### Integral Transforms and the Pull-Push Perspective, I

#### Posted by Simon Willerton

In a series of posts I want to give a flavour of the idea, well known to experts, that integral transforms, given in terms of kernels, can be viewed from a pull-push perspective, and to tie this in to the the idea of enriched profunctors as transforms between presheaf categories. I would like to discuss Fourier transforms, the Legendre transform, the Radon transform, Fourier-Mukai transforms and many other things; I’m not sure how far I will get, however, as I want to try to do this quite gently.

These sorts of things have been discussed here at the Café on various occasions.

This time I will discuss the Fourier transform, or rather Fourier series, in terms of a pull-push operator. Next time I will talk about the composition of kernels and the relevance of the Beck-Chevalley Condition and the Projection Formula (also known as the Frobenius Identity).

The main point this time is the notion of the push-forward of a function. (Category theorists who don’t know the term ‘push-forward’ might find it useful to think of Kan extensions.) I don’t know enough analysis to be able to give the general setting of push-forwards in this situation here, but it is only necessary, for the Fourier transform, to understand push-forwards along projections.

#### The pull-push picture

We can see many examples of correspondences and kernels across mathematics. Provided that the words **space** and **function** are appropriately interpreted, the pull-push picture is roughly as follows.

For spaces $X$ and $Y$ a **correspondence** $Z$ (also called a **span** by category theorists) is a space with maps

$\array{ &&Z\\ &{}^{p}\swarrow&&\searrow^{q}\\ X&&&&Y }$

and a **kernel** is a “scalar-valued” function $\kappa$ on $Z$. In most standard examples either the correspondence or the kernel is trivial, meaning that we consider a constant function on an interesting correspondence $Z$ or else we consider an interesting function $\kappa$ on the trivial correspondence $X\times Y$. This data then gives rise to a **transform** $\mathcal{T}_\kappa$ which is basically a function from functions on $X$ to functions on $Y$.

$\mathcal{T}_\kappa\colon F(X)\to F(Y)$

This is defined via a pull-push formalism. Starting with a function on $X$, we pull it back to a function on $Z$, multiply by the kernel $\kappa$, and push the result down to a function on $Y$.

$\array{ &&F(Z)&\righttoleftarrow \kappa\cdot\\ &{}^{p^\ast}\nearrow&&\searrow^{q_\ast}\\ F(X)&&&&F(Y) }$

So for a function $\phi\in F(X)$ we have

$\mathcal{T}_\kappa(\phi)\coloneqq q_\ast(\kappa\cdot p^\ast\phi).$

The pull-back of a function will usually be obvious: given $f\colon A\to B$ the pull-back $f^\ast\colon F(B)\to F(A)$ is given by composition with $f$, so for a function $\phi\in F(B)$ we have $f^\ast(\phi)\coloneqq \phi\circ f$. The push-forward $f_\ast\colon F(A)\to F(B)$ is usually more subtle, and does not always exist. The push-forward will be adjoint in some sense to the pull-back. This will mean the function spaces $F(A)$ and $F(B)$ will have some notion of inner product $\langle {\cdot},{\cdot}\rangle$ and for functions $\psi\in F(A)$, $\phi\in F(B)$ then

$\langle f^\ast\phi,\psi\rangle_{F(A)}=\langle \phi,f_\ast\psi \rangle_{F(B)}.$

In some cases the transform $\mathcal{T}_\kappa$ will give a bijection between function spaces $F(X)$ and $F(Y)$, sometimes it will not.

Let’s start by looking at the Fourier transform on the circle.

#### Fourier series and the Fourier transform

##### The general picture

The theory of complex Fourier series for the circle can be viewed as a theory of transforming between functions on the circle and functions on the integers. (I will be slightly vague at this point about exactly what I mean by functions.)

I will make two notational decisions here that I might regret. Firstly I will consider the circle $\mathbb{T}$ as the unit complex numbers and so will write the group structure multiplicatively, thus flouting the convention that abelian groups are written additively. Secondly I will not specify what measure I am taking on the circle, this will avoid me having to specify the standard factors at the front of the Fourier transform.

Given a complex-valued function $\phi$ on the circle $\mathbb{T}$ we get the associated Fourier coefficient $a(n)$ for $n\in \mathbb{Z}$ as

$a(n)\coloneqq \int_{w\in \mathbb{T}}w^{-n}\phi(w)\;\text{d}w.$

If I was being less perverse and considered the circle as $\mathbb{R}/2\pi\mathbb{Z}$ then the Fourier coefficient would be in the more familiar form.

