Zeitschrift für Analysis und ihre Anwendungen

Journal for Analysis and its Applications Volume 19 (2000), No. 1, 13–22

Notes on Modular Conjugations of von Neumann Factors

M. Leitz-Martini and M. Wollenberg

Abstract. In this paper we present some results about the characterization of modular conjugations of von Neumann algebras. Further, we show that hyperfinite factors of type II, III_{1}, and III_{λ} have algebraic conjugations which are not modular conjugations.

Keywords: Operator algebras, modular theory, modular conjugation

AMS subject classification: 46 L 10

1. Introduction

Let M be a von Neumann algebra on a separable Hilbert space H and let u be a cyclic and separating vector for M. Then the Tomita operator S, the modular operator Δ, and the modular conjugation J are defined by

S_{o}Mu = M^{∗}u, M ∈ M,  S = S̄_{o},  Δ = S^{∗}S,  S = JΔ^{\frac{1}{2}} .

These modular objects (Δ, J) for (M, u) satisfy the following relations (see, e.g., [3, 14]):

J^{2} = 1 and J is antiunitary  (1)
JΔJ = Δ^{−1}  (2)
Ju = u  (3)
Δu = u  (4)
JMJ = M′.  (5)

Let U be a unitary operator on H. Then v := Uu is cyclic and separating for N := UMU^{∗} and (N , v) has the modular objects

(Δ_{v}, J_{v}) = (UΔU^{∗}, UJU^{∗}).  (6)

M. Leitz-Martini: Math. Inst. Univ., Auf der Morgenstelle 10, D-72076 Tübingen M. Wollenberg: Inst. Math. Univ., Augustusplatz 10/11, D-04109 Leipzig e-mail: martin.leitz@uni-tuebingen.de and wollenbg@mathematik.uni-leipzig.de

ISSN 0232-2064 / $ 2.50 \begin{matrix}c \\ ◯ \end{matrix} Heldermann Verlag Berlin

We call an operator J satisfying (1) a conjugation. A conjugation J satisfying (5) is called an algebraic conjugation for M (in [15: p. 337] it is called a unitary involution of (M, H)).

Now we are interested in the following questions:

(a) Let J be a conjugation on a Hilbert space H. Does there exist a von Neumann algebra M with a cyclic and separating vector u such that J is the modular conjugation for (M, u)?

(b) Let J be a conjugation on H and let Δ be the modular operator for (M, u) where M is a von Neumann algebra on H with the cyclic and separating vector u. Suppose (2) holds. Does there exist a von Neumann algebra M_{o} with a cyclic and separating vector u_{o} such that (Δ, J) are the modular objects for (M_{o}, u_{o})?

(c) Let J be an algebraic conjugation for the von Neumann algebra M. Does there exist a cyclic and separating vector u for M such that J is even the modular conjugation for (M, u)?

The answer to question (a) is yes (at least if the Hilbert space is infinite-dimensional) and very easy to show. For completeness we discuss this simple question in Section 2. In Section 3 we show that question (b) can be answered affirmatively. The answer to question (c) is in general no. This question (and a partial answer without proof) is mentioned in [14: p. 321] in connection with a possible difference between standard von Neumann algebras and hyperstandard von Neumann algebras. We treat question (c) in some detail in Section 4.

The modular theory plays an important role in the theory of von Neumann algebras and in the algebraic approach to quantum physics (see, e.g., [2, 5, 8]). Modular conjugations are used for studying different problems in quantum field theory (see, e.g., [2, 8, 12]). In particular, the characterizations of modular conjugations, which we investigate here, are – for example – important for some inverse problems of the modular operator (see, e.g., [4, 12, 17]).

In this paper we only consider von Neumann algebras which are factors, shortly von Neumann factors, and which are separable, i.e. they can be represented faithfully on a separable Hilbert space.

If M is a von Neumann factor acting on a separable Hilbert space H, then we denote by MC(M) the set of modular conjugations for M. This means that J ∈ MC(M) if and only if there is a cyclic and separating vector u for M such that J is the modular conjugation for (M, u).

We will use the notation unit vector for a vector with norm 1 and antiunitary operator for an operator V which is antilinear, i.e. V (cu) = c̄V u, and satisfies V V^{∗} = V^{∗}V = 1. Further, we denote the set of unitaries of M by U(M).

2. Conjugations and modular conjugations

First we are interested in the question whether a conjugation is always a modular conjugation for some von Neumann factor with a cyclic and separating vector. The answer is known and we will present a simple proof. For this proof we use the following lemma about conjugations.

Lemma 2.1. Let J_{1} and J_{2} be two conjugations acting on a separable Hilbert space H_{o}. Then:

(i) There is a unitary operator U on H_{o} such that J_{2} = UJ_{1}U^{∗} and the spectral projections of U commute with both J_{1} and J_{2}.

(ii) Let A be a positive selfadjoint operator on H_{o} such that J_{i}AJ_{i} = A^{−1} (j = 1, 2). Then the unitary operator U from (i) can be chosen such that additionally UAU^{∗} = A.

Proof. 1. First we note that the relation JW J = W^{∗} for a conjugation J and a unitary operator W implies that J commutes with the spectral projections of W, JE_{W} (Γ)J = E_{W} (Γ) with Γ ⊂ [0, 2π]. This follows from the spectral representation W = ∫_{o}^{2π} e^{iμ}E_{W} (dμ) (see, e.g., [1: p. 46]).

2. We set V = J_{2}J_{1}. The operator V is unitary and 1 = (J_{2})^{2} = V J_{1} · V J_{1} implies that J_{1}V J_{1} = V^{∗}. Thus, by Step 1, the spectral projections E_{V} (Γ) commute with J_{1}. We set U = ∫_{0}^{2π} e^{i \frac{μ}{2}} E_{V} (dμ). The functional calculus gives V = U · U (see, e.g., [1: p. 46]) and JE_{V} (Γ) = E_{V} (Γ)J implies J_{1}UJ_{1} = U^{∗}. From V J_{1} = J_{2} therefore J_{2} = V J_{1} = UUJ_{1} = UJ_{1}U^{∗} follows. Further,

J_{2}UJ_{2} = V J_{1}UV J_{1} = UUJ_{1}UUUJ_{1} = UU · U^{∗}U^{∗}U^{∗} = U^{∗}

because of J_{1}U = U^{∗}J_{1}. The relations J_{i}UJ_{i} = U^{∗} (i = 1, 2) imply, by Step 1 , that the spectral projections of U commute with J_{1} and J_{2}. This shows that U satisfies all desired relations in (i).

3. From J_{i}AJ_{i} = A^{−1} (i = 1, 2) it follows, with V = J_{2}J_{1}, that

V AV^{∗} = J_{2}J_{1}AJ_{1}J_{2} = J_{2}A^{−1}J_{2} = A.

Thus the spectral projections of V commute with A and, by the definition of U above, U commutes with A, too

Now we come to the answer of the mentioned question.

Proposition 2.2. Let H_{o} be a separable Hilbert space with dim H_{o} = m^{2} for a natural number m or dim H_{o} = ∞. Let K be a conjugation on H_{o}. Then there is a von Neumann factor M_{o} on H_{o} with a cyclic and separating vector u_{o} such that K is the modular conjugation for (M_{o}, u_{o}), i.e. K ∈ MC(M_{o}).

Proof. 1. The assumption on the dimension of H_{o} secures that we can identify H_{o} with a tensor product K ¯⊗K. We define M = L(K) ¯⊗1 and u = ∑_{i} λ_{i}e_{i} ⊗ e_{i} where (e_{i})_{i} is a basis in K and λ_{i} > 0 with ∑_{i} λ_{i}^{2} = 1. It is easy to show that u is a cyclic and separating vector for M.

2. Now let (Δ, J) be the modular objects for (M, u). For the two conjugations J and K there exist a unitary operator U such that UJU^{∗} = K (see Lemma 2.1). We put u_{o} = Uu and M_{o} = UMU^{∗}. Then (6) implies that u_{o} is a cyclic and separating vector for the von Neumann factor M_{o} and that the modular objects (Δ_{o}, J_{o}) for (M_{o}, u_{o}) are given by Δ_{o} = UΔU^{∗} and J_{o} = UJU^{∗} = K. This concludes the proof

This proposition says that the dimension requirement is sufficient to secure that a conjugation K on H_{o} is a modular conjugation. It is also easy to see that the assumption on the dimension is necessary. Since without this assumption there is no von Neumann factor M on H_{o} with a cyclic and separating vector.

If we look at the proof of Proposition 2.2 we easily see that one can prove the following stronger result (in the case dim H_{o} = ∞):

Let K be a conjugation on a separable infinite-dimensional Hilbert space. Let N be a separable von Neumann factor of type I_{∞}, II, or III. Then there is a von Neumann factor M_{o} on H_{o} such that M_{o} is isomorphic to N, M_{o} has a cyclic and separating vector u_{o}, and K is the modular conjugation for (M_{o}, u_{o}).

3. Modular conjugations for a modular operator

In this section we consider the question how the set of modular conjugations for a fixed modular operator looks like.

Proposition 3.1. Let (Δ_{o}, J_{o}) be the modular objects for (M_{o}, u_{o}). Let J be a conjugation satisfying JΔ_{o} = Δ_{o}^{−1}J. Then there is a unitary operator W such that W Δ_{o}W^{∗} = Δ_{o}, W J_{o}W^{∗} = J, u := W u_{o} is a cyclic and separating vector for M := W M_{o}W^{∗}, and (Δ_{o}, J) are the modular objects for (M, u).

Proof. By Lemma 2.1 we have for the two conjugations J and J_{o} a unitary operator W such that W J_{o}W^{∗} = J and W Δ_{o}W^{∗} = Δ_{o}. Defining M = W M_{o}W^{∗} and u = W u_{o} the proposition is proved with the help of (6)

Proposition 3.2. Let (Δ_{o}, J_{o}) be the modular objects for (M_{o}, u_{o}). Let u be a unit eigenvector of Δ_{o} to the eigenvalue 1. Let J be a conjugation satisfying JΔ_{o} = Δ_{o}^{−1}J and Ju = u. Then there is a unitary operator W such that W Δ_{o}W^{∗} = Δ_{o}, W J_{o}W^{∗} = J, u = W u_{o}, u is a cyclic and separating vector for M := W M_{o}W^{∗}, and (Δ_{o}, J) are the modular objects for (M, u).

Proof. 1. First we note that according to Proposition 3.1 there is a unitary W_{1} such that W_{1}Δ_{o} = Δ_{o}W_{1}, W_{1}J_{o}W_{1}^{∗} = J, v := W_{1}u_{o} is a cyclic and separating vector for M_{1} := W_{1}M_{o}W_{1}^{∗}, and (Δ_{o}, J) are the modular objects for (M_{1}, v).

2. We have Jv = v and Δ_{o}v = v. According to Lemma 5.2 there is a unitary V_{1} such that V_{1}Δ_{o} = Δ_{o}V_{1}, V_{1}JV_{1}^{∗} = J and u = V_{1}v. Then the pair (V_{1}M_{1}V_{1}^{∗}, V_{1}v) has the modular objects (V_{1}Δ_{o}V_{1}^{∗}, V_{1}JV_{1}^{∗}) = (Δ_{o}, J) because (M_{1}, v) has the modular objects (Δ_{o}, J). Setting W = V_{1}W_{1} we find the desired result

From Propositions 3.1 and 3.2 we see first that the condition JΔ_{o} = Δ_{o}^{−1}J is necessary and sufficient that a conjugation J is a modular conjugation for a given modular operator Δ_{o}. Second we see that each eigenvector u for Δ_{o} and J to the eigenvalue 1 is a cyclic and separating vector for some von Neumann factor M such that (Δ_{o}, J) are the modular objects for (M, u). Third we find that the construction of the von Neumann factor M is given by M = W M_{o}W^{∗} where W depends only on Δ_{o}, J_{o}, J, u_{o}, u and does not depend on the whole algebra M_{o}.

4. Algebraic conjugations and modular conjugations

First we note a simple characterization of the set of modular conjugations for a von Neumann factor.

Proposition 4.1. Let J_{o} be a modular conjugation for (M_{o}, u_{o}). Then

MC(M_{o}) = {J = U^{∗}J_{o}U : U ∈ U(M′_{o})}.

Proof. 1. If J is a modular conjugation for a pair (M_{o}, u_{o}), then J = U^{∗}J_{o}U with some U ∈ U(M′_{o}) (see, e.g., [14: p. 331]).

2. Conversely, let J = U^{∗}J_{o}U with some U ∈ U(M′_{o}). Then (6) gives that J is the modular conjugation for (U^{∗}M_{o}U, U^{∗}u_{o}). But U^{∗}M_{o}U = M_{o} in our case. Thus we get J ∈ MC(M_{o})

Next we characterize the modular conjugations J for M_{o} with the help of the unitary V := JJ_{o}.

Proposition 4.2. Let M_{o} be a von Neumann factor on a separable Hilbert space H with a cyclic and separating vector u_{o}. Let J_{o} be the modular conjugation for (M_{o}, u_{o}) and let J be an algebraic conjugation for M_{o}. Further let V := JJ_{o}. Then:

(i) ad V ∈ aut M_{o}.

(ii) ad V ∈ int M_{o} if and only if J ∈ MC(M_{o}). (iii) If int M_{o} = aut M_{o}, then J ∈ MC(M_{o}).

Proof. (i). From (5) for J and J_{o} it follows that ad V ∈ aut M_{o}.

(ii) Now suppose J is a modular conjugation. Then from Proposition 4.1 we get that J = U^{∗}J_{o}U for some U ∈ U(M′_{o}) and V := JJ_{o} = U^{∗} · J_{o}UJ_{o}. Since J_{o}UJ_{o} ∈ U(M_{o}) and U^{∗} ∈ U(M′_{o}) we find that ad V is from int M_{o}.

Conversely, suppose that ad V is an inner automorphism of M_{o}. Then we can decompose V in the form V = V_{1} · V_{2} where V_{1} ∈ U(M_{o}) and V_{2} ∈ U(M′_{o}). By J^{2} = V J_{o}V J_{o} = 1, we get J_{o}V J_{o} = V^{∗} and find V^{∗} = V_{2}^{∗}V_{1}^{∗} = J_{o}V_{1}V_{2}J_{o} = J_{o}V_{1}J_{o} · J_{o}V_{2}J_{o} with J_{o}V_{1}J_{o}, V_{2}^{∗} ∈ M′_{o} and J_{o}V_{2}J_{o}, V_{1}^{∗} ∈ M_{o}. Now using the uniqueness of such a product from unitaries from M_{o} and M′_{o} (see Lemma 5.1) we obtain J_{o}V_{1}J_{o} = V_{2}^{∗}e^{ic} and J_{o}V_{2}J_{o} = V_{1}^{∗}e^{−ic}. By using the freedom in the decomposition V_{1} · V_{2} we can transform c into zero. Therefore V = J_{o}V_{2}^{∗}J_{o} · V_{2} = V_{2} · J_{o}V_{2}^{∗}J_{o} and J = V J_{o} = V_{2}J_{o}V_{2}^{∗}J_{o} · J_{o} = V_{2}J_{o}V_{2}^{∗}. This proves J has the form of an element of MC(M_{o}) (see Proposition 4.1) and therefore J is a modular conjugation. This proves the other direction of (ii).

(iii) It follows from (i) and (ii)

Remark 4.3. If we restrict ourselves to hyperfinite factors, then the property int M = aut M is only true for type I factors. At present it is not clear if there are (non-hyperfinite) type II factors with int M = aut M. But we see from this proposition that every algebraic conjugation for a type I factor is even a modular conjugation for this factor.

From Proposition 4.2 we get a further characterization of algebraic conjugations.

Proposition 4.4. Let J_{o}, J, V be as in Proposition 4.2. Then:

(i) J = V · J_{o} = W · J_{o} · V with a unitary W such that ad W ∈ int M_{o}.

(ii) V^{2} = W, W^{∗} = J_{o} · W · J_{o} and V W V^{∗} = W.

Proof. (i) From the fact that ad V is an automorphism of M_{o} and from (6) we get that V J_{o}V^{∗} is a modular conjugation for M_{o}. Thus V J_{o}V^{∗} = W J_{o} with ad W ∈ int M_{o} (see Proposition 4.2). This implies J = V J_{o} = W J_{o}V where W has the described properties.

(ii) We get V^{2} = W J_{o}V J_{o} · V = W J_{o}V V^{∗}J_{o} = W. The second property follows from

J_{o}W J_{o} = J_{o}V^{2}J_{o} = J_{o}(JJ_{o} · JJ_{o})J_{o} = J_{o}J · J_{o}J = V^{∗} · V^{∗} = (V^{2})^{∗} = W^{∗}.

Further, using that V^{2} = W we find V W V^{∗} = V V^{2}V^{∗} = V^{2} = W

Before we use this assertion about the structure of algebraic conjugations we construct for a large class of non type I factors M_{o} algebraic conjugations which are not modular conjugations (shortly we call such algebraic conjugations purely algebraic conjugations). The construction uses the flip automorphism and the fact that ”many” factors M_{o} are tensor squares, M_{o} ≅ N ¯⊗N.

Theorem 4.5. Let M_{o} be a separable von Neumann factor with a cyclic and separating vector u_{o}. Suppose M_{o} is isomorphic to N ¯⊗N where N is a von Neumann factor and M_{o} is not a type I factor. Then there is an algebraic conjugation for M_{o} which is not a modular conjugation.

Proof. 1. Since M_{o} is separable the factor N is also separable. Thus N is isomorphic to a von Neumann factor L on a separable Hilbert space and L has a cyclic and separating vector v_{o} (see [3: Proposition 2.5.6]). This implies that L ¯⊗L is a von Neumann factor with the cyclic and separating vector v := v_{o} ⊗ v_{o} (see [9]). Further, M_{o} is isomorphic to L ¯⊗L. Since both von Neumann factors have a cyclic and separating vector they are spatially isomorphic. Thus, without loss of generality, we can assume in the following that M_{o} = N ¯⊗N on H = H_{o} ¯⊗H_{o} and M_{o} has a cyclic and separating vector u = u_{o} ⊗ u_{o}.

2. We consider the flip automorphism α on N ¯⊗N given by α(X ⊗ Y) = Y ⊗ X (X, Y ∈ N). We obtain

(u, α(X ⊗ Y)u) = (u, Y ⊗ Xu) = (u_{o}, Y u_{o})(u_{o}, Xu_{o}) = (u, X ⊗ Y u).

This gives that α leaves invariant the vector state (u, ·u). Thus, for α there exist a unitary operator V_{α} such that ad V_{α} = α, V_{α}u = u, and V_{α} commutes with the modular objects (Δ_{o}, J_{o}) for (M_{o}, u) (see, e.g., [3]). Since M_{o} is not of type I we have that α (the flip) is an outer automorphism (see [11]).

3. Next we show that V_{α}^{2} = 1. Since α^{2} is the identity automorphism we find thatα^{2}(M) = V_{α}^{2}M(V_{α}^{∗})^{2} = M (M ∈ M_{o}) and therefore Y := V_{α}^{2} ∈ M′_{o}. This implies, with V_{α}u = u, Mu = MV_{α}^{2}u = V_{α}^{2}Mu (M ∈ M_{o}). Since u is cyclic for M_{o} we get that V_{α}^{2} = 1, thus V_{α} = V_{α}^{∗}.

4. Now we define an antiunitary operator J = V_{α}J_{o}. Since V_{α} commutes with the modular conjugation J_{o} (see Step 2) we have J^{2} = V_{α}J_{o}V_{α}J_{o} = V_{α}^{2}J_{o}J_{o} = 1, i.e. J is a conjugation. Further,

JM_{o}J = V_{α}J_{o}M_{o}V_{α}J_{o} = J_{o}V_{α}M_{o}V_{α}J_{o} = J_{o}α(M_{o})J_{o} = J_{o}M_{o}J_{o} = M′_{o},

i.e. J is an algebraic conjugation for M_{o}. Since V_{α} = JJ_{o} implements an outer automorphism of M_{o} we obtain from Proposition 4.2 that J is an algebraic conjugation which is not a modular conjugation

The next question is which von Neumann factors are tensor squares. First, we note that not all von Neumann factors have such a structure. For example, finite von Neumann factors of type I_{n} where n is not a square of a natural number are not tensor squares. Second, for some non-hyperfinite type II_{1} factors it is not known if they have outer automorphisms and therefore it is not clear if they are tensor squares (tensor squares of non type I factors have an outer automorphism, the flip automorphism [11]). Third, there are hyperfinite III_{o} factors (even ITPFI factors) which are not tensor squares (see [7]).

The following proposition presents a class of von Neumann factors which are tensor squares. It follows directly from well known results in the literature.

Proposition 4.6. Let M_{o} be a hyperfinite factor. Then:

(i) If M_{o} is neither of type I nor of type III_{o}, then M_{o} is a tensor square, i.e. M_{o} ≅ N ¯⊗N where N is a hyperfinite factor.

(ii) If M_{o} is a type III_{o} factor whose flow of weights has pure point spectrum, then M_{o} ≅ N ¯⊗N where N is a hyperfinite factor.

Proof. 1. First we note that the tensor square N ¯⊗N of a hyperfinite factor N is again a hyperfinite factor. This follows from the fact that such factors are injective (we consider only separable von Neumann factors where injective is the same as hyperfinite) and injectivity is stable under tensor products (see, e.g., [5]), and from the fact that the tensor product of factors is again a factor (see [9]).

2. If N is a hyperfinite factor of type II_{1} (II_{∞}), then N ¯⊗N is again a factor of type II_{1} (II_{∞}) (see [9]), and it is hyperfinite because of Step 1. Since these factors are unique (see, e.g., [5]), we get that the hyperfinite factors of type II_{1} and of type II_{∞} are tensor squares.

3. There exists a hyperfinite type III_{o} factor N such that M = N ¯⊗N is a hyperfinite type III_{1} factor (a hyperfinite type III_{λ} factor, for each λ ∈ (0, 1)) (see [6]). Since the hyperfinite factors of type III_{1} and of type III_{λ}, λ ∈ (0, 1) are unique (see [5]), we obtain that they are tensor squares.

4. If M_{o} is a hyperfinite type III_{o} factor whose flow of weights has pure point spectrum, then there is a hyperfinite type III_{o} factor N with M_{o} ≅ N ¯⊗N (see [6])

Remark 4.7. 1. From Proposition 4.6 and Theorem 4.5 we get that all hyperfinite factors of type II, III_{λ}, III_{1}, and III_{o} (whose flow of weights has pure point spectrum) have purely algebraic conjugations.

2. For the hyperfinite factors described in item 1 we have that they are even tensor squares of themselves, i.e. M ≅ M ¯⊗M (see [6, 10]).

For factors which are not hyperfinite or do not belong to the set of hyperfinite type III_{o} factors described in Proposition 4.6 the existence of purely algebraic conjugations is not clear.

At the end of this section we shortly discuss the question how ”large” the set of purely algebraic conjugations is.

From Proposition 4.2 we see that the set of algebraic conjugations for a von Neumann factor M_{o} is given by

{J = V · J_{o} : ad V ∈ aut M_{o} and V J_{o} = J_{o}V^{∗}}

where J_{o} is a fixed modular conjugation of M_{o}. Thus the set of algebraic conjugations is determined by the set of automorphisms ad V. Now the set of automorphisms of M_{o} can be classified with the help of the outer conjugacy relation. Two automorphisms α and β are said to be outer conjugate if there is a θ ∈ aut M_{o} such that β = θ · α · θ^{−1} modulo int M_{o}. In our case (ad V)^{p} = ad W ∈ int M_{o} and ad V (W) = γW with p = 2 and γ = 1 (see Proposition 4.4). This means the automorphisms ad V defining the purely algebraic conjugations have the outer conjugacy class invariants p(ad V) = 2 and γ(ad V) = 1) (for details about outer conjugacy, see [5: Section V.6]). Thus we get the following result.

Proposition 4.8. Let M_{o} be a hyperfinite factor of type II with a cyclic and separating vector u_{o}. Let J_{o} be the modular conjugation for (M_{o}, u_{o}). Further, let J_{1} = V_{1}·J_{o} and J_{2} = V_{2} · J_{o} be two purely algebraic conjugations for M_{o}. Then ad V_{1} and ad V_{2} are outer conjugate.

Proof. From the considerations before we know that ad V_{1} and ad V_{2} have the same invariants p = 2 and γ = 1. From [5: Section V.6/Theorems 14 and 16] it follows that then ad V_{1} and ad V_{2} are outer conjugate

Remark 4.9. 1. From this proposition and Remark 4.7 we see that for hyperfinite type II factors there exists only one purely algebraic conjugation up to outer conjugacy. One can partly extend this result to such hyperfinite type III factors as described in Remark 4.7 with the help of the classification of the automorphisms of these factors (see [16]). We omit the details.

2. Naturally, if an outer automorphism ad V of M_{o} satisfies p(ad V) = 2 and γ(ad V) = 1, then J := V · J_{o} is in generally not an algebraic conjugation for M_{o}. The relation J^{2} = 1 does not follow from these assumptions.

Acknowledgments. We thank one of the referees for the suggestion to separate a part of the proof of Proposition 2.2 as a lemma (Lemma 2.1). This paper is partly based on the Diploma thesis of the first author.

Notes on Modular Conjugations 21

5. Appendix

In this appendix we prove two simple but useful lemmas.

Lemma 5.1. Suppose U = U_{1} · U_{2} with elements U_{1} ∈ M and U_{2} ∈ M′ which are bounded invertible. Then this decomposition of U in bounded invertible elements from M and M′ is unique up to a bounded invertible element from the center of M. This means if there is another such decomposition U = V_{1} · V_{2} with V_{1} ∈ M and V_{2} ∈ M′, then U_{1} = G · V_{1} and U_{2} = G^{−1} · V_{2} with G ∈ Z(M).

Proof. If U = U_{1} · U_{2} = V_{1} · V_{2} with U_{1}, V_{1} ∈ M and U_{2}, V_{2} ∈ M′, then we get that G := V_{1}^{−1}U_{1} = V_{2}U_{2}^{−1} is from M ∩ M′. Clearly, G has an inverse. This proves the assertion

Lemma 5.2. Let H be a selfadjoint operator on a separable Hilbert space H. Suppose H has the eigenvalue 1. Let u_{1}, u_{2} be two unit eigenvectors of H to the eigenvalue 1. Suppose there is an antiunitary operator J with J^{2} = 1, Ju_{1} = u_{1}, and Ju_{2} = u_{2}. Then there is a unitary operator V_{1} on H with the properties V_{1}H = HV_{1}, V_{1}J = JV_{1}, and V_{1}u_{1} = u_{2}.

Proof. 1. Assume u_{1} = e^{ia}u_{2}. Then we get
u_{1} = Ju_{1} = Je^{ia}u_{2} = e^{−ia}Ju_{2} = e^{−ia}u_{2} = e^{−i2a}u_{1}.

Thus, e^{ia} is 1 or −1. So we can choose V_{1} as 1 or -1.

2. Next assume that u_{1}, u_{2} generate a two-dimensional subspace H_{o} of H. We can choose orthonormal bases {u_{1}, u′_{1}} and {u_{2}, u′_{2}} in H_{o} such that Ju′_{j} = u′_{j} (j = 1, 2). Namely, suppose that {u_{1}, v_{1}} is an orthonormal basis in H_{o} such that Jv_{1} ≠ v_{1}. We have (Jv_{1}, u_{1}) = (Ju_{1}, v_{1}) = (u_{1}, v_{1}) = 0. Further, JH_{o} = H_{o}. This gives that {u_{1}, Jv_{1}} is again an orthonormal basis in H_{o} like {u_{1}, v_{1}}. Thus Jv_{1} = e^{ic}v_{1}.

Now we set u′_{1} = e^{ic/2}v_{1}. Then Ju′_{1} = Je^{i \frac{c}{2}} v_{1} = e^{−i \frac{c}{2}} Jv_{1} = e^{i \frac{c}{2}} v_{1} = u′_{1}. Clearly, u′_{1} is from H_{o} and is orthogonal to u_{1}. This shows {u_{1}, u′_{1}} has the right invariance under J. The same can be done with {u_{2}, u′_{2}}. So our assumption on the bases can be fulfilled. Next we define V_{o} as a partial isometry from H_{o} onto H_{o} with V_{o}u_{1} = u_{2} and V_{o}u′_{1} = u′_{2}.

3. We make the ansatz V_{1} := (1−P_{o})+V_{o} where P_{o} is the orthoprojection onto H_{o}. Clearly, P_{o} commutes with H and V_{o} commutes with H. Thus V_{1}H = HV_{1}. Further, we get V_{1}u_{1} = V_{o}u_{1} = u_{2}.

4. It remains to show V_{1}J = JV_{1}. First from Ju_{j} = u_{j} and J^{2} = 1 it follows JP_{o} = P_{o}J. Second, we have for u ∈ H_{o}

\begin{matrix}JV_{o}u & = JV_{o}(a_{1}u_{1} + a_{2}u′_{1}) = J(a_{1}u_{2} + a_{2}u′_{2}) = ā_{1}u_{2} + ā_{2}u′_{2} \\ V_{o}Ju & = V_{o}J(a_{1}u_{1} + a_{2}u′_{1}) = V_{o}(ā_{1}u_{1} + ā_{2}u′_{1}) = ā_{1}u_{2} + ā_{2}u′_{2}. \end{matrix} \}

This implies JV_{o} = V_{o}J. Thus JV_{1} = V_{1}J. This concludes the proof

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Received 14.04.1999