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On Lisbon integrals
Daniel Barlet, Teresa Monteiro Fernandes
To cite this version:
Daniel Barlet, Teresa Monteiro Fernandes. On Lisbon integrals. Mathematische Zeitschrift, Springer,
2020, �10.1007/s00209-020-02540-6�. �hal-02161740�
ON LISBON INTEGRALS
DANIEL BARLET AND TERESA MONTEIRO FERNANDES
Abstract. We introduce a family of integral transforms, the Lisbon Integrals, which naturally arise in the study of the affine space of unitary polynomials P s (z) where s ∈ C k and z ∈ C , s i identified to the i-th symmetric function of the roots of P s (z). We completely determine the D -module (or system of partial differential equations) the Lisbon Integrals satisfy and prove that they are their unique global solutions of this D -module.
Contents
1. Introduction 2
2. Lisbon Integrals and the differential system they satisfy 2
2.a. Lisbon integrals 2
2.b. The partial differential system 4
2.c. Example: The cases k = 2, 3 8
3. The left action of Γ( C , D C ) on Lisbon integrals 11 4. The D C
k-module associated to Lisbon Integrals 12
4.a. An example 18
References 20
Date: 19/06/19.
2010 Mathematics Subject Classification. 44A99, 32C35, 35A22, 35 A27, 58J15.
1
1. Introduction
The purpose of this article is to show that the completely explicit differential system satisfied by the “Lisbon integrals” (defined in formulas (3) and (4) below) is in fact a D -module obtained from a very simple one by the usual functorial operations on D -modules (inverse image and direct image).
More precisely, the D -module approach shows that the Lisbon integrals are holo- morphic solutions of modules obtained as integral transforms associated to the affine space of unitary polynomials of degree k in the variable z ∈ C with com- plex coefficients. Writing its elements in the form P (s, z ) = P k
i=0 (−1) i s i z k−i with s = (s 1 , · · · , s k ) ∈ C k with the convention s 0 = 1, this affine space (see [2, Ch. I]) is naturally identified to the hypersurface H of C k+1 endowed with the coordinates (s, z) of equation s k = (−1) k−1 P k
i=1 (−1) i s i z k−i . Let j : H ⊂ C k+1 denote the closed embedding and let B H| C
k+1denote the regular holonomic D C
k+1-module of holomorphic distributions supported by H. Since the restriction of the projection
π : C k+1 −→ C k , (s, z) 7→ s
to H is proper (with finite fibers), given a D C
k+1-module L non characteristic for H, following [8] we obtain a complex in D b coh ( D C
k), the integral transformed of L , given by the composition of the usual derived functors of direct image and inverse image for D -modules,
Dπ ∗ (B H| C
k+1⊗ L O
Ck+1
L ) ' Dπ ∗ ( H 1 [H] ( L )) ' π ∗ j ∗ L
Note that B H| C
k+1= H 1 [H] ( O C
k+1) and that for such a module L , we have, thanks to Kashiwara’s equivalence theorem (cf Th 4.1 and Prop 4.2, [4]), H 1 [H] ( L ) ' H 0 Dj ∗ Dj ∗ L ' j ∗ j ∗ L . In this note L is the quotient of D C
k+1by the ideal gen- erated by the partial derivatives in s i , i = 1, · · · , k, hence the sheaf of solutions of L is p −1 O C where p(s, z ) = z.
As a consequence, we show that the Lisbon Integrals are exactly the global solutions of π ∗ j ∗ L .
These computations illustrate the fact that it is not so easy, even in a rather simple situation, to follow explicitly the computations hidden in the “yoga” of D -module theory.
2. Lisbon Integrals and the differential system they satisfy 2.a. Lisbon integrals. For (z 1 , . . . , z k ) ∈ C k denote s 1 , . . . , s k the elementary sym- metric functions of z 1 , . . . , z k . We shall consider in the sequel s 1 , . . . , s k as coor- dinates on C k ' C k
S k , isomorphism given by the standard symmetric function theorem.
We shall denote P s (z) := Q k
j=1 (z − z j ) = P k
h=0 (−1) h s h z k−h with the convention s 0 ≡ 1.
We shall often write P (s, z) instead of P s (z) with no risk of ambiguity.
The discriminant ∆(s) of P s is the polynomial in s corresponding to the symmetric polynomial Q
16i<j6k (z i − z j ) 2 via the symmetric function theorem.
Lemma 2.1. For h ∈ N and f ∈ O ( C ) any entire holomorphic function, let us define, for R ||s||,
(1) ϕ h (s) := 1
2iπ Z
|ζ|=R
f(ζ)ζ h dζ P s (ζ) .
Then ϕ h (s) is independent of the choice of R large enough and defines a holomorphic function on C k . For ∆(s) 6= 0 we have
(2) ϕ h (s) =
k
X
j=1
z j h f (z j ) P s 0 (z j ) where z 1 , . . . , z k are the roots of P s (z).
Proof. The independence on R large enough when s stays in a compact set of C n is clear. For s in the interior of a compact set, P s (ζ) does not vanish on {|ζ| = R}
for R large enough, so we obtain the holomorphy of ϕ h near any point in C k . The formula (2) is given by a direct application of the Residue’s theorem.
In fact, it will be convenient to consider the k functions ϕ 0 , . . . , ϕ k−1 as the com- ponent of a vector valued function Φ :=
ϕ 0 ϕ 1 . . . ϕ k−1
. Defining E(z) :=
1 z . . . z k−1
we obtain
(3) Φ(s) = 1
2iπ Z
|ζ|=R
f(ζ)E(ζ)dζ P s (ζ) .
