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Symmetric tensors on the intersection of two quadrics and Lagrangian fibration

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Abstract

Let X be an n -dimensional (smooth) intersection of two quadrics, and let TX{T^{\rm{*}}}X be its cotangent bundle. We show that the algebra of symmetric tensors on X is a polynomial algebra in n variables. The corresponding map Φ:TXCn{\rm{\Phi }}:{T^{\rm{*}}}X \to {\mathbb{C}^n} is a Lagrangian fibration, which admits an explicit geometric description; its general fiber is a Zariski open subset of an abelian variety, which is a quotient of a hyperelliptic Jacobian by a 2 -torsion subgroup. In dimension 3 , Φ{\rm{\Phi }} is the Hitchin fibration of the moduli space of rank 2 bundles with fixed determinant on a curve of genus 2 .
Moduli 1, e4 (2024) 1–19
doi:10.1112/mod.2024.3
Symmetric tensors on the intersection of two
quadrics and Lagrangian fibration
Arnaud Beauville , Antoine Etesse , Andreas oring , Jie Liu and Claire Voisin
Abstract
Let Xbe an n-dimensional (smooth) intersection of two quadrics, and let TXbe its
cotangent bundle. We show that the algebra of symmetric tensors on Xis a polynomial
algebra in nvariables. The corresponding map Φ : TXCnis a Lagrangian fibration,
which admits an explicit geometric description; its general fiber is a Zariski open subset
of an abelian variety, which is a quotient of a hyperelliptic Jacobian by a 2-torsion
subgroup. In dimension 3, Φ is the Hitchin fibration of the moduli space of rank 2
bundles with fixed determinant on a curve of genus 2.
Contents
1 Introduction 2
1.1Comments .................................. 2
1.2Strategy.................................... 3
1.3Notations ................................... 3
2Thecasen=3 3
3 Definition of Φ 4
4Fibersofϕ6
4.1Odd-dimensionalintersectionof2quadrics ................ 7
4.2 An auxiliary construction . . ........................ 7
4.3ProofofProposition4.1............................ 8
5 Fibres of Φ 9
5.1Results .................................... 9
5.2Proofofthetheorem:lemmas........................ 10
5.3ProofofTheorem5.1 ............................ 11
Received 14 May 2024, accepted 15 May 2024.
2020 Mathematics Subject Classification 70H06 (Primary), 14J45 (Secondary)
Keywords: symmetric tensors, quadrics, Lagrangian fibration, completely integrable systems, Hitchin fibration.
c
The Author(s), 2024. Published by Cambridge University Press on behalf of the Foundation Composition
Mathematica, in partnership with the London Mathematical Society. This is an Open Access article, distributed
under the terms of the Creative Commons Attribution licence (http://creativecommons.org/licenses/by/4.0/),
which permits unrestricted re-use, distribution and reproduction, provided the original article is properly cited.
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Arnaud Beauville et al.
6 Proof of Proposition 5.5 11
7 Symmetric tensors: second approach 13
7.1 The cotangent bundle of a smooth quadric . ............... 13
7.2Explicitdescriptionofsymmetrictensors ................. 14
7.3Thedoublecover............................... 15
7.4 The divisor of s0............................... 16
7.5Proofofpart(a)ofthetheorem ...................... 17
References 18
1. Introduction
Let XPn+2
Cbe a smooth n-dimensional complete intersection of two quadrics, with n2, and
let TXbe its cotangent bundle. The C-algebra H0(TX, OTX) is canonically isomorphic to
the algebra of symmetric tensors H0(X, STX). Recall that TXcarries a canonical symplectic
structure. Our main result is the following theorem:
Theorem 1.1.
(a) The vector space W:= H0(X, S2TX)has dimension n,and the natural map SW
H0(X, STX)is an isomorphism.
(b) The corresponding map:TXW
=Cn,is a Lagrangian fibration.
(c) When Xis general, the general fiber of Φis of the form AZ,where Ais an abelian
variety and codim Z2.
We will give a precise geometric description of the map Φ and of the abelian variety Ain
Sections 4and 5.
1.1 Comments
(1) For n= 2, (a) follows from Theorem 5.1 in [DOL19], while (b) and (c) are proved in [KL22].
The proof is based on the isomorphism TX
=Ω1
X(1). The theorem also follows from the fact that
Xis a moduli space for parabolic rank 2 bundles on P1[Cas15], so Φ : TXC2is identified to
the Hitchin fibration (see [BHK10]).
For n=3, Xis isomorphic to the moduli space of vector bundles of rank 2 and fixed deter-
minant of odd degree [New68]; again, the theorem follows from the properties of the Hitchin
fibration (see Section 2). It would be interesting to have a modular interpretation of Φ for n4.
Note that the Hitchin map for G-bundles is homogeneous quadratic only when Gis SL(2) or a
product of copies of SL(2), so this limits the possibilities of using it.
(2) The map Φ is an example of an algebraically completely integrable system; see Remark
5.1. There is an abundant literature on such systems; see, for instance, [A96].
A classical example, the geodesic flow on an ellipsoid, is discussed in detail in [K80]. The
corresponding Lagrangian fibration takes place on the cotangent bundle of one quadric; it is not
related to our Φ. However, some of the tools we use in Sections 4and 5, in particular the variety
Xand the family of planes F, appear already in [K80] (with a different purpose).
(3) Such a situation is rather exceptional: Most varieties do not admit nonzero symmetric ten-
sors (for instance, hypersurfaces of degree 3[HLS22]); when they do, even for varieties as simple
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Symmetric tensors on the intersection of two quadrics and Lagrangian fibration
as quadrics, the algebra of symmetric tensors is fairly complicated (see, for instance, [BLi24]).
We do not have a conceptual explanation for the particularly simple behavior in our case.
(4) For n= 2 or 3, the generality assumption on Xin (c) is unnecessary. It seems likely that
this is the case for all n, but our method does not allow us to make that conclusion.
1.2 Strategy
We will first treat the case n= 3, which is independent of the rest of this article (Section 2). For
the general case, we will develop two different approaches. In the first one we exhibit a natural
n-dimensional subspace WH0(X, S2TX),fromwhichwededuceama:TXW
=Cn
(Section 3). We then show that Φ has the required properties, which implies (a), (b) and (c) for
general X(5.1). In the second approach (Section 7), we directly prove (a) for all smooth X,by
realizing Xas a double covering of a quadric.
