June 13, 2008

An Exercise in Groupoidification: The Path Integral

Posted by Urs Schreiber

As we have been reminded of by the last entry a while ago some of us had been very busily thinking here about

What is the quantum path integral really?

We were trying to understand this by looking at simple finite combinatorial toy models. I can’t tell how far John Baez and Alex Hoffnung have gotten since then, but I know how far I got. Here is where I am coming from:

Extended quantum field theory of $\Sigma$-model type should work like this:

a) the “classical” data is: a target space $X$ together with a nonabelian differential $n$-cocycle $\nabla$ on it, expressed in terms of a parallel transport $n$-functor.

b) the quantization procedure is, roughly: to each piece $\Sigma$ of parameter space assign the result of forming the “space of sections” of the transgression of $\nabla$ to $Maps(\Sigma,X)$.

It’s comparatively clear that and how this works for $dim \sigma \lt n$: transgression of transport is just forming the inner hom and then taking sections.

The more mysterious part is this: with the really right way of looking at this, it should be true that turning this crank for $dim \Sigma = n$ magically leads to the path integral itself, thus realizing Dan Freed’s old observation that the path integral should be just the top dimension part of a general process which always just transgresses and then takes sections. If this comes out as hoped, one would begin to hope that this provides hints for how we should really be thinking of the mystery of the path integral.

Anyway, I had a bunch of ideas about this but didn’t quite get to the point where I was entirely happy. Now here is something which is simple but looks a bit like progress to me. A simple exercise in Groupoidification. I haven’t really had the time to think it true in its entirety. But that’s one reason more for me to share it.

So I want to look at this pathetically simple setup:

Background/motivation

target space is a category $P_1(X)$ generated from a finite graph.

We fix a finite gauge group $G$ and some representation $\rho : \mathbf{B} G \to \mathrm{Set}$ (where $\mathbf{B}G$ is the one-object groupoid version of $G$). Let’s write $V//G$ for the corresponding action groupoid.

The background field is a functor $\nabla : P_1(X) \to \mathbf{B}G \,.$ A state is a section of this restricted to points, namely a lift of $\nabla|_0 : P_0(X) \to \mathbf{B}G$ through $V \to V//G \to \mathbf{B}G \,.$ So that’s just a choice of element in $V$ over each point.

A bit more interesting, if we transgress to path space by homming the interval category $(a \to b)$ into everything to get $tg \nabla : hom((a \to b), P_1(X)) \to hom(a\to b, \mathbf{B}G) \,.$ Then restricting that functor to objects and taking sections in terms of lifts through $hom(a \to b, V) \to hom(a \to b, V//G) \to hom(a \to b, \mathbf{B}G)$ over objects yields: over each path a choice of element in $V$ over the endpoints, such that they are related by the parallel transport of $\nabla$ along that path.

Everybody still following? But to some extent that is just motivation for the following simple situation that I want to look at:

A span of groupoids

Let’s build a span of finite groupoids this way:

Let $\Gamma_X := \cup_{x \in P_0(X)} V//G$ be something like the groupoid of the graph of sections over points. Big words – I just mean the disjoint union of one copy of the action groupoid of our rep over each point of target space. To be thought of as a groupoid of sections over target space.

Similarly, let $\Gamma_{P X} := \cup_{\gamma \in hom(a\to b, P_1(x))} \Gamma( tg_\gamma \nabla)$ be the disjoint union over all morphisms in $P_1(X)$ of all sections of $\nabla$ over that path: for each $\gamma$ this groupoid is isomorphic to $V//G$ again, but we think of an object now as a flat section over the path $(x,v_1) \stackrel{(\gamma,g = \nabla(\gamma))}{\to} (y,v_2 = v_1\cdot g) \,.$

Now we build a span from these of the form

$\array{ &&& \Gamma_{P X} \\ & {}^{in}\swarrow &&& \searrow^{out} \\ \Gamma_{X} &&&&& \Gamma_{X} } \,.$

Here the functor $in : \Gamma_{P X} \to \Gamma_X$ simply projects out the left end of a path, and the functor $out : \Gamma_{P X} \to \Gamma_X$ the right end. So

$in : ((x,v_1) \stackrel{(\gamma, g = \nabla(\gamma))}{\to} (y,v_2)) \mapsto (x,v_1)$

and

$out : ((x,v_1) \stackrel{(\gamma, g = \nabla(\gamma))}{\to} (y,v_2)) \mapsto (y,v_2) \,.$

Okay, now lets do groupoidily linear algebra and see how bundles of sets over $\Gamma_X$ pull-push through this span.

Let me pick one single point $x$ in target space and one section over it, $v \in V$. Define

$\array{ \{\bullet\} \\ & \searrow \\ && \Gamma_X }$

to be the functor of groupoids which sends the single object of the terminal groupoid to that object $(x,v)$ in $\Gamma_X$.

The pull-push

$\array{ &&&&& in^* \{\bullet\} \\ &&&&& \downarrow \\ \{\bullet\}&& &&& \Gamma_{P X} &&&&& \int in^*\{\bullet\} \\ & \searrow && {}^{in}\swarrow &&& \searrow^{out} && \swarrow \\ &&\Gamma_{X} &&&&& \Gamma_{X} }$

produces first $in^* \{\bullet\} \to \Gamma_{P_X}$: that has precisely one point sitting over each labeled path which starts at $x$ and is labeled there by $v$.

Then it produces $\int in^* \{\bullet\} \to \Gamma_X$: this has over the point $y$ with label $v'$ one point per path $x \to y$ which is labeled by $v$ over $x$ and by $v'$ over $y$.

To see what this means, fix one point $y$ in $X$. Then we get one point over each label $v'$ for each path from $x$ to $y$ labeled by $v$ on the left and by $v'$ on the right.

But since the labels of the paths are sections of the transgressed transport functor over these paths, which just means that these are flat sections of the original transport over these paths, it means that the $v'$ appearing here are of the form $\nabla(g) v$ for $g$ the parallel transport over the given path.

So the “total” space of $\int in^* \{\bullet\}$ over $y \in X$ is the $V$-colored set $\cup_{\gamma : x \to y} \{\bullet_{\nabla(\gamma) v} \} \,,$

where I write $\bullet_v$ for an element colored by $v$.

If $V$ has an additive structure, for instance if it is a vector space, we have the cardinality operation on $V$-colored sets $|\cdot| : Set_V \to V$ and get that the cardinality of the above is

$\sum_{\gamma : x \to y} {\nabla(\gamma) v} \,,$

But that’s the right path integral kernel for propagation from $x$ to $y$ acting on the state $v$ over $x$.

So what?

About all ingredients of the above we have talked before, in one way or another. Lots of ingredients from John’s discussion of groupoidification and John an Jeffrey’s “categorified” quantum mechanics appear. But somehow I feel that I have not before put things together in the picture as above. To me, I had the feeling this clarified some things that had been a bit mysterious to me before:

a) the fact that the path integral should be “taking sections at codimension 0”;

b) the natural connection of a) to groupoidification;

c) the natural and automatic appearance of $V$-colored sets.

But I have to stop here. If Konrad or Hisham read this, or some of the other people waiting for me getting back to them with tasks finished, they’ll be unhappy to see me instead talk about foundational abstract nonsense here. But I needed to relax a bit. :-)

Posted at June 13, 2008 4:58 PM UTC

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Re: An Exercise in Groupoidification: The Path Integral

You no doubt saw my comment here and no doubt could guess what I was thinking :)

I still stand by my old comment here, but extend its (admittedly highly speculative) scope to what you are talking about here as well.

I’m slowly catching up, which means you must be getting slow in your old age ;) Just kidding! :) I am finding that I’m understanding more and more about things discussed here lately, which somehow seems miraculous. Thanks!

Posted by: Eric on June 14, 2008 10:57 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

I’m glad you’re taking time off now and then to ponder what groupoidification might do for physics. I’m pretty frantic right now — catching a plane in two hours — so it’ll take me a while longer to absorb what you’re doing. But, I’ll think about it.

I can’t tell how far John Baez and Alex Hoffnung have gotten since then…

It turned out Alex’s real talents lie in areas not so directly connected to physics — something more like ‘algebra’. So, my project of understanding path integrals will wait until some other student comes along and helps me out.

Alex hasn’t studied much physics, so it was hard for him to tackle the project of drastically reformulating it. But what really impressed me is how I once said “you know, someone told me that Chen spaces should be an example of a concrete quasitopos, whatever that is” — and then he learned enough about sheaves on sites, topoi and quasitopoi to prove this! He had a lot of help from James Dolan, and a bit from me too — but still, most people find this stuff frighteningly abstract, and he seemed to jump right in and enjoy it.

So, right now our plan is for him to fill in the many holes in HDA7 — some of which involve topos theory. Then, he can use this to groupoidify the usual sort of Hecke algebra we get from any Dynkin diagram. This is part of a bigger project to work out all the details of what was sketched in the Geometric Representation Theory seminar earlier this year. Christopher Walker is working on another side of that project: first giving a nice ‘practical’ introduction to groupoidification — “Look, ma! No topoi!” — and then groupoidifying Hall algebras.

Hall algebras and Hecke algebras are both closely connected to quantum groups, so very roughly you could say this project is about groupoidifying quantum groups. For this reason, Alex plans to spend a lot of time in New York this summer talking to Aaron Lauda.

Meanwhile, as a kind of side project, he wants to work on De Rham cohomology for smooth spaces.

He will be at HOCAT in Barcelona from June 30th to July 5th — but alas, you won’t.

Posted by: John Baez on June 14, 2008 6:46 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

I’n not sure if I’ve asked this before, but how does this related to the groupoid programme of Ian Stewart and Martin Golubitsky, as reviewed in Bulletin (new series) of the American Mathematical Society, vol.42, no.1. Jan 2005, pp.99-103, and the wonderful paper:

Ian Stewart and Martin Golubitsky

NONLINEAR DYNAMICS OF NETWORKS: THE GROUPOID FORMALISM

AMERICAN MATHEMATICAL SOCIETY, 2006
May 3, 2006

Abstract:

A formal theory of symmetries of networks of coupled dynamical
systems, stated in terms of the group of permutations of the nodes that
preserve the network topology, has existed for some time. Global network
symmetries impose strong constraints on the corresponding dynamical systems,
which affect equilibria, periodic states, heteroclinic cycles, and even chaotic
states. In particular, the symmetries of the network can lead to synchrony,
phase relations, resonances, and synchronous or cycling chaos.

Symmetry is a rather restrictive assumption, and a general theory of
networks should be more flexible. A recent generalization of the group-theoretic
notion of symmetry replaces global symmetries by bijections between certain
subsets of the directed edges of the network, the ‘input sets’. Now the
symmetry group becomes a groupoid, which is an algebraic structure
that resembles
a group, except that the product of two elements may not be defined. The
groupoid formalism makes it possible to extend group-theoretic methods to
more general networks, and in particular it leads to a complete classification
of ‘robust’ patterns of synchrony in terms of the combinatorial structure of the
network.

Many phenomena that would be nongeneric in an arbitrary dynamical
system can become generic when constrained by a particular network
topology. A network of dynamical systems is not just a dynamical system with
a high-dimensional phase space. It is also equipped with a canonical set of
observables—the states of the individual nodes of the network. Moreover, the
form of the underlying ODE is constrained by the network topology—which
variables occur in which component equations, and how those equations relate
to each other. The result is a rich and new range of phenomena, only a few of
which are yet properly understood.

Posted by: Jonathan Vos Post on June 15, 2008 3:04 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Had a busy weekend, celebrating a friend’s wedding. Now I need to run and think about something else.

I should come back here and phrase Freed-Quinn’s discussion of Dijkgraaf-Witten entirely in the above context. But not now.

One thing to notice is: in the above entry I effectively say that what matters as the groupoid of sections of a bundle for the purpose of quantization is the action groupoid of its transport functor.

As it appears, for the fundamental rep., at the very end of the article with David Roberts.

Posted by: Urs Schreiber on June 16, 2008 7:30 AM | Permalink | Reply to this
Read the post Teleman on Topological Construction of Chern-Simons Theory
Weblog: The n-Category Café
Excerpt: A talk by Constant Teleman on extended Chern-Simons QFT and what to assign to the point.
Tracked: June 17, 2008 6:54 PM

Re: An Exercise in Groupoidification: The Path Integral

I really want to put some effort into following this idea. I have a lot of catching up to do though. I have made an effort to trace the references back, but after a couple levels became overwhelmed.

I have questions that you will probably find to be quite basic so feel free to ignore them. You have better things to do :) It never hurts to ask though!

The first question is from the first line of the Background/Motivation:

target space is a category $P_1(X)$ generated from a finite graph.

First a warning: I intend to try to either understand or reinterpret everything here in terms of diamonds.

So given a 2-diamond, a.k.a. a binary tree, how exactly do you generate a category? Objects are nodes and morphisms are edges. But then you need to define composition. The obvious thing to do would generate morphisms in $P_1(X)$ that do not correspond to edges in $X$. Is that going to be a problem? Should we try to keep track of how many “time steps” are present in a morphism?

In a binary tree, we can label the nodes with a space index $i$ and time index $j$ via $(i,j)$, where the generating edges are of the form

$(i,j) \to (i\pm1,j+1)$.

If $P_1(X)$ is generated the way I think it is, do we distinguish between

$(i,j)\to(i+1,j+1)\to(i,j+2)$

and

$(i,j)\to(i-1,j+1)\to(i,j+2)$

? In other words, do we assume there are two morphisms joining $(i,j)$ and $(i,j+2)$?

Baby steps…

I’ll stop there for now although you can imagine with this beginning there are many more questions in store :O

Best wishes

Posted by: Eric on June 18, 2008 7:42 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Actually, asking the question probably helped me answer it.

Yes. I think the cardinality of

$hom((i,j),(i,j+2))$

is two. For some reason, I thought we might want it to be one, but that is silly. Baby steps…

Posted by: Eric on June 18, 2008 8:08 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Now that I think I understand $P_1(X)$, i.e. a morphism in $P_1(X)$ corresponds to a “path” in $X$ (which might have something to do with the fact that you use a “$P$” to denote it!), I’ll try to take another baby step.

