A characterization of pick bodies

Lemmas concerning operators Let Jf be a finite dimensional Hilbert space and B a linear operator on df... 272] which states that the quotient of a uniform algebra by a closed ideal is is

A CHARACTERIZATION OF PICK BODIES

B COLE, K LEWIS AND J WERMER

ABSTRACT

Let 2 = (z v , z n) be an /i-tuple of distinct points in the open unit disk We define the Pick body @(z)

as the totality of points w = (w v ,w n ) in Cn such that there exists feH 00 with H/l^ < 1 and J(z f ) = w jt

for 1 <_/ < n We discuss the properties of Pick bodies and characterize them among compact subsets of

C We also study related questions concerning certain algebras of operators on Hilbert space.

0 Introduction

We consider Cn as an algebra with coordinatewise multiplication and let || • || denote an algebra norm on Cn The closed unit ball in this norm {zeC n \ \\z\\ ^ 1} is

a compact convex circled subset of Cn with non-void interior

In [4] we studied a particular class of such norms which arise as follows: let A be

a uniform algebra on a compact Hausdorff space X, with maximal ideal space Jt, and

fix an H-tuple of points Ml 5 , Mn in M Let / = {g e A \ g(Mj) = 0,1 ^ j ^ n} Then /is a closed ideal in A The quotient algebra A/Us then a normed algebra isomorphic

as an algebra to Cn under the map [/] H>- (J[M X ), ,J(M n )) forfeA, where [f\ denotes

the coset of/ in A/1 The image of the closed unit ball of A/1 under this map is

n

| 3feA with/(M,) = w p 1 <y ^ n, and ||[/]|| ^ 1}.

In addition to the above mentioned properties valid for any algebra norm on Cn, Si now has the following additional property Let k be any positive integer and let

A* denote the closed unit polydisk in Cn If P is a polynomial in k variables with

||P||A* = max(eA* 1^(01 ^ 1, then for each fc-tuple of elements x',x", ,x w in 2) we have that P(x',x", ,x m ) again is in 9.

This statement follows easily from the definition of @) as the closed unit ball of A/1

and is also a special case of a theorem due to I Craw and (independently) B Cole concerning ^-algebras, which is given in [2, §50, Proposition 5] As defined in [4], a (2-algebra is a Banach algebra which is a quotient of a uniform algebra This suggests the following definition

DEFINITION. A subset K of C" is called hyperconvex if K is compact with non-empty interior so that, for all positive integers k, whenever P is a polynomial

in k variables with ||P||A* ^ 1 and x',x", ,x w are in K, then P{x',x", ,x (k) )eK.

We note that, expressed in terms of coordinates, if x' = (x[, ,x' n ),

x" = (xl ,x" n ) etc., then P(x',x", ,*<*>) = {P(x' x ,xj, ,*<*>), ,P{x' n ,<, ,xf)).

Received 21 October 1991.

1991 Mathematics Subject Classification 30E05.

/ London Math Soc (2) 48 (1993) 316-328

A CHARACTERIZATION OF PICK BODIES 3 1 7

We may obtain an example of a hyperconvex set as follows: we select an n-tuple

of distinct points zl 5 ,z n in the open unit disk in C and put

®(z) = {vveC"13/e//0 0 with/(*,) = " „ 1 ^j ^ n, and \\f\\ n ^ 1},

where z represents the point (z1 5 ,zn) in Cn

We call the set 3>{z) a Pick body It follows at once from the definition that 2)(z)

is hyperconvex

Let A(A) denote the disk algebra and let / be the ideal in A(A) consisting of

9 = {weC n \lfeA(A) withf[ Zj ) = w p 1 ^j^n, and M I L < 1},

where \\[F]\\ denotes the norm of[F] in A(A)/I Then ®(z) = ®.

