Doubling metric spaces and embeddings. Assaf Naor

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1 Doubling metric spaces and embeddings Assaf Naor

2 Our approach to general metric spaces bears the undeniable imprint of early exposure to Euclidean geometry. We just love spaces sharing a common feature with R n. (M. Gromov, 2001).

3 Metric spaces A set X equipped with a mapping that assigns a nonnegative number every two elements (points) x; y 2 X d(x; y) = 0 () x = y d(x; y) = d(y; x) d(x; y) 6 d(x; z) + d(z; y) x d : X X! [0; 1) to such that d(x; y) > 0 d(x; y) d(x; z) y d(z; y) z

4 A rich and unstructured (?) world Shortest-path metrics on graphs

7 Heisenberg group The integer grid in 3-space f(x; y; z) : x; y; z are integersg edges: (x; y; z)» (x 1; y; z) (x; y; z)» (x; y 1; z x)

9 Edit distance X consists of all the strings of letters from an alphabet S (e.g. S={A,C,G,T}). The distance between two strings a,b is defined to be the minimum number of insertions or deletions of letters from S required to transform a to b.

10 Transportation cost/earthmover X consists of all the k-point subsets of the n by n grid in the plane. A =

11 A =

12 B =

15 If A = fa 1 ; : : : ; a k g and B = fb 1 ; : : : ; b k g then their distance is defined to be the minimum over all permutations ¼ of f1; : : : ; kgof the total matched distance: d(a; B) def = min ¼ nx j=1 a j b ¼(j)

18 Nice metric spaces R n The n-dimensional Euclidean space. p > 1 For the space `p, consisting of all infinite sequences of numbers x = (x 1 ; x 2 ; x 3 ; : : :), equipped with the distance d x = (x 1 ; x 2 ; : : :); y = (y 1 ; y 2 ; : : :) def = 1 1X jx j y j j p A j=1 1 p

19 Embeddings and distortion (X; d X ) admits a bi-lipschitz embedding into there exists 1 6 D < 1 and f : X! R n with Denote this by R n 8 x; y 2 X; d X (x; y) 6 kf(x) f(y)k 2 6 Dd X (x; y): X D,! R n : if c R n(x) def = smallest D for which this is possible.

20 Embeddings and distortion (X; d X ) admits a bi-lipschitz embedding into there exists 1 6 D < 1 and f : X! `p with 8 x; y 2 X; d X (x; y) 6 kf(x) f(y)k p 6 Dd X (x; y): Denote this by X D,! `p: `p if c p (X) def = smallest D for which this is possible.

21 Theorem (Bourgain, 1985): If (X; d X ) is a metric space of size n then c 2 (X). log n: This is sharp up to the value of the implicit constant (Linial-London-Rabinovich, 1995). Combining with (Johnson-Lindenstrauss, 1983): c R 10 log n(x). log n:

22 Theorem (Mendel-N., 2007): If (X; d X ) is a metric space of size n then for every 0 < ² < 1 there exists a subset Y ½ X with jy j > n 1 ² and c 2 (Y ). 1 ² : This is sharp (Bartal-Linial-Mendel-N., 2002).

23 The bi-lipschitz embeddability problem into R n Characterize those metric spaces (X; d X ) for which there exists an integer n such that (X; d X ) admits a bi-lipschitz embedding into. R n See [Semmes, 1999] for a survey.

24 Obvious restriction: doubling A metric space (X; d X ) is K-doubling if any ball can be covered by K-balls of half its radius. B(x; r) = fy 2 X : d X (x; y) 6 rg: (X; d X ) is said to be doubling if it is K-doubling for some K>0. ([Bouligand, 1928], [Larman, 1967], [Coifman-Weiss, 1971])

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44 R n is doubling R n Denote the unit ball of by. is a maximal set with for every distinct i,j. B n = B(0; 1) x 1 ; : : : ; x k 2 B n kx i x j k 2 > 1 2

45 R n is doubling R n Denote the unit ball of by. is a maximal set with for every distinct i,j. B n = B(0; 1) x 1 ; : : : ; x k 2 B n kx i x j k 2 > 1 2 def B j = B x j ; 1 4

46 R n is doubling R n Denote the unit ball of by. is a maximal set with for every distinct i,j. B n = B(0; 1) x 1 ; : : : ; x k 2 B n kx i x j k 2 > 1 2 vol (B n ) > > kx vol j=1 kx vol (B j \ B n ) j=1 µ B µ 7 8 x j; 1 8 = k vol (Bn ) 8 n :

47 R n is doubling R n Denote the unit ball of by. is a maximal set with for every distinct i,j. B n = B(0; 1) x 1 ; : : : ; x k 2 B n kx i x j k 2 > 1 2 k 6 8 n

48 R n is doubling R n Denote the unit ball of by. is a maximal set with for every distinct i,j. B n = B(0; 1) x 1 ; : : : ; x k 2 B n kx i x j k 2 > 1 2 B n ½ k 6 8 n k[ j=1 µ B x j ; 1 2

49 Simple consequence: If a metric space admits a bi- Lipschitz embedding into R n then it must also be doubling. If X D,! R n K = e O(n log D) then X is K-doubling with

50 Assouad s embedding theorem Theorem (Assouad, 1983): Suppose that is K-doubling and ² 2 (0; 1). Then (M; d 1 ² X ),! D(K;²) R N(K;²) : (X; d X )

51 Theorem (N.-Neiman, 2010): Suppose that is K-doubling and ² 2 (0; 1). Then (M; d 1 ² X ),! D(K;²) R N(K) : (X; d X ) David-Snipes, 2013: Simpler deterministic proof.

52 (M; d 1 ² X ),! D(K;²) R N(K;²) : Obvious question: Why do we need to raise the metric to the power 1 ²?

53 (M; d 1 ² X ),! D(K;²) R N(K;²) : Obvious question: Why do we need to raise the metric to the power 1 ²? Pansu (1989)-Semmes (1996): The power is needed! 1 ²

54 Heisenberg group The integer grid in 3-space H f(x; y; z) : x; y; z are integersg edges: (x; y; z)» (x 1; y; z) (x; y; z)» (x; y 1; z x)

56 Denote the shortest path metric on H by. d W The ball of radius n centered at 0=(0,0,0) is B n def = x 2 H = Z 3 : d W (x; 0) 6 n ª : By induction one shows that 8 m 2 N; 8x 2 H; jb dw (x; m)j = jb m j ³ m 4

57 8 m 2 N; 8x 2 H; jb dw (x; m)j = jb m j ³ m 4 So, (H; d W ) is a doubling metric space. Theorem (Pansu, Semmes, 1996): The metric space (H; d W ) does not admit a bi- Lipschitz embedding into for any integer n. R n

58 Heisenberg non-embeddability 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). Lafforgue-N. (2012). Li (2013). Li (2014).

59 Heisenberg non-embeddability 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). Lafforgue-N. (2012). Li (2013). Li (2014). H does not embed into `p:

60 More doubling spaces that do not admit a bi-lipschitz embedding into R n Laakso (1997):

61 [Laakso,1997], [Bourdon-Pajot, 1998], [Cheeger, 1999].

62 The Lang-Plaut problem (2001) All known examples of doubling metric spaces that fail to admit a bi-lipschitz embedding into actually fail to admit a bi-lipschitz embedding even into the infinite dimensional space. Question (Lang-Plaut): Does a doubling subset of admit a bi-lipschitz embedding into some? `2 R n R n `2

63 Would this solve the bi-lipschitz embeddability problem in? A metric space that embeds into must be doubling, and is clearly also bi-lipschitz to a subset of. `2 R n R n Conversely, can we recognize intrinsically that a given metric space embeds into? `2

64 Linial-London-Rabinovich (1995) Let d be a metric on f1; : : : ; ng. The metric space (f1; : : : ; ng; d) embeds with distortion D into `2 if and only if for every n by n symmetric positive semidefinite matrix A = (a ij ) with 8i; P n j=1 a ij = 0; nx i;j=1 maxf0; a ij gd(i; j) 2 6 D 2 n X i;j=1 maxf0; a ij gd(i; j) 2

