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I read in various places that up to isomorphism there are only two line bundles ( 1-d vector bundles) over a circle, the trivial one and the mobius strip. On the other hand, when I make a mobius strip from a strip of paper, I note that I can give the paper as many rotations as I want. No rotation of the paper and simply gluing the end pieces of the strip gives a cylinder without tops (which we can think as the trivial line bundle), one rotation gives the mobius strip. But more rotations of the paper are possible and they are not isomorphic to each other; you cannot make one change to the other without tearing the paper.

Now I understand that these are not line bundles as they are not extended to infinity and they are "bounded" together in a way that infinite fibers or vectors would not be.

I am just wondering: Is there a notion of "bundles" here over a circle which extends to more than the trivial and mobius strip? After all, no rotation and one rotation of my strip do match up with the two possible vector bundles (trivial and mobius). Can we define a notion of "bundle" for more rotations of the strip?

EDIT: The comments below basically mean that I have not understood the situation correctly. I will not delete the question since I think others might run into the same misunderstanding.

alireza
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    I believe that the surfaces you describe are homeomorphic to either the cylinder or the Möbius strip. – hjhjhj57 Jan 12 '15 at 03:30
  • @hjhjhj57 I do not think so, not in a way a cup and a torus are. Think about it, one cannot be deformed continuously to the other. – alireza Jan 12 '15 at 03:40
  • I already cut some paper :D and I'm playing with it. Have you ever seen the seatbelt trick used to explain spinors? We may be missing an extra dimension to see the homeomorphism. Let's see what the experts have to say about it. – hjhjhj57 Jan 12 '15 at 03:42
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    You're confusing homeomorphisms and isotopes. The $n$-rotated belts you're describing have two homeomorphism classes, depending on whether $n$ is even or odd; but you cannot transform one into another within $\Bbb R^3$. –  Jan 12 '15 at 03:44
  • No. But this I doubt is a question needing more dimensions to deform; Play and you might see what I mean. I could be wrong of course. – alireza Jan 12 '15 at 03:45
  • @Mike Miller any references? I just cannot see the homeomorphism. – alireza Jan 12 '15 at 03:48
  • @alireza Do you know how to define vector bundles in terms of cocycles? If so, try to realize your new vector bundles in that manner. Then show the vector bundle isomorphisms I mentioned above all in terms of cocycles. Isomorphism of vector bundles is much stronger than homeomorphism. (But given an isomorphism you can write down a homeomorphism.) I might write down something about why you can get non-isotopic embeddings of these homeomorphic vector bundles later. –  Jan 12 '15 at 03:50
  • @Mike Miller I think the difference is that vector bundles on a circle are not the twisted belts. They are infinitely extended and hence are not at some end connected to each other. I think that is the key difference. So while vector bundles are indeed isomorphic, my twisted belt cannot be untwisted to one twist or no twist with any sort of continuous transformations. – alireza Jan 12 '15 at 04:11
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    @alireza no, that's irrelevant. Not all homeomorphisms can be given by continuous deformation, which is what Mike's been talking about with isotopies. The easiest way to see that the two-twist bundle is trivial is to observe that it has a nonzero section. This works whether the bundle is open,or bounded like yours. – Kevin Carlson Jan 12 '15 at 04:25
  • @Kevin Carlson I see. So tell me how two twists can be transformed into none by a homeomorphism, is it the problem of "passing through itself" as hjhjhj57 was thinking? How does it look like. I want to visualize it. – alireza Jan 12 '15 at 04:33
  • It doesn't really look like anything; just introduce coordinates on the double twisted band. Say, draw a line a bit above the central circle, and observe it comes back to itself as does not happen with a single twist. This line is "1", i.e. it will be mapped to 1 in the trivial line bundle. And similarly you can find the image of every other level in the fibers of the trivial bundle. – Kevin Carlson Jan 12 '15 at 05:32
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    RE: your edit. You may not have understood the situation entirely correctly, but nonetheless this is a very good question. I'm glad you asked it! –  Jan 12 '15 at 18:00
  • Possible duplicate of Line bundles of the circle – Alex M. Oct 05 '18 at 17:38

2 Answers2

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I will try to sum up the existing comments and talk about the linking number.

Isomorphism vs. Isotopy

So basically there are two things interplaying here when you talk about line bundles over $S^1$: isomorphisms and isotopies. Let's look at the definitions:

  • Two line bundles $E, E' \xrightarrow{p, p'} S^1$ are said to be isomorphic if there exists a map $f : E \to E'$ that makes the triangle commute ($p' \circ f = p$), such that the induced map on fibers is linear (these two conditions make $f$ into a vector bundle morphism), and moreover $f$ has an inverse which is also a vector bundle morphism.

  • Let two line bundles $E, E' \xrightarrow{p, p'} S^1$ and embeddings $E, E', S^1 \subset \mathbb{R}^3$. Then an isotopy between $E$ and $E'$ is a map $H : [0,1] \times \mathbb{R}^3 \to \mathbb{R}^3$ such that:

    • $H(0, -)$ is the identity of $\mathbb{R}^3$;
    • every $H(t, -)$ is a homeomorphism $\mathbb{R}^3 \to \mathbb{R}^3$;
    • $H(t, -)$ is the identity when restricted on $S^1$;
    • $H(1, E) = E'$ and $p' \circ H(1, -) = p$.

It's easy to see that if $E$ and $E'$ are isotopic in $\mathbb{R}^3$, then they are isomorphic. The converse is not true, as you noted: as band twisted twice is isomorphic to the trivial line bundle, but it is not isotopic in $\mathbb{R}^3$ to the cylinder. The isotopy class depends on the specific embedding in $\mathbb{R}^3$, but the isomorphism class does not.

Classification

Isomorphism classes: Stiefel-Whitned

In fact there are only two isomorphism classes, the trivial one and the one corresponding to the Möbius strip; see this question. Basically (this is a different version of the reasoning in the answers I linked), isomorphism classes of line bundles on $X$ are classified by the first Stiefel-Whitney class(*) $w_1(X) \in H^1(X; \mathbb{Z}/2\mathbb{Z})$, and $H^1(S^1; \mathbb{Z}/2\mathbb{Z}) \cong \mathbb{Z}/2\mathbb{Z}$.

Isotopy classes: Linking number

Now it is actually possible to classify (smooth?) isotopy classes too.

Let $E \to S^1$ be a line bundle embedded in $\mathbb{R}^3$, and let's assume that it is smooth for simplicity (the continuous case is more complicated). Then you can construct two curves out of it:

  • The first one $\gamma_0$ corresponds to the zero section of the line bundle.

  • For the second one, take a tubular neighborhood of the zero section in $E$ and take $\gamma_1$ to be one of the boundary components.

Finally consider the linking number: $$\operatorname{lk}(\gamma_0, \gamma_1),$$ which is well defined up to sign. This number basically corresponds to your intuitive notion of "number of twists": it's zero for the cylinder, one for the Möbius strip, two if you add one more twist... And it classifies smooth isotopy classes of smooth line bundles embedded in $\mathbb{R}^3$.

(*) The reason for that is one of my favorite facts in mathematics (the complex case one is just as amazing): $$\operatorname{Vect}^\mathbb{R}_1(X) = [X, \mathrm{Gr}_1(\mathbb{R}^\infty) ] = [X, \mathbb{RP}^\infty] = [X, K(\mathbb{Z}/2\mathbb{Z}, 1)] = H^1(X; \mathbb{Z}/2\mathbb{Z}).$$

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Najib Idrissi
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  • Aren't you using a restricted meaning for the concept of "isomorphism"? The definitions I've seen give this to mean a bundle map $F : E \to E'$ over a map $f : M \to M$, meaning that $p' \circ F = f \circ p$. You, on the other hand, take $f$ to be the identity. Why? – Alex M. Oct 06 '18 at 15:48
  • @AlexM. It makes no difference. Suppose that you have an isomorphism with your definition: let $p : E \to B$, $p' : E' \to B'$ be two bundles, $(F,f) : (E,B) \to (E',B')$ and $(G,g) : (E',B') \to (E,B)$ two bundle maps (i.e. $p' \circ F = f \circ p$ and $p \circ G = g \circ p'$) which are inverse to each other, i.e. $g \circ f$, $G \circ F$, $f \circ g$, and $F \circ G$ are each equal to the relevant identity map. Then I have a bundle $f \circ p : E \to B'$ (because $f$ is a homeomorphism), and an isomorphism with my definition between $f \circ p$ and $p'$. – Najib Idrissi Oct 06 '18 at 15:51
  • The big picture question is whether you're interested in bundles in general, or bundles over a fixed base space. In my answer the base space is fixed. So I use the definition where bundle morphisms have to cover the identity. – Najib Idrissi Oct 06 '18 at 15:53
  • What I am trying to say is that $E, E' \to M$ may be isomorphic by a map $F$ covering some $f$, but may not be isomorphic by any map covering the identity. If you want to classify bundles, how do you count them - as being in different equivalence classes, or in the same one? The fact that there is an isomorphism between $E$ and $f^* E'$ that covers the identity does not change the fact that $E$ and $E'$ are not isomorphic in your sense. – Alex M. Oct 06 '18 at 16:06
  • @AlexM. I see. In this case there are only two homotopy classes of self-homeos $S^1 \to S^1$ so it's easy to check. But the usual definition of a bundle isomorphism is a morphism that covers the identity. Check your favorite reference book. – Najib Idrissi Oct 06 '18 at 16:18
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Not a PROOF but a non-sufficient simple algebraic reason for twice twisted is same as trivial for further readers: Define an equivalence relation on $\Bbb R^2$ by declaring that $(x', y')\sim(x,y)\iff (x',y')=(x+n,(-1)^ny)$ for some $n \in\Bbb Z$. This defines the Mobius bundle. To see that if we twist the band twice is same as trivial bundle (that is the identification $(x,y)\sim (x+m,y)$ for some $m$), just apply this identification once again, i.e.: (If I am not mistaken by notations) $$\text{twisted }\Bbb R^2\ni(a', b')\sim(a,b)\in\text{twisted }\Bbb R^2\iff (a',b')=(a+n,(-1)^nb)=(x+2n,(-1)^{2n}y)=(x+2n,y)\iff (a', b')\sim(x,y).$$ that says $(z,w)\sim (z',w')\iff (z',w')=(z+2n,w)$ that is the trivial bundle. (first factor is a circle and second is a line)

$\qquad\qquad\qquad\qquad\qquad$Source: WikipediaSource: Wikipedia

C.F.G
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