On Partial Differential Operators Which Annihilate the Roots of the Universal Equation of Degree k

Abstract

The aim of this paper is to study in details the regular holonomic $D-$module introduced in Barlet (Math Z 302 $n^03$: 1627–1655, 2022 arXiv:1911.09347 [math]) whose local solutions outside the polar hyper-surface $\{\Delta (\sigma ).\sigma _k = 0 \}$ are given by the local system generated by the power $\lambda$ of the local branches of the multivalued function which is the root of the universal degree k equation $z^k + \sum _{h=1}^k (-1)^h\sigma _hz^{k-h} = 0$. We show that for $\lambda \in \mathbb {C} {\setminus } \mathbb {Z}$ this D-module is the minimal extension of the holomorphic vector bundle with an integrable meromorphic connection with a simple pole which is its restriction on the open set $\{\sigma _k\Delta (\sigma ) \not = 0\}$. We then study the structure of these D-modules in the cases where $\lambda = 0, 1, -1$ which are a little more complicated, but which are sufficient to determine the structure of all these D-modules when $\lambda$ is in $\mathbb {Z}$. As an application we show how these results allow to compute, for instance, the Taylor expansion of the root near $-1$ of the equation:

$$\begin{aligned} z^k + \sum _{h=-1}^k (-1)^h\sigma _hz^{k-h} - (-1)^k = 0. \end{aligned}$$

near $z^k - (-1)^k = 0$.

1 Introduction

1.1 .   

There are several ways to define an interesting function. Of course, the simplest one is to give its value at each point by an explicit finite formula or as a sum of an infinite series (converging somewhere at least). Another way is to give a functional equation which characterizes it. A third approach is to give a partial differential system which has our function f as its unique solution (up to normalisation).

For instance, the function $f(z) = e^z$ may be defined as

• 1. $f(z) = \sum _{n=0}^\infty \frac{z^n}{n!}$.

• 2. $f(z+z') = f(z).f(z')$ with $f(0) = 1$ and $f(1) = e$.

• 3. $\frac{\partial f}{\partial z} = f$ and $f(0) = 1$.

In general, to increase our understanding of such a function it is useful to dispose of at least two kinds of characterization as above. For instance, in the basic example of $e^z$ the description 3. gives easily the formula 1 and also the functional equation 2. Note that the third approach will often lead to a description of the first kind via the Taylor expansion at least when we dispose of a regular holonomic system defining f which is enough simple and suitably described to allow an inductive explicite computation of the coefficients of the Taylor expansion. However, this means that we are at least able to well describe essentially all partial differential operators which annihilate f.

We shall consider, in this paper, the case of the multivalued function $z(\sigma )^\lambda$ on $\mathbb {C}^k$, with $\sigma := (\sigma _1, \dots , \sigma _k)$ and $\lambda$ a complex parameter, where $z(\sigma )$ is defined as the root of the universal monic polynomial of degree k:

$$\begin{aligned} P_\sigma (z):= \sum _{h=0}^k (-1)^h\sigma _hz^{k-h} \quad \mathrm{with \ the \ convention} \quad \sigma _0 \equiv 1. \end{aligned}$$

It is well-known that the description of this function with the first approach is quite difficult (at least for $k \ge 5$). The definition given above of this multivalued function may be seen as a description of the second kind.

The aim of this paper is to give a description of the third kind which characterizes this multivalued function. More precisely we describe completely the structure of the regular holonomic $\mathcal {D}_N$-module¹$\mathcal {D}_N\big /\mathcal {J}_\lambda$ where $\mathcal {J}_\lambda$ is the left ideal in $\mathcal {D}_N$ which annihilates $z^\lambda (\sigma )$.

The case where $\lambda$ is in $\mathbb {Z}$ is of special interest (for $\lambda = 0$ the left ideal $\mathcal {J}_0$ will be defined in a natural way inside the annihilator of the function 1) and is less simple.

Let me explain how I come to study this question.

In the article [1], we characterize the trace functions F on $N:= \mathbb {C}^k$ as a solutions of a sub-holonomic $\mathcal {D}_N$-module $\mathcal {M}$ given by explicit generators in the Weyl algebra. Recall that an entire function F on N is a trace function when there exists a entire function f on $\mathbb {C}$, such that

$$\begin{aligned} F(\sigma ) = \sum _{j=1}^k f(z_j(\sigma )) \end{aligned}$$

where $z_1(\sigma ), \dots z_j(\sigma )$ are the roots of the polynomial $P_\sigma (z):= z^k + \sum _{h=1}^k (-1)^h\sigma _hz^{k-h}$.

Then, adding the quasi-homogeneity condition $U_0 - \lambda$, where $U_0:= \sum _{h=1}^k h\sigma _h\partial _h$ is the expression in $\sigma _1, \dots , \sigma _k$ of the Euler vector field $\sum _{j=1}^k z_j\partial _{z_j}$, to the ideal annihilating trace functions defines a (regular) holonomic $\mathcal {D}_N$-module $\mathcal {N}_\lambda$ whose local solutions are now given by linear combinations of the branches of the multi-valued function $z(\sigma )^\lambda$. Therefore, the goal of this article to understand the structure of theses regular holonomic $\mathcal {D}_N$-modules for each value of the parameter $\lambda \in \mathbb {C}$.

We give the statements of the main results in detail in Sect. 1.2 below

As an application of the structure theorem for the $\mathcal {D}$-module $\mathcal {N}_1:= \mathcal {D}_N\big /\mathcal {J}_1$ we compute the Taylor series at the point $\sigma ^0 = (0, 0, \dots , -1)$ of the holomorphic function of $\sigma _1, \dots , \sigma _k$ which gives the root of the polynomial:

$$\begin{aligned} z^k + \sum _{h=1}^k (-1)^h\sigma _hz^{k-h} - (-1)^k = 0 \end{aligned}$$

which is near $-1$. The fact that the associated $\mathcal {D}_N$-module corresponding to $\lambda = 1$ is not simple make this computation quite complicate when we use the $\mathcal {D}_N$-module $\mathcal {N}_1$ itself. However, with the remark that $z(\sigma ) - \sigma _1/k$ is a solution of the simple part $\mathcal {N}^\square _1$ of this $\mathcal {D}$-module deduced from the structure Theorem 4.1.9, we obtain a complete explicit computation of the Taylor series at $\sigma ^0$, corresponding to the equation $z^k - (-1)^k = 0$, of the root which is near $-1$.

Of course, this method is valid to compute (with some more numerical complications, but without theoretical difficulty) the Taylor expansion of any (uni-valued) holomorphic branch of the multivalued function $z(\sigma )^\lambda$ near any point $\sigma ^0 \in N$ for any given complex number $\lambda$.

To conclude this introduction, let me remark that we produce in this article an explicit description of the image via Riemann–Hilbert correspondence of the minimal extension of a rather involved local system defined in the complement of a hyper-surface with rather complicated singularities: the discriminant.

1.2 The Results

To make the statements clear we have to precise some notations which will be used (and defined again) in the text.

Notations

• 1. The coordinates on $M:= \mathbb {C}^k$ are $z_1, \dots , z_k$ and their elementary symmetric functions $\sigma _1, \dots , \sigma _k$ are the coordinates on $N \simeq \mathbb {C}^k$. The corresponding quotient map by the symmetric group is denoted by $quot: M \rightarrow N$.

• 2. $D_N$ is the sheaf of holomorphic differential operators on N and $\partial _1, \dots , \partial _k$ are the partial derivative in $\sigma _1, \dots , \sigma _k$.

• 3. For each integer $p \ge -1$ we denote $U_p$ the image by the tangent map of the quotient map quot of the vector field $\sum _{j=1}^k z^{p+1}_j\frac{\partial }{\partial z_j}$. The vector fields $U_{-1}, U_0$ and $U_1$ on N are given by

  $$\begin{aligned} U_{-1}:= & {} \sum _{h=0}^{k-1} (-1)^h(k-h) \sigma _h\partial _{h+1}, \ U_0:= \sum _{h=1}^k h\sigma _h\partial _h, \\ U_1:= & {} \sum _{h=1}^k (\sigma _1\sigma _h - (h+1)\sigma _{h+1})\partial _h. \end{aligned}$$

  with the convention $\sigma _0 \equiv 1$ and $\sigma _{k+1} = 0$.

• 4. For $\lambda$ a complex number, the $D_N$-module $\mathcal {N}_\lambda$ is the quotient of $D_N$ by the left ideal $\mathcal {J}_\lambda$ generated by the following global sections of $D_N$:

  • $A_{i, j}:= \partial _i\partial _j - \partial _{i+1}\partial _{j-1} \quad \textrm{for} \quad i \in [1, k-1] \quad \textrm{and} \quad j \in [2, k]$

  • $\mathcal {T}^m:= \partial _1\partial _{m-1} + \partial _mE \quad \textrm{for} \ m \in [2,k], \ \textrm{where} \ E:= \sum _{h=1}^k \sigma _h\partial _h$

  • $U_0 - \lambda .$

• 5. We denote $\mathcal {O}_N(\star \sigma _k)$ the left $D_N$-module $\mathcal {O}_N[\sigma _k^{-1}]$.The $D_N$-module quotient $\mathcal {O}_N(\star \sigma _k)\big /\mathcal {O}_N$ will be denoted $\underline{H}^1_{[\sigma _k = 0]}(\mathcal {O}_N)$.

• 6. We define $\mathcal {N}_1^*$ as the kernel of the map $\varphi _1: \mathcal {N}_1 \rightarrow \mathcal {O}_N$ sending 1 to $\sigma _1$.

• 7. We define $\mathcal {N}_0^*$ as the kernel of the map $\varphi _0: \mathcal {N}_0 \rightarrow \mathcal {O}_N$ sending 1 to 1.

• 8. We define $\mathcal {N}_{-1}^*$ as the kernel of the map $\varphi _{-1}: \mathcal {N}_{-1} \rightarrow \mathcal {O}_N(\star \sigma _k)$ sending 1 to $\sigma _{k-1}/\sigma _k$.

1.2.1 The Case $\lambda \in \mathbb {C} \setminus \mathbb {Z}$

Theorem 1.2.1

Let $\lambda \in \mathbb {C}\setminus \mathbb {Z}$. Then, $\mathcal {N}_\lambda$ is the minimal extension of the associated meromorphic connection (see 3.3.11). Therefore, $\mathcal {N}_\lambda$ is a simple $\mathcal {D}_N$-module.

Moreover, for each $\lambda \in \mathbb {C} \setminus \mathbb {Z}$ we have an isomorphism $\mathscr {T}_{\lambda +1}:= \square U_{-1}: \mathcal {N}_\lambda \rightarrow \mathcal {N}_{\lambda +1}$ given by right multiplication by $U_{-1}$.

See Theorem 3.3.11 for details.

1.2.2 The Case $\lambda \in \mathbb {N}^*$

Theorem 1.2.2

The diagram below describes the structure of $\mathcal {N}_1$. The torsion $\Theta$ in $\mathcal {N}_1$ is generated by the class of $\partial _kU_{-1}$ and we have an isomorphism $\underline{H}^1_{[\sigma _k = 0]}(\mathcal {O}_N) \rightarrow \Theta$ defined by sending $1/\sigma _k$ to $[\partial _kU_{-1}]$.

The $\mathcal {D}_N$-modules $\Theta \simeq \underline{H}^1_{[\sigma _k = 0]}(\mathcal {O}_N)$ and $\mathcal {N}^\square _1$ are simple $\mathcal {D}_N$-modules.

The following commutative diagram of left $\mathcal {D}_N$-modules has exact lines and columns where the maps i and e are defined by $i([U_{-1}]) = [U_{-1}]$ (so the map i induced by the identity of $\mathcal {N}_1$) and $e([U_{-1}]) = k$:

Moreover, for any $p \in \mathbb {N}^*$ the $D_N$-module $\mathcal {N}_p$ is isomorphic to $\mathcal {N}_{p+1}$ via the map $\mathscr {T}_{p+1}:= \square U_{-1}: \mathcal {N}_{p} \rightarrow \mathcal {N}_{p+1}$ given by right multiplication by $U_{-1}$.

See Theorems 4.1.5 and 4.1.9.

1.2.3 The Case $\lambda = 0$

Theorem 1.2.3

Define $\mathcal {N}_0^{\square }:= \mathcal {D}_NU_1 \subset \mathcal {N}_0^*$. Then, $\mathcal {N}_0^{\square }$ is simple and isomorphic to $\mathcal {N}^\square _1$ via the map induced by the map $\square U_1: \mathcal {N}_1 \rightarrow \mathcal {N}_0$, the quotient $\mathcal {N}_0\big /\mathcal {N}_0^{\square }$ is isomorphic to $\mathcal {O}_N(\star \sigma _k)$ and the quotient $\mathcal {N}_0\big /\mathcal {N}_0^*$ is isomorphic to $\mathcal {O}_N$.

The structure of $\mathcal {N}_0$ is described by the exact sequences of $D_N$-modules:

See Theorem 4.2.4.

1.2.4 The Case $\lambda \in -\mathbb {N}^*$

Theorem 1.2.4

We have the following commutative diagram of $\mathcal {D}_N$-module with exact lines and columns, where the $\mathcal {D}_N$-linear map $\varphi _{-1}: \mathcal {N}_{-1} \rightarrow \mathcal {O}_N(\star \sigma _k)$ is defined by $\varphi _{-1}(1) = \sigma _{k-1}\big /\sigma _k$:

Therefore, $\chi$ is an isomorphism. Moreover, the map $\mathscr {T}_0$ induces an isomorphism of $\mathcal {N}_{-1}^{*}$ onto the simple $\mathcal {D}_N$-module $\mathcal {N}_0^{\square } = \mathcal {D}_NU_1 \subset \mathcal {N}_0^*$. Therefore, $\mathcal {N}_{-1}^*$ is simple.

Moreover, for each $p \in -\mathbb {N}^*$ we have an isomorphism $\mathscr {T}_p: \square U_{-1}: \mathcal {N}_{p-1} \rightarrow \mathcal {N}_{p}$ given by right multiplication by $U_{-1}$.

See Theorem 4.3.8.

Section 2 is devoted to preliminaries, Sect. 3 concludes by the simplicity of $\mathcal {N}_\lambda$ for $\lambda \not \in \mathbb {Z}$ and Sect. 4 studies the cases where $\lambda$ is in $\mathbb {Z}$.

Section 5 gives the application and Sect. 6 is an appendix useful in the study of the characteristic variety of the $D_N$-modules $\mathcal {N}_\lambda$.

2 The $D_N$-Modules $\mathcal {W}$ and $\mathcal {M}$

Notations We fix in the sequel and integer $k \ge 2$. Let $\mathbb {C}[\sigma ]\langle \partial \rangle$ be the Weyl algebra in the variables $\sigma _1, \dots , \sigma _k$. We shall note $N = \mathbb {C}^k$ which is the target of the quotient map:

$$\begin{aligned} quot: M:= \mathbb {C}^k \rightarrow \mathbb {C}^k\big /\mathfrak {S}_k = N \simeq \mathbb {C}^k \end{aligned}$$

by the natural action of the permutation group $\mathfrak {S}_k$ on $\mathbb {C}^k$. We shall note $T_{quot}$ its tangent map.

Then, $\mathcal {D}_N$ denotes the sheaf of holomorphic differential operators on N and we shall use the same notations for modules on $\mathbb {C}[\sigma ]\langle \partial \rangle$ and for the corresponding sheaves of $\mathcal {D}_N$-modules on N.

We denote $\Delta \in \mathbb {C}[\sigma ]$ the discriminant of the polynomial $P_\sigma (z):= z^k + \sum _{h=1}^k (-1)^h\sigma _hz^{k-h}$ and $H_\Delta := \{\sigma \in N\ / \ \Delta (\sigma ) = 0\}$ the corresponding hyper-surface in N.

For basic results on $\mathcal {D}$-modules the reader may consult, for instance, the books [5] or [6].

2.1 The $D-$Module $\mathcal {W}$

In this section, we shall consider the $\mathcal {D}_N$-module $\mathcal {W}:= \mathcal {D}_N\big /\mathcal {A}$ where $\mathcal {A}$ is the left ideal sheaf in $\mathcal {D}_N$ generated by

$$\begin{aligned} A_{i, j}:= \partial _i\partial _j - \partial _{i+1}\partial _{j-1} \quad \textrm{for} \quad i \in [1, k-1] \quad \textrm{and} \quad j \in [2, k] \end{aligned}$$

(1)

Notations Let $\mathcal {D}_N(m)$ be the sub-sheaf of $\mathcal {D}_N$ of partial differential operators of order at most equal to m. Then, let $\mathcal {W}(m)$ be the sub-$\mathcal {O}_N$-module in $\mathcal {W}$ of the classes induced by germs in $\mathcal {D}_N(m)$.

Let $\mathcal {W}(m)$ be the sub$-\mathcal {O}_N$-module in $\mathcal {W}$ of the classes induced by germs in $\mathcal {D}_N(m)$.

As we have $\mathcal {A}(m):= \mathcal {D}_N(m) \cap \mathcal {A} = \sum _{i, j} \mathcal {D}_N(m-2)A_{i, j}$ for each $m \in \mathbb {N}$, the quotient $\mathcal {W}(m) = \mathcal {D}_N(m)\big /\mathcal {A}(m)$ injects in $\mathcal {W}$ and we have

$$\begin{aligned} \mathcal {W} = \cup _{m \ge 0} \mathcal {W}_m. \end{aligned}$$

Note that $\mathcal {A}(1) = 0$ so $\mathcal {W}(1) = \mathcal {D}_N(1)$.

It is clear that the characteristic variety of the $\mathcal {D}_N$-module $\mathcal {W}$ is equal to $N \times S(k)$ in the cotangent bundle $T_N^* \simeq N \times \mathbb {C}^k$ of N, where S(k) is the algebraic cone in $\mathbb {C}^k$ defined by the equations

$$\begin{aligned} \eta _i\eta _j - \eta _{i+1}\eta _{j-1} = 0 \quad \forall i \in [1, k-1] \quad \textrm{and} \quad \forall j \in [2, k]. \end{aligned}$$

(2)

We describe this two-dimensional cone and the corresponding ideal in the appendix (see Sect. 6). We shall use in the present section the following results which are proved in the appendix (Proposition 6.1.5 and Corollary 2.1.2).

Proposition 2.1.1

Let $L_1:= \{\eta _1 = 0\} \cap S(k)$ and $L_k:= \{\eta _k = 0 \} \cap S(k)$. Then, $L_1$ is the line directed by the vector $(0, \dots , 0, 1)$ and $L_k$ the line directed by the vector $(1, 0, \dots , 0)$. The maps $\varphi _1: S(k){\setminus } L_1 \rightarrow \mathbb {C}^*\times \mathbb {C}$ and $\varphi _k: S(k) {\setminus } L_k \rightarrow \mathbb {C}^*\times \mathbb {C}$ which are defined by the formulas

$$\begin{aligned} \varphi _1(\eta ):= (\eta _1, -\eta _2/\eta _1) \quad \textrm{and} \quad \varphi _k(\eta ):= (\eta _k, -\eta _{k-1}/\eta _k) \end{aligned}$$

(3)

are isomorphisms. Therefore, $S(k)\setminus \{0\}$ is smooth and connected surface.

Corollary 2.1.2

The ideal of $\mathbb {C}[\eta ]$ defined by the equations in (2) is prime. Moreover, S(k) is a normal surface.

Notation For $\alpha$ in $\mathbb {N}^k$ define $q:= \vert \alpha \vert := \sum _{h=1}^k \alpha _h$ and $r:= w(\alpha ) = \sum _{h=1}^k h\alpha _h$.

Definition 2.1.3

Let P be a germ of section of $\mathcal {D}_N$. We say that P is bi-homogeneous of type (q ,  r) if we may write

$$\begin{aligned} P = \sum _{\vert \alpha \vert = q, w(\alpha ) = r} a_\alpha \partial ^\alpha \end{aligned}$$

where $a_\alpha$ are germs of holomorphic functions in N.

It is clear that any germ P of section of $\mathcal {D}_N$ has a unique decomposition:

$$\begin{aligned} P = \sum _{q, r} P_{q, r} \end{aligned}$$

where $P_{q, r}$ is a bi-homogeneous germ of section of $\mathcal {D}_N$ of type (q, r). Note that this sum is finite because for a given order q the corresponding type (q, r) has non zero representative only when r is in [q, kq].

Lemma 2.1.4

Let P be a germ of section of $\mathcal {D}_N$ and write the decomposition of P in its bi-homogeneous components as $P = \sum _{q,r} P_{q, r}$. Then, P is a germ of section in $\mathcal {A}$ if and only if for each type (q, r) $P_{q, r}$ is a germ of section in $\mathcal {A}$.

Proof

It is clear that P is in $\mathcal {A}$ when each $P_{q, r}$ is in $\mathcal {A}$. Conversely, assume that P is in $\mathcal {A}$. Then, we may write

$$\begin{aligned} P = \sum _{(i, j)} B_{i, j}A_{i, j} \quad \textrm{with} \quad i \in [1, k-1] \quad \textrm{and} \quad j \in [2, k] \end{aligned}$$

and where $B_{i, j}$ are germs of sections of $\mathcal {D}_N$.

Write $B_{i, j} = \sum _{q, r} (B_{i, j})_{q, r}$ the decomposition of $B_{i, j}$ in its bi-homogeneous components; this gives

$$\begin{aligned} P = \sum _{q, r}\sum _{p \ge 0} \left( \sum _{i+j=p} (B_{i, j})_{q, r}A_{i, j}\right) \end{aligned}$$

where $\sum _{i+j=p} (B_{i, j})_{q, r}A_{i, j}$ is bi-homogeneous of type $(q+2, r+p)$ for each (i, j), such that $i+j=p$. This implies that $P_{q, r}$ is equal to the sum $\sum _{i+j+s = r} (B_{i, j})_{q-2, s}A_{i, j}$. Therefore, each $P_{q, r}$ is a germ of section in $\mathcal {A}$.$\blacksquare$

Lemma 2.1.5

The class of $\partial ^\alpha$ in $\mathcal {W}$ only depends on $q:= \vert \alpha \vert$ and $r:= w(\alpha )$. It will be denoted $y_{q, r}$. Moreover, if $\mathcal {W}_q$ is the sub$-\mathcal {O}_N$-module of $\mathcal {W}$ generated by the $y_{q, r}$ for $r \in [q, kq]$, $\mathcal {W}_q$ is a free $\mathcal {O}_N$-module of rank $kq - q + 1$ with basis $y_{q, q}, y_{q, q+1}, \dots , y_{q, kq}$ and, as $\mathcal {O}_N$-module, we have the direct decompositions:

$$\begin{aligned} \mathcal {W}(m) = \oplus _{q=0}^m \mathcal {W}_q \quad \textrm{and} \quad \mathcal {W} = \oplus _{q \in \mathbb {N}} \ \mathcal {W}_q. \end{aligned}$$

(4)

Remark that the action of $\mathcal {D}_N$ on $\mathcal {W}$ is defined by

$$\begin{aligned} \partial _j(y_{q, r}) = y_{q+1, r+j} \quad \forall j \in [1, k] \quad \forall q \in \mathbb {N} \quad \textrm{and} \quad \forall r \in [q, kq] \end{aligned}$$

(5)

and that $\partial _j\mathcal {W}_q \subset \mathcal {W}_{q+1}$.

Proof

The fact that the class induced by $\partial ^\alpha$ in $\mathcal {W}$ depends only on $\vert \alpha \vert$ and $w(\alpha )$ is a direct consequence of the fact that the class induced by $x^\alpha$ in $\mathbb {C}[x_1, \dots , x_k]\big /IS(k)$ only depends on $q:= \vert \alpha \vert$ and $r:= w(\alpha )$ (see Proposition 6.1.1 in the appendix). Then, it is clear that $y_{q, q}, y_{q, q+1}, \dots , y_{q, kq}$ is a $\mathcal {O}_N-$basis of $\mathcal {W}_q$ looking at the symbols and using the appendix (Sect. 6) over the sheaf of $\mathbb {C}-$algebras $\mathcal {O}_N$.$\blacksquare$

The global polynomial solutions of the $\mathcal {D}_N$-module $\mathcal {W}$ are described by our next lemma.

Definition 2.1.6

For each $q \in \mathbb {N}$ and each $r \in [q, kq]$ define the polynomial

$$\begin{aligned} m_{q, r}(\sigma ):= \sum _{\vert \alpha \vert = q, w(\alpha ) = r} \frac{\sigma ^\alpha }{\alpha !} \end{aligned}$$

(6)

Lemma 2.1.7

Any $m_{q, r} \in \mathbb {C}[\sigma _1, \dots , \sigma _k]$ is annihilated by the left ideal $\mathcal {A}$ in $\mathcal {D}_N$ and if a polynomial $P \in \mathbb {C}[\sigma _1, \dots , \sigma _k]$ is annihilated by $\mathcal {A}$, P is, in a unique way, a $\mathbb {C}-$linear combination of the $m_{q, r}$ for $q \ge 0$ and $r \in [q, kq]$ which gives the bi-homogeneous decomposition of $P(\partial _1, \dots , \partial _k) \in \mathbb {C}[\partial _1, \dots , \partial _k]$ (see Lemma 2.1.4).

Proof

First, we shall verify that each polynomial $m^q_r$ is annihilated by each $A_{i, j}$ for all $i \in [1, k-1]$ and all $j \in [2, k]$. We have for each $(i, j) \in [1, k]^2$:

$$\begin{aligned} \partial _i\partial _j (m_{q, r})(\sigma ) = \sum _{\vert \beta \vert = q-2, w(\beta ) = r-(i+j)} \frac{\sigma ^\beta }{\beta !} = m_{q-2, r-(i+j)(\sigma )} \end{aligned}$$

because $\alpha _i\alpha _j = 0$ implies $\partial _i\partial _j \sigma ^\alpha = 0$. The right hand-side above only depends on $i+j$ for q and r fixed. This is enough to conclude our verification.

Note also that the uniqueness is obvious because of the uniqueness of the Taylor expansion of a polynomial.

Let now $P:= \sum _{\alpha } c_\alpha \frac{\sigma ^\alpha }{\alpha !}$ a polynomial in $\mathbb {C}[\sigma ]$ which is annihilated by the left ideal $\mathcal {A}$ in $\mathcal {D}_N$. We want to show that if $\alpha$ and $\beta$ satisfy $\vert \alpha \vert = \vert \beta \vert$ and $w(\alpha ) = w(\beta )$ we have $c_\alpha = c_\beta$. It is enough to prove this equality when there exist $i \in [1, k-1], j \in [2, k]$ and $\gamma \in \mathbb {N}^k$, such that $\sigma ^\alpha = \sigma _i\sigma _j\sigma ^\gamma$ and $\sigma ^\beta = \sigma _{i+1}\sigma _{j-1}\sigma ^\gamma$ by definition of the equivalence relation² given by $\vert \alpha \vert = \vert \beta \vert$ and $w(\alpha ) = w(\beta )$. In this case the coefficient of $\sigma ^\gamma /\gamma !$ in $\partial _i\partial _j P$ is $c_\alpha$ and in $\partial _{i+1}\partial _{j-1} P$ is $c_\beta$. Therefore, they are equal.$\blacksquare$

It is easy to see that an entire holomorphic function $F: N \rightarrow \mathbb {C}$ is solution of $\mathcal {W}$ if and only if its Taylor series at the origin may be written, for some choice of $c_{q, r} \in \mathbb {C}$:

$$\begin{aligned} F(\sigma ) = \sum _{q, r} c_{q, r}m_{q, r}(\sigma ). \end{aligned}$$

In the same way, a holomorphic germ $f: (N, \sigma ^0) \rightarrow (\mathbb {C}, z^0)$ is solution of $\mathcal {W}$ if and only if its Taylor series may be written in the form

$$\begin{aligned} f(\sigma ^0 + \sigma ) = \sum _{q, r} c_{q, r}m_{q, r}(\sigma ) \end{aligned}$$

with $c_{0, 0} = z^0$.

2.2 The $\mathcal {D}_N$-Module $\mathcal {M}$

Definition 2.2.1

Let $m \in [2, k]$ be an integer and define the second-order differential operators in the Weyl algebra $\mathbb {C}[\sigma ]\langle \partial \rangle$

$$\begin{aligned} \mathcal {T}^m:= \partial _1\partial _{m-1} + \partial _mE \quad \textrm{for} \ m \in [2,k], \ \textrm{where} \quad E:= \sum _{h=1}^k \sigma _h\partial _h \end{aligned}$$

(7)

Then, define the left ideal $\mathcal {I}$ in $\mathcal {D}_N$ as

$$\begin{aligned} \mathcal {I}:= \mathcal {A} + \sum _{m=2}^k \mathcal {D}_N\mathcal {T}^m \end{aligned}$$

(8)

and let $\mathcal {M}$ be the $\mathcal {D}_N$-module

$$\begin{aligned} \mathcal {M}:= \mathcal {D}_N\big /\mathcal {I} \end{aligned}$$

(9)

We shall now recall and precise some results of [1].

Let Z be the complex (algebraic) subspace in $N \times \mathbb {C}^k$ (with coordinates $\sigma _1, \dots , \sigma _k, \eta _1, \dots , \eta _k$) defined by the ideal of $(2, 2)-$minors of the matrix

$$\begin{aligned} \begin{pmatrix} \eta _1&{}\quad -l_{\sigma }(\eta ) \\ \eta _2 &{} \eta _1 \\ . &{}. \\ . &{}. \\ \eta _k &{} \eta _{k-1} \end{pmatrix} \end{aligned}$$

(10)

where $l_\sigma (\eta ):= \sum _{h=1}^k \sigma _h\eta _h$. We shall note $I_Z$ the ideal of $\mathcal {O}_{N \times \mathbb {C}^k}$ generated by these minors and by $p_*I_Z$ its direct image by the projection $p: N\times \mathbb {C}^k \rightarrow N$. For each integer $q \ge 0$ the sub-sheaf $p_*I_Z(q)$ of sections of $p_*I_Z$ which are homogeneous of degree q along the fibers of p is a coherent $\mathcal {O}_N$-module.

Proposition 2.2.2

The complex subspace Z is reduced, globally irreducible and Z is the characteristic cycle of the $\mathcal {D}_N$-module $\mathcal {M}$.

Proof

Let $\vert Z\vert$ be the support of the sub-space Z. The fact that Z is globally irreducible is already proved in [1] Proposition 4.2.6 as Z is conic over N. This implies that Z is reduced as a complex sub-space:

Assume that $I_Z$ is not equal to the reduced ideal $I_{\vert Z\vert }$ of the complex analytic subset $\vert Z\vert$ in $N \times \mathbb {C}^k$. By homogeneity in the variables $\eta _1, \dots , \eta _k$ there exists $q \ge 0$, such that the quotient $\mathcal {Q}(q):= I_{\vert Z\vert } (q)\big / I_Z(q)$ is not $\{0\}$ and then the coherent sheaf $p_*(\mathcal {Q}(q))$ is not $\{0\}$ on N. However, this contradicts the fact that any global section on N of $p_*I_{\vert Z\vert }(q)$ is a global section on N of $p_*I_Z(q)$ which is the content of Proposition 4.2.6 in loc. cit.

To complete the proof that Z is the characteristic cycle of $\mathcal {M}$ it is enough to see that the symbol of any germ P of section in $\mathcal {I}$ vanishes on $\vert Z\vert$. This is obvious by definition of $I_Z$.$\blacksquare$

The following proposition, which is a local version of Theorem 5.1.1 in [1], will be useful. Recall that the Newton polynomial $N_m \in \mathbb {C}[\sigma _1, \dots , \sigma _k]$ is the polynomial corresponding (via the standard symmetric function Theorem) to the symmetric polynomial $\sum _{j=1}^m z_j^m$.

Proposition 2.2.3

Let $\mathcal {I}_+$ the left ideal in $\mathcal {D}_N$ of germs of differential operators P, such that $P(N_m) = 0$ for each Newton polynomial $N_m, m \in \mathbb {N}$. Then, $\mathcal {I}_+ = \mathcal {I}$

Proof

Proposition 4.1.2 in [1] already proves the inclusion $\mathcal {I} \subset \mathcal {I}_+$. To prove the other inclusion we shall argue by contradiction. Therefore, assume that at some point $\sigma ^0$ in N we have $\mathcal {I}_{+, \sigma ^0} {\setminus } \mathcal {I}_{\sigma ^0} \not = \emptyset$ and let P be in $\mathcal {I}_{+, \sigma ^0} \setminus \mathcal {I}_{\sigma ^0}$ with minimal order say q. Thanks to Proposition 4.2.8 in loc. cit. we know³ that the symbol s(P) is in $p_*(I_Z)_{\sigma ^0}$ thanks to the equality $I_Z = I_{\vert Z\vert }$ proved above. Therefore, there exists a germ $P_1$ in $\mathcal {I}_{\sigma ^0} \setminus \{0\}$ with symbol $s(P_1) = s(P)$. Then, the order of $P - P_1$ is strictly less than q. But then, $P - P_1$ is in $\mathcal {I}_{+, \sigma ^0}$ with order strictly less than q and then it is in $\mathcal {I}_{\sigma ^0}$. Contradiction. Therefore, $\mathcal {I} = \mathcal {I}_+$.$\blacksquare$

Notations We note $\mathcal {I}(m)$ the sub-$\mathcal {O}_N$-module generated in $\mathcal {I}$ by classes induced by differential operators of order at most equal to m. Therefore, $\mathcal {I}(m) = \mathcal {I}\cap \mathcal {D}_N(m)$.

Then, we note $\mathcal {M}(m):= \mathcal {D}_N(m)\big /\mathcal {I}(m)$.

For any non zero germ of section P of $\mathcal {D}_N$ we note s(P) its symbol in $\mathcal {O}_N[\eta _1, \dots , \eta _k]$. For $P = 0$, let s(P) be 0.

Recall that we note $p: N\times \mathbb {C}^k \rightarrow N$ the projection.

Lemma 2.2.4

We have $\lim _{m \rightarrow \infty } \mathcal {M}(m) \simeq \mathcal {M}$, where the maps $\mathcal {M}(m) \rightarrow \mathcal {M}(m+1)$ are induced by the obvious inclusions

$$\begin{aligned} \mathcal {D}_N(m) \hookrightarrow \mathcal {D}_N(m+1) \quad \textrm{and} \quad \mathcal {I}(m) \hookrightarrow \mathcal {I}(m+1). \end{aligned}$$

Therefore, $\mathcal {M}$ is equal to the inductive limit $\lim _{m \rightarrow \infty } \mathcal {M}(m)$.

Proof

Beware that the maps $\mathcal {M}(m) \rightarrow \mathcal {M}(m+1)$ are not a priori injective.

There is an obvious map $\lim _{m \rightarrow \infty } \mathcal {M}(m) \rightarrow \mathcal {M}$ which is clearly surjective. The point is to prove injectivity. Let P be a non zero germ at some $\sigma \in N$ of order m, such that its image in $\mathcal {M}_\sigma$ is 0. Then, by definition, there exists germs $B_h, h \in [2, k]$ and $C_{p, q}, (p, q) \in [1, k]^2$ in $D_{N, \sigma }$, such that

$$\begin{aligned} P = \sum _{h=2}^k B_h\mathcal {T}^h + \sum _{p, q} C_{p, q}A_{p, q}. \end{aligned}$$

Let r be the maximal order of the germs $B_h$ and $C_{p,q}$. Then, the equality above shows that P is in $\mathcal {I}(r+2)$. Therefore, the image of P in $\lim _{m \rightarrow \infty } \mathcal {M}(m)$ is zero, as it is already 0 in $\mathcal {M}(r+2)$.$\blacksquare$

Lemma 2.2.5

Let P be a non zero germ of section of the sheaf $\mathcal {I}$. Assume that P has order at most 1. Then, $P = 0$.

Proof

Let $P = a_0 + \sum _{h=1}^k a_h\partial _h$. Recall that for each $h \in [1, k]$ and each $m \in \mathbb {N}$ we have (see Proposition  5.2.1 in [1]):

$$\begin{aligned} \partial _hN_m = (-1)^{h-1}mDN_{m-h} \end{aligned}$$

(11)

where the polynomials

$$\begin{aligned} DN_m:= \sum _{P_\sigma (x_j) = 0} \frac{x_j^{m+k-1}}{P'_{\sigma }(x_j)} \end{aligned}$$

vanish for $m \in [-k+1, -1]$ and $DN_0 = 1$.

Then, the equality $\mathcal {I}_+ = \mathcal {I}$ proved in Proposition 2.2.3 implies that for each integer m, we have

$$\begin{aligned} a_0N_m + \sum _{h=1}^k a_h(-1)^{h-1}mDN_{m-h} = 0, \forall m \in \mathbb {N} \end{aligned}$$

For $m = 0$ this gives $a_0 = 0$; if we have $a_0 = a_1 = \dots = a_p = 0$ for some $p \in [0, k-1]$ then $P[N_{p+1}] = 0$ gives $\sum _{h = p+1}^k a_h(-1)^h(p+1)DN_{p+1-h} = a_{p+1}(p+1)DN_0 = 0$ and then $a_{p+1} = 0$. Therefore, $P = 0$.$\blacksquare$

Notation Define the vector fields on N:

$U_0:= \sum _{h=1}^k h\sigma _h\partial _h \quad$ and $\quad U_{-1}:= \sum _{h=0}^{k-1} (k-h)\sigma _h\partial _{h+1}$ with the convention $\sigma _0 \equiv 1$.

Lemma 2.2.6

Let $q \ge 2$ be an integer, $\alpha \in \mathbb {N}^k$, such that $\vert \alpha \vert = q-2$ and let m be an integer in the interval [2, k]. The class induced by $\partial ^\alpha \mathcal {T}^m$ in $\mathcal {W}$ only depends on the integers q and $r:= w(\alpha ) + m$. This class is given by the formula (with the convention $\sigma _0 \equiv 1$)

$$\begin{aligned} \left[ \partial ^\alpha \mathcal {T}^m\right] = \sum _{h=0}^k \sigma _hy_{q, r+h} + (q-1)y_{q, r}, \end{aligned}$$

(12)

where $y_{q, r}$ is the class induced by $\partial ^\gamma$ in $\mathcal {W}$ for any $\gamma \in \mathbb {N}^k$, such that $\vert \gamma \vert = q$ and $w(\gamma ) = r$ (see Lemma 2.1.5).

Let $\lambda$ be a complex number and let $\beta \in \mathbb {N}^k$, such that $\vert \beta \vert = q-1$ and $w(\beta ) = r$. The class induced by $\partial ^\beta (U_0 - \lambda )$ in $\mathcal {W}$ only depends on $\lambda$ and on the integers q and r. This class is given by

$$\begin{aligned} \left[ \partial ^\beta (U_0 - \lambda )\right] = \sum _{h=1}^k h\sigma _hy_{q, r+h} + (r - \lambda )y_{q-1, r}. \end{aligned}$$

(13)

In addition, the class induced by $\partial ^\beta U_{-1}$ in $\mathcal {W}$, again for $\vert \beta \vert = q-1$ and $w(\beta ) = r$, only depends on the integers q and r. This class is given by

$$\begin{aligned} \left[ \partial ^\beta U_{-1} \right] = \sum _{h=0}^k (k-h)\sigma _hy_{q, r+h+1} + (k(q-1) - r)y_{q-1, r +1} \end{aligned}$$

(14)

where, for $r = k(q-1)$, the last term in (14) is equal to 0 by convention.

Proof

By definition $\mathcal {T}^m = \partial _1\partial _{m-1} + \sum _{h=1}^k \sigma _h\partial _h\partial _m + \partial _m$ which implies

$$\begin{aligned} \partial ^\alpha \mathcal {T}^m = \partial ^\alpha \partial _1\partial _{m-1} + \sum _{h=1}^k \sigma _h\partial _h\partial _m\partial ^\alpha + (q-1)\partial ^\alpha \partial _m \end{aligned}$$

as we have $\partial ^\alpha \sigma _h\partial _h = \sigma _h\partial _h\partial ^\alpha + \alpha _h\partial ^\alpha$ for any $\alpha \in \mathbb {N}^k$ and any $h \in [1, k]$. Now, formula (12) follows from Lemma 2.1.5, proving our first assertion.

As $U_0:= \sum _{h=1}^k h\sigma _h\partial _h$ we have

$$\begin{aligned} \partial ^\beta (U_0 - \lambda ) = \sum _{h=1}^k h\partial ^\beta \sigma _h\partial _h - \lambda .\partial ^\beta = \sum _{h=1}^k h.\sigma _h\partial _h\partial ^\beta + (w(\beta ) - \lambda )\partial ^\beta \end{aligned}$$

which gives Formula (13) using Lemma 2.1.5, and this proves our second assertion. The third one is analogous using the fact that $U_{-1} = \sum _{h=0}^{k-1} (k-h)\sigma _h\partial _{h+1}$ with the convention $\sigma _0 \equiv 1$ and the equalities:

$$\begin{aligned} \partial ^\beta \sigma _h\partial _{h+1}= & {} \sigma _h\partial ^\beta \partial _{h+1} + \beta _h\partial ^{\beta +1_{h+1}-1_h}\\ \sum _{h=0}^{k-1} (k-h)\beta _h= & {} k((q-1) - \beta _k) - (w(\beta ) - k\beta _k) = k(q-1) - r \end{aligned}$$

with the convention $\beta _0 = 0$ and the fact that $\partial ^{\beta +1_{h+1}-1_h}$ induces $y_{q-1,r+1}$.$\blacksquare$

Notations

• 1. Let $V_q \subset \mathcal {W}_q$ be the $\mathcal {O}_N-$sub-module with basis $y_{q, r}$ for $r \in [k(q-1)+1, kq]$. Remark that $V_0 = \mathcal {W}_0 = \mathcal {W}(0)= \mathcal {O}_N$ and $V_1 = \mathcal {W}_1 = \oplus _{h=1}^k \mathcal {O}_N.\partial _h$.

• 2. Let $L_q: \mathcal {W}_q \rightarrow \mathcal {M}(q)$ be the map induced by restriction to $\mathcal {W}_q$ of the quotient map $\mathcal {W}(q) \rightarrow \mathcal {M}(q)$ and $l_q: V_q \rightarrow \mathcal {M}(q)$ its restriction to $V_q$.

Lemma 2.2.7

Fix an integer $q \ge 0$. Then, for any $Y \in \mathcal {W}_q$ there exists $X \in V_q$, such that $L_q(Y - X)$ is in $\mathcal {M}(q-1)$, with the convention $\mathcal {M}(-1) = \{0\}$.

Proof

Remark that for $Y = y_{q, r}$ with $r \in [k(q-1)+1, kq]$ we may choose $X = Y$. Therefore, it is enough to prove the lemma for Y in the sub-module with basis $y_{q, r}$ with $r \in [q, k(q-1)]$.

Note that for $q = 0$ and for $q = 1$ there is nothing more to prove.

For each $q \ge 2$ and $r \in [q, k(q-1)]$ there exists $m \in [2, k]$, such that $r-m$ is in $[q-2, k(q-2)]$, because the addition map $(s, m) \rightarrow s+m$ is surjective⁴ from $[q-2, k.(q-2)]\times [2, k]$ to $[q, k(q-1)]$. Therefore, there exists $\alpha \in \mathbb {N}^k$, such that $\vert \alpha \vert = q-2$ and $w(\alpha ) = r-m$. Then

$$\begin{aligned} \partial ^\alpha \mathcal {T}^m = y_{q, r} + (q-1)y_{q-1, r}+ \sum _{h=1}^k \sigma _hy_{q, r+h} \end{aligned}$$

and the class induced by $y_{q, r}$ in $\mathcal {M}(q)$ is, modulo $\mathcal {M}(q-1)$, in the sub$-\mathcal {O}_N$-module of $\mathcal {M}(q)$ induced by the images of classes of $y_{q, r'}$ with $r' > r$. By a descending induction on $r \in [q, k(q-1)]$ we see that, modulo $\mathcal {M}(q-1)$, the image of $\mathcal {W}_q$ by $L_q$ is equal to $L_q(V_q)$. This implies our statement by induction on q.$\blacksquare$

Note that the previous lemma shows that $l_q(V_q) = L_q(\mathcal {W}_q)$ for each $q \ge 0$.

Proposition 2.2.8

For any $q \in \mathbb {N}$, there is a natural isomorphism of $\mathcal {O}_N$-modules

$$\begin{aligned} \Lambda _q:= \oplus _{p=0}^q l_p: \oplus _{p=0}^{q}\ V_p \longrightarrow \mathcal {M}(q) \end{aligned}$$

(15)

which is compatible with the natural map $\oplus _{p=0}^{q}\ V_p \hookrightarrow \oplus _{p=0}^{q+1}\ V_p$ and the natural map $\mathcal {M}(q) \rightarrow \mathcal {M}(q+1)$.

Proof

For $q = 0$ we have $\mathcal {M}(0) = V_0 = \mathcal {O}_N.y_{0, 0}$ where $y_{0, 0} = 1$. Therefore, $\Lambda _0$ is an isomorphism. For $q = 1$, Lemma 2.2.5 shows that the map $\Lambda _1$ is injective. As it is surjective (we have $V_0 = \mathcal {W}_0$ and $V_1 = \mathcal {W}_1$) the assertion is clear.

Assume that we have proved that $\Lambda _{q-1}$ is an isomorphism of $\mathcal {O}_N$-modules for some $q \ge 2$. We shall prove that $\Lambda _q$ is also an isomorphism.

Consider $Y:= \sum _{p=0}^q Y_p$ with $Y_p \in V_p$ for each $p \in [0, q]$, which is in the kernel of $\Lambda _q$. If $Y_q = 0$ the induction hypothesis allows to conclude that $Y = 0$.

So assume that $Y_q \not = 0$. As $Y_q$ is induced⁵ by a differential operator of the form $\sum _{j=1}^k b_j\partial _k^{q-1}\partial _j$, with $b_j \in \mathcal {O}_N$ for $j \in [1, k]$, we may choose a differential operator $P \in \mathcal {I}$ of order q which induces Y, such that its symbol is equal to $\eta _k^{q-1}\sum _{j=1}^k b_j\eta _j$. This symbol vanishes on Z, and, as $\eta _k$ does not vanish on any non empty open set on Z, we conclude that $\sum _{j=1}^k b_j\eta _j$ vanishes on Z. The injectivity of $\Lambda _1$ implies that $b_1 = \dots = b_k = 0$ showing that $Y_q = 0$ and this contradicts our hypothesis. Therefore, $\Lambda _q$ is injective.

We have already noticed that Lemma 2.2.7 implies the surjectivity of $\Lambda _q$ for $q \ge 2$. Therefore, the proof is complete.$\blacksquare$

Corollary 2.2.9

The $\mathcal {D}_N$-module $\mathcal {M}$ has no $\mathcal {O}_N$-torsion.

Proof

This an easy consequence of the previous proposition giving that each $\mathcal {M}(q)$ is a free $\mathcal {O}_N$-module, because for any $\sigma \in N$, a non zero torsion germ in $\mathcal {M}_\sigma$ has to come from a non zero torsion element in $\mathcal {M}(q)_\sigma$ for some q large enough (may be much more larger than the order of the germ in $D_{N, \sigma }$ inducing this class in $\mathcal {M}_\sigma$) thanks to Lemma 2.2.4.$\blacksquare$

2.3 On Quotients of $\mathcal {M}$

We shall use the description of the characteristic variety of $\mathcal {M}$ to examine the holonomic quotients of $\mathcal {M}$ supported by an irreducible complex subset of N.

Proposition 2.3.1

Let Q be a holonomic quotient of $\mathcal {M}$ which is supported by an analytic subset S of N with empty interior in N. Then, S is a hyper-surface and S is contained in $\{\sigma _k = 0\}\cup \{\Delta (\sigma ) = 0\}$.

Proof

Let $S_0$ be an irreducible component of S, the support of a holonomic quotient Q of $\mathcal {M}$. Let $d \ge 1$ be the co-dimension of $S_0$. Then, near the generic point in $S_0$ the co-normal sheaf of $S_0$ is a rank d vector bundle over $S_0$ which is contained in Z. As the fibres of Z over N have pure dimension 1 we have $d \le 1$ and then $d = 1$ and $S_0$ is a hyper-surface in N. Then, S is also a hyper-surface in N.

Let now $S_0$ be an irreducible component of S which is not contained in $\{\Delta (\sigma ) = 0 \}$. Then, near the generic point in $S_0$ the quotient map $quot: M \rightarrow N$ is an étale cover and this shows that $\mathcal {M}$ locally is isomorphic to the quotient of $D_{\mathbb {C}^k}$ by the let ideal with generators $\frac{\partial ^2}{\partial z_i\partial z_j}$ for $i \not = j$ in [1, k]. Therefore, the characteristic variety of $\mathcal {M}$ is locally isomorphic to $C:= \cup _{j=1}^k N\times \{\mathbb {C}.e_j\}$ where $e_j$ is the $j-$th vector in the canonical basis of $\mathbb {C}^k$. If an irreducible hyper-surface has its co-normal bundle contained in C, it has to be equal to the co-normal of one of the hyperplanes $\{z_j = 0\}$. This means that S is contained in $\{\sigma _k = 0 \}$.

But any hyper-surface contained in $\{\Delta = 0 \}$ is equal to $\{\Delta = 0 \}$. Therefore, the only possible irreducible components of the support of Q are $\{\sigma _k = 0 \}$ or $\{\Delta = 0 \}$.    $\blacksquare$

We shall use the following immediate corollary of this proposition:

Corollary 2.3.2

Let Q be a coherent holonomic quotient of $\mathcal {M}$ which is supported in a closed analytic subset S in N with empty interior in N. If Q vanishes near the generic points of $\{\sigma _k = 0 \} \cup \{\Delta = 0 \}$, then $Q = \{0\}$.$\blacksquare$

Let $k \ge 2$. We shall study the $\mathcal {D}_N$-module $\mathcal {M}$ near the generic point of the hyper-surface $\{\Delta = 0\}$ in N.

Let $z_1^0, z_3^0, \dots , z_k^0$ be $(k-1)$ distinct points in $\mathbb {C}$ and let $r > 0$ a real number small enough in order that the discs $D_1, D_3, \dots , D_k$ with respective centers $z_1^0, z_3^0, \dots , z_k^0$ and radius r are two by two disjoint. Let $\mathcal {U}_0:= D_1\times D_1\times \prod _{j=3}^k D_j$ and $\mathcal {V}$ (equal to $D_1\times D_2$ for $k = 2$) the image of $\mathcal {U}_0$ by the quotient map $quot: \mathbb {C}^k \rightarrow \mathbb {C}^k\big /\mathfrak {S}_k = N$. Note $\mathcal {U}:= q^{-1}(\mathcal {V})$. Then, q induces an isomorphism of $\mathcal {U}_0\big /\mathfrak {S}_2$ onto $\mathcal {V}$.

Remark that for each $\sigma \in \mathcal {V}$ we have exactly two roots $z_1(\sigma ), z_2(\sigma )$ distinct or not which are in $D_1$ and for each $j \in [3, k]$ we have exactly one (simple) root $z_j(\sigma )$ in $D_j$. We have the following holomorphic map on $\mathcal {V}$:

• 1. The map $\tau = (\tau _1, \tau _2): \mathcal {V} \rightarrow \mathbb {C}^2$ given by $\tau _1(\sigma ):= z_1(\sigma ) + z_2(\sigma )$ and $\tau _2:= z_1(\sigma )z_2(\sigma )$ where $z_1(\sigma )$ and $z_2(\sigma )$ are the roots of $P_\sigma$ which are in $D_1$.

• 2. For each $j \in [3, k]$ the map $z_j: \mathcal {V} \rightarrow D_j$ given by the unique (simple) root of $P_\sigma$ in $D_j$.

To be completely clear, these holomorphic maps are defined on $\mathcal {V}$ by the following integral formulas:

$$\begin{aligned}&\tau _1(\sigma ) := \frac{1}{2i\pi } \int _{\partial D_1} \frac{\zeta P'_\sigma (\zeta )d\zeta }{P_\sigma (\zeta )} \\&2\tau _2(\sigma ) = \tau ^2_1 - \nu _2(\sigma ) \quad \textrm{where} \quad \nu _2(\sigma ) := \frac{1}{2i\pi } \int _{\partial D_1} \frac{\zeta ^2P'_\sigma (\zeta )d\zeta }{P_\sigma (\zeta )} \\&\mathrm{and \ for} \ j \in [3, k] \quad z_j(\sigma ) := \frac{1}{2i\pi } \int _{\partial D_j} \frac{\zeta P'_\sigma (\zeta )d\zeta }{P_\sigma (\zeta )} \end{aligned}$$

The following lemma is obvious:

Lemma 2.3.3

The holomorphic map $\Phi : \mathcal {V} \rightarrow \mathbb {C}^2\times \prod _{j=3}^k D_j$ given by $(\tau _1, \tau _2, z_3, \dots , z_k)$, is an isomorphism of $\mathcal {V}$ onto the open set $\mathcal {V}_1:= D_{1,2}\times \prod _{j=3}^k D_j$ where we define $D_{1,2}:= (D_1\times D_1)\big /\mathfrak {S}_2$ as the image of $D_1\times D_1$ by the quotient map by the action of the permutation group $\mathfrak {S}_2$.$\blacksquare$

In the sequel, we shall use the coordinate system on $\mathcal {V}$ given by the holomorphic functions $\tau _1, \tau _2, z_3, \dots , z_k$ on $\mathcal {V}$.

