# For a fixed dominant weight $\lambda$, are almost all dominant weights in the same coset above it?

First some notation as in e.g. the book by Humphreys on Lie Algebras.

Let $$E$$ be an Euclidean space with inner product $$(-,-)$$, and denote $$\langle v,w \rangle = \frac{2(v,w)}{(w,w)}$$. Let $$\Phi$$ be an irreducible root system on $$E$$, so $$\langle \beta, \alpha \rangle \in \mathbb{Z}$$ and $$\alpha - \langle \beta, \alpha \rangle \beta \in \Phi$$ for all $$\alpha, \beta \in \Phi$$. Fix a set of simple roots $$\alpha_1$$, $$\ldots$$, $$\alpha_l$$.

Let $$\Lambda$$ be the set of weights, i.e. the set of $$\lambda \in E$$ such that $$\langle \lambda, \alpha \rangle \in \mathbb{Z}$$ for all $$\alpha \in \Phi$$.

Let $$\Lambda^+$$ be the set of dominant weights, that is, the set of $$\lambda \in \Lambda$$ such that $$\langle \lambda, \alpha_i \rangle \geq 0$$ for all $$i$$.

We have the usual partial order on $$\Lambda$$, by defining $$\mu \preceq \lambda$$ iff $$\lambda - \mu = \sum_{i = 1}^k k_i \alpha_i$$ for some integers $$k_i \in \mathbb{Z}_{\geq 0}$$.

It is well known that for a fixed $$\lambda \in \Lambda^+$$, there are only finitely many $$\mu \in \Lambda^+$$ such that $$\mu \preceq \lambda$$ (See for example 13.2 in Humphreys). But is the following true?

Fix $$\lambda \in \Lambda^+$$. Then for all but finitely many $$\mu \in \Lambda^+$$ with $$\lambda - \mu \in \mathbb{Z}\Phi$$, we have $$\lambda \preceq \mu$$.

If the answer is yes, this could be used to give a different solution to a previous question asked here: link.

• Looking back at my choice of notation in 1972 and earlier, I'd now want to make clearer use of coroots and such rather than relying so much on pointed brackets. – Jim Humphreys May 10 '19 at 20:10
• Since the root lattice has finite index in the weight lattice, the restriction that $\lambda-\mu$ be in the root lattice does not affect the answer. – Victor Protsak May 18 '19 at 1:20
• @VictorProtsak: If $\lambda - \mu \not\in \mathbb{Z}\Phi$, then $\lambda \not\preceq \mu'$ for every dominant weight $\mu'$ in the coset $\mu + \mathbb{Z}\Phi$. So it is needed if $\Lambda \neq \mathbb{Z}\Phi$. – Mikko Korhonen May 18 '19 at 8:03
• Let $P_{\geq 0}$ denote the (real) cone generated by the fundamental weights and $Q_{\geq 0}$ the (real) cone generated by the positive roots (note that these are polar dual cones, with $P_{\geq 0}$ strictly contained in $Q_{\geq 0}$). It should be that for any $v \in P_{\geq 0}$ we have $P_{\geq 0}\setminus (v+Q_{\geq 0})$ is a bounded, nonconvex polytope. – Sam Hopkins May 19 '19 at 1:36
• In fact, I think it's not even important that $v\in P_{\geq 0}$ for this to be true. In other words, the answer to your question is yes, even for not necessarily dominant $\lambda \in \Lambda$. – Sam Hopkins May 19 '19 at 1:43

## 2 Answers

To me the answer seems to be yes.

Let $$\varpi_i$$ be the $$i$$th fundamental dominant weight. Recall first that since $$\Phi$$ is irreducible, for all $$i$$ we have $$\varpi_i = \sum_{j = 1}^l q_{ji} \alpha_j$$ for $$q_{ji} \in \mathbb{Q}$$ with $$q_{ji} > 0$$ for all $$j$$. (This can be seen either case-by-case by inverting the Cartan matrix, or with a general proof as in Exercise 13.8 of Humphreys' book).

Let $$\lambda \in \Lambda^+$$ and write $$\lambda = c_1\alpha_1 + \cdots + c_l\alpha_l$$ for $$c_i \in \mathbb{Q}$$.

Now consider $$\mu \in \Lambda^+$$ such that $$\lambda - \mu \in \mathbb{Z}\Phi$$. Write $$\mu = a_1 \varpi_1 + \cdots + a_l\varpi_l$$ for $$a_i \in \mathbb{Z}_{\geq 0}$$. I claim that there is an $$N > 0$$ depending only on $$\Phi$$ and the $$c_i$$ such that if $$\mu \not\succeq \lambda$$, then $$a_i \leq N$$ for all $$i$$. Consequently the number of such $$\mu$$ is finite.

