Notes of Roberta Ghezzi’s lecture

BV functions and sets of finite perimeter in SR manifolds

Roberta Ghezzi

with Luigi Ambrosio and Valentino Magnani

1. BV functions on manifolds

Definition 1 Let be an open set in . An function on is BV if

To make sense of this definition on a manifold, one merely needs to define divergence, i.e. a volume element suffices, and a class of admissible vector fields.

Definition 2 Let be a smooth oriented manifold, a smooth nowhere vanishing top dimensional form on , a compactly supported vector field on . The divergence of is the function such that

Definition 3 Let be a smooth function which is quadratic on fibers. Let denote the class of vector fields on such that everywhere.

Definition 4 Let be a smooth oriented manifold, a smooth nowhere vanishing top dimensional form on . An function on is BV if, for all , the distributional derivative exists and

Theorem 5 (Characterization) Let be on . Then the distributional derivative exists if and only iff

Theorem 6 (Structure) Let be in BV. Then

1. The map, defined on open sets by

   

   is the restriction of a measure.

2. There is a Borel measurable vector field on such that - almost everywhere, and which achieves the supremum.

Note that a more general theory of BV functions on metric spaces has been started by Ambrosio and Miranda in the early 2000′s. Our class of BV functions is a priori larger than theirs (where the metric is the Carnot-Carathéodory metric associated to ). It coincides with theirs in the sub-Riemannian case (i.e. vectors of finite length form a smooth sub-bundle), to which I will stick from now on.

2. Sets of finite perimeter

Definition 7 A subset has finite perimeter if .

In the Euclidean case, De Giorgi’s rectifiability theorem states

Theorem 8 (De Giorgi) In , if has finite perimeter, then is concentrated on the reduced boundary which is -rectifiable.

The main tool in the proof is the

Theorem 9 (Blow up) In , if has finite perimeter, then at -almost every point (Lebesgue points of ), dilates of converge to the indicator of a half space.

Note that . By homogeneity of , the dilated indicators are bounded in BV. Some weak limit exists, it is of the form where has finite perimeter, and its normal is constante. This implies that is a halfspace.

2.1. Sub-Riemannian case

De Giorgi’s theorem has been extended two step 2 Carnot groups by Franchi, Serapioni and Serra-Cassano. One difficulty If

then for all brackets .

2.2. Blow up

We have been able to generalize their result to sub-Riemannian manifolds with varying geometry.

Theorem 10 (Mostow-Margulis, Bellaïche) In a sub-Riemannian manifold, there is a metric tangent cone at each point, it is isometric to a sub-Riemannian structure on the Lie group generated by the homogeneous degree components of the generating vector fields.

Theorem 11 Let have finite perimeter. Let be a point of the reduced boundary . If the nilpotent aproximation at is a 2-step Carnot group, then blow ups of at converge to vertical half-spaces.

The proof relies on the asymptotic doubling property (general for metric measure spaces satisfying Poincaré inequality).

2.3. Open questions

For higher step groups, Ambrosio-Kleiner-Le Donne prove convergence of a subsequence of dilates to a vertical half-space. They cannot prove rectifiability of the reduced boundary.

There is an example of a set with constant normal in Engel’s group which is not a vertical half-space. Nevertheless, its blow-ups are half-spaces at all but one point. Therefore I still think that there should be a blow-up, a half-space, almost everywhere.

Notes of Genevieve Walsh’s lecture

Right-angled Coxeter groups, polyhedral complexes, and acute triangulations

Joint work with Sang-Hyan Kim.

1. Coxeter groups

Given a simplicial graph , there is an associated right-angled Coxeter group, generated by vertices of , and two generators commute iff the corresponding vertices are joined by an edge.

2. Acute triangulations

An acute triangulation is a combinatorial triangulation of the 2-sphere which can be realized by geodesic triangles with acute angles.

Theorem 1 Let be a triangulation of the 2-sphere. The corresponding Coxeter group is one ended and word hyperbolic if and only if is acute.

This group theory theorem has a purely combinatorial corollary.

