I think I know how the Segerman surface is produced. 1. Consider the horizontal plane, P, in R^3, passing through the origin. 2. Take the unit sphere centred at the origin, and let it intersect P to form a unit circle C. This will form the boundary of our surface. 3. For each angle theta in [0, pi), take the vertical plane Q_theta through the origin with an azimuth angle of theta. Within this plane Q_theta, we draw a (possibly degenerate) circular arc defined by the following properties: -- the centre of the arc lies on the vertical line through the origin; -- the endpoints of the arc are the two intersections of Q_theta with C; -- the angle of approach where the arc meets P is 2*theta; Two of these arcs are degenerate circles: one is the intersection of the x-axis with the disc bounded by C, and the other is the intersection of the y-axis with the *complement* of the disc bounded by C. The latter is the arc which passes through the point at infinity in the one-point compactification of R^3. 4. Take the union of all of these arcs. Now, I claim that this surface (considered as a subset of the one-point compactification of R^3) is homeomorphic to a Moebius strip. Specifically, if you take the usual construction of a Moebius strip from taking a long skinny rectangle and identifying the two short sides in opposite orientations, the short line segments parallel to those two short sides correspond exactly to the arcs in the above construction. The only annoyance is the point at infinity, but that's easily amendable: stereographically project from (R^3 union {infinity}) to S^3, then rotate the 3-sphere slighty so there's no longer a point of the surface on the north pole, and then stereographically project back to R^3. (This is topologically equivalent to Allan's suggestion of inserting Segerman's model into a hole in a sphere.) Best wishes, Adam P. Goucher
Sent: Friday, April 03, 2020 at 5:17 AM From: "Allan Wechsler" <acwacw@gmail.com> To: math-fun <math-fun@mailman.xmission.com> Subject: Re: [math-fun] Two topological visualization questions.
The outer "scalloped" edge is intended to suggest being extended to the point at infinity. (Notice how the scalloped edge is almost planar.) In principle you could deflate that embedding (at the cost of some pretty symmetry). Imagine the surface a sphere with a cloverleaf-shaped hole in it; fit Segerman's model into the hole. Mind, I am still not convinced that the resulting model is a Moebius strip, but I have to confess that it might be.
On Fri, Apr 3, 2020 at 12:04 AM Andy Latto <andy.latto@pobox.com> wrote:
On Thu, Apr 2, 2020 at 9:23 PM Michael Kleber <michael.kleber@gmail.com> wrote:
Here is Henry Segerman's 3d-printable model: https://www.shapeways.com/product/YRBSFUEHB/round-mobius-strip
And now that I look at it a bit more carefully, this isn't the object I'm looking for either! This object has a circular edge, but it also has a second edge, consisting of 4 circular arcs with right angles between them. Maybe this object is meant in some way to suggest the object I'm thinking of, but it isn't literally that object.
Andy
On Thu, Apr 2, 2020 at 8:59 PM Andy Latto <andy.latto@pobox.com> wrote:
I think the surface you describe is a punctured Klein bottle, which is different from a mobius strip, which is a punctured projective plane.
Andy
On Thu, Apr 2, 2020 at 7:38 PM Allan Wechsler <acwacw@gmail.com>
wrote:
I now have a "visualization" that I can understand, although, like
Andy, I
would like to have a 3D model to hold in my hand.
Imagine an ordinary circular disk in a horizontal plane. Punch two circular holes in the disk, and starting with the first hole, build a tube with circular cross section. The tube should run straight downward for a short distance, then bend to run horizontally until it is well clear of the underside of the original disk. Once clear, it can bend upward to cross the plane of the disk without interference from the disk itself, until it is comfortably above the plane of the disk. Now it can bend horizontal again and head for a point above the second hole, and on arrival above the second hole, bend downward to seal with it.
On Thu, Apr 2, 2020 at 6:54 PM Andy Latto <andy.latto@pobox.com> wrote:
On Thu, Apr 2, 2020 at 6:35 PM Fred Lunnon <fred.lunnon@gmail.com> wrote: > > Presumably the "Sudanese Möbius Band" (credited to Sue Goodman > & Dan Asimov) at > > https://en.wikipedia.org/wiki/M%C3%B6bius_strip > > I found these easier to interpret than Gosper's old-tech renderings. > Plainly apparent in the first frame is a caustic line where the surface > intersects itself, as might be expected.
If you're right that this figure has a self-intersecting line, it's not the figure I'm looking for. I want a Mobius strip that is *embedded* not just *immersed*, which means no self-intersections.
