## April 8, 2010

### Symmetric Monoidal Bicategories and (n×k)-categories

#### Posted by Mike Shulman

I just put a note on the arXiv about constructing symmetric monoidal bicategories. The definition of symmetric monoidal bicategory is somewhat imposing, but in many cases all the structure can be “lifted” for free from a symmetric monoidal double category, which is a much easier structure. For instance, the bicategories of spans, cobordisms, and profunctors all inherit their monoidal structures in this way. Here’s the link:

The tricky part is that in a monoidal double category, the coherence isomorphisms such as $X\otimes (Y\otimes Z)\cong (X\otimes Y)\otimes Z$ are vertical isomorphisms, but these are the morphisms that get discarded when we pass to the horizontal bicategory. Thus, we need to be able to “lift” these isomorphisms to horizontal equivalences, in such a way that we can ensure coherence is preserved. The structure we need is essentially that of a proarrow equipment (regarded as a double category), although in this context, with attention focused on the “bicategory of proarrows” rather than the 2-category of arrows, it seems more appropriate to call it a framed bicategory or a fibrant double category. Richard Garner and Nick Gurski proved essentially the same result in section 5 of this paper, using “locally-double bicategories;” the machinery I used also makes braiding and symmetry easy to deal with.

More generally, I suspect that there is a theorem along the following lines. Define an $(n\times k)$-category (read “$n$ by $k$ category”) to be an $n$-category internal to $k$-categories. Such a thing has $(n+1)(k+1)$ different kinds of cells which are naturally arranged in a grid, with $(0,0)$-cells at the bottom left (say), $(n,0)$-cells at the top left, and $(n,k)$-cells at the top right. In particular, a double category is a 1x1-category. The theorem I think should be true is that if you have an $(n\times k)$-category which is “fibrant” in a suitable sense, then by discarding most of those types of cells and considering only the $(n+k+1)$ ones that go up the left edge and across the top edge of the rectangle, you should obtain an ordinary $(n+k)$-category.

The notion of $(n\times k)$-category may seem abstruse, but they actually occur quite naturally. Let’s unpack the definition a bit more. It’s easiest to imagine in the case $k=1$, where an $(n\times 1)$-category is just an internal $n$-category in 1-categories, so it comes with $(n+1)$ different categories $D_0, \dots, D_n$ related by source, target, unit, and composition maps (this is what Batanin calls a monoidal $n$-globular category). Now we are supposed to get an $(n+1)$-category by discarding the morphisms of $D_0, D_1,\dots,D_{n-1}$ and keeping only their objects, but also keeping both the objects and morphisms of $D_n$.

When $n=1$, this just means discarding the vertical arrows in a double category to get a bicategory. When $n=2$, a 2x1-category becomes a little more complicated, but if $D_0$ is the terminal category, then it reduces to simply a monoidal double category. Thus, the actual theorem I mentioned above is a special case of the putative theorem for 2x1-categories. But there are also non-degenerate 2x1-categories, such as the one in which $D_0$ consists of commutative rings and ring homomorphisms, $D_1$ consists of two-sided algebras and algebra homomorphisms, and $D_2$ consists of bimodules and bimodule maps. The underlying tricategory of this consists of commutative rings, two-sided algebras, bimodules, and bimodule maps. I believe that the “conformal nets” of Arthur Bartels, Chris Douglas, and André Henriques are supposed to be a similar structure but which is only “braided” at the bottom level, rather than fully commutative.

We can also write down some naturally occurring 1x2-categories; such a thing consists of 2-categories $D_0$ and $D_1$ related by source, target, unit, and composition maps. Suppose we take $D_0$ to be the 2-category of monoidal categories, monoidal functors, and transformations, and $D_1$ to consist of “bimodules” over these—not in the sense of profunctors, but meaning categories with an associative action on both sides by a pair of monoidal categories. Then we have a 1x2-category, which again has an underlying tricategory consisting of monoidal categories, bimodules, bimodule functors, and bimodule transformations.

Finally, this sort of thing also occurs in $(\infty,n)$-category theory. In particular, a Segal space can be viewed as a (weak) internal category in spaces, i.e. a $(1\times (\infty,0))$-category. The “completion” process which takes a Segal space to a complete Segal space, which is a sort of $(\infty,1)$-category, can then be viewed as a version of the above theorem (bearing in mind that $1 + (\infty,0) = (\infty,1)$). Similarly, an $n$-fold Segal space is like an internal $n$-fold category in spaces, i.e. a $(1\times 1\times \dots\times 1\times (\infty,0))$-category, and “completion” associates to this its corresponding $1+1+\dots+1+(\infty,0) = (\infty,n)$ category.

