the brane coordinates): $$\begin{aligned}
 \label{eqsd2}
  \delta \ddot x_i & = & - \tr |X-x|^2 \delta x_i + \tr \left((X^j-x^j)(X_i-x_i)\right)\delta x_j \nonumber \\
  & & \quad - 2 \overline{\delta x^j} \delta x_j \delta x_i + 2 \overline{\delta x_i} \delta x^j \delta x_j\end{aligned}$$ It follows that the Jacobian matrix of this equation is the same than for $N=1$ with the inertia matrix: $$I = \left(\begin{array}{ccc} \tr(Y_2^2+Y_3^2) & -\tr(Y_1Y_2) & -\tr(Y_1Y_3) \\ -\tr(Y_1Y_2) & \tr(Y_1^2+Y_3^2) & -\tr(Y_2Y_3) \\ -\tr(Y_1 Y_3) & -\tr(Y_2 Y_3) & \tr(Y_1^2+Y_2^2) \end{array} \right)$$ The phase trajectory of $\delta x$ is then wrapped around a 6-torus $\mathbb T^6$.\
Concerning the polarisation, it obeys to $$\begin{aligned}
 \label{pola}
  |\ddot u_i \rangle & = & -\mathrm{trm}\,|X-x|^2 |u_i \rangle - 2 \frac{d\ln \delta x_i}{dt} |\dot u_i \rangle \nonumber \\
  & & \quad -\mathrm{trm} \left([X_i,X^j]-(X^j-x^j)(X_i-x_i) \right) \frac{\delta x_j}{\delta x_i} |u_j\rangle \nonumber \\
  & & \quad +\left(\overline{\delta x^j} \delta x_j - 2 \frac{\overline{\delta x_i}}{\delta x_i} \delta x^j \delta x_j \right) |u_i \rangle \nonumber \\
  & & \quad +\left(2 \frac{\overline{\delta x_i}}{\delta x_i} \delta x^j \delta x_j \langle u_i|u^j \rangle - \overline{\delta x^j}\delta x_j \langle u^j|u_i \rangle\right) |u_j \rangle\end{aligned}$$ where $\mathrm{trm}\, A = A - \tr A$.\
In order to illustrate the deformation of the invariant torus with 