Assume that $P(x,y), Q(x,y) \in \mathbb{R}[x,y]$ are two polynomials. We define a linear map

$D$ on $\mathbb{R}[x,y]$ with $D(U)=PU_{x}+QU_{y}$. In fact $D$ is the differential operator corresponding to the vector field $P\partial_{x}+Q\partial_{y}$.

Are there polynomials $P$ and $Q$ such that the codimension of the range of $D$ is finite but different from $0$ and $1$?

Note This codimension is $0$, $1$ and $\infty$ for $\partial_{x}$, $x\partial_{x}+y\partial_{y}$ and $x\partial_{x}-y\partial_{y}$, respectively.

Motivation It can be easily shown that this codimension is an upper bound for the number of closed orbits of the polynomial vector field $P\partial_{x}+Q\partial_{y}$.

However there is a Painful fact: If an smooth vector field $X$ on $\mathbb{R}^{2}$ has a limit cycle which surrounds a nondegenerate singularity (or a singularity which has a local smooth first integral at a deleted neighborhood of the singularity and is discontinuous at the singularity), then the codimension of the range of the differential operator $D_{X}$ on $C^{\infty} (\mathbb{R}^{2})$ is infinite ! The reason is as follows.

Assume that a limit cycle $\gamma$ is attractor and surounds a source singularity at the origin.

Non degeneracy of the singularity help us to find an open set D around origin whose boundary $S$ is a smooth closed curve and the vector field is transverse to $S$ toward the exterior of $S$. So the flow of $X$ defines a smooth retraction $r:\bar{D}\setminus \{0\} \to S$. In fact $r(x)$ is the intersection point of the orbit of $x$ with $S$ This shows that there is an smooth real valued function $\phi$ on $\mathbb{R}^2 \setminus \{0\}$ such that locally around the origin we have $X. \phi =0$ and $P\circ \phi$ is discontinuous at origin for every polynomial $P(x)$. Such $\phi$ can be constructed locally by $\psi \circ r$ where $\psi:S \to \mathbb{R}$ is an arbitrary non constant smooth function.Now we choose an arbitrary smooth extension of this locally defined $\phi$ to whole punctured plane. This extension is denoted again by $\phi$. Obviously every nontrivial linear combination $\sum \lambda_i \phi^i$ is discontinuos at origin. Because its restriction to every small neighborhood of the origin has the same range(image) as the range of $\sum \lambda_i \psi^i: S \to \mathbb{R}$. However $X.\phi^i$ is a smooth function on whole plane, for every $i$ since it vanish locally around the origin.

Now for every $n$,the following set represents an independent subset of $C^{\infty}(\mathbb{R}^2)/Range(D_X)$

$$\{X.\phi, X.\phi^2, \ldots, X.\phi^n \}$$ To prove this, we strongly use the fact that we have at least one limit cycle.

More precisely: assume that $\sum_{i=1}^n \lambda_i (X.\phi^i)= X.F$ for some smooth function $F\in C^{\infty} (\mathbb{R}^2)$. Then $F-\sum \lambda_i \phi^i$ is a first integral(constant of motion) on the puntured plane. So it is constant on the basian of attraction of the limit cycle. A punctured neighborhood of the origin is contained in the basian of attraction of limit cycle. Thus locally around the origin,$F$ differs $\sum_{i=1}^n \lambda_i \phi^i$ by a constant(In a deleted neighborhood). This is a contradiction because $F$ is smooth at origin but $\sum \lambda_i \phi^i$ is discontinuous at origin.

So unfortunately we encounter with the following two inconsistent and opposite but true results:

1)The codimension of the range of $D_X$ is an upper bound for the number of limit cycles.

2.If the codimension is finite then there is no any limit cycle.

So we should choose an appropriate function algebra, different from $C^{\infty} (\mathbb{R}^{2})$, which is invariant under the differential operator corresponding to an algebraic vector field.


According to the answer of Loic Teyssier we put $X=x^k\partial_x+y\partial_y$. Then $X$ can act on various spaces $C^{\infty}(\mathbb{R}^2),\; C^{\omega}(\mathbb{R}^2)$ or the space of holomorphic functions from $\mathbb{C}^2$ to $\mathbb{C}$. What can be said about the dimension of the cokernels in each of these actions?