For this note that $$\mu \not\succeq \lambda$$ if and only if $$\mu = c_1'\alpha_1 + \cdots + c_l'\alpha_l$$ with $$c_t' < c_t$$ for some $$t$$. Now $$c_t' = \sum_{i = 1}^l a_i q_{ti}$$, so $$c_t' < c_t$$ implies that $$a_iq_{ti} < c_t$$ for all $$i$$. Hence we can take $$N = \max \{c_j/q_{ji}\}_{i,j}$$.

• Can you specify what $c_i'$ is in your argument? I am trhying to understand the details. – Jim Humphreys May 19 '19 at 1:10
• @JimHumphreys: In this case $c_i' = \sum_{j = 1}^l a_j q_{ij}$. (I added this detail to the answer). – Mikko Korhonen May 19 '19 at 7:29

Here's basically the same answer as Mikko Korhonen, but written in a way that's slightly easier for me to understand.

Let me use $$\leq$$ to denote the partial order on all of the vector space $$E$$ with $$u \leq v$$ for $$u,v \in E$$ if and only if $$v-u = \sum_{i=1}^{n}a_i \alpha_i$$ with all $$a_i \geq 0$$ (but not necessarily integral). Considering $$\leq$$ instead of $$\preceq$$ means I don't have to deal with the different classes of the root lattice mod the weight lattice.

Fix any $$v \in E$$. We claim there are only finitely many dominant weights $$\mu \in \Lambda^{+}$$ for which we don't have $$v \leq \mu$$, from which the desired claim obviously follows. Indeed, let $$\omega$$ be a fundamental weight. As Mikko explains (and as is also fundamental to the answer I gave to the linked question), we have $$\omega = \sum_{i=1}^{n} a_i \alpha_i$$ with all $$a_i > 0$$. Hence clearly there is some $$m \geq 0$$ so that $$v \leq m\omega$$. Now let $$M$$ be the maximum over all fundamental weights of such $$m$$. Then writing $$\mu = \sum_{i=1}^{n} c_i \omega_i$$, the only way we could fail to have $$v \leq \mu$$ is if $$c_i < M$$ for all $$i$$. (Here we are using the fact that if $$v \leq u$$ and $$\nu \in \Lambda^{+}$$, then $$v \leq u+\nu$$, which can be seen again from the fact that writing $$\omega = \sum_{i=1}^{n} a_i \alpha_i$$ for any fundamental weight $$\omega$$, we have $$a_i > 0$$.) There are clearly only finitely many $$\mu=\sum_{i=1}^{n} c_i \omega_i \in \Lambda^{+}$$ with $$c_i < M$$ for all $$i$$.

EDIT:

Here is an even more general statement/context. Let $$V$$ be an n-dimensional Euclidean vector space with inner product $$\langle \cdot, \cdot \rangle$$ and let $$v_1,\ldots,v_n \in V$$ be a collection of vectors such that:

• $$v_1,\ldots,v_n$$ form a basis of $$V$$;
• $$\langle v_i , v_j \rangle \leq 0$$ for all $$i \neq j$$ (in other words, the vectors are pairwise non-acute);
• there is no nontrivial decomposition of $$V = V_1 \oplus V_2$$ into orthogonal subspaces $$V_1$$ and $$V_2$$ such that $$v_i \in V_1 \cup V_2$$ for all $$i$$ (this is an irreducibility condition- equivalently it says that if we draw the graph on the $$v_i$$ with $$v_i$$ adjacent to $$v_j$$ if $$\langle v_i, v_j\rangle < 0$$ then that graph will be connected).

Then let $$Q_{\geq 0} := \{ v=\sum_{i=1}^{n}a_iv_i, a_i \geq 0\}$$ be the cone generated by the $$v_i$$. And let $$P_{\geq 0} := \{v \in V\colon \langle v,w\rangle \geq 0 \textrm{ for all w\in Q_{\geq 0}}\}$$ be the dual cone to $$Q_{\geq 0}$$.

Then the claim is that for any $$v\in V$$ we have that $$P_{\geq 0} \setminus (v+Q_{\geq 0})$$ is a bounded subset of $$V$$.

To prove this, observe that any nonzero $$w \in P_{\geq 0}$$ (in particular, any generator of this cone) has all $$a_i > 0$$ when we write $$w=\sum_{i=1}^{n} a_i v_i$$. The reason for this is that the matrix $$M=(\langle v_i, v_j \rangle)$$ is a nonsingular, irreducible $$M$$-matrix, and hence $$M^{-1}$$ (which expresses the coordinates of the generators of $$P_{\geq 0}$$) is a matrix with all entries strictly positive (see e.g. Theorem A of https://core.ac.uk/download/pdf/82640451.pdf). Then we can apply the same argument as above to the generators of the cone $$P_{\geq 0}$$.

(The particular situation above corresponds to the $$v_i$$ being the simple roots and the cone $$P_{\geq 0}$$ being the dominant cone, i.e., cone spanned by the fundamental weights.)