Timeline for Second Betti number of lattices in $\mathrm{SL}_3(\mathbf{R})$
Current License: CC BY-SA 4.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Mar 2, 2023 at 7:21 | answer | added | Uri Bader | timeline score: 8 | |
| Dec 4, 2018 at 0:11 | history | edited | YCor | CC BY-SA 4.0 |
corrected an unclear sentence to an alternative
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| Oct 29, 2018 at 20:27 | vote | accept | YCor | ||
| Oct 29, 2018 at 15:05 | answer | added | Steffen Kionke | timeline score: 18 | |
| Oct 28, 2018 at 16:49 | comment | added | YCor | @Marty cocompact lattices in $\mathrm{SL}_3(\mathbf{R})$ are reasonably classified (they're arithmetic and the ways to produce arithmetic lattices are classified, for instance in Witte-Morris' book arxiv.org/abs/math/0106063, §6.7, at least up to commensurability). | |
| Oct 28, 2018 at 16:38 | comment | added | Marty | Maybe one can compute $b_2$ for the following non-division algebra example. Can't remember where I saw it, but it's a standard kind of thing. Take $F = Q(\sqrt{2})$ and $L = Q(\sqrt[4]{2})$. Take the unitary group $\Gamma = SU_3(A)$ , for the trivial Hermitian form and nontrivial element of $Gal(L/F)$, and $A$ a suitable ring of integers. Then $\Gamma$ will embed into $SL_3(R)$ (at one place of $F$) as a cocompact lattice... not sure about torsion. | |
| Oct 27, 2018 at 23:12 | answer | added | Aurel | timeline score: 3 | |
| Oct 27, 2018 at 18:19 | comment | added | Ian Agol | It seems plausible that surface subgroups represented by sublattices in $O(2,1;\mathbb{R})$ might be homologically non-trivial. If one could find a sublattice of complementary dimension intersecting this generically, then one ought to be able to prove that it becomes homologically non-trivial in a finite-index subgroup using double coset separability arguments. | |
| Oct 27, 2018 at 12:19 | history | asked | YCor | CC BY-SA 4.0 |