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Why does the result of cutting a Möbius strip down the middle lengthwise have two full twists in it? I can account for one full twist--the identification of the top left corner with the bottom right is a half twist; similarly, the top right corner and bottom left identification contributes another half twist. But where does the second full twist come from?

Explanations with examples or analogies drawn from real life much appreciated.

edit: I'm pasting J.M.'s Mathematica code here (see his answer), modified for version 5.2.

twist[{f_, g_}, a_, b_, u_] := {Cos[u] (a + f Cos[b u] - g Sin[b u]), 
      Sin[u] (a + f Cos[b u] - g Sin[b u]), g Cos[b u] + f Sin[b u]};

With[{a = 3, b = 1/2, f = 1/2},
Block[{$DisplayFunction = Identity}, 
g1 = ParametricPlot3D[Evaluate[Append[twist[{f - v, 0}, a, b, u],
  {EdgeForm[], FaceForm[SurfaceColor[Red], SurfaceColor[Blue]]}]],
  {u, 0, 2 Pi}, {v, 0, 2 f}, Axes -> None, Boxed -> False];
g2 = ParametricPlot3D[Evaluate[Append[twist[{f - v, 0}, a, b, u],
  EdgeForm[]]], {u, 0, 4 Pi}, {v, 0, 2 f/3},
  Axes -> None, Boxed -> False];
g3 = ParametricPlot3D[Evaluate[Append[twist[{f - v, 0}, a, b, u],
  {EdgeForm[], FaceForm[SurfaceColor[Red], SurfaceColor[Blue]]}]],
  {u, 0, 2 Pi}, {v, 2 f/3, 4 f/3}, Axes -> None, Boxed -> False,
 PlotPoints -> 105]];
GraphicsArray[{{g1, Show[g2, g3]}}]];
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I tweaked your code a bit; hopefully it generates something that looks more like the output from version 8. –  J. M. Sep 28 '11 at 16:35
    
@J.M. I just saw this now. Thanks. (Is there an SE setting that I can turn on, that'll notify me when someone comments? Had I not come back to this post I wouldn't have known that you commented.) –  sasha Oct 25 '11 at 7:01
    
It supposedly should have shown up in your inbox. There's a red indicator in the upper left of the page that will say how many unread comments have you gotten. –  J. M. Oct 25 '11 at 8:22
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3 Answers

up vote 10 down vote accepted

One twist comes from the two half-twists of the Möbius strip. Another comes from the fact that just after you've made the cut, the resulting half-width strip goes two times around the cut, so it will turn an extra time when you unfold it to a large circle.

Try making an ordinary strip that goes two times around a cylinder and then meets itself, without a Möbius twist. If you remove the cylinder and try to unfold your strip to a circle, it will have one full twist. This twist arises from the fact that the strip's centerline must wind around itself when it goes around the cylinder twice. (In the cut-Möbius case, the direction of this winding depends on the direction the original Möbius strip was twisted, which means that the single twist from the unfolding adds to the two half-twists rather than cancel them out).

Another everyday effect that shows this (in reverse) is to try to wrap a rubber band (an ordinary cylindrical-section rubber band with a flat cross section) twice round a package. It will need to twist in order to do this, even if it can lie flat wrapped once around the package.

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In knot theory they talk about what is going on here as interchanging twist and writhe. A quick search didn't bring up any good references. They were all about DNA. –  Ross Millikan Sep 25 '11 at 22:49
    
Thanks! So it appears that my problem was not with the Möbius strip per se, but with its inherent "looping" which I failed to see. I investigated in this manner: I held out a strip of paper at arm's length (without twists) and brought my right hand in, closer to me, and then out again, as if tracing a circle in a horizontal plane clockwise. I had to shift my fingers around while I did this, but the result was equivalent to a strip of paper with a full-twist. –  sasha Sep 26 '11 at 17:43
1  
BTW, I don't think your answer accounts for possible cancellation of twists--though I think it's clear that the loops in the Möbius strip go in the same "direction" as its twists. –  sasha Sep 26 '11 at 17:55
    
I wondered about that, too, and did figure out the same solution as you note -- but somehow I forgot to state it explicitly in the answer. I have edited to add it now. –  Henning Makholm Sep 26 '11 at 18:00
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Observe that the boundary of a Möbius strip is a circle. When you cut, you create more boundary; this is in fact a second circle.

During this process, the Möbius strip loses its non-orientability. Make two Möbius strips with paper and some tape. Cut one and leave the other uncut. Now take each and draw a line down the middle. The line will come back and meet itself on the Möbius strip; on the cut Möbius strip, it won't.

