Why is shear missing in Helmholtz theorem? Let $F_i$ be a well behaved vector field in $\mathbb{R}^3$ which rapidly vanishes at infinity. (I am using here the index notation.) By the Helmholtz theorem, knowledge of divergence $\partial_i F_i$, and the curl $\partial_iF_j - \partial_j F_i$ of $F_i$ provides enough information to reconstruct the original vector field $F_i$.
I am wondering how is it that one can reconstruct $F_i$ only from divergence and curl, while the complete Jacobian $J_{ij} = \partial_j F_i$ corresponds additional information. Divergence is the trace of $J_{ij}$, curl is skew-symmetric part, while the symmetric part $\partial_j F_i + \partial_i F_j$, sometimes known as shear, does not participate in the Helmholtz theorem. I would like to intuitively understand this behavior.
In the case of a scalar function $f$, one needs to know all of its derivatives $\partial_i f$ in order to reconstruct $f$ via line integral. However, in the case of Helmholtz theorem, we are reconstructing $F_i$ as a volume integral. Does this make any difference?

The Helmholtz theorem in effect tells us that the shear can be computed if you know the curl and the divergence. This much is obvious.
If we just count degrees of freedom (dof), a vector field has 3 dof per point. To specify its curl (3 dof) and divergence (1 dof) means to fix 4 dof per point. Thus from this perspective, the puzzle is not why only div and curl is enough, but how is it that specifying div and curl does not result in an overdetermined problem.
To answer this question, let us look at the divergence and curl separately. 


*

*First, if $\nabla\times F=0$, then $F$ is a conservative field, i.e., there is a scalar field $\phi$ such that $F=\nabla\phi$. So roughly speaking, a curl-free field has 1 dof per point.

*Now if $\nabla\cdot F=0$, then there is a vector potential $A$ such that $F=\nabla\times A$. So it appears that a divergence-free vector field has 3 dof per point. However, we should remember that the curl of a gradient vanishes, meaning that $A$ is determined only up to a gradient. More specifically, if $\psi$ is any scalar field, then $\nabla\times(A+\nabla\psi)=\nabla\times A$. This means that a divergence-free vector field really has 2 dof per point.


In other words, the curl of $F$ "cannot see" the gradient in $F$, and so specifying the curl of $F$ means specifying only 2 "real" degrees of freedom. Then the remaining 1 degree of freedom is fixed when we give the divergence.
