# Right-invariance of a volume form on a compact Lie group

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.

• Why downvote this question? I do not see any reason. Commented Jan 14, 2021 at 10:51

## 1 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 $$G$$.

• You're right -- I should have included the hypothesis that $G$ is connected. I've added this to my correction list. Thanks for pointing it out. Commented Jan 10, 2021 at 17:36