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I want to show that the set of orthogonal matrices, $O(n) = \{A \in M_{n \times n} | A^tA=Id\}$, is an embedded submanifold of the set of all $n \times n$ matrices $M_{n \times n}$.

So far, I have used that this set can be described as $O(n) = f^{-1}(Id)$, where $f: M_{n \times n} \rightarrow Sym_n = \{A \in M_{n \times n} | A^t = A\}$ is given by $f(A) = AA^t$, and that the map $f$ is smooth. Hence I still need to show that $Id$ is a regular point of this map, i.e. that the differential map $f_*$ (or $df$ if you wish) has maximal rank in all points of $O(n)$.

How do I find this map? I tried taking a path $\gamma = A + tX$ in $O(n)$ and finding the speed of $f \circ \gamma$ at $t=0$, which appears to be $XA^t + AX^t$, but don't see how to proceed. Another way I thought of was by expressing everything as vectors in $\mathbb{R}^{n^2}$ and $\mathbb{R}^{\frac{n(n+1)}{2}}$, but the expressions got too complicated and I lost track.

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    FYI, I'm pretty sure this is a worked-out example in Guillemin-Pollack.2012-10-28
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    Thanks. I am using Warner's book for my course, which is in my opinion rather condensed and leaves just a little too much as 'an exercise to the reader', but it is good to hear that there are books out there which have worked out examples.2012-10-28

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I think you have almost done. As you said, it suffices to show that $\mathrm{Id}$ is a regular value of $f$, i.e. for each $A\in O(n)$, $f_*:T_A M_{n\times n}\to T_{\mathrm{Id}}Sym_n$ is surjective, where $T_pX$ denotes the tangent space of $X$ at $p$. Note that $T_A M_{n\times n}$(resp. $T_{\mathrm{Id}}Sym_n$) can be identified with $M_{n\times n}$(resp. $Sym_n$) and, as you have known, $f_*(X)=XA^t+AX^t$. Then you only need to verify that for any $S\in Sym_n$, there exists $X\in M_{n\times n}$, such that $XA^t+AX^t=S$. At least you may choose $X=\dfrac{1}{2}SA$.

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    Thank you for your answer. Next, I wondered if I can use this construction to find the tangent space of $O(n)$ at $Id$. In $\mathbb{R}^{n}$, once you have described a manifold $V$ as a level set of some function $g$, the tangent space at a point $x$ is given by the nullspace or kernel of the function, i.e. $T_x V = ker(Dg(x))$. It seems intuitive to describe the tangent space in this case as the kernel of $f_*(X)$, but I can't seem to find any reference that this is the case.2012-10-28
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    Ok, so my guess is that $\gamma(t) = Id + tX$, with $X \in O(n)$ and $t$ sufficiently small describes all curves through $Id$ in $O(n)$, so the collection of all velocities of these curves is simply $O(n)$? Since the tangent space appears to be defined for vectors in the manifold itself (here $O(n)$), are we therefore excluding curves in $M_{n \times n}$ that are not entirely in $O(n)$?2012-10-28
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    However, if my first reasoning is correct, then we have to consider $f_*: T_Id M_{n \times n} \rightarrow T_Id Sym_n$, which is given by $f_*(X) = X + X^t$. Setting this equal to $0$ gives $X = -X^{-t}$, so the tangent space consists of all anti-symmetric matrices. This seems to agree with what I found here: http://mathworld.wolfram.com/OrthogonalGroup.html2012-10-28
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    @Peter, your last comment is correct, but the second one is incorrect. To consider a tangent vector $X\in T_I O(n)$ as $\gamma'(0)$ for some curve $\gamma$, you should suppose $\gamma(t)=\mathrm{Id}+tX+o(t)\in O(n)$ for small $t$, and $X$ cannot be assumed in $O(n)$ a priori. Then from the identity $\gamma(t)\gamma(t)^t=\mathrm{Id}$ you can obtain $X+X^t=0$.2012-10-29
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    One more question: is there any good way to see that the tangent spaces at these matrix spaces are just the matrix spaces themselves (e.g. $T_{A} M_{n \times n} = M_{n \times n}$)? It seems that the spaces are 'exactly the same' as $R^{n}$ and since $R^{n}$ has this property it seems reasonable. Is there any other way to see use this identification or does it really follow by going over the definition of tangent space?2012-10-29
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    I have no better idea than identifying $M_{n\times n}$ with $\mathbb{R}^{n^2}$.2012-10-29