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For the special linear algebra $\frak{sl}_{m}$ which finite dimensional irreducible representations $V_{\mu}$ have non-trivial zero weight spaces?

For $\frak{sl}_2$ this is clear: $V_{2k\pi}$ for $\pi$ the fundamental weight.

Do we have such a description in higher order, even for $\frak{sl}_3$?

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  • $\begingroup$ @SamHopkins: That is quite good! So for $\mathfrak{sl}_3$ can we describe the root lattice in terms of the fundamental weights $\pi_1$ and $\pi_2$? $\endgroup$ Nov 9, 2023 at 17:18
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    $\begingroup$ Ok, so the $i$-th column in the Cartan matrix gives the coefficients of $\alpha_i$ in terms of the fundamental weights $\pi_j$? $\endgroup$ Nov 9, 2023 at 17:27
  • $\begingroup$ If you put this as an answer I am happy to accept it. $\endgroup$ Nov 9, 2023 at 17:31
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    $\begingroup$ Even more strongly, we can ask what is the action of the the Weyl group $S_m$ on the 0-weight space. One reference is David Gay, jstor.org/stable/44236121. $\endgroup$ Nov 9, 2023 at 19:55

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At the request of the question-asker I am converting my comments to an answer.

In general, for any semisimple Lie algebra $\mathfrak{g}$, is $\mu$ is a (dominant, integral) weight of $\mathfrak{g}$ and $\nu$ is another weight, then the $\nu$ weight space in the irreducible representation $V^{\mu}$ is nonzero if and only if $\mu-\nu$ is a nonnegative integral combination of simple roots.

This means $V^{\mu}$ has a nonzero $0$-weight space if and only if $\mu$ is a nonnegative integral combination of simple roots. Since $\mu$ is dominant (and the cone spanned by the simple roots is contained in the cone spanned by the fundamental weights), the "nonnegative" part is automatic. So what matters is whether $\mu$ belongs to the $\mathbb{Z}$-span of the simple roots, or in other words, the root lattice.

So to check for a particular $\mu$ if it satisfies this condition, you need to ask if it is an integral combination of the simple roots $\alpha_i$, which means you need to understand how the $\alpha_i$ are expressed in the fundamental weights $\pi_j$.

In Type A at least, these coefficients are the columns of the Cartan matrix. For instance, in $\mathfrak{sl}_3$ we have $\alpha_1=2\pi_1-\pi_2$ and $\alpha_2=-\pi_1+2\pi_2$. In other types the same should be more-or-less true, but there might be an issue with duality, i.e., roots vs. co-roots. Since Type A is simply laced, no issue like that arises.

All of this should be in any standard source on the representation theory of simple Lie algebras, e.g., Bourbaki or Humphreys.

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    $\begingroup$ It's probably worth saying what this looks like in terms of the standard partition language. Weights of $\mathfrak{sl}_m$ are often described using $m$-tuples $(\mu_1, \mu_2, \ldots, \mu_m)$ modulo $\mathbb{Z}(1,1,\ldots, 1)$. (Concretely, $\mu$ corresponds to the character $\sum \mu_j t_{jj}$ of the Cartan algebra of diagonal matrices in $\mathfrak{sl}_m$. Since $\sum t_{jj}=0$, we quotient by $\mathbb{Z}(1,1,\ldots, 1)$.) The simple root $\alpha_j$ is $e_j - e_{j+1} \in \mathbb{Z}^n/\mathbb{Z}(1,1,\ldots, 1)$; the fundamental weight $\pi_j$ is $e_1+e_2+\cdots + e_j$. (continued) $\endgroup$ Nov 9, 2023 at 18:30
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    $\begingroup$ In this language, $(\mu_1, \mu_2, \ldots, \mu_m)$ is a positive combination of the $\pi_j$ iff $\mu_1 \geq \mu_2 \geq \cdots \geq \mu_m$, and $(\mu_1, \mu_2, \ldots, \mu_m)$ is an integer combination of the $\alpha_j$ iff $\sum \mu_j \equiv 0 \bmod m$. So representations of of $\mathfrak{sl}_m$ with a $0$-weight space correspond to partitions with at most $m$ parts and size $0 \bmod m$, modulo $(1,1,\ldots,1)$. $\endgroup$ Nov 9, 2023 at 18:32

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