7
$\begingroup$

Suppose that $\Gamma$ is an irreducible lattice in a semi-simple real Lie group $G$ of higher rank (with infinite center!), is every homomorphism $\Gamma \to \mathbb{Z}$ trivial?

The case where $G$ has finite center follows easily from Margulis Normal subgroup Theorem. The simplest example I can think of where this question is relevant is the lift of $SL_2(\mathbb{Z}(\sqrt{2}))$ to the universal covering of $SL_2(\mathbb{R})\times SL_2(\mathbb{R})$.

Also, any reference where a discussion about lattices in semi-simple real Lie group of higher rank with infinite center would be appreciated. I only know of Ch.9 Sec.6 in Margulis Book, where I couldn't find an answer to this question.

Thank you!

Improve this question
$\endgroup$
5
  • 1
    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Apr 23, 2020 at 9:28
  • 1
    $\begingroup$ 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. $\endgroup$
    – YCor
    Commented Apr 23, 2020 at 9:41
  • 1
    $\begingroup$ 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 $\mathbb{Z}^2\rtimes \Gamma \to \mathbb{Z}$. $\endgroup$
    – Ian Agol
    Commented Apr 24, 2020 at 5:48
  • 1
    $\begingroup$ @IanAgol you're using a semidirect product notation for a nontrivial central extension... $\endgroup$
    – YCor
    Commented Apr 24, 2020 at 9:10
  • $\begingroup$ Thanks for the insight @IanAgol and for the references Yves. $\endgroup$
    – shurtados
    Commented Apr 25, 2020 at 3:05

2 Answers 2

Reset to default
5
$\begingroup$

Yes, every homomorphism $\Gamma \to \mathbb{Z}$ is trivial.

We may assume that $G$ is simply connected, thus it decomposes as a product of simple factors. Let's consider two cases:

  1. $G$ has exactly one non-compact simple factor.

  2. $G$ has at least two non-compact simple factors.

In case 1 $G$ has property (T), so also does $\Gamma$ and the result follows. In case 2 the result follows from theorem 0.8 in

Shalom, Yehuda Rigidity of commensurators and irreducible lattices. Invent. Math. 141 (2000), no. 1, 1–54.

Formally, the above theorem applies only for $\Gamma<G$ cocompact, but in fact the proof shows that you need 2-integrability of $\Gamma$ in $G$, which holds by Proposition 7.1 here, see the preceding discussion for the definition.

The above is an edit of an earlier partial answer I gave, based on the answer of Mikael de la Salle. See Mikael's answer and YCor's comments for further details.

Improve this answer
$\endgroup$
18
  • $\begingroup$ Some references about square integrability of lattices are listed in Example 6.8 here. $\endgroup$
    – YCor
    Commented Apr 24, 2020 at 9:06
  • $\begingroup$ Hi @YCor, you have a mistake in this reference. A non-compact lattice in $\text{SL}_2(\mathbb{R})$ cannot be square integrable, see Lemma 5.4. in arxiv.org/pdf/1006.5193.pdf. $\endgroup$
    – Uri Bader
    Commented Apr 24, 2020 at 9:25
  • $\begingroup$ Thanks, fortunately it's the only exception and harmless there. Indeed I relied on a Shalom's (Annals 2000) Theorem 3.6-3.7, proving square integrability of lattice in some Lie groups. Theorem 3.6 says "in $\mathrm{SO}(n,1)$ for $n\ge 4$" and Theorem 3.7 says "if $\mathrm{SO}(n,1)$ is replaced with any other rank one simple Lie group". The proof indeed seems to be done for $\mathrm{SO}(n\ge 4,1)$, then $\mathrm{SO}(3,1)$, and $\mathrm{SU}(n\ge 2,1)$, so only $\mathrm{SO}(2,1)$ should be excluded. It's hard to believe that such an ambiguous formulation survived in the published version. $\endgroup$
    – YCor
    Commented Apr 24, 2020 at 9:49
  • $\begingroup$ @YCor, almost. I disagree regarding $\text{SO}(3,1)$. The computation by the end of section 3 in Shalom's paper that you mentioned shows that for $\text{SO}(n,1)$ there is a $p$-integrable domain for every $p<n-1$, in particular a non-uniform lattice in $\text{SO}(3,1)$ is $p$-integrable for every $p<2$. While I do not know a proof that such a lattice in not 2-integrable, I find the converse unlikely. As I mentioned above, for $n=2$ such a proof exists. It could be a nice research project to settle this for a general $n$. $\endgroup$
    – Uri Bader
    Commented Apr 24, 2020 at 11:34
  • $\begingroup$ Thanks again, I hoped you would correct me if necessary. $\endgroup$
    – YCor
    Commented Apr 24, 2020 at 11:58
2
$\begingroup$

This is a follow-up of Uri's answer. My goal is just to confirm that (for any $p$) the $L^p$-integrability of lattices in a connected semisimple Lie group $G$ follows from the $L^p$-integrability of lattices in $G/Z(G)$. The non-trivial ingredient that is needed is that the central extension $G\to G/Z(G)$ is represented by a bounded $2$-cocycle. The argument (which I think I learned from Nicolas Monod) is at least in Proposition 7.1 of my paper with Tim de Laat https://arxiv.org/abs/1401.3611

The fact that this central extension is represented by a bounded $2$-cocycle follows, for simple Lie groups, from the well-known classical work of Guichardet-Wigner, see also the paper Shtern, A. I. Bounded continuous real 2-cocycles on simply connected simple Lie groups and their applications. Russ. J. Math. Phys. 8 (2001), no. 1, 122–133. The case of semisimple Lie groups follows by decomposing into simple parts.

Improve this answer
$\endgroup$
5
  • 1
    $\begingroup$ Actually that at the QI-level the extension $Z\to G\to G/Z$ is trivial (and hence represented by a bounded 2-cocycle) is essentially immediate, using that $Z$ is discrete and that $G/Z$ has a simply connected closed cocompact (solvable) subgroup (this is why I used $T$ in my sketch of argument as a comment to Uri's answer). $\endgroup$
    – YCor
    Commented Apr 30, 2020 at 13:00
  • $\begingroup$ Oh, I had never realized that. (and I had not read the comments...) Thanks. $\endgroup$
    – Mikael de la Salle
    Commented Apr 30, 2020 at 13:27
  • $\begingroup$ Thanks for the reference Mikael. $\endgroup$
    – shurtados
    Commented May 1, 2020 at 19:27
  • 1
    $\begingroup$ Thanks, Mikael, I should have remembered that you had this figured out. I incorporated this into my answer for the benefit of possible future readers. I hope you don't mind... $\endgroup$
    – Uri Bader
    Commented May 3, 2020 at 8:12
  • $\begingroup$ @UriBader Sure, it it better if your answer is complete. $\endgroup$
    – Mikael de la Salle
    Commented May 4, 2020 at 15:36

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .