k k k

Since A is a curve in O(n), this becomes

(A’1 A¯ k + b · (A’1 A¯ k + b

™x ™ ™x ™

1

= mk

2

k

mk (±¯ k + β) · (±¯ k + β) .

1

= x x

2

k

Using the center-of-mass normalization, this simpli¬es to

mk ’txk ±2 xk + |β|2 .

1

¯ ¯

= 2

k

With a slight rearrangement, this takes the simple form

2

L γ(t) = ’tr ±(t) µ + 1 m |β(t)|2

™ 2

where m = m1 + · · · + mN is the total mass of the body and µ is the positive semi-de¬nite

symmetric n-by-n matrix

mk xk txk .

µ= 1 ¯¯

2

k

It is clear that we can interpret L as a left-invariant Lagrangian on G. Actually, even the

formula we have found so far can be simpli¬ed: If we write µ = R δ tR where δ is diagonal

and R is an orthogonal matrix (which we can always do), then right acting on G by the

element

R0

01

will reduce the Lagrangian to the form

2

L g(t) = ’tr ±(t) δ + 1 m |β(t)|2 .

™ 2

Thus, only the eigenvalues of the matrix µ really matter in trying to solve the equations

of motion of a rigid body. This observation is usually given an interpretation like “the

motion of any rigid body is equivalent to the motion of its ˜ellipsoid of inertia™ ”.

Hamiltonian Form. Let us return to the consideration of the Euler-Lagrange equa-

tions. As we have seen, in expanded form, the equations in local coordinates are

‚2L ‚2L ‚L

(y, y)¨ + i j (y, y)y j ’ i (y, y) = 0.

j

™y ™™ ™

‚pi ‚pj ‚p ‚x ‚x

L.4.11 71

In order for these equations to be solvable for the highest derivatives at every possible set

of initial conditions, the symmetric matrix

‚2L

HL (x, p) = (x, p) .

‚pi ‚pj

must be invertible at every point (x, p).

De¬nition 4: A Lagrangian L is said to be non-degenerate if, relative to every local

coordinate system x: U ’ Rn , the matrix HL is invertible at every point of T U .

For example, if L: T M ’ R restricts to each tangent space Tm M to be a non-

degenerate quadratic form, then L is a non-degenerate Lagrangian. In particular, when L

is a Riemannian metric, L is non-degenerate.

Although De¬nition 4 is fairly explicit, it is certainly not coordinate free. Here is a

result which may clarify the meaning of non-degenerate.

Proposition 4: The following are equivalent for a Lagrangian L: T M ’ R:

(1) L is a non-degenerate Lagrangian.

(2) In local coordinates (x, p), the functions x1 , . . . , xn , ‚L/‚p1 , . . . , ‚L/‚pn have every-

where independent di¬erentials.

(3) The 2-form dωL is non-degenerate at every point of T M, i.e., for any tangent vector

v ∈ T (T M), v dωL = 0 implies that v = 0.

Proof: The equivalence of (1) and (2) follows directly from the Chain Rule and is left

as an exercise. The equivalence of (2) and (3) can be seen as follows: Let v ∈ Ta (T M)

be a tangent vector based at a ∈ Tm M. Choose any local any canonical local coordinate

system (x, p) with m ∈ U and write qi = ‚L/‚pi for 1 ¤ i ¤ n. Then ωL takes the form

dωL = dqi § dxi .

Thus,

v dωL = dqi (v) dxi ’ dxi (v) dqi .

Now, suppose that the di¬erentials dx1 , . . . , dxn , dqi , . . . , dqn are linearly independent

—

at a and hence span Ta (T M). Then, if v dωL = 0, we must have dqi (v) = dxi (v) = 0,

which, because the given 2n di¬erentials form a spanning set, implies that v = 0 Thus,

dωL is non-degenerate at a.

On the other hand, suppose that that the di¬erentials dx1 , . . . , dxn , dq1 , . . . , dqn are

—

linearly dependent at a. Then, by linear algebra, there exists a non-zero vector v ∈ Ta (T M)

so that dqi (v) = dxi (v) = 0. However, it is then clear that v dωL = 0 for such a v, so

that dωL will be degenerate at a.

For physical reasons, the function qi is usually called the conjugate momentum to the

coordinate xi .

L.4.12 72

Before exploring the geometric meaning of the coordinate system (x, q), we want to

give the following description of the L-critical curves of a non-degenerate Lagrangian.

Proposition 5: If L: T M ’ R is a non-degenerate Lagrangian, then there exists a unique

vector ¬eld Y on T M so that, for every L-critical curve γ: [a, b] ’ M, the associated curve

γ: [a, b] ’ T M is an integral curve of Y . Conversely, for any integral curve •: [a, b] ’ T M

™

of Y , the composition φ = π —¦ •: [a, b] ’ M is an L-critical curve in M.

Proof: It is clear that we should take Y to be the unique vector ¬eld on T M which satis¬es

Y dωL = ’dEL . (There is only one since, by Proposition 4, dωL is non-degenerate.)

Proposition 3 then says that for every L-critical curve, its lift γ satis¬es γ (t) = Y (γ(t)) for

™ ¨ ™

all t, i.e., that γ is indeed an integral curve of Y .

™

The details of the converse will be left to the reader. First, one must check that, with φ

™

de¬ned as above, we have φ = •. This is best done in local coordinates. Second, one must

check that φ is indeed L-critical, even though it may not lie entirely within a coordinate

neighborhood. This may be done by computing the variation of φ restricted to appropriate

subintervals and taking account of the boundary terms introduced by integration by parts

when the endpoints are not ¬xed. Details are in the Exercises.

The canonical vector ¬eld Y on T M is just the coordinate free way of expressing the

fact that, for non-degenerate Lagrangians, the Euler-Lagrangian equations are simply a

non-singular system of second order ODE for maps γ: [a, b] ’ M

Unfortunately, the expression for Y in canonical (x, p)-coordinates on T M is not very

nice; it involves the inverse of the matrix HL . However, in the (x, q)-coordinates, it is

a completely di¬erent story. In these coordinates, everything takes a remarkably simple

form, a fact which is the cornerstone on symplectic geometry and the calculus of variations.

Before taking up the geometric interpretation of these new coordinates, let us do a

few calculations. We have already seen that , in these coordinates, the canonical 1-form

ωL takes the simple form ωL = qi dxi .

We can also express EL as a function of (x, q). It is traditional to denote this expression

by H(x, q) and call it the Hamiltonian of the variational problem (even though, in a certain

sense, it is the same function as EL ). The equation determining the vector ¬eld Y is

expressed in these coordinates as

(dqi § dxi ) = dqi (Y ) dxi ’ dxi (Y ) dqi

Y dωL = Y

‚H ‚H

= ’dH = ’ i dxi ’ dqi ,

‚x ‚qi

so the expression for Y in these coordinates is

‚H ‚ ‚H ‚

’i

Y= .

‚qi ‚xi ‚x ‚qi

In particular, the ¬‚ow of Y takes the form

‚H ‚H

qi = ’

xi =

™ and ™ .

‚xi

‚qi

L.4.13 73

These equations are known as Hamilton™s Equations or, sometimes, as the Hamiltonian

form of the Euler-Lagrange equations.

Part of the reason for the importance of the (x, q) coordinates is the symmetric way

they treat the positions and momenta. Another reason comes from the form the in¬nites-

imal symmetries take in these coordinates: If X is an in¬nitesimal symmetry of L and

X is the induced vector ¬eld on T M with conserved quantity G = ωL (X ), then, since

LX ωL = 0,

X dωL = ’d(ωL (X )) = ’dG.

Thus, by the same analysis as above, the ODE represented by X in the (x, q) coordinates

becomes

‚G ‚G

qi = ’ i .

xi =

™ and ™

‚qi ‚x

In other words, in the (x, q) coordinates, the ¬‚ow of a symmetry X has the same Hamilto-

nian form as the ¬‚ow of the vector ¬eld Y which gives the solutions of the Euler-Lagrange

equations! This method of putting the symmetries of a Lagrangian and the solutions of

the Lagrangian on a sort of equal footing will be seen to have powerful consequences.

The Cotangent Bundle. Early on in this lecture, we introduced, for each coordinate

chart x: U ’ Rn , a canonical extension (x, p): T U ’ Rn — Rn and characterized it by a

geometric property. There is also a canonical extension (x, ξ): T — U ’ Rn — Rn where

ξ = (ξi ): T — U ’ Rn is characterized by the condition that, if f: U ’ R is any smooth

function on U, then, regarding its exterior derivative df as a section df: U ’ T — U, we have

‚f

ξi —¦ df = .

