# The dual of a dual in a rigid tensor category

For a rigid tensor category $\cal{C}$, can it happen that, for some $X \in {\cal C}$, we have that $X$ is not isomorphic to $(X^{*})^*$, for $*$ denoting dual? If so, what is a good example.

• What is your definition of "rigid tensor category"? For the definition I know, any object is reflexive (that is, the canonical morphism $X \to (X^*)^*$ is an isomorphism). – Ingo Blechschmidt Jun 20 '18 at 13:13
• Alternatively, you can try to look at the category of representations of a suitable finite-dimensional non-involutive Hopf algebra. Others might know better which particular Hopf algebra could give you an example, if any. – Tobias Fritz Jun 20 '18 at 13:13
• A small addition: it's plausible that my Hopf algebra proposal does not work at all, according to a more general conjecture of Etingof, Nikshych, and Ostrik. – Tobias Fritz Jun 20 '18 at 13:23
• @ Ingo: I'm following nLab. On that site a "rigid tensor category" is one where all objects admit left and right duals. A "pivotal category" is a rigid tensor category equipped with a tensor natural isomorphism $X \to (X^*)^*$. SO I guess I am looking for an example for a rigid tensor category which is not "pivotable". – Max Schattman Jun 20 '18 at 14:03
• Note that if the category is semisimple then you can show that they coincide (unnaturally and unmonoidally) by picking splittings of ev and coev for simples. – Noah Snyder Jun 20 '18 at 16:18

As Tobias said in his answer, a good place to look for examples is in endofunctor categories with composition as the monoidal product, where duals are adjoints. But another way to get a rigid monoidal category of such is to just restrict to the full subcategory of endofunctors $F = F_0$ that fit into some infinite adjoint string

$$\cdots \dashv F_{-2} \dashv F_{-1} \dashv F_0 \dashv F_1 \dashv F_2 \dashv \cdots$$

Since composition preserves adjoints, this subcategory is closed under the monoidal structure, hence is itself a monoidal category; and it is evidently rigid since each of the $F_n$ above also belongs to it (just shift the adjoint string back and forth).

So now we just need to find an endofunctor that fits into such an infinite adjoint string but whose left and right adjoints are not isomorphic. One such "naturally ocurring" example appears in Remark 6.11 of arXiv:1704.08084: if $D$ is a stable derivator, then there is an infinite adjoint string

$$\cdots \dashv G_{-2} \dashv G_{-1} \dashv G_0 \dashv G_1 \dashv G_2 \dashv \cdots$$

connecting $D$ and $D^{}$, where $$ is the arrow category $(0\to 1)$. These are not endofunctors, but composing them in pairs we get an adjoint string of endofunctors:

$$\cdots \dashv G_{-1} G_{-2} \dashv G_{-1} G_0 \dashv G_1 G_0 \dashv G_1 G_2 \dashv G_3 G_2 \dashv \cdots$$

Now $G_1: D \to D^{}$ is the constant diagram functor, so that $G_2$ is evaluation at $0$ and $G_{0}$ is evaluation at $1$. Then $G_3$ sends an object $X$ to the morphism $X\to 1$, and $G_{-1}$ sends it to the morphism $0\to X$. Thus $G_1 G_0(X\to Y) = (Y\to Y)$, while $G_1 G_2(X\to Y) = (X\to X)$ and $G_{-1} G_0(X\to Y) = (0\to Y)$. So as long as $D$ is nontrivial, $G_{-1} G_0$ is an object in a rigid monoidal category that is not isomorphic to its double right dual.

• Ha, I was just looking at that paper as well! It wasn't clear to me how to make them into endofunctors, so this is nice. I think I've found a more elementary example in my corrected answer. – Tobias Fritz Jun 20 '18 at 16:49

TL;DR: $X\cong (X^*)^*$ is not necessarily true, and this seems to be a folklore result, mentioned e.g. in these notes by Müger on p.9 (found by Eduardo Pareja Tobes). However, finding explicit examples of this phenomenon is not so trivial, and Müger in particular does not give one. Here's a general construction.

First, note that it's enough to find a monoidal category together with some object $X$ which has all iterated duals on both sides such that $X\cong (X^*)^*$. For then, we can consider the full subcategory on those objects which have all iterated duals, and this is closed under the monoidal structure due to the strong monoidality of taking duals, $(X\otimes Y)^* \cong Y^*\otimes X^*$.

Second, for any category $C$, it is well-known that the category of endofunctors $\mathrm{End}(C)$ is monoidal with respect to functor composition, and that dual objects correspond to adjoint functors. We therefore only need to find some category $C$ together with an endofunctor $F : C\to C$ which has all iterated adjoints, and such that its left and right adjoint are not isomorphic. There are plenty of examples in this MO question, but let me give a very concrete one for the sake of completeness.

