MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Sign up
Here's how it works:
  1. Anybody can ask a question
  2. Anybody can answer
  3. The best answers are voted up and rise to the top

I can't understand this sentence i the article of Majid "Tannaka-Krein theorem for quasi-Hopf algebras and other results" about the reconstruction of a quasi-algebra (in fact its dual) from a given braided category (C,c) with a forgetful functor F ; at the end it considers the action of Drinfeld-twisting upon quasi-algebra and says: if A (quasi-algebra) corresponds to (C,F,c) then A_twisted corresponds to (C,F,c') for a new choice of braiding c'. i am very disturbed: the associativity constraint is modified too obviously; sorry for my very naive question.

share|cite|improve this question
Why don't you email him? – José Figueroa-O'Farrill Mar 20 '10 at 16:12
You're going to have to give more context if you want an answer. It's not fair to require us to locate this paper and read through it to find your question. Is there a link to the paper online? What page? If not could you excerpt the relevant statement? – Noah Snyder Mar 20 '10 at 16:27
The article, incidentally, is in the collection "Deformation Theory and Quantum Groups with Applications to Mathematical Physics", Contemp. Math. 134, 1991. I'm not sure which particular sentence OP is confused by. – Theo Johnson-Freyd Mar 20 '10 at 17:56
up vote 2 down vote accepted

You might find these notes helpful as a second source to read fro:

There is a careful elaboration of the various variants of Tannaka-Krein construction, and the case of quasi-bialgebras and quasi-Hopf algebras is discussed. So far as I recall, Drinfeld twisting of a quasi bi-algebra just means acting on the original category by a tensor functor which is the identity on objects (and so is only interesting as a tensor functor).

As such, it does have a fairly straightforward effect on the associator $\phi$. If you twist by an invertible element $F\in H\otimes H$, then the associator $\phi^F$ is something like $\phi^F:=B\phi B^{-1}$, where $B:=F(\Delta\otimes\operatorname{id})(F)$, or maybe $F(\operatorname{id}\otimes\Delta)(F)$, I can't remember.

It might help you to note that nothing in your question relies on the braiding; there is a reconstruction theorem for any (quasi)bi-algebra with a fiber functor, and since you're asking about the associator, it isn't too important that there's a braiding lying around.

What leads you to expect some more complicated effect on the associator?

share|cite|improve this answer

You probably know all that I'm going to say, but I'll mention it anyway, for any other interlocutors:

I'm going to adopt Stasheff's recommendation ("Drinfel$'$d's quasi-Hopf algebras and beyond", same volume) that "Drinfel$'$d-twisting" be called "skrooching" (a transliteration and abbreviation of the Russian, which is roughly four syllables.). Also, I'm going to talk about associative algebras and quasicoassociative coalgebras, not the other way around, even though the reconstruction theorem lands in "co" territory: if you really want to do Majid's theorem, you have to think about the convolution multiplication.

Anyway, then the basic Drinfel$'$d idea is that you start with something that I think should be called a "weak bialgebra": an associative unital algebra $A$ and an algebra homomorphism $\Delta : A \to A\otimes A$. This gives a functor $\otimes: A\text{-rep} \times A\text{-rep} \to A\text{-rep}$, which forgets to the usual $\otimes$ on vector spaces.

A quasicoassociative bialgebra is a weak bialgebra along with a distinguished invertible element $\psi \in A^{\otimes 3}$ so that $(\text{id} \otimes \Delta)\circ \Delta = \text{Ad}_\psi \circ (\Delta \otimes \text{id})\circ \Delta$, where $\text{Ad}_\psi$ is the inner automorphism of $A^{\otimes 3}$ by conjugating by $\psi$ — then acting by $\psi$ gives a natural $A\text{-rep}$ isomorphism $(X\otimes Y)\otimes Z \to X \otimes (Y\otimes Z)$ — and there's a pentagon for $\psi$ that matches Mac Lane's pentagon.

Given any invertible element $\varphi \in A^{\otimes 2}$, you can skrooch the weak bialgebra $(A,\Delta)$ by $\varphi$ to get a new weak bialgebra $(A,\Delta^\varphi)$, which is the same algebra and $\Delta^{\varphi} = \text{Ad}_\varphi\circ \Delta$. If $(A,\Delta,\psi)$ is quasicoassociative, then $(A,\Delta^\varphi,\psi^\varphi)$ is, with: $$ \psi^\varphi = \text{Ad}_\phi \psi \quad \text{where} \quad \phi = (\varphi\otimes 1) \cdot (\Delta \otimes \text{id})(\varphi) \in A^{\otimes 3}$$ and $\cdot$ is the multiplication in $A^{\otimes 3}$. You can translate this pretty straightforwardly into the representation theory.

The easier observation is quasitriangularity. A weak bialgebra $(A,\Delta)$ is quasitriangular if it comes equipped with a skrooch $\rho \in A^{\otimes 2}$ such that $\Delta^\rho = \Delta^{\rm op}$. A (quasi)coassociative bialgebra is quasitriangular if additionally $\rho$ satisfies two hexagons. If you skrooch a quasitriangular bialgebra $(A,\Delta,\rho)$ by $\varphi$, then you get a new quasitriangular bialgebra $(A,\Delta^\varphi,\rho^\varphi)$, where $\rho^\varphi = \varphi^{\rm op}\rho\varphi^{-1}$ (multiplication in $A^{\otimes 2}$; $\varphi^{\rm op}$ is the obvious element with the two $A$s in $A^{\otimes 2}$ flipped).`

share|cite|improve this answer

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.