Pythagoras Theorem in Noncommutative Geometry Francesco DAndrea GAP - PowerPoint PPT Presentation

Pythagoras Theorem in Noncommutative Geometry Francesco D’Andrea GAP Seminar, PSU, 21/05/2015

Introduction ◮ The line element in nc geometry “is” the M 2 inverse of the Dirac operator: “ ds ∼ D − 1 ” ds M 1 ds 2 ◮ For a product of Riemannian manifolds M = M 1 × M 2 , with product metric: ds 1 ds 2 = ds 2 1 + ds 2 2 is an “infinitesimal” version of Pythagoras equality. D 2 = D 2 ◮ For a product of noncommutative manifolds (spectral triples): 1 ⊗ 1 + 1 ⊗ D 2 2 which is a sort of “inverse Pythagoras equality”: 1 1 1 ds 2 = + ( ⋆ ) ds 2 ds 2 1 2 See e.g.: A. Connes, Variations sur le th` eme spectral (2007), available on-line. ◮ Can we “integrate” ( ⋆ ) to get some (in)equalities for the distance in ncg? 1 / 13

Cartesian products and the product metric An extended semi-metric on a set X is a map d : X × X → [ 0, + ∞ ] s.t., for all x , y , z ∈ X : i) d ( x , y ) = d ( y , x ) (symmetry); ii) d ( x , x ) = 0 (reflexivity); iii) d ( x , y ) � d ( x , z ) + d ( z , y ) (triangle inequality). It is an extended metric if in addition iv) d ( x , y ) = 0 ⇒ x = y (identity of the indiscernibles), and a metric ‘tout court’ if: v) d ( x , y ) < ∞ ∀ x , y . Let p � 0 and � p = L p -norm. Given ( X 1 , d 1 ) and ( X 2 , d 2 ) , define on X = X 1 × X 2 : � �� �� d ( x , y ) := d 1 ( x 1 , y 1 ) , d 2 ( x 2 , y 2 ) ∀ x = ( x 1 , x 2 ) , y = ( y 1 , y 2 ) ∈ X 1 × X 2 . p , � Exercise. Check that d is a (extended semi-)metric if d 1 and d 2 are. If p = 2 , we get the product metric, denoted d 1 ⊠ d 2 : d 1 ( x 1 , y 1 ) 2 + d 2 ( x 2 , y 2 ) 2 � d 1 ⊠ d 2 ( x , y ) = 2 / 13

From points to states Recall that: A state on a C ∗ -algebra A is a linear map ϕ : A → C which is positive, Definition. i.e. ϕ ( a ∗ a ) � 0 ∀ a ∈ A , and normalized: � ϕ � := sup a � = 0 | ϕ ( a ) | / � a � = 1 . � � • The set S ( A ) := states of A is a convex space. Extreme points are called pure states. � � � � � � • A = C 0 ( X ) ⇒ S ( A ) = x : x ∈ X probability measures on X ∧ pure states = ˆ . Here x ( f ) := f ( x ) . ˆ • If π : A → B ( H ) is a bounded rep. on a Hilbert space, a state ϕ is normal (w.r.t. π ) if � � ∀ a ∈ A , ϕ ( a ) = Tr H ρπ ( a ) , where ρ = positive traceclass operator with trace 1 , called a density matrix for ϕ . � � density matrices on H − → S ( A ) is neither injective nor surjective. Remark. The map 3 / 13

Bipartite systems Let A = A 1 ⊗ A 2 . The marginals ϕ ♭ 1 ∈ S ( A 1 ) and ϕ ♭ 2 ∈ S ( A 2 ) of ϕ ∈ S ( A ) are: ϕ ♭ ϕ ♭ 1 ( a 1 ) := ϕ ( a 1 ⊗ 1 ) , 2 ( a 2 ) := ϕ ( 1 ⊗ a 2 ) , for all a 1 ∈ A 1 and a 2 ∈ A 2 (well defined even if 1 / ∈ A 1 , A 2 ). We call ϕ a product state if ϕ = ϕ ♭ 1 ⊗ ϕ ♭ 2 . Separable states := closed convex hull of product states. Peres’ criterion. ρ = separable density matrix ⇒ ρ T 1 � 0 ( T 1 = transposition on the 1st leg). Example (Two qubits). Non-separable (“entangled”) states exist! Let ρ Bell = 1 � ij e ij ⊗ e ij 2 with { e ij } i , j = 1,2 canonical basis of M 2 ( C ) . ✛ Exercise: check that ρ Bell is a rank 1 projection ( = pure state). prove that ρ T 1 Bell has a negative eigenvalue ( = entangled). 4 / 13

