The following is a question from the second edition of John M. Lee's Introduction to Riemannian Manifolds.
3-9. Suppose $G$ is a compact Lie group with a left-invariant metric $g$ and a left-invariant orientation. Show that the Riemannian volume form $dV_g$ is bi-invariant. [Hint: Show that $dV_g$ is equal to the Riemannian volume form for a bi-invariant metric.]
Since $G$ is compact, it admits a bi-invariant metric $\tilde g$, and with this and the given orientation, we have a volume form $dV_{\tilde g}$. It's easy to see that $dV_g$ and $dV_{\tilde g}$ are both left-invariant, using the fact that the metrics $g, \tilde g$ and the orientation are left-invariant. Since they are both left-invariant and positive with respect to the given orientation, there is a $c > 0$ such that $dV_g = c dV_{\tilde g}$. Then $dV_g$ is equal to the Riemannian volume form corresponding to the bi-invariant metric $c^{2/n}\tilde g$, as the hint suggests.
I am having trouble showing that $dV_g$ is also right-invariant; here is my work thus far. Since for any $\varphi \in G$ the forms $R_\varphi^*(dV_g)$ and $dV_g$ are left-invariant, there is a function $f \colon G \to \mathbb{R}^\times$ such that $R_\varphi^*(dV_g) = f(\varphi) dV_g$. Evaluating both sides at $e$, one obtains $f(\varphi) = \det(\mathrm{Ad}(\varphi^{-1}))$, a continuous homomorphism. Since $G$ is compact, $f(G)$ is a compact subgroup of $\mathbb{R}^\times$, i.e. $f(G) = \{1\}$ or $f(G) = \{\pm 1\}$.
I do not see how to exclude the second case, i.e. if $R_\varphi$ is orientation-reversing for some $\varphi \in G$. Since $f(e) = 1$, $f$ is identically $1$ on the identity component of $G$; this would finish the problem if $G$ were connected, but unfortunately, it might not be. Since I haven't used the fact that $dV_g$ equals the volume form for a bi-invariant metric yet, I feel it must be used here, but I cannot see how.
Some searching reveals this may be related to the idea of left/right-invariant Haar measures and unimodular Lie groups, but my measure theory knowledge is insufficient to understand that material. A small part of me believes that the problem is incorrect without the connectedness hypothesis (e.g. consider the diagonal matrix $A$ with $-1$ and $1$ in $O(2)$, then $f(A) = -1$?), but the errata for the book reveals nothing. Any hints or suggestions on how to proceed with this problem would be appreciated.
Best Answer
The claim is simply false in general. I think, Lee forgot to assume that the group is connected or that the orientation is bi-invariant. As a simple example, as you suggested, consider $G=O(2)$. Then the action of $G$ on itself via conjugation does not preserve any orientation on $G_0=SO(2)$ (since this action contains a reflection). From this, it follows that there is no bi-invariant orientation. From this, it follows that there is no bi-invariant volume form. Of course, there is a bi-invariant measure on $G$, given by a bi-invariant density on $G$.