In all of these cases, you can probably already guess that I think it’s often better to just work with the $(n\times k)$-category as the fundamental structure. However, sometimes you really do want to forget the extra structure and just think about the underlying $(n+k)$-category. For instance, the Baez-Dolan cobordism hypothesis asserts some universal property of the $n$-category (or $(\infty,n)$-category) of cobordisms. If we want to construct a particular invariant of manifolds, i.e. a map out of this $n$-category, then we need to fix an $n$-category as the target. Now one way to construct the $(\infty,n)$-category of cobordisms is via completion of an $n$-fold Segal space, and the target $n$-category will also often come from an $((n-1)\times k)$-category for some $k$. But the universal property belongs only to the $(\infty,n)$-category of cobordisms itself, so the invariant we want only lives (a priori) at the level of $(\infty,n)$-categories. Thus, it’s important to understand the construction which takes an $(n\times k)$-category to an $(n+k)$-category.

Posted at April 8, 2010 6:01 AM UTC

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### Re: Symmetric Monoidal Bicategories and (n x k)-categories

Trivial comment:

(n,0)-cells at the top left

I would have expected the top left to have x-coordinate’ 0
or is this another example of categorical contrariness?

Posted by: jim stasheff on April 8, 2010 1:05 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n x k)-categories

I would have expected the top left to have x-coordinate’; 0

I didn’t really think carefully about naming of the cells; I just wanted to say something so that I could say which cells the operation discards and which it keeps. That said, I do tend to think of the $n$ as going “up” and the $k$ as going “sideways,” I’m not sure why – it never even occurred to me to compare that convention to the one on coordinate axes for graphs.

Posted by: Mike Shulman on April 8, 2010 3:42 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n x k)-categories

A better comparison is to entries in a matrix, which I believe we do often write as (row,column).

Posted by: Mike Shulman on April 9, 2010 6:52 AM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n x k)-categories

There is a nice coincidence of terminology here too. Peter Lumsdaine pointed out to me a little while ago that when you start thinking about $\mathbb{Z}$-categories and $\omega^{op}$-categories, it becomes natural to actually define $L$-categories for any linear order $L$. You can view “$n$-category” as a special case of this if $n$ refers to the linear order $(0 \lt 1 \lt \dots \lt n)$, as it does for instance in defining the simplex category (so perhaps we should technically speaking say $[n]$-category instead!). But if $1$ means $(0\lt 1)$, then $1\times 1$ is the partial order $\array{(1,0) & \overset{}{\to} & (1,1)\\ \uparrow && \uparrow\\ (0,0)& \underset{}{\to} & (0,1)}$ which is exactly what I would expect as an “indexing shape” for the notion of double category, and similarly for $n\times k$, $1\times 1\times 1$, and so on.

The snag is that as far as we could see, it’s not really clear how to define a $P$-category for a general partial order $P$. Or even whether partial orders are the right level of generality. Any thoughts?

Posted by: Mike Shulman on April 8, 2010 4:40 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n x k)-categories

If you allow $L$-categories for any linear order, what do $\mathbb{Q}$-categories and $\mathbb{R}$-categories look like?

Posted by: David Corfield on April 16, 2010 12:23 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n x k)-categories

Like not much that I’ve ever seen before. An $L$-category consists of sets $C_n$ for each $n\in L$, source and target maps $s_{n m}, t_{m n}: C_n \to C_m$ whenever $m\lt n$, satisfying the globularity conditions whenever $k\lt m\lt n$, along with composition operations $\circ_m : C_n\times_{C_m} C_n \to C_n$, and identities, which are associative and satisfy the interchange law whenever $k\lt m\lt n$. So a $\mathbb{Q}$-category has a set of $q$-cells for every rational number $q$, which have sources and targets of “dimension” $r$, and can be composed along common $r$-boundaries, for all $r\lt q$.

Posted by: Mike Shulman on April 16, 2010 9:37 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n x k)-categories

And this definition works also for orders which are not linear, so there are a lot of small examples on which to test it.
Posted by: Marc Olschok on April 17, 2010 8:56 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n x k)-categories

I actually don’t know how to state this definition for a non-linear order. The parts I mentioned above work, but the missing question is:

If $n\lt m$ and you compose two $m$-cells $x$ and $y$ along an $n$-cell, what are the $k$-source and target of the resulting $m$-cell?

In a total order, you can say: if $k\lt n$, then $x$ and $y$ must have the same $k$-source and target by the globularity conditions, so their $m$-composite $x\circ_m y$ must have same source as those. On the other hand, if $n\lt k\lt m$, then the $k$-source of $x\circ_m y$ should be the $m$-composite of the $k$-sources of $x$ and $y$, which are $m$-composable again by the globularity conditions. But if $k$ is incomparable to $n$, then I don’t know what to do.