Define on $\mathcal {V} \simeq \mathcal {V}_1= D_{1, 2}\times \prod _{j=3}^k D_j$ the following partial differential operators in the coordinate system described above:

• $T^2:= \partial _{\tau _1}^2 + \tau _1\partial _{\tau _1}\partial _{\tau _2} + \tau _2\partial _{\tau _2}^2 + \partial _{\tau _2}$

• $B_{i,j}:= \partial ^2\big /\partial z_i\partial z_j$ for $3 \le i < j \le k$

• $C_{1,j}:= \partial ^2\big /\partial \tau _1\partial z_j$ for $j \in [3, k]$

• $C_{2, j}:= \partial ^2\big /\partial \tau _2\partial z_j$ for $j \in [3, k]$

• $V_0:= \tau _1\partial _{\tau _1} + 2\tau _2\partial _{\tau _2} + \sum _{j=3}^k z_j\partial _{z_j}$

• $V_{-1}:= 2\partial _{\tau _1} + \tau _1.\partial _{\tau _2} + \sum _{j=3}^k \partial _{z_j}.$

Proposition 2.3.4

The isomorphism of change of coordinates $\Phi$ on $\mathcal {V}$ given by $\sigma \mapsto (\tau _1, \tau _2, z_3, \dots , z_k)$ has the following properties:

• (i) The image of ideal $\mathcal {I}$ of $\mathcal {D}_N$ restricted to $\mathcal {V}$ by the isomorphism $\Phi$ is the left ideal generated by $T^2$, $B_{i,j}$, $C_{1,j}$ and $C_{2, j}$ in $D_{\mathcal {V}_1}$.

• (ii) The vector field $U_0$ is sent to $V_0$ and the vector field $U_{-1}$ is sent to $V_{-1}$ by this isomorphism.

Proof

We shall use the local version of Theorem 5.1.1 in [1] which is given in Proposition 2.2.3 above.

For $\Omega \subset \mathcal {V}$ it is easy to see that the Fréchet space of trace functions admits as a dense subset the finite $\mathbb {C}-$linear combinations of the Newton functions $\nu _m, m \in \mathbb {N}$ of $z_1(\sigma ), z_2(\sigma )$ and of the functions $z_j^m(\sigma ), m \in \mathbb {N}$ for each $j \in [3, k]$. From the case $k = 2$ for which the left ideal $\mathcal {I}$ is generated by $T^2$ and the fact that each $B_{i,j}, C_{1,j}$ and $C_{2,j}$ kill each $\nu _m$ and each $z_j^m$, we conclude that $\mathcal {I}$ contains the left ideal generated by $T^2$, the $B_{i,j}$, the $C_{1,j}$ and the $C_{2, j}$.

Conversely, if P is in $\mathcal {I}$ it has to kill any $\nu _m$ and each $z_j^m, \forall j \in [3, k]$. Therefore, P has no order 0 term. Modulo the ideal generated by the $B_{i,j}$, the $C_{1,j}$ and the $C_{2, j}$ we may assume that we can write

$$\begin{aligned} P = P_0 +\sum _{m=1}^N \sum _{j=3}^k g_{j,m}\partial _{z_j}^m \end{aligned}$$

where $P_0$ is a differential operator in $\tau _1, \tau _2$ with no order 0 term, and with holomorphic dependence in $z_3, \dots ,z_k$ (but no derivation in these variables) and where $g_{j, m}$ are holomorphic functions on $\mathcal {V}$. Applying P to $z_j^N$, with $j \in [3, k]$, gives that

$$\begin{aligned} \sum _{m=1}^N \ g_{j,m}\frac{N!}{(N-m)!}z_j^{N-m} = 0 \end{aligned}$$

and then $g_{j,m} = 0$ for each $m \in [1, N]$ and each $j \in [3, k]$, because the $g_{j,m}$ are holomorphic functions of $(\tau _1, \tau _2)$. Then, $P = P_0$ and $P(\nu _m) = 0$ implies that $P_0$ is in the left ideal generated by $T^2$ in the $\mathcal {O}_{\mathcal {V}_1}$-algebra generated by $\frac{\partial }{\partial \tau _1}$ and $\frac{\partial }{\partial \tau _2}$. Then, $P_0$ and also P are in our ideal and (i) is proved.

The verification of (ii) is easy and left to the reader.$\blacksquare$

Lemma 2.3.5

For $k=2$ we have for each $n \in \mathbb {N}^*$

$$\begin{aligned} (T^2 - 2n\partial _2)\Delta ^n = \Delta ^nT^2 + 2n(2n + 1)\Delta ^{n-1}. \end{aligned}$$

(16)

Proof

Recall that we have $E:= \sigma _1\partial _1 + \sigma _2\partial _2$ and $T^2 = \partial _1^2 + \partial _2E$ and that

$\Delta = \sigma _1^2 - 4\sigma _2$. Therefore, we have

$$\begin{aligned}&\partial _1\Delta = \Delta \partial _1 + 2\sigma _1 \\&\left[ \sigma _1\partial _1, \Delta \right] = 2\sigma _1^2 \\&\partial _2.\Delta = \Delta \partial _2 - 4 \\&\left[ \sigma _2\partial _2, \Delta \right] = -4\sigma _2 \\&\left[ E, \Delta \right] = 2\sigma _1^2 - 4\sigma _2 \end{aligned}$$

and

$$\begin{aligned}&\left[ \partial _2E, \Delta \right] = \partial _2(\Delta E + 2\sigma _1^2 - 4\sigma _2) - (\partial _2\Delta +4).E = 2\Delta \partial _2 - 4\sigma _1\partial _1 - 4 \\&\partial _1^2\Delta = \partial _1(\Delta \partial _1 + 2\sigma _1) = ( \Delta \partial _1 + 2\sigma _1)\partial _1 + 2\sigma _1\partial _1 + 2 = \Delta \partial _1^2 + 4\sigma _1\partial _1 + 2 \\&\left[ T^2,\Delta \right] = \left[ \partial _1^2, \Delta \right] + [\partial _2E, \Delta ] \\&T^2.\Delta = \Delta T^2 + 4\sigma _1\partial _1 + 2 + 2\Delta \partial _2 - 4\sigma _1\partial _1 - 4 = \Delta T^2 + 2\partial _2\Delta + 8 +2 - 4 \\&(T^2 - 2\partial _2)\Delta = \Delta T^2 + 6 \end{aligned}$$

which proves (16) for $n = 1$.

Assume now that we have proved the formula (16) for $n \ge 1$. Then, we have, using that $\Delta ^n\partial _2 = \partial _2\Delta ^n + 4n\Delta ^{n-1}$

$$\begin{aligned} (T^2 - 2n\partial _2)\Delta ^{n+1}&= \Delta ^nT^2\Delta + 2n(2n + 1)\Delta ^n \\ (T^2 - 2n\partial _2)\Delta ^{n+1}&= \Delta ^{n+1}T^2 + \Delta ^n2\partial _2.\Delta + 6\Delta ^n + 2n(2n + 1)\Delta ^n \\ (T^2 - 2n\partial _2)\Delta ^{n+1}&= \Delta ^{n+1}T^2 + 2\partial _2\Delta ^{n+1} + 8n\Delta ^n + 6\Delta ^n + 2n(2n + 1)\Delta ^n \\ (T^2 - 2(n+1)\partial _2)\Delta ^{n+1}&= \Delta ^{n+1}T^2 + 2(n+1)(2n+3)\Delta ^n \end{aligned}$$

because $2n(2n + 1) + 8n + 6 = 2(n+1)(2n+3)$.$\blacksquare$

Theorem 2.3.6

Let Q be a coherent $\mathcal {D}_N$-module which is a quotient of $\mathcal {M}$ and which is supported by $\{\Delta = 0\}$. Then, $Q = \{0\}$. Moreover, any holonomic quotient of $\mathcal {M}$ has no $\Delta -$torsion.

Proof

Thanks to Corollary 2.3.2 it is enough to prove that such a quotient Q is zero near the generic points of $\{\Delta = 0\}$. Therefore, assume that Q is such a non zero quotient. Using now the result of Proposition 2.3.4, the $\mathcal {D}$-module Q is given near the generic points of $\{\Delta = 0\}$ by the quotient of $\mathcal {D}$ by a left ideal $\mathcal {K}$ which contains $T^2$. Then, there exists an integer $n > 0$, such that $\Delta ^n$ belongs to $\mathcal {K}$. Then, Lemma 2.3.5 implies that $\mathcal {K}$ contains $\Delta ^{n-1}$. Then, by a descending induction on n we obtain that 1 is in $\mathcal {K}$ and this contradicts the non vanishing of Q.

The characteristic variety of a holonomic quotient of $\mathcal {M}$ which is supported in codimension $\ge 1$ in N is contained in the characterisc variety of $\mathcal {M}$ so is contained in the union of $N\times \{0\}$ with the co-normal to $\{\sigma _k = 0\}$ and $\{\Delta = 0\}$ thanks to Proposition 2.3.1. However, Lemma 2.3.5 implies that near the generic point of $\{\Delta = 0\}$ a torsion element in such a quotient vanishes. Therefore, the torsion submodule of a holonomic quotient of $\mathcal {M}$ cannot have the co-normal of $\{\Delta = 0\}$ in its characteristic variety. Then, such a quotient has no $\Delta -$torsion.$\blacksquare$

2.4 Action of $sl_2(\mathbb {C})$ on $\mathcal {M}$

Let B be the sub-$\mathbb {C}-$algebra of the Weyl algebra $\mathbb {C}[\sigma ]\langle \partial \rangle$ generated by the vector fields $U_p, p \ge -1$, where $U_p$ is the vector field on N defined as the image by the differential $T_{quot}$ of the quotient map:

$$\begin{aligned} quot: M:= \mathbb {C}^k \longrightarrow N:= \mathbb {C}^k\big /\mathfrak {S}_k \simeq \mathbb {C}^k \end{aligned}$$

of the vector field $\sum _{j = 1}^k z_j^{p+1}\frac{\partial }{\partial z_j}$.

Theorem 2.4.1

For each $p \ge -1$, we have $\mathcal {I}U_p \subset \mathcal {I}$. Then, the right action of B on $\mathcal {D}_N$ induces a morphism of algebras between B and the algebra of left $\mathcal {D}_N$-linear endomorphisms of $\mathcal {M}$.

Moreover, the right action of B on $\mathcal {M}$ satisfies $[U_p, U_q] = (q-p)U_{p+q}, \forall p, q \ge -1$.

Proof

It will be enough to show that for each integer $p \ge -1$ we have the inclusion $\mathcal {I}U_p \subset \mathcal {I}$. If it is not difficult to prove such an inclusion for $p = -1$ or $p= 0$ by a direct computation of the commutators of $U_p$ with the generators of $\mathcal {I}$, it seems rather difficult to do it for p large, because the coordinates of $U_p$ in the $\mathbb {C}[\sigma ]$ basis $\partial _1, \dots , \partial _k$ of the polynomial vector fields on N seems more and more complicated. Therefore, we shall use the local version of Theorem 5.1.1 in [1] given in Proposition 2.2.3.

Let $P \in \mathcal {I}$, $p \ge -1$ an integer and $m \in \mathbb {N}$. Using the formula $U_p[N_m] = mN_{m+p}$ which is easy to verify on M, we get

$$\begin{aligned} P[U_p[N_m]] = P[mN_{m+p}] = 0 \end{aligned}$$

when P annihilates any Newton polynomial. Then, $PU_p$ also annihilates any Newton polynomial and thanks to Proposition 2.2.3 we conclude that $PU_p$ belongs to $\mathcal {I}$ proving the first assertion.

The verification of the commutation formula

$$\begin{aligned} U_pU_q - U_qU_p = (q-p)U_{p+q} \end{aligned}$$

is easy and left to the reader.$\blacksquare$

Remark The commutation relations

$$\begin{aligned}{}[U_0, U_{-1}] = -U_{-1}, \quad [U_0, U_1] = U_1\quad \textrm{and} \quad [U_1, U_{-1}] = 2U_0 \end{aligned}$$

which are easy to check in M, show that the Lie algebra $\mathcal {L}$ generated by the $U_p$ (with the commutators given by the formula (17)) contains a sub-Lie algebra isomorphic to $sl_2(\mathbb {C})$. The formula (17) shows that $\mathcal {L}$ acts on $\mathcal {M}$ and then induces a structure of $sl_2(\mathbb {C})-$module on $\mathcal {M}$.

3 The $\mathcal {D}_N$-Modules $\mathcal {N}_\lambda$

3.1 Homothety and Translation

Notations Let $\lambda$ a complex number. We define the left ideal

$$\begin{aligned} \mathcal {J}_\lambda := \mathcal {I} + \mathcal {D}_N(U_0 - \lambda ) \end{aligned}$$

in $\mathcal {D}_N$ and let $\mathcal {N}_\lambda$ be the quotient $\mathcal {D}_N\big /\mathcal {J}_\lambda$. We shall denote by $q_\lambda : \mathcal {M} \rightarrow \mathcal {N}_\lambda$ the quotient map.

We shall denote, respectively, by $\mathscr {H}_{\lambda }$ and $\mathscr {T}$ the endomorphisms of left $\mathcal {D}_N$-modules on $\mathcal {M}$ induced, respectively, by the right multiplications by $U_{0} - \lambda$ and $U_{-1}$ (see the Theorem 2.4.1). They satisfy the commutation relation (see loc. cit.)

$$\begin{aligned} \mathscr {H}_{\lambda }\circ \mathscr {T}- \mathscr {T}\circ \mathscr {H}_{\lambda } = - \mathscr {T} \end{aligned}$$

for each $\lambda \in \mathbb {C}$ and $\mathcal {N}_{\lambda }$ is, by definition, the co-kernel of $\mathscr {H}_{\lambda }$.

As $\mathcal {I}.U_{-1} \subset \mathcal {I}$, writing this relation in the form $\mathscr {H}_{\lambda -1}\circ \mathscr {T} = \mathscr {T}\circ \mathscr {H}_{\lambda }$ we see that the right multiplication by $U_{-1}$ induces a left $\mathcal {D}_N$-modules morphism

$$\begin{aligned} \mathscr {T}_\lambda : \mathcal {N}_{\lambda -1} \rightarrow \mathcal {N}_{\lambda } \end{aligned}$$

for each $\lambda$.

Proposition 3.1.1

For each $\lambda \in \mathbb {C}$ we have an exact sequence of left $\mathcal {D}_N$-modules on N

$$\begin{aligned} 0 \rightarrow \mathcal {M} \overset{\mathscr {H}_{\lambda }}{\longrightarrow } \mathcal {M} \overset{q_{\lambda }}{\longrightarrow } \mathcal {N}_{\lambda } \rightarrow 0 \end{aligned}$$

(17)

where $q_{\lambda }$ is the obvious quotient map.

Proof

The quotient map $q_{\lambda }$ is surjective by definition, so the point is to prove that the kernel of $q_{\lambda }$ is isomorphic to $\mathcal {M}$.

This kernel is obviously given by

$$\begin{aligned} \mathcal {J}_{\lambda }\big / \mathcal {I} \simeq \big (\mathcal {I} + D_{N}(U_0 - \lambda )\big )\big /\mathcal {I} \simeq D_{N}(U_0 - \lambda )\big /\mathcal {I}\cap D_{N}(U_{0}- \lambda ) \end{aligned}$$

(18)

The proof will be an easy consequence of the following lemma.

Lemma 3.1.2

Let P be a germ in $D_{N, \sigma }$ for some $\sigma \in N$ such that $P(U_{0} - \lambda )$ is in $\mathcal {I}_{\sigma }$. Then, P is in $\mathcal {I}_\sigma$. Therefore, $\mathcal {I}_\sigma \cap D_{N, \sigma }(U_0 -\lambda ) = \mathcal {I}_\sigma (U_0 -\lambda )$.

Proof

Assume that the lemma is wrong. Then, let $P_{0}$ in $D_{N, \sigma }$ having minimal order among germs P in $D_{N, \sigma }$ satisfying the following properties

• 1. $P(U_{0} -\lambda )$ is in $\mathcal {I}_{\sigma }\cap D_{N, \sigma }(U_{0}- \lambda )$;

• 2. P is not in $\mathcal {I}_{\sigma }$.

Let $\pi$ be the symbol of $P_{0}$ and let g be the symbol of $U_{0}$. We have $\pi g \in p_* (I_Z)_\sigma$. However, we know that g does not vanish on any non empty open set of Z, because $\{g = 0\} \cap Z$ has pure co-dimension 1 in Z (see Lemma 2.2.5 above). Then, $\pi$ vanishes on $(V \times \mathbb {C}^k) \cap Z$ where V is a neighborhood of $\sigma$ in N and, as we have proved that Z is reduced and is the characteristic cycle of $\mathcal {M}$, their exists a germ $P_{1}$ in $\mathcal {I}_{\sigma }$ with symbol equal to $\pi$. Then, $(P_{0}- P_{1})(U_{0}- \lambda )$ satisfies again the properties 1 and 2 and is of order strictly less than the order of $P_{0}$. Therefore, $P_0 - P_{1}$ is in $\mathcal {I}_{\sigma }$ and this contradicts the fact that we assumed that $P_{0}$ is not in $\mathcal {I}_{\sigma }$.$\blacksquare$

End of proof of 3.1.1 The previous lemma shows that for each $\lambda \in \mathbb {C}$

$$\begin{aligned} \mathcal {I} \cap D_{N}(U_{0} - \lambda ) = \mathcal {I}(U_{0} - \lambda ). \end{aligned}$$

Therefore, the right multiplication by $U_{0}- \lambda$ induces an isomorphism of left $\mathcal {D}_N$-modules

$$\begin{aligned} \mathcal {M} \rightarrow D_{N}(U_{0}- \lambda )\big /\mathcal {I}(U_{0}- \lambda ) \end{aligned}$$

and the kernel of $q_{\lambda }$ is isomorphic to $\mathcal {M}$ by the inverse of this isomorphism.$\blacksquare$

Definition 3.1.3

Define the $\mathcal {D}_N$-module $\mathfrak {N}$ as the quotient $\mathcal {D}_N\big /(\mathcal {I} + \mathcal {D}_NU_{-1})$. For each $\lambda \in \mathbb {C}$ then define $\mathscr {T}_\lambda : \mathcal {N}_\lambda \rightarrow \mathcal {N}_{\lambda +1}$ as the $\mathcal {D}_N$-linear map induced by $\mathscr {T}$.

Lemma 3.1.4

For each $\lambda \in \mathbb {C}$ the co-kernel of the $\mathcal {D}_N$-linear map $\mathscr {T}_{\lambda +1}$ is naturally isomorphic to the co-kernel of the $\mathcal {D}_N$-linear map $\tilde{\mathscr {H}}_{\lambda +1}: \mathfrak {N} \rightarrow \mathfrak {N}$ induced by $\mathscr {H}_{\lambda +1}$ and there is also a natural isomorphism of $\mathcal {D}_N$-modules between the kernels of $\tilde{\mathscr {H}}_{\lambda +1}$ and $\mathscr {T}_{\lambda +1}$.

Proof

Consider the commutative diagram of left $\mathcal {D}_N$-modules with exact lines and columns:

where $\mathcal {N}^\square _{\lambda +1}$ is, by definition, the co-kernel of $\mathscr {T}_{\lambda +1}: \mathcal {N}_\lambda \rightarrow \mathcal {N}_{\lambda +1}$. By a simple diagram chasing it is easy to see that $\mathcal {N}^\square _{\lambda +1}$ is also the co-kernel of $\tilde{\mathscr {H}}_{\lambda +1}: \mathfrak {N} \rightarrow \mathfrak {N}$.

A elementary diagram chasing gives also the isomorphism between kernels of $\tilde{\mathscr {H}}_{\lambda +1}$ and $\mathscr {T}_{\lambda +1}$. $\blacksquare$

We shall prove now that for $\lambda \not = 0, 1$ the map $\mathscr {T}_\lambda$ is an isomorphism of left $\mathcal {D}_N$-modules. This implies $\mathcal {N}^\square _\lambda = \{0\}$ for $\lambda \not = 0, 1$.

Lemma 3.1.5

Let $\mathscr {G}_\lambda : \mathcal {N}_{\lambda +1} \rightarrow \mathcal {N}_\lambda$ the left $\mathcal {D}_N$-linear map given by right multiplication by $U_1$. Then, we have for each $\lambda \in \mathbb {C}$

Therefore, for $\lambda \not = 0, 1$ the left $\mathcal {D}_N$-linear map $\mathscr {T}_\lambda$ is an isomorphism.

Proof

We shall use the same argument than in the proof of Theorem 2.4.1 to prove the formulas

$$\begin{aligned} U_1U_{-1} = U_0(U_0 - 1)\quad \textrm{modulo} \quad \mathcal {I} \quad \textrm{and} \quad U_{-1}U_1 = U_0(U_0 + 1) \quad \textrm{modulo} \quad \mathcal {I}\nonumber \\ \end{aligned}$$

(19)

For each $m \in \mathbb {N}$ we have:

$$\begin{aligned} U_1U_{-1}[N_m]&= U_1[mN_{m-1}] = m(m-1)N_m \\ U_0(U_0 - 1)[N_m]&= U_0[(m-1)N_m] = m(m-1)N_m \qquad \mathrm{and \ also} \\ U_{-1}U_1[N_m]&= U_{-1}[mN_{m+1}] = (m+1)mN_m \\ U_0(U_0 + 1)[N_m]&= U_0[(m+1)N_m] = m(m+1)N_m \end{aligned}$$

and this implies Formulas (19).

These give (A) and (B) and the conclusion follows.$\blacksquare$

The following important result shows that adding to the ideal $\mathcal {A}$ the invariance by translation and the homogeneity 1, that is to say considering the left ideal in $\mathcal {D}_N$: $\mathcal {A} + \mathcal {D}_NU_{-1} + \mathcal {D}_N(U_0 - 1)$, we recover the ideal $\mathcal {J}_1 + \mathcal {D}_NU_{-1}$ and $\mathcal {D}_N\big /(\mathcal {J}_1 + \mathcal {D}_NU_{-1})$ is the co-kernel of the map $\mathscr {T}_1: \mathcal {N}_0 \rightarrow \mathcal {N}_1$. Therefore, as a corollary, we shall obtain the equality

$$\begin{aligned} \mathcal {N}^\square _1 = \mathcal {D}_N\big / \mathcal {A} + \mathcal {D}_NU_{-1} + \mathcal {D}_N(U_0 - 1). \end{aligned}$$

(20)

Proposition 3.1.6

For $h \in [2, k]$ we have the equality

and for $h = 1$ the equality

Proof

Recall first that, if we put $E:= \sum _{h=1}^k h\sigma _h\partial _h$ then for any $m \in [2, k]$ we have

$$\begin{aligned} \mathcal {T}^m = \partial _1\partial _{m-1} + \partial _mE = \partial _1\partial _{m-1} + E\partial _m + \partial _m. \end{aligned}$$

For $h \in [2, k]$ we have

$$\begin{aligned} \partial _hU_0 + \partial _{h-1}U_{-1}&= \sum _{q=1}^k q\sigma _q\partial _q\partial _h + h\partial _h + \sum _{q=0}^{k-1} (k-q)\sigma _q\partial _{q+1}\partial _{h-1} + (k-h+1)\partial _h \\&\quad = \sum _{q=1}^k q\sigma _q\partial _q\partial _h + \sum _{q=1}^{k} (k-q)\sigma _{q}A_{h-1, q+1} + k\partial _1\partial _{h-1}\\&\qquad + \sum _{q=1}^{k} (k-q)\sigma _q\partial _q\partial _h + (k+1)\partial _h \\&\quad = kE\partial _h + k\partial _1\partial _{h-1} + k\partial _h + \partial _h + \sum _{q=1}^{k-1} (k-q)\sigma _qA_{h-1, q+1}\\&\quad = k\mathcal {T}^h + \partial _h + \sum _{q=1}^{k-1} (k-q)\sigma _{q}A_{h-1, q+1} \end{aligned}$$

which is $(E_h)$.

For $h = 1$ let us compute $\sum _{q=1}^{k-1} (k-q).\sigma _{q}.\mathcal {T}^{q+1}$:

$$\begin{aligned}&\sum _{q=1}^{k-1} (k-q)\sigma _{q}\mathcal {T}^{q+1} = \sum _{q=1}^{k-1} (k-q)\sigma _{q}\left( \partial _1\partial _{q} + \partial _{q+1}E \right) \\&\sum _{q=1}^{k-1} (k-q)\sigma _{q}\mathcal {T}^{q+1} = \left( \sum _{q=1}^{k-1} (k-q)\sigma _q\partial _q\right) \partial _1 + \sum _{q=1}^{k-1} (k-q)\sigma _{q}\partial _{q+1}E \\&\sum _{q=1}^{k-1} (k-q)\sigma _{q}\mathcal {T}^{q+1} = \big (\sum _{q=1}^{k-1} (k-q)\sigma _q\partial _q\big )\partial _1 + (U_{-1} - k\partial _1)E \\&\quad \quad = k(E - \sigma _k\partial _k)\partial _1 - (U_0 - k\sigma _k\partial _k)\partial _1 + (U_{-1} - k\partial _1)E \\&\quad \quad = kE\partial _1 - k\partial _1E - U_0\partial _1 + U_{-1}E \\&\quad \quad = EU_{-1} - \partial _1(U_0 - 1) \end{aligned}$$

using the commutation relations $[U_{-1}, E] = k\partial _1$, $[E, \partial _1] = -\partial _1$ and $[U_0, \partial _1] = -\partial _1$. Therefore, we obtain the equality $(E_1)$.$\blacksquare$

Remarks

• 1. An interesting way to look at these relations is to compare them with the minors of the $(k+1, 2)$ matrix

  $$\begin{aligned} \begin{pmatrix} \mathscr {T} &{}\quad - (\mathscr {H}_0 - 1) \\ \partial _1 &{} -\mathscr {E} \\ \partial _2 &{} \partial _1 \\ . &{}.\\ . &{}.\\ . &{}. \\ \partial _k &{} \partial _{k-1} \end{pmatrix} \end{aligned}$$

  where, by definition, $\mathscr {E}$ is the right product by E in the Weyl algebra $\mathbb {C}[\sigma ]\langle \eta \rangle$. The relations $(E_h), h \in [1, k]$ may also be seen as the fact that

  $$\begin{aligned} \begin{pmatrix} \partial _1 \\ \partial _2 \\ . \\ . \\ . \\ \partial _k \end{pmatrix}(U_0 - 1) + \begin{pmatrix} -E \\ \partial _1 \\ . \\ . \\ . \\ \partial _{k-1} \end{pmatrix}U_{-1} \end{aligned}$$

  is a global section of $\mathcal {I}^k \subset \mathcal {D}_N^k$.

