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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?

Remarks:

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?

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    $\begingroup$ 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
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    $\begingroup$ 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
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    $\begingroup$ 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
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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

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