Heuristically what does Alberti's *rank-one theorem* imply about the structure of a $\mathrm{BV}$ vector field $\boldsymbol{b}$? 


Is it rigorously fair to say that the level lines of $\boldsymbol{b}$ are all "parallel" and pointing in one direction? Why? 



Meaning of Alberti rank-one theorem
Let $X=\sum\_ja\_j(x)\frac{\partial }{\partial x\_j}$ be a $BV$ vector field in an open subset of $\mathbb R^n.$ Alberti's theorem says that 
$$
DX\_s=(\frac{\partial a\_j }{\partial x\_k})\_{1\le j,k\le n}=S \otimes \eta,
\quad \text{$S(x)$ tangent vector at $x$, $\eta(x)$ cotangent vector at $x$}
$$
i.e for $T$ tangent at $x$,
$
(DX)\_s(x) T= \langle \eta(x), T\rangle S(x).
$
Intuitively, it means essentially that, near each point, you can find a coordinate system such that each coefficient $a\_j$ of your vector field is in fact a function 
$$
a\_j(x', x\_n) \text{ such that}\ \frac{\partial a\_j}{\partial x'} \in L^1,\quad
\frac{\partial a\_j}{\partial x\_n} \text{is a Radon measure},
$$
where $x'$ is $n-1$ dimensional and $x\_n$ has just one dimension.


Let me add a couple of comments. You have a canonical decomposition of $DX$. What matters is the singular part. When I write as above $DX\_s=S\otimes \eta$, it means that for a tangent vector $T$, $$\underbrace{DX\_s}\_{\text{matrix}}T=\langle\eta,T\rangle \underbrace{S}\_{\text{vector}}.$$ You may decide that $\eta=e\_n^\*$ so that $DX\_sT=T\_n S$ and if you know that the divergence of $X$ is absolutely continuous, you have $S\_n=0$, so that you may assume that $S=e\_1$. As a result,
$$
DX\_s e\_n=e\_1,\quad DX\_s e\_k=0 \text{ for $1\le k\le n-1$}.
$$
Since $DX=\bigl(\frac{\partial a\_j}{\partial x\_k}\bigr)$, $j$ line-index, $k$ column-index, you get that only the last column
$$
\frac{\partial a\_j}{\partial x\_n}\quad \text{can be a Radon measure},\quad
\frac{\partial a\_j}{\partial x\_k}, k\le n-1\quad \text{(any $j$) is $L^1$}
$$
With our assumption on the divergence and our choice $S=e\_1$, we have only
$$
\frac{\partial a\_1}{\partial x\_n} \text{singular, other entries $L^1$.}
$$
Of course, you have to keep in mind that $S,\eta$ depend on the point $x$, but through an approximation procedure, you may indeed assume that singularity occurs by differentiating only in one direction (and with null divergence (or ac) is concerning only one off-diagonal coefficient).


