Intergal geometry and the minimax inequality in a nutshell

February 7, 2011

The goal for most of the posts in this blog has been to take out some very simple parts of certain papers/subjects and “blow them up” to a point where anybody (myself included) can understand. Ideally the simple parts should give some inspirations and ideas towards the more general subject. This one is on the same vein. This one is based on parts of professor Guth’s minimax paper.

In an earlier post, we talked about the extremal length where one is able to bound the “largest possible minimum length” (i.e. the “maximum minimum length“) of a family of rectifiable curves under conformal transformation. When combined with the uniformization theorem in for surfaces, this becomes a powerful tool for understanding arbitrary Riemannian metrics (and for conformal classes of metrics in higher dimensions).

However, in ‘real life’ we often find what we really want to bound is, instead, the “minimum maximum length” of a family of curves, for example:

Question: Let \mathbb{D} \subseteq \mathbb{R}^2 be the unit disc. Given any family \mathcal{F} of arcs with endpoints on \partial \ \mathbb{D} and \mathcal{F} foliates \mathbb{D}, then how short can the logest arc in \mathcal{F} possibly be?

In other words, let \mathbb{F} be the collection of all possible such foliations \mathcal{F} as above, what is

\displaystyle \inf_{\mathcal{F} \in \mathbb{F}} \ \sup_{A \in \mathcal{F}} \ \ell(A)?

After playing around a little bit with those foliations, we should expect one of the fibres to be at least as long as the diameter ( i.e. no foliation has smaller maximum length leaf than foliating by straight lines ). Hence we should have

\displaystyle \inf_{\mathcal{F} \in \mathbb{F}} \ \sup_{A \in \mathcal{F}} \ \ell(A) = 2.

This is indeed easy to prove:

Proof: Consider the map f: S^1 \rightarrow S^1 where S^1 = \partial \ \mathbb{D}, f switches the end-points of each arc in \mathcal{F}. It is easy to check that f is a continuous, orientation reversing homeomorphism of the circle (conjugate to a reflection). Let p, q be its fixed points, L_1, L_2 be the two arcs in S^1 connecting p to q.

Let

g: z \mapsto -z

be the antipodal map on S^1.

Suppose p \neq g(q) then one of L_1, L_2 is longer than \pi, say it’s L_1.

Then we have

f \circ g (L_1) \subseteq L_1.

Hence f \circ g has a fixed point m in L_1, i.e. f(m) = -m.

There is a fibre A in \mathcal{F} with endpoints m, -m, the fibre must have length

\ell(A) \geq d(-m,m) = 2.

The remaining case is trivial: if p = g(q) then both L_1 and L_2 gets mapped into themselves orientation-reversingly, hence fixed points still exists.

Establishes the claim.

Instead of the disc, we may look at circles that sweep out the sphere (hence to avoid the end-point complications):

Theorem: Any one-parameter family of circles that foliates S^2 (except two points) must have the largest circle being longer than the equator.

This is merely applying the same argument, i.e. one of the circles needs to contain a pair of antipodal points hence must be longer than the equator.

In order for easier generalization to higher dimensions, with slight modifications, this can be formulated as:

Theorem: For any f: T^2 \rightarrow S^2 having non-zero degree, there is \theta \in S^1 where \ell(f(S^1 \times \{ \theta \}) is larger than the equator.

Hence in higher dimensions we can try to prove the same statement for largest image of a lever k-sphere under f: S^k \times S^{n-k} \rightarrow S^n. However before we do that I would like to highlight some intergal geometry machineries that are new to me but seemingly constantly used in proving those kinds of estimates. We shall get some idea of the method by showing:

Theorem: Let \mathbb{R}P^n be equipped with the round metric. p^k \subseteq  \mathbb{R}P^n be a ‘flat’ k-dimensional plane. Then any k-chain z^k \subseteq \mathbb{R}P^n in the same k dimensional homology class as p^k must have volume at least as large as p^k.

