A multiplicative ergodic theorem for von Neumann algebra valued cocycles
aa r X i v : . [ m a t h . OA ] J un A multiplicative ergodic theorem for von Neumannalgebra valued cocycles
Lewis Bowen ∗ , Ben Hayes † and Yuqing (Frank) Lin ‡ June 25, 2020
Abstract
The classical Multiplicative Ergodic Theorem (MET) of Oseledets is generalizedhere to cocycles taking values in a semi-finite von Neumann algebra. This allows for acontinuous Lyapunov distribution.
Keywords : Multiplicative Ergodic Theorem, Oseledets’ Theorem, Lyapunov exponents
MSC : 37H15
Contents ∗ This author acknowledges support from NSF grant DMS-1500389 and a Simons Fellowship. † This author acknowledges support from NSF grant DMS-1827376. ‡ This author acknowledges support from NSF grant DMS-1500389. ( M, τ ) ⊂ L ( M, τ ) . . . . . . . . . . . . . . . . . . . . . . . . . . . 233.4.3 Extending the adjoint . . . . . . . . . . . . . . . . . . . . . . . . . . 233.4.4 Invertible affiliated operators . . . . . . . . . . . . . . . . . . . . . . . 23 P ∞ ( M, τ ) of bounded positive operators . . . . . . . . . . . . . . 285.2 The measure topology . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 302.3 The space P ( M, τ ) of positive log-square integrable operators . . . . . . . . . 355.4 Continuity of the exponential map . . . . . . . . . . . . . . . . . . . . . . . . 405.5 The semi-finite case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
Here is a version of the classical Multiplicative Ergodic Theorem (MET). Let (
X, µ ) bea standard probability space, f : X → X a measure-preserving transformation, and c : N × X → GL( n, R ) a measurable cocycle: c ( n + m, x ) = c ( n, f m x ) c ( m, x ) ∀ n, m ∈ N , µ − a.e. x ∈ X. Assume the first moment condition: Z log + k c (1 , x ) k dµ ( x ) < ∞ , Then there is a limit operator Λ( x ) := lim n →∞ [ c ( n, x ) ∗ c ( n, x )] / n for a.e. x . Let e λ ( x ) > · · · > e λ k ( x ) be the distinct eigenvalues of Λ( x ). Then λ , . . . , λ k are the Lyapunov exponents . They are invariant in the sense that λ i ( f ( x )) = λ i ( x ) fora.e. x . If m i ∈ N is the multiplicity of λ i then the Lyapunov distribution is the discretemeasure P ki =1 m i δ λ i . 3et W i be the e λ i ( x ) -eigenspace of Λ( x ) and define V i = X j ≥ i W j so that V k ( x ) ⊂ · · · ⊂ V ( x ) = R n is a flag. The V i ( x ) are the Oseledets subspaces . Theyare cocycle-invariant in the sense that V i ( f ( x )) = c (1 , x ) V i ( x ) (for a.e. x ).Finally, for a.e. x ∈ X and every vector v ∈ V i ( x ) \ V i +1 ( x ),lim n →∞ n log k c ( n, x ) v k = λ i ( x ) . This last condition can be expressed without reference to Lyapunov exponents by:lim n →∞ k c ( n, x ) v k /n = lim n →∞ k Λ( x ) n v k /n . Infinite-dimensional generalizations of the MET have appeared in [Rue82, Mn83, Blu16,LL10, Thi87, GTQ15, Sch91]. Each of these assumes the operators c ( n, x ) satisfy a quasi-compactness condition and consequently they trivialize away from the discrete part of thespectrum of the limit operators Λ( x ).On the other hand, one does not expect there to be an unconditional generalizationto infinite dimensions. For example, Voiculescu’s example in [HS09, Example 8.4] showsthere is a bounded operator T : ℓ ( N ) → ℓ ( N ) such that | T n | /n does not converge in theStrong Operator Topology. We could define the cocycle c above by c ( n, x ) = T n to see thatconvergence cannot be guaranteed in the general setting of bounded operators on Hilbertspaces. The purpose of this paper is to establish a new MET in which the cocycle takes values inthe group of invertible elements of a semi-finite tracial von Neumann algebra. To explain inmore detail, let H be a separable Hilbert space, B ( H ) the algebra of bounded operators on H . A von Neumann algebra is a sub-algebra M ⊂ B ( H ) containing the identity (I ∈ M )that is closed under taking adjoints and closed in the weak operator topology. Let M + ⊂ M be the positive operators on M . A trace on M is a map τ : M + → [0 , ∞ ] satisfying4. τ ( x + y ) = τ ( x ) + τ ( y ) for all x, y ∈ M + ;2. τ ( λx ) = λτ ( x ) for all λ ∈ [0 , ∞ ), x ∈ M + (agreeing that 0(+ ∞ ) = 0);3. τ ( x ∗ x ) = τ ( xx ∗ ) for all x ∈ M .We will always assume τ is • faithful, which means τ ( x ∗ x ) = 0 ⇒ x = 0; • normal, which means τ (sup i x i ) = sup i τ ( x i ) for every increasing net ( x i ) i in M + ; • semi-finite, which means for every x ∈ M + there exists y ∈ M + such that 0 < y < x and 0 < τ ( y ) < ∞ .The pair ( M, τ ) is a finite tracial von Neumann algebra if τ (I) < ∞ .The trace τ on M is unique (up to scale) if and only if M has trivial center. Many con-structions considered here depend on the choice of trace but we will suppress this dependencefrom the notation and terminology. Fix a standard (semi-finite) measure space (
Y, ν ) and let M = L ∞ ( Y, ν ). For every φ ∈ M ,define the multiplication operator m φ : L ( Y, ν ) → L ( Y, ν ) , ( m φ f )( y ) = φ ( y ) f ( y ) . The map φ m φ embeds M into the algebra of bounded operators on L ( Y, ν ). We willidentify φ with m φ . Define the trace τ : M + → [0 , ∞ ) by τ ( φ ) = Z φ d ν. With this trace, (
M, τ ) is a semi-finite von Neumann algebra. It is finite if ν ( Y ) is finite.5 .5 Example: the full algebra Let M = B ( H ) be the algebra of all bounded operators on a separable Hilbert space H . Alsolet { ξ i } i ∈ I ⊂ H be an orthonormal basis. Define the canonical trace τ H : M + → [0 , ∞ ] by τ H ( a ) = X i ∈ I h aξ i , ξ i i . It is well-known that the canonical trace does not depend on the choice of orthonormal basis.Moreover, ( B ( H ) , τ H ) is semi-finite. The multiplicative ergodic theorem for cocycles takingvalues in this tracial von Neumann algebra was obtained in Karlsson-Margulis [KM99]. Our first main result, Theorem 1.1 shows the existence of a limit operator. The remainingresults, Theorems 1.2-1.4 are derived from Theorem 1.1 in §
6. We state the result here andafterwards explain the notions of convergence and the notation used, such as GL ( M, τ ) and P . Theorem 1.1.
Let ( X, µ ) be a standard probability space, f : X → X an ergodic measure-preserving transformation, ( M, τ ) a von Neumann algebra with semi-finite faithful normaltrace τ . Let M × ⊂ M be the subgroup of elements of M with bounded inverse in M . Let c : N × X → M × ∩ GL ( M, τ ) be a cocycle in the sense that c ( n + m, x ) = c ( n, f m x ) c ( m, x ) for all n, m ∈ N and a.e. x ∈ X . We assume c is measurable with respect to the StrongOperator Topology on M × .Assume the first moment condition: Z X k log( | c (1 , x ) | ) k dµ ( x ) < ∞ . Then for almost every x ∈ X , the following limit exists: lim n →∞ k log( c ( n, x ) ∗ c ( n, x )) k n = D. is called the drift . Moreover, if D > then for a.e. x , there exists a limit operator Λ( x ) ∈ L ( M, τ ) satisfying • lim n →∞ n d P ( | c ( n, x ) | , Λ( x ) n ) = 0 ; • lim n →∞ | c ( n, x ) | /n → Λ( x ) in ( P , d P ) and in measure; • lim n →∞ n − log | c ( n, x ) | → log Λ( x ) in L ( M, τ ) .Remark . The special case of Theorem 1.1 in which (
M, τ ) = ( B ( H ) , τ H ) was proven in[KM99]. Let N = { x ∈ M : τ ( x ∗ x ) < ∞} . The trace induces an inner-product on N by h x, y i := τ ( x ∗ y ) . Let L ( M, τ ) denote the Hilbert space completion of N with respect to this inner product.For x ∈ M , the left-multiplication operator L x : M → M defined by L x ( y ) = xy extendsto a bounded linear operator on L ( M, τ ). Therefore, we may view M as a sub-algebraof the algebra B (L ( M, τ )) of bounded linear operators on L ( M, τ ). This is the regularrepresentation of M (this is explained in more detail in § x on L ( M, τ ) is affiliated with M if it isclosed, densely defined and commutes with every element in the commutant M ′ = { x ∈ B (L ( M, τ )) : xy = yx ∀ y ∈ M } . A subspace is V ⊂ L ( M, τ ) is essentially dense if forevery ǫ > p ∈ M such that τ (I − p ) < ǫ and p L ( M, τ ) ⊂ V .Essentially dense subspaces are reviewed in § M, τ ) is τ -measurable if its domain of definition is essentially dense. Note that when ( M, τ ) isfinite, then all affiliated operators are τ -measurable. Let L ( M, τ ) denote the algebra of τ -measurable operators affiliated with ( M, τ ). This is a ∗ -algebra, and in fact a completetopological ∗ -algebra with respect to the measure topology. Moreover, the trace τ extendsto τ : L ( M, τ ) + → [0 , ∞ ] where L ( M, τ ) + ⊂ L ( M, τ ) is the cone of positive τ -measurableaffiliated operators. Also if x ∈ L ( M, τ ) + then x − / and log x are well-defined via thespectral calculus. See § § ( M, τ ) consist of those elements x ∈ L ( M, τ ) such that log | x | ∈ L ( M, τ ). Weprove in § ( M, τ ) is a group. Let P = GL ( M, τ ) ∩ L ( M, τ ) + . For x, y ∈ P , define d P ( x, y ) = k log( x − / yx − / ) k . We prove in § P , d P ) is a complete CAT(0)metric space on which GL ( M, τ ) acts transitively by isometries. These aforementionedproperties allow us to apply the Karlsson-Margulis theorem, a special case of which is re-produced in § P ∩ M [AL06]. Example . Continuing with our running example, if M = L ∞ ( Y, ν ) thenthe above-mentioned inner product on M is the restriction of the inner product on L ( Y, ν )to M . Therefore, L ( M, τ ) is naturally isomorphic to L ( Y, ν ). The algebra of affiliatedoperators L ( M, τ ) is identified with the algebra of all complex-valued measurable functionson (
Y, ν ) (mod null sets). The exponential map exp : L ( Y, ν ) → GL ( M, τ ) is a surjectivehomomorphism of groups (where we consider L ( Y, ν ) as an abelian group under addition).The kernel consists of all maps φ ∈ L ( Y, ν ) with essential range in 2 πi Z . The restriction ofexp to the real Hilbert space L ( Y, ν ; R ) is an isometry onto ( P , d P ). Example . Suppose (
M, τ ) = ( B ( H ) , τ H ) is as in § N ⊂ M is the algebra of Hilbert-Schmidt operators and L ( M, τ ) = N . A subspace of L ( M, τ ) isessentially dense if and only if it equals L ( M, τ ). This is because every non-zero projectionoperator has trace at least 1 so if τ (I − p ) < p . So L ( M, τ ) = B ( H ). Remark . Let k · k ∞ denote the operator norm. If the cocycle is uniformly bounded inoperator norm (this means there is a constant K such that k c (1 , x ) k ∞ ≤ K for a.e. x ) then k Λ( x ) k ∞ ≤ K as well. Therefore, Λ( x ) ∈ M for a.e. x . Remark . This theorem is a special case of a more general result (Theorem 6.2) whichremoves the restriction of the cocycle to taking values in bounded operators.
Remark . The reader might wonder whether a stronger form of convergence holds in thetheorem above. Namely, whether convergence log Λ( x ) = lim n →∞ log (cid:0) [ c ( n, x ) ∗ c ( n, x )] / n (cid:1) occurs in operator norm. The answer is ‘no’. We provide an explicit example of this in § M = L ∞ ( Y, ν ). 8 onjecture 1.
Assume the hypotheses of Theorem 1.1. If ( M, τ ) is finite then for a.e. x , log (cid:0) [ c ( n, x ) ∗ c ( n, x )] / n (cid:1) converges to log Λ( x ) almost uniformly in the sense of [Pad67] (theequivalent notion of nearly everywhere convergence was first introduced in [Seg53, Defn 2.3]).This means that for every ǫ > and for a.e. x , there exists a closed subspace S ( x ) ⊂ L ( M, τ ) such that the projection operator p S ( x ) satisfies p S ( x ) ∈ M , τ (I − p S ( x ) ) < ǫ and lim n →∞ log (cid:0) [ c ( n, x ) ∗ c ( n, x )] / n (cid:1) p S ( x ) = log(Λ( x )) p S ( x ) where convergence is in operator norm. One of the main advantages of working with a tracial von Neumann algebra (
M, τ ) is thatif x ∈ M is normal (this means xx ∗ = x ∗ x ) then x has a spectral measure. If M = M n ( C ) isthe algebra of n × n complex matrices, then the spectral measure is the uniform probabilitymeasure on the eigenvalues of x (with multiplicity). To define it more generally, recall thatthere is a projection-valued measure E x on the complex plane such that x = R λ d E x ( λ )[Con90, Chapter IX, Theorem 2.2]. The spectral measure of x is the composition µ x = τ ◦ E x . It is a positive measure with total mass equal to τ (I) (where I is the identity operator).Moreover, if p is any polynomial then τ ( p ( x )) = R p dµ x . Example . If M = L ∞ ( Y, ν ) then every operator φ ∈ M is normal. Thespectral measure of φ is its distribution µ φ defined by µ φ ( R ) = ν ( { y ∈ Y : φ ( y ) ∈ R } )for all measurable regions R ⊂ C . Example . Suppose (
M, τ ) = ( B ( H ) , τ H ). Then every normalHilbert-Schmidt operator is unitarily diagonalizable. In particular, there is an orthonor-mal basis of eigenvectors. Therefore, the spectral measure of a normal Hilbert-Schmidtoperator is discrete.This definition of spectral measure extends to x ∈ L ( M, τ ). In the context of Theorem1.1, we define the
Lyapunov distribution to be the spectral measure µ log Λ( x ) of the log9imit operator log Λ( x ). If M = M n ( C ) is the algebra of n × n complex matrices and τ is theusual trace then this definition agrees with the previous definition.To further justify this definition, we recall the notion of von Neumann dimension. If S ⊂ L ( M, τ ) is a closed subspace and the orthogonal projection operator p S lies in M thenthe von Neumann dimension of S is dim M ( S ) = τ ( p S ). For example, the vN-dimensionof L ( M, τ ) itself is τ (I). This notion of dimension satisfies many desirable properties suchas being additive under direct sums and continuous under increasing and decreasing limits[L¨uc02]. Example . If M = L ∞ ( Y, ν ) and if p ∈ M is a projection operator thenthere is a measurable subset Z ⊂ Y such that p is the characteristic function p = 1 Z andthe range of p is the space of all L -functions with support in Z . The vN-dimension of thisspace is the measure ν ( Z ). Example . Suppose (
M, τ ) = ( B ( H ) , τ H ). Then a projection p ∈ M is Hilbert-Schmidt if and only if its range is finite-dimensional. Moreover, the vN-dimensionof a finite-dimensional subspace is its dimension.Let H t ( x ) = 1 ( −∞ ,t ] (log Λ( x ))(L ( M, τ )) ⊂ L ( M, τ )where 1 ( −∞ ,t ] (log Λ( x )) is defined via functional calculus. Alternatively, H t ( x ) is the rangeof the projection E log Λ( x ) ( −∞ , t ]. This is analogous to the Oseledets subspaces definedpreviously. The following theorem is proven is § Theorem 1.2. [Invariance principle] With notation as above, for a.e. x ∈ X and every t ∈ [0 , ∞ ) , c (1 , x ) H t ( x ) = H t ( f ( x )) , µ log Λ( x ) = µ log Λ( f ( x )) . The
Fuglede-Kadison determinant of an arbitrary x ∈ M is defined by∆( x ) = exp (cid:18)Z ∞ log( λ ) dµ | x | ( λ ) (cid:19) where | x | = ( x ∗ x ) / is a positive operator defined via the spectral calculus. The FK-determinant is multiplicative in the sense that ∆( ab ) = ∆( a )∆( b ) [FK52]. From [HS07] it10ollows the definition of FK-determinant extends to operators in GL ( M, τ ) and thereforecan be applied to the limit operator Λ( x ). Example . If M = L ∞ ( Y, ν ) then the FK-determinant of a function φ ∈ M is exp R log | φ ( y ) | d ν ( y ). Example . Suppose (
M, τ ) = ( B ( H ) , τ H ). If H is finite-dimensional,then the FK-determinant is the absolute value of the usual determinant. If H is infinite-dimensional, then the FK-determinant coincides with the absolute value of the Fredholmdeterminant on operators of the form I + a where a ∈ M is trace-class.The following theorem is proven in § Theorem 1.3.
With notation as above, for a.e. x ∈ X , if τ is finite, then lim n →∞ (∆ | c ( n, x ) | ) /n = ∆Λ( x ) . Assume the notation of Theorem 1.1.
Definition 1.
Given ξ ∈ L ( M, τ ), let Σ( ξ ) be the set of all sequences ( ξ n ) n ⊂ L ( M, τ )such lim n →∞ k ξ − ξ n k = 0. Define the upper and lower smooth growth rates of thesystem ( X, µ, f, c ) with respect to ξ at x ∈ X byGr( x | ξ ) = inf n lim inf n →∞ k c ( n, x ) ξ n k /n : ( ξ n ) n ∈ Σ( ξ ) o Gr( x | ξ ) = inf (cid:26) lim sup n →∞ k c ( n, x ) ξ n k /n : ( ξ n ) n ∈ Σ( ξ ) (cid:27) . The following theorem is proven in § Theorem 1.4.
Assume the hypotheses of Theorem 1.1. Then for a.e. x ∈ X and every ξ ∈ L ( M, τ ) , Gr( x | ξ ) = lim n →∞ k Λ( x ) n ξ k /n = Gr( x | ξ ) . Remark . In § n →∞ k c ( n, x ) ξ k /n > lim n →∞ k Λ( x ) n ξ k /n occurs. 11 onjecture 2. If ( M, τ ) is finite then Theorem 1.4 can be strengthened to: for a.e. x ∈ X there exists an essentially dense subspace H x ⊂ L ( M, τ ) such that for every ξ ∈ H x , lim n →∞ k c ( n, x ) ξ k /n = lim n →∞ k Λ( x ) n ξ k /n . Essentially dense subspaces are reviewed in § . In § x ∈ X there exists an essentially dense subspace H x ⊂ L ( M, τ ) such that for every ξ ∈ H x , lim inf n →∞ k c ( n, x ) ξ k /n = lim n →∞ k Λ( x ) n ξ k /n . As in previous examples, suppose M = L ∞ ( Y, ν ). In §
2, we show that with this choiceof (
M, τ ), Theorem 1.1, along with Conjectures 1 and 2, follows readily from Birkhoff’sPointwise Ergodic Theorem. We also provide explicit examples where the limit operatorΛ( x ) has continuous spectrum, where convergence to the limit operator does not occur inoperator norm, and where there exist vectors ξ satisfying the strict inequalitylim inf n →∞ k c ( n, x ) ξ k /n > lim n →∞ k Λ( x ) n ξ k /n . The section § As above, let (
M, τ ) be a finite von Neumann algebra and let T ∈ M . It is a famous openproblem to determine whether T admits a proper invariant subspace. The main results of[HS09] show that the limit lim n →∞ | T n | /n = Λ exists in the Strong Operator Topology (SOT)and moreover, if H t = 1 [0 ,t ] (Λ)(L ( M, τ )) then H t is an invariant subspace. The spectralmeasure of Λ is the same as the Brown measure of T radially projected to the positive realaxis. Moreover, if the Brown measure of T is not a Dirac mass then there exists a properinvariant subspace.Now suppose that T has a bounded inverse T − ∈ M . Regardless of the dynamics, wemay choose to define the cocycle c by c ( n, x ) = T n . Theorems 1.1 and 1.4 then recover12he main results of [HS09] with the exception that our results say nothing of the Brownmeasure and they only apply to the invertible case. Our methods are completely different.In particular, we do not use [HS09]. We will make use of a general Multiplicative Ergodic Theorem due to Karlsson-Margulisbased on non-positive curvature (see also [Kau87] which seems to be the first paper thatdevelops this geometric approach). To accommodate their cocycle convention (which isdifferent from ours), let us say that a measurable map ˇ c : N × X → G is a reverse cocycle if ˇ c ( n + m, x ) = ˇ c ( n, x )ˇ c ( m, f n x )for any n, m ∈ N (where G is a group).The following is a special case of the Karlsson-Margulis Theorem. Theorem 1.5 ([KM99]) . Let ( X, µ ) be a standard probability space, f : X → X an ergodicmeasure-preserving invertible transformation, ( Y, d ) a complete CAT(0) space, y ∈ Y and ˇ c : N × X → Isom(
Y, d ) a measurable reverse cocycle taking values in the isometry groupof ( Y, d ) , where measurable means with respect to the compact-open topology on Isom(
Y, d ) .Assume that Z X d ( y , ˇ c (1 , x ) y ) dµ ( x ) < ∞ . Then for almost every x ∈ X , the following limit exists: lim n →∞ d ( y , ˇ c ( n, x ) y ) n = D. Moreover, if
D > then for almost every x there exists a unique unit-speed geodesic ray γ ( · , x ) in Y starting at y such that lim n →∞ n d ( γ ( Dn, x ) , ˇ c ( n, x ) y ) = 0 . As remarked in [KM99], this result implies the classical MET as follows. Let P ( n, R ) bethe space of positive definite n × n matrices. Then GL( n, R ) acts on P ( n, R ) by g.p := gpg ∗ .13he tangent space to p ∈ P ( n, R ), denoted T p ( P ( n, R )), is naturally identified with S ( n, R ),the space of n × n real symmetric matrices. Define an inner product on T p ( P ( n, R )) by h x, y i p := trace( p − xp − y ) . This gives a complete Riemannian metric on P ( n, R ). All sectional curvatures are non-positive and so P ( n, R ) is CAT(0). Moreover the GL( n, R ) action is isometric and transitive.Every geodesic ray from I (the identity matrix) has the form t exp( tx ) for x ∈ S ( n, R ).Substitute Y = P ( n, R ) and y = I (the identity matrix) in the Karlsson-Margulis The-orem to obtain the classical multiplicative ergodic theorem.Our proof of Theorem 1.1 follows in a similar way from the Karlsson-Margulis Theorem.In [AL06], Andruchow and Larotonda construct a Riemannian metric on the positive cone P ∞ ( M ) of a finite von Neumann algebra. They prove that it is non-positively curved. Wego over the needed facts from their construction in § P ∞ ( M ) is not metrically complete. We prove that its metric completion cannaturally be identified with P , as mentioned earlier in § ( M, τ ) acts transitivelyand is a subgroup of the isometry group of P . This partially answers a question raised in[CL10, Remark 3.21] which asks to identify the metric completion of the space of positivedefinite operators P ∞ ( M ) with respect to the metric d p ( x, y ) = k log( x / y − x / ) k p (1 ≤ p < ∞ ). We obtain a characterization in the special case p = 2. It is possible that our proofcan be modified to handle the general case; we did not attempt it. Acknowledgements . L. Bowen would like thank IPAM and UCLA for their hospitality.The initial ideas for this projects were obtained while L. Bowen was attending the Quanti-tative Linear Algebra semester at IPAM.
As in § M = L ∞ ( Y, ν ) and define the trace τ on M by τ ( φ ) = R φ d ν . This sectionstudies the MET under the hypothesis that the cocycle c takes values in M . It serves asmotivation and can be read independently of the rest of the paper.This special case might seem trivial and indeed, we will see that the conclusions ofTheorem 1.1 are implied by the Pointwise Ergodic Theorem. However, there are curious14eatures not present in previous versions of the MET. Below we will give examples in whichΛ( x ) has continuous spectrum and examples where | c ( n, x ) | /n converges in L -norm to Λ( x )but not in operator norm. We will also show that growth rates do not necessarily exist forevery vector, but do exist for an essentially dense subspace of vectors. Theorem 2.1.
Assume the hypotheses of Theorem 1.1. In addition, let ( Y, ν ) be a standardprobability measure space, M = L ∞ ( Y, ν ) and let the trace τ be given by τ ( φ ) = R φ d ν ( φ ∈ M + ). Also assume that the cocycle is uniformly bounded: ∃ R > such that R − ≤ | c (1 , x )( y ) | ≤ R for a.e. ( x, y ) . Then the conclusion of Theorem 1.1 follows from the Pointwise ErgodicTheorem.Proof. Define F : X × Y → X × Y F ( x, y ) = ( f ( x ) , y ) ,φ ∈ L ( X × Y, µ × ν ) φ ( x, y ) = log | c (1 , x )( y ) | ,A n ( x, y ) ∈ L ( X × Y, µ × ν ) A n ( x, y ) = 1 n n − X k =0 φ ( F k ( x, y )) = 1 n log | c ( n, x )( y ) | A n ( x ) ∈ L ( Y, ν ) A n ( x )( y ) = A n ( x, y ) . Because we assume R k log( | c (1 , x ) | ) k dµ ( X ) < ∞ , it follows that φ ∈ L ( X × Y, µ × ν )as claimed above.The first conclusion of Theorem 1.1 is: for a.e. x ∈ X , k A n ( x ) k converges as n → ∞ . Itis easier to work with the L -norm in place of the L -norm. This is because if | c (1 , x ) | ≥ x then k A n ( x ) k = Z | A n ( x, y ) | d ν ( y ) = 1 n n − X k =0 Z φ ( F k ( x, y )) d ν ( y ) . So the Pointwise Ergodic Theorem applied to x R φ ( x, y ) d ν ( y ) implies k A n ( x ) k convergesfor a.e. x as n → ∞ . A similar argument holds if | c (1 , x ) | < x . By linearity, this15mplies a.e. convergence of k A n ( x ) k in the general case. Because ( Y, ν ) is a finite measurespace and φ ( x, y ) is essentially bounded, L -convergence implies L -convergence. So for a.e. x ∈ X , k A n ( x ) k converges as n → ∞ .The Pointwise Ergodic Theorem implies A n ( x, y ) converges for a.e. ( x, y ) as n → ∞ .So Fubini’s Theorem implies that: for a.e. x , A n ( x ) converges pointwise a.e. as n → ∞ .Scheffe’s Lemma now implies that for a.e. x , A n ( x ) converges in L ( Y, ν ). This proves thelast conclusion of Theorem 1.1.Let log Λ( x ) denote the limit of A n ( x ). As explained in Example 1,1 n d P ( | c ( n, x ) | , Λ( x ) n ) = 1 n k log( | c ( n, x ) | ) − log(Λ( x ) n ) k = (cid:13)(cid:13)(cid:13)(cid:13) n log( | c ( n, x ) | ) − log(Λ( x )) (cid:13)(cid:13)(cid:13)(cid:13) . Thus n d P ( | c ( n, x ) | , Λ( x ) n ) → n → ∞ .Similarly, d P ( | c ( n, x ) | /n , Λ( x )) = (cid:13)(cid:13)(cid:13)(cid:13) n log( | c ( n, x ) | ) − log(Λ( x )) (cid:13)(cid:13)(cid:13)(cid:13) . So | c ( n, x ) | /n converges to Λ( x ) in ( P , d P ) as n → ∞ (for a.e. x ).To prove | c ( n, x ) | /n converges to Λ( x ) in measure, it suffices to show: for every ǫ > ν (cid:0)(cid:8) y ∈ Y : (cid:12)(cid:12) | c ( n, x )( y ) | /n − Λ( x )( y ) (cid:12)(cid:12) > ǫ (cid:9)(cid:1) tends to zero as n → ∞ (for a.e. x ). This is implied by the fact that n − log | c ( n, x ) | converges to log Λ( x ) in L ( Y, ν ) for a.e. x .Similarly, if M = M n ( C ) ⊗ L ∞ ( Y, ν ) where M n ( C ) denotes the algebra of n × n com-plex matrices, then the non-ergodic version of the classical Multiplicative Ergodic Theoremimplies the conclusions of Theorem 1.1. This example is almost trivial. Let ψ ∈ L ∞ ( Y, ν ) be such that log | ψ | ∈ L ( Y, ν ). Define c ( n, x ) = ψ n . Then the limit operator satisfies Λ( x ) = | ψ | for a.e. x and the spectral measureof Λ is the distribution of | ψ | . In particular, if | ψ | has continuous distribution then Λ( x ) hascontinuous spectrum. 16 .3 Almost uniform convergence and growth rates In this subsection, we prove Conjectures 1 and 2 in the special case M = L ∞ ( Y, ν ) and (
Y, ν )is a probability space.
Theorem 2.2.
Assume hypotheses as in Theorem 1.1. In addition, let ( Y, ν ) be a standardprobability space, M = L ∞ ( Y, ν ) and let the trace τ be given by τ ( φ ) = R φ d ν . In thissetting, Conjecture 1 is true.Proof. Define
F, φ and A n as in § A n ( x, y ) convergesto log Λ( x )( y ) for a.e. ( x, y ). By Fubini’s Theorem, there exists a subset X ′ ⊂ X withfull measure such that for a.e. x ∈ X ′ , A n ( x ) converges pointwise a.e. (as n → ∞ ) tolog Λ( x ). Let ǫ >
0. By Egorov’s Theorem, for every x ∈ X ′ there exists a measurablesubset Z ( x ) ⊂ Y with ν ( Z ( x )) > − ǫ such that A n ( x ) converges uniformly to log Λ( x ) on Z ( x ).Let S ( x ) ⊂ L ( Y, ν ) be the closed subspace of functions that equal zero off of Z ( x ).The projection operator p S ( x ) is identified with the characteristic function 1 Z ( x ) ∈ L ∞ ( Y, ν ).Moreover, τ (I − p S ( x ) ) = ν ( Y \ Z ( x )) < ǫ . Because A n ( x ) = n − log | c ( n, x ) | convergesuniformly to log Λ( x ) on Z ( x ), it follows thatlim n →∞ n − log | c ( n, x ) | p S ( x ) = log(Λ( x )) p S ( x ) in operator norm. Proposition 2.3.
We assume the same hypotheses as Theorem 2.2. In this setting, Con-jecture 2 is true.Proof.
Let ξ ∈ L ( Y, ν ). We first prove lim n →∞ k Λ( x ) n ξ k /n = k Λ( x )1 support ( ξ ) k ∞ .Without loss of generality, we may assume k ξ k = 1. It is a standard exercise thatlim n →∞ k φ k n = k φ k ∞ for φ ∈ L ∞ . So k Λ( x ) n ξ k /n tends to k Λ( x ) k L ∞ ( Y, | ξ | d ν ) as n → ∞ .The latter is the same as k Λ( x )1 support ( ξ ) k L ∞ ( Y,ν ) .By the proof of Theorem 2.2, for every r ∈ N and a.e. x , there exists a measurable subset Z r ( x ) ⊂ Y such that ν ( Z r ( x )) > − /r and n − log | c ( n, x ) ↾ Z r ( x ) | converges uniformly tolog Λ( x ) ↾ Z r ( x ) as n → ∞ (where ↾ means “restricted to”).17et S r ( x ) ⊂ L ( Y, ν ) be the subspace of vectors ξ such that ξ ( y ) = 0 for a.e. y ∈ Y \ Z r ( x ).Let S ( x ) = ∪ r ∈ N S r ( x ) ⊂ L ( Y, ν ). Because ν ( Z r ( x )) > − /r for all r , S ( x ) is essentiallydense.Let ξ ∈ S r ( x ). Then k c ( n, x ) ξ k ≤ k Λ( x ) n ξ k (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) y ∈ Z r ( x )
7→ | c ( n, x )( y ) | Λ( x ) n ( y ) (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) L ∞ ( Z r ( x ) ,ν ) . Since n − log | c ( n, x ) ↾ Z r ( x ) | converges to log Λ( x ) uniformly on Z r ( x ), this implieslim sup n →∞ k c ( n, x ) ξ k /n ≤ lim n →∞ k Λ( x ) n ξ k /n . Similarly, k Λ( x ) n ξ k ≤ k c ( n, x ) ξ k (cid:13)(cid:13)(cid:13)(cid:13)(cid:18) y ∈ Z r ( x ) Λ( x ) n ( y ) | c ( n, x )( y ) | (cid:19)(cid:13)(cid:13)(cid:13)(cid:13) L ∞ ( Z r ( x ) ,ν ) . So lim inf n →∞ k c ( n, x ) ξ k /n ≥ lim n →∞ k Λ( x ) n ξ k /n . This proveslim n →∞ k c ( n, x ) ξ k /n = lim n →∞ k Λ( x ) n ξ k /n = k Λ( x )1 support ( ξ ) k ∞ for all vectors ξ ∈ S r ( x ). Because r ∈ N is arbitrary, the above limits hold for all ξ ∈ S ( x ).Because S ( x ) is essentially dense, this implies Conjecture 2. Remark . The same result holds if M = M n ( C ) ⊗ L ∞ ( Y, ν ) with essentially the same proof.One needs only use the non-ergodic version of Oseledet’s Multiplicative Ergodic Theoreminstead of Birkhoff’s Pointwise Ergodic Theorem.
Theorem 2.4.
There exist standard probability spaces ( X, µ ) , ( Y, ν ) , an ergodic pmp invert-ible transformation f : X → X , a measurable cocycle c : Z × X → M = L ∞ ( Y, ν ) satisfyingthe hypotheses of Theorem 1.1 and a vector ξ ∈ L ( Y, ν ) such that lim n →∞ k c ( n, x ) ξ k /n L ( Y,ν ) = lim n →∞ k c ( n, x ) k /n L ( Y,ν ) > lim n →∞ k Λ( x ) n ξ k /n L ( Y,ν ) = lim n →∞ k Λ( x ) n k /n L ( Y,ν ) . Moreover, we can choose the cocycle so that k c (1 , x ) k ∞ ≤ C for some constant C and a.e. x . Moreover, n − log | c ( n, x ) | does not converge to log Λ( x ) in operator norm (for a.e. x ). roof. Let X = Z be the compact group of 2-adic integers. An element of Z is writtenas a formal sum x = P ∞ i =0 x i i with x i ∈ { , } and the usual multiplication and additionrules. Let µ be the Haar probability measure on X . There is a bijection between X and { , } N ∪{ } given by x ( x , x , . . . ). This bijection maps the measure µ to the ( N ∪ { } )-thpower of the uniform measure on { , } .Define f : X → X by f ( x ) = x + 1. It is well-known that a translation on a compactabelian group is ergodic if and only if every orbit is dense. Thus f is an ergodic measure-preserving transformation. Alternatively, f is the standard odometer which is well-knownto be ergodic.Let ( Y, ν ) be a probability space that is isomorphic to the unit interval with Lebesguemeasure. Let Y = ⊔ ∞ n =1 Y n be a partition of Y into positive measure subsets. We will choosethe partition more carefully later. Define the cocycle c : Z × X → L ∞ ( Y, ν ) by c (1 , x )( y ) = y ∈ Y m for some m and x m = 02 otherwiseThis extends to a cocycle via c ( n, x ) = c (1 , f n − x ) · · · c (1 , f ( x )) c (1 , x ).For every y ∈ Y , Z log c (1 , x )( y ) dµ ( x ) = (1 /
2) log(2) . Since f is ergodic, it follows that the limit operator Λ( x ) defined by log Λ( x ) = lim n →∞ n − log c ( n, x )(where convergence is in L and pointwise a.e.) is the constant function Λ( x ) = √ x . For n, m ∈ N , let S n,m = { x ∈ X : x m = 1 , x n = 0 } . We claim that if x ∈ S n,m and n < m , 0 ≤ l ≤ n + 1, then c ( l, x )( y ) = 2 l ∀ y ∈ Y m .Indeed, note that if x ∈ S n,m the smallest k such that c (1 , f k x )( y ) = 1 must be when k + P mi =0 x i i = 2 m +1 . But 2 m ≤ m X i =0 x i i ≤ m +1 − n − . Therefore, c (1 , f k x )( y ) = c (1 , x + k )( y ) = 2 for all 0 ≤ k ≤ n and c ( l, x ) = c (1 , f l − x ) · · · c (1 , x ) = 2 l . µ ( S n,m ) = 1 /
4. Moreover, if n = n and m = m then S n ,m and S n ,m are independent events. It follows that if T n = S n,n +10 then the events { T n } ∞ n =1 are jointlyindependent and, by Borel-Cantelli, a.e. x is contained in infinitely many of the sets T n .If x ∈ T n then for 0 ≤ l ≤ n + 1 k c ( l, x )1 Y k ( Y,ν ) ≥ k c ( l, x )1 Y n +10 k ( Y,ν ) ≥ ν ( Y n +10 )2 l . We could choose the subsets { Y m } so that ν ( Y m ) ≥ Cm − for some constant C . Withthis choice and x ∈ T n , k c ( l, x )1 Y k ( Y,ν ) ≥ C l / ( n + 10) − . Since a.e. x is contained in infinitely many T n ’s it follows thatlim n →∞ k c ( n, x )1 Y k /n L ( Y,ν ) = 2 . On the other hand, lim n →∞ k Λ( x ) n Y k /n L ( Y,ν ) = √ . This proves the theorem with ξ = 1 Y . By Theorem 2.2, | c ( n, x ) | /n does not converge toΛ( x ) in operator norm (for a.e. x ). Remark . The essential phenomena behind this counterexample is that there is no uniformrate of convergence in the Pointwise Ergodic Theorem. Precisely, while n P n − k =0 log c (1 , f k x )( y )converges to log(2) / y and a.e. x , the convergence is not uniform in y . Throughout these notes, by a tracial von Neumann algebra we mean a pair (
M, τ ) where M is a von Neumann algebra, and τ is a faithful, normal, tracial, state. By a semi-finitevon Neumann algebra we mean a pair ( M, τ ) where M is a von Neumann algebra and τ isa faithful, normal, and semi-finite trace. Throughout, we assume M is a sub-algebra of thealgebra B ( H ) of all bounded operators on a separable Hilbert space H . This implies ( M, τ )has separable pre-dual. We will consider many constructions that depend on the choice oftrace τ but we suppress this dependence from the notation.20 .1 Spectral measures Suppose x is a (bounded or unbounded) self-adjoint operator on H . By the Spectral Theorem([RS80, Theorem VIII.6]), there exists a projection valued measure E x on the real line suchthat x = Z λ d E x ( λ ) . The support of E x is contained in the spectrum of x . The projections of the form E x ( R ) (forBorel sets R ⊂ R ) are the spectral projections of x . If f : R → R is Borel then f ( x ) is aself-adjoint operator on H defined by f ( x ) := Z f ( λ ) d E x ( λ ) . In the case of unbounded x , f ( x ) has the same domain as x . The absolute value of x isdefined by | x | = ( x ∗ x ) / = R √ λ d E x ∗ x ( λ ) and is equal to R | λ | d E x ( λ ).If x is such that all of its spectral projections lie in the von Neumann algebra M , thenthe composition τ ◦ E x is a Borel probability measure on C called the spectral measureof x and denoted by µ x . In particular, if x ∈ M then µ x is well-defined. Example . If M = L ∞ ( Y, ν ) (as in § φ ∈ M is self-adjoint if and only if it is real-valued. The projection-valued measure E φ satisfies: E φ ( R ) isthe projection onto the subspace of L -functions with support in φ − ( R ) (for Borel R ⊂ R ).Moreover, µ φ = φ ∗ ν is the distribution of φ . We will frequently have to use the polar decomposition, see [RS80, Theorem VIII.32]. Werestate it here.
Proposition 3.1.
Let x be a closed densely defined operator on H . Then there is a positiveself-adjoint operator | x | with dom( | x | ) = dom( x ) and a partial isometry u with initial space ker( x ) ⊥ and final space Im( x ) so that x = u | x | (where Im( x ) denotes the image of x ).Moreover | x | and u are uniquely determined by these properties together with the additionalcondition ker( | x | ) = ker( x ) . The expression x = u | x | is called the polar decomposition of x .21 .3 The regular representation For the remainder of this section, fix semi-finite von Neumann algebra (
M, τ ) . Recall from theintroduction that L ( M, τ ) is the Hilbert space completion of N = { x ∈ M : τ ( x ∗ x ) < ∞} with respect to the inner product defined on N by h x, y i = τ ( x ∗ y ) . Let k x k = h x, x i / and k x k ∞ be the operator norm of x (as an operator on H ).For any x, y ∈ M , k xy k ≤ k x k ∞ k y k and k xy k ≤ k x k k y k ∞ . (e.g., [Tak02, V.2, equation (8)]). Therefore, the operator L x : N → N defined by L x ( y ) = xy admits a unique continuous extension from L ( M, τ ) to itself. Moreover, the operator normof L x is bounded by k x k ∞ . In fact, they are equal. This follows, for example, from [Tak02,Corollary I.5.4 and Theorem V.2.22]. Similarly, the map R x : N → N defined by R x ( y ) = yx admits a unique continuous extension to L ( M, τ ) and the operator norm of R x is k x k ∞ .We will identify M with its image { L x : x ∈ M } (viewed as a sub-algebra of the algebraof bounded operators on L ( M, τ )).
Definition 2 (L ( M, τ )) . The commutant of M , denoted M ′ , is the algebra of boundedoperators y on L ( M, τ ) such that xy = yx for all x ∈ M . An unbounded operator x onL ( M, τ ) is affiliated with M if for every unitary u ∈ M ′ , xu = ux . By [Dix81, ChapterI.1 Exercise 10] or [AP16, Proposition 7.2.3], if x is a closed densely defined operator and x = u | x | is its polar decomposition, then x is affiliated with M if and only if u and thespectral projections of | x | are in M . A subspace is V ⊂ L ( M, τ ) is essentially dense iffor every ǫ > p ∈ M such that τ (I − p ) < ǫ and p L ( M, τ ) ⊂ V .Essentially dense subspaces are reviewed in § M, τ ) is τ -measurable if its domain of definition is essentially dense. Note that when( M, τ ) is finite, then all affiliated operators are τ -measurable. Let L ( M, τ ) denote the setof τ -measurable operators. By [Tak03, Theorems IX.2.2, IX.2.5], L ( M, τ ) is closed underadjoint, addition and multiplication and is a ∗ -algebra under these operations.22 .4.1 Domains For x ∈ L ( M, τ ) we write dom( x ) ⊂ L ( M, τ ) for its domain. We remark now that for a, b ∈ L ( M, τ ) the sum a + b is defined as the closure of the operator T with dom( T ) =dom( a ) ∩ dom( b ) and with T ξ = aξ + bξ for ξ ∈ dom( T ) . Similarly ab is defined as the closure of the operator T with domain b − (dom( a )) ∩ dom( b ) and T ξ = a ( bξ ) for ξ ∈ dom( T ) . Thus, for example, the domain of ab is often larger than b − (dom( a )) ∩ dom( b ) . This willoccasionally cause us some headaches, and we will try to remark when it actually presentsan issue. Regardless, this paragraph should be taken as a blanket warning that ab is notliterally defined to be the composition, and a + b is not the literal sum. ( M, τ ) ⊂ L ( M, τ )We can include L ( M, τ ) in L ( M, τ ) as follows. For x ∈ L ( M, τ ) and y ∈ M , define L x ( y ) = R y ( x ) = xy . Then L x is closable but not bounded in general. Let L x denote theclosure of L x . The map x L x defines a linear bijection from L ( M, τ ) into L ( M, τ ). Byabuse of notation, we will identify L ( M, τ ) with its image in L ( M, τ ). While L ( M, τ ) is asubspace of L ( M, τ ), it is not a sub-algebra in general.For x ∈ L ( M, τ ) we set | x | = ( x ∗ x ) / and k x k = (cid:18)Z t dµ | x | ( t ) (cid:19) / ∈ [0 , ∞ ] . Then L ( M, τ ) is identified with the set of all x ∈ L ( M, τ ) which have k x k < ∞ . The anti-linear map x x ∗ on M uniquely extends to an anti-linear isometry J : L ( M, τ ) → L ( M, τ ). By [AP16, Proposition 7.3.3], if x ∈ L ( M, τ ) then the following are equivalent:(1) x is self-adjoint, (2) J x = x , (3) x is in the L -closure of M sa = { y ∈ M : y ∗ = y } . LetL ( M, τ ) sa = { x ∈ L ( M, τ ) :
J x = x } . We say an operator x ∈ L ( M, τ ) is invertible if there exists an operator y ∈ L ( M, τ )such that xy = yx = I where, following our abuse of notation, xy and yx denote the23losures of the compositions of the operators x and y . In this case we write y = x − . LetL ( M, τ ) × ⊂ L ( M, τ ) be the set of invertible affiliated operators x . Lemma 3.2. If ( M, τ ) is semi-finite and x ∈ L ( M, τ ) × has polar decomposition x = u | x | then u is unitary, | x | ∈ L ( M, τ ) × and x ∗ ∈ L ( M, τ ) × with ( x ∗ ) − = ( x − ) ∗ . If ( M, τ ) isfinite then x ∈ L ( M, τ ) is invertible if and only if it is injective.Proof. Because u is a partial isometry, u ∗ u is the orthogonal projection onto ker( u ) ⊥ . If x is invertible, then u is injective, so u ∗ u = I. Similarly, uu ∗ is projection onto the closure ofthe image of x . So if x is invertible then uu ∗ = I. This proves u is unitary.Because x is injective, the equality k xξ k = k| x | ξ k for ξ ∈ dom( x ) implies that | x | isinjective. Thus 1 { } ( | x | ) = p ker( | x | ) = 0 , and so | x | − may be defined as a closeable operatorin L ( M, τ ) . The computation ( x ∗ ) − = ( x − ) ∗ is straightforward.Now suppose ( M, τ ) is finite. Without loss of generality, τ (I) = 1. Suppose x is injective.Because u also injective, u ∗ u = I. Thus 1 = τ ( u ∗ u ) = τ ( u ∗ u ). Because the only projectionwith trace 1 is I (by faithfulness of τ ), u ∗ u = I and u is unitary. As above, | x | is invertible.So | x | − u ∗ is an inverse to x . Given a semi-finite von Neumann algebra (
M, τ ), letGL ( M, τ ) = { a ∈ L ( M, τ ) × : log( | a | ) ∈ L ( M, τ ) } be the log-square integrable general linear group of ( M, τ ). For brevity we will write G = GL ( M, τ ). Although we call this set a group, it is not at all obvious that G is closedunder multiplication. The main result of this section is: Theorem 4.1. G is a subgroup of L ( M, τ ) × . Moreover, for every a ∈ G we have that a ∗ ∈ G and additionally: k log( | a | ) k = k log( | a ∗ | ) k = k log( | a − | ) k . We start with some basic facts about spectral measures.24 roposition 4.2.
Let ( M, τ ) be a semi-finite von Neumann algebra and a ∈ L ( M, τ ) × . Then:1. µ | a | = µ | a ∗ | , µ | a − | = µ | a | − = r ∗ ( µ | a | ) , where r : (0 , ∞ ) → (0 , ∞ ) is the map r ( t ) = t − . Proof. (1): Let a = u | a | be the polar decomposition. Since a ∈ L ( M, τ ) × by Lemma 3.2 wehave that u ∈ U ( M ) (which is the unitary group of M ). Then a ∗ = | a | u ∗ , and | a ∗ | = aa ∗ = u | a | u ∗ . From this, it is easy to see that | a ∗ | = u | a | u ∗ , because ( u | a | u ∗ ) = u | a | u ∗ Thus, forevery Borel E ⊆ [0 , ∞ ) µ | a ∗ | ( E ) = τ (1 E ( | a ∗ | )) = τ (1 E ( u | a | u ∗ )) = τ ( u E ( | a | ) u ∗ ) = τ (1 E ( | a | )) = µ | a | ( E ) . (2): Again, let a = u | a | be the polar decomposition. So a − = | a | − u ∗ . As in (1), it isdirect to show that | a − | = u | a | − u ∗ . The proof then proceeds exactly as in (1), using that r ∗ ( µ | a | ) = µ | a | − (which follows from functional calculus).Because expressions like 1 ( λ, ∞ ) ( | a | )(L ( M, τ )) will show up frequently, it will be helpfulto introduce the following notation. Given a ∈ L ( M, τ ) and E ⊆ [0 , ∞ ) Borel, we let H aE =1 E ( | a | )(L ( M, τ )) . It will be helpful for us to derive an alternate expression for k log( | a | ) k . Note k log( | a | ) k = Z ∞ t d µ | log( | a | ) | ( t )= Z ∞ (cid:18)Z t λ d λ (cid:19) d µ | log( | a | ) | ( t )= 2 Z ∞ λ Z ∞ ( λ, ∞ ) ( t ) d µ | log( | a | ) | ( t ) d λ = 2 Z ∞ λµ | log( | a | ) | ( λ, ∞ ) d λ. (1)Note that we have used Fubini’s Theorem. This is valid if µ | log( | a | ) | ( λ, ∞ ) < ∞ for all λ > µ | log( | a | ) | is sigma-finite. On the other hand, if µ | log( | a | ) | ( λ, ∞ ) = ∞ for some λ > µ | log( | a | ) | is the pushforward of µ | a | under the map t
7→ | log( t ) | . So k log( | a | ) k = 2 Z ∞ λ (cid:2) µ | a | ( e λ , ∞ ) + µ | a | (0 , e − λ ) (cid:3) d λ.
25y Proposition 4.2, k log( | a | ) k = 2 Z ∞ λ [ µ | a | ( e λ , ∞ ) + µ | a − | ( e λ , ∞ )] d λ. (2) Proposition 4.3.
Let ( M, τ ) be a semi-finite von Neumann algebra and a, b ∈ L ( M, τ ) . Given λ , λ , λ ∈ (0 , ∞ ) with λ λ = λ, we have that τ (1 ( λ, ∞ ) ( | ab | )) ≤ τ (1 ( λ , ∞ ) ( | a | )) + τ (1 ( λ , ∞ ) ( | b | )) . Proof.
The proof is almost immediate from [FK86, Lemma 2.5 (vii) and Proposition 2.2]. Toexplain, for a ∈ L ( M, τ ), let e µ t ( a ) be the infimum of k ap k ∞ over all projections p ∈ M with τ (1 − p ) ≤ t . This is the t -th generalized s -number of a . Also let e λ t ( a ) = τ (1 ( t, ∞ ) ( | a | )).So t e λ t ( a ) is the distribution function of a . These invariants are related by [FK86,Proposition 2.2] which states e µ t ( a ) = inf { s ≥ e λ s ( a ) ≤ t } . It follows that e µ t ( a ) ≤ s if and only if e λ s ( a ) ≤ t . In particular, e µ e λ t ( a ) ( a ) ≤ t always holds.[FK86, Lemma 2.5 (vii)] states e µ t + s ( ab ) ≤ e µ t ( a ) e µ s ( b )for any t, s > a, b ∈ L ( M, τ ).Now that the tools above are ready, we return to the proposition we want to prove. Theinequality τ (1 ( λ, ∞ ) ( | ab | )) ≤ τ (1 ( λ , ∞ ) ( | a | )) + τ (1 ( λ , ∞ ) ( | b | )) is equivalent to the statement e λ ts ( ab ) ≤ e λ t ( a ) + e λ s ( b ) . By [FK86, Proposition 2.2], this is true if and only if e µ e λ t ( a )+ e λ s ( b ) ( ab ) ≤ ts. By [FK86, Lemma 2.5 (vii)], e µ e λ t ( a )+ e λ s ( b ) ( ab ) ≤ e µ e λ t ( a ) ( a ) e µ e λ t ( b ) ( b ) . By [FK86, Proposition 2.2] again, e µ e λ t ( a ) ( a ) ≤ t and e µ e λ s ( b ) ( b ) ≤ s . Combining these inequali-ties proves the proposition. 26 roof of Theorem 4.1. The fact that G is closed under inverses and the ∗ operation is obviousfrom Proposition 4.2. Let a, b ∈ G. By Proposition 4.3: k log( | ab | ) k = 2 Z ∞ λ (cid:2) µ | ab | ( e λ , ∞ ) + µ | b − a − | ( e λ , ∞ ) (cid:3) d λ ≤ Z ∞ λ (cid:2) µ | a | ( e λ/ , ∞ ) + µ | a − | ( e λ/ , ∞ ) (cid:3) d λ + 2 Z ∞ λ (cid:2) µ | b | ( e λ/ , ∞ ) + µ | b − | ( e λ/ , ∞ ) (cid:3) d λ = 4 Z ∞ λ (cid:2) µ | a | ( e t , ∞ ) + µ | a − | ( e t , ∞ ) (cid:3) d t + 4 Z ∞ λ (cid:2) µ | b | ( e t , ∞ ) + µ | b − | ( e t , ∞ ) (cid:3) d µ = 2( k log( | a | ) k + k log( | b | ) k ) . We also need the following fact analogous to Proposition 4.3, but whose proof is easier.
Proposition 4.4.
Let ( M, τ ) be a semi-finite von Neumann algebra, and a, b ∈ L ( M, τ ) . Then for all λ , λ > we have that µ | a + b | ( λ + λ , ∞ ) ≤ µ | a | ( λ , ∞ ) + µ | b | ( λ , ∞ ) . Proof.
We use [FK86, Lemma 2.5, (v)], which shows that e µ µ | a | ( λ , ∞ )+ µ | b | ( λ , ∞ ) ( a + b ) ≤ e µ µ | a | ( λ , ∞ ) ( a ) + e µ µ | b | ( λ , ∞ ) ( b ) ≤ λ + λ . Apply [FK86, Proposition 2.2] to the inequality above to obtain e λ λ + λ ( | a + b | ) ≤ µ | a | ( λ , ∞ ) + µ | b | ( λ , ∞ ) . By definition of e λ , e λ λ + λ ( | a + b | ) = µ | a + b | ( λ + λ , ∞ ) so this finishes the proof. Let P = P ( M, τ ) = { x ∈ GL ( M, τ ) : x > } be the positive definite elements of GL ( M, τ ).In § P ∩ M . In § ( M, τ ). By approximating P by P ∩ M (in themeasure topology), we show in § d P (as defined in the introduction) is a metric on P .27oreover, GL ( M, τ ) acts transitively and by isometries on ( P , d P ). In § ( M, τ ) sa → P is a homeomorphism. From this, we concludethat ( P , d P ) is a complete CAT(0) metric space and characterize its geodesics. In § P ∞ ( M, τ ) of bounded positive operators Throughout this section, we assume (
M, τ ) is finite.Let M sa ⊂ M be the subspace of self-adjoint elements and P ∞ = P ∞ ( M, τ ) := { exp( x ) : x ∈ M sa } ⊂ M sa be the positive definite elements with bounded inverse. Note P ∞ = P ∩ M × . This sectionstudies P ∞ equipped with a natural metric, as introduced in [AL06]. The results in thissection are obtained directly from [AL06].Let GL ∞ ( M, τ ) = M × be the group of elements x ∈ M such that x has a bounded inverse x − in M . This group acts on M sa by g.w := gwg ∗ ( ∀ g ∈ GL ∞ ( M, τ ) , w ∈ M sa ) . For w ∈ P ∞ , the tangent space to P ∞ at w , denoted T w ( P ∞ ), is a copy of M sa withthe inner product h· , ·i w defined by h x, y i w := h w − / .x, w − / .y i = τ ( w − / x ∗ w − yw − / ) = τ ( w − xw − y ) . Let k · k w, denote the L -norm with respect to this inner product. In the special case that w = I is the identity, this is just the restriction of the standard inner product to M sa .These inner products induce a Riemannian metric on P ∞ ( M, τ ). The reader might beconcerned that the tangent spaces T w ( P ∞ ) are not complete with respect to their innerproducts. This causes no difficulty in defining the metric on P ∞ but it does mean thatTheorem 1.5 cannot be directly applied to P ∞ .Here is a more detailed explanation of the metric. Let γ : [ a, b ] → P ∞ be a path. TheL -derivative of γ at t is defined by γ ′ ( t ) = lim h → γ ( t + h ) − γ ( t ) h -metric on T γ ( t ) ( P ∞ ). Then the length of γ isdefined as in the finite-dimensional case:length P ( γ ) = Z ba k γ ′ ( t ) k γ ( t ) , dt. Define distance on P ∞ ( M, τ ) by d P ( x, y ) = inf γ length P ( γ ) where the infimum is taken overall piece-wise smooth curves γ with derivatives in M . For this to be well-defined, it needs tobe shown that there exists a piecewise smooth curve between any two points of P ∞ . For anyexp( x ) ∈ P ∞ , the map t exp( tx ) defines a smooth curve from I to exp( x ). A piecewisesmooth curve between any two points can be obtained by concatenating two of these specialcurves. Lemma 5.1.
The action of GL ∞ ( M, τ ) on P ∞ is transitive and by isometries.Proof. The action of GL ∞ ( M, τ ) on P ∞ is by isometries since the Frechet derivative of g at w is the map x ∈ T w ( P ∞ ) g.x = gxg ∗ ∈ T g.w ( P ∞ )and h g.x, g.y i g.w = τ (( g.w ) − ( g.x )( g.w ) − ( g.y ))= τ (( g ∗ ) − w − g − ( gxg ∗ )( g ∗ ) − w − g − ( gyg ∗ ))= τ ( w − xw − y ) = h x, y i w . The action GL ∞ ( M, τ ) y P ∞ is transitive since for any w ∈ P ∞ , w / ∈ GL ∞ ( M, τ ) and w / . I = w. Lemma 5.2. [AL06, Lemma 3.5] For any a, b ∈ P ∞ , d P ( a, b ) = k log( b − / ab − / ) k ≥ k log( a ) − log( b ) k . Theorem 5.3. P ∞ ( M, τ ) is a CAT(0) space.Proof. This follows from [AL06, Lemma 3.6] and [BH99, Chapter II.1, Proposition 1.7 (3)].29 orollary 5.4.
Let x, y ∈ M sa and σ ≥ be a scalar. Then d P ( e σx , e σy ) ≥ σd P ( e x , e y ) . Proof.
Let x ′ , y ′ ∈ M sa and let f ( t ) = d P ( e tx ′ , e ty ′ ). By [AL06, Corollary 3.4], f is convex.Therefore, f ( t ) ≤ tf (1) + (1 − t ) f (0) = tf (1)for any 0 ≤ t ≤
1. Set t = 1 /σ , x ′ = σx , y ′ = σy to obtain f ( t ) = d P ( e x , e y ) ≤ σ − d P ( e σx , e σy ) . This section reviews the measure topology on L ( M, τ ). The results here are probably well-known but being unable to find them explicitly stated in the literature, we give proofs forcompleteness. We will need this material in the next two sections.Let (
M, τ ) be a semi-finite von Neumann algebra. By [Tak03, Theorem IX.2.2], the sets O ε,δ ( a ) = (cid:8) b ∈ L ( M, τ ) : τ (1 ( ε, ∞ ) ( | a − b | )) < δ (cid:9) ranging over a ∈ L ( M, τ ) and ε, δ > ( M, τ ) , and this topology turns L ( M, τ ) into a topological ∗ -algebra (i.e. the productand sum operations are continuous as a function of two variables, as is the adjoint). Weshall call this topology the measure topology . This motivates the following definition. Definition 3.
Let (
M, τ ) be a semi-finite von Neumann algebra. Given a sequence ( a n ) n inL ( M, τ ) , and an a ∈ L ( M, τ ) we say that a n → a in measure if for every ε > τ (1 ( ε, ∞ ) ( | a − a n | )) → n →∞ . Lemma 5.5.
Let ( M, τ ) be a semi-finite von Neumann algebra. Suppose x , x , . . . ∈ L ( M, τ ) and lim n →∞ x n = x ∞ in L ( M, τ ) . Then x n → n →∞ x ∞ in measure. roof. For any ǫ > τ (1 ( ǫ, ∞ ) ( | x n − x | )) ǫ ≤ k x n − x k . Since x n → x in L ( M, τ ), this shows x n → x in measure. Proposition 5.6.
Let ( M, τ ) be a von Neumann algebra with a finite trace. Let C > andlet M C ⊂ M be the set of all elements x with k x k ∞ ≤ C . Then the measure, strong operatorand L topologies all coincide on M C . By Strong Operator Topology we mean with respect toeither of the inclusions M ⊂ B ( H ) or M ⊂ B ( L ( M, τ )) .Proof. By [Dix81, I.4.3. Theorem 2] or [AP16, Corollary 2.5.9 and Proposition 2.5.8], thetopology induced on M C from the SOT on B ( H ) is the same as the topology it inherits fromthe SOT on B (L ( M, τ )).Let x ∈ M C and let ( x n ) n ⊂ M C be a sequence. We will show that if x n → x in one ofthe three topologies then x n → x in the other topologies. After replacing x n with x n − x and C with 2 C if necessary, we may assume x = 0. Also without loss of generality we mayassume τ (I) = 1.Suppose that x n → x n → . For any ε > k x n k = τ ( x ∗ n x n ) ≤ C τ (1 ( ε, ∞ ) ( | x n | )) + ε . Since x n → n →∞ k x n k ≤ ε . Since ε is arbitrary, this shows x n → . Now suppose that x n → . We will show that x n → ξ ∈ L ( M, τ ). If ξ ∈ M then lim sup n →∞ k x n ξ k ≤ lim sup n →∞ k x n k k ξ k ∞ = 0 . In general, for any ξ ∈ L ( M, τ ) and ǫ >
0, there exists ξ ′ ∈ M with k ξ − ξ ′ k < ǫ . Thenlim sup n →∞ k x n ξ k ≤ lim sup n →∞ k x n ξ ′ k + k x n ( ξ − ξ ′ ) k ≤ lim sup n →∞ k x n ( ξ − ξ ′ ) k ≤ Cǫ.
Since ǫ > x n → x n → ∈ L ( M, τ ) and x n I = x n , k x n k →
0. Thisshows x n → . 31ow suppose x n → . Let ε >
0. Then τ (1 ( ε, ∞ ) ( | x n | )) ≤ ε − k x n k . Since k x n k → n →∞ τ (1 ( ε, ∞ ) ( | x n | )) = 0 . So x n → Remark . It is possible that the topology on M inherited from the SOT on B ( H ) is notthe same as topology it inherits from the SOT on B (L ( M, τ )) [AP16, Exercise 1.3].
Definition 4.
If ( µ n ) n is a sequence of Borel probability measures on a topological space X and µ is another Borel probability measure on X then we write µ n → µ weakly if for everybounded continuous function f : X → C , R f d µ n converges to R f d µ as n → ∞ .Recall that C ( R ) denotes continuous functions on R that vanish at infinity while C c ( R ) ⊂ C ( R ) denotes those functions with compact support. Proposition 5.7.
Let ( M, τ ) be a von Neumann algebra with a finite trace. Suppose that ( a n ) n , ( b n ) n ∈ L ( M, τ ) , a n → a in measure, b n → b in measure and b n is self-adjoint for all n . Then:1. µ b n → µ b weakly.2. For all but countably many λ ∈ R we have that µ b n ( λ, ∞ ) → µ b ( λ, ∞ ) .3. For every bounded, continuous f : R → R we have that k f ( b n ) − f ( b ) k → n →∞ .
4. For every continuous f : R → R we have that f ( b n ) → f ( b ) in measure.5. | a n | → | a | in measure.Proof. After scaling if necessary we will assume without loss of generality that τ (I) = 1.(1): Let f ∈ C ( R ) (where C ( R ) is the space of continuous functions that vanish atinfinity). By [Sti59, Corollary 5.4], we know thatlim n →∞ k f ( b n ) − f ( b ) k = 0 . | τ ( x ) − τ ( y ) | ≤ k x − y k for all x, y ∈ M, the above convergence shows thatlim n →∞ Z f dµ b n = lim n →∞ τ ( f ( b n )) = τ ( f ( b )) = Z f dµ b . Now that we know the integrals R f dµ b n converge as n → ∞ for f ∈ C ( R ), it followsthat these integrals converge for all bounded continuous f : R → R because µ b n , µ b are allprobability measures (see, e.g. [Fol99, Exercise 20 of Chapter 7]).(2): This follows from (1) and the Portmanteau theorem.(3): Let R > | φ ( t ) | ≤ R for all t ∈ R . Let ε > , and choose a T > µ b ( { t : | t | ≥ T } ) < ε. Choose a function ψ ∈ C c ( R ) with ψ ( t ) = 1 for | t | ≤ T and sothat 0 ≤ ψ ≤ . Then, k φ ( b n ) − φ ( b ) k ≤ k φψ ( b n ) − φψ ( b ) k + k φ (1 − ψ )( b n ) k + k φ (1 − ψ )( b ) k = k φψ ( b n ) − φψ ( b ) k + (cid:18)Z | φ ( t ) | (1 − ψ ( t )) dµ b n ( t ) (cid:19) / + (cid:18)Z | φ ( t ) | (1 − ψ ( t )) dµ b ( t ) (cid:19) / . By [Sti59, Corollary 5.4] and (1) we thus have thatlim sup n →∞ k φ ( b n ) − φ ( b ) k ≤ (cid:18)Z | φ ( t ) | (1 − ψ ( t )) dµ b ( t ) (cid:19) / < Rε.
Letting ε → φ ∈ C c ( R ) . By [Sti59, Theorem 5.5], it suffices to show that k lim n →∞ φ ( f ( b n )) − φ ( f ( b )) k = 0 . Since φ ◦ f is bounded and continuous, this follows from (3).(5): Since L ( M, τ ) is a topological ∗ -algebra in the measure topology, a ∗ n a n → a ∗ a in themeasure topology. Let g : [0 , ∞ ) → [0 , ∞ ) be the function g ( t ) = √ t. Then | a n | = g ( a ∗ n a n ) . So this follows from (4) with b n = a ∗ n a n . We can give a more refined improvement of Proposition 5.7.4 which we will need later.We first note the following. 33 orollary 5.8.
Let ( M, τ ) be a von Neumann algebra with a finite trace, and let K ⊆ L ( M, τ ) have compact closure in the measure topology. Then for every ε > , there is an R > so that τ (1 ( R, ∞ ) ( | a | )) < ε for all a ∈ K. Proof.
Replacing K with its closure, we may as well assume K is compact. By Proposition5.7 (5) and (1), the map L ( M, τ ) → Prob( R ) sending x µ | x | is continuous if we giveL ( M, τ ) the measure topology, and Prob( R ) the weak topology. So { µ | a | : a ∈ K } iscompact in the weak topology, and thus tight. Tightness means there exists an R > µ | a | ( R, ∞ ) < ε for all a ∈ K. As τ (1 ( R, ∞ ) ( | a | )) = µ | a | ( R, ∞ ) , we are done. Corollary 5.9.
Let ( M, τ ) be a von Neumann algebra with a finite trace. Then the map E : L ( M, τ ) sa × C ( R , R ) → L ( M, τ ) sa given by E ( a, f ) = f ( a ) is continuous if we give L ( M, τ ) sa the measure topology and C ( R , R ) the topology of uniform convergence on compact sets.Proof. Suppose we are given sequences ( f n ) n ⊂ C ( R , R ), ( a n ) n ⊂ L ( M, τ ) sa and f ∈ C ( R ), a ∈ L ( M, τ ) sa with f n → f uniformly on compact sets and a n → a in measure.To prove f n ( a n ) → f ( a ) in measure, fix λ > . Let ε >
R > n ∈ N τ (1 ( R, ∞ ) ( | a n | )) < ε, τ (1 ( R, ∞ ) ( | a | )) < ε. Let g : R → R be a bounded continuous function with g ( t ) = t for | t | ≤ R, and define h : R → R by h ( t ) = f ( t ) − f ( g ( t )) and h n : R → R by h n ( t ) = f n ( t ) − f n ( g ( t )). Then: f n ( a n ) − f ( a ) = f n ( g ( a n )) − f ( g ( a )) + h n ( a n ) − h ( a ) . Then, by Proposition 4.4 we have that: τ (1 ( λ, ∞ ) ( | f n ( a n ) − f ( a n ) | ) ≤ τ (1 ( λ/ , ∞ ) ( | h n ( a n ) | )) + τ (1 ( λ/ , ∞ ) ( | h ( a ) | )) + τ (1 ( λ/ , ∞ ) ( | f n ( g ( a n )) − f ( g ( a )) | )) ≤ τ (1 ( λ/ , ∞ ) ( | h n ( a n ) | )) + τ (1 ( λ/ , ∞ ) ( | h ( a ) | )) + 4 λ k f n ( g ( a n )) − f ( g ( a )) k . λ > , and h = 0 in [ − R, R ] it follows that for all n ∈ N : τ (1 ( λ/ , ∞ ) ( | h n ( a n ) | )) ≤ τ (1 ( R, ∞ ) ( | a n | )) < ε. Similarly, τ (1 ( λ/ , ∞ ) ( | h ( a ) | )) < ε. For the last term: let
T > k g k ∞ ≤ T. Then: k f n ( g ( a n )) − f ( g ( a )) k ≤ k f n ( g ( a n )) − f ( g ( a n )) k + k f ( g ( a n )) − f ( g ( a )) k ≤ k f n ( g ( a n )) − f ( g ( a n )) k ∞ + k f ( g ( a n )) − f ( g ( a )) k ≤ sup t ∈ R : | t |≤ T | f n ( t ) − f ( t ) | + k f ( g ( a n )) − f ( g ( a )) k . We have that sup t ∈ R : | t |≤ T | f n ( t ) − f ( t ) | → n →∞ n → ∞ since f n → f uniformly oncompact sets. We also have that k f ( g ( a n )) − f ( g ( a )) k → k f n ( g ( a n )) − f ( g ( a )) k → n →∞
0. Altogether, we have shown thatlim sup n →∞ τ (1 ( λ, ∞ ) ( | f n ( a n ) − f ( a n ) | )) ≤ ε. Since ε > ε → τ (1 ( λ, ∞ ) ( | f n ( a n ) − f ( a ) | )) → n →∞ . Since this is true for every λ > , we have that f n ( a n ) → n →∞ f ( a ) in measure. P ( M, τ ) of positive log-square integrable operators Definition 5.
Let (
M, τ ) be a tracial von Neumann algebra, and let G = GL ( M, τ ) = { a ∈ L ( M, τ ) : log( | a | ) ∈ L ( M, τ ) } . Set P = P ( M, τ ) = { a ∈ G : a ≥ } . For a, b ∈ P , set d P ( a, b ) = k log( b − / ab − / ) k . This is well-defined by Theorem 4.1. Note that P ∞ = P ∩ M and d P restricted to P ∞ agreeswith the formula above by Corollary 5.2. 35he main result of this section is: Theorem 5.10.
Let ( M, τ ) be a von Neumann algebra with a finite trace. Then,1. d P is a metric.2. The group G acts on P by isometries by g.a = gag ∗ .3. The action G y P is transitive.4. P ∞ is dense in P . To prove this theorem, we will approximate elements of P by elements of P ∞ in themeasure topology and then apply results from the previous section on P ∞ . Because wewill use the Dominated Convergence Theorem, we need some basic facts about operatormonotonicity. Recall that if a, b are operators then by definition, a ≤ b if and only if b − a ∈ L ( M, τ ) is a positive operator (where b − a is defined to be the closure of b − a restricted to dom( b ) ∩ dom( a )). Proposition 5.11.
Let ( M, τ ) be a semi-finite von Neumann algebra.1. Suppose a, b ∈ L ( M, τ ) and | a | ≤ | b | . Then for every λ > we have that µ | a | ( λ, ∞ ) ≤ µ | b | ( λ, ∞ ) .
2. If a, b ∈ L ( M, τ ) are self-adjoint and a ≤ b, then cac ∗ ≤ cbc ∗ for every c ∈ L ( M, τ ) . Proof. (1): This is implied by [BK90, Lemma 3.(i)] or [FK86, Lemma 2.5(iii)].(2): We may write b − a = d ∗ d for some d ∈ L ( M, τ ) . Then cac ∗ − cbc ∗ = ( dc ∗ ) ∗ dc ∗ .The next proposition contains the approximations results we will need. Proposition 5.12.
Let ( M, τ ) be a von Neumann algebra with a finite trace. Suppose that ( a n ) n , ( b n ) n are sequences in G, that a ∈ L ( M, τ ) and that a ± n → a ± , b ± n → b ± in measure.Further assume that there are A , A , B , B ∈ P with A ≤ | a n | ≤ A , B ≤ | b n | ≤ B forall n ∈ N .
1. Then a ∈ G and k log( | a n | ) k → n →∞ k log( | a | ) k . . If a n and b n ∈ P for all n, then d P ( a n , b n ) → n →∞ d P ( a, b ) . Proof. (1): As in (2), k log( a n ) k = 2 Z ∞ λ [ µ | a n | ( e λ , ∞ ) + µ | a − n | ( e λ , ∞ )] dλ. Moreover, since | a n | ≤ A , we have that µ | a n | ( λ, ∞ ) ≤ µ A ( λ, ∞ ) . Let a n = u n | a n | be thepolar decomposition. Since a − n = u − n ( u n | a n | − u − n ), it follows that | a − n | = u n | a n | − u ∗ n . Soby operator monotonicity of inverses, | a − n | ≤ u n A − u ∗ n and thus by Proposition 5.11 (1), µ | a − n | ( e λ , ∞ ) ≤ µ u n A − u ∗ n ( e λ , ∞ ) . Since a n ∈ L ( M, τ ) × we know that each u n is a unitary, so µ | a − n | ( e λ , ∞ ) ≤ µ A − ( e λ , ∞ ).Thus λ [ µ | a n | ( e λ , ∞ ) + µ | a − n | ( e λ , ∞ )] ≤ λ [ µ A ( e λ , ∞ ) + µ A − ( e λ , ∞ )] . Since A , A ∈ P , the right hand side of this expression is in L ( R ) . Since a ± n → a ± in measure, Proposition 5.7 implies that µ | a ± n | ( λ, ∞ ) → µ | a ± | ( λ, ∞ ) forall but countably many λ. So by the Dominated Convergence Theorem, k log( | a | ) k = 2 Z ∞ λ [ µ | a | (( e λ , ∞ )) + µ | a − | (( e λ , ∞ ))] dλ = lim n →∞ Z ∞ λ [ µ | a n | (( e λ , ∞ )) + µ | a − n | (( e λ , ∞ ))] dλ = lim n →∞ k log( | a n | ) k . Moreover, we already saw that2 Z ∞ λ h µ | a n | ( e λ , ∞ ) + µ | a − n | ( e λ , ∞ ) i d λ ≤ Z λ h µ A ( e λ , ∞ ) + µ A − ( e λ , ∞ ) i d λ < ∞ . Thus log( | a | ) ∈ L ( M, τ ) and we have established that k log( | a n | ) k → n →∞ k log( | a | ) k . (2): By definition, d P ( a n , b n ) = k log( b − / n a n b − / n ) k , and as in (1) we have that d P ( a n , b n ) = 2 Z ∞ λ h µ b − / n a n b − / n ( e λ , ∞ ) + µ b / n a − n b / n ( e λ , ∞ ) i d λ. (3)37y Proposition 5.7 (4) we know that b − / n a n b − / n → b − / ab − / in measure, and similarly b / n a − n b / n → ba − b measure. Hence by Proposition 5.7 (5) we know thatlim n →∞ µ b − / n a n b − / n ( e λ , ∞ ) + µ b / n a − n b / n ( e λ , ∞ ) = µ b − / ab − / ( e λ , ∞ ) + µ b / a − b / ( e λ , ∞ ) , (4)for all but countably many λ. Moreover, by Proposition 4.3 we have that µ b − / n a n b − / n ( e λ , ∞ ) ≤ µ b − / n ( e λ/ , ∞ ) + µ a n ( e λ/ , ∞ ) = 2 µ b − n ( e λ/ , ∞ ) + µ a n ( e λ/ , ∞ ) . By operator monotonicity of inverses, we have that b − n ≤ B − and so by Proposition 5.11(1) we have µ b − / n a n b − / n ( e λ , ∞ ) ≤ µ B − ( e λ/ , ∞ ) + µ A ( e λ/ , ∞ ) . (5)Similarly, µ b / n a − n b / n ( e λ , ∞ ) ≤ µ B ( e − λ/ , ∞ ) + µ A − ( e − λ/ , ∞ ) . (6)As in the proof of (1), λ λ h µ B − ( e λ/ , ∞ ) + µ B ( e − λ/ , ∞ ) + µ A − ( e − λ/ , ∞ ) + µ A ( e λ/ , ∞ ) i is in L ( R ) . So by (5),(6), and (4) we may apply the dominated convergence theorem to (3)to see thatlim n →∞ d P ( a n , b n ) = 2 Z ∞ λ (cid:2) µ b − / ab − / ( e λ , ∞ ) + µ b / a − b / ( e λ , ∞ ) (cid:3) d λ = d P ( a, b ) . Proof of Theorem 5.10. (1):We first prove non-degeneracy. So suppose that a, b ∈ P and d P ( a, b ) = 0 . Thenlog( a − / ba − / ) = 0 , and so a − / ba − / = 1 . Multiplying this equation on the left andright by a / proves that b = a. For the triangle inequality, we already know by Corollary 5.2 that d P is a metric whenrestricted to GL ∞ ( M, τ ) = M × . Here M × is the set of elements of M with a bounded inverse.Define f n : [0 , ∞ ) → [0 , ∞ ) by f n ( t ) = n, if t > nt, if n ≤ t ≤ n n , if 0 ≤ t < n .Given a, b ∈ P , set a n = f n ( a ) , b n = f n ( b ) , A = | log( a ) | , B = | log( b ) | and observe that:38 a ± n → a ± , b ± n → b ± in measure, • exp( − A ) ≤ a n ≤ exp( A ) , exp( − B ) ≤ b n ≤ exp( B ) for all n ∈ N . By Proposition 5.12 (2), lim n →∞ d P ( a n , b n ) = d P ( a, b ) . Since d P is a metric when restricted to P ∩ M × , and f n ( P ) ⊆ P ∩ M × , the above equationimplies the triangle inequality for d P . It also implies d P is symmetric. So it is a metric.(2): It is easy to see that (2) is true if g ∈ U ( M ) (where U ( M ) ≤ M × is the group ofunitaries in M ). Every g ∈ G can be written as g = u | g | where u ∈ U ( M ) . Since | g | ∈ P forevery g ∈ G, and (2) is true when u ∈ U ( M ) , it suffices to show (2) for g ∈ P . So we willassume throughout the rest of the proof that g ∈ P . We first show that d P ( gag ∗ , gbg ∗ ) = d P ( a, b ) for a, b ∈ P ∩ M × . Since g ∈ P , as in (1) wemay find a sequence g n ∈ P ∩ M × so that • g ± n → g ± in measure, • g n = f n ( g ) for some f n : [0 , ∞ ) → [0 , ∞ ) • exp( − H ) ≤ g n ≤ exp( H ) for some self-adjoint H ∈ L ( M, τ ) . Since L ( M, τ ) is a topological ∗ -algebra in the measure topology, we have that g n ag n → n →∞ gag in measure. Moreover by Proposition 5.11 (2) k a − k − ∞ exp( − H ) ≤ k a − k − ∞ g n g n ≤ g n ag n ≤ k a k ∞ g n ≤ k a k ∞ exp(2 H ) , and similarly k b − k − ∞ exp( − H ) ≤ g n bg n ≤ k b k ∞ exp(2 H ) . So as in (1) we may apply Proposition 5.12 (2) to see that d P ( gag, gbg ) = lim n →∞ d P ( g n ag n , g n bg n ) . (7)By Lemma 5.1, d P ( g n ag n , g n bg n ) = d P ( a, b ). We thus have that d P ( gag, gbg ) = d P ( a, b ) . We now handle the case of general a, b ∈ P . As in (1), we find may sequences a n , b n ∈ P ∩ M × so that: 39 a ± n → a ± , b n → b ± in measure • exp( − A ) ≤ a n ≤ exp( A ) , exp( − B ) ≤ b n ≤ exp( B ) for some A, B ∈ L ( M, τ ).As in (1), we have that d P ( a, b ) = lim n →∞ d P ( a n , b n ) . (8) d P ( gag ∗ , gbg ∗ ) = lim n →∞ d P ( ga n g ∗ , gb n g ∗ ) . So combining (8) with the first case shows that d P ( gag ∗ , gbg ∗ ) = d P ( a, b ) . (3) Let p, q ∈ P . Then p − / , q / ∈ G = GL ( M, τ ). Moreover,( q / p − / ) · p = q. (4) Let a ∈ P and define a n = f n ( a ) as in (1). Then a n ∈ P ∞ and a n → a in measure.Apply Proposition 5.12 with b n = a = B = B to obtain d P ( a n , a ) → d P ( a, a ) = 0 as n → ∞ . Since a ∈ P is arbitrary, this proves P ∞ is dense in P . This section proves that the exponential map exp : L ( M, τ ) sa → P is a homeomorphismand obtains as a corollary that P is a complete CAT(0) metric space. We also obtain aformula for the geodesics in P . First we need the following estimate which extends the P ∞ case proven earlier. Proposition 5.13.
Let ( M, τ ) be a von Neumann algebra with a finite trace. Then for all a, b ∈ L ( M, τ ) sa , k a − b k ≤ d P ( e a , e b ) . If a and b commute then k a − b k = d P ( e a , e b ) . roof. Define a function f n : R → R by f n ( t ) = n, if t > nt, if | t | ≤ n − n, if t < − n. Set a n = f n ( a ) , b m = f n ( b ) . Then: • e a n → e a , e b n → e b in measure, • exp( − A ) ≤ e a n ≤ exp( A ) , exp( − B ) ≤ e b n ≤ exp( B ) for all n ∈ N . So as in Theorem 5.10 (1) we have that d P ( e a , e b ) = lim n →∞ d P ( e a n , e b n ) . Additionally, it is direct to see from the spectral theorem thatlim n →∞ k a − a n k = lim n →∞ k b − b n k = 0 . So, by Corollary 5.2, d P ( e a , e b ) = lim n →∞ d P ( e a n , e b n ) ≥ lim n →∞ k a n − b n k = k a − b k . Suppose a and b commute. By definition d P ( e a , e b ) = k log( e − b/ e a e − b/ ) k . Since a and b commute, e − b/ e a e − b/ = e a − b . So k log( e − b/ e a e − b/ ) k = k a − b k . Theorem 5.14.
Let ( M, τ ) be a von Neumann algebra with a finite trace. Then the expo-nential map exp : L ( M, τ ) sa → P is a homeomorphism.Proof. By Proposition 5.13, we know that log : P → L ( M, τ ) sa is continuous. So it justremains to show that exp : L ( M, τ ) sa → P is continuous.41uppose that ( a n ) n is a sequence in L ( M, τ ) and a ∈ L ( M, τ ) with k a − a n k → . Let ε > , and for λ >
0, define f λ : R → R by f λ ( t ) = λ, if t > λt, if | t | ≤ λ − λ, if t < − λ. If λ > k a − f λ ( a ) k < ε. Fix such a choice of λ. Since a and f λ ( a ) commute, d P ( e a n , e a ) ≤ d P ( e a , e f λ ( a ) ) + d P ( e f λ ( a n ) , e f λ ( a ) ) + d P ( e a n , e f λ ( a n ) ) (9)= k a − f λ ( a ) k + k a n − f λ ( a n ) k + d P ( e f λ ( a n ) , e f λ ( a ) ) . Since a n → a in L ( M, τ ), a n → a in measure. By Proposition 5.7 (3), lim n →∞ k f λ ( a n ) − f λ ( a ) k = 0 . Furthermore, max( k f λ ( a n ) k ∞ , k f λ ( a ) k ∞ ) ≤ λ for all n ∈ N . By Proposition 5.7 (4), e − f λ ( a n ) / → e − f λ ( a ) / in measure. Since L ( M, τ ) is a topological ∗ -algebra in the measure topology, e − f λ ( a n ) / e f λ ( a ) e − f λ ( a n ) / → e − f λ ( a n ) / e f λ ( a ) e − f λ ( a n ) / ) → e − λ ≤ e − f λ ( a n ) / e f λ ( a ) e − f λ ( a n ) / ≤ e λ . Choose a continuous function φ : R → R with φ ( x ) = log( x ) for all e − λ ≤ x ≤ e λ .Then φ ( e − f λ ( a n ) / e f λ ( a ) e − f λ ( a n ) / ) = log( e − f λ ( a n ) / e f λ ( a ) e − f λ ( a n ) / ). So the claim follows fromProposition 5.7 (4).By Proposition 5.6, the claim above implies log( e − f λ ( a n ) / e f λ ( a ) / e − f λ ( a n ) / ) → ( M, τ ).Since d P ( e f λ ( a n ) , e f λ ( a ) ) = k log( e − f λ ( a n ) / e f λ ( a ) / e − f λ ( a n ) / ) k this shows d P ( e f λ ( a n ) , e f λ ( a ) ) → n →∞ . (10)Since a n → a and f λ ( a n ) → f λ ( a ) in L ( M, τ ), k a n − f λ ( a n ) k → n →∞ k a − f λ ( a ) k . Combiningwith (10), (9) we have shown thatlim sup n →∞ d P ( e a n , e a ) ≤ k a − f λ ( a ) k < ε. ε → d P ( e a n , e a ) → . Corollary 5.15.
Let ( M, τ ) be a von Neumann algebra with a finite trace. Then ( P , d P ) isa complete metric space.Proof. Let ( a n ) be a Cauchy sequence in P . Set b n = log( a n ) . By Proposition 5.13, we knowthat ( b n ) is Cauchy in L ( M, τ ) . By completeness of L ( M, τ ) , there is a b ∈ L ( M, τ ) with k b n − b k → n →∞ . Then a = e b ∈ P , and by Theorem 5.14 we know that a n = e b n → e b = a. Corollary 5.16. If ( M, τ ) is finite then P is CAT(0).Proof. Recall that P ∞ is CAT(0) by Theorem 5.3. By Theorem 5.10 P ∞ is dense in P .Because metric completions of CAT(0) spaces are CAT(0) by [BH99, II.3, Corollary 3.11],this implies P is CAT(0). Corollary 5.17.
Let ( M, τ ) be a von Neumann algebra with a finite trace. Then the measuretopology on P ( M, τ ) is weaker than the d P -topology.Proof. Let ( b n ) n be a sequence in P ( M, τ ) and b ∈ P ( M, τ ) with lim n →∞ d P ( b n , b ) = 0 . Let a n = log b n , a = log( b ). Then k a n − a k → n →∞ , since the logarithm map is continuous.So a n → a in measure. But then by applying the exponential map in Proposition 5.7 (4) wehave that b n → b in measure. Corollary 5.18.
For ξ ∈ L ( M, τ ) sa , the map γ ξ : R → P defined by γ ξ ( t ) = exp( tξ ) is a minimal geodesic with speed k ξ k . Moreover every geodesic γ with γ (0) = I is equal to γ ξ for some ξ . Moreover, for any a, b ∈ P , the unique unit-speed geodesic from a to b is themap γ : [0 , d P ( a, b )] → P defined by γ ( t ) = a / γ ξ ( t ) a / where ξ = log( a − / ba − / ) k log( a − / ba − / ) k = log( a − / ba − / ) d P ( a, b ) . (11)43 roof. For any t > d P (I , γ ξ ( t )) = k log γ ξ ( t ) k = t k ξ k . This proves γ ξ is a minimal geodesic with speed k ξ k . Because P is CAT(0), there is a uniqueunit-speed geodesic between any two points. By uniqueness of geodesics, every geodesic γ with γ (0) = I has the above form.In particular, if a, b ∈ P and ξ is defined by (11) then γ ξ : [0 , d P ( a, b )] → P is a unit-speed geodesic from I to a − / ba − / . Because the action of GL ( M, τ ) on P is by isometries, γ ( t ) = a / .γ ξ ( t ) is a unit-speed geodesic from a = a / . I to b = a / .a − / ba − / . Let (
M, τ ) be a semi-finite tracial von Neumann algebra. Let G = GL ( M, τ ) and P = P ( M, τ ) = exp( L sa ( M, τ )) as before. We want to show that d P ( a, b ) := k log( b − / ab − / ) k is a distance function which makes P into a complete CAT(0) space. Since we have shownthis fact when τ (I) is finite, our approach will often involve reducing to the finite case. Tothis end we first need to identify the following objects.For a finite projection p ∈ M , observe that ( pM p, τ ◦ p ) is a von Neumann algebrawith a finite trace. Let P p = exp(L sa ( pM p, τ ◦ p )) ⊂ L ( pM p, τ ◦ p ). For a, b ∈ P p define d P p ( a, b ) = || log( b − / ab − / ) || L ( pMp,τ ◦ p ) . Since ( pM p, τ ◦ p n ) is finite, Theorem 5.10 implies d P p is a metric and Corollary 5.16 implies P p is complete CAT(0).Also define ˜ P p = exp( p L sa ( M, τ ) p ) ⊂ P ( M, τ ). and define the metric d ˜ P p of ˜ P p to be therestriction of d P to ˜ P p . Proposition 5.19.
The following are true:1. For every projection p ∈ M, the inclusion pM p ֒ → M extends to a ∗ -isomorphismof topological ∗ -algebras ι : L ( pM p, τ (cid:12)(cid:12) pMp ) ∼ = pL ( M, τ ) p. Further µ | ι ( x ) | = µ | x | , so inparticular ι induces an isometry P p → ˜ P p . d P is a metric.3. G acts on P by isometries. . G acts on P transitively.5. P is complete.6. P is CAT(0).7. Let P ∞ = exp( M sa ∩ L ( M, τ ) sa ) . Then P ∞ is dense in P and P ∞ = P ∩ M × .Proof of Proposition 5.19(1). Let j : pM p → M the inclusion map. Tautologically, τ (1 ( λ, ∞ ) ( | j ( x ) | )) = τ (1 ( λ, ∞ ) ( | x | )) , and the above equality implies that j extends to a linear map ι : L ( pM p, τ (cid:12)(cid:12) pMp ) → L ( M, τ )with closed image, and that this map is a homeomorphism onto its image. Moreover j ( pM p ) = pM p, and so passing to closures we have ι ( L ( pM p, τ (cid:12)(cid:12) pMp )) = pL ( M, τ ) p. By uniform continuity and the fact that ι is a ∗ -homomorphism on a dense ∗ -subalgebra(namely, pM p ) it follows that ι is a ∗ -homomorphism. The fact that µ | ι ( x ) | = µ | x | for all x ∈ L ( pM p, τ (cid:12)(cid:12) pMp ) follows from the fact that it is true for x ∈ pM p and Proposition 5.7(5). Since µ | ι ( x ) | = µ | x | , we know k log( | x | ) k = k log( | ι ( x ) | ) k for all x ∈ pM p. So ι ( P p ) = ˜ P p . Moreover, the fact that ι is ∗ -homomorphism and the aboveequality implies that ι : P p → ˜ P p is an isometry.To prove (2) of Proposition 5.19, we take an approximation approach similar to that ofthe finite case. Although we do not have all the tools available in the finite case such asProposition 5.7, we still have enough to work with. We first state the tools that we will beusing. Lemma 5.20.
Suppose x k , x ∈ L sa ( M, τ ) and x k → k →∞ x in measure and f : R → R is aBorel function continuous on the spectrum of x and bounded on bounded subsets of R . Then f ( x k ) → k →∞ f ( x ) in measure. roof. This is implied by [Tik87, Theorem 2.4].
Lemma 5.21.
Suppose x k , x ∈ L ( M, τ ) and x k → k →∞ x in measure. Suppose λ µ | x | ( λ, ∞ ) is continuous at λ . Then µ | x k | ( λ , ∞ ) → k →∞ µ | x | ( λ , ∞ ) .Proof. Let λ µ | x | ( λ, ∞ ) be continuous at λ . By Proposition 4.4, for any 0 < δ < λ and k ∈ N , µ | x | ( λ + δ, ∞ ) ≤ µ | x − x k | ( δ, ∞ ) + µ | x k | ( λ , ∞ ); µ | x k | ( λ , ∞ ) ≤ µ | x k − x | ( δ, ∞ ) + µ | x | ( λ − δ, ∞ ) . Since x k → x in measure (as k → ∞ ), µ | x | ( λ + δ, ∞ ) ≤ lim inf k →∞ µ | x k | ( λ , ∞ )lim sup k →∞ µ | x k | ( λ , ∞ ) ≤ µ | x | ( λ − δ, ∞ ) . Since λ is a point of continuity, this implieslim sup k →∞ µ | x k | ( λ , ∞ ) ≤ µ | x | ( λ , ∞ ) ≤ lim inf k →∞ µ | x k | ( λ , ∞ )which implies the lemma.We now prove that d P satisfies the triangle inequality and symmetry properties; theidentity property is similar to the finite case. We do this by approximating via elementsfrom a “reduced” von Neumann algebra with a finite trace. Notation 1.
For any x ∈ L sa ( M, τ ) and n ∈ N , let p xn = 1 ( −∞ , − /n ) ∪ (1 /n, ∞ ) ( x ) and x n := p xn xp xn = xp xn . Then x n is an increasing sequence converging in measure to x (as n → ∞ ).Because x ∈ L sa ( M, τ ), p xn is a finite projection. Proposition 5.22.
Suppose z, w ∈ P and we can write z = e x · · · e x k , w = e y · · · e y l forsome x i , y j ∈ L sa ( M, τ ) . Let x i,n = p x i n x i p x i n , z n = e x ,n · · · e x k,n y j,n = p y j n y j p y j n , w n = e y ,n · · · e y l,n . Assume z n and w n are positive for all n . Then d P ( z n , w n ) → d P ( z, w ) and d P ( z n , z ) → as n → ∞ . roof. We will just prove d P ( z n , w n ) → n →∞ d P ( z, w ) since the proof that d P ( z n , z ) n →∞ → q n = z − / n w n z − / n and q = z − / wz − / . Because x i,n → x i in measure, the expo-nential map is continuous and bounded on bounded subsets, and L ( M, τ ) is a topological*-algebra, it follows from Lemma 5.20 that z n → z in measure as n → ∞ . Similarly z − n → z − , w n → w and w − n → w in measure as n → ∞ . It follows from Lemma 5.20 that z − / n → z − / in measure. Since L ( M, τ ) is a topological *-algebra in the measure topology, q n → q in measure.Next use Proposition 4.3 and operator monotonicity (Proposition 5.11) to get (for any λ > µ z ± n ( e λ , ∞ ) ≤ k X i =1 µ e ± xi,n ( e λ/k , ∞ ) = k X i =1 µ ± x i,n ( λ/k, ∞ ) (12) ≤ k X i =1 µ | x i,n | ( λ/k, ∞ ) = k X i =1 µ k | x i,n | ( λ, ∞ ) ≤ k X i =1 µ k | x i | ( λ, ∞ ) . (13)A similar calculation shows that µ w ± n ( e λ , ∞ ) ≤ P lj =1 µ l | y j | ( λ, ∞ ).Next use Proposition 4.3 to get (for any λ > µ q n ( e λ , ∞ ) ≤ µ z − / n ( e λ/ , ∞ ) + µ w n ( e λ/ , ∞ )= 2 µ z − n ( e λ/ , ∞ ) + µ w n ( e λ/ , ∞ ) ≤ k X i =1 µ k | x i | (2 λ/ , ∞ ) + l X j =1 µ l | y j | ( λ/ , ∞ )= k X i =1 µ (3 / k | x i | ( λ, ∞ ) + l X j =1 µ l | y j | ( λ, ∞ )where the last inequality follows from (12).Note q − n = z / n w − n z / n . So by a similar computation we obtain µ q − n ( e λ , ∞ ) ≤ k X i =1 µ (3 / k | x i | ( λ, ∞ ) + l X j =1 µ l | y j | ( λ, ∞ ) . Now since each x i ∈ L sa ( M, τ ), by equation (1) we conclude that λ λ k X i =1 µ (3 / k | x i | ( λ, ∞ ) + l X j =1 µ l | y j | ( λ, ∞ ) !
47s integrable. It follows that λ λ ( µ q n ( e λ , ∞ ) + µ q − n ( e λ , ∞ )) is dominated by an integrablefunction, so that by the Dominated Convergence Theorem, Lemma 5.21 and equation (2), k log q n k → k log q k . By definition, k log q n k = d P ( z n , w n ) and k log q k = d P ( z, w ), so thisimplies the proposition.We can now prove d P is a metric. Proof of Proposition 5.19(2).
To prove the triangle inequality in ( P , d P ), suppose e x , e y , e v ∈ P with x, y, v ∈ L sa ( M, τ ). Define x n , y n , v n ∈ M as above. Let p n = p xn ∨ p yn ∨ p vn be thesmallest projection dominating p xn , p yn , p vn . By [Dix81][Part III, Ch. 2, Prop. 5], p n is afinite projection in M . So x n , y n , v n are in the finite von Neumann sub-algebra p n M p n . ByTheorem 5.10(1), d P ( e x n , e v n ) ≤ d P ( e x n , e y n ) + d P ( e y n , e v n ) . By Proposition 5.22, this implies the triangle inequality d P ( e x , e v ) ≤ d P ( e x , e y ) + d P ( e y , e v ) . Similarly, the symmetry d P ( e x , e v ) = d P ( e v , e x ) follows from by taking the limit as n → ∞ in d P ( e x n , e v n ) = d P ( e v n , e x n ). Lastly, if d P ( e x , e y ) = 0 then, by definition, log( e − x/ e y e − x/ ) =0 which implies e − x/ e y e − x/ = 1 which implies e x = e y . Corollary 5.23.
Let
Proj ⊂ M denote the set of finite projections in M . Then ∪ p ∈ Proj P p is dense in P .Proof. For any x ∈ L ( M, τ ) sa and n ∈ N , e x n ∈ ∪ p ∈ Proj P p and lim n →∞ e x n = e x in ( P , d P )by Proposition 5.22.To prove Proposition 5.19(3), we first show that unitary elements in M act by isometrieson P . Then, by polar decomposition, it suffices to consider the action of P on P . Lemma 5.24.
Let x, y ∈ L sa ( M, τ ) and u ∈ M is unitary. Then1. ( ue x u ∗ ) − = ue − x u ∗ ;2. ( ue x u ∗ ) / = ue x/ u ∗ ;3. e uxu ∗ = ue x u ∗ ; . d P ( ue x u ∗ , ue y u ∗ ) = d P ( e x , e y ) .Proof. The first claim is obvious. The second follows from observing that ue x/ u ∗ is positiveand its square is ue x u ∗ .For the third claim, first consider a sequence x k ∈ M converging to x in L . For each x k ,because M is a unital Banach algebra, e ux k u ∗ = ∞ X n =0 ( ux k u ∗ ) n n ! = ∞ X n =0 ux nk u ∗ n ! = ue x k u ∗ . By Lemma 5.5, x k → x in measure as k → ∞ . By Lemma 5.20, ue x k u ∗ → ue x u ∗ in measure, and also e ux k u ∗ → e uxu ∗ . But since ue x k u ∗ = e ux k u ∗ , by uniqueness of limits ue x u ∗ = e uxu ∗ .The last claim now follows using the previous three claims: d P ( ue x u ∗ , ue y u ∗ ) = k log[( ue x u ∗ ) − / ue y u ∗ ( ue x u ∗ ) − / ] k = k log[( ue − x/ u ∗ ) ue y u ∗ ( ue − x/ u ∗ )] k = k log[ ue − x/ e y e − x/ u ∗ ] k = k u log[ e − x/ e y e − x/ ] u ∗ k = τ ( u log[ e − x/ e y e − x/ ] ∗ log[ e − x/ e y e − x/ ] u ∗ ) / = τ (log[ e − x/ e y e − x/ ] ∗ log[ e − x/ e y e − x/ ]) / = k log[ e − x/ e y e − x/ ] k = d P ( e x , e y ) . The first equality is by definition of d P . The second follows from the first two claims above.The third equality uses uu ∗ = 1. The fourth follows from the third item of this lemma. Thefifth is by definition of k · k . The sixth holds because τ is a trace. Proof of Proposition 5.19(3).
By Lemma 5.24 and polar decomposition it suffices to considerthe action of P on P . Let g, a, b ∈ P , where g = e h , a = e x , b = e y , h, x, y ∈ L sa ( M, τ ). Wewant to show that d P ( gag ∗ , gbg ∗ ) = d P ( e x , e y ). As before consider reduced versions h n , x n , y n of h, x, y . Let g n = e h n , a n = e x n , b n = e y n . By Proposition 5.22, d P ( g n a n g ∗ n , g n b n g ∗ n ) → d P ( gag ∗ , gbg ∗ ). 49ow d P ( g n a n g n , g n b n g n ) = d P ( a n , b n ) → d P ( a, b ) (equality because we are again in a vonNeumann algebra with a finite trace and so Theorem 5.10(2) applies, and convergence byProposition 5.22), so it must be that d P ( gag, gbg ) = d P ( a, b ).The proof of Proposition 5.19(4) is the same argument as in the finite case.To prove ( P , d P ) is complete, we first show that exp is a homeomorphism from L ( M, τ ) sa to P . Lemma 5.25.
Let x, y ∈ L ( M, τ ) sa . Then k x − y k ≤ d P ( e x , e y ) .Proof. As above, consider x n = p xn xp xn and y n = p yn yp yn . Let p n = p xn ∨ p yn be the supremum of p xn and p yn . By [Dix81][Part III, Ch. 2, Prop. 5], p n is a finite projection in M . Proposition5.13 applies to ( p n M p n , τ ◦ p n ). So k x n − y n k ≤ d P ( e x n , e y n ).By Proposition 5.22, d P ( e x n , e y n ) → d P ( e x , e y ) as n → ∞ . It remains to show that k x n − y n k → k x − y k as n → ∞ . Now we know that x n − y n → x − y in measure. Furthermorewe can write k x n − y n k , in a similar fashion as equation (2), as 2 R ∞ λµ | x n − y n | ( λ, ∞ )d λ , andby Lemma 5.21, µ | x n − y n | ( λ, ∞ ) → µ | x − y | ( λ, ∞ ), while by Proposition 4.4 and Proposition5.11 µ | x n − y n | ( λ, ∞ ) ≤ µ | x | ( λ, ∞ ) + µ | y | ( λ, ∞ ). So by the Dominated Convergence Theorem k x n − y n k → k x − y k . Lemma 5.26.
Let f : R → R be a bounded continuous function. Suppose there is an openneighborhood O ⊂ R of such that f ( t ) = 0 for all t ∈ O . Suppose a , a , . . . is a sequencein L ( M, τ ) sa that converges to a ∈ L ( M, τ ) sa . Then d P ( e f ( a k ) , e f ( a ) ) → as k → ∞ .Proof of Lemma 5.26. By Lemma 5.5, a k converges to a in measure as k → ∞ . It followsby Lemma 5.20 that f ( a k ) converges to f ( a ) in measure, and also e f ( a k ) converges to e f ( a ) in measure. Since multiplication is jointly continuous with respect to the convergence inmeasure topology, z k := e − f ( a k ) / e f ( a ) e − f ( a k ) / → λ > t ∈ R | f ( t ) | ≤ λ and f ( t ) = 0 for all | t | ≤ /λ . Now since e − λ ≤ e − f ( a k ) / e f ( a ) e − f ( a k ) / ≤ e λ and log is a continuous function on the spectrum of z k ,by Lemma 5.20 log z k → z k converges to 0 in L . Now − λ ≤ log z k ≤ λ is uniformlybounded by 2 λ , log z k is also in L sa ( M, τ ). 50 laim 1. sup k τ (1 (0 , ∞ ) ( | log z k | )) < ∞ .Proof. Note that ker( f ( a k )) ∩ ker( f ( a )) ⊂ ker(log z k ), so ker(log z k ) ⊥ ≤ (ker( f ( a k )) ∩ ker( f ( a ))) ⊥ =span(ker( f ( a k )) ⊥ ∪ ker( f ( a k )) ⊥ ). Equivalently, 1 (0 , ∞ ) ( | log z k | ) ≤ (0 , ∞ ) ( | f ( a k ) | ) ∨ (0 , ∞ ) ( | f ( a ) | ) ≤ (0 , ∞ ) ( | f ( a k ) | ) + 1 (0 , ∞ ) ( | f ( a ) | ). Now τ (1 (0 , ∞ ) | f ( a k ) | ) = τ (1 (1 / λ, ∞ ) | a k | ) ≤ λ k a k k . Since a k is converging to a in L , the right hand side is bounded independently of k . The claimfollows.Since | log z k | ≤ λ , the claim implies that for any ǫ > k log z k k ≤ ǫ µ | log z k | (0 , ǫ ] + (2 λ ) µ | log z k | ( ǫ, ∞ ) ≤ ǫ K + (2 λ ) µ | log z k | ( ǫ, ∞ )where K = sup k τ (1 (0 , ∞ ) ( | log z k | )) is constant. Since ǫ > | log z k | → | log z | in measure (as k → ∞ ), it follows that log z k → . This implies d P ( e f ( a k ) , e f ( a ) ) → k →∞ Proposition 5.27. exp : L ( M, τ ) sa → P is continuous.Proof. We use a strategy similar to that used in the finite trace setting. Suppose ( a k ) is asequence in L ( M, τ ) sa converging to a in L .Let λ >
0. Let f λ : R → R be a continuous nondecreasing function such that for f λ ( x ) = 0 on [ − / λ, / λ ], f λ ( x ) = λ for x > λ , f λ ( x ) = − λ for x < − λ , and f λ ( x ) = x on[ − λ, − /λ ] ∪ [1 /λ, λ ].Because a k and f λ ( a k ) commute, d P ( e a k , e f λ ( a k ) ) = k a k − f λ ( a k ) k . Similarly, d P ( e a , e f λ ( a ) ) = k a − f λ ( a ) k . So two applications of the triangle inequality yield d P ( e a k , e a ) ≤ d P ( e a k , e f λ ( a k ) ) + d P ( e f λ ( a k ) , e f λ ( a ) ) + d P ( e f λ ( a ) , e a ) (14)= k a k − f λ ( a k ) k + d P ( e f λ ( a k ) , e f λ ( a ) ) + k a − f λ ( a ) k ≤ d P ( e f λ ( a k ) , e f λ ( a ) ) + k a k − a k + 2 k a − f λ ( a ) k + k f λ ( a ) − f λ ( a k ) k . By Lemmas 5.25 and 5.26, lim sup k →∞ d P ( e a k , e a ) ≤ k a − f λ ( a ) k k a − f λ ( a ) k = Z ( t − f λ ( t )) d µ a ( t ) , and ( t − f λ ( t )) ≤ t , the Dominated Convergence Theorem implies k a − f λ ( a ) k → λ → ∞ . Combined with the previous inequality, this implies d P ( e a k , e a ) → k → ∞ .The proof that ( P , d P ) is complete now follows from the same argument as in Corollary5.15. This finishes the proof of Proposition 5.19(5). Proof of Proposition 5.19(6).
We use arguments similar to those found in [BH99, Theo-rem II.3.9] in order to apply [BH99, Proposition II.1.11]. A sub-embedding of a 4-tuple( x , y , x , y ) of points in a metric space ( X, d X ) is a 4-tuple of points (¯ x , ¯ y , ¯ x , ¯ y ) in theEuclidean plane E such that d X ( x i , y j ) = k ¯ x i − ¯ y j k ( ∀ i, j ∈ { , } ) and d X ( x , x ) ≤ k ¯ x − ¯ x k and d X ( y , y ) ≤ k ¯ y − ¯ y k . A pair of points x, y ∈ X is said to have approximate mid-points if for every δ > m ∈ X such thatmax { d X ( x, m ) , d X ( m, y ) } ≤ d X ( x, y ) + δ. According to [BH99, Proposition II.1.11], a metric space (
X, d X ) is CAT(0) if and onlyif every 4-tuple of points in X admits a sub-embedding into E and every pair of points x, y ∈ X admits approximate midpoints. We will verify that ( P , d P ) satisfies this condition.Now let a i ∈ P , 1 ≤ i ≤ a i,n = exp( p n log( a i ) p n ) ∈ P p n , where p n = ∨ i =1 p log( a i ) n is as defined in Notation 1. For each n , ( P p n , d P ) is CAT(0)by Corollary 5.16. So each 4-tuple ( a i,n ) i has a sub-embedding in Euclidean space E : a 4-tuple of points (¯ a i,n ) i such that d P n ( a ,n , a ,n ) ≤ k ¯ a ,n − ¯ a ,n k , d P n ( a ,n , a ,n ) ≤ k ¯ a ,n − ¯ a ,n k , d P n ( a i,n , a j,n ) = k ¯ a i,n − ¯ a j,n k for all other i, j .By translation invariance of the standard metric on E , we can assume ¯ a ,n = ¯ a isthe same for all n . We have shown in Proposition 5.22 that for each i, j , d P pn ( a i,n , a j,n ) → d P ( a i , a j ) as n → ∞ . So the sub-embedding condition and triangle inequality show that (¯ a i,n )is contained in a compact set as i and n vary. In particular, by passing to a subsequence ifnecessary, we can assume ¯ a i,n converges to some ¯ a i in E for each i . It follows that (¯ a i ) i =1 isa sub-embedding of ( a i ) i =1 . 52or the approximate midpoint condition, let x, y ∈ P . Let x n , y n ∈ P p n be the reduced ver-sions where p n is now redefined to be p n = p log( x ) n ∨ p log( y ) n . For example, x n = exp( p n log( x ) p n ).Let δ > n such that | d P ( x n , y n ) − d P ( x, y ) | < δ/ , d P ( x, x n ) ≤ δ/ , d P ( y, y n ) ≤ δ/ . Because ( P p n , d P ) is CAT(0), there exists m ∈ P p n withmax { d P ( x n , m ) , d P ( m, y n ) } ≤ d P ( x n , y n ) + δ/ . By the triangle inequality,max { d P ( x, m ) , d P ( m, y ) } ≤ max { d P ( x n , m ) , d P ( m, y n ) } + δ/ ≤ d P ( x n , y n ) + 2 δ/ ≤ d P ( x, y ) + δ. Thus x, y have approximate midpoints. This completes the verification of the conditions in[BH99, Proposition II.1.11].
Proof of Proposition 5.19(7).
That P ∞ is dense in P follows from M sa ∩ L ( M, τ ) sa beingdense in L ( M, τ ) sa and Proposition 5.27 (that exp : L ( M, τ ) → P is continuous). Now if x ∈ P ∞ , then x = e y for y ∈ M sa . Then k e y k ∞ ≤ e k y k ∞ and similarly for x − = e − y , so x ∈ M × . Thus P ∞ ⊂ P ∩ M × . Conversely if x ∈ P ∩ M × then log x ∈ L sa ( M, τ ) ∩ M = M sa ,so x ∈ P ∞ . Corollary 5.28.
