So, after scouring the entirety of the internet, I managed to find one (and, so far, only one) source that actually explains how to invert the $p$-adic mellin transform:

$$\mathscr{M}_{p}\left\{ f\right\} \left(s\right)\overset{\textrm{def}}{=}\frac{p}{p-1}\int_{\mathbb{Q}_{p}^{\times}}\left|\mathfrak{z}\right|_{p}^{s-1}f\left(\mathfrak{z}\right)d\mathfrak{z},\textrm{ }\forall s\in\mathbb{C}$$ where $d\mathfrak{z}$ is the haar probability measure on $\mathbb{Z}_{p}$, and where $s$ is a complex variable. The source in question are these notes from the University of Chicago, specifically, pages 72 and 73. However, being an analyst, the word "(un)ramified" gives me heart palpitations; I'll be honest, I don't know exactly how to interpret equations (4.15) and (4.16) from the notes (pages 72 & 73), nor their accompanying text. I know just enough to know that the integral I wrote above is what the writer meant in writing (4.14).

However, because of the maddening $t$ business in the notes—among other things—I cannot understand how to correctly write down the inversion formula, among other things. Before I ask my questions, let me just say:

i. I have no interest in integrating over anything other than complex-valued functions on $\mathbb{Z}_{p}$. For what I'm trying to learn, all the business about field extensions are needless complications in these notes that I'm trying to do away with as I explain the material to myself.

ii. I have no interest in Representation theory; I'm just an analyst whose work has led him into non-archimedean waters, and would like to know what the rules are for swimming in these circumstances.


Is the correct way of writing (4.15):

$$\mathscr{M}_{p}^{-1}\left\{ F\right\} \left(\mathfrak{z}\right)=\frac{1}{2\pi i}\oint_{p^{-\sigma}\partial\mathbb{D}}\frac{F\left(s\right)}{\left|\mathfrak{z}\right|_{p}^{s}}ds$$ where $p^{-\sigma}\partial\mathbb{D}$ is the circle in $\mathbb{C}$ centered at $0$ of radius $p^{-\sigma}$, and where $\sigma$ is a positive real number.

Or is it: $$\mathscr{M}_{p}^{-1}\left\{ F\right\} \left(\mathfrak{z}\right)=\frac{1}{2\pi i}\int_{\sigma-i\infty}^{\sigma+i\infty}\frac{F\left(s\right)}{\left|\mathfrak{z}\right|_{p}^{s}}ds$$ where the contour is the line $\textrm{Re}\left(s\right)=\sigma$ in $\mathbb{C}$?

Or is it: $$\mathscr{M}_{p}^{-1}\left\{ F\right\} \left(\mathfrak{z}\right)=\frac{1}{2\pi i}\oint_{p^{-\sigma}\partial\mathbb{D}}\left|\mathfrak{z}\right|_{p}^{-s}F\left(-\frac{\ln s}{\ln p}\right)ds$$

Or is it something else, entirely?

Next, as a test-run, I tried to compute and then invert the transform of the constant $\mathbb{Z}_{p}$. Like in the notes, I computed: $$\mathscr{M}_{p}\left\{ \mathbf{1}_{\mathbb{Z}_{p}}\right\} \left(s\right)=\frac{1}{1-p^{-s}}$$ where $\mathbf{1}_{\mathbb{Z}_{p}}$ is the indicator function for $\mathbb{Z}_{p}$. This is the same as the notes, albeit they use $t=p^{-s}$ and write this as $\frac{1}{1-t}$.

However, when I try to use either of the above two attempts at interpreting the inversion formula (4.15), I end up with gobbledygook.

• The first formula I gave yields the constant function $$f\left(\mathfrak{z}\right)=\frac{1}{\ln p}$$

• The second formula yields (using the residue theorem): $$f\left(\mathfrak{z}\right)=\frac{1}{\ln p}\sum_{k\in\mathbb{Z}}\left|\mathfrak{z}\right|_{p}^{-\frac{2k\pi i}{\ln p}}=\frac{1}{\ln p}\sum_{k\in\mathbb{Z}}e^{2k\pi i\textrm{val}_{p}\left(\mathfrak{z}\right)}$$ which is always divergent.

