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math/0105062 | If MATH, then the exact sequence REF reduces to MATH . If MATH, then MATH as well. If MATH and MATH is non-trivial, then MATH for MATH, so REF reduces further to MATH. |
math/0105063 | For the trivial local system MATH associated to MATH, the long exact cohomology sequence of the pair MATH splits into short exact sequences MATH see CITE and CITE. In particular, the MATH-th relative cohomology group MATH is canonically isomorphic to MATH, the MATH-th cohomology of MATH with constant coefficients MATH, see REF . So it suffices to show that the fundamental group of MATH acts trivially on MATH. Let MATH denote the inclusion of the fiber in the total space of the bundle MATH. It is known CITE that the image, MATH, of the cohomology of the total space MATH with (trivial) coefficients in the field MATH is invariant under the action of MATH. Consider the logarithmic forms on MATH defined by MATH denotes the gradient of MATH with respect to the variables MATH. Clearly these forms represent non-trivial classes in MATH. Furthermore, we have MATH . As is well known, the forms MATH generate the cohomology ring of MATH. It follows that MATH is totally nonhomologous to zero in MATH with respect to MATH: The inclusion MATH induces a surjection MATH in cohomology with trivial coefficients MATH. Consequently, the fundamental group MATH acts trivially on the MATH for each MATH, and the representation MATH is trivial. |
math/0105063 | By REF , the result holds at MATH. Therefore it holds for MATH close to MATH. The result follows. |
math/0105063 | Given MATH and MATH, let MATH be an eigenvalue of MATH. Then MATH is an eigenvalue of MATH for every MATH. It follows that the function MATH, MATH is single-valued and has no poles. Thus, MATH is a NAME polynomial in MATH. Write MATH, where MATH is a polynomial. Since MATH is an automorphism for every MATH, we have MATH for all MATH. Thus, MATH is a non-zero constant, and MATH is a unit in MATH. Using REF , we have MATH. |
math/0105063 | Recall from REF that the eigenvalues of MATH are monomials of the form MATH. Since MATH is the linear term in the power series expansion of MATH in MATH, the result follows. |
math/0105063 | Given MATH, the endomorphism MATH of MATH is the linearization of the automorphism MATH of MATH. Recall from REF that MATH. It follows that MATH may be realized as a logarithmic derivative of MATH at MATH. This being the case, we have MATH where MATH. Since MATH, the specialization MATH yields MATH . Thus a NAME connection matrix MATH satisfies MATH, see REF . Now the entries of MATH are linear forms in MATH. Consequently, we have MATH. Therefore, the specialization MATH yields the NAME connection matrix MATH. |
math/0105063 | Let MATH, and consider the corresponding induced bundles over the circle as discussed above. By REF , the NAME connection on the vector bundle MATH is given by the matrix MATH for each MATH. By REF , the eigenvalues of the connection matrix MATH are evaluations at MATH of linear forms with integer coefficients for each MATH. Passage to cohomology yields a connection matrix MATH, corresponding to the cohomology representation MATH, whose eigenvalues satisfy the same condition. |
math/0105063 | For MATH, we have an automorphism MATH of the universal complex MATH by REF . Write MATH and MATH, and consider these maps as matrices with entries in MATH. Then, for each MATH, MATH . Now make the substitution MATH, and denote power series expansions in MATH by MATH and MATH. In this notation, MATH. Comparing terms of degree two in the above equality, we obtain MATH . By REF , we have MATH. By REF , the linearization of MATH is equal to the boundary map of the NAME complex, MATH. Also, REF implies that MATH and MATH. These facts, together with REF, imply that MATH. In other words, MATH is a chain map of the NAME complex. |
math/0105063 | For MATH, the formal connection MATH is a chain map on the NAME complex, which induces upon specialization the NAME connection in NAME algebra cohomology. The result follows. |
math/0105064 | For cancellative MATH and MATH it is obvious. |
math/0105064 | From regularity REF it follows MATH (see also REF ). Multiplying REF on MATH gives MATH. Using second equation in REF for term in bracket we obtain MATH . For MATH similarly, but using REF . By analogy, multiplying REF on MATH we have MATH, and so MATH. For MATH similarly, but using REF . |
math/0105064 | For MATH it follows from REF . Multiplying first equation in REF on MATH we derive MATH, and the applying second equation in REF obtain MATH. For MATH similarly, but using REF . |
math/0105064 | The proof is obvious, if we note that MATH and MATH. |
math/0105064 | Multiplying REF on MATH we have MATH, and using REF we obtain MATH, and so MATH which has shape of the first equation in REF . For MATH similarly using REF we obtain MATH . REF can be modified using REF and then applying REF , then we obtain MATH which coincides with REF . For conjugated equations (second ones in REF - REF ) after multiplication of REF on MATH we have MATH or using REF MATH . By analogy from REF it follows MATH . |
math/0105064 | The first two relations can be resulted easily from REF . The third one follows by induction using REF and MATH . Applying the automorphism REF to REF , one gets REF . |
math/0105064 | CASE: Take any MATH and consider the product MATH. We have MATH. By the definition of MATH, there exists uniquely determined elements MATH and MATH of MATH such that MATH. This defines maps MATH and MATH in a unique fashion. The left multiplication by MATH being linear, so are MATH and MATH. Expanding both sides of the equality MATH in MATH using MATH for MATH, we get MATH . It follows that MATH and MATH. And, MATH. So, MATH, MATH. Therefore, we know that MATH is an algebra endomorphism and MATH is a MATH-derivation. The uniqueness of MATH and MATH follows from the freeness of MATH over MATH. CASE: We need to construct the multiplication on MATH as an extension of that on MATH such that MATH. For this, it needs only to determine the multiplication MATH for any MATH. Let MATH and each row and each column has only finitely many MATH and MATH is the identity of MATH. For MATH, let MATH satisfying MATH. Then MATH; and for MATH, MATH, MATH, thus MATH, MATH in MATH. And, obviously, for MATH, MATH; MATH. |
math/0105064 | Let MATH. Assume MATH. For MATH, obviously, MATH are linear independent. Since MATH and MATH, we have MATH and MATH for any MATH. Thus, MATH. It means that MATH for all MATH, then MATH. Hence MATH. |
math/0105064 | We have MATH. Now, we complete the proof of REF . Let MATH denote the subalgebra generated by MATH and MATH (all MATH) in MATH. From REF , we see that every element of MATH can be generated linearly by some elements in the form as MATH (MATH, MATH). But MATH, so MATH, that is, MATH is surjective. Then by REF , MATH is bijective. It follows that MATH and MATH are linearly isomorphic. Define MATH, then we can extend this formula to define the multiplication of MATH with MATH for any MATH and MATH, MATH. Under this definition, MATH becomes an algebra and MATH is an algebra isomorphism from MATH to MATH. And, MATH for all MATH. Obviously, the inclusion of MATH into MATH is an algebra morphism. |
math/0105064 | CASE: It follows from the fact that MATH is just MATH as a left MATH-module. CASE: Firstly, we can show that MATH, that is, for any MATH, there are MATH,MATH,MATH,MATH such that MATH. Equivalently, we show by induction on MATH that for any MATH, MATH can be in the form MATH for some MATH. When MATH, it is obvious. Suppose that for MATH the result holds. Consider the case MATH. Since MATH is surjective, there is MATH such that MATH. But MATH, we get MATH by the hypothesis of induction for some MATH with MATH. For any MATH and MATH, MATH since MATH is an algebra. Then MATH is a right MATH-module. Suppose MATH for MATH and MATH. Then MATH can be written as an element of MATH by the formula MATH whose highest degree term is just that of MATH, that is, MATH. From REF , we get MATH. It implies MATH. It is a contradiction. Hence MATH is a free right MATH-module. |
math/0105064 | As is well known, the two-variable polynomial algebra MATH is NAME (see for example, CITE). Then MATH is also NAME. For any MATH and MATH, if at least one element of MATH does not equal MATH, MATH is not in the ideal MATH of MATH. So, in MATH, MATH. It follows that MATH is a basis of MATH. Let MATH satisfies MATH and MATH. Then MATH can be extended to an algebra automorphism on MATH and MATH is a weak NAME extension of MATH from MATH and MATH. By REF , MATH is a free left MATH-module with basis MATH. Thus, MATH is a MATH-algebra with basis MATH and MATH run respectively over all non-negative integers, MATH runs over all positive integers-MATH. But, from the definition of the weak NAME extension, we have MATH, MATH, MATH. Thus, we can conclude that MATH and MATH run respectively over all non-negative integers, MATH runs over all positive integers-MATH is a basis of MATH. Let MATH satisfies MATH, MATH, MATH. Then MATH can be extended to an algebra automorphism on MATH. Let MATH satisfies MATH for MATH and MATH. Then just as in the proof of REF, it can be shown that MATH can be extended to a MATH-derivation of MATH such that MATH is a weak NAME extension of MATH. Then in MATH, MATH . From these, we conclude that MATH as algebras. Thus, from REF , MATH is NAME. By REF , MATH is free with basis MATH as a left MATH-module. Thus, as a MATH-linear space, MATH has the basis MATH run over all non-negative integers, MATH runs over all positive integers-MATH. By REF any MATH can be MATH-linearly generated by some elements of MATH, and therefore MATH and MATH generate the same space MATH. |
math/0105064 | The two-variable polynomial algebra MATH is NAME (see for example, CITE). Then MATH is also NAME. For any MATH and MATH, if at least one element of MATH does not equal MATH, MATH is not in the ideal MATH of MATH. So, in MATH, MATH. It follows that MATH is a basis of MATH. Let MATH satisfies MATH and MATH. Then MATH can be extended to an algebra automorphism on MATH and MATH is a weak NAME extension of MATH from MATH and MATH. By REF , MATH is a free left MATH-module with basis MATH. Thus, MATH is a MATH-algebra with basis MATH and MATH run respectively over all non-negative integers, MATH runs over all positive integers-MATH. From the definition of the weak NAME extension, we have MATH, MATH, MATH. So, we conclude that MATH and MATH run respectively over all non-negative integers, MATH runs over all positive integers-MATH is a basis of MATH. Let MATH satisfies MATH, MATH, MATH. Then MATH can be extended to an algebra automorphism on MATH. Let MATH satisfies MATH for MATH and MATH. Then just as in the proof of REF, it can be shown that MATH can be extended to a MATH-derivation of MATH such that MATH is a weak NAME extension of MATH. Then in MATH, MATH . From these, we conclude that MATH as algebras. Thus, from REF , MATH is NAME. By REF , MATH is free with basis MATH as a left MATH-module. Thus, as a MATH-linear space, MATH has the basis MATH where MATH run over all non-negative integers, MATH runs over all positive integers. By REF - REF any MATH can be MATH-linearly generated by some elements of MATH, and therefore MATH and MATH generate the same space MATH. |
math/0105064 | The proof is similar to that of REF for MATH. It suffices to check that MATH and the map MATH satisfying MATH, MATH, MATH, MATH are reciprocal algebra morphisms. |
math/0105064 | It can be shown by direct calculation that the following relations hold valid. MATH . Therefore, through the basis in REF , MATH and MATH can be extended to algebra morphisms from MATH to MATH and from MATH to MATH, MATH can be extended to an anti-algebra morphism from MATH to MATH respectively. Using REF - REF it can be shown that MATH for any MATH or MATH. Let MATH and MATH be the product and the unit of MATH respectively. Hence MATH becomes into a bialgebra. |
math/0105064 | Follows from REF - REF by tedious calculations. For MATH,MATH it is easy, and so we consider MATH, as an example. We have MATH . By analogy, for REF and MATH we obtain MATH . |
math/0105064 | Follows from REF . |
math/0105064 | First we should prove that MATH defines a morphism of algebras from MATH into MATH. We check that MATH . The relations REF - REF are clear from REF . For REF we have MATH . Relation REF is obtained similarly. Next for REF exploiting REF we derive MATH . Then we show that MATH is coassociative MATH . Take MATH as an example. On the one hand MATH . On the other hand MATH which coincides with previous. Proof that the counit MATH defines a morphism of algebras from MATH onto MATH is straithforward and the result has the form MATH . Moreover, it can be shown that MATH for MATH. Further we check that MATH defines an anti-morphism of algebras from MATH to MATH as follows MATH . The first three relations are obvious. For REF using REF we have MATH . For last relation REF using REF - REF we obtain MATH . Therefore, we conclude that MATH has a structure of a bialgebra. |
math/0105064 | Follows from REF - REF . As an example for MATH we have MATH . |
math/0105064 | Follows from REF - REF . For MATH,MATH it is easy, and so we consider MATH, as an example. We have MATH . By analogy, for REF and MATH we obtain MATH . |
math/0105064 | Suppose MATH is a group-like element, that is, MATH. By REF , MATH can be written as MATH. Here and in the sequel, every MATH, MATH and MATH with subscripts is in the field MATH and does not equal zero. Then MATH and MATH . It is seen that if MATH or MATH, MATH is impossible to equal MATH. So, MATH and MATH. We get MATH. Then MATH . If there exists MATH, then MATH possesses the monomial MATH, which does not appear in MATH. It contradicts to MATH. Hence we have only a unique MATH. Similarly, there exists a unique MATH. Thus MATH. Moreover, it is easy to see that MATH, MATH and MATH can not appear simultaneously in the expression of MATH. Therefore, we conclude that MATH, MATH or MATH (no summation) and we have MATH . It follows that MATH run over all non-negative integers-MATH. For any MATH, one can find MATH such that the regularity REF takes place MATH, which means that MATH forms a regular monoid under the multiplication of MATH. |
math/0105064 | Suppose MATH is a group-like element, that is, MATH. By REF , MATH can be written as MATH. Here and in the sequel, every MATH, MATH and MATH with subscripts is in the field MATH and does not equal zero. Then MATH and MATH . It is seen that if MATH or MATH, MATH is impossible to equal MATH. So, MATH and MATH. We get MATH. Then MATH . If there exists MATH, then MATH possesses the monomial MATH, which does not appear in MATH. It contradicts to MATH. Hence we have only a unique MATH. Similarly, there exists a unique MATH. Thus MATH . Moreover, it is easy to see that MATH, MATH and MATH can not appear simultaneously in the expression of MATH. Therefore, we conclude that MATH, MATH or MATH (no summation) and we have MATH . It follows that MATH run over all non-negative integers-MATH. For any MATH, one can find MATH such that the regularity REF takes place MATH, which means that MATH forms a regular monoid under the multiplication of MATH. |
math/0105064 | Let MATH be the linear sub-space generated by MATH for all MATH, MATH, MATH and MATH, and MATH is the linear sub-space generated by MATH. It is easy to see that MATH; MATH, thus, MATH is an ideal; and, MATH is a sub-algebra of MATH. Note that the identity of MATH is MATH. Moreover, MATH is a NAME algebra with the unit MATH, the comultiplication MATH satisfying MATH and the same counit, multiplication and antipode as in MATH. Let MATH be the algebra morphism from MATH to MATH satisfying MATH, MATH, MATH and MATH. Then MATH is, in fact, a NAME algebra isomorphism since MATH for all MATH, MATH, MATH and MATH is a basis of MATH by REF . |
math/0105070 | The dual of the differential of MATH at MATH is the pull - back map MATH, which is clearly injective. It follows that MATH is surjective. Denote by MATH the connected component of MATH that contains MATH and set MATH. The abelian variety MATH is a finite étale cover of MATH, and MATH factors through the induced map MATH, which is a connected étale cover of the same degree as MATH. Since the fibers of MATH are connected by assumption, it follows that MATH is an isomorphism and the fibers of MATH are connected. Let MATH be a general fiber of MATH. By REF, the NAME map of MATH is surjective, and therefore the image of MATH via MATH is a translate of an abelian subvariety of MATH which we again denote by MATH. MATH is contained in MATH and is independent of MATH, since MATH moves in a continuous system. Denote by MATH the quotient abelian variety MATH. The induced map MATH is constant on the general fiber of MATH and thus induces a rational map MATH. Since MATH is an abelian variety, MATH is actually a morphism and MATH generates MATH by construction. It follows that MATH, namely MATH. It follows that MATH is equal to MATH. By the theorem on the dimension of the fibers of a morphism, if MATH has maximal NAME dimension, then MATH also has maximal NAME dimension, and MATH has dimension MATH. Thus MATH, and Corollary MATH of CITE implies that the NAME map of MATH is a birational morphism. So the NAME variety of MATH is isogenous to MATH and, in particular, it does not depend on MATH, since MATH moves in a continuous system. |
math/0105070 | Follows immediately from CITE. |
math/0105070 | Let MATH be the cokernel of MATH. Clearly MATH for all MATH and all MATH. Therefore, MATH by REF , and MATH is an isomorphism. |
math/0105070 | MATH satisfies the I.T., MATH is a line bundle on MATH that satisfies the W.I.T. with index MATH, and MATH. Any line bundle with MATH is negative semidefinite. It is well known that there exists a morphism of abelian varieties MATH such that MATH for some negative definite line bundle MATH on MATH. It follows that MATH and hence MATH are supported on the image of MATH. |
math/0105070 | Compare REF . |
math/0105070 | REF correspond to REF. REF follows from REF. |
math/0105070 | Clearly we may replace MATH by the image of MATH and assume that MATH is onto. In addition, may assume that MATH is normal, since MATH factors through the normalization MATH and MATH for any coherent sheaf MATH on MATH and MATH. By the NAME spectral sequence, to prove the claim it is enough to show that the sheaves MATH are zero for MATH and MATH. Fix an ample divisor MATH on MATH such that: CASE: MATH is generated by global sections; CASE: MATH for MATH. The NAME spectral sequence that computes MATH degenerates at MATH by REF , and we have isomorphisms between MATH and MATH. For MATH, the former group vanishes by REF, since MATH is again nef and big. Now for MATH the sheaf MATH is zero by REF , and thus MATH is also zero. |
math/0105070 | If MATH, then clearly MATH for all MATH. Suppose now that MATH. The sheaf MATH is torsion free by REF. Let MATH be the NAME map of MATH, which is generically finite by assumption. The sheaf MATH is non-zero, supported on MATH. By REF , for every MATH and for every MATH we have isomorphisms MATH. If MATH, then the latter group vanishes by REF, and hence MATH is independent of MATH. In addition, by REF there is MATH such that MATH. So MATH for every MATH. |
math/0105070 | By the definition of NAME factorization, MATH is birational, MATH is finite and MATH. The variety MATH is normal, since MATH is normal, hence MATH is a normal sheaf, as defined above. Denote by MATH the dualizing sheaf of MATH. By REF p. REF, if MATH is the inclusion of the smooth locus MATH of MATH, one has MATH. If we write MATH and denote by MATH the restriction of MATH to MATH, then we have MATH since MATH is a birational morphism of smooth varieties. This identification induces a sheaf map MATH that is an isomorphism outside the image MATH of the singular locus of MATH, which is a closed subset of MATH of codimension MATH. The sheaf MATH is normal, since it is locally free and MATH is smooth, while MATH is normal by the definition of MATH and by the fact that MATH is finite. It follows that MATH is an isomorphism. By REF , there is an isomorphism MATH, where, for a coherent sheaf MATH of MATH and a finite morphism MATH, one defines MATH as the coherent sheaf on MATH corresponding to the MATH-module MATH. Now MATH, by REF , hence MATH. Taking duals, one has MATH. The sheaf MATH is normal and torsion free, since MATH is normal, and thus it is reflexive by REF. Thus we have MATH. In particular, MATH is locally free, namely MATH is flat. |
math/0105070 | REF is REF. For REF , the fact that the components of MATH are translates of abelian subvarieties follows from REF and the fact that the translation is by a torsion point follows from REF. REF follows from REF. REF is REF. |
math/0105070 | We write MATH and MATH. For MATH the claim follows by NAME duality, hence we may assume MATH. We will assume that MATH and show that MATH. We remark that since MATH, one has MATH. Let MATH be the greatest integer such that MATH. Then by REF, MATH is an isolated point of MATH for MATH. By REF and NAME duality, for every MATH the complex: MATH is exact except at the last term. As MATH varies, the above complexes fit together to give a complex of vector bundles on MATH (see CITE, proof of REF , for a similar argument): MATH which is again exact except at the last term, since exactness can be checked fibrewise. For MATH in an appropriate neighborhood of MATH, one has that MATH for MATH, hence MATH. It follows by REF , that for every MATH, the cokernel of the map MATH is MATH-dimensional. Therefore the cokernel of the map of vector bundles MATH is a line bundle MATH with MATH. Let MATH be the complex of vector bundles given by the exact sequence MATH . For any MATH, one can consider the two spectral sequences associated with the hypercohomology of the complex MATH. Since the complex is exact, we have MATH for all MATH, and thus MATH. On the other hand, for MATH, one has that MATH except for MATH and MATH, in which case MATH (recall that MATH by assumption). Comparing the two spectral sequences, one sees that MATH and hence MATH. For MATH the only non zero terms are-MATH and MATH, therefore MATH. In particular, the map MATH is an isomorphism. For MATH the only non zero terms are MATH , MATH and MATH. The differential MATH is an isomorphism and hence MATH that is, MATH. |
