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Let $Y$ be the abelian variety $\mathbb{C}/\mathbb{Z}[i] \times \mathbb{C}/\mathbb{Z}[i] $. Let $X$ be the quotient of $Y$ by action of the group generated by the map $\eta(x,y)=(ix,iy)$. This group generated is of order 4, and is given by $\{e, -e, \eta, -\eta\}$ where $e$ is the identity map.

How can we show that $X$ is in fact a rational surface and has 10 singularities? I do know we have to look at the fixed points by the subgroups generated by $\eta$ but I have very little idea on how to proceed. Any hints given or links to papers describing this particular construction would be greatly appreciated!

Edit: I have been told that I need to look at the fixed points of the orbit, which in this case is simply the 0 class. How can I proceed?

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    $\begingroup$ From where do you know that $X$ has 10 singularities? I could find only $9$ special orbits in $Y$. $\endgroup$ – Alan Muniz Feb 22 at 19:12
  • $\begingroup$ From the paper: arxiv.org/abs/1007.0895 "Normal subgroups in the Cremona group". They claimed that $X$ has 10 singularities, all of which can be resolved within a single blowup. $\endgroup$ – thedilated Feb 23 at 4:30
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Let us call $E=\mathbb{C}/\mathbb{Z}[i]$. Then $Y= E\times E$ and $X=Y/\eta$. The singularities of $X$ come from orbits on $Y$ with nontrivial stabilizer.

Since $\eta$ has order four, a possible stabilizer is generated either by $\eta$ or $\eta^2$. Hence we first look for the points fixed by $\eta^2(x,y) = (-x,-y)$. Moreove the action is diagonal and we may analyse each factor.

Imposing that $-(a+bi) \equiv a+bi \mod{\mathbb{Z}[i]}$ it follows that $$a+bi \mod{\mathbb{Z}[i]}\in \left\{0, \frac{1}{2}, \frac{i}{2}, \frac{1}{2}+\frac{i}{2} \right\} = K\subset E $$ Then the points with nontrivial stabilizer by $\eta$ are in $K\times K$. This gives us ten orbits $$ \left\{\left(0,0\right)\right\}, \left\{\left(0,\frac{1}{2}\right),\left(0,\frac{i}{2}\right)\right\}, \left\{\left(0,\frac{1}{2}+\frac{i}{2}\right)\right\}, \left\{\left(\frac{1}{2},0\right),\left(\frac{i}{2},0\right)\right\}, \left\{\left(\frac{1}{2},\frac{1}{2}\right),\left(\frac{i}{2},\frac{i}{2}\right)\right\}, \left\{\left(\frac{1}{2},\frac{i}{2}\right),\left(\frac{i}{2},\frac{1}{2}\right)\right\}, \left\{\left(\frac{1}{2},\frac{1}{2}+\frac{i}{2}\right),\left(\frac{i}{2},\frac{1}{2}+\frac{i}{2}\right)\right\}, \left\{\left(\frac{1}{2}+\frac{i}{2},0\right)\right\}, \left\{\left(\frac{1}{2}+\frac{i}{2},\frac{1}{2}\right),\left(\frac{1}{2}+\frac{i}{2},\frac{i}{2}\right)\right\}, \left\{\left(\frac{1}{2}+\frac{i}{2},\frac{1}{2}+\frac{i}{2}\right)\right\}. $$

We may use Castelnuovo's criterion to determine the rationality. First, if $S$ is a a smooth surface obtained from $X$ by resolution of singularities, then from $Y= E\times E$ we have -- by Proposition 2.2 here -- $$ q(S) = g(E/\eta|_E) + g(E/\eta|_E) = 0. $$ and using a Kunneth type formula for coherent sheaves we have $$ P_2 = h^0(S,K_S^{\otimes 2}) = h^0(E\times E, K_E^{\otimes 2} \boxtimes K_E^{\otimes 2} )^\eta = h^0(E, K_E^{\otimes 2})^{\eta|_E}\cdot h^0(E, K_E^{\otimes 2})^{\eta|_E} = 0. $$

Therefore $P_2 = q =0$ and $S$ (hence $X$) is rational by Castelnuovo's rationality criterion.

Besides Serrano's paper I'd recomend Harris, Algebraic Geometry and Barth et ali, Compact Complex Surfaces.

Let me know if there is any point that is not clear.

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  • $\begingroup$ Hi Alan. Thank you so much for the amazing and detailed answer. May I also ask the following: the paper claims that every automorphism on $X$ can be lifted to that on $Y$. Why is this true? I know it is not true that you can lift an autmorphism on a rational map to that on an abelian variety in general. $\endgroup$ – thedilated Mar 4 at 11:19
  • $\begingroup$ I could not understand their argument but I believe it has something to do with the Galois Theory of the covering $Y \to X$. $\endgroup$ – Alan Muniz Mar 5 at 19:18

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