# Surjectivity of differential operators with constant coefficients

I would like a proof or a reference (or a counter-example...) for the following fact. Let $$P\in \mathbb{C}[x_1,\ldots ,x_n]$$ and $$D\in \mathbb{C}[\frac{\partial }{\partial x_1} ,\ldots ,\frac{\partial }{\partial x_n}]$$ be nonzero homogeneous polynomials. Then there exists a homogeneous polynomial $$Q\in \mathbb{C}[x_1,\ldots ,x_n]$$ such that $$D(Q)=P$$.

Here is another approach. Let $$R$$ be a non-zero homogeneous polynomial of degree $$n$$. We want to show that the mapping $$Q\mapsto R(\partial)Q$$ is surgective from $$V_{m+n}$$ to $$V_m$$ where $$V_k$$ is the space of homogeneous polynomials of degree $$k$$. Note that $$\langle A,B\rangle=[A(\partial)\bar B](0)$$ is a scalar product on $$V_k$$ (with monomials forming an orthogonal basis). If our mapping is not surjective, then there exists a non-zero polynomial $$S\in V_m$$ such that $$\langle S, R(\partial) Q\rangle=\langle S\bar R,Q\rangle=0$$ for all $$Q\in V_{n+m}$$. But $$S\bar R$$ is a non-zero polynomial, so taking $$Q=S\bar R$$, we get a contradiction.

• Nice, thanks! In fact that works over any field, you just need to view $R(\partial )$ as an element of $V_n^{*}$. – abx Nov 21 '18 at 17:27
• @abx I think the argument needs a field of characteristic 0, or at least characteristic not dividing $(m+n)!$ (here $n$ is not the number of variables) – Mizar Nov 21 '18 at 22:07
• Right, thanks. This is needed for the duality between symmetric powers. – abx Nov 22 '18 at 7:12
• @abx more specifically to this argument, monomials are orthogonal but not orthonormal, and for a monomial $M$, the pairing $\langle M,M\rangle$ vanishes in characteristic $p$ if $p$ is less than or equal to one of the exponents of $M$. – Vladimir Dotsenko Nov 23 '18 at 12:18

Too long for a comment. I want to use a version of the Lojaciewicz theorem of division of distributions by an analytic function (in fact Hörmander's result of division by a polynomial). We may assume that $$P(x) =x^\alpha=x_1^{\alpha_1}\dots x_n^{\alpha_n}$$ a monomial homogeneous with degree $$\vert \alpha\vert=\alpha_1+\dots +\alpha_n$$. Let us replace your notation $$D$$ by an operator $$A(D)=\sum_{\vert \beta\vert =m}a_\beta D^\beta,\quad D=-i\nabla.$$ The question at hand is to find an homogeneous polynomial $$Q$$ such that $$A(D) Q= x^\alpha.$$ By Fourier transformation, it is equivalent to solve $$A(\xi)\widehat Q(\xi)=i^{\vert \alpha\vert}\delta^{(\alpha)},$$ which means divide a derivative of the Dirac mass by an homogeneous ($$A$$ is assumed to be non-zero) polynomial. It is indeed possible by the aforementioned results of division and we find that $$\widehat Q$$ is homogeneous with degree $$-n-\vert \alpha\vert-m$$, so that $$Q$$ is homogeneous with degree $$\vert \alpha\vert+m$$. Moreover $$\widehat Q$$ is supported at the origin, proving that $$Q$$ is a polynomial.

• Interesting, thanks! Though a bit of an overkill... – abx Nov 22 '18 at 7:14

If $$V_a$$ is the space of homogenous polynomials in $$x_1,...,x_n$$ of degree $$a$$ and $$V_a^\vee$$ its dual, then differentiation is a $$GL_n$$-equivariant bilinear map $$V_{d+e}\times V_e^\vee\to V_d$$. Your question is whether, given non-zero $$D\in V_e^\vee$$, the linear map $$V_{d+e}\times\{D\}\to V_d$$ is surjective.

For this you can specialise $$D$$ to define a closed $$GL_n$$-orbit in $$\mathbb P(V_e^\vee)$$. That is, you can take $$D=\partial^e/\partial x_1^e$$ and then the map is obviously surjective.

• I'm not sure I understand the last two sentences, What is obvious (to me) from what was written before is that it is enough to get $x_1^d$ in the image, but that's not what you are saying. Can you elaborate a bit? – fedja Nov 21 '18 at 14:34
• Same as @fedja: why is it enough to check surjectivity for one particular orbit? – abx Nov 21 '18 at 14:38
• In the projective space $\mathbb P(V_e^\vee)$ there is a unique closed orbit, the orbit of the highest weight vector $\partial^e/\partial x_1^e$. This is contained in the closure of every orbit. Let $U$ be the locus of points $D$ in $\mathbb P(V_e^\vee)$ for which $V_{d+e}\times\{D\}$ is surjective. Then $U$ is Zariski open and $GL_n$-invariant, so it is enough to show that $U$ contains $\partial^e/\partial x_1^e$. For this the map is obviously surjective. – inkspot Nov 21 '18 at 17:16