Zehnder E. Lectures on Dynamical Systems: Hamiltonian Vector Fields and Symplectic Capacities
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V.1. Symplectic vector spaces 191
The pullback of a k-form ˛ on R
d
under the linear map AW R
d
! R
d
is the
k-form A
˛ defined by
.A
˛/.v
1
;:::;v
k
/ ´ ˛.Av
1
;:::;Av
k
/
for vectors v
1
;:::;v
k
2 R
d
. The map ˛ 7! A
˛ is linear. Moreover,
A
.˛ ^ ˇ/ D .A
˛/ ^ .A
ˇ/:
If the form ˛ 2
V
d
.R
d
/ is of highest degree, then we claim that its pullback
under the map A 2 L.R
d
/ is the multiplication by the determinant, so that A
˛ D
.det A/˛. Indeed, due to dim
V
d
D 1, the form ˛ can be written as ˛ D a.e
1
^^
e
d
/ μ aˇ with a real number a 2 R. In view of the definition of the determinants
(Leibnitz-formula),
A
ˇ.e
1
;:::;e
d
/ D ˇ.Ae
1
;:::;Ae
d
/ D det A:
Since also A
ˇ has to be of form A
ˇ D const. ˇ, it immediately follows that
A
ˇ D .det A/ˇ and in view of the linearity of the pullback we find indeed that
A
˛ D .det A/˛
for every d -form ˛.
We next turn to the proof of Proposition V.4 which is now very easy.
Proof of Proposition V.4 . We introduce the 2n-form ´ !
0
^^!
0
(the product
taken n times). Since this is a 2n-form on R
2n
, it follows for every linear map
A 2 L.R
2n
/ that on one hand
A
D .det A/ ;
and on the other hand, since A 2 Sp.n/ is a symplectic map,
A
D .A
!
0
/ ^^.A
!
0
/
D !
0
^^!
0
D
so that det A D 1, as claimed in Proposition V.4.
Having this proposition we investigate the spectrum of symplectic linear maps.
Proposition V.5 (Spectral proposition for Sp.n/). Let A 2 Sp.n/ and let .A/ be
the spectrum of A.If belongs to the set .A/, then also
1
;
N
and
N
1
belong to
.A/, and all these eigenvalues have the same algebraic multiplicity. If 1 2 .A/
or 1 2 .A/, then these eigenvalues have even multiplicity.
192 Chapter V. Hamiltonian vector fields and symplectic diffeomorphisms
It follows for the dynamics of the map A 2 Sp.n/ that there exist no attractors!
In view of the spectrum of a symplectic map, A 2 Sp.n/ is organized in groups of
eigenvalues and the following cases can be distinguished.
(a) If jj¤1 and … R then ;
1
;
N
;
N
1
are four different eigenvalues A.
(a)
1
C
N
1
N
1
(b) If jj¤1 and 2 R, then there are two different eigenvalues and
1
.
(b)
1
C
1
(c) If jjD1, then
N
D
1
and there are two eigenvalues and
N
on the unit
circle, respectively one eigenvalue in the cases D˙1.
(c)
1
C
N
D
1
For every eigenspace on which A is contracting, there exists another eigenspace
on which A is expanding.
Proof of Proposition V.5 . If A 2 Sp.n/, then A
T
JA D J and hence JAJ
1
D
.A
1
/
T
. Moreover, det A
1
D det A D det J D 1 and using the rules of determi-
nants we compute, for the characteristic polynomial p./ of the map A,
p./ D det.A / D det.JAJ
1
/ D det..A
1
/
T
/ D det.A
1
/
and
det.A
1
/ D det.A
1
.1 A// D det.1 A/ D .1/
2n
2n
det.A
1
/:
Summing up, the characteristic polynomial satisfies the identity
p./ D
2n
p.
1
/:
V.1. Symplectic vector spaces 193
If 2 .A/, then ¤ 0 and p./ D 0 and from the above formula one concludes
p.
1
/ D 0, so that also
1
is an eigenvalue of A.
If k is the multiplicity of
0
2 .A/ then p./ D .
0
/
k
p
1
./ for a
polynomial p
1
satisfying p
1
.
0
/ ¤ 0 and hence
p./ D
2n
p.
1
/
D
2n
.
1
0
/
k
p
1
.
1
/
D
k
0
2nk
.
1
0
/
k
p
1
.
