paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0009124 | Since MATH is integrable, MATH is NAME equivalent to MATH, CITE. |
math/0009124 | Let MATH be the natural projection MATH. It suffices to show that MATH is isomorphic to MATH. In the notation of the proof of REF , the isomorphism is MATH. Here MATH is the MATH-component of MATH, the form MATH is the symplectic form on MATH, and MATH is extended to a genuine differential form on MATH using the direct... |
math/0009124 | Weak NAME equivalence is just morphism in the weak NAME MATH-category. |
math/0009124 | We focus on the properties of MATH. For MATH the reasoning is identical. To show that MATH is a submersion, we need to verify that this map is onto and so is its derivative MATH. The surjectivity is a pure set-theoretic fact. (Let MATH. Take any MATH and any MATH with MATH. Then MATH.) This argument uses only the assum... |
math/0009124 | Assume first that MATH and the homotopy MATH is just the identity map. Let MATH . This is a compact set. It is easy to see that there exists a neighborhood MATH of MATH in MATH, a small MATH, and an embedding MATH such that MATH, where MATH and MATH. Here and in what follows, we identify MATH with a neighborhood of MAT... |
math/0009124 | We will prove the lemma for MATH. (This implies the lemma for MATH, for MATH can always be extended from MATH to some MATH-cube containing MATH and MATH can also be extended to MATH.) The proof is by induction in MATH. For MATH, that is, MATH, the assertion follows from the assumptions that the fibers of MATH are MATH-... |
math/0009124 | As in the proof for MATH, we argue inductively. For MATH and MATH, the image MATH is comprised of two mapped MATH-spheres contained in the fibers of MATH. Since these fibers are MATH-connected, the filling does exist. Assuming that the lemma is proved for MATH, we establish it for MATH. Let us think of MATH as the unio... |
math/0009124 | We only outline the argument, for it is very similar to the proof of REF . As in that proof we replace MATH by MATH. Arguing inductively, we assume that the lemma is proved for the cubes of dimension less than MATH. As above, we set MATH and MATH, where MATH, and consider the maximal open interval in MATH over which th... |
math/0009124 | Let MATH, where MATH, be a smooth map and let MATH be a smooth homotopy of MATH to a point. We will find a continuous homotopy MATH of MATH to a point. By REF , there exists a lift MATH of MATH. Furthermore, by REF , MATH can be extended to a little closed collar MATH of MATH in MATH. Let MATH be this extension. Denote... |
math/0009131 | REF |
math/0009131 | REF . |
math/0009131 | In view of REF and the fact that MATH and MATH are isomorphisms the three assertions are of course equivalent. REF follows from the identities MATH and MATH by an easy induction. |
math/0009131 | This is REF. |
math/0009131 | First note that we have the identities MATH which together give MATH . Then MATH . |
math/0009131 | The first equality can be found in CITE. The second equality is REF. |
math/0009131 | Because of REF we may assume that MATH is of the form MATH or MATH. We will therefore argue by induction on weight and degree and assume that the assertion holds for all MATH of either less degree or less weight than MATH. (The assertion is certainly trivial for the vacuum, the - up to a scalar factor - unique element ... |
math/0009131 | Let MATH denote the component of MATH of degree MATH. We must show that for each MATH the elements MATH where MATH runs through all partitions of MATH, form a set of generators of the MATH-module MATH. Claim: The restriction map MATH is a surjective and degree preserving ring homomorphism and maps MATH to MATH. In part... |
math/0009134 | compare CITE. |
math/0009134 | We refer to REF for a complete proof. In op.cit. , it is shown under which condition the NAME vector MATH induces permutation on the field MATH, for MATH. |
math/0009134 | It is rather easy to check that the affine REF - folds MATH and MATH have the same number of points over fields containing MATH. Hence, any bijective linear transformation between MATH and MATH will be defined over such fields. Among the critical points of MATH, one finds the points MATH with MATH satisfying the equati... |
