On Defining the Fourier Transform and Performing Changes of Variable on Quotient Subgroups of $\mathbb{Q}/$

Much to my dismay, in my work the more number-theoretic side of harmonic analysis (ex: the fourier transform on the adeles, on the profinite integers, etc.), I have found myself struggling with technicalities that emerge from frustratingly simple issues—so simple (and yet, so technical) that I haven’t been able to find anything that might shine any light on the matter.

Let $ L_{\textrm{loc}}^{2}\left(\mathbb{Q}\right)$ be the space of functions $ f:\mathbb{Q}\rightarrow\mathbb{C}$ which satisfy $ \sum_{t\in T}\left|f\left(t\right)\right|^{2}<\infty$ for all bounded subsets $ T\subseteq\mathbb{Q}$ . Letting $ \mu$ be any positive integer, it is easy to show that any functions which is both $ \mu$ -periodic (an $ f:\mathbb{Q}\rightarrow\mathbb{C}$ such that $ f\left(t+\mu\right)=f\left(t\right)$ for all $ t\in\mathbb{Q})$ and in $ L_{\textrm{loc}}^{2}\left(\mathbb{Q}\right)$ is necessarily an element of $ L^{2}\left(\mathbb{Q}/\mu\mathbb{Z}\right)$ , the complex hilbert space of functions $ f:\mathbb{Q}/\mu\mathbb{Z}\rightarrow\mathbb{C}$ so that: $ $ \sum_{t\in\mathbb{Q}/\mu\mathbb{Z}}\left|f\left(t\right)\right|^{2}<\infty$ $ Equipping $ \mathbb{Q}/\mu\mathbb{Z}$ with the discrete topology, we can utilize Pontryagin duality to obtain a Fourier transform: $ \mathscr{F}_{\mathbb{Q}/\mu\mathbb{Z}}$ . The ideal case is when $ \mu=1$ . There, $ \mathscr{F}_{\mathbb{Q}/\mathbb{Z}}$ is an isometric hilbert space isomorphism from $ L^{2}\left(\mathbb{Q}/\mathbb{Z}\right)$ to $ L^{2}\left(\overline{\mathbb{Z}}\right)$ , where: $ $ \overline{\mathbb{Z}}\overset{\textrm{def}}{=}\prod_{p\in\mathbb{P}}\mathbb{Z}_{p}$ $ is the ring of profinite integers, where $ \mathbb{P}$ is the set of prime numbers, and where $ L^{2}\left(\overline{\mathbb{Z}}\right)$ is the space of functions $ \check{f}:\overline{\mathbb{Z}}\rightarrow\mathbb{C}$ which are square-integrable with respect to the haar probability measure $ d\mathfrak{z}=\prod_{p\in\mathbb{P}}d\mathfrak{z}_{p}$ on $ \overline{\mathbb{Z}}$ .

The first sign of trouble was when I learned that, for any integers $ \mu,\nu$ , the (additive) quotient groups $ \mathbb{Q}/\mu\mathbb{Z}$ and $ \mathbb{Q}/\nu\mathbb{Z}$ are group-isomorphic to one another, and thus, that both have $ \overline{\mathbb{Z}}$ as their Pontryagin dual. From my point of view, however, this isomorphism seems to only cause trouble. In my work, I am identifying $ L^{2}\left(\mathbb{Q}/\mu\mathbb{Z}\right)$ with the set of $ \mu$ -periodic functions $ f:\mathbb{Q}\rightarrow\mathbb{C}$ which are square integrable with respect to the counting measure on $ \mathbb{Q}\cap\left[0,\mu\right)$ . As such, $ \mathbb{Q}/\mu\mathbb{Z}$ and $ \mathbb{Q}/\nu\mathbb{Z} $ cannot be “the same” from my point of view, because $ f\left(t\right)\in L^{2}\left(\mathbb{Q}/\mu\mathbb{Z}\right)$ need not imply that $ f\left(t\right)\in L^{2}\left(\mathbb{Q}/\nu\mathbb{Z}\right)$ .

