# A graph theory question inspired by a question about Morse functions in \$ mathbb {S} ^ 2 \$

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.