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math/9912088 | The first map is an isomorphism by REF , and the second map is an isomorphism by REF . |
math/9912088 | Without loss of generality, we can assume MATH. We need to show that the following diagram is commutative: MATH . In this diagram the only map which has not been defined is MATH. But since MATH it follows that MATH fixes MATH, so we can replace MATH by MATH in REF , and obtain a map which we also denote by MATH. Since ... |
math/9912088 | We start by showing that the construction of MATH is natural. Let MATH and MATH be two finite MATH complexes, and MATH a MATH-equivariant map between them. We want to define a map of sheaves MATH with the properties that MATH and MATH. Consider two MATH-cell decompositions of MATH and MATH, and let MATH be the collecti... |
math/9912088 | Notice that if MATH is a point, translation by MATH induces an isomorphism of the corresponding sheaf MATH on MATH to MATH. Via this isomorphism, the gluing map MATH is the usual restriction of holomorphic functions. Therefore MATH. |
math/9912088 | Recall that we defined MATH as the complex algebraic variety MATH. As in the proof of REF , denote by MATH the algebraic structure sheaf of MATH, and by MATH the analytic structure sheaf. Let MATH. We need to show that the natural map MATH is faithfully flat. We first show it is flat: By the local characterization of f... |
math/9912088 | This follows from MATH and from MATH, where the last isomorphism comes from the NAME formula. |
math/9912088 | First assume that MATH is a line bundle. Then from the definition of the NAME character we get MATH. We have already seen that MATH. Since MATH is finite, MATH is a finite module over MATH. Let MATH be a set of generators for MATH. Choose a set of elements MATH so that MATH . Denote by MATH. The element MATH is in MATH... |
math/9912088 | The proof of this lemma is mostly formal, and just makes intensive use of the identifications and definitions we have made so far. Start by using the exponential map MATH . The element MATH is in the image, so pick MATH such that MATH. Denote by MATH the element corresponding to MATH via the map MATH. Then one can appl... |
math/9912088 | We notice that, by REF , MATH does indeed belong to MATH, which by REF is the stalk of MATH at MATH. Fix MATH a cover of MATH adapted to MATH. Let MATH with MATH and MATH. This implies MATH and also MATH. Denote by MATH and MATH. REF implies that MATH. Now we have to prove that MATH, that is, that MATH. Consider the su... |
math/9912088 | Because of the NAME - NAME sequence, it is enough to verify the isomorphism for ``equivariant points" of the form MATH, with MATH a compact subgroup of MATH. Choose an identification MATH such that via this identification MATH . Then MATH. We use now REF , which is also true if we replace MATH by MATH-theory (because i... |
math/9912088 | By REF the images of the two maps MATH and MATH are the same. But MATH is injective, so we can identify MATH with the image of MATH. This implies that MATH factors through a map MATH, and this yields a splitting MATH. |
math/9912088 | Let MATH. Any compact MATH-manifold admits a decomposition as a finite MATH complex (see for example CITE). Let MATH be such a cell decomposition. We saw that if MATH is a compact subgroup of MATH, MATH. In particular, this implies that MATH. Let MATH, which is again equivariantly formal. By REF there is a natural iden... |
math/9912089 | The second and and the sixth maps are isomorphisms because MATH, and therefore MATH. The properties of the tensor product imply that the third and the fifth maps are isomorphisms. The fourth map comes from translation by MATH, so it is also an isomorphism. Finally, the second map MATH is an isomorphism because CASE: If... |
math/9912089 | We modify slightly the notations used in REF to indicate the dependence on MATH: MATH. Recall that MATH is sent to MATH via the algebra map MATH. If MATH is another additive uniformizer, we saw at the beggining of this Section that there exists a nonzero constant MATH in MATH such that MATH. Choose a square root of MAT... |
math/9912089 | For MATH to be a cohomology theory, we need naturality. Let MATH be a MATH-equivariant map of finite MATH complexes. We want to define a map of sheaves MATH with the properties that MATH and MATH. Choose MATH an open cover adapted to MATH, and MATH an additive uniformizer of MATH. Since MATH is MATH-equivariant, for ea... |
math/9912089 | One notices that MATH, because of the third condition in the definition of an adapted cover. If MATH, a global section in MATH is a collection of sections MATH which glue, that is, MATH. So MATH in MATH, which means that we get an element in MATH, since the MATH's cover MATH. So MATH. But clearly MATH for MATH, and the... |
math/9912089 | First, we show that the construction of MATH is canonical: Let MATH be another identification of MATH. We then have MATH, and MATH is identified with MATH. Since MATH is also identified with MATH, we get a group map MATH. This implies that we have a continuous group map MATH such that MATH. Any such map must be multipl... |
math/9912089 | The difficult part, namely that MATH is holomorphic, is proved in the Appendix, in REF . Consider the usual cup product, which is a map MATH, and extend it by tensoring with MATH over MATH. We obtain a map MATH. The equivariant elliptic NAME class of MATH is MATH, so we have to show that both these classes are holomorp... |
math/9912089 | Recall CITE that the ordinary pushforward is defined as the composition of three maps, two of which are NAME isomorphisms, and the third is a natural one. The existence of the elliptic pushforward follows therefore from the previous corollary. The proof that MATH is a map of MATH-modules is the same as for the ordinary... |
math/9912089 | From the hypothesis, we know that MATH is an isomorphism, because it is multiplication by the invertible class MATH. Also, since MATH is invertible, the localization theorem implies that MATH is an isomorphism. Therefore MATH is an isomorphism. Start with a class MATH on MATH. Because MATH is an isomorphism, MATH can b... |
math/9912089 | Denote by MATH the normal bundle of the embedding MATH. Let us show that, if MATH, then MATH is invertible in MATH. Denote by MATH the nonequivariant NAME roots of MATH, and by MATH the corresponding rotation numbers of MATH (see REF in the Appendix). Since MATH, MATH. Also, the MATH-equivariant NAME class of MATH is g... |
math/9912089 | To simplify notation, we are going to identify MATH with MATH, where MATH is the ``doubled" lattice described in REF. We want now to think of points in MATH as points in MATH, and of MATH as the pullback of MATH on MATH via MATH. Then we call MATH a torsion point if there is an integer MATH such that MATH (notice that ... |
math/9912089 | We first study the class MATH on each connected component of MATH in MATH. We will see that it lifts naturally to a class on MATH. The problem arises from the fact that we can have two connected components of MATH inside one connected component of MATH, and in that case the two lifts will differ by a sign. We then show... |
math/9912089 | The proof follows CITE. Denote by MATH the MATH-sphere with the MATH-action which rotates MATH times around the north-south axis as we go once around MATH. Denote by MATH and MATH its NAME and NAME poles, respectively. Consider a path in MATH which connects MATH with MATH, and touches MATH or MATH only at its endpoints... |
math/9912089 | By lifting the MATH-action to a double cover of MATH, we can make the MATH-action preserve the spin structure. Then with this action MATH is a spin MATH-manifold. At the beginning of this Section, we say that if MATH is a compact spin MATH-manifold, that is, the map MATH is spin, then we have the NAME pushforward, whic... |
math/9912089 | We know that the map MATH when regarded at the level of stalks at zero is the usual equivariant elliptic pushforward in MATH. Now MATH is the elliptic genus of the family. We have MATH, where MATH is the bundle of tangents along the fiber. Since MATH is spin, REF allows us to extend REF to the NAME section MATH. Since ... |
math/9912089 | From the embedding MATH, we have the following isomorphism of vector bundles: MATH . So, in terms of MATH-equivariant elliptic NAME classes we have MATH. Rewrite this as MATH where MATH is the twisted cocycle from REF . Notice that we can extend REF to virtual bundles as well. In other words, we can define MATH to be M... |
math/9912089 | We have MATH, where we write MATH for MATH. Assume MATH has a trivial MATH-action. It is easy to see that MATH. Hence we can write MATH, with MATH, and MATH nilpotent in MATH. We expand MATH in NAME expansion in multiindex notation. We make the following notations: MATH, MATH, and MATH. Now we consider the NAME expansi... |
math/9912089 | CASE: If MATH, MATH is a power series in the equivariant NAME classes MATH. But NAME classes are independent of the orientation, so MATH. CASE: If MATH, then MATH, with MATH. Hence MATH. MATH changes sign when orientation changes sign, while MATH is invariant, because of REF. |
math/9912091 | We note that REF define the affine subspace MATH in the linear space MATH. Since there exists an invariant linear connection on MATH, MATH is non-empty. Clearly MATH is a MATH-module and we need to check that MATH contains a MATH-fixed element. But this is evident since MATH is MATH-invariant and MATH is reductive. Nam... |
math/9912091 | We will prove only the assertion about orbits, leaving the assertion about coverings to the interested reader. Though MATH is not necessarily reductive, the homogeneous space MATH is always reductive, that is, MATH, where MATH. Suppose that MATH is a linear map satisfying REF for MATH, MATH. Then we can extend MATH to ... |
math/9912091 | This is an easy application of REF . |
math/9912091 | Let MATH be an irreducible MATH-submodule in MATH with the highest weight MATH and the highest vector MATH. Let MATH. The operator MATH induces a structure of MATH-module on MATH. The highest weight is MATH and the highest vector is MATH. We notice that if we have MATH for MATH then we have MATH, therefore MATH. |
math/9912095 | In view of the interpretation of MATH as MATH, the proposition follows from the fact that for a meromorphic function MATH on MATH, we have MATH regular at a point MATH if and only if MATH is regular at MATH. |
math/9912095 | Straightforward. |
math/9912095 | Let MATH be a local defining equation for MATH. Write MATH where MATH is the reduced divisor with support MATH. We claim first that the map MATH is a quasiisomorphism if and only if for all MATH the map MATH defined by the commutative diagram MATH given by MATH is a quasi-isomorphism. This follows by considering the co... |
math/9912095 | With notation as in REF, we take MATH, where MATH is a local coordinate. Integrability implies MATH with MATH. Multiplying through by MATH, we conclude that MATH is regular on MATH, which is equivalent to the assertion of the lemma. |
math/9912095 | We must show MATH. Using REF and MATH we reduce to showing MATH. Since MATH, we may do the computation formally locally and replace MATH by MATH. The desired vanishing follows because an exact form has no residues. |
math/9912095 | Again the proof is given in detail in CITE and we omit it. |
math/9912095 | Write MATH. Write the universal element in MATH as a matrix MATH. The assertion that MATH vanishes in the fibre at MATH means MATH for some MATH. Note this is an identity of the form MATH . We first claim that in fact this identity holds already in MATH. To see this, write MATH. Note MATH is a free module on generators... |
math/9912095 | We first consider the case when MATH has a pole of order MATH at infinity, so the MATH in REF. A basis for MATH is given by MATH . MATH and MATH has basis MATH . To compute the NAME connection, we consider the diagram (here MATH and MATH) MATH . One deduces from this diagram the NAME connection MATH . We may choose MAT... |
math/9912095 | The NAME complex (of global sections on MATH) for MATH is the tensor product of the corresponding complexes for the MATH, so REF follows from REF . For REF we have for example, the complex of global sections MATH. In MATH this gives for all MATH . REF follows easily. |
math/9912095 | Let MATH be the relative connection on MATH with respect to the map MATH, and take MATH to be the fibre coordinate for MATH. Write MATH. Then MATH so in MATH we have MATH . It follows that MATH has rank MATH (generated for example, by MATH and MATH), and the other MATH as claimed. |
math/9912095 | We can now calculate the connection MATH on MATH just as before. We have the NAME diagram (MATH). MATH . Here MATH . We get MATH . We can now substitute MATH getting finally MATH . For convenience define MATH . Representing an element in our rank two bundle as a column vector MATH the matrix for the connection on MATH ... |