• 2. Let g and $\gamma$ be the symbols of $U_0$ and $U_{-1}$, respectively. Looking at the symbols in formulas $(E_h), h \in [1, k]$, we obtain (recall that $l_\sigma (\eta )$ is the symbol of E):

  

• 3. For $\lambda \not = 0$ the sheaf of solutions⁶ of $\mathcal {N}_\lambda$ near the generic point in N is the rank k local system with basis $z_j^\lambda$. This is consequence of the the fact that any local trace function⁷F which satisfies $(U_0 - \lambda )[F] = 0$ is the trace of a homogeneous function of degree $\lambda$.

• 4. Thanks to Lemma 3.1.5, the map induced by $\mathcal {T}_\lambda$ on solutions

  $$\begin{aligned} Sol^0(\mathcal {N}_\lambda ) \rightarrow Sol^0(\mathcal {N}_{\lambda -1}) \end{aligned}$$

  sends $z_j^\lambda$ to $\lambda z_j^{\lambda -1}$. This is clearly an isomorphism for $\lambda \not = 0, 1$.

3.2 Characteristic Varieties

Recall that for a differential operator $P \in \mathcal {D}_N$ we note s(P) its symbol which is a section of the sheaf $\mathcal {O}_N[\eta ]$ of homogeneous polynomials in $\eta := (\eta _1, \dots , \eta _k)$.

Proposition 3.2.1

Let $\mathcal {I}$ be a coherent left ideal in $\mathcal {D}_N$ such that its characteristic ideal $I_Z$ is the reduced ideal of an analytic subset Z in $N \times \mathbb {C}^k$. Let $U \in \Gamma (N, \mathcal {D}_N)$ be a differential operator of order q, such that its symbol u does not vanish on any non empty open set in Z. Assume that $\mathcal {I}U \subset \mathcal {I}$. Then, the characteristic ideal of $\mathcal {I} + \mathcal {D}_NU$ is equal to $I_Z + \mathcal {O}_{N\times \mathbb {C}^k}u$.

Moreover, for any $\sigma \in N$ and any germ at $\sigma$ of order $q+r$: $Q = P + BU$ where $P \in \mathcal {I}_\sigma$ and $B \in D_{N, \sigma }$, there exists $P_1 \in \mathcal {I}_\sigma$ of order at most $q+r$ and $B_1 \in D_{N, \sigma }$ of order at most r such that $Q = P_1 + B_1U$.

Proof

First, assume that there exists $P \in \mathcal {I}_\sigma$ and $B \in D_{N, \sigma }$, such that the symbol of $Q:= P + BU$ is not in $I_Z + (u)$. Then, consider such a couple $(P_0, B_0)$ with $B_0$ of order b minimal among all such couples. Then, $P_0$ and $B_0U$ have the same order, because when their orders are different we have $s(Q) = s(P_0)$ or $s(Q) = s(B_0)u$ contradicting the fact that s(Q) is not in $I_Z + (u)$.

In addition, if $P_0$ and $B_0U$ have equal orders which is the order of Q, we have the equality $s(Q) = s(P_0) + s(B_0)u$ contradicting our assumption.

Therefore, the only case left is when $P_0$ and $B_0U$ have the same order $b_0 + q$ which is strictly bigger than the order of Q. In this case we have $s(P_0) + s(B_0)u = 0$ which implies that $s(B_0)u$ vanishes on Z. However, our hypothesis on u implies then that $s(B_0)$ vanishes on Z. As $I_Z$ is reduced and is the characteristic ideal of $\mathcal {I}$ we may find a germ $B \in \mathcal {I}_\sigma$, such that $s(B) = s(B_0)$. Then, write

$$\begin{aligned} Q = P_0 + B_0U = P_0 + BU + (B_0 - B)U. \end{aligned}$$

Since B is in $\mathcal {I}_\sigma$ and $\mathcal {I}U \subset \mathcal {I}$ we have $P_1 = P_0 + BU$ in $\mathcal {I}_\sigma$ and the order of $B_0 - B_1$ is strictly less than b. This contradicts the minimality of b and proves our first assertion.

Assume now that $Q = P + BU$ has order $q+r$ and that B has order $r + s$ with $s \ge 1$. If the order of P and BU are not equal then either P or BU is of order $q+r$ and P and BU have orders at most $q+r$ we are done.

So we may assume that P and BU have the same order $q+r+s$ with $s \ge 1$. Then, the previous considerations will produce $B' \in \mathcal {I}_\sigma$ with $s(B') = s(B)$ and then $P_1:= P + B'U$ and $B_1:= B - B'$ give that $Q = P_1 + B_1U$ with $P_1 \in \mathcal {I}_\sigma$ and $B_1 \in D_{N, \sigma }$ of order at most $q + r + s - 1$. By a descending induction on s this completes our proof, because when $B_1$ has order at most r the order of $P_1$ is at most $q+r$, because we assume that Q has order $q+r$.$\blacksquare$

The following two corollaries are immediate applications of the previous proposition, using Proposition 2.2.2 and Theorem 2.4.1 which allow to verify that needed hypotheses.

Corollary 3.2.2

The characteristic cycle of $\mathcal {N}_\lambda , \forall \lambda \in \mathbb {C}$, is the cycle associated to the ideal $I_Z + (g)$ in $\mathcal {O}_N[\eta ]$ where g is the symbol of $U_0$.

Also the characteristic cycle of $\tilde{\mathcal {M}}$ is the cycle associated to the ideal $I_Z + (\gamma )$ in $\mathcal {O}_N[\eta ]$ where $\gamma$ is the symbol of $U_{-1}$.$\blacksquare$

Corollary 3.2.3

Let $\mathcal {I}$ the left ideal in $\mathcal {D}_N$ that we introduced in Definition 2.2.1 and let $U:= U_0 -\lambda$. Then, for any non zero germ $Q \in \mathcal {I} + \mathcal {D}_NU$ of order $q+1$ there exist a germ $P \in \mathcal {I}$ of order at most $q+1$ and a germ $B \in \mathcal {D}_N$ of order at most q such that $Q = P + BU$. $\blacksquare$

Remark

• 1. We shall be interested mainly by the special case of Corollary 3.2.3. Define for each $q \ge 0$

  $$\begin{aligned} \mathcal {J}_\lambda (q+1) = \mathcal {I}(q+1) + \mathcal {D}_N(q)(U_0 - \lambda ). \end{aligned}$$

  Then, this corollary gives, for each $\lambda \in \mathbb {C}$ and for each $q \in \mathbb {N}^*$ the equality $\mathcal {J}_{\lambda } \cap \mathcal {D}_N(q) = \mathcal {J}_\lambda (q)$. This implies that the natural map

  $$\begin{aligned} \mathcal {N}_\lambda (q) \rightarrow \mathcal {N}_\lambda \end{aligned}$$

  (21)

  is injective

• 2. Note that $\mathcal {J}_\lambda (0):= \mathcal {I}(0) = \{0\}$ as no non zero differential operator of order 0 annihilates the Newton polynomials (in fact $N_0:= k$ is enough !)

• 3. Also the fact that $\mathcal {I}(1) = \{0 \}$ (see Lemma 2.2.5) implies the equality

  $$\begin{aligned} \mathcal {J}_\lambda (1) = \mathcal {O}_N(U_0 - \lambda ). \end{aligned}$$

The irreducible component X. Let $H_\Delta := \{\Delta (\sigma ) = 0 \}$ in N. At the generic point $\sigma$ of this hyper-surface, the polynomial $P_\sigma$ has exactly one double root $\varphi (\sigma )$ and $\varphi : H_\Delta \dashrightarrow \mathbb {C}$ is a meromorphic function which is locally bounded on $H_\Delta$. Then, define the meromorphic map

$$\begin{aligned} \Phi : H_\Delta \dashrightarrow \mathbb {P}_{k-1} \end{aligned}$$

by letting $\Phi _h(\sigma ) = (- \varphi (\sigma ))^{k-h}$ for $h \in [1, k]$ in homogeneous coordinates. Let $X \subset N \times \mathbb {C}^k$ be the $N-$relative cone over the graph of the meromorphic map $\Phi$. This is a $k-$dimensional irreducible subset in $H_\Delta \times \mathbb {C}^k$ and its fiber at the generic point in $H_\Delta$ is the line directed by the vector $\Phi _h(\sigma ), h \in [1, k]$.

We shall consider the following sub-spaces in $N \times \mathbb {P}_{k-1}$ (where s(P) is the symbol of P)

$$\begin{aligned} \mathbb {P}(Z):= & {} \left\{ (\sigma , \eta ) \in N \times \mathbb {P}_{k-1} \ / \ s(P)(\sigma , \eta ) = 0 \quad \forall P \in \mathcal {I}\setminus \{0\} \right\} \\ \mathbb {P}(X):= & {} \left\{ (\sigma , \eta ) \in \mathbb {P}(Z) \ / \ \gamma (\sigma , \eta ):= \sum _{h=0}^{k-1} (k-h).\sigma _h.\eta _{h+1} = 0 \right\} \\ \mathbb {P}(Y):= & {} \left\{ (\sigma , \eta ) \in \mathbb {P}(Z) \ / \ g(\sigma , \eta ):= \sum _{h=1}^k h.\sigma _h.\eta _h = 0 \right\} . \end{aligned}$$

The next proposition will justify our notations in proving that $\mathbb {P}(X)$ is the graph of the meromorphic map $\Phi$ !

Proposition 3.2.4

The subspace $\mathbb {P}(Z)$ is a complex sub-manifold of dimension k which is a $k-$branched covering of N via the natural projection $N \times \mathbb {P}_{k-1} \rightarrow N$. The sub-space $\mathbb {P}(X)$ is reduced and equal to the irreducible component of

$$\begin{aligned} \mathbb {P}(Z) \cap \big ( H_\Delta \times \mathbb {P}_{k-1}\big ) \end{aligned}$$

which is the graph of the meromorphic map $\Phi : H_\Delta \rightarrow \mathbb {P}_{k-1}$ defined above, and $\mathbb {P}(Y)$ is the sum (as a cycle) of   $\mathbb {P}(X)$ with the reduced hyper-surface in $\mathbb {P}(Z)$ defined by the (reduced) divisor $\{\eta _{k-1} = 0\}$ in $\mathbb {P}(Z)$.

Proof

First remark that if $(\sigma , \eta )$ is in Z and satisfies $\eta _k = 0$ then we have $\eta = 0$. Therefore,   $\mathbb {P}(Z)$ is contained in the open set $\Omega _k:= \{\eta _k \not = 0\}$ and, on this open set which is isomorphic to $N \times \mathbb {C}^{k-1}$, we may use the coordinates $\sigma _1, \dots , \sigma _k, \eta _1/\eta _k, \dots , \eta _{k-2}/\eta _k$ and $z:= - \eta _{k-1}/\eta _k$.

Lemma 3.2.5

We have an isomorphism

$$\begin{aligned} \varphi _k: \mathbb {P}(Z) \rightarrow \mathbb {C}^k \quad \mathrm{given \ by} \quad (\sigma , \eta ) \mapsto (\sigma _1, \dots , \sigma _{k-1}, z). \end{aligned}$$

Proof

Remark first that the vanishing of the (2, 2) minors which give the equations of Z (see Formula (10) before Proposition 2.2.2) implies, assuming $\eta _k \not = 0$, that:

$$\begin{aligned} \eta _h/\eta _k = (-z)^{k-h} \quad \textrm{for} \ h \in [1, k-1] \end{aligned}$$

and also, as the symbol of $\mathcal {T}^k$ is equal to $\eta _1\eta _{k-1} + \eta _kl_\sigma (\eta )$, that $l_\sigma (\eta )/\eta _k = -(-z)^k$. But then,

$$\begin{aligned} l_\sigma (\eta )/\eta _k = \sum _{h=1}^k (-1)^{k-h}\sigma _hz^{k-h} = (-1)^k(P_\sigma (z) - z^k) \end{aligned}$$

shows that $P_\sigma (z) = 0$ on $\mathbb {P}(Z)$.

Let us show that the holomorphic map $\psi _k: \mathbb {C}^k \rightarrow \mathbb {P}(Z)$ given by

$$\begin{aligned} \eta _h = (-z)^{k-h} \quad \textrm{for} \ h \in [1, k] \quad \textrm{and} \quad \sigma _k = -\sum _{h=0}^{k-1} (-1)^{k-h}\sigma _hz^{k-h} \end{aligned}$$

with the convention $\sigma _0 \equiv 1$ gives an inverse to $\varphi _k$.

First, we shall verify that $\psi _k$ takes its values in $\mathbb {P}(Z)$. Note that the definition of $\sigma _k$ implies $P_\sigma (z) = 0$. We have for $(\sigma , \eta ) = \psi _k(\sigma ', z)$ the equality:

$$\begin{aligned} l_\sigma (\eta ) = \sum _{h=1}^k \sigma _h\eta _h = \sum _{h=1}^k (-1)^{k-h}\sigma _hz^{k-h} = -(-z)^k \end{aligned}$$

Then, we have to verify that the vectors $(\eta _1, \dots , \eta _{k-1}, 1)$ and $((-z)^k, \eta _1, \dots , \eta _{k-1})$ are co-linear. This is clear as the second one is $\eta _{k-1} = (-z)\eta _k$-times the first one (see again Formula (10)).

To complete the proof, it is enough to check that $\psi _k \circ \varphi _k$ and $\varphi _k \circ \psi _k$ are the identity maps. This is easy verification is left to the reader.$\blacksquare$

End of proof of 3.2.4. In this chart we have

$$\begin{aligned} g(\sigma , \eta )/\eta _k&= \sum _{h=1}^k h\sigma _h\eta _h/\eta _k = \sum _{h=1}^k (-1)^{k-h}h\sigma _hz^{k-h} \\ g(\sigma , \eta )/\eta _k&= (-1)^{k+1}\left( \sum _{h=1}^k (-1)^h(k-h)\sigma _hz^{k-h} - k\sum _{h=1}^k (-1)^h\sigma _hz^{k-h}\right) \end{aligned}$$

and this gives

$$\begin{aligned} g(\sigma , \eta )/\eta _k = (-1)^{k+1}\Big (zP'_\sigma (z) -kz^k - k(P_\sigma (z) - z^k)\Big ) = (-1)^{k+1}zP'_\sigma (z). \end{aligned}$$

We have also

$$\begin{aligned} \gamma (\sigma , \eta )/\eta _k = \sum _{h=0}^{k-1} (-1)^{k-h-1}(k-h)\sigma _hz^{k-h-1} = (-1)^{k+1}P'_\sigma (z). \end{aligned}$$

Therefore, $g = z\gamma$ in this chart,⁸ and the ideal generated by g and $\gamma$ in $\mathbb {P}(Z)$ is generated by $\gamma$ which defined the hyper-surface $\mathbb {P}(X)$.

However, on this hyper-surface we have $P_\sigma (z) = 0$ and $P'_\sigma (z) = 0$, so z is a double root of $P_\sigma$. This implies that $\Delta (\sigma ) = 0$ for $(\sigma , \eta )$ in the analytic subset $\vert \mathbb {P}(X)\vert$.

On a Zariski dense open set in $H_\Delta$ the unique double root of $P_\sigma$ is equal to $\varphi (\sigma )$ which is given by $z = -\eta _{k-1}/\eta _k$ when $(\sigma , \eta ) \in \vert \mathbb {P}(X)\vert$. Therefore, $\vert \mathbb {P}(X)\vert$ contains the graph of the meromorphic map $\Phi$. Moreover, as the projection $\mathbb {P}(Z) \rightarrow N$ is clearly a branched covering (of degree k) and over the generic point in $H_\Delta$ there exists an unique root of multiplicity 2 for $P_{\sigma }$, $\mathbb {P}(X)$ has generic degree 1 over $H_\Delta$. In addition,, because $P''_\sigma (z)$ does not vanish at the generic point in $\mathbb {P}(X)$ (which has to be over the generic point of $H_\Delta$) implies that the hyper-surface $\mathbb {P}(X)$ of $\mathbb {P}(Z)$ is reduced. This is enough to conclude that $\mathbb {P}(X)$ is equal to the graph of $\Phi$.

The previous computation shows also that $\mathbb {P}(Y)$ is the sum of $\mathbb {P}(X)$ with the divisor defined by $\{z = 0\}$ in $\mathbb {P}(Z)$ which is a smooth and reduced hyper-surface given by the equation $\eta _{k-1} = 0$ in $\mathbb {P}(Z)$.$\blacksquare$

The determination of the characteristic cycles of the holonomic $\mathcal {D}_N$-modules $\mathfrak {N}$ and $\mathcal {N}_\lambda$ is an easy consequence of the previous proposition thanks to Proposition  2.2.3.

Corollary 3.2.6

The characteristic cycle of the $\mathcal {D}_N$-module $\mathfrak {N}$ is equal to $\mathbb {P}(X)$. For each complex number $\lambda$ the characteristic cycle of the $\mathcal {D}_N$-module $\mathcal {N}_\lambda$ is equal to $\mathbb {P}(Y) = \mathbb {P}(X) + \big (\mathbb {P}(Z) \cap \{ \eta _{k-1} = 0 \}\big )$.$\blacksquare$

Remarks

• 1. The intersection $\mathbb {P}(Z) \cap \{ \eta _{k-1} = 0 \}$ is equal to $N \times [v]$ where v is the point $(0, \dots , 0, 1) \in \mathbb {P}_{k-1}$ and this intersection is the projectivization of the co-normal to the hyper-surface $\{\sigma _k = 0\}$.

• 2. At the set-theoretical level we have

  $$\begin{aligned} Z \cap \{\gamma = 0 \}= & {} X \cup (N\times \{0\}) \quad \textrm{and}\\ Z \cap \{g = 0\}= & {} X \cup \Big ( \{\sigma _k = 0\} \times \{\eta _1= \eta _2 = \dots = \eta _{k-1} = 0 \} \Big ) \cup (N\times \{0\}). \end{aligned}$$

• 3. Despite the previous results, $g(\sigma , \eta )$ does not belongs to the ideal of $\mathbb {C}[\sigma , \eta ]$ generated by $I_Z$ and $\gamma (\sigma , \eta )$ at the generic point in $N\times \{0\}$. This is consequence of the fact that $I_Z$ does not contain a non zero element in $\mathbb {C}[\sigma , \eta ]$ which is homogeneous of degree 1 in $\eta$, using Corollary 3.2.3.

The following lemma will be useful later on

Lemma 3.2.7

Assume that $f\partial _k^nU_{-1}$ is in $\mathcal {J}_{\lambda , \sigma }$ for some $f \in \mathcal {O}_{N, \sigma }$, some integer $n \ge 1$ and some $\lambda \in \mathbb {C}$. Then, f is in $\sigma _k\mathcal {O}_{N, \sigma }$.

Proof

The fact that $f\partial _k^nU_{-1}$ is in $\mathcal {J}_{\lambda , \sigma }$ implies that $f\eta _k^n\gamma$ vanishes on the characteristic variety of the $\mathcal {D}_N$-module $\mathcal {N}_\lambda$. Therefore, thanks to Corollary 3.2.6$f.\eta _k^n\gamma$ vanishes on C, the co-normal bundle of the hyper-surface $\{\sigma _k = 0 \}$. However, $\eta _k$ and $\gamma$ do not vanish on any non empty open set in C:

this is clear for $\eta _k$ and the restriction of $\gamma$ to C is equal to $\sigma _{k-1}\eta _k$ and $\sigma _{k-1}$ also does not vanish on any non empty open set in C. Therefore, $f \in \mathcal {O}_{N, \sigma }$ has to vanish on C and we conclude that f is in $\sigma _k\mathcal {O}_{N, \sigma }$. $\blacksquare$

3.3 The Case $\lambda \not \in \mathbb {N}$

Notation For each $\lambda \in \mathbb {C}$ and each $q \ge 0$ we shall note

$$\begin{aligned} \mathcal {J}_\lambda (q+1):= \mathcal {I}(q+1) + \mathcal {D}_N(q)(U_0 - \lambda ) \end{aligned}$$

and

$$\begin{aligned} \mathcal {N}_\lambda (q+1):= \mathcal {D}_N(q+1)\big /\mathcal {J}_\lambda (q+1). \end{aligned}$$

For $q = 0$ we note $\mathcal {J}_\lambda (0):= \mathcal {I}(0)$ and $\mathcal {N}_\lambda (0):= \mathcal {O}_N\big /\mathcal {J}_\lambda (0)$.

The goal of this paragraph is to prove the following theorem.

Theorem 3.3.1

For $\lambda \in \mathbb {C}\setminus \mathbb {N}^*$ the $\mathcal {D}_N$-module $\mathcal {N}_\lambda$ has no $\mathcal {O}_N$-torsion.

Proof

This result is a direct consequence of Proposition 3.3.5, thanks to the injectivity for each $q \ge 0$ of the natural map $\mathcal {N}_\lambda (q) \rightarrow \mathcal {N}_\lambda$ (see Remark 1 following Corollary 3.2.3).$\blacksquare$

Definition 3.3.2

For any $\lambda \in \mathbb {C}\setminus \mathbb {N}^*$, for any integer $q \ge 2$ and for any integer $r \in [q, k(q-1)]$ define the following elements in $\mathcal {W}_q$ (see Formulas (12) and (13) in Lemma 2.2.6)

$$\begin{aligned}&\theta _{q, r} := (r - \lambda )\partial ^\alpha \mathcal {T}^m] - (q - 1)\left[ \partial ^\beta (U_0 - \lambda )\right] \end{aligned}$$

(22)

$$\begin{aligned}&\quad \textrm{so} \quad \theta _{q, r} = \sum _{h=0}^k (r - \lambda - (q-1)h)\sigma _hy_{q, r+h}. \end{aligned}$$

(23)

where in Formula (23) we assume that $\alpha \in \mathbb {N}^k$ and $m \in [2, k]$ satisfy $\vert \alpha \vert = q-2$ and $w(\alpha ) = r-m$, and that $\beta \in \mathbb {N}^k$ satisfies $\vert \beta \vert = q-1$ and $w(\beta ) = r$.

Corollary 3.3.3

For any integer $q \ge 1$ the kernel of the quotient map

$$\begin{aligned} l_q: \mathcal {W}(q) \rightarrow \mathcal {N}_\lambda (q) \end{aligned}$$

is equal to the sub-$\mathcal {O}_N$-module generated by $U_0 - \lambda \in \mathcal {W}(1)$ and the elements $\theta _{p, r}$, for each $p \in [2, q]$, and for each $r \in [p, (k-1)p]$.

Proof

We have to prove that if a non zero differential operator P of order $p \le q$ is in $\mathcal {J}_\lambda$ then it may be written as $Q + B(U_0 -\lambda )$ with $Q \in \mathcal {I}$ of order at most p (or $Q = 0$) and B of order at most $p-1$ (or $B = 0$). When $p \ge 2$ this is precisely the statement proved in Proposition 3.2.1. For $p \le 1$ the only P which are in $\mathcal {J}_\lambda (1)$ are in $\mathcal {O}_N(U_0 - \lambda )$ thanks to Remarks 2 and 3 following Corollary 3.2.3. $\blacksquare$

Lemma 3.3.4

Let $\lambda$ be in $\mathbb {C}\setminus \mathbb {N}^*$; for each integer $q \ge 2$ the elements $\theta _{q, r}$ and $y_{q, s}$, with $r \in [q, k(q-1)]$ and $s \in [k(q-1)+1, kq]$ form a $\mathcal {O}_N-$basis of $\mathcal {W}_q$.

Proof

Let $\mathcal {W}_{q, p}$ be the $\mathcal {O}_N$-module of $\mathcal {W}_q$ with basis the $y_{q, r}$ for $r \ge p+1$. Then, we have for $r \in [q, k(q-1)]$

$$\begin{aligned} \theta _{q, r} \in (r- \lambda )y_{q, r} + \mathcal {W}_{q, r+1} \end{aligned}$$

so the determinant of the $k(q-1) - q + 1 + k = kq - q + 1$ vectors $\theta _{q, r}, y_{q, s}$ in the basis $\big (y_{q, r}, r \in [q, kq]\big )$ of $\mathcal {W}_q$ is upper triangular and is equal to $\prod _{r = q}^{k(q-1)} (r - \lambda )$ which is in $\mathbb {C}^*$ as soon as $\lambda$ is not in the subset $[q, k(q-1)]$ of $\mathbb {N}^*$.$\blacksquare$

Proposition 3.3.5

Let $q \ge 1$ be an integer and assume that $\lambda$ is not an integer in $[0, k(q-1)]$. Let $L_q: \mathcal {W}_q \rightarrow \mathcal {N}_\lambda (q)$ be the restriction to $\mathcal {W}_q$ of quotient map $l_q$. This $\mathcal {O}_N$-linear map is surjective and its kernel is the sub-module of $\mathcal {W}_q$ with basis the $\theta ^q_r$ for $r \in [q, k(q-1)]$. Therefore, $\mathcal {N}_\lambda (q)$ is a free $\mathcal {O}_N$-module of rank k.

Proof

Remark first that for $q= 1$ the result is clear as for $\lambda \not = 0$ we have

$$\begin{aligned} \mathcal {N}_\lambda (1) = \oplus _{h=1}^k \mathcal {O}_N\partial _h \end{aligned}$$

thanks to Remark 3 following Corollary 3.2.3 and $\mathcal {W}_1 = \oplus _{h=1}^k \mathcal {O}_N.y_{1, h}$ with $L_1(y_{1, h}) = [\partial _h]$. Therefore, we may assume that $q \ge 2$.

We shall prove first that $\mathcal {N}_\lambda (q)$ is equal to the image of $L_q$ by induction on $q \ge 2$.

Assume that $q = 2$. Then, the image of

$$\begin{aligned} \partial _j(U_0 - \lambda ) - (j-\lambda )\partial _j = \sum _{h=1}^k h\sigma _hy_{2, h+j} \in \mathcal {W}_2 \end{aligned}$$

by $L_2$ is the class of $-(j-\lambda )\partial _j$ in $\mathcal {N}_\lambda$. Therefore, the image of $L_2$ contains the classes of $\partial _1, \dots , \partial _k$ as $\lambda$ is not in [1, k] and also contains the class of 1 as we assume $\lambda \not = 0$ and as the equality $\lambda = \sum _{h=1}^k h\sigma _h\partial _h$ holds in $\mathcal {N}_\lambda$. However, the image of $L_2$ contains obviously the classes of $\partial ^\alpha$ for any multi-index $\alpha \in \mathbb {N}^k, \vert \alpha \vert = 2$. Therefore, our assertion is proved for $q = 2$.

Assume now that $q \ge 3$ and that our assertion is proved for $q-1$. Remark that the image of $L_q$ contains obviously the classes of $\partial ^\alpha$ for each $\alpha \in \mathbb {N}^k, \vert \alpha \vert = q$. We shall use now the following easy formula:

• For any $r \in [q-1, k(q-1)]$ and any $j \in [1, k]$ we have in $\mathcal {N}_\lambda$ the equality

  $$\begin{aligned} \partial _j L_{q-1}(y_{q-1, r}) = L_q(\partial _j y_{q-1,r}) = L_q(y_{q, r+j}). \end{aligned}$$

For any $\beta \in \mathbb {N}^k {\setminus } \{0\}$ with $\vert \beta \vert \le q-1$ we may find $j \in [1, k]$ and $\gamma \in \mathbb {N}^k$, such that $\partial ^\beta = \partial _j\partial ^\gamma$. By our inductive assumption there exists $x \in \mathcal {W}_{q-1}$, such that $L_{q-1}(x) = \partial ^\gamma$. Then, $\partial _jx$ is in $\mathcal {W}_q$ and thanks to the formula above we have

$$\begin{aligned} L_q(\partial _jx) = \partial _jL_{q-1}(x) = \partial _j.\partial ^\gamma = \partial ^\beta \quad \textrm{in } \quad \mathcal {N}_\lambda . \end{aligned}$$

Again, we conclude that the class of 1 in $\mathcal {N}_\lambda (q)$ is in the image of $L_q$ using $\lambda \not = 0$ and the equality $\lambda = \sum _{h=1}^k h\sigma _h\partial _h$ which holds in $\mathcal {N}_\lambda (q)$. This complete the proof of our first statement.

However, it is clear that $\theta _{q,r}$ for $r \in [q, k(q-1)]$ are in the kernel of $L_q$. Therefore, the $\mathcal {O}_N$-free rank k module with basis $\big (y_{q, r}, r \in [k(q-1)+1, kq]\big )$ is surjective via $L_q$ onto $\mathcal {N}_\lambda (q)$. The next lemma completes the proof, as we already know that $\mathcal {N}_\lambda (1)$ is a $\mathcal {O}_N$-free rank k sub-module of $\mathcal {N}_\lambda (q)$ with basis $\partial _1, \dots , \partial _k$. $\blacksquare$

Lemma 3.3.6

Let A be an integral commutative ring and let M be a A-module. Assume that there exists a surjective A-linear map $p: A^k \rightarrow M$ and an injective A-linear map $i: A^k \rightarrow M$. Then, p is an isomorphism.

Proof

Let $j: A^k \rightarrow A^k$ be a $A-$linear map, such that $j\circ p = i$. Therefore, j is injective and the co-kernel C of j is a torsion module. Let $q: A^k \rightarrow C$ be the quotient map and let K be the kernel of p. The restriction of q to K is injective, because if $x \in K$ satisfies $q(x) = 0$ then $x = j(y)$ for some $y \in A^k$ and then $i(y) = p(j(y)) = p(x) = 0$, which implies $y = 0$ and $x=0$. Therefore, K is a sub-module of C and then K is a $A-$torsion module. However, as $K \subset A^k$ we have $K = 0$ and so p is an isomorphism.$\blacksquare$

Lemma 3.3.7

For $\lambda \not \in \mathbb {N}$ we have $\sigma _k\Delta (\sigma )\mathcal {N}_\lambda (2) \subset \mathcal {N}_\lambda (1)$.

The proof will be a simple consequence of the following lemma.

Lemma 3.3.8

Let $y:= (y_{2}, \dots , y_{2k})$ be in $\mathbb {C}[\sigma ]^{2k-1}$ and consider the $\mathbb {C}[\sigma ]-$linear system $(2k-1, 2k-1)$ on $\mathbb {C}[\sigma ]^{2k-1}$ given by the following $\mathbb {C}[\sigma ]-$linear forms:

$$\begin{aligned} L_{q}(y)&:= \sum _{h=0}^{k} h\sigma _{h}y_{q+h} \quad \textrm{for } \quad q \in [1, k] \\ \Lambda _{r}(y)&:= \sum _{h=0}^{k} \sigma _{h}y_{r+h} \quad \textrm{for} \quad r \in [2, k] \end{aligned}$$

Then, the determinant of this linear system is equal to $\sigma _{k}\Delta (\sigma )$ where $\Delta (\sigma )$ is the discriminant of the polynomial $P_{\sigma }(z):= z^k + \sum _{h=1}^{k} (-1)^{h}\sigma _{h}z^{k-h}$.

Proof

Remark first that $\Delta (\sigma )$ is also the discriminant of the polynomial (see the computation below):

$$\begin{aligned} \tilde{P}_{\sigma }(z):= \sum _{h=0}^{k} \sigma _{h}z^{k-h}. \end{aligned}$$

Then, remark also that the resultant of the polynomials $\tilde{P}_{\sigma }(z)$ and $k\tilde{P}_{\sigma }(z) - z(\tilde{P}_{\sigma })'(z)$ coincides with the determinant of the $(2k-1, 2k-1)$ $\mathbb {C}[\sigma ]-$linear system defined in the statement of the lemma. Therefore, it is enough to compute this resultant. It is given by

$$\begin{aligned} R(\sigma )= & {} \prod _{\tilde{P}_{\sigma }(z_{j}) = 0} \big (k\tilde{P}_{\sigma }(z_{j}) - z_{j}(\tilde{P}_{\sigma })'(z_{j})\big ) \\ {}= & {} \sigma _{k}\prod _{P_{\sigma }(-z_{j}) = 0} (-1)^{k-1}P'_{\sigma }(-z_{j}) = \sigma _{k}\Delta (\sigma ) \end{aligned}$$

as $\tilde{P}_\sigma (-z) = (-1)^{k}P_\sigma (z)$ implies $\tilde{P}'_\sigma (-z) = (-1)^{k-1}P'_\sigma (z)$. This conclude the proof $\blacksquare$

proof of 3.3.7 It is enough to prove that for each $(p, q) \in [1, k]^2$ there exist polynomials $a_{h, q}^p(\lambda )$ in $\mathbb {C}[\sigma , \lambda ]$ (in fact affine in $\lambda$), such that

$$\begin{aligned} \sigma _k\Delta (\sigma )\partial _p\partial _q - \sum _{h=1}^k a_{h, q}^p(\lambda )\partial _h \in \mathcal {J}_\lambda . \end{aligned}$$

For $m \in [2, k]$ we have

$$\begin{aligned} \mathcal {T}^m = y_{2, m} + \sum _{h=1}^k \sigma _hy_{2, m+h} + y_{1, m} \in \mathcal {I} \subset \mathcal {J}_\lambda \end{aligned}$$

and for $q \in [1, k]$:

$$\begin{aligned} \partial _q(U_0 - \lambda ) = \sum _{h=1}^k h\sigma _hy_{2, q+h} + (q-\lambda )y_{1, q} \in \mathcal {J}_\lambda . \end{aligned}$$

This gives $(2k-1)$ $\mathbb {C}[\sigma ]-$linear relations between the basis elements $y_{2, r}, r \in [2, 2k]$ of $\mathcal {W}_2$ modulo $L_2^{-1}(\mathcal {N}_\lambda (1))$. However, the determinant of these $2k-1$ vectors in the basis $y_{2,r}$ of $\mathcal {W}_2$ is equal to $\sigma _k\Delta (\sigma )$ thanks to the previous lemma. The conclusion follows, as we know that $L_2: \mathcal {W}_2 \rightarrow \mathcal {N}_\lambda (2)$ is surjective for $\lambda \not \in \mathbb {N}$.$\blacksquare$

Lemma 3.3.9

Assume that on a Stein open set U in N the equality of the sheaves $\mathcal {N}_\lambda (2)_{\vert U} = \mathcal {N}_\lambda (1)_{\vert U}$ is true for some $\lambda \in \mathbb {C}$. Then, we have

$$\begin{aligned} (\mathcal {N}_\lambda )_{\vert U} = \mathcal {N}_\lambda (1)_{\vert U}. \end{aligned}$$

Proof

It is enough to prove the equality $\mathcal {N}_\lambda (q)_{\vert U} = \mathcal {N}_\lambda (1)_{\vert U}$ for any $q \ge 2$, because we know that $\mathcal {N}_\lambda = \cup _{q \ge 0}\ \mathcal {N}_\lambda (q)$. As this is true for $q = 2$ by assumption, we shall prove this equality by induction on $q \ge 2$. Therefore, assume that this equality is proved for some $q \ge 2$ and we shall prove it for $q+1$.

Let $\alpha \in \mathbb {N}^k$, such that $\vert \alpha \vert = q+1$ and write $\partial ^\alpha = \partial _p\partial ^\beta$ for some $p \in [1, k]$ and some $\beta \in \mathbb {N}^k$ with $\vert \beta \vert = q$. By the inductive assumption we may write $\partial ^\beta = \sum _{h=1}^k b_h\partial _h$ in $\mathcal {N}_\lambda (q)$ with $b_h \in \mathcal {O}(U)$, because we know that $\mathcal {N}_\lambda (1) = \oplus _{h=1}^k \mathcal {O}_N\partial _h$ on N. Then, we obtain that $\partial _p\partial ^\beta$ is in $\mathcal {N}_\lambda (2)_{\vert U} = \mathcal {N}_\lambda (1)_{\vert U}$, concluding our induction.$\blacksquare$

Corollary 3.3.10

For each $\lambda \in \mathbb {C}\setminus \mathbb {N}$ there exists a meromorphic integrable connection $\nabla _\lambda : \mathcal {O}_N^k \rightarrow \frac{1}{\sigma _k.\Delta }.\mathcal {O}_N^k\otimes \Omega ^1_N$ with a simple pole on the reduced hyper-surface $\{\sigma _k\Delta (\sigma ) = 0 \} \subset N$, such that the restriction of $\mathcal {N}_\lambda$ to the Stein (in fact affine) open set $U:= \{\sigma _k\Delta (\sigma ) \not = 0 \}$ is isomorphic to the $\mathcal {D}_U$-module defined by $(\mathcal {O}_N^k, \nabla _\lambda )$. Moreover, this isomorphism is the restriction of an injective $\mathcal {D}_N$-linear map

$$\begin{aligned} \mathcal {N}_\lambda \rightarrow \big (\mathcal {O}_N^k(*\sigma _k\Delta (\sigma )), \nabla _\lambda \big ). \end{aligned}$$

Proof

This is an easy consequence of the $\mathcal {O}_N$ isomorphism $\mathcal {N}_\lambda (1) \rightarrow \oplus _{h=1}^k \mathcal {O}\partial _h$ and previous Lemmas 3.3.7 and 3.3.9.$\blacksquare$

We shall conclude this section by the following theorem.

Theorem 3.3.11

Let $\lambda \in \mathbb {C}\setminus \mathbb {Z}$. Then, $\mathcal {N}_\lambda$ is the minimal extension of the meromorphic connection given by $(\mathcal {N}_\lambda (1), \nabla _\lambda )$. Therefore, $\mathcal {N}_\lambda$ is a simple $\mathcal {D}_N$-module.

Proof

To see that $\mathcal {N}_\lambda$ is the minimal extension of the simple pole meromorphic connection $\big (\mathcal {N}_\lambda (1), \nabla _\lambda \big )$ it is enough to prove that $\mathcal {N}_\lambda$ has no torsion, and this is given by Proposition 3.3.5, and no co-torsion, that is to say that there is no non trivial coherent left ideal $\mathcal {K}$ in $\mathcal {D}_N$ containing $\mathcal {J}_\lambda$ and generically equal to $\mathcal {J}_\lambda$ on N. Such an ideal defines a holonomic quotient Q of $\mathcal {N}_\lambda$ which is supported in a closed analytic subset S of N with empty interior in N. As $\mathcal {N}_\lambda$ is a quotient of $\mathcal {M}$, we may apply Corollary 2.3.2 and so it is enough to show that near the generic points of $\{\sigma _k\Delta (\sigma ) = 0\}$ such an ideal $\mathcal {K}$ is equal to $\mathcal {J}_\lambda$ or to $\mathcal {D}_N$.

Near the generic point of $\{\sigma _k = 0\}$ we have $\Delta \not = 0$ and we may use a local isomorphism of N given by a holomorphic section of the quotient map

$$\begin{aligned} quot: M = \mathbb {C}^k \rightarrow \mathbb {C}^k\big /\mathfrak {S}_k = N. \end{aligned}$$

Via such an isomorphism $\mathcal {N}_\lambda$ is the quotient of $D_{\mathbb {C}^k}$ by the left ideal generated by the $\frac{\partial ^2}{\partial z_i \partial z_j}$ for $i \not = j \in [1, k]$ and $\sum _{j=1}^k z_j\frac{\partial }{\partial z_j} - \lambda$. The lemma below allows to conclude this case. For the other case, that is to say near the generic point of $\{ \Delta = 0 \}$, Theorem  2.3.6 completes the proof.

The fact that $\mathcal {N}_\lambda$ is a simple $\mathcal {D}_N$-module is then consequence of the irreductibility of the monodromy representation of its associated meromorphic connection. $\blacksquare$

Lemma 3.3.12

Let $\mathcal {J}_\lambda$ for $\lambda \not \in -\mathbb {N}^*$ be the ideal in $D_{\mathbb {C}^k}$ generated by the differential operators $\frac{\partial ^2}{\partial z_i\partial z_j}$ for $1 \le i < j \le k$ and $\sum _{h=1}^k z_h.\frac{\partial }{\partial z_h} - \lambda$. Let assume that Q is a quotient of the $D_{\mathbb {C}^k}-$module $\mathcal {N}_{\lambda }:= D_{\mathbb {C}^k}\big /\mathcal {J}_\lambda$ in a neighborhood U of the point $(z^0_1, \dots , z^0_k)$ in $\mathbb {C}^k$ where $z_1^0 = 0$ and $z_i \not = z_j$ for $1 \le i < j \le k$, with support in $\{z_1 = 0 \}$. Then, $Q = 0$.

Proof

Assume that $Q \not = 0$ Then $Q = D_U\big /\mathcal {K}$ where $\mathcal {K}$ is a left ideal in $D_U$, such that $\mathcal {J}_\lambda \subsetneq \mathcal {K} \subsetneq D$. Then, restricting the open neighborhood U of $z^0$ if necessary, there exists a positive integer n, such that $z_1^n$ belongs⁹ to $\mathcal {K}$. Then, we have

as $\sum _{h=1}^k z_h\frac{\partial }{\partial z_h} - \lambda \in \mathcal {J}_\lambda \subset \mathcal {K}$ on U. However, $z_1^n \in \mathcal {K}$ implies also, for each $j \in [2, k]$:

again as $\mathcal {J}_\lambda \subset \mathcal {K}$. Combining (a) and (b) we conclude that $z_1^{n-1}$ belongs to $\mathcal {K}$, as we assume $n > 0$ and $\lambda \not \in \mathbb {N}^*$.

By a descending induction on n we conclude that 1 belongs to $\mathcal {K}$ which contradicts our assumption that Q is not 0.$\blacksquare$

Remark Note that the $\mathcal {D}_N$-linear map

$$\begin{aligned} \varphi _{-1}: \mathcal {N}_{-1} \rightarrow \underline{H}_{[\sigma _k = 0]}^1(\mathcal {O}_N):= \mathcal {O}_N[\sigma _k^{-1}]\big /\mathcal {O}_N \end{aligned}$$

defined by $\varphi _{-1}(1):= \sigma _{k-1}/\sigma _k$ is surjective, because $\varphi _{-1}(\partial _{k-1}) = 1/\sigma _k$. This shows that for $p= -1$ the sheaf $\mathcal {N}_{-1}$ has a non zero quotient supported by $\{\sigma _k = 0\}$. Then, using the isomorphism $\mathscr {T}_\lambda : \mathcal {N}_{\lambda -1} \rightarrow \mathcal {N}_\lambda$ for $\lambda \in -\mathbb {N}^*$ to deduce the case $\lambda -1$ from the case $\lambda$ for each $\lambda \in - \mathbb {N}^*$, we see that the sheaf $\mathcal {N}_{-p}$ has a non zero quotient supported by $\{\sigma _k = 0\}$ for any $p \in \mathbb {N}^*$.

4 The $\mathcal {D}_N$-Modules $\mathcal {N}_p, p \in \mathbb {Z}$

4.1 Structure of $\mathcal {N}_p, p \ge 1$

The first important remark is that, thanks to Lemma 3.1.5, it is enough to determine the structure of $\mathcal {N}_1$ as for each $p \ge 2$ the $\mathcal {D}_N$-module $\mathcal {N}_p$ is isomorphic to $\mathcal {N}_1$ via the right multiplication by $U_1^{p-1}$.

4.1.1 Minimality of $\mathcal {N}_1^\square$

Recall that $\mathcal {N}_1^\square$ is the co-kernel of the left $\mathcal {D}_N$-linear map $\mathscr {T}_1: \mathcal {N}_0 \rightarrow \mathcal {N}_1$ defined by the right multiplication by $U_{-1}$.

Thanks to formulas $E_h, h \in [2,k]$ (see Proposition 3.1.6) we obtain that $\mathcal {N}_1^\square$ is the quotient of $\mathcal {D}_N$ by the left ideal $\mathcal {A} + \mathcal {D}_N(U_0 - 1) + \mathcal {D}_NU_{-1}$, because these formulas imply that the partial differential operators $\mathcal {T}^m, m \in [2, k]$ are contained in the ideal $\mathcal {A} + \mathcal {D}_N(U_0 - 1) + \mathcal {D}_NU_{-1}$ and we have $\mathcal {J}_1 = \mathcal {I} + \mathcal {D}_N(U_0 - 1)$ by definition (see Formula (8) for the definition of $\mathcal {I}$ and the beginning of Paragraph 3.1 for the definition of the ideal $\mathcal {J}_\lambda$).

We shall note $\mathcal {N}_1^\square (q):= \mathcal {D}_N(q)\big /\big (\mathcal {J}_1 \cap \mathcal {D}_N(q)\big )$ for each integer $q \ge 0$.

Proposition 4.1.1

For each q the natural map $\mathcal {N}_1^\square (q) \rightarrow \mathcal {N}_1^\square$ is injective.

Proof

The proof will use Proposition 3.2.1 two times: the first time for the left ideal $\mathcal {A}$ and with $U:= U_{-1}$ and the second time for the left ideal $\mathcal {A} + \mathcal {D}_NU_{-1}$ and with $U:= U_0 - 1$. This will give the equalities

$$\begin{aligned} \big (\mathcal {A} + \mathcal {D}_NU_{-1}\big ) \cap \mathcal {D}_N(q)&= \mathcal {A}(q) + \mathcal {D}_N(q-1)U_{-1} \quad \textrm{and} \\ \big (\mathcal {A} + \mathcal {D}_NU_{-1} + \mathcal {D}_N(U_0 - 1) \big ) \cap \mathcal {D}_N(q)&= \mathcal {A}(q) + \mathcal {D}_N(q-1)U_{-1}\\ {}&\quad + \mathcal {D}_N(q-1)(U_0 - 1). \end{aligned}$$

This will conclude the proof.

To apply Proposition 3.2.1 we have to show that the following properties hold:

• (i) The coherence of $\mathcal {A}$ and of $\mathcal {A} + \mathcal {D}_NU_{-1}$.

• (ii) The fact that the characteristic ideals of $\mathcal {A}$ and of $\mathcal {A} + \mathcal {D}_NU_{-1}$ are reduced.

• (iii) The inclusions $\mathcal {A}U_{-1} \subset \mathcal {A}$ and $\big (\mathcal {A} + \mathcal {D}_NU_{-1})(U_0 - 1)\big ) \subset \mathcal {A} + \mathcal {D}_NU_{-1}$.

• (iv) The symbol of $U_{-1}$ does not vanish on any non empty open set of the characteristic variety of $\mathcal {D}_N\big /\mathcal {A}$.

• (v) The symbol of $U_0 - 1$ does not vanish on any non empty open set of the characteristic variety of $\mathcal {D}_N\big /\mathcal {A} + \mathcal {D}_NU_{-1}$.

The point (i) is clear.

The characteristic ideal of $\mathcal {A}$ is the pull-back by the projection $p_2: N \times \mathbb {C}^k \rightarrow \mathbb {C}^k$ of the ideal of the reduced ideal IS(k) of the surface S(k) (see Corollary 2.1.2 in the appendix).

The point (ii) is completed by the following lemma:

Lemma 4.1.2

Let $\gamma (\sigma , \eta ):= \sum _{h=0} (k-h)\sigma _h\eta _{h+1}$ and $g(\sigma , \eta ):= \sum _{h=1}^k h\sigma _h\eta _h$. Then, defined the following ideals in $\mathcal {O}_N[\eta ]$, where $I_1:= (p_2)^*(IS(k))$:

$$\begin{aligned} I_2:= I_1 + (\gamma ) \quad \textrm{and} \quad I_3:= I_2 + ( g). \end{aligned}$$

Then, $I_2$ is reduced and g does not vanish on any non empty open set of the analytic subset $(N \times S(k))\cap \{\gamma = 0\}$.

Proof

To see that $I_2$ is reduced, as $N \times S(k)$ is normal, it is enough to prove that $\{\gamma = 0\}$ defined a reduced and irreducible hyper-surface in $N \times S(k)$. Looking at the chart on the dense open set $\eta _k\not = 0$ of $N \times S(k)$ which is given by the map $(\sigma , \eta ) \mapsto (\sigma , -\eta _{k-1}/\eta _k, \eta _k) \in N \times \mathbb {C}\times \mathbb {C}^*$ (see Paragraph  3.2) we find that $\gamma$ is given in this chart by

$$\begin{aligned} \gamma (\sigma , \eta ) = (-1)^{k-1}P'_\sigma (z)\eta _k \quad \textrm{where} \quad z:= -\eta _{k-1}/\eta _k \end{aligned}$$

using the fact that $\eta _h = (-z)^{k-h}\eta _k$ in this chart. This gives the fact that $\{\gamma = 0\}$ is reduced and irreducible in $N \times S(k)$.

The computation of g in the same chart gives that

$$\begin{aligned} g(\sigma , \eta ) = (-1)^kzP'_\sigma (z)\eta _k - (-1)^kkP_\sigma (z)\eta _k. \end{aligned}$$

and this proves that g does not vanishes identically on any non zero open set in $(N \times S(k)) \cap \{\gamma = 0\}$, because

$$\begin{aligned} (N \times S(k)) \cap \{\gamma = 0\} \cap \{ g = 0\} \subset Z \cap \{\gamma = 0\} \end{aligned}$$

which has dimension k, so co-dimension 2 in $N \times S(k)$.$\blacksquare$

End of proof of 4.1.1 The point iii) is consequence of the following easy formulas:

$$\begin{aligned} A_{p, q}U_{-1}&= U_{-1}A_{p, q} - (k-p-1)A_{p+1, q} - (k-q)A_{p, q+1} \\ A_{p, q}U_0&= U_0A_{p, q} - (p+q)A_{p, q} \\ U_{-1}(U_0 - 1)&= U_0U_{-1}. \end{aligned}$$

The points (iv) and (v) are obvious, because a non zero germ of section of $\mathcal {O}_N[\eta ]$ which is homogeneous of degree 1 in $\eta$ does not vanishes of $N \times S(k)$.$\blacksquare$

Recall that in $\mathcal {W}(q):= \oplus _{p=0}^q \mathcal {W}_p$ we have, for each $\beta \in \mathbb {N}^k$ with $\vert \beta \vert = q-1$ and $w(\beta ) = r-1$ (compare with Formulas (13) and (14), but here $w(\beta ) = r-1$)

$$\begin{aligned} \left[ \partial ^\beta (U_0 - 1)\right] = \sum _{h=1}^k h\sigma _hy_{q, r+h-1} + (r - 2)y_{q-1, r -1}. \end{aligned}$$

(24)

and

$$\begin{aligned} \left[ \partial ^\beta U_{-1} \right] = \sum _{h=0}^k (k-h)\sigma _hy_{q, r+h} + (k(q-1) - r+1)y_{q-1, r} \end{aligned}$$

(25)

Now, note $\beta ^+$ a multi-index with $\vert \beta ^+\vert = q-1$ and $w(\beta ^+) = r$, when $r \not = k(q-1)+1$ and $\beta ^+ = 0$ for $r = k(q-1)+1$.

Then for $r \not = k(q-1) + 1$ we have

$$\begin{aligned} \partial ^{\beta ^+}(U_0 -1) = \sum _{h=0}^k h\sigma _hy_{q, r+h} + (r-1)y_{q-1, r} \end{aligned}$$

with the convention $\sigma _0 \equiv 1$ and $\partial ^{\beta ^+}(U_0 -1) = 0$ for $r = k(q-1)+1$.

Then define for $q \ge 2$ and $r \in [q, k(q-1)]$ the following elements in $\mathcal {W}_q$:

$$\begin{aligned} \tilde{\theta }_{q, r}:= (r-1)\partial ^{\beta } U_{-1}- (k(q-1) - r+1)\partial ^{\beta ^+}(U_0 - 1) \end{aligned}$$

(26)

This gives

$$\begin{aligned} \tilde{\theta }_{q, r}= k\sum _{h=0}^k \big ( (r-1) - h(q-1) \big )\sigma _hy_{q, r+h} \end{aligned}$$

(27)

Remark that for $r = k(q-1)+1$ and $h = k$ the vector $y_{q, r+k}$ is not defined in $\mathcal {W}_q$ and we cannot use the formula (27) to define $\tilde{\theta }_{q, k(q-1)+1}$. But with our convention $\partial ^{\beta ^+}(U_0 -1) = 0$ for $r = k(q-1)+1$, we define the vector

$$\begin{aligned} \tilde{\theta }_{q, k(q-1)+1}:= k(q-1)\partial ^{q-1}_kU_{-1} = k(q-1)\sum _{h= 0}^{k-1} (k-h)\sigma _h.y_{q, k(q-1) + 1 + h}\nonumber \\ \end{aligned}$$

(28)

which is in $\mathcal {W}_q$.

Then, for $q \ge 2$, let $\tilde{\Theta }_q \subset \mathcal {W}_q$ be the sub-$\mathcal {O}_N$-module generated by the elements $\tilde{\theta }_{q, r}, r \in [q, k(q-1]+1]$. Of course $\tilde{\Theta }_q$ is in the kernel of the $\mathcal {O}_N-$linear map

$$\begin{aligned} L_q: \mathcal {W}_q \rightarrow \mathcal {N}_1^\square (q) \end{aligned}$$

induced by the quotient map $\mathcal {W} \rightarrow \mathcal {N}^\square _1$.

For $q = 1$ define $\tilde{\Theta }_1:= \mathcal {O}_NU_{-1}$ and $V_1:= \oplus _{h=2}^k \mathcal {O}_N[\partial _h]$ (where $y_{1, h}:= [\partial _h]$ in $\mathcal {W}_1$).

Lemma 4.1.3

For each $q \ge 1$ we have a direct sum decomposition $\mathcal {W}_q = \tilde{\Theta }_q \oplus V_q$ where $V_q$ is the $\mathcal {O}_N-$sub-module with basis $y_{q, r}$ with $r \in [k(q-1)+2, kq]$.

Proof

For $q = 1$ our assertion is clear. For $q \ge 2$ ( and so $r \ge 2$) the difference $\tilde{\theta }_{q, r} - k(r-1)y_{q, r}$ is a $\mathbb {C}[\sigma ]-$linear combination of the $y_{q, s}$ for $s \ge r+1$, so the matrix of the vectors $\tilde{\theta }_{q, r}$ for $r \in [q, k(q-1)+1]$ and $y_{q, s}, s \in [k(q-1)+2, kq]$ is triangular in the basis $y_{q, t}, t \in [q, kq],$ of $\mathcal {W}_q$ with determinant equal to

$$\begin{aligned} k^{(k-1)(q-1)+1} \prod _{r = q}^{k(q-1)+1} (r-1) = k^{(k-1)(q-1)+1}\frac{(k(q-1))!}{(q-2)!} \end{aligned}$$

which is a positive integer. $\blacksquare$

Lemma 4.1.4

For each $q \ge 1$ the map $l_q: V_q \rightarrow \mathcal {N}^\square _1(q)$ induced by $L_q$ is bijective.

Proof

We shall prove this lemma by induction on $q \ge 1$. First remark that the map $l_1: V_1 \rightarrow \mathcal {N}^\square _1(1)$ is surjective (in fact an isomorphism of free rank $(k-1)$ $\mathcal {O}_N$-modules), because $1 = \sum _{h=1}^k h\sigma _h\partial _h$ and $k\partial _1 = - \sum _{h=1}^{k-1} (k-h)\sigma _h\partial _{h+1}$ in $\mathcal {N}^\square _1$.

So let $q \ge 2$ and assume that $l_{q-1}:V_{q-1} \rightarrow \mathcal {N}^\square _1(q-1)$ is surjective. Then, $\mathcal {N}^\square _1(q-1)$ is contained in the image of $L_q$, because for $r \in [k(q-2)+2, k(q-1)]$ the relation (26) shows that the image of $y_{q-1, r}$ by $l_{q-1}$ is in the image of $L_q$.

Then remark that $L_q$ induces a surjective map onto the quotient $\mathcal {N}_1^\square (q)\big /\mathcal {N}_1^\square (q-1)$ and that $\tilde{\Theta }_q$ is in the kernel of $L_q$. Therefore, $l_q$ is surjective on $\mathcal {N}^\square _1(q)$. Therefore, we have a surjective map $l_q$ of the rank $k-1$ free $\mathcal {O}_N$-module $V_q$ onto $\mathcal {N}^\square _1(q)$ and an injective map of the rank $k-1$ free $\mathcal {O}_N$-module $\mathcal {N}_1^\square (1)$ into $\mathcal {N}_1^\square (q)$. Then, Lemma 3.3.6 gives that $l_q$ is bijective.$\blacksquare$

Theorem 4.1.5

The restriction of $\mathcal {N}^\square _1(1)$ to the Zariski open set $\{\Delta (\sigma ) \not = 0\}$ is a rank $(k-1)$ free¹⁰$\mathcal {O}_N$-module with a simple pole meromorphic connection along $\{\Delta = 0\}$ given by the inclusion $\Delta (\sigma )\mathcal {N}^\square _1(2) \subset \mathcal {N}^\square _1(1)$ (see Lemma 4.1.6 below). Its sheaf of horizontal sections is locally generated by $z_i - z_j$ where $z_h, h \in [1, k],$ are local branches of the multivalued function $z(\sigma )$ defined by $P_{\sigma }(z(\sigma )) \equiv 0$. The $\mathcal {D}_N$-module $\mathcal {N}^\square _1$ is the minimal extension on N of this vector bundle with its integrable regular meromorphic connection. Therefore, it is a simple $\mathcal {D}_N$-module.

Proof

Lemma 4.1.4 gives that the map $l_q: V_q \rightarrow \mathcal {N}^\square _1(q)$ is a isomorphism of $\mathcal {O}_N$-modules for each $q \ge 1$ and Proposition 4.1.1 implies that $\mathcal {N}^\square _1$ is the union of the sheaves $\mathcal {N}^\square _1(q), q \ge 1$. Therefore, the $\mathcal {D}_N$-module $\mathcal {N}^\square _1$ has no $\mathcal {O}_N$-torsion.

Thanks to Lemma 4.1.6 below we have the inclusion $\Delta \mathcal {N}^\square _1(2) \subset \mathcal {N}^\square _1(1)$. This implies that $\mathcal {N}^\square _1(1) \simeq \mathcal {O}_N^{k-1}$ has an integrable meromorphic connection $\nabla _1$ with a simple pole along $\{\Delta (\sigma ) = 0\}$ on N. The fact that $\mathcal {K}_1 = \mathcal {J}_1 + \mathcal {D}_NU_{-1}$ implies that the horizontal sections of $\mathcal {N}^\square _1(1)$ are trace functions (see [1]) which are homogeneous of degree 1 and killed by $U_{-1}$. Therefore, they are $\mathbb {C}-$linear combinations of $z_1(\sigma ), \dots , z_k(\sigma )$, the local branches of the multivalued function $z(\sigma )$ on N defined by $P_\sigma (z(\sigma )) = 0$.

The condition for $\sum _{h=1}^k a_hz_h(\sigma ), a_h \in \mathbb {C},$ to be killed by $U_{-1} \simeq \sum _{h=1}^k \frac{\partial }{\partial z_h}$ is given by $\sum _{h=1}^k a_h = 0$ and then the horizontal sections are linear combinations of the differences $z_i - z_j, \ i, j \in [1, k]$. A basis of horizontal sections is given, for instance, by $z_2(\sigma ) - \sigma _1/k, \dots , z_k(\sigma ) - \sigma _1/k$ (note that $\sum _{j=1}^k (z_j(\sigma ) - \sigma _1/k) \equiv 0$).

The $\mathcal {D}_N$-module $\mathcal {N}^\square _1$ has neither $\mathcal {O}_N$-torsion nor $\mathcal {O}_N$-co-torsion, because its characteristic variety is the union of $N \times \{0\}$ and X (X is defined in Sect. 3.2) and, thanks to Theorem 2.3.6, it has neither $\Delta$-torsion nor $\Delta$-co-torsion as a quotient of $\mathcal {M}$. Therefore, $\mathcal {N}^\square _1$ is the minimal extension of the meromorphic connection $\big (\mathcal {N}^\square _1(1), \nabla _1\big )$ and it is a simple $\mathcal {D}_N$-module, because the monodromy representation of the local system of horizontal sections of $(\mathcal {N}^\square _1(1), \nabla _1)$ is irreducible. $\blacksquare$

Lemma 4.1.6

We have $\Delta \mathcal {N}^\square _1(2) \subset \mathcal {N}^\square _1(1)$.

Proof

In $\mathcal {W}_2\big /\mathcal {W}_1$ the $2k-1$ vectors induced by $\partial _j(U_0 - 1), j \in [2, k]$ and $\partial _hU_{-1}, h \in [1, k]$ are given in the basis $y_{2, r}, r \in [2, 2k]$ of this free $\mathcal {O}_N$-module by the relations:

$$\begin{aligned} A _j&:= \partial _j(U_0 - 1) = \sum _{p=1}^k p\sigma _py_{2, j+p} \\ B_h&:= \partial _hU_{-1} = \sum _{p=0}^{k-1} (k-p)\sigma _py_{2, h+p+1} \end{aligned}$$

with the convention $\sigma _0 = 1$.

Put $\tilde{P}_\sigma (z):= \sum _{p=0}^k \sigma _pz^{k-p}$ and $y_{2, k+p} = z^{k-p}$.

Then $B_k = \tilde{P}_\sigma '(z)$ and $A_k = z\tilde{P}_\sigma '(z) - k\tilde{P}_\sigma (z)$. Therefore, the resultant of $A_k$ and $B_k$ is equal to $(-k)^{k-1}\Delta (\sigma )$. The determinant of the vectors $A_j, j \in [2, k]$ and $B_h, h \in [1, k]$ in the basis $y_{2, r}, r\in [2, 2.k]$ of $\mathcal {W}_2 \simeq \mathcal {W}_2\big /\mathcal {W}_1$ is then equal to $(-k)^{k-1}\Delta (\sigma )$ (compare with Lemma 3.3.8).$\blacksquare$

4.1.2 The Structure Theorem for $\mathcal {N}_p, p\ge 1$

We first examine the case $p = 1$. As already explained in the beginning of this section this will be enough to describe the structure of $\mathcal {N}_p$ for any $p \in \mathbb {N}^*$.

The torsion sub-module of $\mathcal {N}_1$ is described by the following result. Remark that we already know from Theorem 4.1.5 that the torsion sub-module of $\mathcal {N}_1$ is contained in the image of $\mathscr {T}_1: \mathcal {N}_0 \rightarrow \mathcal {N}_1$ as $\mathcal {N}^\square _1 = \mathcal {N}_1\big /Im(\mathscr {T}_1)$ has no torsion.

Proposition 4.1.7

There exists a injective $\mathcal {D}_N$-linear map $\chi : \underline{H}^1_{[\sigma _k = 0]}(\mathcal {O}_N) \rightarrow \mathcal {N}_1$ which sends the class $[1/\sigma _k]$ in $\underline{H}^1_{[\sigma _k = 0]}(\mathcal {O}_N)$ to the class $[\partial _kU_{-1}]$ in $\mathcal {N}_1$. Its image is the torsion sub-module $\Theta$ of $\mathcal {N}_1$.

Proof

Note first that $\underline{H}^1_{[\sigma _k = 0]}(\mathcal {O}_N)$ is given by $\mathcal {D}_N\big /\big (\sum _{h=1}^{k-1} \mathcal {D}_N\partial _h + \mathcal {D}_N\sigma _k\big )$ as the annihilator of $[1/\sigma _k]$ is generated by $\partial _h, h \in [1, k-1]$ and $\sigma _k$. To show that $\chi$ exists it is enough to show that $\partial _h, h \in [1, k-1]$ and $\sigma _k$ annihilate the class $[\partial _kU_{-1}]$ in $\mathcal {N}_1$. The fact that $\partial _h[\partial _kU_{-1}] = 0$ in $\mathcal {N}_1$ for $h \in [1, k-1]$ is a direct consequence of Formulas $(E_h), h \in [2, k]$ which give $[\partial _hU_{-1}] = 0$ in $\mathcal {N}_1$. Then, Formula $(E_1)$ gives the vanishing of the class of $EU_{-1} = \sum _{h=1}^k \sigma _h\partial _hU_{-1}$ in $\mathcal {N}_1$. Therefore, we obtain that $\sigma _k[\partial _kU_{-1}]$ vanishes in $\mathcal {N}_1$ and $\chi$ is well defined. Moreover, as $\underline{H}^1_{[\sigma _k = 0]}(\mathcal {O})$ is a $\mathcal {D}_N$-module with support in $\{\sigma _k = 0 \}$, its image is contained in the torsion sub-module in $\mathcal {N}_1$.

Note that we know that the torsion in $\mathcal {N}_1$ is only $\sigma _k-$torsion thanks to Corollary 3.2.6 and Theorem 2.3.6.

To prove the injectivity of $\chi$, assume that the kernel of $\chi$ is not 0 and consider an element $K:= \sum _{p=1}^m f_p\partial _k^p[1/\sigma _k]$ in this kernel with $f_p \in \mathcal {O}_N\big /(\sigma _k)$ and with m minimal. Then, we have $0 = \chi (K) = [\sum _{p=1}^m f_p\partial _k^{p+1}U_{-1}]$ in $\mathcal {N}_1$. Therefore, $f_m\eta _k^{m+1}\gamma$ is the symbol of an element in $\mathcal {J}_1$. Then, Lemma 3.2.7 implies that $f_m$ is in $\sigma _k\mathcal {O}_N$ contradicting the minimality of m. Therefore, $\chi$ is injective.

To complete the proof we have to show that if P induces a torsion class in $\mathcal {N}_1$ then there exists $Q \in \mathcal {D}_N$, such that $P - Q\partial _kU_{-1}$ is in $\mathcal {J}_1$. As we already know (because $\mathcal {N}^\square _1$ has no torsion) that there exists $P_1 \in \mathcal {D}_N$, such that $\mathscr {T}_1(P_1) = [P_1U_{-1}] = [P]$ in $\mathcal {N}_1$ and as we know that $\partial _hU_{-1} = 0$ for each $h \in [1, k-1]$ we may assume that $P_1$ is in $\mathcal {O}_N[\partial _k]$. However, $\partial _k^nU_{-1}$ is torsion in $\mathcal {N}_1$ for $n \ge 1$, because $\partial _kU_{-1}$ is torsion (see above). Therefore, the only point to prove is that if $fU_{-1}$ is torsion in $\mathcal {N}_1$ for some $f \in \mathcal {O}_N$ then $f = 0$. This a consequence of the following lemma. $\blacksquare$

Lemma 4.1.8

The class of $U_{-1}$ is not in the $\sigma _k-$torsion of $\mathcal {N}_1$.

Proof

Assume that $\sigma _k^nU_{-1}$ is in $\mathcal {J}_1$ for some $n \in \mathbb {N}$. Then, choose n minimal with this property and compute

$$\begin{aligned} \partial _k\sigma _k^nU_{-1} = n\sigma _k^{n-1}U_{-1} + \sigma _k^n\partial _kU_{-1} \in \mathcal {J}_1. \end{aligned}$$

As $\sigma _k\partial _kU_{-1}$ is in $\mathcal {J}_1$ (see above) we obtain that $n = 0$ by minimality of n. However, $U_{-1}$ is not in $\mathcal {J}_1$, because its symbol $\gamma (\sigma , \eta )$ restricted to the co-normal C to the hyper-surface $\{\sigma _k = 0 \}$ is equal to $\sigma _{k-1}$ which does not vanish identically on C. In addition, C is a component of the characteristic variety of $\mathcal {N}_1$ (see Paragraph 3.2). This concludes the proof.$\blacksquare$

Theorem 4.1.9

The diagram below describes the structure of $\mathcal {N}_1$, where $\Theta$ is the torsion sub-module of $\mathcal {N}_1$, where $\varphi _1: \mathcal {N}_1 \rightarrow \mathcal {O}_N$ is the $\mathcal {D}_N$-linear map defined by $\varphi _1(1) = \sigma _1$ and where the isomorphism $\chi : \underline{H}^1_{[\sigma _k = 0]}(\mathcal {O}_N) \simeq \Theta$ is defined by sending $1/\sigma _k$ to $[\partial _kU_{-1}]$.

The $\mathcal {D}_N$-modules $\Theta \simeq \underline{H}^1_{[\sigma _k = 0]}(\mathcal {O}_N)$ and $\mathcal {N}^\square _1$ are simple $\mathcal {D}_N$-modules.

Moreover, we have the direct sum decomposition of left $\mathcal {D}_N$-modules:

$$\begin{aligned} \mathcal {N}_1\big /\Theta = Im(\mathscr {T}_1)\big /\Theta \oplus \mathcal {N}_1^*\big /\Theta = \mathcal {O}_N[U_{-1}] \oplus \mathcal {N}^\square _1. \end{aligned}$$

The following commutative diagram of left $\mathcal {D}_N$-modules has exact lines and columns where the maps i and e are defined by $i([U_{-1}]) = [U_{-1}]$ and $e([U_{-1}]) = k$:

Proof

Note first that the quotient by the torsion sub-module $\Theta$ (which is the image of $\mathcal {D}_N\partial _kU_{-1}$ in $\mathcal {N}_1$; see Proposition 4.1.7) of the image of $\mathcal {N}_0$ (equal to $D_NU_{-1}$ in $\mathcal {N}_1$) is isomorphic to $\mathcal {O}_N$, because its generator $[U_{-1}]$ is killed by $\partial _h, \forall h \in [1, k]$ (see Formulas $(E_h), h \in [1, k]$) and this quotient has no torsion, because $\Theta$ is also the torsion sub-module of $Im(\mathscr {T}_1)$. This gives the exactness of the first line. The exactness of the second line and of the columns are clear.

Note also that $\varphi _1(U_{-1}) = k$, so the upper right square commutes. The commutations of the other squares are obvious.

To show that the map $\theta$ is well defined and is an isomorphism is a simple exercise in diagram chasing which is left to the reader.

The direct sum decomposition of $\mathcal {N}_1\big /\Theta$ is given by the left $\mathcal {D}_N$-linear map

$$\begin{aligned} r: \mathcal {N}_1\big /\Theta \rightarrow Im(\mathscr {T}_1)\big /\Theta \end{aligned}$$

constructed as follows:

Note first that $\varphi _1(U_{-1}) = k$. For $[P] \in \mathcal {N}_1$ let $f:= \varphi _1([P])$. Then, we define

$$\begin{aligned} r([P]):= \left[ (f/k)U_{-1}\right] \in Im(\mathscr {T}_1)\big /\Theta . \end{aligned}$$

As $\Theta$ is in the kernel of $\varphi _1$, this map is well defined on $\mathcal {N}_1\big /\Theta$ and $[P] - r([P])$ is in $ker(\varphi _1) = \mathcal {N}_1^*$ and defines a class in $\mathcal {N}_1^*\big /\Theta$. Remark that Lemma 4.1.8 shows that the kernel of r is equal to $\mathcal {N}_1^* \big /\Theta$, because $\varphi _1$ is injective on $\mathcal {O}_N[U_{-1}]$. This gives the desired splitting, as r induces the identity on $Im(\mathscr {T}_1)\big /\Theta$.$\blacksquare$

4.2 The Structure of $\mathcal {N}_0$

Define $\mathcal {N}_0^*$ as the kernel of the $\mathcal {D}_N$-linear map $\varphi _0: \mathcal {N}_0 \rightarrow \mathcal {O}_N$ given by $\varphi _0(1) = 1$. The sub-module $\mathcal {N}_0^*$ is generated by $\partial _1, \dots , \partial _k$. We shall show that it contains a copy of $\mathcal {O}_N$.

Note that the $\mathcal {D}_N$-module $\mathcal {N}_0$ has no $\mathcal {O}_N$-torsion thanks to Theorem 3.3.1.

Proposition 4.2.1

The kernel of the $\mathcal {D}_N$-linear map $\mathscr {T}_1: \mathcal {N}_0 \rightarrow \mathcal {N}_1$ given by the right multiplication by $U_{-1}$ is $\mathcal {D}_NU_1$ which is contained in $\mathcal {N}_0^*$ and the quotient $\mathcal {N}_0^*\big /\mathcal {D}_NU_1$ is isomorphic to the $\mathcal {D}_N$-module $\underline{H}_{[\sigma _k = 0 ]}^1(\mathcal {O}_N)$.

The proof of this proposition will used the following results from [1] Proposition 5.2.1.

Proposition 4.2.2

For each $m \in \mathbb {Z}, m \ge -k+1$ and for each $\sigma \in N$, such that $\Delta (\sigma ) \not = 0$ define

$$\begin{aligned} DN_{m}(\sigma ):= \sum _{P_{\sigma }(x_{j}) = 0} \frac{x_{j}^{m+k-1}}{P'_{\sigma }(x_{j})}. \end{aligned}$$

(29)

Each $DN_{m}$ is the restriction to the open set $\{\Delta (\sigma ) \not = 0 \}$ of a polynomial of (pure) weight m in $\mathbb {C}[\sigma _{1}, \dots , \sigma _{k}]$ and the following properties are satisfied:

• (i) For $m \in [-k+1, -1], DN_{m} = 0$ and $DN_0 = 1$.

• (ii) For each $m \ge 1, \quad \sum _{h=0}^{k} (-1)^{h}\sigma _{h}DN_{m-h} = 0$ with the convention $\sigma _{0} \equiv 1$.

• (iii) For each $h \in [1, k]$ and each $m \ge 0$ we have

  $$\begin{aligned} \partial _h N_{m} = (-1)^{h-1}mDN_{m-h}.\qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \qquad \blacksquare \end{aligned}$$

We shall use also the following lemma.

Lemma 4.2.3

For any $h \in [2, k]$ we have

Moreover, we have also

Therefore, we have $\partial _hU_1 = -\partial _h$ for $h \in [2, k]$ and $\partial _1U_1 = E$ in $\mathcal {N}_0^*$.

Remark Note that Formulas $(F_h)$ for any $h \in [1, k]$ give that $\partial _hU_1 = 0$ in $\mathcal {N}_{-1}$.

Proof

Thanks to the characterization of trace functions given in [1] it is enough to prove that for each $m \in \mathbb {N}$ we have $\partial _hU_1[N_m] = -\partial _{h-1}(U_0 + 1)[N_m]$ for $h \in [2, k]$ and $\partial _1U_1(N_m) = E(U_0+1)[N_m]$ for all $m \in \mathbb {N}$. This is consequence of the following formulas which use the results of [1] recalled in Proposition 4.2.2 and the equality $U_p[N_m] = mN_{m+p}$ which is valid for each $m \in \mathbb {N}$ and each integer $p \ge -1$, because $U_p$ is the image by the tangent map to the quotient map $quot: M = \mathbb {C}^k \rightarrow \mathbb {C}^k\big /\mathfrak {S}_k = N$ of the vector field $\sum _{j=1}^k z_j^{p+1}\frac{\partial }{\partial z_j}$:

$$\begin{aligned}&\partial _hU_1[N_m] = \partial _h[m.N_{m+1}] = (-1)^{h-1}m(m+1)DN_{m+1-h} \quad \forall h \in [1, k] \quad \forall m \in \mathbb {N} \\&\partial _{h-1}(U_0 + 1)[N_m] = \partial _{h-1}[(m+1)N_m] = (-1)^h m(m+1)DN_{m-h+1} \quad \forall h \in [2, k] \end{aligned}$$

proving Formulas $(F_h)$ for $h \in [2, k]$.

Now $\sigma _h\partial _h[N_m] = (-1)^{h-1}m\sigma _hDN_{m-h}$ gives

$$\begin{aligned} E(U_0+1)[N_m] = m(m+1)\sum _{h=1}^k (-1)^{h-1}\sigma _hDN_{m-h} = m(m+1)DN_m \end{aligned}$$

because for $m \ge 1$ we have $\sum _{h=0}^k (-1)^h\sigma _hDN_{m-h} = 0$ (see ii) above) and also $E[1] = 0$ for $m = 0$. However, for $h = 1$ we have

$$\begin{aligned} \partial _1U_1[N_m] = \partial _1[mN_{m+1}] = m(m+1)DN_{m} \quad \forall m \in \mathbb {N}. \end{aligned}$$

This gives Formula $(F_1)$.$\blacksquare$

Proof of Proposition 4.2.1 Remark first that $U_1$ is in $\mathcal {N}_0^*$ and, thanks to the previous lemma, that $\mathcal {D}_NU_1$ contains $\partial _1, \dots , \partial _{k-1}$ and $\sigma _k\partial _k$. Define the sub-$\mathcal {D}_N$-module $S:= \sum _{h=1}^{k-1} \mathcal {D}_N\partial _h$. Then, we have a natural surjective $\mathcal {D}_N$-linear map $\alpha : S + \mathcal {D}_N\partial _k\big /(S + \mathcal {D}_N\sigma _k\partial _k) \rightarrow \mathcal {N}_0^*\big /\mathcal {D}_NU_1$. However, we have

$$\begin{aligned} S + \mathcal {D}_N\partial _k\big /(S + \mathcal {D}_N\sigma _k\partial _k) \simeq \mathcal {D}_N\partial _k\big /(S \cap \mathcal {D}_N\sigma _k\partial _k) \simeq \mathcal {D}_N\big /(S + \mathcal {D}_N\sigma _k) \end{aligned}$$

thanks to the equality $S \cap \mathcal {D}_N\sigma _k\partial _k = (S + \mathcal {D}_N\sigma _k)\partial _k$. Moreover, the $\mathcal {D}_N$-module

$$\begin{aligned} \mathcal {D}_N\big /(S + \mathcal {D}_N\sigma _k) \simeq \underline{H}_{[\sigma _k = 0 ]}^1(\mathcal {O}_N) \end{aligned}$$

is simple, so $\alpha$ must be an isomorphism.$\blacksquare$

Theorem 4.2.4

Define $\mathcal {N}_0^{\square }:= \mathcal {D}_NU_1 \subset \mathcal {N}_0^*$. Then, $\mathcal {N}_0^{\square }$ is simple and isomorphic to $\mathcal {N}^\square _1$ via the map induced by the map $\square U_1: \mathcal {N}_1 \rightarrow \mathcal {N}_0$, and the quotient $\mathcal {N}_0\big /\mathcal {N}_0^{\square }$ is isomorphic to $\mathcal {O}_N(\star \sigma _k)$.

Proof

The only point which is not already proved above is the fact that right multiplication by $U_1$, $\square U_1: \mathcal {N}_1 \rightarrow \mathcal {N}_0$ has its image in $\mathcal {N}_0^*$ and induces an isomorphism of $\mathcal {N}^\square _1$ to $\mathcal {N}^\square _0 = \mathcal {D}_NU_1 \subset \mathcal {N}_0^*$. However, $U_1$ is in $\mathcal {N}_0^*$ so the first assertion is clear. This map vanishes on the image of $\mathscr {T}_1$, because $U_{-1}U_1 = (U_0+1)U_0 \quad \textrm{modulo} \ \mathcal {I}$ (and also $\mathcal {I}U_1 \subset \mathcal {I}$ see Theorem 2.4.1), so that right multiplication by $U_1$ induces a map which is clearly surjective. As $\mathcal {N}^\square _1$ is simple, this surjective map is an isomorphism.

