paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0006054 | First note that if MATH is torsion-free then MATH is also torsion-free. Hence its support consists is either empty or consists of the whole relative jacobian surface MATH. Moreover, if MATH is not MATH-MATH then the support of MATH is contained in the support of MATH since MATH for all MATH. However, MATH cannot be sup... |
math/0006054 | Suppose that MATH is an isolated point. It cannot belong to a fibre MATH such that MATH is semistable, since the restriction to every nearby fibre is also semistable and the restriction of MATH to the fibres varies holomorphically. So MATH is unstable. But MATH contains all such fibres, since MATH for all MATH. To obta... |
math/0006054 | Suppose that the support of MATH decomposes as MATH, where MATH is the sum of fibres. When this happens, we have a subsheaf MATH of MATH supported on MATH with degree REF. Assume that MATH is the maximal such effective subdivisor of MATH. Let MATH. Then MATH is MATH-MATH and MATH is MATH-MATH. The resulting sequence af... |
math/0006054 | Suppose that MATH is a destabilizing sequence for MATH. We can assume MATH is semistable and MATH has pure dimension REF. Now MATH, for some MATH. Then MATH for all MATH, since MATH. Thus MATH is MATH-MATH by REF . Arguing as in REF , we have MATH hence MATH. So MATH will destabilize MATH unless MATH. For this to be th... |
math/0006054 | The first statement follows from REF , which requires us to show that MATH for all MATH. However, MATH and any map MATH would contradict the stability of MATH. Now suppose that MATH is not MATH-stable and let MATH be the destabilizing subsheaf, so that MATH . Moreover, we may assume that MATH is MATH-stable, thus MATH ... |
math/0006054 | We must first show that MATH is a well defined map of varieties. To see this we use the argument given by NAME and NAME in CITE. The key idea is to observe that MATH coincides (locally in the étale or complex topology) with the projection map from the universal sheaf corresponding to the relative NAME scheme of degree ... |
math/0006054 | By continuity it suffices to prove this when MATH are defined over a point MATH given by an extension: MATH where MATH consists of discrete points and MATH is smooth. Deformations of MATH arising from the fibre of MATH are determined by deformations of MATH along MATH. Then MATH is generated by a non-zero vector in MAT... |
math/0006057 | We prove the identities containing the target counital map, the proofs of their source counterparts are similar. Using REF we compute MATH where MATH stands for the second copy of the unit, proving REF . For REF we have MATH . To prove REF we observe that MATH on the other hand, applying MATH to both sides of MATH, we ... |
math/0006057 | MATH and MATH are coideals by REF , they commute by REF . We have MATH, conversely MATH, therefore MATH. To see that it is an algebra we note that MATH and for all MATH compute, using REF : MATH . The statements about MATH are proven similarly. |
math/0006057 | Let MATH be the convolution of MATH. Then MATH, and MATH. If MATH is another antipode of MATH then MATH . To check that MATH is an algebra anti-homomorphism, we compute MATH for all MATH, where we used REF and easy identities MATH and MATH. Dualizing the above arguments we show that MATH is also a coalgebra anti-homomo... |
math/0006057 | Using results of REF we compute MATH for all MATH. The second identity is proven similarly. Clearly, MATH maps MATH to MATH and vice versa. Since MATH is bijective, and MATH by REF , therefore MATH and MATH are anti-isomorphisms. |
math/0006057 | It is clear that MATH is a homomorphism. If we write MATH with MATH and MATH linearly independent, then MATH. By REF , MATH, that is, MATH is surjective. Since MATH then MATH, therefore, MATH so MATH is bijective. The proof for source subalgebras is similar. |
math/0006057 | For all MATH we compute, using REF : MATH applying MATH to this identity we get MATH. Clearly, MATH, whence MATH is a separability element. The second statement follows similarly. |
math/0006057 | The proof is a straightforward application of REF , and is left as an exercise for the reader. |
