ON THE EXTENDED HAAGERUP TENSOR PRODUCT IN OPERATOR SPACES. 1. Introduction

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1 ON THE EXTENDED HAAGERUP TENSOR PRODUCT IN OPERATOR SPACES TAKASHI ITOH AND MASARU NAGISA Abstract We descrbe the Haagerup tensor product l h l and the extended Haagerup tensor product l eh l n terms of Schur product maps, and show that l h l B(l 2 ) (resp l eh l B(l 2 )) concdes wth c 0 h c 0 B(l 2 ) (resp c 0 eh c 0 B(l 2 )) For C*-algebras A,B, t s shown that A h B = A eh B f and only f A or B s fnte-dmensonal Introducton For Hlbert spaces H and K, we let B(H, K) and K(H, K) denote the bounded operators and the compact operators of H to K An operator space X on H s a subspace of B(H) = B(H, H) whch s endowed wth norms to each n m matrces M n,m (X) over X as a subspace of M n,m (B(H)) = B(H m, H n ) We allow to use the notaton M I,J (B(H)) = B(H J, H I ) for arbtrary ndex sets I and J Let X and Y be operator spaces The Haagerup tensor product of X and Y s the completon of the algebrac tensor product X Y by the norm n u h = nf{ [a,, a n ] t [b,, b n ] u = a b X Y, n N, a X, b Y }, and s denoted by X h Y [] We also recall the extended Haagerup tensor product X eh Y An element u of X eh Y s represented by the followng formal sum: u = I a b, where a = [a ] I M,I (X), b = t [b ] I M I, (Y ) (n other words, a = I a a /2 <, b = I b b /2 <

2 for a X and b Y ) We apprecate ths formal sum as the blnear form on X Y as follows: u(f, g) = I a (f)b (g) for f X, g Y For ths element u X eh Y, ts norm s defned by u eh = nf{ a b u = I a b, a M,I (X), b M I, (Y )} Then we can realze X eh Y as a subspace of the dual operator space (X h Y ) [6] In [7], the authors studed the Schur product on B(H) and used the extended Haagerup tensor product to descrbe the property of Schur product maps Effros and Ruan has shown that X h Y s (completely sometrcally) embedded to X eh Y [6] We wll be concerned wth the dfference between the Haagerup tensor product and the extended Haagerup tensor product, snce t s essental to deal wth Schur product maps derved from (possbly unbounded) operators The Schur product map on B(l 2 ) s a normal l -bmodule map, where l s a maxmal abelan subalgebra of B(l 2 ) and s dentfed wth the bounded sequences on N (cf [7]) As a deep result concernng (normal) bmodule maps, we often refer to the followng theorem by Blecher and Smth n [2]: f M s a von Neumann algebra, then M w h M s completely somorphc to the completely bounded M -bmodule maps of K(H) to B(H) denoted by CB M (K(H), B(H)), where w h concdes wth eh n ths settng In secton 2, we study the dfference between l h l and l eh l from the vew pont of Schur product and characterze them n terms of Schur product maps Moreover we characterze c 0 h c 0 and c 0 eh c 0 n terms of Schur product maps, where c 0 s the complex sequences on N tends to 0 As a result for Schur product maps derved from bounded operators, we show that l h l B(l 2 ) (resp l eh l B(l 2 )) concdes wth c 0 h c 0 B(l 2 ) (resp c 0 eh c 0 B(l 2 )) In secton, we ntroduce some notons (rght-compact, weakly rghtcompact, left-compact, weakly left-compact) for whch dstngush the Haagerup tensor product from the extended Haagerup tensor product for operator spaces As a man result n ths secton, for C*-algebras A, B, t s shown that A h B = A eh B f and only f A or B s fnte-dmensonal 2

