Talk:Möbius strip/Archive 1

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Axeeeeeel. Nice work. One day I'll put some pictures with strip and some outstanding Maple code or something like that.FireJamXRasta 3 Wednesday [2002.02.27]

Wouldn't it be cool to have a small java applet of the moebius strip in 3D?

Yes, (and Maple or Mathematica code too)

I took out the R from the parametrization for three reasons:

It was not explained.
It looked as if it was a parameter, but it was in fact a fixed number representing the radius of the Möbius band.
Not all values of R work (you can get self intersections if R is too small.

I explained the parametrization a bit better. AxelBoldt

Nice Axel. Yes R seems to be a constant and not a parameter. We should investigate for which R Moebius strip is really defined. I would like to say something more: I didn't mean that those presumptions about connection Universe<->Moebius strip come from SF - they come from science world (physics, cosmology) I guess. We should correct that fact somehow. Uh, Axel I don't want to be your student, ha, ha. I still owe to this page another picture of a strip... XJam [2002.03.25]] 1 Monday (0)

I haven't seen any serious cosmology suggesting a Moebius strip universe, but if you find anything, make sure to put it in the article. AxelBoldt

I assume by "generalized Moebius strip" something like a finite volume universe with orientation reversing paths is meant. My understanding of cosmology (very weak admittedly) is that such universes are possible. But for some reason most physicists suppose the universe to be orientable.--C S 10:09, Sep 7, 2004 (UTC)

I have checked 'very briefly' for R. As it seems strip degenerate near 0 and probably R must be positive or non-positive real number. Very intersting - how small should R be to get self intersections. We can also split R to R₁ and R₂. Does any self intersection appear if R₁ = - R₂. (I guess not - but how can you be shure?) Another output picture is coming...
XJam [2002.03.26]] 2 Tuesday (0)

The Klein bottle isn't a 3D analogue, since it's also a surface -- it's more an extension. -- Tarquin

The Klein bottle is actually two Moebius strips glued together along their edges. --C S 10:09, Sep 7, 2004 (UTC)

I have a question regarding

Another equation for a Möbius strip is log(r)*sin(θ/2)=z*cos(θ/2).

I assume this is in cylindrical coordinates (r,θz)? This equation describes an unbounded figure though (you can enlarge r and z beyond all bounds), so I don't see how it can describe a Moebius strip. AxelBoldt, Sunday, June 2, 2002

It is in cylindrical coordinates. It describes an unbounded Moebius strip. If you want a bounded strip, you can take the part inside a torus, or restrict r and z. --phma

moebius in fiction: there's also A Subway Named Moebius, AJ Deutsch.

Wouldn't it be better to have "Boston subway authority" link to MBTA? Counterfit

Am I the only person bothered by the phrase "Mobius strip is a topological object with only one surface"??? What does this mean??? I know what it is meant to mean, (i.e. that it's not orientable), and that this is the way to put it in as comfortable a language as possible, but the way it's phrased this way it seems too inaccurate (or meaningless) to be worth the benefit. Revolver 10 Nov 2003

I wouldn't say that's meaningless. It's implicit in that statement that the Moebius band under consideration is in 3-dimensional Euclidean space. There the band is one-sided. So I interpret "only one surface" to mean "one-sided". --C S 10:09, Sep 7, 2004 (UTC)

Harmless in the intro, I'd say. Anyone reading on is given a clearer idea. Generally speaking the first para of an article has some license to use looser language, and not to define everything with exactitude. There again, the surface link is probably unhelpful there.

Charles Matthews 18:00, 10 Nov 2003 (UTC)

I removed this:

A family of 3D solids that closely relate to Möbius strips are the Sphericons. They are like a Möbius band, but without the hole in the middle. If you make a Möbius band out of a n-sided polygon sectioned strip, rotate it k amount and count the number of sides and edges created, more parallels can be found with the Sphericon.

I'm convinced at this point that the sphericon is that closely related; certainly not topologically.

Charles Matthews 13:48, 7 Apr 2004 (UTC)

I apologise, I think I was wrong. I was regarding a mobius band as not just a 2D rectangle cuved round into a circle and twisted, but also as 3D prisms curved round and twisted. These result in shapes with one side and one edge just as a mobius band does, and I believed they all fell under the same name. At simplest you can have a triangular prism, but a prism with many sides approaches a torus. Similarily, the simplest sphericon is based on a sixty degree apexed cone split and twisted, and a more complex sphericon approaches a sphere. What page would you reference it to?

You could add it to List of polygons, polyhedra and polytopes, perhaps as a 'see also'; also to the list of geometry topics under the 3D shapes.

Charles Matthews 20:01, 9 Apr 2004 (UTC)

Thanks, it's done.

Proberts2003 20:50, 9 Apr 2004 (UTC)

1 Möbius strip with a circular boundary
- 1.1 A nice description
2 Handedness of Möbius strip
3 Pronunciation
4 Should we replce Möbius with Mobius?
5 Humanist symbol
6 Good Article, but...
7 who can help me to add

Möbius strip with a circular boundary

Don't know how to edit this -- the second to last paragraph

A cross-cap is a two-dimensional surface that is topologically equivalent to a Möbius strip. The term 'cross-cap', however, often implies that the surface has been deformed so that its boundary is an ordinary circle. This cannot be done in three dimensions without the surface intersecting itself.

is wrong. This error is repeated in the cross-cap entry as well (word for word!). It is possible to embed a Mobius Band in $\mathbf{R}^3$ with boundary a perfect circle. Here is the idea: let C be the unit circle in the xy plane in $\mathbf{R}^3$ . Now connect opposite points on C, i.e., points at angles $θ$ and $θ + π$ , by an interval which is an arc of a circle. For $θ$ between $0$ and $π / 2$ the arc lies above the xy plane, and for other $θ$ the arc lies below (there are two places where the arc lies in the xy plane).

I don't know what this embedding of the Mobius strip is called. sam Mon Aug 16 17:49:53 CDT 2004

Thanks for pointing this out. I have adapted what you wrote and added it to the page. Charles Matthews 10:20, 7 Sep 2004 (UTC)

Unfortunately, I can't make sense of the given description. I think an improved version is needed. The way I personally visualized this is hard to verbalize, but I found some nice descriptions in George Francis' A Topological Picturebook. They rely extensively on some key pictures however. Because of copyright I can't just scan them in, but maybe I could draw my own versions. At least, for now, we can give Francis' book as a reference (for a doubting Thomas).--C S 11:45, Sep 7, 2004 (UTC)

I'm having some problems with this description as well. It is a little ambigouous. A picture would be worth a thousand words here. If someone can give me a more concrete description, I'd like to plot this in Mathematica. -- Fropuff 17:10, 2004 Oct 20 (UTC)

Perhaps these photos help. I made an actual physical model, with a wire coat-hanger and a tee-shirt. In the first picture, quadrants 1, 3, and 4 are just a single layer. Quadrant 2 inside the circle has three layers; the middle layer connects to the boundary circle, while the top and bottom layers make the corner-shaped pouch in the upper left. The rope demonstrates that there's really a hole. Travelling from the end that goes off the edge (towards the yellow taped end), the rope passes up through the circle twice: it wraps around the circle inside the pouch.

The second picture is just the first one flipped over.

Fropuff, does that help at all? I've been fortunate enough not to have to learn Mathematica, so I can't make a plot of it. dbenbenn | talk 11:49, 29 Jan 2005 (UTC)

Thanks dbenbenn, I appreciate the effort. Though I must admit that I've stared at this picture for quite awhile and I'm afraid I still can't picture it. How can the rope pass twice through the circle? Where exactly is the circle in this picture? -- Fropuff 07:16, 2005 Feb 1 (UTC)

Tonight, I'll cut a "window" in the model and upload a new photograph. Perhaps that'll help. I'll also try highlighting the boundary with the Gimp. dbenbenn | talk 23:43, 3 Feb 2005 (UTC)

Okay, I procrastinated ridiculously, but it's here now. Does that help? dbenbenn | talk 07:24, 19 Feb 2005 (UTC)

— — —

Uh, no. I just read this for the first time, and I'm also baffled by this proposed method of embedding a Mobius Band in $\mathbf{R}^3$ with a perfect circle boundary, without self-intersections. Alas the "shirt & coat hanger model" photos aren't helping me, but perhaps adding some more detail in the paragraph?

Let C be the unit circle in the xy plane in $\mathbf{R}^3$ .
Now connect opposite points on C, i.e., points at angles $θ$ and $θ + π$ ....

Okay, so far so good. I agree if we can connect these opposing points, we'll get an embedding of a Mobius Band. But I don't believe this surface won't intersect itself.

.. by an interval which is an arc of a circle.

Is the author suggesting connecting the opposing points using arcs which extend into the z direction, so the middle of each arc will lie on the z axis?

For $θ$ between $0$ and $π / 2$ the arc lies above the xy plane, and for other $θ$ the arc lies below (there are two places where the arc lies in the xy plane).

If I understand this right, the author is suggesting that:

For $θ$ equals 0, the arc lies in the xy plane
Then as $θ$ increases away from 0, the arc starts going higher and higher above the xy plane
Until some critical $θ$ is reached (maybe $π / 4$ ), when the arc goes highest above the xy plane.
Then as $θ$ continues increasing, approaching $π / 2$ , the arc starts going lower and lower, but is still above the xy plane.
When $θ$ reaches $π / 2$ , the arc lies in the xy plane again.
Then as $θ$ increases away from $π / 2$ , the arc starts going lower and lower below the xy plane.
Until some critical $θ$ is reached (maybe $3π / 4$ ), when the arc goes lowest below the xy plane.
Then as $θ$ continues increasing, approaching $π$ , the arc starts going higher and higher, but is still below the xy plane.
And when $θ$ reaches $π$ , the arc joins with the arc created when $θ$ equalled 0, completing the surface.

If this is what is being suggested, and the middle of each arc lies on the z-axis,

I never said that the center of the arc lies on the z-axis. In fact the arc "topples" over. At angle 0 the arc in in the xy plane and is inside the circle C. At angle 90 degrees the arc is in the xy plane and is outside the circle C. I hope that helps! Sam nead 20:07, 7 September 2005 (UTC)

then it looks like there will have to be self-intersection on the z-axis. During "step 2" in my list above, each arc uses a higher point on the z-axis until an arc with maximum height is reached, so when the arcs try to descend in "step 4", they must use those same z-axis points coming down. Each z-axis point that is used once going up must be used again coming down. That sounds like self-intersection to me.

In fact, even if one doesn't try to guess what is meant to be happening off the xy plane, the description itself says: "there are two places where the arc lies in the xy plane". As I read it, these two places are ( $θ = 0$ ), when we're trying to connect the east and west points of the circle, and ( $θ = π / 2$ ), when we're trying to connect the north and south points of the circle -- yet both arcs must lie in the xy plane. Well, I can't draw a curve connecting the east and west points of a circle, and another curve connecting the north and south points, without those two curves intersecting.

I would love to see better what the author did with his shirt; otherwise, I'm proposing deleting this fragment from the Mobius band page. I think it's fairly well established that embedding a Mobius band in $\mathbf{R}^3$ such that its boundary is a circle must have self-intersection. Anonymous Reader -- UTC 12:24 Sep 6, 2005

— — —

Hmmm.. Thinking about it some more, I think I see how it might be done, if you don't mind that the resulting surface is infinitely large.

In my list of "steps", above, the arc is rising higher and higher in "step 2". But maybe it doesn't reach any maximum, so there is no "step 3" and no subsequent descent in "step 4"; rather, the arc just keeps getting higher and higher, and wider and wider, so that when $θ$ reaches $π / 2$ , the highest point of the arc reaches infinity.

Technically at this point, the arc is in the xy plane again, because it's so huge that the only thing you can see at any scale, is a line going north from the north-point of the circle, and another line going south from the south point of the circle, and we're supposed to believe that these two lines meet at the point at infinity. So that's how the north and south points of the circle can "connect" in the xy plane, without intersecting the perfectly normal horizontal-line-segment that connects the east and west points.

Then as $θ$ increases past $π / 2$ , the extreme point of the arc jumps to negative infinity and continues rising. So instead of "steps 6,7,8", the extreme point of the arc just rises smoothly from negative infinity back to zero, as $θ$ increases from $π / 2$ to $π$ .

