paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9905032 | We follow the argument in REF . Let MATH be an arbitrary function on MATH such that MATH for all but a finite number of MATH's. Form the probability generating functional: MATH . Then, viewing MATH as a diagonal matrix, we get MATH where the last equality is justified by REF applied to the operator MATH. Now, set MATH,... |
math/9905032 | Set MATH, MATH. By the inclusion-exclusion principle we have MATH . This alternating sum is easily seen to be identical to the expansion of MATH by linearity using MATH . |
math/9905032 | Without loss of generality one can assume that the block MATH is a nonnegative diagonal matrix, MATH. Write the blocks MATH and MATH as matrices, too, and let MATH and MATH be their diagonal entries. Since MATH, we have MATH and therefore MATH . |
math/9905032 | Clearly, REF implies REF . To check the converse claim, write MATH in the block form, MATH, where MATH is of finite size and MATH is small. Write all the MATH's in the block form with respect to the same decomposition of the NAME space, MATH. Since MATH weakly, we have convergence of finite blocks, MATH, which implies ... |
math/9905032 | The fact that MATH is holomorphic in MATH for any trace class operators MATH is proved in CITE. The continuity of the map follows from the inequality MATH which holds for any MATH, see CITE. |
math/9905034 | Let MATH be a local parameter on MATH near MATH, so that the sheaf MATH is locally generated by an element MATH which is well defined up to a MATH-th root of unity. Define the map MATH by MATH. It is easy to check that this definition is independent of local parameter, and hence defines a canonical isomorphism. |
math/9905034 | If MATH has no automorphisms, then each term MATH in the MATH-spin structure MATH is locally free (since MATH is smooth) of rank one and has automorphism group MATH, which acts on MATH by multiplication. However, an automorphism of the MATH-spin structure must also be compatible with the structure maps MATH. In particu... |
math/9905034 | In the case of MATH, smooth MATH-spin curves of type MATH and index MATH correspond to the torsion points of MATH of exact order MATH. It is well known that the involution MATH acts without fixed points on the points of exact order MATH, unless MATH is MATH or MATH, in which case the involution fixes all MATH-torsion p... |
math/9905034 | When MATH the bundle MATH is trivial because the sections are disjoint. In the case MATH the result follows from the fact that taking residues gives an isomorphism between MATH and MATH. |
math/9905034 | The map MATH pulls back, via MATH, to give MATH . Since MATH is disjoint from the nodes of MATH, MATH is an isomorphism, even on the boundary strata where MATH fails to be locally free. Consequently, we have MATH . |
math/9905034 | This is a direct consequence of REF and the previous theorem. |
math/9905034 | First, we claim that because MATH is a regular imbedding it has finite Tor dimension; that is, there is an integer MATH such that for every coherent MATH-module MATH, the MATH-modules MATH vanish for MATH. This can be seen as follows. We may assume that MATH is MATH and MATH is MATH for some regular element MATH in a r... |
math/9905034 | The degree of MATH is an integer and is equal to MATH. Thus MATH, and the degree of MATH is strictly negative. Therefore, when MATH is irreducible, MATH has no global sections. When MATH is not irreducible, but MATH is locally free REF at each node, the same argument holds. If MATH is NAME at some nodes, then normaliza... |
math/9905034 | It is clear from the construction of the classes MATH that they satisfy REF (convexity). REF follows from the fact that since MATH, the top-dimensional NAME class MATH of MATH, is simply the product of the top-dimensional classes of MATH and MATH. Now we will show that REF holds. If MATH is the MATH-th root from the un... |
math/9905034 | Let us show first that the conditions of the theorem are necessary. If MATH, then by REF , MATH must vanish; therefore, we can assume that MATH. In this case the dimension MATH (given in REF ) of the class MATH is equal to zero, and since MATH has two connected components MATH and MATH it will be sufficient to find MAT... |
math/9905034 | We only need to check the case MATH. Since the class has dimension MATH, REF gives MATH where MATH are the numbers of even/odd theta characteristics on a smooth curve of genus MATH and the last factor of MATH is the local (orbifold) degree of the map MATH near a generic point of MATH. |
math/9905034 | This follows from the definition of the potential, the dimensions of the cohomology classes, and the dimensions of the moduli spaces MATH. It encodes the fact that intersection numbers between cohomology classes vanish if the classes do not have proper the dimension. |
