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$O(n)$ is the manifold of the orthogonal $n \times n$ matrices.

How do I prove that its dimension is $\displaystyle \frac{n(n-1)}{2}$?

Edit: Thanks to all your answers. I appreciate your help.

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Do you know how to prove that $O(n)$ is a manifold? One easy way is to use the implicit function theorem and to consider the map $A \mapsto A^{T}A$ from the $n \times n$-matrices to the symmetric $n \times n$ matrices and observing that $O(n)$ is the pre-image of the regular value given by the identity matrix, then counting dimensions $n(n-1)/2 = n^2 - n(n+1)/2$. –  t.b. May 23 '11 at 18:39
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A different argument than Theo's would be to observe there's the fibre bundle $$O(n-1) \to O(n) \to S^{n-1}$$. Iterating that bundle and counting the dimensions of the spheres gives you $dim(O(n)) = n-1 + n-2 + \cdots + 3 + 2 + 1 = n(n-1)/2$. –  Ryan Budney May 23 '11 at 18:57
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A more elementary way of phrasing Ryan's comment might be that the columns of an $O(n)$ matrix form an ON basis in $\mathbf{R}^n$, and you have $n-1$ degrees of freedom when choosing the first vector in such a basis (since the only restriction is that it has to be of length one), then $n-2$ degrees of freedom when choosing the second basis vector (since now there is also the restriction that it has to be orthogonal to the first vector), $n-3$ degrees of freedom for the third vector (since it must have unit length and be orthogonal to the first two), etc. –  Hans Lundmark May 23 '11 at 19:32

2 Answers 2

I think the fastest way might be to consider it's Lie Algebra $\mathcal{o}(n)$. Which is the set of skew symmetric matrices. Their dimensions will be the same since $\mathcal{o}(n)$ is the tangent space at any point on the manifold $O(n)$. You get the result you want by counting the dimensions and taking into account the skew symmetry.

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No real objection, but in order to know that the Lie algebra of $O(n)$ consists of the skew-symmetric matrices, you still have to determine the derivative of $A \mapsto A^{T}A$ at the identity, no? So I'm saying it's essentially the same argument as the one I outlined in my comment to the question. –  t.b. May 23 '11 at 18:45
    
You're quite right. I think we posted at about the same time. I didn't read your comment until after my answer. –  Schmitty May 23 '11 at 18:52
    
Sorry, I didn't mean to imply that you were copying my comment :) I just wanted to point out that these two arguments boil down to the same computation, in the end. (you got an upvote from me, by the way). –  t.b. May 23 '11 at 18:57
    
Haha! I was thinking the same thing! Up votes all around! –  Schmitty May 23 '11 at 19:29

One way is to use singular value decomposition. Let $D(n)$ be the space of $n\times n$ diagonal matrices, and let $F\colon O(n) \times D(n) \times O(n) \to \mathbb{R}^{n\times n}$ be the function $F(O_1,D,O_2) = O_1DO_2$. Then $F$ is a local homeomorphism at each point $(O_1,D,O_2)$ for which the diagonal entries of $D$ are all different, so it follows that $$ \dim\;O(n) \,+\, \dim\;D(n) \,+\, \dim\;O(n) \;=\; \dim\;\mathbb{R}^{n\times n} $$ Since $\dim\;D(n) = n$ and $\dim\;\mathbb{R}^{n\times n} = n^2$, you can solve to get $\dim\;O(n) = (n^2 - n)/2$.

If you don't like singular value decomposition, it works just as well to use diagonalization of symmetric matrices. Let $S(n)$ be the space of all $n\times n$ symmetric matrices, and define $G\colon O(n) \times D(n) \to S(n)$ by $G(O,D) = O^T D O$. Again, $G$ is a local homeomorphism whenever the entries of $D$ are all different, so $$ \dim\;O(n) \,+\, \dim\;D(n) \;=\; \dim\;S(n) $$ Since $\dim\;S(n) = (n^2 + n)/2$ and $\dim\;D(n) = n$, we conclude that $\dim\;O(n) = (n^2-n)/2$.

Edit: Of course, the solution using symmetric matrices is very similar to the one that Theo gives above in the comments. The difference is that making the local product structure explicit avoids the use of the Implicit Function Theorem.

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