Vector bundles in Algebraic Geometry Enrique Arrondo. 1. The notion of vector bundle


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1 Vector bundles in Algebraic Geometry Enrique Arrondo Notes(* prepared for the First Summer School on Complex Geometry (Villarrica, Chile 79 December The notion of vector bundle In affine geometry, affine varieties are defined by zeros of polynomials in the sense that a polynomial determines a welldefined function that we can evaluate and decide whether it is zero or not In projective geometry, however, a homogeneous polynomial of positive degree never defines a function, and it only makes sense to say whether the polynomial vanishes or not at some point Moreover, the only regular functions on irreducible projective sets are the constants We want to introduce the language of vector bundles as a method to describe equations of projective sets, generalizing the notion of regular function We start with a couple of examples that illustrate what we have in mind Example 11 Let F K[,, x n ] be a homogeneous polynomial of degree d If we consider f i, the dehomogenization of F with respect of the variable X i, now f i becomes a function on the affine open set U i = { 0} Let us try to analyze why it is not possible to glue all these functions F i to get a function on P n If we write the coordinates in U i as,, 1, +1,, x n xi, then f i = F Therefore the reason why we cannot glue together x d i the functions on U i and U j is that f j is obtained from f i when multiplying by xd i This x d j means that any hypersurface V (F of P n can be described locally by functions f i, but they can be glued together only after a multiplication If now X P n has arbitrary codimension r, one would have a priori the right to expect that X could be defined by exactly r equations This is not true in general, and in fact this happens rarely (if it holds, X is said to be a complete intersection However it is true that good sets X (for instance smooth projective sets enjoy that property locally around each point The question now is: is it possible somehow to glue together those local functions in a similar way as in the previous example? Although the answer is negative in general, we present a positive case in order to give a clearer idea of what we mean Example 12 Let X P 2 be the subset consisting of the points (1 : 0 : 0, (0 : 1 : 0 and (0 : 0 : 1 At each open set U i = { 0}, X U i consists of one point, so its affine (* Part of these notes are extracted from the last (and incomplete sections of [A2], where the reader can find some preliminaries 1
2 ideal is generated by exactly two elements More precisely, the ideals of X U 0, X U 1 and X U 2 are respectively generated by, ;, ; and, For any λ K \ {0} (the reader will understand soon why we do not take just λ = 1, it is possible to relate the two first sets of generators by the expression = ( 2 0 (1 λ 0 λ and the matrin the expression has entries regular in U 0 U 1 and is invertible in that open set Similarly, there is a relation = λ (1 λ 1 0 ( 2 Putting together the last two relations we get = λ 0 (1 λ λ (1 λ( 2 (λλ λ + 1 x2 2 Thus if we want to avoid divisions by zero (and get a matrix similar to the previous ones, we need λλ λ + 1 = 0, ie λ = 1 1 λ Hence, if λ 0, 1, we can glue together the equations The precise definition is the following (in which the reader can replace the category of algebraic varieties with any category of topological spaces: Definition A vector bundle of rank r (or line bundle if r = 1 over an algebraic variety X is an algebraic variety F equipped with a morphism π : F X such that there exists a covering X = i I U i by (Zariski open subsets such that: (i For each i I there is an isomorphism ψ i : π 1 (U i U i K r satisfying that the composition π ψ 1 i : U i K r U i is the first projection (ii For each i, j I there is an (r rmatrix A ij (called transition matrix, or transition function if r = 1 whose entries are regular functions in U i U j satisfying that the composition ϕ ij = ψ j ψ 1 i U i U j : (U i U j K r π 1 (U i U j (U i U j K r takes the form ϕ ij (x, v = (x, A ij (xv 2
3 Remark 13 It is clear that any subpartition of the covering still satisfies conditions (i and (ii, so we will always work, when necessary, with sufficiently fine partitions If only one open set is needed, ie F = X K r and π is the first projection, we say that F is a trivial bundle of rank r Condition (i is saying that, for any x X the set π 1 (x (called the fiber of the vector bundle at the point x, is bijective to K r, and that locally the fibers are glued to produce a trivial vector bundle From this, it is clear that, in condition (ii, the first coordinate of ϕ ij (x, v must be x Thus condition (ii is just saying that the fibers of F glue together in different trivial representations in a linear way In other words, the fibers of the vector bundle have to be regarded as vector spaces It is frequent in the literature to identify a vector bundle with the set of its transition matrices for a suitable partition of the variety This is possible by the following result: Proposition 14 Let X be an algebraic variety with an open covering X = i I U i Let F be a vector bundle over X with transition matrices A ij corresponding to the covering Then: (i A ii is the identity matrix I r (ii A jk A ij = A ik Reciprocally, given a set of matrices {A ij } i,j I satisfying (i and (ii there is a vector bundle over X whose transition matrices are the given matrices Proof: Properties (i and (ii are easy to proof The fact that they allow to reconstruct the vector bundle is a standard and tedious technique of constructing algebraic varieties by glueing several pieces Example 15 Proposition 14 allows to actually construct vector bundles from Examples 11 and 12 In particular, Example 11 yields a line bundle L d over P n with transition functions xd i, in which we can now even allow d to be negative (if the reader is familiar x d j with the identification of vector bundles and locally free sheaves, L d is the line bundle corresponding to the invertible sheaf O P n(d Similarly, Example 12 provides, for any λ 0, 1, a vector bundle F λ over P 2 Exercise 16 Construct a ranktwo vector bundle over the smooth quadric X P 4 of equation x 3 + x by glueing the local equations of the line L X of equations = = = 0 This vector bundle is called the spinor bundle over X, and in fact it is independent on the choice of the line L contained in X Exercise 17 Given a regular map ϕ : Y X and a rankr vector bundle F Y over X, show that the second projection of the fibered product F X Y defines a rankr vector bundle over Y (called pullback of F via ϕ and denoted by ϕ F 3
4 Remark 18 It is possible to define in the natural way the notion of dual vector bundle, direct sum or tensor product of vector bundles, wedge or symmetric power of a vector bundle We leave the reader the task of checking how the transition matrices can be constructed from the transition matrices of the original vector bundles We concentrate in just few examples: Exercise 19 Show that we have the following isomorphisms in P n (following the notation of Example 15: (i L 0 is the trivial line bundle (ii L d L e = Le+d (iii L 1 d = L d (iv Symm d L 1 = Ld (v If n = 2, 2 F λ = L3 Exercise 110 Show that, if F is a vector bundle determined by transition matrices A ij, then the dual vector bundle F is a vector bundle determined by transition matrices (A t ij 1 (where A t ij denotes the transposed matrix of A ij In particular, the dual vector bundle Eλ of the vector bundle E λ obtained from Example 12 has transition matrices: (1 λ x 1 λ λ A 10 = 0 1 λ 1 λ 1 1 x 0 0 λ, A 21 = λ 2 1 2, A 20 = λ 2 Example 111 Let us define the cotangent bundle of P n K, for any field K, in terms of its transition matrices with respect to the open sets U i = { 0} We first define a differential form on U i as a formal expression of the form ω = f 0 d( + + f i 1 d( 1 + f i+1 d( f n d( x n where f 0,, f i 1, f i+1,, f n are regular functions depending on the affine coordinates,, 1, +1,, x n xi To restrict this to any U i U j we write, for any k i: so that we get ω = x j d( x k = d ( ( x k x j ( x j ( f 0 d( x j + +( f 0 1 = x j d( x k x kx j x j d( i x j f i f i+1 x n f n d( + +f n d( x n x j x j
5 In other words, the transition matrix from U i to U j is given by A ij = x j x j x n xi This provides a vector bundle of rank n on P n that we denote by Ω P n (and it is called the cotangent bundle on P n For example, in the case n = 2, we have A 10 = x , A 21 = 0 2 x2 1 2, A 20 = In fact the notion of vector bundle determined by transition matrices does not make any sense unless we have the notion of isomorphism ( determined by always means up to isomorphism The precise definition is: Definition A morphism of vector bundles is a regular map ϕ : F F such that π = π ϕ (ie it sends fibers F x to fibers F x and each induced map ϕ x : F x F s linear If any ϕ s an isomorphism we will say that ϕ is an isomorphism Exercise 112 Express the notion of morphism of vector bundles in terms of transition matrices and show that the matrices A 0 = ( 1 λ 0, A = ( ( λ , A 0 λ 2 = 0 λ define an isomorphism between Ω P 2 and the vector bundle Fλ defined in Example 110 (this shows that all the bundles F λ are isomorphic to T P 2, in particular isomorphic to each other Exercise 113 Let us consider U P n K n+1 to be the subset of pairs (p, v such that v is a vector in the vector line of K n+1 defining the point p P n Show that the first projection U P n endows U with the structure of a line bundle over P n which is isomorphic to L 1 5
6 2 Vector bundles and Grassmannians We will relate here the theory of vector bundles with the theory of Grassmannians We start by recalling briefly the notion of Grassmannian and its main properties (more details can be found in [A1] or [A2] Definition The Grassmannian G(k, n is the set of all kdimensional linear subspaces of P n We will sometimes write G(k, P(V if we want to specify the concrete projective space to which the linear subspaces belong Exercise 21 Prove the following properties of the Grassmannian: (i If P n = P(V is the projective space of lines in the (n + 1dimensional vector space V, show that the map G(k, n P( k+1 V (called Plücker embedding sending the subspace of P(V spanned by independent points [v 0 ],, [v k ] to [v 0 v k ] is welldefined (ie it does not depend on the choice of the k + 1 points and injective (ii If we choose a basis for V and take the corresponding natural basis for k+1 V, let Λ be the subspace spanned by the points of coordinates (a 00 : a 0n,, (a k0 : a kn Then show that the image of Λ under the Plücker embedding has coordinates the maximal minors of the Plücker matrix a 00 a 0n a k0 a kn (iii If, for any choice 0 i 0 < < i k n, we write a 0i0 a 0ik p i0 i k = a ki0 a kik the p i0 i k are called the Plücker coordinates of Λ Show that the image of the Plücker embedding of G(1, 3 is the smooth quadric in P 5 of equation p 01 p 23 p 02 p 13 +p 13 p 12 = 0 [It can be shown in general that the Plücker embedding allows to regard G(k, n as a projective set whose ideal is generated by quadratic equations in the Plücker coordinates] (iv Show that, for any choice 0 i 0 < < i k n, the set U i0 i k := {p i0 i k 0} is the set of all spaces Λ not meeting the linear space 0 = = k = 0 Moreover, the image of U i0 i k := {p i0 i k 0} under the Plücker embedding is isomorphic to an affine space of dimension (k + 1(n k 6
7 Example 22 There is a geometric way of verifying that G(1, 3 has degree two in its Plücker ambient space P 5 As Exercise 21(iv shows, a hyperplane of P 5 intersects G(1, 3 in the set of lines meeting a given line (be careful! not all hyperplanes of P 5 have the same nice interpretation when intersected with G(1, 3 The degree of G(1, 3 will be the number of elements of G(1, 3 obtained when intersecting with four hyperplanes of P 5 If we take these four hyperplanes as before, then the degree of G(1, 3 should be the number of lines in P 3 meeting four given lines L 1, L 2, L 3, L 4 Assume we take L 1 and L 2 meeting in one point p (hence contained in the same plane Π Then the set of lines