Note that in a double category, a (1,1)-cell (i.e. a square) actually has four (0,0)-cells (i.e. objects) in its boundary. That makes me wonder whether instead of indexing things on posets, we are really indexing them maybe on free categories generated by directed graphs.

What happens in a double category, by the way, is that when I compose two (1,1)-cells $x$ and $y$ along a (0,1)-cell, the (1,0)-source and target of the result are the composite of the (1,0)-source and target of $x$ and $y$ along a (0,0)-cell. So it’s not written entirely in terms of $n=(1,1)$, $m=(0,1)$, and $k=(1,0)$; somehow another index (0,0) sneaks in there, so how is it being determined? Is it the meet of (0,1) and (1,0)? I haven’t thought about this much, though, just basically to this point and then I decided I had more important things to do. (-:

Posted by: Mike Shulman on April 18, 2010 12:38 AM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n x k)-categories

Here’s an incredibly important question. Why, in your paper, do you say ‘$n \times k$ category’ but then avoid using the beautiful TeX $\times$ symbol when you speak of, say ‘1x2 categories’?

I really dislike using ‘x’ to mean ‘$\times$’ when prettier fonts are available. If a normal person did this I’d just assume they were being a lazy, ignorant slob. But you’re so perfectionistic that I figure you must have an entire philosophy lurking behind this decision!

I can make up a story about why wrote ‘n x k’ in the title of this blog entry: namely, because you noticed that TeX doesn’t work in the titles, and didn’t notice that HTML does.

(You could have used &times;.)

But in the TeX version of your paper…???

Posted by: John Baez on April 8, 2010 6:29 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and n×k-categories

If a normal person did this I’d just assume they were being a lazy, ignorant slob. But you’re so perfectionistic that I figure you must have an entire philosophy lurking behind this decision!

Hahaha! But can’t we all have our lazy and ignorant days?

I can make up a story about why wrote ‘n x k’ in the title of this blog entry: namely, because you noticed that TeX doesn’t work in the titles, and didn’t notice that HTML does.

And unlike many stories that one might make up, that one is actually true. Thanks for the tip about &times;.

But in the TeX version of your paper…

So, I did have a reason for doing it that way, as you guessed, but it’s nothing so deep as an entire philosophy, and I just now realized that there’s a better solution. The default spacing that TeX puts around \times made “$(1\times 2)$-category” look quite ugly to my eyes, uglier than “1x2-category” (Your Aesthetics May Vary, of course). With “$(n\times k)$-category” I didn’t have much of a choice, though, because even I was unwilling to use a lowercase “x” between two other lowercase letters — and it also didn’t look quite as bad for some reason. Finally, that extra space around \times also meant I basically needed parentheses around $(1\times 2)$, and I preferred to be able to write “1x2-category” without parentheses.

However, this paper was actually written a little while ago, and I just got around to putting it up; and in the interim I’ve learned some more TeX that I should have applied in revising it. Namely, I think that if I write \mathord{\times} instead of \times then the spacing becomes much more palatable, and I could even throw in some \!s.

Thanks for drawing my attention to this; when I update the paper I’ll improve the notation too. Since I am perfectionistic, I do think this is important; concepts with ugly-looking or ugly-sounding names are less likely to catch on than ones with natural and beautiful names.

Posted by: Mike Shulman on April 8, 2010 9:23 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n×k)-categories

Hi Mike,

One comment: I would think “fibrant” were better reserved for the case where one need have horizontal liftings only of vertical isomorphisms. That is all that is necessary for the proofs, and makes some kind of sense, in that such things are the fibrant objects in the Reedy model structure on $\mathbf{Cat}^{\bullet \rightrightarrows \bullet}$ (which, who knows, with a bit of luck and a good wind might lift to $\mathbf{DblCat}$).

Posted by: Richard on April 8, 2010 9:59 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n×k)-categories

So, certainly the notion of “fibrant” depends on the model structure you have in mind. The existence of the putative model structure you mention is at least a little dubious, since it is shown in theorem 7.2 of this paper that the Reedy model structure on $Cat^{\Delta^{op}}$ does not transfer to $DblCat$. But of course one can talk about the transferred fibrations and fibrant objects regardless of whether the model structure exists.

