I'm curious if anyone can see a route to get the Borsuk-Ulam theorem from Gromov's waist inequality. For the sake of notation, here's the inequality:

Let $S^n$ denote the round unit sphere in $\mathbb{R}^{n+1}$. It carries the spherical measure which we denote $vol(\cdot)$ and a Riemannian metric which induces a distance function $d(\cdot,\cdot)$. We write $S^{n-k}$ for the equatorial $(n-k)$-sphere inside $S^n$: $$ S^{n-k} = \{ (x_1, \dots, x_{n+1}) \in S^n : x_i = 0, i>n-k+1\} $$ For a continuous map $f : S^n \rightarrow \mathbb{R}^k$ we write $S_z = \{s \in S^n : f(s) = z\}$ for the level sets of $f$. Write $U_\epsilon(S) = \{\sigma \in S^n : d(\sigma, S) < \epsilon\}$ for an $\epsilon$-neighbourhood of a subset $S$.

Theorem (Gromov): If $f : S^n \rightarrow \mathbb{R}^k$ is continuous then there is a point $z \in \mathbb{R}^k$ such that: $$ vol(U_\epsilon(S_z)) \geq vol(U_\epsilon(S^{n-k})) $$ for all $\epsilon > 0$.

This inequality is quite strong. It implies the spherical isoperimetric inequality. It also implies topological invariance of domain. For more background, see Guth's survey The Waist Inequality in Gromov's Work.

Getting back to my question: If $n=k$ and the level set given by the inequality contains two points then taking $\epsilon = \pi/2$ gives the Borsuk-Ulam theorem. Can one arrange for this to always happen and hence derive Borsuk-Ulam?

  • $\begingroup$ I think Gromov mentions this case in his paper. $\endgroup$ Nov 14, 2013 at 8:24
  • $\begingroup$ @alvarezpaiva: Now that you mention it, I remember reading Gromovs paper and noticing that he is careful enough to write $\operatorname{card}f^{-1}(z)\leq 2$. I edited my answer below. $\endgroup$ Nov 14, 2013 at 10:17

1 Answer 1


Yashar Mermarian writes here that the answer is yes. And the argument he gives is pretty much the same as the one you already started.

Taking $n=k$ and $\epsilon=\pi/2$ Gromov's waist inequality gives a point $z\in \mathbb{R}^n$ such that $$vol(U_{\pi/2}(S_Z))\geq vol(U_{\pi/2}(S^0))=vol(U_{\pi/2}(\{(1,0,\dots,0),(-1,0,\dots,0)\})=vol(S^n).$$ It follows that $U_{\pi/2}(S_Z)$ has to have the same volume as the sphere. This can only be the case when $S_Z$ contains at least $2$ elements, since the volume of $U_\epsilon(S_Z)$ would be to small otherwise. If it contains $2$ elements and not more, these $2$ elements would have to be antipodal, otherwise the volume of $U_\epsilon(S_Z)$ would be to small again. Although Memarian writes there "is no choice" for $S_Z$ but to "pass through antipodal points", I don't really see how one can rule out that $S_Z$ might consist of more than two points.

This gives the following (weaker) version of a Bursuk Ulam type result:

If $f:\; S^n\to\mathbb{R}^n$ is continuous, then $f$ maps two antipodal points in $S^n$ to the same point in $\mathbb{R}^n$ or $f$ maps (at least) three points to the same point in $\mathbb{R}^n$.

two semispheres $U_{\pi/2}$ for two points which are not antipodal.

In the original paper Gromov is more careful. He writes:

If $k=n$, and $\operatorname{card}(f^{-1}(z))\leq 2$, $z\in\mathbb{R}^n$, then [the waist of the sphere theorem] applied to $\pi/2$ amounts to the Borsuk–Ulam theorem: some level $f^{-1}(z)$ of $f\;:S^n\to \mathbb{R}^k$ equals a pair of opposite points.

Of course $f^{-1}(z)$ is just a different notation of $S_Z$. So the important detail is: "and $\operatorname{card}(f^{-1}(z))\leq 2$".

In general $\operatorname{card}(f^{-1}(z))\leq 2$ is not true for arbitrary maps. And even if you perturbe $f$ a little bit this condition won't be fullfilled. Of course any two maps from $S^n$ to $\mathbb{R}^n$ are homotopic, so in particular $f$ will be homotopic to a map with $\operatorname{card}(f^{-1}(z))\leq 2$ for all $z$, but this doesn't really give you anything.

To summarize the answer is: Gromov's Waist Inequality almost implies Borsuk-Ulam.

  • $\begingroup$ In Mermarian's write up, the only statement I take issue with is this: "But the right hand side of the inequality is equal the total volume of the sphere, so there is no choice for $f^{-1}(z)$ but to pass through two diametrically opposite points." There are certainly $S \subset S^n$ such that $vol(U_\epsilon(S)) \geq vol(U_\epsilon(S^0))$ for all $\epsilon$ but $S \cap (-S) = \emptyset$. I guess the problem is: How generic are function $f : S^n \rightarrow \mathbf{R}^n$ such that $|f^{-1}(z)| \leq 2$ for all $z \in \mathbf{R}^n$? $\endgroup$
    – pgadey
    Nov 14, 2013 at 19:06
  • $\begingroup$ I also take issue with this. I guess Mermarian implicitly presume $|f^{-1}(z)|\leq2$. I will add a line about the fact that this is a serious restriction. $\endgroup$ Nov 14, 2013 at 19:07

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

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