Let $\mathbb K$ be a characteristic-$0$ field, $R=\mathbb K[x_1,\ldots, x_n]$ be a polynomial ring, and $W=\mathbb K[x_1,\ldots,x_n,\partial_1,\ldots \partial_n]$ be the Weyl algebra. As usual $W$ acts on $R$ as, \begin{equation} \label{eq:1} x_i \bullet f=x_i f,\quad \partial_i \bullet f=\frac{\partial f}{\partial x_i},\quad \forall f \in R \end{equation} Let $\mathcal D=\{D_1, \ldots, D_k\}$ be a finite subset of $W$. Denote $\mathcal D \bullet R$ be the $\mathbb K$-linear space, \begin{equation} \label{eq:2} \mathcal D \bullet R\equiv {\rm span}_{\mathbb K}\{ D_i \bullet f |1\leq i\leq k,\ \forall f\in R\} \end{equation} The $\mathbb K$-linear quotient space $R/(\mathcal D \bullet R)$ is the object of our interest. We want a practical way (1) to calculate $\dim_{\mathbb K} R/(\mathcal D \bullet R)$, if it is finite, (2) to determine if a given polynomial $F\in R$ is in $(\mathcal D \bullet R)$ or not.

Note that if $D_i$'s contain no differential operator, i.e. if $D_i\in R$, $1\leq i \leq k$, then both questions can be solved by the (commutative) Gr\"obner basis computation of a polynomial ideal. For the general case, is there a way to solve them by computing Gr\"obner basis for an ideal of the Weyl algebra?

  • $\begingroup$ There is a filtration on the Weyl algebra such that the associated graded ring is just a polynomial algebra in $2n$ variables, and so I suspect that the non-commutative analogue of Groebner bases works out very nicely. Have you had any luck doing computations when $|\mathcal D|=1$? $\endgroup$ – Aaron May 31 '16 at 23:39
  • $\begingroup$ Gröbner bases work fine for the Weyl algebra, e.g. Macaulay2 has a good implementation. Also, do you really want $\mathcal{D}$ to be a finite set, or the left ideal in $W$ generated by that finite set? $\endgroup$ – Ketil Tveiten Jun 1 '16 at 8:22
  • $\begingroup$ Thanks Ketil! I tried Macaulay2: Consider the {\it right}-ideal I generated by $\mathcal D$. (Since Macaulay2 deals with left ideals, I used the antihomomorphism of $W$ to switch between the left/right). Let the $\{g_i\}$ be the Gr\"obner basis of $I$, by the division in $W$, $F=\sum_i g_i \cdot q_i + O$. Act this relation on $R$, so $F=(O\bullet 1) +( {\text polynomials\ in\ } \mathcal D\bullet R)$. $(O\bullet 1)$ seems like ``normal form'' of $F$. However, I am not sure how to choose a good weight vector of $W$, so this division works for some cases but failed for others. $\endgroup$ – Yang Zhang Jun 1 '16 at 14:21
  • $\begingroup$ Keith, yes, I want $\mathcal D$ to be a finite set since for my problem, the context is that $D_i$'s are coming from some tangent vector fields of an affine variety. $\endgroup$ – Yang Zhang Jun 1 '16 at 14:31
  • $\begingroup$ In usual (commutative polynomial) Gröbner basis theory, one has a gradation and therefore filtration on the polynomial ring $R$, and an ideal $I$, so therefore $gr\ R$ acts on $init\ I := gr\ I$. A generating set for the latter ideal lifts to a set in $I$ that automatically generates, and gets called a Gröbner basis. In standard treatments $R$ and $gr\ R$ are identified, which is confusing when one wants to deal with situations like yours where the filtration doesn't come from a gradation (and $gr\ W \not\cong W$). $\endgroup$ – Allen Knutson Jun 1 '16 at 22:29

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