1.3 Notations
Throughout this article, Xwill be a smooth complete intersection of two quadrics in Pn+2,with
n2. We denote by TXits cotangent bundle and by PTXits projectivisation in the geometric
sense (not in the Grothendieck sense). If Vis a vector space, we denote by P(V) the associated
projective space V{0}/Cparametrising 1-dimensional subspaces of V.
2. The case n=3
In this section we show how our general results can be obtained in the case n= 3 by interpreting
Xas a moduli space.
As in Section 4.1 below, we associate to Xagenus2curveCsuch that the variety of lines
in Xis isomorphic to JC. Let us fix a line bundle Non Cof degree 1; then Xis isomorphic
to the moduli space Mof rank 2 stable vector bundles on Cwith determinant N[New68].
The cotangent bundle TMis naturally identified with the moduli space of Higgs bundles;that
is, pairs (E, u)withEMand u:EEKCa homomorphism with Tru=0. The Hitchin
map Φ:TMH0(K2
C) associates to a pair (E, u) the section det uof K2
C. It is a Lagrangian
fibration [Hit87].
Let ωH0(K2
C). We assume in what follows that ωvanishes at 4distinct points.LetCω
be the curve in the cotangent bundle TCdefined by z2=ω. The projection π:CωCis a
double covering branched along div(ω), and Cωis a smooth curve of genus 5. Let Pbe the Prym
variety associated to π, that is, the kernel of the norm map Nm : JCωJC; it is a 3-dimensional
abelian variety.
Proposition 2.1. The fibre Φ1(ω)is isomorphic to the complement of a curve in P.
Proof.Recall that the map L→ πLestablishes a bijective correspondence between line bundles
on Cωand rank 2 vector bundles Eon Cendowed with a homomorphism u:EEKCsuch
that u2=ω·IdEor, equivalently, Tru=0 and detu=ω(see, for instance, [BNR89]). To get
(E, u)i
1(ω), we have to impose det E=Nand Estable. Since det πL=Nm(L)K1
C,
the first condition means that Lbelongs to the translate PN:= Nm1(KCN)ofP.
Then the vector bundle πLis unstable if and only if it contains an invertible subsheaf
Mof degree 1; this is equivalent to saying that there is a nonzero map πML;thatis,
L=πM(p) for some point pCω. The condition LPNmeans that M2(π(p)) = KCN,so
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Arnaud Beauville et al.
Mis determined by pup to the 2-torsion of JC. Thus the locus of line bundles LPNsuch
that πLis unstable is a curve.
Let ρ:CP1be the canonical double covering, with BP1its branch locus. Since the
homomorphism S2H0(KC)H0(K2
C) is surjective, the divisor of ωis of the form ρ(p+q), for
some p, q P1;byassumption,wehavep=qand p, q /B.
Proposition 2.2. Let Γbe the double covering of P1branched along B∪{p, q}.There is an
exact sequence
0Z/2JΓP0.
Proof.Let χ:P1P1be the double covering branched along {p, q}.Sincediv(ω)=ρ(p+q),
there is a cartesian diagram of double coverings
Cω
ξ
π
P1
χ
CρP1
which gives rise to two commuting involutions σ, τ of Cω, exchanging the two sheets of πand ξ,
respectively. The field of rational functions on Cωis
C(x, y, z)withy2=f(x),z
2=g(x),
where fand gare polynomials with divf=Band divg={p, q}.Thenσand τchange the sign
of yand z, respectively.
The involution στ is fixed-point free, so the quotient Γ := Cω/στ has genus 3; its field of
functions is C(x, w), with w=yz and w2=f(x)g(x). We have again a cartesian square
Cω
ϕ
π
Γ
ψ
CρP1.
Let αJΓ. We have Nmπϕα=ρNmψα= 0; hence, ϕmaps JΓintoPJCω.Sinceϕis ´etale,
we have Kerϕ=Z/2; since dim JΓ=dimP=3, ϕis surjective.
3. Definition of Φ
Let Ybe a smooth degree dhypersurface in PN,denedbyanequationf= 0. Recall that one
associates to fa section hfof S2Ω1
Y(d), the hessian or second fundamental form of f[GH79]: at
apointyof Y, the intersection of Ywith the tangent hyperplane Hto Yat yis a hypersurface
in Hsingular at y,andhf(y) is the degree 2 term in the Taylor expansion of f|Hat y.
Now let XPn+rbe a smooth complete intersection of rhypersurfaces of degree d;let
VH0(Pn+r,OP(d))
be the r-dimensional subspace of degree dpolynomials vanishing on X. By restricting hf,for
fV,toX, we get a linear map
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Symmetric tensors on the intersection of two quadrics and Lagrangian fibration
VOX−→ S2Ω1
X(d),
which gives at each point xXa linear space of quadratic forms on the tangent space Tx(X).
Note that when d= 2, the corresponding quadrics in P(Tx(X)) can be viewed geometrically as
follows: The projective space P(Tx(X)) can be identified with the space of lines in Pn+rpassing
through xand tangent to X;thenforeachqV, the quadric defined by hq(x) parameterises
the lines passing through xand contained in the quadric {q=0}.
Now we want to consider the ‘inverse’ of the quadratic form hf(x)onTx(X); that is, the form
on T
x(X) given in coordinates by the cofactor matrix. Intrinsically, each fVgives a twisted
symmetric morphism
hf:TX−→ Ω1
X(d),
which induces a twisted symmetric morphism on (n1)-th exterior powers, namely,
n1hf:n1TX−→ n1Ω1
X((n1)d).
We now observe that KX=OX(n1r+dr); hence
n1TX
=Ω1
X(n+1r(d1)) and n1Ω1
X
=TX(n1+r(d1)) ,
so n1hfinduces a symmetric morphism from Ω1
X(n+1r(d1)) to TX((n1)dn1+
r(d1)), hence provides a section
n1hfH0(X, S2TX(d(n+2r1) 2(n+r+ 1))).
Being locally given by the cofactor matrix, n1hfis homogeneous of degree n1inf. Hence,
we have constructed a linear map
α:Sn1V−→ H0(X, S2TX(d(n+2r1) 2(n+r+ 1))) such that α(fn1)=n1hf.
