Tags: Fixed Point Method

Let’s discuss the fixed point problem. We aim to solve a system of equations using a fixed point method.

The Bethe equations are:

\[\xi_i^L = (-1)^{N-1} \prod_{j=1}^{N} \frac{1 + \xi_i \xi_j - \Delta \xi_j}{1 + \xi_i \xi_j - \Delta \xi_i}, \quad i= 1, \dots, N\]

for \(N\leq L \in\mathbb{Z}_{\geq 1}\) and \(\Delta \in \mathbb{R}\).

The idea was to defibe a sequence \(\xi(k) =(\xi_1(k), \dots, \xi_N(k))\) as follows

\[\xi_i(k+1)^L = (-1)^{N-1} \prod_{j=1}^{N} \frac{1 + \xi_i(k) \xi_j(k) - \Delta \xi_j(k)}{1 + \xi_i(k) \xi_j(k) - \Delta \xi_i(k)}, \quad i= 1, \dots, N\]

The problem is that we need to take an \(L^{th}\) root and choose a branch cut. Then, to apply the Fixed Point Theorem, we need to make sure that our sequence is bounded away from the brach cut. This seems to be true in practice for \(\Delta\) near \(0\), but it seems to be fairly involved to prove that its true. In particular, it requires some bounds on \(\Delta\), depending on \(N\) and \(L\), that are not trivial to optimize. I’m thinking of abandoning this approach.

I would like to consider an alternate sequence instead. First, note that we have

\[\begin{align} &(-1)^{N-1} \xi_i^L \prod_{j=1}^{N}(\xi_j^{-1} + \xi_i - \Delta \xi_i\xi_j^{-1})\\ &= \prod_{j=1}^{N} ( \xi_i - (\Delta- (\xi_j^{-1}) )\\ &= \sum_{m=0}^{N} (-1)^{N-m} e_{N-m}(x_1, x_2, \dots, x_n) \xi_i^m\\ &= (-1)^{N-1} e_{N-1}(x_1, \dots, x_N)\xi_i +\sum_{m\neq 1}^{N} (-1)^{N-m} e_{N-m}(x_1, x_2, \dots, x_n) \xi_i^m \end{align}\]

where \(x_j= \Delta - \xi_j^{-1}\) and \(e_{m}\) is the elemtary symmetric polynomial. Thus, we may consider the following sequence

\[\begin{align} &\xi_i(k+1)\\ &= \frac{\left(\xi_i(k)^L \prod_{j=1}^{N}(\xi_j(k)^{-1} + \xi_i(k) - \Delta \xi_i(k)\xi_j(k)^{-1}) + (-1)^N \sum_{m\neq 1}^{N} (-1)^{N-m} e_{N-m}(x_1(k), \dots, x_N(k)) \xi_i(k)^m\right)}{e_{N-1}(x_1(k), \dots, x_N(k))} \end{align}\]

where \(x_j(k) = \Delta - \xi_j(k)^{-1}\).

Let us denote the right-side of the equation above as \(F_i(\xi(k))\) and set \(F=(F_1, \dots, F_N)\). Then, the sequence is given as follows

\[\xi(k+1) = F(\xi(k)).\]

Then, to use the Fixed Point Theorem, we need to show that \(F\) is continuous (in the appropate region) and that the map is a contraction (or, sufficiently, that the norm of the Jacobian of \(F\) is less than one).

Questions

  1. Can we show \(F\) is contnuous? Note that \(F\) is not continuous if \(e_{N-1}(x_1, \dots, x_N) = 0\) or, equivalently, when
\[\frac{1}{x_1} + \cdots + \frac{1}{x_N} =\frac{\xi_1}{1- \Delta \xi_1} + \cdots+ \frac{\xi_N}{1 -\Delta \xi_N} = 0.\]
  1. Can we show that the sequence \(\xi(k)\) stays away from the set described by the equation above? We are just interested in the case \(\Delta \approx 0\). Heuristically, we then have \(x_i \approx 1/\xi_i\) and \(\xi_i's\) are \(L^{th}\) roots of unity. Thus, under the heuristic approximation, the execptional set becomes
\[\xi_1 + \cdots + \xi_N =0.\]
  1. Can we show that the equation above is not true for \(\xi_i's\) are \(L^{th}\) roots of unity? Maybe except for \(N\) a divisor of \(L\) or \(N=L\).

Fixed Point Method