Genus of an Orientable Surface Must Be an Integer

Question

My question comes from Livingston's "Knot Theory." It states:

Prove that the genus of an orientable surface is an integer. (Apply induction on the number of bands, and check the effect of adding an oriented band on the number of boundary components.)

Here is my attempt at the solution:

We consider the base case of the surface $S_0$ produced by attaching $0$ bands to $D$ disks with $C$ boundary components. By a theorem from the textbook, we have that the genus is $$g(S_0)=\frac{1}{2}(2-D+0-C)=1-\frac{D+C}{2}$$ We know that if there are $0$ bands attached, then there is only $1$ disk and thus $1$ boundary component, since otherwise we would have 2 surfaces (we only consider connected surfaces). Then, $$g(S_0)=1-\frac{1+1}{2}=1-1=0$$ For the inductive step, we assume that the result holds for a surface formed by attaching $n$ bands to a collection of disks. We then want to show that the result holds for a surface formed by attaching $n+1$ bands to a collection of disks. Any help on how to do this would be greatly appreciated.

Answer 1

There is a problem with your induction in that it does not handle a space made with more than one disk. One fix is to just use the genus formula anyway, and change the induction to consider all oriented surfaces with finitely many disks and bands, connected or not. $D=C$ so you get an integer in the base case.

There are two options for an additional band.

1. The band is between two distinct boundary components. The additional band joins them into a single boundary component. Adding a band and losing a boundary component adds 2 inside the parentheses for your formula.

2. The band is along the same boundary component. Since the surface is orientable, the band must split the boundary component into two. Adding a band and gaining a boundary component cancels out in your accounting.