properties of riemannian manifolds › ... › sc618 › riemannian_m2.pdf · tensor...

Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold

Properties of Riemannian manifolds

Ravi N [email protected] 1

1Systems and Control Engineering,IIT Bombay, India

March 24, 2017

Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN


Outline

1 Tensor fields

2 Curvature

3 Euler Lagrange equations

4 Conventional robotics

5 Control on a Riemannian manifold



Tensor fields

DefinitionThe C∞ sections of T rs (TM) are called class C∞(r, s) tensor fields on Mand denoted by Γ∞(T rs (TM)).

• A C∞ vector field on M is a class C∞(1, 0) tensor fields on M .

• A C∞ covector field on M is a class C∞(0, 1) tensor fields on M .

• A C∞ Riemannian metric on M is a class C∞(0, 2) tensor field G onM having the property that G(x) is an inner product on TxM .

Associating maps with tensor fields

Given t ∈ Γ∞(T rs (TM)), α1, . . . , αr ∈ Γ∞(T ∗M), andX1, . . . , Xs ∈ Γ∞(TM), we define a function on M

x→ t(x) · (α1(x), . . . , αr(x), X1(x), . . . , Xs(x))

and intepret t as a map

Γ∞(T ∗M)× · · · × Γ∞(T ∗M)× Γ∞(TM)× · · · × Γ∞(TM)→C∞(M)



Differentiating tensors

The Lie derivative of a tensor

• Say α ∈ Γ∞(T rs (TM)). Then what is LXα ?

• Going the natural way, the Lie drivative of α must satisfy

LX〈α, Y 〉 = 〈LXα, Y 〉+ 〈α,LXY 〉

⇒ 〈LXα, Y 〉 = LX〈α, Y 〉 − 〈α,LXY 〉

• Check whether the RHS is C∞(M) linear in Y .

Explicit computation

(LXt)(α1, . . . , αr, X1, . . . , Xs) = LX(t(α1, . . . , αr, X1, . . . , Xs))

−r∑i=1

t(α1, . . . ,LXαi, . . . , αr, X1, . . . , Xs)−s∑i=1

(α1, . . . , αr, X1, . . . ,LXXj , . . . , Xs)



Tensors associated with a Riemannian manifold

DefinitionA tensor T of order r on a Riemannian manifold is a multilinear mapping

T : X (M)× · · · × X (M)→D(M)

• A tensor is a pointwise object.

• The curvature tensor

R : X (M)×X (M)×X (M)×X (M)→D(M)

is defined as

R(X,Y, Z,W ) = 〈R(X,Y )Z,W 〉 X,Y, Z,W ∈ X (M)

• The metric tensor

G : X (M)×X (M)→D(M) G(X,Y ) = 〈X,Y 〉

• The Riemannian connection is not a tensor

∇ : X (M)×X (M)×X (M)→D(M) ∇(X,Y, Z) = 〈∇XY , Z〉

since ∇ is not linear with respect to Y .



Covariantly differentiating tensors

Definition

The covariant derivative ∇ZT of a tensor T of order r relative to Z is atensor of order r given by

∇ZT (Y1, . . . , Yr) = Z(T (Y1, . . . , Yr))− [

r∑i=1

T (Y1, . . . ,∇ZYi, . . . , Yr)]

DefinitionThe covariant differential of a tensor T of order r is a tensor of order (r+ 1)given by

∇T (Z, Y1, . . . , Yr) = ∇ZT (Y1, . . . , Yr)



Relating Lie differentiation and covariant differentiation

Claim

(LXS)(Y1, . . . , Yr) = (∇XS)(Y1, . . . , Yr)−∇S(Y1,...,Yr)X+

r∑i=1

S(Y1, . . . ,∇YiX, . . . , Yr)

Can be proved from the two identities

LX(S(Y1, . . . , Yr)) = ∇X(S(Y1, . . . , Yr))−∇S(Y1,...,Yr)X

LX(S(Y1, . . . , Yr)) = (LXS)(Y1, . . . , Yr) +r∑i=1

S(Y1, . . . ,LXYi, . . . , Yr)



A few other results

ClaimThe covariant differential of the metric tensor is the zero tensor

∇G(Z,X, Y ) = Z 〈X,Y 〉 − 〈∇ZX,Y 〉 − 〈X,∇ZY 〉 = 0

• The tensor ∇ZX can be identified with the vector field ∇ZX since

∇ZX(Y ) = ∇X(Z, Y ) = Z(X(Y ))−X(∇ZY ) = Z 〈X,Y 〉−〈X,∇ZY 〉 =

〈∇ZX,Y 〉

• If S is a tensor,∇kS = ∇(∇k−1S)

• From the above identity,

∇2X(Y,Z) = ∇Y∇ZX −∇∇Y ZX



Contravariant tensors

• On a Riemannian manifold, to each X ∈ X (M) we associate a uniqueone-form θX ∈ X ∗(M) (dual to X ) given by

θX (Y ) = 〈X,Y 〉 ∀Y ∈ X (M)

• For each x ∈M , there are isomorphisms G] : T ∗M→TM andG[ : TM→T ∗M associated with the Riemannian metric G.

• Further,dθX(V,W ) + LXg(V,W ) = 2g(∇VX,W )



Coordinate computation of the Lie derivative of the metric tensor -

• Take t = gijdxidxj and Y = (Y1, Y2) .

•

LX(t(Y1, Y2)) = ∂kgijXkY i1Y

j2 + +gij∂kY

i1X

kY j2 + gijYi1 ∂kY

j2 X

k

•t(LXY1, Y2) + t(Y1,LXY2) = gij [∂kY

i1 − ∂mXiY m1 ]Y j2

+gij [∂kY

j2 X

k − ∂mXjY m2 ]Y i1

•LXt(Y1, Y2) = LX(t(Y1, Y2))− [t(LXY1, Y2) + t(Y1,LXY2)]



Outline

1 Tensor fields

2 Curvature






Curvature of a Riemannian manifold

DefinitionThe curvature R of a Riemannian manifold M is a correspondence thatassociates to every pair X,Y ∈ X (M) a mapping R(X,Y ) : X (M)→X (M)given by

R(X,Y )Z = ∇Y∇XZ −∇X∇Y Z +∇[X,Y ]Z, Z ∈ X (M)

Measures the non-commutativity of the covariant derivative.

Proposition

• R is bilinear in X (M)×X (M)

• The curvature operator R(X,Y ) : X (M)→X (M) is linear.



Curvature identities

Proposition

Bianchi Identity

R(X,Y )Z +R(Y,Z)X +R(Z,X)Y = 0



Outline

1 Tensor fields

2 Curvature






Euler-Lagrange equations on a Riemannian manifold

Preliminaries

• Action functional

AL(γ) =

∫ b

a

L(t, γ′(t))dt

• γ : [a, b]→Q is a twice differentiable curve on M starting at qa andending at qb.

• ν : J × [a, b]→Q denotes a class of such admissible curves, each curvebeing parametrized by s ∈ J .

• Necessary condition for a minimum

d

ds|s=0(AL(ν(s, t)) = 0



Facts

• Two vector fields

Sν(s, t) =d

dsν(s, t) Tν(s, t) =

d

dtν(s, t)

Define XSν ◦ ν = Sν XTν ◦ ν = Tν

• Equality of mixed partials leads to

[XSν , XTν ](ν(s, t)) = 0

•∇SνTν = ∇TνSν (∇XY −∇YX = [X,Y ] = 0)

•d

dtG(Sν , Tν)(s, t) = LTνG(Sν , Tν)(s, t) =

G(∇TνSν , Tν)(s, t) + G(Sν ,∇TνTν)(s, t)

•d

dsG(Tν , Tν)(s, t) = LSνG(Tν , Tν)(s, t) = 2G(∇SνTν , Tν)(s, t)



Outline of the proof

•0 =

d

ds|s=0(AL(ν(s, t))

•

=d

ds|s=0

1

2

∫ b

a

G(Tν(s, t), Tν(s, t))dt

•

=

∫ b

a

G(∇SνTν(s, t), Tν(s, t))dt|s=0

•

=

∫ b

a

G(∇TνSν(s, t), Tν(s, t))dt|s=0

•

=

∫ b

a

G(∇γ′(t)δν(t), γ′(t))dt

•

=

∫ b

a

[d

dtG(δν, γ′(t))−G(δν(t),∇γ′(t)γ′(t))dt]

Leads to∇γ′(t)γ′(t) = 0



Outline

1 Tensor fields

2 Curvature






A robotic manipulator

l1

x

y

z

a

B

A

l2

Figure: A conventional manipulator (Courtesy: Murray, Li and Sastry)



Forward kinematics

How do the joint angles map to the position of the end-effector ?

• Configuration space: Revolute joint - S1, prismatic joint - R1

• Say p revolute joints and m prismatic joints,

• Forward kinematics mapf(·) : S1 × . . .× S1︸︷︷︸

p revolute

×R1 × . . .× R1︸︷︷︸m prismatic

= Q→ SE(3).

• ξi is the velocity vector (translational or rotational) associated withthe ith joint.

• For n joints - the product of exponentials yields

f(θ) = eξ1θ1eξ2θ2 . . . eξnθn︸︷︷︸a product of exponentials

f(0)

• Recall the geometric interpretation - the exponential map takes the Liealgebra to the Lie group.



