In some lecture notes of mine we define a Cartan subalgebra $\mathfrak h$ for semisimple $\mathfrak g$ as an abelian subalgebra of $\mathfrak g$ containing ad-diagonizable elements which are maximal.

It then says that for more general Lie algebras $\mathfrak g$, a Cartan subalgebra is defined as a self-normalising nilpotent subalgebra. It then goes on to say that this is automatically maximal among nilpotent subalgebras.

My question is how would one show this (that it's maximal)? It says automatically but I can't see how it follows. Thanks in advance.


That's a consequence of

Engel's Theorem: If $V\neq 0$ is a finite-dimensional representation of a Lie algebra ${\mathfrak g}$ such that any $X\in{\mathfrak g}$ acts nilpotently on $V$, then $0\neq V^{\mathfrak g} := \{v\in V\ |\ X.v=0\text{ for all } X\in{\mathfrak g}\}$.

Now suppose that ${\mathfrak h}\subsetneq {\mathfrak t}$ are two nilpotent subalgebras of your given algebra ${\mathfrak g}$. Then ${\mathfrak h}$ acts on ${\mathfrak t}/{\mathfrak h}$ by the adjoint action (this is not meant to say that ${\mathfrak t}/{\mathfrak h}$ inherits a Lie algebra structure, and we do not need to assume that ${\mathfrak h}$ is an ideal here), and by the nilpotency of ${\mathfrak t}$, the hypothesis of Engel's Theorem is satisfied. Hence you find some $0\neq\overline{T}\in{\mathfrak t}/{\mathfrak h}$ such that $0=[\overline{H},\overline{T}]=\overline{[H,T]}$ in ${\mathfrak t}/{\mathfrak h}$ for all $H\in{\mathfrak h}$ - in other words, $T\in{\mathfrak t}\setminus{\mathfrak h}$ but $[H,T]\in{\mathfrak h}$ for all $H\in{\mathfrak h}$. Hence $T\in{\mathfrak n}_{\mathfrak g}({\mathfrak h})\setminus{\mathfrak h}$.

  • $\begingroup$ You need H to be an ideal of t to do that, no? $\endgroup$ – Mariano Suárez-Álvarez Jan 8 '15 at 6:26
  • $\begingroup$ @MarianoSuárez-Alvarez: Mh, so far I don't see where this would be necessary: since ${\mathfrak h}\subset{\mathfrak t}$, the bracket with elements from ${\mathfrak h}$ preserves both ${\mathfrak h}$ and ${\mathfrak t}$, hence descends to ${\mathfrak t}/{\mathfrak h}$ - or did I make a mistake somewhere else? $\endgroup$ – Hanno Jan 8 '15 at 7:16
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    $\begingroup$ @LSpice: I am not claiming that ${\mathfrak g}/{\mathfrak h}$ inherits a Lie algebra structure, but only that it admits a natural ${\mathfrak h}$-action by the usual bracket. For this, one only needs that $[{\mathfrak h},{\mathfrak h}]\subset{\mathfrak h}$ and $[{\mathfrak h},{\mathfrak t}]\subseteq {\mathfrak t}$. $\endgroup$ – Hanno Aug 3 '15 at 4:56
  • $\begingroup$ @Hanno, yes, you are right, and I misread. I apologise, and will delete my comment. $\endgroup$ – LSpice Aug 4 '15 at 1:36

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