At one point, I made a sincere effort to learn some basic representation theory, but didn’t get very far. Now you denote a representation as

$\rho: \text{B}G\to\text{Set}$.

This is not like the representations I remember looking at. Having lurked around here long enough, I think you are referring to a more recent understanding of representations? I think I can simply squint and continue, but I think I need to pause and try to get a little bit of a grasp on what $V//G$ is.

How do we think about $V//G$ in this set up?

Posted by: Eric on June 18, 2008 8:27 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

By the way, I’m finding page 9 of this to be very helpful so far. Baby steps…

Posted by: Eric on June 18, 2008 8:40 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Groupoids (and more group actions)

Posted by: Eric on June 18, 2008 11:06 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Ok. I think I understand the basics of action groupoids now thanks to John Armstrong’s awesome article. No questions at the moment.

PS: Sorry for the plethora of comments. Urs encouraged me to ask questions so I am. As I answer them myself I thought I’d say so to avoid anyone taking time to answer a question unnecessarily. Making progress…

Thanks!

Posted by: Eric on June 18, 2008 11:40 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Thanks. Honest, I’ll be back up to speed again soon, in case you’re eager to hear more about linear algebra.

Posted by: John Armstrong on June 19, 2008 12:32 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

My next goal is to understand “transgression”. Tracing back references led me to your “Integration wihout Integration” paper, which I remember seeing but couldn’t spend the time to absorb it. Neat idea!

Now that I think about it a litte bit, it reminds me of one of our old conversation from way back when at the String Coffee Table:

Before the Flood

Were we talking about transgression? I was trying to pull back a 1-form on a base manifold to a 0-form on loop space so that integration on the base manifold was evaluation on loop space. Is it at all related?

*glimmer of hope to actually understand this some day*

Thanks

Posted by: Eric on June 19, 2008 1:01 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Looking at your comment here reminds me of a source of confusion I’ve always had when it comes to loop space, i.e. parameterized vs unparameterized loops.

What you described is what I would call a parameterized loop map

$\mathcal{L}_{param}: \mathcal{L}M \times [0,1]\to M$.

What I was talking about was some funky multivalued map that might be called an unparameterized loop map

$\mathcal{L}_{unparam}: \mathcal{L}M \to M$.

This takes a point in loop space to an unparameterized loop in base space, i.e. a 0-dimensional manifold to a 1-dimensional manifold.

I mention this almost at random, but also because it reminded me of a recent comment you made about unparameterized loops here.

Back then we were struggling because the mathematics/geometry of loop space hadn’t been developed yet. I assume that has changed now, but is there a significant distinction between the geometry of parameterized loop space as opposed to unparameterized loop space? I would be more interested in the latter.

Posted by: Eric on June 19, 2008 2:13 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Okay, let’s start going through this in small steps. I propose some steps, and you let me know what you think.

Step 1) A group as a groupoid.

Given any group $G$, there is a groupoid with a single object and one morphism per element of $G$, with composition of morphisms coming from the product in $G$.

This construction is so trivial that often it is taken as being empty. But it is not and we need the distinction between a monoidal set (for instance a group) and the one-object category it gies rise to (for instance a groupoid) later on.

So I write $\mathbf{B} G = \{ \bullet \stackrel{g}{\to} \bullet | g \in G \}$

for the one-object groupoid version of the group $G$.

Okay?

Step 2) A representation as a functor.

We eventally want to talk about sections of associated vector bundles. That’ll be easy once we realize that a representation of a group, i.e. a group homomorphism

$\rho : G \to GL(V)$

for some vector space $V$ is nothing but a functor

$\rho : \mathbf{B} G \to Vect$

which sends $\rho : (\bullet \stackrel{g}{\to} \bullet) \mapsto (V \stackrel{\rho(g)}{\to} V) \,.$

Okay?

Step 3) An action groupoid from a representation.

The construction of the path integral that we are approaching here will be entirely formulated in terms of spans of action groupoids. So we need to understand action groupoids.

Given a representation $\rho : \mathbf{B} G \to Vect$ as above, we define a groupoid to be denoted $V//G$ as follows: its objects are the elements of $V$ (i.e. the vectors in $V$) and from each element $v$ there starts precisey one morphism per element in $g$ which goes to $\rho(g)(v)$:

$V//G := \{ v \stackrel{g}{\to} \rho(g)(v) | v \in V, g \in G \} \,.$

Composition of morphisms is the obvious one coming from the product in the group.

Okay so far?

As an aside, to be skipped by those who want to stick to the elementary discussion, I mention the more abstract definition of $V//G$ that will eventually provide the big abstract picture to be developed later on:

Over the category of (small) Sets there sits the category of pointed (small) sets $T_{pt} Set$ (forming the “universal Set-bundle”)

$\array{ T_{pt} Set \\ \downarrow \\ Set }$

Using our representation and the forgetful functor $Vect \to Set$ we can pull back this universal Set-bundle to $\mathbf{B}G$:

$\array{ V//G &\to& T_{pt} Vect &\to& T_{pt} Set \\ \downarrow && \downarrow && \downarrow \\ \mathbf{B}G &\stackrel{\rho}{\to}& Vect &{\to}& Set } \,.$

This witnesses $V//G$ as the groupoid incarnation of the $\rho$-associated vector bundle to the universal $G$-bundle. But never mind that.

Step 4) enter $V$-colored sets

Consider the simple situation where there is some finite set $S$ and a map of that set down to the objects of $V//G$:

$s : S \to V//G \,.$

Those objects are just the elements of $V$, so the map $s : S \to V$ defines a “$V$-coloring” of $S$. Each element $a$ of $S$ is sent to some vector $s(a)$ in $V$, so we can think of it as being labeled or colored by $s(a)$.

Since $V$ is a vector space we can add elements in $V$. This allows to define a “cardinality of $V$-colored sets”

$|\cdot| : SetsOver(V) \to V$

by

$|S| = \sum_{a \in S} s(a) \,.$

so our finite set $s : S \to V$ in particular defines a single vector in $V$. (Notice that the empty set maps to the 0-vector.) It is like a single vector equipped with extra information for how that vector was obtained by adding a couple of other vectors.

Okay so far?

Step 5) Finally: vector bundles.

Pick your favorite finite category and allow me to call is $P_1(X)$. That’s because we will think of its objects as points in some space and of its morphisms as paths between these points.

Consider a functor $tra : P_1(X) \to \mathbf{B}G$ which labels each path with a group element, such that composition of paths corresponds to multiplication of group elements.

Such a functor we can regard as a $G$-connection on a trivial $G$-bundle over $X$.

Using our representation, we can turn this into a connection on the associated vector bundle, simply by composing functors:

$\rho_*tra : P_1(X) \stackrel{tra}{\to} \mathbf{B}G \stackrel{\rho}{\to} Vect \,.$

This functor sends each point of $X$ to the vector space $V$, to be thought of as the fiber of a vector bundle over that point, and sends a path labeled by the group element $g$ to the linear map $V \stackrel{\rho(g)}{\to} V$ between the vector spaces over its endpoints.

Okay?

Step 6) Sections of the vector bundle.

we can combine now the idea which led to the action groupoid with the functor $\rho_* tra$ to get something like the action groupoid of $P_1(X)$ acting on $V$ by $tra$ and $\rho$.

this groupoid I’ll call $tra^* V//G\,.$

Its objects are pairs $(x,v)$ consisting of a point $x$ in $X$ and a vector $v \in V$. For each path $x \stackrel{\gamma}{\to} y$ in $P_1(X)$ there is one morphism in $tra^* V//G$ starting at each $(x,v)$ which goes to $(y, \rho(tra(\gamma))(v))$

$(x,v) \stackrel{(x \stackrel{\gamma}{\to} y)}{\to} (y, \rho(tra(\gamma))(v)) \,.$

You can forget the morphisms for the time being, actually. Doing so, we are left with nothing but the set $X \times V$ of objects: there is the vector space $V$ sitting over each point of $X$.

Then again consider a finite set $s : S \to tra^* V//G$ sitting over our groupoid. This is hence the same as nothing but a map of sets $s : S \to X \times V \,.$ But this is nothing but a bundle of $V$-colored sets over $X$! It is one $V$-colored set over each point of $X$.

Do you see that?

Taking our cardinality of $V$-colored-sets, we thus obtain from $S \to X \times V$ an assignment of one vector in $V$ over $x$ for each point $x$ of $X$. That’s a section of our vector bundle.

Okay so far?

Here again a more abstract description, to be skipped if you don’t want to see it.

The groupoid $tra^* V//G$ is the pullback $\array{ tra^* V//G &\to& V//G &\to& T_{pt} Vect &\to& T_{pt} Set \\ \downarrow && \downarrow && \downarrow && \downarrow \\ P_1(X) &\stackrel{tra}{\to}& \mathbf{B}G &\stackrel{\rho}{\to}& Vect &{\to}& Set } \,.$

Sections of our vector bundle $E \to X$ (at the moment being just the trivial vector bundles $E = X \times V$, but that will change as we proceed) correspond to (finite)set-bundles over this groupoid:

$FiniteSetsOver(tra^* V//G) \to \gt \Gamma(E \to X) \,.$

We could also talk about flat sections of $E$, since we have a connection in the game. These would come from full functors down to $tra^* V//G$.

Step 7) Transgression

The next step would be transgression of $tra$ to path space. Then we’d repeat all the above steps there and end up with a span of action groupoids. Pull-pushing sections through that span will realize the path integral.

But before doing that, let me know if you follow all of the steps above. Or else, let me know about what is unclear.

Posted by: Urs Schreiber on June 19, 2008 11:47 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Wow! Thank you Urs. This is awesome.

Step 1) A group as a groupoid. Okay?

Yeah, I think I’m ok with this.

Step 2) A representation as a functor. Okay?

Makes sense.

Step 3) An action groupoid from a representation. Okay?

Yep. John Armstrong’s article was very helpful with this.

Step 4) Enter $V$-colored sets. Okay so far?

Yep, but I’m getting on thin ice. I like the idea though.

Step 5) Finally: vector bundles.

Pick your favourite finite category and allow me to call it $P_1(X)$.

I already picked my $X$ to be a 2-diamond complex, a.k.a. binary tree, and my $P_1(X)$ is a category whose objects are nodes and morphisms are paths on 2-diamonds :)

Consider a functor

(1)$\text{tra}: P_1(X)\to \text{B}G$

which labels each path with a group element, such that composition of paths corresponds to multiplication of group elements.

Couldn’t this functor have been generated from $X$ and an assignment of $G$ to each edge in $X$? Better yet, couldn’t we have just defined elementary (two object) groupoids on each edge of $X$ to begin with?

Okay?

Yep, I think so.

Step 6) Section of the vector bundle. Okay so far?

I was a bit on thin ice regarding $V$-colored sets and that hasn’t improved, so I’m a bit on thin ice here as well, but think I am “ok”. I’m not exactly sure what $S$ is supposed to represent, but can trudge along.

This is hence the same as nothing but a map of sets

(2)$s:S\to X\times V.$

I would have understood this better if you said there is a set $S$ and a map $s:S\to V$ setting over each point of $X$. Is that the same as saying there is a map $s:S\to X\times V$. If so, I think I’m ok.

Step 7) Transgression

Woohoo! This is the part I was looking forward too!

Posted by: Eric on June 19, 2008 6:35 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Eric said

I would have understood this better if you said there is a set $S$ and a map $s:S\rightarrow V$ setting over each point of $X$. Is that the same as saying there is a map $s:S\rightarrow X\times V$.

I don’t think that can be quite right, if I understand what Urs is saying. A map $S\rightarrow X\times V$ will have different (sets of) points of $S$ sitting over different points of $X$, and possibly no points of $S$ over some of the points of $X$. So we get, for each point of $X$, a map from a subset of $S$ to $V$, with the subsets for different points in $X$ being disjoint.

Is that right, Urs?

Posted by: Tim Silverman on June 19, 2008 7:28 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Hmm. I’d think we want the same set $S$ available at each point of $X$. Maybe there was a typo? I’m just guessing because I’m not sure what $S$ is supposed to be.

$S$ seems to provide some “internal structure” to each point of $X$ that leads to a bunch of vectors over each point. Computing the cardinality results in a single vector at each point, i.e. a section of a vector bundle (?).

I’m just thinking out loud…

Posted by: Eric on June 19, 2008 8:03 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Hmmm${}^2$.

Reading it again, I don’t think it was a typo. So it seems that disjoint subsets of $S$ are over distinct points in $X$ (???)

Posted by: Eric on June 19, 2008 10:27 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Is that right, Urs?

Yes, that’s precisely right.

I think I mentioned the “right” way to think about these $V$-colored sets, but let me amplify:

consider again our trivial vector bundle $X \times V$ over $X$, all whose fibers look like the fixed vector space $V$.

A section of this vector bundle is a choice of element $v_x \in V$ for each $x \in X$.

But now, instead of just grabbing these elements out of $V$, we leave them in there and instead add a marker which tells us which element we picked.

In other words, instead of writing $v_x \in V$ we’ll write

$v_x : \{\bullet \} \to V \,.$

We regard the choice of the vector $v_x$ in $V$ as a map from the set with a single element to the vector space $V$.

This identifies the vector $v_x$ in $V$ as a set over $V$. In this case just a one-element set.

But the point then is that this allows us to generalize a bit.

First of all, we could consider the empty set over $V$ and the unique map of sets

$\{\} \to V \,.$

We should agree that this empty set over $V$, i.e. the set $V$ with none of its elements labelled, is a way to talk about the 0-vector in $V$.

Next, if we have a set with more than one element sitting over $V$, for instance the two-element set

$\{a,b\} \to V$ $a \mapsto v_a$ $b \mapsto v_b$

which hence assigns a label to two elements in $V$ (which might coincide, in which case we’d have a single element with two labels) then we agree that we read this as a way to think about the vector obtained by the sum of these two vectors, i.e. $v_a + v_b$ in the above example.

