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math/0105041 | In the local case of quaternion Hermitian NAME MATH, we have to explain how to project the structure of MATH over MATH. The key point is that locally, MATH has admissible basis formed by MATH-parallel (hence integrable) complex structures. Then formulae REF show that MATH is projectable. The foliation being Riemannian, MATH is also projectable. The compatibility of the projected quaternion bundle with the projected metric is clear. To show that the projected structure is quaternion NAME, let MATH be the MATH-form of the projected structure. As the projection is a totally geodesic Riemannian submersion, MATH coincides with the restriction of MATH to basic vector fields on MATH. Hence, it is enough to show that MATH on basic vector fields. But MATH and the result follows from REF . The scalar curvature of MATH is easily computed using NAME formulae. The global case of a hyperhermitian NAME MATH now follows. |
math/0105041 | Fix MATH be a compatible complex structure on MATH. Then MATH is a homogeneous NAME manifold and by REF we have the regularity of both the foliations MATH and MATH. Therefore MATH projects on homogeneous manifolds MATH and MATH. In particular the projections of MATH on MATH are regular Killing vector fields. Then REF assures that REF-dimensional foliation spanned by the projections of MATH is regular. This, in turn, implies that MATH is a homogeneous manifold, thus MATH is regular on MATH. |
math/0105041 | Let first MATH be a regular vector field. Accordingly, MATH is compact locally MATH-Sasakian manifold, NAME with positive scalar curvature. From NAME theorem, its Riemannian universal cover MATH is compact and MATH is finite. Hence, the pull-back MATH (see REF ) is trivial and MATH is globally MATH-Sasakian. Let now MATH be the pull-back of the MATH-bundle MATH: being a flat principal circle bundle over a MATH-Sasakian manifold, REF provides a hyperhermitian NAME structure on MATH. By construction, this one projects on the quaternion Hermitian NAME structure of MATH. In the weaker assumption that MATH has only compact leaves (it is a quasi-regular foliation), the leaves space MATH is a compact orbifold with same Riemannian properties as above. Its universal orbifold covering MATH is a complete Riemannian orbifold with positive NAME curvature. According to REF, the diameter of MATH is finite. Hence MATH is compact and MATH is finite. Now the pull-back of MATH to MATH is again trivial and, as in the manifold case, one shows that MATH is a globally MATH-Sasakian orbifold. The proof then continues as above. Note that the total space MATH is again a manifold. |
math/0105042 | Suppose MATH and MATH as a graded vector space, where MATH is an ideal in MATH and MATH is a subalgebra in MATH. Consider the left MATH-modules MATH . Then the MATH-modules MATH and MATH (respectively, the MATH-modules MATH and MATH) are naturally isomorphic. The induction functor MATH takes MATH to MATH and provides an inclusion of algebras MATH: MATH . CASE: Consider the following action of MATH on MATH: MATH . The action of MATH on MATH introduced above commutes with the left regular action of MATH. Moreover it is well defined on MATH. Along with the action of MATH on MATH it defines the inclusion of algebras MATH. The first statement is obvious. Let MATH, MATH, MATH, MATH. Note that MATH. Thus we have MATH . To prove the third statement note that MATH . The following statement is crucial in the proof of the Theorem. There exists an isomorphism of left MATH-modules MATH . Let MATH, MATH. Denote the element MATH by MATH. Then such elements form a base of the vector space MATH. We calculate the action of MATH on MATH in this base. For MATH we have MATH . For MATH we have MATH . Now recall that MATH is an ideal in MATH and the actions of the subalgebras MATH and MATH on MATH are given by these very formulas. We return to the situation of REF . Note that each module MATH is in fact a MATH-bimodule by REF . The left MATH-modules MATH and MATH are naturally isomorphic to each other. We prove by induction by MATH that the MATH-modules MATH and MATH are naturally isomorphic to each other. Note that each time the induction hypothesis provides the MATH-bimodule structure on MATH: MATH . Thus by the previous Lemma we have MATH . The Lemma is proved. The statement of the Theorem follows immediately from the previous Lemma. |
math/0105042 | It follows from the definition of the functor MATH that MATH is isomorphic to MATH. Now the statement of the Lemma follows form REF . |
math/0105042 | Essentially one has to construct the quasi-inverse functor for MATH on suitably chosen derived categories. Consider the functor MATH as follows: MATH where MATH denotes taking the maximal MATH-semisimple submodule. CASE: The functor MATH is well defined on the category MATH and is left-exact. CASE: The functor MATH is well defined as a functor MATH. Denote the right derived functor of MATH by MATH . Consider also the functor on the category of MATH-bimodules MATH, where MATH denotes the maximal MATH-locally finite submodule in MATH. CASE: The functor MATH is left-exact. CASE: We have MATH. For any MATH we have a natural map MATH . The map MATH is an isomorphism in the category MATH. First we apply the functor MATH to a NAME module MATH. On the level of vector spaces (not regarding the MATH-module structure) we have MATH . We leave to the reader to check that the above module is isomorphic to MATH and that MATH provides this isomorphism. It follows that for any MATH-module MATH from the category MATH the map MATH provides an isomorphism MATH. Now note that for any finite complex of modules from the category MATH is quasi-isomorphic to a finite complex of modules finitely filtered by NAME modules. This assertion implies the statement of the Lemma. Now the statement of the proposition follows from the Lemma since the functor MATH is quasi-inverse for the functor MATH. |
math/0105042 | First note that there exists a canonical map MATH since MATH has a highest weight vector of the weight MATH. Now apply the functor MATH in the derived category to the modules MATH, MATH and to the map MATH. Evidently by the definition of a contragradient quasi-Verma module we have MATH. The complex MATH is concentrated in homological degree MATH. Moreover the character of MATH coincides with the one of MATH. Since the module MATH is simple, we conclude that the map MATH is an isomorphism in the derived category. Thus it is an isomorphism in MATH. It follows that MATH is an isomorphism itself. |
math/0105042 | Denote by MATH the projection MATH. Then by definition of local cohomology we have MATH . Here we used that the map MATH is smooth of relative dimension MATH. Next note that the map MATH is pro-finite, thus we can replace MATH by MATH in the above formula. We obtain the following map MATH . Here we used the canonical adjointness map MATH. |
math/0105042 | All the assertions of the Lemma follow immediately from REF . |
math/0105042 | Denote by MATH the MATH-th infinitesimal neighborhood of MATH in MATH. Let MATH. Then we have a natural filtration MATH of MATH by MATH. The spaces MATH are isomorphic to MATH with the second grading given by the action of the multiplicative group of MATH. |
math/0105042 | First note that MATH . Now consider the map MATH and use REF . The obtained morphism is the one of MATH-modules because the map MATH is equivariant with respect to the action of MATH by left translations. |
math/0105042 | Let us identify the normal bundle to MATH in MATH (respectively, the normal bundle to MATH in MATH) with MATH (respectively, with MATH). By the previous Lemma we have the map of the filtered objects MATH . Now use REF and consider the morphism of associated graded objects MATH . The morphism MATH is an isomorphism of vector spaces. |
math/0105042 | For the sake of simplicity we prove the Theorem only for the weight MATH (and thus MATH). The proof in the general case does not differ much from the one presented below. We start with reformulating the statement of the Theorem. Consider the action of the group MATH on MATH and the locally closed subset MATH. Here MATH denotes the unique zero dimensional NAME cell. The MATH-modules MATH are naturally isomorphic. To prove the Lemma note only that the natural action of the element MATH of the NAME group on MATH takes MATH to the NAME cell MATH. Now consider the map MATH provided by the action of MATH on the Flag variety. On the other hand MATH can be considered as taking quotient of MATH by the free diagonal action of MATH. There exists a natural map of the left MATH-modules MATH . Here the coinvariants in the LHS are taken along the action of MATH on MATH provided by the diagonal action of MATH on MATH. The map in question is constructed using REF . The obtained morphism respects the MATH-module structures on the LHS and Right-hand side since the map MATH is MATH-equivariant. Remark: Note that we have MATH by the NAME formula. The following statement is well-known. There exist natural isomorphisms of the left MATH-modules MATH . Thus REF compared with REF provides a natural MATH-module morphism MATH . The map MATH is an isomorphism of the MATH-modules. First it is known that MATH is equal to the character of the NAME module MATH. Next it is easy to verify that the MATH-module MATH has the same character. The rest of the proof is left to the reader. REF completes the proof of the Theorem. |
math/0105042 | This follows from the existence of the NAME complex for MATH in this case and from the identifications MATH and MATH. |
