Non commutative Khintchine inequalities and Grothendieck s theo

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1 Non commutative Khintchine inequalities and Grothendieck s theorem Nankai, 2007

2 Plan Non-commutative Khintchine inequalities 1 Non-commutative Khintchine inequalities 2

3 µ = Uniform probability on the set { 1, 1} IN Notation ε = (ε n ) { 1, 1} IN Classical Khintchine inequalities : For any 0 < p < there are constants A p > 0 and B p > 0 such that for any sequence x = (x n ) in l 2 we have A p ( xn 2) 1/2 ( xn ε n p dµ(ε) ) 1/p B p ( xn 2) 1/2. (1)

4 If B is a Banach space and x k B, the square function must be replaced by ( x n 2 ) 1/2 ( sup ) n ε 2 1/2 kx k dµ (2) n D 0 B where D = { 1, 1} IN equipped with its usual probability µ and where ε n = D { 1, 1} denotes the n-th coordinate. When B is a Hilbert space, for all x k in B we have n 0 ε kx k 2 = n 0 x k 2 and hence we recover the square function, but in general this is not possible and we must work with (2).

5 Theorem (Kahane) For any 0 < p < q < there is a constant K (p, q) such that for any Banach space B and any finite subset x 1,..., x n in B we have εk x k Lp(B) εk x k Lq(B) K (p, q) εk x k Lp(B). In particular ε k x k L2 (B) is equivalent to ε k x k Lp(B) for any 0 < p <.

6 We will concentrate on the very important example of B-valued martingales of the form M n = n 0 ε kx k with B = S p or more generally B = L p (M, τ) when (M, τ) is a semi-finite non-commutative measure space (due mainly to Lust-Piquard). This had a major influence on all the subsequent developments of non-commutative martingale theory.

7 Theorem Let (M, τ) be a semi-finite generalized measure space, let 2 p < and let (x n ) be a finite sequence in L p (τ). There are constants A p > 0 and B p > 0 such that A p 1/2 ( x n x n + x n xn) L p(τ) In particular, if x n = xn we have ( xn 2 ) ( 1/2 p 2 A p ( IE p ε n x n B p L p(τ) ) 1/p 1/2 ( x n x n + x n xn) IE ε n x n p L p(τ)) 1/p 2 B p L p(τ) ( xn 2 ) 1/2 p..

8 The case p < 2 is different from that of p > 2. We emphasize that even if x n = x n the expression defined as [(x n )] p which replaces ( x 2 n ) 1/2 p cannot be simplified significantly. Theorem Now assume 1 p 2. For any finite sequence (x n ) in L p (τ) we define { ) } 1/2 [(x n )] p = inf ( b 1/2 n b n + ( cn cn) where the infimum runs over all possible decompositions x n = b n + c n with b n, c n in L p (τ). Then α > 0 such that ( α [(x n )] p IE p ε n x n p L p(τ) ) 1/p [(x n )] p. (3) p

9 The dual norm to the norm [(x n )] p is NOT [(x n )] p it is the following norm y = (y n ), y n L p (τ) { ) 1/2 p [y] p = max ( yn yn ) 1/2, ( y n y n We will concentrate on case p = 1, p = so we define { ( ) 1/2 y RC = max yn yn ) } 1/2, ( y n y n. p }.

10 Proof of Non-com Khintchine for p = 1 Very recently (to appear in JFA) Haagerup and Musat found a short proof of the case p = 1 which gives optimal constant for the complex Gaussian case. We will sketch this proof for (ε n ) for which they obtain the constant 3, i.e. [(x n )] 1 L1 3IE εn x n. (τ)

11 Recall B non-commutative L 1, M = B von Neumann algebra. By duality the following are equivalent (for a fixed constant C) Indeed : x B N y M N [(x n )] 1 C ε n x n L1 (B) inf ϕ [εn] ε n y n + ϕ L (M) C y RC [(x n )] 1 and y RC are dual to each other ε n x n L1 (B) and inf ϕ [ε n] ε n y n + ϕ L (M) are dual to each other

12 Assume B non-commutative L 1, i.e. M = B is a von Neumann algebra. Key Lemma Let y = (y n ) 1 n N ; y n M. Assume y RC 1, where : Then, ŷ y RC = max{ y n y n 1/2, y n y n 1/2 }. [ε n ] = {ϕ L (M) ϕ [ε n ] such that ε n ŷ n + ϕ L (M) 3/2 y ŷ RC 1/2. ϕε n = 0, n}

13 Key Lemma implies C 3 Indeed, we have C 3/2 + C/2 and hence C/2 3/2.

14 Proof of Key Lemma (with constant 4 3 for short) : Can assume y n = y n. Let S = ε n y n. Let S t = S1 S t. Clearly S t L (M) t Moreover, if ŷ n = ε n S t we may write a priori S t = ε n ŷ n + ϕ with ϕ [ε n ] CLAIM : y ŷ RC 3/t Assuming this claim, if we choose t = 2 3 then 3/t = 1/2, we obtain y ŷ RC 1/2 and S t L (M) = ε n ŷ n + ϕ L (M) t = 2 3.

