## February 12, 2007

### QFT of Charged n-Particle: Dynamics

#### Posted by Urs Schreiber

Given an $n$-particle (a point, a string, a membrane, etc.) coupled to a $n$-bundle with connection (an electromagnetic field, a Kalb-Ramond field, a 2-gerbe, etc.), what is the corresponding quantum theory?

The answer to this question has a kinematical and a dynamical aspect and up to now I had concentrated on the kinematics:

Here I start talking about dynamics. The right abstract point of view on the dynamics of quantum systems, as far as path integrals are concerned, is certainly that adopted for instance in

E. Lupercio, B. Uribe
Topological Quantum Field Theories, Strings, and Orbifolds
hep-th/0605255.

A parameter space $\mathrm{par}$, describing the shape of the quantum object (the point, the circle, the sphere, etc.) propagates along “worldvolumes” (graphs, surfaces, 3-manifolds, etc.), which are spaces whose boundaries look like $\mathrm{par}$, by pull-push along correspondences of the form $\array{ & \mathrm{worldvol} \\ \multiscripts{^{\mathrm{in}}}{\nearrow}{} && \nwarrow^{\mathrm{out}} \\ \mathrm{par} && \mathrm{par} } \,.$

Here I formulate this in an arrow theoretic way (a little more arrow-theoretic than the discussion in the above text, that is) that fits into the context of kinematics that I discussed before.

While everything is categorical, a crucial point is that, as opposed to the kinematics, the dynamics requires the push-forward of a set at one point. In the absence of a notion of adjoint morphisms of sets, this requires extra structure on our sets: a measure. This is the infamous measure that appears in the path integral.

The following definition is taken from

Recall # that an $n$-particle is something that looks like the ($n-1$)-category $\mathrm{par}$ (the point $\mathrm{par} = \{\bullet\}$, or the interval $\mathrm{par} = \{a \to b\}$, or the sphere, $\mathrm{par} = \Pi_2(S^2)$, etc.) and comes equipped with a choice of maps $(\gamma : \mathrm{par} \to \mathrm{tar}) \in \mathrm{conf} \subset [\mathrm{par},\mathrm{tar}]$ into some $n$-category called “target space” (ordinarily, $\mathrm{tar} = P_n(X)$ are $n$-paths in some “spacetime” $X$).

Each such map is a configuration of the $n$-particle: one of many ways for it to sit in spacetime.

If our $n$-particle, however, does not just want to sit around, we get the same picture just slightly lifted in dimension: we look at $n$-categories whose boundaries look like $\mathrm{par}$, and think of them as inducing interpolations between different configurations of the $n$-particle.

Deinition. A worldvolume (or diagram) for our $n$-particle is an $n$-category $\mathrm{worldvol}$ together with a collection of embeddings $\mathrm{in}_i : \mathrm{par} \to \mathrm{worldvol}$ and $\mathrm{out}_j : \mathrm{par} \to \mathrm{worldvol} \,.$

Given a worldvolume, a space of histories or space of trajectories or space of paths is a choice of subcatgeory $\mathrm{hist} \subset [\mathrm{worldvol},\mathrm{tar}]$ which is compatible with the above choice of configuration space in that $\mathrm{in}_i^* \mathrm{hist} \simeq \mathrm{conf}$ and $\mathrm{out}_j^* \mathrm{hist} \simeq \mathrm{conf} \,.$

For instance, for $n=1$ and $\mathrm{par} = \{\bullet\}$ the single point particle, the simplest worldvolume is the worldline, which is nothing but the interval $\{a \to b\}$ with the only two possible injections $\array{ & \{a \to b\} \\ \multiscripts{^{\mathrm{in}}}{\nearrow}{} && \nwarrow^{\mathrm{out}} \\ \bullet && \bullet } \,.$ In a simple standard example, we would take target space to be $\mathrm{tar} = \mathrm{Moore}(X)$, the category of Moore paths (paths with a parameter length) in $X$ and would take the category $\mathrm{hist}$ of histories to be the discrete category on the set of paths of parameter length $t \in \mathbb{R}$, say.

Not before long, we will need to equip the space of objects of this category with a measure. The canonical choice here would be the Wiener measure.

The interval $\{a \to b\}$ in the above example is special, in that it is a cylinder over parameter space.

Definition. A worldvolume $(\mathrm{worldvol},\mathrm{in},\mathrm{out})$ is a cylinder over parameter space if there is a unique transformation $\array{ & \nearrow \searrow^{\mathrm{in}} \\ \mathrm{par} &\Downarrow& \mathrm{worldvol} \\ & \searrow \nearrow^{\mathrm{out}} } \,.$

Whenever we have a worldvolume which is a cylinder (not only then, but let us concentrate on this for the time being), the unique transformation it comes with induces for us a transformation filling the diagram $\array{ & & \mathrm{hist}\times \mathrm{worldvol} \\ & \multiscripts{^{\mathrm{out}^*}}{\swarrow}{} && \searrow^{\mathrm{in}^*} \\ \mathrm{conf}\times\mathrm{par} & &\;\;\Downarrow^{\mathrm{cyl}}& & \mathrm{conf}\times\mathrm{par} \\ & \multiscripts{^{\mathrm{ev}}}{\searrow}{} && \swarrow^{\mathrm{ev}} \\ && \mathrm{tar} } \,.$

This is noteworthy, because this diagram can be regarded as a correspondence through which we may pull-push states of the $n$-particle.

Recall that a state of the particle charged under the $n$-bundle with parallel transport $\mathrm{tra} : \mathrm{tar} \to \mathrm{phas}$ is a transformation of the form $\array{ & \nearrow \searrow^{1} \\ \mathrm{conf}\times\mathrm{par} &\;\Downarrow^e& \mathrm{worldvol} \\ & \searrow \nearrow^{\mathrm{ev}^*\mathrm{tra}} } \,.$

Given the above $\mathrm{cyl}$-diagram, we may

- first pull such a state back along $\mathrm{in}^*$, “from the incoming $n$-particle to the space of paths”

- then tranport it by composition with $\mathrm{cyl}$ to the other end of the cylinder

- and finally push it forward along $\mathrm{out}^*$ from the space of paths to the outgoing $n$-particle.

The diagrams below illustrate this process, and its appearance in pasting diagrams, in full detail.

Remarkably, while everything in the theory of the charged $n$-particle is completely canonical and based only on abstract arrow theory, here is one point where it seems as if the Dao does not yield an answer, and we need to intervene by hand:

in order to reproduce standard quantum theory, we need to interpret the above push-foward as a push-forward of measure spaces.

If we do this, we add to our definition of space of paths, $\mathrm{hist}$, a measure on this space. Then the required push-forward is integration over the fiber with respect to this measure.

In our above example, with this measure chosen as the ordinary Wiener measure, the map $e \mapsto e'$ which we get from the worldline of length $t$ is precisely the path integral propagator over time $t$ in standard quantum mechanics.

