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?


2 Answers 2


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.

  • $\begingroup$ 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. $\endgroup$
    – Hachino
    Jan 31, 2015 at 6:51
  • $\begingroup$ @Hachino, by virtue of being homogeneous polynomials, all $w_k(0) = 0$, except for $k=1$. $\endgroup$ Jan 31, 2015 at 15:16
  • $\begingroup$ @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. $\endgroup$
    – Hachino
    Feb 1, 2015 at 9:14
  • $\begingroup$ 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 $x\ne 0$. $\endgroup$ Feb 1, 2015 at 13:19
  • $\begingroup$ @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. $C^\infty$). $\endgroup$
    – Bazin
    Feb 2, 2015 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.

  • $\begingroup$ 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. $\endgroup$
    – Bazin
    Jan 27, 2015 at 12:15
  • $\begingroup$ The answer has been thoroughly modified, have a look at it if you want to. :) $\endgroup$
    – Hachino
    Jan 28, 2015 at 8:20
  • $\begingroup$ @Hachino, I think you might find it helpful to work out the Fourier space version of my updated answer. $\endgroup$ Jan 28, 2015 at 9:42
  • $\begingroup$ 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.) $\endgroup$
    – Hachino
    Jan 28, 2015 at 9:47
  • $\begingroup$ 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 $\text{curl}\ u$. $\endgroup$
    – Hachino
    Jan 28, 2015 at 11:15

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