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Let $[a_{i,j}(x_1,\ldots,x_n)]$ be a skew-symmetric $n\times n$ matrix of functions $a_{i,j}\in C^\infty(\mathbb{R}^n)$. The vector field $$v=\sum\left(\dfrac{\partial}{\partial x_i}a_{i,j}\right)\dfrac{\partial}{\partial x_j}$$ is divergence-free.

Prove by induction that for every $n\geq 2$, every $C^\infty$ divergence-free vector field on $\mathbb{R}^n$ is of this form.

Consider $n=2$. Suppose the vector field is $f_1(x_1,x_2)\dfrac{\partial}{\partial x_1}+f_2(x_1,x_2)\dfrac{\partial}{\partial x_2}$. Since the vector field is divergence-free, we have that $\dfrac{\partial}{\partial x_1}f_1(x_1,x_2)+\dfrac{\partial}{\partial x_2}f_2(x_1,x_2)=0$. By this result, there exists a function $g(x_1,x_2)$ whose $x_1$-derivative equals $f_2$ and whose $x_2$-derivative equals $-f_1$. The result follows.

But how about for $n>2$? To use induction, I have to relate a divergence-free vector field of $\mathbb{R}^n$ to a divergence-free vector field of $\mathbb{R}^{n-1}$. It is possible that the following result will help:

Let $v$ be a vector field on $\mathbb{R}^n$. Show that $v$ can be written as a sum $v=f_1\dfrac{\partial}{\partial x_1}+w$ where $w$ is a divergence-free vector field.

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    $\begingroup$ The case $n=3$ is the result that says if $\text{div}(u) = 0$, then $u = \text{curl} v$ for some vector $v$. $\endgroup$ – Stephen Montgomery-Smith Nov 24 '13 at 5:21
  • $\begingroup$ Also the case $n=2$ was answered here: math.stackexchange.com/questions/578709/… - think of $(-h,g)$ as the divergence free vector field, and $\pm \frac12 f$ as the off diagonal entries of the skew matrix. $\endgroup$ – Stephen Montgomery-Smith Nov 24 '13 at 16:00
  • $\begingroup$ @StephenMontgomery-Smith Yes, exactly. I wrote that the case $n=2$ was taken care of by that link. What I'm stuck is jumping from $n-1$ to $n$ for the induction. $\endgroup$ – JJ Beck Nov 24 '13 at 16:01
  • $\begingroup$ Today is very busy for me, so I probably won't have time to think of a solution. But in view of the previous questions you asked, and their relationship to this problem, it makes sense to think "anti-derivative." $\endgroup$ – Stephen Montgomery-Smith Nov 24 '13 at 16:02
  • $\begingroup$ Sorry I didn't see your $n=2$ comment. Hey, here is a way that I find helpful. Instead of working with the functions directly, work with their Fourier transforms. Then the condition that $v$ is divergence free becomes $k \cdot \vat v(k) = 0$, where $k \in \mathbb R^n$ is sometimes called the wavenumber. So in the 3D case, you want to conclude that there exists a vector $v$ such that $\hat u = k \times \hat v$. This reduces the whole problem to vector algebra (e.g. in this case try $v = -k\times \hat u$.) Once you have the algebra worked out, it should tell you where to apply anti-derivs. $\endgroup$ – Stephen Montgomery-Smith Nov 24 '13 at 16:20
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Induction Hypothesis over $k$: Given a smooth divergence free vector field $v:\mathbb R^n \to \mathbb R^n$ such that $\text{div}_{k} v := \sum_{i=1}^k \frac{\partial v_i}{\partial x_i} = 0$, there exists a smooth skew-symmetric matrix $a:\mathbb R^n \to \mathbb R^{n\times n}$ such that $v_j = \sum_{i=1}^k \frac\partial{\partial x_i} a_{ij}$ for $1 \le i \le k$.

The case $k=0$ is trivial.

Suppose it is true for $k-1 \ge 0$. We prove it for $k$.

