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 $K$?

  • $\begingroup$ You mean, a cocompact arithmetic lattice somewhere... probably a nontrivial semisimple Lie group. $\endgroup$ – YCor Jun 23 '16 at 8:17
  • $\begingroup$ @YCor Thanks. Yes. Possibly of higher rank. All examples of cocompact lattices I know, are constructed as (irreducible) lattices in semisimple groups with compact factors. So, projecting them on such a factor gives a required homomorphism. The question is equivalent to asking for an example which is not of this kind. $\endgroup$ – Jarek Kędra Jun 23 '16 at 8:45
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    $\begingroup$ I was thinking of $S(\mathbf{Z})$ when $S$ is the group of norm 1 elements in a central simple algebra $M$ such that $M(\mathbf{Q})$ is a division algebra while $M(\mathbf{R})$ is isomorphic to, say, $M_n(\mathbf{R})$ with $n\ge 3$. Then $S(\mathbf{Z})$ is cocompact in $S(\mathbf{R})$ (because $S$ has $\mathbf{Q}$-rank 0)$. Then one should consider precise statements of superrigidity to show there are no homomorphisms with infinite image in compact Lie groups. $\endgroup$ – YCor Jun 23 '16 at 8:49
  • $\begingroup$ Nice question ! $\endgroup$ – Andreas Thom Jun 23 '16 at 20:42
  • $\begingroup$ Dear Yves and Dave, thanks a lot for your replies!! $\endgroup$ – Jarek Kędra Jun 23 '16 at 20:56

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.

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.

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