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

inverse of the Dirac operator: “ ds ∼ D−1 ”

◮ For a product of Riemannian manifolds

M = M1 × M2, with product metric: ds2 = ds2

1 + ds2 2

M1 M2 ds1 ds2 ds

is an “infinitesimal” version of Pythagoras equality.

◮ For a product of noncommutative manifolds (spectral triples):

D2 = D2

1 ⊗ 1 + 1 ⊗ D2 2

which is a sort of “inverse Pythagoras equality”: 1 ds2 = 1 ds2

1

+ 1 ds2

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?

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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 = Lp-norm. Given (X1, d1) and (X2, d2), define on X = X1 × X2: d(x, y) :=

• d1(x1, y1) , d2(x2, y2)
• p ,

∀ x = (x1, x2), y = (y1, y2) ∈ X1 × X2.

• Exercise. Check that d is a (extended semi-)metric if d1 and d2 are.

If p = 2, we get the product metric, denoted d1 ⊠ d2: d1 ⊠ d2(x, y) =

• d1(x1, y1)2 + d2(x2, y2)2

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From points to states

Recall that: Definition. A state on a C∗-algebra A is a linear map ϕ : A → C which is positive, i.e. ϕ(a∗a) 0 ∀ a ∈ A, and normalized: ϕ := supa=0 |ϕ(a)| /a = 1.

• The set S(A) :=
• states of A
• is a convex space. Extreme points are called pure states.
• A = C0(X) ⇒ S(A) =
• probability measures on X
• ∧
• pure states
• =
• ˆ

x : x ∈ X

• .

Here ˆ x(f) := f(x) .

• If π : A → B(H) is a bounded rep. on a Hilbert space, a state ϕ is normal (w.r.t. π) if

ϕ(a) = TrH

• ρπ(a)
• ,

∀ a ∈ A, where ρ = positive traceclass operator with trace 1, called a density matrix for ϕ.

• Remark. The map
• density matrices on H
• −

→ S(A) is neither injective nor surjective.

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Bipartite systems

Let A = A1 ⊗ A2 . The marginals ϕ♭

1 ∈ S(A1) and ϕ♭ 2 ∈ S(A2) of ϕ ∈ S(A) are:

ϕ♭

1(a1) := ϕ(a1 ⊗ 1) ,

ϕ♭

2(a2) := ϕ(1 ⊗ a2) ,

for all a1 ∈ A1 and a2 ∈ A2 (well defined even if 1 / ∈ A1, A2). 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 ⇒ρT1 0 (T1 = transposition on the 1st leg). Example (Two qubits). Non-separable (“entangled”) states exist! Let ρ Bell = 1 2

• ij eij ⊗ eij

with {eij}i,j=1,2 canonical basis of M2(C).

✛ Exercise:

check that ρ Bell is a rank 1 projection (= pure state). prove that ρT1

Bell has a negative eigenvalue (= entangled).

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Spectral triples

Definition

A spectral triple (A, H, D) is given by:

◮ a complex separable Hilbert space H; ◮ a ∗-algebra A of bounded operators on H; ◮ a (unbounded) selfadjoint operator D on H

s.t. [D, a] is bounded and a(D + i)−1 is a compact operator for all a ∈ A. It is called:

◮ unital if 1B(H) ∈ A ; ◮ even if ∃ a grading γ on H s.t. A is even and

D is odd. Example: the Hodge-Dirac operator M = oriented Riemannian manifold.

◮ A = C∞

0 (M)

◮ H = Ω•(M) = L2-diff. forms ◮ D = d + d∗

γ = (−1)degree It is unital ⇐ ⇒ M is compact. D induces an extended metric on S(A) called spectral distance, as follows: dD(ϕ, ψ) := supa∈As.a.

• ϕ(a) − ψ(a) : [D, a] 1
• ,

∀ ϕ, ψ ∈ S(A) .

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The Wasserstein distance of order 1

Monge d´ eblais et remblais problem:

c Villani, Optimal transport, old and new

dµ1 = distribution of material in a mine (d´ eblais) Same total mass: dµ2 = distribution in the construction site (remblais)

• dµ1 =
• dµ2 = 1 (in suitable units)

Moving a unit material from x to y = T(x) (transport plan) costs dgeo(x, y). The total cost is cT(µ1, µ2) :=

• dgeo(x, T(x))dµ1(x). For ϕi(f) =
• f(x)dµi(x), the Wasserstein dist. is:

W(ϕ1, ϕ2) := inf

T : T∗(µ1)=µ2

cT(µ1, µ2) (†)

◮ On a complete Riem. manifold, (†) is the spectral distance of the Hodge-Dirac operator!

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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): H =

• x,y∈X

x=y

C2 , π(f) =

• x,y∈X

x=y

• f(x)

f(y)

• ,

D =

• x,y∈X

x=y

1 g(x, y)

• 1

1

• ,

γ =

• x,y∈X

x=y

• 1

0 −1

• ,

for all f ∈ C(X).

Proposition. dD(ˆ x, ˆ y) = g(x, y). dD(ϕ, ψ) = Wasserstein distance with unit cost g.

