# 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. May 10, 2019 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. May 18, 2019 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 $\Lambda \neq \mathbb{Z}\Phi$. May 18, 2019 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. May 19, 2019 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 $\lambda \in \Lambda$. May 19, 2019 at 1:43

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. May 19, 2019 at 1:10
• @JimHumphreys: In this case $c_i' = \sum_{j = 1}^l a_j q_{ij}$. (I added this detail to the answer). May 19, 2019 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.)