math/0006057 | REF implies that the MATH-Hopf module structure on MATH is well defined, and it is easy to check that MATH is a well defined homomorphism of MATH-Hopf modules. We will show that MATH is an inverse of MATH. First, we observe that MATH for all MATH, since MATH so MATH maps to MATH. Next, we check that it is both module a... |
math/0006057 | Take MATH in REF and use REF . |
math/0006057 | CASE: Suppose that MATH is semisimple, then since MATH is a left ideal in MATH we have MATH for some idempotent MATH. Therefore, MATH and MATH is a left integral by REF . It is normalized since MATH. REF : if MATH is normalized then MATH is a separability element of MATH by REF . REF : this is a standard result CITE. |
math/0006057 | In the notation of REF the element MATH is a normalized integral in MATH. |
math/0006057 | A straightforward (but rather lengthly) computation shows that MATH is a well defined homomorphism, compare (CITE). Let MATH be a basis of MATH and MATH be the dual basis of MATH, that is, such that MATH for all MATH, then the element MATH does not depend on the choice of MATH and the following map MATH is the inverse ... |
math/0006057 | REF for MATH shows that MATH is a projective generating MATH-module such that MATH. Therefore, MATH and MATH are NAME equivalent. Since MATH is semisimple (as a separable algebra), MATH is semisimple. |
math/0006057 | Define a left unit homomorphism MATH by MATH . It is an invertible MATH-linear map with the inverse MATH. Moreover, the collection MATH gives a natural equivalence between the functor MATH and the identity functor. Similarly, the right unit homomorphism MATH defined by MATH has the inverse MATH and satisfies the necess... |
math/0006057 | We know already that MATH is monoidal. One can check (CITE) that MATH and MATH are MATH-linear. To show that they satisfy the identities MATH take MATH. Using the isomorphisms MATH and MATH identifying MATH, MATH and MATH, for all MATH and MATH we have: MATH which completes the proof. |
math/0006057 | Note that MATH is well-defined, since MATH. To prove the MATH-linearity of MATH we observe that MATH . The inverse of MATH is given by MATH therefore MATH is is an isomorphism. Finally, we check the braiding identities. Let MATH, then MATH . Similarly, we have MATH. The third equality of this computation shows that the... |
math/0006057 | It follows from the first two relations of REF , that MATH . |
math/0006057 | The proof is essentially the same as (CITEEF). |
math/0006057 | Note that MATH for all MATH. Hence, we have MATH for all MATH. Therefore, MATH . The remaining part of the proof follows the lines of (CITEEF). The results for MATH can be obtained by applying the results for MATH to the quasitriangular quantum groupoid MATH. |
math/0006057 | Take MATH. Then we have : MATH therefore MATH, as required. |
math/0006057 | The proof is a straightforward verification that all the structure maps are well-defined and satisfy the axioms of a quantum groupoid, which is carried out in full detail in (CITE). |
math/0006057 | The identities MATH and MATH can be written as (identifying MATH with MATH and MATH with MATH) : MATH . The above equalities can be verified, for example, by evaluating both sides against an element MATH in the third factor (respectively, against MATH in the first factor), see (CITEEF). It is also straightforward to ch... |
math/0006057 | One can use the explicit form of the MATH-matrix REF of MATH and the description of the dual MATH to check that in this case the map MATH from REF is surjective. |
math/0006057 | Since MATH is an invertible central element of MATH, the homomorphism MATH is a MATH-linear isomorphism. The twist identity follows from the properties of MATH : MATH for all MATH. Clearly, the identity MATH is equivalent to the twist property. It remains to prove that MATH for all MATH, that is, that MATH where MATH i... |
math/0006057 | Follows from REF , and REF. |
math/0006057 | Any endomorphism of an irreducible module is the multiplication by a scalar, therefore, we must have MATH. Applying the counit to both sides and using that MATH, we get the result. |
math/0006057 | Here we only need to prove the invertibility of the matrix formed by MATH where MATH are as above, MATH is a basis in the space MATH of characters of MATH (we used above REF for the quantum trace). It was shown in (CITE, REF) that the map MATH is a linear isomorphism between MATH and the center MATH (here MATH). So, th... |
