Complex Multiplication and Noether-Lefschetz Loci of the Twistor Space of a K3 Surface
CCOMPLEX MULTIPLICATION AND NOETHER-LEFSCHETZ LOCI OFTHE TWISTOR SPACE OF A K3 SURFACE F RANCESCO V IGANÒ * Abstract
For an algebraic K3 surface with complex multiplication (CM), algebraicfibres of the associated twistor space away from the equator are again of CMtype. In this paper, we show that algebraic fibres corresponding to points atthe same altitude of the twistor base S (cid:39) P C share the same CM endomor-phism field. Moreover, we determine all the admissible Picard numbers ofthe twistor fibres. A projective (or equivalently, algebraic) complex K3 surface X is said to beCM (complex multiplication) if the endomorphism field K T ( X ) of the Hodgestructure on the transcendental lattice T ( X ) = NS ( X ) ⊥ Q ⊆ H ( X , Q ) is a CMfield, and dim K T ( X ) T ( X ) =
1. Denote by
X → P C the twistor space of theprojective K3 surface X associated with a Kähler class given by an ample class (cid:96) = c ( L ) ∈ H ( X , Q ) (see Section 1.1 for a short description). Despite the tran-scendental nature of the twistor construction, the fibres X ζ that are again alge-braic share some arithmetic properties. In particular, all algebraic fibres awayfrom the equator P C (cid:39) S are CM, and the corresponding CM endomorphismfields share the same totally real maximal subextension [Huy19, Theorem 5.3].In this paper we prove that the CM fields corresponding to fibres at the samealtitude of S (cid:39) P C coincide (Theorem 5.8). This result will be proven by intro-ducing an action of the topological multiplicative group K × T ( X ) on the Noether-Lefschetz locus of the upper-half sphere. Theorem 0.1.
Consider the twistor space
X → P C associated with a projective complexK3 surface X with complex multiplication. Assume that ζ , ζ ∈ P C are two points ofPicard jump at the same altitude and not on the equator. Then X ζ is algebraic if andonly if X ζ is such. If so, then the CM endomorphism fields of these K3 surfaces coincide.Moreover, the set of points of Picard jump at the same altitude of ζ (and ζ ) is countableand dense in the circle at that altitude. * LSGNT (London School of Geometry and Number Theory), Centre for Doctoral Train-ing across Imperial College London, University College London and King’s College London,[email protected] . a r X i v : . [ m a t h . AG ] F e b he Picard number of a K3 surface is the rank of its Néron-Severi group. Huy-brechts proved that, if a fibre X ζ has excessive Picard number (that is, bigger thanthe original Picard number ρ ( X ) ), then ζ lies on the equator of S (cid:39) P C [Huy19,Proposition 3.2]. We prove, in the CM case, that there is only one admissibleexcessive Picard value (Theorem 5.3). Theorem 0.2.
Consider the twistor space
X → P C associated with a projective complexK3 surface X with complex multiplication. If ζ is a point of Picard jump on the equator,then ρ ( X ζ ) = + ρ ( X ) Moreover, the Noether-Lefschetz locus of the equator is dense in the equator. More pre-cisely, the locus of points on the equator whose fibres are algebraic K3 surfaces is dense inthe equator.
Admissible valuesof ρ ( X ζ ) , CM caseOutside theequator ρ ( X ) − ρ ( X ) On theequator ρ ( X ) −
1, 10 + ρ ( X ) Outline.
Section 1 contains some basic information about the geometry ofthe twistor space, Hodge structures of K3 type and CM fields. In Section 2 wedescribe a family of Hodge structures, parameterized by a sphere, attached toa given Hodge structure of K3 type. This construction is, at the level of Hodgestructures, the algebraic equivalent of the twistor space. Afterwards, in Section3, we prove Proposition 3.4 (algebro-equivalent of Theorem 0.2) and other resultscharacterizing points of Picard jump on the equator of the mentioned sphere. Twoactions of the topological multiplicative group K × T ( X ) are introduced in Section4; these will play a key role in the proof of Corollary 4.10 (algebro-equivalent ofTheorem 0.1) and related statements. Finally, the translation into geometric termsis exposed in Section 5, and Theorems 0.1 and 0.2 are proven. Acknowledgements.
I thank François Charles for introducing me to the sub-ject, and both him and Daniel Huybrechts for our prolific discussions and theirsignificant comments. I am also thankful to Paolo Stellari for his useful remarksin the revision of this paper. 2 P RELIMINARIES
We present a short description of the twistor space of a complex K3 surface, andwe recall some facts about Hodge structures of K3 type and CM fields.1.1 T
HE TWISTOR SPACE OF A COMPLEX K3 SURFACE
A complex K3 surface X admits a sphere of different complex structures. Moreexplicitly, one can attach a set of different complex structures I ζ to the underlyingdifferentiable manifold X , so that ( X , I ζ ) is again a K3 surface, and these struc-tures are parameterized by a sphere ζ ∈ S (cid:39) P C . These K3 surfaces can bepatched together into a 3-dimensional complex manifold X , called twistor space (see [HKLR87, Section 3.F], [Hit92] or [Joy00, Chapter 7] for more on the twistorspace and the details of its construction). X comes with a holomorphic map X → P C with the property that, for ζ ∈ P C (cid:39) S , the fibre X ζ is the K3 sur-face ( X , I ζ ) . In fact, this construction is non-canonical, and depends on the choiceof a Kähler class of X (any K3 surface is Kähler, [Siu83]). The original K3 surfacecorresponds to one of the poles of P C (cid:39) S .Even if the original K3 surface X is algebraic (or equivalently, projective), X is no longer algebraic, and has to be thought of as of transcendental nature. Thetwistor space of a K3 surface plays an important role in several situations. Amongall, it provides an example of K3-fibration over a compact base, and it is used ina modern proof of Torelli Theorem (any two K3 surfaces are connected by a finitepath of twistor lines, see [Huy16, Section 7.3]).1.2 H ODGE STRUCTURES OF K3 TYPE By polarized Hodge structure of K3 type we mean the data of a vector space T over Q of dimension r ≥
2, endowed with a symmetric bilinear form ( . ) of signature ( r − ) , and a decomposition T C = T ⊕ T ⊕ T such that the C -linear extension of ( . ) satisfies:1. the subspaces T and T ⊕ T are orthogonal;2. ( . ) is positive definite on P T = ( T ⊕ T ) ∩ T R and T , T ⊆ T C areisotropic;3. complex conjugation on T C preserves T and exchanges T and T ;4. dim C T = σ a C -generator of T . Note that σ generates T , andthe required conditions give ( σ . σ ) = ( σ . σ ) > ( (cid:60) ( σ )) = ( (cid:61) ( σ )) and ( (cid:60) ( σ ) . (cid:61) ( σ )) =
0. The plane P T is considered with the orientation given by thebasis {(cid:60) ( σ ) , (cid:61) ( σ ) } . We will also assume that T is irreducible.3efine K T to be the ring of endomorphism of the Hodge structure T . As T isirreducible, K T is a division algebra. It was pointed out by Zarhin [Zar83] that K T is indeed a number field, endowed with an embedding K T (cid:44) → C . This embeddingis defined in the following way: as ϕ preserves the (
2, 0 ) -part of T , ϕ ( σ ) = ϕ · σ ,where we identify ϕ with a scalar in C . Zarhin proved also that K T is either totallyreal or complex multiplication (CM). Notice that T is naturally a K T -module and,therefore, it becomes a K T -vector space. Definition 1.1. T is said to be:• of totally real type if K T is totally real,• of almost complex multiplication type , or almost CM , if K T is CM, and• of complex multiplication type , or CM , if K T is CM and dim K T T = K T T = K T is totally real, dim K T T ≥ K T T = [ K T : Q ] = r .As the form ( . ) is non-degenerate, we can define the transpose ϕ (cid:48) of ϕ by thecondition ( γ . ϕ ( δ )) = ( ϕ (cid:48) ( γ ) . δ ) ,for any γ , δ ∈ T . ϕ (cid:48) is in fact an element of K T , and corresponds to the complexconjugate of ϕ via the embedding K T (cid:44) → C , that is: ϕ (cid:48) ( σ ) = ϕ · σ (see [Huy19,Remark 2.6] or [Huy16, Chapter 3]). In particular, an element ϕ ∈ K T is an isom-etry for ( . ) if and only if its image in C has unitary norm. We will denote by K T = K T ∩ R the real part of K T , once seen K T as a subfield of C via the pre-scribed embedding. Notice that K T can be characterized by the subfield of K T ofself-transpose endomorphisms.The period field k T of T is defined in the following way. Let γ vary in T . Amongthe periods ( σ . γ ) , at least one is not zero, say for ˜ γ , since σ (cid:54) = ( . ) is non-degenerate. The period field is defined to be the subfield of C generated over Q by the quotients ( σ . γ ) / ( σ . ˜ γ ) . Clearly, it is enough to consider these quotientsonly for γ varying in a Q -basis of T . For a proof of the following result, see[Huy19, Lemma 2.10]. Lemma 1.2.
Assume that T is a polarized irreducible Hodge structure of K3 type withcomplex multiplication (i.e. K T is a CM field and dim Q T = [ K T : Q ] ). Then1. The endomorphism field K T and the period field k T coincide (as subfields of C );2. For any basis { γ i } of T and any σ ∈ T , σ (cid:54) = , the coordinates x i = ( σ . γ i ) satisfy K T = k T = r (cid:77) i = Q · ( x i / x ) . More precisely, if T is irreducible, ( σ . γ ) = γ = SEFUL FACTS ON CM FIELDS
First of all, we recall a result proven by Blanksby and Loxton [BL78].
Theorem 1.3.
If E ⊆ C is a CM field, then E = Q ( α ) , for a primitive element α satisfying | α | = . We now present other elementary properties carried by CM fields.
Lemma 1.4.
Let E ⊆ C be a number field given as a subfield of C , and let e ∈ E, e (cid:54) = .Then the dimension of the Q -vector space ( R · e ) ∩ E does not depend on e, and is equalto dim Q ( E ∩ R ) .Proof. It is enough to notice that ( R · e ) ∩ E = e · ( E ∩ R ) .This Lemma assumes a particular form in the case of a CM field. Corollary 1.5.
Let E ⊆ C be a CM field given as a subfield of C , and let α be a generatorof E satisfying | α | = , n = [ E : Q ] . For any e ∈ E, e (cid:54) = , the dimension of the Q -vectorspace ( R · e ) ∩ E does not depend on e, and is equal to n /2 .Proof. The statement follows from Lemma 1.4, together with the fact that E ∩ R isthe maximal totally real subfield of E , satisfying dim Q ( E ∩ R ) = n /2. Proposition 1.6.
Assume that a number field E is given as a subfield of C . Suppose thatE (cid:42) R and E = E. Then E ∩ S is dense in S .Proof. It is enough to show that there exist elements α ∈ E lying on the circle ofarbitrary small non-zero argument (density follows taking powers of these ele-ments). Let δ be an element of E \ R . For r ∈ Q , define α r = δ + r δ + r ∈ E .Then | α r | = α r has arbitrary small non-zero argument, since lim r → ∞ α r = Corollary 1.7.
If E is a CM field given as a subfield of C , then E ∩ S is dense in S .Proof. A CM field satisfies the hypotheses of Proposition 1.6.
WISTOR SPHERE OF H ODGE STRUCTURES
Associated with T and an abstract class (cid:96) of positive square, there exists a sphereof related Hodge structures. Here we outline its construction, following [Huy19,Section 3].If one wishes to keep in mind the geometric picture, they should think of T as the transcendental lattice of a projective complex K3 surface X , of ( . ) asthe restriction of the cup product on H ( X , Q ) to T , and of (cid:96) as an ample classof X . Altered Hodge structures (or sub-Hodge structures of these) correspond5o transcendental lattices of the K3 surfaces constituting the twistor space of X (compare with Remark 5.1).Fix a positive integer d ∈ Z > . We extend T to the Hodge structure of K3type T ⊕ Q (cid:96) , of dimension r + ( . ) of signature ( r + − ) , by declaring (cid:96) to be of type (
1, 1 ) , orthogonal to T , and to satisfy ( (cid:96) . (cid:96) ) = d . P T ⊕ R (cid:96) ⊆ T R ⊕ R (cid:96) is, therefore, a positive 3-space. Notice that T ⊕ Q (cid:96) is no longer irreducible. The associated twistor base is the conic P (cid:96) = (cid:26) z = [ σ (cid:48) ] ∈ P ( T ⊕ T ⊕ C (cid:96) ) (cid:12)(cid:12)(cid:12) (cid:16) σ (cid:48) . σ (cid:48) (cid:17) = (cid:27) .Points of the conic P (cid:96) define different Hodge structures on T ⊕ Q (cid:96) , in the follow-ing sense. Any z ∈ P (cid:96) defines a Hodge structure of K3 type on T ⊕ Q (cid:96) , that wesay corresponding to z : its (
2, 0 ) -part is the line corresponding to z = [ σ (cid:48) ] , i.e. theline C σ (cid:48) , its complex conjugate the (
0, 2 ) -part, and the (
1, 1 ) -part is given as theorthogonal complement of the former two. Notice that ( . ) is positive definite on ( C σ (cid:48) ⊕ C σ (cid:48) ) ∩ ( T R ⊕ R (cid:96) ) = R (cid:60) ( σ (cid:48) ) ⊕ R (cid:61) ( σ (cid:48) ) ; this follows immediately from thefact that ( . ) is positive definite on P T ⊕ R (cid:96) , which contains R (cid:60) ( σ (cid:48) ) ⊕ R (cid:61) ( σ (cid:48) ) .Mapping z ∈ P (cid:96) to the oriented, positive real plane P z = (cid:104)(cid:60) ( z ) , (cid:61) ( z ) (cid:105) R yields an identification P (cid:96) (cid:39) Gr po ( P T ⊕ R (cid:96) ) with the Grassmannian of oriented,positive planes in P T ⊕ R (cid:96) . The complex conjugate z , i.e. the point (cid:104) σ (cid:48) (cid:105) , corre-sponds to the same plane with reversed orientation: P z = (cid:104)(cid:61) ( z ) , (cid:60) ( z ) (cid:105) R .Indeed, the basis given by {(cid:60) ( z ) , −(cid:61) ( z ) } induces the same orientation as thebasis given by {(cid:61) ( z ) , (cid:60) ( z ) } .We define the period point of T as x = [ σ ] ∈ P ( T C ) . Via the natural inclusion P ( T C ) ⊆ P ( T C ⊕ C (cid:96) ) , we see that both x = [ σ ] and its complex conjugate x = [ σ ] belong to the conic P (cid:96) , as ( σ . σ ) = ( σ . σ ) =
0. Thinking of P z with itsorientation being given as the orthogonal complement of a generator α z of theline P ⊥ z ⊆ P T ⊕ R (cid:96) provides a natural identification P (cid:96) (cid:39) Gr po2 ( P T ⊕ R (cid:96) ) (cid:39) S (cid:96) = (cid:8) α ∈ P T ⊕ R (cid:96) (cid:12)(cid:12) ( α . α ) = (cid:9) .With this identification, x and x correspond to the normalizations of (cid:96) and − (cid:96) .We think of them as the north and south poles of S (cid:96) . The equator of the twistorbase is the circle S (cid:96) = (cid:110) z ∈ P (cid:96) (cid:12)(cid:12)(cid:12) (cid:96) ∈ P z (cid:111) (cid:39) (cid:110) α ∈ S (cid:96) (cid:12)(cid:12)(cid:12) ( α . (cid:96) ) = (cid:111) .Indeed, if (cid:96) ∈ P z then the line orthogonal to P z in P T ⊕ R (cid:96) is orthogonal to (cid:96) , andvice versa. 6f, for z ∈ P (cid:96) , we write z = [ σ (cid:48) = a σ + b σ + c (cid:96) ] , for some a , b , c ∈ C , then z ∈ P (cid:96) if and only if ( σ (cid:48) . σ (cid:48) ) =
0, i.e.2 ab ( σ . σ ) + c d =
0. (1)The only points with c = x = [ σ ] , x = [ σ ] .For all the other points, c (cid:54) = σ (cid:48) , we may assume c =
1. Thefollowing result gives the explicit form of the isomorphism P (cid:96) (cid:39) S (cid:96) . Lemma 2.1.
Pick an element z = [ σ (cid:48) = a σ + b σ + (cid:96) ] ∈ P (cid:96) corresponding to a pointon S (cid:96) different from both poles. Then its image in S (cid:96) is given, in coordinates for the basis {(cid:60) ( σ ) , (cid:61) ( σ ) , (cid:96) } , by the point x ( a , b ) = v ( a , b ) (cid:107) v ( a , b ) (cid:107) , where v ( a , b ) = (cid:32) (cid:60) ( b − a ) , (cid:61) ( b − a ) , (cid:16) aa − bb (cid:17) ( σ . σ ) d (cid:33) and (cid:107) (cid:107) corresponds to (cid:112) ( . ) . Remark 2.2.
