paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0002152 | Let MATH such that such that the functional MATH belongs to MATH. We have to show that MATH and MATH. So we will see that, for any cube MATH, MATH and for any two doubling cubes MATH, MATH . First we will show that REF holds for any doubling cube MATH. In this case the argument is almost the same as the one of CITE. We... |
math/0002152 | This lemma is an easy consequence of the NAME representation theorem. The same argument as the one of CITE works here. |
math/0002152 | Notice that, by REF , MATH for MATH. Now we repeat the arguments in CITE again. We consider the diagram MATH . The map MATH is an inclusion and MATH is the canonical injection of MATH in MATH (with the identification MATH for MATH). By REF , MATH is a closed subspace of MATH. By NAME 's closed range theorem (see CITE),... |
math/0002152 | We assume MATH. The proof for MATH is similar. For some fixed MATH and all MATH, we will prove that there exists some MATH such that for any MATH we have the following good MATH inequality: MATH . It is well known that by this inequality one gets MATH if MATH. We denote MATH and MATH . For the moment we assume MATH. Fo... |
math/0002152 | Assume MATH. The operator MATH is sublinear and it is bounded in MATH and MATH. By the NAME interpolation theorem, it is bounded on MATH, MATH. That is, MATH . We may assume that MATH has compact support. Then MATH and so MATH. Thus MATH, and so MATH. By REF , we have MATH . The proof for MATH is similar: Given MATH, w... |
math/0002152 | We will assume MATH. CASE: Taking into account REF , for MATH-almost all MATH such that MATH, there exists a cube MATH satisfying MATH and such that if MATH is centered at MATH with MATH, then MATH . Now we can apply NAME 's covering theorem to get an almost disjoint subfamily of cubes MATH satisfying REF . CASE: Assum... |
math/0002152 | For simplicity we assume MATH. The proof follows the same lines as the one of CITE. The functions MATH having compact support with MATH are dense in MATH, MATH. For such functions we will show that MATH . By REF , this implies MATH . Notice that if MATH has compact support and MATH, then MATH and MATH. Thus MATH, and t... |
math/0002152 | We have already seen REF . Let us prove REF. Suppose that MATH, for example. Let MATH be a function supported on some cube MATH. Suppose first MATH (this is always the case if MATH). We have MATH . So it is enough to show that MATH . Let MATH be the point (or one of the points) in MATH which is closest to MATH. We deno... |
math/0002152 | Suppose that MATH. Let us see that this implies MATH. The estimates are similar to the ones that we used to show that CZO's bounded on MATH are also bounded from MATH into MATH. Assume, for example MATH. We have to show that if MATH, then MATH . We denote MATH. Then we write MATH . Since MATH is antisymmetric, we have ... |
math/0002152 | Let MATH be a cube concentric with MATH such that MATH, with MATH for some MATH. Then MATH for all MATH, with MATH depending on MATH and MATH. Observe also that if we take MATH so that MATH, then MATH and so MATH . Therefore, MATH . On the other hand, if MATH is big enough, then MATH. Indeed, MATH and so MATH, assuming... |
math/0002152 | Let MATH be two doubling cubes in MATH, with MATH. Let MATH be the first cube of the form MATH, MATH, such that MATH. Since MATH, we have MATH. Therefore, for the doubling cube MATH, we have MATH, with MATH depending on MATH, MATH, MATH and MATH. In general, given MATH, we denote by MATH the first cube of the form MATH... |
math/0002152 | For all MATH, we will show the pointwise inequality MATH where, for MATH, MATH is the non centered maximal operator MATH and MATH is defined as MATH . The operator MATH is bounded on MATH for MATH, and MATH is bounded on MATH for MATH because MATH is bounded on MATH (see CITE). Then the pointwise REF for MATH implies t... |
math/0002154 | Let MATH. Then it is decomposed as MATH with MATH. Consequently MATH can be non-zero only for MATH. Hence MATH with MATH . Since MATH is a (non-zero) scalar we have found MATH. For such a MATH the condition MATH is equivalent to MATH. Sandwiching with MATH and MATH gives the desired intertwining relations for all MATH.... |
math/0002154 | The preservation of addition and the multiplication is a straight-forward corollary of REF . The statement for the conjugates is derived in the same way as CITE. |
math/0002154 | Note that if MATH, MATH and MATH is an isometry, then MATH since MATH. Hence MATH for any MATH. With this it is easy to check that MATH is unitary. The first inclusion of REF together with CITE imply that MATH is an intertwiner from MATH to MATH. With that it is easy to check that MATH (compare the proof of CITE). Next... |
math/0002154 | Thanks to Izumi's result, REF , and due to the first inclusion in REF , all what we have to verify is the relation MATH for all MATH whenever MATH. For MATH there is some isometry MATH with some MATH. Then this is just MATH thanks to naturality. |
math/0002154 | It follows from MATH and the first inclusion in REF that MATH. The elements are clearly linearly independent as MATH are orthonormal isometries in MATH. Now the statement follows since MATH by NAME reciprocity. |
math/0002154 | By REF we have to show that MATH . So first we assume that MATH is in the right-hand side, and such a MATH satisfies MATH in particular for all MATH. So choose an isometry MATH with some MATH. Then, by REF , MATH whereas MATH . Equating these and multiplying by MATH from the left and MATH from the right we obtain, usin... |
math/0002154 | Again by REF , we only need to show that for MATH the linear space of intertwiners MATH satisfying MATH for all MATH is given by MATH whenever MATH and vanishes otherwise. Thus suppose that MATH satisfies MATH for all MATH. Then in particular MATH whenever MATH is an isometry and MATH. Sandwiching this with MATH and MA... |
math/0002154 | Let MATH be an isometry, MATH. Then MATH where we used the BFE MATH. |
math/0002154 | Using REF , we can compute MATH and now the result follows by REF and since MATH. |
math/0002154 | Immediate from CITE. |
math/0002154 | Literally the same as the proof of REF , apart from the simplification in the second half that we now only have to consider MATH. |
math/0002154 | Using once again REF , we first show that if a non-zero MATH satisfies MATH for all MATH, then this implies MATH. So suppose we have such a MATH. Since MATH there will also be some MATH and an isometry MATH. Then the intertwining condition reads MATH whenever MATH and MATH is an isometry. Multiplication with MATH from ... |
math/0002154 | We compute MATH where we used the BFE for the relative braiding CITE, MATH. |
math/0002154 | Analogous to the proof of REF , this is reduced to REF by use of REF . |
math/0002164 | In local coordinates, the formula MATH becomes REF for the tensor MATH on MATH; thus REF are equivalent. The NAME rule for graded NAME algebras shows that MATH . Thus MATH is a differential on MATH if and only if MATH. The bracket MATH is skew-symmetric: MATH . As for the NAME rule, we have MATH . The anomalous term MA... |
math/0002164 | By REF , a weak equivalence MATH of dg NAME algebras induces a bijection MATH; indeed, MATH is filtered by subspaces MATH, and similarly for MATH. It remains to prove that MATH induces bijections MATH for all MATH and MATH. Given MATH, define a dg NAME algebra MATH where MATH is placed in degree MATH. The construction ... |
math/0002164 | It suffices to define MATH on the generators MATH of MATH over MATH. By the hypotheses on MATH, we have MATH so the definition of MATH is forced. By induction on MATH, we see that MATH . It follows that MATH, and hence that, for MATH, MATH . Thus MATH . The sum over MATH reduces to MATH, and the right-hand side to MATH... |
math/0002164 | From the formula MATH it follows that MATH and that MATH for MATH; the lemma follows easily from this formula. |
math/0002164 | We have MATH . Since MATH, this in turn equals MATH . The sum over MATH reduces to MATH, and the lemma follows. |
math/0002164 | It is clear that MATH is compatible with the differential on MATH: MATH . It is also easy to see that MATH is a morphism of graded NAME algebras, since MATH . It only remains to check that MATH is a weak equivalence; this is a variant on the ``exactness of the variational bicomplex." We learned the idea used in the fol... |
