On a class of partial planes related to biplanes

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Problem: Reconstructed statement: characterize the finite biplanes $\mathcal B=(P,\mathcal L,I)$ that occur as the canonical quotient of a regular antipodal plane, i.e. a partial plane with involutions $p\mapsto p^*$, $L\mapsto L^*$, where $p^*$ is the unique point not joined to $p$, $L^*$ is the unique line disjoint from $L$, and $pIL\Rightarrow p^*IL^*$. The quotient identifies $x\sim x^*$. This matches Schleiermacher’s Section 4 context: regular antipodal planes have canonical homomorphic images that are biplanes.

Result: A biplane $\mathcal B$ is such a quotient iff its flags admit an odd quadrangle signing.

Precisely, for every quadrangle

with $p\neq q$, $L\neq M$, define a signing $\varepsilon:I\to\{\pm1\}$. Then $\mathcal B$ is the underlying biplane of a regular antipodal plane iff there exists $\varepsilon$ such that

for every quadrangle.

Equivalently, over $\mathbb F_2$, if $x_{pL}$ is a variable for each flag, the system

for every quadrangle must be solvable. Equivalently again: every mod-2 set of quadrangles in which each flag occurs an even number of times has even cardinality.

Proof. Given such $\varepsilon$, construct

with incidence

The antipodes are $(p,s)^*=(p,-s)$, $(L,t)^*=(L,-t)$. If $p=q$ and $s\neq u$, no line joins $(p,s)$ and $(p,u)$. If $p\neq q$, the two common base lines $L,M$ give exactly one lifted common line, because the two possible incidence conditions have product $-1$. Thus any two non-antipodal points are joined by exactly one line. Dually, any two non-antipodal lines meet in exactly one point. Also incidence is preserved by antipodes:

Hence this is a regular antipodal plane, and quotienting by antipodes recovers $\mathcal B$.

Conversely, suppose a regular antipodal plane $\mathcal A$ quotients to $\mathcal B$. Choose signs on each antipodal pair over every point and line. For each base flag $pIL$, regularity gives a perfect matching between the two lifted points over $p$ and the two lifted lines over $L$; define $\varepsilon(p,L)$ by that matching. If a quadrangle had product $+1$, then suitable lifts of $p$ and $q$ would be incident with lifts of both $L$ and $M$, giving two distinct joining lines in the partial plane, impossible. Hence every quadrangle has product $-1$.

The linear parity formulation is exactly the standard consistency criterion for the above affine system over $\mathbb F_2$.

Verification audit: no extra structure beyond the biplane incidence relation is used; the constructed object satisfies the antipodal uniqueness and regularity axioms; the quotient is exactly the given biplane.

Citation: Problem source: A. Schleiermacher, “On a class of partial planes related to biplanes,” J. Geom. 107, 445–466 (2016), DOI: 10.1007/s00022-016-0321-7. No prior exact resolution is cited here; the proof is the signed two-cover/linear-algebra criterion specialized to this setting.