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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 summands than MATH. By REF and hence MATH also lie in MATH. By the minimality of MATH, MATH must be zero. That is, we must have MATH for each MATH. In other words, MATH lies in the subalgebra MATH and so the ideal MATH is non-zero. |
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 each MATH, then MATH is an element of the desired form and we're done. If not, as in the proof of REF, apply the operation MATH (or MATH in the case MATH) to MATH. From the NAME formula and REF , we have MATH, which is non-zero. Clearly this lies in MATH and by the stability assumption it lies in MATH and hence in MATH. |
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 operations on the generators. We consider the case that MATH and simply make note of the appropriate operations in the case MATH. The interested reader may fill in the details of the latter case. From REF and the NAME formula: MATH we first observe that repeated application of MATH (respectively, MATH) to MATH results in an element of the form MATH since the other terms are eventually killed by MATH (respectively, MATH). By the assumption that MATH is stable under the action of the NAME operations, MATH lies in MATH and hence in MATH. So if at least one of the MATH or MATH is non-zero, applying REF we're done. Hence we may assume that MATH has the form MATH with not all coefficients being zero. As already noted, repeated application of MATH (respectively, MATH) will eventually kill such a MATH. Stopping at the last point before we get zero, we may further assume that MATH has the form MATH with not all coefficients being zero. If every MATH, then MATH lies in MATH and we are done by REF . Hence we may assume that some MATH is non-zero. Applying MATH (respectively, MATH) and using the NAME formula: MATH along with REF , we get MATH . And further applying MATH (respectively, MATH) to this, using REF , we get MATH . Now we apply MATH (respectively, MATH) to this. With the NAME formula and REF , we get MATH . Successively applying MATH, MATH, (respectively, MATH, MATH, ) we eventually conclude that MATH contains an element of the form MATH with some MATH, which is of the desired form. |
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 cohomology ring MATH acts on the spectral sequence with the action on the abutment via MATH. Hence, the periodicity isomorphism follows from that of MATH or MATH on MATH. |
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 and MATH, define a map MATH by MATH. For any pair MATH, we check that MATH is in fact a homomorphism. On the one hand, we have MATH . On the other hand, MATH . Since MATH, these agree and MATH is a homomorphism. |
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) we have MATH. Consider the cobar resolution (compare CITE) MATH which is an injective resolution of MATH over MATH. Further, this is a sequence of rational MATH-modules and maps if MATH is defined to act by the left regular representation on the first MATH factor of each term MATH (compare CITE). Hence MATH acts on the sequence and does so compatibly with respect to its action on MATH. Let MATH be the dual complex (that is, take the MATH-linear dual of each module and reverse the arrows). Then MATH is necessarily a projective resolution of MATH over MATH, which can be considered as a resolution over MATH on which MATH acts compatibly. Finally, since MATH is a closed subgroup scheme of MATH, any projective MATH-module is also a projective MATH-module (compare CITE). So MATH is in fact a projective resolution of MATH over MATH with MATH acting compatibly. Given that such a resolution exists, the same argument as on p. REF (for a semi-direct product of finite groups) shows that there is an isomorphism MATH where the latter cohomology group is a hypercohomology group. That is, the coefficients consist of a cochain complex of modules. Further, this isomorphism is preserved under the natural embedding MATH so that the following diagram commutes: MATH . Finally, the left hand map will be injective if the right hand map is. However, the injectivity of the right hand restriction map in ordinary cohomology can be extended to hypercohomology. For example, one standard proof in ordinary cohomology (compare CITE or CITE) is based on the fact that for a MATH-module MATH, the composite MATH is multiplication by the index MATH which is invertible in MATH, where MATH denotes the transfer map. Hence the composite is an injection and so the restriction map is also. This same argument works with MATH replaced by a cochain complex. Alternately, by the NAME lemma, we may identify MATH and the restriction map MATH is simply the map induced from the canonical map MATH which sends MATH. This module map is split by the map MATH which sends MATH, where MATH is a set of left coset representatives of MATH in MATH. This splitting of the module map induces a splitting of the restriction map in cohomology when MATH is either a module or a complex. |
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 noted in the Introduction, we can identify MATH as the semi-direct product MATH where MATH is the finite group of MATH-points of MATH. By assumption, MATH is unipotent, but MATH may be an arbitrary finite group. However, by REF , we may assume that MATH is in fact a MATH-group, for, if MATH is nilpotent after restriction to the subgroup MATH (for a MATH-Sylow subgroup MATH of MATH), since the restriction map in cohomology is injective, MATH must necessarily be nilpotent. Hence, the group MATH may be assumed to be unipotent. We now proceed by induction on MATH and are trivially done if MATH or MATH, and so assume that the theorem holds for all groups MATH over any field MATH with MATH. As in REF, let MATH denote the map induced by the module map MATH which defines MATH. The next step is to reduce to the case that MATH is elementary abelian. If not, by NAME 's theorem REF , there exists a finite product MATH which is zero in MATH for non-zero MATH. Each MATH can be considered as a non-zero map MATH, which can be extended to MATH. Let MATH denote the kernel of MATH. Since MATH is non-trivial, MATH and so by induction, the restriction of MATH to each MATH is nilpotent. Hence, by REF , MATH is divisible by MATH for each MATH. Consider the canonical projection MATH and the induced map MATH. Since MATH for each MATH, some power of MATH is divisible by MATH and hence is zero. Hence, we may assume that MATH for some MATH (possibly zero). In the trivial case MATH, we have MATH and the claim is precisely REF. In any case, the succeeding argument holds in general (with REF being impossible in that situation). The remainder of the argument consists of three cases depending on MATH and uses arguments similar to those in the above reduction to the case that MATH is elementary abelian. CASE I: MATH . Consider the natural projection MATH and the induced map MATH. By the assumption and REF , we have an isomorphism MATH. Consider the map in degree REF, MATH. By REF , either MATH and we are done or MATH contains some non-zero element of degree REF. Now, the map MATH preserves the action of the NAME algebra (compare CITE, CITE, and also CITE) and hence MATH is a homogeneous ideal stable under the action of the NAME operations. By REF , there exists a non-zero product MATH which lies in MATH. In other words, the image of MATH under MATH is zero in MATH. On the other hand, arguing as above, by REF , we conclude that some power of MATH is divisible by MATH and hence is zero. CASE II: MATH . Let MATH be a representative map preserved by MATH. Since MATH is infinitesimal, the image of MATH lands in some MATH. If MATH, the composition MATH where MATH is the NAME morphism, would give another linearly independent homomorphism preserved by MATH. Hence we must have MATH. Consider the homomorphism MATH and the induced morphism MATH. Specifically, consider the map on MATH (compare REF ): MATH . Again, this is evidently an isomorphism and as before, either MATH and we're done or there exists a non-zero element of degree REF in MATH. In the latter case, by REF , there exists a product MATH which lies in MATH. Each element MATH may again be identified with MATH for some non-zero MATH. Just as above, for each MATH, we consider the homomorphism MATH corresponding to MATH and conclude by REF that MATH is divisible by MATH. On the other hand, consider the composite MATH where the last map is the canonical projection. By REF , MATH is also divisible by MATH. Hence some power of MATH is divisible by MATH and thus zero since MATH is zero in MATH. CASE III: MATH . This case must be further divided into two, based on MATH. As noted above, since MATH is infinitesimal, any MATH has image in some MATH. So there is certainly one non-trivial MATH-preserved map MATH, but there need not be two linearly independent such maps. CASE III REF : MATH . The proof of this case is similar to the proof of REF. Let MATH be a non-trivial representative map preserved by MATH. Since MATH is preserved by the action of MATH, it can be extended to a non-trivial map MATH by MATH. The NAME map MATH induces via