Table Of ContentSYMPLECTIC VECTOR SPACES, LAGRANGIAN SUBSPACES, AND
LIOUVILLE’S THEOREM
CONNER JAGER
Celestial Mechanics Junior Seminar - Nicolas Templier
Abstract. The purpose of this paper is to prove Liouville’s theorem on volume-preserving flows
from a symplectic linear algebra point of view. Along the way, we will prove a series of properties
about symplectic vector spaces that elucidate the usefulness of this approach with regard to the
correspondence between the configuration space and the symplectic phase space. Though only
briefly discussed, this paper motivates how symplectic manifolds provide the natural abstraction
of the phase space of a dynamical system through the simplified linear example: symplectic vector
spaces.
0. Motivation
In the study of celestial mechanics, there arises a natural connection between symplectic manifolds
andthephasespaceofadynamicalsystem. Intuitively, wemaythinkofthetwodefiningproperties
of symplectic manifolds as follows: (1) each point is associated with a vector in the tangent space
of some embedded manifold and a covector in the cotangent space, and (2) it is equipped with a
2-form satisfying certain “natural” constraints. Indeed, in classical mechanics, we generate systems
of objects under force that are naturally symplectic in nature: (1) each object is associated with
a state consisting of a position in some embedded configuration space and momentum, and (2) we
have a set of laws governing how the forces act upon the object (the Hamiltonian function). In
fact, as pointed out in Cohn’s article “Why symplectic geometry is the natural setting for classical
mechanics,” symplectic manifolds provide precisely this desired abstraction of phase spaces.
A fundamental example of this intimate connection is the classic n-body problem, where we seek to
determinetheorbitsofnobjectsinR3 bytheirmutualgravitationalinteractions. Byidentifyingthe
initialphasespace,thatis,identifyingavectorinthesymplecticvectorspaceR6n characterizingthe
initial position and momentum for each of the n bodies, and determining the energy (Hamiltonian)
function, can construct a 6 dimensional phase flow that fully characterizes how the phase space (ie.
position and momentum) evolves through time [C, A]. Understanding the properties of symplectic
manifolds provides a clean way to analyze these dynamical systems.
Though the manifold under which the system of study operates need not be Euclidean, as the n-
bodyproblemelucidates,veryoftenthesysteminvolvesthemotionofparticlesthroughthreespace,
and thus the manifold is Euclidean [C]. As such, this paper will limit its scope to symplectic vector
spaces. This is a very strong simplifying assumption since, as we will later prove, every symplectic
vector space is isomorphic to the R2n and therefore is characterized fully by its dimension.
1
2 CONNERJAGER
1. Symplectic Vector Spaces
It is, of course, important to first establish some basic definitions and principles concerning sym-
plectic vector spaces.
Definition 1.1. A symplectic vector space is a (finite-dimensional) vector space V equipped with
a bilinear form ω (called the symplectic form of V) that is
antisymmetric: ω(u,v) = ω(v,u)
• −
non-degenerate: if u = 0, then there exists a vector v such that ω(u,v) = 0.
•
If we also assume the characteristic of the field of V is = 2, we can deduce that ω(v,v) = 0 for all
vectors v V from the antisymmetry of ω. For the purpose of this paper, we will assume that the
∈
field of V is R. [HZ]
Example 1.1. The canonical example of a symplectic vector space is R2n with the bilinear form
ω (u,v) = Ju,v , where , is the Euclidean inner product (ie. dot product) and J is the 2n 2n
0
· · ×
matrix:
0 I
n
J =
In 0
−
We can easily verify that ω is antisymmetric and non-degenerate:
0
Let u = (u ,...,u ),v = (v ,...,v ).
1 2n 1 2n
•
ω (u,v) = u v + +u v u v u v
0 n+1 1 2n n 1 n+1 n 2n
··· − −···−
ω (v,u) = v u + +v u v u v u = ω (u,v)
0 n+1 1 2n n 1 n+1 n 2n 0
··· − −···− −
Let u = 0 = u = (u ,...,u ) where u = 0 for some i. Thus, ω (u,e ) = u = 0 =
1 2n i 0 i i
• ⇒ | | ⇒
ω (u,e ) = 0 for the standard basis vector e .
0 i i
Definition 1.2. The symplectic orthogonal complement of some subspace E V is defined as
⊆
E := v V : ω(u,v) = 0, u E
⊥
{ ∈ ∀ ∈ }
It is critical to note that in general, E and E need not be complements and E need not itself
⊥ ⊥
be a symplectic vector space. As a counterexample, consider when E = span(v) for some nonzero
vector v V. For any x,y E,x = c v,y = c v and thus
1 2
∈ ∈
ω(x,y) = c c ω(v,v) = 0
1 2
·
(In this case, E E and ω is degenerate on E.)
⊥
⊆
From these observations, we will define special classes of subspaces of a symplectic vector space.
[L]
SYMPLECTIC VECTOR SPACES, LAGRANGIAN SUBSPACES, AND LIOUVILLE’S THEOREM 3
Definition 1.3. Let V be a symplectic vector space. A subspace E V is called
⊆
isotropic if E E
⊥
• ⊆
Lagrangian if E = E
⊥
•
symplectic if ω = 0
E
• |
We will now prove the following theorem that allows gives us an isomorphism between any sym-
plecticvectorspaceandthecanonicalsymplecticspace(R2n,ω0). Wewillmakeuseofthefollowing
propositions. The proofs of these propositions are straightforward, and they are left to the reader
as an exercise (see [L]).
Proposition 1.1. dimV = dimE +dimE
⊥
Proposition 1.2. (E ) = E
⊥ ⊥
Proposition 1.3. Let E V. The following conditions are equivalent:
⊆
(1) E is symplectic
(2) E is symplectic
⊥
(3) E E = 0
⊥
∩ { }
(4) V = E E
⊥
⊕
Theorem 1.1. Given a symplectic vector space (V,ω), there exists a basis e ,...,e ,f ,...,f of V
1 n 1 n
such that
ω(e ,e ) = 0
i j
ω(f ,f ) = 0
i j
ω(e ,f ) = δ
i j ij
Proof. Choose any vector e V. By the non-degeneracy of ω, there exists some vector f V such
∈ ∈
that ω(e,f ) = c = 0. Using the linearity of ω in the second component, we can normalize f to
f = f /c such that ω(e,f) = 1. Suppose e and f were linearly dependent. We could then write
f = λe, but then ω(e,f) = λω(e,e) = 0, a contradiction. Thus, e and f span a subspace E V
⊆
of dimension 2. By our construction, E is symplectic: every non-zero vector v E has a non-zero
∈
component of either e or f, so ω applied to v and f or e, respectively, is non-zero and thus ω is
E
|
non-degenerate. If dimV = 2, we are done; if not, we use Proposition 1.3 and induction to generate
the desired basis on E . Since V = E E and ω(v,w) = 0 for v E,w E , this completes
⊥ ⊥ ⊥
⊕ ∈ ∈
the proof.
We leave it to the reader to show that the desired conditions on the basis hold, and that the matrix
representation of ω with respect to this basis is J. One important consequence of Theorem 1.1 is
that the symplectic vector spaces have even dimension. [HZ]
4 CONNERJAGER
2. Lagrangian Subspaces
The goal of this section is to highlight the importance of Lagrangian subspaces. As we will see,
we can think of the configuration space as an embedded subspace of the symplectic phase space.
This embedded space is Lagrangian. The following theorem, which shows how we can generate a
symplectic vector space from a given vector space, proves this point.
Theorem 2.1. For any vector space V with dual V , the vector space V V is symplectic, with
∗ ∗
⊕
symplectic form
ω((v ,f ),(v ,f )) = f (v ) f (v )
1 1 2 2 2 1 1 2
−
Proof. It suffices to show that ω is non-degenerate and skew-symmetric since the bilinearity is
inherited from the linearity of f ,f . Choose any non-zero (v,f) V V . If v = 0, then there
1 2 ∗
∈ ⊕
must be some linear functional g V such that g(v) = 0. The vector (0,g) V V satisfies the
∗ ∗
∈ ∈ ⊕
non-degeneracy condition:
ω((v,f),(g,0)) = g(f) f(0) = g(f) = 0
−
If v = 0, then f = 0 = f(w) = 0 for some w V. The vector (w,0) V V satisfies the
∗
⇒ ∈ ∈ ⊕
non-degeneracy condition:
ω((v,f),(w,0)) = 0 f(w) = f(w) = 0
− −
Finally, we note that
ω((v ,f ),(v ,f )) = f (v ) f (v ) = ω((v ,f ),(v ,f ))
2 2 1 1 1 2 2 1 1 1 2 2
− −
Corollary 2.1. E = V 0 and E = 0 V are Lagrangian subspaces of the symplectic vector
⊕ ∗n n ⊕ ∗
space V V = E E .
∗
⊕ ⊕
Proof. ( ) Let ω be the symplectic form on V V as defined above. Choose any v V.
