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$n$ by $n$ real orthogonal matrices have $n (n-1)/2$ degrees of freedom. So do the skew-symmetric matrices.

But what about matrices that are both skew-symmetric and orthogonal?

Is the number of such matrices finite for any given $n$? If not, how many degrees of freedom do they have?

We know that such matrices exist only if $n$ is even, in which case they are equal to

$$\bigoplus_{i=1}^{n/2}\begin{bmatrix} 0 & 1\\ -1 & 0\end{bmatrix}$$

up to an orthogonal change of basis. However, the number of their degrees of freedom is still unclear to me.

YuiTo Cheng
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MWB
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1 Answers1

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The space of all such is $G/G_x$ where $G$ is the whole orthogonal group and $G_x$ is the stabilizer in $G$ of your particular example under the action $g\times x=gxg^{-1}$. The dimension of the latter is $>0$, since at the very least orthogonal matrices with $2\times 2$ blocks scalar are in that centralizer. More generally, upon writing out the centralizer condition, it is a copy of $U(n)$ inside $O(2n,\mathbb R)$.

paul garrett
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    There is an analysis here http://scipp.ucsc.edu/~haber/webpage/antiortho.ps. Didn't have time to read it. It's the same approach, though. – fred goodman Feb 20 '18 at 23:40
  • I didn't really understand this answer, I guess because I'm not that familiar with group theory. Does this mean that we have a formula for the degrees of freedom as a function of $n$, and in that case, what is that formula? – HelloGoodbye Oct 13 '21 at 13:42
  • @fredgoodman How do I open the file? – HelloGoodbye Oct 13 '21 at 13:43
  • @HelloGoodby, download the linked file and open it with something that recognizes postscript files, e.g. a ps to pdf converter, or your tex/latex system, or Preview on Apple machines. – fred goodman Oct 14 '21 at 14:23
  • This implies that the number of degrees of freedom is $(\frac{n}{2})^2$ (with $n$ as in the question), I think? – Tuomas Laakkonen Jun 28 '24 at 00:58