math/9905034 | Recall that on the moduli space of stable curves, MATH, the MATH classes (where MATH) satisfy the equation MATH, where MATH is the image of the MATH-th canonical section and MATH. Let MATH be the forgetful morphism MATH, then since the class MATH on MATH is the pullback via MATH of the MATH class on MATH, one can lift ... |
math/9905034 | The proof follows from the facts that MATH forms a NAME, and that MATH and MATH are lifts of the analogous classes on the moduli space of stable curves, and from REF . |
math/9905034 | On MATH, the class MATH can be written in terms of boundary classes as MATH . This equation is obtained from lifting the analogous relation on MATH. The classes MATH can be written similarly by applying an element of the permutation group MATH. The recursion relation follows from this presentation and the restriction p... |
math/9905034 | The proof of the first topological recursion relation arises from the relation on MATH which is obtained from lifting the analogous relation from MATH. The action of MATH yields MATH. This, combined with the restriction properties of the MATH classes, yields the desired result. The second comes from the presentation of... |
math/9905034 | CITE write down a formula (see REF for an explicit proof) for the large phase space potential MATH which in our case is MATH where MATH. The term MATH is equal to the small phase space potential MATH, evaluated at MATH for all MATH, but the small phase potential MATH vanishes. |
math/9905034 | MATH is nonempty if and only if MATH, where MATH and MATH for all MATH. The genus zero correlators are given by MATH where MATH is the (top) NAME class of degree MATH . The class MATH vanishes unless MATH for all MATH by REF . Furthermore, the correlator can only be nonzero if MATH. If MATH then the dimensionality cond... |
math/9905034 | The large phase space, genus zero potential MATH is completely determined by its values on the small phase space by the topological recursion relations. Let MATH denote the small phase space potential, which must satisfy the WDVV equation since MATH yields a NAME. Furthermore, the grading REF shows that the small phase... |
math/9905034 | REF shows that when MATH the class given by REF satisfies the axioms of a virtual class, and REF implies that the large phase space potential of the corresponding MATH-spin NAME is equal to the generating function of tautological intersection numbers on MATH (the large space potential of pure topological gravity). By N... |
math/9905034 | In this case, the conjecture means that the genus zero part MATH of the large phase space potential REF of the MATH-spin NAME REF coincides with the potential MATH of the semiclassical limit of the MATH hierarchy. In genus zero the virtual class MATH exists by REF . From REF it follows that the corresponding potential ... |
math/9905034 | By REF there is a unique formal power series MATH, of the proper grading, satisfying the equations of REF , WDVV, and the genus-zero topological recursion relations. CITE shows by a straightforward computation that any such power series yields a solution of the semiclassical limit of the MATH hierarchy. |
math/9905034 | The proof follows from REF and the fact that the potential of the NAME structure on the base of the versal deformation of the MATH singularity is equal to the potential MATH of the semiclassical limit of the MATH hierarchy (compare CITE). |
math/9905037 | See CITE, CITE. |
math/9905037 | CASE: Let MATH be associated to MATH, let MATH be the NAME morphism (relative to MATH) on the dual plane of MATH, and suppose that MATH is not defined over MATH. Then, since the envelope is defined over MATH and MATH is MATH-rational, MATH would belong to two different components of the envelope, namely MATH and MATH. ... |
math/9905037 | That a set of MATH lines no three concurrent satisfies the bound is trivial. Let MATH be the equation of MATH, let MATH be the factorization of MATH in MATH, and let MATH be the curve given by MATH. For simplicity we assume MATH even, say MATH. Setting MATH, MATH and MATH we have MATH. The singular points of MATH arise... |
math/9905037 | CASE: Let MATH be the equation of MATH over MATH. By REF , MATH admits a factorization in MATH of type MATH with MATH and MATH. Let MATH be the plane curve given by MATH . Then MATH satisfies the hypothesis of REF and it has even degree by REF . From REF and NAME 's theorem, for each line MATH (in the dual plane) corre... |
math/9905037 | CASE: This follows from the proof of REF . CASE: This follows from REF in the appendix. |
math/9905037 | Let MATH. Suppose that it is non-singular and an inflexion point of MATH. Then, from REF and the definition of MATH, we have that MATH is not the tangent line of MATH at MATH, that is, we have that MATH. Now suppose that MATH is either singular or a non-inflexion point of MATH. Then from REF we have MATH and the result... |
