Let $k$ be a field, and let $q\in k^{\times}$. We can then consider the Temperley-Lieb category $TL(q)$. The objects of $TL(q)$ are the non-negative integers, and morphisms are roughly isotopy classes of string diagrams (without crossings) contained in a planar strip up to the relation that any circle evalutes to $-(q^{-1}+q)$. This category is monoidal, and objects have duals. In fact, it has a canonical spherical structure.

It is quite standard that, provided that $q$ admits a square root $q^{1/2}$ in $k$, then $TL(q)$ can be endowed with a braiding. Explicitly, it is given by Kauffman's skein relation. Now, I am fairly sure that the answer to the following question is yes, but I haven't been able to find a reference in the literature.

Are these braidings unique? More precisely, is it true that (the equivalence classes of) braidings on $TL(q)$ are classified by the square roots of $q$ in $k$?

This question was asked on MS first, but wasn't answered there.


1 Answer 1


Let $|$ denote the standard (single-strand) monoidal generator of $\mathrm{TL}(q)$. Since $|$ is a monoidal generator, any braiding $\beta$ on $\mathrm{TL}(q)$ will be fully determined by the map $\beta_{|,|} : | \otimes | \to | \otimes |$. Note that the space of all such maps is $2$-dimensional, spanned by the identity $||$ and a cup-cap-composition $(\overset\cup{\underset\cap{}}) : | \otimes | \to \emptyset \to | \otimes |$ satisfying $(\overset\cup{\underset\cap{}})^2 = -(q+q^{-1})(\overset\cup{\underset\cap{}})$. Thus we can decide that $$ \beta_{|,|} = x (||) + y(\overset\cup{\underset\cap{}})$$ for some scalars $x,y$.

Since the diagrammatics trivializes associators, for $\beta$ is to be braiding, a necessary condition is that $$ | \otimes \cap = \beta_{|,\emptyset} \circ (| \otimes \cap) = (\cap \otimes |) \circ (\beta_{|, ||}) = (\cap \otimes |) \circ (| \otimes \beta_{|, |}) \circ (\beta_{|, |} \otimes |) $$ Here the first equation is unitality of $\beta$, the second equation is naturality, and the third is associativity. Expanding the right-hand side gives $$ = x^2(\cap \otimes |) - xy(q+q^{-1})(\cap \otimes |) + xy (| \otimes \cap) + y^2 (\cap \otimes |) $$ Combining like terms, we find that necessary conditions for $\beta$ to be a braiding are: $$ x^2 - xy(q+q^{-1}) + y^2 = 0, \qquad xy = 1.$$ These equations have four solutions, indexed by the two square roots of $q^{\pm 1}$.

This shows that there are not more braidings than the ones you already know.

  • $\begingroup$ I agree with you, this is pretty much the proof sketch I had. Do you have any idea where this could be done in the literature? $\endgroup$
    – JeCl
    Apr 23 at 14:10
  • $\begingroup$ I don't know a reference. I learned it as an exercise. $\endgroup$ Apr 24 at 11:04

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