# Surjectivity of curl

Let: $\mathbb R^3\ni x\mapsto v(x)\in\mathbb R^3$ be a vector field with null divergence belonging to the Schwartz class such that $$\int_{\mathbb R^3} v(x) dx=0.$$ Is it true that there exists a vector field $w$ in the Schwartz class such that $$\text{curl } w=v\quad?$$ In other words, this is a regularity question for a Poincaré lemma: let $u$ be a closed two-form on $\mathbb R^3$. Then, there exists a one-form $a$ such that $u=da$. If $u$ is smooth, $a$ can be chosen smooth; the above question can be reformulated: if $u$ belongs to the Schwartz class, is it possible to choose $a$ in the Schwartz class?

I think that the answer is Yes.

1st step. Because Fourier transform is an automorphism of the Schwartz class, the problem is equivalent to show that every vector field $v(x)\in{\mathcal S}({\mathbb R}^3)^3$ verifying $x\cdot v(x)\equiv0$ can be divided'', that is can be written $$v(x)=x\wedge w(x), \qquad w\in{\mathcal S}({\mathbb R}^3)^3.$$

2nd step. The jets at the origin. Let $$v(x)\sim \sum_kv^k(x)$$ be the formal power series of $v$ at the origin, with $v_k$ a homogeneous polynomial vector field of degree $k$. Each $v_k$ satifies $x\cdot v_k(x)=0$.

It is an algebraic fact that there exists a polynomial vector field $w_k$, homogeneous of degree $k-1$, such that $v_k=x\wedge w_k$. Choose a $\phi\in{\mathcal D}({\mathbb R}^3)$ be such that $\phi\equiv1$ in $B(0;1)$, and form $$w^0(x)=\sum_k\phi(kx)w_k(x).$$ This is a locally finite series, therefore convergent to a ${\cal C}^\infty$-function away from the origin, compactly supported. It is actually ${\cal C}^\infty$ everywhere, because $w^0$ differs from a ${\cal C}^k$ field by an $O(|x|^{k+1})$. Therefore $w^0$ is in the Schwarz class. Then the jet of $v^0:=x\wedge w^0$ equals that of $v$ at the origin.

3rd step. Away from the origin. Define $v^1:=v-v^0$, which is of Schwartz class and is flat at the origin. Define $$w^1=\frac{x}{|x|^2}\wedge v^1$$ is of Schwarz class and satisfies $v^1=x\wedge w^1$.

Finally, $w=w^0+w^1$ is the solution of the problem.

• Nice one ! Could you please elaborate a bit on why the series defining $w^0$ converges in a neighborhood of the origin ? I'm probably missing something simple, but since the Taylor coefficients at $0$ of $v$ could be anything, convergence is not completely obvious. Jan 31 '15 at 6:51
• @Hachino, by virtue of being homogeneous polynomials, all $w_k(0) = 0$, except for $k=1$. Jan 31 '15 at 15:16
• @IgorKhavkine : This is convergence at $0$, not around $0$. Assume that the dominant coefficient of $v_k$ grows like $(k!^k)$ (which may indeed happen), how come that $w^0$ makes sense ? I trust Denis Serre when he says so, but still, I would like to understand why this holds. Feb 1 '15 at 9:14
• As Denis wrote, the series defining $w^0(x)$ is locally finite, in the sense that $\phi(kx)w_k(x)$ is non-zero only for finitely many $k$ at any $x\ne 0$. Feb 1 '15 at 13:19
• @Denis Serre: I do not understand your Borel-type argument for the smoothness of $w^0$ at the origin. The size of $w_k$ on the sphere could be anything and your argument ("differs from…") would provide smoothness for $\sum_{k\ge 0}\phi(kx)a_k x^k$ for any sequence. Given a sequence $(a_k)_{k\ge 0}$, it is possible to choose a sequence $(\mu_k)_{k\ge 0}$ (depending heavily on the $a_k$) such that $\sum_{k\ge 0}\phi(\mu_k x)a_k x^k$ is smooth (i.e. $C^\infty$). Feb 2 '15 at 21:17

Update : I deleted the previous answer, as it was irrelevant. Here is a second attempt, hopefully better than the first one. Morally speaking, it is often (if not always) possible to choose a Schwartz anticurl.

We will still work in Fourier space, but more symbolically this time. Let

$$\hat{w}(k) = \frac{k \wedge \hat{v}(k)}{|k|^2}$$

be our first guess. We know that it has a singularity near $0$, but if this singularity may be carried by a gradient, that is, if we have a decomposition

$$\hat{w}(k) = \hat{w}_{smooth}(k) + k \hat{f}(k)$$

with some irrelevant $\hat{f}$ and a Schwartz $\hat{w}_{smooth}$, we are done.

Because $v$ is divergence free, a few computations using that $k \cdot \hat{v}(k) = 0$ lead to :

$$\hat{w}(k) = \frac{1}{k_2} \left( \begin{array}{c} \hat{v}_3 \\ 0 \\ - \hat{v}_1 \\ \end{array} \right) + k \left( \frac{k_3}{k_1} \hat{v}_1 - \frac{k_1}{k_2}\hat{v}_3 \right).$$

And indeed, using again that $k \cdot \hat{v}(k) = 0$, you may check that $k \wedge$ (the left term) is, up to a sign we don't care about, $\hat{v}(k)$.

So, a sufficient condition for $\hat{w}_{smooth}$ to be... well, smooth is for both $\hat{v}_1$ and $\hat{v}_3$ to be divisible by $k_2$, which is almost given by the divergence free condition, but not exactly. In particular, such a condition implies immediately that $v$ has zero mean, but is much, much weaker than flatness.

Notive that we may reason symetrically with the other variables, leading to a divisibility condition with a "OR", not an "AND".

If you try to cook up a counterexample by negating one divisibility condition, then surely you will end up with something like

$$\hat{v}(k) = k \wedge e_1$$

(up to a fast decreasing factor to make it Schwartz) where $e_1$ is some fixed vector in $\mathbb{R}^3$, which obviously has a Schwartz anticurl.

If someone is able to prove or disprove that a Schwartz vector field satisfying $k \cdot \hat{v}(k)$ always satisfies the "OR divisibility" condition, this would be a definitive answer.

• Thanks for this very nice counterexample. The vanishing of the mean (i.e. $\hat \nu(0)=0$) that I required in the question above is certainly necessary, but as you have just pointed out is not sufficient. All moments should vanish, this is an elegant addendum to Poincaré Lemma. Jan 27 '15 at 12:15
• The answer has been thoroughly modified, have a look at it if you want to. :) Jan 28 '15 at 8:20
• @Hachino, I think you might find it helpful to work out the Fourier space version of my updated answer. Jan 28 '15 at 9:42
• I don't follow you when you claim that $\text{curl} \ u = v$, in that in Fourier space, your equation reads $\hat{u}(k) = \frac{k \wedge \hat{v}(k)}{|k|^2} \hat{\phi}(k)$. Thus, we have $k \wedge \hat{u}(k) = \hat{v}(k) \hat{\phi}(k)$, which is not $\hat{v}(k)$. (Again, up to a sign somewhere, but who cares.) Jan 28 '15 at 9:47
• Correction : $\hat{u}(k) = (k \wedge \hat{v}(k)) \hat{\left( \frac{\phi}{|\cdot|} \right)}(k)$ and $k \wedge \hat{u}(k) = |k|^2 \hat{v}(k) \hat{ \left( \frac{\phi}{| \cdot |} \right)}(k)$. Which still has few reasons to agree with the $\text{curl}\ u$. Jan 28 '15 at 11:15