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I got stuck on the following problem:

I am looking at the general linear group $GL_n(\Bbb R)$ which is a subset of $\Bbb R^{n^2}$. I want now to prove that $GL_n(\Bbb R)$ is a locally connected topological group, without the fact that it is locally compact.

Any help would be much appreciated! Thanks in advance.

1 Answers1

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This can be done in two steps:

  1. $GL_n(\mathbb R)$, regarded as a subset of $\mathbb R^{n^2}$, is an open subset, as one proves using continuity of the determinant function $\text{det} : \mathbb R^{n^2} \to \mathbb R$
  2. Every open subset of $\mathbb R^{n^2}$ is locally connected, as one proves using the standard "open ball" basis for the topology on $\mathbb R^{n^2}$.
Lee Mosher
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  • Thanks for the answer. I don't see how the continuity helps me out in point 1.... –  Mar 23 '20 at 17:35
  • Continuity of a function is equivalent to saying that the inverse image of an open set is open. $GL_n(\mathbb R)$ is by definition the subset of $\mathbb R^{n^2}$ defined by $\text{det}(M) \ne 0$, and therefore $GL_n(\mathbb R)$ is the inverse image, under the continuous function $\text{det}$, of the open subset $\mathbb R - {0} \subset \mathbb R$. – Lee Mosher Mar 23 '20 at 18:02
  • Thank you a lot! The seconde point is clear. Now I have a seconde question, I know that $GL_n(\Bbb R)$ has two connected components: $GL_n^+(\Bbb R)$ and $GL_n^-(\Bbb R)$ but here again I got stuck the show this. –  Mar 23 '20 at 18:33
  • That's perhaps a good followup question. But, rather than ask it here in a comment to an answer to this post, where few other than you or I will know about it, you should ask it in a new post. – Lee Mosher Mar 23 '20 at 18:41
  • Of course, thanks for everything! –  Mar 23 '20 at 18:42
  • Although, a little searching led me to this question in which your latest question is asked and answered completely. – Lee Mosher Mar 23 '20 at 18:43