3 2 l h l and l eh l Let X and Y be operator spaces and X Y the algebrac tensor product of X and Y For a = [a,, a n ] M,n (X), b = t [b,, b n ] M n, (Y ) and α = [α j ] M n (C), we denote n a b X Y by a b, and n,j= α ja b j by aα b Proposton If u X Y, then u h = nf{ a α b u = aα b X Y, n N, α M n (C), a M,n (X), b M n, (Y )} Proof It follows from u h = nf{ a n b u = a n b} nf{ a α b u = aα b} nf{ aα b u = aα b} nf{ a b u = a b} = u h Proposton 2 If u X h Y, then u h = nf{ a α b α K(l 2 ), a M, (X), b M, (Y ), u = α j a b j (= aα b) converges n X h Y },j= = nf{max λ a b (λ ) c 0, a M, (X), b M, (Y ), u = λ a b converges n X h Y } Proof Suppose that u X h Y wth u h < To prove the frst equalty, t suffces to show that there exst a = [a, a 2, ] M, (X) wth a <, α = [α j ] K(l 2 ) wth α < and b = t [b, b 2, ] M, (Y ) wth b < such that k α j a b j,j= converges to u n X h Y when k tends to Gven ε = u h > 0 Then we can choose a sequence {u n } X Y, whch converges to u, satsfyng that u n h < ε and u n+

4 u n h < 2 n ε (n ), u 0 = 0 If we put t n = u n+ u n, then t turns out k t n u h = u k+ u h 0 (k ) n=0 For t n X Y, there exst v n M,l(n) (X), β n M l(n) and w n M l(n), such that t n = v n β n w n wth β n = (n 0), v n w n < 2 n ε(n ), v 0 w 0 < ε and v n = w n It follows that t n h v n w n < n=0 Then we can choose an ncreasng sequence {c n } R such that c n >, lm c n =, c n v n w n < n n=0 Now we put a() = c v, α = β /c and b() = c w Then we have k k u k+ = v n β n w n = a(n)α n b(n) n=0 n=0 n=0 α 0 = [ a(0) a() a(k) ] α b(0) b(), α k b(k) and [a(0), a(),, a(k)], t [b(0), b(),, b(k)] <, α k 0 (k ) If we defne a n X, b n Y and α K(l 2 ) by the followng relaton: [a(0), a(),, a(k)] = [a, a 2,, a l(0)+l()+ +l(k) ] [b(0), b(),, b(k)] = [b, b 2,, b l(0)+l()+ +l(k) ] α = α k, k=0 then we can get the frst equalty For the above α K(l 2 ), we can take untares u k, v k M l(k) such that λ α k = u k k =0 l()+ λ k =0 l()+2 4 λ k =0 l() v k (k = 0,,2, )

5 If we put U = u k, V = k=0 λ λ v k, Λ = 2 k=0 λ, then we can get Λ = max λ, au M, (X) and a = au, V b M, (Y ) and b = V b, for any a M, (X) and b M, (Y ) By the fact we can get the second equalty aα b = auλv b = (au)λ (V b), By the above proof, we also get the followng fact: X h Y = {aα b α K(l 2 ), a M, (X), b M, (Y )} = { λ a b (λ ) c 0, a M, (X), b M, (Y )} Let H be a separable Hlbert space, {f } a completely orthonormal system of H and {e j },j= a system of matrx unts of B(H) defned by e j ξ = (ξ f j )f, ξ H We can naturally dentfy the bounded sequences l on N wth the maxmal abelan subalgebra of B(H) generated by {e } We denote by CB l (K(H), B(H)) the l -bmodule completely bounded maps of K(H) to B(H) Then there exsts completely sometrc somorphsm between l eh l and CB l (K(H), B(H)) by the followng: for a b l eh l, < a b > CB l (K(H), B(H)) s defned by < a b > (k) = a kb for k K(H) [2] By the l -bmodularty of < x > for x l eh l, there exsts a scalar x j satsfyng that < x > (e j ) = x j e j,, j =,2, 5

