Let $X$ be a projective complex manifold and $G$ a finite group. Assume that $G$ acts on $X$ holomorphically and freely. Is it true that any birational map $\phi \in Bir(X/G)$ lifts to some birational map $\tilde{\phi}\in Bir(X)$? If this is not true, what kind of condition should one impose on the manifold or the action?

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    $\begingroup$ or the anion, conceivably $\endgroup$ – Tom Goodwillie Feb 8 '13 at 22:14
  • $\begingroup$ Sorry, couldn't resist the lame pun on "cation". $\endgroup$ – Tom Goodwillie Feb 8 '13 at 23:26

This is all about the function fields, if I understand the question. Let $L$ be the function field of $X$ and let $K$ be the function field of $X/G$. $L$ is a finite Galois extension of $K$. A typical automorphism of $K$ will not extend to an automorphism of $L$. For example, if $G$ has order $2$ and $L=K(\sqrt f)$ then the automorphism would have to take $f$ to $f$ times a square.

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  • $\begingroup$ Doesn't your example have fixed locus $f=0$? $\endgroup$ – user2013 Feb 11 '13 at 0:40
  • $\begingroup$ Yes. I totally failed to see the word "freely" in the question. $\endgroup$ – Tom Goodwillie Feb 11 '13 at 1:11
  • $\begingroup$ Thank you for the answer, Tom. I now think my claim is positive although I cannot prove it. I will post something once I am convinced with my idea. $\endgroup$ – Koopa Feb 18 '13 at 18:40

It seems to me that the answer is no even if the action of $G$ is free.

In fact, assume that $G$ acts freely and set $Y:= X/G$. Let $f \colon Y \to Y$ be an automorphism and $f_* \colon \pi_1(Y) \to \pi_1(Y)$ the induced automorphism of the fundamental group. The Galois cover $p \colon X \to Y$ is induced by a normal subgroup of $\pi_1(Y)$, that we call $H$, such that $\pi_1(Y)/H = G$. It is immediate to check that

if $f \colon Y \longrightarrow Y$ lifts to $g \colon X \longrightarrow X$, then necessarily $f_*H=H$.

Now let me give an example of a free $G$-action and an automorphism that does not lift.

Let $Y = \mathbf{C}/ \Lambda$, where $\Lambda = \mathbf{Z} \oplus i \mathbf{Z}$. Then it is well-known that the elliptic curve $Y$ has an automorphism $f \colon Y \to Y$ of order $4$, induced by $z \to iz$.

Now set $X = \mathbf{C}/\Gamma$, where $\Gamma = \mathbf{Z} \oplus 2i \mathbf{Z}$. Since $\Gamma$ has index $2$ in $\Lambda$, we have an étale double cover (degree $2$ isogeny) $p \colon X \to Y$.

However, the automorphism $f \colon Y \to Y$ does not lift to $X$. The point is that the lattice $\Gamma$ is not invariant under $f_*$, in fact $f_*(\mathbf{Z} \oplus 2i\mathbf{Z}) = 2\mathbf{Z} \oplus i\mathbf{Z}.$

Remark. A similar construction works for covers of any degree taking $\Gamma_n:= \mathbf{Z} \oplus ni\mathbf{Z}$. This lattice has index $n$ in $\Lambda$, hence there is a degree $n$ cyclic cover $p_n \colon X_n \to Y$, where $X_n = \mathbf{C}/ \Gamma_n$. The same argument used above shows that the automorphism $f \colon Y \to Y$ does not lift to $X_n$.

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