# Equation of motion of a charged string in a twisted torus.

## The flat torus background:

Say we want to study the sigma model of strings (closed strings $S^1= \mathbb{R}/\mathbb{Z}$) on a flat torus (for example $T^3=\mathbb{R}^3/\mathbb{Z}^3$ with a flat metric induced from $\mathbb{R}^3$), so for each fixed time time $\tau$ we have our string represented by a map $\phi = S^1 \rightarrow T^3$. Choosing a coordinate $\sigma$ in the universal cover $\mathbb{R}$ of $S^1$ and $\vec{x}$ in the universal cover $\mathbb{R}^3$ of $T^3$. We view our map $\phi$ as an equivariant map $\vec{x} ( \sigma, \tau)$ subject to the conditions $\vec{x} (\sigma+1, \tau) = \vec{x} + \vec{b}$ for some $\vec{b} \in \mathbb{Z}^3$. The equations of motion are in this case $$(\partial^2_\sigma - \partial_\tau^2) \vec{x}(\sigma, \tau) = 0. \label{1}$$ And this can be solved by Fourier analysis in general as $$\vec{x} (\sigma, \tau) = \vec{x}_0 + \vec{W} \sigma + \vec{P} \tau + \sum_{n \in \mathbb{Z}\setminus \{0\}} \vec{x}^+_n e^{- 2 \pi i n (\sigma + i \tau)} + \sum_{n \in \mathbb{Z} \setminus\{0\}} \vec{x}_n^- e^{-2 \pi i n (\sigma - i \tau)}, \label{2}$$

where $\vec{x}_0$ is a point in $T^3$ so that we think of it as a vector in $\mathbb{R}^3$ well defined modulo $\mathbb{Z}^3$, $\vec{W} \in \mathbb{Z}^3$ is the "winding", $\vec{P} \in \mathbb{R}^3$ is "momentum" and the complex vectors $\vec{x}^\pm_n \in \mathbb{C}^3$ satisfy some equation to ensure the sum is real (we should expand in terms of cosines and sines to avoid this).

In the twisted case the situation is changed as follows, the equations of motion are now given by $$(\partial^2_\sigma - \partial^2_\tau) \vec{x} = \partial_\sigma \vec{x} \times \partial_\tau \vec{x}$$ subject to the same boundary conditions $\vec{x}(\sigma+1, \tau) = \vec{x} \mod (\mathbb{Z}^3)$ and here $\times$ denotes the cross product in $\mathbb{R}^3$ . My question is:

What are the solutions of this differential equation? is there a similar description as in the Fourier case above?

**Granted, the background is useless but perhaps you've seen these equations solved in a similar context and that'll be great. **

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