vertical versus horizontal poincare inequalities...vertical versus horizontal poincare inequalities...

Vertical Versus Horizontal Poincare Inequalities

Assaf Naor

Courant Institute

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Bi-Lipschitz distortion

a metric space and a Banach space.

the infimum over those for which there exists satisfying

(M;dM) (X;k ¢ kX)

cX(M) = D 2 (1;1]

f :M !X

MD,¡!X:

8 x; y 2M; dM(x; y) 6 kf(x)¡ f(y)kX 6DdM(x; y):

The discrete Heisenberg group

• The group generated by subject to the relation stating that the commutator of is in the center:

where

H a; ba; b

ac= ca and bc= cb

c = [a; b] = aba¡1b¡1

Concretely,

H =

8<:

0@1 x z

0 1 y

0 0 1

1A : x; y; z 2 Z

9=;

a =

0@1 1 0

0 1 0

0 0 1

1A and b =

0@1 0 0

0 1 1

0 0 1

1A

c =

0@1 0 1

0 1 0

0 0 1

1A

The left-invariant word metric on corresponding to the generating set is denoted .

fa; a¡1; b; b¡1gH

dW

Denote

Basic facts:

8n 2 N; Bn = fx 2H : dW(x; eH) 6 ng

8k 2 N; dW(ck; eH) ³pk

8m 2 N; jBmj ³m4

Uniform convexity

The modulus of uniform convexity of :

±X(²) = inf

½1¡

°°°°x+ y

2

°°°°X

: kxkX = kykX = 1; kx¡ ykX = ²

¾(X;k ¢ kX)

PnX

i=1

Pn

i=1bkblblRR 10f(x)dx

Pn

i=1

Pn

i=1

Pn

i=1

x

y0

x+y

2

²

1¡°°x¡y

2

°°X¸ ±X(²)

• X is uniformly convex if

• For , X is q-convex if it admits an equivalent norm with respect to which

Theorem (Pisier, 1975). If X is uniformly convex then it is q-convex for some .

is -convex for .

8 ² 2 (0;1); ±X(²) > 0:

q 2 [2;1)

±X(²) & ²q:

q 2 [2;1)

`p maxf2; pg p > 1

Mostow (1973), Pansu (1989), Semmes (1996)

Theorem. The metric space does not admit a bi-Lipschitz embedding into for any .

(H; dW )

Rn n 2 N

Assouad’s embedding theorem (1983)

• A metric space is K-doubling if any ball can be covered by K-balls of half its radius.

(M;dM)

Assouad’s embedding theorem


(M;dM)

Assouad’s embedding theorem (1983)


(M;dM)

Theorem (Assouad, 1983). Suppose that is K-doubling and . Then

Theorem (N.-Neiman, 2010). In fact

David-Snipes, 2013: Simpler deterministic proof.

(M;dM)

² 2 (0;1)

(M;d1¡²M )D(K;²),¡¡¡¡! RN(K;²):

(M;d1¡²M )D(K;²),¡¡¡¡! RN(K):

Obvious question: Why do we need to raise the metric to the power ?

1¡ ²

(M;d1¡²M )D(K;²),¡¡¡¡! RN(K;²):

Obvious question: Why do we need to raise the metric to the power ?

Since in we have

the metric space is O(1)-doubling.

By Mostow-Pansu-Semmes,

1¡ ²

(M;d1¡²M )D(K;²),¡¡¡¡! RN(K;²):

8m 2 N; jBmj ³m4;(H; dW )

(H; dW )

(H; dW) 6,! RN :

Proof of non-embeddability into

By a limiting argument and a non-commutative variant of Rademacher’s theorem on the almost-everywhere differentiability of Lipschitz functions (Pansu differentiation) we have the statement

“If the Heisenberg group embeds bi-Lipschitzly into then it also embeds into via a bi-Lipschitz mapping that is a group homomorphism.”

A non-Abelian group cannot be isomorphic to a subgroup of an Abelian group!

Rn

Rn Rn

Heisenberg non-embeddability

• Mostow-Pansu-Semmes (1996).

• Cheeger (1999).

• Pauls (2001).

• Lee-N. (2006).

• Cheeger-Kleiner (2006).



• Cheeger-Kleiner-N. (2009).

• Austin-N.-Tessera (2010).

• Li (2013).



• Cheeger (1999).

