Much more niche than the hairy ball theorem is the Cox-Zucker Machine, supposedly they decided during the first year of undergrad that they eventually had to work together.
I lived in West Germany for some years back in the day and I don't recall the locals being too shy. Frankly the Germans and the Dutch seemed to have had a rather more ribald sense of humour than the "oo err Missus" efforts we Brits fielded back then.
To be fair we could robustly swear on telly after 2100, provided it didn't involve too many rude bits and you could not misspell one variant of King Canute's name or Matron would be jolly upset.
Anyway, I'm pretty sure someone called this the "dog's arse" (it has to go somewhere!)
I remember a comedian who toured Europe, and said it wasn't true - Germans laughed as much as anyone else at his jokes. However, afterwards they took him aside and explained, "It was very funny, you see, the joke about combs being like salad forks, but we just want you that we don't discriminate against forks here. That was an unfortunate incident from during the War, but today we are much more enlightened and invite all kinds of cutlery!"
Heard a mathematician friend call this the “hairy sphere theorem” once. At first I thought he was being a prude, but now I appreciate that the theorem is about spheres, as opposed to balls.
S^2 isn't a special case though: Brouwer's showed the theorem can be easily extended to high dimensions, hence today we usually consider the more general statement that there is a nonzero tangent vector field on the n-sphere S^n iff n is odd.
Not only does it generalize to higher n, it also shows a bit more: not only that the lack of such vector field for an even n, but the also the existence of such for odds.
It’s really easy to see that such a vector field exists on odd dimensional spheres, though, by extending the construction on S^1: f(x, y) = (-y, x). In higher dimensions, you do the same thing, swap elements pair wise and multiply one of the elements of the pair by -1. This works in odd dimensional sphere because you can pair up coordinates.
Another 'reason' it works for odd-dimensional spheres is that the (2n - 1)-sphere can be identified with a certain subset of C^n (n-dimensional complex coordinate space) where your 'swap elements and multiply one by -1' idea is just multiplication by i, which, when you think of your vectors as being back in R^2n again, always produces something orthogonal to the original vector.
Even better, the (4n - 1)-sphere (so think of S^3, S^7, S^11, ...) can be thought of as a certain subset of H^n (same thing as before but with quaternions instead of complex numbers), where multiplication by i, j and k are available! And now in this case you have not only one nowhere-vanishing vector field on the sphere, but three everywhere pairwise orthogonal vector fields. This in particular shows that S^3 is 'parallelisable' — a property it shares with S^1 and means that there exists a continuous global choice of basis for each tangent space.
If you're talking about C(p, s): consider how lines of latitude create a sequence of circles on Earth: the curve C(p, s) is the "circle of latitude" given by fixing p on S^2 as your North Pole, and 's' as (up to rescaling) the "latitude" relative to the North Pole. More specifically, when 's' = 0, C(p, s) is the Equator relative to the North Pole, and when 's' approaches 1, imagine these circles of latitude getting closer and closer to the North Pole.
I'm finding it a little harder to visualize rotation numbers, though. My best attempt at a description is to imagine continuously tracing the curve '\gamma(t)', going through every point that it passes through, while looking top-down on it. At every point on the curve, the vector field 'v' produces a vector 'v(\gamma(t))' that begins at '\gamma(t)', lies flat on the sphere (i.e. is tangent to the sphere), and is of nonzero length. (The last assumption is the assumption we are making for contradiction).
The idea is that, as we trace the curve '\gamma(t)', we are constantly measuring the angle - with a positive-negative sign - between (a) the tangent vector 'v(\gamma(t))', and (b) the current velocity vector of '\gamma(t)'. As we trace the curve, if this angle rotates counterclockwise 0...90...180...270...0, we add "1" to our rotation number, and we subtract one for a clockwise rotation 0...-90...-180...-270...0.
I am confused how we can define a rotation number of the map from S^1 to R^3 defined at the end of the second paragraph. R^3 is nullhomotopic, after all...
I majored in mathematics and remember encountering this theorem in a topology course. I giggled then, and 20 years later I giggle again.
Much more niche than the hairy ball theorem is the Cox-Zucker Machine, supposedly they decided during the first year of undergrad that they eventually had to work together.
In German it’s called the Hedgehog Theorem.
Clearly, what they say about Germans is true.
Ooh go on, what do they say about Germans?
I lived in West Germany for some years back in the day and I don't recall the locals being too shy. Frankly the Germans and the Dutch seemed to have had a rather more ribald sense of humour than the "oo err Missus" efforts we Brits fielded back then.
