Vertical Versus Horizontal Poincare Inequalities Assaf Naor Courant Institute
Bi-Lipschitz distortion a metric space and (M; d M ) (X; k k X ) a Banach space. c X (M) = the infimum over those D 2 (1; 1] for which there exists f : M! X satisfying 8 x; y 2 M; d M (x; y) 6 kf(x) f(y)k X 6 Dd M (x; y): M D,! X:
The discrete Heisenberg group The group H generated by a; b subject to the relation stating that the commutator of a; b is in the center: where ac = ca and bc = cb c = [a; b] = aba 1 b 1
Concretely, H = 80 1 9 < 1 x z = @ 0 1 ya : x; y; z 2 Z : ; 0 0 1 a = 0 1 1 1 0 @ 0 1 0A and b = 0 0 1 0 1 0 1 0 @ 0 1 1A 0 0 1 c = 0 1 0 1 1 @ 0 1 0A 0 0 1
The left-invariant word metric on H corresponding to the generating set is denoted. d W fa; a 1 ; b; b 1 g
Denote 8n 2 N; B n = fx 2 H : d W (x; e H ) 6 ng Basic facts: 8 k 2 N; d W (c k ; e H ) ³ p k 8 m 2 N; jb m j ³ m 4
Uniform convexity The modulus of uniform convexity of : ± X (²) = inf ½ 1 x + y 2 (X; k k X ) ¾ : kxk X = kyk X = 1; kx yk X = ² X x x+y 2 0 ² y 1 x y 2 X ± X (²)
X is uniformly convex if 8 ² 2 (0; 1); ± X (²) > 0: For q 2 [2; 1), X is q-convex if it admits an equivalent norm with respect to which ± X (²) & ² q : Theorem (Pisier, 1975). If X is uniformly convex then it is q-convex for some q 2 [2; 1). `p is maxf2; pg-convex for p > 1.
Mostow (1973), Pansu (1989), Semmes (1996) Theorem. The metric space (H; d W ) does not admit a bi-lipschitz embedding into for any. R n n 2 N
Assouad s embedding theorem (1983) A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem (1983) A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem (1983) A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem (1983) A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem (1983) A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem (1983) A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem (1983) A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem (1983) A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem (1983) A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem (1983) A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem (1983) A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem (1983) A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem (1983) A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem (1983) A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem (1983) A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem (1983) A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Assouad s embedding theorem (1983) A metric space (M; d M ) is K-doubling if any ball can be covered by K-balls of half its radius.
Theorem (Assouad, 1983). Suppose that is K-doubling and ² 2 (0; 1). Then (M; d 1 ² M ),! D(K;²) R N(K;²) : (M; d M ) Theorem (N.-Neiman, 2010). In fact (M; d 1 ² M ),! D(K;²) R N(K) : David-Snipes, 2013: Simpler deterministic proof.
(M; d 1 ² M ),! D(K;²) R N(K;²) : Obvious question: Why do we need to raise the metric to the power? 1 ²
(M; d 1 ² M ),! D(K;²) R N(K;²) : Obvious question: Why do we need to raise the metric to the power 1 ²? Since in (H; d W ) we have the metric space (H; d W ) 8 m 2 N; jb m j ³ m 4 ; is O(1)-doubling. By Mostow-Pansu-Semmes, (H; d W ) 6,! R N :
Proof of non-embeddability into R n By a limiting argument and a non-commutative variant of Rademacher s theorem on the almosteverywhere differentiability of Lipschitz functions (Pansu differentiation) we have the statement If the Heisenberg group embeds bi-lipschitzly into R n then it also embeds into R n via a bi-lipschitz mapping that is a group homomorphism. A non-abelian group cannot be isomorphic to a subgroup of an Abelian group!
Heisenberg non-embeddability Mostow-Pansu-Semmes (1996). Cheeger (1999). Pauls (2001). Lee-N. (2006). Cheeger-Kleiner (2006). Cheeger-Kleiner (2007). Cheeger-Kleiner (2008). Cheeger-Kleiner-N. (2009). Austin-N.-Tessera (2010). Li (2013).
Heisenberg non-embeddability Mostow-Pansu-Semmes (1996). Cheeger (1999). Pauls (2001). Lee-N. (2006). H does not embed into any uniformly convex space: Cheeger-Kleiner (2006). Cheeger-Kleiner (2007). Cheeger-Kleiner (2008). Cheeger-Kleiner-N. (2009). Austin-N.-Tessera (2010). Li (2013).
Heisenberg non-embeddability Mostow-Pansu-Semmes (1996). Cheeger (1999). Pauls (2001). Lee-N. (2006). Cheeger-Kleiner (2006). Cheeger-Kleiner (2007). Cheeger-Kleiner (2008). Cheeger-Kleiner-N. (2009). Austin-N.-Tessera (2010). Li (2013).
[ANT (2010)]: Hilbertian case c`2(b n ; d W ) ³ p log n:
[ANT (2010)]: Hilbertian case c`2(b n ; d W ) ³ p log n: 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.
