Bound on the sum of intersection number of any projectivized measured foliation with two transverse measured foliations

Question

Let $R$ be a finite Riemann surface (having negative Euler Characteristic) without boundary (may have punctures) and $q$ be a unit area quadratic differential on $R$. We define $\mathcal{MF}_{1}=\{F \in \mathcal{MF} : Ext_{R}(F)=1\}$ and the function

$$I : \mathcal{MF}_{1} \rightarrow [0,+\infty]$$ as $I(F) = i(H(q),F) + i(V(q),F)$ for $F \in \mathcal{MF}_{1}$, where $\mathcal{MF}$ is the set of all measured foliations on $R$ upto Whitehead Equivalence, $Ext_{R}(F)$ is the extremal length of the foliation $F$ on $R$, $i(\cdot,\cdot)$ denotes the intersection number between measured foliations and $H(q), V(q)$ are respectively the horizontal and vertical foliations of $q$. Using Minsky's Inequality ($i(G,F)^{2} \le Ext_{R}(F)Ext_{R}(G)$) it is easy to see that $I(F) \le 2$ for all $F \in \mathcal{MF}_{1}$. Note that for the foliations $H(q)$ and $V(q)$ we have $I(H(q))=1=I(V(q))$. It seems that for a foliation $F \in \mathcal{MF}_{1}$ the more it intersects $H(q)$, the less it intersects $V(q)$. Can it be shown that $I$ is actually bounded by 1? If not, then what should be a counter example. Any help would be appreciated.

Answer 1

I think that there are counterexamples, but I am not an expert.... Here is my attempt.

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Let $S$ be the unit square in the complex plane. We obtain $R$ by gluing opposite sides by translation. So $R$ is the "square torus of side-length one". (If you really insist that $R$ have negative Euler characteristic, then we can remove the image of the origin, making $R$ a punctured surface.) Let $q = dz^2$ be the resulting quadratic differential and note that it has area equal to one. $\newcommand{\Ext}{\mathrm{Ext}_R}$

Let $H$ and $V$ be the horizontal and vertical foliations. These very conveniently have $\Ext(H) = \Ext(V) = 1$. Let $D$ be the foliation whose leaves are the geodesics of slope one.

Claim: $\Ext(D) = 2$.

Proof: We cut $R$ along one leaf of $D$ (running through the origin, if we removed it). This gives a flat right annulus with circumference $\sqrt{2}$ and width (or "height") $1/\sqrt{2}$. So the modulus of the annulus is $1/2$. This is the reciprocal of the extremal length. $\Box$

So the measured foliation $D/\sqrt{2}$ has the correct extremal length: that is, one. We now find that $i(V, D/\sqrt{2}) = i(H, D/\sqrt{2}) = 1/\sqrt{2}$. So their sum is $2/\sqrt{2} = \sqrt{2}$ which is a bit more than one.