Noncommutative hyperbolic geometry on the unit ball of B(H) n

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1 Journal of Functional Analysis 256 (2009) Noncommutative hyperbolic geometry on the unit ball of B(H) n Gelu Popescu Department of Mathematics, University of Texas at San Antonio, One UTSA Circle, San Antonio, TX 78249, USA Received 3 October 2008; accepted 3 February 2009 Available online 20 February 2009 Communicated by N. Kalton Abstract In this paper we introduce a hyperbolic (Poincaré Bergman type) distance δ on the noncommutative open ball [ B(H) n ] := { (X,...,X n ) B(H) n : X X + +X nx n /2 < }, where B(H) is the algebra of all bounded linear operators on a Hilbert space H. It is proved that δ is invariant under the action of the free holomorphic automorphism group of [B(H) n ], i.e., δ ( Ψ(X),Ψ(Y) ) = δ(x,y), X,Y [ B(H) n], for all Ψ Aut([B(H) n ] ). Moreover, we show that the δ-topology and the usual operator norm topology coincide on [B(H) n ]. While the open ball [B(H) n ] is not a complete metric space with respect to the operator norm topology, we prove that [B(H) n ] is a complete metric space with respect to the hyperbolic metric δ. We obtain an explicit formula for δ in terms of the reconstruction operator R X := X R + +X n R n, X:= (X,...,X n ) [ B(H) n], associated with the right creation operators R,...,R n on the full Fock space with n generators. In the particular case when H = C, we show that the hyperbolic distance δ coincides with the Poincaré Bergman Research supported in part by an NSF grant. address: gelu.popescu@utsa.edu /$ see front matter 2009 Elsevier Inc. All rights reserved. doi:0.06/j.jfa

2 G. Popescu / Journal of Functional Analysis 256 (2009) distance on the open unit ball B n := { z = (z,...,z n ) C n : z 2 < }. We obtain a Schwarz Pick lemma for free holomorphic functions on [B(H) n ] with respect to the hyperbolic metric, i.e., if F := (F,...,F m ) is a contractive ( F ) free holomorphic function, then δ ( F(X),F(Y) ) δ(x,y), X,Y [ B(H) n]. As consequences, we show that the Carathéodory and the Kobayashi distances, with respect to δ, coincide with δ on [B(H) n ]. The results of this paper are presented in the more general context of Harnack parts of the closed ball [B(H) n ], which are noncommutative analogues of the Gleason parts of the Gelfand spectrum of a function algebra Elsevier Inc. All rights reserved. Keywords: Noncommutative hyperbolic geometry; Noncommutative function theory; Poincaré Bergman metric; Harnack part; Hyperbolic distance; Free holomorphic function; Free pluriharmonic function; Schwarz Pick lemma 0. Introduction Poincaré s discovery of a conformally invariant metric on the open unit disc D := {z C: z < } of the complex plane was a cornerstone in the development of complex function theory. The hyperbolic (Poincaré) distance is defined on D by δ P (z, w) := tanh z w zw, z,w D. Some of the basic and most important properties of the Poincaré distance are the following: () the Poincaré distance is invariant under the conformal automorphisms of D, i.e., δ P ( ϕ(z),ϕ(w) ) = δp (z, w), z, w D, for all ϕ Aut(D); (2) the δ P -topology induced on the open disc is the usual planar topology; (3) (D,δ P ) is a complete metric space; (4) any analytic function f : D D is distance-decreasing, i.e., satisfies δ P ( f(z),f(w) ) δp (z, w), z, w D. Bergman (see [2]) introduced an analogue of the Poincaré distance for the open unit ball of C n, B n := { z = (z,...,z n ) C n : z 2 < },

3 4032 G. Popescu / Journal of Functional Analysis 256 (2009) which is defined by β n (z, w) = 2 ln + ψ z(w) 2 ψ z (w) 2, z,w B n, where ψ z is the involutive automorphism of B n that interchanges 0 and z. The Poincaré Bergman distance has properties similar to those of δ P (see () (4)). There is a large literature concerning invariant metrics, hyperbolic manifolds, and the geometric viewpoint of complex function theory (see [5,6,45,7] and the references therein). There are several extensions of the Poincaré Bergman distance and related topics to more general domains. We mention the work of Suciu [38,40,39], Foiaş [8], and Andô, Suciu and Timotin [] on Harnack parts of contractions and Harnack type distances between two contractions on Hilbert spaces. Some of their results will be recover (with a different proof) in the present paper, in the particular case when n =. In this paper, we continue our program to develop a noncommutative function theory on the unit ball of B(H) n (see [30,3,33,32,34]). The main goal is to introduce a hyperbolic metric δ on the noncommutative ball [ B(H) n ] := { (X,...,X n ) B(H) n : X X + +X nx n /2 < }, where B(H) denotes the algebra of all bounded linear operators on a Hilbert space H, which satisfy properties similar to those of the Poincaré metric δ P (see () (3)), and which is a noncommutative extension of the Poincaré Bergman metric β n on the open unit ball of C n.the secondary goal is to obtain a Schwarz Pick lemma for free holomorphic functions on [B(H) n ] with respect to the hyperbolic metric. We should mention that the noncommutative ball [B(H) n ] can be identified with the open unit ball of B(H n, H), which is one of the infinite-dimensional Cartan domains studied by L.A. Harris [ 3]. He has obtained several results, related to our topic, in the more general setting of JB -algebras (see also the book by H. Upmeier [42]). We also remark that the group of all free holomorphic automorphisms of [B(H) n ] [34], can be identified with a subgroup of the group of automorphisms of [B(H n, H)] considered by R.S. Phillips [9] (see also [44]). However, the hyperbolic distance δ that we introduce in this paper is different from the Kobayashi distance on [B(H) n ], with respect to the Poincaré distance on D, and also different from the one considered, for example, in []. In [30,3,33,32,34], we obtained several results concerning the theory of free holomorphic (resp. pluriharmonic) functions on [B(H) n ] and provided a framework for the study of arbitrary n-tuples of operators on a Hilbert space H. Several classical results from complex analysis [3, 9,4,37] have free analogues in the noncommutative multivariable setting. To put our work in perspective, we need to set up some notation and recall some definitions. Let F + n be the unital free semigroup on n generators g,...,g n and the identity g 0. The length of α F + n is defined by α :=0ifα = g 0 and α :=k if α = g i g ik, where i,...,i k {,...,n}. If(X,...,X n ) B(H) n,wesetx α := X i X ik and X g0 := I H. Throughout this paper, we assume that E is a separable Hilbert space. A map F :[B(H) n ] B(H) min B(E) is called free holomorphic function on [B(H) n ] with coefficients in B(E) if there exist A (α) B(E), α F + n, such that lim sup k α =k A (α) A (α) /2k and F(X,...,X n ) = X α A (α), k=0 α =k

