8

Consider the group scheme $\mathbb{G}_m = \mathrm{Spec}(\mathbb{Z}[t, t^{-1}])$. An action of $\mathbb{G}_m$ on an affine scheme $X = \mathrm{Spec}(A)$ is a morphism of schemes $$ \sigma: \mathbb{G}_m \times X \to X $$ such that for every scheme $T$, $\sigma_T : \mathbb{G}_m(T) \times X(T) \to X(T)$ is a group action in the standard definition. I want to show that the action $\sigma$ is equivalent to a $\mathbb{Z}$-grading on $A$. I've been trying to do this for a little while but I am stuck. How can I prove/think about this?

The closest I've got to solving this is by using the idea of considering $A$ as a $\mathbb{Z}[t, t^{-1}]$-comodule using the ring homomorphism $$ \Delta: A \to A \otimes \mathbb{Z}[t, t^{-1}] \cong A[t, t^{-1}] $$ dual to the action $\sigma$. This was taking from example 1.6.6 in Hida's book on geometric modular forms, which refers to Jantzen's book on Reps of Algebraic Groups. However, I'm not comfortable with the idea of comodules, including why this is a comodule map. However, I think I see why, if this is a comodule map and satisfies the requesite properties mentioned in Jantzen, then we get a grading. Given a grading, there seems to be an obvious way to define a homomorphism $\Delta$ taking $a \mapsto \sum_{n \in \mathbb{Z}} p_n(a) \otimes t^n$ where $p_n$ is the projection map onto the $n^{th}$ graded piece. Is it true that this gives a group action? Again I'm not that comfortable with this equivalence between group actions and $G$-modules and $\mathbb{Z}[t, t^{-1}]$-comodules. If the above description is the best way to think of this then I will try to understand these better, but I wanted an outside opinion on how to prove/think about this.

ggg
  • 1,433
  • 1
    I think it would be best if you could derive the coalgebra structure on $\mathbb{Z}[t, t^{-1}]$ from the group structure on $\mathbb{G}_m$, and also the comodule structure on $A$ from the group action on $\operatorname{Spec}(A)$. It should be quite easy to verify that anything you come up with is indeed a coalgebra / comodule, there are not many axioms to check. If you want an example to check against, verify that the natural scaling action of $\mathbb{G}_m$ on affine space $\operatorname{Spec}(\mathbb{Z}[x, y])$ gives the usual grading on $\mathbb{Z}[x, y]$. – Joppy Jun 06 '19 at 01:58
  • 1
    If you have access to it, you might want to look at the first few pages of Waterhouse's "Introduction to Affine Group Schemes", which is considerably easier to digest than Jantzen. – Joppy Jun 06 '19 at 03:16
  • Thanks for the suggestion. The section on comodules (3.2) in Waterhouse is indeed clearer than Jantzen. Quick side-question: does everything extend well when we consider general actions on an affine scheme rather than on some vector space over a field? I'll think about your example. – ggg Jun 06 '19 at 16:03
  • Everything is meant to extend nicely, by defining the group schemes over $\mathbb{Z}$, and then taking a product with $\operatorname{Spec} k$ of the desired field $k$. For example, the start of section II in Jantzen assumes that $k$ is an integral domain, rather than just a field - there is some result saying that every split reductive group scheme may be realised over $\mathbb{Z}$. – Joppy Jun 07 '19 at 03:21
  • That's interesting, I didn't think about base changing to a field. I was thinking that if we take our group scheme $G$ (suppose affine for now, $G=\mathrm{Spec}(B)$) and scheme $X = \mathrm{Spec}(A)$ defined over some affine base scheme $S = \mathrm{Spec}(R)$ then an action is a morphism of $S$-schemes $G \times_S X \to X$. This corresponds to a morphism of $R$-algebras $A \to B \otimes_R A$, and we can consider this as a comodule map. Although I've never seen a definition not over a field. – ggg Jun 07 '19 at 15:26
  • I know my above comment may not make total sense. I'm a bit confused about the generality in which I'm working. Specifically, in all the references I'm using (e.g. Waterhouse) they talk about linear representations of an affine group scheme on a module functor taking a $k$-algebra $C$ to the tensor product $M \otimes_k C$, where $M$ is some $k$-module. Actually, Hida works with a general affine base $S= \mathrm{Spec}(R)$, but still only considers action on these types of functors. I see that if $M$ is a finitely generated free (maybe projective is enough) ... – ggg Jun 07 '19 at 15:37
  • $R$-module then such a functor is representable by the affine scheme $\mathrm{Spec}(\mathrm{Sym}(M^\vee))$ so we can translate this to the situation in my previous comment, but I don't think this is general. – ggg Jun 07 '19 at 15:37
  • OK I understand better the correspondence between linear representations of $G$ on a module fuctor $M$ and $k[G]$-comodules now. The part I'm confused about is the extension to the setting of my question where the action is now on a general affine scheme. In Jantzen, it's mentioned that such an action makes the global sections of $X$ naturally into a $k[G]$-comodule, so we should be able to get a grading this way, and I think the other direction should follow from the same arguments as in the more specific case. I will write up my progress in the next day or two. – ggg Jun 07 '19 at 18:56

0 Answers0