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Consider $G = GL_n( \mathbb{Q}_p)$ and $K = GL_n( \mathbb{Z}_p)$, or more favorable replace $ \mathbb{Q}_p$ by a non archimedean field and $\mathbb{Z}_p$ by the ring of integers.

Note that the space $G //K$ is discrete, since $K$ is open and closed in $G$.

Given a Haar measure on $G$, I can prove that there exists a unique (discrete) measure on $G//K$, such that $$ \int\limits_{G} f(g) d g = \sum\limits_{x \in G//K} w(x) \int\limits_{K \times K} f(k_1 x k_2) d k_1 d k_2 .$$

How can $w$ be expressed, if we pick a representative $x = diag( w^{k_1}, \dots, w^{k_n})$ for a uniformizer $w$?

Perhaps easier, but equivalent what is the ratio: $vol_G (K xK)/ vol_G(K)?$

(More out of curiousity: How is the Plancherel measure related to this?)

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You are dealing with the integral formula for the Haar measure on $G$, associated with the Cartan decomposition $G=KAK$. Since the Plancherel measure lives on the dual object $\hat{G}$, I don't see any obvious relation between the two. –  Alain Valette Dec 17 '11 at 22:05

2 Answers 2

up vote 3 down vote accepted

The measure of $KxK$ is a classical computation that may be found in: Macdonald "Symmetric Functions and Hall Polynomials" (Oxford Mathematical Monographs), more precisely in Chapter V: The Hecke ring of ${\rm GL}(n)$ over a local field.

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This is very useful, thank you. I will delete my response as obsolete. The final answer is (2.9) on p.298 of Macdonald's book. –  GH from MO Dec 17 '11 at 18:05
My copy of Mc Donalds (1979) has it on pg. 162! Thanks to both of you. –  Marc Palm Dec 18 '11 at 10:02

Depending on taste, one might also find appealing or helpful the description of this in terms of the Iwahori-Hecke algebra, with affine Weyl group and affine cartan decomposition $G=\bigcup_w BwB$ where $B$ is the Iwahori. Among other features, this does give a way to inductively determine the measure of $BwB$, once the measure of $B$ is normalized, because there is a precise cell-multiplication rule $BwB\cdot BsB=BwsB$ when the length of $ws$ is strictly greater than that of $w$, and $s$ is one of the affine reflections generating $W$. That is, the length in $W$ is equivalent to knowing the measure of the Iwahori coset.

(Inevitably, surely MacDonald's discussion does something equivalent to this, but I don't remember, and I don't have a copy accessible to me.)

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I checked the other answer, because it directly answers my question, but your suggestion seems a lot more suitable for my purposes than the ``naive'' KAK decomposition I was aking about. Do you have a reference for this? Thanks a lot for mentioning this. –  Marc Palm Dec 18 '11 at 10:04
@pm, In my "Buildings and Classical Groups" book, (math.umn.edu/~garrett/m/buildings/book.pdf) chapter 5 exactly treats this: it is simply the BN-pair (with corresponding Bruhat decomposition, etc) attached to the affine building. Probably any book on buildings mentions this at some point. In some sources, the cell multiplication rule is an "axiom". In my chapter 5 I prove that it follows from the building properties. –  paul garrett Dec 18 '11 at 15:01
So just to confirm that I understand what you are saying notationwise: In your notation $B$ is the pullback of the Borel subrgoup $B'$ in $GL(n, o/p)$ along the projection $GL(n,o) \rightarrow \GL(n, o/p)$ and $w$ runs through the normalizer of all diagonal matrices $M$ (=affine Weyl group $MW$?). This is called affine Cartan decomposition? What is $N$ in the $BN$ pair, if $B$ is the Iwahori? ($N=MW$?) I am rather fine with using axioms, but it usually gives me a headache to verify that they hold for specific examples. –  Marc Palm Dec 18 '11 at 16:53
@pm: Yes, in this situation, $B$ can be described as the inverse image in $GL(n,o)$ of the Borel in $GL(n,o/p)$. Yes, the affine Weyl group is normalizer of diagonals, modulo diagonal unit matrices. In my book/notes I verify (as did Bruhat and Tits ages ago, maybe it's in Bourbaki Lie ch. iv-...?) that the affine building for $SL(n,k)$ can be constructed via homothety classes of lattices. Then the building properties prove an abstracted Bruhat decomposition. One can also prove, following Tits, that the Coxeter-group property follows from the building set-up. –  paul garrett Dec 18 '11 at 18:22

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