# How to draw a weight diagram?

Given a weight, say $\omega=3\lambda_1+4\lambda_2$, where $\lambda_1, \lambda_2$ are fundamental weights (type A Lie algebra). How to draw the weight diagram of the irriducible representation with heighest weight $\omega$ (for example, the one in chapter 6 of the book introduction to Lie algebras and representation theory)? How to compute the multiplicity (dimension of weight spaces) quickly (as shown in the diagram)? How can we compute the multiplicity directly (find a basis of the weight space)? How to compute the multiplicity using Weyl formula or other formulas? Thank you.

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You have a lot of questions here, and maybe someone will come along and answer them all systematically for your specific example(though you don't tell us what n is), but let me just say this: take a look at chapter 12 of Fulton and Harris, books.google.com/…, I think that reading through this example will make this more clear. – BBischof Jul 6 '11 at 18:38

Using the Weyl formula is easy using figures like the one you mention: You start building the multiplicity table from the highest weight $\lambda$. It always happens that if you subtract a (legally low) positive integer multiple of a simple root from the highest weight, the multiplicity of the weight space is equal to 1. You can prove this using the Weyl formula as well, but that's overkill (this fact is also Exercise #1 of that chapter). I will show you, how you get that 2 for the weight $\mu=\lambda-\alpha_1-\alpha_2$ using Weyl's formula, and leave the rest to you. This is for type $A_2$, so the orbit of $\delta$ (=half-sum of positive roots) consist of the roots. The element $q=\sum_{w\in W}(-1)^{\ell(w)}\epsilon_{w(\delta)}=\epsilon_{\alpha_1+\alpha_2}-\epsilon_{\alpha_1}-\epsilon_{\alpha_2}+$ corresponding terms involving negative roots.
The key to computing the multiplicity $c(\mu,\lambda)$ of $\mu$ in this irreducible representation is to study the coefficient of $\epsilon_{\mu+\delta}=\epsilon_{\lambda}$ in the product $q*ch_\lambda$. Because $\mu+\delta$ is not in the $W$-orbit of $\lambda+\delta$, Weyl's formula tells us that this coefficient is equal to zero. On the other, this coefficient clearly is $c(\mu,\lambda)-c(\mu+\alpha_1,\lambda)-c(\mu+\alpha_2,\lambda)$, because the weights $\mu$, $\mu+\alpha_1$ and $\mu+\alpha_2$ are the only weights of this module that differ from $\lambda$ by an element of $W(\delta)$. So we get an equation $$c(\mu,\lambda)-c(\mu+\alpha_1,\lambda)-c(\mu+\alpha_2,\lambda)=0.$$ Earlier (when working with the outer layer of weights) we showed that $c(\mu+\alpha_1,\lambda)=c(\mu+\alpha_2,\lambda)=1,$ so we get $c(\mu,\lambda)=2$ as claimed.
This process is recursive: start from the the highest weight, and work towards zero. Use symmetry under $W$ heavily along the way. Or even more simply: Keep the multiplicities as unknowns, and write Weyl's formula as a system of linear equations. Solve that system. When viewed in this order (from the highest dominant weight to the lowest), the matrix of coefficients is triangular, so the system is easy to solve.