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Borsuk-Ulam theorem.

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 DaveHK 30 Jan 2020

I heard this referred to on an In Our Time podcast about Emmy Noether. The way the guest spoke about it it sounded like it was an actual property of our planet but that goes against most of what I know about planetary climate and seasonality.

I wondered if it was more a theoretical thing that made predictions about some idealised sphere with hypothetical properties rather than real world properties.

The Wikipedia article on it doesn't shed much light on it for me.

Here's the bit that caught my interest: 

"The case n=2 is often illustrated by saying that at any moment, there is always a pair of antipodal points on the Earth's surface with equal temperatures and equal barometric pressures."

I guess it's most likely to be the case if those antipodal points are on the equator but that isn't specified.

My money is on Robert Durran to shed the most light on this but all contributions welcome!

 DaveHK 30 Jan 2020
In reply to DaveHK:

This claims to offer a simple proof of the theory but I'm too tired to think hard about it right now: https://brilliant.org/discussions/thread/borsuk-ulam-theorem-the-global-version/

 DaveHK 30 Jan 2020
In reply to DaveHK:

About 10:40 into this: https://m.youtube.com/watch?v=csInNn6pfT4

I think I might be beginning to understand...?

In reply to DaveHK:

Never heard of it but just looked it up and had a quick think.......... The n=1 (circle) case is simply proved: Take two antipodal (opposite) points (A and B) on a circle. Suppose the temperature at A is greater than at B. Now rotate them both round at the same rate until they have change positions. The temperature at A is now lower than at B. So at some point in between, their temperatures must have been the same (because there are no sudden jumps - that's the continuity thing).

I'll now start thinking about n=2.........

 petemeads 30 Jan 2020
In reply to DaveHK:

Brilliant video! I hope he has a version which explains the ham sandwich theorem which is apparently closely related...

In reply to DaveHK:

OK, I think I can explain it reasonably non-technically for n=2:

Again consider two antipodal points A and B (say the north and south pole). Again we can rotate them at the same rate around any circle consisting of opposite lines of longitude (there are infinite such circles). By the same argument as for n=1, there will be antipodal points on each circle with equal temperature. Because of the continuity thing, these points will form a continuous loop around the world. Now consider any pair of antipodal points on this loop (P and Q). Now move P and Q around this loop at rates such that they remain antipodal and therefore always have equal temperature. By the same argument, they will at some time be in positions with equal pressure. So we have found a pair of antipodal points at equal temperature and pressure.

Feeling quite pleased with myself

In reply to DaveHK:

42.

 HansStuttgart 30 Jan 2020
In reply to Robert Durran:

> OK, I think I can explain it reasonably non-technically for n=2:

> Again consider two antipodal points A and B (say the north and south pole). Again we can rotate them at the same rate around any circle consisting of opposite lines of longitude (there are infinite such circles). By the same argument as for n=1, there will be antipodal points on each circle with equal temperature. Because of the continuity thing, these points will form a continuous loop around the world. Now consider any pair of antipodal points on this loop (P and Q). Now move P and Q around this loop at rates such that they remain antipodal and therefore always have equal temperature. By the same argument, they will at some time be in positions with equal pressure. So we have found a pair of antipodal points at equal temperature and pressure.

> Feeling quite pleased with myself


But a problem emerges at the continuous loop. Because I can think of a situation in which I can make one of these loops infinitely small. For example the north pole has a local temperature minimum for all polar angles smaller than 1 deg and the south pole has a local maximum at the same temperature for all polar angles larger than 178.  And it is not guarantied that there is an equal pressure on the loop of polar angles equal to one. There will of course be another (probably larger) that will have this property. But to me the proof needs some method of selection which of the (possibly infinite?) numbers of small loops of antipodal points in temperature will have the antipodal point in pressure.

 petemeads 30 Jan 2020
In reply to HansStuttgart:

Surely the minimum size for a loop is a circumference?

Post edited at 23:19
In reply to HansStuttgart:

> But a problem emerges at the continuous loop

I've just realised that of course that all the antipodal equi-temperature positions of A and B form separate loops for A and B. For example suppose the N.Pole is at 0C and the S. Pole at 10C, all latitudes of 80N are at 5C and all latitudes of 80S are also at 5C. Then the antipodal equi-temperature points form separate loops for A at B (lines of latitude at 80N and 80S).  Need to think again...........

 elsewhere 31 Jan 2020
In reply to DaveHK:

> This claims to offer a simple proof of the theory but I'm too tired to think hard about it right now: https://brilliant.org/discussions/thread/borsuk-ulam-theorem-the-global-version/

That really is brilliantly clear.

 DaveHK 31 Jan 2020
In reply to Robert Durran:

Thanks Robert, think I'm getting there!

