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In the introduction to this paper, the author mentions that any action of a lattice $\Gamma < G$ on a rank one symmetric space $X$ has a fixed point, where $G$ is a higher rank semisimple algebraic group over a local field (let's say the reals, for the purpose of this question), without rank one factors.

I'm looking for either a sketch of a proof of this fact or a reference for it (without using the results in this paper, since it seems it is an older result), at least in the case $X = \mathbb{H}^n$ (maybe the same proof will work in general, I do not know). Thanks in advance!

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  • $\begingroup$ You want to assume that G is simple and not only semisimple. For instance, a (possibly irreducible) lattice in $SL(2,\mathbb R) \times SL(2,\mathbb R)$ admits a non-elementary action on $\mathbb H^2$ by projecting to the first factor. $\endgroup$
    – Nicolast
    Commented May 21, 2021 at 22:33
  • $\begingroup$ True, thanks! Should say "without rank one factors" maybe? Or do we need simplicity of $G$? $\endgroup$
    – Mauro
    Commented May 21, 2021 at 22:35
  • $\begingroup$ Yes, you can say that. Then the statement is essentially Margulis's superrigidity theorem in the particular case of a target group of rank $1$. Perhaps Margulis's proof can be streamlined in that case, but I think it is a very strong statement in any case ! $\endgroup$
    – Nicolast
    Commented May 21, 2021 at 22:37
  • $\begingroup$ I see. I'm slightly confused about how we can apply Margulis' superrigidity here though. The statement of the theorem (at least the one in Zimmer's "Ergodic theory and semisimple groups", perhaps there's other more general statements) requires the homomorphism $\pi : \Gamma \to SO(1,n)$ to have Zariski dense image (which doesn't seem to be obvious here). Then, that would yield an extension to a homomorphism $G \to SO(1,n)$, would this give us the result? $\endgroup$
    – Mauro
    Commented May 21, 2021 at 22:53
  • $\begingroup$ That's the idea. But if it is not Zariski dense what can be the Zariski closure of the image ? Another way to state Margulis superrigidity is that if $\rho$ is a morphism from an irreducible lattice in a higher rank Lie group $G$ into a semisimple Lie group $H$, then there is a $\rho$-equivariant totally geodesic map from the symmetric space of $G$ to that of $H$. Now, you only need to understand why a totally geodesic map from a higher rank irreducible symmetric space to a rank $1$ symmetric space is constant. $\endgroup$
    – Nicolast
    Commented May 21, 2021 at 23:06

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