Geometric intuition for linearizing algebraic group action

Question

I am reading an overview of geometric invariant theory and find myself stuck when we begin linearizing the action of an algebraic group on a variety. The definition given in my notes is that given an action $\sigma: G\times X\rightarrow X$ of an algebraic group $G$ and a line bundle $\pi:L\rightarrow X$, a linearization of $\sigma$ is an extension to an action $\bar{\sigma}: G\times L\rightarrow L$ that commutes with $\sigma$ via the projection $\pi$. I understand the definition, but do not see why we do this/how it helps us construct quotients. Any insight would be appreciated, preferably from a geometric perspective.

Answer 1

Here is some intuition for choosing a linearization as far as I understand. I'm also just learning the subject so take this with a grain of salt but here is the picture I've gotten.

Let us first consider a finite group $G$ acting nicely on an affine variety $X = \operatorname{Spec}A$. Here we want a (presumably affine) quotient $X/G$ to be a space whose points represent the orbits of $G$ acting on $X$ in a nice way. Therefore, by definition, we want the functions on $X/G$ to be functions on $X$ that are $G$-invariant since this will ensure that they are constant on orbits. Well then we just take the natural action of $G$ on $A$ and pick the $G$-invariant functions $A^G$ and define $X/G := \operatorname{Spec}(A^G)$. In this way, things work out nicely.

However, we don't want to be restricted to just quotients of affine varieties by finite groups. Linearizations come into the picture when we try to take more general quotients. They are generalizations of natural action of $G$ on $A$ used above to define the quotient. Here are the two types of problems that linearizations fix.

The first one is pretty obvious. What do we do with a projective variety that isn't determined by an affine coordinate ring? We realize that a projective variety has the next best thing, a homogenous coordinate ring. Explicitly, if $X \hookrightarrow \mathbb{P}^n$ is a projective variety, we can look at the affine cone $\tilde{X}$ in $\mathbb{A}^{n+1}$ and consider the coordinate ring $A(\tilde{X})$. Then we can lift the group action of $G$ on $X$ to a compatible one on $\tilde{X}$. This is a linearization of the group action. The recipe above tells us that the quotient should be $\tilde{X}/G = \operatorname{Spec}A(\tilde{X})^G$. So we guess that maybe we should define the quotient of $X$ by $X/G = \operatorname{Proj}(A(\tilde{X})^G)$. It turns out this definition works pretty well for the same reason the one above does. We are defining it by saying sections should be $G$-invariant since we want the space to be an orbit space.

The above definition of $X/G$ depends on the embedding. Explicitly, once we have an embedding, we can recover $A(\tilde{X})$ directly as the ring

$$ \bigoplus_{i} H^0(X,\mathcal{O}_X(i)) $$

and the linearization is the induced action of $G$ on this ring that is compatible with the original action. More generally, if we have an abstract projective variety $X$, we know that embeddings correspond to a choice of very ample line bundle $L$. Thus to apply the above prescription for a quotient, what we are really doing is picking a very ample line bundle $L$ and a compatible action of $G$ and then defining the quotient to be $\operatorname{Proj}(A^G)$ where

$$ A = \bigoplus_i H^0(X,L^{\otimes i}) $$

where this ring is exactly the projective coordinate ring of the variety under the embedding given by $L$. This is where the definition of linearization comes from and how it helps us give a quotient.

However, this isn't the end the story. We are taking Proj so we are throwing away some points, specifically those points in the irrelevant ideal. This process depends on the ring $A^G$ and the grading on it, thus the choice of linearization actually matters and different linearizations give us different quotients. So really, we should be writing $X//_{L}G$ or something of that sort. A lot of the theory of GIT is built to tell us exactly what we're throwing and what is the modular interpretation of $X//_{L}G$, i.e., what orbits do its points represent.

Finally, I'll give an example to show why we need to throw out points to get a good quotient even in the simplest possible cases.

Consider $\mathbb{C}^*$ acting on the plane $\mathbb{A}^2$ by scaling. First let's look at the topological orbit space and see why the naive quotient is not desirable from an algebraic geometry point of view. The orbits of this action are the copy of $\mathbb{C}^*$ contained in each line through the origin, and the orbit consisting of the single point $0$. Thus, the topological orbit space is a highly nonseperated space where the closure of each point consists of that point plus the orbit of $0$. This is inherently not a nice space and in fact you can show that it doesn't satisfy the nice properties we want quotients to satisfy in the category of schemes (e.g. categorical quotient, geometric quotient, etc). Intuitively, we know that what we should do is throw out the origin and then take the quotient to get the really nice variety $\mathbb{P}^1$.

Now, lets try applying the machinery above about linearizations to take this quotient. Consider the trivial line bundle $L = \mathcal{O}_{\mathbb{A}^2}$ and pick some integer $r$. Let's lift the scaling action of $t \in \mathbb{C}^*$ on $\mathbb{A}^2$ to $L$ by multiplying by $t^{-r}$ on the fibers.

Then for $r < 0$ the invariant sections are trivial so we get an empty quotient so that's not the linearization we want. For $r = 0$ the invariant sections are just the constants since this is the trivial lift of the action and we get a single point as the quotient. This represents the idea that if we don't throw out the origin but still want a good quotient in some sense, we have to identify all the points because of the nonseparatedness discussed above. However, if $r > 0$, we get the graded ring

$$ \bigoplus_k R_{kr} $$

where $R_{kr}$ are the homogeneous polynomials of degree $kr$ in two variables. In particular, for $r = 1$ we get just $\mathbb{C}[x,y]$ with the usual grading and so the quotient with this linearization is given by $\mathbb{A}^2/\mathbb{C}^* = \operatorname{Proj}\mathbb{C}[x,y] = \mathbb{P}^1$ as we wanted.

I hope this helps give you some intuition as to why linearizations come up in GIT and why they're important to keep in mind. Like I said, I'm still learning the subject myself but this is how I picture the role of a linearization. A lot of what I said above and what I know about the subject comes from this really nice article by Richard Thomas: http://arxiv.org/abs/math/0512411