math/9905037 | CASE: Let MATH be an irreducible envelope of MATH and MATH the degree of MATH. If MATH, then MATH so that MATH and the result follows. So let MATH. Then from CITE we have that MATH is NAME classical and REF follows from REF . |
math/9905037 | CASE: As we mentioned in REF , MATH. Since MATH for MATH, REF follows from REF . CASE: If would exist a NAME classical irreducible envelope of MATH, then from REF we would have MATH so that MATH. |
math/9905037 | We can assume MATH. If MATH is not a power of MATH, by the MATH-adic criterion CITE we have MATH and MATH. |
math/9905037 | From the MATH-adic criterion CITE, MATH. Then from REF we have MATH and the result follows. |
math/9905037 | From REF and MATH follow that MATH and so that MATH. From MATH and MATH we have that MATH and it follows the assertion on MATH. The bound on MATH follows from REF . |
math/9905037 | Let MATH; so MATH as MATH. From REF we have that MATH so that MATH. Let MATH. If MATH (so MATH), then from REF and MATH we would have that MATH which is a contradiction for MATH. Let MATH. Then MATH and MATH and MATH. Thus from REF we have MATH, that is, MATH. This is a contradiction since by REF we must have MATH. Thi... |
math/9905037 | If MATH, then from MATH we would have that MATH and so, from REF , that MATH, a contradiction. |
math/9905039 | Indeed, REF shows that the subsheaf MATH of MATH of sections having a norm with moderate growth is a meromorphic bundle (compare CITE). This meromorphic bundle contains MATH because of REF . As both bundles MATH and MATH coincide on MATH, they are equal. If MATH is irreducible, the object MATH is stable: if MATH is a p... |
math/9905039 | Let MATH be the MATH complex associated with the connection MATH and the harmonic metric MATH given by REF (the complete metric on MATH being fixed as above). This complex is equal to the MATH complex associated with MATH and MATH, as MATH and MATH are mutually bounded, hence is isomorphic to MATH, according to REF . L... |
math/9905039 | Let MATH be such that MATH has a formal decomposition. Let MATH and MATH. As MATH, we have an isomorphism MATH. Let MATH be an elementary meromorphic connection in the MATH-variable equipped with a formal isomorphism MATH, given by NAME 's theorem. Put MATH . As MATH and MATH are elementary, this isomorphism comes from... |
math/9905039 | Recall that for MATH and MATH we have MATH . We will use the following identities: MATH . The matrix MATH of MATH in the basis MATH is MATH and the matrix of the metric connection MATH is MATH . The computation gives MATH . In the basis MATH, the curvature MATH has matrix MATH with MATH . The matrix of MATH in the orth... |
math/9905039 | Denote by MATH (respectively, MATH) the matrix of MATH (respectively, MATH) in the basis MATH. We have MATH . Put MATH and MATH, MATH. The matrix MATH of the operator MATH in the basis MATH is MATH. The pseudo-curvature MATH has matrix MATH . Put MATH and MATH. Then MATH . It is thus enough to prove the lemma when MATH... |
math/9905039 | Working in the basis MATH and denoting by MATH the matrix of MATH in this basis, we see that the matrix of the hermitian form MATH is equal to MATH, if MATH denotes the matrix of MATH. As MATH is orthogonal and commutes with MATH and MATH and as MATH, this matrix is equal to MATH. |
math/9905039 | Let MATH be the rank-one MATH-submodule of MATH which is equal to MATH on MATH and which is generated by the section MATH near MATH, for any MATH. The residue formula and the fact that MATH is an integer give MATH . Indeed, if MATH is trivial, this is the usual residue formula for meromorphic differential forms; if MAT... |
math/9905039 | Let MATH with MATH. Let MATH containing MATH. We assume that MATH contains at most one zero of MATH and, if it contains one, this zero belongs to MATH. Let MATH (respectively, MATH) be a one-form (respectively, two-form) on MATH such that MATH and MATH (respectively,). We will show that there exists a function MATH (re... |
math/9905039 | Let us fix a decomposition MATH lifting the formal decomposition, and let MATH be the first projection. One may write MATH where MATH is asymptotic to MATH, and MATH is asymptotic to MATH. Then MATH is equal to MATH, because MATH is a horizontal section of the meromorphic connection MATH; as MATH has a regular singular... |
math/9905041 | Let MATH be the subset MATH in MATH. NAME 's Theorem gives that MATH where MATH is the inward-pointing unit normal to MATH. But for large MATH we have MATH and MATH on MATH, so that the right-hand side of REF is MATH. Since MATH we see that the right-hand side of REF tends to zero as MATH, and this proves REF . The poi... |