On the other hand, the objects which I’m calling “fibrant” are also characterized by a right lifting property, against the inclusions of the walking vertical arrow into the walking companion pair and the walking conjoint pair. So I can imagine that they might also be the fibrant objects in some other model structure, although so far all my efforts to construct such a model structure have been unsuccessful. Moreover, they are precisely the double categories for which $(s,t)\colon D_1 \to D_0\times D_0$ is a Grothendieck fibration, or equivalently (in this case) an opfibration. Since the lifting of isomorphisms only is equivalent to $(s,t)$ being an isofibration, I decided that it was reasonable to call the one “fibrant” and the other “isofibrant” (see Remark 3.22 in the paper).

However, if majority opinion is on the side of calling isofibrant things “fibrant,” then I might be swayed. But the ones with liftings of all vertical arrows don’t otherwise have a good name of the form “XXX double categories,” although they can equivalently be called “framed bicategories” or “proarrow equipments.” I suppose that may be a reason not to introduce another name for them, but neither of those names generalizes well to the case of n×k-categories.

Posted by: Mike Shulman on April 8, 2010 10:27 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n×k)-categories

I am wondering how this paper and that of Garner and Gurski affect *my* life beyond the obvious of providing more reading on the flight to Minnesota this evening.

I am pretty sure that the free symmetric monoidal 2-category with duals on one self-dual object generator is the same thing as the category of generic surfaces in 3-space, and I would be willing to bet that the Baez-Langford proof goes through with “braided” replaced by “symmetric” uniformly. Is a “symmetric monoidal bicategory” a “symmetry monoidal 2-category?” Or am I asking for a spoiler?

Posted by: Scott Carter on April 9, 2010 2:54 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n×k)-categories

I am pretty sure that the free symmetric monoidal 2-category with duals on one self-dual object generator is the same thing as the category of generic surfaces in 3-space

Don’t you need your surfaces to live in some higher-dimensional space in order to get a symmetric monoidal 2-category? The Baez-Langford result is about surfaces in 4-space (I don’t think surfaces in 3-space are even braided), and the Tangle Hypothesis would suggest that surfaces in 5-space would be sylleptic and only surfaces in (6+)-space would be symmetric.

Is a “symmetric monoidal bicategory” a “symmetry monoidal 2-category?”

A bicategory is a weak 2-category; every bicategory is equivalent to a strict 2-category. Likewise, a symmetric monoidal bicategory is the fully weak notion (tensor product associative and commutative only up to equivalence, etc.), while a symmetric monoidal 2-category can exist at many levels of strictness, but the “semistrict” version is most common; any symmetric monoidal bicategory is I think equivalent to a semistrict one. So a symmetric monoidal 2-category is, in particular, a symmetric monoidal bicategory. Does that answer your question?

Posted by: Mike Shulman on April 9, 2010 5:24 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n×k)-categories

If the surfaces were embedded in a higher dimensional space, you can project them into 3-space. And so the symmetric picture becomes the projection of the braided picture. At the level of groups, the permutation group is a quotient of the braid group and so the standard braid generator projects to an X, or an ×.

You wrote, “A bicategory is a weak 2-category; every bicategory is equivalent to a strict 2-category. Likewise, a symmetric monoidal bicategory is the fully weak notion (tensor product associative and commutative only up to equivalence, etc.), while a symmetric monoidal 2-category can exist at many levels of strictness, but the “semistrict” version is most common; any symmetric monoidal bicategory is I think equivalent to a semistrict one. So a symmetric monoidal 2-category is, in particular, a symmetric monoidal bicategory. Does that answer your question?” (old school)

Yes, I think that is exactly what I was questioning.

Posted by: Scott Carter on April 9, 2010 7:58 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n×k)-categories

If the surfaces were embedded in a higher dimensional space, you can project them into 3-space.

You can project into 3-space, but they’ll no longer be embedded, only immersed. When someone talks about “surfaces in 3-space” I generally assume they mean embedded surfaces. If you meant immersed ones, then I agree that you should get a free symmetric monoidal 2-category, as long as you deal with all the technicalities right.

And so the symmetric picture becomes the projection of the braided picture. At the level of groups, the permutation group is a quotient of the braid group and so the standard braid generator projects to an X, or an $\times$

This seems to me like going in the opposite direction. Braided surfaces live naturally in 4-space, symmetric ones in 6-space, and we can’t project from 4-space to 6-space.

Posted by: Mike Shulman on April 9, 2010 8:40 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n×k)-categories

About to go to the airport. I didn’t say “immersed” since they can have branched points. If the surface is embedded in HD, as it is projected it will generally acquire multiple points. I don’t think that we are in significant disagreement. We project from 6 into 3. All of this is related, BTW, to the cobordism group of surfaces.

