Poincaré inequalities and rigidity for actions on Banach ... · For any Hilbert space H =, every...

Poincare inequalities and rigidity for actions on Banachspaces

Piotr Nowak

IMPAN & University of Warsaw

Young GGT 2012

Piotr Nowak (IMPAN & U.Warsaw) Rigidity for actions on Banach spaces Young GGT 2012 1 / 33

Affine actions

An affine map:Av = Lv + b ,

where L is a linear map and b is a fixed vector.

Given a representation π of a group G on a linear space V an affineπ-action is of the form

Agv = πgv + bg,

where g 7→ bg is a map b : G → V such that

bgh = πgbh + bg.

A vector v ∈ V is a fixed point of this action if and only if b is a coboundary:

bg = v − πgv

for every g ∈ G.

H1(G, π) = 1-cocycles/1-coboundariesPiotr Nowak (IMPAN & U.Warsaw) Rigidity for actions on Banach spaces Young GGT 2012 2 / 33

Property (T)

Property (T) was defined by Kazhdan in late 1960’ies.

We use a characterization of (T) due to Delorme – Guichardet as adefinition.

DefinitionA group G has Kazhdan’s property (T) if every action of G by affineisometries on a Hilbert space has a fixed point.Equivalently,

H1(G, π) = 0

for every unitary representation π.


Property (T)

Property (T) is a powerful rigidity property and has several applications:

construction of expanders – Margulis (1973). Finite quotients of aresidually finite group with (T) form a family of expanders;

solution of the Ruziewicz problem – Sullivan, Margulis (1980), basedon work of J. Rosenblatt. The Lebesque measure is the only finitelyadditive rotation-invariant measure on the n-sphere, n ≥ 4;

rigidity for operator algebras and related dynamical systems(Connes–Weiss, Popa,. . . )

. . .

Proving (T) is always non-trivial.Examples:– SLn(Z) has (T) for n ≥ 3,– amenable and, more generally, a-T-menable groups do not have (T).


Generalizing (T) to other Banach spaces

X – Banach space, reflexive (X = X∗∗)

Example: Lp are reflexive for 1 < p < ∞, not reflexive for p = 1,∞.

We are interested in groups G for which the following property holds:

every affine isometric action of G on X has a fixed point

or equivalently,H1(G, π) = 0

for every isometric representation π of G on X .

This is much more more difficult than for the Hilbert space, even whenX = Lp .


Previous results

Only a few positive results are known:

(T)⇐⇒ fixed points on Lp and any subspace, 1 < p ≤ 2

(T) =⇒ ∃ ε = ε(G) such that fixed points always exists on Lp forp ∈ [2, 2 + ε) (Gromov, Fisher–Margulis)(a general argument, ε unknown)

lattices in higher rank simple Lie groups for X = Lp for all p > 1 (Bader –Furman –Gelander – Monod, 2007)

SLn(Z[x1, . . . xk ]) for n ≥ 4; X = Lp for all p > 1 (Mimura, 2010) [both use arepresentation-theoretic Howe-Moore property]

Gromov monsters for X = Lp for all p > 1 (Naor – Silberman, 2010)

Some of these results also extend to Shatten p-class operators and othernon-commutative Lp-spaces


Previous results

Some groups with property (T) admit fixed point free actions on certain Lp .

G = Sp(n, 1), the isometry group of the quaternionic hyperbolicspace, admits fixed point free actions on Lp(G), p ≥ 4n + 2(Pansu 1995)

hyperbolic groups admit fixed point free actions on `p(G) forp ≥ confdim(∂G) (Bourdon and Pajot, 2003)

for every hyperbolic group G there is a p > 2 (sufficiently large) suchthat G admits a metrically proper action by affine isometries on`p(G × G) (Yu, 2006)


Values of p (after Cornelia Drutu)

Consider e.g. a hyperbolic group G with property (T).

There are many natural questions about the above values of p.