• The third formula yields $f\left(\mathfrak{z}\right)=0$, because the integrand: $$\left|\mathfrak{z}\right|_{p}^{-s}F\left(-\frac{\ln s}{\ln p}\right)=\frac{\left|\mathfrak{z}\right|_{p}^{-s}}{1-p^{--\frac{\ln s}{\ln p}}}=\frac{\left|\mathfrak{z}\right|_{p}^{-s}}{1-s}$$ is holomorphic inside the unit disk.

None of these seem right to me, which makes me worry that none of the inversion formulae I've proposed are correct.

As such, I ask:

(1) What is the correct formula for the inversion of the $p$-adic mellin transform?

(2) What is the procedure for evaluating said integral? (Ex. Do I use the residue theorem, but ignore the existence of certain poles—if so, which ones?)

(3) More generally, given an $f:\mathbb{Z}_{p}\rightarrow\mathbb{C}$ so that the integral: $$F\left(s\right)=\int_{\mathbb{Z}_{p}\backslash\left\{ 0\right\} }\left(f\left(\mathfrak{z}\right)\right)^{s}d\mathfrak{z}$$ exists and has an analytic continuation to a meromorphic or entire function of $s\in\mathbb{C}$, how would I go about inverting it to re-obtain $f$? What would be the inversion formula, are there any special cares I should take in computing it (ignoring certain singularities when computing residues, etc.)? And to what extent can I re-obtain $f$ in this way?

To anyone who has read this far: thank you very much for your time!

  • $\begingroup$ I think that the correct answer is somewhere between your first and second formula: $$\mathscr M_p^{-1}\{F\}(\mathfrak z) = \frac1{2\pi i}\int_{\sigma-i\pi}^{\sigma+i\pi}\frac{F(s)}{\lvert\mathfrak z\rvert_p^s}\mathrm ds.$$ I'll check details and post later. However, in the meantime, while your obvious frustration is understandable, I think that it's not necessary to give such an apparently hostile pre- and post-lude to your question; perhaps it suffices to respond to such bad behaviour only after it happens here? $\endgroup$
    – LSpice
    Feb 9, 2020 at 1:56
  • 1
    $\begingroup$ I've deleted the hostility. xD I had intended it more as a safe-guard against someone giving a patronizing "oh, just look it up yourself" type response. Anyhow, thanks for your time. :D $\endgroup$
    – MCS
    Feb 9, 2020 at 2:53

2 Answers 2


In those notes, taking $F = \mathbf Q_p$, the scary term "unramified character" of $\mathbf Q_p^\times$ means a continuous homomorphism $\chi \colon \mathbf Q_p^\times \rightarrow \mathbf C^\times$ that is trivial (equal to $1$) on the units $\mathbf Z_p^\times$. The simplest example of such a character is the $p$-adic absolute value: $x \mapsto |x|_p$. This is continuous on $\mathbf Q_p^\times$ and it is definitely trivial on $\mathbf Z_p^\times$ since those are exactly the $p$-adic number of $p$-adic absolute value 1. A complex power $x \mapsto |x|_p^s$ for $s \in \mathbf C$ is also an unramified character of $\mathbf Q_p^\times$, and he is saying all unramified characters of $\mathbf Q_p^\times$ look like this for some $s$. Why is that?

