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math-ph/0002045 | For every MATH consider the subbundle MATH that consists of the intersection of the subbundle of all elements of MATH of degree at most MATH with either the subbundle MATH or MATH in accordance with the parity of MATH. In the same manner we define MATH, where MATH runs over all indices between MATH and MATH differing f... |
math-ph/0002054 | The sesquilinear form MATH on MATH is separately continuous, and as a consequence of the nuclearity of MATH it is jointly continuous. This gives the norm continuity of MATH. Since the action of MATH on MATH is smooth, the action of MATH on MATH is continuous. As a consequence we obtain the strong continuity of MATH. |
math-ph/0002054 | One can always choose a neighbourhood MATH of MATH and a chart MATH with coordinates MATH, such that MATH and the dual metric tensor is diagonal in the point MATH, that is, MATH . MATH is then a positive matrix. Since the matrix-valued function MATH is continuous, there exists a neighbourhood MATH of MATH, on which it ... |
math-ph/0002054 | We introduce the following notations: MATH . MATH is clearly a MATH-invariant subspace and MATH is dense in MATH. Identifying MATH with MATH we can endow MATH with the locally convex quotient topology. Since MATH is nuclear and MATH is continuous, MATH is a nuclear space (see CITE), and clearly the inclusion map MATH i... |
math-ph/0002054 | We start with the fermionic case. Let MATH be the subspace of MATH generated by the set MATH and let MATH be the unital MATH-subalgebra of MATH generated by the set MATH. Clearly, MATH is equal to the MATH-subalgebra of MATH generated by MATH. The group of time translations MATH is a strongly continuous one parameter g... |
math-ph/0002054 | Let MATH. For each MATH we then have at least for some open neighbourhood MATH of REF MATH . Since MATH is a NAME REF or a ground-state (MATH), MATH is the boundary value of a function MATH which is analytic on the strip MATH and bounded and continuous on MATH. By the NAME reflection principle MATH vanishes on the whol... |
math/0002001 | Let MATH and MATH be the natural projections. Then MATH with MATH, MATH. Since MATH implies MATH (by the NAME result mentioned above), there exists a map MATH extending MATH. Then MATH extends MATH. Fix a function MATH such that MATH and define MATH by MATH. Since MATH, MATH can be extended to a map MATH. Then MATH is ... |
math/0002001 | MATH for each MATH. Observe first that MATH implies MATH, in particular, MATH. For fixed MATH extend MATH to a map MATH (such an extension exists because MATH is a closed subset of MATH, so MATH). Then MATH and MATH define a map from MATH into MATH which is extendable to a map MATH. Obviously, MATH and MATH, so MATH. C... |
math/0002001 | Suppose MATH is closed and MATH is a map. We are going to find a continuous extension MATH of MATH. Let MATH be the cone of MATH with a vertex MATH. Since MATH, there exists a map MATH extending MATH. Then MATH is a zero-set in MATH (because MATH is such a set in MATH) containing MATH. Therefore, we can assume that MAT... |
math/0002001 | Since MATH is a k-space and MATH, we can apply REF . |
math/0002001 | This lemma was proved by CITE for metric spaces MATH. His proof, coupled with CITE, works in our situation as well. |
math/0002001 | Observe first that MATH is paracompact as a closed image of the paracompact MATH. By REF , MATH for any MATH. Then the proof follows from REF with MATH replaced by MATH. |
math/0002001 | We follow the proof of REF . Maintaining the same notations and applying now REF (instead of REF ), it suffices to show that if MATH is a zero-set in MATH, then the formula MATH defines a set-valued map MATH which is strongly lsc and each MATH is aspherical. And this follows from REF . |
math/0002001 | Any MATH is aspherical (because the class MATH contains all finite-dimensional spaces) and a MATH. Suppose MATH is aspherical and MATH. We are going to apply REF in the special case when MATH and MATH being the identity map. In this special REF is true if MATH is replaced by MATH. Indeed, REF becomes trivial; to prove ... |
math/0002001 | First condition was proved by CITE, see also CITE for the case of local contractibility. REF can be obtained by using the arguments of CITE. |