Definition 2.2. We call Φ (sometimes also denoted by Φ f when precision is re- quired) the Lisbon Integral associated to f
It will be also interesting to introduce another type of Lisbon integrals:
(4) Ψ(s) := 1
2iπ Z
|ζ|=R
f (ζ)E(ζ).P s 0 (ζ)dζ P s (ζ)
Ψ will also be noted below by Ψ f when precision is required.
It is easy to see that this is again a vector valued holomorphic function on C k and the Residue’s theorem gives that, for ∆(s) 6= 0, the component ψ h of Ψ is given by:
(5) ψ h (s) =
k
X
j=1
z h j f(z j ).
Proposition 2.3. If f is not identically zero then Φ and Ψ are non zero vector- valued holomorphic functions on C k .
Proof. Suppose that f is non identically zero. Then the statement follows as an
immediate consequence of the non vanishing of the Van der Monde determinant of
z 1 , . . . , z k when these complex numbers are pair-wise distinct.
An example. Take f ≡ 1. Then formula (5) shows that ψ h (s) is the h−th Newton symmetric functions of the roots of the polynomial P s . So it is a quasi-homogeneous polynomial in s of weight h (the weight of s j is j by definition).
Let us show that we have ϕ h (s) ≡ 0 for h ∈ [0, k − 2] and ϕ k−1 (s) ≡ 1 in this case.
For h ∈ [0, k − 2] the formula (1) gives the estimate (with f ≡ 1)
|ϕ h | 6 R h+1 (R − a) k−1
if each root of P s is in the disc {|z| 6 a} when R > a > 0. When R → +∞ this gives ϕ h (s) ≡ 0 for h ∈ [0, k − 2].
For h = k − 1 write
kz k−1 = P s 0 (z) −
k−1
X
h=1
(−1) h k(k − h)s h z k−h−1 . This gives, using the previous case and formula (2), that ϕ k−1 (s) ≡ 1.
2.b. The partial differential system. Let us introduce the (k, k) matrix A asso- ciated to the polynomial P s by
(6) A :=
0 1 0 · · · 0
0 0 1 0 · · · 0
0 · · · 0
0 · · · 0 1
(−1) k−1 .s k · · · (−1) h−1 .s h · · · · s 1
Theorem 2.b.1. The vector value holomorphic function Φ on C k satisfies the fol- lowing differential system
(@) (−1) k+h . ∂Φ
∂s h
(s) = ∂(A k−h Φ)
∂s k
(s) ∀ s ∈ C k and ∀ h ∈ [1, k − 1].
Moreover, this system is integrable 1 and if Φ is a solution of this system, so is AΦ.
The proof of this result will use several lemmas.
Lemma 2.4. Let A be a (k, k) matrix with entries in C [x] and put B := λ ∂A ∂x where λ is a complex number. Let M be the (2k, 2k) matrix given by
M :=
A B 0 A
. Then for each p ∈ N we have
(*) M p =
A p B p 0 A p
1 We shall explain in the proof what we mean here.
where B p := λ ∂(A ∂x
p) .
Proof. As the relation ( ∗ ) is clear for p = 0, 1 let us assume that it has been proved for p and let us prove it for p + 1. We have:
A p B p 0 A p
A B 0 A
=
A p+1 A p B + B p A 0 A p+1
which allows to conclude.
Corollary 2.b.2. For each integer p ∈ [0, k − 1] the following equality holds in the module C k ⊗ C C [s 1 , . . . , s k , z]
(P 2 ) over the C −algebra C [s 1 , . . . , s k , z]
(P 2 ) (7) z p E(z) = A p E (z) + (−1) k−1 P s (z) ∂ (A p )
∂s k E(z) In particular for any entire function on z,
Φ zf = A(s)Φ f
Moreover the following identity in the module C k ⊗ C C [s 1 , . . . , s k , z]
(P 2 ) holds (7-bis) P s 0 (z)E(z) = P s 0 (A)E(z) + (−1) k−1 P s (z) ∂(P s 0 (A))
∂s k
E(z).
Proof. In the basis 1, z, · · · , z k−1 , P s (z), zP s (z), · · · , z k−1 P s (z) of this algebra which is a free rank 2k module on C [s 1 , . . . , s k ], the multiplication by z is given by the matrix M of the previous lemma with A as in (6) and with B := (−1) k−1 ∂s ∂A
k
. This proves formula (7).
As P s 0 (z) = P k−1
h=0 (−1) h (k − h)z k−h−1 does not depend on s k it is enough to sum up the previous equalities with p = k − h − 1 with the convenient coefficients to obtain
the equality (7 − bis).
Lemma 2.5. Let A be a (k, k) matrix with entries in C [x, y]. Assume that for some integer q we have
∂A
∂x A q = ∂A
∂y then we have for any integer h
∂A h
∂x A q = ∂A h
∂y .
Proof. The assertion is our hypothesis for h = 1. Assume the assertion proved for some h > 1. Then Leibnitz rule gives:
∂A h+1
∂x .A q = ∂A h
∂x A q+1 + A h ∂A
∂x A q which implies:
∂A h+1
∂x A q = ( ∂A h
∂y A + A h ∂A
∂y ) = ∂A h+1
∂y
which is our assertion for h + 1.
Corollary 2.b.3. For any h ∈ [1, k] and any p ∈ N the matrix A in (6) satisfies the relation:
(8) (−1) k−h ∂A p
∂s h = ∂A p
∂s k A k−h
Proof. The case p = 1 of (7) is an easy direct computation on the matrix A.
Assume that the assertion is proved for p > 1. Then Leibnitz’s rule gives:
(−1) k−h ∂A p+1
∂s h
= (−1) k−h ∂A p
∂s h
A + A p (−1) k−h ∂A
∂s h
= ∂A p
∂s k A k−h+1 + A p ∂A
∂s k A k−h = ∂A p+1
∂s k A k−h
concluding the proof of (8).
Proof of the theorem 2.b.1 . By derivation inside the integral in (3) we obtain:
∂Φ
∂s h (s) = 1 2iπ
Z
|ζ|=R
f (ζ)E(ζ)(−1) h+1 ζ k−h dζ
P s (ζ) 2 and in particular
∂Φ
∂s k (s) = 1 2iπ
Z
|ζ|=R
f (ζ )E(ζ)(−1) k+1 dζ P s (ζ) 2
Now for h ∈ [1, k − 1] we use the formula of corollary 2.b.2 to obtain:
∂Φ
∂s h = (−1) h+1 A k−h (−1) k−1 ∂ Φ
∂s k + (−1) h+1 (−1) k−1 ∂A k−h
∂s k Φ that is to say:
(@) (−1) k+h ∂Φ
∂s h = ∂(A k−h Φ)
∂s k ∀ h ∈ [1, k]
By the integrability of the system (@) we mean that for any Φ such that (@) holds, then the computation of the partial derivatives ∂s ∂
2Φ
h
∂s
jusing the system (@) gives a symmetric result in (h, j ) for any pair (h, j) in [1, k]. Note that if h or j is equal to k the assertion is trivial.
So consider a couple (h, j) ∈ [1, k − 1] 2 . Thanks to corollary 2.b.3 we have : (−1) h+j . ∂ 2 Φ
∂s j ∂s h = (−1) k−j ∂
∂s k
∂(A k−h Φ)
∂s j
= (−1) k−j ∂
∂s k
∂A k−h
∂s j Φ + A k−h ∂ Φ
∂s j
= ∂
∂s k
∂A k−h
∂s j A k−j Φ + A k−h ∂(A k−j Φ)
∂s k
= ∂ 2
∂s 2 k
A 2k−h−j Φ which is symmetric in (h, j).
To finish the proof of the theorem we have to show that AΦ is a solution of (@)
when Φ is a solution of (@). This is given by the following computation
(−1) k−h ∂(AΦ)
∂s h = (−1) k−h ∂A
∂s h Φ + (−1) k−h A ∂Φ
∂s h
= ∂A
∂s k .A k−h Φ + A ∂A k−h Φ
∂s k = ∂(A k−h (AΦ))
∂s k
which also uses corollary 2.b.3.
Remark. A consequence of our computation on the integrability of the system (@) is the fact that for any solution Φ and any pair (h, j ) ∈ [1, k] the second order partial derivative ∂s ∂
2Φ
j
∂s
honly depends on h + j . This implies that any scalar Lisbon integral ϕ h satisfies
(9) ∂ 2 ϕ h
∂s p ∂s q+1
= ∂ 2 ϕ h
∂s p+1 ∂s q
∀ p, q such that 1 6 p < q 6 k − 1
The following corollary of theorem 2.b.1 describes the analogous system for the the vector functions Ψ defined in (4) which is singular along the discriminant subset {∆(s) = 0} in C k .
Corollary 2.b.4. The vector value holomorphic function Ψ on C k satisfies the fol- lowing differential system:
(@@) (−1) k+h ∂ Ψ
∂s h (s) = ∂(A k−h Ψ)
∂s k (s) + (−1) k (k − h)A k−h−1 P s 0 (A) −1 Ψ(s) for all h ∈ [1, k − 1], which is singular along the discriminant sub-set
∆ := {s ∈ C k / ∆(s) = 0}.
Again, if Ψ is any solution of (@@) then AΨ is also a solution of (@@).
Proof. Define F (z) := P 0 (z)E(z). Then using formula (7) we obtain Ψ(s) = P s 0 (A)Φ(s) and this gives for each h ∈ [1, k − 1]
(−1) k−h ∂Ψ
∂s h
(s) = (−1) k−h A k−h−1 Φ(s) +
k−1
X
p=0
(−1) p (k − p)(−1) k−h s p . ∂(A k−p−1 Φ)
∂s h
(s) and using now the fact that A k−p−1 Φ is solution of (@) we obtain
(−1) k−h ∂Ψ
∂s h (s) = (−1) k−h A k−h−1 Φ(s) +
k−1
X
p=0
(−1) p (k − p)s p . ∂(A 2k−p−h−1 .Φ)
∂s k (s) (−1) k−h ∂Ψ
∂s h (s) = (−1) k−h A k−h−1 Φ(s) + ∂ (A k−h P s 0 (A)Φ)
∂s k (s) (−1) k−h ∂Ψ
∂s h
(s) = (−1) k−h A k−h−1 P s 0 (A) −1 Ψ(s) + ∂(A k−h Ψ)
∂s k
(s)
Since P s 0 (A) commutes with A, the last assertion is easy.
2.c. Example: The cases k = 2, 3. The left ideal of partial differential operators in the Weyl algebra C [s 1 , . . . , s k ]h∂ s
1, . . . , ∂ s
ki which anihilate Lisbon integrals is described in the article [1].
Below we will explicit the cases k = 2 and k = 3.