We have the following exact sequences of $\mathcal {D}_N$-modules which describe the structure of $\mathcal {N}_0$ and $\mathcal {N}_0^*$:

where the map $\square U_1: \mathcal {N}_1 \rightarrow \mathcal {N}_0$ given by right multiplication by $U_1$ induces an isomorphism $\mathcal {N}^\square _1 \rightarrow \mathcal {N}_0^{\square }$, showing that $\mathcal {N}_0^{\square }$ is a simple $\mathcal {D}_N$-module.$\blacksquare$

Note that the local horizontal basis of $\mathcal {N}_0^{\square }$ on the open set $\{\Delta (\sigma )\sigma _k \not = 0\}$ is (locally) generated by the $(Log \, z_i - Log\, z_j)$ and their images by the isomorphism induced by $U_1$ are the $(z_i - z_j)$ which generates a local horizontal basis of $\mathcal {N}^\square _1$ on the open set $\{\Delta (\sigma )\sigma _k \not = 0\}$.

4.3 The Structure of $\mathcal {N}_p$ for $p \in -\mathbb {N}^*$

Again it is enough, thanks to Lemma 3.1.5, to describe the structure of $\mathcal {N}_{-1}$. Define $\mathcal {N}_{-1}^*$ as the kernel of the $\mathcal {D}_N$-linear map

$$\begin{aligned} \varphi _{-1}: \mathcal {N}_{-1} \rightarrow \mathcal {O}_N(\star \sigma _k) \end{aligned}$$

which is given by $\varphi _{-1}(1) = \sigma _{k-1}\big /\sigma _k$. This map is well defined, because the meromorphic function $\sigma _{k-1}\big /\sigma _k = \sum _{j=1}^k 1\big /z_j$ is a local trace function of the open set $\{\sigma _k \not = 0\}$ and so it is killed by $\mathcal {I}$ everywhere as the $\mathcal {D}_N$-module $\mathcal {O}_N(*\sigma _k)$ has no torsion. Moreover, we have, still on the open set $\{\sigma _k \not = 0\}$:

$$\begin{aligned} U_0(\sigma _{k-1}\big /\sigma _k) =( k-1)\sigma _{k-1}\big /\sigma _k -k\sigma _{k-1}\big /\sigma _k = -\sigma _{k-1}\big /\sigma _k. \end{aligned}$$

So $U_0 +1$ is also in the annihilator of $\sigma _{k-1}\big /\sigma _k$ in $\mathcal {O}_N(\star \sigma _k)$. Therefore, the map $\varphi _{-1}$ is well defined. It is surjective, because $\varphi _{-1}(\partial _{k-1}) = 1/\sigma _k$.

Lemma 4.3.1

The symbol of $U_1$ does not vanish identically on X for each integer $k \ge 2$.

Proof

Recall that X is defined in Sect. 3.2. We have $\eta _h = (-z)^{k-h}\eta _k$ on X where $z = -\eta _{k-1}/\eta _k$ is the double root of $P_\sigma$ at the generic point of $H_\Delta$ (recall that X is the closure of the graph of the meromorphic map $H_\Delta \dashrightarrow \mathbb {C}$ given by the double root of $P_\sigma$ at the generic point of $H_\Delta$). As $U_1 = \sum _{h=1}^k (\sigma _1\sigma _h - (h+1)\sigma _{h+1})\partial _h$ (with the convention $\sigma _{k+1} = 0$), we obtain

$$\begin{aligned} s(U_1)&= (-1)^k\eta _k\sum _{h=1}^k (\sigma _1\sigma _h - (h+1)\sigma _{h+1})(-1)^hz^{k-h} \\&= (-1)^k\eta _k\sum _{h=1}^k (-1)^h\sigma _1\sigma _hz^{k-h} + (-1)^k\eta _k\sum _{p=2}^k (-1)^{p-1}(k-p)\sigma _pz^{k-p+1} + \\&\quad - (-1)^kk\eta _k\sum _{p=2}^k (-1)^{p-1}\sigma _pz^{k-p+1}\\ s(U_1)&= (-1)^k\eta _k\sigma _1\big (P_\sigma (z) - z^k\big ) + (-1)^k\eta _kz^2\big (P'_\sigma (z) - kz^{k-1} + (k-1)\sigma _1z^{k-2}\big ) \\&\quad - (-1)^kk\eta _kz\big (P_\sigma (z) - z^k + \sigma _1z^{k-1}\big ) \\&= -(-1)^k\eta _k\sigma _1z^k - (-1)^kk\eta _kz^{k+1} + (-1)^k(k-1)\eta _k\sigma _1z^k \\&\quad + (-1)^kk\eta _kz^{k+1} - (-1)^kk\eta _k\sigma _1z^k \\&= -2(-1)^k\eta _k\sigma _1z^k \end{aligned}$$

as $P_\sigma (z) = P'_\sigma (z) = 0$ on X.$\blacksquare$

So $U_1$ is not zero in any $\mathcal {N}_\lambda$ for any $\lambda \in \mathbb {C}$.

Proposition 4.3.2

The kernel of the $\mathcal {D}_N$-linear map $\mathscr {T}_0: \mathcal {N}_{-1} \rightarrow \mathcal {N}_0$ given by right multiplication by $U_{-1}$ is equal to $\mathcal {D}_NU_1 = \mathcal {O}_NU_1$ in $\mathcal {N}_{-1}$.

Proof

Recall that Formulas $(F_h), h \in [1, k]$ show the equality $\mathcal {D}_NU_1 = \mathcal {O}_NU_1$ in $\mathcal {N}_{-1}$ (see Remark following Lemma 4.2.3). Moreover, we know that $\mathcal {N}_{-1}$ has no $\mathcal {O}_N$-torsion, thanks to Theorem 3.3.1, so $\mathcal {D}_NU_1$ is a sub-module of $\mathcal {N}_{-1}$ which is isomorphic to $\mathcal {O}_N$ as $U_1$ is not zero in $\mathcal {N}_{-1}$, because its symbol does not vanish on X (see Lemma 4.3.1 above).

The end of the proof of this proposition will use the following lemmas:

Lemma 4.3.3

Let a and b be holomorphic function on an open set U in N, such that the function $a\gamma - bg$ is a section on $U \times \mathbb {C}^k$ which vanishes on $Z \cap (U \times \mathbb {C}^k)$. Then, a and b vanishes identically on U

Proof

The first remark is that we have $a\gamma = bg$ on $U \times \mathbb {C}^k$, because the sheaf $p_*(I_Z)$ has no non zero section which homogeneous of degree 1 in $\eta _1, \dots , \eta _k$ (see Lemma 2.2.5). Then, looking at the coefficients of $\eta _1$ and $\eta _2$ in the equality $a\gamma = bg$ gives

$$\begin{aligned} ka = \sigma _1b \quad \textrm{and} \quad (k-1)\sigma _1a = 2\sigma _2b \quad \mathrm{and \ so} \quad (k-1)\sigma _1^2b = 2k\sigma _2.b \end{aligned}$$

which implies $b \equiv 0$ and then $a \equiv 0$ on U.$\blacksquare$

Lemma 4.3.4

Let $P\in \mathcal {D}_N$, such that $PU_{-1} = AU_0 + Q$ with $A \in \mathcal {D}_N$ and $Q \in \mathcal {I}$. Then, A is unique modulo $\mathcal {I}$.

Proof

We have to show that $AU_0 \in \mathcal {I}$ implies that A is in $\mathcal {I}$.

If this is not true, let $A \in \mathcal {D}_N \setminus \mathcal {I}$ be of minimal order, such that $AU_0 \in \mathcal {I}$. We have $s(A)g \in I_Z$ and, as g is generically $\not = 0$ on Z and $I_Z$ is prime (so reduced), there exists $A_1 \in \mathcal {I}$ with $A- A_1$ of order strictly less than the order of A. Then, $(A - A_1)U_0$ is in $\mathcal {I}$.

This contradicts the minimality of A, since $A - A_1$ cannot be in $\mathcal {I}$.$\blacksquare$

Lemma 4.3.5

There exists a natural $\mathcal {D}_N$-linear map $\psi : Ker(\mathscr {T}_0) \rightarrow \mathcal {N}_1$ given by $\psi (P) = [A]$ when $PU_{-1} = AU_0$ modulo $\mathcal {I}$.

Proof

First recall that the right multiplication by $U_{-1}$ induces a $\mathcal {D}_N$-linear map $\mathscr {T}_0: \mathcal {N}_{-1} \rightarrow \mathcal {N}_0$, because we have $\mathcal {I}U_{-1} \subset \mathcal {I}$ and the relation $(U_0 +1)U_{-1} = U_{-1}U_0$. If $P \in \mathcal {D}_N$ induces a germ of section of $Ker(\mathscr {T}_0)$ then the previous lemma shows that if we write $PU_{-1} = AU_0 + Q$ with $Q \in \mathcal {I}$, the image of the germ A in $\mathcal {D}_N\big /\mathcal {I}$ is well defined. Then, we have a $\mathcal {D}_N$-linear map $Ker(\mathscr {T}_0) \rightarrow \mathcal {D}_N\big /\mathcal {I} = \mathcal {M}$ and after composition by the quotient maps $\mathcal {D}_N\big /\mathcal {I} \rightarrow \mathcal {N}_1$ we obtain the desired map.$\blacksquare$

End of proof of Proposition 4.3.2 First remark that $U_1$ is sent to 0 in $\mathcal {N}_1$ by the map $\psi$ because of the relation $U_1U_{-1} = (U_0 -1)U_0 \quad \textrm{modulo} \quad \mathcal {I}$.

So $\psi$ composed with the quotient map $\mathcal {N}_1 \rightarrow \mathcal {N}^\square _1$ induces a map

$$\begin{aligned} \tilde{\psi }: Ker(\mathscr {T}_0)\big /\mathcal {O}_NU_1 \rightarrow \mathcal {N}^\square _1. \end{aligned}$$

We shall prove that this map is injective and not surjective. As $\mathcal {N}^\square _1$ is simple, this will prove that $Ker(\mathscr {T}_0) = \mathcal {O}_NU_1$ completing the proof of Proposition 4.3.2.

We shall first prove the injectivity of $\tilde{\psi }$, so the fact that if $[P] \in Ker(\mathscr {T}_0)$ satisfies $\psi (P) = [A]$ with $[A] = 0$ in $\mathcal {N}^\square _1$ then [P] is a germ of section of the sub-sheaf $\mathcal {D}_NU_1$ of $Ker(\mathscr {T}_0)$.

Let $P \in \mathcal {D}_N$ of minimal order, such that the class of [P] in $Ker(\mathscr {T}_0)\big /\mathcal {D}_NU_1$ is not zero and satisfies $\tilde{\psi }([P]) = 0$. Then, we have

$$\begin{aligned} (P + X(U_0 + 1) + Q_0)U_{-1} = A_0U_0 + Q_1 \quad \textrm{with} \quad Q_0, Q_1 \in \mathcal {I} \quad \textrm{and} \ X \in \mathcal {D}_N. \end{aligned}$$

Then, thanks to the relations $(U_0 +1)U_{-1} = U_{-1}U_0 \ \textrm{modulo} \quad \mathcal {I}$ we obtain

$$\begin{aligned} PU_{-1} = AU_0 + Q_2 \quad \textrm{with} \ Q_2:= Q_1 - Q_0U_{-1} \in \mathcal {I} \quad \textrm{and} \ A = A_0 -XU_{-1}. \end{aligned}$$

Then, our hypothesis implies that there exist $R, S, Q_3 \in \mathcal {D}_N$ with $Q_3 \in \mathcal {I}$, such that $A = R(U_0 -1) + SU_{-1} +Q_3$. Therefore

$$\begin{aligned} (P - S(U_0 +1) - RU_1)U_{-1} = Q_3U_0 \quad \textrm{modulo} \ \mathcal {I} \end{aligned}$$

and $Q_3U_0$ is again in $\mathcal {I}$. So looking at the symbols we find

$$\begin{aligned} s(P - S(U_0 +1) - RU_1)\gamma \in I_Z. \end{aligned}$$

As $\gamma$ is generically $\not = 0$ on Z and $I_Z$ is prime (then reduced) we conclude that there exists $P_1 \in \mathcal {I}$ with symbol $s(P_1) = s(P - S(U_0 +1) - RU_1)$. Therefore, the order of $P - P_1- S(U_0 +1) - RU_1$ is strictly less than the order of P but the class of $P - P_1 -S(U_0+1) - RU_1$ in $Ker(\mathscr {T}_0)\big /\mathcal {D}_NU_1$ is the same than the class induced by P. This contradict the minimality of the order of P; so $Ker(\mathscr {T}_0) = \mathcal {D}_NU_1$ and the map $\tilde{\psi }$ is injective.

To conclude it is now enough to prove that $\tilde{\psi }$ is not surjective, as explained above. Therefore, assume that there exists $P \in \mathcal {D}_N$ with $PU_{-1} = AU_0 + Q$ with $Q \in \mathcal {I}$ and $[A - 1] = 0$ in $\mathcal {N}^\square _1$. This would implies that $PU_{-1} = (1 + T(U_0 - 1) + YU_{-1})U_0$ modulo $\mathcal {I}$ and so we obtain the equality

$$\begin{aligned} (P -Y(U_0+1) -TU_1)U_{-1} = U_0 \quad \textrm{modulo} \quad \mathcal {I}. \end{aligned}$$

So looking at the symbols restricted to Z this gives:

$$\begin{aligned} s\big (P -Y(U_0+1) -TU_1\big )\gamma = g \end{aligned}$$

in $\mathcal {O}_Z$. By homogeneity in $\eta$ this implies that $f:= s\big (P -Y(U_0+1) -TU_1\big )$ is the pull-back of a holomorphic function on an open set in N and this gives a contradiction thanks to Lemma 4.3.3. $\blacksquare$

Proposition 4.3.6

Let $\mathcal {N}_{-1}^{\square }$ be the sub-$\mathcal {D}_N$-module of $\mathcal {N}_{-1}^*$ which is generated by $\partial _1, \dots , \partial _{k-2}$. Then, $\mathscr {T}_0$ sends $\mathcal {N}_{-1}^{\square }$ onto $\mathcal {N}_0^{\square }$ and induces an isomorphism between theses two simple $\mathcal {D}_N$-modules.

Proof

As we know that $\mathcal {N}_0^{\square }$ is equal to $\mathcal {D}_NU_1 \subset \mathcal {N}_0^*$, we first check that the generators of $\mathcal {N}_{-1}^\square$ have their images by $\mathscr {T}_0$ in $\mathcal {D}_NU_1$.

For $h \in [1, k-2]$ Formulas $(E_{h+2})$ and $(F_{h+1})$ imply

$$\begin{aligned} \partial _{h+2}U_1 + \partial _{h+1}(U_0 +1) + \partial _hU_{-1} + \partial _{h+1}(U_0- 1) \in \mathcal {I} \end{aligned}$$

which implies $\partial _hU_{-1} = - \partial _{h+2}U_1$ in $\mathcal {N}_0$.

Note that, as $\mathcal {N}^\square _1$ is obviously generated¹¹ by $\partial _1, \dots , \partial _{k-2}$ its image by the right multiplication by $U_1$ in $\mathcal {N}_0^*$ is generated by $\partial _hU_1, h \in [1, k-2]$ giving a direct proof of the surjectivity of $\mathscr {T}_0:\mathcal {N}_{-1}^{\square } \rightarrow \mathcal {D}_NU_1 = \mathcal {N}_0^{\square }$.

The injectivity of this map is clear thanks to Proposition 4.3.2 and the fact that $\varphi _{-1}(U_1) = -k$ which implies that the sub-modules $\mathcal {D}_NU_1 = \mathcal {O}_NU_1$ and $Ker(\mathscr {T}_0)$ of $\mathcal {N}_{-1}$ have an intersection reduced to $\{ 0 \}$.$\blacksquare$

Proposition 4.3.7

The sub-module $\mathcal {N}_{-1}^{\square }$ defined in the previous proposition is equal to $\mathcal {N}_{-1}^*$.

Proof

By definition $\mathcal {N}_{-1}^*$ is the kernel of the map $\varphi _{-1}: \mathcal {N}_{-1} \rightarrow \mathcal {O}_N(*\sigma _k)$ which sends [1] to $\sigma _{k-1}/\sigma _k$. Then, $\mathcal {N}_{-1}^*$ is generated by the annihilator of $\sigma _{k-1}/\sigma _k$ in $\mathcal {O}_N(*\sigma _k)$. Therefore, $\mathcal {N}_{-1}^*$ is generated by the class in $\mathcal {N}_1$ of

$$\begin{aligned} \partial _1, \dots , \partial _{k-2}, \partial _{k-1}^2, \sigma _{k-1}\partial _{k-1} - 1, \sigma _k\partial _k +1. \end{aligned}$$

We already know that $\partial _1, \dots , \partial _{k-2}$ are in $\mathcal {N}_{-1}^{\square }$ for $h \in [1, k-2]$ by definition of $\mathcal {N}_{-1}^\square$. Moreover, we have $\partial _{k-1}^2 = \partial _k\partial _{k-2} \quad \textrm{modulo} \ \mathcal {I}$ gives that $\partial _{k-1}^2U_{-1} = \partial _k\partial _{k-2}U_{-1}$ belongs to $\mathcal {N}_0^\square$. Then, Proposition 4.3.6 implies that $\partial _{k-1}^2$ is in $\mathcal {N}_{-1}^\square + Ker(\mathscr {T}_0)$. Therefore, write $\partial _{k-1}^2 = e + fU_1$ with $e \in \mathcal {N}_{-1}^{\square }$ and $f \in \mathcal {O}_N$ using Proposition 4.3.2.

Then $\varphi _{-1}(\partial _{k-1}^2) = 0$ implies $\varphi (fU_1) = -kf = 0$, because e is in $\mathcal {N}_{-1}^\square \subset \mathcal {N}_{-1}^*$. Therefore, $f = 0$ and $\partial _{k-1}^2$ is in $\mathcal {N}_{-1}^\square$.

So it is enough to prove that $a:= \sigma _{k-1}\partial _{k-1} - 1$ and $b = \sigma _k\partial _k +1$ are in $\mathcal {N}_{-1}^{\square }$ to complete the proof.

Formula $(E_1)$ gives $EU_{-1} = -\partial _1$ in $\mathcal {N}_0$ and Formula $(F_2)$ gives $\partial _2U_1 = -\partial _1$ in $\mathcal {N}_0$. This implies that $\mathscr {T}_0(E) = \partial _2U_1 \in \mathcal {N}^\square _0$. This implies that E is in $\mathcal {N}_{-1}^{\square } + Ker(\mathscr {T}_0)$.

So write $E = e + fU_1$ with $e \in \mathcal {N}_{-1}^{\square }$ and $f \in \mathcal {O}_N$ using Proposition 4.3.2.

Now

$$\begin{aligned} \varphi _{-1}(E) = E[\sigma _{k-1}/\sigma _k] = 0 \quad \textrm{and} \quad \mathcal {N}^\square _{-1} \subset \mathcal {N}_1^* = ker(\varphi _{-1}). \end{aligned}$$

Therefore, $\varphi _{-1}(fU_1) = f\varphi _{-1}(U_1) = -kf = 0$. This implies $E = e$ is in $\mathcal {N}^\square _{-1}$.

But $a + b = E \quad \textrm{modulo} \ \mathcal {N}_{-1}^{\square }$. Therefore, $a + b$ belongs to $\mathcal {N}_{-1}^{\square }$.

We have also in $\mathcal {N}_{-1}$:

$$\begin{aligned} 0 = U_0 + 1 = (k-1)(\sigma _{k-1}\partial _{k-1} - 1) + k(\sigma _k\partial _k + 1) \quad \textrm{modulo} \ \mathcal {N}_{-1}^{\square } \end{aligned}$$

and this gives $(k-1)a + kb \in \mathcal {N}_{-1}^{\square }$, concluding the proof.$\blacksquare$

Theorem 4.3.8

We have the following commutative diagram of $\mathcal {D}_N$-module with exact lines and columns, where the $\mathcal {D}_N$-linear map $\varphi _{-1}: \mathcal {N}_{-1} \rightarrow \mathcal {O}_N(\star \sigma _k)$ is defined by $\varphi _{-1}(1) = \sigma _{k-1}\big /\sigma _k$:

Therefore, $\chi$ is an isomorphism. Moreover, the map $\mathscr {T}_0$ induces an isomorphism of $\mathcal {N}_{-1}^{*}$ onto the simple $\mathcal {D}_N$-module $\mathcal {N}_0^{\square } = \mathcal {D}_NU_1 \subset \mathcal {N}_0^*$.

Proof

The exactness of the first line is consequence of the equality $\varphi _{-1}(U_1) = -k$. The exactness of the second line is consequence of the surjectivity of $\varphi _{-1}$ which is consequence of the equality $\varphi _{-1}(\partial _{k-1}) = 1/\sigma _k$.

As $\mathcal {Q}$ is the obvious quotient the injectivity of the induced map $\chi$ is easily obtained by a diagram chasing. $\blacksquare$

The local solutions of $\mathcal {N}_{-1}$ are the $1/z_j, j \in [1, k]$ and the local solutions of $\mathcal {N}_{-1}^*$ are the $1/z_j - 1/z_h$ which generate the linear combinations of the $1/z_j$ which are killed by $U_1 = \sum _{j=1}^k z_j^2\frac{\partial }{\partial z_j}$.

Conclusion For each integer $p \ge 2$ define

$$\begin{aligned} \mathcal {N}_p^{\square }:= \mathcal {D}_N\big /\mathcal {I} + \mathcal {D}_N(U_0 - p) + \mathcal {D}_NU_{-1}^{p} \end{aligned}$$

and $\mathcal {N}_{-p}^*:= ker(\varphi _{-p})$, where $\varphi _{-p}: \mathcal {N}_{-p} \rightarrow \mathcal {O}_N(*\sigma _k)$ is given by $\varphi _{-p}(1) = U_{-1}^{p-1}[\sigma _{k-1}/\sigma _k]$. Then, we have the chain of isomorphisms:

where $\mathscr {T}_p:= \square U_{-1}$ is given by right multiplication by $U_{-1}$.

4.4 Some Higher Order Solutions of $\mathcal {N}_p$ for $p \in \mathbb {N}$

Let $N:= \mathbb {C}^{k}$ with coordinates $\sigma _{1}, \dots , \sigma _{k}$ and note $D_{N}$ the sheaf of (holomorphic) differential operators on N and $\mathcal {D}b_{N}^{p, q}$ the sheaf of $(p, q)-$currents on N.

Recall that $\mathcal {D}b_{N}^{p, q}$ is a left $D_{N}-$module and that we have the following theorem due to M. Kashiwara (see [7])

Theorem 4.4.1

For any regular holonomic $D_{N}-$module $\mathcal {N}$ and any integer $j \ge 1$ we have

$$\begin{aligned} Ext_{D_{N}}^{j}(\mathcal {N}, \mathcal {D}b_{N}^{0, p}) = 0. \end{aligned}$$

Note that the case $p \ge 1$ is an obvious consequence of the case $p = 0$ as $\mathcal {D}b_{N}^{0, p}$ is the direct sum of $C_{k}^{p}$ copies of $\mathcal {D}b_{N}^{0, 0}$ as a left $D_{N}-$module.

Corollary 4.4.2

For any regular holonomic $D_{N}-$module $\mathcal {N}$ and any integer $j \ge 0$ we have a natural isomorphism of sheaves of $\mathbb {C}-$vector spaces

$$\begin{aligned} Sol^{j}(\mathcal {N}):= Ext_{D_{N}}^{j}(\mathcal {N}, \mathcal {O}_{N}) \simeq H^{j}\big ((Hom_{D_{N}}(\mathcal {N}, \mathcal {D}b_{N}^{0, \bullet }), \bar{\partial }^{\bullet })\big ). \end{aligned}$$

For instance, if $\mathcal {N}:= D_{N}\big /\mathcal {J}$ is a regular holonomic system (where $\mathcal {J}$ is a coherent left ideal in $D_{N}$), we have a natural isomorphism of sheaves of complex vector spaces, for each j:

$$\begin{aligned} Sol^{j}(D_{N}\big /\mathcal {J}) \simeq \{T \in \mathcal {D}b_{N}^{0, j} / \ / \mathcal {J}.T = 0, \ \bar{\partial }T = 0 \} \Big / \bar{\partial }\big ( \{T \in \mathcal {D}b_{N}^{0, j-1} / \ / \mathcal {J}.T = 0 \}\big ). \end{aligned}$$

Proof

As the Dolbeault–Grothendieck complex $(\mathcal {D}b_{N}^{0, \bullet }, \bar{\partial }^{\bullet })$ is a resolution of $\mathcal {O}_{N}$ by $D_{N}-$modules for which the functor

$$\begin{aligned} \mathcal {N} \mapsto Hom_{D_{N}}(\mathcal {N}, -) \end{aligned}$$

is exact, thanks to the previous theorem, the conclusion follows by degeneracy of the spectral sequence.$\blacksquare$

Proposition 4.4.3

Let $\sigma ^{0}$ be a point the hypersurface $\{\sigma _{k} = 0 \}$ in N and let d be the multiplicity of the root 0 in $P_{\sigma ^{0}}$. Let U be a small open neighborhood of $\sigma ^{0}$ in N on which there exists a holomorphic map $f: U \rightarrow Sym^{d}(\mathbb {C})$ whose value at $\sigma \in U$ is the $d-$tuple of roots of $P_{\sigma }$ which are near by 0.¹²

Then define for $q \in \mathbb {N}$ the distribution on U (given by a locally integrable function)

$$\begin{aligned} T_{q}(\sigma ) = \sum _{j=1}^{d} \ z_{j}^{q}Log\vert z_{j}\vert ^{2} \quad \textrm{where} \ [z_{1}, \dots , z_{d}] = f(\sigma ) \end{aligned}$$

(30)

Then, the current $\bar{\partial }T_{q}$ defines a section on U of the sheaf $Sol^{1}(\mathcal {N}_{q})$, such that its germ at a point $\sigma ^{0}$ is non zero in $Sol^{1}(\mathcal {N}_{q})_{\sigma ^{0}}$.

Proof

Let $pr: H \rightarrow \mathbb {C}$ and $\pi : H \rightarrow N$ are the projections, where

$$\begin{aligned} H:= \{(\sigma , z) \in N\times \mathbb {C}\ / \ P_\sigma (z) = 0\}. \end{aligned}$$

We may assume that the open set $pr(\pi ^{-1}(U))$ is the disjoint union of D with an open set $\Omega$ in $\mathbb {C}$. Then, if we define the locally integrable function $f: D \cup \Omega$ as $f(z) = z^{q}Log\vert z\vert ^{2}$ on D and $f \equiv 0$ on $\Omega$ we have $T_{q}(\sigma ) = \pi _{*}(f)(\sigma ) = \sum _{j=1}^{k} f(z_{j})$ where $z_{1}, \dots , z_{k}$ are the roots of $P_{\sigma }$. It is then easy to verify that $\mathcal {I}T_{q} = 0$ and that $(U_{0} - q)T_{q} = N_{q}(\sigma )$ the q-th Newton function of the $d-$tuple $d(\sigma )$ of roots of $P_\sigma$ which are in D. Therefore, it is holomorphic on U. Then, the $(0, 1)-$current $\bar{\partial }T_{q}$ is $\bar{\partial }-$closed and is killed by $\mathcal {J}_{q}$. Then, thanks to Corollary 4.4.2 it induces a section on U of the sheaf $Sol^{1}(\mathcal {N}_{q})$.