6 Then we can defne an nfnte dmensonal matrx [x] = [x j ],j=, and also dentfy [x] wth a lnear map from c c (N) to l as follows: for ξ = [ξ, ξ 2, ] c c (N), [x]ξ = [ x j ξ j, x 2j ξ j, ], j= where ξ = [ξ, ξ 2, ] c c (N) means that ξ n = 0 for suffcently large n Clearly c c (N) s contaned n l 2 and the mage of c c (N) by [x] s not necessarly contaned n l 2 If [x] can be extended to B(l 2 ) (resp K(l 2 )), then we wrte x (l eh l ) B (resp x (l eh l ) K) We also use the followng notaton: for any subspace S of l eh l, j= S B = (l eh l ) B S, S K = (l eh l ) K S Lemma x l eh l f and only f there exst ξ, η l 2 ( =,2, ) such that sup{ ξ, η } < and x j = (ξ η j ) Proof For x l eh l, note that there exst countable sequences {a } and {b } n l such that [a, a 2, ], t [b, b 2, ] < and x = a b Then we defne ξ = [a (), a 2 (), ] and η = [b (), b 2 (), ] for any N Clearly we have ξ, η l 2, and sup{ ξ, η } max{ [a, a 2, ], t [b, b 2, ] }, < x > (e j ) = = a k e j b k k= a k ()b k (j)e j = (ξ η j )e j k= Conversely, for ξ = [ξ (), ξ (2), ], η = [η (), η (2), ] l 2 wth sup{ ξ, η } < and x j = (ξ η j ), we defne a = [ξ (), ξ 2 (), ] and b = t [η (), η 2 (), ] 6

7 Then we have [a, a 2, ], t [b, b 2, ] sup{ ξ, η } and x = a b Lemma 4 x l h l f and only f there exst β K(l 2 ), ξ, η l 2 ( =,2, ) such that sup{ ξ, η } < and x j = (βξ η j ) Proof By Proposton 2, for gven ε > 0 and x l h l, there exst [a, a 2, ] M, (l ), t [b, b 2, ] M, (l ) and [α j ] K(H) satsfyng [a, a 2, ] t [b, b 2, ] < and [α j ] < x h + ε such that x =,j= α ja b j If ξ = [a (), a 2 (), ], η = [b (), b 2 (), ] and β = [β j ] where β j = α j, then t s clear that sup { x, η } < Thus we have < x > (e j ) = s,t α st a s e j b t = s,t α st a s ()b t (j)e j = (βξ η j )e j Conversely, for gven ξ = [ξ (), ξ (2), ], η = [η (), η (2), ] l 2 and β = [β j ] K(l 2 ), we put a = [ξ (), ξ 2 (), ], b = [η (), η 2 (), ] l and α = [α j ] K(l 2 ) where α j = β j Then we have, for any postve nteger N, [a, a 2,, a N ], t [b, b 2,, b N ] sup{ ξ, η } < For an element x n = [ a a 2 b α ] α n a n b 2 l l, α n α nn b n 7

8 we have x n+k x n = [ ] a a n+k 0 0 α,n+ α,n+k b 0 0 α n+, α n+,n+ b n+k α n+k, α n+k,n+k By the compactness of α and Proposton, lm x n+k x n h = 0 n for any postve nteger k Thus we have that the sequence {x n } converges to x n l h l Snce c 0 s a C*-subalgebra of l, we can see c 0 h c 0 as a subspace of l h l Lemma 5 x c 0 h c 0 f and only f there exst ξ, η l 2 ( =,2, ) such that lm ξ = lm η = 0 and x j = (ξ η j ) Proof By Proposton 2, for gven ε > 0 and x c 0 h c 0, there exst [a, a 2, ] M, (c 0 ), t [b, b 2, ] M, (c 0 ) and [α j ] K(H) satsfyng [a, a 2, ] t [b, b 2, ] < and [α j ] < x h + ε such that x =,j= α ja b j We put ξ = [a (), a 2 (), ], η = [b (), b 2 (), ] and β = [β j ] where β j = α j Then we have and, by the fact a, b c 0, x j = (βξ η j ) lm ξ (j) = lm η (j) = 0 for any j N Ths means that {ξ }, {η } l 2 weakly converges to 0 We can choose β, β 2 K(l 2 ) such that β = β 2 β Then we have lm β ξ = lm β 2 η = 0 and x j = (βξ η j ) = (β ξ β 2 η j ) 8