Let P ={p : H1(G, π) = 0 for every isometric rep. π on Lp

}The only thing we know about P is that it is open.

Question: Is P connected?Piotr Nowak (IMPAN & U.Warsaw) Rigidity for actions on Banach spaces Young GGT 2012 8 / 33

Spectral conditions for property (T)

Based on the work of Garland,

used to prove (T) by Ballmann – Swiatkowski, Dymara – Januszkiewicz,Pansu, Zuk

Theorem (general form)

Let G be acting properly discontinuously and cocompactly on a2-dimensional contractible simplicial complex K and denote by λ1(x) thesmallest positive eigenvalue of the discrete Laplacian on the link of avertex x ∈ K. If

λ1(x) >12

for every vertex x ∈ K then G has property (T).


Link graphs on generating sets

G - group, S = S−1 - finite generating set of G, e < S.

Definition

The link graph L(S) = (V ,E) of S:

vertices V = S,

(s, t) ∈ S × S is an edge ∈ E if s−1t ∈ S.

Example. Let Z2 be generated by (1, 0), (0, 1), (1, 1) and their inverses.


The spectral criterion

Laplacian on `2(S, deg):

∆f(s) = f(s) −1

deg(s)

∑t∼s

f(t)

λ1 denotes the smallest positive eigenvalue

Theorem (Zuk)

If L(S) connected and λ1(L(S)) >12

then G has property (T).


Poincare inequalities

Let Mf =∑

x∈V f(x)deg(x)

#Ebe the mean value of f

Definition (p-Poincare inequality for the norm of X )

X -Banach space, p ≥ 1, Γ = (V ,E) - finite graph. For every f : V → X∑s∈V

‖f(s) −Mf‖pX deg(s)

1/p

≤ κ

∑(s,t)∈E

‖f(s) − f(t)‖pX

1/p

.

The inf of κ for L(S), giving the optimal constant, is denoted κ(p,S,X)

The classical p-Poincare inequality when X = R.

1 κ(1,S,R) ' Cheeger isoperimetric const

2 κ(2,S,R) =√λ−1

1 ;

3 for 1 ≤ p < ∞ we have κ(p,S, Lp) = κ(p,S,R)


The Main Theorem

Given p > 1 denote by p∗ the adjoint index:1p

+1p∗

= 1.

Main TheoremLet X be a reflexive Banach space, G a group generated by S as earlier. Iffor some p > 1

max{

2−1p κ(p,S,X) , 2−

1p∗ κ(p∗,S,X∗)

}< 1

thenH1(G, π) = 0

for any isometric representation π of G on X.

Remark 1. By reflexivity, the same conclusion holds for actions on X∗

Remark 2. The roles of the two constants in the proof are different.


The Main Theorem

Remark 3. In the case

p = 2 and X = Hilbert space

this reduces to the spectral criterion for property (T) in the form given byZuk.

Then p = p∗ soκ(p∗,S,X∗) = κ(p,S,X)

and2−1/pκ < 1 ⇔ λ1 >

12.


Sketch of proof

Difficulty: lack of self-duality when X is not a Hilbert space

For any Hilbert space H∗ = H , every subspace has an orthogonalcomplement

For Y ⊆ X Banach spaces, Y might not have a complement,

Y∗ = X∗/Ann(Y)

[Ann(Y)=all functionals in X∗ which identically vanish on Y ]with the quotient norm ∥∥∥ [y]

∥∥∥Y∗ = inf

x∈Ann(Y)‖y − x‖Y∗

General remark: computing quotients of Banach spaces can be difficult.E.g., every separable Banach space is a quotient of `1(N).


(cochainsπ)∗

X∗ �δ∗

(cocyclesπ)∗

i∗

??

Xδv(s) = v − πsv

- cocyclesπ

cochainsπ

i

?

∩

X∗ is equipped with the adjointrepresentation,

πg = π∗g−1 .

We want to show that δ is onto.

This is equivalent to δ∗ havingclosed range.