From now on, we restrict to the case d=2,r=2, so Xis the complete intersection of two
quadrics, and the previous construction gives a linear map
α:Sn1V−→ H0(X, S2TX).
Using the canonical isomorphism H0(TX, OTX)=H0(X, STX), we deduce from αa
morphism
Φ:TX−→ Sn1V
=Cn.
We have Φ(λv)=λ2Φ(v)forvTX,λC, so Φ induces a rational map
ϕ:PTX Pn1,
whose indeterminacy locus Zis the image of Φ1(0).
Proposition 3.1.
(1) αis injective.
(2) Φ is surjective.
(3) The image of Zby the structure map p:PTXXis a proper subvariety of X.
Proof.Let xbe a general point of X. We claim that the base locus in P(Tx(X)) of the pencil
of quadratic forms {hq(x)}qVis smooth. Indeed, this locus can be viewed as the variety Fxof
lines in Xpassing through x.LetFbe the Fano variety of lines contained in X,andlet
GF×X={(, y)|y}.
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Arnaud Beauville et al.
Then Fand therefore Gare smooth [Reid72, Theorem 2.6], hence Fx, which is the fibre above x
of the projection GX,issmoothsincexis general. It follows that in an appropriate system
of coordinates (k1,...,k
n)ofTx(X), the forms {hq(x)}can be written as
tk2
i+αik2
iwith αidistinct in C,tC.
Then n1hq(x) is given by the diagonal matrix with entries βi:=
j=i
(t+αj)(i=1,...,n).
These polynomials in tare linearly independent; hence, they generate the space of quadratic
forms on T
x(X), which are diagonal in the basis (ki). This linear system has dimension n,soα
is injective; it has no base point, so ϕinduces a finite, surjective morphism P(T
x(X)) Pn1.
Thus, Φ is surjective, and ZP(T
x(X)) = , which gives (2) and (3).
We want to give a geometric construction of the rational map ϕ:PTX Pn1.Apoint
of PTXis a pair (x, H), where xXand His a hyperplane in Tx(X). Restricting the pencil
{hq(x)}qVto Hgives a pencil of quadrics on H, which for general(x, H) contains n1sin-
gular quadrics q1,...,q
n1. The subset {q1,...,q
n1}of P(V) corresponds to a point ϕx,H of
P(Sn1V); namely, the hyperplane in Sn1Vspanned by qn1
1,...,q
n1
n1.
Proposition 3.2. ϕ(x, H)=ϕx,H .
Proof.We can assume that xis general. We have seen that the restriction of ϕto P(T
xX)is
the morphism given by the linear system of quadratic forms W
=Sn1Vspanned by the forms
n1hq(x), for qV;inotherwords,ϕmaps the point Hof P(T
x(X)) to the hyperplane of
forms in Wvanishing at H.
On the other hand, ϕx,H is the hyperplane of Sn1Vspanned by the qn1for those qV
such that hq(x)|His singular; this condition is equivalent to saying that the form n1hq(x)
on T
xXvanishes at H. Therefore, ϕx,H is spanned by quadratic forms vanishing at H, hence
coincides with ϕ(x, H).
Corollary 3.3. codimZ2.
Proof.Suppose that Zcontains a component Z0of codimension 1; since p(Z)=X,wehave
Z0=p1(p(Z0)). We claim that this is impossible; in fact, Zcannot contain a fibre p1(x).
Indeed, its doing this would mean that for qV,theformhq(x) is singular along all hyperplanes
HTx(X); that is, hq(x)hasrankn2. But the rank of hq(x) is the rank of the restriction
of qto the projective tangent subspace to Xat x. Restricting a quadratic form to a hyperplane
lowers its rank by up to two. Since a general qin Vhas rank n+ 3, its restriction to a codimension
2 subspace has rank n1.
4. Fibers of ϕ
In an appropriate system of coordinates (x0,...,x
n+2), our variety Xis defined by the equations
q1=q2=0, with
q1=x2
iq2=μix2
i,with μiCdistinct.
Let Π = P(V)(
=P1) be the pencil of quadrics containing X. We choose a coordinate ton Π so
that the quadrics of Π are given by tq1q2= 0. Then the singular quadrics of Π correspond to
the points μ0,...,μ
n+2.
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Symmetric tensors on the intersection of two quadrics and Lagrangian fibration
The goal of this section is to describe the general fibre of the rational map ϕ:PTX
Sn1Π(
=Pn1). For λ=(λ1,...,λ
n1)Sn1Π, let Cμ,λ denote the hyperelliptic curve y2=
(tμi)(tλj), of genus n. We will then prove the following:
Proposition 4.1. For λgeneral in Sn1Π, the fibre ϕ1(λ)is birational to the quotient of the
Jacobian JCμ,λ by the group Γ:=1JCΓ+,where Γ+
=(Z/2Z)n2is a group of translations
by 2-torsion elements.
4.1 Odd-dimensional intersection of 2 quadrics
We briefly recall here the results of Reid’s thesis ([Reid72]; see also [DR76]). Let YP2g+1 be
a smooth intersection of 2 quadrics, and let Ξ (
=P1) be the pencil of quadrics containing Y.
Let Σ Ξ be the subset of 2g+ 2 points corresponding to singular quadrics, and let Cbe the
double covering of Ξ branched along Σ; this is a hyperelliptic curve of genus g. The intermediate
Jacobian JY of Yis isomorphic to JC (as principally polarized abelian varieties). The variety F
of (g1)-planes contained in Yis also isomorphic to JC, but this isomorphism is not canonical.
In an appropriate system of coordinates, the equations of Yare of the form
x2
i=αix2
i=0,with αiCdistinct;
then Σ = {α1,...,α
2g+2}. The group Γ := (Z/2Z)2g+1 acts on Y(hence also on F) by changing
the signs of the coordinates. Let Γ+Γ be the subgroup of elements that change an even number
of coordinates. Choose an element γΓΓ+; there is an isomorphism F
−→ JC such that γ
corresponds to (-1JC). Then the image of Γ+in Aut(JC) is the group T2of translations by
2-torsion elements of JC, and the image of Γ is T2×{±1JC}[DR76, Lemma 4.5].
