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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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    $\begingroup$ The dimension of $O(n)$ and $SO(n)$ is $n(n-1)/2$. $\endgroup$
    – user20266
    Mar 25, 2012 at 12:26
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    $\begingroup$ You know, technically closed sets can be open ;) You can appeal to connectedness here, though. $\endgroup$ Mar 25, 2012 at 12:42
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    $\begingroup$ yes but $\mathbb R^{n^2}$ is connected so the only clopen subsets are $\mathbb R^{n^2}$ and $\emptyset$ $\endgroup$
    – palio
    Mar 25, 2012 at 12:48
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    $\begingroup$ palio, shouldn't you correct the $n^2$ as pointed out to you in the comments? $\endgroup$ Jun 28, 2014 at 19:04

1 Answer 1

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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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    $\begingroup$ 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 $n(n+1)/2$. $\endgroup$
    – martini
    Mar 25, 2012 at 12:38
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    $\begingroup$ the dimension is not $n^2-1$. With your reasoning this would imply $S_n$ has dimension $1$ $\endgroup$
    – user20266
    Mar 25, 2012 at 12:38
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    $\begingroup$ $S_n(\mathbb{R})$ is the vector space of symetric matrices. For $f :X \rightarrow Y$ a differentiable function between $X$ and $Y$ two finite dimensionnal normed $\mathbb{R}$-vector spaces, we say $y \in Y$ is a regular value of $f$ if for any $x$ such that $f(x) = y$, $Df(x)$ is onto(or,equivalently, that $Df(x)$ has maximal rank). My remark is based on the following theorem : Let $f : U \subset \mathbb{R}^m \rightarrow \mathbb{R}^n$ be a function $\mathcal{C}^k$ on a domain U, and let $y$ be a regular value of $f$. Then $f^{-1}(\{y\})$ is a submanifold of $\mathbb{R}^m$ of dimension $m-n$ $\endgroup$ Mar 25, 2012 at 14:49
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    $\begingroup$ ok i see and why $SO(n)$ is a submanifold of $O(n)$, is it true that being a connected component makes it open in $O(n)$? it seems not true as the case of $\mathbb Q$ above shows $\endgroup$
    – palio
    Mar 25, 2012 at 15:06
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    $\begingroup$ Yes but in this case, since $SO_n(\mathbb{R}) = \det^{-1}(\mathbb{R}_+^*)$ it is open $\endgroup$ Mar 25, 2012 at 15:18

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