math/0105070 | Notice that by REF, the assumptions on MATH imply that MATH is a proper subset of MATH. Write MATH and MATH. As explained in the proof of REF, MATH is a component of MATH and MATH for MATH. By REF, the assumptions on MATH imply that the complex MATH: MATH is exact for all MATH. As we have already seen in the proof of REF , the complexes MATH fit together to give an exact complex MATH of vector bundles on MATH: MATH . Similarly, there is an exact sequence MATH of vector bundles on MATH: MATH . There is a map of complexes MATH induced by MATH . Clearly for MATH there is an isomorphism MATH . Proceeding by induction, assume MATH and MATH for all MATH. NAME by MATH and taking cohomology, one gets: MATH . If the vertical rows are exact, then the required isomorphism follows from the five lemma . Consider the spectral sequence associated to the complex MATH. We have MATH for any MATH, thus MATH. On the other hand for MATH one has MATH if MATH and MATH. If MATH then MATH. For MATH this group is MATH as MATH, while for MATH it is MATH since MATH by assumption. Comparing spectral sequences one sees that the sequence MATH is exact. The exactness of the first vertical row can be shown in the same way. |
math/0105070 | Write as usual MATH, MATH. By REF we have MATH. This implies MATH, hence MATH, since MATH has maximal NAME dimension. In particular we have MATH, implying MATH. |
math/0105070 | The statement is trivial if MATH, hence we may assume MATH. CASE: For all MATH, MATH. Arguing as in the proof of CITE, one can show that if MATH then MATH for all MATH. Since MATH is dense in MATH, by semicontinuity one has that MATH for every MATH. CASE: Let MATH. We have MATH for all MATH . This follows by a procedure analogous to CITE. We illustrate this for MATH. Write MATH and denote by MATH the composition of MATH and of the NAME map of MATH. Fix MATH an ample line bundle on MATH and MATH. By REF , for MATH sufficiently big and divisible we may pick a divisor: MATH . Possibly replacing MATH by an appropriate birational model, we may assume that MATH has normal crossings support and that (as in CITE) MATH . Let MATH that is, MATH is numerically equivalent to the pull back of a nef and big MATH-divisor plus a klt divisor. Then comparing base loci as in CITE MATH where the third and the fifth equality follow from REF. CASE: For all MATH, MATH. Let MATH be the maximum of the function MATH for MATH and let MATH be such that MATH. By REF , MATH for all MATH. Since MATH is dense in MATH, by semicontinuity we have MATH for all MATH, hence MATH for all MATH. |
math/0105070 | This is CITE, Lemma D. We thank NAME and NAME for allowing us to reproduce their proof. Let MATH, and let MATH be the codimension of MATH in MATH. Choose a basis MATH such that MATH are pull backs under MATH of a basis for MATH. In particular we may assume that MATH are conjugate to a basis of the tangent space to MATH. There is a homomorphism: MATH defined by the section MATH. MATH vanishes at a point MATH if and only if MATH is not smooth at MATH. Let MATH be the divisor along which MATH vanishes. There is a diagram: MATH . An easy local computation shows that MATH is torsion free. Fix a general point MATH. By the proof of REF (see also REF ) one sees that MATH. We claim that MATH induces an isomorphism: MATH . Fix a non-zero section MATH. Since MATH is a torsion free sheaf, it suffices to show that at a general poit MATH, MATH lies in the subspace of MATH spanned by MATH. By REF for all MATH, one has MATH. Since MATH span MATH at a general point, MATH must be a multiple of MATH. Consider finally the commutative diagram: MATH . The top row is multiplication by the ramification divisor MATH, therefore MATH. As we have seen above, MATH gives an isomorphism on global sections. By the proof of REF (compare REF ) one sees that the right hand side vertical homomorphism also induces an isomorphism on global sections. Therefore MATH . In other words MATH is a fixed divisor of the linear series MATH. |
math/0105070 | Denote by MATH the restriction to MATH of the NAME line bundle on MATH and by MATH, MATH, the projections of MATH on the MATH-th factor. Set MATH and MATH and denote by MATH the generic rank of MATH, MATH. Denote by MATH the sheaf map induced by the multiplication map MATH on MATH. The sheaf MATH is isomorphic to the trivial bundle MATH. In addition, there exists a nonempty open set MATH such that MATH and MATH are locally free on MATH. We denote by MATH, MATH the vector bundles associated to the restriction of MATH, MATH to MATH. For each MATH there are natural identifications MATH and MATH. In particular we have MATH and MATH. Composing the natural bilinear map MATH with the restriction of MATH, we obtain a morphism MATH. In turn, passing to the projectivized bundles and composing with the projection MATH, this induces a morphism MATH such that the restriction of MATH at the fibers over MATH corresponds to the natural map of linear systems MATH. The map MATH has finite fibers, since each element of MATH can be decomposed as the sum of two effective divisors in a finite number of ways. It follows that the dimension of MATH is greater than or equal to MATH. |
math/0105070 | Denote by MATH the natural inclusion. Pushing forward to MATH, one has an inclusion MATH. By REF, the sheaves MATH and MATH vanish for all MATH, hence for all MATH the NAME spectral sequence gives natural isomorphisms MATH and MATH. Thus for all MATH and MATH we have isomorphisms MATH . By REF , the sheaves MATH and MATH are isomorphic and, in particular, they have the same rank at the generic point of MATH. Thus the degree of MATH is equal to REF, that is, MATH is birational. |
math/0105070 | If MATH, then MATH is a point and the claim is of course true. Therefore we may assume MATH. By REF, we may assume that MATH and MATH are smooth and that the NAME fibrations of MATH and MATH fit in a commutative diagram where MATH and MATH . Consider the induced maps MATH, MATH and MATH. The claim will follow if we show that MATH has connected fibers. Fix MATH ample on MATH. The map MATH is generically finite onto its image by REF, hence MATH is nef and big on MATH. For MATH, the linear system MATH is nonempty. Let MATH be a general member of MATH and let MATH. We may assume that both MATH and MATH have normal crossings support. Define MATH and MATH . CASE: MATH is effective. Let MATH, where the MATH are distinct prime divisors. If the MATH are integers, then MATH. So it is enough to consider the case MATH, that is, MATH. Let MATH be a point such that MATH for MATH and let MATH be local coordinates centered in MATH such that MATH is a local equation for MATH, MATH. Let MATH be a point such that MATH and let MATH be a component of MATH containing MATH. Choose local coordinates MATH around MATH such that MATH is a local equation for MATH. Assume that MATH is a component of MATH for MATH, so that for MATH we have MATH, with MATH and MATH a regular function that does not vanish identically on MATH. A local equation for MATH around MATH is given by the determinant of the Jacobian matrix MATH, which is easily seen to vanish on MATH to order at least MATH. So the coefficient of MATH in MATH is greater than or equal to MATH. CASE: There is a map of sheaves MATH inducing an isomorphism on global sections. Since MATH is of NAME general type, we have MATH for every MATH. After replacing MATH by appropriate birational models, we may assume that MATH where MATH is free and MATH has normal crossings, MATH being the divisor chosen before. We wish to define a new divisor MATH such that MATH. To this end, pick MATH such that MATH has normal crossings support and MATH is smooth, not contained in the support of MATH. Let MATH and MATH. We may write MATH and MATH. For all MATH we have that the multiplicity of MATH along MATH satisfies MATH . We now replace MATH by MATH. It follows that MATH. By REF , there is an injection of sheaves MATH, corresponding to the effective divisor MATH. Pushing forward to MATH, one gets an inclusion MATH. Since MATH, the corresponding map on global sections is an isomorphism. CASE: MATH . By REF , there is an injection of sheaves MATH inducing an isomorphism on global sections. Pushing forward via MATH we obtain an exact sequence of sheaves on MATH . By REF, for all MATH, MATH, MATH . Therefore MATH for all MATH and MATH and we have MATH. By REF , the map MATH is an isomorphism. Therefore we have MATH and hence MATH by REF . MATH is connected. Since MATH is generically finite onto its image, it follows from REF that the generic rank of MATH is equal to the generic rank of MATH. Let MATH be a general fiber of MATH. Recalling that MATH is effective and that MATH, we have MATH. This shows that the generic rank of MATH is equal to MATH. Since MATH is effective, the generic rank of MATH is greater than or equal to the number of connected components of the generic geometric fiber. Therefore, MATH is connected. |
math/0105070 | The assumption implies that MATH is a divisor of MATH, hence MATH is generically finite onto its image. In addition, MATH is of general type, since otherwise, by CITE, MATH would be a pull-back of a divisor from a quotient MATH of MATH of dimension MATH. In that case, if MATH are the pull-backs of the elements of a basis of MATH, then the restriction of MATH to MATH is zero, contradicting the assumptions. Notice that the map MATH is injective for every MATH by the assumption that it is injective for MATH. The proof of REF gives MATH and thus MATH for MATH. |
math/0105070 | Let MATH be a desingularization of MATH. We may assume that MATH factors through MATH and MATH. CASE: MATH is a principal polarization. Notice first of all that, by the injectivity of MATH is a divisor and generates MATH. In particular, MATH has maximal NAME dimension. By REF , to prove that MATH is a principal polarization it is enough to show that: CASE: MATH for all MATH; CASE: MATH for all MATH and MATH. REF follows directly from REF . To prove REF , assume by way of contradiction that MATH contains a point MATH. By REF, and REF , MATH is a proper subset of MATH. Let MATH be a component of MATH, let MATH be a general point and let MATH be a vector that is not tangent to MATH at MATH. By REF ), REF , the complex MATH is exact for MATH. In particular, we have MATH, with equality holding iff the map MATH is zero. Since MATH is of maximal NAME dimension, it is easy to see that there exists MATH such that the map MATH is non-zero. If we denote by MATH the conjugate of MATH, then by REF MATH is not tangent to MATH at MATH. By NAME conjugation and NAME duality with twisted coefficients (compare REF, the map MATH is non-zero. Therefore, MATH. On the other hand, MATH, and MATH as observed above. Thus we have a contradiction and the proof of REF is complete. CASE: The map MATH is birational. Since MATH factors through MATH, it is clear that MATH is an isomorphism. Then REF implies that the maps MATH are isomorphisms for MATH. In particular one has MATH. In addition, for every MATH the map MATH is injective and we have MATH. Now the same argument as in REF can be used to show that MATH for MATH and MATH. Thus for every MATH we have MATH. It follows that for MATH the map MATH is an isomorphism, being a non trivial map of MATH-dimensional vector spaces. So the assumption of REF is satisfied for MATH. For MATH it follows from the fact that MATH is an isomorphism by using NAME conjugation and NAME duality. Thus MATH has degree REF by REF . |
math/0105070 | We write MATH, and MATH. Consider the map MATH. If it is not an isomorphism, then there exist MATH-forms MATH such that MATH on MATH. By the Generalized NAME Theorem (compare for instance REF ) it follows that there exists a fibration MATH REF such that MATH are pull-backs from MATH. Notice that MATH, since MATH has maximal NAME dimension. Then the induced morphism MATH is surjective and is not an isogeny, contradicting the assumption that MATH be simple. Thus MATH is an isomorphism and, by REF , MATH is also an isomorphism for all MATH. In particular, we have MATH. So the claim will follow from REF if we show that MATH for all MATH and MATH. By the generic vanishing theorems REF , the components of MATH are torsion translates of subtori of MATH of codimension at least MATH. Since MATH is simple, it follows that for all MATH the sets MATH consist of finitely many torsion points in MATH. Since MATH for MATH by REF, we have MATH for MATH by REF . |
