Obtaining Weights of Tensor Product representations of Lie Algebra

Question

While studying representations of Simple Lie Algebras the procedure is quite straight forward.

1. Find the Cartan Sub-Algebra
2. Label states with the corresponding eigenvalues
3. Use the remaining generators to construct SU(2) blocks and use them as raising/lowering operators.

This leds to the notion of roots and in general it's straight forward to find the corresponding root structure or decomposition for the classical groups.

Having all this information (root system, Cartan Matrix, and so on) one can in principle use them to construct new representations of the algebra. My question is the following, given all the information regarding the roots of a classical Lie algebra how can I build the weights of representation with fixed symmetry properties. For example say you are in SP(N) and you know all the root system and Cartan Matrix, how do you build the Weights corresponding to the antisymmetric representation?. Is there a systematic way to do so?.

Any bibliography will be appreciated!

Answer 1

It's actually very straightforward. Let's go through a simple example. Suppose $V$ is a $3$-dimensional representation of your Lie algebra, and suppose that $V$ has the basis $$v_1, v_2, v_3, $$ where $v_1$, $v_2$, $v_3$ each lie in a weight space, and the corresponding weights are $w_1$, $w_2$, $w_3$ respectively. (So for any $H$ in the Cartan subalgebra, we have $H(v_i) = w_i(H) v_i$ for $i \in \{ 1, 2, 3 \}$.)

Then the symmetric tensor product $Sym^2 (V)$ is a six-dimensional representation, with basis $$ v_1 \otimes v_1, v_2 \otimes v_2, v_3 \otimes v_3, \tfrac 1 2 (v_1 \otimes v_2 + v_2 \otimes v_1), \tfrac 1 2 (v_2 \otimes v_3 + v_3 \otimes v_2) , \tfrac 1 2 (v_3 \otimes v_1 + v_1 \otimes v_3) $$ These basis vectors each lie in a weight space, and the corresponding weights are $2w_1, 2w_2, 2w_3, w_1 + w_2, w_2 + w_3, w_3 + w_1$ respectively.

[In practice, you may find that some of the weights coincide. If this happens, then it means you have degenerate weight spaces (i.e. a weight space corresponding to a given weight has dimension greater than one) - it's not a problem!]

Returning to our example, the antisymmetric tensor product $\wedge^2 (V)$ is a three-dimensional representation, with basis $$ \tfrac 1 2 (v_1 \otimes v_2 - v_2 \otimes v_1), \tfrac 1 2 (v_2 \otimes v_3 - v_3 \otimes v_2) , \tfrac 1 2 (v_3 \otimes v_1 - v_1 \otimes v_3) $$ and these basis vectors each lie in a weight space, and the corresponding weights are $w_1 + w_2, w_2 + w_3, w_3 + w_1$ respectively.

This method generalises to every example you'll ever encounter, and it is as systematic a method as you can hope for!