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Let $G$ be a reductive group and fix a choice of positive roots inside the associated root system.

My question is about the relationship between the cone spanned by $\mathbb{Z}_{\geq 0}$-linear combinations of these positive roots and the dominant Weyl chamber.

More precisely, suppose $\beta$ is a weight and $w \in W$ is an element of the Weyl group with $w(\beta)$ dominant. What restrictions on $w$ does one impose if $\beta$ is assumed $\geq 0$ (i.e. a $\mathbb{Z}_{\geq 0}$-linear combination of positive roots)?

The following is what I would like. Can one always choose $w$ so that $l(w) < r$ in the above situation?

Here $l(w)$, as usual, denotes the smallest possible length of an expression of $w$ in terms of simple reflections, and $r$ is the number of simple roots.

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    $\begingroup$ The answer to your second question is no; in type $C_2$ take $\beta=s_1s_2\theta = s_1\theta=\alpha_2$ where $\theta$ is the highest root and $\alpha_2$ is the long simple root. $\endgroup$
    – Grant B.
    Commented Aug 27 at 21:33
  • $\begingroup$ Yes, good point! But that shouldn't really contradict what I'm after. I've tried to fix the formulation. $\endgroup$
    – user536506
    Commented Aug 27 at 22:11
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    $\begingroup$ Still doesn't work: in type $G_2$ take $\beta=s_1s_2\theta=\alpha_2$ where $\theta$ is the highest root and $\alpha_2$ is the long simple root. $\endgroup$
    – Grant B.
    Commented Aug 27 at 23:47
  • $\begingroup$ I should have checked the G2 case myself. Thanks alot. $\endgroup$
    – user536506
    Commented Aug 28 at 11:01

1 Answer 1

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Grant B gave a great example in type $G_2$. Here is another example in type $A_3$. The roots are $\{\epsilon_i-\epsilon_j:1\le i\ne j\le 4\}$, with simple roots $\alpha_i=\epsilon_i-\epsilon_{i+1}$. Note that $\sum_{i=1}^4x_i\epsilon_i$ is dominant if and only if $x_1\ge x_2\ge x_3\ge x_4$.

Let $$\beta=\alpha_1+3\alpha_2+6\alpha_3=\epsilon_1+2\epsilon_2+3\epsilon_3-6\epsilon_4.$$ Then the unique element $w\in S_4$ such that $w(\beta)$ is dominant is the permutation $(13)=s_1s_2s_1$, which has length three.


What one can say in type $A$ is the following. The positive weights of type $A_n$ are of the form $$\begin{align*}\beta&=t_1\alpha_1+t_2\alpha_2+\cdots+t_n\alpha_n\\&=t_1\epsilon_1+(t_2-t_1)\epsilon_2+\cdots+(t_n-t_{n-1})\epsilon_n-t_{n}\epsilon_{n+1} \end{align*}$$ for $t_i\in\mathbb Z_{\ge0}$, which are in particular of the form $\sum_{i=1}^{n+1}x_i\epsilon_i$ where $x_1\ge x_n$. So there exists a permutation $w\in S_{n+1}$ such that $w(\beta)$ is dominant and $w(1)<w(n)$. Thus $$\ell(w)\le {n\choose 2}.$$ When $n=2$ this gives $\ell(w)\le 1$ which is compatible with your expectation, but that is a coincidence.

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