Parker Glynn-Adey

Of Waists and Spheres

Posted in Math by pgadey on 2013/11/19

These are some notes that I’m writing up on Gromov‘s waist inequality. We’ll look at some standard material about the Borsuk-Ulam theorem and finish with a nice application of the inequality.

Firstly recall the celebrated Borsuk-Ulam theorem:

Theorem 1 If {f : S^n \rightarrow {\mathbb R}^n} is a continuous function then there is {x \in S^n} such that {f(x) = f(-x)}.

The statement above can be thought of quantitatively. Let {A(f) = \{x \in S^n : f(x) = f(-x) \}}. Borsuk-Ulam asserts that: {|A(f)|\geq 2} for all {f}. This is a little artificial, but it is an interesting re-phrasing of the result in that it makes the statement about a particular measurement of the complexity of {f}. Here is a statement, due to Lyusternik and Shnirel’man, which is equivalent to Borsuk-Ulam and which has an even stronger quantitative feel to it:

Theorem 2 If {S^n} is covered by open sets {U_1, \dots, U_{n+1}} then some {U_i} contains a pair of antipodal points. Moreover, the result holds if we drop the requirement that each {U_i} is open but require instead each set to be either open or closed.

The Borsuk-Ulam theorem is excellent in that it has many restatements and many proofs. I’m going to try and put it in a larger context of maps and their fibers. For more information about Borsuk-Ulam, one can do no better than Using Borsuk-Ulam by Jiri Matoushek.

We now turn our attention to Gromov’s wasit inequality: Let {S^n} denote the round unit sphere in {{\mathbb R}^{n+1}}. It carries the spherical measure which we denote {\mathop{\operatorfont vol}\nolimits(\cdot)} and a Riemannian metric which induces a distance function {d(\cdot,\cdot)}. We write {S^{n-k}} for the equitorial {(n-k)}-sphere inside {S^n}:

\displaystyle  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 3 If {f : S^n \rightarrow {\mathbb R}^k} is continuous then there is a point {z \in {\mathbb R}^k} such that:

\displaystyle  \mathop{\operatorfont vol}\nolimits(U_\epsilon(S_z)) \geq \mathop{\operatorfont vol}\nolimits(U_\epsilon(S^{n-k}))

for all {\epsilon > 0}

The statement above says that there is a point such that Minkowski volume of the level set hitting this point is at least the Minkowski volume of the equator. Consider the case {n=k}. If we are lucky and get that the level set in question has {|S_z| \leq 2} then this inequality implies Borsuk-Ulam. If the equality above holds for all {\epsilon} then we can take {\epsilon = \pi / 2} and obtain that {S_z = \{x, -x\}}. If the two points in {S_z} were not antipodal to begin with then their {\pi/2}-neighbourhoods would not cover and the inequality would be violated.

Here’s another demonstation which shows how powerful the waist inequality is. We’ll derive topological invariance of domain from the waist inequality. Suppose we had a homeomorphism: {f : {\mathbb R}^{k+1} \rightarrow {\mathbb R}^k}. Take {T : S^n \rightarrow {\mathbb R}^{n+1}} a linear map. Note that {T} the fibers of {T} are going to be {(n-k-1)}-dimensional spheres. Now form the composition {h(x) = f \circ T : S^n \rightarrow R^{n}}. Since {f} is injective, the level sets of {h} look like the level sets of {T}. The waist inequality applies to this composition of continuous function. We then that the level sets of {h} are {(n-k-1)}-dimensional and hence have zero {(n-k)}-volume. Taking {\epsilon \rightarrow 0} and normalizing to pick up {n-k} volume, one contradicts the waist inequality.

And to end of this brief tour, here’s a way to a simple but important case of the waist inequality which I’ve mentioned before. The classical isoperimetric inequality on the sphere says: Any curve of length {L} on the sphere bounds a region with area {A} satisfying {A(\mathop{\operatorfont vol}\nolimits_2(S^2) - A) \leq L^2}. Note {A \mapsto \mathop{\operatorfont vol}\nolimits_2(S^2) - A} symmetry of the lefthand side. This means that the inequality always picks the smaller of the two halves of the sphere bounded by our curve. The waist inequality for maps {S^2 \rightarrow {\mathbb R}^1} follows from the isoperimetric inequality. We know there is some {t} such that {\mathop{\operatorfont vol}\nolimits_2(f^{-1}((-\infty, t]) = \mathop{\operatorfont vol}\nolimits_2(f^{-1}((t,\infty))) = \frac{ 1 }{ 2 }\mathop{\operatorfont vol}\nolimits(S^2)}. Let then have that {L = \mathop{\operatorfont vol}\nolimits_1(f^{-1}(t))} satisfies {\frac{ 1 }{ 2 }\mathop{\operatorfont vol}\nolimits(S^2) \left( \mathop{\operatorfont vol}\nolimits(S^2) - \frac{ 1 }{ 2 }\mathop{\operatorfont vol}\nolimits(S^2) \right) \leq L^2} and hence { 2\pi = \mathop{\operatorfont vol}\nolimits_1(S^1) \leq L = \mathop{\operatorfont vol}\nolimits_1( f^{-1}(t))}.

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