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Does a Lie group's group structure (not Lie group structure) determine its topology? Said another way, can you have two Lie groups that are isomorphic as groups but not homeomorphic?

If so, the group isomorphism map will not be continuous (and thus not a Lie group isomorphism), and there will be no natural map between their tangent spaces (Lie algebras).

I suspect you can't (the group does determine the topology), but I don't know how to prove it.

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    $\begingroup$ Should be "determine its homeomorphism type". Determine its topology could mean determine its structure of topological space, which is a stronger, but still reasonable, requirement (for instance, the topology of $\mathbf{R}$ (not only its homeomorphism type) is determined by its field structure, and the topology of a compact semisimple Lie group is determined by its group structure. $\endgroup$
    – YCor
    Oct 27 '20 at 13:12
  • $\begingroup$ @YCor That's true, I hadn't thought of that subtlety. I think you're right that those are both reasonable things to expect to be true. $\endgroup$ Oct 27 '20 at 13:23
  • $\begingroup$ For $n\ge 2$, I think that the group structure determines the topology in $\mathrm{SL}_n(\mathbf{R})$, while in $\mathrm{SL}_n(\mathbf{C})$ it determines the homeomorphism type, but not the topology. $\endgroup$
    – YCor
    Oct 27 '20 at 13:24
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$\mathbb{R}^n$ and $\mathbb{R}^m$ are abstractly isomorphic (assuming the axiom of choice) for $n \neq m$ but not homeomorphic and so not isomorphic as topological groups.

I think this might be the only thing that can go wrong, though; e.g. it seems plausible that for, say, compact semisimple Lie groups an abstract isomorphism must be continuous (hence smooth, hence analytic) but I don't know how to prove it. Some googling turned up these notes which claim that

  • a Haar-measurable homomorphism of Lie groups is automatically continuous, and it's consistent with ZF that every subset of a Lie group is Haar-measurable, so the existence of discontinuous homomorphisms of Lie groups is independent of ZF, and
  • some Lie groups admit no discontinuous automorphisms (but not much is said about which).

Edit: Some more googling turned up Braun, Hofmann, and Kramer's Automatic continuity of abstract homomorphisms between locally compact and Polish groups, which proves very general results about this. Assuming I've parsed it correctly, I think Theorem A implies that a Lie group $G$ with at most countably many connected components has a unique Lie group topology provided that

  • The center $Z(G_0)$ of the identity component of $G$ is finite, and
  • The Lie algebra $\mathfrak{g}$ is a direct sum of Lie algebras $\mathfrak{g}_i$ which are absolutely simple in the sense that $\mathfrak{g}_i \otimes_{\mathbb{R}} \mathbb{C}$ is simple.

(This is equivalent to the claim that any abstract isomorphism from $G$ to another Lie group is automatically continuous.)

It is also apparently an old result of Cartan and van der Waerden that every abstract isomorphism between compact simple Lie groups is automatically continuous.

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    $\begingroup$ For topological groups, we also have examples like $\mathbb{Q}$ or $S^1$ with usual topology and discrete topology. $\endgroup$ Oct 27 '20 at 2:53
  • $\begingroup$ Thanks, Qiaochu! $\endgroup$ Oct 27 '20 at 12:17

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