Today, we shall prove the sharp gap theorem. Let me state it again.

Theorem 1 (Wenger 2008)Let be a geodesic metric space. Suppose

- There exists a geodesic thickening of with ;
- There exists and such that
Then is Gromov hyperbolic.

The assumption on existence of a thickening with quadratic filling is not restricitve. Indeed, we saw last time that Gromov guarantees extistence of such a thickening provided the coarse filling function is everywhere positive. This works for Cayley polyhedra, so we get

Corollary 2Let be a finitely presented group. If there exist and such that for all ,

then is Gromov hyperbolic.

**1. Preparation **

** 1.1. Asymptotic cones **

Asymptotic cones reflect a metric space when we look at it from further and further away.

Let be a metric space. Pick a sequence of base points and a sequence of scales . Define

We want to identify sequences which cannot be distinsguished at resolution , and get a distance on the resulting quotient. For this, we fix once and for all a non principal ultrafilter . This is a finitely additive measure on subsets of which takes only values and , and is on all finite sets. Using the axiom of choice, one can show that such things exist.

Proposition 3Let be a compact Hausdorff space, a sequence of points of . Then there is a unique point such that for all neighborhoods of , . This is denoted by

Based on this fact, for any two sequences , belonging to , the limit exists and defines a pseudo metric on .

Definition 4Let be a non principal ultrafilter on . The-asymptotic coneof is

where two sequences are identified iff their vanishes.

Theorem 5 (Gromov)Let be a geodesic metric space. Then is hyperbolic iff all its asymptotic cones are metric trees.

In fact, one can do with a fix ultrafilter, and all sequences , , or the sequence , and all sequences and ultrafilters .

** 1.2. Lipschitz maps to asymptotic cones **

Proposition 6 (Wenger 2008)Let be a geodesic metric space, a geodesic thickening with quadratic filling function. If is not hyperbolic, there exists an asymptotic cone of , a compact set and a Lipschitz map such that has positive -dimensional Hausdorff measure.

Definition 7Say that a metric space ispurely -unrectifiableif all images of Lipschitz maps from subsets of to have zero -dimensional Hausdorff measure.

Example 1The first Heisenberg group (and its quaternionic and octonionic analogues) have purely -unrectifiable asymptotic cones.

More generally, this holds for Carnot groups whose first layer of the stratification of the Lie algebra does not contain any -dimensional sub-algebra.

We shall not prove this. This depends on Rademacher’s a.e. differentiability theorem.

Corollary 8Let be geodesic and not hyperbolic. Assume that all asymptotic cones of are purely -unrectifiable. Then for all geodesic thickenings of ,

** 1.3. Metric differentiability **

Definition 9Let be an open subset of . Let be Lipschitz. Themetric directional derivativeof in direction is

Theorem 10 (Kirchheim 1994)For a.e. , metric derivatives at exist in all directions. Moreover, there exist a countable family of disjoint compact sets filling almost all of such that on each such compact set , for all there exists such that for all , ,

The estimate in Theorem implies that is a semi-norm. When it is a norm, then is nearly isometric around (-bilipschitz in a small enough ball).

Remark 1Theorem 1 extends to the case where is merely a Borel set.

One merely needs to extend to a neighborhood, with values in .

*Proof:* (Sketch). Existence of .

**Step 1**: Case .

Let be a dense countable subset of . Then is Lipschitz so a.e. differentiable. Set

Then

Furthermore, for any , ,

At a Lebesgue point of continuity of , this integral converges to , and

**Step 2**: There exists for a.e. and all .

Step 1 gives it for a dense countable set of vectors. Triangle inequality implies it for all .

**Step 3**: Prove estimate on .

Write

By Lusin’s theorem, partition into compact subsets on which , , is continuous and on which, by Egorov’s theorem,

converges uniformly to , and this does the job.

I wrote a slightly simplified proof of Theorem 1 as an appendix of my paper on *Characterization of real trees..*.

** 1.4. Area formula **

Theorem 11 (Kirchheim 1994)Let be an open subset of . Let be Lipschitz. Let be Lebesgue measurable. Then

Here, for a semi-norm on , define unless is a norm, in which case is the -dimensional Hausdorff measure, in the Banach space , on the unit cube.

Note that one recovers our definition of area for Lipschitz maps of the -disk.

*Proof:* **Step 1**: Show that the image by of the set of points where is not defined or nor a norm has vanishing -Hausdorff measure.

**Step 2**: Use the fact that is nearly isometric on the pieces of a partition.

**2. Proof of sharp gap theorem **

** 2.1. Sketch of the proof **

By contradiction. Assume has nearly Euclidean filling function, but is not hyperbolic. Then there exists an asymptotic cone and a Lipschitz map with .

By Kirchheim’s theorems 1 and 11, after possibly replacing by a smaller set, there exists a Lebesgue density point at which the metric derivative exists and is a norm, and is nearly isometric around , i.e. -bilipschitz in a small enough ball.

Let be a convex isoperimetric set in . By convex geometry, the isoperimetric profile of a -dimensional Banach space is minimal for Euclidean plane, so

Let be an arc-length parametrization. Choose large enough (depending on and ) and cut in equidistant intervals, leading to vertex set . By assumption, there exist arbitrarily large such that contains a nearly isometric copy of . Connect these vertices by geodesics to get a piecewise geodesic loop in . By assumption, there exists a Lipschitz extension of of area

One would like to extend the -Lipschitz map to a map from to in order to push the filling to .

Not possible without increasing Lipschitz constant by a factor of . Instead, view as a linear subspace of . Use injectivity of to extend into a -Lipschitz map

Then nearly fills . Adding little -dimensional pieces, one gets with and

**Open question** (Busemann): Let be an affine subspace of a Banach space , let be a convex compact subset. Does minimize -dimensional Hausdorff measure among all Lipschitz disks in with same boundary ?

Answer unknown. Fortunately, we can remedy this. Take the Holmes-Thompson area . instead of Hausdorff measure. It has the following properties.

- .
- .
- Affine disks minimize (Dima Burago and Sergey Ivanov 2002).

Therefore we have found in a filling of with

contradicting above three properties. This completes the proof.

Remark 2Properties 2. and 3. is easy for Gromov’s -measure.

If we choose the notion of area in the definition of the filling function , we do not need to resort to Burago-Ivanov’s result. We can then also allow surfaces of arbitrary genus as fillings in the function making it potentially smaller than when only allowing disk fillings and thus making the assumption 2. in the gap theorem even weaker.

** 2.2. Proof of Proposition 6 **

We shall prove a stronger statement that will be needed again for nilpotent groups. We need some preparation. To motivate it, let us attempt to prove

Proposition 12 (Wenger 2008)Let be a geodesic metric space, a geodesic thickening with quadratic filling. If is not hyperbolic, there exists an asymptotic cone of , a compact set and a Lipschitz map such that has positive -dimensional Hausdorff measure.

By Gromov’s hyperbolicity criterion, there exists an asymptotic cone which is not a tree, i.e. contains a non trivial loop. One obtains from it a sequence of uniformy Lipschitz loops in . Fill with a Lipschitz disk . Will (sub)-converge to a disk in ?

Related problem: Plateau’s problem in . Try to solve it by the direct method, i.e. pick a minimizing sequence of disks (now is always the same curve). There could be thin spikes of small area sticking out, so no convergence. People have used the following tricks:

- Douglas and Rado minimize Dirichlet integral instead. Scope limited to -disks.
- Federer and Fleming introduce normal and integral currents, i.e. linear functionals generalizing surfaces. They show that minimizing sequences have subsequences which converge to limit functionals which have nice structure.

** 2.3. Euclidean theory of currents **

**Idea**: Every -dimensional oriented submanifold with boundary of defines a linear functional on compactly supported differential -forms, with the following properties.

- Stokes’ theorem: .
- Duality volume/-norm:

So define *currents* are elements of the dual space to the space of compactly supported -forms normed by -norm.

Banach-Alaoglu’s theorem gives that the unit ball in , i.e. currents of mass , is weak-sequentially compact.

We want to define *integral currents* as the closure in of oriented submanifolds with boundary. Consider the following family of currents:

where is countably -rectifiable. is a density, is an -vector defining an orientation on the (a.e. defined) tangent space of .

Definition 13Anintegral currentis a current such that both and are of the above form.

Theorem 14(Closure theorem). If a sequence of integral currents with bounded mass and bounded boundary mass converges in , then the limit is an integral current.

** 2.4. Metric theory of currents **

**Idea** (de Giorgi): Replace smooth differential form with -tuple of Lipschitz functions .

The formula

suggests generalization

This leads to a solid theory. Here is our generalized Proposition 12.