composition a map (of the same name) MATH. Further, since any MATH-preserved map remains so after composition with MATH, this restricts to a map MATH, and we may identify the kernel of this (restricted) map with MATH (compare CITE). By assumption, the kernel must then be one-dimensional, and so there exists a non-negative integer MATH and homomorphism MATH with MATH and such that the set MATH is a basis for the subspace MATH. Consider the homomorphism MATH and the induced map on cohomology MATH. On MATH the map MATH is the identity on the right factor and on the left factor maps the basis MATH to the basis MATH. Hence, this is an isomorphism and either MATH and we're done or there is a homomorphism MATH for which (by REF ) the kernel of MATH contains an element of the form MATH with MATH and for each MATH, MATH for some non-trivial MATH. By REF and the usual argument, some power of MATH is divisible by MATH and hence is zero. CASE III REF : MATH . Let MATH and MATH be two linearly independent (and non-trivial) homomorphisms which are preserved by the action of MATH. Since they are preserved by MATH, these maps can be extended to maps MATH by MATH and MATH by MATH. Clearly these remain non-trivial and linearly independent maps MATH. Now, the identical argument as in the case MATH in the proof of CITE may be applied with MATH and MATH to imply that MATH is indeed nilpotent. |
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 the NAME ring MATH, we must have MATH for all MATH for some MATH. As the notions of projective and injective are equivalent (see the discussion preceding REF ), it follows that MATH must in fact be projective. |
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 algebras, we necessarily have MATH. Hence, an element of MATH is nilpotent if and only if it is nilpotent when considered as an element of MATH. As nilpotence is detected by certain NAME subalgebras of the form MATH in the latter case, this is also true in the former. So, if MATH is nilpotent upon restriction to all subalgebras MATH, it is nilpotent upon restriction to each MATH and hence, by REF , is itself nilpotent. |
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 Theorem in CITE which was based on the original arguments of CITE. By the assumption on the points of MATH (see the Introduction), we may write MATH as usual, with MATH assumed to be unipotent. The first step is to reduce to the case that MATH is a MATH-group. Let MATH be a MATH-Sylow subgroup of MATH, and let MATH denote the subgroup MATH. To show that MATH is projective, it suffices to show that MATH for all MATH and any rational MATH-module MATH. If MATH is projective over MATH, then MATH for all MATH, and since, by REF , the restriction map MATH is an injection, we also have MATH for all MATH. Thus it suffices to assume that MATH is a MATH-group and MATH is in fact unipotent. Under the assumption that MATH is unipotent, to show that MATH is projective over MATH, it suffices, by REF , to show that MATH. If MATH is any field extension, then MATH (compare CITE). Hence, if it can be shown that MATH for some field extension MATH, then we also have MATH. So it suffices to assume that MATH is algebraically closed. We now proceed by induction on MATH and are trivially done if MATH or MATH, and so assume that the theorem holds for all groups MATH over any field MATH with MATH. As in the proof of REF , the next step is to reduce the problem to the case that MATH is elementary abelian. If MATH is not elementary abelian, by NAME 's theorem REF , there exists a finite product MATH which is zero in MATH for non-zero MATH. Each MATH can be considered as a non-zero map MATH, which can be extended to MATH. Let MATH denote the kernel of MATH. Since MATH is non-trivial, MATH and so by induction MATH is projective upon restriction to MATH for each MATH. Hence, MATH for all MATH and each MATH, and so by REF the action of MATH (where MATH is induced from the canonical projection MATH) induces a periodicity isomorphism MATH for all MATH. Hence, the product MATH acts via an isomorphism on MATH. But, MATH is zero and so we must have MATH. Thus MATH is projective. From now on we may suppose that MATH for some MATH (possibly zero). In the trivial case MATH, we have MATH and the claim is precisely the Theorem in CITE. In any case, the following argument still holds (with REF being impossible in that situation). As in REF, the rest of the proof consists of three steps based on MATH with arguments like the preceding one. CASE I: MATH . Consider the natural projection MATH and the induced map MATH. By assumption and REF , we have an isomorphism MATH. Consider the map in degree REF, MATH. By REF , either MATH and we are done or MATH contains some non-zero element of degree REF. Again, the map MATH preserves the action of the NAME algebra and hence MATH is a homogeneous ideal stable under the action of the NAME operations. By REF , there exists a non-zero product MATH which lies in MATH. In other words, its image under MATH is zero in MATH. On the other hand, applying the above argument to MATH, we conclude that MATH acts isomorphically on MATH and hence MATH. CASE II: MATH . Just as in REF, there is a homomorphism MATH which is either an isomorphism (in which case we're done) or there exists a non-zero element of degree REF in MATH. In the latter case, by REF , there exists a product MATH which lies in MATH. Each element MATH may again be identified with MATH for some non-zero MATH. Just as above, for each MATH, we consider the homomorphism MATH corresponding to MATH and conclude that the action of MATH on MATH gives an isomorphism MATH for all MATH. On the other hand, consider the composite MATH where the last map is the canonical projection. By REF , MATH induces a periodicity isomorphism MATH for all MATH. Hence MATH acts isomorphically on MATH. Since MATH is zero in MATH we must have that MATH. CASE III: MATH . This case is again divided into two sub-cases based on MATH. CASE III REF : MATH . As in REF, we deduce that there is a homomorphism MATH which is either an isomorphism (in which case we're done) or the kernel of MATH contains a non-zero element of degree REF. In the latter case, by REF , the kernel contains an element of the form MATH with MATH for some non-trivial MATH and MATH for some non-trivial MATH for each MATH. As above, the actions of MATH and MATH (for each MATH) induce a periodicity isomorphism MATH for all MATH. Hence, MATH acts isomorphically, but since it's zero we must have MATH. CASE III REF : MATH . Let MATH and MATH be two linearly independent and non-trivial homomorphisms MATH which preserve the action of MATH. These can then be extended to two linearly independent and non-trivial homomorphisms MATH. We now apply the same argument as in CITE to the maps MATH and MATH. For any homorphism MATH, let MATH denote the image of the canonical generator MATH under the induced map in cohomology. Once again, the inductive argument as above shows that both MATH and MATH induce periodicity isomorphisms MATH for all MATH. Moreover, for any MATH, by REF, MATH . If at least one of MATH, MATH is non-zero, by linear independence, the map MATH is non-trivial and hence by REF , the element MATH also induces a periodicity isomorphism MATH. On the contrary, since MATH is by assumption algebraically closed, if MATH is finite-dimensional, an eigenvalue argument implies that for some MATH with not both zero, this map will not be an isomorphism unless MATH. Thus, for finite dimensional modules the proof is complete and we see that it is only necessary to extend to the algebraic closure of MATH. For infinite dimensional MATH, this eigenvalue argument need not work and must be replaced by an infinite dimensional substitute used in CITE and which requires a further field extension. Let MATH be any non-trivial algebraically closed field extension. After base change, the (extended) maps MATH remain linearly independent. Hence, as the inductive arguments apply equally well over MATH, we again conclude that for any MATH with not both zero, the element MATH also induces a periodicity isomorphism MATH . Applying REF (or REF) we conclude that MATH as desired. |