math/9912095 | The essential point is the expansion MATH where MATH runs through permutations of MATH. |
math/9912095 | The differential form MATH on MATH is translation invariant. Indeed, to see this we may trivialize the torsor and take the point MATH to be the identity. Introducing a formal variable MATH with MATH, the group structure is then given by MATH with MATH . Since MATH it follows that the MATH are additive, whence MATH is t... |
math/9912095 | The proof is close to REF . We write (abusively) MATH for the power sum MATH, taken as a function of the elementary symmetric functions MATH. The quadratic monomials MATH all occur with nonzero coefficient in MATH. By construction, MATH with MATH. If we think of MATH as having weight MATH, MATH is pure of weight MATH, ... |
math/9912095 | Let MATH be the blowup. Then MATH is the strict transform of MATH in MATH. Since MATH is the NAME divisor MATH in MATH, it follows that the strict transform of MATH in MATH or MATH is isomorphic to MATH. We consider the structure of MATH locally around the exceptional divisor. We may assume some MATH is invertible and ... |
math/9912095 | Write MATH for the above integrand. Let MATH be some neighborhood of MATH. Then MATH is rapidly decreasing on MATH near MATH, where the size is defined by some metric on the holomorphic MATH forms on MATH. Since the chains are compact, a chain MATH on MATH will be supported on MATH for a sufficiently small neighborhood... |
math/9912095 | Let MATH be a point. We first show MATH . We calculate MATH using the NAME spectral sequence. For MATH, MATH is a smooth, affine quadric of dimension MATH. So MATH away from MATH for MATH, and MATH is a rank MATH local system. The monodromy about MATH and MATH is induced by MATH and MATH, respectively. Both actions giv... |
math/9912097 | Let MATH be a MATH-point of MATH. We have to prove that the Cartesian product MATH is representable by a scheme proper over MATH. For a weight MATH consider the NAME scheme MATH whose MATH points are pairs: (A map MATH; a subsheaf MATH of MATH on MATH such that MATH is locally free of rank MATH and the quotient MATH is... |
math/9912097 | Consider the embeddings MATH, MATH. We take MATH to be the maximal open subset of MATH, such that none of the above MATH's has zeroes on MATH. From the NAME relations, we obtain that on MATH, the other MATH's have no zeroes either. Hence, MATH is the thought-for MATH-bundle on MATH. Let MATH. To each of the points MATH... |
math/9912097 | The proof will use the same idea as the proof of REF, of which the latter is a particular case, since for MATH, MATH. Let MATH be a rerpesentation of MATH; put MATH, MATH. Let MATH denote the stack of rank MATH vector bundles on MATH and let MATH denote the stack classifying pairs MATH where MATH is a rank MATH vector ... |
math/9912097 | The proof is a combination of REF and of the following (obvious) statement: Let MATH be an open sub-stack of finite type. Then the set of MATH for which the intersection MATH is non-empty has the form MATH where MATH is some fixed element of MATH. |
math/9912097 | Let MATH and let MATH denote MATH, viewed as a local system just on MATH. Let MATH denote the NAME acting on MATH and on MATH. To construct the sought-for automorphic function it is sufficient to produce an object MATH together with a NAME structure on it, which is a NAME eigen-sheaf with respect to MATH in a MATH-comp... |
math/9912097 | A MATH-point MATH of MATH consists by definition of the following data: MATH; a pair of MATH-bundles MATH and MATH on MATH identified with one another outside on MATH; a collection of line bundles MATH, each embedded as subsheaf into the corresponding MATH, such that the NAME relations hold. Recall that the fact that t... |
math/9912097 | For MATH let us denote by MATH the map MATH . By definition, for MATH, the sheaf MATH is canonically the same as MATH. We have: CASE: The maps MATH and MATH from MATH to MATH coincide. CASE: For every MATH, the maps MATH and MATH from MATH to MATH coincide. For the proof of REF, observe that for an object MATH, the she... |
math/9912097 | It is easy to see that the assertion of the proposition amounts to the following: Let MATH be as MATH-point of MATH, such that for every MATH-module MATH, the sheaf embedding MATH has no zero along the divisor MATH. Suppose that MATH is a MATH-point of MATH such that the a priori meromorphic maps MATH extend to regular... |
