paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0008168 | Let MATH be a non-zero homogeneous polynomial in MATH. Choose such a MATH that has lowest possible degree and for which the number of summands, MATH, is least. Having made such a choice, it suffices to assume that MATH. From REF and the NAME formula (MATH), we see that MATH . Since MATH, the difference MATH has fewer s... |
math/0008168 | We have MATH. Consider the set MATH. This is evidently an ideal in MATH and moreover homogeneous since MATH was. If MATH is non-zero, by REF , the ideal MATH is also non-zero as well as homogeneous and NAME stable. Applying REF to MATH, the result follows. Let MATH with MATH be a non-zero degree two element in MATH. If... |
math/0008168 | Consider the ideals MATH and MATH. These ideals satisfy the structural hypotheses of REF respectively. By REF contains some non-zero element MATH of degree REF. Generically, MATH has the following form: MATH for some constants MATH which are not all zero. We remind the reader of REF which lists the actions of the NAME ... |
math/0008168 | Let MATH denote either MATH or MATH and MATH be the given homorphism. Since MATH is non-trivial we have a short exact sequence of group schemes over MATH: MATH and a corresponding NAME spectral sequence MATH . By assumption, the spectral sequence collapses to MATH and hence we have an isomorphism MATH where MATH. The c... |
math/0008168 | Any homomorphism MATH determines two homomorphisms MATH by MATH and MATH by MATH. Further, MATH necessarily preserves the action of MATH. Given any MATH and MATH, since MATH is a homomorphism, we have MATH . On the other hand, we also have MATH . Hence, we must have MATH as claimed. Conversely, given homomorphisms MATH... |
math/0008168 | The idea is to identify MATH-cohomology groups with certain MATH-cohomology groups. We first note that there exists a projective resolution MATH of rational MATH-modules which admits a compatible action of the finite group MATH. That is, this is also a complex of MATH-modules and for any MATH, MATH, and MATH (any MATH)... |
math/0008168 | The strategy of the proof is to extend that of REF. For any field extension MATH, since MATH (compare CITE), if MATH is nilpotent after base change, then it is necessarily nilpotent. Hence it suffices to assume that MATH is algebraically closed. In this case, the points of MATH are certainly MATH-rational and so, as no... |
math/0008168 | If MATH is projective over MATH, then it remains so upon restriction to any closed subgroup scheme (compare CITE). Conversely, given a MATH with MATH, if MATH is projective upon restriction, then MATH for all MATH. Hence, by REF , every element MATH for MATH is nilpotent. Since, by CITE, MATH is finitely generated over... |
math/0008168 | For any closed subgroup scheme of the form MATH, the corresponding NAME algebra MATH is isomorphic as an algebra to the restricted enveloping algebra, MATH, of an abelian NAME algebra MATH over MATH with trivial MATH-mapping. Given a MATH-module MATH, let MATH denote the module MATH considered as a MATH-module. As NAME... |
math/0008168 | As previously noted, if MATH is projective over MATH, then it remains so upon restriction to any closed subgroup scheme (compare CITE). Conversely, suppose that all restrictions are projective. The outline of the proof is the same as for the proof of REF with the details modified along the lines of the proof of the The... |
math/0008172 | To prove the theorem, we follow CITE in starting with a single peg, which we denote MATH and playing the game in reverse. The first `unhop' produces MATH and the next MATH . (As it turns out, MATH is the only configuration that cannot be reduced to a single peg without using a hole outside the initial set of pegs. Ther... |
math/0008172 | Our proof of REF is constructive in that it tells us how to unhop from a single peg to any feasible configuration. We simply reverse this series of moves to play the game. |
math/0008172 | Suppose we are given a string MATH where each MATH. Let MATH be a nondeterministic finite automaton (without MATH-transitions) for MATH, where MATH is the set of states in MATH, MATH is the start state, and MATH is the set of accepting states. We then construct a directed acyclic graph MATH as follows: Let the vertices... |
