Poincar´e inequalities for differential forms on Heisenberg groups

Motivation Sub-Riemannian case Questions

Poincar´ e inequalities for differential forms on Heisenberg groups

Pierre Pansu, Univ. Paris-Sud. Joint with A. Baldi, B. Franchi, Bologna

October 17th, 2017

Motivation Sub-Riemannian case Questions

Euclidean Poincar´e inequality Leray’s acyclic covering theorem revisited

Definition

X simplicial complex. Cochains are functions on simplices. When is an`<sup>p</sup> cocycle the coboundary of an`^qcochain ? Set

`<sup>q,p</sup>H<sup>k</sup>(X) ={`^pk-cocycles}/d{`^qk−1-cochains}.

X finite : topological invariant.X infinite : quasi-isometry invariant. Well defined for discrete groups, gives rise to numerical invariants of discrete groups...

Example.X = subdivided line. Then all`<sup>q,p</sup>H<sup>0</sup>(X) = 0,`^∞,1H¹(X) = 0, all other

`^q,pH¹(X)6= 0.

Example.X = tesselated plane. Then`^q,pH¹(X) = 0 if 1 p−1

q ≥1

2. Indeed, Sobolev inequality allows to handle the case of finitely supported cocycles. It states that, for a smooth compactly supported functionuon the plane, ifp<2,

kuk_q≤Ckduk_p.

Questions. Handle infinitely supported cocycles ? Pass from discrete to continuous and backward ?

Motivation Sub-Riemannian case Questions

Euclidean Poincar´e inequality Leray’s acyclic covering theorem revisited

Definition

X simplicial complex. Cochains are functions on simplices. When is an`<sup>p</sup> cocycle the coboundary of an`^qcochain ? Set

`<sup>q,p</sup>H<sup>k</sup>(X) ={`^pk-cocycles}/d{`^qk−1-cochains}.

X finite : topological invariant.X infinite : quasi-isometry invariant. Well defined for discrete groups, gives rise to numerical invariants of discrete groups...

Example.X = subdivided line. Then all`<sup>q,p</sup>H<sup>0</sup>(X) = 0,`^∞,1H¹(X) = 0, all other

`^q,pH¹(X)6= 0.

Example.X = tesselated plane. Then`^q,pH¹(X) = 0 if 1 p−1

q ≥1

2. Indeed, Sobolev inequality allows to handle the case of finitely supported cocycles. It states that, for a smooth compactly supported functionuon the plane, ifp<2,

kuk_q≤Ckduk_p.

Questions. Handle infinitely supported cocycles ? Pass from discrete to continuous and backward ?

Motivation Sub-Riemannian case Questions

Euclidean Poincar´e inequality Leray’s acyclic covering theorem revisited

Definition

X simplicial complex. Cochains are functions on simplices. When is an`<sup>p</sup> cocycle the coboundary of an`^qcochain ? Set

`<sup>q,p</sup>H<sup>k</sup>(X) ={`^pk-cocycles}/d{`^qk−1-cochains}.

X finite : topological invariant.X infinite : quasi-isometry invariant. Well defined for discrete groups, gives rise to numerical invariants of discrete groups...

Example.X = subdivided line. Then all`<sup>q,p</sup>H<sup>0</sup>(X) = 0,`^∞,1H¹(X) = 0, all other

`^q,pH¹(X)6= 0.

Example.X = tesselated plane. Then`^q,pH¹(X) = 0 if 1 p−1

q ≥1

2. Indeed, Sobolev inequality allows to handle the case of finitely supported cocycles. It states that, for a smooth compactly supported functionuon the plane, ifp<2,

kuk_q≤Ckduk_p.

Questions. Handle infinitely supported cocycles ? Pass from discrete to continuous and backward ?

Motivation Sub-Riemannian case Questions

Euclidean Poincar´e inequality Leray’s acyclic covering theorem revisited

Definition

X Riemannian manifold. Set

L^q,pH^k(X) ={L^pclosedk-forms}/d{L^qk−1-formsωsuch thatdω∈L^p}.

Questions. Compute it. IfX is triangulated, doesL^q,pH^k(X) =`^q,pH^k(X) ?