math/9905041 | This is an analogue for MATH of REF . If MATH for MATH and MATH, then by CITE we have MATH where MATH is the volume of the unit sphere MATH in MATH. This is NAME 's representation for MATH. Let MATH be a radius function on MATH. Then MATH, so REF gives MATH . We split this into integrals over the three regions MATH, MA... |
math/9905041 | The theory of weighted NAME spaces on AE manifolds and the Laplacian is developed by CITE, who restrict their attention to the case MATH. In particular, they prove REF for the case MATH, MATH and MATH, CITE. Their proof uses a result equivalent to REF in the case MATH and MATH. By using REF together with the methods of... |
math/9905041 | Let MATH be the NAME form of MATH. Then if MATH is a smooth function on MATH we have MATH . Also, if MATH is a real MATH -form on MATH and MATH it can be shown that MATH where MATH is the NAME star and MATH the volume form of MATH. REF hold on any NAME manifold of dimension MATH. Define a function MATH on MATH by MATH.... |
math/9905041 | Let MATH be an ALE NAME metric on MATH, with NAME form MATH. By REF there exists a closed, compactly-supported, real MATH -form MATH on MATH with MATH in MATH. Define MATH. Then MATH is an exact real MATH -form on MATH. Now the NAME form of MATH on MATH is MATH. So from the definition of ALE NAME metric we see that MAT... |
math/9905041 | Let MATH in MATH. Then there are coordinates MATH on MATH in which MATH acts by MATH . As MATH acts freely on MATH we can take MATH for MATH. Since MATH we know that MATH preserves a complex symplectic form on MATH, and we can choose MATH so that this form is MATH. Thus REF gives MATH for MATH. But as MATH this implies... |
math/9905043 | If MATH is a generic point of MATH then MATH is locally isomorphic to MATH near MATH, and MATH is locally isomorphic to MATH near MATH, as MATH is a local product resolution. But if MATH then MATH, and MATH is a generic point of MATH, so that MATH is also locally isomorphic to MATH near MATH, and MATH is locally isomor... |
math/9905043 | Let MATH be fixed throughout the proof. Above we defined a lot of notation such as MATH and so on, associated to MATH. The corresponding data associated to MATH will be written MATH, etc., in the obvious way. We will express the data for MATH in terms of that for MATH. Define the indexing set MATH by MATH, and for each... |
math/9905043 | We shall use the notation defined in the proof of REF . In addition, let MATH be the Euclidean metric on MATH. Let MATH, so that MATH and MATH. Writing the Euclidean metric MATH on MATH as MATH, REF with MATH replaced by MATH becomes MATH on MATH. Using MATH to pull REF from MATH back to MATH, and substituting MATH sin... |
math/9905043 | Let MATH be smooth with MATH for MATH and MATH for MATH, where MATH is the constant in REF . For each MATH, define a smooth function MATH on MATH by MATH . The idea here is that MATH at distance at least MATH from the pull-back of MATH in MATH, that MATH at distance no more than MATH from the pull-back of MATH, and tha... |
math/9905043 | Let MATH be as in the proof of REF . Then MATH is positive outside a compact set in MATH, because at large distances from MATH we have MATH for some MATH, and so MATH is positive because MATH is positive. Let MATH, and choose a smooth function MATH such that MATH at distance less than MATH from MATH in MATH, and MATH a... |
math/9905043 | Let MATH, and suppose for simplicity that MATH. Then MATH, so by REF there exists a unique MATH such that MATH and MATH as MATH in MATH. Since MATH, we see that MATH on MATH. Suppose that MATH at some point of MATH. Then MATH is non-constant and has a maximum in MATH, since MATH and MATH as MATH in MATH. But this contr... |
math/9905043 | From REF we see that MATH lies in MATH, and hence in MATH. So there exists a unique MATH with MATH. Clearly MATH is smooth, MATH and MATH for some MATH. Also, using REF we can show that MATH for some small MATH, and thus by the proof of the previous theorem we have MATH. Thus MATH satisfies REF , as we want. |
math/9905043 | Suppose for a contradiction that MATH are distinct NAME QALE NAME metrics on MATH in the same NAME class, and let MATH be of the smallest dimension in which this can happen. Clearly MATH, since otherwise we can replace MATH by MATH, which has smaller dimension. Let MATH be asymptotic to metrics MATH on MATH for MATH, a... |
math/9905043 | Let MATH have NAME forms MATH. Then REF shows that MATH, where MATH is a unique function of NAME potential type on MATH. As MATH is the only NAME QALE metric in its NAME class by REF , we see that MATH, and MATH is asymptotic to MATH when MATH and MATH. Using these facts one can show that the MATH satisfy REF . Therefo... |