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. Therefore, for all larger configurations we can ignore the MATH's on each end.) We take the second of these as our example. It has two ends, MATH and MATH. The latter can propagate itself indefinitely by unhopping to the right, MATH . When the former unhops, two things happen; it becomes an end of the form MATH and it leaves behind a space of two adjacent holes, MATH . Furthermore, this is the only way to create a MATH. We can move the MATH to the right by unhopping pegs into it, MATH . However, since this leaves a solid block of MATH's to its left, we cannot move the MATH back to the left. Any attempt to do so reduces it to a single hole, MATH . Here we are using the fact that if a peg has another peg to its left, it can never unhop to its left. We prove this by induction: assume it is true for pairs of pegs farther left in the configuration. Since adding a peg never helps another peg unhop, we can assume that the two pegs have nothing but holes to their left. NAME the leftmost peg then produces MATH, and the original (rightmost) peg is still blocked, this time by a peg which itself cannot move for the same reason. In fact, there can never be more than one MATH, and there is no need to create one more than once, since after creating the first one the only way to create another end of the form MATH or MATH is to move the MATH all the way through to the other side MATH and another MATH created on the right end now might as well be the same one. We can summarize, and say that any configuration with three or more pegs that can be reduced to a single peg can be obtained in reverse from a single peg by going through the following stages, or their mirror image: CASE: We start with MATH. By unhopping the rightmost peg, we obtain MATH. If we like, we then CASE: NAME the leftmost peg one or more times, creating a pair of holes and obtaining MATH. We can then CASE: Move the MATH to the right (say), obtaining MATH. We can stop here, or CASE: Move the MATH all the way to the right, obtaining MATH, or CASE: Fill the pair by unhopping from the left, obtaining MATH. REF simply states that the set of configurations is the union of all of these plus MATH, MATH, and MATH, with as many additional holes on either side as we like. Then MATH is regular since it can be described by a regular expression CITE, that is, a finite expression using the operators MATH and MATH. |
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 of MATH consist of all pairs MATH where MATH and MATH. Draw an arc from MATH to MATH in MATH whenever MATH makes a transition from state MATH to state MATH on symbol MATH. Also, draw an arc from MATH to MATH for any MATH and any MATH. Since MATH, MATH. Then any path from MATH to MATH in MATH consists of MATH arcs of the form MATH to MATH, together with some number MATH of arcs of the form MATH to MATH. Breaking the path into subpaths by removing all but the last arc of this second type corresponds to partitioning the input string into substrings of the form MATH, so the length of the shortest path from MATH to MATH in MATH is MATH, where MATH is the minimum number of pegs to which the initial configuration can be reduced. Since MATH is a directed acyclic graph, we can find shortest paths from MATH by scanning the vertices MATH in order by MATH, resolving ties among vertices with equal MATH by scanning vertices MATH (with MATH) earlier than vertex MATH. When we scan a vertex, we compute its distance to MATH as one plus the minimum distance of any predecessor of the vertex. If the vertex is MATH itself, the distance is zero, and all other vertices MATH have no predecessors and infinite distance. Thus we can find the optimal strategy for the initial configuration by forming MATH, computing its shortest path, using the location of the edges from MATH to MATH to partition the configuration into one-peg subconfigurations, and applying REF to each subconfiguration. Since MATH, this algorithm runs in linear time. |
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 MATH, consider the following game: MATH . Now MATH and MATH are separated by two gaps of the form of REF . Since hopping from MATH and MATH into MATH gives MATH, and since hopping into this gives MATH, and since hopping into MATH gives another word of the same form, we're done. |
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 Lemma for context-free languages CITE by considering the word MATH where MATH, MATH, and MATH where MATH is sufficiently large. Since regular and context-free languages are closed under finite-state transduction and under intersection with a regular language, neither MATH nor MATH is regular or context-free. A more general argument applies to both MATH and the set of MATH-positions MATH. We define MATH similarly to MATH. Now the NAME mapping, which counts the number of times each symbol appears in a word, sends any context-free language to a semilinear set CITE. This implies that the set MATH is eventually periodic. However, it is easy to see that this is MATH . Suppose MATH is eventually periodic with period MATH, and let MATH be sufficiently large that MATH is both in the periodic part of MATH and larger than MATH. Then MATH, but if MATH then MATH, while if MATH then MATH. This gives a contradiction, and since MATH is not eventually periodic neither is its complement. Thus neither MATH nor MATH is regular or context-free. |
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 class for every nim-value. |
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 below. MATH . The horizontal maps above are injective, so we will view them as inclusions. By REF of Subsection REF, there is a local section of MATH that is not zero on MATH. Now, MATH has rank REF. Since MATH, there is also a local section of MATH that vanishes on MATH. So MATH . Now, by REF , MATH . Hence, by analogy with REF , MATH . Since the two squares in the above diagram are commutative, REF imply MATH and hence MATH as divisors on MATH. Both sides of REF are effective because MATH and MATH are distinct for each MATH. Moreover, since MATH is not in the support of MATH and MATH is not in the support of MATH for each MATH, neither MATH nor MATH is in the support of either side of REF . So the right-hand side of REF is a relative NAME divisor on MATH. As both sides of REF are relative NAME divisors and agree on the generic fiber of MATH, they agree everywhere. |
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 on MATH is given by the formula MATH for all MATH and all MATH in the NAME algebra MATH of MATH, where MATH denotes the linear vector field induced by MATH on MATH. Hence MATH where MATH is the restriction of MATH to MATH. Note that MATH is also a moment map for the action of MATH on MATH. Since MATH is compact there exists a MATH invariant complex structure MATH on MATH compatible with MATH. Let MATH denote the unit sphere with respect to the inner product MATH. Since MATH is homogeneous, MATH. Since the action of MATH is linear, MATH. Note that the restriction of MATH to MATH is a contact form and that the restriction of MATH to MATH is the corresponding contact moment map. Therefore the space MATH is the contact quotient MATH and the first part of the lemma follows. The second part of the lemma follows as well since the contact moment map MATH on the contactization MATH is related to the symplectic moment map MATH on MATH by MATH for all MATH (see REF ). |
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 they are locally closed. Similarly REF together with REF imply that a symplectic quotient is locally isomorphic as a partitioned space to a product of an even 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 symplectic quotient are even dimensional manifolds. We next argue by induction on the dimension of contact quotients that contact quotients are stratified spaces. The smallest dimension that a contact quotient can have is one. If it is one dimensional then it is a one dimensional manifold (which need not be connected). Assume now that the dimension of our contact quotient MATH is bigger than one and that any contact quotient MATH with MATH is a stratified space. By REF a neighborhood of a point in MATH is isomorphic, as a partitioned space, to a neighborhood of a point in the contact quotient of a contact vector space. By REF , the contact quotient of a contact vector space is isomorphic to the product of a contact vector space and a cone on the contact quotient of a sphere. By induction, the contact quotient of the sphere is a stratified space. Therefore MATH is a stratified space by definition. Finally, by REF a symplectic quotient is isomorphic to a product of a disk and a cone on the contact quotient of a contact sphere. By the previous paragraph, the contact quotient of the sphere is a stratified space. Therefore a symplectic quotient is also a stratified space. |
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 MATH. Since the projective space MATH is compact and connected, the fibers of the moment map MATH are connected by REF (see CITE and CITE). On the other hand, the restriction MATH is the contact moment map for the action of MATH on the sphere MATH. Since the fibers of MATH are MATH quotients of the fibers of MATH, the connectedness of the fibers of MATH implies that the fibers MATH are connected as well. Therefore MATH is connected. |
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 open stratum. This implies that if a stratum contains a set which is open in MATH then the whole stratum is open as well. Let MATH be a point in MATH and MATH be the stratum containing MATH. By definition there an open neighborhood MATH of MATH in MATH, an open ball MATH in MATH, and a isomorphism MATH of partitioned spaces, where MATH is the link of MATH. If the link is empty, then MATH. Otherwise, by induction MATH contains a unique open dense stratum MATH. Then MATH is open in MATH. Let MATH denote the stratum in MATH with MATH. Then MATH is open in MATH. Therefore MATH is open in MATH, hence MATH. Clearly MATH lies in the closure of MATH. Therefore MATH is in the closure of MATH. This proves that MATH is dense. We now prove that MATH is connected. Suppose not. Choose disjoint open sets MATH and MATH in MATH with MATH. By density, MATH. Because MATH is connected, however, MATH. Choose MATH. Then MATH and so the link MATH of MATH is non-empty. Hence, by induction, there is a unique open stratum MATH in MATH. Choose a neighborhood, MATH of MATH of the form MATH, where MATH is a ball. On the one hand, MATH is connected. On the other, MATH is not. |
