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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

  1. 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.
  2. 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.
  3. 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?

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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.

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  • $\begingroup$ Thanks a lot for your answer. A short question to make sure I understand you. The crucial point, I think, is why $H_i r = \lambda_i r \rightarrow \lambda=( \lambda_1, \lambda_2, \lambda_3, \ldots ) \leftrightarrow \tilde \lambda_1\omega_1+\cdots+ \tilde \lambda_n\omega_n $. In general, the coefficents in the $H$ and in the Dynkin basis are very different, i.e. $\tilde \lambda_i \neq \lambda_i$. For example, the $3$ rep of $SU(3)$ reads in the Dynkin basis: $$ \left( \begin{array}{cc} 1 & 0 \\ -1 & 1 \\ 0 & -1 \\ \end{array} \right)$$ $\endgroup$ – jak Aug 26 '15 at 10:42
  • $\begingroup$ and in the $H$-basis: $\left( \begin{array}{cc} \left\{\frac{1}{2}\right\} & \left\{\frac{1}{2 \sqrt{3}}\right\} \\ \{0\} & \left\{-\frac{1}{\sqrt{3}}\right\} \\ \left\{-\frac{1}{2}\right\} & \left\{\frac{1}{2 \sqrt{3}}\right\} \\ \end{array} \right)$. Are these simply two different choices for the basis of the Cartan subalgebra? $\endgroup$ – jak Aug 26 '15 at 10:43
  • $\begingroup$ I found the answer myself. Yes its just a different basis for the Cartan subalgebra, known as Chevalley Basis $\endgroup$ – jak Aug 26 '15 at 14:31
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The Dynkin basis is the basis where we label the weights in terms of the Eigenvalues of the Cartan generators in the Chevalley basis.

Or in other words

The Dynkin labels are the eigenvalues of the Chevalley generators of the Cartan subalgebra

The $H$-basis and the Chevalley basis are simply two different choices for the basis for the Cartan subalgebra.

The Dynkin basis is often more useful, because there each basis element of the Cartan subalgebra corresponds to a simple root.

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