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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.

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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 $Y\otimes Y$.

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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).

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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

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