. Hermitian operators and isometries on Banach algebras of continuous maps with values in unital C∗-algebras .

Osamu Hatori∗ (Niigata University) ∗ Supported by JSPS KAKENHI Grant Numbers JP16K05172, JP15K04921 «Joint work with Shiho Oi»

RTOTA 3 May, 2018 University of Memphis

1/ 25 General Problem . Problem . Ej : Banach space, C(Xj , Ej ) ⊃ Aj with a Banach space norm When is it true that every surjective isometrya

U : A1 → A2 will be of the form (canonical form) [ ] (Uf )(x) = Vx (f (φ(x)) , x ∈ X2 for continuous (with respect to SOT) map x 7→ Vx from X2 to the space of surjective isometries on E1 to E2, and a homeomorphism φ?

aIn this talk an isometry on a complex Banach space always means a .complex-linear isometry. 2/ 25 The origin of the problem

. The Banach-Stone theorem (1932, 1937 and later) . Suppose that U : C(X1, C) → C(X2, C) is a surjective isometry. ⇐⇒ U(1) is a continous function with the unit modulus

∃φ : X2 → X1 : homeomorphism s.t.

U(f )(x) = U(1)(x)f (φ(x)), x ∈ X2 ∈ C .for every f C(X1, ).

This canonical form is also called a weighted composition operator.

3/ 25 After the Banach-Stone theorem:

. Several surjective isometries are of weighted composition operators Lp spaces (Isometries between Lp[0, 1] have been already studied by Banach in his book, Lamperti (1958) and several variations) Kadison’s theorem for a unital C∗-algebra 1951,1952 (later non-unital case by Paterson and Sinclair 1972) Nagasawa’s theorem for a uniform algebra (simple counterexample for non-unital case) Spaces of analytic functions (de Leeuw-Rudin-Wermer for H1 and for a uniform algebra (1960), Forelli for Hp (1964), and ··· ) Algebras of Lipschitz functions (de Leeuw(1961) and ········· ) ···

4/ 25 Not always of a canonical form.

An example1 of Lip({a, b, c}) with ∥ · ∥ = max{∥ · ∥∞, L(·)}

The Wiener algebra ∑ ∞ ˆ W (T) = {f ∈ C(T): ∥f ∥W = −∞ |f (n)| < ∞}, and it’s analytic part ˆ WA(T) = {f ∈ W (T): f (n) = 0, ∀n < 0}. are isometric by ∑ ∑ ∑ U( ˆf (n)einθ)) = ˆf (ℓ(n))einθ, ˆf (n)einθ ∈ W (T) n∈Z n∈Z n∈Z

for any bijection ℓ : Z → Z+.

· · ·· and many 1N.Weaver, Isometries of noncompact Lipschitz spaces, CMB (1995) 5/ 25 Not always of a canonical form.

An example1 of Lip({a, b, c}) with ∥ · ∥ = max{∥ · ∥∞, L(·)}

The Wiener algebra ∑ ∞ ˆ W (T) = {f ∈ C(T): ∥f ∥W = −∞ |f (n)| < ∞}, and it’s analytic part ˆ WA(T) = {f ∈ W (T): f (n) = 0, ∀n < 0}. are isometric by ∑ ∑ ∑ U( ˆf (n)einθ)) = ˆf (ℓ(n))einθ, ˆf (n)einθ ∈ W (T) n∈Z n∈Z n∈Z

for any bijection ℓ : Z → Z+.

· · ·· and many 1N.Weaver, Isometries of noncompact Lipschitz spaces, CMB (1995) 5/ 25 Not always of a canonical form.

An example1 of Lip({a, b, c}) with ∥ · ∥ = max{∥ · ∥∞, L(·)}

The Wiener algebra ∑ ∞ ˆ W (T) = {f ∈ C(T): ∥f ∥W = −∞ |f (n)| < ∞}, and it’s analytic part ˆ WA(T) = {f ∈ W (T): f (n) = 0, ∀n < 0}. are isometric by ∑ ∑ ∑ U( ˆf (n)einθ)) = ˆf (ℓ(n))einθ, ˆf (n)einθ ∈ W (T) n∈Z n∈Z n∈Z

for any bijection ℓ : Z → Z+.

