Show that $\langle[U,X],V\rangle = -\langle U,[V,X]\rangle$ for bi-invariant metric in Lie group

Question

I know that $\langle U,V \rangle = \langle dR_{x_{t(e)}}U, dR_{x_{t(e)}}V \rangle$ and $\langle U,V \rangle = \langle dL_{x_{t(e)}}U, dL_{x_{t(e)}}V \rangle$ because it is bi-invariant. How do I proceed?

Answer 1

Let $g$ denote the metric of $G$. The right invariance of $g$ guarantees that $$R^*_{\exp (tX)}g=g, \forall t \in \mathbb{R}. $$ On the other hand, $t\to R_{exp(tX)}$ is the flow of X, we have $\mathcal{L}_Xg=0$, where $\mathcal{L}_Xg$ denotes the Lie derivative of $g$ in the direction X. $$(\mathcal{L}_Xg)(U,V)=X(g(U,V))-g(\mathcal{L}_XU,V)-g(U,\mathcal{L}_YZ) \\ =-\langle[X,U],V\rangle-\langle U,[X,V]\rangle$$

where we use the invariance of $g$ at last equal to conclude that $g(U,V)$, seen as function of $G$, is constant in each connected component of $G$.