# Lie algebra representations and tensor product decompositions.

Find the weights for $V_{L_1 - 2L_3}$, where $L_1, L_2, L_3$ are the weights for the standard representation of $\mathfrak{sl}_3 \Bbb{C}$ on $V \cong \Bbb{C}^3$.

In order to find these weights, we look at $\text{Sym}^1 V \otimes \text{Sym}^2 V^{*}$. We know that the weights for $\text{Sym}^1 V$ are just $L_1, L_2, L_3$ and those for $\text{Sym}^2 V^{*}$ are $\{-2L_i, -L_i-L_j\ | i,j=1,2,3, \space i\not=j\}$. So the weights for $V_{L_1 - 2L_3}$ must be the sum of $\{L_1, L_2, L_3\}$ and $\{-2L_i, -L_i-L_j\ | i,j=1,2,3, \space i\not=j\}$, right? Is my answer correct?

• I'm not sure if your answer is correct or not because I'm not sure what $V_{L_{1} - 2L_{3}}$ means in your problem. Is it the irreducible representation of $\mathfrak{sl})_{3}$ with highest weight $L_{1} - 2L_{3}$? If it is the highest weight module, then the general recipe is to take the orbit of the highest weight under the Weyl group and then take the intersection of the convex hull of this orbit with the weight lattice. I can elaborate on what I'm talking about if you clarify your question. – Siddharth Venkatesh Dec 2 '14 at 9:13
• @SiddharthVenkatesh Thanks. Yes it is the representation with highest weight $L_1 - 2L_3$. – Artus Dec 2 '14 at 9:58
• Ok. The answer you obtained is not correct then because $V_{L_{1} - 2L_{3}}$ is not the tensor product of the symmetric power representations. I will elaborate in a little bit on what I meant with my recipe. I also have a slight correction. Where I said weight lattice, it should be intersection with translate of the root lattice. – Siddharth Venkatesh Dec 2 '14 at 10:49

As I mentioned in the comments, the method to find the weights of an irreducible representation with highest weight $\lambda$ is:

1. Take the orbit of $\lambda$ under the Weyl group. Since the set of weights of any representation is preserved by the Weyl group, all these weights are in the weight space of $V_{\lambda}$.

2. For an irreducible representation, the set of weights form an unbroken chain. This comes from the representation theory of $\mathfrak{sl}_{2}$. What this means in practice is that if you take the convex hull of the weights obtained in step 1, then any weight in this convex hull that is an integer multiple of roots away from $\lambda$ must be in the set of weights of $V_{\lambda}.$

So, let's apply this to your particular problem. One important thing to remember is that in $\mathfrak{sl}_{3}$, $L_{3} = -L_{1} - L_{2}$. Now, our Weyl group is $S_{3}$, with two generators $s_{1}, s_{2}$ and the elements are $1, s_{1}, s_{2}, s_{1}s_{2},s_{2}s_{1}, s_{1}s_{2}s_{1}$ where

$$s_{1}(L_{1}) = L_{2}, \; s_{1}(L_{2}) = L_{1},\; s_{1}(L_{3}) = L_{3}$$

$$s_{2}(L_{1}) = L_{1}, \; s_{2}(L_{2}) = L_{3} = -L_{1} - L_{2}, \; s_{2}(L_{3}) = L_{2}$$

You can find this description of the action in Fulton and Harris' book I think but most introductory references will have it. Basically $S_{3}$ acts by permuting $L_{1}, L_{2}, L_{3}$ as you would expect.

Now, you can compute the orbit of $L_{1} - 2L_{3}$ under the action of the Weyl group, this will give the weights:

$$L_{1} - 2L_{3}, L_{2} - 2L_{3}, L_{1} - 2L_{2}, L_{2} - 2L_{1}, L_{3} - 2L_{2}, L_{3} - 2L_{1}.$$

The final step is to take all weights in the convex hull of these six weights (which form a hexagon in the $L_{1}, L_{2}$ plane once you replace $L_{3}$ by $-L_{1}-L_{2}$) that are integer multiples of $L_{1} - L_{2}, L_{2} - L_{3}$ away from $L_{1} - 2L_{3}$. I will leave that computation to you. The easiest way to do it might be to plot the weights on the $L_{1}, L_{2}$ plane.