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Let me recall the definition of a rational $G$-module from M. Brions notes Introduction to actions of algebraic groups (Def. 1.6)

Let $G$ be an affine group scheme over $\mathbb{C}$. A rational $G$-module is an action of $G$ on a (not necesseraly finite dimensional) $\mathbb{C}$-vector space V such that every $v \in V$ is contained in a finite dimensional $G$-stable subspace of $V$ on which $G$ acts algebraically.

Just after this definition Brion states that if $G$ is irreducible and not trivial, then the action of $G$ on the rational function field $\mathbb{C}(G)$ is an example of a $G$-module that is not rational.

The first point I do not understand here is the following: Brion says that $G$ operates on $\mathbb{C}(G)$ via left multiplication. On the coordinate ring $\mathbb{C}[G]$ the group $G$ operates via right multiplication: $(g.f(x)) = f(xg)$ for $g,x \in G$ and $f \in \mathbb{C}[G]$. So I would guess that one would try to extend this action to $\mathbb{C}[G]$. However it is not clear to me why this works in general. So the first question is: How does $G$ act on $C(G)$?

Second, I would like to see a concrete example where this action is not rational, ideally in an easy case such as $G = \mathbb{G}_a$.

As I understand it, it seems that the existence of non-rational $G$-modules would contradict the fundamental lemma on representations (every representation is the finite unions of its finite-dimensional subrepresentations), see e.g. Corollary 4.7 in Milnes book Algebraic Groups. Probably I am misunderstanding something fundamental here. I appreciate clarifications.

skew41
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  • On a quick read it seems that in Milnes book all representations (at least where he states Corollary 4.7) are rational (i.e. comodules over the coordinate algebra). – Tobias Kildetoft May 27 '16 at 08:22
  • I will think a bit more about a more concrete example, but for a very prominent thing that happens in positive characteristic: The category of rational $G$-modules for a simple (or just reductive) algebraic group in positive characteristic has no projectives (i.e. all the projective modules are not rational). – Tobias Kildetoft May 27 '16 at 08:25
  • @TobiasKildetoft If Milne would assume that all representations would be rational, then Corollary 4.7 would not say anything new, right? As I understand it, the affirmation there (the representation is the union of finite dimensional subrepresentations) is the definition of rational. Is not it? – skew41 May 27 '16 at 09:06
  • Not quite, as Milne has defined representations to be comodules over the coordinate ring, so in that setup one needs to prove that this implies the definition used by Brion (which is the result that precedes Corollary 4.7). – Tobias Kildetoft May 27 '16 at 09:09
  • I agree. The result in Proposition 4.6 and Corollary 4.7 in Milne seems to be what Brion uses to define a G-module to be rational. So the aforementioned results seems to say that every G-module is rational. – skew41 May 27 '16 at 09:13
  • No, only when you define $G$-module as $k[G]$-comodule (or what the notation was that Milne used). The point is that not all actions of $G$ define comodule structures of $k[G]$. – Tobias Kildetoft May 27 '16 at 09:17
  • In 4.1 Milne explains how G-representations correspond to k[G]-comodules. So maybe I do not understand properly what a G-module is. Actually it seems that Brion does not define this precisely. He just defines a rational G-action to be a vector space with a linear action of G plus the finiteness condition. So I guess a G-module is just a vector space with G-action. Does he mean something else as Milne by a linear representation of $G$? – skew41 May 27 '16 at 09:31
  • Right, but note that his $G$-representations require that the associated map $G\to GL_V$ is a homomorphism of group-valued functors. This is stronger than just being a homomorphism of groups (which would be all that was needed to define a representation if we did not require it to be rational, since this would basically mean ignoring the geometric structure on $G$). – Tobias Kildetoft May 27 '16 at 09:37
  • So I guess that when Brion speaks of a G-module, he only considers the $k$-rational points, i.e. he wants a morphism $k[G(k)] \otimes_k V \to V$ of $k$-vector spaces that provides a unital and associative $k[G(k)]$-structure on $V$. – skew41 May 27 '16 at 09:41
  • Yes, he seems to use $G$-module to simply mean a representation of $G$ as an abstract group (whereas Milne uses it to mean when Brion calls a rational $G$-module). This is a thing that can cause a lot of confusion when people are not careful (there are plenty of papers on rational $G$-modules which just call them modules because to the authors, the non-rational ones are not of interest for what they are doing. I try to make sure I either always call them rational or at least make a note at the beginning that when I say module, I mean rational module). – Tobias Kildetoft May 27 '16 at 09:45
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    For the case of $\mathbb{C}(G)$ where $G=\mathbb{G}_a$, $\mathbb{C}(G)$ is the field of rational functions $\mathbb{C}(t)$ in an indeterminate $t$, with $g\in G$ acting as $p(t)\mapsto p(t+g)$. This is not a rational representation since if $1/t$ is in a $G$-stable subspace then so is $1/(t+a)$ for every $a\in\mathbb{C}$. But $\left{1/(t+a): a\in\mathbb{C}\right}$ spans an infinite-dimensional subspace of $\mathbb{C}(t)$. – Jeremy Rickard May 27 '16 at 11:29

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Let us for the moment forget all about algebraic geometry and just consider a group $G$ and a $G$-module $V$. The definition of being rational makes sense in this context, so let us just find an example of a non-rational $G$-module which will work whenever $G$ is infinite (so it will also work for whatever algebraic group you choose as long as it is infinite).

It turns out that we can just take the regular $G$-module (note that this term tends to mean something slightly different when we consider $G$ as an algebraic group though). So let $V$ be the vector space with basis $\{v_g\mid g\in G\}$ and define the action of $G$ by letting $gv_x = v_{gx}$.

I claim that this is not a rational $G$-module. To see this note that the submodule generated by any $v_g$ is all of $V$, so none of these can be contained in a finite dimensional $G$-stable subspace (and in fact, this module has no non-trivial finite dimensional submodules).

Tobias Kildetoft
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  • Thank you. It seems that I didn't properly distinguish between (1) actions of algebraic group schemes $G$ on $V$ as morphisms $G \to GL_V$ of group schemes and (2) actions of abstract groups (I guess Brion thinks of $G$ as its $\mathbb{C}$-points) on $V$. You gave an example of the latter that is not rational. The above-cited corollary in Milne's book shows that for representations of algebraic group schemes in the sense of (1) always fulfill the finiteness property that is used in the definition of "rational". So it seems to me that all representations in the sense of (1) are rational. – skew41 May 27 '16 at 11:05
  • @skew41 Yes, precisely (and indeed, this is one possible definition of the action being rational). The distinction quickly becomes hard to find when working in the setup of Milne since the rationality is "hidden" in the functors (but in some sense this also makes everything easier to deal with once you get used to it). – Tobias Kildetoft May 27 '16 at 11:07