Corollary 2 Let be a triangulation of the 2-sphere. Then is acute if and only if it has no separating 3 or 4-cycles.

Indeed, Davis and Moussong show that 3 and 4-cycles produce spheres and tori in the Cayley graph. Conversely, Andreev’s theorem implies that, when there are no 3 and 4-cycles, the triangulation is achieved by a hyperbolic polyhedron . coincides with the group generated by reflections in the faces of , which is one ended and hyperbolic.

3. Proof

From , Davis and Moussong construct a cube complex . It is a subcomplex of the cube . For each simplex in , add the faces parallel to the coordinate planes. One easily sees that a 4-cycle produces a torus in .

Useful: draw when is a triangulation of the 1-sphere. Get a polyherdon combinatorially equivalent to the reflection tiling of a right-angled polygon in hyperbolic plane.

The link of a vertex in is dual to . So is a 3-manifold.

3.1. Hyperbolic acute

Assume that is one ended and hyperbolic. Then admits a -invariant hyperbolic metric. Since is generated by mirror symmetries, it has a fundamental domain which is a right-angled polygon. Pick a point inside , draw geodesics normal to each side. This gives rise to an acute triangulation of the sphere at infinity, which is combinatorially equivalent to .

3.2. Acute hyperbolic

Conversely, from an acute triangulation, we construct a complex on which acts geometrically. It will be made of hyperbolic polyhedra. In order to get , it suffices to check that links are . For this, we shall use a result of C. Hodgson and I. Rivin.

3.3. Gauss images

Hodgson and Rivin introduce the Gauss image of a hyperbolic polyhedron . For a Euclidean polyhedron , for each , take the set of unit normals of supporting planes at . This tessalates the 2-sphere. Similarly, for each vertex , take the polar dual, glue them together. This produces a spherical complex which is not isometric to the 2-sphere.

Theorem 3 (Hodgson-Rivin) A spherical complex is the Gauss image of a hyperbolic polyhedron iff

1. angles around vertices are ,
2. lengths of closed geodesics are

We call such complexes strongly .

3.4. Sequel of proof

Start with an acute triangulation . Construct a Euclidean polyhedron whose Gauss image is as follows. For each vertex of , ake the plane in tangent to the sphere at that vertex, let be the intersection of half-spaces defined by these spaces which contain the unit ball. Then is strongly obtuse. Do the same thing in hyperbolic space from a very small ball, get a polyhedron which is again strongly obtuse. Do as if it were right angled, and build a complex from it. The links are combinatorially equivalent to the octahedron triangulation. According to Theorem 2, triangles in links are larger than octants, so links are .

Question. Do there exist acute triangulations of , ? Examples exist for .

Question. Study the space of acute triangulations of a given combinatorial type. Study the space of polyhedral complexes with a given combinatorial structure.

Notes of Marc Bourdon’s lecture

Some applications of -cohomology to the boundaries of Gromov hyperbolic spaces

Joint work with Bruce Kleiner.

1. cohomology

1.1. Definition

Definition 1 Let be a contractible hyperbolic simplicial complex with bounded geometry. Its -cohomology is the following quotient space,

Theorem 2 (Strichartz, Bourdon-Pajot) If , then admits a measurable radial extension . Furthermore, the map mod constants descends to an injective map on cohomology.

Radial means that is the limit of along rays representing .

1.2. Numerical invariants

It is convenient to stick to continuous functions arising as radial extensions. These form an algebra of functions, from which further invariants can be extracted.

Definition 3

The equivalence relation on is defined by

Define numerical invariants

Theorem 4 (Kleiner, Carrasco) Assume is cocompact. Then

Theorem 5 (Bourdon-Pajot) Assume satisfies the combinatorial -Loewner property for some . Then

I will not define combinatorial Loewner property. It is quasi-Möbius invariant. It follows from the usual, analytic, Loewner property. The converse is conjectured to hold (Kleiner).

Example 1 Boundaries if rank one symmetric spaces, of Fuchsian buildings, are Loewner. Poincaré inequality implies Loewner.

2. New results

2.1. Ways to construct cohomology classes

The classical procedure is

Proposition 6 (Elek, Pansu) If , then .

I will use relative classes. Assume is a subcomplex. Then, by excision,

where denotes the boundary of as a subset of .

I will illustrate this construction in two situations.

2.2. Amalgamated products

Let , be hyperbolic groups, let be a quasi-convex and malnormal subgroup both of and of . Let . Take as (resp. ) the Cayley graph of (resp. ). Then is a disjoint union of copies of the Cayley graph of , indexed by .

Theorem 7 Assume is separable in , i.e. there exist a sequence of finite index subgroups of such that . Let . Then

1. If , then for large enough,

   

2. If , then, for and for large enough, the equivalence relation on is of the form

   

In other words, when replacing with , the cosets get far away from each other, and one can construct functions which are constant on these cosets, while keeping their gradients in .

Example 2 Let and have Sierpinski carpet boundaries with different conformal dimensions. Amalgamate them along peripheral Fuchsian subgroups. The result cannot have Loewner property.

2.3. Polygonal complexes

Theorem 8 Let be a simply connexted 2-complex such that

1. Every 2-cell is isometric to an Euclidean polygon with at least sides.
2. Every pair of 2-cells meet along at most a vertex or an edge.
3. Every edge belongs to at most 2-cells.

(thus is hyperbolic). Then

In many cases, this upper bound is much better than previously known bounds (independent on ), based on area growth for the natural metric.

For this, one first constructs a tree in using geodesic segments in cells orthogonal to edges. Let be a thickening of that tree. The 2-cells in are polygons with two types of edges. It is not hard to compute when the relative cohomology of separates ends. Indeed, push frontier edges to infinity, replacing each cell by an ideal triangle. The result, like a hyperbolic building, admits retractions to hyperbolic plane. Pull back a smooth function on the closed disk. This yields a function which is constant on frontier edges, with controllable .

This construction seems specific to dimension 2.

Remark 1 increases under quasi-isometric embeddings, but does not.

Notes of Matias Carrasco’s lecture

Conformal dimension and canonical splittings of hyperbolic groups

Can one characterize hyperbolic groups whose conformal dimension equals one ? For the definition of conformal dimension, see Haissinsky’s talk.

1. The Kleinian dimension

For Kleinian groups, an other dimension-like invariant arises, the infimum of Hausdorf dimensions of limit sets of fundamental groups of Kleinian manifolds homeomorphic to .

By definition, . Groups with have a nice characterization.

Theorem 1 (Canary-Minsky-Taylor) if and only if is a generalized book of -bundles, i.e. there exists a collection of essential annuli in such that all closures of connected components of their complement are either

1. Solid tori.
2. -bundles.

In particular, there is no rigid vertex in the JSJ decomposition of .

Question. ?

We shall see that the answer is no.

2. Splittings

In general, one understands ideal boundaries of groups which admit quasiconvex splittings: contains one copy of the ideal boundary of each vertex group , a vertex of the splitting tree , whose size decays as tends to infinity, and is compactified by adjoining . We shall first split along finite groups, and then along virtually cyclic groups.

Theorem 2 (Stability under Dunwoody-Stalling splitting) Recall that a Dunwoody-Stalling splitting is a splitting over finite groups with one-ended vertex groups. In this case,

Theorem 3 (Well spread local cut points condition) Let be one-ended. Assume

Then

Recall Bowditch’s JSJ decomposition for hyperbolic groups (see Haissinsky’s talk).

Corollary 4 If all vertex groups in the JSJ decomposition are virtually free, then satisfies condition (WS). Therefore .

Example 1 (of a group with conformal dimension 1 and ) Remove a disk to a 2-torus, take a product with an interval. Then glue to it a solid torus and two -bundles. This satisfies (WS) but it is not a generalized -bundle.

Remark 1 If there exists a rigid vertex groups whose ideal boundary is a circle, then does not satisfy (WS).

It would desirable to fully understand the following family of examples.

Example 2 Glue two surfaces along filling geodesics.