Case k = 2:
Here we use the notations s := s 1 and p := s 2 . In that case, the differential system (@) becomes:
− ∂ Φ 0
∂s = ∂Φ 1 (10) ∂p
− ∂ Φ 1
∂s = −Φ 0 − p ∂Φ 0
∂p + s ∂Φ 1 (11) ∂p
corresponding to the matrix A :=
0 1
−p s
. Differentiating (9) with respect to p we obtain, after substituting via (10)
∂ 2 Φ 0
∂s 2 = −2 ∂Φ 0
∂p − p ∂ 2 Φ 0
∂p 2 − s ∂ 2 Φ 0
∂s∂p and so ∂ 2 Φ 0
∂s 2 + s ∂ 2 Φ 0
∂s∂p + p ∂ 2 Φ 0
∂p 2 + 2 ∂Φ 0
∂p = 0 (])
Differentiating (10) with respect to s we obtain, after substituting via (9)
− ∂ 2 Φ 1
∂s 2 = ∂Φ 1
∂p + p ∂ 2 Φ 1
∂p 2 + ∂Φ 1
∂p + s ∂ 2 Φ 1
∂s∂p and so ∂ 2 Φ 1
∂s 2 + s ∂ 2 Φ 1
∂s∂p + p ∂ 2 Φ 1
∂p 2 + 2 ∂Φ 1
∂p = 0 (]])
Then the second order differential operator of weight −2
(]]]) Θ := ∂ 2
∂s 2 + s ∂ 2
∂s∂p + p ∂ 2
∂p 2 + 2 ∂
∂p anihilates Φ 0 and Φ 1 for any solution Φ of the system (10), (11).
A direct proof that Θ anihilates ϕ m for all m ∈ N. We have, for any entire function f : C → C and for R max{|s|, |p|}:
(a) ϕ m (s, p) = 1
2iπ Z
|ζ|=R
f(ζ) ζ m dζ
ζ 2 − sζ + p
This gives:
∂ϕ m
∂s (s, p) = 1 2iπ
Z
|ζ|=R
f (ζ) ζ m+1 dζ (ζ 2 − sζ + p) 2 (b)
∂ϕ m
∂p (s, p) = − 1 2iπ
Z
|ζ|=R
f (ζ) ζ m dζ (ζ 2 − sζ + p) 2 (c)
∂ 2 ϕ m
∂s 2 (s, p) = 2 1 2iπ
Z
|ζ|=R
f (ζ) ζ m+2 dζ (ζ 2 − sζ + p) 3 (d)
∂ 2 ϕ m
∂s∂p (s, p) = −2 1 2iπ
Z
|ζ|=R
f (ζ) ζ m+1 dζ (ζ 2 − sζ + p) 3 (e)
∂ 2 ϕ m
∂p 2 (s, p) = 2 1 2iπ
Z
|ζ|=R
f (ζ) ζ m dζ (ζ 2 − sζ + p) 3 (f)
Now it is easy to check that (d) + s(e) + p(f ) + 2(c) = 0.
Case k = 3:
The differential system system (@) becomes now:
(@) (−1) k−h ∂ h Φ = ∂ k (A k−h Φ) for h ∈ [1, k − 1]
with k = 3. Here Φ =
Φ 0 Φ 1 Φ 2
and A =
0 1 0 0 0 1 s 3 −s 2 s 1
. Explicitely this gives:
∂ 1 Φ 0 = ∂ 3 Φ 2
∂ 1 Φ 1 = ∂ 3 s 3 .Φ 0 − s 2 Φ 1 + s 1 Φ 2
∂ 1 Φ 2 = ∂ 3 s 1 s 3 Φ 0 − (s 3 − s 1 s 2 )Φ 1 + (s 2 1 − s 2 )Φ 2
− ∂ 2 Φ 0 = ∂ 3 Φ 1
− ∂ 2 Φ 1 = ∂ 3 Φ 2
− ∂ 2 Φ 2 = ∂ 3 s 3 Φ 0 − s 2 Φ 1 + s 1 Φ 2 which implies the equalities
∂ 1 Φ 0 = −∂ 2 Φ 1 = ∂ 3 Φ 2 ∂ 1 Φ 1 = −∂ 2 Φ 2 Then we deduce that:
(∂ 2 2 − ∂ 1 ∂ 3 )(Φ j ) = 0 for j = 0, 1, 2 because
∂ 2 2 Φ 0 = −∂ 2 ∂ 3 Φ 1 = ∂ 3 ∂ 3 Φ 2 = ∂ 3 ∂ 1 Φ 0 ,
∂ 1 ∂ 3 Φ 1 = −∂ 1 ∂ 2 Φ 0 = ∂ 2 ∂ 3 Φ 2 = ∂ 2 2 Φ 1 ,
and ∂ 2 2 Φ 2 = −∂ 2 ∂ 1 Φ 1 = ∂ 3 ∂ 2 Φ 2 .
Lemma 2.6. Let us define:
L := ∂ 1 2 + s 1 ∂ 1 ∂ 2 + s 2 ∂ 2 2 + s 3 ∂ 2 ∂ 3 + 2∂ 2 L 0 := ∂ 1 ∂ 2 + s 1 ∂ 2 2 + s 2 ∂ 2 ∂ 3 + s 3 ∂ 3 2 + 2∂ 3 Then we have:
L Φ 0 = 0 L 0 Φ 1 = 0 et L Φ 2 = 0 and also:
(r) ∂ 3 L − ∂ 2 L 0 ∈ D C
k(∂ 1 ∂ 3 − ∂ 2 2 ) This entails that ∂ 3 L and ∂ 2 L 0 anihilate Φ j for j = 0, 1, 2.
Proof. We have
∂ 1 Φ 1 = Φ 0 + s 3 .∂ 3 Φ 0 − s 2 .∂ 3 Φ 1 + s 1 .∂ 3 Φ 2
= Φ 0 + s 3 .∂ 3 Φ 0 + s 2 .∂ 2 Φ 0 + s 1 .∂ 1 Φ 0 and so ∂ 2 ∂ 1 Φ 1 = −∂ 1 2 Φ 0 =
∂ 2 Φ 0 + s 3 .∂ 2 ∂ 3 Φ 0 + ∂ 2 Φ 0 + s 2 .∂ 2 2 Φ 0 + s 1 .∂ 1 ∂ 2 Φ 0
and so Φ 0 is killed by L as (∂ 1 ∂ 3 − ∂ 2 2 )Φ 0 = 0.
