## November 7, 2010

### 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.

Posted at November 7, 2010 9:19 PM UTC

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

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.

Posted by: David Corfield on November 11, 2010 7:57 PM | Permalink | Reply to this

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

Er, no. I put it on the server four days ago but only just “published” it. I guess it’s date-stamped from then.

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.

Well the way I intended it was that the functions on $X$, $Y$ and $Z$ are also “scalar-valued”.

I’m only presenting one way of formalizing these kinds of constructions, I’m not saying this is the only way to do it – I have the idea of profunctors sitting at the back of my mind as I write, so I’m trying to emphasize the similarity with that.

As I said, the word “function” might need appropriate interpretation, and it can be sensible to allow the distributions (generalized functions) into the picture.

For instance, Schwartz’s Kernel Theorem says, I believe, that for $X$ and $Y$ open subsets of $\mathbb{R}^n$ every continuous linear map from smooth functions on $X$ to distributions on $Y$ can be represented by a kernel which is a distribution on $X\times Y$.

Posted by: Simon Willerton on November 11, 2010 8:31 PM | Permalink | Reply to this

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

Thank you for writing this Simon. This is awesome.

Posted by: Eric on November 11, 2010 11:06 PM | Permalink | Reply to this

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

Posted by: Simon Willerton on November 12, 2010 9:43 AM | Permalink | Reply to this

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

Yes I am enjoying it too.

Posted by: Bruce Bartlett on November 12, 2010 9:15 PM | Permalink | Reply to this

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

I like that adjoint relation. It seems like the fundamental bit.

Kindergarten question…

If $F(A)$ and $F(B)$ each have inner products, does that imply $A$ and $B$ are measure spaces? Does a choice of inner product on $F(A)$ induce a measure on $A$ and vice versa?

Posted by: Eric on November 13, 2010 12:54 AM | Permalink | Reply to this

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

My thinking was that if $A$ and $B$ are a measure spaces, then we can define inner products on $F(A)$ and $F(B)$ as indicated and it will satisfy the relation

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

Now, if instead of telling you $A$ and $B$ are measure spaces, I tell you that $F(A)$ and $F(B)$ are inner product spaces. Could we use that information to deduce $A$ and $B$ are measure spaces? In a way, I’m asking if instead of defining an inner product on $F(A)$ in terms of integration on $A$, can we define integration on $A$ in terms of an inner product on $F(A)$?

This is probably too basic…

Posted by: Eric on November 15, 2010 11:20 PM | Permalink | Reply to this

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

Nice! I enjoyed that.

Just a micro-comment. Near the end, you wrote:

multiplicative characters on $G$; these are not characters in the sense of representation theory, but rather are group homomorphisms to the complex numbers

Actually, the representation-theory meaning of ‘character’ includes multiplicative characters. A multiplicative character is just a one-dimensional representation-theoretic character.

Why? A one-dimensional representation of $G$ is a multiplication-preserving map $G \to GL_1(\mathbf{C}) = \mathbf{C}^\times$. The trace map $GL_1(\mathbf{C}) \to \mathbf{C}$ is the inclusion $\mathbf{C}^\times \hookrightarrow \mathbf{C}$. Hence a one-dimensional character of $G$ is a multiplication-preserving map $G \to \mathbf{C}$, i.e. a multiplicative character.

Incidentally, I think you want to take unitary characters in the last section — otherwise your equation $\xi(g^{-1}) = \overline{\xi(g)}$ won’t hold.

Posted by: Tom Leinster on November 12, 2010 2:18 AM | Permalink | Reply to this

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

So we’re dealing here with the distinction that I mentioned Mackey drew between characters and generalized characters?

The Pontryagin dual group of characters for $\mathbb{Z}$ is $S^1$. But Mackey points out that the generalized characters for $\mathbb{Z}$ are all of the non-zero complex numbers – essentially 1 can be sent to any non-zero complex number, $c$, so

$n \mapsto c^n$

is a generalized character.

Elements of finite groups are forced to land on the complex circle. If you want this for infinite groups you can specify unitary character, as Tom mentioned. Otherwise, the transform is Mackey’s Laplace transform for locally compact abelian groups.

Posted by: David Corfield on November 12, 2010 8:34 AM | Permalink | Reply to this

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

Hmmm. I guess I fluffed the section on general Abelian groups a little.

My comment about multiplicative characters not being the same as characters was emphasized by me because authors often call them just characters (see Character group at wikipedia) and define them as group homomorphisms which confuses me no end. I have just caused confusion the other way now, so will reword it.

I meant to say that they are group homomorphisms to the circle, but didn’t write that. I think I had finite abelian groups in my head and in that case it didn’t make any difference. I will amend it.

Posted by: Simon Willerton on November 12, 2010 12:12 PM | Permalink | Reply to this

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

Hi Simon,

As you mention, profunctors are an example of this situation of a trivial correspondence with an interesting kernel, and these can be interpreted as cocontinuous maps between presheaf categories.

The other case with constant kernel and interesting correspondence is more or less the case of spans, right? This is still closely related to cocontinuous maps between presheaf categories, but the relationship is not quite as nice as for profunctors.

Is there a nice class of maps between presheaf categories in which to interpret the second case?

Or should there be something a little more general than cocontinuous functors to capture both extremes?

Posted by: Alex Hoffnung on November 12, 2010 4:17 AM | Permalink | Reply to this

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

Is there a nice class of maps between presheaf categories in which to interpret the [case of spans]?

It’s a good question but I haven’t really thought about this before and nothing springs to mind.

This is still closely related to cocontinuous maps between presheaf categories, but the relationship is not quite as nice as for profunctors.

Can you say precisely what the not-as-nice relationship is?

Posted by: Simon Willerton on November 13, 2010 11:22 PM | Permalink | Reply to this

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

I can say a bit.

At some point, I mistakenly thought that I should be able to send a span to a cocontinuous functor and then recover the span up to equivalence.

I was attempting to recover the span by converting a cocontinuous functor into a profunctor and then applying the category of elements construction.

This always leaves one with a jointly faithful span, however these are not closed under composition as Mike Shulman pointed out to me.

So the “not as nice” relationship is that I can pull-push a span to a cocontinuous functor, and I can recover from this an interesting kernel, i.e., a profunctor, and from that an interesting span, but not the same interesting span that I started with.

Posted by: Alex Hoffnung on November 14, 2010 12:23 AM | Permalink | Reply to this

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

Is there a nice class of maps between presheaf categories in which to interpret the [case of spans]?

A special case of the results of Gambino-Kock is that the bicategory of spans in a category $C$ with pullbacks is equivalent to the 2-category whose

• objects are the slice categories $C/x$,
• morphisms are functors $C/x \to C/y$ with the property of being isomorphic to $f_! g^*$ for some span $x \xleftarrow{g} z \xrightarrow{f} y$ (“linear polynomial functors”) and
• 2-morphisms are strong natural transformations (transformations compatible with the canonical tensorial strengths of linear polynomial functors).

I’m not sure if that’s exactly what you’re looking for, but it sounds kind of like it.

Posted by: Mike Shulman on November 14, 2010 5:44 AM | Permalink | Reply to this

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

Thanks for pointing out that result and paper! I was hoping for something just like that.

Posted by: Alex Hoffnung on November 14, 2010 7:06 AM | Permalink | Reply to this

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

Mike wrote:

A special case of the results of Gambino-Kock is that the bicategory of spans […]

Do you mean the bicategory of Spans which has as 2-Morphisms Morphisms between Span? I have the feeling that there are two different Span categories in the literature, namely also the bicategory which has as 2-Morphisms (classes) of Spans of Span-Morphisms. In the first case this amount to adjoint some right inverses to the original category $C$ (though not freely see Universal properties of Span ). The second choice means to adjoin something like two- sided-adjoints to $C$, but I am not entirely sure about this case.

morphisms are functors $C/x \to C/y$ with the property of being isomorphic to $f_! g^*$ for some span […]

That is a very nice reformulation of that category. Is there a more intrinsic characterisation of these functors?