Proof Fix w e ®(z) Choose <p e H™, \\ <f> || ^ 1, such that w j = (f>(zj) for each/ Put

A ( 0 = 0 ( ( 1 - 1 A ) O - Then/f ce^(A) and \\[f k ]\\ ^ 1 Hence

It follows from the definition that 9) is closed Therefore w = l i m ^ ^ w

Conversely, let WE Si For k any positive integer, there exists geA(A),

| | g | | ^ l + l/fc, with g(Zj) = Wj for all / Passing to a pointwise convergent subsequence, we obtain gsH™ with HgH^ ^ 1 and g(zj) = w } for ally Hence we3)(z) LEMMA 1 Fix £ = ((19 ,£„) in Cn wi7/i |Cy| < 1 for each] and (, # Cy'/' # / • Let

d® denote the boundary of 3){Q Then the points C = (Cf. -.CS) fo/ong to 5 ^ for k= 1 , , « - 1

The following argument is due to Herbert Alexander If

then there exists geA(A) such that g ( Q = C* f o r y = 1, ,«, and ||g|| < 1 Hence

\g(z)\ < \z k \ for z on the unit circle By Rouche's theorem, then, the functions z k

and z k —g have the same number of zeros in \z\ < 1 But z k —g vanishes at the

n points £ p 1 < / < n, while z* has exactly k zeros This contradiction yields that

>, as desired.

Thus each Pick body ®(Q has the two properties: (i) @(Q is hyperconvex in Cn and (ii) the kth power of C belongs to the boundary of ®(Q for k = 1, ,« — 1 We

shall show that these two properties characterize Pick bodies

(1) K is hyperconvex,

(2) dK contains a point z = (z15 , z n ) with |z^| < 1 for each j and z< ^ z, if i # y , such that z, , zn - 1 all belong to dK.

Then K is a Pick body.

1 Lemmas concerning operators Let Jf be a finite dimensional Hilbert space and B a linear operator on df.

3 1 8 B COLE, K LEWIS AND J WERMER

\\B\\ = ll^""1! = 1 Choose a unit vector <f> in tf with ||5n"VII = 1- Then tne n-tuple

<f>, B^, ,^' 1 ^ is linearly independent.

Proof Note that Wy/f = WBy/f is equivalent to {{I-B*B)\j/, y/) = 0 and since I—B*B is positive this is equivalent to (I—B*B)y/ = 0 Note that

so all the inequalities are equalities Hence we have that (I—B*B)B i <f> = 0 for

j = 0,\, ,n-2.

Now suppose that the w-tuple $, i?0, , i?""10 is linearly dependent; then there

exists k < n — 1 such that B k <j> is a linear combination of <f>, B(f>, ,B k ~ x <j> The space

spanned by these vectors is invariant under B and is contained in the kernel o f / - B*B

by the above Hence B is unitary on this subspace and thus has an eigenvalue of modulus 1 This contradicts our hypothesis and so the n-tuple (f>, B(J>, ,B n ~ l (j> is

linearly independent

Jtf with eigenvalues z x , ,z n Assume that

(3) \ Zj \<\,forl^j^n,

(4) z t *z t ifi*j t

(5) | | * | | - I I * -11 1 = 1.

Then there exists a basis of eigenvectors v lt , v n corresponding to z1? , z n so that

(v<5 v,) = for 1 < i,j ^ n (6)

\.—z i z i Proof Choose 0 e J ^ with ||0|| = 1 and H*""1^ = 1 As in Lemma 2,

it follows that (I—B*B)B 1 <f> = 0, for j = 0 , l , , n - 2 , since for each such j ,

||B(B 1 0)|| = 1 = | | # 0 | | Also by Lemma 2, 0, B<j>, ,B n ~ l <f> are linearly independent.

Hence I—B*B has a null space of dimension at least n— 1 Since B is not unitary, we

must have r a n k ( / - 5 * £ ) = 1

Let v\, ,v' n be any basis of eigenvectors for B, and put T= I—B*B Since rank T= 1, the nxn matrix ((7X,v])) has rank 1 This matrix is also positive

semidefinite Hence, there exist scalars cl s ,cn so that

Since |cj2 = (1 — |zj2) ||v|||2 # 0, we see that c t # 0 So the vectors v( = v'Jc i define a basis of eigenvectors with

and Lemma 3 is proved

Fix z = (z v , z n ) in Cn satisfying (3) and (4) We define the inner product ( , ) , on

Cnb y

n

(MX13 L t i s j0-~ z t^})~ 1 fort,seC n

A CHARACTERIZATION OF PICK BODIES 3 1 9

An elementary calculation gives

C>4 = ^ V l ± ~ ie dO,

and so (, \ is positive definite and makes Cn into an n-dimensional Hilbert space which

we designate by C"

