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§1. The Discrete Version of the Dynamics of a Rigid Body 261
Corollary. The subspaces V
[V
and isotropic with respect to
, V−are Lagrangian:
+
]=[V−,V−] = 0 (26)
+,V+
:
(z)=
1
(Hz, z)=0for z ∈ V±.
2
The first statement follows immediately from the lemma because
V
+
To prove that V
=span
s=1, ...,N
is isotropic with respect toconsider for ϕ ∈ Eμ, ψ ∈ Eν,
±
E
,μs∈ S+,Eμ=Ker(A − μI)2N.
μ
s
(Hϕ, ψ)=[Aϕ, ψ]=μ[ϕ, ψ]+[!ϕ, ψ]=0
where !ϕ =(A − μI)ϕ ∈ E
Note that V
, V−are real subspaces since S+= S+, S−= S−, while E
+
.
μ
are generally complex.
Now we return to the proof of Theorem 2. Let z
in V
; combining these as column vectors of an N × 2N matrix
+
Z
+
of rank N we have from AV
=(z1, ..., zn)=
⊂ V
+
+
AZ+= Z+C
X
Y
+
, ..., znbe any basis
1
+
+
for some real N ×N matrix C+.
To prove (23) we use the relations (26), yielding for any u, v ∈ R
k
N
0=[Z+u, Z+v]=(BZ+u, Z+v)=(MX+u − Y+u, X+v)+(X+u, Y+v).
Assuming v is so chosen that X
= X
Cv = Y+v and we find from the above identity
+
i. e. Y
v =0, hence Z+v =0,i.e. v =0. Therefore det X+=0, proving (23).
+
Thus V
is given by y = W+x.SinceV+lies on the zero energy surface
+
it follows
n
for all x ∈ R
Moreover spec W
proving W
= S+proving Theorem 1.
+
v =0,wesetu = C+v,sothatX+u =
+
0=|Y
2
|Jx|
T
W+= J2, hence ω = JW
+
v|2,
+
−|W+x|2=0
+
−1
is orthogonal.
262 Discrete Versions of Some Classical Integrable Systems
1.5. The Integration of the Discrete Euler Equation
Now we apply our results to finding the solution of (4), following the
procedure which was described in the continuous case by Dubrovin [9, 10].
For the initial data X
−1
= X
X0∈ O(N ) and M1= ω
1
, X1∈ O(N ) we define ω1= X
0
T
J − Jω1. As follows from the previous
1
T
X0=
1
considerations Eqs. (4) define only a correspondence, but if we fix the splitting S = S
∪ S−of the roots of the polynomial Q(ν)(ν2I − μMk− J2),
+
which in fact does not depend on k, then we have a well defined mapping
f
S+,S
:(ωk,Mk) → (ω
−
k+1,Mk+1
). In order to «integrate» this dynamics
consider the spectral curve Γ:
det(M + λJ
We will assume that J
for i = j,andJ
2
=0. For generic M, Γ has a genus g =
i
2
has distinct eigenvalues different from zero: J
2
− μI)=0,M= M1. (27)
(N − 1)(N − 2)
2
eigenvector ψ(λ, μ),
(M + λJ
2
− μI)ψ(λ, μ) = 0 (28)
normalized by the condition
1
ψ
+ ···+ ψN= 1 (29)
is meromorphic on Γ whose poles define a divisor
(see [10, 16]). In the points at infinity Pi∈ Γ,whereμ ≈ λJ
=
+ ···+
1
2
, λ →∞,
i
(i =1, ...,N) we have
This means that ψ
i
(λ, μ) is the basis of the linear space of meromorphic functions
with the pole divisor
i
(Pj)=δ
ψ
, determined by the conditions (30).
i
. (30)
j
In our case M is skew symmetric, therefore Γ has a symmetry σ :Γ→Γ,
2
σ
=id:
σ(λ, μ)=(−λ, −μ). (31)
The divisor
T
(−λ, −μ) by ψ∗(λ, μ) and fix λ ∈ C such that the eigenvalues μ1, ..., μ
ψ
also is not arbitrary because of the following proposition. Denote
of M + λJ2determined by (27) are distinct.