math/0002164 | We must show that if MATH is a symplectic manifold, the inclusion MATH of sheaves of dg NAME algebras is a weak equivalence. By the NAME theorem (in its original sense!), it suffices to consider a convex subset MATH of MATH with its standard symplectic structure, and NAME tensor MATH . Let MATH be the differential asso... |
math/0002164 | Lifts MATH of MATH are characterized by the equations MATH and MATH . Let MATH be a section of MATH such that MATH; there are no obstructions to the existence of MATH, because MATH is NAME. Since MATH, we see that there is a section MATH of MATH such that MATH; again, there are no obstructions to the existence of MATH.... |
math/0002164 | CASE: MATH is a morphism of complexes (that is, MATH): Let MATH be an element of MATH. It is obvious that MATH since MATH and MATH both vanish. As for MATH, we have MATH and MATH . CASE: MATH preserves the NAME bracket: If MATH, we have MATH . On the other hand, MATH . From these formulas, we see that MATH. Finally, it... |
math/0002164 | We must show that MATH vanishes. Rewritten using the operators MATH and MATH, and taking into account that the operators MATH and MATH graded commute with MATH and MATH, and that MATH, we see that this equals MATH which vanishes, since MATH and MATH. |
math/0002164 | We have MATH . The formula follows, since MATH. |
math/0002164 | There is short exact sequence of complexes MATH and hence, for MATH, a long exact sequence MATH . There is an isomorphism between the complex MATH and the NAME complex MATH, obtained by mapping MATH to MATH and MATH to the basis vector MATH of MATH. Likewise, the complex MATH is isomorphic to the reduced NAME complex M... |
math/0002169 | Obviously, REF follows from REF . For MATH the statement was proved by NAME and NAME REF . Now suppose that MATH and REF is minimal. Since MATH is globally generated, NAME 's theorem ensures that a generic map MATH is injective. Then, in the following commutative diagram columns and rows are exact and MATH is locally f... |
math/0002169 | We first prove that if MATH is semistable, then MATH. In fact, if MATH then MATH split as MATH and by weakly admissibility MATH then we have MATH which contradicts the semistability of MATH. Now suppose that MATH is semistable and MATH for some MATH with MATH and let MATH be the largest of such MATH. Since MATH the min... |
math/0002169 | Let MATH be the minimal resolution of MATH where MATH . The stability of MATH ensures the vanishing MATH so that MATH. Then, from REF we easily find the following data: MATH and from REF we have MATH . Now, by splitting MATH as MATH with MATH, the above formula becomes MATH . Since MATH and MATH so MATH . Finally, dist... |
math/0002169 | It can be verified by direct computation from REF that, if MATH has natural pair, then the codimension of MATH is zero. Conversely, let MATH, MATH be two non-negative integers. Since all finite difference MATH are non decreasing functions of MATH, then MATH and by the previous lemma MATH . If MATH, we have MATH and MAT... |
math/0002169 | This is a verification; we outline the main steps of the computation. In the first place, one must ensure that the natural pair MATH is actually associated to vector bundles in MATH. This amount to show that from REF the pair MATH has the appropriate NAME classes and that REF hold. From REF , any pair MATH such that MA... |
math/0002169 | Moduli space of stable rank MATH vector bundles on MATH are smooth. By the previous proposition they can have only one connected component. |
math/0002169 | Let MATH be the admissible pair associated to a vector bundle MATH in MATH. Then, one has MATH if and only if MATH. By semicontinuity of cohomology groups and REF , it is enough to restrict ourselves to the case where MATH is general. So, by REF one has MATH if and only if MATH and the condition MATH is equivalent to M... |
math/0002170 | Let MATH. We prove that MATH for any MATH. Then we conclude that MATH and hence MATH. Since MATH, we obtain MATH. Now REF and the assertion of the Proposition for MATH imply that MATH . Hence, MATH is the symmetrizer of the algebra MATH. Let MATH. We prove that MATH for any MATH. For this we use the relations of the al... |
