Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Put $B_p := \left\{ \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} \in GL_2(Q_p) : a, b, c \in Q_p \right\}$ the subgroup of upper triangular matrices in $GL_2(Q_p)$, $Q_p$ denoting the $p$-adic rationals. I have already figured out that the modularity function is

$$\Delta \begin{pmatrix} a & b \\ 0 & c \end{pmatrix} = (|a|/|c|)^\lambda$$ i.e. if $\mu$ is the Haar measure on $B_p$ and $M$ is a measurable set then for any $x \in B_p$, $$\mu(Mx)=\Delta(x)\mu(M)$$

Does anybody know how to figure out that $\lambda=1$?


Fabian Werner

share|improve this question
I may be saying stupid, but is not the Haar measure of $B_p$ simply the restriction of the Haar measure of $GL_2({\mathbb Q}_p)$? Also, is not the Haar measure on $GL_2({\mathbb Q}_p)$ defined by the $\frac{1}{det}$ formula the way it is on $GL_2({\mathbb R})$ ? –  Ewan Delanoy Nov 29 '12 at 20:02
No! It is not like: G an LCH group, H a subgroup then the Haar measure on H is simply the one on G restricted to H. The problem becomes visible in the above example: Every set of the form $* \times * \times \{0\} \times *$ is a Null set in $Q_p^4$ (with respect to the unique product measure $\mu$ of the usual Haar measure, remark that the cases $\infty \cdot 0$ and $0 \cdot \infty$ are defined as $0$!) So as you say: the measure on $GL_2$ is roughly $(1/det) \cdot \mu$, so $B_p$ is a Null set with respect to that restriction (see next comment). –  Fabian Werner Dec 18 '12 at 10:50
So, let us set $B_N := \{ x \in B_p: |det(x)| \geq p^{-N}\}$, and let $\nu$ be the restriction of the measure on $GL_2$ then we see that $B_p = \cup_{N \in \mathbb{N}} B_N$ and consequently $\nu(B_N) = \int_{GL_2(Q_p)} 1_{B_N} dx/|det(x)| \leq const \int_{GL_2(Q_p)} 1_{B_N} dx = 0$ by the last comment. $|\cdot|$ denotes the p-adic norm. –  Fabian Werner Dec 18 '12 at 11:15

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Browse other questions tagged or ask your own question.