So sets over $V$ can be thought of as

vectors in $V$ together with information about how that vector arises from addition of other vectors

We can forget this extra information and just rememeber the summed-up vector. This operation “forgetting” is a map

$|\dot| : SetsOver(V) \to V$

which is the cardinality operation on $V$.

You can find more about this somewhere in John’s online notes. I’ll try to provide the link when I find the time to look for it.

As an aside, for those interested: the category $SetsOver(V)$ is symmetric bimonoidal: it has two distriuting monoidal structures on it, namely the cartesian product and the disjoint union

$(SetsOver(V),\times, \cup)$

Also $V$ is a monoidal category (a 0-category, though): it comes equipped with the operation of vector addition.

The cardinality operation above is a monoidal functor

$|\cdot| : (SetsOver(V),\cup) \to (C,+)$

in that the disjoint union of sets over $V$ maps to the sum of the corresponding vectors.

If $V$ happens to actually be an algebra (the ground field, in particular) such that it is also bimonoidal $(V,\cdot,+)$ then the cardinality is even a bimonoidal functor

$|\cdot| : (SetsOver(V),\times,\cup) \to (C,\cdot,+)$

The point of this is: sets over $V$ behave just as elements of $V$ do, too, only that sets over $V$ are a bit more “fluffy” than elements in $V$. For one, there are many sets over $V$ which correspond to one and the same element $v \in V$. They correspond to different ways of obtaining $V$ from sums of other vectors of $V$.

Finally, to get back to our situation, consider $X = \{x_1, x_2\}$ to be a “space” consisting of two points and let again $X \times V = \{(x_i,v) | x_i \in X, v \in V\}$ be the trivial $V$-vector bundle over that space.

Here are some examples of sets over $X \times V$ and how they correspond to sections of this vector bundle:

the empty set over $X \times V$

$\{\} \to X \times V$

maps under $|\cdot|$ to the 0-section of the vector bundle.

The one-element set

$\sigma : \{a\} \to X \times V$ with

$\sigma : a \mapsto (x_1,v)$

corresponds to the section which over $x_1$ is the vector $v$ and over $x_2$ is the vector $0$.

Last example: the three-element set

$\sigma : \{a,b,c\} \to X \times V$ with

$a \mapsto (x_1,v_a)$ $b \mapsto (x_1, v_b)$ $c \mapsto (x_2, v_c)$

corresponds to the section which over $x_1$ is the vector $v_a + v_b$ and over $x_2$ is the vector $v_c$.

Does that clarify the idea?

Posted by: Urs Schreiber on June 20, 2008 10:02 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

It seems you want to consider a one-element set

$S = \{\bullet\}$

and the bundle of $V$-colored sets over $X$

$s:S\to X\times V.$

I blabbered about this before, but find that I’m still confused. If there is only one element in $S$, then I don’t see how this one element can map to anything but just one element in $X\times V$, i.e.

$s:\bullet\mapsto (x_\bullet,v_\bullet).$

Every other point besides $x_\bullet$ must be assigned the 0 vector. Are you sure we don’t want

$s:X\times S\to X\times V$

with

$s: (x,\bullet)\mapsto (x,v_\bullet)$

? In this way, one copy of $S$ is available over every point in $X$.

For the two-element set $S=\{a,b\}$, then

$s:X\times S\to X\times V$

allows us to keep track of two vectors over each point of $X$, i.e.

$s:(x,a)\mapsto (x,v_a)$

and

$s:(x,b)\mapsto (x,v_b).$

One way or the other, I think we want to have one copy of our set available at each point, particularly when our set is $\{\bullet\}$. Otherwise $\{\bullet\}$ would seem to lead to a very boring section with only one point having a nonzero vector.

Posted by: Eric on July 1, 2008 11:21 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

I’m gaining conviction that something like this must be correct. I think we want to be able to say that for each $x\in X$ we have a map

$s_x: S\to V.$

I’m not sure what the correct way to express this is, but I don’t think it is

$s:S\to X\times V.$

Posted by: Eric on July 2, 2008 3:25 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Let

$F = hom(S,V)$

be the set of $V$-colored maps. Since $X$ is finite, we have a trivial bundle

$\pi:X\times F\to X.$

Sections are just maps

$s:X\to X\times F$

such that $\pi s(x) = x$.

This is scary. I think I’m starting to understand this stuff! :)

Posted by: Eric on July 2, 2008 3:49 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

OK. Given a section of a $V$-colored bundle

$s:X\to X\times hom(S,V)$

we can reinterpret this as a map

$s:S\to X\times V.$

For every $x\in X$ we have a pair $(x,s_x)\in X\times F$. If we take any element $a\in S$ and feed it to this section, we get an element $(x,v_x)\in X\times V$ for every point $x\in X$, but this means that $s$ can also be thought of as a map

$s:S\to X\times V.$

I think I got it!

Posted by: Eric on July 2, 2008 6:06 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Couldn’t this functor have been generated from $X$ and an assignment of $G$ to each edge in $X$?

I think what you mean is this: you are considering the case that the category I called $P_1(X)$ is the category freely generated by a graph, i.e. the result of taking a graph, declaring objects to be the vertices of the graph and morphisms all finite sequences of matching edges, with composition of morphisms just being concatenation of such sequences of edges.

Then, yes, a functor from such a category to any category $C$ is precisely the same thing as a graph map from your graph to the graph underlying $C$.

That is: such a functor comes precisely from sending each vertex of the graph to an object in $C$ and each single(!) edge in the graph to a morphism in $C$.

For the special case we were talking about, where $C = \mathbf{B}G$, this amounts to assigning one group element to each edge in the graph. Yes.

Better yet, couldn’t we have just defined elementary (two object) groupoids on each edge of $X$ to begin with?

Wait, we don’t want to assign groupoids to edges. At least not at the moment. But maybe you mean that the groupoid freely generated from a graph with just a single edge and two vertices is precisely a two-object groupoid with a single nontrivial morphism (and its inverse). If so, yes.

Posted by: Urs Schreiber on June 20, 2008 9:39 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Step 7) Transgression

Woohoo! This is the part I was looking forward too!

All right. The fun thing is, we’ll give that fancy word now a really simple but profound interpretation:

Again, choose some finite category and allow me to call it $P_1(\Sigma) \,.$

Just as we thought of our finite category $P_1(X)$ as points and paths in some space which from the point of view of physics is “spacetime” or “target space”

$\mathbf{tar} = P_1(X)$

we now think of $P_1(\Sigma)$ as points and paths in a space called “parameter space”

$\mathbf{par} = P_1(\Sigma)$

which models the shape of the things which we want to regard as trajectories in $X$.

So in particular, consider the case where

$P_1(\Sigma) = \{a \to b\}$

is the “interval category”, the cylinder object in the category of categories, the category freely generated from a graph with a single edge.

This category is our model for the worldline (or the “world-interval”, rather) of a particle.

Now, the “space” of all $\Sigma$-shaped trajectories in $X$ is simply

$hom(P_1(\Sigma), P_1(X)) = Funct(P_1(\Sigma), P_1(X))$

the category of functors from $P_1(\Sigma)$ to $P_1(X)$.

An object in this category is nothing but a choice of morphism in $P_1(X)$, hence a path in $X$. A morphism in this category is something like a translation of this path. We won’t need the morphisms in this functor category until much later, so just ignore them for the time being.

We can also map $P_1(\Sigma)$ into our target category $\mathbf{B}G$. And into $V//G$. Into everything in sight, really.

And given a functor $tra : P_1(X) \to \mathbf{B}G$ we canonically get a functor $Funct(P_1(\Sigma),P_1(X)) \to Funct(P_1(\Sigma),\mathbf{B}G)$ simply by postcomposing a functor from $P_1(\Sigma)$ to $P_1(X)$ by $tra$:

$Funct\left(P_1\left(\Sigma\right),P_1\left(X\right)\right) : \left( \left(a \to b \right) \stackrel{\gamma}{\mapsto} \left( x_a \stackrel{\gamma(a\to b)}{\to} y_a \right) \right) \mapsto \left( \left(a \to b \right) \stackrel{\gamma}{\mapsto} \left( x_a \stackrel{\gamma(a\to b)}{\to} y_a \right) \stackrel{tra}{\mapsto} \left( \bullet \stackrel{tra(\gamma(a\to b))}{\to} \bullet \right) \right)$

This functor is oficially denoted

$Funct(P_1(\Sigma),tra) : Funct(P_1(\Sigma),P_1(X)) \to Funct(P_1(\Sigma),\mathbf{B}G) \,.$

But that’s heavy notation for a simple concept. You should at this point take a pen and paper and draw out the action of this functor. It’s really very simple.

And, lo and behold, this functor $Funct(P_1(\Sigma),tra)$ is the transgression of $tra$ to the space of paths.

Let me know if you are happy with this so far.

As an aside, I again end by mentioning the bigger picture which we are developing here:

I’ll be taking the big pullback diagram which encoded our vector bundle with connection

$\array{ tra^* V//G &\to& V//G &\to& T_{pt}Vect &\to& T_{pt}Set \\ \downarrow && \downarrow && \downarrow && \downarrow \\ P_1(X) &\stackrel{tra}{\to}& \mathbf{B}G &\stackrel{\rho}{\to}& Vect &\to& Set }$

and then apply the inner hom

$hom(P_1(\Sigma), --) : Cat \to Cat$

to all of it.

$\array{ hom(P_1(\Sigma),tra^* V//G) &\to& hom(P_1(\Sigma),V//G) &\to& hom(P_1(\Sigma),T_{pt}Vect) &\to& hom(P_1(\Sigma),T_{pt}Set) \\ \downarrow && \downarrow && \downarrow && \downarrow \\ hom(P_1(\Sigma),P_1(X)) &\stackrel{hom(P_1(\Sigma),tra)}{\to}& hom(P_1(\Sigma),\mathbf{B}G) &\stackrel{hom(P_1(\Sigma),\rho)}{\to}& hom(P_1(\Sigma),Vect) &\to& hom(P_1(\Sigma),Set) }$

That is transgresson to the mapping space $Maps(\Sigma,X)$.

Posted by: Urs Schreiber on June 20, 2008 10:45 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Thanks again!

I see that $S$ is more general than I had originally thought and could easily accommodate what I had in mind, i.e. a copy of some set (not $S$) over each point in $X$, if we wanted to (which we probably don’t). Ok! Progress!

If I understand what you are saying, since everything was formulated category theoretically, then we can pass to “path space” by simply “hom”ing everything in sight with an elementary path.

I feel like I just finished a half marathon and that was just the warm up! :)

Posted by: Eric on June 20, 2008 4:33 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

If I understand what you are saying, since everything was formulated category theoretically, then we can pass to “path space” by simply “hom”ing everything in sight with an elementary path.

That’s right! The interested reader is supposed to look at Transgression of $n$-Transport and $n$-Connections

Posted by: Urs Schreiber on June 20, 2008 5:37 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

I feel like I just finished a half marathon and that was just the warm up! :)

We are almost there. I think if you go back over what we did, it is actually remarkably elementary. I mean, part of the problem here is the general problem of math: an idea may be very simple and elegant in the space of ideas, but once it is pushed through the bottelneck of human communication it incarnates as a weird bucnh of symbols from which each reader has to distill again, internally, the original platonic idea in its pure form. Just think through what we did and I think you’ll find that it is just a couple of trivialities.

Posted by: Urs Schreiber on June 20, 2008 5:41 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

On to the next step:

Step 8) Pullback of sets over groupoids

Recall our finite groupoid $tra^* V//G$ of which I asked you to remember only the objects for the moment (so it’s really just a finite set for the time being): its objects are pairs $(x,v)$ with $x$ a point in “target space” and $v \in V$ a vector in our representation space.

Over that groupoid we have sitting some finite set $S$ which defines for us a section of a trivial vector bundle over target space $X$

$\array{ S \\ &\searrow \\ && tra^* V//G }$

(for other readers: here) is a description for how to think of these sets over the groupoid).

But now we also have the “in”-projection from our bigger groupoid with the funny name $hom(\Pi_1(\Sigma,tra^* V//G))$:

$\array{ S &&&& hom(\Pi_1(\Sigma), tra^* V//G) \\ &\searrow && {}^{in}\swarrow \\ && tra^* V//G }$

The objects of that bigger groupoid are pairs consisting of path $x \stackrel{\gamma}{\to} y$ in $X$ and the parallel transport $v \stackrel{\rho(tra(\gamma))}{\to} v ' = \rho(tra(\gamma))(v)$ of an element $v \in V$ over that path.

The projection map $in$ just remember the start point $x$ and that vector $v$ at that point and sends that down to $(x,v) \in tra^* V//G$.

Now, there is a natural operation to get from this setup with the set $S$ sitting over $tra^* V//G$ another set, to be called $in^* S$ – the pullback along $in$ – over $hom(\Pi_1(\Sigma),tra^* V//G)$.

$\array{ &&&& in^* S \\ &&&& \downarrow \\ S &&&& hom(\Pi_1(\Sigma), tra^* V//G) \\ &\searrow && {}^{in}\swarrow \\ && tra^* V//G }$

In words: the pullback set $in^* S$ has one copy of each element $a$ of $S$ sitting over each element that maps by $in$ to the same element in $tra^* V//G$ that $a$ maps to.

In other words: if there is an element $a \in S$ sitting over the pair $(x,v)$, then in $in^* S$ there is one copy of $a$ (meaning: one abstract element) sitting over each pair $(x \stackrel{\gamma}{\to}y, v \stackrel{tra(\gamma)}{\to} rho(tra(\gamma))(v))$ with that same $x$ and $v$ at the left end.

In yet other words: consider the special case that $S = \{\bullet\}$ contains a single element which maps down to the pair $(x,v)$. Then $in^* S$ contains one element per object in $hom(\Pi_1(\Sigma),tra^* V//G)$ whose left end is $(x,v)$.

Let me know if that makes sense. Then we can continue with pushing forward along $out$.

Posted by: Urs Schreiber on June 23, 2008 9:45 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Hi Urs,

I’ve drawn several pages of arrow diagrams at this point. It definitely helps!