9 Conversely suppose that lm ξ = lm η = 0 We may assume that ξ < c for all N Then, for any ε > 0, we can choose a number N such that where ξ < c for all and ξ < ε f > N, { ξ (j) = ξ (j) f j < N 2ξ (j) otherwse Clearly we have lm ξ = 0 Applyng ths argument to {ξ } repeatedly, we can choose = n(0) < n() < n(2) < and {ζ } l 2 such that ζ (j) = 2 k ξ (j) f n(k) j < n(k + ), ζ < c for all and ζ < 2 k f > n(k) We put a = [ζ (), ζ 2 (), ], b = [η (), η 2 (), ] and Then we have a, b, (λ ) c 0, and Thus we have λ = 2 k f n(k) < n(k + ) [a, a 2, ], t [b, b 2, ] sup{ ζ, η } < x = λ a b c 0 h c 0 Lemma 6 x c 0 eh c 0 f and only f there exst ξ, η l 2 ( =,2, ) such that lm ξ = lm η = 0 (weakly) and x j = (ξ η j ) Proof By the fact c 0 eh c 0 l eh l and the proof of Lemma, we can choose ξ, η l 2 ( =,2, ) satsfyng that and sup{ ξ, η } <, x j = (ξ η j ) lm ξ j() = lm η j () = 0 ( =,2, ) j j Ths means that {ξ }, {η } weakly converges to 0 9

10 Conversely, snce ξ, η converges to 0 weakly, they are unformly bounded As n the proof of Lemma, f we put a (j) = ξ j () and b (j) = η j (), then we have a, b c 0, [a, a 2, ] <, t [b, b 2, ] < and x = a b c 0 eh c 0 From the proof of the prevous lemmas, we have another representatons of norms for the (extended) Haagerup tensor product on c 0 c 0 and l l Remark 7 () For x l eh l, x eh = nf{sup ξ η j x j = (ξ η j ), ξ, η j l 2 } j (2) For x l h l, x h = nf{sup ξ η j β x j = (βξ η j ), ξ, η j l 2, β K(l 2 )} j () For x c 0 eh c 0, x eh = nf{sup ξ η j x j = (ξ η j ), ξ, η j l 2, ξ 0, η j 0 weakly} j (4) For x c 0 h c 0, x h = nf{sup ξ η j x j = (ξ η j ), ξ, η j l 2, ξ 0, η j 0 strongly} j Theorem 8 () (c 0 h c 0 ) B = (l h l ) B (2) (c 0 eh c 0 ) B = (l eh l ) B Proof () It s clear that (c 0 h c 0 ) B (l h l ) B Let x (l h l ) B By Lemma 4, there exst β K(l 2 ), ξ, η l 2 ( =,2, ) such that sup{ ξ, η } < and x j = (βξ η j ) We choose β, β 2 K(l 2 ) such that β = β 2 β, that s, and we may assume that x j = (β ξ β 2 η j ), Range(β ) span{β 2 η j j N}, and Range(β 2 ) span{β ξ j j N} 0