Take $C = \mathbb{Z}$ to be the integers with their usual order considered as a category, and let $f : \mathbb{Z}\to\mathbb{Z}$ be the functor $f(x) = 2x$. This has a left adjoint given by $x\mapsto \lceil \frac{x}{2} \rceil$ and a right adjoint given by $x\mapsto \lfloor \frac{x}{2} \rfloor$. Thanks to $$\lceil \frac{x}{2} \rceil = \lfloor \frac{x+1}{2} \rfloor,$$ it's easy to see that we have a further left adjoint given by $x\mapsto 2x+1$. Continuing like this, we obtain an infinite sequence of adjoints, no two of which are the same. In their full glory, the individual adjuntions making up the sequence look like this: $$2x+n \leq y \quad \Leftrightarrow \quad x \leq \lfloor \frac{y-n}{2}\rfloor,$$ $$\lfloor \frac{x-n}{2}\rfloor \leq y \quad \Leftrightarrow \quad x\leq 2y + n + 1,$$ where the third functor $x\mapsto 2x + n + 1$ now looks the same as the first $x\mapsto 2x + n$, but with $n$ increased by $1$. The second equivalence is easiest to see using $\lfloor\frac{x-n}{2}\rfloor = \lceil \frac{x-n-1}{2} \rceil$.

• The adjoint functor theorem only says that every sup-preserving endofunctor has a right adjoint, and that every inf-preserving endofunctor has a left adjoint, so I don't think this works. – Mike Shulman Jun 20 '18 at 15:35
• @MikeShulman: of course, thanks for the correction. I'll see if I can fix this answer soon or will otherwise delete it. – Tobias Fritz Jun 20 '18 at 15:37
• Can you say any more about why you can "continue like this"? – Mike Shulman Jun 20 '18 at 16:57
• @MikeShulman: sure, I've expanded a bit. – Tobias Fritz Jun 20 '18 at 17:11

Another way of showing this isn't true in general (and this is a fairly systematic way to handle this kind of questions) is to look at string diagrams, ie at the free rigid (strict, say) monoidal category generated by one object $X$. It is not hard to see that morphisms in that category can be represented by "planar open oriented tangles", i.e. collections of intervals embedded in a rectangle so that the endpoints of each interval are attached to the top or bottom edge of the rectangle, modulo planar isotopy.

In that category, $Hom(X,X^{**})$ is simply empty. ANother way to say this is that if this statement was always true, then it would be a formal consequence of the axioms of rigid monoidal categories, and one would be able to write down such an isomorphism purely in terms of the duality data, and it's easy to see there is'nt even a candidate map (let alone an isomorphism).

It might be worth mentionning that the statement that does generalize to any rigid category, is the fact that there are canonical isomorphism $${}^*(X^*)\cong X \cong ({}^*X)^*.$$

• Can you say more precisely what "planar open oriented tangles" are, or provide a reference? The reason that I'm asking is because it sounds like the monoid of iso classes of objects in your monoidal category will be $\mathbb{N}$. But the monoid of iso classes of objects in the free rigid monoidal category with a single generator should be something that involves binary strings in letters "$\ell$" and "$r$", corresponding to iterated left and right dualization. – Tobias Fritz Jun 21 '18 at 14:40
• I think it would rather be $\mathbb{Z}$, implicitly using the identification above (the left dual of the right dual of $X$ is $X$). A nice reference is arxiv.org/abs/0908.3347 – Adrien Jun 21 '18 at 15:11
• Right, I meant to say $\mathbb{Z}$, not $\mathbb{N}$. Then taking the left or right dual corresponds to $x\mapsto -x$ in $\mathbb{Z}$, right? In particular, taking the double dual on either side is the identity $x\mapsto x$. Doesn't this give $X^{**} \cong X$? In other words, I'm afraid that what you're describing is not the free rigid monoidal category on a single generator. – Tobias Fritz Jun 21 '18 at 15:32
• No, taking the left/right dual is adding $\pm 1$, – Adrien Jun 21 '18 at 15:38
• I think I see the confusion, sorry: the set of objects is actually finite sequences of integers, not $\mathbb{Z}$, to account for the tensor product. – Adrien Jun 21 '18 at 15:40

Left and right duals are left and right adjoints in the 2-category determined by $\mathcal{C}$; as such there's no reason for them to coincide (they will of course under the presence of a symmetry, I think a braiding is enough too).

Searching around a bit, in these notes by Michael Müger we see on p16 that the left and right duals in the module category of a Hopf algebra do not coincide in general; and this condition is equivalent to X being isomorphic to its double (left or right) dual.

This should be related to Frobenius objects and the like; most likely someone more knowledgeable than me will chime in and say something more meaningful.

• By "do not coincide", do you mean "are not isomorphic"? If so, where exactly do you see that in Müger's notes? Wouldn't that be a very easy disproof of the conjecture mentioned in the comments? – Tobias Fritz Jun 20 '18 at 13:54
• Yup re isomorphic; p9 in those notes. But anyway I guess it's mostly obvious from the duals as adjoints picture. I just saw your comments above; I think that the point of a pivotal category lies in requiring the iso between Id and double dualization being monoidal; see on the structure of modular categories by Müger himself. – Eduardo Pareja Tobes Jun 20 '18 at 14:27