4 G. Popescu / Journal of Functional Analysis 256 (2009) where the series converges in the operator norm topology for any (X,...,X n ) [B(H) n ]. The set of all free holomorphic functions on [B(H) n ] with coefficients in B(E) is denoted by Hol(B(H) n ).LetH (B(H) n ) denote the set of all elements F in Hol(B(H)n ) such that F := sup F(X,...,X n ) <, where the supremum is taken over all n-tuples of operators (X,...,X n ) [B(H) n ] and any Hilbert space H. According to [30] and [33], H (B(H) n ) can be identified to the operator algebra Fn B(E) (the weakly closed algebra generated by the spatial tensor product), where Fn is the noncommutative analytic Toeplitz algebra (see [23,2,25]). We say that a map u :[B(H) n ] B(H) min B(E) is a self-adjoint free pluriharmonic function on [B(K) n ] if u = Re f := 2 (f +f)for some free holomorphic function f. We also recall that u is called positive if u(x,...,x n ) 0 for any (X,...,X n ) [B(K) n ], where K is an infinite-dimensional Hilbert space. In Section, we introduce an equivalence relation H on the closed ball [B(H) n ], and study the equivalence classes (called Harnack parts) with respect to.twon-tuples H of operators A := (A,...,A n ) and B := (B,...,B n ) in [B(H) n ] are called Harnack equivalent (and denote A H B) if and only if there exists a constant c such that c 2 Re p(b,...,b n ) Re p(a,...,a n ) c 2 Re p(b,...,b n ) for any noncommutative polynomial p C[X,...,X n ] M m, m N, with matrix-valued coefficients such that Re p 0. Here M m denotes the algebra of all m m matrices with entries in C. We also use the notation A H B to emphasize the constant c in the inequalities above. The c Harnack parts of [B(H) n ] are noncommutative analogues of the Gleason parts of the Gelfand spectrum of a function algebra (see [0]). In Section, we use several results (see [27 29,36]) concerning the theory of noncommutative Poisson transforms on Cuntz Toeplitz C -algebras (see [4]) and free pluriharmonic functions (see [33,32]) to obtain useful characterizations for the Harnack equivalence on the closed ball [B(H) n ]. On the other hand, a characterization of positive free pluriharmonic functions (see [33]) and dilation theory (see [4]) are used to obtain a Harnack type inequality (see [3]) for positive free pluriharmonic function on [B(H) n ]. More precisely, we show that if u is a positive free pluriharmonic function on [B(H) n ] with operator-valued coefficients in B(E) and 0 <r<, then u(0) r + r u(x,...,x n ) u(0) + r r for any (X,...,X n ) [B(H) n ] r. This result is crucial in order to prove that the open unit ball [B(H) n ] is a distinguished Harnack part of [B(H) n ], namely, the Harnack part of 0. In Section 2, we introduce a hyperbolic (Poincaré Bergman type) metric on the Harnack parts of [B(H) n ]. More precisely, given a Harnack part of [B(H)n ] we define δ : R+ by setting δ(a,b) := ln ω(a,b), A,B,

5 4034 G. Popescu / Journal of Functional Analysis 256 (2009) where { ω(a,b) := inf c>: A H } B. c We prove that δ is a metric on. Consider the particular case when =[B(H) n ] and let δ :[B(H) n ] [B(H) n ] [0, ) be the hyperbolic metric defined above. We prove, in Section 2, that δ is invariant under the action of the group Aut([B(H) n ] ) of all the free holomorphic automorphisms of the noncommutative ball [B(H) n ], i.e., δ ( Ψ(A),Ψ(B) ) = δ(a,b), A,B [ B(H) n], for all Ψ Aut([B(H) n ] ). We mention that the group Aut([B(H) n ] ) was determined in [34]. Using a characterization of the Harnack equivalence on [B(H) n ] in terms of free pluriharmonic kernels, we obtain an explicit formula for the hyperbolic distance in terms of the reconstruction operator. More precisely, we show that δ(a,b) = ln max { C A CB, C B CA }, A,B [ B(H) n], where C X := ( X I)(I R X ) and R X := X R + +X n R n is the reconstruction operator associated with the n-tuple X := (X,...,X n ) [B(H) n ] and with the right creation operators R,...,R n on the full Fock space with n generators. In particular, we show that δ Bn B n coincides with the Poincaré Bergman distance on B n, i.e., δ(z,w) = 2 ln + ψ z(w) 2 ψ z (w) 2, z,w B n, where ψ z is the involutive automorphism of B n that interchanges 0 and z. We mention that similar results concerning the invariance under the automorphism group Aut([B(H) n ] ) as well as an explicit formula for the hyperbolic metric hold on any Harnack part of [B(H) n ]. In Section 3, we study the relations between the δ-topology, the d H -topology (which will be introduced), and the operator norm topology on Harnack parts of [B(H) n ]. We prove that the hyperbolic metric δ is a complete metric on any Harnack part of [ B0 (H) n] := { (X,...,X n ) [ B(H) n] : r(x,...,x n )< }, and that all the topologies above coincide on the open ball [B(H) n ]. In particular, we deduce that [B(H) n ] is a complete metric space with respect to the hyperbolic metric δ and that the δ-topology and the usual operator norm topology coincide on [B(H) n ]. A very important property of the Poincaré Bergman distance β m : B m B m R + is that any holomorphic function f : B n B m is distance-decreasing, i.e., β m ( f(z),f(w) ) βn (z, w), z, w B n. In Section 4, we extend this result and prove a Schwarz Pick lemma for free holomorphic functions on [B(H) n ] with operator-valued coefficients, with respect to the hyperbolic metric on the noncommutative ball [B(H) n ].

6 G. Popescu / Journal of Functional Analysis 256 (2009) More precisely, let F j :[B(H) n ] B(H) min B(E), j =,...,m, be free holomorphic functions with coefficients in B(E), and assume that F := (F,...,F m ) is a contractive ( F ) free holomorphic function. If X, Y [B(H) n ], then we prove that F(X) H F(Y) and δ ( F(X),F(Y) ) δ(x,y), where δ is the hyperbolic metric defined on the Harnack parts of the noncommutative ball [B(H) n ]. This result is used to show that the Carathéodory and the Kobayashi distances, with respect to δ, coincide with δ on [B(H) n ]. The present paper makes connections between noncommutative function theory (see [27,30, 33,34]) and classical results in hyperbolic complex analysis and geometry (see [5 7,3,9,4, 37]). In particular, we obtain a new formula of the Poincaré Bergman metric on B n using Harnack inequalities for positive free pluriharmonic functions on [B(H) n ],aswellasaformulain terms of the left creation operators on the full Fock space with n generators. It would be interesting to see if the results of this paper can be extended to more general infinite-dimensional bounded domains such as the JB -algebras of Harris [], the domains considered by Phillips [9], or the noncommutative domains from [35]. Since our results are based on noncommutative function theory, dilation and model theory for row contractions, we are inclined to believe in a positive answer at least for the latter domains.. Harnack equivalence on the closed unit ball [B(H) n ] In this section, we introduce a preorder relation H on the closed ball [B(H) n ] and provide several characterizations. This preorder induces an equivalence relation H on [B(H) n ], whose equivalence classes are called Harnack parts. Several characterizations for the Harnack parts are provided. We obtain a Harnack type inequality for positive free pluriharmonic functions and use it to prove that the open unit ball [B(H) n ] is a distinguished Harnack part of [B(H) n ], namely, the Harnack part of 0. Let H n be an n-dimensional complex Hilbert space with orthonormal basis e,e 2,...,e n, where n =, 2,...,orn =. We consider the full Fock space of H n defined by F 2 (H n ) := C k H k n, where Hn k is the (Hilbert) tensor product of k copies of H n. Define the left (resp. right) creation operators S i (resp. R i ), i =,...,n, acting on F 2 (H n ) by setting S i ϕ := e i ϕ, ϕ F 2 (H n ), (resp. R i ϕ := ϕ e i, ϕ F 2 (H n )). The noncommutative disc algebra A n (resp. R n )isthe norm closed algebra generated by the left (resp. right) creation operators and the identity. The noncommutative analytic Toeplitz algebra Fn (resp. R n ) is the weakly closed version of A n (resp. R n ). These algebras were introduced in [23] in connection with a noncommutative von Neumann type inequality [43], and have been studied in several papers (see [2,24 26,5,6], and the references therein).

