TITLE: Reps of $Lie(G)$ lift to universal cover of $G$. Reps of $G$ descend to highest weight reps of $Lie(G)$?
QUESTION [5 upvotes]: Let us work over an algebraically closed field of characteristic $0$. Let $G$ be a semisimple, or perhaps reductive algebraic group, so we are working with the Zariski topology. Let $\mathfrak{g}$ be the Lie algebra associated to $G$.
If possible, could we not appeal to Lie groups and/or exponentiation, unless we explain how this relates to the algebraic group setting, with the Zariski topology.

How does one lift a representation from $\mathfrak{g}$ to $G$?
It lifts to a representation of the universal cover of $G$?
In the setting of Algebraic groups, I can't use the exponential to lift representations, since it isn't algebraic?
How does one descend a representation from $G$ to $\mathfrak{g}$?
These descend to highest weight representations?

Essentially I would like to understand the functors:
$$d:Rep(G)\to Rep(\mathfrak{g}),\qquad \int: Rep(\mathfrak{g})\to Rep(G)$$

An example on $\text{SL}(2,\Bbb C)$ would be nice, since the rep theory is simple.

REPLY [2 votes]: While Tobias Kildetoft's answer is technically correct, and there are (in general, many) finite dimensional representations of the Lie algebra of a reductive group that do not lift to the universal covering group, nevertheless for semisimple Lie algebras in characteristic $0$ all finite dimensional representations do in fact lift, and one can write down the lifting in a very explicit fashion. This is a rather long story to write out completely here, so I'll just give some hints and a reference.
Roughly speaking, you start with the action of the Chevalley generators $e_i$ and $f_i$ of the Lie algebra on a given finite dimensional representation $V$; these turn out to be nilpotent on $V$, and hence we may exponentiate them in any characteristic greater than the order of nilpotency to obtain a lift to a certain group (which is always a quotient of the universal covering group). But more is true: there is an integral version of the enveloping algebra that eliminates all the denominators by taking the divided powers $$e_i^{(k)}=\frac{e_i^k}{k!} \quad \text{and} \quad f_i^{(k)}=\frac{f_i^k}{k!}$$ as generators, and every finite dimensional representation of the complex Lie algebra contains a lattice stable by this integral form. This allows you to lift any such representation to the group, and in fact obtain these representations in any characteristic. 
For most of this, you might look at Steinberg's wonderful Lectures on Chevalley groups, available e.g. here
http://www.ms.unimelb.edu.au/~ram/Resources/YaleNotes.pdf
Going from representations of $G$ to those of $\mathrm{Lie}(G)$ is rather less subtle: a representation of $G$ is a morphism of algebraic groups 
$$G \rightarrow \mathrm{GL}(V)$$ for some vector space $V$, and this induces a morphism of Lie algebras
$$\mathrm{Lie}(G) \rightarrow \mathfrak{gl}(V),$$ which is the desired representation.