math/0009134 | If MATH is a point in the singular locus of the variety MATH (compare REF), then the coordinates of MATH must satisfy REF . This implies that for MATH, MATH for MATH and MATH. Therefore, the singular locus of MATH is described by the set of points MATH in the singular locus of the variety MATH . The MATH-tuples MATH ar... |
math/0009134 | It follows from the above isomorphism and from the fact that MATH (independent of the number of the singular points). For smooth, quintic hypersurfaces in MATH one knows that MATH. |
math/0009134 | The vanishing cycle exact sequence applied to a NAME pencil of quintic hypersurfaces in MATH with degenerating fiber MATH and generic fiber MATH, reads as MATH . Here, MATH is the MATH-vector space of vanishing cycles. Because MATH has only double points, the rank of MATH equals the number of double points of MATH. Hen... |
math/0009134 | The equality MATH follows from REF and the fact that the rank-one vector space MATH is generated by the class of a hyperplane section. The equality MATH is a direct consequence of the above lemma and REF (that is, MATH). The equality MATH follows from REF. Finally, MATH is deduced from the difference MATH (compare proo... |
math/0009134 | In order to prove this, we study the behaviour of the critical points of MATH mod MATH. In paragraph REF, we observed that MATH has MATH critical points with value MATH, MATH with value MATH and MATH with value MATH. We list these points in the following table. MATH . One can easily check that these MATH critical point... |
math/0009134 | Suppose MATH. In paragraph REF we showed that the polynomial MATH, with MATH and MATH . NAME polynomials. It follows from REF that the map MATH permutes MATH. Hence MATH assumes each value exactly MATH times and therefore there are MATH solutions to the equation MATH in MATH. From REF it follows that there are no criti... |
math/0009134 | It follows from REF that MATH. The inequality in MATH yields: MATH . Hence, MATH for MATH big enough. Using the trace formula MATH it follows that MATH. For small MATH the MATH are listed in the table MATH. |
math/0009134 | Let MATH denote the vectorspace MATH, and let MATH be the extension of scalars of MATH. Let MATH be the NAME character of MATH with kernel MATH. By REF we have that MATH for all MATH not in MATH. So MATH for all MATH. It follows that MATH and MATH are isomorphic. Let MATH be an isomorphism that intertwines MATH and MAT... |
math/0009134 | We only show that MATH ramifies at MATH and at both the infinite places. For more details on the properties of MATH, as well as for the proof of its uniqueness up to isomorphism, we refer to CITE. Notice that MATH is totally definite (that is, MATH is a division quaternion algebra isomorphic to two copies of the NAME '... |
math/0009134 | We refer to REF , for the proof. The computations written down in op.cit. generalize easily to any quaternion algebra over a number field. |
math/0009134 | Suppose MATH is an element of MATH. Then MATH, MATH, MATH and MATH. Recall the definition of MATH from MATH in terms of harmonic polynomials MATH. From the definition of the MATH it immediately follows that MATH if MATH, MATH, MATH, MATH or MATH, and MATH otherwise. From the definition of MATH it follows that MATH if M... |
math/0009135 | This is immediate from the identities MATH for MATH and MATH. |
math/0009135 | Let MATH and MATH be pointed matroids on MATH and MATH respectively. A pointed strong map MATH arises from a set function MATH mapping MATH to MATH. This function yields a homomorphism of exterior algebras MATH determined by MATH. According to CITE, the image of each circuit of MATH is dependent in MATH. Using REF this... |
math/0009135 | Let us write MATH for MATH. Using the fact that tensor product is a sum in the category of connected graded algebras, together with REF , we obtain a surjective homomorphism MATH. Using REF and CITE, one can show that the domain and target have the same dimension in each degree. Thus the two algebras are isomorphic. |
math/0009135 | Consider the pointed parallel connection MATH. The underlying matroid MATH is MATH, which is precisely MATH. Then, by REF , the MATH. On the other hand, by REF , MATH is also isomorphic to MATH, which again by REF , is isomorphic to MATH. Now, according to REF , we may change the base points of MATH and MATH without af... |
math/0009135 | Suppose MATH is an isomorphism of MATH to MATH. To begin with, we can then assume without loss that MATH and MATH have the same ground set. The isomorphism MATH determines an isomorphism MATH, and MATH . We need only show that MATH . Let MATH. Then, for MATH, MATH . Since the truncations have rank MATH, we also have, f... |
math/0009136 | As noted in REF the equivalence of REF follows form the computation REF - REF there. Namely, since MATH defines MATH, it is nonsingular, so MATH for some real valued function MATH (possibly after replacing MATH by MATH). Then MATH . Therefore, REF holds. In the last equality in REF , we have again used that MATH (see R... |
math/0009141 | As MATH is finite dimensional, the functor MATH where the right MATH-action on MATH is given by MATH is an equivalence of categories CITE. As the antipode of MATH is bijective CITE we have the following equivalences of categories MATH that is, MATH is NAME equivalent to MATH. It follows from the NAME theory that there ... |
math/0009141 | It is obvious that MATH, MATH are algebra maps. For MATH we have MATH and MATH so MATH is coassociative if and only if MATH . Using the pentagon REF , we find that this is equivalent to MATH and this equality holds for any MATH. In a similar way we can prove that MATH is also coassociative. |
math/0009141 | Taking into account the multiplication rule of MATH we have MATH and MATH so we have to prove the equality MATH in MATH. Let's fix the indices MATH, MATH, MATH and MATH and evaluate REF at MATH. REF is then equivalent to MATH which can be obtained by applying the definition of the convolution product and the dual basis... |
math/0009141 | CASE: We shall use the notations introduced above. First we shall prove that MATH (respectively, MATH) are unitary subalgebras in MATH and subcoalgebras of MATH (respectively, MATH). This will follow from the formulas: MATH and MATH for all MATH, MATH. We prove REF, (being similar, REF is left to the reader). MATH that... |
math/0009141 | CASE: We have proved in REF that if the spaces MATH and MATH are nonzero, then MATH is unitary. Conversely, if MATH is unitary, the isomorphisms of vector spaces from REF of REF show us that MATH and MATH are nonzero. CASE: Assume now that MATH is finite dimensional. From the construction of REF we have that MATH. In p... |
math/0009141 | CASE: MATH is a right MATH-invariant; hence, MATH. If we apply MATH to this equality we get that MATH, for any MATH. As MATH is a basis of MATH, we obtain that MATH, for all MATH. If we apply the right MATH-module map MATH to this, we obtain that MATH is a right integral of MATH. CASE: Left to the reader. |
math/0009141 | Let MATH be a basis of MATH and MATH the canonical element. We have to prove that the NAME algebra MATH extracted from REF of REF is isomorphic to MATH, with the initial NAME algebra structure. MATH, MATH is an injective algebra map. We identify MATH . From the construction, MATH is the subalgebra of MATH having MATH a... |
math/0009141 | In REF of REF we proved that there exists a NAME algebra isomorphism MATH for any MATH and MATH, which means that all the NAME algebras associated to the elements of an orbit of the action REF are isomorphic. We shall now prove the converse. First we show that two finite dimensional NAME algebras MATH and MATH are isom... |
math/0009143 | It clearly suffices to deal with the case where MATH and both have mean zero. Let MATH . Take a function in MATH, of mean zero, and consider its NAME expansion MATH where MATH is the partial sum over all frequencies of norm less than MATH. Then the correlation function MATH is a sum of terms MATH . By NAME, the second ... |
math/0009143 | This is the well known Euclidean algorithm: Assume that MATH with MATH. Primitivity is equivalent to MATH being co-prime. Then find an integer MATH so that MATH with MATH. We then arrive at a new vector MATH. Now find an integer MATH so that MATH with MATH to get another vector MATH and so on. This proceeds until we ge... |
