Take the 2-minute tour ×
MathOverflow is a question and answer site for professional mathematicians. It's 100% free, no registration required.

First, why all the coefficients in the characteristic polynomial of L are elliptic functions, since the diagonal entries of the matrix L are the momentums?

second, how to understand the ramification of the Spectral curve over the base elliptic curve? i.e, where are the branched points?

thanks

share|improve this question
    
Could you provide some background and/or a brief introduction for the question? –  Justin Curry Mar 24 '10 at 23:25
add comment

1 Answer

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 Lamé function. Using the Lax equation $\dot{L}=[L,M]$, 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 $$\Gamma(\lambda, \mu)\equiv\det(L(t,\lambda)-\mu I)=\sum_{i=0}^N r_i(\lambda)\mu^i$$ 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.

share|improve this answer
    
Hi, Gjergji thanks! what i am not understanding, for example in case N=2, I have the explicit expression for spetral curve: k^2+k(p1+p2)+(p1*p2-\pi(q1-q2)+\pi(\numda)), where \pi is the weiestrass p function. p1 and p2 are the momentums. q1,q2 positions. so p1+p2 and p1*p2 are elliptic functions of the parameter? –  Jimmy Mar 24 '10 at 22:09
    
So, yes this all agrees with the above. The $r_i$'s are representable as a linear combination of the Weierstrass p function with coefficients that are integrals of your system (independent of the spectral parameter). If you don't have access to the book I mentioned, try finding the paper "Elliptic solutions of the Kadomtsev-Petviashvili equation and integrable systems of particles" by Krichever, section 1 addresses both your questions. –  Gjergji Zaimi Mar 25 '10 at 1:57
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.