How to obtain the root $2 \alpha_6 + \alpha_1 + 2 \alpha_2 + 3 \alpha_3 + 2 \alpha_4 + \alpha_5$ in type $E_6$ root system by Weyl group action?

Question

The Dynkin diagram for $E_6$ is \begin{align} \circ - \circ - & \circ - \circ - \circ \\ & \ | \\ & \ \bullet \end{align} where $\bullet$ corresponds to the simple root $\alpha_6$ and the other vertices correspond to $\alpha_1, \ldots, \alpha_5$ (from left to right) respectively.

The element $2 \alpha_6 + \alpha_1 + 2 \alpha_2 + 3 \alpha_3 + 2 \alpha_4 + \alpha_5$ is a root in type $E_6$ root system.

Let $W$ be the Weyl group in type $E_6$. I would like to find $w \in W$ and some simple root $\alpha_i$ such that $w \cdot \alpha_i = 2 \alpha_6 + \alpha_1 + 2 \alpha_2 + 3 \alpha_3 + 2 \alpha_4 + \alpha_5$. Where $w \cdot \alpha_i$ denotes the Weyl group action. Thank you very much.

Answer 1

The problem can be solved by Sage. I used sage to apply Weyl group action on $\alpha_2$ (which is $\alpha_6$ in my notation):

DynkinDiagram(['E', 6])
W = WeylGroup(RootSystem(['E',6]).root_lattice())
s = W.simple_reflections()
alpha = W.domain().simple_roots()
r1=[]
r2=[]
for i in W:
    t1=i.action(alpha[2])
    if r1.count(t1)==0:
        r1.append(t1)
        r2.append([i.reduced_word(),t1])

Then I choose the $w$ which gives $w \cdot \alpha_6 = 2 \alpha_6 + \alpha_1 + 2 \alpha_2 + 3 \alpha_3 + 2 \alpha_4 + \alpha_5$.

Let $w = s_6 s_3 s_4 s_5 s_2 s_3 s_4 s_1 s_2 s_3$. Then $w \cdot \alpha_6 = 2 \alpha_6 + \alpha_1 + 2 \alpha_2 + 3 \alpha_3 + 2 \alpha_4 + \alpha_5$.