Antisymmetric periodic flux potential

For each smooth periodic scalar $v$, [F2] gives ${\int }_{Y}a(e_j+\mathrm{\nabla }{\chi }_j)\cdot \mathrm{\nabla }v=0$, so ${\mathrm{div}}_y{q}^j=0$ distributionally. Its mean is zero by the definition of $A^{\ast }$ in [F1]. Testing with the real and imaginary parts of $e^{-2\pi im\cdot y}$ therefore gives, for every nonzero Fourier mode $m\in {\mathbb{Z}}^d$, $m\cdot {\widehat{q}}^j(m)=0$, and ${\widehat{q}}^j(0)=0$.

For $m\mathrm{\ne }0$ define ${\widehat{\sigma }}_{ik}^j(m)=\frac{m_k{\widehat{q}}_i^j(m)-m_i{\widehat{q}}_k^j(m)}{2\pi i\mathrm{\mid }m{\mathrm{\mid }}^2},{\widehat{\sigma }}_{ik}^j(0)=0.$ Then ${\widehat{\sigma }}_{ik}^j=-{\widehat{\sigma }}_{ki}^j$ and ${\sum }_{k}2\pi i{m}_k{\widehat{\sigma }}_{ik}^j(m)={\widehat{q}}_i^j(m)-\frac{m_i}{\mathrm{\mid }m{\mathrm{\mid }}^2}{\sum }_k{m}_k{\widehat{q}}_k^j(m)={\widehat{q}}_i^j(m)$ by step 1.1. Also $\mathrm{\mid }{\widehat{\sigma }}^j(m)\mathrm{\mid }\le C_d\mathrm{\mid }m{\mathrm{\mid }}^{-1}\mathrm{\mid }{\widehat{q}}^j(m)\mathrm{\mid }$, so ${\sum }_m(1+\mathrm{\mid }m{\mathrm{\mid }}^2)\mathrm{\mid }{\widehat{\sigma }}^j(m){\mathrm{\mid }}^2\le C_d{\sum }_m\mathrm{\mid }{\widehat{q}}^j(m){\mathrm{\mid }}^2$. By [F3], the symmetric Fourier partial sums and each of their first derivatives are Cauchy in $L^2$, since differentiation multiplies the coefficients by $2\pi i{m}_k$. Their $L^2$ limits satisfy the weak-derivative identity by passing to the limit in integration by parts against smooth tests. Thus the partial sums converge in $H^1$ to a tensor in the periodic $H^1$ closure, with zero mean and the asserted norm bound. The coefficient identity and completeness [F3] give its divergence $q^j$; conjugate symmetry makes it real.