$a(n)\coloneqq \int_{x=0}^{2\pi}e^{-i n x}\phi(x)\;\text{d}x.$

I guess I have written it in this way partially because the standard form is so familiar that I was perhaps trying to shake away the over-familiarity. Regardless of that notational point, what is written above is nothing other than an integral transform as I will now explain. We have the diagram

$\array{ &&\mathbb{T}\times \mathbb{Z}\\ &{}^{p}\swarrow&&\searrow^{q}\\ \mathbb{T}&&&&\mathbb{Z} }$

and on $\mathbb{T}\times\mathbb{Z}$ there is the complex-valued function, or kernel, $\kappa$ defined by $\kappa(w,n)\coloneqq w^{-n}$. I claim that the Fourier coefficients come from the transform of this kernel, so $a=q_\ast(\kappa\cdot p^\ast\phi)$. The interesting part of the construction is understanding what the push-forward of a map is.

##### The push-forward of a projection

As we will see below, the push-forward of functions can not be defined for an arbitrary map between spaces, but we can define it for a projection $f\colon C\times B\to B$.

On a measure space $A$ we have a candidate for an inner product on the complex-valued functions on $A$:

$\langle \phi_1,\phi_2\rangle\coloneqq \int_A\overline{\phi_1(a)}\cdot\phi_2(a)\; \text{d} a$

If $A$ is not compact then this might not be defined for all pairs of functions and we might have to restrict the functions we consider, say to compactly supported ones, or rapidly decaying ones.

Aside:Note that the complex numbers $\mathbb{C}$ have an “internal inner product” $\langle {\cdot},{\cdot}\rangle_{\mathbb{C}}\colon \overline \mathbb{C}\otimes \mathbb{C}\to \mathbb{C}; \quad a\otimes b\mapsto \overline{a}b$ and the inner product of two $\mathbb{C}$-valued functions on a space A is given by $\langle \phi_1,\phi_2\rangle_{F(A)}\coloneqq \int_A\langle{\phi_1(a)},\phi_2(a)\rangle_{\mathbb{C}}\; \text{d} a.$ On the other hand, a closed monoidal category $\mathcal{V}$ has an internal hom $[ {\cdot},{\cdot}]_{\mathcal{V}}\colon \mathcal{V}^{op}\otimes \mathcal{V}\to \mathcal{V}$ and the hom object of two $\mathcal{V}$-valued presheaves on a $\mathcal{V}$-category $\mathcal{A}$ is given by $[ P_1,P_2]_{\mathcal{V}^{\mathcal{A}}}\coloneqq \int^{a\in\mathcal{A}}[{P_1(a)},P_2(a)]_{\mathcal{V}}.$ I won’t say anymore about this for the moment.

Given a map $f\colon A\to B$ we have the pull-back $f^\ast \colon F(B)\to F(A)$ given by precomposition, so $f^\ast \phi\coloneqq \phi\circ f$. We want to define an adjoint map $f_\ast \colon F(A)\to F(B)$. The idea is that this should be ‘integration along the fibre’, so we will want

$f_\ast \phi(b)\;\text{"}\coloneqq\text{"}\;\int_{a\in f^{-1}(b)}\phi(a)\;\text{d} a.$

This comes from a standard philosophy of pushing things forward by “sticking the things in the preimage together”; here sticking together the values of the function means adding them together, or, rather, integrating them. There is a fundamental question of what the measure should be here. Let’s see where this actually comes from. Because we want $f_\ast$ and $f^\ast$ to be adjoint, for all $\psi\in F(A)$ and $\phi \in F(B)$ we need

$\langle f^\ast \phi,\psi\rangle_{F(A)}=\langle \phi,f_\ast \psi \rangle_{F(B)}.$

in other words

$\begin{aligned} \int_A\overline{\phi(f(a))}\cdot\psi(a)\;\text{d} a &= \int_B\overline{\phi(b)}\cdot f_\ast \psi(b)\;\text{d} b \\ &= \int_{b\in B}\overline{\phi(b)}\cdot\biggl( \int_{c\in f^{-1}(b)} \psi(c)\;\text{d} c\biggr)\text{d} b \\ &= \int_{b\in B}\int_{c\in f^{-1}(b)}\overline{\phi(b)}\cdot \psi(c)\;\text{d} c\;\text{d} b. \end{aligned}$

