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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
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
Outline
1 Tensor fields
2 Curvature
3 Euler Lagrange equations
4 Conventional robotics
5 Control on a Riemannian manifold
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
Outline
1 Tensor fields
2 Curvature
3 Euler Lagrange equations
4 Conventional robotics
5 Control on a Riemannian manifold
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics 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)
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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)
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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 .
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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)
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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)
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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 )
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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)]
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
Outline
1 Tensor fields
2 Curvature
3 Euler Lagrange equations
4 Conventional robotics
5 Control on a Riemannian manifold
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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.
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
Curvature identities
Proposition
Bianchi Identity
R(X,Y )Z +R(Y,Z)X +R(Z,X)Y = 0
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
Outline
1 Tensor fields
2 Curvature
3 Euler Lagrange equations
4 Conventional robotics
5 Control on a Riemannian manifold
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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)
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
Outline
1 Tensor fields
2 Curvature
3 Euler Lagrange equations
4 Conventional robotics
5 Control on a Riemannian manifold
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
A robotic manipulator
l1
x
y
z
a
B
A
l2
Figure: A conventional manipulator (Courtesy: Murray, Li and Sastry)
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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.
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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)
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
Outline
1 Tensor fields
2 Curvature
3 Euler Lagrange equations
4 Conventional robotics
5 Control on a Riemannian manifold
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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.
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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)))
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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 .
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
Geodesic spray
• The geodesis spray is defined as
S = vi∂
∂xi− Γkijv
jvk∂
∂vi
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
Pendulum on a cart
A schematic
e
m
M
l
x
. .
Figure: A conventional manipulator
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN
Tensor fields Curvature Euler Lagrange equations Conventional robotics Control on a Riemannian manifold
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
]
Lectures on Riemannian geometry (IIT-B/IIT-GN) March 2017, IIT-GN