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 1


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.

  • $\begingroup$ 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! $\endgroup$ Apr 3, 2021 at 12:50

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