Which branch of mathematics talks about this?

  • #1
Imagine two smooth scalar functions defined on the surface of a sphere. Is it possible to find a smooth coordinate translation that slides the values of the first function over the sphere, until it becomes equal to the second function everywhere?

It seems reasonable that if the number of maxima and minima are different, then we can't stretch one function into another. But if the number of maxima and minima are the same, I'm not sure if it's always possible.

[Edit: Oops ... we need to assume that each maximum/minimum of f1 has the same value as a corresponding maximum/minimum of f2, although in a possibly different location.] Let's assume that as well.

My questions are, which branch(es) of mathematics address(es) this kind of thing? What are some key results that exist for this topic? How does it apply to the previous paragraphs?
 
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  • #2
andrewkirk
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I think it would need to be the case that every contour map of the first function - call it f - is topologically equivalent to the corresponding contour map of the second function - call it g. That is a stronger condition than having the same number of local maxima and minima, with the same function values. Consider a 'sombrero' function that has a single peak that is ringed by a concentric trough and ridge. Such a function may have two local maxima and one local minimum. But it cannot be equivalent to a function that is flat everywhere but with two isolated peaks and one inverted peak, even though that function may have corresponding extrema with identical values.

It would be something like - for every partition ##P## of the range ##R## of the function f, there exists a homotopy (continuous function) ##F_P:S^2\times[0,1]\to S^2## such that ##f## is identical to ##x\mapsto F_P(x,0)## and ##g## is identical to ##x\mapsto F_P(x,1)##.
 
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  • #3
every contour map of the first function...

I don't know the definition of a contour map in topology/analysis, but if it means roughly the same as it does in cartography, then a function would have just one contour map... ? Does the technical meaning allow multiple contour maps to be thought of?
 
  • #4
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I am not aware of any studies on the subject, but it seems that one requirement would be that every value in the range would have to be taken on the same number of times with the same multiplicities by the two functions. And I am not even sure what that means when the "multiplicity" is infinite.
 
  • #5
jim mcnamara
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I think you want to consider Geodesy. (I reserve the right to be wrong...)

A common problem is to calculate surface (great circle) distance distance between two sets of coordinates (latitude, longitude) using the WGS84 datum. Haversines are an example solution. Check out the Law of Haversines here: https://en.wikipedia.org/wiki/Haversine_formula

So routes (a vector "map" of coordinates, which is often a polygon) can be translated across lat and long. Can I ask what you are trying to do? We could guess, which is what I did. So this answer probably should live in the circular file. Good questions help to generate better answers.
 
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  • #6
RPinPA
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So you're trying to find out if two functions of spherical angles ##\theta## and ##\phi## are identical up to rotation of the coordinate axes? That's how I read your question.
 
  • #7
So you're trying to find out if two functions of spherical angles θ\theta and ϕ\phi are identical up to rotation of the coordinate axes? That's how I read your question.

No, not just simple rotation of the axes.

Given two smooth functions ##f1(\theta, \phi)## and ##f2(\theta, \phi)##.
Let ##P=(\theta, \phi)## be any point on the sphere.
Now can we find a smooth displacement function ##g(P)## that moves every point ##P## on the sphere to a new point ##P'=g(P)## such that
##f1(P')=f2(P)## ? So g can be any arbitrary but smooth function, not necessarily a rotation.

Edit: I suspect that's not the correct way to state what I'm asking. But imagine f1 and f2 to represent the variation of some scalar property of the material of a physical sphere (e.g. a grayscale shade that varies smoothly). If we are allowed to slide each point (pixel) over the sphere, can the post-transformation shading profile of f1 be made to match the f2 shading profile, by using a transformation where the displacement function varies smoothly?

[In the grayscale example above, we ignore physical facts like the change in apparent brightness if we bring a lot of bright pixels together]

I think the answer lies in Andrew Kirk's reply, especially the sombrero example. We can't stretch f1 into f2 unless they have the same number of hills, sombreros, valleys, sombreros within sombreros, and so on. However, if the topmost circle of a sombrero has a gradient in f1 and no gradient in f2, they can't be brought to match. An elliptical sombrero can be transformed into a circular sombrero if their topmost ridge contours are both flat, or if they have the same max/ min when moving along the ridge. What happens between maxima and minima may not matter (??) because our arbitrary shifting function can handle such differences. But what is the body of theory that explains this formally, was my question. It's topology, said Andrew. (Thanks, Andrew).
 
