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.

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    $\begingroup$ Why downvote this question? I do not see any reason. $\endgroup$
    – Sumanta
    Commented Jan 14, 2021 at 10:51

1 Answer 1


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 $G$.

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    $\begingroup$ 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. $\endgroup$
    – Jack Lee
    Commented Jan 10, 2021 at 17:36

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