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?



Is this quotient of a threefold known? What are its singularities?
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 $\mathbb{P}^1$.


