2
$\begingroup$

$F$ is a non archimedean field here. To be more precise, I would actually prefer a set of representative in $B(F)$ for the discrete space $B(F) / B(o)Z(F)$?

This can be phrased also as question about lattices in $F^n$, but I would prefer to stay on the group level.

$\endgroup$
3
  • $\begingroup$ B denotes the Borel subgroup of GL(N). $\endgroup$
    – Marc Palm
    Commented Dec 14, 2011 at 12:58
  • 1
    $\begingroup$ To clarify further: I assume $B$ is a Borel subgroup, $B(F)$ its $F$ rational points, and I assume $o$ is supposed to be the ring of integers $\mathcal{O}$ of $F$, and $Z(F)$ the center (i.e. diagonal matrices). Correct? $\endgroup$
    – Max Horn
    Commented Dec 14, 2011 at 14:37
  • $\begingroup$ correct! ... $\endgroup$
    – Marc Palm
    Commented Dec 15, 2011 at 9:20

2 Answers 2

3
$\begingroup$

You somehow want to parametrise the vertices of the building of $G={\rm GL}(n,F)$ : $$ G/F^\times K = BK/F^\times K= B(F)/B({\mathfrak o})Z(F) $$ (by Iwasawa decomposition).

For $n=2$ ou can easily find representatives, but for $n>2$, it's going to be tricky!

I just give some hints. Write $N$ for the unipotent radical of $B$ and $T$ for the diagonal torus so that $B=T\ltimes N$.

-- If $n,n'\in N$ and $t, t'\in T$, then if $nt\sim nt'$ mod $B(O)Z(F)$, one has $t\sim t'$ mod $Z(F)T(O)$. So you may assume that $t$ is of the form

$$ t= {\rm diag}(\varpi^{k_1}, ...,\varpi^{k_n}) $$ where $(k_1 ,...,k_n )$ is well defined modulo the diagonal action of $\mathbb Z$ on ${\mathbb Z}^n$.

-- You have $nt \sim n't$ mod $Z(F)B(O)$ iff $n\sim n'$ mod $tN(0)t^{-1}$.

So for each $t$ as above, you need to find a system of representatives of $$ N(F)/tN(O)t^{-1} $$ For $n=2$, this is easy. For $n>2$, this seems tricky. I've never tried ...

$\endgroup$
3
  • $\begingroup$ Here $O={\mathfrak o}$ is the ring of integers, and $\varpi$ a uniformizer. $\endgroup$ Commented Dec 14, 2011 at 20:14
  • $\begingroup$ That was my initial idea too. I have now succeeded by iterating your argument also for the unipotent radical, seen as an iterated semidirect product. Thanks. $\endgroup$
    – Marc Palm
    Commented Dec 15, 2011 at 9:18
  • $\begingroup$ Nevertheless, it is certainly solved somewhere. The final partition looks like $$ \coprod\limits_{\alpha_1 \in T(F) / T(o)} \coprod\limits_{\alpha_2 \in E_1(F) / E_1(o)^{\alpha_{1}} } \cdots \coprod\limits_{E_n(F) / E_n(o)^{\alpha_1 \alpha_2 \cdots \alpha_{n-1}}},$$ where $E_k$ is the set of upper diagonal matrices with only $k-th$ column nonzero and $H^x = x^{-1} H x$. $\endgroup$
    – Marc Palm
    Commented Dec 15, 2011 at 9:35
2
$\begingroup$

The comments together with Paul's answer emphasize the importance of formulating the question more precisely and with some context (what you've read on the subject, for instance). Though I'm not at all a specialist in buildings, I know some of the complicated history of the subject as it evolved into long and highly sophisticated papers by Bruhat-Tits and others. But your question about general (or equally well special) linear groups, which are split over the prime field, goes back to the foundational papers such as the 1965 Publ. Math. IHES paper by Iwahori and Matsumoto, freely available by a quick author search here, followed by the detailed exercises in Chapter IV of Bourbaki's 1968 treatise Groupes et algebres de Lie where the BN-pair structure (or Tits system) is developed into the basic theory of buildings.

In this early work there is a treatment of the special subgroup structure present in a split (Chevalley) group over a standard $p$-adic field: fixing a Borel subgroup over the finite residue class field, one can lift it to the $p$-adic integers where it becomes an Iwahori subgroup. Such groups are determined up to conjugacy in the ambient algebraic group. Along with a copy of the affine Weyl group, an Iwahori subgroup determines a BN-pair structure and Bruhat decomposition. In turn there are finitely many maximal (proper) "parahoric" subgroups. Their cosets in the big group become the vertices for the resulting building. If you regard the original Borel subgroup as "canonical", this pathway should lead to canonical vertices of the building. As Paul observes, in the case $n=2$ all of this is fairly easy to write down; here the building is just an infinite tree.

Once you get beyond split groups and ordinary $p$-adic extensions of the rationals, a lot more machinery has to be developed in order to work effectively with buildings and subgroup actions on them. But the general linear group, especially in semisimple rank 1, is the natural starting point for combining group theory and combinatorial geometry in a visualizable way.

$\endgroup$

You must log in to answer this question.

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