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Let $g$ be a lie algebra over an algebraically closed field of characteristic $0$. Let $h$ be a Cartan subalgebra of $g$. Let $n$ be an ideal of $g$. Consider the quotient map $q:g \rightarrow g/n$ and let $\overline{h}=q(h)=(h+n)/n$ be the image of $h$ in $g/n$. Prove that $\overline{h}$ is a Cartan subalgebra of $g/n$

I can prove $\overline{h}$ is nilpotent but not that it is self-normalizing. Its clear that if $x+n \in N_{g/n}(\overline{h})$ then $[x,y] \in h+n$ $\forall y \in h$. I need to prove that $x \in h+n$ but don't know how to do this, since I don't see why $[x,y]$ can't be in $n$ and therefore $0$ in $g/n$?

Callum
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Lottie Woods
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  • Indeed this seems a little more intricate than one would think. The proof in Bourbaki goes down a rabbit hole of corollaries to previous results, ultimately relying on a similar result for generalized nullspaces (the image of a generalized nullspace of certain operators under a surjective map compatible with the operator action is again the generalized nullspace). It would be nice to see a more direct proof. – Torsten Schoeneberg Jun 06 '23 at 14:39
  • I feel like it should be fairly easy but can't see how. Is there a way to see the proof you found easily? – Lottie Woods Jun 08 '23 at 21:56

1 Answers1

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As said in a comment, the statement is to be found in Bourbaki's volume on Lie Groups and Algebras (ch. VII § 2 no.1, cor. 2 to prop. 4), but for the non-trivial part of the proof, it refers to earlier propositions which refer to earlier corollaries which rely on a theorem resting on a lemma ...

I would like to see a more direct proof, but here is the idea I distilled out of that. Maybe one really needs the following machinery, because it is what Cartan subalgebras are ultimately for: to decompose the Lie algebra into (generalized) weight spaces.

Definition: For a subalgebra $\mathfrak h \subset g$, and any linear form $\lambda \in \mathfrak h^*$, let $$g^\lambda(\mathfrak h) :=\{ x \in g: \text{ for all } h \in \mathfrak h \text{ there is } n\in \mathbb N \text{ such that } (ad(h)-\lambda(h) \cdot id)^n (x) = 0\}$$ be the generalized weight space to the weight $\lambda$. (Note that in the theory of semsimple Lie algebras, one looks at the more straightforward weight spaces $g_\lambda(\mathfrak h) :=\{ x \in g: \text{ for all } h \in \mathfrak h, (ad(h)-\lambda(h) \cdot id) (x) = 0\}$.)

Fact 1: If $\mathfrak h$ is nilpotent, then $\mathfrak g$ is the full sum of its generalized weight spaces i.e. $$ g = \bigoplus_{\lambda \in \mathfrak h^*} g^\lambda(\mathfrak h).$$ This is a generalization of the well known theorem that commuting endomorphisms are simultaneously diagonalizable. Nilpotency guarantees something Bourbaki calls "almost commutativity", which in turn guarantees this "simultaneous trigonalizability".

Fact 2: If $\mathfrak h$ is nilpotent, then it is self-normalizing if and only if $\mathfrak h = g^0(\mathfrak h)$. (This is rather easy to see.)

Now the proof goes like this: If one has a map between things which have a full decomposition as in Fact 1, and the map respects these decompositions (which is trivially true in our application by it being a Lie algebra homomorphism), then that map can only be surjective if its restriction to each weight space is surjective. In particular in our case, and for $\lambda = 0$, we can conclude

$$ \overline{\mathfrak h} = \overline{g^0(\mathfrak h)} = \overline{g}^0 (\overline{\mathfrak h}) $$

which by Fact 2 tells us $\overline{\mathfrak h}$ is self-normalizing in $\overline{g}$.

Torsten Schoeneberg
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