PROCEEDINGS OF THE AMERICAN MATHEMATICAL SOCIETY Volume 136, Number 2, February 2008, Pages 605–611 S 0002-9939(07)09033-8 Article electronically published on November 1, 2007
FUNCTION REPRESENTATION OF A NONCOMMUTATIVE UNIFORM ALGEBRA
KRZYSZTOF JAROSZ
(Communicated by N. Tomczak-Jaegermann)
Abstract. We construct a Gelfand type representation of a real noncommu- tative Banach algebra A satisfying f 2 = f 2, for all f ∈ A.
1. Introduction A uniform algebra A is a Banach algebra such that (1.1) f 2 = f 2 , for all f ∈ A. The commutative, complex uniform algebras constitute the most classical class of Banach algebras, one which has been intensely studied for decades. It is well known that any such algebra is isometrically isomorphic with a subalgebra of CC (X), the algebra of all continuous complex-valued functions defined on a compact set X and equipped with the sup-norm topology. Hirschfeld and Zelazko˙ [6] proved that the assumption that the algebra is commutative is superfluous—commutativity already follows from (1.1). Algebras of analytic functions serve as standard examples of complex uniform algebras. Commutative, real uniform algebras have also been studied for years [9]. Any such algebra A is isometrically isomorphic with a real subalgebra of CC (X)forsome compact set X;furthermore,X can often be divided into three parts X1,X2, and X3
such that A|X1 is a complex uniform algebra, A|X2 consists of complex conjugates
of the functions from A|X1 ,andA|X3 is equal to CR (X3). Any commutative, real uniform algebra can be naturally identified with a real subalgebra of df CC (X, τ) = f ∈ CC (X):f (τ (x)) = f (x)forx ∈ X ,
2 where τ : X → X is a surjective homeomorphism with τ = idX .Inthecase of real uniform algebras the condition (1.1) no longer implies commutativity; the four-dimensional algebra of quaternions serves as the simplest counterexample. There has been very little study of noncommutative real uniform algebras. Only recently it was proved [1] that any such algebra is isomorphic with a real subalgebra of CH (X), the algebra of all continuous functions defined on a compact set X and taking values in the field H of quaternions. There is no simple theory of analytic functions of quaternion variables. The standard definition
Received by the editors October 31, 2005 and, in revised form, November 24, 2006. 2000 Mathematics Subject Classification. Primary 46H20, 46H05. Key words and phrases. Uniform algebra, function algebra, Banach algebra, quaternions.
c 2007 American Mathematical Society Reverts to public domain 28 years from publication 605
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−1 f (a) = limq→0,q∈H (f (a + q) − f (a)) q is not useful since only linear functions are differentiable (there is an interesting alternative approach to differentiability of such functions, see for example [10, 12], but that approach is however not appli- cable for our purpose). Hence there is no obvious example of a noncommutative real uniform algebra other than the direct sum of the entire algebra CH (X)anda commutative real uniform algebra. We prove here that in fact there is no nontrivial example and any such algebra is roughly equal to such a direct sum; we also obtain a Gelfand type representation of such algebras analogous to CC (X, τ). All algebras are assumed to contain a unit.
2. The results We first need an easily verifiable elementary description of isomorphisms of the four-dimensional real algebra H. Proposition 1. A surjective map T : H → H is linear (over the field of real numbers) and multiplicative if and only if ⎡ ⎤ ⎡ a ⎤ 10 ⎢ b ⎥ (2.1) T (a, b, c, d)= ⎢ ⎥ ,fora + bi + cj + dk ∈ H, 0M ⎣ ⎣ c ⎦ ⎦ d where M is an isometry of 3-dimensional Euclidean space R3 preserving the ori- entation of that space. The space of all such maps forms a 3-dimensional compact connected group, which we will denote by M. We should also notice that if A is a real algebra and F : A → H is a multiplicative functional such that dim (F (A)) = 2, then F (A) is isomorphic with C and the set {T ◦ F : T ∈M}can be identified with the unit sphere in R3. Indeed to define an isometry T on F (A) we just have to decide which unit vector of the form bi+cj+dk is to be the image of the imaginary unit of F (A) .
Definition 1. We say that A ⊂ C (X) separates the points of X if for any x1,x2 ∈ X there is f ∈ A such that f (x1) = f (x2). We say that A ⊂ C (X) strongly separates the points of X if for any x1,x2 ∈ X there is f ∈ A such that f (x1) = f (x2)=0. Definition 2. A real algebra A is fully noncommutative if any nonzero linear and multiplicative functional F : A → H is surjective. Notice that A is not fully noncommutative iff {F (a):a ∈ A} is isomorphic with R or with C for some H-valued multiplicative functional F on A.
Definition 3. Let Φ : M→Hom(X):Φ(T )=ΦT be a homomorphism of the group M onto a group of homeomorphisms of a compact set X. We define df CH (X, Φ) = {f ∈ CH (X):f ◦ ΦT (x)=T ◦ f (x)forx ∈ X, T ∈M}. For a commutative real or complex uniform algebra A the standard way to rep- resent A as a subalgebra of CC (X) is to consider the space X = M (A)ofall complex-valued linear-multiplicative functionals and use the Gelfand transforma- tion: df (2.2) A a −→ aˆ ∈ C (X):ˆa (x) = x (a) .
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For a real algebra A and for any x ∈ M (A), also the complex conjugatex ¯ = τ (x) of x is an element of M (A), hence the representation of A as a subalgebra of ∧ CC (X, τ). An important feature of such a representation is that maps the set of invertible elements of A exactly onto the subset of Aˆ consisting of functions that do not vanish on X. We will construct a similar representation for a noncommutative uniform algebra. Theorem 1. Assume A is a real uniform algebra. Then there is a compact set X and an isomorphism Φ:M→Hom(X) such that A is isometrically isomorphic with a subalgebra Aˆ of CH (X, Φ).Furthermorea ∈ A is invertible if and only if the corresponding element aˆ ∈ Aˆ does not vanish on X. If A is fully noncommutative, then Aˆ = CH (X, Φ) . We have obvious candidates for X, Φ, and the map ∧ :
df X = MH (A) = quaternion-valued real-linear and multiplicative functionals df df (2.3) ΦT (x) = T ◦ x, aˆ (x) = x (a) . ∧ It is clear that the map is a homomorphism of A into CH (X, Φ). We need to show that it is an isometry, that it preserves the set of noninvertible elements, and that Aˆ = CH (X, Φ) in the fully noncommutative case. So without loss of generality we may assume that A is separable, or equivalently countably generated. Before proving the theorem it will be useful to notice that for x ∈ X we may encounter exactly three distinct cases: the set {T ◦ x : T ∈M}may be • equal to the singleton {x} if x (A)=R,or • homeomorphic with the unit sphere in R3 if x (A) is a 2-dimensional com- mutative subalgebra of H isomorphic with C,or • homeomorphic with M if x (A)=H.
Consequently the set MH (A) can be divided into three parts X1,X2,X3:
• X1 = {x ∈ X : T ◦ x = x for all T ∈M}= {x ∈ X : x (A)=R};thatset is equal to the subset of X consisting of the points where all the functions from Aˆ are real valued; • X2 = {x ∈ X :dimx (A)=2}; that set is a union of disjoint copies of the 3 unit sphere in R , and the algebra Aˆ restricted to X1 ∪ X2 is commutative; • X3 = {x ∈ X : x (A)=H}; that set is a union of disjoint copies of M and the algebra Aˆ restricted to X3 is fully noncommutative. That is analogous to the commutative case when a real-linear multiplicative functional x takes only real values if τ (x)=x, or takes all complex values if τ (x) = x and the set M (A) can be divided into just two parts. Put df a ∈ A :ˆa is constant on X1 ∪ X2 AR = and on {T ◦ x : T ∈M} for each x ∈ X3 and AˆR = {aˆ ∈ CR (X):a ∈ A} . We prove the theorem in several short steps; the steps also provide a much more detailed description of the representation of A: (1) the map ∧ defined by (2.3) is injective and A is semisimple;
License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use 608 KRZYSZTOF JAROSZ f ∈ CR (X):f is constant on X1 ∪ X2 (2) AˆR = ;furthermore a = and on {T ◦ x : T ∈M} for each x ∈ X3 aˆ for any a ∈ AR; df ∪ ˆ ˆ| ∪ (3) X 1 X2 is a weak peak set for A, so any function from A X1 X2 = f| ∪ ∈ C (X ∪ X ):f ∈ Aˆ has a norm-preserving extension to a X1 X2 1 2 ˆ ˆ ∪ function in A; M A|X1∪X2 = X1 X2; ˆ (4) {f ∈ CH (X, Φ) : f (x)=0forx ∈ X1 ∪ X 2}⊂A; ˆ ∈ ∈ ˆ (5) A = f CH (X, Φ) : f|X1∪X2 A|X1∪X2 ; ∧ ˆ (6) the map is an isometry and A|X1∪X2 is a complete algebra; (7) any noninvertible element of A is contained in a kernel of a functional from MH (A); Aˆ = CH (X, Φ) if A is fully noncommutative. The first part follows from Theorem 3 of [1]. We provide an independent proof here for completeness. We will need the following special case of the main result of [2]. Theorem 2 (Aupetit-Zemanek). Let A be a real Banach algebra with unit. If lim n an = a for all a in A, then for every irreducible representation π : A → L (E), the algebra π (A) is isomorphic with its commutant Cπ in the algebra L (E) of all linear transformation on E.
Proof. Part 1. Since the commutant Cπ is a normed real division algebra ([3], p. 127) it is isomorphic with R, C, or H.LetX be the set of all irreducible representa- tions of A;forx ∈ X and a ∈ A puta ˆ (x)=x (a) ∈ H.Themap∧ is an isomorphism of A into the algebra of H-valued functions on X.Letσ be the strongest topology on X such that all the functionsa, ˆ a ∈ A are continuous. Assume (X, σ)isnot compact and let x0 be a point from βX\X; the operator A a → aˆ (x0) ∈ H is an irreducible representation on A,sox0 ∈ X, a contradiction. To show that ∧ is injective we need to show that for any 0 = a ∈ A there is an irreducible representation π with π (a) =0.Fix0 = a ∈ A and let Aa be the closed subalgebra of A generated by all the elements of the form q (a), where q is a rational function with real coefficients and with poles outside the spectrum of a.NoticethatAa is a commutative uniform algebra such that radAa = {0} . If −1 −1 b ∈ A ∩ Aa,thenb as given by a rational function is in Aa; that means that −1 ∩ −1 ∩ ⊂ { } ∈ A Aa = Aa . Hence by [11, p. 476], Aa radA radAa = 0 ,soa/radA and, since a was arbitrary, A is semisimple. Part 2. Put 2 R = (fg − gf) ∈ CH (X):f,g ∈ Aˆ .
For arbitrary quaternions wp = ap + bpi + cpj + dpk, p =1, 2wehave 2 2 (w1w2 − w2w1) =(2(b1c2 − b2c1) ij +2(b1d2 − b2d1) ik +2(c1d2 − c2d1) jk) − − − − 2 =4(( b1c2 b2c1) k (b1d2 b2d1) j +(c1d2 c2d1) i) 2 2 2 = −4 (b1c2 − b2c1) +(b1d2 − b2d1) +(c1d2 − c2d1)
∈ CR (X) , so R contains only real-valued functions. Let ∼ be an equivalence relation on X defined by x1 ∼ x2 iff h (x1)=h (x2) for all h ∈R,
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let X0 = X/∼ be the quotient space, and let π : X → X0 be the natural projection. For x ∈ X there is a function in R not vanishing at x if and only if x ∈ X3; furthermore on X3 we have x1 ∼ x2 if and only if ker x1 =kerx2, or equivalently if and only if {T ◦ x1 : T ∈M} = {T ◦ x2 : T ∈M}. By the Stone-Weierstrass Theorem AˆR can be identified with a dense subset of CR (X0). We shall show that in fact it is equal to the entire CR (X0). AR is a commutative real uniform algebra, so it is isometric with a closed subalge- bra of CR (M (AR)); X0 can be