Rational curves on complete intersection Calabi-Yau 3-folds
B. Wang

TL;DR
This paper proves that generic complete intersection Calabi-Yau 3-folds contain rational curves of all degrees, which are immersions with specific normal bundle properties, advancing understanding of their geometric structure.
Contribution
It establishes the existence of rational curves of all degrees on generic complete intersection Calabi-Yau 3-folds and characterizes their normal bundles.
Findings
Existence of rational curves of every degree on generic Calabi-Yau 3-folds.
All such rational curves are immersions.
Normal bundle of these curves is isomorphic to or P^1}(-1)\u00d7or P^1}(-1).
Abstract
We prove the following results. If is a generic complete intersection Calabi-Yau 3-fold, (1) then for each natural number there exists a rational map \par\hspace{1 cc} of , (2) further more all such are immersions satisfying \begin{equation} N_{c(\mathbf P^1)/ X_3}\simeq \mathcal O_{\mathbf P^1}(-1)\oplus \mathcal O_{\mathbf P^1}(-1).
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Taxonomy
TopicsAlgebraic Geometry and Number Theory · Commutative Algebra and Its Applications · Geometry and complex manifolds
Rational curves on complete intersection Calabi-Yau 3-folds
B. Wang
(汪 镔)
Abstract
We prove the following results. If is a generic complete intersection Calabi-Yau 3-fold,
(1) then for each natural number there exists a rational map
of ,
(2) further more all such are immersions satisfying
[TABLE]
Contents
1 Statement
In mirror symmetry there is a general consensus that a “generic” Calabi-Yau 3-fold over should contain and only contain finitely many irreducible rational curves of each degree with respect to the polarization. This is the basis for the calculation of instantons in Mirror symmetry. While we are still in the process of finalizing the precise statement, let us prove it for the complete intersection Calabi-Yau 3-folds in a single projective space.
Theorem 1.1**.**
Let be a generic, complete intersection Calabi-Yau 3-folds over .
Then
(1) admits an irreducible rational curve of each degree,
(2) all such are immersions and the normal bundle is
isomorphic to
[TABLE]
2 Sketch of the proof
2.1 Setting
Throughout the paper rational curves are curves rationally parametrized by . The image is called an irreducible rational curve.
Rational curves on projective varieties has been a topic for many decades. The general theory which has its own technique is not the focus of this paper. Instead we are interested in a specific type of problems with a quite different technique.
Focus 2.1**.**
Which generic complete intersection admits an irreducible rational rational curve of each degree?
Focus 2.2**.**
Once the first question is affirmative, what is the normal sheaf
[TABLE]
Theorem 1.1 tries to answer these two questions in the case of Calabi-Yau 3-folds.
The idea of the work starts from and stays in a down-to-earth setting, which employees linear algebra only. The method first converts the invariant expression of Theorem 1.1 to a variant expression as the content of Theorem 1.1 stays the same. Then it explores the unique linear algebra in the variant setting to reach an algebraic result. At last it converts the algebraic result back to the invariants. In technique the first conversion
[TABLE]
uses classical geometry. The second conversion
[TABLE]
uses Clemens’ deformation idea [1].
Let’s start with this alternative setting. Let
[TABLE]
The open set of represents (but is not equal to)
[TABLE]
Let
[TABLE]
be sections in and
[TABLE]
Let
[TABLE]
In this paper, the Cartesian product
[TABLE]
is also called a complete intersection of type . So
[TABLE]
is a complete intersection Calabi-Yau 3-fold in the usual sense for generic
[TABLE]
Choose distinct points . Let
[TABLE]
In the rest of the paper, we’ll use following conventions in affine coordinates.
(a) or denotes a complex number which is a point in an affine open
set ,
(b) denotes the image
[TABLE]
(c) is a homogeneous polynomial of degree in
variables.
We should note that in these affine coordinates, the incidence relation can be expressed as the composition for all
[TABLE]
Let be a system of affine coordinates for , which determines an isomorphism
[TABLE]
We define a system of polynomials in the variable
[TABLE]
Then the subsets of polynomials
[TABLE]
give a rise to a holomorphic map
[TABLE]
where . We”ll denote
[TABLE]
which will be called the incidence scheme of rational curves on the complete intersection of . Then the differential map is represented by the Jacobian matrix of size
[TABLE]
denoted by , which depends on . However once the points are fixed, the matrix is well-defined and varied algebraically on the entire affine space , where is affine. We call it the Jacobian matrix of the incidence scheme .