Proof. [D, π(f)] =

• x,y∈X

x=y

f(x) − f(y) g(x, y)

• 0 −1

1

• =

⇒ [D, π(f)] ≡ fLip := max

x=y

|f(x) − f(y)| g(x, y)

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Products of spectral triples

In nc geom., the Cartesian product of spaces is replaced by the product of spectral triples. Given two spectral triples (A1, H1, D1, γ1) and (A2, H2, D2) , their product (A, H, D) is A = A1 ⊙ A2 H = H1 ⊗ H2 D = D1 ⊗ 1 + γ1 ⊗ D2

(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 dD the product metric?

◮ dD is an extended metric on S(A). ◮ dD1 ⊠ dD2 extended semi-metric on S(A), and extended metric on S(A1) × S(A2).

Is dD = dD1 ⊠ dD2 on product states? (Or on some subset?)

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Pythagoras

S(A1) S(A2) b c a ϕ♭

2

ψ♭

2

ϕ♭

1

ψ♭

1

ϕ ψ More explicitly. . . , given any ϕ, ψ ∈ S(A), let a := dD(ϕ, ψ) , b := dD1(ϕ♭

1, ψ♭ 1) ,

c := dD2(ϕ♭

2, ψ♭ 2) .

In general, one has

Only unital spectral triples

• √

b2 + c2 a √ 2 √ b2 + c2

• Only product states

If a2 = b2 + c2 (†) we say that ϕ, ψ satisfy Pythagoras equality. Which states/spectral triples satisfy (†) ?

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Examples

Consider the following list of spectral triples/states: (1) two point space A = C2 (arbitrary) product states (2) finite metric space A = CN pure states (3) Riemannian manifold A = C∞

0 (M)

pure states (4) Moyal space A = K (compact operators) coherent states (5) A = MN(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) = (A2, H2, D2) be any spectral triple, ϕ2, ψ2 ∈ S(A2) , and δ↑, δ↓ pure states of C2. If Pythagoras eq. holds for δ↑ ⊗ ϕ2 and δ↓ ⊗ ψ2 in the product (1) × (6) , then it is satisfied by ϕ1 ⊗ ϕ2 and ψ1 ⊗ ψ2 in the product (2) × (6) for all pure states ϕ1, ψ1 of CN.

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Rephrasing the problem

Let A1 + A2 :=

• a = a1 ⊗ 1 + 1 ⊗ a2
• a1 ∈ A1, a2 ∈ A2
• and, for all ϕ, ψ ∈ S(A):

d×

D(ϕ, ψ) =

sup

a∈(A1+A2)sa

• ϕ(a) − ψ(a) : [D, a] 1
• .

Clearly d×

D(ϕ, ψ) dD(ϕ, ψ) .

(⋆)

Theorem.

For any ϕ, ψ one has d×

D(ϕ, ψ) = dD1 ⊠ dD2(ϕ♭, ψ♭) ,

where, as before, ϕ♭

1 and ϕ♭ 2 are the marginals of ϕ and ϕ♭ = (ϕ♭ 1, ϕ♭ 2) .

To demonstrate Pythagoras equality for ϕ, ψ, one has to prove the opposite inequality in (⋆).

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A simple criterion for product states

If 1A ∈ A and π ∈ S(A) π♯ : A → A , π♯(a) = π(a)1A , is an idempotent map. Let ϕ = ϕ1 ⊗ ϕ2 and ψ = ψ1 ⊗ ψ2 . Define a projection Pϕ,ψ : A → A1 + A2 by: Pϕ,ψ := ϕ♯

1 ⊗ Id + Id ⊗ ψ♯ 2 − ϕ♯ 1 ⊗ ψ♯ 2

Then:

Proposition. d×

D(ϕ, ψ) :=

sup

a∈(A1+A2)sa

• ϕ(a) − ψ(a) : [D, a] 1
• =

sup

a∈Asa

• ϕ(a) − ψ(a) : [D, Pϕ,ψ(a)] 1
• Corollary.

If for all a ∈ Asa, [D, Pϕ,ψ(a)] [D, a] , then ϕ, ψ satisfy Pythagoras equality.

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Example: C2 ⊗ C2

S(C2) ≃ [−1, 1], the state corresponding to ϕ ∈ [−1, 1] being: C2 ∋ (a↑, a↓) → 1 − ϕ 2 a↑ + 1 + ϕ 2 a↓ . Write a = a∗ ∈ C2 ⊗ C2 as a = x0 1 ⊗ 1 + x1 2 γ ⊗ 1 + x2 2 1 ⊗ γ + x3 2 vϕ,ψ , x0, . . . , x3 ∈ R, where γ = (1, −1) and vϕ,ψ is a (suitable) basis vector of ker Pϕ,ψ . Then [D, a] = √ 2 |x3| +

• (x1 + x3 ψ2)2 + (x2 + x3 ϕ1)21/2
• x2

1 + x2 2

1/2 = [D, Pϕ,ψ(a)] , for all ϕ1, ψ2 ∈ [−1, 1].

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