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?
 A: 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 $\dim O_+(n)=\dim O_-(n)$.
A: Consider $g=\operatorname{diag}(-1,1,1,1\ldots,1)\in O_-(n)$.  The map $h\mapsto gh$ induces a bijection between $O_+$ and $O_-$.
A: I think it might help you to understand the connection between degrees of freedom and eigenvalues when thinking more geometrically. 
Take $n = 2$. Here every elements of $O_+$ is a rotation around the origin and every element of $O_-$ is a relection in a plane through the origin. Suppose you would actually use this as the definition of $O_+$ and $O_-$ and try to compute how big the sets of rotations and reflections are using your eigenvalue/eigenvector idea...
Reflections are the easy ones. Just by thinking about what it means to be a reflection we see that it has eigenvalue 1 with eigenvectors all lying on the 'mirror' and and eigenvalue -1 lying on the line perpendicular to the mirror. So we have zero degrees of freedom in choosing the eigenvalues but that doesn't mean the set of reflections in mirrors passing through the origin is zero-dimensional because we can still choose the location of the 1-eigenvector. The location of the $(-1)$-eigenvector is then fixed.
In other words we see from a purely geometric perspective that the space $O_-(2)$ itself has the topology of a the projective line (the set of all lines through the origin in the plane) which is 1-dimensional, homeomorphic to a circle. (Perhaps it is more correct to say that it is a circle with its antipodal points identified, which by some quirk of nature is itself homeomorphic to a circle.)
By contrast in the set of rotations the situation concering eigenvalues and eigenvectors is more or less 'reversed'. Rotations seemingly have no eigenvectors which naively could be seen as a reason to believe that $O_+(2)$ is zero-dimensional. But as you noted in this case we have something we did not have in the reflection case: the freedom to choose an eigenvalue. So there still is a degree of freedom. 
Now from a geometric perspective it is hard to understand what an eigenvalue 'is' when there are no eigenvectors, but in the geometric picture there is a different and much easier way to describe the degree of freedom: it is just the angle over which you rotate. It is clear that the set of angles together has the topology of a circle. So both $O_+(2)$ and $O_-(2)$ are circles and hence diffeomorphic. 
One way to talk this back to eigenvalues, though, is to view the entire plane as a one-dimensional complex eigenspace. The rotation is then just the multiplication with the single eigenvalue associated to that space (the second eigenvalue acting on a different, 'invisible' plane we can safely ignore as our interest is in degrees of freedom and we already knew that the second eigenvalue is completely determined by the first eigenvalue that we 'do' see)
