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and the spaces H are ¬nite-dimensional. Hence, we have the following
isomorphism:
,m
,m
H H
HdeRham (M ) HDolbeault(M ) .
+m=k +m=k

For the proof, see for instance [48, 107].

17.2 Immediate Topological Consequences
Let bk (M ) := dim HdeRham (M ) be the usual Betti numbers of M , and let
k
,m
h ,m (M ) := dim HDolbeault (M ) be the so-called Hodge numbers of M .
bk ,m
= +m=k h
Hodge Theorem =’ ,m m,
h =h
Some immediate topological consequences are:
1. On compact K¨hler manifolds “the odd Betti numbers are even:”
a
k
2k+1 ,(2k+1’ )
,m
b = h =2 h is even .
+m=2k+1 =0


2. On compact K¨hler manifolds, h1,0 = 1 b1 is a topological invariant.
a 2

3. On compact symplectic manifolds, “even Betti numbers are positive,” be-
cause ω k is closed but not exact (k = 0, 1, . . . , n).

ωn = d(± § ω n’k ) = 0 .
Proof. If ω k = d±, by Stokes™ theorem,
M M
This cannot happen since ω n is a volume form.
,
4. On compact K¨hler manifolds, the h
a are positive.
,
Claim. 0 = [ω ] ∈ HDolbeault (M ).

Proof.
ω ∈ „¦1,1 =’ ω ∈ „¦ ,
¯
dω = 0 =’ 0 = dω = ‚ω + ‚ω
( +1, ) ( , +1)
¯
=’ ‚ω = 0 ,
¯
,
so [ω ] de¬nes an element of HDolbeault. Why is ω not ‚-exact?
¯
If ω = ‚β for some β ∈ „¦ ’1, , then
¯ n,n
ω n = ω § ω n’ = ‚(β § ω n’ ) =’ 0 = [ω n ] ∈ HDolbeault (M ) .
n,n
2n
But [ω n ] = 0 in HdeRham (M ; C) HDolbeault (M ) since it is a volume form.
101
17.3 Compact Examples and Counterexamples


There are other constraints on the Hodge numbers of compact K¨hler man-
a
,m
ifolds, and ongoing research on how to compute HDolbeault . A popular picture
to describe the relations is the Hodge diamond:
hn,n
hn,n’1 hn’1,n
hn,n’2 hn’1,n’1 hn’2,n
.
.
.

h2,0 h1,1 h0,2
h1,0 h0,1
h0,0
Complex conjugation gives symmetry with respect to the middle vertical, whereas
the Hodge — induces symmetry about the center of the diamond. The middle
vertical axis is all non-zero. There are further symmetries induced by isomor-
phisms given by wedging with ω.
,
The Hodge conjecture relates HDolbeault (M ) © H 2 (M ; Z) for projective
manifolds M (i.e., submanifolds of complex projective space) to codim C =
complex submanifolds of M .

17.3 Compact Examples and Counterexamples

⇐= K¨hler
symplectic a
“ “
almost complex ⇐= complex


smooth even-dimensional orientable


almost complex

symplectic

complex
K¨hler
a




Is each of these regions nonempty? Can we even ¬nd representatives of each
region which are simply connected or have any speci¬ed fundamental group?
¨
102 17 COMPACT KAHLER MANIFOLDS


• Not all smooth even-dimensional manifolds are almost complex. For ex-
ample, S 4 , S 8 , S 10 , etc., are not almost complex.

• If M is both symplectic and complex, is it necessarily K¨hler?
a

No. For some time, it had been suspected that every compact symplectic
manifold might have an underlying K¨hler structure, or, at least, that a
a
compact symplectic manifold might have to satisfy the Hodge relations on
its Betti numbers. The following example ¬rst demonstrated otherwise.

The Kodaira-Thurston example (Thurston, 1976 [99]):

Take R4 with dx1 § dy1 + dx2 § dy2 , and “ the discrete group generated
by the following symplectomorphisms:

(x1 , x2 , y1 , y2 ) ’’ (x1 , x2 + 1, y1 , y2 )
γ1 :=
(x1 , x2 , y1 , y2 ) ’’ (x1 , x2 , y1 , y2 + 1)
γ2 :=
(x1 , x2 , y1 , y2 ) ’’ (x1 + 1, x2 , y1 , y2 )
γ3 :=
(x1 , x2 , y1 , y2 ) ’’ (x1 , x2 + y2 , y1 + 1, y2 )
γ4 :=

Then M = R4 /“ is a ¬‚at 2-torus bundle over a 2-torus. Kodaira [69]
had shown that M has a complex structure. However, π1 (M ) = “, hence
H 1 (R4 /“; Z) = “/[“, “] has rank 3, b1 = 3 is odd, so M is not K¨hler [99].
a

• Does any symplectic manifold admit some complex structure (not neces-
sarily compatible)?

No.

(Fern´ndez-Gotay-Gray, 1988 [37]): There are symplectic manifolds
a
which do not admit any complex structure [37]. Their examples are circle
bundles over circle bundles over a 2-torus.
S1 ’ M

S1 ’ P tower of circle ¬brations

T2

• Given a complex structure on M , is there always a symplectic structure
(not necessarily compatible)?

No.

The Hopf surface S 1 — S 3 is not symplectic because H 2 (S 1 — S 3 ) = 0.
But it is complex since S 1 — S 3 C2 \{0}/“ where “ = {2n Id | n ∈ Z} is a
group of complex transformations, i.e., we factor C2 \{0} by the equivalence
relation (z1 , z2 ) ∼ (2z1 , 2z2 ).
103
17.4 Main K¨hler Manifolds
a


• Is any almost complex manifold either complex or symplectic?

No.

