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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 $n(n-1)/2$.2012-03-25
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    You know, technically closed sets can be open ;) You can appeal to connectedness here, though.2012-03-25
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    yes but $\mathbb R^{n^2}$ is connected so the only clopen subsets are $\mathbb R^{n^2}$ and $\emptyset$2012-03-25
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    palio, shouldn't you correct the $n^2$ as pointed out to you in the comments?2014-06-28

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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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    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 $n(n+1)/2$.2012-03-25
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    the dimension is _not_ $n^2-1$. With your reasoning this would imply $S_n$ has dimension $1$2012-03-25
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    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 $\frac{n(n-1)}{2}$.2012-03-25
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    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?2012-03-25
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    On the other hand the singleton is a connected component of $\mathbb Q$ but it is not opoen in $\mathbb Q$2012-03-25
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    $S_n(\mathbb{R})$ 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$2012-03-25
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    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 shows2012-03-25
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    Yes but in this case, since $SO_n(\mathbb{R}) = \det^{-1}(\mathbb{R}_+^*)$ it is open2012-03-25