In a similar way:
∂ 1 Φ 1 = −∂ 2 Φ 2 = ∂ 3 s 3 .Φ 0 − s 2 .Φ 1 + s 1 .Φ 2
∂ 2 ∂ 1 Φ 1 = −∂ 3 (s 3 .∂ 3 Φ 1 ) − ∂ 2 ∂ 3 (s 2 .Φ 1 ) − s 1 .∂ 2 2 Φ 1
= −∂ 3 Φ 1 − s 3 .∂ 3 2 Φ 1 − ∂ 3 Φ 1 − s 2 .∂ 2 ∂ 3 Φ 1 − s 1 .∂ 2 2 Φ 1 showing that L 0 anihilates Φ 1 .
In a similar way:
∂ 1 Φ 2 = s 1 .Φ 0 + s 1 .s 3 ∂ 3 Φ 0 + Φ 1 + (s 3 − s 1 .s 2 ).∂ 3 Φ 1 + (s 2 1 − s 2 )∂ 3 Φ 2
= s 1 .∂ 1 Φ 1 + Φ 1 − s 2 .∂ 3 .Φ 2 + s 3 .∂ 3 Φ 1
= s 1 .∂ 1 Φ 1 + Φ 1 + s 2 .∂ 2 .Φ 1 + s 3 .∂ 3 Φ 1
∂ 1 2 Φ 2 = −∂ 2 Φ 2 − s 1 .∂ 1 ∂ 2 Φ 2 − ∂ 2 Φ 2 + s 2 .∂ 2 (−∂ 2 Φ 2 ) + s 3 .∂ 3 .(∂ 1 Φ 1 )
= −2∂ 2 Φ 2 − s 1 .∂ 1 ∂ 2 Φ 2 − s 2 .∂ 2 2 Φ 2 − s 3 .∂ 2 ∂ 3 Φ 2 and Φ 2 is anihilated by L .
Let us prove now the relation (r):
∂ 3 L = ∂ 3 ∂ 2 1 + s 1 .∂ 1 ∂ 2 ∂ 3 + s 2 .∂ 1 ∂ 3 2 + s 3 .∂ 2 ∂ 3 2 + 3∂ 2 ∂ 3
∂ 2 L 0 = ∂ 1 ∂ 2 2 + s 3 .∂ 2 .∂ 3 2 + s 2 .∂ 2 2 ∂ 3 + s 1 .∂ 2 3 + 3∂ 2 ∂ 3 this allows to conclude, as
∂ 3 ∂ 1 2 − ∂ 1 ∂ 2 2 = ∂ 1 (∂ 1 ∂ 3 − ∂ 2 2 )
∂ 1 ∂ 2 ∂ 3 − ∂ 2 3 = ∂ 2 (∂ 1 ∂ 3 − ∂ 2 2 )
For the study of differential operators anihilating all the scalar components of the
Lisbon integrals (for any value of k) see [1].
3. The left action of Γ( C , D C ) on Lisbon integrals
For each entire function f of the variable z we shall henceforward denote by Φ f the associated Lisbon integral (previously generically denoted by Φ). Clearly the assignement
f 7→ Φ f
is C -linear and, according to Proposition 2.3, it is injective.
Lemma 3.1. Let f be an entire function on C and let Φ f the corresponding Lisbon integral.
(1) Let h be an entire function of z. Then Φ hf = h(A(s))Φ f . In particular Φ f = f (A(s))Φ 1 .
(2) We have the identity
Φ ∂
z(f ) (s) = −∇Φ f (s) +
k−1
X
h=0
(k − h)s h ∂ s
h+1(Φ f )(s)
where ∇ is the constant (k, k) matrix
0 0 0 . . . 0 1 0 0 . . . 0 0 2 0 . . . 0 0 0 . . . 0 0 . . . 0 k − 1 0
.
Proof. When h is a polynomial on z, statement (1) follows easily in Corollary 2.b.2.
For an arbitrary entire function h, it is a consequence of [2, Lem 3.1.8].
Let us prove (2): Consider the Lisbon integral Φ ∂
zf (s) = 1
2iπ Z
|ζ|=R
f 0 (ζ)E(ζ) dζ P s (ζ) Integration by parts gives, as ∂ z (E)(z) = ∇E(z) :
(*) Φ ∂
z(f ) (s) = −∇φ f (s) + 1 2iπ
Z
|ζ|=R
f(ζ)E(ζ) P s 0 (ζ) P s (ζ) 2 dζ Now, using the equalities P s 0 (ζ ) = P k−1
h=0 (−1) h (k − h)s h ζ k−h−1 and
∂Φ f
∂s h (s) = − Z
|ζ|=R
f(ζ)E(ζ) (−1) h ζ k−h dζ P s (ζ) 2
we obtain the formula of the lemma.
Remark. From the formula (∗) and using the fact that P s 0 (ζ)E(ζ) = P s 0 (A)E(ζ) we obtain also the formula
(**) Φ ∂
z(f) (s) = −∇Φ f (s) + (−1) k ∂(P s 0 (A)Φ f )
∂ s
k(s)
It is not obvious that when Φ is solution of the system (@), then (Φ∂ z )(s) :=
−∇Φ(s) + (−1) k ∂ (P
s∂
0(A)Φ)
sk
(s) (given by the formula ( ∗∗ )) is also a solution of the
same system. A direct verification of this fact is consequence of the formula given
in the lemma below.
