Inverse of left-invariant metric is left-invariant Given a Lie group $G$ of dimension $n$ with a left-invariant metric $g$.
Is it true that $g^{-1}$, that is, the metric in the cotangent bundle (whose matrix form is the inverse of that of $g$) is also left-invariant?
The reason that I think this is not true is that for a particular example, I have $n$ left-invariant 1-forms, and if my computations are correct, some of the inner products (via $g^{-1}$) of these forms are not constants.
 A: Let's denote by ${\flat} : TM \to T^*M$ the musical isomorphism defined by $X^{\flat}(Y) =g\left(X,Y\right)$. Suppose $g$ is left-invariant under some groupe action of $G$, that is
$$
g\left(h\cdot X,h\cdot Y\right) = g (X,Y)
$$
whenever $h \in G, X,Y$ vector fields on $M$. Recall that $G$ has a natural left action on $T^*M$ by
$$
\left(h \cdot \alpha \right)(X) = \alpha\left(h^{-1} X\right).
$$
Then if $X$ and $Y$ are vector fields, and $h \in G$,
$$
\left(h\cdot X\right)^{\flat}(Y) = g\left( h\cdot X,Y\right) = g\left( h\cdot X, h\cdot \left(h^{-1} \cdot Y\right)\right) = g\left(X,h^{-1}\cdot Y\right) = \left(h\cdot X^{\flat}\right)(Y).
$$
It follows that $\left(h\cdot X\right)^{\flat} = h \cdot X^{\flat}$ so that the musical isomorphism is also left invariant.
Recall that the metric on $T^*M$ is defined so that musical isomorphisms are isometries. Denoting $\tilde{g}$ the metric on $T^*M$ it follows that if $X$ and $Y$ are vector fields,
$$
\tilde{g}\left( h \cdot X^{\flat}, h \cdot Y^{\flat}\right) = \tilde{g}\left( \left(h\cdot X\right)^{\flat}, \left(h\cdot Y\right)^{\flat} \right) = g\left( h\cdot X,h\cdot Y\right) = g\left(X,Y\right) = \tilde{g}\left(X^{\flat},Y^{\flat}\right),
$$
that is, if $\alpha$ and $\beta$ are $1$-forms,
$$
\tilde{g}\left(h \cdot \alpha,h\cdot \beta\right) = \tilde{g}\left(\alpha,\beta\right),
$$
and $\tilde{g}$ is left invariant.
Therefore, what is stated above can be applied to $M = G$ and this brings me to the conclusion that you made some error in your calculations.