math/0008172 | Any attempt to cross this gap only creates a larger gap of the same form; for instance, a hop on the left end from MATH yields MATH. Thus the two games cannot interact. |
math/0008172 | Previous steals Next's strategy until Next hops into the gap. Previous then hops into the gap and over Next's peg, leaving a position of the form in REF . The games then separate and Previous can continue stealing Next's strategy, so the nim-value is MATH. To show that this remains true even if Next tries to hop from M... |
math/0008172 | Let MATH be the set of MATH-positions. Since the nim-value of MATH is MATH, the intersection of MATH with the regular language MATH is MATH . To simplify our argument, we run this through a finite-state transducer which the reader can easily construct, giving MATH . It is easy to show that MATH violates the Pumping Lem... |
math/0008172 | Recall that a language is regular if and only if it has a finite number of equivalence classes, where we define MATH and MATH as equivalent if they can be followed by the same suffixes: MATH if and only if MATH. Since MATH if and only if MATH and MATH have the same nim-value by REF , there is at least one equivalence c... |
math/0008176 | Let MATH . Let MATH, MATH and MATH be the respective determinants of the natural maps, MATH . The zero scheme of MATH is MATH and that of MATH is MATH. Let MATH be the zero scheme of MATH. As observed in Subsection REF, we have MATH. By REF , there is a sequence of inclusions MATH, which induces the commutative diagram... |
math/0008178 | Since MATH is a slice, for any MATH, MATH is a single MATH-orbit. |
math/0008178 | Let MATH be the vector subspace of MATH-fixed vectors in MATH; it is a symplectic subspace of MATH. Let MATH denote the symplectic perpendicular to MATH. It is a symplectic MATH-invariant subspace as well. Let MATH as before denote the radial vector field on MATH. Since MATH, the moment map MATH for the action of MATH ... |
math/0008178 | REF together with REF imply that a contact quotient is locally isomorphic as a partitioned space to a product of an odd-dimensional disk and a cone on the contact quotient of a standard contact sphere. It follows that the pieces of the canonical decomposition of a contact quotient are odd dimensional manifolds and that... |
math/0008178 | The map MATH, MATH is a moment map for the action MATH of the circle MATH on MATH, and the sphere MATH is a level set of MATH. Since the action of MATH commutes with the action of MATH, the restriction of the MATH-moment map MATH to MATH descends to a moment map MATH for the action of MATH on the symplectic quotient MA... |
math/0008178 | The proof is an induction on the dimension of MATH. Let MATH be the union of all the open strata in MATH. We show first that MATH is dense. Using density, we then show that MATH is connected and hence consists of a single stratum. Note that a point in a stratified space has an empty link if and only if it lies in an op... |
math/0008178 | Define MATH by MATH. Since the function MATH is nonvanishing the map MATH sends MATH equivariantly and homeomorphicly onto the set MATH. Hence MATH as partitioned spaces (the partitioning is by MATH-orbit type). Therefore it is enough to show that MATH . Next note that the vector space MATH embeds canonically into MATH... |
math/0008178 | The proof is essentially the same as that of REF above. Since we are not reducing as zero, it is a little more delicate. We use a version of the local normal form theorem due to NAME and to NAME and NAME which is proved on pp. CASE: Let MATH be a symplectic manifold with a proper Hamiltonian action of a NAME group MATH... |
math/0008178 | For MATH, define MATH. For all MATH and all MATH, we have MATH and MATH. Hence, MATH is contact in an open neighborhood of MATH. Since MATH is invariant, NAME 's theorem applies. The time REF map of the isotopy MATH exists on some open set MATH of MATH since it exists on a neighborhood of each point of MATH in MATH. Se... |
math/0008180 | We perform an induction on MATH for arbitrary MATH. The case MATH is immediate. For the case MATH, observe that MATH . For the inductive step MATH, we shall rewrite the recursions REF in the following way: MATH . Substitution of these recursions into REF and interchange of summations transform the identity into MATH wh... |
math/0008184 | Taken literally, the equation consists of REF numerical equalities, because there are REF ways to orient the six boundary edges on each side. However, both sides are zero unless three edges point in and three point out. This leaves REF non-zero equations. The equation also has three kinds of symmetry: The right side is... |
math/0008184 | The argument is similar to that for REF . Both sides are zero unless two boundary edges point in and two point out. There is a symmetry exchanging the two sides given by reflecting through a horizontal line and simultaneously reversing all arrows. (Note that the weights of a NAME are not invariant under reflection alon... |
math/0008184 | Diagonal reflection exchanges the two sides. Both sides are zero if an odd number of boundary edges point inward. If two boundary edges point in and the other two point out, then arrow reversal is also a symmetry, because one corner must have inward arrows and the other outward arrows. These facts together imply that a... |
math/0008184 | Invariance of MATH is an illustrative case. We exchange MATH with MATH for any MATH by crossing the corresponding lines at the left side. If the spectral parameter of the crossing is MATH, we can move it to the right side using the NAME equation REF and then remove it: MATH . The argument for symmetry in MATH is exactl... |
math/0008184 | In both cases, the symmetry is effected by reflecting the square ice grid or the alternating-sign matrices through a horizontal line. |
math/0008184 | This lemma is clearer in the alternating-sign matrix model than it is in the square ice model. The partition function MATH is a sum over MATH alternating-sign matrices in which each entry of the matrix has a multiplicative weight. When MATH, the weight of a REF in the southwest corner is REF. Consequently this corner i... |
math/0008184 | Our proof is by the factor exhaustion method CITE. The determinant MATH is divisible by MATH because when MATH, two rows of MATH are proportional. Likewise it is also divisible by MATH. At the same time, the polynomial MATH has degree MATH, so it has no room for other non-constant factors. This determines MATH up to a ... |
math/0008184 | Factor exhaustion. We first view each determinant as a fractional NAME polynomial in MATH. By choosing special values of MATH, we will find enough factors in each determinant to account for their entire width, thus determining them up to a rational factor MATH. (Each determinant is a centered NAME polynomial in MATH wi... |
math/0008184 | Factor exhaustion in both MATH and MATH. If MATH, then MATH is, as written, a sum of MATH rank REF matrices. Therefore the Pfaffian, whose square is the determinant, it is divisible by MATH. The same argument applies to MATH. It also applies to MATH since MATH is symmetric in MATH and MATH. This determines MATH up to a... |
math/0008190 | Working inductively, it will suffice to construct a MATH-scheme MATH that represents MATH and is an affine bundle over MATH. Fix a universal bundle MATH on MATH. For any scheme MATH, an element of MATH determines a map MATH, and, if MATH is the given element of MATH, there is an isomorphism of MATH with MATH compatibly... |
math/0008192 | If MATH is not special, then MATH and the result is obvious. If MATH is special, then it is not contained in MATH (by the definition of an adapted cover), and so MATH is not contained in MATH. In particular, MATH is a unit in MATH. The localization REF gives the result. |
math/0008192 | If the rank MATH is even, this is REF. If the rank of MATH is odd, then we may apply that lemma to MATH. |
math/0008192 | Without loss of generality we may suppose that MATH. If in addition MATH, then MATH. Otherwise, we have MATH so MATH, and so MATH is a unit in MATH. On each component MATH of MATH, there are integers MATH and complex vector bundles MATH over MATH such that MATH . Here MATH acts on MATH fiberwise by the character MATH. ... |
math/0008192 | Choose a cover adapted to MATH and the pair MATH. Suppose that MATH, MATH is special, and MATH is not. Then the diagram MATH commutes (all the arrows are isomorphisms). The left column describes the sheaf MATH, while the right column describes MATH. |
math/0008192 | This is REF. |
