algebraic-topologygeneral-topology
I would like to understand how to prove that the connected sum $\mathbb{R}P^2 \# T^2$ of the projective plane with a torus is homeomoprhic to $\mathbb{R}P^2 \# \mathbb{R}P^2 \# \mathbb{R}P^2$.
I got as far as showing that it must be equivalent to a connected sum of projective planes, how can I argue though that I need precisely three projective planes ?
Thanks for your help!
(P.S. not a homework exercise, this is for me to understand the classification of surfaces).
Here's an answer more in the spirit of the question. All figures should be read from upper left to upper right to lower left to lower right.
Fig 1: A Klein bottle...gets a yellow circle drawn on it; this splits it into two regions, which we reassemble, at which point they're obviously both Mobius bands:
To see that $P^2$ minus a disk is a Mobius band, look at the following. In the upper left is $P^2$, drawn as a fundamental polygon with sides identified. In the upper right, I've removed a disk. The boundary of the now-missing disk is drawn at the lower left as a dashed line, and the two remaining parts of the edge of the fundamental polygon are color-coded to make the matching easier to see. In the lower right, I've morphed things a bit, and if you consider the green-followed-by-red as a single edge, you can see that when you glue the left side to the right, you get a M-band.
You can take as a reference these notes (check corollary 5.3.5.1). The corollary states the following:
If $M$ is a differentiable surface of $\mathbb{R}^3$ connected and compact with Gauss curvature $K \ge 0$ and not identically zero then $M$ is homeomorphic to a sphere.
My understanding is that one requires differentiability to be able to use Gauss curvature.
Now, the ingredients to arrive to this corollary are the following:
a classification of topological surfaces which you cite and which is stated at theorem 5.3.5.1 of the document.
the Gauss-Bonnet theorem which has been mentionned in the comments and appears in the document at theorem 5.3.4.1.
Realize that homeomorphic implies homotopy equivalent.
In fact, the document cites a stronger result by Hadamard in theorem 5.3.6.2:
If $M$ is an ovalid then the Gauss map $\stackrel{\to}{N}: M \to \mathbb{S}^2$ associated with any unitary normal $N$ is a dipheomorphism. In particular, $M$ is dipheomorphic to a sphere.
So as you can see you need quite a bit of machinery to prove the result. However you get to a nice result for ovaloids.
Best Answer
The following picture will help you to understand this problem intuitively. (If you can understand why $\mathbb{R}\mathrm{P}^2 \# \mathbb{R}\mathrm{P}^2 = K$, where $K$ stands for the Klein bottle.)
The figure in the upper right corner is $K \setminus \text{disk} $. It may take some time to think why it looks like this.
Source of the picture: link