Some novel constructions of optimal Gromov-Hausdorff-optimal correspondences between spheres
Saúl Rodríguez Martín


1/f Noise in the Heliosphere: A Target for PUNCH Science
    [
    September 9, 2024
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§ ABSTRACT 
In this article, as a first contribution, we  provide alternative proofs of recent results by Harrison and Jeffs which determine the precise value of the Gromov-Hausdorff (GH) distance between the circle 𝕊^1 and the n-dimensional sphere 𝕊^n (for any n∈ℕ) when  endowed with their respective geodesic metrics.
Additionally, we prove that the GH distance between 𝕊^3 and 𝕊^4 is equal to 1/2arccos(-1/4), thus settling the case n=3 of a conjecture by Lim, Mémoli and Smith.




§ INTRODUCTION


In this article we consider the problem of determining the Gromov-Hausdorff (GH) distances between 𝕊^1 and all other spheres, as well as  the GH distance between 𝕊^3 and 𝕊^4. Let us first recall some definitions. The Hausdorff distance between two subspaces A,B of a metric space (X,d_X) is defined as 

    d_(A,B)=
    max(
    sup_a∈ Ad_X(a,B),
    sup_b∈ Bd_X(b,A)
    ),

where d_X(x,A):=inf_a∈Ad_X(x,a) for all x∈X and A⊆X. If (X,d_X) and (Y,d_Y) are metric spaces, we will write X≅Y whenever X,Y are isometric. Then the GH distance between two metric spaces (X,d_X) and (Y,d_Y) is defined as 

    d_(X,Y)=
    inf{d^Z_H( X',Y');(Z,d_Z) metric space; X',Y'⊆Z; X'≅X;Y'≅Y}.

The GH distance takes values in [0,∞], and it satisfies the triangle inequality (cf. <cit.> Prop. 7.3.16). If X and Y are compact metric spaces, then d_GH(X,Y)=0 iff X and Y are isometric, and the value d_GH(X,Y) is sometimes described as a way to measure how far the spaces X and Y are from being isometric. 



If X is a dense subspace of Y, then d_GH(X,Y)=0.


Since it was introduced by Edwards (<cit.>, 1975) and independently by Gromov (<cit.> 1981), the GH distance has been instrumental in research areas such as the analysis of shapes formed by point cloud data <cit.>, convergence results for sequences of Riemannian manifolds <cit.>, differentiability in metric measure spaces <cit.>, and the robustness of topological invariants of metric spaces when they suffer small deformations <cit.>. 



Since being introduced by Edwards (<cit.>, 1975) and Gromov (<cit.>, 1981), GH distance has been used mostly as an asymptotic measure of the distance between spheres, see Include a few papers here, can take them from my S1SimpCon paper.






Recently, there has been growing interest (see e.g. <cit.>, and for a historical account of these efforts see <cit.>) in computing the exact value of the GH distance between certain simple metric spaces, specifically round spheres 𝕊^n⊆ℝ^n+1 (n∈ℕ:={1,2,…}) equipped with the geodesic metric d_𝕊^n. 
In their paper <cit.>, Lim, Mémoli and Smith provided some upper and lower bounds for d_GH(𝕊^n,𝕊^m) for all n,m∈ℕ and they gave exact values for the pairwise distances between 𝕊^1,𝕊^2 and 𝕊^3. 
Some bounds for the GH distances between spheres were further improved in <cit.>. Most importantly for our purposes, a concrete case of <cit.> implies that  
    2d_GH(𝕊^n,𝕊^n+1)≥ζ_n:=arccos(-1/n+1),
    n∈ℕ,

and in <cit.> it was proved that 

    d_ GH(𝕊^1,𝕊^2n)≥π n/2n+1 and 
    d_ GH(𝕊^1,𝕊^2n+1)≥π n/2n+1.
    n∈ℕ.


In <cit.>, Harrison and Jeffs proved that the inequalities from <Ref> are actually equalities: 


For any integer n≥1, d_ GH(𝕊^1,𝕊^2n)=π n/2n+1.



For any integer n≥1, d_ GH(𝕊^1,𝕊^2n+1)=π n/2n+1.

In this article we prove three main results.

    
  * We give an alternative proof of <Ref>. The construction we use is an immediate generalization of the one in <cit.>. Our proof was found independently of <cit.>, but it is similar to it, as we explain in <Ref>.
    
  * We give a proof of <Ref> which is distinct from (and considerably shorter than) the one in <cit.>. The proof in <cit.> uses an `embedding-projection correspondence' (see <cit.>) to prove <Ref>, while we instead use a certain modification of the alternative  construction we developed for proving <Ref>.
    
    
    
    
  * We also establish the following novel result: 
    
    
    
    The distance 
    d_ GH(𝕊^3,𝕊^4) is 1/2ζ_3.
    
    
    This proves case n=3 of <Ref>, of which cases n=1,2 were proved in <cit.>:

    
    
    For all n∈ℕ we have d_(𝕊^n,𝕊^n+1)
    =1/2ζ_n.
    
    The proof strategy  of <Ref> is also valid for cases n=1,2 of <Ref>, and perhaps it could be adapted to cases n=4,5,6 (see <Ref>).
    




The definition of d_GH is hard to work with; we now recall (cf. 7.3 in <cit.>) an equivalent definition based on correspondences between sets. Recall that a relation between two sets X,Y is a subset of X×Y. We will say a relation R⊆X×Y is a correspondence between X and Y if π_X(R)=X and π_Y(R)=Y, where π_X:X×Y→X and π_Y:X×Y→Y are the coordinate projections.



If (X,d_X) and (Y,d_Y) are metric spaces, we will define the distortion of a nonempty relation R⊆X×Y as 

    dis(R):=sup{|d_X(x,x')-d_Y(y,y')|;(x,y),(x',y')∈ R}∈[0,∞].

In Theorem 7.3.25 of <cit.> it is proved that, if (X,d_X),(Y,d_Y) are metric spaces, 

    d_GH(X,Y)=1/2inf{dis(R);R⊆X×Y correspondence between X and Y}.


So the distortion of any correspondence R between X and Y is an upper bound for  2d_GH(X,Y). 



The graph of a function ϕ:X→Y is a relation, and it will be a correspondence between X and Y iff ϕ is surjective. We then let 

    (ϕ):=sup{|d_X(x,x')-d_Y(ϕ(x),ϕ(x'))|;x,x'∈X}
 be the distortion of the graph of ϕ.




We can slightly relax the definition of correspondence: we say R⊆X×Y is a metric correspondence between (X,d_X) and (Y,d_Y) if the projections π_X(R),π_Y(R) are dense in X, Y respectively. If R is a metric correspondence between X and Y, then the triangle inequality for d_GH and <Ref> imply that 1/2dis(R) is an upper bound for d_GH(X,Y).





Thanks to <Ref>, in order to prove <Ref> it suffices to construct metric correspondences between spheres having adequate distortions. 

All of the correspondences we construct are related to constructions in <cit.>, and they have several points in common.  Firstly, all of them use the helmet trick (<Ref> below), which tells us that, if n,m∈ℕ and H^n_+:={x∈𝕊^n;x_n+1≥0}, then for any correspondence R⊆ H^n_+×𝕊^m there is a correspondence R'⊆𝕊^n×𝕊^m containing R and satisfying (R')=(R). When estimating the distance d_(𝕊^n,𝕊^m), this allows us to use correspondences in H^n_+×𝕊^m, which are easier to construct than correspondences in 𝕊^n×𝕊^m.

Secondly, our constructions all use regular simplices inscribed in 𝕊^n, by which we mean a set of n+2 distinct points p_1,…,p_n+2∈𝕊^n such that d_𝕊^n(p_i,p_j) is the same for all (i,j) with i≠ j. A list of useful properties of such simplices can be found in <Ref>.



  §.§.§ Structure of the paper


In <Ref> we prove <Ref>, which has the easiest proof of our three main results. To do this, we devise a metric correspondence R_2n⊆ H^2n_+×𝕊^1 with distortion 2π n/2n+1. The correspondence R_2n is an immediate generalization of a construction from <cit.>. 



In <Ref> we prove <Ref>. 
To explain the approach we take in that section: given <Ref> and the fact that d_(𝕊^1,𝕊^2n)=π n/2n+1 (proved in 
<Ref>), one could optimistically conjecture that d_(𝕊^1,𝕊^2n+1)=d_(𝕊^1,𝕊^2n)=π n/2n+1. Therefore, a natural approach is using the correspondence from <Ref> to create some correspondence R_2n+1⊆ H^2n+1_+×𝕊^1. And indeed, we start with a natural adaptation of the correspondence R_2n to dimension 2n+1 and after `rotating' it in a small subset B⊆ H^2n+1_+ (shown in <Ref>), we obtain a correspondence R_2n+1⊆ H^2n+1_+×𝕊^1 with distortion 2π n/2n+1. The way in which we rotate the correspondence in the set B⊆ H^2n+1_+ is inspired by the arguments from <cit.>.


In <Ref> we prove <Ref> using a surjective map F:H^4_+→𝕊^3 with distortion ζ_3=arccos(-1/4). The construction of the map F is not particularly complicated, but the author has only found proofs that (F)=ζ_3 using computer assistance, see <Ref>.

Roughly speaking, the map F is obtained as an `interpolation' between two functions F',F”:H^4_+→𝕊^3 described below and depicted in <Ref>.

We obtain F':H^4_+→𝕊^3 by taking points p_1,…,p_n+2 forming a regular simplex inscribed in 𝕊^3≡{x∈𝕊^4;x_n+2=0} and for each x∈ H^4_+ defining F'(x)=p_i, where i∈{1,…,n+2} is chosen so that p_i is as close as possible to x. The map F”:H^4_+→𝕊^3 is obtained by, for each p∈ H^4_+, choosing F”(p) to be a point of 𝕊^3 which minimizes the distance to p. So F(p) is the `projection' of p to 𝕊^3, except if p is the north pole N:=(0,0,0,0,1).


We want our function F to have distortion ζ_3. The map F' has distortion ζ_3, but it does not induce a correspondence. The map F” has the opposite problem: it is surjective but has distortion π, because for points x,x' very close to the north pole,  d_𝕊^3(F(x),F(x'))-d_𝕊^4(x,x') may be as close to π as we want. In the proof of <cit.>, they find d_GH(𝕊^1,𝕊^2) via a surjective map ϕ:H^2_+→𝕊^1 which is equal to (the lower dimensional analogue of) F' for points in the equator and equal to F” in the rest of H^2_+. 
The analogous map in higher dimension, ϕ_n:𝕊^n+1→𝕊^n, has distortion η_n>ζ_n (η_n is defined in <Ref>) for n≥3, so it cannot be used to find d_GH(𝕊^n+1,𝕊^n). 

Our map F is defined as F' for points near the north pole (so that its distortion is not π), F” for points in 𝕊^3 (so that F is surjective) and an interpolation between F' and F” in between. A depiction of the map F can be found in <Ref>.
In <Ref> we define a higher dimensional analogue of F, which we call F_n:𝕊^n+1→𝕊^n.
The map F_n has distortion >ζ_n when n≥7 (see <Ref>), so in that case it cannot be used to prove <Ref>.


Do we have (F_n)=ζ_n for n=4,5,6?


In the rest of the introduction we explain how we use computer assistance to prove some inequalities needed in <Ref>, as well as give some ideas that could be used to prove <Ref>.



In <Ref>, when we prove that the map F:𝕊^4→𝕊^3 has distortion ζ_3, we need to prove inequalities of the form f(x_1,x_2)≥ 0, where x_i are in intervals [a_i,b_i]⊆ℝ. The expression for the function f is sometimes very complicated, and it is not clear how to give a clean proof of the inequality. However, we can prove f(x_1,x_2)≥0 using brute force if two conditions are met:

    
  *  The function f is bounded below by some positive constant ε.
    
  * The function f is uniformly continuous: there is a constant δ (which we can compute explicitly) such that, if |x_1-x_1'|,|x_2-x_2'|<δ, then |f(x_1,x_2)-f(x_1',x_2')|<ε.

Under these conditions we can use a computer program to check that f(x_1,x_2)>ε for all points (x_1,x_2) in some finite set G (a grid) inside [a_1,b_1]×[a_2,b_2] such that every point of [a_1,b_1]×[a_2,b_2] is at distance <δ of some point of G, concluding the inequality.


We have used this method to check three crucial inequalities in <Ref>. The python code can be found in the GitHub repository <cit.>; both files that check the inequality and files that output a 3D plot of the function f(x,y) are included. For the most complicated inequality we have explained in detail how to obtain the uniform continuity constants in <Ref> (see page CheckUnifCont).






 
If the answer to <Ref> is positive, it should be theoretically  provable using the same ideas of <Ref>[One first needs to use mathematical arguments to ensure that conditions <ref> and <ref> of <Ref> are satisfied.]. Indeed, <Ref> reduces to the inequality

    |d_𝕊^n+1(x,x')-d_𝕊^n(F(x),F(x'))|<ζ_nx,x'∈𝕊^n+1, for n=4,5,6.

However, the author has not been able to find a computer program efficient enough to prove dist(F_n)=ζ_n using this strategy in a reasonable amount of time, the main obstacle being that grids in 𝕊^n have too many points for n≥4.




















  
Acknowledgements. Special thanks to Facundo Mémoli for introducing the author to the study of GH distances and providing guidance and extensive feedback
while writing this article. The author gratefully acknowledges support from the
grants BSF 2020124 and NSF CCF AF 2310412. Also thanks to Daniel Hurtado, River Li and Pablo Vitoria for helping improve to some of the arguments in this article. The large language models ChatGPT-4 and Claude 3.5-Sonnet were used to aid in writing part of the python code.




§ NOTATION AND PRELIMINARIES


Throughout most of this article, the metric spaces we will be studying are the unit spheres 𝕊^n with the spherical distance, which will be denoted by d_𝕊^n. That is, considering 𝕊^n:={x∈ℝ^n+1;∑_i=1^n+1x_i^2=1}, for all x,y∈𝕊^n we will have 

    d_𝕊^n(x,y):=arccos(⟨ x,y⟩), where ⟨ x,y⟩:=∑_i=1^n+1x_iy_i.
 Thus distances take values in the interval [0,π]. For any two non-antipodal points x,x'∈𝕊^n we will denote the (unique) geodesic segment from x to x' by [x,x'].  Also, if p,q,r∈𝕊^n+1 are distinct points, then we denote by ∠ pqr∈[0,π] the angle at q of the spherical triangle with vertices p,q,r, as specified by the spherical cosine rule.


[Antipodal sets]
For any subset X⊆𝕊^n⊆ℝ^n+1, we define 

    -X:={-x;x∈X}.

Similarly, for any relation R⊆𝕊^n×𝕊^m, we define

    -R:={(-x,-y);(x,y)∈ R}.



The following is a version of Lemma 5.5 of <cit.> for relations:

Let R⊆𝕊^n×𝕊^m be a relation and let -R={(-x,-y);(x,y)∈ R}. Then the relation 

    -R∪ R⊆𝕊^n×𝕊^m
 has the same distortion as R.



For any k and for any two points x,y∈𝕊^k, we have d_𝕊^k(x,y)=π-d_𝕊^k(x,-y). So for any two pairs (x,y),(x',y')∈𝕊^n×𝕊^m, 

    |d_𝕊^n(x,-x')-d_𝕊^m(y,-y')|=|π-d_𝕊^n(x,x')-(π-d_𝕊^m(y,y'))|=|d_𝕊^n(x,x')-d_𝕊^m(y,y')|.

Thus, any distortion |d_𝕊^n(x,x')-d_𝕊^m(y,y')| between two pairs of points (x,y) and (x',y') of -R∪ R is also attained between two pairs of points of R, which proves dis(-R∪ R)=dis(R).



Fix a point q∈𝕊^n+1. For any nonempty X⊆{x∈𝕊^n+1;⟨ x,q⟩=0}, we define the cone C_qX as the union of geodesic segments 

    C_qX:=⋃_x∈X[x,q].



As proved in Lemma 6.1 of <cit.>, the diameter of a cone C_qX is given by

    diam(C_qX)=max(π/2,diam(X)).


For each n∈ℕ, we can find points p_1,…,p_n+2∈𝕊^n which form a regular simplex in ℝ^n+1. We can associate to them the open Voronoi cells

    V_i:={x∈𝕊^n;d_𝕊^n(x,p_i)<d_𝕊^n(x,p_j) for all j≠ i}, i=1,2,…,n+2.

We will say a subset A⊆𝕊^n is convex when any geodesic segment between two points of A is contained in A (so an open hemisphere is convex, and a closed hemisphere is not). The convex hull of a set A⊆𝕊^n is the intersection of all convex sets containing A.

We will need some properties of regular simplices inscribed in 𝕊^n (some of them are proved in <cit.>;  see also Section 3 of <cit.> for related results):
 Let (p_i)_i=1^n+2 and (V_i)_i=1^n+2 be as above. Then

    
  * d_𝕊^n(p_i,p_j)=ζ_n:=arccos(-1/n+1) for i≠ j.
    
    
  *  For all i, V_i is the convex hull inside 𝕊^n of the set {-p_j;j≠ i}.
    
    
  * The diameter of the Voronoi cells V_i is 
    
    η_n:={[  arccos(-n+1/n+3)         for n odd; arccos(-√(n/n+4))       for n even. ].

    
    
  * The Voronoi cell V_i satisfies
    
    B_𝕊^n(p_i,ζ_n/2)⊆ V_i⊆ B_𝕊^n(p_i,π-ζ_n),

    where B_𝕊^n(x,r) denotes the ball centered at x of radius r in 𝕊^n.






<Ref> are discussed in Remarks 6.4 and 6.5 of <cit.>. To prove <Ref> note that, as p_1,…,p_2n+2 are unit vectors forming a regular simplex centered at 0, we have ∑_i=1^n+2p_i=0. Moreover, by symmetry the scalar products ⟨ p_i,p_j⟩ take a single value κ for any pair (i,j) with i≠ j. This value κ can be obtained from the equation

    0=⟨ p_i,0⟩=
    ⟨ p_i,∑_j=1^n+2p_j⟩
    =
    1+(n+1)κ.

So cos(d_𝕊^n(p_i,p_j))=-1/n+1 for all i≠ j, as we wanted.

Finally we prove <Ref>. The first containment is a consequence of the fact that, if d_𝕊^n(x,p_i)<ζ_n/2, then for any j≠ i we have

    d_𝕊^n(x,p_j)≥ 
    d_𝕊^n(p_i,p_j)-d_𝕊^n(x,p_i)
    >ζ_n-ζ_n/2>d_𝕊^n(x,p_i),

so x∈ V_i. For the second containment, note that B_𝕊^n(p_i,π-ζ_n) is convex (balls of radius <π/2 are convex in 𝕊^n) and contains the points -p_j for all j≠ i. So by <Ref> we have V_i⊆B_𝕊^n(p_i,π-ζ_n), and as V_i is open we also have V_i⊆ B_𝕊^n(p_i,π-ζ_n).












