<< стр. 5(всего 9)СОДЕРЖАНИЕ >>
and the spaces H are п¬Ѓnite-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 =в‡’ ,m m,
h =h
Some immediate topological consequences are:
1. On compact KВЁhler manifolds вЂњthe odd Betti numbers are even:вЂќ
a
k
2k+1 ,(2k+1в€’ )
,m
b = h =2 h is even .
+m=2k+1 =0

2. On compact KВЁhler manifolds, h1,0 = 1 b1 is a topological invariant.
a 2

3. On compact symplectic manifolds, вЂњeven Betti numbers are positive,вЂќ be-
cause П‰ k is closed but not exact (k = 0, 1, . . . , n).

П‰n = d(О± в€§ П‰ nв€’k ) = 0 .
Proof. If П‰ k = dО±, by StokesвЂ™ theorem,
M M
This cannot happen since П‰ n is a volume form.
,
4. On compact KВЁhler manifolds, the h
a are positive.
,
Claim. 0 = [П‰ ] в€€ HDolbeault (M ).

Proof.
П‰ в€€ в„¦1,1 =в‡’ П‰ в€€ в„¦ ,
ВЇ
dП‰ = 0 =в‡’ 0 = dП‰ = в€‚П‰ + в€‚П‰
( +1, ) ( , +1)
ВЇ
=в‡’ в€‚П‰ = 0 ,
ВЇ
,
so [П‰ ] deп¬Ѓnes an element of HDolbeault. Why is П‰ not в€‚-exact?
ВЇ
If П‰ = в€‚ОІ for some ОІ в€€ в„¦ в€’1, , then
ВЇ n,n
П‰ n = П‰ в€§ П‰ nв€’ = в€‚(ОІ в€§ П‰ nв€’ ) =в‡’ 0 = [П‰ n ] в€€ HDolbeault (M ) .
n,n
2n
But [П‰ 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 KВЁhler 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,nв€’1 hnв€’1,n
hn,nв€’2 hnв€’1,nв€’1 hnв€’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 в€— 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 П‰.
,
The Hodge conjecture relates HDolbeault (M ) в€© 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

в‡ђ= KВЁhler
symplectic a
в‡“ в‡“
almost complex в‡ђ= complex

smooth even-dimensional orientable

almost complex

symplectic

complex
KВЁhler
a

Is each of these regions nonempty? Can we even п¬Ѓnd representatives of each
region which are simply connected or have any speciп¬Ѓed fundamental group?
ВЁ
102 17 COMPACT KAHLER MANIFOLDS

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

вЂў If M is both symplectic and complex, is it necessarily KВЁhler?
a

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

The Kodaira-Thurston example (Thurston, 1976 ):

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

(x1 , x2 , y1 , y2 ) в€’в†’ (x1 , x2 + 1, y1 , y2 )
Оі1 :=
(x1 , x2 , y1 , y2 ) в€’в†’ (x1 , x2 , y1 , y2 + 1)
Оі2 :=
(x1 , x2 , y1 , y2 ) в€’в†’ (x1 + 1, x2 , y1 , y2 )
Оі3 :=
(x1 , x2 , y1 , y2 ) в€’в†’ (x1 , x2 + y2 , y1 + 1, y2 )
Оі4 :=

Then M = R4 /О“ is a п¬‚at 2-torus bundle over a 2-torus. Kodaira 
had shown that M has a complex structure. However, ПЂ1 (M ) = О“, hence
H 1 (R4 /О“; Z) = О“/[О“, О“] has rank 3, b1 = 3 is odd, so M is not KВЁhler .
a

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

No.

(FernВґndez-Gotay-Gray, 1988 ): There are symplectic manifolds
a
which do not admit any complex structure . Their examples are circle
bundles over circle bundles over a 2-torus.
S1 в†’ M
в†“
S1 в†’ P tower of circle п¬Ѓbrations
в†“
T2

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

No.

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

вЂў 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 п¬Ѓt KodairaвЂ™s clas-
siп¬Ѓcation of complex surfaces), nor symplectic (as shown by Taubes 
in 1995 using Seiberg-Witten invariants).
вЂў In 1993 Gompf  provided a construction that yields a compact sym-
plectic 4-manifold with fundamental group equal to any given п¬Ѓnitely-
presented group. In particular, we can п¬Ѓnd simply connected examples.
His construction can be adapted to produce nonKВЁhler examples.
a

17.4 Main KВЁhler Manifolds
a

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

Deп¬Ѓnition 17.4 A Stein manifold is a KВЁhler manifold (M, П‰) which
a
iВЇ
admits a global proper KВЁhler potential, i.e., П‰ = 2 в€‚ в€‚ПЃ for some proper
a
function ПЃ : M в†’ R.

