Nonexistence of disjoint compact Baer subplanes (Q1125398)

A closed subplane \({\mathcal Q}=(Q,{\mathcal M})\) in a topological compact connected finite-dimensional projective plane \({\mathcal P}=(P,{\mathcal L})\) is a Baer subplane if the point set \(Q\) of \({\mathcal Q}\) has half the dimension of the point set \(P\) of \({\mathcal P}\). Baer subplanes frequently arise as fixed point sets of automorphisms. In particular, if two Baer subplanes are the fixed point sets of Baer involutions, then it is well known that they have both a point and a line in common. The author shows that this remains true for any two compact Baer subplanes (not necessarily associated with Baer involutions). This result is in marked contrast to the finite case where the point set of every finite desarguesian projective plane may be partitioned into mutually disjoint Baer subplanes. The method of proof is to first show that for a Baer subplane \({\mathcal Q}=(Q,{\mathcal M})\) of a compact connected \(4k\)-dimensional projective plane \({\mathcal P}=(P,{\mathcal L})\), where \(k=1,2,4\), the complement \(P\setminus Q\) is a locally trivial fibre bundle over \({\mathcal M}\) with fibres homotopy equivalent to the sphere of dimension \(k-1\). The fibre map \(\alpha\) sends a point of \(P\setminus Q\) to the unique line to the unique line in \({\mathcal M}\) incident with it. In the next step, using various homology and cohomology groups, Lefschetz duality and the topology of compact Baer subplanes, the author then proves that the fibre bundle associated with a Baer subplane does not admit a cross section. The final step is to construct a cross section to \(\alpha\) from a second disjoint Baer subplane \({\mathcal Q'}=(Q',{\mathcal M'})\) and thus obtain a contradiction. By a result of Salzmann two Baer subplanes that have a line in common also have a point in common so that in an assumed counterexample both point sets \(Q\) and \(Q'\) and line sets \({\mathcal M}\) and \({\mathcal M'}\) are disjoint. By duality one obtains a fibre map \(\beta:{\mathcal L}\setminus{\mathcal M'}\to Q'\) and its restriction to \({\mathcal M}\) yields a cross section to \(\alpha\).