Let $f : S^2 \to \mathbb{R}$ be a continuous map such that $f(-x) = -f(x)$. Consider the set $Z = f^{-1}(0)$. Must $Z$ contain some path from some point to its antipode? Indeed, must $Z$ contain a continuous loop intersecting each "meridian", passing through antipodal points on antipodal meridians?

[I see this often asserted as a step in intuitive arguments for the Borsuk-Ulam theorem, but do not see why rigorously it must be so. Indeed, as a child, I once attempted precisely this argument for the Borsuk-Ulam theorem in a math camp, and was chided for asserting this unsupported lemma; thus it has stuck with me always. Certainly, $Z$ must contain a point on each meridian, but that these points must line up into a continuous path is not obvious to me.]

Update: My original question turned out to have already been answered at How bogus is the glitzy proof of Borsuk-Ulam?, as well as by Loïc Teyssier in the same way below, but a new question with same motivation is: must Z contain two antipodal points in the same connected component? (If so, then the Borsuk-Ulam proof via this approach can still be salvaged; if not, we may write it off as a lost cause)

  • $\begingroup$ For what it's worth, I now find the winding number proof of Borsuk-Ulam much more intuitive and illuminating (given odd map from sphere to R^2, its winding number around 0 as its input traces around the equator must be odd and thus nonzero. But winding numbers around boundaries add as regions combine, and also vanish in the vicinity of non-zeros of the map. Thus, by repeated bisection of input loops, we can constructively zoom in on some point which is a zero of the map.). I'm just curious whether this other approach that everyone goes for can indeed be fleshed out to hold up or not. $\endgroup$ Oct 29 '17 at 23:55
  • 1
    $\begingroup$ I've just noticed essentially identical question: mathoverflow.net/questions/251921/… $\endgroup$ Oct 30 '17 at 1:58
  • 1
    $\begingroup$ I've updated this question, and am now very interested in whether Z must contain antipodal points in the same connected component. $\endgroup$ Oct 30 '17 at 2:44
  • $\begingroup$ As noted, I'm still very interested in the connected component version of this question… Not sure if it's best to keep bumping this question or to make a new one. $\endgroup$ Nov 5 '17 at 23:30

Let $K\subset \mathbb S^2$ be the compact set obtained by modifying the equator of $\mathbb S^2$ around two antipodal points so that it locally looks like the adherence of the graph of $t\neq0\mapsto \sin \frac{1}{t}$. You can make these modifications so that $K$ is symmetric for the antipodal involution. It is compact and not locally connected at the pair of points. In particular no two antipodal points in $K$ can be joined by a continuous path ranging in $K$.

The open set $\mathbb S^2\setminus K$ has two connected components, a "north" (say +) and a "south" (say -) domains, which are in antipodal involution. Let $f$ be the "distance to $K$" function multiplied by $\pm1$ according to the sign chosen for the north and south domains. Then $f$ is continuous, odd and $Z=K$.

Do you mean instead that some point of $Z$ should lie in the same connected component as its antipode?

  • $\begingroup$ This f isn't odd, though, since distances are always nonnegative. Can this f be odd-ized? $\endgroup$ Oct 29 '17 at 23:37
  • $\begingroup$ Ah... you're right of course. I'll edit the answer. $\endgroup$ Oct 29 '17 at 23:40
  • $\begingroup$ Great, thanks! And, sure, I'd also be interested in the connected component version of this, since I suppose Z having two antipodal points in same connected component still allows us to conclude that any other odd function from S^2 to R must have a zero in the same Z as well, and thus obtain Borsuk-Ulam. $\endgroup$ Oct 29 '17 at 23:59
  • $\begingroup$ The zero set can be made to have no simple curves. $\endgroup$ Oct 30 '17 at 1:49

The answer to the connected component version is yes, though the only proof I can come up with seems like more trouble than it would've been worth in math camp homework! Two steps:

  1. If there there are finitely many components of $S^2 \smallsetminus Z$, produce a finite-diameter bipartite tree on which the antipodal map on $S^2$ induces an involution fixing a point.
  2. Approximate any $f$ by functions satsifying step 1.

Most of the work seems to be a part of Step 1 that vaguely sounded like the Jordan curve theorem:

Key Fact. Let $X$ be a connected, compact, locally path-connected metric space with zero first homology. For an open subset $U$, inclusion induces a bijection between the connected components of $\partial U = \bar{U} \smallsetminus U$ and those of $X \smallsetminus U$.

