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math/0002203 | Let MATH denote the curvature form of the NAME metric MATH on MATH . Then by REF, we have MATH where MATH is the complex conjugation of MATH with respect to MATH . MATH restricts to a metric MATH and induces a MATH-decomposition MATH . One obtains MATH where MATH is the second fundamental form of the sub-bundle MATH an... |
math/0002203 | Let MATH be the dual of MATH. We have the projection MATH . Note that MATH as a NAME bundle of the system of NAME bundles corresponding to the dual variation of NAME structures MATH. The monodromy of MATH around MATH is again unipotent. We have the projection MATH where MATH is the q-th subbundle in the extended NAME f... |
math/0002203 | If the fibres of MATH are connected and if MATH is semistable, this is nothing but REF. In general, let MATH be a finite extension of the function field MATH, containing the NAME hull of the algebraic closure of MATH in MATH, and let MATH be the normalization of MATH in MATH. Consider the normalization MATH of MATH, a ... |
math/0002203 | Let MATH and MATH be as in REF, and let MATH be a component of the normalization of MATH. If MATH are the induced morphisms, then MATH . This, for all MATH, implies that MATH. |
math/0002203 | Let MATH be a point. Then MATH is ample, hence MATH is semi-ample. By CITE, the sheaf MATH is nef. Since the same holds true over all MATH, finite over MATH and unramified in MATH, one obtains REF from REF . |
math/0002203 | If MATH, the sheaf MATH is trivial, hence MATH and REF obviously holds true. Hence we will assume MATH. For some MATH there exists an effective divisor MATH, disjoint from MATH with MATH. By REF and by flat base change, we are free to replace MATH by any MATH, finite over MATH and unramified over a neighborhood of MATH... |
math/0002203 | It remains to verify REF. Comparing the first NAME classes of the sheaves in REF one finds MATH. Hence for some invertible sheaf MATH of degree MATH one has an inclusion MATH . Recall that MATH, and that MATH where MATH is the relative fix locus of MATH. In particular, if MATH denotes the effective divisor with MATH, t... |
math/0002203 | To handle both cases at once, define MATH, if MATH and MATH, otherwise. Assume that MATH . Then by REF, MATH is an invertible subsheaf of MATH of degree MATH in particular it is ample. For MATH, we will construct by induction an invertible subsheaf MATH of MATH of degree MATH . If MATH, the sheaf MATH is ample, and by ... |
math/0002203 | The sheaf MATH is nef, hence if REF does not hold true, it is ample. Since MATH is semistable, MATH is compatible with further pullbacks. Replacing MATH by a covering, we may assume that MATH. By REF MATH hence MATH is ample, for a point MATH. REF implies that MATH, obviously a contradiction. |
math/0002203 | Choose an invertible sheaf MATH on MATH of degree MATH. If MATH one finds MATH. By REF MATH hence MATH is ample. REF implies that MATH contradicting the choice of MATH. |
math/0002205 | An abelian variety is simple if and only if its endomorphism ring contains no zero-divisors. Thus, if MATH is simple and MATH is not, there must exist an element of MATH that does not come from MATH. But it follows from the NAME theorem CITE that MATH if MATH. This proves the first statement of the lemma. If MATH is or... |
math/0002205 | Let MATH be a primitive MATH-th root of unity in an algebraic closure of MATH and let MATH. Note that MATH contains MATH. Since MATH is a NAME extension of MATH it is also a NAME extension of MATH, and it follows that MATH and MATH are linearly disjoint over MATH, so that MATH. Let MATH. Since MATH is a NAME extension ... |
math/0002205 | If MATH is an integer in MATH then clearly MATH is a proper subfield of MATH. On the other hand, if there exists some MATH such that MATH then there exists a smallest such MATH, and by REF this MATH lies in MATH. This proves the first statement of the proposition. It is clear that MATH is absolutely simple if and only ... |
math/0002205 | Let MATH be the NAME endomorphism of MATH and let MATH be the field MATH. Because MATH is ordinary, the field MATH is a NAME of degree MATH over MATH. The ordinariness of MATH also implies that MATH is a NAME for every positive integer MATH, and that MATH splits over the degree-MATH extension of MATH if and only if MAT... |
