Von Neumann Algebras.
Vaughan F.R. Jones
1
13th May 2003
1 Supported
in part by NSF Grant DMS9322675, the Marsden fund UOA520,
and the Swiss National Science Foundation.
2
Chapter 1
Introduction.
3
4
Chapter 2
Background and Prerequisites
2.1 Hilbert Space
A Hilbert Space is a complex vector space
C
H with inner product h, i : HxH →
hξ, ηi = hη, ξi, is positive
which is linear in the second variable, satises
hξ, ξi > 0
p
hξ, ξi.
denite, i.e.
||ξ|| =
Exercise 2.1.1
for
ξ 6= 0,
and is complete for the norm dened by test
Prove the parallelogram identity :
||ξ − η||2 + ||ξ + η||2 = 2(||ξ||2 + ||η||2 )
and the Cauchy-Schwartz inequality:
|hξ, ηi| ≤ ||ξ|| ||η||.
Theorem 2.1.2
If C is a closed convex subset of H and ξ is any vector in
H, there is a unique η ∈ C which minimizes the distance from ξ to C , i.e.
||ξ − η 0 || ≤ ||ξ − η|| ∀η 0 ∈ C .
Proof. This is basically a result in plane geometry.
0
Uniqueness is clearif two vectors η and η in C minimized the distance,
ξ
and these two vectors dene a plane and any vector on the line segment
0
between η and η would be strictly closer to ξ .
d be the distance from C to ξ and choose a sequence
ηn ∈ C with ||ηn − ξ|| < d + 1/2n . For each n, the vectors ξ , ηn and ηn+1
dene a plane. Geometrically it is clear that, if ηn and ηn+1 were not close,
some point on the line segment between them would be closer than d to ξ .
To prove existence, let
Formally, use the parallelogram identity:
||ξ −
ηn + ηn+1 2
ξ − ηn ξ − ηn+1 2
|| = ||
+
||
2
2
2
5
= 2(||
ξ − ηn+1 2
ξ − ηn 2
|| + ||
|| − 1/8||ηn − ηn+1 ||2 )
2
2
≤ (d + 1/2n )2 − 1/4||ηn − ηn+1 ||2
Thus there is a constant
2
would be less than d .
Thus
(ηn )
K
such that
||ηn − ηn+1 ||2 < K/2n
C
is Cauchy, its limit is in
and has distance
or
d
||ξ − ηn +η2 n+1 ||2
from
ξ.
Exercise 2.1.3
If
φ ∈ H∗
(the Banach-space dual of
continuous linear functionals from
of
H.
φ 7→
to
C), kerφ
H
consisting of all
is a closed convex subset
with φ(η) =
ξφ is a conjugate-linear isomorphism from H∗ onto H.
Show how to choose a vector
and that
H
ξφ
orthogonal to
ker φ
hξφ , ηi
We will be especially concerned with s eparable Hilbert Spaces where there
is an orthonormal basis, i.e.
hξi , ξj i = 0
all the ξi .
for
i 6= j
Exercise 2.1.4
a sequence
and such that
{ξi }
is
of unit vectors with
is the only element of
{ξi } is
dense in H.
Show that is
linear span of the
0
{ξ1 , ξ2 , ξ3 , ...}
H
orthogonal to
an orthonormal basis for
H
then the
a : H → K is said to be bounded if there is a
||aξ|| ≤ K||ξ|| ∀ξ ∈ H. The inmum of all such K is called
the n orm of a, written ||a||. The set of all bounded operators on H is written
B(H). Boundedness of an operator is equivalent to continuity.
To every bounded operator a between Hilbert spaces H and K, by exercise
∗
2.1.3 there is another, a , between K and H, called the a djoint of a which is
∗
dened by the formula haξ, ηi = hξ, a ηi.
A linear map (operator)
number
K
with
Exercise 2.1.5
Prove that
||a|| =
sup
|haξ, ηi|
||ξ||≤1,||η||≤1
= ||a∗ || = ||a∗ a||1/2 .
The identity map on H is a bounded operator denoted 1.
∗
An operator a ∈ B(H) is called self-adjoint if a = a .
2
∗
An operator p ∈ B(H) is called a projection if p = p = p .
a ∈ B(H) is called positive if haξ, ξi ≥ 0 ∀ξ ∈ B(H). We
say a ≥ b if a − b is positive.
∗
An operator u ∈ B(H) is called an isometry if u u = 1.
∗
∗
An operator v ∈ B(H) is called a unitary if uu = u u = 1.
∗
An operator u ∈ B(H) is called a partial isometry if u u is a projection.
An operator
The last three denitions extend to bounded linear operators between
dierent Hilbert spaces.
6
Exercise 2.1.6
Show that every
a ∈ B(H)
is a linear combination of two
self-adjoint operators.
Exercise 2.1.7
Exercise 2.1.8
A positive operator is self-adjoint.
Find an isometry from one Hilbert space to itself that is not
H = `2 (N) is a ne example. There is an
unitary. The unilateral shift on
obvious orthonormal basis of
just sends the
nth.
Exercise 2.1.9
K
If
H
indexed by the natural numbers and the shift
basis element to the
K
(n + 1)th.
is a closed subspace of
which assigns to any point in
H
H
show that the map
the nearest point in
K
PK : H →
is linear and a
projection.
Exercise 2.1.10
Show that the correspondence
ercise is a bijection between closed subspaces of
⊥
is a subset of H, S
is by denition
⊥
Note that S is always a closed subspace.
If
S
K → PK of the previous exH and projections in B(H).
{ξ ∈ H : hξ, ηi = 0 ∀η ∈ S}.
Exercise 2.1.11
If
K
is a closed subspace then
Exercise 2.1.12
If
u
is a partial isometry then so is
K⊥⊥ = K
u,
is then closed and called the initial domain of
closed and called the nal domain of
u.
u∗ .
and
PK⊥ = 1−PK .
The subspace
the subspace
uH
u∗ H
is also
Show that a partial isometry is the
composition of the projection onto its initial domain and a unitary between
the initial and nal domains.
The commutator
Exercise 2.1.13
If
[a, b]
K
of two elements of
is a closed subspace and
aK ⊆ K
In general
(aK ⊆ K
B(H)
and
if f
is the operator
a = a∗
ab − ba.
then
[a, PK ] = 0.
a∗ K ⊆ K) ⇐⇒ [a, PK ] = 0.
2.2 The Spectral Theorem
The spectrum
σ(a)
Exercise 2.2.1
of
a ∈ B(H)
||a||
{λ ∈ C : a − λ1
(Look up proofs if necessary.)
empty closed bounded subset of
with either
is
or
−||a||
in
C
σ(a).
and that if
7
a
is not invertible}.
Show that
= a∗ , σ(a)
σ(a) is a non⊆ [−||a||, ||a||]
The spectral theorem takes a bit of getting used to and knowing how
to prove it does not necessarily help much. If one cannot see the spectral
decomposition of an operator it may be extremely dicult to obtainexcept
in a small nite number of dimensions where it is just diagonalisation. But
∗
fortunately there is nothing like a course in operator algebras, either C or
von Neumann, to help master the use of this theorem which is the heart of
linear algebra on Hilbert space. The book by Reed and Simon, Methods of
mathematical physics vol. 1, Functional Analysis, contains a treatment of
the spectral theorem which is perfect background for this course.
We will
make no attempt to prove it herejust give a vague statement which will
establish terminology.
The spectral theorem asserts the existence of a projection valued measure
a∗ or more generally when
from the Borel subsets of th σ(a) (when a =
a is normal i.e. [a, a∗ ] = 0) to projections in
λ → E(λ),
B(H),
written symbolically
such that
Z
a=
λdE(λ).
This integral may be interpreted as a limit of sums of operators (necessitating
R
B(H)), as a limit of sums of vectors
aξ = λdE(λ)ξ or simply
R
in terms of measurable functions hξ, aηi =
λdhξ, E(λ)ηi. The projections
E(B) are called the spectral projections of a and their images are called the
spectral subspaces of a.
Given any bounded Borel complex-valued function f on σ(a) one may
R
form f (a) by f (a) =
f (λ)dE(λ).
a topology on
Exercise 2.2.2
If
Exercise 2.2.3
If
µ is a sigma-nite measure on X and f ∈ L∞ (X, µ),
2
2
the operator Mf : L (X, µ) → L (X, µ), Mf g(x) = f (x)g(x), is a bounded
ess−sup
operator with ||Mf || =
(|f (x)|). If f is real valued then Mf is self
x∈X
adjoint. Find σ(f ) and the projection-valued measure E(λ).
measure or
a
dim(H) < ∞
nd the spectrum and projection-valued
(which is a Hermitian matrix).
The example of exercise 2.2.2 is generic in the sense that there is a version
of the spectral theorem which asserts the following. If ξ ∈ H is any vector
a = a∗ ∈ B(H), let K be the closed linear span of the {an ξ : n =
and
0, 1, 2, 3, ...}, then a denes a self-adjoint operator on K and there is a nite
measure µ on the spectrum σ(a) such that (K, a) is isomorphic in the obvious
2
sense to (L (σ(a), µ), multiplication by x). Continuing such an argument by
⊥
restricting to K one obtains a full spectral theorem.
8
Exercise 2.2.4
Show that a self-adjoint operator
a
is the dierence
of two positive operators called the positive and negative parts of
as functions of
a
a,
a+ − a−
obtained
as above.
2.3 Polar decomposition
Exercise 2.3.1
square root
a
1/2
Show that every positive operator
a
has a unique positive
.
Given an arbitrary
Exercise 2.3.2
a ∈ B(H)
we dene
|a| = (a∗ a)1/2 .
Show that there is a partial isometry
and that
u
ker(a)⊥ .
The nal domain of this
u
such that
a = u|a|,
is unique subject to the condition that its domain subspace is
u is Im(a) = ker(a∗ )⊥ .
2.4 Tensor product of Hilbert Spaces.
H and K are Hilbert spaces one may form
H ⊗alg K (in the category of complex vector
If
their algebraic tensor product
spaces). On this vector space
one denes the sesquilinear form
hξ ⊗ η, ξ 0 ⊗ η 0 i = hξ, ξ 0 ihη, η 0 i
and observes that this form is positive denite.
The Hilbert space tensor
H ⊗ K is then the completion of H ⊗alg K. It is easy to see that if
a ∈ B(H), b ∈ B(K), there is a bounded operator a ⊗ b on H ⊗ K dened by
a ⊗ b(ξ ⊗ η) = aξ ⊗ bη .
product
Exercise 2.4.1
Let
L2 (X, H, µ)
be the Hilbert space of measurable square
integrable functions (up to null sets) f : X →
f ∈ L2 (X, µ) let fξ ∈ L2 (X, H, µ) be dened by
the map
ξ ⊗ f 7→ fξ
denes a unitary from
9
H. For each ξ ∈ H and
fξ (x) = f (x)ξ . Show that
H ⊗ L2 (X, µ) onto L2 (X, H, µ).
10
Chapter 3
The denition of a von Neumann
algebra and some basic examples.
3.1 Topologies on B(H)
(i) The norm
or uniform
topology is given by the norm
||a||
dened in the
previous chapter.
(ii)The topology on
B(H)
H is called the
a ∈ B(H) is formed
of pointwise convergence on
strong operator topology. A basis of neighbourhoods of
by the
N (a, ξ1 , ξ2 , ..., ξn , ) = {b : ||(b − a)ξi || < ∀i = 1, · · · , n}
(iii)The weak operator topology is formed by the basic neighbourhoods
N (a, ξ1 , ξ2 , ..., ξn , η1 , η2 , .., ηn , ) = {b : |h(b − a)ξi , ηi i| < ∀i = 1, · · · , n}
Note that this weak topology is the topology of pointwise convergence
on
H
in the weak topology on
H
dened in the obvious way by the inner
product.
The unit ball of
of
B(H)
H
is compact in the weak topology and the unit ball
is compact in the weak operator topology. These assertions follow
easily from Tychono 's theorem.
Exercise 3.1.1
Show that we have the following ordering of the topologies
(strict in innite dimensions).
(weak operator topology) < (strong operator topology) < (norm topology)
11
Note that a weaker topology has less open sets so that is a set is closed in
the weak topology it is necessarily closed in the strong and norm topologies.
We will now prove the von Neumann density or bicommutant theorem
which is the rst result in the subject. We prove it rst in the nite dimensional case where the proof is transparent then make the slight adjustments
for the general case.
Theorem 3.1.2
with
M be a self-adjoint
dim(H) = n < ∞. Then M = M 00 .
Let
subalgebra of
B(H)
containing
1,
M ⊆ M 00 .
00
So we must show that if y ∈ M then y ∈ M . To this end we will amplify
the action of M on H to an action on H⊗H dened by x(ξ⊗η) = xξ⊗η . If we
n
choose an orthonormal basis {vi } of H then H ⊗ H = ⊕i=1 H and in terms of
matrices we are considering the n x n matrices over B(H) and embedding M
Proof. It is tautological that
in it as matrices constant down the diagonal. Clearly enough the commutant
0
of M on H ⊗ H is the algebra of all n x n matrices with entries in M and the
00
second commutant consists of matrices having a xed element of M down
the diagonal.
v be the vector ⊕ni=1 vi ∈ ⊕ni=1 H and let V = M v ⊆ H ⊗ H. Then
M V ⊆ V and since M = M ∗ , PV ∈ M 0 (on H ⊗ H) by exercise 2.1.13. So if
y ∈ M 00 (on H⊗H), then y commutes with PV and yM v ⊆ M v . In particular
y(1v) = xv for some x ∈ M so that yvi = xvi for all i, and y = x ∈ M . Let
Theorem 3.1.3
(von Neumann) Let M be a self-adjoint subalgebra of B(H)
00
containing 1. Then M = M (closure in the strong operator topology).
00
Proof. Commutants are always closed so M ⊆ M .
00
So let a ∈ M and N (a, ξ1 , ξ2 , ..., ξn , ) be a strong neighbourhood of a.
n
n
We want to nd an x ∈ M in this neighbourhood. So let v ∈ ⊕i=1 H be ⊕i=1 ξi
n
and let B(H) act diagonally on ⊕i=1 H as in the previous theorem. Then the
same observations concerning matrix forms of commutants are true. Also M
n
commutes with PM v which thus commutes with a (on ⊕i=1 H). And since
1 ∈ M , av = ⊕aξi is in the closure of M v so there is an x ∈ M with
||xξi − axii || < for all
Corollary 3.1.4
1)
If
i. M = M∗
is a subalgebra of
B(H)
with
1 ∈ M,
TFAE:
00
M =M
2) M is
3) M is
strongly closed.
weakly closed.
Denition 3.1.5
A subalgebra of
B(H) satisfying the conditions of corollary
3.1.4 is called a von Neumann algebra.
12
(A self-adjoint subalgebra of
∗
is called a C -algebra.)
Example 3.1.6
B(H)
which is closed in the norm topology
1) Any nite dimensional *-subalgebra of
B(H)
containing
1.
Example 3.1.7 B(H) itself.
Example 3.1.8
Exercise 3.1.9 Let (X, µ) be a nite measure space and
∞
2
consider A = L (X, µ) as a *-subalgebra of B(L (X, µ)) as multiplication
0
operators as in exercise 2.2.2. Show that A = A , i.e. A is maximal abelian
0
and hence a von Neumann algebra. (Hint: if x ∈ A let f = x(1). Show that
f ∈ L∞ and that x = Mf .)
Example 3.1.10
If
S ⊆ B(H),
we call
(S ∪ S ∗ )00
the von Neumann algebra
generated by S. It is, by theorem 3.1.3 the weak or strong closure of the *algebra generated by
1
and
S.
Most constructions of von Neumann algebras
begin by considering some family of operators with desirable properties and
then taking the von Neumann algebra they generate. But is is quite hard, in
general, to get much control over the operators added when taking the weak
closure and all the desirable properties of the generating algebra may be
lost.(For instance any positive self-adjoint operator
a with ||a|| ≤ 1 is a weak
limit of projections.) However, if the desirable properties can be expressed
in terms of matrix coecients then these properties will be preserved under
weak limits since the matrix coecients of
form
hξ, aη .
a
are just special elements of the
We shall now treat an example of this kind of property which is
at the heart of the subject and will provide us with a vast supply of interesting
von Neumann algebras quite dierent from the rst 3 examples.
2
Let Γ be a discrete group and let ` (Γ) be the Hilbert space of all functions
X
X
f (γ)g(γ). An
f : Γ → C with
|f (γ)|2 < ∞ and inner product hf, gi =
γ∈Γ
` (Γ) is given by the {εγ } where εγ (γ ) = δγ,γ 0 so that
f (γ)εγ in the `2 sense. For each γ ∈ Γ dene the unitary operator uγ
orthonormal basis of
f=
X
γ∈Γ
0
2
γ∈Γ
(uγ f )(γ 0 ) = f (γ −1 ). Note that uγ uρ = uγρ and that uγ (ερ ) = εγρ . Thus
γ 7→ uγ is a unitary group representation called the left regular representation.
The uγ are linearly independent so the algebra they generate is isomorphic
to the group algebra CΓ. The von Neumann algebra generated by the uγ
goes under various names, U (Γ), λ(Γ) and L(Γ) but we will call it vN (Γ) as
it is the group von Neumann algebra of Γ.
by
13
To see that one can control weak limits of linear combinations of the
uγ ,
Γ = Z/nZ.
consider rst the case
With basis
u0 , u1 , u2 , · · · , un−1 , u1
is
represented by the matrix:

0
0

0

0

0
1
1
0
.
.
0
0
0
1
0
.
.
0
0
0
1
0
.
.
.
0
0
1
0
.

.
.

0

0

1
0
which is a matrix constant along the diagonals. Clearly all powers of
u1
and
all linear combinations of them have this property also so that an arbitrary
element of the algebra they generate will have the matrix form (when

a
f

e

d

c
b
b
a
f
e
d
c
c
b
a
f
e
d
d
c
b
a
f
e
e
d
c
b
a
f
n = 6):

f
e

d

c

b
a
(Such matrices are known as circulant matrices but to the best of our knowledge this term only applies when the group is cyclic.) If
Z/nZ were replaced
by another nite group the same sort of structure would prevail except that
the diagonals would be more complicated, according to the multiplication
table of the group.
(γ, ρ) matrix entry of
−1
a nite linear combination of the uγ 's will depend only on γ
ρ. As observed
Now let
Γ be an innite group.
It is still true that the
above, the same must be true of weak limits of these linear combinations,
hence any element of
vN (Γ).
We see that the elements of
vN (Γ)
have matrices (w.r.t. the basis
{(γ, ρ) : γ −1 ρ is constant}.
εγ )
which are constant along the diagonals :
Exercise 3.1.11
Check whether it should be
γ −1 ρ
or
γρ−1
or some other
similar expression.....
Using the number
cγ
at least, any element of
on the diagonal indexed by
vN (Γ)
as a sum
X
c γ uγ .
γ
we can write, formally
It is not clear in what
γ∈Γ
sense this sum converges but certainly
X
cγ uγ
must dene a bounded linear
γ∈Γ
operator. From this we deduce immediately the following:
14
(i) The function
(ii)
γ 7→ cγ
is in
`2 .
(Apply
X
c γ uγ
to
εid .)
γ∈Γ
X
X
XX
(
cγ uγ )(
dγ uγ ) =
(
cρ dρ−1 γ )uγ
γ∈Γ
γ∈Γ
γ∈Γ ρ∈Γ
where the sum dening the coecient of
ρ 7→ cρ and ρ 7→ dρ−1 γ are in `2 .
uγ
on the right hand side con-
verges since
Exactly what functions
γ 7→ cγ
dene elements of
vN (Γ)
is unclear but
an important special case gives some intuition.
Case 1.
Γ = Z.
It is well known that the map
c n εn →
cneinθ denes a unitary V
2
2
−1 ikθ
from ` (Γ) to L (T). Moreover V un V
(e
= V un (εk ) = V ε(k + n) =
einθ eikθ so that V un V −1 is the multiplication operator Meinθ . Standard spec∞
tral theory shows that Meinθ generates L (T) as a von Neumann algebra,
P
P
∞
−1
and clearly if Mf ∈ L (T),
V Mf V =
cn εn where
cn einθ is the
P
Fourier series of
f.
dene elements of
P
We see that, in this case, the functions
vN (Z)
are precisely the Fourier series of
γ 7→ cγ which
L∞ functions.
In case we forget to point it out later on when we are in a better position
to prove it, one way to characterise the functions which dene elements on
vN (Γ) is as all functions which dene bounded operators on `2 (Γ). This is
not particularly illuminating but can be useful at math parties.
At the other extreme we consider a highly non-commutative group, the
n generators, n ≥ 2.
Γ = Fn .
free group on
Case 2.
Just for fun let us compute the centre Z(vN (Γ)) of vN (Fn ), i.e. those
P
cγ uγ that commute with all x ∈ vN (Γ). By weak limits, for
cγ uγ to
be in Z(vN (Γ)) it is necessary and sucient that it commute with every uγ .
This clearly is the same as saying cγργ −1 = cρ
∀γ, ρ, i.e. the function c is
constant on conjugacy classes. But in Fn all conjugacy classes except that
2
of the identity are innite. Now recall that γ 7→ cγ is in ` !! We conclude
that cγ = 0
∀γ 6= 1, i.e. Z(vN (Γ)) = C1.
Note that the only property we used of F∞ to reach this conclusion was
P
that every non-trivial conjugacy class is innite (and in general it is clear
that
Z(vN (Γ))
is in the linear span of the
uγ
with
γ
in a nite conjugacy
class.) Such groups are called i.c.c. groups and they abound. Other examples
S∞ (the group of
P SL(n, Z) and Q ×Q∗ .
include
nitely supported permutations of an innite set),
Unsolved problem in von Neumann algebras:
Is
vN (Fn ) ∼
= vN (Fm )
for
n 6= m
Note that it is obvious that the group algebras
15
(≥ 2)?
CFn
and
CFm
are not iso-
morphic, just consider algebra homomorphisms to
C-but they will not extend
vN (Γ).
to
Denition 3.1.12
A von Neumann algebra whose centre is
C1
is called a
H = K1 ⊗ K2 and let M = B(K1 ) ⊗ 1
M and M 0 are factors.
Show that
factor.
Exercise 3.1.13
Show that
Exercise 3.1.14
Suppose
0
M = 1 ⊗ B(K2 )
so that
B(H)
is a factor.
This exercise is supposed to explain the origin of the term factor as
M 0 come from a tensor product factorisation of H.
M
and
vN (Γ) is of an entirely dierent nature
B(H). To see this P
consider the function tr : vN (Γ) → C dened by
tr(a) = haε1 , ε1 i, or tr( cγ uγ ) = c1 . This may is clearly
linear, weakly
P
∗
2
continuous, satises tr(ab) = tr(ba) and tr(x x) =
γ |cγ | ≥ 0 (when
P
x = γ cγ uγ ). It is called a trace on vN (Γ). If Γ = Z it obviously equals
R 2π
f (θ)dθ under the isomorphism between vN (Z) and L∞ (T).
0
The factor we have constructed as
from
Exercise 3.1.15
tr : B(H) → C
tr(ab) = tr(ba),
dim H < ∞ show that there is a constant K with tr(x) = Ktrace(x).
(ii) There is no non-zero weakly continuous linear map tr : B(H) → C
satisfying tr(ab) = tr(ba) when dim(H) = ∞.
(iii) There is no non-zero linear map tr : B(H) → C satisfying tr(ab) =
tr(ba) and tr(x∗ x) ≥ 0 when dim(H) = ∞.
(iv) (harder) There is no non-zero linear map tr : B(H) → C satisfying
tr(ab) = tr(ba) when dim(H) = ∞.
(i)If
Thus our factors
vN (Γ)
when
Γ
seem to have more in common with
dim H = ∞! They
H in this case.
is a linear map with
is i.c.c.
B(H)
are innite dimensional but
when
dim H < ∞
than when
certainly do not come from tensor product factorisations
of
Let us make a couple of observations about these factors.
1)They contain no non-zero nite rank operators for such an operator
x∗ x of necessary.)
cannot be constant and non-zero down the diagonal. (Take
2)They have the property that tr(a) = 0 ⇒ a = 0 for a positive element
2
(a = |a| by the spectral theorem and all spectral projections of a, hence
a
|a|
,are in the von Neumann algebra generated by a.)
∗
∗
3)They have the property that uu = 1 ⇒ u u = 1 (i.e. they contain no
non-unitary isometries).
∗
∗
Proof. If u u = 1, uu is a projection so
1 − tr(u∗ u) = 0. 16
1 − uu∗
is too and
tr(1 − uu∗ ) =
Exercise 3.1.16
vN (Γ), ab = 1 ⇒ ba = 1.
0, ab = 1 ⇒ ba = 1 in F Γ.
Show that in
any eld of characteristic
Show that if
F
is
As far as I know this assertion is still open in non-zero characteristic. The
above exercise is a result of Kaplansky.
The next observation is a remarkable property of the set of projections.
4) If
Γ = Fn , {tr(p) : p
a projection in
vN (Γ)} = [0, 1].
Proof. It is clear that the trace of a projection is between
0 and 1.
To see
that one may obtain every real number in this interval, consider the subgroup
hai generated by a single non-zero element. By the coset decomposition of Fn
2
the representation of hai on ` (Fn ) is the direct sum of countably many copies
of the regular representation. The bicommutant of ua is then, by a matrix
argument, vN (Z) acting in an amplied way as block diagonal matrices
with constant blocks so we may identify vN (Z) with a subalgebra of vN (Γ).
Under this identication the traces on the two group algebras agree. But as
∞
we have already observed, any element f ∈ L (0, 2π) denes an element of
vN (Z)
whose integral is its trace. The characteristic function of an interval
is a projection so by choosing intervals of appropriate lengths we may realise
projections of any trace.
We used the bicommutant to identify
vN (Z) with a subalgebra of vN (Γ).
It is instructive to point out a problem that would have arisen had we tried
2
to use the weak or strong topologies. A vector in ` (Γ) is a square summable
2
2
sequence of vectors in ` (Z) so that a basic strong neighbourhood of a on ` (Γ)
P∞
2
would correspond to a neighbourhood of the form {b :
i=1 ||(a − b)ξi || < }
P
∞
2
2
for a sequence (ξi ) in ` (Z) with
i=1 ||ξi || < ∞. Thus strong convergence
2
on ` (Z) would not imply strong convergence on (Γ)! This leads us naturally
to dene two more topologies on
Denition
3.1.17
P
∞
i=1
B(H).
The topology dened by the basic neighbourhoods of a,
2
in ` (Z) with
||(a − b)ξi ||2 < } for any and any sequence (ξi )
2
i=1 ||ξi || < ∞, is called the ultrastrong topology on B(H).
{b
P∞:
The topology dened by the basic neighbourhoods
{b :
∞
X
|h(a − b)ξi , ηi i| < }
i=1
for any
>0
and any sequences
∞
X
(ξi ), ηi
in
`2 (Z)
||ξi ||2 + ||ηi ||2 < ∞
i=1
is called the ultraweak topology on
B(H).
17
with
Note that these topologies are precisely the topologies inherited on
if it is amplied innitely many times as
Exercise 3.1.18
B(H) ⊗ 1K
with
B(H)
dim K = ∞.
Show that the ultrastrong topology and the ultraweak topol-
ogy are stronger than the strong topology and weak topology respectively. Show
that the ultrastrong and strong topologies coincide on a bounded subset of
B(H)
as do the weak and ultraweak topologies.
Exercise 3.1.19
Repeat the argument of the von Neumann density theorem
(3.1.3) with the ultrastrong topology replacing the strong topology.
Here are some questions that the inquisitive mind might well ask at this
stage. All will be answered in succeeding chapters.
Question 1) If there is a weakly continuous trace on a factor, is it unique
(up to a scalar multiple)?
Question 2)If there is a trace on a factor
p a projection in
M
is it true that
{tr(p) :
M } = [0, 1]?
Question 3) Is there a trace on any factor not isomorphic to
B(H)?
Question 4) Are all (innite dimensional)factors with traces isomorphic?
Question 5) If
M
is a factor with a trace, is
M0
also one? (Observe that
the commutant of a factor is obviously a factor.)
Question 6) Is
vN (Γ)0
the von Neumann algebra generated by the right
regular representation?
Question 7) If
φ : M → N
is a
Neumann algebras on Hilbert spaces
∗
so that φ(a) = uau for a ∈ M ?
H
18
∗
-algebra isomorphism between von
and
K
is there a unitary
u:H→K
Chapter 4
Elementary properties of von
Neumann algebras.
Throughout this chapter
space
M
will be a von Neumann algebra on a Hilbert
H.
EP1) If
a = a∗
is an element of
bounded Borel functions of
a
are in
M , all the spectral projections and
M . Consequently M is generated by
all
its
projections.
According to one's proof of the spectral theorem, the spectral projections
E(λ)
of
a
are constructed as strong limits of polynomials in
erty that the spectral projections of
a
a.
Or the prop-
are in the bicommutant of
a
may be
an explicit part of the theorem. Borel functions will be in the bicommutant.
EP2) Any element in
M
is a linear combination of
4
unitaries in
M.
x is a linear√combination of 2 self-adjoints,
||a|| ≤ 1, let u = a + i 1 − a2 . Then u is unitary and
Proof. We have seen that any
and if a is self-adjoint,
u+u∗
. 2
a=
EP3)M is the commutant of the unitary group of
M 0 so that an alternative
denition of a von Neumann algebra is the commutant of a unitary group
representation.
This follows from EP2)
Exercise 4.0.20
Show that multiplication of operators is jointly strongly
continuous on bounded subsets but not on all of
Show that
∗ : B(H) 7→ B(H)
B(H).
is weakly continuous but not strongly con-
tinuous even on bounded sets.
The following result is well known.
19
Theorem 4.0.21
{aα } is a net of self-adjoint operators with aα ≤ aβ
for α ≤ β and ||aα || ≤ K for some K ∈ R, then there is a self-adjoint
a with a = limα aα , convergence being in the strong topology. Furthermore
a = lub(aα ) for the poset of self-adjoint operators.
If
a for the limit can be found by weak compactness of
the unit ball. Then haα ξ, ξi is increasing with limit haξ, ξi for all ξ ∈ H and
√
a ≥ aα ∀α. So limα a − aα = 0 in the strong topology. Now multiplication
is jointly strongly continuous on bounded sets so s−lim aα = a. Note that we have slipped in the notation s−lim for a limit in the strong
topology (and obviously w−lim for the weak topology).
If a and (aα ) are as in 4.0.21 we say the net (aα ) is monotone convergent
to a.
Prof. A candidate
EP4)
M
is closed under monotone convergence of self-adjoint operators.
The projections on
B(H)
form an ortholattice with the following proper-
ties:
p ≤ q ⇐⇒ pH ⊆ qH
p ∧ q = orthogonal
projection onto
pH ∩ qH
p⊥ = 1 − p
p ∨ q = (p⊥ ∧ q ⊥ )⊥ = orthogonal
Exercise 4.0.22
Show that
projection onto
pH + qH.
p ∧ q = s−lim n→∞ (pq)n .
The lattice of projections in
B(H) is complete (i.e.
closed under arbitrary
sups and infs) since the intersection of closed subspaces is closed.
EP5) The projections in
M
generate
M
as a von Neumann algebra, and
they form a complete sublattice of the projection lattice of
B(H).
is a setWof projections in M then nite subsets of S are a
p is a net in M satisfying the conditions of 4.0.21.
directed set and F →
p∈F
Thus the strongWlimit of this net exists and is in M . It is obvious that this
Proof.
If
S
strong limit is
p∈S
p,
the sup being in
B(H). Easier proof. For each projection p ∈ M , pH is invariant under each
M 0 . Thus the intersection of these subspaces is also. T
EP6) Let A be a *-subalgebra of B(H). Let W be
ker(a) and K =
a∈A
⊥
W . Then K is invariant under A and if we let B = {a|K : a ∈ A}, then 1K
element of
is in the strong closure of
B
which is thus a von Neumann algebra.
20
Proof. By the above, if
p and q
p ∨ q = 1 − (1 − p) ∧ (1 − q)
generated by p and q . By spectral
are projections
is in the strong closure of the algebra
∗
theory, if a = a the range projection Pker(a)⊥ is in the strong closure of the
algebra generated W
by a so we may apply the argument of the proof of EP5)
P
to conclude that
⊥ is in the strong closure of A. But this is 1K . a∈A ker(a)
Thus if we were to dene a von Neumann algebra as a weakly or strongly
closed subalgebra of
B(H),
it would be unital as an abstract algebra but its
identity might not be that of
B(H)
so it would not be equal to its bicommu-
tant. However on the orthogonal complement of all the irrelevant vectors it
would be a von Neumann algebra in the usual sense.
EP7) If
(M 0 p)0
and
M is a von Neumann algebra and p ∈ M is a projection, pM p =
pM p0 = M 0 p as algebras of operators on pH. Thus pM p and M 0 p
are von Neumann algebras.
0
Proof. Obviously pM p and M p commute with each other on pH. Now
0 0
suppose x ∈ (M p) ⊆ B(pH) and dene x̃ = xp(= pxp) ∈ B(H). Then if
y ∈ M 0 , yx̃ = yxp = ypxp = (xp)(yp) = xpy = x̃y , so x̃ ∈ M and x = px̃p.
0 0
Thus (pM ) = pM p which is thus a von Neumann algebra.
0
If we knew that M p were a von Neumann algebra on pH we would be
done but a direct attempt to prove it strongly or weakly closed fails as we
0
0
would have to try to extend the limit of a net in M p on pH to be in M .
0
0
So instead we will show directly that (pM p) ⊆ M p by a clever extension
0
of its elements to H. By EP2 it suces to take a unitary u ∈ (pM p) . Let
K⊆H
clearly invariant under
for
M pH and let q be projection onto it. Then K is
M and M 0 so q ∈ Z(M ). We rst extend u to K by
X
X
ũ
xi ξi =
xi uξi
be the closure of
xi ∈ M
and
ξi ∈ H.
ũ is an isometry:
X
X
||ũ
xi ξi ||2 =
hxi uξi , xj uξj i
We claim that
i,j
=
X
=
X
hpx∗j xi puξi , uξj i
i,j
hupx∗j xi pξi , uξj i
i,j
= ... = ||
X
This calculation actually shows that
xi ξi ||2
ũ
is well dened and extends to an
isometry of K. By construction ũ commutes with M on
0
0
0
So ũq ∈ M and u = ũqp. Hence (pM p) = M p. 21
M pH,hence
on
K.
Corollary 4.0.23
over the map
M is a factor,pM p is a factor on pH, as is pM 0 . Morex →
7
xp from M 0 to M 0 p is a weakly continuous *-algebra
If
isomorphism.
Proof.
Factoriality follows immediately from EP7.
Continuity and the
*-algebra homomorphism property are obvious so all we have to show is
0
injectivity. But if x ∈ M and xp = 0 then x = 0 on M pH. But as before
the projection onto
M pH
Corollary 4.0.24
If
implies either
Proof. Let
is in
M is a
a = 0 or b = 0.
Z(M ),
hence equal to
factor and
a ∈ M
1.
and
So
x = 0. b ∈ M0
then
ab = 0
p be the range projection of b and apply the previous corollary.
Exercise 4.0.25
Show that if
M
is a von Neumann algebra generated by the
self-adjoint, multiplicatively closed subset S , then pSp generates pM p (if p is
0
a projection in M or M ). Show further that the result fails if S is not closed
under multiplication.
Exercise 4.0.26
Show that if M is a factor and V and W are nite dimen0
sional subspaces of M and M respectively then the map a ⊗ b 7→ ab denes
a linear isomorphism between
with
v∈V
and
V ⊗W
and the space
VW
spanned by all
vw
w ∈ W.
a ∈ M and a = u|a| is the polar decomposition of a then u ∈ M
|a| ∈ M . Proof. By the uniqueness of the polar decomposition both |a|
u commute with every unitary in M 0 .
EP8) If
and
and
EP9) None of the topologies (except
they all are on the unit ball (when
H
|| − ||)
is metrizable on
is separable) and
H
B(H)
but
is separable for all
except the norm topology.
Proof.
First observe that a weakly convergent sequence of operators is
bounded.
This follows from the uniform boundedness principle and 2.1.5
which shows how to get the norm from inner products.
{ηi , i = 1, · · · ∞} be an orthonormal basis of
H and let ei be projection onto Cηi . Consider the family {em + men : m, n =
1,
. Let V a basic ultrastrong neighbourhood of 0 dened by and {ξi :
P· · · ∞}
2
||ξP
writing
i || < ∞} and let
P| − |Vi 2be the corresponding seminorm,
P then
i
i 2
ξi = j ξj ηj we have i,j |ξj | < ∞. Now choose m so that i |ξm | < 2 /4
P i 2
2
2
2
n 2
and n so that
i |ξn | < /4m . Observing that ||en (ξi )|| = |ξi | we have
Here is the cunning trick. Let
|em + men |V ≤ |em |V + m|en |V
22
=
s
X
||em ξ||2 + m
s
X
i
||en ξ||2
i
≤ /2 + /2
so that
en + men ∈ V .
On the other hand no subsequence of
tend even weakly to
0
{em + men : m, n = 1, · · · ∞}
can
since it would have
to be bounded in norm which would force some xed
m to occur innitely
often in the sequence, preventing weak convergence! So by the freedom in
choosing
m
and
n
to force
em + men
to be in
V,
there can be no countable
basis of zero for any of the topologies (except of course the norm).
If we consider the unit ball, however, we may choose a dense sequence ξi of
P
d(a, b) = [ i 2−i ||(a − b)ξi ||2 ]1/2 which is a metric on
the unit ball dening the strong topology. (Similarly for the weak topology.)
unit vectors and dene
We leave non-separability of
B(H)
in the norm topology as an exercise.
EP10) An abelian von Neumann algebra on a separable Hilbert space is
generated by a single self-adjoint operator.
Proof.
Let
{e0 , e1 , e2 , · · ·}
be a sequence of projections that is strongly
dense in the set of all projections in the Abelian von Neumann algebra A.
P∞ 1
the norm topology so a ∈ A. The
Let a =
n=0 3n en . The sum converges
Pin
∞
norm of the self-adjoint operator a1 =
n=1 is obviously at most 1/2 so that
the spectral projection for the interval [3/4, 2] for a is e0 . Continuing in this
00
0
way we see that all the en s are in {a} . This relegates the study of Abelian von Neumann algebras to the spectral
theorem. One can show that any Abelian von Neumann algebra on a sepa∞
rable Hilbert space is isomorphic to either ` ({0, 1, · · · , n}) (where n = ∞
∞
is allowed) or L ([0, 1], dx) or a direct sum of the two. This is up to abstract algebra isomorphism.
To understand the action on a Hilbert space,
multiplicity must be taken into account.
23
24
Chapter 5
Finite dimensional von Neumann
algebras and type I factors.
5.1 Denition of type I factor.
The rst crucial result about factors (remember a factor is a von Neumann
algebra with trivial centre) will be the following ergodic property.
Theorem 5.1.1
there is an
If
x∈M
M
is a factor and
pxq 6= 0.
with
p
q are non-zero projections in M
x can be chosen to be unitary.
and
Moreover
∗
Proof.
that for any unitary u ∈ M , puq = 0. Then u puq = 0 and
W Suppose
W
u∗ pu q = 0. But clearly u∈M u∗ pu commutes with all unitaries u ∈ M
u∈M
so is the identity.
The reason we have called this an ergodic property is because of a pervasive analogy with measure-theoretic dynamical systems (and it will become
much more than an analogy).
serving the measure
µ(X \ A) = 0
µ
A transformation
−1
is called ergodic if T
(A)
T : (X, µ) → (X, µ) pre⊆ A implies µ(A) = 0 or
A ⊆ X . If T is invertible one can then show
n
that there is, for any pair A ⊂ X and B ⊂ X of non-null sets, a power T
n
2
N
of T such that µ(T (A) ∩ B) 6= 0. Or, as operators on L (X, µ), AT B 6= 0
where we identify A and B with the multiplication operators by their charn
acteristic functions. The proof is the samethe union of all T (A) is clearly
invariant, thus must dier from all of X by a set of measure 0.
for a measurable
Corollary 5.1.2
Let
p
u be the partial
pxq 6= 0. Proof. Let
such that
q be non-zero projections
u (=
6 0) in M with uu∗ ≤ p
and
there is a partial isometry
in a factor M . Then
∗
and u u ≤ q .
isometry of the polar decomposition of
25
pxq
for
x
p and q had been swapped
around in the proof, so we
interchanged uu∗ and u∗ u
in the statement.
Denition 5.1.3
p∈M
If
M
is a von Neumann algebra, a non-zero projection
is called minimal, or an atom, if
Exercise 5.1.4
Show that
Denition 5.1.5
p
(q ≤ p) ⇒ (q = 0 or
is minimal in
M
i
q = p).
pM p = Cp.
A factor with a minimal projection is called a type I fac-
tor.
5.2 Classication of all type I factors
We will classify all type I factors quite easily.
We begin with the model,
which we have already seen.
B(H)⊗1 be the constant diagonal matrices on H ⊗K. Its commutant
1 ⊗ B(K) will be our model. It is the algebra of all matrices dening bounded
Let
operators with every matrix entry being a scalar multiple of the identity
matrix on
H.
A matrix with a single
1
on the diagonal and zeros elsewhere
is obviously a minimal projection.
Theorem 5.2.1
spaces
H
and
K
If
M
L, there are Hilbert
uM u∗ = B(H) ⊗ 1.
is a type I factor of a Hilbert space
and a unitary
u:L→H⊗K
with
{p1 , p2 , ...} be a maximal family of minimal projections in M
such that pi pj = 0 for i 6= j . (We assume for convenience that L is separable.
W
W
Our rst claim is that
i pi = 1 so that L = ⊕i pi L. For if 1 −
i pi were
∗
∗
nonzero, by corollary 5.1.2 there would be a u 6= 0 with uu ≤ p1 and u u ≤
W
1 − i pi . By minimality uu∗ is minimal and hence so is u∗ u contradicting
maximality of the pi . Now for each i choose a non-zero partial isometry e1i
∗
∗
∗
∗
with e1i e1i ≤ p1 and e1i e1i ≤ pi . By minimality e1i e1i = p1 and e1i e1i = pi .
P
Then M is generated by the e1i 's, for if a ∈ M we have a =
i,j pi apj the sum
∗
∗
converging in the strong topology, and pi apj = e1i e1i ae1j e1j ∈ p1 M p1 = Cp1 .
P
∗
Thus there are scalars λij so that a =
i,j λij e1i e1j . (The details of the
convergence of the sum are unimportantwe just need that a be in the
Proof. Let
strong closure of nite sums.)
If n is the cardinality
u : `2 (X, p1 L) → L by
of
{pi },
uf =
let
X = {1, 2, ..., n}
X
and dene the map
e∗1i f (i).
i
∗
u is unitary and u e1i u is a matrix on `2 (X, p1 L) with an identity
operator in the (1, i) position and zeros elsewhere. The algebra generated by
2
2
these matrices is B(` (X)) ⊗ 1 on ` (X) ⊗ p1 L and we are done.
Observe that
26
Remark 5.2.2
The importance of being spatial.
We avoided all kinds of problems in the previous theorem by constructing
our isomorphism using a unitary between the underlying Hilbert spaces. In
general given von Neumann algebras
M
and
tively, to construct an isomorphism between
N
M
generated by
and
N
S
and
T
respec-
it suces to construct
(if possible !!!) a unitary u between their Hilbert spaces so that T is contained
∗
in uSu . To try to construct an isomorphism directly on S could be arduous
at best.
5.3 Tensor product of von Neumann algebras.
If
M
is a von Neumann algebra on
K we dene M ⊗ N to be
{x ⊗ y : x ∈ M, y ∈ N }.
Exercise 5.3.1
M ⊗alg N
H
and
N
is a von Neumann algebra on
the von Neumann algebra on
Show that
M ⊗N
H⊗K
generated by when you use \h, make sure
it's enclosed in $ signs
contains the algebraic tensor product
as a strongly dense *-subalgebra.
Denition 5.3.2
units (s.m.u.)
Let
of size
M be a von Neumann algebra . A system of matrix
n in M is a family {eij : i, j = 1, 2, ..., n} (n = ∞
allowed) such that
(i)
e∗ij = eji .
eij ekl = δj,l eil
P
(iii)
i eii = 1.
(ii)
Exercise 5.3.3
{eij ; i, j = 1, ..., n} is
eij generate a type I
Show that if
Neumann algebra
2
B(` ({1, 2, ..., n}))
M,
then the
and that
an s.m.u.
in a von
factor isomorphic to
M
is isomorphic (unitarily equivalent to in this
2
instance) The von Neumann algebra e11 M e11 ⊗ B(` ({1, 2, ..., n})).
5.4 Multiplicity and nite dimensional von Neumann algebras.
Theorem 5.2.1 shows that type I factors on Hilbert space are completely
classied by two cardinalities
n1 =
n2 =
(n1 , n2 )
according to:
rank of a minimal projection in
rank of a minimal projection in
27
M , and
M 0.
We see that the isomorphism problem splits into abstract isomorphism
n2 alone), and spatial isomorphism, i.e. unitary equivalence.
n = n2 . It is abstractly
B(H) with dim H = n. The integer n1 is often called the
(determined by
A type In factor is by denition one for which
isomorphic to
multiplicity of the type I factor.
We will now determine the structure of all nite dimensional von Neumann algebras quite easily. Note that in the following there is no requirement
that
H
be nite dimensional.
Theorem 5.4.1
M
be a nite dimensional von Neumann algebra on
k
the Hilbert space H. Then M is abstractly isomorphic to ⊕i=1 Mni (C) for
some positive integers k, n1 , n2 , ..., nk . (Mn (C) is the von Neumann algebra
of all
n×n
Let
matrices on
n-dimensional Hilbert space.) Moreover there are
u : ⊕i `2 (Xi , Ki ) → H (with |Xi | = ni ) with
Hilbert spaces Ki and a unitary
u∗ M u = ⊕i B(`2 (Xi )) ⊗ 1.
Z(M ) is a nite dimensional abelian von Neumann algebra
. If p is a minimal projection in Z(M ), pM p is a factor on pH. The theorem
follows immediately from theorem 5.2.1 and the simple facts that Z(M ) =
⊕ki=1 pi C where the pi are the minimal projections in Z(M ) (Two distinct
minimal projections p and q in Z(M ) satisfy pq = 0), and M = ⊕i pi M pi . Proof. The centre
The subject of nite dimensional von Neumann algebras is thus rather
simple. It becomes slightly more interesting if one considers subalgebras
M.
N⊆
Let us deal rst with the factor case of this. Let us point out that the
identity of
M
is the same as that of
Theorem 5.4.2
If
M
N.
determined, up to conjugation by unitaries in
k > 0
such that
the subfactor
N
Im factors are all uniquely
M , by the integer (or ∞)
is a type In factor, its type
pM p is a type
mk = n.
Ik factor,
p
being a minimal projection in
and
N1 and N2 be type Im subfactors with generating s.m.u.'s {eij }
{fij } respectively.
PmIf k is the integer (in the statement of the theorem)
for N − 1 then 1 =
1 eii and each eii is the sum of k mutually orthogonal
minimal projections of M , hence n = mk . The same argument applies to
N2 . Build a partial isometry u with uu∗ = e11 and u∗ u = f11 by adding
Proof. Let
and
together partial isometries between maximal families of mutually orthogonal
projections less than
w=
M
e11
and
f11
respectively. Then it is easy to check that
∗
∗
i ej1 uf1j is a unitary with wfkl w = ekl . So wN2 w = N1 . P
n
Now we can do the general (non-factor) case. If N = ⊕i=1 Mki (C) and
= ⊕m
j=1 Mrj (C) and N ⊆ M as von Neumann algebras, let pj be minimal
28
central projections in
p j qi N
is a factor and
where
λij
M
is a subfactor so we may form the
(i, j), pj qi M qi pj
matrix Λ = (λij )
pj qi N ⊆ pj qi M qi pj
by theorem 5.4.2.
and
qi be those of N .
is the integer associated with
Exercise 5.4.3
ei is a minimal
Mr j C.
Example. Let
Then for each
λij dened
factor qi N , λij =
Show that the integer
projection in the
M = M5 (C) ⊕ M3 (C)
and
N
above is the following: if
trace of the matrix
pj ei ∈
be the subalgebra of matrices of
the form:

X
0
0
0
X
0

0
X
0⊕
0
z
z ∈ C and X is a 2×2 matrix. Then N
p1 = 1 ⊕ 0, q1 = 1 ⊕ 0, etc., we have
2 1
Λ=
.
1 1
where
and if
The matrix
Λ
of edges between
0
z
is isomorphic to
M2 (C)⊕C
is often represented by a bipartite graph with the number
i
and
j
being
λij .
The vertices of the graph are labelled by
the size of the corresponding matrix algebras. Thus in the above example
the picture would be:
This diagram is called the Bratteli diagram for
Exercise 5.4.4
sion
N ⊆M
N ⊆ M.
Generalise the above example to show that there is an inclu-
corresponding to any Bratteli diagram with any set of dimen-
sions for the simple components of
N.
5.5 A digression on index.
N ⊆ M are type I factors we have seen that there is an integer k (possibly
∞) such that M is the algebra of k × k matrices over N . If k < ∞, M is
2
thus a free left N-module of rank k . It seems reasonable to call the number
k 2 the index of N in M and write it [M : N ]. This is because, if H < G
are groups and CH ⊆ CG their group algebras, the coset decomposition of
G shows that CG is a free left CH -module of rank [G : H].
If
29
30
Chapter 6
Kaplansky Density Theorem.
6.1 Some simple but telling results on linear
functionals.
We begin with a result about linear functionals of independent interest.
Theorem 6.1.1
Let
V
be a subspace of
B(H)
and let
φ:V →C
be a linear
functional. T.F.A.E.
(i) There are vectors in
H, ξ1 , ξ2 , ..., ξn
φ(x) =
n
X
and
η1 , η2 , ..., ηn
with
hxξi , ηi i
i=1
(ii)
(iii)
φ
φ
is weakly continuous.
is strongly continuous.
⇒ (ii) ⇒ (iii) are obvious, so suppose φ is strongly continuous.
Then let ν1 , ..., νn be vectors such that ||xνi || < 1 ∀i ⇒ |φ(x)| < 1 for
pP
2
x ∈ V . Then there is a constant K > 0 so that |φ(x)| < K
i ||xνi || . Let
ν = ν1 ⊕ ...νn ∈ ⊕i H and let K = (V ⊗ 1)(ν). Then dene φ̃ on V ⊗ 1(ν) by
φ̃(⊕i xνi ) = φ(x). Observe that φ̃ is well-dened and continuous so extends to
K and there is a vector ξ = ⊕ξi ∈ K with φ(x) = φ̃(x⊗1)(ν) = h(x⊗1)(ν), ξi.
Proof. (i)
Exercise 6.1.2
Replace weak and strong by ultraweak and ultrastrong, and
`2 -convergent ones in the previous theorem.
the nite sequences of vectors by
Corollary 6.1.3
If
C
is a convex subset of
sures coincide.
31
B(H),
its weak and strong clo-
Proof.
Two locally convex vector spaces with the same continuous linear
functionals have the same closed convex sets. This is a consequence of the
Hahn-Banach theorem to be found in any text on functional analysis..
Corollary 6.1.4
If
dim H = ∞
the strong and ultrastrong topologies dier
the proof here actually on B(H).
proves that the weak and
ultraweak topologies dier,
and to get the result here
Proof. Let (ξi ) be an orthonormal basis of H and let ω(x)
we need to use the previous
exercise
1
i n2 hxξi , ξi i.
Then ω is ultraweakly continuous but not strongly continuous. For if it were
Pn
weakly continuous it would be of the form
i=1 hxνi , ηi i and ω(p) = 0 where p
=
P
is the projection onto the orthogonal complement of the vector space spanned
by the
νi .
But by positivity
ω(p) = 0
forces
p(ξi ) = 0
for all
i. 6.2 The theorem
In our discussion of
vN (Γ) we already met the desirability of having a norm-
bounded sequence of operators converging to an element in the weak closure
of a *-algebra of operators.
This is not guaranteed by the von Neumann
density theorem. The Kaplansky density theorem lls this gap.
Theorem 6.2.1
Let
A
be a *-subalgebra of
B(H).
Then the unit ball of
is strongly dense in the unit ball of the weak closure
adjoint part of the unit ball of
the unit ball of
A
M
of
A,
A
and the self-
is strongly dense in the self-adjoint part of
M.
Proof. By EP6) we may assume
1∈M
and the worried reader may check
that we never in fact suppose 1 ∈ A. We may further suppose that A is
∗
norm-closed, i.e. a C -algebra. Consider the closure of Asa , the self-adjoint
The * operation is weakly continuous so if xα is a net converging
∗
α
converges to x so the weak closure
x ∈ M , xα +x
2
of Asa is equal to Msa .
∗
Let us now prove the second assertion of the theorem. Let x = x ∈ M ,
part of
A.
to the self-adjoint element
||x|| < 1,
ξ1 , ..., ξn , > 0 dene a strong neighbourhood of x. We must
2t
come up with a y ∈ Asa , ||y|| < 1, with ||(x−y)ξi || < . The function t →
1+t2
is a homeomorphism of [−1, 1] onto itself. So by the spectral theorem we may
2X
choose an X ∈ Msa with ||X|| ≤ 1, so that
= x. Now by strong density
1+X 2
choose Y ∈ Asa with
and
||Y xξi − Xxξi || < ,
Put
y=
2Y
and note that
1+Y 2
and
||
Y
X
ξi −
ξi || < /4.
2
1+X
1 + X2
||y|| ≤ 1.
32
Now consider the following equalities:
y−x=
2Y
2X
−
2
1+Y
1 + X2
1
1
2
2
(Y
(1
+
X
)
−
(1
+
Y
)X)
)
1+Y2
1 + X2
1
1
Y
X
= 2(
(Y − X)
+
(X − Y )
)
2
2
2
1+Y
1+X
1+Y
1 + X2
1
1
2
(Y − X)
+ y(X − Y )x.
=
2
2
1+Y
1+X
2
= 2(
By the choice of
Y,
we see that
||(y − x)ξi || < .
This proves density for
the self-adjoint part of the unit ball.
||x|| ≤ 1. The trick
isto form
aα b α
a b
∈ M ⊗M2 (C). Strong convergence of a net
to
cα dα
c d
is equivalent to strong convergence of the matrix entries so A ⊗ M2 (C) is
aα b α
0 x
strongly dense in M ⊗ M2 (C). Moreover if
→
strongly
∗
cα dα
x 0 aα b α
then bα tends strongly to x. And ||bα || ≤ 1 follows from ||
|| ≤ 1
cα dα
aα b α
0
η
and hbα ξ, ηi = h
,
i. cα dα
ξ
0
Now consider a general
0
x∗
x
0
x ∈ M
with
Corollary 6.2.2
If
M
is a *-subalgebra of
von Neumann algebra i the unit ball of
Proof. The unit ball of
M
M
B(H)
containing
1
then
M
is a
is weakly compact.
is weakly closed since the norm topology is stronger
than the weak topology, and a closed subset of a compact set is compact.
Conversely, if the unit ball of
closed. Let
net
xα
x
M
is weakly compact, then it is weakly
M . By Kaplansky density
||xα || ≤ 1. Hence x ∈ M . be in the weak closure of
weakly converging to
x
with
33
there is a
this proof is wrong!
34
Chapter 7
Comparison of Projections and
Type II1 Factors.
7.1 Order on projections
Denition 7.1.1
p
If
and
q
are projections in a von Neumann algebra M
∗
if there is a partial isometry u ∈ M with uu = p and
we say that p q
u∗ u ≤ q . We say that
isometry
u∈M
≈
Observe that
with
p and q are equivalent, p ≈ q
uu∗ = p and u∗ u = q .
if there is a partial
is an equivalence relation.
Theorem 7.1.2
The relation
is a partial order on the equivalence classes
of projections in a von Neumann algebra .
Proof. Transitivity follows by composing partial isometries. The issue is to
ef
show that
and
f e
imply
e ≈ f.
Compare this situation with sets
and their cardinalities.
Let
u
and
v
satisfy
uu∗ = e, u∗ u ≤ f
and
vv ∗ = f, v ∗ v ≤ e.
Note the
picture:
We dene the two decreasing sequences of projections e0 = e, en+1 =
∗
and f0 = f, fn+1 = u en u. The decreasing property follows by induc∗
tion since p → v pv gives an order preserving map from projections in M
v ∗ fn v
less than
f
to- .
projections in
and
f, v
and
u.
M
Let
and similarly interchanging the roles of e
∞
^
ei and f∞ =
fi . Note that v ∗ f∞ v = e∞ and
less than
∞
^
e∞ =
e
i=0
i=0
f∞ vv ∗ f∞ = f∞ so that e∞ ≈ f∞ . Also e = (e − e1 ) + (e1 − e2 ) + · · · + e∞ and
f = (f −f0 )+(f1 −f2 )+· · ·+f∞ are sums of mutually orthogonal projections.
35
∗
But for each even i, u (ei − ei+1 )u = fi+1 − fi+2 so ei − ei+1 ≈ fi+1 − fi+2 ,
∗
and v (fi − fi+1 )v = ei+1 − ei+2 so one may add up, in the strong topology,
all the relevant partial isometries to obtain an equivalence between
e
and
f.
Note that if we had been dealing with
vN (Γ)
this argument would have
been unnecessary as we could have used the trace:
tr(v ∗ v) ≤ tr(e) = tr(uu∗ ) = tr(u∗ u) ≤ tr(f ) = tr(vv ∗ ) = tr(v ∗ v)
so that
tr(e − v ∗ v) = 0
which implies
e = v∗v.
However in general it is
certainly possible to get a projection equivalent to a proper subprojection of
2
itself. Just take the unilateral shift on B(` (N)) which exhibits an equivalence
1
between
and the projection onto the orthogonal complement of the rst
basis vector. This is analogous to the notion of an innite setone which is
in bijection with a proper subset of itself.
Denition 7.1.3
nite if
p≈q
p in a von Neumann
q < p, p 6= q . Otherwise p
A projection
for some
algebra
M
is called in-
is called nite. A von
Neumann algebra is called nite if its identity is nite, and it is called purely
innite if it has no nite projections other than
0.
A factor is called innite
if its identity is innite.
We will show that purely innite von Neumann algebras exist though it will
not be easy.
Remark 7.1.4
If
Remark 7.1.5
A factor with a trace like
Remark 7.1.6
Every projection in a nite von Neumann algebra is nite.
dim H = ∞
Or, more strongly, if
For if
p ≈ p0 ,
Remark 7.1.7
in
M ⊗ B(H)
if
p0 < p,
and
q
p 6= p0
B(H)
is innite.
vN (Γ)
is nite then
then
p
is nite.
is nite.
p + (q − p) ≈ p0 + (q − p) 6= q .
M is any von Neumann algebra , 1 is an innite projection
dim H = ∞.
If
Theorem 7.1.8
or
p≤q
then
If
M
is a factor and
p, q
q p.
36
are projections in
M,
either
pq
∗
∗
Consider the family of partial isometries u with uu ≤ p, u u ≤ q .
∗
∗
This set is partially ordered by u ≤ v if u u ≤ v v and v = u on the initial
∗
domain u uH of u. This partially ordered set satises the requirements for
∗
∗
Zorn's lemma so let u be a maximal element in it. If u u = q or uu = p we
∗
∗
are done so suppose q − u u and p − uu are both non-zero. Then by 5.1.1
∗
∗
∗
∗
there is a v 6= 0 with v v ≤ q − u u and vv ≤ p − uu . But then u + v is
Proof.
larger than
u
which was supposed maximal.
Exercise 7.1.9
tor
M
Show that two equivalent projections
are unitarily equivalent, i.e. there is a unitary
p and q in a nite facu ∈ M with upu∗ = q .
We see that the equivalence classes of projections in a factor form a totally
ordered set.
It is known that, on a separable Hilbert space, the possible
isomorphism types for this set
We added 0 to the list for
type In , so the comment
below about type III and
type I1 makes sense
are:
{0, 1, 2, ..., n}
1)
2)
3)
4)
n=∞
[0, 1]
[0, ∞]
{0, ∞}
where
is allowed.
type In type II1 type II∞ type III
Strictly speaking this is nonsense as type III is the same as type I1 and
II1 is the same as II∞ . We mean not only the order type but whether
1
is
innite or not.
Observe that the type II1 case certainly exists.
has projections of any trace between
0
and
1.
We saw that
vN (F2 )
By the previous theorem
it is clear that the trace gives an isomorphism between the ordered set of
equivalence classes of projections ant the unit interval. We will proceed to
prove a statement generalising this considerably.
Denition 7.1.10
H
A type II1 factor is an innite dimensional factor
admitting a non-zero linear function
tr : M → C
M
on
satisfying
tr(ab) = tr(ba)
(i)
(ii)
(iii)
tr(a∗ a) ≥ 0
tr
is ultraweakly continuous.
The trace is said to be normalised if
tr(1) = 1.
Denition 7.1.11
In general a linear functional φ on a *-algebra A is called
∗
positive if φ(a a) ≥ 0 (and φ(a ) = φ(a) though this is redundant if A is a
∗
C -algebra), and faithful if φ(a∗ a) = 0 ⇒ a = 0. A positive φ is called a
∗
1 ∈ A and φ(1) = 1.
if φ(ab) = φ(ba).
state if
trace)
A linear functional
37
φ
is called tracial (or a
It is our job now to show that a II1 factor has a unique ultraweakly
continuous tracial state, which is faithful. First a preliminary result on ideals.
Theorem 7.1.12
M.
M is
Let
mann algebra
M = M e.
If
M
be an ultraweakly closed left ideal in a von Neu-
Then there is a unique projection
2-sided,
e
is in
e ∈ M
such that
Z(M ).
M ∩ M∗ is an ultraweakly closed *-subalgebra so it has a largest
projection e. Since e ∈ M, M e ⊆ M. On the other hand if x ∈ M let
x = u|x| be its polar decomposition. Since u∗ x = |x|, |x| ∈ M ∩ M∗ . Hence
|x|e = |x| and x = u|x| = u|x|e ∈ M e. So M = M e.
Uniqueness follows easily since f = xe ⇒ f ≤ e.
∗
Moreover if M is 2-sided, for any unitary u ∈ M , uM = M = uMu =
M e = M ueu∗ so ueu∗ = e by uniqueness. Hence e ∈ Z(M ).
Proof.
Corollary 7.1.13
An ultraweakly continuous positive non-zero trace
Tr
on
a II1 factor is faithful.
M = {x ∈ M : T r(x∗ x) = 0}. Then since x∗ a∗ ax ≤ ||a||2 x∗ x, M
is a left ideal and since T r(ab) = T r(ba), M is a 2-sided ideal. Moreover by
∗
the Cauchy Schwarz inequality T r(x x) = 0 i T r(xy) = 0 ∀y ∈ M . Thus
M is ultraweakly closed, being the intersection of the kernels of ultraweakly
continuous functionals. Thus M = M e for some central projection. And e
must be zero since M is a factor. Proof. Let
Corollary 7.1.14
projection,
pM p
If
M
is a type II1 factor on
is a type II1 factor on
Proof. This is cleara trace on
M
H
and
p∈M
is a non-zero
pH.
restricts to a trace on
pM p which is non-
zero by faithfulness and all the other properties are immediate. Since a minimal projection in
pM p would be minimal in M , pM p is innite dimensional.
The uniqueness of
tr
will follow easily once we have gathered some facts
about projections in a II1 factor.
Theorem 7.1.15
There are non-zero projections in a type II1 factor of ar-
bitrarily small trace.
d = inf {tr(p) : p ∈ M, p2 = p∗ = p 6= 0}. Suppose d > 0. Let
p be a projection with tr(p) − d < d. Then p is not minimal since we have
seen that M is not isomorphic to B(H). So there is a non-zero projection
q < p. But then we have tr(p − q) = tr(p) − tr(q) ≤ tr(p) − d < d. This is a
contradiction. So d = 0. Proof. Let
38
Theorem 7.1.16
positive non-zero
M be a type II1 factor with an ultraweakly continuous
2
∗
trace tr . Then {tr(p) : p ∈ M, p = p = p} = [0, tr(1)].
Let
Proof. For
r ∈ [0, tr(1)]
r}.
is a partially ordered set and if
S = {p : p a projection in M and
W tr(p) ≤
pα is a chain in S , p = α pα ∈ M
and p is in the strong closure of the pα so p is in S . So by Zorn, S has a
maximal element, say q . If tr(q) were less than r , then by 7.1.8, q ≺ p. So
0
choose q ∼
= q, q 0 < p. Applying 7.1.14 to p − q 0 we nd a projection strictly
0
between q and p. Then
S
Corollary 7.1.17
consider
The map
tr
gives an isomorphism between the totally or-
dered set of equivalence classes of projections on a type II1 factor and the
interval
[0, tr(1)].
Proof. By 7.1.16 it suces to show that the equivalence class of a projection is determined by its trace. This is immediate from 7.1.8.
Exercise 7.1.18
a subfactor
Let
N ⊆M
Corollary 7.1.19
M
with
be a type II1 factor . Then for each
N∼
= Mn (C).
n∈N
there is
Any two non-zero ultraweakly continuous normalised traces
on a type II1 factor are equal.
Proof. By the elementary facts it suces to prove that two such traces
and
tr
agree on projections. We may assume one of them, say
tr,
Tr
is positive.
By the previous exercise, 7.1.17, and the uniqueness of the trace on a matrix
algebra,
tr
and
Tr
are equal on projections for which
a projection for which
tr(p)
tr
is rational. Given
is irrational build an increasing sequence
ei
of
subprojections as follows:
ei with tr(ei ) = T r(ei ) and tr(p) −
(p − ei )M (p − ei ) is a type II1 factor so tr and T r agree
on projections in it whose tr is arbitrarily close to tr(p − ei ). So choose in
it a projection ei+1 between ei and p, on which tr and T r agree and with
W
1
. Then tr and T r agree on
tr(p) − tr(ei+1 ) < i+1
i ei which is equal to p by
the faithfulness of tr . Suppose we have already constructed
tr(ei ) < 1/i.
Then
We shall see that a positive trace on a type II1 factor is norm-continuous
and a self-adjoint operator is actually a norm-limit of linear combinations
of its spectral projections so in fact an apparently weaker property than
ultraweak continuity is all we used in the previous corollarynamely that
the trace of the supremum of an increasing net of projections is the supremum
of the traces.
39
Corollary 7.1.20
Let
M
be a von Neumann algebra with a positive ultra-
weakly continuous faithful normalised trace
i
T r = tr
Proof.
tr.
Then
M
is a type II1 factor
for all ultraweakly continuous normalised traces
We just have to show that
Z(M )
is trivial.
T r.
But if it were not,
choose by faithfulness a projection p ∈ Z(M ) with 0 < tr(p) < 1. Dene
1
)tr(xp). Then T r is an ultraweakly continuous normalized
T r(x) = ( tr(p)
trace dierent from tr on 1 − p. Exercise 7.1.21
a
be a non-zero positive self adjoint operator. Show
+
+
that there is a bounded piecewise smooth function f : R → R such that
af (a)
Let
is a non-zero projection.
Exercise 7.1.22
A type II1 factor is algebraically simple.
(Hintuse the
previous exercise to show that a 2-sided ideal contains a projection, then add
projections to obtain the identity.)
7.2 The GNS construction
Thus uniqueness of the trace implies factoriality. This suggests another interesting way to construct a type II1 factor . If
is embedded
a ,→ a ⊗ 1. Iterate this procedure to form an
An of *-algebras with A1 = A and An+1 = An ⊗ A, and
∞
consider the *-algebra A∞ = ∪n An which could also be called ⊗alg,n=1 An . If
we normalise the matrix trace on all matrix algebras so that tr(1) = 1 then
tr(a ⊗ 1) = tr(a) so that tr denes a positive faithful normalised trace on
A∞ . Elements of A∞ can be thought of as linear combinations of tensors
of the form a1 ⊗ a2 ⊗ a3 ⊗ · · · ⊗ 1 ⊗ 1 ⊗ 1 ⊗ · · ·, on which the trace is just
0
the product of the traces of the ai s. We now turn A∞ into a von Neumann
in
A⊗A
A = M2 (C), A
as diagonal matrices:
increasing sequence
algebra .
hx, yi = tr(y ∗ x). Then A∞ is a preHilbert space and let H be its completion. Note that Mn (C) is a von Neu∗ ∗
2
∗
mann algebra so tr(y x xy) ≤ ||x|| tr(y y). This means that the operator
Lx on A∞ , Lx (y) = xy , satises ||Lx (ξ)|| ≤ ||x||||ξ|| (where ||x|| is the operator norm of the matrix x and ||ξ|| is the Hilbert space norm of ξ ) and so
extends uniquely to a bounded operator also written Lx on H. One checks
∗
that (Lx ) = Lx∗ so x → Lx denes a faithful (=injective) representation of
the *-algebra A∞ on H . Let M be the von Neumann algebra on H generated
by the Lx and identify A∞ with a subalgebra of M .
Dene an inner product on
The trace on
1 ∈ A∞
A∞
A∞
by
tr(a) = haξ, ξi where ξ is the element
H. So tr extends to a trace on M which is
is dened by
considered as a vector in
40
ultraweakly continuous, positive and normalised. It is also unique with these
properties by the uniqueness of the trace on the ultraweakly dense subalgebra
A∞
of
M.
If we can show that
tr
is faithful on
M
then it follows that
M
is a
type II1 factor . It is important to note that this does not follow simply from
the faithfulness of
tr
on
A.
In fact it is true but we need to do something to
prove it.
Lx was bounded, the same calculation, with tr(ab) =
tr(ba), would have shown that Rx , right multiplication by x, is also bounded.
Associativity shows that Lx and Ry commute on A∞ , hence on H. Thus M
commutes with Ry for each y ∈ A∞ . Now we can show faithfulness: if
tr(x∗ x) = 0 for x ∈ M then for each a ∈ A∞ we have
When we showed that
||x(a)||2 = ||xRa (ξ)||2 = ||Ra x(ξ)||2 ≤ ||Ra ||2 ||xξ||2 = ||Ra ||2 tr(x∗ x) = 0.
Since
A∞
is dense, this means
x = 0.
So
tr
M
is faithful on
which is thus a
type II1 factor .
Many points are raised by this example.
the properties of the vector
M ξ = H and M 0 ξ = H.
Denition 7.2.1
ξ
The easiest to deal with are
which played a prominent role. We used both
M be a von Neumann algebra on H. A vector ξ ∈ H is
called cyclic for M if M ξ = H and separating for M if
(xξ = 0) ⇒ (x = 0)
for all x ∈ M .
Let
Proposition 7.2.2
rating for
With notation as above,
ξ
is cyclic for
M
i
ξ
is sepa-
M 0.
Proof.(⇒) Exercisein fact done in the discussion of
(⇐) Let
(1 − p)ξ = 0
p
be the projection onto the closure of
so
A∞
M ξ.
above.
Then
p ∈ M 0.
But
p = 1. The construction of
M
from
A∞
is a special case of what is known
as the GNS construction (Gelfand-Naimark-Segal). Given a positive lin∗
ear functional φ satisfying φ(a ) = φ(a) on a *-algebra A we let Nφ be
{x ∈ A : φ(x∗ x) = 0}. We also dene a sesquilinear form h, iφ on A by
hx, yiφ = φ(y ∗ x). This form is positive semidenite but this is enough for the
Cauchy-Schwartz inequality to hold so that
0 ∀y ∈ A}
so that
N
is a subspace and
structure on the quotient
tiplication by
A/N .
N is the same as {x : hx, yiφ =
h, iφ denes a pre-Hilbert space
Under favourable circumstances, left mul-
x , Lx
denes a bounded linear operator on it.
∗
circumstances are provided by C -algebras.
41
Favourable
Exercise 7.2.3
φ is a linear functional on a C ∗ -algebra satisfying φ(a∗ a) ≥
0 show that φ(a ) = φ(a). Moreover if A is unital show that φ is normcontinuous and in fact ||φ|| = φ(1).
If
∗
Remark 7.2.4
It is a standard elementary fact in
∗
always adjoin an identity to a C -algebra.
Proposition 7.2.5
If
A
is a unital
C ∗ -algebra
C ∗ -algebras that
and
φ:A→C
one may
is a positive
linear functional then
φ(y ∗ x∗ xy) ≤ ||x||2 φ(y ∗ y)
Proof. Let φ̃(a)
||x||2 φ̃(1). = φ(y ∗ ay).
Then
φ̃
is positive so by the exercise
φ̃(x∗ x) ≤
φ on a unital
πφ (x) on the
Hilbert space Hφ of the GNS construction via left multiplication: πφ (x)(y) =
xy . Moreover ||πφ (x)|| ≤ ||x|| and πφ (x∗ ) = πφ (x)∗ since hπφ (x)y, zi =
φ(z ∗ xy) = hy, πφ (x∗ )zi. Note that φ(x) = hπφ (x)1, 1i.
It follows immediately that, given a positive linear functional
C ∗ -algebra,
each
x∈A
determines a bounded linear operator
To sum up we have the following:
Denition 7.2.6
on
A,
If
A
is a
C ∗ -algebra
and
φ
is a positive linear functional
the Hilbert space of the GNS construction is written
resentation
πφ
Hφ
and the rep-
by left multiplication is called the GNS representation.
Proposition 7.2.7
If
A
is a
C ∗ -algebra
on
H
haξ, ξi. Then φξ is a positive linear functional
u : Hφξ → Aξ such that uπφξ (a)u∗ = a.
ξ ∈ H, dene φξ (a) =
a 7→ aξ denes a unitary
and
and
Proof. Obvious.
A is actually a von Neumann algebra , πφ (A) will not in general be one
Hφ . However we will see that if φ is ultraweakly continuous then πφ (A)
If
on
will be a von Neumann algebra . For this we need a little preparation.
Lemma 7.2.8
A be a C ∗ -algebra on H containing 1. If ψ is a positive
linear functional on A and ξ ∈ H is a vector with ψ ≤ φξ (i.e. φξ − ψ is
0
positive), then there is a t ∈ A with ψ = φtξ .
Let
(, ) on Aξ by (aξ, bξ) = ψ(b∗ a). CauchySchwarz and ψ ≤ φξ give that |(aξ, bξ)| ≤ ||aξ||||bξ|| so (, ) is well-dened and
∗
there is a bounded positive operator t on Aξ with haξ, tbξi = ψ(b a). But
∗ ∗
∗
0
haξ, tbcξi = ψ(c b a) = hb aξ, tcξi = haξ, btcξi
√ so that t ∈ A on Aξ . If p =
0
pAξ , tp is a positive operator in A and if s = t, ψ(a) = haξ, tξi = hasξ, sξi.
.
Proof. Dene a sesquilinear form
42
Corollary 7.2.9
(on a
C
∗
A
-algebra
Proof. For
ξ
and
on
H)
If
η
ω(a) = haξ, ηi
ν with ω = φν .
are vectors such that
then there is a vector
is positive
a ≥ 0,
haξ, ηi = 1/4(ha(ξ + η), ξ + ηi − ha(ξ − η), ξ − ηi)
≤ 1/4φξ+η (a).
Theorem 7.2.10
If
M
H
is a von Neumann algebra on
and
φ
is a positive
ultraweakly continuous linear functional on M then there are vectors
P
P
with
||νi ||2 < ∞ so that φ(x) = hxνi , νi i for x ∈ M .
Proof. Let
hxξ, ηi
ξi , ηi
M
be as supplied by 6.1.2. Then for
so by the previous corollary we are done.
Corollary 7.2.11
on
νi ∈ H
H ⊗ `2 (N), φ(x) =
φ is an ultraweakly continuous positive linear functional
on the von Neumann algebra M then the GNS representation πφ is ultraweakly
continuous onto a von Neumann algebra on Hφ .
If
φ(x) = hx⊗1(ν), νi on H ⊗`2 (N).
continuous. By 7.2.7 we have that πφ
reduction to M ⊗ 1(ν) is ultraweakly
Proof. We saw in the last theorem that
The map
x 7→ x ⊗ 1
is ultraweakly
is ultraweakly continuous since the
continuous. So the kernel of
πφ
is an ultraweakly closed 2-sided ideal, hence
M e for some e in the centre of M . It follows that πφ is injective on
M (1 − e) and since the norm of an operator x is determined by the spectrum
∗
of x x, the unit ball of the image of M is the image of the unit ball which is
weakly compact so by 6.2.2 we are done. of the form
7.3 Exercises on two projections.
Let
p
space
and
U
q
be projections onto closed subspaces
M = {p, q}00 .
H
and
K
of the Hilbert
respectively. Let
Exercise 7.3.1
and this decomposition is
Exercise 7.3.2
U = (H∩K)⊕(H⊥ ∩K⊥ )⊕(H∩K⊥ )⊕(H⊥ ∩K)⊕W
invariant under p and q .
Show that
W , p and q are in general
p ∧ q = 0, p ∨ q = 1, (1 − p) ∧ q = 0 and (1 − p) ∨ q = 1.
Show that, on
43
position, i.e.
Exercise 7.3.3
a ∈ B(H), 0 ≤ a ≤ 1,
Show that if
is a projection on
H ⊕ H.
Exercise 7.3.4
Let
Exercise 7.3.5
Show that
p a
a(1 − a)
When is it in general position with
p
a(1 − a)
1− a
1 0
?
0 0
a = (p − q)2 and A = {a}00 . Show that a ∈ Z(M ) and
that {a0 +a1 p+a2 q +a3 pq +a4 qp : ai ∈ A} is a *-algebra (which is necessarily
weakly dense in M ).
>From now on suppose
Exercise 7.3.6
tion of
Show that
pM p
p
is abelian, generated by
and
p∼
=q
q
in
pqp.
are in general position.
M.
(Hint,consider the polar decomposi-
pq .)
Exercise 7.3.7
Show there is a
2×2
system of matrix units
(eij ) ∈ M
with
p = e11 .
Exercise 7.3.8
B ⊗ M2 (C) for some
abelianvon Neumann
algebra B generated
1, with p correspond
by b, 0 ≤ b ≤p
b
b(1
−
b)
1 0
p
ing to
and q corresponding to
0 0
b(1 − b)
1−b
Show that
M
is spatially isomorphic to
Now drop the hypothesis that
Exercise 7.3.9
Show that
p
and
q
are in general position.
p∨q−p∼
=q−p∧q
in
M
Alternative approach using group representations.
Exercise 7.3.10
Show that
(Z/2Z) ∗ (Z/2Z) ∼
= Zo(Z/2Z)
(innite dihedral
group).
Exercise 7.3.11
Classify all unitary representations of
Zo(Z/2Z).
(Hint
use the spectral theorem for unitaries.)
Exercise 7.3.12
Observe that
Exercise 7.3.13
Obtain the structure of 7.3.8 using the last 3 exercises.
2p − 1
and
44
2q − 1
are self-adjoint unitaries.
Chapter 8
The Predual
φ on a von Neumann algebra M
∗
is norm continuous so denes an element of M . Our goal in this chapter is
∗
to show that the set of all such φ is a closed subspace M∗ of M and that
the duality between M∗ and M makes M equal to the Banach space dual of
M∗ . We will rst establish this in the special case M = B(H).
An ultraweakly continuous linear functional
8.1 Trace class and Hilbert Schmidt operators.
The material in this section is standard so we will only prove results as it
suits us, otherwise referring any unproved assertions to Reed and Simon.
Lemma 8.1.1
bases of
H,
If
a ∈ B(H) is positive and (ξi ) and (ηi ) are two orthonormal
then
X
X
haξi , ξi i =
i
(where
∞
haηi , ηi i
i
is a possible value for the sum).
Proof. We have
X
haξi , ξi i =
X √
|| aξi ||2
i
i
XX √
=
(
|h aξi , ηj i|2 )
i
j
XX √
=
(
|h aηj , ξi i|2 )
j
i
X √
=
|| aηj ||2
j
45
=
X
haηj , ηj i
j
where every number is positive so the order of the sum is immaterial.
The number
P
i haξi , ξi i of the previous theorem is called the trace of
a,
T race(a).
written
Denition 8.1.2
An element
a ∈ B(H) is said to be of
trace class if
T race(|a|) <
∞.
If
a
is trace class and
(ξi )
is an orthonormal basis, the sum
X
haξi , ξi i
i
converges absolutely and is called the trace,
Theorem 8.1.3
The trace class operators on
of compact operators,
T race(|a|)
|a|1 .
T race(a),
I1 ,
in
denes a norm on
B(H). The
I1 for which
H
of
a.
form a self-adjoint ideal
|a|1
function
dened by
it is complete. Moreover
|a|1 =
||a|| ≤
Proof. The only thing not proved in Reed and Simon is completeness. For
this observe that if
an
is a Cauchy sequence in
| − |1 ,
|| − || so
| − |1 -Cauchy sequence
that limit in | − |1 . So
it is Cauchy in
what we have to do is show that the norm limit of a
(an )
is trace class and that the sequence tends to
suppose
>0
is given. Then for
∞
X
m
and
n
large enough
h|an − am |ξi , ξi i < .
i=1
So for any
N,
N
X
h|an − am |ξi , ξi i < .
i=1
Now if bn tends in norm to b, then |bn | tends in norm to |b| (obviously
∗
bn bn → b∗ b, and approximate the square root function by polynomials on an
interval) so for each xed i,
lim |an − am |ξi = |a − am |ξi .
n→∞
PN
i=1 h|a − am |ξi , ξi i < and letting N tend to ∞ we see that
I1 is a vector space, and also that an → a in | − |1 . So
46
a ∈ I1
since
The trace is independent of the orthonormal basis and if
and
a
is trace class
b ∈ B(H), T r(ab) = T r(ba).
We see that each
h ∈ I1
determines a linear functional
φh
on
B(H)
by
φh (x) = T race(xh).
Denition 8.1.4
for the state
The trace-class matrix as above is called the density matrix
φh .
Proposition 8.1.5
Each
element of
|h|1 .
B(H)∗
is
φh
is ultraweakly continuous and its norm as an
h is compact, choose an orthonormal basis (ξi ) of eigenvectors
|h| with eigenvalues λi and let h = u|h| be the polar decomposition. Then
Proof. Since
of
φh (x) =
∞
X
hxu|h|ξi , ξi i
i=1
so ultraweak continuity is apparent, and
φh (x) ≤
∞
X
||x|| || |h|ξi ||
i=1
= ||x||
∞
X
λi
i=1
= ||x|| |h|1 .
Moreover evaluating
on
u∗
gives
||φh || = |h|1 . are Hilbert spaces, a bounded operator x : H → K is called
P∞
2
x∗ x is trace class, i.e.
i=1 ||xξi || < ∞ for some (hence
any) orthonormal basis (ξi ) of H. The set of all Hilbert-Schmidt operators
2
∗
from H to K is written ` (H, K) and if x is Hilbert-Schmidt, so is x , and x
If
H
φh
and
K
Hilbert-Schmidt if
is compact.
Theorem 8.1.6
a ∈ B(H), b ∈ B(K) and x ∈ `2 (H, K) then bxa ∈
`2 (H, K). If x ∈ ` (H, K) and y ∈ `2 (K, H) then yx is trace class. With the
∗
2
inner product hx, yi = T race(y x), ` (H, K) is a Hilbert space in which the
If
2
nite rank operators are dense.
Proof. See Reed and Simon.
Exercise 8.1.7
Prove all the assertions made above about trace-class and
Hilbert-Schmidt operators.
47
Exercise 8.1.8
∗
K ⊗H
Let
H and K are Hilbert spaces construct
` (H, K) and show that it is unitary.
If
a natural map from
2
to
|x|2
be the Hilbert space norm on Hilbert-Schmidt operators.
Lemma 8.1.9
If
x ∈ `2 (H, K) and y ∈ `2 (K, H) then T race(xy) = T race(yx).
Proof. First note that the result is true if we suppose that
For then let
x = u|x|
|x| is trace class.
be the polar decomposition, choose an orthonormal
of the nal domain of u and extend it to an orthonormal basis of
K. Also extend (u∗ ξi ) to an orthonormal basis of H by vectors in ker(|x|).
basis
(ξi )
Then
X
T race(xy) =
hu|x|yξi , ξi i
i
=
X
h|x|yuu∗ ξi , u∗ ξi i
i
= T race(|x|(yu))
= T race((yu)|x|)
= T race(yx.)
Now suppose only that x is Hilbert-Schmidt. Let
x0 of nite rank with |x − x0 |2 < . Then
>0
be given and choose
|T race(xy) − T race(yx)| = |T race((x − x0 )y) − T race(y(x − x0 ))|
which by Cauchy-Schwartz is
Corollary 8.1.10
B(H),
If
ω
≤ 2|y|2 . is an ultraweakly continuous linear functional on
there is a trace class
h
so that
ω = φh .
P
(ξi ) and (ηi ) in `2 (N, H) so that
ω(x) = i hxξi , ηi i.
P
2
Then if we dene a and b from ` (N) to H by a(f ) =
i f (i)ξi and b(f ) =
P
∗
i f (i)ηi , a and b are Hilbert Schmidt and ω(x) = T race(b xa) which is
T race(xab∗ ) by the previous result. Proof. By 6.1.2 there are
Putting everything together so far, we have identied the image of the
∗
Banach space I1 under the map h 7→ φh with the closed subspace of B(H)
consisting of ultraweakly continuous linear functionals. To close the loop we
only need to show that the Banach space dual of
Theorem 8.1.11
x ∈ B(H)
so that
I1
is
α : I1 → C is linear and bounded
α(a) = φa (x), and ||α|| = ||x||.
If
48
B(H).
for
| − |1 ,
there is an
x of I1
by x(v) = hv, ξiη so one may dene a sesquilinear form on H by (ξ, η) = α(x).
Boundedness of x follows from that of α so there is an appropriate x ∈ B(H).
To show that the norm of x as an element of the dual of I1 is actually ||x||,
suppose ||x|| = 1 and choose a unit vector ξ with ||xξ|| almost equal to 1.
Then T r(hx) is almost 1 if h is the partial isometry which sends v ∈ H to
ξ
. hv, xξi ||xξ||
Proof. This is rather routine. Two vectors
Exercise 8.1.12
ξ
and
η
dene an element
Fill in the missing details in the previous proof.
Now we pass to von Neumann algebras though in fact these results work
for any ultraweakly closed subspace of
Theorem 8.1.13
V
⊥⊥
V
If
is an ultraweakly closed subspace of
φ(x) = 0
x∈V.
in the sense that if
which
φ(V ) = 0
then
B(H).
B(H)
V =
φ for
then
for every ultraweakly continuous
Proof. This is a simple application of the Hahn-Banach theoremif
V
construct an ultraweakly continuous functional which is zero on
non-zero on
and
x. Exercise 8.1.14
orthogonal to
Exhibit a non-zero trace class operator on
`2 (Γ)
which is
vN (Γ).
Theorem 8.1.15
If
V
is an ultraweakly closed subspace of
is canonically the dual Banach space of
V∗
V
is the weak-* topology on
V
B(H)
then it
which is dened as the space
of ultraweakly continuous linear functionals on
topology on
V
x∈
/
V.
Moreover the ultraweak
as the dual of
V∗ .
∗
is a Banach space with dual B and V is a
∗
⊥
weak-* closed subspace of B then V is the dual of B/V
(surjectivity
⊥
of the natural map from V to the dual of V /B
is a result of the previous
Proof. If
B
theorem), so V is a dual space. So we just have to identify the Banach
⊥
∗∗
space B/V
with the space of weak-* continuous (as elements of B ) linear
functionals on
V.
Exercise 8.1.16
V∗
This is a simple exercise. Putting
If
V
B = I1
is an ultraweakly closed subspace of
is a separable Banach space if
H
we are done.
B(H),
is a separable Hilbert space.
49
show that
8.2 Normal maps
There is still a very unsatisfactory feature about the previous section in that
ultraweak continuity and the predual depend on the Hilbert space representation on which a von Neumann algebra acts. In this section we shall show
that ultraweak continuity is actually equivalent to a property called normality which is determined by the purely algebraic structure of a von Neumann
algebra .
Denition 8.2.1
A positive linear map
Φ:A→B
between von Neumann
algebras is called normal if
_
_
Φ( aα ) =
Φ(aα )
α
for any increasing net
Theorem 8.2.2
(aα )
α
of self-adjoint operators in
A.
A positive linear functional on a von Neumann algebra is
normal i it is ultraweakly continuous.
Proof. That ultraweak implies normal is obvious. For the moment we refer
to Dixmier for the other direction and content ourselves to remark that our
proof of the uniqueness of the trace on a II1 factor used only normality (a
self-adjoint operator is a norm limit of linear combinations of its spectral
projections) so that a normal trace is attached to it by its algebraic structure
alone.
Observe that by 6.1.2 any ultraweakly continuous linear functional is a
linear combination of positive ones (by polarisation) so the predual of a von
Neumann algebra is dened entirely by the algebraic structure. In fact the
predual is unique up to isometry.
There is another important structure theorem for whose proof we will
also provisionally simply refer to Dixmier:
Theorem 8.2.3
A
Φ
be a normal homomorphism of von Neumann algebras
is the composition of an amplication
x → x ⊗ id,
a reduction to a closed
invariant subspace, and a unitary equivalence.
Proof. Dixmier.
A unitary equivalence is also called a spatial isomorphism.
50
8.3 A technical lemma.
Let us prove a lemma which shows what the techniques developed so far can
be good for. It will be crucial in our treatment of Tomita-Takesaki theory.
It is a Radon-Nikodym type theorem inspired by one due to Sakai([]).
Lemma 8.3.1
Let
λ ∈ R+
be given and let
φ
be a faithful ultraweakly con-
tinuous state
von Neumann algebra M .
p on a p
∗
∗
|ψ(y x)| ≤ φ(x x) φ(y ∗ y). Then there is an
≤ 1/2)
ψ ∈ M∗ be such that
a ∈ M1/2 (elements of norm
Let
so that
ψ(x) = λφ(ax) + λ−1 φ(xa).
Proof. For
α(a) = θa ,
a∈M
let
θa (x) = φ(λax + λ−1 xa).
is continuous for the topologies of duality
But we know that this topology on
α(M1 )
α(M ).
M
α : M → M∗ ,
between M and M∗ .
Then the map
is the ultraweak topology so that
is a compact convex set. By contradiction suppose that
ψ
is not in
h ∈ M with <(ψ(h)) > D where
h = u|h| = |h∗ |u is the polar decomposition
Then by Hahn-Banach there is an
D = supa∈M1/2 <(θa (h)). But if
h, we have
θu∗ /2 (h) = 1/2(λφ(|h|) + λ−1 φ(|h∗ |))
of
so that
But also
tion.
p
p
1
2D ≥ λφ(|h|) + φ(|h∗ |) ≥ 2 φ(|h|) φ(|h∗ |).
λ
p
p
D < |ψ(h)| = |ψ(u|h|1/2 |h|1/2 )| ≤ φ(|h|) φ(u|h|u∗ ),
51
a contradic-
52
Chapter 9
Standard form of a II1 factor and
II∞ factors.
9.1 Standard form.
In this section
M
will be a von Neumann algebra with an ultraweakly contr and L2 (M, tr) will be abbreviated to
tinuous faithful normalized trace
L2 (M ).
In section 7.2 we learned how to construct a von Neumann algebra from a
C
∗
-algebra and a positive linear functional on it. If we apply this construction
R
∞
to L (X, µ) with trace given by
f dµ, the Hilbert space would be L2 (X, dµ).
2
For this reason, if M is a type II1 factor we write L (M, tr) for the GNS
p
Hilbert space obtained from the trace. In fact one can dene L spaces
p
1/p
for 1 ≤ p ≤ ∞ using the L norm ||x||p = tr(|x|)
. A noncommutative
p
version of the Holder inequality shows that || − ||p is a norm and L (M ) is
∞
1
the completion. We set L (M ) = M and we shall see that L (M ) is the
predual
M∗ .
Let us x on the notation
of
Ω for the vector in L2 (M ) which is the identity
M.
Proposition 9.1.1
If
M
complete metric space for
is a type II1 factor the
|| − ||2
|| − ||-unit
and the topology dened by
ball of
|| − ||2
M
is a
on the
unit ball is the same as the strong (and ultrastrong and *-strong) topology.
xn is Cauchy in || − ||2 then for each a ∈ M , xn a is also since
||xa||2 ≤ ||a|| ||xn ||2 . So we can dene x on the dense subspace M Ω of L2 (M )
by x(aΩ) = limn→∞ xan Ω. Since ||x|| ≤ 1, we have ||xξ|| ≤ ||ξ|| for ξ ∈ M Ω
2
so x extends to a bounded operator on L (M ) which is obviously in M , and
xΩ = x = limn→∞ = limn→∞ xn in || − ||2 .
Proof. If
53
The strong topology is obviously no stronger than
seminorm
|| − ||2
a 7→ ||aΩ||
|| − ||2
since the single
denes the
topology. Moreover
||xaΩ|| ≤ ||x||2 ||a||
shows that
|| − ||2
controls
the strong topology on the unit ball.
Finally note that in the statement of the theorem it does not matter what
representation of
M
is used to dene the strong topology on the unit ball as
the ultrastrong topology does not change under the manipulations that we
used to get the GNS construction from a II1 factor on an arbitrary Hilbert
space.
2
The action of M on L (M, tr) is called the standard form of M . Note
2
that vN (Γ) on ` (Γ) is already in standard form. (We see that we could have
obtained our rst example of a II1 factor by applying the GNS construction
P
γ cγ uγ = ci d.)
0
We now want to determine the commutant M when M is in standard
to the group algebra
with the trace
CΓ
tr(
form.
Denition 9.1.2
tion which is the
Lemma 9.1.3
J : L2 (M ) → L2 (M ) be the antilinear unitary involu2
∗
extension to L (M ) of the map x 7→ x from M to M .
Let
For
x, a
in
M,
and
ξ, η
in
L2 (M )
hJξ, Jηi = hη, ξi
∗
(ii) JxJ(aξΩ) = ax Ω
(i)
Proof.
ξ = aΩ and η = bΩ, hJξ, Jηi = tr(ba∗ ) = hη, ξi.
JxJ(aΩ) = J(xa∗ Ω) = ax∗ Ω.
(i) If
(ii)
Corollary 9.1.4
For
M
on
L2 (M ), JM J ⊆ M 0 .
Proof. Left and right multiplication commute.
Lemma 9.1.5
Proof. Take
For
a ∈ M,
M
on
L2 (M ),
if
x ∈ M 0 , JxΩ = x∗ Ω.
then
hJxΩ, aΩi = hJaω, xΩi
= ha∗ Ω, xΩi
= hΩ, xaΩi
= hx∗ Ω, aΩi.
54
Theorem 9.1.6
For
M
on
L2 (M ), JM J = M 0 .
0
Proof. If we can show that x 7→ hxΩ, Ωi is a trace on M we are done since
2
2
0
by the above results L (M ) can be canonically identied with L (M ) so that
the two
also for
But for
J maps coincide. (Note that Ω is cyclic and
M 0 .) So we would have JM 0 J ⊆ M .
x, y ∈ M 0 ,
hxyΩ, Ωi = hyΩ, x∗ Ωi
separating for
M
hence
= hyΩ, JxΩi
= hxΩ, JyΩi
= hxΩ, y ∗ Ωi
= hyxΩ, Ωi.
We see that the commutant of the left regular representation of
Γ on `2 (Γ)
is the von Neumann algebra generated by the right regular representation
Juγ Jεγ 0 = εγ 0 γ −1 . And more generally the commutant of the left action
2
∗
of M on L (M ) is the
− algebra of right multiplication operators. In
2
particular the commutant of a type II1 factor M on L (M ) is also a type
II1 factor . This is not the case for M on an arbitrary Hilbert space. For
2
instance we could consider M ⊗1 on L (M )⊗H for some innite dimensional
H. Then the commutant of M ⊗ 1 would be JM J ⊗ B(H)innite matrices
over JM J .
since
Denition 9.1.7
A II∞ factor is a factor on the form
type II1 factor and
pM p
with
M
a
dim H = ∞.
Proposition 9.1.8
that
M ⊗ B(H)
Let
M
be an innite factor with a projection
is a type II1 factor . Then
M
p∈M
so
is a II∞ factor.
orthogonal projections in
P {pα } of mutually
P
M with
Put q =
α pα . If 1 −
α pα were ≥ p then
P we could
contradict the maximality of the family {pα }. So write 1 = q +
α pα with
q ≤ p. Since M is innite the set of indices {α}P
is innite so we may choose
P
a bijection with itself minus α0 and write q +
p
p
+
α
α
0
α
α6=α0 pα =
P
1. We conclude that α pα is equivalent to 1 so we may suppose it equal
to 1. We may then construct a system of matrix units by using partial
isometries implementing the equivalences between the pα to obtain the result
Proof. Choose a maximal family
pα ∼
= p ∀α.
from exercise 5.3.3.
55
Note that we used implicitly in the above proof the fact that the supremum of nitely many nite projections is nite. This will be covered in a
series of exercises on projections.
It could conceivably happen that, given a II∞ factor
of the form
pM p
depends on
p
M , the type II1 factor
(obviously only up to equivalence). We now
introduce the trace on a II∞ factor which will make this issue more clear.
If
M
is a type II1 factor , dene the map
of positive elements of
M ⊗ B(H)),
to
tr((xij )) =
[0, ∞]
∞
X
tr
from
(M ⊗ B(H))+
(the set
by
tr(xii )
i=1
where we have chosen a basis of the innite dimensional Hilbert space
identify
M ⊗ B(H)
Theorem 9.1.9
with certain matrices over
H
to
M.
M be as above.
(i) tr(λx) = λtr(x) for λ ≥ 0.
(ii) tr(x + y) = tr(x) + tr(y).
W
(iii) If (aα ) is an increasing net of positive operators with
α aα = a then
W
tr( α aα ) = limα tr(aα ).
∗
∗
(iv) tr(x x) = tr(xx ) ∀x ∈ M .
∗
(v) tr(uxu ) = tr(x) for any unitary u ∈ M and any x ≥ 0 in M .
(vi) If p is a projection in M then p is nite i tr(p) < ∞.
(vii) If p and q are projections with p nite then p q i tr(p) ≤ tr(q).
(viii) pM p is a type II1 factor for any nite projection p.
Proof.
Let
The rst two assertions are immediate.
For (iii), note that the
diagonal entries of positive matrices are ordered as the matrices, and all
numbers are positive in the sums. (iv)Is obvious using matrix multiplication.
√ √ ∗
∗
(v) follows from (iv) via uxu = (u x)( xu ). For (vi), if tr(p) < ∞ but p is
innite, there is a proper subprojection of
p having the same trace as p.
The
dierence would be a projection of trace zero which is clearly impossible.
tr(p) = ∞ then if q is a projection of nite trace, q p and if q ≤ p
then tr(p − q) = ∞ so one may construct an innite sequence of mutually
orthogonal equivalent projections less than p. Using a bijection with a proper
subsequence, p dominates
If
an innite projection so is innite itself. (vii) follows easily as in the case
of a type II1 factor . For (viii) simply observe that
p q
for some
q
tr(p) < ∞
means that
whose matrix is zero except for nitely many
diagonal. And obviously
qM q
is a type II1 factor for such a
56
q. 1's
on the
Corollary 9.1.10
tr
Let
M
be a II∞ factor on a separable Hilbert space and
be the trace supplied by a decomposition II1
⊗ B(H).
Then
tr
denes an
isomorphism of the totally ordered set of equivalence classes of projections in
M
to the interval
[0, ∞].
Proof. Given the previous theorem, we only have to prove that any innite
projection is equivalent to the identity. But if p is innite choose u with
uu∗ = p and u∗ u strictly less than p. Then (u∗ )n u are a strictly decreasing
sequence of equivalent projections so we may write p as an orthogonal sum
P
p = p∞ + ∞
i=1 pi with all the pi equivalent for i ≥ 1. Now write the identity as
p1 (using the decomposition
II1 ⊗ B(H) if necessary). We see that 1 ≤ p. Unlike the II1 case, or for that matter the B(H) case, the trace cannot be
normalised (by tr(1) = 1 in the type II1 factor case or tr(minimalprojection) =
1
in the B(H) case). This allows for the possibility of an automorphism α
of M with tr(α(x)) = λtr(x) for x ≥ 0 and λ > 0, λ 6= 1.
a countable orthogonal sum of projections all
Exercise 9.1.11
Show that the trace on a II∞ factor is unique with proper-
ties (i) to (vi), up to a scalar.
Exercise 9.1.12
α : M → N is a *-homomorphism from a type II1 factor
α is an isomorphism, strongly continuous on the unit ball.
If
onto another, then
57
58
Chapter 10
The Coupling Constant
We want to compare actions of a given II1 factor on (separable) Hilbert
spaces.
We will show that they are parameterized by a single number in
[0, ∞].
Denition 10.0.13
If
M
is a type II1 factor , by
M -module
we will mean a
H together with an ultraweakly continuous unital *-homomorphism
H. Thus M acts on H and we will
action simply as xξ for x ∈ M and ξ in H.
Hilbert space
from
M
to a type II1 factor acting on
write that
In fact the ultraweak continuity condition is superuous.
The identity
M is dened into an M -module.
Given M on H and another Hilbert space K, x 7→ x ⊗ id makes H ⊗ K into
2
an M -module. The GNS representation makes L (M ) into an M -module.
(The notion of M −M bimodule is dened similarly as two commuting actions
2
of M on some Hilbert space, L (M ) being the rst example.) There is an
obvious notion of direct sum of M -modules. We will compare a given M 2
module H with L (M ) by forming the direct sum of it H and innitely many
2
copies of L (M ).
map makes the Hilbert space on which
10.1 Denition of dimM H
Theorem 10.1.1
Let
M
Then there is an isometry
(i.e.
u
is
H a separable M -module.
u : H → L (M ) ⊗ ` (N) such that ux = (x ⊗ 1)u
be a type II1 factor and
2
2
M -linear).
M -module K = L2 (M ) ⊗ `2 (N). Let p = id ⊕ 0 ∈ B(K)
2
2
be the projection onto H and q = 0⊕id be the projection onto L (M )⊗` (N).
0
Both p and q are in M (on K) which is a II∞ factor since q is clearly innite in
Proof. Form the
59
M 0 and if e is a rank one projection in B(`2 (N)) then (0⊕(1⊗e))M (0⊕(1⊗e))
2
is a type II1 factor , being the commutant of M on L (M ).
0
Since q is an innite projection in M , by 9.1.10 there is a partial isometry
0
∗
∗
in M with u u = p and uu ≤ q . Using the obvious matrix notation for
operators on K, let u be represented by
a b
.
c d
Then calculating
u∗ u = p and uu∗ ≤ q gives b∗ b+d∗ d = 0 and aa∗ +bb∗ = 0
so that
u=
for some isometry
Moreover the fact
0
w
0
0
w : H → L2 (M ) ⊗ `2 (N).
that u commutes with M̃
is equivalent to
wx = (x ⊗ 1)w
∀x ∈ M . Corollary 10.1.2
The commutant of a type II1 factor is either a type II1
factor or a type II∞ factor.
Proof. We leave the proof as an exercise.
Proposition 10.1.3
then
∗
uu ∈ M
0
on
u : H → L2 (M ) ⊗ `2 (N) is an M -linear
L (M ) ⊗ `2 (N) and tr(uu∗ ) is independent of u.
If
isometry
2
v were another M -linear isometry then uu∗ = uv ∗ vu∗
tr(uu ) = tr((vu∗ )(uv ∗ )) = tr(vv ∗ ). Proof. If
∗
so by 9.1.9
Observe that if M were replaced by C in the above construction the
∗
number tr(uu ) would be the dimension of H.
Denition 10.1.4
For a type II1 factor (or the n × n matrices) and an M ∗
module H, the number tr(u u) dened by the two previous results is called
dimM H,
or the coupling constant or the
(M ⊗ 1)0
[0, ∞].
Simply by reducing by projections in
whose
M -dimension
is any number in
M -dimension
Trivial examples
dimM L2 (M ) = 1.
2
2
(ii) dimM (L (M ) ⊗ ` (N)) = ∞
(i)
60
of
H.
one obtains Hilbert spaces
10.2 Elementary properties of dimM H
Theorem 10.2.1
With notation as above,
dimM (H) < ∞ i M 0 is a type II1 factor .
(ii) dimM H = dimM K i M on H and M on K
(i)
are unitarily equivalent (=
spatially isomorphic).
(iii) If
Hi
M -modules,
X
dimM (⊕i Hi ) =
dimM Hi .
are (countably many)
i
(iv)
(v)
dimM (L2 (M )q) = tr(q) for any projection q ∈ M .
−1
If p is a projection in M , dimpM p (pH) = trM (p)
dimM (H).
For the next two properties we suppose
with trace
M0
is nite, hence a type II1 factor
trM 0 .
p is a projection in M 0 , dimM p (pH) = trM 0 (p) dimM H.
(dimM H)(dimM 0 H) = 1.
(vi) If
(vii)
Proof.
M -linear isometry u we see that M on H is unitarily
uu∗ L2 (M ) ⊗ `2 (N). This makes (i) and (ii) obvious.
Using an
equivalent to
M
on
To see (iii), choose
M -linear
isometries
ui
from
Hi
to
L2 (M ) ⊗ `2 (N)
and
compose them with isometries so that their ranges are all orthogonal. Adding
P
∗
we get an M -linear isometry u with uu =
ui u∗i . Taking the trace we are
done.
For (iv), choose a unit vector
uu∗
is
JqJ ⊗ e
where
e
ξ ∈ `2 (N)
and dene
u(v) = v ⊗ ξ .
Then
is a rank one projection.
(v) Let us rst prove the relation in the case
H = L2 (M )q
where
q
is a
M with q ≤ p.
pxpΩ 7→ p(xΩ)p is a unitary from L2 (pM p) to pL2 (M )p which inter2
twines the left an right actions of pM p. Hence pM p on pL (M )q is unitarily
2
equivalent to pM p on L (pM p)q . So by (iv), dimpM p (pH) = trpM p (q) =
trM (p)−1 trM (q) = trM (p)−1 dimM H.
2
2
0
Now if H is arbitrary, it is of the form e(L (M ) ⊗ ` (N)) for e ∈ (M ⊗ 1) /
But e is the orthogonal sum of projections all equivalent to ones as in (iv)
with q ≤ p.
projection in
Then
H = e(L2 (M ) ⊗ `2 (N)) so M 0 = e(JM J ⊗ B(`2 (N))e
and p denes the isometry in the denition of dimM (pH). But p is a projection less than e in a II∞ factor so by uniqueness of the trace, dimM (pH) =
tr(M ⊗1)0 (p) = tr(M ⊗1)0 (p)/tr(M ⊗1)0 (e) dimM (H) = trM 0 (p) dimM (H).
(vi) We may suppose
61
L2 (M ), dimM (H) dimM 0 (H) = 1 so by (v) and
2
(vi) the result is true for M -modules of the form L (M )p. Also if one forms
k
K = ⊕i=1 H then dimM ⊗1 (K) = k dim H and dim(M ⊗1)0 K = k −1 dimM 0 by
2
2
k
(v). But any H can be obtained from L (M ) as ⊕i=1 L (M )p for suitable k
and p. (vii) Observe that, on
Example 10.2.2
Γ0 < Γ are icc groups, vN (Γ0 ) acts on `2 (Γ). And if γ ∈
Γ the unitary ρ(γ) of the right regular representation gives a vN (Γ0 )-linear
2
2
−1
unitary between ` (Γ0 ) and ` (Γ0 γ
). Hence by the coset decomposition,
2
dimvN (Γ0 ) (` (Γ)) = [Γ : Γ0 ].
If
N ⊆ M are II1
dimN L2 (M ).
This inspires the following denition: If
[M : N ]
of
N
in
M
is the real number
Exercise 10.2.3
Show that
Example 10.2.4
(Due to Atiyah and Schmidt.)
[M : N ] = 1
implies
factors, the index
N = M.
Discrete series representations of locally compact groups.
Reduction by a nite projection in the commutant of a type II1 factor
occurs in the representation theory of locally compact groups. If a discrete
series representation is restricted to an icc lattice it generates a type II1 factor
. The coupling constant is given by the ratio of the formal dimension and
the covolume of the lattice.
We illustrate in the case of
P SL(2, R)
which is the group of transfor-
az + b
H = {z ∈ C : Im(z) > 0}, z 7→
cz + d
a b
dened by invertible real 2 × 2 matrices
It is well known that there
c d
is a fundamental domain D for the action of the subgroup Γ = P SL(2, Z)
mations of the upper half plane
illustrated below:
DO FIGURE
P SL(2, Z) cover H and are disjoint
apart from boundaries which are of Lebesgue measure 0. Thus if µ is an
2
invariant measure equivalent to Lebesgue measure, L (H, dµ) gives a unitary
representation of Γ which is unitarily equivalent to the left regular repre2
2
sentation tensored with the identity on L (D, dµ), making L (H, dµ) into a
vN (Γ)-module whose vN (Γ) dimension is innite.
dxdy
The measure
is Γ-invariant but we want to vary this procedure
y2
dxdy
slightly. For each n ∈ N consider
. This measure is not invariant but
y 2−n
The set
D
and all its translates under
62
we can make the action of
P SL(2, R) unitary on L2 (H,
a b
c d
f (z) =
dxdy
) by the formula
y 2−n
1
az + b
)
f(
n
(cz + d)
cz + d
(with perhaps an inverse matrix...exercise as usual). This changes noth-
P SL(2, Z) so we obtain (unitarily
dxdy
2
equivalent) vN (Γ)-modules Hn = L (H,
) for each n.
y 2−n
The commutant of vN (Γ) on Hn is a II∞ factor. But as is well known,
2
holomorphic functions form a closed subspace of L functions which is manifestly invariant under P SL2 (R). The ensuing unitary representation is known
to be irreducible and in the discrete series of P SL2 (R). It can be shown to be
0
a nite projection in Γ . Thus we have a concrete example of a vN (Γ)-module
with nite vN (Γ)-dimension or coupling constant.
In general, if G is a locally compact group with Haar measure dg , the
ing as far as how the representation looks to
discrete series representations are precisely those irreducible unitary representations π that are direct summands of the left regular representation on
L2 (G, dg). So if Γ is a discrete subgroup with a fundamental domain D so
that
G
is covered by the
γ(D)
which are disjoint up to measure zero sets,
we may apply the same analysis as above to obtain a
vN (Γ)
module. The
obvious question is to calculate its coupling constant. This turns out to be
quite simple because of a key property of discrete series representations.
H is a Hilbert space aording a discrete
G, then the functions g 7→ hπg ξ, ηi, the so-called
coecients of π are in L (G, dg). We may then imitate the usual procedure
2
for nite or compact groups embedding H in L (G, dg). And the usual Schur
orthogonality of the coecients of a representation yields a number dπ such
that
Z
dπ hπg ξ, ηihη 0 , πg ξ 0 idg = hξ, ξ 0 ihη 0 , ηi.
See [ref robert] for the proof that if
series representation
π
of
2
G
G is compact and Haar measure is normalized so that G has measure 1, dπ
is the dimension of the vector space H. In general dπ depends on the choice
R
of Haar measure but obviously the product of dπ with the covolume
dg
D
2
does not. The coecients give an explicit embedding of H in L (G, dg) and
If
a straightforward calculation of the trace of the projection onto the image of
H in vN (Γ)0 yields immediately the formula
dimvN (Γ) (H) = dπ covolume(Γ).
The detailed calculation from this point of view can be found in [1] pp. 142148.
63
Example 10.2.5
Consider the subfactor
N ⊆ N ⊗ Mn (C) (diagonal emN -linear isometry of L2 (N ) onto
Each matrix unit eij gives an
L2 (N ⊗ Mn (C)) which is an orthogonal direct sum of these
2
equivalent subspaces. Hence [N ⊗ Mn (C) : N ] = n .
bedding).
a subspace of
Proposition 10.2.6
If
M
is a type II1 factor on
(a)
M
has a separating vector if
(b)
M
has a cyclic vector if
H
then
dimM (H) ≥ 1.
dimM (H) ≤ 1.
Proof. Both assertions follow immediately by comparing
H
to
L2 (M )p
or a
direct sum of copies of it.
In fact both conditions in the last proposition are i. For that one needs
L2 (M ). In fact the original denition of the
to control arbitrary vectors in
coupling constant by Murray and von Neumann was as follows.
on
H
be a type II1 factor whose commutant is a type II1 factor .
Let
M
Choose
any nonzero vector ξ ∈ H and let p and q be projections onto the closures
0
0
of M ξ and M ξ respectively. Then p ∈ M and q ∈ M and using the
normalised traces the coupling constant was dened as the ratio
the hard part being to show that this ratio is independent of
ξ.
trM (q)
,
0
trM
(p)
Assuming
this last statement it is trivial to identify the Murray-von Neumann coupling
constant with our
dimM (H) but at this
stage we have nothing to oer in the
way of a simplied proof of why this
number does not depend on
Example 10.2.7
ξ.
(due to M. Rieel) If
(X, µ)
is a measure space and
Γ
is
a countable group acting by measure preserving transformations on (X, µ)
2
so that Γ acts by unitaries uγ on L (X, µ) in the obvious way. We say that
D ⊆ X is a fundamental domain for Γ if X = ∪γ γ(D)
and µ(Dγ(D)) = 0 for all γ ∈ Γ, γ 6= id. (One may clearly suppose the
γ(D) are disjoint by removing a set of measure zero.) In this situation the
∞
Γ
∞
abelian von Neumann algebra L (X) of Γ-invariant L
functions may be
∞
identied with the space L (D).
Now suppose Γ and Λ are two groups acting on X as above with fundamental domains D and E respectively. We may consider the von Neumann
2
∞
Λ 00
algebra MΓ,Λ on L (X, µ) dened as {{uγ : γ ∈ Γ} ∪ L (X) } .
a measurable subset
64
Chapter 11
The Crossed Product
construction.
Perhaps the most useful way of producing von Neumann algebras from others
is the crossed product.
We begin by dening a very general notion about
which there is not a lot to say, but then examine it carefully in special cases.
11.1 Group actions.
Let
M
be a von Neumann algebra and
is a homomorphism
g 7→ αg
from
G
G
a group. An action of
to the automorphism group
G on M
AutM of
M
(where automorphisms may be assumed ultraweakly continuous if necG
essary). The algebra of xed points for the action is denoted M
and is a
von Neumann algebra . A special case of some importance is when the is a
∗
unitary group representation g 7→ ug with ug M ug = M ∀g ∈ G. In that case
∗
0
setting αg (x) = ug xug denes an action of G on M (and M ). We say that
α is implemented by the unitary representation ug . If the ug are
actually in M , we say that the action is inner as an inner automorphism of
M is by denition one of the form Ad u(x) = uxu∗ for u a unitary in M . An
the action
automorphism is called outer if it is not inner.
Actions are not always implementable though the notion depends on the
Hilbert space on which
Exercise 11.1.1
M
acts.
T is a bijection of X
−1
which preserves the measure class of µ (i.e. µ(A) = 0 ⇔ µ(T
(A)) = 0.)
∞
show how T denes an automorphism αT of L (X, µ). Show further that
2
this automorphism is implemented by a unitary u on L (X, µ).
A bijection
subset
A⊆X
T
If
(X, µ)
is a measure space and
T (A) = A
µ(X \ A) = 0.
as above is called ergodic if
implies either
µ(A) = 0
or
65
for a measurable
Proposition 11.1.2
points for
αT
With notation as above
T
is ergodic i the only xed
are constant functions.
f ∈ L∞ and αT (f ) = f . After throwing away a null set we
may assume that f (x) = f (T (x)) for all x ∈ X . Then for every > 0, by
the denition of the essential supremum, µ({x : ||f || − |f (x)| < } =
6 0. But
this set is invariant under T so it is equal to X up to a set of measure 0.
Letting tend to 0 we see that µ({x : |f (x)| =
6 ||f ||}) = 0. So we may assume
ig(x)
f (x) = e
for some measurable g taking values in [0, 2π). Repeating the
argument for g gives f constant almost everywhere.
(⇐) If A is a measurable invariant set then its characteristic function is
∞
xed by α in L
i A is invariant. 0 1
0 −i
1 0
Exercise 11.1.3 Let σx = 1 0 , σy = i 0 and σz = 0 −1
be the Pauli spin matrices. Show that Ad ux , Ad uy and Ad uz dene an
action of the group Z/2Z ⊕ Z/2Z on the two by two matrices which is not
2
implementable for M2 (C) on C .
Proof. (⇒) Let
Exercise 11.1.4
Show that any group action is implementable for a type
II1 factor in standard form and more generally any automorphism group pre-
serving a faithful normal state is implementable in the GNS representation.
Exercise 11.1.5
Show that every automorphism of
Exercise 11.1.6
Show that the automorphism of
B(H)
is inner.
vN (F2 )
coming from the
group automorphism which exchanges the 2 generators is outer.
If
G
is a topological group there are many possible notions of continuity.
The most useful is that of pointwise *-strong convergence, i.e. we assume that
the map
g 7→ α(g)(x)
is *-strong continuous for any
x ∈ M.
Typically many
other notions of continuity will be equivalent to that and even a measurability
assumption can be enough to ensure this continuity.
We will always assume pointwise *-strong continuity when referring to an
action of a topological group.
Exercise 11.1.7
Is the action by translation of
R on L∞ (R) pointwise norm
continuous? pointwise strongly continuous? pointwise *-strong continuous?
Actions of a given group on von Neumann algebras are easy to construct
but actions of a group on a given von Neumann algebra may be hard to come
by.
66
Denition 11.1.8
Exercise 11.1.9
sulting action of
An action of
G
on
M
is said to be ergodic if
M G = Cid.
Show that if G acts preserving µ on (X, µ) then the re∞
on L (X, µ) is ergodic i the only measurable subsets
G
A ⊆ X which satisfy µ(g(A)∆A) = 0 ∀g ∈ G
µ(X \ A) = 0.
(Here A∆B means A \ B ∪ B \ A.)
satisfy either
µ(A) = 0
or
The following question is an intriguing open problem:
Does
SU (3)
have any ergodic action on a type II1 factor ?
It is shown in [] that
SU (2)
has no such action and it is shown in [] that
if a compact group acts ergodically on a von Neumann algebra then that von
Neumann algebra has a faithful normal trace.
11.2 The crossed product
α is an action of the locally compact group G with Haar measure dg on the
von Neumann algebra M on the Hilbert space H. Form the Hilbert space
K = L2 (G, H) = L2 (X) ⊗ H and let G act on K by ug = λg ⊗ 1, λ being the
left regular representation. Further, let M act on K by
If
(x̃f )(g) = αg−1 (f (g))
.
Exercise 11.2.1
of
M
Show that
onto a von Neumann
Exercise 11.2.2
x 7→ x̃ is an ultraweakly continuous *-isomorphism
subalgebra of B(K).
Show that
^
ug x̃u∗g = α
g (x).
Note that this gives another way of making a group action implementable,
at least when it is locally compact.
Denition 11.2.3
M oα G is the
{x̃ : x ∈ M }.
If
M , H, G
and
α
von Neumann algebra on
K
generated by
{ug : g ∈ G}
P
algebraic crossed product.
67
and identify
M
and
M̃ . Note
that nite linear combinations
of M oα G.
g xg ug form a dense *-subalgebra P
Moreover the ug are linearly independent over M in the sense that
g x g ug =
0 ⇒ xg = 0 for each g in the sum. This dense subalgebra could be called the
>From now on we will drop the
˜
are as above, the crossed product
with
M oα G when G is compact or abelian,
but we shall be mostly interested in the case where G is discrete as then we
may replay the matrix element game that we played for vN (Γ) to gain control
There is a well-developed theory of
of weak limits of elements in the algebraic crossed product. (In fact of course
vN (Γ) is the special case of the crossed product when M = C and the action
is trivial.) Indeed we see
G is discrete,
any element of M oα G
P
g 7→ xg so that the sum
x
u
g g g stands for a certain
2
matrix of operators on K = H ⊗ ` (G). Moreover any matrix of this form
which denes a bounded operator on K is in
M oα G. This is because the sum converges pointwise at least on the
dense set of functions of nite support from G to H. In the case where the
immediately as in 3.1.10 that if
denes a function
crossed product is a II1 factor we know that the commutant consists of right
0
multiplication by elements of M oα G so a wealy dense subalgera of (M oα G)
P
g xg ug and
commute. We will return to the general
preserves this dense subspace of vectors and on that subspace
right multiplication by
ug
and
x∈M
case later on.
Moreover the formulae
X
X
(
xg ug )∗ =
αg (xg−1 )ug
and
X
X
X X
(
xg ug )(
y g ug ) =
{
xh αh (yh−1 g)}ug
g
h
are justied by matrix multiplication.
We shall now provide some sucient conditions for
G
is discrete.
Denition 11.2.4
An action
always assuming
G
for which
αg
α
of
G
on
M
M oα G to be a factor
g
in
is an outer action of
G
is called outer if the only
is inner is the identity.
Proposition 11.2.5
G is a discrete group
M oα G is a factor.
If
and
α
M then
P
Proof. If x =
xg ug ∈ Z(M ) then equating coecients in the expression
that x commutes with M gives us yxg = xg αg (y) ∀y ∈ M ,g ∈ G. By the
next lemma this implies xg = 0 for any g 6= 1. Thus x ∈ M and since M is
a factor we are done. on the factor
Lemma 11.2.6
x 6= 0,
Let
α ∈ Aut M
for a factor
yx = xα(y)
Then
M.
Suppose there is an
with
α
is inner.
68
∀ y ∈ M.
x ∈ M,
Proof. If
x
were unitary this would be obvious. So take the adjoint of the
∗
∗
∗
∗
∗
relation to obtain x y = α(y)x ∀y ∈ M . Thus yxx = xα(y)x = xx y
∗
∗
∗
∗
and xx ∈ Z(M ). Similarly x x ∈ Z(M ). But xx and x x always have
∗
∗
the same spectrum so since M is a factor both xx and x x are equal to the
same positive number
are done.
λ.
√
Dividing by
λ
converts
x
into a unitary and we
These two results prompt the following denition.
Denition 11.2.7
α
An automorphism
of a von Neumann algebra
M
is
called free if
∀ y ∈ M ⇒ x = 0.
yx = xα(y)
An action
α
is called free if
αg
is free for every
g 6= id.
α is a free action
Z(M oα G) ⊆ M , in fact that M 0 ∩
The argument of proposition 11.2.5 shows in fact that if
on a von Neumann algebra
M
then
M oα G ⊆ M .
Theorem 11.2.8
gebra
M,
then
α is a free ergodic
M oα G is a factor.
If
action of
G
on a von Neumann al-
Proof. This follows immediately from the preceding remark.
To understand the meaning of freeness for automorphisms of the form
αT we need to make a hypothesis on (X, µ) as otherwise one could envisage
a T which is non-trivial on X but for which αT is the identity. So we will
suppose from now on that (X, µ) is countably separated. This means there
is a sequence Bn of measurable sets with µ(Bn ) > 0 for which, if x 6= y , there
is an n with x ∈ Bn but y ∈
/ Bn . Obviously Rn is countably separated.
Exercise 11.2.9
Show that
αT = id
means that
Tx = x
almost everywhere.
Hint-look at the proof of the next result.
Proposition 11.2.10
If
T
is a transformation of
(X, µ)
then
αT
is free i
µ({x : T (x) = x}) = 0.
Proof. (⇒)If A is any measurable set on which
∞
for all f ∈ L .
T = id
then
χA f = αT (f )χA
(⇐) First throw away any xed points of T . Then suppose f1 αT (f2 ) =
f2 f1 ∀ f2 ∈ L∞ . Let A be the support of f1 . Then since T has no xed
−1
points, A = ∪n (A ∩ Bn \ T
(Bn )).
∞
If f1 were non-zero in L , we could thus choose an n for which µ(A ∩ Bn \
−1
T (Bn )) > 0. Set f2 = χBn . Then for any x ∈ A ∩ Bn \ T −1 (Bn ) we have
69
f1 (x)f2 (x) 6= 0 but f1 (x)f2 (T x) = f1 (x)χBn (T x) = 0 since x ∈
/ T −1 (Bn ).
∞
Thus f1 αT (f2 ) 6= f2 f1 in L . So the measure of A must be zero. We conclude that if Γ is a countable group acting freely and ergodically
on a measure space (X, µ), preserving the class of µ, then the crossed product
L∞ (X, µ)oΓ is a factor.
Note that if Γ is abelian, ergodic implies free.
Exercise 11.2.11
Show that freeness of the action actually proves that
L∞ (X, µ)
is maximal abelian in the crossed product.
The crossed product
M oΓ
when
M
is abelian and
Γ
is discrete is called
the group measure space construction. Here are several examples.
Example 11.2.12 X = Z, Γ = Z
acting by translation,
µ =
counting
measure.
The action is free and ergodic and
Example 11.2.13
L∞ (X, µ)oΓ = B(`2 (Z)).
The irrational rotation algebra-von Neumann algebra ver-
sion.
(X, µ) = (T1 , dθ), Γ = Z generated by the transformation T
e z and α/2π is irrational.
where
T (z) =
iα
Exercise 11.2.14
Use Fourier series to show that this
Example 11.2.15
Let
H
T
(hn )
Γ =
L
n∈N H be
with hn eventually the identity. Put
be a nite abelian group and
the countable group of sequences
is ergodic.
Q
X = G =
G is a
Then
n∈N H (the set of all sequences) with the product topology.
compact group so has a Haar measure µ. Γ acts on X by left
translation. The action is clearly free and ergodic as we shall now argue.
There is a particularly von Neumann algebraic way to view this example
without even constructing the space (X, µ) !
∞
Let A = L (H) = CĤ be the group algebra of the dual group
Ĥ , with its
⊗alg,n∈N A
usual trace. As in section 7.2, form the algebraic tensor product
with product trace
tr.
Then perform the GNS construction with respect to
tr to obtain an abelian von Neumann algebra. It may be identied with
L∞ (G, µ) so the Hilbert space H of the GNS construction is L2 (X, µ). But it
is clear that an orthornormal basis of H is given by nite sequences (χn ) of
elements of Ĥ which dene elements χ1 ⊗ χ2 ⊗ · · · ⊗ 1 ⊗ 1 ⊗ 1 · · · in ⊗alg,n∈N A.
The point is that these basis vectors are eigenvectors for the action of Γ on
L2 (X, µ):
Y
(hn )(χ1 ⊗ χ2 ⊗ · · · ⊗ 1 · · ·) = ( χn (hn )) χ1 ⊗ χ2 ⊗ · · · ⊗ 1 · · · .
n
70
Ergodicity follows easily since the only basis element which is xed by all the
(hn )
is the one with all
Exercise 11.2.16
χn
equal to
1.
H = Z/2Z in this example then the subalgebra of the crossed product generated by ⊗alg,n∈N A and Γ is the algebraic
innite tensor product of M2 (C).
Show that if
Both of the last two examples are special cases of a more general one:
X
is a compact group with its Haar measure and
Γ
is a countable dense
subgroup acting (freely) by left translation. The Peter Weyl theorem shows
that this action is ergodic.
Example 11.2.17
Bernoulli Shift.
Γ is any innite group and A = CZ/2Z we may form the tensor product
indexed by Γ of a copy of A for each γ ∈ Γ. The von Neumann algebra thus
∞
obtained is once again the L
space of the innite product measure space,
this time with the set indexing the product being Γ. As in the previous
2
example we can obtain a basis of L indexed by funcions from Γ to the set
{0, 1} which are almost always 0. These functions are the same as nite
subsets of Γ and the action of Γ on the Hilbert space is by permuting the
If
basis in the obvious way. Ergodicity follows from the fact that the orbit of
any non-empty subset is innite.
One could also chose another trace than the usual one and modify the
orthonormal basis of
A accordingly.
The measures are the obvious ones unless
specied.
We give a few more examples of free ergodic actions without supplying
proofs of ergodicity.
Example 11.2.18 SL(2, Z) acts on T2 = R2 /Z2 via the linear action on R2 .
Example 11.2.19 P SL(2, Z) acts on R ∪ {∞} by linear fractional transformations.
Example 11.2.20 SL(2, Z) acts on R2 by linear transformations.
Example 11.2.21 Q acts on R by translation.
There are two fairly easy ways to see that this action is ergodic. The rst
is to reduce it to a dense subgroup of a compact group by observing that an
L∞ function on R which is invariant under translation by Z denes an L∞
function on the quotient
T.Then
use Fourier series.
The second way is a direct attack which should generalise to show that bullshit
71
translation by any countable dense subgroup of a locally compact group is
∞
ergodic. If f ∈ L (R) is invariant under Q, set things up so that there are
sets
and
A
B
and
B
both of nonzero measure, so that
g(A) ∩ g(B) = ∅.
with intervals of the same width with rational endpoints.
these must intersect
A
The ax+b" group
Example 11.2.23
Same as example 11.2.13 with
CH
QoQ∗
acts on
1−p
for
R
H = Z/2Z
but using a
which is dierent from the usual one. Such a trace
is specied by its values on the minimal projections of
and
Some of
B in non-nul sets. But all these intervals are all
g cannot be invariant up to sets of measure zero.
Example 11.2.22
p
A
and
translates of each other so
normalised trace on
Cover
0 < p < 1.
CH
which we could call
The product measure is not absolutely continous
with respect to Haar measure, and it is not preserved by group translation
so this example is perhaps most easily approached by the von Neumann
L
algebra construction where one can implement the action of
n∈N Z/2Z by
2
unitaries. These unitaries come from ones on L (H) which exchange two
points of unequal weight so they must be correctly scaled.
Exercise 11.2.24
Work out the details of example 11.2.23
In the examples we see four dierent kinds of free ergodic actions:
Type I :
Γ
acts transitively.11.2.12
Γ preserves a nite measure. 11.2.13,11.2.15,11.2.17,11.2.18
II∞ : Γ preserves an innite measure.11.2.20,11.2.21
III : Γ preserves no measure equivalent to µ.11.2.19,11.2.22,11.2.23
Type II1 :
Type
Type
11.3 The type of the crossed product.
We adopt the notations and conventions of the previous section. The map
Em : M oα Γ → M
aid
P
γ∈Γ is destined to play
a big role in the theory. It is called the conditional expectation onto M and
which assigns
to the element
obviously satises the following contitions:
2
EM
= EM .
∗
∗
∗
(ii) EM (x) = EM (x ), EM (1) = 1, EM (x x) = 0if f x = 0.
∗
∗
(iii) EM (x x) ≥ EM (x )EM (x), ||E(x)|| ≤ ||x||.
(iv) EM (axb) = aEM (x)b for a, b ∈ M .
(v) EM is ultraweakly continuous.
(i)
So
EM
is a projection of norm one in the Banach space sense.
condition (iv) says that
EM
is an
M − M -bimodule
72
map.
The
Theorem 11.3.1
Γ acts non-transitively,freely and ergodically, preserving
∞
the nite measure µ then L (X, µ)oΓ is a II1 factor. If Γ preserves the
∞
innite σ -nite measure µ then L (X, µ)oΓ is a II∞ factor unless Γ acts
∞
transitively in which case L (X, µ)oΓ is type I.
If
Proof.
(i) It is clearer to prove a more general statement (in the case where
preserves
µ
and
µ(X) = 1).
ultraweakly continuous trace
action is free and ergodic.
So suppose
tr
Γ
preserves the faithful positive
on the von Neumann algebra
Then we claim
Γ
M = AoΓ
A
and that its
is a type II1 factor
(or a nite dimensional factor). By previous results we need only show that
it has an ultraweakly continous positive trace. So dene
M.
T r = tr ◦ EA
on
Ultraweak continuity and positivity are obvious so by continuity and
T r(auγ buη ) = T r(buη auγ ). For either side of
−1
the equation to be non-zero means η = γ
amd then the left hand side is
−1
−1
tr(aαγ (b)) = tr(αγ (aαγ (b))) = tr(bα (a) ) which is equal to T r(buη auγ ).
(ii) If µ is innite and Γ does not act transitively then there are no atoms
hence there are subsets Y of X of arbitrary positive measure. Let Y have
nite non-zero measure and let ξ be the function ξ(γ) = δγ,id χY . Then
R
hauγ ξ, ξi = ωξ (auγ ) = δid,γ Y a(x)dµ(x).
One easily checks that ωξ ((pauγ p)(pbuη p)) = ωξ ((pbuη p)(pauγ p)) so by
4.0.25 ωξ denes a positive ultraweakly continuous trace on p(AoΓ)p which
is a type II1 factor . But AoΓ is not itself a type II1 factor since A contains
linearity it suces to prove
an innite family of equivalent mutually orthogonal projections. By 9.1.8 we
are done.
(iii) If
Γ
acts transitively then
(X, µ) = (Γ, counting
measure) and the
characteristic function of a set with one element is a minimal projection in
L∞ (X, µ)oΓ .
Exercise 11.3.2
sure
µ
If
Γ
acts ergodically on
(X, µ)
preserving the
σ -nite meaµ.
then any other invariant equivalent measure is proportional to
We now want to show that there are factors that are neither of type I nor
∞
type II . Suppose that M = L (X, µ)oΓ is a type I or II factor. Then it has
tr : M+ → [0, ∞].
X,
µ,by reversing the procedure of theorem
∞
11.3.1 and dening the measure σ(A) to be tr(ξA ) (ξA ∈ L (X, µ) ⊆ M ).
Invariance of the measure σ is no problem. The snag is that tr(χA ) could be
innite for every non-nul set A. We will show that this is not the case. To
a trace
We would like to dene an invariant measure on
absolutely continous with respect to
this end the concept of lower semicontinuity will be useful.
73
Denition 11.3.3
X
If
is a topological space we say that
lower semicontinous if for every
such that
f (u) > f (x) − Exercise 11.3.4
(a)f
−1
there is an
for all
Prove that if
((−∞, K]))
x ∈ X and > 0
u ∈ U.
f
is lower semicontinous then
is closed for every
K ∈ R.
(b)f attains its minimum on any compact subset of
Exercise 11.3.5
from
B(H)
to
R
If
H
f : X → R is
open set U ⊆ X
is a Hilbert space and
ξ ∈ H,
X.
the function
a 7→ ||aξ||
is weakly lower semicontinuous.
Exercise 11.3.6
If
fα
are lower semicontinous then
∨α fα
is lower semicon-
tinous if it exists.
Lemma 11.3.7
Let
M
be a type I or II factor and
T race in type I,as in 9.1.9 in type II∞
∗
each K ≥ 0, M1,K = {x : tr(x x) ≤ K}
tr : M + → [0, ∞]
be
and the trace in type II1 . Then for
is weakly compact.
Proof. Clear in the II1 case. In a decomposition
M ∼
= N ⊗ B(`2 (N))
on
H
N a type II1 factor or C we may assume by 10.2.6 that there is a vector
ξ ∈ e11 H with ωξ a trace on e11 M e11 . So if ξi = ei1 ξ we have, up to a scalar,
with
that
tr(x) =
∞
X
hxξi , ξi i.
i=1
By the previous exercises and weak compactness of the unit ball, we are done.
Proposition 11.3.8
let W (x) be the
P
P x ∈ M1,K
∗
weak closure of the convex set of nite sums {
i λi = 1, λi >
i λi ui xui :
∞
0, ui unitary in L (X, µ)}. Then W (x) ⊆ M1,K and if φ(y) = tr(y ∗ y) for
y ∈ W (x) then φ attains its minimum at a unique point E(x) of W (x).
With notation as above, for
∗
Proof. Note rst that {z ∈ M : tr(z z) < ∞} is a vector space on which
||z|| = tr(z ∗ z) denes a prehilbert space structure. (Since (a + b)∗ (a + b) ≤
2(a∗ a + b∗ b) as operators, and the parallelogram identity passes to the potentially innite sum dening
of
M
tr.)
Moreover
so by lower semicontinuity
φ
W (x)
is a weakly compact subset
attains its minimum at a point which is
unique by two dimensional Euclidean geometry as in 2.1.2.
Proposition 11.3.9
projection in
M
with
M = L∞ (X, µ)oΓ is a type I or II factor
L∞ (X, µ). Let tr be as above and p be a
Suppose that
Γ on
tr(p) < ∞. Then
for a free ergodic action of
E(p) = EL∞ (X,µ) (p)
and
0 < tr(E(p)2 ) ≤ tr(p).
74
E = EL∞ (X,µ) . By the uniqueness of E(p) it commutes with every
∞
∞
unitary in L
so it is in L
by 11.2.11. On the other hand E(y) = E(p)
for all y ∈ W (p) by the bimodule linearity of the conditional expectation
and its ultraweak continuity. So E(E(p)) = E(p) = E(p). But φ(E(p) ≤
φ(p) = tr(p)∞. Finally E(p) = E(p2 ) which is a positive non-zero selfadjoint operator and hence has non-zero trace. Proof. Let
Theorem 11.3.10
rated
Let
Γ
act freely and ergodically on the countably sepa-
σ -nite measure space (X, µ) so that there is no σ -nite Γ-invariant
X absolutely continuous with respect to µ. Then L∞ (X, µ)oΓ is
measure on
a factor not of type I or II.
Proof. If the crossed product were of type I or II, dene the measure
X
ρ
on
By the previous result ρ(A) would have to be nite
∞
2
and non-zero for some A since the L
functionf = E(p) must dominate a
by
ρ(A) = tr(χA ).
χA for some A (e.g. let A be those x with f (x) suciently close
||f ||). But then by ergodicity X = ∪γ∈Γ γ(A) (up to null sets) so that ρ
is σ -nite. It is automatically absolutely continuous wrt µ. Invariance of ρ
−1
under Γ follows from tr(uγ xuγ ) = tr(x) for x ≥ 0. multiple of
to
Denition 11.3.11
A factor not of type I or II is called a type III factor.
Example 11.2.22 provides a type III factor since the subgroup
ergodically so the only possible invariant measure is a multiple
Q acts
of dx by
exercise 11.3.2 and this is not invariant under multiplication!
Note that the above technique works in somewhat greater generality than
actions of groups on measure spaces.
Exercise 11.3.12
M oα Z
Adapt the proofs of the results just obtained to show that
is a type III factor if the action
α
is generated by a single automor-
phism of the II∞ factor scaling the trace by a factor
75
λ 6= 1.
76
Chapter 12
Unbounded Operators and
Spectral Theory
There are many naturally arising examples of unbounded operators, some
of the most fundamental being multiplication by
x
and dierentiation, the
position and momentum operators of quantum mechanics. Our immediate
motivation for studying unbounded operators here is to facilitate the study
of arbitrary von Neumann algebras acting on GNS Hilbert spaces. Here we
establish the necessary preliminaries on unbounded operators. The material
closely follows Reed and Simon [2].
12.1 Unbounded Operators
Denition 12.1.1
subspace
D(T ),
An operator
the domain of
T,
T
on a Hilbert space
H
and a linear map from
consists of a linear
D(T )
to
H.
Example 12.1.2
(i)
Mx ,
multiplication by
x
on
D(Mx ) =
L2 (R).
Z
2
f ∈ L (R) :
2
2
x |f (x)| d x < ∞ .
R
(ii)
T =
d
on
dx
L2 ([0, 1]). D(T ) = C 1 [0, 1].
In order to do some analysis we want to restrict our attention a little so
as not to consider completely arbitrary linear maps.
77
Denition 12.1.3
Let
T
be an operator on
H.
The graph of
T
is
Γ(T ) = {(ξ, T ξ) : ξ ∈ D(T )} ⊂ H ⊕ H.
T
is closed if
Γ(T )
Remark 12.1.4
is closed in
Note that if
H ⊕ H.
T
D(T ) = H
is closed and
then
T
is bounded
by the Closed Graph Theorem.
Lemma 12.1.5
(0, η) ∈ Γ
A linear subspace
Γ ⊂ H⊕H
is the graph of an operator i
η = 0.
implies
Proof. Straightforward.
Many operators are not closed, but can be extended to a closed operator.
Denition 12.1.6
S ⊂ T,
denoted
Let S, T be
Γ(S) ⊂ Γ(T ).
if
Denition 12.1.7
An operator
operators on
Equivalently
T
H. T is an
D(S) ⊂ D(T )
S,
T |D(S) = S .
extension of
and
is preclosed (or closable) if it has a closed
extension.
Lemma 12.1.8
tension
T
Suppose
T
is preclosed. Then
has a smallest closed ex-
T . Γ(T ) = Γ(T ).
Denition 12.1.9 T
Proof of Lemma.
contains
Γ(T )
so
is called the closure of
T.
Take a closed extension
Γ(T ) ⊂ Γ(A). Γ(T )
A
of
T . Γ(A)
is closed and
is the graph of an operator (call it
T)
because:
(0, η) ∈ Γ(T ) ⊂ Γ(A) ⇒ η = A(0) = 0.
T is the smallest closed extension because for all closed extensions A, Γ(T ) =
Γ(T ) ⊂ Γ(A).
Remark 12.1.10
We thus obtain two equivalent denitions of a preclosed
operator:
(i)
(0, η) ∈ Γ(T ) ⇒ η = 0.
(ii) (ξn
∈ D(T ), ξn → 0
and
T ξn
converges)
⇒ T ξn → 0.
Exercise 12.1.11
(i) Dene
S
on
D(S) = C0∞ (R) (innitely dierentiable
0
support), Sf = f . Show that S is preclosed.
L2 (R)
with compact
by
78
functions
(ii) Dene
T
Show that
from
T
to
C
D(T ) = L1 (R) ∩ L2 (R), T (f ) =
by
is not preclosed.
Denition 12.1.12
subspace
L2 (R)
D0 ⊂ D(T )
Suppose
T is a closed
T |D0 = T .
operator. A core for
T
R
R
f.
is a linear
such that
We can perform basic arithmetic operations with (unbounded) operators
as follows: S + T is the operator with domain D(S + T ) = D(S) ∩ D(T ) and
(S + T )ξ = Sξ + T ξ . ST is the operator with domain D(ST ) = {ξ ∈ D(T ) :
T ξ ∈ D(S)} and (ST )ξ = S(T ξ). Of particular importance is the adjoint.
Denition 12.1.13
Let
T
be a densely dened operator on
H.
Let
D(T ∗ ) = {η ∈ H : ∃σ ∈ H such that < T ξ, η >=< ξ, σ > ∀ξ ∈ D(T )}
= {η ∈ H : ∃C > 0 such that | < T ξ, η > | ≤ C||ξ|| ∀ξ ∈ D(T )}.
For
and
ξ ∈ D(T ∗ ) note
∗
dene T ξ = η .
that the element
σ
is unique (by the density of
D(T ))
Remark 12.1.14
Note that if
Exercise 12.1.15
Give an example to show that the domain of the adjoint
S⊂T
need not be dense. [In fact it can be
Proposition 12.1.16
1.
T∗
2.
D(T ∗ )
3. If
Let
T
then
T ∗ ⊂ S ∗.
{0}].
be a densely dened operator. Then
is closed.
T
is dense i
T
is preclosed then
is preclosed. In that case
T = T ∗∗ .
(T )∗ = T ∗ .
(η, σ) ∈ Γ(T ∗ )
< (−T ξ, ξ), (η, σ) >= 0. Hence
Proof. Note that
i
< T ξ, η >=< ξ, σ >
for all
ξ ∈ D(T )
i
Γ(T ∗ ) = {(−T ξ, ξ) : ξ ∈ D(T )}⊥ = (uΓ(T ))⊥ = uΓ(T )⊥ ,
where
u:H⊕H→H⊕H
is the unitary operator
1. Orthogonal complements are closed, hence
79
u(ξ, η) = (−η, ξ).
Γ(T ∗ )
is closed.
Now:
2.
Γ(T ) = (Γ(T )⊥ )⊥ = u∗ Γ(T ∗ )⊥ ,
so
(0, ξ) ∈ Γ(T ) ⇔ (−ξ, 0) ∈ Γ(T ∗ )⊥
⇔ 0 =< (−ξ, 0), (η, T ∗ η) >= − < ξ, η > for
⇔ ξ ∈ D(T ∗ )⊥ .
Hence
T
is preclosed i
D(T ∗ )⊥ = {0}
∗∗
∗ ⊥
Γ(T ) = uΓ(T )
In that case
i
D(T ∗ )
⊥⊥
2
= u Γ(T )
all
η ∈ D(T ∗ )
is dense.
= −Γ(T ) = Γ(T ),
so
T ∗∗ = T .
3.
T ∗ = T ∗ = T ∗∗∗ = (T )∗ .
Denition 12.1.17
T is symmetric if T ⊂ T ∗ . Equivalently
< T ξ, η >=< ξ, T η > for all ξ, η ∈ D(T ). T is self-adjoint if T = T ∗ . A
self-adjoint operator T is positive if < T ξ, ξ >≥ 0 for all ξ ∈ D(T ).
An operator
12.2 Spectral Theory for Unbounded Operators
Denition 12.2.1
Let
T
be a closed operator on
ρ(T ) = {λ|λ1 − T : D(T ) → H
The spectrum of
T
Remark 12.2.2
Note that if
T )−1
is
H.
T
The resolvent of
is
is a bijection}.
σ(T ) = C\ρ(T ).
λ1 − T : D(T ) → H
is a bijection then
(λ1 −
is bounded by the Closed Graph Theorem.
Exercise 12.2.3
The spectrum is highly dependent on the domain.
AC[0, 1] denote the
d
d
, T2 = dx , with
dx
set of absolutely continuous functions on
[0.1].
Let
Let
T1 =
D(T1 ) = {f ∈ AC[0, 1] : f 0 ∈ L2 ([0, 1])}
D(T2 ) = {f ∈ AC[0, 1] : f 0 ∈ L2 ([0, 1]), f (0) = 0}.
Show that
T1
and
T2
Proposition 12.2.4
are closed. Show that
while
σ(T2 ) = ∅.
F a measure2
able, real-valued, a.e. nite function on X . Let D(Mf ) = {g ∈ L (X, µ) :
2
f g ∈ L (X, µ)} and let Mf g = f g . Then Mf is self-adjoint and σ(Mf ) =
ess.range(f ) = {λ ∈ C : µ({x : |λ − f (x)| < }) > 0 ∀ > 0}.
Let
(X, µ)
σ(T1 ) = C
be a nite measure space and
80
Exercise 12.2.5
Prove Prop 12.2.4.
Theorem 12.2.6 (Spectral Theorem - Multiplier Form)
Let
A be a self-
H with dense domain. Then there exists a nite measure
(X, µ), a real-valued a.e. nite function f on X and a unitary operator
u : H → L2 (X, µ) such that uAu∗ = Mf
adfoint operator on
space
Proof. See [2]
Remark 12.2.7 (Borel Functional Calculus)
Note that the Spectral The-
orem allows us to dene a Borel functional calculus for self adjoint operators.
∗
Given a Borel function h on the spectrum of A, dene h(A) = u Mh◦f u.
12.3 Polar Decomposition
Theorem 12.3.1
Let
A : H → K
be a closed, densely dened operator.
Then:
(i)
A∗ A and AA∗ are
(AA∗ )1/2 exist).
(ii) There exists a partial isometry with initial space
nal space
(A∗ A)1/2
and
Range(A∗ A)1/2
and
positive self-adjoint operators (hence
Range(A)
and
A = v(A∗ A)1/2 .
A = uB for some positive B and partial
Range(B) then u = v and B = (A∗ A)1/2 .
(iii) If
(iv) In addition
isometry
v
with initial space
A = (AA∗ )1/2 v .
Proof.
P : Γ(A) → H be
projection onto the rst component. Since A is an operator Ker(P ) =
{0} and hence Range(P ∗ ) is dense in Γ(A) (so P P ∗ H is a core
(i) Since
for
Γ(A)
A).
Let
is closed, it is a Hilbert space.
ξ ∈ H, P ∗ ξ = (η, Aη).
Let
Then, for all
σ ∈ D(A),
< ξ, σ >=< P ∗ ξ, (σ, Aσ) >=< η, σ > + < Aη, Aσ >
⇒ < ξ − η, σ >=< Aη, Aσ >
⇒ Aη ∈ D(A∗ ) and A∗ Aη = ξ − η.
81
D(A∗ A) ⊃ P P ∗ H which is a core for A.
1) = H.
Thus
In addition
Range(A∗ A+
∗
∗
∗
∗
It is easy to see that A A is symmetric, so A A + 1 ⊂ (A A + 1) .
∗
∗
∗
Let ξ ∈ D((A A + 1) ). Since Range(A A + 1) = H there exists
˜.
˜
ξ˜ ∈ D(A∗ A + 1) with (A∗ A + 1)∗ ξ = (A∗ A + 1)ξ(=
(A∗ A + 1)∗ ξ)
∗
∗
∗
Ker((A A + 1) ) = {0} because Range(A A + 1) = H, and hence
ξ = ξ˜ ∈ D(A∗ A+1). Thus (A∗ A+1)∗ = A∗ A+1 and so (A∗ A)∗ = A∗ A.
ξ ∈ D(A∗ A), < A∗ Aξ, ξ >=< Aξ, Aξ >≥ 0 so A∗ A
∗
i.e. σ(A A) ⊂ [0, ∞) (just use the Spectral Theorem).
Finally, for
positive,
is
∗
∗
(ii) As we noted above, D(A A) is a core for A. D(A A) is also a core for
∗
1/2
|A| = (A A) (use spectral theory). Thus AD(A∗ A) = RangeA and
|A|D(A∗ A) = Range|A|. Note that for ξ ∈ D(A∗ A),
|||A|ξ||2 =< A∗ Aξ, ξ >=< Aξ, Aξ >= ||Aξ||2 ,
v : |A|ξ 7→ Aξ , ξ ∈ D(A∗ A), extends to a partial
isometry with initial space |A|D(A∗ A) = Range|A| and nal space
AD(A∗ A) = RangeA.
so that the map
ξ ∈ D(|A|) take ξn ∈ D(A∗ A)
Aξn = v|A|ξn → v|A|ξ and, as A is
For
For
ξ ∈ D(A)
take
ξn ∈ D(A∗ A)
(ξn , |A|ξn ) → (ξ, |A|ξ). Then
closed, ξ ∈ D(A) and Aξ = v|A|ξ .
with
with
∗
(ξn , Aξn ) → (ξ, Aξ).
∗
Then
∗
|A|ξn = v v|A|ξn = v Aξn → v Aξ.
Since
|A|
Hence
is closed,
ξ ∈ D(|A|).
D(A) = D(|A|)
and
A = v|A|.
A∗ = B ∗ u∗ = Bu∗ . A∗ A = Bu∗ uB = B 2 since u∗ u is
projection onto Range(B). By uniqueness of the positive square root
∗
1/2
of a positive operator (Exercise 12.3.3), (A A)
= B . Thus the initial
space of u is Range(|A|) and u|A| = A = v|A| so u = v .
(iii) If
(iv)
A = uB
then
A = v(A∗ A)1/2 so A∗ = (A∗ A)1/2 v ∗ and hence AA∗ = v(A∗ A)1/2 (A∗ A)1/2 v ∗ =
v(A∗ A)v ∗ (Exercise 12.3.3). Thus v implements the unitary equivalence
∗
∗
∗ 1/2
= v(A∗ A)1/2 v ∗ and
of AA |Range(A) and A A|Range(A∗ ) . Hence (AA )
∗
1/2
then A = v(A A)
= (AA∗ )1/2 v .
Remark 12.3.2
∗
D(A A)
Note that it was very important in (i) to establish that
contained a core for A and hence was dense. It was not clear a
D(A∗ A) contained any elements other than 0.
priori that
Exercise 12.3.3
(i) Let
T
be a positive operator. Show that
T 1/2 T 1/2 = T .
(ii) Show that a positive operator has a unique positive square-root.
82
12.4 Unbounded operators aliated with a von
Neumann algebra .
M is a von Neumann algebra on H, an element a ∈ B(H)
au = ua for every unitary in M 0 . This inspires the following.
If
Denition 12.4.1
is in
M
i
T : D(T ) → H is a linear operator on the Hilbert space
H and M is a von Neumann algebra on H we say that T is aliated with
M , written T ηM if, for any unitary u ∈ M 0 ,
If
uD(T ) = D(T )
and
uT ξ = T uξ ∀ξ ∈ D(T ).
Lemma 12.4.2
If
T
is preclosed with closure
Proof. It is clear that
T ηM
i
T
uΓ(T ) = Γ(T )
If
T
T = u|T |
Γ(T )
is a
2×2
M
T
T ηM .
M 0.
But
then
matrix of operators in
is the polar decomposition of
for any bounded Borel function of
if
is a closed operator aliated with
1. The projection onto
2. If
T ηM
for all unitaries in
this property passes to the closure of the graph.
Lemma 12.4.3
then
then
u∈M
M.
and
f (|T |) ∈ M
|T |.
Proof.
1. is obvious from the characterisation of aliation given in the proof of
the previous lemma.
2. follows from uniqueness of the polar decomposition and the bicommutant
theorem.
83
84
Chapter 13
Tomita-Takesaki theory.
In chapter 9 we showed that the GNS construction on
M
using a faithful
Htr with respect
M and its commutant. This is because the map J , which is the extension
to Htr of the * operation on M , is an isometry. So x 7→ JxJ is the extension
∗
to Htr of right multiplication by x . Unfortunately if we use a (normal)
non-tracial state φ the * operation is no longer an isometry and there is no
reason to expect either it or right multiplication by elements of M to have
bounded extensions to Hφ . But as we shall see, the * operation is actually
1/2
preclosed in the sense of unbounded operators and if S = J∆
is the polar
normal trace produces a perfectly symmetric Hilbert space
to
decomposition of its
JM J = M 0 ! Quite remarkably, the operator
M so that a state actually gives rise to a
0
automorphism group of M (and M )!!
closure S , we will show that
1/2
it
−it
∆
will satisfy ∆ M ∆
=
dynamics-a one parameter
We will prove these results using a method of van Daele for which we
have followed some notes of Haagerup ([],[]). But before getting started on
this dicult theory it is essential to do some elementary calculations to see
how it all works out in the
2×2
matrices.
Exercise 13.0.4
form
i
h
Let M be M2 (C). Show that any state φ on M is of the
φ(x) = T race(hx) for some positive h of trace 1. And that φ is faithful
is invertible. Thus with respect to the right basis,
φ(x) = T race(x
for some
1
1+λ
0
0
λ
1+λ
)
λ, 0 ≤ λ ≤ 1.
Exercise 13.0.5
faithful and let
S
With notation as in the previous exercise, suppose
be the * operation on the GNS Hilbert space
85
Hφ .
φ
is
Calculate
1/2
the polar decomposition S = J∆
and show that
z
−z
Show that ∆ M ∆
= M for z ∈ C so that σzφ (x) =
representation of
C
as automorphisms of
M
SM S = JM J = M 0 .
∆z x∆−z = M denes a
which are
∗
-automorphisms i
z ∈ iR.
Exercise 13.0.6
Generalize the above to the
n×n
matrices and in fact any
nite dimensional von Neumann algebra .
13.1 S,F and their graphs.
will be a von Neumann algebra on H and Ω ∈ H
0
a cyclic and separating vector for M and hence M . (The same data as a
Throughout this section
M
faithful normal state.) Let S0 and F0 be the conjugate linear operators with
0
∗
∗
domains M Ω and M Ω dened by S0 (xΩ) = x Ω and F0 (xΩ) = x Ω for
x ∈ M and M 0 respectively.
Lemma 13.1.1
so that
S0
and
In the sense of unbounded operators
F0
F0 ⊆ S0∗
and
S0 ⊆ F0∗
have densely dened adjoints and hence are preclosed.
hS0 (aΩ), a0 Ωi extends to a bounded
0
0 ∗
conjugate linear map on all of H. But hS0 (aΩ), a Ωi = h(a ) Ω, aΩi
0
which is bounded as a function of aΩ by Cauchy-Schwartz. Hence a Ω is
0 ∗
0
∗ 0
∗
in the domain of S0 and S0 (a Ω) = (a ) Ω = F0 (a Ω). Interchanging S0 and
F0 we get the other inclusion.
Proof.
To show
S0∗ (a0 Ω)
Denition 13.1.2
Let
1/2
S = J∆
Let
is dened if
S
S0 = S0−1
unbounded operators.
Thus
∆−1/2 .
The same goes for
∗
see that F = S .
Theorem 13.1.3
M
and
F
be the closures of
be the polar decomposition of
Observe that
∗
F
S
so
1/2
∆
S0
and
respectively.
S.
S2 = 1
2
kernel J
= 1
is injective and
has zero
F0
in the sense of
1/2
and J∆
J =
and its polar decomposition, but we shall now
(Takesaki,[].)
(cΩ, c Ω) where c
Ω ∈ D(c) ∩ D(c∗ ).
set of all
and
S∗ = F , F ∗ = S
and the graph of
S
is the
is a closed densely dened operator aliated with
∗
∗
Proof. Let (ξ, F ξ) be in the graph of F . By the denition of F we know
0 ∗
0
∗
that hξ, (a ) Ωi = ha Ω, F ξi. Now dene operators a and b with domain
M 0 Ω by ax0 Ω = x0 ξ and bx0 Ω = x0 F ∗ ξ . Then a and b are closable for if x0
0
0
and y are in M we have
ha(x0 Ω), y 0 Ωi = hx0 ξ, y 0 Ωi = hξ, (x0 )∗ y 0 Ωi
86
= h(y 0 )∗ x0 Ω, F ∗ ξi = hx0 Ω, y 0 F ∗ ξi = hx0 Ω, b(y 0 Ω)i
a ⊆ b∗
b ⊆ a∗ .
∗
∗
Let c be the closure of a. Then cΩ = aΩ = ξ and c = a ⊇ b so
c∗ Ω = F ∗ ξ . Now by construction the domain of a is invariant under the
0
0
unitary group of M and on it a commutes with the unitaries in M . This
means that c is aliated with M . At this stage we have shown that the graph
∗
∗
of F consists of all (cΩ, c Ω) where c is a closed densely dened operator
∗
aliated with M and Ω ∈ D(c) ∩ D(c ).
∗
We now want to show that the graph of F is contained in the graph of S .
√
c∗ c be its polar decomposition.
This is not hard. Let c be as above and c =
Then if fn (t) = t for 0 ≤ t ≤ n and fn (t) = 0 for t > n we have that
√
√
fn ( c∗ c) → √ c∗ c on any vector in the
√ domain of c, and since c is aliated
∗
with M , fn ( c c) ∈ M so that ufn ( c∗ c)Ω is in the domain of S and tends
√
∗
∗
∗
∗
to ξ . Moreover fn ( c∗ c)u Ω tends to c Ω = F ξ so (ξ, F ξ) is in the graph
of S .
∗
∗
∗
Thus F ⊆ S and we have already observed that S ⊆ F . Hence S = F
∗
and S = F . so that as before
Corollary 13.1.4
and
The polar decomposition of
We now prove a crucial result connecting
Lemma 13.1.5
with
aΩ
in the
F
M
is
J∆−1/2 .
and
M 0.
λ ∈ R+ be given. Then for a0 ∈ M 0
0
−1
domain of F and a Ω = (λS + λ F )aΩ.
Let
there is an
||a0 || ≤ 1 we may apply theorem 8.3.1 to the ψ
ψ(x) = hxΩ, a Ωi and φ(x) = hxΩ, Ωi to obtain the existence of
Proof. Assuming
0
a∈M
dened by
an
a∈M
with
hxΩ, a0 Ωi = λhaxΩ, Ωi + λ−1 hxaΩ, Ωi
= λhxΩ, a∗ Ωi + λ−1 haΩ, x∗ Ωi.
Provided
aΩ is in the domain of F
this equation reads
a0 Ω = (λS +λ−1 F )aΩ.
On the other hand rearranging the equation gives
haΩ, x∗ Ωi = λhxΩ, a0 Ω − λa∗ Ωi
so by Cauchy Schwartz
Corollary 13.1.6
an ∈ M
with
an Ω
aΩ
is in the domain of
For each ξ ∈
→ ξ , ∆1/2 an Ω
F = S ∗. D(∆1/2 ) ∩ D(∆−1/2 ) there is a sequence
→ ∆1/2 ξ and ∆−1/2 an Ω → ∆−1/2 ξ .
87
Proof. Set
η = (S + F )ξ
a0n ∈ M 0 with a0n → η . By
(S + F )an Ω = a0n Ω. But S +
and choose a sequence
the previous lemma there are an ∈ M with
F = J(∆1/2 + ∆−1/2 ) has bounded inverse (in the usual sense of unbounded
−1 0
−1 0
operators) so put ξn = (S + F ) (an Ω). So an Ω = (S + F ) an Ω → ξ .
Moreover
∆1/2 an Ω = ∆1/2 (∆1/2 + ∆−1/2 )−1 Ja0n Ω
+ ∆−1/2 )−1 is bounded by spectral theory. So ∆1/2 an Ω →
(S + F ) (S + F )ξ = ∆1/2 ξ . In the same way ∆−1/2 an Ω → ∆−1/2 ξ .
1/2
and ∆
(∆1/2
1/2
−1
∆
M
We put everything together with a lemma linking
subspace to which many functions of
Lemma 13.1.7
If
ξ
η
and
are in
∆
and
M0
on a dense
can be applied.
D(S) ∩ D(F ), a0 , λ
and
a
as in 13.1.5,
then
λhSaSξ, ηi + λ−1 hF aF ξ, ηi = ha0 ξ, ηi.
Proof. By moving one
S
and
F
to the other side of the inner products, we
η are xΩ and yΩ
respectively, both in D(F ), for x and y in M . But on M Ω, SaS acts by right
∗
∗
∗
multiplication by a so hSaSξ, ηi = hxa Ω, yΩi = hSaΩ, x yΩi. On the other
∗
∗
hand, systematically using F = S we obtain hF aF xΩ, yΩi = hy xΩ, aΩi =
∗
∗
hSx yΩ, aΩi = hF aΩ, x yΩi. Combining these two we see
see by the previous lemma that we may assume
ξ
and
λhSaSξ, ηi + λ−1 hF aF ξ, ηi = h(λSa + λ−1 F a)Ω, x∗ yΩi.
But by 13.1.5 this is
ha0 Ω, x∗ yΩi = ha0 ξ, ηi. 13.2 Proof of the main theorem.
We begin with an easy exercise in countour integration.
Exercise 13.2.1
f
Let
S
is continuous and bounded on
Z
∞
f (0) =
−∞
Hint: Integrate
{z ∈ C : −1/2 ≤ <(z) ≤ 1/2}. Suppose
S and analytic on the interior of S . Then
be the strip
f (z)
sin πz
f (1/2 + it) + f (−1/2 + it)
dt
2coshπt
around rectangular contours in
tending to the boundary of
S.
88
S
Proposition 13.2.2
With notation as in the previous section
Z
∞
∆it Ja0 J∆−it
dt
2coshπt
λ2it
a=
−∞
Proof. Since
1/2
J∆
−1/2
J =∆
we have
J(D(S) ∩ D(T )) = D(S) ∩ D(T )
so
after a little rearrangement the formula of 13.1.7 reads
hJa0 Jξ, ηi = λha∆−1/2 ξ, ∆1/2 ηi + λ−1 ha∆1/2 ξ, ∆−1/2 ηi.
H0 be the dense subspace of all vectors in H which is the union of
ξ[a,b] (∆ for 0 < a < b < ∞. Certainly H0 ⊆ D(S) ∩ D(F ), H0 is invariant
z
z
under J and ∆ for z ∈ C, and moreover for ξ ∈ H0 , z 7→ ∆ ξ is an entire
function of z .
Now let
all
For
ξ, η ∈ H0
dene the analytic function
f (z) = λ2z ha∆−z ξ, ∆z ηi.
f is bounded in the strip S of the
f (1/2 + it) = h∆it ∆1/2 ξ, ηi so that
Then
Also
previous lemma and
f (0) = haξ, ηi.
f (1/2 + it) + f (−1/2 + it) = λ2it h∆it Ja0 J∆−it ξ, ηi.
So by the previous lemma we are done.
Theorem 13.2.3
H and Ω a cyclic
∗
and separating vector for M . Suppose S is the closure of xΩ 7→ x Ω on M Ω.
∗
1/2
Let ∆ = S S , and J be the antiunitary of the polar decomposition S = J∆
.
Let
M
be a von Neumann algebra on
Then
(i)
(ii)
JM J = M 0
∆it M ∆−it = M ∀t ∈ R
Proof. If
a0 ∈ M 0
we know that
Z
∞
λ2it
−∞
.
Conjugating by a unitary
∆it Ja0 J∆−it
dt ∈ M
2coshπt
u ∈ M0
and writing
iθ
λ = e2
we see that the
Fourier transforms of the strongly continuous rapidly decreasing functions
∆it Ja0 J∆−it
∆it Ja0 J∆−it ∗
and u
u are equal. Hence ∆it Ja0 J∆−it ∈ M for
2coshπt
2coshπt
t since it commutes with every unitary u ∈ M 0 .
all real
(Take inner products
with vectors if you are not happy with Fourier transforms of operator valued
functions.)
Putting
0
JM J = M
t=0
JM 0 J ⊆ M
done. we see
and we are
and by symmetry
89
JM J ⊆ M 0 .
Hence
Denition 13.2.4
The operator
J
of the previous result is called the mod-
ular conjugation and the strongly continuous one-parameter group of autoφ
it
−it
morphisms of M dened by σt (x) = ∆ x∆
is called the modular automorphism group dened by
φ.
13.3 Examples.
Example 13.3.1
ITPFI
The acronym ITPFI stands for innite tensor product of nite type I.
These von Neumann algebras are formed by taking the *-algebra
m
O
union
A∞
of tensor products
being diagonal. The state
each
Mnk .
φ
Mnk (C), the inclusion of Am
as the
in
Am+1
k=1
on A∞ is then the tensor product of states on
One may then perform the GNS construction with cyclic and
separating vector
∞
O
M =
Am =
A∞
Mnk (C)
Ω
given by
1 ∈ A∞ ,
to obtain the von Neumann algebra
as the weak closure of
k=1
all the nk are equal to
2
A∞
acting on
Hφ .
The case where
and all the states are the same is called the Powers
factor and the product state the Powers state as it was R.Powers who rst
showed that they give a continuum of non-isomorphic type III factors.
A slight snag here is that we do not know that
M.
on
Ω
denes a faithful state
But if we proceed anyway to construct what have to be
will soon see that the state is indeed faithful, i.e.
Ω
is cyclic for
J and ∆
M 0 Ω.
we
Mn (C), and φh (x) =
Ptrace(xh) where
h is the diagonal
q matrix (density matrix) with hii = µi , µi = 1, µi > 0,
µj
e and ∆n (eij ) = µµji eij (where dependence on h has been
then Jn (eij ) =
µi ji
Recall from exercise 13.0.6 that, for
suppressed).
To diagonalise the modular operators on
Hφ
completely it is most con-
di of the diagonal matrices, with
d1 = 1. Then a basis for the Hilbert space Hφ is formed by tensors ⊗∞
k=1 vk Ω
where vk = 1 for large k , and is otherwise a di or an eij with i 6= j .
We can guess that J is, on each basis vector, the tensor product of the J 's
vincing to choose an orthonormal basis
coming from the matrix algebras. Dening it as such it is clearly an isometry
x ∈ A∞ , JxJ is in
M by the nite dimensional calculation! But the linear span of these JxJΩ
0
is dense so Ω is cyclic for M and hence separating for M . We are hence in
A∞ Ω
on
and thus extends to all of
Hφ .
But then, for any
0
a position to apply Tomita Takesaki theory. Each of the basis elements is in
∞
M Ω so S(⊗∞
k=1 vk Ω) = ⊗k=1 wk Ω where wk is vk if vk is diagonal, and eji if
90
vk = eij .
So
JS
is diagonal and hence essentially self-adjoint. We conclude
that
J(xΩ) = Jm (x)Ω
and
∆(xΩ) = ∆m (x)Ω
and
σtφ
=
∞
O
σ φhk
for
x ∈ Am ,
.
k=1
Example 13.3.2
Let
Γ
Group-measure-space construction.
be a discrete group acting on the nite measure space
(X, µ)
pre-
serving the class of the nite measure µ. The Hilbert space of the crossed
L∞ (X, µ) is L2 (X, µ) ⊗ `2 (Γ) and as we saw in chapter 11 the vector
1 ⊗ εid is a cylic and separating vector Ω for M = L∞ (X, µ)oΓ.
product
µ is preserved by the γ ∈ Γ the Radon Nikodym theorem
1
L
functions hγ so that φ(hγ αγ (y)) = φ(x) where φ(y) =
R
ydµ. We know that, if x ∈ L∞ (X, µ) then S(uγ x) = x∗ uγ −1 . In general
X
we will not be able to completely diagonalise ∆ but the same argument as
in the previous example gives that the domain of ∆ is
XZ
2
|hγ (x)f (x)|2 dµ(x) < ∞}
{f : Γ → L (X, µ) :
Since the class of
guarantees positive
γ
X
on which
(∆f )(γ) = hγ f (γ),
and
(Jf )(γ) = hγ−1/2 f (γ).
We can now nally answer the question as to which sums
∞
elements of M = L (X, µ)oΓ.
Theorem 13.3.3
P
With notation as above, if the function
P
γ
xγ uγ
dene
γ 7→ xγ ∈ L∞ (X, µ)
γ xγ uγ , interpreted as a matrix of operators as in section 11.2,
∞
denes a bounded operator, then that operator is in M = L (X, µ)oΓ.
is such that
P
By 13.2.3 it suces to show that
γ xγ uγ commutes with Jxuγ J
∞
for all x ∈ L (X, µ) and γ ∈ Γ. And for this it suces to check that the
2
commutation holds on functions of the form f ⊗ εγ for f ∈ L . This is just
Proof.
a routine computation.
91
Exercise 13.3.4
Show that example 13.3.1 is in fact a special case of this
∞
group-measure-space example in which L (X, µ) is provided by the tensor
products of the diagonal elements and the group
Γ
is a restricted innite
Cartesian product of cyclic groups, constructed from the non-diagonal
eij 's.
Conclude by the method of 11.2.15 that ITPFI algbras are factors.
This example brings to light a signicant inadequacy of our treatment of
Tomita-Takesaki theory. We would like to treat the case where the measure
of the space is innite.
Although of course we could choose an equivalent
nite measure, this choice may not be natural.
To do this we would have
to consider the theory of weights which are to states as the trace on a II∞
factor is to the trace on a type
II1
factor. We need the same notion in order
to understand the origin of the term modular used above as coming from
the modular function on a non-unimodular locally compact group.
But a
serious treatment of weights would take many pages so we simply refer the
reader to Takesaki's book [3].
Example 13.3.5
If
G
Hecke algebras à la Bost-Connes.
is a nite group let
ug
and
vg
be the unitaries of the left and right
regular representations respectively. If H is a subgroup, the projection pH =
P
1
2
h∈H vh projects from ` (G) onto functions that are right translation
|H|
invariant under
H,
i.e. functions on the quotient space
called quasi-regular representation of
G
on
G/H
G/H .
Thus the so-
is a direct summand of
the left regular representation and we have from EP7 of chapter 4 that the
2
commutant of the action of G on ` (G/H) is pH ρ(G)pH where ρ(G) is the
algebra generated by the right regular representation (of course isomorphic
to
C).
This commutant is spanned by the
functions on
G,
pH vg pH
which, thought of as
are multiples of the characteristic functions of the double
HgH which form the double coset
ρ(G) spanned by these double cosets is
cosets
space
H\G/H .
The subalgebra
the space of H − H bi-invariant
2
functions and we see it is the commutant of G on ` (G/H). It is known as
of
the Hecke algebra for the pair
(G, H)
and has a considerable role to play
in the representation theory of nite groups. A famous example is the case
where
G
is the general linear group over a nite eld and
H
is the group of
upper triangular matrices. The coset space is then the so-called ag variety
and the Hecke algebra in this case leads to a lot of beautiful mathemtatics.
See Bourbaki [].
Nothing could better indicate how dierently things work for innite discrete groups than how the Hecke algebra works. Far from being direct summands, the quasiregular representations can be totally dierent from the left
92
regular representations and can even generate type III factors! These Hecke
algebras give nice examples where the modular operators can be calculated
explicitly.
Denition 13.3.6
A subgroup
H
of the discrete group
G
is called almost
normal if either of the two equivalent conditions below is satised.
(a)
gHg −1 ∩ H
is of nite index in
H
(b) Each double coset of
orbits of
on
G/H
for all
g ∈ G.
is a nite union of left cosets of
H
(i.e. the
are all nite).
G one may construct operators in the commutant
G on `2 (G/H) as follows:
Given an element x of G/H let εx be the characteristic function of x.
2
These functions form an orthonormal basis of ` (G/H). Moreover each vector
εx is cyclic for the action of G hence separating for the commutant. If D is
a double coset of H dene TD by the matrix
−1
(TD )x,y = 1 if y x = D;
0 otherwise.
If
H
H
H
is almost normal in
of the quasiregular representation of
Clearly
TD
is bounded since
mutes with the action of
G.
H
is almost normal and it obviously com-
From the denition we have
TD∗ = TD−1 .
It is also easy to check that
TD TE =
X
nFD,E TF
F
where the structure constants are dened by
nFD,E =
n
#(E/H)
0
if
F ⊆ ED;
otherwise.
TD 's the HeckeHvN (G, H). The
We will call the von Neumann algebra generated by the
H ⊆ G and write it
φ dened on HvN (G, H) by εH is faithful and normal, and
hTD εH , TD0 εH i = 0 unless D = D0 so that the TD 's are orthogonal. It is thus
easy to calculate the operators for the modular theory on Hφ (note that this
2
is not ` (G/H)). We guess as usual that J(TD Ω) = (constant)TD−1 Ω and
by counting cosets in double cosets (or sizes of orbits of H on G/H ) we nd
von Neumann algebra of the pair
vector state
93
(#(D/H))1/2 (#(H\D))−1/2 . Thus as
TD Ω of Hφ so essentially self-adjoint and
that the constant has to be
is diagonal on the basis
∆(TD Ω) =
before
JS
#(H\D)
TD Ω
#(D/H)
with the obvious domain. Thus
σtφ (TD )
=
#(H\D)
#(D/H)
it
TD .
Examples of almost normal subgroups are not hard to nd. The classical
example of Hecke himself is the case where
G = SL(2, Q) and H = SL(2, Z).
In this case the Hecke algebra is abelian. Bost and Connes in [4] examined
ax+b group over the rationals with the subgroup being integer
They showed that HvN (G, H) in this case is a type III factor
the case of the
translations.
and made a connection with prime numbers.
13.4 The KMS condition.
In the examples of the previous section the operators of the modular group
were easy to calculate explicitly, including the domain of
∆.
One can imagine
that this is not always so. If we are particularly interested in the modular
φ
group σt it would be useful to be able to guess it and show that the guess is
right without bothering about the domain of
∆.
The KMS (Kubo-Martin-
Schwinger) condition from quantum statistical mechanics allows us to do just
that. The modular group came from the non-trace-like property of a state
and the KMS condition allows us to correct for that.
Let us do a formal
calculation assuming that the modular group can be extended to complex
numbers (remember that
Ω
is xed by
φ(xy) =
=
=
=
S, J
and
∆):
hyΩ, x∗ Ωi
hyΩ, J∆−1/2 ∆x∆−1 Ωi
h∆x∆−1 Ω, SyΩi
hy∆x∆−1 Ω, Ωi.
We conclude that
φ(xy) = φ(yσiφ (x)).
Thus the trace is commutative provide we operate by the modular group.
94
Exercise 13.4.1
If M is nite dimensional and φ is a faithful state, show
φ
that φ ◦ σt = φ and that for each x and y in M there is an entire function
F (z) with, for t ∈ R,
F (t) = φ(σtφ (x)y)
F (t + i) = φ(yαt (x)).
If
M
and
is innite dimensional we would not expect the function
F (z)
of the
previous exercise to be entire.
Denition 13.4.2
Let
αt
be a strongly continuous one parameter automor-
M , and φ be a faithful normal state
α satises the KMS condition for φ if φ ◦ αt = φ and ,
for each x and y in M , there is a function F , continuous and bounded on
the strip {z : 0 ≤ =m(z) ≤ 1}, analytic on the interior of the strip and such
that for t ∈ R,
phism group of a von Neumann algebra
on
M.
We say that
F (t) = φ(σtφ (x)y)
F (t + i) = φ(yαt (x)).
and
Theorem 13.4.3
If φ is a faithful normal state on a von Neumann algebra
φ
M then σt is the unique one parameter automorphism group satisfying the
KMS condition for φ.
This chapter has been heavily technical so we defer the proof, which is by
approximation on dense subspaces of the domain of
∆ to which
the previous
calculations can be applied, to an appendix. We content ourselves here with
an interesting corollary, identifying a part or
M
on which
φ
behaves as a
trace.
Corollary 13.4.4
1.
φ(ax) = φ(xa)
2.
σtφ (a) = a
For
a∈M
for all
for all
the following are equivalent:
x ∈ M.
t ∈ R.
⇒ 2) Observe that for x ∈ M , hx∗ Ω, aΩi = hΩ, xaΩi = hΩ, axΩi
∗
∗
∗
∗
(by 1). So hSxΩ, aΩi = ha Ω, xΩi so that aΩ ∈ D(S ) and S (aΩ) = Ω . So
∆(aΩ) = aΩ, ∆it aΩ = aΩ and nally σtφ (a) = a for all t ∈ R.
Proof.(1
(2
⇒ 1) φ(σtφ (x)a) = φ(σtφ (xa)) = φ(xa) so that F (t) is constant.
Use the
Schwarz reection principle to create a holomorphic function, constant on
in the union of the strip with its complex conjugate. Thus
the strip and
φ(xa) = φ(ax). 95
F
R,
is constant on
Denition 13.4.5
The von Neumann subalgebra of
M
dened by either of
the conditions of the previous corollary is called the centraliser of the state
φ.
96
Chapter 14
Connes' theory of type III factors.
14.1 The Connes unitary cocycle Radon-Nikodym
theorem.
This result will allow us to extract information from the modular group of a
state which is independent of the state.
Theorem 14.1.1
algebra
M.
Let
φ
and
ψ
be faithful normal states on a von Neumann
Then there is a strongly continous map
unitary group of
M
ut
from
R
to the
so that
σtφ = Adut σtψ
Morevoer
t → ut
∀ t ∈ R.
satises the cocycle condition
ut σtψ (us ) = ut+s .
Φ onM ⊗ M2 (C) by Φ((x)ij ) =
1 0
1
Φ
(φ(x11 ) + ψ(x22 )). The projection p =
is xed by σ
by 13.4.4.
2
0 0
Φ
So σ denes a one parameter automorphism group of pM ⊗ M2 (C)p which
φ
Φ
satises the KMS condition for φ. Hence σt (x ⊗ e11 ) = σt (x) ⊗ e11 . Similarly
0 0
ψ
Φ
Φ
∗
σt (x ⊗ e22 ) = σt (x) ⊗ e22 . Let Vt = σt (1 ⊗ e21 ). Then Vt Vt =
and
0 1
1 0
0 0
Vt∗ Vt =
. Hence Vt =
for some unitary vt ∈ M . Routine
0 0
vt 0
computations give the rest. Proof. We dene the faithful normal state
Corollary 14.1.2
M
If
M
is a factor and
is of type III.
97
σtφ
is outer for any
phi
and
t
then
Proof. By the previous result it suces to exhibit a single faithful normal
state on a type II factor with inner modular group. In the II1 case use the
φ
trace and in the II∞ case choose a faithful normal state
tr ⊗ φ,
on
B(H)
and use
using the KMS condition (if necessary) to very that the modular
group for the tensor product is the tensor product of the modular groups.
Corollary 14.1.3
The subgroup of all
pendent of the faithful normal state
Denition 14.1.4
variant of
M,
t∈R
for which
σtφ
is inner is inde-
φ.
The subgroup of the previous corollary, which is an in-
is called
T (M ).
T (M ) for the Powers factor Rλ where this refers
with
all nk = 2 and all states having the same density
0
We shall now calculate
to the ITPFI
factor
1
1+λ
matrix h =
λ
0
1+λ
.
Theorem 14.1.5
T (Rλ ) =
2π
Z
log λ
.
Proof. By the formula for the modular group
σ φ2π = id
log λ
so
2π
Z
log λ
⊆ T (Rλ ).
For the other direction it suces to show that an automorphism
α
of the
form
α = ⊗∞
k=1 Adu
u is
uk = ⊗k1 u
is outer whenever the unitary
not a scalar.
For this rst dene
and observe that if α = Adv then (uk ⊗
−1
1) v = id on the matrix algebra Ak = ⊗k1 M2 (C). By exercise 5.3.3 this
means that v = uk ⊗ w . Now it is clear from our basis that we can choose
⊗pj=1 xi ⊗ 1Ω with non-zero inner procuct with vΩ. But then xing p and
letting k tend to innity we see that
h(⊗pj=1 xi ⊗ 1)Ω, vΩi =
p
Y
hxi , uih1, uik−p h1, wi.
j=1
is a scalar
|h1, wi| ≤ 1 so we must have
multiple of 1 by the Cauchy-
Rλ
are type III factors ,mutually
The left hand side does not depend on
|h1, ui| = 1
which means that
Schwarz inequality.
u
k
and
We conclude that the Powers factors
non-isomorphic for dierent values of
λ.
98
14.2 Type IIIλ.
The spectrum of the modular operator
∆
is easy to calculate for an ITPFI
µi
as µ varies over
µj
all the density matrices dening the product state. Apart from being closed
factor.
It is simply the closure of the set of all ratios
under the inverse operation this set of non-negative real numbers has no
particular structure and can be altered easily by making changes in nitely
many density matrices which of course do not change the factor.
Denition 14.2.1
If
M
is a von Neumann algebra the invariant
the intersection over all faithful normal states
responding modular operators
φ
S(M )
is
of the spectra of their cor-
∆φ .
Theorem 14.2.2
A factor
Theorem 14.2.3
(Connes-van Daele)
M
is of type III i
0 ∈ S(M ).
S(M ) \ {0}
is a closed subgroup of
the positive real numbers.
There are only three kinds of closed subgroups of
Denition 14.2.4
A factor
M
is called type IIIλ for
R+ .
0≤λ≤1
if
λ = 0 : S(M ) = {0} ∪ {1}
0 < λ < 1 : S(M ) = {0} ∪ {λn : n ∈ Z}
λ = 1 : S(M ) = {0} ∪ R+
Theorem 14.2.5
The Powers factor
Rλ
is of type IIIλ .
In his thesis , Connes showed that every type IIIλ factor for
0<λ<1
is Connes thesis
canonically isomorphic to the crossed product of a type II∞ factor with an
action of
Z
whose generator scales the trace by
λ.
A is a locally compact abelian group with an action α on a von Neumann
M , there is an action α̂ of the Pontryagin dual  on the crossed
product M oα A satisfying
If
algebra
α̂a (x) = x
α̂â (ua ) = â(a)ua
x∈M
ua are the
for
if
unitaries dening the crossed product.
The existence of the so-called dual action
α̂
is trivial proved since it is
2
implemented by the obvious unitary representation of  on L (A).
99
Exercise 14.2.6
P
A is nite consider the projection p = a ua ∈ M oA.
A
Show that pM oAp = M p and thus show that (M oα A)oα̂ Â is isomorphic
to M ⊗ M|A| (C).
If
Observe that the crossed product of a von Neumann algebra M on H by
φ
the modular group σ does not depend, up to isomorphism, on the faithful
normal state φ. This follows from theorem 14.1.1 by dening the unitary
2
on L (R, H) by
V
V f (t) = ut f (t)
where
ψ
is another faithful normal state with unitary one-cocycle
gating the operators that generate
of
M oσφ R
by
V
ut .
Conju-
one obtains the generators
M oσψ R.
Theorem 14.2.7
The crossed product of
M
by the modular group admits a
trace satisfying the properties of 9.1.9
Denition 14.2.8
weights of
The action of
R̂
on
Z(M oσφ R)
is called the ow of
M.
Theorem 14.2.9
(Takesaki duality) The crossed product
(M oσφ R)oσˆφ R̂
is isomorphic to the tensor product
Thus if
M
M ⊗ B(H)
for
H = L2 (R).
is a factor the ow of weights is ergodic.
Theorem 14.2.10
If
M
is a factor of type IIIλ the ow of weights is
M
III1 :
The trivial ow on a one point set if
IIIλ :
2π
The transitive ow on the circle with period λ if
λ < 1.
0<
III0 :
Ergodic non-transitive if
M
is III1 .
M
is of type IIIλ ,
is of type III0 .
Moreover any ergodic non-transitive ow arises as the ow of weights for
some type III0 factor.
100
Chapter 15
Hyperniteness
Denition 15.0.11
A von Neumann algebra
M
on a separable Hilbert space
is called hypernite if there is an increasing sequence
*-subalgebras of
M
which generates
M
An
of nite dimensional
as a von Neumann algebra .
15.1 The hypernite type II1 factor R
The rst main result about hyperniteness was proved by Murray and von
Neumann in [].
We will use
R
to denote the hypernite II1 factor whose
uniqueness they proved.
Theorem 15.1.1
Up to abstract isomorphism there is a unique hypernite
II1 factor.
||x||2 = tr(x∗ x)1/2 on M . It is
not hard to show that a von Neumann subalgebra N of M is strongly dense
in M i it is dense in ||−||2 . Given a subalgebra A of M and a subset S ⊆ M
Sketch of proof. One works with the norm
one says
S ⊆ A
ε
if for each
x∈S
there is a
y∈A
with
||x − y||2 < ε.
The hyperniteness condition then implies:
S ⊆M
M with
For every nite subset
sional *-subalgebra
A
of
and every
S ⊆ A.
ε
101
ε>0
there is a nite dimen-
The next step is to show that the A in the preceeding condition can be
n
n
chosen to be the 2 × 2 matrices for some (possibly very large) n. This part
uses what might be described as II1 factor technique. One begins with
A
and approximates all its minimal projections {ei } by ones whose traces are
n
numbers of the form k/2 . The matrix units of A can be changed a little bit
|| − ||2 so that, together with matrix units conecting projections of trace
1/2n less than the ei , they generate a 2n ×2n matrix algebra containing, up to
ε, the matix units of A. Perturbation of the matrix units will involve results
in
of the form:
u ∈ M satises ||(uu∗ )2 − uu∗ ||2 < v ∈ M with ||v − u||2 < F ()
(for some nice function f with f (0) = 0).
If
then there is a partial isometry
or:
p and q are projections with ||pq||2 < pq = 0 and ||q − q 0 || < F ().
If
then there is a projection
q0
with
0
or:
If
fij
are almost
n×n
(a)
||fij − fji ||2 < (b)
||fij fkl − δj,k fil ||2 < P
||1 − ni=1 f ii||2 < (c)
then there are
only on
n
and
n × n matrix
F (0) = 0.
matrix units, i.e.
units
eij
with
||eij − fij || < F ()
where
F
depends
Such results are proved by a skilful use of the polar decomposition and
spectral theorem.
Thus one shows that in a hypernite type II1 factor one has:
Property * :
2n × 2n
For every nite subset
matrix subalgebra of
M
S⊆M
and every
ε>0
there is a
with
S ⊆ A.
ε
One may now proceed to prove the theorem by choosing a
sequence
nk
nk
2
×2
xk
in
M
matrix algebras
For each
|| − ||2 -dense
and inductively constructing an increasing sequence of
Ak
k = 1, 2, 3, ...,
with the property that
{x1 , x2 , ..., xk }
102
⊆ Ak .
1/k
Ak 's is clearly dense in || − ||2 . This is enough to prove
theorem since the Ak 's can be used to give an isomorphism of M with
∞
type II1 factor ⊗ M2 (C) constructed in section 7.2.
The union of the
the
the
To construct
Ak+1
from
Ak
one needs to arrange for the new algebra
to contain the one already constructed. So to the elements
add matrix units
containing the
xi
eij
eij
Ak+1 .
eij .
In
B
x1 , x2 , ..., xk+1 ,
B almost
Now use property * to obtain a
and the matrix units, with
over the matrix units
close to the
for
small enough to control sums
we know there are approximate matrix units
so by a technical lemma, exact matrix units
fij
close to the
eij .
Now choose a unitary close to the identity which conjugates the
the
eij
fij
to
B to a superalgebra of Ak . This
xi up to epsilon since u is close to
and use this unitary to conjugate
superalgebra is
Ak+1
and it contains the
the identity.
This completes the sketch of the proof. The technique involved is considered standard in von Neumann algebras and the details can be found in
dixmier
.
Corollary 15.1.2
If
S∞
is the group of nitely supported permutations of a
countably innite set then vN (S∞ ) ∼
= ⊗∞ M2 (C).
Proof. The subgroups of
show that
vN (S∞ )
S∞
permuting an increasing sequence of nite sets
is hypernite.
It is surprising at rst sight that the type II1 factor
from an ergodic measure-preserving transformation
T
L∞ (X, µ)oZ obtained
is hypernite. This can
be shown by Rokhlin's tower theorem which asserts that, for each
each
>0
there is a measurable subset
(1)
T i (A) ∩ T j (A) = ∅
(2)
µ(X \ ∪ni=0 T i (A)) < .
1 ≤ i < j ≤ n,
with
and
A
can be combined, with a little perturbation to get the identity, to get a n × n
The unitary
u1
for
A⊆X
n ∈ N and
of the crossed product and the characteristic function of
matrix algebra. Careful application of the tower theorem will allow one to
∞
get any element of L (X, µ), and u1 , in this matrix algebra up to some .
This was rst proved by Henry Dye in who went on to prove that in fact all Dye
groups of polynomial growth give hypernite II1 factors in this way.
The ultimate result in this direction is the celebrated Injective factors
theorem of Connes who showed that hyperniteness for a von Neumann algebra
M
on
H is equivalent to injectivity
which means there is a projection
in the Banach space sense of norm one from
B(H)
onto
M.
This theorem,
whose proof is a great, great tour de force, has a raft of corollaries, many of
103
which were open questions. Let us just mention the fact that it follows easily
that any subfactor of
R.
as
R which is innite dimensional is in fact isomorphic to
vN (Γ), as well as L∞ (X, µ)oΓ is hypernite as soon
It also implies that
Γ
is amenable.
15.2 The type III case.
The complete classication of injective(=hypernite) factors is a triumph
Connes actions
of 20th.
century mathematics.
trace-scaling automorphism of
Connes Injective factors
Connes showed in that there is only one
R ⊗ B(H)
λ 6= 1 up to
0 < λ < 1 there is
for each scaling factor
conjugacy. Together with this shows that for each
λ
with
a unique injective factor of type IIIλ .
krieger
injective
Using results of Krieger in , his thesis and , Connes showed that hypernite type III0 factors are classied by their ow of weights (up to conjugacy of
ows, not orbit equivalence). This means that there is a rather large number
of III0 factors but their classication is in the realm of ergodic theory rather
than von Neumann algebras.
The remaining case of injective type III1 factors was solved by Haagerup
ueIIIone
in . There is just one such factor and a hypernite factor is generically of
type III1 .
104
Chapter 16
Central Sequences.
16.1 Generalities.
Denition 16.1.1
M is a type II1 factor , a central sequence in M is a
(xn ) with limn→∞ ||[xn , a]||2 = 0. A central sequence
if limn→∞ ||xn − tr(xn )id||2 = 0. M is said to have
If
norm bounded sequence
is said to be trivial
property
Γ
if there is a central
∗
∞
The collection of all central sequences is clearly a C -subalgebra of ` (N, M ).
∞
If ω is a free ultralter on N, the subalgebra Iω of ` (N, M ) consisting of
∞
sequences with limn→ω ||xn ||2 = 0 is a 2-sided ideal of ` (N, M ). Note also
∞
that M is embedded in ` (N, M ) as constant sequences.
Denition 16.1.2
With notation as above, the ultraproduct of M along ω is
ω
by Iω . It is written M . The algebra of (ω -)central
0
ω
∞
sequences is the centraliser Mω = M ∩ M of M in ` (N, M ).
the quotient of
`∞ (N, M )
By compactness, the trace on
M
denes a trace on
Mω
by
tr((xn )) = lim tr(xn )
n→ω
and by denition it is faithful on
M ω.
Exercise 16.1.3
in
|| −
∗
Show that the unit ball (in the C norm) of
ω
||2 so that M and Mω are von Neumann algebras.
Mω
is complete
16.2 Central sequences in R
All models for
R
exhibit central sequences in abundance. The most obvious
∞
situation is that of ⊗ M2 (C). Fixing x ∈ M2 (C) we can dene the sequence
105
xn = 1 ⊗ 1 ⊗ 1...x ⊗ 1 ⊗ 1... with the x in the nth. slot in the tensor product.
For large enough n, xn will commute with any element in the algebraic tensor
product so by the obvious (in the II1 case!) inequality ||[xn , a]|| ≤ 2||xn || ||a||2
we see that (xn ) is central and clearly non-trivial if x is not a scalar. Just as
clearly the central sequence algebra is non-commutative as we only need to
choose
(yn )
x
y
and
that do not commute and construct the sequences
as above. In fact it is not hard to show that
Theorem 16.2.1
Proof. If
(xn )
The central sequence algebra
represents an element
Rω
Rω
(xn )
and
is a factor.
is a type II1 factor .
X ∈ Rω ,
16.3 Type II1 factors without property Γ.
Theorem 16.3.1
Let
Γ be an icc group possessing a subset ∆ not containing
α, β and γ such that
the identity and three elements
(a)Γ
= {1} ∪ ∆ ∪ α∆α−1
(b)∆, β∆β
then for
−1
and
γ∆γ −1
are mutually disjoint.
x ∈ vN (Γ),
||x − tr(x)id||2 ≤ 14 max{||[x, uα ]||2 , ||[x, uβ ]||2 , ||[x, uγ ]||2 }.
Proof.Write
x
as
P
ν∈Γ
x ν uν .
We will frequently use the formula
||[x, uρ ]||22 = ||uρ−1 xuρ − x||2 =
X
|xν − xρνρ−1 |2 .
ν∈Γ
x − tr(x)id it suces to prove the result if tr(x) = 0
||x||2 = 1 so that for such an x we must show 1 ≤
14 max{||[x, uα ]||2 , ||[x, uβ ]||2 , ||[x, uγ ]||2 }.
We rst make a simple estimate. If Λ is any subset of Γ and ρ ∈ Γ then
X
X
X
(|xν | + |xρνρ−1 |)||xν | − |xρνρ−1 ||
|
|xν |2 −
|xρνρ−1 |2 | =
By replacing
x
by
and we may suppose
ν∈Λ
ν∈Λ
ν∈Λ
≤
X
(|xν | + |xρνρ−1 |)(|xν − xρνρ−1 |)
ν∈Λ
X
≤ 2||x||2 (
|xν − xρνρ−1 |2 )1/2
ν∈Λ
106
so that if
ρ ∈ {α, β, γ}
we have
|
X
|xν |2 −
ν∈Λ
where
X
|xρνρ−1 |2 | ≤ 2
ν∈Λ
= max{||[x, uα ]||2 , ||[x, uβ ]||2 , ||[x, uγ ]||2 }.
||x||2 = 1:
X
X
1 ≤
|xν |2 +
|xανα−1 |2
Let us now rst overestimate
ν∈∆
≤ 2
ν∈∆
X
2
|xν | + 2.
ν∈∆
Now underestimate it:
1 ≥
X
|xν |2 +
≥ 3
|xβνβ −1 |2 +
ν∈∆
ν∈∆
X
X
X
|xγνγ −1 |2
ν∈∆
2
|xν | − 4.
ν∈∆
y=
1 ≥ 3y − 4
Let
2
ν∈∆ |xν | and eliminate
to obtain
P
y
from the inequalities
1 ≤ 2y + 2 and
≥ 1/14
as desired.
It is easy to come up with groups having subsets as in the previous theorem. For instance if
G = F2 ,
free on generators
g
and
h,
let
∆
be the set
of all reduced words ending in a non-zero power of g . Let α = g , β = h and
γ = h−1 . The same works for more than two generators. We conclude:
Theorem 16.3.2
The type II1 factor
vN (Fn ) for n ≥ does not have property
Γ.
107
108
Chapter 17
Bimodules and property T
109
110
Chapter 18
Subfactors
111
112
Chapter 19
Free probability
113
114
Appendix A
Normal implies ultraweakly
continuous.
115
116
Appendix B
Proof of the KMS condition.
Theorem B.0.3
φ be
a faithful normal state on a von Neumann algebra
M . Then the modular group σtφ is the unique one parameter automorphism
group of
M
Let
which satises the KMS condition for
φ.
Proof. Perform the GNS construction with canonical cyclic and separating
1/2
vector Ω and modular operators S = J∆
. Recall that f (∆)Ω = Ω for any
φ
it
−it
function of ∆ with f (1) = 1. In particular φ(σt (x) = h(∆ x∆
Ω, Ωi so σtφ
preserves
φ.
Now let us check the rest of the KMS condition. We have
φ(σtφ (x)y) = h∆−it yΩ, x∗ Ωi
and
φ(yσtφ (x)) =
=
=
=
hyσtφ (x)Ω, Ωi
hJ∆1/2 σtφ (x∗ )Ω, J∆1/2 yΩi
h∆1/2 yΩ, ∆1/2 ∆it x∗ Ωi
h∆1/2−it yΩ, ∆1/2 x∗ Ωi
ξ = yΩ, η = x∗ Ω and let pn be the spectral projection for ∆ for
±1
the interval [1/n, n] so that pn tends strongly to 1 and ∆
are bounded on
pn Hφ . The functions
Fn (z) = h∆−iz pn ξ, ηi
So let
are then entire and
|Fn (t) − φ(σtφ (x)y)| = |h∆−it (1 − pn )ξ, ηi| ≤ ||(1 − pn )ξ|| ||η||
|Fn (t + i) − φ(yσtφ (x))| = |h∆1/2−it (1 − pn )ξ, ∆1/2 ηi| ≤ ||(1 − pn )∆1/2 ξ|| ||∆1/2 η||.
117
Hence the
Fn are bounded and continuous on the strip {z : 0 < =mz < 1}
and converge uniformly on its boundary. By the Phragmen-Lindelof theorem
we are done.
Now let us prove uniqueness of the modular group with the KMS condition.
αt
Let
be another continous one-parameter automorphism group satisfy-
ing KMS for
φ.
The fact that
αt
φ
preserves
means we can dene a strongly continous
one-parameter unitary group t 7→ ut by ut xΩ = αt (x)Ω. By Stone's theorem
it
it is of the form t 7→ D for some non-singular positive self-adjoint operator
A.
D = ∆.
The goal is to prove that
set of analytic vectors in
MΩ
As a rst step we construct a dense
by Fourier transform. Let
operators of the form
Z
A
be the set of all
∞
fˆ(t)αt (x)dx
−∞
for all
C∞
f
functions
of compact support on
strongly so
Z
R.
The integral converges
∞
f (log(D))xΩ =
fˆ(t)Dit (xΩ)dx
−∞
AΩ. Thus the spectral projections of D are in the strong closure of A
AΩ is dense. Moreover z 7→ Dz xΩ is analytic for x ∈ A since xΩ is
in the spectral subspace of A for a bounded interval. Also AΩ is invariant
z
under D by the functional calculus. To compare with φ dene, for x and y
in A, the entire function
is in
and
F1 (z) = hD−iz yΩ, x∗ Ωi.
Let
F
be the function, analytic inside the strip and continuous and bounded
on it, guaranteed for
for
−1 ≤ =mz ≤ 1
x
and
y
by the KMS condition. Then if we dene
G(z)
by
G(z) =

