# Is automorphism on a compact group necessarily homeomorphism? How about N-dimensional torus？ [closed]

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

• If I remember it correctly, this is true for semisimple Lie groups. Commented Jun 29 at 0:47
• 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. Commented Jun 29 at 3:35
• @RyanBudney this would mean the question is "is every automorphism an automorphism". This is probably not the intended question.
– YCor
Commented Jun 29 at 7:07
• According to the axiom of choice, there are discontinuous group-automorphisms of $\mathbb T$. Commented Jun 29 at 12:19
• @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.
– YCor
Commented Jul 1 at 5:26

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
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.