In this version (as above) the arc at 90 degrees is in the xy plane, is outside of C, but now it is a pair of rays. This is totally fine -- in fact, your version gives a nice Mobius strip in the three-sphere, the one-point compactification of R^3. To turn your version into mine, just apply a Mobius transformation (like a linear map, but more general) which fixes the circle C. Sam nead 20:07, 7 September 2005 (UTC)

Okay. If this infinite-sized-curiousity is what the author intended, I agree that an embedding with no self-intersection is technically possible, I no longer think it should be deleted. Though I still don't understand is how this shirt-photo was supposed to be a "physical model" of a surface with infinite extent! :)

The paragraph in the article doesn't really explain the resulting surface at all -- a plot would be very helpful. Anonymous Reader -- UTC 13:37 Sep 6, 2005

I did the best I could! Sam nead 20:07, 7 September 2005 (UTC)

— — —

Oh wow, you're totally right, Sam Nead! I can see it now, it's finite, smooth, and .. really pretty. Now that I understand what the thing is supposed to look like, I made up a parameterization and created the following plot of the object: Anonymous Reader -- 25 September, 2005

What beautiful pictures!! I have added them to the article.Sam nead 00:58, 28 September 2005 (UTC)

Nice pictures, thanks. I would be very curious to see the parametrization you used to make these. -- Fropuff 02:20, 28 September 2005 (UTC)

Well, I just basically plodded through the verbal description above, making up functions as needed to make the thing look nice.

Let the first parameter be $θ$ running from -π/2 to π/2.

For any given $θ$ we want to connect antipodal points (cos $θ$ ,sin $θ$ ,0) and (-cos $θ$ ,-sin $θ$ ,0) with the arc of a circle that lies in a plane which "tips over" as $θ$ increases away from zero. So I decided that the arc would lie in the (u,v) plane, with $ρ$ indicating the amount of tipping.

$\vec{\mathbf{u}} = \left \{ \cos \theta, \sin \theta, 0 \right \}$

$\vec{\mathbf{v}} = \left \{ \sin \theta \sin \rho, - \cos \theta \sin \rho, \cos \rho \cdot \operatorname{sign} \theta \right \}$

And of course $ρ$ had to be 0 when $θ$ was 0, and π/2 when $θ$ was ±π/2, so it looked best if I just set:

$\rho = \theta \,$

Now within the (u,v) plane, we want a circular arc to connect (-1,0) with (1,0), with the shape of the arc changing as a function of $θ$ . The shape of the arc can be expressed by $τ$ , defined as half the angle subtended by the arc. So if $τ$ is π/2, the arc is a semicircle, connecting (-1,0) with (1,0) through (0,1). For lower $τ$ the arc is shallower and in the limit as $τ$ approaches 0 the arc becomes a straight line through the origin. For higher $τ$ the arc is more than half a circle, and u gets to take on values outside the {-1...1} range.

Let the second parameter be t running from -1 to +1 indicating our position along a given arc.

Then the arc in the (u,v) plane is given by:

$u(t) = \frac{\sin ( \tau t ) }{\sin \tau} \qquad v(t) = \frac{\cos ( \tau t ) - cos \tau }{\sin \tau}$

so the only thing remaining is to select how $τ$ should vary as a function of $θ$ . When $θ$ is 0, we select $τ = 0$ to get a horizontal line connecting the east-west points of the circle. When $θ$ is ±π/2, (and the (u,v) plane is tipped 90 degrees, so (u,v) = (±y,x)), we need $τ$ to exceed π/2 so that the arc will bulge outward and connect the north-south points of the circle without ever going inside the circle. I arbitrarily selected $τ = 3π / 4$ when $θ$ was ±π/2, but a linear relationship between $τ$ and $θ$ didn't look so good. So fiddled with it and ultimately selected:

$\tau = \frac{3\pi}{8} \left ( 1 - \cos \left ( \sqrt{2\pi|\theta|} \right ) \right ) \,$

this being an arbitrary function that starts $τ$ off at 0 for $θ = 0$ and gives it a sort of smooth rapid rise up to its maximum of 3π/4. So I just told my plotting program to compute a grid of values for $θ$ and t, found the (u,v) basis vectors and the u(t),v(t) values to map each point into (x,y,z) coordinates, and plotted the resulting surface.

Alas the resulting surface looked squashed, so I scaled it by 150% along the z axis. :)

--Anonymous Reader 8:35, 29 September 2005 (UTC)

A nice description

After sifting through the above discussion and a lot of tinkering around, I have stumbled upon a very nice desription of the Möbius strip with a circular boundary.

The first step is to embed the Möbius strip in the 3-sphere (thought of as a subset of C²) via the maps

$z_1 = \sin\eta\,e^{i\phi}$

$z_2 = \cos\eta\,e^{i\phi/2}$

Here η runs from 0 to π and φ runs from 0 to 2π. The boundary of the strip is given by $| z 2 | = 1$ , which is clearly a circle on the 3-sphere.

The second step is to map S³ to R³ via a stereographic projection. Stereographic projections map circles to circles and will preserve the circular boundary of the strip. In order to get a finite surface one must project from a point not on embedded Möbius strip in S³. After some trial and error I found that projecting from the point (½,−½,−½,−½) looks quite nice. The resulting surface looks similiar to the one shown above. (I can upload Mathematica graphics if anyone is interested).

Absolutely, I'd love to see it! Your description is a lot cleaner. (I guess it has fewer knobs for sculpting the surface, but that's a good thing. More natural, less hacky. I can't visualize 4D projection in my head (yet!), but the Möbius-Strip-In-4D with which you're starting has an awesome symmetry to it.) -- Anonymous Reader. 8:35, 29 September 2005 (UTC)

Should we replace the description in the article with this one? I think this description is a little cleaner and less ambiguous.

-- Fropuff 05:57, 29 September 2005 (UTC)

Okay, here is a plot of the surface I described above. It was just done with Mathematica so it doesn't look quite as nice as the other one.

-- Fropuff 14:59, 29 September 2005 (UTC)

I implemented your 4D-stereographic-projection idea so I could look at it myself, and find a view of it that looked as good as my first hacky one. Using your equations for the strip in C², I set the projection-point to be that point on the 3-sphere closest to { Re(z₁), Im(z₁), Re(z₂), Im(z₂) } == { X, Y, 1, 0 }, and proceeded to modify (X,Y) to see how this changed the surface.

With (X,Y) == (0, .33), the circle forms the opening of a deep well, that twists like a snail-shell into a slender tube. Striking, but asymmetric and kind of lumpy..

With (X,Y) == (-.33, 0), the film is flat between the circle, and nearly a disc, but with the top and bottom "sides" connected by a tube that is centered and much neater!

I'm really impressed at the variety of different-looking Mobius-Strips-With-Circular-Boundaries that your technique makes possible! Great job! In both cases the "slender tube" bloats up as the magnitudes of X or Y increase, until the "tube" is fatter than the disc. The symmetric one, bloated up at (X,Y) == (-1.5, 0) and with the colors reversed, looks like the closest match to my first manually-made version from above.

I agree that this stereographic-projection-of-a-4D-mobius-strip explanation is a much more precise and powerful description than the arc-that-flops-over paragraph, and it should replace the paragraph in the article. -- Anonymous Reader -- 18 October 2005 (UTC)

The 4D Möbius strip on which this is based is generally thought to have been first discovered by Blaine Lawson in his Stanford Ph.D. thesis (1960's). It has not only tremendous symmetry (I think its symmetry group is O(2)), but it also happens to be *minimal* in S^3. With the help of Doug Lerner, I made an animation of this strip -- stereographically projected into R^3 in the most symmetrical way so as to create a non-compact surface in R^3 (that's technically the Möbius strip minus one point) in 1983; it appears in Siggraph Video Review #17.Daqu 12:50, 17 January 2006 (UTC)

P.S. For historical reasons we called the 4D strip the "Sudanese Möbius band". It can be geometrically described as follows: There is a way to think of the 3-sphere S^3 via an "open book decomposition". That is, choose some great circle, calling it C. There is a circle's worth of great hemispheres all having C as their common boundary, and whose union is all of S^3. Denote these hemispheres by H_t for 0 <= t < 2pi according to the angle H_t makes with H_0. Each hemisphere H_t has a "pole" -- the unique point farthest from its boundary -- which we call P_t. We *also* parametrize C itself by c(t), 0 <= t <= 2pi, where c(t) traverses C at unit speed.

Finally, on each H_t we draw the unique great semicircle S_t passing through both P_t and c(2t). The union of all the S_t's forms the Sudanese Möbius band. (Thus S_(t+pi) = S_t for 0 <= t < pi.)

Note: Sam nead's description sounds like exactly what the Sudanese Möbius band looks like in my Siggraph film of the same name . . . after it is stereographically projected S^3 - {pt} --> R^3 (with projection point being one of the poles P_t above).Daqu 13:27, 17 January 2006 (UTC)

Handedness of Möbius strip

Is the photo at the top of Möbius strip a right or left handed strip? I can't tell from reading the definition in the article. Perhaps someone who knows the convention could improve the description. dbenbenn | talk 23:42, 14 Mar 2005 (UTC)

In topology, a Möbius strip (like any surface or topological space) is defined independent of its surroundings. It's thought of as a thing-in-itself. More accurately, it's thought of as not only a rectangle-with-one-pair-of-opposite-sides-identified-by a-flip, but actually any topological space that's topologically equivalent (aka homeomorphic) to that surface.

Also, a Möbius strip *might* be identified with a subset of another topological space, and of course it's usually depicted in 3-space. There's no standard terminology for the "left-" and "right-" handed versions, but in fact there are many more than just these two! To start with, any odd number of twists, clockwise or counterclockwise, in a strip -- before glueing its ends -- creates a topological Möbius strip. (The logo of the wool industry, and the logo for recycling, are each 3-twist Möbius strips). And to make matters worse, its core circle could also be knotted!Daqu 20:09, 5 May 2006 (UTC)

Pronunciation

The IPA for the article says (pronounced /ˈmøbiʊs/), but isn't the ø a long vowel? 惑乱分からん 18:34, 27 February 2006 (UTC)

I'm not well-versed in the IPA, but the German pronunciation of Möbius (deemed correct by international mathematicians for pronouncing "Möbius strip") has the first vowel pronounced as a hybrid of the (English) u in "but" (ŭ) and the oo in "look" (denoted here by ʊ). More accurately, it's a diphthong composed of the ŭ followed by the ʊ. That is, just as the long i (ī) is the diphthong ah-ee when spoken quickly, the ö is at least very close to ŭ-ʊ (spoken very quickly, with the ŭ extremely short).

The common pronunciation in the U.S. by the *non*-cognoscenti is with long o (ō), but I feel this error should not be encouraged.Daqu 20:42, 5 May 2006 (UTC)

Should we replce Möbius with Mobius?

It would certainly make for easier editing. Klosterdev 20:29, 6 March 2006 (UTC)

If you can't type an "ö" easily, you could always just write "Mobius" and let someone with an "ö" key fix it. The regexp replace user script makes fixing such things particularly easy. —Ilmari Karonen (talk) 21:07, 6 March 2006 (UTC)

There is an entirely standard way to represent a German o-umlaut when you don't have an umlaut symbol: you just put an e after the o. So among mathematicians, at least, an entirely standard alternative to "Möbius" is "Moebius". (But definitely *not* "Mobius".)Daqu 19:27, 7 August 2006 (UTC)

Humanist symbol

The strip is the symbol of the Humanist International and also of many humanist parties (ex: Portuguese Humanist Party [1] ). I think that should be refered in the article. Mário 00:05, 23 March 2006 (UTC)

Good Article, but...

Hi to the contributors - great work. I'd love to nominate this as a Good Article but I fear it would be shot down by reviewers for a lack of references. Anyone think they could add a few? Inline citations would be great too - no harm in asking eh? Cheers SeanMack 16:37, 11 April 2006 (UTC)

Searching the net I came across chemistry papers talking about aromatic compounds with Möbius strip shapes, for example here,

http://www.rsc.org/ej/CC/2000/b002462g.PDF, any chemists know enough about this to do it justice? SeanMack 18:10, 30 May 2006 (UTC)

who can help me to add

Image:MobiusAnime.gif

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Talk:Möbius strip/Archive 1

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Möbius strip with a circular boundary

A nice description

Handedness of Möbius strip

Pronunciation

Should we replce Möbius with Mobius?

Humanist symbol

Good Article, but...

who can help me to add

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