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I have been reading di Francesco's Conformal Field Theory section on Lie algebras (chp 13, pg 491). Following their notation, let $[\cdot,\cdot]$ be the multiplication in the Lie algebra. The Cartan-Weyl basis first chooses the generators $\{H^i\}$ where $[H^i, H^j] = 0$, the maximal abelian subalgebra $H$ of the Lie algebra. The remaining generators $\{E^a\}$ are chosen such that $[H^i, E^a] = \alpha^i E^a$.

Now, they claim that because $H$ is the maximal abelian subgroup, the roots are non-degenerate. I have been trying to prove this claim to no avail. I have tried to assume the contrary and construct a new element of $H$ to contradict the maximality of $H$, but have been unsuccessful. How does one prove the statement that the roots $\alpha^i$ are non-degenerate?

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    $\begingroup$ Would Mathematics be a better home for this question? $\endgroup$
    – Qmechanic
    Oct 23, 2017 at 21:40

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I think I prove this in part (viii) of the discussion of roots starting on page 246 of my lecture notes:

https://courses.physics.illinois.edu/phys509/sp2017/bmaster.pdf

I don't recall explicitly using the maximality of the Cartan algebra, so maybe I fudged somewhere!

Mmm... maybe Im answering the wrong question. I's not the $e_\alpha$ that you want to show unique for a given $\alpha$, but the $\alpha$'s themselves.... but I think it boils down to the same thing.

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  • $\begingroup$ I think it is the same thing. The maximality of the algebra indicates the root vectors are "rank n"-dimensional and should differ by at least one component... $\endgroup$ Oct 23, 2017 at 23:46

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