I'm reading some lecture notes about GIT, and I was wondering the following:
Let $\mathbb{C}^*$ act on $\mathbb{C}^2$ as $t\cdot (x,y)=(tx,t^{-1}y)$. One may show that there are 3 types of orbits:
- the origin $(0,0)$, which is closed and has positive dimensional stabilizer;
- non-closed orbits $x=0$ and $y=0$;
- closed orbits $xy=c$, for $c\in \mathbb{C}\setminus \{0\}$.
By definition, the set of stable point is $\mathbb{C}^2\setminus Z(xy)$. Moreover, there exists a GIT quotient $\pi: \mathbb{C}^2\to \mathbb{C}, (x,y)\mapsto xy$, which is not geoemtric since the preimage of $0$ is the union of $3$ orbits. On the other hand, the restriction $\mathbb{C}^2\setminus Z(xy)\to \mathbb{C}^*$ is geometric, since -among all the other properties- the preimage of every point is a single orbit.
Problem: Suppose I consider the restriction $\mathbb{C}^2\setminus Z(x)\to \mathbb{C}$. By definition, $\mathbb{C}^2\setminus Z(x)$ contain points which are not stable, that is the non-closed orbit $y=0$. However, the preimage of every point is a single orbit; indeed for every point $p\in \mathbb{C}\setminus \{0\}$ it is trivial, while for $0$ the preimage consists of a single orbit.
Question: Is is true that the restriction $\mathbb{C}^2\setminus Z(x)\to \mathbb{C}$ is a geometric quotient? If yes, then there may exist geometric quotients which parametrize also points which are not stable?
Thanks in advance.
Yes, you are right - $\mathbb C^2\setminus Z(x) \to \mathbb C$ is a geometric quotient.
Of course, $\mathbb C^2\setminus Z(y) \to \mathbb C$ is also an equally good geometric quotient. This gives us two different choices for how to enlarge our geometric quotient to include more than just the stable points. (We don't like $(\mathbb{C}^2\setminus Z(xy)) \cup\{(0,0)\}$, since it is not an irreducible quasiprojective variety.)
You might like to read up on 'semistability' and 'the Hilbert-Mumford criterion' (e.g. in section 6 of those notes that you linked). These two different quotients parameterise semistable points (as opposed to just the stable points) and the notion of semistability requires the choice of a 1-parameter subgroup $\lambda \colon G\to \mathbb{C}^*$. In your case $G=\mathbb{C}^*$, and these two geometric quotients correspond to the two non-trivial choices for a 1-parameter subgroup: either $\lambda(t)=t$ or $\lambda(t)=t^{-1}$.