• Pauls (2001).

• Lee-N. (2006).






• Li (2013).

H does not embed into

any uniformly convex space:



• Cheeger (1999).

• Pauls (2001).

• Lee-N. (2006).






• Li (2013).

[ANT (2010)]: Hilbertian case

c`2(Bn; dW) ³plogn:

[ANT (2010)]: Hilbertian case

A limiting argument combined with [Aharoni-Maurey-Mityagin (1985), Gromov (2007)] shows that it suffices to treat embeddings that are 1-cocycles associated to an action by affine isometries. By [Guichardet (1972)] it further suffices to deal with coboundaries. This is treated by examining each irreducible representation separately.

c`2(Bn; dW) ³plogn:

[ANT (2010)]: q-convex case

If is q-convex then

(X;k ¢ kX)

cX(Bn; dW ) &X

µlogn

log logn

¶1=q

:

[ANT (2010)]: q-convex case, continued

Qualitative statement: There is no bi-Lipschitz embedding of the Heisenberg group into an ergodic Banach space X via a 1-cocycle associated to an action by affine isometries.

X is ergodic if for every linear isometry and every the sequence

converges in norm.

T :X!Xx 2X

1

n

n¡1X

j=1

T jx


N.-Peres (2010): In the case of q-convex spaces, it suffices to treat 1-cocycle associated to an affine action by affine isometries.

For combining this step with the use of ergodicity, uniform convexity is needed, because by [Brunel-Sucheston (1972)], ultrapowers of X are ergodic if and only if X admits an equivalent uniformly convex norm.


Conclusion of proof uses algebraic properties of cocycles combined with rates of convergence for the mean ergodic theorem in q-convex spaces.

Li (2013): A quantitative version of Pansu’s differentiation theorem. Suboptimal bounds.

Almost matching embeddability

Assouad (1983): If a metric space is O(1)-doubling then there exists and 1-Lipschitz functions such that for ,

(M;dM)

k 2 N©Áj :M ! Rk

ªj2Z x; y 2M

dM(x; y) 2 [2j¡1;2j] =) kÁj(x)¡Áj(y)k2 & dM(x; y):

Almost matching embeddability

Assouad (1983): If a metric space is O(1)-doubling then there exists and 1-Lipschitz functions such that for ,

So, define by

For the bi-Lipschitz distortion of f is of order

(M;dM)

k 2 N©Áj :M ! Rk

ªj2Z x; y 2M

dM(x; y) 2 [2j¡1;2j] =) kÁj(x)¡Áj(y)k2 & dM(x; y):

f : Bn !O(logn)M

j=1

Rk f(x) =

O(logn)M

j=1

Áj(x):

p 2 [2;1)

(logn)1=p:

Lafforgue-N., 2012

Theorem. For every q-convex space every and every ,

(X;k ¢ kX);n 2 Nf : H!X

n2X

k=1

X

x2Bn

kf(xck)¡ f(x)kqXk1+q=2

.XX

x2B21n

¡kf(xa)¡ f(x)kqX + kf(xb)¡ f(x)kqX

¢:

The proof of this inequality relies on real-variable Fourier analytic methods. Specifically, a vector-valued Littlewood-Paley-Stein inequality due to Martinez, Torrea and Xu (2006), combined with a geometric argument.

For embeddings into one can use the classical Littlewood-Paley inequality instead.

`p

Sharp non-embeddability

If

and

8x; y 2B22n; dW(x; y) 6 kf(x)¡ f(y)kX 6DdW(x; y);

X

x2B21n


¢

. DqjB21nj ³ Dqn4;

n2X

k=1

X

x2Bn


>n2X

k=1

X

x2Bn

dW (xck; x)q

k1+q=2

&n2X

k=1

X

x2Bn

kq=2

k1+q=2³ jBnj logn ³ n4 log n:

so,

n2X

k=1

X

x2Bn


.XX

x2B21n


¢;

n4 logn.X Dqn4 =) D &X (logn)1=q:

cX(Bn; dW) &X (logn)1=q:

Sharp distortion computation

p 2 (1;2] =) c`p (Bn; dW ) ³pplogn:

p 2 [2;1) =) c`p (Bn; dW) ³p (logn)1=p:

The Sparsest Cut Problem

Input: Two symmetric functions

Goal: Compute (or estimate) in polynomial time the quantity

C;D : f1; : : : ; ng£f1; : : : ; ng ! [0;1):

©¤(C;D) = min;6=S(f1;:::;ng

Pn

i;j=1C(i; j)j1S(i)¡ 1S(j)jPn

i;j=1D(i; j)j1S(i)¡ 1S(j)j:

The Goemans-Linial Semidefinite Program

The best known algorithm for the Sparsest Cut Problem is a continuous relaxation called the Goemans-Linial SDP (~1997).