For each w = (w 1 , ,w n ), we denote by P w the operator on C" defined by

P w (t 1 , ,t n ) = (w 1 t 1 , ,w n t n ) We call P w a Pick operator on C? We also set

S = P,

Let e v , e n denote the standard basis on Cn Then, suppressing the subscript z,

we have

\—Z i 2 j

By the definition of S, we also have

Then there exists a unitary map U from Jtf to C? such that S = UBU' 1

Proof Choose vx, ,vn as in Lemma 3 From (6) and (7), it follows that (e(, e t ) = (v<5 v,) for 1 ^ ij < n.

We define an operator U from Jf to C" by setting Uv } = e i fory = l, ,n and

extending linearly Now UBv } = Vz i \ i = z i e i = Se } = SUv } , and so UBU' 1 = S Also,

if 0e^f, </> = Yih v p t h e n

= II E hUvtf = || £tfyf = Et,tk(e},ek) = £ t}fk(v},vk) = || E ^v,||2 = ||0||2.

Hence U is unitary, and we are done.

We now drop the assumption that our Hilbert space 3f? is finite-dimensional, and

we choose an n-tuple of bounded operators E } ,\ ^j < n on #? We assume that

E i E i = ^ if i ^j, E} = E i # 0 for 1 < i,j ^ n a n d E 1 + E 2 + - + E n = I (9)

For each w = (w lt , w n ) in CB, we put T w = E w, ^ and let si = {T w \ weC n } Then

sf is an n-dimensional operator algebra on Jf, and the map T w ^ w is an isomorphism

of sf onto the algebra C".

LEMMA 5 Let 9 = {weC n \ \\TJ ^ 1} Fix z = (z15 ,zn) in Cn such that \z t \ < 1 for 1 ^ 7 < n and z i # z } if i ^j Put T = T z and assume that

\\T\\ = 1 and there exists </>e^ with WT^fiW = \\<f>\\ * 0 (10) Then 2 is the Pick body 2{z).

Proof Assume without loss of generality that ||0|| = 1 We have

3 2 0 B COLE, K LEWIS AND J WERMER

Let "W be the subspace of M spanned by the vectors <j>, T<f>, , T n ~ x (f> Since st is

n-dimensional, T n is a linear combination of /, T, , T n ~ l and so if is invariant under

T We shall show that iV is ^-dimensional and that the restriction of T to if is

unitarily equivalent to the operator 5 on C2

Define f to be the restriction of T to HT By (11),

so \\f1| = IIT71"1!! = 1 Let x be an eigenvector of f with corresponding eigenvalue X Choose j Q such that E } x ^0 Then

Hence X = z } and so |A| < 1 Hence each eigenvalue of f has modulus less than 1 So Lemma 2 applies to f and gives that the w-tuple of vectors <j>, f <j>, , fn"10 is linearly

independent and hence is a basis of #" Since for each k, T k = £^z*£,, the vectors

E 1 (j>, E 2 0, , E n $ span itr Also, for each ;, TE i <f> - ( £ zfc £fc) Erf = z, Ei <f> So the

set JE*! 0, £"2 (j>, , E n (f> is a basis of eigenvectors for T.

Thus Lemma 4 applies to T and so there exists a unitary operator U from T^" to

Czn with S = UTU*.

Fix a in ^ Then ||2J < 1 Let Q be a polynomial with Q(z } ) = a } for

y = 1, ,« If P a denotes the Pick operator introduced above, then

P a = Q(S) = UQ(f)U* = UQ(T Z )U* = UT a U*,

and so HPJ < ||£|| < 1 Hence for all /eCzn, ||Paf||2 < ||r||2, the norms being taken in C" Thus we have

By Pick's theorem, (12) implies that there exists feH™ with H/H^ ^ 1 a n d / ^ ) = a, for ally, and so a e ^ ( z ) Thus 2> c Q)(z).

