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

Pythagoras Theorem in Noncommutative Geometry

Francesco D’Andrea

A B C c a b

Workshop on Noncommutative Geometry and Optimal Transport Besanc ¸on, 27/11/2014

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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Pythagoras

A B C c a b

a2 = b2 + c2

• r

d(B, C)2 = d(A, B)2 + d(A, C)2 (†) Generalization to nc geometry:

• what are points A, B, C, . . .?
• what is d(A, B)?
• what is a right-angle triangle?
• is (†) still valid?

Reference: FD & P . Martinetti, On Pythagoras Theorem for Products of Spectral Triples,

• Lett. Math. Phys. 103 (2013), 469–492.

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What are points?

• Any topological space X is the spectrum of a commutative C∗-algebra A = C0(X) .
• Points are linear maps δx : C0(X) → C , δx(f) := f(x) .

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
• =
• δx : x ∈ X
• .
• Quantum mechanical interpretation:

As.a. :=

• a ∈ A : a = a∗

← → physical observables

• utcome of a measure ←

→ eigenvalue of a (or residual spectrum) ϕ ∈ S(A) associates to any observable a its expectation value

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What is a right-angle triangle?

A(b, 0) B(b, c) C (0, c) Sp(A1) Sp(A2) b c a

Pythagoras revised (here C = (0, 0)): d(δb ⊗ δc, δ0 ⊗ δ0)2 = = d(δb, δ0)2 + d(δc, δ0)2

Dictionary Point A(b, 0)

• pure state δb of A1 = C0(R)

Point (0, c)

• pure state δc of a second copy A2 = C0(R)

Point B(b, c)

• product δb ⊗ δc: state of A1 ⊗A2 ≃ C0(R × R)

Given two separable states ϕ = ϕ1 ⊗ ϕ2 and ψ = ψ1 ⊗ ψ2 on A = A1 ⊗A2 , is d(ϕ, ψ)2 = d(ϕ1, ψ1)2 + d(ϕ2, ψ2)2 ?

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Spectral triples

Definition

A spectral triple 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;

◮ non-degenerate if

Span

• av : a ∈ A, v ∈ H
• ≡ H .

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. Example: the generator of K0(C)

◮ A =

a

• : a ∈ C
• ◮ H = C2

◮ D =

1 1

• γ =

1 0 −1

• Note that this is a non-unital spectral

triple (because 1M2(C) / ∈ A ), even if A ≃ C is a unital algebra.

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What is d(ϕ, ψ)?

A spectral triple induces a distance on S(A): dA,D(ϕ, ψ) := supa∈As.a.

• ϕ(a) − ψ(a) : [D, a] 1
• ,

∀ ϕ, ψ ∈ S(A) .

c Villani, Optimal transport, old and new

Monge d´ eblais et remblais problem: 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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Example: the SNCF spectral triple

Let X ⊂ Rn be a finite∗ subset and x0 a basepoint. Take x0 = 0 to simplify the notations. Let A :=

• f : X → C s.t. f(0) = 0
• ,

ex be the indicator function of x ∈ X and {e±} the canonical basis of C2, H := A ⊗ C2 with orthonormal basis {ex ⊗ e±}x∈X. The representation π of A on H and D are π(f) = f

• D(ex ⊕ e±) =
• x−1ex ⊕ e∓

if x = 0 if x = 0 Then [D, π(f)] = sup

x=0

|f(x)| x 1 ⇒ |f(x) − f(y)| x + y ∀ x, y ∈ X (including 0) The sup is attained on the function fx,y = xex − yey . Thus: dA,D(δx, δy) ≡ dSNCF(x, y) =

• x + y

if x = y if x = y

∗For the compact resolvent condition, which could be dropped if only interested in the metric aspect.

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Optimal public-transport

dgeo(Lyon, Besanc ¸on) = 188 km dSNCF(Lyon, Besanc ¸on) = 720 km

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Optimal control

Other “non-Riemannian” examples arise in minimization problems with constraints.

Soft Moon landing. Falling cat.

total mass = mHtL thrust = -k m'HtL height = hHtL m g c 2012 Wolfram Media, Inc. c

• R. Montgomery, Commun. Math. Phys. 1990

Goal: minimize the amount of fuel. Goal: turn of 180 degrees while falling. Constraint: at height h = 0 the velocity must be v = 0 (avoid crashing). Constraint: while falling, keep the total angular momentum J = 0.∗

∗ Motion constrained on a distribution H ⊂ TM, with M the configuration space.

