# Necessity of diagonalizability of adjoint representation of cartan subalgebra in definition

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?

• $h$ 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? – Torsten Schoeneberg Feb 15 at 8:55
• @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. – user45765 Feb 15 at 16:12
• $(2)$ 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. – Torsten Schoeneberg Feb 15 at 19:20