On the other hand, fix a in Q)(z) Then, for each e > 0, there exists a function /e,4(A), which we may take to be a polynomial, with \\f\\ m ^ 1 +e and^z^) = a } for

all j by the Claim before Lemma 1 Then

./TO =

and so by von Neumann's inequality, ||./(^)|| ^ 1+e Hence ||7^|| < 1+e for all

e > 0, implying that || 7^|| ^ 1, and so a e Si Therefore 2>{z) a S>, and we have shown that Q)(z) = 2.

2 Proof of Theorem 1

In the proof of Theorem 1, we shall use Cole's representation theorem, [2, Theorem 7, p 272] which states that the quotient of a uniform algebra by a closed ideal is isometrically isomorphic to an algebra of operators on a Hilbert space Furthermore, every operator in the algebra achieves its norm on some unit vector

Proof of Theorem 1 By hypothesis K is a hyperconvex subset of Cn, and hence

is the closed unit ball of a Banach algebra 3& which is algebraically isomorphic to Cn and which satisfies the following condition: whenever x',x", ,x {k) are in the closed

unit ball of ^ , then P(x',x", ,x lk) ) is also in the closed unit ball of 38 provided that

P is a polynomial in k variables with ||P||A* < 1.

A CHARACTERIZATION OF PICK BODIES 3 2 1

Proposition 5 in [2, §50] implies that J1 is isometrically isomorphic to

the quotient A/I of a uniform algebra A by a closed ideal / Also there exist

M x , ,M n zM A with / = {feA\j{M^ = 0, j = 1, ,«} By Cole's representation theorem stated above, there exists a further isometric isomorphism of A/1 on an algebra of operators, d, on a Hilbert space, Jf Combining these two maps,

we have an isometric isomorphism x'M -> s/ and

for all Test, there exists </>e3? with ||0|| = 1 and \\T</>\\ = ||r|| (13) Let |/J e A/1 for 7 = 1, ,« be such that /,(Mt) = <5,t, and put £, = T((/J). Then

E x , ,E n satisfy condition (9) For w e Cn, we put 7; = £ , w, £r Thus, r(zfc) = 7J for

A: = 1,2, In particular, with z as in Theorem 1 (2), we have that \\T Z \\ = || T^\\ = 1

in st since ||z|| = Hz""1!! = 1 in @.

By (13) there exists <pe J4? with ||0|| = 1 and WT^^W = 1, and so (10) in Lemma

5 is satisfied Hence, by that lemma, S> = {weC n \\\TJ < \} = 2{z) Since T is

isometric, w is in the closed unit ball of $1 if and only if || £ , w j E } \\ ^ 1 which occurs

if and only if we&>(z), and so the closed unit ball of 38 is Q)(z) Thus K = Q){z) and

Theorem 1 is proved

3 Operator algebras

In this section we give an alternative approach to some of the questions considered

in the previous sections of this paper These results are independent of those obtained earlier and are somewhat more general Theorem 2 implies Theorem 1, and other results generalize Theorem 1 in various ways The price to be paid is that these calculations are less explicit and elementary than those used in the proof of Theorem 1

The main objects of interest here are finite dimensional commutative Banach algebras, principally operator algebras on Hilbert space (which are not assumed to be

self-adjoint) When $t is any semisimple Banach algebra of dimension n, the Gelfand theory identifies the closed unit ball of stf with a subset Q)^ of Cn Specifically, we put

% = {{x{M1), ,x{Mn))\xesf,\\x\\ ^ 1},

where the maximal ideal space of $0 consists of the n distinct points Ml s , M n Since

the Gelfand representation is essentially the map

<D: sf -> Cn with O(x) = (x(Mx), , x(M n )),

we see that O is an isomorphism when Cn is viewed as an algebra under

coordinatewise operations Furthermore, under this map, the closed unit ball of s#

corresponds to ^

The hyperconvex set Si, associated with the Q-algebra s$ = All in the Introduction, is precisely the set S>^ defined above Moreover, as discussed in the

proof of Theorem 1, every hyperconvex set arises in this way Thus, from the Banach algebra point of view, Theorem 1 can be interpreted as a statement about the unit ball

of a finite dimensional semisimple Q-algebra

We now turn our attention to a broader class of Banach algebras: finite dimensional commutative algebras of operators on Hilbert space Since we know that every Q-algebra is in fact an operator algebra, results about operator algebras automatically apply to Q-algebras And consequently, general results about the unit ball of an operator algebra provide information about hyperconvex sets

3 2 2 B COLE, K LEWIS AND J WERMER

For the remainder of this section, we let A denote the disk algebra A(A) and let

n be a positive integer All Banach algebras considered are assumed to contain an

identity

isometrically isomorphic to A/J where / is a closed ideal in A with hull(/) £ int(A) Furthermore, an element xesf is said to be a Pick generator if

{p(x)\p is a polynomial, \\p\\ A ^ 1}

constitutes a dense subset of the closed unit ball of $0.