Why the "as can be expected"? The standard embedding of a mobius strip in R^3, the one you get by giving a strip of paper a half-twist and joining it into a band, has no self-intersections, and the embedding of the boundary into R^3 is homotopic to the embedding of the geometric circle. So you can gradually deform this figure, aways with no self-intersections, into a figure where the edge is a geometric circle. The fact that it feels like there must be a self-intersection shows how difficult the resulting surface (which has no self-intersections) is to visualize.
Andy
> > WFL > > > On 4/2/20, Andy Latto <andy.latto@pobox.com> wrote: > > On Thu, Apr 2, 2020 at 4:55 PM Fred Lunnon < fred.lunnon@gmail.com> wrote: > >> > >> << embed a mobius strip in R^3 >> immerse, perhaps? WFL > > > > No, embedded. You can embed a Mobius strip with edge being homoptopic > > to a geometric circle, so you can embed it with the edge actually > > being a geometric circle. There are illlustrations on the wikipedia > > page for mobius strip, but they aren't helping me visualize it. > > > > I want someone to 3-d print me one of these! > > > > Andy > > > > > >> > >> > >> > >> On 4/2/20, Andy Latto <andy.latto@pobox.com> wrote: > >> > On Thu, Apr 2, 2020 at 3:46 PM James Propp < jamespropp@gmail.com> > >> > wrote: > >> >> > >> >> I'm confused by there first sentence ("there's only one embedding of a > >> >> circle in R^3 up to homotopy"), since a knot isn't homotopic to an > >> >> unknot. > >> > > >> > Sorry; I should have said "the embedding of a circle in R^3 given by > >> > the edge of the most familiar embedding of the mobius strip in R3 is > >> > homotopic to the embedding of a geometric circle in R^3, so... > >> > > >> > So while my argument was completely wrong, the conclusion that you can > >> > embed a mobius strip in R^3 with a geometric circle as boundary is > >> > still true, as is the fact that my efforts to visualize this have > >> > proved completely unsuccessful. > >> > > >> >> > >> >> But I think I understand and sympathize a lot of what follows; in > >> >> particular, I'm pretty sure Klein bottles are easier to grok than > >> >> Boy's surface for nearly everybody. I don't know whether this as a > >> >> mathematical question or a psychological question or both, but I think > >> >> it's > >> >> an interesting one! > >> >> > >> >> Jim > >> >> > >> >> On Thu, Apr 2, 2020 at 3:07 PM Andy Latto < andy.latto@pobox.com> > >> >> wrote: > >> >> > >> >> > Since there's only one embedding of a circle in R^3 up to homotopy, > >> >> > there's an embedding of a mobius strip in R^3 where the edge is a > >> >> > geometric perfect circle. But I find myself unable to visualize such > >> >> > a > >> >> > thing. Has anyone seen a 3-d model of this surface? Second-best > >> >> > thing > >> >> > would be a graphic of such a thing, preferably one that you could > >> >> > rotate in 3 dimensions. > >> >> > > >> >> > I'd also like to better visualize Boy's surface, or any other > >> >> > immersion of RP^2 in R^3. It would also be interesting to have an > >> >> > insight into why immersing a Klein bottle in R^3 is easy, while > >> >> > immersing RP2 is "hard". I don't know of any formal sense in which > >> >> > this is true, but apparently Boy came up with this surface when > >> >> > challenged by Hilbert to prove that immersing RP^2 in R^3 was > >> >> > impossible. > >> >> > > >> >> > Also, are these two questions related? That is, can you immerse a > >> >> > mobius strip in R^3 in such a way that the boundary is a geometric > >> >> > circle, and that the union of this mobius strip and a disk with the > >> >> > same boundary is still an immersion (of RP^2 in R^3)? > >> >> > > >> >> > Andy Latto > >> >> > > >> >> > andy.latto@pobox.com > >> >> > > >> >> > -- > >> >> > Andy.Latto@pobox.com > >> >> > > >> >> > _______________________________________________ > >> >> > math-fun mailing list > >> >> > math-fun@mailman.xmission.com > >> >> > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > >> >> > > >> >> _______________________________________________ > >> >> math-fun mailing list > >> >> math-fun@mailman.xmission.com > >> >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > >> > > >> > > >> > > >> > -- > >> > Andy.Latto@pobox.com > >> > > >> > _______________________________________________ > >> > math-fun mailing list > >> > math-fun@mailman.xmission.com > >> > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > >> > > >> > >> _______________________________________________ > >> math-fun mailing list > >> math-fun@mailman.xmission.com > >> https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > > > > > > -- > > Andy.Latto@pobox.com > > > > _______________________________________________ > > math-fun mailing list > > math-fun@mailman.xmission.com > > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun > > > > _______________________________________________ > math-fun mailing list > math-fun@mailman.xmission.com > https://mailman.xmission.com/cgi-bin/mailman/listinfo/math-fun
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