In the book "Symmetries, Lie Algebras and Representations: A Graduate Course for Physicists" by Jürgen Fuchs,Christoph Schweigert the authors write
"In the description of representations, the Dynkin components of a weight play the role of eigenvalues with respect to the generators $H^i$ of the Cartan subalgebra."
They don't explain this any further, thus does anyone understand why this is the case?
In my understanding there are three commonly used bases for the weight space
- The simple root basis $$ r =(a_1,a_2,a_3,\ldots)=a_1 \alpha_1 + a_2 \alpha_2 + a_3 \alpha_3 +\ldots,$$ where $α_i$ denotes the simple roots.
- The fundamental weight basis (=Dynkin basis), $$ w =(b_1,b_2,b_3,\ldots)=b_1 \omega_1 + b_2 \omega_2 + b_3 \omega_3 +\ldots,$$ where $\omega_i$ denotes the fundamental weights. We can change between these two bases using the Cartan matrix and its inverse.
- The H-basis, where we write each weight or root in terms of the eigenvalue of the Cartan generators $H_i$, when they act on them:
$$ H_i r = \lambda_i r \rightarrow r=( \lambda_1, \lambda_2, \lambda_3, \ldots ) $$
What the authors say in the quote above is basically that the second and third basis are the same.
Is it because we have some freedom in choosing a basis for the Cartan subalgebra? In other words, are the second and third basis in my list simply different choices of a basis for the Cartan subalgebra? If yes, why?
Weights of a representation of a Lie algebra are elements of $\mathfrak{h}^*$, i.e. linear functional from the Cartan Subalgebra $\mathfrak{h}$ to $\mathbb{C}$.
The two bases for $\mathfrak{h}^*$ play different roles in the representation theory of the Lie algebra. The simple roots (or roots in general) have to do with the root space decompostion of $\mathfrak{g}$: if $X\in\mathfrak{g}_{\alpha}$ and $H\in\mathfrak{h}$, then $$[H,X]=\alpha(H)X.$$ The fundamental weights are coordinate functions: $$\omega_i(H^j)=\delta_{ij}.$$
Now, if $r$ is a weight vector of weight $\lambda=\lambda_1\omega_1+\cdots+\lambda_n\omega_n$, then an element $H\in\mathfrak{h}$ acts on $r$ by $H.r=\lambda(H)r$. In particular, since $\lambda(H^i)=\lambda_i$, and the $H^i$ form a basis for $\mathfrak{h}$, we get an identification of (2) and (3) above.