If we take the definition of a real projective space $\mathbb{R}\mathrm{P}^n$ as the space $S^n$ modulo the antipodal map ($x\sim -x$), it is possible to see that $\mathbb{R}\mathrm{P}^1$ is topologically equivalent to the circle. It is equivalent to the upper half of the circle where the two end points are glued together – i.e. another circle.
Is there an intuitive or visual way of understanding the real projective plane, $\mathbb{R}\mathrm{P}^2$? By the same intuitive reasoning, it seems that $\mathbb{R}\mathrm{P}^2$ should be topologically equivalent to the upper half of $S^2$ with the antipodal equivalence on the rim. This is the definition of $\mathbb{R}\mathrm{P}^1$, where the antipodal map doesn't change the topology, so it seems that $\mathbb{R}\mathrm{P}^2$ should simply be the upper half of $S^2$ as well.
This is clearly wrong, however, since $\mathbb{R}\mathrm{P}^2$ cannot be imbedded in $\mathbb{R}^3$. What is the fault in this reasoning, and is there an intuitive or visual way to imagine the real projective plane?
Best Answer
There are some interesting constructive ways to visualize $\mathbb{R}\mathrm{P}^2$. I think the simplest way is as follow:
First, we prepare a Whitney umbrella, which is homeomorphic to the upper half of $S^2$ :
Let's check the points to quotient (be glued).
After gluing, the red and blue paths should match their start points and the end points respectively so that their antipodal points are matched.
In the picture of half $S^2$ their start points are far from each others' start points now, while Whitney Umbrella has the two start points meet together.
Now we can just bend the umbrella and glue the edge:
And get a cross-cap, which is homeomorphic to $\mathbb{R}\mathrm{P}^2$.
I cannot color it with two color since it is not "two sided" now.
Many books suggest Möbius strip for intuition. It is a little bit more difficult but more interesting to visualize that $\pi_1(\mathbb{R}\mathrm{P}^2)$ is isomorphic to $\mathbb{Z}_2$.