math/0105070 | By CITE, MATH is surjective, hence MATH has maximal NAME dimension. By an argument due to NAME and NAME (compare CITE), MATH is birational to MATH. |
math/0105070 | By assumption the NAME map MATH is generically finite and surjective. By a result of NAME (compare CITE) one has MATH for all MATH. By NAME symmetry and NAME duality, one has that MATH. In particular MATH. Since MATH is simple, MATH is MATH-dimensional for MATH. Since MATH, then MATH is also MATH-dimensional and the claim follows from REF . |
math/0105070 | If MATH then this is CITE. If MATH or MATH for some MATH, this is CITE. We will follow the proof of CITE. Assume that MATH is not surjective. By CITE, up to replacing MATH by an appropriate birational model, there is a morphism MATH where MATH is a smooth variety of general type of dimension MATH, such that the NAME map MATH is birational onto its image. We denote by MATH a general fiber of MATH. In particular, notice that, MATH. CASE: There exists an ample divisor MATH on MATH such that, after replacing MATH by an appropriate birational model, there exist an integer MATH and a divisor MATH such that MATH has normal crossings support and MATH where MATH. We remark first of all that the linear system MATH is nonempty, since MATH. Let MATH be an ample MATH-divisor on MATH such that MATH is big. By the proof of CITE, after replacing MATH by an appropriate birational model, for MATH sufficiently big and divisible there exists a divisor MATH such that the restriction of MATH to the general fiber MATH is equal to MATH, where MATH and MATH is a general smooth member. Let MATH be a general member of MATH and set MATH. Let MATH. After replacing MATH by an appropriate birational model, we may assume that MATH has normal crossings support. Then MATH . Therefore MATH . CASE: MATH for all MATH. Let MATH be a divisor as in REF . Define MATH . One has: MATH . In particular MATH, and hence, by REF , MATH is a nonzero constant, independent of MATH. We are going to show that this constant is MATH. Indeed assume that it is equal to MATH and consider the NAME map MATH. By REF we have MATH. If MATH, then MATH by REF, and by REF it follows that MATH is supported on an abelian subvariety of MATH. However, by REF, MATH is a torsion free sheaf, and thus its support is MATH. Since MATH is birational, the support of MATH is MATH, contradicting the fact MATH is of general type. This shows that for every MATH we have MATH REF . We have (compare CITE) that MATH, MATH for all MATH, and if MATH then MATH. It follows that by REF , MATH for all MATH and MATH, and if MATH then MATH. We now apply REF with MATH, MATH, MATH, so that MATH. Recalling that MATH, for MATH we get MATH, the required contradiction. For MATH and MATH, we have MATH for every MATH and REF gives again MATH. Finally, for MATH and MATH, we have MATH and the contradiction follows by considering the morphism of linear series MATH . |
math/0105070 | If MATH is the double cover described in the statement, then the standard formulas for double covers give MATH and, for MATH, MATH if MATH is even and MATH if MATH is odd. Assume now that MATH and MATH. By REF , MATH is surjective and thus MATH has maximal NAME dimension. Let MATH denote the NAME fibration of MATH. We have a commutative diagram: MATH . Moreover, by REF , MATH is connected and there exists an abelian variety MATH isogenous to MATH and birational to MATH. Let MATH . Then MATH is the union of finitely many translates of MATH corresponding to the finite group MATH . For any line bundle MATH on MATH, define compare REF MATH . Since MATH, it follows that MATH for every MATH. By REF , one sees that for all MATH, the dimension of MATH is MATH. Moreover, by REF the only possible MATH-dimensional component of MATH is the origin. The proof is divided into several steps. CASE: MATH. Since MATH, we have MATH. By REF , we have MATH for every MATH, namely MATH. Consider an irreducible component MATH of MATH, where MATH is torsion. Then by CITE MATH has positive dimension, in particular it is nonempty. Thus MATH is also nonempty and REF implies again MATH. CASE: If MATH is an irreducible component of MATH, then MATH. Since MATH, we have also MATH, and hence by REF , MATH for all MATH. Applying REF with MATH and MATH we get MATH. CASE: For every MATH one has MATH. Assume that there is MATH such that MATH. Then we have MATH and thus, by REF , we have MATH for all MATH. Since MATH, then MATH. Let MATH be a point of MATH (such a point exists by CITE). Then from MATH and MATH it follows immediately MATH, against the assumptions. CASE: For any MATH, MATH. Let MATH be such that MATH. By REF , one has MATH for all MATH. Since MATH also MATH. Let MATH be an irreducible component of MATH. Then by REF , one has MATH for all MATH. Since MATH has positive dimension by REF gives MATH, a contradiction. CASE: MATH. Recall that by REF MATH is the only possible MATH-dimensional component of MATH. Let MATH be a positive dimensional component of MATH. Then by REF, MATH, where MATH and MATH is an abelian subvariety of MATH. For every MATH, we have MATH and MATH. Thus REF gives MATH. By REF we have MATH for every MATH. In particular, we have MATH and MATH for all MATH. Applying REF again gives MATH, a contradiction. CASE: MATH is birational. Since MATH is of NAME general type, the same is true for MATH by REF . So if MATH then the claim is true by CITE. Assume MATH. By the commutativity of the diagram at the beginning of the proof, the map MATH is surjective, since MATH is surjective. By a result of NAME and NAME (see also CITE) there is an irreducible component MATH of positive dimension. The divisor MATH is effective since MATH has maximal NAME dimension, hence MATH, contradicting REF . Denote by MATH the composition of the NAME map MATH and of MATH. CASE: For any MATH, there exists a principal polarization MATH on MATH and an effective divisor MATH on MATH such that MATH . See REF in the proof of CITE. Let MATH be any MATH-dimensional irreducible component of MATH. We recall that by REF, MATH is a translate of an abelian subvariety MATH. Let MATH be the dual map of abelian varieties and MATH the induced morphism. By REF , there exists a divisor MATH, vertical with respect to MATH, such that for all MATH, MATH is a fixed divisor of each of the linear series MATH. In other words, we have for all MATH . The divisors in MATH are vertical with respect to MATH. To see this, notice that MATH so MATH. Since MATH, the map MATH factors through the map MATH, which has connected fibers since it is dual to an inclusion of tori. The map MATH has connected fibers by REF , hence MATH has connected fibers. CASE: For general MATH, there exists a line bundle MATH of degree MATH on MATH such that MATH. The divisors MATH are nonempty and vary with MATH. Since MATH is vertical, for MATH general it contains a smooth fiber of MATH, where MATH. So we may set MATH. CASE: MATH. Let MATH be a component of MATH of positive dimension. By REF , MATH has dimension REF. As before, we let MATH and we denote by MATH the corresponding map. Let MATH be a general point. By REF , there is a line bundle MATH of degree REF on MATH such that MATH, and by REF there is a principal polarization MATH on MATH such that MATH. We denote by MATH the pull-back of MATH to MATH. There is an inclusion MATH, and thus the dimension of MATH is MATH. On the other hand, for every MATH we have MATH and MATH. So REF gives MATH. It follows that MATH, and the proof of REF shows that for every divisor D in MATH there is MATH such that MATH can be written as MATH, where MATH and MATH. Let MATH be the moving part of MATH. The line bundle MATH is positive semidefinite, so that there exist a quotient MATH of MATH and an ample line bundle MATH on MATH such that MATH is the pull-back of MATH to MATH and MATH compare REF. Notice that since the continuous system MATH is base point free, every divisor of MATH can be written as MATH for suitable MATH. Moreover, MATH, since MATH. So the general divisor of MATH is reducible, and the same is true for the general divisor of MATH on MATH. Since MATH is ample and has no fixed part, by REF this can only happen if MATH has dimension REF. Since MATH, it follows that MATH, namely the moving part of MATH, and thus also of MATH, is a pull-back from MATH. This condition determines the map MATH uniquely and, taking duals, it determines uniquely MATH. So we have shown that all the positive dimensional components of MATH are translate of the same abelian subvariety MATH of dimension REF. Since by CITE, MATH is generated by the sum of these translates, it follows that MATH has dimension REF, namely MATH. CASE: MATH. By REF , MATH is a smooth curve of genus REF. As we have remarked at the beginning of the proof, MATH intersects each component of MATH in a set of positive dimension. This shows that MATH contains MATH. The statement now follows from REF . MATH is of degree REF. Since MATH and MATH have connected fibers, then MATH is just the cardinality of MATH. If MATH, then there exist elements MATH such that MATH. Since MATH for all MATH by REF , it follows applying REF that MATH for all MATH and similarly that MATH, a contradiction. CASE: MATH where MATH is a principal polarization on MATH and MATH but MATH. By REF , we have MATH for an appropriate REF-torsion element MATH. The sheaf MATH is torsion free by REF and it has rank REF since the general fiber MATH of MATH is birational to an abelian variety and the restriction of MATH to MATH is trivial. Since MATH is a curve, MATH is actually a line bundle. In addition, we have MATH for all MATH. Therefore MATH is a principal polarization. Similarly the sheaf MATH is a line bundle such that MATH and MATH for all MATH, therefore MATH. Let MATH and let MATH. There are inclusions of sheaves MATH . There is a corresponding map MATH . Restricting to a generic fiber MATH of MATH, one has that the above map is given by MATH . Let MATH be a general fiber of MATH, MATH. The above map of sheaves is equivalent to the isomorphism MATH . It follows that MATH is an inclusion of sheaves. We wish to show that MATH is an isomorphism. To this end, by REF , it suffices to show that MATH induces isomorphisms of cohomology groups MATH for all MATH and MATH. For MATH the map above is injective and it is enough to check that the two vector spaces have the same dimension. By REF and by the description of MATH that we have given, it follows that for MATH and any MATH, we have isomorphisms in cohomology. If MATH and MATH is not in MATH, then both vector spaces are MATH (compare REF). We will prove that the above isomorphism holds for all MATH. (The proof for MATH proceeds analogously but is easier). For such a choice of MATH one has: MATH . We observe that MATH by REF, since MATH has maximal NAME dimension and we have seen that MATH is a proper subset of MATH. Let MATH be a subspace complementary to the tangent space to MATH. The assumptions of REF are satisfied for all MATH, hence for all MATH and MATH there are isomorphisms MATH induced by cup product. The required isomorphism is given by the following commutative diagram MATH REF . Conclusion of the proof. If MATH is the NAME factorization of MATH, then by REF , MATH is flat and MATH. The branch locus of MATH is reduced, since MATH is normal, and it is linearly equivalent to MATH, thus it consists of two fibers of MATH. |
math/0105077 | From the NAME exact sequence of the MATH-bundle MATH, it is easy to see that MATH. Hence MATH is equal to MATH and does not depend on the choice of the normal framing MATH. |
math/0105077 | Since MATH and MATH in the diagram above, we see that MATH. Hence MATH. Then the result follows directly from REF . |
math/0105077 | CASE: First assume that MATH, where MATH. Consider the spin structure of MATH corresponding to MATH under the identification in REF . Then by a result of CITE, there exists a framed REF-manifold MATH such that CASE: MATH, CASE: the framing of MATH restricted to the MATH-skeleton MATH of MATH coincides with the given spin structure of MATH, CASE: MATH has a special handlebody decomposition, consisting of one REF-handle and some REF-handles attached to REF-handle simultaneously. Since MATH is a nonclosed spin REF-manifold, it is parallelisable and so it can be immersed into MATH. Furthermore, since MATH has a MATH-complex as its spine, its immersion into MATH can be deformed into an embedding. Take an embedding MATH and set MATH. Now it suffices to show that MATH. Taking the normal framing MATH for MATH corresponding to the normal vector field of MATH, we see that the map MATH restricted to the MATH-skeleton can be written as the following composition: MATH where MATH is the map determined by the differential of MATH using the given framing on MATH. Thus MATH, since MATH. CASE: In the case when MATH cannot be written as MATH with MATH, put MATH with MATH. As in REF , consider the spin structure of MATH corresponding to MATH, and take an embedding MATH of a MATH-manifold MATH with MATH such that MATH has the same properties as in REF . Set MATH. In order to show that MATH, recall that we have fixed a trivialisation MATH of MATH to obtain the map MATH . If we take another trivialisation MATH of MATH, then we obtain another map MATH and with respect to this new trivialisation MATH, we have a new map MATH and a new induced homomorphism MATH which satisfies REF . Furthermore, if we choose MATH so that MATH corresponds to MATH as a spin structure, then we have MATH and hence MATH . Since the restriction of MATH to the MATH-skeleton MATH can be decomposed into the same composition MATH as in REF , by the same reason as in REF we have MATH and hence MATH. Thus MATH is a required embedding. |
math/0105077 | As it has been mentioned in the paragraph just after REF , we have MATH . This means that MATH is surjective. Let us prove the injectivity. For two immersions MATH and MATH, we have MATH . Note that MATH and MATH are homotopic on the MATH-skeleton of MATH. Let MATH be the difference MATH-cochain between MATH and MATH, that is, REF-cochain which assigns MATH to MATH considered as a MATH-cell, and MATH to the other REF-cells. Then we have MATH using the fact that MATH and the following lemma. Two maps MATH and MATH are homotopic if and only if CASE: MATH, and CASE: there exist a MATH-cocycle MATH and a MATH-cochain MATH such that MATH . Note that MATH is the generator of MATH and that MATH (see REF ). Now if MATH, then MATH (which we abbreviate here to MATH) is trivial, hence a coboundary. Conversely, if MATH is a coboundary, then there exists a MATH-cochain MATH such that MATH. Thus, we have MATH . Furthermore, MATH since MATH takes MATH on all MATH-cells of MATH. Therefore, we have MATH . This completes the proof of REF . (Note that this last argument also shows that the map MATH is indeed well-defined.) |
math/0105077 | By CITE, there exists a REF-connected compact oriented spin REF-manifold MATH with MATH such that the spin structures on MATH induced by MATH and MATH coincide with each other. Then by NAME 's theorem, MATH. Thus we have only to show the assertion for MATH instead of MATH. Since MATH is REF-connected, we have the exact sequence MATH where the map MATH is identified with the intersection form of MATH through the NAME duality MATH. By taking an appropriate basis for MATH, we may assume that the matrix representative of MATH is of the form MATH, where MATH is the zero form and MATH is nonsingular. Then MATH can be regarded as a presentation matrix of MATH and its size has the same parity as MATH. Now, since MATH is spin, its intersection form is of even type. Thus all the diagonal entries of MATH are zero, where MATH denotes the reduction modulo two. Now, it is an easy exercise to show that the size of such a matrix has the same parity as the dimension of the MATH vector space MATH which it presents. Thus the result follows. |
math/0105077 | Let MATH and MATH be generic maps of the compact oriented MATH-manifolds MATH and MATH respectively as in REF . First, consider the case where the normal vector fields along MATH in MATH and in MATH are homotopic. Then after a suitable deformation near MATH, we may regard MATH as the image of a generic map of the closed MATH-manifold MATH which is an immersion on a neighbourhood of MATH in MATH. Now we obtain MATH by REF, which claims that for any generic map MATH of a closed oriented MATH-manifold MATH, the equality MATH holds. Thus the result follows. Suppose that the normal vector fields along MATH in MATH and in MATH are not homotopic. We may assume that both normal vector fields are of equal constant length. Let MATH and MATH be formed by the endpoints of vectors of these two normal vector fields. Then we may assume that MATH and MATH intersect each other along a surface MATH which is the image of a surface MATH by the map associating to MATH the endpoint of the normal vector at MATH. Thus, MATH can be deformed, by a small modification near MATH, into the image of a smooth generic map which has only NAME umbrella singular points on MATH and no other singularities nearby. Now we can apply REF again. This completes the proof. |
math/0105077 | Let us first prove that MATH if MATH and MATH are regularly homotopic. Suppose that MATH and MATH are regularly homotopic. Let MATH be the track of a generic regular homotopy MATH between MATH and MATH. Then we can take a normal vector field MATH of MATH, which determines normal vector fields MATH and MATH of MATH and MATH for MATH and MATH respectively. (The argument below follows that of NAME in CITE.) There is a small positive MATH such that on the collars MATH and MATH the maps MATH and MATH, respectively, are immersions. Let us denote these immersions by MATH and MATH respectively. The tangent bundle of the cylinder MATH is MATH, where MATH, MATH is the projection, and MATH is the trivial line bundle. We may assume that the regular homotopy MATH is such that MATH for MATH and MATH for MATH. Now we define the bundle monomorphism MATH as MATH, that is, for a tangent vector MATH, where MATH and MATH, put MATH. This map is monomorphic on each fibre indeed, since MATH is normal to MATH. Furthermore, it coincides with the differentials of the immersions MATH and MATH over the collars. Hence - by the relative version of NAME 's theorem - there is an immersion MATH which coincides with the given immersions on the collars. Let MATH and MATH be generic maps as in REF . Now via the smoothing process as in the proof of REF , we obtain a generic map MATH of the closed oriented MATH-manifold MATH. By REF we obtain that MATH . Now let us show that MATH is injective on the set MATH of regular homotopy classes of immersions having the same NAME invariant MATH. Notice that the analogous map MATH has been shown to be an isomorphism in CITE (see REF in the present paper). Let us consider the effective and transitive group action described in the previous section restricted to just one orbit: MATH . Choose an embedding MATH in MATH (by REF , such an embedding exists). Let us consider the map MATH . This map is bijective by REF . Compose it with the map MATH . The resulting composition MATH coincides with the map MATH. Indeed, for any embedded NAME surface MATH for MATH we have MATH and for a singular NAME surface of MATH one can take the boundary connected sum of the NAME surface MATH for MATH with any singular NAME surface for MATH. Hence the map MATH is injective. |
math/0105077 | Let MATH and MATH be generic maps of the compact oriented MATH-manifolds MATH and MATH respectively as in REF . Then, we obtain a generic map MATH. From the fact that for any generic map MATH of a closed oriented MATH-manifold MATH, the equality MATH holds (see CITE), the result follows. |
math/0105077 | Let MATH be an immersion with trivial normal bundle. By REF , there is an embedding MATH with MATH. Furthermore, by REF , there exists an immersion MATH - unique up to regular homotopy - such that MATH. Let MATH be the track of a generic regular homotopy between MATH and MATH. Let MATH be the algebraic number of triple points of MATH. Take a compact MATH-manifold MATH and a generic map MATH as in REF for MATH, and let MATH be a NAME surface for MATH. Then, we can deform MATH into the image of a generic map of the manifold MATH into MATH bounded by MATH (see REF ). Therefore, by REF , we have MATH . Here we need the following lemma. We have MATH for any normal framing MATH for MATH. We can extend MATH to the whole cylinder MATH using the regular homotopy MATH . We shall denote this extended normal field also by MATH. Since on MATH, the normal field MATH coincides with the homotopically unique normal framing of MATH, we have MATH . (Indeed, recall that for an immersion the invariant MATH was defined as the linking number of the image of the immersion with the curves formed by the double points of the immersion pushed off out of the image of the immersion into the direction defined by MATH. Since MATH is an embedding the equality follows.) Furthermore, one can show that MATH in the same way as in CITE. This completes the proof of REF . Thus, MATH does not depend on the choice of MATH and we obtain MATH . Since the regular homotopy class of MATH is determined by MATH, again by REF , the regular homotopy class of MATH is completely determined by MATH, hence by MATH. This completes the proof of REF . |
math/0105077 | Since MATH has no MATH-torsion, MATH is equal to MATH for every embedding MATH. If we take a NAME surface MATH for an embedding MATH, then clearly MATH. Thus the result follows directly from REF . |
math/0105077 | CASE: It is known that there exists a spin MATH-manifold MATH of signature MATH with MATH, see CITE for example. Furthermore, MATH can be chosen to have a special handlebody decomposition with one MATH-handle and some MATH-handles with even framings (see CITE). As we have seen in the proof of REF , such a MATH-manifold MATH embeds into MATH. Such an embedding, restricted to the boundary MATH, gives a required embedding MATH. CASE: It is shown also in CITE that MATH with any spin structure spin-bounds either the solid torus MATH or the above mentioned MATH. Now let MATH be any NAME surface of any embedding MATH. Then MATH induces a spin structure on MATH, which is induced also from a spin structure either on the solid torus or on MATH. Therefore, either MATH or MATH carries a spin structure, and so by NAME 's theorem its signature is divisible by MATH. By NAME 's additivity these signatures are MATH and MATH respectively. In both cases MATH is divisible by MATH. |
