# Continuous scalar fields on spheres

I've been thinking about this for awhile now (I am trying to find a method of proving the Borsuk-Ulam theorem in 2 dimensions without resorting to the usual, and not so intuitive to non-mathematicians topological arguments). I saw the question Does a continuous scalar field on a sphere have continuous loop of "isothermic antipodes" posted here, but am curious about a stronger statement.

Say you have a continuous map $f$ from $S^2$ to $R$. Let $U$ be the set of points such that $f(x)=f(-x)$ for $-x$ being the antipode of $x$. I;ve been able to show that if $U$ is not the entire sphere, then every pair of antipodal points which are not mapped to the same value by $f$ are disconnected in $S^2-U$ (using an argument similar to one posted in the above question). So; the set $U$ must then contain at least one continuous loop. What I am trying to show is that $U$ must contain a loop $C$ such that $C=-C$ (equivelantly, that there is a loop $C$ in $U$ for which $S^2-C$ is two disjoint "hemispheres" of equal areas).

It seems to me that if this were not the case then you could construct a second map $g$ on $S^2$ such that its set of antipodes mapping to the same value would be completely disjoint from $U$ (since two loops on a sphere which are not "equators" in the sense that they divide the area of the sphere in half do not necessarily need to intersect), and then you could construct a map from $S^2$ to $R^2$ which had no antipodes mapping to the same value by $F=(f,g)$, which would contradict the theorem I know is true but would like to prove.

Maybe I'm going about this in a difficult way, so any suggestions on how to study the nature of the set $U$ would be appreciated.

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Since the question you refer to was answered negatively (such a loop does not necessarily exist), you should not assume that the set $U$ contains at least one continuous loop.

Even if $U$ does contain a continuous loop, it is not necessarily the case that there exists a loop $C$ so that $C = -C$. For example, there exists a function on the sphere whose $U$ region is the union of two antipodal small circles on the sphere connected by a pair of antipodal topologist's sine curves.

Of course, if the function $S^2\to\mathbb{R}$ is sufficiently well-behaved, the set $U$ will be a graph on the sphere. In this case, it is relatively easy to show that $U$ must contain a loop $C$ such that $C = -C$. Specifically, since $U=-U$, the antipode of each vertex of $U$ is again a vertex of $U$. If we choose a path from a vertex to its antipode, then the union of this path with the antipodal path is a closed loop. (This argument works whenever $U$ is path-connected.)

It is possible that the loop created in this fashion intersects itself, since the original path may contain more than one pair of antipodal points. However, it is always possible to shorten the path so that the only pair of antipodal points are the endpoints. Specifically, given a path $\gamma\colon[0,1]\to S^2$ between two antipodal points, let $$a = \max\{t \in [0,1] \mid \gamma(t) = -\gamma(s)\text{ for some }s>t\}$$ and let $$b = \min\{s \in [a,1] \mid \gamma(s) = -\gamma(a)\}$$ Then the restriction of $\gamma$ to $[a,b]$ is a path whose only pair of antipodal points are its endpoints, so the union of $\gamma$ and $-\gamma$ is a simple closed curve $C$ for which $C=-C$.

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Thanks. How would I go about showing that $U$ is path connected? (Or, in cases where $U$ is not path connected; that it contains a connected component with a pair of antipodes?) I am attempting by reasoning that since each pair of antipodes $x, -x$ in $S^2$ which do not map to the same value under $f$ are disconnected in $S^2-U$, that if I assume that $U$ is a collection of disjoint components such that no component contains both a point and its antipode, I will reach a contradiction. But so far have gotten stuck. – Steve Feb 19 '11 at 23:08
I'm not quite sure what you're asking. It's not true in general that $U$ contains a path-connected component with a pair of antipodes, so you're not going to be able to prove it except in specific cases. – Jim Belk Feb 20 '11 at 16:03

While surfing through some math, I came across this thread that seems not to be finally clarified.

So, Steve was asking whether the set $U=\{x\in\mathbb{S}^{2}|\ f(x)=f(-x)\}$ contains a connected component $K$ such that $K=-K$. This is a consequence of the Borsuk-Ulam theorem:

Suppose the contrary, then for any $x\in U$ we have $K(x)\cap K(-x)=\varnothing$, where $K(x)$ is the component of $U$ containing $x$. Now, as $U$ is compact Hausdorff, each component is an intersection of open-closed subsets, so one easily finds an open-closed $V(x)\supset K(x)$ such that $V(x)\cap(-V(x))=\varnothing$. Consider the covering $\{V(x),\ x\in U\}$, since $U$ is compact, take a finite subcover $\gamma=\{V(x_{i} )\},\ i=1,...,n$. Let $\nu$ denote the maximal disjoint cover of $U$ inscribed in $\gamma\cup(-\gamma)$ (sorry...). Then $\nu$ is a disjoint finite symmetric cover of $U$ such that $W\cap(-W)=\varnothing$ for any $W\in\nu$. Thus we may decompose $U$ into the union $U=C\cup(-C)$, where $C$ is compact and $C\cap(-C)=\varnothing$. Let $\varphi:\mathbb{S}^{2}\rightarrow \lbrack0,1]$ be a continuous function such that $\varphi(C)=0$, $\varphi(-C)=1$ (Urysohn's Lemma) and consider finally the map $F=(f,\varphi ):\mathbb{S}^{2}\rightarrow\mathbb{R}^{2}$. It is now easily seen that $F$ has the property $F(-x)\neq F(x)$ for any $x\in\mathbb{S}^{2}$, that contradicts Borsuk-Ulam's Theorem.

As Jim pointed out, it may happen that no two antipodal points of $U$ may be connected by a topological arc in $U$, although there are two antipodal points lying in one and the same component of $U$.

It's interesting, that something more may be said about the set $U=\{x\in \mathbb{S}^{2}|\ f(x)=f(-x)\}$. For example, if $\ U$ is supposed to be a 1-manifold, then it contains an odd number of symmetric components (and consequently, an even number of other components).

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