Proper means that the preimage by ПЃ of a compact set is compact, i.e.,
вЂњПЃ(p) в†’ в€ћ as p в†’ в€ћ.вЂќ
Stein manifolds can be also characterized as the properly embedded ana-
lytic submanifolds of Cn .
вЂў Complex tori
Complex tori look like M = Cn /Zn where Zn is a lattice in Cn . The form
П‰ = dzj в€§ dВЇj induced by the euclidean structure is KВЁhler.
z a
вЂў Complex projective spaces
The standard KВЁhler 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.)
вЂў Products of KВЁhler manifolds
a
вЂў Complex submanifolds of KВЁhler 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
satisп¬Ѓes NewtonвЂ™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

вЂ“ What does a symplectic geometer do with a real function?...
Let (M, П‰) be a symplectic manifold and let H : M в†’ R be a smooth
function. Its diп¬Ђerential dH is a 1-form. By nondegeneracy, there is a unique
vector п¬Ѓeld XH on M such that Д±XH П‰ = dH. Integrate XH . Supposing that
M is compact, or at least that XH is complete, let ПЃt : M в†’ M , t в€€ R, be the
one-parameter family of diп¬Ђeomorphisms generated by XH :
пЈ±
пЈґ ПЃ0 = idM
пЈІ
пЈґ dПЃt
пЈі в—¦ ПЃв€’1 = XH .
t
dt

Claim. Each diп¬Ђeomorphism ПЃt preserves П‰, i.e., ПЃв€— П‰ = П‰, в€Ђt.
t

dв€—
= ПЃв€— LXH П‰ = ПЃв€— (d Д±XH П‰ +Д±XH dП‰ ) = 0.
Proof. We have dt ПЃt П‰ t t
0
dH

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

Deп¬Ѓnition 18.1 A vector п¬Ѓeld XH as above is called the hamiltonian vector
п¬Ѓeld with hamiltonian function H.

Example. The height function H(Оё, h) = h on the sphere (M, П‰) = (S 2 , dОёв€§dh)
has
в€‚
Д±XH (dОё в€§ dh) = dh в‡ђв‡’ XH = .
в€‚Оё

105
106 18 HAMILTONIAN VECTOR FIELDS

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

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

Remark. If XH is hamiltonian, then

LXH H = Д±XH dH = Д±XH Д±XH П‰ = 0 .

Therefore, hamiltonian vector п¬Ѓelds preserve their hamiltonian functions, and
each integral curve {ПЃt (x) | t в€€ R} of XH must be contained in a level set of H:

H(x) = (ПЃв€— H)(x) = H(ПЃt (x)) , в€Ђt .
t

в™¦

Deп¬Ѓnition 18.2 A vector п¬Ѓeld X on M preserving П‰ (i.e., such that LX П‰ = 0)
is called a symplectic vector п¬Ѓeld.

в‡ђв‡’
X is symplectic Д±X П‰ is closed .
в‡ђв‡’
X is hamiltonian Д±X П‰ is exact .
Locally, on every contractible open set, every symplectic vector п¬Ѓeld is hamil-
1
tonian. If HdeRham (M ) = 0, then globally every symplectic vector п¬Ѓeld is hamil-
1
tonian. In general, HdeRham (M ) measures the obstruction for symplectic vector
п¬Ѓelds to be hamiltonian.
в€‚
Example. On the 2-torus (M, П‰) = (T2 , dОё1 в€§ dОё2 ), the vector п¬Ѓelds X1 = в€‚Оё1
в€‚
в™¦
and X2 = в€‚Оё2 are symplectic but not hamiltonian.
To summarize, vector п¬Ѓelds on a symplectic manifold (M, П‰) which preserve
П‰ are called symplectic. The following are equivalent:
вЂў X is a symplectic vector п¬Ѓeld;
вЂў the п¬‚ow ПЃt of X preserves П‰, i.e., ПЃв€— П‰ = П‰, for all t;
t

вЂў LX П‰ = 0;
вЂў Д±X П‰ is closed.
A hamiltonian vector п¬Ѓeld is a vector п¬Ѓeld X for which
вЂў Д±X П‰ is exact,
i.e., Д±X П‰ = dH for some H в€€ C в€ћ (M ). A primitive H of Д±X П‰ 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 П‰0 =
dqj в€§ dpj . The curve ПЃt = (q(t), p(t)) is an integral curve for XH exactly if
пЈ±
пЈґ dqi (t) = в€‚H
пЈґ
пЈґ dt
пЈІ в€‚pi
(Hamilton equations)
пЈґ dp
пЈґ в€‚H
пЈґ i
пЈі (t) = в€’
dt в€‚qi
n
в€‚H в€‚ в€‚H в€‚
в€’
Indeed, let XH = . Then,
в€‚pi в€‚qi в€‚qi в€‚pi
i=1

n n
Д±XH (dqj в€§ dpj ) = [(Д±XH dqj ) в€§ dpj в€’ dqj в€§ (Д±XH dpj )]
Д± XH П‰ =
j=1 j=1
n
в€‚H в€‚H
= в€‚pj dpj + в€‚qj dqj = dH
j=1