11 It s suffcent to show that Assume that lm β ξ = lm β 2 η = 0 lm sup β ξ > 0 Then there exst δ > 0 and a subsequence {n(k)} such that β ξ n(k) > δ for k =,2, Snce sup ξ <, we may also assume that {ξ n(k) } weakly converges to some ξ 0 l 2 By the compactness of β, we have lm β ξ n(k) β ξ 0 = 0 Thus t turns out β ξ 0 0 We can choose j 0 such that Then there exsts K N such that (β ξ 0 β 2 η j0 ) 0 x n(k),j0 = (β ξ n(k) β 2 η j0 ) > (β ξ 0 β 2 η j0 ) for k > K 2 Ths contradcts to [x] = [x j ] B(l 2 ) (2) For x (l eh l ) B, that s, [x] B(l 2 ), we can choose α = [ξ j ] and β = [η j ] n B(l 2 ) such that Remarkng the fact we defne for all Then we have [x] = αβ and α = β = [x] /2 x j = k ξ k η kj, a = [ξ, ξ 2, ], b = [η, η 2, ] l 2 c 0 [a, a 2, ], t [b, b 2, ] [x] /2 < and x = a b c 0 eh c 0 Corollary 9 Let x l eh l and lm sup k x (k),j(k) > 0 for some njecton N k ((k), j(k)) N N Then x does not belong to c 0 h c 0 Moreover, f x satsfes an addtonal condton [x] B(l 2 ), then x does not belong to l h l

12 Example 0 () Let x =,j= ( λ λ j ) t e e j l eh l, where λ s are postve real and t s real Then we have ( λ λ ) t ( λ λ 2 ) t [x] = ( λ 2 λ ) t ( λ 2 λ 2 ) t / B(l 2 ), x j = ( λ ) t = ( λ j 0 λ t 0 λ t and x j = Ths means x / (l eh l ) B, x / c 0 eh c 0 (by Lemma 6) and x l h l (by Lemma 4) (2) Let x = k= e k e k c 0 eh c 0 Snce 0 [x] = 0 B(l 2 ) then we have x / l h l (by Corollary 9) () (l eh l ) K (l h l ) B By Lemma 4, t s clear that (l eh l ) K (l h l ) B We consder the followng nfnte dmensonal matrx: p = j 0 ) Snce p s an nfnte dmensonal projecton, p does not belong to K(l 2 ) If we put ξ = [,0,0,0, ] ξ 2 = ξ = [0,, 0,0 ] 2 ξ 4 = ξ 5 = ξ 6 = [0,0, 2, 0, ]

13 and ξ n = η n (n =,2, ), then ξ n, η n l 2 satsfy Ths means that lm n ξ n = lm n η n = 0 and p = [(ξ η j )] ((l h l ) B) ((l eh l ) K) c φ (4) Let a = b = [, 2,, n, ] c 0 Then x = a b c 0 h c 0 and [x j ] = / B(l 2 ) By the above argument, we can get the followng dagram of nclusons: (c 0 h c 0 ) K (c 0 h c 0 ) B (c 0 eh c 0 ) B = = = (l eh l ) K (l h l ) B (l eh l ) B c 0 h c 0 c 0 eh c 0 l h l l eh l X h Y and X eh Y Let X be an operator space We call X rght-compact (resp leftcompact) f M,I (X) = M,I (X)K(l 2 (I)) (resp M I, (X) = K(l 2 (I)) M I, (X)) If X s rght-compact, then, for any a = [a ] I M,I (X), there exst b = [b ] I M,I (X) and α = [α j ],j I K(l 2 (I)) such that a = bα (a j = b α j ) I We also call X weakly rght-compact (resp weakly left-compact) f we have, for any a = [a ] M,I (X) (resp a = [a ] M I, (X)), that { I a 0} s countable and lm a = 0 Lemma If X s a rght-compact (resp left-compact) operator space, then X s weakly rght-compact (resp weakly left-compact)

14 Proof Snce α = [α j ],j I K(l 2 (I)), we have that {(, j) I I α j 0} s countable and lm j α j = 0 For b = [b ] I M,I (X), {b I} s bounded So we have for any a = bα, that { I a 0} s countable and lm a = 0 As a typcal example of rght-compact operator spaces, we can get the followng: Lemma 2 Let X be an operator space on a Hlbert space H If X pb(h) for some fnte-dmensonal projecton p B(H), then X s rght-compact In partcular, any fnte-dmensonal C*-algebra s left- and rghtcompact Proof We assume that dm ph = n < Let a = [a ] I M,I (X), e, I a a < We can consder a a as an element of M n (C), so we put a a = (α jk ) (j, k =,2,, n) By the postvty of a a, we have 0 sup α jj a a j n I I Ths mples that s countable Remarkng the fact we have I 0 = { I α jj > 0 for some j} a a n sup α jj, j n a a a a n n α jj < We can choose a sequence of postve numbers λ such that λ > 0, λ 0 ( ), a a < Then we have j= a = [a ] I = [ a λ ] I [δ j λ ],j I M I (X)K(l 2 (I)), 4 λ 2 I

