paper
stringlengths 9
16
| proof
stringlengths 0
131k
|
|---|---|
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 from MATH by zero or an even number. CASE: MATH for every MATH. We prove the claim by induction. For MATH a comparison of the definitions shows that MATH for every MATH. In case MATH we obtain MATH for every MATH. To show the general argument recall that the complexified NAME algebra of a real vector space MATH and the complexified exterior algebra of MATH are related by the isomorphisms MATH for every MATH. So there exist induced symbol maps between the components of the NAME bundle and the exterior algebra bundle over MATH, MATH, with kernels MATH. Consequently, every smooth section of the MATH-th component MATH of the NAME bundle can be represented as a finite linear combination of elementary elements of the form MATH, where the central dot denotes the NAME multiplication. To see this apply NAME 's theorem and take the projection MATH of the canonical orthonormal basis MATH of the trivial bundle that houses MATH as a direct summand. By REF the set MATH generates MATH as a MATH-module. There exists a finite atlas of MATH and a partition of unity MATH corresponding to it such that every component MATH of a certain generator MATH can be written as a finite sum of elements of the set MATH. Since all sums are finite we get the desired decomposition property for smooth sections of MATH. Pulling this system of generators back via MATH we obtain it for smooth sections of MATH, too. To show the inclusion MATH we have only to check the canonical elements MATH since the inclusion is supposed to be already established for lower degrees by induction. We have MATH . Conversely, we have to show that MATH for every MATH. By the results of our considerations on the symbol maps and by induction we have to verify the inclusion only for finite sums MATH. We get MATH . This establishes the statement of the first claim. CASE: MATH for any MATH with MATH. Suppose, MATH. By the first step MATH and MATH since MATH is injective on such elementary elements. Consider MATH : MATH . Consequently, MATH for every MATH with MATH. To show the reverse inclusion, let MATH. If MATH is a partition of unity corresponding to the selected atlas then we can assume MATH since MATH. Furthermore, by the discussions in the first part of this proof MATH for some functions MATH. Let MATH with MATH for any MATH and MATH for every MATH. This forces MATH and MATH for every MATH. Furthermore, MATH for every MATH, and by the first step we obtain MATH . Therefore, for any MATH since the latter is an ideal in MATH. Finally, by REF and the first step: MATH . We arrive at MATH for every MATH with MATH, and REF is proved. As the final step we list the following chain of identifications and isomorphisms: MATH . |
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 is positive. |
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 is continuous. We denote the dual of MATH by MATH. It follows that MATH is a NAME triple. We denote the selfadjoint generator of the group MATH by MATH. Clearly, MATH and MATH restricts to a continuous operator MATH. Moreover MATH is invariant under the action of MATH. As a consequence MATH is essentially selfadjoint on MATH. Hence, there exists a complete set of generalized eigenvectors (see CITE) for MATH, that is, a family MATH indexed by a subset MATH, such that MATH . By continuity MATH defines for each MATH a distribution in MATH, such that MATH where MATH is the timelike Killing vector field induced by MATH and MATH the NAME derivative on MATH defined by MATH. For each point MATH there exists an open contractible neighbourhood MATH and a chart mapping MATH to MATH which we can choose to be of the form constructed in REF . The restriction of MATH to MATH is trivial and we can identify MATH with MATH in such a way that MATH for both functions and sections. We consider the distribution MATH in such a chart. We have MATH, where MATH is the adjoint operator. Moreover, REF reads MATH. Note that the principal part of MATH has the form MATH . Replacing the MATH-derivatives by MATH and adding the term MATH we obtain an elliptic second order differential operator MATH with MATH. Hence, MATH is smooth (see for example . CITE) and since MATH has scalar principal part the classical result of CITE (see especially REF ) implies that MATH in each such chart in which MATH vanishes in an open nonvoid set, in particular in each such chart intersecting with MATH. Since MATH is connected this implies MATH on MATH. The set of generalized eigenvectors MATH was complete and therefore MATH equals MATH and the theorem is proved. |
