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Suppose that $C$ is a 2-category, perhaps $C=\rm{Cat}$, the 2-category of small categories, functors, and natural transformations. Let $T$ be an object in $C$.

I form the new 1-category whose objects are morphisms $f\colon A\rightarrow T$ in $C$, and in which a morphism from $f$ to some $f'\colon A'\rightarrow T$ consists of a pair $(\phi,\phi^\sharp)$ where $\phi\colon A\rightarrow A'$ is a 1-morphism in $C$ and $\phi^\sharp\colon\phi\circ f'\rightarrow f$ is a 2-morphism between arrows $A\rightarrow T$. Call this new category the $(C\Uparrow T)$. An obvious variation comes about by reversing the direction of the 2-morphism, i.e. we could take $\phi^\sharp\colon f\rightarrow\phi\circ f'$; perhaps I might call this variation $(C\Downarrow T)$.

What is the high-brow way to refer to these strange slice-categories? How do you locate them within a good understanding of 2-categories? Where are the properties of such things discussed? What is the relation between these strange slices and the usual 2-categorical slices?

Thanks!

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3 Answers 3

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The second definition looks like the 'lax comma category' $C // T$, where a morphism $f \to f'$ is given by a 2-cell $f \to f'\phi$. The defining universal property is the same as for comma objects, except that the 2-cells in the squares are lax natural transformations. Your first definition should be the oplax version.

See Kelly, On clubs and doctrines, LNM 420, or Gray, Adjointness For 2-Categories, LNM 391, who calls these '2-comma categories'.

In more detail, Gray's 2-comma categories come from (strict, I think) 2-functors $A \overset{F}{\rightarrow} K \overset{G}{\leftarrow} B$. An object is a 1-cell $FA \to GB$, a morphism is a square with a 2-cell in, and a 2-cell is given by a pair of 2-cells in $K$ that fit into a commuting cylinder (it's pretty obvious if you draw a picture). In your example, (what I've called) $C // T$ has 2-cells $(\phi,\phi^\sharp) \Rightarrow (\psi,\psi^\sharp)$ given by 2-cells $\alpha \colon \phi \Rightarrow \psi$ such that $\psi^\sharp \circ f'\alpha = \phi^\sharp$. (Again, pictures make it much clearer!) So your slices are actually 2-categories, coming from $C \overset{1}{\rightarrow} C \overset{T}{\leftarrow} \bullet$.

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Welcome to MO, Finn! –  Tom Leinster Feb 27 '10 at 2:03
    
Thanks, Tom. Most of the questions here are way over my head, so imagine my delight/surprise when I saw one I knew the answer to! –  Finn Lawler Feb 28 '10 at 17:26
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Interesting question!

If C is a 1-category then the overcategory C/T can be described as the lax (or oplax, I forget) limit of the diagram • → C in the 2-category Cat, where the arrow is given by the object T of C. The lax limit means we ask for a universal limit cone on the diagram where the triangle is filled with a noninvertible 2-morphism. We can adapt this definition to any object C of any 2-category equipped with a map T from the terminal object.

When C is a 2-category, I believe we need to ask for a "very lax" limit, in which the triangle is not even filled by a (noninvertible) natural transformation, but only a lax (or oplax, depending on which of your constructions you want) natural transformation. As far as I can see, there is no way to perform your constructions starting only with 2Cat as a 3-category and the data of C and T; we need the extra structure of the (op)lax natural transformations in 2Cat. Moreover, there is no 3-category of 2-categories, functors, and lax natural transformations in which to take a lax limit and obtain your constructions. So, these "lax" overcategories are still rather mysterious to me.

I assume by "the usual 2-categorical slices" you mean to require the 2-morphism between $\phi \circ f'$ and $f$ to be invertible, which is the (op?)lax limit of the diagram • → C in 2Cat.

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Thanks Reid. Yes, your assumption in the last paragraph is correct. I don't want to assume that triangles commute after a choice of 2-isomorphism; I want the right to choose a non-invertible 2-morphism. –  David Spivak Feb 26 '10 at 23:41
    
Slightly off-topic question: over the last couple of years I've started hearing people use the term "overcategory", which I gather is a synonym for slice category. Now "slice category" has been around for god knows how many decades, and I guess "overcategory" is much newer, but I have no idea where it comes from. Do you know? –  Tom Leinster Feb 27 '10 at 2:08
    
I think it comes from parts of Alg. Top. and is at least as old a term as 'slice category'. I remember hearing Peter Johnstone, way back, talking about splices and thinking 'aha, that is just another word for the over category' .... but of course, my memory may be faulty on this :-) –  Tim Porter Feb 27 '10 at 7:54
    
Ah, thanks Tim. Somehow I only started hearing it recently. –  Tom Leinster Feb 28 '10 at 1:36
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For what it's worth, the construction you describe features prominently in http://front.math.ucdavis.edu/0807.4146, though that paper does not use 2-categorical language.

More generally, given a 2-category $C$ (which I usually assume is pivotal, though maybe that's not necessary here), one can construct a 1-category $D$ whose objects are 1-morphisms $f:a\to b$, and whose morphisms are "rectangles": the domain $f:a\to b$ along the bottom, the range $f':a'\to b'$ along the top, additional 1-morphisms (of $C$) $g:a\to a'$ and $h:b\to b'$ along the right and left sides, and a 2-morphism of $C$ filling in the rectangle. Composition in $D$ is given by stacking the rectangles vertically. I like to think of the pair $(g, h)$ as the (bi)grading of the morphisms of $D$. What you describe is the case where we restrict $h$ to be an identity 1-morphism (of $C$). In the paper linked to above we put an inner product on $D$ and complete it to a von Neumann algebra (in fact, a factor).

In response to David's comment below:

Modulo some details, a planar algebra is equivalent to a pivotal 2-category whose 2-morphisms are vector spaces and whose 1-morphisms are finitely generated. The standard example is constructed from a pair of factors (irreducible von Neumann algebras) $N\subset M$. From this data we construct a 2-category whose objects are $N$ and $M$, whose 1-morphisms are generated by the two bimodules $_N M_M$ and $_M M_N$, and whose 2-morphisms are intertwinors.
(So for example the 1-morphisms are $M\otimes_N M\otimes_N\cdots\otimes_N M$, thought of as either an $N$-$N$ or $N$-$M$ or $M$-$N$ or $M$-$M$ bimodule.) You can think of the usual planar algebra definition as axiomatizing the "string diagrams" you would draw for this 2-category.

The diagrams in the paper I referred to are rotated 90 degrees from my explanation above. The left and right sides of the rectangles in the paper correspond to the $f$ and $f'$ of your (David's) original question. The tops of the rectangles corresponds to your $\phi$, and the interiors of the rectangles correspond to your $\phi^\sharp$.

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Hi Kevin, I don't know enough about planar algebras to even understand what I'm seeing here perhaps, but a word search reveals that this paper doesn't mention the word "category" or "morphism." Could it possibly be the wrong paper? If it's the right one, would you point me to the construction? –  David Spivak Feb 27 '10 at 12:15
    
It's the right paper. I'll expand the original answer to address the rest of your question. –  Kevin Walker Feb 27 '10 at 15:09
    
Just to check that I've followed the translation correctly, David's T becomes the shading of the bottom of the rectangle? –  Noah Snyder Feb 27 '10 at 22:42
    
Another way to think about the connection between planar algebras and 2-categories is that the planar algebra is a way of studying a 2-category from the point of view of your favorite 1-morphism. This means that certain natural categorical notions translate poorly, but you can talk about say rotation that only make sense once you've chosen a 1-morphism. Since "subfactors" are roughly the same as "Frobenius algebra objects in pivotal categories" you don't lose much by picking that algebra object to always be your favorite object (ie 1-morphism thinking of the monoidal category as a 2-category). –  Noah Snyder Feb 27 '10 at 23:02
    
Noah: Yes, David's T becomes the shading at the bottom of the rectangle. –  Kevin Walker Feb 27 '10 at 23:42
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