The basis {(cid:60) ( σ ) , (cid:61) ( σ ) , (cid:96) } is orthogonal for ( . ) ; nonetheless, it is notnormalized. Remark 2.3.
If we assume a = b , then, if σ (cid:48) (cid:54) = ( σ (cid:48) . σ (cid:48) ) = aa ( σ . σ ) + d > ( σ . σ ) >
0. Thus z = [ σ (cid:48) ] ∈ P (cid:96) forces a (cid:54) = b . Proof of the Lemma.
Consider the plane P z = (cid:104)(cid:60) ( σ (cid:48) ) , (cid:61) ( σ (cid:48) ) (cid:105) R . Let α z be the uniqueelement of P T ⊕ R (cid:96) of norm 1 that is positively orthogonal to P z . We are provingthat v = v ( a , b ) gives the coordinates of a vector in P T ⊕ R (cid:96) that is positivelyaligned to α z , and this will be enough. First of all, we check the orthogonalityrelations. Note that (cid:60) ( σ (cid:48) ) = σ (cid:48) + σ (cid:48) = a + b σ + b + a σ + (cid:96) = (cid:60) ( b + a ) (cid:60) ( σ ) + (cid:61) ( b + a ) (cid:61) ( σ ) + (cid:96) and (cid:61) ( σ (cid:48) ) = σ (cid:48) − σ (cid:48) i = a − b i σ + b − a i σ = (cid:61) (cid:16) a − b (cid:17) (cid:60) ( σ ) + (cid:60) (cid:16) a − b (cid:17) (cid:61) ( σ ) .Recall that ( (cid:60) ( σ ) . (cid:60) ( σ )) = ( (cid:61) ( σ ) . (cid:61) ( σ )) = ( σ . σ ) ( (cid:60) ( σ ) . (cid:61) ( σ )) = ( v . (cid:60) ( σ (cid:48) )) = ( σ . σ ) (cid:16) (cid:60) ( b − a ) (cid:60) ( b + a ) + (cid:61) ( b − a ) (cid:61) ( b + a ) + (cid:16) aa − bb (cid:17)(cid:17) = ( v . (cid:61) ( σ (cid:48) )) = ( σ . σ ) (cid:16) (cid:60) ( b − a ) (cid:61) (cid:16) a − b (cid:17) + (cid:61) ( b − a ) (cid:60) (cid:16) a − b (cid:17)(cid:17) = (cid:60) ( b + a ) (cid:61) ( b + a ) (cid:61) (cid:16) a − b (cid:17) (cid:60) (cid:16) a − b (cid:17) (cid:60) ( b − a ) (cid:61) ( b − a ) (cid:16) aa − bb (cid:17) ( σ . σ ) d which is (cid:16) aa − bb (cid:17) ( σ . σ ) d + (cid:16) a − b (cid:17)(cid:16) a − b (cid:17) > a (cid:54) = b , thanks to Remark 2.3. Remark 2.4.
The equator S (cid:96) is defined by the condition | a | = | b | , or equivalently aa = bb , for z = [ σ (cid:48) = a σ + b σ + (cid:96) ] . Lemma 2.5.
Choose a non-zero element (cid:96) (cid:48) ∈ T ⊕ Q (cid:96) . Then there are exactly two pointsz , z (cid:48) ∈ P (cid:96) such that (cid:96) (cid:48) is orthogonal to z and z (cid:48) , i.e. (cid:96) (cid:48) ∈ P ⊥ z and (cid:96) (cid:48) ∈ P ⊥ z (cid:48) . Moreover,z (cid:48) = z, and z and z correspond to antipodal points on S (cid:96) via the isomorphism P (cid:96) (cid:39) S (cid:96) .Proof. The case (cid:96) (cid:48) = (cid:96) has already been discussed; suppose that (cid:96) (cid:48) / ∈ Q (cid:96) . Theorthogonal complement of (cid:96) (cid:48) in T C ⊕ C (cid:96) does not contain the whole T ⊕ T ⊕ C (cid:96) (since the form ( . ) is non-degenerate on this last vector space). Therefore (cid:96) (cid:48)⊥ ∩ ( T ⊕ T ⊕ C (cid:96) ) is a C -plane in T ⊕ T ⊕ C (cid:96) . Passing to the projectivespaces, (cid:96) (cid:48)⊥ ∩ ( T ⊕ T ⊕ C (cid:96) ) defines a line in P ( T ⊕ T ⊕ C (cid:96) ) , which cutsthe conic P (cid:96) in two distinct points, or one single point (with double multiplicity).On the other hand, the second case is not admissible: if z = [ σ (cid:48) = a σ + b σ + (cid:96) ] satisfies conditions (1) and (2), then z = (cid:104) σ (cid:48) = b σ + a σ + (cid:96) (cid:105) satisfies both equations as well. Besides, as already pointed out in Remark 2.3, a (cid:54) = b , so that z (cid:54) = z . Then z , z are the required points. Lastly, note that z and z areantipodal on the sphere P (cid:96) (cid:39) S (cid:96) ; for, compare with the isomorphism of Lemma2.1. ICARD JUMP ON THE EQUATOR
We focus our attention on the Noether-Lefschetz locus of the equator S (cid:96) ⊆ P (cid:96) , itspoints of Picard jump and their period fields.8.1 E XCESSIVE P ICARD JUMP VALUES
We have already remarked that the original extended Hodge structure of K3 typeon T ⊕ Q (cid:96) is no longer irreducible, being (cid:96) a (
1, 1 ) -class. Given z ∈ P (cid:96) , we de-fine the Picard number ρ z of the Hodge structure corresponding to z to be the Q -dimension of the space of (
1, 1 ) -classes of T ⊕ Q (cid:96) , that is ρ z = dim Q (cid:16) P ⊥ z ∩ ( T ⊕ Q (cid:96) ) (cid:17) .For instance, as the original T is irreducible, we deduce that ρ x = ρ x = Definition 3.1.
We say that a point z ∈ P (cid:96) is of Picard jump if ρ z ≥
1, of excessivePicard jump if ρ z >
1. We call
Noehter-Lefschetz locus the set of points z ∈ P (cid:96) ofPicard jump.We are interested in understanding how points of Picard jump distribute onthe sphere P (cid:96) (cid:39) S (cid:96) , and the possible relative Picard numbers. Huybrechts provedthe following result [Huy19, Proposition 3.2]. Proposition 3.2.
