How to compute an integral?

Question

I am reading the lecture notes. I am trying to understand the prove of Lemma 0.0.1.1 on page 4. From line 3 to line 4 in the proof of Lemma 0.0.1.1., how to prove that $$ \int_{F^{n-1}} \hat{1}_{\mathfrak{p}^{-k}} \ (x) \pi \left( \begin{matrix} 1_{n-1} & x \\ 0 & 1 \end{matrix} \right) v dx = vol(\mathfrak{p}^{-k}) \int_{F^{n-1}} 1_{\mathfrak{p}^{m+k}} \ (x) \pi\left( \begin{matrix} 1_{n-1} & x \\ 0 & 1 \end{matrix} \right) v dx? $$ Where do we use the condition $\mathfrak{p}^m$ is the conductor of $\psi$ in the proof of Lemma 0.0.1.1? Thank you very much.

Answer 1

It is really only a statement about $\def\P#1{{{\mathfrak p}^{#1}}}\hat 1_\P k$, where $$ \hat I_{\P {-k} }(x) = \int_{F^{n-1}}I_{ \P {-k} } (y) \psi ( y^tx) \, dy = \int_{(\P {-k})^{n-1}}\psi ( y^tx) \, dy ,$$ and follows from a general fact about integration over (compact) topological groups (with Haar measure):

Fact: Suppose $\psi: G \mapsto \mathbb C^*$ is a character. Consider $$ \int_G \psi (g) dg.$$ If $\psi$ is not identically one, then the integral is zero. Otherwise, the integral is the volume of group.

Proof - if there exists $h\in G$ such that $\psi(h) \not = 1$, then $$\int_G \psi ( g )\, dg = \int_G \psi ( h g)\, dg = \psi (h) \int_G \psi( g) \, dg. $$ So the integral must vanish. On the other hand, if $\psi \equiv 1$, the integral is the volume.

The conductor $\P m$ is the largest subgroup of $F$ on which $\psi$ is trivial - correct? Therefore the statement follows from the above fact: $\hat I_{ \P {-k}}(x)$ is non-zero if and only if $x \in (\P {k+m})^{n-1}$, and then equal to the volume of $(\P {-k})^{n-1}$, i.e., $\hat I_{ \P {-k}} $is the corresponding indicator function $ I_{ \P {k+m}} $ multiplied by the volume.