**Set-up:** Consider the action of $\mathbb{C}^\*$ on $\mathbb{C}^4$ defined as follows: $(t,(x,y,z,w))=(tx,ty,t^{-1}z,t^{-1}w)$. I know that the affine GIT quotient is equal to $$\phi: \mathbb{C}^4 \to \mathbb{C}^4//\mathbb{C}^\*=Z(XW-YZ),$$ $$(p\_1,p\_2,p\_3,p\_4)\to (p\_1p\_3,p\_1p\_4,p\_2p\_3,p\_2p\_4).$$


**Question:** Consider now the open subset $U=\{p\in \mathbb{C}^4\mid \lim\_{t\to 0}tp \text{ doesn't exists} \}$. I would like to prove that the restriction morphism $\phi:U\to \phi(U)\subset \mathbb{C}^4//\mathbb{C}^\*$ is a geometric quotient.


**My trial:** In order to visualize it, I did the smaller case $\mathbb{C}^\*$ acting on $\mathbb{C}^2$ by $(tx,t^{-1}y)$, and there I had no problems as $U=\{(x,y)\mid y\neq 0 \}$, therefore the orbit $\mathbb{C}^\*\cdot (0,y)$ is mapped to $0$, the orbits $\mathbb{C}^\*\cdot (x,y)$ (with $x,y\neq 0$) are mapped to a point $c\neq 0$ (the product of the coordinates), therefore the preimage of $\phi$ at every point was a single orbit and I was able to conclude.


The problem in this new example is that $U=\{p\in \mathbb{C}^4\mid p\_3\neq 0 \vee p\_4\neq 0\}$, and now when I consider the morphism $\phi$ the following orbits are still mapped to the origin:


* $\mathbb{C}^\*\cdot (0,0,z,0)$
* $\mathbb{C}^\*\cdot (0,0,0,w)$
* $\mathbb{C}^\*\cdot (0,0,z,w)$


therefore $\phi^{-1}(0,0,0,0)$ is not a single orbit, hence it is not a geometric quotient.


Am I doing something wrong? I intuitively thought that, even $U$ does not coincide with the stable points (which would give me a geometric quotient), still throwing away only some bad points would have helped me



Geometric quotients obtained by throwing away limits
Throwing away a subset of codimension two you will not give any more invariants. In particular, points of the form $(p\_1,p\_2,0,0)$ will not be separated.


Nevertheless, the geometric quotient exists. It is just not an affine variety anymore. There are two basically equivalent ways to see this:


1. The ad-hoc way: Since regular functions are not enough, use also rational invariants. Here, $p\_3/p\_4$ and $p\_4/p\_3$ come to mind. They glue to give a morphism
$$
U\to\mathbf P^1:(p\_1,p\_2,p\_3,p\_4)\mapsto [p\_3:p\_4].
$$
Now throw in all invariants which you already have to get a morphism
$$
U\to\mathbb C^4\times\mathbf P^1:(p\_1,p\_2,p\_3,p\_4)\mapsto (p\_1p\_3,p\_1p\_4,p\_2p\_3,p\_2p\_4,[p\_3:p\_4]).
$$
The image of this map is precisely the quotient of $U$. It is given by the equations
$$
q\_{13}q\_{24}=q\_{14}q\_{23}, q\_{13}p\_4=q\_{14}p\_3, q\_{23}p\_4=q\_{24}p\_3
$$
with $q\_{ij}=p\_ip\_j$. So $U/\!/\mathbb C^\*$ is a three dimensional homogeneous quadric with the vertex replaced by a $\mathbb P^1$. One checks easily that it is smooth.
2. The GIT way: All quotients in GIT depend on the choice of a linearized line bundle. On $\mathbb C^4$ there is only the trivial bundle $\mathbb C^4\times\mathbb C$ but one can tamper with the action on it. More precisely let $\mathbb C^\*$ act by the formula
$$
t\cdot(p\_1,p\_2,p\_3,p\_4,u)=(tp\_1,tp\_2,t^{-1}p\_3,t^{-1}p\_4,tu).
$$
Then one gets two more invariants namely $p\_3u$ and $p\_4u$. The invariant ring is graded by putting $u$ in degree $1$ and all other variables in degree $0$. Taking the proj of the invariant ring just amounts to combine $p\_3u$ and $p\_4u$ to $[p\_3u:p\_4u]=[p\_3:p\_4]\in\mathbb P^1$. So the result is the same as above.


