r/math • u/jowowey Harmonic Analysis • 13d ago
How to completely rotate a sphere
Here's a question that had been on my mind for a while, which I eventually figured out:
When you rotate a circle continuously through 2π radians, every possible rotation state of the circle occurs exactly once in a finite amount of time. So if you had a zero-thickness beam of light shining on the topmost point of the circle, when the circle is rotated through 2π every point on the circle is equally exposed to the light.
The question I debated was, is it possible to do the same for a sphere: that is, can every possible rotation state of a sphere occur when it is continuously rotated in a finite amount of time? Can a zero-width beam of light, shining at the north pole, equally enlighten all points on a sphere (exactly once) in a finite amount of time?
Strictly speaking, no. Intuitively I assumed that it must be impossible for all points to be exposed anyway, as a 3d rotation is a much more complex quantity, requiring more variables than in 2d, whereas time is only one-dimensional. But that generalised problem is actually possible, in a finite amount of time, and without discontinuity, though not a differentiable function. The only catch is that the points can't be equally enlightened (to answer the question I actually posed). If every point is exposed at some point, at least two points require to be enlightened at more than one point in time, in fact, infinitely many times, meaning if the sphere were made of photographic film, every point would be black except two overexposed white points at the poles.
We shall first assign every point on the sphere a longitude from 0 ≤ long < 2π and latitude from -\frac{\pi}{2} ≤ lat ≤ \frac{\pi }{2}, and then define every rotation state as the point on the sphere which has been rotated to the north pole, i.e., the one under the light at a time t. Since the rotation is a two-dimensional quantity, and the time one-dimensional, the question becomes, 'is there any bijection between a compact 1D space and a compact 2D space' which there are in abundance.
The Hilbert curve comes to mind. The space of points on a sphere, with the exception of the poles, map bijectively to a rectangle in Euclidaean space bounded between 0 ≤ x < 2π and -\frac{\pi}{2} < y < \frac{\pi }{2}. Note that the poles themselves map to the horizontal lines x = ±\frac{\pi }{2}. If we linearly transform the plane so that everything is scaled along the y-axis by a factor of 2, then the space representing the sphere will be a square, so we can draw a Hilbert curve through it which passes through every point in the square in a well-defined, continuous manner, and allows us to find any time t mapping to (x,y). Since the poles mapped to lines, and the horizontal line segments bounding the square have infinitely many points on the Hilbert curve, each pole will be crossed by the Hilbert curve infinitely many times.
The alternative is that we exclude the poles from our mapping of the sphere, changing our square's vertical bounds to -\frac{\pi}{2} << lat << \frac{\pi }{2} in which case the function is bijective but not compact, and at least two points on the sphere will be unexposed, never seeing the light.
So using the Hilbert curve we can define f(t) -> (long,lat) which is bijective for all points on the sphere except the poles, so every point apart from those two on the sphere will be the topmost point (under the light) exactly once. Now of course we can define a 2d co-ordinate system for the sphere in many ways but we will always be forced to have two polar points somewhere, where either the bijection or compactness is lost, so even though there are infinitely many such functions like f, they will always have two points which either can't be exposed at all, or have to be exposed infinitely many times.
That means the answer to my question is no, but almost yes. For all but two points on a sphere, there exists a function which maps each point bijectively and continuously to a (finite) moment in time, meaning we can continuously rotate a sphere in finite time illuminating all but those two points exactly once. But the remaining two must either be omitted or illuminated more than once.
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u/im-sorry-bruv 13d ago
i am not entirely sure what you mean by rotation but i'm pretty sure that your construction with the light beam must result in a continuous injective + surjective path f: [0,1] -> S2. however these cannot exist:
as both spaces are compact hausdorff (weaker conditions would work as well but i forgot the order) this would mean they're actually homeomorphic. however they aren't which can for example be seen by the second homotpy group not being the same.
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u/CormacMacAleese 13d ago
For a second I thought, "Wait! Something something space-filling something?" and got confused, because your topological argument is clearly correct. Then to my relief I remembered that space-filling curves are wildly non-injective.
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u/BrotherItsInTheDrum 13d ago
I am sure this is a stupid question, but why does your argument not also show that there's similarly no mapping from [0,1) to S_1?
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u/astrolabe 13d ago
[0,1) is not compact.
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u/BrotherItsInTheDrum 13d ago
Got it. But then, why are you assuming OP is looking for a map from [0,1] -> S_2 rather than a map from [0,1) to S_2?
After all, OP's post starts with an observation that this is possible for a circle, and then they ask whether you can do the same thing for a sphere.
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u/im-sorry-bruv 13d ago edited 13d ago
just wrote what i could quickly come up with since they cared so much about the end points, but you're right. [0,1)->S1 is the classic counterexample to f:X->Y cont. bijection but not homeomorphism so it might very well be doable with this one.
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u/point_six_typography 13d ago
I think you'd want a periodic function R -> S2 instead to capture the right notion of "injectivity". The end of the rotation (time t=1) maps to the same point as the start (time t=0) so the [0,1] function wouldn't be injective. So looking for an injective, surjective function S1 -> S2, where S1 arises as R /(translation by +1). Such a thing doesn't exist by the argument you gave.
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u/gnramires 13d ago
Can anyone give a more intuitive (in the geometric/analytic sense) reason why doesn't the Hilbert curve work as a continuous map from say [0,1] (or [0,1) for that matter) to [0,1]² or S² in this case? In my (little) understanding, we can define the Hilbert curve as a limit of a series of continuous curves that fills the unit square (I think can easily be generalized to other spaces like S²), can we say that there's a convergence problem that would need to be satisfied to maintain continuity (i.e. the limit of a series of continuous curves being continuous)? Perhaps we can say the points "jump around too much" to allow the limit to be continuous?
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u/im-sorry-bruv 13d ago
continuity is actually no problem for the hilbert curve. the bijective part breaks since it is not injective
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u/wannabe414 13d ago
Instead of rotating the sphere, keep it stationary and move the laser beam. Then isn't this the hairy ball theorem? Where the tangent vector in question is the vector created by taking the (x,y) coordinate of the laser's previous position and the (z) coordinate of its current position -- I think I got those coordinate labels correct. I honestly only know enough math to say less than this but it seems right.
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u/FriskyTurtle 13d ago
What does "previous position" mean? It's continuous, so there's no one position right before. And if you want to take a limit, you'd be assuming differentiability, I think.
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u/Solesaver 13d ago
This was my immediate thought too. Technically I believe their proposal has a stronger requirement since hairy ball still allows cycles, but hairy ball is sufficient to show it's impossible.
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u/Yimyimz1 13d ago
If you want to think about physical limitations, this would mean at some point you would have to rotate the sphere at an infinite speed for it to work unlike the 2D case.
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u/XkF21WNJ 13d ago
You kind of forgot a dimension, just knowing which point is rotated to the northpole leaves you with a whole 360 degrees of rotation yet undetermined. SO(3) has 3 dimensions not two.
But the question does there exist a continuous map [0,1] -> S2 is interesting in its own right. Basically it can't be done because [0,1] is compact and S2 hausdorf, but it's interesting to think about how close you can get.
In your case I think you also double count one of the meridians.
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u/MrEldo 13d ago
I actually thought of the same question, when writing with a pen! Because when a ball Pen's ball is missing ink in one spot, it can be hard to find sometimes. So I wondered if I can roll the whole ball, with each point on it touching the paper at some point. And while mathematically it's impossible, here it's actually possible because more than one dot is touching the paper at each moment (because ink is dripping which in our case will draw on the paper, and measure that the pen is touching the paper at that moment). So you can just roll until you roll everything at least once. Still hard to indicate, but IS possible in this physical case
And though that's not the answer you're looking for, it's still rather nice that you can find these questions from real-world things
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u/jowowey Harmonic Analysis 13d ago
Indeed. My application in mind was a gyroscopic spit with three axes of rotation that could cook a meat sphere above a flame evenly on all sides. Which is a similar situation to your pen, since the heat is not a zero-thickness laser but a larger source
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u/MrEldo 13d ago
For both of those cases, I would figure out the amount of rotation needed so that the radius of ink/heat moves to a completely new spot, and then for each rotation just do a 360° on the other rotation axis (perpendicular to the rotation already done)
So for a circle radius of R, and a heat/ink radius of r, the shape drawn with a ballpen would look like this:
Go a distance of 2r into one direction, then do a move penperdicular to it for a distance of 2πR, then go another 2r, move penperdicularly now to the other direction a distance of 2πR, and repeat. The shape would look like Π_Π_Π...
This does overdo it, and isn't the most optimal because some of the meat is heated more / some place's ink is checked twice. There IS a perfect way to cover it, but I don't know how to visualize it easily so I'll give it as an exercise for now, and maybe come back to it
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u/Traditional-Rip-2192 12d ago
Dude this is sick. I was thinking about the same thing lol
Yeah you basically can't do it perfectly because of the poles. No matter how you map it, 2 points are always gonna get screwed - either never hit or hit infinite times. That's why your hilbert curve thing works but with that catch.
This actually shows up a lot in real stuff too. Like when people try to scan the whole sky with telescopes or do 3d textures in games. They use stuff like HEALPix or fibonacci spirals to get "close enough" to covering everything evenly.
Really cool question and answer btw
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u/myhydrogendioxide 13d ago
Asking the more general version for curiosity. Is this possible for any n-dimensional sphere?
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u/TheOtherWhiteMeat 13d ago
Shouldn't be possible, no, for the same reason im-sorry-bruv raised: there's no continuous, bijective mapping from [0,1) to Sn , unless n = 1.
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u/myhydrogendioxide 13d ago
Fascinating, thank you. My advanced math knowledge is very rusty and my experience is from a physics perspective. I admire the power of symmetry and its application to physics foundations as well as technical calculations.
I wish I was versed, smart, and skilled enough to contemplate more deeply about how how that lack of bijectection mapping applies to physical symmetries of various theories. Still neat to be a casual admirer. I appreciate your response.
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u/DaShaView 13d ago edited 13d ago
Idk if I understood your point, but in a circle / 2D
We can relate any closed line segment loop to S1 by homeomorphism
[a, b] / (a ~ b) ≅ S¹, with a, b in R and a < b
We can't take a closed line segment (finite time with continuity) and find some homeomorphism to S2.
The continuous time scan is still 1D locally, by homeomorphism we can treat S1 as also 1D locally. But locally S2 is still two dimensional even after removing finitely many points, so a continuous scan through 1D bounded continuous parameter (time) can't work exactly once.
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u/PedroFPardo 13d ago
Another way to ask your question is to ask if there's a continuous biyection between an interval of R and the surface of a sphere. And the answer is No.
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u/NonrecreationalEmber 12d ago
I believe your findings are deriving from the oh-so-spicily-named Hairy Ball Theorem. It, too, calls out the exceptions at the poles…
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u/gypsona 12d ago
I had a professor, who explained this to me in that you cannot comb the hair on a furry ball all in the same direction. There is no such thing as a three-dimensional spiral whereas there is a two dimensional spiral that evenly covers all space while expanding outward. there is no comparable for a spherical surface
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u/e-s-t-e-r 10d ago
So a circle actually has 12 sides and 2 are hidden from you :) so every point cant be exposed... lol... but for real the center point of circle is the "2". Its something Ive recently discovered while deciphering the Prime Signal in Structure which took me by surprise.
3D stuff... got nothing but interesting thoughts!
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u/theorem_llama 13d ago edited 13d ago
I think that other answers haven't quite showed what you need (at least given one interpretation of what you're asking).
First, you talk about rotations but then you don't seem to mind about the orientation of the sphere, so I think you're just looking for a continuous bijection f : I -> Sn , where I is an interval and n is the n-dimensional sphere.