2
i
.The
g+N −1
= J
2
j
N
§1. The Discrete Version of the Dynamics of a Rigid Body 263
Proposition. Let μ
For μ
= μ this product is d ifferent from zero:
= μ be two distinct eigenvalues of M + λJ2,then
∗
(λ, μ)ψ(λ, μ)=0. (32)
ψ
∗
(λ, μ)ψ(λ, μ) =0. (33)
ψ
To prove this consider the product
∗
ψ
(λ, μ)(M + λJ2)ψ(λ, μ)=μψ∗(λ, μ)ψ(λ, μ).
On the other hand
∗
(λ, μ)(M + λJ2)ψ(λ, μ)=
ψ
2
= −((M −λJ
We s e e th at i f μ
μ = μ
, ..., μNform a basis, therefore ψ∗(λ, μ)ψ(λ, μ) can’t be equal to
1
)ψ(−λ, −μ))Tψ(λ, μ)=μψ∗(λ, μ)ψ(λ, μ).
= μ then ψ∗(λ, μ)ψ(λ, μ)=0.Butψ(λ, μ) for all possible
zero.
Corollary. The divisorof the poles of ψ satisfies the equation
+ σ() ≈ B, (34)
where B is the set of branch points of μ as a function of λ, and ≈ means linear
equivalence of divisors.
∗
This equivalence is given by the function F (λ, μ)=ψ
as follows from the proposition. Thus
belongs to the shifted Prym variety
(λ, μ)ψ(λ, μ)
P ⊂ J(Γ). We restrict ourselves to these considerations because the detailed
discussion of the algebraic-geometric aspects of this spectral problem can be
found in the literature (see [21] and references therein). The corresponding
problems of real algebraic geometry are considered in [23].
Now we use the representation (19) for describing the analytic properties
on Γ of ψ
for arbitrary k.
k
Fix some splitting Σ=Σ
ψ(λ, μ)=(ω
∪ Σ−. As follows from (19)
+
− λJ )ψk(λ, μ) (35)
k
264 Discrete Versions of Some Classical Integrable Systems
is an eigenvalue of M
This means that we can define ψ
Notice that ψ
only for ψ
does not satisfy Dubrovin’s normalization (29) which is required
k+1
.
1
One can see from (36) that ψ
P
, ..., PN. In order to find the new zeros consider the hyperbola, deter-
1
k+1
(M
+ λJ2:
+ λJ2)ψ(λ, μ)=μψ(λ, μ).
k+1
as
k+1
ψ
=(ωk− λJ )ψk. (36)
k+1
has N new poles at the «infinities»
k+1
mined by the equation
λμ =1
and the intersection
−1
μ = λ
and
det(M + λJ
which coincides with (11).
So we have 2N points of intersection which we denote Q
−
Q
1
, ..., Q
−
in agreement with the splitting Σ=Σ+∪ Σ−. As follows from
N
the construction of ω
This means that Q
Lemma. For a given splitting Σ=Σ
tion ψ
(36) of the matrix M
k+1
analytical properties, which determine ψ
1. ψ
has a simple pole independing on the initial data M1and the
k+1
poles at the «infinities» P
P
j
2. ψ
has a zero of order k in Q
k+1
In particular, we see that the pole-divisor
∩ Γ. This intersection is described by the equations
2
− λ−1I)=(−λ)−Ndet(I −λM − λ2J2)=0
(see Sect. 3)
k
+
)=0.
i
. Thus we have proven
k+1
∪ Σ−the vector eigenfunc-
+
uniquely:
k+1
i
+ O(λ−1)),λ→∞.
j
+
, ..., Q
1
+
.
N
k+1
of ψ
is connected with
k+1
+
, ..., Q
1
i
: ψ
k+1
(ω
− λiI)ψk(Q
k
+
are the new zeros of ψ
N
+ λJ2has on spectral curve Γ the following
k+1
, ..., PNwith asymptotics in
1
= λk(−Jj)k(δ
+
, ..., Q
1
by the relation
≈
+ U, (38)
k
+
.
