[math-fun] Homology of YGB Sculpture
Consider a Mobius strip, colored symmetrically: Yellow-Green-Blue-Green. (Actually, there are two shades of blue and accents of orange, purple, and magenta, but these extra colors do not contribute to analysis.) If the Mobius strip were non-self-intersecting, a 4*PI cycle could be written as a word, Y1G1B1g1Y2G2B2g2. Instead require that B2 is bisected by the inner Y edge. The path through B2 is obstructed, and the 4*Pi cycle can not be completed, see for example: https://0x0.st/zpTf.JPG https://0x0.st/zpT9.JPG https://0x0.st/zpTp.JPG https://0x0.st/zpTL.JPG Fix the topology by allowing that Y1 and Y2 connect directly to the left and right parts of B2. PROBLEM: How many homology classes does the YGB sculpture have? ANSWER: Three paths are valid from Y1 to Y2: Y1G1B1g1Y2, Y1B2G2Y2, and Y1g2B2Y2. An entire cycle involves one half path plus the reverse of another. Three choose two equals three. Thus the geometry has three homology classes, two of length 7, one of length 6. Explicitly they are: Y1G1B1g1Y2G2B2, Y1G1B1g1Y2B2g2, and Y1B2G2Y2B2g2. A Mobius strip has only one homology class, so this YGB Surface must be something else entirely. Is it known already? If yes, does it also have a vanity name after a famous mathematician? Cheers, Brad
The Arnold Principle. If a notion bears a personal name, then this name is not the name of the discoverer. See also: https://arxiv.org/abs/1609.07779 . Still no luck on the “triple Möbius strip”, or is it a “bifurcated al-Jaziri belt”, or maybe I should try searching for “pretzel-ization of Axion’s Zodiac”? —Brad
On Jul 21, 2019, at 4:28 AM, Brad Klee <bradklee@gmail.com> wrote:
Consider a Mobius strip, colored symmetrically: Yellow-Green-Blue-Green. (Actually, there are two shades of blue and accents of orange, purple, and magenta, but these extra colors do not contribute to analysis.)
If the Mobius strip were non-self-intersecting, a 4*PI cycle could be written as a word, Y1G1B1g1Y2G2B2g2. Instead require that B2 is bisected by the inner Y edge. The path through B2 is obstructed, and the 4*Pi cycle can not be completed, see for example:
https://0x0.st/zpTf.JPG https://0x0.st/zpT9.JPG https://0x0.st/zpTp.JPG https://0x0.st/zpTL.JPG
Fix the topology by allowing that Y1 and Y2 connect directly to the left and right parts of B2.
PROBLEM: How many homology classes does the YGB sculpture have?
ANSWER: Three paths are valid from Y1 to Y2: Y1G1B1g1Y2, Y1B2G2Y2, and Y1g2B2Y2. An entire cycle involves one half path plus the reverse of another. Three choose two equals three. Thus the geometry has three homology classes, two of length 7, one of length 6. Explicitly they are: Y1G1B1g1Y2G2B2, Y1G1B1g1Y2B2g2, and Y1B2G2Y2B2g2.
A Mobius strip has only one homology class, so this YGB Surface must be something else entirely. Is it known already? If yes, does it also have a vanity name after a famous mathematician?
Cheers,
Brad
On Sun, Jul 21, 2019 at 4:28 AM Brad Klee <bradklee@gmail.com> wrote:
Instead require that B2 is bisected by the inner Y edge.
Shouldn't one of the ravenous proof-checkers debate this point? Define W0 = B2Lg2Y1G1B1g1Y2G2B2R, the word associated to the usual Mobius action, bisected at B2. Depending on which yellow edge collides with blue, there are in fact two alternatives. Here is a depiction of the other: https://0x0.st/zfX4.JPG . (It looks kind of ragged, but at least the thread covers the rust and mangle. Sry, can't afford much more.) It is characterized by two non-inverting cycles: B2Lg2Y1 and B2RG2Y2. I did find the following Bridges article with nice drawings: https://pdfs.semanticscholar.org/4d8e/4d5b384e82808b0a87b2afddeab7e97ebfc6.p... The model with two non-inverting cycles is about the same as Figure 1b., after obstruction and YB-reconnection. It appears that changing the angle from 0 to -2*Pi, would reproduce the earlier obstruction in a slightly less contorted figure-eight form, yet homotopy-Eq. to the earlier geometry (after YB-reconnect). Either topology can be drawn in summary as a graph on four nodes. Both figure-eight surfaces are subsets of the 3D-immersed Klein bottle, this time with YB-reconnection along the entire self-intersection loop. (colored computer graphs anyone?) I talked to a Math Grad student studying for Quals from Allen Hatcher. He was quite surprised, and couldn't give a proper name. For now, I am content to call the geometries: "Self-obstructed Möbius strip of the first kind" and "Self-obstructed Möbius strip of the second kind" LOL. Even better, instead of saying "of the first kind": "Self-obstructed Möbius strip with three shortcuts". Under a slightly different integration, the three inverting cycle lengths are 3*Pi, 3*Pi, and 2*Pi, relative to total length 4*Pi. (prob. better to report the length metric in this symmetry.) And just for fun . . . If a character is hoping for a modicum of anonymity, this latest trash-wire, cheap-thread creation can also be worn as a mask: https://0x0.st/zfXJ.JPG https://0x0.st/zfXy.JPG This scene "ici" was shot today in Joëlle Storet's art studio. The paintings and mannequins are her work also. Cheers, Brad
Suspend disbelief, and is it possible that the earlier YGB surface is the connected sum of two separate Möbius loops? Here's one "failed attempt": https://0x0.st/zOWy.JPG https://0x0.st/zOW6.JPG Starting on yellow, if an ant is to take the high road, eventually it will pass two blue bars and return inverted to yellow. Then it may complete the Möbius cycle by traversing the high road again. Too bad, the ant will miss the two red bars on the low road. Similarly if the ant twice takes the low road it will miss the two blue bars entirely. Fortunately, there are four cycles (up to reversal), which pass both blue and red measures. So, if the ant chooses at random, most likely it will turn out to be a well-balanced ant (in the sense of having travelled red and blue... as well as green, yellow, brown and white). The homology count is just four choose two equals six, and all six cycles are measured by 4*Pi. This is enough to ensure the latest work different from the initial YGB. If the maker is more careful and does a "T" join on yellow, it should be possible to recreate a homology with exactly three Möbius cycles. I thought upon this construction the other day and doubted that the colouring could be preserved, but did not write out a rigorous proof. There is one final note on the Klein Bottle upcoming, but humoursly, we have run out of the teal blue thread, thus find ourselves in a lurch, waiting on a shipment from the merchant, ha ha ha. SPQR VOX VERITAS --Brad
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