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• CommentRowNumber1.
• CommentAuthorUrs
• CommentTimeOct 6th 2010

following public demand, I created an entry titled

(If you don’t like this title, please have a look first to see if it makes sense afterwards. If still not, I won’t be dogamitic about this and am open for suggestions for other titles.)

• CommentRowNumber2.
• CommentAuthorzskoda
• CommentTimeOct 6th 2010
I like it.
• CommentRowNumber3.
• CommentAuthorMike Shulman
• CommentTimeOct 6th 2010

I don’t understand what this is saying. A transformation $F\to G$ between $n$-categories $C$ and $D$ is certainly given by an $n$-functor $C\to D^{\Delta^1}$ or equivalently an $n$-functor $C\times \Delta^1\to D$. But where do the $(n-1)$-categories come in?

• CommentRowNumber4.
• CommentAuthorUrs
• CommentTimeOct 6th 2010
• (edited Oct 6th 2010)

But where do the (n−1)-categories come in?

Because for $D$ an $n$-category, that component-functor of the transformation is fixed by its value on the $(n-1)$-skeleton of $C$.

An ordinary natural transformation $\eta : F \Rightarrow G : C \to D$ is in components a 0-functor

$\eta : Obj(C) \to D^{\Delta} \,.$

A pseudonatural transformation between 2-functors between 2-categories is really a 1-functor

$\eta : sk_1 C \to D^{\Delta}$

and so on. Because the naturality condition and the fact that $D$ has no nontrivial $(k \gt n)$-cells fixes the remaining data.

• CommentRowNumber5.
• CommentAuthorMike Shulman
• CommentTimeOct 6th 2010

Well, that’s sort of true, but in general, the “$(n-1)$-skeleton” of a weak $n$-category is not an $(n-1)$-category. E.g. if you discard the 2-cells from a bicategory, you don’t get a category, because composition is not associative.

• CommentRowNumber6.
• CommentAuthorUrs
• CommentTimeOct 6th 2010

Well, that’s sort of true,

Okay. So for strict $\infty$-categories it is precisely true.

For weak categories the description is not as straightforward, but the general principle still holds: the transformation is fixed by its values on $(n-1)$-cells and the naturality condition that it satisfies makes it respect compositions in a natural way. One needs to organize this data properly.

Possibly a way to formalize this is to throw away only all those $n$-morphisms that are not structure morphisms. So for instance I could start with an algebraic quasi-$n$-category and then form the $(n-1)$-skeleton but keeping all the chosen $n$-horn fillers. That should yield an $(n-1)$-category in the sense that it is $(n-1)$-truncated.

For low $n$ one can arrgange things by hand: for 2-categories we can invoke semistrictification to pass to an equivalent model where the 1-morphisms do form a category.

For general $n$-categories I could appeal to Simpson’s conjecture to similarly deduce that the components of a transformation have as domain an $(n-1)$-semicategory (no identity $k$-morphisms). (The transformation will of course also respect these units in some way.)

• CommentRowNumber7.
• CommentAuthorMike Shulman
• CommentTimeOct 6th 2010

Possibly a way to formalize this is to throw away only all those $n$-morphisms that are not structure morphisms. So for instance I could start with an algebraic quasi-$n$-category and then form the $(n-1)$-skeleton but keeping all the chosen $n$-horn fillers. That should yield an $(n-1)$-category in the sense that it is $(n-1)$-truncated.

I don’t see any reason why that would give you an $(n-1)$-truncated thing; it seems to me that the structure isomorphisms might still be nontrivial automorphisms. Nor is it clear to me that it would be invariant under passing to an equivalent $n$-category.

One thing that should be true is that for any algebraic notion of $n$-category, there’s a corresponding notion of “top-level-incoherent $(n-1)$-category” which just has all the operations on $\lt n$-dimensional cells that your notion of $n$-category has. If your notion of $n$-category had strict associativity and interchange, then this would be your $(n-1)$-semicategories.

• CommentRowNumber8.
• CommentAuthorUrs
• CommentTimeOct 6th 2010
• (edited Oct 6th 2010)

I don’t see any reason why that would give you an (n−1)-truncated thing; it seems to me that the structure isomorphisms might still be nontrivial automorphisms.

Right, that was wrong.

One thing that should be true is that for any algebraic notion of $n$-category, there’s a corresponding notion of “top-level-incoherent $(n-1)$-category” which just has all the operations on $\lt n$-dimensional cells that your notion of $n$-category has. If your notion of $n$-category had strict associativity and interchange, then this would be your $(n-1)$-semicategories.

Okay, I’ll go through the entry now, and try to clarify these issues a bit more.

One interesting aspect to notice (which I still need to say in the entry), with an eye towards the motivating “holography for QFT $n$-functors” is that Chris Schommer-Pries famously found (slide 81 of his notes referenced in the entry) that boundary conditions and defect lines in 2d QFT are not in fact given by natural transformations of 2-functors, but by canonical morphisms aka “supernatural transformations”, those that are natural only with respect to the invertible $n$-cells.

This is something that still deserves better conceptual understanding, I think.

• CommentRowNumber9.
• CommentAuthorUrs
• CommentTimeOct 7th 2010

Okay, I edited the entry a bit more. There is now one expository section with elementary details on how natural transformations between strict n-categories work for low $n$.

Then following this is a section with more on the general statement. There is lots of room to further improve this. But I am getting a bit tired. Has been a long day.

• CommentRowNumber10.
• CommentAuthorMike Shulman
• CommentTimeOct 7th 2010

I tried to clarify a bit further and copied my remark from above about (n-1)-dimensional structures.

• CommentRowNumber11.
• CommentAuthorDavid_Corfield
• CommentTimeOct 7th 2010

Mike #10, copied it where?

• CommentRowNumber12.
• CommentAuthorUrs
• CommentTimeOct 7th 2010

Mike #10, copied it where?

Last paragraph of the Formalizations-subsection

• CommentRowNumber13.
• CommentAuthorDavid_Corfield
• CommentTimeOct 7th 2010

Are the boundary QFTs of QFTs involved in S- and T-duality related interestingly?

• CommentRowNumber14.
• CommentAuthorUrs
• CommentTimeOct 7th 2010

Are the boundary QFTs of QFTs involved in S- and T-duality related interestingly?

That’s actually a good question. I don’t really know.

Concretely, since Witten identifies ordinary quantum mechanics on the boundary of the A-model, it would be very interesting to see what happens on the boundary of the mirror-symmetric B-model. Maybe he already wrote about that. I didn’t follow all the latest articles closely enough.

• CommentRowNumber15.
• CommentAuthorDavid_Corfield
• CommentTimeJun 14th 2015

I had an enquiry by email about the status of abstract formulations of holography in view of what there is at duality in physics. The page for this thread is the closest to an abstract general account of holography, I guess. Given that it’s often called a physical duality, could there be any connection between the pages?