Timeline for quantum groups... not via presentations
Current License: CC BY-SA 3.0
12 events
| when toggle format | what | by | license | comment | |
|---|---|---|---|---|---|
| Aug 25, 2012 at 21:52 | vote | accept | André Henriques | ||
| Aug 23, 2012 at 12:55 | comment | added | Theo Johnson-Freyd | If you are unlucky, then at worst you can take the isomorphism class which includes the quantum group constructed from the presentation you gave with some random choice of Cartan. An intermediate amount of luck is that this isomorphism class consists of all fibers that degenerate to the standard one on $\mathrm{Rep}(\g)$ as $q\to 0$; the standard one is $\hom(U\g,-)$. Finally, two orthogonal disclaimers: you may have to sprinkle in associators and the word "quasi"; you may have to take "Hopf duals" from time to time (getting things like "Hopf coalgebroids"). | |
| Aug 23, 2012 at 12:50 | comment | added | Theo Johnson-Freyd | A braided Hopf algebroid is a natural isomorphism between $\Delta$ and $\Delta^{op}$ (with the obvious definition) satisfying some Yang-Baxter type relation. What I believe that you can do now is to look up the correct papers on Tannakian reconstruction for braided Hopf algebras (there must be many write-ups ... Magid or Majid or Etingof or someone), and play the game but consider all fiber functors simultaneously, and you should be able to build the above structure. If you're lucky, then for quantum groups of algebraically closed fields all fibers are isomorphic. | |
| Aug 23, 2012 at 12:43 | comment | added | Theo Johnson-Freyd | The reason the Hopf condition is unnatural is that it required choosing an object set $C_0$. I could have equivalently defined a representation of $C$ to be a linear functor $C \to VECT$. Up to equivalence of categories, all we then have is a linear category $C$ and its monoidal category of representations; but to say the Hopf condition requires that this category be "over" something for which the canonical morphism between internal homs is an iso. So really "bialgebroid" just means "linear category with monoidal structure on its rep'ns" and it is functors of bialgebroids that are Hopf. | |
| Aug 23, 2012 at 12:39 | comment | added | Theo Johnson-Freyd | The Hopf condition is less natural, for a good reason that I'll get back to. There is an obvious tensor product of functors $C_0 \to VECT$, which is the tensor product fiberwise over each object of $C_0$. By construction, the restriction map from $C$-representations to $\hom(C_0,VECT)$ is monoidal. Thus it induces a natural transformation between internal homs in the two different categories, which I'll let you work out. The cleanest way to say the Hopf condition is that this natural transformation is an isomorphism. It is a general fact that Hopf-ness is an open condition for deformations. | |
| Aug 23, 2012 at 12:33 | comment | added | Theo Johnson-Freyd | Recall that a "(quasi) bialgebroid" $C$ is a linear category with object-set $C_0$ and a functor $\Delta: C \to C \times_{C_0} C$, where the codomain is the fibered product of linear categories, and $C_0$ is thought of as a discrete linear category, i.e. different objects have only the $0$ map between them, and all endomorphism algebras are one-dimensional. A representation of $C$ is a functor $F: C_0 \to VECT$ and the natural notion of action; then $\Delta$ determines a functor of two variables on the category of $C$-representations, and I demand it to be (non-strict) monoidal. | |
| Aug 23, 2012 at 12:27 | comment | added | Theo Johnson-Freyd | @Andre: Not really — again I'm going based on intuition from the group case. What I'd expect is that there is a category whose objects are faithful braided monoidal functors from "$\mathrm{Rep}(U_q(\g)$" to some version of VECT; equivalently, the objects of the category are braided coalgebras which are generators of the category. Morphisms are all linear maps, but this category should be some version of "Hopf algebroid" where the comultiplication encodes the various coalgebra structures, e.g. the grouplike elements are the coalgebra homomorphisms. | |
| Aug 23, 2012 at 12:14 | comment | added | Theo Johnson-Freyd | @B. Bischof: That said, most likely even if there are other fiber functors, I can ask for those that degenerate to "the" fiber functor as $q\to 1$, and these should all be isomorphic, I'd expect. | |
| Aug 23, 2012 at 12:13 | comment | added | Theo Johnson-Freyd | @B. Bischof: I believe so, but I could be mistaken. Certainly to have a chance of any two fibers being isomorphic I had better word over an algebraically closed field — otherwise you should expect that the fibers correspond to "Galois actions" of the quantum group on the field. But I'm trusting intuition from the case of actual algebraic groups. | |
| Aug 23, 2012 at 9:48 | comment | added | André Henriques | Thank you Theo for your insightful answer. I'm quite curious about the Hopf algebroid you mentioned in your last paragraph... can you say a bit more about it? | |
| Aug 22, 2012 at 22:57 | comment | added | B. Bischof | To be clear, are you saying there is a Tannakian reconstruction for Uq(g) up to non-unique isomorphism? | |
| Aug 22, 2012 at 14:35 | history | answered | Theo Johnson-Freyd | CC BY-SA 3.0 |