Proof of Key Fact: Mayer-Vietoris and some point-set debris

Given $\epsilon > 0$, let $U_\epsilon$ be the set of points less than $\epsilon$ away from the closure $\bar{U}$ of $U$; and let $V_\epsilon$ be the set less than $\epsilon$ away from $X \smallsetminus U$. The Mayer-Vietoris sequence in reduced homology ends with

$$ H_1(X) \to \tilde{H}_0(U_\epsilon \cap V_\epsilon) \to \tilde{H}_0(U_\epsilon) \oplus \tilde{H}_0(V_\epsilon) \to \tilde{H}_0(X) , $$

which gives a bijection between the path components of $U_\epsilon \cap V_\epsilon$ and those of $V_\epsilon$. Pass to connected components of their closures using:

  1. A connected open set $W$ in a locally path-connected space is path-connected.

    Proof. Each point of $W$ is has a path-connected open neighborhood in $W$; so the path components form a partition of $W$ by open sets.

  2. The closure of a connected open set $W$ is connected.

    Proof. In a partition of $\bar{W}$ by clopen subsets, whichever one contains $W$ contains $\bar{W}$.

Then send $\epsilon$ to $0$. That is, since

$$ \begin{align*} X \smallsetminus U &= \bigcap_{\epsilon > 0} \bar{V}_\epsilon & &\text{and} & \partial U &= \bigcap_{\epsilon > 0} \overline{U_\epsilon \cap V_\epsilon} , \end{align*} $$

the Key Fact is proven if we can biject connected components of $X \smallsetminus U$ with nested sequences of components $C_n$ of $\bar{V}_{1/n}$ (and similarly for $\partial U$):

Lemma. In a compact space $X$, the intersection $C$ of nested closed connected sets $C_1 \supseteq C_2 \supseteq \cdots$ is connected.

Proof. Given disjoint open subsets $U$ and $V$ of $X$ that cover $C$, their union and the sequence $X \smallsetminus C_n$ form an open cover of $X$—which has a finite subcover, so some $C_n$ lies in $U \cup V$. Since $C_n$ is connected, it lies entirely in one of $U$ or $V$; and so does $C$.

Step 1: Turn connectivity data into a tree

Claim. Let $Z$ be the zero locus of a continuous odd $f: S^2 \to \mathbb{R}^2$. If $S^2 \smallsetminus Z$ has finitely many connected components, then the antipodal map on $S^2$ sends some component of $Z$ to itself.

Let $G$ be the graph where:

  • Vertices are connected components of $Z$ and $S^2 \smallsetminus Z$, the sets of which we respectively denote $V_0$ and $V_{\neq 0}$.
  • Two vertices are joined by an edge if their union is a connected set in $S^2$—equivalently, if $K \in V_0$ meets the closure of $U \in V_{\neq 0}$.

$G$ is bipartite with parts $V_0$ and $V_{\neq 0}$, and the assumption of Step 1 is that $V_{\neq 0}$ is finite; so—since $S^2$ is connected—$G$ is connected with finite diameter. By the Key Fact, every vertex in $V_{\neq 0}$ is a cut vertex, which makes $G$ a tree.

Every involution of a finite-diameter tree preserves either an edge or a vertex; and since the antipodal map can't exchange the parts of $G$ or fix any member of $V_{\neq 0}$, it has to fix exactly one vertex in $V_0$. This is the connected component of $Z$ we're after.

Step 2: Infinitely many components in $S^2 \smallsetminus Z$ is okay

Approximate $f$ by a sequence $f_n$ where $f_n$ is equal to $f$ on the connected components of $\{f \neq 0\}$ admitting an open ball of radius $\pi/n$ and zero everywhere else.

The area of an open ball of radius $\pi/n$ on $S^2$ is at least $4\pi/n^2$ (the area of a sphere of circumference $2\pi/n$). So each $f_n$ has no more than $n^2$ components to its nonzero locus; and running them through Step 1 yields a sequence of nested closed connected sets $C_1 \supseteq C_2 \supseteq \cdots$ preserved by the antipodal map.

Then the intersection of the sets $C_n$ is preserved by the antipodal map, in $Z$ (every component of $S^2 \smallsetminus Z$, being open, contains an open ball), and connected (by the Lemma before Step 1).

  • $\begingroup$ Thank you so much for resolving this! I wish I could accept two answers to this question. $\endgroup$ Nov 13 '17 at 0:21

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.