Spectral triples Definition Example: the Hodge-Dirac operator A spectral triple ( A , H , D ) is given by: M = oriented Riemannian manifold. ◮ A = C ∞ 0 ( M ) ◮ a complex separable Hilbert space H ; ◮ H = Ω • ( M ) = L 2 -diff. forms ◮ a ∗ -algebra A of bounded operators on H ; ◮ D = d + d ∗ γ = (− 1 ) degree ◮ a (unbounded) selfadjoint operator D on H It is unital ⇐ ⇒ M is compact. s.t. [ D , a ] is bounded and a ( D + i ) − 1 is a compact operator for all a ∈ A . It is called: ◮ unital if 1 B ( H ) ∈ A ; ◮ even if ∃ a grading γ on H s.t. A is even and D is odd. D induces an extended metric on S ( A ) called spectral distance, as follows: � � d D ( ϕ , ψ ) := sup a ∈ A s.a. ϕ ( a ) − ψ ( a ) : � [ D , a ] � � 1 , ∀ ϕ , ψ ∈ S ( A ) . 5 / 13

The Wasserstein distance of order 1 Monge d´ eblais et remblais problem: � Villani, Optimal transport, old and new c dµ 1 = distribution of material in a mine ( d´ eblais ) Same total mass: � � dµ 2 = distribution in the construction site ( remblais ) dµ 1 = dµ 2 = 1 (in suitable units) Moving a unit material from x to y = T ( x ) (transport plan) costs d geo ( x , y ) . The total cost is � � c T ( µ 1 , µ 2 ) := d geo ( x , T ( x )) dµ 1 ( x ) . For ϕ i ( f ) = f ( x ) dµ i ( x ) , the Wasserstein dist. is: W ( ϕ 1 , ϕ 2 ) := inf c T ( µ 1 , µ 2 ) ( † ) T : T ∗ ( µ 1 )= µ 2 ◮ On a complete Riem. manifold, ( † ) is the spectral distance of the Hodge-Dirac operator! 6 / 13

Finite metric spaces Every finite metric space ( X , g ) can be reconstructed from a canonical even spectral triple. For x ∈ X , let ˆ x ( f ) = f ( x ) the corresponding pure state of A = C ( X ) . Set ( ∀ f ∈ A ): � � f ( x ) 0 C 2 , � � H = π ( f ) = , 0 f ( y ) x , y ∈ X x , y ∈ X x � = y x � = y � � � � 1 0 1 1 0 � � D = , γ = , g ( x , y ) 1 0 0 − 1 x , y ∈ X x , y ∈ X x � = y x � = y for all f ∈ C ( X ) . Proposition. d D ( ˆ y ) = g ( x , y ) . d D ( ϕ , ψ ) = Wasserstein distance with unit cost g . x , ˆ Proof. � � f ( x ) − f ( y ) 0 − 1 | f ( x ) − f ( y ) | � [ D , π ( f )] = = ⇒ � [ D , π ( f )] � ≡ � f � Lip := max g ( x , y ) g ( x , y ) 1 0 x � = y x , y ∈ X � x � = y 7 / 13

Products of spectral triples In nc geom., the Cartesian product of spaces is replaced by the product of spectral triples. Given two spectral triples ( A 1 , H 1 , D 1 , γ 1 ) and ( A 2 , H 2 , D 2 ) , their product ( A , H , D ) is A = A 1 ⊙ A 2 H = H 1 ⊗ H 2 D = D 1 ⊗ 1 + γ 1 ⊗ D 2 (algebraic tensor product) One can also consider the case when both spectral triples are odd, however note that: • any odd spectral triple can be transformed into an even one without changing the metric. Question: is d D the product metric? ◮ d D is an extended metric on S ( A ) . ◮ d D 1 ⊠ d D 2 extended semi -metric on S ( A ) , and extended metric on S ( A 1 ) × S ( A 2 ) . Is d D = d D 1 ⊠ d D 2 on product states? (Or on some subset?) 8 / 13

Pythagoras S ( A 2 ) ψ ψ ♭ 2 a c ϕ ϕ ♭ 2 b S ( A 1 ) ϕ ♭ ψ ♭ 1 1 More explicitly. . . , given any ϕ , ψ ∈ S ( A ) , let b := d D 1 ( ϕ ♭ 1 , ψ ♭ c := d D 2 ( ϕ ♭ 2 , ψ ♭ a := d D ( ϕ , ψ ) , 1 ) , 2 ) . In general, one has Only unital spectral triples � �� � √ √ √ b 2 + c 2 � a � b 2 + c 2 2 � �� � Only product states If a 2 = b 2 + c 2 ( † ) we say that ϕ , ψ satisfy Pythagoras equality. Which states/spectral triples satisfy ( † ) ? 9 / 13

Examples Consider the following list of spectral triples/states: (1) two point space A = C 2 (arbitrary) product states (2) finite metric space A = C N pure states (3) Riemannian manifold A = C ∞ 0 ( M ) pure states (4) Moyal space A = K (compact operators) coherent states (5) A = M N ( C ) SU ( N ) -coherent states Pythagoras equality holds for the products: (1) × (1) ( i ) × ( j ) ∀ i , j = 2, 3 (2) × (4) The proof should still work if (4) is replaced by (5) (we didn’t check). Proposition. Let (6) = ( A 2 , H 2 , D 2 ) be any spectral triple, ϕ 2 , ψ 2 ∈ S ( A 2 ) , and δ ↑ , δ ↓ pure states of C 2 . If Pythagoras eq. holds for δ ↑ ⊗ ϕ 2 and δ ↓ ⊗ ψ 2 in the product (1) × (6) , then it is satisfied by ϕ 1 ⊗ ϕ 2 and ψ 1 ⊗ ψ 2 in the product (2) × (6) for all pure states ϕ 1 , ψ 1 of C N . 10 / 13

Rephrasing the problem Let � � � a 1 ∈ A 1 , a 2 ∈ A 2 � A 1 + A 2 := a = a 1 ⊗ 1 + 1 ⊗ a 2 and, for all ϕ , ψ ∈ S ( A ) : � � d × D ( ϕ , ψ ) = sup ϕ ( a ) − ψ ( a ) : � [ D , a ] � � 1 . a ∈ ( A 1 + A 2 ) sa Clearly d × D ( ϕ , ψ ) � d D ( ϕ , ψ ) . ( ⋆ ) Theorem. For any ϕ , ψ one has d × D ( ϕ , ψ ) = d D 1 ⊠ d D 2 ( ϕ ♭ , ψ ♭ ) , 2 are the marginals of ϕ and ϕ ♭ = ( ϕ ♭ where, as before, ϕ ♭ 1 and ϕ ♭ 1 , ϕ ♭ 2 ) . To demonstrate Pythagoras equality for ϕ , ψ , one has to prove the opposite inequality in ( ⋆ ). 11 / 13

A simple criterion for product states If 1 A ∈ A and π ∈ S ( A ) � π ♯ : A → A , π ♯ ( a ) = π ( a ) 1 A , is an idempotent map. Let ϕ = ϕ 1 ⊗ ϕ 2 and ψ = ψ 1 ⊗ ψ 2 . Define a projection P ϕ , ψ : A → A 1 + A 2 by: P ϕ , ψ := ϕ ♯ 1 ⊗ Id + Id ⊗ ψ ♯ 2 − ϕ ♯ 1 ⊗ ψ ♯ 2 Then: Proposition. � � d × D ( ϕ , ψ ) := sup ϕ ( a ) − ψ ( a ) : � [ D , a ] � � 1 a ∈ ( A 1 + A 2 ) sa � � = sup ϕ ( a ) − ψ ( a ) : � [ D , P ϕ , ψ ( a )] � � 1 a ∈ A sa Corollary. If for all a ∈ A sa , � [ D , P ϕ , ψ ( a )] � � � [ D , a ] � , then ϕ , ψ satisfy Pythagoras equality. 12 / 13