math/0006057 | The proof is a straightforward verification, see CITE for details. |
math/0006057 | The existence and uniqueness of the solution of REF and the fact that MATH satisfies REF are established by induction on the degree of the first component of MATH, see (CITE). |
math/0006057 | CASE: If MATH, then by MATH one has MATH. Therefore, MATH is a dual pair of non-degenerate left integrals. The property MATH is equivalent to that the quasi-basis of MATH satisfies MATH from where MATH and MATH. Furthermore, MATH. Thus, MATH is a NAME integral. The inverse statement is obvious. CASE: The "only if" part... |
math/0006057 | The assumption on MATH is used only to ensure that MATH, once knowing that it is semisimple, so that there is a MATH-basis of matrix units MATH in MATH. MATH : Recall that any trace MATH on MATH is completely determined by its trace vector MATH. Now let MATH be the trace with trace vector MATH. Then MATH is non-degener... |
math/0006057 | Clearly, MATH and MATH verify all the conditions of REF , from where the existence of NAME integral follows. Since MATH is non-degenerate, the scalar product MATH is also non-degenerate. By the equality MATH positivity of MATH follows from REF . |
math/0006057 | The result follows from the fact that MATH is a MATH-quantum groupoid, which is easy to verify using the explicit formulas from REF . |
math/0006057 | Since MATH also implements MATH REF , MATH is central, therefore MATH is also central. Clearly, MATH must commute with MATH. The same Proposition gives MATH, which allows us to compute MATH . REF and the trace property imply that MATH . Since MATH and MATH is central, the above relation means that MATH and, therefore, ... |
math/0006057 | The proof follows from REF . |
math/0006057 | If MATH is such that MATH , then, using CITE, one has MATH therefore, using the properties of MATH and NAME projections, we get MATH so MATH. Similarly for MATH. |
math/0006057 | One can check (CITE) that MATH, hence, since MATH and MATH commutes with MATH, we have MATH which is REF . REF dual to it is obtained similarly, considering the comultiplication in MATH. As a consequence of the ``symmetric square" relations MATH one obtains (CITE) the identity MATH from where it follows that MATH . We ... |
math/0006057 | The above map defines a left MATH-module structure on MATH, since MATH and MATH . Next, using REF we get MATH . By CITE and properties of MATH we also get MATH . Finally, MATH and MATH iff MATH. |
math/0006057 | If MATH is such that MATH for all MATH, then MATH. Taking MATH, we get MATH which means that MATH. Thus, MATH. Conversely, if MATH, then MATH commutes with MATH and MATH therefore MATH. |
math/0006057 | By definition of the action MATH we have : MATH for all MATH, so MATH is a well defined linear map from MATH to MATH. It is surjective since an orthonormal basis of MATH over MATH is also a basis of MATH over MATH (CITEEF). Finally, one can check that MATH is a NAME algebra isomorphism (CITE). |
math/0006057 | Note that MATH for all MATH. By CITE, the tower of basic construction for MATH is MATH therefore, the depth of this inclusion is equal to MATH, where MATH is the smallest positive integer such that MATH. |
math/0006057 | First, let us check that MATH and MATH are indeed maps between the specified lattices. It follows from the definition of the crossed product that MATH is a NAME subalgebra of MATH, therefore MATH is a NAME subalgebra of MATH, so MATH is a map to MATH. To show that MATH maps to MATH, let us show that the annihilator MAT... |
math/0006057 | If MATH is an object of MATH, then every simple subcomodule of MATH is MATH-dimensional. Let MATH REF be one of these comodules, then all other simple subcomodules of MATH are of the form MATH, where MATH, and MATH . Vice versa, MATH with natural MATH bimodule and MATH-comodule structures is a simple object of MATH. |
math/0006057 | First, we observe that MATH for any MATH. Indeed, since both categories are semisimple, it is enough to check that they have the same set of simple objects. All objects of MATH are also objects of MATH. Conversely, since irreducible MATH subbimodules of MATH are contained in the decomposition of MATH for all MATH, we s... |
