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Let $X$ be a variety with a $G$-action by an algebraic group.

My question refers to a motivating example from:

https://web.maths.unsw.edu.au/~danielch/thesis/mbrassil.pdf

Here is the relevant excerpt:

enter image description here

Here the author discusses an example of $X/G$ in order to explain that it is necessary to form $X/G$ as categorical quotient and not the topological one.

We consider this motivating example introduced at page 27:

Take $X:= \mathbb{C}^2$ with action by $G:=\mathbb{C}^\times$ via multiplication $\lambda \cdot (x,y) \mapsto (\lambda x, \lambda y)$.

Obviously the "naive" topological quotient consists set theoretically of the lines $\{(\lambda x, \lambda y) \vert \lambda \in \mathbb{C}^x \}$ and the origin $\{(0,0)\}$.

Topologically the origin lies in the closure of every line.

So the QUESTION is why does this argument already imply that $Y:=X/G$ cannot have the structure of a variety? I don't understand the argument given by the author.

If we denote by $p:\mathbb{C}^2 \to Y$ the canonical projection map and by (continuity?) this map can't separate orbits, why does this already imply that $Y$ doesn't have structure of a variety as stated in the excerpt?

Especially which role does the fact that we can't separate the lines from the origin play (in a purely topological way)? Does it cause an obstruction to form a variety structure on $X/G$?

Remark: I know that there are different ways to deduce that if we define $X/G$ purely topologically then it cannot have a structure of a variety. The most common argument is to introduce the invariant ring $R^G$ and to calculate it explicitly. It seems to me that the given argument is a bit too "elementary" in the sense that the author doesn't explicitly work in this example with the concept of the invariant ring $R^G$.

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    $\begingroup$ This is a good question, but probably not research level. It might be better at MSE. $\endgroup$
    – LSpice
    May 6, 2019 at 1:18
  • $\begingroup$ I imagine the main point is that the topological space underlying a complex variety should be Hausdorff $\endgroup$
    – gcousin
    May 6, 2019 at 2:02
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    $\begingroup$ @gcousin: I'm not sure. We consider the underlying topological space endowed with Zariski topology so unless for finite algebraic sets no algebraic set is ever a Hausdorff space. Or do I oversee an aspect? $\endgroup$
    – user267839
    May 6, 2019 at 2:18
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    $\begingroup$ To simplify the problem: I think essentially the same argument that the author gave in his example would show that if we let act the multiplicative variety $G_m$ on $\mathbb{A}^1$. Then the topological quotient $p: \mathbb{A}^1 \to \mathbb{A}^1/G_m$ should also fail to become a variety. Another idea: If we assume that the topological quotient $\mathbb{A}^1/G_m$ has a variety structure, should all orbits /point be closed? Why? If yes, this would lead to the desired contradiction but I don't see an argument why we can make this assumption. $\endgroup$
    – user267839
    May 6, 2019 at 2:34
  • $\begingroup$ Another remark: I think that in case of $\mathbb{C}^2/\mathbb{C}^*$ the fact that Zariski topology has the Kolmogorov property en.wikipedia.org/wiki/Kolmogorov_space does it's job. indeed there exist no open set which contains a line $(\lambda x, \lambda y)$ but not the origin $\{(0,0)\}$. But in case of $\mathbb{A}^1/G_m$ it seems to be a bit more subtle: the "point"/orbit $\mathbb{A}^1/G_m - \{0\}$ is open so it can be separated from point $\{0\}$ in Kolmogorov's way. So what fails in case $\mathbb{A}^1/G_m$? $\endgroup$
    – user267839
    May 6, 2019 at 3:43

2 Answers 2

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Honestly, I think you are missing the point (pun intended). The issue is not whether there is a scheme structure on the set of orbits or not (every set admits such a structure). The issue is whether the projection morphism is "polynomial".

If there was the structure of a variety on the set of orbits such that the projection map was regular, then the orbits would be distinguished by polynomial maps. As correctly stated by the author you reference (who in turn correctly read Newstead), this can't happen in the examples he (and you) consider since any such morphism would be $G$-invariant and so would identify the origin to the lines (over $\mathbb{C}$ this just follows since polynomials are continuous).

As you say, you can see this via the coordinate ring, namely the ring of invariants, since the only invariants are the constants and so the corresponding variety is a point. But this is the answer to the motivation you seek; the fact that invariant polynomials can't distinguish orbits in your example is good motivation.

The coordinate ring statement confers the fact that the "optimal fit" of an affine variety structure to the orbit set that has the quotient map regular will be when the corresponding coordinate ring (to the affine variety) is exactly the ring of invariants. One then goes on to show that $Spec$ of the ring of invariants is a good categorical quotient (when the ring of invariants is finitely generated, like when $G$ is reductive), and then you patch such affine quotients together to obtain categorical quotients in the quasi-projective setting (with respect to an ample line bundle).

Here are my recommendations to learn more: Preparation for GIT

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The scheme $Spec \mathbb{C}[x,y]^{\mathbb{C}^{\times}}=Spec \mathbb{C}$ is a point, so this approach to giving $X/\mathbb{C}^{\times}$ the structure of a variety does not work. The quotient is $\mathbb{P}^1$ in particular it is not an affine variety. In order to realize this, we need to remove the two axes $\{x=0\}$ and $\{y=0\}$ (one at a time). The set $\mathbb{C}^2/(0,0)=U_1\cup U_2$ where $U_1=Spec \mathbb{C}[x,y,y^{-1}]$ (i.e where $y\neq 0$) and $U_2=Spec \mathbb{C}[x,y,x^{-1}]$ (where $x\neq 0$). The quotients $U_1/\mathbb{C}^{\times}=Spec \mathbb{C}[x,y,y^{-1}]^{\mathbb{C}^{\times}}=Spec \mathbb{C}[x/y]$ and $U_2/\mathbb{C}^{\times}=Spec \mathbb{C}[x,y,x^{-1}]^{\mathbb{C}^{\times}}=Spec \mathbb{C}[y/x]$ glue together to $\mathbb{P}^1$.

If you want to understand the general theory, I'd recommend reading about the GIT construction of quotients. This involves choosing a very ample line bundle $L$ on a variety $X$ such that the group action of $G$ on $X$ lifts to $L$. After removing a certain "bad" subset from $X$ called the locus of non-semistable points $X^{ns}$, the GIT quotient is simply $Proj \left(\bigoplus_{m\geq 0} H^0(X/X^{ns},L^{\otimes m})^{G}\right)$. The group $G$ needs to be reductive for this to work.

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  • $\begingroup$ A point is a variety. A GIT quotient is not "the" quotient. There are more than one. For affine varieties, the point is the quotient. But in the quasi-projective setting there are more than one. Each depends on a line bundle, in this case, a character. The trivial character gives the affine quotient, whereas the identity character gives your construction. See math.stackexchange.com/a/2022872/84231 for a more general description and reference. $\endgroup$
    – Sean Lawton
    May 6, 2019 at 17:01

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