 DaveHK 31 Jan 2020
In reply to petemeads:

> Surely the minimum size for a loop is a circumference?

I think this is Hans' point: According to Robert's explanation one moves around the lines of longitude until one finds the equal temperature antipodes then you move around a line of latitude (at the same rate to remain antipodal) until one finds equal pressure. But the closer to the poles you are the smaller the loop of the latitude line is meaning it's less likely that you will find equal pressure. I think...

In reply to DaveHK:

> I think this is Hans' point: According to Robert's explanation one moves around the lines of longitude until one finds the equal temperature antipodes then you move around a line of latitude (at the same rate to remain antipodal) until one finds equal pressure. 

Not around a line of latitude, but around a closed loop consisting of antipodal points of equal temperature. Unfortunately I overlooked the fact that my antipodal points might be moving round two separate loops.

Anyway, having lain awake a good bit of the night thinking about it, I think I may have resolved this problem (maybe a few details to sort out). I'll post later when I have time.

Looking forward to looking at the brilliant.org link and the video once I've fully convinced myself of my own explanation!

Post edited at 07:36
In reply to DaveHK:

Emmy Noether deserves to be far better known; she was of a time when it was almost impossible for a woman to get the recognition she deserved as a mathematician or scientist. As I understand it her theorems proving a correspondence between symmetries and conservation laws completely underpin the modern approach to theoretical physics.

Post edited at 07:41
 DaveHK 31 Jan 2020
In reply to Robert Durran:

> Not around a line of latitude, but around a closed loop consisting of antipodal points of equal temperature. Unfortunately I overlooked the fact that my antipodal points might be moving round two separate loops.

That makes more sense, I was wondering how you'd conserve temperature moving round like that.

Surely if you can make an infinite number of longitudinal loops each with antipodal points of equal temperature that infinite number of points will form a loop of their own and you can just scoot around that loop (conserving temperature) until you find equal pressure points?

In reply to DaveHK:

> Surely if you can make an infinite number of longitudinal loops each with antipodal points of equal temperature that infinite number of points will form a loop of their own and you can just scoot around that loop (conserving temperature) until you find equal pressure points?

Yes, that was what I thought ( though not conserving temperature - just keeping my two points at antipodal positions of equal temperature). Unfortunately my two points might be scooting round  separate loops, so never change places (see n=1 argument). I described a scenario which results in this happening late last night.

My new argument is that there must be an odd number of such loops and that I can have my antipodal points of equal temperature both scooting around the middle one........ I think!

 HansStuttgart 31 Jan 2020
In reply to Robert Durran:

The question is whether there is always at least one loop with A and B on the same loop?

In reply to HansStuttgart:

> The question is whether there is always at least one loop with A and B on the same loop?

I think there is. I think there will always be an odd number of loops and the middle one of these will do the job.

 HansStuttgart 31 Jan 2020
In reply to Robert Durran:

> I think there is. I think there will always be an odd number of loops and the middle one of these will do the job.


How does it have to be the middle one if I can start with an arbitrary point A and B in the procedure?

The other extreme is easy: start with A and B on the loop we are looking for. Then the correct loop is at the edge instead of at the middle.

Thought example:

we start with A and B at the north and southpole. We define the equator as the major loop we are looking for. We also add rings of antipodal points at 10 deg polar and 170 deg. Now we add another set of loops just like the 10 deg polar but then rotated by 20 degrees. Now if we go from A to B we encounter 5 temperature antipodal points. And the one we are looking for is number 4, not the middle one.

Post edited at 09:33
In reply to HansStuttgart:

> Thought example:

> we start with A and B at the north and southpole. We define the equator as the major loop we are looking for. We also add rings of antipodal points at 10 deg polar and 170 deg. Now we add another set of loops just like the 10 deg polar but then rotated by 20 degrees. Now if we go from A to B we encounter 5 temperature antipodal points. And the one we are looking for is number 4, not the middle one.

"Now we add another set of loops just like the 10 deg polar but then rotated by 20 degrees"

I'm not clear what you mean by this.

In reply to DaveHK:

> This claims to offer a simple proof of the theory but I'm too tired to think hard about it right now: https://brilliant.org/discussions/thread/borsuk-ulam-theorem-the-global-version/

Had a look at this now that I'm pretty confident of my own explanation. I'm not convinced that the "band" of isothermic antipodal poinmts does not have the same problem I had - the antipodal points could in fact be on two separate bands. But maybe I'm missing something.

The video uses the same explanatioon and identical graphics.

 HansStuttgart 31 Jan 2020
In reply to Robert Durran:

Sketches would be easier....

procedure to create a new ring of antipodal equal temperature points. Start with an antipodal point at coordinate theta (polar angle) and phi (azimuthal angle) where C at (theta, phi) and D at (180-theta, phi + 180) have a difference in temperature. Choose the point (C or D) with the lowest temperature and increase its temperature until it is higher than the temperature at the antipodal point. By the continuity of the temperature field this procedure creates a ring of antipodal equal temperature points around C and D.

"Now we add another set of loops just like the 10 deg polar but then rotated by 20 degrees" means do the above procedure at coordinates theta = 20, phi = 0.

And in the procedure to get from A to B in order to find antipodal equal temperature points, follow the line phi = 0, so that this line cuts through the newly created loop.

 alexm198 31 Jan 2020
In reply to Robert Durran:

Absolutely agree. Noether’s theorems are some of the most beautiful results in mathematical physics, in my opinion. I vividly remember being taught about them in a theoretical physics lecture and being blown away. 

 cb294 31 Jan 2020
In reply to DaveHK:

Just skimmed over the thread, so my apologies if this has been posted below and I missed it:

So if I remember correctly from my university maths ages ago the whole thing is a generalized claim about properties of two steady functions f(x) and g(x) that is obvious for the simple case R -> R case: If within some interval of x  f(x) rises from A to B and g(x) drops from B to A, their graphs must cross at some point if neither fuctions is allowed to do sudden jumps.

The most intuitive physical equivalent is that you can stabilize any four legged table that stands on uneven ground with one leg in the air by simply rotating it around its centre (the table top may not end up horizontal, of course).

Simply plot the height the floating leg will have above or below the ground (relative to some absolute height mark) against the angle of rotation as you turn the table, keeping the other three legs fixed to the ground.

Similarly, plot the height of the bumps and dips of the ground relative to the same height mark as you go around the same circle*

After one full turn both curves will end up at their starting height, but the table leg height will have, at some point, been above and at some other point, below the level of the ground.

Since there are no sudden jumps in either the height of the fourth table leg or the ground, the values must be identical for some rotation angle and thus all four legs on the ground

The BU theorem (again IIRC) is the same concept but expanded to higher dimensional functions. The increase in dimensionality is also the reason that you can make the claim that you can find anitpodean points with identical properties.

CB

* the table leg will not describe a true circle, as it moves inwards as it tilts up, but the argument can still be made using the projection of the curve onto the ground and using the height above ground along that curve)

 SenzuBean 31 Jan 2020
In reply to DaveHK:

> This claims to offer a simple proof of the theory but I'm too tired to think hard about it right now: https://brilliant.org/discussions/thread/borsuk-ulam-theorem-the-global-version/

That is really clear and easy.

I tried to 'invert' the problem by thinking about surfaces in the temperature/pressure domain to see if it might become easier (I wondered if there might be an easy way to use the Brouwer fixed-point theorem's graphical formulation), but it didn't get any easier...

If you take any equatorial line around the Earth (every point of which's unique antipode is also present on that equator), then every possible temperature/pressure of this equator can be drawn as a closed loop (that may self-cross) on a 2D temp/pressure graph. If you take rotation of this 'equator' as your other variable (from 0º to 180º), and you extend that 2D plane into 3D with this dimension, then you sweep out a surface in that 3D area. However due to the 180º rotation, the other loop at the end 3D surface must be both perfectly co-linear with the 0º loop and inside-out(if I'm not mistaken), which would mean there must be at least one twist (and thus at least one crossing point) in the surface (this crossing point would be a same temperature/pressure point on a given equator). However I can't quite see how this guaranteed crossing point must at least lie once on some antipodal points.

In reply to SenzuBean:

> That is really clear and easy.

Why do you think it is clear that there will be a single band of isopthermic antipodal pairs of points rather than more than one band with pairs of antipodal points being on different bands?

In reply to HansStuttgart:

> Sketches would be easier....

Indeed. You have completely lost me!

> procedure to create a new ring of antipodal equal temperature points..................

In reply to DaveHK:

Here's my argument:

Colour the sphere so that every point whose antipodal point is at a lower temperature is red, and every point whose antipodal point is at a higher temperature blue. Every point and its antipodal point will therefore be different colours. The boundaries between red and blue regions will consist of points whose antipodal points are the same temperature. Importantly, the antipodal point of any point on a boundary will also be on a boundary, so that each boundary will have a "mirror" antipodal boundary.

Suppose the N. Pole is red and the S. Pole blue. If you travel by any route from the N. Pole to the S. Pole, you must switch colour an odd number of times (since you switch overall from red to blue). Therefore you must cross a boundary an odd number of times. There must therefore be an odd number of boundaries forming closed loops encircling the poles. Some of these loops might touch but that does not affect the argument. There may also be closed loops which do not enclose the poles, but this does not affect the argument either. Suppose there are, say, 9 such closed loops numbered 1 to 9 from north to south. Loops 1 and 9, 2 and 8, 3 and 7, 4 and 6 must be "mirror" antipodal boundaries, but loop 5 must be its own "mirror" antipodal boundary - in other words it consists of pairs of antipodal points at the same temperature. We can now use the n=1 argument to show that one such antipodal pair must also be at the same pressure.

 SenzuBean 01 Feb 2020
In reply to Robert Durran:

> Why do you think it is clear that there will be a single band of isopthermic antipodal pairs of points rather than more than one band with pairs of antipodal points being on different bands?

I think there must be at least one, continuous isothermic band of antipodal points*
(quoting: There will be no existence of gaps between any two points on this band due to the fact that if there exists a gap, then we would be able to swap the thermometers antipodally between AAA and BBB without ever encountering an intersection where the temperature readings meet, which is not possible.)

As long as we have at least one continuous isothermic band, we can then perform the intermediate value trick on that for pressure (this trick will still work if there are multiple bands and we just randomly select one).

* - Some degenerate examples showing multiple would include
- The entire earth is isothermic, so there are infinite possible isothermic bands of antipodes.
- A fat isothermic band around the equator - there are also infinite possible isothermic bands of antipodes in here.

In reply to SenzuBean:

> I think there must be at least one, continuous isothermic band of antipodal points*

Yes, but I remain to be convinced that the argument on brilliant.org proves this!

In reply to DaveHK:

I've been pondering my red and blue coloured sphere and now realise that it gives a really elegant and illuminating explanation through symmetry.

When viewed from above the N. Pole and from above the S. Pole it looks identical except with the colours swapped. So the sphere must consist of two pieces, identical apart from having the colours swapped, which  must fit together with the join between them consisting of a continuous loop of antipodal isothermic points.

Symmetry is always the best

I suspect that this argument ought to extend to higher dimensions.....

Post edited at 07:44
In reply to Robert Durran:

> When viewed from above the N. Pole and from above the S. Pole........

Another nice point is that the poles can be chosen completely arbitrarily and one will always be on each of the symmetrical halfs of the sphere, so we always get the same continuous loop of antipodal isothermic points, even though the other pairs of bands I mentioned in my earlier post will come and go with the choice of poles.

 SenzuBean 01 Feb 2020
In reply to Robert Durran:

> Yes, but I remain to be convinced that the argument on brilliant.org proves this!

Okay having read it again, I think there's likely a small mistake:
Now, repeating the same process with infinitely many sets of antipodal points A and B
I think should be
Now, repeating the same process with infinitely many antipodal walks between antipodal points A and B

I think I glanced over that as the pictures show the correct story (the band forms while point A remains in the same location).
 

 HansStuttgart 01 Feb 2020
In reply to Robert Durran:

clear explanation.

My issue above refers to what you call loops that do not enclose the pole. I was pointing out that it is possible to to cross 9 loops going from north to south where 1 mirrors 9, 2 and 3 are on the same loop that does not enclose the pole and therefore do not have mirror loops that intersect the north to south routes (there are of course mirrors on the antipodal south to north route), 4 mirror 8, 5 mirrors 7, and 6 mirrors itself. So the loop that mirrors itself it only the middle one after we substract the loops that do not enclose the poles. But your mirror boundary argument works in all cases. We just select the one loop whose mirror boundary is itself.

I was also a bit lost yesterday evening in thinking about complicated crossing patterns. But again the mirror boundary argument can identify the right branch at each crossing point.

Thanks!

 HansStuttgart 01 Feb 2020
In reply to Robert Durran:

> I've been pondering my red and blue coloured sphere and now realise that it gives a really elegant and illuminating explanation through symmetry.

> When viewed from above the N. Pole and from above the S. Pole it looks identical except with the colours swapped. So the sphere must consist of two pieces, identical apart from having the colours swapped, which  must fit together with the join between them consisting of a continuous loop of antipodal isothermic points.

> Symmetry is always the best

nice

In reply to SenzuBean:

> Okay having read it again, I think there's likely a small mistake:Now, repeating the same process with infinitely many sets of antipodal points A and B

> I think should be: Now, repeating the same process with infinitely many antipodal walks between antipodal points A and B

> I think I glanced over that as the pictures show the correct story (the band forms while point A remains in the same location).

Yes, I think I took it to be trying to say that but I still don't see how that means there is a single band of isothermic antipodal points, rather than pairs of bands with one of each pair of points on each band.

In reply to HansStuttgart:

> My issue above refers to what you call loops that do not enclose the pole............

I was a bit worried about these, but, because any pole to pole route can simply go around them without crossing any boundaries, I don't think they affect the argument. Anyway, as you say, my symmetry argument makes the issue go away anyway.


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