Completely bounded isomorphisms of operator algebras

4 ALVARO ARIASProposition 4. C (K) is 2-cb-complemented inC (F 1). Moreover, the orthogonalprojection onto `2(K) is completely bounded from C (F 1) onto C (K).We will present the proof of Proposition 4 after the proof of Theorem 2.Proposition 5. C (K) C (F 1) C (F 1) C (F 1). Moreover, the isomorphismsare completely bounded.S S S 1Proof. Decompose K = K 1 K2 where K 1 =j=0 e 12jF ,andK 2 =j=0 e 2j+1F .It is clear that C (K 1) and C (K 2) are completely isometric to C (K). Moreover,Proposition 4 applied to K 1 and K 2 implies that they are cb-complemented inC (F 1)by the orthogonal projection. Therefore they are complemented in C (K)and we haveC (K) =C (K 1) C (K 2) C (K) C (K):Similarly, decompose K = K 3SK4 , where K 3 = e 1 F and K 4 = S 1j=2 e jF ,and apply the previous argument to conclude that C (K) C (F 1) C (K).Proposition 4 tells us that C (F 1)=C (K) Z for some Z. ThenC (F 1 ) C (K) Z C (K) C (K) Z C (K) C (F 1 ) C (K):Proposition 6. C (F k) C (F k) C (F k), for k =2 3 .Proof. Divide the generators of F 1 into 1 k e 1 e 2 , and denote S by F 1the subgroup generated by the's F is isomorphic to F k . Let K =j=0 e jF .The proof of Proposition 4 works and we get that C (K ) is 2-cb complementedin C (F 1) hence, by Proposition 1, it is 2-cb-complemented in C (F k) also. Itis clear that C (K ) C (K ) C (K ) and that C (K ) C (F k) C (K ).Hence the proof of Proposition 5 applies and we get the result.We will present the proof of Theorem 2 for completeness. This is a standardversion of Pe lczynski's decomposition method.Proof of Theorem 2. Proposition 1 tells that C (F k) C (F 1) Y for someY , and that C (F 1) C (F k) Z, for some Z. Then Propositions 5 and 6 giveC (F k ) C (F 1 ) Y C (F 1 ) C (F 1 ) Y C (F 1 ) C (F k ):On the other handC (F 1 ) C (F k ) Z C (F k ) C (F k ) Z C (F k ) C (F 1 ):The rst step for the proof of Proposition 4 is to understand how tonormtheelements in M n (C P P (K)). The typical element inL K looks like:ik e ip i , wherep i 2L i.e., p i =j a ije xj , for some x j 2 F .Whenwe consider operator spaces,we replace the scalars by matrices, so the canonical element ofM n (C (K)) lookslike(1)Xik(I e i )A i for some A i 2 M n (L )

COMPLETELY BOUNDED ISOMORPHISMS OF OPERATOR ALGEBRAS 7We see that the rst term is the norm of T in M n ( C C (L ) ), and the second oneis the norm of T in M n (R R(F ) ). Here R(F )is`2(F ) with the row operatorspace structure.Using the notation of interpolation theory of operator spaces (see [P]) we concludeProposition 9. C (F 1) c:b: [ C C (F 1)] T [ R R(F 1 )]:Remark. If we are interested only in the Banach space structure, we have thatC (F 1) is isomorphic (but probably not completely isomorphic) to C C (F 1).3. Isomorphisms of non-commutative analytic algebras.In this section we will consider only the words consisting of positive powers ofthe generators of F k , and the identity. We denote this set by P k F k ,andlet`2(P k ) be the Hilbert space with orthonormal basis fe x : x 2 P k g. This Hilbertspace is also denoted by F 2 (H k ), the full Fock spaceonk-generators, see [Po]. LetP (or P k to avoid confusion) be the linear span of the basic elements, and considertwo norms on P: The kk 2 -norm, induced by `2(P k ), and the kk 1 -norm, denedaskpk 1 =supfkpqk 2 : q 2P k kqk 2 1g:Notice that p 2P k induces a left multiplication operator on `2(P k )andkpk 1 equalsthe operator norm of that map.Remark. If p 2P k ,thenthekpk 1 -norm does not coincide with the kpk-norm asan element of C (F k). In fact, if Q is the orthogonal projection onto `2(P k ) thenkpk 1 = kQpQk. We always have that kpk 1 kpk and sometimes the inequality isstrict (see Proposition 17).Let A (k) be the closure of P k in the norm topology of B(`2(P k )), and F 1 (k) theclosure in the strong operator topology. These spaces are studied in [Po], where hecalls them non-commutative analogues of the disk algebra and H 1 . When k =1they coincide with the classical denitions.The main results of this section are.Theorem 10. A (k) is completely isomorphic to A (1) when n =2 3 4 :Theorem 11. F 1 (k) is completely isomorphic to F 1 (1) when n =2 3 4 :As in the previous section we will only present the proof of the rst one, the otherone is essentially the same. The proof of Theorem 10 will follows from Propositions12, 13 and 14.Proposition 12. Let n m, and let :A (n) ! A (m) be the formal identity.Then is a complete isometry. Moreover, the orthogonal projection onto `2(P n ) iscompletely contractive from A (m) onto A (n).Proof. Let E P m be the set of all y 2 P m whose rst letter does not start frome 1 e n . It is easy to see that fP n y : y 2 Eg forms a partition of P m . Then`2(P m )= P 1j=1 `2(P n y j ), where E = fy j : j 2 Ng.

8 ALVARO ARIASLet p 2P n and q 2P m , kqk 2 1. Use the previous partition to decompose q asq = P j r je yj , for some r j 2P n and y j 2 E. Then1X1Xkpqk 2 = 2kpr i k 2 = 2i=0 p r i 2kri=0 i k 2 2kr i k 2 sup2 pi r i 2kr i k 22kpk 2 1 :This tells us that kpkA(n) = kpkA(m). Moreover, if p 2 A (n) A (m), we normitwith elements from P n .This is the fact that we use for the complementation. Given p 2P m , write it asp = p 1 + p 2 ,wherep 1 2P n and p 2 2 (P n ) ? .Thenifq 2P n ,wehave that p 1 q 2P nand p 2 q 2 (P n ) ? . Hence,kp 1 k 1 = supq2P nkqk 2The completely bounded part is very similar.kp 1 qk 2 supq2P nkqk 2kpqk 2 kpk 1 :Proposition 13. There exists a subspace ofA (2) completely isometric to A (1)and completely complemented by the orthogonal projection.Proof. Let a b be the generators of P 2 .Let P be the set of all words generatedby ab a 2 b a 3 b a 4 b . P is clearly isomorphic to P 1 . Let E P 2 be the setof all words in P 2 such that no initial segment belongs to P . Thenitiseasytosee that fP y : y 2 Eg forms a partition of P 2 , and the proof is like thatoftheprevious proposition.Proposition 14. A (1) is completely isomorphic to C A (1), where C is thecolumn Hilbert space.Proof. Divide the generators of P 1 into S 1 2 e 1 e 2 , and let P be the1set of words generated by the's. Let K = e j=1 jP P 1 , and let P(K) bethespan of the basic elements in K. Denote the closure of P(K) inthe`2-norm by`2(K), and in the kk 1 -norm by A (K).P The canonical elementofP(K) looks likeik e ip i , for some k 2 N and p 2P .Given any q 2P the e i p i q's are orthogonal. Then we have, 1 X 1e i p i =iksupkqk 21 X ike i p i q2= supkqk 21s Xikkp i qk 2 2 = XSince A (P ) is isometric to A (1) we conclude that A (K) is isometric