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Oct 30, 2018 at 22:24 comment added Sam Hopkins Ok, I think I understand. I did not realize this non-isotropic condition was enough to make the finite field case behave like the real case.
Oct 30, 2018 at 22:24 vote accept Sam Hopkins
Oct 30, 2018 at 22:22 comment added Ilya Bogdanov Sorry, I did not mean to say this is completely trivial;). Surely, two points $(x_1:x_2:x_3:x_4)$ and $(y_1:y_2:y_3:y_4)$ are orthogonal if $\sum x_iy_i=0$. Notice that the points orthogonal to a given non-isotropis point $p$ form a 2-dimensional projective plane which does not contain $p$.
Oct 30, 2018 at 21:20 comment added Sam Hopkins Sorry, maybe the problem is indeed simple: I'm not used to thinking about these kind of configurations. But can I ask what "orthogonal" means in this context?
Oct 30, 2018 at 19:34 history answered Ilya Bogdanov CC BY-SA 4.0