3. A global picture

Above results should be compared to

Theorem 5 (McKay) If the JSJ decomposition of is trivial (no local cut points), then .

McKay’s theorem rules out Sierpinsky curves or Menger curves from the list of possible 1-conformal dimensional ideal boundaries of groups.

Start from a group with . Assume that has no 2-torsion. DS split it, then JSJ split vertex groups, then DS vertex groups again… In the end, only Fuchsian cocompact groups and finite groups may arise. This is why Example 2 is so important.

4. Proof of Theorem 3

The argument is inspired by J. Tyson and J.-M. Wu’s treatment of the Sierpinski gasket.

Thanks to (WS), and by self-similarity, there is a uniform upper bound on the number of points needed to disconnect an -ball into sets of diameter , for all . Fix a large . Split a ball of radius into concentric annuli of width . Each of them contains a bounded number of cut points sufficient to disconnect it from the next layers. Consider the family of curves joining the center to the complement of the ball. Any curve of must pass through at least such cut-points. So make a conformal change of metric where the weight is at cut-points and 0 elsewhere. This will give length to all curves in . On the other hand, the -volume is

which tends to as tends to infinity. This shows that the -modulus of vanishes for all , and thus, that conformal dimension equals 1.

Notes of Gilles Courtois’ lecture

Poincaré inequalities and Ricci curvature

Joint work with G. Besson and S. Hersonsky.

1. Poincaré inequalities

1.1. Definition

Definition 1 Say a metric measure space satisfies a Poincaré inequality if there exist constants , , such that for all , all functions and all balls,

Here, is an average. Tthe length of the gradient is defined as an upper gradient, i.e. the least function such that for all , and any rectifiable path from to ,

Example 1 , ; Riemannian manifolds with nonnegative Ricci curvature, nilpotent Lie groups satisfy Poincaré inequalities. Hyperbolic space, trees do not.

1.2. What are they good for, I

Colding and Minicozzi’s proof of a conjecture of Yau stating that when , the space of harmonic functions of polynomial growth is finite dimensional goes as follows.

Pick very large, . Cover optimally with balls of radius . Then by Bishop-Gromov. Let be a finite dimensional vector space of harmonic functions of growth . Estimate the dimension of as follows. Consider the map , mapping to its averages on balls. One can choose such that for any , there exists such that is injective. Indeed, let be in the kernel. Then

Since is harmonic, the reverse Poincaré inequality holds,

(this does not require any curvature assumption). Combining these inequalities shows that grows fast, which contradicts the assumption that has polynomial growth, unless .

Question. Is the above property (finitely many harmonic functions of polynomial growth) a quasi-isometry invariant ? What about rough isometries ?

There are counter examples for bounded harmonic functions (Lyons-Sullivan).

1.3. What are they good for, II

Theorem 2 (Bonk-Kleiner) Let be a compact, -Ahlfors regular metric space which is homeomorphic to the 2-sphere. Assume satisfies a -Poincaré inequality. Then and is quasi-symmetric to the round 2-sphere.

As a special case, let be the ideal boundary of a hyperbolic group . In general, Poincaré inequality is not satisfied. Otherwise, one could conclude that is virtually a lattice in hyperbolic 3-space, thus solving Cannon’s conjecture.

2. Attempt : a weaker inequality

Question. Does polynomial growth plus a lower bound on Ricci curvature imply a Poincaré inequality ?

Example 2 Let be a compact negatively curved manifold. The horospheres in the universal cover have bounded sectional curvature and polynomial volume growth.

The following result can be extracted from works by Th. Coulhon and L. Saloff-Coste.

Theorem 3 Let be a Riemannian manifold with Ricci curvature bounded below and polynomial growth . Then there exist , , , such that for all , for all , all functions and all balls,

The weakness is the in the exponent. Nevertheless, it is sharp. Indeed, consider the stupid comb-shaped tree (of quadratic growth), and a function with concentrated on one edge. Note that horosphere should not look like that. Indeed, they are kind of quasi-periodic.

The proof is done first for graphs, in which case it is trivial (sometimes called Poincaré’s duplication principle). Then manifolds are approximated by graphs.

Question. Let be a compact hyperbolic manifold. Change the metric. Compare respective horospheres : are they quasi-symmetrically equivalent ?

Question. Let be a compact negatively curved manifold. Do horospheres satisfy two-sided volume bounds

Notes of Sa’ar Hersonsky’s lecture

Combinatorial harmonic coordinates

Uniformizing combinatorial annuli.

1. Perspective

Can a combinatorial structure determine a rigid geometry ? Here are interesting cases where this works.

Theorem 1 (Thurston, Rodin-Sullivan, Schramm-He, Beardon-Stephenson, Colin de Verdière…) Cover a planar domain with small equal circles. Apply Koebe’s theorem, get a piecewise affine map of the unit disk into mapping centers of circles to centers of circles. As size of circles tends to zero, this map converges uniformly to a Riemann mapping. The ratio of radii of corresponding circles converges to the modulus of the derivative of the Riemann mapping.

Our work : We construct flat surfaces starting from combinatorial data. This can be viewed as a discrete uniformization, in the spirit of Schramm and Cannon-Floyd-Parry.

2. Boundary values on graphs

Let be -connected. Split its boundary into wher is the outermost component. Triangulate it. Let a symmetric conductance function. Then Laplacian makes sense,

Harmonic functions satisfy at inner vertices.

The discrete Dirichlet boundary value problem (D-BVP) consists in finding a harmonic function with prescribed boundary values constant at and at .

The solution is used in the following theorem, in specifying the target space (replacement for the unit disk).

3. A warm up

Theorem 2 (Brooks-Smith-Stone-Tutte 1940) Let be an annulus, a positive number. Let be the straight Euclidean cylinder with height and circumference

Then there exists a mapping which associates to each edge of a unique embedded Euclidean rectangle in in such a way that the collection of these rectangles form a tiling of .

This map preserves energy. It seems that Dehn already had the idea of using Kirckhhoff’s laws in 1903, and pointed out difficulties which are still. This was clarified by Cannon-Floyd-Parry (1994) and Benjamini-Schramm (1996). I have an alternate proof.

The difficulty is that the mapping cannot be extended to a homeomorphism

We shall make a change of charts: We view the given triangulaton as a set of initial charts, and we shall improve on it.

4. A new theorem

Theorem 3 Let be the concentric Euclidean annulus with inner and outer radii and period of (see below).

Then there exist a cellular decomposition of and

1. a tiling of by annular shells,
2. a homeomorphism mapping each quadrilateral in onto a single annular shell in , and preserving area.

4.1. What goes into the proof

Fact: the level curves of foliate . No critical points.

4.2. Construction of a combinatorial angle

We define a new function, , on , on the annulus minus a slit, the conjugate function of . It is obtained by summing the normal derivatives of along a suitably chosen PL path, joigning the slit to a vertex.

Properties: Level curves of have no endpoints in the interior, and join to . Any two are disjoint. The intersection number between level curves of and level curves of is 1.

4.3. Constructing a rectangular net

Consider the collection of level sets of containing all vertices of . So and . Do the same for .

Definition 4 A rectangular combinatorial net on i a cellular decomposition of where each 2-cell is a simple quadrilateral, and a pair of functions and which satisfy

for all edges.

Theorem 5 There exists a choice of conductances such that and and their level sets form a rectangular combinatorial net.

5. Higher connectivity

Most of the discussion extends to higher connectivity domains.

Split along singular level sets of . Components need not be annuli (this makes it hard). These level sets have a naturally defined length, in terms of the period of a conjugate function . This allows to pile up model annuli and get a model surface of high connectivity.

Notes of Peter Haissinsky’s lecture

Hyperbolic groups with planar ideal boundaries

I will prove the following theorem.

Theorem 1 Let be a word hyperbolic group whose boundary is homeomorphic to a proper subset of the 2-sphere. Then the following are equivalent.

1. is virually a convex cocompact Kleinian group.
2. .
3. acts cellularly and geometrically on a cube complex.

This allows us to give new proofs of the following results.

Corollary 2 (Bonk-Kleiner) If is homeomorphic to the Sierpinski carpet, then is virtually Kleinian if and only if .

(note that no proof of this longstanding announcement has appeared yet).

Corollary 3 (Markovic) If is homeomorphic to the 2-sphere and acts cellularly and geometrically on a cube complex, then is virtually a uniform lattice of .

1. Tools

1.1. Kleinian groups

Kleinian means discrete group of isometries of . Such a group acts conformally on the 2-sphere, with a limit set and an ordinary set where the action is properly discontinuous. If is convex-cocompact, is a compact 3-manifold with boundary.

I will need Thurston’s hyperbolization theorem in the following form.

Theorem 4 (Thurston) If is a compact 3-manifold with non empty boundary and word hyperbolic fundamental group, then for some Kleinian group .

1.2. Conformal gauge

Gromov, Coornaert, Bowditch. The ideal boundary of a word hyperbolic group carries a quasi-symmetry (in fact, quasi-Möbius) class of metrics whch are Ahlfors-regular

Definition 5 (Bourdon-Pajot) The Ahlfors-regular conformal dimension is the infimum of Hausdorff dimensions of Ahlfors-regular metrics in the gauge.

(1)(2) in our Theorem is due to D. Sullivan.

(1)(3) in our Theorem is due to N. Bergeron and D. Wise.

2. Sketch of proof of (2)(1)

2.1. Planar actions

First treat a special case : planar actions. Assume that the given homeomorphism has the following extra property: for any connected component of the complement of , and any , there exists an other component such that .

The heart of the proof is the following

Proposition 6 Let be one-ended with . Assume admits a planar action. Then the action is conjugate to that of a Kleinian group.

2.2. JSJ decomposition

Theorem 7 (Bowditch) Let be a word hyperbolic group. Then acts on a tree with finite quotient and no edge inversions. Furthermore,

1. Edge stabilizers are virtually cyclic.
2. Vertex stabilizers belong to one of the following classes
   1. Virtually cyclic,
   2. Virtually free and preserving a canonical cyclic order on adjacent edges.
   3. Rigid, i.e. quasi-convex, non elementary, and not in former classes.

The keypoint of reduction from general case to special case (planar action) is the following

Proposition 8 Let be one ended and let be a planar embedding. Then the action of any rigid vertex subgroup of the JSJ decomposition of on extends to a convergence action on the whole 2-sphere.

This provides us with a compact 3-manifold with boundary for each rigid vertex.

The next steps (getting further pieces for non rigid vertices and for edges, gluing them together) require to improve slightly the JSJ decomposition. One needs

1. Elementary vertex groups are cyclic and fix components of the complement of their limit sets.
2. Surface (with boundary) vertex groups act freely.
3. Rigid vertex groups are torsion free.

Proposition 9 If all rigid vertex groups have conformal dimension , then there exists a finite index subgroup with a regular JSJ decomposition.

The proof relies on Agol and Wise’s hierarchical decompositions.

NB: The proof gives a slightly stronger result in case has no 2-torsion: It suffices to assume that Sierpinsky carpet subgroups have conformal dimension .

3. Rigid vertex groups are planar

3.1. Special case : rigid groups

When itself is rigid, i.e. it has a trivial JSJ decomposition, has no cut points, so it is a Sierpinsky carpet. The planar action assumption is automatically satisfied. Indeed, components of the complement are disks bounded by peripheral circles embedded in the carpet. Any self-homeomorphism of the carpet permutes peripheral circles, and thus permutes components of the complement.

3.2. General case

Being a rigid vertex subgroup in the JSJ decomposition of a group is a weaker assumption. Components of the complement of are simply connected, the Riemann mapping onto them extends continuously to the boundary, but the extension may be non injective, in case has cut points. One must get rid of them.

Each time two cut points arise on the boundary of the same component, join them with an arc and pinch it. Show that the resulting space is again a 2-sphere, that components of the complement of the limit set are disks. By Gabai’s theorem, the action extends to these disks.