Why is the Street nerve of the Gray tensor product $[1]\otimes [1]$ isomorphic to $[1]\times [1]$

2012-01-01T06:30:48Z

Recall that given two strict ω-categories $A$ and $B$, their lax Gray tensor product $A\otimes B$ is sent to the Verity-Gray tensor product of their associated complicial sets $N_{\operatorname{Strat}}(A) \otimes N_{\operatorname{Strat}}(B)$ by the stratified nerve of Verity. Further, given any strict ω-category $A$, the underlying simplicial set of the complicial set $N_{\operatorname{Strat}}(A)$ is exactly $N_\omega(A)$, where $N_\omega = N_{\mathcal{O}}$ is the nerve functor associated with the cosimplicial object $\mathcal{O}:\Delta \to \omega\operatorname{-cat}$ where $\mathcal{O}[n]$ is $n$th oriental as defined by Street.

Consider the following case: If we take the lax Gray tensor product of two freestanding 1-cells $[1]\otimes [1]$ and apply the stratified nerve, we obtain a Verity-Gray tensor product $N_{\operatorname{Strat}}([1])\otimes N_{\operatorname{Strat}}([1])$. By the definition of this tensor product, its underlying simplicial set is given simply as $[1]\times [1]$, and therefore, we see that the Street nerve $N_\omega([1]\otimes [1])=[1]\times [1]$.

If we actually take a moment to draw out the strict ω-category $[1]\otimes [1]$, we see that it can be visualized as:

•====•--->•   
|\   \    |  
| \   \   |  
|  \ =>\  |   
|   \   \ |  
v    v   vv
•--->•====•

where the "====" means that we are identifying the vertices on either end.

Also, the second oriental $\mathcal{O}[2]$ is traditionally written as:

 •---->•   
  \    |  
   \=> |  
    \  |   
     \ |  
      vv
       •

but the strict ω-category that this generates can be visualized as:

•====•--->•   
 \   \    |  
  \   \   |  
   \ =>\  |   
    \   \ |  
     v   vv
     •====•

But $[1]\times [1]$ viewed as a simplicial set is just the union of its two nondegenerate $2$-simplices. These two nondegenerate $2$-simplices should correspond to maps of strict ω-categories $\mathcal{O}[2] \to [1]\otimes [1]$. The bottom-left $2$-simplex is obviously given by the map sending $\mathcal{O}[2]$

•===•   
|\   \      
| \   \     
|  \<= \     
|   \   \   
v    v   v
•--->•===•

onto the bottom-left simplex

•   
|\       
| \      
|  \      
|   \      
v    v  
•--->•

by collapsing the 2-cell (note the flipped orientation)

•===•   
 \   \      
  \   \     
   \<= \     
    \   \   
     v   v
     •===•

to an edge.

The top-right $2$-simplex of $[1]\times [1]$ classifies the inclusion of $\mathcal{O}[2]$

•====•--->•   
 \   \    |  
  \   \   |  
   \ =>\  |   
    \   \ |  
     v   vv
     •====•

in $[1]\otimes [1]$

•====•--->•   
|\   \    |  
| \   \   |  
|  \ =>\  |   
|   \   \ |  
v    v   vv
•--->•====• .

The thing I don't understand is why some of the other maps $\mathcal{O}[2]\to [1]\otimes [1]$ classify degenerate 2-faces in $[1]\times [1]$.

For instance, consider either of the maps $\mathcal{O}[2] \to D_2$

sending

•====•--->•   
 \   \    |  
  \   \   |  
   \ =>\  |   
    \   \ |  
     v   vv
     •====•

onto

•===•   
 \   \      
  \   \     
   \ =>\     
    \   \   
     v   v
      •===•

whose restriction to the subobject

•---->•   
 \    |  
  \   |  
   \  |   
    \ |  
     vv
      •

is given by a codegeneracy (collapsing this simplex either to 0 0 2 or 0 2 2).

Then since $D_2$ embeds in $[1]\otimes [1]$, we obtain a map $\mathcal{O}[2] \to [1]\otimes [1]$ that doesn't appear to be degenerate.

However, it follows from the description of $N_\omega([1]\otimes [1])=[1]\times [1]$ that these maps must classify degenerate 2-simplices. Why are the simplices classified by these maps degenerate?

Answer by Harry Gindi for Why is the Street nerve of the Gray tensor product $[1]\otimes [1]$ isomorphic to $[1]\times [1]$

2012-01-03T13:21:50Z

Alright, I figured it out.

Here's the problem: The Verity tensor product of complicial sets is obtained as follows:

$$A\otimes_{\operatorname{Cs}} B = L_{\operatorname{Cs}} (\iota_{\operatorname{Cs}}(A) \otimes_{\operatorname{Strat}} \iota_{\operatorname{Cs}}(B)),$$

where $$L_{\operatorname{Cs}}:\operatorname{Strat} \rightleftarrows \operatorname{Cs}: \iota_{\operatorname{Cs}}$$ is the reflection-inclusion adjunction from the inclusion ${\operatorname{Cs}}\subseteq {\operatorname{Strat}}$.

The problem earlier was that I was only computing $$\iota_{\operatorname{Cs}}(A) \otimes_{\operatorname{Strat}} \iota_{\operatorname{Cs}}(B),$$ then taking the underlying simplicial set of this stratified simplicial set.

So in the example in the question, suppose that we took the stratified tensor product immediately. Then we end up with a stratified simplicial set that is obtained by gluing a thin 2-simplex to a standard one along the edge $0\to 2$. Then apply $L_{\operatorname{Cs}}$ to this. Since $L_{\operatorname{Cs}}$ preserves colimits, we can look at $L_{\operatorname{Cs}}$ applied to each component.

On the thin component, we do nothing, since a thin $2$-simplex is just a commutative triangle of 1-cells, which is, in particular, complicial set.

However, on the standard 2-cell, applying $L_{\operatorname{Cs}}$ gets us something rather more interesting. It gives us the minimal complicial approximation of a standard 2-simplex (standard meaning that it is the stratified 2-simplex whose thin cells are only the degenerate simplices), which is "obviously" the second oriental (where "obviously" means here that I don't know how to prove it and also don't care enough to try). However, this ends up being what we had originally expected!

Why is the Street nerve of the Gray tensor product $[1]\otimes [1]$ isomorphic to $[1]\times [1]$ - MathOverflow

Why is the Street nerve of the Gray tensor product $[1]\otimes [1]$ isomorphic to $[1]\times [1]$

Answer by Harry Gindi for Why is the Street nerve of the Gray tensor product $[1]\otimes [1]$ isomorphic to $[1]\times [1]$