7 4036 G. Popescu / Journal of Functional Analysis 256 (2009) We need to recall from [27] a few facts about noncommutative Poisson transforms associated with row contractions T := (T,...,T n ) [B(H) n ]. Let F + n be the unital free semigroup on n generators g,...,g n, and the identity g 0. We denote e α := e i e ik and e g0 :=. Note that {e α } α F + n is an orthonormal basis for F 2 (H n ). For each 0 <r, define the defect operator T,r := (I H r 2 T T r2 T n T n )/2.The noncommutative Poisson kernel associated with T is the family of operators defined by K T,r : H T,r H F 2 (H n ), 0 <r, K T,r h := r α T,r Tα h e α, h H. k=0 α =k When r =, we denote T := T, and K T := K T,. The operators K T,r are isometries if 0 <r<, and K T K T = I H SOT- lim k α =k T α Tα. Thus K T is an isometry if and only if T is a pure row contraction, i.e., SOT- lim k α =k T α Tα = 0. We denote by C (S,...,S n ) the Cuntz Toeplitz C -algebra generated by the left creation operators. The noncommutative Poisson transform at T := (T,...,T n ) [B(H) n ] is the unital completely contractive linear map P T : C (S,...,S n ) B(H) defined by P T [f ]:=lim r K T,r (I H f)k T,r, f C (S,...,S n ), where the limit exists in the norm topology of B(H). Moreover, we have P T [ Sα S β] = Tα T β, α,β F+ n. When T := (T,...,T n ) is a pure row contraction, we have P T [f ]=K T (I D T f)k T, where D T = T H. We refer to [27,28,36] for more on noncommutative Poisson transforms on C -algebras generated by isometries. For basic results concerning completely bounded maps and operator spaces we refer to [8,20,7]. When T [B(H) n ] is a completely noncoisometric (c.n.c.) row contraction, i.e., there is no h H, h 0, such that

8 G. Popescu / Journal of Functional Analysis 256 (2009) Tα h 2 = h 2 for any k =, 2,..., α =k an F n -functional calculus was developed in [24]. We showed that if f = α F + n a αs α is in F n, then Γ T (f ) = f(t,...,t n ) := SOT- lim r α a α T α r k=0 α =k exists and Γ T : Fn B(H) is a completely contractive homomorphism and WOT-continuous (resp. SOT-continuous) on bounded sets. Moreover, we showed (see [36]) that Γ T (f ) = P T [f ], f Fn, where P T [f ]:=SOT- lim r K T,r (I H f)k T,r, f F n, (.) is the extension of the noncommutative Poisson transform to the noncommutative analytic Toeplitz algebra F n. We introduced in [33] the noncommutative Poisson transform Pμ of a completely bounded linear map μ : R n + R n B(E) by setting (Pμ)(X,...,X n ) := (id μ) [ P(X,R) ], X:= (X,...,X n ) [ B(H) n], where the free pluriharmonic Poisson kernel P(X,R)is given by P(X,R):= Xα R α + I + X α R α, X [ B(H) n], k= α =k k= α =k and the series are convergent in the operator norm topology. We recall that the joint spectral radius associated with an n-tuple of operators (X,...,X n ) B(H) n is given by r(x,...,x n ) := lim X α Xα k We remark that the free pluriharmonic Poisson kernel P(X,R) makes sense for any n-tuple of operators X := (X,...,X n ) B(H) n with r(x,...,x n )<. According to [36], X [B(H) n ] if and only if α =k /2k r(x,...,x n ) and P(rX,R) 0, r [0, ). We say that a free pluriharmonic function u is positive if u(x,...,x n ) 0 for any (X,...,X n ) [B(K) n ] γ and any Hilbert space K. We recall [33] that u 0 if and only if u(rs,...,rs n ) 0 for any r [0, ). In particular, if p is a noncommutative polynomial with operator-valued coefficients, then Re p 0 if and only if Re p(s,...,s n ) 0..

9 4038 G. Popescu / Journal of Functional Analysis 256 (2009) Now, we introduce a preorder relation H on the closed ball [B(H) n ] and provide several characterizations. Let A := (A,...,A n ) and B := (B,...,B n ) be in [B(H) n ]. We say that A is Harnack dominated by B, and denote A H B, if there exists c>0 such that Re p(a,...,a n ) c 2 Re p(b,...,b n ) for any noncommutative polynomial with matrix-valued coefficients p C[X,...,X n ] M m, m N, such that Re p 0. When we want to emphasize the constant c, we write A H c B. Theorem.. Let A := (A,...,A n ) and B := (B,...,B n ) be in [B(H) n ] the following statements are equivalent: and let c>0. Then (i) A H B; c (ii) (P A min id)[q q] c 2 (P B min id)[q q] for any polynomial q = α k S α C (α) with matrix-valued coefficients C (α) M m, and k,m N, where P X is the noncommutative Poisson transform at X [B(H) n ] ; (iii) P(rA,R) c 2 P(rB,R) for any r [0, ), where P(X,R) is the free pluriharmonic Poisson kernel associated with X [B(H) n ] ; (iv) u(ra,...,ra n ) c 2 u(rb,...,rb n ) for any positive free pluriharmonic function u with operator-valued coefficients and any r [0, ); (v) c 2 P B P A is a completely positive linear map on the operator space A n + A n. Proof. First we prove the equivalence (i) (ii). Assume that (i) holds, and let q = α k S α C (α) be an arbitrary polynomial in A n M m. Since the left creation operators are isometries with orthogonal ranges, q q has form Re p(s,...,s n ) for some noncommutative polynomial p C[X,...,X n ] M m, m N, such that Re p 0. Using the properties of the noncommutative Poisson transform, one can see that (i) (ii). Conversely, assume that (ii) holds. Let p C[X,...,X n ] M m, m N, be a polynomial of degree k such that Re p 0. Then Re p(s,...,s n ) is a positive multi-toeplitz operator with respect to R,...,R n acting on the Hilbert space F 2 (H n ) C m, i.e., ( )[ R i I C m Re p(s,...,s n ) ] (R j I C m) = δ ij G, i, j =,...,n. According to the Fejér type factorization theorem of [25], there exists a multi-analytic operator q in B(F 2 (H n ) C m ) such that Re p(s,...,s n ) = q q and ( ) S α I C m q( h) = 0 for any h C m and α >k. Therefore, q is a polynomial, i.e., q = α k S α C (α) for some operators C (α) B(C m ). Note that Re p(a,...,a n ) = (P A min id) [ q q ] c 2 (P B min id) [ q q ] = c 2 Re p(b,...,b n ), which proves (i).

10 G. Popescu / Journal of Functional Analysis 256 (2009) Let us prove that (i) (iii). For each m N, consider R (m) := (R (m),...,r n (m) ), where R (m) i, i =,...,n, is the compression of the right creation operator R i to P m := span{e α : α F + n, α m}. Note that R α (m) = 0 for any α F + n with α m + and, consequently, we have P ( rx,r (m)) = α m r α X α R(m) α + I + α m r α X α R (m) α. Note that R (m) is a pure row contraction and the noncommutative Poisson transform id P R (m) is a completely positive map. We recall that X P(X,R)is a positive free pluriharmonic function with coefficients in B(F 2 (H n )). Hence P(rX,R) 0 for any X [B(H) n ] and r [0, ). Applying id P R (m), we obtain Now, applying (i), we obtain P ( rx,r (m)) = (id P R (m)) [ P(rX,S) ] 0. P ( ra,r (m)) c 2 P ( rb,r (m)) for any m N and r [0, ). Using Lemma 8. from [33], we deduce that P(rA,R) c 2 P(rB,R)for any r [0, ). Therefore (iii) holds. To prove the implication (iii) (iv), assume that condition (iii) holds and let u be a positive free pluriharmonic function with coefficients in B(E). According to Corollary 5.5 from [33], there exists a completely positive linear map μ : R n + R n B(E) such that u(y ) = (Pμ)(Y ) := (id μ) ( P(Y,R) ) for any Y [B(H) n ]. Hence and using the fact that c 2 P(rB,R) P(rA,R) 0, we deduce that c 2 u(rb,...,rb n ) u(ra,...,ra n ) = (id μ) [ c 2 P(rB,R) P(rA,R) ] 0, which proves (iv). Now, we prove the implication (iv) (v). Letg A n + A n M m be positive. Then, according to Theorem 4. from [33], the map defined by u(x) := (P X id)[g], X [ B(H) n], (.2) is a positive free pluriharmonic function. Condition (iv) implies u(ra,...,ra n ) c 2 u(rb,...,rb n ) for any r [0, ). On the other hand, by relation (.2), we have c 2 (P rb id)[g] (P ra id)[g]=c 2 u(rb,...,rb n ) u(ra,...,ra n ) 0 (.3) for any r [0, ). Since g A n + A n M m,wehave (P A id)[g]=lim r (P ra id)[g] and (P B id)[g]=lim r (P rb id)[g],

11 4040 G. Popescu / Journal of Functional Analysis 256 (2009) where the convergence is in the operator norm topology. Taking r in (.3), we deduce item (v). To prove the implication (v) (i),letp C[X,...,X n ] M m, m N, be a noncommutative polynomial with matrix coefficients such that Re p 0. Due to the proprieties of the noncommutative Poisson transform, we have c 2 Re p(b,...,b n ) Re p(a,...,a n ) = c 2 (P B id) [ Re p(s,...,s n ) ] (P A id) [ Re p(s,...,s n ) ]. Since Re p(s,...,s n ) 0 and c 2 P B P A is a completely positive linear map on the operator space A n + A n, we deduce item (i). This completes the proof. We remark that each item in Theorem. is equivalent to the following: P ( ra,r (m)) c 2 P ( rb,r (m)) for any m N, r [0, ), where R (m) is defined in the proof of Theorem.. In what follows, we characterize the elements of the closed ball [B(H) n ] dominated by 0. which are Harnack Theorem.2. Let A := (A,...,A n ) be in [B(H) n ]. Then A H 0 if and only if the joint spectral radius r(a,...,a n )<. Proof. Note that the map X P(X,R) is a positive free pluriharmonic function on [B(H) n ] with coefficients in B(F 2 (H n )) and has the factorization P(X,R)= (I R X ) I + ( I RX ) = ( I RX) [ I RX ( I RX) (I RX ) + I RX ] (I RX ) = ( I R X) [( I X X X nx n) I ] (I RX ), where R X := X R + +Xn R n is the reconstruction operator associated with the n- tuple X := (X,...,X n ) [B(H) n ]. We remark that, due to the fact that the spectral radius of R X is equal to r(x,...,x n ), the factorization above holds for any X [B(H) n ] with r(x,...,x n )<. Now, using Theorem., part (iii) and the factorization obtained above, we deduce that A H 0 if and only if there exists c such that ( ) I rr [( A I r 2 A A r2 A n A ) ] n I (I rra ) c 2 I for any r [0, ). Similar inequality holds if we replace the right creation operators by the left creation operators. Then, applying the noncommutative Poisson transform id P e iθ R we obtain ( I r 2 A A r2 A n A ) n I c 2 ( I re iθ RA)( I re iθ ) R A (.4) for any r [0, ) and θ R.

12 G. Popescu / Journal of Functional Analysis 256 (2009) Assume now that A := (A,...,A n ) [B(H) n ] is such that A H 0. Then r(a,...,a n ). Suppose that r(a,...,a n ) =. Since r(r A ) = r(a,...,a n ), there exists λ 0 T in the approximative spectrum of R A. Consequently, there is a sequence {h m } in H F 2 (H n ) such that h m = and λ 0 h m R A h m m for m =, 2,... Hence and taking r = m in relation (.4), we deduce that ( ( I ) 2 ( RA m R A )h m,h m c 2 h m 2 ) λ 0 R A h m m c ( λ 2 0 h m R A h m + ) 2 m R Ah m 4c2 m 2. (.5) Combining this result with the fact that R A h m, we deduce that ( m) 2 ( ) 2 R A h m 2 4c2 m m 2. Hence, we obtain 2m 4c 2 + for any m =, 2,..., which is a contradiction. Therefore, we have r(a,...,a n )<. Conversely, assume that A := (A,...,A n ) [B(H) n ] has the joint spectral radius r(a,...,a n )<. Note that M := sup r (0,) (I rr A ) exists and, therefore, ( ) I rr [( A I r 2 A A r2 A n A ) ] n I (I rra ) M 2 I for any r (0, ), which, due to Theorem., shows that A H 0. The proof is complete. We mention that in the particular case when n = we can recover a result obtained in []. Since H is a preorder relation on [B(H) n ], it induces an equivalent relation H on [B(H) n ], which we call Harnack equivalence. The equivalence classes with respect to H are called Harnack parts of [B(H) n ].LetA:= (A,...,A n ) and B := (B,...,B n ) be in [B(H) n ]. It is easy to see that A and B are Harnack equivalent (we denote A H B) if and only if there exists c such that c 2 Re p(b,...,b n ) Re p(a,...,a n ) c 2 Re p(b,...,b n ) (.6) for any noncommutative polynomial with matrix-valued coefficients p C[X,...,X n ] M m, m N, such that Re p 0. We also use the notation A H B if A H B and B H A. c c c A completely positive (c.p.) linear map μ X : C (S,...,S n ) B(H) is called representing c.p. map for the point X := (X,...,X n ) [B(H) n ] if μ(s α ) = X α for any α F + n. Next, we obtain several characterizations for the Harnack equivalence on the closed ball [B(H) n ]. The result will play a crucial role in this paper.