Every nonzero $p$-adic number has the form $p^nu$ for some $n \in \mathbf Z$ and $u \in \mathbf Z_p^\times$. For an unramified character $\chi$ of $\mathbf Q_p^\times$, we have $\chi(u) = 1$, so $\chi(p^nu) = \chi(p^n) = \chi(p)^n$. The number $\chi(p)$ is in $\mathbf C^\times$, so we can write $\chi(p)$ as $1/p^s$ for some $s \in \mathbf C$. (This $s$ is not unique, but is well-defined up to adding an integer multiple of $2\pi i/\log p$ on account of looking at the complex solutions $s$ to $p^s = 1$.) Then $\chi(p^nu) = \chi(p)^n = (1/p^s)^n = (1/p^n)^s = |p^nu|_p^s$, so $\chi(x) = |x|_p^s$ for all $x \in \mathbf Q_p^\times$: $\chi$ is the $s$-th power of the basic unramified character $x \mapsto |x|_p$, where $s$ satisfies $\chi(p) = 1/p^s$. When a homomorphism $\mathbf Q_p^\times \rightarrow \mathbf C^\times$ is trivial on $\mathbf Z_p^\times$, it is continuous since it is locally constant (it is constant near 1 and a homomorphism) and is completely determined by its value at $p$.

The value of $\chi(p)$ can be arbitrary in $\mathbf C^\times$: for each $t \in \mathbf C^\times$ set $t = 1/p^s$ for some $s \in \mathbf C$ and define $\chi_t \colon \mathbf Q_p^\times \rightarrow \mathbf C^\times$ by the rule $\chi_t(p^nu) = t^n$ for $u \in \mathbf Z_p^\times$ and $n \in \mathbf Z$. This is a homomorphism, its value at $p$ is $t$, it is trivial on $\mathbf Z_p^\times$ ($\chi_t$ is "unramified"), and it is continuous since it is locally constant. Since $\chi_t(p^nu) = t^n = (1/p^s)^n = (1/p^n)^s = |p^nu|_p^s$, we have $\chi_t(x) = |x|_p^s$ for all $x \in \mathbf Q_p^\times$. This is why he says for each $t \in \mathbf C^\times$ there is a unique unramified character $\chi$ of $\mathbf Q_p^\times$ with $\chi(p) = t$: that $\chi$ is $\chi_t$.

The connected components of the group of characters $\Omega(\mathbf Q_p^\times)$ are entirely determined by how the characters look on $\mathbf Z_p^\times$: two characters of $\mathbf Q_p^\times$ are in the same connected component exactly when they are equal on $\mathbf Z_p^\times$, and by continuity a character $\mathbf Z_p^\times \rightarrow \mathbf C^\times$ is trivial on some neighborhood $1 + p^n\mathbf Z_p$ of 1 (a subgroup!) since $\mathbf C^\times$ has no subgroup in a neighborhood of 1 other than $\{1\}$. Therefore a character on $\mathbf Z_p^\times$ is a homomorphism to $\mathbf C^\times$ on some quotient group $\mathbf Z_p^\times/(1 + p^k\mathbf Z_p)\cong (\mathbf Z/p^k\mathbf Z)^\times$, which is finite. Going the other way, for each homomorphism $(\mathbf Z/p^k\mathbf Z)^\times \rightarrow \mathbf C^\times$ we can lift it to a character $\eta$ on $\mathbf Z_p^\times$ using the composite map $$ \mathbf Z_p^\times \rightarrow \mathbf Z_p^\times/(1+p^k\mathbf Z_p) \cong (\mathbf Z/p^k\mathbf Z_p)^\times \rightarrow \mathbf C^\times $$ (this is automatically continuous since it is a homomorphism and it is trivial on the neighborhood $1 + p^k\mathbf Z_p$ of 1) and then we can multiply this by an unramified character $\chi$ to get a character of $\mathbf Q_p^\times$: $p^nu \mapsto \chi(p)^n\eta(u)$. In other notation, since unramified characters of $\mathbf Q_p^\times$ are just complex powers $|\cdot|_p^s$, each character of $\mathbf Q_p^\times$ is $|\cdot|_p^s\eta$ where $s \in \mathbf C$ and $\eta$ is a character of $\mathbf Z_p^\times$: each connected component can be labeled by the common $\eta$ (restriction to $\mathbf Z_p^\times$) for all characters in that component. (The choice of $s$ for a character really is in $\mathbf C/(2\pi i/\log p)\mathbf Z$, a cylinder, which topologically is the same as $\mathbf C^\times$ using $s + (2\pi i/\log p)\mathbf Z \mapsto 1/p^s$.)

