Let $L$ be a finite dimensional semisimple complex Lie algebra.

Let $M$ be a subalgebra with the property that all elements of $M$ are semisimple, and is maximal w.r.t. this property.

Then $M$ is abelian and its centralizer is itself. So it is maximal abelian subalgebra.

Q.1 Is it always true that a maximal abelian subalgebra should contain only semisimple elements?

Q.2 Is it true that two maximal abelian subalgebras can have different dimensions? (I want to know answer to Q. 2 in which one may make cases - $L$ is emisimple or not semisimple. )

  • $\begingroup$ You may also be interested in this remark on Wikipedia: A refinement of the Cartan decomposition for symmetric spaces of compact or noncompact type states that the maximal Abelian subalgebras $\mathfrak a$ of $\mathfrak p$ are unique up to conjugation by $K$. en.wikipedia.org/wiki/Cartan_decomposition $\endgroup$ – punctured dusk Dec 13 '18 at 17:15

Q.1 Any 1 dimensional Lie subalgebra is abelian so is contained in at least one maximal abelian subalgebra. So taking any non-semisimple element we find a abelian subalgebra with non-semisimple elements and thus there exist maximal abelian algebras with non-semisimple elements.

Q.2 I think this is best illustrated with some examples: $$ D = \left\{ \begin{pmatrix} a_1 & \dots & 0 \\ \vdots & \ddots & \vdots \\ 0 & \dots & a_{2n} \\ \end{pmatrix} : a_i \in \mathbb{C} \right\} $$ $$ A = \left\{ \begin{pmatrix} 0 & 0 \\ B & 0 \\ \end{pmatrix}: B \in \mathfrak{gl}(n,\mathbb{C})\right\} $$ These are both maximal abelian and A has dimension $n^2$ where D has dimension $2n$. In terms of semisimplicity of L these examples fit in to the solvable Lie algebra of lower triangular matrices. But I suspect that it is possible in semisimple Lie algebras as well


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