math/0002205 | Let MATH be the polynomial whose existence is guaranteed by REF . Then MATH satisfies the last four of the five hypotheses of REF ; we will show that it satisfies the first hypothesis as well. First we will consider the case in which MATH. Since MATH we find that the polynomial MATH defined in REF may be written in the... |
math/0002205 | Suppose, to obtain a contradiction, that MATH corresponds to an isogeny class that is not absolutely simple. Then by REF there is a positive integer MATH such that MATH is a proper subfield of MATH. Let MATH be the smallest positive integer with this property. Since MATH is ordinary, the field MATH is a NAME, and its m... |
math/0002205 | Since MATH modulo MATH is irreducible, MATH itself is irreducible in MATH, and since all of its complex roots are real, MATH defines a totally real number field MATH. Let MATH be a root of MATH in MATH. The discriminant of the polynomial MATH is totally negative because the roots of MATH all have magnitude less than MA... |
math/0002205 | This follows from the techniques of CITE. |
math/0002205 | If MATH we can simply choose the appropriate value of MATH from REF , so let us assume that MATH. REF (below) shows that there exist monic degree-MATH polynomials MATH in MATH and MATH in MATH such that MATH is irreducible, such that MATH is a linear times an irreducible and has nonzero constant term, and such that the... |
math/0002205 | For MATH we choose MATH and MATH from REF . For MATH we argue as follows: REF (p. REF) of CITE shows that there exists a monic irreducible polynomial in MATH of degree MATH with zeroes for the first six coefficients after the leading MATH. We take this polynomial for our MATH. The same corollary shows that there is a m... |
math/0002205 | Let MATH be the arithmetic function defined by MATH, where MATH is NAME 's MATH-function, let MATH be the number of isogeny classes of abelian surfaces over MATH, let MATH be the number of isogeny classes of simple ordinary abelian surfaces, and let MATH be the number of isogeny classes of absolutely simple ordinary ab... |
math/0002205 | Combining the estimates for MATH and MATH given in REF, we find that the quantity MATH is less than or equal to MATH so MATH . An easy induction shows that MATH, and certainly MATH, so we have MATH . Since MATH we have MATH, and combining this with the fact that MATH we find that MATH . But MATH is greater than MATH fo... |
math/0002205 | Let MATH and let MATH denote the lattice MATH. If MATH is a point in MATH let MATH denote the ``brick" MATH . Let MATH denote the set of all MATH such that MATH. The proof of REF (see especially p. REF) shows that MATH where MATH is the mesh of MATH (see p. REF), which is MATH. Since the covolume of MATH is MATH, we fi... |
math/0002205 | The lemma follows easily from the well-known exact formula MATH where MATH is the NAME function. |
math/0002216 | Suppose MATH and MATH non-contracting. Let MATH and MATH be two morphisms of strictly positive dimension and MATH. Then MATH therefore MATH cannot be MATH-dimensional. If MATH then MATH if MATH and if MATH for two different reasons. Therefore MATH cannot be MATH-dimensional as soon as MATH. |
math/0002216 | The characterization of REF gives the solution. |
math/0002216 | Let MATH and let MATH be the property : ``for any MATH-category MATH and any MATH, then MATH." One has MATH, MATH and necessarily MATH by definition of a morphism of cuts. Therefore MATH holds. Now suppose MATH proved for some MATH. One has MATH since MATH is a morphism of cuts and MATH . Therefore the natural transfor... |
math/0002216 | The natural map MATH for MATH from MATH to MATH is of the form MATH for MATH with MATH for MATH and MATH. Therefore for MATH, MATH by REF . And by construction of MATH, one obtains MATH. |
math/0002216 | Consider the MATH of CITE. One has MATH where, for the last formula, MATH are other operators which is not important to explicitly define here : the only important thing is that MATH remains MATH-dimensional if the argument is MATH-dimensional. Hence REF . As for REF , it is enough to check it for MATH. And in this cas... |