15 where δ j means Kronecker s symbol Lemma Let {a } be a sequence of bounded operators on a Hlbert space H wth a a < Suppose that there exst sequences {u }, {ξ } of unt vectors n H such that (a u ξ ) > for N Then, for any ε > 0, there exsts a number 0 such that s nfnte { N (a 0 u ξ ) < ε} Proof Let ε > 0 Suppose that {j N (a u j ξ j ) < ε} s fnte for all N If we omt the fnte set we may assume that If we omt agan the fnte set we may assume that { N (a u ξ ) < ε}, (a u ξ ) >, (a 2 u 2 ξ 2 ) >, (a u 2 ξ 2 ) ε { N > 2, (a 2 u ξ ) < ε}, (a u ξ ) ε, (a 2 u ξ ) ε Usng ths argument repeatedly, we may assume that, for any n, Then we have (a u n ξ n ) ε, (a 2 u n ξ n ) ε,, (a n u n ξ n ) ε n n (n )ε 2 (a u n ξ n ) 2 u n 2 a ξ n 2 n (a a ξ n n ξ n ) a a Ths contradcts to the assumpton a a < Therefore we can get a number 0 requred n the statement 5

16 Lemma 4 Let X be an operator space on a Hlbert space H If X s not weakly rght-compact, then there exst a sequence {a } of X, sequences {u }, {ξ } of unt vectors n H and some constant K such that () a a < (2) < a < K () < (a u ξ ) < K (4) (a k u j ξ j ) for k j Kk Proof Snce X s not weakly left-compact, we can choose a sequence {a } of X such that a a < and { a } s not convergent to 0 Then we may assume that < a < K for any and some constant K We choose sequences {u }, {ξ } of unt vectors n H satsfyng (a u ξ ) > for all N Usng Lemma, we can choose a subsequence {n(k)} k= such that (a n(k) u n(j) ξ n(j) ) < for k < j K k If we replace {a n(k) } wth {a }, then we can get the condtons () () and (4) for k < j We consder a sequence {a, u, ξ } of trplets By the calculaton (a u j ξ j ) 2 a ξ j 2 = ( a a ξ j ξ j ) we have a a <, lm (a u j ξ j ) = 0 for any j Choosng a subsequence of {a, u, ξ }, we may assume that (a k u j ξ j ) < K k for k > j Thus we can get the condtons () (4) 6

17 Lemma 5 Let α > β > 0 If sequences {a k }, {b k } of vectors n C m satsfy the followng condtons: then (a k b k ) > α and (a k b l ) < β for k l, sup{ a k (), b k () =,, m, k =,2, } = Proof We assume that sup{ a k (), b k () =,, n, k =,2, } s fnte By the compactness, we can choose a par of convergent subsequences {a n(k) }, {b n(k) } Then we have But ths contradcts to lm k (a n(k) b n(k+) ) = lm k (a n(k) b n(k) ) α lm sup (a n(k) b n(k+) ) β k Theorem 6 Let X and Y be operator spaces Then we have () X h Y = X eh Y f X s rght-compact or Y s left-compact (2) X s weakly rght-compact or Y s weakly left-compact f X h Y = X eh Y Proof () We assume that X s rght-compact For any s X eh Y, there exst a = [a ] M,I (X) and b = t [b ] M I, (Y ) such that s = a b = I a b By the assumpton, there exst c M,I (X) and α K(l 2 (I)) such that a = cα So we have s = a b = cα b X h Y Ths means that X h Y = X eh Y When Y s left-compact, we can also have X h Y = X eh Y by the same argument (2) Let X (resp Y ) be an operator space on H (resp K) We assume that X s not weakly rght-compact and Y s not weakly leftcompact By Lemma 4, we can choose a sequence {a } of X (resp a sequence {b } of Y ), sequences {u }, {ξ } of unt vectors n H (resp sequences {v }, {η } of unt vectors n K) and some constant K satsfyng 7

