I'm studying Humphreys's Introduction to Lie Algebras and Representation Theory, and I'm working on the following problem from the book (Chapter 13, problem 8):

Let $\Phi$ be irreducible. Prove that each $\lambda_i$ is of the form $\sum_jq_{ij}\alpha_J$, where all $q_{ij}$ are positive rational number.

Here, $\lambda_i$ is a fundamental dominant weight, and every $\alpha$ is simple. The book gives a hint, too:

Deduce from a previous exercise that all $q_{ij}$ are nonnegative. From $(\lambda_i,\lambda_i) > 0$ obtain $q_{ij}>0$. Then show that if $q_{ij}> 0$ and $(\alpha_j,\alpha_k)<0$, then $q_{ik}>0$.

I've managed to show the first part of the hint, and I can complete the problem using it, but I can't figure out how to show that, if $q_{ij}>0$ and $(\alpha_j,\alpha_k)<0$, then $q_{ik}>0$.

Any help would be really appreciated.


Assume that $q_{ij} > 0$ and $(\alpha_j, \alpha_k) < 0$. We know from Humphreys' problem 7 that $q_{ik} \ge 0$. If $q_{ik} = 0$, then we have $$0 \le \delta_{ik} (\alpha_k, \alpha_k)/2 = (\lambda_i, \alpha_k) = \sum_{m\neq k} q_{im} (\alpha_m, \alpha_k) \le 0$$ hence $q_{im} = 0$ or $(\alpha_m, \alpha_k) = 0$ for all $m\neq k$. This contradicts our assumptions on $m = j$. Hence, it must be that $q_{ik} > 0$.


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.