∗
⊆ ⊕ ∈
ω((v,0),(v,0)) = 0(v) 0(v) = 0 = (v,0) E
⊥
− ⇒ ∈
( ) Let (w,f) E . For any v V,
⊥
⊇ ∈ ∈
ω((w,f),(v,0)) = 0(w) f(v) = 0 = f(v) = 0 v V = f = 0 = (w,f) E
− ⇒ ∀ ∈ ⇒ ⇒ ∈
Thus, E = E⊥. A similar argument shows E is also Lagrangian.
Corollary 2.1 tells us that not only can we generate a symplectic vector space from a generic vector
space, but the generating vector space will be embedded in the symplectic space as a Lagrangian
subspace. [L]
We will now prove the converse.
Theorem 2.2. Any symplectic vector space (V,ω) contains a Lagrangian subspace E V such
⊆
that V = E E .
∗
⊕
SYMPLECTIC VECTOR SPACES, LAGRANGIAN SUBSPACES, AND LIOUVILLE’S THEOREM 5
Proof. Choose a basis e ,...,e ,f ,...,f of V as in Theorem 1.1. Let E = span(e ,...,e ) and
1 n 1 n 1 n
F = span(f ,...,f ). By construction, E F = V. We must show that E is Lagrangian and that
1 n
⊕
E = F.
∗
Let v = a e + +a e E.
1 1 n n
··· ∈
ω(v,v) = a2ω(e ,e )+ +a2ω(e ,e ) = 0 = E E
1 1 1 ··· n n n ⇒ ⊆ ⊥
Since dimE = dimV dimE = 2n n = n = dimE = E = E .
⊥ ⊥
− − ⇒
Now, we can chose a basis e ,...,e for E such that e (e ) = δ . Noting that ω( ,f ) E , we
∗1 ∗n ∗ ∗i j ij · i ∈ ∗
write
ω( ,f ) = b e + +b e
· i 1 ∗1 ··· n ∗n
ω(e ,f ) = δ = e (e ) = f = ω( ,f ) = e
j i ij ∗i j ⇒ i · i ∗i
Thus, we conclude that f1,...,fn is a basis for E∗ = F = E∗.
⇒
3. The Symplectic Group
Definition 3.1. Let (V,ω) be a symplectic vector space. A linear map T : V V is called a
→
symplectic transformation if
T ω(u,v) := ω(Tu,Tv) = ω(u,v)
◦
As with any linear transformation, we can describe T as a matrix with entries characterizing the
action of T on some basis for V. Such a matrix is called a symplectic matrix.
Using the basis from Theorem 1.1, we can alternatively characterize a symplectic matrix as satis-
fying the property:
ω (Tu,Tv) = J(Tu),Tv = (Tv)tJ(Tu) = vt(TtJT)u = vtJu = ω (u,v)
0 0
= TtJT = J
⇒
Taking the determinant of both sides:
det(TtJT) = det(T)2det(J) = det(T) = 1
⇒ ±
In fact, det(T) = 1 for all symplectic matrices T. To prove this, we will briefly digress to discuss
skew-symmetric matrices.
If A is an n n skew-symmetric matrix (ie. At = A) with n odd, then
× −
det(A) = det(At) = det( A) = ( 1)ndet(A) = det(A) = det(A) = 0
− − − ⇒
However,ifniseven,thedeterminantbecomesmuchmoreinteresting; namely,thedeterminantcan
be written as the square of a polynomial of it’s entries (Cayley, Muir). We define this polynomial
as the Pfaffian. [M]
6 CONNERJAGER
Definition 3.2. If A is an even-dimensional skew-symmetric matrix, the Pfaffian of A is defined
as
n
1
pf(A) = sgn(σ) a
2nn! σ(2i−1),σ(2i)
σ∈S2n i=1
As a consequence of this definition, it follows that pf2(A) = det(A). Another important identity of
the Pfaffian is that for any matrix B, pf(BtAB) = det(B)pf(A) (the proof of this identity can be
verified using the explicit Pfaffian definition, or by using exterior algebra). [M]
Theorem 3.1. If T is a symplectic matrix, det(T) = 1.
Proof. Since J is an even-dimensional (Theorem 1.1) skew-symmetric matrix, pf(J) > 0 and there-
fore
pf(J) = pf(TtJT) = det(T)pf(J) = det(T) = 1
⇒
Theorem 3.1 provides us with an important characterization of symplectic matrices; namely, that
the set of 2n 2n symplectic matrices are a subset of the special linear group:
×
SL2n(R) := all 2n 2n matrices M : det(M) = 1
{ × }
In fact, this subset actually forms a subgroup under matrix multiplication, denoted Sp(n), and is
called the symplectic group. This can be verified as follows:
identity: I Sp(n) since It JI = J trivially.
