Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

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.

share|improve this question

1 Answer 1

up vote 5 down vote accepted

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.

share|improve this answer
    
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
1  
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

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.