Posted by: Thomas Nikolaus on November 14, 2010 4:00 PM | Permalink | Reply to this

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

I don’t understand the two bicategories you mean. The only bicategory of Spans that I’m familiar with has the 2-morphisms being diagrams of the form $\array{ && A\\ & \swarrow && \searrow \\ X && \downarrow && Y\\ & \nwarrow && \nearrow\\ && B}$

Posted by: Mike Shulman on November 17, 2010 2:37 AM | Permalink | Reply to this

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

Mike says to Thomas

I don’t understand the two bicategories you mean.

There is for each $n$ an $(\infty,n)$-category whose objects are $\infty$-groupoids, morphisms are spans, 2-morphisms are spans-of-spans

$\array{ && A \\ & \swarrow &\uparrow& \searrow \\ X &&K&& Y \\ & \nwarrow &\downarrow& \nearrow \\ && B } \,,$

3-morphisms are spans of spans of spans, and so on, up to level $n$, where we just have homotopy classes of morphisms.

In Classification of TFTs this $(\infty,n)$-category is called $Fam_n$, starting around p. 57 . It reappears and features prominently in a context that is the $n$-fold categorification of pull-push that we are discussing here in TQFT from compact Lie groups.

I think in his above comment Thomas was thinking of the homotopy 2-category of (some slight variant of) $Fam_3$.

Posted by: Urs Schreiber on November 17, 2010 10:22 AM | Permalink | Reply to this

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

Okay, thanks. I think using a different name for the categories of “higher spans,” like $Fam_n$, is a good idea to avoid confusion; I’ve only ever seen the one I described referred to as “the bicategory of spans.”

Posted by: Mike Shulman on November 17, 2010 3:33 PM | Permalink | Reply to this

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

Urs is right, that is the category I meant. Let me describe that a little bit more in detail:

1) For a category $C$ we denote the (usual) bicategory of spans by $\mathfrak{Span}(C)$. Then $\mathfrak{Span}$ is a functor $\mathfrak{Cat} \to \mathfrak{BiCat}$. It is clear that $\mathfrak{Span}$ is product preserving.

2) Hence let $B$ be an arbitrary bicategory. Then by applying $\mathfrak{Span}$ to the Hom-categories we obtain another tricategory $\overline{B}$ (in fact the pentagon axiom and everything hold strictly).

3) Now for a one-category $C$ we first form the bicategory $\mathfrak{Span}(C)$ and then apply the construction from 2) to obtain a tricategory $\overline{\mathfrak{Span}(C)}$.

4) The categoy I meant in my last post is then obtained by dividing out the 3-isomorphisms.

Let me say two things about these type of span categories. First of all it is clear that the first three steps can by iterated, hence we get (n+1)-categories of the type:

objects, 1-morphisms,…,n-morphisms are given by spans
n+1-morphisms are span morphisms

Then one can decategorify to an n-category by dividing out 2-isomorphisms. These span categories are important in extended field theory. That is because the extended cobordism categories are categories of cospans of this type. But the disadvantage is that there are not so much functors from $\overline{\mathfrak{Span}(C)}$ to some other bicategory. That is related to the ambidextrous adjoints in $\overline{\mathfrak{Span}(C)}$.

Posted by: Thomas Nikolaus on November 17, 2010 9:36 PM | Permalink | Reply to this

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

Well, it really only acts on categories with pullbacks, of course. And pullback-preserving functors (otherwise you get only an oplax functor after applying Span).

Oh, you are right. Maybe I was a little bit to enthusiasthic. I mainly wanted to make the point that it can be defined by some iterated construction.

Posted by: Thomas Nikolaus on November 18, 2010 7:47 AM | Permalink | Reply to this

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

Then $Span$ is a functor $Cat \to Bicat$

Well, it really only acts on categories with pullbacks, of course. And pullback-preserving functors (otherwise you get only an oplax functor after applying Span).

Hence let $B$ be an arbitrary bicategory. Then by applying $Span$ to the Hom-categories we obtain another tricategory $\bar{B}$.

So I guess you need the hom-categories of $B$ to have pullbacks, and its composition functors to preserve pullbacks. But fortunately, $Span(C)$ has those properties, so I believe your construction. (-:

Posted by: Mike Shulman on November 18, 2010 1:22 AM | Permalink | Reply to this

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

Is there a nice class of maps between presheaf categories in which to interpret the [case of spans]?