Step 1: identify (cochainsπ)∗.


TheoremIf X-reflexive, π – isometric representation. Then

(cochains π)∗ is isometrically isomorphic to (cochains π).

Sketch of proof: we view cochains π as a complemented subspace of alarger Banach space, Y:

cochainsπ ⊕Z = Y,

cochainsπ ⊕Z = Y∗.

Compute to get

(cochainsπ)∗ = Y∗/Z isomorphic to cochainsπ

This is not sufficient – we need an isometric isomorphism.


Orthogonality-type conditions

We need an additional geometric condition.

TheoremIf π is isometric then

‖c − x‖Y = ‖c + x‖Y,

for c ∈ cochains π, x ∈ Z

This is an orthogonality-type condition


This allows to compute i∗δ∗

Step 2. i∗δ∗f = Mf , the mean value operator.

The next step would naturally be to identify (cocyclesπ)∗, but it is not clearhow to do it, except for some cases.E.g., when X is isomorphic to the Hilbert space then we know that everysubspace is complemented and then

(cocyclesπ)∗ ' cocyclesπ.

QuestionAre the cocyclesπ complemented in the cochainsπ?

For now we need to find a way around this.


cocyclesπ ⊂i- cochainsπ

X∗

δ

6

�δ∗

(cocyclesπ)∗

i∗

??

Xδ- cocyclesπ

(cocyclesπ)∗

δ∗

6

��i∗

cochainsπ

i

?

∩

Thm 1. If 21/p∗κ(p∗,S,X∗) < 1then δ∗ i∗ i has closed range.

Thm 1 follows from asequence of inequalities

It implies δ∗ has closed rangeon image of i∗ i

The same argument for theother inequality gives:21/pκ(p,S,X) < 1 thenδ∗

i∗

i has closed range

⇒ i∗

i has closed range

⇒ i∗ i is surjective. �


Examples and Applications


Isometric representations on Lp

We want to apply this to X = Lp , p > 2

Desired outcome: vanishing of cohomology for all Lp ,

p ∈ [2, 2 + c),

where we can say something about c.

This cannot be improved, in the sense that we cannot expect vanishing forall 2 < p < ∞:

1 the assumptions of the theorem are almost never satisfied for psufficiently large

2 the main theorem applies to some hyperbolic groups

Estimating p-Poincare constants is a hard analytic problem whenp , 1, 2,∞.


A2 groups

Cartwright, Młotkowski and Steger defined finitely presented groups Gq

where q = k n for k - prime such that

L(S) = incidence graph of a projective plane over a finite field.

In the 60ies Feit and Higman computed spectra of such incidence graphs:

2−12 κ(2,S,R) =

√(1 −

√q

q + 1

)−1

−→1√

2.

We now want to estimate κ(p,S, Lp) for these graphs.


Estimating the p-Poincare constant

When p ≥ 2, in finite dimensional spaces: ‖f‖`np≤ ‖f‖`n

2≤ n1/2−1/p‖f‖`n

p.

#V = 2(q2 + q + 1),

#E = 2(q2 + q + 1)(q + 1)

deg(s) = q + 1 for every s ∈ S

Similarly for p∗ < 2.

Theorem

For each q=power of a prime we have

H1(Gq, π) = 0

for any isometric representation π of Gq on any Lp for all

2 ≤ p <2 ln

(2(q2 + q + 1)

)ln (2(q2 + q + 1)) − ln

√2(1 −

√q

q + 1

) .


Numerical values of p

We have 2 ≤ p ≤ 2.106 and p → 2 as q → ∞.


Hyperbolic groupsZuk used the spectral conditions to prove that many hyperbolic groups have (T).

Theorem (Zuk)

G in the density model for 1/3 < d < 1/2 is, with high probability, of the form

H -- Γ ⊆f .i. G,

where G is hyperbolic and H has a link graph with 2−1/2κ(2,S,R) < 1.

