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Using the axiom of choice, $\mathbb{R}$ and $\mathbb{R}^2$ are equal-dimensional vector spaces over $\mathbb{Q}$ and so are isomorphic as $\mathbb{Q}$-vector spaces thus as groups.

This is obvious, however I recently began reading Godement's Introduction à la théorie des groupes de Lie and in particular I was reading the chapter on topological groups when I came across this statement:

Given a group $G$, there is at most one topology that makes it into a topological group that is at the same time locally compact and countable to infinity.

(I don't know how to translate "dénombrable à l'infini" better, it means $G$ is a union of countably many compact sets)

Here's my reasoning: let $\phi: \mathbb{R}\to \mathbb{R}^2$ be a group isomorphism. In particular, it is a bijection, and so one can define a topology $\mathcal{T}$ such that $\phi$ is a homeomorphism from $(\mathbb{R}, \mathcal{T})$ to $\mathbb{R}^2$ with the usual topology.

Obviously, $(\mathbb{R}, \mathcal{T})$ is locally compact and countable to infinity.

Moreover, $+: \mathbb{R}\times \mathbb{R}\to \mathbb{R}$, and letting $add$ denote the addition in $\mathbb{R}^2$, because $\phi$ is an isomorphism, we get $+= \phi^{-1}\circ add \circ (\phi\times\phi)$. Therefore, since all these maps are continuous (wrt the usual topology on $\mathbb{R}^2$ and $\mathcal{T}$ on $\mathbb{R}$), so is $+$, and similarly one gets that $x\mapsto -x$ is continuous on $\mathbb{R}$ wrt $\mathcal{T}$.

But then $(\mathbb{R}, \mathcal{T})$ is a locally compact topological group that's countable to infinity: according to Godement's claim $\mathcal{T}$ is the usual topology !

This leads to the absurdity that $\mathbb{R}$ and $\mathbb{R}^2$ are homeomorphic, which is trivially false.

I'm really stuck on this and I don't know where I went wrong. Could anybody please solve my problem?

EDIT : as suggested in the comments, here's a link to a dropbox file with the proof in Godement's book : https://www.dropbox.com/sh/2gxg1jpbdmmcg23/AACP__txcn3o26cw-JR5W5Oea?dl=0 Sorry for the quality of the pictures. And it's in french !

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    $\begingroup$ You have refuted that sentence of Godement. $\endgroup$ Aug 4, 2017 at 20:55
  • $\begingroup$ @LordSharktheUnknown but he proves it and I think uses it later on, so my first guess would be that I made a mistake somewhere (it seems to be quite a big theorem) $\endgroup$ Aug 4, 2017 at 20:56
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    $\begingroup$ "Union of countably many compact sets" = $\sigma$-compact. $\endgroup$ Aug 4, 2017 at 21:10
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    $\begingroup$ A google search for "dénombrable à l'infini godement" finds a googlebooks extract that includes the relevant parts of the book. I agree with you that this result looks unbelievable. In your counterexample, Godement's proof needs us to accept that the graph of a group isomorphism between $\Bbb{R}$ and $\Bbb{R}^2$ is a closed subspace of $\Bbb{R} \times \Bbb{R}^2$. I don't think that can be right. $\endgroup$
    – Rob Arthan
    Aug 4, 2017 at 21:43
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    $\begingroup$ Two topological groups can be both isomorphic as groups and homeomorphic as spaces, but not isomorphic as topological groups. $\endgroup$ Aug 4, 2017 at 22:13

1 Answer 1

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You are correct, and Godement's proof is incorrect. His proof proceeds by letting $G$ and $G'$ be two different $\sigma$-compact locally compact groups with the same underlying group, and considers the diagonal $D\subseteq G\times G'$. He then applies a theorem about $\sigma$-compact locally compact groups to the projection maps $D\to G$ and $D\to G'$. The problem is that the theorem does not apply, since $D$ may not be $\sigma$-compact or locally compact, since it may not be closed in $G\times G'$.

This is exactly what happens in the case of $\mathbb{R}$ and $\mathbb{R}^2$: the "diagonal" in $\mathbb{R}\times\mathbb{R}^2$ is the graph of a group-isomorphism $\mathbb{R}\to\mathbb{R}^2$. A group-isomorphism $\mathbb{R}\to\mathbb{R}^2$ is a horrendously discontinuous map, so its graph is a horrendous non-closed subgroup of $\mathbb{R}\times\mathbb{R}^2$. (Quick proof that the graph of an isomorphism $f:\mathbb{R}\to\mathbb{R}^2$ cannot be closed: since $f$ cannot be $\mathbb{R}$-linear, there exists $x\in \mathbb{R}$ such that $f(x)\neq xf(1)$. But $f(q)=qf(1)$ for all $q\in\mathbb{Q}$, so by approximating $x$ by rationals, we find that $(x,xf(1))$ is in the closure of the graph of $f$.)

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    $\begingroup$ Godement explicitly states that the diagonal $D\subset G\times G'$ is closed but doesn't say why. This is a good lesson in paying attention to innocent-looking statements. $\endgroup$ Aug 4, 2017 at 22:28
  • $\begingroup$ Your answer actually begs a question: is there a simpler proof than the one implicit in Godement's (apparently correct) Corollaire 1 that the graph of a group isomorphism between $\Bbb{R}$ and $\Bbb{R}^2$ cannot be closed in $\Bbb{R}^3$? (That concern is why I provided my observation about this problem with Godement's proof as a comment rather than an answer.) $\endgroup$
    – Rob Arthan
    Aug 4, 2017 at 22:37
  • $\begingroup$ @RobArthan: Suppose $f:\mathbb{R}\to\mathbb{R}^2$ is an isomorphism, and take any $x\in\mathbb{R}$ such that $f(x)\neq xf(1)$. By approximating $x$ by rationals, $(x,xf(1))$ is in the closure of the graph of $f$. By a more complicated elaboration of this idea, you can show the graph of $f$ is actually dense in $\mathbb{R}^3$. $\endgroup$ Aug 4, 2017 at 23:01
  • $\begingroup$ @EricWofsey: thanks. That argument would be a helpful addition to your answer. $\endgroup$
    – Rob Arthan
    Aug 4, 2017 at 23:07
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    $\begingroup$ @Max: Since Godement, sadly, passed away last year, yes, you may want to contact the editorial board of Springer Verlag (Universitext series); just this year they published an English translation of his book. However, first check this edition if the mistake was corrected. Also, consider contacting the translator (at Reims University) instead. $\endgroup$ Aug 9, 2017 at 4:01

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