• R. Montgomery (1993): Gauge theory of the falling cat.

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The Carnot-Carath´ eodory distance

A sub-Riemannian geometry on a manifold M consists of a vector sub-bundle H ⊂ TM with a fiber inner-product g( · , · ) on it. A curve γ : [0, 1] → M is horizontal if ˙ γ(t) ∈ H ∀ t. The Carnot-Carath´ eodory distance is defined as: dCC(x, y) := inf

• g( ˙

γ(t), ˙ γ(t))

1 2 dt ,

where the inf is on all horizontal curves from x to y. A contact manifold (M, H) is a 2n + 1-dimensional manifold M equipped with a codimension 1 vector sub-bundle H ⊂ TM which is completely non-integrable (that is, sections of H generate Vect(M) as a Lie algebra). Locally, H = ker τ where τ ∈ Ω1M satisfies τ(dτ)n = 0 (contact 1-form). Fact: there is a natural sub-Riemannian metric on any contact manifold.

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The Rumin complex

Let τ be a contact for on a 2n + 1-dim. manifold M, and:

• I• = τ, dτ

— differential ideal generated by τ.

• J• = ker(τ ∧) ∩ ker(dτ ∧)

— annihilator.

Theorem (Rumin, 1994) ∃ a 2nd order diff. op. D such that the following is a cochain complex: Ω0/I0

d

− → Ω1/I1

d

− → . . .

d

− → Ωn/In

D

− − − − → Jn

d

− → Jn+1

d

− → . . .

d

− → J2n+1 Its cohomology is the de Rham cohomology of M.

• Observations. For f ∈ C∞(M):
• D is not 1st order, so [D, f] is not bounded =

⇒ no spectral triple;

• [D, f] is bounded relatively to |D|

1 2 =

⇒ bounded Fredholm module;

• [[D, f], f] is bounded, and

[R. Yuncken, Bonn 2014]

dCC(x, y) = sup

f∈C∞(M)

• |f(x) − f(y)| : [[D, f], f] 2
• .

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Back to Pythagoras. . .

Given two separable states ϕ = ϕ1 ⊗ ϕ2 and ψ = ψ1 ⊗ ψ2 on A = A1 ⊗A2 , is d(ϕ, ψ)2 = d(ϕ1, ψ1)2 + d(ϕ2, ψ2)2 satisfied by the spectral distance?

S(A1) S(A2) d(ϕ1, ψ1) d(ϕ2, ψ2) d(ϕ, ψ) ψ2 ϕ2 ψ1 ϕ1 ψ ϕ

A last ingredient is missing: on A = A1 ⊗A2 we want a product metric.

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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 . (†) Here A1 ⊗ A2 is the algebraic tensor product. One can also consider the case when both spectral triples are odd, however note that:

◮ In the Hodge-Dirac example

• C∞

0 (M), Ω•(M), D = d + d∗, γ = (−1)degree

the spectral triple is even whatever is the dimension of M!

◮ In the latter case, (†) corresponds to equipping M = M1 × M2 with the product metric. 13 / 19

Products of manifolds

◮ Let CS(M) be the Hodge-Dirac spectral triple of M. The associated spectral distance is

the Wasserstein distance on M, or the geodesic distance between pure states.

◮ Denote by CS(M1) ⊗ CS(M2) the product of spectral triples CS(M1) and CS(M2).

• Proposition. For M = M1 ×M2 with product metric, CS(M) and CS(M1)⊗CS(M2) induce

the same distance on the space of probability distributions on M.

• Proof. H and D are the same (modulo unitary equivalence), and

A = C∞

0 (M) ≃ C∞ 0 (M1)

⊗ C∞

0 (M2) ⊃ A1 ⊗ A2 = C∞ 0 (M1) ⊗ C∞ 0 (M2)

with ⊗ the projective tensor product. So, for all ϕ, ψ: dA,D(ϕ, ψ) dA1⊗A2,D(ϕ, ψ) . One can prove equality using density arguments.

• Now, 1. every pure state is separable: δx = δx1 ⊗ δx2, x = (x1, x2) ∈ M;
• 2. dgeo satisfies Pythagoras identity [FD & P

. Martinetti, Lett. Math. Phys. 2013]. So:

• Proposition. Pythagoras identity holds for pure states, in the product of Hodge-Dirac spectral

triples of complete oriented Riemannian manifolds.

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Pythagoras identity: a counterexample

Let M = R × R with standard Euclidean metric and let ϕλ ∈ S(C0(R)) be the state: ϕλ(f) := λf(1) + (1 − λ)f(0) , 0 λ 1 . Let W1 resp. W2 be the Wasserstein distance on the 1st resp. 2nd factor of R × R, and W the one on the product. An explicit computation gives: W(ϕλ ⊗ ϕλ, ϕ0 ⊗ ϕ0) = kλ

• W1(ϕλ, ϕ0)2 + W2(ϕλ, ϕ0)2 ,

where kλ = λ + √ 2(1 − λ) .

◮ Pythagoras identity holds when kλ = 1, i.e. λ = 1 (note that ϕ1 is a pure state).

It is also trivially true for λ = 0.

◮ For λ ∈ [0, 1], kλ assumes all possible values in [1,

√ 2]. The same argument works on a torus (with flat metric), providing then a compact example (unital spectral triple).

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Pythagoras inequalities for spectral triples

Here we consider a product of arbitrary (i.e. not necessarily commutative) spectral triples. We shall use the shorthand notation d = dA,D and di = dAi,Di for i = 1, 2.

Theorem

Given two spectral triples (A1, H1, D1, γ1) and (A2, H2, D2): i) For any two separable states ϕ = ϕ1 ⊗ ϕ2 and ψ = ψ1 ⊗ ψ2, d(ϕ, ψ) d1(ϕ1, ψ1) + d2(ϕ2, ψ2) . (⋆) ii) Furthermore, if the spectral triples are unital: d(ϕ, ψ)

• d1(ϕ1, ψ1)2 + d2(ϕ2, ψ2)2 .

Notice that from (⋆) and (a + b)2 = 2(a2 + b2) − (a − b)2 2(a2 + b2) it follows that:

• d1(ϕ1, ψ1)2 + d2(ϕ2, ψ2)2 d(ϕ, ψ)2

√ 2

• d1(ϕ1, ψ1)2 + d2(ϕ2, ψ2)2 .

The example in previous slides proves that the coefficient √ 2 is optimal.

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Pythagoras inequalities for spectral triples

S(A1) S(A2) b c a ψ2 ϕ2 ψ1 ϕ1 ψ ϕ √ b2 + c2 a √ 2 √ b2 + c2

◮ These inequalities are optimal. ◮ The lower bound holds only if the spectral triples are unital.

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The importance of being non-degenerate

A product of two-point spaces Let λ, µ > 0 be two parameters. A first unital spectral triple (A1, H1, D1, γ1) is given by A1 = a b

• : a, b ∈ C
• H1 = C2

D1 = 1 λ 1 1

• γ1 =

1 0 −1

• A second non-unital spectral triple (A2, H2, D2) is

A2 =               a b      : a, b ∈ C          H2 = C4 D2 = 2 µ      1 1 1 1      Let ϕ↑, ϕ↓ be the (only) two pure states of A1 ≃ A2 ≃ C2. Then: d1(ϕ↑, ϕ↓) = λ , d2(ϕ↑, ϕ↓) = µ . Pythagoras inequalities would give d(ϕ↑ ⊗ ϕ↑, ϕ↓ ⊗ ϕ↓)

• λ2 + µ2 . Instead:
• Proposition. The distance d(ϕ↑ ⊗ ϕ↑, ϕ↓ ⊗ ϕ↓) = µ is independent of λ.

For λ, µ > 0, the ratio d/

• d2

1 + d2 2 assumes all possible values in (0, 1).

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The importance of being non-degenerate

A two-sheeted real line Let λ > 0 and (A1, H1, D1, γ1) be the unital spectral triple of C2 as in previous slide. Let the second spectral triple (A2, H2, D2) be given by A2 = f

• : f ∈ C∞

0 (R)

• H2 = L2(R) ⊗ C2

D2 = ∂ / 2 2 −∂ /

• with ∂

/f(x) = if′(x).

• Proposition. For any x, y ∈ R and λ > 0, we have

d(ϕ↑ ⊗ δx, ϕ↓ ⊗ δy) 1 λ−1 d1(ϕ↑, ϕ↓)2 + d2(δx, δy)2 . In particular, if λ > 1: d(ϕ↑ ⊗ δx, ϕ↓ ⊗ δx) = d1(ϕ↑, ϕ↓)

R R      1 = λ ϕ↑ ⊗ δx ϕ↓ ⊗ δx

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