Any advice on how to render $\mathrm{B}G$? Here is the best I could do

$\begin{matrix} \bullet & {}\\ \circlearrowleft & g\in G \end{matrix}$

I’d prefer to have a nice big clockwise loop above the node. All my diagrams are pretty much of this form with arrows going between similar diagrams.

If that is the best we can do for now, it is fine, but thought you might have an idea.

Posted by: Eric on June 25, 2008 6:38 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

I’ve drawn several pages of arrow diagrams at this point. It definitely helps!

Sounds good!

Any advice on how to render $\mathbf{B}G$?

As you have seen, I never bother with restricting to drawing a single copy of the single object of $\mathbf{B}G$. Not only that one runs into the immediate typesetting problem you have now, but it gets way worse once one starts doing something. There is just in general no good point in trying to draw in possibly big $n$-categorical pasting diagrams all elements which are supposed to be equal on top of each other. That quickly leads to absurd typesetting – and reading! – problems much worse than what we have so far.

So, for me the picture of $\mathbf{B}G$ is simply

$\mathbf{B}G = \{ \bullet \stackrel{g}{\to} \bullet | g \in G\} \,.$

I mean, if you have an ordinary formula such as $f(\ln(x)) = \exp( a x)$ you do not try to write the two $x$s on top of each other, either.

That’s my opinion. I am having this kind of discussion a lot with people who wonder why I draw two dots and speak of a single object. But so far nobody could convince me not to do that. :-)

Posted by: Urs Schreiber on June 25, 2008 7:27 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Sounds good to me. For some reason, looking at a bunch of loops seems to help me understand. Maybe a compromise is

$\text{B}G = \{\bullet\righttoleftarrow g | g\in G\}$

or

$\bullet\righttoleftarrow g\in G$

or simply

$\bullet\righttoleftarrow g$

? As long as we understand each other, it is fine :)

By the way, that uses \righttoleftarrow in case you’re curious. I tend to get hung up a bit on notation.

Here is a library of loops:

$\text{B}G = \{\bullet\righttoleftarrow g | g\in G\}$

$\bullet\righttoleftarrow g$

$\rho: \text{B}G \to \text{Vect}$

$\begin{matrix} \bullet\righttoleftarrow g\\ \downarrow \rho \\ V\bullet\righttoleftarrow \rho(g) \end{matrix}$

$V//G$

$V\bullet\righttoleftarrow (v,g)$

$P_1(X)$

$X\bullet\righttoleftarrow \gamma$

$\text{tra}: P_1(X)\to \text{B}G$

$\begin{matrix} X\bullet\righttoleftarrow \gamma \\ \downarrow \text{tra}\\ \bullet\righttoleftarrow \text{tra}(\gamma) \end{matrix}$

$\rho_*\text{tra}: P_1(X)\to\text{Vect}$

$\begin{matrix} X\bullet\righttoleftarrow \gamma \\ \downarrow \text{tra} \\ \bullet\righttoleftarrow \text{tra}(\gamma) \\ \downarrow\rho \\ V\bullet\righttoleftarrow \rho\circ\text{tra}(\gamma) \end{matrix}$

$\text{tra}^* V//G$

$X\times V\bullet\righttoleftarrow (x,v,\text{tra}(\gamma))$

$s: S\to \text{tra}^* V//G$

$\begin{matrix} S\bullet\righttoleftarrow f \\ \downarrow s \\ X\times V\bullet\righttoleftarrow (x_a,v_a,\text{tra}(\gamma_a)) \end{matrix}$

$P_1(\Sigma)$

$\{a,b\}\bullet\righttoleftarrow (a\to b)$

$\text{Funct}(P_1(\Sigma),P_1(X))$

$\begin{matrix} \{a,b\}\bullet\righttoleftarrow (a\to b) \\ \downarrow \gamma \\ X\bullet\righttoleftarrow \gamma(a\to b) \end{matrix}$

or

$\begin{matrix} \gamma(a\to b)\\ \downarrow \\ \gamma'(a\to b) \end{matrix}$

(I’m not too sure about that last one.)

$\text{Funct}(P_1(\Sigma),\text{B}G)$

$\begin{matrix} \{a,b\}\bullet\righttoleftarrow (a\to b) \\ \downarrow \gamma \\ X\bullet\righttoleftarrow \gamma(a\to b) \\ \downarrow\text{tra} \\ \bullet\righttoleftarrow \text{tra} \gamma(a\to b) \end{matrix}$

or

$\begin{matrix} \text{tra}\gamma(a\to b)\\ \downarrow \\ \text{tra}\gamma'(a\to b) \end{matrix}$

$\text{Funct}(P_1(\Sigma),\text{tra})$

$\begin{matrix} \gamma(a\to b) & \stackrel{\text{tra}}{\to} & \text{tra}\gamma(a\to b) \\ \downarrow & {} & \downarrow \\ \gamma'(a\to b) & \stackrel{\text{tra}}{\to} & \text{tra}\gamma'(a\to b) \end{matrix}$

(I’m not sure about that at all.)

$\text{Funct}(P_1(\Sigma)),\text{tra}^* V//G)$

$\begin{matrix} \{a,b\}\bullet\righttoleftarrow (a\to b) \\ \downarrow \\ X\times V\bullet\righttoleftarrow (x_a,v_a,\text{tra}\gamma(a\to b)) \end{matrix}$

(I’m not sure about that.)

Ok! So that should give you some idea of the progress I’m making, i.e. not much! I guess you could say I’m still lingering on Step 7 and Step 7 b. I suppose I wasn’t ready to move on to Step 8 yet, which means I will probably not make much progress on Step 9, but I’m working on it! :)

Posted by: Eric on June 25, 2008 9:35 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

It’s good that you give that list, because it shows that we need to clarify some things before proceeding.

I start at the top of your list:

where you display $\rho : \mathbf{B}G \to Vect$ it is mildly important to realize that the arrow in the middle of

$\rho : (\bullet \stackrel{g}{\to} \bullet) \mapsto (V \stackrel{\rho(g)}{\to} V)$

is a \mapsto arrow, indicating how a map of sets maps one element to another, and not a \to-arrow which indicates a morphism between two objects.

Then, when you display $V//G$ it seems that there is maybe a misunderstanding – maybe by me of your notation:

the groupoid denoted $V//G$ has many objects, each vector $v \in V$ is one of its objects. Then it has a morphism labeled $g$ between any vector $v$ and its image under the action of $g \in G$ on $v$ under the action $\rho(g) : V \to V$. So $V//G$ is

$V//G = \{ v \stackrel{g}{\to} \rho(g)(v) | v \in V, g \in G \} \,.$

A similar comment applies to $P_1(X)$. I’d write that as

$P_1(X) = \{ x \stackrel{\gamma}{\to} y | x,y points in X, \gamma path in X \} \,.$

You write:

I guess you could say I’m still lingering on Step 7 and Step 7 b

I see, I wasn’t aware of that. Let me know what your questions are, then I can try to clarify. I suppose there is a bit of – in principle trivial but potentially distracting – terminology to get straight here, in our communication process.

Posted by: Urs Schreiber on June 26, 2008 12:22 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

where you display $\rho:\text{B}G\to\text{Vect}$ it is mildly important to realize that the arrow in the middle of

$\rho:(\bullet\stackrel{\rho}{\to}\bullet)\mapsto (V\stackrel{\rho(g)}{\to}V)$

is a \mapsto arrow, indicating how a map of sets maps one element to another, and not a \to-arrow which indicates a morphism between two objects.

I’m probably confused, but when I wrote

$\bullet\righttoleftarrow g$

I meant to indicate a category with one object and one morphism for each $g\in G$, i.e. $\bullet\righttoleftarrow g$ was intended to represent all of $\text{B}G$. At one point, I wrote

$\bullet\righttoleftarrow g\in G$

but thought the notation was a bit heavy. Similarly, when I wrote

$V\bullet\righttoleftarrow\rho(g)$

I meant a category with one object $V$, a vector space, and one morphism $\rho(g)$ for each $g\in G$.

Then when I wrote

$\begin{matrix} \bullet\righttoleftarrow g \\ \downarrow \rho \\ V\bullet\righttoleftarrow \rho(g) \end{matrix}$

I meant a functor that maps the one object $\bullet$ in $\text{B}G$ to the one object $V$ in $V\subset\text{Vect}$ and maps each morphism $g$ in $\text{B}G$ to the morphism $\rho(g)$ in $V\subset\text{Vect}$.

I know we’re going WAY back to Step 1 and Step 2, but instead of squinting my eyes and trudging along, I’m trying to understand in terms of arrow diagrams. Not sure if that is misguided.

Do my diagrams, and hence the way I’m thinking about this, make sense? At lead for $\rho$?

Am I committing notation fowls? :)

PS: I might have been mistaken, but I was thinking of $V//G$ as a one-object category with many morphisms, i.e. $V\bullet\righttoleftarrow (v,g)\in V\times G$ as described on John Armstrong’s blog. Did I misunderstand? When you say $V//G$ has many objects, I suspect I did.

Same for $P_1(X)$. I was thinking of it as a one-object category whose object is the set $X$ and whose morphisms are paths in $X$.

That is why all my diagrams looked like loops because everything seemed to consist of one-object categories with functors between them.

Posted by: Eric on June 26, 2008 1:44 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

when I wrote

$\bullet\righttoleftarrow g$

I meant to indicate a category with one object and one morphism for each $g\in G$,

Yes, no problem, I didn’t object to that.

At one point, I wrote

$\bullet\righttoleftarrow g\in G$

but thought the notation was a bit heavy.

Okay, this is a minor point, I don’t want to be pedantic here, even though we might rather want to be starting being conveniently sloppy with notation only after everything is clear.

Then when I wrote

$\begin{matrix} \bullet\righttoleftarrow g \\ \downarrow \rho \\ V\bullet\righttoleftarrow \rho(g) \end{matrix}$

I meant a functor

I am not really happy with that but I won’t be pedandic here. The more important point is the next one:

I might have been mistaken, but I was thinking of $V//G$ as a one-object category

That’s not the case. Unless $V$ happens to be the 0-vector space, the groupoid $V//G$ has infinitely many objects. Each vector $v \in V$ is one of its objects.

It also has many morphisms: from each object there emanates one morphism per element $g \in G$.

as described on John Armstrong’s blog

No, there is a misunderstanding here. John Armstrong gives the same definition I am using here. Notice that he just writes $G$ for what I denote $\mathbf{B}G$. This has a single object. But $V//G$ (which he calls $S//G$) has many.

everything seemed to consist of one-object categories with functors between them

No, please have another look at my descriptions to see that most every groupoid in the game here has more than one object. In fact, the very special fact that $\mathbf{B} G$ has a single object and hence comes from a group is emphasized by the boldface-B notation. That highlights that $\mathbf{B}G$ is a very special groupoid, one which happens to have just a single object. Pretty much every other groupoid that we are talking about has many objects.

Same for $P_1(X)$. I was thinking of it as a one-object category whose object is the set $X$

That’s not right. Rather, $X$ is the set of objects of $P_1(X)$: every element $x \in X$ is one obkject in $P_1(X)$.

I know we’re going WAY back to Step 1 and Step 2, but instead of squinting my eyes and trudging along, I’m trying to understand in terms of arrow diagrams.

That’s good! Please keep going this way. We need to sort this out before proceeding.

Posted by: Urs Schreiber on June 26, 2008 7:27 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

I might have been mistaken, but I was thinking of $V//G$ as a one-object category

That’s not the case. Unless $V$ happens to be the 0-vector space, the groupoid $V//G$ has infinitely many objects. Each vector $v\in V$ is one of its objects.

Uh oh! Now I need to unlearn something. Unlearning is harder than learning! I managed to convince myself through pictures that $V//G$ had one object.

Is this open to interpretation? Is it absolutely unarguable that $V//G$ has one object for each element of $V$?

Let me try to make a case for $V//G$ having one object and the flaw in my argument should be easy to point out.

Warning: I know everything I am about to say is probably wrong and I’m not trying to be difficult or clever. This is really the way I NEED to learn stuff. I don’t know any other way. In other words, I need to propose alternative ways of thinking about things (because I generally don’t understand the standard ways), figure out why I’m wrong, and THEN I will understand what’s going on.

I won’t blame you if you give up on me :)

If I understand, $\text{Vect}$ is a category whose objects are vector spaces $\{V,W,\dots\}$ and whose morphisms $\{F:V\to W,\dots\}$ are maps between them.

When there is only one vector space in sight, we can promote it to a category by simply considering it to be a one-object category whose object is $V$ and whose morphisms are endomorphisms of $V$. This, I would denote by

$V\bullet\righttoleftarrow\text{End}(V)$

Note: My notation is undergoing an evolution. Instead of labeling the morphism by an element, I am labeling it by the set. Hence, I would now draw $\text{B}G$ as

$\bullet\righttoleftarrow G,$

which is a little clearer I think. This is compatible with the picture

$\begin{matrix} \bullet\righttoleftarrow G \\ \downarrow \rho \\ V\bullet\righttoleftarrow \text{End}(V). \end{matrix}$

(Note again the change in notation from before.)

Now, consider two vector spaces $V$ and $W$ and its corresponding two-object category

$\lefttorightarrow\stackrel{V}{\bullet}\leftrightarrows\stackrel{W}{\bullet}\righttoleftarrow ,$

which I will denote by $VW$.

Next, consider a two-object category $\text{B}GG$ which is like $\text{B}G$ except it has two objects and can be depicted as

$G\lefttorightarrow\bullet\stackrel{G}\leftrightarrows\bullet\righttoleftarrow G,$

where there is one copy of $G$ for each collection of morphisms between any two objects (nodes).

Next define a functor

$\rho:\text{B}GG\to VW.$

This is intended to look like a representation.

FINALLY, I want to claim that $VW//G$ is to be understood as the image of $\rho$, i.e.

$VW//G := \rho(VW).$

This is a two-object category where each morphism is labeled by an element of $G$ and is invertible.

Similarly, $V//G$ is to be understood as the image of the functor

\rho:\text{B}G\to V,$

i.e.

