ON THE BIRATIONALITY OF TORIC DOUBLE MIRRORS

Transcription

1 ON THE BIRATIONALITY OF TORIC DOUBLE MIRRORS by ZHAN LI A dissertation submitted to the Graduate School New Brunswick Rutgers, The State University of New Jersey In partial fulfillment of the requirements For the degree of Doctor of Philosophy Graduate Program in Mathematics Written under the direction of Lev A. Borisov And approved by New Brunswick, New Jersey May, 2014

2 ABSTRACT OF THE DISSERTATION On the birationality of toric double mirrors By ZHAN LI Dissertation Director: Lev A. Borisov We prove that generic complete intersections associated to double mirror nef-partitions are all birational. This result solves a conjecture of Batyrev and Nill in [6] under some mild assumptions. This dissertation is based on my paper [18]. ii

3 Acknowledgements I am deeply indebted to my advisor, Professor Lev A. Borisov, for his encouragement, immense knowledge, and constant guidance: in our weekly meetings, he never shows impatience when I stick to trivial questions, instead, he explains to me every details on blackboard or even writes down every tiny things on piece of papers for my record; he is never too busy to talk with; and he is always willing to reply my long s consisting of math questions. I always appreciate his generosity of sharing ideas and even proofs with me. Without his supervision, this dissertation would have never been written. I would like to thank Professor Anders S. Buch who taught me commutative algebra and first course of algebraic geometry. It is his encouragement and kindness that make algebraic geometry less formidable. I would like to thank Professor Xiaojun Huang who has been helping me from my first day at Rutgers. He is guiding me on academics as well as on my life. To Professor Chris Woodward, it is always my pleasure to observe him together with Lev shooting questions to seminar speakers; To Professor Jerrold B. Tunnell for his crystal clear lectures and enlightening seminars; To Professor Simon Thomas for helps on many occasions during my early years. I would like to thank the faculty and staff of Mathematics Department at Rutgers, especially Professor Stephen Miller, Professor Michael Saks, Professor Zhengchao Han, Ms. Lynn Braun, Ms. Demetria Carpenter and Ms. Katie Guarino, for their patience, help and support. I would like to thank Rutgers University for providing excellent academic environment and financial support; to the organizers of the conferences, workshops and seminars I spoke on and attended. To Howard Nuer for many helpful discussions and constant help. iii

4 Many thanks go to my friends: Dr. Jin Wang, Zhuo Feng, Simão Herdade, Edward Chien, Ming Xiao, John Miller, Mo Wang, Hanlong Fang, Liming Sun, Dr. Knight Fu, Sjuvon Chung, Dr. Tian Yang, Xu Dong, Dr. Jingang Xiong, Dr. Jinwei Yang, Professor Yuan Yuan, Yuxiang Qin, Fanxing Hu, Hongbin Wang, Dr. Kaige Zhu... And to many many people I don t know their names, it is their kindness that survives me from those difficult days. To my girlfriend Bai Yixiu for her patient waiting... To my relatives, especially my grandmother Ms. Fu Zhe for her unwavering confidence in me. Finally, I would give all my gratitude to my parents for raising me up, supporting me, understanding me and trusting me. Without them, any math is meaningless. iv

5 Dedication To My Parents. v

6 Table of Contents Abstract ii Acknowledgements iii Dedication v 1. Introduction Background Gorenstein cones and Nef-partitions Relationship between nef-partitions and reflexive Gorenstein cones The main question The main question and its motivation Example The main theorem Results on the decomposition of lattices Construction of the determinantal variety Proof of the main theorem Open questions: D-equivalence and K-equivalence Appendix: -regularity, singularities and Calabi-Yau varieties vi

7 1 Chapter 1 Introduction Mirror symmetry was first discovered in string theory as a duality between families of 3-dimensional Calabi-Yau manifolds. Since its discovery more than twenty years ago, it has drawn much attention from physicists and mathematicians. Batyrev [2] used -regular hypersurfaces in toric varieties associated to reflexive polytopes as a way to construct a large set of mirror pairs. In this case, the mirror pair consists of the family of -regular hypersurfaces associated to a reflexive polytope and the family of - regular hypersurfaces associated to its dual polytope. Borisov [7] generalized Batyrev s construction by considering nef-partitions of reflexive polytopes. A nef-partition of a reflexive polytope corresponds to a decomposition of the boundary divisor into nef divisors. In this case, the mirror pairs are constructed as the family of complete intersections associated to a nef-partition and the family of complete intersections associated to its dual nef-partition. These complete intersections are Calabi-Yau varieties, and their string-theoretic Hodge numbers behave as predicted by mirror symmetry [4]. Compared to hypersurfaces, complete intersections associated to nef-partitions are more complicated. In particular, they may exhibit nontrivial double mirror phenomenon, i.e. two Calabi-Yau varieties X, X may have the same mirror Y. If this is the case, the homological mirror symmetry conjecture [17] implies that the derived categories of coherent sheaves on X, X are equivalent. Indeed, according to the conjecture, the derived categories of X, X are expected to be equivalent to the Fukaya categories of their mirrors, which in this case are the same because X, X are double mirrors. Instead of derived equivalence, Batyrev and Nill [6] asked whether toric double mirrors are birational. We give an affirmative answer to this question in Theorem under some mild assumptions:

8 2 Theorem. Let X, X be toric double mirrors, then there exists a variety D, called the determinantal variety, with morphism X X D such that if X, X and D are all irreducible with dim D = dim X = dim X, then X, X are birational. Now, we describe briefly the content of each chapter: In Chapter 2, we fix the notations used throughout the paper. We give relevant background information on reflexive Gorenstein cones and nef-partitions. At the end of the chapter, we prove Proposition which connects the notions of nef-partitions and reflexive Gorenstein cones. This will be used to reformulate Batyrev and Nill s original question in the language of Gorenstein cones. We also give a constructive proof of the converse of Proposition in Proposition In Chapter 3, we reformulate the question of Batyrev and Nill using reflexive Gorenstein cones. We also discuss the motivation of this question and give an example which motivates our proof. In Chapter 4, we give a proof for the main result Theorem We also discuss the necessity of its assumptions. In Chapter 5, we present some open questions related to the subject. In the appendix, we give the definition of -regularity and discuss its properties. We show that the singularities of -regular intersections are inherited from the ambient toric variety. In particular, the complete intersections considered in the paper are Calabi-Yau varieties with canonical, Gorenstein singularities. This fact is used in the proof of the main theorem.

