Order isomorphisms of operator intervals
∗
Peter Šemrl
Faculty of Mathematics and Physics
University of Ljubljana
Jadranska 19
SI-1000 Ljubljana
Slovenia
[email protected]
Abstract
We develop a general theory of order isomorphisms of operator intervals. In this way we unify and extend several known results, among
others the famous Ludwig’s description of ortho-order automorphisms of
effect algebras and Molnár’s characterization of bijective order preserving
maps on bounded observables. Besides proving several new results, one
of the main contributions of the paper is to provide self-contained proofs
of several known theorems whose original proofs depend on various deep
results from functional analysis, operator algebras, and geometry. At the
end we will show the optimality of the obtained theorems using Löwner’s
theory of operator monotone functions.
AMS classification: 47B49.
1
Introduction
Let H be a Hilbert space and S(H) the set of all bounded linear self-adjoint
operators on H. We say that A ∈ S(H) is positive, A ≥ 0, if hAx, xi ≥ 0 for
every x ∈ H, and A is strictly positive, A > 0, if A is positive and invertible.
The set S(H) is partially ordered by the relation ≤ defined by A ≤ B ⇐⇒
B − A ≥ 0.
In the Hilbert space formalism of quantum mechanics bounded self-adjoint
operators correspond to bounded observables. The physical meaning of A ≤ B is
that the mean value (or, in other words, expectation) of the bounded observable
A in any state is less than or equal to the mean value of B in the same state.
∗ The
author was supported by a grant from ARRS, Slovenia, Grant No. P1-0288.
1
By E(H) we denote the effect algebra on H, that is, the set of all positive
operators on H that are bounded by the identity. More precisely, for any pair
of bounded self-adjoint operators B and C such that C − B is strictly positive
we define an operator interval by [B, C] = {A ∈ S(H) : B ≤ A ≤ C}. Then
E(H) = [0, I].
Elements of E(H) are called effects. They describe yes-no measurements
which can be unsharp (for detailed explanation see [1] and [8]). For every
A ∈ E(H) the orthocomplement A⊥ is defined by A⊥ = I − A. One of the
basic results in mathematical foundations of quantum mechanics is Ludwig’s
characterization of ortho-order automorphisms of effect algebras [9, Section V.5].
It states that for every bijective map φ : E(H) → E(H), dim H ≥ 2, satisfying
A ≤ B ⇐⇒ φ(A) ≤ φ(B),
A, B ∈ E(H),
and
φ(A⊥ ) = φ(A)⊥ ,
A ∈ E(H),
there exists an either unitary or antiunitary operator U on H such that
φ(A) = U AU ∗
for every A ∈ E(H).
Ludwig formulated this theorem only in the case when dim H ≥ 3. His proof
has been clarified in [3] and it was shown in [15] that the statement remains
valid when dim H = 2. It is somewhat surprising that the proof in the twodimensional case is much more difficult than in higher dimensions. The reason
is that one of the main tools in the proof of Ludwig’s theorem is the fundamental
theorem of projective geometry and in order to apply it one needs the dimension
of the underlying Hilbert space to be at least three.
The order automorphisms of S(H) were studied by Molnár [10, 12]. Observe
that when studying order automorphisms of S(H), that is, bijective maps φ :
S(H) → S(H) satisfying
A ≤ B ⇐⇒ φ(A) ≤ φ(B),
A, B ∈ S(H),
there is no loss of generality in assuming that φ(0) = 0. Indeed, the map φ is an
order automorphism if and only if the same is true for the map A 7→ φ(A)−φ(0).
With this harmless normalization Molnár’s result reads as follows: assume that
dim H ≥ 2 and that φ : S(H) → S(H) is an order automorphism satisfying
φ(0) = 0. Then there exists an invertible bounded linear or conjugate-linear
operator T : H → H such that
φ(A) = T AT ∗
for every A ∈ S(H). If we add the linearity assumption then the proof reduces to
a nice exercise. It is remarkable that the real-linearity of the map φ in Molnár’s
theorem is not an assumption but a conclusion.
2
Both Ludwig’s and Molnár’s result belong to a quite active research area
in mathematical physics dealing with bijective maps on quantum structures
preserving certain operations and/or relations that are relevant in the mathematical foundations of quantum mechanics. There is an obvious difference
between these two theorems. In both cases we are dealing with order automorphisms but in Ludwig’s theorem we have the additional assumption that
the orthocomplementation is preserved. It is then tempting to conjecture that
this additional assumption is superfluous, that is, every order automorphism
φ : E(H) → E(H) is of the form φ(A) = U AU ∗ , A ∈ E(H), for some unitary
or antiunitary operator U on H. This turns out to be wrong as can be demonstrated by the following example given in [14]. For any fixed invertible operator
T ∈ E(H), the transformation
A 7→
T2
2I − T 2
−1/2
2
(I − T + T (I + A)
−1
T)
−1
−I
T2
2I − T 2
−1/2
is a bijective map of E(H) onto itself, it preserves order in both directions but
at first glance it does not seem to be of a form that can be easily described.
In order to get a better insight we will first give some examples of order
isomorphisms between operator intervals. Here, of course, a map φ : J1 → J2
between operator intervals J1 and J2 is said to be an order isomorphism if it is
bijective and for every X, Y ∈ J1 we have X ≤ Y ⇐⇒ φ(X) ≤ φ(Y ). Assume
that A, B ∈ S(H) and B − A is a strictly positive operator. Let T ∈ S(H) be
any self-adjoint operator. Then the translation map X 7→ X + T is an order
isomorphism of [A, B] onto [A+T, B+T ]. Further, if T : H → H is any invertible
bounded linear or conjugate-linear map, then the congruence transformation
X 7→ T XT ∗ is an order isomorphism of [A, B] onto [T AT ∗ , T BT ∗ ]. And finally,
if A is strictly positive, then the map X 7→ X −1 is an order anti-isomorphism
of [A, B] onto [B −1 , A−1 ]. Recall that a bijective map φ between two operator
intervals is an order anti-isomorphism if for every pair X, Y from the domain
we have X ≤ Y ⇐⇒ φ(Y ) ≤ φ(X). To verify the last example we need to
recall the well-known fact that if C, D ∈ S(H) are strictly positive operators,
then we have C ≤ D if and only if D−1 ≤ C −1 .
It is now trivial to understand the above “complicated” example. Namely,
the map
−1
A 7→ ξ(A) = I − T 2 + T (I + A)−1 T
−I
is a product of several order isomorphisms and two order anti-isomorphisms:
A 7→ I + A 7→ (I + A)−1 7→ T (I + A)−1 T 7→
I − T 2 + T (I + A)−1 T 7→ (I − T 2 + T (I + A)−1 T )−1 7→
(I − T 2 + T (I + A)−1 T )−1 − I,
3
and is therefore an order isomorphism of E(H) onto [ξ(0), ξ(I)]. Clearly, ξ(0) =
0 and ξ(I) = T 2 (2I − T 2 )−1 . Composing ξ with the suitable congruence transformation we obtain the order automorphism of E(H) given above.
With this insight the next obvious step would be to describe the group of
order automorphisms of E(H) by finding a small set of generators that would
consists of products of simple maps described above. Much to our surprise
it turned out that all order automorphisms of E(H) can be described with a
rather simple formula [19, 20]. We first need to introduce a family of bijective
monotone increasing functions of the unit interval onto itself. For every real
number p < 1 we define such a function fp : [0, 1] → [0, 1] by
fp (x) =
x
,
px + (1 − p)
x ∈ [0, 1].
Then we have the following description of order automorphisms of the effect
algebra: Let dim H ≥ 2. Assume that φ : E(H) → E(H) is an order automorphism. Then there exist real numbers p, q ∈ (−∞, 1) and a bijective linear or
conjugate-linear bounded operator T : H → H with kT k ≤ 1 such that
−1/2
−1/2
φ(A) = fq (fp (T T ∗ ))
fp (T AT ∗ ) (fp (T T ∗ ))
, A ∈ E(H).
Finally, we mention three more results that serve as a motivation for our
research. For the cones of positive operators and strictly positive operators we
will use the interval notation, that is, we will write
[0, ∞) = {A ∈ S(H) : A ≥ 0}
and
(0, ∞) = {A ∈ S(H) : A > 0}.
Then φ is an order automorphism of [0, ∞) if and only if there exists an invertible
bounded linear or conjugate-linear operator T : H → H such that φ(A) = T AT ∗
for every A ∈ [0, ∞) [12], and similarly, φ is an order automorphism of (0, ∞) if
and only if there exists an invertible bounded linear or conjugate-linear operator
T : H → H such that φ(A) = T AT ∗ for every A ∈ (0, ∞) [18]. It was proved in
[13] that (0, ∞) is not order isomorphic to S(H) = (−∞, ∞).
The aim of this paper is to develop a general theory of order isomorphisms of
operator intervals. More precisely, we will answer two questions. First we want
to know which operator intervals are order isomorphic. And then, of course, we
would like to know what is the general form of order isomorphisms for each pair
of operator intervals that are order isomorphic. Giving a complete solution of
these two problems we will extend and unify the results mentioned above. It
will turn out that after some rather simple reductions we will need to consider
only a very small number of special cases. In particular, some of our main
results have been known before. However, in the past different special cases were
treated by quite different ad hoc methods, while we will here present a unified
4
approach. We will show that all structural results for order isomorphisms of
operator intervals can be deduced rather easily from one special case only. The
main contribution of this paper are completely new proof techniques that are
much simpler than the ones used so far in studying this kind of problems. For
example, the original Molnár’s proof of the description of order isomorphisms
of S(H) [10, 12] depends on several deep results from functional analysis and
ring theory: Rothaus’ theorem on automatic linearity of order isomorphisms of
closed convex cones in normed spaces [17], Vigier’s theorem [16, 4.1.1.Theorem],
Kadison’s theorem on order preserving linear bijections between C ∗ -algebras
[7, Corollary 5], and Herstein’s theorem on Jordan homomorphisms onto prime
rings [6]. An alternative proof can be found in [18]. This one is somewhat shorter
but it also depends on a non-trivial tool, that is, on an infinite-dimensional
extension of the Hua’s fundamental theorem of geometry of hermitian matrices.
The proof given in the present paper is much simpler and mostly self-contained
- whenever we will use other results we will give references where rather short
proofs of the statements that we will need are available.
In the next section we will show that our problems can be reduced to a
small number of special cases and then we will formulate our main theorems.
The third section will be devoted to preliminary results. As already mentioned,
the two-dimensional case is somewhat special when dealing with this kind of
problems. The fourth section will deal with this case. In the next section
we will then complete the characterization of order automorphisms of effect
algebras. The case of effect algebras turns out to be the crucial one. Namely,
we will see in the sixth section that all other main results follow rather quickly
from the description of order automorphisms of effect algebras. Finally, in the
last section we will show that our results are optimal.
2
Reduction to special cases and statement of
main results
Let H be a Hilbert space, dim H ≥ 2, and A, B ∈ S(H) such that B − A > 0.
Then we define operator intervals
[A, B] = {C ∈ S(H) : A ≤ C ≤ B},
[A, B) = {C ∈ S(H) : A ≤ C < B},
and
(A, B) = {C ∈ S(H) : A < C < B}.
Similarly, we define
[A, ∞) = {C ∈ S(H) : C ≥ A},
(A, ∞) = {C ∈ S(H) : C > A},
5
and S(H) = (−∞, ∞). The notations (A, B], (−∞, A], and (−∞, A) should
now be self-explanatory.
We will deal with two questions. First we will examine which of the above
operator intervals are order isomorphic. And then we will describe the general
form of all order isomorphisms between operator intervals that are order isomorphic. When studying the second question we do not need to consider all pairs of
order isomorphic operator intervals. Namely, if operator intervals I1 , I2 , J1 , J2
are order isomorphic and the order isomorphisms ϕj : Ij → Jj , j = 1, 2, are
given and we know the description of the general form of order isomorphisms
between I1 and I2 , then we immediately get the general form of order isomorphisms between J1 and J2 . Indeed, each order isomorphism ψ : J1 → J2 is of
the form
ψ = ϕ2 φ ϕ−1
1 ,
where φ : I1 → I2 is any order isomorphism.
A further reduction is possible if we know that certain operator intervals
are order anti-isomorphic. For example, the operator intervals [0, I) and (0, I]
are not order isomorphic. But as they are order anti-isomorphic (the map X 7→
I −X is a bijective map of [0, I) onto (0, I] satisfying X ≤ Y ⇐⇒ φ(Y ) ≤ φ(X),
X, Y ∈ [0, I)) it is enough to know the structure of order automorphisms of the
operator interval [0, I) to get automatically the description of the general form
of order automorphisms of (0, I]. Indeed, a map φ : [0, I) → [0, I) is an order
automorphism if and only if the map X 7→ I − φ(I − X), X ∈ (0, I], is an order
automorphism of the operator interval (0, I].
Clearly, we have the following more general reduction principle: if I1 and
I2 are order isomorphic operator intervals, and J1 and J2 are order isomorphic
operator intervals, and ϕj : Ij → Jj , j = 1, 2, are order anti-isomorphisms, then
each order isomorphism ψ : J1 → J2 is of the form ψ = ϕ2 φ ϕ−1
1 , where φ is an
order isomorphism of I1 onto I2 .
Let A1 , A2 , B1 , B2 ∈ S(H) be any operators with A1 < B1 and A2 < B2 .
Then the bijective map φ : [A1 , B1 ] → [A2 , B2 ] given by
φ(X) = (B2 − A2 )1/2 (B1 − A1 )−1/2 (X − A1 )(B1 − A1 )−1/2 (B2 − A2 )1/2
+A2 ,
X ∈ [A1 , B1 ],
is an order isomorphism. In short, all operator intervals of the form [A, B]
with A < B are isomorphic and in order to know the general form of order
isomorphisms between two such operator intervals it is enough to describe the
general form of order automorphisms of the effect algebra [0, I]. Similarly, all
operator intervals of the form [A, B) with A < B are isomorphic and in order
to know the general form of order isomorphisms between two such operator
intervals it is enough to describe the general form of order automorphisms of
the operator interval [0, I). The reader will now easily formulate analogous
statements for the operator intervals (A, B] and (A, B).
6
It is obvious that any two of the operator intervals [0, I], [0, I), (0, I], and
(0, I) are order non-isomorphic. For example, 0 ∈ [0, I) has the property that
0 ≤ A for every A ∈ [0, I), while there does not exist an operator B ∈ (0, I)
satisfying B ≤ A for every A ∈ (0, I). It follows that [0, I) and (0, I) are not
order isomorphic.
Further, for any A ∈ S(H) the operator interval [A, ∞) is order isomorphic
to [0, ∞) via the translation isomorphism X 7→ X − A. Similarly, for any
A ∈ S(H) the operator interval (−∞, A] is order isomorphic to (−∞, 0]. And
clearly, [0, ∞) and (−∞, 0] are not order isomorphic.
However, the operator intervals [0, ∞) and [0, I) are order isomorphic. The
map φ : [0, I) → [0, ∞) given by
φ(X) = (I − X)−1 − I,
X ∈ [0, I),
is an example of an order isomorphism of [0, I) onto [0, ∞) (one can easily verify
that the map φ is an order isomorphism using the ideas from Introduction and
the simple fact that the map X 7→ −X is an order anti-isomorphism of interval
J onto the interval −J). Similarly, (−∞, 0] is order isomorphic to (0, I].
Observe next that the map φ(X) = I − X −1 is an order isomorphism of
(0, I) onto (−∞, 0). And similarly, the map φ(X) = (I − X)−1 − I is an order
isomorphism of (0, I) onto (0, ∞).
Finally, we have the following result obtained by Molnár in [13].
Theorem 2.1 Operator intervals (−∞, ∞) and (0, ∞) are not order isomorphic.
To summarize, each operator interval J is isomorphic to one of the following
operator intervals: [0, I], [0, ∞), (−∞, 0], (0, ∞), and (−∞, ∞). And any two
of these operator intervals are order non-isomorphic. For any given operator
interval J one can easily construct an order isomorphism between J and the
one of the above five operator intervals that is order isomorphic to J by using
the ideas above.
The operator intervals [0, ∞) and (−∞, 0] are obviously order anti-isomorphic.
Hence, to understand the structure of all order isomorphisms between any two
order isomorphic operator intervals it is enough to describe the general form
of order automorphisms of the following four operator intervals: [0, I], [0, ∞),
(0, ∞), and (−∞, ∞).
As we shall see later the crucial step will be to describe the general form
of order automorphisms of the effect algebra [0, I]. The structure of order isomorphisms on other operator intervals will be deduced rather quickly from this
special case. Of course, due to the example given in Introduction one can not
expect that the description of order isomorphisms of effect algebras will be as
simple as in the case of some other operator intervals. Still, a rather simple
description was given in [19]. Before formulating this statement let us recall
7
that by fp , p < 1, we denote a bijective monotone increasing function of the
unit interval [0, 1] onto itself defined by
fp (x) =
x
,
px + 1 − p
x ∈ [0, 1].
Theorem 2.2 Let dim H ≥ 2. Assume that φ : [0, I] → [0, I] is an order
automorphism. Then there exist real numbers p, q, 0 < p < 1, q < 0, and a
bijective linear or conjugate-linear bounded operator T : H → H with kT k ≤ 1
such that
−1/2
−1/2
φ(A) = fq (fp (T T ∗ ))
fp (T AT ∗ ) (fp (T T ∗ ))
, A ∈ [0, I].
Conversely, for any pair of real numbers p, q, 0 < p < 1, q < 0, and any bijective
linear or conjugate-linear bounded operator T : H → H with kT k ≤ 1 the map φ
defined by the above formula is an order isomorphism of the effect algebra [0, I].
This statement is a slight improvement of one of the main results in [19]. A
much simpler proof will be given here leading to a much better insight into the
structure of order automorphisms of effect algebras. In particular, it is possible
to give alternative descriptions of order authomorphisms of effect algebras. For
example, we can describe them in the following way.
Theorem 2.3 Let dim H ≥ 2. Assume that φ : [0, I] → [0, I] is an order
automorphism. Then there exist a negative real number p and an invertible
bounded linear or conjugate-linear operator T : H → H such that
φ(A) =
fp (I + (T T ∗ )−1 )1/2 I − (I + T AT ∗ )−1 (I + (T T ∗ )−1 )1/2 ,
A ∈ [0, I].
Conversely, for any real number p < 0 and any invertible bounded linear or
conjugate-linear operator T : H → H the map φ defined by the above formula is
an order isomorphism of the effect algebra [0, I].
The simple proof we will present gives us many additional information on
the group of order automorphisms of the effect algebra. In particular, we have
the following statement on the transitivity of the group of order automorphisms
of [0, I].
Theorem 2.4 Let dim H ≥ 2. Assume that φ : [0, I] → [0, I] is an order
automorphism. Then
φ( (0, I) ) = (0, I).
Moreover, for every pair of effects A, B ∈ (0, I) there exists an order automorphism ψ : [0, I] → [0, I] such that
ψ(A) = B.
8
The following three theorems will be obtained as rather straightforward consequences of the above descriptions of groups of order automorphisms of effect
algebras. These statements have been known before [10, 12, 18]. We will present
here proofs that are much simpler and shorter than the original ones. The converse statements of the three theorems below are obviously true.
Theorem 2.5 Let dim H ≥ 2. Assume that φ : [0, ∞) → [0, ∞) is an order
automorphism. Then there exists a bounded bijective linear or conjugate-linear
operator T : H → H such that
φ(A) = T AT ∗
for every A ∈ [0, ∞).
Theorem 2.6 Let dim H ≥ 2. Assume that φ : (0, ∞) → (0, ∞) is an order
automorphism. Then there exists a bounded bijective linear or conjugate-linear
operator T : H → H such that
φ(A) = T AT ∗
for every A ∈ (0, ∞).
Theorem 2.7 Let dim H ≥ 2. Assume that φ : (−∞, ∞) → (−∞, ∞) is an
order automorphism. Then there exist a bounded bijective linear or conjugatelinear operator T : H → H and S ∈ S(H) such that
φ(A) = T AT ∗ + S
for every A ∈ S(H).
When defining intervals [A, B], [A, B), (A, B], and (A, B) we have assumed
that A < B. What happens if we deal with the weaker assumption that A ≤ B?
It is clear that if A ≤ B but A 6< B, then the intervals [A, B), (A, B], and (A, B)
are empty sets. Thus, the question makes sense only for the interval [A, B].
Assume first that B − A ≥ 0 is an injective operator that is not invertible
(obviously, this can happen only when the underlying Hilbert space H is infinitedimensional). Then, of course, the interval [A, B] is nonempty. We know that
all intervals of the form [C, D], where C, D ∈ S(H) satisfy C < D, are order
isomorphic. In particular, they are order isomorphic to [0, I]. On the other
hand, [A, B] is order isomorphic to [0, B − A] = [0, T ]. Here, T is a positive
injective non-invertible operator. Then the question is whether [0, I] and [0, T ]
are order isomorphic. The answer is in the affirmative.
Theorem 2.8 Let H be an infinite-dimensional Hilbert space and T ∈ S(H) a
positive injective non-invertible operator. Then the map φ : [0, I] → [0, T ] given
by
φ(A) = T 1/2 AT 1/2 , A ∈ [0, I],
is an order isomorphism of [0, I] onto [0, T ].
9
It is clear that φ maps [0, I] into [0, T ] and that for any pair A, B ∈ [0, I] we
have A ≤ B ⇒ φ(A) ≤ φ(B). We need to prove the converse implication, that
is, for any pair A, B ∈ [0, I] the inequality φ(A) ≤ φ(B) yields that A ≤ B, and
we need to show that φ is bijective.
Thus, if [A, B] is an operator interval with B − A ≥ 0 injective but not
invertible, then [A, B] is order isomorphic to the effect algebra. Hence, any two
operator intervals [A, B] and [C, D] where both B − A and D − C are positive
injective (they can be invertible or non-invertible) are order isomorphic and since
we know the general form of order automorphisms of the effect algebra and we
also have explicit formulas for order isomorphisms of [0, I] onto [A, B] and order
isomorphism of [0, I] onto [C, D], we actually have a complete description of
order isomorphisms of [A, B] onto [C, D].
But what happens if A ≤ B and B −A is not injective? This is a special case
of a more general question. Namely, as above we only need to know whether
[0, I] and [0, T ] are order isomorphic. Here, T = B − A is a positive operator.
We denote by Im T and Ker T the image of T and the kernel of T , respectively.
Then H is the orthogonal direct sum H = Im T ⊕ Ker T . If we denote K = Im T
and S = T|K , then clearly, the operator interval [0, T ] on the Hilbert space H
is order isomorphic to the operator interval [0, S] on the Hilbert space K. But
we already know that the interval [0, S] is order isomorphic to [0, I], where I
stands for the identity operator on the Hilbert space K.
We will now deal with an even more general question. Is it possible that
an operator interval on a Hilbert space H is isomorphic to an operator interval
on a Hilbert space K? Clearly, if H and K are isomorphic (their orthonormal
bases have the same cardinality), then we can identify H with K, and in this
special case the question is trivial. Hence, the problem is what happens if the
orthonormal bases of H and K have different cardinalities?
We already know that each operator interval J on H is isomorphic to one
of the following operator intervals: [0, I]H , [0, ∞)H , (−∞, 0]H , (0, ∞)H , and
(−∞, ∞)H . Here, (−∞, 0]H = {A ∈ S(H) : A ≤ 0} and other notations
should be self-explanatory. Similarly, each operator interval on K is isomorphic
to one of the following operator intervals: [0, I]K , [0, ∞)K , (−∞, 0]K , (0, ∞)K ,
and (−∞, ∞)K . So, the above question reads as follows: Assume that H and K
are non-isomorphic Hilbert spaces. Is any of operator intervals [0, I]H , [0, ∞)H ,
(−∞, 0]H , (0, ∞)H , and (−∞, ∞)H order isomorphic to any of operator intervals
[0, I]K , [0, ∞)K , (−∞, 0]K , (0, ∞)K , and (−∞, ∞)K ? The first guess that this
cannot happen turns out to be true.
Theorem 2.9 Let H and K be Hilbert spaces such that the cardinalities of the
orthonormal bases of H and K are different. Assume that J1 is an operator
interval on H and J2 an operator interval on K. Then J1 and J2 are not order
isomorphic.
10
3
Preliminary results
We will start with some rather trivial observations.
Lemma 3.1 Let H be a Hilbert space and A ∈ E(H). Then the following are
equivalent:
• A = 0 or A is a rank one operator;
• For every pair of operators B, C ∈ E(H) we have: B, C ≤ A ⇒ B ≤
C or C ≤ B.
Proof. Assume first that A = 0 or A is of rank one. Then A = tP for some rank
one projection P and some nonnegative real number t ≤ 1. It is rather easy to
verify that then B, C ∈ E(H) and B, C ≤ A yield that B = sP and C = rP for
some real numbers r and s, r, s ∈ [0, t], and hence, B and C are comparable.
Assume now that A is neither the zero operator nor a rank one operator.
A straightforward consequence of the spectral theorem for bounded self-adjoint
operators is the existence of a projection P of rank two and a positive real
number c such that cP ≤ A. Let Q be a rank one projection satisfying Q ≤ P .
Then for B = cQ ∈ E(H) and C = c(P − Q) ∈ E(H) we have B, C ≤ A but
neither B ≤ C nor C ≤ B is true.
2
Lemma 3.2 Let H be a Hilbert space and A, B ∈ S(H) operators satisfying
A ≤ B. Assume that H1 ⊂ H is a closed subspace that is invariant for both A
and B. Suppose further that the restrictions of A and B to H1 coincide, that
is, A|H1 = B|H1 . Let C ∈ S(H) be an operator such that A ≤ C ≤ B. Then H1
is invariant for C and we have
A|H1 = C|H1 .
Proof. The assumptions yield that with respect to the orthogonal direct sum
decomposition H = H1 ⊕ H1⊥ the operators A and B have the following matrix
representation
A1 0
A1 0
A=
and B =
0 A2
0 B2
with A2 ≤ B2 . We have 0 ≤ C − A ≤ B − A. Let the matrix representation of
C − A with respect to the same decomposition of H be
D E
C −A=
.
E∗ F
We have
0
0
0
D
≤
0
E∗
E
F
11
≤
0
0
0
B2 − A2
and we need to show that D = 0 and E = 0. The above chain of inequalities
implies first that hDx, xi = 0 for every x ∈ H1 , and consequently, D = 0. Then
we apply the first of the above two inequalities to conclude that also E = 0, as
desired.
2
Lemma 3.3 Let A and R be positive 2×2 matrices. Assume that A is invertible
and R of rank one. Then the following two conditions are equivalent:
• R ≤ A and there exists a positive real number ε such that sR 6≤ A for
every real s, 1 < s < 1 + ε;
• tr (A−1 R) = 1.
Proof. The condition that R ≤ A and there exists a positive real number ε such
that sR 6≤ A for every real s, 1 < s < 1 + ε, is equivalent to A−1/2 RA−1/2 ≤ I
and there exists a positive real number ε such that sA−1/2 RA−1/2 6≤ I for every
real s, 1 < s < 1 + ε. Since A−1/2 RA−1/2 is a rank one hermitian matrix, it is
of the form A−1/2 RA−1/2 = tP , where t is a real scalar and P is a projection
of rank one. Moreover, t = tr (A−1/2 RA−1/2 ) = tr (A−1 R). Using the obvious
fact that we have tP ≤ I and there exists a positive real number ε such that
stP 6≤ I for every real s, 1 < s < 1 + ε, if and only if t = 1, we complete the
proof.
2
Lemma 3.4 Let P and Q be 2 × 2 projections of rank one. Then there exists
a 2 × 2 unitary matrix U such that
1 1 w
1 1 u
and U QU ∗ =
UPU∗ =
2 u 1
2 w 1
for some complex numbers u and w of modulus one.
Proof. Recall that a 2 × 2 hermitian rank one matrix is a projection if and only
if its trace equals one. Since each projection is positive both diagonal entries
must be nonnegative. Thus, P and Q are simultaneously unitarily similar to
p
a
eit a(1 − a)
1 0
P0 =
and Q0 = −it p
,
0 0
a(1 − a)
1−a
e
respectively, where 0 ≤ a ≤ 1 and t ∈ R. Replacing P 0 and Q0 by W P 0 W ∗ and
W Q0 W ∗ , where W is an appropriate diagonal unitary matrix, we may, and we
will assume that t = 0.
All we need to do is to find a unit vector z ∈ C2 such that
hP 0 z, zi = hQ0 z, zi =
12
1
.
2
One can take
1
z=√
2
1
i
to complete the proof.
2
For k = 1, 2, . . . we denote by Pk ⊂ E(H) the subset of all projections of
rank k.
Lemma 3.5 Let H be a Hilbert space and A ∈ E(H) be of rank one. Then the
following are equivalent:
• A ∈ P1 ;
• If B ∈ E(H) satisfies A ≤ B then either A = B or B is not of rank one.
Proof. Let us assume first that A ∈ P1 . Then with respect to the orthogonal
direct sum decomposition H = Im A ⊕ Ker A, where Im A and Ker A denote the
image and the kernel of A, respectively, we have
1 0
A=
.
0 0
If B ∈ E(H) satisfies A ≤ B then
1 0
1
≤B≤
0 0
0
0
,
I
and by Lemma 3.2 we have
1
B=
0
0
B1
for some bounded self-adjoint operator B1 acting on Ker A. Now, if B1 = 0,
then A = B, and otherwise B is not of rank one.
If A is of rank one but is not a projection, then A = sP for some projection
P of rank one and some real s ∈ (0, 1). Take any t ∈ (s, 1) and B = tP to
complete the proof.
2
Lemma 3.6 Let H be a Hilbert space, k an integer, k ≥ 2, and A ∈ E(H).
Then the following are equivalent:
• A ∈ Pk ;
• There exist P1 , . . . , Pk ∈ P1 such that:
1. P1 , . . . , Pk ≤ A,
13
2. for every Q ∈ Pk−1 there exists an integer j, 1 ≤ j ≤ k, such that
Pj 6≤ Q, and
3. for every C ∈ E(H) satisfying P1 , . . . , Pk ≤ C we have A ≤ C.
Proof. Assume that A is a projection of rank k. Then we can find pairwise
orthogonal rank one projections P1 , . . . , Pk such that P1 + . . . + Pk = A. It is
trivial to verify that they satisfy the first two properties. If C ∈ E(H) satisfies
P1 , . . . , Pk ≤ C then Pj ≤ C ≤ I, j = 1, . . . , k, together with Lemma 3.2 yield
that C = A + C1 with C1 satisfying 0 ≤ C1 ≤ I − A. In particular, A ≤ C.
Assume now that there exist P1 , . . . , Pk ∈ P1 such that all three properties
above are satisfied. From the second property we conclude that the linear span
of images of Pj , j = 1, . . . , k, is k-dimensional. Denote this linear span by H1 .
Applying the first property, A ≤ I, and Lemma 3.2 we conclude that there are k
linearly independent vectors x1 , . . . , xk ∈ H1 such that Axj = xj , j = 1, . . . , k.
It follows that H1 is invariant under A and the restriction of A to H1 acts
like the identity. In other words, with respect to the orthogonal direct sum
decomposition H = H1 ⊕ H1⊥ we have
I 0
A=
0 A1
for some A1 with A1 ≥ 0. If we set
I
C=
0
0
0
then P1 , . . . , Pk ≤ C, and therefore by the third property we have A ≤ C.
Hence, A1 = 0, and consequently, A ∈ Pk , as desired.
2
In the next few lemmas we will always assume that H is a Hilbert space,
dim H ≥ 2, and φ : E(H) → E(H) an order automorphism with the additional
property that
1
1
I = I.
φ
2
2
Let us start with some trivial observations that follow easily from the results
obtained so far. Clearly, there is only one element T of E(H) with the property
that T ≤ A for every A ∈ E(H). This is the zero operator. Thus, φ(0) = 0,
and similarly, φ(I) = I. Using Lemma 3.1 we see that φ maps the set of rank
one effects onto itself. This fact together with Lemma 3.5 yields φ(P1 ) = P1 . A
straightforward application of Lemma 3.6 then gives φ(P2 ) = P2 . By induction
we see that φ(Pk ) = Pk , k = 1, 2, . . . Let k be an integer ≥ 2. Since A ∈ E(H)
is of rank k if and only if for every Q ∈ Pk−1 we have A 6≤ Q and A ≤ P for
some P ∈ Pk , we conclude that φ maps the set of effects of rank k onto itself.
14
Further, if A ∈ E(H) is of rank k and P ∈ Pk such that A ≤ P , then A and P
have the same images. Hence, such a P is uniquely determined. It follows that
two finite rank effects A and B have the same images if and only the same is
true for φ(A) and φ(B).
In particular, for every projection P of rank one there exists a bijective
monotone function fP : [0, 1] → [0, 1] (and hence, fP is continuous) such that
φ(tP ) = fP (t)φ(P )
for every t ∈ [0, 1]. Let P ∈ P1 . Then for every t ∈ [0, 1] we have t ≤ 1/2 if
and only if tP ≤ (1/2)I which is equivalent to φ(tP ) ≤ φ((1/2)I), or in other
words, fP (t)φ(P ) ≤ (1/2)I. Hence, for every P ∈ P1 we have
1
1
= .
fP
2
2
By observing that for any two projections P and Q we have P ≤ Q if and
only if φ(P ) ≤ φ(Q) we can improve one of the above statements by concluding
that for every pair of finite rank effects A, B we have Im A ⊂ Im B if and only
if Im φ(A) ⊂ Im φ(B).
The above simple facts will be frequently used in the proofs that follow. We
will call two projections P and Q orthogonal, P ⊥ Q, if P Q = 0 (and then, of
course, we have also QP = 0).
Lemma 3.7 Let P, Q ∈ P1 be orthogonal. Then
φ(P ) ⊥ φ(Q)
and for all pairs of real numbers t, s ∈ [0, 1] we have
φ(tP + sQ) = fP (t)φ(P ) + fQ (s)φ(Q).
Proof. We will first show that
φ
1
1
(P + Q) = R
2
2
for some R ∈ P2 . Indeed, because of bijectivity of φ we have φ(P ) 6= φ(Q).
Since (1/2)P ≤ (1/2)(P + Q) ≤ (1/2)I we have (1/2)φ(P ) ≤ φ((1/2)(P + Q)) ≤
(1/2)I, and similarly, (1/2)φ(Q) ≤ φ((1/2)(P + Q)) ≤ (1/2)I. Thus, by Lemma
3.2, 1/2 is an eigenvalue of φ((1/2)(P + Q)) with multiplicity at least two. But
this effect has rank two, and therefore, it is of the form (1/2)R for some R ∈ P2 .
The set (P + Q)E(H)(P + Q) consists of all effects whose image is contained
in the image of P + Q. And, of course, this set can be in a natural way identified
with E2 , the set of 2 × 2 hermitian matrices, whose eigenvalues belong to the
unit interval [0, 1], in such a way that P corresponds to E11 and Q corresponds
15
to E22 . Here, as usual, Eij stands for the matrix whose all entries are zero
but the (i, j)-entry that is equal to one. And because φ(P + Q) is a rank two
projection whose image coincide with the image of φ((1/2)(P + Q)), we have
φ(P + Q) = R, and again, RE(H)R can be identified with E2 , too.
In order to prove our lemma we need to consider only the restricition of
φ to (P + Q)E(H)(P + Q) as a bijection from (P + Q)E(H)(P + Q) onto
RE(H)R. This restriction will be again denoted by φ and after the above
mentioned identifications of both the domain and the codomain with E2 we
actually have an order automorphism
φ : E2 → E2
satisfying
φ
1
I
2
=
1
I,
2
and we need to show that φ(E11 ) ⊥ φ(E22 ) and that for all pairs of real numbers
t, s ∈ [0, 1] we have
φ(tE11 + sE22 ) = f1 (t)φ(E11 ) + f2 (s)φ(E22 ),
where fj = fEjj , j = 1, 2.
Because
1
1/2
I≤
0
2
0
≤I
1
and φ(tI) = tI for t = 1/2 and t = 1, we know that the eigenvalues of
1/2 0
C=φ
0 1
belong to [1/2, 1]. Now, φ(E22 ) ≤ C ≤ φ(I) = I, and hence, by Lemma 3.2, one
eigenvalue of C is 1 and its eigenspace coincides with the image of φ(E22 ). If
the other eigenvalue of C was strictly larger than 1/2, then we would be able to
find for every projection S ∈ E2 of rank one a real number tS > 1/2 such that
tS S ≤ C. But then we would be able to find for every projection T ∈ E2 of rank
one a real number tT > 1/2 such that
1/2 0
tT T ≤
,
0 1
a contradiction. Thus, the other eigenvalue of C must be 1/2, and moreover,
the corresponding eigenspace coincides with the image of φ(E11 ).
After replacing φ by a map A → U φ(A)U ∗ , where U is an appropriate 2 × 2
unitary matrix, we may, and we will assume that
1/2 0
1/2 0
φ
=
.
0 1
0 1
16
It follows that φ(E11 ) = E11 and φ(E22 ) = E22 . In particular, we have proved
the first of the desired conclusions, that is, φ(E11 ) ⊥ φ(E22 ).
We further know that
φ(tE11 ) = f1 (t)E11
and φ(tE22 ) = f2 (t)E22 ,
t ∈ [0, 1].
We apply the fact that for every s ∈ [0, 1] and every positive real ε we have
1 0
0 0
1 0
1 0
,
≤
≤
0 0
0 s
0 s
0 1
and
to conclude that
0
0
0
1
6
≤
s+ε
0
1
0
0
s
φ
=
1
0
0
s
0
.
f2 (s)
Let now t, s be an arbitrary pair of real numbers from the unit interval.
From
0
0
t 0
1
0
≤φ
=
0 f2 (s)
0 s
0 f2 (s)
we infer that
φ
t
0
0
s
∗
=
0
0
,
f2 (s)
where ∗ stands for some real number from the unit interval. Applying
t 0
t 0
t+ε 0
t 0
≤
and
6≤
0 0
0 s
0
0
0 s
we finally conclude that
φ
t
0
0
s
=
f1 (t)
0
0
,
f2 (s)
as desired.
2
Lemma 3.8 Assume that φ(P ) = P for every P ∈ P1 . Then φ(A) = A for
every A ∈ E(H) of rank one.
Proof. We know that for every P ∈ P1 there exists a bijective monotone increasing function fP : [0, 1] → [0, 1] such that φ(tP ) = fP (t)P for every t ∈ [0, 1].
According to our assumptions we have fP (1/2) = 1/2 for every rank one projection P . All we need to do is to show that fP is the identity function for every
P ∈ P1 .
17
So, let P be an arbitrary projection of rank one and we want to prove that fP
is the identity of the unit interval. We choose and fix Q ∈ P1 that is orthogonal
to P . We denote fP = f . We know that φ(P + Q) is a projection of rank
two and since φ(P ) = P ≤ φ(P + Q) and similarly, Q ≤ φ(P + Q), we have
φ(P + Q) = P + Q. In the rest of the proof we will be interested only in effects
belonging to (P + Q)E(H)(P + Q). This set is invariant under φ. We will
identify it with E2 in such a way that Q corresponds to E11 and P to E22 . The
restriction of φ to (P + Q)E(H)(P + Q) ≡ E2 will be again denoted by φ.
Take any real p ∈ (0, 1/2). Then there exists a unique a ∈ (0, 1) such that
p=
a−1
,
a−2
and then Lemma 3.3 and straightforward calculation show that
p
1 p a
a(1 − a)
1 0
≤
0 p
a(1 − a)
1−a
2
and there exists a positive real number ε such that
p
1 p a
a(1 − a)
1
6≤
s
0
a(1 − a)
1−a
2
0
p
for every real s, 1 < s < 1 + ε. The hermitian matrix
p
a
a(1 − a)
p
a(1 − a)
1−a
has rank one and trace one, and is therefore a projection of rank one. Hence,
p
p
1 p a
1 p a
a(1 − a)
a(1 − a)
φ
=
,
a(1 − a)
1−a
a(1 − a)
1−a
2
2
and therefore,
p
1 p a
a(1 − a)
1
≤
0
a(1 − a)
1−a
2
0
f (p)
and there exists δ > 0 such that
p
1 p a
a(1 − a)
1
6≤
s
0
a(1 − a)
1−a
2
0
f (p)
for every real s, 1 < s < 1 + δ, which again by Lemma 3.3 yields that
f (p) =
a−1
.
a−2
Hence, f (p) = p for every real p, 0 < p < 1/2. In a similar way we prove that
f (p) = p holds true also for any p between 1/2 and 1. Hence, f (p) = p for all
p ∈ [0, 1], as desired.
18
2
Let H be a Hilbert space with dim H ≥ 3. The famous Uhlhorn’s generalization of Wigner’s unitary-antiunitary theorem [21] was formulated in [12, p.13]
in the following way: if φ : P1 → P1 is a bijective map such that for every pair
P, Q ∈ P1 we have
P Q = 0 ⇐⇒ φ(P )φ(Q) = 0,
then there exists an either unitary or antiunitary operator U : H → H such that
φ(P ) = U P U ∗ for every P ∈ P1 . Let us just mention that this statement is
a rather straightforward consequence of the fundamental theorem of projective
geometry, see [4], where also a very short proof of the fundamental theorem of
projective geometry is given.
Lemma 3.9 Assume that dim H ≥ 3. Then there exists an either unitary or
antiunitary operator U : H → H such that
φ(A) = U AU ∗
for every A ∈ E(H).
Proof. The inverse of φ has the same properties as φ. Hence, Lemma 3.7 tells
that we can apply Uhlhorn’s theorem to find an either unitary or antiunitary
operator U : H → H such that φ(P ) = U P U ∗ for every P ∈ P1 . The map
A 7→ U ∗ φ(A)U has the same properties as φ and moreover, it acts like the
identity on P1 . Replacing φ by this map and applying Lemma 3.8 we conclude
that φ(A) = A for every A ∈ E(H) of rank one. It is known that every effect
A ∈ E(H) is equal to the supremum of all rank one effects R ∈ E(H) satisfying
R ≤ A, see [2, Corollary 3]. Consequently, we have φ(A) = A for all A ∈ E(H).
2
We will say that a continuous bijective function f : [0, 1] → [0, 1] is strictly
operator monotone on the unit interval if for every pair A, B ∈ E(H) we have
A ≤ B ⇐⇒ f (A) ≤ f (B).
In particular, if f : [0, 1] → [0, 1] is strictly operator monotone on the unit
interval, then the map φ : E(H) → E(H) defined by φ(A) = f (A), A ∈ E(H),
is an order automorphism.
Lemma 3.10 For every real number p < 1 the function fp : [0, 1] → [0, 1]
defined by
x
, x ∈ [0, 1],
fp (x) =
px + 1 − p
is strictly operator monotone on the unit interval. Moreover, the set of functions
{fp : p < 1} with the operation of functional composition is a group. We have
fp ◦ fq = fp+q−pq
19
and
p .
fp−1 = f p−1
Proof. The verification of the fact that {fp : p < 1} with the operation ◦ is a
commutative group with the identity f0 is straightforward (note that 1 − (p +
q − pq) = (1 − p)(1 − q) > 0, and hence p + q − pq < 1). It has been proved
already in [19] that the functions fp , p < 1, are strictly operator monotone. As
the proof is very short we will repeat it here for the sake of completeness. All we
need is to use the arguments from Introduction to check that for every positive
real number r the map
1
−1
δ(A) = r(r + 1) (rI + A) −
I
r+1
is an order anti-automorphism of E(H) onto [δ(I), δ(0)] = [0, I]. Clearly, δ(A) =
gr (A), where gr (x) = r 1−x
r+x , x ∈ [0, 1]. In order to complete the proof we observe
that
x
gs (gr (x)) =
, x ∈ [0, 1],
px + (1 − p)
s−r
with p = s(1+r)
∈ (−∞, 1), s, r > 0, and moreover, for every p ∈ (−∞, 1) we
s−r
can find positive real numbers r, s such that p = s(1+r)
.
2
4
The two-dimensional case
As already mentioned in Introduction, the two-dimensional case is the most
difficult and some additional work is needed to get the same result as we have
obtained for higher dimensional spaces in the previous section.
Lemma 4.1 Let φ : E2 → E2 be an order automorphism such that φ((1/2)I) =
(1/2)I. Then:
• If P, Q ∈ P1 are orthogonal, then fP = fQ .
• For every pair P, R ∈ P1 we have: tr (P R) = 1/2 ⇐⇒ tr (φ(P )φ(R)) =
1/2.
Proof. As the inverse of φ has the same properties as φ it is enough to show
that tr (P R) = 1/2 yields tr (φ(P )φ(R)) = 1/2.
So, let P , Q, and R be any projections of rank one such that P ⊥ Q and
tr (P R) = 1/2. Without loss of generality we may assume that P = E11 and
Q = E22 . By Lemma 3.7 we know that φ(P ) ⊥ φ(Q), and after composing
φ with an appropriate unitary similarity transformation, we may, and we will
assume that φ(Ejj ) = Ejj , j = 1, 2.
20
We denote fEjj = fj , j = 1, 2. Using the fact that a rank one projection has
trace one we conclude that
1 1 z
R=
2 z 1
for some complex number z of modulus one. We need to show that f1 = f2 and
that φ(R) has both diagonal entries equal to 1/2.
Set
1 1 −z
S=
2 −z 1
Clearly, S ∈ P1 and R ⊥ S. Using Lemma 3.7 once more we conclude that
φ(R) ⊥ φ(S). It follows that there exists a, 0 < a < 1, and a complex number
u with |u|2 = a(1 − a) such that
a
u
1 − a −u
φ(R) =
and φ(S) =
u 1−a
a
−u
(note that the possibilities a = 0 or a = 1 cannot occur because of bijectivity of
φ). Moreover, since fR (1/2) = fS (1/2) = 1/2, we have
1 1 z
1 a
u
φ
=
4 z 1
2 u 1−a
and
1 1
φ
4 −z
−z
1
1 1−a
=
2 −u
−u
.
a
We know that for every real p from the unit interval we have
1 0
1
0
φ
=
.
0 p
0 f2 (p)
Applying Lemma 3.3 we see that
1 0
(1/2)R ≤
0 1/3
and s(1/2)R 6≤
1
0
0
1/3
for every s > 1. And the same is true with S instead of R. Hence,
1 a
u
1 0
≤
,
0 q
2 u 1−a
where q = f2 (1/3) and there exists ε > 0 such that
1 a
u
1 0
6≤
s
0 q
2 u 1−a
21
for every s ∈ (1, 1 + ε). By Lemma 3.3 we have
a 1−a
+
=1
2
2q
and similarly,
a
1−a
+
= 1,
2
2q
which immediately yields that a = 1/2. Hence,
1 1 w
1 1
φ(R) =
and φ(S) =
2 w 1
2 −w
−w
1
for some complex number w of modulus one.
For every p ∈ (0, 1) there exists a unique r ∈ (0, 1) such that
p 0
1 0
rR ≤
,
0 1
0 p
and for every s > 1 we have
0
1
,
1
0
0
.
p
srR 6≤
p
0
1 1
2 w
w
f (p)
≤ 1
1
0
Hence,
φ(rR) = fR (r)
and there exists ε > 0 such that
1 1 w
f (p)
6≤ 1
fR (sr)
w
1
0
2
0
1
,
1
0
0
1
,
1
0
0
f2 (p)
0
f2 (p)
for every real s, 1 < s < 1 + ε. Applying the fact that fR is strictly monotone
increasing and Lemma 3.3 we conclude that f1 (p) = f2 (p), as desired.
2
Lemma 4.2 Let φ : E2 → E2 be an order automorphism such that φ((1/2)I) =
(1/2)I. Then there exists a bijective strictly increasing function f : [0, 1] → [0, 1]
such that
φ(tP ) = f (t)φ(P )
for every P ∈ P1 and every t ∈ [0, 1].
Proof. Let P and Q be any two projections of rank one. We need to show that
fP = fQ . According to Lemma 3.4 we can assume with no loss of generality
that
1 1 u
1 1 w
P =
and Q =
2 u 1
2 w 1
22
for some complex numbers u and w of modulus one. We further know that
φ(E11 ) is a projection of rank one and after composing φ with an appropriate
unitary similarity transformation we may, and we will assume that φ(E11 ) =
E11 . By the previous lemma there exists a bijective strictly increasing function
g : [0, 1] → [0, 1] such that
φ(tE11 + sE22 ) = g(t)E11 + g(s)E22
for every pair t, s ∈ [0, 1]. Moreover, we have
1 1
1 1 u0
and φ(Q) =
φ(P ) =
2 u0 1
2 w0
w0
1
for some complex numbers u0 and w0 of modulus one.
Fix p ∈ (0, 1). Then it is possible to find t, s ∈ (0, 1) such that pP ≤
tE11 + sE22 but qpP 6≤ tE11 + sE22 for any q > 1. Lemma 3.3 yields that then
also pQ ≤ tE11 + sE22 but qpQ 6≤ tE11 + sE22 for any q > 1. Applying φ to
these inequalities and then using Lemma 3.3 once more we easily conclude that
fP = fQ , as desired.
2
Lemma 4.3 Let φ : E2 → E2 be an order automorphism such that φ((1/2)I) =
(1/2)I. Let f : [0, 1] → [0, 1] be the function from the previous lemma. Then
there exists an increasing sequence (rn ) ⊂ [1/2, 1] such that
• lim rn = 1, and
• for every positive integer n we have f (rn ) = rn .
Proof. Assume without loss of generality that φ(E11 ) = E11 . Then we know
that
1 0
1
0
φ
=
, p ∈ [0, 1],
0 p
0 f (p)
and
φ
t 1
2 1
1
1
=
f (t) 1 u
,
u 1
2
for some complex number u of modulus one.
By Lemma 3.3, we have
2p
t=
p+1
if and only if
t 1
2 1
1
1
≤
1
0
23
0
p
t ∈ [0, 1],
and there exists ε > 0 such that
t 1
s
2 1
1
1
6
≤
1
0
0
p
for every s ∈ (1, 1+ε). Applying φ and then using Lemma 3.3 again we conclude
that for all pairs of real numbers t, p ∈ (0, 1) we have
t=
2f (p)
2p
⇐⇒ f (t) =
.
p+1
f (p) + 1
Take r1 = 1/2 and define
rn+1 =
2rn
.
rn + 1
Clearly, (rn ) is an increasing sequence with lim rn = 1. Using
2p
2f (p)
f
=
p+1
f (p) + 1
we get inductively that f (rn ) = rn , n = 1, 2, . . .
2
Lemma 4.4 Let φ : E2 → E2 be an order automorphism such that φ((1/2)I) =
(1/2)I. Then there exists an increasing sequence (an ) ⊂ [0, 1] such that
• lim an = 1, and
• for every positive integer n and every pair P, Q ∈ P1 we have
tr (P Q) = an ⇐⇒ tr (φ(P )φ(Q)) = an .
Proof. Set
an =
2rn − 1
,
rn
n = 1, 2, . . . ,
where (rn ) is the sequence from the previous lemma. Then clearly, lim an = 1,
and since φ and its inverse have the same properties, we only need to show
that if P, Q ∈ P1 satisfy tr (P Q) = an for a given positive integer n, then
tr (φ(P )φ(Q)) = an .
We can assume with no loss of generality that P = E11 . After composing φ
with an appropriate unitary similarity transformation, we may further assume
that φ(E11 ) = E11 . But then
1 0
1 0
=
.
φ
0 12
0 21
24
We further know that
an
Q=
u
u
1 − an
for some complex number u with |u|2 = an (1 − an ). Denote
b
w
φ(Q) =
,
w 1−b
where 0 ≤ b < 1 and w is a complex number with |w|2 = b(1 − b). The proof
will be completed when we verify that b = an .
By Lemma 3.3,
an
u
1 0
rn
≤
u 1 − an
0 21
and
srn
an
u
u
1 − an
6≤
1
0
0
1
2
for every s > 1. Using by now the standard arguments we conclude that
b=
2rn − 1
,
rn
as desired.
2
Lemma 4.5 Let φ : E2 → E2 be an order automorphism such that φ((1/2)I) =
(1/2)I. Then there exists a 2 × 2 unitary matrix U such that either
φ(P ) = U P U ∗
for every P ∈ P1 , or
φ(P ) = U P t U ∗
for every P ∈ P1 .
Proof. All we need to verify is that
tr(P Q) = tr(φ(P )φ(Q)),
P, Q ∈ P1 .
Then the conclusion follows directly from the non-bijective version of Wigner’s
theorem (for a short proof see [5]).
If we identify projections of rank one with their images, that is, with onedimensional subspaces of C2 , then it is an elementary linear algebra exercise
to see that tr (P Q) determines the angle 6 (P, Q) between images of P and Q.
Recall that the angle ϕ ∈ [0, π/2] between the images of P and Q is defined by
cos ϕ = |hx, yi|,
25
where x and y are unit vectors belonging to image of P and image of Q, respectively.
Indeed, let P and Q be two projections of rank one. Then up to a unitary
similarity we can write them as
1 0
1
P =
=
[1 0]
0 0
0
and
Q=
cos2 ϕ
cos ϕ sin ϕ
cos ϕ
=
[ cos ϕ sin ϕ ]
cos ϕ sin ϕ
sin2 ϕ
sin ϕ
for some uniquely determined ϕ ∈ [0, π/2]. Clearly,
tr (P Q) = cos2 ϕ
and
1
cos ϕ
,
= cos ϕ.
0
sin ϕ
In the rest of the proof we will need some simple well-known facts. For
the sake of completeness we will mention them and briefly sketch some of the
proofs. If we take sequences (Pn ), (Qn ) ∈ P1 , then obviously tr (Pn Qn ) tends
to 1 if and only if the angle 6 (Pn , Qn ) tends to zero. If P and Q are as above,
then I − P = E22 is the unique projection orthogonal to P and we have
π
= 6 (P, I − P ) = 6 (P, Q) + 6 (Q, I − P ).
2
It is well-known and rather easy to verify that for arbitrary rank one projections
P , Q, and R we have the triangle inequality
6
(P, Q) ≤ 6 (P, R) + 6 (R, Q).
Indeed, we may assume that P and Q are as above and then we can write
cos2 ψ
eit cos ψ sin ψ
R = −it
e cos ψ sin ψ
sin2 ψ
for some ψ ∈ [0, π/2] and some t ∈ R. If ψ ≥ ϕ, then 6 (P, Q) ≤ 6 (P, R) and
we are done. Otherwise we easily compute that
cos ϕ
cos ψ
6
cos( (R, Q)) = , −it
sin ϕ
e sin ψ = | cos ϕ cos ψ + eit sin ϕ sin ψ|
≤ | cos ϕ cos ψ + sin ϕ sin ψ| = cos(ϕ − ψ),
26
and therefore, 6 (R, Q) ≥ ϕ − ψ, as desired. It is straightforward to see that
if ϕ1 , ϕ2 ∈ [π/2] and ϕ1 + ϕ2 = 6 (P, Q) for some P, Q ∈ P1 , then there exists
R ∈ P1 such that 6 (P, R) = ϕ1 and 6 (R, Q) = ϕ2 . And finally, it is similarly
easy to check that for arbitrary P, Q ∈ P1 with 6 (P, Q) = ϕ and any ψ ∈ [0, π/2]
satisfying (1/2)ϕ ≤ ψ and ψ + ϕ ≤ π/2 we can find R such that
6
(P, R) = 6 (R, Q) = ψ.
The previous lemma tells that there is a sequence of angles (ϕn ) ∈ [0, π/2]
such that
• lim ϕn = 0, and
• for a given positive integer n and for every pair P, Q ∈ P1 we have
6 (P, Q) = ϕn ⇐⇒ 6 (φ(P ), φ(Q)) = ϕn .
It is now easy to verify that for every pair P, Q ∈ P1 we have
6
(φ(P ), φ(Q)) ≤ 6 (P, Q).
In other words, φ is a contraction with respect to the distance d(P, Q) = 6 (P, Q).
Indeed, there is nothing to prove if ϕ = 6 (P, Q) = 0 or if ϕ = π/2. Hence,
assume that 0 < ϕ < π/2 and take any ε > 0. Then we can find a positive
integer n such that ϕn < min{ε, π/2 − ϕ, ϕ}. Let m be a positive integer such
that (m − 1)ϕn ≤ ϕ < mϕn . Clearly, m ≥ 2.
We can find inductively a finite sequence of rank one projections P =
P0 , P1 , . . . , Pm−2 such that 6 (Pj−1 , Pj ) = ϕn , j = 1, . . . , m − 2, and 6 (Pj , Q) =
ϕ − jϕn . In particular,
ϕn ≤ 6 (Pm−2 , Q) < 2ϕn
and also 6 (Pm−2 , Q)+ϕn ≤ ϕ+ϕn < π/2, and therefore we can find Pm−1 ∈ P1
such that
6 (Pm−2 , Pm−1 ) = 6 (Pm−1 , Pm ) = ϕn ,
where we have denoted Pm = Q.
It follows that
6 (φ(P ), φ(Q)) = 6 (φ(P0 ), φ(Pm ))
≤ 6 (φ(P0 ), φ(P1 )) + . . . + 6 (φ(Pm−1 ), φ(Pm )) = mϕn
= (m − 1)ϕn + ϕn < 6 (P, Q) + ε.
As ε was an arbitrary positive real number we see that 6 (φ(P ), φ(Q)) ≤ 6 (P, Q),
as desired.
In the next step we will prove that we have actually
6
(φ(P ), φ(Q)) = 6 (P, Q),
27
P, Q ∈ P1 .
Let P, Q be any projections of rank one. Then
π
= 6 (P, I − P ) = 6 (P, Q) + 6 (Q, I − P ).
2
Applying
π
= 6 (φ(P ), I − φ(P )) ≤ 6 (φ(P ), φ(Q)) + 6 (φ(Q), I − φ(P ))
2
= 6 (φ(P ), φ(Q)) + 6 (φ(Q), φ(I − P ))
together with the fact that φ is a contraction we conclude that φ preserves all
angles. It follows that tr(P Q) = tr(φ(P )φ(Q)) for all P, Q ∈ P1 . This completes
the proof.
2
Corollary 4.6 Let H be a Hilbert space with dim H ≥ 2. Let φ : E(H) → E(H)
be an order automorphism such that φ((1/2)I) = (1/2)I. Then there exists an
either unitary or antiunitary operator U : H → H such that
φ(A) = U AU ∗
for every A ∈ E(H).
Proof. The case when dim H ≥ 3 has been already proved in the previous
section. In the case when dim H = 2 we know by the previous lemma that
there exists an either unitary or antiunitary operator U : H → H such that
φ(P ) = U P U ∗ for every P ∈ P1 . Applying Lemma 3.8 we conclude that
φ(A) = U AU ∗ for every rank one operator A ∈ E(H). Exactly the same
arguments as in the proof of Lemma 3.9 show that we have φ(A) = U AU ∗ for
every A ∈ E(H).
2
5
Order automorphisms of effect algebras
We will need the following statement that was proved by Busch and Gudder [2,
Theorem 3]. The proof is rather short and self-contained.
Lemma 5.1 Let H be a Hilbert space, A ∈ E(H), and P ∈ E(H) a projection
of rank one. Then there exists a positive real number t > 0 satisfying tP ≤ A if
and only if Im P ⊂ Im A1/2 .
Corollary 5.2 Let H be a Hilbert space and φ : E(H) → E(H) an order
isomorphism. Then
φ( (0, I) ) = (0, I).
28
Proof. We are interested only in the case when dim H ≥ 2, but this statement
is obviously true also in the trivial case when dim H = 1. Assume from now on
that dim H ≥ 2.
Suppose that A ∈ E(H) and A > 0. Then Im A = H, and consequently,
Im A1/2 = H. It follows from Lemma 5.1 that for every rank one projection
P there exists a positive real number tP such that tP P ≤ A (of course, one
can verify this directly without using Lemma 5.1). By Lemmas 3.1 and 3.5, φ
maps the set of projections of rank one bijectively onto the set of projections of
rank one. It follows easily that for every rank one projection Q there exists a
positive real number tQ such that tQ Q ≤ φ(A). Hence, by Lemma 5.1 we have
Im (φ(A))1/2 = H, and therefore, Im φ(A) = H. This yields that φ(A) > 0.
In the same way we show that A < I implies that φ(A) < I. Hence,
φ( (0, I) ) ⊂ (0, I), and since φ−1 is also and order automorphism, we have
φ( (0, I) ) = (0, I).
2
Lemma 5.3 Let R ∈ (0, I). Then there exist real numbers p, q, 0 < p < 1,
q < 0, and an operator S ∈ (0, I) such that
1
I = R,
ξ
2
where ξ : E(H) → E(H) is an order automorphism defined by
ξ(A) = fq (fp (S 2 ))−1/2 fp (SAS)(fp (S 2 ))−1/2 , A ∈ E(H).
Proof. A slightly weaker version of this lemma has been proved in [19]. The
same proof works also for this stronger version. We will repeat it here (with
minor changes) for the sake of completeness.
Let a, b be real numbers such that 0 < a ≤ b < 1 and σ(R) ⊂ [a, b]. If S
is an operator belonging to (0, I), 0 < p < 1, q < 0, and ξ is defined as above,
then
1
ξ
I = fq (g(S 2 )),
2
where g : (0, 1) → (0, 1) is a real function defined by
fp 21 t
g(t) =
, t ∈ (0, 1).
fp (t)
It is easy to verify that p > 0 implies that g is a strictly
increasing bijection of
1
. The function h = fq ◦ g
open interval (0, 1) onto the open interval 12 , 2−p
is
bijection of open interval (0, 1) onto the open interval
a strictly increasing
1
1
2−q , 2−p+q(p−1) . We have
lim
q→−∞
1
= 0,
2−q
29
and once the real number q is fixed we also have
1
= 1.
p→1 2 − p + q(p − 1)
lim
So we can find p, q, 0 < p < 1, q < 0, such that the function h is a strictly
increasing bijection of open interval (0, 1) onto an open interval (α, β) ⊂ (0, 1)
such that [a, b] ⊂ (α, β). Then h−1 (R) = K ∈ (0, I). Let S be the unique
positive square root of K. Then ξ((1/2)I) = h(S 2 ) = h(K) = R, as desired.
2
Proof of Theorem 2.2. By Corollary 5.2 we have φ( (0, I) ) = (0, I). In particular,
φ((1/2)I) = R for some R ∈ (0, I). Let ξ be as in Lemma 5.3. Set
ψ = ξ −1 ◦ φ : E(H) → E(H).
Then ψ is an order automorphism of E(H) satisfying ψ((1/2)I) = (1/2)I.
Corollary 4.6 yields that there exists an either unitary or antiunitary operator U : H → H such that
ψ(A) = U AU ∗
for every A ∈ E(H). It follows from ψ = ξ −1 ◦ φ that
φ(A) = ξ(U AU ∗ )
= fq (fp (S 2 ))−1/2 fp (SU AU ∗ S)(fp (S 2 ))−1/2 ,
A ∈ E(H).
Set SU = T and note that S 2 = T T ∗ to conclude that
−1/2
−1/2
φ(A) = fq (fp (T T ∗ ))
fp (T AT ∗ ) (fp (T T ∗ ))
for all A ∈ E(H), as desired.
To prove the converse we observe that T T ∗ ≤ I and then the map A →
fp (T AT ∗ ), A ∈ [0, I], is an order isomorphism of [0, I] onto [0, fp (T T ∗ )]. It is
now trivial to conclude the proof using the fact that A 7→ fq (A), A ∈ [0, I], is
an order automorphism of the effect algebra.
2
Lemma 5.4 Let R ∈ (0, I). Then there exist a real number p < 0 and an
operator S ∈ (0, ∞) such that
1
I = R,
ξ
2
where ξ : E(H) → E(H) is an order automorphism defined by
ξ(A) = fp (S + S 2 )1/2 S −1 − (S + A)−1 (S + S 2 )1/2 , A ∈ E(H).
30
Proof. The map X 7→ S + X is an order isomorphism of [0, I] onto [S, S +
I] ⊂ (0, ∞). Hence, the map X 7→ (S + X)−1 is an order anti-isomorphism
of [0, I] onto [(S + I)−1 , S −1 ], which further yields that the transformation
X 7→ −(S + X)−1 is an order isomorphism of [0, I] onto [−S −1 , −(S + I)−1 ].
Using Lemma 3.10 one can now easily conclude that ξ is an order automorphism
of E(H).
Clearly,
1
I = fp (I + S)(I + 2S)−1 .
ξ
2
Lemma 3.10 tells that for every p < 0 there is a real q, 0 < q < 1, such that fq
is the inverse of fp . Then
1
−1
I
.
(I + S)(I + 2S) = fq ξ
2
Let R ∈ (0, I). In order to complete the proof we need to find q, 0 < q < 1, and
S ∈ (0, ∞) such that
(I + S)(I + 2S)−1 = fq (R).
The spectrum of R is contained in some closed subinterval of (0, 1), σ(R) ⊂ [a, b],
0 < a ≤ b < 1.
It is easy to see that there exists q ∈ (0, 1) such that fq (a) > 1/2. Then
clearly, fq ([a, b]) ⊂ ((1/2), 1). Hence, the spectrum of fq (R) is contained in
((1/2), 1), and consequently, the spectrum of
S = (I − fq (R))(2fq (R) − I)−1
is contained in the set of positive real numbers. A direct calculation verifies
that (I + S)(I + 2S)−1 = fq (R).
2
Proof of Theorem 2.3. As in the proof of Theorem 2.2 we see that there exists
an either unitary or antiunitary operator U : H → H such that
U AU ∗ = ξ −1 (φ(A))
for every A ∈ E(H). Here, ξ is as in Lemma 5.4. It follows that
φ(A) = ξ(U AU ∗ )
= fp (S + S 2 )1/2 S −1 − (S + U AU ∗ )−1 (S + S 2 )1/2 ,
A ∈ E(H).
Hence,
φ(A) =
fp (S + S 2 )1/2 S −1/2 I − S 1/2 (S + U AU ∗ )−1 S 1/2 S −1/2 (S + S 2 )1/2 =
31
fp (I + S)1/2 I − (I + S −1/2 U AU ∗ S −1/2 )−1 (I + S)1/2 .
Set T = S −1/2 U and observe that T T ∗ = S −1 in order to complete the proof of
one direction.
All we need to do to prove the converse is to verify that the map A 7→
Ih − (I + T AT ∗ )−1 , Ai∈ [0, I], is an order isomorphism of the effect algebra onto
−1
0, I + (T T ∗ )−1
.
2
Proof of Theorem 2.4. The first part has been already proved, see Corollary
5.2. Assume that A, B ∈ (0, I). By Lemma 5.3 (or by Lemma 5.4) one can find
order automorphisms ξ1 , ξ2 of E(H) such that
1
1
ξ1
I = A and ξ2
I = B.
2
2
Set ψ = ξ2 ◦ ξ1−1 to complete the proof.
2
6
The proofs of the main results
We have already proved the structural theorems for order automorphisms of
effect algebras. In this section we will present the proofs of all other main
results. We will postpone the proof of Theorem 2.1 and start with the proof of
Theorem 2.5. We need one more lemma.
Lemma 6.1 Let A, B, C, D ∈ S(H) satisfy A < B and C < D and assume that
φ : [A, B] → [C, D] is an order isomorphism. Let P ∈ P1 and let ε = sup{t ≥
0 : A + tP ≤ B} > 0. Then there exist real numbers a, b, b > 0, and a rank one
projection Q ∈ P1 such that
φ(A + tP ) = C +
t
Q
at + b
for every t ∈ [0, ε].
Proof. We first observe that there exists a unique order automorphism ψ :
[0, I] → [0, I] such that for every X ∈ [A, B] we have
φ(X) =
(D − C)1/2 ψ (B − A)−1/2 (X − A)(B − A)−1/2 (D − C)1/2 + C.
32
Hence, we need to verify that
(D − C)1/2 ψ t(B − A)−1/2 P (B − A)−1/2 (D − C)1/2 =
t
Q,
at + b
t ∈ [0, ε],
for some real numbers a, b with b positive and some rank one projection Q.
Clearly, for every projection R1 of rank one and for every bounded invertible
linear or conjugate-linear operator S : H → H there exists a rank one projection
R2 and a positive real number c such that
S(tR1 )S ∗ = ctR2 ,
t ≥ 0.
Moreover, for every r < 1 we have
fr (tP ) =
t
P,
rt + (1 − r)
t ∈ [0, 1],
and the compositum of two real functions of the form
t 7→
t
at + b
with b positive is a function of the same form. It is now easy to complete the
proof using Theorem 2.2.
2
Proof of Theorem 2.5. Let C > 0. Then φ(C) ≥ 0. Take any B > φ(C). By
the bijectivity of φ there exists D such that φ(D) = B. Hence, φ(D) > φ(C),
and therefore D ≥ C > 0. It follows that for every A ≥ D we have A > 0 and
φ(A) > 0.
Take a positive integer n such that nI ≥ D. Then φ(nI) > 0. Replacing φ
by
A 7→ nφ(nI)−1/2 φ(A)φ(nI)−1/2 , A ∈ [0, ∞),
we may, and we will assume that
φ(nI) = nI.
Choose and fix a projection P of rank one. Let A be any positive operator
with A ≥ D. Applying the previous lemma to the map φ : [0, A] → [0, φ(A)]
we see that there exist real numbers aA , bA with bA positive and a rank one
projection QA such that
φ(tP ) =
t
QA
aA t + bA
for every t ∈ [0, c], where c = sup{t ≥ 0 : tP ≤ A}. It is trivial to verify that
t
QA ≡ Q is independent of the choice of A, A ≥ D. If two functions t 7→ at+b
and
33
t
t 7→ ct+d
that are monotone increasing in a small neighborhood of 0 coincide
on an interval [0, δ] with δ > 0, then we have a = c and b = d. It follows that
also the real constants aA ≡ a and bA ≡ b are independent of the choice of A,
A ≥ D. Hence, we have
φ(tP ) =
t
Q,
at + b
t ≥ 0.
Before we continue we observe that if P 0 and Q0 are rank one projections and
φ(tP 0 ) = g(t)Q0 , t ≥ 0, for some real function g : [0, ∞) → [0, ∞), then clearly
g must be a bijection of [0, ∞) onto itself. The verification of this simple fact is
left to the reader.
t
maps the set of all nonnegative real
If a > 0, then the function t 7→ at+b
numbers bijectively onto the interval [0, 1/a), a contradiction. Similarly we see
that the possibility a < 0 cannot occur. Hence, a = 0.
It follows that for every P ∈ P1 there exist a unique QP ∈ P1 and a unique
real number cP such that φ(tP ) = cP tQP , t ≥ 0. For a given rank one projection
P we have tP ≤ nI if and only if t ≤ n. Hence, φ(nP ) = nQ, and consequently,
cP = 1 for every P ∈ P1 .
The map P 7→ QP is a bijection of P1 onto itself. By [2, Corollary 3], each
positive operator A ∈ S(H) is the supremum of all positive rank one operators
R satisfying R ≤ A. It follows that φ([0, I]) = [0, I] and φ((1/2)I) = (1/2)I.
By Corollary 4.6 there exists a unitary or antiunitary operator U on H such
that φ(P ) = U P U ∗ for every P ∈ P1 . Replacing φ by A 7→ U ∗ φ(A)U we can
assume with no loss of generality that φ(P ) = P for every rank one projection
P . But then φ(R) = R for every rank one R ≥ 0. Applying the fact that every
positive operator A is the supremum of all positive rank one operators that are
below A, we finally conclude that φ(A) = A for every A ∈ [0, ∞).
2
We will need the following consequence of Lemma 6.1.
Corollary 6.2 Let A, B, C, D ∈ S(H) satisfy A < B and C < D and assume
that φ : [A, B] → [C, D] is an order isomorphism. Assume further that F ∈
S(H) satisfies A < F < B and C < φ(F ) < D. Let P ∈ P1 and let b = sup{t ≥
0 : F + tP ≤ B} and a = inf{t ≤ 0 : F + tP ≥ A}. Then there exist real
numbers m, n, n > 0, and a rank one projection Q such that
φ(F + tP ) = φ(F ) +
t
Q
mt + n
for every t ∈ [a, b].
Proof. We apply Lemma 6.1 for the operator intervals [F + aP, B] and [φ(F +
aP ), D] and for real numbers t ∈ [0, b − a] to conclude that there exist real
34
numbers c, d with d > 0 and a rank one projection Q such that
φ(F + aP + tP ) = φ(F + aP ) +
t
Q
ct + d
for every nonnegative real t ≤ b − a. Choosing t = −a we get
φ(F + aP ) = φ(F ) +
a
Q,
d − ac
and consequently, for every t ∈ [a, b] we have
φ(F + tP ) = φ(F + aP + (t − a)P ) = φ(F ) +
a
t−a
Q +
Q
d − ac
c(t − a) + d
t
Q,
mt + n
where m and n are real numbers with n > 0.