To be fair we could robustly swear on telly after 2100, provided it didn't involve too many rude bits and you could not misspell one variant of King Canute's name or Matron would be jolly upset.
Anyway, I'm pretty sure someone called this the "dog's arse" (it has to go somewhere!)
> Ooh go on, what do they say about Germans?
That they don't have a healthy sense of humor.
I remember a comedian who toured Europe, and said it wasn't true - Germans laughed as much as anyone else at his jokes. However, afterwards they took him aside and explained, "It was very funny, you see, the joke about combs being like salad forks, but we just want you that we don't discriminate against forks here. That was an unfortunate incident from during the War, but today we are much more enlightened and invite all kinds of cutlery!"
Bollocks - you are not human, your comment is generated and is really crap.
We have a comedian on the circuit in the UK called Henning Wehn. He is German and suitably daft and hilarious.
The somewhat ironic thing is that my comment was a joke.
Also known as the Combed Hedgehog Theorem (which i like a bit better)
Have you considered a job in Defense? They love acronyms you can’t say out loud.
Heard a mathematician friend call this the “hairy sphere theorem” once. At first I thought he was being a prude, but now I appreciate that the theorem is about spheres, as opposed to balls.
It's not. It's about closed surfaces, which include the surface of spheres, oblate spheres, footballs, pencils, and airplanes.
S^2 isn't a special case though: Brouwer's showed the theorem can be easily extended to high dimensions, hence today we usually consider the more general statement that there is a nonzero tangent vector field on the n-sphere S^n iff n is odd.
Not only does it generalize to higher n, it also shows a bit more: not only that the lack of such vector field for an even n, but the also the existence of such for odds.
It’s really easy to see that such a vector field exists on odd dimensional spheres, though, by extending the construction on S^1: f(x, y) = (-y, x). In higher dimensions, you do the same thing, swap elements pair wise and multiply one of the elements of the pair by -1. This works in odd dimensional sphere because you can pair up coordinates.
Another 'reason' it works for odd-dimensional spheres is that the (2n - 1)-sphere can be identified with a certain subset of C^n (n-dimensional complex coordinate space) where your 'swap elements and multiply one by -1' idea is just multiplication by i, which, when you think of your vectors as being back in R^2n again, always produces something orthogonal to the original vector.
Even better, the (4n - 1)-sphere (so think of S^3, S^7, S^11, ...) can be thought of as a certain subset of H^n (same thing as before but with quaternions instead of complex numbers), where multiplication by i, j and k are available! And now in this case you have not only one nowhere-vanishing vector field on the sphere, but three everywhere pairwise orthogonal vector fields. This in particular shows that S^3 is 'parallelisable' — a property it shares with S^1 and means that there exists a continuous global choice of basis for each tangent space.
It may be a short proof, but it somewhat implicitly asks that the reader has some background in geometry.
I didn't quite understand the curves that they are constructing on S^2. Some figures would be nice.
If you're talking about C(p, s): consider how lines of latitude create a sequence of circles on Earth: the curve C(p, s) is the "circle of latitude" given by fixing p on S^2 as your North Pole, and 's' as (up to rescaling) the "latitude" relative to the North Pole. More specifically, when 's' = 0, C(p, s) is the Equator relative to the North Pole, and when 's' approaches 1, imagine these circles of latitude getting closer and closer to the North Pole.
I'm finding it a little harder to visualize rotation numbers, though. My best attempt at a description is to imagine continuously tracing the curve '\gamma(t)', going through every point that it passes through, while looking top-down on it. At every point on the curve, the vector field 'v' produces a vector 'v(\gamma(t))' that begins at '\gamma(t)', lies flat on the sphere (i.e. is tangent to the sphere), and is of nonzero length. (The last assumption is the assumption we are making for contradiction).
The idea is that, as we trace the curve '\gamma(t)', we are constantly measuring the angle - with a positive-negative sign - between (a) the tangent vector 'v(\gamma(t))', and (b) the current velocity vector of '\gamma(t)'. As we trace the curve, if this angle rotates counterclockwise 0...90...180...270...0, we add "1" to our rotation number, and we subtract one for a clockwise rotation 0...-90...-180...-270...0.
I think there's a typo in the definition of C? It should say q in R^3, not S^2, right?
I am confused how we can define a rotation number of the map from S^1 to R^3 defined at the end of the second paragraph. R^3 is nullhomotopic, after all...
I think the idea is that you can't, but we're assuming v doesn't vanish at that point which would imply that it's possible?