This question arose from my attempt to solve the problem posed in this fascinating question about the gradient of a soft function in $ mathbb {S} ^ 2 $.

Fix a finite set, $ P $, of an even number of points $ p_1, ldots, p_N in mathbb {S} ^ 2 $. Leave $ C (p_i) = {q in mathbb {S} ^ 2: q cdot p_i = 0 } $ Be set of large circles with axis of $ p_i $.

We make the following assumptions

- $ C (p_i) neq C (p_j) $ for $ p neq j $
- $ C (p_i) cap C (p_j) cap C (p_k) = emptyset $ for different $ p_i, p_j, p_k $.

Now let $ V = {q in mathbb {S} ^ 2: q in C (p_i) cap C (p_j), mbox {some} p_i neq p_j } $. This is the set of {vertices}.

Leave

$$

Gamma = bigcup_ {p in P} C (p) $$

and let $ E $ be the set of components of $ Gamma backslash V $. This is the set of edges. Clearly, $ (V, E) $ It is a flat graph. Finally, let's go $ F $ be the set of the components of $ mathbb {S} ^ 2 backslash Gamma $. This is the set of faces. We can think of $ (F, E) $ as the dual chart for $ (V, E) $.

Keep in mind that the antipodal map $ A: mathbb {S} ^ 2 to mathbb {S} ^ 2 $ Induces an involution in the sets. $ V $, $ E $ Y $ F $ which we will also denote by $ A $.

Now suppose we have a map $ sigma: F to {0, ldots, N } $ with the property that

- $ sigma (f) + sigma (A (f)) = N $
- Yes $ f_1 $ Y $ f_2 $ they are adjacent faces, then $ | sigma (f_1) – sigma (f_2) | = 1 $ (that is to say., $ sigma $ Jump through one through an edge.

Look at the construction ensuring that each vertex. $ v $ It limits four edges and thus one has four faces. $ f_1 $, $ f_2 $, $ f_3 $ Y $ f_4 $ meet in $ v $, then, until the re-labeling $ sigma (f_1) = sigma (f_2) $, $ sigma (f_3) = sigma (f_2) -1 $ Y $ sigma (f_4) = sigma (f_2) + 1 = sigma (f_3) + 2 $.

Now let $ F_e = {f in F: sigma (f) mbox {even} } $ and y $ F_o = {f in F: sigma (f) mbox {odd} }. $

Leave

$$

F_o & # 39; = {f in F_o: sigma (f) geq 3 mbox {y} sigma (A (f)) geq 3 }.

$$

As $ N $ it's even, we have $ A (F_e) = F_e $ Y $ A (F_o) = F_o $.

by $ F_o & # 39; & # 39; $ some unspecified subset of $ F_o & # 39; $, we are interested in the following problem:

**Issue:** Determine if $ F_o & # 39; & # 39; cup F_e $ contains a connected path (in the dual graph $ (F, E) $) connecting antipodal faces.

When $ N = 2 $ or $ N = 4 $, $ F_o & # 39; $ it is empty and it is easy to see any path between the antipodal points in $ F_e $ must go through $ F_o $ – That is, there is no route as requested by the problem. I have no idea what can happen when $ N = 6 $. My suspicion is that for $ F_o & # 39; & # 39; $ big enough one can have such a way.