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.
