Which pictures do children draw with straight lines?

Edray Goins gave a great colloquium today about his work on dessins d’enfants.  And in this talk there was a picture that surprised me.  It was one of the ones on the lower right of this poster.  Here, I’ll put in a screen shot:

Let me tell you what you’re looking at.  You are looking for elliptic curves E admitting a Belyi map f: E -> P^1, which is to say a map ramified only over 0,1, and infinity.  For each such map, the blue graph is f^{-1}([0,1]), the preimage of the line segment joining o and 1 in P^1(R).

In four of these cases, the graph is piecewise linear!  I didn’t know there were examples like this.  Don’t know if this is easy, but:  for which Belyi maps (of any genus, not just genus 1) is f^{-1}([0,1]) a union of geodesics?

Tagged ,

4 thoughts on “Which pictures do children draw with straight lines?

  1. dalitt says:

    It seems to me that this happens if an invariant differential on the elliptic curve descends to a real meromorphic differential on P^1\{0,1,infinity}, regular along (0,1). The point is that you want a tangent vector to your path, evaluated at the invariant differential, to have locally constant argument; this is evidently true if your differential is the pullback of a real differential.

  2. Yes, I think you can *make* a singular flat metric which is geodesic by thinking of each triangle as Euclidean- if you make them regular triangles, then you get a sextic differential (since angles are multiples of pi/3 = 2pi/6). You can also ask what kind of abelian differentials you get from this by lifting to the natural degree 6 cover. I think this is a great question!

  3. Jordan: if you draw the dessin in the correct geometry (sphere, Euclidean plane, or hyperbolic plane), a dessin is always geodesic. The whole point is that it is obtained by a (geodesic) triangulation. So I would guess that the two non-geodesic ones are with triangle groups with negative curvature.

  4. Like Jayadev Athreya said, a dessins d’enfant is fundamentally taking a number of equilateral triangles (literally, like Polydrons), assembling together according to the gluing pattern, and then uniformizing the result. This is a lot clearer to see in your pictures if you draw the inverse image of the whole real axis, not just the segment [0,1].

    To get these extended pictures (combinatorially), take the bipartite graph that is drawn above, add a dot of a third color (purple) in the center of each face, and connect it to all the red and blue dots around the face.

    On three out of the four straight-line diagrams above, you get a tiling of a certain torus by equilateral triangles this way. (The pictures are drawn with a shear to draw the torus inside of a square; you need to undo that for the triangles to appear equilateral.) In these cases uniformization does nothing: the assembly of equilateral triangles is already flat.

    The remaining straight-line picture, in the middle of the bottom row, becomes a tiling by 45-45-90 triangles. It isn’t as obvious that uniformization makes that flat, but the Schwarz reflection principle proves it. (For that matter, the Schwarz reflection principle is a way to see that the description by assembling equilateral triangles is correct: take the Riemann mapping from the upper half-plane to an equilateral triangle, and extend by reflection.)

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out /  Change )

Google+ photo

You are commenting using your Google+ account. Log Out /  Change )

Twitter picture

You are commenting using your Twitter account. Log Out /  Change )

Facebook photo

You are commenting using your Facebook account. Log Out /  Change )


Connecting to %s

%d bloggers like this: