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§3. Reduction of a Hamiltonian System with Symmetries 81
The tangent space at A is given by
T
ψ−1(A)={[B, A] | B ∈ A}.
A
The differential form is given by
ω([B
,A], [B2,A]) = tr(A[B1,B2]).
1
One verifies that this form is nondegenerate, skew symmetric and closed, thus
defining a symplectic structure on O(A).
Consider any two group actions ϕ
: M → M and ϕg: M → M of G.One
g
calls a map τ: M → M equivariant if
τϕ
= ϕgτ.
g
If ϕ
is an exact symplectic group action then the moment map ψ: M →A∗is
g
equivariant with respect to ϕ
(This theorem holds under more general assumptions, that is not only if ϕ
: M → M and the coadjoint representation of A.
g
g
exact symplectic.)
Thus the image of {ϕ
(p) | g ∈ G} for fixed p ∈ M under the moment
g
map ψ is given by the orbit
∗
{Ad
(g)μ | g ∈ G} where μ = ψ(p).
is
The isotropy group G
of μ is defined as the subgroup
μ
G
= {g ∈ G | Ad∗(g)μ = μ}.
μ
of G.
e) The reduced phase space. Given an exact symplectic group action
: M → M we construct the moment map ψ : A→A∗and for a fixed
ϕ
g
∗
μ ∈A
the set
−1
(μ)={p ∈ M | ψ(p)=μ}.
ψ
This is the subset of M of fixed values of the integral. In the example 2 it
consists of those states for which the angular momentum vector is fixed. On this
set the group G
acts, which in this example consists of all rotations leaving the
μ
angular momentum vector μ fixed. If μ =0this is a one-dimensional rotation
group SO(2). To eliminate this angle of rotation we consider the quotient
−1
set ψ
(μ)/Gμwhich corresponds to elimination of the «ignorable» angle of
rotation.
82 Various Aspects of Integrable Hamiltonian Systems
Under appropriate assumptions (e. g. that ψ−1(μ) is a manifold, Gμis
compact, that G
acts on ψ−1(μ) fixed point free) we introduce the «reduced
μ
phase-space»
M = ψ
Thus M is the base space of the bundle π : ψ
(μ)/Gμ.
−1
(μ) → M.
−1
This reduced phase space M is a symplectic manifold again and the sym-
plectic structure is given by a two form ω which is defined as follows: If
−1
j : ψ
to ψ
(μ) → M is the injection map then j∗ω denotes the pullback of ω
−1
(μ). This form is invariant under Gμand therefore defines ω via
∗
π
ω = j∗ω.
In other words, for V
Moreover, if H is a Hamiltonian invariant under ϕ
, V2∈ Tψ−1(μ) letVk=(dπ)Vk,then
1
ω(V
,V2)=ω(V1,V2).
1
then the «reduced flow»
g
is given by a HamiltonianH defined by
H ◦π = H ◦j.
In other words H ◦ j, the restriction of H to ψ
−1
(μ) is invariant under Gμand
therefore defines a functionH on M.
We summarize: If H = H (ϕ
) is invariant under the group action ϕ
g
then (under appropriate assumptions) the phase space can be reduced to M =
−1
= ψ
(μ)/Gμwhich is a symplectic manifold and the Hamiltonian reduces toH
defined by restriction.
We do not go into the precise specification of the assumptions since their
validity can be easily established in the examples which we discuss.
f) Examples. To illustrate this reduction one may consider the Examples
1 and 2, which we leave to the reader.
XAMPLE 3. Consider the group action of G = R
E
(x, y) → (x + sy, y)
2n
on R
, which preserves θ = y, dx. It has the Hamiltonian
g
ψ =
1
2
|x|
.
2
§3. Reduction of a Hamiltonian System with Symmetries 83
Taking μ =
is G = R. To describe ψ
1
the manifold ψ−1(μ) is the unit sphere and the isotropy group
2
−1
(μ)/G we choose the point of minimal distance in
the line x + sy to lie at s =0, i. e. we choose x, y =0. Thus
−1
(μ)/G {(x, y) ∈ R2n, |x| =1, x, y =0}
ψ
which is the cotangent bundle T
∗Sn−1
one form y, dx projects into the natural 1-form of T
of the unit sphere. One verifies that the
∗Sn−1
.
As an example we consider the Hamiltonian
1
2
(|x|
H =
|y|2−x, y2)(∗)
2
which is invariant under the above group action. The corresponding vector field
2
restricts on T
or
∗Sn−1
˙x = |x|
˙y = −|y|
to
˙x = y, ˙y = −|y|
y −x, yx
2
x + x, yy
¨x = −|˙x|
2
x
2
x
which is clearly the geodesic flow on the sphere. Thus the geodesic flow appears
as the reduced system to H =
1
(|x|2|y|2−x, y2). Its orbits are circles on the
2
sphere.
E
XAMPLE 4. The above Hamiltonian (∗) is invariant under the group action
(x, y) → (ax + by, cx + dy),ad− bc =1
of SL(2, R). Its Lie algebra is generated by the Hamiltonians |x|
and we ask for the reduced space corresponding to
2
=1, x, y =0, |y|2=1,
|x|
−1
which defines ψ
(μ) in this case. We need the isotropy group Gμwhich is
easily determined to be given by
ab
cd
=
cos ϕ sinϕ
−sin ϕ cos ϕ
.
2
, |y|2, x, y,
84 Various Aspects of Integrable Hamiltonian Systems
Note that ψ−1(μ) is the unit tangent bundle of S
n−1
and the orbits of G
are precisely the geodesics in the sphere. Thus the reduced space is the orbit
manifold of geodesics on the unit tangent bundle. The reduced flow is constant
in this case.
In the next section we will describe a more complicated example of a
reduced phase space under the coadjoint action of the unitary group extended to
the cotangent bundle of the Lie algebra. This will lead for appropriate choice
of μ to the inverse square potential of Calogero.
Actually similar constructions are applicable to the Toda lattice (the group
being given by the upper triangular matrices) and the Korteweg – de Vries equation as was shown recently by M. Adler [6].
References
[1] V. I. Arnold, Mathematical Methods in Classical Mechanics, Moscow, 1974,
(to appear in English translation), in particular Appendix 5.
[2] J. Marsden and A. Weinstein, Reduction of symplectic manifolds with sym-
metries, Reports on Math. Physics 5, 1974, 121–130.
[3] J. Marsden, Applications to Global Analysis in Mathematical Physics, Pub-
lish or Perish, Inc, 1974, in particular Chap. 6.
[4] J. M. Souriau, Structure des syst`emes dynamiques, Dunod, Paris, 1970.
μ
[5] A. A. Kirillov, Elements of the Theory of Representations, Springer, 1976.
[6] M. Adler, On a trace functional for formal pseudodifferential operators and
symplectic structure of the Korteweg – de Vries equation , preprint, Univ.
Wise., 1978.
§ 4. The Inverse Square Potential
a) In connection with his quantum theoretical work Calogero [4] was
led to consider the n-body problem on the line. He was led to the conjecture
that the corresponding classical problem was integrable which was subsequently
proven ([5]). Recently these results were derived most elegantly from a reduction
process as it was discussed above applied to the coadjoint representation of the
unitary group [1].
§4. The Inverse Square Potential 85
We formulate the results. Consider n distinct points on the real line with
coordinates
x
1<x2
<... <xn,
and define the potential
U(x)=
(xk− xj)−2.
k<j
We consider the Hamiltonian system with the Hamiltonian
H =
n
1
2
y
+ bU (x)+
k
2
k=1
n
2
a
2
k=1
2
x
.
k
This system is integrable and the solutions are algebraic functions in a sense to
be defined.
We write the differential equations in the form
˙x
= yk, ˙yk= −bU
k
x
k
− a2x
k
¨xk=2b
Using an interesting observation due to Perelomov
(xk− xj)−3− a2xk. (1)
j=k
1
one can reduce this system
to the case a =0, b =1by the following elementary transformation.
(t)=c cos(at)Xk(T ) where T =tan(at). (2)
x
k
Indeed, one finds
¨x
+ a2xk=
k
ca
(cos at)
X
k
3
2
wheredenotes differentiation with respect to T . Hence, if Xksatisfies the
above equation with a =0, b =1we have
¨x
+ a2xk=
k
Thus with b = a
2
ca
· 2
(cos at)
2c4
3
the transformation (2) takes the system (1) into that with
(Xk− Xj)−3= c4a2· 2
j=k
(xk− xj)−3.
j=k
a =0, b =1. This may lead to complex values of c — which need not disturb
us since all solutions have complex continuation.
1
I owe this information to Calogero.
86 Various Aspects of Integrable Hamiltonian Systems
b) In the following we will restrict ourselves to the case a =0, b =1
and show that this system possesses n integrals in involution, and that for the
solutions x
(t) are algebraic functions of t. To formulate the results it is good
k
to introduce the matrix
, ..., yn)+i(zkl)(3)
1
where z
L(x, y) = diag(y
=(xk− xl)−1for k = l, zkk=0,andthexkalways distinct.
kl
We will prove the
Theorem. The Hamiltonian system (1) with a =0, b =1has the n rational
integrals
1
=
F
k
tr(L(x, y))
k
which are an involution. The solutions x
k
,k=1, ..., n (4)
= xk(t) are the n distinct eigenvalues
k
of the matrix
diag(x
(0), ..., xn(0)) + tL(x(0),y(0)), (5)
1
hence algebraic functions.
Corollary. The symmetric functions
n
p
x
k
k=1
are polynomials of degree p in t. Hence, using the transformation (2) one
finds that for arbitrary a, b the above expression is a trigonometric polynomial
(±iat)
in e
period 2πa
of degree p. In particular, for a>0 all solutions are periodic of
−1
.
c) To prove these results we consider the cotangent bundle of the unitary
group U(n). We represent its element in the form of a pair (X, Y ) of Hermitian
matrices and with the 1-form
θ =tr(YdX).
The symplectic structure is defined by the two form
ω =tr(dY ∧ dX ).
In this symplectic space the Hamiltonian system associated with a function H =
= H(X, Y ) is given by
˙
X = H
,˙Y = −H
Y
X
§4. The Inverse Square Potential 87
where H
is the matrix with the entries (∂H/∂Ykj) etc. The Poisson bracket is
Y
given by
{F, G} =tr(F
Therefore it is clear that any two functions F
XGY
− FYGX).
= F1(Y ), F2= F2(Y ) inde-
1
pendent of X are in involution. Thus any such Hamiltonian is integrable. For
us
1
=
G
p
will be relevant. The system belonging to G
p
). (6)
tr(Y
p
is even linear and has the solution
2
X(t)=X(0) + tY (0),Y(t)=Y (0). (7)
d) All these systems defined by G
ϕ
:(X, Y ) → (U−1XU, U−1YU).
U
are invariant under the group action
p
The corresponding infinitesimal mapping is
(X, Y ) → i([X, A], [Y, A])
where A is Hermitian, i. e. iA belongs to the Lie algebra of U (n). This vector
field
˙
X = i[X, A],˙Y = i[Y, A]
has the Hamiltonian
!
ψ
= itr(A[X, Y ]) = tr(iA[iX, iY ]).
A
This defines the moment map as
ψ :(X, Y ) → [Y, X]
the right-hand side being interpreted as an element of the dual of the Lie algebra.
The elements of [Y, X] represent the relevant integrals; [Y, X] is the analogue
of the angular momentum vector.
−1
To construct the reduced space ψ
(μ)/Gμwe choose — with Kazhdan,
Kostant and Sternberg the element μ as the matrix
⎛
⎞
0 i i ... i
μ = −
⎜
i 0 i ... i
⎜
⎜
ii0 ... i
⎜
⎜
.....
⎜
⎝
i ... . 0 i
⎟
⎟
⎟
⎟
⎟
⎟
⎠
(8)
i ... . i 0
88 Various Aspects of Integrable Hamiltonian Systems
so that ψ−1(μ) is given by the pairs of Hermitian matrices (X, Y ) for which
[X, Y ]=−μ. (9)
The complete set of solutions is given by the following.
Proposition 1. The most general solution of (9) with μ given by (8) is of
the form
X = V
Y = V
−1
diag(x1, ..., xn)V,
−1
L(x, y)V,
(10)
where x
, ..., xnare distinct real numbers, y =(y1, ..., yn) ∈ Cnand V a
1
unitary matrix satisfying
⎛
⎞
1
⎜
⎟
1
⎜
⎟
Ve= λe where e =
⎜
⎝
.
⎟
.
.
⎠
.
1
Moreover, the group of the unitary matrices V agrees with the isotropy group G
of μ.
Corollary. The quotient space ψ(μ)/G
=(x
, ..., xn), y =(y1, ..., yn) and to any (X, Y ) ∈ ψ(μ) is by (10)
1
is parametrized by x =
μ
associated a unique pair of matrices
(K(x),L(x, y)) where K (x) = diag(x
, ..., xn).
1
The corresponding differential form is computed as
n
ω =tr(dL(x, y) ∧dK(x)) =
P
ROOF OF THE PROPOSITION.
dyk∧ dxk.
k=1
We consider the more general equation
i
2
|v|
[X, Y ]=i(v
kvl
) −
δ
kl
n
μ
where v =(v1, ..., vn) is any complex vector =0. The second term is
determined so that the trace of the right-hand side vanishes. For v
=1the
k
equation goes over into (9).
§4. The Inverse Square Potential 89
We choose a unitary transformation U which diagonalizes X. Note that
−1
under X → U
form where v → U
matrix X = diag(x
we conclude that |v
|v
| =1,orvk= e
k
XU, Y → U−1YU the right-hand side goes over into a similar
−1
v. Therefore we can analyze the equation for a diagonal
, ..., xn)=K(x).From
1
1
2
[K(x),Y]
|2is independent of k. If we specialize to |v|2= n we get
k
iθ
k
. Applying the unitary transformation,
=0=i|vk|2−
kk
U =diag(e
iθ
1
, ..., e
|v|
n
iθ
n
)
we achieve that v = e and the diagonal form of X is not destroyed. Thus we
have reduced our system to the case X = K(x), v = e.From
[K(x),Y]
we conclude that the x
If we set Y
= ykwe have the solution
kk
k
=(xk− xl)Ykl= i for k = l
kl
are distinct and
i
=
Y
kl
xk− x
for k = l.
l
X = K(x),Y= L(x, y)
for the system (9).
To find the most general solution we have to investigate those unitary
transformations U which leave the equation invariant. Since
−1
U
(vkvl)U =(wkwl) where w = U−1v
and since this matrix determines the vector W onlyuptoafactorλ on |λ| =1we
see that the set of U commuting with (v
) is given by those unitary matrices
kvl
for which Uv = λv. This proves the remaining assertions of the proposition.
e) Thus to any (X, Y ) ∈ ψ
π(X, Y )=(K(x),L(x, y)) = (U
and to any function H(X, Y ) on ψ
a function
h(x, y)=π
−1
(μ) we can associate a unique pair
−1
XU, U−1YU); U ∈ Gμ,
−1
(μ) invariant under U(n) we can associate
∗
H = H(K (x),L(x, y)).
90 Various Aspects of Integrable Hamiltonian Systems
This projection takes the symplectic structure of ψ−1(μ)/Gμinto the standard
n
form
dyk∧ dxk.
k=1
In particular, the functions (6) which are in involution go over into
1
p
tr L
p
(x, y)
F
=
p
which are also in involution. If we observe that
1
2
tr L
(x, y)=
2
is the Hamiltonian of the inverse square potential then it is clear that the F
1
2
|y|
+ U (x)
2
are
p
rational integrals in involution of this system, proving one part of our theorem.
−1
0
-flow are
2
)
−1
t
V0we
The other part is equally easily derived. The solutions of the G
given by (7) which we restrict now to ψ
unitary matrices V
(V
wherewesetK
such that
t
−1
V
tKt
,VtLtV
t
= K(x(t)), Lt= L(x(t),y(t)). Hence with Ut= V
t
−1
)=(V0(K0+ tL0)V
t
−1
(μ). By the proposition there exist
−1
,V0L0V
0
have
−1
(U
KtUt,U
t
The equality of the second component shows again that trL
integrals, the equality of the first component shows that the x
of K
+ tL0, proving the theorem.
0
f) Our derivation was based on the fact that the functions tr Y
to F
=trLp(x, y). Similarly, tr Xpis mapped into tr Kp=
p
−1
LtUt)=(K0+ tL0,L0).
t
p
t
(t) are eigenvalues
k
p
n
k=1
p
=trL
0
are mapped
p
x
, but here
k
are
it is evident that these functions are in involution. Nevertheless, it is useful to
consider the map
(X, Y ) → (−Y, X)
which carries one set of functions into the other. This mapping is clearly sym-
−1
plectic, preserving ψ
there exists a U ∈ G
(U
(μ). Projecting this mapping to ψ−1(μ)/Gμwe see that
such that
μ
−1
K(x)U, U−1L(x, y)U )=(−L(ξ, η),K(ξ))
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