Where does the "easy" definition of a weak n-category fail? - MathOverflow most recent 30 from http://mathoverflow.net 2013-06-19T11:20:28Z http://mathoverflow.net/feeds/question/4897 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/4897/where-does-the-easy-definition-of-a-weak-n-category-fail Where does the "easy" definition of a weak n-category fail? Aleks Kissinger 2009-11-10T17:51:56Z 2009-11-10T19:16:38Z <p>Okay, I'm going to ask a naiive question that surely has an interesting answer. So, a first approximation of defining a (small) weak n-category probably goes something like this. Take a pre-n-category C of all the cells, source and target maps that do the right thing (i.e. are globular), and a composition defined for each r in {0,...,n}.</p> <p>For C, define a family of coherent sets $(\Sigma_1, \Sigma_1, \ldots)$ as a family of sets $\Sigma_r$ of r-cells in C such that</p> <ol> <li>$f : a \rightarrow b \in \Sigma_r \Rightarrow \exists f' : b \rightarrow a \in \Sigma_r$</li> <li>$f, f' : a \rightarrow b \in \Sigma_r \Rightarrow \exists \alpha : f \rightarrow f' \in \Sigma_{r+1}$</li> </ol> <p>Now, suppose C admits such a family of coherent sets and all r-cells have associators, uniters, and interchangers in $\Sigma_{r+1}$, one might be tempted to say C is an $\infty$-category. If for all $r \geq n+1$, $\Sigma_r$ is only identites, one might say this defines an n-category.</p> <p>So, the reason I say "one might be tempted to say" is that, if it were that easy, someone much smarter than me would have done it already. :) So, where does the above recipe fail? Or is this definition unsatisfactory because it doesn't express the structure using a finite generating set of commutative diagrams (cf. Mac Lane's coherence etc.)?</p> http://mathoverflow.net/questions/4897/where-does-the-easy-definition-of-a-weak-n-category-fail/4906#4906 Answer by Mike Shulman for Where does the "easy" definition of a weak n-category fail? Mike Shulman 2009-11-10T19:16:38Z 2009-11-10T19:16:38Z <p>If I understand correctly what you're getting at, I think the reason this fails is because for n>2, <em>not</em> every diagram of constraints can be expected to commute (even up to higher constraints) in a weak n-category. For example, a braided monoidal category can be regarded as a weak 3-category with one 0-cell and one 1-cell, but then the "double twist" is a constraint isomorphism which is not equal to the identity (also a constraint isomorphism).</p> <p>One way to get around this is, instead of talking about diagrams of constraints <em>in</em> some particular n-category, to talk about "formal" diagrams of constraints, i.e. diagrams of constraints in a <em>free</em> n-category. I think that when one makes your idea precise using this corrected approach, one will end up with something very similar to Batanin's higher-operadic definition of weak n-category.</p>