Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Let $\lambda$ be a weight and $P^+$ be the set of all positive roots. Define $wt(\lambda) = \{ w(\mu) \mid w\in W, \mu \in P^+, \mu \leq \lambda \}$, where $W$ is the Weyl group of a Lie algebra. So here $wt(\lambda)$ is a set of weights. In some other place, the notation $\sum_{\mu} dim(V_{\mu})e^{wt(\mu)}$ is used. I am confused with this since $wt(\mu)$ is a set. What does $e^{wt(\mu)}$ mean?

share|improve this question
For notational issues, it is generally a good idea to include a reference to where you saw the notation used. Because in context it could be any of (a) inconsistent notation between different authors (b) inconsistent notation by the same author (c) typographical error. –  Willie Wong Jul 5 '11 at 15:08
@Willie, The lecture notes I read does not have a electric version. But it seems the notation is used often. –  LJR Jul 5 '11 at 15:12

1 Answer 1

up vote 2 down vote accepted

This is mostly educated guessing. Standard caveats apply.

It looks like the author wants $wt(\lambda)$ to be the set of weights of an irreducible module with highest weight $\lambda$. In other words I am gambling that there is a typo, and $\mu$ should really range over the set of dominant weights subject to $\mu\le\lambda$, as opposed to the set of positive roots. It is hard to think of a setting, where one would want to use this kind of a definition to describe a set of roots. It is not impossible, but unlikely, because if $\mu$ were meant to be a root as opposed to a weight, the set $wt(\lambda)$ would be (in the simple case) either empty, consist of the short roots, or of all the roots, but there would be a lot of redundancy there. I need a little bit more context to tell, whether that is a live possibility.

The latter formula looks a little bit suspicious, too. Apparently the author wants to write a formula for the formal character of this representation, but I can't be sure, whether he/she wants to use the symmetry under the Weyl group or not. If in that summation $\mu$ ranges over the dominant weights only, then I would hazard a guess that $e^{wt(\mu)}$ really means $\sum_{\mu'\in W(\mu)}e^{\mu'}$, i.e. a formal sum over the orbit of $\mu$ under the Weyl group action. If the summation is over the set of all the weights, then I would expect to see $e^\mu$ only. It is a bit confusing, so I really think that the common malady of haste combined with lack of proofreading is what you are witnessing. Lecture notes are not always subject to the same level of doublechecking than a textbook would. Don't I know it :-(

Here $e^\mu$ denotes an element of the group ring of the additive free abelian group of weights, but you probably knew that.

[Edit] I do not see a way of interpreting $wt(\mu)$ the same way in both these occurrences.[/Edit]

share|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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