Note that J in the above definition is of the form bA where b is a finite Blaschke product and that the Pick algebra A/J has the Pick generator [idj where idA is the identity map on A (the complex coordinate function) Moreover, s/ = A/J is semisimple if and only if b has simple zeros, and this is exactly the situation considered in the Introduction; so we see that Pick bodies arise as S>^ where si a

semisimple Pick algebra

We shall prove a slight modification of a theorem due to D Sarason which gives

a representation of Pick algebras as singly generated algebras of operators on a Hilbert space Sarason's theorem uses if00 in place of A and an arbitrary inner function in place of b Our proof will also work in this situation; however, we need

the disk algebra version in what follows

We review some basic facts concerning Hardy spaces, denoted by H p Let m

denote Lebesgue measure on the unit circle Let S be the shift operator of multiplication by z on H 2 : Sf{z) = zj{z) for feH 2 A function beH 2 is called inner if

\b\ = 1 a.e [m] and a function geH 2 is called outer if g is a cyclic vector for S, that

is, the set of all pg, p a polynomial, is dense in H 2 A theorem of Beurling states that

every closed subspace of H 2 invariant under S has the form bH 2 , where b in an inner

function, and every function/ei/2 can be factored a s / = bg with b inner and g outer Also, for every heH 1 , there exists keH 2 with \h\ = \k\ 2 a.e [m].

Following Sarason [8], we fix an inner function b, let Jf b = H 2 Q bH 2 , denote by Pjr b the orthogonal projection from H 2 to J^, and put T b = P jCb S\ :t - b , where S is

multiplication by z on H 2 The orthogonal projection from L 2 to H 2 is denoted by P.

ForfeA, define AS) on H 2 by AS)g =fg for geH 2 and define <b b (f) = P^JiS) \ Xb

Note that O6 is a contractive map from A to $8(JQ, the space of operators on

JfT b Since bH 2 is invariant under 5, we have that tf b is invariant under S*; so 7? = 5 * 1 ^ and p(T b )*y = p(S)*y for ye JT b Hence for x,yeJr b ,

(p(T b )x,y) = (x,p(T b )*y) = (x,p(S)*y) = (p(S)x,y) = (P Xb p(S)x,y).

That is, p(T b ) = 0> b (p) This shows that O6 is multiplicative on polynomials, and so by norm continuity of multiplication, <D6 is an algebra homomorphism From here on we shall write <P6(/) as J[T b ).

Not only do we have \\J{T b )\\ ^ \\f\\,feA, but since b(T b ) = 0, we have

iiyroii = MTj+bmgmw ^ \\f+bg\\

for all geA, so ||/(7;)|| < | | / + M | | , the coset norm of/in A/bA The content of the

next theorem is that this is always an equality Sarason [8] used this result to prove Pick's theorem

A CHARACTERIZATION OF PICK BODIES 3 2 3

= ||/TO || • In particular, the algebra st generated by T b is a Pick algebra and has Pick generator T b

Proof By the Hahn-Banach Theorem, there exists a complex measure ji with

\\f+bA\\ = jfdfi, ||//|| < 1, and jbgd/i = 0 for all geA We shall show that there

exist x,yeki,, \\x\\, \\y\\ ^ 1, with ffdfi = ffxydm This will prove the theorem

since we shall have

\\f+bA\\ = jfd M = (fx,y) = (P Kb fx,y) = <J[T b )x,y) ^ \\f[T b )\\ ^ | | / + M | |

By the Riesz Theorem, there exists heH 1 with bdfi = hdm There exists geH 2 with \h\ = |g|2, so dfi = u \g\ 2 dm, where u is unimodular By Beurling's theorem, we

may also assume that g is outer This implies that ug _l_ bH 2 since for all keA we have

0 = jbkdfi = J bku \g\ 2 dm = (bkg, ug) and the set of all bkg for keA is dense in bH 2

It follows that y = P(ug) 1 bH 2 and so yeX b Also,

Put x = P Xb g; then ||x|| < 1 also.

