Timeline for Does the dual of an object with trivial symmetry also have trivial symmetry?
Current License: CC BY-SA 3.0
9 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Oct 26, 2011 at 19:10 | vote | accept | Martin Brandenburg | ||
| Oct 25, 2011 at 20:37 | comment | added | Martin Brandenburg | @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. | |
| Oct 25, 2011 at 16:33 | comment | added | Buschi Sergio | See 2.5.4.2 pag.46 on "CAtegories TAnnakiennes" Lnm 265 (Neantro Saavedra Rivano) | |
| Oct 25, 2011 at 16:26 | history | edited | David Jordan | CC BY-SA 3.0 |
added 2 characters in body
|
| Oct 25, 2011 at 16:25 | comment | added | David Jordan | 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. | |
| Oct 25, 2011 at 16:16 | history | edited | David Jordan | CC BY-SA 3.0 |
added 167 characters in body; deleted 63 characters in body
|
| Oct 25, 2011 at 16:01 | comment | added | David Jordan | 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. | |
| Oct 25, 2011 at 14:01 | comment | added | Martin Brandenburg | Thanks. Why does this composition agree with $S_{X,X}$? Also, do you have a reference for the general fact $S_{U^*,V^*}=S_{U,V}^*$? | |
| Oct 25, 2011 at 13:29 | history | answered | David Jordan | CC BY-SA 3.0 |