In order to verify that we have Φ (z.f)
0= Φ f +Φ zf
0corresponding to the usual relation
∂ z z − z∂ z = 1 the following lemma is useful.
Lemma 3.2. We have the following identity:
∇A − A∇ + Id = (−1) k−1 ∂ k (A)P s 0 (A), ∀ s ∈ C k .
Proof. It is an easy computation to obtain that ∇A − A∇ + Id is the matrix which have all lines equal to 0 excepted its last one which is given by (x 1 , . . . , x k ) with x h = (−1) k−h hs k−h for h ∈ [1, k] with s 0 ≡ 0. On the other hand, the matrix (−1) k−1 ∂ k A has only a non zero term at the place (k, 1) which equal to 1, so it is quite easy to see that (−1) k−1 (∂ k A)A p has only a non zero term at the place (k, p + 1) with value (−1) k−1 . According to the computation of (−1) k−1 ∂ k (A)P s 0 (A)
we conclude the desired formula.
4. The D C
k-module associated to Lisbon Integrals
We shall begin by recalling some basic facts on D -modules and fixing our notations.
For a morphism of manifolds f : Y → X, we use the notation of [7]
f d := t f 0 : T ∗ X × X ×Y −→ T ∗ Y and
f π : T ∗ X × X ×Y −→ T ∗ X the associated canonical morphisms of vector bundles.
We recall that a conic involutive submanifold V of the cotangent bundle T ∗ Z of a manifold Z (real or complex) is regular if the restriction ω| V of the canoni- cal 1-form ω on T ∗ Z never vanishes outside the 0−section. Recall also that if (x 1 , · · · , x n , ξ 1 , · · · , ξ n ) are canonical symplectic coordinates on T ∗ Z, then ω(x, ξ) = P n
i=1 ξ i dx i .
Let us fix some k ∈ N , k > 2. In C k+1 = C k × C we consider the coordinates (s 1 , · · · , s k , z). As in the previous sections we set
P (s, z) = z k +
k−1
X
h=1
(−1) h s h z k−h Obviously
P (s 1 , · · · , s k , z) = 0 ⇐⇒ s k = (−1) k−1
k−1
X
h=0
(−1) h s h z k−h , where s 0 = 1. We note s = (s 1 , · · · , s k ) and s 0 := (s 1 , · · · , s k−1 ).
Let H be the smooth hypersurface of C k+1 given by the zeros of P (s, z).
Let us denote by L the D C
k+1-module with one generator u defined by the equations
∂u/∂s 1 = · · · = ∂u/∂s k = 0. Such module is an example of a so called partial de Rham systems, which have the feature, among others, that their characteristic varieties are non singular regular involutive. In our case we have
Char L = {(s, z); (η, τ ) ∈ C k+1 × C k+1 such that η = 0}.
Since H ⊂ C k+1 is defined by the equation P (s, z) = (−1) k s k +
k−1
X
h=0
(−1) h s h z k−h = 0, T H ∗ C k+1 is the subbundle of T ∗ C k+1 | H described by
{(s, z ; η, τ ), (s, z) ∈ H, ∃ λ ∈ C such that (η, τ ) = λdP (s, z )}.
This means that P (s, z ) = 0 and that their exists λ ∈ C with η h = λ.(−1) h .z k−h for each h ∈ [1, k] and that τ = λ.P 0 (s, z). Hence as η k = λ(−1) k and this implies:
Char L ∩ T H ∗ C k+1 ⊂ T ∗
C
k+1C k+1 (as usual, T ∗
C
k+1C k+1 denotes the zero section of T ∗ C k+1 ), in other words H is non characteristic for L . Let us denote by j : H , → C k+1 the closed embedding. By Kashiwara’s classical results (which can be found in [5]) it follows that the induced system Dj ∗ L by L on H is concentrated in degree zero and N := H 0 Dj ∗ L is a D C
H-coherent module whose characteristic variety is exactly
j d j π −1 Char( L ).
Recall that
Dj ∗ L := O H
⊗ L j
−1O
Ck+1
j −1 L and in this non-characteristic case
' j −1 O C
k+1P (s 0 , s k , z) O C
k+1⊗ j
−1O
Ck+1
j −1 L We have
N := j −1 ( D C
k+1P D C
k+1+ D C
k+1∂ s
1+ · · · + D C
k+1∂ s
k) ' j −1 ( O C
k+1/P O C
k+1) < ∂ z >
which is isomorphic as a D H -module to
O H < ∂ z >' D H
D H ∂ s
1+ · · · + D H ∂s k−1
where ∂ s
istands for the derivation ∂/∂s i on O H and ∂ z as a derivation on O H is the class of ∂ z in the quotient above.
In particular N is sub-holonomic and it is a partial de Rham system similarly to L . Let us now determine the image of N under the morphism π : C k ' H → C k given by (s 0 , z) 7→ (s 0 , s k (s 0 , z)). Clearly π is proper surjective with finite fibers.
Recall that one denotes by D C
k←H the transfer (π −1 D C
k, D H )-bimodule (π −1 D C
k⊗ π
−1O
Ck
π −1 Ω ⊗
−1C
k) ⊗ π
−1O
Ck
Ω H
Recall also that, according to the properness and the fiber finiteness of π, we have Dπ ∗ N ' H 0 Dπ ∗ N = π ∗ ( D C
k←H ⊗ D
HM )
where we abusively use the notation π ∗ for the direct image functor in the categoy of
D -modules in the two left terms and for the direct image functor for sheaves in the
right term. According to [5, Th. 4.25 and 4.27] (see also the comments in loc.cit.
before Theorem 4.27), one knows that Dπ ∗ N is concentrated in degree zero and that Char H 0 Dπ ∗ N = π π π d −1 Char N .
So we may henceforward denote for short π ∗ N := Dπ ∗ N without ambiguity.
Let ∆ as above be the zero set of the discriminant of P , which can also be defined as the image by π of the subset of H defined by {P 0 (s, z) = 0}.
Since π d is given by the k × k matrix
Id 0
∂s k /∂s 0 ∂s k /∂z T
we conclude that Char π ∗ N is the image by π π of the set
{(s 0 , z); (η 0 , τ ) ∈ C k × C k / η j = −(−z) k−j .τ ∀ j ∈ [1, k − 1]}
so it is given by the set
{(s, η) ∈ C k × C k / ∃ z ∈ C such that P s (z]) = 0 and with η j = (−z) k−j η k ∀ j ∈ [1, k−1]}
Then Char π ∗ N is an involutif analytic subset of T ∗ C k with codimension k − 1, hence π ∗ N is a subholonomic D C
k+1-module. Denoting by p the projection: T ∗ C k → C k , for any s such that ∆(s) 6= 0, the Implicit Function Theorem implies that
p −1 (s) ∩ Char π ∗ N is a cone of l distinct complex lines l j = C
(−z j ) k−1 (−z j ) k−2
· · · (−z j )
1
, where
z j runs over the set of l distinct roots of P (s, z)
Proposition 4.1. The D C
k−module π ∗ N is the quotient of D k C
k' D C
k⊗ C C k by the action of
A h := ∂ s
h⊗ Id C
k+ (−1) k−h−1 ∂ s
k⊗ A(s) k−p for j ∈ [1, k − 1].
Moreover the action of z and ∂ z on π ∗ N deduced from the action of D H on N 2 are given respectively by
A 0 := 1 ⊗ A(s) and B := 1 ⊗ ∇ + (−1) k−1 ∂ s
k⊗ P s 0 (A(s))
where we put P s 0 (z) := (∂ z (P s (z)) and ∇ :=
0 0 . . . 0 1 0 . . . 0 0 2 0 . . . . . . . 0 . . . k − 1 0
2 Note that z and ∂ z commute with ∂ s
hfor h ∈ [1, k − 1] in D H .
Proof. Our goal is to explicit π ∗ N and to check that it coincides with the D C
k- module associated to the system (@) in Theorem 2.b.1.
In a first step we explicit the transfer-module D C
k←H := π −1 D C
k⊗ π
−1O
Ck
(π −1 Ω ⊗
−1C
k⊗ π
−1O
Ck
Ω H ) as a (π −1 D C
k, D H )-bimodule.
The next step is to determine the cokernel of
α : ( D C
k←H ) k−1 −→ D C
k←H
(u 1 , · · · , u k−1 ) 7→
k−1
X
i=1
u i ∂ s
iThe last step is to apply π ∗ .
Let us denote for short σ := ω H ⊗ ω ⊗
−1C
k:= ds 1 ∧ · · · ∧ ds k−1 ∧ d z ⊗ (ds 1 ∧ · · · ∧ ds k−1 ∧ ds k ) ⊗
−1the generator of the line bundle π −1 Ω ⊗
−1C
k⊗ π
−1O
Ck
Ω H .
Recall that O H = O C
k+1/ J , where J is the ideal generated by P (s, z). Hence O H
is a π −1 O C
k-free module with rank k since each section a(s 0 , z) of O H is equivalent, by Weierstrass Division Theorem, to a unique polynomial P k−1
j=0 a j (s 0 , s k )z j modulo P (s, z), for some sections a j of O C
k.
Hence D C
k←H is a left π −1 D C
k-free module of rank k generated by the k−sections (1 ⊗ z j σ) j=0,···,k−1 . Since the right action of each operator in D H is π −1 D C
k-linear, it is sufficient to calculate each (1 ⊗ z j σ)∂ s
i, i = 1, · · · , k − 1, j = 0, · · · , k − 1.
Now recall that H is defined in C k+1 by the equation s k = (−1) k−1 P k−1
h=0 (−1) h s h z k−h with the convention s 0 = 1 and so s 1 , . . . , s k−1 , z are global coordinates on H. Then we have in H
∂s k
∂s h = (−1) k−h−1 z k−h and ∂s k
∂z = (−1) k−1 P s 0 (z) where P s 0 (z) does not depend on s k .
Let F := 1 ⊗ E(z)σ denote the basis (1 ⊗ z j σ), j ∈ [0, k − 1] of the free rank k left π −1 ( D C
k)−module D C
k←H . Recall that, according to [5, Rem. 4.18], in view of the generators described above, the action of D H in D C
k←H is defined by the following formulas, where we consider F as a k−vector and use the usual matrix product
F θ(s 0 ) = θ(s 0 )F where θ ∈ O H does not depend on z (0)
F z = A(s)F (1)
− F ∂ s
h= ∂ s
hF + (−1) k−h−1 ∂ s
k(A(s) k−h F ) ∀ h ∈ [1, k − 1]
(2)
− F ∂ z = ∇F + (−1) k−1 ∂ s
kP s 0 (A(s))F (3)
where we have used the equalities zE(z) = A(s)E(z) and ∂ z (E(z)) = ∇E(z).
