# Technical lemma on root systems, reduced to linear algebra

Update: I have posted the case of $G = SL(n)$ as a different question here.

This is a technical lemma I am currently stuck at. Any suggestions about how to proceed are welcome.

Let $G$ be a split semisimple group over a number field. Let $B, T$ be a chosen minimal parabolic, torus respectively, and $\theta$ be an automorphism of $G$ of finite order, preserving $B$ and $T$. By the theory of algebraic groups, we can associate to $(G, B, T)$, a root datum $(X^*, \Delta, X_*, \Delta^\vee )$, where $X^*$ (resp. $X_*$) is the lattice of characters (cocharacters) and $\Delta$ (resp. $\Delta^\vee$) the simple roots (coroots). The Weyl group $W$ acts on the root space $V^* = X^* \otimes \mathbb R$ and co-root space $V_*$ as usual and the action of $\theta$ descends to a permutation of the finite sets $\Delta, \Delta^\vee$. For every $\beta \in \Delta$, let $\varpi_\beta \in \hat\Delta$ be the corresponding weight so that $\hat\Delta$ is a basis of $V^*$ dual to $\Delta$.

Question: Fix $w \in W, w \neq 1$. Does there exist a cone $\Omega$ inside the positive Weyl chamber of $V^*$ such that if $\lambda \in \Omega$ and $\lambda - \theta w \lambda = \displaystyle\sum_{\beta \in \Delta} {d_\beta} \beta$, then $d_\beta > 0$?

(Edit: To clarify, by positive Weyl chamber, I mean a positive span of weights, not roots.)

This is not true per se but I would like the coordinates $d_\beta$ positive only in some directions: Let $Q(w)$ be the smallest standard (i.e., containing $B$) parabolic subgroup containing a representative of $w$. By a correspondence between standard parabolic subgroups and subsets of $\Delta$, we have the subset $\Delta^{Q(w)}$ of $\Delta = \Delta^G$ corresponding to $Q(w)$. I need $d_\beta>0$ whenever $\beta \in \Delta^{Q(w)}$.

Example: Let $G = SL(n)$ with usual Borel and torus and let $\theta(x) = {}^Tx^{-1}$. We can identify $V^*$ with the subspace of $\mathbb R^n$ of vectors with coordinates adding to zero so that $\Delta = \{ \beta_1 = (1, -1, 0, \cdots, 0), \cdots, \beta_{n-1} \}$ and $\{ \varpi_\beta \in \hat\Delta\}$ is the dual basis (of weights). We have $\theta(\beta_i) = \beta_{n-i}$. In the case of $SL(3)$ taking $\lambda = c_1 \varpi_1+ c_2 \varpi_2$ and $w$ to be the permutation $(1,2)$ gives $$\lambda - \theta w\lambda = \left(\frac{c_1 - c_2}{3}\right)\beta_1 + \left(\frac{2c_1 + c_2}{3}\right)\beta_2$$ so we can define the cone to be $c_1 > c_2 > 0$. I have written a Python code and verified this till $SL(7)$ but unable to prove so far for general $G$.

EDIT: When $\theta = 1$ we can take $\Omega$ to be the full positive Weyl chamber; see Bourbaki's Lie Groups and Lie Algebras Chapter 6 $\S 1.5$ (Edit: 1.6, not 1.5) Proposition 18 for a proof based on induction on $\ell(w)$.

EDIT': I am elaborating on the case $\theta = 1$. Fix a root $\gamma \in \Delta^{Q(w)}$ and $\lambda$ in the positive Weyl chamber. The coefficient of $\gamma$ in $\lambda - w \lambda$ is given by $\langle \lambda - w \lambda, \varpi_\gamma^\vee \rangle = \langle \lambda, (1 - w^{-1}) \varpi_\gamma^\vee \rangle$ and by Bourbaki's aforementioned lemma, the vector $(1 - w^{-1}) \varpi_\gamma^\vee$ is a non-negative combination of co-roots.

• Is the cone allowed to depend on $w$? Jan 16, 2017 at 20:22
• Yes. (Otherwise it's not true.) Jan 16, 2017 at 21:12
• The phrase "over a number field" isn't needed here, is it? The split assumption seems to be enough, since your question concerns just the Chevalley structure theory of such groups over arbitrary fields. Jan 17, 2017 at 17:00
• That's right but I included it anyways. Jan 17, 2017 at 17:04
• I would appreciate any comments / hints / ideas / thoughts from Prof. Humphreys or Loren. Jan 17, 2017 at 17:07

Maybe it's helpful to add a longer comment, in community-wiki format. The original question is not well-formulated, I think, as shown in the later convoluted remarks on the case $\theta =1$. It's probably better here to follow Bourbaki (Chapter VI), since the essential problem concerns just an irreducible root system in a vector space $V$ having as basis a fixed choice of simple roots. This choice then determines a partition of the complement in $V$ of reflecting walls for the various roots into Weyl chambers. One of these (call it $C$) is the dominant Weyl chamber (called "positive" in the question) defined by $\langle \lambda, \alpha_i^\vee \rangle >0$ for all simple $\alpha_i$ with $\lambda \in V$. Its closure $\overline{C}$ is then a fundamental domain for the action of the Weyl group $W$.
The fundamental weights $\varpi_i$ lie in the walls of $C$ and are at acute (or right) angles. On the other hand, the simple roots $\alpha_i$ are at obtuse (or right) angles and determine a "positive root cone" (call it $D$) consisting of positive linear combinations of simple roots. Then $D$ contains $C$ because each $\varpi_i$ is a positive $\mathbb{Q}$-linear combination of the $\alpha_i$. But $D$ is usually larger than $C$.
We are given an automorphism $\theta$ of the Dynkin diagram, for example the one of order 2 for type $A_\ell$ when $\ell \geq 2$ (coming from $\mathrm{SL}_{\ell+1}$) which switches $\alpha_i$ and $\alpha_{\ell-i+1}$. The question then concerns $(*) \:\lambda - \theta w \lambda$ for a fixed $w \in W$ and any dominant weight $\lambda$ (say in $\overline{C}$). Bourbaki's Prop. 18 says for $\theta =1$ that $\lambda - w \lambda$ lies in $\overline{D}$, but of course usually not in $\overline{C}$.
First write $\lambda$ as a $\mathbb{Z}^+$-linear combination of the $\varpi_i$, say with coefficients $c_i$. Then $(*)$ is a $\mathbb{Q}$-linear combination of simple roots. Maybe the intended question for arbitrary $\theta$ is whether there is a cone inside the $\mathbb{Q^+}$-span of $D$ (or $\overline{D}$) consisting of those elements $(*)$ defined by conditions on the $c_i$ such as those in the example. (A fixed denominator may occur, coming from the index of the root lattice in the weight lattice.) Of course, the cone need not lie in $\overline{C}$, as shown by the case $\theta =1$. Anyway, the version I've stated seems likely to have a positive answer, but I don't know how to prove it.