Volume form on compact Lie group

Question

Suppose $G$ is a compact Lie group, $\omega \in \Omega^n(G)$ is a left-invariant volume form and let $i:G \rightarrow G$ be the inversion $i(g)=g^{-1}$. I have previously shown that $r^*_g \omega$ is left-invariant as well, where $r_g$ is multiplication on the right and so $r^*_g \omega=f(g)\omega$. Moreover, since this $f$ is a homomorphism, by compactness, it follows that $f(g)=1$.

I also need to prove that $i^* \omega = \omega$ or $-\omega$. I think I need to use the fact that $i^* \omega$ is also left-invariant. I can see this, because
$$r^*_h i^* \omega=(i \circ r_h)^* \omega=(l_{h^{-1}}\circ i)^* \omega=i^* l_{h^{-1}}^* \omega=i^* \omega$$
since $\omega$ is left-invariant.
How can I use this to show $i^* \omega = \omega$ or $-\omega$?

Answer 1

Let $X_1, \cdots, X_n \in T_e G$. Using $i_* X = -X$ (see here), we have

$$ (i^* \omega)_e (X_1, \cdots, X_n) = \omega_e (i_* X_1, \cdots, i_*X_n) = (-1)^n \omega_e (X_1, \cdots, X_n),$$

thus $(i^*\omega)_e =(-1)^n \omega_e$. Since both $\omega, i^*\omega$ are left invariant, we have $i^*\omega = (-1)^n \omega$.