TITLE: A semisimple Lie algebra has finitely many representation of a given finite dimension.
QUESTION [5 upvotes]: Let $\mathfrak{g}$ be a finite dimensional, semisimple Lie algebra, with Cartan subalgebra $\mathfrak{h}$. I want to show that $\mathfrak{g}$ has finitely many representations of dimension $n$, up to isomorphism, where $n \in \mathbb{N}$.
I know that finite dimensional representations of $\mathfrak{g}$ are the direct sum of their weight spaces. But since $\mathfrak{h}$ is finite dimensional, also $\mathfrak{h}^*$ is, thus there are only finitely many ways we can decompose a representation $V$, of dimension $n$, as the sum of its weight spaces.
I think that two representations with same root space decomposition are isomorphic,
thus concluding the argument, but I'm not quite sure how to prove it.

REPLY [4 votes]: For finite-dimensional repns, we know there is a unique highest weight (and weight space). By irreducibility, this weight space generates the whole repn. In particular, there is a chain of lowering operators (actions by elements in the negative weight spaces) taking the highest weight vector to the lowest weight vector (which we know has the negative of the weight of the highest). Thus, the dimension of the whole is certainly at least the number of steps it takes to get from highest to lowest weight, which is something like $\sum_i |\lambda_i|$ where (in some normalization) the $\lambda_i$ are the coordinates of the highest weight $\lambda$. Since highest weights are "integral" and "dominant", there are only finitely many with $\sum_i |\lambda_i|$ below a given bound.