13 4042 G. Popescu / Journal of Functional Analysis 256 (2009) Theorem.3. Let A := (A,...,A n ) and B := (B,...,B n ) be in [B(H) n ] the following statements are equivalent: and let c>. Then (i) A H B; c (ii) the noncommutative Poisson transform satisfies the inequalities c 2 (P B min id) [ q q ] (P A min id) [ q q ] c 2 (P B min id) [ q q ] for any polynomial q = α k S α A (α) with matrix-valued coefficients A (α) M m, and k,m N; (iii) the free pluriharmonic kernel satisfies the inequalities c 2 P(rB,R) P(rA,R) c2 P(rB,R) for any r [0, ); (iv) for any positive free pluriharmonic function u with operator-valued coefficients and any r [0, ), c 2 u(rb,...,rb n ) u(ra,...,ra n ) c 2 u(rb,...,rb n ); (v) c 2 P B P A and c 2 P A P B is a completely positive linear map on the operator space A n + A n, where P X is the noncommutative Poisson transform at X [B(H) n ] ; (vi) there are representing c.p. maps μ A and μ B for A and B, respectively, such that c 2 μ B μ A c 2 μ B. Proof. The first five equivalences follow using Theorem.. It remains to show that (i) (vi). Assume that A H B. Then according to item (v), P A P c c 2 B and P B P c 2 A are completely positive linear maps on the operator space A n + A n. Using Arveson s extension theorem, we find some completely positive linear maps ϕ and ψ on the Cuntz Toeplitz algebra C (S,...,S n ) which are extensions of P A P c 2 B and P B P c 2 A, respectively. Define the c.p. maps μ A,μ B : C (S,...,S n ) B(H) by setting μ A := c2 ( c 2 c 4 ϕ + ψ ) and μ B := c2 ( c 2 c 4 ψ + ϕ ). (.7) Note that for any f A n + A n,wehave P A [f ]=ϕ(f) + c 2 P B[f ]=ϕ(f) + c 2 [ψ(f)+ c 2 P A ]. Solving for P A [f ], we obtain P A [f ]=μ A (f ). Similarly, we obtain P B [f ]=μ B (f ). Since P A [S α ]=A α and P B [S α ]=B α for any α F + n, it is clear that μ A, μ B are representing c.p.

14 G. Popescu / Journal of Functional Analysis 256 (2009) maps for A := (A,...,A n ) and B := (B,...,B n ), respectively. Now, since c> and using relation (.7), it is a routine to show that c 2 μ B μ A c 2 μ B. (.8) Conversely, assume that (vi) holds for some c>, and let p(s,...,s n ) := α q S α M (α) be a polynomial such that Re p 0. Since μ A, μ B are representing c.p. maps for A := (A,...,A n ) and B := (B,...,B n ), respectively, relation (.8) implies c 2 Re p(b,...,b n ) Re p(a,...,a n ) c 2 Re p(b,...,b n ). This shows that A H c B and completes the proof. We remark that the first five equivalences in Theorem.3 remain true even when c. The next result is a Harnack type inequality for positive free pluriharmonic functions on [B(H) n ]. Theorem.4. If u is a positive free pluriharmonic function on [B(H) n ] with operator-valued coefficients in B(E) and 0 r<, then for any (X,...,X n ) [B(H) n ] r. u(0) r + r u(x,...,x n ) u(0) + r r Proof. Notice that the free pluriharmonic Poisson kernel satisfies the relation P(rX,R)= r k RX k + I + r k( RX) k, 0 r<, k= for any X := (X,...,X n ) [B(H) n ], where R X := X R + +Xn R n is the reconstruction operator. Since R,...,R n are isometries with orthogonal ranges, we have RX R X = ni= X i Xi I and therefore R X = n i= X i Xi /2 <. Let U be the minimal unitary dilation of R X on a Hilbert space M H F 2 (H n ), in the spirit of Sz.-Nagy Foiaş. Then we have RX k = P H F 2 (H n ) Uk H F 2 (H n ) for any k =, 2,... Since R X is a strict contraction, U is a bilateral shift which can be identified with M e iθ I G for some Hilbert space G, where M e iθ is the multiplication operator by e iθ on L 2 (T). Consequently, there is a unitary operator W : L 2 (T) G M such that where k= P(rX,R)= P H F 2 (H n )[ W ( M(r) IG ) W ] H F 2 (H n ), (.9) M(r) := r k M k e iθ + I + r k M k e iθ. k= k=

15 4044 G. Popescu / Journal of Functional Analysis 256 (2009) For each f L 2 (T),wehave Since k= M(r)f, f = f 2 2 k= r k e iθk = f 2 r 2 2 2r cos θ + r 2. r + r r 2 2r cos θ + r 2 + r r, we have r +r +r I M(r) r I. Hence and due to (.9), we deduce that r + r I P(rX,R) + r r I (.0) for any X [B(H) n ] and 0 r<. Since Y P(Y,R) is a pluriharmonic function on [B(H) n ], hence continuous, we deduce that (.0) holds for any X [B(H) n ]. On the other hand, since u is a positive pluriharmonic function on [B(H) n ], one can use [33] (see Corollary 5.5) to find a completely positive linear map μ : R n + R n B(E) such that u is the noncommutative Poisson transform of μ, i.e., u(y,...,y n ) = (id μ) [ P(Y,R) ], Y := (Y,...,Y n ) [ B(H) n]. Hence, u(rx,...,rx n ) = (id μ)[p(rx,r)] for any X := (X,...,X n ) [B(H) n ].Now, using inequalities (.0) and the fact that μ is a completely positive linear map, we obtain u(0) r + r u(rx,...,rx n ) u(0) + r r, X [ B(H) n]. This completes the proof. We recall that if f A n M m and (A,...,A n ) [B(H) n ], then, due to the noncommutative von Neumann inequality [23] (see also [24]), it makes sense to define f(a,...,a n ) B(H) min M m. In this case we have f(a,...,a n ) f. Our next task is to determine the Harnack part of 0. First, we need the following technical result. Lemma.5. Let A := (A,...,A n ) and B := (B,...,B n ) be in [B(H) n ] such that A H B, and let {f k } k= be a sequence of elements in A n M m, m N, such that f k for any k N. Then lim fk (A,...,A n ) = if and only if lim fk (B,...,B n ) =. k k Proof. Assume that lim k f k (A,...,A n ) =. Then there is a sequence of vectors {h k } k= H Cm such that

