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The Cartan subalgebras of a reductive Lie algebra are abelian.

Are there non-reductive Lie algebras with abelian Cartan subalgebras?

In fact, the elements of a Cartan subalgebra of a reducitive Lie algebra are semisimple, so a weaker question is:

Are there non-reductive Lie algberas with abelian Cartan subalgebras all of whose elements are semisimple?

N.B.: I asked this earlier at math.SE.

flag	asked Feb 18 2012 at 2:42 Mariano Suárez-Alvarez 30.6k●7●68●153

Pasha Zusmanovich · Answer 1 · 2012-04-04 07:40:54Z UTC

If $H$ is a Cartan subalgebra of a Lie algebra $L$, and $A$ is an associative commutative algebra, then $H \otimes A$ is a Cartan subalgebra of $L \otimes A$. Specializing this to the case $L$ classical simple, we get another positive answer to the first question (but not to the second one). This example can be, probably, varied, by taking subalgebras of $L \otimes A$, or adding "tails" of derivations, etc.

Jim Humphreys · Answer 2 · 2012-02-18 19:38:17Z UTC

Maybe it should be emphasized that these questions deal with finite dimensional Lie algebras over a field of characteristic 0; in prime characteristic there are further complications. BR has addressed the first question and observed that a subalgebra of a reductive Lie algebra containing a given Borel subalgebra (in other words a parabolic subalgebra) will have the same type of abelian Cartan subalgebras found in the big algebra.

The follow-up question is more problematic, since the notion of "semisimple" element in a Lie algebra isn't always intrinsically defined. In fact, there is an intrinsic Jordan decomposition precisely when the Lie algebra equals its derived algebra (which for instance excludes nonzero abelian Lie algebras): this is proved by building on the classical theorem for semisimple Lie algebras, in a paper by Cagliero and Szechtman here. (The intrinsic nature of the Jordan decomposition involves good behavior under linear representations. Leaving that aside, parabolic subalgebras of a given semisimple Lie algebra do contain "semisimple" and "nilpotent" elements relative to the big algebra but not relative to their own representations taken in isolation.)

Anyway, I'm not sure right away how much can be said about Cartan subalgebras in a non-reductive Lie algebra which equals its derived algebra. Examples?

BR · Answer 3 · 2012-02-18 04:59:44Z UTC

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The subalgebra of diagonal matrices in the (non-reductive) standard Borel subalgebra ${\mathfrak b}$ of ${\mathfrak gl}_2$ is an abelian Cartan subalgebra of ${\mathfrak b}$, since it is abelian (hence nilpotent) and its own normalizer (directly calculate the bracket of an element $X$ of ${\mathfrak b}$ with a diagonal matrix $A$ to see that the result is diagonal only when $X$ was in fact diagonal or $A$ was in the center).

answered Feb 18 2012 at 4:59

BR
2,307●1●6●14

	Indeed, the subalgebra of diagonal matrices is also a Cartan subalgebra of $gl_n(F)$ for any field $F$ and $n>0$. – Salvatore Siciliano Feb 18 2012 at 16:27
	This method works whenever $\mathfrak b$ is a subalgebra of a reductive Lie algebra $\mathfrak g$ and $\mathfrak b$ contains a Cartan subalgebra $\mathfrak a$ of $\mathfrak g$ (Since $\mathfrak a$ is its own normalizer in $\mathfrak g$, it will be its own normalizer in $\mathfrak b$). – BR Feb 18 2012 at 16:32
	BR: Yes, you are right. – Salvatore Siciliano Feb 18 2012 at 16:46
	Even simpler, an abelian Lie algebra is a Cartan subalgebra of itself. – anon Feb 18 2012 at 18:52
	But an abelian Lie algebra is reductive! – BR Feb 18 2012 at 18:54

Lie algebras with abelian Cartan subalgebras.

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