0
$\begingroup$

This is the definition of cartan subalgebra define in Brian Hall, Lie groups, Lie algebras and representations, 2nd Edition, Chpt 7, Sec 2, Def. 7.10.

I am assuming the ground field is $C$ or it does not make sense to talk about diagonalizability.

If $g$ is complex semisimple lie algebra(here he assumed semisimple=reductive+triviality of center), and $h\subset g$ is an ideal s.t. $(1)\forall H_i\in h$, $[H_i,H_j]=0$

$(2)\forall G\in g, [G,h]=0\implies G\in h$

$(3)$ For all $H\in h, ad_H$ is diagonalizable.(i.e. adjoint representation of $h$ is diagonalizable.)

It is clear that $(1)$ and $(2)$ are required to find maximal amount of joint eigenvalues of $ad_H$. However, since $[H_i,H_j]=0$ and $ad:g\to gl(g)$ is lie algebra homomorphism, certainly it suffices to demand only 1 particular $H$ s.t. $ad_H$ diagonalizable.

$\textbf{Q:}$ Do I always need diagonalizable condition? What is the counter example that I do need? Since there is requirement for lie algebra homomorphism for $ad$, there will be constraint on the structure of $ad$. Furthermore, if I do need diagonalizability, can I just use one per reasoning above?

$\endgroup$
  • 1
    $\begingroup$ $h$ is a subalgebra, not an ideal. And every subalgebra contains $H=0$ which is diagonalisable. So e.g. $\pmatrix{0&*\\0&0} \subset \mathfrak{sl}_2(\Bbb C)$ satisfies 1 and 2, but not 3, and is not a Cartan subalgebra. Is that the counterexample you look for or do I misunderstand the question? $\endgroup$ – Torsten Schoeneberg Feb 15 at 8:55
  • $\begingroup$ @TorstenSchoeneberg So this is not equivalent to standard cartan subalgebra with nilpotency and $h$ being normaliser of itself? Clearly this definition implies nilpotency and $h$ self normalization. The converse fails by $(3)$ then. $\endgroup$ – user45765 Feb 15 at 16:12
  • $\begingroup$ $(2)$ means self-centralising, not necessarily self-normalising, as the example shows. If you replace "$=0$" with "$\in h$" in $(2)$ (and, as said before, replace "ideal" with "subalgebra") then indeed $(1)$ and $(2)$ together would suffice to describe a Cartan subalgebra in your setting. $\endgroup$ – Torsten Schoeneberg Feb 15 at 19:20

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.