16 G. Popescu / Journal of Functional Analysis 256 (2009) h k = and lim f k (A,...,A n )h k =. k Let {α k } k= D be such that α k ask. According to Theorem.5 from [3], the inverse noncommutative Cayley transform Γ of α k f r is in A n M m and Re Γ [α k f k ] 0. Therefore, g k := Γ [α k f k ]:=(I α k f k ) (I + α k f k ) A n M m (.) and Re g k 0 for all k N. Since A H B, Theorem. implies the existence of a constant c such that c 2 Re g k(a,...,a n ) Re g k (B,...,B n ) c 2 Re g k (A,...,A n ). (.2) For each k N, we define the vectors y k := [ I α k f k (A,...,A n ) ] h k and x k := [ I α k f k (B,...,B n ) ] y k. (.3) Note that due to relations (.) and (.3), we have [ gk (B,...,B n )y k,y k = I + αk f k (B,...,B n ) ] x k, [ I α k f k (B,...,B n ) ] x k Consequently, we deduce that Similarly, we obtain = x k 2 α k 2 f k (B,...,B n )x k 2 + i Im αk f k (B,...,B n )x k,x k. Re g k (B,...,B n )y k,y k = xk 2 α k 2 f k (B,...,B n )x k 2. Re g k (A,...,A n )y k,y k = hk 2 α k 2 fk (A,...,A n )h k 2. Hence and using the second inequality in (.2), we deduce that x k 2 α k 2 f k (B,...,B n )x k 2 c 2 ( h k 2 α k 2 f k (A,...,A n )h k 2). Consequently, since h k =, lim k f k (A,...,A n )h k = and α k ask,we deduce that x k 2 α k 2 fk (B,...,B n )x k 2 0, as k. (.4) Now, suppose that lim k f k (B,...,B n ). Due to the noncommutative von Neumann inequality we have f k (B,...,B n ) f k. Passing to a subsequence, we can assume that there is t (0, ) such that f k (B,...,B n ) t< for all k N. Due to relation (.4) and the fact that 0 ( t 2) x k 2 x k 2 α k 2 f k (B,...,B n )x k 2,

17 4046 G. Popescu / Journal of Functional Analysis 256 (2009) we deduce that x k 0ask. Using relation (.3), we obtain y k =[I α k f k (B,..., B n )] x k. Consequently, and using again relation (.4), we have [ I α k f k (A,...,A n ) ] h k = yk 0, as k, (.5) for any sequence {α k } k= D with the property that α k ask.let{β k } k= D be another sequence with the same property and such that α k + β k 0ask. Then, due to (.5), we have lim 2hk (α k + β k )f k (A,...,A n )h k = 0. k Taking into account that f k (A,...,A n )h k for all k N, we deduce that h k 0 as k, which contradicts the fact that h k = for all k N. Therefore, we must have lim k f k (B,...,B n ) =. The converse follows in a similar manner if one uses the first inequality in (.2). The proof is complete. Now, we have all the ingredients to determine the Harnack part of 0 and obtain a characterization in terms of the free pluriharmonic Poisson kernel. Theorem.6. Let A := (A,...,A n ) be in [B(H) n ]. Then the following statements are equivalent: (i) A H 0; (ii) A [B(H) n ] ; (iii) r(a,...,a n )< and P(A,R) ai for some constant a>0, where P(A,R) is the free pluriharmonic Poisson kernel at A. Proof. First, we prove the implication (i) (ii). Assume that A H 0 and A =. For each k N define f k := S S n A n M n. Then f k (A,...,A n ) = A =. Applying Lemma.5, we deduce that 0 = f k (0), as k, which is a contradiction. Therefore A <. Now, we prove that (ii) (i). LetA := (A,...,A n ) be in [B(H) n ] and let r := A <. According to Theorem.4, we have u(0) r + r u(x,...,x n ) u(0) + r r for any positive free pluriharmonic function u on [B(H) n ] with operator-valued coefficients in B(E). Applying now Theorem.3, we deduce that A H 0.

18 G. Popescu / Journal of Functional Analysis 256 (2009) To prove that (ii) (iii), leta [B(H) n ]. Note that r(a,...,a n ) A <, which implies the operators I R A and I A A A na n 0 are invertible. Hence, and using the fact that P(A,R)= ( I R A) [( I A A A na n) I ] (I RA ), one can easily deduce that there exists a>0 such that P(A,R) ai. Therefore (iii) holds. It remains to show that (iii) (i). Assume that (iii) holds. Due to Theorem.2, we have A H 0 and P(X,S)= Xα S α + I + X α S α, X [ B(H) n], k= α =k k= α =k where the series are convergent in the operator norm topology. On the other hand, since P(A,S) ai, applying the noncommutative Poisson transform id P rr, we obtain P(rA,R)= (id P rr ) [ P(A,S) ] ai = ap(0,r) for any r [0, ). Due to Theorem., equivalence (i) (iii), wehave0 H A. Therefore, item (i) holds. The proof is complete. We remark that, when n =, we recover a result obtained by Foiaş[8]. 2. Hyperbolic metric on the Harnack parts of the closed ball [B(H) n ] In this section we introduce a hyperbolic (Poincaré Bergman type) metric δ on the Harnack parts of [B(H) n ], and prove that it is invariant under the action of the group Aut([B(H)n ] ) of all the free holomorphic automorphisms of the noncommutative ball [B(H) n ]. We obtain an explicit formula for the hyperbolic distance in terms of the reconstruction operator and show that δ Bn B n coincides with the Poincaré Bergman distance on B n, the open unit ball of C n. Given A,B [B(H) n ] in the same Harnack part, i.e., A H B, we introduce { ω(a,b) := inf c>: A H } B. (2.) c Lemma 2.. Let be a Harnack part of [B(H) n ] properties hold: and let A,B,C. Then the following (i) ω(a,b) ; (ii) ω(a,b) = if and only if A = B; (iii) ω(a,b) = ω(b,a); (iv) ω(a,c) ω(a,b)ω(b,c).

19 4048 G. Popescu / Journal of Functional Analysis 256 (2009) Proof. The items (i) and (iii) are obvious due to relations (.6) and (2.). If ω(a,b) =, then A H B and, due to Theorem., part (iii), we have P(rA,R)= P(rB,R) for any r [0, ). Applying this equality to vectors of the form h, h H, we obtain r α A α h e α = r α Bα h e α. α α Hence, we deduce that A i = B i for any i =,...,n. Therefore, (ii) holds. Note that, due to Theorem.3, we have and ω(a,b) 2 P(rB,R) P(rA,R) ω(a,b)2 P(rB,R) ω(b,c) 2 P(rC,R) P(rB,R) ω(b,c)2 P(rC,R) for any r [0, ). Consequently, we deduce that ω(a,b) 2 ω(b,c) 2 P(rC,R) P(rA,R) ω(a,b)2 ω(b,c) 2 P(rC,R) for any r [0, ). Applying again Theorem.3, we have ω(a,c) ω(a,b)ω(b,c). This completes the proof. Now, we can introduce a hyperbolic (Poincaré Bergman type) metric on the Harnack parts of [B(H) n ]. Proposition 2.2. Let be a Harnack part of [B(H) n ] and define δ : R+ by setting Then δ is a metric on. δ(a,b) := ln ω(a,b), A,B. (2.2) Proof. The result follows from Lemma 2.. In [34], we showed that any free holomorphic automorphism Ψ of the unit ball [B(H) n ] which fixes the origin is implemented by a unitary operator on C n, i.e., there is a unitary operator U on C n such that Ψ(X,...,X n ) = Ψ U (X,...,X n ) := [X X n ]U, (X,...,X n ) [ B(H) n]. The theory of noncommutative characteristic functions for row contractions (see [22,29]) was used to find all the involutive free holomorphic automorphisms of [B(H) n ]. They turned out to be of the form

