Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

Let $G$ be a finite group with identity element $e$, and $C[G]$ the ring of complex-valued functions on $G$, with pointwise addition and multiplication. Then $C[G]$ is naturally a Hopf algebra, with comultiplication, counit, and antipode given by $$[\Delta(f)](g_1, g_2) = f(g_1 \cdot g_2), [\epsilon(f)] = f(e), [S(f)](g) = f(g^{-1}),$$ for all $f \in C[G]$, $g, g_1, g_2 \in G$. Here we identify $C[G] \otimes_C C[G]$ with $C[G \times G]$ in the usual way.

If $\chi \in Z^2(G, C^\times)$ is a normalized (meaning $\chi(g,1) = \chi(1,g) = 1$) two-cocycle, then one can "twist" the coalgebra structure on $C[G]$, defining new comultiplication and antipode by $$\Delta_\chi(f) = \chi \cdot \Delta(f) \cdot \chi^{-1}, S_\chi f = U (S f) U^{-1},$$ where $U = \sum_i \chi_i^{(1)} (S \chi_i^{(2)})$.

This, according to Theorem 2.3.4 of Shahn Majid's "Foundations of quantum group theory," produces a new Hopf algebra structure. It's typically called $C_\chi[G]$ (though my notation differs slightly from Majid's). Here, I think that we are viewing $\chi$ -- a priori a function from $G \times G$ to $C^\times$ -- as an element $\sum_i \chi_i^{(1)} \otimes \chi_i^{(2)} \in C[G] \otimes_C C[G]$, identifying complex-valued functions on $G \times G$ with elements of $C[G] \otimes C[G]$ as usual, and for some reason $U$ is invertible in the ring $C[G] \otimes C[G]$.

Now for the question...

The Hopf algebra $C_\chi[G]$ obtained through this process is still a commutative Hopf algebra over $C$; only the antipode and comultiplication were changed. So $Spec(C_\chi[G])$ is an affine group scheme over $C$, whose $C$-points are in bijection with the elements of $G$.

So... what is this group scheme?! Or have I messed something up in this construction? It seems very odd to me, since the cocycle should -- group theoretically -- produce a central extension of $G$ by $C^\times$, and I don't know how such a thing would be encoded in a group scheme whose $C$-points are in bijection with $G$. Any references for this group scheme?

share|improve this question
    
I don't understand, multiplication in $\mathbb C[G]\bigotimes\mathbb C[G]$ is commutative so $\Delta_\chi$ looks like it is equal to $\Delta$. –  Torsten Ekedahl Jan 25 '11 at 19:18
    
@Torsten: Of course... I think you are right. I'll put it down as an answer. It's just weird that Majid makes something of this example in his book -- it threw me off completely. –  Marty Jan 25 '11 at 19:55
    
Note that you can twist the coalgebra structure (just not by the formula above), it won't be a Hopf algebra though. –  Torsten Ekedahl Jan 26 '11 at 5:12

4 Answers 4

Suppose that $H$ is a Hopf algebra over a field $k$.

There are two dual notions that we can use in order to deform $H$, one is Drinfeld's twists and the other is Hopf 2-cocycle.

(Remark: Hopf 2-cocycles also are related with Hopf-Galois theory, since there exists a biyective correspondence between Hopf 2-cocycle and cleft Galois object (see Hopf bigalois extensions), and for finite dimensional Hopf algebras, the classification of Hopf 2-cocycles is the same as the classification of Galois object.)

A 2-cocycles over $H$ (also called Hopf 2-coycles) is a linear map $\sigma:H\otimes H\to k$, invertible w.r.t the convolution product, and such that $\sigma(a_1, b_1)\sigma(a_2b_2, c) = \sigma(b_1, c_1)\sigma(a, b_2c_2).$ Using Hopf 2-cocycles you can get a new Hopf twisting the multiplication: $a\cdot_\sigma b = \sigma(a_1,b_1)a_2b_2\sigma^{-1}(a_3,b_3)$ and using the original comultiplication. As an example, for the group algeba $kG$, a Hopf 2-cocycle is the same as an usual 2-cocycle, where $G$ acts trivialy on $k^*$. In this case each twisting is again $kG$.

Now, for the Hopf algebra $k^G$ all is diferent, Hopf 2-cocycles are more complicated and interesting (see Movshev and C.G and Medina for a classification). In this case the deformations are not trivial, and you get examples of non-commutative and non-cocommutative Hopf algebras (See C.G. and Natale for concrete examples of non-trivial deformations).

share|improve this answer

The Hope algebra arising from this "twist" is just the original Hopf algebra, since multiplication is commutative, so $\Delta = \Delta_\chi$ and $S = S_\chi$. So it's a pretty dumb question, after all.

Then again, it makes me think that I'm not twisting the Hopf algebra correctly. After all, one can twist the multiplication in the group algebra $CG$ to obtain a (still cocommutative) Hopf algebra $C_\chi G$ algebra that's honestly different from $CG$. And one can take the dual, to obtain $(C_\chi G)^\ast$. That's probably what I want to do, and I should probably make it a new question.

share|improve this answer
    
The fact that Majid(?) denotes by $C\left[G\right]$ what most of the world calls $C^G$ isn't useful either... –  darij grinberg Jan 25 '11 at 20:16
    
Anyway, if you want a nontrivial twist, then you must either twist the multiplication of a non-cocommutative bialgebra, or twist the comultiplication of a non-commutative bialgebra. Either way, I think the result is (in general) neither commutative nor cocommutative, so there shouldn't be an algberaic group interpretation. Or do I get it wrong? –  darij grinberg Jan 25 '11 at 20:22
    
Well, you can twist the multiplication of the noncommutative (if $G$ is nonabelian) bialgebra $CG$ as I mentioned above -- just put $g_1 \cdot_\chi g_2 = \chi(g_1, g_2) \cdot g_1 \cdot g_2$. This produces $C_\chi G$, which is still cocommutative. Similarly, I suppose that one can twist the comultiplication on $C[G]$ -- probably not by the formula I had in my original question. But probably matching the dual of $C_\chi G$ as I put in this answer. –  Marty Jan 25 '11 at 20:25
    
Are you sure this twist will be a bialgebra? The twist that I know goes like this: $x\cdot_{\sigma} y=\sigma(x_{(1)},y_{(1)})x_{(2)}y_{(2)}\sigma^{-1}(x_{(3)},y_{(3)})$. If you do this to a cocommutative algebra, it doesn't change. –  darij grinberg Jan 25 '11 at 20:41
    
@Grinberg: Argh! Too many possibilities. What you describe seems like the twist dicsussed in Prop. 2.3.8 of Majid's book. That would seem to twist the function algebra $C[G] = C^G$, since $\sigma$ is a normalized two-cocycle for the group if and only if its a counital 2-cocycle for the function algebra. So this explains how to obtain a not-necessarily commutative twist of the function algebra. Very good, I think -- and it explains why you don't get the function algebra of a group scheme. –  Marty Jan 25 '11 at 21:08

This is wrong for silly reasons...

Since you are using complex coefficients and the group is finite, all $2$-cocycles are coboundaries. So your twisted group algebra is simply isomorphic to the original group algebra: you don't get anything new.

share|improve this answer
    
I'm confused, aren't the coefficients in the multiplicative group and not the additive group (see ncatlab.org/nlab/show/Drinfel'd+twist for a formula that's certainly written multiplicatively). –  Noah Snyder Jan 25 '11 at 18:41
    
Urgh, you are right! –  Mariano Suárez-Alvarez Jan 25 '11 at 18:43

In one case you will always get a group isomorphic to the group you started with, namely if your cocycle is symmetric. Actually in http://arxiv.org/abs/1007.1412 Theorem 3.2 proves that all symmetric cocycles on a cocommutative, compact quantum group are trivial. A quantum group is to be understood in the von Neumann algebraic sense there, but for finite groups this point of view is equivalent to the Hopf algebraic $\mathbb{C}G$.

share|improve this answer

Your Answer

 
discard

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.