Timeline for What is the opposite category of the category of modules (or Hopf algebra representations)?
Current License: CC BY-SA 2.5
6 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jun 26, 2010 at 5:36 | comment | added | Ryan Reich | There is actually a lot of separation in the Tannakian duaity theorem. You can prove any number of pieces almost independently: that a fiber functor is ind-representable; that the representing object is a coalgebra; that when the source is abelian then the fiber functor gives an equivalence with comodules over this coalgebra; that the tensor structure makes it an algebra; that the rigidity makes it a Hopf algebra. Then there are the recognition theorems for properties of the corresponding group.... | |
| Jun 26, 2010 at 5:18 | comment | added | Boyarsky | Aha, so your point is that "dual representation" involves both linear duality and group inversion, by separating these two ingredients it's all trivial in the end (and so of course is sure to be useless, sad to say). Thanks for the clarification. | |
| Jun 26, 2010 at 4:26 | comment | added | Ryan Reich | Maybe I should clarify: I am not swapping the multiplication with the comultiplication, but replacing each operation individually with its opposite operation with the arguments (or co-arguments) reversed. | |
| Jun 26, 2010 at 4:25 | comment | added | Ryan Reich | The opposite of a commutative ring is still commutative, no? It's just that I've chosen an unusual isomorphism between the two. | |
| Jun 26, 2010 at 4:17 | comment | added | Boyarsky | But then we have to give up commutativity of multiplication in Hopf algebras. Does the Tannakian stuff work in that generality, and is it useful? (I though the Tannakian stuff always spits out affine algebraic groups, albeit not necessarily of finite type, so the coordinate ring is a commutative ring. Maybe I am ignorant.) | |
| Jun 26, 2010 at 4:02 | history | answered | Ryan Reich | CC BY-SA 2.5 |