Theorem 15 (Wenger)Let be a geodesic metric space, a geodesic thickening with quadratic filling. Assume that

Every asymptotic cone of has quadratic isoperimetric inequality for integral -currents.

Cornulier: Is there a converse ? Answer: Gromov, using his coarse filling, proves that quadratic filling in all cones implies -filling in .

As written, the last sentence of Section 2.1 is a bit misleading. It would be more accurate to write:

If we choose the mass* notion of area in the definition of the filling function FA_0^{X,\infty}(r) we do not need to resort to Burago-Ivanov’s result. We can then also allow surfaces of arbitrary genus as fillings in the function FA_0^{X,\infty}(r) making it potentially smaller than when only allowing disk fillings and thus making the assumption 2. in the gap theorem even weaker.

Dear Stefan,

Thanks for your comment. I edited the file as you suggested.

Can you please have a look at the notes of your first lecture ? The statement of Gromov’s thickening lemma, to be found at the end, looks suspicious to me. I tend to believe that the thickening improves the filling function for small values of r and should not change it for larger values of r. So it cannot state that FA^{X,Y}(r)\preceq r^2 for all r.

Sincerely,

Dear Pierre,

Thanks for making the changes. I will look at all of lecture 1 and also lecture 2 today and will write again.

As for the proposition about thickenings: yes, the thickening improves the filling function to quadratic on small scale (r<1) and has growth on the large scale bounded by that of Gromov's coarse isoperimetric function. One can even find a thickening Y such that FA_0^Y has growth at most that of Gromov's coarse function at large scales and such that FA_0^Y(r) is bounded by Cr^2 for 0<r<1, where C is a suitable constant. Note that FA_0^{X,Y} is bounded by FA_0^Y so this is even a bit better than what you wrote in the notes now. We will need this somewhat better statement when trying to produce uniformly compact fillings of given curves in lecture 3. The proof of the proposition (which is not so difficult) can be found in my paper on the sharp isoperimetric constant. I was unaware of a proof by Gromov. Is it the one you told about in the Filling Riemannian Manifolds paper?

Dear Pierre,

Thank you for taking notes. Here are, as promised, some minor corrections to the notes of lecture 2.

– Paragraph after Theorem 1: it should read “provided the coarse filling function has at most quadratic growth.”

– In Section 1.1, after Proposition 3: in the definition d_\omega(\hat{x}, \hat{y}) insert \lim_\omega

– In Section 1.1, after Theorem 5: write r_n instead of r-n

– In Section 1.2, Proposition 6: “geodesic thickening with quadratic filling function”

– In Section 1.2, in Definition 7: complete the sentence as follows: “… if all images of Lipschitz maps from subsets of R^2 to Z have zero 2-dimensional Hausdorff measure.”

– In Section 1.2, after Example 1: “… for Carnot groups whose first layer of the stratification of the Lie algebra does not contain any 2-dimensional sub-algebra.”

– In Section 1.3, Proof of Theorem 10:

– In step 1, the first inequality should read $$\liminf_{r\to0}\frac{d(\psi(t+r), \psi(t))}{|r|} \geq \liminf_{r\to0}\frac{|\gamma_n(t+r) – \gamma_n(t)|}{|r|}= |\dot{\gamma}_n(t)|.$$

– In the equation that follows, the term “\sup_n\int_s^t|\dot{\gamma}_n(\tau)|d\tau” should be between inequality signs

– In the equation that follows, at the very end insert “| = a(t)”

– In step 3, in the equation, replace w by v-w at the end of the first line

– In Section 2.1, end of second line in first paragraph: H^2(\psi(K))>0

– In Section, 2.1, middle of proof:

– in the equation, replace $\sim length(\partial I)$ by $\sim \frac{1-\varepsilon}{4\pi}length(\partial I)^2.$

– on the next line, complete the sentence “One would like to” to “One would like to extend the (1+\delta)-Lipschitz map \psi^{-1} to a map from L_\infty(X_1) to V in order to push the filling \phi to V.”

– In Section 2.1, in the equation before the Open question, replace the first occurrence of \gamma by \bar{\gamma}

– Section 2.2, title: “Proof of Proposition 6”, not Proposition 14

Thank you again. With kind regards,

Stefan