math/9905046 | CASE: The first part is proven in CITE. As for the second part, note the following equivalences, which hold for each MATH using REF MATH . CASE: Using MATH in REF we obtain especially that a MATH-submodule of the form MATH is a behavior. Thus, write MATH with some appropriate MATH. Then MATH is a behavior (see also CIT... |
math/9905046 | follows from REF , and REF . |
math/9905046 | CASE: By REF we have MATH. Without loss of generality we may assume that MATH is column-reduced, that is, MATH is the sum of the column degrees of MATH. From CITE we obtain matrices MATH such that MATH. Indeed, the parameter ord-MATH in CITE is equal to the degree, compare CITE. Setting MATH and using REF we obtain the... |
math/9905047 | The proof of this theorem follows from the definitions of mean curvature and NAME map. A complete proof can be found in CITE, and the original proof in CITE. |
math/9905047 | Because of the above observations we see that the condition MATH implies the existence of a MATH function MATH such that MATH and which makes the second variation of area REF negative. Namely MATH is the initial point of a REF-parameter family of minimal surfaces with boundary equal to MATH, such that each element in t... |
math/9905047 | The proof of the three assertions in the claim follows easily from the fact that the surfaces being considered are embedded and stable, and that MATH is the homology boundary of these surfaces. |
math/9905047 | First notice that the theorem certainly holds if MATH consists of the union of the two planar regions bounded by MATH and MATH. Hence we need to prove the theorem for connected MATH. The proof will be divided in five steps. CASE: For MATH sufficiently close to MATH, the components of MATH away from the crossing points ... |
math/9905047 | This follows easily from the description of MATH, and from the observation that a cell decomposition of MATH can be computed via a cell decomposition of the unique varifold MATH determined by MATH, and corresponding to MATH. Such a cell decomposition is equivalent to one that has: CASE: - number of vertices equal to MA... |
math/9905047 | Since it is not ambiguous, after an appropriate choice of MATH, we will drop the MATH indicating dependence on MATH, in this proof. Let us consider the foliation MATH, and consider the surface MATH which is obtained by deforming MATH via the NAME vector field MATH, namely the set of all the points MATH, as MATH varies ... |
math/9905047 | Let us consider MATH . Then for each MATH, if MATH is a NAME vector field, one has, as observed in paragraph REF, MATH . Let us choose MATH. Because of our hypotheses, MATH induces a MATH such that REF above is satisfied, namely MATH which implies MATH . Moreover REF above implies MATH which yields the conclusion: MATH... |
math/9905047 | Let MATH be the NAME vector field on MATH, and without loss of generality let us suppose that the maximum of MATH is attained at some point belonging to MATH (otherwise the proof of the lemma is still valid, as one can easily see: this hypothesis takes care of the ``worst possible case"). Moreover let MATH be a bump fu... |
math/9905047 | Suppose that the assertion stated in the theorem is false. Then there would exist a sequence MATH, corresponding to which there would be a sequence of unstable minimal surfaces MATH with boundary MATH, and described according to REF. Hence there would exist a sequence of NAME vector fields MATH defined on MATH, all hav... |
math/9905047 | The theorem will be proven by contradiction. Let MATH and MATH be two distinct families (for MATH sufficiently close to MATH) of minimal surfaces with the same boundary MATH, existing for all MATH, and having the same limit varifold MATH. Let MATH be the unbounded connected component of the region MATH of space given b... |
math/9905049 | The inverse images of MATH and MATH in MATH are represented by algebraic spaces (respectively, schemes), because MATH and MATH are representable (respectively, strongly representable). But these inverse images are open substacks which cover MATH. |
math/9905049 | Since MATH is finitely presented over the ground scheme, we may assume that MATH is obtained by base change from a stack of finite type over MATH. Hence to obtain a cover we may assume that MATH is of finite type over MATH. Also, since the morphism MATH is finite and surjective we can assume MATH is reduced. By working... |
math/9905049 | Let MATH be a point in MATH. Replacing MATH by an étale cover if necessary, we may assume the points of MATH all have residue field equal to the residue field of MATH. Then, for any MATH, MATH, the induced maps MATH are isomorphisms. Thus the composite MATH gives an automorphism of the completed local ring MATH. By ass... |