In my current work, I am dealing with a functional equation of the form:$ $ \sum_{n=0}^{N-1}g_{n}\left(t\right)f\left(\frac{a_{n}t+b_{n}}{d_{n}}\right)=0$ $ where $ N$ is an integer $ \geq2$ , where the $ g_{n}$ s are known periodic functions, where $ f$ is an unknown function in $ L^{2}\left(\mathbb{Q}/\mathbb{Z}\right)$ (i.e., $ f\left(t\right)\in L_{\textrm{loc}}^{2}\left(\mathbb{Q}\right)$ and $ f\left(t+1\right)=f\left(t\right)$ for all $ t\in\mathbb{Q})$ , and where $ a_{n},b_{n},d_{n}$ are integers with $ \gcd\left(a_{n},d_{n}\right)=1$ for all $ n$ . For brevity, I’ll write: $ $ \varphi_{n}\left(t\right)\overset{\textrm{def}}{=}\frac{a_{n}t+b_{n}}{d_{n}}$ $ Because $ \varphi_{n}\left(t+1\right)$ need not equal $ \varphi_{n}\left(t\right)+1$ , the individual functions $ f\circ\varphi_{n}$ , though periodic and in $ L_{\textrm{loc}}^{2}\left(\mathbb{Q}\right)$ , are not necessarily going to be of period $ 1$ . Letting $ p$ denote the least common multiple of the periods of $ g_{n}$ and the $ f\circ\varphi_{n}$ s, I can view the functional equation as existing in $ L^{2}\left(\mathbb{Q}/p\mathbb{Z}\right)$ , and as such, I hope to be able to simplify it by applying the fourier transform.

Letting $ e^{2\pi i\left\langle t,\mathfrak{z}\right\rangle }$ denote the duality pairing between elements $ t\in\mathbb{Q}$ (or $ \mathbb{Q}/\mu\mathbb{Z}$ ) and $ \mathfrak{z}\in\overline{\mathbb{Z}}$ , the idea is to multiply the functional equation by $ e^{2\pi i\left\langle t,\mathfrak{z}\right\rangle }$ , sum over an appropriate domain of $ t$ (ideally, $ \mathbb{Q}/p\mathbb{Z}$ ), make a change of variables in $ t$ to move one of the $ \varphi_{n}\left(t\right)$ s out of $ f$ and into $ e^{2\pi i\left\langle t,\mathfrak{z}\right\rangle }$ , pull out terms from this character, and then invert the fourier transform to return to $ L^{2}\left(\mathbb{Q}/p\mathbb{Z}\right)$ with a vastly simpler equation. My main difficulty can be broken into three parts:

(1) Is taking the least common multiple of the periods to reformulate the functional equation as one over $ L^{2}\left(\mathbb{Q}/p\mathbb{Z}\right)$ legal?

(2) I know that the fourier transform $ \mathscr{F}_{\mathbb{Q}/\mathbb{Z}}:L^{2}\left(\mathbb{Q}/\mathbb{Z}\right)\rightarrow L^{2}\left(\overline{\mathbb{Z}}\right)$ is given by:$ $ \mathscr{F}_{\mathbb{Q}/\mathbb{Z}}\left\{ f\right\} \left(\mathfrak{z}\right)\overset{\textrm{def}}{=}\sum_{t\in\mathbb{Q}/\mathbb{Z}}f\left(t\right)e^{2\pi i\left\langle t,\mathfrak{z}\right\rangle }$ $ and that the inverse transform is: $ $ \mathscr{F}_{\mathbb{Q}/\mathbb{Z}}^{-1}\left\{ \check{f}\right\} \left(t\right)\overset{\textrm{def}}{=}\int_{\overline{\mathbb{Z}}}\check{f}\left(\mathfrak{z}\right)e^{-2\pi i\left\langle t,\mathfrak{z}\right\rangle }d\mathfrak{z}$ $ However, I am at a loss as to what formula to use for $ \mathscr{F}_{\mathbb{Q}/p\mathbb{Z}}$ and its inverse, and for two reasons. On the one hand, because $ \mathbb{Q}/\mathbb{Z}$ and $ \mathbb{Q}/p\mathbb{Z}$ are group-isomorphic, what is to stop me from using the same formula for their fourier transforms? On the other hand, if I use a modified formula—say:$ $ \mathscr{F}_{\mathbb{Q}/p\mathbb{Z}}\left\{ f\right\} \left(\mathfrak{z}\right)=\sum_{t\in\mathbb{Q}/p\mathbb{Z}}f\left(t\right)e^{2\pi i\left\langle \frac{t}{p},\mathfrak{z}\right\rangle }$ $ does the fact that $ \mathscr{F}_{\mathbb{Q}/p\mathbb{Z}}\left\{ f\right\} \left(\mathfrak{z}\right)\in L^{2}\left(\overline{\mathbb{Z}}\right)$ then mean that I can recover $ f$ by applying $ \mathscr{F}_{\mathbb{Q}/\mathbb{Z}}^{-1}$ , or do I have to also modify it in order to make everything consistent? Knowing the correct formula for the fourier transform on $ L^{2}\left(\mathbb{Q}/p\mathbb{Z}\right)$ and its inverse is essential.

(3) I would like to think that performing a change-of-variables for a sum of the form:$ $ \sum_{\mathbb{Q}/r\mathbb{Z}}f\left(\alpha t+\beta\right)$ $ (where $ \alpha,\beta,r\in\mathbb{Q}$ , with $ \alpha\neq0$ and $ r=\frac{p}{q}>0$ ) would be a relatively simple matter, but, that doesn’t appear to be the case. For example, if $ t\in\mathbb{Q}\rightarrow f\left(\alpha t+\beta\right)\in\mathbb{C}$ is not a $ r$ -periodic function, then this sum is not well-defined over the quotient group $ \mathbb{Q}/r\mathbb{Z}$ . Worse yet—supposing $ f$ is $ r$ -periodic take a look at this: write elements of $ \mathbb{Q}/r\mathbb{Z}$ in co-set form: $ t+r\mathbb{Z}$ , where $ t\in\mathbb{Q}$ . Then, make the change-of-variable $ \tau=\alpha t+\beta$ . Consequently, the set of all $ \tau$ is:$ $ \alpha\left(\mathbb{Q}/r\mathbb{Z}\right)+\beta=\left\{ \alpha\left(t+r\mathbb{Z}\right)+\beta:t\in\mathbb{Q}\right\} =\left\{ \tau+\alpha r\mathbb{Z}:t\in\mathbb{Q}\right\}$ $ . Here is where things get loopy.

(1) Since $ \alpha,\beta\in\mathbb{Q}$ with $ \alpha\neq0$ , the map $ \varphi\left(t\right)\overset{\textrm{def}}{=}\alpha t+\beta$ is a bijection of $ \mathbb{Q}$ . As such, I would think that:$ $ \left\{ \tau+\alpha r\mathbb{Z}:t\in\mathbb{Q}\right\} =\left\{ \tau+\alpha r\mathbb{Z}:\varphi^{-1}\left(\tau\right)\in\mathbb{Q}\right\} =\left\{ \tau+\alpha r\mathbb{Z}:\tau\in\varphi\left(\mathbb{Q}\right)\right\}$ $ and hence:$ $ \alpha\left(\mathbb{Q}/r\mathbb{Z}\right)+\beta=\left\{ \tau+\alpha r\mathbb{Z}:\tau\in\mathbb{Q}\right\} =\mathbb{Q}/\alpha r\mathbb{Z}$ $ Using this approach, I obtain:$ $ \sum_{t\in\mathbb{Q}/r\mathbb{Z}}f\left(\alpha t+\beta\right)=\sum_{\tau\in\mathbb{Q}/\alpha r\mathbb{Z}}f\left(\tau\right)$ $

(2) Since $ r=\frac{p}{q}$ , decompose $ \mathbb{Z}$ into its equivalence classes mod $ q$ :$ $ \left\{ \tau+\alpha r\mathbb{Z}:\tau\in\mathbb{Q}\right\} =\bigcup_{k=0}^{q-1}\left\{ \tau+\alpha r\left(q\mathbb{Z}+k\right):\tau\in\mathbb{Q}\right\}$ $ and so:$ $ \sum_{t\in\mathbb{Q}/r\mathbb{Z}}f\left(\alpha t+\beta\right)=\sum_{k=0}^{q-1}\sum_{\tau\in\mathbb{Q}/\alpha rq\mathbb{Z}}f\left(\tau+\alpha rk\right)$ $ On the other hand, for each $ k$ :$ $ \left\{ \tau+\alpha r\left(q\mathbb{Z}+k\right):\tau\in\mathbb{Q}\right\} =\left\{ \tau+\alpha rk+\alpha rq\mathbb{Z}:\tau\in\mathbb{Q}\right\}$ $ and, since $ \tau\mapsto\tau+\alpha rk$ is a bijection of $ \mathbb{Q}$ , the logic of (1) would suggest that:$ $ \left\{ \tau+\alpha r\left(q\mathbb{Z}+k\right):\tau\in\mathbb{Q}\right\} =\left\{ \tau+\alpha rq\mathbb{Z}:\tau\in\mathbb{Q}\right\}$ $ for all $ k$ . But then, that gives:$ $ \sum_{k=0}^{q-1}\sum_{\tau\in\mathbb{Q}/\alpha rq\mathbb{Z}}f\left(\tau+\alpha rk\right)=\sum_{t\in\mathbb{Q}/r\mathbb{Z}}f\left(\alpha t+\beta\right)=q\sum_{\tau\in\mathbb{Q}/\alpha rq\mathbb{Z}}f\left(\tau\right)$ $ which hardly seems right.

Quotient metrics in length spaces

If I collapse a (say, closed) set in a length space, I obtain a length space: is there some literature on this?

We consider length spaces as defined by Gromov and others. [However the case of a Riemannian distance already leads to interesting examples of what I am writing]. If $ X$ is one such space, it has a distance $ d$ which can be recovered by the length of curves. Suppose $ X$ is path connected. Let $ c:[a,b]\to X$ be a curve, define $ |c'(t)|=\lim_{\epsilon\to0}\sup_{|u-t|,|v-t|\le\epsilon}\frac{d(c(u),c(v))}{|u-v|}$ and $ $ L(c)=\int_a^b|c'(t)|dt, $ $ and set $ $ D(x,y)=\inf\{L(c) \text{ $ c$ is a curve having endpoints $ x$ and $ y$ }\}. $ $ We are interested in cases where $ d=D$ and where the distance $ D=d$ is realized by lengths of geodesics (i.e. the hard work has been already done).

Now, let $ E\subseteq X$ be closed and define a distance $ D_X$ on $ X\setminus E\cup\{E\}$ ($ X$ with $ E$ collapsed to a point): $ $ L_E(c)=\int_a^b|c'(t)|\chi_E(c(t))dt, $ $ and set $ $ D_E(x,y)=\inf\{L_E(c) \text{ $ c$ is a curve having endpoints $ x$ and $ y$ }\}, $ $ if $ x,y\in X\setminus E$ , and $ D_E(x,E)=D(x,E)=\inf_{y\in E}D(x,y)$ .

The space $ X\setminus E\cup\{E\}$ is clearly a length space w.r.t. the distance $ D_E$ .

The reason I find these objects interesting is that, if $ E$ is the smooth boundary of an open subset of $ X$ , a nice Riemannian manifold, then the points of $ E$ play the role of the unit vectors in the tangent space of a point; this meaning that they can parametrize those geodesics leaving $ E$ which, at least locally, minimize the distance from $ E$ . Even in the case of the Euclidean plane one obtains interesting pictures.

I would be very surprised if no one had developed this viewpoint in the past.

Show convex function is increasing in both variables of difference quotient using alternative definition of convexity

Let $ \phi$ be a function that satisfies $ $ \frac{\phi (t) – \phi (s)}{t – s} \leq \frac{\phi (u) – \phi (t)}{u – t}$ $

where $ s < t < u$ .

Is it possible to directly use this definition of convexity to prove that $ \phi$ ‘s difference quotients are increasing in each variable, i.e., $ $ \frac{\phi (u) – \phi (s)}{u – s} \leq \frac{\phi (u) – \phi (t)}{u – t}$ $ and $ $ \frac{\phi (t) – \phi (s)}{t – s} \leq \frac{\phi (u) – \phi (s)}{u – s}$ $

where again $ s < t < u$ .