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    $\begingroup$ Here's an idea. If you can take the standard monomial basis, then you can write down the operator as an infinite matrix. The codimension is then the number of zero rows. The matrix has a block structure given by the degree of monomials and the particular form of $D$ shows that you can control the behaviour of coefficients in your matrix with respect to this block structure. In particular, the degree can drop at most by one. ... $\endgroup$ Commented Jul 25, 2014 at 20:50
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    $\begingroup$ ... Thus it may be possible to decouple the problem to a finite initial part, where you could try to construct matrix with zero rows, and the remaining infinite part where you would have to show that your choice of coefficients doesn't lead to zero rows. (Also, tags weyl-algebra and d-modules may be appropriate.) $\endgroup$ Commented Jul 25, 2014 at 20:50

2 Answers 2


The question admits a positive answer when one looks at it for the ring of formal power series $\mathbb R[[x,y]]$. For instance the vector field $x^k\partial_x+y\partial_y$ has a formal cokernel of dimension $k$. Yet when restricted to polynomials the property disappears: the polynomial cokernel is infinite-dimensional (you can never reach monomials of the form $x^{k-1}y^m$ for $m\in\mathbb N$ nor $x^n$ with $n<k$, and those are the only problems).

Somehow I believe the answer should also be positive in the polynomial case. But I have no example offhand.

  • $\begingroup$ I think something is missing in your answer. I think the codimension is $\infty$ for $x\partial_{x}$ or $(x^{2}+y^{2})\partial_{x}$ while this codimension is zero for $\partial_{x}$. Am I mistaken? $\endgroup$ Commented Jul 18, 2014 at 21:40
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    $\begingroup$ No, you're right. I was too quick. I'll modify the answer. $\endgroup$ Commented Jul 18, 2014 at 21:42
  • $\begingroup$ thank you for your edited version. Your new version is a motivation to consider the derivational operator on the space of real analytic function(Instead of polynomial functions). However I am thinking to consider some other functional space which is safe from the danger of infinite dimensionality. so the first subject which arose me is **the space of holomorphic sections ** of a holomorphic bundles on $\mathbb{C}P^{2}$. I learned from the MS theisis of Saeid Zakeri that a polynomial vec. field gives us a holomorphic line bundle. But... $\endgroup$ Commented Jul 30, 2014 at 16:02
  • $\begingroup$ ...But I do not know what can I do with this budle and its holomorphic section. I try to learn some related material from the book Principle of algebraic geometry". I look forward to hear your opinion on this "holomorphic" idea. Is there a reasonable way to continue? As another condidate I would like to consider the space of real analytic functions on $\mathbb{R}^{2}-S$ whith smooth flat extension at $S$, $S$ is singularities. I Have some psychological reasons for this consideration, but I am not sure that this idea works well. Thanks for your attention $\endgroup$ Commented Jul 30, 2014 at 16:08
  • $\begingroup$ Dear Loic, it seems that the link of your talk is temporarely unavailable. Can you write in a paragraph, the summary of your talk in Toulous? I had read your talk in that link but may be other participants wish to read the abstract of your talk?If you have a saved version of that abstract, may you copy past that abstract as a revision of the following post? mathoverflow.net/questions/271336/… $\endgroup$ Commented Jan 30, 2018 at 6:10

This is not an answer.

I've tried some experiments in Maple and I believe that the answer to your question is negative. I've looked at the infinite matrix given by $D$ as I've suggested in my comments.

A column in this infinite matrix is given by coefficients of $D(x^iy^j) = iP(x,y)x^{i-1}y^j + jQ(x,y)x^iy^{j-1}$ which means that the entries in the matrix look like $\lambda p_{ab} + \mu q_{cd}$ for integral $\lambda$ and $\mu$. If one such an entry is zero, then it implies vanishing of infinitely many entries in the matrix, which are given by multiples of this specific $(\lambda, \mu)$ coming from some higher powers $x^{ki}y^{kj}$.

For the corner cases $x^i$ and $y^j$ one recovers multiples $P$ and $Q$, the coefficients are just more spread out throughout the rows. Since the matrix is band diagonal (the girth given by degrees of $P$ and $Q$), it seems that there is a slim that the codimension could be finite and bigger than 1.


If you want to consider various function spaces then it may be fruitful to reinterpret the statement in the theory of $\mathcal{D}$-modules, see for example chapter $8$ of "Homological Algebra" by S. I. gelfand and Yu. I Manian. It will also give you more literature to study the broader scope of your problem. I've took the liberty to add appropriate tags to your question. Please take the following lines with a grain of salt as it has been quite long since I've dealt with $\mathcal{D}$-modules.

