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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.

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    $GL_n(\Bbb R)$ is an open subset of $\Bbb R^{n\times n}$, so the tanget spaces can be identified with $n\times n$ matrices. If you have a smooth subgroup $G\subset GL_n$ then its tangent space at $\Bbb1$ can be identified with a subspace of $T_{\Bbb1}GL_n$, so also with a system of matrices. – 2017-01-27

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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 $c$).

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    Thanks, but I'm still struggling a bit. Concretely, if $A$ was a skew symmetric matrix representing a tangent vector and f was a smooth map to $\mathbb{R}$, how would you evaluate $Af$? – 2017-01-27
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    @icaidm One way to (explicitly) do that is to calculate the $t$- derivate I wrote down in the second equation. By the chain rule it's $df_{A_0}c^\prime(t)|_{t=0} = df_{A_0} V$, which corresponds to the action of $V$ on $f$. – 2017-01-27
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    @icaidm actually the statement as such is rather trivial. To see this is a linear map just apply the procedure to $f+\lambda g$ where $f,g$ are smooth functions on $M$ and $\lambda$ is a scalar, a real number in this case. This applies to vectors at a given point as well as to vector fields on $M$. You should not try to overthink this. The simple statement that this is a linear map has important structural consequences, though. – 2017-01-27
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    @icaidm If $A$ is a skew symmetric matrix and $f:O(n)\to\mathbb{R}$ is a smooth function, then, we view $A$ as a tangent vector of $O(n)$ at $I$ by the formula$$Af=\left.\frac{d}{dt}\right|_{t=0}f(\exp(tA)).$$ (Or you could replace $\exp(tA)$ by any smooth curve $c(t)$ on $O(n)$ which has $c'(0)=A$. It will always give the same number.) – 2017-01-28