In this section, we prove that a general section 𝑠 ∈ 𝐻0(𝐹 ,O (1,1)) gives rise to two non-isomorphic Calabi–Yau threefolds 𝑋 = 𝑍(𝑝∗𝑠) and 𝑌 = 𝑍(𝑞∗𝑠), this result will be stated in Theorem 7.2.6. Be-fore proving the theorem, we will discuss some auxiliary results. In (BCP20), an argument to show that every 𝑋e⊂ X25 is contained in just one pair of Grassmannians has been explained. Using similar ideas, we will prove an analogous result for the boundary ¯X25 of the family, namely that every Calabi–Yau threefold in ¯X25 is contained in just one Grassmannian.
Lemma 7.2.1. Let 𝑋 be a Calabi–Yau threefold described as the zero locus of a section ofQ∨
2(2). Then the following equalities hold for every 𝑡 ≥0:
𝐻0(𝐺(2, 𝑉5),Q2(−𝑡))=𝐻0(𝑋 ,Q2|𝑋(−𝑡)); (7.2.1)
𝐻0(𝐺(2, 𝑉5),∧2Q2(−𝑡))=𝐻0(𝑋 ,∧2Q2|𝑋(−𝑡)). (7.2.2)
In particular,𝐻0(𝑋 ,Q2|𝑋) '𝑉5and
𝐻0(𝑋 ,Q2|𝑋(−𝑡)) =𝐻0(𝑋 ,∧2Q2|𝑋(−𝑡)) =0 for𝑡strictly positive.
Proof. Let us consider the following short exact sequence which comes from tensoring the ideal sheaf sequence of 𝑋 with Q2:
0−→ I𝑋/𝐺(2,𝑉5) ⊗ Q2(−𝑡) −→ Q2(−𝑡) −→ Q2|𝑋(−𝑡) −→0 (7.2.3) Given this sequence, we need to show the vanishing of the first two degrees of cohomology for I𝑋/𝐺(2,𝑉5)⊗ Q2. To do this, we consider the sequence obtained tensoring with Q2 the Koszul resolution of the ideal sheaf of 𝑋:
0−→ Q2(−5−𝑡)−→ Q−𝜃 2⊗ Q2∨(−3−𝑡) −→
−→ Q2⊗ Q2(−2−𝑡)−→ I−𝜙 𝑋/𝐺(2,𝑉5) ⊗ Q2(−𝑡) −→0
(7.2.4)
The bundles Q∨
2(−5−𝑡) and Q2⊗ Q∨
2(−3−𝑡) have no cohomology in degree smaller than six: this follows from the isomorphisms
Q∨2(−5−𝑡) ' ∧2Q2∨⊗ (∧3Q2∨)⊗(4+𝑡) Q2⊗ Q∨2(−3−𝑡) ' (∧3Q2∨)⊗(2+𝑡) ⊗ ∧2Q∨2 ⊗ Q∨2
(7.2.5) and by a Borel–Weil–Bott computation. This, in turn, proves that 𝐻<6(𝐺(2, 𝑉5),ker(𝜙)) = 0 in (7.2.4). Similarly, one finds that Q2 ⊗ Q2(−2−𝑡) has no cohomology in degree smaller than four, due to Q2⊗ Q2(−2−𝑡) ' (∧2Q∨
2)⊗(2+𝑡). Therefore𝐻0(𝐺(2, 𝑉5),I𝑋/𝐺(2,𝑉5) ⊗ Q2) =0 and 𝐻1(𝐺(2, 𝑉5),I𝑋/𝐺(2,𝑉5) ⊗ Q2) = 0. This, together with (7.2.3), proves our claim (7.2.1). The second equality follows from a totally
analogous computation, namely it involves the tensor product of the ideal sheaf sequence with the wedge square of Q2. Lemma 7.2.2. Let 𝑋 be a Calabi–Yau threefold described as the zero locus of a section ofQ∨
2(2). Then the restrictionQ∨
2(2) |𝑋 is slope-stable.
