Let us recall some basic facts about lattice theory for 𝐾3 surfaces.
For Calabi–Yau manifolds 𝑌 of dimension 𝑛 ≥ 2 there exists an iso-morphism 𝑓 : Pic(𝑌) −→ 𝐻2(𝑌 ,Z) as a consequence of the condition 𝐻𝑘(𝑌 ,O) =0 for 1 ≤ 𝑘 ≤ 𝑛−1 and the exponential sequence. However, for 𝐾3 surfaces, 𝑓 is not surjective: hence the exponential sequence yields a natural embedding of the Picard group as a submodule of 𝐻2(𝑌 ,Z).
Let us see this embedding from a Hodge-theoretical point of view:
we call an integral Hodge structure of 𝐾3 type a freeZ-module𝑉 with a direct sum decomposition of its complexification
𝑉 ⊗C= Ê
𝑝+𝑞=2
𝑉𝑝,𝑞
such that:
1. 𝑉𝑝,𝑞 =𝑉𝑞, 𝑝 2. dim𝑉2,0=1
3. 𝑉𝑝,𝑞 =0 for |𝑝−𝑞| > 2
Moreover, we say a Hodge structure 𝑉 of weight 𝑛 is polarizable if there exists a map
𝑉 ⊗𝑉 −→Q(−𝑛)
which fulfills the following requirements (see (Huy16, Definition 1.6) for additional details):
1. it is a morphism of Hodge structures, where we consider Q(−𝑛) a Hodge structure of weight 2𝑛whose only nontrivial direct sum-mand is the one of bidegree (𝑛, 𝑛).
2. its R-linear extension is a bilinear form defined by a symmetric matrix 𝑁 via 𝛼, 𝛽↦−→ (𝛼, 𝑁 𝛽).
Clearly, the Z-module 𝐻2(𝑌 ,Z) for a 𝐾3 surface 𝑌 can be seen as a 𝐾3-type polarizable Hodge structure of weight 2 where the polarization is given by the intersection product, and the embedding of the Picard group in 𝐻2(𝑌 ,Z) is a morphism of Hodge structures since the Picard group is given by the Hodge classes 𝐻1,1(𝑌) ∩𝐻2(𝑌 ,Z).
Observe that the intersection pairing gives 𝐻2(𝑌 ,Z) the structure of a lattice, we can thus identify the Picard sublattice 𝑃𝑖 𝑐(𝑌) ⊂ 𝐻2(𝑌 ,Z) observing that the restriction of the intersection pairing to the Picard group is nondegenerate (Huy16, Proposition 2.4).
Another important sublattice of 𝐻2(𝑌 ,Z) is the transcendental lat-tice𝑇𝑌 defined Hodge-theoretically as the minimal sub-Hodge structure containing 𝐻2,0(𝑌) such that 𝐻2(𝑌 ,Z)/𝑇𝑌 is torsion-free. Moreover, the transcendental lattice of a polarizable Hodge structure of 𝐾3 type
is itself a polarizable irreducible Hodge structure of 𝐾3 type, and 𝑇𝑌 =Pic(𝑌)⊥ (Huy16, Lemma 2.7).
The following theorem (Orl03, Proposition 4.2.3) provides a practi-cal tool to understand whether two 𝐾3 surfaces are derived equiva-lent:
Theorem 8.1.1. Let𝑌,𝑌ebe complex algebraic𝐾3surfaces. Then𝐷𝑏𝑌 ' 𝐷𝑏𝑌eif and only if there exists a Hodge isometry
𝑓 :𝑇𝑌 −→𝑇
e𝑌.
The following result is a consequence of Theorem 8.1.1 and the Hodge structure of a hyperplane section of a roof, as we described it in Chapter 6.
Proposition 8.1.2. LetE andEebe vector bundles of rank𝑟 on rational homogeneous bases 𝐵 and 𝐵e. Let 𝑋 ' P(E∨) ' P(Ee∨) be a roof of dimension𝑟 +2and (𝑌 ,𝑌e)the Calabi–Yau pair associated to 𝑋 defined by a hyperplane section 𝑆 ∈ 𝐻0(𝑋 ,L). Then 𝑌 and 𝑌e are derived equivalent.
Proof. Given the dimension and rank of the Mukai pairs, 𝑌 and 𝑌e are 𝐾3 surfaces. By the fact that 𝐵 and 𝐵eare rational homogeneous, Equation 6.2.5 provides an isometry of transcendental lattices 𝑇𝑌 '𝑇
𝑌e. This, in turn, by Theorem 8.1.1, proves that 𝑌 and 𝑌e are derived
equivalent.