math/0009143 | One can easily check that homogeneous quasi-morphisms on MATH which vanish on parabolics are NAME functions with respect to metric MATH. Let MATH be such a quasi-morphism with MATH. Take MATH sufficiently small and consider any sequence MATH with MATH. It follows that MATH for some MATH and all MATH. Applying REF , we ... |
math/0009143 | The proof is divided into REF steps. CASE: We claim that every hyperbolic matrix MATH is conjugate in MATH to a matrix MATH with MATH. Indeed, assume that MATH, where MATH . One calculates that MATH, where MATH. The discriminant MATH of the quadratic form MATH equals MATH. Since MATH is hyperbolic, MATH is irrational. ... |
math/0009144 | Since MATH, we can find a unitary automorphism MATH of MATH over MATH which is equal to MATH on MATH and MATH, and equal to MATH on MATH. Pulling MATH back by MATH, we reduce to the case MATH, so that the caloron configuration is now a loop in MATH. But this space is contractible, so the result follows. |
math/0009144 | The space of calorons on a given framed bundle, with given boundary data MATH is contractible. It is easy to see that any continuous path joining MATH to MATH gives rise to a norm-continuous path of NAME operators between the NAME spaces in MATH. Since each of these is NAME by REF , it follows that the index is constan... |
math/0009146 | By the proof of REF we have that for any MATH, MATH, where, as before, MATH is the image of MATH as a subspace of MATH, and in fact, by REF and since MATH is stable, it results MATH. Thus we have, up to a change of basis, MATH where MATH and MATH are subspaces of MATH of dimension MATH. It is easily checked that MATH a... |
math/0009146 | The exactness of REF is proven in (CITE, pag. REF). In order to prove the exactness of REF , we proceed by mimicking the proof of the existence of the NAME complex given in CITE. Let MATH and let MATH be the projection onto the first space. The morphism MATH defines a section MATH, given by MATH where the MATH's are th... |
math/0009146 | Let MATH and let us consider a resolution of MATH: MATH where each MATH is a direct sum of line bundles. Let us split the sequence into short exact sequences: MATH then, applying the functor MATH, we get: MATH for any MATH such that MATH. Thus MATH for any MATH. Moreover MATH if MATH and thus by REF , MATH is reflexive... |
math/0009146 | The injective map MATH induces the exact sequence: MATH where MATH is a MATH-rank sheaf such that MATH: thus, dualizing this sequence and observing that, by REF , MATH, it results MATH, in fact by REF all these sheaves are torsion-free. Moreover MATH and therefore REF follows. |
math/0009146 | By REF , we can suppose MATH . Moreover the same technique used above can be applied to prove the thesis for all MATH. Thus it suffices to show that MATH. It is easily checked that by dualizing the sequence REF we get the sequence MATH . Thus we just need to prove that if MATH is such that MATH then MATH: this is a dir... |
math/0009146 | The first statement of REF easily implies that the map MATH is injective. Since MATH is torsion-free, so is MATH. Let now MATH be a torsion-free sub-sheaf of rank MATH. then MATH: since MATH REF , it results MATH, that is, MATH. Thus MATH is MATH-stable. NAME, let MATH be a non-stable matrix. Then, by REF , we can writ... |
math/0009146 | Let us prove first that any MATH is uniquely determined by a morphism of sequences: MATH . This is a direct consequence of the fact that, by the vanishing of MATH and MATH we get the exact sequence: MATH . Thus if MATH is simple, then the only automorphisms of MATH are the homotheties, that implies REF . |
math/0009147 | Let MATH . We want to show that MATH. It is easy to see that MATH is a closed subset and that it is closed under addition and conjugation. Let MATH and MATH. Then MATH so MATH. Hence MATH is also closed under multiplication. So it is a MATH-subalgebra of MATH. Since MATH for each MATH. So MATH. |
math/0009147 | First notice that for each MATH . From this we get that for each MATH . By REF we see that there is only a finite number of different sets MATH. So we can for each MATH choose finite sets MATH and MATH such that MATH . From this we get MATH . So MATH for each MATH. We can therefore define MATH by MATH where MATH is as ... |