math/0105042 | We prove the Theorem providing an explicit description of the commutative square in terms of local cohomology. Note that in the MATH case MATH. Consider the diagonal action of the group MATH on MATH. Here MATH acts on the first factor of the product via the right translations. It acts on the second factor via the projection on its NAME factor MATH. We have MATH. Consider the local cohomology spaces MATH . We have CASE: MATH CASE: MATH . Consider now the projection MATH. There exist natural maps of MATH-modules CASE: MATH CASE: MATH . Both statements of the lemma follow from REF . The maps from the above Lemma provide the isomorphisms: MATH . Now all the four maps in the square in question are constructed, it remains to check that the square commutes. This follows from the functoriality of NAME complex (via the inclusion MATH). The surjectivity of the vertical arrows follows from the one in the MATH case and from the exactness of the induction functor. |
math/0105042 | We prove the Theorem providing an explicit description of the commutative square in terms of local cohomology. The proof goes along the lines of the one of REF . Consider the action of the group MATH on MATH and in particular the orbit of the one dimensional NAME cell MATH under the action. When twisted by the action of MATH the set MATH coincides with the NAME cell MATH. Again, like in the proof of REF , consider the diagonal action of the group MATH on MATH. Consider the local cohomology spaces MATH . We have MATH . Consider now the projection MATH. There exist natural maps of MATH-modules CASE: MATH CASE: MATH . Both statements of the lemma follow from REF . The maps from the above Lemma provide the isomorphisms: MATH . Now applying the functor MATH to the isomorphisms from the above Lemma we obtain isomorphisms as follows: MATH . It remains to check that the square in question commutes. We use the same arguments as in the proof of REF . |
math/0105042 | Note that any two neighboring elements MATH and MATH differ by right multiplication by a simple reflection and their lengths differ by REF. Now apply REF . |
math/0105042 | The statement of the Lemma follows from the general combinatorics of the NAME group. |
math/0105042 | We begin the proof of the Theorem with generalizing the complex MATH. In fact the corresponding statement is contained in REF. Recall that the Flag variety MATH as well as the NAME cells are defined over MATH. Denote the corresponding objects by MATH and MATH. The standard line bundles MATH are also defined over MATH. The following statement is due to NAME. CASE: There exists a complex of free MATH-modules MATH with MATH . CASE: Higher cohomology groups of the complex vanish and MATH. CASE: For any finite field MATH we have MATH. Remark: The complex from the above Proposition is called the global NAME complex for MATH on MATH. Note also that MATH equals the contragradient NAME complex for the regular dominant integral weight MATH. We prefer to work with the contragradient dual complex MATH. The duality makes sense since MATH consists of MATH-free modules graded by the torus action so that the grading components are of finite rank over MATH. For a given element of the NAME group MATH (and the fixed reduced expression of MATH via simple reflections MATH) we have the chain of maps MATH given by the components of the differentials in MATH. All the arrows in the chain are embeddings. The corresponding statements over all the fields MATH are proved above. It follows that the assertion of the Lemma holds since the property MATH is a closed one. The MATH-modules MATH do not depend on the choice of the reduced expression for MATH. This also follows from the corresponding statement for the finite fields. Let us now sum up what we have got already. We have a number of families over MATH of submodules in MATH that is viewed as a quasicoherent sheaf over MATH. Each family is enumerated by an element of the NAME group. We know also that over the generic point of MATH the fiber of the family enumerated by MATH is a submodule in the fiber of the family enumerated by MATH if and only if MATH follows MATH in the NAME order. This is a well-known property of the NAME resolution. But the property MATH is a closed property in a flat family. Thus we obtain the statement as follows. For any finite field MATH we have and a pair of elements MATH of the NAME group such that MATH follows MATH in the NAME order we have MATH as submodules of MATH. This essentially finishes the proof of the first part of the Theorem. We leave the proof of the second part to the reader. |
math/0105042 | First one checks that for a regular dominant integral weight MATH the module MATH is isomorphic to MATH and thus the algebra MATH acts on it. Next note that all the modules MATH can be viewed as submodules in MATH, thus they are acted by MATH as well. Finally the components of the differentials in MATH are just the inclusions MATH for MATH, such that MATH and MATH follows MATH in the NAME order. |
math/0105042 | Formally note that both complexes look as follows. As a MATH-module each of them is a direct some of certain MATH-modules enumerated by the NAME group, and the characters of the corresponding modules in the direct sums coincide. The following statement is the key one in the proof of the Theorem. The specialization of the quasi-Verma module MATH into MATH is isomorphic to the MATH-quasi-Verma module MATH. Again we use the technique of semiinfinite cohomology with few comments and with references to CITE. Note that the algebra MATH posesses a triangular decomposition MATH on the level of MATH-vector spaces. This decomposition comes as the specialization of the one of MATH into the field MATH. We have CASE: The semiregular modules MATH . CASE: We have the isomorphism of algebras MATH. CASE: For any MATH-free module MATH we have an isomorphisms of the MATH-modules MATH . Thus we are to compare MATH and MATH. Let us spell out the definition of the last module in down-to-earth terms. The MATH-module MATH is isomorphic to MATH . Here MATH denotes the restricted dual module for MATH. There exists a natural map of MATH-modules MATH . Below we cheat a little forgetting about MATH in the definition of MATH. Note that we have a canonical identification MATH . Now we use the following statement. There exists a natural MATH-module map MATH . Thus we obtain a canonical element MATH . This way we obtain the required map MATH. The map MATH is an isomorphism of MATH-modules. Now the statement of REF follows from REF . It remains to compare the differentials in the complexes MATH and MATH. To do this note that the components of the differentials in the complexes MATH and MATH are obtained by the same algorythm. Namely we have two lattices of submodules in MATH. Both lattices are enumerated by the NAME group and the inclusions of submodules in the lattices agree with the NAME order on MATH. In both cases for MATH, such that MATH and MATH follows MATH in the NAME order, we obtain the component of the differential in the corresponding complex by taking the inclusion map in the corresponding lattice. Now the statement below follows form REF and from the construction of the quasi-BGG complex from the beginning of the present section. The lattices of submodules MATH in MATH coincide. The Lemma completes the proof of REF . |
math/0105043 | Choose MATH so large that if MATH then MATH for any MATH . Then, MATH when MATH and MATH when MATH . We define two subsets of the MATH axis: MATH . These sets are clearly non-empty (for example, MATH) and open. They are disjoint because MATH cannot have a local maximum in the region MATH or a local minimum in the region MATH . Hence there is a MATH and this corresponds to a solution which is bounded on MATH . This proves REF . |
math/0105043 | Suppose that there are two bounded solutions, say MATH and MATH and let MATH . Then MATH and since MATH is positive definite and MATH cannot be bounded on MATH. This proves the result. |
math/0105043 | Existence of these solutions is a simple application of the shooting technique from REF . It is easy to show that if MATH then at MATH where MATH has a local minimum, MATH . We consider solutions MATH with MATH . For MATH in this region and close to MATH decreases and crosses MATH transversely. For MATH in this region and close to MATH increases and crosses MATH, again transversely. It easily follows that there is a solution MATH such that MATH for all MATH . Similarly, a second solution is found with MATH . Uniqueness of both of these solutions follows as in REF . |