• 2n ∈ 2n 2n
closure: Let T,U Sp(n).
• ∈
(TU)tJ(TU) = Ut(TtJT)U = UtJU = J = TU Sp(n)
⇒ ∈
inverses: Let T Sp(n). Noting J 1 = Jt = J,
−
• ∈ −
J = TtJT = JT 1 = TtJ = T 1 = J 1TtJ
− − −
⇒ ⇒
(T 1)tJT 1 = (JtTJ)J(J 1TtJ) = Jt(TJTt)J = JtJJ = J
− − −
so T 1 is symplectic.
−
A further characterization of symplectic matrices is available in the Appendix.
4. Proving Liouville’s Theorem
We are now equipped to prove Liouville’s Theorem in the context of symplectic vector spaces. Let
R6n be the phase space of generalized momentums and positions, ie.
z = (p,q) R6n
∈
where p and q are the generalized momentum and position, respectively, of the phase state z. With
the canonical symplectic form, this phase space is a symplectic vector space. The identifications
SYMPLECTIC VECTOR SPACES, LAGRANGIAN SUBSPACES, AND LIOUVILLE’S THEOREM 7
proved in Section 2 also shows us that the configuration space R3n of all possible generalized
positions q R3n is an embedded Lagrangian subspace of the phase space. Given a Hamiltonian
∈
function H : R6n R, we define the Hamiltonian vector field Φ : R6n R6n as the map
→ →
∂H ∂H
(p,q) ,
→ − ∂q ∂p
We can equivalently write Φ(z) = J H(z) where J is our matrix as defined in Example 1.1.
− ∇
[L, F]
Definition 4.1. We define the phase flow as the group of transformations
g : (p(0),q(0)) (p(t),q(t))
t
→
with (p(t),q(t)) satisfying the Hamiltonian equations. This constraint on (p(t),q(t)) means
∂g
t
=Φ( g )
t
∂t
In other words, the phase flow must be consistent with the Hamiltonian vector field. [A]
Let G be the Jacobian matrix associated with g :
t t
∂g
t
G :=
t
∂z
Theorem 4.1. G is a symplectic matrix; that is,
t
GTJG = J
t t
for all t R.
∈
Proof. Let Xt = GTt JGt. We want to show that Xt = J for all t ∈ R. First, we note that since g0
is the identity map, G = I and X = J.
0 0
Now, consider the derivative of X with respect to time. Using the product rule, we compute
t
X˙ = G˙TJG +GTJG˙
t t t t t
Using Young’s Theorem (1), the phase flow constraint (2), and the chain rule (3):
∂ ∂g ∂ ∂g ∂ ∂ ∂ ∂
G˙ = t =1 t =2 Φ(g ) = ( J H(g )) =3 J H(g ) g = JH G
t t t t t zz t
∂t ∂z ∂z ∂t ∂z ∂z − ∇ − ∂z∇ · ∂z −
where H is the Hessian matrix of second derivatives of H with respect to z. It is important to
zz
note that H is symmetric (as a consequence of Young’s Theorem):
zz
∂2H ∂2H
(H ) = = = (H ) for all i,j
zz i,j zz j,i
∂z ∂z ∂z ∂z
i j i j
Thus,
G˙T = GTHT ( J)T = GTH J
t t zz − t zz
and therefore
X˙ = GTH JJG +GTJ( J)H G = GTH G +GTH G = 0
t t zz t t − zz t − t zz t t zz t
This means Xt is constant, and therefore Xt = X0 = J for all t R. [F, FKM, A]
∈
8 CONNERJAGER
EquippedwithTheorem5.1,Liousville’sTheoremremarkablyfollowsasasimplecorollary. [FKM]
Corollary 4.1. (Liousville’s Theorem) The phase flow gt is volume preserving.
Proof. Choose any region D R6n. By the change of variable theorem in multivariable calculus
⊂
(*),
∂g
t 3.1
volume(gtD) = dz =∗ det dz = detGt dz = dz = volume(D)
∂z | |
gtD D D D
Another interesting consequence is that g forms a group. [A]
t
Corollary 4.2. G = gt : t R is a group under function composition.