math/0105077 | Let MATH and MATH be the embeddings described above in REF ; that is, they have NAME surfaces-MATH and MATH of signatures MATH and MATH respectively. From the fact that-MATH is a bijection, there is an immersion MATH such that MATH. Let us assume that MATH is regularly homotopic to an embedding. Then the following sequence of equalities hold: MATH for an integer MATH, which is a contradiction. Here the fourth equality MATH follows from the obvious remark that for a NAME surface of the connected sum MATH one can choose the boundary connected sum of the NAME surfaces of the embeddings MATH and MATH. The NAME invariant MATH is a multiple of MATH by the result of CITE recalled in REF. |
math/0105077 | Let MATH be any immersion with NAME invariant MATH. Then MATH is not regularly homotopic to an embedding again by the result of CITE. We show that for any embedding MATH the connected sum MATH is regularly homotopic to an embedding. For this purpose we first compute the MATH-invariant of MATH and then produce an embedding with the same MATH-invariant. Since the MATH-invariant determines the regular homotopy class completely, we obtain that MATH is regularly homtopic to the embedding. The following sequence of equalities hold: MATH . Here the first equlaity holds, since we can form the boundary connected sum of the NAME surface MATH of MATH with that of MATH even if the latter is a ``singular" one (that is, it is a generic map with boundary MATH, not necessarily an embedding). The number MATH is an integer by REF . Let MATH be an embedding with NAME invariant MATH, where MATH if MATH is even and MATH if MATH is odd. Now we see that if MATH is even, then MATH. If MATH is odd, then MATH. |
math/0105077 | Otherwise, restricting MATH to the closure of an appropriate neighbourhood of a component of MATH, we would get a contradiction to the integrality of the invariant MATH and REF . |
math/0105078 | The condition of MATH-hyperbolicity for MATH implies that for any geodesic MATH (finite or infinite) the closest-points projection MATH is coarsely contracting in this sense: If MATH and MATH then the ball MATH has MATH-image whose diameter is bounded by a constant MATH depending only on MATH. (This is an easy exercise in the definitions - see for example, CITE. Indeed this condition for all geodesics MATH implies MATH-hyperbolicity CITE). In the rest of the proof, let a ``dotted path" be a sequence MATH with MATH, and let its ``length" be MATH. Let MATH. If MATH is a dotted path in MATH outside a MATH-neighborhood of MATH then the contraction property of the previous paragraph implies MATH . Now suppose that MATH has endpoints within MATH of MATH and assume also that MATH. Let MATH be a segment of MATH whose endpoints MATH are within MATH of MATH, respectively, but whose interior is outside the MATH-neighborhood of MATH. On the concatenation of geodesics from MATH to MATH, across MATH to MATH and back to MATH, select a sequence of points spaced at most REF apart and apply MATH, obtaining (via the coarse-Lipschitz property) a dotted path MATH in MATH satisfying MATH by the coarse-Lipschitz property. Since MATH and MATH are distance MATH from the respective endpoints MATH and MATH of MATH (by coarse idempotence) and MATH, the contraction property of MATH implies that MATH and MATH are within MATH of MATH and MATH respectively. It follows that MATH. Thus, together with REF we obtain MATH and hence MATH . Now if we choose MATH so that MATH, we obtain an upper bound MATH . This bounds by MATH the maximum distance from a point in MATH to MATH. Since this applies to every excursion of MATH from the MATH-neighborhood of MATH, we conclude that MATH is MATH-quasi-convex. Now let MATH be any point. We have the bound MATH. Let MATH be a nearest point to MATH. We have MATH by coarse idempotence. Now applying coarse NAME to the path from MATH to MATH, whose length is at most MATH, we find that MATH is at most MATH. Finally by the triangle inequality we obtain a bound on MATH. |
math/0105078 | Let MATH be an arc of MATH - there may be one or two such arcs. Let MATH. Deform MATH fixing endpoints to a path MATH with MATH orthgonal paths from MATH to MATH, and MATH running along MATH. The lengths of MATH are at most MATH, and MATH. REF allows us to deform MATH, fixing endpoints, to MATH with MATH and MATH in the MATH-Margulis tube of MATH, so that MATH where we write MATH, MATH, MATH, and MATH. Since MATH, we conclude that MATH can be deformed to an arc MATH, so that MATH with MATH depending on the previous constants. This is shorter than MATH by at least MATH, so we can shorten the curve MATH by at least this much, concluding that MATH . On the other hand MATH since MATH, so we obtain an upper bound on MATH. |
math/0105078 | The width of the collar of MATH is the function MATH described in the Appendix. Lift MATH to a MATH-neighborhood MATH of a geodesic lift MATH of MATH in MATH. For MATH, MATH lifts to a geodesic arc MATH connecting MATH to the circle at infinity. Let MATH denote the length of its orthogonal projection to MATH. This is largest when MATH is tangent to MATH, so let us consider this case. The arcs MATH and its projection to MATH form opposite sides of a quadrilateral with three right angles and an ideal vertex, whose two finite sides have lengths MATH and MATH (see REF ). Hyperbolic trigonometry CITE gives us MATH . On the other hand we have from the definitions in REF that MATH where MATH satisfies MATH . From this we obtain an expression for MATH as a function of MATH, and one can deduce that this quantity is bounded. Indeed the maximum is obtained at MATH, and we have MATH . In other words, MATH travels less than REF times around the annulus, as measured by its orthogonal projection. We deduce that MATH and MATH intersect at most REF times in MATH. The bound REF follows. |
math/0105078 | Let MATH be the compactified annular lift, and let MATH and MATH be the lifts of MATH and MATH to MATH. Let MATH denote the lift of MATH to MATH, and let MATH be the component of MATH which is a homeomorphic lift of MATH. Let MATH be one of the components of MATH. There is a uniform upper bound MATH to the MATH-length of the shortest arc MATH connecting MATH to any component MATH of MATH contained in MATH. This comes from a standard area bound: Since MATH is bounded below by MATH (see REF), an embedded collar of radius MATH around MATH has area at least MATH. Since the area of MATH is fixed by the NAME theorem, MATH is uniformly bounded. The first self-tangency of this collar yields a bound for MATH. Since by REF the choice of MATH cannot change the twisting number in the lemma by more than REF, we may assume that MATH has an initial segment which is this shortest arc MATH. Similarly we may assume that MATH is orthogonal to MATH in the metric MATH. The NAME condition in the complement of MATH implies that the MATH-length of MATH is bounded by MATH. Thus the number of essential intersections that MATH can have with MATH in the interval MATH is bounded by MATH, since between any two such there is a segment of MATH that projects orthogonally (in MATH) to an entire boundary component of MATH. The remainder of MATH is contained in a convex set whose boundary MATH is the lift of a boundary component of MATH. Since MATH is a convex subsurface both in MATH and MATH, MATH still bounds a convex set in the metric MATH. Thus MATH can have at most one essential intersection with MATH in this convex set. |
math/0105078 | Since the MATH end is degenerate, there is a sequence of pleated surfaces MATH eventually contained in every neighborhood of the end. Choose MATH. Then the geodesic representatives of MATH must also eventually exit MATH. In particular they must converge to MATH in MATH. The same discussion applies with MATH replaced by MATH. Since MATH is homeomorphic to MATH and MATH is compact, any pleated surface in the proper homotopy class of MATH which is sufficiently far in a neighborhood of MATH (or MATH) can be deformed to infinity, through cusp-preserving maps, without meeting MATH. Choosing MATH high enough, then, insures that MATH and MATH homologically encase MATH and that MATH. Furthermore MATH implies that MATH, so that by REF there is a MATH-good homotopy (with MATH depending on MATH) between MATH and any map in MATH. This homotopy has bounded tracks outside the NAME tubes of MATH, if any. Thus we may choose MATH high enough that this homotopy also avoids MATH, so that MATH satisfies the conclusions of the lemma. |
math/0105078 | Let MATH be a vertex that is in MATH. If MATH meets MATH then (see REF) MATH, and in particular MATH . It follows from REF r@DisplaY NAME REF that MATH where MATH depends only on the topological type of MATH. Thus, since MATH are the vertices of a geodesic, the possible values of MATH lie in an interval of diameter at most MATH, which we call MATH. In other words, MATH. It remains to notice that, by the choice of MATH and the MATH-good homotopy property, if any part of a block MATH meets MATH then one of the boundaries must meet MATH, and hence MATH or MATH are in MATH. |
math/0105078 | By NAME 's REF , MATH is naturally identified with the NAME boundary of MATH. By the definition of the NAME boundary, there is a neighborhood MATH of MATH such that, if MATH are in MATH then any geodesic MATH in MATH connecting them must lie outside a MATH-ball of MATH. In particular every vertex of MATH must be at least distance REF from MATH, and hence must intersect MATH essentially. REF states that in such a situation MATH for a constant MATH depending only on the topological type of MATH, and in particular MATH. Now consider a sequence MATH converging to MATH in MATH, with MATH. After restricting to a subsequence if necessary, the MATH converge in the NAME topology to a lamination that includes the support of MATH, which means that eventually MATH. Thus MATH. |
math/0105078 | We begin by describing a decomposition of MATH into controlled pieces in a way that is determined by the MATH. In the following, MATH always denotes a permutation of MATH. CASE: Suppose that the three ``triangle inequalities" MATH all hold. Then there is a unique triple MATH satisfying MATH . (where we use the convention MATH). In fact we simply set MATH. Now define three ``bands" MATH in MATH as follows: MATH is the (closed) MATH-neighborhood of the edge MATH of MATH which is the common perpendicular of MATH and MATH. See REF . The two segments MATH and MATH cover all of MATH and meet in their common boundary point. The interiors of MATH are disjoint as a consequence of the triangle inequality and the fact that MATH. The closure MATH of the complement MATH is a ``triangle" with the following properties: CASE: Each edge MATH has curvature MATH, which is in fact given by MATH. The curvature vector points outward. CASE: Adjacent edges meet at angle REF. From this and the NAME theorem we have MATH which implies the upper bound MATH . There is also a lower bound MATH obtained as follows: Extend each MATH to a line of constant curvature in MATH. Each bounds a region MATH that contains MATH, so that the three regions have disjoint interiors. Since MATH, through any point MATH passes the boundary of a horoball contained in MATH. One can check that for any three horoballs in MATH with disjoint interiors, and any point MATH, the distance from MATH to at least one of the horoballs is at least MATH (the extreme case is where all three are tangent). The midpoint of MATH meets a horoball in MATH and is a distance at most MATH from horoballs in MATH and MATH. Thus we have REF . The geometry of the band MATH is easy to describe: it can parametrized