Riemann surfaces, separation of variables and classical and quantum integrability.

O. Babelon 1 and M. Talon1

September 2002

Abstract: We show that Riemann surfaces, and separated variables immediately provide classical Poisson commuting Hamiltonians. We show that Baxter's equations for separated variables immediately provide quantum commuting Hamiltonians. The construction is simple, general, and does not rely on the Yang–Baxter equation.

1  Introduction.

We know since Liouville that integrability means commuting Hamiltonians. It is the primary role of Lax matrices and the Yang-Baxter equation to provide non trivial such Hamiltonians. In the classical theory, additional benefits are the spectral curve Γ and the ability to separate variables by considering g = genus (Γ) points on it [1].
In the quantum theory, the analog construction is Sklyanin's method of separation of variables and Baxter's equations [2, 3]. Despite the beauty of this result, the route from a Yang–Baxter defined quantum integrable model to the separated variables is usually long and difficult, especially in the non hyperelliptic case.
Here, we show that we can reverse the strategy. We start from separated variables and consider Baxter's equations as equations for the Hamiltonians. We then prove that these Hamiltonians commute under very general hypothesis.
By its generality, its simplicity and its close analogy to the classical case, this result could provide a good starting point to build a theory of quantum integrable systems.

2  The main theorem.

Consider a curve in C2
Γ (λ,μ) ≡ R0(λ,μ) +
g
Σ
j=1
Rj(λ,μ) Hj =0       (1)
where the Hi are the only dynamical moduli, so that R0(λ,μ) and Ri(λ,μ) do not contain any dynamical variables. If things are set up so that Γ is of genus g and there are exactly g Hamiltonian Hj (see below for realizations of this setup), then the curve is completely determined by requiring that it passes through g points (λi, μi), i=1,⋯, g. Indeed, the moduli Hj are determined by solving the linear system
g
Σ
j=1
Rjii) Hj + R0ii) = 0,   i=1,⋯ ,g       (2)
whose solution is
H = − B−1 V       (3)
where
H =




H1
Hi
Hg





,    B =




R111) Rg11)
 
R1ii) Rgii)
 
R1gg) Rggg)





,    V =




R011)
R0ii)
R0gg)







Here, of course, we assume that generically detB ≠ 0.

Theorem 1   Suppose that the variables (λi, μi) are separated i.e. they Poisson commute for ij:
{ λij} = 0,    { μij}=0,    { λij} = pi, μi) δij       (4)
Then the Hamiltonians Hi, i=1⋯ g, defined by eq.(3) Poisson commute
{ Hi , Hj } = 0
Proof. Let us compute
B1 B2 { (B−1V)1,(B−1V)2 } = { B1, B2 } (B−1V)1 (B−1V)2
    − { B1, V2 } (B−1V)1 − { V1, B2 } (B−1V)2 + { V1, V2 }
Taking the matrix element i,j of this expression, we get
( B1 B2 { (B−1V)1,(B−1V)2 })ij =
δij
 
Σ
k,l
{ Bik, Bil } (B−1V)k (B−1V)l
   
− δij
 
Σ
k
{ Bik, Vi } (B−1V)k − δij
 
Σ
l
{ Vi, Bil } (B−1V)l + δij{ Vi, Vi } =0
where δij occurs because the variables are separated.

It can hardly be simpler. The only thing we use is that the Poisson bracket vanishes between different lines of the matrices, and then the antisymmetry. We did not even need to specify the Poisson bracket between λi and μi. The Hamiltonian are in involution whatever this Poisson bracket is. This is the root of the multihamiltonian structure of integrable systems.

Can we make it quantum ? Let us consider a set of separated variables
[ λij ] = 0,    [ μij] =0,    [ λij] = pi, μi) δij
We want Baxter's equation, so we start from the linear system
 
Σ
j
Rjii) Hj + R0ii) = 0       (5)
Here the Hj are on the right, and in Rjii), R0ii), we assume some order between λii, but the coefficients in these functions are non dynamical. Hence we start from the linear system
B H = −V       (6)
We notice that we can define unambiguously the left inverse of B. First, the determinant D of B is well defined because it never involves a product of elements on the same line. The same is true for the cofactor Δij of the element Bij (we include the sign (−1)i+j in the definition of Δij). Define
Bij−1 ≡ (B−1)ij = D−1 Δji
We have
(B−1 B)ij =
 
Σ
k
D−1 Δki Bkj
But Δki does not contain any element Bkl, hence the product Δki Bkj is commutative, and the usual construction of the inverse of B is still valid. Since the left and right inverse coincide in an associative algebra with unit, we have the identities
(BB−1)ij=
 
Σ
k
Bik Bkj−1 =
 
Σ
k
Bik D−1 Δjk = δij       (7)
We write the solution of eq.(6) as
H = − B−1 V       (8)


Theorem 2   The quantities Hi defined by eq.(8), which solve Baxter's equations eqs.(5), are all commuting
[ Hi , Hj ] = 0
Proof. Using that Vk and Vl commute, [ Vk,Vl ] = 0, we compute
[ Hi, Hj ] =
 
Σ
k,l
  [ Bik−1 Vk , Bjl−1 Vl ]  
    (9)
  =
 
Σ
k,l
  [Bik−1, Bjl−1 ] Vk VlBik−1 [Bjl−1, Vk ] Vl + Bjl−1 [Bik−1, Vl ] Vk
 
Using
[A−1, B−1 ] =A−1 B−1[A,B]B−1A−1= B−1 A−1[A,B]A−1B−1
so that
[ Bik−1, Bjl−1 ] =
 
Σ
rs,r's'
Bir−1Bjr'−1 [Brs, Br's'] Bs'l−1Bsk−1
  =
 
Σ
rs,r's'
Bjr'−1 Bir−1 [Brs, Br's']Bsk−1 Bs'l−1
the first term can be written
 
Σ
k,l
  [Bik−1, Bjl−1 ] VkVl
=
Σ
1
2
Bir−1Bjr'−1 [Brs, Br's'] ( Bs'l−1 Bsk−1 + Bs'k−1 Bsl−1 ) VkVl
  =
Σ
1
2
Bjr'−1 Bir−1 [Brs, Br's'] ( Bsk−1Bs'l−1 + Bsl−1 Bs'k−1 ) VkVl
Using that [Brs, Br's'] = δrr'[Brs, Brs'] and is therefore antisymmetric in ss', and setting
Kss' =
 
Σ
k,l
  ( Bs'l−1 Bsk−1 + Bs'k−1 Bsl−1Bsl−1 Bs'k−1Bsk−1 Bs'l−1 ) VkVl
we get
 
Σ
k,l
[Bik−1, Bjl−1 ] VkVl
=
 
Σ
rss'
1
4
Bir−1Bjr−1 [Brs, Brs'] Kss'
  =
 
Σ
rss'
1
4
Bjr−1 Bir−1 [Brs, Brs']Kss'
  =
 
Σ
rss'
1
8
[Bir−1, Bjr−1] [Brs, Brs']Kss'
The last two terms in eq.(9) are simpler, we get
 
Σ
k,l
  Bjl−1 [Bik−1, Vl ] VkBik−1 [Bjl−1, Vk ] Vl =
 
Σ
rsk
[Bir−1, Bjr−1] [Brs, Vr] Bsk−1 Vk
The quantities Hi will commute if
[Bir−1, Bjr−1] = 0,    ∀ i,j,r       (10)
This is true as shown in the next Lemma.

The condition eq.(10) says that the elements on the same column of B−1 commute among themselves. In a sense this is a condition dual to the one on B. It is true semiclassically because
{ Bir−1, Bjr−1 } =
 
Σ
a,a',b,b'
Bia−1 Bja'−1 { Bab, Ba'b' } Bbr−1 Bb'r−1 =
 
Σ
a,b,b'
Bia−1 Bja−1 { Bab, Bab' } Bbr−1 Bb'r−1=0
where in the last step we use the antisymmetry of the Poisson bracket. We show that it is also true quantum mechanically.
Lemma 1   Let B be a matrix whose elements commute if they do not belong to the same line
[ Bik , Bjl ] = 0    if   ij
Then the inverse B−1 of B is defined without ambiguity and moreover elements on a same column of B−1 commute
[ Bir−1 , Bjr−1 ] = 0
Proof. We want to show that
ΔriBjr−1 = ΔrjBir−1
denote by βi(r) the vector with components Bki, kr. Then we have (with j>i) (blue symbols are omitted):
ΔriBjr−1 = (−1)r+i β1(r) ∧ β2(r) ∧ ⋯ βi(r) ∧ ⋯ βj(r) ∧ ⋯βg(r) Bjr−1
  = (−1)r+i+gjβ1(r) ∧ β2(r) ∧ ⋯ βi(r) ∧ ⋯ βj(r) ∧ ⋯βg(r) ∧ βj(r) Bjr−1
  =
(−1)r+i+gj+1β1(r) ∧ β2(r) ∧ ⋯ βi(r) ∧ ⋯ βj(r) ∧ ⋯βg(r)
 
Σ
kj
βk(r) Bkr−1
  = (−1)r+i+gj+1β1(r) ∧ β2(r) ∧ ⋯ βi(r) ∧ ⋯ βj(r) ∧ ⋯βg(r) ∧ βi(r) Bir−1
  = (−1)r+jβ1(r) ∧ β2(r) ∧ ⋯ βi(r) ∧ ⋯ βj(r) ∧ ⋯βg(r) Bir−1
  = ΔrjBir−1
In the above manipulations, we never have two operators Bij on the same line so we can use the usual properties of the wedge product. Moreover it is important that the line r is absent in the definition of β(r). Remark that this equation can also be written ΔriD−1 Δrj= ΔrjD−1 Δri which is a Yang–Baxter type equation.

With this Lemma, we have completed the proof of our theorem. It is remarkable that, again, only the separated nature of the variables λi, μi is used in this construction, but the precise commutation relations between λi, μi does not even need to be specified. This is the origin of the multi Hamiltonian structure of integrable systems, here extended to the quantum domain.

3  Choosing the right number of dynamical moduli.

Let us explain how one can set up things in order that the number of dynamical moduli is equal to the genus of the Riemann surface. To understand the origin of the conditions we will write, let us explain first what happens in the setting of general rational Lax matrices described in [4, 9]. Quite generally, a Lax matrix L(λ) depending rationally on a spectral parameter λ, with poles at points λ k can be written as
L(λ) = L0 +
 
Σ
k
Lk(λ)  
        (11)
where L0=Diag(a1,⋯,aN) is a constant diagonal matrix and Lk(λ) is the polar part of L(λ ) at λ k, ie. Lk(λ)=Σr=−nk−1 Lk,r (λ −λ k)r. In order to have a good phase space to work with, we assume that Lk(λ) lives in a coadjoint orbit of the group of N× N matrix regular in the vicinity of λ =λ k, i.e.
Lk = (gk Ak gk−1)
Here Ak(λ ) is a diagonal matrix with a pole of order nk at λ =λ k, and gk has a regular expansion at λ =λ k. The notation () means taking the singular part at λ =λ k. This singular part only depends on the singular part (Ak) and the first nk coefficients of the expansion of gk in powers of (λ −λ k). The matrix (Ak) is an orbit invariant which specifies the coadjoint orbit, and is not a dynamical variable. It is in the center of the Kirillov bracket which as shown in [4] induces the Poisson bracket eq.(4), with pii)=1, on the separated variables. The physical degrees of freedom are contained in the first nk coefficients of gk(λ ). Note however that since Ak commutes with diagonal matrices one has to take the quotient by gkgk dk where dk(λ ) is a regular diagonal matrix, in order to correctly describe the dynamical variables on the orbit. The dimension of the orbit of Lk is thus N(N−1)nk so that L(λ ) depends on Σk N(N−1)nk degrees of freedom. Finally, the form and analyticity properties of L(λ ) are invariant under conjugation by constant matrices. To preserve the normalization, L0, at ∞ these matrices have to be diagonal (if all the ai's are different). Generically, these transformations reduce the dimension of the phase space by 2(N−1), yielding:
dim M =(N2N)
 
Σ
k
nk −2(N−1)
The spectral curve is
Γ   :   R(λ, μ ) ≡ det( L(λ )−μ 1 ) =(−μ)N +
N−1
Σ
q=0
rq(λ) μq =0  
        (12)
The coefficients rq(λ) are polynomials in the matrix elements of L(λ ) and therefore have poles at λ k. The curve is naturally presented as a N–sheeted covering of the λ-plane. We call μj(λ) the N branches over λ. Using the Riemann–Hurwitz formula, we can compute the genus of Γ [4]:
g=
N(N−1)
2
 
Σ
k
nkN +1 =
1
2
dim M


It is important to observe that the genus is half the dimension of phase space. So the number of action variables occurring as independent parameters in the eq.(12) should also be equal to g. Let us verify it. Since rj(λ) is the symmetrical function σj1,⋯,μN), it is a rational function of λ . It has a pole of order jnk at λ =λ k. Its value at λ = ∞ is known since μj(λ)→ aj. Hence it can be expressed on jΣk nk parameters namely the coefficients of all these poles. Altogether we have 1/2 N(N+1) Σk nk parameters. They are not all independent however. Above λ =λ k the various branches can be written:
μj(λ)=
nk
Σ
n=1
cn(j)
(λ −λ k)n
+regular       (13)
where all the coefficients c1(j),⋯,cnk(j) are fixed and non–dynamical because they are the matrix elements of the diagonal matrices (Ak), while the regular part is dynamical. This implies on rj(λ) that the coefficients of its nk highest order pole terms are fixed. Summing over j, we get Nnk constraints and we are left with 1/2 N(N−1) Σk nk parameters, that is g+N−1 parameters. It remains to take the quotient by the action of constant diagonal matrices. The generators of this action are the Hamiltonians Hn=(1/nresλ =∞Tr  (Ln(λ ))  dλ , i.e. the term in 1/λ in Tr (Ln(λ )). Setting
μj(λ) = aj +
bj
λ
+ ⋯       (14)
around the point Qj = (∞,aj), we have Hn = Σj ajn−1 bj . After Hamiltonian reduction these quantities are to be set to fixed (non–dynamical) values. So, both ai (by definition) and bi are non dynamical. On the functions rj(λ ) this implies that their expansion at infinity starts as rj(λ ) = rj(0)+ rj(−1)/λ + ⋯, with rj(0) and rj(−1) non dynamical. Hence when the system is properly reduced we are left with exactly g action variables.

The constraints eqs.(13, 14) can be summarized in a very elegant way [5, 9]. Introduce the differential δ with respect to the dynamical moduli. Then our constraints mean that the differential δ μ dλ is regular everywhere on the spectral curve because the coefficients of the various poles being non dynamical, they are killed by δ:
δ μ  dλ = holomorphic
Since the space of holomorphic differentials is of dimension g, the right hand side of the above equation is spanned by g parameters which are the g independent action variables we were looking for. Notice that these action variables are coefficients in the pole expansions of the functions rj(λ), and thus appear linearly in the equation of Γ. Hence eq.(12) can be written in the form eq.(1). Clearly, these considerations can be adapted by considering more general conditions such as
δ μ
μn
 
dλ
λm
holomorphic


4  Examples.

Let us show how well known models fit into our scheme. For the hyperelliptic ones, things are so simple that we can directly check the commutation of the Hamiltonians.

4.1  Neumann model.

The spectral curve can be written in the form [9]:
μ2 =
N−1
Π
i=1
(λ−bi)
N
Π
i=1
(λ−ai)
=
P(λ)
Q(λ)
      (15)
Performing the birational transformation s= μ Q(λ), we get:
s2=Q(λ)P(λ)       (16)
which is an hyperelliptic curve of genus g=N−1. The polynomial Q(λ) is non dynamical. We have (N−1) independent dynamical quantities, namely the (N−1) symmetrical functions of the bi, coefficients of P(λ). We have
δ μ  dλ =
δ P(λ)
2 μ Q(λ)
dλ =
δ P(λ)
2 s
dλ = holomorphic
Asking that a curve of the form eq.(15) passes through the g points (λii) determines the polynomial P(λ). The solution of Baxter's equations
Pi ) = Qi) μi2
simplifies in this case because the matrix B depends only on the λi. It is equivalent to Lagrange interpolation formula:
P(λ) = P(0)(λ) + P(2)(λ)
with
P(0)(λ) =
 
Π
i
(λ − λi),    P(2)(λ) =
 
Σ
j
Sj (λ) Qjj2,   Sj(λ) =
 
Π
kj
(λ − λj)
 
Π
kj
j − λk)
Introducing the canonical commutation relations
[ μj , λk ] = i ℏ δjk
so that
[ μj, fj) ] = iℏ ∂λj fj),   [ μj2, fj) ] = 2 iℏ ∂λj fj) μj + (iℏ)2λj2 fj)
We can check that [ P(λ), P(λ') ] =0 is a consequence of ∂λjP(0)(λ) = − Πkj ( λj − λk) Sj(λ), and the identities
Sj(λ) ∂λjnSi(λ') − Sj(λ') ∂λjnSi(λ) =0,    ∀ n >0
These identities follow from the remark that, if we define the translation operators tj λi = λj + σ δij, then
Sj(λ) tj Si(λ') − Sj(λ') tj Si(λ) =
 
Π
ki,j
(λ − λk)
 
Π
ki,j
(λ' − λk)
 
Π
kj
j − λk)
 
Π
ki
i − λk)
i − λj) (λ − λ')       (17)
is independent of σ.

4.2  Toda Chain.

The spectral curve can be written in the form [8, 9]:
μ + μ−1 = 2 P(λ)           (18)
where 2P(λ)=λn+1 − Σi=1n+1 pi λn+⋯ is a polynomial of degree (n+1). The spectral curve is hyperelliptic since it can be written as
s2 = P2(λ) − 1,    with   s = μ −P(λ)           (19)
The polynomial P2(λ) is of degree 2(n+1) so the genus of the curve is g = n. The number of dynamical moduli is g=n in the center of mass frame Σi=1n+1 pi =0. We have
δ μ
μ
 dλ =
2 δ P(λ)
μ − μ−1
dλ =
δ P(λ)
s
dλ =  holomorphic
Asking that the curve eq.(18) passes through the n points (λii), we get Baxter's equations.
2 Pi) = μi + μi−1
Their solution is again given by Lagrange interpolation formula:
2P(λ) = P(0)(λ) + P(1)(λ)
where
P(0)(λ) = (λ +
 
Σ
i
λi)
n
Π
i=1
(λ − λi),    P(1)(λ) =
 
Σ
i
Si(λ) (μi + μi−1)
The polynomial P(0)(λ) is of degree n+1, vanishes for λ = λi and has no λn term. Let the commutation relations of the separated variables be given by:
μj λj = q λj μj ,    μj λi = λi μj,    ij
Then again [ P(λ), P(λ') ] =0 as a result of eq.(17), where tj is interpreted as tj λj = q λj, and the facts that
Sj(λ) tj± 1 P(0)(λ') − Sj(λ') tj± 1 P(0)(λ) = P(0)(λ') Sj(λ) − P(0)(λ) Sj(λ')
   
=
 
Π
kj
(λ − λk )
 
Π
kj
(λ' − λk )
 
Π
kj
j − λk)
(λ + λ' +
 
Σ
ij
λi)(λ'−λ)


4.3  A non–hyperelliptic model.

We consider the model studied in [2, 6, 7] . The spectral curve can be written in the form:
R(λ,μ) ≡ μN + t(1)(λ) μN−1 + ⋯ t(N)(λ) = 0       (20)


The polynomials t(k)(λ) are such that degree  t(k)(λ) ≤ kn −1 and degree  t(N)(λ) = Nn −1 for some integer n. The genus of this curve is
g =
1
2
(N−1)(Nn−2)
Assuming that there is no singular point at finite distance, the homomorphic differentials are
ωkl =
μl λk
μR(λ,μ)
dλ,   0 ≤ l < N−1,   0 ≤ k < (Nl−1)n −1
We have
δ μ
dλ
λ
= −
N
Σ
k=1
δ t(k)(λ) μNk
μR(λ, μ)
dλ
λ
This will be holomorphic if δ t(1)(λ) = 0 and
δ t(k)(λ) =δ H1(k) λ + ⋯ + δ H(k−1)n −1(k)λ(k−1)n −1
Baxter's equations and the commutation of the Hamiltonians where proved in this case, starting from the definition of the quantum model through it Lax matrix and the Yang–Baxter equation. Our approach gives a very simple proof of this result.

5  Conclusion.

We have shown that starting from the separated variables, one can give an easy definition of a quantum integrable system. The next step is to reconstruct the Lax matrix and the original dynamical variables of the model. While this is a well understood problem in the classical theory [1, 9], (it is the essence of the classical inverse scattering method), its quantum counterpart will require a deeper understanding of the quantum affine Jacobian [10].

Acknowledgements. We thank D. Bernard and F. Smirnov for discussions.

References

[1]
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[2]
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[3]
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[4]
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[5]
I.M. Krichever, D.H. Phong, On the integrable geometry of soliton equations and N=2 supersymmetric gauge theories. J. Diff. Geom. 45 (1997) p.349–389.

[6]
F.A. Smirnov, Separation of variables for quantum integrable models related to Uq(^slN). math-ph/0109013

[7]
F.A. Smirnov, V. Zeitlin, Affine Jacobians of spectral Curves and Integrable Models. math-ph/0203037

[8]
L. Faddeev, L. Takhtajan, Hamiltonian Methods in the Theory of Solitons. Springer 1986.

[9]
O.Babelon, D. Bernard, M. Talon, Introduction to Classical Integrable Systems. Cambridge University Press (to appear).

[10]
F.A. Smirnov, Dual Baxter equations and quantization of Affine Jacobian. math-ph/0001032

1
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