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So, the Lax operator $L(\lambda)$ is given by $$L(t,\lambda)_{ij}=p_i \delta_{ij}+(1-\delta_{ij})f_{ij}\Phi(q_i-q_j,\lambda)$$ with lambda the spectral parameter, and $\Phi$ the LamÃ© function. Using the Lax equation $\dot{L}=[M,L]$, \dot{L}=[L,M]$, which is equivalent to$[L,\frac{\partial}{\partial t}+M]=0$, if a matrix$A(t,\lambda)$satisfies $$\left(\frac{\partial}{\partial t}+M(t,\lambda)\right)A(t,\lambda)=0$$ 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$\det(L-\mu I)$, (and so the spectral curve) is independent of time. Now, the equation of the spectral curve is $$\Gamma:\quad\det(L(t,\lambda)-\mu I)=0$$ Writing $$\det(L(t,\lambda)-\mu \Gamma(\lambda, \mu)\equiv\det(L(t,\lambda)-\mu I)=\sum_{i=0}^N r_i(\lambda)\mu^i$$ Your first question is why are the$r_i(\lambda)$'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$r_i$'s are meromorphic, all you need is a gauge transformation to get rid of this singularity. Note that $$L(t,\lambda)=G(t,\lambda)\bar{L}(t,\lambda)G^{-1}(t,\lambda)$$ with $$G=\left(\delta_{ij}e^{\zeta(\lambda)q_i(t)}\right)_{1\le i,j\le N}$$ where$\zeta$is the Weierstrass zeta function, does the job. So each$r_i(\lambda)$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$\frac{\partial \Gamma(\lambda,\mu)}{\partial \lambda}$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. 2 edited body So, the Lax operator$L(\lambda)$is given by $$L(t,\lambda)_{ij}=p_i \delta_{ij}+(1-\delta_{ij})f_{ij}\Phi(q_i-q_j,\lambda)$$ with lambda the spectral parameter, and$\Phi$the Lame LamÃ© function. Using the Lax equation$\dot{L}=[M,L]$, it is clear that the spectral curve is independent of time. Now, the equation of the spectral curve is $$\Gamma:\quad\det(L(t,\lambda)-\mu I)=0$$ Writing $$\det(L(t,\lambda)-\mu I)=\sum_{i=0}^N r_i(\lambda)\mu^i$$ Your first question is why are the$r_i(\lambda)$'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$r_i$'s are meromorphic, all you need is a gauge transformation to get rid of this singularity. $$L(t,\lambda)=G(t,\lambda)\bar{L}(t,\lambda)G^{-1}(t,\lambda)$$ with $$G=\left(\delta_{ij}e^{\zeta(\lambda)q_i(t)}\right)_{1\le i,j\le N}$$ where$\zeta$is the Weierstrass zeta function, does the job. Look at "Introduction to classical integrable systems" by O. Babelon, D. Bernard, M. Talon for more details. 1 So, the Lax operator$L(\lambda)$is given by $$L(t,\lambda)_{ij}=p_i \delta_{ij}+(1-\delta_{ij})f_{ij}\Phi(q_i-q_j,\lambda)$$ with lambda the spectral parameter, and$\Phi$the Lame function. Using the Lax equation$\dot{L}=[M,L]$, it is clear that the spectral curve is independent of time. Now, the equation of the spectral curve is $$\Gamma:\quad\det(L(t,\lambda)-\mu I)=0$$ Writing $$\det(L(t,\lambda)-\mu I)=\sum_{i=0}^N r_i(\lambda)\mu^i$$ Your first question is why are the$r_i(\lambda)$'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$r_i$'s are meromorphic, all you need is a gauge transformation to get rid of this singularity. $$L(t,\lambda)=G(t,\lambda)\bar{L}(t,\lambda)G^{-1}(t,\lambda)$$ with $$G=\left(\delta_{ij}e^{\zeta(\lambda)q_i(t)}\right)_{1\le i,j\le N}$$ where$\zeta\$ is the Weierstrass zeta function, does the job. Look at "Introduction to classical integrable systems" by O. Babelon, D. Bernard, M. Talon for more details.