[ANT (2010)]: q-convex case If is q-convex then µ 1=q log n c X (B n ; d W ) & X : log log n (X; k k X )
[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 x 2 X the sequence n 1 1 X T j x n j=1 converges in norm. T : X! X
[ANT (2010)]: q-convex case, continued 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.
[ANT (2010)]: q-convex case, continued 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 (M; d M ) is O(1)- doubling then there exists k 2 N and 1-Lipschitz functions such that for, Áj : M! R kª j2z x; y 2 M d M (x; y) 2 [2 j 1 ; 2 j ] =) ká j (x) Á j (y)k 2 & d M (x; y):
Almost matching embeddability Assouad (1983): If a metric space (M; d M ) is O(1)- doubling then there exists k 2 N and 1-Lipschitz functions Áj : M! R kª such that for x; y 2 M, j2z d M (x; y) 2 [2 j 1 ; 2 j ] =) ká j (x) Á j (y)k 2 & d M (x; y): So, define f : B n! O(log n) M by R k f(x) = O(log n) M Á j (x): j=1 j=1 For order p 2 [2; 1) (log n) 1=p : the bi-lipschitz distortion of f is of
Lafforgue-N., 2012 Theorem. For every q-convex space every f : H! X and every n 2 N, n 2 X X kf(xck ) f(x)k k 1+q=2 x2b n X. X k=1 x2b 21n q X (X; k k X ); kf(xa) f(x)k q X + kf(xb) f(x)kq X :
The proof of this inequality relies on realvariable 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 `p one can use the classical Littlewood-Paley inequality instead.
Sharp non-embeddability If and 8x; y 2 B 22n ; d W (x; y) 6 kf(x) f(y)k X 6 Dd W (x; y); X x2b 21n kf(xa) f(x)k q X + kf(xb) f(x)kq X. D q jb 21n j ³ D q n 4 ; n 2 X k=1 & X q X kf(xck ) f(x)k > k 1+q=2 x2b n n 2 X X k=1 x2b n n 2 X k=1 kq=2 k 1+q=2 ³ jb nj log n ³ n 4 log n: X d W (xck ; x)q k 1+q=2 x2b n
n 2 X X kf(xck ) f(x)k k 1+q=2 x2b n X. X k=1 x2b 21n q X kf(xa) f(x)k q X + kf(xb) f(x)kq X ; so, n 4 log n. X D q n 4 =) D & X (log n) 1=q : c X (B n ; d W ) & X (log n) 1=q :
Sharp distortion computation p 2 (1; 2] =) c`p (B n ; d W ) ³ p p log n: p 2 [2; 1) =) c`p (B n ; d W ) ³ p (log n) 1=p :
The Sparsest Cut Problem Input: Two symmetric functions C; D : f1; : : : ; ng f1; : : : ; ng! [0; 1): Goal: Compute (or estimate) in polynomial time the quantity (C; D) = min ;6=S(f1;:::;ng P n i;j=1 C(i; j)j1 S(i) 1 S (j)j P n i;j=1 D(i; j)j1 S(i) 1 S (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 (C; D): (log n) 1 2 +o(1)
nx Minimize C(i; j)kv i v j k 2 2 over all, subject to the constraints and i;j=1 v 1 ; : : : ; v n 2 R n nx i;j=1 D(i; j)kv i v j k 2 2 = 1; 8 i; j; k 2 f1; : : : ; ng; kv i v j k 2 2 6 kv j v k k 2 2 + kv k v j k 2 2:
The link to the Heisenberg group Theorem (Lee-N., 2006): The Goemans-Linial SDP has an integrality gap of at least. c`1(b n ; d W )
Cheeger-Kleiner-N., 2009: There exists a universal constant c>0 such that c`1(b n ; d W ) > (log n) c : 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.
Khot-Vishnoi (2005): The Goemans-Linial SDP has integrality gap at least. (log log n) c
How well does the G-L SDP perform? Conjecture: c`1(b n ; d W ) & p log n: 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. p log n [Arora-Rao-Vazirani, 2004] and the best known integrality gap lower bound is [Kane-Meka, 2013]. e cp log log n
Vertical perimeter versus horizontal perimeter Conjecture: For every smooth and compactly supported f : R 3! R; Ã Z 1. 0 Z R 3! 1 2 µz 2 dt jf(x; y; z + t) f(x; y; z)jdxdydz R t 2 µ 3 @f (x; y; z) @x + @f (x; y; z) + x@f (x; y; z) @y @z dxdydz: Lemma: A positive solution of this conjecture implies that c`1(b n ; d W ) & p log n:
Theorem (Lafforgue-N., 2012): For every p>1, Ã Z 1 µz! 2=p 1=2 jf(x; y; z + t) f(x; y; z)j p dt dxdydz 0 R t 2 3 µz p @f. p µ (x; y; z) @x + @f p 1=p (x; y; z) + x@f (x; y; z) @y @z dxdydz : R 3
Equivalent form of the conjecture Let A be a measurable subset of R 3. For t>0 define ³ (x; ª v t (A) = vol y; z) 2 A : (x; y; z + t) =2 A : Then Z 1 0 v t (A) 2 t 2 dt. PER(A) 2 :
µz 1 Proof of the vertical versus horizontal Poincare inequality Equivalent statement: Suppose that q-convex and f : R 3! X is smooth and compactly supported. Then 0 Z. X µz R 3 R 3 q kf(x; y; z + t) f(x; y; z)kx µ @f @x t 1+q=2 (x; y; z) q X + 1 q dxdydz @f (x; y; z) + x@f @y @z (X; k k X ) (x; y; z) q X is 1 q dxdydz :
Proof of the equivalence: partition of unity argument + classical Poincare inequality for the Heisenberg group.
a = 0 1 1 1 0 @ 0 1 0A and b = 0 0 1 0 1 0 1 0 @ 0 1 1A 0 0 1 d f d² ²=0 = d d² ²=0 f 00 @@ 1 x z 1 0 0 1 ya @ 1 ² 0 11 0 1 0AA 0 0 1 0 0 1 00 11 1 x + ² z @@ 0 1 yaa = @f @x : 0 0 1
a = 0 1 1 1 0 @ 0 1 0A and b = 0 0 1 0 1 0 1 0 @ 0 1 1A 0 0 1 d f d² ²=0 = d d² ²=0 f 00 @@ 1 x z 1 0 0 1 ya @ 1 0 0 11 0 1 ² AA 0 0 1 0 0 1 00 11 1 x z + ²x @@ 0 1 y + ² AA = @f @y + x@f @z : 0 0 1
The Poisson semigroup P t (x) = 1 ¼(t 2 + x 2 ) : Q t (x) = @ @t P t(x) = x2 t 2 ¼(t 2 + x 2 ) 2 :
Vertical convolution For à 2 L 1 (R); à f(x; y; z) = Z R Ã(u)f(x; y; z u)du 2 X:
Heisenberg gradient r H f = µ @f @x ; @f @y + x@f @z : R 3! X X: Proposition: µz 1 Z q kf(x; y; z + t) f(x; y; z)kx dxdydz 0 R 3 t 1+q=2 µz 1 1. t q 1 kq t r H fk q L q (R 3 ;X X) dt q : 0 1 q
Littlewood-Paley By Martinez-Torrea-Xu (2006), the fact that X is q-convex implies µz 1 1 t q 1 kq t r H fk q L q (R 3 ;X X) dt q 0. kr H fk Lq (R 3 ;X X) : So, it remains to prove the proposition.
By a variant of a classical argument (using Hardy s inequality and semi-group properties), µz 1 0. Z R 3 µz 1 q kf(x; y; z + t) f(x; y; z)kx 0 t 1+q=2 t q 2 1 kq t fk q L q (R 3 ;X) dt 1 q : dxdydz 1 q
So, we need to show that µz 1 0. t q 2 1 kq t fk q L q (R 3 ;X) dt 1 q µz 1 0 t q 1 kq t r H fk q L q (R 3 ;X X) dt 1 q :
Key lemma: For every t>0, kq t f Q 2t fk Lq (R 3 ;X). p t kq t r H fk Lq (R 3 ;X X)
The desired estimate µz 1 0. t q 2 1 kq t fk q L q (R 3 ;X) dt 1 q µz 1 0 t q 1 kq t r H fk q L q (R 3 ;X X) dt 1 q Follows from key lemma by the telescoping sum 1X Q t f = (Q 2 m 1 t Q 2 m t f) : m=1
Proof of key lemma Since P 2t = P t P t we have Q 2t = P t Q t. R 3 H h 2 R 3 So, by identifying with, for every, Q t f(h) Q 2t f(h) = Q t f(h) P t Q t f(h) Z = P t (u) Q t f(h) Q t f(hc u ) du: R
For every s>0 consider the commutator path s : [0; 4 p s]! R 3 ; s (µ) = 8 a >< µ if 0 6 µ 6 p s; a ps b µ p s if p s 6 µ 6 2 p s; a ps b ps a >: µ+2p s if 2 p s 6 µ 6 3 p s; a ps b ps a ps b µ+3p s if 3 p s 6 µ 6 4 p s:
So, s (0) = 0 = e H s (4 p s) = and h a ps ; b p i s = [a; b] s = c s : Hence, Q t f(h) Q t f(hc u ) = Z 4 p u 0 d dµ Q t f hc u u (µ) dµ:
By design, d dµ Q t f (hc u u (µ)) is one of @ a Q t f(hc u u (µ)) = Q t @ a f(hc u u (µ)) or @ b Q t f(hc u u (µ)) = Q t @ b f(hc u u (µ)); where @ a = @ x and @ b = @ y + x@ z. We used here the fact that since Q t is convolution along the center, it commutes with @ a ; @ b :
We saw that Q t f(h) Q 2t f(h)) Z = P t (u) Q t f(h) Q t f(hc u ) du = Z R R P t (u) Z 4 p u 0 d dµ Q t f hc u u (µ) dµdu: Now the key lemma follows from the triangle inequality and the fact that Z 1 0 p upt (u)du ³ p t: