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I am not an expert at all in the subject of Lie groups, lattices, arithmetic groups and rigidity. But, lately I am interested in Margulis superrigidity theorem, which in most versions can be stated as follows:

Theorem. Let $G$ and $G'$ be semisimple connected real center-free Lie groups without compact factors with $\mathrm{rk}(G)\geq 2$, $\Gamma < G$ be an irreducible lattice, and $\pi: \Gamma \to G'$ a homomorphism with $\pi(\Gamma)$ being Zariski dense in $G'$. Then $\pi$ extends to a rational epimorphism $\pi':G\to G'$.

Here "$H$ is without compact factors" means, for $H$ center-free, that if we write $H=\prod_{i=1}^kS_i$ with $S_i$ simple, then each $S_i$ is non-compact, or equivalently has positive (real) rank; the (real) rank $\mathrm{rk}(H)$ of $H$ is the sum of the ranks of all $S_i$. Irreducibility of a lattice $\Gamma$ means that its projection in $H/S_i$ has a dense image for all $i$.

Questions:

Do you have examples of Lie groups and lattices for which Margulis theorem applies, and also groups for which the theorem does not holds. In particular, do this theorem apply for $G=G'=\mathrm{PSL}(n,\mathbb{R})$ and $\Gamma=\mathrm{PSL}(n,\mathbb{Z})$. Does this implies that $\mathrm{Out}(\mathrm{PSL}(n,\mathbb{Z}))$ is finite?

Thank you all.

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  • $\begingroup$ If $G$ semisimple is written as $S_1\dots S_n$ where $S_i$ are its simple normal subgroups, "no compact (simple) factor" means that none of the $S_i$ is compact. Probably MathStackExchange seems more appropriate. $\endgroup$
    – YCor
    Commented Dec 8, 2017 at 13:10
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    $\begingroup$ For $n\ge 3$, $Out(PSL_n(\mathbf{Z}))$ is finite. More generally $Out(\Gamma)$ is finite for every irreducible. lattice $\Gamma\subset G$, $G$ semisimple not locally isomorphic to $SL_2(\mathbf{R})$. This follows from Mostow rigidity (which is anterior to the stronger Margulis' superrigidity): every automorphism of $\Gamma$ extends to an automorphism $G$. Since $Out(G)$ is finite, a finite index subgroup of $Aut(\Gamma)$ consists of the normalizer $N$ of $\Gamma$ in $G$. Since $N$ contains $\Gamma$ with finite index, $Out(\Gamma)$ finite follows. $\endgroup$
    – YCor
    Commented Dec 8, 2017 at 16:10
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    $\begingroup$ (For $n\ge 2$ $Out(PSL_2(\mathbf{Z}))$ is also finite, as mentioned by Igor, but for reasons essentially unrelated to (super)rigidity: but this does not hold for its torsion-free finite index subgroups) $\endgroup$
    – YCor
    Commented Dec 8, 2017 at 16:12
  • $\begingroup$ @YCor: Why does $N$ contain $\Gamma$ as a finite index subgroup? $\endgroup$
    – Luis Jorge
    Commented Dec 8, 2017 at 16:26
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    $\begingroup$ PS: there are several questions in your question; some but not all rather belong on MathSE. The question why (super)rigidity implies finiteness of Out is reasonable here. Possibly you could reedit accordingly. $\endgroup$
    – YCor
    Commented Dec 8, 2017 at 16:33

2 Answers 2

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Actually, the way it is stated, the "Theorem" is false. Take $G=PSL_3(\mathbb R)$, $G'=SL_3(\mathbb R)$, and $\Gamma ' \subset G'$ a torsion free lattice. Then the map $G'\rightarrow G$ sends $\Gamma '$ isomorphically to a lattice $\Gamma \subset G$. But the map $\Gamma \simeq \Gamma '$ does not extend to a rational epimorphism $PSL_3(\mathbb R) =G \rightarrow G'=SL_3(\mathbb R)$. It is not just that $G,G'$ are center freee but are of adjoint type (over $\mathbb C$, they should be centre free).

(the modified question has been answered by many people in the answers and comments).

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Compact factors:

$$SO(3) \times SO(3).$$

No compact factors $SL(3, \mathbb{R}).$

$SL(3, \mathbb{R})$ - theorem applies. $SL(2, \mathbb{R})$ -- does not (rank 1). $Out(PSL(2, \mathbb{Z})$ is finite, however - it consists of the Dyer automorphism (which is an involution) and the identity, see my answer to this question.

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  • $\begingroup$ Than you for your answer. I was tinking on finiteness of $PSL(n,\mathbb{Z})$ for all $n\geq 2$, I already edited my question. Do you know if all that outer automorphisms groups are finite? do youknow any good reference? $\endgroup$
    – Luis Jorge
    Commented Dec 8, 2017 at 15:58
  • $\begingroup$ It is known that $SL(n,\mathbb{R})$ is diffeomorphic to $SO(n)\times \mathbb{R}^m$ with $m=n(n-1)/2$. That was actually the reason I got confused in the first place, I mean, it seems like $SL(n,\mathbb{R})$ does have a compact factor $\endgroup$
    – Luis Jorge
    Commented Dec 8, 2017 at 16:33
  • $\begingroup$ Yes "compact factor" is somewhat ambiguous. It should be "compact simple factor in this case", but this remains ambiguous. I explained the intended meaning in my first comment to your question. $\endgroup$
    – YCor
    Commented Dec 8, 2017 at 17:29
  • $\begingroup$ I see, so, the "word" factor should be understood as group theoretic factors, not as direct factors. I guess the decomposition of $SL(n,\mathbb{R})=SO(n)\times \mathbb{R}$ is not a contradiction siince $\mathbb{R}$ is not considered simple as a Lie group. $\endgroup$
    – Luis Jorge
    Commented Dec 8, 2017 at 18:01
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    $\begingroup$ It's not the point. First, $SL_n(\mathbf{R})=SO(n)\times \mathbf{R}$ is senseless since they are not even homeomorphic (different dimension). It's true that $SL_n(\mathbf{R})$ is homeomorphic to $SO(n)\times \mathbf{R}^{n(n+1)2-1}$. But it's not a group isomorphism. $\endgroup$
    – YCor
    Commented Dec 8, 2017 at 22:04

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