Posted at February 12, 2007 5:00 PM UTC

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### Re: QFT of Charged n-Particle: Dynamics

I just wrote a long reply to this post talking in a pretty general way about categories of spans, functors giving configuration spaces and state spaces, and the pull-push procedure for turning spans of configuration spaces into operators on state spaces, and then mistakenly hit the “cancel” button rather than “preview”. Ugh: never edit anything in a web browser. The bottom line can be summed up thus:

You have a span of configuration spaces - one for a cobordism, and one each for its input and output boundaries. Then you have a “state space” which is something like a space of functions on these configuration spaces (or, if the configuration spaces are categories, a category of presheaves). You want to take a function from input to output. You do this by pulling back the function (or presheaf) to the middle of the span (your space of histories) - then pushing forward onto the output. The result is a path integral.

There are a bunch of variations depending on two things: (1) what category $C$ the configuration spaces live in, and (2) what kind of algebraic structure $A$ the functions take values in. In particular, if configuration spaces are literally spaces, and functions are literally complex-valued function, then you find you need a measure to do the pushforward - which is then integration. But there are other cases where no extra structure is needed and the Tao remains un-intervened with - basically, if $A$ has infinitary “sums”. For example, as John has discussed in the “quantization and cohomology” seminar recently, if $A$ is one of these idempotent algebras using “min” instead of summation, you are taking an infemum rather than an integral - so no measure is needed. In a different vein, if we categorify things, and we’re taking presheaves valued in a category $A$ with infinite colimits, then again everything works internal to $A$.

Some questions that occur to me, then: what combinations of $A$ and $C$ work automatically? What sorts of $A$ need what kinds of extra structure on $C$ to make the push-forward part work? How “canonical” are the choices? Etc.

Posted by: Jeff Morton on February 13, 2007 10:42 AM | Permalink | Reply to this

### Re: QFT of Charged n-Particle: Dynamics

Thanks a lot for the comment!

the pull-push procedure for turning spans of configuration spaces into operators on state spaces

I know how hard it is to lose a long comment one has typed into the computer. It feels awful.

But if you ever find the energy to write up your ideas and observations on this issue again, I’d sure be interested in looking at it.

I should maybe point out that what I tried to accomplish in the above entry is to see how this general pull-push business fits into the framework where I allow everything to remember its $n$-categorirical structure, and where I take (nontrivial) $n$-bundles into account by way of the transport $n$-functors.

I find that “states” and “sections” are most naturally thought of as transformations between $n$-functors. And so one has to (at least I had to) think about what it means to pull-push such transformations.

In particular, since for a nontrivial $n$-bundle, we don’t just pull a section up the correspondence and then multiply it with a number, but instead we have to properly “parallel transport” it over the corresponding cobordism. This is naturally accomplished by the pull-push in terms of those pasting diagrams which I drew, and this is supposed to be my little contribution to this general idea of path-integration by pull-push.

Incidentally, I was just made aware of slides by Dan Freed where this pull-push quantization is also discussed.

then you find you need a measure to do the pushforward

Yes. This put me off for quite a while.

As you know, also due to Freed is the idea that, somehow, doing the path integral is the same kind of operation as constructing the space of states.

But maybe it is not really the same, after all, but just analogous.

Forming the space of states is an honest push-forward to a point, in the sense that it is the adjoint functor to a canonical pull-back functor.

Doing the path integral, on the other hand, is a push-forward of measures.

I was hoping for a while that, with enough thinking, I would find a point of view that unifies both these procudures. I did come up with one observation which might be half-way what we need. Or maybe not.

In case anyone is interested, I will describe that in a separate post.

if $A$ is one of these idempotent algebras using “min” instead of summation, you are taking an infemum rather than an integral - so no measure is needed.

All right, but my concern would be: measure or not, it seems we need to define by hand (without some abstract nonsense doing it for us) what we want to understand under the push-forward part of the pull-push through the correspondence.

Maybe that’s just the way it is. Or maybe we are missing something.

Posted by: urs on February 13, 2007 11:15 AM | Permalink | Reply to this

### Lebesgue integral as collection of 0-sections?

I wrote:

Maybe that’s just the way it is. Or maybe we are missing something.

The following is potentially silly. But I like to share it anyway. Maybe some expert out there can recognize a fraction of a good idea and can help me out with some comments.

So this is the issue, from my perspective:

Let $X$ be some set and let $\mathrm{tra} : \mathrm{Disc}(X) \to \mathrm{Vect}_\mathbb{C}$ be an assignment of a vector space to each point of the set (to be thought of as assigning the fibers of a vector bundle to points).

Let $1 : \mathrm{Disc}(X) \to \mathrm{Vect}$ be the tensor unit in the space of all such assignments, i.e. that which sends each point to $\mathbb{C}$ itself.

Then the space of sections of $\mathrm{tra}$ is $\mathrm{Hom}(1,\mathrm{tra}) \,.$

Since the functor $1$ may be regarded as the pullback of a line bundle over a point to $X$, we may regard this space of sections equivalently as the push-forward of $\mathrm{tra}$ to a point. (I have talked about that here, but I guess it is obvious to anyone who ever thought about it.)

Now decategorify once. Instead of vector spaces, let us assign numbers to points, i.e. consider 0-functors $\mathrm{tra} : X \to \mathbb{C} \,.$

($\mathbb{C}$ is the 0-category of complex 0-vector spaces).

I believe that it is not really important that we are dealing with complex number here. So let me just replace this by

$\mathrm{tra} : X \to \mathbb{R} \,,$ for simplicity.

Okay. Now, what would be a section of this??

Well, it seems that, by the above reasoning, the space of sections would be $\mathrm{Hom}(c,\mathrm{tra}) \,,$ where $\mathrm{c} : X \to \mathbb{R}$ is any constant function on $X$ – the pullback of some function on the point.

If we take this notion of section seriously, maybe there is an interesting picture arising:

namely, it is clear that a 0-functor $\mathrm{tra} : X \to \mathbb{R}$ has

- precisely one section when it is itself a constant function

- precisely no section otherwise.

The reason is, of course, that the only morphisms between functions are equalities. So $\mathrm{Hom}(c,\mathrm{tra})$ is either of the two -1-categories: equal or non-equal.

But let’s really take the concept of sections seriously in our context: then we should really say that a 0-functor $\mathrm{tra}$ has, unless it is constant, no global sections.

It does, however, always have local sections: it has precisely one section over each subset of $X$ on which it is constant.

So, we need to conclude that one thing that changes for 0-vector transport is that, while an $n$-vector transport always has global sections, a 0-vector transport need not have any global sections.

This may be a useful point of view, since local sections are still very much a concept following the Dao: they are given to us – we don’t invent them.

So then the question is this: what is the “space of states” that we are looking for, in the case that our bundle does not admit a global section?

The answer to this question might be of more general value: instead of only quantizing systems that coupled to vector bundles, we might want to think about what it would mean to quantize a particle that couples to a principal bundle.

