MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

Let $C$ be a symmetric monoidal category. I am interested in objects $X \in C$ such that the symmetry

$S_{X,X} : X \otimes X \cong X \otimes X$

is equal to the identity. There are many examples of such objects, e.g. invertible sheaves. My first question is: How would you call such an object?

Now assume that $X$ has a dual $Y$, i.e. we have morphisms $e: Y \otimes X \to 1$ and $c : 1 \to X \otimes Y$ such that the triangular identities are satisfied.

Question. Assuming $S_{X,X}$ is the identity, can we conclude that $S_{Y,Y}$ is the identity? If not, does it suffice to assume that $e$ (and thus $c$) is an isomorphism?

Edit: I am still interested how objects with $S_{X,X}=\mathrm{id}$ are called in the literature or which terminology you would suggest.

share|cite|improve this question
up vote 5 down vote accepted

I believe the answer to your question is yes, without a further assumption that e is an isomorphism. The symmetry S_{Y,Y} can be obtained from the symmetry S_{X,X}

as follows

$Y\otimes Y \xrightarrow{c\circ c} Y\otimes Y \otimes X\otimes X \otimes Y \otimes Y \xrightarrow{id_Y^{\otimes 2} S_{X,X}\otimes id_Y^{\otimes 2}} Y\otimes Y \otimes X\otimes X \otimes Y \otimes Y \xrightarrow{e\circ e}Y\otimes Y$.

Here, $c\circ c$ is shorthand for $(id_Y^{\otimes 2}\otimes c \otimes id)\circ(id_Y^{\otimes 2}\otimes c)$, and similarly for $e\circ e$.

In pictures, all I'm doing (which I would draw if I knew an easy way) is:

Take $Y \otimes Y$ up, and then bend them around to the right and back down (they become X's on the downward strand, apply $S_{X,X}$, then bend the X's back around and up to the right (where they become Y's again.

here is a pdf of the computation

I am just really repating a proof here that $S_{U^*,V^*}=S_{U,V}^*$, which holds for the braiding in any rigid braided monoidal categetory.

Since $S_{X,X}$ is the identity, you will get a diagram which is recognizable as the identity for $Y\otimes Y$.

share|cite|improve this answer
Thanks. Why does this composition agree with $S_{X,X}$? Also, do you have a reference for the general fact $S_{U^*,V^*}=S_{U,V}^*$? – Martin Brandenburg Oct 25 '11 at 14:01
I imagine it's in Kassel's book quantum groups, among other places. To see that this composition agrees with $S_{Y,Y}$ (that's what you meant I think), you use the naturality of the braiding. So you can rewrite this as a morphism where you do $c\circ c$ and then the $e\circ e$ (which cancel), and then the braiding. Let me add a picture. – David Jordan Oct 25 '11 at 16:01
picture added. should have done so in the first place. Note that the linked pdf has the proof in general that S_{U^*,V^*}=S_{U,V}^* implicit, since there was no need for both the original slots to be equal. – David Jordan Oct 25 '11 at 16:25
See pag.46 on "CAtegories TAnnakiennes" Lnm 265 (Neantro Saavedra Rivano) – Buschi Sergio Oct 25 '11 at 16:33
@Buschi: Does this really help here? In my version, this is just a trivial reformulation that two objects are inverse to each other. @David: Thanks a lot! Meanwhile I've also found the relevant section in Kassel's book. It will take a while to digest it. – Martin Brandenburg Oct 25 '11 at 20:37

Brandenburg, I think that the answere is yes:

From the theory of adjunctions given $(F_k, G_k, \epsilon_k, \eta_k): \mathcal{C}\to \mathcal{C}$ for $k=1, 2$ (Maclane CWM notations), and given a natural morphism $\phi: F_2\circ F_1 \to F_1\circ F_2$ there exist a natural morphisms $\widetilde{\phi}: G_1\circ G_2 \to G_2\circ G_1$ defined as :

$G_1G_2\xrightarrow{\eta_2 G_1G_2} G_2F_2G_1G_2 \xrightarrow{G_2\eta_1 F_2 G_1G_2} G_2G_1F_1F_2G_1G_2$

$\xrightarrow{GG\phi F_2 G_1G_2} G_2G_1F_2F_1G_1G_2 \xrightarrow{GGF\epsilon_1 G} G_2G_1F_2G_2\xrightarrow{GG\epsilon_2} G_2G_1$

Considering the case $(F_1, G_1, \epsilon_1, \eta_1)= (F_2, G_2, \epsilon_2, \eta_2)$ and indicate it as
$(F, G, \epsilon, \eta)$.

By naturality, we have $GF\epsilon\ast \eta FG= \eta\ast \epsilon $, then $GGF\epsilon\ast G\eta FG= G\eta\ast G\epsilon $, then $GGF\epsilon G\ast G\eta FGG= G\eta G\ast G\epsilon G $.

Let $\phi=1$, then $\widetilde{\phi}= GG\epsilon\ast GGF\epsilon G\ast G\eta FGG\ast \eta GG = GG\epsilon\ast G\eta G\ast G\epsilon G \ast \eta GG =$

$=G(G\epsilon\ast \eta G)\ast (G\epsilon \ast \eta G)G =1_G\ast1_G=1_G $.

Now we use this proof for a 2-category with a only one object, (essentially a strict monoidal category), and then to a bicategory with one object (essentially a monoidal category).

share|cite|improve this answer
I realized that mine answer is repetitive (see the Jordan answere above), because don't show that $\widetilde{\phi}$ is the symmetry . – Buschi Sergio Dec 7 '12 at 21:31

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

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