$

$V//G := \rho(V),$$

where there is only one object$

V$and each morphism is labeled by an element of$G$and is invertible, where

$

$g:V\to V$$

is defined by

$

$g: v\mapsto \rho(g)v.$$

What you described as$

V//G$would correspond to what I would call

$

$Explode(V)//G,$$

where$

Explode(V)$is a category with one object for each element of$V$with corresponding exploded endomorphisms, i.e. for each$v\in V$there is one morphism$v\to F(v)$for each$F\in\text{End}(V)$.

PS: For some reason I am not fond of the idea of "exploding" nice categories, i.e. taking a nice category and creating a new category whose objects correspond to elements of the previous objects. I think my one-object$

V//G$might somehow be equivalent to your infinite object$Explode(V)//G$(what you called$V//G$). Posted by: Eric on June 26, 2008 8:04 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Is this open to interpretation? No. Is it absolutely unarguable that $V//G$ has one object for each element of $V$? Yes. If you want to denote by $V//G$ the action groupoid of $G$ acting on $G$. If you want the notation to mean something else I can’t stop you from doing so, but then we’ll have a problem here with steps 1)to oo) When there is only one vector space in sight, we can promote it to a category by simply considering it to be a one-object category whose object is $V$ and whose morphisms are endomorphisms of $V$. This, I would denote by I would denote this by $\mathbf{B} End(V)$ by the way. I want to claim that $VW//G$ is to be understood as the image of $\rho$ If you claim this based on the generally accepted definition then it is wrong, unfortunately. You can of course declare the notation to be such. Posted by: Urs Schreiber on June 26, 2008 8:49 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Morally, what is $V//G$ supposed to represent? I thought it was supposed to represent a vector space $V$ and an invertible $G$-action on $V$ (not sure if that is the right term). If that is morally correct, I can think of two ways to realize this. One way is apparently the standard way as outlined in John Armstrong’s blog. I could very well be (and am likely) wrong, but I think there is another way to realize this moral concept. The reason why I am looking for an alternative is that I am not too happy about “exploding” $V$ and considering elements of $V$ to be objects of some category. I think I can illustrate the concept with a simple example. Consider a set $S = \{a,b\}$ with two maps $E:S\to S$ and $T:S\to S$ defined by $E: a\mapsto a, b\mapsto b$ $T: a\mapsto b, b\mapsto a$ which can be depicted $S\bullet\righttoleftarrow \{E,T\}.$ Now consider a two element group $G = \{e,t\}$, where $e$ is the identity and $t t = e$. This can be depicted as $\bullet\righttoleftarrow \{e,t\}.$ We have the obvious representation $\begin{matrix} \bullet\righttoleftarrow \{e,t\} \\ \downarrow \rho \\ S\bullet\righttoleftarrow \{\rho(e),\rho(t)\} \end{matrix}$ where $\rho(e) = E$ and $\rho(t) = T$. I think that contains everything needed to define $V//G$. To be sure, let’s now consider the exploded (i.e. “standard”) version. The exploded version consists of a category with two objects $\lefttorightarrow\stackrel{a}{\bullet}\leftrightarrows\stackrel{b}{\bullet}\righttoleftarrow$ There are four morphisms, two for each object, $(a,e): a\to a$ $(a,t): a\to b$ $(b,e): b\to b$ $(b,t): b\to a$ This exploded version is clearly equivalent (at least morally) to the unexploded $S\bullet\righttoleftarrow G$ The advantage is that the unexploded version consists of a category with only one object while the exploded version consists of a category with $|S|$ objects and $|S|*|G|$ morphisms, where $|\cdot|$ denotes cardinality. I think this line of thought carries over to $V$ where there are infinitely many objects in the exploded version. Posted by: Eric on June 26, 2008 10:30 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Morally, what is V//G supposed to represent? Ah. Good question if that was unclear. $V//G$ is the weak quotient of $V$ by $G$. You probably know the ordinary quotient $V/G$ (just a single slash). This is the set of equivalence classes of elements in $V$, where two elements $v, v' \in V$ are considered equivalent if and only if there is a $g \in G$ such that $v' = \rho(g)(v)$. So $V/G$ is obtained by considering elements “on the same $G$-orbit” as equal. On the other hand, $V//G$ is obtained from the same idea, but considering elements on the same $G$-orbit just as isomorphic. We still rememeber that they are different, but make them “essentially the same” by throwing in a isomorphism for each group element relating them. Therefore, if we pass to the set $[V//G]$ of isomorphism classes of objects in $V//G$, i.e. the set of all connected sub-groupoids of $V//G$, then this is the ordinary quotient: $[V//G] = V/G \,.$ In other words, the quotient $V/G$ of the vector space by the group is the decategorification of the action groupoid $V//G$. This is what $V//G$ is morally speaking. Posted by: Urs Schreiber on June 26, 2008 11:37 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral I suspect that either I am totally wrong or that I am only partially wrong (in both cases I’m wrong!) and threw you off with some incorrect statements. Instead of claiming $V//G$ to be an image of $\rho$, which is probably not spoken correctly, I’ll try to explicitly construct it with the same example as my previous post. Again, let $S = \{a,b\}$ and $G = \{e,t\}$. In the unexploded version, we have just one object $S$ and two morphisms $(S,e): a\mapsto\rho(e)a (=a), b\mapsto\rho(e)b (=b)$ $(S,t): a\mapsto\rho(t)a (=b), b\mapsto\rho(t)b (=a).$ In the exploded version, we have two objects $a$ and $b$ and four morphisms $(a,e): a\mapsto a$ $(a,t): a\mapsto b$ $(b,e): b\mapsto b$ $(b,t): b\mapsto a$ In this way, I HOPE it is clear that I am not trying to propose something completely different, but rather am trying to interpret the moral meaning of $V//G$ with an alternative (but equivalent) realization. Posted by: Eric on June 26, 2008 10:52 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral an alternative (but equivalent) realization. No, unfortunately they are not equivalent. What you are describing is indeed $im(\rho) \subset \mathbf{B}Aut(V)$. If the representation is faithful, then this is equivalent (as a category) to $\mathbf{B}G$. But not to $V//G$. Instead, $\mathbf{B}G$ plays the role of the classifying space of the universal $G$-bundle and $V//G$ plays the role of the total space of the vector bundle associated by $\rho$ to the universal $G$-bundle. The bundle projection corresponds to the faithful and surjective functor $V//G \to \mathbf{B}G$ which forgets the elements $v \in V$ and just remembers the group elements that act. $V$ itself by the way, regarded as a category with $V$ as its space of objects and only identity morphisms is the fiber of this map: $V \hookrightarrow V//G \to \mathbf{B}G \,.$ This is how all these things fit together. What you are describing is essentially the thing on the far right. But $V//G$ is the thing in the middle. Posted by: Urs Schreiber on June 26, 2008 11:52 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Ok. Thanks for your patience. I will need to chew on this a bit more, but think I am making some progress. It would probably help if I had a better handle on bundles, but my differential geometry education became hazy at around that point. I know (or I have seen) the standard definitions and have seen them discussed for years, but haven’t worked them directly enough to have an intuitive understanding. These last posts of yours seem particularly helpful so I’ll work on understanding them. Thanks! Posted by: Eric on June 27, 2008 1:11 AM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral I should say that at this point I didn’t mean to imply that getting into these statements about bundles in detail is necessary or even helpful. I mentioned it mainly to indicate that the different categories (groupoids all of them) $\mathbf{B}G$, $V//G$ (and also $V$, regarded as a category with just identity morphisms) are all different and non-equivalent in general but fit together nicely into one nice story. You should of course feel free to let this discussion here make you explore all kinds of directions, but do we want to continue with going through the steps 1) - 11)? Let me know up to which step you feel comfortable and from where on we need more discussion. Posted by: Urs Schreiber on June 27, 2008 11:05 AM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Hi Urs, I thought I had made some real progress when I drew all my diagrams, i.e. the library of loops. Now I know that almost all of my diagrams were incorrect, so I feel like I’m starting from scratch again. I think I want to try to complete the library of diagrams before proceeding. Each step should, I think, admit a nice picture. I will work on it (time permitting) and try again. At least my first two pictures were “ok”. Now I hope to provide a diagram for $V//G$. Thanks Posted by: Eric on June 27, 2008 2:56 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral provide a diagram for $V//G$ I wish I had a better idea of what it is you are looking for and find missing from the discussion so far. The good way to think of $V//G$ pictorially is, quite literally, as the orbits of the $G$-action on $V$, i.e. the decomposition of $V$ int all those elements that are connected by the $G$-action. Another way to say this is that the connected components of the groupoid $V//G$ (all the parts that hang together through morphisms) are the orbits. Which is just another way of saying that the decategorification of the weak quotient $V//G$ is the ordinary quotient $V/G$. Posted by: Urs Schreiber on June 27, 2008 4:19 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral With this post, you might have managed to penetrate my thick skull :) Let me try to rephrase what you have been trying to tell me to make sure I’ve got it. $V//G$ can be thought of as a bi-directed graph (i.e. for each directed edge, there is an opposite directed edge) representing a bunch of closed paths. The nodes correspond to elements of $V$. There are $N$ generators $g_i\in G$ for $N$ orbits. For each node $v\in V$ there is one edge pair $v\leftrightarrows\rho(g_i)v$ for each generator $g_i\in G$. Question: There are $N$ closed paths passing through each node $v\in V$. One for each generator $g_i\in G$. Should we distinguish the nodes for each orbit, i.e. should we make $N$ copies of $v$ so that we have disjoint closed paths? Or should we allow the closed paths to intersect? Thanks! Posted by: Eric on June 27, 2008 5:29 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral $V//G$ can be thought of as a bi-directed graph In the sense that every category has an underlying oriented graph (simply the category with the composition operation forgotten) and that for a groupoid in this graph all oriented edges happen to occur in pairs with opposite orientation. Yes. representing a bunch of closed paths. I am not sure what you mean by “closed paths” here. Maybe the answer to that one is “No”. :-) The nodes correspond to elements of $V$. Yes, the vertices of the graph underlying a category are the objects of that category. There are $N$ generators $g_i \in G$ for $N$ orbits. No, this does not sound right. An “orbit” of a group action is an element $v$ of the space $V$ that the group acts on together with all other elements that can be reached from $v$ by acting with all the available group elements. $Orbit(v) = \{ \rho(g)(v) | \forall g \in G \} \,.$ So orbits are labelled by elements of $V$, not by elements of $G$. This label is highly non-unique. Any two objects $v_1, v_2$ which are related by the action of a group element $\exists g \in G : v_2 = \rho(g)(v_1)$ sit in and hence label, in this sense, the same orbit. In $V//G$ we remember which group elements relate which two points on one orbit by connecting them with an arrow which is labeled by that group element. Posted by: Urs Schreiber on June 27, 2008 6:03 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Oops! Sorry. I confused “orbit” with “cycle”, i.e. a cyclic subgroup, because I was thinking of an orbit in the planetary since. In other words, I was thinking of carrying $v$ around the “orbit” by repeatedly multiplying it by the generator $g_i\in G$. $Cycle(g_i,v) = \{\rho(g_i)^j v|\forall j\in Z\}.$ Then I was thinking that (finite) $G$ could be decomposed into $N$ cyclic subgroups with generators $g_i\in G$. In which case, I THINK we would have $Orbit(v) = \Union_{g_i} Cycle(g_i,v)$ I guess what I was trying to do (without realizing it) was to find a directed graph that generates $V//G$. Posted by: Eric on June 27, 2008 7:20 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral $V$ itself by the way, regarded as a category with V as its space of objects and only identity morphisms is the fiber of this map: $V\embedsin V//G\to BG.$ Does that means that $\pi^{-1}(\bullet) = V$, the objects of $V$, and for each morphism $g\in BG$, $\pi^{-1}(g)$ is an identity morphism for each $v\in V$? Posted by: Eric on June 27, 2008 11:21 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Hi Urs, I think I am confused because given a group $G$, a representation $\rho$, and a vector space $V$ and you told me to consider a category whose objects are the elements of $V$, there are two such categories that might come to mind. One is a category whose objects are elements of $V$ and whose morphisms are all identity morphisms. Another category is a category whose objects are elements of $V$ and whose morphisms correspond to the action of $G$ (via $\rho$) on that object, i.e. $v\stackrel{g}{\to}v',$ where $v' = \rho(g) v$. This latter category is the only category I can conceive of that should correspond to $V//G$. Is that right? Is $V//G$ the second category described above? If not, I am hopelessly lost… Posted by: Eric on June 27, 2008 11:50 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral One is a category whose objects are elements of $V$ and whose morphisms are all identity morphisms. This is the category I am just calling $V$. If you want to be more explicit you can call it $Disc(V)$, the “discrete” category over the set $V$. Another category is a category whose objects are elements of $V$ and whose morphisms correspond to the action of $G$ (via $\rho$) on that object Yup, as I said, this is $V//G$. Exercise: spell out what the sequence of functors $V \to V//G \stackrel{p}{\to} \mathbf{B}G$ does. Determine $p^{-1}(\bullet)$. Posted by: Urs Schreiber on June 27, 2008 11:59 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Yup, as I said, this is $V//G$. HALLELUJAH!! PRAISE CARTAN!! :D Ok. This means that I am simply failing to explain myself clearly, which is not surprising. Exercise: spell out what the sequence of functors $V\to V//G\stackrel{p}{\to}BG$ does. Ok. The first functor $i: V\to V//G$ takes each object $v\in V$ and maps it to the same $v\in V//G$ since the objects of the two categories coincide. It maps each (identity) morphism in $V$ to the identity morphism $v\stackrel{e}{\to}v$ in $V//G$, where $e$ is the identity element in $G$. This functor is faithful and essentially surjective (trying out new wikipedia terms, which is always dangerous!). The second functor $p:V//G\to BG$ maps every object $v$ in $V//G$ to the single object $\bullet$ in $BG$. It maps every morphism $v\stackrel{g}{\to}v'$ in $V//G$ to the morphism $g$ in $BG$. This functor is fully faithful but not essentially surjective. Determine $p^{-1}(\bullet)$. The inverse image is the colection of objects of $V//G$, i.e. the elements of $V$. Posted by: Eric on June 28, 2008 12:30 AM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Great, now we are making progress. I am glad we could sort that out. What you say is correct. Just one minor quibble with your newly acquired Wikipedia knowledge: the functor $p : V//G \to \mathbf{B}G$ is essentially surjective (even plain surjective) and is faithful. But it is not full! (In general, except in the entirely degenerate case of a trivial group acting on a single element.) For it to be full, it had to be true that for any two elements $v,v' \in V$ there is a morphism $v \stackrel{g}{\to} v'$ for all $g \in G$. But that can’t be, as you can quickly see. There is this theorem that two categories are equivalent precisely if there exists an essentially surjective and fully faithful functor between them. So we see, once again, that $V//G$ is not equivalent to $\mathbf{B}G$. Okay, so how should we proceed? Maybe you want to check if you can spell out the pullback of the sequence of functors you just looked at along $tra$, i.e. $\cdots \to tra^*(V//G) \to P_1(X) \,.