And in the case where the vector bundle in question is associated to that principal bundle, probably we would want the two procedures to agree.

Anyway, back to our 0-sections of 0-bundles:

we found that there is precisely one 0-section over every subset of $X$ over which $\mathrm{tra}$ is constant.

Maybe we should try to “collect” all these local sections together, somehow.

I wish I knew a good way to make this “somehow” precise, conceptually. I don’t.

But I do notice that, incidentally, the Lebesgue integral, which we want to see in the quantum theory anyway, seems to suggest itself as doing just this:

notice how the Lebesgue integral is built on the idea that, a priori, the only thing that can directly be integrated are constant functions: the indicator functions of subsets of $X$!

It seems, that if we just translate the words that describe the Lebesgue integral into the langue of sections above, we find that the Lebesgue integral prescription says:

“count the number of all local sections of your 0-functor, then add them all up”

This is as far as my observation goes.

I thought this might suggest that, maybe, the Lebesgue integral can, in a precise way, be regarded as an expression like $\mathrm{Hom}(1,\mathrm{tra}) \,,$ applied to the $n=0$-case.

Does my motivation here make sense to anyone? Does somebody maybe see how to make this work in a precise way?

I am really following John’s principle here: for a definition to be a good defintion, it needs to hold in all degenerate cases without further qualification (without us intervening “by hand”). Here the definition in question is that of “space of sections of an $n$-vector bundle”. And the degenerate case is $n=0$.

Posted by: urs on February 13, 2007 12:17 PM | Permalink | Reply to this

### Re: Lebesgue integral as collection of 0-sections?

$\mathbb{C}$ is the 0-category of complex 0-vector spaces

Run the argument for this past me again. Is it true for all versions of $n$-vector space that this is so for $n = 0$?

An extra test for consistency might be to think what $(-1)$-vector spaces would be.

Posted by: David Corfield on February 13, 2007 1:02 PM | Permalink | Reply to this

### Re: Lebesgue integral as collection of 0-sections?

$\mathbb{C}$ is the 0-category of complex 0-vector spaces

Run the argument for this past me again. Is it true for all versions of $n$-vector space that this is so for $n=0$?

Here I am thinking of $n$-vector spaces as module $(n-1)$-categories for the canonical 1-dimensional $(n-1)$-vector space.

So:

let $K$ be some field. Ordinary $K$-vector spaces are nothing but $K$-modules: $1\mathrm{Vect} \simeq \mathrm{Mod}_K \,.$ Notice the triviality that $K$ is the canonical 1-dimensional 1-vector space over itself.

Next, it turns out that $1\mathrm{Vect}$ is itself

- abelian

- monoidal .

So we may regard it as a “2-field” – or rather: a 2-ring.

So we say that 2-vector spaces are 2-modules for this 2-ring: $2\mathrm{Vect} \simeq \mathrm{Mod}_{1\mathrm{Vect}} \simeq \mathrm{Mod}_{\mathrm{Mod}_K} \,.$ Moreover, $1\mathrm{Vect} \simeq \mathrm{Mod}_K$ is the canonical 1-dimensional 2-vector space over itself.

Now, $2\mathrm{Vect}$ is itself abelian, and monoidal. Hence it qualifies as a 3-ring. The 3-category of its 3-vector spaces is $3\mathrm{Vect} \simeq \mathrm{Mod}_{\mathrm{Mod}_{\mathrm{Mod}_K}} \,.$

So this is the pattern for $n \geq 1$. Extending it to $n=0$ means regarding $K$ itself as the 0-category of 0-vector spaces.

I haven’t thought about what it would mean to consider $n=-1$ here!

But since there are only two -1 categories, there is not much of a choice.

Does it make sense to consider $K$ as a 0-module for the -1-category true??

Posted by: urs on February 13, 2007 1:42 PM | Permalink | Reply to this

### Re: Lebesgue integral as collection of 0-sections?

But since there are only two −1 categories, there is not much of a choice.

I’d be very careful about this! This is like saying that your only choices for the 0category of 0vspaces are the cardinal numbers, since these are the only 0categories. But in fact, you have both more and fewer choices; you can’t pick the cardinal number 0 (and maybe some others are unavailable, I don’t know), but you have several choices when the cardinal number is 20 (R, C, and several others).

This is because a field (better, rig) is a 0category equipped with extra structure, just as a 2rig is a 1category equipped with extra structure, and so on. Thus a −1rig (which is really where your possibilities lie) is a −1category equipped with extra structure —and man, that could be anything!

(More precisely, the 0category of −1rigs, whatever those may be, will define −1rigs as -1categories with extra structure if it comes with a faithful 0functor to the 0category of −1categories. And every 0functor is faithful! So the 0category of −1rigs could, a priori, be any set whatsoever.)

The last time that I tried to figure out what a −1rig is, I don’t think that I got anywhere. Even if you get a satisfactory answer to this, I don’t think that any of these will be the −1rig of complex −1vspaces; it seems to me that you have to start with C at the 0vspace level to get anywhere. But I may be wrong about that.

In a similar vein, it’s not really fair to say that −1categories are truth values, although I’ve been following your lead in pretending that it is fair. The truth values are both the −1groupoids and the 0posets, and the sets are both the 0groupoids and the 0categories, but the concept of ncategory —unlike (n + 1)­poset and ngroupoid— does break down at n = −1 for a legitimate reason (because its definition calls for one exceptional level, which there is no room to accomodate). Perhaps the concept of nrig is similarly inapplicable.

Posted by: Toby Bartels on February 13, 2007 10:14 PM | Permalink | Reply to this

### Re: Lebesgue integral as collection of 0-sections?

More precisely, the 0category of −1rigs, whatever those may be, will define −1rigs as -1categories with extra structure if it comes with a faithful 0functor to the 0category of −1categories. And every 0functor is faithful! So the 0category of −1rigs could, a priori, be any set whatsoever.

I understand. Thanks for pointing that out!

What I don’t understand yet is what you say afterwards:

The truth values are both the -1groupoids and the 0posets, and the sets are both the 0groupoids and the 0categories, but the concept of $n$category —unlike $(n+1)$­poset and $n$groupoid— does break down at $n=-1$ for a legitimate reason (because its definition calls for one exceptional level, which there is no room to accomodate).

Why is that? Didn’t you define (-1)-categories as the $\mathrm{Hom}$-objects of 0-categories?

Why does it matter whether this is in the world of categories or of groupoids?

What is the difference between 0-categories and 0-groupoids?? :-)

And how do you define a 0-poset?

By the way, all this reminds me of one other idea that looks like it might potentially be a way to solve the issue about 0-push-forward discussed above:

Since we are encountering problems at $n=0$, maybe we would want to shift everything up one level.

Following ideas used in Jeffrey Morton’s discussion of categorified quantum mechanics, we might want to pass from some rig $R$ to $R$-sets, i.e. to sets $S \to R$ equipped with maps to $R$.

This way we pass from the 0-category $R$ to the 1-category $\mathrm{Set}/R$.