18 that a a <, b b <, < a, b < K, < (a u ξ ), (b η v ) < K and (a k u j ξ j ), (b k η j v j ) < K k for k j We defne s X eh Y, ϕ k X and ψ k Y as follows: s = Then we have a b, ϕ k ( ) = ( u k ξ k ), ψ k ( ) = ( η k v k ) s(ϕ k, ψ k ) = = ϕ k (a )ψ k (b ) (a u k ξ k )(b η k v k ) (a k u k ξ k )(b k η k v k ) k (a u k ξ k )(b η k v k ) 9 > 8, K2 and, for j k, s(ϕ j, ψ k ) = = ϕ j (a )ψ k (b ) (a u j ξ j )(b η k v k ) (a u j ξ j )(b η k v k ) K + j K + < k K2 8

19 Suppose that X eh Y = X h Y, then s belongs to X h Y We can choose m t = x y X h Y and s t h < Snce ϕ j = ψ k =, that s, Then we have and, for j k, s(ϕ j, ψ k ) t(ϕ j, ψ k ) <, s(ϕ j, ψ k ) m ϕ j (x )ψ k (y ) < m ϕ k (x )ψ k (y ) > 7 m ϕ j (x )ψ k (y ) < 4 Ths contradcts to the boundedness of { ϕ k (x ), ψ k (y ) m, k N} by Lemma 5 We are done Remark 7 The row Hlbert space H r s rght-compact and s not weakly left-compact and the column Hlbert space H c s left-compact and s not weakly rght-compact Then t s clear that (cf[5]) H r h H c = H r eh H c, H c h H r H c eh H r Corollary 8 Let A and B be C*-algebras Then the followng assertons are equvalent: () A h B = A eh B, (2) A or B s fnte dmensonal Proof We have already shown that every fnte-dmensnal C*-algebra s rght-compact and left-compact n Lemma 2 It s suffcent to show that every nfnte-dmensonal C*-algebra s nether weakly rghtcompact nor weakly left-compact 9

20 Suppose that A s nfnte dmensonal Snce the maxmal abelan *-subalgebras n A s nfnte dmensonal, there exst self-adjont elements {a n } A such that a n = and a a j = 0 f j Then we have a 2 = a a = a a < and { a } does not converge to 0 Ths means that A s nether weakly rght-compact nor weakly left-compact References [] D P Blecher and V I Paulsen, Tensor products of operator spaces, J Funct Anal 99, (99), pp [2] D P Blecher and R R Smth, The dual of the Haagerup tensor product, J London Math Soc 45, (992), pp [] E G Effros and A Kshmoto, Module maps and Hochschld-Johnson cohomology, Indana Math J 6, (987), pp [4] E G Effros and Z -J Ruan, A new approach to operator spaces, Canad Math Bull 4, (99), pp 29 7 [5] E G Effros and Z -J Ruan, Self-dualty for the Haagerup tensor product and Hlbert space factorzatons, J Funct Anal 00, (99), pp [6] E G Effros and Z -J Ruan, Operator convoluton algebras: An approach to Quantum groups, preprnt [7] T Itoh and M Nagsa, Schur products and module maps on B(H), Publ RIMS Kyoto Unv 6, (2000), pp [8] R R Smth, Completely bounded module maps and the Haagerup tensor product, J Funct Anal 02, (99), pp Department of Mathematcs, Gunma Unversty, Gunma 7-850, Japan E-mal address: toh@edugunma-uacjp Department of Mathematcs and Informatcs, Chba Unversty, Chba , Japan E-mal address: nagsa@mathschba-uacjp 20