math/0009147 | Let MATH. By REF MATH for each MATH, and since MATH, we have MATH. So MATH . Hence MATH is a partial isometry. Since the left NAME cover graph is left-resolving, we have that if MATH either MATH or MATH. If MATH by REF , and if MATH . So MATH for MATH. By REF MATH . So the partial isometries MATH satisfy the NAME relat... |
math/0009147 | We will prove the statement by induction over the length of MATH. First assume that MATH. Then the statement follows directly from the definition of the left NAME cover graph. Assume next that we have proved the statement for MATH, and that MATH. Let MATH. If MATH, then MATH. So there exists a unique path MATH such tha... |
math/0009147 | Observe that MATH for all MATH if MATH, and that MATH for all MATH if MATH. So MATH if MATH and MATH if MATH. Since MATH is generated by MATH and MATH for MATH there exists a MATH-homomorphism MATH from MATH to MATH sending MATH to MATH. For each MATH define MATH by MATH . Since MATH we have that MATH for each MATH. Si... |
math/0009147 | According to REF there exists a MATH-homomorphism MATH such that MATH, and according to REF there exists a MATH-homomorphism MATH such that MATH and MATH. We have that MATH where we for the last equality use that MATH, and that MATH if there does not exists an edge with range MATH and label MATH. So MATH, and since MAT... |
math/0009148 | The first assertion is proved using CITE and the fact that the full dimensional cones of the NAME fan of MATH are exactly the cones corresponding to monomial initial ideals of MATH. The second assertion is easily proved by noticing that MATH is a monomial ideal none of whose generators contain the variable MATH. |
math/0009148 | First suppose that MATH does not have minimum negative support. Then there is MATH such that MATH is strictly contained in MATH. This means that MATH for MATH, and MATH. Say MATH. If MATH, then MATH does not hold. If MATH, then MATH does not hold. If MATH, then MATH does not hold, and if MATH then MATH does not hold. A... |
math/0009148 | This follows from the same arguments that proved REF . |
math/0009148 | To see that our candidate for standard pair satisfies the criterion of REF, we have to show that the only integer point in a certain polytope is the origin. This follows exactly from the same arguments of REF if MATH is a monomial ideal. Otherwise, we shrink MATH so that the same arguments will work when we use the wei... |
math/0009148 | Fix MATH. Let MATH such that MATH if MATH; MATH otherwise. Observe that MATH for MATH. For MATH we define, following CITE: MATH . Following the proof of REF, we see that, for MATH there exists a positive integer MATH such that MATH contains a nonzero integer vector MATH. It must satisfy MATH. The reason for this is tha... |
math/0009148 | We compute canonical series with respect to the weight vector MATH, as in CITE, assuming that MATH is a monomial ideal. We will deal with the case when MATH is not monomial later. The logarithm-free canonical solutions of MATH are of the form MATH where MATH is a fake exponent of minimum negative support. The fact that... |
math/0009148 | It is clear that the functions described above belong to the kernel of MATH. Suppose first that MATH is a logarithm-free solution of MATH that is constant with respect to MATH. We compute canonical series with respect to the weight vector MATH. If this cannot be done (that is, if MATH is not a monomial ideal) we replac... |
math/0009148 | That MATH is the fake exponent of MATH corresponding to MATH follows from the fact that MATH (and that we have only modified the third coordinate of MATH). Now we have to show that MATH has minimum negative support. We know that MATH for MATH, so that MATH has at least three integer coordinates. If it has exactly those... |
math/0009148 | Suppose there is a solution MATH of MATH such that MATH. We will obtain a contradiction. We proceed as in the part of the proof of REF where we show that the functions MATH are logarithm-free. The first step is to use REF to write MATH for every MATH. We apply REF , with the goal of showing that MATH has no terms in MA... |