For $x \in \mathbf Q_p^\times$, written as $p^nu$ where $n \in \mathbf Z$ and $u \in \mathbf Z_p^\times$, write $u$ as $u_x$ to indicate its dependence on $x$. Then for each character $\eta$ of some $(\mathbf Z/p^k\mathbf Z)^\times$ and $s \in \mathbf C$, we get a character $\chi$ of $\mathbf Q_p^\times$ by $\chi(x) = |x|_p^s\eta(u_x\bmod p^k)$. (Note that $\eta(p)$ makes no sense.) All characters of $\mathbf Q_p^\times$ look like this.

Prop. 4.6 is about continuous functions $\mathbf Q_p^\times \rightarrow \mathbf C$ with compact support. Note that your test-run example of the characteristic function of $\mathbf Z_p$, viewed as a function on $\mathbf Q_p^\times$ by taking $0$ out of its domain, does not have compact support in $\mathbf Q_p^\times$: the set $\mathbf Z_p - \{0\}$ is not compact in $\mathbf Q_p^\times$ just as $(0,1]$ and $[-1,1] - \{0\}$ are not compact in $\mathbf R^\times$. Therefore this is not a good example for a test-run for Prop. 4.6 (unlike Prop. 4.7).

For a better choice of test-run for Prop. 4.6, let $\xi_A$ be notation for the characteristic function of a set $A$ (1 if the variable is in $A$ and 0 otherwise). For $a \in \mathbf Q_p^\times$ and $n \in \mathbf Z$ chosen large enough so that $|a|_p > 1/p^n$, set $\phi = \xi_{a + p^n\mathbf Z_p}$: this is the characteristic function of the ball $a + p^n\mathbf Z_p$, which is a subset of $\mathbf Q_p^\times$ since we can't have $a + p^nx = 0$ for $x \in \mathbf Z_p$, as $|p^nx|_p \leq 1/p^n < |a|_p$. (If you don't like the characteristic function of a general ball in $\mathbf Q_p$ not containing $0$, consider the special case $a = 1$: $\xi_{1 + p^n\mathbf Z_p}$ for $n \geq 1$.)

Letting $|a|_p = 1/p^m$, so $m<n$, the Mellin transform $(M\phi)(\chi)$ of a character $\chi$ of $\mathbf Q_p^\times$ is the following integral $$ (M\phi)(\chi) = \int_{\mathbf Q_p^\times} \xi_{a + p^n\mathbf Z_p}(x)\chi(x)d^\times x, $$ where I write $d^\times x$ for the (multiplicative) Haar measure on $\mathbf Q_p^\times$ that you write as $p/(p-1) dx/|x|_p$. It is the Haar measure on $\mathbf Q_p^\times$ that gives the compact open subgroup $\mathbf Z_p^\times$ measure 1. After some calculation that I omit (tell me if you can work this out), we get $$ (M\phi)(\chi) = \chi(a)\int_{1 + p^{n-m}\mathbf Z_p} \chi(y)d^\times y, $$ which is $\chi(a)$ times the integral of the multiplicative character $\chi$ over the compact multiplicative group $1 + p^{n-m}\mathbf Z_p$. The value of this is determined by whether or not $\chi$ is trivial on $1 + p^{n-m}\mathbf Z_p$: if $\chi \not\equiv 1$ on $1 + p^{n-m}\mathbf Z_p$ then $$ (M\phi)(\chi) = 0, $$ and if $\chi \equiv 1$ on $1 + p^{n-m}\mathbf Z_p$ then $$ (M\phi)(\chi) = \chi(a)\int_{1 + p^{n-m}\mathbf Z_p} d^\times y = \frac{\chi(a)}{[\mathbf Z_p^\times:1 + p^{n-m}\mathbf Z_p]} = \frac{\chi(a)}{p^{n-m-1}(p-1)}. $$ There are only finitely many connected components containing the $\chi$ where $\chi \equiv 1$ on $1 + p^{n-m}\mathbf Z_p$ since such $\chi$ are determined on $\mathbf Z_p^\times$ (not on $\mathbf Q_p^\times$!) by their values on $\mathbf Z_p^\times/(1 + p^{n-m}\mathbf Z_p)$, which is a finite group (so it has only finitely many homomorphisms to $\mathbf C^\times$). Thus $M\phi$ is a "polynomial" in the sense defined above Prop. 4.6.

