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Let $\mathfrak{g}$ be a Lie algebra and $\mathfrak{h}$ is a Cartan subalgebra. We know when the lie algebra is semi simple then the killing form, $\kappa(X,Y) = Tr(X^{ad}Y^{ad})$, is non degenerate in $\mathfrak{g}$ and more importantly in $\mathfrak{h}$. So it is a bilinear form analogous to the metric providing an isomorphism between $\mathfrak{h}$ and $\mathfrak{h}*$ and allowing us to take the inner product across these spaces.

One thing that isn't clear fromt the literature I've read (and from my Universities lecture notes) is when we are able to to put the killing form in the form $\kappa_{ij} = -\kappa \delta_{ij}$ and so we can treat the inner product as the normal euclidean dot product. Some notes I have read say this is only true for compact groups, others I have read say you can always do it and some don't even mention it (my lecture notes).

glS
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SheldonCooper
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    The question is equivalent to https://math.stackexchange.com/q/3953914/96384. – Torsten Schoeneberg Jan 11 '21 at 00:20
  • @Torsten: is it? I think the other question is equivalent to asking for an Ad-invariant positive or negative-definite form, whereas in this question we specifically want the form to be the Killing form. It takes an additional small argument here, I think. – Qiaochu Yuan Jan 11 '21 at 00:23
  • Well, at least on each absolutely simple summands (NB, https://math.stackexchange.com/q/3931433/96384), the Killing form is the unique $ad$-invariant non-degnerate bilinear form up to scaling. Further, since the tacit assumption here seems to be that the ground field is $\mathbb R$, for the purposes of both questions, such scaling comes down to flipping between a Euclidean product and its negative, which the OP actually does themselves in this question. – Torsten Schoeneberg Jan 11 '21 at 04:21

2 Answers2

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Call a semisimple Lie algebra $\mathfrak{g}$ compact if its Killing form is negative definite. Then the simply connected Lie group $G$ with Lie algebra $\mathfrak{g}$ has its adjoint representation on $\mathfrak{g}$ preserving the Killing form; its image lies in the Lie subgroup of $\text{Aut}(\mathfrak{g})$ preserving the Killing form, which is an orthogonal group $O(\mathfrak{g})$ and in particular which is compact. It follows that the adjoint form $G/Z(G)$ of $G$ (the image of the Killing form) is compact, and hence if the center of $G$ is finite that $G$ is compact. (I think this is always true with these hypotheses but I'm not sure how to prove it.)

The semisimple Lie algebras which arise in this way are exactly the compact real forms of the semisimple complex Lie algebras, e.g. $\mathfrak{su}(n), \mathfrak{so}(n), \mathfrak{sp}(n)$ and the exceptionals.

In the noncompact case the Killing form may have indefinite signature; for example, the Killing form on $\mathfrak{sl}_2(\mathbb{R})$ has signature $(2, 1)$, and the adjoint representation induces an exceptional isomorphism $\mathfrak{sl}_2(\mathbb{R}) \cong \mathfrak{so}(2, 1)$. This is the split analogue of the exceptional isomorphism $\mathfrak{su}(2) \cong \mathfrak{so}(3)$, and both of these isomorphisms complexify to the exceptional isomorphism $\mathfrak{sl}_2(\mathbb{C}) \cong \mathfrak{so}_3(\mathbb{C})$.

Qiaochu Yuan
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    Actually, conversely in the noncompact case the Killing form is always indefinite. Indeed in some (and every) copy of $\mathfrak{sl}_2$ it restrict to a nonzero invariant bilinear form, which has to be indefinite. – YCor Jan 10 '21 at 22:55
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It is true for compact Lie algebra (Lie algebras of compact simply connected groups) for which Killing is negative definite.

https://en.wikipedia.org/wiki/Killing_form#Connection_with_real_forms

Tsemo Aristide
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