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2-cocycle twists of Hopf algebras

Let $H$ be a Hopf algebra over a field $k$. Then a (left, unital) 2-cocycle on $H$ is a map $$ f: H \otimes H \to k$$ such that $$ f(x_{(1)},y_{(1)})f(x_{(2)} y_{(2)}, z) = f(y_{(1)}, z_{(1)}) f(x, y_{(2)} z_{(2)}) $$ (in Sweedler notation) and $$ f(x,1) = \varepsilon(x) = f(1,x) $$ for $x,y,z \in H$. Also one usually wants that $f$ is invertible for the convolution product, i.e. that there is some functional $$ \bar{f} : H \otimes H \to k $$ such that $$ f(x_{(1)}, y_{(1)}) \bar{f}(x_{(2)}, y_{(2)}) = \varepsilon (x) \varepsilon (y) = \bar{f} (x_{(1)}, y_{(1)}) f(x_{(2)}, y_{(2)}). $$

The cocycle can be used to twist the multiplication of $H$ as follows: $$ x \cdot_f y = f(x_{(1)}, y_{(1)}) x_{(2)} y_{(2)} \bar{f}(x_{(3)}, y_{(3)}), $$ The cocycle condition ensures that $\cdot_f$ is associative, and the fact that $f$ is unital means that the old identity element is still the identity element for $\cdot_f$.

You can also twist the antipode of the Hopf algebra in such a way as to get a new Hopf algebra structure on $H$, where the comultiplication and counit are the same as before. You have to check that the original comultiplication and counit are still algebra homomorphisms with respect to $\cdot_f$.

Twisting the algebra structure only

You can modify this construction in such a way as to obtain only an algebra instead of a Hopf algebra. This is done by defining $$ x \cdot_f y = f(x_{(1)}, y_{(1)}) x_{(2)} y_{(2)}. $$ Again, $f$ being unital means that the identity element is the same.

The braided setting

All of this can be done in exactly the same way whenever $H$ is a Hopf algebra object in a braided monoidal category. All the maps can be interpreted as string diagrams, and the proofs that $\cdot_f$ is associative and that the comultiplication and counit are algebra maps for the twisted multiplication go through in exactly the same way by manipulating the string diagrams.

My question is:

Does anybody know a reference for 2-cocycle twists of a Hopf algebra object in a braided monoidal category? I've looked in Majid's book, and he does talk a great deal about Hopf algebras in braided categories (what he calls braided groups), and also about 2-cocycle deformations, but I don't think he puts the two concepts together. Please correct me if I'm wrong. Thanks!

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1 Answer 1

up vote 6 down vote accepted

The two concepts - twisting a Hopf algebra one-sided to an algebra and two-sided to a new Hopf algebra - are actually intimately connected and play an important role in several areas of current research. The former are known (as always, there's restrictions on the equivalence, e.g. in infinite dimension) as Galois objects, the latter as Doi-twist.

A Galois object over a Hopf algebra $H$ is an $H-$comodule algebra $_HA$ with a nondegeneragy condition (can be bijective); mostly (if a cleaving exists, s.a.) this means $A=H^\sigma$ is a one-sided 2-cocycle deformation. It can be completed to a Bigalois object $_HA_L$ (a second compatible comodule structure on the other side) between $H$ and what turns out to be the (two-sided Hopf algebra) Doi-twist $L=H_\sigma$. These can be multiplied by the contensor product if they fit together:

$$_HA_L\;\;and\;\;_LB_M\rightarrow\;_H(A\Box_L B)_M$$

This forms the Bigalois gruppoid (gruppoid because you dont stick with one Hopf algebra $H$, but "hop accross" the Doi twists $L,M,...$ and can only multiply "fitting ends"), which is the best analogon you have to a cohomology group (which is no group for the reason above!) and usually much more organized to actually calculate with...the basics can be found e.g. in Schauenburg's paper or more elaborate in Susan Montgomory's "Hopf algebras and their actions on rings". There are many worked out examples (e.g. for generalized Taft algebras), stuff like a Küneth-formula etc.pp.

On the other hand, Doi-twists are a very "mild" operation, that do not change the module-category....that's way the question "what Doi twists are there of $H$?" often appears as the question for quasi-isomorphisms $H\rightarrow L$ - this may also help googeling ;-)

Now directly to the question...

Though I've never seen it, I'd definitely agree that you get no problem, as all the concepts (both approaches) just use the braided category structure ("braided" doesn't mean it's nontrivial, just that you got the structure map)...or as you say, are expressable in string diagrams.

Are you interested in a specific use of these deformations? Do you have a specific braided category in mind?

...and a good source of examples?

My favourite examples of braided categories "in-use" are the Yetter-Drinfel'd modules $V$ over a finite group $G$, which means: $G$-graded, $G$-action, such that $g.V_h=V_{ghg^{-1}}$. Of course this taste is influenced by it being my field of work ;-) They include the "super- and color-spaces" ($\mathbb{Z}_2,\mathbb{Z}_3$), but of course mus more...

In these categories you have much-studied braided Hopf algebras resembling (but maybe finite-dimensional!) a universal enveloping of a Lie algebra, the Nichols algebras. These are glued to the groupring itself to form new quantum groups by a Radford/Majid-construction, and often deformed by a Doi twist to get a complete classification (Andruskiewitsch/Schneider)...well, usually AFTER the relations are found it is proven, that it's really a Doi twist...

I think one could use these known Doi twists of the Hopf algebra to write down deforming 2-cocycles just on the Nichols algebra in the braided sense. I've tried this for the "generic cases" (linking- and root-relations), but there once you get rid of the groupring, they're trivial (because in the deformed relations some $1-g$ appears...I could even imagine that's why they're the "easyer" ones??). However I've found several exceptional later-worked-out cases in literature (e.g. selflinkings Daniel Didt's Dissertation p. 46-49) where this is no longer the case. Here you have "real" Nichols-with-Nichols-to-Nichols relations and in my opinion these deformations have to correspond to nontrivial braided 2-cocycles on the Nichols algebra part alone...(?)

If there's still interest in the question (it's been almos a year) I'm sure we could work something out :-)

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