The second homotopy group of the sphere: π₂(S²) in classical and digital topology
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- Опубліковано 27 лис 2023
- A very gentle introduction to the second higher homotopy group of the 2-sphere, plus some digital topology.
The paper is: "A Second Homotopy Group for Digital Images" by Gregory Lupton, Oleg Musin, Nicholas A. Scoville, P. Christopher Staecker, Jonathan Treviño-Marroquín, at arxiv here: arxiv.org/abs/2310.08706
Talk given at the Fairfield University math department research seminar, Nov 17, 2023.
For a slightly faster and more technical version of the second part of this talk, see my JMM talk here: ua-cam.com/video/Lk-HW-jjRL8/v-deo.html
Super lecture! I discovered your channel through your reviews of antique calculating devices, and although I’m not a mathematician at all (never did more than differetial equations and some analysis in undergrad), your lecture was easy to understand and thoroughly enjoyable!
So accessible! Thank you kindly!
41 minutes on homotopy? Wrap it up!
Well that was absolutely fascinating! Going to read the paper for sure. Makes me want to build some sort of puzzle game to break down nonagrams into T and T inverses or something. Great work!
damn, I'm only like halfway through a Mathematical Fundamentals course rn (it's a course in my uni that teaches you surface level Linear Algebra and Analysis) and I was able to follow the whole thing! this is awesome, i love this :D
Thank you! I love getting into new topics, especially when it's as accessible as this!
stellar video, thanks!
Beautiful visual proof
I thought this lecture was both fascinating and very well explained. Nice job!
Followed for the mechanical calculators, stayed for mathematical ideas that might actually be relevant for my research
and the really nice theorem at the end
Marvelous. A hint, for a wrapper for which inside-out is obvious, take a bread-bag. Or wrto napkin, a tea towel with humorous saying or souvenir image on one side.
I mean this is what I had been thinking this whole time. Glad you made a video about it.
Sometimes you just gotta state the obvious.
for my future reference
00:00 - 15:15 recap of higher homotopy groups
Great lecture!
Awesome talk! The filling trick is really elegant! Just FYI, it looks like you have a typo at 29:31 at the bottom left (there's a light blue next to a dark blue).
Yes thanks! I meant to have the third column from the left doubled, and everything else like it was originally. But I totally wrecked the right side of the picture! Sorry-
Great talk! Very accessible to someone not in the field. In digital media and data we often have 3D or higher dimension data structures. E.g., voxels, gridded volumetric data, or other multidimensional arrays (I think machine learning enthusiasts sometimes call them "tensors" or "data tensors"). Would there be interest in investigating the properties of these higher-dimensional digital structures?
Yes- one of the very basic goals of "digital topology" theories is to find ways to deduce the topological "shape" of discretized data. The goals are similar to "topological data analysis", which is a much more mainstream developed field, but the techniques are totally different.
(commented at 24:00) How does continuity imply the "no color adjacent to opposite" property, since we have diagonal connections? Looking at some minimal triangle in the napkin ABC so there's an edge for each pair, what prevents a coloring like A,B,C \mapsto white,light blue,black? I thought about this for a bit but couldn't figure it out, so wondering if there's a simple explanation for what I'm missing.
Enjoying the video so far at this halfway point.
Edit: The final trick to reduce the function is very beautiful! It's incredible that you can make such a big transformation with 3 really natural operations.
Thanks for watching carefully- the continuity condition says that two adjacent vertices in the domain must map to adjacent (or equal) vertices in the codomain. "Adjacent" means that there is an edge in the graph (possibly diagonal) connecting them. So (A,B,C) \mapsto (white,light blue,black) is not allowed because A is adjacent to C, but their images are white and black, which are not adjacent.
@@ChrisStaecker Oh, very silly overlooking on my part. Thanks for the response!
I hope I've just misunderstood, but in your large example of the (digital cover?), you've got an almost 5x5 red square near the bottom right, I believe in that square there is a black diagonally adjacent to a white, which I understood to be illegal according to the graph rules outlined earlier in the talk. Is it the case that diagonally adjacent opposite mappings are allowed or illegal? I guess I should read the paper to find out.
Really cool fill method to find T and T^-1! I noticed you subdivided by 5 and that corresponds to the sidelength of the T (map?) on this surface. Is there a relationship between those values? Is T the elementary (mapping?) Between all digital spheres or just this octahedron? I would guess it would increase in size but I'm unsure how, for larger digital spheres. There's lots of different caveats I can think of that I don't know how to answer atm.
You're right! This is a mistake in the picture! This example was made by hand just for this talk (it's not in the paper), and I tried my best to make sure it was legal. If I give this talk again I will need to change that white (the one diagonally adjacent to the black) to a light blue- then everything will be fine.
The subdivision by 5 vs 5x5 shape of T are related, I guess. It turns out that after the second fill, the pink points will be isolated, and also that they will only occur at the "corner" points of the 5x5 blocks resulting from the subdivision. Since the subdivision was 5x5, this means that the isolated pink points are guaranteed to be at least 5 pixels away from each other, which means that when I do the final fill I get separate blocks of T's and T^{-1}'s. If the subdivision was only 3x3 for example, after my last fill I will mostly get T's and T^{-1}'s, but some may be overlapping and then you'd have to do some extra steps to separate them.
For non-octahedral shapes things would look very different. The important property of T is that it "wraps" all the way from the bottom to the top of the octahedron. If that octahedron is replaced by a bigger thing, the simplest possible T will have to be very large just to make it all the way around. And then the step I said where we proved all 5x5 blocks are either T, T^{-1}, or c- this step would be a lot more complicated.
Why is a tetrahedron not the simplest "digital sphere"?
Using the standard definitions, the tetrahedron does not surround any “missing point”, so when you consider it in this theory it looks like a solid. If you want something sphere-like, it needs to have some empty space inside.
sure it's beyond me but i'm at least curious what "A-theory(?)" is, they've kinda just chosen the worst possible name for their field to trying to search up
Yes it's a uniquely bad name for SEO. It started in the late 90s, I guess before anybody was thinking about that. If you want to read an introduction, try "Perspectives on A-homotopy theory and its applications" by Barcelo and Laubenbacher, 2005. It's very similar to what I'm doing here, but a slightly different version of the homotopy relation makes many of the basic results different.
Missed opportunity to make a Christmas pudding
I wouldn't call it "fill in with red" but instead "propagate the red" (?)
if the set of (p+n) of positive and negative sums really isomorphic to Z then how is the cancellation happening in pixel land?
@@user-gu2fh4nr7h I skipped over this detail, but you can show directly that if you take the +1 and -1 blocks from 34:40 and put them next to each other, then you can change 1 point at a time to make it cancel fully (all red).
In the paper we called it "flood" rather than "fill". To me "propagate" has the connotation that you are taking all the existing reds and extending them outwards to make more red. But that's not really what you're doing. When you do a fill with red, it can create new red pixels where there were none before. But of course anybody can call it whatever they want!
@@ChrisStaecker I see. Thank you for your reply.