The scheme , which could be reducible with multiple dimensions,111For instance, it contains the components of multiple covering maps of . is the alternative to various moduli spaces of rational curves and maps. In this paper we show a methodology in the calculation of the Jacobian matrix at a point of corresponding to an irreducible rational curve on generic complete intersections. The components containing such points always have the smallest dimension as expected. They correspond to the components that have actual fundamental classes instead of virtual fundamental classes in Mirror symmetry. The methodology is rooted in a specific pattern of the Jacobian matrix modeled on the Vandermonde matrices.
In general we are interested in the following questions,
(1) what is the dimension of ?,
(2) is it reduced?
(3) is it irreducible, and if not, what is the structure of each component?
Answers to these questions would solve some technical problems in Mirror symmetry, that have been left out of the physics’ formulation. Thus it is an alternative to patch some holes in the general mathematical theory. The complete answers to these questions are out of scope of this paper. Here we are going to use linear algebra to explore the dimension of for some complete intersections.
2.2 Existence
In the first part, we prove the existence of irreducible rational curves on a generic complete intersection Calabi-Yau 3-fold . We avoid a direct construction.
(I) First we’ll use the proven existence of irreducible rational curves of arbitrary degrees on a single generic hypersurface of lower degrees (which are Fano). After its extension to a special complete intersection in by joining more Fano hypersurfaces, we use linear algebra to glue the block matrix for each Fano hypersurface to obtain the non-degeneracy of Jacobian matrix . This will show the existence of an irreducible rational curve of each degree on a special complete intersection Calabi-Yau 3-fold with the non-degenerate Jacobian at the point . 222 For curves of a fixed and small degree, computer software has been used to find its Jacobian. See p. 295, [2].
(II) Applying Clemens’ deformation idea, this smooth is deformed to generic as a different irreducible rational curve . The requirement for such a deformation is the non-degeneracy of Jacobian matrix .
2.3 Rigidity
The second step is to show the normal bundle of all such irreducible curves are split as in (0.1). This rigidity is determined by the dimension of the incidence scheme . First we have the general study of a uniruled projective variety to deduce that a free rational curve on it has the unobstructed deformation on its generic hypersurfaces. This no-trivial result follows from [7] or [8]. Then we notice that the filtration of complete intersections
[TABLE]
corresponds to another filtration of subvarieties
[TABLE]
where the subscript means the analytic neighborhood around the point with . Then we notice all for are Fano, therefore uniruled. Applying above deformation result for rational curve on uniruled varieties, we obtain a recursive formula
[TABLE]
( or equivalently ). Applying the Calabi-Yau condition (2.5) we obtain that
[TABLE]
(or ).
Then the dimension forces to be an immersion and furthermore to be rigid.
We organize the rest of the paper as follows. In section 3, we prove the existence, and in section 4 we prove the rigidity. Appendix covers a particular technique in linear algebra for the existence.
3 Existence of rational curves
Starting from this section we give technical proofs. This section focuses on the existence.
We resume all notations introduced in section 1. Recall are sections in such that and . Let
[TABLE]
In particular, if
[TABLE]
is generic, is a smooth complete intersection Calabi-Yau 3-fold.
In this subsection, we prove part (1) of Theorem 1.1. It asserts the existence of an irreducible rational curve of each degree on the generic .
Theorem 3.1**.**
There exist irreducible rational curves of each degree on the following generic complete intersection Calabi-Yau 3-folds.
[TABLE]
[TABLE]
[TABLE]
[TABLE]
The case (3.5) has been proved by H. Clemens [1] and S. Katz ([3]). Let’s consider other 3 cases. Suppose that in all 3 cases, there exist a special complete intersection, denoted by sections and a smooth rational curve of degree with for all such that the corresponding Jacobian matrix at has full rank. Then we divide the coordinates of to two parts, and . The partial derivatives with respect to form the maximal minor block in the Jacobian matrix . Thus is a maximal non-degenerate block. The size of is therefore
[TABLE]
Let the remaining coordinates of be , and corresponding block matrix in is . Hence are the affine coordinates of and . Since evaluated at the special complete intersection and smooth is non-degenerate, by the implicit function theorem in complex analysis, there exists analytic functions near , ( are locally free sections) such that
[TABLE]
where . In geometric term, this means that there exists a smooth rational curve of degree on each same type of complete intersection (free choice of ) near the special one . Since these deformed complete intersections are all generic in Zariski-topology, we complete the proof of the existence in these cases. So in the following subsections we are going to find such a special complete intersection in each case.
3.1 Complete intersection of type in
Proposition 3.2**.**
There is a smooth rational curve of each degree on a special complete intersection of type in such that at is non-degenerate, where is the normalization of .
Proof.
Let be homogeneous coordinates of . Let the subspace defined by . First we consider a generic quadric in . By [4], Hilbert scheme at a smooth rational curve
[TABLE]
of degree is non-empty and smooth with the expected dimension. Then is smooth at with expected dimension. Let be the corresponding Jacobian matrix of in (as defined in the introduction). By Lemma A.2, it has full rank. In the following we use the “gluing” technique for block matrices to extend the Jacobian matrix to . Assume . It is extended to a smooth rational curve in as follows
[TABLE]
which is isomorphic to . We define the special complete intersection of type as follows. Let
[TABLE]
be four quadrics in . Then we have for all .
Let be the affine coordinates for and copies in . Let be the hyperplane of , whose elements vanish at [math]. Let be the affine coordinates for in the components of . The we’ll show these variables, that are not all variables of , are . This amounts to show that the Jacobian matrix has full rank. Thus we first write down the blocks of Jacobian matrix. We let
[TABLE]
Then one of maximal minor blocks of the Jacobian matrix ( in ) at is formed by the block matrices
[TABLE]
We can verify that in (3.10) satisfy all conditions in Lemma A.2. Furthermore all conditions in Lemma A.6 for are also satisfied. By the Lemmas is non-degenerate. We complete the proof. ∎
3.2 Complete intersection of type in
Proposition 3.3**.**
There is a smooth rational curve of each degree on a special complete intersection of type in such that at is non-degenerate, where is the normalization of .
Proof.
Let be homogeneous coordinates of . Let be the subspace defined by . By [4], there exists a generic
[TABLE]
such that it contains a smooth rational curve
[TABLE]
We define a new smooth rational curve of degree to be
[TABLE]
We define hypersurfaces as
[TABLE]
where is a generic quadric with respect to . Let be the affine coordinates for copies in . Let
[TABLE]
We write down the block matrices for the Jacobian matrix. Let
[TABLE]
Next we write down the Jacobian matrix directly as
[TABLE]
Then we verify satisfy all conditions in Lemmas A.2, A.3. Then we can apply Lemma A.5. We obtain that Jacobian matrix of the incidence scheme has full rank. We complete the proof of this case. ∎
3.3 Complete intersections of and types in
Proposition 3.4**.**
There is a smooth rational curve of each degree on a special complete intersection of type in such that at is non-degenerate, where is the normalization of .
Proof.
Let be homogeneous coordinates of . Let be the subspace defined by . By [4], there exists a generic such that it contains an irreducible rational curve
[TABLE]
of degree . Next we define a smooth rational curve of degree in as
[TABLE]
We also define two cubic hypersurfaces
[TABLE]
where is a generic quadric with respect to . Let be the affine coordinates for copies in (all variables of ). Let
[TABLE]
Let
[TABLE]
for , . Then we directly write down the Jacobian of the incidence scheme at .
[TABLE]
Similarly conditions of Lemma A.3 are met. By Lemma A.4, it has full rank. We complete the proof in this case. ∎
Proposition 3.5**.**
There is a smooth rational curve of each degree on a special complete intersection of type in such that at is non-degenerate, where is the normalization of .
Proof.
As in the case 1, are homogeneous coordinates of . Let be the subspace covered by the coordinates
[TABLE]
Let be the subspace covered by coordinates
[TABLE]
By the Mori’s result [6]. there is a smooth rational curve of degree on a generic quartic . Let
[TABLE]
be generic.
Then the Jacobian matrix
[TABLE]
has full rank. 333 The number of points is unusual. This exceptional case holds only for quartic surface in . Let’s extend it to . Define a new quartic by setting in and new rational curve by setting
[TABLE]
where is the Mori’s curve. Next we add one generic point
[TABLE]
and extend the coordinates of
[TABLE]
to
[TABLE]
These add the a new row –the differential of to the Jacobian , Hence we obtain a new Jacobian matrix in at ,
[TABLE]
where
[TABLE]
is the differential 1-form . Hence in has full rank. To summarize it, we found a special quartic in containing a smooth rational curve of the given degree and its Jacobian matrix is of full rank. 444 We added one more feature to Mori’s existence. That is the non-degeneracy of the Jacobian matrix.
Next we extend it one more time to . Let be the extension of original to . Let and the
[TABLE]
be the smooth curve in . Thus lies on the complete intersection of . We add new set of generic points in ,
[TABLE]
Then we can write down the Jacobian matrix in this case. It is equal to
[TABLE]
where
[TABLE]
is the Jacobian matrix at of many functions (in ),
[TABLE]
and is the block with respect to coordinates in last two components of
[TABLE]
By Lemma A.2, has full rank. Hence the Jacobian matrix also has full rank. We complete the proof.∎
3.4 Quintic in
This has been proved in [3]. We’ll repeat the same construction in [1], but continue with our method. By the Mori’s result [6]. there is a smooth rational curve of degree on a generic quartic in . Now we extend it to by setting
[TABLE]
and the new quintic
[TABLE]
where is a generic linear polynomial and is a generic quartic. By [9], the corresponding Jacobian matrix has full rank.
Corollary 3.6**.**
Let be a generic complete intersection Calabi-Yau 3-fold. Then it contains an irreducible rational curve of each degree .
Proof.
is projectively embedded as one of the complete intersections in Theorem 3.1. Therefore the corollary follows.
∎
Remark In above arguments for the existence, all rational curves are smooth. But non-smooth and irreducible rational curves on a generic complete intersection Calabi-Yau 3-fold do exist.
4 Rigidity of rational curves
4.1 Rational curves with unobstructed deformation
In this subsection we let be an arbitrary smooth projective variety over . We’ll prove a general assertion in Theorem 4.3 about uniruled variety. For a parametrized rational curves on , the notion of having unobstructed deformation is weaker than being a free morphism. Precisely let .
Definition 4.1**.**
If the normal sheaf denoted simply by has the vanishing first cohomology, i.e.
[TABLE]
then we say has unobstructed deformation on .
Since is equivalent to , then Definition 4.1 is equivalent to the following splitting
[TABLE]
On the other hand, we define
Definition 4.2**.**
If is generated by global sections, we say is a free morphism, and is uniruled.
This is a special case of the more general definition in [5].
It is clear that being free is equivalent to the splitting
[TABLE]
So being free is stronger than having unobstructed deformation.
Theorem 4.3**.**
Let be a very ample line bundle on , and . Let
[TABLE]
where is generic with an . Let
[TABLE]
be a birational morphism onto an irreducible rational curve . If is free on , then has unobstructed deformation on , i.e.
[TABLE]
Proof.
of Theorem 4.3: By the assumption of the theorem, we have a polarization of such that
[TABLE]
is a smooth subvariety of dimension , and . Let
[TABLE]
be generic. Let .
We denote
[TABLE]
By the genericity of , scheme-theoretically
[TABLE]
Let be generic in .
We have a non-commutative diagram of exact sequences
[TABLE]
Let’s define and analyze the diagram. Including all zero spaces, there are 5 rows and 6 columns. Second and third rows are parts of the long exact sequences from the short exact sequences of sheaves over ,
[TABLE]
[TABLE]
The third column is the part of the long sequence of the short exact sequence of sheaves over ,
[TABLE]
The isomorphism in the fourth column is from the adjuntion formula.
The homomorphism is well-defined only if has unobstructed deformation on . Let’s see this in the following. Because is free, therefore has unobstructed deformation on . So . Therefore the third column exact sequence in (4.9) splits, i.e.
[TABLE]
We define to be the restriction of to the subspace, i.e.
[TABLE]
(But may not be zero.). In the sense of this splitting, the map also splits as
[TABLE]
Next we go further to use a construction to prove that is the zero map due to the global generation. This is just a specification of
[TABLE]
inside of
[TABLE]
By (4.13), there are a Zariski open set and a trivial subbundle
[TABLE]
over such that
[TABLE]
and
[TABLE]
Let
[TABLE]
So is a subspace of . It generates a sheaf over (in general could be zero, so did .). Because
[TABLE]
are generated by global section, the sheaf over satisfies
[TABLE]
and (4.17) extends to
[TABLE]
We then define to be the restriction of to (this is the same as the definition before. But this time, the subspace is uniquely identified.). By the condition (4.20), is the zero map.
