2

I would like to have a way of generating points on an n-sphere so that the points are 'symmetric' in the sense that the distribution of points 'looks the same' from any of the points and such that the points are densely covering the sphere.

Here is a more precise definition of what I want:

For some $\epsilon>0$, find a finite set $S\subseteq\mathbb{R}^{n+1}$ st:
1: $\forall a\in S, |a|=1$ and $|S|<∞$
2: $\forall a,b\in S, \exists$ orthogonal $A\in\mathbb{R}^{(n+1)\times(n+1)}$ st. $S=\{As:s\in S\}$ and $a=Ab$
3: $\forall x\in\mathbb{R}^n$, if $|x|=1, \exists s\in S$ st. $|x-s|<\epsilon$

In more plain language:
1: Find a finite set of points on the n-sphere embedded in $\mathbb{R}^{n+1}$,
2: for which I can take any point in the set to any other via an orthogonal map (symmetry),
3: and such that the points are 'densely packed', so that any point on the sphere is within $\epsilon$ of a point in the set

In general, I would like a way of constructing such a set, call it $S_{n,\epsilon}$, for a given $n$ and $\epsilon$. Ideally, the number of points in $S_{n,\epsilon}$ should scale like $O((\frac{1}{\epsilon})^n)$.

Motivation:

I want to solve PDEs on a sphere with a discretization which doesn't introduce any global artifacts that are due to how the function was oriented on the grid. For example, if you place your points at fixed latitudes, even if you space them out well, there may still be artifacts that come about due to where the 'north pole' was placed. If there is a better way of doing this than the one I proposed above where I find a collection of points with sufficient 'symmetry', please let me know, however I am still interested in knowing if there is a solution the problem I gave above because I find it interesting in and of itself, without any reference to PDEs.

Thanks

Thanks to @ErikTowers for pointing me towards some helpful Wikipedia articles

Regarding Orthogonality
Here is the precise definition of orthogonal I am using:
https://en.wikipedia.org/wiki/Orthogonal_matrix
That is, QQ^T=I
Notice, for example, that flipping about a plane (1 negative and 2 positive eigenvalues) is an orthogonal map.
Consider the vertices of this object:
https://upload.wikimedia.org/wikipedia/commons/5/55/Uniform_tiling_532-t012.png
I believe I can take any vertex to any adjacent vertex via a flip about the plane that goes through the centre of the edge between the vertices, and is orthogonal to this edge. Because I can take any vertex to any neighbouring vertex, and because there is a path between any two vertices, I can take any vertex to any other via a sequence of orthogonal maps, which it self is a single orthogonal map.

regarding embedded 2-spheres It is not clear to me that a tessellation of an n-sphere induces a tessellation of the embedded 2-spheres, as I believe @EricTowers is claiming (if this is not the claim, my apologies, please correct me). If we take an n-sphere and let an affine plane slice through it, we get a new lower dimensional sphere. Say this new sphere has a number of tessellation points on it from the higher dimensional sphere. I don't think this new sphere must itself have these points form a tessellation. Yes, there must be an orthogonal map taking any tessellation point on the new sphere to any other, however this orthogonal map may take points in the new sphere to points outside of it that are in the old sphere, and vice versa.

Mathew
  • 1,894
  • 2
    In symbols, condition 2 requires that the rigid rotations that take any point to any other point fix the set $S$. In words, you have said the rigid rotations exist with no fixed set requirement. Having $S$ fixed is infeasible except for an enumerated set of unsatisfactory exceptions. Such a set $S$ would be a vertex set of a tesselation of the sphere and these have been enumerated: https://en.wikipedia.org/wiki/Lists_of_uniform_tilings_on_the_sphere,plane,_and_hyperbolic_plane#Spherical_tilings(r_=_2) . You want denser sets that are on that list, and they don't exist. – Eric Towers Nov 20 '21 at 02:58
  • @EricTowers, thanks a lot for giving me this link. I glanced at the list and it seems to only cover the case of n=2. For n=1, what I am asking for is trivially possible. It seems that for n=2, there are only finitely many such sets S. How about n>2? – Mathew Nov 20 '21 at 04:14
  • On the lists there are the vertices of Platonic solids and certain other tesselating solids (most of which require reflections that break your $S$-invariance requirement). There are also $n$ points evenly spaced on a single great circle for any positive integer $n$. – Eric Towers Nov 20 '21 at 04:20
  • I see, so we still have the same result for higher dimensions then. Thanks again – Mathew Nov 20 '21 at 04:27
  • @EricTowers If you could find somewhere that states this result (the wikipedia article doesn't seem to actually state that this is only possible for a finite number of sets) then I would accept that as an answer. – Mathew Nov 20 '21 at 05:16
  • The second answer here gives such a proof. – Eric Towers Nov 20 '21 at 06:31
  • my constraints don't require that item 2 holds in answer 2 on the given question: "Only one type of polygon is permitted and all instances are congruent" For example, my constraints will allow for this shape: https://upload.wikimedia.org/wikipedia/commons/5/55/Uniform_tiling_532-t012.png Also this only addresses the 2-sphere and I am interested in all n-spheres. – Mathew Nov 20 '21 at 07:08
  • 1
    Incorrect. Rotation of one blue square vertex to an adjacent blue square vertex pushes the target vertex onto a non-vertex in a red or yellow region, violating that the rotation takes vertices to vertices (i.e. the image of each vertex is another vertex). The situation in $n$-spheres is worse -- each $n$- sphere has multiple embedded $2$-spheres and you are requiring a tesselation simultaneously on all of these embedded $2$-spheres. – Eric Towers Nov 20 '21 at 07:10
  • @EricTowers I am worried that we are using two different definitions of orthogonal. I have updated the question to show the specific definition I am using (the one on wikipedia), and I have explained why I think the red-yelllow-blue tiling I linked to does in fact obey my requirements. Hopefully you can explain to me where I have gone wrong in my reasoning, assuming we are using the same definition for orthogonal. Also, it is not clear to me that a tessellation of an n-sphere induces a tessellation of a 2-sphere, I will adress this concern in the question body. – Mathew Nov 20 '21 at 20:22

0 Answers0