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Let $p_1,...,p_7\in\mathbb{P}^{2}$ be seven general points in the projective plane $\mathbb{P}^{2}$ over the complex numbers.

Let $f$ be an automorphism of $\mathbb{P}^{2}$ inducing a permutation of $\{p_1,...,p_7\}$.

Does this imply that $f$ is the identity on $\mathbb{P}^{2}$ ?

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  • $\begingroup$ (out of curiosity) Why seven? What is known about the cases with fewer points? $\endgroup$
    – j.c.
    Commented Sep 26, 2017 at 19:11

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If you fix a general quadrilateral, you fix the coordinates of every point in the affine chart of that quadrilateral. See Bumcroft, Modern Projective Geometry, chapter 3, section I or Hartshorne, Foundations of Projective Geometry. If you permute the order of points in the quadrilateral, you move around which points lie in the affine chart, covering the whole plane.

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  • $\begingroup$ This doesn't answer the question quite. You can see that your $f$ generates a finite abelian group, since some power fixes 4 points. So then $f$ generates a finite cyclic subgroup of $PU(3)$, being compact so lying in a maximal compact. One can then diagonalize $f$. I hope this helps. $\endgroup$
    – Ben McKay
    Commented Sep 26, 2017 at 15:44

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