math/0008192 | What must be shown is that MATH has no pole at the origin. Suppose it does; we shall show that MATH is a special point. Choose a lift of MATH to MATH; we may call it MATH in view of REF . We have MATH . If this has a pole at the origin then MATH has a zero at MATH for some MATH or a pole at MATH for some MATH. Let us t... |
math/0008192 | We have MATH . The iterated transformation REF gives MATH where MATH for MATH. In view of REF , it remains to observe that MATH while for MATH we have MATH . |
math/0008192 | The argument is similar to the proof in the odd case, REF . Once again we have MATH . The result follows from REF , and the equations MATH . |
math/0008192 | The first part is clear, since MATH. The third part follows from the multiplicative property of the NAME isomorphism REF (and so of its associated euler class), together with the construction REF of MATH. For REF, let MATH, and consider the multiplicative orientation MATH given by the MATH function. A point MATH is spe... |
math/0008192 | Let MATH be a divisor on MATH. Recall that the line bundle MATH associated to MATH is trivial precisely when MATH the second sum is taken in the group structure of the elliptic curve MATH. For the divisors we are considering, REF is always satisfied, since as a group MATH and MATH . It follows that MATH is trivial prec... |
math/0008192 | Introduce formal roots MATH and MATH so that MATH . CASE: MATH implies that MATH . On the other hand, this class is given by half the degree-four component of MATH . Examining the coefficient of MATH gives REF, and examining the coefficient of MATH gives REF follows from the equation MATH by a similar argument. |
math/0008192 | We treat the case that MATH is even; the case that MATH is odd is similar. Let MATH. Introduce formal roots MATH and MATH so that MATH . Then MATH is given by minus half the degree-four component of MATH . The coefficient of MATH is MATH . The characteristic class restriction REF implies that this quantity is zero; the... |
math/0008192 | We have MATH . The third equation uses REF . |
math/0008192 | Recall that MATH . Substituting the equation MATH (and similarly for MATH) into REF gives MATH . It follows that MATH and that this quantity is divisible by MATH. |
math/0008192 | The first NAME class of the bundle MATH is MATH . The restriction of MATH to MATH has NAME class MATH . Adding to this zero in the form (from REF) MATH gives MATH . |
math/0008192 | First observe that if MATH then REF shows that MATH has a MATH root, with the property that MATH . The restriction of MATH to MATH has NAME class MATH . Adding to this zero in the form (from REF) MATH gives MATH . |
math/0008194 | Since the MATH's are all fixed for MATH, MATH is fixed. Since MATH are onto, one may choose regular values MATH respectively of these two maps MATH and MATH such that MATH. Let MATH such that MATH. Then MATH is a smooth point (The representation corresponding to MATH is irreducible) and the map MATH is regular at MATH.... |
math/0008195 | CASE: We first show that MATH is a right comodule mapping, i. CASE: MATH for MATH. Recall that the coproduct on MATH is given by MATH, see CITE. Since MATH one has MATH and consequently, MATH . Hence MATH which proves REF . CASE: We show that MATH is a right module map. Since MATH is an algebra map of MATH we have for ... |
math/0008195 | CASE: The braiding MATH is compatible with the tensor product of NAME bimodules in the sense that the identity MATH is fulfilled for all NAME bimodules MATH, MATH, and MATH, see CITE. Iterating this yields MATH . CASE: In what follows we skip the space indices MATH and MATH to simplify the notations. Since MATH we obta... |
math/0008195 | We carry out the proof of the first equation. The proof of the second one is analogous. By REF , MATH . Note that the endomorphisms MATH, MATH, , MATH do not act in the first component. So we can separate the last summand. Applying REF we continue MATH . This finishes the proof. |
math/0008195 | Using the definition of MATH and MATH, the identity MATH and REF we get (the not underlined terms remain unchanged) MATH . The last equation follows by REF. |
math/0008195 | Let MATH, where MATH is given by REF. Further let MATH be the corresponding dual basis elements in the quantum tangent space of MATH. Since the MATH-metric is bicovariant, the complex matrices MATH and MATH are elements of the vector space MATH, where MATH denotes the corepresentation corresponding to the right coactio... |