Let $$ f_1(x_1,\dots,x_n) = \int_0^{x_1} \frac{\partial}{\partial x_k} v_k(\xi,x_2,x_1,\dots,x_n) \, d\xi .$$ Then $$ \frac{\partial}{\partial x_1}(v_1+f_1) + \frac{\partial}{\partial x_2}v_2 + \dots + \frac{\partial}{\partial x_{k-1}}v_{k-1} = 0 .$$

By the inductive hypothesis, there is a skew symmetric matrix $a_{ij}$ such $$ v_1 + f_1 = \sum_{i=1}^{k-1} \frac{\partial}{\partial x_i} a_{i1} $$ $$ v_j = \sum_{i=1}^{k-1} \frac{\partial}{\partial x_i} a_{ij} \quad \text{ for $2 \le j \le k-1$}$$

We define $$ f_2(x_1,\dots,x_n) = \int_0^{x_1} v_k(\xi,x_2,\dots,x_{k-1},0,x_{k+1},\dots,x_n) \, d\xi - \int_0^{x_k} f_1(x_1,\dots,x_{k-1},\xi,\dots,x_n) \, d\xi .$$ Then $$ \frac\partial{\partial x_1} f_2 = v_k(x_1,\dots,x_{k-1},0,\dots,x_n) - \int_0^{x_k} \frac\partial{\partial x_k} v_k(x_1,\dots,x_{k-1},\xi,\dots,x_n) \, d\xi = -v_k $$ and $$ \frac\partial{\partial x_k} f_2 = - f_1 $$ Now extend the matrix $a$ by setting $a_{k1} = -a_{1k} = f_2$, and $a_{kj}=a_{jk} = 0$ for $2 \le j \le k$. Then $$ \sum_{i=1}^{k} \frac{\partial}{\partial x_i} a_{i1} = v_1 + f_1 + \frac\partial{\partial x_k} f_2 = v_1, $$ $$ \sum_{i=1}^{k} \frac{\partial}{\partial x_i} a_{j1} = v_j \quad \text{ for $2 \le j \le k-1$}, $$ and $$ \sum_{i=1}^{k} \frac{\partial}{\partial x_i} a_{ik} = - \frac\partial{\partial x_1} f_2 = v_k. $$

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  • $\begingroup$ Hmm.. I think the inductive hypothesis only holds if you know $v_1-f_1,v_2,\ldots,v_{n-1}$ are functions of $x_1,x_2,\ldots,x_{n-1}$, but here they're also functions of $x_n$. $\endgroup$ – JJ Beck Nov 25 '13 at 4:16
  • $\begingroup$ That's why you fix $x_n$ so that you can apply it separately for each $x_n$. $\endgroup$ – Stephen Montgomery-Smith Nov 25 '13 at 4:35
  • $\begingroup$ Why will the resulting matrix have non-zero entries only in the first column or the first row? I can't see that... $\endgroup$ – JJ Beck Nov 25 '13 at 5:12
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    $\begingroup$ Because at the $n$th induction step, all I added was entries in the $(1,n)$ and $(n,1)$ positions to the prior $(n-1)\times(n-1)$ matrix. $\endgroup$ – Stephen Montgomery-Smith Nov 25 '13 at 5:15
  • $\begingroup$ For 100 points, I'll put in some more details. I have to pick up my kids right now. So I'll get to it later. $\endgroup$ – Stephen Montgomery-Smith Nov 26 '13 at 21:24
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Well, since induction is mentioned, I'd like to give a try. Note that $$\nabla\cdot v=\sum_{i,j}^n\partial_i(\partial_ja_{ij})=\sum_{i,j}^{n-1}\partial_i(\partial_ja_{ij})+\sum_i^{n-1}\partial_i(\partial_na_{in})+\sum_j^{n-1}\partial_n(\partial_ja_{nj})=0$$ Wow... It seems easier than what I've thought.

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  • $\begingroup$ I'm confused - we're asked to show that the $a_{ij}$ exist, so how can you assume their existence in the first place? Or maybe I misunderstood your argument. $\endgroup$ – JJ Beck Nov 25 '13 at 4:40
  • $\begingroup$ @JJBeck As I mentioned here, Poincare lemma gives the existence of such $A$. $\endgroup$ – Shuchang Nov 25 '13 at 4:43
  • $\begingroup$ Hmm okay. I appreciate your answer, but unfortunately it's way beyond the material I'm studying, so its use is limited for me. $\endgroup$ – JJ Beck Nov 25 '13 at 5:25
  • $\begingroup$ This question is equivalent to the Poincare lemma. Treat vector field as one form, then divergence can be thought of as $d^*$ with the standard metric. $\endgroup$ – user99914 Nov 25 '13 at 8:37

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