Proof. Consider a subobject F ⊂ Q∨
2|𝑋(2). Then, since 𝐺(2, 𝑉5) has Picard number one, we have 𝑐1(F ) =O (𝑡) for some𝑡 and this leads to the injection
0−→ O −→ ∧𝑟Q∨2|𝑋(2𝑟−𝑡) (7.2.6) where 𝑟 is the rank of F, which can be either one or two. To have F as a destabilizing object for Q∨
2|𝑋(2), 𝑡 must satisfy the following
On the other hand, for the injection in (7.2.6) to exist it means that
∧𝑟Q∨
2|𝑋(2𝑟 − 𝑡) has global sections. Let us now consider the case 𝑟 =1. ThenQ∨
2|𝑋(2−𝑡) has sections only for𝑡 ≤1, but for such values the inequality 7.2.7 cannot be satisfied. We can prove that the same happens for 𝑟 =2: in fact,∧2Q∨
2|𝑋(4−𝑡) ' Q2(3−𝑡) has sections only for 𝑡 ≤ 3, but the inequality 7.2.7 cannot be fulfilled with these values
of 𝑡.
Let us suppose 𝑋 is contained in two Grassmannians𝐺1and𝐺2, where the latter is the image of the former under an isomorphism of P9. Since both the restrictions of the normal bundlesN𝑖|𝑋 =N𝐺
𝑖/P9|𝑋 =Q∨
2𝑖(2) |𝑋
are stable with the same slope, every morphism between them must
be either zero or an isomorphism. Below we furthermore prove that the isomorphism class of the normal bundle determines the Grassman-nian. Combining these two facts will give us the uniqueness of the Grassmannian containing 𝑋.
Lemma 7.2.3. Let 𝑋 be a Calabi–Yau threefold described as the zero locus of a section ofQ∨
2(2). Then the following isomorphism holds:
𝐻0(P9,O (1)) ' 𝐻0(𝑋 ,O𝑋(1)) (7.2.8) Proof. The claim follows by proving separately the following claims:
𝐻0(P9,O
P9(1)) ' 𝐻0(𝐺(2, 𝑉5),O𝐺(2,𝑉5)(1)) (7.2.9) 𝐻0(G(2, 𝑉5),O𝐺(2,𝑉5)(1)) ' 𝐻0(𝑋 ,O𝑋(1)) (7.2.10) Let us begin by verifying Equation 7.2.10. A twist of the Koszul resolution of 𝑋 ⊂𝐺(2, 𝑉5) yields:
0−→ O (−4) −→ ∧2Q2(−3) −→ Q2(−1) −→ O (1) −→𝑖∗O𝑋(1) −→0 (7.2.11) where𝑖 is the embedding of 𝑋 in𝐺(2, 𝑉5). The desired isomorphism is a consequence of the vanishing of cohomology of the first three bundles.
To prove the validity of Equation 7.2.9 we follow basically the same argument applied to the Pfaffian resolution of 𝐺(2, 𝑉5), yielding the following exact sequence on P9:
0−→ O (−4) −→𝑉5⊗ O (−2) −→𝑉5⊗ O (−1) −→
−→ OP9(1) −→ 𝑗∗O𝐺(2,𝑉5)(1) −→0
(7.2.12) where 𝑗 is the embedding of 𝐺(2, 𝑉5) in P9. Again the first three bundles have no cohomology. For both computations, the vanishings
can be computed by the Borel–Weil–Bott theorem (or, in the first case,
by Lemma 7.2.1).
Lemma 7.2.4. Consider a Calabi–Yau threefold 𝑋 ∈ X25such that 𝑋 is contained in two translates𝐺1, 𝐺2 of𝐺(2, 𝑉5) inP9. IfN𝐺
1|P(∧2𝑉5)|𝑋 ' N𝐺
2|P(∧2𝑉5)|𝑋, then𝐺1 =𝐺2.
Proof. Let E be a globally generated rank three vector bundle on 𝑋 such that𝐻0(𝑋 ,E) =𝑉5. Then it defines a unique morphism 𝑓 : 𝑋 −→
The proof is concluded by observing that sinceN𝐺
1|P(∧2𝑉5)|𝑋 ' N𝐺
2|P(∧2𝑉5)|𝑋,
one has 𝑓1= 𝑓2, and hence 𝐺1=𝐺2.
Corollary 7.2.5. If 𝑋 ⊂ P9 is a Calabi–Yau threefold from the family
X¯25, then 𝑋 is contained as a zero locus of a vector bundle in a unique Grassmannian𝐺(2,5)in its Plücker embedding.
Proof. Suppose that 𝑋 is contained in two Grassmannians𝐺1, 𝐺2such that for each of them we have an exact sequence:
0→ N𝑋|𝐺𝑖 → N𝑋| isomorphism it induces an isomorphism N𝐺
1|P9|𝑋 ' N𝐺
2|P9|𝑋 and we conclude by Lemma 7.2.4. If it is trivial it lifts to an isomorphism N𝑋|𝐺1 ' N𝑋|𝐺2 which again gives an isomorphism N𝐺
1|P9|𝑋 ' N𝐺
2|P9|𝑋
and permits us to conclude again by Lemma 7.2.4.
Now we are ready to prove the main theorem of this chapter.
Theorem 7.2.6. Let𝐹 be the partial flag manifold𝐹(2,3, 𝑉5), let 𝑝and
4. Because of Lemma 7.1.1, we deduce that if there ex-ists an isomorphism mapping 𝑋 to 𝑌, then it is given by a map 𝑓 : 𝐺(2, 𝑉5) → 𝐺(3, 𝑉5). Recall that such a map is determined by a linear isomorphism from𝑇𝑓 :𝑉5→𝑉∨
5.
Thus, because of Corollary 7.2.5, 𝑋 and 𝑌 are isomorphic only if there exists 𝑓 :𝐺(2, 𝑉5) →𝐺(3, 𝑉5) such that 𝑋 is 𝑓-dual to 𝑋. This, by Corollary 7.1.8 translates to the fact that a section 𝑠𝑋 ∈ H𝐹 from Lemma 7.1.5 defining 𝑋 on 𝐹 satisfies 𝑀−1
𝑓 𝑆𝑇𝑀𝑓 = 𝜆𝑆 for 𝑆 being the matrix associated to the section 𝑠𝑋 and some constant𝜆. But since 𝑆 and 𝑆𝑇 are similar matrices then multiplication by 𝜆 must then pre-serve the spectrum of 𝑆. The proof amounts now to find a matrix 𝑆 corresponding to an element of H𝐹 with spectrum that is not fixed by multiplication with 𝜆≠ 1 and such that the equation
𝑆𝑇𝑀− 𝑀 𝑆=0
has no solutions among matrices 𝑀 of the form 𝑀 = ∧2𝑇, and then expand by openness to the general element ofH𝐹. This is done via the following script in Macaulay2 (GS19):
R=QQ[a_1..a_25]
S=matrix{
{ 1 ,0,0,0,0,0,0,0,0,0}, {0, 2 ,0,0,0,0,0,0,0,0}, {0,0, 0 ,0,0,0,0,0,0,0}, {0,0,0, 0 ,0,0,0,0,0,0}, {0,1,0,0, 0 ,0,0,0,0,0}, {0,0,0,0,0, 1 ,0,0,0,0}, {0,0,0,0,0,0,-1 ,0,0,0}, {0,0,0,0,0,0,0,-1 ,0,0}, {0,0,0,0,0,0,0,0,-1 ,0},
{0,0,0,0,0,0,0,0,0,-1 }}
T=genericMatrix(R,5,5) M=exteriorPower(2,T)
Sol=ideal flatten(transpose(S)*M-M*S) saturate(Sol, ideal det T)
Here we chose a matrix 𝑆 satisfying the equations defining H𝐹 = 𝐻0(𝐼𝐹∨|𝑃∨)⊥ as in Remark 7.1.6.
This implies that a general hyperplane section 𝑠 of the flag variety 𝐹 yields two Calabi–Yau threefolds 𝑋 and 𝑌 which are dual, but not projectively isomorphic. By the fact that the studied manifolds have Picard number one we conclude that they are not birational (OR17,
proof of Theorem 4.1).
The proof above being very explicit has the advantage that it permits to construct concrete examples of pairs of Calabi–Yau varieties in our family which are dual but not birational. We can however perform a more conceptual proof, which is more suitable to generalization and allows to estimate the expected codimension of the fixed locus of our duality.