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math/0009085 | The image MATH of MATH are the coefficients of the NAME series MATH . But MATH equals to the resultant of the two polynomials MATH and MATH. (This is a less known form of the resultant MATH which can be obtained by multiplying the NAME by a matrix obtained from the coefficients of the NAME series of MATH see CITE.) On the other hand the resultant is equal to the product of differences of the roots of the two polynomials. |
math/0009085 | This is a straightforward generalization of the pullback property of the ordinary NAME dual. |
math/0009085 | The unfolding map MATH is transversal so we can apply REF . |
math/0009092 | The fact that the sets MATH, MATH and MATH are globally invariant follows immediately from the descriptions REF - REF. The étale algebra MATH corresponding to a finite MATH-set MATH may be defined as the algebra MATH where MATH is the algebra MATH and where MATH acts simultaneously on MATH and MATH. In the generic case, let us consider MATH . Then MATH sends MATH on MATH and acts trivially on MATH, MATH and MATH. We may describe similarly the actions of MATH and MATH. The action of MATH on MATH in the generic case is given by the table MATH . This implies that if MATH is not a cube in MATH, then MATH is isomorphic to MATH as a MATH-set. Then the corresponding étale algebra is MATH . Similarly if MATH is a cube in MATH, then we may decompose MATH into two orbits and we see that the corresponding étale algebra is MATH . The proofs for MATH and MATH are similar. |
math/0009092 | By REF, we have in MATH the relations MATH which proves that the natural projection from MATH to MATH is surjective. Moreover one has the relations MATH which gives a homomorphism of MATH-modules MATH and the exact sequence of the lemma. |
math/0009092 | By REF , we have MATH . Thus it is enough to prove that if MATH is an arbitrary étale algebra over MATH corresponding to a MATH-set MATH and if MATH is a prime such that MATH is not ramified at MATH, then MATH . This well-known assertion follows from the fact that the components of MATH are in bijection with the orbits of MATH in MATH, and the degree of each component is the length of the corresponding orbit. This proves REF . But this also shows that MATH which implies REF . |
math/0009092 | By a result of NAME (see CITE), MATH . REF implies that MATH . |
math/0009092 | Let us consider the set of quadruples MATH in MATH such that MATH . If MATH then MATH, MATH and MATH. Therefore, for any MATH as above, MATH and MATH. But for any triple MATH in MATH, there exists exactly one MATH verifying REF if MATH mod MATH and exactly three of them if MATH mod MATH. If MATH and MATH belongs to MATH then REF implies that MATH. For any triple MATH with MATH in MATH, MATH in MATH and MATH in MATH there exists exactly three MATH in MATH which satisfy REF. We get that MATH . Let us now turn to the set of MATH in MATH such that MATH . We decompose this set as follows MATH . As above we have the formula MATH where for MATH we use the equality MATH . On the other hand, MATH and MATH . We conclude: MATH . |
math/0009092 | To compute MATH we shall use its original definition CITE: MATH where the NAME measure on the affine hyperplane MATH is defined by the MATH-form MATH which is characterized by the relation MATH (where MATH is the linear form defined by MATH on MATH and MATH is the form corresponding to the natural NAME measure on MATH). More explicitely, let MATH be a basis of MATH and MATH be the dual basis. Write MATH with MATH. Let MATH be the projection of MATH on MATH along MATH. Then MATH . When MATH is given by REF the NAME group MATH is MATH and the orbits of its action on REF lines are MATH . In the basis MATH, a basis of MATH is given by MATH . The effective cone MATH is generated by the classes MATH, which in the basis MATH are given by MATH . Therefore, this cone is generated by the elements MATH and MATH and MATH is given as the volume of the domain MATH that is, as the volume of the segment MATH and MATH. If MATH is given by REF then MATH is isomorphic to MATH and the orbits of the NAME action on the MATH lines are MATH . A basis of MATH is given by MATH and the cone MATH is generated by MATH that is, by MATH (since MATH). Thus MATH is the volume of the domain given by MATH . Using the description above, MATH is the volume of MATH . Therefore MATH. If MATH is given by REF , then MATH and the orbits of the NAME action on the MATH lines are given by MATH . A basis of the NAME group is given by MATH . The effective cone MATH is generated by MATH . Since MATH and MATH, we get that MATH is generated by MATH . The anticanonical class is given by MATH . Thus MATH is the volume of the domain MATH that is, of the domain MATH in MATH given by MATH . We compute its volume as follows: decompose MATH into cones with appex MATH and supported by the faces not containing this point. Thus we consider the following faces of MATH: MATH . One has MATH . The area of MATH is the volume of the domain MATH and we get MATH. For MATH we have the equations MATH . We get MATH. For MATH we have the same equations and the same area. For MATH we have the equations MATH . We find MATH. The face MATH is given by the same equations and MATH. Finally MATH . |
math/0009097 | We first blow up the singular point (the origin) of the threefold MATH in REF . The resulting threefold, denoted by MATH, is isomorphic to the total space of the restriction of the tautological line bundle on MATH to the surface MATH. The exceptional divisor MATH is isomorphic to the zero section of this line bundle, which is isomorphic to MATH. An isomorphism MATH is given by MATH . The normal bundle MATH is isomorphic to the restriction of the tautological line bundle to MATH, which is MATH, under the above isomorphism. Hence we can contract either of the MATH-factor in MATH to obtain a smooth threefold. In the following we will show that we can choose the contraction so that the proper transforms of the MATH and the MATH-axes intersect and the proper transforms of the MATH and the MATH-axes intersect. Let MATH be the obvious projection. Then under the isomorphism REF the map MATH is of the form MATH where MATH is the variable representing points on the total space of the tautological line bundle. Note that MATH has homogeneous degree MATH, as MATH and MATH have homogeneous degree MATH and MATH respectively. Hence the proper transforms of the MATH, MATH, MATH and the MATH-axes intersect MATH at MATH respectively. Therefore if we contract the first MATH-factor the resulting threefold MATH satisfies the required property. |
math/0009097 | We prove this Lemma by induction on MATH. For MATH, there is nothing to prove. We now assume that the Lemma holds for MATH. Namely, we have found coverings MATH of MATH that satisfy the required property. Following the construction, MATH is a small resolution of MATH. Using the explicit description of MATH, we see that MATH are smooth MATH-invariant open subsets of MATH. Clearly, we can choose a unique coordinate MATH of MATH so that the projection MATH is given by MATH and the projection MATH is given by the formula REF in the statement of the Lemma. A routine check shows that for MATH the MATH-action on MATH and the transition function from MATH to MATH is exactly as shown in the statement of the Lemma. It remains to check the case where MATH and MATH. We now look at the chart MATH. Let MATH be the composite of the desingularization MATH with the obvious projection. Then MATH where MATH is the tautological line bundle (degree MATH) on MATH and MATH is the total space of MATH. We now cover MATH by MATH and MATH. We choose trivializations MATH and MATH so that the transition function from MATH to MATH is given by MATH . We let MATH be MATH and let MATH. Then MATH form a covering of MATH. Next for MATH we choose a coordinate MATH and for MATH we choose MATH . We now check that this choice of coordinates satisfy the requirement of the Lemma. We first check the transition functions. The transition function from MATH to MATH is determined by the transition function from MATH to MATH. Thus we have MATH . This is the requirement in REF for MATH. It remains to work out the transition function from MATH to MATH. We let MATH be a general point and let MATH be the point defined by the relation REF with MATH replaced by MATH and MATH replaced by MATH. Then MATH . Clearly, to prove REF we only need to check that the images of MATH and MATH in MATH coincide and their images in MATH also coincide. By definition, the image of MATH in MATH is MATH and the image of MATH in MATH is MATH . They coincide using the relations mentioned above. On the other hand the image of MATH in MATH is MATH and its image in MATH, using REF and the induction hypothesis, is MATH . The image of MATH in MATH is MATH . They coincide as well. This proves the relation (MATH). REF concerning the group action can be checked directly and will be omitted. |
math/0009097 | In case MATH lies over MATH, then there is nothing to prove. Now assume MATH lies over MATH. Let MATH be the integer so that MATH has MATH irreducible components. Then by REF , there is an open neighborhood MATH of MATH and two morphisms MATH so that MATH is equivalent to MATH via an arrow. Thus to prove the Lemma it suffices to consider the case where MATH. Now we assume, MATH and MATH. Let MATH be the MATH-isomorphism and let MATH be the associated morphisms. We define a functor MATH from the category of MATH-schemes to sets that associates to any MATH-scheme MATH the set of all morphisms MATH such that with MATH the induced morphism (that is, the first projection) then MATH . (For the notation here see REF .) By REF schemes developed by NAME coupled with the trivialization of the stabilizers (see REF ), MATH is represented by a subscheme MATH of MATH. We now show that MATH is non-empty and dominates a neighborhood of MATH. Since MATH, the restriction of the isomorphism MATH to fibers over MATH induces an isomorphism MATH that commutes with MATH. Then it is induced by an element MATH by REF . Hence MATH. Let MATH be the maximal ideal of the local ring MATH. By the previous discussion, we have MATH so that MATH factor through MATH, where MATH is the inclusion. Let MATH with MATH the inclusion morphism. Suppose for some MATH we have MATH extending MATH that factor through MATH. We will show that MATH extends to MATH satisfying similar condition. Since MATH is smooth, we can extend MATH to MATH. Let MATH be the canonical immersion. We consider the problem of extending the MATH-morphism MATH to a MATH-morphism MATH . This is a typical extension problem. Since MATH and MATH are two such extensions, by deformation theory they are related by an element in MATH, where MATH . We claim that MATH is isomorphic to MATH and is generated by the group action of MATH. We let MATH be the immersion and MATH be the composite of MATH with the projection. Then we have the exact sequence MATH and its induced exact sequence MATH . At the origin MATH, we have a canonical isomorphism MATH. We let MATH be the MATH-th component. The dual of MATH, which is MATH, defines a homomorphism MATH. Now let MATH say given by a homomorphism MATH. Then it must be a linear combination of MATH, say equal to MATH. Now let MATH be the extension class in MATH of the exact sequence REF and let MATH be the homomorphism induced by MATH. We claim that MATH. Indeed, let MATH be the immersion representing the tangent vector MATH. Then MATH is an infinitesimal smoothing of the MATH-th nodal divisor of MATH whenever MATH, and consequently the exact sequence MATH does not split since MATH. On the other hand, this exact sequence is exactly defined by the extension class MATH. In case MATH, then MATH, violating the non-splitness of REF . This proves MATH. As a consequence MATH . Now let MATH be any element in MATH. Then MATH by the above identity. Hence the restriction of MATH to MATH, denoted by MATH, is a vector field of MATH. Since MATH lies in the kernel, we have MATH. Furthermore, the restriction of MATH to the two (or one if MATH or MATH) distinguished divisors is tangential to these divisors and must satisfy the compatibility condition MATH. Now let MATH be the subspace of MATH consisting of the vector fields whose restrictions to the two distinguished divisors are tangential to the two divisors. Then MATH is canonically isomorphic to the direct sum MATH, where MATH is generated by the vector field generated by the group action MATH. Let MATH be the component of MATH in MATH. Then the compatibility condition translates to MATH. Adding the fact that MATH, we conclude that MATH for MATH. Therefore, MATH must be a vector field tangential to the fibers of MATH and vanishes on the two distinguished divisors. By REF , such MATH is generated by the group action MATH. If we apply this to the two extensions MATH and MATH, we see that there is a morphism MATH with MATH factor through MATH such that with MATH, MATH . Since MATH is arbitrary, this implies that the projection MATH dominates a neighborhood of MATH. We let MATH be the projection. To complete the proof of the Lemma, we need to find a neighborhood MATH of MATH, so that MATH is an open neighborhood of MATH. First, by REF , for any MATH the set MATH has at most one point. Applying REF again, we see that the choice of MATH in the proof is unique. Hence the formal completion of MATH along MATH is isomorphic to the formal completion of MATH along MATH, under MATH. Hence MATH is étale over a neighborhood of MATH. Combining the one-to-one and étale property, we conclude that there is a neighborhood MATH of MATH so that MATH is an open immersion. Let MATH be the morphism given by the definition of the functor MATH and let MATH be the inclusion. Then MATH defines an effective arrow between MATH and MATH. This proves the Lemma. |
