# Using fundamental polygons to prove $PR^2\# T^2 = PR^2\# PR^2 \# PR^2$.

## Background

Let $$P^2$$ denote the real projective plane, and $$T^2$$ the torus. These generate a monoid where the operation is the connected sum. This monoid is abelian, so for notational brevity, we write $$nX+mY$$ to denote $$X\# Y \# Y \# \cdots$$ (where $$X$$ appears $$n$$ times and so on).

The following two closely related results are well known:

Every orientable closed surface is of the form $$nT^2$$ for some $$n \in \{0,1,2,\ldots\}$$. (The empty sum is the sphere.) Every non-orientable closed surface is of the form $$mP^2$$ for some $$m \in \{1,2,3,\ldots\}$$.

Every closed surface is of the form $$nT^2+mP^2$$, for some $$n \in \{0,1,2,\ldots\}$$, and $$m \in \{0,1,2\}$$.

These two facts can be related through the following result:

$$T^2 + P^2 = 3P^2$$.

There are many visual proofs of this fact in which $$T^2 + M^2$$ is shown to be $$K^2 + M^2$$ (where $$K^2$$ is the Klein bottle and $$M^2$$ is the Möbius strip.) These visualisations (e.g. see here) often follow this process:

1. Cut out a disk from a torus. Observe that this hole can be "stretched" so that the remaining surface looks like a cylinder with a small region of the ends identified.

2. The connected sum of the torus and Möbius strip is then visually shown to be the "same" as cutting out two disks from the mobius strip and gluing the ends of a cylinder to these two holes.

3. One end of the cylinder can now be "twisted" by dragging it along the Möbius strip until it is again close to the other end of the cylinder. This is in fact the connected sum of a Klein bottle with a Möbius strip.

If one first establishes that $$2P^2 = K^2$$, this is almost believable - although it is necessary to be careful since $$M^2$$ is not $$P^2$$.

## My question

I decided to try and "prove" these facts by using fundamental polygons. That is, I wanted to draw a sequence of pictures that can be directly formalised by defining the spaces as quotients of the subsets of $$\mathbb R^2$$ represented by my images. I was able to achieve this for $$2P^2 = K^2$$ as shown below. However, despite hours of drawing I just couldn't figure out how to show that $$T^2 + P^2 = K^2 + P^2$$ in a sufficiently convincing way. I attempted to use the intuition from the visualisation shown above. How would one "draw a proof" that $$T^2 + P^2 = K^2 + P^2$$ where every step is clear?

## Example for $$2P^2 = K^2$$:

• See pages 19 to 21 of courses.maths.ox.ac.uk/node/view_material/4169 – Gerry Myerson Oct 30 '18 at 8:58
• Have you had a look at that link, Harambe? – Gerry Myerson Nov 1 '18 at 10:51
• Ah yes I had a look just then, it solves the problem. Thank you! – Harambe Nov 3 '18 at 19:43
• You might consider writing it up to post as an answer here, since that link may go away some time, and it would be nice to have the details here. – Gerry Myerson Nov 4 '18 at 2:51
• You could do that today. – Gerry Myerson Nov 5 '18 at 8:48