N
where U = P
+ ···+ PN− Q
1
k+1
+
−··· −Q
1
+
,
N
k
§1. The Discrete Version of the Dynamics of a Rigid Body 265
For given ψ
M
k+1
as ω
T
k+1
one can reconstruct ψ
k+1
J −Jω
. To find the solution of (2) Xk∈ O(N ):
k+1
X
k
−1
= ωkω
k−1
by using the formula (14), and
k+1
...ω1X
0
−1
,
one can use again Eq. (36). Indeed, from (36) follows that
= ωkΦk, (39)
Φ
k+1
where Φ
For ψ
is N × N matrix with ψk(0,μi) as a column. This means that
k
= ωkω
Φ
k+1
one can write the explicit formulas in terms of Prym’s θ-functions as
k+1
...ω1Φ1and X
k−1
−1
k
=Φ
k+1
−1
−1
Φ
X
1
. (40)
0
it was done, for example, by Bobenko in [24]. But here we restrict ourselves to
the example of O(3) (see below).
We summarize the result of this section in the following theorem.
Theorem 3. The discrete Euler Eq. (4) corresponds to the shifts on the
Prym variety P ⊂ J (Γ) (34) by the vector U = P
depending on the splitting Σ=Σ
∪ Σ−. If such splitting is fixed the general
+
+···+PN−Q
1
+
−··· −Q
1
+
N
solution of (4) and (2) can be expressed as some abelian function on P in the
points z
= z0+ kU.
k
1.6. Explicit Formulas for the Discrete Dynamics of the 3-Dimensional
Rigid Body
,
We consider here Eqs. (2), (4) for N =3. In this case the solution can be
expressed by elliptic functions. The spectral curve Γ (27) has the equation
or
where H = J
(λJ
2
M
3
det
2
− μ)(λJ
1
2
+ J
12
2
− μM12M
λJ
1
2
2
−M
−M
M
13
12
13
2
− μ)(λJ
2
2
+ J
1
2
λJ
−M
M
2
− μM
2
2
23
λJ
23
2
− μ)+Hλ −M2μ =0, (41)
3
.
13
=0
23
2
− μ
3
In the new variables
x = μ/λ, y = λ,
(41) has the form
2
y
Q(x)=H −M2x
266 Discrete Versions of Some Classical Integrable Systems
with Q(x)=(x −J
2
)(x − J
1
2
)(x − J
2
2
). After another change of variables
3
w = Q(x)y = Q(μ/λ)λ
we obtain the standard form of the elliptic curve
2
w
= R(x)=(x −J
2
)(x − J
1
2
)(x − J
2
2
)(H −M2x). (42)
3
The involution σ:(λ, μ) → (−λ, −μ) in these variables is σ(w, x)=(−w, x)
and the Prym variety coincides with J (Γ) ≈ Γ. The «infinities» P
correspond to the branch points x = J
point x = H/M
zero 0 on Γ.Letx
numbers J
2
corresponds to the point λ = μ =0, so we choose it as the
<x2<x3<x4be the ordered roots of R(x),i.e. the
2
, J
1
1
2
2
, J
, H/M2(notice that min{Ji} H/M2 max{Ji}).
2
3
2
, x = J
1
2
, x = J
2
2
. The fourth branch
3
, P2, P
1
The elliptic integral
(x, w)
z =
H/M
gives the equivalence of Γ and C/Z
x
3
dx
; τ
=2
#
R(x)
x
2
is real, τ2is purely imaginary.
1
dx
#
R(x)
2
x
2
dx
+ Z
τ
1
,whereτ1=2
τ
2
#
x
1
R(x)
, τ
=
2
The equation μλ =1in the variables x, w has the form
2
(x − J
it determines on Γ the set of six points, which we denote as Q
−
−
, Q
Q
according to the splitting Σ=Σ+∪ Σ−(see Fig. 1).
2
3
)(x − J
1
This figure depicts the situation, corresponding to J
sufficiently small M
2
and H, when all roots of (43) and P (λ) are real numbers.