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 group on MATH, and with MATH we can apply REF above. It follows that MATH is dense in MATH and hence MATH is norm dense in MATH. In the bosonic case let MATH be the complexification of MATH and take the complexification of MATH as a scalar product on MATH. We complete this space and obtain a NAME space MATH. We complexify the real vector bundle MATH and obtain the complex vector bundle MATH with a canonical action of MATH. We can extend the map MATH to a map MATH which maps from the section of MATH to MATH. By construction MATH has dense range. Since MATH is invariant under the action of the time translations MATH we get a strongly continuous unitary action of the group of time translations on MATH such that MATH is equivariant. Let MATH be the complex subspace of MATH generated by the set MATH . We see that all the assumptions for REF are fulfilled and hence MATH is dense in MATH. Using REF one therefore concludes that the MATH-subalgebra of MATH generated by the set MATH is strongly dense in MATH. |
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 whole real axis. Therefore MATH for all MATH. Hence, MATH is invariant under the action of MATH. |
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 the required extension of MATH. |
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. CASE: MATH is strongly lsc. Let MATH and MATH be compact. We have to find a neighborhood MATH of MATH in MATH such that MATH for every MATH. Let MATH, MATH. Since MATH is compact and MATH is a MATH-space, by the NAME theorem, each MATH is compact and MATH is evently continuous. This easily implies that the set MATH is open in MATH and, obviously, MATH. Because MATH is closed, there exists a neighborhood MATH of MATH in MATH with MATH. Then, according to the choice of MATH and the definition of MATH, MATH for every MATH. CASE: Each MATH is MATH. For a fixed MATH take an arbitrary map MATH, where MATH. We are going to show that MATH can be extended continuously to a map from MATH into MATH (we identify MATH with the boundary of MATH). Since MATH is a k-space (as a product of a compact space and a k-space), the map MATH, MATH, is continuous (see CITE). Because MATH for every MATH, we can extend MATH to a map MATH, MATH. Then, we have a closed subset MATH of MATH and a map MATH defined by MATH and MATH. Since MATH and MATH are zero-sets in MATH and MATH, respectively, both MATH and MATH are zero-sets in MATH, so is MATH. Note that MATH because MATH is closed in MATH. Therefore, by REF , MATH extends to a map MATH. Now, let MATH be the union of the sets MATH, MATH and MATH. We define the map MATH by MATH, MATH and MATH. Obviously, MATH is closed in MATH. Since MATH, there exists an extension MATH of MATH. To finish the proof of REF , observe that MATH generates the map MATH, MATH. Moreover, MATH for any MATH and MATH. So, MATH is a map from MATH to MATH which extends MATH. |
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 MATH is a zero-set in MATH. Next, define the set-valued map MATH, MATH (a similar idea was earlier used by NAME and NAME). By REF , MATH is a strongly lsc map with each MATH being a MATH-set. Since MATH, we can apply REF to obtain a continuous selection MATH for MATH. Then MATH, defined by MATH, is continuous on every compact subset of MATH and because MATH is a k-space, MATH is continuous. Since MATH, we have MATH for all MATH and MATH, MATH. Finally, if MATH denotes the natural retraction, then MATH is the required continuous extension of MATH. |
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 REF we don't need to apply REF because the set MATH is homeomorphic either to MATH if MATH or MATH otherwise, we need that any map from MATH into MATH is extendable to map from MATH into MATH, MATH. In order to apply REF , it remains only to check that MATH for all MATH. And that is true because MATH CITE. |
