Mulase and Penkava studied Ribbon Graphs and came up with a constructive proof of the correspondence between two kinds of moduli: a Riemann surface moduli space and a Ribbon Graph moduli space. The original theorem is due to Penner, Thurston and others, and relies on the study of Grothendieck’s Children’s Drawings.

One works with a category of Ribbon Graphs. An object is a collection of vertices and edges and an incidence map i. Arrows are pairs of arrows that form a commuting square with the two incidence maps. Vertices are always at least trivalent, but we then add bivalent vertices at the centre of each edge to create half edges. A cyclic ordering on half-edge vertices gives an orientation to the ribbon edges. By definition, a boundary of a graph is a sequence of directed edges which cycles back on itself. Then Euler’s relation holds,

v – e + b = 2 – 2g

where g is the genus of the surface represented by the ribbon graph.

### Like this:

Like Loading...

*Related*

## CarlB said,

November 19, 2006 @ 10:21 pm

Another lesson, this one a little easier to understand.

I’ve made two important discoveries. First, the people who write the subtitles for Venezuelan television do not have as good a vocabulary for Spanish swear words as I learned growing up in New Mexico. Because of this, I have doubts about my ability to learn the Spanish language this way.

Secondly, I’ve discovered that I can cancel the twin obstructions of television and the internet. One puts on a movie, and then uses the internet to find the plot synopsis on Wikipedia. One can then spend ones time on physics until the start of the next movie.

## Kea said,

November 19, 2006 @ 11:51 pm

Hi Carl

Brilliant! I usually don’t turn on the TV (when I’m near one, which is rare, but usually in a hotel of some sort) because it costs money. I saw a nice Mongolian movie the other day, but I’m not getting out much. Can’t wait for the Maths workshop next week.

## Kea said,

November 20, 2006 @ 4:46 am

Cool stuff about Fool’s Gold, Carl. You know much more than me about symmetry. Don’t tell Baez too much. Of course I agree that geometry is the true path. I wouldn’t be talking to you otherwise, would I?