20

Imagine bending a square sheet of paper and gluing two opposite edges together. The most natural way to do this results in a cylinder with open ends. Call this object $C_0$. If one twists the sheet one half turn ($180^{\circ}$) and glues the edges, one obtains a Mobius band. Call that object $C_1$. It's easy to imagine object $C_n$ with $n$ twists.

Notice that $C_n$ is homeomorphic to $C_m$ whenever $m-n$ is even, since the gluing identifications are the same. In particular, $C_0$ and $C_2$ are homeomorphic. Thus, topologically, $C_0$ and $C_2$ are equivalent. But intuitively, they strike me as different: one cannot continuously deform $C_0$ into $C_2$ without ripping the paper or some such thing.

My (soft) question: what's the best way to understand this difference between $C_0$ and $C_2$?

I'm clearly thinking of $C_0$ and $C_2$ as embeddings of $S^1\times[0,1]$ into $\mathbb{R}^3$. I think one can describe the difference between $C_0$ and $C_2$ as a difference between their embedding maps -- they're not homotopic in some way. (I'm not sure exactly how to formulate the non-equivalence of the embeddings.) But does one even need to refer to embeddings to capture the difference between $C_0$ and $C_2$? Intuitively, it seems to me the difference is largely, if not completely, independent of the space in which they're embedded. I'm inspired by the beautiful way manifold theory is built up without ever needing to describe the manifold as embedded in a higher dimensional space. Can one distinguish between $C_0$ and $C_2$ without embedding them, or at least showing that the distinction is independent of the space in which they're embedded?

Will Nelson
  • 5,161

2 Answers2

18

Perhaps the equivalence relation you want is "isotopic", which briefly means "homotopic through homeomorphisms". More precisely, you want to know whether there is an isotopy from $C_0$ to $C_2$, which by definition is a continuous function $H : C_0 \times [0,1] \to \mathbb{R}^3$ such that $H(x,0)=x$, $H | C_0 \times \{t\}$ is a homeomorphism onto its image for each $t \in [0,1]$, and $H(C_0)=C_2$.

The boundary of $C_m$ when $m$ is even is a 2-component link which I'll denote $L_m$. As it turns out, given unequal even integers $m,n$ the links $L_m,L_n$ are not isotopic, and therefore the twisted cylinders $C_m,C_n$ are not isotopic. The reason is that for any 2-component link there is an isotopy invariant called the linking number. One can compute for each even number $m$ that the linking number of $L_m$ equals $m/2$ (see the picture on that web page which shows $C_8$).

Lee Mosher
  • 120,280
  • Very interesting, thanks! Just a note: The Wikipedia definition of linking numbers seems to require immersion in $\mathbb{R}^3$. I still wonder if one can dispense with the that space and just say something like, "$S^1\times[0,1]$ has an "immersion group" of $\mathbb{Z}$, where the space in which it's embedded is left unspecified. Different members of this hypothetical "immersion group" correspond to distinct classes of immersions. Intuitively, for example, I think the "immersion group" for $S^2$ would be trivial. But perhaps one cannot make sense of this "immersion group" concept at all... – Will Nelson Aug 06 '17 at 02:05
  • Regarding the definition of linking number of two closed curves, what is required for the definition to make sense is that the images of the closed closed curves are disjoint from each other, and what is required for invariance of linking number is that they stay disjoint under homotopy. The individual components of the link need not be immersed. But if you are studying embeddings of $S^1 \times [0,1]$ and the restricted embedding of $S^1 \times {0,1}$, that all seems irrelevant. – Lee Mosher Aug 06 '17 at 13:07
  • 1
    Regarding the "immersion group", this idea was very thoroughly studied by Morris Hirsch in his 1958 dissertation on immersions of manifolds. Sadly for your idea of the immersions of $S^1$, in general the collection Imm(M, K) of regular homotopy classes of immersions of $M$ into $K$ depends on both $M$ and $K$. For instance, if $M = S^1$, the group $Imm(S^1, \Bbb R^2)$ is just $\Bbb Z$, measured by turning number (cf. the Whitney-Graustein theorem), but $Imm(S^1, \Bbb R^3) = \Bbb Z/2\Bbb Z$, while $Imm(S^1, T^2)$ is considerably more complicated. – John Hughes Aug 06 '17 at 17:15
7

To add a little to what @LeeMosher has said: you're thinking of $C_0$ and $C_2$ as the images of two distinct embeddings. But the domains of these embeddings are, of course, homeomorphic.

I just wanted to mention that in addition to the idea of isotopy that Lee presents (which seems like it might be what you're after), there's a slightly weaker idea that works in the category of smooth manifolds and that's regular homotopy of immersions, in which you are allowed to move around the image in any smooth enough way (you'll have to look at the definition for details), but this "moving around" is allowed to include stuff passing through other stuff. In the plane, for instance, you could have two circles next to each other, the left one being smaller, OR you could have two concentric circles. Both of these situations represent embeddings (or immersions) of a pair of circles in the plane. They're not isotopic, but they are regularly homotopic: you just slide the left (smaller) circle to the right until it sits inside the larger one; at each stage, you have an immersion of a "pair of circles" manifold.

The actual definition contains some interesting subtleties, worth looking at carefully, and the great introductory paper on the subject is this:

Whitney, Hassler (1937). "On regular closed curves in the plane". Compositio Mathematica. 4: 276–284.

In the case of your two embeddings of the surface, they're not only homotopic, as @AndrewD.Hwang notes in a comment, they're regularly homotopic. But as Lee notes, they're not isotopic, because of linking numbers, etc. [See below for slight correction to this statement.]

Post-comment addition: I was mistaken in my claim above, having misread the definition of $C_2$ --- I thought it was two full twists. Calling that item $C_4$, what's true is that $C_4$ and $C_0$ are regularly homotopic, but $C_2$ is not regularly homotopic to either of them.

This is tied to the fact that $\pi_1(SO(3)) = \Bbb Z / 2 \Bbb Z$: if you imagine $C_0$ as a wide, but not tall, cylinder, then you can draw a centerline $\gamma$ on it, going around the circle. The tangent vector $T = \gamma'(t)$ at each point of this centerline can be length-adjusted to be a unit vector, and an outward normal $N$ at each point gives a second, perpendicular, unit vector. A third vector, $S = N \times T$, at each point gives you a triple $(T, N, S)$, which you can think of as a $3 \times 3$ matrix with perpendicular unit-length columns, i.e, an element of $SO(3)$. Since you get one of these triples at each point $\gamma(t)$, you end up with a map $t \mapsto M(t) \in SO(3)$, with $M(0) = M(1)$, i.e. an element of $\pi_1(SO(3))$. For the standard cylinder $C_0$, the resulting path is the generator of $\pi_1(SO(3))$.

For $C_1$, you don't actually get a loop in $\pi_1(SO(3))$, because the starting and ending directions of $N$ are opposite.

For $C_2$, you DO get a loop in $\pi_1(SO(3))$, but it turns out to be the non-generator in $\pi_1(SO(3))$.

Under regular homotopy, this element of $\pi_1(SO(3))$ is invariant, and thus $C_0$ and $C_2$ cannot be regularly homotopic maps. But $C_0$ and $C_4$ can be, and indeed are, and thisis the basis for some rather clever equipment used to centrifuge blood during platelet pheresis, which (when I used to donate platelets) was done by extracting blood from one vein, running it through a long plastic tube with a collecting bag hanging off it, and then back into another location on the vein. The whole tube-and-bag assembly was fresh for each patient, but in the middle of the thing, they needed to spin the blood to separate the platelets into the collecting bag. This was done with a very clever assembly that works because of the $\pi_1(SO(3))$ trick.

John Hughes
  • 93,729