Summary This question is about covering a sphere with great circles so that the resulting density is uniform and that if we for each circle associate a normal vector, then the magnitude of the sum of them would have an upper limit of 1/2. This has a bearing in a classical model of the electron that yields, when classically calculating, the correct excitation energy for hydrogen.

Definition Define a covering of a unit sphere, $S^2$, in $E^3$ as a covering of type $G(\mu_o)$, where $\mu_o$ is as a positive measure on the set of orthogonal 3 dimensional matrices, $O(3)$ s.t. $\mu_o(O(3)) = 1$, Let $s_z$ be the great circle with normal $\hat z$ and $\mu_z$ it's the uniform measure on $s_z$ such that $\mu_z(s_z) = 1$. Now the elements of $O(3)$ maps a great circles to all great circles and we can define a map $F: O(3)\times s_1 \rightarrow S^2$ with $F(O,p) = O \,p$. Then for the covering to be of type $G$ we define that the induced measure through $F(O,p)$ e.g. via the $\mu_o(dO)\mu_1(dp)$ and achange of variables, should be the uniform measure $\nu$ on $S^2$ such that $\nu(S^2) = 1$. We will assume the normal Borell sets e.g. the measure theory that you get from the eucledian metric, for all geometrical objects included in the definition.

Also define $\text{total}(G)$ as the vector sum of all normals,

$$ \text{total}(G) = \int_{O \in O(3)} O \, \hat z \,d\mu_o $$

We will consider a genralization of the above:

Definition Let a covering of $^2$ inbedded in $E^3$ be of typ $G^2(\mu,\mu_z)$ if we add to the definition of $G$ that $\mu_z$ is a probability meassure on the great circle $s_z$. Fo this case we should consider the set of orthogonal matrixes module a rotation with normal in the $\hat z$ direction. e.g. we consider $O_1,O_2$ in the same equivalence class if we can write $O_1=O_2R_z$ with $R_z$ a rotation with normal in the $\hat z$ direction.

Note due to the triangle inequality and that orthogonal matrices maps the unit sphere to the unit sphere we have, $$ |\text{total}(G)| \leq \int_{O(3)} |O \, \hat z| \,d\mu_o = \int_{O(3)} \,d\mu_o = 1 $$

and therefore there is a supremum for the values that the magnitude of $total(G)$ can take. Hence this is a well formed conjecture

Conjecture A There exists at least one covering of type $G$ and furthermore $\sup_G |\text{total}(G)| = 1/2$ and there is an example that attains that supremum.

Conjecture B There exists at least one covering of type $G^2$ and furthermore $\sup_{G^2} |\text{total}(G^2)| = 1/2$ and there is an example that attains that supremum.

The example below shows a non trivial example with $|\text{total}(G)| = 1/2$

We also see that we can't allow general measures on each circle because we can take the optimal example below and redistribute mass from great circles with normal not pointing to the north pole to the one pointing to the north pole and yield a $\text{total}(G)$ larger than $1/2$.

Question The question is if the conjecture A is true under the definitions above or with some reasonable restrictions or even generalizations like asking if Conjecture B is true.

Commentary

The text below is a commentary for the actual question you can stop now.

Alternate Definition We could have defined a covering of type $G(\mu)$ as follows in stead. Let $\mu$ be a probability measure on $S^2$ inbedded in $E^3$ and associate to each great circle the normal pointing to the half sphere. This is a one to two corespondance between a great circle and a point on the sphere of $S^2$ imbedded in $E^3$. Now for each set $A\subset S^2$ then for each $p\in A$ there is a set $B_p \subset S^2$ e.g. a subset of great circles which cover $p$ such that we can defined the measure $\nu$ on $S^2$ through $$ \nu(A) = \mu_2(\bigcup_{p\in A} B_b) $$ Then for being of type $G$ we constrain $\mu$ so that the measure $\nu$ is a uniform probability measure on $S^2$. Note that $B_p$ is a great circle with normal pointing to $p$. If $\mu$ has a density $f_{\mu}$ one would then assume that should be $$ \int_{q \in B_p} f_{\mu}(q) dq = 1 / 4\pi $$

