In chapter 1 of do Carmo's Riemannian Geometry the author makes the following claim:
Let $G$ be a Lie group with identity element $e$ and let $\langle \cdot, \cdot \rangle_e$ be a positive blinear form on $\mathfrak g$ such that
$$
\langle[U, X], V \rangle = – \langle U, [V, X] \rangle \quad \forall U, V, X \in \mathfrak g \quad \quad (1)
$$
Then the riemannian metric induced on $G$ by the relation
$$
\langle u, v \rangle_x = \langle d(L_{x^{-1}})_x(u), d(L_{x^{-1}})_x(v) \rangle_e, \quad x \in G, \ u, v \in TxG
$$
is bi-invariant.
He says that the proof is easy but will be omitted. I am not finding it easy. I have tried using the definitions and showed that the metric is left-invariant:
$$
\langle d(L_x)_y (u), d(L_x)_y(v) \rangle_{xy} = \langle d(L_{(xy)^{-1}})_{xy} d(L_x)_y(u), d(L_{(xy)^{-1}})_{xy} d(L_x)_y(v) \rangle_e = \langle d(L_{y^{-1}x^{-1}} \circ L_x)_y(u), d(L_{y^{-1}x^{-1}} \circ L_x)_y(v) \rangle
= \langle d(L_{y^{-1}})_y(u), d(L_{y^{-1}})_y(v) \rangle_e = \langle u, v \rangle_y.
$$
I am having trouble showing that it is also right-invariant. Also, I have no clue on where to use $(1)$.
Any hints will be the most appreciated.
Thanks in advance and kind regards.
Best Answer
I would not call the proof easy, but it just needs general principles. An outline is as follows: Since you transport an inner product on the Lie algebra $\mathfrak g$ to all points of $G$, left invariance is clear (and this is verified in the computation in your question). If a metric is invariant under both left and right translation, then it is invariant under conjugations, which easily implies that the corresponding inner product on $\mathfrak g$ is $Ad$-invariant, i.e. $\langle Ad(g)(X),Ad(g)(Y)\rangle=\langle X,Y\rangle$ for all $g\in G$ and $X,Y\in\mathfrak g$. It is also easy to show that this $Ad$-invariance is equivalent to bi-invariance of the induced metric (basically since right translations can be obtained from conjugations and left translations). Finally, if $G$ is connected (which probably should be added to the assumptions) then invariance of the inner product under the adjoint representation of $G$ is equivalent to invariance under the corresponding infinitesimal representation of $\mathfrak g$. This reads as $\langle ad(Z)(X),Y\rangle +\langle X,ad(Z)(Y)\rangle=0$ for all $X,Y,Z\in\mathfrak g$, which (after small rearrangements) is exactly the condition you have.