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Is automorphism on a compact group necessarily homeomorphism? I don't think so,but I think it is possible on the N-dimensional torus.

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    $\begingroup$ If I remember it correctly, this is true for semisimple Lie groups. $\endgroup$ Commented Jun 29 at 0:47
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    $\begingroup$ What do you mean by automorphism? You assume you have a compact group so I think most would presume you are talking about continuous group isomorphisms that have continuous inverses. $\endgroup$ Commented Jun 29 at 3:35
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    $\begingroup$ @RyanBudney this would mean the question is "is every automorphism an automorphism". This is probably not the intended question. $\endgroup$
    – YCor
    Commented Jun 29 at 7:07
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    $\begingroup$ According to the axiom of choice, there are discontinuous group-automorphisms of $\mathbb T$. $\endgroup$ Commented Jun 29 at 12:19
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    $\begingroup$ @RyanBudney It is frequent that new users do not interact after asking their first question, this one seems to also be one of them. Second, questions around the theme "is every abstract homo/iso-morphism between these given [topological] groups continuous" is a very recurrent question here and at MathSE, so this interpretation (which is the literal one) is quite likely. $\endgroup$
    – YCor
    Commented Jul 1 at 5:26

1 Answer 1

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Suppose that $G$ is a compact Hausdorff group. Let's call $G$ semisimple if it is connected and perfect (i.e. $G=[G,G]$). Then, according to Theorem B in

Braun, O.; Hofmann, Karl H.; Kramer, L., Automatic continuity of abstract homomorphisms between locally compact and Polish groups, Transform. Groups 25, No. 1, 1-32 (2020). ZBL1483.22001.

every abstract automorphism of a semisimple compact Hausdorff group is automatically a self-homeomorphism. (A homomorphism of topological groups is called abstract if it is a morphism in the category of groups.) For compact semisimple Lie groups this was proven (independently) much earlier by Cartan and van der Waerden.

Without the semisimplicity assumption, this result is false as Qiaochu Yuan correctly noted in his comment to the other (now deleted) answer. In particular, it is false for tori (assuming the Axiom of Choice). To be more specific, consider $G=S^1=U(1)$. Using the axiom of Choice, we split $G$ as the direct product $F\times H$, where $F$ is the group of roots of unity. (The subgroup $H$, is abstractly isomorphic to the direct sum of continuum of copies of $\mathbb Q$.) Both subgroups $F$ and $H$ are dense in $G$. Now, take the abstract automorphism $\phi: G\to G$ which is identity on $H$ and is the inversion $z\mapsto z^{-1}$ on $F$. It is clear that $\phi$ is discontinuous.

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  • $\begingroup$ Is the result in this full generality of non-compact, semisimple Lie groups really so recent? $\endgroup$
    – LSpice
    Commented Jun 29 at 17:03
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    $\begingroup$ It is not quite true for noncompact semisimple Lie groups, you have to take field automorphisms into account. It is true for absolutely simple groups (I forgot the author, about 80 years ago). @LSpice $\endgroup$ Commented Jun 29 at 17:17
  • $\begingroup$ For absolutely simple groups, I believe it's due to Borel and Tits - Homomorphisms “abstraits” de groupes algébriques simples (although, as you say, one still needs to allow field automorphisms). $\endgroup$
    – LSpice
    Commented Jun 29 at 17:50
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    $\begingroup$ @LSpice: Ok, for absolutely simple groups, this is due to Freudenthal, Annals of Math, 1941. Borel and Tits prove a much more general result. $\endgroup$ Commented Jun 29 at 18:54

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