 F (z) − F1 (z)
if
=mz ≥ 0;
F (z) − F1 (z)
if
=mz ≤ 0.

G is analytic for −1 < =mz < 1, hence
equal to 0, and since both F and F1 are continous on the strip, φ(yσt (x)) =
F (t + i) = F1 (t + i) = hD1−it yΩ, x∗ Ωi. In particular putting t = 0 we get
Since
F
and
F1
agree on the real line
hDyΩ, x∗ Ωi =
=
=
=
φ(yx)
hxΩ, y ∗ Ωi
hJ∆1/2 x∗ Ω, J∆1/2 yΩi
h∆1/2 yΩ, ∆1/2 x∗ Ωi
118
So
∆1/2 yΩ
is in the domain of
∆1/2
and
∆yΩ = DyΩ.
z
agree on AΩ. But multiplication by the function e + 1 is
∞
a linear isomorphism of Cc so by functional calculus (D +1)AΩ = AΩ which
is thus dense. Since D + 1 is invertible by spectral theory, any (ξ, (D + 1)ξ)
Thus
D
and
∆
D + 1 can be approximated by (An Ω, (D + 1)An Ω). Thus D is
AΩ, and both ∆ and D are self-adjoint extensions
restriction of D to this domain. Thus D = ∆. in the graph of
essentially self-adjoint on
of the
119
120
Bibliography
[1] F.M. Goodman, P. de la Harpe, and V.F.R. Jones, Coxeter graphs and
towers of algebras, Springer-Verlag, 1989.
[2] M. Reed and B. Simon, Methods of Modern Mathematical Physics I:
Functional Analysis, Academic Press, 1972.
[3] M. Takesaki, Theory of Operator Algebras II. EMS series on Operator
Algebras, 2002.
[4] J.-B. Bost and A.Connes, Hecke algbebras, type III factors and phase
transitions with spontaneous symmetry breaking in number theory, Selecta Math. (New Series),1, (1995), 411-457.
121