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 by MATH and that MATH is a slice through MATH for the action of MATH on MATH. Because MATH and MATH is injective, MATH if and only if MATH and MATH. Therefore MATH . Hence MATH. Therefore by REF holds. |
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 and a corresponding equivariant moment map MATH. Let MATH be a point. Let MATH denote its isotropy group. Let MATH and let MATH denote the isotropy group of MATH. Define MATH where MATH denotes the symplectic perpendicular to the tangent space to the orbit MATH in MATH; it is a symplectic vector space with a linear symplectic action of MATH. Let MATH, MATH and MATH denote the NAME algebras of MATH, MATH and MATH, respectively. Let MATH denote the annihilator of MATH in MATH, and MATH the annihilator of MATH in MATH. Choose a MATH-equivariant splitting MATH; let MATH and MATH denote the corresponding injections. There exists a MATH-invariant neighborhood MATH of MATH in MATH, a MATH-invariant neighborhood MATH of the zero section on the vector bundle MATH and a MATH-equivariant diffeomorphism MATH such that MATH where CASE: MATH is the coadjoint representation and CASE: MATH is the moment map for the action of MATH on MATH. Define MATH by MATH. The map MATH sends MATH homeomorphicly and MATH-equivariantly onto MATH. Hence MATH as spaces partitioned by MATH-orbit types. Therefore it is enough to prove that for some sufficiently small MATH-invariant neighborhood MATH of MATH in MATH as partitioned spaces for some MATH-invariant neighborhood MATH of REF in MATH, where the left hand side is partitioned by MATH-orbit types and the right hand side is partitioned by MATH-orbit types. Note that the canonical embedding of the vector space MATH into MATH, MATH makes MATH into a slice at MATH for the action of MATH on MATH. The vector space MATH is also a slice at MATH for the action of MATH on MATH. Therefore MATH. It follows from REF that for any subgroup MATH of MATH . Conversely if MATH, we may assume that MATH. Then REF holds. Therefore the partition of the left hand side of REF by MATH-orbit types is the same as its partition by MATH-orbit types. Recall that the tangent space to the coadjoint orbit through MATH is canonically isomorphic to the annihilator of MATH in MATH: MATH. Since MATH, the vector bundle MATH is the normal bundle of the orbit MATH in MATH. Consider the map MATH given by MATH. It is the exponential map for a flat MATH-invariant metric on MATH. The differential MATH is an isomorphism at the point MATH. Therefore MATH is an open embedding on a sufficiently small neighborhood MATH of MATH. In particular MATH. We now factor MATH as follows: let MATH . We have MATH. Let MATH. Then MATH. Consequently by REF MATH as partitioned spaces, where the left hand side is partitioned by MATH-orbit types, the right hand side is partitioned by MATH-orbit types and where MATH. But we have seen that the partition of MATH by MATH-orbit types and by MATH orbit types is the same partition. The Proposition now follows. |
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. Set MATH and MATH. |
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 where MATH denotes the right hand side of REF. Now we use the induction hypothesis. As it turns out, factorization of powers of MATH from MATH yields the same power for MATH and MATH, whence we can group these terms together. After several steps of simplification we arrive at the following identity: MATH where MATH for MATH even and MATH for MATH odd. The sum can be evaluated by means of the very - well - poised MATH summation formula CITE: MATH . The sum we are actually interested in does not extend to infinity, so we rewrite is as follows: MATH where MATH denotes the summand in REF. Now, replacing MATH by MATH, MATH by MATH, MATH by MATH, MATH by MATH and MATH by MATH in the summand of REF gives MATH times the fraction MATH, which cancels. So we obtain after some simplification: MATH . Substitution of this evaluation in REF and simplification yield for both cases MATH even REF and odd REF the same equation MATH which, of course, is true. This finishes the proof. |
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 the left side rotated by REF degrees, all arrows may be reversed, and both sides may be rotated by REF degrees if the variables MATH, MATH, and MATH are cyclically permuted. By the three symmetries, REF of the non-zero equations are tautological, and the other REF are all equivalent. One of REF non-trivial equations is MATH . In algebraic form, the equation is MATH . Cancelling a factor of MATH, expanding, and cancelling terms yields MATH which is implied by the condition MATH. |
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 alone.) Under this symmetry REF of REF non-zero equations are tautological, and the other REF are equivalent. One of these is: MATH . Algebraically, the equation reads: MATH . All terms of the equation match or cancel when MATH and MATH. |
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 all cases of the equation are null or tautological. |
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 exactly the same for all of the square ice grids without NAME. If the grid has diagonal boundary with corner vertices, we can bounce the crossing off of it using REF . If the grid has NAME boundary on the right, we exchange MATH with MATH by crossing the MATH line over the two lines above it. We let the spectral parameters of these two crossings be MATH and MATH. We move both crossings to the right using the NAME equation, then we bounce them off of the NAME using the reflection equation REF : MATH . Also if the grid has a NAME on the right, we establish covariance under MATH by switching the lines with these two labels and eating the crossing using the fish equation REF . The same arguments establish symmetry in the coordinates of MATH. All of the arguments used in combination establish the claimed properties of MATH. |
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 is forced to be REF and the left column and bottom row are forced to be REF, as in REF . The sum reduces to one over MATH ASMs. The only discrepancy between MATH and MATH is the weights of the forced entries, which the lemma lists as factors. f:forcedASM entries forced by MATH . The argument in the other determinant cases is identical. The argument in the Pfaffian cases is only slightly different: All QTSASMs have zeroes in the corners, and the specialization MATH instead forces a REF next to each corner and zeroes the first two rows and columns from each edge. Likewise the specialization MATH forces a REF next to each corner of an OSASM or an OOSASM and a REF in the third row entry bottom of a UOSASM, and several rows and columns of zeroes in each of these cases. |
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 constant, which can be found inductively by setting MATH. The determinant MATH is argued the same way. The NAME MATH and MATH are also argued the same way; here the constant factor can be found by setting MATH. |
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 with fractional exponents. The notion of width make sense for these.) We will derive this factor by a separate method. For example, if MATH, then MATH is divisible by MATH because MATH . Evidently MATH is a sum of MATH rank REF matrices at this specialization, so its determinant has a MATH-fold root at MATH. Likewise MATH also has rank MATH and MATH also divides MATH. All four of the determinants have this behavior. In each case, the singular values of MATH can be read from the product formulas for the determinants. The only detail that changes is the form of each rank REF term, which is summarized in REF . t:terms Details of factor exhaustion for REF Finally the MATH-independent factor MATH can be found by examining the coefficient of the leading power of MATH, or equivalently, taking the limit MATH. For example MATH as MATH with MATH . In this case MATH is given by MATH in REF . This happens in each case, although for the matrix MATH it is slightly more convenient to specialize to MATH. The best extra value of MATH in all four cases is given in REF . |
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 factor MATH depending only on MATH. This factor can be determined by taking the limit MATH: MATH . In other words, after rescaling rows and columns, MATH has a block matrix limit: MATH . (The bullet MATH is the exponent above that is different for different rows and columns.) This establishes that the leading coefficient of MATH as a polynomial in MATH is MATH which in turn determines MATH. The Pfaffian MATH is argued the same way. To find the factor MATH which is independent of MATH and MATH, we take the limit MATH. In this limit MATH reduces to a special case of MATH in REF . |
