Consider a rigid braided monoidal category, with braiding $\beta_{x,y} : x \otimes y \cong y \otimes x$, and every object has a dual such that $\epsilon_x : 1 \to a \otimes a^*, \bar\epsilon_x : a^* \otimes a \to 1$ satisfying the zig-zag identities. Now we have the Reidemeister moves, e.g. $$ (c \otimes \beta_{a,b}) \circ \beta_{a\otimes b, c} = \beta_{b \otimes a, c} \circ (\beta_{a,b}\otimes c) $$ saying that braiding the two string with another, and then braiding the two, is the same as first braiding the two and then with the other. This is Reidemeister III. Similarly $\beta_{a\otimes a^*, b} \circ (b \otimes \epsilon_a) = \epsilon_a \otimes b$ is Reidemeister II, and $(\bar\epsilon_{a} \otimes a) \circ (a^* \otimes \beta_{a,a}) \circ (\epsilon_a\otimes a) = \mathrm{id}_a$ is Reidemeister I. This is rather like the Morse link presentation of knots and links.

Is there any reference that develops such an idea? How can the theory of quandles be integrated in this picture? I'm guessing that quandles either behave like modules over a knot category, or we have a joint generalization of quandles and knot categories.

  • $\begingroup$ I think a relevant keyword is "tangle hypothesis", which is about a higher dimensional, higher-categorical generalization of what you're asking about. The relationship between knots and braided monoidal categories of dualizable objects should be a low-dimensional special case. $\endgroup$ Commented May 17, 2023 at 8:06
  • 2
    $\begingroup$ Your last relation which is supposed to translate to Reidemeister I does not hold in general: rigid and braided is not enough to get knot invariants. The keyword is "ribbon category" though I would be surprised you've heard of braided monoidal categories and not of ribbon ones (so maybe I just misunderstand your question). If you indeed haven't heard of those then yes, there is an enormous amount of litterature around that idea. $\endgroup$
    – Adrien
    Commented May 17, 2023 at 12:11
  • $\begingroup$ @Adrien I did gloss over that but did not think about the connection. If you could turn the comment into an answer (perhaps also addressing subtleties like orientation and framing) I would accept it! $\endgroup$
    – Trebor
    Commented May 18, 2023 at 10:28
  • $\begingroup$ This isn't an answer to the question in the body of your message but to the one in the title. There are many knot categories, i.e. categories where the objects are knots. Usually the arrows are cobordisms of the knots with various constraints, usually that the cobordism is an embedded product. Rather than talk about the underlying category usually this avenue of study is called "knot concordence". $\endgroup$ Commented May 18, 2023 at 21:48

1 Answer 1


To expand on my comment, this connection is indeed well-known and the key concept is that of ribbon category. A standard textbook reference is Turaev, Quantum Invariants of Knots and 3-Manifolds.

Braided and rigid is not enough to get links invariants, because RI will not hold in general (and in fact pretty much never). A nice exposition of that issue can be found in Selinger, A survey of graphical languages for monoidal categories (https://arxiv.org/abs/0908.3347) (in that reference autonomous means rigid and tortile means ribbon).

Any ribbon category provides an invariant of framed, oriented links for each object $X$. If $X$ is simple then the ribbon element $\theta_X$ acts as a multiple of the identity so that you can renormalize to get an invariant of oriented links. The choice of an isomorphism $X\cong X^*$, provided one exists, gets rid of the orientation.

If you want non trivial invariants of oriented links on the nose, you'd need that $\theta_X=id_X$ and $\theta_{X\otimes X} \neq id_{X\otimes X}$. While this isn't impossible (tautologically, the category of oriented tangles is a ribbon category which satisfies this for example) this is a pretty unnnatural condition.


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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