As noted here , a Lie group $G$ admits a bi-invariant metric if and only if $G$ is the cartesian product of a compact (Lie) group and a vector space $\mathbb{R}^n$.

The question:

For which Lie groups that posses a bi-invariant metric, this metric is unique up to scalar multiple?


1) Every Lie group posses a left invariant metric. Just take any inner product on $T_eG$ and translate it to all the other tangent spaces via the differential of left translation.

This implies there are many left invariant metrics on any Lie group $G$ (which are not scalar multiples of one another). This follows from the fact this is true in the level of linear algebra. (Just take any two inner products on $T_eG$ which are not scalar multiple of one another).

2) As noted by Daniel Fischer, in the case of an abelian group, since right-invariance & left-invariance coincide, remark 1 above implies existence of many different bi-invariant inner products.

As a corollary, it follows that any direct products of an abelian Lie group of (dimension >1) and a non-abelian Lie Group also have non-unique metrics.

Hence, we must exclude from the search direct product of compact groups and a vector space. This leaves out the case of compact nonabelian groups which are not direct products of abelian & non-abelian groups. For which of these Lie groups the uniqueness hold?

  • 1
    $\begingroup$ A bi-invariant metric is a (Riemannian) metric such that the differentials of all left and all right translations are isometries? Then any inner product on $T_eG$ induces a bi-invariant metric for abelian $G$, so in general, it's not unique up to scalar multiples. $\endgroup$ – Daniel Fischer Jun 10 '15 at 11:32
  • $\begingroup$ You are clearly right. The non-trivial case is for non-abelian groups. I will edit the question. $\endgroup$ – Asaf Shachar Jun 10 '15 at 11:41
  • 1
    $\begingroup$ Not to be a spoil-sport, but somebody should exclude direct products of an abelian Lie group of dimension $> 1$ and a non-abelian Lie Group. $\endgroup$ – Daniel Fischer Jun 10 '15 at 13:08
  • $\begingroup$ You are right. This is an immediate corollary to the case of abelian group. But I am not sure how many Lie groups does that leave us? $\endgroup$ – Asaf Shachar Jun 10 '15 at 13:23
  • 1
    $\begingroup$ For semisimple lie groups the killing form is the only ad invariant bilinear form modulo constant, I think that it's also the case when G is non abelian compact and $H^3(G)=\mathbb Z $. $\endgroup$ – k76u4vkweek547v7 Jan 31 '16 at 17:28

The answer to your question comes from Schur's lemma, in the theory of group representations, applied to the adjoint representation of the Lie group.

The bi-invariant metric restricts to an Ad-invariant scalar product on the Lie algebra. Schur's lemma says that, in an irreducible representation, there is only one invariant symmetric bilinear form, up to a scalar multiple, (this is just one version of Schur's lemma, which has many other uses).

A Lie group is simple if and only if its adjoint representation is irreducible. Therefore, the answer to you question is : simple Lie group.

What if a compact Lie group is not simple. Well, then, it is a direct product of simple Lie groups and of a torus (afterwards, there can be a quotienting by a discrete subgroup). The cone of bi-invariant metrics is made up of positive linear combinations of the bi-invariant metrics of the factors in this direct product. -- Salem


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.