Theorem (Arora, Lee, N., 2005). The Goemans-Linial SDP outputs a number that is guaranteed to be within a factor of

of

(logn)12+o(1)

©¤(C;D):

Minimize

over all ,

subject to the constraints

and

nX

i;j=1

C(i; j)kvi ¡ vjk22

v1; : : : ; vn 2 Rn

nX

i;j=1

D(i; j)kvi ¡ vjk22 = 1;

8 i; j; k 2 f1; : : : ; ng;kvi ¡ vjk22 6 kvj ¡ vkk22 + kvk ¡ vjk22:

The link to the Heisenberg group

Theorem (Lee-N., 2006): The Goemans-Linial SDP has an integrality gap of at least .

c`1(Bn; dW)

Cheeger-Kleiner-N., 2009: There exists a universal constant c>0 such that

Cheeger-Kleiner, 2007, 2008: Non-quantitative versions that also reduce matters to ruling out a certain more structured embedding.

Quantitative estimate controls phenomena that do not have qualitative counterparts.

c`1(Bn; dW) > (logn)c:

Khot-Vishnoi (2005): The Goemans-Linial SDP has integrality gap at least . (log logn)c

How well does the G-L SDP perform?

Conjecture:

Remark: In a special case called Uniform Sparsest Cut (approximating graph expansion) the G-L SDP might perform better. The best known performance guarantee is [Arora-Rao-Vazirani, 2004] and the best known integrality gap lower bound is

[Kane-Meka, 2013].

c`1(Bn; dW) &plogn:

.plogn

ecplog logn

Vertical perimeter versus horizontal perimeter

Conjecture: For every smooth and compactly supported

Lemma: A positive solution of this conjecture implies that

f : R3 ! R;ÃZ 1

0

µZ

R3jf(x; y; z + t)¡ f(x; y; z)jdxdydz

¶2dt

t2

! 12

.Z

R3

µ¯̄¯̄@f@x

(x; y; z)

¯̄¯̄+¯̄¯̄@f@y

(x; y; z) + x@f

@z(x; y; z)

¯̄¯̄¶dxdydz:

c`1(Bn; dW) &plogn:

Theorem (Lafforgue-N., 2012): For every p>1,

ÃZ 1

0

µZ

R3jf(x; y; z + t)¡ f(x; y; z)jpdxdydz

¶2=pdt

t2

!1=2

.p

µZ

R3

µ¯̄¯̄@f@x

(x; y; z)

¯̄¯̄p

+

¯̄¯̄@f@y

(x; y; z) + x@f

@z(x; y; z)

¯̄¯̄p¶

dxdydz

¶1=p

:

Equivalent form of the conjecture

Let A be a measurable subset of For t>0 define

Then

R3.

vt(A) = vol³©

(x; y; z) 2 A : (x; y; z + t) =2 Aª´

:

Z 1

0

vt(A)2

t2dt . PER(A)2:

Proof of the vertical versus horizontal Poincare inequality

Equivalent statement: Suppose that is q-convex and is smooth and compactly supported. Then

(X;k ¢ kX)f : R3 !X

µZ 1

0

Z

R3

kf(x; y; z + t)¡ f(x; y; z)kqXt1+q=2

dxdydz

¶ 1q

.X

µZ

R3

µ°°°°@f

@x(x; y; z)

°°°°q

X

+

°°°°@f

@y(x; y; z) + x

@f

@z(x; y; z)

°°°°q

X

¶dxdydz

¶ 1q

:

Proof of the equivalence: partition of unity argument + classical Poincare inequality for the Heisenberg group.