1
/:
One reads off that also the eigenvalue
1
0
has multiplicity k.
Let
1
;
1
1
;:::;
k
;
1
k
be all eigenvalues ¤˙1 of A listed according to their
multiplicities. If l is the multiplicity of 1 2 .A/, then m ´ 2n 2k l is the
multiplicity of 1 2 .A/. Since det A is the product of all 2n eigenvalues of A,we
have
1 D .
1
1
1
/:::.
k
1
k
/.1/
l
;
so that l must be an even integer and consequently also m.
Finally, since the characteristic polynomial has real coefficients, we conclude
that with an eigenvalue also its complex conjugate
N
is an eigenvalue. The proof
of Proposition V.5 is complete.
Definition. If .V
1
;!
1
/ and .V
2
;!
2
/ are two symplectic vector spaces, then a linear
map A 2 L.V
1
;V
2
/ is called symplectic,ifA
!
2
D !
1
, explicitly if
!
2
.Au; Av/ D !
1
.u; v/ for all u; v 2 V
1
:
Remark. Linear symplectic maps are injective. Indeed, if u belongs to the kernel
of A, then
!
1
.u; v/ D A
!
2
.u; v/ D !
2
.Au; Av/ D !
2
.0; Av/ D 0 for all v 2 V
1
;
and therefore u D 0, since !
1
is nondegenerate. In particular, dim V
1
dim V
2
,if
there exists a linear symplectic map A W V
1
! V
2
.
Proposition V.6. If .V
1
;!
1
/ and .V
2
;!
2
/ are two symplectic vector spaces of the
same dimension, then there exists a linear isomorphism A 2 L.V
1
;V
2
/ satisfying
A
!
2
D !
1
.
Such a map A is called a symplectic isomorphism. It can be easily verified that
the inverse A
1
is also symplectic.
Proof of the proposition. If we choose symplectic bases fe
i
;f
j
g in .V
1
;!
1
/ and
fOe
i
;
O
f
j
g in .V
2
;!
2
/ and define A W V
1
! V
2
by Ae
j
DOe
j
and Af
j
D
O
f
j
for all j ,
then it follows that A
!
2
D !
1
.
194 Chapter V. Hamiltonian vector fields and symplectic diffeomorphisms
Corollary V.7. All symplectic vector spaces of the same dimension are symplecti-
cally isomorphic. The only invariant is the dimension!
In view of the proof of Proposition V.2 we can choose e
1
and Oe
1
freely in the
symplectic bases. In the case of V
1
D V
2
we therefore find, in view of the proof
of Proposition V.6 , a symplectic isomorphism A W V
1
! V
1
satisfying Ae
1
DOe
1
.
This proves the following statement.
Proposition V.8. The symplectic group Sp.n/ acts transitively on R
2n
nf0g.
In the same way one sees that Sp.n/ acts transitively on the set of all symplectic
subspaces of R
2n
of fixed dimension 2k. We choose for a symplectic subspace
E a symplectic basis and complete it to a symplectic basis of R
2n
by choosing
a symplectic basis in the symplectic complement E
?
. In the same way one can
proceed with a second symplectic subspace F of the same dimension. By mapping
the corresponding basis vectors onto each other, we haveindeed defined a symplectic
map A 2 Sp.n/ satisfying A.E/ D F .
Literature. A presentation of symplectic algebra can be found, for instance, in
Appendix A in the book [24] by R. Cushman and L. Bates.
V.2 The exterior derivative d
This section introduces Cartan’s formalism of differential forms. Proceeding at first
locally we consider an open subset U R
n
. The spaces
k
.U / of the smooth
differential forms on U are defined in the following way.
(1) The space
0
.U / of the 0-forms on U is the set C
1
.U; R/ of the smooth
functions.
(2) A 1-form on the set U is a map U !
V
1
.R
n
/ D .R
n
/
,
x 7! ˛.x/ D
n
X
j D1
˛
j
.x/dx
j
;˛
j
2 C
1
.U; R/; dx
j
D e
j
:
The space of the 1-forms on U is denoted by
1
.U /.
(3) Analogously, a k-form on U is a map U !
V
k
.R
n
/,
x 7! ˛.x/ D
X
1i
1
<i
2
<<i
k
n
f
i
1
:::i
k
.x/ dx
i
1
^^dx
i
k
;
where f
i
1
:::i
k
2 C
1
.U; R/ are smooth functions. The space of k-forms on
U is denoted by
k
.U /.