65 Schoenberg, 1934 A metric space (X; d X ) is isometric to a subset of Hilbert space if and only if for every the matrix d(x i ; x j ) 2 is negative semidefinite on the subspace of mean-zero vectors: nx nx c j = 0 =) c i c j d(x i ; x j ) 2 6 0: j=1 j=1 x 1 ; : : : ; x n 2 X

66 Bi-Lipschitz dimensionality reduction Equivalent quantitative version of the Lang-Plaut question: Does every K-doubling subset X ½ `2 with distortion D(K) into R N(K)? embed This is known to be possible if one weakens the bi- Lipschitz requirement (e.g. nearest neighbor preserving embeddings, Indyk-N., 2007.)

67 Bi-Lipschitz dimensionality reduction Theorem (Johnson-Lindenstrauss, 1983): For every n points x 1 ; : : : ; x n 2 `2 and every ² 2 (0; 1), (fx 1 ; : : : ; x n g; kx yk 2 ) 1+²,! R C(²) log n : Most natural variants for important open problems. `p; p 6= 2 remain major In other metrics, e.g. earthmover, also wide open.

68 Failure of the Lang-Plaut problem? Theorem (Lafforgue-N., 2013): For every p>2 there exists a doubling subset of `p that does not admit a bi-lipschitz embedding into for every integer n. R n The construction relies on the Heisenberg group. Conjecture: The same construction refutes the Lang-Plaut problem in as well. `2

69 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 :

70 The Goemans-Linial Semidefinite Program The best known algorithm for the Sparsest Cut Problem is a continuous relaxation (~1997) called the Goemans-Linial Semidefinite program (G-L SDP). Theorem (Arora, Lee, N., 2005). The Goemans-Linial SDP outputs a number that is guaranteed to be within a factor of (log n) 1 2 +o(1) from (C; D):

71 Minimize nx i;j=1 C(i; j)kv i v j k 2 2 over all and v 1 ; : : : ; v n 2 R n, subject to the constraints 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 kv j v k k kv k v j k 2 2:

72 How well does the G-L SDP perform? Khot-Vishnoi (2005): The Goemans-Linial SDP must make an error of at least (log log n) c on some inputs.

73 The link to the Heisenberg group Theorem (Lee-N., 2006): The Goemans-Linial SDP makes an error of at least a constant multiple of c 1 (B n ; d W ) on some inputs. Motivated by Assouad s theorem.

74 How well does the G-L SDP perform? Theorem (Cheeger-Kleiner-N., 2009): There exists a universal constant c>0 such that c 1 (B n ; d W ) > (log n) c :

75 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. p log n [Arora-Rao-Vazirani, 2004] and the best known impossibility result is e cp log log n [Kane-Meka, 2013]. c 1 (B n ; d W ) & p log n:

76 Conjecture: For every compactly supported), Ã Z 1. 0 Z R 3 f : R 3! R (smooth and µz 2 dt jf(x; y; z + t) f(x; y; z)jdxdydz R t 2 µ (x; y; (x; y; z) + x@f (x; dxdydz: Lemma: A positive solution of this conjecture implies that c 1 (B n ; d W ) & p log n:! 1 2

77 Theorem (Lafforgue-N., 2012): For every p>1, Ã Z 1 0. p µz µz R 3 2 p! 1 2 jf(x; y; z + t) f(x; y; z)j p dt dxdydz R t 2 3 µ (x; y; p (x; y; z) + x@f (x; 1 p dxdydz

78 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 : Previously known for p=2 [Austin-N.-Tessera, 2010].

79 Geometric form of the conjecture Let A be a measurable subset of Then R 3. For t>0 define ³ (x; ª v t (A) def = vol y; z) 2 A : (x; y; z + t) =2 A : Z 1 0 v t (A) 2 t 2 dt. PER(A) 2 :

Vertical Versus Horizontal Poincare Inequalities. Assaf Naor Courant Institute

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]