math/9912097 | The fact that MATH is contained in MATH follows from the definitions. Therefore, to prove the proposition we have to show that every MATH-point MATH of MATH is contained is some MATH. The latter amounts to showing that there exists a MATH-bundle MATH and an isomorphism MATH, such that the composition MATH extends to a ... |
math/9912097 | Point REF follows immediately from REF. Moreover, by the same reason, for a perverse sheaf MATH on MATH, the sheaf MATH is perverse. Therefore, to prove point REF, it is enough to show that whenever MATH is irreducible, MATH is irreducible as well. First of all, since the map MATH is smooth (and has connected fibers), ... |
math/9912097 | The assertion of the proposition is in fact an easy corollary of the following fact proven in CITE: Let MATH be a semi-simple group and let MATH and MATH be two MATH-bundles on MATH. Then for any MATH, the restrictions MATH and MATH become isomorphic. Let MATH and MATH be two MATH-points of MATH and let MATH and MATH d... |
math/9912097 | of REF Let us first prove the "if" part of the proposition. We will fix MATH and let us denote by MATH the fiber product MATH, where MATH is mapped to MATH by means of the projection MATH. Let also MATH (respectively, MATH, MATH) denote the relative version of the stack MATH (respectively, MATH, MATH) introduced in REF... |
math/9912097 | The fact that MATH is representable is evident, since both the source and the target are representable over MATH. Hence, for a MATH-point of MATH, its preimage under MATH is a set, rather than a category, and to prove that MATH is a locally closed embedding, one has to show that this set consists of at most one element... |
math/9912097 | Let MATH be a MATH-point of MATH. To prove the proposition, it is enough to construct an irreducible stack MATH with a map MATH such that REF MATH is non-empty. CASE: MATH. Let MATH be the set of points where MATH has singularities and let MATH be the corresponding defects. Let us choose elements MATH in such a way tha... |
math/9912097 | For every collection of non-negative integers MATH, MATH, consider the corresponding map MATH . REF implies that the function on MATH corresponding to the sheaf MATH equals MATH . By summing up along the fibers of the projection MATH, we derive the formula of REF. |
math/9912097 | Assume that MATH. In particular, the corresponding point MATH belongs to MATH. By definition, we must have that for MATH the map MATH is regular on MATH, which implies that MATH. The latter inequality means exactly that MATH. Conversely, let us assume that MATH. Let MATH be a MATH-module of the form MATH, where MATH is... |
math/9912097 | Let MATH be a point of MATH, where MATH is a MATH - bundle on MATH and MATH is an isomorphism between MATH and MATH defined outside of MATH. We attach to it a point of MATH as follows: The corresponding MATH-bundle is induced from MATH, that is, MATH and the corresponding MATH-bundle is MATH. Now, the data of MATH is o... |
math/9912097 | As we have seen above, the fibers of the map MATH are never empty. Therefore, it is enough to prove that MATH is dense in MATH. Let MATH be a MATH-point of MATH. As in the case of REF, it is enough to construct an irreducible stack MATH with a map MATH, whose image contains MATH and such that MATH. To simplify the nota... |
math/9912097 | Set MATH and for a a MATH - bundle, we will denote by a subscript MATH the corresponding induced MATH - bundle. CASE: Let MATH and MATH be two MATH - points of MATH. First we will show that MATH is MATH - equivalent to another MATH - point MATH, such that MATH and the induced isomorphism MATH is regular on the whole of... |
math/9912097 | Let MATH and MATH be two MATH-points of MATH. Let us denote by MATH (respectively, MATH) the corresponding map. Consider the complexes MATH over MATH. We will normalize (by making a cohomological shift and NAME 's twist) in such a way that their highest (respectively, lowest) cohomology is MATH in degree MATH. To prove... |
math/9912097 | It is enough to prove that for each MATH, every irreducible subquotient of MATH is a constant sheaf on MATH. Therefore, it suffices to show that MATH is a constant function on MATH for all MATH. For that purpose, consider two points MATH and for an extension MATH, let MATH and MATH denote the corresponding MATH - point... |
math/9912097 | We have: MATH . As the projection MATH is proper, it follows from REF that it is enough to show that the sheaf MATH on MATH is ULA with repect to the projection MATH. However, since MATH is a fibration (locally trivial in the smooth topology), the needed assertion follows from REF. |
math/9912097 | Let MATH be equal to MATH. Consider the non-symmetrized NAME stack MATH, defined as MATH, where MATH. Let MATH, MATH and MATH denote the projections from MATH to MATH and MATH, respectively. The map MATH is an étale covering. Therefore, it is enough to show that the complex MATH lives in non-positive cohomological degr... |
math/9912097 | It is clear that if MATH is a ``good" perverse sheaf on MATH, then so is MATH. Therefore, to prove the theorem we have to show that the restriction of MATH to MATH lives in negative cohomological degrees if MATH is ``good". In other words, we must show that for every MATH, the sheaf MATH lives in negative cohomological... |
math/9912097 | Consider the following general set-up: Let MATH be a map of algebraic stacks with MATH smooth. Let MATH be an open substack such that the map MATH is smooth as well. Assume, in addition, that the sheaves MATH and MATH are ULA with respect to the map MATH. Now let MATH be another algebraic stack mapping to MATH. Let MAT... |
math/9912097 | A point of the stack MATH is by definition a triple of bundles MATH plus a collection of embeddings MATH which satisfy the NAME relations. We set: MATH, where for MATH, the map MATH is the composition: MATH . |
math/9912097 | By definition, for MATH, MATH which, according to the projection formula, can be rewritten as MATH . However, the maps MATH and MATH from MATH to MATH coincide and so do the maps MATH and MATH from MATH to MATH. Hence, REF can be identified, using once again the projection formula, with MATH which, by REF is the same a... |
math/9912097 | Consider the fiber product MATH . It classifies the data of MATH, with MATH, MATH and MATH. As in REF, there are two projections MATH and MATH from this stack to MATH that ``forget" MATH and MATH, respectively. By applying the projection formula, we obtain that for MATH, MATH and MATH, MATH can be written as: MATH wher... |
math/9912097 | Since MATH is the only fundamental weight, MATH is a evidently a closed substack in the stack classifying triples MATH as above. To show that this closed embedding is an isomorphism, one has to show that any map MATH satisfies the NAME relations, which is evident. To show that MATH is smooth we argue as follows: consid... |
math/9912097 | By construction, we have a natural map MATH, which corresponds to the open embedding MATH. Its cone is by REF. However,the latter complex is the same as the pull-back under MATH of MATH (compare REF). This implies our assertion. |
math/9912097 | Indeed, let us consider the fiber product MATH. We have the evaluatuion map MATH and let us denote by MATH the pull-back of the NAME sheaf on MATH under this map, tensored by MATH. To construct the isomorphism MATH, it is enough to show that both MATH and MATH can be canonically identified with the MATH - direct image ... |
math/9912097 | Consider the fiber product MATH, where MATH is the map MATH. By definition, this stack classifies quintuples MATH, where MATH are rank MATH-bundles, MATH is an embedding of coherent sheaves such that MATH is a skyscraper sheaf at MATH, MATH is a line bundle and MATH is an arbitrary map. We have a natural map MATH, whic... |
math/9912097 | Assume that REF does not hold, that is, that there exists a root MATH of MATH such that MATH. Let MATH denote the corresponding map. Then it is easy to see that MATH commutes with the image of MATH. Hence MATH contains MATH and therefore it is not commutative. To prove REF let us note that MATH. Hence, if MATH is commu... |
math/9912097 | (of REF .) Assume that MATH is not commutative. Let MATH denote its NAME algebra. We claim that there exists a nilpotent element MATH, such that MATH for a scalar MATH. Indeed, MATH defines an automorphism of MATH (it is obvious that MATH and hence also MATH is normalized by MATH). Therefore, the above lemma shows that... |
math/9912100 | We may assume both, MATH and MATH to be semi-stable. The natural inclusion MATH induces an inclusion MATH, for all MATH. Hence if MATH is not birationally isotrivial, REF is satisfied. |
math/9912100 | It is easy to find a finite cover MATH and an isomorphism MATH of polarized manifolds. In fact, there exists a coarse moduli space MATH of polarized manifolds, and NAME and NAME constructed a finite cover of MATH which carries a universal family (see CITE, p. REF). Of course one may assume MATH to be NAME with group MA... |
math/9912100 | Let us write again MATH. Then MATH. If MATH is not étale, there exists some MATH such that MATH contains a multiple point. This remains true, if we replace MATH by any finite cover MATH. In particular, in order to prove the first part of REF, we may assume that MATH, for MATH the image of a section of MATH. The same ca... |
math/9912100 | We may assume, that MATH is a normal crossing divisor and that the MATH-invariant defines a morphism in a neighborhood of MATH. We choose MATH to be ramified over MATH of order divisible by the multiplicities of the components of MATH, and such that MATH has a stable reduction MATH. REF allow to assume that REF holds t... |
math/9912100 | By the canonical bundle formula CITE, MATH is a subsheaf of MATH. Hence MATH is an invertible subsheaf of MATH, and since both coincide outside of a finite number of points, they are the same. The second equality follows from REF. |
math/9912100 | For a Jacobian fibration MATH write MATH. By REF we can find a covering MATH, étale over MATH, such that the conditions in REF are satisfied for suitable models of both, MATH and MATH. We will drop the MATH and assume MATH. The family MATH is birational to the semistable family MATH. If MATH is not birationally isotriv... |
math/9912100 | REF follows from REF and from the well-known uniqueness of the decomposition of MATH (see for example REF ). For REF let MATH be a representative of a class MATH, with MATH. Since MATH is numerically equivalent to MATH, for some MATH, one obtains from the NAME formula MATH . Therefore MATH or MATH. Since MATH in MATH, ... |
math/9912100 | In what follows we will work locally in MATH, but by abuse of notations we will write MATH and MATH. Consider first the case where the reduced general fibre of MATH is a smooth elliptic curve. Let us assume for a moment that MATH has ramification order MATH, the multiplicity of MATH. Then the normalization MATH is étal... |
math/9912100 | Both sheaves are subsheaves of MATH, hence in order to show that the inclusion MATH factors through MATH, we can argue locally in a neighborhood of MATH. Using the notation from REF, one has (locally in a neighborhood of MATH) a diagram MATH and REF holds true. |
math/9912101 | Let MATH and MATH be three vector fields on MATH. Then: MATH so that MATH is compatible with MATH. From the definition of MATH: MATH and this shows that: MATH and the result follows. |
math/9912101 | Let MATH be the orthonormal basis of MATH for MATH which diagonalizes MATH, and let MATH be the associated eigenvalues. We need to prove the upper bound above with MATH replaced by MATH, because MATH is skew-adjoint, and MATH is an orthonormal basis of MATH for MATH. According to the previous proposition and to REF , i... |
math/9912101 | We only have to prove the second assertion, concerning the curvature. Let MATH and MATH be the area elements associated to the metrics MATH and MATH on MATH. By the NAME REF : MATH . Let MATH be an orthonormal moving frame on MATH, and let MATH be its connection REF-form, that is: MATH . Then: MATH . But MATH is an ort... |
math/9912101 | A direct computation shows that, for MATH and MATH : MATH because MATH is torsion-free. |
math/9912101 | By definition: MATH . |
math/9912101 | From REF : MATH so, if MATH and MATH: MATH so that: MATH . But MATH and MATH, and it follows that: MATH . Take the scalar product (for MATH) with MATH and then with MATH, and use the symmetry of MATH with respect to MATH to obtain the result. |
math/9912101 | According to the previous proposition: MATH . Let MATH, call MATH the restriction of MATH to the NAME of REF in MATH. A simple compactness argument shows that there exists a constant MATH (which does not depend on MATH) such that: MATH where MATH is the maximum of the sectional curvatures of MATH at MATH. Therefore, is... |
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