§ DISTANCE FROM 𝕊^1 TO EVEN DIMENSIONAL SPHERES

This section is devoted to proving <Ref> by constructing a correspondence between 𝕊^1 and 𝕊^2n with distortion 2π n/2n+1. This construction generalizes the one used in Appendix D of <cit.> to find d_GH(𝕊^1,𝕊^2).


Let us start with some notation:

    𝕊^2n:={x∈ℝ^2n+1;∑_i=1^2n+1x_i^2=1}


    H^2n_+:={x∈𝕊^2n;x_2n+1≥0}


    𝕊^2n-1:={x∈𝕊^2n;x_2n+1=0}


Let p_1,…,p_2n+1∈𝕊^2n-1 be the vertices of a regular simplex in ℝ^2n×{0}⊆ℝ^2n+1 inscribed in 𝕊^2n-1. For i=1,…,2n+1, consider the Voronoi cells 

    V^2n-1_i:={
    x∈𝕊^2n-1;d_𝕊^2n(x,p_i)<d_𝕊^2n(x,p_j) for all j≠ i
    },


    V^2n_i:={
    x∈ H^2n_+;d_𝕊^2n(x,p_i)<d_𝕊^2n(x,p_j) for all j≠ i
    }.






Note that V^2n_i is obtained by taking a cone (see <Ref>) from V^2n-1_i with respect to the point (0,…,0,1)∈ℝ^2n+1, so by Propositions <ref> and <ref><ref> we have

    diam(V^2n_i)=diam(V^2n-1_i)=arccos(-2n/2n+2)=arccos(-n/n+1)


Also, letting q_1,…,q_2n+1 be the vertices of a regular 2n+1-gon inscribed in 𝕊^1, we define the Voronoi cells

    W_i:={
    y∈𝕊^1;d_𝕊^1(y,q_i)<d_𝕊^1(y,q_j) for all j≠ i
    },

which are intervals of length 2π/2n+1.

We will need the fact that the diameter of V^2n_i is  at most 2π n/2n+1:

For all positive x∈[1,∞) we have arccos(-x/x+1)≤2π x/2x+1.

The following elegant proof of <Ref> is due to Pablo Vitoria.[This proposition plays a similar role to <cit.>.]


Note that for any x∈(1,∞) we have

    arccos(-x/x+1)≤2π x/2x+1-x/x+1≥cos(2π x/2x+1)x/x+1≤cos(π/2x+1)

Changing variables to y=π/2x+1 (so that x/x+1=2π/π+y-1), it will be enough to check that cos(y)≥2π/π+y-1 for all y∈[0,π/3]. But this inequality follows from the facts that:

    
  * The functions cos(y) and 2π/π+y-1 agree at y=0,π/3.
    
  * In the interval [0,π/3] the function cos(y) is concave while 2π/π+y-1 is convex.

 




Consider the following relation R_2n⊆ H^2n_+×𝕊^1:

    R_2n:=(⋃_i=1^2n+1V^2n_i×{q_i})
    ⋃(⋃_i=1^2n+1{p_i}× W_i)

Note that the projection of R_2n onto its first coordinate is dense in H^2n_+, because it is the union of the Voronoi cells V_i^2n for i=1,…,2n+1. Similarly, the projection of R_2n to its second coordinate is dense in 𝕊^1. Thus, R_2n∪-R_2n is a metric correspondence between 𝕊^2n and 𝕊^1. So thanks to the Helmet trick (<Ref>), we have

    d_GH(𝕊^2n,𝕊^1)≤1/2dis(R_2n∪-R_2n)=1/2dis(R_2n).


So to prove <Ref> it will be enough to prove that dis(R_2n)≤2π n/2n+1. That is, we need to prove that if (x,y) and (x',y') are in R_2n, then 

    |d_𝕊^2n(x,x')-d_𝕊^1(y,y')|≤2π n/2n+1.

To prove <Ref> we will divide the analysis into 6 cases. 

    
  * x,x'∈ V^2n_i,y=y'=q_i for some i∈{1,…,2n+1}. In this case 
    
    |d_𝕊^2n(x,x')-d_𝕊^1(y,y')|=d_𝕊^2n(x,x')≤diam(V^2n_i)=arccos(-n/n+1)≤2π n/2n+1.

    
    
  * x∈ V^2n_i,x'∈ V^2n_j, y=q_i,y'=q_j for some i≠ j. In this case we have d_𝕊^1(y,y')∈[π/2n+1,2π n/2n+1], which implies <Ref>.
    
    
  * x∈ V^2n_i,x'=p_i,y=q_i,y'∈ W_i for some i. Then we have
    
    d_𝕊^1(y,y')=d_𝕊^1(q_i,y')≤π/2n+1

    
    d_𝕊^2n(x,x')=d_𝕊^2n(x,p_i)<π-ζ_n (see Proposition <ref><ref>),

    so |d_𝕊^2n(x,x')-d_𝕊^1(y,y')|≤max(π/2n+1,π-ζ_n)<π/2.
    
    
  * x∈ V^2n_i,x'=p_j,y=q_i,y'∈ W_j for some i≠ j. Then we have
    
    d_𝕊^1(y,y')=d_𝕊^1(q_i,y')>π/2n+1

    
    d_𝕊^2n(x,x')=d_𝕊^2n(x,p_j)>ζ_2n-1/2=1/2arccos(-1/2n) (see Proposition <ref><ref>),

    but for n≥1 we have 1/2arccos(-1/2n)≥π/2n+1, because we have equality for n=1 and for n≥2 we have 1/2arccos(-1/2n)≥π/4≥π/2n+1. So both d_𝕊^2n(x,x') and d_𝕊^1(y,y') are in the interval [π/2n+1,π], thus |d_𝕊^2n(x,x')-d_𝕊^1(y,y')|≤2π n/2n+1 in this case.
    
    
    
  * x=x'=p_i,y,y'∈ W_i for some i. Then d_𝕊^2n(x,x')=0 and d_𝕊^1(y,y')<2π/2n+1 so we are done. 
    
    
  * x=p_i,x'=p_j, y∈ W_i, y'∈ W_j for some i≠ j. Then d_𝕊^2n(x,x')=ζ_n∈[π/2n+1,2π n/2n+1], which implies <Ref>.
    









Our construction is similar to that of <cit.>. Both are extensions of the idea from <cit.> of considering finite subsets 𝒫,𝒬 of 𝕊^2n,𝕊^1 of the same cardinality and using a bijection between 𝒫 and 𝒬 to construct a correspondence between 𝕊^2n and 𝕊^1. The main difference is that in our case we need the helmet trick (as the sets V_i^2n are Voronoi cells inside H_+^2n, not 𝕊^2n) and that we use the vertices of a 2n+1-simplex as centers of the Voronoi cells in 𝕊^2n, while in <cit.> they use an orthonormal basis of vectors, together with their antipodals. However, in both our construction and the one in <cit.>, the correspondence between 𝕊^1 and 𝕊^2n has distortion 2π n/2n+1 for the same reason: there are pairs of points in 𝕊^n+1 which are arbitrarily close to each other  which are mapped to points at distance 2π n/2n+1 in 𝕊^1. 







§ DISTANCE FROM 𝕊^1 TO ODD DIMENSIONAL SPHERES


This section is devoted to proving <Ref>. 



 §.§ An optimal correspondence between 𝕊^2n+1 and 𝕊^1


Let us start with some notation which we will use in this section:

    𝕊^2n+1={x∈ℝ^2n+2;∑_j=1^2n+2x_j^2=1}


    𝕊^2n:={x∈𝕊^2n+1;x_2n+2=0}


    𝕊^2n-1:={x∈𝕊^2n+1;x_2n+1=x_2n+2=0}


    H^2n+1_+:={x∈𝕊^2n+1;x_2n+2≥0}


    H^2n_+:={x∈𝕊^2n;x_2n+1≥0}


    H^2n+1_++:={x∈𝕊^2n+1;x_2n+1≥0,x_2n+2≥0}.


    H^2n+1_-+:={x∈𝕊^2n+1;x_2n+1≤0,x_2n+2≥0}.

These sets are illustrated in <Ref>. 




Now, for a dense subset D of H_+^2n+1, we will define a map Φ:D→𝕊^1 with distortion 2π n/2n+1 and such that, if G_Φ is the graph of Φ, then G_Φ∪ -G_Φ⊆𝕊^2n+1×𝕊^1
is a metric correspondence between 𝕊^2n+1 and 𝕊^1. By the helmet trick (<Ref>) this will establish <Ref>.

We describe the map Φ in detail after <Ref>; we first give a more informal description of it.


Let N=(0,…,0,1)∈ℝ^2n+2 be the north pole of 𝕊^2n+1. We will denote points x∈ H_+^2n+1∖{N} as (p,α)∈𝕊^2n×[0,π/2), where p∈𝕊^2n⊆𝕊^2n+1 is the point of 𝕊^2n closest to p and α∈[0,π/2) is the geodesic distance from p to 𝕊^2n.


Now, note that the correspondence R_2n from the <Ref> was the union of the graphs of two maps, one map f:𝕊^2n→{q_0,…,q_2n}⊆𝕊^1 and one map g:𝕊^1→{p_0,…,p_2n}⊆𝕊^2n. Then, the restriction of our map Φ to 𝕊^2n will be just the map f: we let Φ(p,0)=f(p) for all p∈𝕊^2n.