$ Posted by: Urs Schreiber on June 28, 2008 1:06 AM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Are we ready for Step 9, by the way? I am afraid I have to go and watch soccer now with the guys here at the Hausdorff institute, but maybe you can meanwhile guess how step 8 should work: Given sets $A$ and $B$ and a map $\array{ A \\ & \searrow \\ && B }$ and finally given a finite set $S$ sitting over $A$ $\array{ S \\ \downarrow \\ A \\ & \searrow \\ && B }$ we want to naturally “push $S$ down to $B$” by doing a “fiber integration”. Can you see how to do that? If so, you can write the description of step 9 for me while I take a little break… Posted by: Urs Schreiber on June 25, 2008 7:34 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Okay, it seems then as if we are ready to contiunue. Step 7 b) Spelling out the transgression of $tra^* V//G$ We need to get a good idea for what the groupoid with the fancy name $hom(P_1(\Sigma), tra^* V//G)$ is like. Actually, as already mentioned before, I shall restrict attention for the time being to just the spaces of objects of our various groupoids. The morphism will come in in step 11+, as indicated below. An object of $hom(P_1(\Sigma), tra^* V//G)$ is a pair, consisting of a path $x \stackrel{\gamma}{\to} y$ and a vector $v \in V$ which we, however, think of as giving as a full morphism $v \stackrel{\rho(tra(\gamma))}{\to} v'$ in $V//G$ which describes the parallel transport of $v$ along $\gamma$. So we think of the objects of $hom(P_1(\Sigma), tra^* V//G)$ as pairs consisting of a path $x \stackrel{\gamma}{\to} y$ in $X$ and the parallel transport $v \stackrel{\rho(tra(\gamma))}{\to} v'$ of some vector $V$ along this path. The important point is that there are two obvious maps from this collection of objects down to the objects of the groupoid $tra^* V//G$ which, recall, were pairs consisting of a point of $x$ and a vector in $V$. These two maps $\array{ && hom(P_1(\Sigma), tra^* V//G) \\ & {}^in\swarrow && \searrow^{out} \\ tra^* V//G &&&& tra^* V//G }$ simply project out the beginning and the end, respectively, of the path and the parallel transport $in : ((x,v) \stackrel{(\gamma,\rho(tra(\gamma)))}{\to} (y,v') ) \mapsto (x,v)$ $in : ((x,v) \stackrel{(\gamma,\rho(tra(\gamma)))}{\to} (y,v') ) \mapsto (y,v') \,.$ Okay? Step 8) Pull back a $V$-colored set along the map “$in$ […] Step 9) Push-forward a $V$-colored set along the map “$out$ […] Step 10) There it is – the path integral […] Step 11 +) The canonical path integral measure from the Leinster measure/ groupoid cardinality […] […] Posted by: Urs Schreiber on June 20, 2008 5:35 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Step 7b) Spelling out the transgression of $\text{tra}^* V//G$. Okay? I probably couldn’t recite it on the spot, but I think I can follow each of the steps. So far so good! I’m a little hazy on the details for Step 8 through Step 11+ though :) I can see where its going though thanks to having a look at Professor Baez’ slides on Spans in Quantum Theory. Googling for the slides I found two other articles that look interesting too for the uninitiated (like me!) Jeffrey Morton, Spans and Vector Spaces - pt 1 John Armstrong, Spans and Cospans Great stuff! Posted by: Eric on June 20, 2008 7:28 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral I’m a little hazy on the details for Step 8 through Step 11+ though :) I hope it is clear that these are still to be described by me. I was only trying to indicate what part of the road still lies ahead. But steps 8 and 9 are simple enough to guess how they might work. Maybe you want to try. More later. Posted by: Urs Schreiber on June 22, 2008 6:05 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Urs said: I hope it is clear that these are still to be described by me. I was only trying to indicate what part of the road still lies ahead. But steps 8 and 9 are simple enough to guess how they might work. Maybe you want to try. Sure definitely. I had slipped into Californese, which is a dialect of English, except it contains high doses of sarcasm. If I can (time and brain cells willing) I will try to fill in the rest of the blanks. If anyone else cares to contribute, please feel free! I think this is a really cute problem and might make a nice tutorial that introduces many concepts with a concrete problem. I hope this is the way QM is taught in 10 years (or less)! By the way, is this what you call “arrow theory”? It seems you simply try to explain every single simple concept in terms of arrow diagrams. The phynance world calls for now though… Posted by: Eric on June 23, 2008 2:53 AM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Urs said: Step 6) Sections of the vector bundle. [snip] Then again consider a finite set (1)$s:S\to\text{tra}^* V//G$ sitting over our groupoid. This is hence the same as nothing but a map of sets (2)$s:S\to X\times V.$ But this is nothing but a bundle of $V$-colored sets over $X$! It is one $V$-colored set over each point of $X$. I wonder if this should be Then again consider a finite set (3)$s:X\times S\to\text{tra}^* V//G$ sitting over our groupoid. This is hence the same as nothing but a map of sets (4)$s:X\times S\to X\times V,$ where (5)$s(x,a) = (x,s(a)).$ But this is nothing but a bundle of $V$-colored sets over $X$! It is one $V$-colored set over each point of $X$. Otherwise we run into the issue of disjoint subsets of $S$ that Tim pointed out. Posted by: Eric on June 19, 2008 11:40 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Err… you know what I mean :) Given a $V$-colored set (1)$s:S\to V$ I think we want to consider a $V$-colored section (2)$1_X\otimes s:X\times S\to X\times V.$ The cardinality operator applied pointwise gives rise to a section of the vector bundle. Something like that! :) Posted by: Eric on June 19, 2008 11:56 PM | Permalink | Reply to this Read the post Eli Hawkins on Geometric Quantization, I Weblog: The n-Category Café Excerpt: Some basics and some aspects of geometric quantization. With an emphasis on the geometric quantization of duals of Lie algebras and duals of Lie algebroids. Tracked: June 20, 2008 5:14 PM Read the post Category Algebras Weblog: The n-Category Café Excerpt: On an interpretation of the category algebra of a category in terms of pull-push of finite sets through correspondence spaces. Tracked: June 26, 2008 9:59 PM Exploding a Category Given a category $C$ with objects $ob(C)$ and morphisms $hom(C)$, it is often the case that each object $oC\in ob(C)$ itself contains many elements $\{c\in oC\}$. There is probably a standard term for what I’m going to describe, but I am going to call it “exploding C”. Exploding a category $C$ results in a new category $Explode(C)$. The objects are given by $ob[Explode(C)] = \Union_{oC\in ob(C)} oC.$ In other words, the objects of $Explode(C)$ consist of all the elements of all the objects of $C$. The morphisms of $Explode(C)$ are the obvious ones inherited from the morphisms of $C$, i.e. given a morphism $oC\stackrel{mC}{\to}oC'$ in $C$, we have the obvious morphism $(c\in oC)\stackrel{mExplodeC}{\to}(c'\in oC')$ in $Explode(C)$. I hope I’m not introducing a notation clash with something else, but now given a vector space $V$, a group $G$, and a representation $\rho$, I want to define a category $V\nearrow G.$ There is one object $V$ in $V\nearrow G$ and one morphism $g:V\to V$ for each element in $G$. This has a nice pictorial depiction via $V\bullet\righttoleftarrow G.$ I am only talking about this here, because I claim that the action groupoid $V//G$ is given by $V//G = Explode(V\nearrow G).$ This is what I meant in my “Library of Loops”. I was confusing $V\nearrow G$ with $V//G$. Since $V//G$ can be easily constructed from $V\nearrow G$ via exploding, I’m still a little hazy as to why we need to consider $V//G$ instead of simply working with $V\nearrow G$. Posted by: Eric on June 28, 2008 1:26 AM | Permalink | Reply to this Re: Exploding a Category Here is another example of an exploded category… Consider a category $X\nearrow\Gamma$ with one object, a set $X$, and one morphism $\Gamma: X\to X$ for each “path map” in $X$. This category may be depicted as $X\bullet\righttoleftarrow \Gamma.$ Just like I confused $V\nearrow G$ with $V//G$ in the Library of Loops, where it should have been $V//G = Explode(V\nearrow G),$ I also confused $X\nearrow\Gamma$ for $P_1(X)$, where it should have been $P_1(X) = Explode(X\nearrow\Gamma).$ Posted by: Eric on June 28, 2008 6:01 AM | Permalink | Reply to this Re: Exploding a Category I wrote this last comment in a rush, which is not usually a great idea, just as grandma arrived for the weekend. One thing I noticed right at the last minute was a matho (like a typo). I had written “paths”, but what I really had was a something I called (in a rush) a “path map”. I’m still groggy because our daughter Sophia slept with us last night, which means she slept great, but I was kicked in the face all night :) Anyway, I wanted to talk about $\Gamma$ a little bit. If we begin with a finite (or at least countable) set $X$ and construct a directed graph $DX$ by assigning directed edges to pairs of points in $X$, then $\Gamma$ is constructed by promoting the directed edges to a collection of morphisms $T$. This collection represents the possible “one time step” moves available from any point in $X$. Next, we can generate the collection of “two time step” paths $T^2$ by gathering all composable paths $T^2 = T\circ T.$ Similarly, $T^n = \underset{n times}{T\circ\dots\circ T}.$ Then $\Gamma$ is the collection of all such maps, i.e. $\Gamma = \{T^i | i\in N\},$ which I called “path maps”, but are kind of like flows generated by discrete vector fields. The category $P_1(X)$ is then obtained by exploding the category (monoid?) of these discrete flows on $X$, i.e. $P_1(X) = Explode(X\nearrow\Gamma).$ Posted by: Eric on June 28, 2008 3:47 PM | Permalink | Reply to this Re: Exploding a Category Concerning “explosion”: yes this is the heart of the idea of the action groupoid (albeit in different terminology). Here is another way to think about the same thing, which will become useful later on: Let $Set$ be the category of finite sets (objects are finite sets, morphisms are maps between sets). And let $Set_*$ be the category of finite pointed sets: objects are finite sets with one element singled out and morphisms are maps between finite sets which map chosen elements to chosen elements. There is a canonical functor $\array{ Set_* \\ \downarrow \\ Set }$ which simply forgets the chosen element. Now take your category $V \nearrow G$ (which I might have denoted $im(\rho)$). It has a canonicaly embedding into $Set$ $V\nearrow G \hookrightarrow Set \,.$ So we can put the pieces together and consider the diagram $\array{ && Set_* \\ && \downarrow \\ V\nearrow G &\hookrightarrow& Set } \,.$ Given a diagram of this sort, we are tempted to wonder if we can put something in its top left corner such that we get a commuting square $\array{ (?) &\to& Set_* \\ \downarrow && \downarrow \\ V\nearrow G &\hookrightarrow& Set } \,.$ Certainly there are some boring choices to do so. So we want to lok for the maximally non-boring one. This we can take to be the choice $(?)$ such that any other possible choice $(?')$ has a map to $(?)$ $(?') \to (?)$ such that all possible triangles and squares in the game here commute. Such a $(?)$ is called the pullback of the above diagram. This pullback is your $Explode(V\nearrow G)$ $\array{ Explode(V\nearrow G) &\to& Set_* \\ \downarrow && \downarrow \\ V\nearrow G &\hookrightarrow& Set } \,.$ The functor to the right $Explode(V \nearrow G) \to Set_*$ sends the object $(v \in V)$ to the set $V$ with chosen element $v$ $v \mapsto (V,v) \,.$ Posted by: Urs Schreiber on June 29, 2008 9:44 AM | Permalink | Reply to this Re: Exploding a Category Great! Just one quick question and I can return to the Steps. Can you confirm that $V//G = Explode(V\nearrow G)$ and $P_1(X) = Explode(X\nearrow\Gamma)$ ? If those two are correct, then I think I understand. However, if $V//G = Explode(V\nearrow G)$ then it seems to me we can also write $V\nearrow G \sim Implode(V//G).$ Imploding a category is as obvious as exploding one. Each object in a category $C$ maps to an element of the single object in $Implode(C)$ in such a way that it remembers the morphisms so that morphisms in $C$ become strands of an endomorphism in $Implode(C)$. I think there may be many choices in how to implode a category, but I think they all should be related some how so that the different choices are equivalent in some way so that, given a category $M$ with one object, we have $M \sim Implode\circ Explode(M).$ Does that make any sense? Posted by: Eric on June 29, 2008 3:33 PM | Permalink | Reply to this Re: Exploding a Category Can you confirm that $V//G = Explode(V \nearrow G)$ and $P_1(X) = Explode(X \nearrow G)$ Concerning $V//G$: yes. Concerning $P_1(X)$: here it depends on what you mean. Originally in the discussion here I said something like “let $P_1(X)$ be your favorite finite category modelling points and paths in some space”. So I didn’t really specify what precisely it should be. But, yes, you are free to let $P_1(X)$ be ofthe form $X//\Gamma$, where $\Gamma$, in the case you like to consider, is a group acting on $X$ which you imagine to be generated from some collection of elements of which you think as “possible one-time-step movements” of the points in $X$. Concerning implosion: this works under some assumptions. You are using here the fact that you already know that $V//G$ comes from “exploding” a group action. Suppose I hand you an arbitrary groupoid $Gr$ and asked you to implode it. What would you do? I think there is a theorem that you can always do it if you have in addition the information of a faithful functor to a one-object groupoid $\mathbf{B}G$. If I recall correctly in the Geometric Representation Theory seminar John taught that every groupoid with such a faithful functor is equivalent to an action groupoid. But I’d need to check that more carefully than I have time for right this moment. Posted by: Urs Schreiber on June 29, 2008 4:00 PM | Permalink | Reply to this Re: Exploding a Category Great. Thanks again for your patience. It seemed to me that, for the things we were considering, that it shouldn’t matter whether we worked with $V\nearrow G$ or $V//G$. This is what I meant by “morally equivalent”. As long as I understand how they are related, I am ok. If you next tell me something like, “Remember how before it didn’t matter much whether we considered $V\nearrow G$ or $V//G$ since they were morally equivalent? For what we are going to do next, the differences are important so we really do need to consider $V//G$ on its own.” I’ll be ok with it. Progress! For now, family calls :) Posted by: Eric on June 29, 2008 5:31 PM | Permalink | Reply to this Re: Exploding a Category Concerning this notion of “morally equivalent” in the sense that one thing contains the information to get another thing: that may be true. In that sense most everything we have been saying here is “morally equivalent” to simply the ordinary concept of a representation of a group. We could drop the entire formalism here if we adopt the point of view that it is just the ordinary concept of a group action. But the point of the game called “arrow theory” we are playing here is that we want to extract the various aspects contained in the concept of a representation in a “concept invariant way”. While you are right that (as I understand your statement now) $V//G$ does not give you any information you didn’t already have with having the representation, it does allow you to construct certain things just using arrows – such as (eventually, if we get there) the path integral! :-) You should think of $V//G$ as a first step in an “information processing” operation which leads us from input data (group, rep, vector space, vector bundle, connection) to output data (path integral) in a way that uses just diagrams. The point of the diagrams being that they’ll allow us to vastly generalize once we understand everything in terms of just them. Does that sound sensible? Posted by: Urs Schreiber on June 29, 2008 5:48 PM | Permalink | Reply to this Re: Exploding a Category Does that sound sensible? That sounds more than just sensible. It sounds exciting! That is what I thought we were trying to do, but could only vaguely see the big picture. Now I’m starting to understand the individual pieces. Posted by: Eric on June 29, 2008 6:11 PM | Permalink | Reply to this Library of Loops Urs has managed to force some knowledge into my head, which is no small fete. Thanks! Here is a summary of what I’ve learned so far: $BG$ $\bullet\righttoleftarrow G$ $BEnd(V)$ $V\bullet\righttoleftarrow End(V)$ $\rho: BG\to BEnd(V)$ $\begin{matrix} \bullet\righttoleftarrow G \\ \downarrow\rho \\ V\bullet\righttoleftarrow End(V) \end{matrix}$ $V\nearrow G$ $V\bullet\righttoleftarrow G$ $X\nearrow\Gamma$ $X\bullet\righttoleftarrow\Gamma$ $flo: X\nearrow\Gamma\to BG$ $\begin{matrix} X\bullet\righttoleftarrow\Gamma \\ \downarrow flo \\ \bullet\righttoleftarrow G \\ \end{matrix}$ $\rho_* flo: X\nearrow\Gamma\to BEnd(V)$ $\begin{matrix} X\bullet\righttoleftarrow\Gamma \\ \downarrow flo \\ \bullet\righttoleftarrow G \\ \downarrow \rho \\ V\bullet\righttoleftarrow End(V) \end{matrix}$ That is about as far as I dare go for now. This is slightly different than what you’ve outline so far so I hope you don’t mind the modifications. I believe we can recover what you had by inserting “Explode” in several places, e.g. $V//G = Explode(V\nearrow G)$ $P_1(X) = Explode(X\nearrow\Gamma)$ $tra = Explode(flo).