math/0006057 | First, let us show that there is a bijective correspondence between right relative MATH . NAME modules and MATH-modules. Indeed, every right MATH . NAME module MATH carries a right action of MATH. If we define a right action of MATH by MATH then we have MATH for all MATH and MATH which shows that MATH and MATH act on M... |
math/0006057 | In this case MATH and inclusion MATH is connected, so that MATH (note that MATH is biconnected). |
math/0006063 | We assume that MATH is defined on a coordinate chart MATH, and take MATH. Since MATH, the last condition of the definition of a formal integral takes the form MATH . Equating to zero the coefficient at MATH, of the l.h.s. of REF we get MATH, which can be rewritten as a recurrent equation MATH . Since MATH is a nondegen... |
math/0006063 | The condition that MATH is a nondegenerate critical point of the function MATH directly follows from the fact that MATH is a potential of the non-degenerate MATH -form MATH. REF of formal integral are trivially satisfied. It remains to check REF . Replace the pair MATH by the equivalent pair MATH. Put MATH. For MATH RE... |
math/0006063 | REF of formal integral are satisfied. It remains to check REF . Let MATH be a vector field on MATH. Denote by MATH the corresponding NAME derivative. We have MATH. Applying MATH we obtain that MATH, which concludes the proof. |
math/0006063 | Using integral representations REF one gets for MATH . Changing variables MATH in REF gives MATH . In order to apply the method of stationary phase to the integral in REF the following preparations should be made. Using REF express the phase function of the integral in REF as follows: MATH . In order to find the critic... |
math/0006063 | It was already shown that the phase function MATH of integral REF satisfies the conditions required in the method of stationary phase. Thus REF can be applied to REF . We get that MATH, where MATH is a formal integral at the point MATH associated to the pair MATH and MATH is a nonvanishing formal function on MATH. It f... |
math/0006065 | Clearly, we may assume that a group into which we embedd the amalgam MATH is generated by MATH and MATH. For MATH odd this means the group itself is of exponent MATH, and for MATH even of exponent MATH. |
math/0006065 | Since MATH, MATH, and MATH are of finite exponent, they are the direct product of their MATH-parts, that is, MATH, MATH, and MATH. Clearly, an embedding of MATH provides embeddings for MATH for each MATH; conversely, if we have embeddings into MATH for each MATH, then MATH will give an embedding for MATH. |
math/0006065 | Let MATH be the order of MATH (MATH if MATH is not torsion); then consider the group MATH. Denote the generator of the cyclic groups by MATH, and mod out by the subgroup generated by MATH. Since the MATH are central, the subgroup is normal and this works out. |
math/0006065 | Let MATH be the order of MATH, with MATH if MATH is not torsion. Consider the group MATH and denote the generators of the cyclic groups by MATH and MATH. Then mod out by the subgroup generated by MATH. Again, since the MATH are central, this subgroup is normal. If MATH is of odd exponent MATH, then MATH is generated by... |
math/0006065 | Since MATH and MATH are commutators in MATH, they are central. Therefore: MATH . Therefore, MATH, as claimed. |
math/0006065 | Suppose that MATH satisfies the conditions, and let MATH be any overgroup of MATH with MATH. Let MATH and MATH be such that MATH for some MATH, MATH. Let MATH, MATH. We want to verify that MATH. Note that since MATH and MATH are MATH-th powers modulo the commutator in MATH, then there is an extension of MATH in MATH wi... |
math/0006065 | CASE: As per the remark above, if MATH is cyclic or trivial, then MATH is absolutely closed in MATH, and hence also in MATH. If MATH is not cyclic, then let MATH and MATH be elements of MATH which project to distinct cyclic summands of MATH. Then note that a product MATH is a MATH-th power in MATH if and only if MATH a... |
math/0006065 | Note that if MATH and MATH lie in MATH, then so do MATH and MATH, so both commutator brackets lie in the dominion of MATH. Expanding the bracket bilinearly, we have MATH . Since MATH, and MATH, the last three terms on the right hand side lie in MATH, so the left hand side lies in MATH if and only if MATH lies in MATH, ... |
