aa r X i v : . [ m a t h - ph ] S e p Jakˇsi´c-Last Theorem for Higher Rank Perturbations
Anish Mallick ∗ The Institute of Mathematical Sciences,Chennai-600113, India
September 27, 2018
Abstract
We consider the generalized Anderson Model ∆ + P n ∈N ω n P n , where N is a countable set, { ω n } n ∈N are i.i.d random variables and P n are rank N < ∞ projections. For these models weprove theorem analogous to that of Jakˇsi´c-Last on the equivalence of the trace measure σ n ( · ) = tr ( P n E H ω ( · ) P n ) for n ∈ N a.e ω . Our model covers the dimer and polymer models. In this paper we address the nature of spectral measure for generalized Anderson type models with singlesite potentials of higher rank or a constant randomness over several neighboring collection of sites . Thebasic setup of the problem is the following. We have a self adjoint operator A on separable Hilbertspace H , and rank N projections { P n } n ∈N where N is countable or a finite set. Given an absolutelycontinuous measure µ on R , we define the set of operators H ω = A + X n ∈N ω n P n (1.1)for { ω n } n ∈N ∈ R N distributed identically and independently following the distribution µ . This definesa map from measure space (Ω , B , P ) (product measure space ( R N , B ( R N ) , ⊗ µ )) to the set of essentiallyself adjoint linear operators on H . We are interested in the spectral measure of this set of operators.In the case of Anderson tight-binding model, we have the Hilbert space l ( Z d ) on which we have theoperator ∆ defined by (∆ u )( n ) = X | n − m | =1 u ( m ) ∀ u ∈ l ( Z d ) , n ∈ Z d and the collection of rank one projection {| δ n ih δ n |} n ∈ Z d . Prior works [9, 11, 21, 22] proved simplicity ofspectrum for such models using the property that {| δ n ih δ n |} n ∈ Z d are rank one.Similar to the tight-binding model, we have the random Schr¨odinger operators, defined by( H ω f )( x ) = − d X i =1 ∂ f∂x i ( x ) + X p ∈ Z d ω p G ( x − p ) f ( x ) ∀ x ∈ R d , f ∈ C ∞ c ( R d )where G is a compactly supported function on [0 , d . Simplicity of the singular spectrum for this modelis still an open problem.These models are also considered on graphs (for example Bethe lattices and one dimensional strips).In the case of Bethe lattice and Bethe strips absolute continuous spectrum was shown to exists [7, 13, 14].All these models show localization at high disorder [1] and so have pure point spectrum also.Multiplicity of these spectra are not well understood for projection valued perturbation. Knownresults study rank one perturbation [9, 11] and cyclicity [22, 21]. In work by Naboko, Nichols and Stolz[15], such a problem is handled for pure point part of the spectrum. Sadel and Schulz-Baldes [18] werelooking at quasi-one-dimensional stochastic dirac operator, and provided conditions for singular and ∗ [email protected] n ∈ N and ω ∈ Ω define H n,ω as the cyclic subspace generated by H ω defined by (1.1) and the vector space P n H (this vector spaceis isomorphic to C N ), and set Q ωn : H → H n,ω as the canonical projection. Let E H ω be the spectralprojection measure for the operator H ω ; set Σ ωn ( · ) = P n E H ω ( · ) P n (which is now a matrix valued measure)and set σ ωn ( · ) = tr (Σ ωn ( · )) as the trace measure (these are finite measures). Let P ωac be the orthogonalprojection onto the absolutely continuous spectral subspace H ac ( H ω ). For n, m ∈ N , define E n,m = { ω ∈ Ω | Q ωn P m has same rank as P m } (1.2)We will be working with the following set M = { n ∈ N | σ ωn is not equivalent to Lebesgue measure for a.e ω } The reason for confining oneself in this set is a theorem of F. and M. Riesz [17], which implies that the Borel transform of any complex measure which is zero in C + has to be absolutely continuous withrespect to Lebesgue measure . But one can prove that the total variation measure need to be equivalent toLebesgue measure, see [11, Theorem 2.2] for a proof. So confining to M we get that the Borel transformof non-zero measure in M can never be identically zero in C + , and so one can use results about boundaryvalues of analytic functions like [2, 3]. We state the main theorem: Theorem 1.1.
For any N ∈ N , let { P n } n ∈N be collection of rank N projections such that P n ∈N P n = I ,and µ be a absolutely continuous measure on R . Let { H ω } ω ∈ Ω be a family of operator defined as in (1.1) ,then1. For n, m ∈ M we have P ( E n,m ) ∈ { , } .2. Let n, m ∈ M such that P ( E n,m ∩ E m,n ) = 1 . For a.e ω ∈ Ω , the restrictions P ωac H ω | H n,ω and P ωac H ω | H m,ω are unitary equivalent.3. Let n, m ∈ M such that P ( E n,m ∩ E m,n ) = 1 , for a.e ω ∈ Ω the measures σ ωn and σ ωm are equivalentas Borel measures. Remark 1.2.
Two examples for which the condition P ( E n,m ) = 1 can be verified are:1. Consider l ( Z ) with the operator H ω = ∆ + P n ∈ Z ω n P n where P n = P N − k =0 δ Nn + k , we have P ( E n,m ) = 1 for each n, m ∈ Z . This is because for n , n ∈ Z , (cid:10) δ n , ( H ω ) | n − n | δ n (cid:11) = 1 andhence all cyclic subspaces intersect with each other non-trivially. If we considered l ( Z d ) and someenumeration of its basis { δ n k } k ∈ Z and define P n as before, again we can prove P ( E n,m ) = 1 .2. Consider the Hilbert space ⊕ Ni =1 l ( Z ) , and ∆ as adjacency operator on each space separately. Set π i : Z → Z be surjective map for i = 1 , · · · , N , and define P n = P Ni =1 δ iπ i ( n ) , where { δ in } n ∈ Z isbasis for each l ( Z ) . Then for this case also P ( E n,m ) = 1 .In case the measure µ has compact support on R and A is bounded, none of the σ ωn can have full supporton R , and so M = N similar to the rank case of Jakˇsi´c and Last [11]. The approach to gain information about the spectral measure is by using the matrix valued function: P n ( H ω − z ) − P n : P n H → P n H for z ∈ C + . Since we will be working with n ∈ M , it is enough to look at the above matrices. These aretermed Matrix valued Herglotz functions or Birman-Schwinger operators. Birman-Schwinger principlewas developed for compact perturbations in [4, 19] and some notable applications can be found in[5, 12, 20].We will be working with the above as Matrix valued Herglotz functions whose properties can befound in [8]. By combining theorem A.2 (see [8, Theorem 5.4] for proof) and A.3 we obtain conditionsin terms of lim ǫ ↓ P n ( H ω − E − ιǫ ) − P n .Second and third part of the theorem 1.1 are consequences of perturbations by two projections, andthe first part is because of Kolmogorov 0-1 law. Lemma 3.4 is the primary step for the first part ofthe main theorem. It tells us that the event E n,m ( Q ωn P m has same rank as P m ), is independent of2ny other perturbation, whence Kolmogorov 0-1 law applies. For second part, whenever the condition issatisfied, we have to show that for E in a full measure set, the density of the measure has same rank forboth indices; this is done in corollary 3.6. For the last part, the second part of the theorem 1.1 helps byasserting that absolute continuous parts are equivalent and for the singular part we only need to considerthe lowest (Hausdorff) dimensional part. This is the case because we are using Poltoratskii’s theorem[16], and lowest dimensional part of the spectrum contributes maximum rate of growth to the Herglotzfunction as its argument approaches the boundary of C + . Corollary 3.8 gives the equivalence for thelowest dimensional parts of the measure.Before attempting to handle the problem, it is important to note that the set of perturbations wherethe procedure may not be applicable is a measure zero set . Lemma 2.1 gives such a statement, and alsotells us that for almost all perturbation, the measure of singular part (w.r.t to Lebesgue measure) is zero. Following lemma is a result concerning the zero sets of polynomials. This lemma helps in the proof ofour main theorem by ensuring that for almost all perturbation the set where singular part lie is measurezero.
Lemma 2.1.
For a σ -finite positive measure space ( X, B , m ) , and a collection of measurable functions a i : X → C , define the function f ( λ, x ) = 1 + P Nn =1 λ n a n ( x ) . The set defined by Λ f = { λ ∈ C | m { x ∈ X | f ( λ, x ) = 0 } > } (2.1) is countable.Proof. The proof is by induction on degree of f (as a polynomial of λ ). We will use the notation: S λ = { x ∈ X | f ( λ, x ) = 0 } (2.2)By definition the sets S λ are measurable.Base case of induction is N = 1, so f ( λ, x ) = 1 + λa ( x ). Clearly for λ = λ ∈ C we have S λ ∩ S λ = φ . Since, if x ∈ S λ ∩ S λ then1 + λ a ( x ) = 0 & 1 + λ a ( x ) = 0 ⇒ λ = − a ( x ) = 1 λ ⇒ λ = λ but we assumed λ = λ . Since ( X, m ) is σ -finite, we have a countable collection { X i } i ∈ N such that ∪ i X i = X and for each i we have m ( X i ) < ∞ . Now for each λ ∈ C and n ∈ N define S λ,n = S λ ∩ X n ,so we have ∪ n S λ,n = S λ , and ∪ λ ∈ Λ f S λ,n ⊂ X n . We have X λ ∈ Λ f m ( S λ,n ) = m ( ∪ λ ∈ Λ f S λ,n ) ≤ m ( X n ) < ∞ , so only for countably many λ ∈ Λ f we have m ( S λ,n ) = 0. Set Λ n = { λ ∈ Λ f | m ( S λ,n ) > } , we haveΛ f = ∪ n ∈ N Λ n , but since countable union of countable set is countable, we get Λ f countable. Thiscompletes base case.Now assume the induction hypothesis, i.e for measurable functions a i : X → C , and f ( λ, x ) =1 + P Nn =1 λ n a n ( x ), the set Λ f is countable.We have to show for f ( λ, x ) = 1+ P N +1 n =1 λ n a n ( x ), the set Λ f is countable. First we define the relation ∼ for elements of Λ f ; for µ, ν ∈ Λ f we define µ ∼ ν if there exists { λ i } ki =1 such that λ = µ , λ k = ν and m ( S λ i ∩ S λ i +1 ) > i = 1 , · · · , k −
1. For µ ∈ Λ f we have µ ∼ µ because m ( S µ ) > ∼ is reflexive. If µ ∼ ν for µ, ν ∈ Λ f , then we have a sequence { λ i } ki =1 such that λ = µ and λ k = ν and m ( S λ i ∩ S λ i +1 ) >
0, hence choosing ˜ λ i = λ k − i +1 we get ν ∼ µ and so ∼ is symmetric. If µ ∼ ν and ν ∼ η , then we have sequences { α i } pi =1 and { β i } qi =1 such that α = µ , α p = β = ν and β q = η , sodefining the sequence { λ i } p + qi =1 defined as λ i = α i for i ≤ p and λ i = β i − p for i > p we get µ ∼ η givingtransitivity of ∼ . So ∼ is a equivalence relation on Λ f , and can break the set Λ f into equivalence classesindexed by ˜Λ = Λ f / ∼ , where we view [ λ ] ∈ ˜Λ as [ λ ] = { µ ∈ Λ f | µ ∼ λ } and define S [ λ ] = ∪ µ ∈ [ λ ] S µ .3irst we will show for any [ λ ] ∈ ˜Λ, the set [ λ ] is countable. Let λ ∈ Λ f , so we have the m ( S λ ) = 0. Wewill restrict to subspace S λ , on this space f ( ν, x ) can be written as f ( ν, x ) = λ ( λ − ν ) (cid:16) P Nn =1 ˜ a n ( x ) ν n (cid:17) (since λ is a solution). So we have the new function ˜ f ( ν, x ) = 1 + P Nn =1 ˜ a n ( x ) ν n , and by our assumption(induction hypothesis) we get Λ ˜ f is countable. For any ν ∈ Λ f with m ( S λ ∩ S ν ) = 0 implies ν ∈ Λ ˜ f , sofor fixed λ ∈ Λ f the set of ν ∈ Λ f such that m ( S λ ∩ S ν ) = 0 is countable.Next choose λ ∈ Λ f , and set A = { λ } , and define A i = ∪ β ∈ A i − { ν ∈ Λ f | m ( S ν ∩ S β ) = 0 } ∀ i ∈ N by previous step each A i are countable. So ∪ ∞ i =0 A i is countable. By definition of ∼ we have [ λ ] = ∪ ∞ i =0 A i .Now we will prove ˜Λ is countable. By definition m ( S [ λ ] ) > λ ] ∈ ˜Λ, and for [ λ ] = [ µ ] ∈ ˜Λ wehave m ( S [ λ ] ∩ S [ ν ] ) = 0. For n ∈ N define S [ λ ] ,n = S [ λ ] ∩ X n , then we have X n ∈ ˜Λ m ( S [ λ ] ,n ) = m ( ∪ [ λ ] ∈ ˜Λ S [ λ ] ,n ) ≤ m ( X i ) < ∞ From last step only countably many [ λ ] can have m ( S [ λ ] ,n ) >
0. Call ˜Λ n = { [ λ ] ∈ ˜Λ | m ( S [ λ ] ,n ) > } (which are countable); for any [ λ ] ∈ ˜Λ we have0 < m ( S [ λ ] ) ≤ X n ∈ N m ( S [ λ ] ,n )So [ λ ] ∈ ˜Λ for some n ∈ N we have m ( S [ λ ] ,n ) >
0, hence ˜Λ = ∪ n ∈ N ˜Λ n ; giving us ˜Λ is countable.Since Λ f = ∪ [ λ ] ∈ ˜Λ [ λ ] and both the sets are countable we get the countability of Λ f . Remark 2.2.
It should be clear that above result holds for a function of the type f ( λ, x ) = P Nn =0 a n ( x ) λ n on the set { x ∈ X | a ( x ) = 0 } . One should note that one cannot extend the result for whole of X .We can view f ( λ, x ) = λ N (cid:16)P Nn =0 a N − n ( x ) (cid:0) λ (cid:1) n (cid:17) , and so the result also holds on the set { x ∈ X | a N ( x ) = 0 } . Corollary 2.3.
For a σ -finite positive measure space ( X, B , m ) and a collection of functions a i : X → C , b i : X → C , define the function f ( λ, x ) = P Ni =1 a i ( x ) λ i P Ni =1 b i ( x ) λ i , then the set Λ f = { λ ∈ C | m { x ∈ X | f ( λ, x ) = 0 } 6 = 0 } (2.3) is countableProof. Set g ( λ, x ) = 1+ P Nn =1 a n ( x ) λ n , then { ( x, µ ) ∈ X × C | f ( λ, x ) = 0 } ⊆ { ( x, µ ) ∈ X × C | g ( λ, x ) = 0 } .So by lemma 2.1 we get the desired result.We will need the spectral averaging result (see[6, Corollary 4.2] for proof): Lemma 2.4.
Let E λ ( · ) be the spectral family for the operator A λ = A + λP , where A is self adjointoperator, and P is a rank N projection. Then for M ⊂ R such that | M | = 0 (Lebesgue measure), wehave P E λ ( M ) P = 0 for Lebesgue almost all λ . This lemma guarantees us that we can omit any Lebesgue measure zero set from any analysis thatfollows. Following lemma from [9, Proposition 2.1] will be used extensively, as it guarantees the existenceof limits, almost surely. We denote H φ to be the cyclic subspace generated by A and φ ∈ H . Lemma 2.5.
Let A be a self adjoint operator on a separable Hilbert space H with φ, ψ ∈ H such that H φ H ψ . Then for a.e E ∈ R (Lebesgue) the limit lim ǫ ↓ (cid:10) φ, ( A − E − ιǫ ) − ψ (cid:11) = (cid:10) φ, ( A − E − ι − ψ (cid:11) exists and is non-zero. We note that the limit always exists a.e E , and it is non-zero if and only if H φ H ψ . We will needPoltoratskii’s theorem [16]. 4 heorem 2.6. For any complex valued Borel measure µ on R and for f ∈ L ( R , dµ ) , with Boreltransform F µ ( z ) = R dµ ( z ) x − z lim ǫ → F fµ ( E + ιǫ ) F µ ( E + ιǫ ) = f ( E ) for a.e E with respect to µ -singular. A proof can be found in [10]. This theorem will be used for proof of equivalence of measure for thesingular part in lemma 3.7 and corollary 3.8.
In this section we will be working with ( A, H , { P i } i =1 ), where A is a self adjoint operator on the Hilbertspace H , and { P i } i =1 are three rank N projections. We will work with the case that the measures tr ( P i E A ( · ) P i ) are not equivalent to lebesgue measure (hence using Riesz theorem [17], the Borel transformof these measures are non-zero on the upper half plane). Define A µ = A + µP , G ij ( z ) = P i ( A − z ) − P j and G µij ( z ) = P i ( A µ − z ) − P j for i, j = 1 , , z ∈ C + , and will use the notation g ( E + ι
0) := lim ǫ ↓ g ( E + ιǫ )for E ∈ R (whenever the limit exists). Using the relation A − − B − = B − ( B − A ) A − = A − ( B − A ) B − , we have G µ ( z ) = G ( z )( I + µG ( z )) − (3.1)( I + µG ( z ))( I − µG µ ( z )) = I (3.2) G µij ( z ) = G ij ( z ) − µG i ( z )( I + µG ( z )) − G j ( z ) ( i, j ) = (1 ,
1) (3.3)For any E ∈ R such that G ( E + ι
0) exists and finite, and f : (0 , ∞ ) → C be such that lim ǫ → f ( ǫ ) = 0look at lim ǫ ↓ f ( ǫ ) G µ ( E + ιǫ ) using equation (3.2)lim ǫ ↓ f ( ǫ )( I − µG µ ( E + ιǫ ))( I + µG ( E + ιǫ )) − f ( ǫ ) I = 0( I + µG ( E + ι (cid:18) lim ǫ ↓ f ( ǫ ) G µ ( E + ιǫ ) (cid:19) = 0So we get range (cid:18) lim ǫ ↓ f ( ǫ ) G µ ( E + ιǫ ) (cid:19) ⊆ ker ( I + µG ( E + ι ⊆ ker ( ℑ G ( E + ι ℑ G ( E + ι ≥ P H = ker ( ℑ G ( E + ι ⊕ ker ( ℑ G ( E + ι ⊥ with range ( ℑ G ( E + ι ker ( ℑ G ( E + ι ⊥ , so on ker ( ℑ G ( E + ι ⊥ we have ℑ G ii ( E + ι >
0. Thisfact will be used in identifying appropriate subspaces. We will need some preliminary results before weattempt to prove our main results. The Following lemma relates the invertibility of the matrices G µ ( z )with the ranks of Q P and P . Lemma 3.1.
Let A be a self-adjoint operator on a Hilbert space H and P and P be two projections ofrank N . Let H i denote the cyclic subspace generated by A and P i H and Q i : H → H i be the canonicalprojection onto that subspace, for i = 1 , . If Q P has same rank as P , then P ( A − z ) − P is almostsurely invertible for a.e z ∈ C + .Proof. Let φ ∈ P H \ { } . Since Q P has same rank as P , we have 0 = Q φ ∈ H (if it is zero, then ker ( Q ) ∩ P H = { } and so rank ( Q P ) < rank ( P )), so there is ψ ∈ P H and f ∈ L ( R , dµ ψ ) suchthat Q φ = f ( A ) ψ . So0 = h Q φ, Q φ i = h ψ, f ∗ ( A ) Q φ i = h ψ, f ∗ ( A ) φ i = Z ¯ f ( x ) dµ ψ,φ ( x )since Q commutes with any functions of A . So the measure µ ψ,φ is non-zero, hence the Borel transform Z dµ ψ,φ ( x ) x − z = (cid:10) ψ, ( A − z ) − φ (cid:11) ,
5s almost surely non-zero on C + .So for each vector φ ∈ P H there exists a ψ ∈ P H such that (cid:10) ψ, ( A − z ) − φ (cid:11) is non-zero, in otherwords P ( A − z ) − P is an injection, and since P ( A − z ) − P is an n × n matrix we get invertibility. Remark 3.2.
The lemma above also assures that for almost all E the matrix valued function P ( A − E − ι − P is invertible.For some z ∈ C + , the invertibility of P ( A − z ) − P gives us Q P has same rank as P . So bylooking at det( G mn ( z )) we can obtain a statement about non-orthogonality of the subspace { H i } i =1 , . Choose a basis of P i H , then G ij ( z ) is a matrix in that basis. We can write S = { E ∈ R | Entries of G ij ( E + ι
0) exists and are finite ∀ i, j = 1 , , } (3.5)Then by lemma 2.5 we know that S has full measure. Define S ij = { E ∈ S | G ij ( E + ι
0) is invertible } ∀ i, j = 1 , , S ij has full measure whenever Q i P j has same rank as P j . Remark 3.3.
On the set S , the limit G ( E + ι exists, and since det( I + µG ( E + ι P Ni =1 a i ( E ) µ i , using lemma 2.1 for almost all µ the matrix I + µG ( E + ι is invertible on a set of fullmeasure. Lemma 3.4.
Let A be self adjoint operator on Hilbert space H and { P i } i =1 be rank N projections.Define A µ = A + µP , G ij ( z ) = P i ( A − z ) − P j and G µij ( z ) = P i ( A µ − z ) − P j . If G ( E + ι is invertiblefor a.e E , then G µ ( E + ι is also invertible for a.e ( E, µ ) .Proof. From equations (3.1) and (3.3) and remark 3.3 we get G µ ( E + ι
0) = G ( E + ι − µG ( E + ι I + µG ( E + ι − G ( E + ι G µ ( E + ι G ( E + ι P a n ( E ) µ n det( I + µG ( E + ι µ the matrix G ( E + ι
0) is invertible on a set of fullmeasure.Next lemma provide the relation between the absolute continuous component of the measure.
Lemma 3.5.
On Hilbert space H we have two rank N projections P , P and a self adjoint operator A .Set A µ = A + µP , G ij ( z ) = P i ( A − z ) − P j and G µij ( z ) = P i ( A µ − z ) − P j ; set S and S as (3.5) , (3.6) .Define V µE,i = ker ( ℑ G µii ( E + ι ⊥ for each E ∈ S ∩ { x ∈ R | lim ǫ ↓ G µ ( x + ιǫ ) exists and finite } . Assume S has full measure. Then fora.e µ ( G ( E + ι − : V µE, → V µE, is injective, and ( I + µG ( E + ι V E, → V µE, is isomorphism.Proof. From the equation (3.3) and (3.2) we get G µ ( z ) = G ( z ) − µG ( z ) G ( z ) + µ G ( z ) G µ ( z ) G ( z )For E ∈ S ∩ { x ∈ R | lim ǫ ↓ G µ ( x + ιǫ ) exists and finite } , let v ∈ V µE, , and set φ = ( G ( E + ι − v ,observe (every quantity in RHS below exists and finite so limit can be taken)lim ǫ ↓ h φ, ( ℑ G µ ( E + ιǫ )) φ i = lim ǫ ↓ [ h φ, ( ℑ G ( E + ιǫ )) φ i − µ h φ, ℑ ( G ( E + ιǫ ) G ( E + ιǫ )) φ i µ h φ, ( ℑ G ( E + ιǫ ) G µ ( E + ιǫ ) G ( E + ιǫ )) φ i (cid:3) Since ℑ G µ ( E + ι
0) is positive matrix, looking at h φ, ( ℑ G µ ( E + ι φ i is enough.If h φ, ( ℑ G ( E + ι φ i = 0 which implies ( ℑ G ( E + ι φ = 0 so using (A.3) we have G ( E + ι φ = G ∗ ( E + ι φ , and solim ǫ ↓ h φ, ( ℑ G µ ( E + ιǫ )) φ i = µ h G ( E + ι φ, ( ℑ G µ ( E + ι G ( E + ι φ i− µ h φ, ℑ ( G ( E + ι G ( E + ι φ i = µ h v, ( ℑ G µ ( E + ι v i So φ ∈ V µE, and hence G ( E + ι − gives the injection.For the other assertion, let v ∈ V E, observe h v, ( I + µG ( E + ι v i = k v k + µ ( h v, ℜ G ( E + ι v i + ι h v, ℑ G ( E + ι v i )since h v, ℑ G ( E + ι v i 6 = 0, so the above equation cannot be zero for any µ ∈ R . So on V E, theoperator ( I + µG ( E + ι φ = ( I + µG ( E + ι v , observelim ǫ → h φ, ( ℑ G µ ( E + ιǫ )) φ i = lim ǫ → (cid:10) φ, ℑ ( G ( E + ιǫ )( I + µG ( E + ιǫ )) − ) φ (cid:11) = (cid:10) ( I + µG ( E + ι − φ, ( ℑ G ( E + ι I + µG ( E + ι − φ (cid:11) = h v, ( ℑ G ( E + ι v i 6 = 0This gives the isomorphism ( I + µG ( E + ι V E, → V µE, .This only gives the injection between the absolutely continuous spectral subspaces. One cannotexpect more from this setting. By a second perturbation we obtain an isomorphism, which is attainedin the next corollary. Corollary 3.6.
Let A be self adjoint operator on Hilbert space H , and P , P are two rank N projections.Set A µ = A + µ P + µ P and G ij ( z ) = P i ( A − z ) − P j , G µ ,µ ij ( z ) = P i ( A µ ,µ − z ) − P j for i, j = 1 , and define the vector space V µ ,µ E,i = ker ( ℑ G µ ,µ ii ( E + ι ⊥ for each E ∈ S ∩ { x ∈ R | lim ǫ ↓ G µ ,µ ii ( x + ιǫ ) exists and finite for i = 1 , } . Assume S , S have fullmeasure. Then for a.e µ , µ the two vector space V µ ,µ E, and V µ ,µ E, are isomorphic.Proof. This is just application of lemma 3.5. For E in full measure set we have V µ ,µ E, ֒ → V µ ,µ E, where the map is ( G µ , ( E + ι − . Lemma 3.4 tells us G µ , ( E + ι
0) is also invertible for almost all µ . Now we can do the same thing other way around: V µ ,µ E, ֒ → V µ ,µ E, Since we are working in finite dimensional spaces ( V µ ,µ E,i are finite dimensional), injection in bothdirection tells us that they are isomorphic.The next lemma is similar to lemma 3.5, but for the singular part. The conclusion is for subspaceswhere growth of the Herglotz function is maximum or equivalently its associated measure has lowest(Hausdorff) dimension. We will use the fact that a matrix valued measure Σ n ( · ) = P n E A ( · ) P n isabsolutely continuous with respect to the trace measure σ n ( · ) = tr (Σ n ( · )) and so lim ǫ ↓ σ n ( E + ιǫ ) Σ n ( E + ιǫ ) = M ( E ) is L w.r.t σ n -singular ( σ n ( z ) , Σ n ( z ) are corresponding Borel transform). Lemma 3.7.
On Hilbert space H we have two rank N projections P , P and a self adjoint operator A .Set A µ = A + µP , G ij ( z ) = P i ( A − z ) − P j and G µij ( z ) = P i ( A µ − z ) − P j . Set f E ( ǫ ) = tr ( G µ ( E + ιǫ )) − and E ∈ R be such that f E ( ǫ ) ǫ ↓ −−→ , define ˜ V µE,i = ker (cid:18) lim ǫ ↓ f E ( ǫ ) G µii ( E + ιǫ ) (cid:19) ⊥ ssume S defined as (3.6) has full measure, then for E ∈ S such that f E ( ǫ ) ǫ ↓ −−→ defined as in (3.5) the map ( G ( E + ι − : ˜ V µE, → ˜ V µE, is injective. So the measure σ µ (where σ µi ( · ) = tr (cid:0) P i E A µ ( · ) P i (cid:1) ) is absolutely continuous with respect to σ µ -singular.Proof. Using i, j = 2 in (3.3), we have G µ ( z ) = G ( z ) − µG ( z ) G ( z ) + µ G ( z ) G µ ( z ) G ( z )Since we are working with E ∈ S , the limits for G ij ( E + ι
0) exists for i, j = 1 ,
2. For φ, ψ ∈ P H wehave h ψ, G µ ( E + ιǫ ) φ i = h ψ, G ( E + ιǫ ) φ i − µ h ψ, G ( E + ιǫ ) G ( E + ιǫ ) φ i + µ h ψ, G ( E + ιǫ ) G µ ( E + ιǫ ) G ( E + ιǫ ) φ i lim ǫ ↓ f E ( ǫ ) h ψ, G µ ( E + ιǫ ) φ i = µ lim ǫ ↓ f E ( ǫ ) h ψ, G ( E + ιǫ ) G µ ( E + ιǫ ) G ( E + ιǫ ) φ i = µ (cid:28) ψ, G ( E + ι (cid:18) lim ǫ ↓ f E ( ǫ ) G µ ( E + ιǫ ) (cid:19) G ( E + ι φ (cid:29) And now using (3.4) and (A.3) we have (cid:28) ψ, G ( E + ι (cid:18) lim ǫ ↓ f E ( ǫ ) G µ ( E + ιǫ ) (cid:19) G ( E + ι φ (cid:29) = (cid:28) ψ, G ( E + ι ∗ (cid:18) lim ǫ ↓ f E ( ǫ ) G µ ( E + ιǫ ) (cid:19) G ( E + ι φ (cid:29) From above if φ = G ( E + ι − v for v ∈ ˜ V µE, , then φ ∈ ˜ V µE, , giving us that the map G ( E + ι − isinjection.Finally lim ǫ ↓ tr ( G µ ( E + ιǫ )) tr ( G µ ( E + ιǫ )) = tr (cid:18) G ( E + ι ∗ (cid:18) lim ǫ ↓ f E ( ǫ ) G µ ( E + ιǫ ) (cid:19) G ( E + ι (cid:19) where RHS is L for σ µ -singular by lemma 2.6 (Poltoratskii’s theorem). Corollary 3.8.
Let A be self adjoint operator on Hilbert space H , and P , P are two rank N projections.Set A µ = A + µ P + µ P , G ij ( z ) = P i ( A − z ) − P j and G µ ,µ ij ( z ) = P i ( A µ ,µ − z ) − P j for i, j = 1 , .Let E ∈ S ∩ S (defined as in (3.6) ) and tr ( G µ ,µ ii ( E + ιǫ )) − ǫ ↓ −−→ for either i = 1 , , then ˜ V µ ,µ E,i = ker (lim ǫ ↓ tr ( G µ ,µ ii ( E + ιǫ )) − G µ ,µ ii ( E + ιǫ )) ⊥ i = 1 , are isomorphic. In particular the singular part of trace measure associated with G µ ,µ ii are equivalent toeach other.Proof. Define ˜ V µ ,µ E,i,j = ker (lim ǫ ↓ tr ( G µ ,µ jj ( E + ιǫ )) − G µ ,µ ii ( E + ιǫ )) ⊥ This is exactly like corollary 3.6. By action of lemma 3.7 we have V µ ,µ E, , ֒ → V µ ,µ E, , & V µ ,µ E, , ֒ → V µ ,µ E, , where first is given by G ,µ ( E + ι − and second is given by G µ , ( E + ι − which are a.e (with respectto perturbation µ , µ ) invertible because of lemma 3.4. Because of the second conclusion of the previouslemma 3.7 we have lim ǫ ↓ tr ( G µ ( E + ιǫ )) tr ( G µ ( E + ιǫ )) exists for a.e tr ( P E A µ ( · ) P )-singular , lim ǫ ↓ tr ( G µ ( E + ιǫ )) tr ( G µ ( E + ιǫ )) exists for a.e tr ( P E A µ ( · ) P )-singular . So as a vector space V µ ,µ E,i,j = V µ ,µ E,i,i = V µ ,µ E,i for a.e tr ( P i E A µ ( · ) P i )-singular. Since we have injectionboth direction and finite dimensionality of the spaces involved, we get the isomorphism.8 .1 Proof of Main theorem Proof.
The notation we will use is G ωnm ( z ) = P n ( H ω − z ) − P m ∀ n, m ∈ N and for some p ∈ M we will denote H ωµ,p = H ω + µP p and G ω,µ,pnm ( z ) = P n ( H ωµ,p − z ) − P m ∀ n, m ∈ M
1. For n, m ∈ M , let ω ∈ E n,m , using lemma 3.1 we get G ωnm ( z ) is almost surely invertible. For any p ∈ N , we have H ωµ,p , and using lemma 3.4 we get G ω,µ,pnm ( z ) is also almost surely invertible foralmost all µ . So we get, if ω ∈ E n,m , then ˜ ω ∈ E n,m (˜ ω is defined by ω n = ˜ ω n ∀ n ∈ M \ { p } ) or inother words, E n,m is independent of the ω p for any p ∈ M . We can repeat the procedure and showthat E n,m is independent of { ω p i } Ki =1 for p i ∈ M . So we can use Kolmogorov 0-1 law to concludethat P ( E n,m ) ∈ { , } .2. For any n ∈ M , we have ( H ω , H n,ω ) is unitary equivalent to ( M id , L ( R , Σ ωn , C N )) (see theoremA.3). For m ∈ M such that P ( E n,m ∩ E m,n ) = 1, we have to show (Σ ωn ) ac is equivalent to (Σ ωm ) ac .Using (5) of theorem A.2 we have d (Σ ωn ) ac ( E ) = 1 π ℑ G ωnn ( E + ι dE For ω ∈ E n,m , we can write the operator H ˜ ω = H ω + µ P n + µ P m , and using corollary 3.6 we get V ˜ ωn are isomorphic to V ˜ ωm , where V ˜ ωi = ker (cid:0) P i ( H ˜ ω − E − ι − P i (cid:1) ⊥ Since ℑ G ωnn ( E + ι
0) = ℑ (cid:0) P n ( H ω − E − ι − P n (cid:1) , the isomorphism gives the equivalence. By proofof part (1), we know E n,m is independent of ω n and ω m , so the result holds for a.e ω .3. For n, m ∈ M such that P ( E n,m ∩ E m,n ) = 1. Let ω ∈ E n,m , define H ˜ ω = H ω + µ n P n + µ m P m (almost always ˜ ω ∈ E n,m ), then corollary 3.8 gives the equivalence of the trace measure for singularpart. As for absolute continuous part, second part of the theorem gives the equivalence. Acknowledgement
I would like to thank M. Krishna for discussions and helpful suggestions. This work is partially supportedby IMSc Project 12-R&D-IMS-5.01-0106.
A Appendix
For A ∈ M n ( C ), we have the decomposition A = ℜ A + ι ℑ A , where both ℜ A and ℑ A are self adjoint. For A , A , A , A ∈ M n ( C ) such that A ∗ A − A ∗ A = I and A ∗ A = A ∗ A , A ∗ A = A ∗ A .Define ˜ A = ( A − A A )( A − A A ) − then we have ℑ ˜ A = 12 ι ( ˜ A − ˜ A ∗ )= 12 ι (cid:16) ( A − A A )( A − A A ) − − (cid:0) ( A − A A ) − (cid:1) ∗ ( A − A A ) ∗ (cid:17) = (cid:0) ( A − A A ) − (cid:1) ∗ ℑ A (cid:0) ( A − A A ) − (cid:1) (A.1)This is equivalent to mobius transforms for complex numbers. These kind of transform will play impor-tant role for determining Σ ωn .Following lemma gives some of the properties of the off-diagonal terms of Herglotz matrices.9 emma A.1. For A ij ∈ M n ( C ) i, j = 1 , , define A = (cid:18) A A A A (cid:19) with the property that ℑ A ≥ . Then for u, v ∈ C n (cid:12)(cid:12)(cid:12)(cid:12)(cid:28) u, A − A ∗ ι v (cid:29)(cid:12)(cid:12)(cid:12)(cid:12) ≤ h u, ( ℑ A ) u i h v, ( ℑ A ) v i (A.2)It’s proof follows same steps as 2 × ℑ A ) v = 0 ⇒ A v = A ∗ v, & ( ℑ A ) u = 0 ⇒ A u = A ∗ u (A.3)also ℑ tr ( A ) = 0 ⇒ A = A ∗ , & ℑ tr ( A ) = 0 ⇒ A = A ∗ (A.4)We will use Matrix valued Herglotz function to work, following theorem from [8, Theorem 5.4] listsome of the useful properties. Proof of these statements are similar to that of scalar case. Theorem A.2.
Let M : C + → M n ( C ) be a matrix-valued Herglotz function, then1. M ( z ) has finite normal limits, i.e M ( E ± ι
0) = lim ǫ → M ( E + ιǫ ) for a.e E ∈ R .2. If for each diagonal element M ii ( z ) , ≤ i ≤ n of M ( z ) has zero normal limit on a fixed subset of R which has positive Lebesgue measure, then M ( z ) = C where C is a constant self-adjoint n × n matrix with vanishing diagonal elements.3. There exists a matrix-valued measure Σ on the bounded Borel subset of R satisfying Z h v, d Σ( x ) v i (1 + x ) − ∀ v ∈ C n such that the Nevanlinna, respectively, Riesz-Herglotz representation M ( z ) = C + Dz + Z R d Ω( x ) (cid:0) ( x − z ) − − x (1 + x ) − (cid:1) ∀ z ∈ C + C = M ( ι ) , D = lim η ↑∞ ιη M ( ιη ) holds.4. The Stieltjes inversion formula for Σ is π lim ǫ ↓ Z λ λ dλ ℑ ( M ( λ + ιǫ )) = Σ( { λ } ) + Σ( { λ } )2 + Σ(( λ , λ ))
5. The absolute continuous part of the measure is given by d Σ ac ( λ ) = 1 π ℑ ( M ( λ + ι dλ
6. Any poles of M ( z ) are simple and are located on real axis. We will use the given version of spectral theorem.
Theorem A.3.
Let A be a self adjoint operator on Hilbert space H and P be an rank N projection.Let the vector space P H has basis { δ n } Nn =1 and define the cyclic subspace generated by A and δ n by H n for all n = 1 , · · · , N , and define the subspace H P = N X i =1 H i also let Σ A denote the spectral projection of A , then ( L ( R , P Σ A P, P H ) , id ) and ( H P , A ) are unitarilyequivalent, where id is multiplication by identity. roof. We have a basis of P H given by { δ n } Nn =1 . So define the map U : L ( R , P Σ A P, C N ) → H P as U ( f , · · · , f N ) N X i =1 f i ( A ) δ i the map is injection because 0 = k U ( f , · · · , f n ) k = N X i,j =1 h f i ( A ) δ i , f j ( A ) δ j i = N X i,j =1 Z ¯ f i ( x ) f j ( x ) dµ ij ( x )where µ ij ( · ) is the measure h δ i , Σ A ( · ) δ j i , so the last equation tells us Z h f ( x ) , dP Σ A P ( x ) f ( x ) i = 0where f ( x ) = ( f ( x ) , · · · , f N ( x )), and ( P Σ A ( · ) P ) ij = h δ i , Σ A ( · ) δ j i . So the map U is injection. The mapis surjection because for φ ∈ H P by definition we can find { f i } Ni =1 such that f i ∈ L ( R , µ ii , C ) such that φ = P Ni =1 f i ( A ) δ i , and so ( f , · · · , f N ) maps to φ . Finally we have to show U M = AU , by definition( M f )( x ) = xf ( x ) U ( M f ) = N X i =1 ( Af i ( A )) δ i = A N X i =1 f i ( A ) δ i = A ( U f )completing the proof.
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