# The Unapologetic Mathematician

## Braidings on Span 2-Categories

Now that we can add a monoidal structure to our 2-category of spans, we want to add something like a braiding.

So, what’s a braided monoidal 2-category? Again, we need some data:

• For any objects $A$ and $B$ an equivalence $R_{A,B}:A\otimes B\rightarrow B\otimes A$ called the “braiding” of $A$ and $B$.
• For any 1-morphism $f:A\rightarrow A'$ and object $B$, a 2-isomorphism $R_{f,B}:R_{A',B}\circ(f\otimes B)\Rightarrow(B\otimes f)\circ R_{A,B}$ called the braiding of $f$ and $B$.
• For any object $A$ and morphism $g:B\rightarrow B'$, a 2-isomorphism $R_{A,g}:R_{A,B'}\circ(A\otimes g)\Rightarrow(g\otimes A)\circ R_{A,B}$ called the braiding of $A$ and $g$.
• For any objects $A$, $B$, and $C$, 2-isomorphisms $\tilde{R}_{(A|B,C)}:(B\otimes R_{A,C})\circ(R_{A,B}\otimes C)\Rightarrow R_{A,B\otimes C}$ and $\tilde{R}_{(A,B|C)}:(R_{A,C}\otimes B)\circ(A\otimes R_{B,C})\Rightarrow R_{A\otimes B,C}$ called the “braiding coherence 2-morphisms”.

We define the braiding to be an equivalence rather than an isomorphism because we don’t want to ask it to be exactly invertible. That is, there will be another 1-morphism $R^*_{A,B}:B\otimes A\rightarrow A\otimes B$ with 2-isomorphisms $R^*_{A,B}\circ R_{A,B}\cong1_{A\otimes B}$ and $R_{A,B}\circ R^*_{A,B}\cong1_{B\otimes A}$. The 2-isomorphisms braiding objects with 1-morphisms act in the place of naturality relations, allowing us to pull 1-morphisms back and forth through the object braiding. And then the hexagon identities from the definition of a braiding in a 1-category are now weakened to the braiding coherence 2-morphisms.

So, where are we going to find this structure for our bicategory $\mathbf{Span}(\mathcal{C})$? Well, let’s assume we have a braiding $\beta_{A,B}$ on $\mathcal{C}$. Then we’ll just define $R_{A,B}$ to be the span $(A\otimes B)\stackrel{1_{A\otimes B}}{\leftarrow}(A\otimes B)\stackrel{\beta_{A,B}}{\rightarrow}(B\otimes A)$. This is actually a 1-isomorphism, using the obvious inverse.

As we saw before with the tensorator, it turns out that we can set all of the $R_{f,B}$ and the $R_{A,g}$ to be the appropriate identities, as well as all the braiding coherence relations. Requiring the the monoidal product on $\mathcal{C}$ preserve pullbacks turns out to be an extremely powerful condition!

And now what are the conditions that make this data into a braided monoidal 2-category?

1. For 1-morphisms $f:A\rightarrow A'$ and $g:B\rightarrow B'$, we have
$(\bigotimes_{g,f}\circ R_{A,B})\bullet((g\otimes A')\circ R_{f,B})\bullet(R_{A',g}\circ(f\otimes B))=$
$((B'\otimes f)\circ R_{A,g})\bullet(R_{f,B'}\circ(A\otimes g))\bullet(R_{A',B'}\circ\bigotimes_{f,g}^{-1})$
2. For any 1-morphisms $f:A\rightarrow A'$ and $f':A\rightarrow A'$, 2-morphism $\alpha:f\Rightarrow f'$, and object $B$, we have
$((B\otimes\alpha)\circ R_{A,B})\bullet R_{f,B}=R_{f',B}\bullet(R_{A',B}\circ(\alpha\otimes B))$
3. For any 1-morphisms $g:B\rightarrow B'$ and $g':B\rightarrow B'$, 2-morphism $\beta:g\Rightarrow g'$, and object $A$, we have
$((A\otimes\beta)\circ R_{A,B})\bullet R_{A,g}=R_{A,g'}\bullet(R_{A,B'}\circ(\beta\otimes A))$
4. For any 1-morphisms $f:A\rightarrow A'$ and $f':A'\rightarrow A''$ and object $B$, we have
$((B\otimes f')\circ R_{f,B})\bullet(R_{f',B}\circ(f\otimes B))=R_{f'\circ f,B}$
5. For any 1-morphisms $g:B\rightarrow B'$ and $g':B'\rightarrow B''$ and object $A$, we have
$((A\otimes g')\circ R_{A,g})\bullet(R_{A,g'}\circ(A\otimes g))=R_{A,g'\circ g}$
6. For any objects $A$, $B$, and $C$, and 1-morphism $f:C\rightarrow C'$, we have
$R_{A,B\otimes f}\bullet(\tilde{R}_{(A|B,C')}\circ(A\otimes B\otimes f))=$
$((B\otimes f\otimes A)\circ\tilde{R}_{(A|B,C)})\bullet((B\otimes R_{A,f})\circ(R_{A,B}\otimes C))\bullet((B\otimes R_{A,C'})\circ\bigotimes_{R_{A,B},f})$
7. There are five more conditions like the last one, with the 1-morphism in other slots and different associations of the three terms.
8. For any objects $A$, $B$, $C$, $D$, we have
$\tilde{R}_{(A|B,C\otimes D)}\bullet((B\otimes\tilde{R}_{(A|C,D)})\circ(R_{A,B}\otimes C\otimes D))=$
$\tilde{R}_{(A|B\otimes C,D)}\bullet((B\otimes C\otimes R_{A,D})\circ(\tilde{R}_{(A|B,C)}\otimes D))$
9. There are two more conditions like the last one, corresponding to different ways of associating the four terms.
10. For any objects $A$, $B$, and $C$, we have
$(\tilde{R}_{(A|B,C)}\circ(A\otimes R_{B,C}))\bullet R_{A,R_{B,C}}^{-1}\bullet((R_{B,C}\otimes A)\circ\tilde{R}_{(A|B,C)}^{-1})=$
$((C\otimes R_{A,B})\circ\tilde{R}_{(A,B|C)}^{-1})\bullet R_{R_{A,B},C}\bullet(\tilde{R}_{(A,B|C)}\circ(R_{A,B}\otimes C))$.
11. $R_{\cdot,\cdot}$, $\tilde{R}_{(\cdot|\cdot,\cdot)}$, and $\tilde{R}_{(\cdot,\cdot|\cdot)}$ are each the identity whenever one of their slots is filled with the unit object $\mathbf{1}$.

Now in our currrent situation, almost all the structure here is trivial, and we have proofs to match! Specifically, all of the 2-morphisms that show up here are just the identities for various 1-morphisms. That is, they move between different ways of writing the same 1-morphism. For example, the first condition just reduces to saying that the composite of three applications of the identity 2-morphism on the span $R_{A',B'}\circ g\circ f$ are the same as three other applications of the identity on the same span. And on and on they go, identities on identities, and there’s ultimately nothing to do here.

So the upshot is that if we have a braiding on $\mathcal{C}$, then $\mathbf{Span}(\mathcal{C})$ is a braided monoidal 2-category. Dually, $\mathbf{CoSpan}(\mathcal{C})$ gets this structure if $\mathcal{C}$ is a braided monoidal category with pushouts preserved by the monoidal product.

October 12, 2007 - Posted by | Category theory

1. John,

braid diagrams and movie diagrams so that we can understand this in terms of the FREE STRICT BRAIDED MONOIDAL 2-CAT
with duals. And when you find the duals, jump up and down.

Scott

Comment by Scott Carter | October 12, 2007 | Reply

2. Scott, I’m unsure whether these will be at all useful or not, especially because so much of the structure is trivial. But when I try to sew these posts into a paper I’ll definitely try to put them into surface diagrams.

Alternately, I’m looking at what happens when I relax the requirement that the monoidal structures preserve pullbacks. Unfortunately, it gets… weird. We can talk about it more when I come by Mobile.

As for the duals, though, they’re on their way…

Comment by John Armstrong | October 12, 2007 | Reply

3. Hello,

The data that you require for the braiding of 2 categories are same as Kapranov and Voevodsky’s on “2-Categories and Zamolodchikov Tetrahedra Equations”. In this paper, they also define 2 2-morphisms and call them Z-systems. Basically, they represent the interplay between R_{A,B}, R_{B,C}, R_{A,C} for any 3 objects A,B,C.

At the same time, Lawrence Breen, in the 8th chapter of his book, titled “On the classification of 2-gerbes and 2-stacks”, suggests to add the data of Z-systems to the data that you require for braiding.

Now, my questions are;
Why is this difference in definitons? and What do these Z-system data add new to the braiding?

Thank You,

Emin

Comment by Emin | January 26, 2008 | Reply

4. Emin, you’re right, they are the same as Kapranov and Voevodsky’s definition. I’ve got a reprint of that paper that Kapranov gave to me right here in fact, though I’m actually pulling the definitions out of Baez and Langford’s Higher Dimensional Algebra IV: 2-Tangles.

As for Breen’s definition, I honestly don’t know. I’ve not seen that definition, nor have I read the book in question, and so I’m not even sure what the additional structure consists of.

Comment by John Armstrong | January 26, 2008 | Reply

5. I ought to pull out the Kapranov-Voevodsky paper, but are you sure they’re absolutely the same? My memory from back in the day is that KV omitted an axiom for the braiding which was later pointed out by others (e.g., Sjoerd Crans), and perhaps also axioms for the unit data — by the time Baez and Langford came out, it had all gotten sorted out. (Could the missing braiding axiom be what is denoted $(\bullet \otimes \bullet) \otimes (\bullet \otimes \bullet)$ in Baez-Langford? There’s a coherence axiom which just about everyone misses the first time around; it might be this.)

If I’m mistaken, I’d like to hear!

Comment by Todd Trimble | January 26, 2008 | Reply

6. Not absolutely the same, no.. I suppose I’m fudging equality here a bit in casual conversation. KV overlooked a couple things that were caught by the time of BL, but those are just conditions similar to the long list they already had. What Emin is suggesting is a whole new sort of data, which in my mind changes the definition far more than just adding in a coherence condition.

Comment by John Armstrong | January 26, 2008 | Reply

7. Okay, data as opposed to equations — I read too quickly. So wait, I’m confused: there are no data in Baez-Langford which correspond to data in things called Z-systems?

Comment by Todd Trimble | January 26, 2008 | Reply

8. I’m really not sure what a “Z-system” is. All the data BL require is listed above.

Comment by John Armstrong | January 26, 2008 | Reply

9. Hello again,

Sorry for my late reaction and thank you for pointing me out HDA series. I guess I found some answers to my questions in HDA1 which led to more questions. I think what is called as Z-systems by Breen is called Yang-Baxter hexagons in definition6 of HDA1 and it is required that the 2-morphisms, denoted S^{+}_{A,B,C} and S^{-}_{A,B,C}, defining these hexagons should match. However this condition is left out in the list of data for braiding in HDA4. I don’t know why.

Comment by Emin | January 31, 2008 | Reply

10. Emin, first off this is a coherence condition, not data. Secondly, this coherence condition is exactly my condition 11 in the list above.

Comment by John Armstrong | January 31, 2008 | Reply