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Is there a compact matrix group $G\subseteq GL(n,\mathbb{R})$ such that $|G|$ is countable infinite?

the_fox
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yao4015
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  • What topology are we using? Would $$\left<\begin{pmatrix}2&0\0&2\end{pmatrix}\right>$$ satisfy your conditions? It would form a closed set in the product topology of $\mathbb{R^{n^2}}.$ – Melody Dec 03 '18 at 09:15
  • @Melody But it is not bounded. – Paul Frost Dec 03 '18 at 09:17
  • @Paul Frost True. Nevermind. – Melody Dec 03 '18 at 09:17
  • Where did you find this problem? What are your thoughts on the problem? What did you try and where did you get stuck? – Servaes Dec 03 '18 at 09:31
  • If $G$ exists, then it must be contained in the orthogonal group $O(n)$. To see this, consider the continuous function $\det : M(n,\mathbb{R}) \to \mathbb{R}$. Since $G$ is compact, $\det(G)$ is a compact subset of $\mathbb{R}$ which does not contain $0$. Assume there exists $A \in G$ such that $A \notin O(n)$. Then $d = \lvert \det(A) \rvert \ne 1$. Note that $A^n \in G$ and $\lvert \det(A^n) \rvert = d^n$. This shows that if $d > 1$, then $\det(G)$ is not bounded, and if $d < 1$, then $\det(G)$ is not closed. – Paul Frost Dec 03 '18 at 09:36
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    There is no countable infinite compact Hausdorff group. See this: https://mathoverflow.net/questions/4939/is-there-a-compact-group-of-countably-infinite-cardinality – freakish Dec 03 '18 at 09:38
  • @Servaes $d^n \to 0$ and $0 \notin \det(G)$. – Paul Frost Dec 03 '18 at 09:40
  • @PaulFrost There is a problem with your argument : there exists matrices $A$ such that $\det(A)=1$ and $A\not\in O(n)$. In fact $\ker\det = \operatorname{SL}_n(\mathbb{R})$. – Roland Dec 03 '18 at 09:43
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    @Servaes Of course. But we know that $\det(G)$ is a compact subset of $\mathbb{R}$, so it must contain $\lim d^n = 0$. – Paul Frost Dec 03 '18 at 09:43
  • @Roland You are right. So my argument only shows that $G$ must be contained in the set of matrices with $\det(A) = \pm1$. – Paul Frost Dec 03 '18 at 09:49
  • @freakish You should give an official answer, even it is essentially a reference. – Paul Frost Dec 03 '18 at 09:53

1 Answers1

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So the answer is negative. In general there is no countably infinite and compact Hausdorff group. See this:

https://mathoverflow.net/questions/4939/is-there-a-compact-group-of-countably-infinite-cardinality

To add a bit to the answer (so that it isn't completely naked ;) ): every compact subgroup of $GL_n(\mathbb{R})$ is obviously contained in the preimage $\det^{-1}(\{-1,1\})$ which is the largest compact subgroup of $\mathbb{R}^*$. But we can do better. The orthogonal group $O(n)$ is a maximal subgroup of $GL_n(\mathbb{R})$ among compact subgroups (see this). Therefore every compact subgroup of $GL_n(\mathbb{R})$ is a subconjugate of $O(n)$ which is a consequence of the Cartan-Iwasawa-Malcev theorem.

In particular if you wish to analyze compact subgroups of $GL_n(\mathbb{R})$ then it is enough to look at subgroups of $O(n)$.

freakish
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