Union of Cartan subalgebras

Question

Let $L$ be a semisimple Lie algebra over the field $F$ of characteristics zero. Let $H_1,H_2$ be any two of its Cartan subalgebras (that exist, are maximal toral, self-normalizing, Abelian). Let $\mathfrak{A}(H_1,H_2)$ be minimal Lie subalgebra of $L$ that contains $H_1$ and $H_2$. I can not find in the literature anything about $\mathfrak{A}(H_1,H_2)$. I tried concluding something myself, but I got nowhere. Is there any book or paper that object $\mathfrak{A}(H_1,H_2)$ was studied? Thankies in advance!

What I tried (this is new idea) is using Theorem 4.1.3 in

Veeravalli S. Varadarajan: $\textit{Lie Groups, Lie Algebras and their Representations}$, Springer-Verlag, 1984.

as there is is clearly said that all Cartan subalgebras are $\mathfrak{h}_X$ (his notations, kind of generalized centralizer) but as I mentioned in previous paragraph these objects $\mathfrak{h}_X$ are harder to control then centralizers and I do not even know the answer to (simpler as it turns out) question:

If we take two elements $x,y \in L$, and denote $C_L(x),C_L(y)$ its centralizer and with $\mathfrak{C}(x,y)$ the minimal subalgebra of $L$ that contains $C_L(x)$ and $C_L(y)$ then what do we know in general about $\mathfrak{C}(x,y)$?

I would appreciate answers in forms of general lemmas from books or papers rather than some examples, thankies!

Answer 1

You didn't want examples, but that's what I'm gonna give, for I haven't been doing research related to Lie algebras for ages, and wouldn't know what buzzwords to use to search the publications.

As examples I proffer pairs of Cartan algebras that behave kinda extremally in the sense of your question. This was inspired by the comment Callum made.

So consider the simple Lie algebra $L=\mathfrak{so}_m(\Bbb{C})$ where $m=2\ell+1$ is an odd integer, $\ell\ge2$. I define it as the Lie algebra of antisymmetric $m\times m$ matrices. Let $E_{ij}$ be the $m\times m$ matrix with its sole non-zero entry a $1$ at position $(i,j)$. Let $S_{ij}:=E_{ij}-E_{ji}$ be its antisymmetric counterpart. The matrices $S_{ij},1\le i<j\le m$ span $L$, and somewhat redundantly we also have the matrices $S_{ji}=-S_{ij}$.

It is easy to verify that the elements $S_{ij}$ are all $\mathrm{ad}$-semisimple. As the matrices $S_{12}$, $S_{34}$, $S_{56}$, $\ldots$, $S_{2\ell-1;2\ell}$ commute, and there are $\ell$ of them, they span a Cartan subalgebra $H_1$.

Another Cartan subalgebra $H_2$ is spanned by $S_{23}$, $S_{45}$, $S_{67}$, $\ldots$, $S_{2\ell;2\ell+1}$. Observe that $H_1$ and $H_2$ intersect trivially.

A third Cartan subalgebra $H_3$ I will use is almost the same as $H_2$. I just replace the first generating matrix $S_{23}$ with $S_{12}$. So the subspaces $H_2$ and $H_3$ have a codimension one intersection.

On with the subalgebras the pairs generate. A tool I will use is that whenever $i,j,k$ are three distinct indices, then $S_{ij},S_{jk}$ and $S_{ki}$ span a copy of the Lie algebra $\mathfrak{so}_3(\Bbb{C})$, and we have the familiar commutator relations $$[S_{ij},S_{jk}]=S_{ik},\quad [S_{jk},S_{ki}]=S_{ji},\quad [S_{ki},S_{ij}]=S_{kj}.$$ In other words, we get the third (up to sign) as the commutator of the other two of any such triple.

• The smallest Lie algebra $\mathfrak{A}(H_1,H_2)$ containing both $H_1$ and $H_2$ is then actually all of $L$. This follows easily from the above observation. We get $S_{13}$ as the commutator of $S_{12}\in H_1$ and $S_{23}\in H_2$. Similarly for the other matrices $S_{i;i+2}$, $i=1,2,\ldots,m-2$. By induction on $d$ we then similarly get all the other matrices of the form $S_{i;i+d}$ with a difference $d$ in the indices as commutators of matrices already known to be in $\mathfrak{A}(H_1,H_2)$.
• On the other hand $\mathfrak{A}(H_2,H_3)$ is much smaller. We easily see that all we need to add to the subspace sum $H_2+H_3$ is the commutator $S_{13}=[S_{12},S_{23}]$. In other words $\mathfrak{A}(H_2,H_3)$ is a direct sum of the algebra $\langle S_{12},S_{23},S_{13}\rangle\simeq \mathfrak{so}_3(\Bbb{C})$ and a toral subalgebra spanned by $S_{45},S_{67},\ldots,S_{2\ell;2\ell+1}$.

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It is easy to imagine that something similar might happen in the general case also. The intersection $H_1\cap H_2$ becomes uninteresting, but the rest generate something else. It would not suprise me if something like the above Levi decomposition above would always come out, but I'm too ignorant to be sure.