15 So it remains to prove CLAIM : y ŷ RC = ( (y n ŷ n ) 2 ) 1/2 3/t this follows from Lemma (P. JFA1978) y RC 1 ( S 4 ) 1/4 3 1/4 y RC 3 1/4 Indeed, y n ŷ n = ε n (S S t ) = ε n S1 S >t therefore (yn ŷ n ) 2 (y n ŷ n ) 2 + ϕ 2 = S 2 1 S >t t 2 S 4 and hence (y n ŷ n ) 2 S 2 1 S >t t 2 S 4 t 2 3 = 1/4

16 GT Non-commutative Khintchine inequalities In 1956, Grothendieck published the fundamental theorem of the metric theory of tensor products", now known as Grothendieck s theorem (GT in short), or Grothendieck s inequality, it has played a major role both in Banach space theory and in C -algebra theory. PROBLEM : extend to operator spaces"

17 A, B C -algebras with subspaces E A, F B. Previous versions : Any bounded linear map T : A B factors boundedly through a Hilbert space as : A T 1 H T 2 B More precisely : H comes from GNS-Hilbert spaces constructed using states f, g on A, B, and : < x, y >= f (x y) << x, y >>= g(yx )

18 Commutative case (Grothendieck 1956) (Banach spaces) K, L compact sets A = C(K ) B = C(L) T : A B There are P and Q probabilities on K and L such that A J P T L 2 (P) L b 2 (Q) J Q B J P : C(K ) L 2 (P) J Q : C(L) L 2 (Q) canonical inclusions T K G T < K C G < K IR G < (unknown exact value of) K G

19 Key inequality (Grothendieck s inequality) : For any finite sequence (a i, b i ) in C(K ) C(L) Ta i, b i K G T a i 2 1/2 bi 2 1/2 (a, b) C(K ) C(L) ( Ta, b K G T ) 1/2 ( a 2 dp ) 1/2 b 2 dq

20 Non-commutative case (Banach spaces) f, g states on A, B, we define L 2 (f ) and L 2 (g) as completion of A and B respectively for scalar products < x, y >= [f (x y)+f (yx )]/2 << x, y >>= [g(x y)+g(yx )]/2 Then we obtain analogous factorisation with A J f T L 2 (f ) L b 2 (g) J g B T K T. Key inequality : For any finite sequence (a i, b i ) in A B Tai, b i K T [a a i i + a i ai ]/2 1/2 [b b i i + b i bi ]/2 1/2 Reference : P. JFA 1978, Haagerup Adv. Math. 1985

21 Equivalent form of Key inequality : (K T ) 1 Tai, b i { a i a i 1/2 + ai ai 1/2} { b i b i 1/2 + bi b i 1/2} = a 1/2 i a i b 1/2 i b i + a 1/2 i a i bi bi 1/2 + ai a i 1/2 b i b i 1/2 + ai a i 1/2 bi b i 1/2.

22 New (operator space) version of non-com GT completely bounded linear maps T : A B and completely bounded factorizations. cf. paper with D. Shlyakhtenko, Inv. Math (2002). Previously Conjectured by Effros-Ruan and Blecher

23 OS version of GT (under suitable assumption, and say in the separable case) Any completely bounded T : A B factors completely boundedly through the direct sum of two very simple building blocks" : the row and column Hilbert operator spaces, as : A T 1 R C T 2 B New feature (without Banach Space analogue) : same factorization valid for c.b. maps T : E F when E, F are EXACT operator spaces.

24 Equivalently, any T : E F decomposes as a sum T = T 1 + T 2 with T 1 factoring through R and T 2 factoring through C. So we will prove K T : E F inf {γ R (T 1 ) + γ C (T 2 )} K T cb. T =T 1 +T 2 Equivalently, this gives a factorization" for jointly CB bilinear forms ϕ : E F C as a sum ϕ = ϕ 1 + ϕ 2 with ϕ 1 (E h F ) and ϕ 2 (F h E). Left Open : Extension to the case of a B(H)-valued bilinear form E F B(H).

25 Exact operator spaces Let d SK (E) = inf{d cb (E, F) F K } By a simple perturbation argument, it can be shown that d SK (E) = inf{d cb (E, F) F M N, N dim E}. An operator space X is exact iff d SK (X) def = sup{d SK (E) E X, dim E < } <. We call this number the exactness constant of X.

26 The terminology exact" (due to Kirchberg) comes from the following FACT : Consider the short exact sequence {0} K B(l 2 ) B(l 2 )/K {0}. An operator space X is exact IFF this sequence remains exact after tensoring by X, more precisely if {0} K min X B(l 2 ) min X (B(l 2 )/K ) min X {0} is an exact sequence.

27 Theorem Theorem Let E A, F B be exact operator spaces (A, B C -algebras). Let C = d SK (E)d SK (F). Any c.b. map T : E F with T cb = 1 satisfies the KEY INEQUALITY : (a n, b n ) A B and w n > 0 ( weights") Tan, b n C [ w 2 n ana 1/2 n + w 2 n a n an [ w 2 n bnb 1/2 n + w 2 n b n bn 1/2] 1/2]

28 Theorem Consequently : Tan, b n 2C a n an 1/2 b 1/2 n b n + a 1/2 n a n bn bn 1/2. Conversely any map satisfying such an inequality must be c.b. with T cb 2C. So we obtain a characterization of c.b. maps from E to F.

29 Corollary Recall E A, F B. Any T CB(E, F ) extends" to a mapping T CB(A, B). Meaning : T viewed as bilinear form on A B extends T viewed as bilinear form on E F. A B E F T T Case of C -algebras : If E = A and F = B enough to assume one of them exact, or T suitably approximable by finite rank maps. NOTE : Very recently (announced in June 2007), Haagerup and Musat showed that this holds without any restriction on C -algebras A, B. C

30 Corollary T : A B with T cb 1, f 1, f 2, g 1, g 2 states on A, B respectively such that (a, b) A B Ta, b 2 3/2 [(f 1 (aa )g 1 (b b)) 1/2 + (f 2 (a a)g 2 (bb )) 1/2 ]. Corollary Assume E A is exact. Then a map u : E OH is c.b. IFF there is a constant K and states f, g such that a A ua 2 K f (a a) 1/2 g(aa ) 1/2. Note : Not true if assumption exact" is removed... If OH is replaced by (C, R) θ the characterization becomes a A ua 2 K f (a a) 1 θ g(aa ) θ.

31 Corollary T : A B with T cb 1, f 1, f 2, g 1, g 2 states on A, B respectively such that (a, b) A B Ta, b 2 3/2 [(f 1 (aa )g 1 (b b)) 1/2 + (f 2 (a a)g 2 (bb )) 1/2 ]. Corollary Assume E A is exact. Then a map u : E OH is c.b. IFF there is a constant K and states f, g such that a A ua 2 K f (a a) 1/2 g(aa ) 1/2. Note : Not true if assumption exact" is removed... If OH is replaced by (C, R) θ the characterization becomes a A ua 2 K f (a a) 1 θ g(aa ) θ.

32 Corollary A map u : A OH is c.b. IFF there is a constant K and states f, g such that a A ua 2 K f (a a) 1/2 g(aa ) 1/2. More generally, u : A (C, R) θ is c.b. IFF there is a constant K and states f, g such that a A ua 2 K f (a a) 1 θ g(aa ) θ.

33 Corollary Reformulation as a factorization (already mentioned) Any T CB(A, B ) can be factorized completely boundedly through an operator space of the form R C (or more generally R I C J for abstract sets I, J in the non separable case). More precisely there are c.b. maps T 1 : A R C T 2 : R C B such that T = T 2 T 1 and T 1 cb T 2 cb 2 3/2 T cb.

34 Applications In classical Banach space theory, GT has several well known consequences. Our result automatically allows to transfer some of these to the operator space setting. One of them is Corollary For a Banach space E {E L 1 and E L 1 } E Hilbert Proof One can assume (unessential) E reflexive for simplicity. We simply factorize through H the map L quotient E Id E L 1 and conclude that the identity of E factors through a subspace of a quotient of H, hence E Hilbert

35 If we now follow the analogy we find : Corollary For an operator space E {E Non.Com.L 1 and E Non.Com.L 1 } E SQ(R C) where SQ(R C) denotes the class of all quotients of subspaces of R C. Thus the single Hilbert space is now replaced by a class of spaces namely the class of all subspaces of quotients of R C denoted by SQ(R C). Note that this class is self-dual : indeed, (R C) R C C R R C and for any X,

