Let $G$ be a Lie group.

Suppose $G$ admits a a bi-invariant volume form. Is it true that $G$ admits a bi-invariant (Riemannian) metric?

Since we know that $G$ admits a bi-invariant metric if and only if it is a cartesian product of a compact group and a vector space $\mathbb{R}^n$, this question is equivalent to the following:

Are there any Lie groups which are not products (in the form mentioned above), which admit a bi-invariant volume?

  • $\begingroup$ All semisimple groups do, en.wikipedia.org/wiki/Haar_measure. $\endgroup$ – Moishe Kohan Nov 27 '16 at 22:23
  • $\begingroup$ Thanks. Is there an easy way to see these measures can be induced by volume forms? (a-priori the Haar measure only respects the topology, so to show it is induced by a volume form, we need (for a start) to know something about its "smoothness"). $\endgroup$ – Asaf Shachar Nov 27 '16 at 23:37
  • $\begingroup$ Yes, read about the modular function in the same wikipedia article (use the left-invariant measure induced by a left-invariant Riemannian metric); keep in mind that connected semisimple Lie groups have only trivial characters. $\endgroup$ – Moishe Kohan Nov 28 '16 at 0:19
  • $\begingroup$ There is a more direct argument: For a connected semisimple group any one-dimensional representation is trivial. Hence the action of $G$ on $\Lambda^{\dim(\mathfrak g)}\mathfrak g^*$ induced by the adjoint action is trivial. Thus any non-zero element in there extends to a bi-invariant volume form on $G$. $\endgroup$ – Andreas Cap Nov 29 '16 at 14:54

The existence of a biinvariant volume form does not imply the existence of a biinvariant Riemannian metric. For instance, each noncompact connected semisimple Lie group $G$ admits such a form but never a biinvariant metric (the latter part I will skip).

Here is a self-contained argument. Start with a left-invariant Riemannian metric on a connected semisimple Lie group $G$, e.g. $G=PSL(n, {\mathbb R})$. This metric defines a left-invariant volume form $\omega$ on $G$. Now, let us prove that $\omega$ is also right-invariant. Given $g\in G$, consider its adjoint representation $Ad_g$ on ${\mathfrak g}$. Let $det: GL({\mathfrak g})\to {\mathbb R}^*$ be the determinant defined by the form $\omega$ at $T_{1}G={\mathfrak g}$. Then the composition $det\circ Ad$ is a homomorphism $G\to {\mathbb R}^*$, a character. Since $G$ is connected and semisimple, it has only trivial characters, hence $Ad$ preserves the volume form $\omega$ at $1\in G$. From this and left-invariance of $\omega$, you see that $\omega$ is right-invariant.


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