Background: $ \phi$ is convex and using the usual definition of convexity the proof is fairly direct. So I’m curious if there’s a direct proof from this definition of convexity.

Maximal unramified quotient of $E[p]$ for the action of $G_{\mathbb{Q}_p}$

Let $ E$ be an elliptic curve with good and ordinary reduction at an odd prime $ p$ . Suppose $ E[p]$ denotes the $ p$ -torsion points of $ E$ and $ G_{\mathbb{Q}_p} := \text{Gal}(\overline{\mathbb{Q}_p}/\mathbb{Q}_p)$ .

In the article `Selmer group and congruences (page 6)’, Greenberg says that one can characterize $ \widetilde{E}[p]$ as the maximal unramified quotient of $ E[p]$ for the action of $ G_{\mathbb{Q}_p}$ where $ \widetilde{E}$ denotes the reduction of $ E$ in $ \mathbb{F}_p$ .

This is so because $ p$ is assumed to be odd and therefore the action of the inertia subgroup of $ G_{\mathbb{Q}_p}$ on the kernel of the reduction map $ \pi: E[p] \longrightarrow \widetilde{E}[p]$ is nontrivial.

It will be every helpful if someone can explain how `$ p$ being odd’ is playing a role in proving the non trivial action of the inertia subgroup on the kernel of the reduction map $ \pi$ ?

Extenstion functor on quotient map

Let $ \rho$ be the half sum of positive roots in $ \Phi^+$ , $ M_x$ be the Verma module with highest weight $ x\cdot(-2\rho)$ and $ L_w$ be the simple highest weight module with highest weight $ w\cdot(-2\rho)$ .

I want to show $ \mathrm{Ext}^i_{\mathcal{O}}(M_x,L_w)\neq 0\implies\mathrm{Ext}^i_{\mathcal{O}}(M_x,M_w)\neq 0$ .

My attempt:

Let $ T(B):=\mathrm{Hom}_{\mathcal{O}}(M_x,B)$ . Denote by $ R^iT$ the $ i$ -th right derived functor of $ T$ . Then $ R^iT(B)=\mathrm{Ext}_{\mathcal{O}}^i(M_x,B)$ .
Let $ \pi_w:M_w\to L_w$ be the quotient map.

Apply functor $ R^iT$ to this, we get $ R^iT(\pi_w):\mathrm{Ext}_{\mathcal{O}}^i(M_x,M_w)\to \mathrm{Ext}_{\mathcal{O}}^i(M_x,L_w)$ .

If I can show $ R^iT(\pi_w)$ is a nonzero map, then I am done.

My question:

  1. How to show $ R^iT(\pi_w)$ is a nonzero map?

  2. How to show $ \mathrm{Ext}^i_{\mathcal{O}}(M_x,L_w)\neq 0\implies\mathrm{Ext}^i_{\mathcal{O}}(M_x,M_w)\neq 0$ in case (1) is false?

Quotient of Three Dimensional Torus by Permutation on Coordinates

The Mobius Strip can be realized as a quotient of $ T = (S^1)^2$ via the identifications $ (x,y) \tilde (y,x)$ .

I tried to generalized this concept to a higher dimension, and consider the quotient of $ (S^1)^3$ by the action of the symmetric group $ S_3$ on the coordinates.

I was able to compute the homology of this space: $ H_n = \mathbb{Z}$ for $ n = 0,1$ , and 0 otherwise (with reduced homology being 0 at $ n=0$ as well).

Even with this information I wasn’t able to identify said space in any other way. Is it well known, or, can it be described in any other fashion? What can be said about higher dimensions?

Topology of the automorphic quotient

The MO question https://mathoverflow.net/questions/331549/compactness-of-the-automorphic-quotient-and-genericity made me realise that I don’t really understand the topology on the adèlic points of an algebraic group $ G$ . Indeed, the way I realised this is by making the wrong comment https://mathoverflow.net/questions/331549/compactness-of-the-automorphic-quotient-and-genericity#comment826950_331549. To save your following the link, the question is about consequences of compactness of $ G(F)\backslash G(\mathbb A)$ . Now I thought that, for each place $ v$ ,

  1. the group $ G(F_v)$ is a closed subgroup of $ G(\mathbb A)$ ; and,

  2. since $ G(F)$ is embedded diagonally in $ G(\mathbb A)$ , it intersects $ G(F_v)$ trivially; so

  3. $ G(F_v)$ sits as a closed subset of $ G(F)\backslash G(\mathbb A)$ .