36 Ingredients of the proof Kirchberg s ideas on Exactness Kirchberg s theorem that there is only one C -norm on C (IF ) B(H) The analysis of Voiculescu s free circular elements (free analog of Gaussian random variables) as generalized by Shlyakhtenko This last ingredient replaces the Gaussian isometric embedding in the classical GT

37 Gaussian Embeddings (g n ) independent Gaussian random variables, standard (mean zero, variance 1) on (Ω, A, P) J : l 2 L p (P) J(x) = x n g n because Jx p = x l2 g 1 p Jx dist = x l2 g 1 Hence x g 1 1 p Jx is an isometric embedding l 2 L p (P) (0 < p < ) In particular, if E = l 2 : we have both E L 1 and E L 1 CONVERSELY : Grothendieck s isomorphic charact. of Hilbert space : {E L 1 and E L 1 } E Hilbert

38 Problem : what remains for operator spaces? Does OH embed in a non-commutative L p -space for p <? Note : Using operator space duality such a non-commutative L 1 -space actually becomes an operator space, then using complex interpolation any non-commutative L p -space (1 < p < ) becomes an operator space.

39 Some preliminary answers The above Gaussian embedding produces a different space than OH in the operator space setting The embedding OH non commutative L p is impossible for 2 < p < (Junge, 1996) The spaces R and C (and hence their direct sum R C) embed in a non-commutative L 1 -space, namely the Schatten class S 1, the predual of B(l 2 ). Indeed their natural embedding in B(l 2 ) is completely complemented", i.e. there is a c.b. projection from B(l 2 ) onto either R or C. One can show that OH is completely isometric to a subspace of a quotient of R C. Thus to embed OH into a non-commutative L 1 -space it suffices to embed any quotient of R C.

40 Some preliminary answers The above Gaussian embedding produces a different space than OH in the operator space setting The embedding OH non commutative L p is impossible for 2 < p < (Junge, 1996) The spaces R and C (and hence their direct sum R C) embed in a non-commutative L 1 -space, namely the Schatten class S 1, the predual of B(l 2 ). Indeed their natural embedding in B(l 2 ) is completely complemented", i.e. there is a c.b. projection from B(l 2 ) onto either R or C. One can show that OH is completely isometric to a subspace of a quotient of R C. Thus to embed OH into a non-commutative L 1 -space it suffices to embed any quotient of R C.

41 Some preliminary answers The above Gaussian embedding produces a different space than OH in the operator space setting The embedding OH non commutative L p is impossible for 2 < p < (Junge, 1996) The spaces R and C (and hence their direct sum R C) embed in a non-commutative L 1 -space, namely the Schatten class S 1, the predual of B(l 2 ). Indeed their natural embedding in B(l 2 ) is completely complemented", i.e. there is a c.b. projection from B(l 2 ) onto either R or C. One can show that OH is completely isometric to a subspace of a quotient of R C. Thus to embed OH into a non-commutative L 1 -space it suffices to embed any quotient of R C.

42 Some preliminary answers The above Gaussian embedding produces a different space than OH in the operator space setting The embedding OH non commutative L p is impossible for 2 < p < (Junge, 1996) The spaces R and C (and hence their direct sum R C) embed in a non-commutative L 1 -space, namely the Schatten class S 1, the predual of B(l 2 ). Indeed their natural embedding in B(l 2 ) is completely complemented", i.e. there is a c.b. projection from B(l 2 ) onto either R or C. One can show that OH is completely isometric to a subspace of a quotient of R C. Thus to embed OH into a non-commutative L 1 -space it suffices to embed any quotient of R C.

43 Some preliminary answers The above Gaussian embedding produces a different space than OH in the operator space setting The embedding OH non commutative L p is impossible for 2 < p < (Junge, 1996) The spaces R and C (and hence their direct sum R C) embed in a non-commutative L 1 -space, namely the Schatten class S 1, the predual of B(l 2 ). Indeed their natural embedding in B(l 2 ) is completely complemented", i.e. there is a c.b. projection from B(l 2 ) onto either R or C. One can show that OH is completely isometric to a subspace of a quotient of R C. Thus to embed OH into a non-commutative L 1 -space it suffices to embed any quotient of R C.

44 Theorem (Junge s Theorem Invent. 2005) There is a von Neumann algebra M and a subspace of the predual E M that is completely 2-isomorphic to OH (i.e. d cb (E, OH) 2). Actually the same is true for any subspace of a quotient of R C. Moreover, OH embeds into a non-commutative L p for any 1 p 2. Open question Can 2 be replaced by 1 (completely isometric)?