CP2 #CP2 #CP2 is almost complex (proved by a computation with charac-
teristic classes), but is neither complex (since it does not ¬t Kodaira™s clas-
si¬cation of complex surfaces), nor symplectic (as shown by Taubes [95]
in 1995 using Seiberg-Witten invariants).
• In 1993 Gompf [46] provided a construction that yields a compact sym-
plectic 4-manifold with fundamental group equal to any given ¬nitely-
presented group. In particular, we can ¬nd simply connected examples.
His construction can be adapted to produce nonK¨hler examples.
a


17.4 Main K¨hler Manifolds
a

• Compact Riemann surfaces
As real manifolds, these are the 2-dimensional compact orientable mani-
folds classi¬ed by genus. An area form is a symplectic form. Any compat-
ible almost complex structure is always integrable for dimension reasons
(see Homework 10).
• Stein manifolds

De¬nition 17.4 A Stein manifold is a K¨hler manifold (M, ω) which
a

admits a global proper K¨hler potential, i.e., ω = 2 ‚ ‚ρ for some proper
a
function ρ : M ’ R.

Proper means that the preimage by ρ of a compact set is compact, i.e.,
“ρ(p) ’ ∞ as p ’ ∞.”
Stein manifolds can be also characterized as the properly embedded ana-
lytic submanifolds of Cn .
• Complex tori
Complex tori look like M = Cn /Zn where Zn is a lattice in Cn . The form
ω = dzj § d¯j induced by the euclidean structure is K¨hler.
z a
• Complex projective spaces
The standard K¨hler form on CPn is the Fubini-Study form (see Home-
a
work 12). (In 1995, Taubes showed that CP2 has a unique symplectic
structure up to symplectomorphism.)
• Products of K¨hler manifolds
a
• Complex submanifolds of K¨hler manifolds
a
Part VII
Hamiltonian Mechanics
The equations of motion in classical mechanics arise as solutions of variational
problems. For a general mechanical system of n particles in R3 , the physical path
satis¬es Newton™s second law. On the other hand, the physical path minimizes
the mean value of kinetic minus potential energy. This quantity is called the
action. For a system with constraints, the physical path is the path which
minimizes the action among all paths satisfying the constraint.
The Legendre transform (Lecture 20) gives the relation between the varia-
tional (Euler-Lagrange) and the symplectic (Hamilton-Jacobi) formulations of
the equations of motion.


18 Hamiltonian Vector Fields

18.1 Hamiltonian and Symplectic Vector Fields

“ What does a symplectic geometer do with a real function?...
Let (M, ω) be a symplectic manifold and let H : M ’ R be a smooth
function. Its di¬erential dH is a 1-form. By nondegeneracy, there is a unique
vector ¬eld XH on M such that ±XH ω = dH. Integrate XH . Supposing that
M is compact, or at least that XH is complete, let ρt : M ’ M , t ∈ R, be the
one-parameter family of di¬eomorphisms generated by XH :
±
 ρ0 = idM

 dρt
 —¦ ρ’1 = XH .
t
dt

Claim. Each di¬eomorphism ρt preserves ω, i.e., ρ— ω = ω, ∀t.
t

d—
= ρ— LXH ω = ρ— (d ±XH ω +±XH dω ) = 0.
Proof. We have dt ρt ω t t
0
dH

Therefore, every function on (M, ω) gives a family of symplectomorphisms.
Notice how the proof involved both the nondegeneracy and the closedness of ω.

De¬nition 18.1 A vector ¬eld XH as above is called the hamiltonian vector
¬eld with hamiltonian function H.

Example. The height function H(θ, h) = h on the sphere (M, ω) = (S 2 , dθ§dh)
has

±XH (dθ § dh) = dh ⇐’ XH = .
‚θ

105
106 18 HAMILTONIAN VECTOR FIELDS


Thus, ρt (θ, h) = (θ + t, h), which is rotation about the vertical axis; the height
™¦
function H is preserved by this motion.

Exercise. Let X be a vector ¬eld on an abstract manifold W . There is a unique
vector ¬eld X on the cotangent bundle T — W , whose ¬‚ow is the lift of the ¬‚ow of
X; cf. Lecture 2. Let ± be the tautological 1-form on T — W and let ω = ’d± be
the canonical symplectic form on T — W . Show that X is a hamiltonian vector
™¦
¬eld with hamiltonian function H := ±X ±.

Remark. If XH is hamiltonian, then

LXH H = ±XH dH = ±XH ±XH ω = 0 .

Therefore, hamiltonian vector ¬elds preserve their hamiltonian functions, and
each integral curve {ρt (x) | t ∈ R} of XH must be contained in a level set of H:

H(x) = (ρ— H)(x) = H(ρt (x)) , ∀t .
t

™¦

De¬nition 18.2 A vector ¬eld X on M preserving ω (i.e., such that LX ω = 0)
is called a symplectic vector ¬eld.

⇐’
X is symplectic ±X ω is closed .
⇐’
X is hamiltonian ±X ω is exact .
Locally, on every contractible open set, every symplectic vector ¬eld is hamil-
1
tonian. If HdeRham (M ) = 0, then globally every symplectic vector ¬eld is hamil-
1
tonian. In general, HdeRham (M ) measures the obstruction for symplectic vector
¬elds to be hamiltonian.

Example. On the 2-torus (M, ω) = (T2 , dθ1 § dθ2 ), the vector ¬elds X1 = ‚θ1

™¦
and X2 = ‚θ2 are symplectic but not hamiltonian.
To summarize, vector ¬elds on a symplectic manifold (M, ω) which preserve
ω are called symplectic. The following are equivalent:
• X is a symplectic vector ¬eld;
• the ¬‚ow ρt of X preserves ω, i.e., ρ— ω = ω, for all t;
t

• LX ω = 0;
• ±X ω is closed.
A hamiltonian vector ¬eld is a vector ¬eld X for which
• ±X ω is exact,
i.e., ±X ω = dH for some H ∈ C ∞ (M ). A primitive H of ±X ω is then called a
hamiltonian function of X.
107
18.2 Classical Mechanics


18.2 Classical Mechanics

Consider euclidean space R2n with coordinates (q1 , . . . , qn , p1 , . . . , pn ) and ω0 =
dqj § dpj . The curve ρt = (q(t), p(t)) is an integral curve for XH exactly if
±
 dqi (t) = ‚H

 dt
 ‚pi
(Hamilton equations)
 dp
 ‚H
 i
 (t) = ’
dt ‚qi
n
‚H ‚ ‚H ‚

Indeed, let XH = . Then,
‚pi ‚qi ‚qi ‚pi
i=1

n n
±XH (dqj § dpj ) = [(±XH dqj ) § dpj ’ dqj § (±XH dpj )]
± XH ω =
j=1 j=1
n
‚H ‚H
= ‚pj dpj + ‚qj dqj = dH
j=1