by the rectangle MATH with the metric MATH . Where MATH is identified with MATH. CASE: Suppose that one of the opposite triangle inequalities holds, for example, MATH . We then let MATH, MATH, and MATH . So that MATH. We then have bands MATH and MATH as before, and MATH is defined as follows (see REF ): Let MATH be the common perpendicular of MATH and its opposite edge MATH. In MATH let MATH be the closure of the complement of MATH and MATH. This has length MATH. Join each MATH to a point MATH with a curve equidistant from MATH. We obtain a foliated rectangle, MATH (if MATH then MATH is a single segment). Similarly to the other bands, we can describe the metric in MATH by REF , but on a rectangle of the form MATH, where MATH. The closure of MATH is now two triangles MATH, with angles MATH, MATH and MATH. Let MATH and let MATH, for MATH. Note that the curvature of MATH is MATH. Upper and lower bounds on MATH and MATH can be derived from REF , by observing that the union of MATH with its reflection across MATH is a triangle of the same type as MATH in REF . For MATH we obtain REF exactly, whereas for MATH we get MATH and MATH . Let MATH be the larger of MATH. Then MATH, and since MATH must then be the larger of MATH we have a bound for MATH: MATH CASE: Now consider two hexagons MATH satisfying the bound MATH. Suppose first that both MATH and MATH are in REF . Note immediately that we also have MATH. If MATH for each MATH, then MATH. In this case we have upper bounds of MATH on both MATH and MATH. The bilipschitz map MATH can be constructed separately on each piece of the decomposition. Map each band MATH to MATH using an affine stretch on the parameter rectangles. The metric described in REF gives a uniform bilipschitz bound on this map. Note that the assumption that both MATH and MATH are bounded away from REF bounds the bilipschitz constant in the MATH direction, and the fact that their difference is bounded gives a bound in the other direction (since the MATH factor gives an exponential scaling). The triangles MATH and MATH are also in uniform bilipschitz correspondence, since in fact they vary in a compact family of possible figures (due to the length and curvature bounds). If at least one MATH we must take a bit more care. Let us consider a limiting case where at least one MATH and the rest are no smaller than MATH. These cases are illustrated in REF . If MATH then MATH and MATH is an ideal triangle. Recall that we are interested only in bounds on the complements of the collars, and this is in this REF right-angled ``hexagon" with three of the sides equal to horocyclic arcs of definite length. If MATH but MATH, then MATH has two geodesic edges meeting in an ideal vertex, and a third leg MATH satsifying the bounds REF . To this leg is attached the band MATH. If MATH and MATH then we obtain a triangle MATH with one geodesic leg and two curved legs to which are attached the bands. Note that the length bounds on the curved legs imply a length bound on the straight leg. If we now take a general MATH with some MATH, consider the family of shapes obtained by taking those MATH to REF, and fixing the rest. These must converge to one of the three cases above, and the bands corresponding to MATH remain fixed. Thus we can find some uniformly bilipschitz mapping taking MATH to the limiting case. Two hexagons MATH and MATH both near one of the boundary cases are therefore near to each other, via the composition of the two maps. Now suppose our two hexagons are in REF . Again we obtain bounds MATH for MATH and MATH. Assume first MATH for MATH. The bands MATH again have uniformly bilipschitz maps to MATH and MATH, respectively, coming from the affine stretches of the parameter rectangles. The same is true for MATH, using the parametrization MATH with metric REF , together with the bounds REF . The cases where some MATH are handled as before, by comparing to the boundary cases where some MATH. We leave the details to the reader. It is also possible for MATH to be in REF and MATH to be in REF . However, this can only happen if both of them are near their respective boundary cases, and one can check that these boundary cases are in fact in the overlap of REF , and hence can be compared with each other. What we have shown is that MATH and MATH admit a homeomorphism which is bilipschitz outside the collars of those MATH that are sufficiently short. For the long MATH we observe that, as in REF , the length of the boundary of MATH in MATH is at most MATH, and the collar width MATH has been chosen so that it is at most half the width of the largest embedded collar. It follows from this that the map can be adjusted to take all collars of MATH to collars of MATH, with a bounded change in the bilipschitz constant. The additive distortion in length of all subarcs of MATH follows from the construction. |
math/0105078 | Fix a generator MATH of MATH, and let MATH. Let MATH be the translation distance of MATH. Suppose first that MATH is loxodromic. If MATH with MATH and MATH then one can obtain see for example, CITE MATH . Suppose that MATH and MATH. Then, since the function MATH is decreasing, we have MATH, yielding an upper bound on MATH. Let MATH be the closest point to MATH (so MATH), and let MATH be the point on MATH with MATH. Suppose that MATH and hence MATH. We claim that MATH where MATH depends only on MATH. To see this, note that the distance from MATH to MATH along the MATH-equidistant surface to MATH is at most a constant MATH, since we can project the geodesic MATH to this surface with bounded distortion. On the other hand the distance from MATH to MATH along the MATH-equidistant surface is MATH, where MATH is the complex translation length of MATH. A little algebra yields REF . It follows that there exists MATH depending only on MATH such that MATH for all MATH. When MATH is parabolic the same discussion works using coordinates adapted to its parabolic fixed point, replacing equidistant surfaces to MATH with horospheres. Now for any arc MATH, with endpoints MATH, let MATH and MATH be the orthogonal geodesics from MATH and MATH, respectively, to MATH when MATH, and to MATH when MATH. The above discussion bounds the lengths of MATH and MATH uniformly. Let MATH be the other endpoints of MATH and MATH. Consider first the case that MATH. Let MATH be the geodesic segment MATH on MATH. Write MATH as in the statement of the lemma. If MATH then MATH, and it is immediate that MATH, which is what we wished to prove. Suppose MATH, so that MATH, and let us prove REF . Identify MATH with MATH so that the generator MATH of MATH acts by MATH. Let MATH. We may assume MATH. Let MATH be the closest point to MATH. Then MATH and MATH are distance at most MATH from MATH and MATH, respectively. We have MATH and MATH by the triangle inequality and the fact that MATH is arclength-preserving. We have by REF . We have MATH since MATH, and we have MATH by the triangle inequality, so REF yields: MATH . Now rearranging and applying REF we have MATH . Now MATH and MATH, so REF together bound MATH. Together with the upper bound REF on MATH, we find that MATH is in fact constrained to within bounded distance of the point in MATH at distance MATH from MATH (as opposed to the point on the other side of MATH at the same distance). This together with the case of MATH yields the general case, by concatentation. Now consider the case that MATH. Writing MATH, let MATH be the concatenated chain of geodesic segments connecting the sequence of MATH. The length estimate on MATH is immediate. |
math/0105079 | This is proved using essentially the same argument as in CITE. The natural map MATH induces the coaction MATH which uses an isomorphism MATH that follows from the flatness condition. |
math/0105079 | To deal with the multiplicative structure we need to modify the original construction given in REF. We remind the reader that we are working in the derived homotopy category MATH. Let MATH be an free MATH-resolution of MATH. Using freeness, we can choose a map of complexes MATH which lifts the multiplication on MATH. For each MATH let MATH be a wedge of sphere MATH-modules satisfying MATH. Set MATH and choose a map MATH inducing MATH in homotopy. If MATH is the homotopy fibre of MATH then MATH and we can choose a map MATH for which the composition MATH induces MATH in homotopy. Next take MATH to be the cofibre of MATH. The map MATH has a canonical extension to a map MATH. If MATH is the homotopy fibre of MATH then MATH and we can find a map MATH for which the composite map MATH induces MATH in homotopy. We take MATH to be the cofibre of MATH and find that there is a canonical extension of MATH to a map MATH. Continuing in this way we construct a directed system MATH whose telescope MATH is equivalent to MATH. Since we can assume that all consecutive maps are inclusions of cell subcomplexes, there is an associated filtration on MATH. Smashing this with MATH we get a filtration on MATH and an associated spectral sequence converging to MATH. The identification of the MATH-term is routine. Recall that MATH and therefore MATH are MATH ring spectra. Smashing the directed system of REF with itself we obtain a filtration on MATH, MATH where the filtrations terms are unions of the subspectra MATH. Proceeding by induction, we can realize the multiplication map MATH as a map of filtered MATH-modules so that on the cofibres of the filtration terms of REF it agrees with the pairing MATH. We have constructed a collection of maps MATH. Using these maps and the multiplication on MATH we can now construct maps MATH which induce the required pairing of spectral sequences. |
math/0105079 | As in the discussion preceding REF , making use of a NAME resolution we obtain MATH . The generators have bidegree MATH, so the differentials MATH are trivial on the generators MATH for dimensional reasons. Together with multiplicativity, this shows that spectral sequence collapses, giving MATH where the generator MATH has degree MATH and is represented by MATH. For each MATH, MATH with MATH. Under the coproduct, MATH is primitive for degree reasons. By comparing the two NAME Spectral Sequences we find that MATH can be chosen to be the image of MATH under the evident ring homomorphism MATH, which is actually a morphism of NAME algebroids over MATH. Hence MATH is coaction primitive in MATH. For MATH, we construct the NAME operation MATH using the composition MATH to induce a map MATH, then use the NAME resolution to determine the Universal Coefficient Spectral sequence MATH which collapses at its MATH-term. Further details on the construction of these operations appear in CITE. |
math/0105079 | An explicit splitting as in REF is obtained using the multiplication map MATH which induces a homomorphism of MATH-modules MATH. |
math/0105079 | Working throughout in the derived category MATH, the proof follows that of CITE, with MATH replacing the sphere spectrum MATH. The canonical NAME resolution of MATH is built up in the usual way by splicing together the cofibre triangles in the following diagram. MATH . The algebraic identification of the MATH-term proceeds as in CITE. |
math/0105079 | Smashing MATH with the cofibre sequence REF and taking homotopy, we obtain an exact triangle MATH . As multiplication by MATH induces the trivial map in MATH-homology, this is actually a short exact sequence of MATH-modules, MATH which clearly splits, so MATH is MATH-projective. In the evident diagram of exact triangles MATH the map MATH is an isomorphism, so MATH is also an isomorphism by the Five Lemma. |