2
)(x − J
2
2
) − x(H −M2x) = 0; (43)
3
+
+
, Q
1
2
<H/M2<J
1
+
, Q
, Q
2
3
2
2
<J
2
3
−
1
and
The condition that P (λ) has no purely imaginary roots, which is sufficient for
the solvability of Eq. (6) (see Sects. 1.2 and 1.4), is equivalent to the absence
of negative roots for Eq. (43). This leads to a certain restriction on the integrals
2
and H.TheshiftU is given by
M
+
+
U = P
+ P2+ P3− Q
1
1
− Q
2
− Q
+
= −(Q
3
+
1
+ Q
+
2
+ Q
+
),
3
3
,
because of P
+ P2+ P3=0.
1
§1. The Discrete Version of the Dynamics of a Rigid Body 267
Fig. 1
Using the results of the previous section one can easily express ψ
k+1
(z) in
terms of the classical elliptic σ-function and the initial position of the poles ψ
(ζ
1,ζ2,ζ3
ψ
×
). For example, the function
1
(z)=(−J1)k×
k+1
k
σ
(z −Q
+
)σk(z −Q
1
+
)σk(z −Q
2
+
)σ(z −(ζ1+ ζ2+ ζ3) − kU − P1)
3
σ(z −ζ1)σ(z −ζ2)σ(z −ζ3)σk(z −P1)σ
k−1
(z −P2)σ
k−1
(z −P3)
(44)
has all analytic properties of the first component of ψ
(z) (see the lemma in
k+1
Sect. 1.5) and therefore coincides with it. We can write now explicit formulas
, Mkand Xkas it was explained in the previous section.
for ω
k
Omitting nonessential multipliers in (44) we define a matrixΦ
− ζ − kU −Pi)
σ(z
= ±
j
σ(zj− Pi)
+
)σ(z −Q
2
∞
2
H/M
#
dx
R(x)
(Φ
where
ζ = ζ
+ ζ2+ ζ3, zi(i =1, 2, 3) correspond to λ =0in (41):
1
z
=0x =
i
=(−Ji)kfk(zj)
k+1)ij
f(z)=
M
+
σ(z −Q
)σ(z −Q
1
σ(z −P1)σ(z −P2)σ(z −P3)
H
,z
2
2, 3
, (45)
+
)
3
,
(x = ∞).
k+1
as
:
1
268 Discrete Versions of Some Classical Integrable Systems
Finally we have from (40)
X
k
T
=Φ
k+1
−1
T
Φ
X
,ωk=Φ
0
1
k+1
−1
Φ
k
(46)
withΦ
defined by (45).
k
In the continuous limit for M = εM
w
(w
)2
=(
2
)
ε
=(x − J
2
)(x − J
1
, ε → 0 the curve Γ remains
c
2
)(x − J
2
2
)(Hc− M
3
2
c
x)s,
but Eq. (43) becomes
(x − J
2
)(x − J
1
2
)(x − J
2
2
) − ε2x(Hc− M
3
2
x)=0.
c
When ε =0we arrive at
2
)(x − J
1
±
, Q
tend to P1, P2, P3when ε → 0 and the shift U =
3
This means that Q
=ΣQ
i
− ΣQ
+
→ 0. One should observe that the continuous limit corresponds
i
(x − J
±
±
, Q
1
2
to the special splitting Σ,where: Σ
2
)(x − J
2
contains all roots in the right half plane
+
2
)=0.
3
of P(λ) of (13) Sect. 1.2, which becomes here
(λ
2J2
1
− 1)(λ2J
2
− 1)(λ2J
2
2
− 1) + Hλ4− Mλ2=0.
3
As was shown before this continuous limit coincides with the classical problem
about the force free motion of a rigid body, the explicit solution for which in
terms of elliptic functions were found by Jacobi [25]. The comparison of this
formula and the continuous limit of (46) may be complicated.
§ 2. The Discrete Dynamics on Stiefel Manifolds and the
Heisenberg Chain with Classical Spins
In this section we generalize the previous results and consider the functional
S =
tr(XkJX
k
) on the sequences X =(Xk) of n + N matrices Xk,
k+1
1 n N satisfying the condition
T
X
X
= In. (1)
k
k
N
The rows of such matrices are orthogonal unit vectors in R
form the Stiefel manifold V
. The function of the interaction=tr(XJYT)
n, N
. Such matrices
§2. The Discrete Dynamics on Stiefel Manifolds 269
is a symmetric bilinear function, invariant under the action of O(n): X → sX ,
Y → sY , s ∈ O(n). We show that some results of Sect. 1, which correspond to
the case n = N, can be generalized for n<N.
For n =1our results agree with the solution of this problem proposed
in [1]. Notice that for n =1, N =3we have the problem about the stationary
states of a Heisenberg chain with classical spins (see [1, 5–7] and Introduction).
2.1. The Equation of the Dynamics and Isospectral Deformations
The variation of S under the condition (1) leads to the equation
where Λ
k
J + X
X
k+1
T
=Λ
is an (n × n) matrix multiplier. We assume J to be symmetric
k
J =ΛkXk, (2)
k−1
and nondegenerate.
Theorem 4. The Eqs. (2) are equivalent to the equation of isospectral
deformations
L
(λ)=Ak(λ)Lk(λ)A
k+1
with L
A
(λ)=J2+ λMk− λ2X
k
(λ)=J −λX
k
so that
L
=(J + λX
k
L
T
k
k+1
X
k−1
= A
. Moreover,
T
k−1
T
k+1
T
X
k−1
Xk)(J −λX
(−λ)A
k+1
k−1
T
k
(λ)=Ak(λ)A
The proof follows by a direct calculation. The determinant of L
−1
(λ)(3)
k
, Mk= X
X
)=A
k−1
T
XkJ − JX
k−1
T
(−λ)Ak(λ), (4)
k
T
(−λ). (5)
k
T
X
k
has the
k
k−1
form
P (λ)=detL
(λ)=detA
k
T
(−λ)detAk(λ)=
k
=detJ
2
det(In− λ2(X
J−1Xk)2), (6)
k−1
where we used the Weinstein– Aronszajn formula, see [10]. Therefore P(λ) is
an even polynomial in λ of degree 2n. The factorization (4) determines the
splitting Σ=Σ
Σ
= {λ:detAT(−λ)=0}. This splitting satisfies the conditions
−
∪ Σ−of the set Σ of zeros of Σ+= {λ :detA(λ)=0},
+
,
Σ+=Σ+, Σ−=Σ−, Σ+= −Σ−. (7)
270 Discrete Versions of Some Classical Integrable Systems
If such a splitting is fixed and all zeros of P(λ) are distinct then for given X
X
the matrix X
k
Consider the «eigenvectors» ψ
∈ Σ+,
to λ
i
can be defined as follows.
k+1
i
L
k+1(λi)ψi
of L
k+1=Ak
=0,
T
(λ)A
(−λ), corresponding
k
k−1
and combine these vectors to an N × n matrix ψ as columns. Then because
Σ
= {λ:detA
+
or
with Λ = diag(λ
then
(λ)=0} we have from (5),
k+1
(J −λ
Jψ −X
, ..., λn), λi∈ Σ+.Ifthen × n matrix Xkψ is invertible
1
T
X
k+1
T
X
Xk)ψi=0
i
k+1
T
XkψΛ=0 (8)
k+1
= JψΛ−1(Xkψ)
−1
(9)
is reconstructed uniquely.
Leaving the further discussion of the general case 1 n N we now turn
to a detailed treatment of the case n =1in the next section. In this connection
we want to mention the work of Adams, Harnad, and Previato [26], in which the
matrices introduced in [10] are generalized.
2.2. Discrete Version of the Neumann System and the Heisenherg Chain
With Classical Spins
,
For n =1we have the functional
where x =(x
J = diag(J
1
S(x)=
), xkbelong to the unit sphere S
k
, ..., Jn).ForN =3we have unit vectors in R3which can be
k∈Z
(xk,Jx
), (10)
k+1
N−1
in RN: |xk| =1and
interpreted as classical spins. In this case the functional S (10) defines the energy
of a spin chain in the Heisenberg model (see [1, 6–8]). The equations of the
stationary configuration have the form
Jx
k+1
+ Jx
k−1
= λkx
k
or
x
k+1
+ x
= λkJ−1xk. (11)
k−1
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