TITLE: Homomorphisms from higher rank lattices with infinite center to $\mathbb{Z}$
QUESTION [7 upvotes]: 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!

REPLY [2 votes]: 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.