math/0006065 | If MATH, the result is easy: the group described is a special amalgamation base of MATH (since everything there is a special amalgamation base), but not a special amalgamation base in MATH or MATH for MATH, as it is abelian but not cyclic. So we may assume that MATH. Consider the group MATH . Then MATH is isomorphic to... |
math/0006065 | We prove the MATH case first. One implication is trivial. So suppose that MATH is not absolutely closed, and let MATH be any nil-REF overgroup of MATH. Let MATH. We want to show that MATH is also not absolutely closed. If MATH, then there is nothing to prove. So we may assume that MATH is properly contained in MATH. By... |
math/0006065 | Clearly MATH and MATH. We show that MATH. Suppose that MATH does not satisfy REF . If MATH, then we take an element MATH. We let MATH be a MATH-overgroup in which MATH is a commutator, using REF , and we let MATH be a MATH-overgroup in which MATH is not central, say MATH. Then the amalgam MATH cannot be embedded in a M... |
math/0006065 | Note that for any MATH, MATH, as every MATH-th power is central and of exponent MATH, and every commutator is central and of exponent MATH. Clearly MATH. Suppose MATH fails to satisfy MATH. If MATH, let MATH be central of exponent MATH, but not in MATH. Since MATH is of exponent MATH and central, we can extend MATH to ... |
math/0006067 | To prove the theorem, we follow CITE in starting with a single peg, which we denote MATH and playing the game in reverse. The first `unhop' produces MATH and the next MATH . (As it turns out, MATH is the only configuration that cannot be reduced to a single peg without using a hole outside the initial set of pegs. Ther... |
math/0006067 | 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/0006067 | Suppose we are given a string MATH where each MATH. Let MATH be a nondeterministic finite automaton (without MATH-transitions) for MATH, where MATH is the set of states in MATH, MATH is the start state, and MATH is the set of accepting states. We then construct a directed acyclic graph MATH as follows: Let the vertices... |
math/0006074 | Though MATH-modules MATH fail to be MATH-modules CITE, one can use the fact that the sheaves MATH are projections MATH of sheaves of MATH-modules. Let MATH be a locally finite open covering of MATH and MATH the associated partition of unity. For any open subset MATH and any section MATH of the sheaf MATH over MATH, let... |
math/0006074 | A fiber bundle MATH is a strong deformation retract of MATH. Then, the first isomorphism in REF follows from the NAME - NAME theorem (REF ; REF ), while the second one is a consequence of the well-known NAME theorem. |
math/0006074 | The isomorphism REF also follows from the facts that MATH is a strong deformation retract of MATH and that MATH is the pull-back onto MATH of the sheaf MATH on MATH REF . |
math/0006074 | The proof is obvious. The complex REF is exact due to the NAME lemma, and is a resolution of the constant sheaf MATH on MATH since MATH are sheaves of MATH-modules. Then, REF complete the proof. |
math/0006074 | The exact sequence REF is a resolution of the pull-back sheaf MATH on MATH. Then, by virtue of REF , we have a cohomology isomorphism MATH . REF completes the proof. |
math/0006074 | Due to the relation MATH the horizontal projection MATH provides a homomorphism of the NAME complex REF to the complex MATH . Accordingly, there is a homomorphism MATH of cohomology groups of these complexes. REF show that, for MATH, the homomorphism REF is an isomorphism (see the relation REF below for the case MATH).... |
math/0006074 | Being nilpotent, the vertical differential MATH defines a homomorphism of the complex REF to the complex MATH and, accordingly, a homomorphism of cohomology groups MATH of these complexes. Since MATH, the result follows. |
math/0006074 | Let the common symbol MATH stand for the coboundary operators MATH and MATH of the variational bicomplex. Bearing in mind the decompositions REF , it suffices to show that, if an element MATH is MATH-exact with respect to the algebra MATH (that is, MATH, MATH), then it is MATH-exact in the algebra MATH (that is, MATH, ... |
math/0006074 | By taking a smooth partition of unity on MATH subordinate to the cover MATH and passing to the function with support in MATH, one gets a smooth real function MATH on MATH which is REF on a neighborhood of MATH and REF on a neighborhood of MATH in MATH. Let MATH be the pull-back of MATH onto MATH. The exterior form MATH... |
