Branching form of the resolvent at threshold for discrete Laplacians - PowerPoint PPT Presentation

Branching form of the resolvent at threshold for discrete Laplacians Kenichi ITO (Kobe University) joint work with Arne JENSEN (Aalborg University) 9 October 2016

Introduction: Discrete Laplacian ◦ Thresholds generated by critical values For any function u : Z d → C define △ u : Z d → C by d � � ∑ for n ∈ Z d . ( △ u )[ n ] = u [ n + e j ] + u [ n − e j ] − 2u [ n ] j = 1 The operator H 0 = − △ is bounded and self-adjoint on H = ℓ 2 ( Z d ) . It has spectrum σ ( H 0 ) = σ ac ( H 0 ) = [ 0, 4d ] , and thresholds τ ( H 0 ) = { 0, 4, . . . , 4d } . 1

H = L 2 ( T d ) , T = R / ( 2π Z ) , and define the Fourier transform Let � F : H → � H and its inverse F ∗ : � H → H by ( F u )( θ ) = ( 2π ) − d/2 ∑ e − i nθ u [ n ] , n ∈ Z d ∫ ( F ∗ f )[ n ] = ( 2π ) − d/2 T d e i nθ f ( θ ) d θ, and then F H 0 F ∗ = Θ ( θ ) = 2d − 2 cos θ 1 − · · · − 2 cos θ d . Since ∂ j Θ ( θ ) = 2 sin θ j , the critical points of signature ( p, q ) are { } θ ∈ { 0, π } d ; # { θ j = 0 } = p, # { θ j = π } = q γ ( p, q ) = . Hence the critical values 0, 4d are thresholds of elliptic type, and 4, . . . , 4 ( d − 1 ) are those of hyperbolic type. 2

◦ Purpose: Asymptotic expansion of resolvent Q. Can we compute an asymptotic expansion of the resolvent R 0 ( z ) = ( H 0 − z ) − 1 ∼ ?? as z → 4q for q = 0, 1, . . . , d ? Note that the resolvent R 0 ( z ) has a convolution kernel: ∫ e i nθ k ( z, n ) = ( 2π ) − d R 0 ( z ) u = k ( z, · ) ∗ u ; Θ ( θ ) − z d θ. T d A. Yes. By localizing around γ ( p, q ) ⊂ T d and changing variables the situation reduces to that for an ultra-hyperbolic operator. • As far as we know, an explicit asymptotics of R 0 ( z ) around a threshold seems to have been open except for 0 and 4d . 3

Ultra-hyperbolic operator (a model operator) Consider an ultra-hyperbolic operator on R d : � = ∂ 2 1 + · · · + ∂ 2 p − ∂ 2 p + 1 − · · · − ∂ 2 p + q ; p, q ≥ 0, d = p + q. The operator H 0 = − � is self-adjoint on H = L 2 ( R d ) with D ( H 0 ) = { u ∈ H ; � u ∈ H in the distributional sense } . It has spectrum  [ 0, ∞ ) if ( p, q ) = ( d, 0 ) ,  σ ( H 0 ) = σ ac ( H 0 ) = (− ∞ , 0 ] if ( p, q ) = ( 0, d ) ,  R otherwise , and a single threshold τ ( H 0 ) = { 0 } . 4

Using the Fourier transform F : H → H and its inverse F ∗ : H → H , we can write F H 0 F ∗ = Ξ ( ξ ) = ξ ′ 2 − ξ ′′ 2 ; ξ = ( ξ ′ , ξ ′′ ) ∈ R p ⊕ R q . The only critical point is ξ = 0 , and the associated critical value, or a threshold 0 is said to be 1. of elliptic type if ( p, q ) = ( d, 0 ) or ( 0, d ) ; 2. of hyperbolic type otherwise. Q ′ . Can we compute an asymptotic expansion of the resolvent R 0 ( z ) = ( H 0 − z ) − 1 ∼ ?? as z → 0 ? A ′ . Yes. In particular, square root, logarithm and dilogarithm branchings show up, depending on parity of ( p, q ) . 5

◦ Square root, logarithm and dilogarithm We always choose branches of √ w and log w such that Im √ w > 0 for w ∈ C \ [ 0, ∞ ) , − π < Im log w < π for w ∈ C \ (− ∞ , 0 ] , respectively. In addition, let us set for w ∈ C \ [ 1, ∞ ) ∫ w Li 1 ( λ ) Li 1 ( w ) = − log ( 1 − w ) , Li 2 ( w ) = d λ, λ 0 which have the Taylor expansions: For | w | < 1 ∞ w k ∞ w k ∑ ∑ Li 1 ( w ) = Li 2 ( w ) = k , k 2 . k = 1 k = 1 6

◦ Elliptic operator in odd dimensional space Theorem. Let d be odd, ( p, q ) = ( d, 0 ) , and γ > 0 . Then ∫ ρ d − 1 e ( ρx ) 2 ( √ z ) d − 2 e ( √ zx ) − 1 k γ ( z, x ) = i π d ρ, ρ 2 − z 2 � Γ ( γ ) Γ ( r ) = { r e i θ ∈ C ; θ ∈ [ 0, π ] } where � . ◦ Elliptic operator in even dimensional space Theorem. Let d be even, ( p, q ) = ( d, 0 ) , and γ > 0 . Then � γ 2 � 2 ( √ z ) d − 2 e ( √ zx ) Li 1 k γ ( z, x ) = − 1 z √ √ λx ) − ( √ z ) d − 2 e ( √ zx ) ∫ γ 2 λ ) d − 2 e ( + 1 ( d λ. 2 λ − z 0 7

Proposition. 1. The function e ( ζ ) is even and entire in ζ ∈ C d , and (− 1/4 ) | α | ∑ 2 e α ζ 2α ; e ( ζ ) = e α = α ! Γ ( | α | + d/2 ) . 2 d π d/2 α ∈ Z d + 2. For any z ∈ C the function e ( √ zx ) satisfies the eigenequation (− △ − z ) e ( √ zx ) = 0 ; △ = � d,0 . Here a branch of √ z does not matter, since e ( ζ ) is even. 8

◦ Hyperbolic case with odd-even or even-odd signature Theorem. Let ( p, q ) be odd-even or even-odd, and γ > 0 . Then 2 ( √ z ) d − 2 ψ + ( √ zx ) + χ γ ( z, x ) , k γ ( z, x ) = i π where ∫ ∫ τ d − 1 ψ + ( τx ) τ d − 1 ψ − ( τx ) χ γ ( z, x ) = − 1 d τ + 1 d τ τ 2 − z τ 2 + z 2 2 Γ ( γ ) Γ ( γ ) ∫ + 1 h + ,γ ( λ, x ) d λ λ − z 4 i 2 � Γ ( γ 2 ) with ∫ γ f ± ( σ/τ, τx ) h ± ,γ ( τ 2 , x ) = τ d − 2 d σ. σ − γ 9

◦ Hyperbolic case with even-even signature Theorem. Let ( p, q ) be even-even, and γ > 0 . Then � γ 4 � 2 ( √ z ) d − 2 ψ + ( √ zx ) Li 1 k γ ( z, x ) = − 1 + χ γ ( z, x ) , z 2 where � ∫ γ 2 � ( √ τ ) d − 2 ψ + ( √ τx ) − ( √ z ) d − 2 ψ + ( √ zx ) ∫ 0 χ γ ( z, x ) = 1 − d τ 2 τ − z − γ 2 0 ∫ γ 2 + 1 h + ,γ ( λ, x ) d λ λ − z 2 − γ 2 with ∫ γ f ± ( σ/τ, τx ) h ± ,γ ( τ 2 , x ) = τ d − 2 d σ − τ d − 2 ψ ± ( τx ) . σ τ 10

◦ Hyperbolic case with odd-odd signature Theorem. Let ( p, q ) be odd-odd, and γ > 0 . Then � � γ 2 � � �� − γ 2 4 ( √ z ) d − 2 φ + ( √ zx ) k γ ( z, x ) = − 1 Li 2 − Li 2 + χ γ ( z, x ) , z z where √ √ λx ) − ( √ z ) d − 2 φ + ( √ zx ) � γ 2 � ∫ γ 2 λ ) d − 2 φ + ( χ γ ( z, x ) = 1 ( log d λ λ − z 4 λ 0 √ √ λx ) − ( √ z ) d − 2 φ + ( √ zx ) � � ∫ 0 λ ) d − 2 φ + ( − γ 2 + 1 ( log d λ 4 λ − z λ − γ 2 ∫ γ 2 h + ,γ ( λ, x ) + 1 d λ 2 λ − z − γ 2 ∫ γ f ± ( σ/τ, τx ) − φ ± ( τx ) with h ± ,γ ( τ 2 , x ) = τ d − 2 d σ . σ τ 11

◦ Properties of φ ± ( ζ ) and ψ ± ( ζ ) The functions φ ± ( ζ ) and ψ ± ( ζ ) are entire in ζ ∈ C d , and   ∑  ∑ ζ 2α  , φ ± ( ζ ) = f ± ,α,a α ∈ Z d a ∈ J α + � �   i 2 | α | − 2 | a | + d − 2 − 1 f ± ,α,a ∑ ∑ ζ 2α   , ψ ± ( ζ ) = 2 | α | − 2 | a | + d − 2 α ∈ Z d a ∈ I α \ J α + where { � } � a ∈ Z 2 2 | α ′ | + p − 1, 2 | α ′′ | + q − 1 I α = + ; 0 ≤ a ≤ , { } J α = a ∈ I α ; | a | = | α | + ( d − 2 ) /2 , f ± ,α,a = ( ± 1 ) a ′ ( ∓ 1 ) a ′′ e ′ � 2 | α ′ | + p − 1 �� 2 | α ′′ | + q − 1 � α ′ e ′′ α ′′ . a ′ a ′′ 2 2 | α | + d − 2 12

◦ Properties of φ ± ( ζ ) and ψ ± ( ζ ) , continued The functions φ ± ( ζ ) and ψ ± ( ζ ) satisfy 1. φ ± ( ζ ) = 0 and ψ ± ( ζ ) = ψ ± (− ζ ) if ( p, q ) is odd-even or even- odd; 2. φ ± ( ζ ) = 0 and ψ ± ( ζ ) = − i d − 2 ψ ∓ ( i ζ ) if ( p, q ) is even-even; 3. φ ± ( ζ ) = i d − 2 φ ∓ ( i ζ ) and ψ ± ( ζ ) = 0 if ( p, q ) is odd-odd. In addition, for any z ∈ C (− � ∓ z ) φ ± ( √ zx ) = 0, (− � ∓ z ) ψ ± ( √ zx ) = 0. 13

◦ Outline of the results for ultra-hyperbolic operator Theorem. 1. If ( p, q ) is odd-even or even-odd, there exist op- erators F ( z ) , G ( z ) analytic at z = 0 such that R 0 ( z ) = F ( z ) √ z + G ( z ) . 2. If ( p, q ) is even-even, there exist operators F ( z ) , G ( z ) analytic at z = 0 such that � 1 � R 0 ( z ) = F ( z ) Li 1 + G ( z ) . z 3. If ( p, q ) is odd-odd, there exist operators F ( z ) , G ( z ) analytic at z = 0 such that � � 1 � � �� − 1 R 0 ( z ) = F ( z ) Li 2 − Li 2 + G ( z ) . z z 14

◦ “Very rough” strategy for proof The resolvent has a limiting convolution expression ∫ for u ∈ S ( R d ); ( R 0 ( z ) u )( x ) = lim R d k γ ( z, x − y ) u ( y ) d y γ →∞ ∫ e ixξ k γ ( z, x ) = ( 2π ) − d ξ ′ 2 − ξ ′′ 2 − z d ξ. | ξ ′ | + | ξ ′′ | <γ It suffices to expand the kernel k γ ( z, x ) , since it contains all the singular part of R 0 ( z ) . If we move on to the spherical or hyperbolic coordinates, a sin- gular part of k γ ( z, x ) takes, more or less, the standard form ∫ γ a ( ρ ) I = ρ 2 − z d ρ. 0 There could appear only the following three types of a ( ρ ) : 15

• If a ( ρ ) = 2b ( ρ 2 ) with b analytic, then ∫ γ ∫ b ( ρ 2 ) b ( ρ 2 ) ( ρ − √ z )( ρ + √ z ) d ρ = i πb ( z ) I = √ z − ρ 2 − z d ρ. − γ | z | = γ, Im z ≥ 0 • If a ( ρ ) = 2ρb ( ρ 2 ) with b analytic, then ∫ γ ∫ γ ∫ γ b ( λ ) b ( z ) b ( λ ) − b ( z ) I = λ − z d λ = λ − z d λ + d λ λ − z 0 0 0 • If a ( ρ ) = 2ρb ( ρ 2 )( log ρ 2 ) with b analytic, then ∫ γ ∫ γ ∫ γ b ( λ ) log λ b ( z ) log λ [ b ( λ ) − b ( z )] log λ I = d λ = d λ + d λ λ − z λ − z λ − z 0 0 0 16