One case in which this will work in is the case where $f$ is a projection, so $A$ is of the form $C\times B$ and $f\colon C\times B\to B$ is the projection onto $B$. Then, provided $A=C\times B$ has the product measure of $B$ and $C$, we can define $f_\ast \colon F(C\times B) \to F(B)$, the push-forward of $f$, by

$f_\ast \phi(b)\coloneqq \int_{a\in C\times \{b\}}\phi(a)\;\text{d} a=\int_{c\in C}\phi(c,b)\;\text{d} c.$

Of course, if $C$ is non-compact then there still might be problems with this if $\phi$ does not decay sufficiently rapidly. I will not worry about that by not saying what the domain and range of the pull-back and push-forward maps are.

##### Push-forward is not always defined as a function

Before getting back to the Fourier transform, it is worth having a quick digression. It is important to note that the push-forward, as an adjoint to pull-back, cannot always be defined in the world of functions. One simple example is that of an inclusion. Suppose that $i\colon X\hookrightarrow Y$ is an inclusion of say compact measure spaces, then the pushforward $i_\ast 1$ of the constant function $1$ on $X$ would, for all $\phi\in F(Y)$, have to satisfy the following:

$\int_X \overline{\phi(i x)} \;\text{d}x= \int_{Y} \overline{\phi(y)} \cdot f_\ast 1(y)\;\text{d}x\;\;\text{d}y$

In other words, the push-forward $f_\ast 1$ would have to be a delta-function supported $X$. In general, this is not going to be possible with a function but might be possible if we allow the push-forward to be some generalized function such as a distribution.

##### Fourier coefficients via a transform

Now we can finally see that the Fourier coefficients can be viewed in this pull-push formalism. If $\phi$ is a function on the circle and $n\in \mathbb{Z}$ then

$\begin{aligned} \mathcal{T}_\kappa\phi (n) &\coloneqq \bigl(q_\ast (\kappa\cdot p^\ast \phi)\bigr)(n) =\int_{w\in \mathbb{T}}\kappa(w,{n})\cdot\phi(p(w,n))\;\text{d} w \\ &=\int_{w\in \mathbb{T}}w^{-n}\phi(w)\;\text{d} w =a(n), \end{aligned}$

as required.

A fundamental property of the Fourier transform on the circle is that it is, essentially, invertible. To look at this from the pull-push, integral transform point of view we will need to see how kernels compose. That’s something for next time. However, let’s finish this time by mentioning how the above construction generalizes to other abelian groups.

##### The Fourier transform for an abelian group

Suppose that $G$ is a locally compact abelian group, say equipped with the Haar measure, then the dual group $\widehat G$ is defined to be the group of **multiplicative unitary characters** on $G$ which means the group homomorphisms to the unit complex numbers.

$\widehat{G}\coloneqq \{\xi\colon G\to \mathbb{T} \mid \xi\; \text{ is a homomorphism}\}.$

[I am occasionally slightly confused by the fact that these are sometimes just called characters when I am used to characters not being multiplicative, but only conjugation invariant, for non-abelian groups.]

The kernel that we want to use will be the function $\kappa\colon G\times \widehat G\to \C$ given by $\kappa(g,\xi)\coloneqq \xi(g^{-1})$, which can also be written as the conjugate: $\kappa(g,\xi)\coloneqq \overline{\xi(g)}$. Using this kernel on the trivial correspondence

$\array{ &&G\times \widehat{G}\\ &{}^{p}\swarrow&&\searrow^{q}\\ G&&&&\widehat{G} }$

gives the transform $\mathcal{T}_\kappa$ from functions on $G$ to functions on $\widehat{G}$, so we get

$\mathcal{T}_\kappa\phi(\xi)\coloneqq \int_{g\in G}\xi(g^{-1})\cdot\phi(g)\;\text{d} g$

and this can actually be written as $\langle \xi,\phi\rangle$.

Next time we’ll start with the question of how to compose kernels and we’ll consider the inverse Fourier transform.

## Re: Integral Transforms and the Pull-Push Perspective, I

Was this really posted 4 days ago? I never noticed.

So what can be said of the values taken by the functions in $F(X)$, etc.? And what of the “scalar-valued” $\kappa$?

Is it enough to say the scalars must act on the codomain of the functions on $X$, $Y$, $Z$? The latter must have an inner product you say.