Remark. The gradient vector ¬eld of H relative to the euclidean metric is
n
‚H ‚ ‚H ‚
H := + .
‚qi ‚qi ‚pi ‚pi
i=1

‚ ‚
If J is the standard (almost) complex structure so that J( ‚qi ) = and
‚pi
‚ ‚
J( ‚pi ) = ’ ‚qi , we have JXH = H. ™¦

The case where n = 3 has a simple physical illustration. Newton™s second
law states that a particle of mass m moving in con¬guration space R3 with
coordinates q = (q1 , q2 , q3 ) under a potential V (q) moves along a curve q(t) such
that
d2 q
m 2 = ’ V (q) .
dt
Introduce the momenta pi = m dqi for i = 1, 2, 3, and energy function H(p, q) =
dt
1
|p|2 + V (q). Let R6 = T — R3 be the corresponding phase space, with co-
2m
ordinates (q1 , q2 , q3 , p1 , p2 , p3 ). Newton™s second law in R3 is equivalent to the
Hamilton equations in R6 :
± dq 1 ‚H
i
 = pi =

dt m ‚pi
2
 dpi d qi ‚V ‚H
 =m 2 =’ =’ .
dt dt ‚qi ‚qi

The energy H is conserved by the physical motion.
108 18 HAMILTONIAN VECTOR FIELDS


18.3 Brackets
Vector ¬elds are di¬erential operators on functions: if X is a vector ¬eld and
f ∈ C ∞ (X), df being the corresponding 1-form, then

X · f := df (X) = LX f .

Given two vector ¬elds X, Y , there is a unique vector ¬eld W such that

LW f = LX (LY f ) ’ LY (LX f ) .

W is called the Lie bracket of vector ¬elds X and Y and denoted W = [X, Y ],
since LW = [LX , LY ] is the commutator.
Exercise. Check that, for any form ±,

±[X,Y ] ± = LX ±Y ± ’ ±Y LX ± = [LX , ±Y ]± .

Since each side is an anti-derivation with respect to the wedge product, it su¬ces
to check this formula on local generators of the exterior algebra of forms, namely
™¦
functions and exact 1-forms.

Theorem 18.3 If X and Y are symplectic vector ¬elds on a symplectic mani-
fold (M, ω), then [X, Y ] is hamiltonian with hamiltonian function ω(Y, X).

Proof.
= L X ±Y ω ’ ± Y L X ω
±[X,Y ] ω
= d±X ±Y ω + ±X d±Y ω ’±Y d±X ω ’±Y ±X dω
0
0 0
= d(ω(Y, X)) .


A (real) Lie algebra is a (real) vector space g together with a Lie bracket
[·, ·], i.e., a bilinear map [·, ·] : g — g ’ g satisfying:
(a) [x, y] = ’[y, x] , ∀x, y ∈ g , (antisymmetry)
∀x, y, z ∈ g .
(b) [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 , (Jacobi identity)
Let
χ(M ) = { vector ¬elds on M }
sympl
(M ) = { symplectic vector ¬elds on M }
χ
ham
(M ) = { hamiltonian vector ¬elds on M } .
χ

Corollary 18.4 The inclusions (χham (M ), [·, ·]) ⊆ (χsympl (M ), [·, ·]) ⊆ (χ(M ), [·, ·])
are inclusions of Lie algebras.

De¬nition 18.5 The Poisson bracket of two functions f, g ∈ C ∞ (M ; R) is

{f, g} := ω(Xf , Xg ) .
109
18.4 Integrable Systems


We have X{f,g} = ’[Xf , Xg ] because Xω(Xf ,Xg ) = [Xg , Xf ].

Theorem 18.6 The bracket {·, ·} satis¬es the Jacobi identity, i.e.,

{f, {g, h}} + {g, {h, f }} + {h, {f, g}} = 0 .

Proof. Exercise.

De¬nition 18.7 A Poisson algebra (P, {·, ·}) is a commutative associative
algebra P with a Lie bracket {·, ·} satisfying the Leibniz rule:

{f, gh} = {f, g}h + g{f, h} .

Exercise. Check that the Poisson bracket {·, ·} de¬ned above satis¬es the
™¦
Leibniz rule.
We conclude that, if (M, ω) is a symplectic manifold, then (C ∞ (M ), {·, ·})
is a Poisson algebra. Furthermore, we have a Lie algebra anti-automorphism

C ∞ (M ) ’’ χ(M )
H ’’ XH
{·, ·} ’[·, ·] .

18.4 Integrable Systems

De¬nition 18.8 A hamiltonian system is a triple (M, ω, H), where (M, ω)
is a symplectic manifold and H ∈ C ∞ (M ; R) is a function, called the hamilto-
nian function.

Theorem 18.9 We have {f, H} = 0 if and only if f is constant along integral
curves of XH .

Proof. Let ρt be the ¬‚ow of XH . Then
d
—¦ ρt ) = ρ— LXH f = ρ— ±XH df = ρ— ±XH ±Xf ω
dt (f t t t
— —
= ρt ω(Xf , XH ) = ρt {f, H} = 0


A function f as in Theorem 18.9 is called an integral of motion (or a ¬rst
integral or a constant of motion). In general, hamiltonian systems do not
admit integrals of motion which are independent of the hamiltonian function.
Functions f1 , . . . , fn on M are said to be independent if their di¬erentials
(df1 )p , . . . , (dfn )p are linearly independent at all points p in some open dense
subset of M . Loosely speaking, a hamiltonian system is (completely) integrable
if it has as many commuting integrals of motion as possible. Commutativity is
with respect to the Poisson bracket. Notice that, if f1 , . . . , fn are commuting
110 18 HAMILTONIAN VECTOR FIELDS


integrals of motion for a hamiltonian system (M, ω, H), then, at each p ∈ M ,
their hamiltonian vector ¬elds generate an isotropic subspace of Tp M :
ω(Xfi , Xfj ) = {fi , fj } = 0 .
If f1 , . . . , fn are independent at p, then, by symplectic linear algebra, n can be
at most half the dimension of M .
De¬nition 18.10 A hamiltonian system (M, ω, H) is (completely) integrable
if it possesses n = 1 dim M independent integrals of motion, f1 = H, f2 , . . . , fn ,
2
which are pairwise in involution with respect to the Poisson bracket, i.e., {f i , fj } =
0, for all i, j.

Example. The simple pendulum (Homework 13) and the harmonic oscillator
are trivially integrable systems “ any 2-dimensional hamiltonian system (where
™¦
the set of non-¬xed points is dense) is integrable.

Example. A hamiltonian system (M, ω, H) where M is 4-dimensional is inte-
grable if there is an integral of motion independent of H (the commutativity
condition is automatically satis¬ed). Homework 18 shows that the spherical
™¦
pendulum is integrable.
For sophisticated examples of integrable systems, see [10, 61].
Let (M, ω, H) be an integrable system of dimension 2n with integrals of
motion f1 = H, f2 , . . . , fn . Let c ∈ Rn be a regular value of f := (f1 , . . . , fn ).
The corresponding level set, f ’1 (c), is a lagrangian submanifold, because it is
n-dimensional and its tangent bundle is isotropic.
Lemma 18.11 If the hamiltonian vector ¬elds Xf1 , . . . , Xfn are complete on
the level f ’1 (c), then the connected components of f ’1 (c) are homogeneous
spaces for Rn , i.e., are of the form Rn’k — Tk for some k, 0 ¤ k ¤ n, where Tk
is a k-dimensional torus.

Proof. Exercise (just follow the ¬‚ows).
Any compact component of f ’1 (c) must hence be a torus. These compo-
nents, when they exist, are called Liouville tori. (The easiest way to ensure
that compact components exist is to have one of the fi ™s proper.)
Theorem 18.12 (Arnold-Liouville [3]) Let (M, ω, H) be an integrable sys-
tem of dimension 2n with integrals of motion f1 = H, f2 , . . . , fn . Let c ∈ Rn
be a regular value of f := (f1 , . . . , fn ). The corresponding level f ’1 (c) is a
lagrangian submanifold of M .
(a) If the ¬‚ows of Xf1 , . . . , Xfn starting at a point p ∈ f ’1 (c) are complete,
then the connected component of f ’1 (c) containing p is a homogeneous
space for Rn . With respect to this a¬ne structure, that component has
coordinates •1 , . . . , •n , known as angle coordinates, in which the ¬‚ows
of the vector ¬elds Xf1 , . . . , Xfn are linear.
111
18.4 Integrable Systems


(b) There are coordinates ψ1 , . . . , ψn , known as action coordinates, comple-
mentary to the angle coordinates such that the ψi ™s are integrals of motion
and •1 , . . . , •n , ψ1 , . . . , ψn form a Darboux chart.

Therefore, the dynamics of an integrable system is extremely simple and the
system has an explicit solution in action-angle coordinates. The proof of part
(a) “ the easy part “ of the Arnold-Liouville theorem is sketched above. For the
proof of part (b), see [3, 28].
Geometrically, part (a) of the Arnold-Liouville theorem says that, in a neigh-
borhood of the value c, the map f : M ’ Rn collecting the given integrals of
motion is a lagrangian ¬bration, i.e., it is locally trivial and its ¬bers are
lagrangian submanifolds. The coordinates along the ¬bers are the angle coordi-
nates.11 Part (b) of the theorem guarantees the existence of coordinates on Rn ,
the action coordinates, which (Poisson) commute with the angle coordinates.
Notice that, in general, the action coordinates are not the given integrals of
motion because •1 , . . . , •n , f1 , . . . , fn do not form a Darboux chart.




11 The name “angle coordinates” is used even if the ¬bers are not tori.
Homework 13: Simple Pendulum

This problem is adapted from [52].

The simple pendulum is a mechanical system consisting of a massless rigid
rod of length l, ¬xed at one end, whereas the other end has a plumb bob of mass
m, which may oscillate in the vertical plane. Assume that the force of gravity is
constant pointing vertically downwards, and that this is the only external force
acting on this one-particle system.

(a) Let θ be the oriented angle between the rod (regarded as a point mass)
and the vertical direction. Let ξ be the coordinate along the ¬bers of T — S 1
induced by the standard angle coordinate on S 1 . Show that the function
H : T — S 1 ’ R given by

ξ2
+ ml(1 ’ cos θ) ,
H(θ, ξ) =
2ml2
V
K

is an appropriate hamiltonian function to describe the spherical pendu-
lum. More precisely, check that gravity corresponds to a potential energy
V (θ) = ml(1 ’ cos θ) (we omit universal constants), and that the kinetic
1
energy is given by K(θ, ξ) = 2ml2 ξ 2 .
(b) For simplicity assume that m = l = 1.
Plot the level curves of H in the (θ, ξ) plane.
Show that there exists a number c such that for 0 < h < c the level curve
H = h is a disjoint union of closed curves. Show that the projection of
each of these curves onto the θ-axis is an interval of length less than π.
Show that neither of these assertions is true if h > c.
What types of motion are described by these two types of curves?
What about the case H = c?
(c) Compute the critical points of the function H. Show that, modulo 2π in θ,
there are exactly two critical points: a critical point s where H vanishes,
and a critical point u where H equals c. These points are called the
stable and unstable points of H, respectively. Justify this terminology,
i.e., show that a trajectory of the hamiltonian vector ¬eld of H whose
initial point is close to s stays close to s forever, and show that this is not
the case for u. What is happening physically?




112
19 Variational Principles
19.1 Equations of Motion
The equations of motion in classical mechanics arise as solutions of variational
problems:
A general mechanical system possesses both kinetic and potential
energy. The quantity that is minimized is the mean value of kinetic
minus potential energy.