Posted by: Scott Carter on April 9, 2010 9:47 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n×k)-categories

Of course, in addition to self-intersections produced by projection into lower dimensions, there are also branch points corresponding to the 1-cells and 2-cells in question—but that’s intrinsic to the topology of the “surface” rather than having anything to do with how it sits it n-space. If the word “immersed” is only ever used for non-singular manifolds, then I guess my usage was wrong. Regardless, yes, I think we are saying the same thing now.

Posted by: Mike Shulman on April 9, 2010 10:17 PM | Permalink | Reply to this

### Stallings; Re: Symmetric Monoidal Bicategories and (n×k)-categories

I’m just trying to visualize this for finite graphs, in the context of Theorem (Stallings): For any immersion f from a finite graph D to G there is a finite-sheeted covering space D’ of G that extends f. More precisely, there is an embedding of D into D’ and the restriction of the covering map to D coincides with f.

Posted by: Jonathan Vos Post on April 10, 2010 2:09 AM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n×k)-categories

Well, actually branch points do have a lot to do with how the surface sits in higher dimensional space! For example take a closed (necessarily non-orientable) surface of odd Euler characteristic, C. If it is embedded in 4-space, then its projection into 3-space will have 2C branch points. Massey’s theorem (Whitney’s conjecture) says that the lifts provide local information about the normal Euler class. For example, a projective plane in 4-space will have either 2 positive or 2-negative branch points. (I am speaking about the DIAGRAM rather than the projection).

Meanwhile Boy’s surface has no branch points; it does not lift to an embedding in 4-space, but does lift to an embedding in 5-space.

In general, the branch points of a diagram of a surface can be used to compute the normal Euler class because one can push the surface off of itself, and compute an intersection number in terms of some local linking number.

Sorry that this is so far from the thread.

Posted by: Scott Carter on April 11, 2010 7:47 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n×k)-categories

I was using “branch points” to refer to aspects of the intrinsic topology of the surface, and “self-intersections” to refer to aspects of how it immerses into some ambient space. Of course if you just draw a picture of an immersed singular surface, it is not clear which are which; you need to be given the singular surface by itself, along with its immersion as a separate datum, in order to tell the difference.

Posted by: Mike Shulman on April 12, 2010 5:14 AM | Permalink | Reply to this

### Whitney, Massey, Cohen; Re: Symmetric Monoidal Bicategories and (n×k)-categories

In differential topology, after the Whitney immersion theorem, Massey went on to prove that every n-dimensional manifold is cobordant to a manifold that immerses in R^(2n - a(n)) where a(n) is the number of 1’s that appear in the binary expansion of n. This was proved by Ralph Cohen (1985).

a(n)=a([n/2])+n; also denominators in expansion of 1/sqrt(1-x) are 2^a(n); also 2n - number of 1’s in binary expansion of 2n.

Posted by: Jonathan Vos Post on April 12, 2010 2:15 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n×k)-categories

Ah, nice idea. I was just going to ask if you had thought about this in relation to Cartesian Bicategories II, but see that you do so in the note. Their cartesian structure (living on MapB), which they show becomes a symmetric monoidal bicategory, always seemed more natural to me at the double level.

Posted by: Geoff Cruttwell on April 10, 2010 4:53 AM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n×k)-categories

Yes, the whole theory of cartesian bicategories has always seemed to me like an overly complicated way of getting at something which is much easier to express in double-categorical language. But possibly I’m missing something.

Posted by: Mike Shulman on April 10, 2010 5:13 AM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n×k)-categories

Anything to be gained from the definition of (cartesian) monoidal $(\infty,1)$-category

No, I don’t think so. A cartesian bicategory is very different from a cartesian monoidal bicategory. It’s supposed to capture examples like $Rel(C)$ and $Span(C)$, whose monoidal structure is the cartesian product of objects in the category $C$, but is not a cartesian product in the bicategory in question. This is what creates all the complication. Hence why it is useful to consider them as double categories, where the morphisms in $C$ are included as basic data and the cartesian monoidal property can easily be expressed.

Posted by: Mike Shulman on April 10, 2010 4:17 PM | Permalink | Reply to this

### Re: Symmetric Monoidal Bicategories and (n×k)-categories

the whole theory of cartesian bicategories has always seemed to me like an overly complicated way of getting at something which is much easier to express

Anything to be gained from the definition of (cartesian) monoidal $(\infty,1)$-category?

With Thomas Nikolaus we have come to believe that the following simple definition is right: a symmetric monoidal $(\infty,1)$-category is a dendroidal set that has # all inner dendroidal horn fillers and all outer corolla horn fillers.

That’s at least a simple, tractable definition.

Posted by: Urs Schreiber on April 10, 2010 1:35 PM | Permalink | Reply to this

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