9 3 Chapter 2 Background 2.1 Gorenstein cones and Nef-partitions We fix the following notations throughout the paper. Let M = Z d be a lattice of rank d, and N = Hom Z (M, Z) be its dual lattice with pairing, : M N Z. Let M R := M Z R, and N R := N Z R be the R-linear extensions. The pairing between M, N can be extended to, : M R N R R. Let M = Z s M be the lattice extended from M, and N = Z s N be its dual lattice with pairing: M N Z (a 1,, a s ; m) (b 1,, b s ; n) a i b i + m, n, where the integer s should be obvious from the context. The purpose of introducing notations M, N will become clear in a moment: if a nefpartition lives in M (or N), then the corresponding reflexive Gorenstein cone will live in M (or N). Sometimes we also use lattice M 1 and its dual lattice N 1. The convention is as follows: we always use M (or N) to denote the lattice where polytopes live, if the cones come from nef-partitions, we use M (or N) to denote the lattice where they live. However, when talking about general cones which do not come from nef-partitions, we use M 1 (or N 1 ) to denote the lattice where they live. Let S M R be a set, we use Conv(S) to denote its convex hull. If M R with the origin 0 in the interior is a lattice polytope (i.e. the convex hull of a finite set of lattice points), then := {y N R x, y 1, x } is its dual polytope. We use Vert( ) to denote the set of vertices of a lattice polytope, and l( ) to denote the set of its lattice points, i.e. l( ) = M.

10 4 Definition Let be a lattice polytope with the origin 0 as an interior point. If the dual polytope is also a lattice polytope, then is called reflexive polytope. Definition (See [5]) A d-dimensional rational polyhedral cone K (M 1 ) R is called a Gorenstein cone, if it is generated by lattice points which are contained in an affine hyperplane {x (M 1 ) R x, n = 1} for some n N 1. This n is uniquely determined if dim K = rank M 1, and this is the only case considered in the paper. We denote this unique element by deg, and call it the degree element. By definition, deg must live in K N 1, where K := {y (N 1 ) R x, y 0, x K} is the dual cone of K. In general, K is a Gorenstein cone does not imply K is a Gorenstein cone. However, if this is the case, we arrive at the notion of reflexive Gorenstein cone. Definition (See [5]) A Gorenstein cone K is called reflexive Gorenstein cone if K is also a Gorenstein cone. Let deg K, deg K be the degree elements in K, K respectively, then deg, deg is called the index of this pair of dual reflexive Gorenstein cones. We will see in a moment how reflexive Gorenstein cones relate to nef-partitions. Before doing this we should briefly recall the notion of nef-partition. In the projective toric variety defined by a reflexive polytope, a nef-partition is equivalent to a decomposition of the boundary divisor into a summation of nef divisors. On the other hand, there exists a purely combinatorial definition of nef-partition without invoking toric variety constructions. For simplicity, we use this combinatorial definition here. The readers can find its equivalent form and its motivation in Borisov s original paper [7]. Definition If the Minkowski sum of s lattice polytopes s i is a reflexive polytope, and the origin 0 i (0 may not be an interior point) for each i, then { i i = 1,..., s} is called a length s nef-partition of the convex hull Conv( s i). Nef-partitions arise in pairs [7]: if we fixed a nef-partition { i i = 1,..., s} with i M R, then there exists a dual nef-partition { i i = 1,..., s} with i N R. The

11 5 relations between them are ( i ) = Conv( s i ) ( i ) = Conv( s i ). Furthermore, they satisfy the property min i, j δ ij, and w j Vert( j ) {0}, the minimum value can be achieved, that is min x, w j = δ ij. x i 2.2 Relationship between nef-partitions and reflexive Gorenstein cones From a nef-partition, one can construct a reflexive Gorenstein cone [5]. On the other hand, from a reflexive Gorenstein cone associated to a nef-partition, if we have a decomposition of the degree element deg, we can construct another nef-partition. Now we will give a precise statement of the above relations, which appeared in a slightly different form in [6]. In fact, we will prove a general result. Let K, K be full dimensional reflexive Gorenstein cones in (M 1 ) R, (N 1 ) R, with degree elements deg, deg in K, K respectively. Suppose the index is deg, deg = s and with e i N 1 K, e i 0. deg = e i, Let S = {x K x, deg = 1} S i = {x K x, e i = 1, x, e j = 0, j i} T = {y K deg, y = 1}. Because K is a Gorenstein cone, any vertex v of S is a lattice point. Thus v, e i are nonnegative integers which add up to 1. Hence, there exists precisely one e i such that

12 6 v, e i = 1. On the other hand, for any e j, because e j 0 and K is a full dimensional cone, there exists at least one vertex w of S such that w, e j = 1. Using these facts, one can show that {e 1,..., e s } must be part of a Z-basis of N 1. Let Ann(e 1,..., e s ) := {m M 1 m, e i = 0, i, 1 i s} be a sublattice of M 1 (we also use Ann(e) for simplicity if no confusion arises), and Span Z {e 1,..., e s } := Ze i be a sublattice of N 1. From the fact that {e 1,..., e s } is part of a Z basis, it follows that the pairing between M and N induces a pairing Ann(e 1,..., e s ) (N 1 / Span Z {e 1,..., e s }) Z, which identifies Ann(e 1,..., e s ) and N 1 / Span Z {e 1,..., e s }. Proposition Under the above notations, the lattice polytope is a reflexive polytope. S i deg Ann(e 1,..., e s ) R Proof. We will show that the dual polytope of s S i deg is exactly T (N 1 / Span Z {e 1,..., e s }) R, where T is the image of T under the projection (N 1 ) R (N 1 / Span Z {e 1,..., e s }) R. First, we show that T has 0 as an interior point. Because deg is in the interior of K, 1 s deg is also in the interior of K, and thus in the interior of T. This property is kept under the projection map T T. The image of 1 s deg is 0 in (N 1 / Span Z {e 1,..., e s }) R, and thus 0 is in the interior of T. Second, we show 0 s S i deg is an interior point. Let deg = i I λ iv i, where λ i R and v i Vert(S) be vertices of S. We have 1 = deg, e 1 = i I λ i v i, e 1 = i I 1 λ i,

13 7 where I 1 = {i I v i, e 1 = 1}. This implies i I 1 λ i v i S 1. By continuing this procedure, one can show deg s S i. If w Vert(T ) is a vertex of T, then s S i, w cannot always be zero. Indeed otherwise, all the S i would be contained in a facet of K, which is impossible. Thus, for any w Vert(T ), there exists v s S i such that v deg, w 0. If s S i deg did not have 0 as an interior point, then R 0 ( s S i deg) (M 1 ) R. We have already showed that T had 0 as an interior point, so R 0 T = (N 1 / Span Z (e 1,..., e s )) R. In particular, there exists a vertex w of T, and thus a vertex w T, such that s S i deg, w < 0, a contradiction. Next, we show that s S i deg Ann(e 1,..., e s ) R is a reflexive polytope with dual T (N 1 / Span Z (e 1,..., e s )) R. Because min S i deg, T = min S i deg, T = min S i, T deg, T 0 1 = 1, we have T ( s S i deg). We only need to show the other inclusion T ( s S i deg). Let y ( s S i deg) such that there exists x s i S i deg with x, y = 1 (this y corresponding to some boundary point of the dual polytope of s i S i deg). We will show for this y, y T. Then it follows for arbitrary y ( s S i deg), y T. Let θ i = min x Si x, y and set y = y s θ ie i. We claim y K. Indeed, K = s t is i with t i 0, and we have min K, y = min = t i S i, y min t i ( S i, y ) θ i e i = S i, θ j e j = j=1 min t i S i, y θ i e i (t i (min S i, y θ i )) 0. Finally, we will show y T and this will imply y T. By the assumption on y, we have min s S i deg, y 1, and there exists x s S i deg, such that x, y = 1. Let x = s x i deg with x i S i, then we must have x i, y = θ i. Indeed, otherwise there exists k such that x k, y > θ k, and all the others satisfy x i, y θ i.

14 8 Thus ( ) 1 = min S i deg, y = min S i, y deg, y ( ) ( ) = θ i deg, y = x i, y deg, y, a contradiction. We have deg, y = deg, y and this implies y T. θ i e i = deg, y θ i = 1 The converse is proved in [6] Theorem 2.6. We will give a direct proof by constructing the dual cone K explicitly. Proposition Let 1,..., s M R be lattice polytopes such that the Minkowski sum s i has dimension dim(m R ) and s i m be a reflexive polytope for some m M. Let M = Z s M, then the associated cone in M R K = {(a 1,..., a s ; a i i ) a i 0} is a reflexive Gorenstein cone of index deg, deg = s. Proof. Let = ( s i m), and for any vertex w j Vert( ), we set m ij = min x i x, w j. Then we claim K N R is generated by the lattice points {(m 1j, m 2j,..., m sj ; w j ) N w j Vert( )} { (0,..., 1,... 0 ; 0) N 1 i s}. }{{} 1 at the i-th position Suppose these lattice points generate a cone C, then it is straightforward to check C K. The difficult part is to show K C. Let (a 1,..., a r ; t) K, then we must have a i + min i, t 0 for all i. Subtracting a non-negative combination of the (0,..., 1,..., 0; 0) if necessary, we have a i +min i, t = 0 for all i. In this case, if one can show (a 1,..., a s; t) C, then adding

15 9 back those non-negative combination of the (0,..., 1,..., 0; 0), we have (a 1,..., a s ; t) C. By the above argument, we can assume without loss of generality, a i +min i, t = 0 for all i. Moreover, if t = 0, then we are done. If t 0, then one can multiply t by a positive real number λ, such that λt lands on the boundary of. In this case, we still have λa i + min i, λt = 0, and if one can show (λa 1,..., λa s ; λt) C, then certainly (a 1,..., a s ; t) C. Thus, we can reduce to the case when t is on the boundary of, particularly, it is on some facet F v. Here, v is a vertex of s i m such that v, F v = 1, where we have used the 1 1 correspondence between vertices and facets in dual reflexive polytopes. Let v = s v i m, with v i a vertex of i, and t = j λ jt j, λ j 0, j λ j = 1, with t j vertices of F v. Then because 1 + m, t = min i, t = min i, t = v i, t, we have min i, t = v i, t = j λ j v i, t j = j λ j min i, t j. The last equation uses the fact that t j F v, and because ( ) 1 = min i m, t j = min i, t j m t j = ( ) v i m, t j = v i, t j m, t j, we must have i, t j = v i, t j. Putting everything together, we have ( min 1, t,..., min s, t, t) = j λ i ( min 1, t j,..., min 1, t j ; t j ). This proves the claim K C. In order to show K is also a Gorenstein cone, let deg = (1, 1,..., 1; m). By using the property min s i m, v j = 1, it is straightforward to show that for the vertex v j of, deg, (m 1j, m 2j,..., m rj ; v j ) = 1

16 10 and deg, ( 0,..., 1,..., 0 ; 0) = 1. }{{} 1 at the i-th position Thus we finish the argument K is a reflexive Gorenstein cone. Because deg = (1, 1,..., 1; 0), the index is deg, deg = s. The above theorem can be applied to the case of nef-partitions, where s i itself is a reflexive polytope with dual polytope ( s i) = Conv( s i). Because 0 i, and min i, j = δ ij, we can write the reflexive Gorenstein cones associated to this pair of nef-partitions in a symmetric way K = {(a 1,..., a s ; a i i ) (M) R a i 0} K = {(b 1,..., b s ; b i i ) (N) R b i 0}. This result can also be proved directly as in [5]. The following is our key construction which is used to reformulate the question from polytopes to reflexive cones. Now we start off with a nef-partition { i 1 i s}, and let K be the reflexive Gorenstein cone associated to it as above with the degree element deg K. If deg = s ẽi, with ẽ i 0, ẽ i K N, then we can similarly define S i as in Proposition ( s In this case, S i deg) is a reflexive polytope in Ann(ẽ 1,..., ẽ s ). Without loss of generality, we can assume ẽ i = ( 0,..., 1,..., 0 ; p }{{} i ) Z s N. 1 at the i-th position We claim that there exists a lattice isomorphism φ : Ann(ẽ 1,..., ẽ s ) M defined by restricting to the projection p : Z s M M. In fact, if φ(x) = 0, then x = (a 1,..., a s ; 0), but x Ann(ẽ 1,..., ẽ s ) implies that i, a i = 0, thus φ is injective.

17 11 The surjectivity comes from the fact that for m M, if we let a i = m, p i, then (a 1,..., a s ; m) Ann(ẽ 1,..., ẽ s ) maps to m under φ. Under this isomorphism, we can identify Ann(ẽ 1,..., ẽ s ) with M. Let i = p( S i ), one can verify directly that s s Conv( i ) = Conv( i ). Moreover, since φ(deg) = 0, and by Proposition 2.2.1, ( s S i deg) is a reflexive polytope in Ann(e 1,..., e s ) R. Hence, ( s i ) is a reflexive polytope in M. Because ( 0,..., 1,..., 0 ; 0) }{{} S i, we have 0 i, and this implies { i 1 i s} is another 1 at the i-th position nef-partition of Conv( s i) (see Definition 2.1.1). Remark One cannot exhaust all the nef-partitions of length s of Conv( s i) using the above construction (i.e. first construct reflexive Gorenstein cone K, K, then decompose deg = s ẽi, and finally construct i ). For example, any subsets of the vertices of an octahedron will give a nef-partition, but some subsets cannot be obtained from the above construction. However, the above process will give exactly the combinatorial data for toric double mirrors (details see Theorem 3.1.3). Next, we give the geometry meaning of this construction. Let X(Σ) be the toric variety defined by the fan Σ := {0} {R 0 θ θ Conv( i i ) is a face}, and L i = D ρ, Li = ρ Vert( i )\{0} ρ Vert( i )\{0} D ρ be the nef divisors corresponding to { i }, { i } respectively, where D ρ is the torus invariant divisor associated to the primitive element ρ. The following result gives a characterization of the nef-partitions obtained from reflexive Gorenstein cones as above. Proposition The nef-partition { i 1 i s} of Conv ( s i) is obtained from the same reflexive Gorenstein cone if and only if the corresponding divisors { L i 1 i s} and {L i 1 i s} are pairwise linearly equivalent.

18 12 Proof. Suppose deg = s ẽi = s e i. Without loss of generality, we can assume ẽ i e i = p i N. Then one can check that L i L i is exactly the principle divisor (X p i ) on X(Σ). On the other hand, suppose L i, L i are linearly equivalent divisors for each i, then there exists p i N such that L i L i = (X p i ). one can check that ẽ i = e i + (0; p i ) satisfies the requirement. One may ask what is the relation between associated Gorenstein cones of these double mirror nef-partitions. As one can imagine, they are all isomorphic. In fact, let ẽ i Z s M as before, then η i = ( 0,..., 1,..., 0 ; 0) Z }{{} s M must be in 1 at the i-th position S i by definition. Let S i = S i η i, then we have (1) 0 S i Ann(ẽ 1,, ẽ s ); (2) s S i is reflexive; (3) S i has the same image as S i under the aforementioned projection. Also, because Ann(ẽ 1,, ẽ s ) is isomorphic to M under the same projection, we can identify { S i } with the nef-partition i. Hence we only need to show the claim for { S i } in lattice Ann(ẽ 1,, ẽ s ). This is straightforward to check, because Zη i + Ann(ẽ 1,, ẽ s ) = Z s M, and the reflexive Gorenstein cone associated to S i is K = r i (η i + S i ) = r i S i = K, r i 0. We will prove the birationality for the -regular complete intersections associated to nef-partitions (i.e. double mirror nef-partition) which are obtained from above.

19 13 Chapter 3 The main question 3.1 The main question and its motivation After establishing the relation between reflexive Gorenstein cones and nef-partitions, we are ready to state the question asked in [6] more explicitly. Let us repeat the construction in the last part of Section 2 in order to extract the main ingredients. Let M be a reflexive polytope, Conv( s i) =, and { i i = 1,..., s} be a nef-partition of. Let M = Z s M, N = Z s N, and K M R be the reflexive Gorenstein cone associated to this nef-partition. The dual cone of K is K N R and deg K is the degree element. Then deg = s e i with e i = ( 0,..., 1,..., 0 ; 0) gives back the original nef-partition { }{{} i }. If there exists 1 at the i-th position another decomposition deg = s ẽi with ẽ i 0, ẽ i K N, then we can associate to it { i } which gives another nef-partition of. Whenever one has a polytope, there is a family of Laurent polynomials associated to it. Let l( i ) be the set of lattice points in i, then the family of Laurent polynomials associated to i is f i = v l( i ) c v X v C[M], where c v is a complex coefficient only depends on the vertex v. Here we abuse notations, using v to represent the lattice point as well as its coordinate in M. For example, if v = (a 1,..., a n ) M, then X v = x a 1 1 xan n C[M]. In the same fashion, j produces a family of Laurent polynomials f j = v l( j ) c v X v C[M].

20 14 Remark We should emphasize that for the same vertex v, v 0, the coefficient c v is the same in all Laurent polynomials. However, the coefficient of the origin, c 0 (i.e. the constant term) might be different in different Laurent polynomials. We abuse notations to avoid writing c 0,j in place of c 0. We can take the zero locus of all f i in (C ) d = Spec(C[M]), and denote this variety by X ( i ). To be precise X ( i ) (C ) d is defined by: X ( i ) : f 1 = f 2 = = f s = 0, and similarly, X ( i ) (C ) d is defined by X ( i ) : f1 = f 2 = = f s = 0. Remark From toric variety point of view, this construction can be stated as follows. Let X := X(Σ( )) be the projective toric variety associated to the polytope s i, T X be the big torus. Let L i be the line bundle associated to the dual nef-partition { i 1 i s}. Generic global sections in H 0 (X, L i ) can be identified with Laurent polynomials with Newton polytopes i. In particular, for 1 i s, f i = v l( i ) c vx v H 0 (X, L i ). Let (f i ) 0 be the zero locus of f i, then X ( i ) = T (f 1 ) 0 (f s ) 0. We will return to this point of view in the appendix. The following question was asked by Batyrev and Nill in [6] Question 5.2: (Nef-partition version) Are the Calabi-Yau complete intersections X ( i ) and X ( i ) birational to each other? We can reformulate this question in terms of reflexive Gorenstein cones as follows. Let S = {x K x, deg = 1}, S i = {v K v, deg = v, ẽ i = 1}. Because deg = s ẽi, for each lattice point v in S, that is v l( S), there exists a unique

21 15 i, such that v, ẽ i = 1. We have a disjoint union l( S) = s l( S i ). One can define a Laurent polynomial in C[M] by setting: g i = v l( S i ) c v X v. For any lattice point w i such that w i, ẽ i = 1, w i, ẽ j = 0, i j, X wi g i is a Laurent polynomial in C[Ann(ẽ 1,..., ẽ r )]. We can similarly define an intersection X (ẽi ) (C ) d = Spec(C[Ann(ẽ 1,..., ẽ s )]) by X (ẽi ) : X w1 g 1 = X w2 g 2 = = X ws g s = 0 This intersection does not depend on the choice of w i, because any other choice will differ by a factor X w, w C[Ann(ẽ 1,..., ẽ s )] and this will not affect the zero loci defined in (C ) d. Similarly, we can construct S i and g i associated to the decomposition deg = r e i, and an intersection X (ei ) (C ) d = Spec(C[Ann(e 1,..., e s )]) by X (ei ) : X w 1 g1 = X w 2 g2 = = X w s g s = 0. We can compare the equations defined by these intersections with the equations defined the intersections above by nef-partitions. Because the lattice isomorphism φ : Ann(ẽ 1,..., ẽ r ) M sends S i deg to i, we can identify g i C[Ann(ẽ 1,..., ẽ r )] with f i C[M] up to a factor X v i, v i C[M]. Hence, X (ẽi ) and X ( i ) are isomorphic varieties. The same thing is true for X (ei ) and X ( i ) as well. The importance of the above construction is explained in the following theorem. Theorem The complete intersections X (ẽi ) and X (ei ) are toric double mirror in the sense that they both mirror to the same family. We abuse the notations: X (ẽi ) here means the family parameterized by the coefficients c v, and the same for X (ei ).