math/0105043 | In proving that the solutions of REF are periodic, we give a second proof of their existence, for the case MATH. Let MATH. First observe that MATH while MATH. Therefore MATH decreases monotonically to MATH. If MATH, then MATH and so MATH initially increases. However, for sufficiently small positive values of MATH, MATH has a first zero, which we denote by MATH. This means, in turn, that MATH must have a first zero at MATH. For small MATH . Further, MATH . From MATH we conclude that for small MATH on MATH . Hence, MATH . Clearly, MATH approaches MATH as MATH tends to MATH from above. Extend the function MATH to larger MATH by continuity, as long as possible. That is, MATH is the continuous function such that MATH and, for MATH sufficiently close to MATH, MATH is the first positive zero of MATH. Then MATH remains the first zero of MATH until either REF MATH (since we can then no longer use the implicit function theorem to solve MATH for MATH) or REF there exists a first MATH such that MATH and MATH (since then MATH ceases to be the first positive zero of MATH). But if REF occurs at some first MATH, and MATH is the first zero of MATH with MATH then MATH . This implies that MATH to the left of MATH, a contradiction. If REF occurs we get the same contradiction at MATH . If REF occurs at MATH and MATH then MATH and MATH . This means that MATH on either side of MATH and MATH could not be the first zero of MATH . Hence MATH is continuous and remains the first zero of MATH as long as MATH . On the other hand, if MATH, then the first maximum of MATH is larger than MATH. Therefore, MATH must vary continuously as MATH increases from MATH until it eventually takes on the value MATH for some MATH. For any such MATH, MATH gives a periodic solution of REF with the properties that MATH for MATH and MATH for MATH. From REF , it follows that such a MATH is unique and the corresponding solution MATH coincides with the unique bounded solution with MATH and MATH in REF . The reflection MATH of this periodic solution gives a second periodic solution, which corresponds to the unique bounded solution with MATH and MATH in REF . The proof that MATH and MATH are the minimal and maximal bounded solutions of MATH on MATH follows from that of REF . To obtain the third periodic solution, MATH again let MATH the smallest root of MATH . We saw that if MATH then MATH for all MATH and all MATH and MATH decreases monotonically to MATH, (in finite time). If MATH so that MATH then MATH and again MATH decreases below MATH and tends to MATH . On the other hand, if MATH is sufficiently small in magnitude then MATH and it is easy to show that MATH increases to cross zero at some first MATH with MATH on MATH . Lowering MATH from MATH the function MATH is continuous as long as MATH. If MATH then MATH since MATH cannot have a local minimum at MATH . This implies that MATH . Since MATH is not defined for MATH, it must increase beyond MATH as MATH decreases from zero. Thus there is some first (largest) MATH with MATH . Let MATH . An easy symmetry argument shows that MATH for all MATH showing that MATH is MATH-periodic, with a maximum at MATH. Suppose, therefore, that MATH satisfies MATH for some first MATH . By the anti-symmetry of MATH around MATH we can assume that MATH . Then MATH . If MATH then MATH so MATH becomes negative. In either case MATH on some interval to the right of MATH. From the graphs of MATH it is clear that if MATH on MATH and MATH before MATH then after that point, MATH at least until MATH . This implies that MATH on MATH . The uniqueness of MATH follows from REF below. Assume that MATH. If MATH and MATH in MATH, then MATH in MATH. First, using the NAME technique we get MATH . Assume that the lemma is false. Then, there is a first MATH such that MATH in MATH and MATH at MATH. Then, MATH. Evaluating REF at MATH shows that the left side is nonpositive and the right side is positive. This contradiction proves the lemma and completes the proof of REF . |
math/0105043 | From REF it suffices to show that MATH for MATH. If MATH, then REF implies MATH for all MATH so that MATH has a jump at MATH, a contradiction. Hence we have MATH. Assume now that the corollary is false. Since MATH in MATH, there is a MATH such that MATH and MATH. The NAME technique shows that MATH . If MATH for some largest MATH then MATH in MATH a contradiction. We again get that MATH has a jump at MATH and this proves the corollary. |
math/0105043 | Suppose that MATH and let MATH . It will be helpful to consider the graph of the set of solutions of MATH shown in REF . Starting from MATH we consider the solution MATH as MATH is lowered. Let MATH denote the first intersection of this solution with MATH . Then near MATH, MATH is defined and continuous. Further, MATH on MATH as long as MATH . The solution cannot be tangent to the line MATH until we reach a value of MATH say MATH where MATH . Suppose that MATH . Since MATH it follows that MATH and we easily calculate that MATH while MATH . But this implies that MATH at some earlier point, a contradiction. Therefore MATH. A similar argument shows that there is a MATH such that MATH and MATH. If MATH then MATH (or MATH is not defined, which is easily handled), while if MATH then MATH. Therefore a simple continuity argument shows that there are three values of MATH such that MATH, and this give three MATH-periodic solutions of REF. Further, this argument extends by continuity to values of MATH close to MATH . That is, keeping MATH and MATH fixed, we still have MATH or undefined, and MATH or undefined. This proves REF . |
math/0105043 | We define two functions, MATH and MATH as follows: MATH A key observation is that no solution can intersect the curve MATH tangentially from above, or the curve MATH tangentially from below. This is because MATH when MATH when MATH and no two distinct solution graphs can be tangent to each other. Also, no solution can be tangent to either of the lines MATH from between these two lines. Further, we will need two other functions, MATH and MATH, where MATH . These are defined by MATH . As in the figure below, the function MATH can be described as the function whose graph is the line MATH except at downward bumps when this line meets a MATH with MATH even. Also, the graph of MATH is the line MATH except for upward bumps when this line meets the graph of MATH for some odd MATH . MATH-The functions MATH and MATH are reflections and translations of MATH and MATH . Recall that in REF we required that MATH . No solution can be tangent to MATH from above or to MATH from below. Also, no solution can be tangent to MATH from above or to MATH from below. CASE: Graphs of MATH and MATH . Assuming that the sequence MATH is chosen, we shall define a sequence of closed intervals MATH with the following properties: (See REF ). MATH . For each MATH implies that MATH intersects MATH and no other MATH with MATH . MATH . Let MATH and for MATH let MATH denote the first MATH such that MATH . Then either MATH and MATH or MATH and MATH . MATH . Since the intervals will be closed, their intersection will be non-empty, and a point MATH will have the properties stated in the Theorem. We do not prove that the intersection is only one point. We will be able to do this for sufficiently small MATH. We will choose MATH as a subinterval of MATH . Suppose, first, that MATH is even. For MATH close to MATH intersects MATH before MATH, and therefore before MATH could intersect MATH . Let MATH denote the first intersection of MATH with MATH . Then MATH is continuous in some maximal subinterval MATH of MATH . However, MATH for solutions MATH with MATH close to MATH do not intersect MATH at all. (Instead they decrease monotonically to below MATH and then to MATH.) The nontangency of MATH with MATH implies that MATH . We do not know if MATH is monotone increasing. However, there must be an interval MATH such that MATH, and MATH in MATH on MATH . As long as MATH is continuous, it must remain the first intersection of MATH with MATH because of the nontangency of MATH with MATH . Further, while MATH is continuous the solution MATH cannot intersect MATH in MATH because of non-tangency with MATH. A similar argument is used to construct MATH if MATH is odd. In this case we have MATH and MATH. We now assume that a (decreasing nested) sequence of closed intervals MATH has been constructed with properties MATH for MATH . We wish to construct MATH . We will consider the case where MATH is even and MATH is odd. We will also assume that MATH while MATH. We will show that we can construct MATH so that MATH and MATH, or we could make a different choice which would result in MATH . (Hence, for any given sequence MATH there will be many solutions with the properties MATH . This does not by itself imply a reduced sensitivity to initial conditions, because these solutions may be separated from each other. This is discussed in a separate section below. ) The assumption that MATH is even means that the function MATH is a downward pointing spike and MATH while taking MATH odd means that MATH is an upward pointing spike. CASE: Several solutions used in the shooting process, with the scale MATH . Solutions shown are MATH-and MATH together with MATH and MATH . To define MATH in the case where MATH is even, let MATH . Then MATH for MATH , and MATH . The last inequality is true because MATH does not intersect MATH, MATH-an upward pointing spike, and MATH. (If we wished to have MATH we would set MATH . In this case MATH for MATH , and MATH . Slight changes are necessary if MATH is odd.) Consider values of MATH near to MATH . Since MATH and MATH we can define MATH continuously by the equations MATH . Then MATH will be continuous in some maximal interval around MATH . However, as MATH increases from MATH must tend to infinity, since it is not defined at MATH where the solution MATH reaches MATH at MATH and never decreases below MATH after that. Let MATH . The interval MATH will be a subinterval of MATH . Observe that MATH crosses MATH from above, at its maximum point, at MATH . This is also an intersection of MATH with MATH . Let the intersection of this solution with MATH be denoted by MATH, and extend MATH as a function of MATH continuously for MATH, MATH being defined as the intersection of MATH with MATH as long as it remains continuous. The solution MATH may possibly intersect MATH at earlier points, but MATH is defined uniquely by requiring that MATH . We see that as MATH increases from MATH the function MATH must eventually be undefined, since it is not defined at MATH . So, it must increase (not necessarily monotonically), and there must be some closed subinterval of MATH in which MATH lies in the interval MATH and moves from the left end of this interval to the right end (not necessarily monotonically) as MATH increases. This subinterval of MATH is chosen for the interval MATH. We define it unambiguously by choosing the subinterval with the given properties which lies nearest to MATH . The possibility of an infinite set of subintervals with these properties is precluded by bounds on the variables for a given MATH. The construction of MATH shows that it satisfies REF. A similar construction will give MATH in the case where MATH and MATH are both even. In this case we can obtain MATH by decreasing MATH from MATH and observing that the intersection of MATH with MATH must tend to MATH before we reach MATH. The case where MATH is odd is also handled similarly. At each step we can obtain MATH if MATH is even and MATH if MATH is odd. This completes the induction step and the proof of REF . |
