Maria Valtorta
72 pages

Maria Valtorta

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Maria Valtorta THE NOTEBOOKS 1944 Maria Valtorta THE NOTEBOOKS 1944 Translated from the Italian by David G. Murray CENTRO EDITORIALE VALTORTIANO Contents Preface to the English Edition 13 JAN. 1 Contemplation of the Holy Face of Jesus 19 2 Vision of the Apostolic College The Resemblance Between Jesus and John 3 Jesus as the Good Samaritan 24 4 Those Kissed by God and the Falsifiers 25 Three Crosses: Three Victim Souls 27 5 Vision of the Dormition of Mary 28 6 Epiphany: The Meaning of Gold, Incense, and Myrrh 31 7 The Order of Human Life: Spirit and Flesh 33 8 Satanism as Opposed to Deification through Christ 35 9 The Father's Condemnation of
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Publié par
Nombre de lectures 26
Langue English


JULY 31|AUGUST 27, 1988
1. Introduction
This is a course about K3 surfaces and several related topics. I want to begin by
working through an example which will illustrate some of the techniques and results
we will encounter during the course. So consider the following problem.
3Problem. Find an example of C XP ,where C is a smooth curve of genus 3
and degree 8 andX is a smooth surface of degree 4.
Of course, smooth surfaces of degree 4 are one type of K3 surface. (For those who
don’t know, a K3 surface is a (smooth) surfaceX which is simply connected and has
trivial canonical bundle. Such surfaces satisfy (O ) = 1, and for every divisor DX
on X, DD is an even integer.)
We rst try a very straightforward approach to this problem. Let C be any smooth
curve of genus 3, and let Z be any divisor on C of degree 8. (For example, we may
takeZ to be the sum of any 8 points onC.) I claim that the linear systemjZj de nes
an embedding of C. This follows from a more general fact, which I hope you have
seen before.
Theorem. LetC be a smooth curve of genusg, and letZ be a divisor onC of degree
d> 2g. Then the linear systemjZj is base-point-free, and de nes an embedding of
Proof. First suppose that P is a base point of Z. We have an exact sequence
0!O (Z−P)!O (Z)!O (Z)! 0C C P
0 0and the assumption that P is a base point implies H (O (Z−P)) = H (O (Z)).C C
But deg(Z−P)=d−1> 2g−2 and deg(Z)=d> 2g−2 so that both of these divisors
0Z and Z−P are non-special. By Riemann-Roch, h (O (Z−P)) =d− 1−g +1 =C
0d−g +1 =h (O (Z)), a contradiction.C

Similarly, suppose that P and Q are not separated by the linear systemjZj.(We
include the case P =Q, where we suppose that the maximal ideal of C at P is not
embedded.) Then in the long exact cohomology sequence associated to
0!O (Z−P−Q)!O (Z)!O (Z)! 0C C P+Q
0 0we must have that the map H (O (Z))! H (O (Z)) is not surjective. ThisC P+Q
1implies that H (O (Z−P−Q))=(0). ButC
1 0 H (O (Z−P−Q)) H (O (K −Z +P +Q))=C C C
and deg(K −Z +P +Q)= 2g− 2−d +2 < 0 so this divisor cannot be eectiveC
(again a contradiction). Q.E.D.
I have included this proof because later we will study the question: on a surface,
when does a linear system have base points, and when does it give an embedding?
To return to our problem, we have a curveC of genus 3 and a divisorZ of degree
8. This divisor is non-special and gives an embedding (since 8> 2 3). By Riemann-
0h (O (Z)) = 8−3+1 = 6C
5sojZj maps C intoP .
3The theory of generic projections guarantees that we can project C intoP from
5 3P in such a way as to still embedC. So we assume from now on: CP is a smooth
curve of degree 8 and genus 3. (The linear systemjO (1)j is not complete.)C
0 0We now consider the map H (O 3(k))! H (O (k)) for various degrees k.WeC
(k +1)(k +2)(k +3)0h (O 3(k)) =
0h (O (k)) = 8k−3+1 = 8k− 2C
(by Riemann-Roch).
0 0It is easy to see that for k 3we have h (O 3(k))<h (O (k)) while for k 4C
0 0we have h (O 3(k))>h (O (k)). This means that for k 3, the linear system cutC
out on C by hypersurfaces of degree k is incomplete, while for k 4there must be
a hypersurface of degree k containing C.Infact,for k =4 we have
3h (O (4)) = 35
0h (O (4)) = 30C
4so that there is at least aP of quartic hypersurfaces X containing C. By Bertini’s
4theorem, the generic X is smooth away from C (the base locus of thisP ). So we
3have almost solved our problem: we have constructed CXP with C smooth,
but X is only known to be smooth away from C.
Where do we go from here? We could continue to pursue these methods of projec-
3 4tive geometry inP . For example, we might consider a generic pencil inside ourP

0 0of quartics: the base locus of this pencil is C[C with C another curve of degree
08. (C = C since the arithmetic genus would be wrong). We could try varying the
0pencil and showing that the induced family of divisorsC\C onC has no base point.
But the arguments are very intricate, and I’m not sure if they work! So we will try
a dierent approach.
In this second approach, we construct X rst instead of C. Whatwe wantis
aK3 surface X together with two curves H and C on X.(H is the hyperplane
3section from the embeddingXP ). We want C to have genus 3,jHj to de ne an
3embeddingXP ,andC should have degree 8 under this embedding. To translate
these properties into numerical properties of H and C on X, notice that for any
curveDX we have
deg(K )=deg((K +D)j )= deg(Dj )=DDD X D D
(using the adjunction formula, and the fact that K is trivial). Thus, g(D)=X
The numerical versions of our properties are: C C =4 (so that g(C)= 3),
3HH =4 (so thatwe get a quartic XP )and HC =8 (so that C will have
3degree 8 inP ). In addition, we wantC to be smooth andjHj to de ne an embedding
3inP .
The rst step is to check that the topology of a K3 surface permits curves with these
numerical properties to exist. The topological properties I have in mind concern the
2intersection pairing onH (X;Z). Each curveD on X has a cohomology class [D]2
2H (X;Z), and the intersection number for curves coincides with the cup product
2 2 4 H (X;Z)⊗H (X;Z)!H (X;Z) Z:=
Poincar e duality guarantees that this is a unimodular pairing (i. e. in a basis, the
matrix for the pairing has determinant1). In addition, the signature of the pairing
(the number of +1 and−1 eigenvalues) can be computed as (3,19) for a K3 surface.
2The pairing is also even: this means xx2 2Z for all x2 H (X;Z). These three
properties together imply that the isomorphism type of this bilinear form overZ is
2 3(−E ) U whereE is the unimodular even positive de nite form of rank 8, and8 8
U is the hyperbolic plane. Let denote this bilinear form; is called the K3 lattice.
We will return to study these topological properties in more detail (and de ne the
terms!) in section 11.
2 In our example, we need a submodule of H (X;Z) generated by 2 elements h=
and c such that the matrix of the pairing on these 2 elements is
The fact that such a submodule exists is a consequence of

