# Homomorphisms from higher rank lattices with infinite center to $\mathbb{Z}$

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!

• 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.
– YCor
Apr 23 '20 at 9:28
• 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.
– YCor
Apr 23 '20 at 9:41
• 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 24 '20 at 5:48
• @IanAgol you're using a semidirect product notation for a nontrivial central extension...
– YCor
Apr 24 '20 at 9:10
• Thanks for the insight @IanAgol and for the references Yves. Apr 25 '20 at 3:05

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 $$\Gamma cocompact, but in fact the proof shows that you need 2-integrability of $$\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.

• Some references about square integrability of lattices are listed in Example 6.8 here.
– YCor
Apr 24 '20 at 9:06
• 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. Apr 24 '20 at 9:25
• 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.
– YCor
Apr 24 '20 at 9:49
• @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 $n$. Apr 24 '20 at 11:34
• Thanks again, I hoped you would correct me if necessary.
– YCor
Apr 24 '20 at 11:58

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.

• 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).
– YCor
Apr 30 '20 at 13:00