${\mathcal{D}}_{i+j-1}$, and we define $\{\delta, \delta'\} \in \operatorname{gr}^{i+j-1} {\mathcal{D}}$ to be the principal symbol of $[d,d']$. It is easy to check that $\{\delta, \delta'\}$ does not depend on the choices of $d$ and $d'$. The unique biadditive extension $$\{,\}: \operatorname{gr}{\mathcal{D}}\times \operatorname{gr}{\mathcal{D}}\rightarrow \operatorname{gr}{\mathcal{D}}$$ is called the *Poisson bracket* on $\operatorname{gr}{\mathcal{D}}$, and, in particular, is a biderivation. An ideal $I \subset \operatorname{gr}{\mathcal{D}}$ is called *involutive* if it is closed under the Poisson bracket, that is, if $\{I,I\} \subset I$.

[@gabber Thm. I]\[gabber\] Let $M$ be a finitely generated left ${\mathcal{D}}$-module. Let $\operatorname{gr}M$ be the associated graded $\operatorname{gr}{\mathcal{D}}$-module of $M$ with respect to a chosen good filtration on $M$. Let $J = \sqrt{\operatorname{Ann}_{\operatorname{gr}{\mathcal{D}}} \operatorname{gr}M} \subset \operatorname{gr}{\mathcal{D}}$ be the *characteristic ideal* of $M$ (as stated earlier, $J$ does not depend on the choice of good filtration). Then $J$ is involutive.

[@essen Lemma 1.12]\[primeinvol\] With the notation of Theorem \[gabber\], let $\mathfrak{p}$ be a minimal prime ideal over $J$. Then $\mathfrak{p}$ is again involutive.

As $J$ is a radical ideal in a Noetherian ring, we can write $J$ as an intersection $J = \mathfrak{p}_1 \cap \cdots \cap \mathfrak{p}_t$ of prime ideals of $\operatorname{gr}{\mathcal{D}}$, and we may assume that each $\mathfrak{p}_i$ is minimal over $J$. If 