Timeline for Frobenius isomorphism for Hopf algebras
Current License: CC BY-SA 3.0
11 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Jan 29, 2012 at 15:17 | vote | accept | user18951 | ||
| Jan 24, 2012 at 8:38 | comment | added | darij grinberg | @unknown: If there was a closed formula, there would be a functor from the category of finite-dimensional Hopf algebras (with all Hopf algebra morphisms, not just isomorphisms). I think this could be easily led to a contradiction, although I don't see how exactly. | |
| Jan 24, 2012 at 0:58 | answer | added | Ralph | timeline score: 1 | |
| Jan 23, 2012 at 23:34 | comment | added | user18951 | No, I just wanted to know if there is a closed formula for the Frobenius isomorphism that only uses the structure maps noted in the question. But according to Mariano's comment there seems to be little hope. | |
| Jan 23, 2012 at 22:56 | comment | added | darij grinberg | @unknown: Do you want a functor? If so, from which category? | |
| Jan 23, 2012 at 21:11 | comment | added | Mariano Suárez-Álvarez | It is a theorem that a finite dimensional Hopf algebra has (up to scalars) a unique integral in $H^*$; the space it spans can be described as the coinvariant subspace in $H^*$ for some Hopf module structure (this is explained in standard placed, like Sweedler's book or Susan Montgomery's)... It is not obvious what you mean by express, so it is hard to answer your question—I do doubt there is a universal expression for the isomorphism in the category generated by the structure maps of the Hopf algebra, though. | |
| Jan 23, 2012 at 20:19 | comment | added | user18951 | But the existence of such an integral is guaranteed by the Hopf Algebra axioms and the finite dimensionality, isn't it ? So isn't it possible to express this integral (and hence the isomorphism) in terms of the Hopf algebra's structure maps ? | |
| Jan 23, 2012 at 19:51 | comment | added | darij grinberg | I don't think this can be done using left integrals of $H$ itself. But if $\lambda$ is a left integral of $H^{\ast}$, then the linear map $H\to H^{\ast},\ a\mapsto a\lambda$ is an $H$-left linear isomorphism. | |
| Jan 23, 2012 at 19:36 | comment | added | user18951 | How would the isomorphism look then ? | |
| Jan 23, 2012 at 19:30 | comment | added | Mariano Suárez-Álvarez | Do you consider cointegrals to be part of The Hopf Algebra™? They are uniquely determined (up to scalars) from equations which can be written down in terms of the Hopf algebra structure maps, just as the antipode is... If you do, then yes. | |
| Jan 23, 2012 at 18:56 | history | asked | user18951 | CC BY-SA 3.0 |