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It seems to me, I could be wrong, that the toral sub algebra goes against the following rules:

For a semisimple Lie algebra:

If the killing form is nondegenerate the Lie algebra is semi simple-> the toral algebra is nilpotent

If the lie algebra is abelian the killing form is degenerate-> contradicting the previous statement.

Can anyone explain why? And if I am correct, the toral sub algebra of a semisimple Lie algebra is not an ideal.

dylan7
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  • Sorry, what is the contradiction here exactly? If the Lie algebra is abelian then the toral algebra is as well, so it's nilpotent. –  Aug 05 '14 at 21:09
  • What I see as a contradiction is for a semi-simple Lie algebra,not necessarily abelian, it's toral sub algebra is abelian. In addition, the killing form of the toral sub algebra is non-degenerate. Since the killing form is non- degenerate by Cartan's criterion the toral sub algebra should be semi-simple and not abelian, but the toral sub algebra is nilpotent and abelian. What I am asking is how can the toral sub algebra have a non-degenerate killing form, be abelian, nilpotent, and not semi-simple? – dylan7 Aug 05 '14 at 21:27
  • The Killing form of the toral subalgebra is not nondegenerate, indeed the Killing form of any abelian Lie algebra $L$ is identically zero because $\operatorname{ad}(x)$ is the zero map for any $x \in L$. Perhaps you're confusing the Killing form of the torus with the Killing form of the big algebra restricted to the torus; they're different. – Matthew Towers Aug 05 '14 at 22:45
  • Ah yes, that must be it. Could you possibly explain the difference between the killing form restricted to a sub-algebra and the killing form of that sub-algebra? – dylan7 Aug 06 '14 at 01:51
  • You have two Lie algebras in play: one called $L$ and another, $T$, its toral subalgebra, with Killing forms $\kappa_L$ and $\kappa_T=0$ respectively. The restriction of $\kappa_L$ to $T$ means the bilinear form $T\times T \to \mathbb{C}$ (or whichever field you work over) sending $(s,t)$ to $\kappa_L(s,t)$. This is not necessarily the same as $\kappa_T$, and in fact is nondegenerate if $L$ is semisimple. – Matthew Towers Aug 06 '14 at 21:27
  • Ok that makes sense thank you. And another question regarding root space decomposition. If I have a root alpha of an element of the toral sub-algebra, that is by definition equal to the killing form of the basis element of the toral associated with that root with that element you are applying the root functional too. In other words alpha(h)=k(basis element of the toral associated with alpha,h), shouldn't this equal one of the eigenvalues of h? When I try it on sl2 with the adj of its cartan basis element, k(diag(2,0,-2),diag(4,0,-4)), shouldn't this pop out 4? – dylan7 Aug 06 '14 at 21:45

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