math/0009097 | This is straightforward and will be left to the readers. |
math/0009097 | Clearly, any MATH can be expressed as above. We now show that such expression is unique. Let MATH be the MATH-sub-module spanned by all MATH satisfying MATH. Clearly, the quotient MATH-module MATH is a free MATH-module with basis MATH. Let MATH be the quotient homomorphism of MATH-modules. Assume MATH has an expression as in REF , then MATH in MATH. Since MATH is a free MATH-module and MATH is a basis, all MATH and MATH are zero for MATH. Since MATH is arbitrary, this implies that all MATH and MATH are zero. This proves the Lemma. |
math/0009097 | Let MATH be the maximal idea and let MATH be the normal form of MATH. Since MATH is non-degenerate, the homomorphism MATH induced by MATH and MATH must be of the form (possibly after exchanging MATH and MATH) MATH for some positive integers MATH and MATH. Hence in REF , all MATH for MATH and MATH for MATH are in the maximal ideal MATH. We let MATH be the ideal generated by the listed terms for all MATH. We first show that MATH. After expressing MATH in its normal form MATH we have MATH and MATH . Because MATH is the image of MATH in MATH, all MATH and MATH for MATH are zero. Now let MATH be the largest integer so that MATH for MATH. Then by considering the relation MATH, we obtain MATH . By assumption, all terms in the above expression except MATH are in MATH. Hence MATH. Since MATH, MATH. This shows that there is no such MATH and hence all MATH are in MATH. Similarly, by considering the relation MATH we can show that all MATH. Now let MATH and let MATH be the associated ideal of MATH in the quotient ring MATH. Let MATH be the maximal ideal of MATH. We now show that MATH. Note that in the expression of MATH and MATH, terms MATH and MATH are in MATH for all MATH. We let MATH. Note that MATH for MATH. Now let MATH be the smallest integer so that MATH for all MATH. Hence MATH. Using MATH, we obtain MATH which implies MATH . Hence MATH. By minimality of MATH we have MATH for all MATH. For similar reason, we can prove MATH for all MATH. To show MATH, we use the relation MATH, which combined with MATH and MATH implies MATH. Now assume MATH for MATH, then the above inclusion implies MATH . Since MATH and MATH, we have MATH and hence MATH. For similar reason, MATH for all MATH. Combined, this proves that MATH. Since MATH is NAME, this is possible only if MATH. This proves MATH. We next show MATH. Assume MATH. Then MATH implies that MATH . Note that all terms in this expression except MATH belong to MATH. This is impossible. Hence MATH. We now look at the normal form of MATH. Since MATH for all MATH, there are MATH so that MATH. On the other hand, since MATH, there are MATH so that MATH for MATH. Using MATH, we see that there are MATH and MATH in MATH so that MATH. Further by our choice of MATH both MATH and MATH are units in MATH. It remains to show that there is a unit MATH so that MATH and MATH. First, since MATH, which is in the ideal MATH, there is a MATH so that MATH. Hence MATH in MATH. Because MATH is flat over MATH, this implies that MATH. Hence MATH is a unit in MATH. This completes the proof of the Lemma. |
math/0009097 | We let MATH, where MATH runs through all positive integers, be indeterminants. We consider the ring MATH, where MATH and MATH mean MATH and MATH respectively. We let MATH be the MATH-adic completion of MATH. We let MATH and its inverse be MATH where MATH and MATH are elements in MATH. We consider the ideal MATH that is generated by all coefficients of MATH in the normal forms of MATH and MATH . We now investigate these generators of MATH, or equivalently the relations in MATH. First note that the coefficient of MATH in MATH is MATH. Hence we obtain elements MATH in MATH for all MATH. Similarly, the coefficient of MATH in MATH is MATH, which by definition is in MATH. In the following we will call them the canonical relations of MATH and MATH, respectively. We let MATH be the ideal generated by MATH and all MATH. We next find relations of MATH. First, by expanding MATH in power series and using MATH, we can express MATH canonically as MATH, where MATH. Then the coefficient of MATH in MATH become MATH . We let MATH be the ideal generated by MATH and this set of elements for all MATH. Since MATH, modulo MATH we can substitute MATH and all MATH by elements in MATH using relations in MATH and then substitute all MATH in MATH by MATH. This way we obtain MATH where MATH and MATH. Note that here we have used the fact that MATH is complete with respect to the ideal MATH. Repeating this procedure, we eventually obtain an element MATH, where MATH. It remains to find relation in MATH. By using the coefficient of MATH in MATH, we obtain an element MATH where MATH. We then let MATH be the ideal generated by MATH and this element. Then again modulo MATH, which amounts to repeatedly replace MATH, MATH and MATH by relations in MATH, we obtain an element MATH, where MATH. By iteration and taking the limit, we obtain a canonical element MATH for some MATH. It is clear from this construction that MATH is a unit since MATH is a unit. This way we have found elements MATH and MATH for MATH and a unit MATH so that MATH . From this set of generators, we see that the canonical homomorphism MATH is an isomorphism. Since MATH is a sub-ideal of MATH, we have quotient homomorphism MATH. We let MATH be the kernel of MATH. We will show that the ideal MATH satisfies the property specified in the statement of the Lemma. First, from the construction it is direct to check that MATH satisfies the base change property. Now let MATH and MATH be as in the Lemma so that MATH has pure MATH-contact. Then it follows from our construction that the ideal MATH. Therefore MATH factor through MATH since MATH. This proves one direction of the Lemma. Now assume MATH is a homomorphism so that it factor through MATH. Then by the base change property we have MATH. It follows from the construction of MATH that MATH does have pure MATH-contact. This completes the proof of the Lemma. |
math/0009097 | It suffices to prove the case where MATH is reduced and irreducible, which we assume now. Let MATH be induced by MATH. In case MATH is dominant over MATH, then REF implies that MATH as topological spaces and hence the Lemma holds. Now assume MATH factor through MATH. Since MATH is reduced, MATH factor though MATH and then MATH as schemes. Let MATH and MATH be the decomposition of MATH provided by REF . Let MATH be the projection. We let MATH be the composite MATH. As before, for any MATH we denote by MATH the restriction of MATH to the fiber over MATH and denote by MATH the fiber of MATH over MATH. Then the left and the right contact order of MATH along MATH define two functions MATH . It follows from the definition that MATH if the following two conditions hold: CASE: For any MATH, MATH; CASE: The preimage MATH is identical to MATH. Applying the usual semi-continuity theorem, we see that both MATH are upper semi-continuous. Hence REF combined implies that MATH is a constructible set. Therefore, to show that MATH is closed it suffices to show that if MATH is a discrete valuation domain with residue field MATH and quotient field MATH, and if MATH extends to MATH, then the extension MATH factor through MATH. Now we let MATH be a morphism so that MATH factor through MATH. We let MATH be the pull back family over MATH via MATH. By abuse of notation, we still denote by MATH, MATH and MATH the decomposition data of the MATH-curve MATH. Let MATH be the generic point of MATH. Then by assumption REF hold for MATH. Now let MATH be any element. As before, for MATH or MATH we let MATH and MATH be the restrictions of the respective families to the fiber over MATH. We let MATH be the pull-back divisor where MATH is viewed as a divisor in MATH. Since MATH is non-degenerate, MATH . Clearly, the equality in REF holds for MATH if MATH. Hence REF is equivalently to that the equalities in the above two inequalities hold. Because the function of contact order is upper semi-continuous, for the closed point MATH, MATH . On the other hand, since MATH is a flat family of non-degenerate morphisms, we have MATH . Combined with the inequality in REF for MATH and MATH and the equality in REF for MATH and MATH, we see that the equality in REF for MATH and MATH must also hold. For similarly reason, the equality in REF also hold for MATH and MATH. Hence the two contact order functions in REF must be constant functions. Therefore, REF hold at MATH and hence MATH factor through MATH. This shows that MATH is closed in MATH. |
math/0009097 | The existence of such a parameterization follows immediately from the deformation of nodes CITE. This also follows from the local parameterization constructed in CITE. Now let MATH and MATH be two such parameterizations of MATH along MATH. Because MATH and MATH define the same divisor in MATH corresponding to the subscheme where the nodes are not smoothed, there is a unit MATH such that MATH. Hence MATH and hence MATH . It follows from the deformation of nodes that there are units MATH and MATH in MATH such that MATH . Using the above relation, we obtain MATH. Hence MATH for some MATH satisfying MATH. We now demonstrate that if we choose MATH, MATH and MATH appropriately, then MATH will be zero. We first write MATH in its normal form MATH . Since MATH, by the uniqueness of the normal form we have MATH and MATH for MATH while MATH. We let MATH be its normal form. Since MATH is a unit, MATH must be a unit. We now let MATH. Then since MATH, we can replace MATH in REF by MATH. Then the new error term MATH, which obeys MATH, belongs to the ideal MATH. Hence without loss of generality we can assume the MATH, MATH and MATH are chosen appriori that the MATH in the normal form of MATH is zero. On the other hand, since MATH for MATH, we can replace MATH by MATH so that REF still hold. Hence we can assume appriori that all MATH. Lastly, we let MATH. Clearly, MATH and MATH since MATH. Thus MATH with MATH is what we want. The uniqueness of MATH can be proved following same strategy. Since a more general version will be proved in CITE, we will omit the proof here. |
math/0009097 | Clearly, should MATH be represented by a subscheme in MATH, then set-theoretically it must be MATH. Hence it suffices to give MATH a closed scheme structure so that it represents the functor MATH. As before, we let MATH be MATH endowed with the reduced scheme structure and MATH be the formal completion of MATH along MATH. We let MATH be any connected component of MATH endowed with the open scheme structure and let MATH be MATH endowed with the reduced scheme structure. Our first objective is to show that MATH admits a canonical scheme structure (as closed scheme of MATH) so that it represents the functor MATH. We continue to use the notation developed so far. For instance MATH and MATH are the decomposition data of the restriction of MATH to MATH along the divisor MATH. We let MATH be the étale morphism so that MATH. We let MATH and MATH be the projections. Let MATH be any closed point and let MATH be the set of points in MATH. Note that MATH are exactly the preimage set MATH. Hence each MATH is associated with an integer MATH that is the contact order of MATH at MATH. For each MATH between MATH and MATH, we pick an affine open neighborhood MATH of MATH. We let MATH and let MATH be the pull back of MATH to MATH. We let MATH be MATH. Then the multi-section MATH defines a section MATH. Without loss of generality, we can assume MATH admits a parameterization of the formal neighborhood of its nodes along MATH. By shrinking MATH if necessary, we can assume that there is an admissible chart MATH of MATH near MATH so that MATH is entirely contained in MATH. Therefore, following the construction associated to REF , by shrinking MATH if necessary we have an ideal sheaf MATH so that it defines the subscheme of MATH that parameterizes all morphisms in the family MATH that are pre-deformable and have contact order MATH along the nodes MATH. To obtain the ideal sheaf of MATH that defines the desired closed subscheme structure of MATH we need to descend the ``intersection" of MATH since MATH is defined by the requirement that all nodes are pre-deformable. As before, we denote by MATH the number of elements in MATH for MATH. Since MATH is connected and MATH is étale and finite, MATH is independent of the choice of MATH. We consider the fiber product, denoted MATH, of MATH-copies of MATH over MATH. We denote by MATH accordingly the fiber product of MATH copies of MATH over MATH. Note that points in MATH are MATH with not necessarily distinct MATH for some MATH. We let MATH be the subset of those points MATH so that all MATH are distinct. Since MATH is finite and étale, and that each fiber of MATH has cardinal MATH, MATH is an open subset of MATH. Then since MATH is homeomorphic to MATH, MATH is an open subset of MATH. We let MATH be MATH endowed with the open scheme structure. Then the obvious projection MATH is étale and finite. Now let MATH be the morphism induced by the MATH-th projection of MATH, which is étale. Now consider any point MATH. To each MATH between MATH and MATH we let MATH be the neighborhood of MATH and MATH the ideal sheaf just constructed. We let MATH be an open neighborhood of MATH so that MATH for each MATH. We then let MATH be the ideal sheaf of MATH that is the pull back of MATH under MATH, and let MATH be the intersection MATH. This ideal sheaf defines a close subscheme of MATH, denoted by MATH. Now let MATH be the composite MATH. It may happens that MATH and MATH are different as subsets of MATH. However, by the proof of the closedness of MATH, the connected component of MATH containing MATH coincides with the connected component of MATH containing MATH. Hence by shrinking MATH if necessary, we can assume MATH is connected. Now let MATH be the family over MATH that is the pull-back of the family MATH via MATH. Since for MATH the preimage MATH has exactly MATH points, it is direct to check that the subscheme MATH represents the functor MATH. Now by repeating this procedure, we can cover MATH by open subsets MATH with ideal sheaves MATH, for MATH, such that the closed subscheme MATH defined by the ideal sheaf MATH represents the functor MATH. By the universal property of these schemes, whenever MATH the ideal sheaves MATH and MATH coincide over MATH. Thus they patch together to form an ideal sheaf MATH. Again, by the universal property, this ideal sheaf forms a descent data for the faithfully flat étale morphism MATH. Hence it descends to an ideal sheaf MATH which defines a closed subscheme MATH. It follows from the previous construction that MATH represents the functor MATH. We now show that the functor MATH is represented by a closed subscheme. Since each connected component MATH of MATH has a closed scheme structure that represents the functor MATH, the functor MATH is represented by a closed subscheme of MATH. We denote this subscheme by MATH. It remains to show that there is a closed scheme structure on MATH so that it represents the functor MATH. As before, we let MATH be the MATH-th standard parameter of MATH. We let MATH be the subsheaf of elements annihilated by some power of MATH. Since MATH is noetherian, there is an integer MATH so that MATH is zero. Similarly, we let MATH be the subsheaf of MATH consisting of elements annihilated by some power of MATH. Then MATH is canonically isomorphic to MATH. As an ideal sheaf, MATH defines a closed subscheme MATH. Clearly, MATH is flat over MATH near MATH. Hence by REF , the restriction of the family MATH to MATH is pre-deformable along MATH. Consequently, the ideal sheaf of MATH is contained in MATH. Using the natural isomorphism MATH, this ideal sheaf defines an ideal sheaf MATH of MATH. By our construction the support of the subscheme defined by MATH coincides with MATH. We define the closed scheme structure of MATH to be the one defined by MATH. It is straightforward to check that with this closed scheme structure, MATH represents the functor MATH. As to MATH, it is clear that it is represented by the intersection of the closed subschemes MATH. This completes the proof of the Theorem. |
math/0009097 | Let MATH be any integer between MATH and MATH. We let MATH be the group action given in REF . For any closed purely one-dimensional subset MATH of MATH we define the relative automorphism group MATH to be the set of pairs MATH where MATH is an automorphism leaving MATH and MATH fixed and MATH is an element in MATH such that MATH for all MATH. Let MATH be any automorphism. Then MATH induces a permutation of MATH for each MATH. As a result, it defines a homomorphism MATH where MATH is the permutation group of MATH. Clearly, its kernel is MATH . Since MATH is an ordinary stable morphism, MATH and MATH are finite. Hence MATH is finite if and only if MATH are finite for MATH. Now assume MATH is a connected component of MATH so that MATH is not a single point set and that MATH is infinite. Then the following must hold: MATH must be contained in a fiber of MATH; MATH must be a smooth rational curve; MATH contains no marked points and MATH consists of exactly two nodal points of MATH, one lies in MATH and the other lies in MATH. Further if we let MATH be the image of MATH then MATH is a branched covering ramified at MATH. We call such connected components trivial components. Clearly, if MATH is not a trivial component, then MATH is finite. On the other hand, if MATH is a trivial component then MATH is surjective. Hence MATH is finite if and only if not all connected components of MATH are trivial. To complete the proof of the Lemma, it suffices to show that MATH if and only if all connected components of MATH are trivial. Let MATH be any connected component of MATH. In case MATH, then since MATH is very ample we can assume without loss of generality that MATH, and hence MATH. In case MATH, then MATH must be contained in a fiber of MATH. Since MATH and MATH is connected, MATH can not be a single point. Hence MATH is a fiber of MATH. Because MATH is pre-deformable, MATH contains at least two nodal points of MATH, one lies in MATH and the other lies in MATH. Hence MATH. Moreover, the equality holds if and only if MATH, MATH contains no marked points and contains exactly two nodal points of MATH. Again since MATH is pre-deformable, it is easy to see that this is possible only if MATH is a trivial component. This shows that any connected component MATH of MATH has MATH and the equality holds if and only if MATH is a trivial component. Now we assume MATH is stable. Then to each MATH between MATH and MATH at least one connected component of MATH is non-trivial. Then combined with the fact that the weight of any connected components of MATH are non-negative, we have MATH. Conversely, if for some MATH between MATH and MATH the weight MATH, then all connected components of MATH are trivial. Since to each component MATH of MATH the automorphism group MATH surjects onto MATH, MATH will surjects onto MATH as well. Thus MATH, and hence MATH is infinite. This proves the Lemma. |
math/0009097 | The proof of these two Lemmas are standard and will be omitted. |
math/0009097 | Following the definition of stacks (compare CITE), it suffices to show the following: CASE: For any scheme MATH over MATH and two stable morphisms MATH and MATH, the functor MATH which associates to any morphism MATH the set of isomorphisms in MATH between MATH and MATH is a sheaf in the étale topology; CASE: Let MATH be a covering of MATH (over MATH) in the étale topology. Let MATH and let MATH be isomorphisms in MATH satisfying the cocycle condition. Then there is a MATH with isomorphism MATH so that MATH . We first prove REF . Let MATH, MATH and MATH, be represented by MATH. To each morphism MATH we define MATH to be the set of all arrows from MATH to MATH. This defines a functor MATH . Clearly it is a pre-sheaf. To show that this is a sheaf it suffices to show that if MATH is an étale open covering of MATH, then MATH is an equalizer diagram. This means that if MATH is an element in the middle term above, then MATH for all MATH if and only if there is a MATH so that its restriction to MATH is MATH. By definition, each MATH consists of two isomorphisms MATH and MATH and a commutative diagram MATH . REF makes MATH a descent data for an isomorphism between MATH and MATH and makes MATH a descent data for an isomorphism between MATH and MATH. Since MATH is an étale covering, by Descent Lemma there is an isomorphism MATH and MATH restricting to MATH and MATH respectively. Hence MATH is the element so that MATH. This proves REF . The proof of REF is similar. It amounts to construct a family MATH over MATH, a family MATH in MATH and a MATH-morphism MATH, which form an element MATH, so that MATH and the identity map is MATH. This is again a standard application of Descent Lemma and will be omitted. |
math/0009097 | We let MATH be represented by MATH. Let MATH and MATH be the standard choices of relative ample line bundles on MATH and on MATH. It follows from the uniqueness of these line bundles that whenever MATH is a morphism and MATH is a pair of isomorphisms shown below that makes the diagram MATH commutative then MATH . Hence MATH is isomorphic to the functor associating to MATH the set of pairs of isomorphisms of polarized projective schemes over MATH satisfying the obvious compatibility condition. It is a routine application of NAME 's results on the representability of the NAME scheme and related functor CITE, see also CITE, that the functor MATH is represented by a scheme quasi-projective over MATH. This completes the proof of the Lemma. |
math/0009097 | To prove that MATH is unramified over MATH, we only need to check the case where MATH is itself a closed point. Namely MATH, in which case MATH is either empty or is isomorphic to MATH. Since MATH, it is well-known that MATH is unramified over MATH if there are no vector fields MATH of MATH and vector fields MATH of MATH (whose push-forward to MATH is trivial) so that they are compatible via MATH. Such vector field does not exist because MATH is stable. Hence MATH is unramified over MATH. Further, since MATH is of finite type over MATH, it is quasi-finite over MATH. This proves the Lemma. |
math/0009097 | Without lose of generality we can assume that MATH is represented by MATH with MATH the associated morphism so that MATH. We will prove the case where MATH does not factor through MATH. The other case is similar and will be left to readers. We let MATH and MATH be the generic and the closed point of MATH, as stated in the Lemma. Let MATH, which is isomorphic to MATH by assumption. Let MATH be the composite MATH. Since MATH does not factor through MATH, MATH is an ordinary stable morphism. By the property of stable morphisms, possibly after a base change MATH the morphism MATH extends to a family of stable morphisms MATH over MATH. For simplicity by replacing MATH with MATH we can assume MATH. We consider the induced morphism MATH where MATH is the canonical projection. By our assumption, the restriction of MATH to MATH is isomorphic to MATH. Therefore by the property of stable morphisms MATH is the stabilization of MATH for MATH. In particular, there is a unique contraction morphism MATH so that MATH . Now let MATH and let MATH be any irreducible component contracted under MATH. Following the same argument in the proof of REF , one can show that MATH, MATH is contained in MATH for some MATH and MATH is contained in a fiber of MATH. There are two cases we need to distinguish. One is when MATH is a point. Then that MATH is contracted by MATH implies MATH contains at least two nodal points of MATH. On the other hand, when MATH is not a point, then MATH is a fiber of MATH and hence because MATH is stable and hence pre-deformable, the component MATH contains at least two nodal points of MATH as well. Because this is true for every irreducible components in MATH contracted under MATH, by the property of stable contraction, an irreducible component MATH of MATH is contracted under MATH if and only if it is isomorphic to MATH, it contains exactly two nodal points but no marked points and MATH. Then since MATH is stable, such MATH must be a trivial component of MATH. Conversely, any trivial component of MATH obviously will be contracted under MATH. This shows that the morphism MATH contracts exactly all trivial components of MATH. Consequently, if we let MATH be the union of all trivial components in MATH, then the contractions MATH and MATH define an isomorphism MATH extending the isomorphism MATH. To proceed, we assume MATH without lose of generality. Note that MATH is a resolution of MATH , which induces a morphism MATH. We consider MATH . Clearly no irreducible components of MATH are mapped entirely to the nodal divisor of MATH unless MATH and in which case some non-trivial components of MATH will be mapped to and only to the MATH-th nodal divisor of MATH. Let MATH be the morphism induced by MATH. Because MATH does not factor through MATH, there are isomorphisms MATH and MATH fitting into the commutative diagram MATH . By REF , MATH is induced by the MATH-action on MATH via a morphism MATH. Let MATH be the uniformizing parameter of MATH and let MATH be defined by MATH where all MATH are units in MATH and MATH are integers. Here we follow the convention MATH . Now let MATH be an irreducible component of MATH whose image under MATH is contained in the first irreducible component of MATH. Then MATH is not contained in MATH, MATH is a non-trivial component and MATH is also contained in the first component of MATH. Now let MATH be a non-trivial irreducible component of MATH whose image under MATH is contained in the second irreducible component of MATH. (Here we assume MATH since otherwise there is nothing to prove.) We let MATH, where MATH is defined in REF , be a curve so that MATH is a closed point in MATH in general position and the composite MATH is a branched covering ramified at MATH. We pick an affine open MATH so that the image of MATH under MATH is contained in MATH. Let MATH be the formal completion of MATH along MATH, let MATH be the formal completion of MATH along MATH and let MATH be the induced morphism. Then the morphism MATH factor through MATH, which we denote by MATH . Here MATH is the scheme defined before REF . We let MATH be the morphism induced by MATH. Because of REF , if we let MATH be the second projection of MATH and let MATH be the group action morphism, then MATH . To utilize this identity, we use the open subset MATH of MATH constructed in REF . Since MATH is contained in the second irreducible component of MATH, if we let MATH be the coordinate chart of MATH introduced in REF , then MATH . Because MATH is in the smooth locus of MATH, all MATH are units in MATH except MATH, which is in the maximal ideal MATH of MATH. On the other hand, because MATH is flat, there is a MATH so that MATH is given by MATH. Therefore using the formula in REF , MATH for some MATH in the quotient field of MATH. Hence if we let MATH be the projection, then MATH maps MATH . Because this composition specializes to MATH by assumption, all MATH are regular and are in the maximal ideal MATH. Furthermore, we know MATH does not specialize to the first nodal point of MATH, therefore MATH is not in MATH. Since MATH is a unit in MATH, MATH is a unit in MATH and MATH is in the maximal ideal MATH, this implies MATH. Similarly, we can pick a non-trivial irreducible component MATH of MATH so that its image under MATH is contained in the second component of MATH. Then because MATH is not contained in the first nodal divisor of MATH, a parallel argument shows that MATH. Therefore, MATH. Repeating this argument, using the fact that no irreducible components of MATH are mapped entirely to the first MATH nodal divisors of MATH under MATH, we can show that all MATH are zero. Therefore, MATH extends to MATH. Now we prove that the arrow MATH extends to an arrow MATH from MATH to MATH. Let MATH be the stable contraction of MATH as a morphism to MATH. Then MATH is MATH, which is isomorphic to MATH via the arrow MATH. We consider the morphism MATH . By assumption, its restriction to MATH is identical to the restriction of MATH to MATH via MATH. On the other hand, both MATH and MATH are stable extensions of a stable morphism from MATH to MATH. Hence by the uniqueness of the extensions of ordinary stable morphisms, the isomorphism MATH extends to an isomorphism MATH and the morphism MATH is identical to MATH. This shows that MATH maps no irreducible components of MATH to the nodal divisors of MATH. We claim that MATH must be equal to MATH. Otherwise, all irreducible components of MATH that are mapped to MATH (under MATH) are trivial components, contradicting to REF . Once we have MATH, then MATH and hence MATH and MATH follows immediately. This defines an arrow MATH that is an extension of MATH. This proves the Lemma. |
math/0009097 | Let MATH be represented by MATH. We will prove the case where MATH does not factor though MATH and leave the other case to the readers. Since MATH does not factor through MATH, we can assume without loss of generality that MATH is represented by MATH. Then MATH is an ordinary stable morphism over MATH. By the property of stable morphism, possibly after a base change MATH, the morphism MATH extends to a MATH-family of ordinary stable morphism MATH. Again by replacing MATH with MATH we can assume MATH. Let MATH and MATH be the closed and the generic points of MATH. In case MATH sends MATH to MATH, then MATH is already a family of stable morphisms in MATH. Now assume MATH is mapped to MATH. Let MATH be any MATH-morphism. Then since MATH is canonically isomorphic to MATH, MATH induces a morphism MATH. We say MATH admits a partial extension to MATH if after a base change MATH the morphism MATH extends to a family of quasi-stable morphisms over MATH so that the associated MATH maps the closed point of MATH to MATH. Here we say MATH is quasi-stable if MATH is finite, where the automorphism group is the set of pairs MATH such that MATH is an automorphism of the domain of MATH and MATH such that MATH. (As before, we let MATH be the restriction of MATH to the fiber MATH.) It is clear that the extension MATH is a quasi-stable extension to MATH. On the other hand, by the proof of REF any quasi-stable extension MATH satisfies MATH. Now we let MATH be the largest possible integer so that there is a quasi-stable extension MATH. We show that MATH defines an extension of MATH. Let MATH be a quasi-stable extension as in REF via MATH. If MATH is non-degenerate, then REF implies that MATH is pre-deformable. Hence MATH as desired. Now assume MATH is degenerate. Let MATH be an irreducible component of MATH so that MATH is contained, say, in the MATH-th nodal divisor of MATH. We distinguish the case where MATH from the case where MATH. We first consider the case MATH. We let MATH be a uniformizing parameter of MATH and let MATH be defined by MATH for an integer MATH. Let MATH be the MATH-th one-parameter subgroup of MATH. We consider MATH . As in the proof of REF , we can pick a curve MATH covering MATH so that MATH is flat, MATH and is in general position of MATH. Further we can find an integer MATH so that MATH specializes to a point in the MATH-th irreducible component of MATH and is away from the nodal divisors of MATH. A simple analysis using the covering MATH of MATH shows that if we let MATH be the extension of REF to a family of stable morphisms, then MATH is still quasi-stable. Further, some irreducible component of MATH is mapped entirely to the MATH-th nodal divisor of MATH. This shows that we can assume without loss of generality that MATH maps an irreducible component MATH of MATH to MATH. We now show that there is a quasi-stable extension of MATH to MATH. Let MATH and MATH be as before so that MATH. We let MATH and MATH be the standard embedding associated to MATH. Namely, they are induced by the embedding MATH that keep the last coordinate REF. Let MATH be the composite of MATH with MATH. Then the same technique as before shows that we can find an integer MATH so that if we let MATH be defined by MATH, then the extension (possibly after a base change MATH) of MATH to a family of ordinary stable morphisms MATH is quasi-stable. This contradicts to the maximal assumption of MATH. Hence MATH must already be a family of non-degenerate morphism. Since MATH is flat, it must be pre-deformable and thus in MATH, extending MATH. This proves the Lemma. |
math/0009097 | The fact that the moduli stack MATH is separate over MATH follows from REF and that it is proper over MATH follows from REF . It remains to show that it is algebraic. Namely, MATH admits an étale cover by a scheme of finite type. We now show that it admits an étale covering by a quasi-projective scheme. Let MATH be the moduli stack of stable morphisms from pointed curves to MATH of type MATH, and let MATH be the substack of MATH consisting of all pre-deformable morphisms that are stable in the sense of REF . Clearly, MATH is a locally closed substack of MATH. Since members of MATH are stable morphisms to MATH, there is a natural morphism MATH. Because of REF , there is an integer MATH so that MATH is surjective. Hence MATH is surjected onto by a quasi-projective scheme since each MATH does. We now show that we can find a quasi-projective scheme MATH and a surjective étale morphism MATH. Let MATH be any closed point in MATH. Since REF is surjective, MATH is contained in the image of MATH for some MATH. Now let MATH be a chart so that MATH is contained in the image of MATH in MATH. To obtain a chart of MATH that contains MATH, we need to investigate the MATH-action on MATH. Since MATH is a MATH-variety, MATH admits a natural MATH-action. By the definition of stability (compare REF ), the MATH action on MATH has only finite stabilizer. Let MATH be the morphism induced by the group action. Namely, MATH. Then there is an open subset MATH that contains MATH so that MATH lifts to a morphism MATH so that MATH. Let MATH be a point so that MATH. Since the stabilizer of MATH (of the MATH-action) is finite, MATH is a smooth morphism near MATH. Hence we can find a locally closed subscheme MATH containing MATH so that MATH is étale at MATH. By shrinking MATH if necessary, we can assume MATH is étale near MATH. Therefore, the composite MATH is étale. Since MATH is bounded, we can find a finite number of étale charts of MATH, each quasi-projective, so that their union covers MATH. This proves that MATH is an algebraic (that is, NAME) stack. |
math/0009097 | The proof is straightforward and will be omitted. |
math/0009097 | Clearly, the composite of MATH with MATH is trivial. Hence by REF MATH which is isomorphic to MATH. This proves the Corollary. |
math/0009097 | The proof is similar to that of REF and will be omitted. |
math/0009097 | The proof is straightforward and will be omitted. |
math/0009097 | The proof is parallel to that of MATH and will be omitted. |
math/0009097 | The proof is straightforward and will be omitted. |
math/0009097 | The proof that the morphism MATH is finite and étale is straightforward, and will be left to readers. We now check that the degree of MATH is as stated. Let MATH be any element. It follows from the discussion preceding to the statement of this Proposition that there is a MATH so that the decomposition of MATH along the divisor MATH is the MATH-decomposition of MATH. Because MATH is stable, for all other MATH the decomposition types of MATH along MATH will be different from MATH (that is, MATH). We now let MATH be represented by MATH, after fixing an ordering on MATH so that the topological type of MATH is MATH. Thus MATH and MATH. Now let MATH be any permutation. We let MATH be the same morphism MATH except that the distinguished marked points of MATH are reordered according to MATH. This way MATH . Clearly, it is in the domain of MATH if and only if MATH. To derive the degree formula, we need to investigate the automorphism group MATH of MATH and the automorphism group MATH of MATH. First, the automorphism group MATH is naturally isomorphic to MATH. Because MATH is derived from gluing MATH and MATH along MATH and MATH, elements in MATH induce permutations of MATH. Hence we obtain a natural homomorphism of groups MATH whose kernel is isomorphic to MATH. Further the image of MATH lies in the subgroup MATH and the coset MATH is exactly the set MATH. This shows that MATH consists of exactly MATH distinct elements. Hence the degree of MATH is MATH . This proves the Proposition. |
math/0009103 | The main point is that induction takes projectives to projectives. All modules in this proof are assumed to be direct sums of simple finite-dimensional MATH-modules. The category of locally finite MATH-modules has enough projectives. This may be checked as follows. In the category of semisimple locally finite MATH-modules, every module is projective and injective. Therefore, if MATH is any such MATH-module, then MATH is projective in the category of locally finite MATH-modules. In particular, if MATH is a locally finite MATH-module, then MATH will be a quotient of the standard projective MATH, and therefore the category of locally finite MATH-modules has enough projectives. Finally, if MATH, then there will exist a projective cover MATH, therefore we can get MATH as a quotient MATH, where the last arrow is the canonical homomorphism MATH, which is given by MATH. This proves REF . For REF , we first note that, in this case, a locally finite MATH-module will in fact be a direct sum of finite-dimensional MATH-modules, and, therefore, if MATH is projective in the category of finite-dimensional MATH-modules, then MATH is projective in this slightly larger category. Then induction makes MATH a projective MATH-module. |
math/0009103 | Let MATH be an extension of MATH by MATH, where MATH; then we need to show that the centre of MATH acts semisimply. Let MATH be the action of MATH. Let MATH be the NAME vector field, and consider MATH. The NAME decomposition gives MATH, where MATH (respectively, MATH) is semisimple (respectively, nilpotent). Furthermore, MATH and MATH are respectively semisimple and nilpotent, they commute, the decomposition is unique, and each of them maps MATH into MATH. Note that MATH is semisimple, hence MATH is semisimple. Then MATH, the difference of two commuting semisimple endomorphisms, so MATH also acts semisimply on MATH, therefore MATH restricted to MATH is zero. Therefore we see that the nilpotent part MATH of the action of the NAME vector field is a nilpotent MATH-module endomorphism of MATH. Since MATH is nilpotent, we have MATH and MATH. If MATH, the only morphism MATH is zero. |
math/0009103 | The condition on MATH ensures that all the NAME modules involved are MATH-typical. By the remarks above, in MATH there is a projective resolution MATH of MATH, with MATH. The MATH-module MATH is typical, therefore semisimple, so it is a direct sum of NAME modules. We can calculate MATH as the cohomology of the complex MATH . Now, MATH . Therefore, if we substitute MATH into the above complex, we get MATH so there are no coboundaries, and we claim that every MATH gives a cocycle. Indeed, MATH, where MATH, and it is easy to verify that there are no nonzero MATH-module maps MATH: define the height of a weight MATH to be MATH. Then the height of any MATH such that MATH is greater than or equal to MATH, while MATH. This weight calculation shows that MATH simply does not occur among the MATH, and we are done. Since all the NAME modules are typical, we can calculate the decomposition of MATH into a direct sum of NAME modules as the decomposition of MATH into a direct sum of MATH-modules. In particular, for MATH as in REF , the multiplicity of MATH is nonzero. |
math/0009103 | Consider the cohomology MATH as MATH varies. The complex computing this cohomology is finite-dimensional, and shifting the weights by MATH does not change the dimension of the components. We can therefore view it as a complex with fixed terms with a differential that depends polynomially on MATH. By REF , MATH for generic values of MATH. By semicontinuity, MATH can only increase at special values. |
math/0009103 | Consider a generalised weight module MATH, so that MATH as a vector space (recall that, for MATH, the subspace MATH is simply MATH in this case). We have the following simple fact: if MATH, then MATH. An immediate consequence is that the MATH, defined above, are submodules. |
