0
$\begingroup$

The longest word in type $A_3$ Weyl group written as a matrix is \begin{align} w_0=s_1s_2s_1=\left(\begin{array}{cccc} 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & 1 & 0 & 0\\ 1 & 0 & 0 & 0 \end{array}\right). \end{align}

What is the longest word in type $C_2$ Weyl group written in terms of a matrix? In this case, $w_0 = s_1 s_2 s_1 s_2$. How to write $s_1, s_2$ as matrices? Thank you very much.

Improve this question
$\endgroup$
2
  • $\begingroup$ Do you mean "what is the matrix, in terms of a basis of simple roots, of the transformation of the root system induced by …"? In this case, the matrices of simple reflections very nearly come from the Cartan matrix. Although you must usually specify your numbering of simple roots, of course here it doesn't matter. If $\alpha_1$ is the short root and $\alpha_2$ the long, then $s_1 = \begin{pmatrix} -1 & 2 \\ 0 & 1 \end{pmatrix}$ and $s_2 = \begin{pmatrix} 1 & 0 \\ 1 & -1 \end{pmatrix}$. $\endgroup$
    – LSpice
    Jul 7, 2018 at 19:48
  • 4
    $\begingroup$ It's scalar multiplication by $-1$. $\endgroup$
    – Ben Webster ♦
    Jul 7, 2018 at 20:22

2 Answers 2

Reset to default
2
$\begingroup$

To answer these questions we have great computer algebra systems such as Sage

W = WeylGroup("C2")
W.long_element()

You can run short snippets of code like these on SageCell

Improve this answer
$\endgroup$
0
$\begingroup$

The Chevalley basis of $sp_4$ is generated by \begin{align} & e_1=\left(\begin{array}{cccc} 0 & 1 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & -1 & 0 \end{array}\right), \ e_2 = \left(\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0 \end{array}\right), \\ & f_1 = \left(\begin{array}{cccc} 0 & 0 & 0 & 0\\ 1 & 0 & 0 & 0\\ 0 & 0 & 0 & -1\\ 0 & 0 & 0 & 0 \end{array}\right), f_2 = \left(\begin{array}{cccc} 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \end{array}\right). \end{align} We have \begin{align} \overline{s}_i = \exp(-e_i)\exp(f_i)\exp(-e_i). \end{align} Therefore \begin{align} & \overline{s}_1 = \left(\begin{array}{cccc} 0 & 1 & 0 & 0\\ -1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & -1 & 0 \end{array}\right), \quad \overline{s}_2 = \left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & 0 & 0 & 1\\ 0 & 0 & 1 & 0\\ 0 & -1 & 0 & 0 \end{array}\right), \\ & \overline{s}_1 \overline{s_2} \overline{s}_1 \overline{s}_2 = \left(\begin{array}{cccc} 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \end{array}\right), \quad s_1 s_2 s_1 s_2 = \left(\begin{array}{cccc} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ 1 & 0 & 0 & 0\\ 0 & 1 & 0 & 0 \end{array}\right). \end{align}

Improve this answer
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.