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In the paper "Normal Subgroups in the Cremona Group", under remark 5.1 they stated that for any generic set $\Sigma \subset \mathbb{P}^2_\mathbb{C}$ of $k$ points, and $h$ is an automorphism of $\mathbb{P}^2_\mathbb{C}$, then $h$ is the identity as soon $h(\Sigma)\cap \Sigma$ contains at least 5 points.

Can anyone be kind enough to show how do I prove it or are there any papers proving this result?

Thank you very much

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In other words, if $\Sigma$ is generic, there does not exist a non-identity automorphism $h$ for which $|\Sigma\cap h(\Sigma)|\geqslant 5$. Denote $\Sigma=\{p_1,\dots,p_k\}$ and assume that $h(p_i)=p_{f(i)}$ for $i=1,2,3,4,5$ and a certain (we may supposed that fixed) injective function $f:\{1,2,3,4,5\}\rightarrow \{1,2,\dots,k\}$. Consider two cases.

1) Assume that some $f(i)$ is greater than 5, say, $f(5)\notin \{1,2,3,4,5\}$. Then if we fix all points except $p_{f(5)}$, $h$ is uniquely determined by $h(p_i)=p_{f(i)}$ for $i=1,2,3,4$, and the event $h(p_5)=p_{f(5)}$ generically does not hold.

2) $f$ is a bijection of the set $\{1,2,3,4,5\}$, but not identity permutation (if $f$ is identity, then $h$ is identity).

2.1) If $f$ has a fixed point, say $f(5)=5$, again $h$ is determined by $h(p_i)=p_{f(i)}$ for $i=1,2,3,4$ (and $h$ is not identity), and the event that $p_5$ is a fixed point of $h$ generically does not hold.

2.2) $f$ is a cycle of length 5 or a product of disjoint cycles of lengths 3 and 2. The 5-tuple $\{p_1,p_2,p_3,p_4,p_5\}$ is projectively equivalent to the 5-tuple $\{p_{f(1)},p_{f(2)},p_{f(3)},p_{f(4)}, p_{f(5)}\}$. Take your favourite projective invariant of a 5-tuple, like the cross ratio of the points $p_1,p_2,p_3p_4\cap p_1p_2,p_3p_5\cap p_1p_2$ and verify that generically it changes under the permutation $f$ (it suffices to find one specific 5-tuple of points for which it changes).

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  • $\begingroup$ Hi. Thank you very much for your reply. May I clarify a few things? For case 1) why is $h(p_i)=p_{f(p_i)}$ uniquely determined for $i=1,2,3,4$ and how does this suggest $h(p_5)=p_{f(p_5)} $ does not hold generically? $\endgroup$
    – Soby
    Commented Jan 13, 2019 at 9:50
  • $\begingroup$ $h$ is uniquely defined by the images of four points, this is the basic fact about projective transformations of the plane $\endgroup$ Commented Jan 13, 2019 at 10:15
  • $\begingroup$ oh right how silly of me to look over that. Thank you so much! $\endgroup$
    – Soby
    Commented Jan 13, 2019 at 11:24

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