How to compute the dimension of a representation of a Lie algebra from its weights?

Question

For some "short" representations of Lie algebras is simple to find their dimensionality by looking at how many weights appear in them. The $(1,0)$ of $A_2$ is immediate to see that it has 3 weight vectors so it must be called a triplet. Less obvious if one takes the $(1,1)$ of $A_2$, which has 7 distinct weight vectors, but somehow $(0,0)$ counts twice. Looking at other small rank algebras, like $G_2$ or $B_2$ the "plot thickens", as the there can easily be a big difference between the distinct weight vectors and the dimension of the representation, e.g. $(2,0)$ of $G_2$ has 19 distinct but is $d=27$.

So, what is the way to figure how many times a weight vector must be counted to determine the dimension of the representation of a generic Lie algebra?

Counting different weights in representations of Lie algebras is about An I am asking for any Lie algebra

Answer 1

Of course there are plenty of people more competent than me on this topic. However, I can suggest you Chapter VI of the wonderful text "Introduction to Lie Algebras and Representation Theory" of J. E. Humphreys. If I well remember, if $V(\lambda)$ is the irreducible representation of highest dominant integral weight $\lambda$ of a simple Lie algebra $L$, then the corresponding set of weights $\Pi(\lambda)$ is invariant under the action of the Weyl group. Conjugate weights correspond to weight spaces of same dimension. Thus, in principle you can compute $\text{dim}\, V(\lambda)$ by computing the cardinality of such orbits and by computing the dimension of the weight spaces for each conjugation class.

Answer 2

Ah sorry, I forgot to properly answer to your question! In the case of above, the weight space $V_\lambda$ associated to the highest dominant integral weight $\lambda$ is one dimensional, but this is not true for the other weights! The result you are looking for is the Freudenthal multiplicity formula. The dimension of all the $L$-module $V(\lambda)$ is given by the Weyl dimension formula.

Answer 3

As mentioned by Lorenzo, the easiest way to compute the dimension of an irreducible representation $V(\lambda)$ is the Weyl dimension formula, which expresses the dimension as a product over the positive roots.

There are some particularly nice cases when the dimension can be read directly from the weights. Namely, your case of the $V(\varpi_1)$ representation for $A_2$ is an example of a minuscule representation. These have all weights being in the same Weyl group orbit, and thus each has dimension one so that number of weights is the dimension of the representation. For $A_n$, the minuscule representations are the fundamental representations. For other types, lists of minuscule highest weights are easily found in the literature.

There are other examples of representations where each weight space is one dimensional; for example, for $A_n$ and the representation $V(k \varpi_1)$ for any positive integer $k$. There is a nice paper of Berenstein and Zelevinsky, "When is the multiplicity of a weight equal to 1?", which is in this direction.

There are some combinatorial ways which you can get at weight multiplicities exactly, by constructing a representative for each vector--namely, the theory of crystals or crystal bases is such an example. But in terms of ease of computation, the Frudenthal formula for multiplicity or the Weyl dimension formula will be most applicable.