math/0008195 | We prove REF. Let MATH be a presentation of MATH with MATH, MATH. By REF, since MATH and MATH are MATH-module homomorphisms and since MATH, we obtain MATH . In the fourth equation we used REF, in the fifth REF and in the last one REF. Now let MATH, MATH, MATH. By REF we have MATH . In the last equation we used MATH for... |
math/0008195 | CASE: An easy computation shows that indeed MATH. Since MATH and MATH, the first part of REF follows. Let us verify the second part of REF. The first sum is direct by the grading. The second sum is direct, since matrix elements of inequivalent irreducible corepresentations are linearly independent. CASE: Since MATH and... |
math/0008195 | For MATH define the following rational functions of MATH and MATH: MATH where MATH. It follows from CITE that for the quantum groups MATH and MATH and for MATH we have MATH, where MATH in the MATH case and MATH in the MATH case. Note that we have to replace MATH in CITE by MATH according to our definition of MATH. Obvi... |
math/0008195 | CASE: Set MATH and fix MATH, MATH. Let us prove that MATH is finite dimensional. The space MATH is finite dimensional since MATH is. Suppose that MATH. Since MATH, we deduce that MATH. Because MATH is finite dimensional, the assertion follows. Similarly, MATH. For MATH let MATH and MATH denote the linear bases of MATH ... |
math/0008195 | MATH . Fix MATH. Since MATH is a NAME algebra homomorphism and MATH, we have MATH . Moreover, MATH. Hence MATH. MATH . Since MATH and MATH are orthogonal subspaces with respect to the pairing of MATH and MATH, we have MATH for all MATH and MATH. Furthermore, MATH. Hence MATH. MATH . MATH is uniquely determined by REF, ... |
math/0008195 | CASE: First it is to show that there exists a NAME algebra automorphism MATH which satisfies REF. To do this we prove that MATH preserves the relations of the NAME algebra MATH. It is easily shown that MATH and MATH. Moreover, the MATH-antisymmetric tensor satisfies MATH. We show that the algebra homomorphism MATH, giv... |
math/0008198 | Let MATH denote the projection on the second factor, let MATH denote the projection to MATH, and, by abuse of notation, let MATH also denote the projection to MATH. By assumption, we have MATH for each MATH. Since MATH is MATH-simple, we find that MATH is a line bundle on MATH for each MATH; therefore, we get a line bu... |
math/0008198 | Because MATH by a MATH-equivariant map, we may find, locally along MATH, some MATH-invariant sections MATH and MATH of MATH that generate MATH. These sections induce a MATH-invariant isomorphism MATH. Thus, locally along MATH we get a MATH-invariant map MATH. Choosing a local MATH-invariant generator of MATH gives an e... |
math/0008198 | We have an exact sequence MATH . Now MATH, since it is (after applying NAME duality) isomorphic to MATH of a sheaf supported on a finite collection of points. Consequently, it is enough to show that MATH: then any equivariant extension pulls back from an extension of MATH by MATH, which can only be locally free if alre... |
math/0008198 | MATH . The NAME spectral sequence shows that this is just MATH furthermore, the restriction map is evidently just the map MATH which is given fiberwise by the map MATH (where MATH is the zero point in MATH) arising from evaluation at infinity. Since the unique MATH-invariant section of MATH under our normalization of t... |
math/0008198 | Given MATH, let MATH. Then the extension in REF determines data of MATH and the inclusion MATH, and REF shows that this data in turn determines MATH as a framed bundle. Now the inclusion MATH is MATH-equivariant, hence the quotient MATH is also MATH-equivariant, as desired. |
math/0008198 | The map MATH when restricted to MATH extends to a map MATH . One gets a sheaf MATH, whose restriction to MATH is canonically isomorphic to MATH. One clearly gets a framing MATH from the framing MATH. Restricting MATH to MATH and MATH to MATH gives exactly the input data for the Extension Lemma, and so determines a uniq... |
math/0008198 | We will show that the images of MATH and MATH in MATH and MATH, respectively, are identified under the given isomorphism. Locally along MATH we may trivialize MATH, and then over an open set MATH we may replace MATH, MATH by MATH, MATH respectively. Choose MATH so that MATH and MATH are free modules over MATH; we may c... |
