Universal enveloping algebra vs algebra of continuous functions

Question

I'm a physicist studying quantum groups, but this question is about the usual classical Lie groups (though it is related to the language that is used to describe the deformation to quantum groups, thus the quantum-group tag).

Suppose I have a finite-dimensional Lie algebra $\mathfrak{g}$ (you can assume it to be compact and/or semisimple, whatever suits your needs) and its universal cover Lie group $G$.

I can always make the universal enveloping algebra $U(\mathfrak{g})$ into a Hopf algebra by defining:

• The co-product is defined as: $$\Delta(1) = 1 \otimes 1$$ $$\Delta(t_a) = t_a \otimes 1 + 1 \otimes t_a$$ where $\{t_a\}$ are the set of generators of $\mathfrak{g}$, or, equivalently, the basis of the order-1 subspace of $U(\mathfrak{g})$. This definition is then extended to the rest of $U(\mathfrak{g})$ through bialgebraic properties.
• The trace is defined as $$ \varepsilon(1) = 1$$ $$ \varepsilon(t_a) = 0$$ and extended to the rest of $U(\mathfrak{g})$ through bialgebraic properties.
• The antipode is defined as $$ S(1) = 1$$ $$ S(t_a) = - t_a $$ and extended to the rest of $U(\mathfrak{g})$ through bialgebraic properties.

Question: please mark the following claims as "true" or "false", and for those which are true, please provide a draft of the proof.

1. The group-like elements of the Hopf algebra $U(\mathfrak{g})$ (those which satisfy $\Delta(g) = g \otimes g$) form a Lie group, which is exactly the universal cover Lie group $G$.

2. The Hopf algebra $U(\mathfrak{g})$ is to some extent (which?) equivalent to the commutative algebra of functions over $G$. (I've seen this claim made many times in one form or another, see e.g. here, but I never actually understood this. How can a commutative algebra be equivalent to a noncommutative one? Anyway, looks like this duality is taken as the definition of the quantum group as a deformed Hopf algebra, so I assume it has to be valid in the classical case, too).

Answer 1

In the meantime, I have posted an answer to the question Duality between universal enveloping algebra and algebras of functions you are citing. So, for the second question I suggest to check the corresponding literature. But the idea is that there is a pairing $\langle\cdot,\cdot\rangle\colon U({\frak g})\otimes O(G)\to\mathbb C$, where $O(G)$ is the algebra generated by the coordinate functions on $G$. If you interpret elements of $\frak g$ as tangent vectors of $G$ at identity, then elements of $U({\frak g})$ indeed act on $O(G)$. Namely by $$\langle x_1\cdots x_n,f\rangle=\left.{\partial^n\over \partial t_1\cdots\partial t_n}\right|_{t_1,\dots,t_n=0}f(\exp(t_1x_1)\cdots\exp(t_nx_n)),$$ where $x_1,\dots,x_n\in{\frak g}$.

As for the first question, the answer is no. More precisely, the only group-like element in $U({\frak g})$ is the identity. To gain some insight, let's maybe look on the motivation behind group-like elements.

Let $\Gamma$ be a finite discrete group, then the group algebra $\mathbb C\Gamma$ forms a Hopf algebra with respect to the coproduct $\Delta(g)=g\otimes g$. In addition, for any Hopf algebra $A$, it holds that, if $g,h\in A$ are group-like, then also the product $gh$ is group-like. So, the set of all group-like elements forms a group – it somewhat cuts of exactly the part of $A$, which looks like $\mathbb C\Gamma$.

For instance, if $G$ is a Lie group or a compact group, so $O(G)$ forms a Hopf algebra, then the group-like elements of $O(G)$ are exactly all the one-dimensional representations of $G$. We know, that those indeed form a group. By the way, if $G$ is compact and abelian, then all the irreducible representations are one-dimensional and the group they form $\hat G$ is exactly the Pontryagin dual of $G$.

But what you are suggesting is that you would like to discover somehow the group algebra $\mathbb CG$ inside $U({\frak g})$. But in order to do that, you would have to somehow exponentiate the elements of ${\frak g}$ to get the elements of $G$, which is not possible inside $U({\frak g})$ (since you have only finite products and linear combinations here).

There is even a classification statement by Cartier–Gabriel (see Theorem 3.8.2 here) saying that every cocommutative Hopf algebra is of the form $A=U({\frak g})\rtimes \mathbb C\Gamma$. So, the property of being primitive (i.e. $\Delta(x)=1\otimes x+x\otimes 1$) and being group-like (i.e. $\Delta(x)=x\otimes x$) are rather complementary than compatible. Every cocommutative Hopf algebra consists of a part that looks like group algebra $\mathbb C\Gamma$ (generated by group-likes) and of a part that looks like an enveloping algebra $U({\frak g})$ (generated by primitives).