Hi

In Mixed Hodge modules Saito computes the Verdier specialisation of a D-modules with respect to a monomial $g = x_1^{m_1}\ldots x_n^{m_n}$. This is a very nice result as I find such explicit computations are quite rare in the litterature but I'm having problems understanding his proof.

Let $X$ be a polydisc, $D_i = \{x_i = 0\}$, $D_I = \bigcap_{i\in I} D_i$. Consider $M$ a regular holonomic algebraic quasi-unipotent right D-module with characteristic variety contained in the union of the conormal bundles to the $D_I$. For $\nu = (\nu_1,\ldots,\nu_n)$, set $$ M^\nu := \langle u\in M ~|~ u(x_i\partial_i - \nu_i)^p = 0~~ \textrm{for $p\gg 0$} \rangle $$ We have $\bigoplus_{\nu \in \mathbb{Q}^n} M^\nu = M$.

Let $g:X\to S$, $S$ an open disk, given by $g=x_1^{m_1}\ldots x_n^{m_n} = x^m$ and $i_g:X \to X\times S$ its graph. The D-module $\tilde{M} = (i_g)_* M$ is $M[\partial_t]$ with the action of $D_{X\times S}$ given by:

$$
(u\otimes \partial_t^j)x_i = ux_i\otimes \partial_t^j, \quad
(u\otimes \partial_t^j)\partial_i = u\partial_i\otimes \partial_t^j - u(\partial_ig)\otimes \otimes \partial_t^{j+1}
$$
$$
(u\otimes \partial_t^j)t = ug\otimes \partial_t^j + ju \otimes \partial_t^{j-1} , \quad
(u\otimes \partial_t^j)\partial_t = ux_i\otimes \partial_t^{j+1}
$$

**Theorem (3.4 p. 280)**: The V-filtration of $\tilde{M}$ along $t = 0$ is generated over $D_X$ by
$$
M^\nu \otimes 1 ~~\textrm{with $\nu_i \leq m_i\alpha$} \quad \textrm{if $\alpha< 0$}
$$
$$
M^\nu \otimes \partial_t^j ~~\textrm{with $\nu_i \leq m_i(\alpha-j)$} \quad \textrm{in general}
$$

It is enough to check that the filtration defined in the theorem satifies the properties of the V-filtration. Saito says:

"We have $$ (u\otimes \partial_t^j)x_i\partial_i = (u\otimes \partial_t^j)(N_i + \nu_i - m_i(s-j) ) \qquad \forall u \in M^\nu $$ where $s = t\partial_t$ and $(u\otimes \partial_t^j)N_i = u(x_i\partial_i-\nu_i)\otimes \partial_t^j$ if $u\in M^\nu$. Thus we get that $s-\alpha$ is nilpotent on $Gr^V_\alpha \tilde{M}$ and $V_\alpha \tilde{M}$ ar $V_0D_{X\times S}$-submodules."

I don't understand this last statement. $Gr^V_\alpha \tilde{M}$ is generated over $D_X$ by the $u \otimes \partial_t^j \in V_\alpha \tilde{M}$ with at least one $i$ so that $\nu_i = m_i(\alpha-j)$. For such $\nu$, the above equation reads

$$
(u\otimes \partial_t^j)x_i\partial_i = (u\otimes \partial_t^j)(N_i - m_i(s-\alpha))
$$
Why would this imply that $s-\alpha$ is nilpotent?

**Edit**: I think I have found the answer. Consider $u\otimes \partial_t^j$ with $\nu_i \leq m_i(\alpha-j)$ forall $i$. Set $I := \langle i ~|~ \nu_i = m_i(\alpha-j),~ m_i\neq 0\rangle$. Then
$$
(u\otimes \partial_t^j)(\prod_{i\in I} x_i\partial_i) = (u \prod_{i\in I}x_i \otimes \partial_t^j) (\prod_{i\in I}\partial_i)
$$

is in $V_{<\alpha} \tilde{M}$. So, by applying the relation above, we have
$$
(u\otimes \partial_t^j)\prod_{i\in I} (N_i - m_i(s-\alpha)) =
(u\otimes \partial_t^j)(\prod_{i\in I} x_i\partial_i) = (u \prod_{i\in I}x_i \otimes \partial_t^j) (\prod_{i\in I}\partial_i) = 0
$$
in $Gr^V_\alpha \tilde{M}$. So
$$
(u\otimes \partial_t^j)(s-\alpha)^{|I|} = (u\otimes \partial_t^j)(\sum_{k=0}^{k=|I|-1} \sigma_k(N_i/m_i)(s-\alpha)^k) )
$$
The operator on the right hand side is a polynomial in the $N_i$'s with coefficient in $Q[s-\alpha]$ and no constant term. So if $p_i$ is the nilpotency order of $N_i$ operatating on $u\otimes \partial_t^j$ and we elevate the relation to the power $\sum p_i$ we find that $(u\otimes \partial_t^j)(s-\alpha)^{|I|(\sum p_i)} = 0$. So $s-\alpha$ is indeed nilpotent.