math/0002170 | Let MATH. Using the relations REF - REF we obtain MATH . The other cases can be shown similarly. |
math/0002180 | This is a direct corollary of the general category-theoretic REF , under `NAME below. Alternatively, a direct argument can be used, as follows. Given a colax symmetric monoidal functor MATH we must define a functor MATH. On objects, take MATH. To define MATH on morphisms, first note the following: CASE: for each MATH, ... |
math/0002180 | REF is immediate from the second half of the proof of REF . An easy induction, again using this half of the proof, shows that MATH equals MATH so by the comments in REF , we have REF . MATH . |
math/0002180 | CASE: From the definition of MATH in the proof of REF , it is easy to show that MATH, and similarly MATH. Hence MATH . Also, MATH is the unique map from MATH to MATH. CASE: Similarly, an easy induction on MATH shows that MATH where MATH is the map defined in REF . But by the comments in REF , MATH is a homotopy monoid ... |
math/0002180 | For each MATH, let MATH denote the standard MATH-simplex MATH with its MATH vertices collapsed to a single point, and this point declared the basepoint. Informally, MATH can be described as follows: CASE: MATH, for example, MATH is a single point, MATH, and MATH looks like MATH CASE: MATH is defined on morphisms by the... |
math/0002180 | Take a homotopy monoidal category MATH in MATH, and construct from it a monoidal category as follows. CASE: MATH. CASE: What we want to define is a functor MATH what we actually have are functors where MATH is the unique map MATH in MATH. So for each MATH and MATH, choose (arbitrarily) a pseudo-inverse MATH to MATH, an... |
math/0002180 | Let MATH be a monoidal transformation. Let MATH and MATH be the (arbitrarily-chosen) functors used in the construction of MATH, and MATH the natural transformations, and similarly MATH etc.for MATH. We now construct a monoidal functor from MATH to MATH. The functor part is MATH. For the rest of the structure we need is... |
math/0002180 | Let MATH be the choices for MATH, and MATH those for MATH. Then MATH and MATH are isomorphic objects of MATH, so their images in MATH under the functor of REF are also isomorphic. In other words, MATH in MATH. MATH . |
math/0002180 | There is a REF-category MATH whose objects and REF-cells are the same as those of REF category MATH, and whose REF-cells are homotopy classes of homotopies. An equivalence in this REF-category is just a homotopy equivalence. (All homotopies mentioned here must respect basepoints.) Moreover, cartesian product MATH and t... |
math/0002180 | First we show that MATH is a MATH-fold homotopy comonoid, that is, a homotopy MATH-algebra in MATH. Let MATH be the colax monoidal functor of REF , exhibiting MATH as a homotopy comonoid. Also let MATH be the MATH-fold smash product, MATH . Observe that since MATH distributes over MATH, MATH naturally becomes a multi-m... |
math/0002180 | Simply apply the axioms for a class of equivalences REF to the commuting diagrams in the definition of monoidal transformation REF . MATH . |
math/0002183 | Since MATH admits a finite MATH-invariant measure, by REF , we have that MATH. Let MATH . Then MATH, MATH, and MATH. Therefore MATH admits a finite MATH-invariant measure, and hence MATH is closed. Hence replacing MATH by MATH, we may assume that MATH is normal in MATH. Now without loss of generality we can replace MAT... |
math/0002183 | Take any MATH. By REF , there exists a compact set MATH such that the set MATH is infinite. Therefore there exists MATH such that MATH. Therefore there exists MATH such that MATH. Put MATH. Then MATH. Put MATH . Then MATH. Suppose there exists a closed subgroup MATH of MATH containing MATH such that MATH is closed. The... |
math/0002183 | Let MATH be such a subgroup. Then the claim follows from replacement of MATH by MATH. |
math/0002183 | Let MATH denote a locally finite MATH-invariant NAME measure on MATH. Since MATH is a proper map, the projected measure MATH on MATH is a locally finite and MATH-invariant. Now the claim follows from the uniqueness, up to constant multiple, of the locally finite MATH-invariant measures on MATH. |