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

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

The case where n = 3 has a simple physical illustration. NewtonвЂ™s second
law states that a particle of mass m moving in conп¬Ѓguration 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 = в€’ 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 в€— R3 be the corresponding phase space, with co-
2m
ordinates (q1 , q2 , q3 , p1 , p2 , p3 ). NewtonвЂ™s second law in R3 is equivalent to the
Hamilton equations in R6 :
пЈ± dq 1 в€‚H
i
пЈґ = pi =
пЈІ
dt m в€‚pi
2
пЈґ dpi d qi в€‚V в€‚H
пЈі =m 2 =в€’ =в€’ .
dt dt в€‚qi в€‚qi

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

18.3 Brackets
Vector п¬Ѓelds are diп¬Ђerential operators on functions: if X is a vector п¬Ѓeld and
f в€€ C в€ћ (X), df being the corresponding 1-form, then

X В· f := df (X) = LX f .

Given two vector п¬Ѓelds X, Y , there is a unique vector п¬Ѓeld W such that

LW f = LX (LY f ) в€’ LY (LX f ) .

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

Д±[X,Y ] О± = LX Д±Y О± в€’ Д±Y LX О± = [LX , Д±Y ]О± .

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

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

Proof.
= L X Д±Y П‰ в€’ Д± Y L X П‰
Д±[X,Y ] П‰
= dД±X Д±Y П‰ + Д±X dД±Y П‰ в€’Д±Y dД±X П‰ в€’Д±Y Д±X dП‰
0
0 0
= d(П‰(Y, X)) .

A (real) Lie algebra is a (real) vector space g together with a Lie bracket
[В·, В·], i.e., a bilinear map [В·, В·] : g Г— g в†’ g satisfying:
(a) [x, y] = в€’[y, x] , в€Ђx, y в€€ g , (antisymmetry)
в€Ђx, y, z в€€ g .
(b) [x, [y, z]] + [y, [z, x]] + [z, [x, y]] = 0 , (Jacobi identity)
Let
П‡(M ) = { vector п¬Ѓelds on M }
sympl
(M ) = { symplectic vector п¬Ѓelds on M }
П‡
ham
(M ) = { hamiltonian vector п¬Ѓelds on M } .
П‡

Corollary 18.4 The inclusions (П‡ham (M ), [В·, В·]) вЉ† (П‡sympl (M ), [В·, В·]) вЉ† (П‡(M ), [В·, В·])
are inclusions of Lie algebras.

Deп¬Ѓnition 18.5 The Poisson bracket of two functions f, g в€€ C в€ћ (M ; R) is

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

We have X{f,g} = в€’[Xf , Xg ] because XП‰(Xf ,Xg ) = [Xg , Xf ].

Theorem 18.6 The bracket {В·, В·} satisп¬Ѓes the Jacobi identity, i.e.,

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

Proof. Exercise.

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

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

Exercise. Check that the Poisson bracket {В·, В·} deп¬Ѓned above satisп¬Ѓes the
в™¦
Leibniz rule.
We conclude that, if (M, П‰) is a symplectic manifold, then (C в€ћ (M ), {В·, В·})
is a Poisson algebra. Furthermore, we have a Lie algebra anti-automorphism

C в€ћ (M ) в€’в†’ П‡(M )
H в€’в†’ XH
{В·, В·} в€’[В·, В·] .

18.4 Integrable Systems

Deп¬Ѓnition 18.8 A hamiltonian system is a triple (M, П‰, H), where (M, П‰)
is a symplectic manifold and H в€€ C в€ћ (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 ПЃt be the п¬‚ow of XH . Then
d
в—¦ ПЃt ) = ПЃв€— LXH f = ПЃв€— Д±XH df = ПЃв€— Д±XH Д±Xf П‰
dt (f t t t
в€— в€—
= ПЃt П‰(Xf , XH ) = ПЃt {f, H} = 0

A function f as in Theorem 18.9 is called an integral of motion (or a п¬Ѓrst
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 diп¬Ђerentials
(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, П‰, H), then, at each p в€€ M ,
their hamiltonian vector п¬Ѓelds generate an isotropic subspace of Tp M :
П‰(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 .
Deп¬Ѓnition 18.10 A hamiltonian system (M, П‰, 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 вЂ“ any 2-dimensional hamiltonian system (where
в™¦
the set of non-п¬Ѓxed points is dense) is integrable.

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

Proof. Exercise (just follow the п¬‚ows).
Any compact component of f в€’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 вЂ™s proper.)
Theorem 18.12 (Arnold-Liouville ) Let (M, П‰, H) be an integrable sys-
tem of dimension 2n with integrals of motion f1 = H, f2 , . . . , fn . Let c в€€ Rn
be a regular value of f := (f1 , . . . , fn ). The corresponding level f в€’1 (c) is a
lagrangian submanifold of M .
(a) If the п¬‚ows of Xf1 , . . . , Xfn starting at a point p в€€ f в€’1 (c) are complete,
then the connected component of f в€’1 (c) containing p is a homogeneous
space for Rn . With respect to this aп¬ѓne structure, that component has
coordinates П•1 , . . . , П•n , known as angle coordinates, in which the п¬‚ows
of the vector п¬Ѓelds Xf1 , . . . , Xfn are linear.
111
18.4 Integrable Systems

(b) There are coordinates П€1 , . . . , П€n , known as action coordinates, comple-
mentary to the angle coordinates such that the П€i вЂ™s are integrals of motion
and П•1 , . . . , П•n , П€1 , . . . , П€n 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) вЂ“ the easy part вЂ“ 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 в†’ Rn collecting the given integrals of
motion is a lagrangian п¬Ѓbration, i.e., it is locally trivial and its п¬Ѓbers are
lagrangian submanifolds. The coordinates along the п¬Ѓbers 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 П•1 , . . . , П•n , f1 , . . . , fn do not form a Darboux chart.

11 The name вЂњangle coordinatesвЂќ is used even if the п¬Ѓbers are not tori.
Homework 13: Simple Pendulum

This problem is adapted from .

The simple pendulum is a mechanical system consisting of a massless rigid
rod of length l, п¬Ѓxed 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 Оё be the oriented angle between the rod (regarded as a point mass)
and the vertical direction. Let Оѕ be the coordinate along the п¬Ѓbers of T в€— S 1
induced by the standard angle coordinate on S 1 . Show that the function
H : T в€— S 1 в†’ R given by

Оѕ2
+ ml(1 в€’ cos Оё) ,
H(Оё, Оѕ) =
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 (Оё) = ml(1 в€’ cos Оё) (we omit universal constants), and that the kinetic
1
energy is given by K(Оё, Оѕ) = 2ml2 Оѕ 2 .
(b) For simplicity assume that m = l = 1.
Plot the level curves of H in the (Оё, Оѕ) 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 Оё-axis is an interval of length less than ПЂ.
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 2ПЂ in Оё,
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 п¬Ѓeld 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
п¬Ѓeld F ; let x(t), a в‰¤ t в‰¤ b, be its path of motion in R3 . NewtonвЂ™s second law
states that
d2 x
m 2 (t) = F (x(t)) .
dt
Deп¬Ѓne the work of a path Оі : [a, b] в€’в†’ R3 , with Оі(a) = p and Оі(b) = q, to be
b
dОі
F (Оі(t)) В·
WОі = (t)dt .
dt
a

Suppose that F is conservative, i.e., WОі depends only on p and q. Then we
can deп¬Ѓne the potential energy V : R3 в€’в†’ R of the system as
V (q) := WОі
where Оі is a path joining a п¬Ѓxed base point p0 в€€ R3 (the вЂњoriginвЂќ) to q. NewtonвЂ™s
second law can now be written
d2 x в€‚V
m 2 (t) = в€’ (x(t)) .
dt в€‚x
In the previous lecture we saw that
в‡ђв‡’
NewtonвЂ™s second law Hamilton equations
in R3 = {(q1 , q2 , q3 )} in T в€— 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
NewtonвЂ™s second law in conп¬Ѓguration space R3 is equivalent to solving in
phase space for the integral curve T в€— R3 of the hamiltonian vector п¬Ѓeld with
в™¦
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 в€‚V
m 2 =в€’ ,
dt в€‚x
where x(t) is the position of earth at time t, and V (x) = const. is the gravi-
|x|
в™¦
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 NewtonвЂ™s second laws. Variational principles are due mostly to DвЂ™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 conп¬Ѓguration of
this system is described by a vector in conп¬Ѓguration space R3n

x = (x1 , . . . , xn ) в€€ R3n

with xi в€€ R3 describing the position of the ith particle. If V в€€ C в€ћ (R3n ) is the
potential energy, then a path of motion x(t), a в‰¤ t в‰¤ b, satisп¬Ѓes

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

Consider this path in conп¬Ѓguration space as a map Оі0 : [a, b] в†’ R3n with
Оі0 (a) = p and Оі0 (b) = q, and let

P = {Оі : [a, b] в€’в†’ R3n | Оі(a) = p and Оі(b) = q}

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

Deп¬Ѓnition 19.1 The action of a path Оі в€€ P is
2
b
mi dОіi
AОі := в€’ V (Оі(t)) dt .
(t)
2 dt
a

Principle of least action.
The physical path Оі0 is the path for which AОі is minimal.
NewtonвЂ™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 Оі0 : [a, b] в†’ M , with Оі0 (a) = p and Оі0 (b) = q, as being вЂњtheвЂќ path which
minimizes AОі among all those hypothetical paths Оі : [a, b] в†’ R3n with Оі(a) = p,
Оі(b) = q and satisfying the rigid constraints Оі(t) в€€ 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 в†’ R be a smooth function.
115
19.3 Variational Problems

If Оі : [a, b] в†’ M is a smooth curve on M , deп¬Ѓne the lift of Оі to T M to be
the smooth curve on T M given by

Оі : [a, b] в€’в†’ T M
Лњ
Оі(t), dОі (t)
t в€’в†’ .
dt

The action of Оі is
b b
dОі
в€—
AОі := (Лњ F )(t)dt =
Оі F Оі(t), (t) dt .
dt
a a

For п¬Ѓxed p, q в€€ M , let

P(a, b, p, q) := {Оі : [a, b] в€’в†’ M | Оі(a) = p, Оі(b) = q} .

Problem.
Find, among all Оі в€€ P(a, b, p, q), the curve Оі0 which вЂњminimizesвЂќ AОі .
First observe that minimizing curves are always locally minimizing:

Lemma 19.2 Suppose that Оі0 : [a, b] в†’ M is minimizing. Let [a1 , b1 ] be a
subinterval of [a, b] and let p1 = Оі0 (a1 ), q1 = Оі0 (b1 ). Then Оі0 |[a1 ,b1 ] is minimiz-
ing among the curves in P(a1 , b1 , p1 , q1 ).

Proof. Exercise:
Argue by contradiction. Suppose that there were Оі1 в€€ P(a1 , b1 , p1 , q1 ) for
which AОі1 < AОі0 |[a1 ,b1 ] . Consider a broken path obtained from Оі0 by replacing
the segment Оі0 |[a1 ,b1 ] by Оі1 . Construct a smooth curve Оі2 в€€ P(a, b, p, q) for
which AОі2 < AОі0 by rounding oп¬Ђ the corners of the broken path.
We will now assume that p, q and Оі0 lie in a coordinate neighborhood
(U, x1 , . . . , xn ). On T U we have coordinates (x1 , . . . , xn , v1 , . . . , vn ) associated
в€‚ в€‚
with a trivialization of T U by в€‚x1 , . . . , в€‚xn . Using this trivialization, the curve

Оі : [a, b] в€’в†’ U , Оі(t) = (Оі1 (t), . . . , Оіn (t))

lifts to
dОі1 dОіn
Оі : [a, b] в€’в†’ T U ,
Лњ Оі (t) =
Лњ Оі1 (t), . . . , Оіn (t), (t), . . . , (t) .
dt dt

Necessary condition for Оі0 в€€ P(a, b, p, q) to minimize the action.
Let c1 , . . . , cn в€€ C в€ћ ([a, b]) be such that ci (a) = ci (b) = 0. Let ОіОµ : [a, b] в€’в†’
U be the curve

ОіОµ (t) = (Оі1 (t) + Оµc1 (t), . . . , Оіn (t) + Оµcn (t)) .

For Оµ small, ОіОµ is well-deп¬Ѓned and in P(a, b, p, q).
116 19 VARIATIONAL PRINCIPLES

b dОіОµ
Let AОµ = AОіОµ = dt. If Оі0 minimizes A, then
F ОіОµ (t), dt (t)
a

dAОµ
(0) = 0 .
dОµ
b
dAОµ в€‚F dОі0 в€‚F dОі0 dci
(0) = Оі0 (t), (t) ci (t) + Оі0 (t), (t) (t) dt
dОµ в€‚xi dt в€‚vi dt dt
a i
b
в€‚F d в€‚F
(. . .) в€’
= (. . .) ci (t)dt = 0
в€‚xi dt в€‚vi
a i

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

в€‚F dОі0 d в€‚F dОі0
Оі0 (t), (t) = Оі0 (t), (t) E-L
в€‚xi dt dt в€‚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
в€‚F dОі0
= 0 в‡ђв‡’ the curve Оі0 sits on the critical set of F .
Оі0 (t), (t)
в€‚xi dt

For generic F , the critical points are isolated, hence Оі0 (t) must be a con-
stant curve.
Case 2: Suppose that F (x, v) depends aп¬ѓnely on v:
n
F (x, v) = F0 (x) + Fj (x)vj .
j=1

n
в€‚F0 в€‚Fj dОіj
LHS of E-L : (Оі(t)) + (Оі(t)) (t)
в€‚xi в€‚xi dt
j=1
n
d в€‚Fi dОіj
RHS of E-L : Fi (Оі(t)) = (Оі(t)) (t)
dt в€‚xj dt
j=1

The Euler-Lagrange equations become
n
в€‚F0 в€‚Fi в€‚Fj dОіj
в€’
(Оі(t)) = (Оі(t)) (t) .
в€‚xi в€‚xj в€‚xi dt
j=1
nГ—n matrix
117
19.5 Minimizing Properties

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

n
dОіj в€‚F0
(t) = Gji (Оі(t)) (Оі(t))
dt в€‚xi
i=1

is a system of п¬Ѓrst order ordinary diп¬Ђerential 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
в€‚ 2F
Legendre condition: det =0.
в€‚vi в€‚vj
в€’1
в€‚2F
Letting Gij (x, v) = в€‚vi в€‚vj (x, v) , the Euler-Lagrange equations become

d2 Оіj в€‚2F
в€‚F dОі dОі dОіk
в€’
= Gji Оі, Gji Оі, .
dt2 в€‚xi dt в€‚vi в€‚xk dt dt
i i,k

This second order ordinary diп¬Ђerential equation has a unique solution given
initial conditions
dОі
Оі(a) = p and (a) = v .
dt

19.5 Minimizing Properties

Is the above solution locally minimizing?
в€‚2F
0, в€Ђ(x, v), i.e., with the x variable frozen, the
Assume that в€‚vi в€‚vj (x, v)
function v в†’ F (x, v) is strictly convex.
Suppose that Оі0 в€€ P(a, b, p, q) satisп¬Ѓes E-L. Does Оі0 minimize AОі ? Locally,
yes, according to the following theorem. (Globally it is only critical.)

Theorem 19.3 For every suп¬ѓciently small subinterval [a1 , b1 ] of [a, b], Оі0 |[a1 ,b1 ]
is locally minimizing in P(a1 , b1 , p1 , q1 ) where p1 = Оі0 (a1 ), q1 = Оі0 (b1 ).

Proof. As an exercise in Fourier series, show the Wirtinger inequality: for
f в€€ C 1 ([a, b]) with f (a) = f (b) = 0, we have
2
b b
ПЂ2
df
|f |2 dt .
dt в‰Ґ
(b в€’ a)2
dt
a a

Suppose that Оі0 : [a, b] в†’ U satisп¬Ѓes E-L. Take ci в€€ C в€ћ ([a, b]), ci (a) =
ci (b) = 0. Let c = (c1 , . . . , cn ). Let ОіОµ = Оі0 + Оµc в€€ P(a, b, p, q), and let
A Оµ = A ОіОµ .
118 19 VARIATIONAL PRINCIPLES

dAОµ
E-L в‡ђв‡’ dОµ (0) = 0.
b
d2 AОµ в€‚2F dОі0
(0) = Оі0 , ci cj dt (I)
dОµ2 в€‚xi в€‚xj dt
a i,j
b
в€‚2F dОі0 dcj
+2 Оі0 , ci dt (II)
в€‚xi в€‚vj dt dt
a i,j
b
в€‚2F dОі0 dci dcj
+ Оі0 , dt (III) .
в€‚vi в€‚vj dt dt dt
a i,j

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

2
dc
в‰Ґ KIII
III
dt L2 [a,b]

в‰¤ KI |c|2 2 [a,b]
|I| L

dc
|II| в‰¤ KII |c|L2 [a,b]
dt L2 [a,b]

where KI , KII , KIII > 0. By the Wirtinger inequality, if b в€’ a is very small, then
III > |I|, |II|. Hence, Оі0 is a local minimum.
Homework 14: Minimizing Geodesics

This set of problems is adapted from .

Let (M, g) be a riemannian manifold. From the riemannian metric, we get
a function F : T M в†’ R, whose restriction to each tangent space Tp M is the
quadratic form deп¬Ѓned by the metric.
Let p and q be points on M , and let Оі : [a, b] в†’ M be a smooth curve joining
p to q. Let Оі : [a, b] в†’ T M , Оі (t) = (Оі(t), dОі (t)) be the lift of Оі to T M . The
Лњ Лњ dt
action of Оі is
2
b b
dОі
в€—
A(Оі) = (Лњ F ) dt =
Оі dt .
dt
a a

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

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

3. Let П„ : [a, b] в†’ [a, b] be a smooth monotone map taking the endpoints of
[a, b] to the endpoints of [a, b]. Prove that
2
b
dП„
dt в‰Ґ b в€’ a ,
dt
a

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

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

119
120 HOMEWORK 14

5. Let Оі0 : [a, b] в†’ M be a curve joining p to q. Suppose that Оі0 is action-
minimizing, i.e., suppose that

A(Оі0 ) в‰¤ A(Оі)

for any other curve Оі : [a, b] в†’ M joining p to q. Prove that Оі0 is also arc-
length-minimizing, i.e., show that Оі0 is the shortest geodesic joining p
to q.

