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t tв€’1
&%

2. Let (P2 , П‰FS ) be 2-(complex-)dimensional complex projective space equipped
with the Fubini-Study form deп¬Ѓned in Section 5.1. The T2 -action on P2 by
(eiОё1 , eiОё2 ) В· [z0 : z1 : z2 ] = [z0 : eiОё1 z1 : eiОё2 z2 ] has moment map

|z1 |2 |z2 |2
1
Вµ[z0 : z1 : z2 ] = в€’ , .
|z0 |2 + |z1 |2 + |z2 |2 |z0 |2 + |z1 |2 + |z2 |2
2
81
5.8. SYMPLECTIC TORIC MANIFOLDS

T

(в€’ 1 , 0)
t(0, 0) E
2
t
dВ В В В В
В
dВ В В В
В
dВ В В
В
dВ В
В
dВ
dt
(0, в€’ 1 )
2

The п¬Ѓxed points get mapped as

[1 : 0 : 0] в€’в†’ (0, 0)
1
[0 : 1 : 0] в€’в†’ в€’ 2 , 0
1
[0 : 0 : 1] в€’в†’ 0, в€’ 2 .

Notice that the stabilizer of a preimage of the edges is S 1 , while the action
is free at preimages of interior points of the moment polytope.

Exercise 32
Compute a moment polytope for the T 3-action on P 3 as
(eiОё1 , eiОё2 , eiОё3 ) В· [z0 : z1 : z2 : z3 ] = [z0 : eiОё1 z1 : eiОё2 z2 : eiОё3 z3 ] .

Exercise 33
Compute a moment polytope for the T 2-action on P 1 Г— P 1 as
(eiОё , eiО· ) В· ([z0 : z1 ], [w0 : w1 ]) = ([z0 : eiОё z1 ], [w0 : eiО· w1 ]) .

By Proposition 4.17, symplectic toric manifolds are optimal hamiltonian
torus-spaces. By Theorem 4.15, they have an associated polytope. It turns out
that the moment polytope contains enough information to sort all symplectic toric
manifolds. We now deп¬Ѓne the class of polytopes which arise in the classiп¬Ѓcation.

Deп¬Ѓnition 5.11 A Delzant polytope в€† in Rn is a polytope satisfying:
вЂў simplicity, i.e., there are n edges meeting at each vertex;
вЂў rationality, i.e., the edges meeting at the vertex p are rational in the sense
that each edge is of the form p + tui , t в‰Ґ 0, where ui в€€ Z n;
82 LECTURE 5. SYMPLECTIC REDUCTION

вЂў smoothness, i.e., for each vertex, the corresponding u 1 , . . . , un can be cho-
sen to be a Z-basis of Z n.

Examples of Delzant polytopes in R2 :

d d
d d
d d
d d

The dotted vertical line in the trapezoidal example is there just to stress that it is a
picture of a rectangle plus an isosceles triangle. For вЂњtallerвЂќ triangles, smoothness
would be violated. вЂњWiderвЂќ triangles (with integral slope) may still be Delzant.
The family of the Delzant trapezoids of this type, starting with the rectangle,
correspond, under the Delzant construction, to the so-called Hirzebruch surfaces.
в™¦
Examples of polytopes which are not Delzant:

В„
d
rr В„d
rr В„d
r
rr В„d
r В„В

The picture on the left fails the smoothness condition, since the triangle is not
в™¦
isosceles, whereas the one on the right fails the simplicity condition.

DelzantвЂ™s theorem classiп¬Ѓes (equivalence classes of) symplectic toric manifolds
in terms of the combinatorial data encoded by a Delzant polytope.

Theorem 5.12 (Delzant ) Toric manifolds are classiп¬Ѓed by Delzant poly-
topes. More speciп¬Ѓcally, the bijective correspondence between these two sets is given
by the moment map:

{toric manifolds} в†ђв†’ {Delzant polytopes}
(M 2n , П‰, Tn , Вµ) в€’в†’ Вµ(M ) .

In Section 5.9 we describe the construction which proves the (easier) existence
part, or surjectivity, in DelzantвЂ™s theorem. In order to prepare that, we will next
give an algebraic description of Delzant polytopes.
83
5.9. DELZANTвЂ™S CONSTRUCTION

Let в€† be a Delzant polytope in (Rn )в€—1 and with d facets.2 Let vi в€€ Z n,
i = 1, . . . , d, be the primitive3 outward-pointing normal vectors to the facets of в€†.
Then we can describe в€† as an intersection of halfspaces
в€† = {x в€€ (Rn )в€— | x, vi в‰¤ О»i , i = 1, . . . , d} for some О»i в€€ R .

Example. For the picture below, we have
= {x в€€ (R2 )в€— | x1 в‰Ґ 0, x2 в‰Ґ 0, x1 + x2 в‰¤ 1}
в€†
= {x в€€ (R2 )в€— | x, (в€’1, 0) в‰¤ 0 , x, (0, в€’1) в‰¤ 0 , x, (1, 1) в‰¤ 1} .
в™¦


В
В
В
(0, 1)
В  в€’в€’ в€’в†’
r
d В  v3 = (1, 1)
d В
d В
' dВ
в€’в€’в†’
в€’в€’ d
v2 = (в€’1, 0)
d
d
dr
r
(0, 0) (1, 0)

в€’в€’в†’
в€’в€’
v1 = (0, в€’1)
c

5.9 DelzantвЂ™s Construction
Following [14, 24], we prove the existence part (or surjectivity) in DelzantвЂ™s the-
orem, by using symplectic reduction to associate to an n-dimensional Delzant
polytope в€† a symplectic toric manifold (Mв€† , П‰в€† , Tn , Вµв€† ).
1 Although we identify Rn with its dual via the euclidean inner product, it may be more clear
to see в€† in (Rn )в€— for DelzantвЂ™s construction.
2 A face of a polytope в€† is a set of the form F = P в€© {x в€€ R n | f (x) = c} where c в€€ R

and f в€€ (Rn )в€— satisп¬Ѓes f (x) в‰Ґ c, в€Ђx в€€ P . A facet of an n-dimensional polytope is an (n в€’ 1)-
dimensional face.
3 A lattice vector v в€€ Z n is primitive if it cannot be written as v = ku with u в€€ Z n, k в€€ Z

and |k| > 1; for instance, (1, 1), (4, 3), (1, 0) are primitive, but (2, 2), (4, 6) are not.
84 LECTURE 5. SYMPLECTIC REDUCTION

Let в€† be a Delzant polytope with d facets. Let vi в€€ Z n, i = 1, . . . , d, be the
primitive outward-pointing normal vectors to the facets. For some О»i в€€ R, we can
write
в€† = {x в€€ (Rn )в€— | x, vi в‰¤ О»i , i = 1, . . . , d} .
Let e1 = (1, 0, . . . , 0), . . . , ed = (0, . . . , 0, 1) be the standard basis of Rd . Consider

Rd в€’в†’ Rn
ПЂ:
в€’в†’ vi .
ei

Lemma 5.13 The map ПЂ is onto and maps Z d onto Z n.