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 with the framings by MATH. But then, because MATH-framed bundles on MATH are rigid, we find that MATH as a functor over MATH is isomorphic to the functor taking MATH to the set of isomorphism classes of pairs MATH consisting of a bundle MATH on MATH together with an isomorphism MATH of MATH with MATH. We will refer to such a pair as a MATH-framed bundle. Because the statement of the proposition is local on MATH, we may assume that MATH is an affine scheme that is the spectrum of a local ring MATH. For simplicity, write MATH and MATH. The ``change of rings" spectral sequence (see REF) MATH yields the exact sequence of terms of low degree MATH . Note that MATH is surjective since the next term in the sequence is MATH, which vanishes because MATH is one-dimensional. Using MATH one may check that there is a canonical element MATH of MATH such that MATH is exactly the MATH-subtorsor of MATH that classifies REF-extensions MATH for which MATH is a locally free MATH-module. Now, REF , together with Cohomology and Base Change, implies that the MATH-module MATH is projective, hence free. One can easily construct, moreover, a universal REF-extension over MATH (using, for example, an affine subspace of the NAME cocycles that maps isomorphically to MATH to furnish gluing data). Because the exact sequence REF and the element MATH are functorial under pullback along morphisms of affine schemes MATH, this universal REF-extension induces a functorial bijection between the set MATH (the inverse image of the canonical element under the base-changed map MATH) and the set of isomorphism classes of pairs MATH consisting of a vector bundle MATH on MATH and a framing MATH. Consequently MATH is represented as a functor over MATH by the torsor over MATH defined by MATH, proving the proposition. |
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. Let MATH be the characteristic series of the orientation MATH, so that MATH for MATH a complex line bundle. Let MATH be the roots of the total NAME class of MATH. Since MATH is compact, the MATH are nilpotent. It suffices to check that MATH is a unit modulo nilpotents in MATH. We have MATH where the equivalences are modulo nilpotents. The result follows, since the MATH are non-zero and MATH is a unit in MATH. The cocycle REF is easy, because as usual the equation needs only to be verified when at most one of MATH, MATH, and MATH is special. |
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 take the first case for definiteness. By assumption, the zeros and poles of MATH are contained in MATH, so MATH and MATH . By REF , MATH is a special point. |
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 special if and only if MATH, and for such MATH we have MATH. For ordinary MATH we have MATH. It follows that MATH a function which vanishes at precisely the points MATH such that MATH. Thus in the gluing REF, a trivialization of MATH corresponds to a section of MATH which vanishes at MATH; this is a description in terms of cocycles of the ideal sheaf MATH. The isomorphism MATH of REF gives the result. |
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 precisely when MATH, that is when MATH. It is well-known and easy to check that the product of sigma functions REF descends to a trivialization of MATH, and it is also easy to check that it coincides with the trivialization of REF . |
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 claims of the lemma follow. |
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. It will then be sufficient to fix MATH and show that the map MATH is regular at MATH. Again, since both MATH are regular values for the maps MATH is a regular value for the map MATH . This implies that the differential MATH is surjective. |
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 . Since MATH is surjective, MATH is a right module map. By REF, MATH is a left-coinvariant REF-form. Hence MATH is a left comodule map. Therefore MATH is an isomorphism of NAME bimodules. In particular, MATH and MATH. CASE: We prove the last assertion. Since MATH defines the right action on MATH, we obtain from REF that MATH, i. CASE: MATH. Since MATH, MATH, MATH, and MATH are bimodule maps it suffices to prove the statement for MATH and MATH. By REF MATH . |
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 obtain MATH, MATH. It follows that MATH. Hence MATH maps MATH to MATH and therefore it defines a mapping MATH. Moreover, by REF it coincides with MATH. Since MATH defines a braiding as well and MATH, the proof for MATH is analogous. |
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 coaction on MATH. By NAME 's lemma these matrices have to be zero (see also the proof of REF ). Using REF and MATH the assertion follows. |
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 all MATH which follows from the centrality of MATH. Inserting both parts into REF we obtain REF. Let us prove REF. Similarly to the preceding equation one shows that MATH. Let MATH with left-coinvariant elements MATH. Using REF we get MATH . This gives REF. Since MATH is central, MATH follows from REF. |
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, by REF we obtain the equation MATH . The proof of the second part of REF is analogous. |
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. Obviously, MATH for MATH. Later we will see that MATH for MATH and not only for MATH. We prove REF . Set MATH. In the remark to REF it was noted that MATH where MATH, MATH. On the other hand, computing the limit MATH directly from REF one gets MATH . Using the formulae MATH, MATH, and MATH from CITE, we obtain REF from REF. From REF it follows that MATH for MATH and MATH if and only if MATH. Suppose that MATH for some MATH. Since MATH is an algebraic function of MATH and MATH is a transcendental root, MATH. In particular MATH . Hence MATH. Let us prove REF . We will show that MATH for arbitrary MATH. For this purpose we prove that MATH is an eigenvector for MATH if MATH is and we compute the corresponding eigenvalue. Suppose that MATH for a complex number MATH. Since MATH and MATH we have MATH . Hence MATH. Acting from the right by MATH gives MATH. For MATH we thus have MATH. Since MATH is grouplike, MATH, and MATH, we obtain by REF the following formulae for MATH: MATH . Since MATH corresponds to the weight MATH and MATH we have for MATH . Next we will show that for MATH, MATH . For MATH set MATH. How to obtain MATH from MATH? In cases MATH shift MATH to the right by MATH, then add one box MATH for each MATH. In cases MATH remove the box MATH and then shift the remainder by MATH to the right. Since boxes with MATH have negative sign and MATH we obtain by REF MATH . In the last line we used MATH. Since there exists MATH such that MATH and MATH , from REF we obtain MATH. Comparing REF yields MATH. We will show that MATH is regular. Suppose that MATH for some MATH. Inserting MATH into REF we get MATH. Since MATH is a rational function of MATH and MATH is a transcendental root of it, MATH. In particular MATH. Set MATH. Further we have MATH. Since MATH the limit MATH gives MATH. Let us define MATH for MATH. By REF, MATH for MATH and MATH if and only if MATH. It is easy to check that MATH for MATH. Hence MATH, MATH, and MATH if and only if MATH for some MATH. We conclude that MATH. Inserting this into equation REF we obtain MATH, where equality holds on the left hand side if and only if MATH and MATH for some MATH and on the right hand side if and only if MATH. Hence MATH and MATH is regular. Finally consider the case when MATH and MATH is a primitive MATH root of unity (MATH). Let MATH. Observe that MATH by REF and since MATH. Hence MATH for MATH. Therefore the if part of the assertion holds. Moreover, for the only if part we can assume that MATH. Since MATH is a transcendental root of the algebraic function MATH, we conclude that MATH has to be identically zero. Therefore MATH has to be zero, and by REF it follows that MATH. Then MATH. Further, one can compute that MATH (see REF). This sum is positive except for the case MATH. Moreover, the latter is equivalent to MATH for some MATH. Together with MATH we get the assertion. CASE: There exists an isomorphism of NAME bimodules MATH and MATH, see CITE. In particular MATH and consequently MATH. By REF and the definition of MATH one obtains MATH where MATH. Paying attention to the MATH-admissible parameter MATH, the choice of the coinvariant REF-form MATH, and the definition of the MATH-metric it follows from the remark after REF that MATH, where MATH . Suppose that MATH. Since MATH for MATH we may assume MATH. Since MATH is a rational function with transcendental root MATH, MATH. In particular MATH . Using the inequality MATH, MATH, from the proof of REF it is easily seen that for MATH in the case MATH and in the case (MATH and MATH) we have MATH. Thus, by REF, MATH or (MATH and MATH). In the latter case MATH where equality holds if and only if MATH, MATH. Inserting this into REF yields MATH. Indeed we have MATH in these cases. |