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  • #8
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Consider an easier problem. Suppose we have functions f and g, where g is invertible on the range of f. Then ##g^{-1}(f(x))## is a well defined function from the sphere to itself. Pick a point, ## x_f##, on the sphere. Let ##x_g = g^{-1}(f(x_f))##. Then ##g(x_g)=f(x_f)##. Consider the line ##(1-\lambda)x_f + \lambda x_g, 0\le\lambda\le1##. It is a straight line inside the sphere, starting at a point, ##x_f##, on the sphere and ending at the point, ##x_g)## where f and g map to the same value, f(x). Let P denote the radial projection operator from points in ##R^3## onto the sphere. Then ##P((1-\lambda)x_f + \lambda x_g), 0\le\lambda\le1## is a line on the sphere.

Keeping in mind that ##x_g## is a function of ##x_f##, this should be useful in defining a smooth transformation from the points on the sphere to the associated points on the sphere where the values of ##f## and ##g## are identical. In this simpler problem, it may have the properties that you want. But since your problem does not have an invertible function, I don't know how it can be extended to your problem.
 
  • #9
Andrew Kirk's sombrero has helped clarify things a lot.

But I think one concern at the back of my head is like this. Imagine that we start at one point on the sphere where the functions are nearly equal, and the local topology is fairly simple. We spread outwards from this point, aligning various sombreros, hills and so on across the two functions. This alignment may require sliding as well as local rotations (swirling) movements. But when we approach the antipodal point of our starting point, we may find a sort of singularity that we can't reconcile. Or perhaps not.
 
  • #10
andrewkirk
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This is quite an intriguing problem. It would be nice if it turned out that it hasn't been explored before, as it's so hard to find a problem that hasn't been explored somewhere, sometime. I find I keep on thinking about it in my spare moments. Here are a few random thoughts.

Firstly, I wonder whether the requirement for the transformation to be smooth may be too restrictive. The sorts of transformations I've been thinking of are mostly piecewise, and hence would not be smooth at the boundaries between pieces. I suppose that may be fixable by using bump functions to smooth things out at boundaries. In that case the transformation would be smooth but not analytic. But in order to keep things simple at the outset, I would keep the restriction to just single differentiability, and see where we get with that.

I think one could try to work one's way towards a general solution by starting simple. A simple start would be to find a general method of finding a differentiable transformation that can relocate n isolated point extrema to n new locations, without worrying about matching values at places other than the extrema. I am confident that, provided the domain is path-connected we will be able to do that, but specifying a general method for doing so appears a little laborious (I have a plan in my head but it's not ready for committing to a post).

Having done that, perhaps we would try to extend it to relocating not just isolated points but also a finite number of non-closed curves, with the idea that those curves would be contours (isolines) of the function. I'm pretty sure we could do that in all cases.

The next step would be when topology starts to come in - relocating finite collections of points, non-closed curves and closed curves in the domains. In this case, I think the two collections would need to be topologically equivalent. I would expect it to be provable that a transformation exists if and only if that topological equivalence holds.

After that I would imagine trying to allow the collections of curves to be countably infinite, and then cardinality C, defined as the cardinality of ##\mathbb R##, and see what can be proven. Since a function of this type can be completely specified by a collection of contour maps of cardinality C, that should give us what we want. Often one can infer results from a continuum from results from a countable set if uniform continuity is assumed in strategic places.

I think we'd need to rule out pathological cases such as space-filling curves, in order to get anywhere. I think the differentiability requirement may be enough to achieve that. If not we might need to require a higher order of differentiability.
 
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  • #11
andrewkirk
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I've thought of a way to turn it into a purely topological problem.

We make a definition that, given a differentiable manifold ##M##, two differentiable functions ##f,g:M\to\mathbb R## are topographically equivalent if there exists a map ##\phi:M\to M## such that ##\phi## is a bijective diffeomorphism and ##f\circ \phi = g##.