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7  
This seems to explain why the cut Möbius strip is a single strip, but not why it is twisted twice. –  Henning Makholm Sep 25 '11 at 22:33
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I seem to be terribly late (I'm quite slow as a typist), but hopefully this answer is complementary rather than redundant.

cut Möbius strip

That the Möbius strip has a single "twist" is apparent in the color clash that appears when one attempts to color one "face" red and the opposite face blue. This color clash does not occur in the cut strip, as one requires two twists to finish a circuit through the surface.

Mathematically, consider the following parametrization of the Möbius strip:

$$\begin{align*}x&=\left(r+(2-v)\cos\frac{u}{2}\right)\cos\,u\\y&=\left(r+(2-v)\cos\frac{u}{2}\right)\sin\,u\\z&=(2-v)\sin\frac{u}{2}\end{align*}$$

where $0 \leq v \leq 2$ and $0 \leq u \leq 2\pi$. The normal vector to this surface is

$$\begin{pmatrix} \frac12\left(r\sin\frac{u}{2}-r\sin\frac{3u}{2}-2(v-2)\sin\,u\sin^2\frac{u}{2}\right) \\ \frac12\left((v-2)(\sin^2 u+\cos\,u)-2r\sin\frac{u}{2} \sin\,u\right) \\ \left(r-(v-2)\cos\frac{u}{2}\right)\cos\frac{u}{2} \end{pmatrix}$$

One can verify that the normal vector expressions at $u=0$ and $u=2\pi$ are not equal, but the normal vector expressions at $u=0$ and $u=4\pi$ do agree. This means that if you trace a pencil through the surface of the usual Möbius strip (that is, fix $v$ and vary $u$), after exactly one turn through the surface, your pencil should end up at the spot exactly under your original starting point. For the "cut" Möbius strip (same parametric equation, but $0 \leq v \leq \frac23$ and $0 \leq u \leq 4\pi$), two turns ($2\times2\pi$) will return your pencil to its starting point.


Here is sundry Mathematica code:

twist[{f_, g_}, a_, b_, u_] := {Cos[u] (a + f Cos[b u] - g Sin[b u]), 
  Sin[u] (a + f Cos[b u] - g Sin[b u]), g Cos[b u] + f Sin[b u]}

With[{a = 3, b = 1/2, f = 1/2},
 g1 = ParametricPlot3D[
   twist[{f - v, 0}, a, b, u], {u, 0, 2 Pi}, {v, 0, 2 f}, 
   Axes -> None, Boxed -> False, Mesh -> None, 
   PerformanceGoal -> "Quality", PlotStyle -> FaceForm[Red, Blue]];
 g2 = ParametricPlot3D[
   twist[{f - v, 0}, a, b, u], {u, 0, 4 Pi}, {v, 0, 2 f/3}, 
   Axes -> None, Boxed -> False, Mesh -> None, 
   PerformanceGoal -> "Quality", PlotPoints -> 85, 
   PlotStyle -> FaceForm[Red, Blue]];
 g3 = ParametricPlot3D[
   twist[{f - v, 0}, a, b, u], {u, 0, 2 Pi}, {v, 2 f/3, 4 f/3}, 
   Axes -> None, Boxed -> False, Mesh -> None, 
   PerformanceGoal -> "Quality", PlotPoints -> 85, 
   PlotStyle -> Opacity[1/10, Blue]];
 GraphicsGrid[{{g1, Show[g2, g3]}}]]
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1  
+1 Your visualization helped me understand what was being asked, as you can see from the (now deleted) answer of mine, I did not quite get it the first time. –  Sasha Sep 26 '11 at 5:24
1  
Thanks. :) I must credit Andres Caicedo though; his answer there has stuck firmly to my memory... –  J. M. Sep 26 '11 at 5:42
    
The coloring argument was helpful, thank you. I tried your code in Mathematica 5.2 (it's all I have at the moment) without luck. I see a few unfamiliar options there, so likely compatability issues... –  sasha Sep 26 '11 at 18:03
    
@J.M. Thank you for providing the link, that was also helpful besides being a great read. –  sasha Sep 26 '11 at 18:04
    
@sasha: 5.2, huh? I did that bit in 8, so yes, there are some incompatibilities. Sorry about that. You'll have to replace GraphicsGrid[] with GraphicsArray[], remove the Mesh, PerformanceGoal, and PlotStyle options, and replace the first argument twist[{f - v, 0}, a, b, u] with Evaluate[Append[twist[{f - v, 0}, a, b, u], {EdgeForm[], FaceForm[Red, Blue]}]]. I don't have the computer with 5.2, so I can't test to see if there are other stuff I forgot. –  J. M. Sep 26 '11 at 18:11
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