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 the construction from the proof of CITE. For every MATH and MATH fix points MATH and neighborhoods MATH of MATH in MATH such that: CASE: MATH contracts in MATH to MATH for all MATH and MATH; CASE: the cover MATH refines MATH; CASE: the cover MATH star-refines MATH for any MATH, that is, MATH refines MATH. Observe that we have corresponding covers MATH, MATH and MATH of MATH such that MATH refines MATH and MATH star-refines MATH, MATH. For every MATH and MATH we fix a contraction map MATH with MATH. Since MATH is a MATH-space (as a closed subset of MATH), there is a sequence of disjoint open families MATH in MATH such that the restriction of each MATH on MATH refines MATH and MATH covers MATH. Further, let MATH be the nerve of MATH and MATH a barycentric map. We are going to define a map MATH such that the family MATH refines MATH. Then the map MATH will be the required MATH-approximation of MATH. Any simplex MATH from MATH, where MATH, can be ordered such that MATH (this is possible because MATH, so the numbers MATH are different). By MATH, for any MATH there exists MATH with MATH. We define MATH by MATH, MATH. Using the contractions MATH, as in CITE, we can define by induction maps MATH such that the restriction of MATH on MATH is MATH, MATH, and for any simplex MATH we have CASE: MATH. So, we obtain a map MATH and, by MATH, MATH and MATH are MATH-close, where MATH. Indeed, if MATH and MATH for some simplex MATH, then MATH and MATH. According to MATH, the last two inclusions imply that both MATH and MATH belong to MATH. So, MATH and MATH are MATH-close and, since MATH, they are also MATH-close. Therefore, MATH is an approximate absolute neighborhood extensor for the class MATH. To complete the proof we state the following result which was actually proved in CITE but not explicitely formulated: If MATH is a class of metrizable spaces such that MATH for every MATH, then any approximate absolute neighborhood extensor for MATH is a MATH. Since MATH is closed with respect to multiplication by MATH, we have MATH. |
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 there is a neighborhood MATH such that any map from a closed subset of a (strongly) countable-dimensional metrizable space MATH into MATH extends to a map from MATH into MATH. Hence, MATH yields that the identity map of MATH is MATH. So, it remains to prove REF . Let MATH be a MATH map with MATH metrizable. We need the following result of CITE: There exists a MATH-soft map from a MATH-compact strongly countable-dimensional metrizable space onto the NAME cube. Here, a map MATH is called MATH-soft if for every strongly countable-dimensional metrizable space MATH, its closed subset MATH and any two maps MATH, MATH such that MATH there exists a map MATH extending MATH with MATH. Using the NAME result, for every cardinal MATH we can construct a strongly countable-dimensional metrizable space MATH of weight MATH and a MATH-soft map MATH (see CITE for a similar reduction), where MATH denotes the NAME space of weight MATH. Embedding MATH into MATH for some MATH and considering the restriction MATH of MATH onto MATH, we obtain a strongly countable-dimensional metrizable space MATH and a MATH-soft map MATH. Let MATH. We are going to show that MATH satisfies the hypotheses of REF . To this end, let MATH be a neighborhood of MATH. Since MATH is MATH, there exists a neighborhood MATH such that every map from a strongly countable-dimensional metrizable space MATH into MATH extends to a map from MATH into MATH. Then MATH for some neighborhood MATH of MATH in MATH because MATH is closed. Now consider MATH and MATH. Since MATH is strongly countable-dimensional, there exists a map MATH extending the restriction MATH. Finally, using that MATH-soft, we can lift MATH to a map MATH such that MATH is the identity. Therefore, MATH is contractible in MATH and, by REF , MATH. |
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 system of REF is actually mainly the first two equations, which are quite simple, plus a series of special cases when MATH. The commutation relations between the MATH and the MATH are independent of the normalization of the latter, so that they impose no restriction on the MATH. Writing down the explicit expression for the MATH-matrix REF and using the fact that MATH for MATH, one finds that the relation REF becomes: MATH finally, when MATH, MATH . These relations can be checked one by one with a lengthy but straightward calculation by substituting the explicit REF for MATH and commuting MATH through it. More particularly, in the case MATH both sides of the equations are identically REF, because both MATH and MATH are lower-triangular matrices. In the case MATH we first use REF again to commute MATH with MATH. Then we need REF to commute MATH with MATH if MATH, while we need to apply REF to the expressions of the type MATH to write them in terms of MATH and MATH if MATH. In the case MATH we need REF to commute MATH through MATH and MATH. In the particular REF follows from the MATH commutation relation REF . |
