Is this quotient of a threefold known? What are its singularities?

Question

Assume $G$ is the Klein four group $G=\{1,\sigma_1,\sigma_2,\sigma_3\}$.

Let $G$ act on $X=\mathbb{A}^2\times\mathbb{P}^1$ via:

$$\sigma_1\cdot(x,y,[\lambda:\mu])=(-x,y,[\lambda:-\mu]) \text{ and } \sigma_2 \cdot ((x,y,[\lambda:\mu])=(x,-y,[\mu:\lambda]).$$

Is the quotient $X/G$ some well known variety? What are its singularities? Can one understand the fiber over $(0,0)$ of the map $X/G \rightarrow \mathbb{A}^2/G$ induced from the projection on the first factor?

I tried to do some computations: remove $0,\infty \in \mathbb{P}^1$, let $z$ be the coordinate on the rest. Then I tried to find $\mathbb{C}[x,y,z,\frac{1}{z}]^G$ and found a lot of invariants with a lot of relations, for example: $x^2$, $y^2$, $(z+\frac{1}{z})x$, $(z-\frac{1}{z})xy$, $z^2+\frac{1}{z^2}$ etc. But I don't see what variety this ring of invariants describes. Maybe it is well known?

Answer 1

I denote by $\mathbb{A}^1_x$ the affine line given by $y=0$ in $\mathbb{A}_2$ and by $\mathbb{A}^1_y$ the line $x=0$ in $\mathbb{A}_2$.

Then, the fixed locus of $\langle \sigma_1 \rangle$ is $\mathbb{A}^1_y \times [1:0] \cup \mathbb{A}^1_y \times [0:1]$.

The fixed locus of $\langle \sigma_2 \rangle$ is $\mathbb{A}^1_x \times \left([1,1] \cup [-1:1] \right)$.

The fixed locus of $\sigma_3 = \sigma_1 \times \sigma_2$ is $(0,0) \times [1,i] \cup (0,0) \times [1,-i]$.

There are no points of $X$ which are fixed by the whole $G$.

Let us denote by $\pi$ the quotient map : $\mathbb{A}^2 \times \mathbb{P}^1 \rightarrow (\mathbb{A}^2 \times \mathbb{P}^1)/G$.

As a consequence of the above discussion on the stablizers, one deduces that $(\mathbb{A}^2 \times \mathbb{P}^1)/G$ looks like:

_$\mathbb{A}^1 \times (\mathbb{A}^2/\mathbb{Z}_2)$ locally around any point of $\pi(\mathbb{A}^1_y (\times [1:0] \cup [0:1]) \cup \mathbb{A}^1_x \times ([1:1] \cup [1,-1])$. Put simply, the quotient variety has surface ordinary double point singularities along: $\pi(\mathbb{A}^1_y (\times [1:0] \cup [0:1]) \cup \mathbb{A}^1_x \times ([1:1] \cup [1,-1])$.

_the quotient variety has singularity of type $\mathbb{A}^3/\mathbb{Z}_2$ locally around $\pi((0,0) \times([1,i] \cup [1,-i]))$,

_the quotient variety is smooth everywhere else.

The ordinary double points can be resolved by blowing up the singular locus, that is blowing up $\pi(\mathbb{A}^1_y (\times [1:0] \cup [0:1]) \cup \mathbb{A}^1_x \times ([1:1] \cup [1,-1])$ (which is smooth).

The singular points $\pi(50,0) \times [1:i])$ and $\pi((0,0) \times [1:-i])$ can also be resoved by blowing them up. But note that the variety $X/G$ is not Gorenstein at the points $\pi((0,0) \times [1:i])$ and $\pi((0,0) \times [1,-i])$.

As far as the fiber over $(0,0)$ of the map $X/G \rightarrow \mathbb{A}^2/G$ is concerned, I think that it is the quotient of $\mathbb{P}^1$ by $G$ which acts as its restriction on $\mathbb{P}^1$. The quotient map is a $4:1$ cover ramified in $4$ points (namely $[1:0],[0:1],[1:i],[1:-i]$). The Hurwitz formula shows that the quotient is $\mathbb{P}^1$.

Answer 2

For whatever it's worth, the invariant ring $\mathbb{C}[x,y,z,z^{-1}]^G$ can be obtained algorithmically by embedding $\mathbb{A}^2 \times (\mathbb{P}^1 \setminus \{0,\infty\})$ equivariantly into $\mathbb{A}^4$ and then applying standard algorithms to the action on $\mathbb{A}^4$. The result is that $\mathbb{C}[x,y,z,z^{-1}]^G$ is generated by

$f_1 = x^2, f_2 = y^2, f_3 = (z+z^{-1}) x, f_4 = (z - z^{-1}) x y, f_5 = z^2 + z^{-2}$, and $f_6 = (z^2 - z^{-2}) y$.

Just the invariants you obtained (except maybe $f_6$). The relations can also be computed. The result are the following six relations:

$-f_3 f_6 + f_4 f_5 + 2 f_4, -f_1 f_6 + f_3 f_4, -f_1 f_5 - 2 f_1 + f_3^2, f_2 f_5^2 - 4 f_2 - f_6^2, f_2 f_3 f_5 - 2 f_2 f_3 - f_4 f_6, f_1 f_2 f_5 - 2 f_1 f_2 - f_4^2$.

I'm not sure this is enlightening in any way. At least one consequence is that making $f_5^2 - 4$ invertible yields a localized polynomial ring. So also removing the points $[1:\pm 1]$ and $[1:\pm i]$ from $\mathbb{P}^1$ yields a nonsingular quotient.