Local compactness of a subspace of a locally compact metric space

$\mathbb{R}^{m^2}$ is a locally compact metric space with the Euclidean metric. Is it true that the set of all orthogonal $m \times m$ matrices (considered as points in $\mathbb{R}^{m^2}$) is locally compact in the subspace topology?

Thanks, Phanindra

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1 Answer

The orthogonal group is compact (this does not depend on whether we are looking at the subspace topology on $\textrm{O}_n$). Now firstly it is a closed subset of $M_n(\Bbb{R})$ - to prove this note that the map $f : M_n(\Bbb{R}) \to M_n(\Bbb{R})$ that sends a matrix $A$ to $AA^T - I$ is continuous with kernel precisely $\textrm{O}_n$. Furthermore the columns of an orthogonal matrix being orthonormal imposes the condition that there is a constant $C$ such that $|a_{ij}|\leq C$ for all $1 \leq i,j \leq n$. The result you are asking for now follows because each orthogonal matrix has an open neighbourhood (namely the whole of $\textrm{O}_n$) about it that is compact.

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Thanks for your answer. From your reply, $O_n$ is a compact subset of $M_n$. Considering our new space as $O_n$ with the subspace topology, I wanted to know if it is locally compact. Precisely, I wanted to show that if $A \in O_n$ then there exists a compact set $K$ (in the subspace topology) such that $A \in K \subset O_n$. I am unable to see how your argument proves this. – jpv Sep 17 '12 at 11:28
@jpv Take $K=O_n$. – user38268 Sep 17 '12 at 11:29
Thanks. I understand it now. – jpv Sep 18 '12 at 4:04