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 even (in the latter case the sum of course runs over MATH). It is easy to show the commutation relations (for MATH) MATH and, for MATH even, MATH which follow from REF . To show now the relation REF MATH we consider first the case MATH, excluding the cases MATH odd and MATH, and MATH even and MATH. By using REF MATH, REF we obtain respectively the identities MATH . If MATH is even and MATH we obtain MATH using respectively the identities REF . We proceed similarly in the case MATH when MATH is even. The calculation is analogous for the other cases MATH. Summing up, for MATH the MATH and MATH-commute, so that there is no restriction on the normalization constants MATH. We now consider the cases MATH. It follows from REF that MATH . We use these two relations as initial steps to show by induction that MATH . In fact MATH . Assuming that REF holds for MATH, the first two terms in the last line are opposite and therefore cancel, and the third gives REF for MATH, as claimed. We now consider the commutators MATH with MATH. For MATH even we find the claim MATH by a straightforward calculation. In all other cases we proceed as follows, MATH as claimed. With the choice REF for the normalization constants MATH, the algebra generated by MATH is symmetric, that is, is invariant, with respect to the following transformation MATH . Notice that MATH is an involution, MATH. Because of REF under MATH the MATH-commutation relations REF and the MATH-commutation relations REF are exchanged. REF and the MATH relations REF are exchanged as well, while the MATH relations REF are invariant. We can immediately check that for MATH odd under MATH where we have used REF . In the case of even MATH the same calculation holds for the terms with MATH, but we have to treat the term with MATH separately MATH . The relation REF expressing MATH in terms of MATH is invariant as well. To see this we first express MATH and MATH in REF through MATH and MATH respectively, then we use REF to move MATH to the left. In this way we are able to rewrite REF in the form MATH . Taking into account REF , under REF becomes MATH that is, we recover REF . Again, in the case of even MATH the special case MATH has to be treated separately, but due to REF MATH and REF , it is easily checked that REF is invariant in this case, too. This transformation is useful, because it enables us to get REF by applying MATH to REF and then using REF of the MATH-matrix and REF . For MATH odd and MATH even, MATH . In the particular case that MATH is even and MATH from MATH and REF of the MATH-matrix, it is easily seen that REF still holds. This concludes the proof of REF . |
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 the MATH-image of REF ] also for MATH, which we had not proved yet. |
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 holds a fortiori. REF for MATH is a consequence of a result by NAME (CITE; see also CITE). More precisely, when MATH, his result implies that MATH where MATH means that the quotient tends to MATH as MATH, and MATH is the NAME zeta function. We sketch the proof for MATH below: we suppose that MATH, so MATH by REF , and compute MATH . Thus MATH, and the inequality MATH is tight by the saddle-point principle when the right-hand side is minimized. This is done by choosing MATH, whence as claimed MATH. Finally, we show that REF yields the same asymptotics when MATH as when MATH. Clearly MATH for all MATH, where for two power series MATH and MATH the inequality MATH means that MATH for all MATH. It thus suffices to consider the case MATH. For this purpose define MATH and compare the series developments of MATH and MATH in MATH. From MATH it follows that MATH so both series have the same odd-degree coefficients, and thus MATH. Their exponentials then have the same asymptotics; more precisely, MATH for all MATH, so MATH termwise, and MATH. |
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 by REF is of dimension MATH. |
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 alternative CITE either MATH contains a free group on two generators (contradicting the assumption on the growth of MATH) or MATH is virtually solvable. By the results of NAME and NAME every virtually solvable group is either of exponential growth or is virtually nilpotent CITE. The asymptotic growth is invariant under taking finite-index subgroups, and the growth of a nilpotent group is polynomial of degree MATH, as was shown by CITE. |
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 it is not a submodule. We shall write MATH for the action of MATH on MATH, and MATH for that of MATH on MATH. Observe that if MATH then MATH. Thus MATH is always divisible by MATH because if MATH for some MATH then MATH and MATH . Write MATH in base MATH. For some MATH and MATH in MATH, we may write MATH . Then by induction MATH as in the second summand MATH is divisible by MATH. This proves the first claim of the lemma. Next, we prove MATH by showing that MATH. As above, write MATH in base MATH. If MATH, we have MATH, and by induction MATH for some MATH, MATH and MATH. Then MATH where the last inclusions hold by REF and induction. Finally, if MATH, note that MATH where MATH is invariant under MATH; thus MATH . |
