# Does every $C_4$-free bipartite graph lies in some finite projective plane?

A projective plane $$Π$$ is a 3-tuple $$(P,L,I)$$ where $$P$$ and $$L$$ are sets, and $$I$$ is a relation between $$P$$ and $$L$$, such that:

• For every two elements $$p_1$$, $$p_2\in P$$, there exists a unique element $$l \in L$$ such that $$p_1 I l$$ and $$p_2 I l$$.

• For every two elements $$l_1$$, $$l_2\in P$$, there exists a unique element $$p \in L$$ such that $$p I l_1$$ and $$p I l_2$$.

A finite projective plane is a projective plane where $$P$$ and $$L$$ are finite.

Identify projective planes with the bipartite graph with two parts $$P$$ & $$L$$ where $$p\in P$$ is connected to $$l \in L$$ iff $$pIl$$.

Such graphs do not have $$C_4$$ as a subgraph: Suppose there's a $$C_4$$ subgraph formed by the vertices $$p_1,l_1,p_2,l_2$$ where $$p_1$$ and $$p_2$$ are both connected to $$l_1$$ and $$l_2$$. Then the element $$l$$ where $$p_1 I l$$ and $$p_2 I l$$ is not unique, thus violating the rules. Their induced subgraphs has no $$C_4$$s, either.

Q: Is the converse true, i.e. if $$G$$ is a $$C_4$$-free bipartite graph, is there a projective plane $$Π$$ where $$G$$ is an induced subgraph of $$Π$$?

One cannot expect the conjecture above holding for finite-field planes: Let $$Γ$$ be the Desagures graph. Let $$e$$ be an edge of $$Γ$$(The Desagures graph is edge-transitive, so all edges are the same) .

The Desagures theorem states, if the configuration $$Γ＼e$$ can be found in a finite-field plane, then the edge $$e$$ is also present there. So $$Γ＼e$$ is not an induced subgraph of any finite-field plane.

Known: All graphs with 13 vertices or less are subgraphs of finite-field projective planes, as checked by computer.

By the Lefschetz principle, if $$G$$ is a subgraph of the incidence graph of $$\mathbb{RP}^2$$, then $$G$$ is the subgraph of an incidence graph of some finite-field plane. So trees satisfy the conjecture.