{ ∈ }
Proof.
closure: Let g ,g G. For any (p(0),q(0)) V,
t s
• ∈ ∈
g g (p(0),q(0)) = g (p(s),q(s)) = (p(s+t),q(s+t)) = g (p(0),q(0))
t s t t+s
= g g = g G
t s t+s
⇒ ∈
identity: g g = g g = g = g = id = g
0 t t 0 0+t t 0
• ⇒
inverses: g g = g g = g = g 1g = g G
t t t t 0 − t t
• − − ⇒ − ∈
associativity: function composition is associative
•
One can easily show that G˜ = Gt : t R is also a group, and the Jacobian is a group homomor-
{ ∈ }
phism between G and G˜. Also, G˜ is a subgroup of the symplectic group Sp by Theorem 5.1, so the
map
∂
G ∂z Sp
−→
is a symplectic group representation of G.
5. Extensions
While symplectic vector spaces provide a clean way to analyze dynamical systems in Euclidean
space, the restriction to Euclidean space is not always sufficient. The complete abstraction of a
phase space over an arbitrary manifold requires symplectic manifolds. For further reference of
how the properties of symplectic vector spaces and their connection with the phase space may be
generalized to symplectic manifolds, consult [A].
SYMPLECTIC VECTOR SPACES, LAGRANGIAN SUBSPACES, AND LIOUVILLE’S THEOREM 9
6. Appendix: Characterizing Symplectic Matrices
Consider an arbitrary, 2n 2n matrix A. For A to be symplectic, it must satisfy the matrix
×
equation:
AtJA = J
Since there are 4n2 entries in A, this matrix equation represents a system of 4n2 equations.
6.1. 2x2 Symplectic Matrices. Let
a b
A =
c d
We set up our system of four equations as:
a c 0 1 a b 0 1
=
b d 1 0c d 1 0
− −
ac ac = 0
−
ac ac ad bc 0 1 ad cb = 1
− − = −
bc ad bd bd 1 0 ⇔ bc ad = 1
− − − − −
bd bd = 0
−
reduce to the condition that ad cb = det(A) = 1 = Sp(1) = SL(2).
− ⇒
6.2. 4x4 Symplectic Matrices. Let
a b c d
e f g h
A =
i j k l
m n o p
This time, we will set up sixteen equations:
a e i m 0 0 1 0 a b c d
b f j n 0 0 0 1 e f g h
AtJA =
c g k o 1 0 0 0 i j k l
−
d h l p 0 1 0 0m n o p
−
0 bi+aj fm+en ci+ak gm+eo di+al hm+ep
− − − − − −
bi aj +fm en 0 cj +bk gn+fo dj +bl hn+fp
= − − − − − −
ci ak+gm eo cj bk+gn fo 0 dk+cl ho+gp
− − − − − −
di al+hm ep dj bl+hn fp dk cl+ho gp 0
− − − − − −
0 0 1 0
0 0 0 1
=
1 0 0 0
−
0 1 0 0
−
10 CONNERJAGER
A quick inspection of the matrix AtJA is skew-symmetric, which means the off-diagonal equations
are redundant. Similarly, the diagonal is already satisfied. This leaves us with (16 4)/2 = 6
−
equations, which are, as it turns out, independent (a proof of this requires differential geometry
beyond the scope of this talk). This means that there are 16 6 = 10 free variables, and Sp(2)
−
forms a 10 dimensional submanifold of R16.
6.3. Higher Dimensional Symplectic Matrices. Though the characterization of 2x2 symplec-
tic matrices is clear, as the 4x4 example shows, higher dimensional symplectic matrices do not have
such an obvious form. In these cases, it is more helpful to consider A in n n block form:
×
X Y
A =
Z W
Xt Zt 0 I X Y Xt Zt Z W
AtJA = n =
Yt Wt In 0Z W Yt Wt X Y
− − −
XtZ ZtX XtW ZtY 0 I
n
= − − =
YtZ WtX YtW WtY In 0
− − −
Thus, the conditions to be symplectic reduce to
(1) XtZ = ZtX = (XtZ)t = XtZ is symmetric
⇒
(2) YtW = WtY = (YtW)t = YtW is symmetric
⇒
(3) XtW ZtY = I
n
−
We can eliminate the fourth equation by noting
WtX YtZ = I (WtX YtZ)t = It XtW ZtY = I
− n ⇔ − n ⇔ − n
Now we can count the number of non-redundant equations obtained. Conditions (1) and (2) put
constraints on the off-diagonal entries only, which results in 1n(n 1) equations each. Condition
2 −
(3) constrains all n2 entries of XtW ZtY. Thus, there are at total of
−
1
n2+2 n(n 1) = 2n2 n
2 − −
equations that characterize the symplectic matrix. Again, we state without proof that these equa-
tions are independent. Therefore, the symplectic group Sp(n) forms a (2n2+n)-dimensional sub-
manifold of R4n2.