math/0009148 | Pick any solution MATH of MATH whose derivative with respect to MATH lies in MATH. Write MATH as in REF : MATH for MATH. Here we must have MATH, since MATH lies in MATH. Suppose that MATH. Look at the logarithm-free function MATH . If MATH, the term MATH has logarithms, so it must be cancelled with terms coming from MA... |
math/0009148 | By REF , an element MATH of the solution space of MATH such that MATH lies in MATH is of the form MATH with MATH a logarithm-free function with integer exponents, no term MATH, and MATH in the kernel of MATH. Notice that once the MATH are fixed, MATH is unique with those MATH, since the difference of two such functions... |
math/0009148 | By contradiction, suppose there is a solution MATH of MATH such that MATH where MATH is a linear combination of the functions from REF . We can write MATH where MATH is the sum of the terms in MATH whose exponents and MATH differ by an integer vector, and MATH is the sum of the terms in MATH whose exponents and MATH di... |
math/0009148 | In REF we built one function in MATH for each function in a basis of MATH (which we knew from REF ). Moreover, REF provided at least two linearly independent functions for MATH. REF shows that all of these functions are linearly independent. Therefore MATH and this implies that: MATH where MATH is the image of MATH. Th... |
math/0009148 | Pick MATH, and MATH as in Construction REF. We can choose MATH small enough so that the numbers MATH satisfy the conditions of Construction REF for all MATH. Call MATH and MATH . Then REF implies that MATH for all MATH. Now the proof of REF implies that MATH. This concludes the proof. |
math/0009155 | The automorphism group of MATH is a semidirect product of MATH and the outer automorphism group of the root system corresponding to MATH. It is easy to verify that, in the cases at hand, a nontrivial outer automorphism acts by sending MATH to MATH. From this, the result is clear. |
math/0009155 | By NAME duality, MATH. Since MATH, where MATH is nef and big, and MATH, MATH for every root MATH. A similar statement holds for MATH. Let MATH be a root. Then MATH, and so by the NAME theorem, MATH. By the first part of the lemma, MATH and so MATH. Thus MATH if and only if MATH for some effective curve MATH. In this ca... |
math/0009155 | In any case MATH has degree zero. Suppose that MATH is not effective. Then MATH for all MATH. Consider the long exact cohomology sequence arising from MATH . Since MATH is NAME dual to MATH, it is zero for all MATH. Thus MATH is an isomorphism for all MATH. It follows that MATH is effective if and only if MATH, if and ... |
math/0009155 | Let MATH be a vector bundle on MATH such that MATH for every irreducible MATH-curve MATH. We shall show that MATH is trivial in some analytic neighborhood of the exceptional curve MATH. If MATH are the MATH-curves, and MATH, it follows easily from the fact that the dual graph of MATH is a union of contractible componen... |
math/0009155 | First we claim: With MATH as in the statement of REF , CASE: MATH; CASE: Let MATH be MATH-curves and let MATH. Then MATH . CASE: By NAME, MATH only depends on the topological type of MATH. Thus we may replace MATH by MATH. But MATH and as we have seen, MATH for every root MATH. Hence MATH. CASE: The bundle MATH has a f... |
math/0009155 | First, we claim that MATH is topologically trivial. It suffices to check that MATH is topologically trivial. The surjection MATH induces a MATH-bundle MATH, and the corresponding line bundle is easily seen to be MATH, which is trivial. Hence MATH lifts to a MATH-bundle, which is automatically topologically trivial sinc... |
math/0009155 | By REF , it suffices to prove the result for one choice of MATH, which we may assume to be a smooth elliptic curve. By CITE, there exists a MATH-bundle MATH of the form MATH, such that REF MATH, and REF MATH is a lift to MATH of the bundle MATH. Let us show that we can find a MATH-bundle MATH such that MATH. To make th... |
math/0009155 | It is easy to check via the exact sequence MATH that the obstruction to lifting MATH to a MATH-bundle vanishes if and only if the bundle MATH is trivial. Moreover, using the above sequence and the fact that there is a subgroup of MATH isomorphic to MATH which surjects onto MATH any lift of MATH to MATH is unique up to ... |