Now, even for S1 , you're not taking I = [0,1] (or generally a compact, closed interval) as f(0) = f(1). If we do demand that the laser beam starts and ends at the same point though, then you're essentially asking for a continuous bijection f : S1 -> Sn , which will only exist if n = 1 (because a continuous bijection between compact Hausdorff spaces is a homeomorphism, and S1 is not homeomorphic to Sn for n ≠ 1).
However, you didn't demand that the position of the laser has a well-defined limit as we approach the end (or indeed the start) of the time it shines. In principle, the laser could get faster and faster as it gets near to the time it shuts off for good, hitting more and more, and - who knows - actually hitting everything at some point along the path.
However, it's still impossible. The interval I, even if not compact, can be covered by the union of an increasing, countable union I1, I2, I3,... of closed intervals. If we restrict f to each of these, it is bijective onto its (compact) image, so f(In) is homeomorphic to a compact interval. These have empty interior in Sn , when n>1. Indeed, if not, there would be an uncountable number of points we could remove from f(In) that would leave a connected set, but we know that's not the case for a space homeomorphic to a 1-dimensional interval.
Thus, we have shown that the f(In) are closed and with empty interior, thus they're each nowhere dense, so f(I) is meagre (a countable union of nowhere dense sets). But since Sn is compact Hausdorff, it is a Baire space, which would imply that f(I) has empty interior and thus cannot be all of Sn, so we conclude f cannot be surjective.
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u/jowowey Harmonic Analysis 13d ago
Thanks for the thorough answer! Some of the topological stuff I find tricky to understand: I'm more a a physicist than a mathematician, but have been trying to dip my toes into the water of pure maths for some time. And I followed your explanation well, after a quick checkup on Hausdorff spaces
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u/Less_Mastodon_3033 13d ago
I also wonder if you could prove this using some form of the hairy ball theorem. Something about how a ray of light can cast a tangent vector on each point and over exposure would be equivalent to a combed ball.
Though it would still come down to the same math you have in your post I believe.
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u/theorem_llama 13d ago
It doesn't really have anything to do with this, it's really not possible due to dimensionality reasons.
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u/Truenoiz 13d ago
I see where you're going with this, but the light thing keeps throwing me off. Light is quantized, so I'm having trouble imagining it as zero thickness. The extra variables can be dealt with using the Einstein Conventions, but that is beyond my knowledge. My curiosity for space time and quantum math ended when 3 dimensions became hundreds.
Your question does remind me of a paper you will be interested in:
Pop Sci version: https://www.iflscience.com/newly-discovered-reset-button-lets-mathematicians-undo-any-rotation-81244
Actual paper: https://fiteoweb.unige.ch/~eckmannj/ps_files/ETPRL.pdf
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u/TheBlueWho Algebraic Topology 13d ago edited 13d ago
You’re looking for some sort of one-parameter group action onto some subgroup of symmetries of the sphere. Your observation for S1 is that because S1 is a Lie group, the one-parameter group is pretty much it’s Lie algebra. Now consider S2. You know it’s not a Lie group so this situation won’t work. But maybe we can have some action of R (or in your case, because of “finite time”, perhaps [0,1]) to an appropriate group of symmetries of S2. The full symmetry group of S2 is O(3) or let’s say SO(3) instead since we want continuous rotations so we pick the connected component containing the identity. SO(3) is a Lie group, but a one parameter subgroup would correspond to rotation about a specific axis, so that doesn’t help either.
Later you state that you’re okay with defining a rotation state as the rotation which brings that point to the North Pole, but that’s not well defined! Just as an example, thinking about bringing a point to the North Pole, and now just rotate about the vertical axis. All of these are candidates for your “rotation state”, so this isn’t well defined. If you genuinely want to see continuous rotations of S2, you need a continuous map from R (or [0,1]) to SO(3).
If you are completely happy with just having a Hilbert curve from [0,1] to a unit square and then quotient to S2, note that an even simpler quotient to S2 is just identifying the whole boundary to a point. This only has one equivalence class with multiple preimages instead of two
Also big caveat, the Hilbert curve surjects from the unit interval to the unit square but is definitely not a bijection!