Eigenvalues of sums of hermitian matrices and the cohomology of - PowerPoint PPT Presentation

Eigenvalues of sums of hermitian matrices and the cohomology of Grassmannians Edward Richmond University of British Columbia January 16, 2013 Richmond (UBC) The hermitian eigenvalue problem January 16, 2013 1 / 27

Outline The eigenvalue problem on hermitian matrices 1 Schubert calculus of the Grassmannian 2 Majorized sums and equivariant cohomology 3 Outline of proof 4 Richmond (UBC) The hermitian eigenvalue problem January 16, 2013 2 / 27

The eigenvalue problem on hermitian matrices The eigenvalue problem on hermitian matrices: Consider the sequences of real numbers α := ( α 1 ≥ α 2 ≥ · · · ≥ α n ) β := ( β 1 ≥ β 2 ≥ · · · ≥ β n ) γ := ( γ 1 ≥ γ 2 ≥ · · · ≥ γ n ) . Question: For which triples ( α, β, γ ) do there exist n × n hermitian matrices A, B, C with respective eigenvalues α, β, γ and A + B = C ? Richmond (UBC) The hermitian eigenvalue problem January 16, 2013 3 / 27

The eigenvalue problem on hermitian matrices Motivation: Functional analysis: decomposing self-adjoint operators on Hilbert spaces Frame theory (sensor networks, coding theory, and compressed sensing) Invariant theory of representations Richmond (UBC) The hermitian eigenvalue problem January 16, 2013 4 / 27

The eigenvalue problem on hermitian matrices Example: Let n = 2 . � � � � � � α 1 0 a c a + α 1 c + = 0 α 2 c ¯ b ¯ c b + α 2 Then � ( a − b ) 2 + | c | 2 β = a + b ± 2 � ( a − b + α 1 − α 2 ) 2 + | c | 2 γ = α 1 + α 2 + a + b ± 2 Solution for the n = 2 case We get that matrices A + B = C exist if and only if the triple ( α, β, γ ) satisfies α 1 + β 1 ≥ γ 1 α 1 + α 2 + β 1 + β 2 = γ 1 + γ 2 and α 1 + β 2 ≥ γ 2 α 2 + β 1 ≥ γ 2 . Richmond (UBC) The hermitian eigenvalue problem January 16, 2013 5 / 27

The eigenvalue problem on hermitian matrices What about n = 3 ? Theorem: Horn ’62 There exist 3 × 3 matrices A + B = C with eigenvalues ( α, β, γ ) if and only if ( α, β, γ ) satisfies α 1 + α 2 + α 3 + β 1 + β 2 + β 3 = γ 1 + γ 2 + γ 3 and up to α ↔ β symmetry α 1 + β 1 ≥ γ 1 α 1 + α 2 + β 1 + β 2 ≥ γ 1 + γ 2 α 1 + β 2 ≥ γ 2 α 1 + α 2 + β 1 + β 3 ≥ γ 1 + γ 3 α 1 + β 3 ≥ γ 3 α 1 + α 2 + β 2 + β 3 ≥ γ 2 + γ 3 α 2 + β 2 ≥ γ 2 α 1 + α 3 + β 1 + β 3 ≥ γ 2 + γ 3 . Richmond (UBC) The hermitian eigenvalue problem January 16, 2013 6 / 27

The eigenvalue problem on hermitian matrices What about the general case? Conjecture: Horn ’62 There exist n × n matrices A + B = C with eigenvalues ( α, β, γ ) if and only if ( α, β, γ ) satisfies n n n � � � α i + β j = γ k i =1 j =1 k =1 and for every r < n we have a “certain” collection of subsets I, J, K ⊂ [ n ] := { 1 , 2 , . . . , n } of size r where � � � α i + β j ≥ γ k . i ∈ I j ∈ J k ∈ K Example: If n = 4 and r = 2 , then ( I, J, K ) = ( { 1 , 3 } , { 2 , 4 } , { 3 , 4 } ) corresponds to the linear inequality α 1 + α 3 + β 2 + β 4 ≥ γ 3 + γ 4 . Richmond (UBC) The hermitian eigenvalue problem January 16, 2013 7 / 27

Schubert calculus of the Grassmannian Schubert calculus of the Grassmannian: Fix a basis { e 1 , . . . , e n } of C n and consider the Grassmannian Gr( r, n ) := { V ⊆ C n | dim V = r } . For any partition λ := ( λ 1 ≥ · · · ≥ λ r ) where λ 1 ≤ n − r define the Schubert subvariety X λ := { V ∈ Gr( r, n ) | dim( V ∩ E n − r + i − λ i ) ≥ i ∀ i ≤ r } where E i := Span { e 1 , . . . , e i } . For example X ∅ = Gr( r, n ) X Λ = { E r } where Λ := ( n − r, . . . , n − r ) � In general, codim C ( X λ ) = | λ | := λ i . i Richmond (UBC) The hermitian eigenvalue problem January 16, 2013 8 / 27

Schubert calculus of the Grassmannian Denote the cohomology class of X λ by σ λ ∈ H 2 | λ | (Gr( r, n ) , C ) . Additively, we have that Schubert classes σ λ form a basis of � H ∗ (Gr( r, n )) ≃ C σ λ . λ ⊆ Λ Define the Littlewood-Richardson coefficients c ν λ,µ as the structure constants � c ν σ λ · σ µ = λ,µ σ ν . ν ⊆ Λ Facts: c ν λ,µ ∈ Z ≥ 0 . If c ν λ,µ > 0 , then | λ | + | µ | = | ν | . Richmond (UBC) The hermitian eigenvalue problem January 16, 2013 9 / 27

Schubert calculus of the Grassmannian The LR-rule for computing c ν λ,µ : For any partition λ = ( λ 1 ≥ λ 2 · · · ≥ λ r ) , we have the associated Young diagram λ 1 λ 2 λ 3 ... λ r If λ ⊆ ν, then we can define the skew diagram ν/λ by removing the boxes of the Young diagram of λ from the Young diagram of ν. (4 , 3 , 2) / (2 , 2) = We have c ν λ,µ = # { LR skew tableaux of shape ν/λ with content µ } . Richmond (UBC) The hermitian eigenvalue problem January 16, 2013 10 / 27

Schubert calculus of the Grassmannian Appearances of Littlewood-Richardson coefficients c ν λ,µ : The number of points in a finite intersection of translated Schubert varieties | g 1 X λ ∩ g 2 X µ ∩ g 3 X ν ∨ | = c ν λ,µ . Let V λ be the irreducible, finite-dimensional representation of GL r ( C ) of highest weight λ. Then � ⊕ c ν V λ ⊗ V µ ≃ V λ,µ . ν ν Let s λ denote the Schur function indexed by λ. Then � c ν s λ · s µ = λ,µ s ν . ν Richmond (UBC) The hermitian eigenvalue problem January 16, 2013 11 / 27

Schubert calculus of the Grassmannian Let Λ = ( n − r, . . . , n − r ) . There is a bijection between � �� � r { Partitions λ ⊆ Λ } ⇔ { Subsets of [ n ] of size r } . Consider the Young diagram of λ ⊆ Λ . 9 7 8 6 5 3 4 2 1 Example: n = 9 and r = 4 with λ = (5 , 3 , 3 , 1) . We identify λ with vertical labels on the boundary path. (5 , 3 , 3 , 1) ↔ { 2 , 5 , 6 , 9 } . For any subset I = { i 1 < · · · < i r } ⊆ [ n ] , let φ ( I ) := { i r − r ≥ · · · ≥ i 1 − 1 } denote the corresponding partition. Richmond (UBC) The hermitian eigenvalue problem January 16, 2013 12 / 27

Schubert calculus of the Grassmannian Solution to the eigenvalue problem: Theorem: Klyachko ’98 Let ( α, β, γ ) be weakly decreasing sequences of real numbers such that n n n � � � α i + β j = γ k . i =1 j =1 k =1 Then the following are equivalent. There exist n × n matrices A + B = C with eigenvalues ( α, β, γ ) . For every r < n and every triple ( I, J, K ) of subsets of [ n ] of size r such that the Littlewood-Richardson coefficient c φ ( K ) φ ( I ) ,φ ( J ) > 0 , we have � � � α i + β j ≥ γ k . i ∈ I j ∈ J k ∈ K Richmond (UBC) The hermitian eigenvalue problem January 16, 2013 13 / 27

Schubert calculus of the Grassmannian Question: Is Klyachko’s solution the same as Horn’s conjectured solution? Theorem: Knutson-Tao ’99 Klyachko’s solution is the same to Horn’s solution. Let λ, µ, ν ⊆ Λ be partitions such that | λ | + | µ | = | ν | . The following are equivalent: The Littlewood-Richardson coefficient c ν λ,µ > 0 . There exist r × r matrices A + B = C with respective eigenvalues λ, µ, ν. Richmond (UBC) The hermitian eigenvalue problem January 16, 2013 14 / 27

Schubert calculus of the Grassmannian Example: Consider the Grassmannian Gr(2 , 4) . We have ( σ ) 2 = σ + σ . For c > 0 , we have , � 1 � 1 � 2 � � � 0 0 0 + = . 0 0 0 0 0 0 For c > 0 , we have , � 1 � 0 � 1 � � � 0 0 0 + = . 0 0 0 1 0 1 But for c = 0 , we have , � 1 � a � 2 � � � 0 c 0 + = . 0 1 c ¯ b 0 2 Not possible with eigenvalues (2 , 0) ! Richmond (UBC) The hermitian eigenvalue problem January 16, 2013 15 / 27

Majorized sums and equivariant cohomology Generalizations of the eigenvalue problem: Lie groups of other types (Berenstein-Sjamaar ’00): Let O λ denote an orbit of G acting on the dual Lie algebra g ∗ . Question: For which ( λ, µ, ν ) is ( O λ + O µ ) ∩ O ν � = ∅ ? Find a minimal list of inequalities (Knuston-Tao-Woodward ’04). c ν λ,µ > 0 (redundant list) ⇒ c ν λ,µ = 1 (minimal list) Find a minimal list in any type (Belkale-Kumar ’06, Ressayre ’10). Replace A + B = C with majorized sums A + B ≥ C (Friedland ’00, Fulton ’00). Richmond (UBC) The hermitian eigenvalue problem January 16, 2013 16 / 27

Majorized sums and equivariant cohomology Definition: We say a matrix A ≥ B ( A majorizes B ) if A − B is semi-positive definite (i.e. A − B has nonnegative eigenvalues). Consider the sequences of real numbers α := ( α 1 ≥ α 2 ≥ · · · ≥ α n ) β := ( β 1 ≥ β 2 ≥ · · · ≥ β n ) γ := ( γ 1 ≥ γ 2 ≥ · · · ≥ γ n ) . Question: For which triples ( α, β, γ ) do there exist n × n hermitian matrices A, B, C with respective eigenvalues α, β, γ and A + B ≥ C ? Richmond (UBC) The hermitian eigenvalue problem January 16, 2013 17 / 27