When does a monoidal functor between ribbon categories preserve cups and caps, but not necessarily braidings?

Imagine a monoidal functor between ribbon categories (i.e. monoidal, with a braiding, a twist and compatible left and right duals). An important example would be the restriction functor from the representation category of one quasitriangular quantum group $G$ to a quantum subgroup, $H$.

Are there functors that don't preserve the braiding (this would be a very strong assumption), but preserve the duals?

I would call such a functor a "rigid functor", but apparently this term isn't being used much. Such functors would preserve all unbraided planar diagrams, but not necessarily the braided ones.

Are there easy-to-check criteria for this in the case of quantum groups? Has it been studied (e.g. with a list of known examples)?

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Wow, such diverse answers... it's a shame I can't accept multiple answers. –  Turion Feb 11 at 18:54

For Rep($U_q(g)$), the monoidal and pivotal structures can be defined with coefficients depending only on $q$, while the braiding requires fractional powers of $q$, say $q^{1/m}$, where $m$ depends on the Lie algebra $g$. So one source of examples would be to replace $q^{1/m}$ with $\xi q^{1/m}$, where $\xi^m = 1$. This changes the braiding but doesn't affect the monoidal structure or the pivotal (duality) structure. In other words, the "identity" functor is monoidal and preserves duals, but does not preserve the braiding.

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Does this change effect the twist functor? or can that be taken to be the same in both cases? –  Chris Schommer-Pries Feb 7 at 8:18
For most (all?) choices of $\xi$, the twist functor will change. –  Kevin Walker Feb 7 at 13:45
When I look at the triangular element of $U_q SL(2)$ in Kassel's book, I find the $q^{1/2}$, but don't the duals depend on the triangular element as well? –  Turion Feb 11 at 18:54
I think that one can write out bases and relations for the tensor category structure on $Rep(U_q(sl_2))$ using only powers of $q$, but that the braiding requires $q^{1/2}$. Check out Kuperberg's "spider" paper (arxiv.org/pdf/q-alg/9712003v1.pdf) and compare the tensor category relation on pages 12-14 with the braidings on page 16. –  Kevin Walker Feb 11 at 19:10
Ah, in other words, the duals only depend on the spherical element in the algebra, which can be written in terms of $q$? –  Turion Feb 12 at 15:49

The categories of (finite dimensional) vector spaces and super vector spaces are rigid and symmetric monoidal and so admit ribbon structure where the twist is trivial. The forgetful functor from super vector spaces to vector spaces is monoidal and preserves duals, but is not braided monoidal.

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Assuming that by "monoidal" you mean "strong monoidal", i.e. a functor $F$ comes with a natural isomorphism $f_{X,Y} : F(X\otimes Y) \cong F(X) \otimes F(Y)$ (and an isomorphism $i : F(1) \cong 1$, and these isomorphisms should intertwine the unit and associativity morphisms on the target category with the images under $F$ of the unit and associativity morphisms in the source), then every monoial functor is rigid, in the following sense:

Let $X$ have a left dual $X^*$, i.e. we have equipped the pair $(X,X^*)$ with a cap $X^* \otimes X \to 1$ and a cup $1 \to X\otimes X^*$, such that the zig-zag equations hold. Then $F(X)$ has $F(X^*)$ as its left dual. Indeed, $i \circ F(\text{cap}) \circ f_{X^\ast,X}^{-1} : F(X^\ast) \otimes F(X) \cong F(X^\ast\otimes X) \to F(1) \cong 1$ and $f_{X,X^\ast} \circ F(\text{cup}) \circ i$ are a cap and cup between $F(X^*)$ and $F(X)$. (The zig-zag equations do require that the isomorphism $f,i$ intertwine the various associators and unitors.)

In particular, suppose that the source of your monoidal functor is rigid. Then the essential image of your functor lands within the largest rigid subcategory of the target. Moreover, suppose both source and target are rigid. The previous paragraph implies that there is a canonical (and unique) natural isomorphism $F(X^*) \cong F(X)^*$ intertwining the cups and caps.

It should be emphasized, however, that the axioms of rigid category do not require an isomoprhism between the left dual $X^\ast$ and the right dual $^\ast X$. (Now I wish I had chosen different left/right conventions. C'est la vie.) In many examples, these are isomorphic, and indeed in many examples one can choose a natural isomorphism $X^\ast \cong {^\ast X}$. For example, in any braided category, the braiding determines such an isomorphism. But actually the braiding determines many such isomorphisms, and there's no best way to choose one, and many errors have been made (e.g. by me) by pretending that $X^\ast = {^\ast X}$ in some canonical way. (In a symmetric monoidal category, there is a canonical isomorphism $X^\ast \cong {^\ast X}$. But it's not necessarily compatible with other structure you might have around.)

Long before thinking about quantum groups, a good example to work through is the category of supervector spaces (over a field in which $2$ is invertible). As a monoidal category, this is the category of $\mathbb Z/2$ modules, but the braiding is different. Namely, the braiding is uniquely determined (up to the usual words) by demanding that on the one-dimensional nontrivial representation $S$, the braiding $\beta_{S,S} : S\otimes S \to S\otimes S$ is multiplication by $-1$. There are isomorphisms $S\otimes S \cong 1$, with which you can present $S$ as its own dual. The canonical isomorphism (determined by the supervector space braiding) $S \cong ^\ast S \cong S^\ast \cong S$ with these most natural choices is then multiplication by $-1$, whereas the canonical isomorphism for the usual braiding is multiplicaiton by $+1$. Note that the identity monoidal functor from supervect to $(\mathbb Z/2)$-rep is not a braided functor.

The next example to think through is the "quantum group" related to the reductive algebraic group $\mathbb G_m$. As a monoidal category, this is the category of $\mathbb Z$-graded vector spaces. The braiding is determined by declaring that on the generating line $L$ (i.e. the one-dimensional vector space in grading $1$), we have $\beta_{L,L} : L\otimes L \to L\otimes L$ is multiplication by $q \in \mathbb K^\times$. This category has a skeletalization with one invertible object $[n] = L^{\otimes n}$ for each $n\in \mathbb Z$; general objects are formal direct sums of these. This skeletalization is a strict monoidal category (generically skeletalizations of monoidal categories have nontrivial associators). In this skeletalization, the braiding $\beta_{[m],[n]} : [m+n] = [m] \otimes [n] \overset\sim\to [n]\otimes[m] = [m+n]$ is multiplication by $q^{mn}$. This example shows all of the problems that come from being too cavalier about the difference between isomorphism and equality.

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