Doubling metric spaces and embeddings. Assaf Naor

Doubling metric spaces and embeddings Assaf Naor

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).

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

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

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)

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.

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

A =

B =

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)

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 = 0 @ 1 1X jx j y j j p A j=1 1 p

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.

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.

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:

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).

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.

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])

Doubling

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

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

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 :

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

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

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

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 )

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.

(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?

(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

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)

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

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, 1989 + Semmes, 1996): The metric space (H; d W ) does not admit a bi- Lipschitz embedding into for any integer n. R n

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).

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

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

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

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

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

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

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.)

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.

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

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 (~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):

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 2 2 6 kv j v k k 2 2 + kv k v j k 2 2:

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.

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.

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 :

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:

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 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:! 1 2

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 p @f µ (x; y; z) @x + @f p (x; y; z) + x@f (x; y; z) @y @z 1 p dxdydz

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].

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 :