math/0002001 | First step is to show that MATH is an approximate absolute neighborhood extensor for the class MATH, that is, if MATH is a metrizable MATH-space, MATH closed and MATH a map, then for every open cover MATH of MATH there is a neighborhood MATH of MATH in MATH and a map MATH such that MATH is MATH-close to MATH. We follow... |
math/0002001 | Since every metrizable (strongly) countable-dimensional space has property MATH, REF implies REF . Standard arguments show that every metrizable MATH which is a MATH for the class of metrizable (strongly) countable-dimensional spaces has the following property MATH: For every MATH and its neighborhood MATH in MATH ther... |
math/0002007 | We prove the first formula. We find MATH . The proof of the second formula is completely analogous. |
math/0002007 | We prove this in the first case. MATH . The proof in the second case is completely analogous. |
math/0002007 | We prove the claim in the first case (in the second the proof is completely analogous). For both MATH . |
math/0002007 | MATH . The proof is completely analogous for MATH. |
math/0002007 | We prove the claim in the first case. For both MATH we have MATH . The proof is completely analogous for MATH. |
math/0002007 | As a first step in the proof that REF is satisfied by the MATH of REF , we calculate the commutation relations between the MATH and the MATH: MATH . To obtain these relations we have used REF . In the case MATH we also need REF , and for even MATH, REF . It will be noticed that although complicated in appearence the sy... |
math/0002007 | It is interesting to note that the commutation relations REF between the MATH are the same as those REF satisfied by the MATH, because MATH. As REF is equivalent to REF , so will REF be equivalent to MATH where the quantities MATH are defined by the equation MATH for MATH in the case MATH odd, and for MATH in the case ... |
math/0002007 | To prove REF , we use REF MATH that is, the `NAME for MATH. By repeated application of the `NAME it is an immediate result that for any polynomial MATH . In particular the projectors MATH, MATH, MATH are of this form. If we write MATH explicitly using REF , this yields the MATH-relations REF [which are equivalent to th... |
math/0002010 | We first suppose MATH, and prove REF following CITE. Let MATH and MATH. Clearly MATH as MATH for all MATH; we now prove that MATH. Define MATH viewed as a complex analytic function in the half-plane MATH. We have MATH, from which MATH. Now applying the NAME residue formula, MATH for all MATH, so MATH . For MATH, REF ho... |
math/0002010 | Let MATH be a finite generating set for MATH; choose a generating set MATH for MATH. Apply REF with MATH to obtain MATH. Clearly MATH for all MATH. |
math/0002010 | Fix a generating set MATH. The identities MATH show that MATH so MATH is generated over MATH by MATH and elements of the form MATH for all MATH and MATH. Now MATH modulo MATH, so MATH is spanned by the MATH . All these elements are in the vector subspace MATH of MATH spanned by products of at most MATH generators, and ... |
math/0002010 | By the previous corollary, MATH for some MATH. Consider the MATH-completion MATH of MATH. As NAME algebras, MATH and MATH coincide, so MATH. By NAME 's criterion MATH is an analytic pro-MATH-group CITE and thus is linear over a field. Since MATH is residually-MATH it embeds in MATH so is also linear. By the NAME altern... |
math/0002010 | If MATH were nilpotent it would be finite, as it is finitely generated and torsion; since it is infinite REF yields the left inequality. The right inequality was proven by the first author in CITE, using purely combinatorial techniques. |
math/0002010 | That MATH is a MATH-group was observed by NAME and follows from the fact that MATH is a MATH-nilalgebra. Let MATH be the natural map MATH. Then MATH is generated by MATH and more generally MATH, so by REF there is a MATH such that the estimate MATH holds for any system MATH of generators of MATH. |
math/0002010 | We proceed by induction on MATH in lexicographic order. For MATH the claim holds trivially; suppose thus MATH. In order to prove MATH, it suffices to check that for all MATH we have MATH, as the MATH form an ascending tower of subspaces. During the proof we will consider MATH as a subspace of MATH; beware though that i... |
math/0002010 | We first suppose MATH. Then MATH; it is easy to check that MATH modulo MATH, so all generators of MATH are of order MATH. Further, MATH and MATH, so the quotient MATH is the elementary abelian group MATH, and MATH is an isomorphism in that case. For MATH it suffices to note that both sides of the isomorphism are direct... |
