Apropos the discussion during Gopakumar's talk about special points on M(g,n) corresponding to vertices of the triangulation. I think that works as follows.
For each n-tuple of points on a fixed compact surface the collection of all quadratic differentials with double poles at these points with given local expansion is an affine space over the space of all quadratic differentials. The Strebel differential is a solution of some extremisation problem in this affine space.
Thus the zeroes of the Strebel differential cannot be chosen for a given Riemann surface. In general, there are 4g-4+2n zeroes which are simple. However, there will be special cases where the multiplicity of the zeroes are higher. All the way up to multiplicity 4g-4+2n. The special "points" are those where this multiplicity is attained.
These special points are (as far as I know) not special points on M(g,n) since the choice of the Strebel differential depends upon the choice of n positive real constants and so the multiplicity and number of zeroes could change if the constants are changed.
Consider the quadratic differential on P^1 given by (dz)2/(z2-z) and take a Belyi map X -> P^1. This will give a quadratic differential on X with double poles at all the points lying over infinity and zeroes at precisely the points lying over 0 and 1 --- except points where the ramification degree is two.
Now it is reasonable to ask whether:
- This is a Strebel differential.
- all the vertices of the Harer triangulation arise in this way.
Vertices of the Harer triangulation also correspond to special values for the "residues" of the quadratic differential at the poles.
I think (dz)2/z2 (instead of (dz)2/(z2-z)) is a more likely candidate to give a Strebel differential via a Belyi map.
Yes, but in that case the graph will have nothing to do with the "dessin" of the Belyi cover.
Here is how you define the "dessin" of a special Belyi cover. Recall that a special Belyi cover is one where the ramification order over 1 is exactly 2.
Let X -> P^1 be a special Belyi cover. Then let the vertices of a cell decomposition be given by the inverse images of zero and the open edges be given by the inverse images of (0,1] and the open cells be given by the inverse images of the complement of [0,1] in P^1.
This gives a cell decomposition of X (called the "dessin" of X) which is very similar to the one given by a Strebel differential whose poles are at the points lying over infinity and zeroes are at the points lying over 0. The points lying over 1 are the midpoints of the edges joining various vertices. (Note that given a "dessin" it is easy to construct a Belyi cover).
Of course, for this to work, as you point out, the points lying over 0 must be ramified. By composing with a suitable map (say z-> z^2), this can always be arranged.
However, there are still a lot of parameters in the choice of differential. The inverse image of a(dz)^2/z - b(dz)^2/(z-1) for all choices of a and b gives a quadratic differential which has double poles are points lying over infinity and zeroes at points lying over 0. Moreover, there are multiple zeroes for this quadratic differential so it does not lie on a vertex of the triangulation.