For a divergence free smooth vector field $v : \mathbb{R}^3 \to \mathbb{R}^3$, how to find the commutator form of the matrix $A=(\partial_i v_j)$?

Question

The question is as in the title.

I know that a traceless matrix can be written as a commutator of two matrices.

Then, let $v : \mathbb{R}^3 \to \mathbb{R}^3$ be a divergence-free smooth vector field. That is, $\nabla \cdot v=0$.

Then, the matrix $A(x)=\bigl[\partial_i v_j(x) \bigr]$ is smooth as a mapping from $\mathbb{R}^3$ into $M_{3 \times 3}(\mathbb{R})$ and tracelss for each $x$.

Then, what does any two matrices $B(x)$ and $C(x)$ that satisfy $A(x)=B(x)C(x)-C(x)B(x)$ look like?

This is quite new and original for me..Could anyone please help me?

Answer 1

You might want to first carry out a unitary transformation $A(x)\mapsto U(x)A(x)U^\top(x)$, such that all diagonal elements are zero.$^\ast$
Then $A(x)=BC(x)-C(x)B$ with $$B=\begin{pmatrix} 1&0&0\\ 0&2&0\\ 0&0&3 \end{pmatrix},\;\; C_{ij}(x)=\begin{cases} 0&\text{if}\;i=j\\ \frac{A_{ij}(x)}{B_{ii}-B_{jj}}&\text{if}\;\;i\neq j. \end{cases} $$

---

$^\ast$ This is always possible for a traceless $A$ (proof).

For a $3\times 3$ matrix the unitary has the two-parameter form $$U(x)=\begin{pmatrix} \cos\alpha(x)&\sin\alpha(x)&0\\ -\sin\alpha(x)&\cos\alpha(x)&0\\ 0&0&1 \end{pmatrix} \begin{pmatrix} 1&0&0\\ 0&\cos\beta(x)&\sin\beta(x)\\ 0&-\sin\beta(x)&\cos\beta(x) \end{pmatrix} .$$ You can solve first for $\alpha(x)$, $$(\partial_1 v_1) \cos ^2\alpha+(\partial_2v_2) \sin ^2\alpha+ (\partial_1v_2+\partial_2 v_1)\sin\alpha\cos\alpha=0,$$ and then for $\beta(x)$.