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Is the following proposition true?



> 
> Given a multidifferential operator $D$ on $\mathbb{R}^n$ with constant coefficients, i.e. for all functions $f\_1,\dots,f\_k \in C^\infty(\mathbb{R}^n)$ we have
>  \begin{align\*}
> D(f\_1,\dots,f\_k)(x) := \sum\_{\alpha\_1,\dots,\alpha\_k} c\_{\alpha\_1 \dots \alpha\_k} \partial^{\alpha\_1} f\_1(x) \dots \partial^{\alpha\_k} f\_k(x)
> \end{align\*}
>  where we sum over all multiindices $\alpha\_1,\dots,\alpha\_k \in \mathbb{N}\_0^n$ and only finitely many $c\_{\alpha\_1 \dots \alpha\_k} \in \mathbb{R}$ are nonzero. 
> 
> 
> Then 
>  \begin{align\*}
> \int\_{\mathbb{R}^n} D(f\_1,\dots,f\_k)(x) dx = 0 \quad \forall f\_1,\dots,f\_k \in C^\infty\_c(\mathbb{R}^n)
> \end{align\*}
>  if and only if $D = \sum\_{i=1}^n \partial\_i E\_i$ for some multidifferential operators $E\_1,\dots,E\_n$ with constant coefficients.
> 
> 
> 




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This looks a lot like you could just hit it with the Poincaré lemma for compactly supported forms, but the Poincaré lemma only gives me that whenever I fix the functions $f\_1,\dots,f\_n$, I get that $D(f\_1,\dots,f\_n)$ is the derivative of another function. It seemingly does not allow me to get a single "integrating" differential operator for all functions. 


Looking into the proof for the compactly supported Poincaré lemma, it seems not directly clear to me that the fibrewise integration procedure will again give me a differential operator in $f\_1,\dots,f\_n$.


Also, is there something simple one can say about the shape of $D$ when we replace $
\int\_{\mathbb{R}^n} D(f\_1,\dots,f\_k)(x) dx = 0 $ with $
\int\_{\mathbb{R}^n} p(x\_1,\dots,x\_n) D(f\_1,\dots,f\_k)(x) dx = 0 $ for some fixed polynomial $p$? Or a general fixed (smooth) function?


(Tagging this with homological algebra because I imagine this statement could be the vanishing of some cohomology group if one stares at it for long enough.)



Multidifferential operators with vanishing integrals
Denote the space of differential operators $D$ you consider by $\mathcal D^k$, and the subspace where $c\_{\alpha\_1\dots\alpha\_k} = 0$ for $\alpha\_1\neq 0$ by $\mathcal D^k\_{lin}$, i.e. which are $C^\infty(\mathbb R^n)$-linear in $f\_1$. The functional
$$
I:\mathcal D^k\to \operatorname{Hom}(C^\infty\_c(\mathbb R^n)^{\otimes k},\mathbb R),D\mapsto \left(f\_1\otimes\dots,f\_k\mapsto \int\_{\mathbb R^n} D(f\_1,\dots,f\_k)\,\mathrm dx\right)
$$
is injective when restricted to $\mathcal D^k\_{lin}$: By definition, $D(f\_1,\dots,f\_k) = f\_1\widetilde D(f\_2,\dots,f\_k)$ for some $\widetilde D\in \mathcal D^{k-1}$, and we can always find some $f\_2,\dots,f\_k$ such that $\widetilde D(f\_2,\dots,f\_k)(0)> 0$ (for instance, choose polynomials dual to some term of highest degree). Taking $f\_1$ to be a bump function supported in a small neighbourhood of $0$ then shows that $I(D)(f\_1\otimes\dots\otimes f\_k) > 0$.


Now consider the divergence operator 
$$
\operatorname{div}:\mathcal D^k\otimes\mathbb R^n\to\mathcal D^k, D^i\otimes e\_i\mapsto \sum\_{i=1}^n \partial\_i D^i
$$
and denote its image by $\mathcal D^k\_0$. By partial integration, $I\circ\operatorname{div} = 0$, so by the previous result $\mathcal D^k\_{lin}\cap \mathcal D^k\_0 = 0$. On the other hand, $\mathcal D^k = \mathcal D^k\_{lin} + \mathcal D^k\_0$, which we can prove by induction on the degree $d$ of the differential operator hitting $f\_1$: $d = 0$ is just the definition of $\mathcal D^k\_{lin}$, and you can always decrease this degree by formal partial integration, which corresponds to adding a divergence. It follows that $\mathcal D^k = \mathcal D^k\_{lin} \oplus \mathcal D^k\_0$ and by injectivity of $I|\_{\mathcal D^k\_{lin}}$ we obtain $\ker I = \mathcal D^k\_0$.


