Projective planes
I'm working with the following definition of projective plane here: we have a set $P$ of points, and a collection $L$ of subsets of $P$ called "lines" with the following special properties:
- For every two points there is a unique line containing them both
- Every two lines intersect in exactly one point
(so this is a bit of an intermediate position between the definition where "points", "lines" and their incidence relation are completely abstract concepts and the one where the projective plane is defined as a very concrete set described in terms of elements of a field)
This question is about finite projective planes, but my intuition comes from the ordinary projective plane over $\mathbb{R}$. I need your help because somehow this intuition seems not to carry over correctly to the finite case and I don't understand why.
In the finite case I say that $P$ is non-degenerate if all lines have the same cardinality.
I believe there is a more general way of defining non-degenerateness that does not require the notion of 'cardinality of a line' to make sense, but I forgot what it was and this one is fine for my purposes.
The Question
Let $(P, L)$ be a non-degenerate projective plane. By a fake line in $P$ I mean some subset $f$ of $P$ with same cardinality as all the lines but which is not in $L$ itself.
The fake line $f$ tries to convince it fellow lines that it is one of them. Each actual line $l$ has only one way of deciding if $f$ is one of its peers or not: by seeing if it indeed has exactly one point in common with $f$. If a a line $l$ intersects $f$ in exactly 1 point (rather than 0 or 2, or some other number) we say that $l$ is fooled by $f$
Questions:
- What is the maximum number of lines a fake line $f$ can fool?
- What is the maximum number of points $p$ such that $f$ can fool all lines through $p$?
Intuition
One of the first cases I considered was in the real projective plane, where I take $f$ to be the graph $y = x^3$ together with the point at infinity that is the intersection of all lines parallel to the $y$-axis. (This seems the most natural choice given how the graph becomes steeper and steeper as it gets further and further from the origin.)
Now here the answer to question 1 and 2 seem to be 'the vast majority of them' and 'half a line worth of points'.
To wit: the lines $y = ax + b$ are fooled by $f$ whenever $a \leq 0$ (which in particular means that of the 'line at infinity' (the set of points $p_a$ that are the intersection of all lines with slope $a$) half of the points has the property sought after in question 2). Of the lines $y = ax + b$ with $a < 0$ we have that most of them (those with $b$ "sufficiently large") are also fooled, hence my optimism in question 1. Finally none of the lines perpendicular to the $x$-axis are fooled, so 'half of the points at infinity' seems to be the final answer to question 2.
Now contrast this with the finite case. In the triangle, the smallest non-degenerate plane, every set of two points is a line and the question is moot. In the Fano-plane (with seven points and seven lines), every set of three non-colinear points fools exactly 3 lines that do go through one point, so the answers to question 1 and 2 are 3 and 1 respectively.
I tried some things in projective planes of order 4, 5 and 7 but never got an answer to question 1 that reached half the number of lines. Also I looked at curve $y = x^3$ over $\mathbb{F}_p$ for $p \equiv 2 \mod 3$ (so that, as in the real case, $x \mapsto x^3$ is injective) and found that (quite different from the real case) there is no number $a$ such that all lines $y = ax + b$ are fooled!
Of course $y = x^3$ is only one way to pick $f$ and there might be others that perform better, but still I feel that in the finite case there is some force working against me that does not exist over $\mathbb{R}$. Maybe something pigeon-hole related? All thoughts are welcome.