Theorem (James [12]). Given an even symmetric bilinear formL over the integers
such that L has signature (1;r− 1) and the rank r of L is 10, there exists a
submodule of isomorphic to L.
1(Later we will study re nements of this theorem in which the rank is allowed to
be larger.)
2So there is no topological obstruction in our case. Moreover, later in the course
we will partially prove the following:
Fact. Given a submodule L of on which the form has signature (1 ;r− 1) with
r 20, there exists a (20−r)-dimensional family of K3 surfacesfXg,eachequippedt
2 with an isomorphism H (X;Z) in such a way that elements of L correspond=t
to cohomology classes of line bundles on X.Moreover,for t generic, these are thet
only line bundles on X,that is, the Neron-Severi group NS(X ) [together with itst t
intersection form] is isomorphic toL.
I said we would partially prove this: what we will not prove (for lack of time) is the
global Torelli theorem and the surjectivity of the period map for K3 surfaces. This
3fact depends on those theorems.
To return to our problem: we now have a K3 surface X and two line bundles
O (H),O (C) with the correct numerical properties, which generate NS(X). ForX X
2 0 any line bundleL onX,wehaveH (L) H (L ) (sinceK is trivial), from which= X
follows the Riemann-Roch inequality:
L L0 0 h (L)+h (L ) +2
(since (O )=2). Thus, ifL L− 2eitherL orL is e ective.X
In our situation we conclude:H is e ective and C is e ective. Replacing H by
−H if necessary we may assume H is e ective. (The choice of C is then determined
by HC =8.)
4To nish our construction, we need another fact which will be proved later.
2Fact. Let H be an e ective divisor on a K3 surface X with H 4. IfjHj has
base points or does not de ne an embedding, then there is a curve E on X with
2 2 1E =−2or E = 0. Moreover, when it does de ne an embedding H (O (H)) = 0X
2and H (O (H)) = 0.X
5In our situation, we wish to rule out the existence of such an E.Wehave E
2 2 2mH +nC sinceH andC generate NS(X). Thus, E =4m +16mn+4n is always
1Need cross-reference.
2Need cross-reference.
3For statements of the theorems and further discussion, see section 12.
4Need cross-reference: section 5?
5The symbol denotes linear equivalence.

