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

Question

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.

Answer 1

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 $N = \max \{c_j/q_{ji}\}_{i,j}$.

Answer 2

The answer is apparently no. This is probably seen most easily in type $G_2$, where the root lattice equals the weight lattice and there is therefore just one coset to consider. As in Bourbaki's planche, the first fundamental weight $\varpi_1 = 2\alpha_1 + \alpha_2$, while $\varpi_2 = 3\alpha_1 + 2\alpha_2$. If my arithmetic is correct, $\alpha_1 = 2\varpi_1 - \varpi_2$ and $\alpha_2 = -3 \varpi_1 + 2\varpi_2$.

Taking $\lambda$ to be any fixed dominant weight, it follows readily that infinitely many dominant weights $\mu$ will fail to be above $\lambda$ in the standard partial ordering.

ADDED: Though I might be oversimplifying matters, the intention was to provide a counterexample in the case of $G_2$. In view of the answers by Mikko and Sam, I am more doubtful about my own answer but need to understand theirs in detail before going deeper. Stay tuned. (In any case, the weights $\lambda + k\alpha_1$ are not usually dominant when $k \in \mathbb{Z}^+$.)

Answer 3

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.)