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!



Homomorphisms from higher rank lattices with infinite center to $\mathbb{Z}$
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](https://arxiv.org/pdf/1401.3611.pdf), see the preceding discussion for the definition.




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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.