naturally identified with a closed subset of M (AR) . Assume X0 = M (AR), and let y0 ∈ M (AR) \X0. Since M (AR) is equipped with the weak * topology, there are a1, ..., an ∈ AR such thata ˆ1 (y0)=... =ˆan (y0)=0 {| | }≥ ∈ df 2 2 ∈ and max aˆj (y) : j =1, ..., n 1 for any y X0. Hencea ˆ0 =ˆa1 + ... +ˆan AR is strictly positive on X0 = X/∼, and since A is semisimple, a0 is an invertible element of AR ⊂ A contrary to the fact that it is in the kernel of the functional y0. The contradiction shows that X0 = M (AR) and consequently AR = CR (X0). n n Assume there is a ∈ AR such that a =1> aˆ .Since aˆ →0 while a =1 it follows that the range of ∧ is not complete; that contradiction shows that ∧ is an isometry on AR. Part 3. We recall that a subset K of the maximal ideal space of a function algebra is called a weak peak set iff for any open neighborhood U of K there is a function f in that algebra such that f (x)=1= f > |f (x)| for any x ∈ K and any x ∈/ U.SinceCR (X0) AR ⊂ A it follows that X1 ∪ X2 is a weak peak set. Assume a ∈ A with sup {|aˆ (x)| : x ∈ X1 ∪ X2} =1. Since M consists of isometries, 1 { ◦ ∈M} ∈ max{1,|aˆ|} is constant on the sets T x : T for x X3 as well as on the set ∪ 1 ∈ ˆ · 1 ∈ ˆ X1 X2. Hence max{1,|aˆ|} A soa ˆ max{1,|aˆ|} A; that function coincides with aˆ on X ∪ X and has norm 1. It is important to notice for further reference that 1 2 ∧ · 1 ≤ 1 since is an isometry on AR we have a max{1,|a|} a max{1,|a|} = a , · 1 so not only is a max{1,|a|} a norm-preserving extension of a with respect to the spectral norm but also with respect to the original norm of the algebra A. ∪ The maximal ideal space of A|X1∪X2 is equal to X1 X2 since any linear and
multiplicative functional on A|X1∪X2 gives a functional on A as well. Part 4. Fix x0 ∈ X3 and let f1,f2,f3,f4 ∈ A be such that f1 (x0)=1,f2 (x0)= i, f3 (x0)=j, f4 (x0)=k. Let U0 be an open neighborhood of x0 such that |fp (x) − fp (x0)| < 1/4forx ∈ U0 and p =1, 2, 3, 4. Let g0 ∈ CH (X, Φ) and df −1 assume the support of g0 is contained in V0 = π (π (U0)). For any x ∈ U0, the numbers f1 (x) ,f2 (x) ,f3 (x) ,f4 (x) can be seen as linearly independent vectors in 4 H R , so there are unique real-valued functions hp defined on U0 such that 4 (2.4) g0 (x)= hp (x) fp (x)forx ∈ U0. p=1 −1 Let x ∈ U0 and let x = T ◦ x be another point of π (π (x)), where T ∈M.By the definition of CH (X, Φ) we have 4 4 g0 (x )=T ◦ g0 (x)= hp (x) T ◦ fp (x)= hp (x) fp (x ) . p=1 p=1 −1 Hence the functions hp,p=1, 2, 3, 4 are constant on sets of the form π (π (x)) and consequently can be naturally extended to V0; furthermore, they can be extended
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to the entire set X by assigning value zero outside V0. Since the coefficients with respect to a fixed basis in R4 are continuous functions of a vector in R4, the functions hp,p=1, 2, 3, 4belongtoAR ⊂ A. Hence g0 ∈ A. −1 We proved that for any point x0 in X there is a neighborhood V0 of π (π (x)) −1 such that A| = CH (X, Φ) .LetV = π (π (U )) ,s ∈ S be an open cover of V0 |V0 s s X consisting of such sets. Since A is separable and X3 is a subset of its dual space equipped with the weak * topology, X3 is metrizable and consequently paracompact. Let s∈S αs = 1 be a locally finite partition of unity in CR (X3) subordinated to the ∈ ∈ cover π (Us) ,s S of X0 ([4], page 375). For any g CH (X, Φ) such that g (x)=0 ∈ ∪ ◦ ◦ ⊂ for x X1 X2,wehaveg (x)= s∈S αs π (x) g (x) where suppαs πg Vs,so g ∈ A. ∈ ∈ ˆ ∈ Part 5. Assume f CH (X, Φ) is such