# Classifying Surfaces (CliffsNotes Version)

/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:

## Step 1: Find the Euler Characteristic

*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:

## 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**

*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

## 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: $\mathbb{R}P^2 \;\#\; \mathbb{R}P^2\; \#\; \cdots\; \#\; \mathbb{R}P^2$

Voila!

*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:

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.