7
$\begingroup$

Is there an invariant, which encodes the failure of the Bruhat decomposition to hold for a reductive group with coefficients in a local ring like the p-adic integers or the ring $\mathbb{Z}/\mathfrak{p}^r$?

Example $G =GL_2$: Fix a Borel subgroup $B$, e.g. the upper diagonal matrixes. The coset space $B \backslash G$ (or $G/B$) are isomorphic to the projective line. However, the Bruhat decomposition $G = BWB$, where $W$ is the Weyl group, does not hold for the group of $R$ points, where $R$ is not a field. Can we describe $B\backslash G/B$ as a variety over $\mathbb{Z}$ here?

$\endgroup$

1 Answer 1

10
$\begingroup$

Bruhat decomposition over $\mathbf Z/p^r\mathbf Z$ is precisely the problem we looked at in this paper. We defined several invariants of double cosets, and classified the pairs $(n,k)$ for which, when $G=GL_n(\mathbf Z/p^k\mathbf Z)$, the cardinality of $B\backslash G/B$ does not depend on $p$. Unfortunately, the general question seems to involve wild classification problems.

$\endgroup$
5
  • $\begingroup$ Thanks for the fast response. So, I guess the quotient $B\GL_n/B$ is pretty nasty and not an algebraic variety? $\endgroup$
    – Marc Palm
    Commented Mar 9, 2011 at 10:02
  • $\begingroup$ We did $n=3$ explicitly. At least there $p$ enters the picture in a very controlled way. In general, it seems like there should some hidden structure which we do not understand. $\endgroup$ Commented Mar 9, 2011 at 10:28
  • 2
    $\begingroup$ There have been earlier attempts by people working in algebraic K-theory to exploit Bruhat decomposition over rings and ideals, including early work by M.R. Stein; but that work is probably not close enough to the precise question asked here. (By the way, it's useful to note that the cited paper is available at arXiv, since our library and many others long ago found the journal unaffordable.) $\endgroup$ Commented Mar 9, 2011 at 17:58
  • $\begingroup$ Thanks for the helpful comment Jim. I would be happy to send a copy of the published version to anyone who requests it. $\endgroup$ Commented Mar 10, 2011 at 1:02
  • 4
    $\begingroup$ Despite the difficulty of the problem as posed, there is still the Iwahori version, that $G(\mathbb Z_p)=JWJ$ (disjoint union!), where $J$ is Iwahori, $W$ is the spherical Weyl group. The disjointness of the union sometimes reduces what is otherwise reasonably perceived as fairly chaotic. $\endgroup$ Commented Jul 16, 2011 at 17:30

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .