# When is this braiding not a symmetry?

Given a topological space $X$ instead of forming the fundamental groupoid $\pi(X)$ which is the category whose objects are the points and morphisms the homotopy classes of paths one can also form the fundamental 2-groupoid which is the bicategory with objects = points, 1-morphisms = paths and 2-morphisms = homotopy classes of homotopies between paths. Composition of 1-morphisms is the "standard" composition $a\otimes b$ of paths given by $(a\otimes b)(t) = a(2t)$ for $0 \leq t \leq \frac{1}{2}$ and $(a\otimes b)(t) = b(2t-1)$ for $\frac{1}{2} \leq t \leq 1$.

By restricting to loops around a fixed point $e \in X$ one therefore gets a bicategory with one object, i.e., a monoidal category. When moreover $X$ has the structure of a topological monoid it is easy to see that the structure map $\mu: X \times X \rightarrow X$ induces a braiding $\gamma_{a,b}: a\otimes b \stackrel{\sim}{\rightarrow} b\otimes a$ on this monoidal category.

My question is: Is there an (elementary) example where this braiding is not a symmetry? One obvious necessary condition is that $X$ has to be noncommutative in order to give such an example.

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## 1 Answer

If $X=\Omega Y$, that braided monoidal category (indeed groupoid) classifies the homotopy type of $P_3Y$, the $3$-type of $Y$. Such $3$-type is completely determined by the map $\eta^*\colon \pi_2(Y)\rightarrow \pi_3(Y)$ defined by precomposition with the Hopf map $\eta\colon S^3\rightarrow S^2$. This map is quadratic, i.e. $\eta^* (x)=\eta^* (-x)$ and the map defined by $\eta^* (x|y)=\eta^* (x+y)-\eta^* (x)-\eta^* (y)$ is bilinear. One can recover this map from the monoidal category, essentially $\eta^*(x)$ corresponds to the braiding $\gamma_{x,x}\colon x\otimes x\cong x\otimes x$. The category is symmetric if and only if $\eta^*$ is a homomorphism. This does not always happen, since any quadratic map between abelian groups $A\rightarrow B$ can be realized by some appropriate $Y$. For instance, if you take $Y=S^2$ the quadratic map is $\eta^* \colon \mathbb{Z}\rightarrow \mathbb{Z}$ is $\eta^* (n)=n^2$, hence you get an example.

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Very nice! Do you know a good reference for the relation between braided monoidal categories and homotopy 3-types? –  Nicolas Schmidt Mar 1 '13 at 23:48
Thanks Nicolas, unfortunately this is a typical example of theory which is "well known to experts". Despite being easy and beautiful there's no monograph covering for instance what you ask. See section 3 of Berger's "Double loop spaces, braided monoidal categories and algebraic 3-type of space" and references therein. –  Fernando Muro Mar 2 '13 at 0:03
Isn't it true that when X is itself a topological monoid (and not just a loop space) the quadratic map eta* is in fact a homomorphism? In any case I think one can use the monoid product to show that the braiding is its own homotopy inverse whenever X is a topological monoid. –  Carl Futia Mar 2 '13 at 0:18
@Carl, that's not true, for any loop space has the homotopy type of a topological group. –  Fernando Muro Mar 2 '13 at 1:57
@Fernando, thanks, I see that now - the Moore loops and the ordinary loop space are of the same homotopy type. –  Carl Futia Mar 2 '13 at 4:10