Twist of a group Hopf-algebra - MathOverflow

Twist of a group Hopf-algebra

2011-01-25T17:49:47Z

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 $$[\Delta(f)](g_1, g_2) = f(g_1 \cdot g_2), [\epsilon(f)] = f(e), S(f) = f(g^{-1}),$$ 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?

Answer by Mariano Suárez-Alvarez for Twist of a group Hopf-algebra

2011-01-25T18:36:17Z

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.

Answer by Marty for Twist of a group Hopf-algebra

2011-01-25T20:00:25Z

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.

Answer by César Galindo for Twist of a group Hopf-algebra

2011-01-28T11:54:19Z

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).

Answer by Sven Raum for Twist of a group Hopf-algebra

2011-01-28T12:42:07Z

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 $\mathbb{C}G$.