math/0002216 | Using the freeness of MATH, the construction in the proof of REF yields the MATH-functor MATH from MATH to MATH. The hypotheses stated in CITE were too strong indeed. If moreover the shell is fillable in the above sense, one concludes still as in the proof of REF . |
math/0002216 | One proves by induction on MATH the following property MATH : `` For any MATH-simplex MATH of the globular simplicial nerve of any MATH-category MATH, the map MATH from MATH to MATH induces a MATH-functor and moreover an element of MATH." Let MATH be a MATH-simplex of the globular nerve of MATH. Then MATH is a MATH-fun... |
math/0002216 | Let MATH and MATH be two morphisms of cuts from MATH to MATH. One proves by induction on MATH that MATH and MATH from MATH to MATH coincide. For MATH, MATH. The only natural transformation from MATH to itself is MATH, therefore MATH. Suppose MATH proved for some MATH. Then for any MATH, and for any MATH, MATH . Now wit... |
math/0002216 | The natural transformation MATH from MATH to MATH can be lifted to a natural transformation MATH from MATH to MATH since the cut MATH is regular. Since MATH and since MATH is one-to-one in positive degree, there is at most one solution for this lifting problem. MATH . Let MATH. For MATH, the natural transformation MATH... |
math/0002216 | It is a consequence of the naturality of MATH and MATH and of REF . |
math/0002216 | The equality MATH implies MATH, implies by REF that for MATH, MATH therefore MATH. Conversely, if for MATH, MATH, then MATH and therefore MATH. |
math/0002216 | Consider the MATH operators of REF . If MATH, then MATH is negative. So MATH is also negative by REF and determines a unique element MATH such that MATH. It is clear that these operators MATH induces the identity map on the reduced globular complex by REF . Since MATH is also negative, then by REF , MATH for some seque... |
math/0002216 | One has MATH with MATH a thin MATH-cube and MATH a thin MATH-cube. The proof made in CITE shows that MATH and MATH are in the image of MATH. Indeed, the existence of MATH and MATH comes from the vanishing of some globular nerve. Therefore MATH and MATH where MATH is a thin MATH-simplex and MATH a thin MATH-simplex. Thi... |
math/0002216 | For MATH, it is obvious. For MATH, one has MATH where MATH means the simplicial homology of the simplicial nerve of the MATH-category MATH. |
math/0002216 | The only non-trivial part is the first assertion. Let MATH be the property : ``for any non-contracting MATH-category MATH and any MATH-functor MATH from MATH to MATH, the set map MATH from the set of faces of MATH to MATH is constant." There is nothing to check for MATH. For MATH, if MATH is a MATH-functor from MATH to... |
math/0002216 | One has MATH by REF . So it suffices to prove the vanishing of MATH as soon as MATH contains morphisms in strictly positive dimension to prove the theorem. Let MATH and MATH be two MATH-morphisms of MATH such that MATH contains other morphisms than MATH and MATH. Then in particular it contains some MATH-morphisms from ... |
math/0002216 | By proceeding as in REF , we see that it suffices to prove that MATH for any pair MATH of MATH-morphisms of MATH and for MATH. However, MATH is non-empty if and only if MATH with our conventions and in this case, MATH . Therefore MATH by REF . |
math/0002216 | This is due to the fact that for MATH, the natural map MATH is induced by the set map MATH from MATH to MATH REF . |
math/0002216 | We have to prove that REF implies REF . Let us consider MATH : ``for any non-contracting MATH-category MATH and any MATH-functor MATH and MATH from MATH to MATH such that MATH and such that the image of MATH is a subset of MATH, then MATH yields a MATH-functor from MATH to MATH and MATH for any MATH." Property MATH is ... |
math/0002216 | Equalities MATH, MATH, MATH, MATH are consequences of REF . Equality MATH is proved right above. The verification of MATH is straightforward. |
math/0002219 | Let MATH be a countable dense set in MATH. Using REF we note that for any MATH and any MATH it follows that MATH . Assuming now that for all MATH the condition MATH is satisfied we can choose a countable subset of MATH, MATH and observe that MATH where MATH is defined just like MATH with the difference that the family ... |
