Proof: The completeness of the ¬‚ow of Y follows immediately from the fact that the in-

’1

tegral curve of Y which passes through v ∈ T M must stay in the compact set EL EL (v) .

(Recall that EL is constant on all of the integral curves of Y .) Details are left to the reader.

I now turn to the proof of the recurrence property. Let E0 = EL (v). By hypothesis,

’1

the set C = EL [E0 ’ 1, E0 + 1] is compact, so the µL -volume of the open set W =

’1

EL (E0 ’1, E0 +1) (which lies inside C) is ¬nite. It clearly su¬ces to prove the recurrence

property for any open neighborhood U of v which lies inside W , so let us assume that

U ‚ W.

Let φ: W ’ W be the di¬eomorphism φ(w) = ¦(T, w). This di¬eomorphism is

clearly invertible and preserves the µL -volume of open sets in W . Consider the open sets

U k = φk (U) for k > 0 (integers). These open sets all have the same µL -volume and

hence cannot be all disjoint since then their union (which lies in W ) would have in¬nite

µL -volume. Let 0 < j < k be two integers so that U j © U k = …. Then, since

U j © U k = φj (U) © φk (U) = φj U © φk’j (U) ,

it follows that U © φk’j (U) = …, as we wished to show.

This theorem has the amazing consequence that, whenever one has a non-degenerate

Lagrangian with a proper energy function, the corresponding mechanical system “recurs”

in the sense that “arbitrarily near any given initial condition, there is another initial

condition so that the evolution brings this initial condition back arbitrarily close to the

¬rst initial condition”. I realize that this statement is somewhat vague and subject to

misinterpretation, but the precise statement has already been given, so there seems not to

be much harm in giving the paraphrase.

L.4.16 76

Exercise Set 4:

Symmetries and Conservation Laws

1. Show that two Lagrangians L1 , L2 : T M ’ R satisfy

EL1 = EL2 and dωL1 = dωL2

if and only if there is a closed 1-form φ on M so that L1 = L2 + φ. (Note that, in this

equation, we interpret φ as a function on T M.) Such Lagrangians are said to di¬er by a

“divergence term.” Show that such Lagrangians share the same critical curves and that

one is non-degenerate if and only if the other is.

2. What does Conservation of Energy mean for the case where L de¬nes a Riemannian

metric on M?

3. Show that the equations for geodesics of a rotationally invariant metric of the form

I = E(r) dr2 + 2 F (r)dr dθ + G(r) dθ2

can be integrated by separation of variables and quadratures. (Hint: Start with the con-

servation laws we already know:

™™ ™

E(r) r2 + 2 F (r)r θ + G(r) θ2 = v0

2

™

™

F (r)r + G(r) θ = u0

™

™

where v0 and u0 are constants. Then eliminate θ and go on from there.)

4. The de¬nition of ωL given in the text might be regarded as somewhat unsatisfactory

since it is given in coordinates and not “invariantly”. Show that the following invariant

description of ωL is valid: The manifold T M inherits some extra structure by virtue of

being the tangent bundle of another manifold M. Let π: T M ’ M be the basepoint

projection. Then π is a submersion: For every a ∈ T M,

π (a): Ta T M ’ Tπ(a)M

is a surjection and the ¬ber at π(a) is equal to

π ’1 π(a) = Tπ(a) M.

It follows that the kernel of π (a) (i.e., the “vertical space” of the bundle π: T M ’ M

at a) is naturally isomorphic to Tπ(a) M. Call this isomorphism ±: Tπ(a) M ’ ker π (a) .

˜

Then the 1-form ωL is de¬ned by

ωL (v) = dL ± —¦ π (v) for v ∈ T (T M).

Hint: Show that, in local canonical coordinates, the map ± —¦ π satis¬es

‚ ‚ ‚

±—¦π ai + bi i = ai .

‚xi ‚pi

‚p

E.4.1 77

5. For any vector ¬eld X on M, let the associated vector ¬eld on T M be denoted X .

Show that if X has the form

‚

X = ai i

‚x

in some local coordinate system, then, in the associated canonical (x, p) coordinates, X

has the form

i

i‚ j ‚a ‚

X =a +p .

‚xi ‚xj ‚pi

6. Show that conservation of angular momenta in the motion of a point mass in a central

force ¬eld implies Kepler™s Law that “equal areas are swept out over equal time intervals.”

Show also that, in the n = 2 case, employing the conservation of energy and angular

momentum allows one to integrate the equations of motion by quadratures. (Hint: For

the second part of the problem, introduce polar coordinates: (x1 , x2 ) = (r cos θ, r sin θ).)

