Why is the Street nerve of the Gray tensor product $[1]\otimes [1]$ isomorphic to $[1]\times [1]$ - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-21T00:57:57Z http://mathoverflow.net/feeds/question/84672 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/84672/why-is-the-street-nerve-of-the-gray-tensor-product-1-otimes-1-isomorphic-to Why is the Street nerve of the Gray tensor product $[1]\otimes [1]$ isomorphic to $[1]\times [1]$ Harry Gindi 2012-01-01T06:30:48Z 2012-01-03T13:32:25Z <p>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.</p> <p>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 $N_\omega([1]\otimes [1])=[1]\times [1]$. </p> <p>If we actually take a moment to draw out the strict ω-category $[1]\otimes [1]$, we see that it can be visualized as:</p> <pre><code>•====•---&gt;• |\ \ | | \ \ | | \ =&gt;\ | | \ \ | v v vv •---&gt;•====• </code></pre> <p>where the "====" means that we are identifying the vertices on either end.</p> <p>Also, the second oriental $\mathcal{O}[2]$ is traditionally written as:</p> <pre><code> •----&gt;• \ | \=&gt; | \ | \ | vv • </code></pre> <p>but the strict ω-category that this generates can be visualized as:</p> <pre><code>•====•---&gt;• \ \ | \ \ | \ =&gt;\ | \ \ | v vv •====• </code></pre> <p>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 $\mathcal{O}[2]$</p> <pre><code>•===• |\ \ | \ \ | \&lt;= \ | \ \ v v v •---&gt;•===• </code></pre> <p>onto the bottom-left simplex </p> <pre><code>• |\ | \ | \ | \ v v •---&gt;• </code></pre> <p>by collapsing the 2-cell (note the flipped orientation)</p> <pre><code>•===• \ \ \ \ \&lt;= \ \ \ v v •===• </code></pre> <p>to an edge. </p> <p>The top-right $2$-simplex of $[1]\times [1]$ classifies the inclusion of $\mathcal{O}[2]$ </p> <pre><code>•====•---&gt;• \ \ | \ \ | \ =&gt;\ | \ \ | v vv •====• </code></pre> <p>in $[1]\otimes [1]$</p> <pre><code>•====•---&gt;• |\ \ | | \ \ | | \ =&gt;\ | | \ \ | v v vv •---&gt;•====• . </code></pre> <p>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 $[1]\times [1]$. </p> <p>For instance, consider either of the maps $\mathcal{O}[2] \to D_2$</p> <p>sending </p> <pre><code>•====•---&gt;• \ \ | \ \ | \ =&gt;\ | \ \ | v vv •====• </code></pre> <p>onto </p> <pre><code>•===• \ \ \ \ \ =&gt;\ \ \ v v •===• </code></pre> <p>whose restriction to the subobject </p> <pre><code>•----&gt;• \ | \ | \ | \ | vv • </code></pre> <p>is given by a codegeneracy (collapsing this simplex either to 0 0 2 or 0 2 2).</p> <p>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. </p> <p>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? </p> http://mathoverflow.net/questions/84672/why-is-the-street-nerve-of-the-gray-tensor-product-1-otimes-1-isomorphic-to/84802#84802 Answer by Harry Gindi for Why is the Street nerve of the Gray tensor product $[1]\otimes [1]$ isomorphic to $[1]\times [1]$ Harry Gindi 2012-01-03T13:21:50Z 2012-01-03T13:32:25Z <p>Alright, I figured it out.</p> <p>Here's the problem: The Verity tensor product of complicial sets is obtained as follows:</p> <p>$$A\otimes_{\operatorname{Cs}} B = L_{\operatorname{Cs}} (\iota_{\operatorname{Cs}}(A) \otimes_{\operatorname{Strat}} \iota_{\operatorname{Cs}}(B)),$$</p> <p>where $$L_{\operatorname{Cs}}:\operatorname{Strat} \rightleftarrows \operatorname{Cs}: \iota_{\operatorname{Cs}}$$ is the reflection-inclusion adjunction from the inclusion ${\operatorname{Cs}}\subseteq {\operatorname{Strat}}$. </p> <p>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.</p> <p>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.</p> <p>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. </p> <p>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!</p>