= φ(F ) +
2
Proof of Theorem 2.6. Replacing φ by the map X 7→ φ(I)−1/2 φ(X)φ(I)−1/2 ,
X ∈ (0, ∞), we can assume that φ(I) = I. Then clearly,
φ( [I, ∞) ) = [I, ∞)
and φ( (0, I] ) = (0, I].
The map ψ(X) = φ(I + X) − I, X ∈ [0, ∞), is an order automorphism of
[0, ∞), and by Theorem 2.5 there exists a bounded invertible linear or conjugatelinear operator T : H → H such that
φ(A) = T AT ∗ + (I − T T ∗ )
for every A ∈ [I, ∞).
−1
Similarly, the map τ (X) = φ (I + X)−1
− I, X ∈ [0, ∞), is an order
automorphism of [0, ∞), and using the substitution A = (I +X)−1 and Theorem
2.5 we arrive at
−1
φ(A) = I + S A−1 − I S ∗
, A ∈ (0, I],
for some bounded invertible linear or conjugate-linear operator S : H → H.
Let P be any rank one projection. Then for every t ≥ 0 we have
φ(I + tP ) = I + ctQ,
where Q is the rank one projection and c the positive real number satisfying
T P T ∗ = cQ.
35
When t ≤ 0 the calculation is just a little bit longer. For −1 < t ≤ 0 we
have
−1
.
φ(I + tP ) = I + S (I + tP )−1 − I S ∗
From
(I + tP )−1 = I −
t
P
1+t
we get
−1
dt
,
φ(I + tP ) = I −
R
1+t
where R is the rank one projection and d the positive real number such that
SP S ∗ = dR. Hence,
dt
φ(I + tP ) = I +
R,
1 + t − dt
and if we write b =
1−d
d ,
then
φ(I + tP ) = I +
t
R,
bt + (1 + b)
−1 < t ≤ 0.
By now standard arguments yield the existence of self-adjoint operators A ∈
(0, I] and B ∈ [I, ∞) such that A < I < B and φ(A) < I < φ(B). Applying
Corollary 6.2 with order isomorphism φ : [A, B] → [φ(A), φ(B)] and F = I we
see that Q = R and
φ(I + tP ) = I + ctQ
for all real numbers t, t > −1. It follows that b = 0, and consequently, c = 1.
In particular, for every rank one projection P the rank one operator T P T ∗
is a projection of rank one which yields that T is either a unitary or antiunitary
operator. After replacing φ by X 7→ T ∗ φ(X)T , X ∈ (0, ∞), we have
φ(A) = A,
A ∈ [I, ∞),
and for every rank one projection P and for every real t, t > −1, we have
φ(I + tP ) = I + tP.
In the rest of the proof we use similar arguments as before. We will give here
just the main ideas. First, it is easy to check that φ(A) = A for every A ∈ (0, I].
For an arbitrary A ∈ (0, ∞) we can find positive real numbers a, b such that
0 < a < 1 < b and aI ≤ A ≤ bI. Since φ(aI) = aI and φ(bI) = bI we see that
the restriction of φ to the operator interval [aI, bI] is an order automorphism
which acts like the identity on the subintervals [aI, I] and [I, bI]. Applying the
canonical order isomorphism between the effect algebra [0, I] and [aI, bI] and the
structural theorem for order automorphisms of [0, I] we see that the restriction
of φ to [aI, bI] is the identity map. In particular, φ(A) = A, as desired.
36
2
Now we are ready to prove Theorem 2.1. A short proof has been already
given in [13] but we were able to find even a shorter and simpler proof.
Proof of Theorem 2.1. We will prove the statement by contradiction. So, assume
there exists an order isomorphism ϕ : (0, ∞) → (−∞, ∞). With no loss of
generality we can assume that ϕ(I) = 0, since otherwise we replace ϕ by A 7→
ϕ(A) − ϕ(I), A ∈ (0, ∞). We can find a positive integer m such that N =
ϕ(mI) > 0. Set M = ϕ((m + 1)I). Clearly, the map ψ : (−∞, ∞) → (−∞, ∞)
defined by
ψ(A) = M 1/2 N −1/2 AN −1/2 M 1/2 , A ∈ S(H),
is an order automorphism, and consequently, the map φ : (0, ∞) → (0, ∞)
defined by
φ(A) = ϕ−1 (ψ(ϕ(A))), A ∈ (0, ∞),
is an order automorphism, too. It is straightforward to check that φ(I) = I,
and therefore, by Theorem 2.6, we have φ(A) = U AU ∗ , A ∈ (0, ∞), for some
unitary or antiunitary operator U on H. This contradicts the obvious fact that
φ(mI) = (m + 1)I.
2
Proof of Theorem 2.7. We replace φ by the map X 7→ φ(X) − φ(0). Thus, we
may assume that φ(0) = 0 and then
φ( [0, ∞) ) = [0, ∞)
and φ( (−∞, 0]) = (−∞, 0].
By Theorem 2.5 we see that there exists a bounded invertible linear or conjugatelinear operator T : H → H such that φ(A) = T AT ∗ for every A ≥ 0. Replacing
φ again, this time by X 7→ T −1 φ(X)(T ∗ )−1 , X ∈ S(H), we can assume with no
loss of generality that φ(A) = A for every A ∈ [0, ∞).
Using the fact that (−∞, 0] is order anti-isomorphic to [0, ∞) we conclude
that there exists a bounded invertible linear or conjugate-linear operator S :
H → H such that φ(A) = SAS ∗ for every A ≤ 0.
Using Corollary 6.2 and similar arguments as in the proof of Theorem 2.6
we see that for every projection P of rank one we have
φ(tP ) = tP,
t ∈ R.
In particular, −SP S ∗ = −P for every projection P of rank one. This yields that
S = zI for some complex number z of modulus one. It follows that φ(A) = A
for every A ∈ (−∞, 0]. As in the proof of Theorem 2.6 we show that actually
φ(A) = A for every A ∈ S(H).
2
37
For the proof of the next theorem we need a simple lemma.
Lemma 6.3 Let H be a Hilbert space and K ⊂ H a dense linear subspace.
Assume that L : K → H is a linear operator such that for every u ∈ K we have
hLu, ui ∈ R and
0 ≤ hLu, ui ≤ kuk2 .
Then there exists a bounded linear self-adjoint operator L̃ : H → H such that
L̃ ∈ [0, I] and L̃|K = L.
Proof. For every pair u, v ∈ K we have
hLu, vi =
1
(hL(u + v), u + vi − hL(u − v), u − vi)
4
i
+ (hL(u + iv), u + ivi − hL(u − iv), u − ivi) .
4
It follows that
|hLu, vi| ≤
1
(ku + vk2 + ku − vk2 + ku + ivk2 + ku − ivk2 )
4
≤ (kuk + kvk)2 .
Thus, for every u, v ∈ K with kuk, kvk ≤ 1 we have
|hLu, vi| ≤ 4,
and consequently, L : K → H is bounded. Hence, there exsits a unique extension L̃ : H → H. By continuity we have
hL̃u, ui ∈ [0, kuk2 ]
for every u ∈ H, and therefore, L̃ is self-adjoint, and moreover, it belongs to
the effect algebra [0, I].
2
Proof of Theorem 2.8. As already mentioned in the second section all we need
to do is to prove that for any pair A, B ∈ [0, I] the inequality φ(A) ≤ φ(B)
yields that A ≤ B, and we need to show that φ is bijective.
If φ(A) ≤ φ(B), that is, T 1/2 AT 1/2 ≤ T 1/2 BT 1/2 , then for every u ∈ H we
have
hAT 1/2 u, T 1/2 ui ≤ hBT 1/2 u, T 1/2 ui.
Hence, hAv, vi ≤ hBv, vi for every v ∈ Im T 1/2 , and since Im T 1/2 is dense in H
we have A ≤ B.
To prove that φ is surjective we choose and fix any C ∈ [0, T ]. By T −1/2 we
denote the bijective linear map from Im T 1/2 onto H. We claim that Im C ⊂
38
Im T 1/2 . Indeed, if u ∈ Im C, then clearly, u ∈ Im C 1/2 , and therefore, by
Lemma 5.1 we know that there exists a positive real number t such that
tP ≤ C ≤ T,
where P is a rank one projection onto the span of u. Applying Lemma 5.1 once
more we see that u ∈ Im T 1/2 , as desired.
Hence, T −1/2 CT −1/2 is a well-defined linear operator from Im T 1/2 into H.
Let u be any vector from Im T 1/2 . Then u = T 1/2 v for some v ∈ H and we have
hT −1/2 CT −1/2 u, ui = hT 1/2 T −1/2 CT −1/2 T 1/2 v, vi = hCv, vi ≥ 0.
Moreover,
hT −1/2 CT −1/2 u, ui = hCv, vi ≤ hT v, vi = hT 1/2 v, T 1/2 vi = hu, ui.
We apply the previous lemma with L = T −1/2 CT −1/2 and K = Im T 1/2 .
Denote L̃ = A. For every u ∈ H we have
T 1/2 AT 1/2 u = T 1/2 A|Im T 1/2 T 1/2 u = T 1/2 (T −1/2 CT −1/2 )T 1/2 u = Cu.
2
Proof of Theorem 2.9. Assume on the contrary that there exists an order isomorphism φ : J1 → J2 . Using the same ideas as before we can find A, B ∈ J1
such that A < B and φ(A) < φ(B). Then, of course, the restriction of φ to
[A, B] is an order isomorphism of [A, B] onto [φ(A), φ(B)]. But these two intervals are order isomorphic to effect algebras. Thus, we can assume without loss
of generality that φ is an order isomorphism from E(H) onto E(K).
We claim that φ((1/2)IH ) ∈ (0, I)K . The verification goes through in exactly
the same way as in the case of order automorphisms of E(H). Using Lemma 5.3
or Lemma 5.4 we see that there exists an order isomorphism φ : E(H) → E(K)
such that φ((1/2)I) = (1/2)I. Using the same arguments as in the proof of
Lemma 3.7 we see that both φ and its inverse map orthogonal pairs of rank
one projections into orthogonal pairs of rank one projections. In particular,
φ maps every maximal set of pairwise orthogonal rank one projections on H
bijectively onto a pairwise orthogonal rank one projections on K, contradicting
the assumption that the cardinalities of orthonormal bases of H and K are not
the same.
2
7
Final remarks
1. Theorems 2.2 and 2.3 look somewhat mysterious. There are two formulas
describing the general form of order automorphisms of effect algebras.
39
They look quite different and it is not at all obvious that they describe
the same class of transformations. Let us explain this in more detail. The
map given in the conclusion of Theorem 2.2 is of the form
φ1 (A) = ξ1 (U AU ∗ ),
A ∈ E(H),
where U is a unitary or antiunitary operator on H and ξ1 is an automorphism of E(H) given by
ξ1 (A) = fq (fp (S12 ))−1/2 fp (S1 AS1 )(fp (S12 ))−1/2 , A ∈ E(H),
with 0 < p < 1, q < 0, and S1 ∈ (0, I), while the map given in the
conclusion of Theorem 2.3 is of the form
φ2 (A) = ξ2 (U AU ∗ ),
A ∈ E(H),
where U is a unitary or antiunitary operator on H and ξ2 is an automorphism of E(H) given by
ξ2 (A) = fr (S2 + S22 )1/2 S2−1 − (S2 + A)−1 (S2 + S22 )1/2 , A ∈ E(H),
for some strictly positive S2 ∈ S(H) and some r < 0. We can rewrite ξ2
as follows:
ξ2 (A) =
−1/2
−1/2
−1/2 −1
−1/2
= fr (S2 + S22 )1/2 S2
I − (I + S2
AS2
)
S2
(S2 + S22 )1/2
−1/2
−1/2 −1
= fr (I + S2 )1/2 I − (I + S2
AS2
)
(I + S2 )1/2 .
Recall that each effect A is the supremum of all rank one effects that are
below A. Thus, the behaviour of an order automorphism φ : E(H) →
E(H) is completely determined by its behaviour on the set of rank one
effects.
If T ∈ S(H) is invertible, then for every rank one projection P there exist
a unique positive real number c and a unique rank one projection Q such
that the map X 7→ T XT sends tP into ctQ for every real t. Further, for
p < 1 the order automorphism X 7→ fp (X) of E(H) sends tP into fp (t)P
for every rank one projection P and every real t ∈ [0, 1], and finally, the
t
P for
map X 7→ I − (I + X)−1 from [0, ∞) onto [0, I) maps tP into t+1
every rank one projection P and every t ≥ 0.
We will first consider the transformation ξ1 . By the previous paragraph we
see that for every rank one projection P there exist positive real numbers
c1 , c2 and a rank one projection Q such that
ξ1 (tP ) = fq (c2 fp (c1 t))Q,
40
t ∈ [0, 1].
A straightforward calculation shows that we have
ξ1 (tP ) =
t
Q,
at + b
t ∈ [0, 1],
for some real numbers a, b with b > 0. We further know that ξ1 (P ) = Q,
and therefore, a = 1 − b < 1, and
ξ1 (tP ) = fa (t)Q.
Let the image of P be spanned by a nonzero vector x. We will find
out what Q is. We know it is a projection of rank one, and therefore,
all we need to know is which vector spans the image of Q. The transformation X 7→ fp (S1 XS1 ) sends P into a rank one effect whose image is spanned by S1 x. After applying the congruence transformation
X 7→ (fp (S12 ))−1/2 X(fp (S12 ))−1/2 we arrive at a rank one effect whose image is spanned by fp (S12 )−1/2 S1 x. Because the map X 7→ fq (X) sends
every rank one effect into a rank one effect with the same image, we conclude that the image of Q is spanned by
−1/2 −2 −1/2
fp (S12 )−1/2 S1 x = S12 (pS12 + (1 − p)I)−1
S1
x
= (pS12 + (1 − p)I)1/2 x.
In other words, the image of Q is spanned by
1/2
p
S12
x.
I+
1−p
Next, we will calculate ξ1 ((1/2)I). We have
1
1 2
ξ1
I = fq fp (S12 )−1 fp
S1
.
2
2
A straightforward calculation gives us
!
fp x2
1 + bx
=
,
fq
fp (x)
bx + 2 − q
where
b=
Note that the real function p 7→
(0, 1) onto (0, ∞).
x ∈ [0, 1],
p
.
1−p
p
1−p
is a bijection of the open interval
Hence,
ξ1
1
I
2
= (bS12 + I)(bS12 + (2 − q)I)−1 .
41
Since S1 belongs to the operator interval (0, I), the operator bS12 = T
is strictly positive. And, of course, for every strictly positive operator T
there exists p, 0 < p < 1, and S1 ∈ (0, I) such that
bS12 =
p
S 2 = T.
1−p 1
Now we can describe the action of the map ξ1 in an alternative way. As
above we denote bS12 = T . Thus, we are given a strictly positive operator
T and a negative real number q. The map ξ1 can be described as follows:
• ξ1 sends (1/2)I into (T + I)(T + (2 − q)I)−1 = R.
• For every nonzero vector x ∈ H the rank one projection Px whose
image is the linear span of x is mapped into the rank one projection
P(T +I)1/2 x whose image is the linear span of (T + I)1/2 x.
• For every nonzero x ∈ H we set
sx = sup{t ∈ [0, 1] : tP(T +I)1/2 x ≤ R}.
There exists a unique ax < 1 such that fax (1/2) = sx .
• If A is a rank one effect, then it is of the form A = tPx for some t ∈
[0, 1] and some nonzero x ∈ H. We have ξ1 (A) = fax (t)P(T +I)1/2 x .
• We now use the fact that every nonzero effect A is the supremum of
all rank one effects that are below A. Hence,
ξ1 (A) = sup{ξ1 (R) : R ∈ E(H)1 and R ≤ A}.
Here, E(H)1 denotes the set of all effects of rank one.
We will now show that the map ξ2 acts in the same way. As in the case of
ξ1 it is straightforward to verify that for every rank one projection P there
exists a unique rank one projection Q and a unique real number a < 1
such that
ξ2 (tP ) = fa (t)Q
for every t ∈ [0, 1]. Similarly as above we see that if the image of P is
spanned by a nonzero vector x, then the image of Q is spanned by
−1/2
(I + S2 )1/2 S2
x = (I + S2−1 )1/2 x.
And finally,
ξ2
1
I
2
= fr
−1 !!
1 −1
(I + S2 ) I − I + S2
2
= (I + S2 )(I + (2 − r)S2 )−1 = (I + S2 )S2−1 (S2−1 )−1 (I + (2 − r)S2 )−1
42
= (I + S2−1 )(S2−1 + (2 − r)I)−1 .
Hence, we see that ξ1 = ξ2 if and only if
p
S 2 = S2−1
1−p 1
and q = r
and in this case the behaviour of this map on rank one effects (that
uniquely determines the behaviour on the whole effect algebra) is well
understood. It should be mentioned here once more that both sets
p
S12 : 0 < p < 1 and S1 ∈ (0, I)
1−p
and
−1
S2 : S2 ∈ S(H) and S2 > 0
coincide with the set of strictly positive operators. Thus, the classes of
transformations that appear in the conclusions of Theorems 2.2 and 2.3
are indeed equal.
2. It is clear that the statement of Theorem 2.2 remains true if we replace
the request that 0 < p < 1 and q < 0 with the weaker condition that
p, q < 1.
3. We know that the map A 7→ A−1 − I, A ∈ (0, I), is an order antiisomorphism of (0, I) onto (0, ∞) and its inverse A 7→ (I + A)−1 is an
order anti-isomorphism of (0, ∞) onto (0, I). This together with Theorem
2.6 yields that every order automorphism φ : (0, I) → (0, I) is of the form
φ(A) = I + T (A−1 − I)T ∗
−1
,
A ∈ (0, I),
where T is some bounded invertible linear or conjugate-linear operator on
H. By Theorem 2.2 and Corollary 5.2, for every pair of real numbers p, q,
0 < p < 1, q < 0, and every bijective linear or conjugate-linear bounded
operator T : H → H with kT k ≤ 1 the map
−1/2
−1/2
φ(A) = fq (fp (T T ∗ ))
fp (T AT ∗ ) (fp (T T ∗ ))
, A ∈ (0, I),
is also an order automorphism of (0, I). As in the first item we have two
formulas that look very different. But, of course, our theorems tell that
each map of the second form can be written also in the first form. A
detailed explanation can be given in a similar manner as in the first item
and is left to the reader.
4. The next problem we will discuss is whether our results are optimal. Let
us consider the optimality in the case of Theorem 2.5. The first question
43
is whether we get the same conclusion under the weaker assumption that
φ : [0, ∞) → [0, ∞) is a bijective map preserving order in one direction
only. Recall that φ preserves order in one direction if for every pair A, B ∈
[0, ∞) we have
A ≤ B ⇒ φ(A) ≤ φ(B).
Here we need some facts about operator monotone functions. For example,
the square root is operator monotone on [0, ∞), while the square is not
operator monotone. More precisely, for every A, B ∈ [0, ∞) we have
A ≤ B ⇒ A1/2 ≤ B 1/2 ,
but the converse implication is false. It follows that
A 7→ A1/2 ,
A ∈ [0, ∞),
is a bijective map of [0, ∞) onto itself which preserves order in one direction, but is not an order automorphism. As the theory of operator
monotone functions is highly non-trivial there is no much hope to get a
nice description of bijective maps on [0, ∞) that preserve order in one direction only. Moreover, we have other examples of such maps that are not
of the standard form as in the conclusion of Theorem 2.5. For example,
define ψ : [0, ∞) → [0, ∞) by ψ(0) = 0,
1
ψ(A) = 1 − n A
2
whenever rank A = n, and ψ(A) = A whenever Im A is infinite-dimensional
(in this case we say that rank A = ∞). It is well-known (and easy to verify) that for any pair of positive operators C, D satisfying C ≤ D we have
rank C ≤ rank D. Using this fact it is easy to verify that ψ is a bijective
map on [0, ∞) preserving order in one direction, but it is not an order
automorphism.
5. The next question when considering the optimality of Theorem 2.5 is
whether we can relax the bijectivity assumption. Clearly, it is enough to
assume that φ is surjective to get the same conclusion. Namely, each map
φ preserving order in both directions is injective. Indeed, if φ(A) = φ(B),
then A ≤ B and B ≤ A, and consequently, A = B.
Can we get a nice description of maps φ : [0, ∞) → [0, ∞) satisfying
A ≤ B ⇐⇒ φ(A) ≤ φ(B) in the absence of the surjectivity assumption?
The answer is negative. To see this assume that H is infinite-dimensional.
Then H can be identified with H ⊕ H, and the operators on H ⊕ H can be
represented by 2×2 operator matrices. The map φ : [0, ∞)H → [0, ∞)H⊕H
defined by
A
0
A 7→
, A ∈ [0, ∞),
0 ϕ(A)
44
preserves order in both directions whenever ϕ : [0, ∞) → [0, ∞) preserves
order in one direction. We have seen in the previous item that at present
there is no much hope to characterize bijective maps from [0, ∞) onto itself
preserving order in one direction only. But we have here even a much more
general situation; the only request on ϕ is that it preserves order in one
direction and no bijectivity is assumed. Thus, we have a lot of additional
examples. In particular, we have even a lot of linear maps that preserve
order in one direction (in the presence of the linearity assumption such
maps are called positive) but do not preserve order in both directions just recall the non-trivial theory of linear positive and completely positive
maps on matrix and operator algebras.
6. We will conclude the paper by a short remark on strictly operator monotone functions. Let I, J ⊂ R be intervals. A continuous bijective function
f : I → J is said to be strictly operator monotone if for every pair of
self-adjoint bounded operators A, B whose spectra belong to the interval
I we have
A ≤ B ⇐⇒ f (A) ≤ f (B).
It is well-known that the theory of operator monotone functions, that
is, functions satisfying the weaker condition A ≤ B ⇒ f (A) ≤ f (B), is
highly non-trivial. Under the stronger condition of preserving order in
both directions the problem becomes rather easy and can be treated in an
elementary way.
Let us start with strictly operator monotone functions f : [a, b] → [c, d]
between two closed intervals. Of course, if we want to have a full understanding of such functions it is enough to check what are all strictly
operator monotone functions of the unit interval. So, let f : [0, 1] → [0, 1]
be strictly operator monotone. Then φ(A) = f (A), A ∈ E(H), is an order
automorphism of the effect algebra. We know that for every projection
P of rank one there exist a rank one projection Q and p < 1 such that
φ(tP ) = fp (t)Q, t ∈ [0, 1]. On the other hand, φ(tP ) = f (tP ) = f (t)P .
We conclude that the set of all strictly operator monotone functions of the
unit interval onto itself is the set of all functions {fp : p < 1}.
In a similar way we see that the set of all strictly operator monotone
functions f : [0, ∞) → [0, ∞) is the set of all functions of the form f (t) =
ct, where c > 0. Just slightly less easy is the question what are all strictly
operator monotone functions from [0, ∞) onto [0, 1). Let f : [0, ∞) →
[0, 1) be strictly operator monotone. Again, the map φ : [0, ∞) → [0, I)
defined by φ(A) = f (A), A ∈ [0, ∞), is an order isomorphism between
operator intervals [0, ∞) and [0, I). Then the map
ψ(A) = (I − φ(A))
45
−1
− I,
A ∈ [0, ∞),
is an order automorphism of [0, ∞). Thus, it is of the form ψ(A) = T AT ∗
for some bounded invertible linear or conjugate-linear operator T : H →
H. If P is a rank one projection, then
ψ(tP ) = tT P T ∗ = ctQ,
t ≥ 0,
where c is the positive real number and Q the rank one projection such
that T P T ∗ = cQ. On the other hand,
ψ(tP ) = (I − f (t)P )−1 − I =
f (t)
P,
1 − f (t)
t ≥ 0.
It follows that strictly operator monotone functions f : [0, ∞) → [0, 1) are
exactly all functions of the form
f (t) =
t
,
t+p
t ≥ 0,
where p is any positive number.
The general form of strictly operator monotone functions between other
intervals can be found using similar simple arguments.
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47