math/0105043 | The proof is a refinement of the proof of REF . The notation is different, however, for in the statement of REF all of the MATH are included, not just those picked out by some index set MATH . As before, MATH will be the endpoints of the graphs of MATH. The functions MATH and MATH and MATH are the same as before. Also, MATH will denote the first intersection of MATH with MATH . Rather than carry out a formal induction, we will show how to construct a solution corresponding to a sequence beginning with MATH or MATH or MATH . Hence we want a solution which misses MATH intersects MATH misses MATH and MATH intersects MATH, and does not intersect MATH . It will then be apparent how to deal with the general case. In the notation of REF we are starting with MATH . If MATH then we proceed in the same way as in the proof of REF . The induction step there going from MATH to MATH produced an interval MATH in which the solutions MATH do not cross MATH in MATH . Further, the solutions in this interval will not intersect MATH or MATH and will intersect the two endpoints of MATH when MATH is at the endpoints of MATH. The refinement of this interval according to whether MATH or MATH will be similar to the way we handle the case MATH so we turn to that case now. When MATH we follow the proof of REF up to the definition of MATH . However now we set MATH . Then for MATH intersects MATH to the right of the first point where MATH . We now let MATH be the intersection of MATH with MATH defined by continuity from the point MATH . Set MATH and MATH . These are defined because MATH does not intersect MATH after MATH and because MATH varies between MATH and MATH as MATH moves from MATH to MATH . Then MATH and for MATH intersects MATH and also intersects MATH in the interval MATH . The solution MATH intersects MATH at its minimum point, at MATH and from there increases to cross MATH before MATH . (This is shown by a comparison with MATH as in the proof of REF .) The solution MATH intersects MATH twice, intersects MATH at its maximum point and then decreases to cross MATH before MATH . Following the same procedure as in the proof of REF , we lower MATH from MATH and find an interval MATH such that MATH and for MATH intersects MATH in MATH and also intersects MATH . Other cases, and succeeding steps, being similar, this completes the proof of REF . |
math/0105043 | We already know that MATH and MATH which exist by REF , are increasing in MATH . We give two proofs of the existence of MATH and MATH. The first is a little shorter, while the second shows in addition that MATH and MATH are increasing on MATH . Proof REF: We will find a solution MATH with the property that MATH, MATH and MATH intersects MATH . (None of the three solutions MATH and MATH have all of these properties.) We assume in REF that MATH . Then in the construction used in the proof of REF we find MATH such that MATH on MATH and MATH . We also find MATH such that MATH and necessarily, MATH . If we decrease MATH from MATH we come to a first (largest) MATH, say MATH where MATH first intersects MATH at MATH . At that point, MATH . Somewhere between MATH and MATH there must be a MATH with MATH . This solution must also intersect MATH and this gives the desired fourth periodic solution (beyond those constructed in REF .) The fifth is obtained by translation and reflection. From REF , it follows that MATH . Here is a second proof: If MATH, then MATH and MATH for MATH, and MATH in MATH. MATH-Also, MATH intersects MATH in MATH, and it follows that for MATH and sufficiently close to MATH is also positive on MATH and again MATH intersects MATH at some point in MATH . Let MATH . Then MATH since MATH does not intersect MATH at all. By continuity, and since no MATH is tangent to MATH intersects MATH in MATH and MATH in MATH. The first intersection of MATH with MATH is either at MATH or in MATH. In the first case MATH which implies MATH and so MATH just before MATH, a contradiction. Hence MATH first intersects MATH in MATH. Then by the definition of MATH, it follows that MATH at some first MATH. We claim that MATH. If MATH. then MATH for otherwise, MATH and this implies that MATH just after MATH, a contradiction. So MATH which gives MATH just before and after MATH, again a contradiction. This shows that MATH so that MATH is a periodic solution which satisfies MATH and MATH in MATH . Further, MATH intersects MATH in MATH, which implies that MATH. Let MATH. From REF , it again follows that MATH . |
math/0105043 | Let MATH . Since MATH, it follows that MATH is well defined. Then for MATH, MATH and then REF gives MATH. Multiply this inequality by MATH and integrate over MATH to give MATH and hence, for MATH, MATH . This implies that MATH and MATH (recall that MATH). If MATH, then MATH, which proves REF for MATH. Assume that MATH and write MATH where MATH and MATH . It follows from REF that MATH and therefore REF follows from the definition of MATH and the assumption that MATH. |
math/0105043 | Consider the case MATH even. The branch MATH is strictly negative in MATH . Further, given a sufficiently small MATH depending on MATH and MATH the quantity MATH is negative and the quantity MATH is positive. It follows MATH can be chosen so that if MATH for some MATH and MATH then MATH crosses MATH as MATH increases in MATH, while if MATH then MATH crosses MATH in MATH as MATH decreases. Similarly, if MATH then MATH will cross zero before MATH leaves MATH either forward or backward depending on the sign of MATH . The case MATH odd is similar. This proves REF . |
math/0105043 | It is easy to check that there are MATH and MATH such that if MATH and MATH, then MATH for all MATH. Assume for contradiction that there are MATH and MATH such that MATH at the point MATH. Assume first that MATH. Let MATH. Since MATH in MATH, MATH and MATH, it follows that MATH, MATH and MATH for MATH. Then for MATH we find that MATH and so for MATH we obtain MATH . Assume that MATH. Evaluating REF at MATH and MATH and using MATH gives MATH . Since MATH and MATH, it follows that the above inequality does not hold if MATH is small. Therefore, MATH can be chosen independent of MATH so that for MATH, MATH if MATH. Then let MATH and MATH. Apply the same argument for MATH as above to get the same contradiction if MATH. If MATH then letting MATH, we see that the right side of REF goes to MATH, contradicting the boundedness of the left side of REF . The other assertions of REF can be proved similarly. CASE: Since MATH is the minimal solution of REF , it follows from REF that to prove MATH it suffices to show that MATH in MATH. We can choose MATH so that for MATH, MATH where MATH. Assume that there is a MATH and MATH such that MATH. Then, as above, by considering MATH for increasing MATH if MATH and for decreasing MATH if MATH, we obtain that MATH if MATH respectively, a contradiction. This proves REF , and REF can be proved similarly. |
math/0105043 | The inequalities in REF follow from REF of the above lemma. To show REF we let MATH, which satisfies MATH . Further, MATH and MATH for all MATH and MATH. Multiply REF by MATH to get MATH . Since MATH, MATH and the derivatives of MATH are all bounded with the bounds independent of MATH, it follows that the right-hand side of REF is bounded by a constant, say, MATH over MATH, that is, MATH for MATH and MATH. Hence REF holds for MATH. Since MATH and MATH are MATH-periodic and even functions, we see that REF holds for all MATH. |
math/0105043 | CASE: Let MATH. Since MATH and MATH in MATH, it follows that MATH in MATH. From REF we see that MATH in MATH. We first assume that MATH at MATH. Then REF holds for this MATH with MATH, and so for MATH, MATH . Since MATH, and MATH if MATH is small, it follows that for MATH, MATH . Let MATH. Since MATH for MATH, REF follows immediately from REF for MATH. Assume that MATH. Let MATH. From what we just proved, we can assume that MATH. Since MATH is decreasing in MATH, MATH in MATH, and MATH in MATH, it suffices to show that REF holds at MATH. Integrating MATH over MATH gives MATH which implies that REF holds at MATH. The proof of the inequality for MATH is similar and therefore is omitted. |
math/0105043 | To show the first part of lemma, from REF it suffices to show that for MATH small, MATH in MATH. Assume that this is false. Then there is a sequence MATH with MATH and a sequence of MATH with MATH for some MATH such that MATH. Then by a phase plane argument we obtain that MATH will reach MATH by time MATH as MATH, contradicting that MATH is bounded. |
math/0105043 | Let MATH. Then MATH satisfies MATH . Then REF follows easily by a phase plane argument. To show REF , we take MATH so large that MATH. Then from REF it follows that if MATH is sufficiently small, then MATH. Therefore, if MATH is small enough to satisfy MATH, then, since MATH in MATH, MATH for MATH. Hence REF follows from REF . |
