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I understand that the Sylvester Gallai theorem doesn't hold for the projective complex plane. Can anyone explain why does Kelly's proof: Here doesn't hold for the complex projective plane?

A counter example for the theorem could be the following SG configuration from basic AG course: The 9 inflection points of a cubic in $\mathbb{P}^2$, it is a well known theorem that every line that passes through 2 of them must pass through a 3rd one, and a simple application of Bezout's theorem shows that each line intersecting a cubic curve will meet it in at most 3 points.

Thank you very much in advance!

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I think the main problem is that lines in the complex projective plane aren't one dimensional (topologically), but instead they are topologically spheres, being isomorphic to complex projective one space.

Thus when the proof talks about points in a line being on one side or the other of each other, it doesn't make sense in the complex case. Since complex lines are topologically two dimensional, so they don't have a betweenness relation.

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  • $\begingroup$ For the same reason I imagine that the SG theorem will not trivially hold for the complex affine plane. Is there a known theorem or counter example for the complex affine case? I could not find online. $\endgroup$ – kindasorta Feb 6 at 19:41
  • $\begingroup$ Gah, of course not! Clearly one could take the same counter example as for the projective plane, remove a single line from it, it is then isomorphic to the affine complex plane but has an SG configuration of points (9 flexes) not all on the same line. $\endgroup$ – kindasorta Feb 6 at 19:54
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    $\begingroup$ @kindasorta Yes, I think you can get a counterexample by taking the projective counterexample and looking at an affine open subset which contains all 9 points. Every line should then still pass through the same 3 points as before. In other words delete any line from the projective plane that doesn't pass through any of the points in the configuration. $\endgroup$ – jgon Feb 6 at 19:58

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