Example. Suppose that a point-particle of mass m moves in R3 under a force
¬eld F ; let x(t), a ¤ t ¤ b, be its path of motion in R3 . Newton™s second law
states that
d2 x
m 2 (t) = F (x(t)) .
dt
De¬ne the work of a path γ : [a, b] ’’ R3 , with γ(a) = p and γ(b) = q, to be
b

F (γ(t)) ·
Wγ = (t)dt .
dt
a

Suppose that F is conservative, i.e., Wγ depends only on p and q. Then we
can de¬ne the potential energy V : R3 ’’ R of the system as
V (q) := Wγ
where γ is a path joining a ¬xed base point p0 ∈ R3 (the “origin”) to q. Newton™s
second law can now be written
d2 x ‚V
m 2 (t) = ’ (x(t)) .
dt ‚x
In the previous lecture we saw that
⇐’
Newton™s second law Hamilton equations
in R3 = {(q1 , q2 , q3 )} in T — R3 = {(q1 , q2 , q3 , p1 , p2 , p3 )}

where pi = m dqi and the hamiltonian is H(p, q) = 2m |p|2 +V (q). Hence, solving
1
dt
Newton™s second law in con¬guration space R3 is equivalent to solving in
phase space for the integral curve T — R3 of the hamiltonian vector ¬eld with
™¦
hamiltonian function H.

Example. The motion of earth about the sun, both regarded as point-masses
and assuming that the sun to be stationary at the origin, obeys the inverse
square law
d2 x ‚V
m 2 =’ ,
dt ‚x
where x(t) is the position of earth at time t, and V (x) = const. is the gravi-
|x|
™¦
tational potential.



113
114 19 VARIATIONAL PRINCIPLES


19.2 Principle of Least Action

When we need to deal with systems with constraints, such as the simple pendu-
lum, or two point masses attached by a rigid rod, or a rigid body, the language
of variational principles becomes more appropriate than the explicit analogues
of Newton™s second laws. Variational principles are due mostly to D™Alembert,
Maupertius, Euler and Lagrange.
Example. (The n-particle system.) Suppose that we have n point-particles
of masses m1 , . . . , mn moving in 3-space. At any time t, the con¬guration of
this system is described by a vector in con¬guration space R3n

x = (x1 , . . . , xn ) ∈ R3n

with xi ∈ R3 describing the position of the ith particle. If V ∈ C ∞ (R3n ) is the
potential energy, then a path of motion x(t), a ¤ t ¤ b, satis¬es

d 2 xi ‚V
mi 2 (t) = ’ (x1 (t), . . . , xn (t)) .
dt ‚xi

Consider this path in con¬guration space as a map γ0 : [a, b] ’ R3n with
γ0 (a) = p and γ0 (b) = q, and let

P = {γ : [a, b] ’’ R3n | γ(a) = p and γ(b) = q}

be the set of all paths going from p to q over time t ∈ [a, b]. ™¦

De¬nition 19.1 The action of a path γ ∈ P is
2
b
mi dγi
Aγ := ’ V (γ(t)) dt .
(t)
2 dt
a



Principle of least action.
The physical path γ0 is the path for which Aγ is minimal.
Newton™s second law for a constrained system.
Suppose that the n point-masses are restricted to move on a submanifold M
of R3n called the constraint set. We can now single out the actual physical
path γ0 : [a, b] ’ M , with γ0 (a) = p and γ0 (b) = q, as being “the” path which
minimizes Aγ among all those hypothetical paths γ : [a, b] ’ R3n with γ(a) = p,
γ(b) = q and satisfying the rigid constraints γ(t) ∈ M for all t.

19.3 Variational Problems

Let M be an n-dimensional manifold. Its tangent bundle T M is a 2n-dimensional
manifold. Let F : T M ’ R be a smooth function.
115
19.3 Variational Problems


If γ : [a, b] ’ M is a smooth curve on M , de¬ne the lift of γ to T M to be
the smooth curve on T M given by

γ : [a, b] ’’ T M
˜
γ(t), dγ (t)
t ’’ .
dt

The action of γ is
b b


Aγ := (˜ F )(t)dt =
γ F γ(t), (t) dt .
dt
a a

For ¬xed p, q ∈ M , let

P(a, b, p, q) := {γ : [a, b] ’’ M | γ(a) = p, γ(b) = q} .

Problem.
Find, among all γ ∈ P(a, b, p, q), the curve γ0 which “minimizes” Aγ .
First observe that minimizing curves are always locally minimizing:

Lemma 19.2 Suppose that γ0 : [a, b] ’ M is minimizing. Let [a1 , b1 ] be a
subinterval of [a, b] and let p1 = γ0 (a1 ), q1 = γ0 (b1 ). Then γ0 |[a1 ,b1 ] is minimiz-
ing among the curves in P(a1 , b1 , p1 , q1 ).

Proof. Exercise:
Argue by contradiction. Suppose that there were γ1 ∈ P(a1 , b1 , p1 , q1 ) for
which Aγ1 < Aγ0 |[a1 ,b1 ] . Consider a broken path obtained from γ0 by replacing
the segment γ0 |[a1 ,b1 ] by γ1 . Construct a smooth curve γ2 ∈ P(a, b, p, q) for
which Aγ2 < Aγ0 by rounding o¬ the corners of the broken path.
We will now assume that p, q and γ0 lie in a coordinate neighborhood
(U, x1 , . . . , xn ). On T U we have coordinates (x1 , . . . , xn , v1 , . . . , vn ) associated
‚ ‚
with a trivialization of T U by ‚x1 , . . . , ‚xn . Using this trivialization, the curve

γ : [a, b] ’’ U , γ(t) = (γ1 (t), . . . , γn (t))

lifts to
dγ1 dγn
γ : [a, b] ’’ T U ,
˜ γ (t) =
˜ γ1 (t), . . . , γn (t), (t), . . . , (t) .
dt dt

Necessary condition for γ0 ∈ P(a, b, p, q) to minimize the action.
Let c1 , . . . , cn ∈ C ∞ ([a, b]) be such that ci (a) = ci (b) = 0. Let γµ : [a, b] ’’
U be the curve

γµ (t) = (γ1 (t) + µc1 (t), . . . , γn (t) + µcn (t)) .