6. Show that, among all curves joining p to q, Оі0 minimizes the action if and
only if Оі0 is of constant velocity and Оі0 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 Christoп¬Ђel equations for a geodesic

d2 Оі k dОі i dОі j
(О“k в—¦ Оі)
+ =0,
ij
dt2 dt dt
where the О“k вЂ™s (called the Christoп¬Ђel symbols) are deп¬Ѓned in terms of
ij
the coeп¬ѓcients of the riemannian metric by

1 в€‚g i в€‚g j в€‚gij
О“k = k
в€’
g + ,
ij
2 в€‚xj в€‚xi в€‚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 п¬Ѓxed point set is the plane spanned by p and q, and show
that this isometry induces on S n an isometry whose п¬Ѓxed 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 в†’ R, F = F (v1 , . . . , vn ),
be a smooth function. Let p в€€ V , u = n ui ei в€€ V . The hessian of F is the
i=1
quadratic function on V deп¬Ѓned by

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

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

Deп¬Ѓnition 20.1 The function F is strictly convex if (d2 F )p 0, в€Ђp в€€ 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) в†’ +в€ћ as p в†’ в€ћ in V .

Proof. Homework 15.

Deп¬Ѓnition 20.3 A strictly convex function F is stable when it satisп¬Ѓes con-
ditions (a)-(d) in Theorem 20.2.

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

20.2 Legendre Transform

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

F : V в€’в†’ R , F (v) = F (v) в€’ (v) .

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

в‡ђв‡’
F is strictly convex F is strictly convex.

121
122 20 LEGENDRE TRANSFORM

Deп¬Ѓnition 20.4 The stability set of a strictly convex function F is

SF = { в€€ V в€— | F is stable} .

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

Proof. Homework 15.
Homework 15 also describes a suп¬ѓcient condition for SF = V в€— .

Deп¬Ѓnition 20.6 The Legendre transform associated to F в€€ C в€ћ (V ; R) is
the map
LF : V в€’в†’ V в€—
в€—
Vв€— .
p в€’в†’ dFp в€€ Tp V

Theorem 20.7 Suppose that F is strictly convex. Then

LF : V в€’в†’ SF ,

i.e., LF is a diп¬Ђeomorphism onto SF .

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

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

Deп¬Ѓnition 20.8 The dual function F в€— to F is

F в€— : SF в€’в†’ R , F в€— ( ) = в€’ min F (p) .
pв€€V

Lв€’1 = LF в€— .
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 в†’ R a function on T M .
Оів€—F .
Problem. Minimize AОі = Лњ
At p в€€ M , let
Fp := F |Tp M : Tp M в€’в†’ R .
Assume that Fp is strictly convex for all p в€€ M . To simplify notation, assume
в€—
also that SFp = Tp M . The Legendre transform on each tangent space

в€—
LFp : Tp M в€’в†’ Tp M
123
20.3 Application to Variational Problems

is essentially given by the п¬Ѓrst derivatives of F in the v directions. The dual
в€— в€—
function to Fp is Fp : Tp M в€’в†’ R. Collect these п¬Ѓberwise maps into

в€’в†’ T в€— M ,
L: L|Tp M
TM = L Fp , and

H : T в€—M в€—
в€’в†’ R , H|Tp M = Fp .
в€—

Exercise. The maps H and L are smooth, and L is a diп¬Ђeomorphism. в™¦
Let
Оі : [a, b] в€’в†’ M be a curve, and
Оі : [a, b] в€’в†’ T M
Лњ its lift.

Theorem 20.10 The curve Оі satisп¬Ѓes the E-L equations on every coordinate
chart if and only if L в—¦ Оі : [a, b] в†’ T в€— M is an integral curve of the hamiltonian
Лњ
vector п¬Ѓeld XH .

Proof. Let
(U, x1 , . . . , xn ) coordinate neighborhood in M ,
(T U, x1 , . . . , xn , v1 , . . . , vn ) coordinates in T M ,
(T в€— U, x1 , . . . , xn , Оѕ1 , . . . , Оѕn ) coordinates in T в€— M .

