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In Humphreys Lie Algebra text, before performing root space decomposition it is required to pick a maximal toral subalgebra of the semisimple lie algebra in question. By maximal, he means not properly contained in any other toral subalgebra.

Why must the toral subalgebra $\mathfrak{h}$ be maximal? What fails if we just find an arbitrary semisimple element $x$ of the lie algebra and set $\mathfrak{h}={\rm span}\{x\}$?

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  • $\begingroup$ Just read on and the maximality condition will get used, and this ought to be clearly noted. Since roots are linear forms on $\frak h$, what is likely to go wrong when taking a non maximal subalgebra is that you lack elements to tell the roots apart (so roots spaces might have higher dimension than$~1$.) $\endgroup$ Oct 7, 2016 at 19:15
  • $\begingroup$ @MarcvanLeeuwen I definitely agree that we can end up with root spaces of higher dimension than $1$, but why is that a problem? $\endgroup$
    – qftey
    Oct 7, 2016 at 19:28
  • $\begingroup$ @MarcvanLeeuwen So maximality is needed to show that the subalgebra is self-centralising. Now I'm investigating why it not being self-centralising is bad. $\endgroup$
    – qftey
    Oct 7, 2016 at 19:56
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    $\begingroup$ Then $\mathfrak{h}$ would not be a Cartan subalgebra. You could take, say, the case of type $A_n$ to see what this would mean. $\endgroup$ Oct 9, 2016 at 14:23
  • $\begingroup$ The question is an exact duplicate of math.stackexchange.com/q/1554766/96384 -- and I shamelessly advertise my answer there for all the things that can happen in the weight decomposition for non-maximal toral $\mathfrak{h}$. Summary: In general the non-zero weights of $\mathfrak{h}$ acting on the LA don't form a root system anymore (and the weight spaces have higher dimensions). The most striking example for all that of course is Dietrich Burde's $\mathfrak{h}=0$. $\endgroup$ Apr 28, 2018 at 17:25

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What we need often is that the maximal total subalgebra is self-normalising, i.e., that it agrees with the Cartan subalgebra in the semisimple case. If we do not have this, then we do not have a Cartan subalgebra. A trivial example would be to take $\mathfrak{h}=0$, as $0$ is a toral subalgebra (it contains no nonzero nilpotent elements).

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