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math/0104145 | We consider the case of odd length MATH; the case of even length is analogous. Setting MATH, MATH, we observe that MATH for suitable coefficients MATH and MATH. Let MATH be the coefficient of MATH in MATH (extending as usual to MATH), and let MATH be the coefficient of MATH in MATH. We find: MATH . If we instead expand MATH and MATH in terms of MATH and MATH, we obtain coefficients: MATH . We need a relation that is a nonnegative linear combination of the coefficients MATH, positive at MATH, as well as a nonnegative linear combination of the coefficients MATH for MATH larger than the bound. Now, consider the relation MATH . On MATH, this has coefficient: MATH which to first order is positive for MATH. Similarly, on MATH, this has coefficient: MATH which is positive for MATH except in a neighborhood of MATH and MATH. The construction of a positive perturbation is straightforward. |
math/0104145 | For each MATH, let MATH be the relation corresponding to the polynomial MATH above; that is: MATH . Thus the coefficient of MATH in MATH is MATH . Also, define coefficients MATH by MATH . We will consider relations of the form MATH with MATH. We first claim that for any such relation, the coefficients of MATH are nonnegative when MATH and MATH is sufficiently large. Indeed, in a neighborhood of MATH, this follows from the estimate of REF ; the terms for MATH are MATH, while the term for MATH is MATH and positive. In the other region, we use the MATH estimate of REF , in which the main term is positive and of order MATH. In the remaining neighborhood of MATH, we use REF . We thus have: MATH where MATH . In particular, if MATH is bounded above the largest zero of MATH then we have positivity for MATH; when MATH, the error term is uniformly of smaller order than the main term. Similarly, MATH with MATH . Thus given any choice of MATH such that MATH if MATH is the largest zero of MATH we have the asymptotic bound MATH . To construct a good relation, we will need some further properties of the NAME polynomials, all classical results: The polynomials MATH are the unique monic polynomials such that MATH . Furthermore, we have the three-variable generating function: MATH convergent for MATH. Finally REF : MATH . Now, given MATH, we can compute MATH using orthogonality; we find: MATH using the same formula to define MATH for MATH gives MATH. But then MATH can be computed as MATH where MATH . Using the three-variable generating function, we find: MATH where MATH and we must satisfy the additional requirement MATH precisely the additional hypothesis above. But then MATH . Since MATH the problem reduces to the following. For an integer MATH, how small can we make the largest zero of a polynomial MATH of degree MATH subject to the condition MATH . Let MATH be the smallest zero of MATH, and consider the polynomials MATH . By NAME, we compute MATH . We thus have the following integrals: MATH . Now, consider the polynomial MATH for MATH small. This certainly satisfies the negative integral condition; on the other hand, its only zero is that near MATH. Thus when MATH, we obtain a bound of the form: MATH . Since MATH we are done. |
math/0104149 | In this case obviously MATH. If MATH, then MATH is easily seen to equal zero. If MATH, then MATH. By REF it then follows that MATH is an invariant. |
math/0104149 | It follows from the computation above that both sides of REF have the same partial derivatives. It is also easy to see that MATH and the theorem follows. |
math/0104149 | Denote the right hand side in REF by MATH and the summand by MATH. Since MATH is odd by assumption, we have MATH. Thus, we can write MATH where we sum over all MATH satisfying the condition. Note that MATH . From this MATH . One checks that the sum in the square brackets is MATH if MATH satisfies either MATH or MATH, and is MATH if MATH . Thus MATH . The case where MATH or MATH does not occur, since by REF |
math/0104149 | Let MATH . Consider MATH. If we apply condition REF to the vector MATH with MATH and MATH we see immediately that MATH has to satisfy MATH . Therefore, MATH with MATH. Now it is immediately checked that this function satisfies REF only if MATH is odd. |
math/0104154 | Let MATH and MATH be the two sheaves induced from the two choices of isomorphism. First, if the automorphism MATH is given by multiplication by an element of MATH, then MATH and MATH are clearly isomorphic (the difference in gluing data is a NAME coboundary). Consequently, we may assume that for each MATH dividing MATH and such that MATH is locally free at MATH, that is, MATH and MATH are congruent to MATH, an isomorphism MATH has been fixed, and the automorphism MATH preserves the homomorphisms MATH induced from MATH. It is straightforward to see CITE that the automorphisms of MATH that preserve MATH are elements of MATH, acting on MATH in the obvious way. Moreover, if either MATH or MATH is not zero, or if MATH and MATH are zero, but the normalization of MATH at MATH is connected, then automorphisms of MATH preserving MATH must, in fact, lie in the image of the diagonal MATH . Thus they are all induced by multiplication by elements of MATH, and consequently induce isomorphic sheaves: MATH. Therefore, the only potential ambiguity arises when MATH, the normalization of MATH at MATH is disconnected, and MATH corresponds to an element MATH of MATH. In this case, we have MATH, and MATH, and thus MATH. Likewise, since MATH and MATH are non-zero, we have MATH. But now the automorphisms MATH and MATH applied to MATH on the two connected components of MATH, respectively, give an isomorphism MATH, as desired. |
math/0104154 | Near a point MATH we have MATH. NAME and NAME 's local description of MATH clearly applies to any line bundle on MATH. Moreover MATH is MATH-equivariant, and thus passes to MATH, as well as inducing the usual MATH operator on the coarse moduli MATH. |
math/0104157 | Let MATH. On a neighbourhood MATH of MATH, there exists a contact form MATH, and REF-forms MATH such that MATH. Pull these forms back to MATH; we will shrink MATH when necessary. Since MATH both restrict to be zero along the fibres of MATH, they cannot be independent modulo MATH. Therefore we can arrange, by adding multiples of MATH, that MATH. (Note that now MATH is no longer the pullback of a form on MATH.). Since MATH, then MATH. Since MATH, then MATH. Since MATH span an integrable system, then there will be a smooth REF-form MATH on MATH that MATH. (Since MATH is one-dimensional at each point, MATH is nonzero on MATH.) By adding multiples of MATH, we can arrange that MATH, giving REF . Because MATH, there must be a nonzero REF-form MATH on MATH such that MATH. We can similarly arrange that MATH. |
math/0104157 | The structure REF enable us to apply the NAME normal form theorem CITE to system MATH. This gives MATH, in terms of the forms defined here. Since MATH, then MATH, and so there exists some function MATH such that MATH along the paths in MATH. |
math/0104157 | Structure REF show that we may first pass to the sub-bundle where MATH and then move along the fibres in a direction dual to MATH to pass to the sub-bundle where MATH. Once there, these equations show that MATH and MATH restrict to have the above form. Of course, REF shows that MATH. |
math/0104157 | Let MATH be any vector field on MATH tangent to the fibres of the projection MATH. Since MATH is annihilated by MATH and MATH, MATH . |
math/0104157 | Let MATH be the quotient of MATH by the foliation by integral curves of the system MATH. Each contact curve in MATH has a unique lift to MATH as an integral curve of MATH. Clearly, arclength is measured along these lifts by the integral of MATH modulo MATH. However, the form MATH on MATH does not descend to be well-defined on MATH, as shown by MATH . However, computing MATH shows that MATH and this, together with MATH, shows that REF-form MATH is well-defined on MATH. Now arclength with respect to the metric may be measured on the integral curves of MATH by the Lagrangian MATH. We will apply the NAME formalism CITE to investigate which of these are geodesics for MATH. Then, we will try to find conditions under which these curves coincide with the projections of the paths in MATH under MATH. Let MATH on MATH. Then one finds that the two-form MATH is of full rank on MATH, except where MATH. Accordingly, let MATH on MATH. Now one computes that MATH . Let MATH be the rank four Pfaffian system on MATH spanned by the four one-forms on the right in REF: MATH . According to the NAME formalism, integral curves of MATH project to be extremal curves for MATH on MATH. These coincide with the projections of the paths in MATH if and only if, in a neighbourhood MATH of each point of MATH, there is a local diffeomorphism MATH such that MATH coincides with MATH, the Pfaffian system on MATH which defines the paths. (The diffeomorphism would follow from the identification of paths with geodesics on MATH.) The form MATH belongs in MATH if and only if MATH. Then, by REF, MATH showing that MATH if and only if MATH. |
math/0104158 | The homomorphism MATH has inverse MATH where MATH is a finitely generated MATH-module such that MATH is free. The definition of the inverse does not depend on the choice of MATH and plainly MATH so we need only check that the `exact sequence relations' REF are respected. Suppose we are given an exact sequence REF . Choose finitely generated MATH-modules MATH and MATH such that MATH and MATH are free. Then MATH is free and there is an exact sequence of endomorphisms: MATH . |
math/0104158 | CASE: Let MATH denote the augmentation map MATH. We shall prove that the sequence MATH is split exact. The composite MATH is the identity map so MATH is surjective and split. We have only to show that an element MATH of MATH which becomes zero in MATH can be written MATH for some MATH. We may certainly write MATH with MATH for each MATH and with MATH invertible. Now MATH so MATH where MATH. Since the diagonal entries of MATH are invertible and all other entries are in MATH, we can reduce MATH by elementary row operations to a diagonal matrix with entries in MATH. Thus MATH where MATH is the product of the diagonal entries. CASE: Taking determinants gives a homomorphism to the group of units MATH . Every element of MATH can be written in the form MATH so the restriction of MATH to MATH is inverse to the canonical map MATH. |
math/0104158 | We aim to show MATH. In MATH we have MATH so MATH . |
math/0104158 | We define a family of rings MATH . There is an obvious surjection MATH for each MATH and MATH is the direct limit of the system MATH . By REF we have MATH so it suffices to prove that for each MATH . We shall see that MATH is sensitive enough to distinguish these two endomorphism classes. Indeed, MATH and in particular the coefficient of MATH is MATH where in each of the `other terms' two or more occurrences of MATH intersperse MATH copies of MATH. Since MATH for all MATH, one may perform a cyclic permutation of the letters in each term to obtain MATH the `other terms' disappearing by the defining relations MATH of MATH. Now the coefficient of MATH in MATH is MATH so it remains to prove that MATH in MATH. We shall argue by contradiction. Let MATH denote the alphabet MATH. If MATH then there is an equation in MATH: MATH where MATH for MATH and MATH are such that MATH for MATH as in REF above. Let MATH denote the MATH-module generated by the cyclic permutations of MATH and let MATH be the MATH-module generated by all other words in MATH . Each basic commutator MATH is either in MATH or in MATH and MATH so by REF MATH where MATH. We have reached a contradiction (for example put MATH) and the proof of REF is complete. |