For those readers not familiar with this I just say that each $R$-set maps to an element of $R$ by taking its $R$-cardinality, i.e. by adding up all the $R$-labels of all elements of the set.

Disjoint union of $R$-sets corresponds to addition of $R$-cardinalities and cartesian product of $R$-sets to the product of $R$-cardinalities.

So $R$-sets behave exactly as $R$ under $R$-cardinality. An $R$-set is like a blown-up element of $R$, having “inner structure”.

The good thing about this is that we then do have all the canonical push-forward technology available, that I was talking about above.

In particular, for $R = \mathbb{C}$, a complex 0-vector bundle on $X$ could then be conceived as a “pseudo-0-functor” $\mathrm{tra} : X \to \mathrm{Set}/\mathbb{C}$ which assigns to each point of $X$ a $\mathbb{C}$-set over it, which, by taking $\mathbb{C}$-cardinality, maps simply to a complex number.

Now, this “pseudo-0-functor” may be “pushed forward to the point” in the canonical way: we simply get the coproduct over all the fibers, i.e. simply the disjoint union of all these $\mathbb{C}$-sets $\oplus_{x \in X} \mathrm{tra}(x) \,.$ This is nice, because the $\mathbb{C}$-cardinality of this is … nothing but the “integral” (sum) of the $\mathbb{C}$-cardinalities over every point! So this would be exactly what we are after!

Hm, now that I used the term “pseudo-0-functor” to describe a map that sends elements of a set to objects of a category, I am reminded of that other insight # which says that we should think of

bundles with connection

in general not as 1-functors from paths to fiber morphsims

but as pseudofunctors from paths to morphisms of the automorphism 2-group of the fibers.

Now it seems I am saying that the best way to think about 0-bundles is by looking at them as pseudo-0-functors! That fits in nicely, at least in a vague way.

Hm…….

Posted by: urs on February 14, 2007 7:54 AM | Permalink | Reply to this

### Re: Lebesgue integral as collection of 0-sections?

now that I used the term “pseudo-0-functor”…

I should have thought of this before!

Let me first adjust my notation. Following def 21 (p. 44) in Jeffrey’s text I should write

$\mathbb{C}\mathrm{Set}$ for the category of sets over $\mathbb{C}$.

We need to understand what happes to our world once we replace the 0-category $\mathbb{C}$ everywhere with the 1-category $\mathbb{C}\mathrm{Set}$.

I guess that’s essentially what Jeffrey Morton has done! But here I would like to regard this not as a categorification (of quantum mechanics), but just as a necessary refinement of our ordinary concepts, following the rules of Schreier theory.

To put it in my own ideosyncratic terms:

the background field for an $n$-particle is, naively, an $n$-functor $\mathrm{tra} : P_n(X) \to \mathrm{something} \,.$ But according to higher Schreier theory, we have to be prepared to find that this is really an $n$-pseudo-functor!

The one central example for this that I was aware of is if we want to couple a 2-particle to a non-fake-flat principal 2-bundle.

But now I seem to have argued that a much more basic example of this need for pseudo-$n$-transport is the 0-particle (what string theorists would call the D(-1)-brane or “D-instanton”).

The 0-particle has a parameter space which is a $-1$-category. As such it is a little hard to distinguish from the single 1-particle which looks like $\{\bullet\}$.

But the 0-particle couples to a background field which is just a 0-functor $\mathrm{tra} : X \to \mathrm{something} \,.$

Now, if we take $\mathrm{something} = \mathbb{C}$ and don’t allow ourselves to intervene as the Dao determines the quantum theory of the 0-particle, then we’ll find that this quantum theory either does not exist (when $\mathrm{tra}$ is not constant) or that it is a little more than boring (when it is constant): in that case the 0-particle would have precisely a single quantum state only (not a 1-dimensional vector space of states, just a single state).

Since we are both reluctant to intervene by hand, as well as reluctant to accept such a sadly boring outcome, we recall that higher Schreier theory at least vaguely suggests that the above failure is possibly due to the fact that we did secretly and wrongly intervene already in the first place, by restricting $\mathrm{tra}$ to be a strict 0-functor. We shouldn’t! We should allow it to be a “pseudo-0-functor”, namely a functor that maps elements of the “target space” 0-category to objects in a category which is allowed to have nontrivial morphisms.

If this is true, then it should affect our expectations for what goes on for $n \geq 0$.

So if a 0-particle couples a pseudo-0-functor $\mathrm{tra}_0 : X \to \mathbb{C}\mathrm{Set}$ then a 1-particle should probably couple to a pseudo-1-functor $\mathrm{tra}_1 : P_1(X) \to {}_{\mathbb{C}\mathrm{Set}}\mathrm{Mod}$ instead of just to a strict 1-functor $\mathrm{tra}_1 : P_1(X) \to {}_{\mathbb{C}}\mathrm{Mod} \simeq \mathrm{Vect}_\mathbb{C} \,.$

If so, we need to understand what ${}_{\mathbb{C}\mathrm{Set}}\mathrm{Mod}$ is like!

Luckily, we have just learned enough general nonsense about enriched category theory to do this easily:

We are now in the world of categories enriched over $V = \mathbb{C}\mathrm{Set} \,.$

In the comment linked to above I talk at great length about how “nice” examples for modules for $V = \mathrm{Vect}$ are obtained from categories $\mathrm{Mod}_A$ of modules over $V$-enriched categories $A$.

Precisely the same argument goes through for our present purpose, simply by exchanging $\mathrm{Vect} \leftrightarrow \mathbb{C}\mathrm{Set}$.

So, we find that a “nice” “$\mathbb{C}\mathrm{Set}$-vector space” (with “nice” exactly in the above sense) is precisely a category of $\mathbb{C}\mathrm{Set}$-presheaves

(1)$W = \mathbb{C}\mathrm{Set}^{C^\mathrm{op}}$

for $C$ any $\mathbb{C}\mathrm{Set}$-enriched category!

We should then consider 1-particles coupled to background fields that are given by pseudo-1-functors $\mathrm{tra} : P_1(X) \to {}_{\mathbb{C}\mathrm{Set}}\mathrm{Mod} \,.$

Following just abstract nonsense, we know how to compute the space of states of such a 1-particle. But now, still following just abstract nonsense without putting in anaything else by hand, this should also allow us to get propagation by the path integral. By taking $\mathbb{C}$-cardinality of everything in sight, the result should then reproduce ordinary quantum mechanics.

(This is potentially solved in Jeffrey’s paper, I’d need to re-read that and remind myself of the details.)

Posted by: urs on February 14, 2007 8:59 AM | Permalink | Reply to this

### CSet-Vector spaces

Above I began convincing myself that I should be interested in considering $\mathbb{C}Set$-vector bundles with connection, i.e. pseudo-1-functors $\mathrm{tra} : P_1(X) \to {}_{\mathbb{C}\mathrm{Set}}\mathrm{Mod} \,.$

As always, I immediately restricted myself to those $\mathbb{C}\mathrm{Set}$-modules that are of the form $W = [C^\mathrm{op},\mathbb{C}\mathrm{Set}]$ for some category $C$.

On these functor categories, $\mathbb{C}\mathrm{Set}$ acts by “pointwise multiplication in $\mathbb{C}\mathrm{Set}$”.

To better see what this means, compare this to ordinary vector spaces with a choice of basis $B$.

These are 0-functors (functions) from the set $B$ to the set $\mathbb{C}$ (I’ll stick to working over the complex numbers, but it’s not relevant.)

$\mathrm{dim}_\mathbb{C}(V) = |B| \;\; \Rightarrow \;\; V \simeq [B,\mathbb{C}] \,.$

So what I called “nice” $V$-vector spaces previously, are really $V$-vector spaces with a choice of basis, in a sense.

But notice, crucially, that I am not restricting my basis category $C$ to be a 0-category. If I did so, i.e. would find the “Kapranov-Voevodsky”-version of $\mathbb{C}\mathrm{Set}$-vector spaces, namely those of the form $W = [\mathrm{Disc}(B),\mathbb{C}\mathrm{Set}] \,.$

As I don’t tire of emphasizing, we have no need, in general, to restric ourselves to $\mathrm{KV}-\mathbb{C}\mathrm{Set}-\mathrm{Vect} \hookrightarrow {}_{\mathbb{C}\mathrm{Set}} \mathrm{Mod} \,.$

You might recall that I offerered the First $n$-Café-Millenium Prize to draw attention to this. .

Anyway. I want to undestand this 2-category ${}_{\mathbb{C}\mathrm{Set}}\mathrm{Mod}$ of $\mathbb{C}\mathrm{Set}$-vector spaces.

To do so, I should first better understand $\mathbb{C}\mathrm{Set}$ itself, and its role as a category we are enriching over.

First, we need a monoidal structure on $\mathbb{C}\mathrm{Set}$. We want this to be cartesian product of sets, combined with the product in $\mathbb{C}$.

In other words, for $s : S \to \mathbb{C}$ and $t : T \to \mathbb{C}$ two $\mathbb{C}$-sets, their product should be $(s : S\to \mathbb{C}) \otimes (t : T\to \mathbb{C}) := ( S \times T \stackrel{s\times t}{\to} \mathbb{C}\times \mathbb{C} \stackrel{\cdot}{\to} \mathbb{C} ) \,.$

Now comes the first potentially interesting point: what is then the closed structure on $\mathbb{C}\mathrm{Set}$?

Clearly, we expect $\mathrm{hom}(s : S \to \mathbb{C}, t : T \to \mathbb{C})$ to be the $\mathbb{C}$-set which as a set is essentially just $\mathrm{Hom}(S,T)$.

But which $\mathbb{C}$-labels $h : \mathrm{Hom}(S,T) \to \mathbb{C}$ do we put on these maps?

Let me write $\{ A_{c}, B_{c_2}, \cdots \}$ for a $\mathbb{C}$-set, where the $\mathbb{C}$-labels are in the subscript.

Then in particular morphisms

(1)$\{A_{c_1}\} \otimes \{B_{c_2}\} \to \{C_{c_3}\}$

exist if and only if $c_3 = c_1 \cdot c_2 \,.$

First assume that $c_2 \neq 0$. Then it follows that $\mathrm{hom}(\{B_{c_2}\},\{C_{c_3}\})$ must be the 1-element set, with the label of the single element being $\frac{c_3}{c_2} \,.$

But what happens if $c_2 = 0$? A short reflection shows that then the internal hom object $\mathrm{hom}(\{B_{c_2}\},\{C_{c_3}\})$ cannot exist! Because if also $c_3 = 0$ we would find that $\mathrm{Hom}_{\mathbb{C}\mathrm{Set}}( \{A_{c_1}\}\otimes \{B_{c_2}\} , \{C_{c_3}\} )$ is the 1-element set, for any value of $c_1$, but no matter what we would choose for $\mathrm{hom}(\{B_{c_2}\},\{C_{c_3}\}) \,,$ the cardinality of the set $\mathrm{Hom}_{\mathbb{C}\mathrm{Set}}( \{A_{c_1}\} , \mathrm{hom}(\{B_{c_2}\},\{C_{c_3}\}) )$ would depend on the label of $c_{1}$. Hence there would not be a natural isomorphism $\mathrm{Hom}_{\mathbb{C}\mathrm{Set}}( \{A_{c_1}\}\otimes \{B_{c_2}\} , \{C_{c_3}\} ) \simeq \mathrm{Hom}_{\mathbb{C}\mathrm{Set}}( \{A_{c_1}\} , \mathrm{hom}(\{B_{c_2}\},\{C_{c_3}\}) ) \,.$

Therefore, with the above choice of tensor product (which we need if we want to have any chance to recover what we want to recover after applying the $\mathbb{C}$-cardinality map) is not closed!

What to do now? I need closedness because I want to enrich over $\mathbb{C}\mathrm{Set}$, which in turn I need in order to talk about $\mathbb{C}\mathrm{Set}$-modules in a sensible way!

At the moment, I can see only one conclusion that this is forcing upon as:

We need to quit using $\mathbb{C}$ and start using $\mathbb{C}^\times$.

While I’d have to see where this leads us, it might be potentially interesting: this will imply that our quantum phases are non-vanishing complex numbers. Which indeed they are supposed to be.

Well, they are even supposed to be unimodular complex numbers, i.e. elements of $U(1) \subset \mathbb{C}^\times \subset \mathbb{C}$.

And indeed, that’s what Jeffrey Morton considers: $U(1)\mathrm{Set}$ instead of $\mathbb{C}\mathrm{Set}$.

Maybe I am just doing nothing but finding (another?) argument for why we indeed are forced to use $U(1)\mathrm{Set}$.

But for the moment I should probably continue with $\mathbb{C}^\times\mathrm{Set}$.

Okay. In that case, we now know that $\mathbb{C}^\times\mathrm{Set}$ is indeed closed. The internal Hom-object is $\mathrm{hom}_{\mathbb{C}^\times\mathrm{Set}}( \mathbf{S}, \mathbf{T} ) = ( h : \mathrm{Hom}(S,T) \supset H \to \mathbb{C}^\times ) \,,$ where the set $H$ contains all those maps $S \to T$ that induce a fixed quotient on the labels of their image and preimage elements. By means of $h$ these maps are labeled by exactly that quotient in $\mathbb{C}^\times$.

There is maybe a more elegant way to say this. But anyway.

Unless I made a mistake, I now have a closed monoidal category $V := \mathbb{C}^\times\mathrm{Set} \,.$ I guess I should also check completeness and cocompleteness. But right now I won’t.

So next I want to understand ${}_{\mathbb{C}^\times\mathrm{Set}}\mathrm{Mod} \,.$

The category of $\mathbb{C}^\times\mathrm{Set}$-vector space has the following bimonoidal structure.

It has a coproduct $[C,\mathbb{C}^\times\mathrm{Set}] \oplus [D,\mathbb{C}^\times\mathrm{Set}] \simeq [C \cup D,\mathbb{C}^\times\mathrm{Set}] \,,$ inherited from $\mathrm{Cat}$, which corresponds to taking the disjoint union of bases of vector spaces as a new generating basis.

Similarly for the product $[C,\mathbb{C}^\times\mathrm{Set}] \times [D,\mathbb{C}^\times\mathrm{Set}] \simeq [C \times D,\mathbb{C}^\times\mathrm{Set}] \,,$ which corresponds to the tensor product of vector spaces.

(Really, this is to be read as the product of $\mathbb{C}^\times\mathrm{Set}$-enriched categories. But I guess that does not make a difference.)

Good. So given a 1-functor $\mathrm{tra} : P_1(X) \to \mathrm{Vect}_{\mathbb{C}^\times\mathrm{Set}} \,,$ and assuming it couples to a 1-particle $\mathrm{par} = \{\bullet\}$ whose configurations are given by $\mathrm{conf} = \mathrm{Disc}(\mathrm{Obj}(P_1(X))) \simeq \mathrm{Disc}(X) \,.$ Then, a state of this 1-particle is a morphism $\psi : 1_* \to \mathrm{tra}_* \,,$ where $\mathrm{tra}_*$ is the restriction of $\mathrm{tra}$ to $\mathrm{Disc}(X)$ and $1_*$ is the tensor unit in the category of such functors, which means it is the functor that sends every point of $X$ to the $\mathbb{C}^\times\mathrm{Set}$-vector space $[\{\bullet\},\mathbb{C}^\times\mathrm{Set}]$ (where the category $\{\bullet\}$ really denotes the 1-object $\mathbb{C}^\times\mathrm{Set}$-enriched category whose objects of morphisms is $I_{\mathbb{C}^\times\mathrm{Set}} = \{\bullet_1\}$.)

Of course $[\{\bullet\},\mathbb{C}^\times\mathrm{Set}] \simeq \mathbb{C}^\times \mathrm{Set} \,.$

Is the space of all these states $[1_*,\mathrm{tra}_*]$ equivalent to $\oplus_{x \in X} \mathrm{tra}(x)$ or is it not?

The answer to that should go in a separate comment.

Posted by: urs on February 15, 2007 1:32 PM | Permalink | Reply to this

### What is a (−1)category? (Was: Lebesgue integral as collection of 0-sections?)

Urs wrote in part:

I understand. Thanks for pointing that out!

Good, that’s probably the important thing.

Then Urs went on to write:

What I don’t understand yet is what you say afterwards

OK, I’ll talk about that now.

Didn’t you define (−1)-categories as the Hom-objects of 0-categories?

You could do that, and certainly there is no better alternative definition, and then you get that a (−1)­category is a truth value. But a 0category should be a category enriched over (−1)­categories. But in fact, a category enriched over (−1)­categories is a poset, not simply a set! So something’s wrong here.

Compare this situation:

And how do you define a 0-poset?

They’re the Hom-objects of 1posets, of course. That is, they’re truth values. But now if I recreate a 1poset as a category enriched over 0posets, I get the correct result!

Note that both above use the correct rule: An (n + 1)­category is a category enriched (in a suitably weak sense) over the ∞category of ncategories; and an (n + 1)­poset is a category enriched over the ∞category of nposets. (To be sure, I myself invented the term ‘nposet’, so it’s a little unfair to use it by analogy to make a terminological point, but I think that you’ll agree that the concept of nposet is reasonable.)

In contrast, an (n + 1)­groupoid is a groupoid enriched over the ∞category of ngroupoids. So now I’ll answer this question:

What is the difference between 0-categories and 0-groupoids?

There’s no difference of course; both of these are just sets. So (−1)­groupoids are defined to be the Hom-objects of 0groupoids, that is (once again) truth values. But now if I recreate 0groupoids as groupoids enriched over (−1)­groupoids, I get the correct answer back.

So we can start with (−1)­groupoids as truth values (or even further back, with (−2)­groupoids as trivial) and get all higher levels of groupoids by applying a single rule (given above). And we can start with 0posets as truth values (or even further back, with (−1)­posets as trivial) and get all higher levels of posets by applying a single rule (given above). But to get higher levels of categories by applying a single rule (the same rule as for higher posets), we can only go back to 0categories!

Here is another way to see the ideas involved, starting from the most general concept of ∞category (which is the same as ∞poset, incidentally). An ∞category is an ∞groupoid if all morphisms at every level are invertible. It’s an nposet if all parallel jmorphisms are equivalent for j ≥ n. It’s an ngroupoid if it’s both an ∞groupoid and an (n + 1)­poset (so the numbering is a bit off between these). Finally, it’s an ncategory if it’s an (n + 1)­poset and every (n + 1)­morphism is invertible.

In other words, an ∞category is an ngroupoid iff parallel jmorphisms are equivalent for j ≥ n + 1 (in which case it follows that all jmorphisms are invertible for jn + 1) and additionally all jmorphisms are invertible for 0 < j ≤ n + 1. But it’s an ncategory iff parallel jmorphisms are equivalent for j ≥ n + 1 and additionally all jmorphisms are invertible for jn + 1, with no restriction for jn + 1. This special rule for jn + 1 is the tricky bit, which doesn’t behave well when n := −1.

Or more simply: The concept of (−1)category doesn’t really make sense, because it calls on the concept of invertibility of 0morphisms, which doesn’t really make sense. Sure, you can simply drop that clause, in which case you get truth values, but it doesn’t fit all the patterns. A similar thing may occur (I don’t really know) for the concept of (−1)rig.

Urs continued to write:

Since we are encountering problems at n = 0, maybe we would want to shift everything up one level.

I expect that you’ll have much better success with this approach; it looks very promising to me. Eventually, it might even tell you what a −1rig really is; but even if not, you probably won’t need to know!

Posted by: Toby Bartels on February 14, 2007 11:01 PM | Permalink | Reply to this

### Re: What is a (−1)category? (Was: Lebesgue integral as collection of 0-sections?)

I wrote:

What I don’t understand yet is what you say afterwards

Toby replied:

OK, I’ll talk about that now.

and did so.

Thanks, that was very illuminating!

It’s really good that somebody (you, that is) takes the time to think about issues like that. It’s the kind of issue which may appear too irrelevant to bother with on first sight, but it’s not.

But in fact, a category enriched over (−1)­categories is a poset, not simply a set!

Okay. Maybe for the record and for the benefit of those following this I’ll recall that

a category enriched over

$-1\mathrm{Cat} = (\{\mathrm{false}, \mathrm{true}\},\otimes = \mathrm{and}, I = \mathrm{true})$

is a category whose Hom-objects “are” either the empty set or the 1-element set (where saying “are” really means applying the canonical monoidal isomorphism $-1\mathrm{Cat} \simeq \{\{\}, \{1\}\} \subset \mathrm{Set}$), therefore these are categories with at most one morphism for any ordered pair of objects, hence these are posets.