math/0009155 | Every set of MATH disjoint lines can be completed to a set of MATH disjoint lines. On the other hand, such a set defines a blowdown to MATH and hence a diagonal basis MATH with MATH and MATH. The NAME group acts transitively on such bases, by REF , and hence on the set of all sets of MATH disjoint lines. A similar argu... |
math/0009155 | For MATH, this is clear by the remarks before the statement. For MATH, the adjoint bundle MATH, where MATH is the set of roots. The result then follows since MATH. |
math/0009158 | Recall first the relation between the cylindrical metric, and the euclidean metrics MATH and MATH (near MATH and MATH, respectively): MATH where MATH . Here we are assuming that MATH is a smooth point, or if not, we pass to a uniformizing chart centred at MATH. It will be clear that MATH-equivariance is preserved throu... |
math/0009158 | Use the conformal isometry MATH given by MATH and the conformal invariance of MATH. Because of the conformal weights, we have if MATH is a section of MATH so that a solution MATH in MATH, homogeneous of degree MATH, translates into an exponential solution with factor MATH on the cylinder. In particular the constant sol... |
math/0009158 | If MATH, MATH, then from REF , MATH has a phg expansion where the index set is just the positive integers. Translating to MATH and remembering the conformal weight, we get MATH which satisfies MATH and MATH near MATH. Hence by elliptic regularity, the MATH-null space of MATH on MATH agrees with the standard null-space ... |
math/0009158 | CASE: Note first by the discussion in REF that the MATH-norm of MATH is MATH for some MATH. Taking MATH, MATH as MATH. To check that the map is smooth, it evidently suffices to show that the nonlinear terms define a smooth map. Note first by the properties of MATH that MATH where MATH is independent of MATH. Hence from... |
math/0009158 | Combine the equations in the form MATH where MATH is a polyhomogeneous section which is a linear combination of MATH, MATH and MATH, MATH and MATH . CASE: Interior regularity. That MATH is MATH in MATH now follows from standard regularity results, which are applicable because MATH is already MATH, where MATH CITE. CASE... |
math/0009158 | Combine REF . The estimate on MATH follows from the fact that the MATH-norm of MATH in REF is of the same order of magnitude as MATH and the estimate REF. The completeness of the family of metrics constructed comes directly from the implicit function theorem. |
math/0009158 | Combine REF with REF . |
math/0009158 | Take MATH to be the conformal blow up at MATH of MATH. Take MATH to be the MATH-manifold obtained by gluing MATH to MATH by identifying MATH with MATH by MATH. Then MATH, viewed as a MATH-manifold, has boundary equal to MATH, while the boundary of MATH is MATH. Applying REF to MATH and MATH now gives the conclusion, in... |
math/0009158 | It is clear that REF combined with REF gives a NAME conformal structure MATH on MATH. However it is obvious that MATH is NAME, so by CITE, there is a NAME representative of the conformal class MATH and this must necessarily be scalar-flat. |
math/0009158 | Recall first the NAME metric MATH CITE, a weakly asymptotically euclidean scalar-flat NAME metric on the blow-up MATH of MATH at the origin. The complement of the exceptional divisor in MATH is biholomorphic to MATH, so we can construct the multiple blow-up MATH, with its standard complex structure, by gluing a copy of... |
math/0009158 | If MATH is NAME, we can identify MATH with the composite MATH of NAME operators as above. To prove the proposition, it is enough to note the NAME formulae MATH which hold whenever MATH and MATH. (The verification of these is left to the reader.) Suppose MATH; set MATH, so that MATH. If MATH then MATH is invertible, so ... |
math/0009158 | Recall the formula MATH from the previous proposition. Let MATH and let MATH. Then MATH as MATH, and so MATH and letting MATH we conclude as before that MATH is parallel. Since MATH at MATH, moreover, MATH. Hence MATH. We make the same argument as above using the NAME formula. This time the boundary term is MATH but we... |
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