In fact, if p∈ H^2n_+ (so p is in the northern hemisphere of 𝕊^2n), we let Φ(p,α)=f(p) for all α∈[0,π/2). That determines Φ(x) for all x∈ H^2n+1_++. 
For points of H^2n+1_-+, that is, points of the form (p,α) where p is in the southern hemisphere of 𝕊^2n, we define Φ(p,α)=f(p)· e^i·min(α,π/2n+1)∈𝕊^1. 
So for fixed p, as α increases from 0 to π/2n+1, Φ(p,α) follows a unit speed geodesic in 𝕊^1 from f(p) to f(p)· e^i π/2n+1, and then Φ(p,α) is equal to f(p)· e^iπ/2n+1 for all α∈[π/2n+1,π/2). 
We include a depiction of the map Φ in <Ref> in the case when 2n+1=3. 
















Now we will define the map Φ in detail. Let p_0,…,p_2n be the vertices of a regular simplex in ℝ^2n×{(0,0)} inscribed in 𝕊^2n-1, and for j=0,…,2n we define the Voronoi cells

    V^2n-1_j:={x∈𝕊^2n-1;d_𝕊^2n+1(x,p_j)<d_𝕊^2n+1(x,p_k) for all k≠ j}


    V^2n_j:={x∈ H^2n_+;d_𝕊^2n+1(x,p_j)<d_𝕊^2n+1(x,p_k) for all k≠ j}


    V^2n+1_j={x∈ H^2n+1_++;d_𝕊^2n+1(x,p_j)<d_𝕊^2n+1(x,p_k) for all k≠ j}

Note that V_j^2n is obtained from V_j^2n-1 by performing the cone operation described in <Ref> (and removing the point (0,…,1,0)∈ℝ^2n+2), and V_j^2n+1 is similarly obtained from V_j^2n. Thus Propositions <ref> and <ref><ref> imply that, for all j=1,…,2n+1,

    diam(V_j^2n+1)
    =diam(V_j^2n)
    =diam(V_j^2n-1)
    =arccos(-2n/2n+2).




In this section, we identify 𝕊^1 with {z∈ℂ;|z|=1}, and for k=0,…,2n we let q_k=e^2π ik/2n+1 denote the (2n+1)-th roots of unity. Note that the map f:𝕊^2n→𝕊^1 given by

    [            f:          𝕊^2n             →          𝕊^1;;                      V_j^2n             ↦           q_j for j=0,…,2n.;                     -V_j^2n             ↦          -q_j for j=0,…,2n. ]

has distortion at most 2π n/2n+1: indeed, f is obtained from applying the helmet trick to the map 

    [          f_+:        H^2n_+             →          𝕊^1;;                      V_j^2n             ↦           q_j for j=0,…,2n, ]

and the map f_+ has distortion at most 2π n/2n+1 because when seen as a relation, it is 

    ⋃_j=0^2nV^2n_j×{b_j},

which is contained in the relation R_2n with distortion 2π n/2n+1 that we used in <Ref>.


Now, using <Ref>, we will repeatedly use the fact that for any fixed p∈𝕊^2n, the curve γ:[0,π/2]→𝕊^2n+1;γ(t)=(p,t) is a unit speed geodesic. Finally, we define 

    D=(⋃_j=0^2nV^2n+1_j)∪(⋃_j=0^2n-V^2n+1_j).

And consider the map

    [                     Φ:                      D                      →                   𝕊^1;                       ;                                         (p,α)                      ↦                    q_k          if p∈ V_k^2n.;                                         (p,α)                      ↦ -q_k· e^imin(α,π/2n+1)          if p∈-V_k^2n.;                        ]

For simplicity we will write Φ(p,α) instead of Φ((p,α)) for the image of the point (p,α). Also note that the condition p∈ V_k^2n is equivalent to (p,α)∈ V_k^2n+1.


The relation G_Φ∪-G_Φ⊆𝕊^2n+1×𝕊^1 is a metric correspondence (as defined in <Ref>) between 𝕊^2n+1 and 𝕊^1.



Firstly, the domain D of Φ is dense in H^2n+1_+. So D∪-D, which is the projection of G_Φ∪-G_Φ to 𝕊^2n+1, is dense in 𝕊^2n+1. Secondly, the image of Φ is the following set I (see <Ref> for the case 2n+1=5)

    I:=⋃_k=0^2nI_k⊆𝕊^1,

where I_k is the following interval of length π/2n+1 having q_k as one endpoint:

    I_k:={e^ix;x∈[π(2k-1)/2n+1,2π k/2n+1]} for k=0,1,…,2n.


So I∪-I, which is the projection of G_Φ∪-G_Φ to 𝕊^1, is the entire 𝕊^1, concluding the proof that G_Φ∪-G_Φ is a metric correspondence between 𝕊^2n+1 and 𝕊^1.





Note that the restriction of Φ to 𝕊^2n is just the function f from <Ref>, which has distortion ≤2π n/2n+1. That is, for any (p,0),(p',0)∈ D we have 

    |d_𝕊^1(Φ(p,0),Φ(p',0))-d_𝕊^2n+1((p,0),(p',0))|≤2π n/2n+1.





Also note that Φ maps most points of D to {q_0,…,q_2n}; letting A=Φ^-1({q_0,…,q_2n}), the set B:=D∖ A of points of D mapped by Φ outside of {q_0,…,q_2n} is essentially the set of points of H^2n+1_-+ at distance <π/2n+1 of 𝕊^2n: 

    B=D∖ A
    =⋃_j=0^2n{(p,α)∈-V^2n+1_j;α<π/2n+1}.












 §.§ Proof that Φ has distortion 2π n/2n+1

We want to prove that, for any two points (p,α) and (p',α') in D⊆ H^2n+1_+, we have

    |d_𝕊^1(Φ(p,α),Φ(p',α'))-d_𝕊^2n+1((p,α),(p',α'))|≤2π n/2n+1.

That means that neither of the following two inequalities can happen:

    d_𝕊^1(Φ(p,α),Φ(p',α'))-d_𝕊^2n+1((p,α),(p',α'))>2π n/2n+1.



    d_𝕊^2n+1((p,α),(p',α'))-d_𝕊^1(Φ(p,α),Φ(p',α'))>2π n/2n+1.












  §.§.§ Inequality <ref>

We will assume that (p,α),(p',α')∈ D satisfy <Ref>
and obtain a contradiction.

Firstly, we may assume without loss of generality that Φ(p,α)∈ I_0 (I_0 is defined in <Ref>; see <Ref>). This, along with the fact that d_𝕊^1(Φ(p,α),Φ(p',α'))≥2n/2n+1π=π-π/2n+1, implies that Φ(p',α') is either in I_n or in I_n+1. We can in fact assume Φ(p',α')∈ I_n+1, swapping (p,α) and (p',α') if not. So we have

    Φ(p,α)∈ I_0  (interval between -q_n and q_0).


    Φ(p',α')∈ I_n+1 (interval between -q_0 and q_n+1).


Note that, as d_𝕊^1(Φ(p,α),Φ(p',α'))>2π n/2n+1, the point Φ(p',α') cannot be exactly q_n+1, as the entire interval I_0 lies at distance ≤2π n/2n+1 from q_n+1. This implies that (p',α')∈ B, that is, α'<π/2n+1 and (p',α')∈ H^2n+1_-+. Now consider the function 

    [                                          h_1:                                    [0,π/2n+1]                                             →                                            ℝ;;                                                                                           t                                             ↦ d_𝕊^1(Φ(p,α),Φ(p',t))-d_𝕊^2n+1((p,α),(p',t)). ]

Then h_1 is decreasing: this is because we have Φ(p',t)=e^i(π+t), so d/dtd_𝕊^1(Φ(p,α),Φ(p',t))=-1, while the function t↦ d_𝕊^2n+1((p,α),(p',t)) is 1-Lipschitz, owing to t↦(p',t) being a unit speed geodesic. So we conclude that 

    d_𝕊^1(Φ(p,α),Φ(p',0))-d_𝕊^2n+1((p,α),(p',0))≥ d_𝕊^1(Φ(p,α),Φ(p',α'))-d_𝕊^2n+1((p,α),(p',α'))>2π n/2n+1.


This implies that d_𝕊^2n+1((p,α),(p',0))<π/2n+1, so we have α<π/2n+1. We also have d_𝕊^2n+1((p,0),(p',0))≤π/2 (if not, d_𝕊^2n+1((p,α),(p',0)) would be more than π/2), so we can easily deduce (e.g. by the cosine rule) that 


    d_𝕊^2n+1((p,0),(p',0))≤ d_𝕊^2n+1((p,α),(p',α')).



Now, we divide the analysis into 2 cases according to the quadrant to which (p,α) belongs:

    
  * Suppose (p,α)∈ H^2n+1_++.
    Then Φ(p,α)=Φ(p,0), so by <Ref> we have
    
    d_𝕊^1(Φ(p,0),Φ(p',0))-d_𝕊^2n+1((p,0),(p',0))>2π n/2n+1,

    contradicting <Ref>.



    
  * Suppose (p,α)∈ H^2n+1_-+. We also know that α∈[0,π/2n+1]. However, the function
    
    [                                         h_2:                                   [0,π/2n+1]                                            →                                           ℝ;;                                                                                         t                                            ↦ d_𝕊^1(Φ(p,t),Φ(p',0))-d_𝕊^2n+1((p,t),(p',0)) ]

    is increasing because d/dtd_𝕊^1(Φ(p,t),Φ(p',0))=1 and the function d_𝕊^2n+1((p,t),(p',0)) is 1-Lipschitz, due to t↦(p,t) being unit speed geodesic. So we have

    h_2(π/2n+1)≥ h_2(α)>2π n/2n+1.

however, this implies that d_𝕊^2n+1((p,π/2n+1),(p',0))<π/2n+1, which is impossible because (p,π/2n+1) is at distance π/2n+1 from 𝕊^2n, and  𝕊^2n    contains the point (p',0).