And how do you define a 0-poset?

They’re the Hom-objects of 1posets, of course. That is, they’re truth values. But now if I recreate a 1poset as a category enriched over 0posets, I get the correct result!

Okay.

What is the difference between 0-categories and 0-groupoids?

There’s no difference of course; both of these are just sets. So (−1)­groupoids are defined to be the Hom-objects of 0groupoids, that is (once again) truth values.

Okay.

But now if I recreate 0groupoids as groupoids enriched over (−1)­groupoids, I get the correct answer back.

This is your central point. Let me see if I understand that.

A groupoid enriched over (-1)groupoids is a poset in which all morphisms are invertible.

But a poset with all morphisms invertible is a disjoint union of codiscrete categories.

Right?

Such a category is equivalent to the the discrete category over its set of connected components.

It is this equivalent discrete category (= 0-category = set) which you are thinking of when saying that a groupoid enriched in (-1)groupoids is a set.

Right?

Posted by: urs on February 15, 2007 9:43 AM | Permalink | Reply to this

### Re: What is a (−1)category? (Was: Lebesgue integral as collection of 0-sections?)

But a poset with all morphisms invertible is a disjoint union of codiscrete categories.

Surely a poset with all morphisms invertible is just a set. Antisymmetry holds in a poset, i.e., $a R b$ and $b R a$ implies that $a = b$.

Is there not something a little dodgy going on in this discussion? When Toby says:

But a 0category should be a category enriched over (−1)­categories.

Why should a 0category be a type of (1)category? Why take the 1-level to be the template?

Posted by: David Corfield on February 15, 2007 10:41 AM | Permalink | Reply to this

### Re: What is a (−1)category? (Was: Lebesgue integral as collection of 0-sections?)

But a poset with all morphisms invertible is a disjoint union of codiscrete categories.

Surely a poset with all morphisms invertible is just a set. Antisymmetry holds in a poset, i.e., $a R b$ and $b R a$ implies that $a = b$.

Hm, true. But antisymmetry is not implied to hold in a $\{\mathrm{true},\mathrm{false}\}$-enriched category. I thought we were using “poset” in the sense of “category with at most one morphism from any object to any other”.

Posted by: urs on February 15, 2007 10:50 AM | Permalink | Reply to this

### Re: What is a (−1)category? (Was: Lebesgue integral as collection of 0-sections?)

Yes. Perhaps I’m not quite getting Toby’s view from infinity. He said:

An $(n + 1)$­category is a category enriched (in a suitably weak sense) over the $\infty$category of $n$categories

Let’s try some recursion here. So

An $(n + 1)$­category is a category [i.e, a category enriched (in a suitably weak sense) over the $\infty$category of $0$categories] enriched (in a suitably weak sense) over the $\infty$category of $n$categories [which are categories enriched (in a suitably weak sense) over the $\infty$category of $(n - 1)$categories].

At each step of the recursion I seem to have put in a replacement for the template term ‘category’ as well as any mention of ‘$n$-category’.

Do I ever know what a category is?:

A (1)­category is a category enriched (in a suitably weak sense) over the $\infty$category of $0$categories

Posted by: David Corfield on February 15, 2007 11:18 AM | Permalink | Reply to this

### Re: What is a (−1)category? (Was: Lebesgue integral as collection of 0-sections?)

Do I ever know what a category is?:

As I understood Toby’s program, for categories he would give his recursion formula and in addition the definition of 0catgories.

For groupoids, though, he would give the recursion formula and the definition of a -2groupoid.

By the way, that’s one more thing that requires using equivalences:

We have $-2Grpd = \{\{\mathrm{true}\},\otimes = \mathrm{and}, I = \mathrm{true} \} \,.$

This implies that a -2Grpd-enriched groupoid as a codiscrete category.

Either its set of objects is empty, or it is not empty. But all codiscrete categories with non-empty set of objects are equivalent to the discrete category on a single element.

Hence there are, up to equivalence (isomorphism, here) just two -2Grpd-enriched groupoids, called $\mathrm{true}$ and $\mathrm{false}$.

Posted by: urs on February 15, 2007 11:58 AM | Permalink | Reply to this

### Re: What is a (−1)category? (Was: Lebesgue integral as collection of 0-sections?)

Do I ever know what a category is?:

A (1)­category is a category enriched (in a suitably weak sense) over the ∞category of 0categories

These statements were never intended as definitions, merely as statements of fact (theorems, if you will). I did put in some defintions later on in the comment; they are in boldface. All that is missing is a definition of ∞category.

However, if you know what an ∞category is and you know what a ‘category enriched over the Cartesian ∞category V’ is (and similarly what an enriched groupoid is), then you can take these as definitions. (Except that I'm not sure if the general theory of enriching —in a suitably weak sense— over an ∞category has ever been worked out, even in the Cartesian case. And we would have to prove that the result at each stage is Cartesian. But morally this should work!)

Posted by: Toby Bartels on February 17, 2007 12:00 AM | Permalink | Reply to this

### Re: What is a (−1)category? (Was: Lebesgue integral as collection of 0-sections?)

David wrote in part:

Toby wrote:

But a 0category should be a category enriched over (−1)­categories.

Why should a 0category be a type of (1)category? Why take the 1-level to be the template?

I'm simply using the ‘category enriched over’ operation to move 1 level up. Indeed, part of the point is that the hierarchy of ncategories is not as well behaved as the hierarchies of ngroupoids and of nposets, so I certainly wouldn't want to take an element of (only) that hierarchy as template! But the process of moving from sets to categories is the template for the iteration process in the hierarchies of ncategories and of nposets (as the process of moving from sets to groupoids is the template for the iteration process in the hierarchy of ngroupoids), so I do use it.

Posted by: Toby Bartels on February 16, 2007 11:58 PM | Permalink | Reply to this

### Re: What is a (−1)category? (Was: Lebesgue integral as collection of 0-sections?)

Urs wrote in part:

Toby wrote:

But in fact, a category enriched over (−1)­categories is a poset, not simply a set!

But now if I recreate a 1poset as a category enriched over 0posets, I get the correct result!

But now if I recreate 0groupoids as groupoids enriched over (−1)­groupoids, I get the correct answer back.

This is your central point. Let me see if I understand that.

Well, I think that all three statements above together constitute the central point. (But you seem to have understood the first two just fine. ^_^)

A groupoid enriched over (-1)groupoids […] is a disjoint union of codiscrete categories. […] Such a category is equivalent to the the discrete category over its set of connected components.

Right. By default, I'm only ever interested in things up to equivalence.

Just as you said (in a later comment here) that a poset is simply a “category with at most one morphism from any object to any other”, so you should say that a set is simply a “groupoid with at most one morphism from any object to any other”. Whereas, if you think the latter is really a setoid (that is a set equipped with an equivalence relation), then you should say that the former is really a proset (that is a preordered set, a set equipped with a reflexive and transitive relation). But I wouldn't do that, because my philosophy is that (in general) equality of objects in a category is meaningless (unless as a synonym for isomorphism).