meeting L 1 and L 2 will consist of the set of lines passing through p or contained in Π If we also assume L 3 and L 4 to meet in a point p and to be contained in a plane Π, then there are only two lines meeting L 1, L 2, L 3, L 4 : the line joining p and p and the line which is the intersection of Π an Π Of course, in principle this proof is not conclusive, since in order to compute the degree one needs to choose general hyperplanes, while we took very special ones The fact that all this actually works is part of the socalled Schubert calculus (more details can be found in [A1] or better in [KL] Exercise 23 Use Schubert calculus as in the previous example to show that G(1, 4 has degree 5 in P 9 Exercise 24 In a similar (and more general way to Exercise 113, consider the subset U G(k, n K n+1 consisting of pairs (Λ, v such that v is a vector in the (k + 1 dimensional linear subspace of K n+1 defining Λ Prove that the second projection over G(k, n defines on U a structure of vector bundle of rank k + 1, called the tautological subbundle of the Grassmannian Definition A section of a vector bundle π : F X is a regular map s : X F such that π s = id X (ie the image of any point s a vector in the fiber F x We will denote by H 0 (X, F the set of sections of F The zero locus of a section s is the set of points x for which s(x is the zero vector in F x Exercise 25 Show that any section of the trivial bundle is given by s : X X K with s(x = (x, f(x, where f is a regular function on X Conclude that the zero locus of any section of any vector bundle is a closed subset with the Zariski topology Exercise 26 Express the definition of section in terms of the transition matrices of the vector bundle and conclude that the vector bundle F λ constructed in Example 12 has a section whose zero locus is {(1 : 0 : 0, (0 : 1 : 0, (0 : 0 : 1} Similarly, the spinor bundle of Exercise 16 has a section whose zero locus is the line L Exercise 27 Show that H 0 (P n, L d is naturally isomorphic to the vector space of homogeneous polynomials of degree d in K[,, x n ] [More generally, it is also true that 7
8 H 0 (G(k, n, Symm d U is canonically isomorphic to the space of homogeneous polynomials of degree d in K[,, x n ], but I do not know of any simple proof; any suggestion will be welcome] Definition Given a vector bundle F on an algebraic set X (we will assume from now on X to be an irreducible projective variety and a linear subspace V H 0 (X, F, we define the evaluation map as the map ev V : X V F defined by ev V (p, s = s(p When V = H 0 (X, F, we usually write ev V = ev F If ev V is surjective we say that F is generated by the sections of V If ev F is surjective, we will say that F is generated by its global sections (or globally generated, or spanned Proposition 28 Let F be a vector bundle of rank r on X that is generated by the sections of V H 0 (F Then there is a natural regular map ϕ V : X G(r 1, P(V such that ϕ V (U = F and H 0 (G(r 1, P(V, U = V Proof: We define the map as follows Since, by hypothesis, the evaluation map ev V : X V F is surjective, its dual F X V is injective This means that, for any x X, the fiber F s an rdimensional subspace of V We thus define ϕ V (x to be the corresponding (r 1dimensional subspace of P(V To see that ϕ V is a regular map we use coordinates We first choose a basis s 1,, s n of V and take coordinates λ 1,, λ n with respect to it Since the regularity of the map can be checked locally, we restrict to open sets U X on which F U is isomorphic to U K r On such open set U, any section s i is represented by regular functions f i1,, f ir on U In particular, the evaluation map ev V restricted to U can be described by λ 1 (x, λ n f 11 (x f n1 (x f 1r (x f nr (x Hence the dual map is given (with respect to the dual bases by the transpose matrix f 11 (x f 1r (x f n1 (x f nr (x In particular, ϕ V (x is the linear subspace of V generated by the points whose coordinates are the columns of the above matrix Therefore the Plücker coordinates of the subspace are the maximal minors of the matrix, which are all regular functions on U This shows that ϕ V is a regular map Finally, since U is the subbundle of G(r 1, P(V V consisting of the pairs (Λ, v such that v is a vector in the subspace of V determined by Λ, it follows that ϕ V (U is the 8 λ 1 λ n
