Can someone show that the roots and the Cartan subalgebra are dual vector spaces?

I don't see how simple roots acting on non-corresponding indices of a Cartan basis produce 0 and a simple root evaluated on its corresponding Cartan basis element equals 1.

According to the definition of dual spaces: $\beta (t_\beta)=1$ and $\beta (t_\alpha)=0$, where ${\alpha,\beta } $ (simple roots) corresponds to ${t_\alpha, t_\beta} $ (Cartan basis) respectively.

If this was true then the inner product of two simple roots would be zero. Since the inner product of roots uses the restriction of the killing form to the Cartan, and so does the root evaluation on a Cartan element.

So $(\alpha,\beta)=k (t_\alpha, t_\beta)_{H\times H}= \alpha (t_\beta)=\beta (t_\alpha)$

Where $H$ is the max Cartan subalgebra.

Since $\beta$ is be being evaluated on an uneven indice ($ t_\alpha $), as above, $\beta (t_\alpha) $ is 0, which is not true for most root systems. In most root systems the inner product of simple roots is less than 0. Also $\beta (t_\beta)$ is not always 1.

So I am confused how the dual space axioms apply here.

In addition, I see how the Cartan basis is orthonormal or at least orthogonal under its own killing form, but not under the Lie algebra's killing form restricted to the Cartan, which is non-degenerate.

I am probably confusing a bunch of stuff.

Help is really appreciated

  • 1
    $\begingroup$ You might be confusing dual spaces and their dual bases. It's entirely possible for two vector spaces to be dual to each other while the bases that you happen to have in mind for those two spaces are not dual bases. $\endgroup$ – Andreas Blass Aug 14 '14 at 16:26
  • $\begingroup$ How are dual bases chosen? Does it depend on bilinear form you are using to define the operation of the functional on the original vector space? $\endgroup$ – dylan7 Aug 14 '14 at 21:46

First you have to realize that the roots are linear functions ($V$) that take elements in some vector space $V$ and produce (bijectively) elements in $\mathbb C$ (complex plane, however we can restrict it to real plane). I encourage to you read the following notes of Robert N. Cahn. In particular pages 15 and 19. There you will find in detail a very clear answer to your question:



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