Algebraic Topology 4: Brouwer Fixed Point Theorem & Borsuk-Ulam
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- Опубликовано: 24 дек 2024
- Playlist: • Algebraic Topology
We use the fundamental group to prove the Brouwer Fixed Point Theorem which states that any continuous map from the disk to itself leaves at least one point fixed. We also prove Borsuk-Ulam which tells us that there are two points on the opposite sides of the earth with the exact same temperature and barometric pressure.
Presented by Anthony Bosman, PhD.
Learn more about math at Andrews University: www.andrews.ed...
In this course we are following Hatcher, Algebraic Topology: pi.math.cornel...
those are very good! having a blast here binge watching this series
Yes. This series is wonderfully written and entertaining.
Esteemed sir, you are one of the finest teacher i have ever seen. Please provide the new lectures of this series. Complete the book of hatcher please.
15:44 how do we know r * h_t is well-defined? What property it follows from?
wait, can we say h(s) = -h(s + 1/2) and h is continuous so h takes 0 somewhere, but h = g \circ l, which means g takes 0 somewhere, then we also get a contradiction??
Don't understand how you can use a unit sphere with it's central point and a straight line from a point on the surface to an anipodal point as an ideal to homotopic other paths across through S2. Is their any correspondence with describing bent spacetime using relationship to orthogonal axes?
33:25 alternative proof of Borsuk-Ulam
There is a slight error/fudging in terminology in the discussion of constructivism. Constructivists follow the principle of constructivity/principle of constructivism which states that a proof of "a or b" is a proof of a or a proof of b similarly for existential quantifiers a proof of “there exists n such that P(n)” is the witness n such that P(n) holds. Surprisingly, there are some constructive systems that allow LEM as long as every proof is still constructive. However Intuitionistic systems reject LEM, often making the negation of LEM an axiom.This is still different from the philosophy that Brouwer pushed for, Intuitionism, which was in opposition to Formalism.
Also worth mentioning that almost every computer scientist is a constructivist by defult, so if you meet them at a confence instead just ask them what the next buzz word to add to your grant proposal should be
I think your definition of f in the antipodal points theorem is wrong; it should be S2 -> R, not R2, right? The example you gave was temperature, which is decidedly in R.
Shouldn't the g in your proof also be R2 -> S1? S2 doesn't even fit in R2.
He's proving it for S1 -> R not S2 -> R2. That's what allows him to use the intermediate value theorem. The higher dimensional proofs take a little more work.
What happens if the winding number helix is observed from an S1 in a constant force field.
The winding number is a topological property and independent of the embedding. Since all elements of the fundamental group are loops the work done along them is always zero in a conservative vector field but that's more of a physics thing and requires a metric.
@@abebuckingham8198 not really grasped much of this yet but surely a metric is implied when there is any homo....(?) with R1.
It seems I have a very bad intuition regarding continuous functions. :/ For example, I really can't convince myself that D1 without origin is homeomorphic to S1. Still there are arbitrarily close points in D1 without origin that map to antipodal (far away) points on S1.
Do you mean D2 with origin removed? S1 is not homeomorphic to D2 minus origin, you can disconnect S1 by removing two points, not so for D2. They are homotopic by deformation retract though
@@MrMetrizable yeah, sorry, I meant D2. Still, I don't know what "homotopic by deformation retract" means. :/
I’m not very knowledgeable about topology, but I guess your concept of continuity is a slightly inaccurate. Your statement on arbitrarily close points in D2-{0} mapping to antinodal points is true, but I guess that what matters for continuity is that given a point x in D2-{0}, there is some neighborhood that gets mapped to an arbitrary small neighbourhood of the image f(x).
Intuitively, for a point (x=epsilon,y=0), your map looks continuous on an open circle of radius epsilon or smaller.
(The map around 18:00 shows that these spaces are homotopic, not homeomorphic, as that would require a unique inverse)
I think I understand where your confusion is coming from. When you say "there are arbitrarily close points that..." what you're doing is starting with two points that you have subjectively deemed to be "close", and then you're looking where they end up. That's backwards. You don't get to pick two points beforehand.
Instead, what you should do is pick one point, not two, and then ask "is there a neighbourhood of points around this one point that will stay close to it". With the punctured disk, you can do that. You just need to make sure that your neighbourhood doesn't include the origin.
@@deadeaded Thank you very much. I think I confused the idea with the one of uniform continuity.
Brilliant proof.
Proofs via the lifting technique are always a bit weird I mean couldn't one just formulate the Theorem directly with respect to the lifted space, give it a meaning of its own right, and then obtain the Theorem about the original space as a corollary via projection?
I'm not exactly sure what you have in mind, but I believe the order here is important to give the contradiction. Recall the contraction is produced by showing that the equivalence class of the loop (in S^1) is both the trivial [w_0] and non-trivial. We show that it is not trivial by lifting it, showing that it must be equivalent to [w_m] for odd m and hence cannot be [w_0].
In your proof of Brouwer's thm, you forgot to talk about the case where the image of a point of the circle (border) by f is a point of the circle.
I might be missing something but f(x)=x doesn't matter if the image is or is not on the border. So this still implies a retraction so I don't see a problem with f(x) being on the border.
I will not forget this quote: "The good news is, while there still are constructivist mathematicians, everyone recognizes them as a big kookie"
Formally it contains all it's Limit points...
i'd like to point out that you might have accidentally misrepresented constructive math and constructive mathematicians. in constructed math, you don't say "the law of excluded middle is false", or "proof by contradiction not a real proof". rather, you just don't *use* the law of excluded middle. As a matter of fact, in constructive math you can prove that the law of excluded middle is not *not* true, which is done by contradiction. there is only a specific kind of proof by contradiction which is not allowed, and that is a proof where you suppose "not A", arrive at a contradiction, and then conclude "therefore, A". this kind of proof relies on the law of excluded middle, or equivalently, double negation elimination. however, another kind of proof by contradiction *does* still hold, where you suppose "A is true", arrive at a contradiction, and then conclude "not A". as a matter of fact, this is how falsehood is defined; "A implies contradiction" is equivalent to "not A".
Also, calling constructive mathematicians kooky is uncalled for. you wouldn't call a mathematician who would rather not use the Axiom of Choice that either.
you might want to watch this video on the topic, as it points out wonderfully what constructive math is and isn't about:
ruclips.net/video/21qPOReu4FI/видео.htmlsi=kG_StLZwHi3zhr8q
I was being lighthearted - your point is well taken.
It is a little kooky you gotta admit