9 subbundle of X V consisting of the pairs (x, v such that v is in the subspace of V determined by ϕ V (x By the definition of ϕ V, this is equivalent to say v F x, so that it immediately follows that ϕ V (U is precisely F Moreover, we know from Exercise 27 that H 0 (G(r 1, P(V, U is the space of linear forms on P(V, ie V Lemma 29 Fix H 1,, H m V linearly independent linear forms and identify them with m sections of the vector bundle U on G(k, n Then the locus on which H 1,, H m have rank at most s is the set of kspaces Λ meeting V (H 1,, H m in dimension at least k s Proof: For a kspace Λ, the rank of H 1,, H m at Λ is the dimension of their span, considered as linear forms on Λ Hence the rank is at most s if and only if the intersection in V of < H 1,, H m > and the (n kdimensional subspace Λ of equations of Λ has dimension at least m s This is equivalent to say that the span of < H 1,, H m > and Λ has dimension at most n k + s, ie Λ V (H 1,, H m has dimension at least k s Proposition 210 For any linear space Λ P n of dimension k and any l k, s the Schubert variety Ω of kspaces of P n meeting Λ in at least dimension l has codimension (l + 1(n + l k k in G(k, n Proof: We consider the incidence diagram I G(l, k G(k, n of pairs (Λ, Λ where Λ is a subspace of Λ contained also in Λ Clearly Ω is the image of I under the second projection Since this projection is birational onto its image, it is enough to check that I has dimension (k + 1(n k (l + 1(n + l k k This is true because G(l, k has dimension (l + 1(k l, while the fibers of the first projection from I are isomorphic to G(k l 1, n l 1, so that they all have dimension (k l(n k Observe that the result of the above proposition is symmetric in k and k, ie the Schubert variety of k spaces of P n meeting a kspace Λ in at least dimension l has also codimension (l + 1(n + l k k in G(k, n The way of remembering the above result is that, in the language of the previous lemma, the locus in which m sections of the (k + 1 rank vector bundle U on G(k, n have rank at most s has codimension (m s(k + 1 s in G(k, n The point now is to prove the same result, at least in a general situation, when we replace G(k, n with any arbitrary variety and U by any vector bundle on it Theorem 211 Let V H 0 (X, F be a set of sections generating X Then, for a general choice of m sections in V, the locus on which they have rank at most s has codimension (r s(m s in X 9
10 Proof: Consider the morphism ϕ V : X G(r 1, dim V 1 By Proposition 28, the locus X s on which m sections in V has rank at most s is the pullback by ϕ V of the locus on which m sections of U has rank at most s Hence, if we take the sections to be independent, by Lemma 29, the locus X s is the pullback of the Schubert variety of (r 1spaces whose intersection with the subspace of dimension k = dim V 1 m (defined by the m sections has dimension at least r 1 s Consider the incidence variety I X G(k, dim V 1 of pairs (x, Λ, such that ϕ V (x and Λ meet in dimension at least r 1 s From the above considerations, the statement to prove is equivalent to the fact that the general fiber of the second projection from I has codimension (r s(m s in X To prove this, it is enough to prove that I has dimension equal to dim G(k, dim V 1 + dim X (r s(m s But this is so because the fiber at any x of the first projection from I is the set of k spaces meeting the (r 1space ϕ V (x in dimension at least r 1 s and, by Proposition 210, this has codimension (r s ( (dim V 1 + (r 1 s (r 1 k, ie dimension dim G(k, dim V 1 (r s(m s 10
11 3 Chern classes and Porteous formula In this section we will need to assume the existence of a good intersection ring on any (smooth ndimensional projective variety X (the standard and sophisticated construction for this can be found in [F] This means that we have a graded ring A(X = n d=0 Ad (X, where each A d (X is made out of classes of linear combination (with integral coefficients of irreducible subvarieties of codimension d Two classes will be equivalent if and only if there intersection product with all subvarieties of complementary dimension coincide Example 31 As a first example, Bézout Theorem can be reinterpreted by saying that A(P n = Z[h]/(h n+1, where h is the class of a hyperplane (and thus h i is the class of a linear subspace of codimension i In fact, the class of an irreducible subvariety of codimension i and degree d is dh i, and its intersection with another irreducible subvariety of codimension i and degree d (hence with class d t i is, if the intersection is proper, a subvariety of codimension i + i and degree dd, so its class is dd h i+i Remark 32 The whole intersection theory is based on the existence of the Chern classes of vector bundles, whose main properties we recall here: (i If L is a line bundle, c 1 (L is the class of any divisor associated to it (in particular c 1 (L 1 L 2 = c 1 (L 1 + c 1 (L 2 and c 1 (L = c 1 (L (ii Any vector bundle of rank r on X has Chern classes c i (F A i (X for i = 1, min{r, n} We will call c(f = 1 + c 1 (F t + + c r (F t r the Chern polynomial of F (we will use the convention c 0 (F = 1 and c i (F = 0 if i > r or i > n (iii For any exact sequence 0 F F F 0 we have c(f = c(f c(f (iv If F has m r independent sections such that the locus on which they have rank at most m 1 is a subvariety Z of codimension r m + 1 (the expected one, according to Theorem 211 then c r m+1 (F is the class represented by Z (to be completely honest, one should consider also a possible coefficient for the class of Z, since it could appear with some multiplicity (v If ϕ : Y X is a morphism and F is a vector bundle on X, then, for any i = 1,, rk(f we have c i (ϕ F = ϕ (c i (F, where we also write ϕ for the natural graded homomorphism A(X A(Y (vi (Splitting principle For any computation on Chern classes one can assume the vector bundle to split (since this is very ambiguous we illustrate it in the next example Example 33 Assume that we have vector bundles F and L of respective ranks one and two and we want to express the Chern classes of F L in terms of the Chern classes of F 11
12 and L If we assume F = L 1 L 2, then properties (ii and (iii easily imply c 1 (F = c 1 (L 1 + c 1 (L 2 c 2 (F = c 1 (L 1 c 1 (L 2 Similarly, since F L = (L 1 L (L 2 L, it follows (using now property (i: c 1 (F L = c 1 (L 1 L + c 1 (L 2 L = c 1 (L 1 + c 1 (L 2 + 2c 1 (L = c 1 (F + 2c 1 (L and c 2 (F L = c 1 (L 1 Lc 1 (L 2 L = (c 1 (L 1 + c 1 (L(c 1 (L 2 + c 1 (L = = c 1 (L 1 c 1 (L 2 + (c 1 (L 1 + c 1 (L 2 c 1 (L + c 1 (L 2 = c 2 (F + c 1 (F c 1 (L + c 1 (L 2 What the splitting principle asserts is that the formulas c 1 (F L = c 1 (F + 2c 1 (L and c 2 (F L = c 2 (F + c 1 (F c 1 (L + c 1 (L 2 are still valid even if F does not split Remark 34 Given any vector bundle F, there is always a line bundle L such that F L is generated by its global sections Hence Theorem 211 and Remark 32(iv allow to compute the Chern classes of F L Using the splitting principle as in Example 33, we can thus compute the Chern classes of F Exercise 35 Given a vector bundle F of rank two, use the splitting principle to write the Chern classes of its third symmetric power Symm 3 F in terms of the Chern classes of F Example 36 As a first application of the theory of Chern classes, we will compute the degree of the Segre variety P 1 P n in P 2n+1 Recall that this Segre can be regarded as the set of matrices of rank one inside the space of matrices of size 2 (n + 1 (all the matrices should be considered up to multiplication by a nonzero constant Hence the equations of the Segre variety are the 2 2 minors of the matrix ( x0 x n x n+1 x n+2 n+1 In other words, the Segre variety is the locus on which the sections (,,, x n, (x n+1, x n+2,, n+1 of L n+1 1 have rank one By Remark 32(iv, its class in A(P 2n+1 is given by c n (L n+1 1 which, using now Remark 32(iii, is the coefficient of t n in the expansion of (1 + c 1 (L 1 t n+1 By Remark 32(i, c 1 (L 1 = ht, where h is the class of the hyperplane section of P 2n+1 (recall Example 31, so that the class of the Segre variety is h n = (n + 1h n Therefore, the Segre variety has degree n + 1 ( n+1 n 12
13 Example 37 Consider on G(k, n the vector bundle U of rank r = k+1 Take m k+1 linearly independent sections of U, thus defining a linear subspace Λ P n of dimension n m Then Lemma 29 implies that the locus on which the m sections of U have rank at most m 1 is the Schubert variety Ω of kspaces meeting the Λ in dimension at least k m + 1 This means that c k m+2 (U is represented by the class of Ω When m = k+1 we get that c 1 (U is represented by the set of kspaces meeting a given subspace of codimension k + 1 (and this is the class of a hyperplane section of the Grassmannian, as Exercise 21(iv shows; if instead m = 1, we get that c k+1 (U is represented by the set of kspaces contained in a given hyperplane Exercise 38 Use the fact that the set of lines contained in a cubic surface in P 3 can be described as the zero locus of a section of Symm 3 U (see Exercise 27 to conclude that one expects a cubic surface to contain 27 lines Example 39 Schubert calculus can be reinterpreted in terms of Chern classes For example, we have seen in Example 22 that the intersection of two hyperplanes in the Plücker space of P 5 yields the set of points passing through a point plus the set of lines contained in a plane In terms of Chern classes, we are saying (using Example 37 that the class of the set of lines passing through one point is c 1 (U 2 c 2 (U In other words, using now Lemma 29, the locus on which three sections of U have rank at most one has class c 1(U c 2 (U c 0 (U c 1 (U This equality generalizes and, in terms of Chern classes, says that the class of the locus on which m independent sections of U have rank at most s is c k+1 s (U c k+2 s (U c k+m 2s (U c k s (U c k+1 s (U c k+m 2s+1 (U c k m+2 (U c k m+3 (U c k+1 s (U We can now prove the following: Theorem 310 (PorteousGiambelli Let F be a globally generated vector bundle of rank r over X Then, for a general choice of m sections of F, the class of the locus X s on which these sections have rank at most s is c r s (F c r s+1 (F c r s+m s 1 (F c r s 1 (F c r s (F c r s+m s (F c r m+1 (F c r m+2 (F c r s (F Proof: Write V = H 0 (X, F and consider the map ϕ V : X G(r 1, P(V given in Proposition 28 We thus have that X s is the inverse image by ϕ V of the locus on which 13
14 m sections of U have rank at most s, and this locus has class, by Example 39, c r s (U c r+1 s (U c r 1+m 2s (U c r 1 s (U c r s (U c r+m 2s (U c r m+1 (U c r m+2 (U c r s (U Since X s has codimension (r s(m s in X, by Theorem 211, it follows that its class is the pullback by ϕ V of the above determinant Since ϕ V (U = F by Proposition 28, the result follows at once from Remark 32 Remark 311 The actual PorteousGiambelli result is much stronger It says that if we have a morphism of vector bundles F F with m = rk(f and r = rk(f, then the locus X s on which the morphism has rank at most s either is empty or has codimension (m s(r s Moreover, when the equality occurs, the class of X s in A(X is c r s c r s+1 c r s+m s 1 c r s 1 c r s c r s+m t c r m+1 c r m+2 c r s where c i represents the coefficient of t i when expanding the quotient c(e c(f in the variable t as a formal series Exercise 312 Use Porteous formula to compute the degree of the Segre variety P 2 P 2 in P 8 14
15 References [A1] E Arrondo, Subvarieties of Grassmannians, Lecture Note Series Dipartimento di Matematica Uni Trento, 10 (1996, arrondo/trentopdf [A2] E Arrondo, Introduction to projective varieties, unpublished lecture notes available at arrondo/projvarhtml [KL] Kleiman, SL Laksov, D, Schubert calculus, Amer Math Monthly 79 (1972, [F] W Fulton, Intersection Theory, Springer 1984 (2nd edition in 1998 Departamento de Álgebra Facultad de Ciencias Matemáticas Universidad Complutense de Madrid Madrid, Spain 15
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