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My question is related to lemma 5.12 in ["Normal Subgroups in the Cremona Group"]. Let $h$ be a birational transformation of a projective surface $X$ and $[D']\in N^1(X)$ be an ample class. We define $[D]=[D']/\sqrt{[D']\cdot[D']}$ so that $[D] \in \mathbb{H}_{\overline{\mathcal{Z}}}$ where $\mathbb{H}_{\overline{\mathcal{Z}}}$ is the hyperboloid of one sheet derived from the (completed) Picard Manin space $\overline{\mathcal{Z}}(X)$ of $X$.

In the last line of the proof, the author stated that the induced isometry $h_\ast$ by $h$ on $\overline{\mathcal{Z}}(X)$ fixes the ample class $[D]$, and since $[D]$ is ample then $h$ is an automorphism of $X$.

My question how does $h_\ast$ fixing an ample class $[D]$ allow us to deduce that $h$ must be an automorphism on $X$? Any insight or resources proving the result will be greatly appreciated.

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    $\begingroup$ Replacing $D$ by a multiple, you can assume that $D$ is very ample, i.e. defines an embedding $\varphi _{D}: X\hookrightarrow \lvert D\rvert^*$, with $\lvert D\rvert=\mathbb{P}(H^0(X,\mathcal{O}_{X}(D)))$. Since $h_{*}$ fixes $[D]$, it induces an automorphism of the linear system $\lvert D\rvert$ which preserves $\varphi _{D}(X)$ and induces $h$ on $X$. Therefore $h$ is an automorphism. $\endgroup$
    – abx
    Commented Jan 23, 2019 at 10:29
  • $\begingroup$ @abx: Does $h_*$ fix the linear system $|D|$, or just the numerical class of $D$? $\endgroup$ Commented Jan 23, 2019 at 10:35
  • $\begingroup$ @Francesco Polizzi: Good point. I was assuming that the surface $X$ is regular, since I think this is the case considered in the quoted paper. $\endgroup$
    – abx
    Commented Jan 23, 2019 at 18:25
  • $\begingroup$ @abx Thank you very much for the input! May I also ask the following question: later on the authors in the proof of proposition 5.13 that if one has $h\in Bir(X)$ s.t. $h_\ast$ preserves the axis of $g_\ast$, then $h_\ast [D']$ is an ample class and so $h$ is an automorphism of $X$. May I know how did they arrive at the image of $[D']$ under $h_\ast$ is ample and how this implies that $h$ is an automorphism? $\endgroup$
    – Soby
    Commented Jan 29, 2019 at 13:44

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