20 G. Popescu / Journal of Functional Analysis 256 (2009) Ψ λ = Θ λ (X,...,X n ) := λ λ (I K ) n λ i X i [X X n ] λ, for some λ = (λ,...,λ n ) B n, where Θ λ is the characteristic function of the row contraction λ, and λ, λ are certain defect operators. Moreover, we determined the group Aut([B(H) n ] ) of all the free holomorphic automorphisms of the noncommutative ball [B(H) n ] and showed that if Ψ Aut([B(H) n ] ) and λ := Ψ (0), then there is a unitary operator U on C n such that i= Ψ = Ψ U Ψ λ. The following result is essential for the proof of the main result of this section. Lemma 2.3. Let A := (A,...,A n ) and B := (B,...,B n ) be in [B(H) n ] and let Ψ Aut([B(H) n ] ). Then A H c B if and only if Ψ(A,...,A n ) H c Ψ(B,...,B n ). Proof. Let p(s,...,s n ) := S α M (α), M (α) M m, α k be a polynomial in A n M m such that Re p(s,...,s n ) 0. Due to the results from [34], if Ψ Aut([B(H) n ] ), then Ψ(S,...,S n ) = (Ψ (S,...,S n ),...,Ψ n (S,...,S n )) is a row contraction with entries Ψ i (S,...,S n ), i =,...,n, in the noncommutative disc algebra A n. Since the noncommutative Poisson transform at Ψ(S,...,S n ) is a c.p. linear map P Ψ(S,...,S n ) : C (S,...,S n ) B(H), we deduce that [ Q(S,...,S n ) := Re α k Ψ α (S,...,S n ) M (α) ] = ( P Ψ(S,...,S n ) id )[ Re p(s,...,s n ) ] 0. If we assume that A H c B, then Theorem., part (v) implies Q(A,...,A n ) = (P A id) [ Q(S,...,S n ) ] c 2 (P B id) [ Q(S,...,S n ) ] = c 2 Q(B,...,B n ), where P A and P B are the noncommutative Poisson transforms at A and B, respectively. Therefore, we have Re p ( Ψ (A,...,A n ),...,Ψ n (A,...,A n ) ) c 2 Re p ( Ψ (B,...,B n ),...,Ψ n (B,...,B n ) ), which shows that Ψ(A,...,A n ) H c Ψ(B,...,B n ).

21 4050 G. Popescu / Journal of Functional Analysis 256 (2009) Conversely, assume that Ψ(A,...,A n ) H Ψ(B,...,B n ). Applying the first part of the proof, c we deduce that Ψ [ Ψ(A,...,A n ) ] H c Ψ [ Ψ(B,...,B n ) ]. Since Ψ Ψ = id on the closed ball [B(H) n ], we deduce that A H B. The proof is complete. c Here is the main result of this section. Theorem 2.4. Let δ :[B(H) n ] [B(H) n ] [0, ) be the hyperbolic metric. Then the following statements hold. (i) If A,B [B(H) n ], then δ(a,b) = ln max { C A C, C B C }, where C X := ( X I)(I R X ) and R X := X R + + X n R n is the reconstruction operator associated with the right creation operators R,...,R n and X := (X,...,X n ) [B(H) n ]. (ii) For any free holomorphic automorphism Ψ of the noncommutative unit ball [B(H) n ], B A δ(a,b) = δ ( Ψ(A),Ψ(B) ), A,B [ B(H) n]. (iii) δ Bn B n coincides with the Poincaré Bergman distance on B n, i.e., δ(z,w) = 2 ln + ψ z(w) 2 ψ z (w) 2, z,w B n, where ψ z is the involutive automorphism of B n that interchanges 0 and z. Proof. Let A,B [B(H) n ]. Due to Theorem.6, we have A H B. In order to determine ω(a,b), assume that A H B for some c. According to Theorem.3, we have c c 2 P(rB,R) P(rA,R) c2 P(rB,R) for any r [0, ). Since A < and A <, we can take the limit, as r, in the operator norm topology, and obtain c 2 P(B,R) P(A,R) c2 P(B,R). (2.3)

22 G. Popescu / Journal of Functional Analysis 256 (2009) We recall that the free pluriharmonic kernel P(X,R), X := (X,...,X n ) [B(H) n ], has the factorization P(X,R)= C X C X, where C X := ( X I)(I R X ). Note also that C X is an invertible operator. It is easy to see that relation (2.3) implies C A C B C BC A c2 I and C B C A C AC B c2 I. Therefore, d := max { C A C, C B C } c, B A which implies d ω(a,b). On the other hand, since C B CA d and C ACB d, we have Hence, we deduce that which is equivalent to C A C B C BC A d2 I and C B C A C AC B d2 I. d 2 C B C B C A C A d 2 C B C B, d 2 P(B,S) P(A,S) d2 P(B,S), where S := (S,...,S n ) is the n-tuple of left creation operators. Applying the noncommutative Poisson transform id P rr, r [0, ), and taking into account that it is a positive map, we deduce that d 2 P(rB,R) P(rA,R) d2 P(rB,R) for any r [0, ). Due to Theorem.3, we deduce that A H B and, consequently, ω(a,b) d. d Since the reverse inequality was already proved, we have ω(a,b) = d, which together with (2.2) prove part (i). To prove (ii), let Ψ Aut([B(H) n ] ).IfA,B [B(H) n ], then, due to Theorem.6, we have A H B. Applying Lemma 2.3, the result follows. Now, let us prove item (iii). Let z := (z,...,z n ) B n. Due to part (i) of this theorem, we have δ(z,0) = ln max { C z, C }, where C z := ( z 2 ) /2 (I n i= z i R i ). First, we show that n I z i R i = + z 2. (2.4) i= z

23 4052 G. Popescu / Journal of Functional Analysis 256 (2009) Indeed, since R,...,R n are isometries with orthogonal ranges, we have ( n n ) /2 z i R i = z i 2 = z 2. i= i= Consequently, n I z i R i + z 2. (2.5) i= Note that, due to Riesz representation theorem, we have sup w=(w,...,w n ) B n + z i w i = + z 2. (2.6) i= On the other hand, due to the noncommutative von Neumann inequality [23], we have + n z i w i I z i R i (2.7) i= i= for any (w,...,w n ) B n. Combining relations (2.5), (2.6), and (2.7), we deduce (2.4). Consequently, we have ( ) C /2 + z 2 z =. (2.8) z 2 Note also that ( I ) n n n 2 z i R i + z i R i + z i R i + i= i= i= = + z 2 + z = z 2. Consequently, we have ( ) /2 + z 2 C z. (2.9) z 2 Due to relations (2.8) and (2.9), we have ( ) /2 + z 2 ω(z,0) =. (2.0) z 2

