# What subspaces of n-tuples of rational functions can be the solution space to a system of differential equations?

Let $K=\mathbb{C}(x)$ denote the field of rational functions (in 1 variable), and let $K^n$ denote an $n$-dimensional $k$-vector space (with basis). For some integer $m$, let $$\delta_{11}, \delta_{12},...\delta_{1n};\delta_{21},\delta_{22}...\delta_{2n};...;\delta_{m1},\delta_{m2},...\delta_{mn}$$ be $mn$-many differential operators in $x$ with coefficients in $K$, so that $$\delta_{11}(f_1)+...+\delta_{1n}(f_n)=0$$ $$\delta_{21}(f_1)+...+\delta_{2n}(f_n)=0$$ $$...$$ $$\delta_{m1}(f_1)+...+\delta_{mn}(f_n)=0$$ defines a system of $m$-many differential equations on $(f_1,...f_n)\in K^n$. Let $S\subseteq K^n$ denote the space of solutions to this system of differential equations; by homogeneous linearity it is a $\mathbb{C}$-subspace of $K^n$.

Usually, people are interested starting with a system of equations and finding the solution space. I have a weak inverse question. When is a given $\mathbb{C}$-subspace $V\subseteq K^n$ the solution space to such a system of differential equations? I don't care (yet) about finding the system of equations or how many equations there are, only whether they exist.

For $n=1$, the answer is appealingly simple. Either the system is degenerate, and the solution space is all of $K$; or it is not, and the solution space is finite $\mathbb{C}$-dimensional. Therefore, an infinite $\mathbb{C}$-dimensional proper subspace of $K$ is not the solution space to any system of equations, and every finite $\mathbb{C}$-dimensional subspace is the solution space of some system (I believe, I have not checked).

I am interested in similar results for larger $n$. I suspect that there are similar small or everything' type results, but I don't know what a good guess for what small might be. Note that any $K$-subspace of $K^n$ is a solution space, with defining equations given by a matrix over $K$ with that space as the kernel.

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Just to be clear: you're assuming that each differential operator $\delta_{jk}$` is linear too, right? Being able to write each operator as a polynomial in $x$ and $d/dx$ is helpful. Nice question, in any case. I'll think about the $n=1$ case myself. If you're interested in "small or everything" type results, may I ask if you have connections to model theory in mind? It sounds like you're trying to prove that something is strongly minimal. – Jerry Gagelman Aug 24 '10 at 8:57
Yup, all the operators are linear. My interests are actually a little bit more banal. Solution subspaces which are also the image of a different set of different operators can be used to create interesting examples of ideals in higher Weyl algebras, and I am trying to construct new examples. – Greg Muller Aug 24 '10 at 13:54