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Main question: Suppose that $\mathfrak g$ is a complex Lie algebra over $\mathbb C$ and let $J$ be a semisimple element of $\mathfrak g$ (meaning that $\mathrm{ad}_J$ is a diagonalizable operator on $\mathfrak g$). Is there a Cartan subalgebra $\mathfrak h \subseteq \mathfrak g$ which contains $J$?

Some motivation: for any Cartan subalgebra $\mathfrak h \subseteq \mathfrak g$ all elements of $\mathfrak h$ are semisimple. Moreover there exists a condition on an element $J$ of $\mathfrak g$ stronger than semisimplicity, called regularity (cf. Serre's book on semisimple Lie algebras) which guarantees existence of a unique Cartan subalgebra $\mathfrak h \subseteq \mathfrak g$ such that that $J \in \mathfrak h$. Here I ask if existence (but with no uniqueness) can be inferred under weaker assumptions. In fact, if the answer to the main question is negative, I would like to ask a slightly more general question.

Generalization: Let $\mathfrak g$ be as above and let $S$ be the union of all Cartan subalgebras of $\mathfrak g$. Is it possible to describe explicitly the set $S$? Let me just mention that $S$ is clearly dense, as it contains the non-empty, Zariski open set of all regular elements of $\mathfrak g$.

Blazej
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    No, there are regular elements that belong to no Cartan subalgebra. For instance, in $\mathfrak{sl}_2$ all nonzero elements are regular, but nilpotent ones belong to no Cartan subalgebra. Actually in general, since elements of Cartan subalgebras are semisimple, the conclusion is that the union of Cartan subalgebras is the set of semisimple elements. – YCor Aug 07 '19 at 13:56
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    The set of regular semisimple elements is, I think, Zariski-open, hence dense. – YCor Aug 07 '19 at 13:59
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    Dear @YCor, I think the problem here is that there is more than one notion of regularity used widely in the literature. I refer to the following notion: $x \in \mathfrak g$ is said to be regular if the multiplicity of $0$ among the eigenvalues of $x$ is equal to the rank of $r$. It is proven e.g. in the Serre's book, and I think also in Bourbaki, that with this definition every regular element belongs to a unique Cartan subalgebra. It is obvious that the set of regular elements is Z-open and nonempty, and that no nilpotent element is regular. – Blazej Aug 07 '19 at 20:54
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    For the reference of people looking at this question in the future: @YCor seems to have referred the definition of regularity of $x$ which demands merely that the dimension of the centralizer of $x$ is equal to $r$, the rank of $\mathfrak g$. Already for $\mathfrak{sl}(2)$ this is a strictly weaker condition. – Blazej Aug 07 '19 at 21:02
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    Thanks for the definition, which is indeed different. Also the conclusion of my first comment is a priori only correct for semisimple Lie algebras. – YCor Aug 07 '19 at 21:02
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    In an arbitrary Lie algebra (finite-dimensional, characteristic zero) the set of regular elements (in the sense of Bourbaki) is Zariski open, so is dense. – YCor Aug 07 '19 at 21:03
  • I think we came to agreement that the set $S$ in my question is dense, since it contains a Zariski open subset. I'm curious if $S$ itself is Zariski open. – Blazej Aug 07 '19 at 21:05
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    Obviously not, since $0\in S$ but $0$ is usually not in the interior of $S$. – YCor Aug 07 '19 at 21:05
  • In general there's no inclusion between Bourbaki-regular and centralizer-regular. Indeed, in a nilpotent Lie algebra, all elements are Bourbaki-regular, but if it is not abelian, central elements are not centralizer-regular. – YCor Aug 07 '19 at 21:13
  • Good point. I was concentrating on the sempisimple case. – Blazej Aug 07 '19 at 21:24
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    OK; the semisimple case is clear-cut (union of Cartan subalgebras = semisimple elements) — by the way it's confusing if you write a hypothesis in the title and not in the question. – YCor Aug 07 '19 at 21:27

2 Answers2

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Yes, and a more general statement is true even over general fields of characteristic $0$, according to Bourbaki's take on Cartan subalgebras (in book VII, §2 of the volume on Lie Groups and Lie Algebras). Namely, proposition 10 says that for an abelian subalgebra $\mathfrak{a} \subseteq \mathfrak{g}$ consisting of semisimple elements,

$$\lbrace \text{Cartan subalgebras of } \mathfrak{g} \text{ containing } \mathfrak{a} \rbrace = \lbrace \text{Cartan subalgebras of } \mathfrak{z}_{\mathfrak{g}}(\mathfrak{a}) \rbrace $$

($\mathfrak{z}_{\mathfrak g} =$ centraliser). But every Lie algebra has Cartan subalgebras (see e.g. corollary 1 to theorem 1 loc. cit.), in particular so does $\mathfrak{z}_{\mathfrak{g}}(\mathbb{C} J)$ in your question.

Torsten Schoeneberg
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  • This answers the main problem, so I will eventually accept this answer. Before I do this, may I ask if you know anything about the second question? – Blazej Aug 06 '19 at 09:16
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Oh, I just saw that in your question title, you restrict to semisimple Lie algebras. Well there's a much easier argument for this case:

Cartan subalgebras of semisimple Lie algebras can equivalently be characterised as the maximal toral ones, where a subalgebra is called "toral" iff it is abelian and consists of semisimple elements. (E.g. Humphreys actually uses this as definition; cf. my answer to Are there common inequivalent definitions of Cartan subalgebra of a real Lie algebra?; a proof for equivalence of these definitions is an exercise in Bourbaki, and done in prop. 3.1.5 of my thesis.) With that, it's just a usual maximality argument in finite dimension.

Again, this works over any field of characteristic $0$ (just that "semisimple" does not necessarily mean diagonalisable, but, well, semisimple, i.e. diagonalisable over an algebraic closure).

Torsten Schoeneberg
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