Classifying Surfaces (CliffsNotes Version)
The Basic Idea
My goal for today is to provide a step-by-step guideline for classifying closed surfaces. (By 'closed,' I mean a surface that is compact and has no boundary.) The information below may come in handy for any topology student who needs to know just the basics (for an exam, say, or even for other less practical (but still mathematically elegant) endeavors) so there won't be any proofs today. Given a polygon with certain edges identified, we can determine the surface that it represents in just three easy steps:
- Step 1: Find the Euler characteristic
- Step 2: Determine if it's orientable or non-orientable
- Step 3: Calculate its genus
In the next section, I'll fill in the details. We'll use the following surface as our working example:
From English to Math
Step 1: Find the Euler Characteristic
Let's call our surface $M_g$, where $g$ is its genus (which we have yet to determine). You recall that the Euler characteristic $\chi(M_g)$ is given by $$\chi(M_g) = \#\text{vertices} \;-\; \#\text{edges} \;+\;\#\text{faces} .$$ Determining the number of edges is easy - you simply count the number of unique edges in the polygon. In our case, the edges are labeled $a,b,c$ and $d$. Four letters implies four edges. Also, there is exactly one face (the octagon's blue shaded region). Hence $$e=4 \qquad \text{and} \qquad f=1.$$ Counting vertices is the only tricky part, and it boils down to "chasing arrows" according to how the edges are glued together. The following video clip shows how this is done with our blue octagon:
Since our surface has 2 vertices, we conclude that the Euler characteristic is $$\chi(M_g)=v-e+f\;=\;2-4+1=-1.$$
Step 2: Determine Orientability
Next, we need to determine if our surface is orientable or non-orientable. In sum,
- any surface that does NOT contain a Möbius band is orientable.
- any surface that DOES contain a Möbius band is non-orientable
(This really follows from the fact that a projective plane is the same as a Möbius band and disk glued together along their boundaries. Remember our discussion on the projective plane? And speaking of Möbius bands, have you ever seen Vihart's tale of Wind and Mr. Ug?) So how do we determine if our surface contains a Möbius band? It's actually quite easy! You just sort of "eyeball it":
Since we found a Möbius band (Yes, yes, we actually found two, but that's okay. It's enough to find at least one.) our surface is non-orientable. As a side note, notice that in our polygon's surface symbol, $abca^{-1}dc^{-1}db$, the inverse of neither $b$ nor $d$ is present. You'll also notice that two the Möbius bands we found above are associated with edges $b$ and $d$. This is not a coincidence.
Step 3: Calculate the genus
Finally, it remains to determine our surface's genus g. This step is straightforward. We simply use the fact that
$$\chi(M_g)=\begin{cases} 2-2g, &\text{if $M_g$ is orientable};\\ 2-g, &\text{if $M_g$ is non-orientable.} \end{cases}$$ In our example, $M_g$ is non-orientable and $\chi(M_g)=-1$, hence $$g=2-\chi(M_g)=2-(-1)=3.$$
The Classification Theorem
We now have all the information needed to determine our surface! We simply apply the Classification Theorem:
The Classification Theorem: Any closed surface is homeomorphic to one of the following:
- a sphere
- a connected sum of tori
- a connected sum of projective planes
We can summarize this along with our observations about orientability as follows:
- An orientable surface of genus g is a connected sum of g tori* : $T \;\# \;T \;\# \;\cdots \;\# \;T$
- A non-orientable surface of genus g is a connected sum of $g$ projective planes: $\mathbb{R}P^2 \;\#\; \mathbb{R}P^2\; \#\; \cdots\; \#\; \mathbb{R}P^2$
In our example, we found that $M_g$ is a non-orientable surface of genus 3. Hence it is a connected sum of three projective planes (a.k.a. cross caps), $\mathbb{R}P^2\;\#\;\mathbb{R}P^2\;\#\;\mathbb{R}P^2$.
Voila!
As an aside, it's helpful to note that $$\boxed{\mathbb{R}P^2\;\#\;\mathbb{R}P^2\;\#\;\mathbb{R}P^2 \cong T\;\#\; \mathbb{R}P^2.}$$ This tells us that any surface that is a connected sum of both tori and projective planes can be written as a connected sum of projective planes only. To see why the above holds, we have the following two claims:
Claim 1: $\mathbb{R}P^2\;\#\;\mathbb{R}P^2\cong K$
Proof: Recall that "$\mathbb{R}P^2\;\#\;\mathbb{R}P^2$" means: "remove a disk from both copies of $\mathbb{R}P^2$ and glue the remaining spaces together." But from previous work we know that $\mathbb{R}P^2-$disc $=$ Möbius band! In other words, $\mathbb{R}P^2\;\#\;\mathbb{R}P^2$ is precisely the space consisting of two Möbius bands glued together along their boundary. But this is precisely a Klein bottle!
Claim 2: $T\;\#\;\mathbb{R}P^2\cong K\;\#\;\mathbb{R}P^2$.
Proof: This is a direct consequence of the following observation: $$T\;\#\;M\cong K\;\#\;M$$ where $M$ is a Möbius band. The drawing below, borrowed from The Shape of Space by Jeffrey Weeks, illustrates this wonderfully:
It follows that $T\;\#\;\mathbb{R}P^2\cong K\;\#\;\mathbb{R}P^2$ since we can glue a disc to the left and right hand sides of $T\;\#\;M\cong K\;\#\;M$ (and, as we mentioned in the proof of Claim 1, gluing a disc to a Möbius band gives us a projective plane.)
Combining Claims 1 and 2 we see that $ T\;\#\; \mathbb{R}P^2 \cong K \;\#\; \mathbb{R}P^2 \cong \mathbb{R}P^2\;\#\;\mathbb{R}P^2\;\#\;\mathbb{R}P^2$ as stated above.
* Note, a sphere has genus 0.