  §.§.§ Second inequality

We will assume that (p,α),(p',α')∈ D satisfy <Ref>
and obtain a contradiction.

<Ref> implies that d_𝕊^1(Φ(p,α),Φ(p',α'))<π/2n+1, so Φ(p,α) and Φ(p',α') are in the same subinterval of 𝕊^1 of length π/2n+1. We can assume that this interval is I_0, so it has  q_0 and -q_n as endpoints. Also note that 

    α+α'   =(π/2-d_𝕊^2n+1(e_n+2,(p,α)))+(π/2-d_𝕊^2n+1(e_n+2,(p',α')))
       ≤π-d_𝕊^2n+1((p,α),(p',α'))
    <π/2n+1.


So both α,α' are 
less than π/2n+1. We consider three cases:

    
  * Both (p,α) and (p',α') are in H^2n+1_++. Then (p,α),(p',α') are both in V^2n+1_0. So by <Ref> and <Ref> we have
    
    d_𝕊^2n+1((p,α),(p',α'))≤diam(V^2n+1_0)=arccos(-2n/2n+2)<2n/2n+1π,

    contradicting <Ref>.



    
  * Both (p,α) and (p',α') are in H^2n+1_-+. Then (p,α),(p',α') are both in -V_n^2n+1, so as in the previous case we have d_𝕊^2n+1((p,α),(p',α'))<2n/2n+1π,
    contradicting <Ref>.





    
    
  * We have (p,α)∈ H^2n+1_++ and (p',α')∈ H^2n+1_-+. So Φ(p,α)=q_0 and, as α'<π/2n+1, we have Φ(p',α')=-q_ne^iα'. Now, the function

    [                                          h_3:                                    [0,π/2n+1]                                             →                                            ℝ;;                                                                                           t                                             ↦ d_𝕊^2n+1((p,α),(p',t))-d_𝕊^1(Φ(p,α),Φ(p',t)). ]

    is increasing, because t↦ d_𝕊^2n+1((p,α),(p',t)) is 1-Lipschitz (due to the fact that t↦(p',t) is a unit speed geodesic) and d/dtd_𝕊^1(Φ(p,α),Φ(p',t))=-1.

    So we have that h_3(π/2n+1)≥ h_3(α)>2n/2n+1π. Which implies that d_𝕊^2n+1((p,α),(p',π/2n+1))>2n/2n+1π. This, however, contradicts the fact that, if e_n+2=(0,…,0,1)∈ℝ^2n+2, then
    
    d_𝕊^2n+1((p,α),(p',π/2n+1))   ≤ d_𝕊^2n+1((p,α),e_n+2)+d_𝕊^2n+1(e_n+2,(p',π/2n+1))
       ≤π/2+(π/2-π/2n+1)=2n/2n+1π.







§ DISTANCE FROM 𝕊^3 TO 𝕊^4

In this section we prove that d_GH(𝕊^3,𝕊^4)
=1/2ζ_3 by constructing a surjective function F_n:𝕊^n+1→𝕊^n and proving that for n=3 it has distortion ζ_3. The construction we utilize is inspired in the proof of <cit.>.
The author suspects that F_n also has distortion ζ_n for more values of n (e.g. n=4,5, see <Ref>), but he has verified that F_n has distortion >ζ_n for n≥7 (see <Ref>). Let us first introduce some notation. 


    
  * We identify 𝕊^n with 𝕊^n×{0}⊆𝕊^n+1⊆ℝ^n+2, and we let e_n+2 be the north pole (0,…,0,1)∈ℝ^n+2. 
    
  * For any x,x'∈𝕊^n+1 such that x'≠-x and λ∈[0,1], we let λ x⊕(1-λ)x' denote the point z in the geodesic segment [x,x'] such that d_𝕊^n+1(z,x')=λ d_𝕊^n+1(x,x'). 
    
  * Let σ:𝕊^n+1∖{e_n+2,-e_n+2}→𝕊^n be the projection to 𝕊^n; that is, for each x in the domain, σ(x) will be the point of 𝕊^n closest to x.
    
  * p_1,…,p_n+2 will be the vertices of a fixed regular simplex inscribed in 𝕊^n.
    
  * For each x∈𝕊^n+1 let α(x):=d_𝕊^n+1(x,e_n+2).
    
  * For each i=1,…,n+2, let 
    
    V_i:=
        {x∈𝕊^n;d_𝕊^n+1(p_i,x)<
        d_𝕊^n+1(p_j,x) for all j≠ i}.

    
  * For each i=1,…,n+2 let N_i be C_e_n+2V_i∖{e_n+2}. Note that each N_i is convex.


Note that ⋃_i=1^n+2N_i is dense in {x∈𝕊^n+1;x_n+2≥0}. We define the function 
    F_n:⋃_i=1^n+2N_i→𝕊^n
 by 

    F_n(x)=(1-f(α(x)))p_i⊕ f(α(x))σ(x),  if x in N_i,


where f(x):=max(0,x+1-π/2), see <Ref>.





Now, it would be enough to prove that dist(F_n)=ζ_n in order to prove that d_GH(𝕊^n,𝕊^n+1)≤ζ_n/2. Indeed, if dist(F_n)=ζ_n, then the function

    F_n':(⋃_i=1^n+2N_i)⋃(-⋃_i=1^n+2N_i)
    →𝕊^n
 defined by F_n'(x)=-F_n'(-x)=F_n(x) for all x∈⋃_i=1^n+2N_i also has distortion ζ_n by <Ref>, and its graph is a metric correspondence (as defined in <Ref>) between 𝕊^n and 𝕊^n+1. Thus, d_GH(𝕊^n,𝕊^n+1)≤ζ_n/2, as we wanted.


For simplicity we will write d instead of d_𝕊^n or d_𝕊^n+1 during the remainder of <Ref>, and we write F:H^4_+→𝕊^3 instead of F_3. 

In the file  of the GitHub repository <cit.> we have included a python program which selects a finite set S of random points in 𝕊^n+1 and computes the maximum value of |d(x,x')-d(F_n(x),F_n(x'))|, for x,x' in S. 



The distortion of F being at most ζ_n means that for all x,x' in ⋃_i=1^n+2N_i we have 

    |d(x,x')-d(F_n(x),F_n(x'))|≤ζ_n.




Depending on whether x,x' are in one cone N∈{N_1,…,N_n+2} or in different ones N,N' and depending on whether d(x,x')-d(F_n(x),F_n(x')) is positive or negative, <Ref> turns into four different inequalities, which give names to the following subsections. 




To experimentally check whether the distortion of F was  ζ_n, we used a python program (see  in <cit.>) which chooses a finite set S⊆𝕊^n+1 of random points in the sphere (choosing the coordinates of the points according to a Gaussian distribution) and computes the distortion |d(x,x')-d(F_n(x),F_n(x'))| for all x,x'∈ S. 
Thanks to Daniel Hurtado for optimizing the program so that it could handle sets S with hundreds of thousands of points.

This program was also useful for determining a function f:[0,π/2]→[0,1] which, when substituted in <Ref>, would lead to F_n having distortion ζ_n. For example, if instead of the function f from <Ref> we chose f(x)=2x/π (this was actually what the author tried first), then we would have  (F_3)>ζ_3.







 §.§ x,x' in the same cone N; d(F(x),F(x'))≤ d(x,x')+ζ_3.

We prove the stronger inequality d(F(x),F(x'))≤ d(x,x')+π/2[This inequality holds for all n, not only n=3. An alternative way to prove it is checking that the function F_n is Lipschitz when restricted to each Voronoi cell N_i; the inequality easily follows.].

Let p∈{p_1,…,p_n+2} be the center of the cone N and suppose that d(F(x),F(x'))>d(x,x')+π/2 for the sake of contradiction. 

Consider <Ref> below, in which the segments [p,σ(x)] and [p,σ(x')] have length <π/2. 



As d(F(x),F(x'))>π/2, the angle ∠ F(x)pF(x') is >π/2 (see <Ref>). Thus,

    d(σ(x),σ(x'))≥ d(F(x),F(x'))≥ d(x,x')+π/2.

where the first inequality follows from the spherical cosine rule. So by the triangle inequality, d(x,σ(x))+d(x',σ(x'))≥ d(σ(x),σ(x'))-d(x,x')≥π/2. That is, α(x)+α(x')=π-d(x,σ(x))+d(x',σ(x'))≤π/2. Which leads to contradiction:

    d(F(x),F(x'))≤ d(F(x),p)+d(F(x'),p)≤ f(α(x))d(σ(x),p)+f(α(x'))d(σ(x'),p)
    ≤π/2(f(α(x))+f(α(x')))≤π/2,

where the last inequality uses that the condition α(x)+α(x')≤π/2 implies f(α(x))+f(α(x'))≤1 (see <Ref>).









 §.§ x,x' in the same cone N; d(x,x')≤ d(F(x),F(x'))+ζ_3.

Suppose that d(x,x')> d(F(x),F(x'))+ζ_3 for the sake of contradiction. Note that by <Ref> applied to the triangle with vertices A=e_n+2,B=x and C=x', the angle ∠ xe_n+2x' is >π/2, so applying again <Ref> to the triangle with vertices A=e_n+2,B=σ(x),C=σ(x'), with the points B',C' from <Ref> being x,x' respectively, we have d(σ(x),σ(x'))>d(x,x')>d(F(x),F(x'))+ζ_3. Thus,

    (π-ζ_3)(π/2-α(x))+(π-ζ_3)(π/2-α(x'))
    ≥ d(p,σ(x))(1-f(α(x)))+
    d(p,σ(x'))(1-f(α(x')))
    
    ≥ d(F(x),σ(x))+d(F(x'),σ(x'))≥ d(σ(x),σ(x'))-d(F(x),F(x'))>η_3,

where the first inequality is due to the definition of F and <Ref> <Ref>, the second one is due to the definition of F and the third one follows from the triangle inequality. The inequality above implies α(x)+α(x')≤π-ζ_3/π-ζ_3<1.76. That cannot happen because 

    α(x)+α(x')=d(e_n+2,x)+d(e_n+2,x')>d(x,x')>ζ_3>1.82.