Thus
[TABLE]
Now we consider the third row, an exact sequence in (4.9). Since is a generic hypersurface of , we can apply Theorem 1.1, [8] (or [7]) to obtain that . The exactness of the sequence implies that is surjective. So is . Next we shift the focus to the second row in (4.9). We apply again to obtain that . Using the exact sequence
[TABLE]
we obtain that
[TABLE]
∎
4.2 Free morphism
In this subsection, we continue the existence result — the existence of irreducible rational curves of each degree on a generic complete intersection Calabi-Yau 3-folds. From now on we change to be a complete intersection as follows,
[TABLE]
for the fixed set of generic . So , which is equal to , is a smooth 4-dimensional complete intersection and Fano. As before for the fixed set of ,
[TABLE]
is a Calabi-Yau 3-fold contained in the Fano 4-fold .
We would like to show a generic rational curve on is free on , in another word, such rational curves cover . Thus we assume
Assumption 4.4**.**
* of exists.*
We’ll work in a neighborhood of each component of the incidence schemes around .
Let’s resume all the notations in section 2. We let be fixed (where ). Then the incidence scheme is also fixed. Let be a component of containing . Let be an irreducible component of the scheme
[TABLE]
containing . Let
[TABLE]
be the projection. Let . So
[TABLE]
Let be a generic point. Let be the closure of the open scheme,
[TABLE]
The equation means that is regular at [math].
We would like to show that
Proposition 4.5**.**
The open scheme is a rational map from to and dominates .
Proof.
Let be an integer satisfying
[TABLE]
Let be a component of containing . Define
[TABLE]
to be the closure of
[TABLE]
for each .
First we consider the case . Suppose is onto , where . Let have coordinates . Then the dimension of the fibre of over the generic point satisfies
[TABLE]
Hence after imposing equations,
[TABLE]
we obtain that
[TABLE]
for all . Let
[TABLE]
be the projection. Then the correspondence sends onto (not a map). Next we show is a rational map.
Applying Assumption 4.4, there is a
[TABLE]
Then generic must be in because is an open set. Hence the correspondence is a rational map. By the induction on the index for
[TABLE]
we showed that the irreducible component
[TABLE]
dominates , through the rational evaluation map
[TABLE]
To prove Proposition 4.5 it suffices to extend above proof to , but the situation is slightly different.
Let
[TABLE]
be generic. By the result of above argument, dominates . So for a generic ,
[TABLE]
For the fixed , we define to be closure of
[TABLE]
where is an open set of space of hypersurface satisfying that the Jacobian matrix of the functions in
[TABLE]
at has full rank. By the inequality (4.35), we count the dimension to obtain the is non-empty. Hence
[TABLE]
is non-empty for generic . This shows that the correspondence is onto . Now let’s show it is a rational map. Notice a generic is a generic element in
[TABLE]
Hence it suffices to show there is one regular point for . By Assumption 4.4, there is a such that is in , and for some (not all ). Therefore is regular at ONE point. With the same reason as above, is a rational map. Thus dominates . This completes the proof. ∎
Corollary 4.6**.**
Let be generic. Then is a free morphism on .
Proof.
By Proposition 4.5, there is a Zariski open set of , such that the projection
[TABLE]
is smooth. Hence its differential is onto. Let . Then the pull-back of the tangent bundle
[TABLE]
does not have negative summand, i.e. . Hence a gives a free morphism . ∎
4.3 Unobstructed deformation
Notice that there is a degree rational curve for a generic . By corollary 4.6, it is a free morphism in , we apply Theorem 4.3. Then the second row of (4.9) implies a recursive formula
[TABLE]
Since
[TABLE]
[TABLE]
Now we consider it from a different point of view. Because is a birational map to its image, there are finitely many points where the differential map
[TABLE]
is a zero map. Assume its vanishing order at is . Let
[TABLE]
Let such that
[TABLE]
The sheaf morphism is injective and induces a composed morphism of sheaves
[TABLE]
It is easy to see that the induced bundle morphism is injective. Let
[TABLE]
Then
[TABLE]
On the other hand, three dimensional automorphism group of gives a rise to a 3-dimensional subspace of
[TABLE]
By (4.41), . Over each point , spans a one dimensional subspace. Hence
[TABLE]
where are some positive integers. This implies that
[TABLE]
Then
[TABLE]
Since , by the formula (4.45), . By the definition of , . Hence is an immersion.
Now is a vector bundle. By the the Calabi-Yau condition,
[TABLE]
where is non-positive. By Theorem 4.3, . Hence . Since is generic, the same proof is valid for all . Together with the existence in section 3, we complete the proof of Theorem 1.1.