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 and MATH, respectively. Then we have MATH . Set MATH. By the left covariance of MATH it follows that MATH . Hence MATH for all MATH. By NAME 's lemma we obtain MATH for MATH. Using right covariance, in a similar way we get MATH for all MATH. Again by NAME 's lemma MATH for MATH. Suppose now that MATH for a fixed MATH and all MATH. By the above arguments MATH for all MATH, MATH, i. e. for all MATH. Since the NAME functional is regular, i. CASE: MATH for all MATH implies MATH and MATH for all MATH implies MATH, and since the pairing MATH is non-degenerate, the pairing MATH is also non-degenerate, compare CITE. Therefore MATH. NAME in the second component can be proved similarly. CASE: Suppose that MATH and MATH. Because of REF, the MATH-symmetry of MATH, and since MATH, we obtain MATH . The proof is complete. |
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, since MATH. We show that MATH is well-defined. Let MATH and MATH be linear bases of MATH and MATH, respectively, and let MATH and MATH be the corresponding dual bases of MATH and MATH, respectively. By assumption there exist MATH such that MATH. Suppose that MATH. Then we have MATH and consequently MATH for all MATH. Using this fact, we conclude that MATH . Hence MATH is well-defined. |
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, given by MATH maps the generating relations appearing in the definition of the NAME algebra MATH to those of the NAME algebra MATH. Here we assume that MATH has both opposite multiplication and opposite comultiplication. By the above identity for the matrix MATH we have MATH . The right hand side generates the relations of the bialgebra MATH. Similarly one shows consistency with the MATH-determinant relation. Finally we have MATH and MATH. Hence MATH is a homomorphism of bialgebras. Since both MATH and MATH are NAME algebras, MATH is a NAME algebra homomorphism. Since the antipode is a NAME algebra map of MATH, MATH is a NAME algebra automorphism. Its inverse MATH is given by MATH. Next we show that MATH. Let MATH be a primitive idempotent such that MATH. For MATH we write MATH and MATH. Let us show that MATH is a projector equivalent to MATH. NAME MATH and MATH are called equivalent, if there exist MATH such that MATH and MATH. For this let MATH and MATH denote the algebra automorphism and algebra antiautomorphism of MATH defined by MATH, respectively. By the theory of NAME algebras it is easy to see that MATH and MATH map each twosided ideal of MATH into itself. In particular, the image of a primitive idempotent is an equivalent primitive idempotent. By induction on MATH we will show that MATH . It is well known that for MATH there exist MATH and MATH such that MATH. Since MATH and MATH can be written in terms of MATH, MATH, and MATH, and since MATH, we have by induction assumption MATH . The character of MATH is MATH CITE. Since MATH is an equivalent idempotent, the corresponding characters of the corepresentations MATH and MATH coincide CITE. We conclude that MATH . Since MATH is a NAME algebra homomorphism, matrix elements of irreducible corepresentations are mapped into each other. Hence MATH. CASE: We show that MATH. Since MATH is a coalgebra homomorphism, MATH is multiplicative on the subalgebra of MATH generated by the matrix elements MATH, MATH. We compute MATH on the generators of MATH. MATH where MATH in the MATH-case and MATH in the MATH-case. Since both MATH and MATH are representations of MATH, we obtain MATH. By CITE we have MATH . Multiplying this relation by MATH and noting that MATH, we get MATH . Multiplying the latter by MATH and using the identity MATH we obtain MATH . By REF, MATH, and MATH, we then have MATH . This completes the proof of REF . Note that MATH is bijective since MATH. By REF , MATH is a well-defined MATH-module map from MATH to MATH. Similarly, MATH, MATH is a well-defined MATH-module map inverse to MATH. CASE: Consider now higher order forms. Let MATH and MATH denote MATH for MATH and MATH for MATH. Since MATH is a MATH-bimodule map, we can extend MATH to an algebra map MATH. We prove that MATH in MATH. Since both MATH and MATH are MATH-bimodule maps and MATH is a free left MATH-module with basis MATH, it suffices to prove this equation for left-coinvariant elements. Let MATH. By REF we have MATH . Hence MATH commutes with the antisymmetriser MATH, MATH. The same is true for MATH. Consequently, MATH is a well-defined algebra map and MATH. Now let us prove REF . Let MATH. By REF and the last part of REF , MATH. Since MATH also satisfies the assumptions of REF , it follows immediately that MATH is inverse to MATH. |
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 bundle MATH on MATH. There is a restriction map MATH since MATH canonically via MATH, we find that MATH canonically. Because each MATH is isomorphic to MATH, the induced map of line bundles MATH is surjective and therefore is an isomorphism (see REF). The inverse image of the identity section MATH now gives a section of MATH; moreover, the unicity of our construction guarantees commutativity of the necessary diagram (see REF) to make the pulled-back morphism MATH an equivariant structure on MATH. This completes the proof. |
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 expression of this map as MATH say with MATH. Because MATH is MATH-equivariant, we find that both MATH and MATH are MATH-invariant. But now if the image of MATH vanishes in a fiber MATH of MATH at a point MATH, then MATH vanishes along MATH, hence MATH has torsion supported along MATH. However, this quotient is torsion-free by construction, and we may conclude that MATH fails to vanish at any point MATH. Therefore MATH is locally free at every point MATH, completing the proof. |
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 already MATH. So, we must compute MATH. Since MATH and MATH has discrete support, we may replace MATH by MATH for this computation. Now, locally along MATH the quotient MATH has generators of the form MATH with MATH a uniformizer in the MATH direction and MATH a parameter in the fiber direction. Here MATH acts by MATH hence, using short exact sequences in the first variable of MATH, we may reduce to considering MATH, where MATH acts on MATH with nonpositive weight MATH. If we resolve MATH locally by MATH we find that MATH is an invariant quotient of MATH . But any MATH-invariant section of MATH vanishes to order at least MATH along MATH and thus maps to zero in MATH. |
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 the MATH action is the one that vanishes to order MATH at MATH (and hence is nonzero at infinity), the map MATH is an isomorphism, and the conclusion follows. |
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 unique extension of MATH to MATH. Finally NAME gluing now pieces together MATH from MATH and its extension across MATH. |
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 choose such a MATH (assuming we have chosen MATH sufficiently small) because MATH and MATH are locally free near MATH. CASE: MATH is the intersection of MATH and MATH in MATH. CASE: MATH is the intersection of MATH and MATH in MATH. By the choice of MATH it is enough to show that MATH . So, write an element of the intersection as MATH . Then MATH . Therefore, all coefficients of MATH are divisible by MATH, and MATH after all; consequently, MATH as desired. This proves the claim. REF , together with the commutative REF , gives an isomorphism MATH so that the diagram MATH commutes. Then by REF, the inclusions MATH induce an isomorphism of the MATH-modules MATH and MATH over the open set MATH of MATH. The isomorphisms so obtained locally along MATH arise from the restrictions of a single diagram MATH over MATH, so they are compatible on intersections of the open sets of the form MATH and consequently induce an isomorphism MATH over all of MATH. |
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 bundle on MATH, and MATH is an ideal of finite colength. Then the restriction to a generic fiber MATH of MATH gives MATH hence an injection MATH; consequently, MATH. Moreover, since MATH is locally free near MATH and MATH is saturated, one has MATH and so one gets MATH . It follows that MATH since MATH is stable. In particular, one has MATH and MATH; consequently, MATH is stable for any polarization MATH for which MATH. Because MATH is equivalent to MATH, we get MATH in homology. So MATH is already stable for any polarization MATH for which MATH. Now, suppose MATH generically, where MATH. If MATH is destabilizing, then MATH generically, for otherwise, as before, the restrictions of MATH to both MATH and a generic fiber MATH are of lower degree than the restrictions of the quotient MATH, proving that MATH is not destabilizing. Therefore, in this case MATH, where MATH is the canonically defined rank one subsheaf of MATH; thus, we may assume that MATH. Writing MATH and MATH where MATH and MATH are ideals of finite colength and MATH and MATH are line bundles on MATH, we get MATH and similarly MATH . So for fixed MATH, and MATH, by choosing MATH sufficiently large one gets MATH. The lemma that follows proves that one can choose the coefficient MATH sufficiently large relative to MATH to work uniformly for all sheaves MATH of fixed NAME classes MATH and MATH. |