Example Now take the the uniform measure $u$ on $O(3)$. Take $s_1$ as the loop with normal $\hat z$ Then take a mapping $M : O(3) \rightarrow O(3)$ so that $M(O)=O$, if $O\hat z$ is on the upper sphere, or the mirroring in the plane defined by actual great circle, $\text{Mir}$ then for this $O$, $M(O) = \text{Mir}\, O$ if $O\hat z$ point to the lower sphere. This means that the image of $M(O)\hat z$ is on the upper sphere. This will create a measure on $O(3)$ such that the resulting density is uniform. also, with $H$ be the upper half sphere, then $$ \text{total}(G) = 2\int_{n\in H} n \,dS $$

Using spherical coordinates, then by symmetry we will have only a z component and it is $$ |\text{total}(G)| = \int_{\theta=0}^{\pi/2} \cos(\theta) \sin(\theta) d\theta =\frac{1}{2}\int_{\theta=0}^{\pi/2} \sin(2\theta) d\theta = 1/2 $$

Explanation Some explanation is needed for the non matematician following this question. So I will go through the definition again with commentary.

"Define a covering of a unit sphere, $S^2$, in $R^3$ as a covering of type $G(\mu_o)$,"

We will need to have a weighted sum of great circles. We want to add a weighted set of great circles and pure abstractly you need a probability measure on all the great circles e.g. a positive measure that's $\mu_o$.

"where $\mu_o$ is as a positive measure on the set of orthogonal 3 dimensional matrises, $O(3)$ s.t. $\mu_o(O(3)) = 1$, "

$O(3)$ is the set of all orthogonal matrices which are linear maps that maps $S^2$ to $S^2$ and is the result of rotations and mirroring along a plane through origo. Now what we will see below. Each great circles can be identified analytically with the help of an Orthogonal matrix so a weighted sum of Orthogonal matrices is a weighted sum of great circels

"Let $s_z$ be the great circle with unit normal $\hat z$ and $\mu_z$ the uniform measure on $s_z$ such that $\mu_z(s_z) = 1$."

The idea is to take a base great circle e.g $s_z$ and then generate all great circles by moving each point $v\in s_z$ to a new great circle. It turns out that orthogonal matrices e.g. rotations and mirrorings maps great circles to great circles on $S^2$. And not only this because you can map a unit normal to any other unit normal through a rotation each great circle on S^2 can be generated for some $O\in O(3)$ hence. Also because every $O\in O(3)$ can be written as $O=\exp(W)$ with $W^t=-W$ e.g. the transpose of $W$ is itself times -1. This means that you have $O=O(x,y,z)$ and a measure is no different to any other measure in the eucledian space.

"Now the elements of $O(3)$ maps a great circle to all great circles and we can define a map $F: O(3)\times s_1 \rightarrow S^2$ with $F(O,p) = O \,p$."

so each point $p\in S^2$ has a set of great circles going through this point and hence there is a set $A \in O(3)\times s_z$ so that for each $O\times v \in A$ we have $p = F(O,v) = O v$ Now this means that if we assume that $F$ is differentiable then we can define what the density of the induced measure through differential analysis.

"Then for the covering to be of type $G$ we define that the induced measure through $F(O,p)$ e.g. via the $\mu_o(dO)\mu_1(dp)$ and a change of variables, should be the uniform measure $\nu$ on $S^2$ such that $\nu(S^2) = 1$."

we make the measure in the great circles independent of the measure int the loop be saying that the weight of p should be the weight of specific great circle multiplied with the weight of a point on the specific great circle or equivalently the weight of a specific point on $s_z$. Now going to the differentiable continuum there is a mathematical machinery of changing variables in the integral e.g. that can calculates the area scaling and resulting measure needed given F. We have

$$ \nu(A) = \int_A d\nu = \int_{F^{-1}(A)} d\mu_o d\mu_z $$

  • I'm very honest with this question but I'm puzzled about the downvotes which was fair in earlier versions. Please comment what needs to be added or rewritten and I will change the formulation. The question is clear though objectively but perhaps not in my writing. – Stefan Jun 11 '17 at 8:29
  • A minor note on formatting. There exist appropriate symbols in LaTeX/MathJax for many standard functions, so if you prefix their names with backslash: \sin or \exp instead of just sin or exp, they will be rendered in upright font with spacing appropriate for a function name (see $\sin$ or $\exp$) instead of italics, looking like multiplication of three variables (like $sin$ and $exp$). – CiaPan Aug 17 '17 at 5:32
  • Thanks!, I addeed slashed math functions as suggested – Stefan Aug 17 '17 at 12:30

Consider https://en.wikipedia.org/wiki/Funk_transform http://www.wikiwand.com/en/Funk_transform

If we start from the alternative definition and let $\nu$ be the uniform measure of the unit sphere and let the measure $\mu$ be as in the definition. Assum that we restrict $\mu$ to have a $L^2$ density hence we have the densities $f_{\nu},f_{\mu}$. Now, $\mu$ being a measure for a covering of type $G$, means that $\mu$ can be associated to a measure on the sphere for the normals of the great circles. Now all great circles that passes through a point on the sphere has normals along the great circle that is 90 degrees tilted from this point and hence the integral of the density, $f_{\mu}$, along a great circle is constant modulo a set of measure zero that is we are saying that the Funck transform evaluates to a constant measure on the sphere. Now furthermore assume thet the $f_{\mu}$ density is in $L^2$ then it can be expanded in spherical harmonics e.g. $$ f_{\mu} = \sum_{n=0}^\infty \sum_{k=-n}^n \hat f_\mu(n,k)Y_n^k $$ $L^2$ wise. using that $Y^0_0$ is the constant density and using the links in the beginning of this answer and applying the Funk transform to $f_{\mu}$, we have:

$$ C Y_0^0 = f_{\nu} = F f_{\mu} = \sum_{n=0}^\infty \sum_{k=-n}^n P_n(0)\hat f_\mu(n,k)Y_n^k $$ $L^2$ - wise, with $P_n(0)$ nonzero for the even terms.

No the orthogonality of $Y_n^k$ means that all even terms, but $n=0,k=0$ $P_n(0) = 0$ Hence $\mu$ must have a uniform density + an odd function. Now for any measure $\mu$ on we can create the associative measure in the upper sphere $\mu_2$. by flipping normals on the negative half sphere to the upper half sphere and by doing that $\mu_2$ represent a new covering of type $G(\mu_2)$ such that $\text{total}(G(\mu)) \leq \text{total}(G(\mu_2))$ Now because for all functions, $f$, on the sphere, $$ f = f_{\text{even}}+f_{\text{odd}} = (f(p) + f(-p))/2 + (f(p) - f(-p))/2, $$ and that the odd part goes to zero in the Funk transform, we can consider the even function for $\mu_2$, that is $f_{even} = (f_{\mu_2}(p) + f_{\mu_2}(-p))/2$ which by the argument above must be of constant density. Now because $f_{\text{even}}$ is basically a union of two half sphere functions with and intersection of measure zero, so can be ignored $L^2$ wise and hence $f_{\mu_2}$ is also equivalent to a constant density on the upper sphere $L^2$ wise. Hence according to the argument above and the example in the question Conjecture A holds for the restriction of $\mu_2$ with a square integrable density. Note that the calculation of the $\text{total}(G)$ is the same for measures with densities that are equivalent $L^2$ wise hence the validity to consider $L^2$ spaces.

To try to tackle Conjecture B, we could relax the definition and consider that each great circle has a desity of it's own. Define a $n$-move by having $\{p_{i}\}_{i=1}^n$ points on the sphere and great circles $\{l_{i}\}_{i=1}^n$ so that $(l_n,l_1)$ intersects at $p_1$, $(l_1,l_2)$ intersects at $p_2$, ..., $(l_{n-1},l_n)$ intersects at $p_n$, then the move is to move a unit mass, $d$, from $l_{n}$ to $l_1$ at point $p_1$, a unit mass $d$ from $l_1$ to $l_2$ at point $p_2$ ... a unit mass $d$ from $l_{n-1}$ to $l_n$ at point $p_n$. This will preserve all measures and redistribute mass on each specific circle and if the conjecture holds before the move then it will hold after the move. So we have

Conjecture C If we for each great circle $l$ has a probability measure $\mu_{l}$ in stead of the uniform measure and if by applying a sequece of $n$-moves to the system results in a unifom meassure on each great circle then for such coverings, $$ \sup_G |\text{total}(G)| = 1/2 $$

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