The other day I was explaining orientability to someone and we were walking through some of the statements about orientability on the Wikipedia page on the topic. While I was able to satisfy his curiosity, one statement on that page (which I did not even attempt to delve into with him) has been nagging me since then:

"For example, a torus in $K^2\times S^1$ can be one-sided and a Klein bottle in the same space can be two-sided."

Because this statement bothered me (since it runs counter to normal intuition about orientable surfaces in Euclidean spaces), I have been thinking about it more over the last few days. I have been able to determine which copies of these submanifolds should have the stated properties and convince myself how the non-orientability of the ambient space $K^2\times S^1$ allows for the submanifolds in question to twist back on themselves in unusual ways, but nevertheless I still cannot form a decent picture of what this really means.

The real issue with my understanding what is going on with these submanifolds seems to be that although these phenomenon occur in a non-orientable space, this space can itself be embedded in an orientable space and so it seems that these odd tori and Klein bottles should therefore embed in an orientable space as well and so I should have some chance of visualizing these phenomena when I project down to $\mathbb{R}^2$ or $\mathbb{R}^3$

Question: Does anyone have a good picture or other approach to help visualize what a one-sided torus or two-sided Klein bottle looks like?

So while it may be too much to hope for a projection that accurately reflects the sidedness of these creatures, I am hoping someone may have a decent projection of either of these creatures to the plane or 3-space that shows some manifestations of their odd behaviour in their ambient space. Or, barring an actual picture, perhaps someone who has thought about this more has some other way of thinking about them which at least gives a better intuitive sense of how to look at them in their ambient space and 'see' (whatever that may mean when you think about them) these counterintuitive features.

  • $\begingroup$ "runs counter to normal intuition about orientable surfaces" - he he. Puns. $\endgroup$ – Reinstate Monica Sep 21 '17 at 2:21

Jeff Weeks' "Shape of Space" book has several good examples of two sided non-orientable surfaces embedded in non-orientable manifolds. That book is written in a manner that is easy to approachable for an outsider, so it might provide a natural next direction for a conversation that started from discussing the wikipedia article. In particular, there is a excellent explanation of the difference between non-orientablity vs. two-sidedness (with pictures) in that book.


2-sided Klein bottle: $KB\times \{0\}$ inside $KB\times [-1,1]$.

1-sided circle: The central core of a Möbius band.

1-sided torus: Previous example cross another circle.


Recall the following well-known picture of $K^2$ as a square with one pair of edges identified and the other pair identified with a twist.

klein bottle picture from wikipedia

Similarly, you can draw a picture of $K^2\times S^1$ as a cube with opposite faces identified, except that one pair of faces is identified with a twist as well (here the twist means that the faces are identified after a reflection through a line).

By thinking about various surfaces in this model, you can find the creatures you are searching for. The two-sided Klein bottle is just an obvious copy of the above picture of the Klein bottle - it's any slice of the cube such that the induced identifications of the edges from the face identifications looks like the picture above.

The one-sided torus is a different square slice of the cube, which we can identify from Kevin Walker's answer. The intersection of the one-sided torus with the picture of the two-sided Klein bottle above is the circle represented by a vertical line running from one blue edge to the other.


(I tried THREE times to post my answer as a comment. It's too hard. Let me do it here.)

@ARupiński, you must be fighting imaginary demons, because there is not much to visualize, this question is simple (not research :-). Let   $S$   be the complex unit circle. First consider the Klein bottle:   $K2\times\{1\}\subseteq K_2\times S$.   Just agree that the normal vector to   $(p\ 1)$   is   $(p\ i)$   for every $p\in K_2$. As you see (it's trivial) you got a smooth field of normal non-zero vectors for your entire Klein bottle   $K2\times\{1\}$.   That's one side of your Klein bottle. The other side is given by normal vectors   $(p\ −i)$.   Your Klein bottle copy is $2$−sided. (Sorry for the imperfect notation for the normal vectors).

Now the challenge is not in seeing things but in writing them down (I wish I could do it without too much pain). Let me start with a convenient (for this text) definition of the Klein bottle. Let   $\mathbb C$   be the complex plane. Let

$$M'\ \ :=\ \ \{z\in \mathbb C\ :\ \frac 12\le |z|\le 1\}$$

Identify   $z\ \ w\in M'\ \Leftrightarrow\ \ w=-z$   and   $|z|=|w|=\frac 12$. The result is a Mobius strip, call it   $M$.   That's one half of your Klein bottle. The boundary of   $M$   is the unit circle. Points obtained from the identification of complex numbers   $z$   of module   $\frac 12$   form the so called equator   $E$   of   $M$.   The whole Klein bottle   $K_2$   is the so-called double of   $M$. We only need to know at this point that in the topological space   $K_2$   equator   $E$   is contained in the topological interior of   $M$.

Consider the torus   $T := E\times S\subseteq K_2\times S$.   At each point   $([z]\ 1)\in E\times S\subseteq K_2\times S$,   where   $|z|= \frac 12$   (i.e. $[z]\in E$) consider two normal vectors:   $(z\ 0)$   and   $-(z\ 0)$ -- these two vectors are on the locally opposite sides of the torus in the   $K_2\times S$ &nbs[; space. Let's move one of the said vectors around, say vector   $(\frac 12\ 0)$   starting at   $([\frac 12]\ 0)\in E\times S$.   Moving in our vector field along the equator we will get back to the same point   $([\frac 12]\ 1)$   of the equator, but we will end up this time with the opposite vector, namely with   vector   $-(\frac 12\ 0)$. This shows that the locally different sides are the same globally.

REMARK   Sorry to do such a lousy job of explaining. If you feel that this mini-exposition is worthy of improving, and if you feel like improving it, then please just go ahead and edit this text.


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