Yes: there is an equivalence between any topos/$\infty$-topos $\mathbf{H}$ and the category of etale geometric morphisms into it and geometric over-morphisms

$\mathbf{H} \simeq ((\infty,1)Topos/\mathbf{H})_{et} \,.$

So a span $X \leftarrow A \to Y$ in the former corresponds to a span

$\array{ \mathbf{H}/X &\stackrel{\overset{i^*}{\to}}{\underset{i_*}{\leftarrow}}& \mathbf{H}/A &\stackrel{\overset{o^*}{\leftarrow}}{\underset{o_*}{\to}}& \mathbf{H}/Y \\ & \searrow\nwarrow & \downarrow\uparrow & \swarrow\nearrow \\ && \mathbf{H} }$

in the latter. Accordingly, a colimit-preserving functor $\mathbf{H}/X \to \mathbf{H}/Y$ comes from a kernel $A$ if there exists such a diagram such that it factors like $o_! i^*$.

Posted by: Urs Schreiber on November 15, 2010 1:52 PM | Permalink | Reply to this

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

Why don’t people study interesting correspondences and interesting kernels combined? It is that the effects of interestingness work independently, in some sense, so they are studied separately?

Posted by: David Corfield on November 12, 2010 8:57 AM | Permalink | Reply to this

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

I’m not saying that people don’t study them; you probably want to think about them if you’re manipulating kernels, we should see a bit of this next time, and if you allow suitably general kernels (such as distributions) then often you can transfer kernels to the trivial correspondence. It just seems that all of the interesting ones naturally live in one of the two camps.

As an example you can think of the correspondence $X\leftarrow X \rightarrow X$ with the trivial (i.e., unit-valued) kernel. This gives rise to the identity transform $F(X)\to F(X)$. This kernel can be pushed-forward to the Dirac delta function on the trivial correspondence $X\leftarrow X\times X \rightarrow X$ giving the same identity transform.

More generally, for a correspondence $X\leftarrow Z \rightarrow Y$ there is clearly a map $Z\to X\times Y$ and one can in many cases push forward the constant function $1$ to get a possibly generalized kernel on $X\times Y$ with the same transform. Again, I’ll mention this next time.

In the slightly different context of sheaves in algebraic geometry see this comment by David Ben-Zvi.

Posted by: Simon Willerton on November 12, 2010 9:40 AM | Permalink | Reply to this

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

Here is a simple precise way to see that nontrivial kernels are equivalent to nontrivial correspondences:

Represent a function object/sheaf/$\infty$-stack/whatever on an object $X$ by a morphism

$\psi : \Psi \to X \,,$

For instance in the category $FinSet$ that’s a map of finite sets which we think of as representing the $\mathbb{N}$-valued function

$|\psi|(x) := |\Psi_x| \,.$

Then the linear action of a correspondence

$A = \left( \array{ && A \\ & {}^{\mathllap{i}}\swarrow && \searrow^{\mathrlap{o}} \\ X &&&& Y } \right)$

on $\psi$ is given by producing the pullback

$A : \psi \mapsto \left( \array{ && i^* \Psi \\ & \swarrow && \searrow^{i^* \psi} \\ \Psi && && A \\ &_{\psi}\searrow& & {}^{\mathllap{i}}\swarrow && \searrow^{\mathrlap{o}} \\ && X &&&& Y } \right)$

regarded as a morphism

$A \psi : i^* \Psi \stackrel{i^* \psi}{\to} A \stackrel{o}{\to} Y \,.$

Now, pullbacks satisfy the pasting law, so that you can equivalently compute this pullback in two stages as

$\array{ i^* \Psi &\to& p_1^* \Psi &\to& \Psi \\ \downarrow && \downarrow && \downarrow^{\mathrlap{\psi}} \\ A &\stackrel{(i,o)}{\to}& X \times Y &\stackrel{p_1}{\to}& X } \,.$

Here the first pullback may be thought of as pulling back to the correspondence-with-kernel

$\array{ && A \\ && \downarrow \\ && X \times Y \\ & \swarrow && \searrow \\ X &&&& Y }$

which first produces

$\array{ p_1^* \Psi && & A \\ & \searrow & \swarrow \\ && X \times Y \\ & {}^{\mathllap{p_1}}\swarrow && \searrow \\ X &&&& Y } \,.$

Then we form the “tensor product” of $p_1^* E$ with $A$ up there – which is nothing but forming their fiber product over $X \times Y$, hence is $i^{*} \Psi$ by the pasting law.

For the context of $FinSet$ you see that $(|A_{x,y}|)$ is a $|X|$-by-$|Y|$-matrix with entries in natural numbers and the function

$|A \psi | : y \mapsto | (i^* \Psi)_y | = \sum_{x \in X} |A_{x,y}| \cdot |\psi|(x)$

is indeed the result of applying the familiar linear map given by this matrix on $|\psi|$.

So

• Every nontrivial correspondence is equivalently a trivial correspondence with kernel, and vice versa.

This argument uses nothing but the existence of pullbacks. Being so basic, it works in vast generality.

For instance, if we take the ambient category to be a sheaf topos $\mathcal{T}$, then morphisms $\Psi \to X$ are objects in the over-topos $\mathcal{T}/X$, which is the little topos of $X$, hence they are sheaves on $X$.

In this large class of cases the pull-push morphism that we are looking at is the combination of base change geometric morphisms

$\mathcal{T}/X \stackrel{\overset{i^*}{\to}}{\underset{i_!}{\leftarrow}} \mathcal{T}/A \stackrel{\overset{o_!}{\to}}{\underset{o^*}{\leftarrow}} \mathcal{T}/Y \,.$

All functors here are left adjoints, hence preserve all colimits, hence are indeed linear with respect to the above notion of functions and linear maps. Some people would call these Lawvere distributions.

Since “there exists $\infty$-category theory” we can also assume the ambient context to be an $\infty$-topos $\mathbf{H}$. Then without doing anything further, all of the above becomes a statement about pull-push of $\infty$-stacks over any site.

This has nothing in particular to do with working over a site for algebraic geometry. This is general abstract.

Posted by: Urs Schreiber on November 12, 2010 4:51 PM | Permalink | Reply to this

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

Urs said:

Represent a function object/sheaf/∞-stack/whatever on an object $X$ by a morphism $\psi\colon \Psi \to X$

How do I do that for a complex-valued, ungeneralized function on the circle?

I can’t see how your construction would work in that case because, if you take $A$ to be the correspondence $\mathbb{T}\leftarrow \mathbb{T} \to \mathbb{T}$ then the transform is the identity transform, and your construction would give an ungeneralized function on the trivial correspondence $\mathbb{T}\leftarrow\mathbb{T}\times \mathbb{T}\to \mathbb{T}$ which represents the identity transform, yet we know that there isn’t such a function.

Posted by: Simon Willerton on November 12, 2010 6:22 PM | Permalink | Reply to this

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

yet we know that there isn’t such a function.

There is such a distribution, though.

The comparison with the un-de-categorified setup shows that in the decategorified picture you need to deal with distributional entities – the images of the skyscraper sheaves.

That’s why people speak of the Lawvere distributions that I mentioned.

$\delta_x : \mathcal{T}/X \to Set$

(also known by other names…)

Posted by: Urs Schreiber on November 12, 2010 7:06 PM | Permalink | Reply to this

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

How do I do that for a complex-valued, ungeneralized function on the circle?

What I said is the story as it works in any $(n,r)$-category with pullbacks. I intentionally said function object instead of function. There is a general abstract category theoretic construction here of which ordinary linear algebra with ordinary functions is supposed to be the decategorification.

I mentioned how to do $\mathbb{N}$-valued functions by working in a sheaf topos and then decategorifying by using cardinality of sets.

(And of course all of this (except maybe the topos/$\infty$-topos generalization) is just repeating what John Baez has been exposing here at some length in the past: you are to think of ordinary linear algebra as the decategorification of the base-change yoga that I recalled.)

To get values in other number systems after decategorificaton, you can go to an $n$-sheaf $n$-topos and apply $\infty$-groupoid cardinality.

That gives at least real numbers. I seem to remember that John mentioned cases on non-tame groupoids whose cardinality we can interpret even as complex numbers, but I forget and can’t search for it right now.

Posted by: Urs Schreiber on November 12, 2010 6:53 PM | Permalink | Reply to this

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

To get values in other number systems after decategorification, you can go to an $n$-sheaf $n$-topos and apply $\infty$-groupoid cardinality.

That gives at least real numbers.

There’s something I’ve never really understood in this picture. Am I supposed to consider the categorification of let’s say a real valued function on a space $X$ to be a stalk-map from points to (tame) $\infty$-groupoids? Can I think of these stalks as (nice) topological spaces, and the total space to be the union of these stalks? Would this be essentially just a space sitting over another space?

I seem to remember that John mentioned cases on non-tame groupoids whose cardinality we can interpret even as complex numbers…

There was also that $U(1)$-phased approach. Looking back, I see Urs in Sept. 2007 claiming that what he had written here

…has the chance of being the deepest observation I’ll ever come up with in my life.

I wonder what his verdict is now.

Gosh, there’s so much material locked up in old posts, and so much forgotten. I see I thought I understood enough about phased-groupoids back then to ask questions, such as here and here.

Now we’ve all gone $\infty$-ish, I suppose we could have phased $\infty$-groupoids.

Posted by: David Corfield on November 13, 2010 9:07 PM | Permalink | Reply to this

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

Am I supposed to consider the categorification of let’s say a real valued function on a space $X$ to be a stalk-map from points to (tame) ∞-groupoids?

Not sure when you are supposed to do that, but you can do that.

Can I think of these stalks as (nice) topological spaces, and the total space to be the union of these stalks? Would this be essentially just a space sitting over another space?

Sure. Specifically $\infty$-sheaves over a paracompact topological space are equivalent to spaces $Y \to X$ over $X$ (in generalization of the similiar 1-categorical statement that sheaves on $X$ are equivalent to étale spaces over $X$, which are spaces over $X$ whose fiber is a homotopy 0-type).

There was also that $U(1)$-phased approach

Right, this option we always have:

we pick some $n$-vector space $V$ and replace all our bases $X$, $Y$, etc. by total spaces of $V$-bundles, say trivial ones $X \times V$ etc, for a start. Then a “function object” $E \to X \times V$ may over each point of $X$ be thought of as a formal linear combination of elements in $V$.

For instance with $V = \mathbb{C}$ this can be used to get complex-valued functions under decategorification, using the cardinality

$|-| : FinSet/\mathbb{C} \to \mathbb{C} \,.$

Finally you write:

I wonder what his verdict is now.

After a good idea is before the next refined idea. I am now working this out in an improved context with Domenico Fiorenza, now under the headline $\infty$-Chern-Simons theory.

Posted by: Urs Schreiber on November 15, 2010 4:45 PM | Permalink | Reply to this

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

So if $Y$ over paracompact $X$ is a categorification of a real function on $X$, haven’t I lost all that interesting structure given by reals as metric space, etc. I can say two real functions are close to one another, or that a sequence of them converges, but I don’t have these in the categorified case. Or do I?

It’s a return to what I was after back here, an idea of distance between groupoids.

I suppose there’s closeness between spaces in Postnikov or Whitehead tower fashions.

Posted by: David Corfield on November 15, 2010 5:54 PM | Permalink | Reply to this

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

Hi Urs,

I seem to be getting a bit confused. Early on when thinking about the spans that should show up in groupoidification, Jim Dolan was throwing around the idea that we might only consider the jointly faithful spans. This was in part an attempt to get the spans of groupoids and generalized bundle picture to line up nicely with cocontinuous functors between presheaf toposes.

Maybe more explicitly, the problem was that in geometric representation theory people usually model elements of some algebra by spans of sets, possibly with an action of some group. This fits nicely into the picture of cocontinuous functors between presheaf toposes, but not so nicely with spans of groupoids, since the kernels for spans of groupoids are just groupoid bundles, not profunctors.

So I thought that it was easy to think of an interesting correspondence as a boring correspondence with an interesting kernel, but not the other way around.

I also thought that this was the statement David BZ was making here.

Am I misinterpreting what you or David, or maybe both, have said?

Posted by: Alex Hoffnung on November 13, 2010 12:48 AM | Permalink | Reply to this

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

the problem was that in geometric representation theory people usually model elements of some algebra by spans of sets, possibly with an action of some group. This fits nicely into the picture of cocontinuous functors between presheaf toposes, but not so nicely with spans of groupoids, since the kernels for spans of groupoids are just groupoid bundles, not profunctors.

The kernels for spans of groupoids are $(2,1)$-profunctors!

Generally, the kernels for spans of $\infty$-groupoids are $(\infty,1)$-profunctors: morphisms in $Pr(\infty,1)Cat$.

Posted by: Urs Schreiber on November 15, 2010 1:41 PM | Permalink | Reply to this

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

“The kernels for spans of groupoids are $(2,1)$-profunctors!”

Great, that is what I hoped. I am still confused though by your comment where it looked like the interpretation for groupoids would have the kernels as bundle groupoids.

By the way, I would guess that kernels for spans of categories would also be $(2,1)$-profunctors. Do you know of cases which use more general profunctors? $(3,1)$-profunctors?

Posted by: Alex Hoffnung on November 16, 2010 12:13 AM | Permalink | Reply to this

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

The kernels for spans of groupoids are (2,1)-profunctors!

Great, that is what I hoped. I am still confused though by your comment where it looked like the interpretation for groupoids would have the kernels as bundle groupoids.

Not sure what you mean, probably we have some terminology mixup here. I’ll try to say again what the situation is, for the case that you seem to be interested in:

given a span $X \stackrel{i}{\leftarrow} A \stackrel{o}{\to} Y$ of $\infty$-groupoids (remember: this includes spans of $n$-groupoids for all $n$), pull-push gives a colimit-preserving functor

$o_! i^* : [X,\infty Grpd] \to [Y,\infty Grpd] \,.$

Being a colimit preserving $\infty$-functor between presheaf-categories, this is an $\infty$-profunctor

$o_! i^* : X ⇸ Y \,.$

Moreover, this pull-push operation is equivalently rewritten as the pull-tensor-push operation

$(p_2)_! ((i,o)\cdot(-)) p_1^* \,.$

You write:

By the way, I would guess that kernels for spans of categories would also be $(2,1)$-profunctors.

If you let them act on the $(2,1)$-presheaf $(2,1)$-categories over these categories, yes.

Do you know of cases which use more general profunctors? (3,1)-profunctors?

This question is a bit like asking: “Do you now examples of functors?” What specifically would you like to see?

Posted by: Urs Schreiber on November 16, 2010 1:19 AM | Permalink | Reply to this

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

This question is a bit like asking: “Do you now examples of functors?” What specifically would you like to see?

Sorry, this part was silly, but also misguided. I had it in my head that your use of $(2,1)$ in $(2,1)$-profunctors was more significant than just the fact that I asked about the case of groupoids. Oops.

I’m still trying to make sure I am clear on the rest of the conversation.

Posted by: Alex Hoffnung on November 16, 2010 2:10 AM | Permalink | Reply to this

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

Alex,

it just occurs to me: maybe I caused a misunderstanding by silently switching between $\infty Grpd/X$ and $PSh_{(\infty,1)}(X)$? These are equivalent (for $X$ an $\infty$-groupoid).

I have added a remark with further pointers on that to integral transforms on sheaves – function spaces. (Scroll down just a little bit to the second example in that section).

Posted by: Urs Schreiber on November 16, 2010 1:59 AM | Permalink | Reply to this

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

it just occurs to me: maybe I caused a misunderstanding by silently switching between $\infty Grpd/X$ and $PSh(\infty,1)(X)$?

Maybe. Let me just point out the spot that made me think your kernels were arrows into the correspondence, rather than arrows out of the correspondence:

Here the first pullback may be thought of as pulling back to the correspondence-with-kernel

$\array{ && A \\ && \downarrow \\ && X \times Y \\ & \swarrow && \searrow \\ X &&&& Y }$

I think here the ambient category was meant to be $FinSet$.

Anyway, of the parts which I think I partially understand, this is causing me the most confusion.

Posted by: Alex Hoffnung on November 16, 2010 3:32 AM | Permalink | Reply to this

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

Let me just point out the spot that made me think your kernels were arrows into the correspondence, rather than arrows out of the correspondence:

Yes, I was describing “function objects” and hence kernels (which are “function objects” on $X \times Y$) generally as objects in over-$\infty$-toposes $\mathbf{H}/X$, $\mathbf{H}/{X \times Y}$, $\mathbf{H}/Y$ for $\mathbf{H}$ a big ambient $\infty$-topos.