Proof. The set {e1 , . . . , ed } is a basis of Z d. The set {v1 , . . . , vd } spans Z n for the
following reason. At a vertex p, the edge vectors u1 , . . . , un в€€ (Rn )в€— , form a basis
for (Z n)в€— which, by a change of basis if necessary, we may assume is the standard
basis. Then the corresponding primitive normal vectors to the facets meeting at
p are symmetric (in the sense of multiplication by в€’1) to the ui вЂ™s, hence form a
basis of Z n.
Therefore, ПЂ induces a surjective map, still called ПЂ, between tori:
ПЂ
Rd /(2ПЂZ d) в€’в†’ Rn /(2ПЂZ n)

Td Tn
в€’в†’ в€’в†’ 1 .

The kernel N of ПЂ is a (d в€’ n)-dimensional Lie subgroup of T d with inclusion
i : N в†’ Td . Let n be the Lie algebra of N . The exact sequence of tori
i ПЂ
1 в€’в†’ N в€’в†’ Td в€’в†’ Tn в€’в†’ 1

induces an exact sequence of Lie algebras
i ПЂ
0 в€’в†’ n в€’в†’ Rd в€’в†’ Rn в€’в†’ 0

with dual exact sequence
ПЂв€— iв€—
0 в€’в†’ (Rn )в€— в€’в†’ (Rd )в€— в€’в†’ nв€— в€’в†’ 0 .
i
Now consider Cd with symplectic form П‰0 = dzk в€§ dВЇk , and standard
z
2
hamiltonian action of Td given by

(eit1 , . . . , eitd ) В· (z1 , . . . , zd ) = (eit1 z1 , . . . , eitd zd ) .

The moment map is П† : Cd в€’в†’ (Rd )в€— deп¬Ѓned by
1
П†(z1 , . . . , zd ) = в€’ (|z1 |2 , . . . , |zd |2 ) + constant ,
2
85
5.9. DELZANTвЂ™S CONSTRUCTION

where we choose the constant to be (О»1 , . . . , О»d ). The subtorus N acts on Cd in a
hamiltonian way with moment map

iв€— в—¦ П† : Cd в€’в†’ nв€— .

Let Z = (iв€— в—¦ П†)в€’1 (0) be the zero-level set.
Claim 1. The set Z is compact and N acts freely on Z.
We postpone the proof of this claim until further down.
Since iв€— is surjective, 0 в€€ nв€— is a regular value of iв€— в—¦ П†. Hence, Z is a
compact submanifold of Cd of (real) dimension 2d в€’ (d в€’ n) = d + n. The orbit
space Mв€† = Z/N is a compact manifold of (real) dimension dim Z в€’ dim N =
(d + n) в€’ (d в€’ n) = 2n. The point-orbit map p : Z в†’ Mв€† is a principal N -bundle
over Mв€† . Consider the diagram
j
в†’ Cd
Z
pв†“
Mв€†

where j : Z в†’ Cd is inclusion. The Marsden-Weinstein-Meyer theorem guarantees
the existence of a symplectic form П‰в€† on Mв€† satisfying

p в€— П‰в€† = j в€— П‰0 .

Since Z is connected, the compact symplectic 2n-dimensional manifold (Mв€† , П‰в€† )
is also connected.
Proof of Claim 1. The set Z is clearly closed, hence in order to show that it is
compact it suп¬ѓces (by the Heine-Borel theorem) to show that Z is bounded. Let
в€† be the image of в€† by ПЂ в€— . We will show that П†(Z) = в€† .

Lemma 5.14 Let y в€€ (Rd )в€— . Then:

yв€€в€† в‡ђв‡’ y is in the image of Z by П† .

Proof. The value y is in the image of Z by П† if and only if both of the following
conditions hold:
1. y is in the image of П†;
2. iв€— y = 0.
Using the expression for П† and the third exact sequence, we see that these
conditions are equivalent to:

1. y, ei в‰¤ О»i for i = 1, . . . , d;
86 LECTURE 5. SYMPLECTIC REDUCTION

2. y = ПЂ в€— (x) for some x в€€ (Rn )в€— .

Suppose that the second condition holds, so that y = ПЂ в€— (x). Then

ПЂ в€— (x), ei в‰¤ О»i , в€Ђi
y, ei в‰¤ О»i , в€Ђi в‡ђв‡’
в‡ђв‡’ x, ПЂ(ei ) в‰¤ О»i , в€Ђi
в‡ђв‡’ x, vi в‰¤ О»i , в€Ђi
в‡ђв‡’ xв€€в€†.

Thus, y в€€ П†(Z) в‡ђв‡’ y в€€ ПЂ в€— (в€†) = в€† .
Since we have that в€† is compact, that П† is a proper map and that П†(Z) = в€† ,
we conclude that Z must be bounded, and hence compact.
It remains to show that N acts freely on Z.
Pick a vertex p of в€†, and let I = {i1 , . . . , in } be the set of indices for the n
facets meeting at p. Pick z в€€ Z such that П†(z) = ПЂ в€— (p). Then p is characterized
by n equations p, vi = О»i where i ranges in I:

в‡ђв‡’
p, vi = О»i p, ПЂ(ei ) = О»i
ПЂ в€— (p), ei = О»i
в‡ђв‡’
в‡ђв‡’ П†(z), ei = О»i
в‡ђв‡’ i-th coordinate of П†(z) is equal to О»i
1
в€’ 2 |zi |2 + О»i = О»i
в‡ђв‡’
в‡ђв‡’ zi = 0 .