Example.X =Rⁿ. ThenL^q,pH^k(X) = 0 if 1<p≤q<∞and 1 p−1

q =1 n. Proof. Let ∆ =d^∗d+dd^∗. Then ∆ has a pseudo-differential inverse which commutes withd.T=d^∗∆⁻¹has a homogeneous kernel of degreen−1, hence is bounded L^p→L^qprovided ¹_p−_q¹=¹_n (Calderon-Zygmund). Finally 1 =dT+Td.

Example.X = ball inRⁿ. ThenL^q,pH^k(X) = 0 if 1<p≤q<∞and 1 p−1

q≤ 1 n. Proof(Iwaniec-Lutoborsky). Poincar´e’s homotopy formula provides a homotopyT, 1 =dT+Td, which has a homogeneous kernel of degreen−1. It does not require forms to be globally defined. H¨older⇒qcan be lessened. Works for convex sets.

Motivation Sub-Riemannian case Questions

Euclidean Poincar´e inequality Leray’s acyclic covering theorem revisited

Definition

X Riemannian manifold. Set

L^q,pH^k(X) ={L^pclosedk-forms}/d{L^qk−1-formsωsuch thatdω∈L^p}.

Questions. Compute it. IfX is triangulated, doesL^q,pH^k(X) =`^q,pH^k(X) ?

Example.X =Rⁿ. ThenL^q,pH^k(X) = 0 if 1<p≤q<∞and 1 p−1

q =1 n. Proof. Let ∆ =d^∗d+dd^∗. Then ∆ has a pseudo-differential inverse which commutes withd.T=d^∗∆⁻¹has a homogeneous kernel of degreen−1, hence is bounded L^p→L^qprovided ¹_p−_q¹=¹_n (Calderon-Zygmund). Finally 1 =dT+Td.

Example.X = ball inRⁿ. ThenL^q,pH^k(X) = 0 if 1<p≤q<∞and 1 p−1

q≤ 1 n. Proof(Iwaniec-Lutoborsky). Poincar´e’s homotopy formula provides a homotopyT, 1 =dT+Td, which has a homogeneous kernel of degreen−1. It does not require forms to be globally defined. H¨older⇒qcan be lessened. Works for convex sets.

Motivation Sub-Riemannian case Questions

Euclidean Poincar´e inequality Leray’s acyclic covering theorem revisited

Definition

X Riemannian manifold. Set

L^q,pH^k(X) ={L^pclosedk-forms}/d{L^qk−1-formsωsuch thatdω∈L^p}.

Questions. Compute it. IfX is triangulated, doesL^q,pH^k(X) =`^q,pH^k(X) ?

Example.X =Rⁿ. ThenL^q,pH^k(X) = 0 if 1<p≤q<∞and 1 p−1

q =1 n. Proof. Let ∆ =d^∗d+dd^∗. Then ∆ has a pseudo-differential inverse which commutes withd.T=d^∗∆⁻¹has a homogeneous kernel of degreen−1, hence is bounded L^p→L^qprovided ¹_p−_q¹=¹_n (Calderon-Zygmund). Finally 1 =dT+Td.

Example.X = ball inRⁿ. ThenL^q,pH^k(X) = 0 if 1<p≤q<∞and 1 p−1

q≤1 n. Proof(Iwaniec-Lutoborsky). Poincar´e’s homotopy formula provides a homotopyT, 1 =dT+Td, which has a homogeneous kernel of degreen−1. It does not require forms to be globally defined. H¨older⇒qcan be lessened. Works for convex sets.

Motivation Sub-Riemannian case Questions

Euclidean Poincar´e inequality Leray’s acyclic covering theorem revisited

Proposition (Leray)

Vanishing of L^q,pH^·of all simplices suffices to prove that L^q,pH^·=`^q,pH^·for bounded geometry triangulated manifolds.

Example. (Ui) covering by stars of vertices.ωclosed 1 form onX.ω|U_i=dui, ui−uj=κij is constant on simplexUi∩Uj, it is a 1-cocycle of the triangulation.