The OP asks what branches of mathematics deals with the question of whether two scalar fields ##f,g## on a sphere are topographically equivalent.

For a differentiable map ##f:S^2\to R## define the ##f##-contour topology ##T_f## on ##S^2## to be the topology generated by the sub-base ##B_f## consisting of all connected components of pre-images ##f^{-1}((a,b))## of open intervals ##(a,b)## in ##\mathbb R##.

That sub-base may seem intimidatingly big, given ##a,b## can be any real numbers. Fortunately, we can show that the topology it generates is the same as that generated by a much smaller sub-base ##B'_f##, in which the intervals ##(a,b)## are required to be of the form ##(k\cdot 2^{-n}, (k+1)\cdot 2^{-n})## for some integers ##k,n##. Then for any interval ##(a',b')## with arbitrary real ##a',b'##, we define ##C## to be the set of pairs of integers ##(k,n)## such that ##a'\le k\cdot 2^{-n} < (k+1)\cdot 2^{-n}\le b'## and we have:
$$(a',b') = \bigcup_{(k,n)\in C} (k\cdot 2^{-n}, (k+1)\cdot 2^{-n}) $$
Hence:
$$ f^{-1}((a',b')) =
f^{-1}\left( \bigcup_{(k,n)\in C} (k\cdot 2^{-n}, (k+1)\cdot 2^{-n}) \mathbb R \right)
= \bigcup_{(k,n)\in C} f^{-1}\left((k\cdot 2^{-n}, (k+1)\cdot 2^{-n}) \mathbb R \right)
$$
so any element of ##B_f## is a union of elements of ##B'_f##, so the topology generated by ##B_f## must be contained in that generated by ##B'_f##. Since ##B'_f\subset B_f## (because ##f## is continuous), the reverse containment also holds, so the two topologies must be identical.

The elements of the sub-base ##B'## are like connected components of isolines (aka contours, aka level sets), eg a closed contour ring of height 250m around a peak of height 280m, except that they are \textit{bands} rather than lines. For instance, a band ringing the peak that covers all ground that has height between 240m and 260m. These bands can be made as narrow as we like, and thereby approximate a contour line with arbitrarily-close accuracy.

I hypothesise that functions ##f,g## are topographically equivalent iff there exists a map ##\phi:S^2\to S^2## that is a homeomorphism when its domain and range have the topologies ##T_f## and ##T_g## respectively, and that ##\phi## will also be a homeomorphism when its domain and range have the metric topologies, so that it is the homeomorphism required under the definition of `topographically equivalent'.

Some constraints may be needed to rule out cases where ##\phi## is a homeomorphism in the contour topology but not a diffeomorphism in the metric topology of ##S^2##. My instinct is that that would only fail in pathological cases. If so, the constraints would be minimal and intuitive, such as requiring that all contour lines have finite length.

If the hypothesis is correct, the question can be translated from one of topography to one of topology, with perhaps a little bit of calculus on manifolds thrown in (for the diffeomorphism).
 
  • #12
Andrew, thanks so much for taking the time!
I don't know enough math to really understand your post, but I hope I can learn as I go forward.

I guess the answer to my question is, the relevant mathematics is -- a lot of calculus and a lot of topology, intertwined in a complex topological knot :smile:
 
  • #13
WWGD
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If I understood correctly, this seems to be about whether the two paths are isotopic?
 
  • #14
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Imagine two smooth scalar functions defined on the surface of a sphere. Is it possible to find a smooth coordinate translation that slides the values of the first function over the sphere, until it becomes equal to the second function everywhere?
Such a transformation is called a conformal transformation.

Generalized for functions not only on the sphere but on any smooth manifold, such transformations are usually possible if you compactify the region, e.g. as in the simplest case with the Riemann sphere.
My questions are, which branch(es) of mathematics address(es) this kind of thing? What are some key results that exist for this topic? How does it apply to the previous paragraphs?
Beyond topology, the branches of mathematics that deal with this topic more specifically are conformal geometry, the theory of complex manifolds and the theory of Riemann surfaces.
 

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