2 abdivisible by 4, soE =−2 is impossible. Moreover, if a rank 2 quadratic form ( )bc
2ab 48represents 0 then− det ( )= b −ac is a square.Inourcase,− det ( )= 48 isbc 84
2not asquare, so E = 0 is impossible.
We conclude thatjHj is very ample. Then HC = 8 implies−C is not e ective,
so that C must be e ective. Furthermore, jCj is then very ample, so this linear
system contains a smooth curve (which we denote again by C). By Riemann-Roch,
0 HH 3h (O (H)) = +2 = 4, sojHj maps X intoP as a smooth quartic surface, andX 2
3we have CXP as desired.
2. K3 surfaces and Fano threefolds
We will use in this course a de nition of K3 surfaces which is slightly dierent from
the standard one. Namely, for various technical reasons which will appear later, it is
convenient to allow K3 surfaces to have some singular points called rational double
6points. These will be the subject of a seminar later on; if you are not familiar with
them, I suggest that you ignore the singularities for the moment and concentrate on
smooth K3 surfaces.
(We do not use the term \singular K3 surface" to refer to these surfaces, because
that term has a dierent meaning in the literature: it refers to a smooth K3 surface
with Picard number 20. Cf. [27].)
Here is a convenient denition of rational double points: a complex surface X
has rational double points if the dualizing sheaf ! is locally free, and if there is aX
f resolution of singularities : X! X such that ! = ! =O (K ). For thoseX e e eX X X
unfamiliar with the dualizing sheaf, what this means is: for every P 2 X there is
a neighborhood U of P and a holomorphic 2-form = (z ;z )dz ^dz de ned1 2 1 2
on U−fPg such that () extends to a nowhere-vanishing holomorphic form on
−1 (U).
The structure of rational double points (sometimes called simple singularities)is
well-known: each such point must be analytically isomorphic to one of the following:
2 2 n+1A (n 1): x +y +z =0n
2 2 n−1D (n 4): x +yz +z =0n
2 3 4E : x +y +z =06
2 3 3E : x +y +yz =07
2 3 5E : x +y +z =08
fand the resolutionX!X replaces such a point with a collection of rational curves
of self-intersection−2 in the following con guration:
A (n curves)n
D (n curves)n
6See Appendix ??.
To return to the de nition of K3 surfaces: a K3 surface is a compact complex
1 analytic surfaceX with only rational double points such thath (O )=0 and! =X X
O .(If X is smooth, the dualizing sheaf ! is the line bundle associated to theX X
canonical divisorK , so this last condition says that the canonical divisor is trivial.)X
fIf X is a K3 surface and : X ! X is the minimal resolution of singularities
(i. e. the one which appeared in the de nition of rational double point) then it turns
1 1out that establishes an isomorphism H (O ) H (O ), and also we have ! ==X e eX X
f ! =O =O .Thus,thesmoothsurface X is alsoaK3 surface.X X eX
We will concentrate on smooth K3 surfaces for quite a while, and only return to
7singular ones in several weeks. I have included the singular case today so that we
don’t have to change the de nitions later.
Here are some of the basic facts about smooth K3 surfaces. Let X be a smooth
K3 surface.
1 2 0 0(1) (O ) = 2, because h (O )=0 andh (O )=h (O (K )) =h (O )=1.X X X X X X
2 2(2) c (X) = 0. [Remember that c (X)=K K :]X X1 1
(3) Therefore, using Noether’s formula
2c (X)+c (X)=12(O )2 X1
1 1we nd that c (X) = 24. Sinceb (X)=2h (O )or 2h (O )− 1, we see that2 1 X X
b (X) = 0. So the Betti numbers must be: b =1, b =0, b = 22, b =0,1 0 1 2 3
b = 1 giving 24 as the topological Euler characteristic.4
[For those who haven’t studied compact complex surfaces, I remind you
1that in the case of algebraic surfaces we always have b (X)=2h (O ). Ko-1 X
daira proved that for non-algebraic complex surfaces this equality can only
1 1 8fail by 1, i. e.,b =2h (O )or 2h (O )−1.](Cf.[BPV,TheoremII.6]. )1 X X
(4) For any line bundleL on X, the Riemann-Roch theorem
(L)=(O )+X
L L0 1 0 h (L)−h (L)+h (L )=2+
2 0 0 sinceh (L)=h (L (K )) =h (L ).X
0 0 In particular, ifL L− 2thenh (L)+h (L )> 0, i. e., eitherL orL
is e ective.
2(5) The intersection form onH (X;Z) has a very explicit structure mentioned in
the introduction. We postpone discussion of this structure until section 11.
7Need cross-reference: section 8?
8Per A. G.
(6) Let D be an irreducible reduced e ective divisor on X. Then the adjunction
formulaK =(K +D)j yieldsD X D
deg(K )=deg(Dj )=DD:D D
In particular,
g(D)= (DD)+1:
0(7) h (D)=g(D)+1.
But even more is true: if we consider the exact sequence (when D is smooth)
0!O !O (D)!O (D)! 0X X D
1 0 0since h (O )=0wehavethat H (O (D))!H (O (D)) is surjective. Thus sinceX X D
O (D) O (K ), the global sections of the line bundleO (D) induce the canonical=D D D X
map onD.SowhenD is not hyperelliptic, it must be embedded byO (D)and thisX
embedding coincides with the canonical embedding of D.
gTo say this another way, if X is embedded in P by the complete linear system
g−1 g−1jO (D)j, then a hyperplane sectionD =X\P is canonically embedded inP X
gP . In brief: \a hyperplane section of a K3 surface is a canonical curve".
This property can be considered as motivating the denition of a K3 surface. That
is, we require K 0 so that the normal bundleO (D) of a hyperplane sectionX D
1agrees with the canonical bundleO (K ), and we require h (O ) = 0sothat theD D X
0 0rational mapH (O (D))!H (O (D)) is surjective.X D
We now ask: What kind of threefold has a K3 surface as its hyperplane section?
1 0If Y is such a threefold, we must have h (O ) = 0 to guarantee that H (O (X))!Y Y
0H (O (X)) is surjective, and by adjunctionX
K =(K +X)jX Y X
so that we need K =−X. (Note that by the Lefschetz hyperplane theorem (cf.Y
1[9]), we have h (O )= 0 so X is aK3surface.) That is, Y is embedded by itsX
anti-canonical linear systemjO (−K )j. Thisiscalleda Fano threefold.Y Y
We can ask the same question for higher dimension, of course.
De nition. A Fano variety is a complex projective varietyY withO (−K )ample.Y Y
A Fano variety has index r if r is the maximum integer such that−K rH forY
some ampleH onY.The coindex of Y is de ned to be c =dim(Y )−r+1.
The linear systemjHj is called the fundamental system of the Fano variety.
Lemma. IfjHj is very ample and we choose r general hyperplanesH;:::;H , then1 r
Y\H \\ H is a variety of dimension c− 1 with trivial canonical bundle.1 r

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