Are rational representations of an (affine) algebraic group the same as modules over some ring?

Question

Given an algebraic group $G$, are the rational representations of $G$ in natural correspondence with modules (of some type?) of a ring associated to the group? Certainly rational representations give rise to representations of $U(\mathfrak g)$, the universal enveloping algebra of the Lie algebra, but my intuition from Lie groups and also the fact that $G\times \mathbb{Z_2}$ has the same Lie algebra as $G$ mean that this shouldn't be the "right" ring.

Answer 1

When $G$ is an affine group scheme, its coordinate ring $k[G]$ is naturally a Hopf algebra. The group unit gives the bialgebra counit, the group multiplication gives the bialgebra comultiplication, and the group inversion gives the Hopf antipode.

Rational representations of $G$ are precisely comodules over the Hopf algebra $k[G]$.

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Added after Tobias' comment

It is true that in characteristic zero, the distribution algebra of $G$ is the same as the universal enveloping algebra $U(\mathfrak{g})$ where $\mathfrak{g} = \operatorname{Lie}(G)$, and hence every $G$-module $V$ becomes a $\mathfrak{g}$-module in a canonical way. However, there are many more $\mathfrak{g}$-modules than $G$-modules.

Take for example $G = \operatorname{SL}_2$, with Lie algebra $\mathfrak{sl}_2$.

1. The first problem is that $\mathfrak{sl}_2 \cong \mathfrak{pgl}_2$ as Lie algebras, while $\operatorname{SL}_2 \not\cong \operatorname{PGL}_2$ as algebraic groups, and so since not every $\operatorname{PGL}_2$-module lifts back to an $\operatorname{SL}_2$-module, we should expect that there are more $U(\mathfrak{sl}_2)$-modules than $\operatorname{SL}_2$ modules. This is the same kind of difference you're seeing between $G$ versus $G \times \mathbb{Z}_2$.

2. There are infinite-dimensional irreducible $\mathfrak{sl}_2$-modules, such as the Verma modules $M(\lambda)$ for any non-integral weight $\lambda$. These are clearly not $\operatorname{SL}_2$-modules, which are unions of finite-dimensional modules.

3. Since $\operatorname{SL}_2$ is a semisimple group, over a field of characteristic zero its representations are all semisimple. However, $U(\mathfrak{sl}_2)$-modules are not semisimple in general, for instance take the Verma modules $M(\lambda)$ for $\lambda$ a dominant integral weight.