math/0009103 | First, define a parity MATH by MATH for some MATH; this is well-defined, by our hypothesis. It extends linearly to a function MATH. Now, suppose MATH is a generalised weight module whose support is contained in a single MATH-coset MATH. Shift the parity function to MATH by setting MATH, where MATH is fixed arbitrarily. Consider the linear map MATH, uniquely defined by requiring that, if MATH, then MATH. Finally, note that, if MATH, then MATH, so MATH. Therefore MATH commutes with the action of MATH, and MATH breaks up into a direct sum of two submodules, determined by this new parity. |
math/0009103 | We define a relation MATH on the set of highest weights, such that MATH implies the existence of a finite-dimensional indecomposable having both MATH and MATH as subquotients, and, consequently, that MATH and MATH are in the same MATH-block. Finally, we show that if MATH, that is, if MATH, then we can get from MATH to MATH with a finite number of intermediate steps. If MATH, then all NAME modules in MATH are simple. In that case, by REF , there exists a nonsplit extension of MATH by MATH as long as MATH. For the atypical case MATH, REF still ensures that there is an indecomposable module with subquotients MATH and MATH, and hence both MATH and MATH are subquotients, for MATH such that MATH. The relation MATH is now defined as follows: set MATH if MATH for some MATH. We have established that MATH implies that MATH and MATH are in the same block. Finally, it is clear that the closure of MATH to an equivalence relation on MATH has equivalence classes which are exactly the cosets MATH. |
math/0009103 | Let MATH be a weight such that MATH, MATH is simple, MATH for every root MATH of MATH, and all the MATH are also simple. For example, any sufficiently dominant weight, that is, MATH, with MATH, will do. Then, by REF , there exists a nontrivial extension of MATH by MATH. Therefore, the MATH-quiver of each block contains a subquiver consisting of a vertex MATH with arrows from it to MATH for each root MATH of MATH. Not counting multiplicities, there are MATH such roots, namely, MATH with MATH and MATH. Since MATH, we always have MATH, and the resulting quiver is already wild (q.v. REF ): MATH . |
math/0009103 | Suppose MATH is NAME dense. First of all, MATH and therefore MATH by REF . Next, applying the NAME Theorem, produce a basis MATH of MATH over MATH, and write MATH as MATH, with the MATH. Applying MATH to MATH gives MATH and, since the MATH are linearly independent, each MATH and therefore MATH. This shows that MATH and that MATH . |
math/0009103 | For any MATH, we have MATH, therefore MATH acts by some scalar MATH on MATH. It therefore also acts by multiplication by MATH on the finite-dimensional quotient MATH of MATH. Furthermore, if MATH and MATH are in the same block, then MATH. However, by REF , any block MATH is already a NAME dense subset of MATH. |
math/0009104 | As usual we note that the components of the total NAME class are universal polynomials in the NAME classes MATH and we may let MATH be the universal bundle. Then as we may think of the coefficients in the universal polynomials as rational numbers we can note that the NAME classes of degree REF to MATH of MATH vanish if and only if the NAME polynomials of the same degrees do, that is, if and only if MATH vanishes in the same degrees. Thus the first part followed from the NAME formula (compare CITE) MATH . Furthermore, if the NAME classes of degree REF to MATH vanish for a bundle MATH, then it is clear that MATH, as can be seen from NAME 's formula. Hence in degree MATH we have MATH is MATH and using the NAME formula again gives the desired formula. |
math/0009104 | By twisting MATH by a line bundle on MATH so that it is trivial along the zero section we may assume that it is of order MATH on MATH. Denoting the class of MATH in MATH by MATH we then either have that MATH has support of codimension MATH if MATH or is zero if MATH. Indeed, from the relation MATH and the fact that MATH is nilpotent we get that MATH is divisible by MATH and that element is supported in codimension MATH. Now in the first case the image under MATH of the support has codimension MATH on MATH and so we may safely remove it and may assume that MATH in MATH. Consider now the NAME bundle MATH on MATH, where MATH denotes the dual abelian variety. By base change MATH is the (derived) pullback along the zero-section of MATH of the sheaf MATH. We have that MATH and so MATH. Now, a fibrewise calculation shows that MATH has support along the inverse of the section of MATH corresponding to MATH. As that section is everywhere disjoint from the zero section the pullback of MATH along the zero section is REF and thus MATH. Using REF and applying the total NAME class then gives MATH which in turn gives the lemma. |
math/0009104 | This follows directly from the structure of MATH and NAME 's prime number theorem. |
math/0009104 | For any prime MATH larger than MATH we can apply REF on the cover obtained by adding a line bundle everywhere of order MATH. Projecting down to MATH again and using that that cover has degree MATH gives MATH. We then finish by using REF (and noting that the factor MATH causes no trouble as by using several primes we see that no prime MATH can divide the smallest annihilating integer). |
math/0009104 | The proof is almost the same as that of REF only that now we consider instead the cover given by putting a full level MATH-structure on MATH. This time we therefore get that MATH, where MATH is some irrelevant positive integer. Using again REF we conclude. |
math/0009104 | The first part follows immediately because MATH is a symplectic vector bundle. As for the second part we may assume that the characteristic is MATH as the case of positive characteristic follows from the characteristic MATH case by specialisation. In that case we may further reduce to the case of the base field being the complex number. We may also prove the annihilation of MATH in MATH-adic cohomology for a specific (but arbitrary) prime MATH. Now, the existence of NAME connection on MATH means that it has a discrete structure group. More precisely, the fundamental group of the algebraic stack MATH is MATH and MATH is the vector bundle associated to the representation of it given by the natural inclusion of MATH in MATH. This complex representation is obviously defined over the rational numbers so we may apply CITE with field of REF. We thus conclude that MATH is killed MATH, where MATH is defined as MATH and MATH is the image of the NAME group of the field of definition of the cyclotomic character. However, as the base field is MATH this image is all MATH and the result follows from the definition of MATH. |
math/0009104 | The MATH-adic part implies the integral cohomology part so we may pick a prime MATH different from the characteristic of the base field and look at MATH in MATH-adic cohomology. What we want to show is that if MATH is odd and MATH and MATH is the largest integer such that MATH, then MATH has order at least MATH and similarly for MATH. We will do this by defining a map MATH, such that inverse image of the MATH to MATH has order MATH. Now, MATH is isomorphic to MATH and hence an element in it has order MATH if and only if its reduction modulo MATH is non-zero. As MATH injects into MATH it is enough to show that the pullback of MATH is non-zero in MATH. Assume first that MATH is odd. Consider now any NAME cover MATH with NAME group MATH which is ramified at MATH, MATH and MATH with ramification group of order MATH, MATH, and MATH respectively (the existence of such a cover follows directly from NAME theory). By the NAME formula the genus of such a covering fulfills the relation MATH, that is, MATH. The action of MATH on MATH has to contain a copy of the irreducible representation given by the action of MATH'th roots of unity as it must be faithful and as MATH has dimension MATH it must be equal to it. Furthermore, this action gives a map MATH. If now MATH is the natural generator (the NAME of the generator of MATH corresponding to the identity map MATH) then the total NAME class of it is MATH which equals MATH. This gives the non-triviality when MATH and when MATH we simply add a principally polarised factor on which MATH acts trivially. When MATH we may assume that MATH as the lower bound to be proven for MATH is implied by the one for MATH. We then make essentially the same construction, a NAME cover with group MATH of MATH ramified at three points with ramification groups of order MATH, MATH, and MATH respectively of genus MATH. The rest of the argument is identical to the odd case. |
math/0009104 | This follows immediately from REF . |
math/0009104 | The top NAME class of MATH is MATH. |
math/0009104 | The class MATH vanishes up to torsion on MATH; for dimension reasons MATH is then represented by a multiple of the fundamental class of MATH. |
math/0009104 | Consider (for MATH) the space MATH of products of a principally polarized abelian variety of dimension MATH and an elliptic curve. It is the image of MATH in MATH under a morphism to MATH which can be extended to a morphism MATH. Since MATH is the affine MATH-line we find a rational equivalence between the cycle class of a generic fibre MATH with MATH a generic point on the MATH-line and a multiple of the fundamental class of the boundary MATH. |
math/0009112 | The main formula we need is MATH which follows easily from the properties stated of the BGG operators. Since MATH, we have MATH this last because by the MATH assumption, MATH annihilates MATH. If MATH, then MATH annihilates MATH too, and the Right-hand side is zero. Combining that with the formula for MATH gives the first result. If MATH, then the Right-hand side is MATH, and two applications of the formula for MATH give the second result. |
math/0009112 | We reduce to the well-studied case (see CITE) of a single flag manifold. Let MATH; since MATH it suffices to consider the map MATH. And MATH, so (omitting the MATH) we're studying the fibers of the composite MATH, as already done in CITE. |
math/0009112 | Let MATH be the set of flags in relative position MATH to MATH, MATH to MATH, and MATH to MATH where MATH are three flags in generic relative position. Then by codimension count (and the usual appeal to NAME 's transversality theorem) the set MATH is finite. However, since none of MATH care what MATH is (since by assumption none of them have a descent at MATH), the set MATH is a union of MATH's. These two facts are only compatible if MATH is empty. |
math/0009112 | Let MATH be the set of flags in relative position MATH to MATH, MATH to MATH, and MATH to MATH where MATH are three flags in generic relative position. Then as in the previous proof, the set MATH is a union of MATH's, reflecting the ambiguity in MATH. If we change one of MATH to have a descent at MATH, each of these MATH's is cut down to a single point. But it doesn't matter which of MATH gets this new descent. |
math/0009112 | Since the second argument has no descents, any place MATH that MATH has a descent and MATH does not gives us an opportunity to cycle a descent from the first argument to the third, replacing MATH. This modification keeps the sum of the lengths MATH and neither causes nor breaks the condition MATH. So we can reduce to the case that any descent in MATH occurs at a descent of MATH. If MATH: then each ascent in MATH occurs at a descent of MATH. By our reduction above, this means that MATH has no ascents. So we're looking at MATH which by assumption is MATH. Conversely if MATH: then no column MATH has three ascents (the NAME problem MATH is not dc-trivial). By our reduction, this means that MATH has no ascents. So MATH. By the assumption on the total length, MATH, so MATH as desired. |
math/0009112 | Let MATH be MATH with the numbers in the MATH-th and MATH-th positions switched, decreasing the number of inversions by exactly one (and so that MATH). In particular every number in MATH physically between the MATH-th and MATH-th positions is not numerically between MATH and MATH. We want to show that MATH unless MATH, in which case MATH. First, we treat the case MATH. If MATH, then neither MATH nor MATH has a descent at MATH. So we can cycle the descent from the second argument of MATH into the third, making them MATH. Now REF tells us that this MATH. Whereas if MATH or MATH (still in the case MATH), then none of MATH, MATH, or MATH have a descent at MATH, and therefore MATH vanishes as it's supposed to. Now take the case MATH. Then since MATH has only one fewer inversion than MATH, MATH must have the same descent-pattern as MATH. Now we reduce (much as in REF ) by cycling descents between the first and third arguments, in order to move the positions MATH and MATH closer together. We can do this descent-cycling in the MATH column as long as MATH, and the MATH column as long as MATH. If MATH are both on the same side of the MATH divide, they can be brought next to each other (by for example, just moving one of them). If MATH are on opposite sides of the divide, we can at least get MATH up to MATH, and MATH down to MATH. Either way we reduce to the MATH case and therefore get the same answer as Monk's rule. |
math/0009115 | There are obvious isomorphisms MATH and MATH, compatible with the first NAME class map, so it suffices to consider the case MATH is a quotient stack, and we are reduced to CITE. |
math/0009115 | A finite cover by a scheme MATH exists by CITE (or in the case of primary interest - MATH a NAME stack - by CITE). We may suppose MATH normal, hence finite flat over MATH away from a closed substack MATH of MATH which is empty or of codimension MATH. This implies (compare CITE) that MATH is a quotient stack, and REF applies. |