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 sums of MATH terms on each of which the lemma for MATH can be applied. |
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 conjugates of MATH, MATH or MATH. Consider the sequence of quotients MATH for MATH. 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 | 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 isomorphism are direct sums of MATH terms on each of which the lemma for MATH can be applied. |
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 for all MATH, and then REF tells us that MATH for MATH. When MATH we have MATH and the induction can continue. |
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 therefore a pro-MATH-group and coincides with the closure of MATH in MATH. The closure of a branch group is again a branch group, as is observed in CITE. The criterion of just-infiniteness for profinite branch groups is the same as the one for discrete branch groups given in CITE; it is that MATH (respectively MATH) be finite, where MATH and MATH are the closures of MATH and MATH. Now MATH, the last inequality following from a computation in CITE. The same inequalities hold for MATH, and this proves the just-infiniteness of MATH. The group MATH has finite width for both versions of REF . This is clear for MATH-width, because the discrete and pro-MATH . NAME algebras MATH and MATH are isomorphic. The finiteness of MATH-width follows from the inequalities MATH which again are consequences of the congruence property of MATH. Finally, MATH does not belong to the list of groups given in Conjecture REF: it is neither solvable, because MATH isn't, nor MATH-adic analytic, by NAME 's criterion CITE (its NAME algebra MATH would have a zero component in some dimension). The other groups in the list of Conjecture REF are hereditarily just-infinite groups, that is, groups every open subgroup of which is just-infinite CITE. Profinite just-infinite branch groups are never hereditarily just-infinite, as is shown in CITE. |
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 MATH generates MATH, this implies that MATH. |
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 for MATH; this isotropy group is the required MATH. |
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. Assume that MATH. Note that MATH at each MATH. Since MATH is connected and MATH is a variety, MATH must have dimension MATH at each point. The set of points where MATH is not smooth is a compact, invariant REF-dimensional subvariety, that is, a finite set of fixed points, hence empty. Since MATH consists of REF orbits, MATH must be a compact, smooth invariant REF-manifold without boundary, that is, each component of MATH is a NAME curve. Since MATH is the union of invariant NAME curves, any component of MATH that meets MATH is a component of MATH. |
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 of MATH on MATH is fixed point free. Therefore we may assume that MATH, so that MATH. By REF there exists a closed normal subgroup MATH of codimension one with MATH. Let MATH be a REF-parameter subgroup transverse to MATH at the identity; then MATH. Because MATH is supersoluble, there is a codimension one subalgebra MATH containing the NAME algebra MATH of MATH. Because MATH is simply connected and solvable MATH is the NAME algebra of a closed subgroup MATH of dimension MATH, and MATH. By the induction hypothesis there exists MATH. Then MATH. Therefore MATH. We now have MATH, a contradiction. |
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 remains to show that if MATH, then MATH. Using induction on MATH and the following identity, MATH we conclude the theorem in the case when MATH is free as a MATH-module. In the case when MATH is not free as a MATH-module, we consider the localization of MATH with respect to a prime ideal MATH of MATH. By REF, MATH is an NAME MATH-algebra. Consider the injective (by flatness of MATH as a MATH-module) map MATH . By REF, MATH which by our discussion on the free NAME case is isomorphic to MATH. Thus, it is sufficient to show that the inclusion of REF MATH is surjective. The following lemma proves this which completes the theorem. |