Fix now $\tau \in U$ and assume that the germ at $\tau$ of the previous section vanishes. Then, there exists on an open polydisc V with center $\tau$ in U and a $(0, 0)-$current S satisfying

$$\begin{aligned} \mathcal {I}S = 0, \quad (U_{0} - q)S = 0 \quad \textrm{and} \quad \bar{\partial }S = \bar{\partial }T. \end{aligned}$$

Then, we may write $S = T- F$ where F is holomorphic on V. However, then F satisfies also $\mathcal {I}F = 0$ and $(U_{0}- q).F(\sigma ) = N_{q}(d(\sigma ))$ for all $\sigma \in V$. The first equation implies that F is a global trace function on V (up to shrink V around $\tau$ if necessary) and using Lemma 3.1.2 in [1] we see that, up to a locally constant function on $D\cup \Omega$, $(U_{0} - q).F$ is the trace of a holomorphic function h define by $h(z) = z^{q}$ on D and 0 on $\Omega$. However, if $F = Trace(g)$ where g is holomorphic on $D \cup \Omega$ this implies

$$\begin{aligned} z\frac{\partial g}{\partial z}(z) = h(z) + \kappa (z) \end{aligned}$$

where $\kappa$ is constant equal to k on D. Therefore, on D the meromorphic function $G:= g\big /z^{q}$ satisfies

$$\begin{aligned} G'(z) = 1/z + \kappa /z^{q+1}. \end{aligned}$$

This is clearly impossible for $q \ge 1$. For $q= 0$ this gives that $G = g$ is constant and so is $F = Trace(g)$. But then, $U_{0}F = d$ is impossible for $d \ge 1$. This shows that at each point $\sigma ^{0}$ of the hyper-suface $\{ \sigma _{k} = 0 \}$ in N the germ induced by $\bar{\partial }T_q$ in $Sol^{1}(\mathcal {N}_{q})_{\sigma ^{0}}$ is not zero. Therefore, the support of the sheaf $Sol^{1}(\mathcal {N}_{q})$ contains this hyper-surface for each $q \in \mathbb {N}$.$\blacksquare$

Remark The exact sequence

$$\begin{aligned} 0 \rightarrow \mathcal {M} \overset{(U_{0}-q)}{\rightarrow } \mathcal {M} \rightarrow \mathcal {N}_{q} \rightarrow 0 \end{aligned}$$

gives a long exact sequence

$$\begin{aligned} 0 \rightarrow Sol^{0}(\mathcal {N}_{q}) \rightarrow Sol^{0}(\mathcal {M}) \overset{U_{0}-q}{\rightarrow } Sol^{0}(\mathcal {M}) \overset{\partial }{\rightarrow } Sol^{1}(\mathcal {N}_{q}) \rightarrow Sol^{1}(\mathcal {M}) \rightarrow \dots \end{aligned}$$

and it is clear that the germ at the origin of the Newton polynomial $N_{q}$ in $Sol^{0}(\mathcal {M})$ is not in the image of $U_{0}- q$. Our computation above shows that the image of the germ of $N_{q}$ at the point 0 is mapped by the connector $\partial$ to the germ in $Sol^{1}(\mathcal {N}_{q})_{0}$ which is constructed above.

The reader will find the computation of the conjugate of the $D_N$-modules $\mathcal {N}_\lambda$ for each $\lambda \in \mathbb {C}$ in the article [4], so the computation of the sheaves:

$$\begin{aligned} Hom_{D_N}(\mathcal {N}_\lambda , Db_N^{0,0}). \end{aligned}$$

5 An Application

5.1 A Taylor Expansion

We shall consider now the universal monic degree k equation near the point $\sigma ^0$ defined by $\sigma ^0_1 = \sigma ^0_2 = \dots = \sigma ^0_{k-1} = 0$ and $\sigma ^0_k = -1$. We shall denote by $z(\sigma )$ the root of $P_{\sigma ^0 + \sigma }(z) = 0$ which is near the (simple) root $-1$, for $\sigma$ small enough, of the equation $P_{\sigma ^0 + \sigma }(z) = z^k -(-1)^k = 0$ when $\sigma$ is small enough. Define

$$\begin{aligned} F(\sigma ^0 + \sigma ):= z(\sigma ) - \sigma _1/k:= \sum _{\alpha \in \mathbb {N}^k} C_\alpha \frac{\sigma ^\alpha }{\alpha !} \end{aligned}$$

(31)

the Taylor expansion at the point $\sigma ^0$ of $z(\sigma ) - \sigma _1/k$ which a solution near $\sigma ^0$ of the $\mathcal {D}_N$-module $\mathcal {N}^\square _1$ (see the Theorem 4.1.5).

The reader may compare the computation below with [8].

An easy consequence of the results in the paragraph 4.1 is the following theorem.

Theorem 5.1.1

The following differential operators annihilate the function F in a neighborhood of $\sigma ^0$, where we note $\partial _h$ for the partial derivative relative to $\sigma _h$.

• 1. $A_{i,j}:= \frac{\partial ^2}{\partial i\partial j} - \frac{\partial ^2}{\partial _{i+1}\partial _{j-1}} \quad \forall i \in {[}1, k-1] \ \textrm{and} \ \forall j \in [2, k]$.

• 2. $\hat{U}_0 - 1:= \sum _{h=1}^k h\sigma _h\partial _h - k\partial _k - 1.$

• 3. $U_{-1}:= \sum _{h=0}^{k-1} (k-h)\sigma _h\partial _{h+1}$ with the convention $\sigma _0 \equiv 1$

Proof This is consequence of the fact that F is a solution in an open neighborhood of $\sigma ^0$ of the regular holonomic system $\mathcal {N}_1^{\square } \simeq \mathcal {D}_N\big /\mathcal {A} + \mathcal {D}_N(U_0 - 1) + \mathcal {D}_N(U_{-1})$. Remark that the operator $\hat{U}_0$ is the expression of $U_0$ in the coordinates centered at $\sigma ^0$. The other operators have in these coordinates the same expression than in the usual coordinates centered at the origin.$\blacksquare$

Corollary 5.1.2

The coefficients $C_\alpha$ is the expansion (1) only depend on the integers $q:= \vert \alpha \vert = \sum _{h=1}^k \alpha _h$ and $w(\alpha ):= \sum _{h=1}^k h\alpha _h$ so we may rewrite the expansion (1) with the convention $C_{q, r} = 0$ when $r \not \in [q, kq]$:

$$\begin{aligned} F(\sigma ^0 + \sigma ) = \sum _{q, r} C_{q, r}m_{q, r}(\sigma ) \end{aligned}$$

(32)

where for $q \in \mathbb {N}$ and $r \in [q, kq]$ we define the polynomial $m_{q, r} \in \mathbb {C}[\sigma ]$ by the formula

$$\begin{aligned} m_{q, r}(\sigma ):= \sum _{\vert \alpha \vert = q, w(\alpha ) = r} \frac{\sigma ^\alpha }{\alpha !} \end{aligned}$$

Proof

This is obvious consequence of the description of the holomorphic functions which are annihilated by the differential operators $A_{i,j}$ for all $i \in [1, k-1]$ and $j \in [2, k]$ (see the paragraph 2.1) which generate the left ideal $\mathcal {A}$ in $\mathcal {D}_N$.$\blacksquare$

Proposition 5.1.3

We have the following formulas, with the conventions $m_{q, r} = 0$ for $r \not \in [q, kq]$ (in particular for $q < 0$ or $r < 0$):

• 1. $(\hat{U}_0 -1)(m_{q, r}) = (r-1)m_{q, r} - km_{q-1, r-k}\quad \forall q \ge 0, \ \forall r \in [q,kq]$

• 2. $U_{-1}(m_{q, r}) = (kq - r + 1)m_{q, r-1} + km_{q-1, r-1} \quad \forall q \ge 0, \ \forall r \in [q,kq]$.

Proof

The first formula is a direct consequence of the easy formulas

$$\begin{aligned} U_0(\sigma ^\alpha /\alpha !) = w(\alpha )\sigma ^\alpha /\alpha ! \quad \textrm{and} \quad \partial _k(\sigma ^\alpha /\alpha !)= \sigma ^\beta /\beta ! \end{aligned}$$

when $\alpha _k \ge 1$, with $\beta + 1_k = \alpha$ and

$$\begin{aligned} \partial _k(\sigma ^\alpha /\alpha !)= 0 \quad \textrm{when} \quad \alpha _k = 0. \end{aligned}$$

The second formula is little more tricky:

For $h \in [2, k-1]$ we have

$$\begin{aligned} \sigma _h\partial _{h+1}(\sigma ^\alpha /\alpha !) = \beta _h\sigma ^\beta /\beta ! \quad \textrm{when} \quad \alpha _{h+1} \ge 1 \end{aligned}$$

with $\beta + 1_{h+1} = \alpha + 1_h$, and

$$\begin{aligned} \sigma _h\partial _{h+1}(\sigma ^\alpha /\alpha !) = 0 \quad \textrm{when} \quad \alpha _{h+1} = 0. \end{aligned}$$

Moreover, for any $\beta$ with $\vert \beta \vert = q-1$ and $w(\beta ) = r-1$ and for each $h \in [2, k]$ there exists exactly one $\alpha$ if $\beta _h \not = 0$ with   $\sigma _h\partial _{h+1}(\sigma ^\alpha /\alpha !) = \beta _h\sigma ^\beta /\beta !$, and it satisfies $\vert \alpha \vert = q$ and $w(\alpha ) = r$, and no such $\alpha$ exists if $\beta _h = 0$. This means that that $\sigma _h\partial _{h+1}(m_{q, r})$ contains $\sigma ^\beta /\beta !$ with the coefficient $\beta _h$.

For $h = 1$ the situation is simpler: $\partial _1(\sigma ^\alpha /\alpha !) = \sigma ^\beta /\beta !$ when $\alpha _1 \ge 1$ with $\beta + 1_1 = \alpha$, and $\partial _1(\sigma ^\alpha /\alpha !) = 0$ when $\alpha _1 = 0$.

Then for each $\beta$ with $\vert \beta \vert = q-1$ and $w(\beta ) = r-1$ there exists a unique $\alpha$, such that $\partial _1(\sigma ^\alpha /\alpha !) = \sigma ^\beta /\beta !$ and it satisfies $\vert \alpha \vert = q$ and $w(\alpha ) = r$. Therefore, we conclude that

$$\begin{aligned} U_{-1}(m_{q, r})&= \sum _{\vert \alpha \vert = q, w(\alpha ) = r} \sum _{h=0}^k (k-h)\sigma _h\partial _{h+1}\left( \frac{\sigma ^\alpha }{\alpha !}\right) \\&= km_{q-1, r-1} + \sum _{\vert \beta \vert = q, w(\beta ) = r-1}\sum _{h=1}^k (k-h)\beta _h\frac{\sigma ^\beta }{\beta !}\\&= km_{q-1, r-1} + \big ( k(q-1) - (r-1)\big )m_{q, r-1} \end{aligned}$$

concluding the proof.$\blacksquare$

Taking in account Eqs. 2 and 3. of Theorem 5.1.1 ( Eq. 1. are used already in the Corollary 5.1.2), we obtain:

Corollary 5.1.4

The coefficients $C_{q, r}$ of the Taylor expansion (1) satisfies the relations:

The formula (B) gives, for each $r \ge 2$ and each $s \in \mathbb {N}$, such that $0 \le s \le \frac{(k-1).r}{k}$

Formula (A) gives for each $r \ge 1$:

Moreover, we have

Proof

Looking at the coefficient of $m_{q,r}$ for $q \ge 1$ and $r \in [q, kq]$ in the equality $(\hat{U}_0 - 1)[F] \equiv 0$ gives the gives (A). Looking at the coefficient of $m_{q, r-1}$ for $q \ge 1$ and $r \in [q+1, kq]$ in $U_{-1}[F] \equiv 0$ gives (B).

The formula $(B^*)$ is a direct consequence of the formula (B) with $q:= r-s$ by an easy induction on $s \in [0, \frac{(k-1).r}{k}]$.

Using formula $(B^*)$ with $r' = r+k$ and $s = k-1$ we obtain

$$\begin{aligned} C_{r+1, r+k} = \frac{(-1)^{k-1}C_{r+k, r+k}}{\prod _{p=0}^{k-2} (r+p - (r-1)/k)} \end{aligned}$$

Combining this formula with the formula (A) with $q = r \ge 1$ which gives

$$\begin{aligned} C_{r+1, r+k} = \frac{r-1}{k}C_{r, r} \end{aligned}$$

and we obtain the formula $(A^*)$.

Formula (C) is a direct consequence of $(A^*)$ for $r = 1$ with an induction on $a \ge 1$ when $r = 1+ ak$.$\blacksquare$

We shall see below that the vanishing of $C_{1+ak, 1+ ak}$ is also valid for $a = 0$ giving $C_{q, 1+ ak} = 0$ for any $q \ge 1$ and any a, such that $q \le 1+ ak \le kq$.

Remark It is enough to compute $C_{0, 0}$ and $C_{1, h}$ for each $h \in [1, k]$ to determine all coefficients $C_{q, r}$ in (1) with $r \in [q, kq]$:

The formula (B) determines $C_{h,h}, h \in [2, k]$ from $C_{1, h}$ with $r = h$ and $s = h-1$. Then, the formulas $(A^*)$ and (C) gives $C_{r, r}$ for any $r \ge 1$. Then, the formula (B) completes the computation of $C_{q, r}$ for any $q \ge 0$ and any $r \in [q, kq]$.

Lemma 5.1.5

We have the following values:

$$\begin{aligned}&C_{0, 0} = -1 \\&C_{1, 1} = 0 \\&C_{1, h} = 1/k \quad \textrm{for} \ h \in [2, k]. \end{aligned}$$

Proof

The value of $C_{0,0}$ is $F(\sigma ^0)$ which is $-1$ by definition of F. The values of $C_{1, h}$ is the derivative $\partial _h(F + \sigma _1/k)(\sigma ^0)$, because we have $m_{1, h} = \sigma _h$ for $h \in [1, k]$. Therefore, it is enough to make an order 1 expansion of F at $\sigma ^0$ to compute the values of the $C_{1, h}, h \in [1, k]$. This is given by the following computation at the first order of $P_{\sigma ^0 + \sigma }(z(\sigma )) \equiv 0$, where we define

$$\begin{aligned} z(\sigma ) = -1 + \sum _{j=1}^k c_j\sigma _j + o(\vert \vert \sigma \vert \vert ) \end{aligned}$$

which gives $c_1 = C_{1, 1} + 1/k$ and $c_j:= C_{1, j}$ for $j \in [2, k]$ and then:

$$\begin{aligned}{} & {} \left( -1 + \sum _{h=1}^k c_h\sigma _h\right) ^k + \sum _{j=1}^k (-1)^j\sigma _j\left( -1 + \sum _{h=1}^k c_h\sigma _h\right) ^{k-j} - (-1)^k = o(\vert \vert \sigma \vert \vert ) \\{} & {} \qquad (-1)^{k-1}k\left( \sum _{h=1}^k c_h\sigma _h\right) + \sum _{j=1}^k (-1)^{k}\sigma _j = o(\vert \vert \sigma \vert \vert ) \quad \mathrm{and \ so} \\{} & {} \qquad c_j = 1/k \quad \forall j \in [1, k] \\ \end{aligned}$$

Then, $C_{1, 1} = 0$ and $C_{1, j} = c_j = 1/k$ for each $j \in [2, k]$.$\blacksquare$

6 Appendix: The Surface S(k)

6.1 Study of S(k)

For $k \ge 2$ an integer and $\alpha \in \mathbb {N}^k$ define

• the length of $\alpha$ given by $\vert \alpha \vert := \sum _{h=1}^k \alpha _h$;

• the weight of $\alpha$ given by $w(\alpha ):= \sum _{h=1}^k h.\alpha _h$.

We shall say that $\alpha$ and $\beta$ are equivalent, noted by $\alpha \sharp \beta$, when $\vert \alpha \vert = \vert \beta \vert$ and $w(\alpha ) = w(\alpha )$.

Remark that for any $\gamma \in \mathbb {N}^k$ we have $(\alpha + \gamma ) \sharp (\beta + \gamma )$ if and only if $\alpha \sharp \beta$.

Let A be a $\mathbb {C}-$algebra which is commutative, unitary and integral. In the algebra $A[x_1, \dots , x_k]$ let IS(k) be the ideal generated by the polynomials $x_p.x_q - x_{p+1}.x_{q-1}$ for all $p \in [1, k-1]$ and $q \in [2, k]$.

We shall say that the two monomials $x^\alpha$ and $x^\beta$ in $A[x_1, \dots , x_k]$ are equivalent when $\alpha$ and $\beta$ are equivalent. In this case we shall also write $x^\alpha \sharp x^\beta$.

Remark that for any $p \in [1, k-1]$ and any $q \in [2, k]$ $x_p.x_q$ is equivalent to $x_{p+1}.x_{q-1}$.

For a monomial $m:= x^\alpha$ we define its length by $l(m):= \vert \alpha \vert$ and its weight $w(m):= w(\alpha )$.

Our first result is the following characterization of the elements in IS(k).

Proposition 6.1.1

Two monomials $x^\alpha$ and $x^\beta$ in $A[x_1, \dots , x_k]$ are equivalent if and only if $x^\alpha - x^\beta$ is in IS(k).

The proof of this proposition will need a preliminary lemma and the next definition.

Definition 6.1.2

We shall say that a monomial m in $A[x_1, \dots , x_k]$ is minimal when it has one of the following forms:

• 1. there exists p, q in $\mathbb {N}$, such that $m = x_1^px_k^q$;

• 2. there exists p, q in $\mathbb {N}$ and $j \in [2, k-1]$ such that $m = x_1^px_jx_k^q$.

Remark Any monomial (minimal or not) is not in the ideal IS(k), because the point $x_1 = x_2 = \dots = x_k = 1$ is not in $\vert S(k)\vert$ the common set of zeros in $A^k$ of the generators of IS(k) and any monomial does not vanish at this point.

Lemma 6.1.3

For each $\alpha \in \mathbb {N}^k$ there exists an unique minimal monomial $x^{\mu (\alpha )}$, such that $x^\alpha \sharp x^{\mu (\alpha )}$. Moreover, for each $\alpha$ $x^\alpha - x^{\mu (\alpha )}$ is in IS(k)

Proof

Let us begin by proving the uniqueness assertion.

We have to show that two minimal monomials which are equivalent are equal. If both are in case 1. (so $m:= x_1^px_k^q$) this is obvious as the length is equal to $l(m) = p + q$ and the weight is $w(m) = p + kq$ and then $(k-1).q = w(m) - l(m)$ proving the uniqueness of q and then of p.

If both are in case 2. let $m:= x_1^px_jx_k^q$ and $m' = x_1^{p'}x_{j'}x_k^{q'}$ then we have

$$\begin{aligned} l(m)&= p + 1 + q = l(m') = p' + 1 + q' \quad \textrm{and} \\ w(m)&= p + j + kq = w(m') = p' + j' + kq' \quad \mathrm{which \ imply} \\ j- j'&= (k-1)(q' - q) \quad \textrm{with} \quad \vert j - j'\vert \in [0, k-3].\\ \textrm{So} \quad j&= j' \quad \mathrm{and \ then} \quad q = q' \quad \textrm{and} \quad p = p'. \end{aligned}$$

If $m = x_1^px_jx_k^q$ and $m' = x_1^{p'}x_k^{q'}$ we have

$$\begin{aligned} (m)&= p + 1 + q = l(m') = l(m') = p' + q' \quad \textrm{and} \\ w(m)&= p + j + kq = w(m') = p' + kq' \quad \mathrm{which \ imply} \\ j - 1&= (k-1)(q' - q) \quad \textrm{with} \quad j \in [2, k-1] \quad \mathrm{and \ this \ is\ impossible}. \end{aligned}$$

The assertion of existence is clear for $\vert \alpha \vert = 0, 1$. We shall prove the existence of $\mu (\alpha )$ by an induction on the length $\vert \alpha \vert$ of $\alpha$.

Assume that the lemma is proved for all $\beta \in \mathbb {N}^k$ with length $1 \le \vert \beta \vert < \vert \alpha \vert$. Then, write $x^\alpha = x_rx^\beta$ for some $r \in [1, k]$. By the induction hypothesis we know that there exists a minimal monomial $x^{\mu (\beta )}$ with $x^\beta \sharp x^{\mu (\beta )}$. Then, we obtain that $x^\alpha \sharp x_rx^{\mu (\beta )}$. If $x^{\mu (\beta )} = x_1^px_k^q$, then $x_rx_1^px_k^q$ is minimal for any choice of $r \in [1, k]$. If $x^{\mu (\beta )} = x_1^px_jx_k^q$ then remark that we have $x_rx_j \sharp x_1x_{r+j-1}$ for $r+j-1 \le k$ and $x_rx_j \sharp x_kx_{r+j-k}$ for $r+j \ge k+1$ and this allows to conclude the induction.

Remark that if, in the induction above, we assume that $x^\beta - x^{\mu (\beta )}$ belongs to IS(k) we obtain that $x^\alpha - x^{\mu (\alpha )}$ is also in IS(k); for instance in the case $x^{\mu (\beta )} = x_1^px_jx_k^q$

$$\begin{aligned} x^\alpha - x^{\mu (\alpha )}&= x_r(x^\beta - x^{\mu (\beta )}) + (x_rx_j - x_1x_{r+j-1})x_1^px_k^q \quad \textrm{for} \quad r+j \le k+1, \\ x^\alpha - x^{\mu (\alpha )}&= x_r(x^\beta - x^{\mu (\beta )}) + (x_rx_j - x_kx_{r+j-k})x_1^px_k^q \quad \textrm{for} \quad r+j \ge k+2. \end{aligned}$$

The other cases are analogous.$\blacksquare$

Proof of the proposition 6.1.1 The previous lemma gives that $x^\alpha \sharp x^\beta$ implies $x^\alpha - x^{\mu (\alpha )}$ and $x^\beta - x^{\mu (\alpha )}$ are in IS(k), so also $x^\alpha - x^\beta$. Conversely, assume that $x^\alpha - x^\beta$ is in IS(k). As the ideal IS(k) is homogeneous (in the sense of length) if $l(\alpha ) \ne l(\beta )$ we conclude that both $x^\alpha$ and $x^\beta$ are in IS(k). This contradicts the remark following Definition 6.1.2.

In a similar way the ideal IS(k) is quasi-homogeneous in the sense of the weight w. Therefore, if $w(\alpha ) \not = w(\beta )$ then $x^\alpha$ and $x^\beta$ are in IS(k) which is again impossible. Therefore, $x^\alpha - x^\beta$ is in IS(k) implies that $\alpha \sharp \beta$.$\blacksquare$

Corollary 6.1.4

For any $q \in \mathbb {N}$ and any $r \in [q, kq]$ there exists a minimal monomial $\mu _{q, r}$ (necessarily unique), such that $\vert \mu _{q, r}\vert = q$ and $w(\mu _{q, r}) = r$.

Proof

The assertion is clear for $q = 0, 1$. Therefore, let us prove it by induction on q. Therefore, let $q \ge 2$ and let $r \in [q, kq]$, and assume that we know that $\mu _{q', r'}$ exists for any $q' \le q-1$ and any $r' \in [q', kq']$. If r is in $[q, k(q-1)+ 1]$, then $r-1$ is in $[q-1, k(q-1)]$ and $\mu _{q-1, r-1}$ exists. Therefore, $\mu _{q, r}:= x_1m_{q-1, r-1}$ is the solution.

If r is in $[k(q-1)+1, kq]$ then $r-k$ is in $[q-1, k(q-1)]$and, because for $q \ge 2$ we have $kq -2k +1 \ge q-1$ and also $r-k \le kq-k \le k(q-1)$, $\mu _{q-1, r-k}$ is defined and $\mu _{q, r}:= x_k\mu _{q-1, r-k}$ is the solution.$\blacksquare$

Proposition 6.1.5

Let $L_1:= \{\eta _1 = 0\} \cap S(k)$ and $L_k:= \{\eta _k = 0 \} \cap S(k)$. Then, $L_1$ is the line directed by the vector $(0, \dots , 0, 1)$ and $L_k$ the line directed by the vector $(1, 0, \dots , 0)$. The maps $\varphi _1: S(k){\setminus } L_1 \rightarrow \mathbb {C}^*\times \mathbb {C}$ and $\varphi _k: S(k) {\setminus } L_k \rightarrow \mathbb {C}^*\times \mathbb {C}$ which are defined by the formulas

$$\begin{aligned} \varphi _1(\eta ):= (\eta _1, -\eta _2/\eta _1) \quad \textrm{and} \quad \varphi _k(\eta ):= (\eta _k, -\eta _{k-1}/\eta _k) \end{aligned}$$

(33)

are isomorphisms. Therefore, $S(k)\setminus \{0\}$ is smooth and connected surface.

Proof of the proposition 6.1.5 Consider the holomorphic map

$$\begin{aligned} \psi _1: \mathbb {C}^2 \rightarrow S(k) \quad (\zeta _0, \zeta _1) \mapsto x_h:= \zeta _0(-\zeta _1)^{h-1} \ \forall h \in [1,k]. \end{aligned}$$

It induces the inverse to the map $\varphi _1$ on $\zeta _0 \not = 0$ and the map $\psi _k$ defined by

$x_h = (-\zeta _0)^{k-h}.\zeta _1 \quad \forall h \in [1, k]$ gives the inverse of $\varphi _k$ on $\zeta _1 \not = 0$.$\blacksquare$

Corollary 6.1.6

The ideal IS(k) of $\mathbb {C}[x]$ is prime. Moreover, $(S_k)$ is a normal surface.

Proof of the corollary 2.1.2 The only point which is not a direct consequence of the previous proposition is the normality of S(k). However, as the blow-up of the maximal ideal at the origin in S(k) gives a desingularization of S(k) with the rational curve¹³ over the origin in S(k). Therefore, is a rational singular point and S(k) is normal.$\blacksquare$