# Reps of $Lie(G)$ lift to universal cover of $G$. Reps of $G$ descend to highest weight reps of $Lie(G)$?

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.

1. How does one lift a representation from $\mathfrak{g}$ to $G$?
2. It lifts to a representation of the universal cover of $G$?
3. In the setting of Algebraic groups, I can't use the exponential to lift representations, since it isn't algebraic?
4. How does one descend a representation from $G$ to $\mathfrak{g}$?
5. 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.
• Presumably, $G$ is a Lie group. How is one supposed to treat this topic "without appealing to Lie groups"? – Omnomnomnom Jul 25 '17 at 11:10
• @Omnomnomnom See significant restructure if you wish :). – user462339 Jul 25 '17 at 11:35
• Indeed, most representations of the Lie algebra do not lift, since all the weights would need to be integral to have any chance (there are no non-integral weights for an algebraic group). – Tobias Kildetoft Jul 25 '17 at 14:02
• @Omnomnomnom Algebraic groups (over any field) also have Lie algebras attached, and if we take a semisimple group of some Dynkin type over the complex numbers, then they have the same one as the Lie group of that type. – Tobias Kildetoft Jul 25 '17 at 14:26
• @user462339 interesting; I never realized you could do all this completely algebraically. – Omnomnomnom Jul 25 '17 at 14:31

While I know you specifically asked for the example $SL_2$, I will take a smaller example for now as everything is a bit easier to explain there.

Let $G = \mathbb{C}^{\times}$ be the group of non-zero complex numbers under multiplication. This is an affine variety with coordinate algebra $k[x,x^{-1}]$. The Lie algebra of $G$ is $1$-dimensional, so in particular it is abelian.

Now, all irreducible algebraic representations of $G$ are $1$-dimensional since it is a torus, so such an irrep is given by a homomorphism from $G$ to itself. It is a nice exercise to show that any such homomorphism which is also algebraic is given by $z\mapsto z^n$ for some $n\in\mathbb{Z}$. If we work through the definitions, we see that the representation of the Lie algebra associated to this is the one given by multiplication by $n$ (since the Lie algebra is $1$-dimensional, a representation is given by a scalar).

But it is now also easy to construct representations of the Lie algebra that do not come from any representation of $G$: Just take ones given by multiplication by something which is not an integer.

The above was a very small example to illustrate how things go in general, though everything becomes more tricky when the representations are no longer $1$-dimensional.

A bit more about the general situation: As illustrated, weights for algebraic groups are always integral, whereas this is of course not the case for Lie algebras, so this puts a restriction on which representations of the Lie algebra can come from the algebraic group (when we have some torus that allows us to make sense of weights). Further, representations of algebraic groups have a build-in finiteness condition which implies that the corresponding representations of the Lie algebra are locally finite (i.e. any finite subset is contained in a finite dimensional submodule), so in particular it is not possible to have infinite dimensional irreducible representations, even though these abound when we study semisimple Lie algebras.

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.

• Very good answer. There is also another, more roundabout, way to "lift" the representations, though it is more accidental than the one you describe. Given an irrep for the Lie algebra, the highest weight will be integral so it lifts to a rep for the maximal torus of the group and thus to the Borel. Inducing this from the Borel to the group gives the desired lift. – Tobias Kildetoft Aug 3 '17 at 15:16
• Wonderful! Thank you for adding this! – user462339 Aug 4 '17 at 8:44