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Symmetric tensors on the intersection of two quadrics and Lagrangian fibration

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Abstract

Let X be a n-dimensional (smooth) intersection of two quadrics, and let T*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 F: T*X -- > C^n is a Lagrangian fibration, which admits an explicit geometric description; its general fiber is a Zariski open subset of an abelian variety, quotient of a hyperelliptic Jacobian by a 2-torsion subgroup. For n = 3 F is the Hitchin fibration of the moduli space of rank 2 bundles with fixed determinant on a curve of genus 2.
arXiv:2304.10919v1 [math.AG] 21 Apr 2023
SYMMETRIC TENSORS ON THE INTERSECTION OF TWO QUADRICS AND LAGRANGIAN
FIBRATION
A. BEAUVILLE, A. ETESSE, A. H ¨
ORING, J. LIU, AND C. VOISIN
ABS TR ACT. Let Xbe a 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, quotient of a hyperelliptic Jacobian by a 2-
torsion subgroup. In dimension 3,Φis the Hitchin fibration of the moduli space of rank 2bundles with fixed
determinant on a curve of genus 2.
1. INTRODU CTI ON
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. a) The vector space W:= H0(X, S2TX)has dimension n, and the natural map SWH0(X, STX)
is an isomorphism.
b) The corresponding map Φ : TXW
=Cnis a Lagrangian fibration.
c) When Xis general, the general fiber of Φis of the form ArZ, where Ais an abelian variety and
codim Z2.
We will give a precise geometric description of the map Φand of the abelian variety Ain § 4 and 5.
1.1. Comments. 1) For n= 2, a) follows from Theorem 5.1 in [DO-L], while b) and c) are proved in [K-L].
The proof is based on the isomorphism TX
=1
X(1). The Theorem also follows from the fact that Xis
a moduli space for parabolic rank 2 bundles on P1[C], so that Φ : TXC2is identified to the Hitchin
fibration (see [B-H-K]).
For n= 3,Xis isomorphic to the moduli space of vector bundles of rank 2 and fixed determinant
of odd degree [N]; again the Theorem follows from the properties of the Hitchin fibration (see §2). It
would be interesting to have a modular interpretation of Φfor 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 Φis an example of an algebraically completely integrable system see for instance [V], and
Remark 5.1. Such a situation is rather exceptional: most varieties do not admit nonzero symmetric ten-
sors (for instance, hypersurfaces of degree 3[H-L-S]); when they do, even for varieties as simple as
quadrics, the algebra of symmetric tensors is fairly complicated. We do not have a conceptual explanation
for the particularly simple behaviour in our case.
J. Liu is supported by the National Key Research and Development Program of China (No. 2021YFA1002300), the NSFC grants
(No. 12001521 and No. 12288201) and the CAS Project for Young Scientists in Basic Research (No. YSBR-033).
1
2 A. BEAUVILLE, A. ETESSE, A. H ¨
ORING, J. LIU, AND C. VOISIN
3) 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 conclude.
1.2. Strategy. We will first treat the case n= 3 , which is independent of the rest of the paper 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), from which we deduce a map TXW
=Cn 3). We then show that
Φhas the required properties, which implies a), b) and c) for general X(5.1). In the second approach
7) we prove directly a) for all smooth X, by realizing Xas a double covering of a quadric.
1.3. Notations. Throughout the paper Xwill be a smooth complete intersection of two quadrics in Pn+2 ,
with n2. We denote by TXits cotangent bundle and by PTXits projectivization in the geometric
sense (not in the Grothendieck sense). If Vis a vector space, we denote by P(V)the associated projective
space Vr{0}/Cparametrising one-dimensional subspaces of V.
2. THE CA SE n= 3
In this section we show how our general results can be obtained in the case n= 3 by interpretating X
as a moduli space.
As in 4.1 below, we associate to Xa genus 2 curve C, such 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[N]. The cotangent bundle TMis naturally
identified with the moduli space of Higgs bundles, that is pairs (E, u)with EMand u:EEKC
a homomorphism with Tr u= 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 [H].
Let ωH0(K2
C).We assume in what follows that ωvanishes at 4distinct points. Let Cωbe 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 : J CωJ C ; it is a 3-dimensional abelian variety.
Proposition 2.1. The fiber Φ1(ω)is isomorphic to the complement of a curve in P.
Proof : Recall that the map L7→ πLestablishes a bijective correspondence between line bundles on
Cωand rank 2 vector bundles Eon Cendowed with a homomorphism u:EEKCsuch that
u2=ω·IdE, or equivalently, Tr u= 0 and det u=ω(see for instance [B-N-R]). To get (E, u)in Φ1(ω)
we have to impose moreover det E=Nand Estable. Since det πL= Nm(L)K1
C, the first condition
means that Lbelongs to the translate PN:= Nm1(KCN)of P.