math/0002219 | We consider the following three cases. If MATH is a reflexive space we can choose according to REF a reflexive space MATH with an FDD MATH which contains MATH. If the dual MATH is separable we can use again a result in REF and choose a space MATH having a shrinking FDD MATH. In the general case we choose MATH to be a M... |
math/0002219 | Let MATH be a subspace of dimension MATH, admitting a continuous projection MATH, so that MATH. Let MATH. By assumption we find a sequence MATH converging to MATH. Let MATH be the (finite dimensional) vector space generated by MATH and choose a basis of MATH of the form MATH. We represent each vector MATH as MATH and p... |
math/0002219 | We first choose a decreasing sequence of finite codimensional spaces MATH in MATH so that for each MATH the equivalences MATH , and, if MATH is separable, MATH , of REF hold. Then we choose the space with an FDD MATH as in REF . We note that trivially REF of REF implies REF . Since the conclusion of REF implies that ev... |
math/0002219 | We put MATH . It is easy to see that MATH is closed in the pointwise topology on MATH, since MATH is closed with respect to the product of the discrete topology on MATH. By the infinite version of NAME 's theorem (compare REF ) we deduce that one of the following two cases occurs. MATH . If the first alternative occurs... |
math/0002219 | Fix MATH. We define MATH where each MATH is given as follows. MATH will be the completion of MATH under the norm MATH . By a segment we mean a sequence MATH with MATH, MATH, for some MATH. Thus a segement can be seen as an interval of a branch (with respect to the usual partial order in MATH), while a branch is a maxim... |
math/0002219 | We choose an appropriate sequence MATH depending upon MATH and the basis constant MATH of MATH. MATH is chosen by the lemma for MATH and MATH. We choose MATH by the lemma for MATH and MATH and so on. If MATH the lemma yields for MATH, MATH with MATH and MATH with MATH. We then let MATH and for MATH, MATH. Thus MATH and... |
math/0002219 | We first show that MATH embeds into MATH for some sequence MATH of finite dimensional spaces. Then to obtain the MATH estimate we adapt an averaging argument similar to the one of CITE. Applying REF to the set MATH we find a reflexive space MATH with an FDD MATH with basis constant MATH which isometrically contains MAT... |
math/0002219 | The implications REF follow from REF and its proof. If we prove REF then REF will follow. Assume that MATH, MATH, MATH and MATH are given as in the statement of REF . By passing to the renorming MATH, MATH, for MATH we can assume without loss of generality that MATH is isometric to a subspace of MATH. We will show that... |
math/0002219 | Let MATH denote the set of all closed subsets of MATH for which MATH holds. For MATH we denote the MATH-neighborhood by MATH and observe the following equivalences MATH . Thus MATH. If MATH then MATH is compact and is contained in the open covering MATH. Thus there exists a finite MATH so that MATH and thus MATH which ... |
math/0002219 | Since there exists MATH so that the unit vector basis of MATH is in MATH (see CITE) we have that MATH must be this unit vector basis. In turn this condition (see CITE or CITE) implies that MATH contains an isomorph of MATH (MATH if MATH) and so MATH. Let MATH, a reflexive space with an FDD MATH. The condition on MATH y... |
math/0002221 | Let MATH be the least integer such that MATH. For MATH we have MATH. Then, MATH . |
math/0002221 | We will show that MATH is of weak type MATH. By similar arguments, one gets that MATH is bounded from MATH into MATH. In this case, one has to use a version of the NAME decomposition in the lemma above suitable for complex measures. For simplicity we assume MATH. Let MATH and MATH. Let MATH be the almost disjoint famil... |
math/0002224 | As MATH preserves MATH and MATH, it (and all its multiples by a constant) preserves the Riemannian metric defined by MATH (we replace, if necessary, MATH with MATH, MATH with MATH, such that MATH is positive definite) with respect to which MATH is Killing and MATH. MATH is then identified to a MATH-invariant, anti-symm... |
