3
$\begingroup$

Let $\mathfrak{g}$ be a semisimple lie algebra. Let $\alpha_1,.., \alpha_n$ form a simple root system, and let the corresponding system of fundamental weights be $\lambda_1,..,\lambda_n$. I have been told that $\frac{1}{2} \sum_{\alpha \in \Delta^+} \alpha = \lambda_1 + ... + \lambda_n$.

However, I have difficulty in seeing why this can be true. For example, in the sl(3) case, the positive roots are $e_1 - e_2, e_2 - e_3, e_1 - e_3$, where $e_i \in \mathfrak{h}^*$ is the functional which pulls out the $i$'th diagonal element. $\frac{1}{2} \sum_{\alpha \in \Delta^+} \alpha = e_1 - e_3$. However, in this case, if we take the simple root system $\alpha_1 = e_1 - e_2, \alpha_2 = e_2 - e_3$, then $\lambda_i = e_1 + .. + e_i$, and so $\lambda_1 + \lambda_2 = 2e_1 + e_2$. Where does my reasoning go wrong in the sl(3) case?

And why is it true that the sum of fundamental weights is equal to half the sum of positive roots?

$\endgroup$
3
$\begingroup$

Nothing is wrong with your reasoning for sl(3). Remember that in sl(3), there is the additional relation $e_1 + e_2 + e_3 = 0$, because the trace of a matrix in sl(3) is 0. So $e_1 - e_3 = 2e_1 + e_2$.

$\endgroup$
  • $\begingroup$ ah great thank you! I can look up the proof myself - what you clarified was the main point of confusion! $\endgroup$ – 902 Feb 19 '18 at 18:00

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.