Let $\Gamma$ be a cocompact arithmetic lattice in a semisimple algebraic group. Does it admit a homomorphism $\Gamma \to K$ with infinite image into a compact real Lie group $K$?



Does every cocompact lattice admit a homomorphism (with infinite image) into a compact Lie group?
No, there are cocompact lattices for which no such homomorphism exists. This is Warning 16.4.3 on page 330 of my book [Introduction to Arithmetic Groups](https://arxiv.org/src/math/0106063v6/anc/IntroArithGrps-FINAL.pdf).


Assume $\Gamma$ is irreducible, $G$ has no compact factors, and $\mathrm{rank}\_{\mathbb{R}} G \ge 2$. Modulo finite groups, the Margulis Arithmeticity Theorem tells us there is a semisimple algebraic group $S$, defined over $\mathbb{Q}$, such that $\Gamma = S(\mathbb{Z})$. One version of the Margulis Superrigidity Theorem (namely, Corollary 16.4.1 in my book) tells us (modulo finite groups) that any homomorphism from $\Gamma$ to $\mathrm{GL}(n,\mathbb{R})$ must extend to be defined on all of $S(\mathbb{R})$. If $S(\mathbb{R})$ has no compact factors (that is, if $G$ has no compact factors and $\Gamma = G(\mathbb{Z})$), this implies that $\Gamma$ has no infinite-image homomorphism to any compact, connected Lie group $K$.


@YCor's construction using norm 1 elements in a division algebra does indeed give an example, since $\Gamma = S(\mathbb{Z})$ and $S(\mathbb{R}) \cong \mathrm{SL}(n, \mathbb{R})$ has no compact factors.


