Coordinates for Laminations: geometric versus shear

Question

Let $S$ be an orientable surface with a triangulation T. A lamination $\ell$ is a simple closed curve on $S$, up to isotopy. We will assume that $\ell$ is drawn in such a way that it intersects the edges of $T$ in the minimal number of points. We can associate to $\ell$ two sets of "coordinates".

The geometric coordinates are defined to be the $|T|$-tuple of numbers $geom(\ell,t)=$ {number of points in $\ell \cap t$}, where $t \in T$ is an edge of the triangulation.

The shear coordinates are another $|T|$-tuple of numbers $shear(\ell,t)$ defined as follows. For every edge $t \in T$ consider the quadrilateral $q$ formed by the two triangles with edge $t$, and the collection of arcs $\ell \cap q$. For each such arc associate a number -1,0,1 depending on how the arc crosses $q$: if it starts and ends at consecutive edges of $q$ associate 0, otherwise if it starts and ends at opposite edge that forms with $t$ a "S" (resp. a "Z") associate -1 (resp. "1"). A more precise definition can be found e.g. in (https://arxiv.org/abs/math/0510312).

My question is whether and how it's possible to move from the two sets of "coordinates", e.g. how to build the $|T|$-tuple $shear(\ell,t)$ from $geom(\ell,t)$.

Answer 1

Yes, it is possible to go back and forth between the two coordinate systems (and a good thing too, as otherwise they would not be coordinates!). My solution goes through a third coordinate system called "normal coordinates".

Suppose that $\gamma$ is a collection of properly embedded arcs in a triangle $T$ which avoids the corners $x, y, z$ of $T$. Suppose that each arc of $\gamma$ meet each edge of $T$ in at most one point. (Thus $\gamma$ forms no bigons with the edges of $T$.) Then we can partition the arcs of $\gamma$ into three collections $\gamma_x, \gamma_y, \gamma_z$ where an arc in $\gamma_x$ (say) separates $x$ from $y$ and $z$. The three numbers $|\gamma_x|, |\gamma_y|, |\gamma_z|$ are called the normal coordinates of $\gamma$ in $T$.

Suppose that $S$ is a surface and $\Delta$ is a triangulation of $S$. Suppose that $\alpha$ is a simple closed multi-curve in $S$ transverse to the skeleta $\Delta^{(k)}$. Suppose that $\alpha$ meets the edges of $\Delta^{(1)}$ minimally, up to isotopies that do not cross the vertices of $\Delta^{(0)}$. (That is, $\alpha$ has no bigons with the edges of $\Delta^{(1)}$.) Then for every triangle $T \in \Delta^{(2)}$, we can compute the normal coordinates of $\alpha$ in $T$.

Here is a sequence of hints for moving between the various coordinate systems.

• To go from normal to intersection coordinates - we simply add. That this is well-defined (independent of which triangle you use) is called the matching equalities for normal coordinates.

• To go from intersection to normal coordinates - this can be done triangle by triangle. Doing this discovers the mod two condition on intersection coordinates as well as the triangle inequality.

• To go from normal to shear coordinates - this is done edge by edge. Suppose that $e \in \Delta^{(1)}$ is an edge. Let $T$ and $T'$ be the adjacent triangles. Let $Q = Q(e)$ be their union; this is the quadrilateral about $e$. If $T = T'$ then we learn that shear coordinates require the surface $S$ to be orientable (unlike the other coordinate systems) and do not allow peripheral curves (that is, components of $\alpha$ homotopic into a neighbourhood of a vertex).

• To go from shear to normal coordinates - this is done vertex by vertex. That is, for a vertex $v$, let $\{T_i\}$ be the triangles in cyclic order about $v$. Note that a triangle may appear as many as three times in this list, in different rotations. Let $P = P(v)$ be the union of these triangles; this is the polygon about $v$. From the sequence of signed shears about $v$ we build a weighted train track, with stops, in $P$. This gives the normal coordinates for the $T_i$ at the corner $v$. Doing this discovers the lack of spiralling in shear coordinates as well as a rule of signs - about every vertex (with some shearing) both signs appear.