4.2 An auxiliary construction
We consider the projective space P2n+1 equipped with the system of homogeneous coordinates
x0,...,x
n+2;y1,..., y
n1
and the affine space An1equipped with the affine coordinates λ1,...,λ
n1.Let
XP2n+1 ×An1
be the complete intersection of the two quadrics with equations
Q1=Q2=0 with Q1=
n+2
i=0
x2
i+
n1
j=1
y2
j,Q
2=
n+2
i=0
μix2
i+
n1
j=1
λjy2
j.
The second projection, XAn1,gives a family of complete intersections of two quadrics Xλof
dimension 2n1 parameterised by An1. Note that Xis the intersection of Xwith the subspace
Pn+2 P2n+1 defined by y1=...=yn1=0.
Let p:FAn1be the family of (n1)-planes contained in the Xλ;thatis
F={(P, λ)|λAn1,P(n1)-plane Xλ}.
For λgeneral, the fibre Fλis isomorphic to the Jacobian of the hyperelliptic curve Cμ,λ (4.1).
Let (P, λ) be a general point of F.ThenPPn+2 is a point xof X.Letπ:P2n+1  Pn+2 be
the projection (xi,y
j)→ (xi). Since the πdifferentials of Qiand qicoincide at x, the differential
πmaps Tx(P)Tx(X)intoTx(X). Since Pis general, πTx(P) is a hyperplane in Tx(X); this
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Arnaud Beauville et al.
will follow from the proof of Proposition 4.2, (1) below, where we explicitly construct pairs (P, λ)
with this property.
Therefore, we have a rational map
ψ:F PTX(P, λ)→ (x=PPn+2
Tx(P)) .
The symmetric group Sn1acts on P2n+1 by permuting the yjand acts on the group (Z/2Z)n1
by changing their signs; this gives an action of the semi-direct product G:= (Z/2Z)n1Sn1.
We make Gact on An1through its quotient Sn1, by permutation of the λi. This induces an
action of Gon Xand therefore on F, which is compatible via pwith the action on the base. The
map ψis invariant under this action; hence, it factors through the quotient F/G.Bypassingto
the quotient, we get a map p:F/G An1/Sn1.
Proposition 4.2. (1) ψinduces a birational map ψ:F/G  PTX.
(2) There is a commutative diagram
F/G ψ
p
PTX
ϕ
An1/Sn1σ
An1Pn1
where pis deduced from pand where σis the isomorphism given by symmetric functions.
Proof.(1) Let (x, H)PTX; we want to describe the pairs (P, λ) such that PPn+2 ={x}
and πTx(P)=H. The latter condition says that via the decomposition
Tx(P2n+1)=Tx(Pn+2)Ker π,
Tx(P) identifies with the graph of a linear map
α:HKer π.
Using the basis (
∂y1,...,
∂yn1)ofKerπ,wehaveα=(α1,...,α
n1), where the αiare linear
forms on H. The condition PXλimplies that the hessians hQ1(x)andhQ2(x) vanish on Tx(P),
which gives
hq1(x)|H=
i
α2
ihq2(x)|H=
i
λiα2
i.(1)
This is a simultaneous diagonalisation of the quadratic forms hq1(x)|Hand hq2(x)|H; when they
are in general position, this determines the λiup to permutation and the αiup to sign and
permutation, which proves (1).
(2) Let (P, λ)F,andlet(x, H):=ψ(P, λ).AccordingtoProposition3.2,ϕ(x, H)isgiven
by the (n1)-uple of quadrics qΠ such that the form hq(x)|His singular. Using (α1,...,α
n1)
as coordinates on H, we see from (1) that this (n1)-uple is given by (λ1,...,λ
n1), which
proves (2).
4.3 Proof of Proposition 4.1.
Let λbe a general element of An1. Let us denote by Γ the subgroup (Z/2Z)n1of G.From
Proposition 4.2 and the cartesian diagram
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Symmetric tensors on the intersection of two quadrics and Lagrangian fibration
F/Γ
p
F/G
p
An1An1/Sn1
we see that the fibre ϕ1(λ) is birational to the quotient Fλ/Γ. By (4.1) there is an isomorphism
Fλ
−→ JCμ,λ suchthatΓactsonJCμ,λ as 1JΓ+,wher
+is a group of translations by
2-torsion elements. This proves the proposition.
5. Fibres of Φ
5.1 Results
We keep the settings of the previous section. Recall that our parameter λlives in An1Sn1Π
=
Pn1.Forλin An1, we denote by ˜
λa lift of λin Cnfor the quotient map Cn{0}→Pn1.
Theorem 5.1. Assume that Xis general. For λAn1general, the fibre Φ1(˜
λ)is isomorphic
to AZ,where:
Ais the abelian variety quotient of JCμ,λ by a 2-torsion subgroup, isomorphic to
(Z/2Z)n2;
Zis a closed subvariety of codimension 2in A.
Corollary 5.2. For every smooth complete intersection of two quadrics XPn+2,the
fibration Φ:TXCnis Lagrangian.
Proof.Assume first that Xis general. The symplectic form on TXis ,whereηis the Liouville
form. By Theorem 5.1 and Hartogs’ principle, the pull-back of ηto a general fibre of Φ is the
restriction of a 1-form on an abelian variety, hence is closed. This implies the result.
Let p:XBbe a complete family of smooth intersection of two quadrics in Pn+2.The
constructions of §3 can be globalised over B: we have a rank 2 vector bundle Vover Bwhose
fibre at a point bBis the space of quadratic forms vanishing on Xb. We get a homomorphism
Sn1VpTX/B, which thus gives rise to a morphism Φ:T(X/B)Sn1Vover Bwhich
induces over each point bBour map Φ. There is a natural Liouville form ηon T(X/B):
Since dηvanishes on a general fibre of Φ, it vanishes on all fibres.
Corollary 5.3. Assume that Xis general. The multiplication map SH0(X, S2TX)
H0(X, STX)is an isomorphism.
(We will give in Section 7a proof that is valid with no generality assumption.)
Proof.Theorem 5.1 implies that every function on a general fibre of Φ is constant; hence,
the pull-back Φ:H0(Cn,OCn)H0(TX, OTX) is an isomorphism. The right-hand space is
canonically isomorphic to H0(X, STX); hence, we get an algebra isomorphism C[t1,...,t
n]
−→
H0(X, STX). By construction, the tiare mapped to elements of H0(X, S2TX), so the Corollary
follows.
Remark 5.4. Let V1,...,V
nbe the Hamiltonian vector fields on TXthat are associated to the
components of Φ. For λgeneral in Cn, let us identify Φ1(λ)toAZas in the theorem. Then
by Hartogs’ principle the Vilinearise on A; that is, they extend to a basis of H0(A, TA). In
principle, this allows to write explicit solutions of the Hamilton equations for Φiin terms of
theta functions.
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Arnaud Beauville et al.
5.2 Proof of the theorem: lemmas
We fix a general point λAn1. We denote by Fothe open subset of Fwhere the rational
map ψis well-defined and denote by Fo
λits intersection with the fibre Fλ.Sinceλis general,
the complement of Fo
λin Fλhas codimension 2. The rational map ψinduces a morphism
ψo:FoPTX; we denote by ψo
λits restriction to Fo
λ.LetZPTXbe the indeterminacy
locus of ϕ(§3), and let Fbad
λ:= (ψo
λ)1(Z)Fo
λ.
Proposition 5.5. Fbad
λhas codimension 2in Fλ.
We postpone the proof of Proposition 5.5 to the next section; here we show how it implies
Theorem 5.1.
Let 0XTXbe the zero section, and let q:TX0XPTXbe the quotient map. Let
ϕo:PTXZPn1be the morphism induced by ϕ.Wethushave q1(˜
λ)) = (ϕo)1(λ),
and the restriction
qλ
1(˜
λ)(ϕo)1(λ)
is an ´etale double cover, with Galois involution ιinduced by (1TX).
We put Foo
λ:= Fo
λFbad
λand consider the restriction
ψo
λ:Foo
λ(ϕo)1(λ)ofψo.
Lemma 5.6. The fibre Φ1(˜
λ)is Lagrangian, and has a trivial tangent bundle.
Proof.The ´etale double cover qλinduces by fibred product an ´etale double cover
π:
Foo
λFoo
λ
such that ψo
λlifts to a morphism ˜
ψo
λ:
Foo
λΦ1(˜
λ).
By Proposition 5.5, the complement of Foo
λin Fλhas codimension 2, so πextends to
an ´etale double cover
FλFλ,where
Fλis an abelian variety or the disjoint union of two
abelian varieties. The morphism ˜
ψo
λ:
Foo
λΦ1(˜
λ) is generically of maximal rank. Again by
Proposition 5.5, the holomorphic 1-forms on
Foo
λare closed; hence by pull-back, the same holds
for the holomorphic 1-forms on Φ1(˜
λ). As in the proof of Corollary 5.2,thisimpliestha
1(˜
λ)
is Lagrangian. The second assertion is a basic property of Lagrangian fibres.
Lemma 5.7 The morphism ψo
λlifts to a morphism ˜
ψo
λ:Foo
λΦ1(˜
λ).
Proof.It suffices to show that the double covering π:
Foo
λFoo
λsplits.
Assuming the contrary,
Fλis an abelian variety. By Lemma 5.6 H01(˜
λ),Ω1) has dimen-
sion n. It follows that the pull-back ( ˜
ψo
λ):H01(˜
λ),Ω1)H0(
Foo
λ,Ω1) is bijective. Since the
Galois involution of the double covering πacts trivially on holomorphic 1-forms, the same holds
for the Galois involution ιof the double covering qλ
1(˜
λ)(ϕo)1(λ).
Now we observe that the 1-forms on Φ1(˜
λ) are ‘pure’; that is, they extend to any smooth
projective compactification of Φ1(˜
λ). This follows from the fact that this holds after pull-back
to
Foo
λ. But the quotient Φ1(˜
λ) is isomorphic to a Zariski open subset of ϕ1(λ), which, by
Proposition 4.1, has no nonzero holomorphic 1-forms, so any Zariski open subset has no nonzero
closed pure holomorphic 1-forms. This contradiction proves the lemma.
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Symmetric tensors on the intersection of two quadrics and Lagrangian fibration
5.3 Proof of Theorem 5.1
Lemma 5.7 gives a factorisation,
ψo
λ:Foo
λ
˜
ψo
λ
−−Φ1(˜
λ)qλ
(ϕo)1(λ).
By Proposition 4.1,ψo
λinduces a birational morphism,
ψo
λ,Γ:Foo
λ/Γ−→ (ϕo)1(λ)
it follows that for some subgroup ΓΓ of index 2, the morphism ˜
ψo
λ:Foo
λΦ1(˜
λ)factors
through a birational morphism,
˜
ψo
λ,Γ:Foo
λ/Γ−→ Φ1(˜
λ).
By Lemma 5.6, the cotangent bundle of Φ1(˜
λ) is trivial. Therefore, the cotangent bundle of
Foo
λ/Γis generically generated by its global sections. This implies that Γacts trivially on
holomorphic 1-forms and, hence, is the subgroup Γ+of Γ generated by translations, isomorphic
to (Z/2Z)n2;thusFλ/Γis an abelian variety A.
To simplify notation, we write Ao:= Foo
λ/Γand u:= ˜
ψo
λ,Γ. The rational map u1:
Φ1(˜
λ) Ais everywhere defined (e.g. [BL92, Theorem 4.9.4]), so we have two morphisms
Aou
Φ1(˜
λ)u1
−−A
whose composition is the inclusion AoA. Since the tangent bundles of Aand Φ1(˜
λ)
are trivial, the determinant of Tu:TAouTΦ1(˜
λ)is a function on Ao, hence constant by
Proposition 5.5. Therefore, uis ´etale and birational, hence an open embedding. This implies
that every function on Φ1(
λ) is constant (because its restriction to Aois constant). Then the
previous argument shows that u1is also an open embedding, hence Φ1(˜
λ) is isomorphic to an
open subset of Acontaining Ao.Thisprovesthetheorem.
6. Proof of Proposition 5.5
We keep the notations of Section 4.2. Recall that we have coordinates (x0,...,x
n+2;y1,...,y
n1)
on P2n+1 and subspaces Pn+2 and Pn2in P2n+1 defined by y= 0 and x=0.
Let q1(x)=q2(x) = 0 be the equations defining Xin Pn+2,andletRbe the vector space of
quadratic forms in y=(y1,...,y
n1). We define an extended family XeP2n+1 ×R2as
Xe={((x, y); (r1,r
2)) P2n+1 ×R2|q1(x)+r1(y)=q2(x)+r2(y)=0}.
The fibre Xe
rat a point r=(r1,r
2)ofR2is the intersection in P2n+1 of the two quadrics
q1(x)+r1(y)=q2(x)+r2(y) = 0. Let Gbe the Grassmannian of (n1)-planes in P2n+1;we
define as before
Fe:= {(P, r)G×R2|PXe
r}
and the extended rational map ψe:Fe PTX, which maps a general PXe
rto the pair
(x, H), with {x}=PPn+2 and H=πTx(P).
We observe that a general pair r=(r1,r
2)ofR2is simultaneously diagonalisable, so the
restriction of ψeto Fe
rcoincides, for an appropriate choice of the coordinates (yi), with the map
ψλthat we want to study.
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Arnaud Beauville et al.
Proposition 6.1. Assume that Xis general.
1.Let ΓFebe the locus of points (P, r)such that either dim PPn+2 >0, or PPn2=
.Then Γhas codimension 2in Fe.
2.There exists no divisor in FeΓthat dominates R2and that is mapped to the base locus
ZPTXby ψe.
We claim that this implies Proposition 5.5. Indeed, as just explained above, it suffices to
prove the analogue of Proposition 5.5 for ψe. Next, it is clear that the indeterminacy locus of
ψeiscontainedi,soψeis well-defined on FeΓ. By Proposition 6.1,(1),itnowsuces
to prove the analogue of Proposition 5.5 for the restriction of ψeto FeΓ. This is exactly the
statement of Proposition 6.1, (2).
Proof of P roposit ion 6.1: (1) Let Qbe the vector space of quadratic forms on P2n+1 of the form
q(x)+r(y) for some quadratic forms qand r. For each pair of integers (k, l)withk0, l≥−1,
let Gk,l be the locally closed subvariety of (n1)-planes PGsuch that
dim(PPn+2)=kdim(PPn2)=l.
(We put, by convention, l=1ifPPn2=.) Let
FQ:= {(P, (Q1,Q
2)) G×Q
2|Q1|P=Q2|P=0}
and
FQ
k,l := FQ(Gk,l ×Q
2).
The general fibre of the projection FQ→Q
2is an abelian variety, and we recover Feby restrict-
ing FQto pairs of quadratic forms of the form (q1(x)+r1(y),q
2(x)+r2(y)), with (q1(x),q
2(x))
fixed. Because we assume Xgeneral, the pair (q1(x),q
2(x)) is general in Q2. It thus suffices to
prove the result for the larger family FQ; that is, to show that FQ
k,l has codimension 2inFQ.
This is done by a dimension count. For PG,letϕPbe the restriction map Q→
H0(P, OP(2)). The fibre of the projection FQGis the vector space (KerϕP)2.ForPgeneral,
ϕPis surjective: This is the case, for instance, if Piscontainedinthe(n+ 2)-plane in P2n+1
defined by yi=xi(i=1,...,n1). However, ϕPis not surjective for PGk,l because the forms
r(y)|Pare singular along PPn+2 and the forms q(x)|Pare singular along PPn2; this implies
that the subspaces PPn+2 and PPn2are apolar for all forms in ImϕP. Therefore, the corank
of ϕPis (k+1)(l+ 1), and there is equality when Pis contained in the subspace defined by
x0=...=xn+1k=y1=...=yn2l= 0, hence for Pgeneral in Gk,l. Thus our assertion follows
from:
codim(FQ
k,l,FQ)=codim(Gk,l,G)2(k+1)(l+1)
=k(k+1)+(l+1)(l+4)2(k+1)(l+1)
=(kl)(kl1) + 2(l+1)
2ifk1orl0.
(2) The base locus ZPTXhas codimension 2 (Corollary 3.3). Note that ψeis well-
defined in FeΓ. If Dis a codimension 1 subvariety in FeΓ, with ψe(D)Z,themapψe
does not have maximal rank along D. This contradicts the following lemma:
Lemma 6.2. ψehas maximal rank on FeΓ.
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Symmetric tensors on the intersection of two quadrics and Lagrangian fibration
Proof.Let (x, H)beapointofPTX; we view Has a hyperplane in the projective tangent
space to xat X. The fibre of ψe:FeΓPTXat (x, H)isthelocus
(ψe)1(x, H)={(P, r1,r
2)G×R2|PPn+2 ={x},PPn2=(P)=H, (2)
(qi+ri)|P=0 (i=1,2)}.(3)
The equations (2) define a smooth, locally closed subvariety Gx,H of G.LetPGx,H ,andlet
χP:RH0(P, OP(2)) be the restriction map. We will show below that the image of χPis the
space of quadratic forms on Pthat are singular at x.Sincetheformsqi|Pare singular at x,this
implies that the solutions of (3) form an affine space over (KerχP)2. Therefore, (ψe)1(x, H)
admits an affine fibration over Gx,H, hence is smooth.
Clearly the quadrics in ImχPare singular at x. To prove the opposite inclusion, choose
the coordinates (xi)sothatx=(1,0,...,0). Since PPn+2 ={x}, there exist linear forms
1,...,
n+2 in the yjso that Pis defined by xi=i(y)fori=1,...,n+ 2. Then a quadratic
form on P2n+1 singular at xcan be written as a form in x1,...,x
n+2;y1,...,y
n1; hence, its
restriction to Pis in ImχP. This proves the lemma and, hence, also the proposition.
7. Symmetric tensors: second approach
7.1 The cotangent bundle of a smooth quadric
We consider a smooth quadric QPn+1 defined by an equation q= 0. Its cotangent bundle
PTQparameterises pairs (x, P )withxQand Pa(n1)-plane tangent to Qat x.Thus,we
get a morphism γfrom PTQto the Grassmannian Gof (n1)-planes in Pn+1,whichisthe
morphism defined by the linear system |OPTQ(1)|. It is birational onto its image, but contracts
the subvariety CPTQthat consists of pairs (x, P ), such that Pis tangent to Qalong a line
Qwith x;thenγ1(P) consists of the pairs (x, P )withx.
Let hqH0(Q, S2Ω1
Q(2)) be the hessian form of q(§3). Choosing coordinates (xi) such that
q(x)=x2
i,wehavehq=(dxi)2(note that this is, up to a scalar, the unique element of
H0(Q, S2Ω1
Q(2)) invariant under Aut(Q)). Then hq(x) is non-degenerate at each point xof Q,
so hqinduces an isomorphism Ω1
Q(1)
−→ TQ(1), hence also S2Ω1
Q(2)
−→ S2TQ(2). The image
in H0(Q, S2TQ(2)) of hqby this isomorphism is hq=2
j. We will view hqas an element of
H0(PTQ, OPTQ(2) pOQ(2)), where p:PTQQis the projection.
Proposition 7.1. The divisor of hqis C.The projection p|C:CQis a smooth quadric
fibration, and Cis a prime divisor for n3.
Proof.Let xQ; the hyperplane tangent to Qat xcuts out a cone over the smooth quadric
QxP(Tx(Q)) defined by hq(x) = 0 (Section 3). The isomorphism Tx(Q)
−→ T
x(Q)givenby
hq(x) carries Qxinto the dual quadric Q
xin P(T
x(Q)). On the other hand, a point yp1(x)
corresponds to a hyperplane HyP(Tx(Q)), and ybelongs to Cif and only if Hyis tangent to
Qx;thatis,ifyQ
x. This proves the equality C=div(hq) and thus, that the fiber of p|C:CQ
at xis Qx, which is smooth and connected if n3.
Remark 7.2 The variety Cis an example of a total dual VMRT [HLS22]. For the proof of the
theorem, we will combine this tool with the birational transformation of PTXdefined by a
double cover. (Compare with [AH23]).
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Arnaud Beauville et al.
We will have to consider the following situation: Let Qbe another quadric in Pn+1, such that
the intersection B:= QQis a smooth hypersurface in Q. The surjection TQNB/Q gives a
section of PTQover B, hence an embedding s:BPTQ.
Lemma 7.3. The image s(B)is not contained in C.
Proof.Let xB.Thepoints(x)inP(T
x(Q)) corresponds to the hyperplane image of Tx(B)in
Tx(Q); we must therefore show that this hyperplane is not tangent to the quadric Qx:= hq(x). In
terms of projective space, this means that the projective tangent space to Qat xis not tangent,
at a smooth point y, to the cone cut out on Qby the projective tangent space to Qat x.
Suppose that this is the case, with y=(y0,...,y
n+1). We can assume that Qis defined
by αix2
i=0, with αiCdistinct. Then the (projective) tangent space to Qat x,givenby
(αixi)ξi= 0, must coincide with the tangent space to Qat y,givenbyyiξi= 0. This implies
y=(α0x0,...,α
n+1xn+1). Thus the point xmust satisfy
x2
i=αix2
i=α2
ix2
i=0.
If these relations hold for all xin B, the quadric α2
ix2
i= 0 must belong to the pencil spanned
by Qand Q. This means that there exist scalars λ, μ, ν such that
λα2
i+μαi+ν=0 for all i,
which is impossible since the αiare distinct. Therefore, there exists xBsuch that
s(x)/C.
7.2 Explicit description of symmetric tensors
We keep the notation of the previous sections: XP=Pn+2 is defined by q1=q2=0, and with
q1=
n+2
i=0
x2
i,q
2=
n+2
i=0
μix2
iwith μiCdistinct.
We put i:=
∂xi
·We have an exact sequence
0TXTP|X
(dq1,dq2)
−−−−−−→ OX(2)20,
where dqimaps the restriction of a vector field Von Pto V·qi. This gives the exact sequence
of symmetric tensors
0S2TXS2TP|X
(dq1,dq2)
−−−−−−→ TP|X(2)2,(4)
where dqi(V1V2)=(V1·qi)V2+(V2·qi)V1for V1,V
2in H0(X, TP|X).
Proposition 7.4. The quadratic vector fields si:=
j=i
(xijxji)2
μjμi
in H0(X, S2TP|X)belong
to the image of H0(X, S2TX).
Proof.According to the exact sequence (4), we have to prove dq1(si)=dq2(si)=0.
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Symmetric tensors on the intersection of two quadrics and Lagrangian fibration
We have (xijxji)·q1= 0; hence, dq1(si)=0 and dq2(xijxji,x
ijxji)=4(μj
μi)xixj(xijxji). Hence, using xjj= 0 and q1|X=0,
dq2(si)=4x2
i
j=i
xjj4xi(
j=i
x2
j)i=0,which proves the proposition.
In the rest of this article, we will consider the sito be elements of H0(X, S2TX).
7.3 The double cover
Let p0:Pn+2  Pn+1 be the projection (x0,...,x
n+2)→ (x1,...,x
n+2). The image p0(X)is
the smooth quadric Qin Pn+1 defined by
n+2
i=1
(μiμ0)x2
i=0.
The restriction π:XQof p0is a double covering that is branched along the subvariety BQ
defined by
n+2
i=1
x2
i=
n+2
i=1
μix2
i=0.
It is a smooth complete intersection of 2 quadrics in Pn+1. The ramification locus RXof π
(isomorphic to B) is the hyperplane section x0=0 of X.
The tangent map of π:XQgives a morphism,
τ:TXπTQ,
whichisanisomorphismoutsideofR. Consider the normal exact sequence
0TRTX|RNR/X 0.
The involution ι:(x0,...,x
n+2)→ (x0,x
1,...,x
n+2)actsonTX|R; this splits the exact
sequence, giving a decomposition
TX|R=TRNR/X
into eigenspaces for the eigenvalues +1 and 1. Let ρ:TX|RTRbetheprojectiononthefirst
summand. We deduce from ρa sequence of homomorphisms
hk:H0(X, SkTX)−→ H0(X, SkTX|R)Skρ
−−H0(R, SkTR).
Since ι0=0and ιj=jfor j>0, we have
h2(s0)=0 and h2(si)=
j>0
j=i
(xijxji)2
μjμi
for i>0(5)
in other words, h2maps s1,...,s
n+2 to the elements ˆs1,...,ˆsn+2 of H0(R, S2TR) constructed
in Proposition 7.2.1 applied to R.
Let πPTQbe the pull-back under πof the projective bundle PTQQ. The homo-
morphism τ:TXπTQgives rise to the birational map g:πPTQ PTX. Following the
geometric description of the tangent map as an elementary transformation of vector bundles in
the sense of Maruyama in [Mar72]and[Mar73, Corollary 1.1.1], one has a commutative diagram
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Arnaud Beauville et al.
Γ
μν
πPTQg
p
PTX
q
X
(6)
where pand qare the canonical projections; νPTXis the blow-up along the subspace
PTRPTXdefined by the projection ρ;andμπPTQis the blow-up of the image B
of the embedding BπPTQdeduced from the surjective homomorphism πTQπNB/X.
Let Eμbe the exceptional divisor of μ.By[Mar73, Theorem 1.1], there is an isomorphism
μOπPTQ(1) OΓ(Eμ)
=νOPTX(1) (7)
as well as the equality
νEμ=qR. (8)
7.4 The divisor of s0
We now consider the divisor CPTQdefined in (7.1) and the cartesian diagram
πPTQπ
PTQ
XπQ.
Set C:= π1(C). The projection CXis again a smooth quadric fibration, so Cis smooth
and connected for n3.
Recall that we have defined the element s0:=
n+2
j=1
(x0jxj0)2
μjμ0
H0(X, S2TX)(7.2). We
will now view s0as an element of H0(PTX, O(2)).
Proposition 7.5 Assume n3. We have gC=div(s0).
Proof.We first show that gC∈|OPTX(2)|.ByProposition7.1 we have C
|OπPTQ(2) pOX(2)|.Using(7), (8) and the projection formula, we get the linear
equivalences
νμC2νμ(c1(OπPTQ(1) pR)) 2(c1(OPTX(1)) + qR)2qR=c1(OPTX(2)) .
Thus, it is enough to prove that νμCis irreducible. Since Cis irreducible and μis the blow-up
along BπPTQ, it suffices to show that Bis not contained in C. If this is the case, then
we have π(B)π(C)=C.Butπ(B)=s(B), where s:BPTQis the embedding defined
by the surjective homomorphism TQNB/Q. Then the result follows from Lemma 7.3.
Since gCand div(s0) are linearly equivalent effective divisors and gCis irreducible, it
suffices to show that their restrictions to PT
x(X) coincide at a general point xX.
Fix a point x=(x0,...,x
n+2)XRso that x0= 0. Then the tangent map (x):
Tx(X)Tπ(x)(Q) is an isomorphism; in diagram (6), the maps μ, ν and grestricted over the
fibres at xare all isomorphisms. Let us show that Cand (div(s0)) define the same quadric
in P(Tπ(x)(Q)).
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Symmetric tensors on the intersection of two quadrics and Lagrangian fibration
Note that CP(T
x(X)) = CP(T
π(x)(Q)) is the quadric defined by the element hqof (7.1).
In the coordinates (zi) defined by zi=(μiμ0)1/2xi, the equation of Qis
n+2
j=1
z2
j=0, so
hq=
n+2
j=1
(
∂zj
)2=
n+2
j=1
2
j
μjμ0
·
On the other hand, since π(x0,...,x
n+2)=(x1,...,x
n+2), we have (0)=0and(j)=j
for j>0; hence,
(s0)=x2
0
n+2
j=1
2
j
μjμ0
·
Since x0= 0, this proves the proposition.
7.5 Proof of part (a) of the theorem
Suppose now that n3. Consider the double cover π:XQand the ramification divisor RX.
The restriction maps hkdefined in Section 7.3 yield a homomorphism of graded C-algebras
h:S(X):=H0(X, STX)−→ H0(R, STR)=:S(R).
Proposition 7.6 The kernel Iof his the ideal generated by s0.
Proof.Since Iis a homogeneous ideal, it suffices to prove that every homogeneous element
sIcan be written as s=ss0for some element sS(X).
Choose an element sIof degree k. This element corresponds to an effective Cartier divisor
Gin the linear system |OPTX(k)|. Recall the commutative diagram (6):
Γ
μν
πPTQg
p
PTX
q
X
Choose ˆ
G:= μνGπPTQ.By(7), ˆ
Gbelongs to the linear system |OπPTQ(k)|.
Here is the key observation: Since sI, the divisor ˆ
GπPTQcontains pR. Indeed, since
(πTQ)|Ris invariant under ι, the homomorphism τ|Rfactors as
τ|R:TX|R
ρ
TR−→ (πTQ)|R.
Therefore, we have a commutative diagram,
H0(X, SkTX)hk
Skτ
H0(R, SkTR)
H0(X, SkπTQ)H0(R, Sk(πTQ)|R)
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https://doi.org/10.1112/mod.2024.3
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Arnaud Beauville et al.
and Skτ(s) vanishes on R.But ˆ
Gis the divisor of Skτ(s), viewed as a section of OπPTQ(k);
hence, ˆ
Gcontains pR.
Now we want to show that the divisor CπPTQis a component of ˆ
GpR. Recall from
(7.1)thatCis the union of the lines that are contracted by the morphism γ:PTQG
and that c1(OPTQ(1)) ·= 0. Thus the curves := πcover Cand satisfy c1(OπPTQ(1)) ·
= 0. On the other hand, the divisor RXis a hyperplane section, so pR·=R·p>0.
Therefore,
(ˆ
GpR)·<0,
so Cis a component of ˆ
G.Thus,gCis a component of G.SincegC=div(s0)by
Proposition 7.5, this proves the proposition.
The following proposition implies part (a) of our main theorem:
Proposition 7.7. Assume n2. For any choice of indices 0i1<...<i
nn+2, the homo-
morphism C[t1,...,t
n]S(X), which maps tjto sij,with deg(ti)=2, is an isomorphism of
graded C-algebras.
Proof.We argue by induction on n. The statement for n= 2 follows from [DOL19,Theorem
5.1], except for the fact that any two of the sigenerate H0(X, S2TX). Up to the permut-
ing of the coordinates, it suffices to prove that s0and s1are linearly independent. But
h2:H0(X, S2TX)H0(R, S2TR) maps s0to zero and maps si(for i>0) to the corresponding
elements ˆsiof H0(R, S2