Anyway, all the terms are correct (whatever your philosophy) up to equivalence, which is all that really matters.

Posted by: Toby Bartels on February 16, 2007 11:54 PM | Permalink | Reply to this

### Re: What is a (−1)category?

While I’m pontificating, I might add that I not only find the concept of ncategory lacking (in general —obviously specific cases are fine if it happens to be what you really want); but I also think that the numbering makes more sense for nposet than for ngroupoid (and ncategory).

So the basic concepts should be nposet and nset. In more familar terms, an nposet is an ncategory in which parallel nmorphisms are always equivalent (usually called ‘equal’ in this context); and an nset is an (n − 1)­groupoid. (I already referred to the fact that a 2set is a groupoid here.) Of course, for n := ∞, the level shift (and the notion of top-level morphism) disappears, so an ∞poset is an ∞category, and an ∞set is an ∞groupoid. And for n := −1 (the lowest that I understand), both (−1)­posets and (−1)­sets are simply truth values.

For that matter, I think that ngroupoids are the really interesting concept philosophically (you can blame Jim Dolan for this, whether or not he actually agrees), particularly for refining the notion of identity. And the basic idea behind ‘categorification’ should be something like enrichment over or internalisation in groupoids, rather than over/in categories, unless you’re categorifying something that already has enough noninvertibility to suggest that the categorified version needs an extra level of that. Even the notion of categorified category is best identified with (2,1)­category (that is a 2category whose 2morphisms are all invertible) rather than 2category (although that answer isn’t too bad; fortunately, categorification isn’t a precise concept, so mutliple answers are OK).

Is that enough idiosyncratically radical opinion? ^_^

Posted by: Toby Bartels on February 17, 2007 12:40 AM | Permalink | Reply to this

### Re: What is a (−1)­category?

I wrote in small part:

And for n := −1 (the lowest that I understand), both (−1)­­posets and (−1)­­sets are simply truth values.

Whoops! Of course I should have said ‘trivial’ here (instead of ‘simply truth values’), consistent with my previous comments. (For n := 0, on the other hand, both 0posets and 0sets are simply truth values.)

Incidentally, the reason that we start with −1 here (rather than 0, the most natural number starting place) is essentially the same reason that one begins with (−1)­simplices, the (−1)­sphere, —and (−1)­morphisms for that matter! I’d kind of like to adjust that numbering system too, but it’s very firmly entrenched. One thing at a time!

Posted by: Toby Bartels on February 17, 2007 9:06 PM | Permalink | Reply to this

### Re: What is a (−1)category?

Is that enough idiosyncratically radical opinion?

We can never get enough.

I’m intrigued by what you say about $n$groupoids. What do you mean by having ‘enough noninvertibility’? Groups don’t have it, but monoids do, so act accordingly? Sets and functions do, but sets and bijections don’t? Can you also have ‘enough dualisability’?

Posted by: David Corfield on February 18, 2007 9:10 AM | Permalink | Reply to this

### Re: What is a (−1)category?

David wrote:

We can never get enough [idiosyncratically radical opinion].

OK, then, here’s some more! ^_^

(Warning: I haven’t thought all of this through before!)

What do you mean by having ‘enough noninvertibility’? Groups don’t have it, but monoids do, so act accordingly? Sets and functions do, but sets and bijections don’t? Can you also have ‘enough dualisability’?

Well, as categorification is not a precise concept, I don’t expect these notions to be precise either. But let’s take ‘enough dualisability’ as an example.

If I teach you the concept of compact category (that is a category with duals for objects) and you try to categorify this to compact bicategory, should it have duals only for objects or should it have duals for morphisms too? (so the hom categories will be compact). It’s not clear to me just what you should do, and I don’t know if anybody has tried.

It’s probably even worse to try to category the concept of dagger category (that is a category with duals for morphisms). But dagger compact categories (that is a category with compatible duals for both) is easy: obviously objects, morphisms, and 2morphisms should now have duals. (The details might not be so simple, but it’s definitely what we should try to do).

And I think that it’s probably no coincidence that dagger compact categories are far more useful than mere dagger categories and mere compact categories (not that these are useless by themselves, but they just don’t pack the same punch; they are like loops or semigroups compared to groups or monoids). Ultimately, this is because there should be a good notion of ∞category with duals (as in the Tangle Hypothesis) that these are merely low-dimensional versions of. But who cares about an ∞category with duals for (say) 5morphisms?

So maybe my point is that concepts need to be treated completely to be worthy of categorification. The concept of category is useful, not least because sets form a category. But one level up, groupoids form (not just a 2category but in fact) a (2,1)category. The concept of 2category is really trying to fill a gap between (2,1)category and 3poset; its justification is that categories form a 2category, but categories already weren’t quite as important as either sets or groupoids (being otherwise just filling a gap between groupoids and 2posets), and this just gets worse as we climb the ladder. In contrast to ncategory, nposet is a complete notion; ncategories only do things half way at the top level.

So categories are missing one level of noninvertibility (since 2morphisms are nontrivial, yet must be invertible), just as compact categories are missing one level of dualisability (since morphisms may be noninvertible yet need not have duals). More complete are 2posets (which may have noninvertible 2morphisms) and dagger compact categories (whose morphisms have duals too).

You might ask about 2posets with duals (at all levels, including 2morphisms). Shouldn’t these be even more complete than dagger compact categories? (since 2posets are more complete than mere categories). But in fact, these are the same as dagger compact categories! (Duals for jmorphisms are automatically inverses when (j + 1)­morphisms are trivial.) So —and I’d never thought of this before— I hereby admit the hierarchy of ncategories with duals to that select class of complete concepts heretofore inhabited by the hierarchies of nposets and of ngroupoids (or nsets) but not by the hierarchy of ncategories.

Does that make any sense?

Posted by: Toby Bartels on February 20, 2007 2:06 AM | Permalink | Reply to this

### Re: QFT of Charged n-Particle: Dynamics

Okay: I’ve tried twice now to reply to this post. The first time I accidentally erased my long detailed reply (never edit in a web browser!), and the second, shorter and sketchier version, suffered a server error. I’ll try again later.

Posted by: Jeff Morton on February 13, 2007 10:44 AM | Permalink | Reply to this

### Re: QFT of Charged n-Particle: Dynamics

Well, reporting server errors is certainly much better than experiencing real ones!

Posted by: Jeff Morton on February 13, 2007 10:46 AM | Permalink | Reply to this

### meta

never edit in a web browser!

I have the habit of copy-pasting my comments to another text editor each time I hit “review”.

suffered a server error

Not sure what this was, but from time to time (very rarely though), I do get a server arrow from golem, too. Usually everything can be recovered by letting the browser go back.

Posted by: urs on February 13, 2007 11:26 AM | Permalink | Reply to this
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