Maximal euler characteristic of surfaces bounding two fixed curves

Question

Let $\gamma_0$ and $\gamma_1$ be two simple closed curves in a closed surface $S$.

What is the maximum Euler characteristic of a compact properly embedded surface $\Sigma \subset S\times [0,1]$ such that $\partial \Sigma = \gamma_0 \times \{0\} \cup \gamma_1 \times \{1\}$?

Of course, in order for such a surface $\Sigma$ to exist, the two curves $\gamma_0$ and $\gamma_1$ must represent the same class in $H_1(S,\mathbb Z_2)$. Note that $\Sigma$ may be non-orientable.

If $2n$ is the minimum geometric intersection number of $\gamma_1$ and $\gamma_2$, it is easy to construct a $\Sigma$ with $\chi(\Sigma)\geqslant \chi(S) - n$. Is there a converse estimate of this kind? Do we have $\chi(\Sigma) \leqslant -n$ when $S$ is a torus?

Answer 1

Here is a graph $\mathcal{G}(S,x)$ associated to a surface and homology class (similar to the 1-skeleton of the curve complex): For a fixed class in $x\in H_1(S;\mathbb{Z}_2)$, consider isotopy classes of embedded multicurves representing the homology class $x$ (one may assume no components are parallel and there are no trivial components). Make this collection the vertices of the graph $\mathcal{G}(S,x)$. Connect two vertices $A, B$ to be adjacent in the graph $\mathcal{G}(S,x)$ if $A\cup B$ are disjointly embedded, and after removing all parallel curves of $A\cup B$, the remaining components bound a pair of pants or a twice-punctured projective plane (this second can only happen if $S$ is non-orientable). I'm not sure if such a complex has been defined before, but there is a somewhat analogous complex defined in the integral homology case by Bestvina, Bux, and Margalit. Also, this is related to a technique of Hatcher-Thurston to undertand surfaces in two-bridge knot complements.

I claim that the maximal Euler characteristic of a surface bounding $\gamma_0\times 0 \cup \gamma_1\times 1$ is the negative of the distance between $\gamma_0$ and $\gamma_1$ in $\mathcal{G}(S,x)$. Put the product metric on $S\times [0,1]$, and make $\Sigma$ into a minimal surface with respect to this metric (Theorem 6.12 of Hass-Scott). Then $S\times t, t\in [0,1]$ gives a foliation by totally geodesic surfaces, and by the maximum principle, they can be tangent to $\Sigma$ in only saddle tangencies (see an argument of Hass, we will assume things are perturbed to be generaic). Thus, for all but finitely many $t$, $S\times t$ meets $\Sigma$ in a finite collection of curves, giving rise to a vertex of $\mathcal{G}(S,x)$. As one passes through a tangency point $S\times t_0$ (assuming things are generic), the intersection changes by a saddle move, giving a surface in $S$ of Euler characteristic $-1$ bounding the curves before and after the tangency. There can never be a closed trivial curve occurring, because this would give rise to a center tangency. Thus, each saddle tangency gives an edge between the adjacent vertices of $\mathcal{G}(S,x)$, and therefore $\Sigma$ gives a path between $\gamma_0$ and $\gamma_1$ in $\mathcal{G}(S,x)$. Conversely, any such path gives rise to a surface.

Of course, there will be many geodesics connecting $\gamma_0$ and $\gamma_1$ in $\mathcal{G}(S,x)$, given by any Morse function on $(\Sigma, \gamma_0,\gamma_1)$ with only index 1 critical points, and I don't expect the distance function to be easy to compute (probably one should use normal surface theory to compute it).

Answer 2

For orientable surfaces of genus at least $2$, pretty sharp bounds are obtained in the unpublished PhD thesis of Ingrid Irmer, which is available here.

Answer 3

Let $a$, $b$ and $x$ be curves on $S$ such that

• $a$ and $x$ intersect once,
• $b$ and $x$ intersect once, and
• $a$ and $b$ have a large algebraic intersection number.

(I assume that $S$ is orientable.) Then there is a surface $\Sigma_a$ connecting $x$ and $x+2a$ with $\chi(\Sigma_a) = -1$ and a similar surface $\Sigma_b$ connecting $x$ and $x+2b$. Combining $\Sigma_a$ and $\Sigma_b$ we get a surface of Euler characteristic $-2$ connecting $x+2a$ and $x+2b$. Thus the geometric intersection number of two curves does not give a lower bound on the complexity of a surface joining them.

(In your original question you ask if $\chi(\Sigma) \le n$, but $n$ is positive and $\chi(\Sigma)$ is negative, so I assume you meant $|\chi(\Sigma)|$.)