Kinetic energy

• Kinetic energy = sum of the kinetic energy of each link. For the ithlink

Ti(θ, θ) =1

2(V bbli)

TMi(Vbbli) =

1

2θT [Jbbli(θ)

TMiJbbli(θ)]θ

• Total kinetic energy

T (θ, θ) =n∑i=1

1

2θTM(θ)θ

• Manipulator inertia matrix

M(θ) =

n∑i=1

Jbbli(θ)TMiJ

bbli(θ)



Potential energy and the Lagrangian

• Potential energy

V (θ) =

n∑i=1

mighi(θ)

• Lagrangian - L(θ, θ) =.

1

2

n∑i,j=1

Mij(θ)θiθj − V (θ)

• Equations of motion for each link

n∑j=1

Mij(θ)θj +n∑

j,k=1

Γkij θj θk +∂V

∂θi(θ) = τi i = 1, . . . , n



Interpretation of the equations of motion

• Christoffel symbols corresponding to the inertia matrix M(θ).

Γkij =1

2[∂Mij(θ)

∂θk+∂Mik(θ)

∂θj− ∂Mkj(θ)

∂θi]︸︷︷︸

arise out of the notion of a covariant derivative

• Coriolis matrix

Cij(θ, θ) =n∑k=1

Γkij θk

• Coriolis and centrifugal forces

C(θ, θ)θ

• Equations of motion are often expressed as

M(θ)θ + C(θ, θ) +N(θ, θ) = τ



Geometric viewpoint

• The dynamic equations are described on the manifold TQ.

• An element of TQ is a two tuple (θ, θ).

• The vector field on TQ is given by[θ

θ

]=

[θ

−M−1(θ)[C(θ, θ)θ +N(θ, θ)]

]• The Lagrangian is a map L : TQ→ R.

• The KE is a metric on the tangent space TqQ induced by the inertiamatrix M(q) as

< θi, θj >4=

1

2

n∑i,j=1

Mij(θ)θiθj

• The inertia matrix M(q) is symmetric and positive definite andM − 2C is skew-symmetric



Outline

1 Tensor fields

2 Curvature






The Riemannian framework

Basic notions

• The manifold Q has a metric G called the Riemannian metric thatsatisfies properties mentioned before.

• G now defines two operators: G] and G[.

G](q) : T ∗q Q→TqQ G[(q) : TqQ→T ∗q Q

〈vq, wq〉 = [G[vq, wq]∀vq, wq ∈ TqQ

• The gradient and the Hessian of a function, f : Q→R:

gradf = G](df) Hessf(q) · (vq, wq) =

⟨vq,

G∇wqgradf

⟩• Please note that often in our use in problems, we interchangeably use

gradf and df . This is because we are working with the natural basis inRn.



Simple mechanical systems

Equations of motion

• γ(t) : [ti, tf ]→Q denotes the trajectory of the mechanical system fromtime ti to time tf .

• Equation of motion (autonomous system)

G∇γ(t)γ(t) = −gradV (γ(t))︸︷︷︸

Potential forces

+ G](Fdiss(γ(t)))︸︷︷︸Dissipative forces

• With state-feedback control

G∇γ(t)γ(t) = −gradV (γ(t)) + G](Fdiss(γ(t))) + G](F (t, γ(t)))



Computation

Coordinate computation

• Assume there are no potential forces or dissipative forces. Further,assign coordinates to γ(t) as x(t). Then, the earlier equation reduces to

∇γ(t)γ(t) = 0 (1)

• Christoffel symbols Γkij

∇γ(t)γ(t) = 0⇒ xk(t) + Γkij(γ(t))xi(t)xj(t) = 0 k = 1, . . . , n.

• Compute the Christoffel symbols from the Riemannian metric as

Γkij =1

2Gkl(∂Gil

∂xj+∂Gjl∂xi

− ∂Gij∂xl

)

where Gkl stands for the inverse of .



Geodesic spray

• The geodesis spray is defined as

S = vi∂

∂xi− Γkijv

jvk∂

∂vi



Pendulum on a cart

A schematic

e

m

M

l

x

. .

Figure: A conventional manipulator



Pendulum on a cart

• Configuration space Q = R× S1.

• The Lagrangian L : TQ→ R is given by

L =1

2(αθ2 − 2β cos θθs+ γs2) +D cos θ

where (s, θ) ∈ (R, S1) and α, β, γ are parameters dependant on m,M, land g.

• The Riemannian metric based on the kinetic energy is

G(s, θ) ((s, θ), (s, θ))︸︷︷︸vq=wq

=1

2

[s θ

] [ γ −β cos θ−β cos θ α

] [s

θ

]


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