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I am trying to understand Borel subgroups of $Sp(4,\mathbb{C})$. I think that the following is a Borel subgroup of $Sp(4, \mathbb{C})$: the subset of $Sp(4, \mathbb{C})$ of all lower triangular matrices $X=\left(\begin{array}{cccc} x_{1,1} & 0 & 0 & 0\\ x_{2,1} & x_{2,2} & 0 & 0\\ x_{3,1} & x_{3,2} & x_{3,3} & 0\\ x_{4,1} & x_{4,2} & x_{4,3} & x_{4,4} \end{array}\right)$ such that $X^T J X = J$, where $J = \left(\begin{array}{cccc} 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1\\ -1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0 \end{array}\right)$. Since $X^T J X = J$, we have $$ X^T J X - J = 0 = \\ \left(\begin{array}{cccc} 0 & x_{1,1}\, x_{3,2} + x_{2,1}\, x_{4,2} - x_{2,2}\, x_{4,1} & x_{1,1}\, x_{3,3} + x_{2,1}\, x_{4,3} - 1 & x_{2,1}\, x_{4,4}\\ x_{2,2}\, x_{4,1} - x_{2,1}\, x_{4,2} - x_{1,1}\, x_{3,2} & 0 & x_{2,2}\, x_{4,3} & x_{2,2}\, x_{4,4} - 1\\ 1 - x_{2,1}\, x_{4,3} - x_{1,1}\, x_{3,3} & - x_{2,2}\, x_{4,3} & 0 & 0\\ - x_{2,1}\, x_{4,4} & 1 - x_{2,2}\, x_{4,4} & 0 & 0 \end{array}\right). $$ There are $5$ independent equations above. Therefore $\dim B^- = 10 - 5 = 5$, where $B^-$ is the above Borel subgroup.

But we also have $B^- = U^- T$, where $T$ is the torus. Since $T^T J T = J$, $\dim T=2$. The dimension of $U^-$ is the length of the longest word in $C_2$ Weyl group which is $4$. Therefore $\dim B^- = 4 + 2 = 6 \neq 5$. Is $\dim B^- = 5$ or $6$? Thank you very much.

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    $\begingroup$ To compute it, it's easier to first work in the Lie algebra where everything is linear, and then exponentiate. I don' remember if J has the right form to make it lower triangular; possibly the antidiagonal matrix $(1,1,-1,-1)$ is better. $\endgroup$
    – YCor
    Commented Jul 7, 2018 at 18:53
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    $\begingroup$ I'm not sure about your matrix equations, but let me confirm that the dimension of the Borel subgroup of type $C_n$ is $n^2+n$ (the full flag variety has dimension $n^2$), so $6$ when $n=2$. $\endgroup$
    – Gro-Tsen
    Commented Jul 7, 2018 at 19:28

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