math/0009115 | Clearly, REF implies REF , and REF implies REF . As MATH is injective, REF implies MATH lies in the image of the boundary homomorphism MATH of the NAME sequence. If MATH, then MATH is isomorphic to a MATH-quotient of the principal bundle on MATH associated to MATH. So, REF implies REF . Given a line bundle as in REF , we can identify MATH with a MATH-quotient of the associated principal bundle MATH. Now MATH is the principal bundle of a class MATH with MATH for some MATH prime to MATH, and REF holds. |
math/0009115 | We follow the program of NAME Intersection Theory CITE: MATH is a MATH-quotient of MATH minus the zero section, so scheme approximations to MATH are complements of zero sections of line bundles over MATH: MATH for MATH sufficiently large. The result now follows from the excision sequence MATH . |
math/0009115 | We compactify MATH by setting MATH, the (complex) projectivization of the NAME sum of MATH and a trivial complex line bundle. Now we have MATH and MATH, complex line bundles over MATH, with MATH. The NAME sequence in cohomology gives MATH . Let MATH denote the tautological complex line bundle on MATH; then the projective bundle theorem dictates MATH via MATH. Since MATH is trivial on MATH and MATH, the leftmost map in REF is MATH . The result is now immediate. |
math/0009115 | Let MATH denote the MATH twist of the tautological line bundle on MATH. The complement of the zero section in MATH is homotopy equivalent to MATH, so by standard arguments, if MATH is a complex line bundle on MATH with MATH, then the complement, over MATH, of the zero section of MATH is homotopy equivalent to MATH. Now the result follows from REF , plus the general fact that MATH and MATH are torsion free. |
math/0009115 | We may assume MATH is reduced. Consider the components of the regular locus MATH. For each MATH, if we let MATH denote the restriction of MATH over MATH, then we claim the image of MATH contains MATH. Indeed, this follows by REF when MATH is NAME locally trivial; otherwise by REF combined with REF , MATH induces an isomorphism of NAME groups in dimension MATH. Now the desired result follows by comparing the excision sequences for MATH and MATH. |
math/0009115 | Denote by MATH the NAME number of MATH; then MATH is strictly less than the second NAME number MATH of MATH. Let MATH be the NAME group of MATH. Without loss of generality, we may suppose the image of MATH in MATH is a primitive lattice element. Then we can write MATH where MATH has rank MATH. Recall that the hypotheses dictate that MATH is topologically (but not algebraically) the projectivization of a rank REF complex vector bundle on MATH. To compute the obstruction map for MATH, we use the constructions and notations of the proof of REF applied to MATH and MATH. In particular, MATH denotes a topological complex rank REF vector bundle such that MATH, and hence MATH . Consider the following claim: the NAME number of MATH is MATH and the MATH-span of the NAME group of MATH is the MATH-span of MATH together with the element MATH, for some MATH. Given this, then, in MATH we have MATH since for any MATH, MATH. Recall that we have MATH. Now MATH, and by REF , MATH, so the image of MATH under the obstruction map REF is nonzero. The obstruction map for MATH is thus completely determined by its vanishing on elements of MATH and its nonvanishing on MATH. It remains to verify the claim. MATH is an extension of MATH by the free group generated by MATH. So MATH is an extension of MATH by some class which pulls back to MATH on MATH. But MATH differs from MATH (the pullback of MATH) by MATH, which lies in MATH. So, indeed, the NAME number of MATH is MATH, and for some MATH, MATH is algebraic on MATH. |
math/0009119 | REF follows from REF : MATH . Now we prove REF : MATH as claimed. Here we have used REF , the definitions and REF . For the proof of REF , we first observe that, if MATH, then MATH . Using that MATH is a coalgebra map, REF , we conclude that MATH . The proof of REF has no difference with the proof of the analogous statement in CITE. |
math/0009119 | We have MATH where we used REF. |
math/0009119 | It follows from results of CITE, CITE, CITE, CITE and CITE that MATH is the positive part of the so-called NAME kernel corresponding to the NAME matrix MATH. See CITE for details. The presentation by generators and relations follows from the considerations in the last paragraph of p. REF and the first paragraph of p. REF referring to CITE. The statement about the basis is CITE. |
math/0009119 | CASE: Let us first assume that the braiding is symmetric, that is MATH for all MATH. By CITE we can assume moreover that the NAME matrix MATH is connected. From our assumptions on the orders of the MATH we then conclude that the braiding has the form MATH for all MATH where MATH is a root of unity of order MATH. See CITE. Hence the Theorem follows directly from REF . CASE: In the case of an arbitrary braiding we know from REF that there exists a finite abelian group MATH satisfying: CASE: The braiding MATH of MATH can be realized from a NAME module structure over MATH that we continue denoting by MATH, compare REF . CASE: There exists a cocycle MATH with corresponding MATH such that the braiding of MATH is symmetric. Let MATH be the isomorphism having the same meaning as in REF. CASE: The braiding of MATH is given in the basis MATH by a matrix MATH such that MATH and the order of MATH is again odd for all MATH and MATH. If MATH, MATH denote the canonical maps, then we have a commutative diagram MATH . Clearly, MATH; if MATH is a set of generators of the ideal MATH with MATH then by REF MATH is a set of generators of the ideal MATH. By the symmetric REF , we know the generators of MATH. Let us denote MATH. Then by REF , we have MATH and MATH, MATH where MATH are non-zero scalars. This implies the first claim of the Theorem. The second follows in a similar way. |
math/0009119 | CASE: As in the proof of REF we first assume that the braiding is symmetric. If MATH, then MATH and hence the corresponding NAME relation REF says that MATH. Thus, we can easily reduce to the connected case. In such case, MATH as before and the Theorem is shown in CITE. CASE: In the general case, we change the group as in the proof of REF . The isomorphism MATH respects the NAME relations up to non-zero scalars by REF . Also, it maps subcoalgebras stable under the action of the group to subcoalgebras by REF . We conclude from REF that MATH is a subcoalgebra of MATH. |
math/0009119 | If MATH then MATH (otherwise MATH) and MATH (otherwise MATH). If MATH then MATH (otherwise MATH) and MATH (otherwise MATH). Assume that MATH. Then MATH . Then MATH divides MATH and analogously, MATH divides MATH. So that MATH by the assumptions on the order of MATH and MATH; by symmetry, MATH. Assume that a vertex MATH is linkable to MATH and MATH. Then MATH, so MATH. |
math/0009119 | . Clearly, MATH is an algebra map if and only if REF holds for all MATH, MATH. It follows also easily that REF holds when MATH, or MATH, or MATH and MATH. Next, let MATH and MATH be arbitrary elements; one can then check that REF holds for MATH and MATH if it holds for all the possibilities MATH and MATH; MATH and MATH; MATH and MATH; MATH and MATH. From this observation and the hypothesis the Lemma follows. |
math/0009119 | By induction on the number of connected components. Here is the first step: REF is true if the NAME diagram corresponding to MATH is connected. Let MATH be a NAME module over MATH with MATH and pick MATH. By REF and the formulas for the biproduct, there exists a unique algebra map MATH such that MATH, MATH. Also, by REF again, there are algebra maps MATH, MATH such that MATH, MATH. Let MATH, MATH, MATH, MATH; then MATH is an algebra map by REF. It is clear now that MATH is an isomorphism with inverse MATH; thus MATH is a NAME algebra and has the desired dimension by the dimension formula in REF . For the rest of this proof we assume: there exists MATH such that MATH, respectively, MATH, if MATH and MATH, respectively, MATH. Let MATH. Let MATH, where the order of MATH is the least common multiple of MATH and MATH, MATH. Let MATH be the unique character of MATH such that MATH, MATH, MATH. This is well defined because MATH divides MATH for all MATH. CASE: MATH, with generators MATH, , MATH (instead of the MATH's) and MATH; CASE: MATH, with generators MATH (instead of the MATH's) and MATH. Note that the linking datum of MATH is empty since MATH is connected. By the recurrence hypothesis, MATH and MATH. CASE: For each MATH, MATH, there exists a unique character MATH such that MATH, MATH. CASE: Let MATH be an arbitrary linking datum. For each MATH, MATH, there exists a unique MATH - derivation MATH such that MATH, MATH. CASE: There exists a unique NAME algebra map MATH such that MATH. CASE: We have to show that MATH preserves the relations REF. This is clear for REF. We check REF: let MATH such that MATH. Then MATH by REF. So that relations REF hold and REF is proven. CASE: This is equivalent to: there exists an algebra map MATH such that MATH, MATH. Then MATH is of the form MATH and MATH is the desired derivation. So, we need to show that the relations REF hold for the matrices in REF. This is evident for REF. For REF it amounts to MATH, which follows from REF when MATH. For REF the argument is clear. Finally, the left hand side of REF for MATH, is REF, whereas the right-hand side also vanishes since MATH by REF again. CASE: It is enough to verify that MATH, MATH satisfy the defining relations REF for MATH. Indeed, this will automatically imply that MATH is a NAME algebra map. Note that REF are empty since the NAME diagram of MATH is connected. For REF, it is enough to verify that the equalities hold when applied to the generators MATH, MATH since both sides are algebra maps. This is now not difficult; for instance MATH . The first relations in REF for MATH hold since MATH divides MATH for all MATH. For REF we need again to verify only on generators, since both sides are skew-derivations; this verification is in turn straightforward. The left-hand side of the NAME relations REF is a skew-derivation by CITE; again we are reduced to see that MATH, MATH, MATH. Write MATH, where MATH is a root of REF. Then MATH . Similarly, MATH since MATH is a homogeneous polynomial in MATH, MATH of positive degree. Finally, relations REF follow from the next Lemma. Let MATH be a finite dimensional pointed NAME algebra generated as an algebra by group-like elements and a family MATH, MATH, of MATH-primitives, for some MATH. Let MATH be the algebra presented by generators MATH and MATH with exactly the same relations as for MATH except for REF; it is a NAME algebra via REF. Let MATH. Assume there exists a NAME algebra map MATH such that MATH and MATH satisfy MATH for all MATH. Then MATH for all MATH. There exists a NAME algebra projection MATH such that MATH and MATH for all MATH. Let MATH be the subalgebra of MATH generated by MATH, MATH, and MATH, MATH. We claim that MATH for all MATH. Clearly, this implies the Lemma. By REF , we know that MATH is a NAME subalgebra of MATH. We have to prove that MATH for MATH a monomial in the group-likes of MATH and the MATH's. We do this by induction on the length of the monomial. We first check the case of length REF. Here we show more generally that MATH for all MATH, MATH and MATH or MATH of the form MATH, with MATH group-like, MATH, and MATH. Note that each element in MATH is a linear combination of such MATH's since MATH. The case when MATH is a group-like is clear. Let MATH, with MATH group-like, MATH, and MATH. Then MATH, and MATH where we used REF and MATH. Assume then that MATH where MATH and MATH are monomials satisfying the claim. Since MATH and MATH are NAME algebra maps, we have MATH. We are ready now to conclude the proof of the Theorem. Consider the cocycle MATH obtained as in REF from the map MATH constructed in REF . Consider the NAME algebra MATH; it has dimension MATH. We claim that the group-like elements MATH are central in MATH for all MATH. By definition of MATH, we have to show for all MATH, MATH and MATH . Since MATH in MATH for all MATH, MATH, it is enough to check REF on generators of MATH and MATH. This in turn follows easily from the definitions. Let MATH be the quotient of MATH by the central NAME subalgebra MATH with quotient map MATH. Then MATH by a result of the second author CITE. Next we claim the existence of a surjective algebra map MATH such that MATH for MATH, MATH, MATH. Again we have to verify the relations REF. Up to REF these relations already hold in MATH. For REF, it is enough to show that MATH . A tedious computation shows that the left-hand side is equal to MATH. Since MATH, we have MATH. Hence the claim follows if we choose MATH for all MATH, MATH. On the other hand, we have algebra maps MATH, MATH given by MATH, MATH, MATH, MATH, MATH, MATH. Here we use that MATH divides MATH for all MATH. Let MATH be defined by MATH for all MATH, MATH. We claim that MATH is an algebra map. By REF , we have to verify MATH, for all generators. This is a straightforward task; for the case MATH and MATH we need again the condition MATH. Since clearly MATH factorizes through MATH, MATH is an isomorphism and the Theorem follows. |
math/0009119 | CASE: Assume that MATH, MATH for some MATH. Substituting MATH and MATH in MATH and using MATH we conclude that MATH . Changing the rôles of MATH and MATH we obtain in the same way MATH . First assume that MATH. In particular, MATH and MATH or MATH. If MATH, respectively, MATH, then we get from REF, respectively, REF, that MATH, respectively, MATH, which is not possible. Next assume that MATH. If MATH then MATH divides MATH by REF. The only possibility is MATH and MATH; but this was excluded in the hypothesis. If MATH then MATH divides MATH by REF and MATH divides MATH by REF; but this contradicts our general assumptions on the MATH's. Finally, if MATH and MATH then MATH and MATH. We discuss the different possible values of MATH. If MATH or MATH, by REF and since MATH is odd we see that MATH or REF, cases excluded by hypothesis. If MATH then MATH. By REF, MATH divides MATH; this discards everything except MATH. But in this last case, MATH and MATH divides REF by REF, a contradiction. Finally, MATH is impossible by analogous arguments. CASE: Assume that MATH. We consider first the case MATH. Evaluating at MATH, we get MATH; hence MATH. Evaluating at MATH, we get then MATH. Since MATH we finally obtain MATH . The possible values of MATH are REF or REF, where REF, respectively, REF, is only possible if MATH is of type MATH, MATH or MATH, respectively, MATH. This contradicts the hypothesis. We consider finally the case MATH; so that MATH. Since MATH is connected, there is a sequence MATH of elements in MATH such that MATH for all MATH, MATH. Then MATH by substituting MATH, then MATH and so on. Note that MATH since MATH and MATH. Hence MATH . The possible values of the sum in REF are MATH or MATH. Hence REF contradicts our assumptions in REF . |
math/0009119 | It is known that MATH, see for instance CITE. REF then follows from REF . To prove REF , let us assume that MATH. By REF again, MATH, for some MATH. We can choose MATH when MATH or else if MATH. That is, MATH is a linking datum for MATH, MATH and MATH; and REF hold. |
math/0009119 | Let us fix MATH. Let MATH be the algebra presented by generators MATH, MATH, MATH and relations REF; it is a NAME algebra via REF. Let MATH and let MATH be the subalgebra of MATH generated by MATH, MATH, and MATH, MATH. By REF , we know that MATH is a NAME subalgebra of MATH. Note that MATH is a graded NAME algebra with trivial coradical. By the choice of the MATH's in MATH and REF , we see there is a well-defined NAME algebra map MATH such that MATH, MATH. The image of MATH under this map is a finite dimensional pointed NAME algebra; it has a trivial coradical by CITE and therefore it is trivial. This implies the Lemma. |
math/0009119 | By REF , there exists a linking datum MATH and a surjective NAME algebra map MATH, where MATH. But MATH by REF ; hence MATH is an isomorphism. |
math/0009119 | By CITE we can realize the braiding over a suitable finite abelian group MATH and twist with a REF-cocycle MATH such that the resulting braiding MATH is symmetric with elements of odd order, has the same diagonal elements and is of NAME type with the same NAME matrix MATH as MATH. We can now conclude from CITE that MATH is of NAME (see CITE). Let MATH be the NAME module over MATH with braiding MATH. Since MATH and MATH have the same dimension, MATH is finite-dimensional. Then MATH is of finite NAME type by CITE. |
math/0009119 | Define MATH, MATH. In both cases we have to show MATH. We assume that MATH is not REF. Let MATH, MATH, with MATH for all MATH. Then action and coaction on MATH are given by MATH, MATH and MATH, MATH for all MATH. Hence MATH, MATH are linearly independent since MATH or MATH. (If both equalities would hold, then MATH, and the NAME type condition MATH would give MATH, hence MATH and MATH which is impossible.) The braiding MATH of REF-dimensional NAME module with basis MATH is given by MATH . Then MATH. We claim that MATH is of NAME type and satisfies the relative primeness condition, that is there are integers MATH such that MATH . In both cases MATH is relatively prime to MATH, because of the hypothesis on MATH. This shows REF. We now prove REF . Then MATH, and it suffices to find an integer MATH relatively prime to MATH with MATH. First assume that MATH. Since MATH is relatively prime to MATH, it is enough to consider the MATH-th power of REF. Since MATH by the NAME condition for MATH, we have to solve MATH. Since MATH is of finite NAME type, the possible values of MATH are -REF, -REF, -REF, -REF, -REF (-REF, -REF respectively, -REF, -REF only occur if the type is MATH or MATH respectively, MATH); the possible values of MATH are -REF, -REF, -REF, (-REF, respectively, -REF only occur if the type is MATH or MATH respectively, MATH). Hence MATH and MATH are relatively prime to MATH by assumption, and the claim follows. (Note that MATH is never MATH). If MATH, then by connectedness there is a sequence MATH of elements in MATH such that MATH for all MATH, MATH. Then as in the proof of REF , MATH . Since the possible values of MATH are REF, -REF, -REF, the MATH-th power of REF leads to the congruence MATH, and the claim again follows. Assume REF , in particular MATH. If MATH, we get a contradiction since the algebra generated by MATH is finite-dimensional, hence MATH by CITE. If MATH, REF is solvable since MATH is odd. Thus we have shown that MATH is of NAME type and satisfies the relative primeness condition. Hence MATH is of finite NAME type by REF . In both cases MATH is a solution of REF, and MATH. Hence the possible values of MATH are REF, -REF, -REF, -REF, and we see that MATH. This contradicts our assumption, and we have shown the NAME relation MATH. |
math/0009119 | We fix a connected component MATH. Let MATH be the NAME submodule of MATH with basis MATH, and MATH the quotient of MATH modulo the NAME relations of all elements MATH with MATH in I. Let MATH. The map MATH factorizes over MATH, since the NAME relations hold in MATH. By REF the subalgebra MATH of MATH generated by the powers of the root vectors MATH, MATH, is a braided NAME subalgebra. As a coalgebra, MATH is pointed and has trivial coradical. Hence MATH is a finite-dimensional pointed and graded NAME subalgebra of MATH in MATH with trivial coradical. We have to show the root vector relation MATH, MATH in MATH, or equivalently that MATH is one-dimensional, that is MATH. Assume MATH. Since MATH is in MATH, there are MATH with MATH. By CITE, we conclude MATH. But this is a contradiction, since for all MATH, MATH implies MATH. For, MATH is the MATH-span of all monomials MATH . For any MATH there are natural numbers MATH, MATH, such that MATH, where MATH are the simple roots. By REF, MATH, where MATH. Hence for all MATH, MATH . It remains to show that MATH. Let MATH. Since the braiding is of NAME type, MATH and MATH . Hence, since all the MATH have order N, MATH . Therefore we obtain MATH . |
math/0009119 | Let MATH be the dual NAME algebra of MATH in the braided sense (see for example CITE). MATH is a graded braided NAME algebra in MATH with MATH, for all MATH. By assumption there are MATH, MATH, with MATH for all MATH, and a basis MATH of MATH with MATH for all i. Let MATH in MATH be the dual basis of MATH. Then MATH with MATH and MATH for all MATH. Thus MATH is a NAME module over MATH with the same braiding as MATH. By CITE, MATH is generated by MATH if and only if MATH. Hence by duality, MATH is generated by MATH, since MATH. It is easy to see that MATH. Hence there are canonical surjections of graded braided NAME algebras MATH . Here MATH is the tensor algebra, the elements MATH are primitive and of degree one, and both maps are the identity on MATH. The kernel MATH of the first map is a homogeneous ideal generated by elements of degree MATH, a coideal and stable under the action and coaction of MATH. Since MATH, where MATH is the largest ideal with the same properties as MATH, there is a canonical surjection MATH. The MATH satisfy the NAME relations REF by REF , and then the root vector relations REF by REF . Therefore it follows from the description of MATH in REF that MATH. This means MATH, hence by duality that MATH is generated by MATH. |
math/0009119 | Let MATH be a finite-dimensional pointed NAME algebra with coradical MATH and let MATH be the diagram of MATH. Then MATH is a NAME module of finite NAME type by CITE. Hence the claim follows from REF . |
math/0009119 | Let MATH be a finite-dimensional pointed NAME algebra with coradical MATH and diagram MATH and MATH as defined above. Since MATH and MATH are braided NAME algebras over MATH of the same dimension, and MATH, we can iterate this process and after finitely many steps we obtain a graded braided NAME algebra MATH over MATH with MATH and MATH. By a result of CITE using CITE which follows from CITE, the number of isomorphism classes of NAME modules MATH over the fixed group MATH with finite-dimensional MATH is finite. Thus we can take for MATH the product of the largest such dimension with the order of MATH. |
math/0009120 | The proof of the Nodal Domain Theorem is based upon deriving a contradiction from Hypothesis W: MATH has MATH weak nodal domains, and Hypothesis S: MATH has MATH strong nodal domains, respectively. We call the domains MATH and define MATH for MATH. None of the functions MATH is identically zero. Since they have disjoint supports their linear span has dimension MATH. It follows that there exist constants MATH such that MATH is non-zero and satisfies MATH for MATH. Without loss of generality we can assume MATH, where MATH denotes the standard scalar product on MATH. Therefore we have MATH . Under hypothesis W we know that MATH . Under hypothesis S we have MATH since the last eigenvalue that is equal to MATH has index MATH. It will be convenient to introduce MATH and to define MATH by MATH so that MATH for all MATH. CASE: Assuming hypotheses W or S, we have MATH. Proof. We have MATH . Summing over the vertex set yields MATH where MATH by symmetrizing. A term of the remainder MATH vanishes if MATH or MATH. If MATH and MATH, that is, MATH, then MATH and MATH lie in the same nodal domain and thus MATH, and the corresponding contribution to MATH vanishes as well. The only remaining terms are those for which MATH and MATH. So we see that MATH. Thus we have MATH. MATH . Under hypothesis S, REF , and REF lead to the desired contradiction, proving the second part of the theorem. Under hypothesis W, REF , and REF imply MATH. Since MATH is by construction orthogonal to all eigenvectors MATH, MATH, a simple variational argument implies MATH . For the second step of the proof of the Weak Nodal Domain Theorem we exploit the fact that the remainder MATH as a consequence of equ. CASE: We proceed with a unique continuation result for the function MATH. CASE: If hypothesis W holds, MATH, MATH, MATH, and MATH then MATH. Proof. If MATH, MATH, MATH, and MATH then MATH (otherwise MATH), and hence MATH. From MATH, MATH, and MATH we conclude that MATH and hence MATH. Now assume that MATH. Define MATH. Then MATH . We have MATH where MATH. Note that MATH by the assumptions of the lemma. Suppose for definiteness that MATH is a positive nodal domain. Then MATH satisfies MATH since otherwise one would have to adjoin MATH to MATH. Thus MATH is a connected set on which MATH. Therefore it is contained in the single (negative) nodal domain MATH. Therefore MATH . The terms in the sum are all negative, thus MATH. The same argument of course works when MATH is a negative nodal domain. MATH . We say that MATH is adjacent to MATH if there are MATH and MATH, MATH. Note that adjacent nodal domains must have opposite signs. Now consider a collection MATH of nodal domains such that MATH. Then there exists a nodal domain MATH, MATH, that is adjacent to some MATH, MATH; otherwise MATH would not be connected. Now we are in the position to prove the first part of the theorem. We assume hypothesis W and thus the conclusions of REF . Since MATH there exists an index MATH for which MATH. If MATH is a nodal domain adjacent to MATH then REF implies MATH. Since the graph MATH is connected by assumption, we conclude in a finite number of steps that MATH for all MATH. Hence MATH. This, however, contradicts the fact that MATH. |
math/0009123 | This is REF. |
math/0009123 | This is REF. |
math/0009123 | Every nontrivial representation of MATH in characteristic MATH has dimension MATH. Indeed, first assume that MATH. Then in the ``generic" case" of MATH such a representation must have dimension MATH, thanks to the already cited REF. If MATH then MATH and the smallest dimension is MATH, according to the Tables in CITE. Second, if MATH then this assertion follows from REF on p. REF; see also the Tables in CITE. Suppose MATH is a central extension of MATH. In addition, assume that either MATH or MATH is a double cover of MATH, that is, MATH is a central cyclic subgroup of order MATH in MATH. Suppose MATH is a finite-dimensional MATH-vector space and MATH is an absolutely irreducible faithful representation of MATH over MATH. Then MATH . Clearly, MATH defines an absolutely irreducible projective representation of MATH in MATH over MATH. Assume first that MATH. Then in the ``generic" case every absolutely irreducible nontrivial projective representation of MATH in characteristic MATH must have dimension MATH (see REF). If MATH then the Proposition follows from the Tables in CITE. Second, suppose MATH. Then MATH. All faithful absolutely irreducible representations of MATH in characteristic zero have dimension MATH (CITE, p. REF). This proves the Proposition in the case when MATH and also when MATH is a trivial double cover, that is, is isomorphic to a product of MATH and a cyclic group of order MATH. If MATH is a nontrivial double cover of MATH then it has precisely two non-isomorphisc MATH-dimensional absolutely irreducible representations in characteristic MATH (up to an isomorphism) CITE. However, none of them is defined over MATH. Indeed, each character of MATH of degree MATH takes on a value, whose square is MATH (REF on p. REF; CITE, p. REF). Assume that MATH is supersingular. Our goal is to get a contradiction. We write MATH for the MATH-adic NAME module of MATH and MATH for the corresponding MATH-adic representation. It is well-known that MATH is a free MATH-module of rank MATH and MATH (as NAME modules). Let us put MATH . Clearly, the natural homomorphism MATH defining the NAME action on the points of order MATH is the composition of MATH and the (surjective) reduction map modulo MATH . This gives us a natural (continuous) surjection MATH whose kernel consists of elements of MATH. We have assumed that the MATH-module MATH is absolutely simple. This implies that the MATH-module MATH is also absolutely simple. Here the structure of MATH-module is defined on MATH via MATH . The absolute simplicity of the MATH-module MATH means that the natural homomorphism MATH is surjective. By NAME 's Lemma, this implies that the natural homomorphism MATH is also surjective (see CITE, p. REF). Let MATH be the MATH-Tate module of MATH. It is well-known that MATH is the MATH-dimensional MATH-vector space and MATH is a MATH-lattice in MATH. Clearly, the MATH-module MATH is also absolutely simple. The choice of polarization on MATH gives rise to a non-degenerate alternating bilinear form (NAME form) CITE MATH . Since MATH contains all MATH-power roots of unity, MATH is MATH-invariant and therefore is MATH-invariant. This means that MATH is a subgroup of the corresponding symplectic group MATH. We have MATH . There exists a finite NAME extension MATH of MATH such that all endomorphisms of MATH are defined over MATH. We write MATH for the MATH-algebra MATH of endomorphisms of MATH. Since MATH is supersingular, MATH . Recall CITE that the natural map MATH is an embedding. Dimension arguments imply that MATH . Since all endomorphisms of MATH are defined over MATH, the image MATH commutes with MATH. This implies that MATH commutes with MATH and therefore consists of scalars. Since MATH is a finite group. Since MATH is a subgroup of finite index in MATH, the group MATH is also finite. In particular, the kernel of the reduction map modulo MATH consists of elements of finite order and, thanks to the NAME Lemma, MATH has exponent MATH or MATH. In particular, MATH is commutative. We have MATH . Since MATH consists of semisimple elements and rank of MATH is MATH, the group MATH is isomorphic (``conjugate") to a multiplicative subgroup of MATH. Since the exponent of MATH is either MATH or MATH, the group MATH is isomorphic to a multiplicative subgroup of MATH. Hence MATH is a MATH-vector space of dimension MATH. This implies that the adjoint action MATH is trivial, since every nontrivial representation of MATH in characteristic MATH must have dimension strictly greater than MATH. This means that MATH lies in the center of MATH. Since the MATH-module MATH is faithful and absolutely simple, MATH consists of scalars. This implies that either MATH or MATH. In other words, either MATH or MATH is a double cover. In both cases MATH is an absolutely irreducible representation of MATH of dimension MATH over MATH. But by REF applied to MATH and MATH, MATH . This gives us the desired contradiction. This ends the proof of REF and therefore of REF . |
math/0009124 | The first and the second properties of MATH comprise the definition of this vector field. The third property is equivalent to that MATH is a MATH-module; the fourth property is clear. Finally, the last property follows from the first (see, for example, the proof of CITE). We leave out the details of the proof for the sake of brevity. |
math/0009124 | The first assertion is clear by REF . If MATH is homogeneous, the representations MATH are equivalent to each other for all MATH. Furthermore, by applying diffeomorphisms fixing MATH, we conclude from REF that the operators MATH, MATH, are conjugate to each other for all MATH and, hence, have the same eigenvalues. In particular, the operators MATH and MATH have the same eigenvalues for all constants MATH. Since MATH, these eigenvalues must be zero. This implies that MATH, when MATH is Hermitian or Euclidean. Hence, MATH. |
math/0009124 | REF is a particular case of REF . Let us outline a direct proof, leaving entirely aside the question of homogeneity. Denote by MATH the natural projection. Fixing the direct product structure on MATH, we embed MATH into MATH as a NAME subalgebra. (Here we use the fact that the symplectic structure on MATH is independent of MATH.) Clearly, for a NAME vector bundle MATH over MATH, the pull-back MATH is a NAME vector bundle over MATH, where the action of MATH on MATH is given by the natural flat connection on MATH along the fibers of MATH. It is easy to see that deformation equivalent vector bundles have deformation equivalent pull-backs. Let us now construct the correspondence inverse to the pull-back. Let MATH be a NAME vector bundle over MATH. The restriction MATH to a fiber MATH inherits the structure NAME vector bundle. Namely, for MATH and MATH, we set MATH, where MATH is an arbitrary extension of MATH to a section of MATH. To verify that this bracket is well defined, we need to show that if MATH is a section of MATH vanishing along MATH and MATH, then MATH vanishes along MATH. To this end note that locally MATH, where MATH are some functions on MATH vanishing along MATH and MATH are some sections. Hence, MATH . Clearly, every term on the right hand side vanishes along MATH and hence so does the left hand side. By REF , MATH is a trivial vector bundle with the NAME vector bundle structure arising from a flat connection. In other words, we obtain a flat MATH-fiberwise connection MATH on MATH. Let MATH be the space of flat sections of MATH. The spaces MATH form a vector bundle MATH over MATH. The sections of MATH can be identified with a subspace of MATH formed by sections which MATH-horizontal. It is easy to see that MATH is invariant under the bracket with elements of MATH and thus inherits the structure of a NAME vector bundle via the embedding MATH. Therefore, MATH. It is clear that the correspondence MATH preserves deformation equivalence. This completes the proof of the proposition. |
math/0009124 | The argument is similar to the proof of REF with NAME 's theorem used to introduce a local (in MATH) direct product structure on MATH. |
math/0009124 | In what follows we assume that the vector bundle MATH is trivial. The proof of the general case requires only superficial changes. In the calculations below we use the sign convention of CITE. Fix a trivialization MATH and thus identify sections of MATH with functions MATH. Let MATH be a NAME vector field on MATH. Since MATH is homogeneous, there exists a function MATH such that MATH. Combining REF with REF, we obtain by a straightforward calculation that MATH for any MATH. Let now MATH be a MATH-cocycle on MATH. Consider the MATH-vector field MATH on MATH defined by MATH . We complete the proof by showing that MATH. This equality is verified by the following (also straightforward) calculation. First recall that MATH where we set MATH . The NAME - NAME REF guarantees that MATH . Hence, MATH . Obviously, MATH . By plugging this into REF, we eliminate the second and the third double sum. As a result of these cancelations, we obtain MATH . By REF, we have MATH. Therefore, MATH . The first two terms in this formula add up to zero because MATH is a cocycle on MATH. Hence, MATH . |
math/0009124 | The argument is by ``pushing the non-linearity of MATH to infinity". Pick a trivialization of MATH so as to identify MATH with MATH as ordinary, not NAME, vector bundles. With the trivialization fixed, the structure of a NAME vector bundle on MATH is given by the MATH-valued vector field MATH on MATH satisfying the NAME - NAME REF as described in REF. Using the structure of a linear space on MATH, we write MATH where MATH is the value of MATH at the origin. The constant MATH-valued vector field MATH on MATH also satisfies the NAME - NAME equation. In fact, MATH determines the NAME vector bundle MATH associated with the representation MATH. Note that the NAME - NAME equations for MATH and MATH imply that MATH where MATH is the NAME structure on MATH. To prove the theorem it suffices to connect MATH and MATH by a family of vector fields MATH, MATH, satisfying the NAME - NAME equation (see REF ). Denote by MATH the dilation by factor MATH, that is, MATH for MATH and MATH. Let us show that MATH satisfies the NAME - NAME equation. To this end we first observe that MATH since MATH is a constant vector field. Hence MATH. Likewise, MATH, for MATH is linear. Therefore, MATH . Here the first equality follows from that MATH satisfies the NAME - NAME equation and the last one is a consequence of REF. Finally, observe that MATH extends smoothly in MATH to MATH with value MATH and hence MATH, MATH, is the required homotopy between MATH and MATH. |
math/0009124 | The algebroid MATH is isomorphic to the action algebroid of the coadjoint action on MATH. The latter integrates to the action algebroid MATH. By REF , there is a one-to-one correspondence between NAME and NAME-equivariant vector bundles over MATH. As is shown in CITE (or in CITE for closed balls centered at MATH), every NAME vector field on MATH is Hamiltonian. Hence, a NAME vector bundle over MATH is automatically homogeneous. |
math/0009124 | The argument is similar to the proof of NAME 's theorem on the formal normal form of a vector field and to NAME 's proof of the formal linearization theorem for NAME structures, CITE. Fix a (formal) trivialization of MATH. We write MATH as a formal power series on MATH and assume that this formal power series contains only a zero order term and terms of degree greater than or equal to MATH: MATH . Here, as in the proof of REF , the zero order term MATH is associated with the representation of MATH on MATH and MATH is a MATH-valued vector field on MATH whose coefficients are homogeneous polynomials of degree MATH. As usual, the dots denote terms of degree higher than MATH. The vector field MATH can be interpreted as a cochain on MATH with coefficients in MATH. Denote by MATH the differential MATH, where MATH is turned into a MATH-module via MATH. The NAME - NAME REF implies that MATH that is, MATH is a cocycle. Our goal is to find a formal change of trivialization MATH of MATH which eliminates the MATH-th order term MATH. We look for MATH in the form MATH, where MATH is a homogeneous MATH-valued polynomial on MATH of degree MATH. (In the Hermitian case, MATH is MATH-valued and MATH.) Under any formal change of trivialization MATH, the MATH-valued vector field MATH transforms to MATH. Hence, MATH . In other words, to eliminate MATH, we need to find MATH with MATH. Since MATH is semisimple, MATH and, by REF, MATH does exist. To eliminate all non-constant terms in the power series MATH, we argue inductively. Let MATH be a change of trivialization with MATH and MATH a change of trivialization eliminating the second order terms in the resulting power series, etc. Then the composition MATH eliminates all terms up to degree MATH. Furthermore, the terms of any given degree in the expansion of MATH stabilize as MATH. Hence, MATH is well defined as a formal power series. This is the required change of trivialization. |
math/0009124 | Let us set MATH and MATH. By REF, to establish REF, it suffices to show that MATH and MATH are anti-isomorphic. The required anti-isomorphism MATH is MATH where MATH is the symplectic structure on MATH. Combined with REF, this proves REF. Moreover, it is clear that the bijection REF induces a bijection on the level of homotopy equivalence classes thus giving REF. Furthermore, REF preserves the classes of Hermitian or Euclidean vector bundles. To prove REF, we will show that the bijection REF sends homogeneous vector bundles to homogeneous ones. We need the following fact, which is implicitly contained in CITE and explicitly in CITE: Let MATH be NAME vector fields on MATH. Then there exists a Hamiltonian vector field MATH on MATH and a NAME vector field MATH on MATH such that MATH . Furthermore, MATH. Let MATH be a homogeneous NAME vector bundle over MATH and let MATH be a NAME vector bundle over MATH which corresponds to MATH under REF. Denote by MATH the vector bundle MATH over MATH. Let MATH be a NAME vector field on MATH. Denote by MATH the (local) flow of MATH in some time MATH, by MATH its lift to MATH, and by MATH the (local) flow of MATH in time MATH on MATH. The flow MATH lifts to MATH. Indeed, for MATH and MATH in MATH such that MATH, we set MATH . (Here we use the fact that MATH.) It is easy to see that MATH is indeed a lift of MATH via automorphisms of the representation MATH of MATH, which covers, in the obvious sense, the lift MATH. By REF , MATH projects to the flow MATH of MATH on MATH. To complete the proof of REF we only need to show that MATH has a well-defined projection MATH which is a lift of MATH to MATH. Let MATH and MATH be two points in the same MATH-fiber. Then for MATH and MATH we also have MATH. Our goal is to show that the following diagram is commutative: MATH where the vertical arrows are the identifications MATH . These identifications can also be interpreted in terms of holonomy. Connect MATH and MATH by a path MATH in the MATH-fiber. Then MATH is just the holonomy along MATH with respect to the natural flat MATH-fiberwise connection on MATH. On the other hand, MATH can also be thought of as a MATH-path and hence as a MATH-path (see REF ). Moreover, the holonomy MATH is equal to the holonomy MATH from REF . Similarly, MATH. Since MATH is an automorphism of MATH as of a MATH-representation, MATH commutes with holonomy along MATH-paths. The anti-isomorphism MATH defined by REF interchanges (up to a sign) the structures of MATH- and MATH-representations on MATH. Therefore, MATH also commutes with holonomy along MATH-paths. In particular, MATH which completes the proof. |
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