Now we show that jfd/n = (fx,y) We have

If dfi = jfu \g\ 2 dm = (fg, ug) = (P(fg), ug)

since fg e H 2 So

(P(fg),ug) = (fg,P(ug)) = (fg,y) = (f{S)g,y) = (g,f[S)*y) = (P^

since yeJf b and Jf b is invariant under S* Finally,

(P^Asry) = (x,AS)*y) = (f{S)x,y) = {fx,y)

and we are done

Our methods rely on a result in [3], and so for completeness we include a proof

with ||31 = 1, rank(7- T*T) = 1, and \\T m x\\ -• 0 as m -• oo for all xeJf, then T is unitarily equivalent to the backward shift S* on H 2 restricted to an S*-invariant subspace.

Proof For two vectors u, v in 2tf, let u ® v be the operator defined by (u <g> v)w

= (w,v)u The hypotheses on Timply that there exists teJf with I—T*T= t®t.

Yoxxetf,

\\Tnx\\2-\\Tn+1 x\\2 = (Tnx,Tnx) -(T*TTnx,Tnx)

and so £»\(T n x, t)\ 2 = ^o(\\T n x\\ 2 -\\T n+1 x\\ 2 ) = \\x\\ 2 since \\T n x\\^0 Hence, the

function defined by f x (z) = f^{T n x, t)z n belongs to H 2 and satisfies | | / J = ||x|| Let

Ube the isometry Jf^H 2 given by U(x) =f x Since/(Tx) = S*fx , we have UT=S*U

on Jf, and so the subspace Jf = XJ{3^) is invariant under S* Viewing U as a unitary map onto Jf, it follows that UTU* = 5*1^.

3 2 4 B COLE, K LEWIS AND J WERMER

The next result provides the key link between operator algebras and quotients of the disk algebra

subalgebra of @{tf) Let Test so that \\T\\ = 1, r a n k ( / - T * r ) = 1, and T has

no eigenvalue of modulus 1 Then, $0 is an n-dimensional Pick algebra with Pick generator T.

Proof Since r a n k ( / - T*T) = r a n k ( 7 - TT*) and since the spectral radius of T,

and hence T*, is less than 1, the hypotheses of Lemma 6 are satisfied by T* So T*

is unitarily equivalent to S* restricted to an invariant subspace which, by Beurling's theorem, has the form Jf b for some inner function b Consequently, $0 is unitarily equivalent to a commutative subalgebra si of 88{X b ) containing T b , and, under this

equivalence, T b corresponds to T Also, since tf has finite dimension, b is a finite

Blaschke product

For the vector <j> = P x 1, {<j>, T b $, , T b ~ x 0} is a basis for X b , and therefore

R s j / w i t h R<f> = 0 implies that R = 0 It follows that dim(j/) = n and j / i s the algebra

generated by T b So, by Proposition 1, s/ is a Pick algebra with Pick generator T.

A Banach algebra s& satisfies the von Neumann Inequality of order 1 if, whenever

xestf with ||x|| < 1, ||/?(x)|| < maxj/jl for every polynomial/? Von Neumann showed

that this condition is satisfied if $$ is a subalgebra of &(&) for a Hilbert space 3f?\ see [4] for more details The next result is used in this paper only when sd x is such an algebra

LEMMA 7 Let s^ be a commutative Banach algebra satisfying the von Neumann

Inequality of order 1, and let s^be a Pick algebra Let Q> be a norm 1 homomorphism from s% into «#£, and let xes^ with \\x\\ ^ 1 so that O(x) is a Pick generator for s^ Then, <f> induces an isometric isomorphism between j^/kerCO) and s^ In particular, if

O is one-to-one, then O establishes an isometric isomorphism between s^ an

Proof Let <£>(*) = y For every polynomial p, <X> maps p(x) to p(y) Restricting

attention to polynomials in the unit ball of the disk algebra, the hypotheses guarantee

that O maps the unit ball of s^, onto a dense subset of the unit ball of s% The

assertion now follows

II T k \\ = \ for 0 < k < n, and T has no eigenvalue of modulus 1, then I— T*T has rank

1 and dim(^f) = n.