V.2. The exterior derivative d 195
Example. If f 2 C
1
.U; R/, then the derivative df is the 1-form
x 7! df .x/ D
n
X
j D1
@f
@x
j
.x/ dx
j
:
By means of the exterior product we can define an algebraic structure on the
space of differential forms. If ˛ 2
k
.U / and ˇ 2
l
.U /, then the product
˛ ^ ˇ 2
kCl
.U / is defined pointwise as
.˛ ^ ˇ/.x/ ´ ˛.x/ ^ ˇ.x/ 2
V
kCl
.R
n
/:
By convention, f ^ ˛ D f˛if ˛ 2
k
.U / and f 2
0
.U /. The product of
differential forms is associative and satisfies
˛ ^ ˇ D .1/
deg ˛ deg ˇ
.ˇ ^ ˛/
where deg ˛ D k if ˛ 2
k
.U /.
Next we define the pullback of forms.IfU R
n
and V R
m
are open sets
and if ' W U ! V is a C
1
-mapping, the pullback
'
W
k
.V / !
k
.U /; k 1
is the linear map defined as
.'
˛/.x/.v
1
;:::;v
k
/ ´ ˛.'.x//
d'.x/v
1
;:::;d'.x/v
k
for all forms ˛ 2
k
.V / and vectors v
j
2 R
k
.Ifk D 0, then '
W
0
.V / !
0
.U / is by definition the composition
.'
f /.x/ D f B '.x/:
The linear map '
is an algebra homomorphism, it satisfies
'
.˛ ^ ˇ/ D .'
˛/ ^ .'
ˇ/:
Next, we turn to the definition of the exterior derivative. It is a sequence of
linear maps
d W
k
.U / !
kC1
.U /; k 0;
which are defined in the following way.
(1) If f 2
0
.U / is a smooth function, then df 2
1
.U / is the ordinary
derivative, so that
df .x/ D
n
X
j D1
@f
@x
j
.x/ dx
j
2 L.R
n
; R/ D .R
n
/
:
196 Chapter V. Hamiltonian vector fields and symplectic diffeomorphisms
(2) If ˛ 2
k
.U / for an integer k 1 is represented by
˛.x/ D
X
1i
1
<i
2
<<i
k
n
f
i
1
:::i
k
.x/ dx
i
1
^^dx
i
k
with functions f
i
1
:::i
k
2 C
1
.U; R/; then we define the form d˛ 2
kC1
.U /
by
d˛.x/ D
X
1i
1
<i
2
<<i
k
n
df
i
1
:::i
k
.x/ ^ dx
i
1
^^dx
i
k
D
X
1i
1
<i
2
<<i
k
n
n
X
j D1
@f
@x
j
.x/ dx
j
^ dx
i
1
^^dx
i
k
„ ƒ‚ …
2
V
kC1
.R
n
/
:
Proposition V.9. d
2
D 0, where d
2
denotes the composition d B d W
k
.U / 7!
kC2
.U /.
Proof [Proposition of Schwarz]. Since d is a linear map, it suffices to look at the
k-form
˛ D fdx
i
1
^^dx
i
k
D fdx
I
in
k
.U /, where we have abbreviated dx
I
´ dx
i
1
^^dx
i
k
. By definition of
the exterior derivative,
d˛.x/ D
n
X
j D1
@f
@x
j
.x/ dx
j
^ dx
i
1
^^dx
i
k
D
n
X
j D1
@f
@x
j
.x/ dx
j
^ dx
I
;
and hence
d.d˛/.x/ D
n
X
sD1
n
X
j D1
@
2
f
@x
j
@x
s
.x/ dx
s
^ dx
j
^ dx
I
:
In view of
@
2
f
@x
j
@x
s
.x/ D
@
2
f
@x
s
@x
j
.x/ and the skew symmetry dx
j
^dx
s
Ddx
s
^dx
j
the terms cancel each other so that d.d˛/.x/ D 0, as claimed.