Since further discussion (for example, paulgarrett’s answer) showed that the conclusion is wrong, obviously one of the steps is wrong—but I’m embarrassed to say I don’t know which one. Is it the first? If so, is there any easier way to see why than “because there’s a definition of closed set that $ G(F_v)$ doesn’t satisfy”?

Compactness of the automorphic quotient and genericity

Let $ G$ be a reductive group defined over a field $ F$ . Let $ \mathbf{A}$ denote the ring of adeles of $ F$ . My question is:

Assuming the automorphic quotient $ [G]=G(F) \backslash G(\mathbf{A})$ is compact, can we say that all the (non-character) automorphic representations of $ G$ are tempered? generic?

I do not precisely understand the relations between the notions of tempered and generic representations beyond the very specific $ \mathrm{GL}(n)$ case, so that any reference about these matters is also welcome.

Motivation Quotient of Algebraic Variety

Let $ X$ be a variety with a $ G$ -action by an algebraic group on it.

My question refers to a motivating example from

https://web.maths.unsw.edu.au/~danielch/thesis/mbrassil.pdf

Here the relevant excerpt:

enter image description here

Here the author discusses an example of $ X/G$ in order to explaine that it is neccessary to form $ X/G$ as categorical quotient and not the topological one.

We consider following motivating example introduced at page 27:

Here we take $ X:= \mathbb{C}^2$ with action by $ G:=\mathbb{C}^x$ via multiplication $ \lambda \cdot (x,y) \mapsto (\lambda x, \lambda y)$ .

Obviously the “naive” topological quotient consists set theoretically of the lines $ \{(\lambda x, \lambda y) \vert \lambda \in \mathbb{C}^x \}$ and the origin $ \{(0,0)\}$ .

Topologically the origin lies in the closure of every line.

So the QUESTION is why does this argument already imply that $ Y:=X/G$ cannot have a strucure of a variety? I don’t understand the argument given by the author.

If we denote by $ p:\mathbb{C}^2 \to Y$ the canonical projection map and by (continuity?) this map can’t separate orbits, why does this already imply that $ Y$ doesn’t have structure of a variety as stated in the excerpt?

Especially which role does here play the fact that we can’t separate the lines from the origin (in pure topologically way)? Does it cause an obstacle in order to form a variety structure on $ X/G$ ?

Remark: I know that there are different ways to deduce that if we define $ X/G$ pure topologically then it cannot have a structure of a variety. The most common argument is to introduce the invariant ring $ R^G$ and to calculate it explicitely here. But the main issue of this question is it made me curious that the given argumentation seems to be a bit more “elementary” in sense that he doesn’t explicitely work in this example with the concept of the invariant ring $ R^G$ .

Why the null space of quotient map is $U$?

I am reading the textbook Linear Algebra Done Right Chapter 3 section E on Products and Quotients of Vectors Spaces.

It tried to prove the dimension of a quotient space is equal to $ \text{dim }V/U = \text{dim }V -\text{dim }U$ .

Before that it defines the quotient map $ \pi$ as follow:

Suppose $ U$ is a subspace of $ V$ . The quotient map $ \pi$ is the linear map $ \pi:V \to V/U$ defined by $ $ \pi(v) = v+U$ $ for $ v \in V$ .

I can understand that the range of $ \pi$ is $ v+U$ which is $ V/U$ according to the definition of $ V/U$ .

But I don’t understand why the null space of $ \pi$ is $ U$ . The book said it is due to the proof like this following:

Suppose $ U$ is a subspace of $ V$ and $ v,w \in V$ . Then the following are equivalent: $ $ v-w \in U$ $ $ $ v+U=w+U$ $ $ $ (v+U) \cap (w+U) \neq \emptyset$ $