##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 \begin{equation} (\partial^2_\sigma - \partial_\tau^2) \vec{x}(\sigma, \tau) = 0. \label{1} \end{equation} And this can be solved by Fourier analysis in general as \begin{equation} \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} \end{equation} 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 \begin{equation} (\partial^2_\sigma - \partial^2_\tau) \vec{x} = \partial_\sigma \vec{x} \times \partial_\tau \vec{x} \end{equation} 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. **