Horns problem, and Fourier analysis Jacques Faraut Symmetries in - PowerPoint PPT Presentation

Horn’s problem, and Fourier analysis Horn’s problem, and Fourier analysis Jacques Faraut Symmetries in Geometry, Analysis, and Spectral Analysis, on the occasion of Joachim Hilgert’s 60th birthday Paderborn, July 26, 2018

Horn’s problem, and Fourier analysis Horn’s problem, and Horn’s conjecture A and B are n × n Hermitian matrices, and C = A + B . Assume that the eigenvalues α 1 ≥ · · · ≥ α n of A , and the eigenvalues β 1 ≥ · · · ≥ β n of B are known. Horn’s problem : What can be said about the eigenvalues γ 1 ≥ · · · ≥ γ n of C = A + B ? Weyl’s inequalities (1912) γ i + j − 1 ≤ α i + β j for i + j ≤ n + 1 , γ i + j − n ≥ α i + β j for i + j ≥ n + 1 . Horn’s conjecture (1962) The set of possible eigenvalues γ 1 , . . . , γ n for C = A + B is determined by a family of inequalities of the form � � � γ k ≤ α i + β j , k ∈ K i ∈ I j ∈ J for certain admissible triples ( I , J , K ) of subsets of { 1 , . . . , n } . Klyachko has proven Horn’s conjecture, and described these admissible triples (1998) .

Horn’s problem, and Fourier analysis n = 3, α = (3 . 5 , 1 . 4 , − 4 . 9), β = (2 , − 0 . 86 , − 1 . 14). Weyl’inequalities gives a 1 ≤ γ 1 ≤ b 1 a 2 ≤ γ 2 ≤ b 2 a 3 ≤ γ 3 ≤ b 3 In the plane x 1 + x 2 + x 3 = 0 , these inequalities determine a hexagon.

Horn’s problem, and Fourier analysis α + β α ε 23 ε 13 ( ε ij = e i − e j ) ε 12

Horn’s problem, and Fourier analysis One observes that the vertices of this hexagon are the points α + σ ( β ) ( σ ∈ S 3 ). This is a special case of the following Theorem (Lidskii-Wielandt) The set H ( α, β ) of possible γ = ( γ 1 , . . . , γ n ) satisfies H ( α, β ) ⊂ α + C ( β ) , where C ( β ) is the convex hull of the points σ ( β ) ( σ ∈ S n ).

Horn’s problem, and Fourier analysis We consider Horn’s problem from a probabilistic viewpoint. The set of Hermitian matrices X with spectrum { α 1 , . . . , α n } is an orbit O α for the natural action of the unitary group U ( n ): X �→ UXU ∗ ( U ∈ U ( n )). Assume that the random Hermitian matrix X is uniformly distributed on the orbit O α , and the random Hermitian matrix Y uniformly distributed on O β . What is the joint distribution of the eigenvalues of the sum Z = X + Y ? This distribution is a probability measure on R n that we will determine explicitly.

Horn’s problem, and Fourier analysis Orbits for the action of U ( n ) on H n ( C ) Spectral theorem : The eigenvalues of a matrix A ∈ H n ( C ) are real and the eigenspaces are orthogonal. The unitary group U ( n ) acts on H n ( C ) by the transformations X �→ UXU ∗ For A = diag ( α 1 , . . . , α n ), consider the orbit O α = { UAU ∗ | U ∈ U ( n ) } , α = ( α 1 , . . . , α n ) ∈ R n . By the spectral theorem � � O α = X ∈ H n ( C ) | spectrum ( X ) = { α 1 , . . . , α n }

Horn’s problem, and Fourier analysis Orbital measures The orbit O α carries a natural probability measure: the orbital measure µ α , image of the normalized Haar measure ω of the compact group U ( n ) by the map U �→ UAU ∗ . For a function f on O α , � � f ( UAU ∗ ) ω ( dU ) . f ( X ) µ α ( dX ) = O α U ( n ) A U ( n )-invariant measure µ on H n ( C ) can be seen as an integral of orbital measures: it can be written � � �� � f ( U diag ( t 1 , . . . , dt n ) U ∗ ) ω ( dU ) f ( X ) µ ( dX ) = ν ( dt ) , R n H n ( C ) U ( n ) where ν is a S n -invariant measure on R n , called the radial part of µ .

Horn’s problem, and Fourier analysis If µ is a U ( n )-invariant probability measure, and X a random Hermitian matrix with law µ , then the joint distribution of the eigenvalues of X is the radial part ν of µ . Assume that the random Hermitian matrix X is uniformly distributed on the orbit O α , i.e. with law µ α , and Y uniformly distributed on O β , i.e. with law µ β , then the law of the sum Z = X + Y is the convolution product µ α ∗ µ β , and the joint distribution of the eigenvalues of Z is the radial part ν α,β of the measure µ = µ α ∗ µ β . Hence the problem is to determine this radial part ν α,β .

Horn’s problem, and Fourier analysis Fourier-Laplace transform For a bounded measure µ on H n ( C ), � e tr ( ZX ) µ ( dX ) . F µ ( Z ) = H n ( C ) If µ is U ( n )-invariant, then F µ is U ( n )-invariant as well, and hence determined by its restriction to the subspace of diagonal matrices. For Z = diag ( z 1 , . . . , z n ), T = diag ( t 1 , . . . , t n ), define � e tr ( ZUTU ∗ ) ω ( dU ) . E ( z , t ) := U ( n ) Then F µ α ( Z ) = E ( z , α ).

Horn’s problem, and Fourier analysis If µ is U ( n )-invariant, � F µ ( Z ) = R n E ( z , t ) ν ( dt ) , where ν is the radial part of µ . Taking µ = µ α ∗ µ β , � E ( z , α ) E ( z , β ) = R n E ( z , t ) ν α,β ( dt ) .

Horn’s problem, and Fourier analysis This is the product formula of the spherical functions for the Gelfand pair ( G , K ). G = U ( n ) ⋉ H n ( C ) , K = U ( n ) . The group G acts on H n ( C ) by the transformations � � g · X = UXU ∗ + A g = ( U , A ) . The spherical functions are given by ϕ z ( g ) = E ( z , t ) , where t 1 , . . . , t n are the eigenvalues of the matrix g · 0. They satisfy the functional equation: � ϕ z ( g 1 Ug 2 ) ω ( dU ) = ϕ z ( g 1 ) ϕ z ( g 2 ) ( g 1 , g 1 ∈ G ) . K With the identification ϕ z ( g 1 ) = E ( z , α ) , ϕ z ( g 2 ) = E ( z , β ) , the functional equation becomes � E ( z , α ) E ( z , β ) = R n E ( z , t ) ν α,β ( dt ) .

Horn’s problem, and Fourier analysis Harish-Chandra-Itzykson-Zuber formula A is an Hermitian matrix with eigenvalues α 1 , . . . , α n , and B with eigenvalues β 1 , . . . , β n . � � e α i β j � 1 e tr ( AUBU ∗ ) ω ( dU ) = δ n ! V n ( α ) V n ( β ) det 1 ≤ i , j ≤ n U ( n ) V n is the Vandermonde polynomial: for x = ( x 1 , . . . , x n ), � V n ( x ) = ( x i − x j ) i < j and δ n = ( n − 1 , n − 2 , . . . , 2 , 1 , 0) , δ n ! = ( n − 1)!( n − 2)! . . . 2!

Horn’s problem, and Fourier analysis Heckman’s measure Consider the projection q : H n ( C ) → D n onto the subspace D n of real diagonal matrices. Horn’s theorem The projection q ( O α ) of the orbit O α is the convex hull of the points σ ( α ) q ( O α ) = C ( α ) := Conv ( { σ ( α ) | σ ∈ S n } ) ( σ ( α ) = ( α σ (1) , . . . , α σ ( n ) )) Heckman’s measure is the projection M α = q ( µ α ) of the orbital measure µ α . It is a probability measure on R n , symmetric, i.e. S n -invariant, with compact support: support ( M α ) = C ( α ).

Horn’s problem, and Fourier analysis Fourier-Laplace transform of a bounded measure M on R n : � � R n e ( z | x ) M ( dx ) M ( z ) = The Fourier-Laplace transform of Heckman’s measure M α is the restriction to D n of the Fourier-Laplace transform of the orbital measure µ α : for Z = diag ( z 1 , . . . , z n ), � M α ( z ) = F µ α ( Z ) Therefore � M α ( z ) = E ( z , α ), and by the Harish-Chandra-Itzykson-Zuber formula, � e z i α j � 1 � M α ( z ) = δ n ! V n ( z ) V n ( α ) det 1 ≤ i , j ≤ n

Horn’s problem, and Fourier analysis Define the skew-symmetric measure � δ n ! η α = ε ( σ ) δ σ ( α ) V n ( α ) σ ∈ S n ( ε ( σ ) is the signature of the permutation σ ). Fourier-Laplace of η α : � � δ n ! δ n ! ε ( σ ) e ( z | σ ( α )) = e z i α j ) 1 ≤ i , j ≤ n η α ( z ) = � V n ( α ) det V n ( α ) σ ∈ S n By the Harish-Chandra-Itzykson-Zuber formula η α ( z ) = V n ( z ) � � M α ( z ) . Proposition � � − ∂ M α = η α V n ∂ x

Horn’s problem, and Fourier analysis � � ∂ Elementary solution of V n ∂ x Proposition Define the distribution E n on R n � �� � � E n , ϕ � = ϕ t ij ε ij dt ij n ( n − 1) 2 R + i < j ( ε ij = e i − e j ) Then � � ∂ V n E n = δ 0 . ∂ x Proof: An elementary solution of the first order differential operator ∂ ∂ ∂ x i − ∂ x j is the Heaviside distribution � ∞ � Y ij , ϕ � = ϕ ( t ε ij ) dt 0 Hence � ∗ E n = Y ij i < j � � ∂ is an elementary solution of V n . ∂ x

Horn’s problem, and Fourier analysis Theorem M α = ˇ E n ∗ η α ϕ ( x ) = ϕ ( − x ), � ˇ ( ˇ E n , ϕ � = � E n , ˇ ϕ ) Heckman’s measure M α is supported by the hyperplane x 1 + · · · + x n = α 1 + · · · + α n . Next figure is for n = 3, drawn in the plane x 1 + x 2 + x 3 = α 1 + α 2 + α 3 .

Horn’s problem, and Fourier analysis ( α 2 , α 1 , α 3 ) ( α 1 , α 2 , α 3 ) ε 23 ε 13 ( α 3 , α 1 , α 2 ) ( α 1 , α 3 , α 2 ) ε 12 ( α 3 , α 2 , α 1 ) ( α 2 , α 1 , α 3 )

Horn’s problem, and Fourier analysis The radial part ν α,β Recall X is a random Hermitian matrix on O α with law µ α , and Y on O β with law µ β . The joint distribution of the eigenvalues of Z = X + Y is the radial part ν α,β of µ α ∗ µ β . Theorem 1 1 = δ n ! V n ( x ) η α ∗ M β ν α,β n ! � 1 1 V n ( x ) = ε ( σ ) δ σ ( α ) ∗ M β . n ! δ n ! V n ( α ) σ ∈ S n The sum has positive and negative terms. However ν α ; β is a probability measure on R n . The measure ν α,β is symmetric (invariant by S n ).

Horn’s problem, and Fourier analysis This theorem can be seen as a special case of a result by Graczyk and Sawyer (2002). A similar result, but slightly different, is given by R¨ osler (2003).