Maths in a Twist [pdf] | Hacker News Reader

Maths in a Twist [pdf]

ima.org.uk

40 pointsColinWright27 comments

By just stretching, I also think it's a Mobius strip: http://i.imgur.com/sshZhxs.png

The only way to get the arrows to connect right is if three of the square's corners are the same point.

The cutting solution is a bit more easy to follow, in my opinion...

Since it's Colin himself posting this, I guess that he linked to where he intended; but, if you're interested in this article, then you may want to look at more of "Mathematics Today", the publication in which this article appears: http://www.ima.org.uk/activities/publications/mathematics_to... .

MAYBE SPOILER

For the last problem you can bend the sheet such as the two arrows point in to the same direction(up) and are in the same side, that is a rectangle. On the side of the arrows(left of the rectangle), the bottom half is an arrow pointing up and the upper half is an arrow also pointing up.

Then you bend the paper such that the lower side meets the center and the wrap around the upper side around the previous bend. I got a cilinder

If I've followed you correctly, I don't think you get a cylinder.

You are right. In the last step, the top arrow has to be bent outwards and then joined against the bottom arrow, instead of overlapping like I previously did/imagined. The result is a figure with one surface and one edge, like the one NateyJay points out: its a mobius strip.

After trying the last puzzle for a while, I think I've concluded it's a torus - is that right?

Nope.

Added in edit: I'm not going to answer any other questions or suggestions - I'd love to see the working behind the answers. Why should it be a Torus? Why should it be a Klein Bottle? Why should it be what you claim? Show me your working!

FWIW, as and when I get the chance I will upvote every guess that shows its working.

A Mobius strip.

Working?

This is not the best wording, but it's what I have for now: Label the top, left, bottom, and right edges of the square T, L, B, and R, respectively (so T and L are the ones which must be glued to each other, in suitable fashion). Cut the square down the \ diagonal; this produces two pieces: an L\B triangle and a T\R triangle. Mark the \ sides of these pieces as having to be ultimately glued to each other in the appropriate fashion, and now re-attach them in the way in which L and T are already indicated to be glued (in the process, flipping one of them over). Now one has the equivalent of the M diagram from before, and thus a Mobius strip.

I also find a Mobius strip, I made the same reasoning

I believe the bottom of L (BL) must coincide with the left of T (LT) and the top of L (TL) must coincide with the right of T (RT); however, musn't all four points coincide in order to properly reglue the \ diagonal? Wouldn't this appear as a "crimped Möbius strip" (if it is even possible)?

Another way to get to this, with perhaps wrong reasoning:

1. Deform the figure until the non-arrowed edges are a disc, and the arrowed edges are next to one another, so you have something like a Pac man figure where the arrowed edges are the mouth.

2. This figure clearly has 1 edge (the outer edge) but only 1 side -- to get to the 'back' side, go through the mouth

Since all the matters topologically is number of edges and sides, this is identical to a Mobius strip.

  > Since all the matters topologically
  > is number of edges and sides, ...

That turns out not to be the case. A torus with a hole and a sphere with a hole both have two sides and one edge.

A Klein Bottle with a hole also has one side and one edge, and is not a Möbius strip.

So that's not enough, and might not be right.

Yeah I think a problem with this article is that hints at what you can do in topology, but it doesn't really explicitly tell you what all the rules of the game are.

I guess to some extent that's the difference between an article intended for a general audience, and an actual proper course on point-set topology. There are limits to what you can cover.

Speaking of topology, check out this numberphile video from a couple days ago: https://www.youtube.com/watch?v=k8Rxep2Mkp8 https://www.youtube.com/watch?v=6Qpfv5y-7WU

Cliff Stoll is a treasure.

It's a Mobius strip: http://i.imgur.com/1MVkbta.png

Interesting. Are you sure a single point can be stretched out to a line? Certainly lines can be stretched or shrunk, but having a point become a line seems a bit of a stretch (if you'll pardon the pun).

It's a nice argument, but is it valid ... ?

It isn't valid. If it were, the cylinder and the cone would be identical.

I'm pretty sure it's valid as long as you put the line back to a point at the end.

Hmmm, actually now that I think about it I'm starting to believe it.

The idea is that the red point is being identified with the entire line, not stretched. My math is a little rusty, but I think you can make a diagram like http://i.imgur.com/PchSkyI.png and use the fact that Q1 and Q2 are quotient maps to define g and show it's a homeomorphism.

The argument goes something like: there's a universal property on quotient spaces X with quotient map Q which says that if F(a) = F(b) for all a and b in X such that a and b are identified (a ~ b), there's a unique continuous G such that F = G ∘ Q.

So let's take a and b in the top left space. We have a homeomorphism f such that a ~ b if and only if f(a) ~ f(b). Since quotients take equivalent elements to equal elements, Q2(f(a)) = Q2(f(b)). Therefore there's a unique continuous g1 such that Q2 ∘ f = g1 ∘ Q1. We can apply the the same reasoning to Q1 ∘ f^-1 to get a unique continuous g2 such that Q1 ∘ f^-1 = g2 ∘ Q2. Furthermore, g1 and g2 are inverses: g1 ∘ g2 ∘ Q2 = g1 ∘ Q1 ∘ f^-1 = Q2 ∘ f ∘ f^-1 = Q2, and Q2 is surjective. Therefore g = g1 is a homeomorphism.

This argument is almost there. The theorem you mentioned about the quotient requires that you are taking a topological quotient on the left of your diagram, which prescribes a specific topology for the quotient space[1] (which you have not demonstrated is the topology you are using). Let Q1: X -> Y. This is the topology such that the open sets for Y are precisely those for which their preimage via Q1 is open in X.

If you think this sounds a lot like continuity, you're right, but it adds one crucial detail: continuity only requires that whatever open sets we have for Y, their preimage is open (so for example you can make Y have the topology of a point, and Q1 would still be continuous). For quotient maps, we require that any set in Y with an open preimage be open, which fixes exactly one topology. This makes sense, because otherwise the quotient could pick from a laundry list of (non-isomorphic) topologies which keep the quotient map continuous, breaking the topological invariance of the quotient map.

[1]: https://en.wikipedia.org/wiki/Quotient_space_(topology)

I think that can't be because a torus doesn't an edge, and the figure clearly still has one

Correct. If it helps, the given diagram clearly has exactly one edge and exactly one surface.

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