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 show that MATH is in the image of MATH. Note, MATH for some MATH. Let MATH be a finite set of generators of MATH as a MATH-module. MATH . By induction hypothesis, for each MATH, MATH, there exists a MATH in MATH, such that MATH. Let, MATH. Then, MATH, for all MATH, MATH. Let MATH. For, MATH and MATH a generator, consider MATH. Now, MATH. Since MATH, MATH, which implies MATH. But, by induction hypothesis, MATH. Hence, MATH. Hence, for any MATH, MATH. Thus, MATH. Hence, MATH. This proves the lemma. |
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 . Hence the theorem. |
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 the variables MATH and the fact that MATH complete the claim. Let MATH be the MATH -subalgebra generated by MATH and MATH. Claim : MATH is a simple MATH-module. By assumption, the characteristic of the field is REF. Hence the claim follows from the fact that given a MATH, there exists a MATH such that MATH and MATH. Note that MATH. Now fix MATH. Then by the NAME Density Theorem, we can find MATH such that MATH. If MATH, then clearly, MATH. Now, by the first claim, MATH. Hence the proposition. |
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 by mapping MATH and MATH. Also, MATH by REF , page REF. Thus, we have a surjection MATH. Now use the fact that MATH is simple to complete the corollary. |
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 fact that MATH and the fact that MATH commutes with all the other generators, to assume that in the expression of MATH, the generators MATH and MATH do not appear. Similarly, as MATH, we can assume that MATH is a polynomial in MATH. Now use the fact that MATH to conclude that there is a non zero scalar in MATH and hence MATH. |
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. Similarily, MATH. To see this claim, it is enough to notice that MATH and since MATH is a decreasing sequence and MATH, we get MATH . A similar estimate can be established for MATH which verifies the claim. To complete the proof, let MATH, MATH. For MATH, we can write MATH as: MATH and using the claim above, MATH. A similar estimate would give MATH and combining these two estimates, we get MATH . This shows that MATH . The proof is complete. |
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. Choose MATH such that MATH. From REF , MATH . Reapplying REF , on MATH, MATH and MATH, one would get a sequence of projections MATH with MATH for every MATH, MATH (as n tends to infinity). As above, on can choose MATH such that MATH and MATH . The induction is clear, repeating the argument above would give a decreasing sequence of projections MATH so that for every MATH, MATH and MATH . If for every MATH, we set MATH then MATH belongs to MATH and MATH which shows that MATH and since MATH, the desired conclusion follows. |
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 construct sequences MATH in MATH and sequences MATH in MATH so that for every MATH, MATH . Let MATH be the diagonal sequence obtained from MATH. For every MATH, MATH is a subsequence of MATH so MATH and MATH . We note that MATH so for every MATH, MATH which implies that MATH . Taking the limit as MATH goes to MATH, MATH . The other inequality is trivial. To check that MATH, it is plain that MATH . The proof of the proposition is complete. |
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 of MATH ( which we will denote again by MATH) and MATH such that MATH . Choose a subsequence MATH so that MATH . Set MATH and MATH for all MATH. Claim: The set MATH is uniformly integrable in MATH. To see this claim, we will first prove the following intermediate lemma: Let MATH, MATH and MATH are uniformly integrable subsets of MATH. We will show that MATH is uniformly integrable. Assume the opposite. There exists MATH such that MATH . From this, there would exists MATH with MATH and such that MATH . In fact, for each MATH, if we denote by MATH the partial isometry in MATH so that MATH, then MATH . Note that for MATH, MATH so MATH and by REF , MATH which shows that MATH . Let MATH. It is clear that MATH and using REF , we conclude that there exists MATH , MATH such that MATH. In particular: MATH . Now choose a subsequence MATH so that there exists MATH satisfying MATH . Since MATH for all MATH, MATH for MATH and therefore MATH and taking the limit as MATH, REF imply MATH. This is a contradiction since MATH. The proof of the lemma is complete. To complete the proof of the theorem, assume that MATH is not uniformly integrable. Using REF , MATH . Again, choose a subsequence MATH so that MATH . Claim: MATH. To see this claim, we note that since MATH, MATH so MATH and therefore MATH and since MATH, MATH and the claim