math/0105043 | We will need the following lemma, which is similar to REF. Let MATH be a solution of REF with MATH. If MATH and MATH are two successive minima (maxima) of MATH with MATH, then MATH. We only show the case that MATH and MATH are successive maxima of MATH and the other case can be proved similarly. Assume that MATH reaches its maximum between MATH and MATH at MATH. Then multiply REF by MATH and integrate over MATH, where MATH and MATH, to yield MATH . Here we use that MATH is decreasing in MATH and MATH and MATH are the inverse function of MATH for MATH and MATH respectively. It then follows that MATH before MATH, which implies that MATH as required. For MATH, recall that MATH are the three solutions of MATH. For given MATH, let MATH . We also need the fact, easily proved, that if MATH is the homoclinic solution of MATH defined earlier, for MATH then the minimum values MATH are decreasing with respect to MATH . Also, for each MATH let MATH be the solution of MATH such that MATH . Let MATH. Choose MATH . When MATH is periodic, and so has MATH local maxima in some interval MATH . For sufficiently small MATH still has at least MATH local maxima in MATH . Hence MATH has at least MATH maxima in MATH . Suppose that these are at MATH . Further, from REF we see that MATH for MATH . If any of the maxima MATH for MATH are such that MATH then because MATH is increasing in MATH we must have MATH in MATH if the solution exists out to MATH and so MATH could not have any more maxima. This shows that for MATH the maxima of MATH must lie in the interval MATH . However, if MATH at some MATH in MATH then MATH so this could not be a maximum. Hence the first MATH maxima of MATH must lie in MATH . The solution MATH must have a minimum at MATH and further minima MATH. Also MATH, by REF . Therefore, MATH for MATH . Now decrease MATH from MATH . Since MATH at a maximum, the maxima are continuous in MATH and so remain above MATH as long as they exist. If MATH has no local minima. But as long as there are, say, MATH local maxima in MATH with MATH and with MATH these maxima and their intervening minima are separated, in that MATH for MATH . This means that the number of maxima cannot decrease until one crosses MATH . This must happen successively for each maximum, which proves REF . |
math/0105043 | In the proof we shall suppress the dependence of MATH on MATH. We first show that MATH. Assume that this is false. Since MATH is bounded, there is a sequence MATH with MATH such that as MATH, the corresponding MATH approaches MATH. For simplicity we assume that MATH. Let MATH. Then MATH satisfies MATH . Let MATH be the solution of MATH . Since MATH, it follows from the continuity of solutions with respect to parameters that for any given MATH if MATH is defined on MATH, then MATH uniformly for MATH. Assume that MATH. Then there is a MATH such that MATH for MATH and MATH. It follows that for MATH sufficiently small MATH crosses MATH, which is impossible. Hence MATH. Then MATH is a periodic function with a period MATH and so MATH has MATH maximum in the interval MATH. Hence by continuity, for MATH sufficiently small, MATH also has MATH maximum in the interval MATH , which implies that MATH has MATH maxima in MATH for MATH sufficiently small, contradicting the assumption on MATH. Therefore, MATH and so REF and the first part of REF follow. The rest of REF can be proved similarly. We next show that MATH. Suppose not. Since MATH, there is a sequence MATH with MATH such that MATH for some MATH. Again, for simplicity, we assume that MATH. Since MATH as MATH, it follows from REF that MATH. We suppose now that MATH. Let MATH. Then MATH satisfies MATH . For MATH (as in REF ), let MATH and MATH . Observe that MATH if MATH is sufficiently small. Hence, MATH is defined and MATH; MATH and so MATH is well defined. Assume that MATH is so small that MATH. Then on MATH we have MATH and so MATH. We see that MATH and MATH. Therefore, by the definition of MATH, it follows that MATH, which is impossible since MATH for all MATH. We assume now that MATH. Note that MATH. We shall show that MATH. For if this is false, then the boundedness of MATH implies that there is a sequence MATH with MATH as MATH such that MATH. It then follows by arguments similar to those above that MATH has more than MATH maxima in MATH for large MATH, which is a contradiction. We now claim that MATH for MATH sufficiently small, where MATH is defined in the statement of this Theorem MATH. Assume that this is false. Let MATH. Then in MATH, MATH and MATH. Since MATH and MATH just to the right of MATH we see that MATH and MATH in MATH. Then, as in MATH(where MATH was slightly different), we again get MATH . We assume now that MATH. Since MATH, we use MATH to obtain MATH . This estimate gives a lower bound on MATH as MATH tends to zero. Evaluating REF at MATH and MATH and letting MATH shows that the right side of REF goes to infinity, while the left side is bounded. This contradiction proves the claim if MATH . We now assume that MATH. We observe that MATH . Then evaluating REF at MATH and MATH and letting MATH, the same contradiction will be obtained. This shows that MATH as claimed. In particular, MATH and so MATH. Continuing with our proof that MATH and under the assumption that this is false and MATH let MATH. Then MATH satisfies MATH . We shall show that MATH, where MATH is the homoclinic solution of MATH with MATH and MATH . Suppose that MATH. Since MATH and MATH is bounded, which is easily verified, there is sequence of values of MATH and corresponding solutions MATH with corresponding MATH such that MATH approaches a number MATH with MATH. We first assume that MATH. Let MATH be the periodic solution, with period MATH to MATH . Since MATH, MATH approaches MATH as MATH uniformly in compact intervals of MATH and it follows that for sufficient small MATH, MATH oscillates more than MATH times in MATH, and so MATH has more than MATH maxima in MATH, a contradiction. Hence we can assume that MATH and again assume that MATH solves REF . Then there is a MATH such that MATH for MATH and MATH, which implies that for MATH sufficiently small, MATH, again a contradiction. Therefore, as MATH, MATH and MATH goes to MATH uniformly in any compact interval. Since MATH is homoclinic to MATH, it follows from continuity that after MATH, MATH will return to the any given neighborhood of MATH before MATH for some MATH independent of MATH. Then, since MATH, MATH increases and stays close to MATH till MATH, which implies that MATH and so MATH for MATH small, contradicting REF . This shows that MATH for MATH. Then, MATH is the heteroclinic solution connecting MATH as MATH to MATH as MATH. So, as MATH the point MATH reaches a point as close to MATH as we like, and then REF implies that MATH and MATH. For MATH, again by REF . Further, MATH remains close to MATH until MATH is close to MATH for otherwise there would be similar contradictions to those obtained above. Hence, MATH must, after its last minimum, follow MATH until close to MATH and then , since its last maximum is above MATH, follow MATH. The bound on MATH follows in the same way as the bound for MATH. For the final statement in the Theorem, we use REF . This proves the theorem for MATH . The proof for MATH is similar. |
math/0105043 | From REF we have MATH . CASE: Let MATH. It follows from REF that for MATH, MATH from which the assertion in REF follows. CASE: From REF , we first have MATH. Evaluate REF at MATH to give MATH so that MATH . This implies that MATH, as desired. |
math/0105043 | Let MATH denote the solution MATH of REF satisfying MATH and MATH. Choose MATH and MATH. By a phase plane argument, there is a MATH such that for MATH, the solution MATH has at least MATH minima and MATH maxima in MATH with all its first MATH minima lying between MATH and MATH and all its first MATH maxima lying between MATH and MATH. Denote the MATH-th minimum and maximum by MATH and MATH. We consider the change in these minima as MATH decreases from MATH. REF shows that all the minima in MATH lie below the line MATH, and so they can neither pass through MATH nor disappear at the middle branch MATH of MATH. Since MATH does not have any minimum at all, it follows that as we decrease MATH to MATH, all the minima will disappear by crossing the lower branch of MATH before MATH. Let MATH . Then MATH and MATH . For MATH small, the MATH-th maximum MATH of MATH exists. Clearly MATH . As before, if MATH, we get MATH for all MATH small, contradicting the definition of MATH. Hence MATH . Therefore, as we raise MATH from MATH, MATH has to move toward to MATH under the lower branch MATH of MATH. This maximum MATH cannot disappear by merger with another minimum in MATH, since this minimum would have to lie above MATH contradicting REF . Since MATH, there must be a point of discontinuity of the MATH-th maximum before reaching MATH-and so there is a MATH such that MATH is continuous in MATH, MATH and MATH. Then MATH is the desired solution. The asymptotic formulas stated in this theorem can be proved in a similar way to that of REF . |
math/0105043 | To prove REF we consider the variational equation and initial conditions satisfied by MATH . These are MATH . We will also be concerned with MATH which satisfies MATH . We observe that MATH when MATH . Multiplying MATH by MATH and MATH by MATH subtracting, integrating by parts and using the initial conditions on MATH and MATH we obtain MATH . If MATH is a solution with MATH which remains in MATH on MATH then MATH on MATH and MATH . Because MATH over the interval MATH, and therefore MATH in this interval, MATH implies that MATH grows exponentially large. More precisely, there are positive numbers MATH and MATH independent of MATH such that MATH in MATH . Also, MATH as MATH . Therefore, for small MATH the right side of MATH is positive as long after MATH as MATH is positive up to MATH . We now show that MATH . If not, then MATH on MATH since MATH when MATH, and MATH in MATH. Therefore MATH. From REF it follows that MATH, contradicting the assumption that MATH remains in MATH . Hence, MATH . This implies that MATH on MATH for if not, then MATH would have a minimum in this interval, and this minimum would lie above MATH contradicting REF . Now suppose in MATH that MATH somewhere in MATH . Then the right side of MATH is positive, while the left side is negative. Hence, MATH on MATH . It is possible that MATH becomes negative somewhere in MATH . Indeed, numerically this is seen to happen. However, we will show that MATH . Suppose that MATH . Then MATH, for otherwise MATH would be positive and increasing on MATH, because MATH there. Thus MATH is positive but decreasing on MATH . From MATH we obtain MATH . Since MATH we get a contradiction from REF in the statement of the theorem, and so MATH could not remain negative on MATH. Therefore, at MATH we have both MATH and MATH and since MATH on MATH both remain positive out to MATH . This proves the lemma. In particular, this applies to the solution MATH . It follows by standard stability arguments CITE that the solution MATH is a stable attractor for the problem MATH on MATH. To prove that MATH is the only MATH-periodic solution which remains in MATH it is convenient to truncate the nonlinearity in MATH by replacing MATH by MATH for all MATH and similarly, if MATH replace MATH in MATH with MATH . This means that all solutions MATH exist on MATH and we can consider MATH to be defined continuously for all negative MATH. We also note that if MATH is a solution which does leave the region MATH then from the point where MATH continuous to increase and cannot satisfy MATH . Recall that there is a unique MATH such that MATH is MATH-periodic and lies entirely below MATH this is the ``minimal" bounded solution, and its graph does not lie in MATH . In considering the possibility of a second MATH-periodic solution, besides MATH which lies in MATH we need only consider MATH . The solution MATH remains in the interior of MATH on MATH and the same is true for MATH if MATH is sufficiently small. However as we raise or lower MATH from MATH we reach values where MATH leaves MATH in MATH . Let MATH be the maximal interval containing MATH such that MATH remains in MATH on MATH for all MATH . MATH is well-defined and closed because MATH is a closed set. If MATH then MATH does not remain in MATH on MATH . : We claim that MATH leaves MATH at MATH and MATH exits MATH at MATH . If not, then one of these solutions is tangent to the boundary of MATH at some MATH . For example, a tangency could occur at MATH . But then, no matter what the slope of MATH is at this point, a phase plane argument shows that for sufficiently small MATH must cross MATH on one direction or the other, and so nearby solutions also leave MATH before MATH contradicting the definition of MATH or MATH . Similar considerations apply at any other possible tangent point in MATH . Consider the case of MATH . By REF , MATH . Hence, for MATH and close to MATH and we cannot, as we lower MATH find a lower MATH where MATH remains in MATH and MATH for at the first such point we would have MATH a contradiction to REF . Similar remarks apply to MATH completing the proof of REF . The proof that if MATH then MATH on MATH and MATH shows that in some neighborhood of MATH . Now, if there is a MATH with MATH, then choose the smallest such MATH. By REF we get MATH at MATH and MATH, a contradiction because these are adjacent zeros of MATH. This contradiction can be reached similarly in the interval MATH, completing the proof of the uniqueness of MATH among solutions with period MATH which remain in MATH . To extend the uniqueness and stability statements to larger intervals we note that starting with MATH, the same analysis allows us to show inductively that MATH remains positive, and MATH is positive at any multiple of MATH . The arguments about uniqueness and stability can then be extended to MATH completing the proof of REF . |
math/0105043 | The corollary follows immediately from the Theorem, the proof of which is more easily understood by reference to REF . We use the original scaling MATH . For any MATH there are MATH and MATH such that MATH on the set MATH . Hence, for sufficiently small MATH no periodic solution can intersect the region MATH at a point where MATH, since then MATH would be forced below MATH before MATH . So any periodic solution must have lie above MATH on MATH . The argument used to prove REF can then be used to show that for sufficiently small MATH is the only periodic solution MATH with a maximum at MATH . This completes the proof of the theorem and Corollary. |
math/0105043 | We have seen that MATH for large negative MATH and also MATH just below MATH . The existence of at least five solutions for MATH follows from the remarks above by showing that if MATH and if MATH, then MATH . Let MATH, and as above, MATH . We see that MATH while MATH satisfies REF . Therefore we obtain, for any MATH . For sufficiently small MATH on MATH . This follows by a slight modification of the proof of REF . Let MATH . Suppose MATH . Then for small enough MATH and MATH there is a MATH such that if MATH then any solution which intersects MATH must decrease monotonically, in at least one direction, to below MATH within a time which is bounded over MATH . By continuity, this is also true for MATH sufficiently close to MATH . As in the proof of REF , no solution other than MATH can have a positive maximum at MATH and not intersect MATH proving the result. Also, if MATH then MATH . Therefore MATH . Suppose that MATH . Then for MATH slightly lower than MATH . Let MATH . Then MATH is well-defined and MATH. Suppose that MATH for some MATH . Since different solutions cannot be tangent, we must have MATH for some MATH but this contradicts the definition of MATH . Therefore MATH and MATH . But in this case, we can lower MATH further, until we find a MATH with MATH . This contradicts the definition of MATH . A similar argument shows that MATH on MATH. Therefore, when MATH . Then MATH and REF show that MATH . Thus, for MATH just above MATH there are at least five solutions (using symmetry). Further, there cannot be a decrease to fewer than five as MATH increases further, for at any point where MATH we would again get MATH . To complete the proof of REF it is convenient to let MATH . Numerically, it appears there is only one MATH-periodic solution between MATH and MATH . Then reflection and translation of MATH-and MATH by the transformation MATH give the additional two asymmetric solutions. Our construction implies that MATH . From REF and the way we define MATH we have MATH for MATH and in this range and close enough to MATH This is because when MATH on MATH . Suppose, however, that for some MATH for some MATH . Let MATH . By the same argument as above we show that MATH and MATH . Hence there is a MATH with MATH . But this contradicts the definition of MATH . This proves that MATH . The proof that MATH is similar, and the construction of MATH and MATH by reflection and translation implies the remaining order relations, namely, MATH . This completes the proof of REF . |
math/0105044 | We prove first the sufficiency. The hypothesis is that for all MATH and for all MATH the relation REF holds. We claim that for any MATH the inequality MATH is true. The ellipticity condition follows then from REF by the choice MATH, for each MATH. Indeed, we have the chain of REF . In order to prove the claim note that MATH implies MATH A straightforward computation which uses the relations MATH gives MATH . The sufficiency part is therefore proven. For the necessity part choose first in the ellipticity condition MATH, MATH. For MATH we obtain MATH, which means the NAME convexity of MATH. (For MATH we obtain MATH, interesting but with no use in this proof.) Next, suppose that MATH and choose MATH for each MATH. The ellipticity condition gives: MATH . Take MATH and get REF, but only for MATH. The expression from the left of REF makes sense for any MATH. By continuity with respect to MATH we prove the thesis. |
math/0105044 | This is a straightforward consequence of the NAME inequality REF MATH and of the NAME convexity of MATH. |
math/0105044 | We have to check the conditions from REF , which gives sufficient conditions on the function MATH in order to satisfy the inequality we are trying to prove. These conditions are: CASE: for any MATH and any symmetric positive definite MATH. This is satisfied by definition of MATH. CASE: for any MATH and any MATH . From the definition of MATH and REF we find that MATH satisfies this condition too. |
math/0105044 | To any MATH we associate it's polar decomposition MATH. For any function MATH such that MATH we shall use the (similar) notation MATH . With the notations from the theorem, we have from the isotropy of MATH, REF that MATH is NAME convex. From REF we obtain the chain of REF . The chain of inequalities continues by using the convexity REF (suppose that MATH): MATH . Now, I claim that the matrix MATH is negative definite. Then, from REF , we find that MATH . We use now the nonincreasing REF to finish the chain of REF Let us see, finally, why the matrix MATH is negative definite. The function MATH is well known polyconcave, hence all the functions MATH satisfy the inequality: MATH . Take now any vector MATH, MATH. Remember that MATH is a symmetric matrix which admits the decomposition MATH hence MATH . Therefore MATH . Use the inequality given by polyconcavity to deduce the claim. |
math/0105044 | Because MATH is NAME convex and for almost any MATH we have the inequality MATH . Use the convexity hypothesis to obtain the desired inequality. |
math/0105045 | Assume that MATH for all MATH. Then, as MATH is a MATH-cover of MATH, so is MATH. In particular, MATH is a large cover of MATH. Now fix any MATH such that MATH for all but finitely many MATH, and let MATH. Then MATH is finite and contains the set of MATH's such that MATH and MATH. |
math/0105045 | REF is immediate. To prove REF , we can apply MATH to MATH-subcovers of the given covers. REF is similar to REF . CASE: Assume that MATH, MATH, are given. Apply MATH to choose finite subsets MATH, MATH, such that MATH. By MATH, there exists a subset MATH of MATH such that MATH. Then for each MATH is a finite (possibly empty) subset of MATH, and MATH. To prove REF , observe that the resulting cover MATH contains an element of MATH, and as MATH is closed under taking supersets, MATH as well. It is clear that reverse inclusion (and therefore equality) hold in REF - REF when MATH. |
math/0105045 | We prove REF . Clearly MATH implies MATH and MATH. On the other hand, by applying the Cancellation Laws REF and then REF we have that MATH . |
math/0105045 | CASE: By the Cancellation Laws REF, MATH. In CITE it was proved that MATH. Thus, MATH, which by the Cancellation Laws is a subset of MATH. The other direction is immediate. REF is similar. |
math/0105045 | We will prove REF (the proof of REF is identical). Assume that MATH satisfies MATH. If MATH is countable then it satisfies all of the properties mentioned in this paper. Otherwise let MATH, MATH, be distinct elements in MATH. Assume that MATH, MATH, are open MATH-covers of MATH. Define MATH. Then MATH is an open MATH-cover of MATH, and thus contains a MATH-cover MATH of MATH. Let MATH be the induced quasiordering. If MATH has a least element, then MATH contains a MATH-cover of MATH. Write MATH. Let MATH be a least element in MATH. Consider the subsequence MATH consisting of the elements MATH such that MATH. Since MATH-covers are large, this sequence is infinite. For all MATH we have MATH, thus MATH for all but finitely many MATH. Since MATH for all MATH, we have that for all but finitely many MATH, MATH. There are two cases to consider. CASE: For some MATH is a least element in MATH. Then MATH contains a MATH-cover MATH of MATH. In this case, for all MATH belongs to all but finitely many members of MATH, thus MATH is finite for each MATH, and MATH is a MATH-cover of MATH. CASE: For each MATH there exists MATH with MATH. For each MATH, MATH is a MATH-cover of MATH, thus MATH belongs to all but finitely many members of MATH. Since MATH does not belong to any of the members in MATH, MATH must be finite. Thus, MATH is a MATH-cover of MATH. |
math/0105045 | See CITE for the clopen version of this theorem (a straightforward usage of REF ). The proof for the NAME case is similar. |
math/0105045 | By REF , MATH. In CITE we defined MATH to be the collection of sets for which every countable clopen MATH-cover contains a MATH-cover, and showed that MATH. But MATH. |
math/0105045 | According to CITE, no continuous image of an analytic set is a tower. In particular, towers are not analytic subsets of MATH. Since NAME images of analytic sets are again analytic sets, we have that every analytic set satisfies MATH. |
math/0105045 | By the Cancellation Laws and REF , MATH. |
math/0105045 | MATH: Assume that MATH is a NAME function, and let MATH. Assume that MATH is centered, and consider the collection MATH where MATH for each MATH. If the set MATH is infinite, then MATH is a pseudo-intersection of MATH and we are done. Otherwise, by removing finitely many elements from MATH we get that MATH is a MATH-cover of MATH. Setting MATH for each MATH, we have that MATH is a NAME MATH-cover of MATH, which thus contains a MATH-cover MATH of MATH. Let MATH, and define a cover MATH of MATH by MATH . Then MATH is a MATH-cover of MATH, and by REF , MATH is linearly quasiordered by MATH. MATH: Assume that MATH is a MATH-cover of MATH. By REF , MATH is centered. Let MATH be a large restriction of MATH which is linearly quasiordered by MATH, and define MATH as in MATH. Then MATH. Thus all elements in MATH are infinite (that is, MATH is a large cover of MATH), and MATH is linearly quasiordered by MATH (that is, MATH is a MATH-cover of MATH). Then MATH is a MATH-cover of MATH. |
math/0105045 | MATH: For each MATH, the collection MATH, where MATH, MATH, is an open MATH-cover of MATH. Assume that MATH is a NAME function from MATH to MATH. By standard arguments we may assume that MATH for all MATH and MATH. Then the collections MATH, MATH, are NAME MATH-covers of MATH. By MATH, there exist finite sets MATH, MATH, such that MATH is a MATH-cover of MATH. Let MATH. Note that for each MATH, MATH. Define MATH by MATH . For all MATH, as MATH is a large cover of MATH, there exist infinitely many MATH such that MATH (that is, MATH). Let MATH be the linear quasiordering of MATH induced by the MATH-cover MATH. Then for all MATH, either MATH or MATH. In the first case we get that for all but finitely many MATH, and in the second case we get the same assertion with MATH and MATH swapped. This shows that MATH satisfies the excluded middle property. MATH: Assume that MATH, MATH, are NAME covers of MATH which do not contain a finite subcover. Replacing each MATH with the NAME set MATH we may assume that the sets MATH are monotonically increasing with MATH. Define a function MATH from MATH to MATH so that for each MATH and MATH: MATH . Then MATH is a NAME map, and so MATH satisfies the excluded middle property. Let MATH be a witness for that. Then MATH is a MATH-cover of MATH: For each MATH we have that MATH, thus MATH is a large cover of MATH. Moreover, for all MATH, we have by the excluded middle property that at least one of the assertions MATH or MATH is possible only for finitely many MATH. Then the first assertion implies that MATH, and the second implies MATH with respect to MATH. |
math/0105045 | MATH, and according to CITE and CITE, MATH. |
math/0105045 | By REF , MATH, thus by REF , MATH. The proof for the NAME case is similar. |
math/0105045 | By REF , MATH. Thus, our theorem will follow from the inclusion MATH once we prove that MATH. To this end, consider a family MATH of size MATH which does not satisfy the excluded middle property, and consider the monotone MATH-covers MATH, MATH, of MATH defined in the proof of REF . Then, as in that proof, we cannot extract from these covers a MATH-cover of MATH. Thus, MATH does not satisfy MATH. |
math/0105045 | Let MATH denote the property that every clopen MATH-cover contains a MATH-cover. Then MATH. By REF , MATH. |
math/0105045 | By REF , MATH . The proof for the NAME case is the same. |
math/0105045 | MATH. |
math/0105045 | By REF , we have that MATH. The proof that MATH is not as elegant and requires a back-and-forth usage of REF . Assume that MATH, and let MATH, MATH, be sets satisfying MATH. Let MATH be a countable open MATH-cover of MATH. Then MATH is linearly quasiordered by MATH. Since each MATH satisfies MATH, for each MATH contains a MATH-cover of MATH, that is, MATH has a pseudo-intersection. By REF , MATH has a pseudo-intersection, that is, MATH contains a MATH-cover of MATH. |
math/0105045 | By REF , MATH, and according to CITE, MATH. By REF , we get that MATH . On the other hand, by REF we have MATH . |
math/0105045 | By REF, assuming CH there exists a set of reals MATH of size continuum such that all subsets of MATH satisfy MATH. (Recall that MATH.) As MATH is closed under taking NAME (continuous is enough) images, we may assume that MATH. For MATH, write MATH for the translation of MATH by MATH. As MATH and only MATH many out of the MATH many subsets of MATH are NAME, there exists a subset MATH of MATH which is not MATH neither MATH. By REF and CITE, for such a subset MATH the set MATH does not satisfy MATH. Set MATH and MATH. Then MATH and MATH satisfy MATH, and MATH does not satisfy MATH. By REF , MATH satisfies MATH and therefore it does not satisfies MATH. But by REF , the set MATH satisfies MATH. |
math/0105045 | Had it satisfied this property, we would have by REF that MATH, contradicting CITE. |
math/0105045 | We modify the aforementioned construction so to make sure that the resulting NAME set MATH does not satisfy the excluded middle property. As we do not need to use any group structure, we will work in MATH rather than MATH. Assume that MATH is an infinite set of natural numbers, and MATH. Then the sets MATH are meager subsets of MATH. For each MATH, the sets MATH are nowhere dense in MATH. Now, MATH, and MATH. Consider an enumeration MATH of MATH which uses only even ordinals. At stage MATH for MATH even, let MATH be the set defined in CITE, and let MATH be the union of MATH and the two meager sets MATH . Then MATH is a union of less than MATH many meager sets. Choose MATH. In step MATH of the construction let MATH be defined as in CITE, and let MATH be the union of MATH with the meager sets MATH . Now choose MATH. Then MATH and MATH witness that MATH does not avoid middles in the resulting set MATH. Consequently, MATH does not satisfy MATH. The proof that MATH satisfies MATH is as in CITE. |