math/0002183 | Since the fibers of MATH have finite invariant measures, we have that MATH has a finite MATH-invariant measure. Therefore MATH has a finite MATH-invariant measure. Again MATH has a finite MATH-invariant measure. Now the claim follows from REF . |
math/0002183 | For any MATH, there exists a closed connected subgroup MATH of MATH such that MATH has a finite MATH-invariant measure, MATH, and MATH. Therefore MATH. Now MATH. Hence MATH has a finite MATH-invariant measure. Therefore MATH is closed. Let MATH be the one-parameter unipotent subgroup of MATH such that MATH. Since MATH,... |
math/0002183 | Due to REF , without loss of generality we can pass to the quotient MATH and assume that MATH. Now arguing as in the proof of REF , we conclude that MATH contains a neighbourhood of MATH in MATH. Now since MATH contains a neighbourhood of MATH in MATH, the claim follows. MATH . |
math/0002185 | We may assume that MATH. Since MATH is exact, it follows from NAME 's theorem CITE that the inclusion map of MATH into MATH is nuclear. Thus, there are MATH and unital completely positive maps MATH and MATH such that MATH . Let MATH be the corresponding linear functional defined as in REF . Since MATH is weak-MATH dens... |
math/0002185 | CASE: We follow the proof of REF. We give ourselves a finite subset MATH and MATH. By the assumption, there is MATH satisfying the conditions in REF for MATH, MATH and MATH. If MATH for MATH, then we put MATH. We define MATH by MATH for MATH. It is not hard to check that MATH has the desired properties. CASE: Let MATH ... |
math/0002186 | For MATH denote by MATH the ring of germs at MATH of real-analytic functions on MATH. For MATH near MATH, we can think of an element MATH in MATH as a MATH-tuple MATH satisfying the condition in REF . Hence the subsheaf MATH coincides with the sheaf of relations MATH . Since the sheaf MATH is coherent by REF, it follow... |
math/0002186 | Define MATH, MATH, where MATH and MATH are the integer valued functions defined by REF respectively. Given MATH, choose MATH and MATH as in REF . Now consider the set of vector valued real-analytic functions MATH (as in REF ) defined in MATH. For each subset of MATH functions in this set, we take all possible MATH mino... |
math/0002186 | We may assume MATH. Since MATH is CR at MATH, there is a neighborhood of MATH in MATH such that the piece of MATH in that neighborhood is contained as a generic submanifold in a complex submanifold of MATH (called the intrinsic complexification of MATH) of complex dimension MATH (see for example, CITE). By a suitable c... |
math/0002186 | Since MATH is generic, and hence CR, we can choose a frame MATH of real-analytic MATH vector fields on MATH near MATH, spanning the space of all MATH tangent vectors to MATH at every point near MATH. (Here MATH, where MATH is the codimension of MATH.) We write MATH, where MATH, MATH, are real valued vector fields. We p... |
math/0002186 | We take normal coordinates MATH vanishing at MATH (see for example, REF), that is, we assume that MATH is given by MATH near MATH, where MATH is a germ at MATH in MATH of a holomorphic MATH-valued function satisfying MATH . We may choose a frame MATH spanning the space of all MATH vector fields on MATH of the form MATH... |
math/0002186 | Denote by MATH the dimension of the span in MATH of the MATH, MATH. By the assumption, there exists a MATH minor MATH, extracted from the components of the MATH, which does not vanish identically. Note that MATH. Let MATH, MATH, be such that MATH. Then for MATH and MATH as in the lemma, it follows that MATH, where MATH... |
math/0002186 | We write MATH. Let MATH be a local defining function for MATH. Then MATH is a local defining function for MATH in a neighborhood of MATH in MATH. By the definition of MATH-equivalence, we obtain MATH for any local parametrization MATH of MATH at MATH. By the second identity in REF , the holomorphic function MATH vanish... |
math/0002186 | We first observe that every MATH-equivalence between MATH and MATH induces a linear isomorphism between MATH and MATH. Since MATH and MATH are MATH-equivalent for every MATH, this implies MATH. To complete the proof of REF , we argue by contradiction. We assume that MATH is CR at MATH but that MATH is not. If MATH is a... |
math/0002186 | Observe that MATH and MATH are generic submanifolds of MATH. We write MATH. Let MATH be a local defining function for MATH. Then MATH is a local defining function for MATH in a neighborhood of MATH in MATH. By the definition of MATH-equivalence, we obtain MATH for any local parametrization MATH of MATH at MATH. We choo... |
math/0002186 | Set MATH, where MATH are defined by REF respectively. Let MATH, MATH a real-analytic submanifold of MATH and MATH. We may assume that MATH and MATH have the same dimension, since otherwise there is nothing to prove. Let MATH be fixed. If, for some integer MATH, MATH and MATH are not MATH-equivalent, then we can take MA... |
math/0002186 | For MATH as above and MATH, define a germ MATH at MATH of a holomorphic function on MATH by MATH . We use the consequence of the chain rule that any partial derivative of a composition of two holomorphic maps can be written as a polynomial expression in the partial derivatives of the components. Then it follows from th... |
math/0002186 | For convenience we use the notation MATH so that the equation of MATH near MATH is given by MATH, or equivalently, by MATH. We first differentiate the identity REF in MATH. Using the chain rule we obtain the identity in matrix notation MATH where MATH. (Observe that MATH in REF depends on the map MATH). The invertibili... |
math/0002186 | We prove the theorem by induction on MATH. We start first with the case MATH and assume that MATH is a MATH-equivalence between MATH and MATH with MATH. By conjugating REF with MATH replaced by MATH we obtain MATH with MATH and MATH as in REF . If we observe that MATH, we conclude that the second term on the right hand... |
math/0002186 | Suppose that MATH is a formal equivalence. If MATH is convergent, it is clear that MATH is also convergent. Conversely, if MATH is convergent, then the first term on the right hand side of REF is a convergent power series in MATH by composition. Since MATH is a MATH-equivalence for every MATH, the remainder term is MAT... |
math/0002186 | We use REF for MATH and substitute MATH for MATH, where MATH is given by REF . The corollary easily follows by taking MATH and MATH. |
math/0002186 | We shall take MATH to be the NAME number of MATH at MATH. In the notation of REF we take MATH, MATH and set MATH where MATH is defined by REF . Here MATH and MATH. Observe that MATH is a linear automorphism of MATH. It follows from REF that MATH. Furthermore it follows from REF that REF also holds. Hence we can apply R... |
math/0002186 | We continue to work with normal coordinates near the origin MATH for MATH. We fix a local parametrization of MATH at MATH of the form MATH . Let MATH be a defining function for MATH near MATH and MATH be given by REF . For MATH, MATH and MATH, we consider the functions MATH . It follows from the properties of MATH, the... |
math/0002186 | Let MATH, MATH, be given. It follows from the definition of the ring MATH that MATH can be viewed as an element of MATH. By REF , given MATH, there exists MATH such that if MATH and MATH are as in the corollary, there exist MATH, MATH satisfying REF . We may now apply REF with MATH (and hence MATH) to conclude that the... |
math/0002187 | If MATH has nonzero square, then, since MATH, MATH. But MATH is a free MATH module generated by MATH. So if MATH has no MATH-torsion, then MATH must be MATH. And then, of course, MATH, as well. Thus MATH. Since MATH, the sequence collapses. Now suppose MATH. Then for each generator MATH, there is a MATH which is linear... |
math/0002187 | Suppose first that MATH acts on MATH. By REF , MATH is a free MATH module on MATH generators. Recall that MATH, where MATH and MATH generate MATH. Since MATH is a polynomial ring, it contains no zero-divisors, so it makes sense to localize at the set MATH consisting of all of the nonzero elements. We check easily that ... |
math/0002187 | If each component of MATH contains at least three spheres, then each sphere intersects its neighbor geometrically once, and the claim follows. If some component contains exactly two spheres, then each intersects the other twice. One of them might, a priori, represent a multiple of two in MATH. But REF cited above impli... |