Conversely, pick partition of unityχi. Given cocycleκ, setui=P

jχjκij. Then dui−duj= 0 onUi∩Uj, hence defines a closed 1-form.

Theorem (Pansu)

Proposition merely requires even weaker analytic information : it suffices that a closed form on ball of radius 2 have a primitive on unit ball with controlled norms.

”Loss on domain is allowed”.

Motivation Sub-Riemannian case Questions

Euclidean Poincar´e inequality Leray’s acyclic covering theorem revisited

Proposition (Leray)

Vanishing of L^q,pH^·of all simplices suffices to prove that L^q,pH^·=`^q,pH^·for bounded geometry triangulated manifolds.

Example. (Ui) covering by stars of vertices.ωclosed 1 form onX.ω|U_i=dui, ui−uj=κij is constant on simplexUi∩Uj, it is a 1-cocycle of the triangulation.

Conversely, pick partition of unityχi. Given cocycleκ, setui=P

jχjκij. Then dui−duj= 0 onUi∩Uj, hence defines a closed 1-form.

Theorem (Pansu)

Proposition merely requires even weaker analytic information : it suffices that a closed form on ball of radius 2 have a primitive on unit ball with controlled norms.

”Loss on domain is allowed”.

Motivation Sub-Riemannian case Questions

Rumin’s complex Heisenberg Poincar´e inequality Back to`p,q-cohomology

To handle Carnot groupsG, need homogeneous Laplacian : Rumin ? Forms onG split into several weights under dilations. Letd0be the weight 0 (algebraic) part ofd. Pick complements ofker(d0) andim(d0) and defined₀⁻¹. Then powers of 1−d₀⁻¹d−dd₀⁻¹stabilize to a projector ΠEonto a subcomplexE.

Π0= 1−d₀⁻¹d0−d0d₀⁻¹projects to a subspaceERof forms where less weights occur.

Set

dR= Π0◦d◦ΠE.

ThenRumin’s complex(ER,dR) is homotopic to the de Rham complex.

Example. Heisenberg groupH¹.

E_R¹consists of horizontal 1-forms,dR:E_R⁰→E_R¹is the horizontal gradient. E_R²consists of vertical 2-forms. ΠE extends a horizontal 1-formαin such a way thatdΠEαis vertical (unique choice). HencedRα=dΠEαinvolves second derivatives.

Motivation Sub-Riemannian case Questions

Rumin’s complex Heisenberg Poincar´e inequality Back to`p,q-cohomology

To handle Carnot groupsG, need homogeneous Laplacian : Rumin ? Forms onG split into several weights under dilations. Letd0be the weight 0 (algebraic) part ofd. Pick complements ofker(d0) andim(d0) and defined₀⁻¹. Then powers of 1−d₀⁻¹d−dd₀⁻¹stabilize to a projector ΠEonto a subcomplexE.

Π0= 1−d₀⁻¹d0−d0d₀⁻¹projects to a subspaceERof forms where less weights occur.

Set

dR= Π0◦d◦ΠE.

ThenRumin’s complex(ER,dR) is homotopic to the de Rham complex.

Example. Heisenberg groupH¹.

E_R¹consists of horizontal 1-forms,dR:E_R⁰→E_R¹is the horizontal gradient.

E_R²consists of vertical 2-forms. ΠE extends a horizontal 1-formαin such a way thatdΠEαis vertical (unique choice). HencedRα=dΠEαinvolves second derivatives.

Motivation Sub-Riemannian case Questions

Rumin’s complex Heisenberg Poincar´e inequality Back to`p,q-cohomology

More generally, for Heisenberg groupHⁿ,ERhas exactly one weight in each degree, hencedR is homogeneous under Heisenberg dilations. It has order 1, except in degree nwhere it has order 2.

One can make choices in a contact invariant manner : (ER,dR) is invariantly defined for contact manifolds.

In presence of a sub-Riemannian metric, adjoints are defined. Let ∆R:=d_R^∗dR+dRd_R^∗ be replaced with ∆R:= (d_R^∗dR)²+dRd_R^∗ord_R^∗dR+ (dRd_R^∗)²in degreesnandn+ 1.