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 isomorphically to a component of a moduli stack MATH with the appropriate degrees. Suppose now that MATH satisfies the four requirements in the statement of REF . Assume, without loss of generality, that MATH with MATH. One gets the canonical short exact sequence MATH . Then MATH implies that MATH . Moreover, MATH . If MATH then MATH; combining this with our assumption that MATH, we see that all terms of REF are nonnegative and that the second term increases linearly with MATH, which gives a bound on MATH since MATH is fixed. Similarly, if MATH then MATH, and consequently all terms of REF are nonnegative with the first term increasing linearly with MATH; thus we again obtain a bound on MATH. This proves the lemma. |
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 stack of stable pairs and they prove the representability of their stack by a quasiprojective variety, the representability of MATH follows. In what follows, we choose MATH, the NAME polynomial of MATH. Then the NAME polynomial of the pair MATH as defined by NAME - NAME is exactly MATH . Suppose, then, that MATH is a rank one subsheaf. If MATH, then the NAME polynomial associated to the pair MATH is MATH the NAME polynomial of MATH. Since MATH is stable, we have MATH as desired. If, on the other hand, MATH is not contained in MATH, then MATH and MATH as desired. NAME - NAME prove in CITE that MATH is smooth whenever the hyper-Ext obstruction group MATH is zero for all pairs MATH in MATH. Since the complex MATH is quasi-isomorphic to MATH in our setting, one obtains MATH . Now by NAME duality, MATH . This completes the proof of the corollary. |
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 projection MATH, we have MATH . Then any endomorphism of MATH that vanishes at infinity is zero; hence any MATH is zero on the generic fiber of MATH and consequently is zero on all of MATH. Similarly, MATH and hence MATH. Suppose next that MATH with MATH. Then, as in REF, we get a canonical exact sequence MATH . Now MATH . If MATH, this group is zero, while if MATH, then, since MATH is semistable of degree zero, we have either MATH or MATH, and in either case the second assumption of the theorem implies that this group is zero. Now if the canonical exact sequence is nonsplit, the above vanishing implies that MATH is simple, hence MATH as desired. If, on the other hand, the canonical exact sequence splits, then MATH and MATH are line subbundles of the polystable bundle MATH, and so MATH and either MATH or MATH. This implies the MATH vanishing just as before. To prove that MATH in this case, we must compute also MATH since MATH. Consequently, MATH in this case. But the two sheaves in this formula are of rank one, hence MATH and the restriction map along MATH or along MATH is just the identity of MATH. As a result MATH here as well. As a consequence of the vanishing of MATH for every MATH-point MATH of MATH, we find that every object of MATH over MATH is rigid. But then by semicontinuity every object of MATH is rigid as well, and by REF, the stack MATH is in fact represented by a locally finitely presented algebraic space. The deformation theory arguments used in CITE to obtain a smoothness criterion for moduli spaces of stable framed sheaves are arguments purely about the smoothness of the moduli functor, and hence carry over mutatis mutandis to our setting; in particular, if MATH for all pairs MATH, then MATH is smooth. But since MATH induces an isomorphism MATH we have that the complex MATH is quasi-isomorphic to MATH concentrated in degree zero, and so MATH for all pairs MATH. As a result, MATH is smooth. Next, we need to prove that MATH is separated; so, writing MATH, suppose that MATH is a family of MATH-framed torsion-free sheaves parametrized by MATH, where MATH is the field of fractions of a DVR MATH. We proceed in several steps. Suppose MATH is an affine scheme and MATH is a MATH-framed family parametrized by MATH. Then MATH is isomorphic to MATH; furthermore, this isomorphism is completely determined by MATH. Suppose we have shown that MATH is isomorphic to MATH. By REF, the obstruction to uniqueness of an extension of MATH to a bundle over MATH is a class in MATH which vanishes whenever MATH by the assumptions of the theorem. So MATH . Moreover, since MATH whenever MATH, we see that the isomorphism MATH is uniquely determined by MATH. The same computation as in the proof of this claim gives MATH . Suppose MATH and MATH are MATH-framed families parametrized by MATH that are equipped with isomorphisms MATH . Restricting to MATH and identifying each of MATH, MATH uniquely with MATH, we obtain an automorphism MATH of MATH whose restriction to MATH is the identity. The automorphism MATH is the restriction of the identity automorphism of MATH. This follows immediately from the description in REF . Now, restrict to MATH. We then have sheaves of MATH-modules MATH, MATH where MATH, together with embeddings MATH moreover, the isomorphisms MATH identify MATH and MATH with subsheaves of MATH, and in addition the restrictions of their images to MATH coincide. By the Uniqueness Lemma, then, the inclusions MATH have the same image and thus induce an isomorphism of MATH and MATH over MATH. Because this is compatible with the isomorphism given over MATH, by the Extension Lemma we obtain an isomorphism MATH . This completes the proof that MATH is separated. We now indicate why MATH is in fact a scheme. Suppose MATH is a MATH-valued point of MATH. Let MATH denote a polarization of MATH, and, given a MATH-module MATH, write MATH . Choose MATH sufficiently large that MATH is globally generated and MATH then the same will hold for all MATH occurring in pairs MATH that lie in an open neighborhood of MATH in MATH. We saw above that MATH for any MATH and that the isomorphism is uniquely determined by MATH. For MATH sufficiently large, this isomorphism determines an injection MATH fix such a MATH. We next construct a scheme representing a neighborhood of MATH in MATH. There is a NAME MATH parametrizing subspaces of MATH that are of CITE dimension equal to the dimension of MATH. Over MATH, in addition, there is a canonical subbundle MATH of MATH; from this family MATH on MATH, one obtains a relative NAME MATH together with a universal subsheaf MATH of MATH on MATH: this NAME parametrizes subsheaves of the bundles MATH REF on MATH with NAME polynomials coinciding with that of MATH . We have a surjective map MATH and there is a locally closed subvariety MATH of MATH parametrizing those MATH on MATH for which CASE: MATH is contained in MATH, CASE: MATH is torsion-free, and CASE: the induced map MATH is an isomorphism. Over MATH we have a universal diagram MATH . The map MATH that is induced by the family MATH represents a neighborhood of MATH in MATH. Suppose MATH is a MATH-flat family in MATH for which CASE: the NAME polynomial of each MATH REF coincides with that of MATH, CASE: each MATH is globally generated, CASE: each MATH, and CASE: each MATH injects into MATH under the canonical homomorphism. Then the family of REF determines a map MATH. This functor gives exactly the inverse of the map MATH, proving the claim. This proves that MATH is locally representable by schemes, and so is itself a scheme. In fact, MATH is a quasiprojective variety and its image in MATH is MATH-invariant; hence, in particular (see REF), MATH may be covered by MATH-invariant quasi-affine open subschemes, and thus, by CITE, MATH admits a NAME decomposition. |
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, the map MATH has in its image the identity element MATH. Furthermore, this map is injective: fiberwise we need only check the following. MATH is injective. But this follows from the conjunction of CASE: MATH is injective for MATH, since MATH is torsion-free; and CASE: MATH for MATH, and consequently MATH . So the inverse image of MATH in MATH is uniquely determined. As in the proof of REF , uniqueness also guarantees commutativity of the diagram required for the map MATH to define a MATH-equivariant structure. |
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 subsheaves (necessarily torsion-free) of MATH; since their composites MATH and MATH are zero, the image of neither can have rank greater than REF, but also MATH and so MATH, MATH must be rank one torsion-free subsheaves of MATH for which MATH . Letting MATH gives the desired splitting. |
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 local MATH-algebra is trivial; this, in turn, follows if MATH . This MATH reduces to MATH; however the NAME complex computing MATH has no MATH-invariants in MATH: this is a consequence of the description in REF . |
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 MATH. |
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. The last two isomorphisms result from the equisingularity of MATH along 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 that MATH is a surjection. All that remains to prove the first statement is to show that MATH is an injection. But in the commutative square: MATH the left homomorphism is an isomorphism, and the lower one is an injection - it is an isomorphism in degrees MATH and MATH vanishes in higher degrees. Finally, the second statement of the theorem follows from the long exact sequence for MATH of the pair MATH. |
