Let's say that the Coulomb gauge can be expressed as a free divergence in space.
$$ \nabla \cdot \mathbf{A}(x) = 0 $$
Now, I understand that I can Fourier transform it and the divergence $X\mapsto \nabla\cdot X$ becomes the cross $\tilde{X}\mapsto k\times\tilde{X}$ and so, from an answer from physics SE
the component $\tilde{X}_\parallel$ of $\tilde{X}$ along the ray joining the origin and $P$ is the part that contributes to the divergence of $X$, and only this part can contribute to the divergence
What I fail to see is however the projection operator (formula 55.4 defined at page 335 of this draft by Mark Srednicki)
$$ A_i(x) \mapsto \biggl(\delta_{ij}-\frac{\Delta_i\Delta_j}{\Delta^2}\biggr)A_j(x) $$
Even assuming that the above formula is obtained by
multiplying $\widetilde{A}_i(k)$ by the matrix $\delta_{ij} − k_ik_j/\mathbf{k}^2$ and then Fourier transforming back to position space.
I can't figure out from where does the matrix $\delta_{ij} − k_ik_j/\mathbf{k}^2$ come from?
Intuitively I guess it is the component in the plane normal to $\mathbf{k}$ and it comes from expressing the cross product in matrix form: am I correct or at least close to the meaning of the expression?
My thesis is that the derivation of the projection operator from the Coulomb gauge is mathematically wrong.
Indeed a recent paper clearly shows how to derive the projection operator from Maxwell equations without applying the Coulomb gauge fixing.
Combining static magnetism equations
$$ \nabla \cdot \vec B = 0 $$ $$ \nabla \times \vec B = \vec j $$
in terms of the vector potential
$$ \nabla(\nabla \cdot \vec A) - \nabla^2 \vec A= \vec j $$
and considering the Fourier transform
$$ − \vec k [ \vec k \cdot \vec {\widetilde A} (\vec k) ] + k^2 \vec {\widetilde A} (\vec k) = \vec {\widetilde j} (\vec k) $$
can be simply rewritten by defining the projection operator as
$$ \delta_{ab} - \frac {k_a k_b} {k^2} $$
More generally, from this pdf (Eq. B.1 and B.2, page. 16-17), to project a vector field into a surface defined by $\phi = constant $, without adding additional structure, we must construct a projection operator using only the functions $\Phi_\alpha$ and cosymplectic form, $J$:
$$ \delta^a_b - { \{ \Phi_\alpha,\Phi_\beta \} }^{-1} \frac {\partial \Phi_\alpha} {\partial z^b} \frac {\partial \Phi_\beta} {\partial z^c} J^{ac} $$