18
$\begingroup$

Recall that a monoidal category $\mathcal C$ is rigid if every object $X\in \mathcal C$ has both left and right duals, i.e. objects $X^l$ and $X^r$ with maps $X^l \otimes X \to \mathbf 1 \to X \otimes X^l$ and $X \otimes X^r \to \mathbf 1 \to X^r \otimes X$ satisfying certain equations. It is a fundamental fact about monoidal categories that "having a left dual" and "having a right dual" are properties of an object, not data: given $X$, the objects $X^l$ and $X^r$, if they exist, are uniquely determined up to unique isomorphism. As such, for a monoidal category itself to be rigid is also property --- the only data is the data of being monoidal.

This should remind you of groups. Given a monoid $G$, an element $x\in G$ is invertible if there are elements $x^l$ and $x^r$ and equalities $x^l x = 1 = x x^r$. (Undergraduate exercise: $x^l = x^r$.) Being invertible is property. A monoid $G$ is a group if every element therein is invertible. Thus this too is a property.

In some cases (e.g. algebraic geometry) you don't always want to think about the elements of a group $G$. Fortunately, there is a very nice way to say when a monoid is a group that does not directly refer to invertibility of elements. Let $G$ be a monoid and consider the map $G \times G \to G\times G$ (a map of underlying spaces, not of groups) that takes $(x,y)$ to $(x,xy)$. The monoid $G$ is a group iff this map is an isomorphism (of underlying spaces).

Is there a similar characterization of when a monoidal category is rigid? Something like "consider the map $\mathcal C \times \mathcal C \to \mathcal C \times \mathcal C$, and ask that it be a left adjoint"?

$\endgroup$
  • $\begingroup$ There are many close connections between groups and rigid monoidal categories; for example, a rigid monoidal category whose only morphisms are identities is a group. I learned from wikipedia about pregroups, which are monoidal categories whose underlying category is a partially ordered set; their sole use seems to be in linguistics and computer science. $\endgroup$ – Theo Johnson-Freyd Dec 6 '15 at 6:46
  • 3
    $\begingroup$ Have you looked at Appendix D in Gaitsgory's 1-affineness paper (arxiv.org/abs/1306.4304)? He discusses the setting of presentable categories. $\endgroup$ – Pavel Safronov Dec 6 '15 at 12:23
  • 1
    $\begingroup$ @მამუკაჯიბლაძე Right, the problem is that sending an object of a rigid category to its dual is contravariant. It is a bit like the problem of defining "Hopf": a bialgebra is an algebra object among coalgebras, but the Hopf condition requires going down to the level of underlying vector spaces (existence of antipode is equivalent to the linear map $\Delta \circ m : H\otimes H \to H \otimes H$ being invertible; but this map is neither a map of algebras nor a map of coalgebras). $\endgroup$ – Theo Johnson-Freyd Dec 6 '15 at 16:38
  • 1
    $\begingroup$ @მამუკაჯიბლაძე In the Hopf case as well, the antipode map is a map to the "dual" (co)algebra, where "dual" here means in the Morita bicategory. Similarly, for a category $\mathcal C$, the opposite category $\mathcal C^\op$, which is the recipient of $X \mapsto X^*$, is the dual object in the "Morita" bicategory of categories, profunctors, and natural transformations. I don't know if that is a useful similarity or not. $\endgroup$ – Theo Johnson-Freyd Dec 6 '15 at 16:41
  • 1
    $\begingroup$ @PavelSafronov In D.1.3 of Appendix D in Gaitsgory's paper he says "... it is easy to show that O is rigid in the sense of Sect. D.1.1 if and only if every compact object of O admits both left and right monoidal duals". However right now I am only able to see that this implies "weak rigidity" as in Bakalov-Kirillov's Lectures on Tensor Cateogries and Modular Functors. Do you understand the argument here? $\endgroup$ – Chris Schommer-Pries Dec 7 '15 at 13:04
8
$\begingroup$

The notion of rigidity can be defined for any pseudomonoid in a monoidal proarrow equipment. See for instance Dualizations and Antipodes by Day, McCrudden, and Street (although they work only with monoidal bicategories and don't make the equipments explicit).

$\endgroup$
  • 6
    $\begingroup$ +1, although if you have some time available, this deserves to be expanded. $\endgroup$ – Todd Trimble Dec 7 '15 at 3:31
  • $\begingroup$ @ToddTrimble yeah, this is a terser answer than I would prefer to give, but unfortunately I don't have the time available at the moment. $\endgroup$ – Mike Shulman Dec 7 '15 at 16:03
  • 2
    $\begingroup$ Note that the relevant proposition in the DMS paper is not right, and was later corrected in "Duals Invert" by López Franco-Street-Wood. $\endgroup$ – Chris Schommer-Pries Dec 8 '15 at 10:57
  • $\begingroup$ @ChrisSchommer-Pries Thank you for the warning! $\endgroup$ – Theo Johnson-Freyd Dec 9 '15 at 2:00
  • 1
    $\begingroup$ The full statement is nice, and simple enough, so it's worth recording here: a monoidal category has duals for objects just when, in Prof, together with its right adjoint, it forms a Frobenius pseudomonoid. $\endgroup$ – Jamie Vicary Mar 13 '16 at 0:20
2
$\begingroup$

Yes, I think it is possible in the following way. Write $|\mathcal C|$ for the set of objects of a monoidal category $\mathcal C$. Then, there is a functor:

$$F \colon |\mathcal C| \times \mathcal C \to |\mathcal C| \times \mathcal C$$

which assigns to a pair $\langle x, y\rangle$ a pair $\langle x, x \otimes y\rangle$, i.e.:

$$F(\langle x, y\rangle) = \langle x, x \otimes y\rangle$$

Monoidal category $\mathcal C$ has right duals if $F$ has a right adjoint $G$ and, moreover, this right adjoint preserves tensor products --- for every object $a \in \mathcal C$: $$G({-}, a \otimes {=}) \approx \langle {-}, a \otimes {=} \rangle \circ G$$

$\endgroup$
  • $\begingroup$ I would argue that this version isn't very "categorical", because of having to take the underlying set of objects. $\endgroup$ – Theo Johnson-Freyd Dec 6 '15 at 16:44
  • $\begingroup$ @TheoJohnson-Freyd, I don't know what you mean by "categorical" here. However, if you are not comfortable with the notion of "underlying set of objects", then you may take "the underlying set of object up to equivalence". In fact, you may substitute $|\mathcal C|$ with the underlying groupoid of $\mathcal C$ as well... $\endgroup$ – Michal R. Przybylek Dec 6 '15 at 16:49
  • 1
    $\begingroup$ I certainly don't have a precise definition of "categorical". At best, I'd like a description of rigidity that can be translated into any bicategory - I mean, there is a perfectly good notion of "monoidal object" in any monoidal bicategory, and at best I could say which of those are "rigid". That is too much to ask, of course - for comparison, in a general monoidal category, certainly it doesn't make sense to ask when a monoid is a group; but if the category has some good properties (e.g. the monoidal structure is the Cartesian product) then it does. In any case, an important part is to use... $\endgroup$ – Theo Johnson-Freyd Dec 7 '15 at 1:45
  • $\begingroup$ ...only data "available" to the bicategory. $\endgroup$ – Theo Johnson-Freyd Dec 7 '15 at 1:46
  • 2
    $\begingroup$ I believe these structures only give "weak rigidity" and not rigidity itself. See for example Bakalov-Kirillov "Lectures on tensor categories and modular functors". $\endgroup$ – Chris Schommer-Pries Dec 7 '15 at 12:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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