d

d²

¯̄¯̄²=0

f

0@0@1 x z

0 1 y

0 0 1

1A0@1 ² 0

0 1 0

0 0 1

1A1A

=d

d²

¯̄¯̄²=0

f

0@0@1 x+ ² z

0 1 y

0 0 1

1A1A =

@f

@x:

a =

0@1 1 0

0 1 0

0 0 1

1A and b =

0@1 0 0

0 1 1

0 0 1

1A

d

d²

¯̄¯̄²=0

f

0@0@1 x z

0 1 y

0 0 1

1A0@1 0 0

0 1 ²

0 0 1

1A1A

=d

d²

¯̄¯̄²=0

f

0@0@1 x z + ²x

0 1 y + ²

0 0 1

1A1A =

@f

@y+ x

@f

@z:

a =

0@1 1 0

0 1 0

0 0 1

1A and b =

0@1 0 0

0 1 1

0 0 1

1A

The Poisson semigroup

Pt(x) =1

¼(t2 + x2):

Qt(x) =@

@tPt(x) =

x2 ¡ t2

¼(t2 + x2)2:

Vertical convolution

For Ã 2 L1(R);

Ã ¤ f(x; y; z) =Z

RÃ(u)f(x; y; z ¡ u)du 2 X:

Heisenberg gradient

Proposition:

rHf =

µ@f

@x;@f

@y+ x

@f

@z

¶: R3 !X ©X:

µZ 1

0

Z

R3


dxdydz

¶ 1q

.µZ 1

0

tq¡1 kQt ¤ rHfkqLq(R3;X©X)dt

¶ 1q

:

Littlewood-Paley

By Martinez-Torrea-Xu (2006), the fact that X is q-convex implies

So, it remains to prove the proposition.

µZ 1

0

tq¡1 kQt ¤ rHfkqLq(R3;X©X)dt

¶ 1q

. krHfkLq(R3;X©X) :

By a variant of a classical argument (using Hardy’s inequality and semi-group properties),

µZ 1

0

Z

R3


dxdydz

¶ 1q

.µZ 1

0

tq2¡1 kQt ¤ fkqLq(R3;X)

dt

¶ 1q

:

The desired estimate

Follows from key lemma by the telescoping sum

µZ 1

0


dt

¶ 1q

.µZ 1

0

tq¡1 kQt ¤ rHfkqLq(R3;X©X) dt

¶ 1q

Qt ¤ f =

1X

m=1

(Q2m¡1t ¡Q2mt ¤ f) :

Proof of key lemma

Since we have .

So, by identifying with , for every ,

P2t = Pt ¤Pt Q2t = Pt ¤Qt

R3 H h 2 R3

Qt ¤ f(h)¡Q2t ¤ f(h)= Qt ¤ f(h)¡ Pt ¤Qt ¤ f(h)

=

Z

RPt(u)

¡Qt ¤ f(h)¡Qt ¤ f(hc¡u)

¢du:

For every s>0 consider the commutator path

°s(µ) =8>><>>:

aµ if 0 6 µ 6ps;

apsbµ¡

ps if

ps 6 µ 6 2

ps;

apsbpsa¡µ+2

ps if 2

ps 6 µ 6 3

ps;

apsbpsa¡

psb¡µ+3

ps if 3

ps 6 µ 6 4

ps:

°s : [0;4ps]! R3;

So, and

Hence,

°s(4ps) =

haps; b

psi= [a; b]s = cs:

°s(0) = 0 = eH

Qt ¤ f(h)¡Qt ¤ f(hc¡u)

=

Z 4pu

0

d

dµQt ¤ f

¡hc¡u°u(µ)

¢dµ:

By design, is one of

or

where and .

We used here the fact that since is convolution along the center, it commutes with

ddµQt ¤ f (hc¡u°u(µ))

@aQt ¤ f(hc¡u°u(µ)) =Qt ¤ @af(hc¡u°u(µ))

@bQt ¤ f(hc¡u°u(µ)) =Qt ¤ @bf(hc¡u°u(µ));@a = @x @b = @y +x@z

Qt

@a; @b:

We saw that

Now the key lemma follows from the triangle inequality and the fact that

Qt ¤ f(h)¡Q2t ¤ f(h))

=

Z

RPt(u)

¡Qt ¤ f(h)¡Qt ¤ f(hc¡u)

¢du

=

Z

RPt(u)

Z 4pu

0

d

dµQt ¤ f

¡hc¡u°u(µ)

¢dµdu:

Z 1

0

puPt(u)du ³

pt:

vertical versus horizontal poincare inequalities...vertical versus horizontal poincare inequalities...

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