Expressing properties of graded algebras in terms of the $\mathbb{G}_m$action

Question

Let us fix a base ring $k$. The category of $\mathbb{Z}$-graded $k$-algebras is equivalent to the category of $\mathbb{G}_m$ equivariant affine $k$-schemes. The following 2 properties often come up when constructing the $Proj$ of a graded ring: Let $A$ be a graded $k$-algebra.

1. The zeroth graded piece is the base ring $A_0 =k$
2. The algebra $A_\bullet$ is positively graded. i.e. $A_j=0$ for all $j<0$.
3. The canonical multiplication homomorphism $Sym^\bullet A_1 \to A_\bullet$ is surjective.

I'm trying to find equivalent characterization of the above properties which don't use the aforementioned equivalence. In other words I want the above 3 properties expressed in terms of properties of $\mathbb{G}_m$-equivariant schemes (orbits, stabilizers, fixed points etc.).

Answer 1

Let $V$ be an affine variety with a $\mathbb{G}_m$-action. Then you should think of $A_0$ as a good model for $V/\mathbb{G}_m$. So being connected says something like $\mathbb{G}_m$ acts transitively- at least from the point of view of functions. Being positively graded is saying something like the action of $\mathbb{G}_m$ extending to an action of the monoid scheme $\mathbb{A}^1$ (under multiplication). The last condition is more involved: The module $A_1$ is the set of $\mathbb{G}_m$-equivariant functions to $\mathbb{A}^1$ where we use the scaling action on $\mathbb{A}^1$. (So it's kind of like "linear" functions on $V$). This is a vector space over $k$, call its dual $W$ and view it as a variety. There's a natural equivariant map $V \to W$ given by evaluation (that's why we took the dual of the vector space of functions). We're asking that this equivariant map be an embedding. So, while $V$ is not a vector space with the standard linear action of $\mathbb{G}_m$, it can be embedded in one in some natural way.

Answer 2

Expanding on Dylan Wilson's answer, condition (2) is indeed just a continuation of $\mathbb{G}_m$-action to an $\mathbb{A}^1$-action, since $\mathcal{O}(\mathbb{A}^1) \hookrightarrow \mathcal{O}(\mathbb{G}_m): \mathbb{k}[t] \hookrightarrow \mathbb{k}[t, t^{-1}]$. Condition (1) is trickier, since the action of $\mathbb{G}_m$ on $V$ will have inseparable orbits --- see the action on $\mathbb{A}^n$. Thus we can't think of it as transitivity (not of the group action at least, the monoid $\mathbb{A}^1$ action is transitive in a sense that any two points can be mapped to a same point). Instead we should look at it together with condition (2), which implies that the action of $\mathbb{G}_m$ has fixed points: for any $x \in V$ the point $0\cdot x \in V$ will be fixed. Here $0 \in \mathbb{A}^1$ and its action exists by (2). Condition (1) then reduces to the statement that the $\mathbb{G}_m$-action has a unique fixed point. On the level of functions it is given by a graded morphism $\mathcal{O}(V) \to \mathbb k$ as graded algebras. Condition (3) means that the action is free if we remove that single fixed point. We can see it because the action on $A_1^*$ is a free action apart from $0 \in A_1^*$ and the embedding $V \hookrightarrow A_1^*$ is $\mathbb{G}_m$-equivariant, as a morphism of graded algebras.

The simplest example where you can see all those statements is the obvious action on $\mathbb{A}^n$. On the other hand, consider the obvious action of $\mathbb{G}_m$ on itself. It corresponds to the natural grading on $\mathbb{k}[t,t^{-1}]$. The reasons (2) fails here is obvious. Since there is no morphism $\mathbb{k}[t,t^{-1}] \to \mathbb{k}$, the action has no fixed points. In this case (1) indeed reduces to transitivity. Instead of (3) we have that $Sym(A_1 \oplus A_{-1}) \to A_\bullet$ is surjective. This also implies that the action is free, for the same reasons as above.