24 G. Popescu / Journal of Functional Analysis 256 (2009) Now, we consider the general case. For each w B n,letψ z be the corresponding involutive automorphism of [B(H) n ]. We recall (see [34]) that Ψ w (0) = w and Ψ w (w) = 0. Due to part (ii) of this theorem and relation (2.0), we have δ(z,w) = δ ( Ψ w (z), Ψ w (w) ) = δ ( Ψ w (z), 0 ) = ln ω ( Ψ w (z), 0 ) = 2 ln + Ψ z(w) 2 Ψ z (w) 2. Since, according to [34], Ψ w is a noncommutative extension of the involutive automorphism of B n that interchanges 0 and z, i.e., Ψ w (z) = ψ w (z) for z B n, item (iii) follows. The proof is complete. Corollary 2.5. For any X, Y [B(H) n ], δ(x,y) ln X Y ( X )( Y ). Proof. According to Theorem.6 and Proposition 2.2, [B(H) n ] is the Harnack part of 0 and δ is a metric on the open ball [B(H) n ]. Therefore δ(x,y) δ(x,0) + δ(0,y). Theorem 2.4, part (i) implies δ(x,0) = ln max { C X, C }, where C X := (I X )(I R X ) and R X := X R + +X n R n. Since R,...,R n are isometries with orthogonal ranges, we have X and On the other hand, since X <, we have I R X + R X = + X (I RX ) + RX + R X 2 + X = + X + X 2 + = X. 2 + XX + XX 2 + = X 2.

25 4054 G. Popescu / Journal of Functional Analysis 256 (2009) Now, one can easily see that and C X = (I RX ) ( I ) X ( + X ) X ( ) + X /2 X C X X X. Note also that, due to the fact that I XX XX,wehave X X ( ) + X /2. X Therefore δ(x,0) = ln max { C X, CX } ln X X. Taking into account that δ(x,y) δ(x,0) + δ(0,y), the result follows. The proof is complete. In what follows we prove that the hyperbolic metric δ, on the Harnack parts of [B(H) n ], is invariant under the automorphism group Aut([B(H) n ] ), and can be written in terms of the reconstruction operator. First, we need the following result. Lemma 2.6. Let A := (A,...,A n ) and B := (B,...,B n ) be in [B(H) n ]. Then the following properties hold. (i) A H B if and only if ra H rb for any r [0, ) and sup r [0,) ω(ra,rb) <. In this case, ω(a,b) = sup ω(ra,rb) and δ(a,b) = sup δ(ra,rb). r [0,) r [0,) (ii) If A H B, then the functions r ω(ra,rb) and r δ(ra,rb) are increasing on [0, ). Proof. Assume that A H B and let a := ω(a,b). Due to relation (.6), it is clear that A H B.By a Theorem.3, we deduce that a 2 P(rB,R) P(rA,R) a2 P(rB,R) for any r [0, ), which shows that ra H a rb for any r [0, ), and sup r [0,) ω(ra,rb) a.

26 G. Popescu / Journal of Functional Analysis 256 (2009) Conversely, assume that sup r [0,) ω(ra,rb) <. Denote c r := ω(ra,rb) for r [0, ), and let c := sup r [0,) c r <. Since ra H rb and c r c,wehave cr c 2 P(trB,R) cr 2 P(trB,R) P(trA,R) cr 2 P(trB,R) c2 P(trB,R) for any t,r [0, ). Due to Theorem.3 we deduce that A H B. Thus ω(a,b) c. Since the c reverse inequality was already proved above, we have a = c. The second part of item (i) is now obvious. To prove (ii), let s,t [0, ) be such that s<t. Applying part (i), we have ω(sa,sb) sup ω(rta,rtb) ω(ta,tb). r [0,) Hence, we deduce item (ii). The proof is complete. Theorem 2.7. Let A := (A,...,A n ) and B := (B,...,B n ) be in [B(H) n ] such that A H B. Then (i) δ(a,b) = δ(ψ (A), Ψ (B)) for any Ψ Aut([B(H) n ] ); (ii) the metric δ satisfies the relation { δ(a,b) = ln max sup r [0,) C ra C rb, sup r [0,) C rb C }, where C X := ( X I)(I R X ) and R X := X R + +X n R n is the reconstruction operator associated with the right creation operators R,...,R n and X := (X,...,X n ) [B(H) n ]. ra Proof. Due to Lemma 2.6 and Theorem 2.4, the result follows. 3. Metric topologies on Harnack parts of [B(H) n ] In this section we study the relations between the δ-topology, the d H -topology (which will be introduced), and the operator norm topology on Harnack parts of [B(H) n ]. We prove that the hyperbolic metric δ is a complete metric on certain Harnack parts, and that all the topologies above coincide on the open ball [B(H) n ]. First, we need some notation. Denote [ B0 (H) n] := { (X,...,X n ) [ B(H) n] : r(x,...,x n )< }, where r(x,...,x n ) is the joint spectral radius of (X,...,X n ). Note that, due to Theorems.2 and.6, we have [ B(H) n ] [ B 0 (H) n] { = X [ B(H) n] : X H } 0.

27 4056 G. Popescu / Journal of Functional Analysis 256 (2009) If A,B are in [B 0 (H) n ], then A H 0 and B H 0. Consequently, there exists c such that, for any f A n B(E) with Re f 0, where E is a separable infinite-dimensional Hilbert space, we have Re f(a) c 2 Re f(0) and Re f(b) c 2 Re f(0). Hence, we deduce that Re f(a) Re f(b) 2c 2 Re f(0). Therefore, it makes sense to define the map d H :[B 0 (H) n ] [B 0(H) n ] [0, ) by setting d H (A, B) := sup { u(a) u(b) : u Re ( A n B(E) ),u(0) = I, u 0 }. Proposition 3.. For any A,B [B 0 (H) n ], d H (A, B) = sup Re p(a) Re p(b), where the supremum is taken over all polynomials p C[X,...,X n ] M m, m N, with Re p(0) = I and Re p 0. Proof. Let f A n B(E) be such that Re f 0 and (Re f)(0) = I. According to [25], f has a unique formal Fourier representation f = S α C (α), C (α) B(E). α F + n Moreover, lim r f r = f in the operator norm topology, where f r = k= α =k r α S α C (α) is in A n B(E) and the series is convergent in the operator norm. Consequently, for any ɛ>0, there exist r ɛ [0, ) and N ɛ N such that p rɛ,n ɛ f < ɛ 2, (3.) where p rɛ,n ɛ := N ɛ k=0 α =k r α ɛ S α C (α). Define the polynomial q ɛ,rɛ,n ɛ := +ɛ (p r ɛ,n ɛ +ɛi) and note that (Re q ɛ,rɛ,n ɛ )(0) = I. On the other hand, due to (3.), we have Re p rɛ,n ɛ Re f < ɛ 2, which, due to the fact that Re f 0, implies Re q ɛ,r ɛ,n ɛ 0. Now, notice that q ɛ,rɛ,n ɛ f + ɛ p r ɛ,n ɛ f + ɛ I + f, + ɛ which together with relation (3.) show that f can be approximated, in the operator norm, with polynomials q = N k=0 S α D (α), D (α) B(E), such that Re q 0 and (Re q)(0) = I. Consider now an orthonormal basis {ξ,ξ 2,...} of E and let E m := span{ξ,...,ξ m }. Setting q m := P F 2 (H n ) E m q F 2 (H n ) E m = N S α P Em D (α) Em, k=0