math/0006078 | Since MATH is a separability element of MATH, MATH commutes with MATH. The assertion about MATH follows similarly. |
math/0006078 | Define a MATH-linear homomorphism MATH by MATH . This map is MATH-linear, since MATH for all MATH. The inverse map MATH is given by MATH . The collection MATH gives a natural equivalence between the functor MATH and the identity functor. Indeed, for any MATH-linear homomorphism MATH we have : MATH . Similarly, the MATH... |
math/0006078 | We know already that MATH is monoidal, it remains to prove that MATH and MATH are MATH-linear and satisfy the identities MATH . Take MATH. Using the axioms of a quantum groupoid, we have MATH therefore, MATH is MATH-linear. To check the MATH-linearity of MATH we have to show that MATH, that is ,that MATH . Since both s... |
math/0006078 | Note that MATH is well-defined, since MATH. To prove the MATH-linearity of MATH we observe that MATH . The inverse of MATH is given by MATH where MATH. Therefore, MATH is an isomorphism. Finally, one can verify that the braiding identities MATH are equivalent to the relations of REF , exactly in the same way as in the ... |
math/0006078 | Since we have MATH, the first line is a consequence of the relation MATH, the second line follows from MATH and MATH. The last two identities are proven similarly. |
math/0006078 | It follows from the first two relations of REF , that MATH . |
math/0006078 | First, using the same argument as in CITE, XI, REFEF, we can show that MATH. Next, using REF , we obtain MATH . Let MATH denote multiplication MATH in MATH. Set MATH and MATH. It follows from the above relations that MATH . On the other hand, REF implies that MATH . Therefore, MATH and MATH. |
math/0006078 | Note that MATH for all MATH, by REF . Hence, we have MATH for all MATH. Therefore, using the axioms of a quantum groupoid, we get MATH . The remaining part of the proof follows the lines of (CITEEF). The results for MATH can be obtained by applying the results for MATH to the quasitriangular quantum groupoid MATH. |
math/0006078 | See (CITEEF). |
math/0006078 | Take MATH. Then we have MATH . Therefore MATH, as required. |
math/0006078 | From the observation that MATH we have that the restriction of MATH to MATH is a linear map onto MATH. On the other hand, REF allows to identify MATH with the dual vector space to MATH, from where MATH and the result follows. |
math/0006078 | Associativity of multiplication in MATH and hence in MATH can be verified exactly as in (CITEEF). Let us check that MATH is an ideal. We have : MATH where MATH and we used the identity MATH. Similarly, one checks that MATH therefore for all MATH we have MATH, so MATH is an ideal. We also compute MATH and similarly MATH... |
math/0006078 | The identities MATH and MATH can be written as (identifying MATH with MATH and MATH with MATH): MATH . The above equalities can be verified by evaluating both sides on an element MATH in the third factor (respectively, on MATH in the second factor), see (CITEEF). To show that MATH is an intertwiner between MATH and MAT... |
math/0006078 | First, observe that for any pair of dual bases MATH and MATH as above and all MATH and MATH the element MATH belongs to MATH. Indeed, MATH for all MATH and, likewise, MATH . Next, we compute MATH where we used the identities MATH that follow from REF and from REF of a quantum groupoid. Therefore, MATH . Thus, we conclu... |
math/0006078 | Since MATH is an invertible central element of MATH, the homomorphism MATH is a MATH-linear isomorphism. The twist identity MATH follows from the properties of MATH: MATH for all MATH. Clearly, the identity MATH is equivalent to the twist property. It remains to prove that MATH for all MATH, that is, that MATH where MA... |
math/0006078 | Follows from REF , and REF. |
math/0006078 | Since the trace of an endomorphism MATH, in terms of the canonical element MATH, is MATH, the definition of MATH gives: MATH where we used REF defining MATH and MATH. |
math/0006078 | An endomorphism of an irreducible module is multiplication by a scalar, therefore, we must have MATH. Applying the counit to both sides and using that MATH, we get the result. |
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