Suppose you have a compact connected Lie group $G$, with an inner product on its Lie algebra $\mathfrak{g}$ defined by the Killing form. I've seen it written that the adjoint representation $Ad: G\rightarrow GL(\mathfrak{g})$ in fact takes values in $SO(\mathfrak{g})$.
Why is this? Many thanks.
This is because $Ad(g)$ acts by change of basis in any representation of $\mathfrak{g}$. Since the Killing form is the trace form on the adjoint representation and trace is independent of the choice of basis, we get that $Ad(g)$ is an element of $SO(\mathfrak{g})$.