follows. From the claim above and REF , there exists MATH such that MATH . Observe that since MATH, MATH . Taking the limit ( as MATH) together with REF would imply MATH. This is a contradiction since MATH. By setting MATH and MATH, the proof for the MATH-finite case is complete. For the general case, let MATH be a NAME algebra (not necessarily MATH-finite) and MATH in MATH as in the theorem. Fix an orthogonal family of cyclic projections MATH in MATH such that MATH ( see for instance, CITE ) . There exists a countably decomposable projection MATH such that for all MATH, MATH. For each MATH and MATH, set MATH and MATH. Claim: MATH is finite (hence MATH is countable). To see this, assume that MATH is infinite. Then there exists an infinite sequence MATH such that MATH for all MATH. If MATH is a finite subset of MATH, then MATH . So MATH (a constant independent of MATH) which shows that MATH is a weakly unconditionally NAME (w.u.c.) series in MATH but since MATH does not contain any copies of MATH, MATH is unconditionally convergent and hence MATH see for instance REF. This is in contradiction with the assumption MATH for all MATH. We proved that MATH is finite. It is clear that MATH so it is at most countable. The claim is verified. Similarly, if MATH then MATH is at most countable. Let MATH ; MATH is at most countable and if MATH then MATH is the union of a countable family of disjoint cyclic projections in MATH so MATH is countably decomposable in MATH CITE. The construction of MATH implies that MATH for all MATH. The lemma is proved. To conclude the proof of the theorem, consider the NAME algebra MATH. Since MATH is countably decomposable, MATH is MATH-finite. Let MATH be the map that takes MATH to MATH. The map MATH is bounded and is weak-MATH to weak-MATH continuous so there exists a map MATH so that MATH. Let MATH be the restriction map. The operators MATH and MATH can be interpolated and since MATH (respectively, MATH) is isometrically isomorphic to MATH respectively, MATH for MATH, (see CITE), we get a bounded linear map MATH. Similarly, if one considers the inclusion map MATH and MATH as above, then by interpolation, we obtain a map MATH. Apply the MATH-finite case to the sequence MATH in MATH to get a decomposition MATH with MATH and MATH satisfying the conclusion of the theorem. It is enough to consider the decomposition: MATH . The proof is complete. |
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 to MATH and MATH. There exists a null sequence MATH of positive numbers such that: MATH for all MATH but since MATH, we get that MATH . This concludes that MATH is asymptotically isometric to MATH. CASE: The sequence MATH is equivalent to MATH. As above, one can find a block so that both the corresponding block for MATH and MATH are asymptotically isometric to MATH. Set MATH. It can be easily seen that MATH is equivalent to an asympotically isometric copy of MATH in MATH. |
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 integrable. The space MATH being reflexive implies that MATH is relatively weakly compact in MATH. We conclude the proof as in the scalar case. |
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 composition. If we identify MATH with both MATH (via MATH) and MATH (via MATH), then it is easily seen that the multiplication map MATH coincides with the evaluation map MATH and this proves the assertion. |
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 trivially on the right MATH. Also the action of MATH on MATH is on MATH the same as the natural (pullback) action. By chasing the diagram we see then that the vector bundle isomorphism above is equivariant with respect to the natural MATH action on both sides. Since we have an an induced isomorphism: MATH and MATH and MATH are both eigenbundles with respect to the trivial character, the lemma follows. |
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 simple. |
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 polystable. On the other hand, by REF the NAME transform of any line bundle is simple, so MATH must be stable. |
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 (compare REF) and so MATH where as usual MATH is the line bundle in Pic-MATH that corresponds to MATH and the last isomorphism follows from REF. Thus we get MATH and the idea is to compute this bundle in a different way, by using the behavior of the NAME transform under isogenies. More precisely, by applying REF we get the isomorphisms: MATH . The second isomorphism follows from REF, while the fourth follows from REF and the NAME formula at level MATH. The outcome is the isomorphism MATH . Now MATH is simple since MATH is simple, so the previous isomorphism implies the stronger fact that MATH . |
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 MATH which in particular implies (recall semistability) that MATH . The important fact that the last index of summation is MATH follows from the well-known ``symmetry" of the NAME formula, which is: MATH . |
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 . The desired isomorphism follows now from the see-saw principle(see for example, CITE, I. REF). |
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. The image of MATH in MATH is nonzero and this implies that MATH for MATH generic. |
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 change theorem, that MATH for all MATH and so by the projection formula we also get MATH for all MATH. The NAME spectral sequence then gives MATH and MATH for all MATH. The first isomorphism implies that there is an exact sequence: MATH . But MATH is globally generated, which means that MATH is surjective. This implies the vanishing of MATH, which by the second isomorphism is equivalent to MATH. |
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 this easily gives the desired vanishing for MATH. On the other hand from REF we know that MATH, which is clearly not globally generated. This shows that the bound is optimal. |
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. Restriction to the fiber gives the following long exact sequence on cohomology: MATH . As in the proof of REF, this sequence can be written in terms of the cohomology of MATH: MATH . The assumption on MATH ensures the fact that the map MATH in the sequence above is nonzero, and as a result MATH . On the other hand we use again REF that MATH is isomorphic to MATH. The additional hypothesis that MATH says then that MATH which is a contradiction. |
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 choice of a theta characteristic MATH is symmetric, this is the same as the global generation of the usual NAME product MATH. As an aside, recall from REF that this would imply the surjectivity of all the multiplication maps MATH for all MATH. The global generation of this NAME product is in turn another application of the general cohomological REF for vector bundles on abelian varieties. We first prove that MATH also has a nice decomposition when pulled back by an isogeny, namely this time by multiplication by MATH. Denote by MATH the NAME transform MATH, so that MATH. Then we have the following isomorphisms: MATH where the second one is obtained by the correspondence between the NAME product and the tensor product via the NAME transform, as in REF. Next, as in the previous sections, we look at the behaviour of our bundle when pulled back via certain isogenies (compare REF): MATH . In REF we proved that MATH and by plugging this into REF we obtain MATH . Finally we apply MATH to both sides of the isomorphism above and use the behavior of the NAME transform of a line bundle when pulled back via the isogeny that it determines (see REF). Since MATH is a direct summand in MATH, we obtain the desired decomposition: MATH . This allows us to apply a trick analogous to the one used in the proof of REF. Namely REF implies that MATH . Thus if we denote by MATH the vector bundle MATH we clearly have: MATH . NAME 's REF immediately gives then that MATH is globally generated for MATH, since MATH . CASE: We will show the vanishing of MATH. By the top sequence in the diagram preceding the theorem, it is enough to prove that MATH . First we prove the vanishing of MATH. The key point is to identify the pull-back of MATH by the étale cover MATH in the diagram MATH described in REF. In the pull-back sequence MATH we can identify MATH with MATH and MATH with MATH. In other words we have the exact sequence MATH which shows that the following isomorphism holds (compare REF): MATH . Finally we obtain the isomorphism MATH . Certainly by the argument mentioned earlier the surjectivity of the multiplication map MATH is also equivalent to MATH. The required vanishing is then an easy application of the NAME formula. The next step is to prove the vanishing of MATH. From the projection formula we know that MATH since obviously MATH for all MATH. The NAME spectral sequence reduces then our problem to proving the vanishing MATH, which is basically equivalent to the surjectivity of the multiplication map MATH . This is the content of REF . |
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 back by multiplication by MATH. Now: MATH hence the required vanishings are obvious as long as MATH and MATH. |
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. |