8.2 Non isomorphic 𝐾 3 pairs
Let us consider a roof 𝑋 where the bases of its vector bundles are smooth quadrics 𝑄 and𝑄e, denote by 𝐿 the line bundle which restricts to O (1) on each fiber of both the projective bundle structures of 𝑋. Call 𝑀 the zero locus of a general section of 𝐿 such that the associated Calabi–Yau pairs have dimension two (see Chapter 4 for details). Both 𝑄 and 𝑄e have cohomology generated by algebraic classes, hence we have an isometry(e𝑗∗◦e𝑞∗|𝑇
𝑌e)−1◦𝑗∗◦𝑞∗|𝑇𝑌 between transcendental lattices 𝑇𝑌 '𝑇
e𝑌 for each associated Calabi–Yau pair (𝑌 ,𝑌e) (in fact K3 pair in this case). Moreover, if 𝑌 and 𝑌e are general, by (Ogu01, proof of Lemma 4.1) 𝑇𝑌 and 𝑇
𝑌e admit no self-isometries different from ±Id.
Hence, to prove that𝑌 and 𝑌eare not isomorphic it is enough to prove that none of the isometries ±(e𝑗∗ ◦
e𝑞∗|𝑇
𝑌e
)−1◦ 𝑗∗ ◦𝑞∗|𝑇𝑌 extends to an isometry between𝐻2(𝑌 ,Z) and𝐻2(e𝑌 ,Z). We will prove it by studying the action of the constructed isometry on the discriminant groups of the associated transcendental lattices. For that we will consider the following notation. Given a lattice 𝑅, let us call 𝑑𝑅 the discriminant group defined by the exact sequence
0−→ 𝑅−→HomZ(𝑅,Z) −→𝑑𝑅 −→0. (8.2.1) Recall that if𝑌 is a K3 surface of Picard number one, with 𝐿 denoting the generator of its Picard group, then h𝐿i = (𝑇𝑌)⊥ ⊂ 𝐻2(𝑌 ,Z) and since 𝐻2(𝑌 ,Z) is unimodular 𝑑𝑇𝑌 ' 𝑑h𝐿i ' Z/𝐿2Z and there is a distinguished generator of 𝑑𝑇𝑌 corresponding to [ 𝐿
𝐿2] under the canon-ical identification 𝑑𝑇𝑌 ' 𝑑h𝐿i. Similarly, 𝑌e is a K3 surface of Picard number one and if we denote by e𝐿 the generator of its Picard group,
we have [e𝐿
e𝐿2
] representing the generator of 𝑑𝑇
e𝑌 associated to the em-bedding𝑇
e𝑌 ⊂ 𝐻2(e𝑌 ,Z).
Note that under the generality assumption for 𝑌 ,𝑌e (using (Ogu01, proof of Lemma 4.1)) each of the lattices 𝑇𝑌, 𝑇
𝑌e can be identified with𝑇𝑀 in a unique way up to ±Id, and this identification is given by
±𝑗∗◦𝑞∗|𝑇𝑌 and±e𝑗∗◦ e𝑞∗|𝑇
𝑌e. Furthermore, 𝐻𝑘(𝑀 ,Z) is unimodular and hence 𝑑𝑇𝑌 and 𝑑𝑇
e𝑌 admit canonical identifications with 𝑑𝐻2𝑟
𝑎𝑙 𝑔(𝑀 ,Z).
On the other hand, by Theorem 6.1.1 and Lemma 6.2.1 both 𝐻2(𝑌 ,Z) and 𝐻2(e𝑌 ,Z) admit Hodge isometric embeddings into 𝐻2𝑟(𝑀 ,Z) ex-tending the embeddings of the transcendental lattices. We conclude that under our identifications [ 𝐿
𝐿2] = ±[𝑗∗𝑞∗𝐿
This is checked in each of the two known cases by the following Lemma.
Lemma 8.2.1. Let 𝑋 be a roof of type 𝐺†
2 or 𝐷4, 𝑀 ⊂ 𝑋 a general hyperplane and𝑌 ,𝑌e the associated pair of 𝐾3 surfaces of degree 12.
Then there is a unique isometry of transcendental lattices 𝑇𝑌 ' 𝑇
𝑌e up to ±Id and this isometry descends to an isomorphism of discriminant groups which maps [𝐿
12] to±7[e𝐿
12].
Proof. By the discussion before the Lemma we just need to compare slightly in each of the two cases. Let us first illustrate the proof for the roof of type 𝐺†
2, which is easier because of the simpler structure of the cohomology ring of the quadric, which is odd dimensional.
By abuse of notation let us denote by 𝐿 ∈ 𝐻2(𝑄 ,Z) the hyperplane class of 𝑄, its restriction to 𝑌 which is the polarization, as well as its pullback to 𝑋 together with its restriction to 𝑀. Fix 𝜉 ∈ 𝐻2(𝑋 ,Z) the To see that, first observe that these are generators 𝐻6
𝑎𝑙 𝑔(𝑋 ,Z) which after restriction to 𝑀 define a sublattice of 𝐻6
𝑎𝑙 𝑔(𝑀 ,Z).
Now, given the Grothendieck relation on 𝑋 as a projective bundle on 𝑄 (Laz04a, page 310):
𝜉3−5𝐿 𝜉2+9𝐿2𝜉−12Π =0 (8.2.2) we can write the intersection form:
Π 𝐿2𝜉 𝐿 𝜉2
Π 0 1 5
𝐿2𝜉 1 10 32 𝐿 𝜉2 5 32 82
It follows that the sublattice hΠ, 𝐿2𝜉 , 𝐿 𝜉2i ⊂ 𝐻6
𝑎𝑙 𝑔(𝑀 ,Z) has rank 3 and discriminant 12. In light of Theorem 6.1.1 and the intersection form above, also 𝐻6
𝑎𝑙 𝑔(𝑀 ,Z) has rank 3 and discriminant 12. We conclude that hΠ, 𝐿2𝜉 , 𝐿 𝜉2i is the whole 𝐻6
𝑎𝑙 𝑔(𝑀 ,Z) proving the claim.
Knowing that (𝑗∗𝑞∗𝐿) ·𝐿2𝜉 =(𝑗∗𝑞∗𝐿) ·Π =0 and 𝑗∗𝑞∗𝐿 is an effective primitive class in 𝐻6
𝑎𝑙 𝑔(𝑀 ,Z) we get:
𝑗∗𝑞∗𝐿 =𝐿 𝜉2−5𝐿2𝜉+18Π (8.2.3) which, from the relation
𝜉 =𝐿+e𝐿 gives
𝑗∗𝑞∗𝐿=7e𝐿 𝜉2−23e𝐿2𝜉+42eΠ. Now by the same argument repeated for𝑌ewe have
e𝑗∗
e𝑞∗e𝐿=e𝐿 𝜉2−5e𝐿2𝜉+18eΠ. We conclude that
1
12(𝑗∗𝑞∗𝐿−7e𝑗∗
𝑞e∗e𝐿) =e𝐿2𝜉−7eΠ∈𝐻6
𝑎𝑙 𝑔(𝑀 ,Z).
Let us now focus on the roof of type 𝐷4. Here the 𝐾3 surfaces are zero loci ofS∨(1). The cohomology ring of a six dimensional quadric is slightly more complicated, since there exist two disjoint families of maximal isotropic linear spaces Π1,Π2. They satisfy the following relations in the cohomology ring:
𝐿3= Π1+Π2, Π1· 𝐿 = Π2·𝐿 , Π21= Π22=0, Π1·Π2=1. (8.2.4)
By the same argument as above, we can construct a basis of the middle cohomology 𝐻8(𝑀 ,Z) given by the classes Π1𝐿 ,Π1𝜉 ,Π2𝜉 , 𝐿2𝜉2, 𝐿 𝜉3. The Grothendieck relation is
𝜉4−6𝐿 𝜉3+14𝐿2𝜉2−14Π1𝜉−16Π2𝜉+12Π1𝐿 =0 (8.2.5) which yields the following intersection matrix:
Π1𝐿 Π1𝜉 Π2𝜉 𝐿2𝜉2 𝐿 𝜉3
Π1𝐿 0 0 0 1 6
Π1𝜉 0 0 1 6 22 Π2𝜉 0 1 0 6 22
𝐿2𝜉2 1 6 6 44 126
𝐿 𝜉3 6 22 22 126 308
As for the non homogeneous roof, we can compute the representation of 𝑗∗𝑞∗𝐿 in terms of the basis above:
𝑗∗𝑞∗𝐿 =𝐿 𝜉3−6𝐿2𝜉2+14Π1𝜉+14Π2𝜉−30Π1𝐿
=𝐿 𝜉3−6𝐿2𝜉2+14𝐿3𝜉−15𝐿4.
(8.2.6)
where the second equality follows from the relations 8.2.4. By substi-tuting the expression 𝐿 =𝜉−e𝐿 in Equation 8.2.6 we find
𝑗∗𝑞∗𝐿 =−6𝜉4+29e𝐿 𝜉3−54e𝐿2𝜉2+46e𝐿3𝜉−15e𝐿4 which by Equation 8.2.5 can be rewritten as
𝑗∗𝑞∗𝐿=−7e𝐿 𝜉3+30e𝐿2𝜉2−38eΠ1𝜉−50eΠ2𝜉+42eΠ1e𝐿 .
On the other hand, we can apply the same argument which leads to Equation 8.2.6 in order to get:
e𝑗∗
e𝑞∗e𝐿 =e𝐿 𝜉3−6e𝐿2𝜉2+14eΠ1𝜉+14eΠ2𝜉−30eΠ1e𝐿 . Finally, we have:
1
12(𝑗∗𝑞∗𝐿+7e𝑗∗
e𝑞∗e𝐿) =e𝐿2𝜉2−5eΠ1𝜉−4eΠ2𝜉+14eΠ1e𝐿 ∈ 𝐻8
𝑎𝑙 𝑔(𝑀 ,Z).
As a result of the discussion above, we get the following:
Corollary 8.2.2. Let 𝑋 be a roof of type𝐺†
2 or 𝐷4, 𝑀 ⊂ 𝑋 a general hyperplane and𝑌 ,𝑌e the associated pair of 𝐾3 surfaces of degree 12.
Then𝑌 and𝑌eare not isomorphic.