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

в€’в†’ T в€— U
L: TU
в€‚F
в€’в†’ (x, Оѕ)
(x, v) where Оѕ = LFx (v) = (x, v) .
в€‚v

(This is the deп¬Ѓnition of momentum Оѕ.)
в€—
H(x, Оѕ) = Fx (Оѕ) = Оѕ В· v в€’ F (x, v) L(x, v) = (x, Оѕ) .
where

Integral curves (x(t), Оѕ(t)) of XH satisfy the Hamilton equations:
пЈ±
пЈґ dx = в€‚H
пЈґ (x, Оѕ)
пЈґ
пЈІ dt в€‚Оѕ
H
пЈґ dОѕ
пЈґ в€‚H
пЈґ
пЈі =в€’ (x, Оѕ) ,
dt в€‚x
whereas the physical path x(t) satisп¬Ѓes the Euler-Lagrange equations:

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

dx
t в†’ (x(t), Оѕ(t)) satisп¬Ѓes H в‡ђв‡’ tв†’ x(t), (t) satisп¬Ѓes E-L .
dt
124 20 LEGENDRE TRANSFORM

The п¬Ѓrst line of H is automatically satisп¬Ѓed:

dx в€‚H dx
(x, Оѕ) = LFx (Оѕ) = Lв€’1 (Оѕ) в‡ђв‡’
= Оѕ = L Fx
в€—
Fx
dt в€‚Оѕ dt

в€‚F
= в€’ в€‚H (x, Оѕ).
Claim. If (x, Оѕ) = L(x, v), then в€‚x (x, v) в€‚x

This follows from diп¬Ђerentiating both sides of H(x, Оѕ) = Оѕ В· v в€’ F (x, v) with
respect to x, where Оѕ = LFx (v) = Оѕ(x, v).

в€‚H в€‚H в€‚Оѕ в€‚Оѕ в€‚F
В·vв€’
+ = .
в€‚x в€‚Оѕ в€‚x в€‚x в€‚x
v

Now the second line of H becomes
d в€‚F dОѕ в€‚H в€‚F
=в€’ в‡ђв‡’
(x, v) = (x, Оѕ) = (x, v) E-L .
dt в€‚v dt в€‚x в€‚x
by the claim
since Оѕ = LFx (v)
Homework 15: Legendre Transform
This set of problems is adapted from .
1. Let f : R в†’ R be a smooth function. f is called strictly convex if
f (x) > 0 for all x в€€ 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) в†’ +в€ћ as x в†’ В±в€ћ.
(d)
The function f is stable if it satisп¬Ѓes 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 в†’ R a smooth function.
The function F is said to be strictly convex if for every pair of elements
p, v в€€ V , v = 0, the restriction of F to the line {p + xv | x в€€ R} is strictly
convex.
The hessian of F at p is the quadratic form
d2
d 2 Fp : v в†’ F (p + xv)|x=0 .
dx2
Show that F is strictly convex if and only if d2 Fp is positive deп¬Ѓnite for
all p в€€ 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) в†’ +в€ћ as p в†’ в€ћ.
(d)
3. As in exercise 2, let V be an n-dimensional vector space and F : V в†’
R a smooth function. Since V is a vector space, there is a canonical
в€—
V в€— , for every p в€€ V . Therefore, we can deп¬Ѓne a map
identiп¬Ѓcation Tp V
LF : V в†’ V в€— (Legendre transform)
by setting
в€—
Vв€— .
LF (p) = dFp в€€ Tp V
Show that, if F is strictly convex, then, for every point p в€€ V , LF maps a
neighborhood of p diп¬Ђeomorphically onto a neighborhood of LF (p).

125
126 HOMEWORK 15

4. A strictly convex function F : V в†’ R is stable if it satisп¬Ѓes the four
equivalent conditions of exercise 2. Given any strictly convex function F ,
we will denote by SF the set of l в€€ V в€— for which the function Fl : V в†’ R,
p в†’ F (p) в€’ l(p), is stable. Prove that:
(a) The set SF is open and convex.
(b) LF maps V diп¬Ђeomorphically onto SF .
(c) If l в€€ SF and p0 = Lв€’1 (l), then p0 is the unique minimum point of
F
the function Fl .
Let F в€— : SF в†’ R be the function whose value at l is the quantity
в€’min Fl (p). Show that F в€— is a smooth function.
pв€€V
The function F в€— is called the dual of the function F .
5. Let F be a strictly convex function. F is said to have quadratic growth
at inп¬Ѓnity if there exists a positive-deп¬Ѓnite quadratic form Q on V and
a constant K such that F (p) в‰Ґ Q(p) в€’ K, for all p. Show that, if F
has quadratic growth at inп¬Ѓnity, then SF = V в€— and hence LF maps V
diп¬Ђeomorphically onto V в€— .
6. Let F : V в†’ R be strictly convex and let F в€— : SF в†’ R be the dual
function. Prove that for all p в€€ V and all l в€€ SF ,
F (p) + F в€— (l) в‰Ґ l(p) (Young inequality) .

7. On one hand we have V Г— V в€— T в€— V , and on the other hand, since
V = V в€—в€— , we have V Г— V в€— V в€— Г— V T в€—V в€— .
Let О±1 be the canonical 1-form on T в€— V and О±2 be the canonical 1-form on
T в€— V в€— . Via the identiп¬Ѓcations above, we can think of both of these forms
 << стр. 5(всего 9)СОДЕРЖАНИЕ >>