3
$\begingroup$

On a introductory Topology course, I learned how to make some connected sums of spaces informally. For example, in the figure below, we have a draft for $\mathbb{R}P^2\ \# \ \mathbb{R}P^2$, where $\mathbb{R}P^2$ is the real projective plane.

First, we take two copies of $\mathbb{R}P^2$, as shown in $(i)$. Then, we ''glue'' them making ''holes'', one of which contains $A$. The sequences of arrows $p$-$q$ and $r$-$s$ can be simplified, and the holes create a new arrow $c$, given us $(ii)$. I regarded the holes in the copies as pieces of the squares containing the lower left vertices. The border of these pieces are the arrows $c$. Then, after some transformations displayed in $(iii)$, we get $K$, the Klein bottle.

enter image description here

Then, I tried the same method to prove that $\mathbb{R}P^2\ \# \ K$ and $\mathbb{R}P^2\ \# \ \mathbb{T}^2$ are homeomorphic, but that becomes more complex since the possibilities of cuts and transformations increase. I'd like to know if someone could give me a suggestion. I could find the homology groups of theses spaces, but I think it doesn't suffice. Thank you!

$\endgroup$
1
  • 1
    $\begingroup$ Well, I found out that the page math.stackexchange.com/questions/358724/… has the same question. But I'm not suppose to know Euler characteristic on my course and I couldn't understand the hint of the picture presented there. $\endgroup$
    – rgm
    Feb 2, 2017 at 12:06

1 Answer 1

2
$\begingroup$

In these notes: https://people.maths.ox.ac.uk/hitchin/hitchinnotes/Geometry_of_surfaces/Chapter_1_Topology.pdf, Nigel Hitchin produces the following solution:

enter image description here

where the first figure represents $\mathbb RP^2\ \# \ \mathbb T^2$. Pasting the last two quadrilaterals by the sides with the right triangles, we obtain $\mathbb RP^2\ \# \ \mathbb K$.

$\endgroup$
1
  • $\begingroup$ Could you explain what happens from the bottom left figure to the upper right one? Which parts are reattached together? $\endgroup$
    – user770533
    Dec 1, 2021 at 12:21

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .