So, the Lax operator L(lambda) is given by
L(t,lambda)ij=pideltaij+(1−deltaij)fijPhi(qi−qj,lambda)
with lambda the spectral parameter, and Phi the Lamé function. Using the Lax equation dotL=[L,M], which is equivalent to [L,fracpartialpartialt+M]=0, if a matrix A(t,lambda) satisfies left(fracpartialpartialt+M(t,lambda)right)A(t,lambda)=0
and is normalized, A(0,lambda)=1 it follows that
L(t,lambda)A(t,lambda)=A(t,lambda)L(0,lambda)
Hence, it is clear that det(L−muI), (and so the spectral curve) is independent of time. Now, the equation of the spectral curve is
Gamma:quaddet(L(t,lambda)−muI)=0
Writing Gamma(lambda,mu)equivdet(L(t,lambda)−muI)=sumNi=0ri(lambda)mui
Your first question is why are the ri(lambda)'s elliptic functions. Note that the matrix elements of L are already doubly periodic, but they have an essential singularity at lambda=0. To show that the ri's are meromorphic, all you need is a gauge transformation to get rid of this singularity. Note that
L(t,lambda)=G(t,lambda)barL(t,lambda)G−1(t,lambda)
with G=left(deltaijezeta(lambda)qi(t)right)1lei,jleN
where zeta is the Weierstrass zeta function, does the job. So each ri(lambda) will be a combination of the Weierstrass wp function and its derivatives, with the coefficients being integrals of the system. For each set of initial values of these integrals, the spectral curve is an N-sheeted covering of the base elliptic curve. The branch points will coincide with the zeros of fracpartialGamma(lambda,mu)partiallambda on Gamma.
Look at "Introduction to classical integrable systems" by O. Babelon, D. Bernard, M. Talon, and the paper of Krichever I mention in the comments for more details.
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