Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 156 / 2

The arcs $\alpha$ and $\beta$ bound a bigon if there are subarcs with common endpoints whose union is the boundary of an embedded disc and whose interiors are disjoint. They are in minimal position if $|\alpha \cap \beta |$ is the least possible intersection count among proper arcs isotopic to $\alpha$ and $\beta$ relative to their ends.

Let $\tilde{\alpha }$ and $\tilde{\beta }$ be lifts to the compactified hyperbolic plane $\overline{{\mathbb{H}}^2}$. Suppose they have two common points. Choose two consecutive common points along one lift. If both lie in ${\mathbb{H}}^2$, the intervening subarcs contain an innermost embedded disc. The covering projection is injective on its interior; otherwise a nontrivial deck translate would produce a still smaller such disc. Its projection is a bigon between $\alpha$ and $\beta$, contrary to the hypothesis.

The same innermost-disc argument works when one corner is on the circle at infinity, after deleting a sufficiently small horoball about that corner. The only possible obstruction would identify the ideal corner by both a hyperbolic deck transformation associated with an essential return and a parabolic deck transformation stabilizing the relevant puncture. A hyperbolic and a parabolic element of the surface group cannot have that fixed point in common, so the truncated disc again projects to a bigon. If both common points are ideal, truncate at both ends and apply the same argument. Therefore any pair of lifts intersects in at most one point of $\overline{{\mathbb{H}}^2}$.

If two arcs bound a bigon, pushing one side across its disc is an isotopy relative to the ends that removes the two corners. Such representatives cannot be in minimal position.

Conversely, compare $\alpha$ with an isotopic representative having the fewest intersections with $\beta$, and lift the isotopy to $\overline{{\mathbb{H}}^2}$. At the first stage where the original excess intersections disappear, two lifted arcs enclose an innermost disc. Part b rules out escape through a puncture or repeated intersections at the ideal boundary, so this disc projects injectively to a bigon on $S$. Thus absence of a bigon implies minimal position. This proves the bigon criterion for essential simple proper arcs.