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• CommentRowNumber1.
• CommentAuthorEric
• CommentTimeApr 14th 2010

The other night, I was drawing some more pictures and drew something like this:

$\array{\bullet &\rightarrow &\bullet &&\\\downarrow &&&\searrow&\\\bullet &&&&\bullet\\&\searrow&&&\downarrow\\&&\bullet&\rightarrow&\bullet}.$

With this, you can play a game “fill the shape with directed triangles”. Here is one way to fill the shape:

$\array{\bullet &\rightarrow &\bullet &&\\\downarrow &\mathllap{{}^\seArrow}{\nearrow}&\Downarrow &\searrow&\\\bullet &&\rightarrow&&\bullet\\&\searrow&\swArrow&\underset{\seArrow}{\swarrow}&\downarrow\\&&\bullet&\rightarrow&\bullet}.$

There are a bunch of other ways to fill the shape with directed triangles. Do the coherence laws of a 2-category basically say that all the ways to fill the shape with directed triangles are the same?

That would be kind of cool because you’d be basically trying to define a surface and the coherence laws would tell you it doesn’t matter how you triangulate the surface.

As far as shapes, does the exchange law say that if there are two downward 2-morphisms connecting parallel morphisms as in the following shape

$\bullet\stackrel{\curvearrowright}{\to}\bullet\stackrel{\curvearrowright}{\to}\bullet$

then this is equal to the shape

$\bullet\stackrel{\curvearrowright}{\to}\bullet$

with a single downward arrow connecting the parallel morphisms?

• CommentRowNumber2.
• CommentAuthorHarry Gindi
• CommentTimeApr 14th 2010
• (edited Apr 14th 2010)

The first diagram you drew is the diagram for a dinatural transformation. So my guess is "no". It is also important to note that in a general 2-category you have to use globular sets rather than simplicial sets to represent your stuff. (Of course, there are ways to make simplicial sets do this job, but it gets pretty combinatorial (c.f. Street's definition of an ω-category).

This is because all of the higher morphisms in a simplicial set (oops! quasicategory) are isomorphisms, so in particular, it doesn't matter which direction you draw them.

In a general bicategory, this is completely not the case. Your diagrams work for (n,1)-categories, but that's it.

• CommentRowNumber3.
• CommentAuthorTim_Porter
• CommentTimeApr 14th 2010

@Harry OOPS! `all of the higher morphisms in a simplicial set are isomorphism'!!! That does not make sense as you state it.

• CommentRowNumber4.
• CommentAuthorMike Shulman
• CommentTimeApr 14th 2010

@Eric, I’m not even sure what is meant by “the same.” If you fill the shape with directed triangles in two different ways, then you’re drawing two different “pasting diagrams” in the language of 2-categories. In a particular 2-category K, you can compose a pasting diagram if its 1-edges are labeled by 1-morphisms in K and its 2-cells are labeled by 2-morphisms in K (and if the directions of everything are compatible). But if you have two different pasting diagrams, even if they have the same “boundary,” then there is no a priori reason why you should be able to label them with 2-cells that have anything to do with each other, or even whether you could label them both at all in a given 2-category. Likewise, in your second question, the notion of “equality of shapes” has no meaning that I can think of.

@Harry, dinatural transformations don’t seem at all relevant to me, since the hexagon there has no 2-cells in it (and doesn’t even live in a 2-category). Not every hexagon is the same as every other hexagon. And invertibility also doesn’t really seem relevant, unless you are worried about the arrows pointing in the wrong direction to be composed with each other.

• CommentRowNumber5.
• CommentAuthorHarry Gindi
• CommentTimeApr 14th 2010
• (edited Apr 14th 2010)

@Mike: 2-morphisms in a quasicategory are just given by filling a triangle. If you look at Eric's filled in diagram, it makes sense as a diagram in a quasicategory, but in a bicategory, the top left 2-cell he's given doesn't make sense. In a bicategory, obviously 2-cells have to go between 1-cells, but his arrow is between a 0-cell and a 1-cell.

@Tim: Fixed!

• CommentRowNumber6.
• CommentAuthorMike Shulman
• CommentTimeApr 14th 2010
• (edited Apr 14th 2010)

@Harry, it makes perfect sense as a “pasting diagram” in a 2-category. If you read any papers on 2-categories or bicategories, there will be lots of such pasting diagrams with 2-cells that go from (the composite of) a string of 1-cells to another such. Although really it should be angled to the left, so that it goes from a 1-cell to a composite of two 1-cells. Likewise the lower right arrow should be angled to the left.

• CommentRowNumber7.
• CommentAuthorHarry Gindi
• CommentTimeApr 14th 2010
• (edited Apr 14th 2010)

Absolutely! I was saying that the direction of the 2-arrow matters unless we're in a quasicategory. The 2-arrow doesn't say "this diagram commutes". It is literally a morphism between (in this case) the composition of two arrows and a third arrow. What I'm saying is that if we take out all of the extra "stuff" (in particular, replacing the two 1-cells with their composition, the diagram will be a globular diagram.

I'm pretty sure we're talking about the same thing.

• CommentRowNumber8.
• CommentAuthorMike Shulman
• CommentTimeApr 14th 2010

Okay – but the globular/simplicial issue is irrelevant. If you’re in a (2,1)-category, then you can compose 2-arrows any which way, while in a general 2-category the direction matters, but it’s irrelevant whether the categories are globular or simplicial or opetopic or whatever.

• CommentRowNumber9.
• CommentAuthorHarry Gindi
• CommentTimeApr 14th 2010
• (edited Apr 14th 2010)

Absolutely. I apologize for being unclear.

Edit: Although I don't really know what a simplicial bicategory actually looks like. When I think of bicategories, they are always globular for all intents and purposes.

• CommentRowNumber10.
• CommentAuthorMike Shulman
• CommentTimeApr 14th 2010

A simplicially-shaped (2,1)-category can be defined as a quasicategory which is “2-truncated” in a suitable sense. If you want noninvertible 2-cells, then you have to get more delicate, as in Street’s definition of $\omega$-category you referred to. The word “bicategory” usually does mean a globular shape, but we can still draw arbitrary pasting diagrams because we have composites. You can make that formal by saying things about computads.

• CommentRowNumber11.
• CommentAuthorHarry Gindi
• CommentTimeApr 14th 2010

Yes, then we were exactly in agreement, but I was unclear. Sorry for the confusion again!

• CommentRowNumber12.
• CommentAuthorDavidRoberts
• CommentTimeApr 15th 2010

Getting back to Eric's question,

Do the coherence laws of a 2-category basically say that all the ways to fill the shape with directed triangles are the same?

if by this you mean "all the ways to fill the shape with associators, unitors (and etc - note these are all invertible, however) are the same?" Then the answer is 'yes'. Note that given an arbitrary 'shell' (the outside 1-arrows) one cannot always fill it in. But when you can, all ways of doing it (with only the structural arrows, associators etc) are equal.

Unless I've completely misunderstood things. In my private 2-Klein geometry-related project there were a lot of such computations, and it was like playing with a puzzle in which you had an infinite supply of pieces of generally similar shapes just with different edges to fit into a frame.

• CommentRowNumber13.
• CommentAuthorMike Shulman
• CommentTimeApr 15th 2010

Note that given an arbitrary ’shell’ (the outside 1-arrows) one cannot always fill it in. But when you can, all ways of doing it (with only the structural arrows, associators etc) are equal.

Whether or not this is true depends on what the words mean. If you are talking about ways of filling in a diagram ’schema’ in the ’theory’ of 2-categories, or equivalently in a free 2-category, then yes, all ways of filling in any fillable shell with constrant data are equal. But in a particular 2-category, it might happen that there are particular shells which could be “filled with constraint data” in multiple inequivalent ways. For instance, you can have a bicategory in which there exists a composable string of arrows $x \overset{f}{\to} y \overset{g}{\to} z\overset{h}{\to} w$ such that $h \circ (g\circ f) = (h\circ g)\circ f$, and yet the associator $\alpha_{f,g,h}$ is not equal to the identity. In such a case, you have a “shell” that can be “filled with constraint data” in two different ways, namely by the identity (i.e. by no constraint data at all) and by the associator.

I think this is one of the subtlest points in the study of coherence theorems.

• CommentRowNumber14.
• CommentAuthorEric
• CommentTimeApr 15th 2010

Thanks David. I was hoping my question wasn’t quite as vacuous as the earlier responses started to make me fear :)

This really is like a puzzle, but I don’t have the rules completely straight yet.

I was just playing with one kind of puzzle piece, but your comment makes me think there are a lot more.

The puzzle piece I’m playing with is a filled directed triangle

$\array{\bullet&&\longrightarrow&&\bullet\\&\searrow&\Uparrow&\nearrow&\\&&\bullet&&}$

For now, I want the rules of the game to be such that the direction matters, but we can play another game later where it doesn’t matter, e.g. with $n$-groupoids.

To be continued…

I started to write down some rules, but was surely doing something wrong. What would be appropriate rules for pasting these together?

For now, I’m being sucked back into the vortex that is reality…

• CommentRowNumber15.
• CommentAuthorHarry Gindi
• CommentTimeApr 15th 2010
• (edited Apr 15th 2010)

That triangle

$\array{\bullet&&\longrightarrow&&\bullet\\&\searrow&\Uparrow&\nearrow&\\&&\bullet&&}$

is not a “commuting triangle” unless the 2-cell is an isomorphism. Also, the 2-cell is a morphism between the straight arrow and the composition of the other two arrows.

• CommentRowNumber16.
• CommentAuthorDavidRoberts
• CommentTimeApr 15th 2010

@Mike,

In the example you gave, I wouldn't count the identity 2-arrow as a legitimate option for a filler, as it isn't one of the structural arrows that come with the definition of a bicategory. I really mean only the associator and the left and right unitors, and if one has weakly invertible 1-arrows, the 2-arrows that are part of the data of those (either having had them given, or using say a result about weak inverses being part of an adjoint equivalence, but this is perhaps too much freedom).

I thought (always a problem) that MacLane's coherence theorem is that if one cooks up a diagram using these structural arrows then it commutes (as a diagram of 2-arrows). Would check but I've gotta go and do the grocery shopping :) (excuses, excuses)

• CommentRowNumber17.
• CommentAuthorEric
• CommentTimeApr 15th 2010

I had a fresh look at Street’s AOS paper on the train ride home and now see my brain was trying to reproduce what I had read there many years ago, but not doing a very good job of it :)

I feel my category kung fu is a little better this time around, so hopefully I can understand a little more if it now.

• CommentRowNumber18.
• CommentAuthorHarry Gindi
• CommentTimeApr 15th 2010
• (edited Apr 15th 2010)

My kung fu:

The game of chess is like a sword fight. You must think first before you move…
Toad style is immensely strong, and immune to nearly any weapon, when it’s properly used, it’s almost invincible.

=D

• CommentRowNumber19.
• CommentAuthorMike Shulman
• CommentTimeApr 15th 2010

@David, I was not calling the identity a structural arrow, but rather the composite of zero structural arrows. You can also cook up examples where neither composite has length zero, however. Take the same category as before, but now consider on the one hand the associator $(f\circ g) \circ h \cong f\circ (g\circ h)$, and on the other hand the composite of the associator with itself, which makes sense in this particular category since $(f\circ g) \circ h = f\circ (g\circ h)$. Since the associator is invertible, it is only idempotent if it is the identity, but it isn’t; thus we have two different parallel composites of structural arrows, both of length $\ge 1$, which are unequal.

• CommentRowNumber20.
• CommentAuthorDavidRoberts
• CommentTimeApr 16th 2010

@Mike

Ok, good. I wasn't particularly disagreeing with you, but I needed to explain myself. I should probably explain that my technique wasn't applied in the case Eric was talking about: I got the wrong end of the stick. I was pasting together coherence diagrams, not the structural arrows, to show composites of structural arrows are equal.

@Eric

sorry if what I have said has given you wonky ideas - I freely admit my higher-category-fu is not the best.

• CommentRowNumber21.
• CommentAuthorDavidRoberts
• CommentTimeApr 16th 2010

Added a little more explanation to coherence theorems around Mike’s example about the non-identity associator. Perhaps a concrete example of this that is worth mentioning (and I learned this from John B a few years back) is a skeletal 2-group that is not strict. By definition we have $(f\circ g) \circ h = f\circ (g\circ h)$ where here $f,g,h$ are arbitrary elements of a such a 2-group, but the associator is not (generally) the identity.

• CommentRowNumber22.
• CommentAuthorEric
• CommentTimeApr 16th 2010

Simple question. The exchange law IS a coherence law, right?

• CommentRowNumber23.
• CommentAuthorDavidRoberts
• CommentTimeApr 16th 2010

No: it’s a consequence of the fact horizontal composition in a 2-category (weak or strict) is a functor (recall that the hom-objects are categories). The exchange law in a Gray-category only holds up to a higher arrow, and this arrow satisfies a coherence law. In general (to my mind), a coherence law is something that is obeyed by morphisms that replace equalities. eg: throw out associativity, and so the associator has to obey a coherence law.

• CommentRowNumber24.
• CommentAuthorHarry Gindi
• CommentTimeApr 16th 2010

Unbiased definitions are so much easier to work with (but so much harder to write out).

• CommentRowNumber25.
• CommentAuthorUrs
• CommentTimeApr 16th 2010
• (edited Apr 16th 2010)

a coherence law is something that is obeyed by morphisms that replace equalities

I would (want to) say that a coherence law in an algebraic n-category like a Nikolaus algebraic quasi-category (see model structure on algebraic fibrant objects) is the reflection of the fact that

1. the underlying simplicial set is an (n+1)-coskeleton

2. therefore fillers between boundaries of (n+2)-simplices consisting of chosen (n+1)-fillers exist uniquely.

The existence of these unique fillers is the coherence law.

For instance for bicategories with n=2 the chosen associators are certain fillers of the shape of 3-simplices in the bicagtogory’s nerve that have been chosen. The nerve of the bicategory being 3-coskeletal means that every boundary of a 4-simplex in the nerve and so in particular every boundary of a 4-simplex consisting of these kinds of chosen 3-fillers, has a unique filler.

• CommentRowNumber26.
• CommentAuthorMike Shulman
• CommentTimeApr 16th 2010

@Urs, that seems to me a pretty strange way to explain it! The prototypical and intuitive (for me) notion of “algebraic n-category” is not an algebraic quasi-category, but rather something more Batanin-like. But I think the general point is valid that a coherence law in an n-category is something that would be a constraint datum in an (n+1)-category, but which becomes an identity due to trucatedness. From this point of view, the exchange law in a 2-category is a coherence law, since it becomes a constraint datum in a weak 3-category.

@Harry #24, why do you say they are harder to write out? I often find them a lot easier. For instance, to define a biased monoidal category, you have to guess or prove that the pentagon is “enough.” But to define an unbiased one, the “correct” axioms are staring you in the face, and they all tend to have much more regular shapes.

• CommentRowNumber27.
• CommentAuthorUrs
• CommentTimeApr 16th 2010
• (edited Apr 16th 2010)

that seems to me a pretty strange way to explain it!

Is it that strange, or is it rather that it is non-traditional?

It would seem to me that if your notion of n-category comes with a notion of simplicial nerve, then this is a powerful way to pretty much explain the statement of coherence laws. it also shows that what makes the coherence laws appear intricate is comes from fact that relating an algebraic n-category to its simplicial nerve may involve some intricate combinatorics. The law itself, in a sense, is not intricate.

• CommentRowNumber28.
• CommentAuthorMike Shulman
• CommentTimeApr 16th 2010

Maybe there are people nowadays who think of an n-category as fundamentally being its simplicial nerve, and for them I guess that would be an explanation. But to me, a simplicial nerve is something constructed from an n-category, or a way to model an n-category, and the fact that something happens in that constructed object or model is not an explanation of something which occurs at a more fundamental level in the n-category itself. The simplicial version may of course provide additional insight or technical convenience. But for this question, I don’t see that introducing simplices actually provides any insight or simplification, rather it makes things more intricate to my mind – isn’t it even simpler to just say that every constraint (n+1)-cell in an n-category becomes an identity, hence a coherence law?

• CommentRowNumber29.
• CommentAuthorUrs
• CommentTimeApr 16th 2010
• (edited Apr 16th 2010)

Sure, simplices shouldn’t have anything intrinsic to do with this. It’s just that for them I happen to know exactly what n-coskeletal means. For other shapes I can guess, but might have to think more.

For me that’s a general bottom line: of course there is nothing that makes simplicial techniques intrinsically better than other techniques. It’s just a fact of life that these happen to be better understood than almost all other techniques. So it helps to see what happens there, even if the more natural description might eventually look different.

• CommentRowNumber30.
• CommentAuthorMike Shulman
• CommentTimeApr 17th 2010

@David #21, I added a mention of 2-groups to coherence theorem, but I removed your parenthetical because I didn’t understand what it meant, maybe you can explain? The statement $a_{x,y,z}^{-1}a_{x,y,z} \neq a_{x,y,z}$ is precisely equivalent to $a_{x,y,z} \neq id$, so why is one more trivial than the other?

• CommentRowNumber31.
• CommentAuthorDavidRoberts
• CommentTimeApr 17th 2010

My objection above about id not being a structure morphism is trivially removed by saying that it is a composite of structure morphisms (and inverses of such) in a trivial way. But the page is ok as it stands: I was just being cantankerous yesterday.