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Is there a standard algorithm for finding the double coset representatives of $H_1$ \ $G/H_2$, where the groups are finite of Lie type?

Specifically, I need to compute the representatives when $G=Sp_4(\mathbb{F}_q)$ (I'm using $J$ the anti diagonal with top two entries $1$, and the other two $-1$), $H_1$ is the parabolic with $4=2+2$, and $H_2=SL_2(\mathbb{F}_q)\ltimes H$, where $H$ is the group of matrices of the form: $$\begin{bmatrix} 1&x&y&z \\\\ 0&1&0&y \\\\ 0&0&1&-x \\\\ 0&0&0&1 \end{bmatrix}$$ which is isomorphic to the Heisenberg group, and $SL_2$ is embedded in $Sp_4$ as: $$\begin{bmatrix} 1&&& \\\\ &a&b& \\\\ &c&d& \\\\ &&&1 \end{bmatrix}$$

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Quick comment: I have managed to do the example I needed with a very long manual sage computation. Still, the question remains. –  simplequestions Jun 3 '11 at 12:31

1 Answer 1

Many such questions yield to using Bruhat decompositions, and often succeed over arbitrary fields (which shows how non-computational it may be). Let P be the parabolic with Levi component GL(2)xSL(2). Your second group misses being the "other" maximal proper parabolic Q only insofar as it misses the GL(1) part of the Levi component. Your double coset space fibers over $P\backslash G/Q$. It is not trivial, but is true that P\G/Q is in bijection with $W_P\backslash W/W_Q$, with W the Weyl group and the subscripted version the intersections with the two parabolics. This is perhaps the chief miracle here. Since the missing GL(1) is normalized by the Weyl group, the fibering is trivial. Then some small bit of care is needed to identify the Weyl group double coset elements correctly (since double coset spaces do not behave as uniformly as "single" coset spaces). In this case, the two smaller Weyl groups happen to be generated by the reflections attached to the two simple roots, and the Weyl group has a reasonable description as words in these two generators.

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