$ Posted by: Eric on June 28, 2008 4:51 PM | Permalink | Reply to this Waking the Baby Maybe I’m feeling a bit romantic at the moment, but I sense a very beautiful story unfolding here. On July 21, 2007, I exclaimed: I’m heartbroken! ;) [snip] I’m pretty sure this does have an arrow theoretic formulation (since it is built on arrows!), but you know I’m not well enough equipped to flow with the Dao (even though I feel it ;)). To which Urs tantalizingly replied: I’m heartbroken! ;) You shouldn’t. I may have a treat for you soon. Let’s not wake up the baby too often before it has grown up sufficiently. One problem with this is babies generally wake YOU up, not the other way around ;) The baby is jumping up and down and wants out of the crib! :) This current thread is one thing. This one is another, where you said: … on a causal lattice 2-path 2-category $P_2(X)$ which could just as well have said … the 2-category generated from a 2-diamond complex… Then, you recently pointed to Week 17 and Week 18 of Quantization and Cohomology. The scent from this trail led me to ask a question on June 12, 2008: I was just looking at the figure on page 47 and was wondering if anything is lost be constraining the morphisms to $\gamma:(t,x)\to(t+1,x\pm 1)$? I was trying to be discreet (har har), but I think you saw right through me :) Since I am blessed with an ability to “sense” things and can see beautiful pictures in the fog just out of my reach, but at the same time cursed with an inability to reach them on my own, let me tell you what I see in the fog and I’m sure you are just steps away from grabbing it. My dream for a discrete theory of physics is about to be realized in a satisfactory manner. When the story is written, I hope I will be mentioned in a footnote as being an energetic side kick :) The key is the concept (as clear to me in this fog I live in as it will ever be) that Groupoidification is discretization This will tie together everything I’ve felt in my gut but was unable to enunciate. While I’m being romantic and discussing what I see in this fog, let me reach further to where things are even less clear, but no less beautiful… The continuum limit is the classical limit. In your birthday present post, you said: Originally I thought that passing to this “field algebra” might correspond to passing to the continuum limit. But in private communication with somebody I was told that possibly this just corresponds to passing to an unbounded lattice. I could be way off base because I’m reaching far back into the fog, but I think that part of the difficulty here is that to truly pass to the continuum limit will require “forgetting” something. I don’t know what it is, but whatever gets forgotten will contain whatever was “quantum”. Something is broken with the continuum and to fix it, you will need to break the Dao. That, or give up on the continuum, which is my vote. See?! This is what happens when you begin teaching me the Dao and it gets mixed with romance (and a dose of lack of sleep for good measure) :) Posted by: Eric on June 30, 2008 6:33 AM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral So, where are we? Do you know what the groupoid $tra^* V//G$ is? Can you spell out the functor $tra^* V//G \to P_1(X)$ which is in the game? And the two functors $in : tra^* V//G \to V//G$ and $out : tra^* V//G \to V//G$ ? I think I gave all the information. Somewhere. Let me know what is clear and what is not. Posted by: Urs Schreiber on June 30, 2008 7:14 AM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Do you know what the groupoid $tra^*V//G$ is? I can repeat what you’ve said. The objects of $tra^* V//G$ are the elements of $X\times V$ the morphisms transport $(x,v)$ to $(x',v')$, where $x' = \gamma(x)$ and $v' = \rho tra(\gamma) v.$ I’m a little hung up on the notation though. I’m pretty sure $tra^* V//G$ can be obtained from exploding something, so to understand it better, it would help me to find an unexploded version. Stealing a look at Wikipedia and your diagrams for Step 6 here, it seems an alternative notation for $tra^* V//G$ might be $P_1(X) \;\times_{\text{B}G}\; V//G.$ If that is the case, it makes me want to denote the unexploded version as $X\nearrow\Gamma \;\times_{\text{B}G} \; V\nearrow G.$ Ok! Actually, this notation helps a lot and makes your next question trivial Can you spell out the functor $P_1(X) \;\times_{\text{B}G}\; V//G\to P_1(X)$ Yep. It is just projection $\pi_1$ :) On objects, it is $\pi_1(x,v) = x.$ On morphisms, it just grabs the underlying path $x\stackrel{\gamma}{\to}x'$, which I suppose can be made explicit by $\pi_1(\gamma,\rho tra(\gamma)) = \gamma.$ And the two functors $in:P_1(X) \;\times_{\text{B}G}\; V//G\to V//G$ and $out:P_1(X) \;\times_{\text{B}G}\; V//G\to V//G$ ? I THINK that $out$ is the second projection. On objects, it is $out(x,v) = v$ and on morphisms, it is $out(\gamma,\rho tra(\gamma)) = \rho tra(\gamma).$ I THINK that $in$ on objects is $in(x,v) = v$ and sends each morphisms to the identity morphism $in(\gamma,\rho tra(\gamma)) = \rho(1).$ Based on what you’ve said, the above for $in$ and $out$ may not be correct because I almost expected $in(x,v) = v$ and $out(x,v) = \rho tra(\gamma) v,$ but I couldn’t think of what the morphisms would be in this case. I hope this makes up for my previous stream of consciousness :) Posted by: Eric on June 30, 2008 9:12 AM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral I’m pretty sure $tra *(V//G)$ can be obtained from exploding something, Indeed. That’s what the notation says: it is a pullback of the action groupoid. Though you will have to generalize your notion of explosion now, since $P_1(X)$ no longer has a single object, as $\mathbf{B}G$ did. it seems an alternative notation for $tra^* V//G$ might be $P_1(X) \times_{\mathbf{B}G} V//G$ That’s right. You also got the functor $tra^* V//G \to P_1(X)$ right, okay, so that is clear. Good. Concerning the rest, you made the mistake of not noticing that I made a mistake. The in and out maps are not defined on $tra^* V//G$ but on $hom(P_1(\Sigma),tra^* V//G)$ ! Sorry for that. They go $in : hom(P_1(\Sigma),tra^* V//G) \to tra^* V//G$ and $out : hom(P_1(\Sigma),tra^* V//G) \to tra^* V//G \,.$ In order for us to continue, you need a clear idea of how this works. Let me know if you have. Posted by: Urs Schreiber on June 30, 2008 9:51 AM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral I am not quite sure why, but I think the objects of $hom(P_1(\Sigma),P_1(X)\;\times_{\text{B}G}\;V//G)$ are the morphisms of $P_1(X)\;\times_{\text{B}G}\;V//G,$ which are of the form $(\gamma,\rho tra(\gamma)): (x,v) \to (\gamma(x),\rho tra(\gamma) v).$ Then, on objects $in(\gamma,\rho tra(\gamma)) = (x,v)$ and $out(\gamma,\rho tra(\gamma)) = (\gamma(x),\rho tra(\gamma) v).$ Posted by: Eric on June 30, 2008 10:36 AM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral I am not quite sure why, but I think the objects of $hom(P_1(\Sigma),P_1(X) \times_{\mathbf{B}G}V//G)$ are the morphisms of $P_1(X) \times_{\mathbf{B}G} V//G$ Yes, iff $P_1(\Sigma)$ is the interval, i.e. iff $P_1(\Sigma) := \{a \to b\}$ is the category with two objects and precisely one nontrivial morphism which goes from one object to the other (or maybe also including the inverse of that morphism, which makes no difference in the following as long as $P_1(X)$ is taken to be groupoid). The reason is that $hom(P_1(\Sigma),P_1(X) \times_{\mathbf{B}G}V//G)$ is the functor category also denoted $Funct(P_1(\Sigma),P_1(X) \times_{\mathbf{B}G}) V//G$ whose - objects are functors $F : P_1(\Sigma) \to P_1(X) \times_{\mathbf{B}G} V//G$ - morphisms are natural transformations $\eta : F \Rightarrow F'$ between these. A functor from the interval category $P_1(\Sigma) = \{a \to b\}$ into any other category $C$ is exactly a choice of morphism in $C$. A morphism in $Funct(P_1(\Sigma),P_1(X) \times_{\mathbf{B}G}) V//G$ between two such functors is a chopice of morphisms in $C$ connecting the two endpoints of the two morphisms in $C$ corresponding to these two functors, such that the obvious square comutes. Do you see which square that is? If so, it should be clear hopw the same game works for more general $P_1(\Sigma)$. But lets concentrate on the interval for now. Then, on objects $in(\gamma, \rho(tra(\gamma))) = (x,\nu)$ and $out(\gama,\rho tra(\gamma)) = (\gamma(x),\rho tra(\gamma)\nu)$ Yes. Assuming that when writing $\gamma(x)$ you mean the target object of $\gamma$. It’s not really correct to write that $\gamma(x)$, though I can see why you did that. In full beauty we have: $in : \left( \array{ (x,v) &\stackrel{(\gamma,tra(\gamma))}{\to}& (y,\rho(tra(\gamma))(v)) \\ \downarrow^{(\gamma_a,g_a)} && \downarrow^{(\gamma_b,g_b)} \\ (x',v') &\stackrel{(\gamma',tra(\gamma'))}{\to}& (y',\rho(tra(\gamma))(v')) } \right) \mapsto \left( \array{ (x,v) \\ \downarrow^{(\gamma_a,g_a)} \\ (x',v') } \right)$ and $out : \left( \array{ (x,v) &\stackrel{(\gamma,tra(\gamma))}{\to}& (y,\rho(tra(\gamma))(v)) \\ \downarrow^{(\gamma_a,g_a)} && \downarrow^{(\gamma_b,g_b)} \\ (x',v') &\stackrel{(\gamma',tra(\gamma'))}{\to}& (y',\rho(tra(\gamma))(v')) } \right) \mapsto \left( \array{ (y,\rho(tra(\gamma))(v)) \\ \downarrow^{(\gamma_b,g_b)} \\ (y',\rho(tra(\gamma))(v')) } \right) ,.$ As soon as we are happy with this state of affairs, the next step would be to understand is how to pullback a set over $P_1(X) \times_{\mathbf{B}G} V//G$ back along $in$. Did I say how that works? Posted by: Urs Schreiber on June 30, 2008 11:53 AM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral A morphism in $Funct(P_1(\Sigma),P_1(X)\;\times_{\text{B}G}\;V//G )$ between two such functors is a chopice of morphisms in $C$ connecting the two endpoints of the two morphisms in $C$ corresponding to these two functors, such that the obvious square comutes. Do you see which square that is? Whenever someone says “obvious”, I get nervous because nothing is obvious to me :) Your “full beauty” diagram provided a hint though. Would a morphism be something like: $(\gamma_a \stackrel{\sigma}{\to} \gamma_b, g_b g_a^{-1}): (\gamma_a, g_a) \to (\gamma_b, g_b)$ ? Oh wait! As I was about to submit this, a light bulb flashed. This should be transport around a loop, i.e. the boundary of surface connecting the paths. Let me modify the notation a little bit: $\gamma_{xy} = x\stackrel{\gamma}{\to}y$ $g_{xy} = \rho tra(\gamma_{xy})$ Then a morphism would be something like $\sigma:(\gamma_{xy},g_{xy})\to(\gamma_{x'y'},g_{x'y'}),$ $x' = \sigma(x),$ $y' = \sigma(y),$ and $g_{x'y'} = g_{yy'} g_{xy} g^{-1}_{xx'}.$ In other words, we need to consider transport along paths $x\to\x'$ and $y\to y'$. As soon as we are happy with this state of affairs, the next step would be to understand is how to pullback a set over $P_1(X)\;\times_{\text{B}G}\; V//G$ back along in. Did I say how that works? You did describe something about it, but I think notation was maybe deceptively simple. It is burdensome, but I think it might be more clear to be explicit $\begin{matrix} & & S\;\times_{(P_1(X)\;\times_{\text{B}G}\;V//G)}\; hom(P_1(X),P_1(X)\;\times_{\text{B}G}\;V//G) & & \\ & \pi_1 \swarrow & & \searrow \pi_2 \\ S & & & & hom(P_1(\Sigma),P_1(X)\;\times_{\text{B}G}\;V//G) \\ & s\searrow & & \swarrow in & \\ & & P_1(X)\;\times_{\text{B}G}\;V//G & & \end{matrix}$ If that is correct, I think I am beginning to see what’s going on. By the way, this pullback trick is kind of neat. Posted by: Eric on June 30, 2008 7:26 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral By the way, I think I understand your shorthand pullback notation now. Going back to the pullback of the action groupoid, we have $\begin{matrix} & & P_1(X)\;\times_{\text{B}G}\;V//G & & \\ & \pi_1 \swarrow & & \searrow \pi_2 & \\ P_1(X) & & & & V//G \\ & tra \searrow & & \swarrow \pi & \\ & & \text{B}G & & \end{matrix}$ To form a pullback, we need two categories and two functors mapping from these two categories to a common third category. The explicit notation keeps track of all three categories, but it is sufficient to specify one of the original two categories and the OTHER functor. If that is the case, then I think for each pullback there are two possible shorthand notations. In this example, we would have $P_1(X)\;\times_{\text{B}G}\;V//G = tra^* V//G = \pi^* P_1(X).$ Is that right? If so, then I am ok with your shorthand notation now. Posted by: Eric on June 30, 2008 7:45 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Your “full beauty” diagram provided a hint though. More than that: it provided the answer! The squares on the left of that diagram are the morphisms from the top horizontal edge to the lower horizontal one. In case it is not clear yet, this is the time that we have to get natural transformations sorted out. Have a look at the beginning of that Wikipedia entry. The first square you see there is the kind of square that we are talking about. This should be transport around a loop, Not totally unrelated yes. Maybe rather flat transport around a loop, at least as long we are talking 1-categories. Concerning the pullback of the set $S$ If that is correct, I think I am beginning to see what’s going on. As a general abstract diagram this is correct, yes. I mean, it’s the right notation. You need to feel comfortable with understanding what in detail that set with the long name on the top is, for instance in cases where the original set had just a single element. So suppose the original set $S$ has a single element which is mapped to, say, the object $(x,v)$. Then, in words: what is the pulled back set like? Over which morphisms does it have elements sitting? When that is clear, the case where $S$ has more than one element is an easy consequence. Posted by: Urs Schreiber on June 30, 2008 10:41 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral I’ll give this a try… The objects in $in^* S$ are of the form $(a,\gamma,\rho tra(\gamma)).$ If we follow this object around the left side of the diagram, we get $(a,\gamma,\rho tra(\gamma))\mapsto (x_a,v_a).$ If we follow this object around the right side of the diagram, we get $(a,\gamma,\rho tra(\gamma))\mapsto \text{"Source of } (\gamma,\rho tra(\gamma)) \text{"}.$ Since the diagram commutes, we have $(a,\gamma,\rho tra(\gamma)) = \text{"Morphism in}\; tra^* V//G \;\text{whose source is} \; (x_a,v_a) \text{"}.$ Therefore, for each $a\in S$, there is one object $(a,\gamma,\rho tra(\gamma))$ in $in^* S$ for each morphism in $tra^* V//G$ that begins at $(x_a,v_a)$. Posted by: Eric on June 30, 2008 11:57 PM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Therefore, for each $a \in S$, there is one [element] in $in^* S$ for each morphism in $tra^*V//G$ that begins at $(x_a, v_a)$. Yup. All right, that was Step 8. Now comes: Step 9) Push-forward along $out$. I am really too tired now, will have to get back to this tomorrow. But I suppose you already know how it works anyway. Think “fiber-integration”: the motto is “collect everything sitting over the same object downstairs”. Posted by: Urs Schreiber on July 1, 2008 12:06 AM | Permalink | Reply to this Re: An Exercise in Groupoidification: The Path Integral Ok! Let me try this. I should tell you that this is NOT easy for me. My brain hurts :) We want to consider this diagram $\begin{matrix} \stackrel{in^* S}{\bullet} & \stackrel{\pi_2}{\rightarrow} & \stackrel{hom(P_1(\Sigma),tra^* V//G)}{\bullet} & \stackrel{out}{\rightarrow} & \stackrel{tra^* V//G}{\bullet} \\ & & in \downarrow & & \downarrow i_2 \\ & & \stackrel{tra^* V//G}{\bullet} & \stackrel{i_1}{\rightarrow} & \stackrel{out_* (tra^* V//G)}{\bullet} \end{matrix}$ Again, if we start with an element $(a,\gamma,\rho tra(\gamma))\in in^* S$ and trace it down and to the right through the diagram we get $\text{"Morphism with source}\; (x_a,v_a)\text{"}.$ Tracing first to the right and then down we get $\text{"Morphism with target}\; (x_b,v_b)\text{"}.$ Since the diagram commutes, then elements in $out_*(tra^* V//G)$$are morphisms that share the same source and target. If I understand, then elements in$ in^* S$are kind of like freely wriggling strands all tied down at the same start point, but whose ends are all over the place. Pushing forward through$out$ ties down the end points as well so we end up with a bunch of strands connecting the same two points.

