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From nCatlab https://ncatlab.org/nlab/show/braiding :

Any braided monoidal category has a natural isomorphism

$$B_{x,y} \;\colon\; x \otimes y \to y \otimes x $$

called the braiding.

A braided monoidal category is symmetric if and only if $B_{x,y}$ and $B_{y,x}$ are inverses (although they are isomorphisms regardless).

This all makes sense, but I'm struggling to think of an instance where you would want to work with an asymmetric braiding. It's plain to me that they can exist, but ... are there any useful examples?

I got to this page from https://ncatlab.org/nlab/show/associative+unital+algebra where it was stating

Moreover, if $(\mathcal{C}, \otimes , 1)$ has the structure of a symmetric monoidal category $(\mathcal{C}, \otimes, 1, B)$ with symmetric braiding $\tau$, then a monoid $(A,\mu, e)$ as above is called a commutative monoid in $(\mathcal{C}, \otimes, 1, B)$ if in addition... [diagram here]

I was also wondering if this was necessary, the symmetry in the braiding. If the braiding was asymmetric, but $\mu \circ B_{x,y} = \mu = \mu \circ (B_{y,x})^{-1}$, we can still make sense of the multiplication being commutative. It seems we could make useful statements about the algebra, even with a strange braiding like. Are there any examples of this, either?

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    $\begingroup$ One very important category which realizes asymmetrical braiding (kind of its canonical example) is $\mathrm{Tang}_2$ the category of 2-tangles. The importance of asymmetrical braiding comes into the picture in full force when one has to deal with fusion categories, which are becoming very very important as people dig deeper into them. Specially for physical purposes, braiding is operation of exchanging particles. If you search Anyons you'll see a lot of extremely non-trivial exchange properties. $\endgroup$ – Hamed Jul 8 '16 at 5:16
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It's plain to me that they can exist, but ... are there any useful examples?

Absolutely. Perhaps the historical motivating example is categories of representations of quantum groups, which can be used to build knot and link invariants such as the Jones polynomial. If you try to play this sort of game with a symmetric monoidal category it becomes very boring: for example, in a braided monoidal category, if $V$ is an object, then $V^{\otimes n}$ naturally acquires an action of the braid group $B_n$. If the braiding is symmetric, this action factors through the symmetric group, but in general it can be very interesting (and that's good, because this sort of thing helps us understand braids).

Braided monoidal categories also naturally arise in homotopy theory in the following way: if $X$ is a pointed space, then its double loop space $\Omega^2 X$ is naturally a "braided monoid," or $E_2$ algebra, and this means that its fundamental groupoid $\Pi_{\le 1}(\Omega^2 X)$ is naturally a braided monoidal groupoid. Every braided monoidal groupoid all of whose objects are invertible arises in this way, and they can be classified using group cohomology.

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It is not that you want to work with an non-symmetric braiding. Sometimes you have to work with one because the category you have in hand is simply a non-symmetric braided category.

The canonical example is the category of braids. You want to study braids, and they are a braided monoidal category, but alas it is not a symmetric category.

There are many other examples, and as soon as you embark upon learning about the subject you'll see that they are both important and useful. From Yetter-Drinfeld modules over a Hopf algebra to representations of quantum groups to all sorts of weird things.

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  • $\begingroup$ Thanks! I can't say I understand really ... any of these. Since I was interested in this in the context of associative algebras, are any of these algebras? $\endgroup$ – Alex Meiburg Jul 8 '16 at 6:15
  • $\begingroup$ Well, these non-symmetric monoidal categories I mentioned are of modules over certain associative algebras. $\endgroup$ – Mariano Suárez-Álvarez Jul 8 '16 at 6:28
  • $\begingroup$ A very readable exposition of all this is Christian Kassel's amazing book on Quantum Groups. $\endgroup$ – Mariano Suárez-Álvarez Jul 8 '16 at 6:29
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Let $A$ be an associative algebra. You probably know about the center $$Z(A) = \{ a \in A \mid \forall b \in A, ab = ba \},$$ the set of elements that commute with all the others. It's easy to show that this is a commutative subalgebra of $A$. It's a rather interesting construction to study.

Now consider a monoidal category $\mathcal{C}$, morally an "associative algebra in categories". It turns out that some notion of center exists for $\mathcal{C}$, called the Drinfeld center. Morally, it is still "the elements that commute with all the others", i.e. the objects $X$ such that for all $Y$, $X \otimes Y \cong Y \otimes X$.

The problem is that as usual in category theory, we're looking for natural stuff, and so it's better to consider pairs $(X, \Phi)$ where $X$ is an object and $\Phi$ is "a way of commuting $X$ with all the other elements", i.e. a natural transformation $X \otimes - \to - \otimes X$ – a so-called half-braiding. It turns out that given some $X$ that "commutes with all the other objects", there may be multiple different ways of making it commute with all the other objects. A morphism between two pairs is the obvious things, and this forms a category, call it $\mathscr{Z}(\mathcal{C})$, the Drinfeld center of $\mathcal{C}$.

Then it's a theorem of Drinfeld, Majid, and Joyal–Street* that $\mathscr{Z}(\mathcal{C})$ is canonically a braided monoidal category, and not a symmetric monoidal category as one might expect. If $\mathcal{C}$ is a category with one object, i.e. a monoid, then we recover the usual notion of center of a monoid, and this is symmetric monoidal. So going from monoids to categories, we've lost (or gained?) something: the center of a category is not symmetric anymore but braided.


* If I understand correctly, it was developed independently by Drinfeld and by Joyal–Street, but Drinfeld initially didn't publish anything about it and it appeared in a paper of Majid who attributed it to a private communication by Drinfeld. I'd be happy to hear a more detailed historical account of this, TBH.

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