I have not yet addressed your question about the Mellin inversion formula. Consider this a partial answer so far. I will save what I have written and come back to this later.

  • $\begingroup$ Woo! Thanks so much! :D $\endgroup$
    – MCS
    Feb 9, 2020 at 3:58
  • $\begingroup$ Some questions: (1) does any of this simplify (even slightly) if we assume that the functions we are transforming are defined only on $\mathbb{Z}_{p}$ (if so, how)? (2) How can we write a generic $\eta$ here "explicitly"? (Ex: like how, for a character $\xi$ on $\mathbb{Z}_{p}$, the "explicit" form of $\xi$ is $\xi\left(\mathfrak{z}\right)=e^{-2\pi i\left\{ t\mathfrak{z}\right\} _{p}}$ for some $t\in\mathbb{Z}\left[\frac{1}{p}\right]/\mathbb{Z}$, treated here as a rational number in $\left[0,1\right)$). I can't know if I can do the integral or not until I see it written explicitly. xD $\endgroup$
    – MCS
    Feb 9, 2020 at 4:21
  • $\begingroup$ So far, I've got $\chi\left(x\right)=\left|x\right|_{p}^{s}\times?$ as the formula for the $\chi$ in the integral. Need to know what the ? is. $\endgroup$
    – MCS
    Feb 9, 2020 at 4:31
  • $\begingroup$ I am using $\xi_A$ in my post for the characteristic function of a set $A$, so please don't use $\xi$ in a comment to mean a character on $\mathbf Z_p$ (or $\mathbf Q_p$). $\endgroup$
    – KConrad
    Feb 9, 2020 at 4:45
  • $\begingroup$ So far, what I've got for the integral is: $\int_{\mathbb{Q}_{p}^{\times}}\left[\mathfrak{z}\overset{p^{n}}{\equiv}a\right]\left|\mathfrak{z}\right|_{p}^{s-1}\eta\left(\mathfrak{z}\right)d\mathfrak{z}=\frac{1}{p^{ms}}\int_{\frac{a}{p^{m}}+p^{n-m}\mathbb{Z}_{p}^{\times}}\left|\mathfrak{z}\right|_{p}^{s-1}\eta\left(p^{m}\mathfrak{z}\right)d\mathfrak{z}$ where $\left|a\right|_{p}=p^{-m}>p^{-n}$, and where the $\eta$ is the one you mentioned above. $\left[\mathfrak{z}\overset{p^{n}}{\equiv}a\right]$ is the iverson bracket for $a+p^{n}\mathbb{Z}_{p}$ (1 if true, 0 if false). $\endgroup$
    – MCS
    Feb 9, 2020 at 5:30


The inversion of $F(s)$, in your notation is

$$f(x) = \frac{\mathrm{ln}(p)}{2 \pi i} \int_{ \{ s=ir \ : r \in (-\pi/\mathrm{ln}(p), \pi/\mathrm{ln}(p)] \}} F(s)|x|^{-s} ds = \frac{\mathrm{ln}(p)}{2 \pi i} \int_{ \{ s=ir \ : r \in (-\pi/\mathrm{ln}(p), \pi/\mathrm{ln}(p)] \}} F(s)p^{\mathrm{val}(x)s} ds$$

Here if $x=p^nu$ where $u \in \mathbb{Z}_p^{\times}$ is a unit $\mathrm{val}(x)=n$. You can evaluate this integral in the normal way, you don't ignore poles. It seems like the closest answer you got was the constant $\frac{1}{\mathrm{ln}(p)}$, but this is incomplete because $f(x)$ is not supported on $\mathbb{Q}_p^{\times}$ but rather on $\{ \mathrm{val}(x) \geq 0 \} $. You can see this because in the series expansion of the rational function $\frac{1}{1-p^{-s}} = 1 + p^{-s}+p^{-2s} + \cdots $ the terms consist of polynomials in the variable $p^{-s}$ and each integral $\int_{ \{ |s|=1 \} } p^{-ns}|x|^{-s}$ is non-zero only when the valuation of $x$ is $n$ for $n \geq 0$. Finally the $\mathrm{ln}(p)$ terms just comes to account for the fact that we are parametrizing the complex circle by $r \mapsto p^{-ir}$ so as to give complex circle measure $1$.


