sub-riemannian curvature: a variational approachrifford/ihp_sr/slides/barilari.pdf · in riemannian...

Motivation Affine control systems and geodesic cost Geodesic growth vector, ample geodesics and curvature Applications: Laplacian

Sub-Riemannian curvature: a variational approach

Davide Barilari (Université Paris Diderot - Paris 7)

IHP Workshop “Geometric Analysis on Sub-Riemannian manifolds”

October 2, 2014

Davide Barilari (Univ. Paris Diderot) SR Curvature October 2, 2014 1 / 38


References

This is a joint work with

Andrei Agrachev (SISSA, Trieste)

Luca Rizzi (CMAP, Ecole Polytechnique)

→ References:

- A. Agrachev, D.B., L.Rizzi, The curvature: a variational approach.ArXiv preprint 2013, 114 p.

- D.B., L. Rizzi, Comparison theorems for conjugate points in sub-Riemanniangeometry, Arxiv preprint 2014.



Outline

1 Motivation

2 Affine control systems and geodesic cost

3 Geodesic growth vector, ample geodesics and curvature

4 Applications: Laplacian of the sub-Riemannian distance

5 Applications: SR geodesic dimension and MCP



Introduction

In Riemannian geometry the curvature tensor plays a major role in many aspects:

Ricci bounds

comparison theorems (volume, distance, Laplacian, Hessian etc.)

asymptotic and estimates of the heat kernel

Obstruction 1. In sub-Riemannian geometry no canonical affine connection

Recently, many efforts to generalize the notion of Ricci/curvature bounds

via optimal transport techniques

Lott, Villani - SturmAmbrosio, Gigli, Savaré (and coauthors)

via gen. curvature dimension inequalities / heat equation techniques

Garofalo, Baudoin (and coauthors)Thalmaier, Grong

via measure contraction properties / other geometric techniques

Julliet - Rifford - VassilevAgrachev, Lee - Lee, Li, Zelenko



Introduction II

In this talk I will be interested in looking for a definition/generalization of thenotion of (sectional) curvature in sub-Riemannian geometry.

This curvature is based on two main tools

distance (→ or more generally a cost)

geodesics (→ or more generally minimizers for the cost)

From these two ingredients we define a “directional curvature”.

Naive idea:

→ The curvature can be found in the asymptotic of the distance along geodesics

Obstruction 2. Non-smoothness of the distance in sub-Riemannian geometry



Riemannian sectional curvature

Let (M , d) be a Riemannian manifold and x ∈ M .

Let γv , γw two geodesics with unit tangent vector v ,w ∈ TxM .

x

γv (t) γw(s)

C (t, s) := 12d

2(γv(t), γw (s))

is smooth in (0, 0)bc

bc bc

We expect to recover the scalar product and the curvature from the asymptoticexpansion of C (t, s).



Asymptotic expansion

C (t, s) ≃1

2(t2 + s2)− 〈v ,w〉 ts

︸︷︷︸

Euclidean

+1

4∂ttssC (0, 0)t2s2 + . . .

Theorem (Loeper, Villani)

−3

2∂ttssC (0, 0) =

⟨R∇(v ,w)v ,w

⟩(= Sec(v ,w))

Remarks

→ In SR geometry the geodesic is not uniquely associated with its tangentvector.

→ The function C (t, s) is not smooth at (0,0).



Lemma: differential of d2(x0, ·)

Let x0 ∈ M and define

fx0(·) =1

2d2(x0, ·)

x0

γ : [0,T ] → M b

x

γ′v(T )

v

fx0 smooth at x ⇐⇒ ∃ ! minimal non conjugate geodesic γ joining x0 and x .

∇fx0(x) = Tγ′v(T )

Remark. If we define ct(x) := − 1

2td2(x , γv (t)) then ∇ct(x0) = v (indep. on t).

∇x0

(

−1

2td2(·, γv (t))

)

= −1

t∇fγv (t)(x0) = −

1

t(−tv) = v



Outline

1 Motivation







Affine optimal control problems

[Dynamic] Let us consider a smooth affine control system on a manifold M

x = X0(x) +

k∑

i=1

uiXi (x), x ∈ M , u ∈ Rk .

we call Dx = spanxX1, . . . ,Xk the distribution.

u ∈ L∞([0,T ],Rk) → γx0,u(·) the solution with x(0) = x0 (admissible curve)

[Cost] Given a smooth function L : M × Rk → R we define the cost at time T as

the functional

JT (u) :=

∫ T

0

L(γu(t), u(t))dt,

Definition

For two given points x , y ∈ M and T > 0, we define the value function

ST (x , y) = inf JT (u) | u admissible, γu(0) = x , γu(T ) = y ,



Assumptions

(A1) The weak bracket generating condition is satisfied

Liex(adX0)

iXj , i ∈ N, j = 1, . . . , k= TxM ,

[the vector field X0 is not included in the generators of the Lie algebra]

(A2) The Lagrangian L : M × Rk → R is of Tonelli-type, i.e.

(A2.a) L has strictly positive Hessian in u, for all x ∈ M.(A2.b) L has superlinear growth, i.e. L(x , u)/|u| → +∞ when |u| → +∞.

(A1) does not ask that the linear part is bracket-generating.

→ when X0 = 0 it is the classical Hormander condition.

(A2) ensures that the Hamiltonian is well defined and smooth



Geometric framework

This setting includes control problems associated with many different geometricstructures such as

Riemannian structuressub-Riemannian structuressmooth Finsler structures (or smooth sub-Finsler).

The (sub-)Riemannian case corresponds to the case when

the system is driftless (X0 ≡ 0) and bracket generating (A1)

k < n (k = n corresponds to Riemannian)

the cost is quadratic L(x , u) = 1

2|u|2 (A2)

→ The cost is induced by a scalar product such that X1, . . . ,Xk are orthonormal.

ST (x , y) =1

2Td2(x , y)

In particular

ct(x) = −1

2td2(x , γ(t)) = −St(x , γ(t))



Main idea: microlocal approach

Fix x0 ∈ M and a minimizing trajectory γ = γx0,u(t).

ct : M → R, t > 0

ct(x) = −St(x , γ(t))

x0

γ(t)

ct : x 7→ −St(x , γ(t))

b

x

The curvature appears in the asymptotic expansion of the second derivative of thecost along a minimizing trajectory.

We need some assumptions on γ to guarantee

the smoothness of ct (→ strongly normal)

the existence of the asymptotic. (→ ample)



End point map

Definition

Fix a point x0 ∈ M and T > 0. The end-point map at time T is the map

Ex0,T : U → M , u 7→ γx0,u(T ),

For every x0, x1 ∈ M one has

ST (x0, x1) = infE

−1x0,T

(x1)JT = infJT (u) |Ex0,T (u) = x1

In general Ex0,T is smooth but

DuEx0,T is not surjective

the set E−1

x0,T(x1) is not a smooth manifold.



Lagrange multipliers rule

A necessary condition for constrained critical point of infE−1x0,T

(x1)JT .

Theorem

Assume u ∈ U is a constr. crit. point, with x = Ex0,T (u). Then (at least) one ofthe two following statements hold true

(i) ∃λT ∈ T ∗x M s.t. λT · DuEx0,T = DuJT , λT · DuEx0,T = DuJT ,

(ii) ∃λT ∈ T ∗x M s.t. λT · DuEx0,T = 0 (here λT 6= 0).

→ A control u (reps. the associated trajectory γu) is called

normal in case (i),abnormal in case (ii).

A priori a single control can be associated with two covectors such that both (i)and (ii) are satisfied.

In what follows we will call geodesic any normal extremal.

