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I'm reading Serre's book about semisimple complex Lie algebras and having problem to understand one part of a proof. The theorem states, that the restriction of the Killing form $B$ of a semisimple Lie algebra $\mathfrak{g}$ to a Cartan subalgebra $\mathfrak{h}$ is nondegenerate. $\mathfrak{h}$ can be viewed a nilspace of $\operatorname{ad} x$ (with $x$ some element in $\mathfrak{g}$).

Within the proof he uses that the nilspaces of $\operatorname{ad} x - \lambda$ and this of $\operatorname{ad} x -\mu$ are orthogonal with respect to $B$ when $\lambda + \mu$ is not $0$. I have no idea how I would calculate that or show that it holds.

Also in the end when we have $\mathfrak{g}$ writen as a orthogonal sum, why does it follow from that that the restriction is nondegenerate. I have read that the restiction of a nondegenerate form is nondegenerate iff the intersection of $\mathfrak{h}$ and its orthogonal complement is $\{0\}$. Is this the reason in this case?

Here is the theorem and proof:

Theorem 3. Let $\mathfrak h$ be a Cartan subalgebra of a semisimple Lie algebra $\mathfrak g$. Then:

(a) $\mathfrak h$ is abelian.

(b) The centralizer of $\mathfrak h$ is $\mathfrak h$.

(c) Every element of $\mathfrak h$ is semisimple (cf. Sec. II.5).

(d) The restriction of the Killing form of $\mathfrak g$ to $\mathfrak h$ is nondegenerate.

(d) By Corollary 2 to Theorem 2, there is a regular element $x$ such that $\mathfrak h = \mathfrak g_x^0$. Let

$$\mathfrak g = \mathfrak g_\alpha^0 \oplus \sum_{\lambda \ne 0} \mathfrak g_x^\lambda$$

be the canonical decomposition of $\mathfrak g$ with respect to $x$ (cf Prop. 2). If $B$ denotes the Killing form of $\mathfrak g$, then a simple calculation shows that $\mathfrak g_x^\lambda$ and $\mathfrak g_x^\mu$ are orthogonal with respect to $B$ provided that $\lambda + \mu \ne 0$. We therefore have a decomposition of $\mathfrak g$ into mutually orthogonal subspaces

$$\mathfrak g = \mathfrak g_x^0 \oplus \sum_{\lambda \ne 0} \mathfrak g_x^\lambda \oplus \mathfrak g_x^{-\lambda}.$$

Since $B$ is nondegenerate, so is its restriction to each of these subspaces, giving (d) since $\mathfrak h = \mathfrak g_x^0$.

LSpice
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1 Answers1

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Here's a proof. Use the standard grading of $\mathfrak{g}$ by weights, i.e., for $\alpha\in\mathrm{Hom}(\mathfrak{h},\mathbf{C})$, one has $\mathfrak{g}_\alpha=\{x\in\mathfrak{g}:\forall h\in\mathfrak{h},[h,x]=\alpha(h)x\}$. In particular $\mathfrak{g}_0=\mathfrak{h}$.

If $\alpha$ is a nonzero weight, we have $$\langle \mathfrak{h},\mathfrak{g}_\alpha\rangle=\langle \mathfrak{h},[\mathfrak{h},\mathfrak{g}_\alpha]\rangle=\langle [\mathfrak{h},\mathfrak{h}],\mathfrak{g}_\alpha\rangle=0.$$

So if $x\in\mathfrak{h}=\mathfrak{g}_0$ is in the kernel of the Killing form restricted to $\mathfrak{h}$, then it is in the kernel of the Killing form on $\mathfrak{g}$, and hence $x=0$. The result is proved.

Edit: let's not take for granted that $\mathfrak{h}$ is abelian. For $h\notin\mathrm{Ker}(\alpha)$, we have $$\langle h,x\rangle=\frac1{\alpha(h)}\langle h,\alpha(h)x\rangle=\frac1{\alpha(h)}\langle h,[h,x]\rangle=\frac1{\alpha(h)}\langle [h,h],x\rangle=0.$$ If $h\in\mathrm{Ker}(\alpha)$, just pick $h'\notin \mathrm{Ker}(\alpha)$. Then $h+h',h'\notin\mathrm{Ker}(\alpha)$. So $\langle h,x\rangle=\langle h+h',x\rangle-\langle h',x\rangle=0$.

YCor
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  • This proves the main result. But the sub-result can be proved in the same way, writing, for $\alpha,\alpha+\beta\neq 0$, $\langle\mathfrak{g}\alpha,\mathfrak{g}\beta\rangle=\langle [\mathfrak{h},\mathfrak{g}\alpha],\mathfrak{g}\beta\rangle=\langle\mathfrak{h},[\mathfrak{g}\alpha,\mathfrak{g}\beta]\rangle\subset\langle\mathfrak{h},\mathfrak{g}_{\alpha+\beta}\rangle=0$ – YCor Dec 14 '23 at 10:22
  • For comparison, Serre seems to use the opposite notation, writing $\mathfrak g^\alpha$ where you write what I think of as the more standard $\mathfrak g_\alpha$. – LSpice Dec 14 '23 at 14:46
  • Thank you. About the last part that ⟨ h, g_{\alpha+\beta} ⟩ =0, why does this hold. Is it because [h,h]=0? In the book this is shown by using the theorem, so it wouldn't make sense to use it in the proof. Or is there a different explanation, I am not seeing? –  Dec 14 '23 at 14:56
  • @paula That $\langle h,g_{\alpha+\beta}\rangle=0$ is proved in the answer (well, with $\alpha$ in the role of $\alpha+\beta$). – YCor Dec 14 '23 at 15:01
  • But don't we use [h,h]=0 there or why is ⟨[h,h],g_α⟩=0? –  Dec 14 '23 at 15:06
  • @paula yes, I'm using (a) to prove (d). – YCor Dec 14 '23 at 16:07
  • Hm, Serre does it the other way around so he later uses d) to show a) and b). Do you know if there is a way to calculate the Killing Form of the nilspaces, without using this property? –  Dec 14 '23 at 16:20