 Translation invariants of Fourier-mode phases

Let a translation symmetry with displacement $s\in {\mathbb{R}}^d$ act on nonzero complex Fourier modes byEach phase changes by $q_j\cdot s$. Therefore a real linear combination of locally unwrapped phases, ${\sum }_j{n}_j{\theta }_j$, is unchanged exactly when ${\sum }_j{n}_j{q}_j=0$. This is a kernel of a linear map calculation for the matrix whose columns are the wavevectors.

A globally defined expression on angular phases requires more care: $e^{i{\sum }_j{n}_j{\theta }_j}$ is independent of the representatives chosen modulo $2\pi$ precisely when every $n_j$ is an integer. The globally defined translation-invariant characters therefore satisfy both $n\in {\mathbb{Z}}^m$ and ${\sum }_j{n}_j{q}_j=0$. An arbitrary real vector in the matrix's kernel of a linear map gives a local unwrapped invariant, but need not give a single-valued function of the angular phases.

For example, take four planar wavevectors $q_1=(1,0)$, $q_2=(0,1)$, $q_3=(1,1)$ and $q_4=(1,-1)$. The two independent angular combinations are ${\theta }_3-{\theta }_1-{\theta }_2$ and ${\theta }_4-{\theta }_1+{\theta }_2$, each interpreted modulo $2\pi$. Translations can set the first two phases to zero; the other two then equal these combinations.

More generally, when four planar wavevectors span a common reciprocal lattice of rank two, their translation action has a closed two-dimensional torus as its image and the quotient of the four-phase torus is two-dimensional. Real rank two alone guarantees only two local independent combinations; without a commensurability condition, the translation image can be dense in a larger torus. At zero amplitude, a phase is undefined, so smooth centre manifold equations should first use real parts and imaginary parts of the complex amplitudes.