Is a Mobius Strip homeomorphic to a torus/ring?

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Is a Mobius Strip homeomorphic to a torus/ring?

Postby Girl-With-A-Math-Fetish » Mon Aug 11, 2014 1:35 am UTC

I say ring because of the "literal" one-sided object, but torus because of the physical paper-strip form.
But is it? Or is it its own class of homeomorphism?
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Re: Is a Mobius Strip homeomorphic to a torus/ring?

Postby z4lis » Mon Aug 11, 2014 1:56 am UTC

If by a ring, you just mean a circle, then the Mobius strip isn't homeomorphic to it because one is two dimensional and the other is one dimensional. It is, however, homotopy equivalent to a circle since you can collapse the strip onto the center circle.

It's also not homeomorphic to a torus. The best way (easiest to understand and most computable) I can think of off the top of my head to show this is homology. Since the Mobius strip is homotopy equivalent to the circle, its homology with Z coefficients will be a copy of Z in dimension 0 and 1. On the other hand, the torus has a copy of Z in dimension 0 and 2, and two copies of Z in dimension one. In more human terms, the Mobius strip is connected, has "one loop", and is not an orientable surface, while the torus is connected, has "two loops", and is orientable.

I should also mention that the Mobius strip may or may not have a boundary, but what I said above holds for either case.
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Re: Is a Mobius Strip homeomorphic to a torus/ring?

Postby jestingrabbit » Mon Aug 11, 2014 2:10 am UTC

Trying to guess homeomorphism classes from first principle is generally unwise, its just too tricky. Mathematicians have developed things called 'invariants' to test for isomorphism. Basically, you can prove that some things are preserved by any homeomorphism. So, lets talk about the three things that you have brought up, and try and work out whether there are any invariants that separate them.

So, we have the mobius strip, the torus, an "the ring". The torus is probably the most straightforward of them all. Its like the skin of a doughnut. One of the properties of that skin is that if you took a tiny little circle around a part of that skin, you would have something homeomorphic with a disc in R^2. Another way to say this is that "the torus is a boundary less two dimensional manifold". Okay, that's cool, what about the other two?

I assume by a "ring" you mean a cylinder without its top and bottom. Now, for most places on that shape, you have the "locally euclidean" property ie "if you took a tiny little circle around a part of that skin, you would have something homeomorphic with a disc in R^2". So, you can say that its a two dimensional manifold. But, there are places where that fails, where, when you take a little circle around them, you don't get a full disc, you always get a half a disc instead, no matter how small you make the circles. So, we have that "the ring is a two dimensional manifold with boundary". More than that, we can see that the boundary comes in two parts, each homeomorphic to a circle.

Now we come to the mobius strip. Is it a two dimensional manifold? does it have a boundary? how many pieces does the boundary have?

So, there are two invariants going on here. Whether a thing is a n-dimensional manifold is preserved by continuous functions, for any n ie the homeomorphism class of a two manifold contains only two manifolds. The number of connected components of the boundary of a manifold is also an invariant. You should be able to answer you own questions using these invariants.

Edit: straight to homology? really?
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Re: Is a Mobius Strip homeomorphic to a torus/ring?

Postby z4lis » Mon Aug 11, 2014 3:17 am UTC

The intimidating part of homology is the details. The idea that there's a precise mathematical way to measure how many "holes" or "loops" a surface has, however, isn't so intimidating. Niether is the number of connected pieces a space has. Isn't homology also one of the easiest ways to carefully show that dimension is actually a topological invariant? I can't remember off the top of my head.
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Re: Is a Mobius Strip homeomorphic to a torus/ring?

Postby Moole » Mon Aug 11, 2014 3:54 am UTC

z4lis wrote:The intimidating part of homology is the details. The idea that there's a precise mathematical way to measure how many "holes" or "loops" a surface has, however, isn't so intimidating. Niether is the number of connected pieces a space has. Isn't homology also one of the easiest ways to carefully show that dimension is actually a topological invariant? I can't remember off the top of my head.


I should probably preface this post noting that I don't know much about topology, except that someone once taught me about fundamental groups and covering spaces, but the only way I've heard to define dimension in a topological space is as "A space of dimension n has that any open cover thereof has a refinement in which no point belongs to more than n+1 sets." This is obviously an invariant under any bijective continuous mapping (since the open cover and refinement are preserved).