Proof: Let Gr(\mathbb{R}P^n, n-k) be the set of all (n-k)-planes in \mathbb{R}P^n (i.e. the Grassmannian).

There is a standard way to associate a measure \mu on Gr(\mathbb{R}P^n, n-k):

Let \lambda be the Haar measure on SO(n+1), fix some Q \in Gr(\mathbb{R}P^n, n-k).

Since SO(n+1) acts on \mathbb{R}P^n, for open set S \subseteq Gr(\mathbb{R}P^n, n-k), we set

\mu(S) = \lambda( \{ T \in SO(n+1) \ | \ T(Q) \in S \}).

–The measure of a collection of planes is the measure of linear transformations that takes the given plane to an element of the set.

Now we are able to integrate over all (n-k)-planes!

For almost all Q \in Gr(\mathbb{R}P^n, n-k), since P is k-plane, we have | Q \cap P | = 1. ( not 1 only when they are ‘parallel’ )

Since [z] = [p] in H_k(\mathbb{R}P^n, \mathbb{Z}_2), for almost all Q, z intersects Q at least as much as P does. We conclude that for almost all Q, \ | z \cap Q | \geq 1.

Fact: There exists constant C such that for any k-chain \Sigma^k \in \mathbb{R}P^N,

\mbox{Vol}_k(\Sigma^k) = \mathbb{E}(|\Sigma \cap Q |).

The fact is obtained by diving the chain into fine cubes, observe that both volume and expectation are additive and translation invariant. Therefore we only need to show this for infinitesimal cubes (or balls) near 0. We won’t work out the details here.

Hence in our case, since for almost all Q we have | z \cap Q | \geq 1, the expectation \mathbb{E}(|z \cap Q |) \geq 1.

We therefore deduce

\mbox{Vol}_k(z) = \mathbb{E}(|z \cap Q |) \geq 1.

Establishes the theorem.

Remark: I found this intergal geometry method used here being very handy: in the old days I always try to give lower bounds on volume of stuff by intersecting it with planes and then pretend the ‘stuff’ were orthogonal to the plane, which is the worst case in terms of having small volume. An example of such bound can be found in the knot distorsion post where in order to lower bound the length we look at its intersection number with a family of parallel planes and then integrate the intersection.

This is like looking from one particular direction and record how many times did a curve go through each height, of course one would never get the exact length if we know the curve already. What if we are allowed to look from all directions?

I always wondered if we know the intersection number with not only a set of parallel planes but planes in all directions, then are there anything we can do to better bound the volume? Here I found the perfect answer to my question: by integrating over the Grassmannian, we are able to get the exact volume from how much it intersect each plane!

We get some systolic estimates as direct corollaries of the above theorem, for example:

Corollary: \mbox{Sys}_1(\mathbb{R}P^2) = \sqrt{\pi/2} where \mathbb{R}P^2 carries the round metric with total volume 1.

Back to our minimax problems, we state the higher dimensional version:

Wish: For any C^1 map f: S^k \times S^{n-k} \rightarrow S^n where S^n carries the standard round metric, there exists some \theta \in S^{n-k} with

\mbox{Vol}_k(f(S^k\times \{\theta\})) \geq \mbox{Vol}_k(E^k)

where E^k \subseteq S^n is the k-dimensional equator.

But what we have is that there is a (small) positive constant c(n,k) s.t. \mbox{deg}(f) \neq 0 implies

\displaystyle \sup_{\theta \in S^{n-k}} \mbox{Vol}_k(f(S^k \times \{\theta\})) \geq c(n,k) \mbox{Vol}_k(E^k)

(shown by an inductive application of the isomperimetric inequality on S^N, which is obtained from applying intergal geometry methods)

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One Response to “Intergal geometry and the minimax inequality in a nutshell”


  1. […] Intergal geometry and the minimax inequality in a nutshell […]


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