math/0002224 | We denote by MATH, respectively, MATH, the NAME connection, respectively, the NAME REF connection associated to the positive MATH . NAME vector field MATH. We have MATH . The claimed result now follows from a straightforward computation. |
math/0002224 | We have to prove that, if MATH is the NAME curvature of the NAME surface MATH, then it satisfies MATH iff MATH is constant (we omit the indices referring to MATH, as we only use the metric, the NAME connection and the operator MATH on MATH in this proof). First we prove the following fact: if MATH, then MATH . We need ... |
math/0002224 | The almost-complex structure MATH on MATH is defined as follows: MATH, and MATH, MATH being the unitary, oriented, generator of MATH. The product metric is, then, given by its NAME form MATH . It is easy to prove that MATH is integrable and that MATH such that the NAME form of the Hermitian metric MATH, is, by definiti... |
math/0002224 | Consider a NAME structure on MATH, with the usual notations. see REF We suppose, with no loss of generality, that MATH is primary, thus MATH is a MATH-bundle over a complex curve MATH, such that the NAME vector field MATH is tangent to the fibers, which all are of equal length MATH (and we can suppose MATH). We want to... |
math/0002225 | The claimed identities are conformally invariant : for REF it is obvious, and the conformal invariance of REF follows from REF ; to see that for REF , let MATH be a MATH-orthonormal oriented basis of MATH, such that MATH is a MATH-orthonormal basis on MATH giving the orientation as above. Then MATH, and, if we take MAT... |
math/0002225 | The proof is different in the cases MATH and MATH; one of the reasons is that conformal flatness reduces, in higher dimensions, to the vanishing of the NAME tensor, while in dimension REF it is a higher-order condition more difficult to handle. The first step (common to all cases) is to prove that a small deformation (... |
math/0002226 | The proof is trivial. |
math/0002226 | It is follows from the obvious relations: MATH . |
math/0002226 | On account of REF we rewrite the equation MATH, as an operator equality in the space MATH: MATH . Substituting REF we get MATH . It is not hard to prove that MATH . Then from REF we have MATH . Substituting REF we get REF . |
math/0002226 | Using REF we rewrite the differential REF in the form of the operator equality in the space MATH: MATH . In view of REF is equivalent to the following operator equality in the space MATH: MATH . It is obvious from REF that it is sufficient to prove that the right-hand side of REF vanishes. From REF it follows that the ... |
math/0002234 | Let MATH be an isometric dilation of MATH, and REF hold. Suppose that MATH is a subspace of MATH, and MATH, is an isometric dilation of MATH. Since for any MATH and MATH we have MATH, and MATH is invariant under MATH. Therefore, MATH and MATH. Thus, MATH is a minimal isometric dilation of MATH. For the rest of this Pro... |
math/0002234 | Let MATH be a unitary dilation of a linear pencil MATH, and REF hold. Suppose that MATH is a subspace of MATH, and MATH, is a unitary dilation of a linear pencil MATH. In the same way as in REF we show that MATH is invariant under MATH and MATH for all MATH. Therefore, MATH and MATH. Thus, MATH is a minimal unitary dil... |
math/0002234 | By REF the equality in REF is true. Since MATH is a minimal uniform unitary dilation of MATH (see our remark in the beginning of this Section), by REF we have REF. Since REF holds, we get MATH and by REF MATH is a minimal unitary dilation of a pencil MATH (of course, a dilation of a dilation of MATH is again a dilation... |
math/0002235 | By NAME 's approximation theorem CITE, one can find a convergent power series mapping MATH such that MATH and for all MATH, MATH in MATH. Let MATH so that if MATH then MATH. Since the radius of convergence of the MATH, MATH, is at least MATH, the radius of convergence of any MATH, MATH, is at least MATH. Thus, the fami... |
math/0002235 | We reproduce here the arguments of CITE. Write MATH where MATH is a MATH complex-analytic matrix such that MATH; that is, MATH. By assumption, we know that we have MATH. This implies that one can find an integer MATH such that if MATH is any formal power series which agrees up to order MATH with MATH then MATH. For thi... |
math/0002235 | We write the expansion MATH . For the sake of clarity, we shall first give the proof of the Proposition in the case MATH. The case MATH. We restrict all the identities given by REF to the MATH-dimensional subspace MATH . This gives, for any multiindex MATH, MATH where MATH is given by REF . Observe that for each multii... |
math/0002235 | First observe that MATH is a formal power series solution of the analytic system in the unknown MATH, MATH . Thus, an application of NAME 's approximation theorem CITE gives, for any positive integer MATH, a MATH-tuple of convergent power series MATH solution in MATH of REF , and which agrees up to order MATH with MATH... |
math/0002235 | Restricting the identity REF to the set MATH where MATH, MATH, are the NAME sets mappings defined by REF , we obtain MATH . Here, MATH is the formal power series defined by REF . We would like to apply REF to the formal REF with MATH, MATH, MATH, MATH, MATH, MATH, MATH and MATH . For this, one has to check that any der... |
math/0002235 | By the NAME expansion REF and by REF , we obtain that all the MATH are convergent in a common neighborhood MATH of MATH. Since MATH is holomorphically nondegenerate, by CITE, there exists MATH, MATH, MATH, such that MATH . Since MATH is a formal biholomorphism, this implies that MATH as a formal power series in MATH. P... |
math/0002235 | We again use the notations of REF. As in the proof of REF , we have, using the expansion REF , MATH . Thus, we know, by REF , that for any multi-index MATH, MATH is convergent in some neighborhood MATH of MATH in MATH. We choose MATH, MATH, of generic maximal rank equal to MATH in a neighborhood MATH of MATH in MATH (s... |
math/0002235 | The implication MATH is clear. The other implication is equivalent to the following proposition. |
math/0002235 | Let MATH be as in the proposition. We can assume that, near MATH, MATH where each MATH. To this complex analytic set MATH, as is customary, we associate the following ideal of MATH defined by MATH . By the NAME property, we can assume that MATH is generated by a family MATH. Furthermore, by the CITE, for MATH, there ex... |
math/0002235 | By REF , there exists a convergent power series mapping MATH such that for each MATH, MATH is convergent, MATH. Then, the graph of MATH is formally contained in the complex analytic set MATH . Let MATH be the decomposition of MATH into irreducible components. For any positive integer MATH, one can find, according to th... |
math/0002237 | It is clear that MATH is a lattice and that MATH is convex in MATH. We now check that MATH is complete. Let MATH. We will show that MATH exists. NAME we may assume that MATH, MATH, MATH. Let MATH. We may also assume that MATH is not the least upper bound of MATH, so let MATH be some upper bound. Fix MATH such that MATH... |
math/0002237 | See CITE and CITE. (The ``moreover" part is not stated there, but following the proof it is easy to see that the lattice MATH constructed in CITE satisfies MATH and will be complete.) |
math/0002237 | We let MATH, MATH, and MATH as above. Let MATH for MATH and MATH for MATH. It remains to show that the map MATH is an orthocomplement for MATH. Clearly this map is well defined and an involution, and it agrees with the map MATH on MATH. Also, we have MATH for all MATH. Now let MATH, without loss of generality MATH. We ... |
math/0002237 | Since every ortholattice can be embedded into a complete ortholattice (the NAME completion; see, for example, CITE, CITE) we may assume that MATH is complete. Let MATH. Let MATH be horizontal sum of MATH and MATH, that is, assume that MATH and MATH have the same least and greatest elements (but are otherwise disjoint),... |
math/0002237 | Choose a cardinal MATH of cofinality MATH such that there is a transfinite enumeration (not necesasrily REF) MATH of all functions MATH. Define an increasing transfinite sequence MATH of MATH-lattices satisfying CASE: MATH. CASE: If MATH, then MATH CASE: For every MATH there is a lattice polynomial MATH with coefficien... |