math/9905054 | Assume without loss of generality that MATH is connected. Since an open surface is a MATH - space, the inclusion MATH is homotopic to a point if and only if the homomorphism MATH is trivial CITE. But MATH is also an open surface. Representing MATH as the union of an increasing chain of compact surfaces with boundary we... |
math/9905054 | It suffices to show that there exist pairwise disjoint closed embedded discs MATH such that MATH is a component of MATH and MATH. Since MATH is compact it has only finitely many connected components which we denote by MATH; let MATH denote the boundary components of MATH. Now fix some MATH. Since MATH is open and MATH ... |
math/9905054 | Assume on the contrary that MATH is a regular value of MATH. Then there exists a segment MATH which consists of regular values of MATH and such that MATH. By definition of MATH, the set MATH contains a non - contractible circle. Since the gradient flow of MATH takes MATH into MATH we conclude that MATH contains a non -... |
math/9905054 | It suffices to show that MATH contains a non - contractible curve from MATH. Write MATH, where MATH are compact surfaces with boundary such that supp-MATH and MATH for all MATH. If some boundary component of some MATH is non - contractible in MATH we are done. Assume therefore that all of them are contractible. Then RE... |
math/9905054 | We are going to apply REF with MATH. If MATH is not contractible in MATH then it contains a curve from MATH. This curve lies either in MATH, which implies MATH, or in MATH, in which case MATH. Suppose now that MATH is contractible in MATH. Then MATH cannot contain a curve from MATH. This means that every curve from MAT... |
math/9905054 | When MATH this is proved in CITE. The case of a general open surface of infinite area can be reduced to this one as follows. Assume without loss of generality that MATH consists of just one disc of area MATH. Let MATH be the closed standard disc of area MATH. Since MATH has infinite area, it is an easy consequence of t... |
math/9905054 | Let us decompose the flow MATH into two commuting flows as follows. Fix any MATH, and choose a smooth function MATH satisfying the following properties: CASE: MATH if MATH CASE: MATH if MATH CASE: MATH if MATH CASE: MATH if MATH or MATH . Define the new Hamiltonians MATH and MATH, and denote their flows by MATH and MAT... |
math/9905054 | The first equalities are proved in CITE, and the second ones in CITE. |
math/9905054 | Since MATH, it suffices to show that MATH . Fix an arbitrary MATH. Choose MATH to be a non - contractible circle on MATH such that MATH. REF imply that MATH. Analogously, MATH. Thus MATH, for every MATH. Thus we get the desired inequality. |
math/9905054 | The proof goes along the lines of CITE, and we only give a sketch here. The argument is devided into three steps. CASE: Choose a compact connected submanifold with boundary MATH whose interior contains both MATH and MATH. Let us perform the following surgery on MATH. We remove the complement to MATH and attach to each ... |
math/9905058 | Let MATH be open and MATH vanish on MATH. We have to show that MATH . Let MATH be any point in MATH. As well-known, there exists a neighborhood MATH of MATH and vector fields MATH, MATH on MATH such that MATH and MATH where MATH depends only on the dimension of MATH. One has, using the fact that MATH is a REF-cocycle M... |
math/9905058 | Any differential operator on MATH is indeed determined by its values on the subspace MATH. |
math/9905058 | : straightforward. |
math/9905058 | Check that for MATH, REF coincides with REF , the result follows then by induction. |
math/9905058 | This follows immediately from REF and the fact that the sequence REF is split when restricted to MATH, see CITE. Let us also give an elementary proof. Recall that a REF-cocycle on MATH with values in MATH is a coboundary if it is of the form MATH for some MATH. Moreover, REF-cocycle vanishes on MATH if and only if MATH... |
math/9905058 | Since the sequence REF is split, one has MATH. If in addition MATH, then by REF MATH and the sequence REF is split. |
math/9905061 | The proof is direct and is left to the reader. |
math/9905061 | Fix MATH as in the hypothesis of the theorem, and let MATH be the set of branches associated with MATH. Let MATH. Clearly MATH is a theory in MATH that satisfies REF 's Compactness Theorem. It follows from REF that there exists a MATH-saturated structure MATH such that: MATH . By the definition of MATH in MATH it follo... |
math/9905061 | Assume, in order to get a contradiction, that there exists MATH such that for every integer MATH there is a normed structure MATH satisfying: MATH . We can now invoke REF to obtain that there exists a normed structure MATH such that MATH and MATH, but this is a contradiction with the hypothesis. |
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