Proposition V.10. If ˛ 2
k
.U / and ˇ 2
l
.U /, then
d.˛ ^ ˇ/ D .d˛/ ^ ˇ C .1/
k
˛ ^ .dˇ/:
Proof [Product rule]. To abbreviate, we set dx
I
D dx
i
1
^^dx
i
k
and dx
J
D
dx
j
1
^^dx
j
l
. Due to the linearity of the exterior derivative d , it suffices to
consider the two forms ˛ D fdx
I
2
k
.U / and ˇ D gdx
J
2
l
.U /. In view of
V.2. The exterior derivative d 197
the definitions,
˛ ^ ˇ D .f g/dx
I
^ dx
J
;
d.˛ ^ ˇ/ D d.f g/dx
I
^ dx
J
;
.d˛/ ^ ˇ D gdf ^ dx
I
^ dx
J
:
Due to dg ^ dx
j
Ddx
j
^ dg,
˛ ^ .dˇ/ D fdx
I
^ dg ^dx
J
D .1/
k
fdg ^ dx
I
^ dx
J
and using the product rule d.f g/ D fdgC gdf the proposition follows.
Proposition V.11. If ' W U R
n
! V R
m
is a C
1
-map, then the pullback of
' commutes with the exterior derivative d , that is,
'
B d D d B '
:
Explicitly, if ˛ 2
k
.V /, then '
.d˛/ D d.'
˛/.
Proof [Chain rule, Proposition V. 9 , Proposition V.10]. (1) We first consider the spe-
cial case in which ˛ D f is an element of
0
.U /. Then in view of the definition
of the pullback of 1-forms, the chain rule and the definition of the pullback of
functions, we obtain
'
.df /.x/ D df .'.x//d'.x/
D d.f B '/.x/
D d.'
f /.x/:
(2) Next we consider the 1-form ˛ D df 2
1
.U /. In view of Proposition V. 9 ,
we have '
.d˛/ D '
.d
2
f/ D '
.0/ D 0. On the other hand, by step (1), we
obtain d.'
df / D d.d.'
f//D d
2
.'
f/and hence also d.'
˛/ D 0, again in
view of Proposition V.9 .
(3) The claim of the general case now follows by induction using Proposi-
tion V.10 and using that the maps d and '
are linear and using the formula
'
.˛ ^ ˇ/ D .'
˛/ ^ .'
ˇ/.
The following proposition is a consequence of the definition of the exterior
derivative.
Proposition V.12. The exterior derivative d is a local operator; if U
0
U are
open sets, then .d˛/j
U
0
D d.˛jU
0
/. Therefore, the value d˛.x/ only depends on
an arbitrarily small neighborhood of x.
198 Chapter V. Hamiltonian vector fields and symplectic diffeomorphisms
Next we introduce the interior product i
X
˛ of a vector field X with a k-form
˛.IfX W U R
n
! R
n
is a C
1
-vector field, then the linear map
i
X
W
k
.U / !
k1
.U /
is defined in the following way. If f 2
0
.U /, then we set i
X
f D 0. For the
k-form ˛ 2
k
.U / of degree k 1 we define
.i
X
˛/.x/
v
1
;:::;v
k1
´ ˛.x/
X.x/; v
1
;:::;v
k1
;v
j
2 R
n
:
We collect some of the properties of the interior product in the following propo-
sition whose proof is a simple exercise left to the reader.
Proposition V.13. The interior product i
X
has the following properties:
(i) i
XCY
D i
X
C i
Y
,
(ii) i
X
B i
Y
Di
Y
B i
X
,
(iii) i
X
B i
X
D 0,
(iv) i
X
.˛ ^ˇ/ D .i
X
˛/ ^ ˇ C .1/
k
˛ ^.i
X
ˇ/ for ˛ 2
k
.U / and ˇ 2
l
.U /.
V.3 The Lie derivative L
X
of forms
Let X W U R
n
! R
n
be a C
1
-vector field on the open set U and let '
t
be the
flow of X satisfying
´
d
dt
'
t
.x/ D X.'
t
.x//; t 2 I
x
;
'
0
.x/ D x; x 2 U;
where I
x
is an open interval around 0. The Lie derivative L
X
of forms is the linear
map
L
X
W
k
.U / !
k
.U /; k 0
defined by
.L
X
˛/.x/ ´
d
dt
.'
t
/
˛.x/
tD0
:
From '
tCs
D '
t
B '
s
D '
s
B '
t
we deduce
d
dt
.'
t
/
˛ D .'
t
/
d
ds
.'
s
/
˛
ˇ
ˇ
ˇ
sD0
D .'
