Is there any example of a continuous endomorphism of a Lie group that has recurrent points with non-compact orbit closure?
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3$\begingroup$ Yes I am more interested in the case where $G$ is connected. The map $f(z) = z^2$ is a surjective endomorphism of $\mathbb{C}^*$ or of $S^1$, but it is not an automorphism, or I am missing something? $\endgroup$– Mauro PatrãoCommented Aug 16, 2016 at 23:16
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1$\begingroup$ Oh, yes you're right. So, I mean, we can reduce to the case of an endomorphism that is a self-covering (surjective with discrete kernel, actually necessarily finite), based on the fact that an endomorphism with dense image always has this form. $\endgroup$– YCorCommented Aug 16, 2016 at 23:30
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1$\begingroup$ If $G$ is semisimple and linear we can assume that $f$ is surjective. In fact, there is an integer $m$ such that $f$ restricted to $H = f^m(G)$ is surjective and $H$ is semisimple and thus closed, since $G$ is linear. $\endgroup$– Mauro PatrãoCommented Aug 17, 2016 at 2:12
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1$\begingroup$ Assuming that $G$ is linear semisimple and $f$ surjective, every recurrent point of $f$ has compact orbit closure. In fact, there is an integer $n$ such that $f^n$ is an inner automorphism and thus a restriction to $G$ of a linear map and $G$ is a closed subset. The claim follows since the orbit closure of every recurrent point of a linear map is a torus. $\endgroup$– Mauro PatrãoCommented Aug 17, 2016 at 2:23
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1$\begingroup$ @YCor: If $G$ is solvable and $T$ is the maximal torus in its center, then $G/T$ is solvable and simply-connected and an endomorphism $f$ induces a endomorphism $\phi$ on $G/T$. Thus there exists a recurrent point of $f$ with a non-compact orbit closure if and only if there exists a recurrent point of $\phi$ with a non-compact orbit closure. Hence, if $G$ is solvable, we can assume that $G$ is simply-connected. $\endgroup$– Mauro PatrãoCommented Aug 17, 2016 at 12:50
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