# Finite dimensional irreducible representations of $\frak{sl}_m$ with non-trivial zero weight spaces

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 $$\frak{sl}_3$$?

• @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 $\pi_2$? Nov 9, 2023 at 17:18
• Ok, so the $i$-th column in the Cartan matrix gives the coefficients of $\alpha_i$ in terms of the fundamental weights $\pi_j$? Nov 9, 2023 at 17:27
• If you put this as an answer I am happy to accept it. Nov 9, 2023 at 17:31
• 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. Nov 9, 2023 at 19:55

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.
• 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) Nov 9, 2023 at 18:30
• 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 $(1,1,\ldots,1)$. Nov 9, 2023 at 18:32