For the incompressible Navier-Stokes equations: $$u_t - \Delta u + (u \cdot \Delta)u+\nabla p = 0\\\nabla \cdot u = 0$$ where $u$ is velocity field and $p$ is pressure. For vorticity $\omega=\nabla \times u$, we have the vorticity formulation:$$\omega_t -\Delta \omega + (u \cdot \nabla)\omega=(\omega \cdot \nabla)u$$
Then we know the vorticity $\omega_{ij}=\partial_i u_j - \partial_j u_i$ satisfies $$\partial_{t}\omega_{ij} -\Delta \omega_{ij} + \partial_i (u_k\omega_{kj})-\partial_j (u_k \omega_{ki})=0$$
My questions is: how to derive $\hat {\partial_i u_j}=\frac{\xi _i \xi_j}{|\xi|^2}\hat{\omega_{kj}}$ from above, where $i,j=1,2,3,...n$. We also assume Einstein summation convention.
This question is old but I write this answer for completeness. It is nothing but converting derivatives to polynomials using the Fourier transform and the fact that $u$ is divergence free.
From $\newcommand{\del}{\partial}w_{ij}=\del_i u_j - \del_j u_i$ we have $\hat w_{kj}=\xi_k \hat u_j - \xi_j \hat u_k$ up to a constant (probably $-2\pi i$). Contracting with $\xi_j$ gives $$\xi_j w_{kj} = - |\xi|^2 \hat u_k$$ since $\xi_j \hat u_j = \widehat{\nabla \cdot u}= 0$. Now multiply by $\xi_i$ and divide by $|\xi|^2$ (assuming this is legal) and the result is proven, up to a multiplicative constant which depends on convention.