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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.

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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 $k[G]$.


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.

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    $\begingroup$ I am a bit rusty in this, but I think the algebra of distributions might be closer to what the OP is after (since then the representations really are modules). But I don't recall precisely how good the connection is (whether it suffices to require the modules to be locally finite to get representations, or if more is needed). $\endgroup$ – Tobias Kildetoft Feb 27 '20 at 9:48
  • $\begingroup$ @TobiasKildetoft You're right, that's probably closer. I've added some thoughts there to my answer $\endgroup$ – Joppy Feb 27 '20 at 11:04
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    $\begingroup$ Another thing of note here (I really ought to write something more detailed myself, but I feel like it has been too long since I worked with it, so I would end up leaving out some too important details due to forgetting them). In some sense, there is no way the representations of an algebraic group can be (just) the modules over some algebra, since the category of rationals reps of a reductive group has either no projective modules or nothing but projective modules. $\endgroup$ – Tobias Kildetoft Feb 27 '20 at 14:48
  • $\begingroup$ This is exactly the sort of thing I was looking for. Do you mind elaborating on this? Also, every object being projective is fine for the module category over a ring - does this happen for any groups that are not finite? Are there examples where this is still not the module category over a ring? $\endgroup$ – Ashwin Trisal Feb 28 '20 at 9:58

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