For µ small, γµ is well-de¬ned and in P(a, b, p, q).
116 19 VARIATIONAL PRINCIPLES


b dγµ
Let Aµ = Aγµ = dt. If γ0 minimizes A, then
F γµ (t), dt (t)
a

dAµ
(0) = 0 .

b
dAµ ‚F dγ0 ‚F dγ0 dci
(0) = γ0 (t), (t) ci (t) + γ0 (t), (t) (t) dt
dµ ‚xi dt ‚vi dt dt
a i
b
‚F d ‚F
(. . .) ’
= (. . .) ci (t)dt = 0
‚xi dt ‚vi
a i

where the ¬rst equality follows from the Leibniz rule and the second equality
follows from integration by parts. Since this is true for all ci ™s satisfying the
boundary conditions ci (a) = ci (b) = 0, we conclude that

‚F dγ0 d ‚F dγ0
γ0 (t), (t) = γ0 (t), (t) E-L
‚xi dt dt ‚vi dt
These are the Euler-Lagrange equations.

19.4 Solving the Euler-Lagrange Equations

Case 1: Suppose that F (x, v) does not depend on v.

The Euler-Lagrange equations become
‚F dγ0
= 0 ⇐’ the curve γ0 sits on the critical set of F .
γ0 (t), (t)
‚xi dt

For generic F , the critical points are isolated, hence γ0 (t) must be a con-
stant curve.
Case 2: Suppose that F (x, v) depends a¬nely on v:
n
F (x, v) = F0 (x) + Fj (x)vj .
j=1

n
‚F0 ‚Fj dγj
LHS of E-L : (γ(t)) + (γ(t)) (t)
‚xi ‚xi dt
j=1
n
d ‚Fi dγj
RHS of E-L : Fi (γ(t)) = (γ(t)) (t)
dt ‚xj dt
j=1

The Euler-Lagrange equations become
n
‚F0 ‚Fi ‚Fj dγj

(γ(t)) = (γ(t)) (t) .
‚xi ‚xj ‚xi dt
j=1
n—n matrix
117
19.5 Minimizing Properties


‚Fj
‚Fi
If the n — n matrix ’ has an inverse Gij (x), then
‚xj ‚xi

n
dγj ‚F0
(t) = Gji (γ(t)) (γ(t))
dt ‚xi
i=1

is a system of ¬rst order ordinary di¬erential equations. Locally it has a
unique solution through each point p. If q is not on this curve, there is no
solution at all to the Euler-Lagrange equations belonging to P(a, b, p, q).

Therefore, we need non-linear dependence of F on the v variables in order to
have appropriate solutions. From now on, assume that the
‚ 2F
Legendre condition: det =0.
‚vi ‚vj
’1
‚2F
Letting Gij (x, v) = ‚vi ‚vj (x, v) , the Euler-Lagrange equations become

d2 γj ‚2F
‚F dγ dγ dγk

= Gji γ, Gji γ, .
dt2 ‚xi dt ‚vi ‚xk dt dt
i i,k

This second order ordinary di¬erential equation has a unique solution given
initial conditions

γ(a) = p and (a) = v .
dt

19.5 Minimizing Properties

Is the above solution locally minimizing?
‚2F
0, ∀(x, v), i.e., with the x variable frozen, the
Assume that ‚vi ‚vj (x, v)
function v ’ F (x, v) is strictly convex.
Suppose that γ0 ∈ P(a, b, p, q) satis¬es E-L. Does γ0 minimize Aγ ? Locally,
yes, according to the following theorem. (Globally it is only critical.)

Theorem 19.3 For every su¬ciently small subinterval [a1 , b1 ] of [a, b], γ0 |[a1 ,b1 ]
is locally minimizing in P(a1 , b1 , p1 , q1 ) where p1 = γ0 (a1 ), q1 = γ0 (b1 ).

Proof. As an exercise in Fourier series, show the Wirtinger inequality: for
f ∈ C 1 ([a, b]) with f (a) = f (b) = 0, we have
2
b b
π2
df
|f |2 dt .
dt ≥
(b ’ a)2
dt
a a

Suppose that γ0 : [a, b] ’ U satis¬es E-L. Take ci ∈ C ∞ ([a, b]), ci (a) =
ci (b) = 0. Let c = (c1 , . . . , cn ). Let γµ = γ0 + µc ∈ P(a, b, p, q), and let
A µ = A γµ .
118 19 VARIATIONAL PRINCIPLES


dAµ
E-L ⇐’ dµ (0) = 0.
b
d2 Aµ ‚2F dγ0
(0) = γ0 , ci cj dt (I)
dµ2 ‚xi ‚xj dt
a i,j
b
‚2F dγ0 dcj
+2 γ0 , ci dt (II)
‚xi ‚vj dt dt
a i,j
b
‚2F dγ0 dci dcj
+ γ0 , dt (III) .
‚vi ‚vj dt dt dt
a i,j

‚2F
Since ‚vi ‚vj (x, v) 0 at all x, v,

2
dc
≥ KIII
III
dt L2 [a,b]


¤ KI |c|2 2 [a,b]
|I| L


dc
|II| ¤ KII |c|L2 [a,b]
dt L2 [a,b]

where KI , KII , KIII > 0. By the Wirtinger inequality, if b ’ a is very small, then
III > |I|, |II|. Hence, γ0 is a local minimum.
Homework 14: Minimizing Geodesics

This set of problems is adapted from [52].


Let (M, g) be a riemannian manifold. From the riemannian metric, we get
a function F : T M ’ R, whose restriction to each tangent space Tp M is the
quadratic form de¬ned by the metric.
Let p and q be points on M , and let γ : [a, b] ’ M be a smooth curve joining
p to q. Let γ : [a, b] ’ T M , γ (t) = (γ(t), dγ (t)) be the lift of γ to T M . The
˜ ˜ dt
action of γ is
2
b b


A(γ) = (˜ F ) dt =
γ dt .
dt
a a




1. Let γ : [a, b] ’ M be a smooth curve joining p to q. Show that the arc-
length of γ is independent of the parametrization of γ, i.e., show that if we
reparametrize γ by „ : [a , b ] ’ [a, b], the new curve γ = γ —¦ „ : [a , b ] ’
M has the same arc-length.


2. Show that, given any curve γ : [a, b] ’ M (with dγ never vanishing), there
dt
is a reparametrization „ : [a, b] ’ [a, b] such that γ —¦ „ : [a, b] ’ M is of
constant velocity, that is, | dγ | is independent of t.
dt



3. Let „ : [a, b] ’ [a, b] be a smooth monotone map taking the endpoints of
[a, b] to the endpoints of [a, b]. Prove that
2
b
d„
dt ≥ b ’ a ,
dt
a

d„
with equality holding if and only if = 1.
dt



4. Let γ : [a, b] ’ M be a smooth curve joining p to q. Suppose that, as s
goes from a to b, its image γ(s) moves at constant velocity, i.e., suppose
that | dγ | is constant as a function of s. Let γ = γ —¦ „ : [a, b] ’ M be a
ds
reparametrization of γ. Show that A(γ ) ≥ A(γ), with equality holding if
and only if „ (t) ≡ t.




119
120 HOMEWORK 14




5. Let γ0 : [a, b] ’ M be a curve joining p to q. Suppose that γ0 is action-
minimizing, i.e., suppose that

A(γ0 ) ¤ A(γ)

for any other curve γ : [a, b] ’ M joining p to q. Prove that γ0 is also arc-
length-minimizing, i.e., show that γ0 is the shortest geodesic joining p
to q.


6. Show that, among all curves joining p to q, γ0 minimizes the action if and
only if γ0 is of constant velocity and γ0 minimizes arc-length.


7. On a coordinate chart (U, x1 , . . . , xn ) on M , we have

gij (x)v i v j .
L(x, v) =

Show that the Euler-Lagrange equations associated to the action reduce
to the Christo¬el equations for a geodesic

d2 γ k dγ i dγ j
(“k —¦ γ)
+ =0,
ij
dt2 dt dt
where the “k ™s (called the Christo¬el symbols) are de¬ned in terms of
ij
the coe¬cients of the riemannian metric by

1 ‚g i ‚g j ‚gij
“k = k

g + ,
ij
2 ‚xj ‚xi ‚x

(g ij ) being the matrix inverse to (gij ).


8. Let p and q be two non-antipodal points on S n . Show that the geodesic
joining p to q is an arc of a great circle, the great circle in question being
the intersection of S n with the two-dimensional subspace of Rn+1 spanned
by p and q.
Hint: No calculations are needed: Show that an isometry of a riemannian
manifold has to carry geodesics into geodesics, and show that there is an isom-
etry of Rn+1 whose ¬xed point set is the plane spanned by p and q, and show
that this isometry induces on S n an isometry whose ¬xed point set is the great
circle containing p and q.
20 Legendre Transform

20.1 Strict Convexity

Let V be an n-dimensional vector space, with e1 , . . . , en a basis of V and
v1 , . . . , vn the associated coordinates. Let F : V ’ R, F = F (v1 , . . . , vn ),
be a smooth function. Let p ∈ V , u = n ui ei ∈ V . The hessian of F is the
i=1
quadratic function on V de¬ned by

‚2F
2
(d F )p (u) := (p)ui uj .
‚vi ‚vj
i,j


d2
Exercise. Show that (d2 F )p (u) = ™¦
dt2 F (p + tu)|t=0 .

De¬nition 20.1 The function F is strictly convex if (d2 F )p 0, ∀p ∈ V .

Theorem 20.2 For a strictly convex function F on V , the following are equiv-
alent:

(a) F has a critical point, i.e., a point where dFp = 0;
(b) F has a local minimum at some point;
(c) F has a unique critical point (global minimum); and
(d) F is proper, that is, F (p) ’ +∞ as p ’ ∞ in V .

Proof. Homework 15.

De¬nition 20.3 A strictly convex function F is stable when it satis¬es con-
ditions (a)-(d) in Theorem 20.2.

Example. The function ex + ax is strictly convex for any a ∈ R, but it is stable
only for a < 0. The function x2 + ax is strictly convex and stable for any a ∈ R.
™¦


20.2 Legendre Transform

∈ V — , let
Let F be any strictly convex function on V . Given

F : V ’’ R , F (v) = F (v) ’ (v) .

Since (d2 F )p = (d2 F )p ,

⇐’
F is strictly convex F is strictly convex.


121
122 20 LEGENDRE TRANSFORM


De¬nition 20.4 The stability set of a strictly convex function F is

SF = { ∈ V — | F is stable} .

Theorem 20.5 The set SF is an open and convex subset of V — .

Proof. Homework 15.
Homework 15 also describes a su¬cient condition for SF = V — .

De¬nition 20.6 The Legendre transform associated to F ∈ C ∞ (V ; R) is
the map
LF : V ’’ V —

V— .
p ’’ dFp ∈ Tp V

Theorem 20.7 Suppose that F is strictly convex. Then

LF : V ’’ SF ,

i.e., LF is a di¬eomorphism onto SF .

The inverse map L’1 : SF ’’ V is described as follows: for l ∈ SF , the
F
value L’1 ( ) is the unique minimum point p ∈ V of F = F ’ .
F

Exercise. Check that p is the minimum of F (v) ’ dFp (v). ™¦

De¬nition 20.8 The dual function F — to F is

F — : SF ’’ R , F — ( ) = ’ min F (p) .
p∈V


L’1 = LF — .
Theorem 20.9 We have that F


Proof. Homework 15.


20.3 Application to Variational Problems

Let M be a manifold and F : T M ’ R a function on T M .
γ—F .
Problem. Minimize Aγ = ˜
At p ∈ M , let
Fp := F |Tp M : Tp M ’’ R .
Assume that Fp is strictly convex for all p ∈ M . To simplify notation, assume

also that SFp = Tp M . The Legendre transform on each tangent space


LFp : Tp M ’’ Tp M
123
20.3 Application to Variational Problems


is essentially given by the ¬rst derivatives of F in the v directions. The dual
— —
function to Fp is Fp : Tp M ’’ R. Collect these ¬berwise maps into

’’ T — M ,
L: L|Tp M
TM = L Fp , and

H : T —M —
’’ R , H|Tp M = Fp .





Exercise. The maps H and L are smooth, and L is a di¬eomorphism. ™¦
Let
γ : [a, b] ’’ M be a curve, and
γ : [a, b] ’’ T M
˜ its lift.

Theorem 20.10 The curve γ satis¬es the E-L equations on every coordinate
chart if and only if L —¦ γ : [a, b] ’ T — M is an integral curve of the hamiltonian
˜
vector ¬eld XH .

Proof. Let
(U, x1 , . . . , xn ) coordinate neighborhood in M ,
(T U, x1 , . . . , xn , v1 , . . . , vn ) coordinates in T M ,
(T — U, x1 , . . . , xn , ξ1 , . . . , ξn ) coordinates in T — M .

On T U we have F = F (x, v).
On T — U we have H = H(u, ξ).

’’ T — U
L: TU
‚F
’’ (x, ξ)
(x, v) where ξ = LFx (v) = (x, v) .
‚v

(This is the de¬nition of momentum ξ.)

H(x, ξ) = Fx (ξ) = ξ · v ’ F (x, v) L(x, v) = (x, ξ) .
where

Integral curves (x(t), ξ(t)) of XH satisfy the Hamilton equations:
±
 dx = ‚H
 (x, ξ)

 dt ‚ξ
H
 dξ
 ‚H

 =’ (x, ξ) ,
dt ‚x
whereas the physical path x(t) satis¬es the Euler-Lagrange equations:

‚F dx d ‚F dx
E-L x, = x, .
‚x dt dt ‚v dt
dx
Let (x(t), ξ(t)) = L x(t), dt (t) . We want to prove:

dx
t ’ (x(t), ξ(t)) satis¬es H ⇐’ t’ x(t), (t) satis¬es E-L .
dt
124 20 LEGENDRE TRANSFORM


The ¬rst line of H is automatically satis¬ed:

dx ‚H dx
(x, ξ) = LFx (ξ) = L’1 (ξ) ⇐’
= ξ = L Fx

Fx
dt ‚ξ dt

‚F
= ’ ‚H (x, ξ).
Claim. If (x, ξ) = L(x, v), then ‚x (x, v) ‚x

This follows from di¬erentiating both sides of H(x, ξ) = ξ · v ’ F (x, v) with
respect to x, where ξ = LFx (v) = ξ(x, v).

‚H ‚H ‚ξ ‚ξ ‚F
·v’
+ = .
‚x ‚ξ ‚x ‚x ‚x
v

Now the second line of H becomes
d ‚F dξ ‚H ‚F
=’ ⇐’
(x, v) = (x, ξ) = (x, v) E-L .
dt ‚v dt ‚x ‚x
by the claim
since ξ = LFx (v)
Homework 15: Legendre Transform
This set of problems is adapted from [53].
1. Let f : R ’ R be a smooth function. f is called strictly convex if
f (x) > 0 for all x ∈ R. Assuming that f is strictly convex, prove that
the following four conditions are equivalent:
(a) f (x) = 0 for some point x0 ,
(b) f has a local minimum at some point x0 ,
(c) f has a unique (global) minimum at some point x0 ,
f (x) ’ +∞ as x ’ ±∞.
(d)
The function f is stable if it satis¬es one (and hence all) of these condi-
tions.
For what values of a is the function ex + ax stable? For those values of a
for which it is not stable, what does the graph look like?
2. Let V be an n-dimensional vector space and F : V ’ R a smooth function.
The function F is said to be strictly convex if for every pair of elements
p, v ∈ V , v = 0, the restriction of F to the line {p + xv | x ∈ R} is strictly
convex.
The hessian of F at p is the quadratic form
d2
d 2 Fp : v ’ F (p + xv)|x=0 .
dx2
Show that F is strictly convex if and only if d2 Fp is positive de¬nite for
all p ∈ V .
Prove the n-dimensional analogue of the result you proved in (1). Namely,
assuming that F is strictly convex, show that the four following assertions
are equivalent:
(a) dFp = 0 at some point p0 ,
(b) F has a local minimum at some point p0 ,
(c) F has a unique (global) minimum at some point p0 ,
F (p) ’ +∞ as p ’ ∞.
(d)
3. As in exercise 2, let V be an n-dimensional vector space and F : V ’
R a smooth function. Since V is a vector space, there is a canonical

V — , for every p ∈ V . Therefore, we can de¬ne a map
identi¬cation Tp V
LF : V ’ V — (Legendre transform)
by setting

V— .
LF (p) = dFp ∈ Tp V
Show that, if F is strictly convex, then, for every point p ∈ V , LF maps a
neighborhood of p di¬eomorphically onto a neighborhood of LF (p).


125
126 HOMEWORK 15


4. A strictly convex function F : V ’ R is stable if it satis¬es the four
equivalent conditions of exercise 2. Given any strictly convex function F ,
we will denote by SF the set of l ∈ V — for which the function Fl : V ’ R,
p ’ F (p) ’ l(p), is stable. Prove that:
(a) The set SF is open and convex.
(b) LF maps V di¬eomorphically onto SF .
(c) If l ∈ SF and p0 = L’1 (l), then p0 is the unique minimum point of
F
the function Fl .
Let F — : SF ’ R be the function whose value at l is the quantity
’min Fl (p). Show that F — is a smooth function.
p∈V
The function F — is called the dual of the function F .
5. Let F be a strictly convex function. F is said to have quadratic growth
at in¬nity if there exists a positive-de¬nite quadratic form Q on V and
a constant K such that F (p) ≥ Q(p) ’ K, for all p. Show that, if F
has quadratic growth at in¬nity, then SF = V — and hence LF maps V
di¬eomorphically onto V — .
6. Let F : V ’ R be strictly convex and let F — : SF ’ R be the dual
function. Prove that for all p ∈ V and all l ∈ SF ,
F (p) + F — (l) ≥ l(p) (Young inequality) .

7. On one hand we have V — V — T — V , and on the other hand, since
V = V —— , we have V — V — V — — V T —V — .
Let ±1 be the canonical 1-form on T — V and ±2 be the canonical 1-form on
T — V — . Via the identi¬cations above, we can think of both of these forms

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