22 16 Proof. By the toric mirror construction in [2] [3], the mirror of X (ẽi ) is a family of generic complete intersections defined by divisors { L i 1 i s} in the toric variety X(Σ) (see the notations above Proposition 2.2.4). Likewise, the mirror of X (ei ) is a family of generic complete intersections defined by divisors {L i 1 i s} in X(Σ). By Proposition 2.2.4, { L i 1 i s}, {L i 1 i s} consist of pairwise linearly equivalent divisors and hence they defined the same family of complete intersections which is the mirror of both X (ẽi ) and X (ei ). Viewing the original question from this perspective, we can ask: (Reflexive Gorenstein cone version) Are the toric double mirror X (ei ), X (ẽi ) birational? We give an affirmative answer to this question in Theorem under some technical assumptions. 3.2 Example In this section, we will illustrate the basic idea of the proof by an explicit example. Let {u 1,..., u 15 } be a basis of Z 15, and we consider a sublattice M Z 15 which is defined by M := { 15 l i u i Z 15 5 l i = The rank of M is 13, it contains a cone K = Z 15 0 M which is defined by nonnegativity 10 i=6 l i = 15 1 of all l i. The 125 generators of rays of K are given by u i1 + u i2 + u i3 l i }. with 5j 4 i j 5j, and let c ijk C denote coefficients. Suppose {v 1,..., v 15 } is the dual basis of {u 1,..., u 15 }, then the dual lattice M is the quotient of Z 15 : M = Z 15 / Span Z { 5 10 v i v i, i=6 5 v i The dual cone K is the image of Z 15 0 in M, and its rays are generated by v i, 1 i The degree elements deg, deg are given by 15 u i and 5 v i respectively. 1 v i }.

23 17 in K : There are three different ways of decomposing deg as a summation of lattice points deg = 5 10 v i, deg = v i, deg = This gives three different complete intersections in P 4 P 4. i=6 15 For deg = 5 v i, the equations of this decomposition can be expressed as 1 j,k 5 1 j,k 5 1 j,k 5 c 1jk x 1 y j z k = 0 c 2jk x 2 y j z k = 0 c 5jk x 5 y j z k = 0. Here [x 1,, x 5 ] are homogenous coordinates of P 4, and similarly for y j, z k. As explained before, we can multiply each equation a factor in order to make it well defined in M Ann(v 1,..., v 5 ). Hence, let f i (y, z) = x 1 i 1 j,k 5 c ijk x i y j z k =. 1 j,k 5 1 v i. c ijk y j z k = 0, 1 i 5. This can be viewed as five bidegree (1, 1) equations in P 4 P 4. Similarly, for deg = 10 i=6 v i and deg = 15 1 v i we have defining equations: g j (x, z) = h k (x, y) = 1 i,k 5 1 i,j 5 c ijk x i z k = 0, 1 j 5, c ijk x i y j = 0, 1 k 5. Our question thus becomes whether these three complete intersections are birational for generic choice of c ijk. Let X 1 be the variety defined by f i = 0, 1 i 5. Let A 1 (z) be 5 5 matrix ( 5 ) A 1 (z) = c ijk z k k=1 ij, 1 i, j 5,

24 18 then f i = 0, 1 i 5 can be written as a matrix equation A 1 (z) y 1. y 5 = 0. Notice that ([y 1,..., y 5 ], [z 1,..., z 5 ]) P 4 P 4 satisfy f i = 0, 1 i 5 if and only if det(a 1 (z)) = 0 in P 4. Let D 1 denote the variety defined by det(a 1 (z)) = 0. For generic coefficients, one can show X 1 and D 1 are birational. Similarly, the variety X 2 defined by g j = 0, 1 j 5 can be written as (x 1,, x 5 ) A 2 (z) = 0 where ( 5 ) A 2 (z) = c ijk z k, 1 i, j 5. k=1 ij Let D 2 be the variety defined be det(a 2 (z)) = 0. The same argument as above shows that X 2 is birational to D 2. On the other hand, D 1 and D 2 are the same varieties, and hence X 1, X 2 are birational. We notice that despite drastically different defining equations, the three complete intersections are all birational. This example suggests us to look at the determinantal variety defined by a common matrix of different nef-partitions. However, it is not very clear how to construct this common matrix at present stage. Besides that, there are following more pressing issues: (1) the dimension of Span R {ẽ 1 e 1,..., ẽ s e s } might be smaller than s 1 which leads to considering the intersection of several determinantal varieties; (2) nonsaturatedness might occur, which forces us to work in auxiliary lattices; (3) in order to show the birationality, we have to take into account of the singularities of the complete intersection. This leads us to consider -regular intersections.

25 19 Chapter 4 The main theorem 4.1 Results on the decomposition of lattices Let be a reflexive polytope, { i 1 i s} be a nef-partition of, and { i 1 i s} be its dual nef-partition. In the following, we assume dim = dim M R. Because i, we have dim( s i) = dim M R. We use Span R {p 1,..., p s } to denote the vector space spanned by p i N R, 1 i s. The following lemma is crucial for our argument. Lemma Let p i i. If s p i = 0, and dim(span R {p 1,..., p s }) = s r, then there exist disjoint sets I k {1,..., s}, 1 k r, such that r k=1 I k = {1,..., s} and for each k, we have i I k p i = 0. Proof. Suppose l is the maximum number such that there exist l nonempty disjoint sets I j, 1 j l satisfying I 1 Il = {1,, s} and j, i I j p i = 0. Because these l equations are linearly independent, we have s r = dim(span R {p 1,..., p s }) s l, and hence l r. All we need to show is l = r. Otherwise, suppose l < r, then there must exist at least one equation 1 i s a ip i = 0, which is not a linear combination of i I j p i = 0. Hence, there must exist an index

26 20 j, such that for i I j, a i are not identically the same. Suppose a m is a minimal element in {a i i I j }. After reindexing the set, we can assume j = 1 and m = 1. Let C be a sufficiently large number, then 0 = a i p i a 1 b i p i 1 i s 2 i s i I 1 p i + C i I 2 I l p i = satisfies b i > 0 when i I 2 I l, and b i 0 when i I 1. Moreover, there exists at least one element t I 1 such that b t > 0 (because a i are not identically the same for i I 1 ). Let S = {i b i 0} be the index set corresponding to nonzero coefficients. Set P = i S p i = i S (1 cb i)p i with c sufficiently big such that i, (1 cb i ) < 0. When k / S, we have k, p i 0 i S k, i S(1 cb i )p i 0. Hence k, P = 0 for k / S. In the following, we will show P = 0. Otherwise, there exists v M R such that v, P > 0. Because M R = s R 0 i, we can chose v = 1 i s v i with v i i. Then we have v, P = v i + v i, P = v i, p j + i, P. i S i/ S i S j / S i/ S v We use the assumption s j=1 p j = 0, and thus P = j / S p j in the second equation. However, i S v i, j / S p j 0, and i/ S v i, P = 0 because k, P = 0 for k / S. This contradiction implies P = i S p i = 0. Because I 1 S and I 1 S, the index set I 1 := I 1 S must satisfy I 1 I 1. Since I 2 I l S, we have p j = P j I 1 i I 2 I l p i = 0. But this implies p j = p j = 0 j I 1 j I 1 \I 1

27 21 which gives a further decomposition of I 1. This is a contradiction to the maximality of l. Remark Under the notation of Lemma 4.1.1, we observe that for each k, dim(span R {p i i I k }) = #(I k ) 1. Let M = Z s M, and K M R be the reflexive Gorenstein cone associated to a nef-partition { 1,..., s } in M R as it is in Proposition This nef-partition corresponds to deg = s e i K, where e i = ( 0,..., 1,..., 0 ; 0). If deg }{{} = 1 at the i-th position s ẽi with ẽ i 0, ẽ i N K, then we can assume without loss of generality that Note that ẽ i = ( 0,..., 1,..., 0 }{{} ; p i ), p i N i. 1 at the i-th position dim(span R {e 1,..., e s, ẽ 1,..., ẽ s }) = s + dim(span R {p 1,..., p s }), hence, if dim(span R {p 1,..., p s }) = s r, by Lemma there exists disjoint index sets I k, 1 k r, such that r k=1 I k = {1,..., s}. For each k, we have i I k p i = 0, with dim(span R {p i i I k }) = #(I k ) 1. Let n k = #(I k ) from now on, and let Ann(e) := Ann(e 1,..., e s ) = {m M m, e i = 0, 1 i s}. If rank M = d, then Ann(e) is a sublattice of M with rank d, and Ann(e, ẽ) : = Ann(e 1,..., e s, ẽ 1,..., ẽ s ) = {m M m, e i = m, ẽ i = 0, 1 i s} a sublattice of M with rank d + r s. For our convenience, we use {(k1), (k2),..., (kn k )} as the index set of I k, and reindex the corresponding elements. For example p i = 0 i I k

28 22 becomes under the new indexing. n k p ki = 0 Because dim(span R {p k1,..., p knk }) = n k 1, we can choose {p 12,..., p 1n1,..., p r2,..., p rnr } as a R-linearly independent set. Another important fact of {p i 1 i s} is that they form a saturated sublattice in N, that is, the abelian group N/ ( s k=1 Zp i) is torsion free. The following combinatorial proof is due to Borisov. Lemma The sublattice s Zp i N is saturated. Proof. Suppose otherwise, there exists n = s a ip i with a i Q such that n N but n s Zp i. Furthermore, we can assume i, 0 a i < 1. Recall that i, p i i, hence a i p i i. By the property of nef-partition, we have n s i = ( i ). i If n 0, then there exists a lattice m s i i such that 1 n, m < 0. Because n is a lattice, we have n, m = 1. On the other hand, the set {m s i i n, m = 1} must contain some vertices of s i i and hence some vertices of i due to nef-partition, without loss of generality, i we can assume m k. Then use the property that min i, j δ ij (see the discussion after Definition 2.1.4), we have this is a contradiction. n s Zp i. 1 = n, m = a i p i, m a k > 1, Thus n = 0, but this contradicts our initial assumption on Using the above two lemmas, we can decompose the lattice to fulfil our purpose.

29 23 Lemma The lattice Ann(e) M can be decomposed as follows: Ann(e) = Ann(e, ẽ) Z[w 12 ] Z[w 1n1 ] Z[w r2 ] Z[w rnr ]. Where w ki M satisfies the following requirements (where by our indexing, w ki starts from w k2 ): 1. w ki, ẽ k1 = 1, w ki, ẽ ki = 1 for i w ki, ẽ lj = 0 for all ẽ lj ẽ k1, ẽ ki. 3. w ki, e lj = 0 for all e lj. Proof. First, if we already have w ki satisfying the given properties, then by definition, we have Ann(e, ẽ) Z[w 12 ] Z[w 1n1 ] Z[w r2 ] Z[w rnr ] Ann(e) as a sublattice. On the other hand, m Ann(e), we set m k m, ẽ ki w ki, ki i 2 then by definition, one can check m k m, ẽ ki w ki ki i 2 Ann(e) Ann(ẽ 12,, ẽ 1n1,, ẽ r2,, ẽ rnr ). Using the fact that k, t I k e t = t I k ẽ t, we have m k m, ẽ ki w ki Ann(e, ẽ). ki i 2 Thus, we only need to show the existence of w ki. Let lattice map θ : M Z s r

30 24 be defined by m ( m, p 12,..., m, p 1n1,..., m, p r2,..., m, p rnr ). We claim that θ is a surjective lattice map. Because of the saturatedness (Lemma 4.1.3), {p 12,..., p 1n1,..., p r2,..., p rnr } forms part of Z-basis of N. It follows that θ is surjective. We can choose m such that m, p ij = 0 j 2 except m, p ki = 1, and set w ki = (0, 0,..., 0; m) M, then w ki satisfies the required properties. Now let L = Span Z {w 12,, w 1n1,, w r2,, w rnr } M we have Ann(e) = Ann(e, ẽ) L. Because of the above decomposition of lattices, we have a corresponding decomposition of toric varieties: Spec(C[Ann(e)]) = Spec(C[Ann(e, ẽ)]) Spec(C[L]). For any closed point in Spec(C[Ann(e)]) with coordinate x, we will write x = (y, ω) with y Spec(C[Ann(e, ẽ)]), ω Spec(C[L]) respectively. 4.2 Construction of the determinantal variety The main ingredient in the proof of Theorem is a determinantal variety D which serves as a bridge to connect two complete intersections. We will show how to construct this variety in Spec(C[Ann(e, ẽ)]) which heavily relies on Lemma Now, let S i,j = {v K v, deg = 1, v, e i = v, ẽ j = 1}

31 25 be a polytope, and g i,j = c v X v v l(s i,j ) be the Laurent polynomial associated to S i,j with coefficients c v C. Let u ki M satisfy: 1. u ki, e ki = u ki, ẽ k1 = 1 2. u ki, e lj = 0 for all e lj e ki 3. u ki, ẽ lj = 0 for all ẽ lj ẽ k1. We point out that unlike those w ki constructed before, u ki starts from u k1 for each k. The existence of u ki follows from the similarly reason as in Lemma 4.1.4, and we do not repeat it here. Next, we proceed to the construction of the determinantal variety D. Let A k (y) be the n k n k matrix with entries in C[M], A k (y) X u k1g k1,k1 X u k1 w k2 g k1,k2 X u k1 w knk g k1,knk X u k2g k2,k1 X u k2 w k2 g k2,k2 X u k2 w knk g k2,knk =... X u kn k g knk,k1 X u kn k w k2 g knk,k2 X u kn k w knk g knk,kn k Notice that the first column is not constructed identically as the rest. The reason for writing the matrix A k (y) as a function of y is that every entry of this matrix is in C[Ann(e, ẽ)], as one can verify. Thus, according to the above decomposition Spec(C[Ann(e)]) = Spec(C[Ann(e, ẽ)]) Spec(C[L]), we use y to represent the corresponding coordinates in Spec(C[Ann(e, ẽ)]). Next, we define n k 1 matrix w k = (1, X w k2,, X w kn k ) t, where t means the transpose of a matrix. And also define the 1 n k matrix u k = (X u k1, X u k2,..., X u kn k ).

32 26 We claim that the condition A k (y) w k = 0 is exactly the same as X u k1g k1. X u kn k g knk = 0. Indeed, recall (Section 3) by definition, we have g ki = c v X v v l(s ki ) where S ki = {v K v, deg = 1, v, e ki = 1} (Notice: this is not the same as S k,i defined before). Because of the relation i I k e i = i I k ẽ i, for any v l(s ki ), v, i I k e i = 1 implies v, i I k ẽ i = 1, thus there exists kj, such that v l( S kj ), where S kj = {v K v, deg = 1, v, ẽ kj = 1}. This means v l(s ki,kj ), and in particular, we have a disjoint union l(s ki ) = l(s ki,kj ). kj I k Hence, g ki = kj I k g ki,kj, and this justifies the claim. On the other hand, u k A k (y) = 0 is exactly the same as ( X w k1 g k1,, X w kn k g knk ) = 0, where g kj = v l( S kj ) c vx v = ki I k g ki,kj because of the disjoint union l( S ki ) = l( S ki,kj ). ki I k Let D k := {det A k (y) = 0}

33 27 in Spec(C[Ann(e, ẽ)]). Let r D = k=1 with its reduced induced subscheme structure. This D will serve as a bridge to prove the birationality of two complete intersections. D k Remark We have det A k (y) 0 for generic coefficients because we always have ( 0,..., 1,..., 0 ; 0) S }{{} i,i. 1 at the i-th position Thus by the definition of determinant, after choosing generic coefficients, these elements will give a nonzero summand in det A k (y), hence, D k is a hypersurface in Spec(C[Ann(e, ẽ)]), and dim D k = d + r s 1, with d = rank M. We need the following lemma about the degree of a morphism and points in the generic fibres. This proof below is indebted from the discussion with Professor Qing Liu. Lemma Let f : X Y be a dominant morphism of varieties over C. Suppose [K(X) : K(Y )] = n. Then there exists a dense open subset U of Y such that f 1 (y) consists of n (distinct) points for all y U. In particular, if f is a dominant, injective morphism, then [K(X) : K(Y )] = 1, so X, Y are birational. Proof. First we can reduce it to the case when f is a finite morphism. In fact, the problem is local in Y, hence we can assume Y = Spec(A) to be affine. Let X X be a non empty open affine subset, it suffice to prove the result for X Y. This is because [K(X) : K(Y )] <, then Y \ f(x\x ), hence the open set f 1 (Y \ f(x\x )) X. Now we assume X = Spec(B). The dominant morphism f corresponds to an injective homomorphism A B. Write k(b) = k(a)[t] where k(b), k(a) are quotient fields of B, A and t annihilates a polynomial P (T ) k(a)[t ] of degree n (theorem of primitive element). Replacing A by a localization A a with a A such that the element t becomes integral over A (also

34 28 localizing B correspondingly). As B is a finitely generated algebra over A, localizing further A, we can suppose A B A[t] (because each element of B belong to some A a [t], it is enough to inverse a common denominator for a system of generators of B over A). As B and A[t] have the same field of fractions and B is finite over A, localizing A again, we have B = A[t] = A[T ]/(P (T )). The discriminant of P (T ) belongs to A (we may need to localize A for this) and is non-zero because P (T ) is separable in k(a)[t ]. Let U be the principal open subset D( ) Y. Then for any y Y, the fiber f 1 (y) is given by the algebra k(y)[t ]/( P (T )) where k(y) = C denotes the residue field at y and P (T ) k(y)[t ] is the canonical image of P (T ). Its discriminant is (y) 0, so it has n (distinct) roots. 4.3 Proof of the main theorem In this section, we will show that X (ei ) and X (ẽi ) are both birational to the determinantal variety D. In fact, we will show that the morphism X (ei ) to D induced by the projection from Spec(C[Ann(e)]) to Spec(C[Ann(e, ẽ)]) gives the birational morphism, and similarly for X (ẽi ) to D. We recall our setup first: Let M, N be rank d lattices, and let K, K be reflexive Gorenstein cones associated to a length s nef-partition. Let deg = s e i = s ẽi where e i, ẽ i K N, e i, ẽ i 0 as before. Again, without loss of generality, we assume 1 i s, e i = ( 0,..., 1,..., 0 }{{} ; 0), ẽ i = ( 0,..., 1,..., 0 }{{} ; p i ) N. 1 at the i-th position 1 at the i-th position By Lemma 4.1.1, we have a decomposition of {p 1,..., p s } into subsets I k = {p k1,..., p knk }, for each 1 k r. We define the intersections X (ei ), X (ẽi ) as in Section 3. With this notation, we have the following birationality result: Theorem For generic coefficients, if X (ei ), X (ẽi ), D are irreducible with dim D = dim X (ei ) = dim X (ẽi ), then the complete intersections X (ei ) and X (ẽi ) are birational.

