# Existence of a semisimple elliptic subalgebra

Let $$\mathfrak{g}$$ be the Lie algebra of a compact Lie group $$G$$. Denote by $$\mathbb{C} \mathfrak g = \mathbb{C} \otimes \mathfrak g$$ the complexification of $$\mathfrak g$$. If necessary, I would not mind making some assumptions regarding $$\mathfrak g$$, like assuming it is simple or semisimple.

Definition: A subalgebra $$\mathfrak h \subset \mathbb C \mathfrak g$$ is called elliptic if $$\mathfrak h + \overline {\mathfrak h} = \mathbb{C}\mathfrak g$$.

The name elliptic is used because there is a natural differential operator associated with the subalgebra $$\mathfrak h$$ that is elliptic if and only if $$\mathfrak h$$ is elliptic.

My question is: does it exists a proper semisimple elliptic subalgebra $$\mathfrak h \subset \mathbb C \mathfrak g$$?

EDIT: I know that, for example, if the dimension of $$\mathfrak g$$ is 3, then any proper elliptic Lie algebra will have dimension 2 and such subalgebra will never be semisimple. So, not all Lie algebras $$\mathbb C \mathfrak g$$ will have semisimple proper elliptic Lie algebras. I would not mind assuming some extra hypothesis on $$\mathfrak g$$. Actually, if I can find any example of semisimple proper elliptic Lie algebra I will be very happy.

Thank you very much for any help.

• Are you sure that you wrote what you meant to ask? Maybe you want a direct sum rather than the sum? Otherwise, you can always take take ${\mathfrak h}= {\mathbb C}{\mathfrak g}$. Or, at least, you probably want ${\mathfrak h}$ to be a proper subalgebra. Then consider ${\mathfrak g}= so(3)$. All proper subalgebras in $sl(2, {\mathbb C})$ are solvable. – Moishe Kohan Jan 12 at 14:40
• I am only interested in proper elliptic subalgebras. Thank you for pointing out that. – Max Reinhold Jahnke Jan 12 at 18:24
• What's the relationship between solvable and elliptic subalgebras? – Max Reinhold Jahnke Jan 12 at 18:26
• A solvable Lie algebra cannot be semisimple. – Moishe Kohan Jan 12 at 18:33
• Interesting. I've thought about it a bit but, besides that the semisimple case immediately reduces to the simple one, I have no clue. Except that in low dimensions, one can exclude the existence of such $\mathfrak{h}$ by dimension arguments ($2dim\mathfrak{h} \ge dim\mathfrak{g}_{\Bbb C}$), an extreme case given by @MoisheCohen's answer. I mean even in the case of type $A_n$, just the classification of semisimple subalgebras of $\mathfrak{sl}_n(\Bbb C)$ (which is kind of equivalent to representation theory) is by no means trivial; and how those subalgebras behave under conjugation ... – Torsten Schoeneberg Jan 12 at 20:21

No, such subalgebra need not exist. For instance, take $${\mathfrak g}= o(3)$$, $${\mathbb C}{\mathfrak g}\cong sl(2, {\mathbb C})$$. However, $$sl(2, {\mathbb C})$$ contains no proper semisimple subalgebras (simply because 3 is the smallest dimension of a complex simple algebra).
• @MaxReinholdJahnke: I will add an example later. My guess is that $o(3)$ is the only exception to the existence problem. – Moishe Kohan Jan 13 at 1:05
• Type $A_2$ i.e. $\Bbb C \mathfrak{g} = \mathfrak{sl}_3(\Bbb C)$ is impossible as well, as the dimension of $\mathfrak{h}$ would need to be between $4$ and $8$; the only semisimple complex Lie algebra which satisfies this is $\mathfrak{sl}_2 \oplus \mathfrak{sl}_2$, which I'm pretty sure does not occur as subalgebra of $\mathfrak{sl}_3$. – Torsten Schoeneberg Jan 13 at 2:42
• @TorstenSchoeneberg: Sorry, I miscalculated, I do not have an example. I checked Levi factors of maximal parabolic subgroups in $SL(n)$ and determined that they cannot work. One only needs to check maximal semisimple subgroups which were computed by Dynkin around 1955. My current guess is that examples simply do not exist. – Moishe Kohan Jan 16 at 1:33