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, that is, L=πM(p)for some point
pCω. The condition LPNmeans M2(πp) = KCN, so Mis determined by pup to the 2-torsion
of J C . Thus the locus of line bundles LPNsuch that πLis unstable is a curve.
Let ρ:CP1be the canonical double covering, and BP1its branch locus. Since the homomor-
phism S2H0(KC)H0(K2
C)is surjective, the divisor of ωis of the form ρ(p+q), for some p, q P1;
by assumption we have p6=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.
SYMMETRIC TENSORS ON THE INTERSECTION OF TWO QUADRICS AND LAGRANGIAN FIBRATION 3
Proof : Let χ:P1P1be the double covering branched along {p, q}. Since div(ω) = ρ(p+q), there is
a cartesian diagram of double coverings
Cω
ξ//
π
P1
χ
Cρ
//P1
which gives rise to two commuting involutions σ, τ of Cω, exchanging the two sheets of πand ξrespec-
tively. The field of rational functions on Cωis
C(x, y, z)|y2=f(x), z2=g(x)
where fand gare polynomials with div f=Band div g={p, q}. Then σand τchange the sign of y
and zrespectively.
The involution στ is fixed point free, so the quotient Γ := Cω/hστ ihas 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
Cρ
//P1.
Let αJΓ. We have Nmπϕα=ρNmψα= 0, hence ϕmaps JΓinto PJCω. Since ϕis ´etale, we
have Ker ϕ=Z/2; since dim JΓ = dim P= 3 ,ϕis surjective.
3. DE FIN ITI ON OF Φ
Let Ybe a smooth degree dhypersurface in PN, defined by an equation f= 0. Recall that one
associates to fa section hfof S21
Y(d), the hessian or second fundamental form of f[G-H]: at a point yof
Y, the intersection of Ywith the tangent hyperplane Hto Yat yis a hypersurface in Hsingular at y,
and hf(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,
to X, we get a linear map
VOX S21
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; then for each qV, the quadric defined by hq(x)parameterizes the lines passing through
xand contained in the quadric {q= 0}.
Now we want to consider the “inverse” of the quadratic form hf(x)on Tx(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 1
X(d)
4 A. BEAUVILLE, A. ETESSE, A. H ¨
ORING, J. LIU, AND C. VOISIN
which induces a twisted symmetric morphism on (n1)-th exterior powers, namely
n1hf:^n1TX ^n11
X((n1)d).
We now observe that KX=OX(n1r+dr), hence
^n1TX
=1
X(n+ 1 r(d1)) ,^n11
X
=TX(n1 + r(d1)) ,
so that n1hfis in fact a symmetric morphism from 1
X(n+1 r(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 n1in f, hence we have
constructed a morphism
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 in
Pn+2 . The previous construction gives a morphism
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)for vTX,λC, so Φinduces a rational map
ϕ:PTX99K 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. Let Fbe the Fano variety of lines contained in X, and let
GF×X={(ℓ, y)|y}.
Then Fand therefore Gare smooth [R, Theorem 2.6], hence Fx, which is the fiber above xof the pro-
jection GX, is smooth since xis general. It follows that, in an appropriate system of coordinates
(k1,...,kn)of Tx(X), the forms {hq(x)}can be written
tXk2
i+Xαik2
iwith αidistinct in C, t C.
Then n1hq(x)is given by the diagonal matrix with entries βi:= Y
j6=i
(t+αj) (i= 1,...,n). These poly-
nomials in tare linearly independent, hence they generate the space of quadratic forms on T
xXwhich
are diagonal in the basis (ki). This linear system has dimension n, so αis injective; it has no base point,
so ϕinduces a finite, surjective morphism P(T
xX)Pn1. Thus Φis surjective, and ZP(T
xX) = ,
which gives 2) and 3).
SYMMETRIC TENSORS ON THE INTERSECTION OF TWO QUADRICS AND LAGRANGIAN FIBRATION 5
We want to give a geometric construction of the rational map ϕ:PTX99K Pn1. A point of PTX
is a pair (x, H ), where xXand His a hyperplane in Tx(X). Restricting the pencil {hq(x)}qVto H
gives a pencil of quadrics on H, which for (x, H )general contains n1singular quadrics q1,...,qn1.
The subset {q1,...,qn1}of P(V)corresponds to a point ϕx,H of P(Sn1V) namely the hyperplane
in Sn1Vspanned by qn1
1, . . . , qn1
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;
in other words, ϕmaps the point Hof P(T
xX)to the hyperplane of forms in Wvanishing at H.