math/0002187 | Recall that MATH while for MATH odd, MATH . Let MATH denote the projection MATH. The NAME spectral sequence of the pair MATH has MATH . On the other hand, MATH is a free MATH-space, so MATH is canonically isomorphic to MATH. Since MATH is a relative manifold pair, NAME duality gives a commutative diagram: MATH . But MA... |
math/0002187 | From the homology spectral sequence of the covering MATH, we obtain a short exact sequence MATH. As we have already seen, MATH is multiplication by MATH in each factor. It follows that MATH is MATH-torsion-free. So in fact all classes in MATH with MATH are mortal. In particular, MATH must vanish, so MATH, so MATH. Supp... |
math/0002187 | For convenience, we continue to assume the action is not pseudofree, or fixed-point-free in the case of MATH. As above, let MATH denote the singular set with two homologically redundant spheres removed. For simplicity of notation, we assume MATH is connected; if it isn't, our argument will carry through on each piece. ... |
math/0002187 | We begin with a slight variant of the plumbing construction of the previous section: Let MATH, and label the spheres consecutively around MATH as MATH. Let MATH denote the ``north pole" of MATH. Choose orientations for each of the MATH, and let MATH denote the corresponding fundamental class. Finally, choose an orienta... |
math/0002188 | By NAME duality if MATH is a root of MATH then MATH is also a root of MATH. Hence if we write MATH it follows that MATH . Let us set for brevity MATH. From REF we get: MATH hence MATH . By REF we have MATH which yields MATH as desired. |
math/0002188 | Given any MATH let MATH be the geodesic with initial condition MATH. Let MATH be the parallel transport along MATH from MATH to MATH. Let MATH be an orthonormal basis of MATH by eigenvectors of the symmetric transformation MATH. Let MATH be the parallel vector field along MATH with initial condition MATH. Given MATH we... |
math/0002189 | Take REF and write MATH as: MATH . The integral converges if: MATH, and in this case, it converges to the constant MATH, independent of MATH. Absorbing this into the main constant, the result for MATH is created. In the case MATH, REF is MATH . Thus again, the error is bounded by a constant, this time, dependent on MAT... |
math/0002190 | Given equations are already satisfied on the disk for the vector fields MATH, MATH. Further at the points of the disk we define transversal to this disk vector fields MATH, MATH, MATH, in such a way that all the union of MATH vectors forms a basis at each point and also that REF is satisfied. Upon constructing the need... |
math/0002190 | Since the commutators of vector fields along MATH are determined by their MATH-prolongations outside MATH, we write the general form for a MATH-prolongation of the vector field MATH: MATH where MATH is the submodule of the module of vector fields, consisting of the vector fields vanishing on the submanifold MATH to the... |
math/0002190 | Perturb the almost complex structure MATH in a neighborhood of the disk MATH so that it coincides with the standard integrable structure MATH near the boundary of this neighborhood: for every MATH there exists such an almost complex structure MATH that MATH in a small neighborhood of the disk MATH and MATH in a neighbo... |
math/0002190 | For a small neighborhood MATH of the point MATH the estimates of REFa of the paper CITE imply existence of a number MATH, dependent only on the almost complex structure MATH and the neighborhood MATH, such that for every MATH, MATH, MATH, and MATH there exists a pseudoholomorphic disk MATH such that MATH, MATH. Setting... |
math/0002190 | The inequality MATH is equivalent to the statement of REF because MATH, where the lower bound is considered over all mappings MATH, such that MATH . |
math/0002190 | The inequality MATH is evident because MATH, where the lower bound is taken over all pseudoholomorphic mappings MATH, MATH, and the norm is count with respect to the NAME metric. Let us prove the reverse. We follow the NAME 's proof CITE. Let MATH be a smooth curve from a point MATH to a point MATH such that MATH. Due ... |
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