# Intuition behind the relation of commutative Hopf algebra and Groups

I've heard that a commutative Hopf algebra can be thought an algebra construction over the space of functions on a group to the ground field. The product should be the pointwise multiplication, Coproduct maps $f$ to the functions on 2 variables $(x,y)↦f(xy)$. Counit is evaluation $f↦f(1)$. Antipode maps f to $x↦f(x^−1)$. Everything is fine so far. Here's my question:

1) How it is related this construction with the other canonical group algebra $\mathbb CG$ where the coproduct is given by $$\triangle\left(g\right)=\left(g\otimes g\right),$$ with $g\in G$. Is it the same construction with different notation? How can I show it's the same?

2) I also don't get if there's an equivalence between commutative Hopf algebras and Groups and to which extent (does the group have to be finite or it can be a Lie Group)?

3) how this is related if it is with the definition of Quantum Groups as non-commutative nor co-commutative Hopf Algebras?

Please if someone can do a systematic and intuitive argument for 1 and for 2 which I suppose to be strictly related. Thank you

• Remember that the group algebra is a cocommutative but not in general a commutative Hopf algebra. So the commutative one is the dual of the group algebra, except you need the group to be finite for this to give something of the same size. – Tobias Kildetoft Mar 5 '18 at 9:55
• In general, a commutative Hopf algebra is the same as an affine group scheme. – Tobias Kildetoft Mar 5 '18 at 9:59
• In addition, the study of quantum groups is not the study of group schemes. Typical quantum groups arising in physics are the so-called Drinfeld-Jimbo algebras. They are not commutative nor co-commutative. There is no definition of a quantum group in general though. Though algebraists and functional analysts have a strong opinion on what a quantum group should be. – Mathematician 42 Mar 5 '18 at 10:02
• if possible could you clarify the problems express in point 1? I don't get the relation if any with this definition of Group Hopf Algebra and the other one – Dac0 Mar 5 '18 at 10:05
• For finite groups, the two are duals of each other. – Tobias Kildetoft Mar 5 '18 at 10:19

Suppose that $G$ is a finite group. We can build the usual hopf algebra associated to the vector space $\mathbb{C}[G]$, with the usual group algebra structure $$\eta: 1 \to e, \quad m: g \otimes h\mapsto gh$$ the coalgebra structure $$\varepsilon: g \mapsto 1, \quad \Delta: g \mapsto g \otimes g$$ where all of the maps above are defined on the basis of group elements.
Now, we can take the dual Hopf algebra $\mathbb{C}[G]^\vee$, which as a vector space are the functions on $G$: $\mathbb{C}[G]^\vee = \{f: G \to \mathbb{C}\}$, and has new operations being the transpose of all the old operations. Now check what happens to all of the Hopf algebra structure. The unit $\eta$ dualises to the new counit $\eta^\vee: \mathbb{C}[G]^\vee \to \mathbb{C}^\vee \cong \mathbb{C}$, defined by $\eta^\vee(f) = (f \circ \eta)(1)$, which on a function $f$ will be $\eta^\vee(f) = f(\eta(1)) = f(e)$. So the new counit maps a function to its value over the identity.
Similarly, you can dualise the other maps, and see what happens. Should should find that the new coproduct $m^\vee$ is a kind of convolution, taking an indicator function $f_g$ to the element $\sum_{xy = g} f_x \otimes f_y$, the new unit $\varepsilon^\vee$ takes $1$ to the constant function on $G$ with value $1$, and the new multiplication $\Delta^\vee$ is pointwise multiplication of functions.