math/0104158 | The `only if' part is easy. Conversely, suppose MATH is invertible and MATH as above. The equation MATH implies that MATH is invertible if and only if MATH is invertible. But applying MATH to REF we learn that MATH is invertible and hence that MATH. Thus MATH and MATH are invertible over MATH. |
math/0104158 | To prove MATH, we need only note that MATH by REF above. Conversely, to prove that MATH it suffices to show that every matrix MATH has an inverse with entries in MATH so that there is a commutative diagram MATH . Recall that MATH is the augmentation given by MATH for all MATH. Multiplying MATH by MATH if necessary we can assume that MATH. Each diagonal entry of MATH has an inverse in MATH so, after elementary row operations (which are of course invertible), MATH becomes a diagonal matrix where each diagonal entry MATH has MATH (compare the proof of REF above). By the definition of MATH, each MATH has an inverse in MATH. |
math/0104158 | It suffices to show that MATH has the form MATH where MATH and MATH have entries in MATH and MATH is a linear matrix MATH with MATH invertible. For then MATH. Note first that if MATH and MATH then MATH whenever the left-hand sides make sense (compare CITE). Hence, we need only treat the cases where REF MATH has entries in MATH and REF MATH with MATH. If MATH has entries in MATH then by repeated application of the equation MATH in which MATH, MATH, MATH and MATH denote matrices, some stabilisation MATH can be expressed as a product of linear matrices. Each linear matrix MATH can be written MATH and REF imply that MATH is of the required form MATH. The case MATH is similar but slightly easier; we repeatedly apply REF to express (a stabilisation of) MATH as a product of inverses of linear matrices in MATH and then apply REF . |
math/0104158 | Observe first that MATH is isomorphic to the NAME localization MATH inverting the MATH matrices MATH. Since MATH the homomorphism MATH factors uniquely through MATH as indicated by the broken arrow MATH in the commutative diagram above. Explicitly, MATH. Now if MATH then MATH is invertible so by REF MATH is invertible. Thus MATH induces a map MATH which, by the universal properties of MATH and MATH, is inverse to MATH. |
math/0104158 | One can check that the canonical map MATH is universal among MATH-inverting homomorphisms. The details are left to the reader. |
math/0104158 | The canonical maps MATH induce maps MATH satisfying MATH for MATH. We aim to prove that any other system of maps MATH with MATH for MATH factors uniquely through MATH: MATH . Suppose MATH is a generator of MATH where MATH. MATH is the image MATH of some matrix MATH so we can define MATH. To show MATH is well-defined there are two things to check: CASE: If MATH is an alternative choice with MATH then we require MATH. Indeed, there exists MATH such that MATH, MATH and MATH. Hence MATH. CASE: We must check that MATH respects the defining relations of MATH. CASE: A matrix MATH is the image of some matrix MATH so MATH. CASE: Suppose MATH for some invertible matrix MATH. For large enough MATH we can choose MATH to represent MATH and MATH respectively. Since MATH there exists MATH such that MATH where MATH and MATH. Thus MATH and MATH. CASE: If MATH is the zero matrix, MATH. Uniqueness of MATH follows from the fact that every class MATH in MATH is an image of a class MATH. |
math/0104160 | In order to simplify the proof, we assume MATH for MATH. Let MATH be a set of representatives of MATH containing MATH. First, we assume MATH for MATH. By a conjugation of MATH, we assume MATH for MATH. Namely, if MATH for MATH, we set MATH such that MATH for MATH and MATH for MATH. It is easy to see that MATH for MATH. Note that MATH, and for MATH, MATH is generated by MATH as a MATH-module. Since MATH is MATH-module automorphism and MATH-cocycle is a MATH-bilinear map, we have MATH for MATH by the direct calculation. Next, we assume MATH for MATH such that MATH is the identity map. For MATH, we set MATH, where MATH. Then we have MATH for MATH. Since MATH and MATH on MATH, we have MATH for MATH. |
math/0104160 | By the definition of MATH, MATH fixes any element of MATH. In particular, MATH fixes the NAME element and the vacuum vector. Note that the fusion rules of tensor products of modules are the tensor products of the fusion rules of those modules. Let MATH. Let MATH: MATH be a map such that MATH for MATH. Let MATH, MATH be elements of MATH. By the definition of fusion rules, we have MATH for MATH, MATH, where MATH is the vertex operator of MATH and MATH is fusion rules for MATH. Therefore we have MATH. |
math/0104160 | By REF , MATH acts by MATH on MATH and acts by MATH on MATH. By CITE, the centralizer of MATH in the Monster simple group is MATH. Since MATH commutes with MATH, we have MATH. Since MATH preserves the space MATH for any MATH, we have MATH. It is well known that there are the four types of Monster elements MATH, MATH, MATH and MATH contained in MATH. Since MATH is an order MATH element, MATH is a MATH element of the Monster. |
math/0104160 | By direct calculation, we have REF . Since MATH contains the all MATH element, the numbers of elements of MATH whose MATH-th coordinate is MATH, and whose MATH-th coordinate is MATH are equal. By the definitions of MATH, we have MATH respectively. Therefore we have REF . |
math/0104162 | In the exceptional groups, we have verified this directly. In the classical groups, we must show that for any MATH and MATH with MATH that we have MATH where the MATH-value is computed with respect to the appropriate subalgebras for MATH, MATH, and MATH. In type MATH, MATH. For any MATH with MATH, we must use both the MATH-value formula in type MATH for MATH and in type MATH for MATH. It is clear that MATH minus various MATH. It turns out that that for each MATH either MATH or MATH will be subtracted in the formula, but not both, as we now show. There are four possibilities for each MATH. If both MATH and MATH belong to MATH or both belong to MATH, then MATH will have a term MATH or MATH, but not both, since the formulas for MATH and MATH select every other MATH to subtract. If on the other hand, MATH belongs to MATH and MATH belongs to MATH, then again MATH will have a term MATH or MATH, but not both. This is because the parity of the position of MATH in MATH will be the same as the parity of the position of MATH in MATH (as the position of MATH in MATH is an even number, namely MATH). But the formula for MATH and MATH choose to subtract parts whose positions have opposite parity. Hence, only one can be selected in the formula for MATH. The result is the same if MATH belongs to MATH and MATH belongs to MATH. Thus in either of the four cases, the effect is to subtract a number which is less than or equal to MATH since MATH, hence the inequality MATH. In types MATH and MATH, a similar argument holds, except we look at the consecutive parts MATH and MATH of MATH (the part MATH will never be subtracted in the formulas for the MATH-value). One or the other, but not both, of these parts will be subtracted in the formula for MATH demonstrating the inequality. |
math/0104162 | We are assuming that MATH is of adjoint type (the result is still valid for any MATH). In the exceptional groups we verified the results directly using the tables for MATH-values and NAME representations in CITE, the tables of conjugacy classes in MATH in CITE, and knowledge of the MATH-values for distinguished orbits given above. In the classical groups we need to do some work in order to understand which local systems appear in the set MATH. We illustrate the situation in type MATH. Let MATH have partition MATH. As before, MATH is the number of parts of MATH of size MATH. So MATH is even whenever MATH is even as we are in type MATH. We associate to MATH a symbol as in REF. It consists of MATH elements. Let MATH. Each odd MATH contributes to the symbol the interval of length MATH and each even MATH contributes to the symbol the MATH numbers MATH each repeated twice in weakly increasing order. Consider the elementary MATH-group with basis consisting of elements MATH, one for each odd MATH with MATH. Then MATH is the subgroup of this group consisting of elements expressible as a sum of an even number of basis elements (this is because we are working in the special orthogonal group and not in the full orthogonal group). Representations MATH of MATH are thus specified by their values (of MATH) on the MATH's. The representations MATH for which MATH is non-zero are those with the property that MATH . If MATH is even, MATH may have the value MATH or MATH on MATH. To determine the set MATH it is necessary to compute the MATH-value of each non-zero representation MATH (see REF). We have to convert between two different notations for symbols. This is a bit of a pain, but the work is greatly simplified since we are only interested in when the MATH-value of MATH equals the MATH-value of MATH. We find that the set MATH consists of the following MATH. Above we listed those odd MATH with MATH odd (which was denoted MATH above) in decreasing order as MATH (note that MATH must be odd). If MATH, we must have MATH and MATH. Let MATH be odd and MATH even. If MATH for some MATH, then MATH may take values MATH on MATH. If, on the other hand, MATH, then MATH. Since MATH is abelian, each element forms its own conjugacy class. We need to relate our two descriptions of conjugacy classes in MATH. Given MATH write MATH where MATH (the usual classical description). Then set MATH and define MATH by MATH. Then MATH exactly corresponds to the conjugacy class of MATH (see CITE). Using our previous notation of MATH, MATH, and MATH, we see that MATH consists of those conjugacy classes where MATH when MATH is even and MATH is even when MATH is odd. By REF this is exactly the condition that the conjugacy class MATH belongs to MATH, showing that MATH. Finally, the statement that MATH is constant on the cosets on MATH is an easy computation similar to the one done to find which classes belonged to MATH. A similar proof holds in type MATH and MATH which we omit. |
math/0104162 | We show that the representation MATH constructed above is independent of MATH (it is clearly independent of MATH, since NAME representations only depend on the orbit through MATH). One of REF duality is that if MATH where MATH is a NAME subalgebra of MATH, then MATH . The notation on the right-side is NAME induction. According to CITE and the validity of REF , one therefore has that MATH where MATH is the NAME group of MATH. Now let MATH be a maximal torus in MATH. Then MATH is a NAME subalgebra of MATH and MATH. Moreover MATH is of the form MATH for a semisimple element MATH and the image of MATH in MATH necessarily belongs to MATH. By the transitivity of truncated induction applied to the sequence MATH, we see that the representation MATH is the same whether we work with respect to MATH or MATH, that is, whether we work with MATH or MATH. The main result of CITE is that the pair MATH (up to simultaneous conjugation by elements in MATH) is determined by (and determines in the case when MATH is of adjoint type) the conjugacy class MATH. In other words, MATH (up to simultaneous conjugation by elements in MATH) is independent of the choice of MATH and thus the construction of MATH is independent of the choices made for MATH and MATH. |
math/0104162 | This is shown by following NAME 's version of Shoji's algorithm (see REF). |
math/0104162 | We give the proof in type MATH, the other types being similar. Assume MATH, but MATH. List all the even parts of MATH in decreasing order as MATH. Then there exists a MATH such that MATH since MATH. Let MATH be the partition obtained from MATH by replacing the part MATH by MATH and the part MATH by MATH and leaving all other parts of MATH unchanged. This is the basic MATH-collapsing move (see CITE) and it suffices to show that MATH. The dual partition MATH equals MATH except that MATH and MATH. Now write MATH where MATH and MATH consist respectively of the odd and even parts of MATH. Given a part MATH of MATH, recall that we are calling MATH the position of MATH in MATH. Now MATH will occur as a part of either MATH or MATH. We refer to its position in whichever partition MATH or MATH it occurs in as its parity position. We use similar language for MATH. Assume first that MATH is even. Then it is possible to show that the parity position of MATH is odd and the parity position of MATH is also odd. These parts will each contribute a root subsystem of type MATH in the definition of MATH or MATH, respectively. On the other hand, if MATH is odd the parity position of these parts is both even and they will each contribute a root subsystem of type MATH in the definition of MATH or MATH, respectively. Since there are no even parts of MATH between MATH and MATH, we have MATH for MATH. Since the same equalities hold for MATH, we see that these parts (which come in pairs) contribute the same terms to MATH and MATH. Finally consider MATH. If MATH is even, then the parity position of both MATH and MATH is even, so they contribute a term MATH in the definition of MATH or MATH, respectively. And if MATH is odd, then they both have odd parity position and their contribution is MATH. Hence a basic collapsing move does not affect the attached representation and the result is proved. |
math/0104162 | We prove the first isomorphism. We noted above that MATH and so in type MATH, MATH since by the previous proposition we can omit the MATH-collapse on the right side. To prove the desired identity we must study the odd and even parts of MATH (in type MATH) and MATH (in type MATH). These partitions are the same except that the latter has an extra part equal to MATH at the end. Now because MATH the definition of the subsystem in REF for MATH in type MATH and for MATH in type MATH coincide (with the extra part in MATH playing no role at all). We now prove the second isomorphism. The first isomorphism implies that MATH, where the left side is in type MATH and the right in type MATH. Since MATH, we have MATH and thus MATH. The last equality holds since in type MATH we can omit the MATH-collapse. |
math/0104162 | We may choose MATH representing MATH so that MATH has semisimple rank equal to the rank of MATH. Then MATH is specified by the pair of partitions MATH. Our first step is to compute the NAME representation MATH of MATH associated to MATH. The pair of partitions associated to MATH is MATH . By REF , the associated NAME representation MATH of MATH is MATH. Consider this now as a representation of MATH in type MATH, MATH in type MATH, and MATH in type MATH (there is no change in type MATH). Then by applying REF in types MATH and MATH and REF again in type MATH, this representation can be described as MATH . These representations possess REF as they possess REF in each simple component. Hence we can apply truncated induction up to MATH. Then by transitivity of induction we claim that we arrive at the representation MATH where the first is a representation of MATH, the second of MATH, and the third of MATH. In type MATH, we use the fact that MATH belongs to MATH. Therefore in all types if the multiplicity of MATH in MATH is odd, then the multiplicity of MATH in MATH or MATH, respectively, is even. Then the validity of REF follows from the definition of MATH in REF . The proof is completed by applying REF in MATH. |
math/0104162 | We verified this case-by-case (NAME already did this in his work with his original map although the details are not recorded). In fact, we will try to exhibit canonical elements MATH of MATH which map bijectively to MATH. These are denoted by a star REF in the tables for the exceptional groups and we now explain their construction in the classical groups. Assume MATH is of type MATH where MATH and MATH where MATH is even or odd depending on MATH. Consider MATH. We ask whether MATH belongs to MATH, or MATH depending on whether MATH is MATH, MATH, or MATH, respectively. In other words, we ask whether MATH is special (note the funny situation in type MATH). If not, we may uniquely write MATH where MATH is distinguished of type MATH, MATH, or MATH, and where MATH belongs to MATH, or MATH, for some MATH, depending on whether MATH is MATH, MATH, or MATH. We now show that MATH has the property that MATH. This is because MATH and the process of forming MATH amounts to taking MATH and diminishing some of its parts; however, parts not congruent to MATH will only be diminished by an even number, so the resulting partition remains of the same type as MATH. Hence MATH for some MATH. It follows that MATH since MATH belonged to MATH, or MATH depending on MATH. It is also true that MATH is itself special in types MATH and MATH. We can now define MATH. In type MATH we set MATH; in type MATH we set MATH; and in type MATH we set MATH. Now because MATH is special (and MATH is special in types MATH and MATH), we have MATH in type MATH; MATH in type MATH; and MATH in type MATH. The second equality in type MATH follows since applying NAME duality twice (to MATH in this case) is the identity on special orbits. Thus in all types MATH where the last equality holds since MATH. We conclude in all types that MATH has the property that MATH. |
math/0104162 | Again we checked this on a case-by-case basis. In the exceptional groups, this amounts to a quick glance at the tables which follow. In the classical groups, it requires attention to the computations in the proof of REF , together with the explicit description of the canonical quotient. We omit the details. |
math/0104166 | REF is proved in CITE. An alternative proof (in the finite generation case) is given in CITE. To see REF , fix arbitrarily a unimodular element MATH such that MATH meets MATH in its relative interior. Since MATH is normal we have MATH. Complete MATH to a free basis MATH of MATH. If a natural number MATH is big enough then all the rays MATH intersect MATH in its interior. This is so because the radial directions of the elements MATH approximate the direction of MATH as MATH. Therefore, for MATH big enough we get the desired basis MATH. REF follows similarly, using the following observation: for any element MATH and MATH there exists a natural number MATH such that MATH, yielding the inclusion MATH. REF is proved in CITE. NAME suggested the following alternative short proof based on REF . By REF we can additionally assume that MATH is normal. Consider the dual cone MATH in the dual space MATH. It is again a pointed convex polyhedral cone of the same dimension MATH. By REF there is a free monoid MATH such that MATH. Then we have the induced embedding MATH where MATH. |
math/0104166 | Fix any representation MATH. By REF there is a basis MATH of the group MATH such that MATH. For a natural number MATH consider the monoid MATH the intersection being considered in MATH. Using the fact that the radial directions of the elements MATH approximate that of MATH when MATH we obtain the desired filtered union representation MATH . |
math/0104166 | Assume MATH for some MATH. Since REF generate the additive monoid MATH there exists MATH such that MATH . But then MATH . |
math/0104166 | Assume MATH for some MATH. Fix any element MATH. There exists MATH such that the element MATH satisfies the condition (in the multiplicative notation): MATH . By REF we have MATH. Now REF applies because of the equalities in MATH: MATH . |
math/0104166 | Since any monoid is a filtered union of finitely generated monoids and MATH-functors commute with filtered inductive limits Conjecture REF reduces to the case of finitely generated monoids. We want to show that MATH assuming the equalities as in the lemma, where MATH is any finitely generated monoid with trivial MATH. By REF (the latter being applied to the extension MATH) one has MATH. By REF the extension MATH yields the natural isomorphisms MATH. Put MATH and MATH. Let MATH be any facet (codimension REF face). Then the ideal MATH is an ideal of the bigger ring MATH as well. Using the same arguments as above we arrive at the equalities MATH where MATH. Next we do the same reduction with respect to another facet of the polytope MATH, and so on. By the same token we get a sequence of rings MATH such that MATH for the corresponding enumeration of the facets MATH, MATH where MATH. Thereafter we handle the codimension REF faces of MATH, and so on. Finally, by annihilating all non-trivial monomials, we shall descend to the coefficient ring MATH. (When we treat MATH-dimensional faces, i. e. the vertices of MATH, we follow the convention - an interior of a point is the point itself.) Finally, we obtain a sequence of rings MATH REF such that MATH. |
math/0104166 | CASE: It follows easily from the results in CITE that for a ring extension MATH, such that MATH the homomorphism MATH is injective. Let MATH be a natural number. Consider the endomorphism MATH. We fix a basis MATH of MATH which is a subset of MATH. Then MATH is a free MATH-module of rank MATH. In particular, one easily concludes (using transfer maps for MATH) that any element MATH, which is mapped to zero in MATH, is of MATH-torsion. By the infinite iteration, involving periodically all the elements MATH, we conclude that every element of MATH where MATH, is annihilated by some positive power of MATH. In view of the equality MATH and REF 's results, the `interior excision' arguments (as in the proof of REF ) for MATH-groups show the implication CASE: (MATH acts nilpotently on MATH) MATH (MATH is of MATH-torsion). We are done because the same is true for another natural number MATH, coprime to MATH. CASE: Since for seminormal monoids MATH it is well known that MATH provided MATH is a seminormal domain, the claim follows from REF . CASE: One only needs to observe that the limit ring MATH is a ``face" ring of the type we had in the proof of REF . This is so because the limit contains no non-zero nilpotent elements, corresponding to monomials from MATH. Therefore, the similar arguments as in that proof apply. |
math/0104166 | Consider a maximal ideal MATH. We claim that there are only two possibilities: either MATH or MATH. In fact, if MATH then for any MATH there exists a natural number MATH such that MATH (as in the proof of REF ). Therefore, MATH which yields MATH. Since the only maximal ideal of MATH containing MATH is MATH the claim follows. Pick MATH. We want to show that MATH is an isomorphism. This can be checked locally on MATH. At MATH both the source and the target localize to MATH. If MATH is another maximal ideal then we know that MATH. Therefore, MATH is a further localization of the identity mapping MATH. |
