Geometric quotients obtained by throwing away limits

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

1 Answer

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.

• Dear @Friedrich Knop, thanks a lot for the exhaustive answer. Before accepting it, can I ask you if there are some references where I can look about it? Thanks again for your patience! Apr 3, 2021 at 12:50