Summing up:
• For g ∈ O H represented by P k−1
r=0 g r (s)z r , the (k × k) matrix G of the π −1 D C
k-linear morphism defined by g on D C
k←H is given by
G := P k−1
r=0 g r (s)A(s) r
• Let us consider the D C
k-linear morphism α : ( D k C
k) (k−1) → D k C
kdefined by the following k − 1 (k, k)−matrices
A h := ∂ s
h⊗ Id C
k+ (−1) k−h−1 ∂ s
k⊗ A(s) k−h
Let Φ be in O k C
k. In view of the relation (2), the map (1⊗z h σ) 7→ Φ h , for h ∈ [0, k−1]
will induce an element of H om D
Ck
(π ∗ N , O C
k), that is to say a solution of π ∗ N , if and only if we have ∂ s
h(Φ) = (−1) k−h ∂ s
k(A k−h Φ), since the generator of π ∗ N is anihilated by the action of ∂ s
hfor each h ∈ [1, k − 1].
In conclusion:
π ∗ N ' coker α
by the finitness of the fibers of π.
Remark 4.2. N is naturally endowed with a structure of right Γ( C ; D C )−module.
By functoriality π ∗ N is also a right Γ( C ; D C )−module and its structure coincides with the induced by the right D C action defined by (1) and (3) on D C
k←H ⊗ D
HN , since it commutes with each ∂ s
i, for i = 1, · · · , k − 1. Therefore we obtain a natural left action of Γ( C , D C ) on H om D
Ck
(π ∗ N , O C
k).
We also conclude, according to Lemma 3.1:
Proposition 4.3. The left action of Γ( C ; D C ) defined by the above Remark 4.2 on H om D
Ck
(π ∗ N , O C
k) coincides with the left action of Γ( C ; D C ) on Lisbon integrals.
If Φ is a solution of π ∗ N , replacing in the formula (3) the second term thanks to the equality obtained for φ after applying (2), we also derive a right action of ∂ z which is given by the formula
−Φ∂ z = ∇Φ +
k−1
X
h=0
(k − h)s h ∂ h+1 Φ
Our next goal is to conclude in Proposition 4.5 below that there are no global holomorphic solutions of π ∗ N other than those of the form Φ f , for some holomorphic function f only depending on z. Since j is non-characteristic we have an isomorphism
j −1 R H om D
Ck+1
( L , O C
k+1) ' R H om D
H( N , O H )
According to Theorem 4.33 (2) of [5], making X = H, Y = C k , f = π, N = O C
kin loc.cit, we obtain
Theorem 4.4. For any open subset Ω of C k we have an isomorphism functorial on N compatible with restrictions to open subsets
RΓ(π −1 (Ω); R H om D
H( N , O H )) ' RΓ(Ω; R H om D
Ck
(π ∗ N , O C
k))
Recall that this isomorphism uses as a tool the "trace morphism": π ∗ O H → O C
kconstructed in [5, Prop. 4.34].
Since for any open subset Ω and any D H -module P , Γ(Ω; ·) and H om D
H( P , ·) are left exact functors, since if Ω is a Stein open set and if P admits a global resolution by free D H -modules of finite rank, then R H om D
H( P , O ) is represented by a complex in degrees > 0 with Γ(Ω, ·)-acyclic entries, we conclude that
H 0 (RΓ(H; R H om D
H( N , O H ))) = Γ(H; H om D
H( N , O H )) and H 0 (RΓ( C k ; R H om D
Ck
(π ∗ N , O C
k))) = Γ( C k ; H om D
Ck
(π ∗ N , O C
k)) therefore Theorem 4.4 entails a C -linear isomorphism
T : Hom D
H( N , O H ) ' Hom D
Ck
(π ∗ N , O C
k) Proposition 4.5. The correspondence
f 7→ Φ(f) := Φ f defines a C -linear isomorphism
Φ : Γ( C ; O C ) −→ Γ( C k ; H om D
Ck
(π ∗ N , O C
k)) = Hom D
Ck
(π ∗ N , O C
k) Moreover, this isomorphism is also Γ( C , D C )-left linear.
Proof. The last statement is clear from Proposition 4.3 and Lemma 3.1.
The remaining of the statement is equivalent to prove that Φ : f 7→ Φ f defines an isomorphism Hom D
H( N , O H ) → Hom D
Ck
(π ∗ N , O C
k).
We already know that Φ is injective. It remains to prove that Φ is surjective. For each f ∈ Γ( C , O C ), we introduce the regular holonomic D C -module M f (a regular flat holomorphic connection on C ) of which the constant sheaf C f in degree zero is the complex of holomorphic solutions. We denote by N f the regular holonomic D H -module (a regular flat holomorphic connection on H)
N f := N M f
It is clear that N f is a quotient of N , and, by the left exactness of π ∗ , π ∗ N f is a quotient of π ∗ N . Moreover, according to Proposition 4.3 and Lemma 3.1, Φ f belongs to Hom D
Ck
(π ∗ N f , O C
k). According to Theorem 4.4, for each f we have a C -linear isomorphism T f : Hom D
H( N f , O H ) ' Hom D
Ck
(π ∗ N f , O C
k) which makes this last one a one dimensional C -vector space. Moreover, by left exactness of Hom and the exactness of π ∗ , we have monomorphisms Hom D
Ck
(π ∗ N f , O C
k)⊂ Hom D
Ck
(π ∗ N , O C
k) and, by functoriality, we have T (f) = T f (f ).
We shall use the following result:
Lemma 4.6. Suppose that f 6= 0. Then Hom D
Ck
(π ∗ N f , O C
k) is a one dimensional
C -vector space generated by Φ f .
Proof. The result follows by Proposition 2.3 since Φ f is a non zero element of Hom D
Ck
(π ∗ N f , O C
k) hence it is a generator as a C -vector space.
Let us now end the proof of Proposition 4.5. Clearly Hom D
H( N , O H ) = P
f Hom D
H( N f , O H ) and, according to Lemma 4.6, for each f , Hom D
Ck