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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

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