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How can we show that $SO(n)$ is an $n^2$-manifold. It would be tempting to say that $SO(n)$ is an open set of $\mathbb R^{n^2}$ but this is not the case since $SO(n)$ is given as the intersection of preimages of singletons. But singletons are closed in $\mathbb R$ hence $SO(n)$ is closed in $\mathbb R^{n^2}$.

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The dimension of $O(n)$ and $SO(n)$ is $n(n-1)/2$. – user20266 Mar 25 '12 at 12:26
You know, technically closed sets can be open ;) You can appeal to connectedness here, though. – Alexei Averchenko Mar 25 '12 at 12:42
yes but $\mathbb R^{n^2}$ is connected so the only clopen subsets are $\mathbb R^{n^2}$ and $\emptyset$ – palio Mar 25 '12 at 12:48
palio, shouldn't you correct the $n^2$ as pointed out to you in the comments? – Rudy the Reindeer Jun 28 '14 at 19:04
up vote 7 down vote accepted

Let $f : M_n(\mathbb{R}) \longrightarrow S_n(\mathbb{R})$ defined by $f(A) = ~^tAA$

$O_n(\mathbb{R}) = f^{-1}(\{I_n\})$. Then check that $I_n$ is a regular value of $f$.

So $O_n(\mathbb{R})$ is a submanifold of $M_n(\mathbb{R})$, it's dimension is $\dim(M_n(\mathbb{R}) - \dim(S_n(\mathbb{R})) = n^2 - \frac{n(n+1)}{2} = \frac{n(n-1)}{2}$ . Since $SO_n(\mathbb{R})$ is a connected component of $O_n(\mathbb{R})$ it is a submanifold to.

Ask me if you want more details

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As @Thomas writes above, the dimension of $O(n)$ is $n^2 - n(n+1)/2 = n(n-1)/2$ as the dimension of $S_n(\mathbb R)$ (symmetric matrices, right?) is $n(n+1)/2$. – martini Mar 25 '12 at 12:38
the dimension is not $n^2-1$. With your reasoning this would imply $S_n$ has dimension $1$ – user20266 Mar 25 '12 at 12:38
Sorry I wrote without thinking. To make sure we are right, one can compute the tangent space at any point of $P \in O_n(\mathbb{R})$ : it is $P A_n(\mathbb{R})$ which is od dimension $\frac{n(n-1)}{2}$. – Selim Ghazouani Mar 25 '12 at 12:58
I don't understand the argument: what is $S_n(\mathbb R)$ and what is a regular value for your map $f$? i this Sard theorem? – palio Mar 25 '12 at 14:18
On the other hand the singleton is a connected component of $\mathbb Q$ but it is not opoen in $\mathbb Q$ – palio Mar 25 '12 at 14:43

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