A geodesic γ : [0,T ] → M is called strongly normal if for all s ∈ [0,T ] therestriction γ|[0,s] is not abnormal.Davide Barilari (Univ. Paris Diderot) SR Curvature October 2, 2014 12 / 38


Hamiltonian and PMP for normal

H(λ) = maxu∈Rk

(〈λ,Xu(x)〉 − L(x , u)) , λ ∈ T ∗M , x = π(λ).

where Xu(x) = X0(x) +∑

uiXi (x)

Theorem (PMP, normal case)

Let (u(t), γu(t)) be a normal geodesic. Then there exists a Lipschitz curveλ(t) ∈ T ∗

γu(t)M such that

λ(t) =−→H (λ(t), u(t)),

Any normal geodesic starting at x0 is characterized by its initial covectorλ0 ∈ T ∗

x0M and we write

γ(t) = Expx0(t, λ0)



Distance from a minimizer

Fix x0 ∈ M and a strongly normal geodesic γ(t) = Expx0(t, λ0) and definethe geodesic cost

ct : M → R, t > 0

ct(x) = −St(x , γ(t))

x0

γ(t)

ct : x 7→ −St(x , γ(t))

b

x

Theorem

There exist ε > 0 and an open set U ⊂ (0, ε)×M such that (t, x0) ∈ U

(i) The geodesic cost function (t, x) 7→ ct(x) is smooth on U .

(ii) For every (t, x) ∈ U there exists a unique minimizer connecting x and γ(t)

Moreover dx0ct = λ0 for any t ∈ (0, ε).

Then dx0 ct = 0 → d2x0ct : Tx0M → R is a well defined family of quadratic forms.



Riemannian case

Fix any Riemannian geodesic γ ⇒ well defined quadratic form d2x0ct : Tx0M → R

Define the family of operators Qγ(t) : Tx0M → Tx0M by

d2

x0ct(v) = 〈Qγ(t)v , v〉, v ∈ Tx0M , t ∈ (0, ǫ)

smooth family of symmetric operators for t ∈ (0, ǫ)

singular at t = 0

Regularity theorem (Riemannian case)

The family of symmetric operators has a second order pole at t = 0 and

Qγ(t) =1

t2I+

1

3R∇(γ, ·)γ + O(t)

→ R∇(γ, ·)γ can be though as “directional curvature operator” in the direction γ



Interpretation for ct(x) in Riemannian geometry

Recall that, for a Riemannian structure

ct(x) := −St(x , γ(t)) = −1

2td2(x , γ(t))

Lemma

For a Riemannian structure, the derivative of the geodesic cost function is

ct(x) =1

2‖W t

x0,γ(t)−W t

x,γ(t)‖2 −

1

2

where W tx,y ∈ TyM is the tangent vector at time t of the unique minimizer

connecting x with y in time t

It encodes the curvature without using parallel transport.



Outline

1 Motivation







Geodesic growth vector

Let γ be smooth admissible curve. Let T ∈ Vec(M) an admissible extension of γ

Geodesic flag

F iγ(t) = span[T , . . . , [T

︸︷︷︸

j≤i−1 times

,X ]]|γ(t) | X ∈ Γ(D), j = 0, . . . , i − 1

These subspaces define a natural flag

F1

γ(t) ⊂ F2

γ(t) ⊂ . . . ⊂ Tγ(t)M

F1γ(t) is the distribution Dγ(t).

Does not depend on the choice of TDepends on the germ of γ and of the distribution at γ(t) (→ microlocal)

Geodesic growth vector

Gγ(t) = k1(t), k2(t), . . ., ki (t) = dimF iγ(t)



Ample geodesic

Definition (Ample geodesic)

A geodesic γ is ample at t if ∃m = m(t) > 0 s.t.

F1

γ(t) ⊂ . . . ⊂ Fmγ (t) = Tγ(t)M

An ample geodesic is equiregular if its growth vector does not depend on t.

is a “microlocal Hörmander condition”.

ample at t = 0 ⇒ strongly normal

Theorem (The generic normal SR geodesic is ample at t = 0)

Let M be a sub-Riemannian manifold. For every x0 ∈ M there exists a non-emptyopen Zariski Ax0 ⊂ T ∗

x0M such that, for all λ ∈ Ax0 , the geodesic γ(t) = exp(tλ)

is ample at t = 0.



Hamiltonian scalar product on Dx

In the Riemannian case we used the scalar product to transform quadraticforms in operators.

In the sub-Riemannian case we have a scalar product on 〈·, ·〉 on Dx

→ restriction of our quadratic forms d2x0ct to Dx0

This is a scalar product induced by the Hamiltonian H . In the general case

L is Tonelli (i.e. d2uL non degenerate)

the second derivative of H |T∗

x M at λ defines a scalar product on Dx

- d2

λHx : T∗

x M → TxM self-adjoint linear map

- Im(d2

λHx ) = Dx for every λ ∈ T∗

x M.

〈v1|v2〉λ :=⟨

ξ1, d2

λHx (ξ2)

⟩

, d2

λHx (ξi ) = vi .

Scalar product 〈·, ·〉λ on Dx , depending on λ.

if H is (quadratic) sub-Riemannian, then it is the SR scalar product (not depon λ!).



Main result

Consider the restriction d2x0ct∣∣Dx0

: Dx0 → R to the distribution.

→ the scalar product 〈·, ·〉λ on Dx0 let us to define a family of symmetricoperators

Qλ(t) : Dx0 → Dx0 , d2

x0ct(v) = 〈Qλ(t)v , v〉λ , v ∈ Dx0

Theorem

Assume the geodesic is ample. Then Qλ(t) has a second order pole at t = 0 and

Qλ(t) ≃1

t2Iλ +

1

3Rλ + O(t), for t → 0

where

Iλ ≥ I > 0

Rλ is the curvature operator. We def Ric(λ) := trace(Rλ).

→ Iλ and Rλ are operators defined on Dx0 (symmetric wrt 〈·, ·〉λ).Davide Barilari (Univ. Paris Diderot) SR Curvature October 2, 2014 21 / 38


Riemannian and Finsler

Riemannian

any geodesic is ample with trivial growth vector Gγ = n

Iλ = I

Rλ = R∇(γ, ·)γ

Finsler

any geodesic is ample with trivial growth vector Gγ = n

Iλ = I

Rλ = RFγ - Flag curvature operator of Finsler manifolds



The Heisenberg case

The Heisenberg group R3 = (x , y , z) with standard left-invariant structure

X = ∂x −y

2∂z , Y = ∂y +

x

2∂z

Every (non trivial) geodesic is

ample and equiregular

with geodesic growth vector G = (2, 3).

If one fix two geodesics γλ(t), γη(s) corresponding to two covectors λ, η

C (t, s) := 1

2d2(γλ(t), γη(s)) is not C 2 at zero!

Still we can determine the main expansion

Qλ(t) ≃1

t2Iλ +

1

3Rλ + O(t), for t → 0



The Heisenberg case

We compute it on the orthonormal basis v := γ(0) and v⊥ := γ(0)⊥ for Dx0 .

The matrices representing Iλ and Rλ in the basis v , v⊥ of Dx0 are

Iλ =

(1 00 4

)

, Rλ =2

5

(0 00 h2

z

)

, (1)

where λ has coordinates (hx , hy , hz) dual to o.n. basis X ,Y + Reeb Z

anisotropy of the different directions on Dx0

curvature is always zero in the direction of γ

curvature of lines hz = 0 is zero.

curvature of lift of circles hz 6= 0 is not bounded (nor constant)

→ trace(Iλ) = 5 ↔ is related to MCP(0,5).



Contact 3D case

Every (non trivial) geodesic is ample and equiregular with geodesic growth vectorG = (2, 3).

The matrices representing Iλ and Rλ in the basis γ, γ⊥ of Dx0 are

Iλ =

(1 00 4

)

, Rλ =2

5

(0 00 rλ

)

, (2)

whererλ = h2

z + κ(h2

x + h2

y) + 3χ(h2

x − h2

y )

and λ has coordinates (hx , hy , hz) dual to o.n. basis X ,Y + Reeb Z .

χ and κ are two functions on M (χ = κ = 0 in Heisenberg)

In high dim contact case it is computed/computable (→ in preparation)

Using generalization of Jacobi fields



Young Diagram for equiregular geodesics

Assume now that the geodesic is also equiregular: [← no dependence on t ]

Define di := ki − ki−1 = dimF iγ − dimF i−1

γ

→ By equiregularity one gets d1 ≥ d2 ≥ . . . ≥ dm,

d1

d2

dm−1

∑mi=1

di = n = dimM

dm

n1

n2

nk

Denote nj := length of the j-th row (j = 1, . . . , k)



The operator Iλ

Being an operator on a Euclidean space Iλ : Dx0 → Dx0 has well definedeigenvalues and trace.

