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I am trying to answer the following question:

Which Lie groups have a Lie algebra admitting an $\text{Ad}$-invariant inner product?

First of all, all compact Lie groups satisfy this condition because one can average an arbitrary inner product on the algebra over the group.

This means that all Lie groups that have the same Lie algebra as some compact group also satisfy the condition. Indeed, the group is then locally isomorphic to a compact group, and taking "the same" inner product on the Lie algebra, we see that this inner product is invariant under all $\text{Ad}_g$ for $g$ in some neighbourhood of the identity, but this neighbourhood generates the whole group, from which the claim follows. Another approach would be to show that an inner product is $\text{Ad}$-invariant precisely if it is $\text{ad}$-invariant (like here, although there may be an issue with connectedness of the group) and conclude that the Lie algebra completely characterizes whether or not an invariant inner product exists.

I suspect these are all the Lie groups that answer the question, because Wikipedia claims this (the second property, at "this property characterizes compact Lie algebras"). However, the reference given is just "SpringerLink", which is of no use.

Could somebody confirm that the answer to the above question is "all Lie groups with a compact Lie algebra"?

I have consulted multiple sources, but none of them completely establish the result I'm looking for. For example, Lie groups beyond an introduction by Knapp establishes what I already know (that it works for groups with a compact Lie algebra), and establishes the converse only in case the inner product is actually the Killing form.

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This reference seems to have a lot about when a Lie group admits a bi-invariant metric. Might be of some use: – muzzlator Mar 3 '13 at 17:26
For a simple Lie algebra there is a unique ad-invariant bilinear form on the Lie algebra, up to constants, essentially from the definition of "simple" and the fact that (by Jacobi identity) ad is a Lie algebra hom, so any other is a multiple of Killing. – paul garrett Mar 3 '13 at 18:18
@muzzlator - The reference you gave has a very nice and explicit answer to the question. A Lie group has a Lie algebra that admits an Ad-invariant inner product precisely if the Lie group admits a bi-invariant metric (this is pretty straightforward), and those Lie groups are exactly direct products of compact groups with $\mathbb{R}^n$. If you post this as an answer, I'll accept it. – Daan Michiels Mar 3 '13 at 21:19
Ah, it's good you found the answer. I'm now reading it a bit more carefully to see how it is proved. You did all the work so I think you should post it as the answer and accept it so that people who may search for it in the future will find it. – muzzlator Mar 3 '13 at 21:28
@muzzlator - I posted the answer. I'll wait until tomorrow to accept it. – Daan Michiels Mar 3 '13 at 21:57
up vote 7 down vote accepted

The answer can be found in the reference given by muzzlator. Combining proposition 18.3 on page 511 and theorem 18.8 on page 514, we see that the following are equivalent:

  • The Lie algebra of the Lie group $G$ admits an $\text{Ad}$-invariant inner product,
  • The Lie group $G$ admits a bi-invariant metric,
  • The Lie group $G$ is the cartesian product of a compact group and a vector space $\mathbb{R}^n$.

The equivalence between the first two statements is relatively straightforward, and a proof is given in the reference linked to above. The equivalence between the last two statements is stated in the above reference, but not proved (though the proof is sketched). One refers to lemma 7.5 of Milnor's paper Curvatures of Left Invariant Metrics on Lie Groups (1976) for a proof.

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