35 29 Proof. When s = 1, then X (ei ) = X (ẽi ), so nothing needs to be proved. Now we assume s 2. From the discussion after Lemma 4.1.4, we have Ann(e) = Ann(e, ẽ) L M. For any closed point x X (ei ), we can write x = (y, ω) Spec(C[Ann(e)]) with y Spec(C[Ann(e, ẽ)]), and ω Spec(C[L]) respectively. We claim that there exists a morphism π: π : X (ei ) D defined by x y. Indeed, by Lemma 4.1.4, we have a lattice decomposition of Ann(e) in M Ann(e) = Ann(e, ẽ) Z[w 12 ] Z[w 1n1 ] Z[w r2 ] Z[w rnr ]. By the construction of A k (y), the following matrix equation A 1 (y) A 2 (y)... w 1 w 2 = 0. A r (y) gives the variety X (ei ), where w k = (1, X w k2,, X w kn k ) t. w r Hence, for a closed point (y, ω) X (ei ) with y Spec(C[Ann(e, ẽ)]) and ω Spec(C[L]), we have for all k, A k (y)w k = 0. Because w k 0, we must have det(a k (y)) = 0. Hence, for all k, y lives in D k, and thus y D = r k=1 D k. This shows that the natural projection Spec(C[Ann(e)]) Spec(C[Ann(e, ẽ)]) maps X (ei ) to D. We denote this morphism by π. Next, we show that π is generically injective, that is, π is injective on a nonempty open subset of X (ei ). Roughly speaking, the proof rests on the fact that a Calabi-Yau variety cannot be uniruled. We show that if π is not generically injective, then X (ei ) is

36 30 a uniruled variety. However, we can construct a compactification X (ei ) of X (ei ) which is a projective, Calabi-Yau variety with canonical, Gorenstein singularities. Put these facts together, and we get a contradiction. The details are given in the follows: Suppose π is not generically injective. By a theorem of Chevalley ([10] Chapter II Ex.3.22(e)), there exists a nonempty open set V π(x (ei )) such that over V, the fibres have the same dimension h. Let y V, and let (X (ei )) y be the fibre over y. We claim that there exists a birational morphism θ y : (X (ei )) y P h. In fact, let Ω y = {ω Spec(C[L]) (y, ω) (X (ei )) y }, and let W y = { λ i ω i ω i Ω y, λ i C} i I,#(I)< be the affine subspace of Spec(C[L]) generated by Ω y, where I is some finite index set. We claim that Ω y W y is a dense open subvariety. To see this, notice by the matrix equation A 1 (y) A 2 (y)... w 1 w 2 = 0,. A r (y) w r if λ i 0 and λ i ω i Spec(C[L]), i I,#(I)< i I,#(I)< then we have (y, λ i ω i ) (X (ei )) y. i I,#(I)< However, the closed points in W y which satisfy i I,#(I)< λ i 0 and i I,#(I)< λ i ω i (C ) s r form an open variety, and this justifies the claim.

37 31 Now, dim Ω y = h implies dim W y = h, and the natural morphism θ y : (X (ei )) y W y P h is a birational morphism. Moreover, if π is not generically injective, then h 1, i.e. the general fibres of π are positive dimensional (one might need to pass to some small open subvariety of V in order to make the fibre contains distinct closed points). Then we can construct a birational morphism π 1 (V ) V P h (y, ω) (y, θ y (ω)). This shows that X (ei ) is a ruled variety and, in particular, a uniruled variety. In the appendix (cf. Remark , Proposition ), we construct a compactification X (ei ) of X (ei ), such that X (ei ) is a projective, Calabi-Yau variety with canonical, Gorenstein singularities. Let X (ei ) be a desingularization of X (ei ). It is also a uniruled variety. Because X (ei ) is a Calabi-Yau variety with canonical singularities, the canonical divisor K (ei ) of X (ei ) is K (ei ) = c j E j, c j 0, where E j are the exceptional divisors. Hence, H 0 ( X (ei ), O(K (ei ))) 0. However, because X (ei ) is a smooth, proper uniruled variety over C, we have H 0 ( X (ei ), O(K (ei ))) = 0 ( [15] IV Corollary 1.11). This is a contradiction, and hence π is generically injective. Let U X (ei ) be an open set where π U is injective. Because for generic coefficients, X (ei ) is smooth of dimension d s (Proposition ), π(u) is a constructible subset of D with dimension d s. This is the same dimension as D by assumption. Thus π is dominant as well. By Lemma 4.2.2, X (ei ) is birational to D. A similar argument can be used to show that X (ẽi ) is birational to D as well. We sketch the argument below:

38 32 First, by the proof of Lemma 4.1.4, one has a decomposition of lattices Ann(ẽ) = Ann(e, ẽ) Z[u 12 u 11 ] Z[u 1n1 u 11 ] Z[u r2 u r1 ] Z[u rnr u r1 ]. where u ki is as defined in Section 4. We can view u ki u k1 as w ki when i 2 because it satisfies the required relation of Lemma (with e ki, ẽ ki switched), and this is enough for the existence of the decomposition. Correspondingly, we have a decomposition of the torus: Spec(C[Ann(ẽ)]) = Spec(C[Ann(e, ẽ)]) (C ) s r. We can similarly define X (ẽi ) D as before, and for the same reason, this is a birational morphism. Hence X (ei ) and X (ẽi ) are both birational to D, and this completes the proof. Remark It is necessary in our argument for D to be irreducible. When we consider the case s = 2, with S 2,1 =, we see that D is a union of zero loci of g 1,1 and g 2,2, where g i,i = v l(s i,i ) c vx v, i = 1, 2. By the proof of the theorem, we see that X (ei ) is birational to the zero locus of g 2,2, but X (ẽi ) is birational to the zero locus of g 1,1. A priori, one cannot expect that the two loci be birational. There is a result due to Batyrev and Borisov ([3] Theorem 3.3) which asserts that X (ei ) is irreducible if the nef-partition is 2-independent. This means there exists no integer n > 0 nor any subset of the nef-partition { k1,..., kn } { 1,..., s } such that dim( k1 + + kn ) n. Remark It is reasonable to require that dim D = dim X (ei ) = d s. Indeed D = r D i is a variety in Spec(C[Ann(e, ẽ)]) = (C ) d (s r) defined by the intersection of r hypersurfaces. Thus D is expected to have dimension d s for generic choice of coefficients.