7. In the example of the motion of a rigid body, show that the Lagrangian on G is always

non-negative and is non-degenerate (so that L de¬nes a left-invariant metric on G) if and

only if the matrix µ has at most one zero eigenvalue. Show that L is degenerate if and

only if the rigid body lies in a subspace of dimension at most n’2.

8. Supply the details in the proof of Proposition 5. You will want to go back to the

integration-by-parts derivation of the Euler-Lagrange equations and show that, even if the

variation “ induced by h does not have ¬xed endpoints, we still get a local coordinate

formula of the form

‚L ‚L

FL,“ (0) = y(b), y(b) hk (b) ’ k y(a), y(a) hk (a)

™ ™

‚pk ‚p

for any variation of a solution of the Euler-Lagrange equations. Give these “boundary

terms” an invariant geometric meaning and show that they cancel out when we sum over

a partition of a (¬xed-endpoint) variation of an L-critical curve γ into subcurves which lie

in coordinate neighborhoods.)

9. (Alternate to Exercise 8.) Here is another approach to proving Proposition 5. Instead of

dividing the curve up into sub-curves, show that for any variation “ of a curve γ: [a, b] ’ M

(not necessarily with ¬xed endpoints), we have the formula

b

FL,“ (0) = ωL V (b) ’ ωL V (a) ’ dωL γ (t), V (t) + dEL V (t) dt

¨

a

™

where V (t) = “ (t, 0)(‚/‚s) is the “variation vector ¬eld” at s = 0 of the lifted variation

™

“ in T M. Conclude that, whether L is non-degenerate or not, the condition γ dωL +¨

dEL γ(t) = 0 is the necessary and su¬cient condition that γ be L-critical.

™

E.4.2 78

10. The Two Body Problem. Consider a pair of point masses (with masses m1 and

m2 ) which move freely subject to a force between them which depends only on the distance

between the two bodies and is directed along the line joining the two bodies. This is what

is classically known as the Two Body Problem. It is represented by a Lagrangian on the

manifold M = Rn — Rn with position coordinates x1 , x2 : M ’ Rn of the form

m1 m2

|p1 |2 + |p2 |2 ’ V (|x1 ’ x2 |2 ).

L(x1 , x2 , p1 , p2 ) =

2 2

(Here, (p1 , p2 ) are the canonical ¬ber (velocity) coordinates on TM associated to the

coordinate system (x1 , x2 ).) Notice that L has the form “kinetic minus potential”. Show

that rotations and translations in Rn generate a group of symmetries of this Lagrangian

and compute the conserved quantities. What is the interpretation of the conservation law

associated to the translations?

11. The Sliding Particle. Suppose that a particle of unit weight and mass (remember:

“geometric units” means never having to state your constants) slides without friction on

a smooth hypersurface xn+1 = F (x1 , . . . , xn ) subject only to the force of gravity (which

is directed downward along the xn+1 -axis). Show that the “kinetic-minus-potential” La-

grangian for this motion in the x-coordinates is

‚F i 2

(p1 )2 + · · · + (pn )2 + ’ F (x1 , . . . , xn ).

1

L= p

2 ‚xi

Show that this is a non-degenerate Lagrangian and that its energy EL is proper if and

only if F ’1 (’∞, a] is compact for all a ∈ R.

Suppose that F is invariant under rotation, i.e., that

F (x1 , . . . , xn ) = f (x1 )2 + · · · + (xn )2

for some smooth function f. Show that the “shadow” of the particle in Rn stays in a ¬xed

2-plane. Show that the equations of motion can be integrated by quadrature.

Remark: This Lagrangian is also used to model a small ball of unit mass and weight

“rolling without friction in a cup”. Of course, in this formulation, the kinetic energy

stored in the ball by its spinning is ignored. If you want to take this “spinning” energy

into account, then you must study quite a di¬erent Lagrangian, especially if you assume

that the ball rolls without slipping. This goes into the very interesting theory of “non-

holonomic systems”, which we (unfortunately) do not have time to go into.

12. Let L be a Lagrangian which restricts to each ¬ber Tx M to be a non-degenerate

(though not necessarily positive de¬nite) quadratic form. Show that L is non-degnerate

as a Lagrangian and that the Legendre mapping „L : T M ’ T — M is an isomorphism of

vector bundles. Show that, if L is, in addition a positive de¬nite quadratic form on each

¬ber, then the new Lagrangian de¬ned by

1

˜ 2

L = L+1

is also a non-degenerate Lagrangian, but that the map „L : T M ’ T — M, though one-to-one,

˜

is not onto.

E.4.3 79

Lecture 5:

Symplectic Manifolds, I

In Lecture 4, I associated a non-degenerate 2-form dωL on T M to every non-degenerate

Lagrangian L: T M ’ R. In this section, I want to begin a more systematic study of the

geometry of manifolds on which there is speci¬ed a closed, non-degenerate 2-form.

Symplectic Algebra.

First, I will develop the algebraic precursors of the manifold concepts which are to

follow. For simplicity, all of these constructions will be carried out on vector spaces over

the reals, but they could equally well have been carried out over any ¬eld of characteristic

not equal to 2.

Symplectic Vector Spaces. A bilinear pairing B: V — V ’ R is said to be skew-

symmetric (or alternating) if B(x, y) = ’B(y, x) for all x, y in V . The space of skew-

symmetric bilinear pairings on V will be denoted by A2 (V ). The set A2 (V ) is a vec-

tor space under the obvious addition and scalar multiplication and is naturally identi¬ed

with Λ2 (V — ), the space of exterior 2-forms on V . The elements of A2 (V ) are often called

skew-symmetric bilinear forms on V. A pairing B ∈ A2 (V ) is said to be non-degenerate if,

for every non-zero v ∈ V , there is a w ∈ V for which B(v, w) = 0.

De¬nition 1: A symplectic space is a pair (V, B) where V is a vector space and B is a

non-degenerate, skew-symmetric, bilinear pairing on V .

Example. Let V = R2n and let Jn be the 2n-by-2n matrix

0n In

Jn = .

’In 0n

For vectors v, w ∈ R2n , de¬ne

B0 (x, y) = tx Jn y.

Then it is clear that B0 is bilinear and skew-symmetric. Moreover, in components

B0 (x, y) = x1 y n+1 + · · · + xn y 2n ’ xn+1 y 1 ’ · · · ’ x2n y n

so it is clear that if B0 (x, y) = 0 for all y ∈ R2n then x = 0. Hence, B0 is non-degenerate.

Generally, in order for B(x, y) = tx A y to de¬ne a skew-symmetric bilinear form on

Rn , it is only necessary that A be a skew-symmetric n-by-n matrix. Conversely, every

skew-symmetric bilinear form B on Rn can be written in this form for some unique skew-

symmetric n-by-n matrix A. In order that this B be non-degenerate, it is necessary and

su¬cient that A be invertible. (See the Exercises.)

L.5.1 80

The Symplectic Group. Now, a linear transformation R: R2n ’ R2n preserves B0 ,

i.e., satis¬es B0 (Rx, Ry) = B0 (x, y) for all x, y ∈ R2n , if and only if tR Jn R = Jn . This

motivates the following de¬nition:

De¬nition 2: The subgroup of GL(2n, R) de¬ned by

Sp(n, R) = R ∈ GL(2n, R) | tR Jn R = Jn

is called the symplectic group of rank n.

It is clear that Sp(n, R) is a (closed) subgroup of GL(2n, R). In the Exercises, you are

asked to prove that Sp(n, R) is a Lie group of dimension 2n2 + n and to derive other of its

properties.

Symplectic Normal Form. The following proposition shows that there is a normal

form for ¬nite dimensional symplectic spaces.

Proposition 1: If (V, B) is a ¬nite dimensional symplectic space, then there exists a

basis e1 , . . . , en , f 1 . . . , f n of V so that, for all 1 ¤ i, j ¤ n,

j

B(ei , f j ) = δi , and B(f i , f j ) = 0

B(ei , ej ) = 0,

Proof: The desired basis will be constructed in two steps. Let m = dim(V ).

Suppose that for some n ≥ 0, we have found a sequence of linearly independent vectors

e1 , . . . , en so that B(ei , ej ) = 0 for all 1 ¤ i, j ¤ n. Consider the vector space Wn ‚ V

which consists of all vectors w ∈ V so that B(ei , w) = 0 for all 1 ¤ i ¤ n. Since the ei

are linearly independent and since B is non-degenerate, it follows that Wn has dimension

m ’ n. We must have n ¤ m ’ n since all of the vectors e1 , . . . , en clearly lie in Wn .

If n < m ’ n, then there exists a vector en+1 ∈ Wn which is linearly independent

from e1 , . . . , en . It follows that the sequence e1 , . . . , en+1 satis¬es B(ei , ej ) = 0 for all

1 ¤ i, j ¤ n + 1. (Since B is skew-symmetric, B(en+1 , en+1 ) = 0 is automatic.) This

extension process can be repeated until we reach a stage where n = m ’ n, i.e., m = 2n.

At that point, we will have a sequence e1 , . . . , en for which B(ei , ej ) = 0 for all 1 ¤ i, j ¤ n.

Next, we construct the sequence f 1 , . . . , f n . For each j in the range 1 ¤ j ¤ n,

consider the set of n linear equations