math/0105079 | Let MATH be a regular sequence generating MATH. Using the notation MATH, we recall from CITE that MATH . For MATH a non-zero divisor, the MATH-free resolution MATH corresponds to a MATH-cell structure on MATH with one cell in each of the dimensions MATH and MATH. There is an associated cofibre sequence MATH for which the induced long exact sequence in MATH-homology shows that MATH is MATH-free. The dual MATH is equivalent to MATH, hence MATH is essentially self dual. For a MATH-module spectrum MATH in MATH, there are two exact triangles and morphisms between them, MATH . The identifications MATH and the Five Lemma imply that MATH . REF now implies that each of the spectra MATH satisfies REF . |
math/0105079 | By REF , MATH with generators MATH which are primitive with respect to the coproduct of this NAME algebroid. The determination of MATH is now standard and the differentials are trivial for degree reasons. |
math/0105079 | The proof is by induction on MATH. Assuming that MATH exists with the asserted properties, we will define a suitable map MATH which induces a fibre sequence of the form MATH for which MATH as a MATH-module. If MATH is a MATH-module which is a MATH module spectrum, REF provides a Universal Coefficient Spectral Sequence MATH . Since MATH is MATH-free, this spectral sequence collapses to give MATH . In particular, for MATH, MATH . By REF , there is an element MATH corresponding to an element MATH inducing a fibre sequence as in REF. It still remains to verify that MATH as a MATH-module. For this, we will use the resolutions MATH and MATH. These free resolutions give rise to cell MATH-module structures on MATH and MATH. By CITE, the MATH-module MATH admits a cell structure with cells in one-one correspondence with the elements of the obvious tensor product basis of MATH. Hence there is a resolution by free MATH-modules MATH . There are morphisms of chain complexes MATH where MATH is the obvious inclusion and MATH is a chain map lifting MATH which can be chosen so that MATH . The effect of the composite MATH on the generator MATH turns out to be MATH while the elements of form MATH with MATH are annihilated. The composite homomorphism MATH is a cocycle. There is a morphism of exact sequences MATH where the cohomology class MATH represents the extension of MATH-modules on the bottom row. It is easy to see that MATH, hence this class also represents the extension of MATH-modules MATH . There is a diagram of cofibre triangles MATH and applying MATH we obtain a spectral sequence converging to MATH whose MATH-term is the homology of the complex MATH where the MATH are essentially the maps used to compute MATH in CITE. By REF , this complex is exact except at the ends, where we have MATH. As a result, this spectral sequence collapses at MATH giving the desired form for MATH. |
math/0105079 | This follows from REF . |
math/0105079 | This follows from REF . |
math/0105079 | This is immediate from REF since MATH . |
math/0105079 | REF is proved by an easy induction on MATH. REF is a consequence of REF . |
math/0105079 | The first statement is easy to verify. By REF , to simplify notation we may as well replace MATH by MATH and so assume that MATH is a commutative regular quotient of MATH. Using the NAME complex MATH, we see that MATH is the homology of the complex MATH with differential MATH. Since the sequence MATH remains regular in MATH, this complex provides a free resolution of MATH as a MATH-module (this is false if the sequence MATH is not finite). Hence we have MATH . To calculate MATH we may use the NAME Spectral Sequence of CITE, MATH . By the first part, the MATH-term is MATH . Hence the natural homomorphism MATH is an isomorphism. |
math/0105084 | REF is contained in REF. REF can be proved using the same idea as in the proof of CITE. |
math/0105084 | CASE: Write MATH and let MATH . Taking the boundary we get the desired result because the cycle MATH cancels with MATH by skew-symmetry. CASE: This is similar to REF if we set MATH . |
math/0105084 | By REF we only need to show MATH which follows easily from a substitution MATH if MATH. If MATH then REF is trivial. The second equation follows from the obvious substitution MATH. |
math/0105084 | To make the proof explicit we will carry it out in a series of steps. Throughout the proof we will use MATH repeatedly without stating it explicitly. As NAME pointed out to the author the major difficulty is to guarantee that all the cycles we use lie in the ``admissible world". Due to its length and pure computational feature we put the proof of admissibility of all the cycles appearing in this paper in the online supplement CITE except for one cycle in REF where we spell out all the details to provide the readers the procedure how we do the checking in general. CASE: Construction of MATH. Let MATH, MATH and MATH . Let MATH and MATH. Then taking MATH we can write MATH by REF because all of the following cycles are admissible and negligible MATH where MATH. Here for the last cycle we need to use the fact that MATH . Next by using the transformation MATH we get MATH . Here for any two rational functions MATH and MATH of one variable we set MATH . CASE: The key reparametrization and a simple expression of MATH. We first observe that under the involution MATH we have MATH . Next if applying both MATH and MATH (denoted by MATH) then we get MATH . By REF MATH . We end this step by showing that MATH is admissible. Note that MATH . We have MATH where the last equation comes from the two solutions of MATH: MATH . By non-degeneracy assumption and MATH it suffices to show the following cycles are admissible: MATH . CASE: MATH is admissible. Because MATH we have MATH . Moreover, by non-degeneracy assumption we see that by REF MATH . Thus both MATH and MATH are clearly admissible by non-degeneracy assumption. CASE: MATH is admissible. This follows from the above proof for MATH. CASE: MATH is admissible. This also follows from the proof for MATH because MATH when MATH by REF . Some admissible cycles for decomposition of MATH. In order to decompose MATH we define the following admissible cycles MATH . We now use REF to remove the coefficients in front of MATH and MATH in MATH and MATH, REF to remove the factor MATH from the fourth coordinate of MATH, and REF to remove the coefficient MATH in front of the third coordinates of MATH, and finally get: MATH . It is not too hard to verify that all the cycles appearing in the above are admissible. Now we can break up the fourth coordinate of MATH according to REF and get MATH . Also by REF we find that MATH . However, the conditions in REF are not all satisfied by MATH. To decompose the third coordinates of MATH we combine MATH and MATH and use REF to get MATH . Here the properties of substitutions MATH and MATH play important roles. Another important thing is that we can write MATH in two ways such that the third coordinate of one of these (that is, MATH) is mapped to the third coordinate of MATH (that is, MATH) under MATH and vice versa. Hence MATH . On the other hand, we can easily see that MATH . Therefore we have the following simple expression of MATH by REF: MATH where MATH . CASE: Decomposition of MATH. Let MATH, MATH, MATH where MATH and MATH. Then we can apply REF and easily get MATH where MATH . We now apply REF to MATH with MATH and MATH to get MATH . CASE: Computation of MATH. Set MATH . Throwing away the appropriate admissible and negligible cycle we have MATH . Then further disregarding some admissible and negligible cycles we get MATH where MATH is defined by REF, MATH and MATH . Here we removed the coefficient MATH in front of the MATH and MATH in front of MATH in MATH by REF . Then we can take MATH, MATH and MATH in REF and get MATH . CASE: Decomposition of MATH. Put MATH which satisfies MATH . Then it follows from REF that MATH where all of the cycles MATH are admissible. This breakup is the key step in the whole paper. CASE: Computation of MATH. To ease the reading of the proof in this section we first summarize our approach here. We very much like to be able to use REF but unfortunately the terms MATH, MATH and MATH cannot be fixed for all the terms because we have to stay inside the ``admissible world". Nevertheless, luckily enough for us, most of the cycles we are going to use have more than one ``realization" so that we can apply REF to obtain the desired results. REF will be crucial to our computation. We begin by setting MATH and MATH such that MATH. By REF we get MATH are all admissible. Then repeatedly applying REF we have MATH . Again by applying REF we get MATH . Therefore MATH . CASE: Computation of MATH. We could use a similar process as in REF to do the computation. But we can get around this by the following argument. Define the substitutions MATH . Let MATH . Then an easy computation shows that MATH . Hence by first splitting off the MATH in front of the first two coordinates of MATH and MATH respectively, then removing some other admissible and negligible cycles we find MATH . CASE: Final decomposition of MATH into MATH's. Putting REF to REF together we see that MATH . We will first simplify the terms in the above expression. Set MATH . Define the admissible cycles MATH . Claim. We have MATH . Proof of the Claim. First by the involution we quickly find MATH . For MATH defined by REF we can remove coefficient MATH from the first two coordinates then apply MATH to get MATH . Here we have sequentially added two admissible and negligible cycles. For the first term in REF, from REF and MATH we find MATH . For the second term in REF, we can first remove the coefficient MATH in MATH and then delete one admissible and negligible cycle to get MATH . This completes the proof of our claim. CASE: Final computation of MATH. Let's compute each MATH separately. Throughout this computation we will repeatedly invoke REF without explicitly stating it. MATH . By REF . MATH . Using MATH we find MATH . MATH . Using substitution MATH we get MATH . MATH . By definition and using substitution MATH we get MATH . Putting the above together we now complete the proof the theorem in the case that none of the terms in NAME 's relations is equal to MATH or MATH. |
math/0105087 | This is a special case of CITE; see also CITE. |
math/0105087 | Fix a natural number MATH relatively prime to MATH and consider MATH, the universal curve of genus MATH with principal NAME structure of level MATH. If MATH, then the final paragraphs of CITE imply that this family has full MATH-monodromy. Indeed, CITE shows that it suffices to verify the statement for the analogous family over MATH, and CITE provides this proof. If MATH is relatively prime to MATH, then consider the moduli space MATH. On one hand, the MATH-torsion of the Jacobian of MATH has full MATH-monodromy. On the other hand, the forgetful map MATH is finite; therefore, MATH has full MATH-monodromy, too. For any MATH as in the statement of the lemma, there is an étale base change MATH so that MATH admits a level MATH structure. Then MATH is the pullback of MATH by the classifying map MATH. Moreover, the sheaf of MATH-torsion on MATH, MATH is the pullback of the universal MATH-torsion: By the versality assumption, MATH has dense image in MATH. We have seen above that MATH has monodromy group MATH. Thus, as long as MATH, MATH has full MATH-monodromy; a fortiori, MATH does, too. |
math/0105087 | In view of the preceding discussion, this is an immediate application of REF. |
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