Let $G$ be an algebraic group acting on a scheme $X$. Then $f: X \to Y$ is called a categorical quotient if it is constant on $G$-orbits and every $X \to Z$ constant on $G$-orbits factors through it in a unique fashion. We call $f$ a 'good' categorical quotient if:

1) $f$ is a surjective open submersion (i.e. $Y$ has the quotient topology).

2) for any open $U \subset Y$, the induced map $\mathcal O_U \to (\mathcal O_{f^{-1}(U)})^G$ is an isomorphism.

Does anyone know an example of a 'bad' categorical quotient (by which I mean...well...a not good one).

• I don't think you mean open immersion. Oct 16, 2009 at 21:06
• Correct; it's sub- not im- Oct 16, 2009 at 22:52

Note that if $f: X \to Y$ is a categorical quotient in the category of schemes which is stable under base change by open immersions, then the second condition (ie. $\mathcal O_Y \to (f_* \mathcal O_X)^G$ is an isomorphism) is automatically satisfied.

In the paper "Examples and counterexamples for existence of categorical quotients" by A'Campo-Neuen and Hausen, there is an example 4.6 of a categorical quotient $f: X \to \mathbb A^1$ such that $f^{-1}(\mathbb A^1-0) \to \mathbb A^1 - 0$ is not a categorical quotient. I haven't checked but I believe this should also give an example where condition (2) fails.

I don't know of example of a categorical quotient where condition (1) fails.

• Sorry, but this paper consist of quite many examples. Where is precisely the example mentioned above? Mar 8, 2016 at 9:59

Let me explain slightly modified example 4.6 by A'Campo-Neuen and Hausen.

Suppose that $\mathbb C^*$ acts on $X=\mathbb P^1_{x_1:x_2} \times \mathbb A^2_{y, z}$ (indices denote corresponding coordinates): $$t \cdot ((x_1 \colon x_2); \; y; \; z)=((tx_1\colon x_2); \; ty; \; t^{-1}z).$$

There are two rational invariants, $x_1z/x_2$ and $yz$, and one can check that $X \to \mathbb A^1_{yz}$ is a categorical quotient. But the restriction $$Y=\pi^{-1}(\mathbb A^1_{yz}-0) \to (\mathbb A^1_{yz}-0)$$ is not a categorical quotient, because now there is $$Y \to \mathbb P^1_{x_1z:x_2} \times (A^1_{yz}-0).$$