that f|X1∪X2 A|X1∪X2 .Letg A − ∈{ ∈ be such that g|X1∪X2 = f|X1∪X2 .Sincef g f CH(X, Φ) : f(x)=0for X1 ∪ X2} by the previous step, f = g +(f − g) ∈ A. Part 6. Put I = {a ∈ A :ˆa (x)=0forx ∈ X1 ∪ X2} and let B = A/I be the quotient algebra. Notice that B is a commutative Banach algebra with M (B)= X1 ∪ X2. Assume there is a + I ∈ B such that a + I = a > 2and aˆ ≤1. By 2 Part 3 we may assume that aˆ = aˆ|X1∪X2 . Suppose that a + I < 1sothat there is g ∈ I with a2 + g < 1. By Part 3 again we may assume thata ˆ2 +ˆg is smaller than 1 on X and consequently g = gˆ < 2. We have a2 + g < 1, g < 2, a2 =4, which is impossible. The above proves that 1 b2 ≥ b 2 , for all f ∈ B. 4 So B is isomorphic with a complete uniform algebra which must be equal to ˆ ∧ A|X1∪X2 ; by the same arguments as at the end of Part 3 we conclude that is in fact an isometry. ˆ ∪ Part 7. This is an obvious consequence of Part 5 since M A|X1∪X2 = X1 X2.
Corollary 1. Assume A is a fully noncommutative closed subalgebra of CH (Y ). Then A = CH (Y ) if and only if A strongly separates the points of Y . We do not assume here that A contains all constant functions; if we assume that the constant functions i, j, k are in A, then the above version of the Stone- Weierstrass Theorem easily follows from the following fact: f ∈ A implies that Re f =(f − ifi − jfj − kfk)/4 ∈ A ([7]). For a fully noncommutative real uniform algebra A, its maximal ideal space MH (A) consists of disjoint copies of M and is locally homeomorphic with a Carte- sian product of M with another set. Quite often we also have a global decomposi- tion: MH (A)=Y ×M. However, such a global decomposition cannot always be achieved, not even its analog in the commutative case. In the commutative case when A ⊂ CC (X, τ) and we can divide X into three parts X1,X2, and X3 such that
A|X1 is a complex algebra, A|X2 consists of complex conjugates of the functions from { ∈ } A|X1 ,andA|X3 is equal to CR (X3) . It is easy to select X3 = x X : τ (x)=x ; we may also easily select one point from each equivalence class x1 ∼ x2 iff τ (x1)= x2 to get X1 and we can form X2 from the remaining points. However we may not
be able to make the sets X1,X2 closed, so A|X1 ,A|X2 may not be uniform algebras;
License or copyright restrictions may apply to redistribution; see https://www.ams.org/journal-terms-of-use FUNCTION REPRESENTATION OF A NONCOMMUTATIVE UNIFORM ALGEBRA 611 1 it 1 see for example A = f ∈ CC S : f e = f ei(t+π) for 0 ≤ t ≤ π where S is a unit circle. This is the reason we represent a real commutative uniform algebra as a subalgebra of CC (X, τ) rather than, what may seem more appealing, a direct
sum of A|X1 ,A|X2 , and A|X3 . We proved that the algebra H has the following property: Any subalgebra A of CH (X)=CR (X) ⊗ H that strongly separates the points of X andsuchthat{f (x):x ∈ X} = H for all x ∈ X is equal to CH (X). We also know (Stone-Weierstrass Theorem) that the algebra R has the same property. It would be interesting to find other algebras with the same property. Acknowledgment The author would like to thank S. H. Kulkarni for his hospitality and for several stimulating discussions during the author’s visit at the Indian Institute of Technol- ogy. References
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Department of Mathematics and Statistics, Southern Illinois University, Edwards- ville, Illinois 62026-1653 E-mail address: [email protected] URL: http://www.siue.edu/~kjarosz
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