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 we can do this, MATH can be extended to MATH by placing a MATH on all vertices and edges outside of MATH, and so MATH, as claimed. Assume that we have chosen MATH so that the lines MATH for MATH are all distinct. Pick a vertex MATH and an adjacent edge MATH. We will construct a degree two section MATH satisfying REF above and for which MATH. The section MATH we want is the product of these sections over all choices of MATH and MATH. If MATH, then MATH is the upper vertex of an edge MATH. Since MATH, there are nonzero vectors MATH, MATH with MATH. The section which is MATH on MATH and MATH everywhere else does the trick. Note that MATH, so MATH lies in a plane in MATH. Now suppose MATH, so MATH. Let MATH be the section guaranteed by REF . It extends to a section MATH which is zero on all of MATH. We can assume that MATH has been chosen so MATH. Thus we can find MATH so that MATH, and putting MATH gives the required section. |
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. Then from REF , MATH with MATH, so that MATH; that is, MATH is hyperelliptic. |
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 and only if MATH is defined over MATH. It this were not the case in our situation, then MATH would be contained in the intersection of two distinct quadrics, contradicting the hypothesis MATH by the NAME theorem. |
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 homogeneous equation MATH . Then MATH because of MATH. Furthermore, MATH and MATH are related in the function field over MATH of MATH by MATH. In addition, the valuation at MATH of the functions MATH are respectively MATH . Hence, MATH. CASE: MATH by REF and the hypothesis MATH. So from the inequalities MATH and REF we obtain REF . CASE: We have from REF that MATH if and only if MATH. Now, as MATH, the last condition is equivalent to MATH and the result follows. CASE: If MATH, from the proof of REF we get MATH. An easy computation shows then that MATH is non-singular at MATH if and only if MATH. Therefore MATH, and the result follows from REF . |
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 MATH, and MATH. If MATH, then MATH is generated by MATH and MATH; CASE: There exist lines MATH and MATH such that MATH, and MATH; If MATH is non-singular, then CASE: No two tangent hyperplanes of MATH at different points coincide. To simplify our notation we shall identify MATH and MATH, according to REF . CASE: Since MATH is non-degenerate, MATH is irreducible. Then MATH is a cone if and only if MATH is singular, as this case can only occur when MATH has just one singular point. Suppose that MATH is non-singular. Then from REF , MATH for each MATH. Note that there exists MATH such that MATH; in fact, otherwise REF would yield MATH; but then MATH would be odd, a contradiction. Hence MATH by REF . Let MATH. We have MATH, as MATH (compare REF ), and hence the plane MATH generated by MATH and MATH is well defined. Then MATH, and the intersection divisor of MATH and MATH becomes MATH where MATH is a divisor on MATH of degree MATH with MATH, and MATH. In addition, REF assures the existence of a line MATH such that MATH . Actually, the line MATH is defined over MATH. In fact, MATH is defined over MATH by REF , and MATH implies that MATH. MATH . Proof of REF . If there exists MATH, then MATH as MATH is defined over MATH. Thus MATH, and hence MATH. It follows MATH, but this is a contradiction. If MATH, then MATH and MATH . Proof of REF . Since MATH, we have MATH by REF . Now if MATH, then MATH by MATH and REF . Also, MATH because MATH by REF . Therefore the plane MATH is generated by the lines MATH and MATH, and hence MATH. Let MATH be the line defined by MATH, and MATH. From REF , we infer that MATH and so MATH, but this is a contradiction. MATH . Proof of REF . Suppose on the contrary that MATH. Equivalently, MATH by REF . Then MATH. In fact, using a similar argument to that in the proof the previous claim, one can show that MATH together with MATH implies MATH in contradiction with REF . Hence, for each MATH, the divisor MATH in REF may also be written as MATH in such a way that REF holds true, MATH, and MATH. Notice that MATH is generated by MATH and MATH where MATH is any point of MATH. Now let MATH such that MATH. Then MATH, otherwise MATH by MATH. This yields that MATH must divide the number of MATH-rational points of MATH, which is a contradiction because MATH is an odd number. So far we have shown that each MATH gives rise to a plane MATH, to a line MATH, and to a divisor MATH such that REF hold with MATH being the sum of MATH-rational points. Notice that MATH. Let MATH be chosen in such a way that MATH and that MATH . Clearly, MATH is MATH-rational, and thus MATH as in the proof of REF . Therefore MATH where MATH is a divisor on MATH of degree MATH such that MATH. MATH, and MATH for each MATH . Proof of REF . Let MATH. Suppose on the contrary that MATH for some MATH. Then MATH contains MATH which is different from MATH as MATH. Hence MATH is generated by MATH and MATH. These lines also generate MATH and so MATH contradicting MATH. Finally suppose on the contrary that MATH. Replacing MATH by MATH, the above argument shows that MATH, whence MATH follows, again a contradiction. Therefore, to each MATH we have associated two lines MATH and MATH such that both REF hold where MATH is a divisor of degree MATH, MATH, and MATH. As it is well-known, MATH has just two families of lines contained in MATH and any two lines of the same family are disjoint. This implies again that MATH must be a multiple of MATH, contradicting the MATH-maximality of MATH. CASE: As MATH is even, from REF there exists MATH such that MATH. Suppose that MATH. From MATH, we find that MATH and so MATH. Therefore MATH by REF , and hence MATH. Now, since MATH is MATH-rational by REF , we have MATH, and hence MATH is generated by MATH and MATH; in particular MATH and thus MATH by REF , a contradiction. Therefore MATH must be MATH-rational and hence MATH must have a singularity at MATH by REF . Then MATH and MATH by REF and the assumption of MATH being even. |
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 MATH in the previous argument yields again a contradiction. In fact, REF reads currently MATH, while MATH by the MATH-maximality of MATH and the lower bound MATH for MATH which has been shown in the proof of REF . Taking MATH into account, this gives MATH, whence MATH follows for MATH; a contradiction. |
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 completes the proof. |
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 REF we would have MATH which implies MATH; whence MATH and so MATH. If MATH, then MATH; otherwise MATH and hence MATH by REF , and this would imply MATH. Finally, assume that MATH. Then from REF MATH and MATH. If MATH, then MATH, whence MATH. Then MATH and MATH. We claim that MATH. In fact, otherwise MATH, MATH and hence MATH, MATH by REF which yields MATH, a contradiction. |
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 then MATH with MATH according to REF . By REF , MATH. Thus, we have to analyze three cases. As before, MATH stands for the MATH-th NAME non-gap at MATH. Recall that MATH by REF . CASE: MATH. From REF , MATH. We have that MATH, since otherwise MATH and hence MATH, a contradiction. Thus MATH. However this leads again to a contradiction since, by REF , MATH does not belong to MATH. CASE: MATH odd, MATH. From REF , MATH does not belong to MATH, a contradiction. CASE: MATH even, MATH. Arguing as in REF we have either MATH or MATH. In the former case, MATH and thus REF implies MATH. Since this is not admitted currently, the latter case can only occur. Then MATH and MATH. Now, as MATH, from REF MATH follows. Since a similar result to REF holds, namely MATH is a MATH-order for MATH, and the set of MATH-orders is MATH we conclude that MATH; whence MATH. Finally from REF , MATH. On the other hand, MATH, a contradiction. CASE: MATH . In virtue of REF , we have MATH. By MATH, REF shows that MATH and that MATH is the canonical linear series, where MATH is a base-point-free REF-dimensional linear series of degree MATH. Let MATH and assume MATH according to REF . If MATH, from REF the result follows. Otherwise, MATH, and we have two possibilities according as MATH is odd or even REF . CASE: MATH is odd MATH, MATH. A similar property to REF holds, namely MATH for any MATH-order MATH. Hence MATH is not in MATH, a contradiction. CASE: MATH is even, MATH, MATH. From REF in the proof of REF , we have MATH. Notice that the degree MATH of the above linear series MATH is coprime to MATH. Then by the well known NAME 's inequality for the genus MATH applied to MATH and the linear series corresponding to MATH we obtain MATH, a contradiction. The implication MATH is a special case of CITE while MATH is trivial. |
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 point MATH CITE satisfies both MATH and MATH. Conversely, from REF we obtain both MATH and MATH. Hence the result from CITE. |
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