So, I guess the only uncertainty is if an n-manifold has dimension n. I'd proceed as follows, assuming the manifold is compact (because I'm sure, that since dimension is essentially a local property and manifolds are presumably locally compact, that extends easily): given an open cover of a compact n-manifold, one could find a refinement thereof such that each open set had a continuous mapping onto Rn - and a finite subcover thereof. Now, consider every intersection of n+2 or more sets in the cover. We seek to "repair" such intersections. Note that any given such intersection can be projected into Rn and handled exactly as if it were a set in Rn, thus by some proof which I'm sure is either hand-wavey-or-tedious, constructing a refinement affecting only that area where each point is a member of at most n+1 sets. If we work backwards making such repairs (i.e. first repairing those where the most sets intersect), eventually, everything with be well, since we assumed a finite subcover.

That seems pretty easy to me. But then again, I don't know what homology is and the Wikipedia page was not very helpful, instead sending me down the rabbit hole to go see what a chain complex or simplical homology is and then try to figure out how that relates to holes. Indeed, I would say that the idea of homology is intimidating in the details (and the intuitive definition is a bit shaky too - everything's nice enough for the 0th and 1st and 2nd homology groups, but beyond that, my brain seems to fizzle. It seems like the kth homology group, from Wikipedia's examples, is just the free abelian group on the set of k-dimensional submanifolds identified by homotopy... but I'm not sure that's even well-defined, and it sounds like it's probably also incorrect)
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Re: Is a Mobius Strip homeomorphic to a torus/ring?

Postby jestingrabbit » Mon Aug 11, 2014 6:51 am UTC

z4lis wrote:Isn't homology also one of the easiest ways to carefully show that dimension is actually a topological invariant?


I'd probably just demonstrate that R^n has non intersecting embeddings of S^(n-1), but R^(n-1) doesn't.

Moole wrote:"A space of dimension n has that any open cover thereof has a refinement in which no point belongs to more than n+1 sets."


The only problem you're gonna have with this is spaces with fractional dimension, but we can pretty easily ignore those, as was the fashion for hundreds of years.
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Re: Is a Mobius Strip homeomorphic to a torus/ring?

Postby z4lis » Mon Aug 11, 2014 4:13 pm UTC

I did some searching for proofs that Euclidean spaces of different algebraic dimension had different topological dimension. There's a book called Dimension Theory by Hurewicz and Wallman that defines a space to have dimension <= n at a point p if the point has arbitrarily small neighborhoods whose boundaries have dimension <= n-1. They then show the result for Euclidean space via one of the forms of the Brouwer fixed point theorem, which they give a proof of independent of defining the degree of a map via its image in homology. I didn't read the proof they gave since I don't have time right now, but it doesn't look like it uses too much machinery.
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Re: Is a Mobius Strip homeomorphic to a torus/ring?

Postby doogly » Mon Aug 11, 2014 7:11 pm UTC

z4lis wrote:The intimidating part of homology is the details. The idea that there's a precise mathematical way to measure how many "holes" or "loops" a surface has, however, isn't so intimidating. Niether is the number of connected pieces a space has. Isn't homology also one of the easiest ways to carefully show that dimension is actually a topological invariant? I can't remember off the top of my head.

This just means it's easy to fool yourself into thinking you understand it.
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Re: Is a Mobius Strip homeomorphic to a torus/ring?

Postby jestingrabbit » Mon Aug 11, 2014 9:47 pm UTC

z4lis wrote:defines a space to have dimension <= n at a point p if the point has arbitrarily small neighborhoods whose boundaries have dimension <= n-1.


That's a really nice way to escape the whole fractional dimensions mess. Really nice.
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Re: Is a Mobius Strip homeomorphic to a torus/ring?

Postby Moole » Tue Aug 12, 2014 3:14 am UTC

Having slept on it, I'm thinking homology might make sense. At least simplical homology seems to make sense. Maybe someone who knows more than me can tell me if this is correct (spoilered since it's not really all that relevant to the original thread):

Spoiler:
The 0th homology group is the quotient group of the free abelian group generated by points with no boundary by the group generated by boundaries of line segments. So, the set of points with no boundary is the set of points and the boundary of line segments is the set of sums p - q for path-connected p and q. Thus, any path connected points p and q are in the same homology class, since their difference is 0 (the identity) in the group quotient. So the 0th homology group is the free abelian group on the number of path-connected components.

The 1st homology group is (sums of line segments with no boundary)/(sums of boundaries of images of triangles). I suppose a useful thing is that, if we take a curve c1 from a to b and curve c2 one from b to c, then the map from the triangle with vertices (0,0), (1,0), (1,1) to the space, then the function f

If x+y<1 then f(x,y) = c1(x+y)
If x+y>=1 then f(x,y) = c2(x+y-1)

has boundary which may be traversed to giving three edges: a traversal of c1 then c2, and the inverses of c1 and c2. Thus, c1 + c2 is in the same homology class as their "concatenation". From here, it's clear that any sum of line segments with a zero boundary is a loop (assuming a path-connected domain. Otherwise, its the sum of loops in disconnected components - note the sum of two loops in a connected component can be reduced to a single loop with a path, traversed both ways, connecting them). Clearly, if two loops are homotopic, then the homotopy may be used to define a square (= sum of two triangles) where the top edge is the first loop and the bottom edge is the second loop and the left and right edges are the base point for the loops. The boundary of this would be two trivial loops (from left and right side, which are constant), in homotopy class of 0 obviously, and then the first loop minus the second one. So, homotopic loops are in the same homology class.

Beyond that, I don't see much more that can be said towards the general case - but, "homology classes have representatives which are sums of loops and homotopy is a sufficient but not necessary condition for homology equivalence" seems good enough. But it's helpful to calculate the homology group of the n-punctured plane, for instance. Knowing that the fundamental group there is the free group on n elements, one quickly sees that any element of a homotopy class can be seen as a sequence of traversals forwards or backwards around the ith point (represented by the ith generator or its inverse) - but, as far as homology is concerned, this is just the sum of the loops around the ith point, because in homology land, everything is magically abelian. However, given any sum of these loops, there is a well-defined notion of how many times it wraps around a given hole (e.g. add a point at infinity. Draw a path from it to the holes. Count the number of positive oriented/negatively oriented crossings of the loops as it heads for the hole), so the homology group must be the free abelian group on the n generators of the fundamental group.

And somehow, my mind is happy to wrap itself around that in higher dimension. (So please tell me it's not wrong). Though, why is it not defined as, instead of using simplexes, using spheres/disks? I suppose it's a bit more complicated to come up with the definition, but it's clearly equivalent and seems like it'd be more elegant and clear. Also, I saw things about "singular simplex homology" - but couldn't tell if this yields a different set of homology classes or what, since the Wikipedia page didn't specify what makes a simplex singular.


jestingrabbit wrote:
z4lis wrote:Isn't homology also one of the easiest ways to carefully show that dimension is actually a topological invariant?


I'd probably just demonstrate that R^n has non intersecting embeddings of S^(n-1), but R^(n-1) doesn't.

Moole wrote:"A space of dimension n has that any open cover thereof has a refinement in which no point belongs to more than n+1 sets."


The only problem you're gonna have with this is spaces with fractional dimension, but we can pretty easily ignore those, as was the fashion for hundreds of years.


Is there even a well-defined notion of fractional dimension in topology? I mean, the Koch Snowflake, for instance, has Hausdorff dimension of log(4)/log(3), but it has a continuos bijection with [0,1], which has Hausdorff dimension 1. This would seem to mean Hausdorff dimension, at least, is a property of a metric space, not a topological one - so I feel totally legitimate in ignoring them unless some other definition exists - especially since the notion I've heard will always give an integer dimension or infinity.
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Re: Is a Mobius Strip homeomorphic to a torus/ring?

Postby jestingrabbit » Tue Aug 12, 2014 8:20 pm UTC

Moole wrote:Is there even a well-defined notion of fractional dimension in topology? I mean, the Koch Snowflake, for instance, has Hausdorff dimension of log(4)/log(3), but it has a continuos bijection with [0,1], which has Hausdorff dimension 1. This would seem to mean Hausdorff dimension, at least, is a property of a metric space, not a topological one - so I feel totally legitimate in ignoring them unless some other definition exists - especially since the notion I've heard will always give an integer dimension or infinity.

I think you're right, you need a metric space to have a hausdorff dimension. You can definitely use topological dimension to classify homeomorphism classes, but that is the one about coverings with overlap.
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Re: Is a Mobius Strip homeomorphic to a torus/ring?

Postby Forest Goose » Wed Aug 13, 2014 11:47 pm UTC

There are some spaces where the topological structure has a canonical metric/fractal structure. See http://matwbn.icm.edu.pl/ksiazki/fm/fm141/fm14135.pdf - just figured it might be of interest.
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Re: Is a Mobius Strip homeomorphic to a torus/ring?

Postby jannii » Mon Oct 27, 2014 11:18 am UTC

If by a ring, you just mean a circle, then the Mobius strip isn't homeomorphic to it because one is two dimensional and the other is one dimensional. It is, however, homotopy equivalent to a circle since you can collapse the strip onto the center circle.



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