math/0008198 | Suppose MATH is a saturated rank one subsheaf of MATH, with quotient MATH. Suppose first that the restriction of MATH to a generic fiber of MATH is of the form MATH. Write MATH, where MATH is the generic fiber degree of MATH (that is, the degree of the restriction of MATH to a generic fiber of MATH), MATH is a line bun... |
math/0008198 | We may assume without loss of generality that MATH has degree MATH or MATH and that the restriction of MATH to a generic fiber of MATH has degree MATH: these are determined by MATH, and by using twists by powers of MATH and pullbacks of line bundles on MATH, we may move any component of a moduli stack MATH isomorphical... |
math/0008198 | Suppose that MATH is a MATH-framed pair and MATH is stable. We will prove that then the pair MATH is a stable pair in the sense of NAME - NAME (CITE, CITE) for the ample divisor MATH chosen in the lemma above and the auxiliary datum of the polynomial MATH; because our stack is then an open substack of their moduli stac... |
math/0008198 | We may assume that MATH, since if we begin with MATH then for any point MATH, MATH and MATH will satisfy the assumptions. We begin by proving that, for any MATH-valued point MATH of MATH, we have MATH . In this computation, we will use repeatedly that MATH . Suppose first that, if MATH denotes the generic fiber of the ... |
math/0008198 | Let MATH denote the projection to MATH, MATH denote the projection to MATH and, by abuse of notation, also the projection MATH to MATH. We have MATH by assumption, and so MATH is a vector bundle on MATH with fiber over MATH equal to MATH. There is a natural restriction map MATH . Now MATH via MATH, and, by assumption, ... |
math/0008198 | By REF , the map MATH is injective, so, because MATH and the image of MATH in this group contains every MATH, we must have MATH . Consider the elements MATH and MATH in MATH that are the inverse images of the elements MATH and MATH, respectively, of MATH. The elements MATH and MATH are idempotents whose images are subs... |
math/0008198 | CASE: Suppose MATH is an open set and MATH, for which MATH . Then by the construction of limits for orbits in REF , one has MATH as desired. CASE: Since the components of the fixed-point set of MATH in MATH are smooth, the conclusion follows, provided any MATH-equivariant deformation of MATH parametrized by an NAME loc... |
math/0008200 | If MATH, we define MATH to be the minimal open set with the same vertices as MATH. If MATH is a complete subgraph of MATH, then restriction gives an isomorphism MATH. The proposition now follows immediately, since the MATH-sheaf can be recovered from the sheaf on MATH as follows: MATH and MATH is given by restriction M... |
math/0008200 | The first statement is clear. The second follows by an easy induction from the following statement: if MATH is a pure MATH-module on MATH, and MATH splits as a direct sum of two pure sheaves, then this splitting can be extended to MATH (see REF for the analogous result for toric varieties). |
math/0008200 | We have MATH, where MATH is the inclusion. According to CITE, the functors MATH and MATH can only increase weights. |
math/0008200 | We have MATH, where MATH is the inclusion. The middle expression vanishes for weights less than the degree, while the right one vanishes for weights greater than the degree (the functor MATH can only decrease weights, and hypercohomology of mixed NAME modules on proper varieties preserves weights). |
math/0008200 | REF implies that the connecting homomorphism MATH vanishes. The equivariant case then follows from the nonequivariant case, since for MATH large enough we have MATH and MATH. |
math/0008200 | For any MATH, let MATH be a MATH-invariant affine neighborhood of MATH, and let MATH be the composition of restriction and localization homomorphisms MATH . It is a surjection, using REF . So it is enough to find a homomorphism MATH with MATH. Such a MATH is given by the composition MATH where MATH is any point in MATH... |
math/0008200 | The claim that MATH is a free module follows from the collapsing of the spectral sequence MATH. This in turn happens because the intersection cohomology of the varieties MATH is pure, by REF . A similar argument shows that MATH is free. REF shows that MATH. Since MATH is projective, we have MATH, and so REF implies tha... |