math/0002010 | Direct computation; see also CITE, where different notations are used. |
math/0002010 | Let MATH be such that it acts like MATH on MATH. Then MATH . Conjugating MATH by elements of MATH yields all cyclic permutations of the above vector, so as MATH is normal in MATH it contains MATH. Likewise, let MATH act like MATH on MATH. Then MATH using MATH, we obtain MATH, so by the same conjugation argument MATH. |
math/0002010 | First compute MATH; it is of index MATH in MATH, with quotient generated by MATH. Compute also MATH of index MATH in MATH, with quotient generated by MATH. This gives the basis of an induction on MATH and MATH. Assume that MATH. Note that the hypothesis of REF is satisfied; indeed MATH can even be chosen among the conj... |
math/0002010 | Write MATH or MATH for some MATH. Then these claims follow immediately, using REF , from MATH . |
math/0002010 | First compute MATH and MATH. This gives the basis of an induction on MATH and MATH. Assume MATH. Consider the sequence of quotients MATH for MATH. We have MATH by REF . REF tell us that MATH; in particular MATH for all MATH, and then REF tells us that MATH for MATH. When MATH we have MATH and the induction can continue... |
math/0002010 | We first suppose MATH. Then MATH it is easy to check that MATH, so all generators of MATH are of order MATH. Further, all commutators of generators belong to MATH, so the quotient MATH is the elementary abelian group MATH, and MATH is an isomorphism in that case. For MATH it suffices to note that both sides of the isom... |
math/0002010 | Direct computation. |
math/0002010 | Let MATH be such that it acts like MATH on MATH. Then MATH so by a conjugation argument MATH. Likewise, let MATH and MATH act like MATH and MATH on MATH. Then MATH . Using MATH, we obtain MATH, so again by a conjugation argument MATH. |
math/0002010 | First compute MATH, of index MATH in MATH, and MATH, with MATH, MATH and MATH. This gives the basis of an induction on MATH and MATH. Assume that MATH. Note that the hypothesis of REF is satisfied for MATH, as it holds for MATH. Consider the sequence of quotients MATH for MATH. REF tell us that MATH; in particular MATH... |
math/0002010 | Write MATH, MATH or MATH for some MATH. Then these claims follow immediately, using REF , from MATH . |
math/0002010 | First compute MATH and MATH. This gives the basis of an induction on MATH and MATH. Assume MATH. Consider the sequence of quotients MATH for MATH. We have MATH by REF . REF tell us that MATH; in particular MATH for all MATH, and then REF tells us that MATH for MATH. When MATH we have MATH and the induction can continue... |
math/0002010 | MATH has the congruence property. This is well known for MATH (see for instance CITE); while for MATH the subgroup MATH contains MATH and enjoys the property that every subgroup of finite index in MATH contains MATH for some MATH; see CITE. The profinite completion of MATH with respect to its subgroups MATH is therefor... |
math/0002013 | By REF and compactness of MATH, it suffices to prove the following: For all convergent sequences MATH in MATH and MATH in MATH, with MATH, we have MATH. Being a connected locally compact group, MATH is generated by a compact neighborbood MATH of the identity. Then MATH uniformly for MATH, so MATH for all MATH. Since MA... |
math/0002013 | Choose a homeomorphism MATH (the circle). Let MATH be the action defined by MATH. Because MATH is solvable, by a result of NAME REF there exists a homeomorphism MATH of MATH conjugating MATH to the rotation group MATH. Since MATH is abelian and acts transitively on MATH, all points of MATH have the same isotropy group ... |
math/0002013 | Since there are no orbits of dimension MATH, MATH is a compact set comprising the points MATH such that MATH. It is easy to see that MATH is a local analytic variety. If MATH then the map MATH is a continuous field of tangent lines to MATH, tangent to MATH at boundary points. The existence of such a field implies that ... |
math/0002013 | Suppose not; then MATH is a closed REF-cell. Since there are no fixed points, MATH is an orbit, hence a component of MATH. Every component of MATH bounds a unique REF-cell in MATH, and there are only finitely many such REF-cells. Let MATH be one that contains no other. Then MATH is invariant under MATH, and the action ... |
math/0002014 | The central elements of the MATH-module MATH are homomorphisms given by right multiplication by elements of MATH. Hence the result. |
math/0002014 | For any MATH-bimodule MATH we have, MATH where MATH denotes the MATH-centre of MATH (analogously defined for MATH). Hence the proposition. |
math/0002014 | Since MATH is commutative, MATH (respectively, MATH), if and only if MATH (respectively, MATH), for MATH. The proposition follows immediately once we notice that if MATH is given by a matrix MATH, then MATH is given by the matrix MATH for MATH. |
math/0002014 | We first prove the theorem in the case when MATH is free over MATH with basis MATH. By REF , any MATH if and only if all the MATH. It remains to show that if all the MATH, then MATH. Let MATH. For any MATH, define MATH as MATH. For each MATH, define MATH (and hence in MATH) given by MATH. Then, we have MATH. Thus, it r... |
math/0002014 | We prove both the statement by induction on MATH. Let MATH be given. There is a MATH in MATH, such that MATH. Thus, MATH is mapped to MATH under MATH. So, the result is proved for MATH. Assuming that the proposition is proved for MATH (which implies that MATH), let MATH be such that MATH for every MATH. It is enough to... |
math/0002014 | The theorem above shows that MATH. The ring MATH is free as a left MATH-module. By REF the corollary is proved. |
math/0002014 | It is clear to see that MATH for MATH and MATH. Thus, MATH implies that MATH. Now use REF to complete the lemma. |
math/0002014 | Let MATH. For each MATH let MATH . Note that MATH. For MATH, we see that MATH . Hence, MATH. Now, by the dual basis lemma, MATH . Hence, MATH . Since MATH, we have MATH. Hence, REF gives MATH . Hence the theorem. |
math/0002014 | The lemma follows from the fact that, for MATH, and MATH, we have MATH. |
math/0002014 | We show that MATH is the identity on MATH. Clearly, MATH. Let any MATH. The MATH as defined in REF are in MATH. Referring to REF , the claim is proved. Now we show that MATH is the identity on MATH. Again, it is obvious that MATH. For the reverse inclusion, use the fact that MATH . For MATH and MATH, we see that MATH .... |
math/0002014 | Let MATH REF for some MATH such that MATH for some MATH and MATH. It is enough to show that there exists a MATH such that MATH for MATH. If MATH then MATH. Else, MATH for some MATH. By referring to the note REF, we have the lemma. |
math/0002014 | Immediate once we see that MATH. |
math/0002014 | Let MATH denote the MATH -span of monomials in MATH . Claim : Let MATH be such that MATH. Then MATH. When MATH then MATH and hence the claim. Assume that we have proved the claim for MATH and fix MATH such that MATH for some MATH. It is enough to show that for MATH, we have MATH. Note that MATH. Similar argument for th... |
math/0002014 | By the above proposition, MATH is generated by MATH and left multiplication by elements of MATH. Since MATH, we have MATH are inner (A more general statement is true, due to NAME REF : All derivations on a NAME algebra are inner). That is, MATH. By REF , we have a surjection MATH. Note that MATH is isomorphic to MATH b... |
math/0002014 | By the properties following REF we see that MATH and MATH generate MATH and MATH for all MATH. The previous proposition completes the theorem. |
math/0002014 | Let MATH. Then MATH can be written as a MATH -linear combination of monomials of the form MATH where MATH where MATH a multiindex. Let MATH be an ideal in MATH. Let MATH. As MATH and the fact that MATH commutes with all the other generators, we can assume that MATH does not appear in the expression of MATH. Now use the... |
math/0002016 | MATH and if MATH as elements of MATH, MATH and therefore MATH. |
math/0002016 | Set MATH. It is clear that MATH is unitary and MATH. It follows that MATH and hence MATH. |
math/0002016 | Let MATH be the polar decomposition of MATH. Then MATH . For the second inequality, MATH . |
math/0002016 | Fix MATH with MATH and define the sequence of projections as follows: MATH . Clearly MATH is a subprojection of the support of MATH so MATH. Also MATH, and since MATH and MATH are commuting operators, MATH and therefore MATH so MATH and MATH are verified. Claim: Let MATH and MATH, for every MATH with MATH, MATH. Simila... |
math/0002016 | The sequence MATH will be constructed inductively. Let MATH be a sequence in the open interval MATH such that MATH and MATH be a faithful state in MATH. By REF , one can choose a sequence of projections MATH with MATH for every MATH, MATH (as n tends to MATH) satisfying the conclusion of REF for MATH, MATH and MATH. Ch... |
math/0002016 | Set MATH and let MATH be a subset of the open interval MATH such that MATH. Since MATH, one can choose a sequence MATH in MATH and MATH such that MATH . A further subsequence MATH can be chosen so that MATH . Set MATH. It is clear that MATH and as above a sequence MATH can be chosen so that MATH . Inductively, one can ... |
math/0002016 | We will assume first that MATH is MATH-finite. Without loss of generality, we can and do assume that MATH for all MATH and MATH is not uniformly integrable. We will show that there exist a sequence MATH in MATH and MATH such that the bounded set MATH is uniformly integrable in MATH. By REF , there exists a subsequence ... |
math/0002016 | Note that MATH is a NAME algebra so subspaces of MATH are subspaces of preduals of NAME algebras. The proof then follows the argument used in CITE using REF . Details are left to the readers. |
math/0002016 | Let MATH be a sequence in MATH that is equivalent to MATH. After taking subsequences, either MATH or MATH is equivalent to MATH. Let assume that MATH is equivalent to MATH. We have two cases. CASE: The sequence MATH is weakly convergent. By taking normalized blocks, we can assume that MATH is asymptotically isometric t... |
math/0002016 | Let MATH be a sequence equivalent to the MATH basis. Since MATH is reflexive, the sequence MATH can not be uniformly integrable (see for instance CITE). Apply the classical NAME subsequence decomposition to the sequence MATH in MATH to get a pairwise disjoint sequence of measurable sets MATH such that MATH is uniformly... |
math/0002017 | The fact that MATH for all MATH and all MATH implies by base change that MATH is locally free with fiber MATH . We also have a natural isomorphism MATH obtained as follows. One one hand by NAME we naturally have MATH and on the other hand the automorphism MATH of MATH induces an isomorphism MATH so MATH is obtained by ... |
math/0002017 | By the push-pull formula (see CITE, III. REF) and REF we obtain: MATH . |
math/0002017 | Both properties can be checked up to finite covers (see for example, REF) and they are obvious for MATH. |
math/0002017 | Using the usual identification between Pic-MATH and Pic-MATH, we have to show that MATH for all MATH-Pic-MATH and all MATH. But MATH is a direct summand in MATH and so it is enough to have the vanishing of MATH. This is obvious by the formula in REF. By REF we have: MATH . The last statement follows also from REF. |
math/0002017 | By REF we have MATH . Since MATH is a theta characteristic, MATH is symmetric and so MATH. We get MATH . Note now that the diagram giving REF is equivariant with respect to the MATH, the group of MATH-torsion points of MATH, if we let MATH act on MATH by MATH, by twisting on MATH, by translation on the left MATH and tr... |
math/0002017 | By definition and the projection formula one has MATH where the last isomorphism is an application of REF. Now the following commutative diagram: MATH shows that MATH, which is exactly the statement of the lemma. |
math/0002017 | This follows from a direct computation of the number of endomorphisms of MATH. By REF and the NAME formula at level MATH, MATH. Then: MATH . On the other hand, since MATH is a NAME cover with NAME group MATH, we have the formula MATH. Combined with REF , this gives MATH . The two relations imply that MATH, so MATH is s... |
math/0002017 | Let's begin by fixing a polarization on MATH, so that stability will be understood with respect to this polarization. Consider the isogeny defined by MATH . If MATH is the index of MATH, it follows from REF that MATH, where MATH. As we already mentioned in the proof of REF, by REF this already implies that MATH is poly... |
math/0002017 | By the duality REF , it is enough to show that MATH. But MATH is just MATH, which is symmetric since MATH is a theta characteristic. So what we have to prove is MATH . Since MATH is NAME with NAME group MATH, we have (see for example, REF ): MATH . But translates commute with tensor products via the NAME transform (com... |
math/0002017 | We know that MATH, so by REF we obtain MATH . As in the previous proposition, since the NAME group of MATH is MATH, we have MATH . Moreover the isomorphisms above show as before that MATH has to be semistable with respect to an arbitrary polarization. On the other by REF we have MATH . This gives us the isomorphism MAT... |
math/0002017 | We will actually prove the following equality: MATH . The statement will then follow from the same symmetry of the NAME formula MATH mentioned in the proof of REF. To this end we can use REF to obtain MATH. But on one hand MATH while on the other hand by REF MATH as required. |
math/0002017 | We will compare the restrictions of the two line bundles to fibers of the projections. First fix MATH. We have MATH where MATH is the map given by twisting with MATH. But by REF one has MATH . On the other hand obviously MATH . Let's now fix MATH such that MATH. Using the same REF we get MATH and we also have MATH . Th... |
math/0002017 | All the ingredients necessary for proving this have been discussed above: by REF, the bundles MATH and MATH are isomorphic and stable. This means that MATH is essentially the unique nonzero morphism between them, and it must be an isomorphism. |
math/0002017 | This follows basically from the proof of REF . One can see in a completely analogous way that: MATH . Now if we consider MATH to be the preimage of MATH by MATH, this shows that MATH is semistable and so MATH is semistable. |
math/0002017 | The proof goes like in REF and we repeat it here for convenience: choose MATH a nonempty open subset of MATH on which MATH is trivial. By NAME 's duality theorem REF we know that MATH, so there exists a nonzero section MATH. Choose now a MATH such that MATH and consider an embedding of MATH in MATH passing through MATH... |
math/0002017 | By REF we have MATH . But from the proof of REF we know that MATH and so MATH. To compute the slope of MATH, first notice that by the proof of REF we know that MATH . But MATH, so MATH . Combining the two equalities we get MATH . |
math/0002017 | Start with MATH and consider MATH. By MATH we have the exact sequence on MATH: MATH . The global generation of MATH implies that there exists a section MATH such that MATH. We would like to lift MATH to some MATH, so it is enough to prove that MATH. The fibers of MATH are reduced, so MATH. REF implies, by the base chan... |
math/0002017 | The trick is to write MATH as MATH. Denote MATH by MATH. NAME 's REF says in our case that MATH is globally generated as long as the condition MATH is satisfied (in fact under this assumption MATH will be globally generated for every ample line bundle MATH). Arguing as usual, MATH is implied by MATH. We have MATH and t... |
math/0002017 | The first part follows by puting together REF in our particular setting. To prove that MATH is not globally generated, let us begin by assuming the contrary. Then the restriction MATH of MATH to any of the fibers MATH is also globally generated. Choose in particular a line bundle MATH in the support of MATH on MATH. Re... |
math/0002017 | CASE: This is not hard to deal with when MATH is a multiple of MATH, since we know from REF that MATH decomposes in a particularly nice way. To tackle the general case though, we have to appeal to REF. Concretely, we have to see precisely when the skew NAME product MATH is globally generated, and since by the initial c... |
math/0002017 | Since MATH is ample, by a standard argument the assertion is true if the multiplication maps MATH are surjective for MATH. For MATH this is proved in the theorem and the case MATH is similar but easier. |
math/0002017 | This follows by REF, which says that on a generic curve the multiplication map MATH on MATH is surjective for MATH. See also REF for a survey of results in this direction. |
math/0002017 | By repeated use of the projection formula, from the commutative diagram MATH we see that it is enough to prove the surjectivity of the multiplication map MATH on MATH. This is an application of the cohomological REF going back to CITE. What we need to check is that MATH . It is again enough to prove this after pulling ... |
math/0002017 | This is a consequence of REF and CITE, where it is proved that on MATH is surjective for MATH. |
math/0002018 | Assume that MATH is strictly semistable. Then it has a NAME filtration: MATH such that MATH are stable for MATH and MATH. By assumption there exist MATH such that MATH and so if we denote MATH this is a proper subset for every MATH. It is clear that any MATH satisfies MATH. |
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