Generally, an object in $\mathbf{H}/X$ we may think of as a sheaf on $X$. $\mathbf{H}/X$ is the little topos incarnation of $X$. Specifically, for the special case that $\mathbf{H} = \infty Grpd$ we have that these over-$\infty$-toposes are equivalent to the $\infty$-presheaf toposes $[X, \infty Grpd]$, etc. with the equivalence given by the $(\infty,0)$-Grothendieck construction, the fully groupoidal case.

So in that case, the colimit-preserving pull-push functor $o_! i^* : [X, \infty Grpd] \to [Y, \infty Grpd]$ is a colimit-preserving functor between presheaf categories, hence a profunctor from $X$ to $Y$.

This is what I kept talking about.

But here is what I guess you are talking about: the function object $(A \to X \times Y) \in \infty Grpd/(X \times Y)$ corresponds under the Grothendieck construction equivalence to a functor

$\tilde A : X \times Y \to \infty Grpd \,.$

And since $Y$ is an $\infty$-groupoid, this is equivalently a functor

$\tilde A : X \times Y^{op} \to \infty Grpd \,.$

So this is also a profunctor in the game. This is the kernel $A$ seen on the right hand of the equivalence $\infty Grpd / (X \times Y) \simeq [X\times Y^{op}, \infty Grpd]$.

Posted by: Urs Schreiber on November 16, 2010 9:09 AM | Permalink | Reply to this

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

Hi Urs,

Thanks for the patient explanations and all of the links. Things are becoming more clear now.

Posted by: Alex Hoffnung on November 16, 2010 4:18 PM | Permalink | Reply to this

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

Here the first pullback may be thought of as pulling back to the correspondence-with-kernel

I think here the ambient category was meant to be $FinSet$.

No. The pull-push I described works in great generality: for any ambient $\infty$-topos H and any span of morphisms

$X\stackrel{i} {\leftarrow} A \stackrel{o}{\to} Y$

in $\mathbf{H}$ there is the colimit preserving (hence “linear”) pull-push-functor

$o_! i^* : \mathbf{H}/X \to \mathbf{H}/A \to \mathbf{H}/Y$

and this is equivalently rewritten as the pull-tensor-push operation

$(p_2)_! (A \cdot (-))p_1^*$

where the “tensor”-operation is itself the product in $\mathbf{H}/(X \times Y)$.

I spelled that out for the case of the 1-topos $\mathbf{H} = FinSet$ only as an example.

Posted by: Urs Schreiber on November 16, 2010 9:24 AM | Permalink | Reply to this

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

Hi David,
One particular example of ‘interesting correspondence space + interesting kernel’ is discussed in the paper

“T-Duality: Topology Change from H-flux” - P. Bouwknegt, J. Evslin, V. Mathai
http://arxiv.org/abs/hep-th/0306062

I’m sure the next post in this series will talk about Fourier-Mukai transforms, but the above paper spins this idea in a more geometrical setting. The span is given by the fibre product $E \times_M E'$ of two particular circle bundles $E,E'$ over a common base $M$. There is extra data in the form of a bundle gerbe on the total space of each bundle, as well as a bundle gerbe on the correspondence space.

The integral transform in this case is a map

$T_! : K^\bullet(E,H) \to K^{\bullet+1}(E',H')$

of the twisted $K$-theories of these bundles twisted by their respective bundle gerbes.

The kernel of this transform is given by tensoring with the Poincare line bundle on the correspondence space.

If you dont want to stray too far from geometry, this paper presents a good aspect of the (twisted) Fourier-Mukai idea.

Posted by: Ryan Mickler on November 15, 2010 2:30 PM | Permalink | Reply to this

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

For ease of reference I have started a page

integral transforms on sheaves

that summarizes some general abstract statements on the analogy between $Pr(\infty,1)Cat$ and vector spaces, linear maps and matrices.

This is not meant to be an exhaustive discussion. But maybe a start.

Posted by: Urs Schreiber on November 15, 2010 1:36 PM | Permalink | Reply to this

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