Proof Choose a unit vector <p e 34? with || J1""101| = 1 From Lemma 2, we know

that </>, T(f>, , T n ~ x 0 is a basis for 2tf Repeating the argument of the first paragraph

of the proof of Lemma 3, we conclude that rank(7— T*T) = 1.

Observe that, if T belongs to a finite-dimensional subalgebra of @}(3ff), then the spectrum of T coincides with the set of eigenvalues For such a T with || T\\ = 1, Thas

no eigenvalue of modulus 1 if and only if its spectral radius is less than 1 which occurs

if and only if ||!T*|| < 1 for some k.

A CHARACTERIZATION OF PICK BODIES 3 2 5

on a Hilbert space tf with dim(j^) < « If there exists Tesi with \\P\\ = 1 for

0 <y < n and if T has no eigenvalue of modulus 1, then si is an n-dimensional Pick

algebra with Pick generator T.

Proof First we show that we may assume that there exists a unit vector <j> e 2tf

with ir""1^!! = 1 By the Gelfand-Naimark-Segal Theorem [1, Corollary to

Theorem 1.7.2], there exists a Hilbert space «#", a unit vector y/eJtf", and a representation n:@(J{?)->#(JT) with WniT^y/W = WT^W = 1 Define

by T(R) = R © n(R) for Resi Then T is an isometric algebra homomorphism, x{T) satisfies all of the original hypotheses, and \\x{T) n ~\Q © ^)|| = ||0 © y/\\ = 1.

The subspace X = {R(j) \ R e jaf} is invariant under d and has dimension at most n The algebra J / = st\ x and the operator f = T\^ satisfy the hypotheses of Lemma 8 and Proposition 2; so si is an n-dimensional Pick algebra with Pick generator f Since the restriction homomorphism Q>:si -^ si satisfies the conditions of Lemma 7 with T in the role of x, <X> is an isometric isomorphism, and the proof is complete When si is semisimple, the above theorem shows that 2^ is a Pick body So, we

are led to a strengthened form of Theorem 1 Note that z( ^ z^ if/ # j:is a consequence

of the theorem, not an assumption This follows since, by the conclusion of Theorem

2, T must have n distinct eigenvalues.

z = (z15 ,zn) with \zt \ < 1 for each j and with z n ~ x edK Then K is a Pick body.

We now identify all Pick generators for a Pick algebra Let Aut(A) denote the set

of conformal automorphisms of the unit disk

generator for si if and only if x = [y/]for ^eAut(A) In particular, if x x , x 2 are both Pick generators for si, then x 2 = y/(x^) for some

Proof Let x = [a] be a Pick generator for A/J; so ||[fl]|| < 1 Observe that since

2, the coset x cannot contain a constant function We write / = bA for a finite Blaschke product b.

Let y = [y/ 0 ] where y/ 0 = idA, and note that ||_y|| ^ 1 Select polynomials g n with

||gj| ^ 1 so that g n (x)-*y in A/J, and select f n e[a] with ||/J| < 1 + 1/n Using a

normal families argument, we obtain /, g, h, keH^ so that ||/|| < 1, ||g|| < 1,

f=a + bh, and gof= y/ 0 + bk Clearly / cannot be constant; so the composition

T = go/is well defined as a holomorphic map on int(A) with values in A Since T leaves

fixed the zeros of b and has derivative 1 at each non-simple zero, and since the sum

of the multiplicities of the zeros of b is at least 2, Schwarz's Lemma implies that

T = y/ 0 Hence, / e Aut(A) and x = [f], as desired.

Since, for ^eAut(A), {po y/\p is a polynomial, \\p\\ < 1} is a dense subset of the closed unit ball of A, the converse is clear.

To prove the last statement of the lemma, suppose that x lf x 2 are both Pick

generators Then, x x = [y/^ and x 2 = [^2] for y/ x , ^2eAut(A) So, x 2 = ^(xx) for

y/ = y/ o y/^ 1 e Aut(A).