Posted by: Eric on July 1, 2008 1:09 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Hm, this does not quite seem right. Here is the intended prescription:

Let $\array{ S \\ \downarrow \\ C }$ be your finite set $S$ sitting over some groupoid $C$. Let’s assume for the moment (I mentioned that before) that $C$ is just a finite set itself, i.e. let’s forget its morphisms for a moment.

Then consider a functor (so really a map of sets) $C \stackrel{out}{\to} D$ i.e. the situation $\array{ S \\ \downarrow \\ C &\stackrel{out}{\to}& D }$

then we want to find a set to be denoted $\array{ \int_{out} S \\ \downarrow \\ D } \,.$

And the prescription is this, in words: over each element $d \in D$ the set $\int_{out} S$ is the disjoint union of elements of $S$ over all $c \in C$ which are mapped by $out$ to $d$.

I am in a hurry, but I hope that is understandable. If so, you should try to say what, using that prescription, the set $\int_{out} in^* S$ in our context is like (forgetting all the morphisms in our groupoids for the moment).

Posted by: Urs Schreiber on July 1, 2008 7:15 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Thanks. I guess I don’t know what you mean by “push forward”. I guessed that you meant “pushout”. When you have two functors with a common source category and distinct target categories, you can (according to Wikipedia and John Armstrong’s blog :)) push this out to a new category. I may have even gotten that wrong, but I was trying to pushout

$hom(P_1(\Sigma),tra^* V//G)$

along $in$ and $out$, which seemed like a sensible thing to do.

Now I’ll think about

$\int_{out} in^* S.$

Posted by: Eric on July 1, 2008 3:05 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

I guessed that you meant “pushout”.

Ah. No, it’s not a pushout. I haven’t really told you about the full abstract way to think about it, I am afraid:

when we come to the full truth, our sets over a groupoid $\array{ S \\ \downarrow \\ C }$ will be replaced by the fiber-assigning transport functors $\array{ Set \\ \uparrow \\ C }$ which send a given object to the subset of $S$ sitting above it.

From that point of view, the pullback that we considered in Step 8 is just ordinary pullback of functors, obtained by precomposition. And the push-forward that we are talking about now will be Kan extension, which will be governed by the Leinster measure.

Maybe at this point it might be helpful to look at section 1.4 “Quantization and categorical coends” of On $\Sigma$-models and differential cohomology. You might encounter some by-now-old friend there if you think of the set-valued functors appearing there in the above way.

Posted by: Urs Schreiber on July 1, 2008 3:17 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Ok. I’ll try again…

For every element $a\in S$ there is an element in $in^* S$ of the form

$(a,\gamma,\rho tra(\gamma))$

which is essentially a morphism in $tra^* V//G$ whose source is $(x_a,v_a)$.

Consider the diagram

$\begin{matrix} \stackrel{in^* S}\bullet & & \stackrel{\int_{out} in^* S}{\bullet} \\ \pi_2 \downarrow & & \downarrow \\ \stackrel{hom(P_1(\Sigma),tra^* V//G)}{\bullet} & \stackrel{out}{\to} & tra^* V//G \end{matrix}$

And the prescription is this, in words: over each element $d\in D$ the set $\int_{out} S$ is the disjoint union of elements of $S$ over all $c\in C$ which are mapped by out to $d$.

We can adapt it to our diagram by saying:

For each element $(x,v)\in tra^* V//G$, the set

$\int_{out} in^* S$

is the disjoint union of elements of $in^* S$ over all $(x',v')\in hom(P_1(\Sigma),tra^* V//G)$ which are mapped by $out$ to $(x,v)$.

Ok. I’ve said it, but it is not yet clear to me how this arises from staring at the diagram.

I’ll think about it…

Posted by: Eric on July 2, 2008 3:10 AM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Hi Urs,

You might have noticed that while I’ve been trying to understand groupoidification, the global economy has gone on the skids. That is no coincidence. I better get back to work and put things back on track. It has been fun! ;)

Posted by: Eric on July 3, 2008 3:06 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

I better get back to work and put things back on track.

Sure. Sorry for not getting back to you earlier. Drop a message whenever you have some time to waste on the vain games of non-financial math.

Posted by: Urs Schreiber on July 3, 2008 6:10 PM | Permalink | Reply to this

Re: An Exercise in Groupoidification: The Path Integral

Hey! You have nothing to apologize for! I know you are extremely busy and I truly appreciate you taking time to walk me through this stuff, which is probably totally trivial for others. I learned a lot!

When the smoke settles a bit, I hope we can come back to this. I really do want to understand it and I think we are close to the finish line.

Posted by: Eric on July 3, 2008 6:54 PM | Permalink | Reply to this
Read the post Block on L-oo Module Categories
Weblog: The n-Category Café
Excerpt: On Jonathan Block's concept of modules over differential graded algebras.
Tracked: June 30, 2008 11:45 PM
Read the post Groupoidification from sigma-Models?
Weblog: The n-Category Café
Excerpt: On constructing sigma-models from differential nonabelian cocycles and how this gives rise to methods of groupoidification.
Tracked: December 30, 2008 4:24 PM

Re: An Exercise in Groupoidification: The Path Integral

I have started turning this into an $n$Lab entry:

Posted by: Urs Schreiber on August 18, 2009 8:17 PM | Permalink | Reply to this

Parallel Transport over Path Space

I’m posting this here since I can’t find any reference to it on this blog. I mean the paper by Chatterjee, Lahiri and Sengupta posted last June. The list of references certainly includes the usual suspects. Did my use of the blog’s search function fail to find a discussion?

Posted by: jim stasheff on March 11, 2010 10:42 PM | Permalink | Reply to this

Re: Parallel Transport over Path Space

Did my use of the blog’s search function fail to find a discussion?

I think we did not have such a discussion here.

Posted by: Urs Schreiber on March 11, 2010 10:57 PM | Permalink | Reply to this

Re: Parallel Transport over Path Space

Is it too late? Even if it has been superseded by stuff here, it would be good to have that pointed out.

Posted by: jim stasheff on March 12, 2010 1:19 PM | Permalink | Reply to this

Re: Parallel Transport over Path Space

I suppose the paper in question is this one:

BTW, I stumbled upon two books that seem to provide some mathematical treatment of (some aspects of?) path integrals, namely

• Albeverio, Sergio A.; Hoegh-Krohn, Raphael J.; Mazzucchi, Sonia: “Mathematical theory of Feynman path integrals. An introduction.” (There is a second recently published version, see ZMATH)

and

• Cartier, Pierre; DeWitt-Morette, Cecile: “Functional integration: action and symmetries.” (see ZMATH)

I would love to take a look at those, and at this thread and the associated nLab entries of course, but if I do a push on my need-to-read stack right now I will get a stack overflow exception thrown into my face. So I am more interested in a “tell me what all that is about and how are the different aspects connected, while I drink a cup of coffee” kind of summary :-)

P.S.: Thanks to the friendly elves who fixed the comment bug!

Posted by: Tim van Beek on March 16, 2010 11:02 AM | Permalink | Reply to this

Re: Parallel Transport over Path Space

BTW, I stumbled upon two books that seem to provide some mathematical treatment of (some aspects of?) path integrals,

Thanks. I moved your references to

Posted by: Urs Schreiber on March 16, 2010 2:24 PM | Permalink | Reply to this

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