$ Nov 4, 2015 at 21:46
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    $\begingroup$ If a lattice is a discrete cocompact subgroup, the Bruhat-Tits building for $\mathrm{SL}_n(\mathbf Q_p)$ having dimension $n-1$, it cannot be more than that. Such a lattice is known (at least when $n\geq 3$) to be arithmetic, and Borel and Serre have produced, for an arithmetic subgroup, a module $M$ such that $H_{n-1}(\Gamma,M)\neq 0$. References are to be found in the books on buildings and cohomology of groups by Kenneth Brown ... $\endgroup$
    – few_reps
    Nov 4, 2015 at 22:00
  • $\begingroup$ A lattice means here "discrete subgroup of finite covolume"; since in SL($n$,$\mathbf{Q}_p$) they are all cocompact, the argument of few_reps shows that it's $\le n-1$ and actually it's indeed very probably $n-1$ (e.g., because they always contain some copy of $\mathbf{Z}^{n-1}$ acting cocompactly on a flat). $\endgroup$
    – YCor
    Nov 4, 2015 at 23:45
  • $\begingroup$ @YCor : I made as if the definition of the cohomological dimension used in the OP is the virtual cohomological dimension for which Borel-Serre dualizing module give a definitive answer ... but it asks for the group to be arithmetic and this is hard to prove (I never studied the proof but it doesn't seem trivial). So your argument would certainly be more natural, (and as definitive since virtual cohomological dimension is monotone with respect to inclusions) but ... how do you exhibit such a $\mathbf Z^{n-1}$ ? $\endgroup$
    – few_reps
    Nov 4, 2015 at 23:55
  • $\begingroup$ I don't know if it uses arithmeticity, or just some general fact about cocompact actions on CAT(0) complexes. $\endgroup$
    – YCor
    Nov 5, 2015 at 0:53

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