45 Note : The above M is not semi-finite... Theorem (P. Bull. LMS 2004) If OH M, then M cannot be semi-finite. More precisely, if M is semi-finite and E M then any factorization through E of the identity OH E OH must be compact. Remark. In (IMRN 2004) I give a characterization of the (infinite dimensional) quotients of R C that embed into M with M semi-finite. There are precisely 7 isomorphic types! These are obtained by direct sum from simply R, C and G = span{g n n IN} L 1 (Ω, A, P) the span of standard Gaussians (g n ) as before. R, C, R C, G, R G, C G, R C G.

46 Note Moreover, where by definition G = span{g n n IN} span{ε n n IN} G R + C R + C = (R C)/{(x, t x)}. Let (e n ) denote the biorthogonal basis to (g n ) in G. Then for any y = (y n ) K N (or B(H) N ) { ( ) 1/2 y n e n = max yn yn ) } 1/2, ( y n y n.

47 Key ingredient : Shlyakhtenko s generalization of Voiculescu s FREE GAUSSIAN VARIABLES" Let H be any Hilbert space. Let H n = H 2 2 H (n times). We denote by F (H) the full Fock space over H, i.e. F (H) = C H H 2 H n We denote by V (for vacuum") the unit in C viewed as an element of F (H). For any h in H, we denote by l(h): F (H) F (H) the left creation operator, defined by : l(h)v = h and for any x in H n with n > 0 : l(h)x = h x. We will assume that H admits an orthonormal basis which can be split in two parts with equal cardinality, {e n n I} and {e n n I} so that the union {e n n I} {e n n I} is an orthonormal system.

48 Key ingredient : Shlyakhtenko s generalization of Voiculescu s FREE GAUSSIAN VARIABLES" Let H be any Hilbert space. Let H n = H 2 2 H (n times). We denote by F (H) the full Fock space over H, i.e. F (H) = C H H 2 H n We denote by V (for vacuum") the unit in C viewed as an element of F (H). For any h in H, we denote by l(h): F (H) F (H) the left creation operator, defined by : l(h)v = h and for any x in H n with n > 0 : l(h)x = h x. We will assume that H admits an orthonormal basis which can be split in two parts with equal cardinality, {e n n I} and {e n n I} so that the union {e n n I} {e n n I} is an orthonormal system.

49 We will denote l n = l(e n ) and l n = l(e n) Then we define, for any weights w n > 0, the generalized circular elements as follows : (1) c n = w n l n + wn 1 l n These are bounded operators on F (H).

50 When w n = 1, we recover circular elements -also called Free Gaussian"- in Voiculescu s sense. (2) f n = l n + l n If we use the symmetric Fock space instead of the full one, then we recover the standard (complex) Gaussians (but now these are unbounded operators of course) (3) g n = l n + l n

51 Let M be the von Neumann algebra generated by one of these three Gaussian systems. In (1) in general M is not semi-finite In (2) it is semi-finite, actually even finite with trace τ(t ) = TV, V. In (3) M L (Ω, A, P) again finite (actually commutative!) Our problem is to produce a quotient map Q : M OH or for any subspace S R C a quotient map Q : M S and then use an S admitting OH as a quotient.

52 In case (3) what works is essentially the orthogonal projection up to a normalization constant c! Q(x) = c g n, x e n then we obtain for any λ l 2 ( λ n 2 ) 1/2 = inf{ x x M Q(x) = λ} showing : l 2 is isometric to a quotient of M = L.

53 The corresponding operator space analogue would be the following : we need to find a Q such that for any (finite) family of coefficients (λ i ) in B(H) we have 1/2 λ n λ n inf { X X M B(H) [Q Id](X) = } e n λ n Then if we use either (2) ( Free Gaussians ") or (3) (standard Gaussians) we find inf { X } max { λ n λ n 1/2, λ nλ n 1/2} which produces G as a quotient of M.

54 So what works is finally to use (1) : we then find in that case { inf { X } max wn 2 λ n λ n 1/2, wn 2 λ nλ n 1/2} which gives us a subspace S R C spanned by or equivalently after rescaling (w n e 1n w 1 n e n1 ) (w 2 n e 1n e n1 ) but it is easy to see (using the rotational invariance of each of R and C) that it is essentially the form of the generic subspace S R C.

55 Conclusion. Any subspace S R C is a quotient of M, and hence, by duality, any quotient of R C is a subspace of M (and actually of M ). Finally : any E in SQ(R C) embeds in M!