# strictifying tricategories

Every tricategory is equivalent to a Gray-categories. However any Gray-category is not equivalent to a 3-category. As far as I know, this is similar to the fact that braided monoidal categories are not equivalent to symmetric monoidal categories. However the difference between braided and monoidal categories lies in just one axiom: the fact that $B_{x,y} \star B_{y,x}$ is the identity.

So my question is: could you add such an axiom to tricategories in order to make them 'strictifyable' into 3-categories? And if the answer is yes, what isn't it a part of the definition of a tricategory?

I think such an axiom may look something like $$\omega_{f,g} \star_2 \omega_{g,f} = 1_{f \star_1g}$$ whenever the left-hand-side term is defined, that is when $f$ and $g$ have the same identity as sources and targets.

Side note: of course one could also add this axiom to Gray categories. However I think it cannot be directly encodded into the Gray tensor product, making the definition a bit awkward.

• In what sense is a symmetric monoidal category like a 3-category? – Dimitri Chikhladze Feb 25 '15 at 14:50
• I do not mean that a symmetric monoidal category is like a 3-category. Only that a symmetric monoidal category is to a braided monoidal category what a 3-category is to a gray category. To be more precise, one object, one arrow tricategories are precisely braided monoidal categories (whith braiding induced by the interchange isomorphism), and one object, one arrow 3-categories are precisely strict symmetric monoidal categories. – Maxime Lucas Feb 25 '15 at 15:09
• I see, I guess you rely on the fact that every symmetric monoidal category can be strictified to a strict symmetric monoidal category. So in a sense the question asks whether there is a 3-categorical analogue of a symmetric monoidal category. – Dimitri Chikhladze Feb 25 '15 at 16:05
• Yes, it is an other way to see it. – Maxime Lucas Feb 25 '15 at 16:13
• @Dimitri: a braided monoidal category is a (weak) 3-category with one object and one morphism. A symmetric monoidal category is an object like this satisfying an additional constraint, one description of which is that it's deloopable to a (weak) 4-category with one object, one morphism, and one 2-morphism. (This is special to the low category number we're working with: in general being symmetric is a structure on top of being braided, not a property.) – Qiaochu Yuan Feb 25 '15 at 19:31

Because we want things that aren't strictifiable. It's known that if a space has strictifiable fundamental $n$-groupoid, then its Whitehead brackets up to $\pi_n$ vanish; in particular, $S^2$ has a nontrivial Whitehead bracket $\pi_2 \times \pi_2 \to \pi_3$, so its fundamental $3$-groupoid is not strictifiable (alternatively, the fundamental groupoid of $\Omega^2 S^2$, as a braided monoidal groupoid, is not equivalent to a symmetric monoidal groupoid), and we want fundamental $3$-groupoids for the homotopy hypothesis. See, for example, Homotopy types of strict 3-groupoids by Simpson.
• @skysurf3000: yes, but that's special to bicategories. It's special to $n \le 2$ that there aren't any interesting Whitehead brackets, and in fact strict $2$-groupoids can model all homotopy $2$-types. But that just stops being true for $n \ge 3$. – Qiaochu Yuan Feb 26 '15 at 17:11
• It might be worth pointing out explicitly that the converse of what you said about Whitehead products does not hold: having them vanish does not guarantee that an n-groupoid is strictifiable. The simplest example is probably $P_3QS^2$, the $3$-type of $QS^2 = \mathrm{colim}\; \Omega^n S^{n+2}$. – Omar Antolín-Camarena Mar 21 '15 at 4:36