t
/
.L
X
˛/:
V.3. The Lie derivative L
X
of forms 199
In order to illustrate the Lie derivative in the special case k D 0 we take the smooth
function f 2
0
.U / and compute using the chain rule
L
X
f.x/ D
d
dt
.'
t
/
f.x/
ˇ
ˇ
ˇ
tD0
D
d
dt
f B '
t
.x/
ˇ
ˇ
ˇ
tD0
D df B '
t
.x/
d
dt
'
t
.x/
ˇ
ˇ
ˇ
tD0
D df .x/ X.x/
so that
L
X
f D i
X
.df /:
The following formula for forms of all degrees will prove extremely useful later on.
Proposition V.14 (Cartan’s formula). The Lie derivative satisfies
L
X
D i
X
B d C d B i
X
:
Explicitly, L
X
˛ D i
X
.d˛/ C d.i
X
˛/ for all differential forms ˛ 2
.U /.
Proof [Propositions V. 9 , V.10, V.11 and V.13]. Since the exterior product ˛ ^ ˇ is
bilinear, the Lie derivative satisfies the product rule
L
X
.˛ ^ ˇ/ D .L
X
˛/ ^ ˇ C ˛ ^ .L
X
ˇ/:
The linear map D ´ i
X
B d C d B i
X
satisfies, due to Proposition V.13 and
Proposition V.10, the same product rule
D.˛ ^ ˇ/ D .D˛/ ^ ˇ C ˛ ^ .Dˇ/:
In view of the linearity of the maps under consideration it therefore suffices to verify
the desired formula L
X
D D for the forms ˛ D f 2
0
.U / and ˛ D df 2
1
.U /.
(1) If ˛ D f 2
0
.U / is a smooth function, then we obtain in view of the
previous calculation and using i
X
f D 0,
L
X
f D i
X
df D i
X
df C d.i
X
f/D .i
X
B d C d B i
X
/f;
as desired.
(2) If ˛ D df 2
1
.U / we compute using the formula d B '
D '
B d
200 Chapter V. Hamiltonian vector fields and symplectic diffeomorphisms
(Proposition V.11),
L
X
df .x/ D
d
dt
.'
t
/
df .x/
ˇ
ˇ
ˇ
tD0
D
d
dt
d Œ.'
t
/
f .x/
ˇ
ˇ
ˇ
tD0
(/
D d
d
dt
.'
t
/
f
ˇ
ˇ
ˇ
tD0
.x/
D dŒi
X
df .x/
V.9
D .d B i
X
C i
X
B d /df .x/I
in step ./ we have used the linearity of d . In the last step we made use of d
2
D 0.
The proof of the proposition is complete.
The differential form ˛ 2
k
.U / is called closed if d˛ D 0 and is called exact
on U if ˛ D dˇ for a differential form ˇ in
k1
.U /. In view of d
2
D 0, an exact
form is necessarily a closed form. Locally, the converse also holds true, as the next
result shows.
Theorem V.15 (Poincaré lemma). If V R
n
is a starlike open set, then a k-form
on V is closed, precisely if it is exact. For all ˛ 2
k
.V / of degree k 1 we have
d˛ D 0 () ˛ D dˇ for some ˇ 2
k1
.V /:
Proof [Cartan’s formula, Proposition V. 9 ]. As already pointed out, an exact form
is closed in view of d
2
D 0. In order to prove that on V a closed form is exact we
make use of the starlike character of the open set. Without loss of generality we
may assume (by translation) that V is starlike with respect to the origin 0 in R
n
,so
that if x 2 V , then also x 2 V for every 2 Œ0; 1. We introduce the linear vector
field X.x/ D x for some real number >0. Its flow is equal to '
t
.x/ D e
t
x
and therefore leaves the open set V invariant for all t 0. The linearized flow is
equal to d'
t
.x/v D e
t
v for all v 2 R
n
. Hence '
t
.x/ ! 0 and d'
t
.x/v ! 0
exponentially as t !1. This allows us to define the linear map
H W
k
.V / !
k1
.V /
by
H˛ ´ i
X
Z
0
1
.'
t
/
˛dt
:
In view of Cartan’s formula L
X
D i
X
B d C d B i
X
we obtain
d.H˛/ D L
X
Z
0
1
.'
t
/
˛dt
i
X
B d
Z
0
1
.'
t
/
˛dt
: