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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?

Marty
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  • 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 05:12

4 Answers4

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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).

César Galindo
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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.

Mariano Suárez-Álvarez
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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.

Marty
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  • 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
  • I THINK that what I describe is the twist discussed in Majid's Prop 2.3.4 (except that I don't care about $\mathcal R$, while Majid doesn't care about bialgebras which are not quasitriangular Hopf) up to dualization. What you describe seems more like a crossed product than like a twist. – darij grinberg Jan 25 '11 at 21:18
  • There are too many duals for me to follow. Are you twisting the multiplication in the function algebra $C[G]$ or the group algebra $CG$ (to use Majid's notation)? I'll try to work it out by hand to see what works... – Marty Jan 25 '11 at 23:12
  • Twisting multiplication in $C[G]$ doesn't change anything; twisting multiplication in $C^G$ makes it neither coco- nor commutative (at least I don't see a reason why it should). Make your choice ;) – darij grinberg Jan 26 '11 at 00:33
  • I'll take the latter. Thanks for the advice! – Marty Jan 26 '11 at 00:55
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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 $\mathbb{C}G$.

Sven Raum
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