Cremona transformation defined by 7 point correspondences

63 Views Asked by At

I believe it is true that given 7 generic point correspondences $x_i \mapsto y_i$ it should be possible to construct a (unique) quadratic Cremona transformation $f$ such that $f(x_i)=y_i$. This would make sense because a quadratic Cremona transformation should have 14 degrees of freedom (12 for the 6 exceptional points and two more for homography [essentially scalars $a_1,a_2,a_3$ that you could throw in front of the standard quadratic Cremona $(x_1:x_2:x_3) \mapsto (a_1x_2x_3:a_2x_1x_3:a_3x_1x_2)$] without changing the exceptional points) and it works for the examples I have tested, but I haven't been able to find a proof of it anywhere or prove it myself. Does anyone have a proof of this or is anyone able to recommend a text that contains this result? Thanks!