‚xi

I will leave to the reader the task of showing that (x, ξ) is indeed a coordinate system

on T — U.

It is a remarkable fact that the cotangent bundle π: T — M ’ M of any smooth manifold

—

carries a canonical 1-form ω de¬ned by the following property: For each ± ∈ Tx M, we

de¬ne the linear function ω± : T± T — M ’ R by the rule ω± (v) = ± π (±)(v) . I leave to

the reader the task of showing that, in canonical coordinates (x, ξ : T — U ’ Rn — Rn , this

canonical 1-form has the expression

ω = ξi dxi .

The Legendre transformation. Now consider a smooth Lagrangian L: T M ’ R

as before. We can use L to construct a smooth mapping „L : T M ’ T — M as follows: At

each v ∈ T M, the 1-form ωL (v) is semi-basic, i.e., there exists a (necessarily unique) 1-

form „L (v) ∈ Tπ(v) M so that ωL (v) = π — „L (v) . This mapping is known as the Legendre

—

transformation associated to the Lagrangian L.

L.4.14 74

This de¬nition is rather abstract, but, in local coordinates, it takes a simple form.

The reader can easily check that in canonical coordinates associated to a coordinate chart

x: U ’ Rn , we have

‚L

(x, ξ) —¦ „L = (x, q) = xi , i .

‚p

In other words, the (x, q) coordinates are just the canonical coordinates on the cotangent

bundle composed with the Legendre transformation! It is now immediate that „L is a local

di¬eomorphism if and only if L is a non-degenerate Lagrangian. Moreover, we clearly have

—

ωL = „L (ω), so the 1-form ωL is also expressible in terms of the canonical 1-form ω and

the Legendre transform.

What about the function EL on T M? Let us put the following condition on the

Lagrangian L: Let us assume that „L : T M ’ T — M is a (one-to-one) di¬eomorphism onto

its image „L (T M) ‚ T — M. (Note that this implies that L is non-degenerate, but is stronger

than this.) Then there clearly exists a function on „L (T M) which pulls back to T M to

be EL . In fact, as the reader can easily verify, this is none other than the Hamiltonian

function H constructed above.

The fact that the Hamiltonian H naturally “lives” on T — M (or at least an open subset

thereof) rather than on T M justi¬es it being regarded as distinct from the function EL .

There is another reason for moving over to the cotangent bundle when one can: The

vector ¬eld Y on T M corresponds, under the Legendre transformation, to a vector ¬eld Z

on „L (T M) which is characterized by the simple rule Z dω = ’dH. Thus, just knowing

the Hamiltonian H on an open set in T — M determines the vector ¬eld which sweeps out

the solution curves! We will see that this is a very useful observation in what follows.

Poincar´ Recurrence. To conclude this lecture, I want to give an application of

e

the geometry of the form ωL to understanding the global behavior of the L-critical curves

when L is a non-degenerate Lagrangian. First, I make the following observation:

Proposition 6: Let L: T M ’ R be a non-degenerate Lagrangian. Then 2n-form µL =

(dωL )n is a volume form on T M (i.e., it is nowhere vanishing). Moreover the (local) ¬‚ow

of the vector ¬eld Y preserves this volume form.

Proof: To see that µL is a volume form, just look in local (x, q)-coordinates:

µL = (dωL )n = (dqi § dxi )n

= n! dq1 § dx2 § dq2 § dx2 § · · · n

§ dqn § dx .

By Proposition 4, this latter form is not zero.

Finally, since

LY (dωL ) = d(Y dωL ) = ’d dEL = 0,

it follows that the (local) ¬‚ow of Y preserves dωL and hence preserves µL .

Now we shall give an application of Proposition 6. This is the famous Poincar´

e

Recurrence Theorem.

L.4.15 75

Theorem 2: Let L: T M ’ R be a non-degenerate Lagrangian and suppose that EL is a

proper function on T M. Then the vector ¬eld Y is complete, with ¬‚ow ¦: R — T M ’ T M.

Moreover, this ¬‚ow is recurrent in the following sense: For any point v ∈ T M, any open

neighborhood U of v, and any positive time interval T > 0, there exists an integer N > 0

so that ¦(T N, U ) © U = ….