I would think intuitively that a proof for the rectangle and the box would start like this:
Take a fractal pattern like the Koch Snowflake but do not put in all the infinite lower levels.
I would believe that a series of boxes and rectangles would form, possibly depending on symmetry; say, having an even number of thorns. Then add more and more lower layers towards a geometric limit of infinite sublayers.
For such a symmetric fractal the boxes and rectangles might expand out towards a circle or converge into a circle as the thorns go infinitely down. This seems a lot like your circular geometric proof for the Basel theorem on the complex polar plane. And here symmetry might have a role to play (the usual symmetry/entropy interaction).
Further thoughts—I can see two limiting cases using the Koch Snowflake:
1) large boxes around the inside with the asymptote a circle
2) small box that diminish to dots with the asymptote a circle
I always struggled with topology. Now I have a much better introduction to it. Thank you, and thanks for all the wonderful videos—they are the stuff of legend. Congratulations! Keep doing what you're doing, and I'm definitely restacking this with a note of my own! I
Oh, with respect to getting ones head around that that embedding of a mobius strip such that the edge lies on a plane but the strip doesn't intersect itself:
The edge of your twisted paper strip describes kind of a folded figure 8 in space, where the crossing doesn't intersect itself (by the width of the strip). Grab one of the two the two tips of that figure 8 - which lie next to one another because the 8 is folded - and pull it up and over to unfold the 8.
The unfolded 8 has 4 side points, and two of them are next to the bit under the crossing point. Grab one of them and pull down and across to turn the 8 into an 0.
We can model the angle of the lines with a number 0-π, and create a sort of doubled surface in 4d space where one of the coordinates gets π/2 added, or just use as space (w mod π/2,x,y,z≥0).
We get problems if two different pairs of points have the same midpoint and angle, but this only happens in nonconvex loops. (Come to think of it, the current solution has this problem too.)
Problem is that if 2 2d manifolds intersect in 4d space, you can always shift the sheet a bit to make the intersection go away. Need to show that if a möbius strip embedded in 4d space intersects itself, then trying to make that intersection go away just shifts the crease. Also need to deal with the fact that the edge of the strip necessarily winds around the cylindrical space.
You could map the angle rather than the length onto 3-d space (mod π/2) ad look for intersections, but all that would show is that points in the interior have at least one pair of pairs at right angles. Come to that, *every* point in the interior has a continuous loop of such points.
Grant,
I would think intuitively that a proof for the rectangle and the box would start like this:
Take a fractal pattern like the Koch Snowflake but do not put in all the infinite lower levels.
I would believe that a series of boxes and rectangles would form, possibly depending on symmetry; say, having an even number of thorns. Then add more and more lower layers towards a geometric limit of infinite sublayers.
For such a symmetric fractal the boxes and rectangles might expand out towards a circle or converge into a circle as the thorns go infinitely down. This seems a lot like your circular geometric proof for the Basel theorem on the complex polar plane. And here symmetry might have a role to play (the usual symmetry/entropy interaction).
Further thoughts—I can see two limiting cases using the Koch Snowflake:
1) large boxes around the inside with the asymptote a circle
2) small box that diminish to dots with the asymptote a circle
I always struggled with topology. Now I have a much better introduction to it. Thank you, and thanks for all the wonderful videos—they are the stuff of legend. Congratulations! Keep doing what you're doing, and I'm definitely restacking this with a note of my own! I
Oh, with respect to getting ones head around that that embedding of a mobius strip such that the edge lies on a plane but the strip doesn't intersect itself:
The edge of your twisted paper strip describes kind of a folded figure 8 in space, where the crossing doesn't intersect itself (by the width of the strip). Grab one of the two the two tips of that figure 8 - which lie next to one another because the 8 is folded - and pull it up and over to unfold the 8.
The unfolded 8 has 4 side points, and two of them are next to the bit under the crossing point. Grab one of them and pull down and across to turn the 8 into an 0.
We can model the angle of the lines with a number 0-π, and create a sort of doubled surface in 4d space where one of the coordinates gets π/2 added, or just use as space (w mod π/2,x,y,z≥0).
We get problems if two different pairs of points have the same midpoint and angle, but this only happens in nonconvex loops. (Come to think of it, the current solution has this problem too.)
Problem is that if 2 2d manifolds intersect in 4d space, you can always shift the sheet a bit to make the intersection go away. Need to show that if a möbius strip embedded in 4d space intersects itself, then trying to make that intersection go away just shifts the crease. Also need to deal with the fact that the edge of the strip necessarily winds around the cylindrical space.
You could map the angle rather than the length onto 3-d space (mod π/2) ad look for intersections, but all that would show is that points in the interior have at least one pair of pairs at right angles. Come to that, *every* point in the interior has a continuous loop of such points.
Meh. Too hard.