4
$\begingroup$

I saw a video (Who cares about topology) that explained the inscribed square problem. The problem say's that you have a simple closed loop in the plane, prove that there's at least one square such that all vertices are in the loop.

But I didn't understand a part of it.

In the video they say that if you take an ordered pair of points that are in the loop, you can map it to a specific point in a torus.

Then he show's that the same can be say for unordered pairs, but in a Möbius strip.

But, by that logic, Isn't possible to just glue two mobius strips (A and B) by their side, such that the pair of points (a,b) is in A, and (b,a) is on B.

What's the problem with that?

Why a torus and not a Klein bottle?

$\endgroup$
5
  • $\begingroup$ Torus is nothing but $S^1\times S^1$, i.e. the product of two circles. Hence an ordered pair of points on the loop $(a,b)$, which is an element of $S^1\times S^1$, can be mapped into the torus. If you glue the points $(a,b)$ and $(b,a)$ in $S^1\times S^1$ you get a Möbius strip. $\endgroup$
    – Levent
    Jan 24 '18 at 21:54
  • $\begingroup$ Thanks, it took me a while to understand what you meant, but i think I finally got it, Because a simple closed curve is similar to a circle, and you represent all parts of the "circle" against all parts of the same "circle" a.k.a circle x circle Yet, i dont unserstant why a torus = $S^1$x$S^1$, but i'll google it. $\endgroup$ Jan 25 '18 at 14:20
  • $\begingroup$ A simple closed curve is homeomorphic to $S^1$ and a torus is by definition $S^1\times S^1$. $\endgroup$
    – Levent
    Jan 25 '18 at 14:31
  • $\begingroup$ Yeah, I think i got it. $\endgroup$ Jan 25 '18 at 14:33
  • $\begingroup$ Still don't know what went wrong so I end up with a klein bottle (Probably broke some topological law or something because im doing all by intuition) $\endgroup$ Jan 25 '18 at 14:34
1
$\begingroup$

A simple closed curve is homeomorphic to $S^1$ (Circle).

So, for the pair of points $(a, b)$ it's possible to draw a circle, where a point in this circle corresponds to $a$ (or a point to the curve), and for each point in this circle, there's another circle that represents the points of $b$ (The same curve).

This mean that you have a shape that's $S^1$x$S^1$, and that shape is the torus.

Again, to see why the torus, but more visually.

enter image description here

So, you chose first a point from the purple circle ($a$), and then another point from red circle ($b$), thinking that both of those circles are the same than the simple closed curve.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.