Tangent Vectors of Orthogonal Group Are Skew Symmetric matrices

Question

The title is a result I've been given, but I don't really understand how it works. I thought tangent vectors were linear functionals on the space of smooth maps from $M$ to $\mathbb{R}$, I don't see how a skew symmetric matrix acts on such a functional.

Answer 1

If $c:(-a,a)\rightarrow O(n)$ is a smooth curve (if you have trouble with the abstract manifold $(O(n)$ you can think of $O(n) $ embedded in the space of $n\times n$-matrices, which is a vector space), then $c^Tc=Id $ is the identity matrix, so $$\frac{d}{dt} c^T c= c^Tc^\prime + (c^\prime)^T c=0$$ If $c(0)=Id$ you see that for $A$ in the tangent space of the orthogonal group at the identity $A=-A^T$ holds, so these are skew symmetric matrices.

A tangent vector acts on functions by taking the derivative in the direction of that vector. If you, say, want to take the derivative of $f(A)$ in some point $A_0$ in direction $V$ you can think of this as follows: take a curve $c$ passing through $A_0$ at $t=0$ with tangent vector $V$ in that point (such a curve always exists) and calculate $$\frac{d}{dt}f( c(t))|_{t=0} $$

(it can be shown that the result of this calculation does not depend on the choice of such $c$).