· · ·· and many 1N.Weaver, Isometries of noncompact Lipschitz spaces, CMB (1995) 5/ 25 Not always of a canonical form.

An example1 of Lip({a, b, c}) with ∥ · ∥ = max{∥ · ∥∞, L(·)}

The Wiener algebra ∑ ∞ ˆ W (T) = {f ∈ C(T): ∥f ∥W = −∞ |f (n)| < ∞}, and it’s analytic part ˆ WA(T) = {f ∈ W (T): f (n) = 0, ∀n < 0}. are isometric by ∑ ∑ ∑ U( ˆf (n)einθ)) = ˆf (ℓ(n))einθ, ˆf (n)einθ ∈ W (T) n∈Z n∈Z n∈Z

for any bijection ℓ : Z → Z+.

· · ·· and many 1N.Weaver, Isometries of noncompact Lipschitz spaces, CMB (1995) 5/ 25 A problem on isometries between spaces of vector-valued maps

e C(Xj , Ej ) ⊃ Bj with a Banach space norm

The problem is : f f If U : B1 → B2 is a surjective isometry, then

E1 ∼ E2? No!

Is X1 homeomorphic to X2? No! What is the form of U? ??? ..... Hence the problem is interesting for me.

6/ 25 A problem on isometries between spaces of vector-valued maps

e C(Xj , Ej ) ⊃ Bj with a Banach space norm

The problem is : f f If U : B1 → B2 is a surjective isometry, then

E1 ∼ E2? No!

Is X1 homeomorphic to X2? No! What is the form of U? ??? ..... Hence the problem is interesting for me.

6/ 25 A problem on isometries between spaces of vector-valued maps

e C(Xj , Ej ) ⊃ Bj with a Banach space norm

The problem is : f f If U : B1 → B2 is a surjective isometry, then

E1 ∼ E2? No!

Is X1 homeomorphic to X2? No! What is the form of U? ??? ..... Hence the problem is interesting for me.

6/ 25 A problem on isometries between spaces of vector-valued maps

e C(Xj , Ej ) ⊃ Bj with a Banach space norm

The problem is : f f If U : B1 → B2 is a surjective isometry, then

E1 ∼ E2? No!

Is X1 homeomorphic to X2? No! What is the form of U? ??? ..... Hence the problem is interesting for me.

6/ 25 A problem on isometries between spaces of vector-valued maps

e C(Xj , Ej ) ⊃ Bj with a Banach space norm

The problem is : f f If U : B1 → B2 is a surjective isometry, then

E1 ∼ E2? No!

Is X1 homeomorphic to X2? No! What is the form of U? ??? ..... Hence the problem is interesting for me.

6/ 25 A problem on isometries between spaces of vector-valued maps

e C(Xj , Ej ) ⊃ Bj with a Banach space norm

The problem is : f f If U : B1 → B2 is a surjective isometry, then

E1 ∼ E2? No!

Is X1 homeomorphic to X2? No! What is the form of U? ??? ..... Hence the problem is interesting for me.

6/ 25 In this talk :

E = a unital C∗-algebra, C(X, C), the algebra of marices

∥ − ∥ { ∈ F(x) F(y) E ∞} Lip(K , E) = F C(K , E): L(F) = supx≠ y d(x,y) < with ∥F∥∞ + L(F) ′ C1(K, E) for K = [0, 1], T with ∥F∥∞ + ∥F ∥∞ A ⊗ E for a uniform algebra A with the supremum norm ··· with the Banach algebra norm

Compair : ∥ · ∥∞ + L(·) with max{∥ · ∥∞, L(·)} for E = C or several Banach spaces Difficulty comes from: the extreme points of the dual space?

7/ 25 In this talk :

E = a unital C∗-algebra, C(X, C), the algebra of marices

∥ − ∥ { ∈ F(x) F(y) E ∞} Lip(K , E) = F C(K , E): L(F) = supx≠ y d(x,y) < with ∥F∥∞ + L(F) ′ C1(K, E) for K = [0, 1], T with ∥F∥∞ + ∥F ∥∞ A ⊗ E for a uniform algebra A with the supremum norm ··· with the Banach algebra norm

Compair : ∥ · ∥∞ + L(·) with max{∥ · ∥∞, L(·)} for E = C or several Banach spaces Difficulty comes from: the extreme points of the dual space?

7/ 25 In this talk :

E = a unital C∗-algebra, C(X, C), the algebra of marices

∥ − ∥ { ∈ F(x) F(y) E ∞} Lip(K , E) = F C(K , E): L(F) = supx≠ y d(x,y) < with ∥F∥∞ + L(F) ′ C1(K, E) for K = [0, 1], T with ∥F∥∞ + ∥F ∥∞ A ⊗ E for a uniform algebra A with the supremum norm ··· with the Banach algebra norm

Compair : ∥ · ∥∞ + L(·) with max{∥ · ∥∞, L(·)} for E = C or several Banach spaces Difficulty comes from: the extreme points of the dual space?

7/ 25 In this talk :

E = a unital C∗-algebra, C(X, C), the algebra of marices

∥ − ∥ { ∈ F(x) F(y) E ∞} Lip(K , E) = F C(K , E): L(F) = supx≠ y d(x,y) < with ∥F∥∞ + L(F) ′ C1(K, E) for K = [0, 1], T with ∥F∥∞ + ∥F ∥∞ A ⊗ E for a uniform algebra A with the supremum norm ··· with the Banach algebra norm

Compair : ∥ · ∥∞ + L(·) with max{∥ · ∥∞, L(·)} for E = C or several Banach spaces Difficulty comes from: the extreme points of the dual space?

7/ 25 In this talk :

E = a unital C∗-algebra, C(X, C), the algebra of marices

∥ − ∥ { ∈ F(x) F(y) E ∞} Lip(K , E) = F C(K , E): L(F) = supx≠ y d(x,y) < with ∥F∥∞ + L(F) ′ C1(K, E) for K = [0, 1], T with ∥F∥∞ + ∥F ∥∞ A ⊗ E for a uniform algebra A with the supremum norm ··· with the Banach algebra norm

Compair : ∥ · ∥∞ + L(·) with max{∥ · ∥∞, L(·)} for E = C or several Banach spaces Difficulty comes from: the extreme points of the dual space?

7/ 25 In this talk :

E = a unital C∗-algebra, C(X, C), the algebra of marices

∥ − ∥ { ∈ F(x) F(y) E ∞} Lip(K , E) = F C(K , E): L(F) = supx≠ y d(x,y) < with ∥F∥∞ + L(F) ′ C1(K, E) for K = [0, 1], T with ∥F∥∞ + ∥F ∥∞ A ⊗ E for a uniform algebra A with the supremum norm ··· with the Banach algebra norm

Compair : ∥ · ∥∞ + L(·) with max{∥ · ∥∞, L(·)} for E = C or several Banach spaces Difficulty comes from: the extreme points of the dual space?

7/ 25 When E = C(Y )

By identifying C(K , C(Y )) with C(K × Y ) by the correspondence

F ∈ C(K , C(Y )) −→ F(x) ∈ C(Y ), ∀x ∈ K [ ] −→ (x, y) 7→ (F(x))(y) ∈ C(K × Y ); we may assume that

Lip(X, C(Y )) ⊂ C(X × Y ), C1(K, C(Y )) ⊂ C(K × Y ),

8/ 25 A surjective isometry on Lip(K , C(Y )) is a weighted composition operator

. Theorem (H. and Oi) . U : Lip(K1, C(Y1)) → Lip(K2, C(Y2)) is a surjective isometry with respect to ∥ · ∥ = ∥ · ∥∞ + L(·) ⇐⇒

∃f ∈ C(Y2) : unimodular function

∃φ : K2 × Y2 → K1 : continuous map s.t. φ(·, y): K2 → K1 is a

surjective isometry for each y ∈ Y2

∃τ : Y2 → Y1 : homeomorphism

∈ × . (UF)(x, y) = f (y)F(φ(x, y), τ(y)), (x, y) K2 Y2

9/ 25 A surjective isometry on Lip(K , C(Y )) is a weighted composition operator . Theorem (H. and Oi) . U : Lip(K1, C(Y1)) → Lip(K2, C(Y2)) is a surjective isometry with respect to ∥ · ∥ = ∥ · ∥∞ + L(·) ⇐⇒

∃f ∈ C(Y2) : unimodular function

∃φ : K2 × Y2 → K1 : continuous map s.t. φ(·, y): K2 → K1 is a

surjective isometry for each y ∈ Y2

∃τ : Y2 → Y1 : homeomorphism

∈ × . (UF)(x, y) = f (y)F(φ(x, y), τ(y)), (x, y) K2 Y2

Note that the composition part has a peculiar form of type BJ.

10/ 25 One can check easily the sufficiency for a surjective isometry. To prove the neccessity;

(1) Prove U1 = 1 ⊗ f for an f ∈ C(Y2) with |f | = 1 on Y2: A proof2 is much involved with a measure theoretic argument applying Choquet theory. (2) Applying a theorem of Jarosz3 to show that (1 ⊗ ¯f )U is an algebra isomorphism, hence it is a composition operator; hence U = (1 ⊗ f )× composition operator

(3) Show that the composition operator above is of a peculiar form by applying Lumer’s method.

2O.Hatori and S.Oi, Isometries on Banach algebras of vector-valued maps, to appear 3Isometries in semisimple, commutative Banach algebra, PAMS 1983 11/ 25 One can check easily the sufficiency for a surjective isometry. To prove the neccessity;

(1) Prove U1 = 1 ⊗ f for an f ∈ C(Y2) with |f | = 1 on Y2: A proof2 is much involved with a measure theoretic argument applying Choquet theory. (2) Applying a theorem of Jarosz3 to show that (1 ⊗ ¯f )U is an algebra isomorphism, hence it is a composition operator; hence U = (1 ⊗ f )× composition operator

(3) Show that the composition operator above is of a peculiar form by applying Lumer’s method.

2O.Hatori and S.Oi, Isometries on Banach algebras of vector-valued maps, to appear 3Isometries in semisimple, commutative Banach algebra, PAMS 1983 11/ 25 One can check easily the sufficiency for a surjective isometry. To prove the neccessity;

(1) Prove U1 = 1 ⊗ f for an f ∈ C(Y2) with |f | = 1 on Y2: A proof2 is much involved with a measure theoretic argument applying Choquet theory. (2) Applying a theorem of Jarosz3 to show that (1 ⊗ ¯f )U is an algebra isomorphism, hence it is a composition operator; hence U = (1 ⊗ f )× composition operator

(3) Show that the composition operator above is of a peculiar form by applying Lumer’s method.

2O.Hatori and S.Oi, Isometries on Banach algebras of vector-valued maps, to appear 3Isometries in semisimple, commutative Banach algebra, PAMS 1983 11/ 25 One can check easily the sufficiency for a surjective isometry. To prove the neccessity;

(1) Prove U1 = 1 ⊗ f for an f ∈ C(Y2) with |f | = 1 on Y2: A proof2 is much involved with a measure theoretic argument applying Choquet theory. (2) Applying a theorem of Jarosz3 to show that (1 ⊗ ¯f )U is an algebra isomorphism, hence it is a composition operator; hence U = (1 ⊗ f )× composition operator

(3) Show that the composition operator above is of a peculiar form by applying Lumer’s method.

2O.Hatori and S.Oi, Isometries on Banach algebras of vector-valued maps, to appear 3Isometries in semisimple, commutative Banach algebra, PAMS 1983 11/ 25 Lumer’s method

Suppose that H ∈ B(B), B is a complex Banach space.

H is a Hermitian operator ⇐⇒ ∥ exp(itH)∥ = 1, ∀t ∈ R

Lumer’s method involves the notion of Hermitian operators and the fact that UHU−1 must be Hermitian if H is Hermitian and U is a surjective isometry: if we know the properties of Hermitian operators, then we get an information on surjective isometries.

12/ 25 Lumer’s method

Suppose that H ∈ B(B), B is a complex Banach space.

H is a Hermitian operator ⇐⇒ ∥ exp(itH)∥ = 1, ∀t ∈ R

Lumer’s method involves the notion of Hermitian operators and the fact that UHU−1 must be Hermitian if H is Hermitian and U is a surjective isometry: if we know the properties of Hermitian operators, then we get an information on surjective isometries.

12/ 25 By a unified approach for Lip(K , C(Y )); a similar result for (1) C1([0, 1], C(Y ))

. Theorem (H. and Oi) . 1 1 Let U : C ([0, 1], C(Y1)) → C ([0, 1], C(Y2)). ′ Then U is a surjective isometry with respect to ∥F∥ = ∥F∥∞ + ∥F ∥∞ ⇐⇒

∃f ∈ C(Y2) : unimodular function

∃φ :[0, 1] × Y2 → [0, 1] : continuous map s.t. for each y ∈ Y2, φ(t, y) = t, t ∈ [0, 1] or φ(t, y) = 1 − t, t ∈ [0, 1]

∃τ : Y2 → Y1 : homeomorphism

∈ × . (UF)(t, y) = f (y)F(φ(t, y), τ(y)), (t, y) [0, 1] Y2

13/ 25 By a unified approach for Lip(K , C(Y )); a similar result for (2) C1(T, C(Y ))

. Theorem (H. and Oi) . 1 1 Let U : C (T, C(Y1)) → C (T, C(Y2)). ′ Then U is a surjective isometry with respect to ∥F∥ = ∥F∥∞ + ∥F ∥∞ ⇐⇒

∃f ∈ C(Y2) : unimodular function

∃φ : T × Y2 → T : continuous map and a continuous map u : Y2 → T

s.t. for each y ∈ Y2 φ(z, y) = u(y)z, z ∈ T or φ(z, y) = u(y)z,¯ z ∈ T

∃τ : Y2 → Y1 : homeomorphism

∈ T × . (UF)(z, y) = f (y)F(φ(z, y), τ(y)), (z, y) Y2

14/ 25 Non commutative case

How about a surjective isometry

U : Lip(K1, E1) → Lip(K2, E2)

∗ for unital C -algebras Ej ’s.

15/ 25 E = Mn(C) : the algebra of matrices . Theorem (Oi 1) . U : Lip(K1, Mn(C)) → Lip(K2, Mn(C)) is a surjective isometry with respect to ∥ · ∥ = ∥ · ∥∞ + L(·) such that U(1) = 1 ⇐⇒ ∃V : unitary matrix

∃φ : K2 → K1 : surjective isometry s.t. ∗ [U(F)](x) = VF(φ(x))V , x ∈ K2, F ∈ Lip(K1, Mn(C)) or t ∗ ∈ ∈ C . [U(F)](x) = V [F(φ(x))] V , x K2, F Lip(K1, Mn( ))

4 4S. Oi, Hermitian operators and isomeries on algebras of matrix-valued Lipschitz maps, submitted 16/ 25 A proof of Theorem (Oi 1) depends on Theorem (Oi 2) about Hermitian operators on Lip(K , Mn(C)) and Lumer’s method. . Theorem (Oi 2) . Let T : Lip(K , Mn(C)) → Lip(K , Mn(C)). T is a Hermitian operator ⇐⇒ ∃H : Hermitian matrix,

∃D: ∗-derivation on Mn(C) s.t. b T = M1⊗H + iD, where M1⊗H is the multiplication operator by 1 ⊗ H and b D : Lip(K , Mn(C)) → Lip(K , Mn(C)) is defined by

b ∈ ∈ C . [D(F)](x) = D(F(x)), x K , F Lip(K , Mn( )).

17/ 25 Problems

1 1 U : C (K, Mn(C)) → C (K, Mn(C)) −→ a similar statement to Theorem (Oi) is expected

U : Lip(K1, Mn(C)) → Lip(K2, Mn(C)) without assuming U(1) = 1 −→ ??? ∗ In general, B(H) or unital C -algebras instead of Mn(C) with or without assuming U(1) = 1 −→ ???

18/ 25 A ⊗ B

A : uniform algebra on the maximal ideal space M B : Banach space A ⊗ B is the uniform closure in C(M, B) of the algebraic tensor product A ⊗ B. A ⊗ B is a Banach space with the supremum norm. If A = C(X), then the decomposition of unity asserts that

A ⊗ B = C(X, B).

19/ 25 Let B be a complex-Banach space. H(B) denotes the space of all Hermitian operators on B. . Theorem (H. and Oi 1) . Suppose that T : A ⊗ B → A ⊗ B is a Hermitian operator. Then ∃ϕ : M → H(B) with respect to the SOT s.t. ∈ ∈ ⊗ . [T (F)](x) = [ϕ(x)](F(x)), x M, F A B.

Theorem (H. and Oi 1) is a generalization with an alternative proof for the case where A = C(X) by Fleming and Jamison5.

5R.J.Fleming and J.E.Jamison, Hermitian operators on C(X, E) and the Banach-Stone theorem, Math. Z. (1980) 20/ 25 . Theorem (H. and Oi 2) . Suppose that Ej is a factor (the center is trivial). Let Iso(E1, E2) be the group of all Jordan ∗-isomorphisms from E1 onto E2. Suppose that

U : A1 ⊗ E1 → A2 ⊗ E2 is a surjective isometry such that U(1) = 1. =⇒

∃φ : M2 → M1 : homeomorphism

∃V : M2 → Iso(E1, E2) : continuous with respect to SOT s.t. ∈ ∈ ⊗ . [U(F)](x) = V (x)[F(φ(x))], x M2, F A1 E1.

Note that a Jordan ∗-isomorphism between unital C∗-algebras is a surjective isometry. A proof of Theorem (H. and Oi 2) is an application of Theorem (H. and Oi 1) for a unital C∗-algebra and Lumer’s method.

21/ 25 Problem

U : A1 ⊗ E1 → A2 ⊗ E2 without assuming U(1) = 1 −→ ???

∗ If Aj = C(Xj ), then Aj ⊗ Ej = C(Xj × Ej ) is a unital C -algebra. By the theorem of Kadison6 we have that U(1) is a unitary element in

C(X2, E2). Hence we have...

6R.V.Kadison, Isometries on operator algebras, Ann. Math. (1951) 22/ 25 . Corollary . U : C(X1, E1) → C(X2, E2) is a surjective isometry ⇐⇒

∃φ : X2 → X1 : a homeomorphism

∃V : X2 → Iso(E1, E2) : continuous with respect to SOT

∃u ∈ C(X2, E2) : a unitary element s.t. ∈ ∈ . [U(F)](x) = u(x)V (x)[F(φ(x))], x M2, F C(X1, E1).

It is not the case when Ej is not a factor. There is a compact Hausdorff spaces X1 is not homeomorphic to X2, but X1 × [0, 1] and X2 × [0, 1] are homeomorphic to each other; hence C(X1, C([0, 1])) and

C(X2, C([0, 1])) are isometrically isomorphic.

23/ 25 . Corollary . U : C(X1, E1) → C(X2, E2) is a surjective isometry ⇐⇒

∃φ : X2 → X1 : a homeomorphism

∃V : X2 → Iso(E1, E2) : continuous with respect to SOT

∃u ∈ C(X2, E2) : a unitary element s.t. ∈ ∈ . [U(F)](x) = u(x)V (x)[F(φ(x))], x M2, F C(X1, E1).

It is not the case when Ej is not a factor. There is a compact Hausdorff spaces X1 is not homeomorphic to X2, but X1 × [0, 1] and X2 × [0, 1] are homeomorphic to each other; hence C(X1, C([0, 1])) and

C(X2, C([0, 1])) are isometrically isomorphic.

23/ 25 We say that a Banach space B have a Banach-Stone property if for any pair X1 and X2 of compact Hausdorff spaces and an isometry U of

C(X1, B) onto C(X2, B) there exists a homeomorphism φ : X2 → X1, a continuous function x 7→ V (x) from X2 into the space of bounded linear operators on B (given its strong topology) such that for each x, V (x) is an isometry on B and

[U(F)](x) = V (x)F(φ(x)), x ∈ X2, F ∈ C(X1, B). Behrends7 proved that if the centralizer of B is of one dimensional, then B has a Banach-Stone property. Fleming and Jamison8 also proved the reuslt by emphasizing that B is a complex-Banach space. Corollary gives a special case with an alternative proof. 7E. Behrends, M-structures and the Banach-Stone theorem, LNM vol.736 (1979), On the Banach-Stone theorem, Math. Ann. (1978) 8R.J.Fleming and J.E.Jamison, Hermitian operators on C(X, E) and the Banach-Stone theorem, Math. Z. (1980) 24/ 25 Thank you!

25/ 25