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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).

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  • $\begingroup$ Let me briefly answer your second question. The correct statement is that the (commutative) space of function $C^\infty(G)$ on $G$ embeds into the dual $U(\mathfrak{g})^*$ of the universal enveloping algebra $U(\mathfrak{g})$. It is clear from the cocommutativity of $U(\mathfrak{g})$ that taking dual is essential. Roughly, this can be seen by observing that $\mathfrak{g}$ acts on $C^\infty(G)$ as derivation via left invariant vector fields, and $U(\mathfrak{g})$ acts as left invariant differential operators. I guess this is explained in more detail in arxiv.org/abs/hep-th/9111043. $\endgroup$ – Henry Jul 27 '18 at 3:35

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