Does the dual of an object with trivial symmetry also have trivial symmetry? - MathOverflow most recent 30 from http://mathoverflow.net 2013-05-22T08:03:14Z http://mathoverflow.net/feeds/question/79068 http://www.creativecommons.org/licenses/by-nc/2.5/rdf http://mathoverflow.net/questions/79068/does-the-dual-of-an-object-with-trivial-symmetry-also-have-trivial-symmetry Does the dual of an object with trivial symmetry also have trivial symmetry? Martin Brandenburg 2011-10-25T12:25:36Z 2012-12-07T15:58:52Z <p>Let $C$ be a symmetric monoidal category. I am interested in objects $X \in C$ such that the symmetry</p> <p>$S_{X,X} : X \otimes X \cong X \otimes X$</p> <p>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?</p> <p>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.</p> <p><strong>Question</strong>. 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?</p> <p>Edit: I am still interested how objects with $S_{X,X}=\mathrm{id}$ are called in the literature or which terminology you would suggest.</p> http://mathoverflow.net/questions/79068/does-the-dual-of-an-object-with-trivial-symmetry-also-have-trivial-symmetry/79077#79077 Answer by David Jordan for Does the dual of an object with trivial symmetry also have trivial symmetry? David Jordan 2011-10-25T13:29:10Z 2011-10-25T16:26:02Z <p>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}</p> <p>as follows</p> <p>$Y\otimes Y \xrightarrow{c\circ c} Y\otimes Y \otimes X\otimes X \otimes Y \otimes Y \xrightarrow{id_Y^{\otimes 2} S_{X,X}\otimes id_Y^{\otimes 2}} Y\otimes Y \otimes X\otimes X \otimes Y \otimes Y \xrightarrow{e\circ e}Y\otimes Y$.</p> <p>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 $e\circ e$.</p> <p>In pictures, all I'm doing (which I would draw if I knew an easy way) is:</p> <p>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.</p> <p>here is a <a href="http://math.mit.edu/~djordan/2011-10-25-Note-11-11.pdf" rel="nofollow">pdf of the computation</a></p> <p>I am just really repating a proof here that $S_{U^<em>,V^</em>}=S_{U,V}^*$, which holds for the braiding in any rigid braided monoidal categetory.</p> <p>Since $S_{X,X}$ is the identity, you will get a diagram which is recognizable as the identity for $Y\otimes Y$.</p> http://mathoverflow.net/questions/79068/does-the-dual-of-an-object-with-trivial-symmetry-also-have-trivial-symmetry/115723#115723 Answer by Buschi Sergio for Does the dual of an object with trivial symmetry also have trivial symmetry? Buschi Sergio 2012-12-07T15:58:52Z 2012-12-07T15:58:52Z <p>Brandenburg, I think that the answere is yes:</p> <p>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 :</p> <p>$G_1G_2\xrightarrow{\eta_2 G_1G_2} G_2F_2G_1G_2 \xrightarrow{G_2\eta_1 F_2 G_1G_2} G_2G_1F_1F_2G_1G_2$</p> <p>$\xrightarrow{GG\phi F_2 G_1G_2} G_2G_1F_2F_1G_1G_2 \xrightarrow{GGF\epsilon_1 G} G_2G_1F_2G_2\xrightarrow{GG\epsilon_2} G_2G_1$ </p> <p>Considering the case $(F_1, G_1, \epsilon_1, \eta_1)= (F_2, G_2, \epsilon_2, \eta_2)$ and indicate it as<br> $(F, G, \epsilon, \eta)$.</p> <p>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 $GGF\epsilon G\ast G\eta FGG= G\eta G\ast G\epsilon G$.</p> <p>Let $\phi=1$, then $\widetilde{\phi}= GG\epsilon\ast GGF\epsilon G\ast G\eta FGG\ast \eta GG = GG\epsilon\ast G\eta G\ast G\epsilon G \ast \eta GG =$</p> <p>$=G(G\epsilon\ast \eta G)\ast (G\epsilon \ast \eta G)G =1_G\ast1_G=1_G$.</p> <p>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).</p>