I've been recently looking into extending the Grothendieck construction to the case of a strict $\omega$-functor $F$ from a strict $\omega$-category $X$ to $\omega-\operatorname{Cat}$. By some recent developments (a paper of Dimitri Ara and Georges Maltsiniotis on defining the lax/oplax join and slice for strict $\omega$-categories), it has become pretty much a breeze to define the Grothendieck construction for such an $\omega$-functor as the oplax colimit of $F$ in $\omega-\operatorname{Cat}$, as it is simply defined to be the initial object of the oplax slice of $\omega-\operatorname{Cat}$ under $F$.

However, in a private conversation, I was informed of Scott Dyer's thesis, where he argues that in dimensions higher than $2$, things go rather bad and we end up in a situation without enough Cartesian lifts.

Here's the trouble: Dyer's thesis is pretty abstruse, and he uses a lot of nonstandard notation. It's also quite long. I'm wondering if anyone here might be familiar enough with it to save me some time reading through it only to find out that I'm barking up totally the wrong tree:

1.) Is Dyer's Grothendieck integral exactly the same construction as the aforementioned oplax colimit?

2.) Assuming the first question is true, has anyone come up with a loosening of either the definition of cartesianness for higher cells or the definition (as far as one exists) of a strict n-fibration. If fibrations in dimension n>2 are not the correct characterization of the image of the Grothendieck construction, has anyone come up with an alternative one since then?

Edit: Here's the paper

  • $\begingroup$ I did skim the section where he discusses fibrations in passing. What looked strange to me was that his higher dimensional lifts did not look like modification cylinders but instead looked looked like smushed-at-one-end versions. $\endgroup$ – Harry Gindi Aug 6 '17 at 11:10
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    $\begingroup$ Even better is a human-readable reference and a non-pdf landing page: Dyer, Scott W., "The Strict Higher Grothendieck Integral" (2015). Dissertations, Theses, and Student Research Papers in Mathematics. 67. digitalcommons.unl.edu/mathstudent/67 $\endgroup$ – David Roberts Aug 6 '17 at 11:21
  • $\begingroup$ Changed the link out to yours. Thanks David. $\endgroup$ – Harry Gindi Aug 6 '17 at 11:24
  • $\begingroup$ No worries! :-) $\endgroup$ – David Roberts Aug 6 '17 at 11:25
  • $\begingroup$ Sorry for repeatedly bumping, MO. I'm a little rusty at posting here, and I don't think there's a way to make a non-bumping edit. I just found another typo. Sorry again =\ $\endgroup$ – Harry Gindi Aug 6 '17 at 11:28

On glancing at the paper, I think that Dyer is just confused. If anything, the place where something new and different happens is $n=2$, not $n=3$, but that case was explained by Hermida.

Dyer's proposition 5.3.2 says (in the case $k=n=2$ that he proves) that every 2-cell factors as a "composition" of a cartesian arrow with a vertical one (replacing his terminology "canonical" and "flat" with the standard words "cartesian" and "vertical"). However, this is not a literal single composition in a hom-category, which would be ill-typed for the same reasons he complains about in the $k=3$ case. As can be seen from his proof of Prop. 5.3.2, the true statement is that every 2-cell factors as a "pasting composite" of a cartesian arrow and a vertical one; more specifically it is an ordinary composite of a whiskering of a cartesian arrow and a whiskering of a vertical one. The whiskering of the cartesian arrow is in fact another cartesian arrow, but the whiskering of the vertical arrow is no longer vertical.

It seems to me that the same thing is happening in the case $k=3$. In the pictures at the bottom of his p98, if you whisker $(b,\chi)$ by $\tilde{l}$ on the bottom, and whisker $\tilde{\mu}$ by $(b,x)$ on the top, then their domains and codomains will match up, and their composite should be $(\mu,\chi)$ as desired.

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  • $\begingroup$ Phew. This was what I was getting as well in my micro-version computations. I found his choice of terminology impenetrable. Gonna leave this open a bit longer so I don't have to switch accepted answers, but I'll probably accept tomorrow. Thanks! $\endgroup$ – Harry Gindi Aug 7 '17 at 0:22

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