to C A (1).Moreover, the elements in A (k) are normed by elements in `2(P ).We willnow see that A (K) is complemented in A (1). Let p 2P 1 and decomposeit as p = p 1 + p 2 , where p 1 2P(K) and p 2 2 (P(K)) ? . If q 2P ,thenp 1 q 2P(K) and p 2 q 2 (P(K)) ? . Hence, kp 1 qk 2 kpqk 2 ,andkp 1 k 1 kpk 1 .As in the proof of Proposition 5 it is clear that A (K) is isomorphic to its square,and then isomorphic to A (1).The completely bounded part follows in the same way after we replace the scalarsby matrices.ikp i p i2:1

COMPLETELY BOUNDED ISOMORPHISMS OF OPERATOR ALGEBRAS 9Proof of Theorem 10. By Proposition 12 and 13 we have thatA (k) =A (1) Yfor some Y . Since A (1) A (1) A (1), we haveA (k) =A (1) Y A (1) A (1) Y A (1) A (k):On the other hand, A (1) =A (k) Z, forsomeZ. If Q : A (1) ! A (k) isthatprojection and I : C ! C is the identity onC, I Q decomposes C A (1) =[C A (k)] [C Z] because Q is completely bounded. Hence,A (1) C A (1)=[C A (k)] [C Z] A (k) [C A (k)] [C Z] A (k) [C A (1)] A (k) A (1):4. Applications to Von Neumann's inequalityFix k for this section and let P k be the positive words generated by e 1 e k .As in the previous section, F 2 (H k )= `2(P k ) is the full Fock space on H k ,a k-dimensional Hilbert space A (k) andF 1 (k) have the same meaning.In [Po] G. Popescu proved that if T 1 T k 2 B(`2) are suchthatk P ik T iT i k1 (i.e., k[T 1 T k ]k1) then any p(e 1 e k ) 2 A (k) satises(4) kp(T 1 T k )kkpkA(k) :When k = 1, this is the classical Von Neumann's inequality.In this section we prove that A (k) andF 1 (k) contain many complementedHilbertian subspaces. Hence we can easily compute kpkA(k) whenever p belongsto one of those subspaces, and use kpkA(k) in Popescu's inequality (4). (See [AP]for more examples and connections with inner functions).We start with the following elementary lemma.Lemma 15. Let p = P i a ie xi q = P j b je yj 2Pbe such that x i y j = x i 0y j 0 ifand only if x i = x i 0 and y j = y j 0, (that is, we cannot have any cancellation), thenkpqk 2 = kpk 2 kqk 2 .P PProof. We have that pq =i j a ib j e xiy j. Since all the x i y j -terms are dierent,the e xiy j's are orthogonal. Hence,kpqk 2 =s XiXjja i b j j 2 =s Xija i j 2 s Xjjb j j 2 = kpk 2 kqk 2 :Remark. The lemma P P extends to the M n -case just as easily. If T =i A i e xi 2 M n (P) and q =j b j e yj 2 `n2(P), P Pthen Tq =i j A ib j e xiy jandqP PkTqk 2 =i j kA ib j k 2.2

10 ALVARO ARIASProposition 16. Let W n P k be the set of all words in P k having n-letters,and let X n = spanfe x : x 2 W n gA (k). Then X n is completely isometric toC n k, the column Hilbert space of the same dimension. Moreover, X n is completelycomplemented inA (k).Proof. Let P p 2 X n and q 2P k , kqk 2 1. Since F P 2 1(H k )=m=0 X m,wewriteq as q =m r m, where r m 2 X m . Notice that pr m 2 X n+m and hence all of themare orthogonal. Moreover, if x 1 x 2 2 X n , y 1 y 2 2 X m and x 1 y 1 = x 2 y 2 ,thenitisnecessary that x 1 = x 2 and y 1 = y 2 . This implies that there is no cancellation inpr m and hence, by the previous lemma, kpr m k 2 = kpk 2 kr m k 2 . Therefore,vu vutX uut 1 1Xkpr m k 2 = 2kpqk 2 =m=0m=0kpk 2 2 kr mk 2 2 = kpk 2kqk 2 and kpk 1 = kpk 2 .The completely P bounded case is very similar. A canonical element ofM n0(X n )looks like T =ik A i e xi , where A i 2 M n0and e xi 2 X n . A canonical elementof `n02 (X P P m) looks like q =j b j e yj , where b j 2 `n02and e yj 2 X m . ThenTq =iPj A ib j e xi e yj 2 `n02 (X n+m) and all of those terms are orthogonal toeach other. Then the proof proceeds as those of section 2. The complementationpart is very easy: If p 2 X n ,thenkpk 1 = kpe 0 k 2 .Multiplication from the left byany one of the e i 's in A (k) orF 1 (k) is an isometry(i.e., kpk 1 = ke i pk 1 ). However, multiplication from the right doesnothave tobelike that.Proposition 17. Let p 2 A (k ; 1) A (k). Then kpe k k 1 = kpe k k 2 = kpk 2 .P P Proof. Let p 2P k;1 , p =i a ie xi where x i 2 P k;1 ,andq 2P k , q =j b je yjwhere y j 2 P k .Thenpe k q = X iXja i b j e xi e k e yj :Since e xi e k e yj = e xi 0 e k e yj 0 i x i = x i 0 and y j = y j 0, then Lemma 15 applies andwe have that kpe k qk 2 = kpk 2 ke k qk 2 = kpk 2 kqk 2 .We conclude with the following two applications of the previous propositions.1. Let p(e 1 e 2 e k ) 2Pbe a non-commutative homogeneous polynomial ofdegree n i.e., p(e 1 P e k )= n p(e 1 e k ), (or p 2 X n with the notation ofProposition 16). If kik T iT i k1, thenkp(T 1 T 2 T k )kkpk 2 :2. Let T 1 T 2 2 B(`2) besuch that k[T 1 T 2 ]k1 (i.e., kT 1 T 1 + T 2T 2k1) andlet p(t) be a polynomial in one variable. The classical Von Neumann's inequalitystates that kp(T 1 )kkpk 1 . Therefore, using the Banach algebra properties ofB(`2) we get thatkp(T 1 )T 2 ksupt2T jp(t)j

COMPLETELY BOUNDED ISOMORPHISMS OF OPERATOR ALGEBRAS 11but if we apply Proposition 17 to Popescu's inequality wegetkp(T 1 )T 2 ks ZT jp(t)j2 dm(t):Acknowledgment. The author thanks Gelu Popescu for useful discussions.References[AP] A. Arias and G. Popescu, Factorization and reexivity on Fock spaces, preprint.[BP] D. Blecher and V. Paulsen, Tensor products of operator spaces, J.Funct. Anal. 99 (1991),262{292.[CS] E. Christensen and A. M. Sinclair, Completely bounded isomorphisms of injective von Neumannalgebras, Proceedings of the Edinburgh Math. Soc. 32 (1989), 317{327.[E]E. Eros, Advances in quantized functional analysis, Proceedings of the International Congressof Math., Berkeley (1986), 906{916.[ER] E. Eros and Z. J. Ruan, A new approach to operator spaces, Canadian Math. Bull. 34(1991), 329{337.[FP] A. Figa-Talamanca and M. Picardello, Harmonic analysis of free groups, lecture notes inpure and applied mathematics, vol. 87, Marcel Dekker, 1983.[HP] U. Haagerup and G. Pisier, Bounded linear operators between C -algebras, Duke Math. 71(3) (1993), 889{925.[PV] M. Pimsner and D. Voiculescu, K-groups of reduced crossed products by free groups, J.Operator Theory 8 (1982), 131{156.[P]G. Pisier, The operator Hilbert space OH, complex interpolation and tensor norms (toappear).[Po] G. Popescu, Von Neumann inequality for (B(H) n ) 1 , Math. Scand. 68 (1991), 292{304..[RW] A. G. Robertson and S. Wassermann, Completely bounded isomorphisms of injective operatorsystems, Bull. London Math. Soc. 21 (1989), 285{290.[S] S. Sakai, C -algebras and W -algebras, Springer-Verlag (1971).Division of Mathematics, Computer Science and Statistics, The University ofTexas at San Antonio, San Antonio, TX 78249, U.S.A.E-mail address: arias@ringer.cs.utsa.edu