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Problem

Let $n$ be a positive integer and let $\phi$ be a root system of type $A_n$. Let $\Delta = \{ \alpha_1, .. , \alpha_n \}$ be a base, such that the Dynkin diagram is a string enumerated from $1$ to $n$ (left to right). Let $w= s_n \circ s_{n-1} .. .. \circ s_2$, where $s_i$ is the simple reflection with respect to the simple root $\alpha_i$ with $i= 1... n$. Lastly, define a partial order relation given by $\alpha > \beta$ iff $\alpha - \beta$ is a linear combination of positive roots.

  1. Prove that $\theta = w(\alpha_1)$ satisfies $\theta \geq \alpha_i$ for every $i$.
  2. Find for which $i \in \{ 1,..,n\}$ it is true that $\theta - \alpha_i \in \phi$.
  3. Prove that $\theta$ is in the fundamental Weyl chamber.

Attempt at a solution

  1. So I know that $A_n$ is the root system of $\mathfrak{sl}(n+1)$. Furthermore, if we consider $H$, the maximal toral subalgebra of $\mathfrak{sl}(n+1)$ (the diagonal matrices in $\mathfrak{sl}(n+1)$), it is true that the Weyl group $W \subseteq GL(H^*)$ is isomorphic to $S_{n+1}$. So, I know that the generators of $S_{n+1}$, the transpositions, correspond to the simple reflections $s_i$, which are the generators of the Weyl group. That means that $\theta = \alpha_n$ (is this correct?) Therefore I know that $\alpha_n \geq \alpha_i$, because $\alpha_i$ are simple roots, therefore positive roots.

  2. Because of the root system of $\mathfrak{sl}(n+1)$, I know that $\theta - \alpha_i \in \phi$ for $i= 1, .., n-1$. Now for the case $i=n$: I know that one way to prove that $\theta - \alpha_n \in \phi$ would be proving that $(\theta, \alpha_n) > 0$. Suppose now that $(\theta, \alpha_n) \leq 0$. That is impossible because $(\alpha_n, \alpha_n) > 0$.

  3. $\theta$ is in the fundamental Weyl chamber because $(\alpha_n, \alpha_i) > 0$ for every $i$. In fact, it can't be negative, because that would imply that $\alpha_n + \alpha_i \in \phi$, which is not for the root system of $\mathfrak{sl}(n+1)$.

Do you think it is correct? Thanks in advance.

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    $\begingroup$ You should not need anything about Lie algebras to answer this. I think that neither $\theta=\alpha_n$ (so your arguments in all three points fall flat), nor is $\alpha_n \ge \alpha_i$. Maybe doe the calculations explicitly for $n=2,3$, that should already show something. $\endgroup$ Commented Feb 5, 2020 at 0:46
  • $\begingroup$ I think I'm missing how the simple reflections act on Dynkin diagrams, that's way I tried to convert them to permutations $\endgroup$
    – cip
    Commented Feb 5, 2020 at 7:28
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    $\begingroup$ There are no order relations between the simple roots. For example, $\alpha_1>\alpha_2$ and $\alpha_2>\alpha_1$ are both false. It is a partial order only. Instead, try and prove that $s_2(\alpha_1)=\alpha_1+\alpha_2$, $s_3(\alpha_1+\alpha_2)=\alpha_1+\alpha_2+\alpha_3$ and continue that chain up to $\theta$. The first step you can verify visually looking at a picture of the root system $A_2$. $\endgroup$ Commented Feb 5, 2020 at 8:55
  • $\begingroup$ So $\theta= \alpha_1 + \alpha_2 + ...+ \alpha_n$? In that case, I know that $\theta - \alpha_i$ is a linear combination of $n-1$ simple roots, therefore positive roots and that would prove the first point. For the second point, I know that $(\theta, \alpha_i) >0$ because $(\theta, \alpha_i) = (\alpha_{i-1}, \alpha_i) + (\alpha_i, \alpha_i)$ (all the others are orthogonal). That also proves that $\theta$ is in the fundamental Weyl chamber. Is this now correct? I'm only missing why $(\alpha_{i-1}, \alpha_{i})$ is positive: I know why it isn't zero, but why can't it be negative? $\endgroup$
    – cip
    Commented Feb 5, 2020 at 10:07
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    $\begingroup$ You are getting there (so +1). When calculating $(\theta,\alpha_i)$ you also need to observe that $(\alpha_{i+1},\alpha_i)=-1$ (whenever applicable). Also observe that for $\theta$ to be in the fundamental chamber you only need $(\theta,\alpha_i)\ge0$. In fact, there is equality here for all but two values of $i$. $\endgroup$ Commented Feb 6, 2020 at 19:56

1 Answer 1

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After discussion in the comments, you seem to be closer to a solution. Maybe the following hints suffice.

Realise that the Dynkin diagram precisely tells you what each $(\alpha_i, \alpha_j)$ is (up to scaling). One standard scaling is $(\alpha_i, \alpha_i) = 2$, $(\alpha_i, \alpha_{i-1}) = -1$ ($\color{red}{!}$), and $(\alpha_i, \alpha_j)=0$ if $j \neq i \pm1$. With this information, you should be able to compute $(\theta, \alpha_i)$ for all $i$, but if you're doing it right (which you don't quite in your latest comment), you should notice that the answer is slightly different in the case $i \in \lbrace 1,n \rbrace$ than in the case $2 \le i\le n-1$.

The same case distinction should apply to the answer to question 2. Which linear combinations of the $\alpha_i$ are actually roots? Notice e.g. that in $A_{17}$, $\alpha_8+\alpha_9 +\alpha_{10} + \alpha_{11}$ is a root, but $\alpha_2 + \alpha_5$ and $\alpha_3+\alpha_4 +\alpha_{16}$ and $\alpha_{9}+ \alpha_{14}+\alpha_{15}$ are not.

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