Dimension of connected components of $O(n)$ The special orthogonal group $SO(n)=O_+(n)$ is the subgroup of the orthogonal group $O(n)$ containing matrices with determinant $+1.$ The other connected component of $O(n)$, call it $O_-(n)$, has matrices with determinant -1.
I always thought $O_+(n)$ and $O_-(n)$ had the same dimension and were in fact isomorphic. However the following consideration shaked this belief when $n$ is even.
The eigenvalues of $g\in O(2m)$ are either real or come in complex conjugate pairs. If all eigenvalues are complex, then the determinant is positive and $g\in O_+(2m)$. The only chance for $g$ to belong to $O_-(2m)$ is if some eigenvalues are real and an odd number of them are $-1$. This stringent condition seems to imply that $O_-(2m)$ has lower dimension than $O_+(2m)$.
Is that true? In even dimension almost all orthogonal matrices are in fact special?

Actually, if $M\in O_-(n)$, then$$\begin{array}{ccc}O_+(n)&\longrightarrow&O_-(n)\\N&\mapsto&MN\end{array}$$is a diffeomorphism. So, yes $\dim O_+(n)=\dim O_-(n)$.