Assume that T is a polarized irreducible Hodge structure of K3 type.Then, for the twistor base P (cid:96) (cid:39) S (cid:96) , one has:1. the set (cid:110) z ∈ P (cid:96) (cid:12)(cid:12)(cid:12) ρ z ≥ (cid:111) is countable and dense (in the classical topology);2. the set (cid:110) z ∈ P (cid:96) (cid:12)(cid:12)(cid:12) ρ z > (cid:111) is at most countable and contained in the equator S (cid:96) . The fact that the Noether-Lefschetz locus is countable is a particular instanceof a more general fact (see [Huy16, Chapter 6, Proposition 2.9]). However, there isa very simple argument that applies in this case. Assume that z is of Picard jump,and let (cid:96) (cid:48) ∈ T ⊕ Q (cid:96) be a (
1, 1 ) -class for the Hodge structure induced by z . Then (cid:96) (cid:48)⊥ identifies a line in P ( T ⊕ T ⊕ C (cid:96) ) , that cuts the conic P (cid:96) in two points: z and z . Thus, (cid:96) (cid:48) is a (
1, 1 ) -class only for finitely many z ∈ P (cid:96) , and therefore theNoether-Lefschetz locus is countable, as T ⊕ Q (cid:96) is.9ssume that z = [ a σ + b σ + c (cid:96) ] . The condition defining (cid:96) (cid:48)⊥ can be explicitlytranslated as a ( σ . (cid:96) (cid:48) ) + b ( σ . (cid:96) (cid:48) ) + ( (cid:96) . (cid:96) (cid:48) ) =
0. (2)
Lemma 3.3.
Assume that (cid:96) (cid:48) ∈ T ⊕ Q (cid:96) , (cid:96) (cid:48) / ∈ Q (cid:96) and choose z to be one of the two pointsorthogonal to (cid:96) (cid:48) . Then the point z is contained in the equator S (cid:96) if and only if (cid:96) (cid:48) ∈ T.Proof.
Write z = [ σ (cid:48) ] . Assume that z ∈ S (cid:96) , i.e. (cid:96) ∈ P z = (cid:104)(cid:60) ( σ (cid:48) ) , (cid:61) ( σ (cid:48) ) (cid:105) R ⊆ (cid:68) σ (cid:48) , σ (cid:48) (cid:69) C . Then (cid:96) = a σ (cid:48) + b σ (cid:48) for some a , b ∈ C ; hence ( (cid:96) . (cid:96) (cid:48) ) = ( σ (cid:48) . (cid:96) (cid:48) ) = (cid:96) (cid:48) ∈ T . Conversely, assume that (cid:96) (cid:48) ∈ T , i.e. ( (cid:96) . (cid:96) (cid:48) ) =
0. Write σ (cid:48) = a σ + b σ + c (cid:96) . After rescaling σ , we may assume that ( σ . (cid:96) (cid:48) ) =
1, so that ( σ . (cid:96) (cid:48) ) = ( σ . (cid:96) (cid:48) ) = ( σ (cid:48) . (cid:96) (cid:48) ) = a ( σ . (cid:96) (cid:48) ) + b ( σ . (cid:96) (cid:48) ) + c ( (cid:96) . (cid:96) (cid:48) ) =
0, i.e. b = − a .Therefore, the condition (1) − a ( σ . σ ) + c d = c (cid:54) = c =
1) and a ∈ R (since d > ( σ . σ ) > σ (cid:48) = a σ − a σ + (cid:96) and (cid:96) = σ (cid:48) + σ (cid:48) = (cid:60) ( σ (cid:48) ) and the proof is concluded, as (cid:96) ∈ P z .If the original Hodge structure T has CM, all points of Picard jump on theequator share the same Picard number. Proposition 3.4.
Suppose that T is a polarized irreducible Hodge structure of K3 typeand CM, and consider its sphere of related Hodge structures. Assume that z ∈ S (cid:96) is apoint of Picard jump on the equator. Then ρ z = r /2 . Admissible valuesof ρ z , CM caseOutside theequator 0, 1On theequator 0, r /2 Proof.
Firstly, note that P ⊥ z ∩ ( T ⊕ Q (cid:96) ) is all contained in T . This follows fromLemma 3.3 applied to any non-zero element (cid:96) (cid:48) ∈ P ⊥ z ∩ ( T ⊕ Q (cid:96) ) . Hence P ⊥ z ∩ ( T ⊕ Q (cid:96) ) = P ⊥ z ∩ T . The key remark is that P ⊥ z ∩ T admits an action of the realpart K T = K T ∩ R of the endomorphism field K T . To show this, choose β ∈ K T (cid:96) (cid:48) ∈ P ⊥ z ∩ T . Write z = [ σ (cid:48) = a σ + b σ + (cid:96) ] . Then, using that ( (cid:96) . (cid:96) (cid:48) ) =
0, weobtain ( β ∈ K T is self-adjoint for ( . ) ) ( σ (cid:48) . β ( (cid:96) (cid:48) )) = ( β ( a σ + b σ + (cid:96) ) . (cid:96) (cid:48) )= a ( β · σ . (cid:96) (cid:48) ) + b ( β · σ . (cid:96) (cid:48) ) + = β · (cid:16) a ( σ . (cid:96) (cid:48) ) + b ( σ . (cid:96) (cid:48) ) + (cid:17) = β · ( σ (cid:48) . (cid:96) (cid:48) ) = P ⊥ z ∩ T is a K T -vector space. As [ K T : Q ] = r /2 (we are in theCM case by assumption, so that [ K T : K T ] = [ K T : Q ] = r ), dim K T P ⊥ z ∩ T can only take three different values: 0, 1, 2. We immediately exclude the casedim K T P ⊥ z ∩ T = z is a point of Picard jump. Assume by contradictionthat dim K T P ⊥ z ∩ T =
2. In fact, this equality would imply P ⊥ z ∩ T = T . Thiswould force σ (cid:48) to belong to C (cid:96) , or equivalently to have zero T C -part (indeed theform ( . ) in non-degenerate on T , and therefore on T C ). On the other hand, thisis a contradiction, since any non-zero σ (cid:48) ∈ C (cid:96) does not satisfy (1). In conclusion,dim K T P ⊥ z ∩ T =
1, or equivalently dim Q P ⊥ z ∩ T = [ K T : Q ] = r /2, i.e. ρ z = r /2. Remark 3.5.
If the original T has CM, the only case where there are no points ofexcessive Picard jump (i.e. ρ z ≤ z ∈ P (cid:96) ) is r = Remark 3.6.
The hypothesis on T of being of CM type plays a fundamental rolein order to ensure the validity of the property ρ z = r /2 for all points of jumpon the equator. It is not difficult to construct explicit examples of irreduciblepolarized Hodge structures of K3 type T for which the excessive Picard valuesare multiple. In addition, not even the sole assumption that K T is CM, withoutsupposing dim Q T = [ K T : Q ] , is enough to guarantee the result. Remark 3.7.
However, if T is not assumed to be of CM type, one may argue as inthe proof of Proposition 3.4 to deduce that ρ z (for z point of Picard jump on theequator) is divisible by [ K T : Q ] and strictly smaller than r .Admissible valuesof ρ z , non-CM caseOutside theequator 0, 1On theequator 0, d such that [ K T : Q ] | d and d < r The following result enriches Proposition 3.2. To prove it, it is not necessaryto assume that T has CM, or that K T is CM. However, in Section 4.2 we will givea better description of the distribution of points of Picard jump on the equator inthe CM case. 11 roposition 3.8. Assume that T is a polarized irreducible Hodge structure of K3 type,and consider its sphere of related Hodge structures. Then the set of points of Picard jumpon the equator is dense in the equator (for the classical topology). In particular, this set iscountable (not finite).Proof.
