TITLE: Regular holonomic D-modules as generalisation of regular singular points
QUESTION [5 upvotes]: I'm trying to understand why the definition of a regular holonomic D-module is a good generalisation of the usual definition of a regular singular point for a differential equation. More precisely, here is the definition of regular holonomic D-module I want to use 

A D-module $M$ on a complex manifold $X$ is holonomic if $\text{char}(M)$ has the same dimension as $X$. Moreover, it is regular if there is a good filtration $\{M_j\}_{j\in \mathbb{Z}}$ such that $$f\cdot\text{gr}(M)=0$$ for all $f\in \text{gr}(D_X)$ vanishing on $\text{char}(M).$

Now I'd like to test this definition on an example. I consider a Bessel equation $$P(f) :=(z^2 \partial_z^2 + z\partial_z  + (z^2-4))(f)=0.$$ The point $z=0$ is regular singular thanks to the $z^2$. Of course if the leading coefficient is replaced by $z^3$ it becomes irregular singular. Now I consider the D-module $$M := D_{\mathbb{C}}/D_{\mathbb{C}} P$$ naturally attached to this equation. We should normaly be able to prove that this D-module on $\mathbb{C}$ is regular holonomic. (And not regular if we put $z^3$ as a leading coefficient) First the characteristic variety is given by the null-set of $(z,w)\mapsto z^2w^2$ and so $$\text{char}(M) = \mathbb{C} \cup T_0^{*}\mathbb{C}$$ which has dimension $1$, so the module is holonomic. Now as a candidate for the good filtration, Simon Wadsley proposes to take $$M_n=D_n.(1+D_\mathbb{C}P).$$
Now I'm getting lost, how can I prove that $f\cdot\text{gr}(M)=0$ for all $f\in \text{gr}(D_X)$ vanishing on $\text{char}(M) \,?$ It doesn't make much sense to me.
Thank you for any help.

REPLY [3 votes]: I'm going to show (eventually) that your proposed filtration does not work.
Let $I$ be an ideal of $D_\Bbb{C}$, and let $M=D_\Bbb{C}/I$. Then, letting $u$ be the image in $M$ of $1$ (so $\operatorname{ann}_{D_\Bbb{C}}(u) = I$),

the filtration $M_j = D_j u$ of $M$ is a good filtration;
the filtration $I_j = D_j \cap I$ of $I$ is a good filtration;
the maps in the exact sequence of $D_\Bbb{C}$-modules $$ 0\to I \to D_\Bbb{C} \to M \to 0$$ are strictly filtered (recall that a map $f\colon M\to N$ is strictly filtered if $f(M)\cap N_j = f(M_j)$ for all $j$).

By 3., the sequence $$0 \to \operatorname{gr} I \to \operatorname{gr}D_\Bbb{C} \to \operatorname{gr} M \to 0$$ is still exact, so
$ \operatorname{gr}M \cong \operatorname{gr}D_\Bbb{C}/\operatorname{gr} I$, and the characteristic variety of $M$ is the zero locus of $\operatorname{gr} I$.
Case $I$ is principal:
Suppose $I= D_\Bbb{C} P$ is principal. Then (prove this!) $\operatorname{gr} I$ is generated by the principal symbol $\sigma(P)$ of $P$. Hence, $\operatorname{gr} M \cong \operatorname{gr} D_\Bbb{C}/(\sigma(P))$.
In particular, if $P = z^2 \partial_z^2 + z\partial_z  + z^2-4$, then $\sigma(P)=z^2\partial_z^2$, and therefore $\operatorname{gr} M \cong \operatorname{gr} D_\Bbb{C}/(z^2\partial_z^2)$. However, the ideal of $\operatorname{Ch}(M)$ is $(z\partial_z)$, which means that this filtration that we chose is not the one whose existence is claimed by the proof.