Corollaries 5.17 and 5.18 also hold for ( M, τ ) semi-finite.Proof. Using Proposition 5.19, the proofs are similar to the finite case.
This subsection proves a generalization of Theorem 1.1. We first need a lemma.53 emma 6.1.
Let ( M, τ ) be a semi-finite tracial von Neumann algebra. For any a, b ∈ P and σ ≥ , d P ( a σ , b σ ) ≥ σd P ( a, b ) . Proof.
First, assume τ is a finite trace. Let x = log a , y = log b . Recall that M sa ⊂ M is the set of self-adjoint elements in M . Because M sa is dense in L ( M, τ ) sa , there exist x n , y n ∈ M sa with x n → x and y n → y in L ( M, τ ) sa as n → ∞ . Thus d P ( a σ , b σ ) = d P ( e σx , e σy ) = lim n →∞ d P ( e σx n , e σy n ) ≥ lim n →∞ σd P ( e x n , e y n ) = σd P ( e x , e y ) = σd P ( a, b )where the second and third equalities follow from continuity of the exponential map (Theorem5.14) and the inequality follows from Corollary 5.4.Next we consider the general semi-finite case. Let x n , y n be the reduced versions of x, y ∈ L sa ( M, τ ) as in Notation 1. Then d P ( a σ , b σ ) = d P ( e σx , e σy ) = lim n →∞ d P ( e σx n , e σy n ) ≥ lim n →∞ σd P ( e x n , e y n ) = σd P ( e x , e y ) = σd P ( a, b ) . Where the second and second-to-last equalities follows from Proposition 5.22 and theinequality follows from the above finite case.We can now prove a slight generalization of Theorem 1.1 by expanding the range of thecocycle.
Theorem 6.2.
Let ( X, µ ) be a standard probability space, f : X → X an ergodic measure-preserving transformation, ( M, τ ) a semi-finite von Neumann algebra with faithful normaltrace τ . Let c : N × X → GL ( M, τ ) be a cocycle: c ( n + m, x ) = c ( n, f m x ) c ( m, x ) ∀ n, m ∈ N , µ − a.e. x ∈ X. Let π : GL ( M, τ ) → Isom( P ) be the map π ( g ) x = gxg ∗ where Isom( P ) is the group ofisometries of P . Suppose π ◦ c is measurable with respect to the compact-open topology on Isom( P ) and Z X L ( c (1 , x ) ∗ c (1 , x )) dµ ( x ) = Z X k log( | c (1 , x ) | ) k dµ ( x ) = Z X d P (1 , | c (1 , x ) | ) dµ ( x ) < ∞ . hen for almost every x ∈ X , the following limit exists: lim n →∞ L ( c ( n, x ) ∗ c ( n, x )) n = D. Moreover, if
D > then for a.e. x , there exists Λ( x ) ∈ L ( M, τ ) with Λ( x ) ≥ such that log Λ( x ) := lim n →∞ log (cid:0) [ c ( n, x ) ∗ c ( n, x )] / n (cid:1) ∈ L ( M, τ ) exists for a.e. x and lim n →∞ n d P (Λ( x ) n , | c ( n, x ) | ) = 0 . Proof.
We will use Theorem 1.5. So let (
Y, d ) = ( P , d P ). By Corollaries 5.15 and 5.16 for thefinite case and Proposition 5.19 for the semi-finite case, ( P , d P ) is a complete CAT(0) metricspace. Let y = I ∈ Y . Observe that the map N × X → GL ( M, τ ) , ( n, x ) c ( n, x ) ∗ is a reverse cocycle. Also d P ( y , c (1 , x ) ∗ .y ) = k log( c (1 , x ) ∗ c (1 , x )) k = L ( c (1 , x ) ∗ c (1 , x )) . So Z X d P ( y , c (1 , x ) ∗ .y ) dµ ( x ) < ∞ . Theorem 1.5 implies: for almost every x ∈ X , the following limit exists:lim n →∞ d P ( y , c ( n, x ) ∗ .y ) n = lim n →∞ L ( c ( n, x ) ∗ c ( n, x )) n = D. Moreover, if
D > x there exists a unique unit-speed geodesic ray γ ( · , x ) in P starting at I such thatlim n →∞ n d P ( γ ( Dn, x ) , c ( n, x ) ∗ .y ) = 0 . By Corollaries 5.18 for the finite case and 5.19 for the semi-finite case, γ ( t, x ) = exp( tξ ( x ))for some unique unit norm element ξ ( x ) ∈ L ( M, τ ) sa . Let Λ( x ) = exp( Dξ ( x ) / n →∞ n d P (Λ( x ) n , c ( n, x ) ∗ c ( n, x )) = 0 . n →∞ n k log(Λ( x ) − n c ( n, x ) ∗ c ( n, x )Λ( x ) − n ) k = lim n →∞ n L (Λ( x ) − n c ( n, x ) ∗ c ( n, x )Λ( x ) − n ) = 0 . Observe thatlim n →∞ (cid:13)(cid:13) log Λ( x ) − log (cid:0) [ c ( n, x ) ∗ c ( n, x )] / n (cid:1)(cid:13)(cid:13) ≤ lim n →∞ d P (Λ( x ) , [ c ( n, x ) ∗ c ( n, x )] / n ) ≤ lim n →∞ n d P (Λ( x ) n , c ( n, x ) ∗ c ( n, x )) = 0where the first inequality follows from Proposition 5.13 for the finite case and Lemma 5.25for the semi-finite case. The second inequality follows from Lemma 6.1. This concludes theproof.In order to show that Theorem 6.2 implies Theorem 1.1, we need to show how SOT-measurability of the cocycle c in Theorem 1.1 implies that π ◦ c is measurable with respectto the compact-open topology.We will need the next few lemmas to clarify the measurability hypothesis on the cocycle.The next lemma is probably well-known. Lemma 6.3.
Let ( Y, d ) be a metric space. Then the pointwise convergence topology on theisometry group Isom(
Y, d ) is the same as the compact-open topology.Proof. It is immediate that the pointwise convergence topology is contained in the compact-open topology. To show the opposite inclusion, let K ⊂ Y be compact, O ⊂ Y be openand suppose g ∈ Isom(
Y, d ) is such that gK ⊂ O . Let g n ∈ Isom(
Y, d ) and suppose g n → g pointwise. It suffices to show that g n K ⊂ O for all sufficiently large n .Because K is compact, there are a finite subset F ⊂ K and for every x ∈ F , a radius ǫ x > B ( x, ǫ x ) ⊂ Y is the open ball of radius ǫ x centered at x then gK ⊂ ∪ x ∈ F B ( gx, ǫ x ) ⊂ O. By compactness again, there exist 0 < ǫ ′ x < ǫ x such that K ⊂ ∪ x ∈ F B ( x, ǫ ′ x ) . g n → g pointwise, there exists N such that n > N implies d ( g n x, gx ) ≤ ǫ x − ǫ ′ x for all x ∈ F . Therefore, g n K ⊂ ∪ x ∈ F B ( g n x, ǫ ′ x ) ⊂ ∪ x ∈ F B ( gx, ǫ x ) ⊂ O as required.Let ˜ H ∈ { H , L ( M, τ ) } be one of the two Hilbert spaces under consideration. We use˜ H − SOT to denote the Strong Operator Topology with respect to the embedding of M in B ( ˜ H ). Similarly, ˜ H − W OT refers to the Weak Operator Topology. If we write SOT orWOT without the ˜ H -prefix then the default assumption is that we have chosen ˜ H = H . Ofcourse, it is possible that H = L ( M, τ ), so SOT by itself refers to both cases.Some of the results of the next Theorem appear in [Joh93].
Theorem 6.4.
Suppose ( M, τ ) is σ -finite, semi-finite and H is separable. Then1. the operator norm M → R , T
7→ k T k ∞ is SOT-Borel;2. a subset E ⊂ M is H -SOT-Borel if and only if it is L ( M, τ ) -SOT-Borel;3. the inverse operator norm M × → R , T
7→ k T − k ∞ is SOT-Borel;4. a subset E ⊂ M is SOT-Borel if and only if it is WOT-Borel;5. the adjoint M → M , T T ∗ is SOT-Borel;6. the multiplication map M × M → M , ( S, T ) ST is SOT-Borel;7. the map P ∩ M × → M defined T log T is SOT-Borel;8. the map M ∩ L ( M, τ ) , T
7→ k T k is SOT-Borel;9. for any x, y ∈ P ∩ M × the map M × ∩ GL ( M, τ ) → R defined by T d P ( T xT ∗ , y ) isSOT-Borel.Proof. (1), (4), (5), (6) follow from [Con00, Proposition 52.2(c) and Proposition 52.5].(2) Let M C = { T ∈ M : k T k ∞ ≤ C } . By [Dix81, I.4.3. Theorem 2] or [AP16, Corollary2.5.9 and Proposition 2.5.8], the H -SOT-topology on M C is the same as the L ( M, τ )-SOT-topology on M C . Since the operator norm is SOT-Borel by (1), this implies that the H -SOT-Borel sigma-algebra is the same as the L ( M, τ )-SOT-Borel sigma-algebra.573) Let H be a countable dense subset of { h ∈ H : k h k = 1 } . Then for every invertible T, we have k T − k − ∞ = inf h ∈ H k T ( h ) k . Since T
7→ k T ( h ) k is SOT-continuous for every h ∈ H , this proves that T
7→ k T − k ∞ isBorel.(7) Note that P ∩ M is SOT-Borel since, by (5), M sa is SOT-Borel and P ∩ M = { a ∈ M sa : h ah, h i ≥ ∀ h ∈ H } . For
D >
0, let M D = { T ∈ P : k T k ∞ ≤ D and k T − k ∞ ≤ D } . By items (1) and (3), M D is SOT-Borel. So it suffices to show that the map M D → M givenby T log T is SOT-continuous. Fortunately, it was proven in [Kap51, Corollary on page232] that if h : R → R is any continuous bounded function then the map on self-adjointoperators given by a f ( a ) is strongly continuous. Since log is bounded on [ D − , D ], thisimplies T log T is SOT-continuous on M D .(8) Let p , p , . . . be a sequence of pairwise-orthogonal finite projections p i ∈ M withlim n →∞ P ni =1 p n = I in L ( M, τ )-SOT. Then for any T ∈ L ( M, τ ), P ni =1 p i T converges to T in L ( M, τ ).Now h p i T, p j T i = 0 for all i = j and P ∞ i =1 p i T = T (where convergence is in L ( M, τ )).Therefore, k T k = ∞ X i =1 k p i T k = ∞ X i =1 k T ∗ p i k where the last equality follows from the tracial property of τ . So it suffices to prove that forany fixed finite projection p ∈ M , the map T
7→ k T ∗ p k is SOT-Borel. This follows fromitem (5) which states that the adjoint map is SOT-Borel (since p ∈ L ( M, τ )).(9) By definition, d P ( T xT ∗ , y ) = k log( y − / T xT ∗ y − / ) k . So this item follows from the previous items.
Corollary 6.5.
Suppose ( M, τ ) is σ -finite and semi-finite. Then π : M × → Isom( P ) isBorel as a map from M × with the SOT to Isom( P ) with the pointwise convergence topology. roof. By definition of the pointwise convergence topology, it suffices to show that for every x, y ∈ P , the map T d P ( T xT ∗ , y ) is SOT-Borel. By (8) of Proposition 5.19, M × ∩ P isdense in P . So it suffices to show that for every x, y ∈ M × ∩ P , the map T d P ( T xT ∗ , y )is SOT-Borel. This is item (9) of Theorem 6.4. Corollary 6.6.
The hypotheses of Theorem 1.1 imply the hypotheses of Theorem 6.2. Inparticular, Theorem 1.1 is true.
This section proves Theorem 1.3. So we assume (
M, τ ) is finite. Following [HS07, Definition2.1], we let M ∆ be the set of all x ∈ L ( M, τ ) such that Z ∞ log + ( t ) dµ | x | ( t ) < ∞ . For x ∈ M ∆ , the integral R ∞ log t dµ | x | ( t ) ∈ [ −∞ , ∞ ) is well-defined. The Fuglede-Kadison determinant of x is∆( x ) := exp (cid:18)Z ∞ log t dµ | x | ( t ) (cid:19) ∈ [0 , ∞ ) . For the sake of context, we mention that by [HS07, Lemma 2.3 and Proposition 2.5], if x, y ∈ M ∆ and ∆( x ) > x − ∈ M ∆ , ∆( x − ) = ∆( x ) − , xy ∈ M ∆ , and ∆( xy ) = ∆( x )∆( y ). Proposition 6.7.
Suppose τ is a finite trace. Then GL ( M, τ ) ⊂ M ∆ . Moreover, ∆ : P → (0 , ∞ ) is continuous.Proof. Let x ∈ GL ( M, τ ). By definition, log | x | ∈ L ( M, τ ). Since τ is a finite trace,L ( M, τ ) ⊂ L ( M, τ ). Thus log | x | ∈ L ( M, τ ) and therefore log + | x | ∈ L ( M, τ ). So x ∈ M ∆ .Now let ( x n ) n ⊂ P and suppose lim n →∞ x n = x ∈ P . By Proposition 5.13, log | x n | converges to log | x | in L ( M, τ ). Therefore, log | x n | converges to log | x | in L ( M, τ ). But thetrace τ : L ( M, τ ) → C is norm-continuous. So τ (log | x n | ) → τ (log | x | ). Since exp : R → R is continuous and ∆( x ) = exp( τ (log | x | )), this finishes the proof.Theorem 1.3 follows immediately from Proposition 6.7 and Theorem 1.1.59 .3 Growth rates In this subsection, we prove Theorem 1.4. The proof uses Theorem 1.1 as a black-box. Theextra ingredients needed to prove the theorem are general approximation results for powersof a single operator. These results will also be needed in later subsections to prove Theorem1.2.
Definition 6.
Let a ∈ L ( M, τ ) be a positive operator and ξ ∈ dom( a ) ⊂ L ( M, τ ). By theSpectral Theorem there exists a unique positive measure ν on C such that ν ([0 , ∞ )) = k ξ k and for every bounded, Borel function f : [0 , ∞ ) → C , h f ( a ) ξ, ξ i = Z f ( s ) d ν ( s ) . Moreover, for a Borel function f : [0 , ∞ ) → C we have that ξ ∈ dom( f ( a )) if and only if R | f ( s ) | d ν ( s ) < ∞ , and h f ( a ) ξ, ξ i = R f d ν if ξ ∈ dom( f ( a )) . The measure ν is called the spectral measure of a with respect to ξ . Let ρ ( ν ) ∈ [0 , ∞ ] be the smallest number suchthat ν is supported on the interval [0 , ρ ( ν )]. Lemma 6.8.
Let a ∈ L ( M, τ ) sa , ξ ∈ T ∞ n =1 dom( a n ) and let ν be the spectral measure of a with respect to ξ . Then ρ ( ν ) = lim n →∞ k a n ξ k /n ∈ [0 , ∞ ] . Moreover, ξ ∈ [0 ,t ] ( a )( L ( M, τ )) if and only if lim n →∞ k a n ξ k /n ≤ t (for any t ∈ [0 , ∞ ] ).Proof. Since ξ ∈ dom( a n ), R s n d ν ( s ) < ∞ for every n ∈ N . Thus k a n ξ k /n = h a n ξ, ξ i / n = (cid:18)Z s n d ν ( s ) (cid:19) / n . (15)It is a standard measure theory exercise that the limit of (cid:0)R s n d ν ( s ) (cid:1) / n as n → ∞ existsand equals ρ ( ν ).Now suppose that t > n →∞ k a n ξ k /n ≤ t. Then, by the above comment wehave that ν is supported on [0 , t ] . Thus: k ξ − [0 ,t ] ( a ) ξ k = Z | − [0 ,t ] ( s ) | d ν ( s ) = 0 . So ξ ∈ [0 ,t ] ( a )(L ( M, τ )) . ξ ∈ [0 ,t ] ( a )(L ( M, τ )) . Thenlim n →∞ k a n ξ k /n = lim n →∞ k a n [0 ,t ] ( a ) ξ k /n = lim n →∞ (cid:18)Z t s n d ν ( s ) (cid:19) / n ≤ t. Lemma 6.9.
Suppose ( M, τ ) is semi-finite. Let C > and suppose x n , x ∈ M C , where M C = { x ∈ M ∩ L ( M, τ ) : k x k ∞ ≤ C } and x n → x in measure as n → ∞ . Then x n → x in SOT.Proof. As in Proposition 5.6 we can assume that x = 0. To begin, let ξ ∈ M ∩ L ( M, τ ).We want to show that x n ξ → ( M, τ ).Let p n,k = 1 (1 /k, ∞ ) ( | x n | ) and x n,k = x n p n,k . By the triangle inequality, k x n ξ k ≤ k x n,k ξ k + k ( x n − x n,k ) ξ k ≤ k x n,k k k ξ k ∞ + k ( x n − x n,k ) k ∞ k ξ k ≤ Cµ x n,k (0 , ∞ ) / k ξ k ∞ + (1 /k ) k ξ k . Since x n → n →∞ µ x n,k (0 , ∞ ) / = µ x n,k (1 /k, ∞ ) / → n →∞ k .Therefore, lim sup n →∞ k x n ξ k ≤ (1 /k ) k ξ k . Since k is arbitrary, this proves x n ξ → n →∞ ( M, τ ).Now suppose ξ ∈ L ( M, τ ) and let ǫ >
0. Because M ∩ L ( M, τ ) is dense in L ( M, τ ),there exists ξ ′ ∈ M ∩ L ( M, τ ) with k ξ − ξ ′ k ≤ ǫ . Solim sup n →∞ k x n ξ k ≤ lim sup n →∞ k x n ξ ′ k + k x n ( ξ − ξ ′ ) k ≤ lim sup n →∞ k x n ( ξ − ξ ′ ) k ≤ lim sup n →∞ k x n k ∞ k ξ − ξ ′ k ≤ Cǫ.
Since ǫ > x n ξ → n →∞ ( M, τ ). Since ξ is arbitrary, thisproves x n → Lemma 6.10.
For n ∈ N , let a, b n ∈ L ( M, τ ) sa and ξ, ξ n ∈ L ( M, τ ) with ξ ∈ dom( a ) , ξ n ∈ dom( b n ) . Assume: b n → a in measure, • k ξ n − ξ k → as n → ∞ .Let ν be the spectral measure of a with respect to ξ and let ν n be the spectral measure of b n with respect to ξ n . Then ν n → ν weakly as n → ∞ .Proof. Let f ∈ C ( R ) be bounded. By Lemma 5.20, f ( b n ) → f ( a ) in measure. Since k f ( b n ) k ∞ , k f ( a ) k ∞ ≤ k f k ∞ = sup x ∈ R | f ( x ) | < ∞ , Lemma 6.9 implies that f ( b n ) → f ( a ) inthe Strong Operator Topology. Hence, (cid:12)(cid:12)(cid:12)(cid:12)Z f ( s ) d ν n ( s ) − Z f ( s ) d ν ( s ) (cid:12)(cid:12)(cid:12)(cid:12) = |h f ( b n ) ξ n , ξ n i − h f ( a ) ξ, ξ i|≤ |h f ( b n ) ξ n , ξ n i − h f ( b n ) ξ n , ξ i| + |h f ( b n ) ξ n , ξ i − h f ( b n ) ξ, ξ i| + |h f ( b n ) ξ, ξ i − h f ( a ) ξ, ξ i|≤ k f k ∞ k ξ n − ξ k ( k ξ n k + k ξ k ) + |h f ( b n ) ξ, ξ i − h f ( a ) ξ, ξ i| . Since k ξ n − ξ k → f ( b n ) → f ( a ) in the SOT, this shows R f ( s ) d ν n ( s ) → R f ( s ) d ν ( s ) as n → ∞ . Since f is arbitrary, ν n → ν weakly. Definition 7. If a ∈ L ( M, τ ) and ξ ∈ L ( M, τ ) \ dom( a ), then let k aξ k = + ∞ .The next definition generalizes Definition 1. Definition 8.
Given ξ ∈ L ( M, τ ), let Σ( ξ ) be the set of all sequences ( ξ n ) n ⊂ L ( M, τ )such lim n →∞ k ξ − ξ n k = 0. Given a sequence c = ( c n ) n ⊂ L ( M, τ ) and ξ ∈ L ( M, τ ), definethe upper and lower smooth growth rates of c with respect to ξ byGr( c | ξ ) = inf (cid:8) lim inf n →∞ k c n ξ n k /n : ( ξ n ) n ∈ Σ( ξ ) , ξ n ∈ dom( c n ) (cid:9) Gr( c | ξ ) = inf (cid:8) lim sup n →∞ k c n ξ n k /n : ( ξ n ) n ∈ Σ( ξ ) , ξ n ∈ dom( c n ) (cid:9) . Lemma 6.11.
For n ∈ N , let c n ∈ L ( M, τ ) , a ∈ L ( M, τ ) with a ≥ and ξ ∈ L ( M, τ ) .Let c = ( c n ) n . Assume: • | c n | /n → a in measure as n → ∞ . • ξ ∈ T ∞ n =1 dom( a n ) .Then Gr( c | ξ ) = lim n →∞ k a n ξ k /n = Gr( c | ξ ) . roof. Let ν be the spectral measure of a with respect to ξ . Let ρ ( ν ) ≥ ν is supported on [0 , ρ ( ν )]. By Lemma 6.8, ρ ( ν ) = lim n →∞ k a n ξ k /n .It is immediate that Gr( c | ξ ) ≤ Gr( c | ξ ) . So it suffices to show ρ ( ν ) ≤ Gr( c | ξ ) andGr( c | ξ ) ≤ ρ ( ν ).We first show ρ ( ν ) ≤ Gr( c | ξ ). Let b n = | c n | /n . Let ( ξ n ) n ∈ Σ( ξ ) with ξ n ∈ dom( b n ). Byhypothesis, b n → a in measure. Let ν n be the spectral measure of ξ n with respect to b n . ByLemma 6.10, ν n → ν weakly. Along with (15) and Fatou’s Lemma we have for every m ∈ N : k a m ξ k /m = (cid:18)Z s m d ν ( s ) (cid:19) / m = (cid:18) m Z λ m − ν ( λ, ∞ ) d λ (cid:19) / m ≤ (cid:18) m lim inf n →∞ Z λ m − ν n ( λ, ∞ ) d λ (cid:19) / m = (cid:18) lim inf n →∞ Z s m d ν n ( s ) (cid:19) / m ≤ lim inf n →∞ (cid:18)Z s n d ν n ( s ) (cid:19) / n = lim inf n →∞ k c n ξ n k /n . So sup m k a m ξ k /m ≤ lim inf n →∞ k c n ξ n k /n . By Lemma 6.8, ρ ( ν ) = lim n →∞ k a n ξ k /n ≤ lim inf n →∞ k c n ξ n k /n . Since ( ξ n ) n is arbitrary, this shows ρ ( ν ) ≤ Gr( c | ξ ).Next we will show Gr( c | ξ ) ≤ ρ ( ν ). So let ε >
0. Choose a continuous function f :[0 , ∞ ) → [0 ,
1] such that f ( t ) = 1 for all t ∈ [0 , ρ ( ν )] and f ( t ) = 0 for all t ≥ ρ ( ν ) + ε . Let ξ n = f ( b n ) ξ .We claim that ξ n → ξ in L ( M, τ ). First observe that h f ( a ) ξ, ξ i = Z f d ν = Z ν = k ξ k . Since k f ( a ) k ∞ ≤
1, we must have f ( a ) ξ = ξ .Next, let ν ′ n be the spectral measure of b n with respect to ξ . Note k ξ n − ξ k = k (1 − f ( b n )) ξ k = h (1 − f ( b n )) ξ, (1 − f ( b n )) ξ i = h (1 − f ( b n )) ξ, ξ i = Z (1 − f ) d ν ′ n .
63y Lemma 6.10, ν ′ n → ν weakly. So R (1 − f ) d ν ′ n → R (1 − f ) d ν as n → ∞ . Since ν is supported on [0 , ρ ( ν )] and 1 − f = 0 on [0 , ρ ( ν )], it follows that R (1 − f ) d ν ′ n → n → ∞ . This proves that k ξ n − ξ k → n → ∞ . Thus, ( ξ n ) n ∈ Σ( ξ ).Let ν n be the spectral measure of b n with respect to ξ n . We claim that d ν n = f d ν ′ n . Tosee this, let g : [0 , ∞ ) → R be a continuous bounded function. Then Z g d ν n = h g ( b n ) ξ n , ξ n i = h g ( b n ) f ( b n ) ξ, f ( b n ) ξ i = h f ( b n ) g ( b n ) f ( b n ) ξ, ξ i = Z gf d ν ′ n . Since g is arbitrary, this proves d ν n = f d ν ′ n .By Lemma 6.10, ν n → ν weakly. SoGr( c | ξ ) ≤ lim sup n →∞ k c n ξ n k /n = lim sup n →∞ h| c n | ξ n , ξ n i / n = lim sup n →∞ h b nn ξ n , ξ n i / n = lim sup n →∞ (cid:18)Z t n d ν n ( t ) (cid:19) / n = lim sup n →∞ (cid:18)Z t n f ( t ) d ν ′ n ( t ) (cid:19) / n ≤ ρ ( ν ) + ε. The last inequality occurs because f ( t ) = 0 for all t > ρ ( ν ) + ε . Since ε is arbitrary,Gr( c | ξ ) ≤ ρ ( ν ). Corollary 6.12.
Let
X, µ, f, M, τ, c, Λ be as in Theorem 6.2. Then for a.e. x ∈ X andevery ξ ∈ L ( M, τ ) with ξ ∈ ∩ n dom( a n ) , lim n →∞ k Λ( x ) n ξ k /n = Gr( c ( x ) | ξ ) = Gr( c ( x ) | ξ ) where c ( x ) = ( c ( n, x )) n . In particular, Theorem 1.4 is true.Proof. Apply Lemma 6.11 with a = Λ( x ), c n = c ( n, x ). Theorem 1.4 now follows fromCorollary 6.6. In this section, we review the notion of an essentially dense subspace. This is used in thelast section to prove Theorem 1.2. 64 efinition 9.
Let (
M, τ ) be a semi-finite von Neumann algebra. A linear subspace V ⊆ L ( M, τ ) is called right-invariant if V x ⊆ V for all x ∈ M. We say that a right-invariantsubspace D of L ( M, τ ) is essentially dense if for every ε > , there is a projection p ∈ M so that τ ( I − p ) ≤ ε, and D ⊇ pL ( M, τ ) . If H is a closed subspace of L ( M, τ ) and W ⊆ H is a right-invariant subspace, we say that W is essentially dense in H if there exists D essentially dense in L ( M, τ ) such that D ∩ H = W .By definition of L ( M, τ ), if a ∈ L ( M, τ ), then dom( a ) is essentially dense. It is anexercise to check that the intersection of countably many essentially dense subspaces isessentially dense.If H ⊆ L ( M, τ ) is closed and right-invariant then the orthogonal projection onto H , denoted p H , is in M as a consequence of the Double Commutant Theorem.Technically, our definition of essentially dense in H is different from the one in [L¨uc02,Definition 8.1]. The next lemma shows that they are in fact equivalent. Lemma 6.13.
Let ( M, τ ) be a semi-finite tracial von Neumann algebra, let H ⊆ L ( M, τ ) be a closed, right-invariant subspace, and let W ⊆ H be a right-invariant subspace. Thenthe following are equivalent:1. W is essentially dense in H ,
2. there is an increasing sequence of projections p n ∈ M so that p n → p H in the StrongOperator Topology, τ ( p H − p n ) → , and W ⊇ p n L ( M, τ ) . Proof. (2) implies (1): Let D = W + (I − p H ) L ( M, τ ) , then clearly D ∩ H = W. Let q n = I − p H + p n .Then D ⊇ q n L ( M, τ ) . Since p n (I − p H ) = 0 , we also have that q n is an orthogonalprojection. Also τ (I − q n ) →
0. Thus D is essentially dense.(1) implies (2): Write W = D ∩ H , where D is essentially dense in L ( M, τ ) . By as-sumption, for every n ∈ N , we find a projection f n in M so that τ (I − f n ) ≤ − n and D ⊇ f n L ( M, τ ) . Set q n = V ∞ m = n f m , and p n = p H ∧ q n . Observe that τ (I − q n ) = τ ∞ _ m = n I − f m ! ≤ ∞ X m = n τ (I − f n ) ≤ − n +1 , τ is normal.By definition, p H − p n = p H − p H ∧ q n is the orthogonal projection onto H ∩ range( q n ) ⊥ ,while I − q n is the orthogonal projection onto range( q n ) ⊥ . It follows that p H − p n ≤ I − q n .Therefore, τ ( p H − p n ) ≤ − n +1 → n → ∞ .Because q n is increasing in n , it also true that p n is increasing in n . So if p ∞ = sup n p n then p ∞ is the SOT-limit of p n as n → ∞ . By definition of least upper bound, p ∞ ≤ p H andsince τ ( p H − p ∞ ) ≤ τ ( p H − p n ) →
0, the fact that τ is faithful implies p ∞ = p H . So p n → p H in the Strong Operator Topology. For each n ∈ N , we have that p n L ( M, τ ) = H ∩ q n L ( M, τ ) ⊆ H ∩ f n L ( M, τ ) ⊆ H ∩ D = W. Lemma 6.14.
Let ( M, τ ) be a tracial von Neumann algebra, and let H ⊆ L ( M, τ ) be aclosed and right-invariant subspace, fix an a ∈ L ( M, τ ) .
1. We have that dim M ( a ( H ∩ dom( a ))) ≤ dim M ( H ) , with equality if ker( a ) ∩ H = { } .2. We have that ( a − ( H )) ⊥ = ( a ∗ )( H ⊥ ∩ dom( a ∗ )) . Proof.
Throughout, let p be the orthogonal projection onto H . (1): Let ap = v | ap | be the polar decomposition. Then v ∗ v = p ker( ap ) ⊥ , vv ∗ = p Im( ap ) . Clearly ker( ap ) ⊇ (1 − p )( L ( M, τ )) so v ∗ v = p ker( ap ) ⊥ ≤ p. So: dim M ( H ) = τ ( p ) ≥ τ ( v ∗ v ) = τ ( vv ∗ ) = dim M ( a ( H ∩ dom( a ))) . If ker( a ) ∩ H = { } , then in fact v ∗ v = p ker( ap ) ⊥ = p so we have equality throughout.(2): This is [Sti59, Lemma 3.4]. In this section, we prove Theorem 1.2. 66 emma 6.15.
For n ∈ N , let c n ∈ L ( M, τ ) and a ∈ L ( M, τ ) with a > . Let T n = a − n | c n | a − n , and S n = T / nn . Suppose S n → I in measure and | c n | /n → a in measure. For t ∈ [ −∞ , ∞ ) , let V t = (cid:26) ξ ∈ L ( M, τ ) : lim inf n →∞ n log k c n ξ k ≤ t (cid:27) H t = (cid:26) ξ ∈ L ( M, τ ) : lim inf n →∞ n log k a n ξ k ≤ t (cid:27) = 1 ( −∞ ,t ] (log a )( L ( M, τ )) . Then there exists an essentially dense subspace D of L ( M, τ ) such that D ∩ V t = D ∩ H t . In particular, we have that H t = V t . Proof.
Choose a decreasing sequence ( ε k ) k of positive real numbers tending to zero. Since S n → I in measure, there is an increasing sequence ( n k ) k X k µ S nk (1 + ε k , ∞ ) < ∞ . By functional calculus, µ T nk ((1 + ε k ) n k , ∞ ) = µ S nk (1 + ε k , ∞ ). So X k µ T nk (cid:0) (1 + ε k ) n k , ∞ (cid:1) < ∞ . Let r k ∈ M denote orthogonal projection onto R k := a − n k [0 , (1+ ε k ) nk ] ( T n k )(L ( M, τ )). Let D = ∞ [ l =1 ∞ \ k = l (cid:16) dom( a n k ) ∩ R k ∩ dom( c n k ) (cid:17) . We claim that D is essentially dense.Because dom( a n k ) and dom( c n k ) are essentially dense,dom( a n k ) ∩ dom( c n k ) ∩ R k is essentially dense in R k . So there exist projections p k ∈ M satisfying • p k L ( M, τ ) ⊆ dom( a n k ) ∩ R k ∩ dom( c n k ) • τ ( r k − p k ) ≤ µ T nk ((1 + ε k ) n k , ∞ ) . l ∈ N , set q l = V ∞ k = l p k . Then for every l ∈ N , we know that D ⊇ q l L ( M, τ ) and τ (I − q l ) ≤ τ (I − r l ) + τ ( r l − q l ) ≤ ∞ X k = l µ T nk (cid:0) (1 + ε k ) n k , ∞ (cid:1) → l →∞ . So we have shown that D is essentially dense.Now suppose that ξ ∈ D . Without loss of generality, k ξ k = 1 . Then:lim inf n →∞ k c n ξ k /n ≤ lim inf k →∞ k c n k ξ k /n k = lim inf k →∞ h| c n k | ξ, ξ i / n k = lim inf k →∞ (cid:10) a − n k | c n k | a − n k a n k ξ, a n k ξ (cid:11) nk = lim inf k →∞ h T n k a n k ξ, a n k ξ i nk . By choice of D , we have that a n k ξ ∈ [0 , (1+ ε k ) nk ] ( T n k )(L ( M, τ )) . Solim inf n →∞ k c n ξ k /n ≤ lim inf k →∞ (1 + ε k ) h a n k ξ, a n k ξ i nk = lim inf k →∞ (1 + ε k ) k a n k ξ k /n k = lim n →∞ k a n ξ k /n , where the last equality holds by Lemma 6.8. So by Lemma 6.11,lim inf n →∞ k c n ξ k /n = lim n →∞ k a n ξ k /n . Thus D ∩ V t = D ∩ H t . Since Lemma 6.11 also shows that V t ⊆ H t , it follows that V t = H t since essentially dense subspaces are dense.Lemma 6.8 also shows that H t = 1 (0 ,e t ] ( a )(L ( M, τ )) = 1 ( −∞ ,t ] (log a )(L ( M, τ )) . Corollary 6.16.
Let
X, µ, f, M, τ, c, Λ be as in Theorem 6.2. For x ∈ X , let H t ( x ) = (cid:26) ξ ∈ L ( M, τ ) : lim inf n →∞ n log k Λ( x ) n ξ k ≤ t (cid:27) = 1 ( −∞ ,t ] (log Λ( x ))( L ( M, τ )) . Then the Oseledets subspaces and Lyapunov distributions are invariant in the following sense.For a.e. x ∈ X , c (1 , x ) H t ( x ) = H t ( f ( x )) ,µ log Λ( f ( x )) = µ log Λ( x ) . In particular, Theorem 1.2 is true. roof. For x ∈ X , let V t ( x ) = (cid:26) ξ ∈ L ( M, τ ) : lim inf n →∞ n log k c ( n, x ) ξ k ≤ t (cid:27) . By the cocycle equation, c ( n, f ( x )) c (1 , x ) = c ( n + 1 , x ). So c (1 , x ) V t ( x ) ⊆ V t ( f ( x )) . Con-versely, using that c (1 , x ) − = c ( − , f ( x )) , we can apply the same logic to see that V t ( x ) ⊇ c (1 , x ) − V t ( f ( x )) . Hence c (1 , x ) V t ( x ) = V t ( f ( x )) . Applying the “in particular” part of Lemma 6.15 shows that c (1 , x ) H t ( x ) = H t ( f ( x )) . To prove invariance of the Lyapunov distribution, let s < t be real numbers. Fix x ∈ X and define φ ∈ M by φ = ( p H t ( f ( x )) − p H s ( f ( x )) ) c (1 , x ) . Because the kernel of p H t ( f ( x )) − p H s ( f ( x )) (viewed as a self-map of H t ( f ( x ))) is H s ( f ( x )), thekernel of φ restricted to H t ( x ) is c (1 , x ) − H s ( f ( x )) = H s ( x ). Thus, the restriction of φ to H t ( x ) ∩ H s ( x ) ⊥ is one-to-one. Moreover, φ ( H t ( x ) ∩ H s ( x ) ⊥ ) = H t ( f ( x )) ∩ H s ( f ( x )) ⊥ . Because injective elements of M preserve von Neumann dimension (by Lemma 6.14),dim M ( H t ( x ) ∩ H s ( x ) ⊥ ) = dim M ( H t ( f ( x )) ∩ H s ( f ( x )) ⊥ ) . By definition, µ log Λ( x ) ( s, t ] = dim M ( H t ( x ) ∩ H s ( x ) ⊥ ). Since s < t are arbitrary, this shows µ log Λ( f ( x )) = µ log Λ( x ) .Theorem 1.2 now follows from Corollary 6.6. References [AL06] E. Andruchow and G. Larotonda. Nonpositively curved metric in the positive coneof a finite von Neumann algebra.
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