$n$ 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.
Answer
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 $O(2n,\mathbb R)$.
No comments:
Post a Comment