math/0008200 | We can find a diagonal linear action of MATH on some affine space MATH, and an equivariant embedding MATH. The tangent space MATH will be generated by a subset of the coordinate directions. Take the linear span of the remaining coordinates and intersect with MATH. |
math/0008200 | The result for MATH is clear from the construction of MATH. Since in REF we showed that MATH we can apply the first part of REF . |
math/0008200 | The moment map gives an embedding MATH of the vertices of MATH into MATH so that if MATH and MATH are joined by an edge MATH, then MATH is a nonzero vector in MATH. If we choose a linear projection MATH which does not kill MATH, then letting MATH provides the required section. |
math/0008200 | Let MATH be the union of MATH with the set of upper vertices of edges in MATH, and let MATH be any extension of MATH to MATH. We will construct an element MATH so that CASE: MATH comes from an element MATH which is a product of MATH-good linear factors, and CASE: for any vertex MATH and any adjacent edge MATH, MATH. If... |
math/0008202 | If MATH is hyperelliptic, MATH at a general point MATH. Then from REF , MATH and so MATH. On the other hand MATH and maximality of MATH yields MATH. From these computations REF follows. Let MATH such that MATH. Then from REF we have MATH and so either MATH is hyperelliptic or MATH. Finally, let MATH such that MATH. The... |
math/0008202 | The set of MATH-Weierstrass points of MATH coincides with the set of MATH-rational points, and MATH for MATH; compare REF . Hence the result follows from REF taking into account the maximality of MATH. |
math/0008202 | CASE: Let MATH and set MATH. From REF , MATH and MATH. Then, as MATH and MATH, it is easy to see that there are at least REF positive NAME non-gaps in MATH and so MATH. CASE: See CITE. CASE: If MATH lies on MATH, then MATH also lies on MATH, where MATH is the NAME collination on MATH relative to MATH. Clearly MATH if a... |
math/0008202 | Set MATH, MATH. Let MATH be MATH-rational functions on MATH, such that MATH. Up to a projective collineation in MATH, we can assume MATH. Let MATH be coordinates in MATH such that each MATH is the pull-back via MATH of MATH restricted to MATH. Then MATH and MATH is given by MATH; see CITE. Let the quadric MATH have hom... |
math/0008202 | General properties of quadrics of a MATH-dimensional projective space over a finite field can be found in CITE. Here we will use the following properties: Let MATH be a non-singular point of MATH and denote by MATH the tangent plane of MATH at MATH. CASE: If MATH, then MATH; CASE: Let MATH and MATH be lines such that M... |
math/0008202 | From REF , MATH. Suppose on the contrary that MATH. If MATH, REF becomes MATH, while MATH-maximality of MATH implies MATH as MATH for every MATH. But then MATH contradicting the hypothesis on MATH. If MATH, then MATH by the MATH-adic criteriom CITE and MATH. Replacing the ramification divisor MATH by the NAME divisor M... |
math/0008202 | The hypothesis on MATH together with REF implies that MATH . Suppose MATH. If MATH, from REF MATH and MATH. Thus MATH so that MATH; that is, MATH. If MATH and MATH, from REF we have MATH and MATH. Then MATH, a contradiction. Finally, if MATH and MATH, then REF together with REF gives MATH . Hence MATH, and this complet... |
math/0008202 | Suppose first that MATH. According to REF we have only three possibilities, namely MATH and MATH. The first one cannot actually occur by MATH; from the second one MATH, MATH, MATH follow, while the third one gives MATH, MATH, and MATH. Suppose next that MATH. Then MATH by MATH. Moreover, MATH; otherwise MATH and from R... |
math/0008202 | CASE: Suppose on the contrary that MATH is a MATH-maximal curve of genus MATH with MATH. Since MATH, we have MATH by REF . Hence, REF implies that MATH and that MATH is the canonical linear series on MATH. Then MATH where the MATH's are MATH-orders, and the MATH's are non-negative integers such that MATH. We choose the... |
math/0008202 | CASE: Let MATH be the unique point over MATH. It is straightforward to check that MATH. Hence MATH by REF . Conversely, from REF we have MATH and so MATH by REF . Now, the result follows from CITE. CASE: We have MATH, where MATH is the linear series cut out by lines on MATH CITE and hence every MATH-rational inflexion ... |
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