Appendix A Vandermonde type of matrices
In the appendix, we give proofs of lemmas dealing with Vandermonde type of matrices that are products of diagonal matrices and Vandermonde matrices. The main purpose is to use rather elementary technique to glue matrices of smaller sizes, the Vandermonde type of matrices, to obtain a matrix of larger size (which is a Jacobian matrix of a complete intersection). We assume that the linear algebra has the ground field .
First we introduce and study Vandermonde type of matrices, that are smaller matrices as entries in the large block matrices. Let
[TABLE]
be a polynomial, where is the variable, and Let
[TABLE]
In the following in order to define matrices, we use the affine expression in subsection 2.1. Let be a diagonal matrix
[TABLE]
Definition A.1**.**
[TABLE]
and
[TABLE]
We call them Vandermonde type of matrices of orders [math] and respectively.
Remark The Jacobian matrices of incidence schemes (we are interested in ) are all made of Vandermonde type of matrices.
Lemma A.2**.**
If are two relatively prime polynomials with distinct zeros, and degrees , then for generic
[TABLE]
the square matrix
[TABLE]
is non-degenerate.
Proof.
It suffices to prove it for a special . Since has degree and it is relatively prime to , we can choose to be the zeros of and is non-zero at all points of . Then
[TABLE]
Then its determinant is
[TABLE]
Since are types of Vandermonde matrices, the distinct and nonvanishing imply the non-degeneracy. Therefore is non-degenerate.
∎
Lemma A.3**.**
Let and be two vectors not on the same line through the origin. They are also pair wisely relatively prime, and all zeros are distinct. Let
[TABLE]
be generic. Then the square matrix ( ),
[TABLE]
is non-degenerate.
Proof.
By the linear algebra in Proposition 2.10, [7], we obtain that the matrix ,
[TABLE]
has full rank for any with distinct points in .
Then we rewrite as a block matrix,
[TABLE]
where
[TABLE]
and
[TABLE]
Then for generic , is column-equivalent to
[TABLE]
By choosing the points of to be the zeros of and remains invertible, we obtain that
[TABLE]
Thus we have a specialization
[TABLE]
which is non-degenerate. Thus for generic ,
[TABLE]
is non-degenerate, so is .
∎
We’ll use two different types matrices in Lemmas A. 2, A.3 as building blocks to obtain matrices of larger sizes. We’ll show that non-degeneracy of large matrices is the result of that of the smaller, block matrices.
Lemma A.4**.**
Let be matrices. Assume
(1)
[TABLE]
(2)
[TABLE]
and
[TABLE]
have full rank,
(3) the columns of
[TABLE]
span the columns of
[TABLE]
Then
[TABLE]
is non-degenerate.
Proof.
In the following, we apply the column and row operations to the matrix. They will not change its rank. Applying the row operations on the matrix we obtain it is row-equivalent to
[TABLE]
By the condition (3), it is column-equivalent to
[TABLE]
By the condition (2), it is further column equivalent to
[TABLE]
By the condition (2), we obtain it has full rank.
∎
Lemma A.5**.**
Consider the block matrix
[TABLE]
where the entries in first row are matrices and all the rest are matrices. We assume
(1)
[TABLE]
(2)
[TABLE]
,
[TABLE]
and
[TABLE]
have full rank,
(3) the columns of
[TABLE]
span the columns of
[TABLE]
then the matrix has full rank.
Proof.
The proof uses row and column operations which are the same as those in Lemma A.4. So in the following we just list the equivalent matrices in the reduction.
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
The last matrix has full rank.
∎
Lemma A.6**.**
Consider the block matrix
[TABLE]
where are non degenerate square matrices of the same size. If
[TABLE]
is non-degenerate, so is .
Proof.
As before we perform row and column operations on to obtain
[TABLE]
[TABLE]
[TABLE]
[TABLE]
[TABLE]
By the assumption,
[TABLE]
is non-degenerate, so is .
∎
The reference list from the paper itself. Each links out to its DOI / PubMed record.
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- 3[3] S. Katz , On the finiteness of rational curves on quintic threefolds , Comp. Math. 60, pp 151-162 (1996)
- 4[4] J. Harris, M. Roth and J. Starr , Rational curves on Hypersurfaces of low degree , J. Reine Angew. Math., 571 (2004), pp. 45-96.
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