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!

$\endgroup$
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
    Apr 23 '20 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
    Apr 23 '20 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
    Apr 24 '20 at 5:48
  • 1
    $\begingroup$ @IanAgol you're using a semidirect product notation for a nontrivial central extension... $\endgroup$
    – YCor
    Apr 24 '20 at 9:10
  • $\begingroup$ Thanks for the insight @IanAgol and for the references Yves. $\endgroup$
    – shurtados
    Apr 25 '20 at 3:05
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.

$\endgroup$
18
  • $\begingroup$ Some references about square integrability of lattices are listed in Example 6.8 here. $\endgroup$
    – YCor
    Apr 24 '20 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
    Apr 24 '20 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
    Apr 24 '20 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
    Apr 24 '20 at 11:34
  • $\begingroup$ Thanks again, I hoped you would correct me if necessary. $\endgroup$
    – YCor
    Apr 24 '20 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.

$\endgroup$
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
    Apr 30 '20 at 13:00
  • $\begingroup$ Oh, I had never realized that. (and I had not read the comments...) Thanks. $\endgroup$ Apr 30 '20 at 13:27
  • $\begingroup$ Thanks for the reference Mikael. $\endgroup$
    – shurtados
    May 1 '20 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
    May 3 '20 at 8:12
  • $\begingroup$ @UriBader Sure, it it better if your answer is complete. $\endgroup$ May 4 '20 at 15:36

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.