math/0104166 | First we observe that for any ring MATH and its two comaximal elements MATH . REF implies the long exact sequence MATH . To prove the lemma it is enough to show that MATH is injective. Let MATH. We write MATH, where MATH and MATH. We have the equality MATH. Since MATH and MATH are comaximal in MATH the observation above yields the exact sequence MATH . By the Fundamental Theorem CITE we have the embeddings MATH with MATH the intersection of their images. Thus, the exact sequence above implies MATH . Now the proof is completed by REF which says that for any functor MATH with transfers from rings to abelian groups the homomorphisms of the type MATH, MATH monic, are always injective. |
math/0104166 | If MATH is affine then MATH for some affine toric variety MATH, whose polyhedral cone is strictly convex, and some torus MATH. Using the Fundamental Theorem, the desired equality is easily derived in the affine case from Conjecture REF. Next we reduce the general case to the affine case as follows. Let MATH be given by a fan MATH. Pick a maximal cone MATH and let MATH denote the fan determined by the remaining maximal cones. Denote by MATH and MATH the open toric subvarieties of MATH, corresponding to MATH and MATH. REF yields the long exact sequence MATH . By passing to the inductive limits we get the exact sequence (in the self explanatory notation) MATH . MATH is again a toric variety. Moreover, MATH and MATH are open toric subvarieties covering MATH so that the underlying fans are products of the corresponding fans by the same ray. Writing up the corresponding NAME sequences one obtains MATH, provided the same equality holds for MATH and MATH. Iterating the process we will produce toric varieties whose fans contain less and less top dimensional cones. But for single cones we are already done. Conversely, using the same homotopy trick as in REF in the proof REF one can show that the equalities MATH for affine toric varieties MATH, whose cones are pointed, are in fact equivalent to Conjecture REF. |
math/0104166 | That the conjecture implies the claim is clear. Next observe that the mentioned inclusions are equivalent to the surjectivity of the homomorphisms MATH for all pyramidal extensions MATH and all sequences MATH. Let MATH be a finitely generated normal monoid with trivial MATH. Consider any sequence MATH of normal submonoids of MATH, satisfying the condition - for any MATH either MATH is a pyramidal extension or MATH. (Observe that MATH, MATH.) Then our condition yields surjectivity of the homomorphisms MATH. Such a sequence of monoids will be called admissible. According to REF for any rational interior point MATH and its any neighborhood MATH there is an admissible sequence of monoids MATH such that MATH for all sufficiently big indices MATH. By REF there exists a rational simplex MATH such that MATH is a free submonoid of MATH. These observations altogether show that (assuming the pyramidal descent) the homomorphism MATH is surjective. But then MATH is a surjection as well. Since MATH is a filtered union of free monoids NAME 's theorem implies MATH. By REF we are done. |
math/0104166 | Assume MATH is a regular ring. By REF MATH satisfies excision. On the other hand the exact sequence MATH and REF show that MATH for MATH. (Here we use the MATH-homotopy invariance of MATH.) Since all faces of MATH are also simplices the same arguments as in the proof of REF provide the desired process of ``annihilating" the interior monoids. |
math/0104166 | We will only treat the case MATH. The general case makes no difference. CASE: Here we prove that MATH. Consider an object MATH. By REF MATH can be represented by a matrix MATH such that MATH. By NAME 's theorem CITE on vector bundles on projective lines (proved by elementary methods for arbitrary fields in CITE) there are MATH and MATH such that MATH for some MATH. We have MATH . By REF we are done. CASE: We prove that MATH is closed under extensions in MATH for arbitrary MATH, MATH. By the previous step the reduction modulo MATH shows that it is enough to prove the following Claim. Assume we are given a commutative diagram of free MATH-modules MATH with exact rows, where MATH and MATH are defined over MATH and MATH and MATH are defined over MATH. Assume, further, MATH, MATH and MATH are MATH isomorphisms determined correspondingly by matrices of the type MATH . Then MATH. We introduce the following notation: REF the matrices of MATH and MATH (with respect to the standard bases) are correspondingly MATH, MATH, MATH, MATH and MATH, MATH, MATH, MATH, REF the matrices of MATH and MATH are correspondingly MATH, MATH, MATH, MATH and MATH, MATH, MATH, MATH. The commutative diagram above implies MATH and MATH . Fix any index MATH. If there exists MATH such that MATH then REF MATH implies MATH. Now consider the case when MATH for all MATH. Then our diagram splits as follows MATH where MATH are the exact sequences of free MATH-modules, obtained by deleting the MATH-th direct summand in MATH, and the mappings MATH establish an isomorphism between the MATH-sheaves MATH and MATH for some MATH. But in this case MATH. Therefore, in all cases we have MATH. Using REF MATH and dual arguments we get MATH. CASE: By REF we have the filtered union representation MATH . Therefore, if we knew that the embeddings MATH, MATH, induce isomorphisms on MATH-groups we could conclude that the embedding MATH also induces isomorphisms on MATH-groups. Fix a nonpositive integral number MATH and consider the two exact functors MATH given correspondingly by MATH and MATH . The exact ``NAME sequence" (where we use matrix theoretical notation) MATH implies the exact sequence of exact functors MATH, where MATH is the identity embedding and MATH is the identity endofuntor of MATH. By CITE we get MATH for the corresponding MATH-group homomorphisms. It remains to show that MATH, the right hand side denoting the identity endomorphism of the corresponding MATH-group of MATH. But this equality is derived by literally the same arguments once we observe the exact ``NAME sequence" MATH of endofunctors of MATH. CASE: Because of the filtered union representation MATH the previous step shows that we only need that the embeddings MATH, MATH, induce isomorphisms on MATH-groups. Assume MATH. Consider an object MATH and fix a matrix MATH satisfying the conditions in REF (with respect to MATH and MATH). We let MATH be the corresponding numbers, where MATH. Without loss of generality we can assume MATH and MATH for some MATH (none of the values MATH and MATH being excluded). Consider the matrix MATH . We have the following (non-functorial) commutative diagram with exact rows MATH where MATH and the matrices at the vertical rows refer to the corresponding maps. Observe that the matrices at the second and third vertical rows satisfy REF so that we get objects of polarization MATH (one uses that MATH). Therefore, any object MATH admits a ``co-resolution" MATH whose second and third terms belong to MATH. In other words, for the dual categories MATH any object MATH admits a resolution MATH where MATH. Since MATH-theories of dual exact categories are the same it suffices to show that the embedding MATH induces isomorphisms on MATH-groups. In view of what has been said above we see that Resolution Theorem CITE applies once we show the implication: if MATH is exact in MATH and MATH then MATH. Returning to the original categories, we claim that for any exact sequence MATH in MATH the following implication holds MATH . Introducing the notation as in Claim in REF this implication rewrites as follows: MATH . First observe that for any MATH there exists MATH such that MATH. In fact, assuming to the contrary that such MATH does not exist, the mapping MATH (and, therefore, MATH too) would not be surjective, which is excluded. By REF MATH we have MATH . In particular, MATH. This completes the proof. |
math/0104166 | Let MATH denote the image of MATH under the homomorphism MATH, induced by MATH. Changing MATH by MATH we can without loss of generality assume MATH. Denote by MATH the image of MATH in MATH. Then MATH with respect to the natural augmentation MATH. CASE: By REF (and continuity of algebraic MATH-functors) there are six polarized monoids MATH,MATH,MATH such that CASE: MATH and MATH, CASE: MATH, for MATH. CASE: MATH. We will assume that MATH is the image of MATH. Next we show that one can also assume CASE: MATH. Since MATH are finitely generated monoids there is an index MATH such that MATH for any natural number MATH. Denote these six polarized monoids respectively by MATH, MATH,MATH. They are naturally isomorphic to MATH. Clearly, the same REF - REF are satisfied for them. In particular, if we show that MATH, where MATH, then we achieve the validity of all REF - REF for the isomorphic system of monoids MATH,MATH,MATH with MATH. By REF there is an intermediate monoid MATH, where MATH is a normal submonoid for which MATH is trivial. In view of the inclusions MATH it suffices to show that MATH, By REF this is equivalent to the inclusion MATH. The hypothesis of the theorem implies that the augmentation MATH, MATH induces the isomorphism MATH. Now the composite mapping MATH factors through MATH. Finally, MATH and by REF we are done. CASE: We let MATH denote the non-affine toric variety corresponding to MATH in the sense of REF. By REF (and REF ) we have the long exact sequence MATH . Hence by REF there exists MATH mapping to MATH. For any natural number MATH we have the commutative diagram of exact categories and exact functors MATH where MATH refers to the diagonal embeddings and the number of factors is MATH. Passing to MATH-groups we get the commutative diagram MATH . Therefore, for any natural number MATH there is an element MATH mapping to MATH under the homomorphism MATH. CASE: As in REF there exists an index MATH such that MATH for any natural number MATH. By REF for any natural number MATH there exists MATH such that the element MATH is in MATH. In particular, by REF we have MATH, where the mapping between the MATH-groups is induced by the composite exact functor MATH . Here MATH is the scheme associated with MATH. By REF MATH where the mapping between the MATH-groups is induced by the `upper route' in the commutative diagram of exact categories and exact functors MATH . That the `lower route' in MATH is in fact possible follows from MATH and REF . Here we have used the following notation: MATH and the scheme MATH is defined by the push-out diagram MATH while MATH refers to the full subcategory of those locally free coherent sheaves on MATH whose ranks are non-negative multiples of MATH. The other members of the diagram are defined similarly with respect to the corresponding polarized monoids, and the exact functors indicated are the ones induced by the appropriate scalar extensions, restrictions of vector bundles to open subschemes and category embeddings. CASE: Fix arbitrarily a natural number MATH. We are going to study the lower route in the diagram MATH. The monoids MATH,MATH,MATH are naturally isomorphic to MATH,MATH,MATH so that they satisfy the obvious analogues of REF - REF . In order to simplify the notation we will below omit the subindex MATH. Thus, the mentioned lower route looks as follows MATH . By REF there is an element MATH mapping to MATH. Clearly, MATH where MATH is the same as in REF . Choose a nonzero object MATH. By REF this object up to isomorphism has the form MATH for some MATH, where MATH is the matrix of MATH in the standard basis of MATH. Since MATH the injective stabilization estimate for MATH CITE CITE implies MATH. It is also clear that MATH. Therefore, by REF there exist MATH and MATH such that MATH, the equality being considered in MATH. We have MATH, where by REF (namely, MATH) the inclusion MATH holds. Therefore, by REF we arrive at Claim A below. First one notation. For MATH we put MATH and MATH . Claim A. Any object of MATH is mapped to an object of MATH which up to isomorphism (within MATH) is of the type MATH for some MATH. CASE: Consider any morphism in MATH of the form MATH . We let MATH and MATH denote the matrices of MATH and MATH in the corresponding standard bases. Then we see that there are systems MATH and MATH such that MATH . (Compare with REF MATH and REF MATH.) By REF (namely, MATH) we get the inclusions MATH for all MATH and MATH. Claim B. Any short exact sequence in MATH of the type MATH is mapped to a split short exact sequence in MATH. By REF any such a sequence reduces modulo MATH to a split sequence in MATH. Let MATH denote the exact functor, obtained by reduction modulo MATH, and let MATH be a monomorphism that splits MATH. We have the scalar extension functor MATH. The composite MATH is isomorphic to the identity functor on MATH. We can assume MATH . Claim B is proved once we show that the composite MATH is mapped to an automorphism of MATH under the functor MATH. It is easily observed that the positive and negative components of MATH have correspondingly entries from MATH and MATH. But it then follows from REF (namely, MATH), the inclusion MATH and the equality MATH that both positive and negative components of MATH are defined over the local ring MATH, that is the corresponding matrices with respect to the standard bases have entries from this local ring. Since these matrices are invertible modulo MATH they are invertible themselves. CASE: Let MATH denote the full subcategory whose objects are MATH, MATH (in the selfexplanatory notation). This is an additive category. We equip it with an exact structure with respect to the class of split short sequences - any additive category carries such an exact structure, as shown by the NAME embedding REF in the category of contravariant additive functors with values in abelian groups. By the previous steps we have the natural commutative triangle of exact categories and exact functors MATH . Consider a morphism MATH. For the corresponding matrices we have the equality MATH . Since MATH is polarized (and MATH) we get: CASE: the entries of MATH and MATH belong to MATH, CASE: the entries of MATH belong to MATH, CASE: the entries of MATH belong to MATH. Conversely, any matrix over MATH of size MATH, whose entries satisfy the three conditions above, is a positive component of a unique morphism in MATH. Consider the subring MATH where CASE: MATH, CASE: MATH, CASE: MATH. Consider the full subcategory MATH of objects of the type MATH. By the mentioned description of morphisms in MATH we have the natural equivalence of categories MATH - the category of left free MATH-modules (after thinking of elements of MATH as the corresponding endomorphisms of MATH). Clearly, MATH is a cofinal subcategory of the exact category MATH in which all exact sequences split by definition. Therefore, by CITE we have MATH. (Recall MATH; for the NAME groups we have MATH and MATH.) We have the exact functor MATH given by picking up the ``positive" part in the objects and morphisms in MATH. In other words, we mean the composite functor MATH, where the third functor is given by restriction to an open subscheme and the fourth functor corresponds to a scalar extension. CASE: Consider the two subrings MATH and MATH of the matrix ring MATH. Put MATH. We regard MATH a central multiplicative set in MATH through the diagonal embedding. It is easily observed that MATH. (For an essentially equivalent equality see CITE.) The inclusion MATH implies MATH. Therefore, we have the commutative diagram (with obvious morphisms) MATH . By REF the boundary of MATH, which consists of MATH-algebra homomorphisms, is a NAME square. Hence the long exact sequence MATH . By REF we have the natural commutative diagram MATH . Also, using MATH, we have the homomorphisms with the same composite MATH . Therefore, using the element MATH (see REF ) and the sequence MATH, we see that MATH for MATH, i. CASE: MATH. CASE: For a natural number MATH put MATH. Clearly, MATH for MATH. Then REF implies MATH for MATH. It is easily observed that all the polarized monoids that have shown up above work equally well for the elements MATH for any natural number MATH. Therefore, for all MATH and all MATH we have MATH. By REF (namely, MATH and MATH) we have the inclusion MATH . Therefore, the polarized monoid MATH is the desired one for MATH. |
math/0104166 | MATH is a nilpotent ideal. So by NAME 's result CITE we have the isomorphism of abelian groups MATH, the right hand side being a finite dimensional MATH-vector space. |
math/0104166 | Since MATH we have MATH for MATH and this yields the desired inclusion. |
math/0104166 | We can gradually approximate MATH by appropriate pyramidal rational cones MATH, MATH such that MATH. Since MATH is a dense subset, we can also keep the condition MATH satisfied, where MATH is the facet `close' to MATH and MATH is the extremal ray of MATH not in MATH. Then MATH is the desired bipyramidal cone for MATH whose desired decomposition into pyramidal cones is MATH . |
math/0104170 | By the assumption that MATH we have that MATH for all MATH. Using the projection formula and the NAME spectral sequence we have MATH . This proves the first part. For every stable vector bundle MATH of rank MATH and degree MATH, MATH . Consequently, the projection formula gives MATH for MATH, and we have by relative NAME duality MATH . Finally, using the projection formula and the NAME spectral sequence, it follows that MATH . MATH for MATH. Thus, MATH as asserted in the proposition. |
math/0104170 | Note that for any MATH . From CITE and CITE we know that for generic MATH, the two vector bundles MATH and MATH are non-isomorphic and stable. Hence MATH . This implies that MATH . So REF gives MATH and the proof is complete. |
math/0104170 | Combining REF and the second part of REF it follows that MATH . From REF it follows immediately that MATH and hence the proof is complete. |
math/0104170 | Since MATH, we have MATH . Consequently, MATH. |
math/0104170 | Let MATH. From REF we obtain an isomorphism MATH . From REF we have MATH . Combining this observation with REF we get MATH . |
math/0104170 | CASE: The fact that MATH is NAME is standard (see CITE). The bundle MATH of rank MATH and degree MATH fails to be MATH-stable if and only if it has a subbundle MATH of rank MATH and degree MATH such that MATH, that is, MATH . By considering the extensions MATH we can estimate the codimension of MATH and show that it is at least MATH (compare the proof of CITE). Note that, since MATH, REF implies that MATH. Given that MATH, we see that MATH only if MATH, MATH or MATH, MATH. These are exactly the cases that were excluded in the introduction. CASE: MATH consists of all pairs MATH for which the bundle MATH in REF is MATH-stable. As in REF , this is a NAME subset. It follows at once from REF that, if MATH is MATH-stable, then MATH is stable. So, if MATH, it follows from REF that MATH can be identified with the projective space MATH. Using the universal projective bundle on MATH, we see that MATH is a projective fibration over MATH (not necessarily locally trivial). Suppose now that MATH belongs to the complement of MATH in MATH. This means that the bundle MATH in REF is not MATH-stable and therefore possesses a subbundle MATH satisfying REF. If MATH, this contradicts the stability of MATH. So there exists an exact sequence MATH with MATH a subbundle of MATH of rank MATH and degree MATH. Moreover, since MATH is a subbundle of MATH, MATH must contain the line MATH. For fixed MATH, these conditions determine a subvariety of MATH of dimension at most MATH . Since MATH, a simple calculation shows that the codimension is at least the number MATH given by REF. As in REF , this gives the required result. |
math/0104170 | Restricting REF to MATH gives MATH . This must coincide with the universal sequence REF up to tensoring by some line bundle lifted from MATH. The result follows. |
math/0104170 | It is sufficient to show that MATH has trivial cohomology. By REF MATH and the result follows. |
math/0104170 | This follows at once from REF . |
math/0104170 | The combination of REF yields MATH . Using REF , the NAME spectral sequence for the map MATH gives-MATH . It is known that MATH CITE. Therefore, MATH . |
math/0104170 | As in the proof of REF we conclude that MATH . Now MATH is just the non-singular part of the moduli space of semistable bundles of rank MATH and determinant MATH, and the latter space is complete and normal. So MATH. |
math/0104170 | Identifying MATH with MATH we have MATH . The proposition follows from a result of CITE that says MATH . For MATH the isomorphism is given by the obvious inclusion of MATH in MATH and therefore globalises to give MATH. Similarly for MATH the isomorphism is given by the infinitesimal deformation map of MATH regarded as a family of bundles over MATH parametrised by MATH; this globalises to MATH. |
math/0104170 | By REF MATH is injective, so its image has dimension MATH. On the other hand, by REF , the NAME tangent space of MATH also has dimension MATH at every point of the image of MATH. It follows that MATH is smooth of dimension MATH at every point of the image of MATH. Hence, by NAME 's Main Theorem, MATH is an isomorphism onto an open subset of MATH. Finally MATH is complete and MATH is connected and separated (since MATH is stable), so MATH is an isomorphism. Let MATH be the polarisation on MATH given by the determinant line bundle CITE. Let MATH denote the principal polarisation on MATH defined by a theta divisor. We wish to show that the isomorphism MATH takes MATH to a nonzero constant scalar multiple (independent of the curve MATH) of MATH. Take any family of pairs MATH, where MATH is a connected non-singular projective curve of genus MATH and MATH is a line bundle on MATH of degree MATH, parametrized by a connected space MATH. Consider the corresponding family of moduli spaces MATH (respectively, NAME MATH) over MATH, where MATH runs over the family. Using the map MATH an isomorphism between these two families is obtained. The polarisation MATH (respectively, MATH) defines a constant section of the second direct image over MATH of the constant sheaf MATH over the family. It is known that for the general curve MATH of genus MATH, the NAME group of MATH is MATH. Therefore, for such a curve, MATH takes MATH to a nonzero constant scalar multiple of MATH. Since MATH is connected, if MATH contains a curve with MATH, then MATH takes MATH to the same nonzero constant scalar multiple of MATH for every curve in the family. Since the moduli space of smooth curves of genus MATH is connected, the proof is complete. |
math/0104170 | This follows at once from the theorem. |
math/0104170 | This follows at once from REF and the classical NAME theorem. |
math/0104175 | Let MATH be a prime ideal of MATH such that MATH. To show that MATH, it suffices to show that MATH. By the going-up property for integral extensions, fix prime ideals MATH and MATH of MATH such that MATH, MATH and MATH. To show that MATH, it suffices to show that MATH because then MATH so that MATH and MATH. Suppose that MATH. We first observe that, for every prime MATH of MATH such that MATH, we have MATH. This is due to the fact that, by integrality MATH, and the catenary and equidimensional assumptions imply that MATH and similarly for MATH. In particular, MATH for MATH. Thus MATH . This clearly gives a contradiction. |
math/0104175 | CASE: Pass to the ring MATH to assume that the residue field of MATH is infinite. Let MATH, MATH and MATH. Then MATH is a local ring with maximal ideal MATH. If MATH contains a field, then the same is true of MATH. If the sequence MATH is a reductive system of parameters of MATH, then the same sequence is a reductive system of parameters of MATH since for some MATH, MATH . Multiplicities are preserved by REF at the beginning of this section. That is, MATH, MATH and MATH. Both rings MATH and MATH are analytically unramified by REF . Thus, if the result holds in MATH, then MATH as desired. CASE: Pass to the completion MATH to assume that MATH is complete and equidimensional with infinite residue field. Let MATH be a prime ideal of MATH that is minimal over MATH such that MATH, and similarly for MATH. Since MATH is analytically unramified, MATH. The fact that the extension MATH is flat therefore implies that MATH by REF at the beginning of this section. Similarly, MATH. If MATH contains a field, then the same is true of MATH. If the sequence MATH is a reductive system of parameters of MATH, then the same sequence is a reductive system of parameters of MATH, as in the previous step. If the result holds for MATH, then it holds for MATH, as in the previous step. CASE: Pass to the quotient MATH for a suitably chosen minimal prime MATH, to assume that MATH is a complete domain with infinite residue field, and that either MATH itself contains a field or MATH satisfies REF in the statement of the theorem. To make this reduction, it suffices to verify that there is a minimal prime MATH of MATH contained in MATH such that MATH. First, we show how this gives the desired reduction. Let MATH with prime ideals MATH, MATH and MATH. If MATH contains a field, then MATH contains a field since MATH is a quotient of MATH. If the sequence MATH is a reductive system of parameters of MATH, then the same sequence is a reductive system of parameters of MATH, as in REF . Since MATH is equidimensional, MATH. And since MATH, it follows that MATH and MATH. Therefore, as in REF , we may pass to MATH. Now we prove that such a minimal prime MATH exists. Let MATH. Suppose that MATH for every MATH such that MATH. (This supposition includes the hypothetical possibility that no MATH is contained in MATH.) By REF , MATH for every MATH contained in MATH, and similarly for MATH contained in MATH. The Associativity Formula then implies that MATH . This clearly contradicts the assumption that MATH. Thus, there is a minimal prime MATH of MATH such that MATH and MATH, as claimed. CASE: We prove the result assuming that MATH is a complete domain with infinite residue field. If MATH contains a field, let MATH be a reductive system of parameters of MATH, MATH a coefficient field of MATH, MATH, and MATH. If MATH has mixed characteristic, then by assumption there is a reductive system of parameters MATH of MATH. By the NAME Structure Theorem there is a complete discrete valuation ring MATH contained in MATH such that the induced map on residue fields is an isomorphism. In this case, let MATH and MATH. Note that, in either case MATH is a regular local ring contained in MATH such that the extension MATH is module-finite. The maximal ideal of MATH is MATH and the extension ideal MATH is a reduction of MATH. Furthermore, the induced maps on residue fields is an isomorphism. Therefore, MATH . If MATH and MATH, then REF implies that MATH, and NAME 's Intersection Theorem implies that MATH . This completes the proof. |
math/0104175 | Fix a nonzero element MATH and without loss of generality, assume that MATH and MATH. Let MATH with maximal ideal MATH and prime ideals MATH and MATH. If MATH, then MATH. REF implies that MATH which gives a contradiction. |
math/0104175 | Let MATH and MATH be the prime ideals of MATH such that MATH and MATH, and let MATH. The aforementioned result of NAME implies that MATH is an element of MATH. Passing the the regular local ring MATH we have MATH which implies that MATH is in the ideal MATH, as desired. |
math/0104175 | By passing to the localization MATH, we assume without loss of generality that MATH is regular local with maximal ideal MATH. We prove the result in this case by induction on MATH. When MATH, the result is trivial. When MATH, this is exactly REF . Assume now that MATH. Fix a prime ideal MATH minimal over MATH. Then by NAME 's Intersection Theorem, MATH. Since MATH is MATH-primary, MATH and it follows that MATH. Therefore, by induction MATH as desired. |
math/0104175 | By passing to the completion MATH-of the ring MATH, we may assume without loss of generality that MATH is complete with infinite residue field. The only nontrivial observation is that, if MATH, then the associated graded rings of both MATH and MATH are MATH. If MATH does not divide MATH in MATH, then the same is true in MATH. The remainder of the proof is similar to that of REF . The only difference is that, in the application of REF we must verify that MATH is part of a reductive system of parameters of MATH. Note that the associated graded ring of MATH is MATH. The fact that MATH is infinite and MATH does not divide MATH implies that MATH is part of a homogeneous system of parameters of degree REF for MATH. As noted in the discussion preceding REF , this implies that MATH is part of a reductive system of parameters of MATH. Thus, REF applies. |
math/0104175 | Since MATH is regular, MATH is generated by elements MATH that form part of a regular system of parameters of MATH. Complete this to a regular system of parameters MATH for MATH. The fact that MATH is a regular sequence and MATH is NAME implies that MATH for all MATH. It suffices to show that MATH for all MATH. To see that this is sufficient, observe that, because MATH is generated by part of the regular sequence that generates MATH, MATH even when MATH. The desired sufficiency now follows by induction on MATH. By passing to the ring MATH as in the proof of REF , we may assume without loss of generality that the residue field of MATH is infinite. We prove that MATH by induction on MATH. If MATH or MATH then MATH is principal. Since MATH is prime, it is straightforward to show that, in either of these cases, MATH. Now assume that MATH. Since MATH is infinite, we choose an infinite sequence MATH such that the residues MATH of the MATH modulo MATH are all distinct. Let MATH and for each MATH choose a prime ideal MATH containing MATH and MATH such that MATH. The quotient MATH with prime ideals MATH and MATH satisfies the induction hypothesis. Suppose that MATH is a nonzero element of MATH. Then the residue of MATH in MATH is in MATH so that MATH for every MATH. The associated graded ring of MATH is the polynomial ring MATH where MATH is the initial form of MATH. If MATH, then the fact that MATH is in the ideal MATH for every MATH implies that the initial form MATH of MATH in MATH is divisible by the initial form of MATH. That is, MATH is divisible by MATH for every MATH. Since the MATH are all distinct in MATH, the MATH all generate distinct prime ideals in MATH. Thus, MATH has an infinite number of distinct, irreducible factors, contradicting the fact that MATH is a unique factorization domain. |
math/0104175 | By assumption, MATH and MATH are NAME modules such that MATH. By CITE Corollary to REF, we know that MATH. By REF , MATH and it follows that MATH. |
math/0104175 | By passing to the ring MATH as in our previous arguments, we may assume without loss of generality that MATH has infinite residue field. The ordinary power MATH is a homogeneous ideal, so that its MATH-primary component MATH is also homogeneous. Similarly, MATH is also homogeneous. Therefore, to check the desired inclusion, we need only check homogeneous elements. Let MATH be a homogeneous element of MATH. Write MATH as a product MATH where each MATH is a nonconstant, irreducible, homogeneous polynomial of degree MATH. Because MATH and MATH are homogeneous and MATH, it is straightforward to show that MATH is an element of either MATH or MATH. We assume without loss of generality that MATH. Then MATH. By NAME 's Intersection Theorem applied to the ring MATH, our assumptions imply that MATH can not be in both MATH and MATH. Let MATH and MATH for MATH. Then, MATH, MATH, and each MATH. If each MATH (this is automatic if either MATH or MATH) then MATH as desired. Thus, we may assume without loss of generality that MATH is irreducible and not divisible by MATH. In particular, MATH has positive degree and some monomial term of MATH has unit coefficient. This implies that, in the associated graded ring of the localization MATH, the initial form of MATH has a monomial term with unit coefficient. Therefore, by REF MATH. |
math/0104179 | According to CITE, the polynomial MATH is an isotopy invariant of MATH-curves. Therefore so is MATH. The property `REF' is a consequence of the fact that MATH is a multiple of MATH for any MATH-curve diagram MATH, which is not hard to see. The property `REF' follows from the connected sum formula MATH for any MATH-curve diagrams MATH and MATH CITE. Finally, the property `REF' is a consequence of the fact MATH CITE. |
math/0104179 | Let MATH be a MATH-curve diagram. Suppose MATH, MATH, MATH and MATH are the diagrams with one crossing of MATH, say MATH, replaced by the local diagrams of REF , respectively. Then we have MATH where MATH and MATH . In the following computations, we omit the subscript MATH from MATH and MATH. Using REF , we obtain MATH . Suppose MATH, MATH, MATH and MATH are the diagrams obtained from MATH by a repeated application of the skein relation REF , which are identical except at one place where they differ as indicated in REF . The local orientations of MATH and MATH shown in REF are induced from the orientation of MATH in which all edges are oriented from one vertex to the other. Using REF again, we have MATH for any nonzero integer MATH. For a set MATH of MATH distinct crossings of MATH, we consider the sum MATH . For MATH, let MATH and MATH. For a MATH-sign MATH, let MATH, MATH, MATH and MATH be the MATH-signs obtained from MATH by appending MATH, MATH, MATH and MATH, respectively. Let MATH and MATH. For each MATH-sign MATH, with MATH, we define MATH for MATH. Applying REF to the right hand side of REF , we obtain MATH . If MATH, then MATH where MATH means a duplication of MATH's so that the length of the whole `sign' is MATH. Using REF with a MATH-sign MATH, we ontain MATH . Repeated applications of this to REF lead us to MATH . This shows that MATH is divisible by MATH, since each MATH is divisible by MATH. Therefore, after the substitution MATH, the power series MATH has no terms of degree less than MATH. This proves the first part of the theorem. Let MATH be the coefficient of MATH in MATH for a MATH-curve MATH. To show that MATH is not finite-type for any integer MATH, we use the MATH-curves MATH with MATH and MATH shown in REF . Since MATH is trivial, we have MATH. Inductively, we can compute the maximal and the minimal degrees of MATH. For MATH, they are MATH and MATH, respectively. For MATH, let MATH be a set of MATH distinct crossings of MATH. The following computation shows that MATH is not finite-type. MATH . Let MATH. Since the minimal degree of MATH is MATH, that of MATH is MATH. For a positive integer MATH, let MATH be a connect sum of MATH copies of MATH. Since the minimal degree of MATH is MATH, we see that MATH if and only if MATH. This leads us to MATH for any MATH. Considering the mirror images and the maximal degrees, we also obtain MATH for any MATH. This proves the second part of the theorem. |
math/0104179 | The first part can be proven in a similar way to that of REF , using the skein relation REF . It is easily seen that MATH has only even degree terms. The maximal degree and the minimal degree of MATH are MATH and MATH, respectively, for MATH, and those of MATH are MATH and MATH, respectively. The second part can be proven in a similar way to that of REF , using MATH instead of MATH. |
math/0104182 | A MATH-manifold MATH arises by identifying points on MATH if and only if there is a torsion free subgroup MATH of MATH with fundamental region MATH and MATH isometric to the quotient MATH. Such a MATH (which is isomorphic to MATH) may be replaced by any of its conjugates in MATH, as these will yield quotients isometric to MATH. Conjugacy classes of subgroups of MATH of index MATH correspond to transitive actions of MATH on MATH, the subgroups arising as the stabilisers of points. These actions in turn correspond to homomorphisms MATH with transitive image. A subgroup MATH is torsion free if and only if it intersects trivially the conjugacy class in MATH of each MATH. This happens precisely when MATH has no fixed points among MATH. Finally, MATH forms a fundamental region for the action of MATH on MATH exactly when MATH forms a transversal (a non-redundant list of coset representatives) for MATH in MATH. Equivalently, MATH and MATH. The first follows immediatly as MATH is torsion free, and the second, since MATH is a subgroup, when the index of MATH in MATH is equal to the order of MATH. |
math/0104184 | Assume in contradiction that MATH. Then MATH and there exist a monomial MATH and a monomial MATH such that MATH appears in the support of MATH. In particular, MATH. We can write MATH where each MATH and MATH. Then MATH with MATH. Let MATH be of degree MATH. If MATH then MATH and MATH, which is not possible. Hence MATH. Write, without loss of generality, MATH where MATH and MATH. Then MATH and clearly, MATH. If MATH with MATH and MATH then MATH for all MATH and MATH which implies that MATH . So MATH . If MATH is different from each MATH, then MATH as MATH. It follows that MATH, and thus MATH, a contradiction. Hence we may assume that MATH for a fixed MATH. Then MATH as MATH . Moreover MATH for all MATH with MATH, because MATH. Then again we obtain that MATH, the final contradiction. |
math/0104184 | Just apply the P-B-W Theorem. |
math/0104184 | Observe first that an element MATH is in MATH if and only if MATH . Indeed, MATH if and only if MATH for all MATH. It is easy to see that MATH . Since this holds for all MATH, it follows that MATH if and only if MATH satisfies MATH. Now let MATH be positive integers with MATH. Then the equality MATH implies MATH . If this holds for all MATH with MATH then each factor in brackets has to be REF, which gives MATH. So MATH, a contradiction which proves the lemma. |
math/0104184 | We assume that all monomials MATH, are ordered as in REF . If some MATH, then MATH for all MATH, so that we can write MATH where MATH with MATH . We say that such monomial has length MATH so the length of MATH is REF. Take an arbitrary ordering on the set of all monomials MATH and, using it, order the set of monomials MATH of a fixed length MATH lexicographically from left to right. Write MATH with MATH where each MATH and MATH if MATH. Among all monomials MATH consider those which have maximal length and denote the corresponding subset of indices by MATH. Let MATH be the smallest monomial in MATH and set MATH . For each MATH and MATH, write MATH, where MATH and set MATH. We consider separately the cases MATH and MATH. So suppose first that MATH. By REF , there exist MATH with MATH such that MATH so setting MATH and MATH we have that MATH . In particular, MATH as MATH . Taking MATH we will show that MATH. First we calculate MATH for arbitrary MATH. Set MATH. Because of the choice of MATH, MATH for all MATH, MATH. Hence MATH . We have MATH . Since MATH, we see that the first monomial MATH is still ordered in the ordering defined in REF . Taking a monomial MATH from REF , not equal to REF observe that its smallest term is greater than MATH, because for MATH either MATH or MATH but MATH. The monomial MATH may not be ordered. If it is not, then after ordering MATH we obtain the ordered permutation of MATH which is bigger than REF and some other shorter summands. Thus we can write MATH where the first monomial coincides with REF . Let MATH be the monomial REF with MATH. So MATH is the first monomial in REF when taking MATH . Observe that MATH as MATH . It is easy to see that MATH does not cancel out in MATH . Indeed, it follows from REF that it is enough to check that MATH does not coincide with REF when MATH. But this obviously follows from the choice of MATH. Observe that MATH and that MATH has degree MATH. Notice also that MATH does not belong to MATH. This is because MATH and in view of MATH, no MATH can be a multiple of some MATH . Now consider the case when MATH . Suppose first that there exists MATH such that MATH . Let MATH be the least such index. By REF , there exist MATH with MATH such that MATH . As before , we take MATH and set MATH . Then MATH and MATH . Let MATH be an index with similar property: MATH for some MATH . Suppose that MATH is the least such index. We have that MATH . We may still need to reorder these monomials, but after doing so we will get only shorter summands. A routine check shows that among the monomials of maximal length in REF the smallest one is the first summand. Let MATH be the first monomial of REF when taking MATH then (MATH becomes MATH). We claim that MATH does not cancel out in MATH . We see that it can not happen in MATH for every MATH with the above property of existence of MATH . Now, if MATH does not have such a property, then the last entries of all MATH's involved in MATH are all MATH's and MATH's. Thus when acting on MATH by MATH-we reduce the number of MATH's, while in MATH it is either the same as in MATH (case MATH), or even greater (case MATH). Hence, MATH does not cancel out in such MATH . Finally, if MATH, then all monomials of MATH which appear after reordering are shorter than MATH . Thus, MATH belongs to the support of MATH, as MATH . Since MATH . Moreover, the degree of MATH is MATH . Suppose now that MATH and MATH for all MATH . Let MATH be the smallest monomial in MATH . If again MATH and MATH for all MATH then we take the next smallest monomial and continue until we get the first monomial MATH with some MATH. Let MATH be the minimal such index. Let also MATH be the maximal index with MATH . Set MATH. By REF we choose MATH with MATH such that MATH . Take MATH . Then MATH where MATH has the same meaning as before. For MATH we let MATH be the maximal index with MATH . We have MATH . Among the monomials of maximal length in REF the first summand is the smallest one. Now let MATH be the first monomial of REF with MATH (then MATH becomes MATH). Since MATH it follows that MATH does not cancel out in MATH for all MATH . If we restrict our attention only to MATH then we are in conditions of previous cases. Let MATH be the smallest monomial among the monomials MATH of maximal length with MATH . In view of the previous considerations let MATH be the first monomial in REF or REF with MATH. Now, if MATH then MATH can not be cancelled out in MATH and it really belongs to the support of MATH as MATH . Let MATH . Taking larger MATH if necessary we can guarantee by REF that MATH and MATH where, as usually, MATH . Then MATH and since MATH is the smallest one among all monomials of maximal length which appear in MATH it does not cancel out. |
math/0104184 | Let MATH. We can assume that MATH is a weight vector and that MATH where MATH, MATH and all the monomials MATH are linearly independent. Note that MATH is an invariant of MATH since MATH is a weight vector. We will denote MATH and will prove the statement by induction on MATH. The base of induction MATH is trivial. Clearly for appropriate MATH and MATH. Hence we can apply our induction hypothesis and conclude that MATH for some MATH . |
math/0104184 | Since MATH we can assume without loss of generality that MATH for all MATH . Use induction on MATH . Let MATH . Then MATH, MATH . Apply MATH . Since MATH and MATH we conclude that MATH . Suppose now that MATH and MATH if MATH . We can also assume that MATH for some MATH since otherwise application of MATH completes the proof. Consider such MATH. Assume that MATH and consider MATH where MATH and MATH. Suppose that MATH, for sufficiently large number of pairs MATH. Then MATH is a constant for all those pairs MATH. Therefore MATH implying that MATH which is a contradiction. Thus we can assume that MATH for a large number of pairs MATH and some MATH . We can also choose MATH and MATH in such a way that MATH for all MATH . Fix MATH and MATH that satisfy conditions above and apply MATH: MATH . If MATH then the first term in the sum above is zero and we can complete the proof by induction on MATH . Assume now that MATH. Denote by MATH . If MATH for all MATH then MATH, with MATH here the same as in MATH, and therefore MATH . Thus we assume that MATH for some MATH . Consider MATH where MATH for MATH . The same argument as above shows that there exists a sufficient number of pairs MATH such that MATH for some MATH and for all such indices MATH, MATH . First note that among all pairs MATH there are some for which MATH . Indeed, MATH as soon as MATH . Without loss of generality we can assume that MATH and MATH are both divisible by MATH for some integer MATH. Set MATH, MATH and assume that MATH for sufficiently many integers MATH. Hence MATH and MATH . We conclude that there exist infinitely many pairs MATH satisfying MATH and MATH as long as MATH . Finally, if MATH then MATH which is impossible since MATH for all MATH . Hence, we can choose MATH and MATH such that MATH and MATH for at least one MATH, and in this case MATH . Apply MATH: MATH . We complete the proof by induction on MATH . The same arguments work in the case MATH. The lemma is proved. |
math/0104184 | Follows from REF . |
math/0104187 | We compute the NAME numbers via NAME cohomology, using the formula in REF . Consider the complex: MATH . Since MATH, it follows that MATH and MATH. The exact sequence MATH induces long exact sequences MATH . By tensoring with MATH and taking global sections, we get the exact sequence MATH . This proves the first assertion in the proposition. We have a similar exact sequence: MATH . Therefore MATH is the dimension over MATH of the cokernel of MATH . Using again the exact sequence MATH, by tensoring with MATH and taking a suitable part of the long exact sequence, we get: MATH . Since MATH, we have MATH and therefore MATH. From the above exact sequence we see that MATH, which proves the second assertion of the proposition. |
math/0104187 | If MATH, then for a general set MATH of MATH points, MATH is a general line bundle on MATH of degree MATH. Since in this case MATH is a general line bundle of degree MATH and MATH, REF says that MRC holds for every MATH if and only if for every MATH such that MATH and for a general line bundle MATH, either MATH or MATH. Since MATH, we have MATH. It follows immediately that MATH if and only if MATH. The first statement of the corollary follows now from the fact that if MATH is a vector bundle on a curve and MATH is a point, then MATH implies MATH and MATH implies MATH. The last statement follows from NAME duality and the isomorphism MATH. |
math/0104187 | CASE: Let MATH be a very ample line bundle of degree MATH. By REF and NAME duality, it is easy to see that IGC holds for MATH if: CASE: MATH for MATH general and REF MATH for MATH general. REF is a simple consequence of the filtration REF . More precisely, if MATH are the points of a general hyperplane section of MATH, from the exact sequence MATH we conclude that it would be enough to prove: MATH for MATH general. Now for every MATH, MATH is a general line bundle of degree MATH, so MATH. On the other hand MATH, so clearly MATH. For REF one needs a different argument. By twisting the defining sequence of MATH: MATH by MATH general and taking cohomology, we see that REF holds if and only if the map MATH is injective, or dually if and only if the cup-product map MATH is surjective. We make the following: Claim. MATH is a base point free pencil. Assuming this for the time being, one can apply the Base Point Free Pencil Trick (see CITE III REF) to conclude that MATH . But MATH is a general line bundle of degree MATH and so MATH. By NAME this means MATH. On the other hand MATH, MATH and MATH, so MATH must be surjective. We are only left with proving the claim. Since MATH is general, MATH, and so we easily get: MATH . Also, for every MATH, MATH is general, hence still noneffective. Thus: MATH . This implies that MATH is base point free. CASE: Here we follow an argument in CITE leading to the required nonvanishing statement. First note that it is clear from REF that for every MATH with MATH there is an inclusion MATH where MATH are general points on MATH. This immediately implies that MATH where MATH and MATH are general effective divisors on MATH of degree MATH. On the other hand we use the fact (see for example, REF. D) that every line bundle MATH can be written as a difference MATH which means that MATH . Now by NAME duality: MATH so that REF easily implies that MATH does not satisfy MRC for MATH as long as MATH. A simple computation gives then the stated conclusion. |
math/0104187 | The argument is similar (and in fact simpler) to the proof of REF . In this case, again by interpreting REF se REF IGC holds if and only if MATH . This is in turn equivalent to the surjectivity of the multiplication map: MATH which is again a quick application of the Base Point Free Pencil Trick. |
math/0104187 | We show that for every MATH, MATH, if MATH is general. Since MATH is hyperelliptic, we have a degree two morphism MATH and if MATH, then MATH. Therefore the morphism MATH defined by MATH is the composition of the NAME embedding MATH with MATH. Note that we have MATH. Since on MATH we have the exact sequence: MATH we get MATH. Therefore for every MATH, we have MATH . Now if MATH is general, then MATH is a general line bundle of degree MATH and so MATH. |
math/0104187 | From REF we see that if MATH is nonhyperelliptic, then the class of MATH is given by: MATH . On the other hand, as MATH is associated to the vector bundle MATH, if it is a divisor, then its cohomology class is MATH (recall that MATH has the same class as MATH). As in this case both MATH and MATH are divisors, in order to finish the proof of the proposition it is enough to prove the first statement. To this end, we follow almost verbatim the argument in REF . Namely, the filtration REF implies that for every MATH there is an inclusion: MATH where MATH are general points on MATH. This means that MATH for all general effective divisors MATH of degree MATH and MATH of degree MATH, which gives the desired inclusion. |
math/0104187 | CASE: This is certainly well known (compare CITE Ch.V.D), and we do not reproduce the proof here. CASE: Assume now that MATH is nonhyperelliptic, so that MATH is birational onto its image. For simplicity we will map everything to the Jacobian of MATH, so fix a point MATH and consider the commutative diagram: MATH where MATH is the MATH-th cartesian product of the curve and the maps are either previously defined or obvious. We will in general denote by MATH the fundamental class of the compact variety MATH. Since MATH clearly has degree MATH, and since MATH is birational by REF , we have: MATH . This means that it is in fact enough to prove that MATH, and note that MATH is the same as the class MATH, where MATH is the usual NAME map: MATH . The last statement is known as NAME 's formula (see for example, CITE I REF). We are now done by the following lemma, which essentially says that adding or subtracting points is the same when computing cohomology classes. |
math/0104187 | For simplicity, in this proof only, we will use additive divisor notation, although we actually mean the associated line bundles. Consider the auxiliary maps: MATH and the addition map: MATH . Then one has: MATH where MATH is the isomorphism MATH . Now the statement follows from the more general fact that MATH induces an isomorphism on cohomology. This is in turn a simple consequence of the fact that the involution MATH induces the identity on MATH. |
math/0104187 | As the arguments involved are standard we will just sketch the proof. Start by considering, for a given MATH, the NAME scheme MATH of curves in MATH with NAME polynomial MATH and MATH the open subset corresponding to smooth connected nondegenerate curves. Let MATH be the universal family over MATH, which is smooth of relative dimension MATH, and MATH the relative cotangent bundle. By base change there is an exact sequence MATH where MATH is a vector bundle on MATH such that if MATH corresponds to a curve MATH (in a suitable embedding), then MATH. The usual deformation theory arguments show that MATH is smooth and has dimension MATH. Moreover, the universal family MATH defines a surjective morphism MATH whose fibers are irreducible and have dimension MATH. It is immediate to see from this that MATH is connected. Fix now MATH such that MATH. Consider MATH and MATH as above and let MATH, where MATH is the composition of the inclusion MATH and the projection onto the second factor. We consider also the closed subset of MATH: MATH . It is clear by definition that MATH if and only if MATH. In particular, REF implies that MATH. MATH is the degeneracy locus of a morphism between two vector bundles of the same rank. Indeed, if MATH is a hyperplane, MATH, and MATH, then MATH is the degeneracy locus of MATH . Note that these are both vector bundles of rank MATH (we use base change and the fact that by REF, for every smooth curve MATH, the bundle MATH is semistable). We therefore conclude that MATH is a divisor on MATH. On the other hand, it is easy to see that the set MATH is closed. Moreover, REF (see also REF ) shows that MATH, and if MATH, then MATH. Let MATH be the set of irreducible components of MATH which are not included in MATH. Using the fact that MATH has irreducible fibers, all of the same dimension, it is easy to see that if MATH, then MATH is closed in MATH, that it is in fact a divisor, and MATH. Moreover, the locus of curves in MATH for which MATH is not a divisor is MATH, which proves the proposition. |
math/0104187 | Note that the pull-back of divisors induced by the projection to the first MATH components induces injective homomorphisms MATH. From the universality of the scheme structure on MATH it follows that the computation of MATH is independent of MATH. Therefore we may assume that MATH is large enough, so that in defining the scheme structure of MATH as above, we may take MATH. We introduce also the notation MATH. As a degeneracy locus, the class of MATH is given by MATH. It is clear that we have MATH if MATH and via MATH, we have MATH. Since MATH is the kernel of the evaluation map for MATH, we get MATH . Before starting the computation of MATH and MATH, we record the following well-known formulas for NAME classes. Let MATH be a vector bundle of rank MATH on a variety MATH and MATH. CASE: MATH. CASE: MATH. CASE: MATH, if MATH and MATH. CASE: MATH, if MATH and MATH. From the previous discussion and REF , we get MATH from which we deduce MATH . In order to compute MATH, we apply the NAME formula for MATH and MATH (see CITE). Note that the varieties are smooth and MATH is smooth and proper. Moreover, since we assume MATH, we have MATH, for MATH. Therefore we get MATH . From this we deduce MATH . We compute now each of the classes involved in the above equation. In order to do this we need to know how to make the push-forward of the elementary classes on MATH. We list these rules in the following: With the above notation, we have CASE: MATH. CASE: MATH. CASE: MATH. CASE: MATH. CASE: MATH. CASE: MATH. CASE: MATH. The proof of REF is analogous to that of the relation MATH (see CITE REFE). The other formulas are straightforward. Using REF for MATH, we deduce that MATH . Applying REF , we get MATH . From the above formula for MATH and REF , we get MATH . REF gives MATH . From the defining exact sequence of MATH we compute MATH . Using now REF , we deduce MATH . Using these formulas and equation MATH, we finally obtain MATH . Since the class of MATH is equal with MATH, we deduce the statement of the proposition. |
math/0104187 | Note that for MATH our assertion is just the statement of REF . Thus we can assume MATH. As mentioned above MATH is reduced, and from REF we see that MATH. We get that MATH is effective, and in fact MATH, where MATH is an effective divisor on MATH and MATH is the projection. Moreover, the map MATH is injective (compare CITE), hence MATH. Since the NAME compactification of MATH has boundary of codimension MATH (see for example, CITE, pg. REF) this implies MATH, that is, MATH. Therefore MATH is a divisor in MATH and the identification MATH holds for every nonhyperelliptic curve MATH. |
math/0104187 | Since the variety MATH introduced above is smooth, it suffices to show that the projection MATH given by MATH, is generically injective. We pick a component MATH of MATH whose general point corresponds to a marked curve MATH together with two different base-point-free MATH's on MATH, both containing MATH and MATH in different fibers. Clearly MATH and if we show that MATH, then we are done. For a general point in MATH we denote by MATH the induced MATH-sheeted maps. We may assume that MATH and MATH . The product map MATH is birational onto its image and MATH will have points of multiplicity at least MATH and MATH at MATH and MATH respectively. If MATH we set MATH, where MATH and MATH are pullbacks of the rulings on MATH, and MATH, MATH are the exceptional divisors. We denote by MATH the NAME variety of curves MATH homologous to MATH. The discussion above shows that MATH lies in the closure of the image of the rational map MATH obtained by projecting MATH onto the two factors. Thus MATH . On the other hand an argument identical to that in REF , shows that since MATH is a regular surface, every irreducible component MATH of MATH having dimension MATH is of the expected dimension provided by deformation theory, that is, MATH. Therefore MATH. |
math/0104187 | We write MATH uniquely as a combination of MATH, tautological classes MATH and MATH with MATH and boundary divisors. To express MATH in MATH we use that MATH and that MATH for MATH . Moreover we have that MATH is boundary in all cases except that MATH (compare REF ). The conclusion follows immediately. |
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