Timeline for Homomorphisms from higher rank lattices with infinite center to $\mathbb{Z}$
Current License: CC BY-SA 4.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Apr 30, 2020 at 12:44 | answer | added | Mikael de la Salle | timeline score: 2 | |
| Apr 25, 2020 at 3:09 | vote | accept | shurtados | ||
| Apr 25, 2020 at 3:05 | comment | added | shurtados | Thanks for the insight @IanAgol and for the references Yves. | |
| Apr 25, 2020 at 2:59 | vote | accept | shurtados | ||
| Apr 25, 2020 at 3:01 | |||||
| Apr 24, 2020 at 9:10 | comment | added | YCor | @IanAgol you're using a semidirect product notation for a nontrivial central extension... | |
| Apr 24, 2020 at 7:22 | answer | added | Uri Bader | timeline score: 5 | |
| Apr 24, 2020 at 5:48 | comment | added | Ian Agol | In the case of $\Gamma=SL_2(Z[\sqrt{2}])$, the central extension of $SL_2(\mathbb{R})$ will be induced by the holomorphic 2-form on $\mathbb{H}^2$. I think this defines a holomorphic Hilbert modular form on $(\mathbb{H}\times\mathbb{H})/\Gamma$ of weight $(2,0)$, and it should give a non-trivial 2nd cohomology class on this Hilbert modular surface for each factor. So I think that the extension by $\mathbb{Z}\times\mathbb{Z}$ should be non-trivial, even for any finite-index subgroup. Hence it should lie in the kernel of a homomorphism of $\mathbb{Z}^2\rtimes \Gamma \to \mathbb{Z}$. | |
| Apr 23, 2020 at 18:41 | history | edited | shurtados | CC BY-SA 4.0 |
added 2 characters in body
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| Apr 23, 2020 at 9:41 | comment | added | YCor | Assume for simplicity that that the center is virtually cyclic (we can boil down to this case), so one has to prove that $\Gamma\times\mathbf{Z}$ and $\tilde{\Gamma}$ are not virtually isomorphic, where $\tilde{\Gamma}$ denotes the lattice and $\Gamma$ is its projection modulo the center. One approach would be to prove that $\Gamma\times\mathbf{Z}$ and $\tilde{\Gamma}$ are not IME (integrably measure equivalent). A result in this direction (for cocompact lattices in $\mathrm{SL}_2(\mathbf{R})$) is due to Das-Tessera. | |
| Apr 23, 2020 at 9:28 | comment | added | YCor | It's true if $G$ has a noncompact simple factor with Property T. So the remaining case is that when $G$ is a product of $\ge 2$ rank-1 groups without Property T, as in your specific example. | |
| Apr 23, 2020 at 8:29 | history | asked | shurtados | CC BY-SA 4.0 |