A detailed proof of the above by M.Kotowski-M.Kotowski.Vanishing of cohomology for all isometric representations on Lp is passed on toquotients and through finite index subgroups, just as (T) is.

Theorem

Let G, Γ and φ be as above, and let L(S) = (V,E) denote the link graph of Γ.Then H1(G, π) = 0 for every isometric representation π of G on Lp for

2 ≤ p < min{p0, p0

∗},

where

p0 =ln degω − ln(2#E)

12

ln(degω#E

)− ln κ(2,S,R)

and p0 =ln(#V degω) − ln 2

12

ln(#V degω) − ln κ(2,S,R)

.


1. Conformal dimension

Definition (Pansu)

G hyperbolic, dV -any visual metric on ∂G.

confdim(∂G) = inf{dimHaus(∂G, d) : d quasi-conformally equiv. to dV

}.

confdim(∂G) is a quasi-isometry invariant of G,

extremely hard to compute or even estimate.

Theorem (Bourdon-Pajot, 2003)

G hyperbolic acts without fixed points on `p(G) for p ≥ confdim(∂G)


1. Conformal dimension

How to estimate conformal dimension of ∂G?

TheoremFor a random hyperbolic group as before (1/3 < d < 1/2),

confdim(∂G) ≥ min{p0, p0

∗},

where

p0 =ln degω − ln(2#E)

12

ln(degω#E

)− ln κ(2,S,R)

and p0 =ln(#V degω) − ln 2

12

ln(#V degω) − ln κ(2,S,R)

.

For d < 1/16 estimates of confdim(∂G) were given by John Mackay.


2. Harmonic analysis: uniformly bounded representations

Let π be a uniformly bounded representation on a Hilbert space: ahomomorphism π : G → B(H), such that

‖π‖ = supg∈G‖πg‖ < ∞

Change the norm on H to

‖v‖′ = supg∈G‖πgv‖.

Let the new Banach space E be H with ‖ · ‖′.

Then π is an isometric representation on E, which is reflexive.

Poincare constants satisfy

κ(2,S,E) ≤ ‖π‖κ(2,S,R).


2. Harmonic analysis: uniformly bounded representations

General question: given a group G with property (T), does H1(G, π) vanishalso for uniformly bounded representations π?

Yes for lattices in higher rank simple Lie gps (Bader-Furman-Gelander-Monod)No for Sp(n, 1) (Shalom)

Conjecture (Shalom)

For every hyperbolic group there is a uniformly bounded π such thatH1(G, π) , 0.

TheoremIn the density model (1/3 < d < 1/2),

H1(G, π) = 0

with high probability for representation satisfying ‖π‖ <√

2.


Applications to actions on the circle

Navas studied rigidity properties of diffeomorphic actions on the circle.

Vanishing of cohomology for Lp for p > 2 improves the differentiabilityclass in his result.

Corollary

Let q be a power of a prime number and Gq be be the corresponding A2

group. Then every homomorphism h : G → Diff1+α+ (S1) has finite imagefor

α >

12

ln(2(q2 + q + 1)(q + 1)) − ln(2) − ln

√

1 −√

qq + 1

ln(q2 + q + 1) + ln(q + 1)

.

Here, α is strictly less than12

, improving for these groups the original

differentiability class.


Eigenvalues of the p-Laplacian

As mentioned before, in the case when X is isomorphic to the Hilbertspace, one of the Poincare consants can be dropped in the assumptionsof the Main Theorem.

In particular, when X is finite dimensional we can use this and the MainTheorem to find a lower bound on the eigenvalue of the p-Laplacian on theCayley graph of a finite quotient, similarly as using the Kazhdan constantto bound the expanding constant of a finite quotient from below.

This is closely related to previous joint work with R. I. Grigorchuk, in whichwe used metric properties bound eigenvalues of the p-Laplacian.


References

P. W. NowakPoincare inequalities and rigidity for actions on Banach spacesarXiv:1107.1896


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