Thanks to Lemma 2.1, a point on the equator S (cid:96) ⊆ P (cid:96) corresponds via theisomorphism P (cid:96) (cid:39) S (cid:96) to the normalization of the vector v ( a , b ) = (cid:16) (cid:60) ( b − a ) , (cid:61) ( b − a ) , 0 (cid:17) .Lemma 2.1 also gives a condition on z to belong to S (cid:96) , namely | a | = | b | . Write R to denote these absolute values; then a = Re i θ , b = Re i τ for some θ , τ ∈ R .Moreover, (1) gives 2 R e i ( θ + τ ) ( σ . σ ) = − d ,from which we deduce that θ + τ ≡ π π . Hence, b = − Re − i θ = − a . Then v ( a , b ) = (cid:16) − R cos ( − θ ) , − R sin ( − θ ) , 0 (cid:17) .Therefore, in order to prove the statement, it is enough to show that the set ofpossible complex arguments assumed by the periods ( σ . γ ) , γ ∈ T is dense inthe circle { z ∈ C | | z | = } (the point of Picard jump corresponding to such a γ would be determined by a = Re i θ such that θ (or π + θ ) is the opposite of theargument of ( σ . γ ) ). However, the Q -vector space P of the periods ( σ . γ ) , γ ∈ T is not contained in an R -line of C ; indeed, if this were the case, then the periodfield k T = Q (( σ . γ i ) / ( σ . γ )) (for a basis { γ i } of T ) would be contained in R ,contradiction. As a consequence, P is dense in C , and then the set of the complexarguments of elements of P \ { } is dense in the circle. CTIONS ON N OETHER -L EFSCHETZ LOCI
In this section, assume that T is a polarized irreducible Hodge structure of K3type, and consider again the associated twistor base P (cid:96) (cid:39) S (cid:96) . We define twoactions: first, an action of the multiplicative group K × T on the Noether-Lefschetzlocus of the upper-half sphere, deprived of the north pole; then, an action of themultiplicative group K × T / ( K T ) × on the Noether-Lefschetz locus of the equator.We treat in particular the situation arising when we assume that T is of CM type:the jumping loci outside and on the equator are homogeneous under the actionsof K × T and K × T / ( K T ) × , respectively.One may define these two actions together as a unique action on the whole P (cid:96) ;however, we prefer to keep them separated to analyse the different behaviours onpoints outside and on the equator S (cid:96) .4.1 O UTSIDE THE EQUATOR
Denote by U the upper-half sphere of P (cid:96) (cid:39) S (cid:96) (defined by the condition | a | > | b | on z = [ σ (cid:48) = a σ + b σ + (cid:96) ] , see Lemma 2.1), and by Q the set of points of Picard12ump in U (that is, the Noether-Lefschetz locus of U ), deprived of the north pole x . Notice that a point z ∈ Q satisfies ρ z =
1, by Proposition 3.2, and hence isorthogonal to a unique (up to a rational scalar) non-zero element (cid:96) (cid:48) ∈ P ⊥ z ∩ ( T ⊕ Q (cid:96) ) . Denote by Q + the set of z ∈ Q for which the corresponding (cid:96) (cid:48) is such that ( (cid:96) (cid:48) . (cid:96) (cid:48) ) > Remark 4.1.
Unless r =
2, the form ( . ) is not positive definite on T . Thus, Q + (cid:40) Q if r > z ∈ P (cid:96) , we define its altitude as the last coordinate of the vectordefined by Lemma 2.1, namely the third coordinate of the corresponding point in S (cid:96) for the basis {(cid:60) ( σ ) , (cid:61) ( σ ) , (cid:96) } of P T ⊕ R (cid:96) .Recall that d = ( (cid:96) . (cid:96) ) ∈ Z > . Choose an element (cid:96) (cid:48) = γ + ( m / d ) (cid:96) ∈ T ⊕ Q (cid:96) , (cid:96) (cid:48) / ∈ T , (cid:96) (cid:48) / ∈ Q (cid:96) , so that the two corresponding orthogonal points are neither thepoles nor on the equator (see Lemma 2.5 and Lemma 3.3). Denote by z ∈ Q the only point of Picard jump in the upper-half sphere orthogonal to (cid:96) (cid:48) . Takinga scalar multiple of (cid:96) (cid:48) identifies the same z ; we may assume m = d , so that (cid:96) (cid:48) = γ + (cid:96) . For an element A ∈ K × T , we define z = A ∗ z , where z ∈ Q isthe only point of Picard jump in the upper-half sphere orthogonal to (cid:96) (cid:48)(cid:48) = γ + (cid:96) ,where γ = A ( γ ) . This association defines an action of K × T on Q . Proposition 4.2.
The action defined above is free. Moreover, if T has CM, the action istransitive.Proof. If A ∈ K × T is different from the identity morphism, (cid:96) (cid:48) and (cid:96) (cid:48)(cid:48) are linearlyindependent over Q , so that they cannot be orthogonal to the same point in Q (thanks to Proposition 3.2); thus, the action is free. The transitivity of the actionin the CM case follows from dim K T T = Remark 4.3. If T does not have CM, the action is no longer transitive (asdim K T T > Proposition 4.4.
Under the same assumptions as above, suppose that z = [ σ (cid:48) = a σ + b σ + (cid:96) ] ∈ Q and additionally that A ∈ K × T . Then1. If | A | = , say A = e i θ , then z = A ∗ z is represented by the element σ (cid:48)(cid:48) = aA ( σ ) + bA ( σ ) + (cid:96) = aA σ + bA σ + (cid:96) = ae i θ σ + be − i θ σ + (cid:96) . In particular, A acts on Q by a rotation of angle − θ along the (cid:96) -axis, and z andz have the same altitude; . If | A | (cid:54) = , then z and z = A ∗ z do not have the same altitude. In particular, if | A | < the altitude of z is greater than the one of z , and vice versa;3. If A ∈ K T = K T ∩ R , then z and z = A ∗ z lie on the same meridian of P (cid:96) (cid:39) S (cid:96) . Remark 4.5.
Recall that the homeomorphism P (cid:96) (cid:39) S (cid:96) of Lemma 2.1 sends z =[ σ (cid:48) = a σ + b σ + (cid:96) ] to the normalization of the vector v ( a , b ) = (cid:32) (cid:60) ( b − a ) , (cid:61) ( b − a ) , (cid:16) aa − bb (cid:17) ( σ . σ ) d (cid:33) .This allows us to deduce that the tangent of the angle between a vector of a point S (cid:96) and the plane of zero-altitude is given, up to sign, by ( σ . σ ) d | a + b | .Indeed, it is suffices to notice that aa − bb = ( a + b )( a − b ) ,since ab ∈ R by (1). Proof of the Proposition.
Firstly, assume that | A | =
1. We are showing that z =[ σ (cid:48)(cid:48) ] actually belongs to the conic P (cid:96) , i.e. ( σ (cid:48)(cid:48) . σ (cid:48)(cid:48) ) =
0. For, it is enough to observethat the (1), that here takes the form2 ( aA ) (cid:16) bA (cid:17) ( σ . σ ) + d = | A | = z ∈ P (cid:96) . If | a | > | b | then | aA | > (cid:12)(cid:12)(cid:12) bA (cid:12)(cid:12)(cid:12) , so that z stillbelongs to the upper-half sphere. To prove the first statement, it suffices to show14hat ( σ (cid:48)(cid:48) . (cid:96) (cid:48)(cid:48) ) = (cid:96) (cid:48)(cid:48) = A ( γ ) + (cid:96) . Recall that, since | A | = A isan isometry for ( . ) ; therefore we obtain ( σ (cid:48)(cid:48) . (cid:96) (cid:48)(cid:48) ) = (cid:16) aA ( σ ) + bA ( σ ) + (cid:96) . A ( γ ) + (cid:96) (cid:17) = a ( A ( σ ) . A ( γ )) + b ( A ( σ ) . A ( γ )) + d = a ( σ . γ ) + b ( σ . γ ) + d = ( σ (cid:48) . (cid:96) (cid:48) ) = z and z , it is enough to compare the tangents of the angles between the vectorscorresponding to z and z , as done in Remark 4.5. Remark 4.6.
One could expect, for a general A ∈ K × T not necessarily of norm one,the point z = A ∗ z to be given by z = (cid:104) σ (cid:48)(cid:48) = aA − ( σ ) + bA − ( σ ) + (cid:96) = aA − σ + bA − σ + (cid:96) (cid:105) ,so that ( σ (cid:48)(cid:48) . (cid:96) (cid:48)(cid:48) ) = z / ∈ P l , so this guess is not true. It is indeed more complicated to deduce anexplicit expression for z when A does not have norm one.We continue the proof of Proposition 4.4. Assume now that | A | (cid:54) =
1. Evenwithout computing explicitly z we can prove that z has different altitude from z . Define B = A − and consider σ (cid:48)(cid:48) = aB σ + bB σ + (cid:96) .It corresponds to a point in P ( T ⊕ T ⊕ C (cid:96) ) orthogonal to (cid:96) (cid:48)(cid:48) . Then 0 =( σ (cid:48)(cid:48) . (cid:96) (cid:48)(cid:48) ) = (cid:16) σ (cid:48)(cid:48) . (cid:96) (cid:48)(cid:48) (cid:17) , where σ (cid:48)(cid:48) = bB σ + aB σ + (cid:96) .Note that σ (cid:48)(cid:48) , σ (cid:48)(cid:48) are linearly independent over C since a (cid:54) = b (see Remark 2.3).Therefore, the line in P ( T ⊕ T ⊕ C (cid:96) ) passing through these points is exactlythe line of elements orthogonal to (cid:96) (cid:48)(cid:48) ; among these points there are z and z ,given by the intersection with the conic P (cid:96) . The general point of this line has theform z = (cid:104) λσ (cid:48)(cid:48) + µσ (cid:48)(cid:48) (cid:105) . Since λ + µ = P (cid:96) , we mayassume λ + µ =
1, and z = (cid:104) λσ (cid:48)(cid:48) + ( − λ ) σ (cid:48)(cid:48) (cid:105) .15ssume that z ∈ P (cid:96) . We claim that λ ∈ R . Firstly, note that λ (cid:54) = σ -coefficient and the σ -coefficient of λσ (cid:48)(cid:48) + ( − λ ) σ (cid:48)(cid:48) would have thesame norm, forcing z to be on the equator S (cid:96) , contradiction. Since z ∈ P (cid:96) , then0 = (cid:16) λσ (cid:48)(cid:48) + ( − λ ) σ (cid:48)(cid:48) (cid:17) = λ ( σ (cid:48)(cid:48) . σ (cid:48)(cid:48) ) + λ (cid:16) − λ )( σ (cid:48)(cid:48) . σ (cid:48)(cid:48) (cid:17) + ( − λ ) (cid:16) σ (cid:48)(cid:48) . σ (cid:48)(cid:48) (cid:17) .Note that ( σ (cid:48)(cid:48) . σ (cid:48)(cid:48) ) = (cid:16) σ (cid:48)(cid:48) . σ (cid:48)(cid:48) (cid:17) . Therefore, also the point z (cid:48) identified by z (cid:48) = (cid:104) ( − λ ) σ (cid:48)(cid:48) + λσ (cid:48)(cid:48) (cid:105) is in the intersection of the conic P (cid:96) with the line of elements orthogonal to (cid:96) (cid:48)(cid:48) .Since λ (cid:54) = z , and thus z (cid:48) = z , where z = (cid:20) λσ (cid:48)(cid:48) + (cid:16) − λ (cid:17) σ (cid:48)(cid:48) (cid:21) .Hence, λ = λ .As noticed in Remark 4.5, to discuss the difference of altitude we are interestedin the quantity | a + b | for the new coefficients. We have λσ (cid:48)(cid:48) + (cid:16) − λσ (cid:48)(cid:48) (cid:17) = (cid:16) λ a + ( − λ ) b (cid:17) B σ + (cid:16) λ b + ( − λ ) a (cid:17) B σ + (cid:96) and, using λ ∈ R , (cid:12)(cid:12)(cid:12)(cid:16) λ a + ( − λ ) b (cid:17) B + (cid:16) λ b + ( − λ ) a (cid:17) B (cid:12)(cid:12)(cid:12) = | a + b | (cid:12)(cid:12)(cid:12) B (cid:12)(cid:12)(cid:12) = | a + b || A | − .This relation tells us exactly that, for points in the upper-half sphere, acting by anelement of K × T of norm smaller than one increases the altitude, while the actionof elements of K × T of norm greater than one decreases the altitude.16inally, we prove the third statement. If A ∈ R then B = A − = A − ∈ R ,too. Arguing as above, we deduce that z is represented by the element λσ (cid:48)(cid:48) + ( − λ ) σ (cid:48)(cid:48) = (cid:16) λ a + ( − λ ) b (cid:17) B σ + (cid:16) λ b + ( − λ ) a (cid:17) B σ + (cid:96) for some λ ∈ R . Set˜ a = (cid:16) λ a + ( − λ ) b (cid:17) B , ˜ b = (cid:16) λ b + ( − λ ) a (cid:17) B .For this element, the quantity ˜ b − ˜ a is equal to (cid:0) ( λ − ) b + ( − λ ) a (cid:1) B = ( b − a )( λ − ) B .As ( λ − ) B ∈ R , the angle determined by (cid:60) (cid:16) ˜ b − ˜ a (cid:17) and (cid:61) (cid:16) ˜ b − ˜ a (cid:17) is the sameas the angle determined by (cid:60) ( b − a ) and (cid:61) ( b − a ) , so that z and z lie on thesame meridian. Remark 4.7.
Of course, one can state a similar result for points in the lower-halfsphere, as the lower-half may be obtained by conjugating the upper-half sphere.Huybrechts proved that, for a point z ∈ Q + orthogonal to (cid:96) (cid:48) , the correspond-ing polarized irreducible Hodge structure of K3 type, namely T (cid:48) = (cid:96) (cid:48)⊥ , is of CMtype, and the real parts K T and K T (cid:48) of K T and K T (cid:48) , respectively, coincide [Huy19,Proposition 3.8]. Moreover, he gave an explicit description of K T (cid:48) [Huy19, Corol-lary 3.10]: it is the quadratic extension of K T = K T (cid:48) described by X + γ X + δ = γ = m ( α + α − ) , δ = m − d ( σ . σ ) ( α + α − − ) , K T = Q ( α ) , d = ( (cid:96) . (cid:96) ) , m = ( (cid:96) . (cid:96) (cid:48) ) and σ is such that ( σ . (cid:96) (cid:48) ) = Proposition 4.8.
Same hypotheses as in Proposition 4.4. Suppose that T has CM.Assume, moreover, that z ∈ Q + and that A ∈ K × T satisfies | A | = . Thenz = A ∗ z ∈ Q + and the polarized irreducible Hodge structures of K3 type corre-sponding to the points z and z , namely (cid:96) (cid:48)⊥ and (cid:96) (cid:48)(cid:48)⊥ , have the same CM field.Proof. To prove that z ∈ Q + , it is enough to notice that ( (cid:96) (cid:48)(cid:48) . (cid:96) (cid:48)(cid:48) ) = ( A ( γ ) + (cid:96) . A ( γ ) + (cid:96) )= ( A ( γ ) . A ( γ )) + ( (cid:96) . (cid:96) )= ( γ . γ ) + ( (cid:96) . (cid:96) )= ( (cid:96) (cid:48) . (cid:96) (cid:48) ) > A is an isometry for ( . ) and z ∈ Q + . In [Huy19, Corollary 3.10] a factor 2 in the expression of δ is missing.
17e now prove that the coefficients γ , δ are the same for the two points z , z ,so that K T (cid:48) = K T (cid:48)(cid:48) . For both points m = d , since we fixed the (cid:96) -part of (cid:96) (cid:48) , (cid:96) (cid:48)(cid:48) ; thus γ is the same. Note that σ depends on (cid:96) (cid:48) : it is chosen such that ( σ . (cid:96) (cid:48) ) =
1. On theother hand, if we choose such a σ for (cid:96) (cid:48) , then A ( σ ) = A σ satisfies ( A ( σ ) . (cid:96) (cid:48)(cid:48) ) = ( A ( σ ) . A ( γ ) + (cid:96) ) = ( A ( σ ) . A ( γ )) = ( σ . γ ) = ( σ . (cid:96) (cid:48) ) = A is an isometry. Hence, we may choose A ( σ ) = A σ for (cid:96) (cid:48)(cid:48) . Then (cid:16) A σ . A σ (cid:17) = (cid:16) A σ . A σ (cid:17) = AA ( σ . σ ) = ( σ . σ ) so that δ is the same as well. Remark 4.9.
It is possible to construct examples to show that not all the degree-2CM extensions of K T are realized as the CM endomorphism fields of some po-larized irreducible Hodge structures of K3 type T (cid:48) = (cid:96) (cid:48)⊥ . Moreover, the samedegree-2 CM extension of K T may occur for infinitely many different T (cid:48) = (cid:96) (cid:48) ,even for an Euclidean dense subset of z ∈ Q + (again, (cid:96) (cid:48) is chosen orthogonal to z ∈ Q + ).Endow K × T with the topology induced as a subspace of C × by the fixed embed-ding K T (cid:44) → C . Notice that the proof of Proposition 4.4 also puts in evidence thecontinuity of the action of K × T on Q . We summarize all the results in the followingcorollary. Corollary 4.10.
Assume that T has CM. Then: • the topological group K × T acts freely, transitively and continuously on Q . Havingfixed an element z ∈ Q , this action induces a homeomorphism between K × T and Q , and this homeomorphism induces a homeomorphism between C × and U , andbetween C and U ∪ { x } ; • the subgroup of K × T given by the elements of norm one acts on Q by rotation alongthe (cid:96) -axis; • all points in Q at the same altitude of z are obtained from z by acting with anelement of K × T of norm one. Besides, given a point z ∈ Q , there exist countablymany points at the same altitude, and they are dense in the corresponding circle of U at that altitude; • if z ∈ Q + , all the other points of Picard jump at the same altitude are in Q + , andthe Hodge structures (cid:96) (cid:48)⊥ and (cid:96) (cid:48)(cid:48)⊥ corresponding to these points have the same CM; • finally, Q + is dense in U .Proof. The only statements left to prove are the ones concerning the density ofpoints of Q at the same altitude and the density of points of Q + . The first asser-tion follows from the fact that, for a CM field E , the set of points in E ∩ S is densein S , according to Corollary 1.7. For the second assertion: let γ ∈ T such that18 γ . γ ) >
0, so that ( γ + (cid:96) . γ + (cid:96) ) = ( γ . γ ) + d > (cid:96) (cid:48) = γ + (cid:96) . Con-sider an element in Q × · ( S ∩ K T ) ⊆ K × T , say A = λα , for λ ∈ Q × and α ∈ K × T satisfying | α | =
1. Then (cid:96) (cid:48)(cid:48) = A ( γ ) + (cid:96) satisfies ( (cid:96) (cid:48)(cid:48) . (cid:96) (cid:48)(cid:48) ) = ( A ( γ ) . A ( γ )) + d = λ ( γ . γ ) + d > Q + in U follows, therefore, from the density of Q × · ( S ∩ K T ) in C × ( S ∩ K T is dense in S thanks to Proposition 1.6).4.2 O N THE EQUATOR
Now we focus on the equator S (cid:96) . In the case P (cid:96) \ S (cid:96) we had at our disposal aclean way to choose one of the two points orthogonal to a certain (cid:96) (cid:48) ∈ T ⊕ Q (cid:96) :picking the one in the upper-half sphere (subject to the condition | a | > | b | ). Herethis choice is not so clear; therefore, we will consider pairs of antipodal points on S (cid:96) . Define R to be the set of points of Picard jump on the equator modulo therelation {±} . Pick an element (cid:96) (cid:48) = γ ∈ T , γ (cid:54) =
0, and consider z = [ σ (cid:48) = a σ + b σ + (cid:96) ] on S (cid:96) which is orthogonal to γ . By our choice of R , here we meanat the same time z and z . For A ∈ K × T , we define z = A ∗ z to be the (pairof) point orthogonal to (cid:96) (cid:48)(cid:48) = γ = A ( γ ) . This defines an action of K × T on R .This action is no longer free: indeed, as explained in the proof of Proposition 3.4,two non-zero elements γ , δ in T are orthogonal to the same point if there existsa β ∈ ( K T ) × = K × T ∩ R such that β ( γ ) = δ . Then, ( K T ) × is contained in thekernel of this action, or equivalently in the stabilizer of each point. This inducesan action of the quotient group K × T / ( K T ) × on R . If T has CM, the action of thequotient is free, i.e. ( K T ) × is exactly the stabilizer of each point: this follows fromthe fact that ρ z = r /2 for points of Picard jump on the equator (Proposition 3.4).However, it will be clear later that the action of K × T / ( K T ) × is free even if T is notof CM type. Notice that the quotient group is trivial if T is of totally real type.Again, if T has CM then the action is transitive (as dim K T T = z = A ∗ z . Pick σ satisfying ( σ . (cid:96) (cid:48) ) = z = [ σ (cid:48) = a σ + b σ + (cid:96) ] , (2) gives a + b =
0, i.e. b = − a . On the other hand, werewrite (1) as − a ( σ . σ ) + d =
0, which forces a ∈ R . Hence, σ (cid:48) = a σ − a σ + (cid:96) .Assume that A = Re i θ ; then z = [ σ (cid:48)(cid:48) ] , where σ (cid:48)(cid:48) = ae i θ σ − ae − i θ σ + (cid:96) .Indeed z ∈ P (cid:96) holds since (1) is satisfied, and ( σ (cid:48)(cid:48) . A ( γ )) = ( ae i θ σ . A ( γ )) + ( − ae − i θ σ . A ( γ ))= ae i θ (cid:16) A ( σ ) . γ (cid:17) − ae − i θ (cid:16) A ( σ ) . γ (cid:17) = ae i θ A − ae − i θ A = P (cid:96) (cid:39) S (cid:96) defined in Lemma 2.1, we see that A = Re i θ acts on R by a rotation of angle − θ . Note that if A ∈ ( K T ) × then θ = θ = π ,and z = z (in R ), as already mentioned.19e add a topological flavour to this discussion. Endow K × T with the topologyinduced by the topology of C × under the fixed embedding K T (cid:44) → C , and ( K T ) × with the subspace topology. Then the topological quotient group K × T / ( K T ) × is asubspace of the topological quotient group P ( R ) = C × / R × . Moreover, R is asubspace of S (cid:96) / {±} and the action of K × T / ( K T ) × on it is given by rotation. Remark 4.11. If K T is a CM field, then K × T / ( K T ) × is dense in P ( R ) . To deducethis, it is enough to show that the elements of norm one inside K T are dense inthe circle: this is the content of Corollary 1.7. Remark 4.12.
We may define the equivalent of Q + also in this setting: let R + bethe set of (pair of) points z ∈ S (cid:96) ⊆ P (cid:96) that are orthogonal to an element of T ofpositive self-intersection. If γ ∈ T is of positive self intersection, denote by z (one of) its orthogonal. For an element A ∈ K × T , | A | = γ = A ( γ ) is of positive self-intersection as well. Then, by acting with elements of S ∩ K × T on z , we see that R + is dense in S (cid:96) / {±} if K T is a CM field.In retrospect, we have proven the following result. Proposition 4.13.