On the other hand, ϕx,H is the hyperplane of Sn1Vspanned by the qn1for those qVsuch that
hq(x)|His singular; this condition is equivalent to say 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 3.1. codim Z2.
Proof : Suppose Zcontains a component Z0of codimension 1; since p(Z)6=X, we have Z0=p1(p(Z0)) .
We claim that this is impossible, in fact Zcannot contain a fiber p1(x). Indeed this would mean that for
qV, the form hq(x)is singular along all hyperplanes HTxX, that is, hq(x)has rank n2. 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. FIBE RS OF ϕ
In an appropriate system of coordinates (x0,...,xn+2 ), our variety Xis defined by the equations
q1=q2= 0, with
q1=Xx2
i, q2=Xµix2
iwith µiCdistinct.
Let Π = P(V) (
=P1)be the pencil of quadrics containing X. We choose a coordinate ton Πso that
the quadrics of Πare given by tq1q2= 0 . Then the singular quadrics of Πcorrespond to the points
µ0,...,µn+2 .
The goal of this section is to describe the general fiber of the rational map ϕ:PTX99K Sn1Π (
=Pn1).
For λ= (λ1,...,λn1)Sn1Π, let Cµ,λ denote the hyperelliptic curve y2=Q(tµi)Q(tλj), of
genus n. We will prove:
Proposition 4.1. For λgeneral in Sn1Π, the fiber ϕ1(λ)is birational to the quotient of the Jacobian J Cµ,λ by
the group Γ := 1JC} × Γ+, 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 ([R],
see also [D-R]). Let YP2g+1 be a smooth intersection of 2 quadrics, and let Ξ (
=P1)be the pencil of
quadrics containing Y. Let ΣΞbe the subset of 2g+ 2 points corresponding to singular quadrics,
and let Cbe the double covering of Ξbranched 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 Fof (g1) -planes contained in Yis also isomorphic to JC , but this isomorphism is not
canonical.
6 A. BEAUVILLE, A. ETESSE, A. H ¨
ORING, J. LIU, AND C. VOISIN
In an appropriate system of coordinates, the equations of Yare of the form
Xx2
i=Xαix2
i= 0 with αiCdistinct;
then Σ = {α1,...,α2g+2}. The group Γ := (Z/2Z)2g+1 acts on Y(hence also on F) by changing the
signs of the coordinates. Let Γ+Γbe the subgroup of elements which change an even number of
coordinates. For an appropriate choice of the isomorphism F
J C , the image of Γ+in Aut(JC)is
the group T2of translations by 2-torsion elements of J C , and the image of Γis T2× 1J C }[D-R,
Lemma 4.5].
4.2. An auxiliary construction. We consider the projective space P2n+1 equipped with the system of
homogeneous coordinates
x0,...,xn+2;y1, . . . , yn1
and the affine space An1equipped with the affine coordinates λ1,...,λn1. Let
XP2n+1 ×An1
be the complete intersection of the two quadrics with equations
Q1=Q2= 0 with Q1=
n+2
X
i=0
x2
i+
n1
X
j=1
y2
j, Q2=
n+2
X
i=0
µix2
i+
n1
X
j=1
λjy2
j.
The second projection XAn1gives a family of complete intersections of two quadrics Xλof dimen-
sion 2n1parameterized 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, that is
F={(P, λ)|λAn1, P (n1)-plane Xλ}.
For λgeneral, the fiber Fλis isomorphic to the Jacobian of the hyperelliptic curve Cµ,λ (4.1).
Let (P, λ)be a general point of F. Then PPn+2 is a point xof X. Let π:P2n+1 99K Pn+2 be
the projection (xi, yj)7→ (xi). Since the differentials of Qiand qicoincide at x, the derivative πmaps
Tx(P)Tx(X)into Tx(X). Since Pis general, πTx(P)is a hyperplane in Tx(X) this will follow
from the proof of Proposition 4.2 1) below, where we construct explicitely pairs (P, λ)with this property.
Therefore we have a rational map
ψ:F99K PTX(P, λ)7→ (x=PPn+2 , πTx(P)) .
The symmetric group Sn1acts on P2n+1 by permuting the yj, and the group (Z/2Z)n1by 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 there-
fore on F, compatible via pwith the action on the base. The map ψis invariant under this action, hence
factors through the quotient F/G. By passing to the quotient we get a map p:F/G An1/Sn1.
Proposition 4.2. 1) ψinduces a birational map ψ:F/G 99K PTX.
2) There is a commutative diagram
F/G ψ
//
p
PTX
ϕ
An1/Sn1σ
//An1Pn1
SYMMETRIC TENSORS ON THE INTERSECTION OF TWO QUADRICS AND LAGRANGIAN FIBRATION 7
where pis deduced from p, and 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)of Ker π, we have α= (α1,...,αn1), where the αiare linear forms on
H. The condition PXλimplies that the hessians hQ1(x)and hQ2(x)vanish on Tx(P), which gives
(1) hq1(x)|H=X
i
α2
i, hq2(x)|H=X
i
λiα2
i.
