I am reading Humphrey's intro to Lie algebra, self-teaching, and have a few questions regarding root space decomp.

1) If I understand this correctly, the toral sub algebra of L represents all semisimple adj reps of L. These are simultaneously diagonalizable, meaning they all must be scalar multiples of one of their members adj L forms, since their adj forns in L are diagonal matrices. The duel basis of this toral subalgebra represents all functionals that when acting on a member of the toral algebra pop out one of the eigenvalues of the adj L matrix the toral members adj forms are all multiples of. The thing I don't understand, unless I am wrong about what I just said, is what is the meaning of the basis of the toral subalgebra, the duel basis of the functionals. I assume it is the element of the toral subalg whose trace of the product with another element of the toral subalg (in adj form) pops out an eigenvalue of the matrices that are simultaneously diagonalizable, meaning an eigenvalue of them two.

2) I don't understand how the toral subalgebra's killing form can be nondegenerate and the subalgebra can be abelian. Since the restriction of the adj rep to the toral subalgebra is zero for all it's members, thus the trace of the prod of their adj forms is zero. But it can't be nondegenerate since L is semisimple.

Thanks for the help


1 Answer 1


In part 2, it is not the Killing form of the maximal toral subalgebra (call it $H$) which is nondegenerate, but the restriction of the Killing form of the whole algebra $L$ to $H$. That is, you look at $Tr(ad_\color{red}L(x)\circ ad_\color{red}L(y))$, for $(x,y) \in H\times H$, not at $Tr(ad_\color{red}H(x)\circ ad_\color{red}H(y))$. The latter, which describes the Killing form of $H$, would indeed be $=0$ as for any abelian Lie algebra.

For part 1, I don't quite understand what you are asking. "What is the meaning of the basis of the toral subalgebra, the du[a]l basis ..." -- do you mean "how is it defined?", or "what is its use?", or something else?


You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .