Explicit orthogonal representations demonstrate non-trivial cohomology in SL(N,Z), suggesting the absence of Kazhdan's property T.
We construct explicit finite-dimensional orthogonal representations π N of SLN(Z) for N ∈ \3,4\ all of whose invariant vectors are trivial, and such that HN - 1(SLN(Z),π N) is non-trivial. This implies that for N as above, the group SLN(Z) does not have property (TN-1) of Bader–Sauer and therefore is not $(N-1)$-Kazhdan in the sense of De Chiffre–Glebsky–Lubotzky–Thom, both being higher versions of Kazhdan’s property T.
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Brück et al. (2025) studied this question.
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