 §.§ x,x' in different cones N≠ N'; d(x,x')≤ d(F(x),F(x'))+ζ_3.

We prove d(x,x')≤ d(F(x),F(x'))+π/2 (which holds for all F_n, not only F=F_3).

Suppose for the sake of contradiction that d(x,x')> d(F(x),F(x'))+π/2. We will denote α=α(x),α'=α(x'), p,p'∈{p_1,…,p_n+2} will be the centers of the cones N,N' and 
    H':={x∈𝕊^m;d(p',x)<d(p,x)}  and  
    H:={x∈𝕊^m;d(p,x)<d(p',x)}.
 We divide the analysis into 2 cases.


    
  * α,α'>π/2-1. As σ(x)∈ H and σ(x')∈ H', by <Ref> and using that d(p,p')=ζ_3>π/2 we get 
        
    d(F(x),H')≥ζ_3/2(π/2-α)

        
    d(F(x'),H)≥ζ_3/2(π/2-α').

        Note that the geodesic from F(x) to F(x') passes through the boundary ∂ H=∂ H' at some point q_0, thus
        
    d(F(x),F(x'))=d(F(x),q_0)+d(F(x'),q_0)≥ d(F(x),H')+d(F(x'),H)≥ζ_3/2(π-α-α').

And now, using the triangle inequality, 

    α+α'≥ d(x,x')
    ≥ζ_3+d(F(x),F(x'))
    ≥ζ_3+ζ_3/2(π-α-α'),

So α+α'≥ζ_31+π/2/1+ζ_3/2≥π/21+π/2/1+π/4>2.26, which contradicts <Ref>.




  * α≤π/2-1. Then F(x)=p, so d(F(x),F(x'))≥ d(p,H')=ζ_3/2≥π/4, so d(x,x')≥ d(F(x),F(x'))+π/2≥3π/4, contradicting d(x,x')≤α+α'≤π-1.














 §.§ x,x' in different cones N≠ N'; d(F(x),F(x'))≤ d(x,x')+ζ_3.

This is the inequality which is most difficult to prove, and it fails when instead of n=3 we have n≥7; such a failure of the inequality can be reached when σ(x)=-p'[σ(x) cannot be exactly -p', but it can be arbitrarily close to it.], σ(x')=-p and α=α'=π-1/2, as in <Ref>. Then we have d(F(x),F(x'))=π and by the cosine rule applied to the triangle with vertices e_n+2,x,x', we have

    d(x,x')=arccos(cos(π-1/2)^2-sin(π-1/2)^2/n+1)
    n→∞⟶arccos(cos(π-1/2)^2)∼1.339.

One can then check that for n≥7, d(x,x')+ζ_n<π=d(F(x),F(x')).







In order to proceed, suppose for the sake of reaching a contradiction that d(F(x),F(x'))>d(x,x')+ζ_3. As in the previous case, we need to introduce some notation: let p,p'∈{p_1,…,p_n+2} be the centers of the cones N,N' respectively, and for brevity we will denote

    κ   :=d(σ(x),σ(x'))
    α   :=d(x,e_n+2)
    α'   :=d(x',e_n+2)


We will give a computer assisted proof. We first prove that κ∈[0.7,1.4] and α+α'∈[π-2,2] in <Ref>. Then we give some upper bounds for d(F(x),F(x')) 
(the functions 𝒰_1,𝒰_2,𝒰_6 introduced below) and a lower bound for d(x,x') (the function ℒ defined below) which only depend on α+α' and κ. Then we check that that (min(𝒰_1,𝒰_2,𝒰_6))(α+α',κ) cannot be bigger than ℒ(α+α',κ)+ζ_3 for all (α+α',κ) in the rectangle [0.7,1.4]×[π-2,2], concluding the proof.





    
  * α,α' are both >π/2-1.







  §.§.§ Lower bound for d(x,x')




    d(x,x')≥ℒ(α+α',κ)   :=
    arccos(cos(α+α')+sin(α+α'/2)^2(1+cos(κ)))
       =
    arccos(cos(α+α')1-cos(κ)/2+1+cos(κ)/2).


Note that the function ℒ(x,y) is defined for all x,y∈ℝ.


By the spherical cosine rule applied to the triangle with vertices x,e_n+2,x' we have

    d(x,x')   =arccos(cos(α)cos(α')+sin(α)sin(α')cos(κ))
       =arccos(cos(α+α')+(1+cos(κ))sin(α)sin(α'))
       ≥arccos(cos(α+α')+sin(α+α'/2)^2(1+cos(κ))).

The last inequality above is Jensen's inequality applied to the function x↦ln(sin(x)) in the interval (0,π).











  §.§.§ Bounds for α+α' and κ


We need some bounds on α+α' and κ for some of our following arguments to work.


 We have

    sin(α+α'/2)cos((π-ζ_3)(π/2-α+α'/2))≤√(3)/2√(2).

This implies α+α'<2.




Note that d(x,F(x))+d(x',F(x')) has to be >ζ_3 due to the triangle inequality and the fact that |d(F(x),F(x'))-d(x,x')|>ζ_3. This cannot happen if α,α' are very close to π/2; let us quantify this observation.

The triangle with vertices x,σ(x),F(x) has a right angle at σ(x) and sides d(x,σ(x))=π/2-α and d(σ(x),F(x))≤(π-ζ_3)(π/2-α), so we see by the cosine rule that

    d(x,F(x))
    ≤
    A(α):=
    arccos(cos(π/2-α)cos((π-ζ_3)(π/2-α)))

But by <Ref> below, the function A(α) is convex in the interval (π/2-1,π/2). So for any fixed value of α+α' we will have

    ζ_3≤
    d(x,F(x))+d(x',F(x'))
    ≤
    A(α)+A(α')≤2A(α+α'/2)

So A(α+α'/2)>ζ_3/2. So cos(A(α+α'/2))≤cos(ζ_3/2)=√(3)/2√(2).










 We have κ≥0.7.



We will need <Ref> below (which does not need κ>0.7 in its proof).

First we prove that κ cannot be very small: Suppose π-ζ_3+κ<arccos(-1/16)(so κ<0.315 approximately). Then we will prove that d(F(x),F(x'))≤ζ_3, contradicting d(F(x),F(x'))> d(x,x')+ζ_3. 
To prove it we first check that d(p,F(x'))≤ζ_3, or in general that p is at distance <ζ_3 of all points of the segment [p',σ(x')].

Note that the triangle with vertices p,p',σ(x') has sides d(p,p')=ζ_3, d(p',σ(x'))≤π-ζ_3 and d(p,σ(x'))≤π-ζ_3+κ. We can assume due to <Ref> that d(p',σ(x'))=π-ζ_3 and d(p,σ(x'))=π-ζ_3+κ, so the angle ∠ pp'σ(x') is <π/2 due to the cosine rule (equality is reached for π-ζ_3+κ=arccos(-1/16)), thus the maximum distance from p to points of the segment [p',σ(x')] is reached at p'. 

Now, knowing that d(p,F(x'))<ζ_3, d(p,σ(x))≤π-ζ_3 and d(F(x'),σ(x))≤κ+π-ζ_3, we repeat the same reasoning to obtain that the distance from F(x') to any point of the segment [p,σ(x)] (and in particular F(x)) is ≤ζ_3.

So we can assume κ∈(0.3,0.7). Now, by <Ref> we have

    d(x,x')≥ℒ(α+α',κ)≥ℒ(π-2,κ)=arccos(cos(2)cos(κ)-1/2+1+cos(κ)/2).


We also give an upper bound for d(F(x),F(x')) depending only on κ:
letting C(κ) be given by <Ref>, we have d(x,F(x'))≤ C(κ) by <Ref>, and note that 

    d(F(x'),σ(x))
    ≤ d(F(x'),σ(x'))+d(σ(x'),σ(x))
    ≤π-ζ_3+κ.

Thus, by <Ref> and <Ref> we conclude that

    d(F(x),F(x'))
    ≤𝒰_3(κ):=
    arccos(
    -4/√(15)√(C(κ)^2+cos(π-ζ_3+κ)^2-1/2C(κ)cos(π-ζ_3+κ))).

However, no value κ∈(0.3,0.7) satisfies ℒ(π-2,κ)+ζ_3≤𝒰_3(κ) (see the graph of 𝒰_3(κ)-ℒ(π-2,κ)-ζ_3 below), so we are done. 









We have κ∈[0.7,1.4] and α+α'∈[π-2,2].



Clearly α+α'>π-2, as we are in the case where α,α'≥π/2-1. The facts that α+α'<2 and κ≥0.7 follow from <Ref>.

In order to deduce that κ<1.4 one can consider the following upper bounds for d(F(x),F(x')) obtained from the triangle inequality:

    d(σ(x),σ(x'))+(d(F(x),σ(x))+d(F(x'),σ(x')))≤𝒰_6(α+α',κ):=
    κ+(π-ζ_3)·(π/2-α)+(π-ζ_3)·(π/2-α')=κ+(π-ζ_3)·(π-(α+α')).


    d(p,p')+d(p,F(x))+d(p',F(x'))≤𝒰_7(α+α',κ):=ζ_3+(π-ζ_3)((α+α')-π+2).

If d(F(x),F(x'))>d(x,x')+ζ_3, then we have min(𝒰_6,𝒰_7)(α+α',κ)-ℒ(α+α',κ)-ζ_3>0. 
However, one can check numerically (see <Ref> and file  in <cit.>) that both inequalities cannot happen if κ≥1.4.
















  §.§.§ Upper bounds for d(F(x),F(x'))







First we obtain an upper bound for d(F(x),F(x')) which is useful when κ is small. We  need the following lemma. 










We have

    cos(d(p,F(x')))≥ C(κ):=-√(1-2cos(ζ_3-κ)+16cos^2(ζ_3-κ))/√(15).



We prove that arccos(C(κ)) is an upper bound for the distance from p to any points on the segment [σ(x'),p']. Note that d(p,p')=ζ_3, d(p',σ(x'))≤π-ζ_3 and d(p,σ(x'))≤π-ζ_3+κ≤π. By <Ref> we can assume d(p,p')=ζ_3, d(p',σ(x'))=π-ζ_3 and d(p,σ(x'))=π-ζ_3+κ. So by <Ref>, the from p to all points of the segment [σ(x'),p'] is bounded above by the function C(κ) from <Ref>.







Letting κ_2:=κ+(π-ζ_3)(π/2-α+α'/2), if π-ζ_3+arccos(C(κ))+κ_2<2π, then 

    d(F(x),F(x'))≤arccos(
    -4/√(15)√(C(κ)^2+cos(κ_2)^2-1/2C(κ)cos(κ_2))).

If on the other hand, π-ζ_3+arccos(C(κ))+κ_2≥2π, then d(F(x),F(x'))≤2π.




Notice that, swapping x,x' if necessary, we have d(F(x'),σ(x))<κ_2. So the triangle with vertices  F(x'),σ(x),p has side lengths d(p,F(x'))≤arccos(C(κ)), d(p,σ(x))≤π-ζ_3 and d(F(x'),σ(x))≤κ_2. If the sum of these three lengths is <2π then by <Ref> and <Ref> we conclude.














    d(F(x),F(x'))
    ≤𝒰_2(α+α',κ):=
    (π-ζ_3)(α+α'/2-π/2+1)
    +
    G(κ,α+α'/2-π/2+1),

where

    G(κ,β'):=arccos(-1/4cos((π-ζ_3)β')+√(15)/4sin((π-ζ_3)β')16cos(π-ζ_3+κ)+1/15),



Assume that α'≥α. We will use the inequality 

    d(F(x),F(x'))≤ d(F(x),p)+d(p,F(x'))≤(π-ζ_3)(α-π/2+1)+d(p,F(x')).

Now, to find an upper bound for d(p,F(x')) we first consider a point q_0∈𝕊^4 such that d(p',q_0)=π-ζ_3 and d(p,q_0)=π-ζ_3+κ. We let P_0:=(α'-π/2+1)q_0+(π/2-α')p'.


d(p,F(x'))<d(p,P_0).


We first note that the points x and x' are fixed, so κ,α' will remain constant during this proof. Let β'=α'-π/2+1∈[0,1]. 

For each a∈[0,π-ζ_3] and b∈[0,π-ζ_3+κ] with a+b≥ζ_3=d(p,p') we consider a point q=q(a,b)∈𝕊^4 given by d(p',q)=a and d(p,q)=b (so q could be σ(x') for adequate a,b), and let F(a,b)=β'q+(1-β')p'. We will be done if we prove that the values a_0,b_0 which 
maximize d(p,F(a,b)) are a_0=π-ζ_3, b_0=π-ζ_3+κ, so d(p,F(a_0,b_0))=d(p,P_0). 


Let us first prove b_0=π-ζ_3+κ. It is not hard to see that for any fixed value of a, d(p,F(a,b)) is increasing with b. Thus, if ζ_3+a_0>π-ζ_3+κ (so that the triangle inequality in the triangle with vertices p,p',q(a,b) holds), we necessarily have b_0=π-ζ_3+κ. But[Note that a can take the value π-2ζ_3+κ, because κ>0.7.]

    d(p,F(a_0,b_0))≥ d(p,F((π-ζ_3+κ)-ζ_3,π-ζ_3+κ))=π-ζ_3+κ.

 Thus, a_0=d(p',q(a_0,b_0))≥ d(p',F(a_0,b_0))≥ d(p,F(a_0,b_0))-ζ_3=π-2ζ_3+κ, concluding the proof that b_0=π-ζ_3+κ.

So it will be enough to check that d(p,F(a,b_0)) is increasing with a. This can be proved using convexity arguments, but in this special case we can compute d(p,F(a,b_0)) explicitly: first note that applying the cosine rule twice, we obtain

    cos(∠ pp'F(a,b_0))
    =
    cos(∠ pp'q(a,b_0))
    =
    cos(b_0)-cos(a)cos(ζ_3)/sin(a)sin(ζ_3).


    cos(d(p,F(a,b_0)))   =cos(ζ_3)cos(aβ')+
    sin(ζ_3)sin(aβ')cos(b_0)-cos(a)cos(ζ_3)/sin(a)sin(ζ_3)
       =
    -cos(aβ')/4+
    sin(aβ')/sin(a)(cos(b_0)+cos(a)/4).

So, computing the derivative with respect to a, we obtain 

    ∂/∂ acos(d(p,F(a,b_0)))
    =
    β'sin(aβ')/4+sin(aβ')/sin(a)·-sin(a)/4+∂/∂ a(sin(aβ')/sin(a))(cos(b_0)+cos(a)/4).

Using that β'∈[0,1] and cos(b_0)<-1/4, one readily checks that both the sum of the first two terms above and the third term above are negative for all a∈(0,π). So d(p,F(a,b_0)) increases with a, as we wanted.



Now, by the cosine rule (using d(p,p')=ζ_3) we have that

    cos(∠(p,p',q_0))=cos(π-ζ_3+κ)+1/16/15/16=16cos(π-ζ_3+κ)+1/15.



Thus, by <Ref> and the cosine rule and letting β'=α'-π/2+1,

    d(p,F(x'))
       ≤ G(κ,β'):=
    d(p,P_0)=
       arccos(-1/4cos((π-ζ_3)β')+√(15)/4sin((π-ζ_3)β')16cos(π-ζ_3+κ)+1/15).


Letting the RHS in <Ref> be G(κ,β')=d(p,(α-π/2+1)q+(π/2-α)p'), our upper bound is d(F(x),F(x'))≤(π-ζ_3)β+G(κ,β'). 
But note that for fixed α+α' with α'≥α, this upper bound will be minimized for α=α'=α+α'/2 (this follows from the fact that, due to the definition of P_0, for any fixed κ the function G(κ,β') is (π-ζ_3)-Lipschitz in β'), completing the proof of <Ref>.














    d(F(x),F(x'))≤arccos(
    -4/√(15)√(C(k)^2+cos(ζ_3-k)^2+1/2C(k)cos(ζ_3-k))).

Letting k_2:=k+(π-ζ_3)(α+α'/2-π/2+1), we also have

    d(F(x),F(x'))≤arccos(
    -4/√(15)√(C^2+cos(k_2)^2-1/2Ccos(k_2))).




We consider the triangle σ(x)pF(x'). In order to find the upper bound for d(F(x),F(x')), we find an upper bound for the maximum distance D_2 from F(x') to any point of [p,σ(x)]. We know that d(p,σ(x))≤π-ζ_3, d(p,F(x'))≤arccos(C)∈[ζ_3,π] and d(σ(x),F(x'))≤π-ζ_3+k. For the second inequality use instead that, swapping x,x' if necessary, d(p,F(x'))≤ k_2.

Reasoning similarly as above[In this case we can first note that for fixed values of d(p,F(x')) and d(σ(x),F(x')), D_2 is maximized for maximal d(p,σ(x)), so we can assume d(p,σ(x))=π-ζ_3.], we can check that D_2 is maximized when the distances are d(p,σ(x))=π-ζ_3, d(p,F(x'))=arccos(C) and d(σ(x),F(x'))=π-ζ_3+k, so by <Ref> we are done. Wait, what if d(p,F(x'))+d(σ(x),F(x'))>π. Then the maximum is just π. So UBdF1 is a function defined in pieces I guess. Same with UBdF3. Formalize this proposition better.










  §.§.§ Concluding the proof of the inequality in <Ref>




Thus, we have a lower bound ℒ:[π-2,2]×[0.7,1.4]→ℝ for d(x,x') and three upper bounds 𝒰_1,𝒰_2,𝒰_6:[π-2,2]×[0.7,1.4]→ℝ for d(F(x),F(x')) in terms of α+α' and κ (see <Ref> for the definition of 𝒰_6). Letting 

    (x, y):=
    min(𝒰_1,𝒰_2,𝒰_6)(x,y)-ℒ(x,y)-ζ_3,

it would be enough to check that (x, y)<0 for all (x,y)∈ R:=[π-2,2]×[0.7,1.4]. 

This is checked numerically in the file  in <cit.>. Indeed, as explained in <Ref>, it is enough to check that (x,y)<-0.08 for all points (x,y) in a grid in the rectangle R with coordinates spaced by d=10^-5, and then check that if two points (x,y) and (x',y') in R satisfy |x-x'|,|y-y'|≤10^-5/2, then |(x,y)-(x',y')|<0.08. This follows from the facts that:

    
  * If |x-x'|,|y-y'|≤10^-5/2, then |ℒ(x,y)-ℒ(x',y')|<10^-5. This follows from the definition of ℒ(x,y), which is the distance between two points of 𝕊^n+1 at distance x/2 of e_n+2 whose geodesics to e_n+2 form an angle of y).

    
  * If |x-x'|,|y-y'|≤10^-5/2, then |𝒰_1(x,y)-𝒰_1(x',y')|<0.07. Similarly with 𝒰_2,
    𝒰_6. This is easy to see for 𝒰_2,
    𝒰_6 (in those cases we can change 0.07 by 10^-4). To check it for 𝒰_1 one can use that the function κ↦ C(κ) is 1-Lipschitz, that C(κ)^2+cos(κ_2)^2-1/2C(κ)cos(κ_2)≥3C(κ)^2/4≥0.15 for all κ∈[0.7,1.5] and the uniform continuity constants of x↦√(x),x↦arccos(x).







  * Either α or α' are <π/2-1. We will assume α'<π/2-1, so F(x')=p'. We can assume α>π/2-1, lest d(F(x),F(x')) be ζ_3. In this case we will also prove the inequality numerically using upper and lower bounds for d(F(x),F(x')) and d(x,x') respectively.

Note that, similarly as in <Ref>, we can deduce that, if β=α-π/2+1, 

    d(F(x),F(x'))=d(F(x),p')
    ≤ G(κ,β).
 
We also have a couple of lower bounds for d(x,x') in this case.
We also know that d(x,x')≥ℒ(α+α',κ)≥ℒ(α,κ):=ℒ_1(α,κ). 
We also know that d(x,x')≥ℒ_2(α,κ), where ℒ_2(α,κ)=α when κ≥π/2 (as in that case, we can assume α'=0, which reduces d(x,x') and does not change d(F(x),F(x'))) and when κ≤π/2, ℒ_2(α,κ) is just the distance from x' to the geodesic e_n+2x, that is, 

    ℒ_2(α,κ)=arcsin(sin(α)sin(κ)).



However, it can be checked numerically (see <Ref> and file  in <cit.>) that for no values α∈[π/2-1,π/2] and κ∈[0.3,π] can we have G(κ,α-π/2+1)>max(ℒ_1(α,κ),ℒ_2(α,κ)), so we are done.






§ SPHERICAL GEOMETRY LEMMAS USED IN <REF>


We will use the notation d=d_𝕊^n+1 in this section.

 
Let T be a spherical triangle with vertices A,B,C and opposite side lengths a,b,c respectively. If b,c≤π/2 and the angle α at A is ≤π/2, then a≤π/2.



By the spherical cosine rule, 
cos(a)=cos(b)cos(c)+sin(b)sin(c)cos(α)≥0.




Let T be a spherical triangle with vertices A,B,C and opposite side lengths a,b,c respectively. If b,c≤π/2 and a≥π/2, then we have α>a≥π/2, where α is the angle at a. If, moreover, B',C' are points in the sides AB and BC respectively and a'=d(B',C'), then a'≤ a.



For the first part, by <Ref> we have α≥π/2, so by the spherical cosine rule, 

    cos(a)=cos(b)cos(c)+sin(b)sin(c)cos(α)≥cos(α).

For the second part, note that if in the formula above we change b,c by some smaller values b',c', then cos(a) increases.







Let p,p' be two points in 𝕊^m (m≥2) at distance a, and consider the hemispheres H'={x∈𝕊^m;d(p',x)<d(p,x)},
H={x∈𝕊^m;d(p,x)<d(p',x)}. Then for any point x∉H' and any λ∈[0,1], 
    d(λ p+_S(1-λ)x,H')≥λa/2.




Note that for any point q∉H', d(H',q) is just d(O_H',q)-π/2, where O_H' is the center of H'. So the inequality above reduces, by Jensen's inequality applied to the function h:[0,1]→ℝ;λ↦ d(λ p+_S(1-λ)x,H') (note h(0)≥0 and h(1)=a/2), to proving that for any point O and any geodesic γ(t), the function t↦ d(O,γ(t)) is convex in the intervals where d(γ(t),O)≥π/2. To prove this we can assume that 𝕊^m=𝕊^2⊆ℝ^3 and O=(0,0,-1), so we just have to prove that for any k∈[0,1], the function arcsin(ksin(t)) is convex in [0,π], which can be seen from computing its second derivative with respect to t.




If we let ρ=π-ζ_3=arccos(1/4), the function A(t)=arccos(cos(t)·cos(ρ t)) is convex in the interval (0,1).[Thanks to River Li for his help in the proof that A(t) is convex; see this https://math.stackexchange.com/a/4911668/807670MathStackExchange answer.]



Let c=cos(t),s=sin(t),c_1=cos(ρ t),s_1=sin(ρ t) during this proof (note that c,c_1,s,s_1>0 for t∈(0,1)).
The second derivative of A(t) is given by the following formula

    A”(t)=-1/(1-c^2c_1^2)^3/2(2ρ ss_1-cc_1ρ^2s^2-cc_1s_1^2).

So we just need to prove that ∀ t∈(0,1), 2ρ ss_1-cc_1ρ^2s^2-cc_1s_1^2>0. This follows from the following two inequalities:

    
  * ρ ss_1>cc_1ρ^2s^2; this is equivalent to s_1>cc_1ρ s, which is true because for all t∈(0,1), tan(ρ t)>ρ t≥ρsin(t)≥ρsin(t)cos(t).
    
  * ρ ss_1>cc_1s_1^2; in fact we have ρ ss_1>s_1^2, as ρsin(t)>sin(ρ t) for all t∈(0,1).






 
Let u,v,w be the vertices of a spherical triangle in 𝕊^n with the sides opposite to u,v,w having lengths x_1,x_2,x_3 respectively (so x_1+x_2+x_3≤2π). Then the maximum distance α between u and any point of the geodesic passing through v,w satisfies

    cos(α)
    =-√(cos(x_2)^2+cos(x_3)^2-2cos(x_1)cos(x_2)cos(x_3))/sin(x_1).



Note that α∈[π/2,π] is π minus the angle between u and the plane containing v and w. Letting a=⟨ v,w⟩=cos(x_1),b=⟨ u,w⟩=cos(x_2) and c=⟨ u,v⟩=cos(x_3), we have

    π-α=arcsin(|(u,v,w)|/√(1-a^2))
       =
    arcsin√([ 1 c b; c 1 a; b a 1 ])/√(1-a^2)
       =arcsin(√(1+2abc-a^2-b^2-c^2)/√(1-a^2)).

So

    cos(α)
    =
    -√(b^2+c^2-2abc)/√(1-a^2)
    =-√(cos(x_2)^2+cos(x_3)^2-2cos(x_1)cos(x_2)cos(x_3))/sin(x_1).








Fix p∈𝕊^2 and a≤ b numbers in [0,π]. For any points q,r∈𝕊^2 with d(p,q)=a and d(p,r)=b let t=d(q,r). Then the function f(t) which gives the maximal distance from p to any points in the geodesic segment qr is well defined and increasing. More concretely,

    
  * If a+b<π, then f:[b-a,b+a]→ℝ is given by f(t)=b.
    
  * If a+b>π, then f:[b-a,2π-a-b]→ℝ is given by f(t)=b for all t<arccos(cos(a)/cos(b)). For t≥arccos(cos(a)/cos(b)) f is increasing, with 

    cos(f(t))
    =-√(cos(a)^2+cos(b)^2-2cos(t)cos(a)cos(b))/sin(t).





If b<π/2 then the closed ball centered at p of radius b is convex, so it contains the segment qr. If b≥π/2 and a+b<π, then consider a closed hemisphere H centered at some point at distance b-π/2 of p. Then H contains some point r at distance b of p, and any segment between r and points at distance a of p is entirely contained in H, proving that f(t)≤ b.

Finally, assume that a+b>π, and fix some point q at distance a of p. The point r can be any point in the circle C of points at distance b of p; as r moves from the point in C closest to q to the the point in C furthest to q, t=d(q,r) increases from b-a to 2π-a-b. 





If cos(t)≥cos(a)/cos(b), then by the cosine rule the angle ∠ prq at r is at most π/2, so f(t)=d(p,r)=b. If cos(t)<cos(a)/cos(b), then ∠ prq is obtuse, and the maximal distance from p to points of the segment qr is given by <Ref>. Note that as r moves further from q, the angle between 0⃗p⃗ and the plane containing the origin, q and r decreases, so f is increasing.






Let a,b,c∈[0,π] satisfy a≤ b+c,b≤ a+c,c≤ a+b. We consider spherical triangles with vertices x,y,z in 𝕊^2, with d(y,z)≤ a,d(x,z)≤ b,d(x,y)≤ c. If a+b+c≥2π, then the antipodal point -x may lie in the segment yz. If a+b+c<2π, then the maximal distance from x to any point of the segment yz is reached when d(y,z)= a,d(x,z)=b and d(x,y)=c.




First note that for any a,b,c as above with a+b+c<2π there are spherical triangles with sides a,b,c. If a+b+c=2π, then the union of the three sides of the triangle is a great circle. So the only nontrivial case is a+b+c<2π.

Let f(α,β,γ) be the maximal distance from x to any point of the segment yz, where d(y,z)= α,d(x,z)=β and d(x,y)=γ.
<Ref> proves that for fixed values of α,β, the function f(α,β,γ) increases with γ. It is also not hard to check that for fixed values of β,γ (so we can consider x,z fixed and y in a radius α circumference around z), f(α,β,γ) is increasing with respect to α, and similarly, for fixed α,γ the function f(α,β,γ) increases with β. So among all values α≤ a,β≤ b and γ≤ c, f(α,β,γ) will be maximized when α=a,β=b,γ=c, and we are done.
















amsalpha


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