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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?

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  • $\begingroup$ Basically, it tells you that the singular part of the (distributional) differential, which is a nasty matrix valued measure, behaves as in the simple case of "jump points" (see Definition 3.67 in the book by Ambrosio, Fusco and Pallara): namely, if you blow up your vector field at a.e. point with respect to this singular measure (rescaling it suitably in domain and target) what you see in the limit is a vector field depending only on, say, the first component, up to rotations. $\endgroup$
    – Mizar
    Commented Nov 22, 2018 at 22:17
  • $\begingroup$ @Mizar Thank you. Do you mind elaborating a bit in an answer on that remark? That is, what do you mean when you say "if you blow up your vector field at a.e. point with respect to this singular measure (rescaling it suitably in domain and target) what you see in the limit is a vector field depending only on, say, the first component, up to rotations"? $\endgroup$
    – user123457
    Commented Nov 23, 2018 at 15:46
  • $\begingroup$ Just to mention that Alberti's result has been generalized by G. De Philippis & F. Rindler in Ann. of Math. 184 (2016), pp 1017-1039. They prove that if a vector-valued bounded measure $\mu$ satisfies a homogeneous PDE $A\mu=0$, where $A$ has constant coefficients, then the singular part $\mu^s$ takes $|\mu|^s$-almost everywhere values in the wave cone of $A$. Alberti's case occurs when $\mu$ is matrix-valued and $A$ is the row-wise curl. $\endgroup$
    – Denis Serre
    Commented Dec 11, 2018 at 19:37
  • $\begingroup$ @DenisSerre Thank you. Heuristically, what does it mean that the singular part of the derivative of a BV function has almost everywhere values in the wave cone of the row-wise curl? $\endgroup$
    – user123457
    Commented Dec 11, 2018 at 19:49
  • $\begingroup$ @Rene. It is hard to develop an answer to this question within a comment. I suggest that you have a look at the introduction of De Philippis & Rindler's paper. In the case covered by Alberti, the wave cone is that of rank-one matrices $a\otimes b$ with $a\cdot b=0$. $\endgroup$
    – Denis Serre
    Commented Dec 11, 2018 at 20:00

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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).

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  • $\begingroup$ I have no more heuristic explanations. If you want to enter the details, I would recommend the paper MR2124585,Transport equations with partially BV velocities, by N. Lerner and the L. Ambrosio article quoted there. $\endgroup$
    – Bazin
    Commented Dec 12, 2018 at 14:14
  • $\begingroup$ After going through your reference, I'm still unclear about the fact that "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).". Could you add some details on it? $\endgroup$
    – user123457
    Commented Apr 16, 2019 at 11:50
  • $\begingroup$ Let me put it that way. You must try to prove your theorem on your vector field $X$ (say uniqueness of weak solutions) for the particular case where $DX$ is as above, which is not so easy. After this, you enter a more technical area, in which you are not stricto sensu reduced to that case, but you can argue via an approximation procedure. The key argument for the uniqueness of weak solutions is a commutator estimate between the vector field and a smoothing operator, and this argument leaves plenty of room for approximation. $\endgroup$
    – Bazin
    Commented Apr 17, 2019 at 11:58

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