0
$\begingroup$

This is a very basic question:

I don't see the following:

If I cut a surface with boundary along non-contracible cycles into components with genus zero, how can those components have an unbounded number of boundary cycles?

Thank you

  • 0
    Yes, sorry I was just about to write it. It is in the second paragraph of the Introduction. The authors write there: "Cutting a surface along non-contractible cycles decomposes the surface into components with genus zero, but those components may have an unbounded number of boundary cycles."2011-11-24

1 Answers 1

1

Okay, I see what you're referring to now. The authors are addressing the issue that if you only cut along non-contractible curves, you could be cutting off annuli. This happens if your curve is "parallel" to a boundary curve. So if you repeatedly cut along curves that are parrallel to the boundary, you'll create possibly an endless list of annuli without ever simplifying the original surface.

If $S_{g,b}$ is a connected surface of genus $g$ with $b$ boundary components, and you cut it along a curve $C$, there are two possibilities:

  • $S_{g,b}$ is cut into a connected surface of the form $S_{g-1,b+2}$
  • $S_{g,b}$ is cut into two connected surfaces of the form $S_{g_1,b_1}$ and $S_{g_2,b_2}$ where $g_1+g_2=g$, and $b_1+b_2 = b+2$.

In particular, if you cut along a boundary-parallel curve you cut $S_{g,b}$ into an $S_{0,2}$ and an $S_{g,b}$.

The first case is when the curve $C$ is non-separating, the 2nd case is when $C$ is separating.

  • 0
    Ok thanks. The first part of your answer i understand, you refer to the 2-cell embedded graph encode with as a rotation system, right? Do you know why it is important that this graph is weighted? But what about the second part. How do you store the actual surface in a data structure. And I think later in the paper this is not used anymore.2011-11-24