Then ∆R is maximally hypoelliptic, henced_R^∗∆⁻¹_R has a smooth (away from the origin), homogeneous kernel.

Proposition (Coifman-Weiss, Koranyi-Vagi 1971)

TR:=d_R^∗∆⁻¹_R is bounded L^p to L^q provided1<p<q<∞and _p¹−¹

q =_2n+2¹ (resp.

2

2n+2 in degree n+ 1).

Again, 1 =dRT+TdR.

Motivation Sub-Riemannian case Questions

Rumin’s complex Heisenberg Poincar´e inequality Back to`p,q-cohomology

Need local version.

Theorem (Baldi-Franchi-Pansu)

Closed L^pforms defined on the Heisenberg 2-ball have dR-primitives on the unit ball which are L^q, provided1<p,q<∞and ¹_p−¹

q ≤ ¹

2n+2 (resp. _2n+2² in degree n+ 1).

Proof. The sub-Riemannian analogue of Poincar´e’s homotopy formula is not compatible with Rumin’s construction.

Instead, use Rumin’s homotopy ΠE and then apply Iwaniec-Lutoborsky. Since ΠEis differential, this requires a preliminary smoothing homotopy.

LetK_Rbe the kernel ofT_R. WriteK_R=K¹+K²whereK1has small support andK2

is smooth. ThenTR=T¹+T²,

1 =dRT¹+T¹dR+S

whereSis smoothing.T¹and thereforeS map forms defined on ball of radius 2 to forms defined on unit ball.T¹mapsL^p toL^qlikeTR.Swins the derivatives that ΠE

looses.

Motivation Sub-Riemannian case Questions

Rumin’s complex Heisenberg Poincar´e inequality Back to`p,q-cohomology

Need local version.

Theorem (Baldi-Franchi-Pansu)

Closed L^pforms defined on the Heisenberg 2-ball have dR-primitives on the unit ball which are L^q, provided1<p,q<∞and ¹_p−¹

q ≤ ¹

2n+2 (resp. _2n+2² in degree n+ 1).

Proof. The sub-Riemannian analogue of Poincar´e’s homotopy formula is not compatible with Rumin’s construction.

Instead, use Rumin’s homotopy ΠE and then apply Iwaniec-Lutoborsky. Since ΠEis differential, this requires a preliminary smoothing homotopy.

LetK_Rbe the kernel ofT_R. WriteK_R=K¹+K²whereK1has small support andK2

is smooth. ThenTR=T¹+T²,

1 =dRT¹+T¹dR+S

whereSis smoothing.T¹and thereforeS map forms defined on ball of radius 2 to forms defined on unit ball.T¹mapsL^p toL^qlikeTR.Swins the derivatives that ΠE

looses.

Motivation Sub-Riemannian case Questions

Rumin’s complex Heisenberg Poincar´e inequality Back to`p,q-cohomology

Leray’s method requires control onkdR(χω)kp in terms ofkωkpandkdRωkp. This fails whend_R is second order. The way around this is

Theorem (Baldi-Franchi-Pansu)

Let1<p,q<∞and ¹_p−_q¹≤_2n+2¹ (resp. _2n+2² in degree n+ 1). On contact manifolds with C^k+1-bounded geometry, global smoothing homotopies

1 =dRT+TdR+S are defined, where T :W^s,p→W^s+1,qand S:W^s,p→W^s+k,q,

−k≤s≤0.

Corollary

Rumin’s complex can be used to compute`^q,p-cohomology of contact manifolds with bounded geometry for this range of p,q.

Corollary

`^q,pH^·(Hⁿ) = 0provided1<p<q<∞and ¹_p−¹

q≥ ¹

2n+2 (resp. _2n+2² in degree n+ 1).

Motivation Sub-Riemannian case Questions

Questions.

1 Sobolev inequality (i.e. for compactly supported forms) ? Yes.

2 Sharpness ? Probably yes.

3 Cases whenp= 1 andq=∞? Work in progress with Baldi and Franchi.

4 Other Carnot groups ? Work in progress with Rumin, but sharp intervals are rarely attained.