This is a simultaneous diagonalization 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, and let (x, H ) := ψ(P, λ). According to Proposition 3.2, ϕ(x, H )is given by the
(n1)-uple of quadrics qΠsuch that the form hq(x)|His singular. Using (α1,...,αn1)as coordinates
on H, we see from (1) that this (n1)-uple is given by (λ1,...,λn1), which proves 2).
4.3. Proof of Proposition 4.1. Let λbe a general element of An1. Let us denote by Γthe subgroup
(Z/2Z)n1of G. From Proposition 4.2 and the cartesian diagram
F/Γ//
p
F/G
p
An1//An1/Sn1
we see that the fiber ϕ1(λ)is birational to the quotient Fλ/Γ. By (4.1) Fλis isomorphic to J Cµ,λ ,
and one can choose the isomorphism so that Γacts on J Cµ,λ as 1J} × Γ+, where Γ+is a group of
translations by 2-torsion elements. This proves the Proposition.
5. FIBE RS OF Φ
5.1. Results. We keep the settings of the previous section. Recall that our parameter λlives in
An1Sn1Π
=Pn1. For λin An1, we denote by ˜
λa lift of λin Cnfor the quotient map
Cnr{0} Pn1.
Theorem 5.1. Assume that Xis general. For λAn1general, the fiber Φ1(˜
λ)is isomorphic to ArZ,
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 5.1. For every smooth complete intersection of two quadrics XPn+2 , the fibration Φ : XCnis
Lagrangian.
8 A. BEAUVILLE, A. ETESSE, A. H ¨
ORING, J. LIU, AND C. VOISIN
Proof : Assume first that Xis general. The symplectic form on TXis , where ηis the Liouville form.
By the Theorem and the Hartogs principle, the pull back of ηto a general fiber of Φis 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 globalized over B: we have a rank 2 vector bundle Vover Bwhose fiber at a point bBis
the space of quadratic forms vanishing on Xb. We get a homomorphism Sn1VpTX/B , which 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 fiber of Φ, it vanishes
on all fibers.
Corollary 5.2. Assume that Xis general. The multiplication map SH0(X, S2TX)H0(X, STX)is an
isomorphism.
(We will give in § 7 a proof valid with no generality assumption.)
Proof : The Theorem implies that every function on a general fiber of Φis 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,...,tn]
H0(X, STX). By construction
the tiare mapped to elements of H0(X, S2TX), so the Corollary follows.
Remark 5.1.Let V1,...,Vnbe the Hamiltonian vector fields on TXassociated to the components of Φ.
For λgeneral in Cn, let us identify Φ1(λ)to ArZas in the Theorem. Then by Hartogs’ principle the
Vilinearize on A that is, they extend to a basis of H0(A, TA). This allows in principle to write explicit
solutions of the Hamilton equations for Φiin terms of theta function.
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 by Fo
λits intersection with the fiber Fλ. Since λis
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. Let ZPTXbe the indeterminacy locus of
ϕ 3), and let Fbad
λ:= (ψo
λ)1(Z)Fo
λ.
Proposition 5.1. Fbad
λhas codimension 2in Fλ.
We postpone the proof of the Proposition to the next section, and first show how it implies Theorem
5.1.
Let 0XTXbe the zero section, and let q:TXr0XPTXbe the quotient map. Let
ϕo:PTXrZPn1be the morphism induced by ϕ. We have q1(˜
λ)) = (ϕo)1(λ), and the
restriction
qλ: Φ1(˜
λ)(ϕo)1(λ)
is an ´etale double cover, with Galois involution ιinduced by (1TX).
We put Foo
λ:= Fo
λrFbad
λ, and consider the restriction
ψo
λ:Foo
λ(ϕo)1(λ)of ψo.
Lemma 5.1. The fiber Φ1(˜
λ)is Lagrangian, and has trivial tangent bundle.
Proof : The ´etale double cover qλinduces by fibered product an ´etale double cover
π:f
Foo
λFoo
λ
SYMMETRIC TENSORS ON THE INTERSECTION OF TWO QUADRICS AND LAGRANGIAN FIBRATION 9
such that ψo
λlifts to a morphism ˜
ψo
λ:f
Foo
λΦ1(˜
λ).
By Proposition 5.1, the complement of Foo
λin Fλhas codimension 2, so πextends to an ´etale
double cover f
FλFλ, where f
Fλis an abelian variety or the disjoint union of two abelian varieties. The
morphism ˜
ψo
λ:f
Foo
λΦ1(˜
λ)is generically of maximal rank. Again by Proposition 5.1, the holomorphic
1-forms on f
Foo
λare closed, hence by pull back the same holds for the holomorphic 1-forms on Φ1(˜
λ).
As in the proof of Corollary 5.1, this implies that Φ1(˜
λ)is Lagrangian. The second assertion is a basic
property of Lagrangian fibers.
Lemma 5.2. The morphism ψo
λlifts to a morphism ˜
ψo
λ:Foo
λΦ1(˜
λ).
Proof : It suffices to show that the double covering π:f
Foo
λFoo
λsplits.
Assume the contrary, so that f
Fλis an abelian variety. By Lemma 5.1 H01(˜
λ),1)has dimension
n. It follows that the pull back (˜
ψo
λ):H01(˜
λ),1)H0(f
Foo
λ,1)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: Φ1(˜
λ)(ϕo)1(λ).
Now we observe that the 1-forms on Φ1(˜
λ)are “pure”, that is, extend to any smooth projective
compactification of Φ1(˜
λ): this follows from the fact that this holds after pull back to f
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 that any Zariski open set has no nonzero closed pure holomorphic
1-forms. This contradiction proves the Lemma.
5.3. Proof of Theorem 5.1. Lemma 5.2 gives a factorization
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 ΓΓof index 2, the morphism ˜
ψo
λ:Foo
λΦ1(˜
λ)factors through
a birational morphism
˜
ψo
λ,H:Foo
λ/Γ Φ1(˜
λ).
By Lemma 5.1, 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,
hence is the subgroup Γ+of Γgenerated by translations, isomorphic to (Z/2Z)n2; thus Fλ/Γis an
abelian variety A.
To simplify notation, we put Ao:= Foo
λ/Γand u:= ˜
ψo
λ,H. The rational map u1: Φ1(˜
λ)99K Ais
everywhere defined (see e.g. [B-L, Theorem 4.9.4]), so we have two morphisms
Aou
Φ1(˜
λ)u1
A
whose composition is the inclusion Ao֒A. Since the tangent bundles of Aand Φ1(˜
λ)are trivial, the
determinant of T u :TAouTΦ1(˜
λ)is a function on Ao, hence constant by Proposition 5.1. Therefore u
is ´etale and birational, hence an open embedding. This implies that every function on Φ1(e
λ)is constant
(because its restriction to Aois constant). Then the previous argument shows that u1is also an open
embedding, so that Φ1(˜
λ)is isomorphic to an open subset of Acontaining Ao. This proves the Theorem.
10 A. BEAUVILLE, A. ETESSE, A. H ¨
ORING, J. LIU, AND C. VOISIN
6. PROO F OF PRO POSIT ION 5.1
We keep the notations of (4.2). Recall that we have coordinates (x0,...,xn+2;y1,...,yn1)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 , and let Rbe the vector space of quadratic
forms in y= (y1,...,yn1). We define an extended family XeP2n+1 ×R2by
Xe={(x, y); (r1, r2)P2n+1 ×R2|q1(x) + r1(y) = q2(x) + r2(y) = 0}.
The fiber Xe
rat a point r= (r1, r2)of R2is 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:Fe99K PTX, which maps a general PXe
rto the pair (x, H)
with {x}=PPn+2 ,H=πTx(P).
We observe that a general pair r= (r1, r2)of R2is simultaneously diagonalizable, so the restriction
of ψeto Fe
rcoincides, for an appropriate choice of the coordinates (yi), with the map ψλthat we want
to study. Thus Proposition 5.1 will follow from the following Proposition:
Proposition 6.1. Assume that Xis general.
1) Let ΓFebe the locus of points (P, r)such that either dim PPn+2 >0, or PPn26=. Then Γ
has codimension 2in Fe.
2) There exists no divisor in FerΓwhich dominates R2and is mapped to the base-locus ZPTXby ψe.
Proof : 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)with k0,l 1, let Gk,l be the locally closed subvariety
of (n1)-planes PGsuch that
dim(PPn+2) = k , dim(PPn2) = l .
(We put by convention l=1if PPn2=.) Let
FQ:= {(P, (Q1, Q2)) G×Q2|Q1|P=Q2|P= 0},
FQ
k,l := FQ(Gk,l ×Q2).
The general fiber of the projection FQQ2is an abelian variety, and we recover Feby restricting FQ
to pairs of quadratic forms of the form (q1(x) + r1(y), q2(x) + r2(y)) . It thus suffices to prove the result
for the larger family FQ, that is, to show that FQ
k,l has codimension 2in FQ.
This is done by a dimension count. For PG, let ϕPbe the restriction map QH0(P, OP(2)). The
fiber of the projection FQGis the vector space (Ker ϕP)2. For Pgeneral, ϕPis surjective: this is
the case for instance if Pis contained in the (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 PPn2
are apolar for all forms in Im ϕP. 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
SYMMETRIC TENSORS ON THE INTERSECTION OF TWO QUADRICS AND LAGRANGIAN FIBRATION 11
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)
2if k1or l0.
2) The base locus ZPTXhas codimension 2(Corollary 3.1). Note that ψeis well-defined in
FerΓ. If Dis a divisor in FerΓwith ψe(D)Z, the map ψehas not maximal rank along D. This
contradicts the following Lemma:
Lemma 6.1. ψehas maximal rank on FerΓ.
Proof : Let (x, H)be a point of TX; we view Has a hyperplane in the projective tangent space to xat
X. The fiber of ψe:FerΓPTXat (x, H)is the locus
(ψe)1(x, H) = {(P, r1, r2)G×R2|PPn+2 ={x}, P Pn2=, π(P) = H, (2)
(qi(x) + ri(y))|P= 0 (i= 1,2)}.(3)
The equations (2) define a smooth, locally closed subvariety Gx,H of G. Let PGx,H , and let χP:RH0(P, OP(2))
be the restriction map. We will show below that the image of χPis the space of quadratic forms on P
which are singular at x. Since the forms qi|Pare singular at x, this implies that the solutions of (3) form
an affine space over (Ker χP)2. Therefore (ψe)1(x, H )admits an affine fibration over Gx,H , hence is
smooth.
Clearly the quadrics in Im χPare singular at x. To prove the opposite inclusion, choose the coordinates
(xi)so that x= (1,0,...,0). Since PPn+2 ={x}, there exist linear forms 1,...,ℓn+2 in the yjso that
Pis defined by xi=i(y)for i= 1,...,n+ 2. Then a quadratic form on P2n+1 singular at xcan be
written as a form in x1,...,xn+2;y1,...,yn1, hence its restriction to Pis in Im χP. This proves the
Lemma, hence also the Proposition.
7. SY MME TRIC TE NSO RS:S ECO ND A PPR OAC H
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 PTQparameterizes pairs (x, P )with xQand Pa(n1) -
plane tangent to Qat x. Thus we get a morphism γfrom PTQto the grassmannian Gof (n1) -planes
in Pn+1 , which is the morphism defined by the linear system |OPTQ(1)|. It is birational onto its image,
but contracts the subvariety CPTQconsisting of pairs (x, P )such that Pis tangent to Qalong a
line Q, and x: then γ1(P)consists of the pairs (x, P )with x.
Let hqH0(Q, S21
Q(2)) be the hessian form of q(§3). Choosing coordinates (xi)such that q(x) = Px2
i,
we have hq=P(dxi)2(note that this is, up to a scalar, the unique element of H0(Q, S21
Q(2)) invari-
ant under Aut(Q)). Then hq(x)is non-degenerate at each point xof Q, so hqinduces an isomorphism
1
Q(1)
TQ(1), hence also S21
Q(2)
S2TQ(2). The image in H0(Q, S2TQ(2)) of hqby this iso-
morphism is h
q=P2
j. We will view h
qas an element of H0(PTQ, OPTQ(2) pOQ(2)) , where
p:PTQQis the projection.
Proposition 7.1. The divisor of h
qis C. The projection p|C:CQis a smooth quadric fibration, and Cis a
prime divisor for n3.
12 A. BEAUVILLE, A. ETESSE, A. H ¨
ORING, J. LIU, AND C. VOISIN
Proof : Let xQ; the hyperplane tangent to xat Qcuts down a cone over the smooth quadric
QxP(Tx(Q)) defined by hq(x) = 0 3). The isomorphism Tx(Q)
T
x(Q)given by hq(x)car-
ries 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, that is yQ
x.
This proves the equality C= div(h
q). Thus the fiber of p|C:CQat xis Qx, which is smooth, and
connected if n3.
Remark 7.1.The variety Cis an example of a total dual VMRT [H-L-S], for the proof of the Theorem we
will combine this tool with the birational transformation of PTXdefined by a double cover, cf. [A-H].
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 7.1. The image s(B)is not contained in C.
Proof : Let xB. The point s(x)in P(T
x(Q)) corresponds to the hyperplane image of Tx(B)in Tx(Q);
we must 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 to the cone QPTx(Q)at a
smooth point yof Q.
Suppose this is the case, with y= (y0,...,yn+1 ). We can assume that Qis defined by Pαix2
i= 0,
with αiCdistinct. Then the (projective) tangent space to Qat x, given by P(αixi)ξi= 0, must
coincide with the tangent space to Qat y, given by Pyiξi= 0. This implies y= (α0x0,...,αn+1 xn+1 ).
Thus the point xmust satisfy Xx2
i=Xαix2
i=Xα2
ix2
i= 0 .
If these relations hold for all xin B, the quadric Pα2
ix2
i= 0 must belong to the pencil spanned by Q
and 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. Wekeep the notation of the previous sections: XP=Pn+2
is defined by q1=q2= 0, with
q1=
n+2
X
i=0
x2
i, q2=
n+2
X
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 an exact sequence of symmetric
tensors
(4) 0S2TXS2TP|X
(dq1,dq2)
TP|X(2)2,
where dqi(V1V2) = (V1·qi)V2+ (V2·qi)V1for V1, V2in H0(X, TP|X).
SYMMETRIC TENSORS ON THE INTERSECTION OF TWO QUADRICS AND LAGRANGIAN FIBRATION 13
Proposition 7.2. The quadratic vector fields si:= X
j6=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 .
We have (xijxji)·q1= 0 , hence dq1(si) = 0, and (xijxji)2·q2= 2(µjµi)xixj(xijxji),
hence, using Pxjj= 0 and q1|X= 0:
dq2(si) = 2x2
iX
j6=i
xjj(2xii)X
j6=i
x2
j= 0 ,which proves the Proposition.
Fron now on we will consider the sias elements of H0(X, S2TX).
7.3. The double cover. Let p:Pn+2 99 K Pn+1 be the projection (x0,...,xn+2 )7→ (x1,...,xn+2 ). The
image p(X)is the smooth quadric Qin Pn+1 defined by
n+2
X
i=1
(µiµ0)x2
i= 0 .
The restriction π:XQof pis a double covering, branched along the subvariety BQdefined by
n+2
X
i=1
x2
i=
n+2
X
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
which is an isomorphism outside of R. Consider the normal exact sequence
0TRTX|RNR/X 0.
The involution ι: (x0,...,xn+2 )7→ (x0, x1,...,xn+2)acts on TX|R; this splits the exact sequence,
giving a decomposition
TX|R=TRNR/X
into eigenspaces for the eigenvalues +1 and 1. Let ρ:TX|RTRbe the projection on the first
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
(5) h2(s0) = 0 and h2(si) = X
j>0
j6=i
(xijxji)2
µjµi
for i > 0 ;
in other words, h2maps s1, . . . , sn+2 to the elements ˆs1,...,ˆsn+2 of H0(R, S2TR)constructed in Propo-
sition 7.2 applied to R.
14 A. BEAUVILLE, A. ETESSE, A. H ¨
ORING, J. LIU, AND C. VOISIN
Let πPTQbe the pull back under πof the projective bundle PTQQ. The homomorphism
τ:TXπTQgives rise to a birational map g:πPTQ99K PTX. Following the geometric descrip-
tion of the tangent map as an elementary transformations of vector bundles in the sense of Maruyama
[M1],[M2, Corollary 1.1.1], one has a commutative diagram
(6) Γ
µ
{{ν
""
πPTQg//
p
##
PTX
q
||
X
where pand qare the canonical projections, ν: Γ PTXis the blow-up along the subspace PTRPTX
defined by the projection ρ,µ: Γ πPTQis the blow-up of the image Bof the embedding
B ֒πPTQdeduced from the surjective homomorphism πTQπNB /X .
Let Eµbe the exceptional divisor of µ. By [M2, Theorem 1.1], there is an isomorphism
(7) µOπPTQ(1) OΓ(Eµ)
=νOPTX(1) ,
as well as the equality
(8) νEµ=qR .
7.4. The divisor of s0.We now consider the divisor CPTQdefined in (7.1), and the cartesian dia-
gram
πPTQπ
//
PTQ
Xπ//Q .
Put C:= π′−1(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
X
j=1
(x0jxj0)2
µjµ0
H0(X, S2TX)(7.2). We will view
s0as an element of H0(PTX, O(2)) .
Proposition 7.3. Assume n3. We have gC= div(s0).
Proof : We first show that gC |OPTX(2)|. By Proposition 7.1 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, we have
π(B)π(C) = C. But π(B) = s(B), where s:B ֒PTQis the embedding defined by the
surjective homomorphism TQNB/Q . Then the result follows from Lemma 7.1.
Since gCand div(s0)are linearly equivalent effective divisors and gCis irreducible, it suffices to
show that their restrictions to PT
xXcoincide for a general point xX.
SYMMETRIC TENSORS ON THE INTERSECTION OF TWO QUADRICS AND LAGRANGIAN FIBRATION 15
Fix a point x= [x0,...,xn+2]XrR, so that x06= 0 . Then the tangent map T π (x) : Tx(X)Tπ(x)(Q)
is an isomorphism; in the diagram (6), the maps µ, ν and grestricted over the fibers at xare all isomor-
phisms. Let us show that Cand T π (div(s0)) define the same quadric in P(Tπ(x)(Q)).
Now CP(T
x(X)) = CP(T
π(x)(Q)) is the quadric defined by the element h
qof (7.1). In the
coordinates (zi)defined by zi= (µiµ0)1/2xi, the equation of Qis
n+2
X
j=1
z2
j= 0, so
h
q=
n+2
X
j=1
∂zj2=
n+2
X
j=1
2
j
µjµ0
·
On the other hand, since π(x0,...,xn+2 ) = (x1,...,xn+2), we have T π(0) = 0 and T π(j) = jfor
j > 0, hence
T π(s0) = x2
0
n+2
X
j=1
2
j
µjµ0
·
Since x06= 0, this proves the Proposition.
7.5. Proof of part a) of the Theorem. Suppose now that n3. Consider the double cover π:XQ
and the ramification divisor RX. The restriction maps hkdefined in (7.3) yield a homomorphism of
graded C-algebras
h:S(X) := H0(X, STX) H0(R, STR) =: S(R).
Proposition 7.4. 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).
Fix an element sIof degree k. It corresponds to an effective Cartier divisor Gin the linear system
|OPTX(k)|. Recall the commutative diagram (6)
Γ
µ
{{ν
""
πPTQg//
p
##
PTX
q
||
X
Put ˆ
G:= µνGπPTQ. By (7), ˆ
Gbelongs to the linear system |OπPTQ(k)|.
Here comes 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)
16 A. BEAUVILLE, A. ETESSE, A. H ¨
ORING, J. LIU, AND C. VOISIN
so that 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 (7.1) that Cis
the union of the lines which are contracted by the morphism γ:PTQG, so that c1(OPTQ(1))·= 0 .
Thus the curves := π′∗ cover C, and 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. Since gC= div(s0)by Proposition 7.3,
this proves the Proposition.
The following Proposition implies part a) of our main Theorem:
Proposition 7.5. Assume n2. For any choice of indices 0i1< . . . < inn+ 2, the homomorphism
C[t1,...,tn]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 [DO-L, Theorem 5.1], except
the fact that any two of the sigenerate H0(X, S2TX). Up to permuting of the coordinates, it suffices to
prove that s0and s1are linearly independent. But h2:H0(X, S2TX)H0(R, S2TR)maps s0to zero
and si, for i > 0, to the corresponding elements ˆsiof H0(R, S2TR); this implies our assertion.
Assume n3. By the induction hypothesis, the homomorphism C[t1,...,tn1]S(R)which maps
tito ˆsiis an isomorphism of graded C-algebras (with deg(ti) = 2 ). It follows that his surjective, and that
(s0,...,sn1)form a basis of H0(X, S2TX)and generate the C-algebra S(X). Thus we have a surjective
homomorphism u:C[t0,...,tn1]S(X), with u(ti) = si.
In particular, the Krull dimension of S(X)is at most n. On the other hand, the ring S(X)is a domain
and s0is neither zero nor a unit. Thus, by Krull’s Hauptidealsatz, the Krull dimension of S(X)is equal
to n, hence uis an isomorphism. By permutation of the coordinates we get the same result for any choice
of nelements in {s0,...,sn+2}, hence the Proposition.
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UNI VER SI T ´
ECˆ
OT E D’AZU R, C NRS L ABO RAT OIRE J .-A . DI EU DO NN ´
E, PARC VA LR OS E , F- 06 108 N I CE C EDE X 2, FR AN C E
Email address:arnaud.beauville@univ-cotedazur.fr
INS TI TUT MAT H ´
EM ATI QUE DE TOUL OU S E, U NIVE RS IT ´
EPAUL SABAT I ER , 118 RTE D E NA RB ON N E, 31400 TO U LO US E
Email address:antoine.etesse@ens-lyon.fr
UNI VER SI T ´
ECˆ
OT E D’AZU R, C NRS L ABO RAT OIRE J .-A . DI EU DO NN ´
E, PARC VA LR OS E , F- 06 108 N I CE C EDE X 2, FR AN C E
Email address:Andreas.Hoering@univ-cotedazur.fr
INS TI T UT E O F MAT H EM ATIC S, A CAD EM Y O F MATHE MAT I CS A ND SY S TE MS SC I EN C E, C HI N ES E AC AD EM Y O F SC IE NCE S, B E I-
JI NG , 100190, CH IN A
Email address:jliu@amss.ac.cn
SOR BO N NE UN IV ERS IT ´
E AN D UN IV ERS IT ´
EPARI S CI T ´
E, CN RS , IMJ -PRG , F-75 005 PAR I S, F RAN CE
Email address:claire.voisin@imj-prg.fr
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