# Lifting of automorphism of rational surface to that on abelian variety

The paper I am referencing is "Normal Subgroups of the Cremona Group." https://arxiv.org/abs/1007.0895. In theorem 5.14, at the bottom of page 52, the author stated for the abelian surface $$Y= \mathbb{C}/\mathbb{Z}[i] \times \mathbb{C}/\mathbb{Z}[i]$$, we consider the quotient $$X= Y/ \Gamma$$ where $$\Gamma$$ is the group of order 4 generated by the map $$(x,y)\mapsto (ix, iy)$$ on $$Y$$. Moreover one can show that $$X$$ is rational.

In general the group of automorphism of $$X$$ cannot be lifted to that on $$Y$$ since there are more automorphisms on a rational surface than on an abelian one. However, the paper stated that in this case we can do so since the fundamental group $$\Lambda$$ of $$X\backslash \text{Sing}(X)$$ is the group $$\mathbb{Z}/4\mathbb{Z} \ltimes (\mathbb{Z}[i]\times \mathbb{Z}[i])$$. They further say that $$\mathbb{Z}[i]\times \mathbb{Z}[i]$$ is invariant under all automorphism of $$\Lambda$$ and used this to conclude that we can lift an aut on $$X$$ to that of $$Y$$.

How did they arrive at this conclusion? Would be grateful if anyone can help to explain what that paragraph meant.

## This question had a bounty worth +100 reputation from thedilated that ended 23 hours ago. Grace period ends in 10 minutes

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Denote $$X\backslash\text{Sing}(X)$$ by $$X_0$$ and its preimage in $$Y$$ as $$Y_0$$. Note that $$Y_0$$ is the Galois cover of $$X_0$$ corresponding to the normal subgroup $$\mathbb{Z}[i]\times\mathbb{Z}[i]$$ inside $$\Lambda.$$ For any automorphism $$f$$ of $$X_0$$, the pullback of $$Y_0$$ along $$f$$ is the cover corresponding to the image of $$\mathbb{Z}[i]\times\mathbb{Z}[i]$$ under the automorphism of $$\Lambda$$ induced by $$f$$. By the reasoning you mentioned, this implies that the pullback of $$Y_0$$ along $$f$$ is again $$Y_0$$ and so we get an automorphism of $$Y_0.$$
It remains to show that any automorphism of $$Y_0$$ extends to an automorphism of $$Y$$. Here is one of many ways to show this. It is a theorem (see e.g. Theorem 3.2 in Milne's notes) that any rational map from a nonsingular variety to an abelian variety extends to a regular map. So we know that our automorphism of $$Y_0$$ comes from a map $$Y\rightarrow Y$$. This latter map is necessarily birational (because it restricts to $$f$$), quasi-finite, and proper, so by normality of $$Y$$ it must be an isomorphism.