Can you "drink" from the projective real plane? and from the Klein bottle?

Question

My geometry II lecturer in the session regarding quotient topology on his notes to the course has elaborated on the following "playful" example:

We have seen that $P^n(\mathbb{R})\simeq S^n/\pi_{\pm}$, the $n$-sphere with identification of antipodal points, this means in particular that for $n=2$ we obtain that the projective real plane is homeomorphic to the "usual" sphere in $\mathbb{R}^3$ where we can choose one representative for each equivalence class (consisting of two points, those diametrically opposite).
Say we choose all the points in the lower hemisphere $z\lt0$, and half of it's equatorial border, with one extreme inculded and one not.
Therefore, in this model, the projective real plane looks like a mug!
It's coffee break time, and we would like to use this mug to drink our coffee, this is anyway not possible for multiple reasons:

• This mug has no handle, this as we know, doesn't really prevent us from drinking from it.
• It also has no border! the points of the half equator are in fact identified with the other "missing" half equator.
• What if we tried to create a border by cutting a disk out of $P^2(\mathbb{R})$? What's left is a Möbius strip, which is clearly not fitting to drink coffee from!

Hence, no coffee break.

I found this example particularly interesting for it's bizzare nature, arguing how some mathematical properties prevent us from "using" an abstract object to perform a concrete action.

I thought it could be fun to make a similar arguement in regards to the Klein bottle, after all, the name suggests that we should be able to drink from it! But can we? Well, clearly not. I wanted to write down a short list similar to the preceding one regarding properties of the Klein bottle which makes it impossible to drink from, I came up with the following:

• Once again, it has no border. Since the Klein bottle can be represented as the unit square $I\times I/\sim$ with the identification of $(0,y)\sim(1,y)$ and $(x,0)\sim(1-x,1)$.
• If we try to cut it in half, we end up with two mobius strips, which doesn't help with our task.
• What if we considered an immersion of the Klein bottle in the Euclidean Space? for example the parametrization of the standard "bottle shape" $3$-dimensional immersion? Now we may actually be able to drink from it, although not very comfortably, but this feels like cheating, in the end, the bottle shouldn't actually have self intersections as it does in the $3$-dimensional visualization of it.

This is as far as I went, since I also tried to read some articles on the Klein bottle but I've felt like I lack the knowledge to investigate deeper into the topic.

Are the aspects I listed "informally" correct for what they argue about? Are there interesting points that I could add regarding the impossibility of drinking from a Klein bottle?

Answer 1

Both statements:

1. "we obtain that the projective real plane is homeomorphic to the "usual" sphere in ${\mathbb R}^3$, where we can choose one representative for each equivalence class (consisting of two points, those diametrically opposite)."

and

2. "If we wish to choose only one representative for every equivalence class, then we could say that a topological model for $P^2({\mathbb R})$ is the half hemisphere with half equatorial border, where we include one extreme and one not of such equatorial border."

are wrong. There is no subset of $S^2$ (or even of ${\mathbb R}^3$) which is homeomorphic to the projective plane. What your lecturer described are subsets $E\subset S^2$ such that the restriction (to $E$) of the quotient map $q: S^2\to P^2({\mathbb R})$ is a continuous bijection $$ q: E\to P^2({\mathbb R}). $$ But a continuous bijection, in general, need not be a homeomorphism. This is exactly what happens here. As a simpler example, consider $S^1$, obtained as the quotient of $[0,1]$ where we identify the end-points. The quotient map $[0,1]\to S^1$ restricts to a continuous bijection $[0,1)\to S^1$. But $[0,1)$ is not a topological model for $S^1$.

At the same time, many other statements made in your post such as "it has no border" (for both projective plane and Klein bottle) and "if you remove a disk from $P^2({\mathbb R})$, the result is [homeomorphic to] a Moebius band" are absolutely true.

On the other hand, one cannot assign a mathematical or physical meaning to "drinking coffee from the projective plane" (or from the Klein bottle). Contemplating such things might be fun, but is meaningless and does not get you any closer to understanding topology of these two surfaces.

Lastly, regarding the "standard" immersion (not embedding) of the Klein bottle in ${\mathbb R}^3$: The (slightly thickened) images of such immersions can be made (I own one): One can buy such glass models, for instance, on Amazon. One can attempt to poor a liquid into a model. The result is that the liquid gets "inside" the bottle and one can never fully wash and dry the bottle afterwards. I do not recommend doing this.