Applications of Grothendieck's inequality to linear symplectic geometry
aa r X i v : . [ m a t h . S G ] M a y Applications of Grothendieck’s inequality to linearsymplectic geometry
Efim Gluskin, Shira TannyMay 19, 2020
Abstract
Recently in symplectic geometry there arose an interest in boundingvarious functionals on spaces of matrices. It appears that Grothendieck’stheorems about factorization are a useful tool for proving such bounds.In this note we present two such applications.
Linear symplectic geometry concerns a non-degenerate anti-symmetricbilinear form, called a symplectic bilinear form , and matrices that preservethis form, called symplectic matrices . Note that such forms exist only oneven dimensional linear spaces. A standard example for this setting is theform defined on R n by ( u, v )
7→ h u, J v i where J (or J n ) is the 2 n × n matrix corresponding to multiplication by i under the identification R n ∼ = C n , J := (cid:18) − n × n n × n (cid:19) . A matrix S preserves this bilinear form if and only if it satisfies the followingrelation: S T J S = J. (1)The space of such matrices is denoted by Sp(2 n ).More generally, given an even dimensional Euclidean vector space E , wedenote by J or J E a linear orthogonal transformation whose square equalsto minus the identity on E , J = − E . In this case, matrices satisfying therelation (1) with respect to J E preserve the form ( u, v )
7→ h u, J E v i and thespace of such matrices is denoted by Sp( E ; J E ).In [1, Appendix A], Buhovsky, Logunov and Tanny proved that thereexists a constant c ( n ), depending on the dimension, such that for any finite1ollection of vectors v , . . . , v N ∈ R n , N X i,j =1 | h v i , J v j i | ≤ c ( n ) · max | s i | , | t j |≤ * N X i =1 t i v i , J N X j =1 s j v j + , (2)where the constant c ( n ) grows exponentially in n . In this paper it is shownthat, using Grothendieck’s inequality [2], the growth of the constant c ( n )in the above inequality can be improved to be √ n . This result is stated inCorollary 4 below, and in Theorem 2 in a more general setting. Example 3shows that the growth of c ( n ) ∝ √ n is sharp. The second main resultconcerns the orbit of a finite collection of vectors under the action of thegroup of symplectic matrices. Grothendieck’s inequality can be used to provea sharp upper-bound for the minimal sum of norms of given vectors underthe action of symplectic matrices, as stated in Theorem 5 below.We use some basic facts and notations from operator theory. Denote by ℓ Np the space R N equipped with the norm: k u k ℓ Np := N X i =1 | u i | p ! p , ≤ p < ∞k u k ℓ N ∞ := max ≤ i ≤ N | u i | , for u = ( u , . . . , u N ) ∈ R N . The space of linear operators from R n to R m is denoted by L ( R n , R m ). We identify an operator A ∈ L ( R n , R m ) with itsmatrix A = ( a ij ) mi =1 , nj =1 . For such a matrix, one denotes by A T ∈ L ( R m , R n )its transpose. For a vector Λ = ( λ , . . . , λ n ) ∈ R n we denote by D Λ thediagonal matrix corresponding to Λ, namely, ( D Λ ) ii = λ i and ( D Λ ) ij = 0for i = j .For 1 ≤ p, q ≤ ∞ one denotes by L ( ℓ np , ℓ mq ) the linear space L ( R n , R m )equipped with the operator norm from ℓ np to ℓ mq : k A k L ( ℓ np ,ℓ mq ) = max k u k ℓnp =1 k Au k ℓ mq . For A ∈ L ( R n , R m ) let λ ≥ λ ≥ · · · ≥ λ n − be the sequence of all eigenval-ues λ j = λ j ( A T A ) of the operator A T A with multiplicities. The j -singularvalue of A is defined as s j ( A ) := (p λ j ( A T A ) , j = 0 , . . . , n − , , j ≥ n. s ( A ) coincides with the operator norm k A k L ( ℓ n ,ℓ m ) . The Hilbert-Schmidt norm of A ∈ L ( R n , R m ) is defined as k A k HS := vuut ∞ X j =0 s j ( A ) . (3)If ( a ij ) mi =1 , nj =1 is the matrix representing A then k A k HS = vuut m X i =1 n X j =1 a ij . (4)For an operator A ∈ L ( R n , R m ) such that rank( A ) = k , one has s j ( A ) =0 for all j ≥ k . Consequently, k A k HS = vuut k − X j =0 s j ( A ) ≤ √ k · s ( A ) = p rank( A ) · k A k L ( ℓ n ,ℓ m ) (5)The following reformulation of Grothendieck’s theorem [2] is a specialcase of Theorem 2.1 in [3], where the compact sets S and T are finite. Theorem (see [3]) . Let ( a ij ) mi =1 , nj =1 be an m × n real matrix. Then, thereexist vectors Λ = ( λ , . . . , λ m ) and Λ = ( λ , . . . , λ n ) with non-negativeentries λ i ≥ , λ i ≥ and Euclidean norms bounded by 1, k Λ k ℓ m ≤ , k Λ k ℓ n ≤ , and there exists an m × n matrix B such that A = D Λ BD Λ (6) and k B k L ( ℓ n ,ℓ m ) ≤ K · k A k L ( ℓ n ∞ ,ℓ m ) (7) where K is an absolute constant. The smallest value of the constant K is called Grothendieck’s constant andis denoted by K G . Its exact value is still unknown, Grothendieck him-self proved that π/ ≤ K G ≤ sinh( π/ Lemma 1.
For any A ∈ L ( R n , R n ) there exists a vector with strictly positivecoordinates Λ = ( λ , . . . , λ n ) ∈ R n , λ i > for all ≤ i ≤ n , such that k Λ k ℓ n ≤ and k D − AD − k L ( ℓ n ,ℓ n ) ≤ K G · k A k L ( ℓ n ∞ ,ℓ n ) . (8)3 roof. Let Λ = ( λ , . . . , λ n ), Λ = ( λ , . . . , λ n ) and B be the vectors andmatrix from Grothendieck’s theorem. Let us define λ i := 1 √ √ − √ √ n + max { λ i , λ i } ! > . Then, for Λ = ( λ , . . . , λ n ), it is clear that k Λ k ℓ n ≤ k D − D Λ j k L ( ℓ n ,ℓ n ) ≤√
3. Inequality (8) easily follows.The following result provides an asymptotically sharp bound for theconstant c ( n ) from (2). Theorem 2.
For any N × N matrix A = ( a ij ) Ni,j =1 , N X i,j =1 | a ij | ≤ K G · √ rank A · k A k L ( ℓ N ∞ ,ℓ N ) . (9) Proof.
Let Λ = ( λ , . . . , λ n ) be the vector from Lemma 1 corresponding tothe matrix A . Denoting B := D − AD − , b ij = λ i λ j a ij , we have N X i,j =1 | a ij | = N X i,j =1 | b ij | · | λ i λ j | . By the Cauchy-Schwarz inequality, N X i,j =1 | b ij | · | λ i λ j | ≤ N X i,j =1 b ij · N X i,j =1 λ i λ j ≤ N X i,j =1 b ij = k B k HS , and by (5) one has k B k HS ≤ √ rank B · k B k L ( ℓ N ,ℓ N ) . Since B is a multipli-cation of A by invertible matrices, rank B = rank A . Moreover, by Lemma 1the L ( ℓ N , ℓ N )-norm of B is bounded by 3 K G ·k A k L ( ℓ N ∞ ,ℓ N ) . Overall we obtain N X i,j =1 | a ij | ≤ k B k HS ≤ √ rank B · k B k L ( ℓ N ,ℓ N ) ≤ √ rank A · K G · k A k L ( ℓ N ∞ ,ℓ N ) A in Theo-rem 2 is sharp. Example 3.
Let k ≤ N and take A = (cid:18) U k × k
00 0 (cid:19) where U = ( u ij ) ij is a k × k orthogonal matrix whose entries satisfy | u ij | ≤ C √ k . (10)For example, when k = 2 m , one can consider the following matrix, relatedto the discrete Fourier transform, which is given in a block form by U = 1 √ m (cid:18)(cid:20) cos( πjℓm ) − sin( πjℓm )sin( πjℓm ) cos( πjℓm ) (cid:21)(cid:19) mj,ℓ =1 . The above matrix satisfies condition (10) for C = √ A , the orthogonality of U implies that k A k L ( ℓ N ∞ ,ℓ N ) ≤ k and P Nj,ℓ =1 a jℓ = k . Therefore, N X j,ℓ =1 | a jℓ | ≥ P Nj,ℓ =1 a jℓ max j,ℓ | a jℓ | ≥ k √ kC ≥ C · k A k L ( ℓ N ∞ ,ℓ N ) · √ rank A, where the middle inequality follows from the assumption (10). Corollary 4.
For any finite collection of vectors v , . . . , v N in R n , N X i,j =1 | h v i , J v j i | ≤ K G · √ n · max | t i | , | s j |≤ * N X i =1 t i v i , J N X j =1 s j v j + . (11) Proof.
Consider the matrix A = ( a ij ) Ni,j =1 defined by a ij := h v i , J v j i . Then,rank A ≤ n and k A k L ( ℓ N ∞ ,ℓ N ) = max | t i | , | s j |≤ N X i,j =1 t i s j h v i , J v j i . Applying Theorem 2 to the matrix A gives the desired inequality.5he next result gives an upper-bound for the infimum of the sum ofnorms of vectors v , . . . , v N ∈ R n under the action of Sp(2 n ), by means ofthe rank and the L ( ℓ N ∞ , ℓ N )-norm of the matrix ( h v i , J v j i ) Ni,j =1 . We remarkthat this matrix is invariant under the action of Sp(2 n ) on the vectors { v i } Ni =1 (as follows easily from (1)). Theorem 5.
Let v , . . . , v N ∈ R n and consider the N × N matrix definedby A := ( h v i , J v j i ) Ni,j =1 . Then, inf S ∈ Sp(2 n ) N X i =1 k Sv i k ℓ n ! ≤ K G · rank A · k A k L ( ℓ N ∞ ,ℓ N ) . (12) Proof.
By homogeneity, we may assume that k A k L ( ℓ N ∞ ,ℓ N ) ≤
1. Denoting by V the 2 n × N matrix whose columns are the vectors { v i } Ni =1 , we can write A = V T J n V. (13)We split the proof into cases, with respect to the rank of A :Case 1: Assume that rank A = 2 n , then V must be of full rank, which meansthat the vectors { v i } Ni =1 span R n . By Lemma 1, there exists a vectorΛ = ( λ , . . . , λ N ) ∈ R N such that λ i > i , k Λ k ℓ N ≤
1, and suchthat the matrix B := D − AD − satisfies k B k L ( ℓ N ,ℓ N ) ≤ K G . Since J is an anti-symmetric operator, it follows from (13) that B T = D − V T J T n V D − = − B, namely, B is an anti-symmetric matrix of rank 2 n . By the spectraltheorem there exists a 2 n × N matrix Q , which is a part of an N × N orthogonal matrix, such that B = Q T RQ , where R is a 2 n × n matrixof the form R = − µ µ − µ n µ n , for some 0 < µ i ≤ K G . Denoting M := ( √ µ , √ µ , √ µ , √ µ , . . . , √ µ n , √ µ n ) ∈ R n ,
6e have R = D M R D M where R = −
11 0 . . . 0 −
11 0 , and is equal, up to a change of order of the basis elements, to J .Namely, R = P T J P , where P is a permutation matrix. We concludethat A = D Λ BD Λ = W T J W, where W is the 2 n × N matrix defined by W := P · D M · Q · D Λ . By (13) we obtain V T J V = A = W T J W. (14)Let us show that ker V = ker W . Indeed, v ∈ ker V if and only iffor any u ∈ R n , h u, V v i = 0. Since J is invertible, this is equivalentto h u, J V v i = 0 for all u ∈ R n . Moreover, V is of full rank andso its image is R n . Therefore, the latter condition is equivalent to (cid:10) w, V T J V v (cid:11) = h V w, J V v i = 0 for all w ∈ R N . We conclude thatker V = ker V T J V . Arguing the same for W and using (14) yieldsker V = ker W .Now, as both matrices V, W are of full rank and have the same kernel,there exists a 2 n × n matrix S such that W = SV.
Plugging this back in (14) yields V T S T J SV = V T J V, which implies (since V is of full rank) that S T J S = J and hence S ∈ Sp(2 n ). Finally, let us bound the sum of norms of the vectors { Sv i } Ni =1 . Denoting by { e i } Ni =1 the standard basis of R N , we have Sv i = W e i . In addition, k W e i k ℓ n = k P · D M · Q · D Λ e i k ℓ n ≤ k P k L ( ℓ n ,ℓ n ) · k D M k L ( ℓ n ,ℓ n ) · k Qe i k ℓ n · | λ i | . P is an orthogonal matrix, k P k L ( ℓ n ,ℓ n ) = 1. In addition, k D M k L ( ℓ n ,ℓ n ) = max i √ µ i ≤ p K G . Therefore, N X i =1 k Sv i k ℓ n ≤ p K G · N X i =1 k Qe i k ℓ n · | λ i |≤ p K G · N X i =1 k Qe i k ℓ n ! · N X i =1 λ i ! ≤ p K G · N X i =1 k Qe i k ℓ n ! ≤ p K G · √ n, where the last inequality follows from the fact that Q is a part of an N × N orthogonal matrix. Since 2 n = rank A , this concludes the proofof the theorem for this case.Case 2: Assume that rank A < n . Set E := span { v , . . . , v N } ⊂ R n . De-note by ℓ the dimension of the kernel of the restriction of the bilin-ear form h· , J ·i to E . By a well known fact from symplectic linearalgebra, dimE − ℓ =: 2 k is even, we have k + ℓ n , and more-over there exists a linear symplectic matrix T ∈ Sp(2 n ) such that T ( E ) = span { e , . . . , e k , e n +1 , . . . , e n + k + ℓ } . Since both sides of theinequality (5) are invariant under the action of Sp(2 n ) on vectors v , . . . , v N , we may assume without loss of generality that E = span { e , . . . , e k , e n +1 , . . . , e n + k + ℓ } . Denote E = span { e , . . . , e k , e n +1 , . . . , e n + k } ,E = span { e n + k +1 , . . . , e n + k + ℓ } ,E = span { e k +1 , . . . , e k + ℓ } ,E = span { e k + ℓ +1 , . . . , e n , e n + k + ℓ +1 , . . . , e n } . We have the orthogonal decompositions E = E ⊕ E and R n = E ⊕ E ⊕ E ⊕ E . Consider the orthogonal projections π i : R n → E i for i = 0 , , ,
3. For any u, v ∈ E we have h v, J u i = h π v, J π u i . A defined by a ij = h v i , J v j i does not changewhen we replace { v i } by { π v i } . Thusrank A = rank h· , J ·i | E = dim E = 2 k and we may apply Case 1 to the vectors { π v i } in ( E , J | E ). We con-clude that there exists a matrix S : E → E such that S T J | E S = J | E and N X i =1 k S π v i k ℓ k ≤ p K G · √ k. For any given ε > S ε := S π + επ + 1 ε π + π . One can check that S ε satisfies relation (1) and therefore belongs toSp(2 n ). Finally, the sum of norms of { S ε v i } is bounded as follows: N X i =1 k S ε v i k ℓ n = N X i =1 k S ε π v i + S ε π v i k ℓ n ≤ N X i =1 k S π v i k ℓ n + N X i =1 k ε · π v i k ℓ n ≤ p K G · √ k + ε N X i =1 k v i k ℓ n . Taking ε → S ∈ Sp(2 n ) N X i =1 k Sv i k ℓ n ≤ p K G · √ k = p K G · √ rank A. Example 6.
Let us show that for any k ≤ n there exist vectors v , . . . , v N ∈ R n with dim span { v , . . . , v N } ≤ k such that for any matrix S ∈ Sp(2 n ), N X j =1 k Sv j k ℓ n ≥ C · k · k A k L ( ℓ N ∞ ,ℓ N ) , A := ( h v i , J n v j i ) Ni,j =1 . Notice that it is enough to consider the casewhere k = 2 m is even. In this case, take the vectors v j to be zero for j > k =2 m and { v , . . . , v m } = { e , . . . , e m , e n +1 , . . . , e n + m } . The correspondingmatrix is A = (cid:18) J m
00 0 (cid:19) and (cid:16)P Nj =1 k v j k ℓ n (cid:17) = k = rank A · k A k L ( ℓ N ∞ ,ℓ N ) . Let S ∈ Sp(2 n ) and set w j := Sv j for all 1 ≤ j ≤ N . Denote by W the 2 n × N matrix whosecolumns are the vectors w j . Then, it follows from (1) that A = W T J n W .Let ˜ W be the 2 n × m matrix whose columns are the first 2 m columns of W , then ˜ W T J n ˜ W = J m .Given a k × ℓ matrix P whose columns are the vectors { p , . . . , p ℓ } ⊂ R k ,we denote by Π( P ) := Q ℓi =1 k p i k ℓ k the product of the Euclidean norms of thecolumns of P . A generalized version of Hadamard’s inequality states that forany pair of k × ℓ matrices P and Q , the determinant of P T Q is bounded bythe product of ℓ -norms of the columns of P and Q , det( P T Q ) ≤ Π( P ) · Π( Q ).Applying this to P = ˜ W , Q = J n ˜ W , we have1 = det( J m ) = det( ˜ W T · J n ˜ W ) ≤ Π( ˜ W ) · Π( J n ˜ W ) = Π( ˜ W ) . Using the inequality of arithmetic and geometric means we conclude that1 ≤ m Y j =1 k w j k ℓ n ≤ m m X j =1 k w j k ℓ n , and so N X j =1 k w j k ℓ n = m X j =1 k w j k ℓ n ≥ m = k = q rank A · k A k L ( ℓ N ∞ ,ℓ N ) . Remark 7.
In the above example we actually proved the following strongerstatement: For any collection of vectors v , . . . , v N ∈ R n that satisfy V T J n V = (cid:18) J m
00 0 (cid:19) =: A for some m ≤ n , where V is the matrix whose columns are { v j } Nj =1 , we have N X j =1 k v j k ℓ N ≥ m = q rank( A ) · k A k L ( ℓ N ∞ ,ℓ N ) . (15)10 cknowledgements. The authors are grateful to the Creator for giving them the understand-ing presented in this paper. We also thank Lev Buhovsky for useful dis-cussions and comments. Shira Tanny extends her special thanks to LevBuhovsky and Leonid Polterovich for their mentorship and guidance.S.T. was partially supported by ISF Grant 2026/17 and by the LevtzionScholarship.
References [1] L. Buhovsky, A. Logunov, and S. Tanny. Poisson brackets of partitionsof unity on surfaces. accepted to Commentarii Mathematici Helvetici ,2019.[2] A. Grothendieck.
R´esum´e de la th´eorie m´etrique des produits tensorielstopologiques . Soc. de Matem´atica de S˜ao Paulo, 1956.[3] G. Pisier. Grothendiecks theorem, past and present.
Bulletin of theAmerican Mathematical Society , 49(2):237–323, 2012.11 fim Gluskin,
School of Mathematical SciencesTel Aviv UniversityRamat Aviv, Tel Aviv 69978IsraelE-mail: [email protected]