Let $\mathcal{O}$ denote the space of polynomials in two variables and let $\mathcal{D}$ be the corresponding Weyl algebra, i.e. the space of differential operators in two variables with coefficients from $\mathcal{O}$. Finally, let $M =\mathcal{D}/\mathcal{D}D$ denote the quotient of $\mathcal{D}$ by the left ideal generated by $D$. If you apply the left exact hom-functor $\mathrm{Hom}(_,\mathcal{O})$ on the short exact sequence $$ 0 \to \mathcal{D} \xrightarrow{Q \mapsto QD} \mathcal{D} \to M \to 0 $$ you will obtain the following long exact sequence $$ 0 \to \mathrm{Hom}_\mathcal{D}(M,\mathcal{O}) \to \mathrm{Hom}_\mathcal{D}(\mathcal{D},\mathcal{O}) \to \mathrm{Hom}_\mathcal{D}(\mathcal{D},\mathcal{O}) \to \mathrm{Ext}^1_\mathcal{D}(M,\mathcal{O}) \to \mathrm{Ext}^1_\mathcal{D}(\mathcal{D},\mathcal{O}) \to \ldots $$

The vector space $\mathrm{Hom}_\mathcal{D}(\mathcal{D},\mathcal{O})$ is actually isomorphic to $\mathcal{O}$ via $\varphi \mapsto \varphi(1)$ and if you unwind the definition of the hom-functor and use this isomorphism, you'll see that the first space $\mathrm{Hom}_\mathcal{D}(M,\mathcal{O})$ is actually isomorphic to $\mathrm{Ker}(D)$ -- the kernel of $D$ acting on the space of polynomials $\mathcal{O}$. Since $\mathrm{Ext}^1_\mathcal{D}(\mathcal{D},\mathcal{O}) = 0$ we obtain exact sequence $$ 0 \to \mathrm{Ker}(D) \to \mathcal{O} \xrightarrow{f \mapsto D(f)} \mathcal{O} \to \mathrm{Ext}^1_\mathcal{D}(M,\mathcal{O}) \to 0, $$ which tells us that $\mathrm{Ext}^1_\mathcal{D}(M,\mathcal{O})$ is actually the cokernel of $D$. All of this goes through for a general $\mathcal{D}$-module $N$ instead of $\mathcal{O}$, the only change is that the kernel is now taken in the space $N$. Thus your question can be reinterpreted/generalized as:

Is there a $\mathcal{D}$-module $N$ such that $\mathrm{Ext}^1_\mathcal{D}(\mathcal{D}/\mathcal{D}D,N)$ is finite-dimensional of dimension greater that one?

There are several computer packages that can give you the cokernel (or ext) of any differential operator over a Weyl algebra. I wouldn't bet on them being able to handle any "nonalgebraic" modules like the space of real analytic functions, but you can try to experiment with some extensions of $\mathcal{O}$.

There are two articles by Coutinho that investigate on the topic: Extensions of modules over Weyl algebras and Modules of codimension one over Weyl algebras).

It would seem that the corollary 2.6 of the former article answers your question in negative since $\mathcal{O} = \mathcal{D}/(\mathcal{D}\partial_x + \mathcal{D}\partial_y)$ is a holonomic module and thus it is singular. But I have no idea whether there can exists $D$ such that $\mathcal{D}/\mathcal{D}D$ is nonsingular.

  • $\begingroup$ So, you mean the answer should be positive (according to your experiment), not negative, right? $\endgroup$ Commented Jul 27, 2014 at 9:57
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    $\begingroup$ OP asks for existence of $P$ and $Q$ such that the codimension of the range of $D$ is finite and bigger than one. I tend to believe that no such polynomials exists. $\endgroup$ Commented Jul 27, 2014 at 10:11
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    $\begingroup$ Ok, I asked because your last sentence was unclear to me regarding the question. Thanks for the clarification. Yet, the example I gave shows that in the formal case the question has a positive answer, and the "matrices" for formal power series are not that different from that for polynomials... In fact they are the same, the difference being on the finite support on the series given as arguments, which is not something easily read in the "matrices" (if I'm not mistaken). $\endgroup$ Commented Jul 27, 2014 at 12:21
  • $\begingroup$ @VítTuček thank you very much for your help and your attention on my question. Your computation which shows that this codimension is not different from 0,1, $\infty$ is a motivation for some other consideration(Consideration of some other space as domain of derivational operator. I explained about this new space in my two recent comments to Loic Teyssier's answer. Any way do you think that there is a complete proof that the codimension is only 0, 1 and $\infty$ when we restrict to polynomial space? Can your computations be extended to a complete proof? $\endgroup$ Commented Jul 30, 2014 at 16:15
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    $\begingroup$ @AliTaghavi You are welcome. $\endgroup$ Commented Nov 16, 2016 at 11:39

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