math/0002237 | Let MATH, MATH small. Let MATH be the sub(ortho)lattice generated by MATH. Since MATH is MATH-power closed, we can find a sublattice MATH and an isomorphism MATH. For MATH let MATH be defined by MATH . Note that MATH . Define a partial function MATH as follows: MATH and MATH undefined if MATH. Note that MATH is monoton... |
math/0002241 | Let MATH denote the set of all countable subsets of the indexing set MATH. Consider the following relation MATH . We need to verify the following three properties of the above defined relation. Existence. If MATH, then there exists MATH such that MATH. Proof. First of all let us make the following observation. Claim. F... |
math/0002241 | Let MATH be a complemented subspace of the sum MATH. Choose a continuous linear map MATH such that MATH for each MATH. Let us agree that a subset MATH is called MATH-admissible if MATH. For a subset MATH, let MATH. CASE: If MATH is a MATH-admissible, then MATH. Proof. Indeed, if MATH, then there exists a point MATH suc... |
math/0002241 | For countable MATH results follows from CITE and CITE. Let now MATH is uncountable and MATH be a complemented subspace of a locally convex direct sum MATH. By REF , MATH is isomorphic to a locally convex direct sum MATH, where MATH is a complemented subspace of the countable sum MATH where MATH for each MATH. According... |
math/0002242 | Let MATH denote the set of all countable subsets of the indexing set MATH. Consider the following relation MATH where MATH and MATH denote canonical projections onto the corresponding subproducts. We need to verify the following three properties of the above defined relation. Existence. If MATH, then there exists MATH ... |
math/0002242 | Let us first of all set up a notation. For a subset MATH, where MATH is an indexing set with MATH, let MATH . Let also for MATH and MATH denote canonical projections onto the corresponding subproducts. Let MATH be a complemented subspace of the product MATH. Choose a continuous homomorphism MATH such that MATH for each... |
math/0002242 | MATH. By CITE and CITE, MATH is an injective object of the category MATH for any set MATH. Obviously (see, for instance, CITE) product of an arbitrary collection of injective objects of the category MATH is also an injective object of this category. Consequently, the NAME space MATH, MATH, as a complemented subspace of... |
math/0002243 | The first thing to notice is that MATH, where MATH is a co-closed MATH-form. Since MATH and MATH are solutions to the NAME we have MATH . Therefore MATH . This last statement and the fact MATH implies that MATH . Since MATH are solutions to the NAME equations we have MATH multiplying by MATH both sides of the equality ... |
math/0002243 | This is a consequence of REF , the expression for MATH (see REF ) and that MATH, where MATH and MATH are the NAME operators of MATH and MATH, respectively. |
math/0002243 | The proof of this proposition can be carry out following the steps in the proof of REF pg. REF, replacing the expression for MATH with MATH. |
math/0002243 | The fact that MATH satisfies the NAME follows from REF . We just need to show that MATH. In order to do this, we will use the metric MATH as the background metric. MATH . To prove that MATH we need to recall that MATH where MATH. The computation is very similar to the one above. |
math/0002243 | Let MATH. The existence of such metric is equivalent to finding a neighborhood MATH of MATH, and a NAME form MATH in the same NAME class of MATH, such that MATH. It is well known that there exist a MATH-neighborhood MATH of MATH and a function MATH such that MATH, where MATH and MATH denotes the distance (using the NAM... |
math/0002243 | It is easy to see that MATH, as it is to see that MATH is a solution of REF with function MATH replacing MATH. The first equation in REF tell us that MATH is a holomorphic section on MATH. Using NAME 's Theorem we can extend this to a holomorphic section on MATH. All the analysis done in proving REF can be carry out if... |
math/0002243 | Since all the analysis done in proving REF can be carry out if we replace the strictly positive function MATH in REF by a non-negative function MATH whose zero set has measure zero, existence is a consequence of REF and uniqueness is obtained using REF . |
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