Isotropic local laws for random matrices Antti Knowles University of Geneva With Y. He and R. Rosenthal
Random matrices Let H ∈ C N × N be a large Hermitian random matrix, normalized so that � H � ≍ 1 . Some motivations: � Quantum mechanics: Hamilton operator of a disordered quantum system (heavy nuclei, itinerant electrons in metals, quantum dots, ...). � Multivariate statistics: sample covariance matrix. Goal: Analysis of eigenvalues λ 1 � λ 2 � . . . � λ N and eigenvectors u 1 , u 2 , . . . , u N ∈ S N − 1 of H .
The key questions (1) Global eigenvalue distribution. Asymptotic behaviour of the empirical � N 1 distribution i =1 δ λ i . N λ i +1 λ i λ 1 (2) Local eigenvalue distribution. Asymptotic behaviour of individual eigenvalues. Examples: distribution of the gaps λ i − λ i +1 or the largest eigenvalue λ 1 . (3) Distribution of eigenvectors. Localization / delocalization of the eigenvectors. Distribution of the components � v , u i � .
Global and local laws The Green function G ( z ) . . = ( H − zI ) − 1 is the right tool to address the questions (1) – (3). Writing z = E + i η ∈ C + , we have N Im 1 1 η � N Tr G ( z ) = ( λ i − E ) 2 + η 2 N i =1 λ i E η Observation: η = Im z is the spectral resolution. � Global law: control of G ( z ) for η ≍ 1 . � Local law: control of G ( z ) for η ≫ 1 N . To answer the questions (2) and (3), one needs a local law.
Deterministic equivalent of Green function 1 One usually needs control of G ( z ) as a matrix, and not just of N Tr G ( z ) . Goal: There is a deterministic matrix M ( z ) , the deterministic equivalent of G ( z ) , such that G ( z ) − M ( z ) is small for η ≫ 1 N with high probability. What does “ G − M small” mean? Canonical notions of smallness (operator topologies): control of (i) |� v , ( G − M ) w �| , (ii) | ( G − M ) v | , (iii) � G − M � for all deterministic v , w ∈ S N − 1 . In fact, already for H = GUE it is easy to see that (ii) and (iii) blow up. Control of (i) is the strongest one can hope for (isotropic control of G ).
Example: Wigner matrices The entries ( H ij . . 1 � i � j � N ) are independent and satisfy E | H ij | 2 = 1 E H ij = 0 , N . The deterministic equivalent is M ( z ) = m ( z ) I where √ � 2 4 − x 2 1 m ( z ) . . = d x 2 π x − z − 2 is the Stieltjes transform of the semicircle law.
Some history Local law η ≫ 1 N for Wigner matrices: (a) Tr( G − M ) [Erd˝ os, Schlein, Yau; 2009] (b) ( G − M ) ij [Erd˝ os, Yau, Yin; 2010] (c) � v , ( G − M ) w � [Knowles, Yin; 2011] More general models (sparse random matrices, covariance matrices, deformed matrices, ...): [Ajanki, Erd˝ os, Knowles, Kr¨ uger, Lee, Schnelli, Yau, Yin, . . . ] Two key steps in all proofs: � Deterministic step: stability of self-consistent equation. Identify M as the solution of a self-consistent equation Π( M ) = 0 . Prove that Π( Q ) ≈ 0 = ⇒ Q ≈ M . � Stochastic step: derivation of the self-consistent equation. Prove that Π( G ) ≈ 0 with high probability.
Derivation of the self-consistent equation: folklore Use Schur’s complement formula to write 1 G ii = , k,l � = i H ik G ( i ) − z − H ii − � kl H li and large deviation estimates to show, with high probability, H ii ≈ 0 and kl H ki ≈ 1 kk ≈ 1 H ik G ( i ) H ik G ( i ) G ( i ) � � � � kl H li ≈ G kk . N N k,l � = i k � = i k � = i k Average over i . Works very well for Wigner matrices and some generalizations theoreof. Problems: (i) Matrix entries have to be independent. (ii) Expectation of H has to be diagonal. (iii) Does not give control of � v , ( G − M ) w � . This requires an additional, difficult, step.
Alternative approach [He, K, Rosenthal; 2016] New way to derive self-consistent equations, overcoming all of the above problems: (i) Admits a very general relationship between matrix entries and the independent random variables. (Can also handle models with short-range correlations, [Erd˝ os, Kr¨ uger, Schr¨ oder; 2017].) (ii) Completely insensitive to the expectation of H . (iii) Yields control of � v , ( G − M ) w � from the outset. Key idea: instead of working on entire rows and columns (Schur’s formula), work on individual entries (resolvent/cumulant expansion).
Resolvent / cumulant expansion Resolvent expansion in individual entries: . = ( H ( ij ) − zI ) − 1 . ( H ( ij ) ) kl . G ( ij ) ( z ) . . = 1 { i,j }� = { k,l } H kl , Starting point: trivial identity I + zG = HG . Then write � E ( HG ) ii = E H ij G ji j � � G ( ij ) − G ( ij ) jj H ji G ( ij ) − G ( ij ) ji H ij G ( ij ) � = E + · · · H ij ji ii ji j 1 � N G ( ij ) jj G ( ij ) = − E + · · · ii j 1 � = − E N G jj G ii + · · · . j Note: resolvent expansion is used twice: G �→ G ( ij ) �→ G .
The resulting algebra is beautifully summarized by the cumulant expansion ℓ 1 C k ( h ) . � k ! C k +1 ( h ) E [ f ( k ) ( h )] + R ℓ , . = ∂ k t | t =0 log E [e th ] . E [ h · f ( h )] = k =0 [Khorunzhy, Khoruzhenko, Pastur; 1996] Performs essentially the same as the resolvent expansion but more tidily. In applications, h = H ij and f ( h ) is a polynomial of resolvents. For example, the previous resolvent calculation is replaced by ℓ �� � k � 1 ∂ � � � E ( HG ) ii = E [ H ij G ji ] = k ! C k +1 ( H ij ) E + · · · G ji ∂H ij j j k =0 1 � � 1 ∂ � � = N E G ji + · · · = N E [ − G jj G ii − G ji G ij ] + · · · ∂H ij j j j | G ij | 2 = 1 Second term small by Cauchy-Schwarz and Ward identity � η Im G ii .
Sketch of results Illustrative model: general mean-field model with independent entries. The entries ( H ij . . 1 � i � j � N ) are independent and satisfy Var( H ij ) = O ( 1 N ) . Split H = W + A where A . . = E H , and define the map Π z ( M ) . S ( M ) . . = I + zM + S ( M ) M − AM , . = E [ WMW ] . Then for z ∈ C + the equation Π z ( · ) = 0 has a unique solution M ( z ) with positive imaginary part – the deterministic equivalent of G for this model. We prove that for all η ≫ 1 N � � Im M � 1 |� v , Π( G ) w �| � + Nη . Nη (Optimal in bulk and at edges.) This deals with the stochastic step – derivation of self-consistent equation. Conclude proof of local law by the deterministic step – stability analysis of self-consistent equation [Lee, Schnelli; 2013], [Ajanki, Erd˝ os, Kr¨ uger; 2016].
How to start the proof Let P vw . . = � v , Π( G ) w � , where Π( G ) = I + zM + S ( M ) M − AM = WG + S ( G ) G . By Markov’s inequality, it suffices to estimate E | P vw | 2 p = E ( WG ) vw P p − 1 vw P p ( S ( G ) G ) vw P p − 1 vw P p � � � � + E . vw vw Apply the cumulant expansion to the first term by writing � ( WG ) vw = v i W ij G j w . i,j The leading term from k = 1 , ��� � � P p − 1 vw P p − E v i G jj G i w , vw i,j ( S ( G ) G ) vw P p − 1 � vw P p � cancels the term E . Everything else has to be vw estimated – main work!
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