It is helpful to state things first in terms of abstract harmonic analysis on locally compact abelian (=LCA) groups might make things clearer.

For any LCA group $A$ let $\widehat{A}$ be the group of it's unitary characters. If $f$ is a function on $A$ it's Fourier transform is $\chi \mapsto \widehat{f}(\chi) = \int_{A} f(x)\chi^{-1}(x)dx$ with $dx$ a fixed Haar measure on $A$. My Pontryagin duality is a unique measure $d\chi$ on the LCA group $\widehat{A}$ dual to $dx$ satisfying the Fourier inversion

$$ f(e) = \int_{\widehat{A}} \widehat{f}(\chi) d\chi $$

$$ f(y) = \int_{\widehat{A}} \widehat{f}(\chi)\chi(y) d\chi $$

$\mathbb{Q}_p^{\times}$ is your abelian group, under multiplication. It's Pontryagin dual is the space of unitary complex characters $ \widehat{\mathbb{Q}_p} =\{ \chi: \mathbb{Q}_p^{\times} \to \mathbb{T} \}$ with $\mathbb{T}$ the complex circle. For any smooth compactly supported function on $\mathbb{Q}_p^{\times}$ it's abstract Fourier transform is the Mellin transform $\widehat{f}(\chi) = \int_{\mathbb{Q}_p^{\times}} f(x) \chi(x) dx$

Because $\mathbb{Q}_p^{\times} = \mathbb{Z}_p^{\times} \times p^{\mathbb{Z}}$ it's unitary dual is $\widehat{\mathbb{Z}_p^{\times}} \times \widehat{\mathbb{Z}}$. The unramified characters are the characters which are trivial on $\mathbb{Z}_p^{\times}$, i.e. they are restrictions to $\{ 1 \} \times \widehat{\mathbb{Z}}$. The unitary characters of $\mathbb{Z}$ can be identified with unit norm complex numbers $\mathbb{T} = \{ |z|=1 \}$. The complex parameter refers to the exponent $s$ in the unitary character determined by $ | \cdot |^{s}:p \mapsto |p|_{p}^s=p^{-s}$ (or however you want to normalize it, but this is natural for number theory). To conform with Ngo's noration let us make a change of variables $t = p^{-s}$ and call this unitary character $\chi_t$. So we should think of the function "$\frac{1}{1-p^{-s}}$" more precisely as a function $L$ on the space of unitary characters:

$$\chi_t \mapsto L(\chi_t) = \frac{1}{1-\chi_t(p)}$$ if $\chi_t$ is an unramified chatacter $$ \eta \mapsto L(\eta) = 0 $$ If $\eta$ is not unramified.

The dual measure $d\chi_{t}$ is always a Haar measure on the unitary dual, which in our case is $\mathbb{T}$, so is a multiple of $dt/t$. Explicitly it looks like $\frac{\mathrm{ln}(p)}{2\pi i} \frac{dt}{t}$ (because the contour is over a parameter between $\frac{-\pi}{\mathrm{ln}(p)} $ and $\frac{\pi}{\mathrm{ln}(p)}$).

We can write $L(\chi_t) = \chi_{t}(e) + \chi_{t}(p) + \chi_{t}(p^2) + \cdots = \chi_{t}(e) + l_p \chi_{t}(e) + l_{p^2} \chi_{t}(e) + \cdots$ where $l_p$ is translation by $p$. Strictly speaking this is not a polynomial function on the space of unitary characters but a rational one, but formally we can do everything for the terms in the sum and then take a limit.