Theorem

Assume the geodesic is ample and equiregular. Then

spec Iλ = n21, . . . , n2

k

tr Iλ = n21+ . . .+ n2

k .

In particular we can rewrite

tr Iλ = d1 + 3d2 + 5d3 + . . .+ (2m− 1)dm

Notice that k1 = dimDx0 is the dimension of the distribution.

In the Riemannian we recover Iλ is the identity.

In the Heisenberg (and any 3D contact) trIλ = 5 for every geodesic.

looks like an “Hausdorff dimension” with odd coeff.



Why “a variational approach”?

Two main reasons:Tonelli Lagrangian → include also Finsler case, and not restrict to(sub)-Riemannian situationsDrift → To have more space to find constant curvature model

In our setting, even Heisenberg has no constant curvature(Rλ not constant with respect to λ)

Linear-Quadratic control system in Rn

x = Ax + Bu, x ∈ Rn, u ∈ R

k , JT (x(·)) =1

2

∫ T

0

|u|2 − 〈Qx , x〉 dt.

with A,B in normal form*, has constant curvature equal to Q.

(*) A, B are in Brunowsky normal form [i.e. systems of the form x (n) = u]

(**) controllability (Kalman) → any geodesic is ample and equiregular

dim Fi = rank(B,AB, . . . ,AiB).

(***) spec(Iλ) = Kronecker indices of Brunovsky normal form

In the paper with L. Rizzi we use these models of “constant curvature” to findDavide Barilari (Univ. Paris Diderot) SR Curvature October 2, 2014 28 / 38


Outline

1 Motivation







SR case and sub-Laplacian operator

A sub-Riemannian manifold M with o.n. frame Dx = spanxX1, . . . ,Xk

horizontal gradient ∇f =∑k

i=1Xi (f )Xi

Given a smooth measure µ we can introduce a sub-Laplacian operator

∆µ = divµ ∇ =

k∑

i=1

X 2

i + (divµXi )Xi

it is essentially self-adjoint on C∞0(M) w.r.t. µ

if the structure is Riemannian, this is the Laplace-Beltrami operator

Corollary (of the Main Theorem)

∆ct∣∣x0

≃tr Iλt2

+1

3Ric(λ) + O(t), for t → 0

→ does not depend on µ since ct has critical point at x0.→ it is just the trace of the main expansion ∆ct

∣∣x0

= trace(d2x0ct)



Application: sub-Laplacian of SR distance

We want a formula for the sub-Laplacian of the distance from a geodesic.

ft(·).=

1

2d2(·, γ(t)) = −tct(·), t ∈ (0, ε)

ft has no critical point at x0 → the volume should appear

Theorem

Assume dim Dx locally constant at x0 and γ(t) = expx0(tλ) equiregular.Then there exists a smooth n-form ω defined along γ, such that for any smoothvolume µ s.t. µγ(t) = eg(t)ωγ(t), we have

∆µft |x0 = trIλ − g(0)t −1

3Ric(λ)t2 + O(t3),

Theorem

Let γ be an equiregular geodesic with initial covector λ ∈ T ∗x0M . Then there

exists a smooth n-form ω defined along γ, such that for any smooth volume µ s.t.µγ(t) = eg(t)ωγ(t), we have

1Davide Barilari (Univ. Paris Diderot) SR Curvature October 2, 2014 30 / 38


Outline

1 Motivation







Sub-Riemannian homotheties

M is a complete, connected, orientable sub-Riemannian manifold

µ smooth volume form on M .

Denote Σx0 is the open and dense set where the function f = 1

2d2(x0, ·) is smooth.

The sub-Riemannian homothety is the map φt : Σx0 → M

φt(x) = γx0,x(t) = π e(t−1)~H(dx f).

Fix Ω ⊂ Σx0 be a bounded, measurable set, with 0 < µ(Ω) < +∞

let Ωx0,t.= φt(Ω).

The map t 7→ µ(Ωx0,t) is smooth

µ(Ωx0,t) → 0 for t → 0.

? At which order µ(Ωx0,t) ∼ t?

the order does not depend on the smooth measure µ!



x0

x

Ωφt(x)

b

b

b

Ωx0,t = φt(Ω)

Figure : Sub-Riemannian homothety of the set Ω with center x0.



Riemannian vs sub-Riemannian

Two basic examples

(R) For a Riemannian structure, it is well known that

µ(Ωx0,t) ∼ tdimM , for t → 0,

[here f (t) ∼ g(t) means f (t) = g(t)(C + o(1)) for t → 0 and C > 0]

(SR) In the Heisenberg group it follows from [Juillet, 2009] that

µ(Ωx0,t) ∼ t5, for t → 0,

5 is neither the topological (dim = 3) nor the metric dimension (dimH = 4).

→ The exponent is a different dimensional invariant, associated with behavior ofgeodesics based at x0.



Geodesic dimension

Definition

Let γ(t) = Exp(t, λ) be ample (at t = 0) with Gγ = k1, k2, . . . , km. Then wedefine

Nλ.=

m∑

i=1

(2i − 1)(ki − ki−1),

and Nλ.= +∞ if the geodesic is not ample.

→ If λ is associated with an equiregular geodesic γ then Nλ = trace Iλ.

The function λ 7→ Nλ is constant a.e. on T ∗x0M , assuming its minimum value.

Nx0

.= minNλ|λ ∈ T ∗

x0M < +∞.

We call Nx0 the geodesic dimension of M at x0.



Main result

For every x0 ∈ M we have the inequality Nx0 ≥ dimM and the equality holdsif and only if the structure is Riemannian at x0.

If the distribution is equiregular at x0 we have Nx0 > dimH M from Mitchell’sformula.

Theorem

Let µ be a smooth volume. For any bounded, measurable set Ω ⊂ Σx0 , with0 < µ(Ω) < +∞ we have

µ(Ωx0,t) ∼ tNx0 , for t → 0. (3)

In general this number can depend on the point

If the structure is left invariant is it does not.

Nx0 = 5 in 3D contact and Nx0 = 2n+ 3 in (2n+ 1)-dim contact manifolds



An open question: MCP on Carnot Groups

We say that an n-dim manifol M satisfies MCP(0,D) if

µ(Ωx0,t) ≥ tDµ(Ω), for t ∈ [0, 1]. (4)

this is a global estimate!

Riemannian with Ric ≥ 0 satisfy MCP(0, n)

[Julliet, 2009] the Heisenberg group satisfies MCP(0, 5)

[Rifford, 2013] Any ideal Carnot group satisfies MCP(0,D) for some integerD ≥ Q + n − k .

It is consistent with our theorem since N ≥ Q + n − k .

Conjecture: Carnot groups satisfy MCP(0,N )?

For Step 2 free Carnot groups Q + n− k ∼ k2 and N ∼ k3

at the moment → MCP(0,14) for the (3,6) Carnot group. (w/ L.Rizzi)



Free step 2 Carnot groups

These are Carnot groups with growth vector (k , n) with

n = k +k(k − 1)

2

The generic geodesic is ample and equiregular with geodesic growth vector:

k

k − 1

2

1

k2

(k − 1)2

12

→ spec(Iλ) = 12, 22, . . . , k2 and N = k(k+1)(2k+1)6

∼ k3



Final remarks

→ Final remarks on MCP

What about non nilpotent situations?

Higher order terms in µ(Ωx0,t)? Bounds on curvature?

→ Open questions

What about other kind of comparison? (Laplacian eigenvalues, etc.)

Comparison for solution of heat equation? (the function ct appears in theheat kernel expression)

pt(x , y) =1

tn/2exp

(

−d2(x , y)

4t

)

(C + O(t))

study of the heat kernel along geodesics



THANK YOU FOR YOUR ATTENTION!


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