Lie algebra of $SO(V)$

Question

As picture below ,why the Lie algebra of $SO(V)$ is obtained by differentiating the relation $\langle Av,A\omega \rangle =\langle v,\omega\rangle$ ?

And how to calculate the differentiating of $\langle Av,A\omega \rangle =\langle v,\omega\rangle$ ?

Answer 1

Here's a proof in synthetic differential geometry using infinitesimals. You can easily convert the proof into a standard one using limits, and derivatives, but this method is perfectly valid (via the embedding of smooth manifolds into the Dubuc topos, for instance), and is a bit more intuitive, in my opinion.

Anyway, let $\epsilon$ be a nilsquare infinitesimal (i.e. $\epsilon^2$=0). A matrix $X \in \mathfrak{gl}(V)$ belongs to $\mathfrak{o}(V)$ if and only if $I + \epsilon X \in O(V)$. We know this matrix is already in $GL(V)$, so we just need to ensure that it preserves the inner product. Well this is straightforward: \begin{align} \langle (I + \epsilon X)v, (I + \epsilon X)w \rangle & = \langle v , w \rangle + \epsilon \langle Xv, w \rangle + \epsilon \langle v, Xw \rangle + \epsilon^2 \langle Xv, Xw \rangle \\ & = \langle v , w \rangle + \epsilon \langle Xv, w \rangle + \epsilon \langle v, Xw \rangle \end{align} We have just expanded and used $\epsilon^2=0$ to eliminate the last term. Thus, it is easy to see that $I + \epsilon X \in O(V)$, and hence $X \in \mathfrak{o}(V)$, if and only if $$\langle Xv, w \rangle + \langle v, Xw \rangle = 0.$$