Thus it is enough to Mellin-invert the function $\widehat{f_0}: \chi_t \mapsto \chi_{t}(e) = 1$ and deduce the rest of it by translating. Using the Fourier inversion formula and the fact that $\widehat{f_0}$ is supported on $\mathbb{Z}_p^{\times}$-invariant characters implies that $f_0$ is $\mathbb{Z}_p^{\times}$-invariant. Thus it is enough to know it's value on $\mathbb{Z}_p^{\times}$ and it's translates $p^n\mathbb{Z}_p^{\times}$.

$$f_0(e)=\mathrm{const} \int_{ |t|=1 } \chi_t(e) \frac{dt}{t} = \mathrm{const} \int_{ |t|=1 } 1 \frac{dt}{t}$$

Thus $f_0(\mathbb{Z}_p^{\times})=1$. Arguing by translation by either $p$ or $p^{-1}$ we obtain

$$f_0(p) = \mathrm{const} \int_{ |t|=1 } \chi_t(p)\frac{dt}{t} = \int_{ |t|=1 } dt=0$$ or

$$f(p^{-1}=\mathrm{const} \int_{ |t|=1 } \chi_t(p^{-1})\frac{dt}{t} = \int_{ |t|=1 } t^{-2}dt =0.$$

So $f_0 = 1_{ \mathbb{Z}_p^{\times}}$.

Arguing similarly with $\chi_{t} \mapsto l_p(\chi)(e)$ produces a function $f_{1}$ such that $l_p(f)(\mathbb{Z}_p^{\times}) =1$ and is invariant under and supported in $\mathbb{Z}_p^{\times}$. In other words it is the characteristic function of the coset $p \mathbb{Z}_p^{\times}$. And so on for the rest of the terms. Adding them all up yields $ \sum_n f_n = 1_{\mathbb{Z}_p - \{ 0 \}}$ as exepcted.

  • $\begingroup$ I am fine with pontryagin duality. The moment anyone breathes a word about group representations, I run screaming for the hills. $\endgroup$
    – MCS
    Feb 9, 2020 at 4:13
  • $\begingroup$ You completely lose me by paragraph 3. I don't know the formula for a "character of $\mathbb{Q}_{p}^{\times}$ that is invariant under $\mathbb{Z}_{p}^{\times}". I also don't understand your notation from that point onward, and cannot follow any of it. $\endgroup$
    – MCS
    Feb 9, 2020 at 4:27
  • $\begingroup$ I know the algorithm for computing $\int_{\mathbb{Z}_{p}}f\left(\mathfrak{z}\right)e^{-2\pi i\left\{ t\mathfrak{z}\right\} _{p}}d\mathfrak{z}$ when $f$ is a schwartz-bruhat function on $\mathbb{Z}_{p}$ and when $t\in\mathbb{Z}\left[\frac{1}{p}\right]/\mathbb{Z}$. I don't know how to do what you are describing. At this point, I just want to know the formula-rules for computing these p-adic mellin integral transforms and their inverses. $\endgroup$
    – MCS
    Feb 9, 2020 at 5:45
  • $\begingroup$ I tried to clarify my answer: the unramified characters are trivial on the unit group $\mathbb{Z}_p^{\times}$, and so are determined by their values at $p$. This introduces the complex variable. The inversion done by the contour integral, however you choose to parametrize it and keep in mind to use the Haar measure on the unit circle. $\endgroup$
    – wskrsk
    Feb 9, 2020 at 10:02
  • $\begingroup$ I don't understand your notation, you aren't parameterizing the integrals, you aren't parameterizing the characters, and you've said absolutely nothing about the heart of my question: giving me the correct formula for the inverse transform and explaining how (if at all) one can use the residue theorem to evaluate it. Although I appreciate you trying to explain the ideas behind the computations, I have no room for that in my head at the moment. Right now, I need formulae and algorithms to memorize. I can focus on understanding the reasoning behind them once I know how to do them. $\endgroup$
    – MCS
    Feb 9, 2020 at 20:39

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