Assume that T has CM. Then: • the topological group K × T / ( K T ) × acts freely, transitively and continuously by ro-tations on the topological space R ; • having fixed an element z ∈ R , this action induces a homeomorphism betweenK × T / ( K T ) × and R , and this homeomorphism passes to the topological completions,inducing a homeomorphism between P ( R ) and S (cid:96) / {±} ; • R is dense in S (cid:96) / {±} , or equivalently: the set of points of Picard jump of theequator is (countable and) dense in the equator; • finally, R + is countable and dense in S (cid:96) / {±} . EOMETRIC INTERPRETATION
Suppose that X is a complex projective K3 surface, and let (cid:96) = c ( L ) be the firstChern class of an ample line bundle. We call period field and endomorphism fieldof X the period field and the endomorphism field, respectively, of the transcen-dental lattice T ( X ) ⊆ H ( X , Q ) , which is a polarized irreducible Hodge structureof K3 type. We say that a complex projective K3 surface X has CM if T ( X ) hasCM, that is: its dimension as a K T ( X ) -vector space is 1.We can construct the twistor space X → P C associated with ( X , L ) , wherewe put in evidence the considered ample line bundle. The fibre over a point ζ ∈ P C is a K3 surface of complex structure x I + x J + x K , if ζ and ( x , x , x ) S (cid:96) (cid:39) P (cid:96) , an element of which corresponds toa class z = [ σ (cid:48) ] ∈ P (cid:96) ; S (cid:39) P C , parametrizing the complex structures x I + x J + x K of the fibres of the geometric twistor space. We want to underline therelation existing between these spheres. We may give explicitly an isomorphism P (cid:96) (cid:39) P C that respects the following property: the Hodge structure determinedby z = [ σ (cid:48) ] ∈ P (cid:96) on T ⊕ Q (cid:96) corresponds to the Hodge structure determined by σ ζ on the same vector space, for ζ ∈ P C corresponding to z ∈ P (cid:96) . Equivalently, onemay ask σ (cid:48) and σ ζ to differ only by a a complex scalar. The isomorphism P C (cid:39) P (cid:96) is defined by ζ (cid:55)→ (cid:104) σ ζ = σ − ζ σ + ζ (cid:96) (cid:105) = (cid:20) ζ σ − ζ σ + (cid:96) (cid:21) (sending ∞ to [ σ ] ). This follows from the explicit form of σ ζ given in [HKLR87,Section 3.F]. It is remarkable that σ ζ is a linear combination of σ , σ and (cid:96) only.Composing these isomorphisms, we get S (cid:96) (cid:39) P (cid:96) (cid:39) P C (cid:39) S .We claim that this composition is nothing but the permutation ( x , y , z ) (cid:55)→ ( x , x , x ) = ( z , x , y ) on the sphere. For, a point ζ ∈ P C is sent, towards theleft, to (cid:104) ζ σ − ζ σ + (cid:96) (cid:105) ∈ P (cid:96) . Set a = ( ζ ) , b = − ζ /2. Then b − a = − ζ − ζ = − ζ ( ζζ + ) = ζζ + ζζ · ( − ζ ) and aa − bb = ζζ − ζζ = ζζ + ζζ · ( − ζζ ) .Therefore the vector v ( a , b ) = (cid:32) (cid:60) ( b − a ) , (cid:61) ( b − a ) , (cid:16) aa − bb (cid:17) ( σ . σ ) d (cid:33) ,that determines the point in S (cid:96) (Lemma 2.1), is positively aligned to the vector (cid:16) − (cid:60) ( ζ ) , − (cid:61) ( ζ ) , 1 − ζζ (cid:17) ,where we use that ( σ . σ ) = ( (cid:96) . (cid:96) ) = d , as already noticed. However, the imageof ζ in S via the stereographic projection is positively aligned with (cid:16) − ζζ , − (cid:60) ( ζ ) , − (cid:61) ( ζ ) (cid:17) ,and this proves the claim.Therefore, we may talk about equator and points at the same altitude for S (cid:96) and S interchangeably (even if the altitude of S (cid:96) corresponds, in fact, to the firstcoordinate of S ). Points at the same altitude correspond, therefore, to complexstructures x I + x J + x K on X with the same coefficient x . The equator corre-spond to complex structures x J + x K , for which the I -part is missing.21 emark 5.1. For a point ζ ∈ P C , let z ∈ P (cid:96) its correspondent under the isomor-phism P (cid:96) (cid:39) P C . We have the equality ρ z + ρ ( X ) − = ρ ( X ζ ) .To prove this, look at the Néron-Severi group of the K3 surface X ζ . For an element γ ∈ H ( X ζ , Q ) = H ( X , Q ) , being in NS ( X ζ ) is equivalent to being orthogonal tothe (
2, 0 ) -form σ ζ = σ − ζ σ + ζ (cid:96) . Of course the orthogonal complement of theclass (cid:96) in the original Néron-Severi group, (cid:96) ⊥ ⊆ NS ( X ) , is always orthogonal to σ ζ (and this corresponds to the addend ρ ( X ) − H ( X , Q ) is T ⊕ Q (cid:96) , and is in direct sum with it. Lying in T ⊕ Q (cid:96) and, at the same time, being orthogonal to σ ζ means belonging to the space P ⊥ z ∩ ( T ⊕ Q (cid:96) ) ,whose dimension is, by definition, ρ z . Therefore, the formula above holds.In the following, when talking about the twistor space X → P C associatedwith a projective complex K3 surface X , we imply the choice of (cid:96) = c ( L ) asKähler class for the construction of X . The analogous of Proposition 3.2, provenby Huybrechts, is the following result. Corollary 5.2.
Consider the twistor space
X → P C associated with a projective complexK3 surface X. If ρ ( X ζ ) > ρ ( X ) , then ζ is contained in the equator S ⊆ S (that is: ζ ismapped to the equator S ⊆ S under stereographic projection P C (cid:39) S ). Proposition 3.4 and Proposition 4.13 yield:
Corollary 5.3.
Consider the twistor space
X → P C associated with a projective complexK3 surface X with complex multiplication. If ζ is a point of Picard jump on the equator,then ρ ( X ζ ) = + ρ ( X ) Moreover, the Noether-Lefschetz locus of the equator is dense in the equator. More pre-cisely, the locus of points on the equator whose fibres are algebraic K3 surfaces is dense inthe equator.
Admissible valuesof ρ ( X ζ ) , CM caseOutside theequator ρ ( X ) − ρ ( X ) On theequator ρ ( X ) −
1, 10 + ρ ( X ) Proof.
Only the equation is left to discuss. If r = dim Q T , then r + ρ ( X ) =
22 and ρ z = r /2 = ( − ρ ( X )) /2, if z corresponds to ζ via the isomorphism P (cid:96) (cid:39) P C .Therefore, the formula above follows from Remark 5.1.22 emark 5.4. The only case where no points of excessive jump appear is under theassumption of maximal Picard number, i.e. ρ ( X ) =
20. This answers [Huy19, Re-mark 5.2] in the CM case. Also, it is a generalization of [Huy19, Remark 5.5]: notonly if ρ ( X ) <
20 there is no fibre such that ρ ( X ζ ) =
20, but the set of admissiblevalues of ρ ( X ζ ) is also very constrained. Remark 5.5.
A K3 surface X is projective, or equivalently algebraic, if and onlyif there exists a line bundle L with L >
0, where L denotes the self-intersectionof this line bundle. For a proof, see [BHPV04, Theorem IV.6.2]. For twistor fibres,we see how the geometric property “being algebraic” agrees with the algebraicrequirement ( (cid:96) (cid:48) . (cid:96) (cid:48) ) > Remark 5.6. If ρ ( X ) =
20, each fibre X ζ corresponding to a point of Picard jump(both on and outside the equator) has Picard number 20 and, therefore, it is au-tomatically algebraic and of CM type. However, when ρ ( X ) <
20, thanks to Re-mark 4.1, there are non-algebraic fibres corresponding to points of Picard jumpoutside the equator.Corollary 4.10 yields the following results.
Proposition 5.7.
Consider the twistor space
X → P C associated with a projective com-plex K3 surface X with complex multiplication. Then the locus of ζ ∈ P C such that X ζ is algebraic is dense (for the classical topology) in P C .Proof. The density of the locus follows from the density of Q + in U . Theorem 5.8.
Consider the twistor space
X → P C associated with a projective complexK3 surface X with complex multiplication. Assume that ζ , ζ ∈ P C are two points ofPicard jump at the same altitude and not on the equator. Then X ζ is algebraic if andonly if X ζ is such. If so, then the CM endomorphism fields of these K3 surfaces coincide.Moreover, the set of points of Picard jump at the same altitude of ζ (and ζ ) is countableand dense in the circle at that altitude. Remark 5.9.
For ζ , ζ ∈ P C , being at the same altitude and outside the equatormeans corresponding to complex structures on X having the same non-zero I -component. 23 EFERENCES [BHPV04] W. Barth, K. Hulek, C. Peters, and A. Van de Ven.
Compact complexsurfaces . Second. Vol. 4. Springer-Verlag, Berlin, 2004, pp. xii+436.[BL78] P. Blanksby and J. Loxton. “A note on the characterization of CM-fields”. In:
J. Austral. Math. Soc. Ser. A
Comm. Math. Phys.
Lectures on K3 surfaces . Vol. 158. Cambridge Studiesin Advanced Mathematics. Cambridge University Press, Cambridge,2016, pp. xi+485.[Huy19] D. Huybrechts.
Complex multiplication in twistor spaces . 2019. arXiv: .[Joy00] D. Joyce.