Let $G$ be a Lie group and $Q$ be a biinvariant metric.
If $h$ is any positive definite scalar product on $T_eG$ then we have a left invariant metric $h$ on $G$ :
$$ h_g (dL_g X, dL_g Y) = h(X,Y)$$
Then we have a $Q$-symmetric $P\in Gl(T_eG)$ which is called a metric tensor : $$ h(X,Y) = Q(PX,Y)$$
I want to know $P$ more details.
Question 1 : For instance, is it a algebra homomorphism ?
And note that ${\rm ad}\ (Z)^\ast = -P^{-1}\circ {\rm ad}\ (Z)\circ P$. Here ${\rm ad}\ (Z)^\ast$ is adjoint operator :
$$ Q(P\ {\rm ad}\ (Z)^\ast\ X,Y)= h(X,[Z,Y])=-Q([Z,PX],Y)$$
Question 2 : And I want to prove
$$(\ast)\ {\rm ad}\ (Y)^\ast\ X = -[Y,PX]$$ where for all $Y\in V ,\ X\in T_eK$.
Here $V = \{ Y\in T_eG |\ PY=Y\}$, $[T_eK,V]\subset V$ and $h$ is ${\rm Ad}_GK$-invariant.
(cf. Lemma 22 in Eschenburg's book "Freie isometrische Aktionen auf kompakten Lie-Gruppen mit positive gekrummten Orbitraumen" : ($\ast$) is use to prove Lemma 22, whose statement is as follows :
Under above assumption, we have $$ h(R(X,Y)X,Y) = 1/4 |[PX,Y]|^2$$ )