# Exercise on root system of type $A_n$

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.

• 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. Commented Feb 5, 2020 at 0:46
• I think I'm missing how the simple reflections act on Dynkin diagrams, that's way I tried to convert them to permutations
– cip
Commented Feb 5, 2020 at 7:28
• 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 $A_2$. Commented Feb 5, 2020 at 8:55
• 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?
– cip
Commented Feb 5, 2020 at 10:07
• 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 $i$. Commented Feb 6, 2020 at 19:56

## 1 Answer

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.