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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?

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    $\begingroup$ You might want to look at the Witten zeta function (which I learnt of in the work of Uri Onn) which is a generating series for these dimensions. For $\mathfrak{sl}_2$ one recovers the Riemann zeta function. $\endgroup$
    – Geordie Williamson
    Commented Feb 5, 2023 at 20:46
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    $\begingroup$ but to your question, of course one needs to assume semi-simple (as you do). Then I think it should be clear (??) from Weyl dimension formula that there are finitely many of a fixed dimension, but I'm not sure of any reasonable bound. $\endgroup$
    – Geordie Williamson
    Commented Feb 5, 2023 at 20:48

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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.

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    $\begingroup$ Making the change of variables $x=n_1+1$, $y = n_2+1$, you are looking for solutions to $xy(x+y) = 2d$. This has come up several times before on MO -- see mathoverflow.net/questions/230712 mathoverflow.net/questions/51193 mathoverflow.net/questions/50661 $\endgroup$
    – David E Speyer
    Commented Feb 6, 2023 at 4:43
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    $\begingroup$ In particular, in the comments at mathoverflow.net/questions/230712 , alpoge raises a good point: We can probably find a $d$ such that the elliptic curve $xy(x+y) = 2d$ has positive rank. So there are infinitely many rational solutions to $xy(x+y) = 2d$. Let $M$ be the product of the denominators of $N$ of those rational solutions. Then we get $N$ integer solutions to $xy(x+y) = 2dM^3$. (Too bad alpoge didn't actually write down an example with positive rank though.) $\endgroup$
    – David E Speyer
    Commented Feb 6, 2023 at 4:47
  • $\begingroup$ Using the lower bound in the paper joro cites in his answer to mathoverflow.net/q/230740, it follows that there is a positive constant $c$ such that there are infinitely many positive integers $m$ such that $xy(x+y)=m$ has at least $c(\log m)^{1/2}$ integer solutions $(x,y)$. Since $m$ has to be even (else there are no integer solutions $(x,y)$), we only have to make sure that are enough solutions with $x$ and $y$ positive. But since $(x,y)$ is a solution if and only if $(x, -x-y)$ and $(-x-y,y)$ are solutions, there are at least $c/3(\log m)^{1/2}$ solutions $(x,y)$ with $x,y>0$. $\endgroup$
    – Robert Bryant
    Commented Jun 10, 2023 at 14:22

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