Incidence identities and exchange rigidity in every projective plane

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These statements hold in every projective plane, Desarguesian or not. They are elementary and are recorded here with complete author proofs by Plasma AI, without independent review. They are the all-plane inputs of the bounded-blocking proof sketch.

Setting. $\Pi$ is a projective plane of order $q\ge2$ with $N=q^2+q+1$ points and lines. For an integer-valued function $z$ on points, write $\partial z(\ell)=\sum_{P\in\ell}z(P)$ for its line sums and $|z|=\sum_Pz(P)$. For a point set $S$ of size $s$, the load of a line is $x_\ell=\partial 1_S(\ell)=|S\cap\ell|$.

Moments, pencils and variance

Moments. Counting incidences and ordered pairs of distinct points of $S$,

$$\sum_\ell x_\ell=(q+1)s,\qquad \sum_\ell x_\ell(x_\ell-1)=s(s-1).$$

Pencil identity. For every integer function $z$ and every point $P$,

$$\sum_{\ell\ni P}\partial z(\ell)=q\,z(P)+|z|.$$

Proof. The $q+1$ lines through $P$ all contain $P$, and every other point lies on exactly one of them. $\square$

For $z=1_S$ this reads $\sum_{\ell\ni P}x_\ell=s+q\,1_S(P)$. The map $z\mapsto\partial z$ is injective: if $\partial z=0$, the identity gives $qz(P)=-|z|$ for every $P$, so $z$ is a constant $c$ with $(q+N)c=0$. A load function therefore determines its point set. Conversely, an integer function $x$ on lines with $\sum_\ell x_\ell=(q+1)s$ whose pencil sums take only the values $s$ and $s+q$ is the load function of the set $S$ of points with pencil sum $s+q$. Summing the pencil sums over all points gives $(q+1)^2s=Ns+q|S|$, so $|S|=s$. Then $x$ and $\partial1_S$ have the same pencil sums, and they are equal because the pencil-sum map on line functions is injective: it is the map $\partial$ of the dual plane.

Variance identity. With $\mu=(q+1)s/N$ the mean load,

$$\sum_\ell(x_\ell-\mu)^2=sq\Bigl(1-\frac sN\Bigr).$$

Proof. By the moments, $\sum_\ell x_\ell^2=s^2+sq$. Since $(q+1)^2=N+q$, $N\mu^2=s^2+s^2q/N$. Subtract. $\square$

So every point set, however it is built, has load variance

$$v=\frac{sq}N\Bigl(1-\frac sN\Bigr)=\mu\,\frac q{q+1}\Bigl(1-\frac sN\Bigr),$$

asymptotic to its mean when $s=O(q)$. This is an equality of second moments, not a Poisson approximation or a constraint on tail shape. If every load lies in $[1,C]$, then $v\le(\mu-1)(C-\mu)$, the largest variance of a distribution on $[1,C]$ with mean $\mu$. Written out, this is exactly the quadratic inequality of L27, whose proof sums $(x_\ell-1)(C-x_\ell)\ge0$. Any construction with a bounded cap must therefore realize a load distribution with variance close to its mean on a bounded range; a concentrated distribution is impossible.

Exchange rigidity

Lemma (exchange footprint). Let $z\ne0$ be an integer function on the points of $\Pi$ with $|\partial z(\ell)|<q$ for every line, and let $F=\{\ell:\partial z(\ell)\ne0\}$. Then $|F|\ge q+1$. If $|F|=q+1$ and $|\partial z(\ell)|<q/2$ for every line, then $F$ is the pencil of a point $P$ and $z=k\,1_{\{P\}}$ for a nonzero integer $k$.

Proof. By the pencil identity, $\sum_{\ell\ni P}\partial z(\ell)\equiv|z|$ (mod $q$) at every point $P$. Since $\partial$ is injective, $F$ is nonempty.

Suppose $|F|\le q$. The lines of $F$ cover at most $q(q+1)<N$ points, so some point lies on no line of $F$, and the congruence there gives $|z|\equiv0$. A line $\ell\in F$ meets the other at most $q-1$ lines of $F$ in at most $q-1$ of its $q+1$ points, so some point of $\ell$ lies on no other line of $F$. The congruence at that point gives $\partial z(\ell)\equiv|z|\equiv0$ (mod $q$), so $\partial z(\ell)=0$ because $|\partial z(\ell)|<q$. This contradicts $\ell\in F$. Hence $|F|\ge q+1$.

Suppose $|F|=q+1$. Each $\ell\in F$ meets the other $q$ lines of $F$ in at most $q$ points, so again some point of $\ell$ lies on no other line of $F$, and $\partial z(\ell)\equiv|z|$ (mod $q$). If some point lay on no line of $F$, then $|z|\equiv0$ and the previous contradiction recurs. So the $q+1$ lines of $F$ cover every point, and by the lemma below they form the pencil of a point $P$. All values $\partial z(\ell)$ with $\ell\in F$ are congruent to $|z|$ modulo $q$ and lie in $(-q/2,q/2)$, so they equal one integer $k$, and $k\ne0$ because $F$ is the support. Then $\partial z=k\,\partial 1_{\{P\}}$, since $\partial 1_{\{P\}}$ is the indicator of the pencil of $P$, and injectivity gives $z=k\,1_{\{P\}}$. $\square$

Lemma (covering pencils). A set of $q+1$ lines covering every point is the pencil of a point.

Proof. Dually, let $B$ be a set of $q+1$ points meeting every line. Take a point $P\notin B$. The $q+1$ lines through $P$ partition the other points, and each meets $B$, so each contains exactly one point of $B$. Hence every line through a point outside $B$ meets $B$ exactly once. For distinct $X,Y\in B$, the line $XY$ meets $B$ twice, so it contains no point outside $B$; thus $XY\subseteq B$ and, comparing sizes, $B=XY$. Dualizing, the lines form a pencil. $\square$

Consequence for exchanges. Let $R\subseteq S$ and $A$ disjoint from $S$, not both empty, with every line meeting $A$ and $R$ in fewer than $q/2$ points. The exchange $S\mapsto(S\setminus R)\cup A$ changes the load of every line in $F$ for $z=1_A-1_R$. It changes at least $q+1$ loads, and at least $q+2$ unless it inserts or removes a single point, the case $z=\pm1_{\{P\}}$.

This is a bound on changed line loads, not on the number of changed points. A single insertion has point support one and line support $q+1$; swapping two distinct points has point support two and line support $2q$. A bounded-point absorber is not excluded merely because many lines see its change. Its legality must account for those lines simultaneously. The small-change hypothesis also matters: replacing a full line by another full line changes only those two loads, by $q$ and $-q$. The stronger, but state-specific, large-deletion obstruction uses saturated pencils of a maximal arc; it does not follow just from the footprint bound.

A modular form of the pencil identity

Lemma. Let $m\ge2$ be coprime to $q$, and let $W=\{\ell:x_\ell\not\equiv1\pmod m\}$. For every point $P$,

$$\sum_{\ell\ni P,\ \ell\in W}(x_\ell-1)\equiv s-q-1+q\,1_S(P)\pmod m.$$

Hence either every point of $S$ or every point outside $S$ lies on a line of $W$.

Proof. Subtract $q+1$ from the pencil identity for $1_S$; lines outside $W$ contribute multiples of $m$. The right sides for points in and outside $S$ differ by $q\not\equiv0$, so at least one of them is nonzero modulo $m$, and a point on no line of $W$ has left side $0$. $\square$

If $W$ were empty, both alternatives would fail unless $S$ or its complement were empty; but then every load is $0$ or $q+1$, neither congruent to $1$, so $W$ is all lines. Hence no point set has all its line intersections congruent to $1$ modulo a number coprime to $q$. For a set with exactly the intersection numbers $1$ and $n$ this gives $\gcd(n-1,q)>1$, a weak form of the classical restriction that $n-1$ divides $q$. For a set with all loads in $[1,4]$, where L27's argument gives $t_2+t_3=-\tfrac12\bigl(s^2-(4q+5)s+4N\bigr)\le(24q+9)/8$ for the numbers $t_i$ of lines with $i$ points, the case $m=3$ with $3\nmid q$ says that the lines with two or three points cover $S$ or cover its complement. In the second case at least $(N-s)/(q-1)$ of them are needed, since each contains at most $q-1$ points outside $S$.