For a semisimple complex Lie algebra $\frak{g}$ it is well known that irreducible finite-dimensional representation are not characterised by their dimension.


More formally, let us define an equivalence relation on dominant weights by $\lambda ~ \mu$, for $\lambda, \mu \in \mathcal{P}^+$, is it holds that
$$
\mathrm{dim}(V\_{\lambda}) = \mathrm{dim}(V\_{\mu}).
$$
As just mentioned, classes can in general have more than one element. Is there an upper bound on the number of elements a class can have, or can one find classes with arbitrarily many elements?



Number of representations of a semisimple Lie algebra of any given dimension
For $\mathfrak{sl}\_2\times \mathfrak{sl}\_2$, the number of irreps of dimension $n$ is the number of factorizations $n=n\_1n\_2$ (you tensor the irreps of the two $\mathfrak{sl}\_2$'s), so there's no upper bound.


For $\mathfrak{sl}\_3$, the Weyl dimension formula says that these dimensions are $\frac{1}{2}(n\_1+1)(n\_2+1)(n\_1+n\_2+2)$. For a fixed dimension, $\frac{1}{2}(n\_1+1)(n\_2+1)(n\_1+n\_2+2)=d$.  

Completing the square in one variable, this is $(n\_1+1+\frac{1}{2}(n\_2+1))^2=\frac{2d}{n\_2+1}+(\frac{1}{2}(n\_2+1))^2$
which implies that $n\_1+1=-\frac{1}{2}(n\_2+1)\pm\frac{1}{n\_2+1}\sqrt{\frac{2d}{n\_2+1}+(\frac{1}{2}(n\_2+1))^2}$. Seems pretty unlikely that one can find a $d$ that gets you a bunch of integral RHSs for n\_2 integral, but maybe I'm missing something.


