Let $S$ be a hyperbolic surface of genus $g \geq 2$.
A *discrete* geodesic lamination on $S$ is a set of disjoint, simple, closed geodesics.


Let $L\_{1}$ and $L\_{2}$ be two discrete geodesic laminations on $S$. We say that $L\_1$ and $L\_2$ fill $S$, if $S \setminus (L\_{1} \cup L\_{2})$ is a disjoint union of open topological disks.
Or equivalently, let $i(\alpha,L\_{i})$ be the intersection number of the closed curve $\alpha$ with the leaves of the lamination $L\_{i}$, then for any closed curve $\alpha$, $i(\alpha,L\_{1}) + i(\alpha,L\_{2}) > 0$ (including the closed curves that belong to $L\_1$ and $L\_2$, and recall that if $\alpha \in L\_{i}$, then $i(\alpha,L\_{i}) = 0$).


**My question is**: If we take two discrete geodesic laminations $L\_1$ and $L\_2$ on $S$, such that $L\_{1}$ and $L\_{2}$ have no common leaf. Can we add leaves to $L\_1$ and leaves to $L\_2$, to obtain two laminations $L^{'}\_{1}$ and $L^{'}\_{2}$ (then $L\_{i} \subset L^{'}\_{i}$) such that $L^{'}\_{1}$ and $L^{'}\_{2}$ fill $S$ ?


Since every geodesic lamination can be completed to $3g-3$ closed curve (then if $\alpha \notin L$, $i(\alpha,L) > 0$), **the question remains to the following:** Let $\alpha$,$\beta\_{1}$,...,$\beta\_{k}$ be disjoint, simple, closed geodesics in $S$. Can we find a simple closed geodesic $\gamma$ such that $i(\gamma,\alpha) >0$ and $\forall j$,$i(\gamma,\beta\_{j}) = 0$.



Pair of laminations that fill on a closed surface
The answer is "no". For consider the case where $L\_1 = L\_2$ is a single simple closed geodesic.




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Now you've added the hypothesis that $L\_1$ and $L\_2$ have no common leaf, the answer becomes "yes". Here is a sketch.


Let $X\_i = S - L\_i$. If $X\_1$ (say) is a union of pants then $L\_1$ is a pants decomposition and so is maximal. Suppose not. Then we pick a lamination $\lambda\_1 \subset X\_1$ which restricts in each non-pants component $X' \subset X\_1$ to a *filling lamination*. That is, a closed subset of $X'$ which is a union of simple geodesics, none of which are loops, so that $X' - \lambda\_1$ is a union of disks and peripheral annuli. Furthermore $\lambda\_1$ admits a transverse measure (so has no leaves that spiral about a component of $L\_1$). Filling laminations exist by Thurston's theory of pseudo-Anosov maps.


We do the same for $L\_2$: that is, we pick a lamination $\lambda\_2$ that lives in and fills all non-pants components of $X\_2$. Now define $\Lambda\_i = L\_i \cup \lambda\_i$


**Exercise**: Since $L\_1$ and $L\_2$ share no leaves the same holds of $\Lambda\_1$ and $\Lambda\_2$.


We deduce that there is a definite angle $\epsilon$ so that, at every point of $\Lambda\_1 \cap \Lambda\_2$, the two laminations intersect with an angle of at least $\epsilon$. In particular $\Lambda\_1 \cup \Lambda\_2$ fills $S$.


We now choose sufficiently close Hausdorff approximations of $\Lambda\_i$ by pants decompositions $L\_i'$. We deduce that $L\_i'$ contains $L\_i$. Also $L\_1'$ and $L\_2'$ fill, as desired.


