1
$\begingroup$

Let $G$ be $SL(n, F)$ for a non-archimedean local field $F$ with Iwahori subgroup $I$. Let $\mu$ be the Haar measure of $G$.

What are the properties of the function $w\mapsto \mu(IwI)/\mu(I)$ for $w$ being an element of the normalizer $N$ of the maximal Levi subgroup of $G$?

Let $d i$ denote the Haar measure on $I$. We have that $$ \int\limits_G f(g) d \mu(g) = \sum\limits_{x \in N / N \cap I} ( \mu(IwI)/\mu(I) )^{-1} \int\limits_I \int\limits_I f(i_1 w i_2) d i_1 d i_2 .$$

I would like to have a reference for the above fact and perhaps also the value of $\mu(IwI)/\mu(I) $ at least for $n=2$ and also for similar situation of $PGL(n)$.

I suspect that the fact that $(I,N)$ is a $BN$ pair and that the Iwahori decomposition is available for $I$ should give everything and also give some nice properties for the function $w\mapsto \mu(IwI)/mu(I)$.

Perhaps one suggestion: Naively, one could argue that the function is multiplicative by the following computation $$ \mu(Iw_1w_2I)/\mu(I) = \mu(Iw_1Iw_2I)/\mu(Iw_1I) \cdot \mu(I w_1 I) / \mu( I )$$ $$ = \mu(Iw_2I)/\mu(I) \cdot \mu(I w_1 I) / \mu( I ),$$ which is wrong. What is the correct relation? $\geq$, if $w_1$ satisfies $w_1^2 = 1$?

$\endgroup$
1
  • $\begingroup$ For the motivation, you can have a look at section 6.2.6 in Abramenko-Brown "Buildings" or chapter 5 in Garrett "Buildings", where the decisive properties of a BN pair are described. $\endgroup$
    – Marc Palm
    Commented Mar 15, 2012 at 17:45

2 Answers 2

1
$\begingroup$

It is well-known that $\mu(IwI)/\mu(I)=q^{l(w)}$, where $l(w)$ is the length of $w$ with respect to the affine simple reflections corresponding to $I$. The multiplicative formula holds so long as $l(w_1w_2)=l(w_1)+l(w_2)$.

Take a look at Iwahori and Matsumoto's original paper on the subject (vol. 25 of IHES Publications) or page 43 of my PhD thesis.

$\endgroup$
2
  • 1
    $\begingroup$ Right to the point. That is exactly what I was hoping for. Beautiful;) $\endgroup$
    – Marc Palm
    Commented Mar 16, 2012 at 8:35
  • $\begingroup$ Glad to have been of use. $\endgroup$ Commented Mar 16, 2012 at 9:09
1
$\begingroup$

The set $IwI/I$ is in bijection with the chambers $gI$ in the affine building $G/I$ at distance $w$ from the chamber $I$. In particular, when $w=s_i$ for $s_i$ a lift of a standard generator of the associated affine Weyl group (so $\langle I, IwI \rangle = I \cup IwI$), then $\mu(IwI)/\mu(I)$ equals the cardinality of the residue field.

In the special that the building is a tree (i.e. $n=2$) and that the product $w_1w_2$ is reduced (i.e. the chamber $w_2I$ is on a geodesic from the chamber $I$ to the chamber $w_1w_2I$) you can cook up a 'multiplicative formula', which I'll leave to you.

$\endgroup$

You must log in to answer this question.

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