Hence, those zвЂ™s are points whose coordinates in the set I are zero, and whose
other coordinates are nonzero. Without loss of generality, we may assume that
I = {1, . . . , n}. The stabilizer of z is

(Td )z = {(t1 , . . . , tn , 1, . . . , 1) в€€ Td } .

As the restriction ПЂ : (Rd )z в†’ Rn maps the vectors e1 , . . . , en to a Z-basis
v1 , . . . , vn of Z n (respectively), at the level of groups, ПЂ : (Td )z в†’ Tn must be
bijective. Since N = ker(ПЂ : Td в†’ Tn ), we conclude that N в€© (Td )z = {e}, i.e.,
Nz = {e}. Hence all N -stabilizers at points mapping to vertices are trivial. But
this was the worst case, since other stabilizers Nz (z в€€ Z) are contained in sta-
bilizers for points z which map to vertices. This concludes the proof of Claim 1.

Given a Delzant polytope в€†, we have constructed a symplectic manifold
(Mв€† , П‰в€† ) where Mв€† = Z/N is a compact 2n-dimensional manifold and П‰в€† is the
reduced symplectic form.
Claim 2. The manifold (Mв€† , П‰в€† ) is a hamiltonian Tn -space with a moment map
Вµв€† having image Вµв€† (Mв€† ) = в€†.
87
5.9. DELZANTвЂ™S CONSTRUCTION

Proof of Claim 2. Let z be such that П†(z) = ПЂ в€— (p) where p is a vertex of в€†, as
in the proof of Claim 1. Let Пѓ : Tn в†’ (Td )z be the inverse for the earlier bijection
ПЂ : (Td )z в†’ Tn . Since we have found a section, i.e., a right inverse for ПЂ, in the
exact sequence
i ПЂ
в€’в†’ Td в€’в†’ Tn
1 в€’в†’ N в€’в†’ 1 ,
Пѓ
в†ђв€’

the exact sequence splits, i.e., becomes like a sequence for a product, as we obtain
an isomorphism
(i, Пѓ) : N Г— Tn в€’в†’ Td .
The action of the Tn factor (or, more rigorously, Пѓ(Tn ) вЉ‚ Td ) descends to the
quotient Mв€† = Z/N .
It remains to show that the Tn -action on Mв€† is hamiltonian with appropriate
moment map.
Consider the diagram
j Пѓв€—
П†
в†’ Cd в€’в†’ (Rd )в€— О· в€— вЉ• (Rn )в€— в€’в†’ (Rn )в€—
Z
pв†“
Mв€†

where the last horizontal map is simply projection onto the second factor. Since
the composition of the horizontal maps is constant along N -orbits, it descends to
a map
Вµв€† : Mв€† в€’в†’ (Rn )в€—
which satisп¬Ѓes
Вµв€† в—¦ p = Пѓ в€— в—¦ П† в—¦ j .
By Section 5.5 on reduction for product groups, this is a moment map for the
action of Tn on (Mв€† , П‰в€† ). Finally, the image of Вµв€† is:

Вµв€† (Mв€† ) = (Вµв€† в—¦ p)(Z) = (Пѓ в€— в—¦ П† в—¦ j)(Z) = (Пѓ в€— в—¦ ПЂ в€— )(в€†) = в€† ,

because П†(Z) = ПЂ в€— (в€†) and Пѓ в€— в—¦ ПЂ в€— = (ПЂ в—¦ Пѓ)в€— = id.
We conclude that (Mв€† , П‰в€† , Tn , Вµв€† ) is the required toric manifold correspond-
ing to в€†.

Exercise 34
Let в€† be an n-dimensional Delzant polytope, and let (M в€† , П‰в€† , T n, Вµв€† ) be the
associated symplectic toric manifold. Show that Вµ в€† maps the п¬Ѓxed points of
T n bijectively onto the vertices of в€†.
88 LECTURE 5. SYMPLECTIC REDUCTION

Exercise 35
Follow through the details of DelzantвЂ™s construction for the case of в€† = [0, a] вЉ‚
Rв€— (n = 1, d = 2). Let v(= 1) be the standard basis vector in R. Then в€† is
described by
x, в€’v в‰¤ 0 and x, v в‰¤ a ,
where v1 = в€’v, v2 = v, О»1 = 0 and О»2 = a.

v
ta
T

t0
c
в€’v

The projection
ПЂ
R2 в€’в†’ R
e1 в€’в†’ в€’v
e2 в€’в†’ v
has kernel equal to the span of (e1 + e2 ), so that N is the diagonal subgroup
of T 2 = S 1 Г— S 1 . The exact sequences become
i ПЂ
T2 S1
1 в€’в†’ N в€’в†’ в€’в†’ в€’в†’ 1
t в€’в†’ (t, t)
tв€’1 t2
(t1 , t2 ) в€’в†’ 1

i ПЂ
R2
0 в€’в†’ в€’в†’ в€’в†’ в€’в†’ 0
R
n
x в€’в†’ (x, x)
(x1 , x2 ) в€’в†’ x 2 в€’ x1

ПЂв€— iв€—
(R2 )в€—
Rв€— nв€—
0 в€’в†’ в€’в†’ в€’в†’ в€’в†’ 0
x в€’в†’ (в€’x, x)
(x1 , x2 ) в€’в†’ x 1 + x2 .

The action of the diagonal subgroup N = {(eit , eit ) в€€ S 1 Г— S 1 } on C2 ,
(eit , eit ) В· (z1 , z2 ) = (eit z1 , eit z2 ) ,
has moment map
1
(iв€— в—¦ П†)(z1 , z2 ) = в€’ 2 (|z1 |2 + |z2 |2 ) + a ,
with zero-level set
(iв€— в—¦ П†)в€’1 (0) = {(z1 , z2 ) в€€ C2 : |z1 |2 + |z2 |2 = 2a} .
Hence, the reduced space is a projective space:
(iв€— в—¦ П†)в€’1 (0)/N = P 1 .

Example. Consider
(S 2 , П‰ = dОё в€§ dh, S 1 , Вµ = h) ,
89
5.9. DELZANTвЂ™S CONSTRUCTION

where S 1 acts on S 2 by rotation. The image of Вµ is the line segment I = [в€’1, 1].
The product S 1 Г— I is an open-ended cylinder. By collapsing each end of the
в™¦
cylinder to a point, we recover the 2-sphere.

Exercise 36
Build P 2 from T 2 Г— в€† where в€† is a right-angled isosceles triangle.

Exercise 37
Consider the standard (S 1 )3 -action on P 3:
(eiОё1 , eiОё2 , eiОё3 ) В· [z0 : z1 : z2 : z3 ] = [z0 : eiОё1 z1 : eiОё2 z2 : eiОё3 z3 ] .
Exhibit explicitly the subsets of P 3 for which the stabilizer under this action
is {1}, S 1 , (S 1 )2 and (S 1 )3 . Show that the images of these subsets under the
moment map are the interior, the facets, the edges and the vertices, respec-
tively.

Exercise 38
What would be the classiп¬Ѓcation of symplectic toric manifolds if, instead of
the equivalence relation deп¬Ѓned in Section 5.8, one considered to be equivalent
those (Mi , П‰i , T i, Вµi ), i = 1, 2, related by an isomorphism О» : T 1 в†’ T 2 and a
О»-equivariant symplectomorphism П• : M 1 в†’ M2 such that:
(a) the maps Вµ1 and Вµ2 в—¦ П• are equal up to a constant?
(b) we have Вµ1 = в—¦ Вµ2 в—¦ П• for some в€€ SL(n; Z)?

Exercise 39

(a) Classify all 2-dimensional Delzant polytopes with 3 vertices, i.e., trian-
gles, up to translation, change of scale and the action of SL(2; Z).
Hint: By a linear transformation in SL(2; Z), we can make one of the angles in
the polytope into a square angle. How are the lengths of the two edges forming
that angle related?

(b) Classify all 2-dimensional Delzant polytopes with 4 vertices, up to trans-
lation and the action of SL(2; Z).
Hint: By a linear transformation in SL(2; Z), we can make one of the angles
in the polytope into a square angle. Check that automatically another angle
also becomes 90o .

(c) What are all the 4-dimensional symplectic toric manifolds that have four
п¬Ѓxed points?

Exercise 40
Let в€† be the n-simplex in Rn spanned by the origin and the standard basis
vectors (1, 0, . . . , 0), . . . , (0, . . . , 0, 1). Show that the corresponding symplectic
toric manifold is projective space, M в€† = P n.

Exercise 41
Which 2n-dimensional toric manifolds have exactly n + 1 п¬Ѓxed points?
Appendix A

Prerequisites from
Diп¬Ђerential Geometry

A.1 Isotopies and Vector Fields
Let M be a manifold, and ПЃ : M Г— R в†’ M a map, where we set ПЃt (p) := ПЃ(p, t).
Deп¬Ѓnition A.1 The map ПЃ is an isotopy if each ПЃt : M в†’ M is a diп¬Ђeomor-
phism, and ПЃ0 = idM .
Given an isotopy ПЃ, we obtain a time-dependent vector п¬Ѓeld, that is, a
family of vector п¬Ѓelds vt , t в€€ R, which at p в€€ M satisfy
d
q = ПЃв€’1 (p) ,
vt (p) = ПЃs (q) where t
ds s=t
i.e.,
dПЃt
= vt в—¦ ПЃt .
dt
Conversely, given a time-dependent vector п¬Ѓeld vt , if M is compact or if the
vt вЂ™s are compactly supported, there exists an isotopy ПЃ satisfying the previous
ordinary diп¬Ђerential equation.
Suppose that M is compact. Then we have a one-to-one correspondence
{isotopies of M } в†ђв†’ {time-dependent vector п¬Ѓelds on M }
ПЃt , t в€€ R в†ђв†’ vt , t в€€ R
Deп¬Ѓnition A.2 When vt = v is independent of t, the associated isotopy is called
the exponential map or the п¬‚ow of v and is denoted exp tv; i.e., {exp tv : M в†’
M | t в€€ R} is the unique smooth family of diп¬Ђeomorphisms satisfying
d
exp tv|t=0 = idM and (exp tv)(p) = v(exp tv(p)) .
dt

91
92 APPENDIX A. PREREQUISITES FROM DIFFERENTIAL GEOMETRY

Deп¬Ѓnition A.3 The Lie derivative is the operator
d
(exp tv)в€— П‰|t=0 .
Lv : в„¦k (M ) в€’в†’ в„¦k (M ) Lv П‰ :=
deп¬Ѓned by
dt
When a vector п¬Ѓeld vt is time-dependent, its п¬‚ow, that is, the corresponding
isotopy ПЃ, still locally exists by PicardвЂ™s theorem. More precisely, in the neigh-
borhood of any point p and for suп¬ѓciently small time t, there is a one-parameter
family of local diп¬Ђeomorphisms ПЃt satisfying
dПЃt
= vt в—¦ ПЃt and ПЃ0 = id .
dt
Hence, we say that the Lie derivative by vt is
d
(ПЃt )в€— П‰|t=0 .
Lvt : в„¦k (M ) в€’в†’ в„¦k (M ) Lvt П‰ :=
deп¬Ѓned by
dt

Exercise 42
Prove the Cartan magic formula,
Lv П‰ = Д±v dП‰ + dД±v П‰ ,
and the formula
dв€—
ПЃt П‰ = ПЃ в€— L v t П‰ , ()
t
dt
where ПЃ is the (local) isotopy generated by vt . A good strategy for each formula
(a) Check the formula for 0-forms П‰ в€€ в„¦ 0 (M ) = C в€ћ (M ).
(b) Check that both sides commute with d.
(c) Check that both sides are derivations of the algebra (в„¦в€— (M ), в€§). For
instance, check that
Lv (П‰ в€§ О±) = (Lv П‰) в€§ О± + П‰ в€§ (Lv О±) .

(d) Notice that, if U is the domain of a coordinate system, then в„¦ вЂў (U ) is
generated as an algebra by в„¦0 (U ) and dв„¦0 (U ), i.e., every element in
в„¦вЂў (U ) is a linear combination of wedge products of elements in в„¦ 0 (U )
and elements in dв„¦0 (U ).

We will need the following improved version of formula ( ).

Proposition A.4 For a smooth family П‰t , t в€€ R, of d-forms, we have
dв€— dП‰t
ПЃ t П‰t = ПЃ в€— L v t П‰t + .
t
dt dt

Proof. If f (x, y) is a real function of two variables, by the chain rule we have
d d d
f (t, t) = f (x, t) + f (t, y) .
dt dx dy
x=t y=t
93
A.2. SUBMANIFOLDS

Therefore,
dв€— dв€— dв€—
ПЃ П‰t = ПЃ П‰t + ПЃ П‰y
dt t dx x dy t
x=t y=t

dП‰y
ПЃв€— L v x П‰ t by ( ) ПЃв€—
x t dy
x=t y=t

dП‰t
= ПЃ в€— L v t П‰t + .
t
dt

A.2 Submanifolds

Let M and X be manifolds with dim X < dim M .

Deп¬Ѓnition A.5 A map i : X в†’ M is an immersion if di p : Tp X в†’ Ti(p) M is
injective for any point p в€€ X.
An embedding is an immersion which is a homeomorphism onto its image. 1
A closed embedding is a proper2 injective immersion.

Exercise 43
Show that a map i : X в†’ M is a closed embedding if and only if i is an
embedding and its image i(X) is closed in M .
Hint:
вЂў If i is injective and proper, then for any neighborhood U of p в€€ X, there
is a neighborhood V of i(p) such that f в€’1 (V) вЉ† U .
вЂў On a Hausdorп¬Ђ space, any compact set is closed. On any topological
space, a closed subset of a compact set is compact.
вЂў An embedding is proper if and only if its image is closed.

Deп¬Ѓnition A.6 A submanifold of M is a manifold X with a closed embedding
i : X в†’ M .3

Notation. Given a submanifold, we regard the embedding i : X в†’ M as an
inclusion, in order to identify points and tangent vectors:

Tp X = dip (Tp X) вЉ‚ Tp M .
p = i(p) and
1 Theimage has the topology induced by the target manifold.
2A map is proper if the preimage of any compact set is compact.
3 When X is an open subset of a manifold M , we refer to it as an open submanifold.
94 APPENDIX A. PREREQUISITES FROM DIFFERENTIAL GEOMETRY

A.3 Tubular Neighborhood Theorem
Let M be an n-dimensional manifold, and let X be a k-dimensional submanifold
where k < n and with inclusion map
i:X в†’M .
At each x в€€ X, the tangent space to X is viewed as a subspace of the tangent
space to M via the linear inclusion dix : Tx X в†’ Tx M , where we denote x = i(x).
The quotient Nx X := Tx M/Tx X is an (n в€’ k)-dimensional vector space, known
as the normal space to X at x. The normal bundle of X is
N X = {(x, v) | x в€€ X , v в€€ Nx X} .
The set N X has the structure of a vector bundle over X of rank n в€’ k under the
natural projection, hence as a manifold N X is n-dimensional.

Exercises 44
Let M be Rn and let X be a k-dimensional compact submanifold of R n .
(a) Show that in this case Nx X can be identiп¬Ѓed with the usual вЂњnormal
spaceвЂќ to X in Rn , that is, the orthogonal complement in R n of the
tangent space to X at x.
(b) Given Оµ > 0 let UОµ be the set of all points in Rn which are at a distance
less than Оµ from X. Show that, for Оµ suп¬ѓciently small, every point p в€€ U Оµ
has a unique nearest point ПЂ(p) в€€ X.
(c) Let ПЂ : UОµ в†’ X be the map deп¬Ѓned in the previous exercise for Оµ
suп¬ѓciently small. Show that, if p в€€ UОµ , then the line segment (1 в€’ t) В·
p + t В· ПЂ(p), 0 в‰¤ t в‰¤ 1, joining p to ПЂ(p) lies in UОµ .
(d) Let N XОµ = {(x, v) в€€ N X such that |v| < Оµ}. Let exp : N X в†’ Rn be
the map (x, v) в†’ x + v, and let ОЅ : N XОµ в†’ X be the map (x, v) в†’ x.
Show that, for Оµ suп¬ѓciently small, exp maps N X Оµ diп¬Ђeomorphically
onto UОµ , and show also that the following diagram commutes:

E
exp
N XОµ UОµ

d В
d В
ОЅd В ПЂ
В‚
d В
X
(e) Suppose now that the manifold X is not compact. Prove that the as-
sertion about exp is still true provided we replace Оµ by a continuous
function
Оµ : X в†’ R+
which tends to zero fast enough as x tends to inп¬Ѓnity. You have thus
proved the tubular neighborhood theorem in Rn .

In general, the zero section of N X,
i0 : X в†’ N X , x в†’ (x, 0) ,
95
A.3. TUBULAR NEIGHBORHOOD THEOREM

embeds X as a closed submanifold of N X. A neighborhood U0 of the zero section
X in N X is called convex if the intersection U0 в€© Nx X with each п¬Ѓber is convex.

Theorem A.7 (Tubular Neighborhood Theorem) Let M be an n-dimensio-
nal manifold, X a k-dimensional submanifold, N X the normal bundle of X in M ,
i0 : X в†’ N X the zero section, and i : X в†’ M inclusion. Then there exist a
convex neighborhood U0 of X in N X, a neighborhood U of X in M , and a diп¬Ђeo-
morphism П• : U0 в†’ U such that

П• E U вЉ†M
N X вЉ‡ U0
d
s В

d В
d В  commutes.
i0 d В i
d В
X
Outline of the proof.

вЂў Case of M = Rn , and X is a compact submanifold of Rn .

Theorem A.8 (Оµ-Neighborhood Theorem)
Let U Оµ = {p в€€ Rn : |p в€’ q| < Оµ for some q в€€ X} be the set of points at a
distance less than Оµ from X. Then, for Оµ suп¬ѓciently small, each p в€€ U Оµ has
a unique nearest point q в€€ X (i.e., a unique q в€€ X minimizing |q в€’ x|).
ПЂ
Moreover, setting q = ПЂ(p), the map U Оµ в†’ X is a (smooth) submersion with
the property that, for all p в€€ U Оµ , the line segment (1 в€’ t)p + tq, 0 в‰¤ t в‰¤ 1, is
in U Оµ .

Here is a sketch. At any x в€€ X, the normal space Nx X may be regarded as
an (n в€’ k)-dimensional subspace of Rn , namely the orthogonal complement
in Rn of the tangent space to X at x:

{v в€€ Rn : v вЉҐ w , for all w в€€ Tx X} .
Nx X

We deп¬Ѓne the following open neighborhood of X in N X:

N X Оµ = {(x, v) в€€ N X : |v| < Оµ} .

Let
в€’в†’ Rn
exp : NX
в€’в†’ x + v .
(x, v)
Restricted to the zero section, exp is the identity map on X.
96 APPENDIX A. PREREQUISITES FROM DIFFERENTIAL GEOMETRY

Prove that, for Оµ suп¬ѓciently small, exp maps N X Оµ diп¬Ђeomorphically onto
U Оµ , and show also that the diagram
exp E UОµ
NXОµ
d В
d В
d В  commutes.
ПЂ0 d В ПЂ
В‚
d В
X
вЂў Case where X is a compact submanifold of an arbitrary manifold M .
Put a riemannian metric g on M , and let d(p, q) be the riemannian distance
between p, q в€€ M . The Оµ-neighborhood of a compact submanifold X is
U Оµ = {p в€€ M | d(p, q) < Оµ for some q в€€ X} .
Prove the Оµ-neighborhood theorem in this setting: for Оµ small enough, the
following assertions hold.
вЂ“ Any p в€€ U Оµ has a unique point q в€€ X with minimal d(p, q). Set q = ПЂ(p).
ПЂ
вЂ“ The map U Оµ в†’ X is a submersion and, for all p в€€ U Оµ , there is a unique
geodesic curve Оі joining p to q = ПЂ(p).
вЂ“ The normal space to X at x в€€ X is naturally identiп¬Ѓed with a subspace of
Tx M :
Nx X {v в€€ Tx M | gx (v, w) = 0 , for any w в€€ Tx X} .
Let N X Оµ = {(x, v) в€€ N X | gx (v, v) < Оµ}.
вЂ“ Deп¬Ѓne exp : N X Оµ в†’ M by exp(x, v) = Оі(1), where Оі : [0, 1] в†’ M is the
geodesic with Оі(0) = x and dОі (0) = v. Then exp maps N X Оµ diп¬Ђeomorphi-
dt
cally to U Оµ .
вЂў General case.
When X is not compact, adapt the previous argument by replacing Оµ by an
appropriate continuous function Оµ : X в†’ R+ which tends to zero fast enough
as x tends to inп¬Ѓnity.

Restricting to the subset U 0 вЉ† N X from the tubular neighborhood theorem,
ПЂ0 в€’1
we obtain a submersion U0 в€’в†’ X with all п¬Ѓbers ПЂ0 (x) convex. We can carry this
п¬Ѓbration to U by setting ПЂ = ПЂ0 в—¦ П•в€’1 :
U0 вЉ† NX U вЉ†M
is a п¬Ѓbration =в‡’ is a п¬Ѓbration
ПЂ0 в†“ ПЂв†“
X X
This is called the tubular neighborhood п¬Ѓbration.
97
A.4. HOMOTOPY FORMULA

A.4 Homotopy Formula

Let U be a tubular neighborhood of a submanifold X in M . The restriction iв€— :
d d
HdeRham (U) в†’ HdeRham (X) by the inclusion map is surjective. As a corollary
of the tubular neighborhood п¬Ѓbration, iв€— is also injective: this follows from the
homotopy-invariance of de Rham cohomology.

Corollary A.9 For any degree , HdeRham (U) HdeRham (X).

At the level of forms, this means that, if П‰ is a closed -form on U and iв€— П‰ is
exact on X, then П‰ is exact. We will need the following related result.

Proposition A.10 If a closed -form П‰ on U has restriction iв€— П‰ = 0, then П‰ is
exact, i.e., П‰ = dВµ for some Вµ в€€ в„¦dв€’1 (U). Moreover, we can choose Вµ such that
Вµx = 0 at all x в€€ X.

Proof. Via П• : U0 в€’в†’ U, it is equivalent to work over U0 . Deп¬Ѓne for every 0 в‰¤ t в‰¤ 1
a map
U0 в€’в†’ U0
ПЃt :
(x, v) в€’в†’ (x, tv) .
This is well-deп¬Ѓned since U0 is convex. The map ПЃ1 is the identity, ПЃ0 = i0 в—¦ПЂ0 , and
each ПЃt п¬Ѓxes X, that is, ПЃt в—¦ i0 = i0 . We hence say that the family {ПЃt | 0 в‰¤ t в‰¤ 1}
is a homotopy from i0 в—¦ ПЂ0 to the identity п¬Ѓxing X. The map ПЂ0 : U0 в†’ X is
called a retraction because ПЂ0 в—¦ i0 is the identity. The submanifold X is then
called a deformation retract of U.
A (de Rham) homotopy operator between ПЃ0 = i0 в—¦ ПЂ0 and ПЃ1 = id is a
linear map
Q : в„¦d (U0 ) в€’в†’ в„¦dв€’1 (U0 )
satisfying the homotopy formula

Id в€’ (i0 в—¦ ПЂ0 )в€— = dQ + Qd .

When dП‰ = 0 and iв€— П‰ = 0, the operator Q gives П‰ = dQП‰, so that we can take
0
Вµ = QП‰. A concrete operator Q is given by the formula:
1
ПЃв€— (Д±vt П‰) dt ,
QП‰ = t
0

where vt , at the point q = ПЃt (p), is the vector tangent to the curve ПЃs (p) at s = t.
The proof that Q satisп¬Ѓes the homotopy formula is below.
In our case, for x в€€ X, ПЃt (x) = x (all t) is the constant curve, so vt vanishes
at all x for all t, hence Вµx = 0.
98 APPENDIX A. PREREQUISITES FROM DIFFERENTIAL GEOMETRY

To check that Q above satisп¬Ѓes the homotopy formula, we compute
1 1
ПЃв€— (Д±vt dП‰)dt ПЃв€— (Д±vt П‰)dt
QdП‰ + dQП‰ = +d
t t
0 0

1
ПЃв€— (Д±vt dП‰ + dД±vt П‰ )dt ,
= t
0
Lvt П‰

where Lv denotes the Lie derivative along v (reviewed in the next section), and we
used the Cartan magic formula: Lv П‰ = Д±v dП‰ + dД±v П‰. The result now follows from
dв€—
ПЃ t П‰ = ПЃ в€— Lvt П‰
t
dt
and from the fundamental theorem of calculus:
1
dв€—
ПЃt П‰ dt = ПЃв€— П‰ в€’ ПЃв€— П‰ .
QdП‰ + dQП‰ = 1 0
dt
0

A.5 Whitney Extension Theorem
Theorem A.11 (Whitney Extension Theorem) Let M be an n-dimensional
manifold and X a k-dimensional submanifold with k < n. Suppose that at each
p в€€ X we are given a linear isomorphism Lp : Tp M в†’ Tp M such that Lp |Tp X =
IdTp X and Lp depends smoothly on p. Then there exists an embedding h : N в†’ M
of some neighborhood N of X in M such that h|X = idX and dhp = Lp for all
p в€€ X.

The linear maps L serve as вЂњgermsвЂќ for the embedding.
Sketch of proof for the Whitney theorem.
Case M = Rn : For a compact k-dimensional submanifold X, take a neigh-
borhood of the form

U Оµ = {p в€€ M | distance (p, X) в‰¤ Оµ} .

For Оµ suп¬ѓciently small so that any p в€€ U Оµ has a unique nearest point in X, deп¬Ѓne a
projection ПЂ : U Оµ в†’ X, p в†’ point on X closest to p. If ПЂ(p) = q, then p = q +v for
some v в€€ Nq X where Nq X = (Tq X)вЉҐ is the normal space at q; see Appendix A.
Let
h : U Оµ в€’в†’ Rn
p в€’в†’ q + Lq v ,
where q = ПЂ(p) and v = p в€’ ПЂ(p) в€€ Nq X. Then hX = idX and dhp = Lp for p в€€ X.
If X is not compact, replace Оµ by a continuous function Оµ : X в†’ R+ which tends
to zero fast enough as x tends to inп¬Ѓnity.
99
A.5. WHITNEY EXTENSION THEOREM

General case: Choose a riemannian metric on M . Replace distance by rieman-
nian distance, replace straight lines q + tv by geodesics exp(q, v)(t) and replace
q + Lq v by the value at t = 1 of the geodesic with initial value q and initial velocity
Lq v.
Appendix B

Prerequisites from Lie
Group Actions

B.1 One-Parameter Groups of Diп¬Ђeomorphisms
Let M be a manifold and X a complete vector п¬Ѓeld on M . Let ПЃt : M в†’ M , t в€€ R,
be the family of diп¬Ђeomorphisms generated by X. For each p в€€ M , ПЃt (p), t в€€ R,
is by deп¬Ѓnition the unique integral curve of X passing through p at time 0, i.e.,
ПЃt (p) satisп¬Ѓes
пЈ±
пЈґ ПЃ0 (p) = p
пЈІ
пЈґ dПЃt (p)
пЈі = X(ПЃt (p)) .
dt
Claim. We have that ПЃt в—¦ ПЃs = ПЃt+s .

Proof. Let ПЃs (q) = p. We need to show that (ПЃt в—¦ ПЃs )(q) = ПЃt+s (q), for all t в€€ R.
Reparametrize as ПЃt (q) := ПЃt+s (q). Then
Лњ
пЈ±
пЈґ ПЃ0 (q) = ПЃs (q) = p
Лњ
пЈІ

пЈіПЃ
пЈґ dЛњt (q) dПЃt+s (q)
= = X(ПЃt+s (q)) = X(Лњt (q)) ,
ПЃ
dt dt
i.e., ПЃt (q) is an integral curve of X through p. By uniqueness we must have ПЃt (q) =
Лњ Лњ
ПЃt (p), that is, ПЃt+s (q) = ПЃt (ПЃs (q)).

Consequence. We have that ПЃв€’1 = ПЃв€’t .
t

In terms of the group (R, +) and the group (Diп¬Ђ(M ), в—¦) of all diп¬Ђeomorphisms
of M , these results can be summarized as:

101
102 APPENDIX B. PREREQUISITES FROM LIE GROUP ACTIONS

Corollary B.1 The map R в†’ Diп¬Ђ(M ), t в†’ ПЃt , is a group homomorphism.

The family {ПЃt | t в€€ R} is then called a one-parameter group of diп¬Ђeo-
morphisms of M and denoted

ПЃt = exp tX .

B.2 Lie Groups

Deп¬Ѓnition B.2 A Lie group is a manifold G equipped with a group structure
where the group operations

G Г— G в€’в†’ G G в€’в†’ G
and
a в€’в†’ aв€’1
(a, b) в€’в†’ a В· b

are smooth maps.

Examples.

вЂў S 1 regarded as unit complex numbers with multiplication, represents rota-
tions of the plane: S 1 = U(1) = SO(2).

вЂў U(n), unitary linear transformations of Cn .

вЂў SU(n), unitary linear transformations of Cn with det = 1.

вЂў O(n), orthogonal linear transformations of Rn .

вЂў SO(n), elements of O(n) with det = 1.

вЂў GL(V ), invertible linear transformations of a vector space V .

в™¦

Deп¬Ѓnition B.3 A representation of a Lie group G on a vector space V is a
group homomorphism G в†’ GL(V ).
1 The operation will be omitted when it is clear from the context.
103
B.3. SMOOTH ACTIONS

B.3 Smooth Actions
Let M be a manifold.
Deп¬Ѓnition B.4 An action of a Lie group G on M is a group homomorphism
П€ : G в€’в†’ Diп¬Ђ(M )
g в€’в†’ П€g .
(We will only consider left actions where П€ is a homomorphism. A right action
is deп¬Ѓned with П€ being an anti-homomorphism.) The evaluation map associated
with an action П€ : G в†’ Diп¬Ђ(M ) is
evП€ : M Г— G в€’в†’ M
(p, g) в€’в†’ П€g (p) .
The action П€ is smooth if ev П€ is a smooth map.

Example. If X is a complete vector п¬Ѓeld on M , then
ПЃ : R в€’в†’ Diп¬Ђ(M )
t в€’в†’ ПЃt = exp tX
в™¦
is a smooth action of R on M .
Every complete vector п¬Ѓeld gives rise to a smooth action of R on M . Con-
versely, every smooth action of R on M is deп¬Ѓned by a complete vector п¬Ѓeld.
{complete vector п¬Ѓelds on M} в†ђв†’ {smooth actions of R on M }

в€’в†’
X exp tX

dП€t (p)
в†ђв€’
Xp = П€
dt t=0

Let G be a Lie group. Given g в€€ G let
Lg : G в€’в†’ G
a в€’в†’ g В· a
be left multiplication by g. A vector п¬Ѓeld X on G is called left-invariant if
(Lg )в€— X = X for every g в€€ G. (There are similar right notions.)
Let g be the vector space of all left-invariant vector п¬Ѓelds on G. Together
with the Lie bracket [В·, В·] of vector п¬Ѓelds, g forms a Lie algebra, called the Lie
algebra of the Lie group G.
104 APPENDIX B. PREREQUISITES FROM LIE GROUP ACTIONS

Exercise 45
Show that the map
в€’в†’ Te G
g
X в€’в†’ Xe
where e is the identity element in G, is an isomorphism of vector spaces.

Any Lie group G acts on itself by conjugation:
G в€’в†’ Diп¬Ђ(G)
П€g (a) = g В· a В· g в€’1 .
g в€’в†’ П€g ,
The derivative at the identity of
П€g : G в€’в†’ G
a в€’в†’ g В· a В· g в€’1
is an invertible linear map Adg : g в€’в†’ g. Here we identify the Lie algebra g with
the tangent space Te G. Letting g vary, we obtain the adjoint representation (or
adjoint action) of G on g:

Exercise 46
Check for matrix groups that
d
Adexp tX Y = [X, Y ] , в€ЂX, Y в€€ g .
dt t=0

Hint: For a matrix group G (i.e., a subgroup of GL(n; R) for some n), we have
Adg (Y ) = gY g в€’1 , в€Ђg в€€ G , в€ЂY в€€ g
and
[X, Y ] = XY в€’ Y X , в€ЂX, Y в€€ g .

Let В·, В· be the natural pairing between gв€— and g:
В·, В· : gв€— Г— g в€’в†’ R
(Оѕ, X) в€’в†’ Оѕ, X = Оѕ(X) .
Given Оѕ в€€ gв€— , we deп¬Ѓne Adв€— Оѕ by
g

Adв€— Оѕ, X = Оѕ, Adgв€’1 X , for any X в€€ g .
g

g
action) of G on gв€— :
Adв€— : G в€’в†’ GL(gв€— )

We take g в€’1 in the deп¬Ѓnition of Adв€— Оѕ in order to obtain a (left) representation,
g
i.e., a group homomorphism, instead of a вЂњrightвЂќ representation, i.e., a group anti-
homomorphism.
105
B.5. ORBIT SPACES

Exercise 47

B.5 Orbit Spaces
Let П€ : G в†’ Diп¬Ђ(M ) be any action.

Deп¬Ѓnition B.5 The orbit of G through p в€€ M is {П€g (p) | g в€€ G}. The stabi-
lizer (or isotropy) of p в€€ M is the subgroup Gp := {g в€€ G | П€g (p) = p}.

Exercise 48
If q is in the orbit of p, then Gq and Gp are conjugate subgroups.

Deп¬Ѓnition B.6 We say that the action of G on M is . . .

вЂў transitive if there is just one orbit,

вЂў free if all stabilizers are trivial {e},
вЂў locally free if all stabilizers are discrete.

Let в€ј be the orbit equivalence relation; for p, q в€€ M ,

pв€јq в‡ђв‡’ p and q are on the same orbit.

The space of orbits M/ в€ј = M/G is called the orbit space. Let

ПЂ : M в€’в†’ M/G
p в€’в†’ orbit through p

be the point-orbit projection.
Topology of the orbit space:
We equip M/G with the weakest topology for which ПЂ is continuous, i.e.,
U вЉ† M/G is open if and only if ПЂ в€’1 (U) is open in M . This is called the quotient
topology. This topology can be вЂњbad.вЂќ For instance:
Example. Let G = R act on M = R by

t в€’в†’ П€t = multiplication by et .

There are three orbits R+ , Rв€’ and {0}. The point in the three-point orbit space
corresponding to the orbit {0} is not open, so the orbit space with the quotient
в™¦
topology is not Hausdorп¬Ђ.
106 APPENDIX B. PREREQUISITES FROM LIE GROUP ACTIONS

n
Example. Let G = C \{0} act on M = C by

О» в€’в†’ П€О» = multiplication by О» .

The orbits are the punctured complex lines (through non-zero vectors z в€€ Cn ),
plus one вЂњunstableвЂќ orbit through 0, which has a single point. The orbit space is
nв€’1
{point} .
M/G = C P
nв€’1
The quotient topology restricts to the usual topology on C P . The only open set
containing {point} in the quotient topology is the full space. Again the quotient
topology in M/G is not Hausdorп¬Ђ.
However, it suп¬ѓces to remove 0 from Cn to obtain a Hausdorп¬Ђ orbit space:
nв€’1
C P . Then there is also a compact (yet not complex) description of the orbit
space by taking only unit vectors:

C \{0} = S 2nв€’1 /S 1 .
nв€’1
= Cn \{0}
CP

в™¦
Appendix C

Variational Principles

C.1 Principle of Least Action

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.
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

107
108 APPENDIX C. VARIATIONAL PRINCIPLES

In Lecture 4 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 hamiltonian
в™¦
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 gravita-
|x|
в™¦
tional potential.

When we need to deal with systems with constraints, such as the simple pen-
dulum, 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
(t) = в€’
mi (x1 (t), . . . , xn (t)) .
dt2 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 C.1 The action of a path Оі в€€ P is
2
b
mi dОіi
AОі := в€’ V (Оі(t)) dt .
(t)
2 dt
a
109
C.2. VARIATIONAL PROBLEMS

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.

C.2 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.
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 C.2 Suppose that Оі 0 : [a, b] в†’ M is minimizing. Let [a1 , b1 ] be a subin-
terval of [a, b] and let p1 = Оі0 (a1 ), q1 = Оі0 (b1 ). Then Оі0 |[a1 ,b1 ] is minimizing 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.
110 APPENDIX C. VARIATIONAL PRINCIPLES

We 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 with a trivialization
в€‚ в€‚
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).
b
Let AОµ = AОіОµ = a F ОіОµ (t), dОіОµ (t) dt. If Оі0 minimizes A, then
dt

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 fol-
lows 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.
Example. 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. On a coordinate chart (U, x1 , . . . , xn ) on M ,
we have
gij (x)v i v j .
F (x, v) =
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 action
Лњ Лњ dt
of Оі is
2
b b
dОі
в€—
A(Оі) = (Лњ F ) dt =
Оі dt .
dt
a a
111
C.3. SOLVING THE EULER-LAGRANGE EQUATIONS

It is not hard to 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 the
ij
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 ). в™¦

C.3 Solving the Euler-Lagrange Equations
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