math/0105045 | Fix a bijective enumeration MATH. Let MATH, MATH, be as in the definition of MATH-covers. For each MATH let MATH be such that MATH. We claim that MATH. As MATH is a MATH-cover of MATH, it is infinite; fix a bijective enumeration MATH of MATH. For each MATH define MATH, and MATH. For each MATH define MATH by: MATH . Then each MATH is a subset of MATH. Each MATH is infinite: For each MATH choose MATH, MATH. Then MATH for all MATH. As MATH is a MATH-cover of MATH, there exists MATH such that MATH. Consider the (unique) MATH such that MATH. Then MATH; therefore MATH, and in particular MATH. As MATH, we have that MATH. The sets MATH are linearly quasiordered by MATH: Assume that MATH. We may assume that MATH. As MATH, we have that MATH for all but finitely many MATH. This shows that MATH is a MATH-cover of MATH. Now, MATH is an extension of MATH by at most countably many elements. It is easy to see that an extension of a MATH-cover by countably many open sets is again a MATH-cover, see CITE. |
math/0105045 | MATH refines MATH. |
math/0105045 | MATH: The proof for this is similar to the proof of MATH in REF . MATH: Assume that MATH is a MATH-cover of MATH. Replacing each member of MATH with all finite unions of Basic clopen subsets of it, we may assume that all members of MATH are clopen (to unravel this assumption we will use the fact that MATH is closed under de-refinements). Thus, MATH is continuous and MATH is centered. Consequently, MATH is linearly refinable, that is, MATH is a MATH-cover of MATH. |
math/0105045 | We make the needed changes in the corresponding proof from CITE. MATH: Assume that MATH, MATH, are open covers of MATH which do not contain finite subcovers. For each MATH, replacing each member of MATH with all of its basic clopen subsets we may assume that all elements of MATH are clopen, and thus we may assume further that they are disjoint. For each MATH enumerate MATH. As we assume that the elements MATH, MATH, are disjoint, we can define a function MATH from MATH to MATH by MATH . Then MATH is continuous. Therefore, MATH satisfies the weak excluded middle property. Let MATH, and for each MATH, MATH be such that MATH is linearly quasiordered by MATH. For each MATH set MATH . We claim that MATH is a MATH-cover of MATH. We will use the followig property. CASE: For each finite subset MATH of MATH and each MATH, MATH. Let MATH be a bijective enumeration of MATH, and let MATH be such that for each MATH, MATH. For each MATH set MATH. We have the following. MATH is a subset of MATH: Assume that MATH, where MATH. Then MATH, therefore MATH. MATH is infinite: Assume that MATH where MATH. For each MATH choose MATH, and set MATH. Then for all MATH. Choose MATH. By property (MATH), MATH, therefore MATH. But MATH, thus MATH, a contradiction. As the sets MATH are linearly quasiordered by MATH, so are the sets MATH. MATH: Since MATH is continuous, MATH also satisfies MATH. Consider the basic open covers MATH defined by MATH. Then there exist finite MATH, MATH, such that either MATH for some MATH, or else MATH is a MATH-cover of MATH. The first case can be split into two sub-cases: If there exists an infinite set MATH such that MATH, then for each MATH the set MATH is finite, and we can define MATH so that MATH for each MATH, and we are done. Otherwise MATH for only finitely many MATH, therefore we may replace each MATH satisfying MATH with MATH, so we are in the second case. The second case is the interesting one. MATH is a MATH-cover of MATH - fix a bijective enumeration MATH of MATH and witnesses MATH, MATH, for that. Define MATH for each MATH. Then the subsets MATH of MATH, MATH, are infinite and linearly quasiordered by MATH. This shows that MATH is linearly refinable. |
math/0105045 | It is easy to see that MATH is the minimal size of a linearly refinable family MATH which has no pseudo-intersection. We will show that MATH. Assume that MATH is a linearly refinable family of size less than MATH, and let MATH be a linear refinement of MATH. As MATH, MATH has a pseudo-intersection, which is in particular a pseudo-intersection of MATH. |
math/0105045 | Clearly MATH. MATH: This can be proved like MATH in REF . MATH: Assume that MATH, MATH, are NAME MATH-covers of MATH. We may assume that these covers are pairwise disjoint. Define a function MATH so that for each MATH and MATH: MATH . Then MATH is a NAME map, and so MATH satisfies the weak excluded middle property. Let MATH and MATH, MATH, be witnesses for that. Set MATH. For each MATH set MATH. Then MATH is infinite and MATH for each MATH, and the sets MATH are linearly quasiordered by MATH. |
math/0105045 | By REF , MATH. On the other hand, MATH implies MATH, whose critical cardinality is MATH. |
math/0105045 | Fix MATH. Let MATH be an enumeration of MATH, and let MATH, MATH, witness that MATH is a MATH-cover of MATH. For each MATH define MATH . As the sets MATH are infinite and linearly quasiordered by MATH, the sets MATH are also infinite and linearly quasiordered by MATH. Moreover, for each MATH and each MATH, MATH, and therefore MATH for each MATH; thus MATH, as required. |
math/0105045 | We will prove the assertion for MATH; the proof for the remaining assertion is similar. Fix MATH. In CITE it is proved that for each open MATH-cover MATH of MATH there exists an open MATH-cover MATH of MATH such that the MATH-cover MATH of MATH refines MATH. Assume that MATH is a sequence of open MATH-covers of MATH. For each MATH choose an open MATH-cover MATH of MATH such that MATH refines MATH. Apply MATH to extract elements MATH, MATH, such that MATH. By REF , MATH is a MATH-cover of MATH. For each MATH choose MATH such that MATH. Then by REF , MATH is a MATH-cover of MATH. |
math/0105045 | By REF , we have that the assertion holds for MATH and MATH. We will use the following result which is analogous to REF and can be proved similarly. The following inclusions hold: CASE: MATH. CASE: MATH. Consequently, MATH. As in REF , we get the remaining assertions follow from this and REF , as MATH is closed under taking countable supersets. |
math/0105045 | We will prove REF ; REF is similar. By REF , we have the following implications MATH . |
math/0105045 | Fix a linear quasiordering MATH of MATH, and assume that MATH are NAME MATH-covers of MATH. Define a NAME function MATH from MATH to MATH by: MATH . MATH is bounded, say by MATH. Now define a NAME function MATH from MATH to MATH by: MATH . Note that MATH is a NAME image of MATH in MATH, thus it is bounded, say by MATH. It follows that the sequence MATH is large, and is a MATH-cover of MATH. |
math/0105045 | MATH: The set MATH is a linear quasiordering of MATH. MATH: If MATH satisfies MATH, then so does its continuous image MATH. Thus, MATH satisfies MATH. MATH: MATH is closed under taking NAME subsets. |
math/0105045 | The property MATH satisfies the assumptions of REF . |
math/0105046 | The estimate is immediate for the finitely many charts covering the interior of MATH, so we need only consider NAME charts near the boundary. In background coordinates, MATH can be written MATH . Pulling back to MATH, we obtain MATH . Since MATH and MATH are smooth functions bounded above and below together with all derivatives on MATH, it suffices to estimate MATH. Our choice of background coordinates ensures that the eigenvalues of MATH are uniformly bounded above and below by a global constant. Uniform estimates on the derivatives of MATH follow from the fact that MATH . Finally, an easy computation yields uniform NAME estimates for the MATH-th derivatives of MATH. (See REF below for a sharper estimate.) |
math/0105046 | We only need to show that there exist points MATH such that MATH cover MATH and MATH are uniformly locally finite, for then we can choose finitely many additional charts for the interior without disturbing the uniform local finiteness. By the preceding lemma, there are positive numbers MATH such that each set MATH contains the MATH-geodesic ball of radius MATH about MATH, and each set MATH is contained in the geodesic ball of radius MATH. Let MATH be any maximal collection of points in MATH such that the open geodesic balls MATH are disjoint. (Such a maximal collection exists by an easy application of NAME 's lemma.) If MATH is any point in MATH, by the maximality of the set MATH, MATH must intersect at least one of the balls MATH nontrivially, which implies that MATH by the triangle inequality. Therefore the sets MATH cover MATH. To bound the number of sets MATH that can intersect, it suffices to bound the number of geodesic balls of radius MATH around points MATH that can intersect. Let MATH be arbitrary and suppose MATH for some MATH. By the triangle inequality again, MATH. Since MATH has bounded sectional curvature, standard volume comparison theorems see, for example, CITE yield uniform volume estimates MATH . Since the sets MATH are disjoint for different values of MATH, there can be at most MATH such points MATH. |
math/0105046 | All of these claims are local, so we fix one background coordinate chart MATH and do all of our computations there. Let MATH be any nonnegative integer less than or equal to MATH. Applying the one-variable version of NAME 's formula to MATH, we obtain MATH . The integral above is easily shown to define a function of MATH that is in MATH up to the boundary and agrees with MATH on MATH; therefore the last term in REF is MATH in general, and if MATH, it is MATH if and only if MATH vanishes on MATH. REF follows easily from this, and REF, and REF follow from REF is an immediate consequence of the definition, and REF follows by setting MATH in REF and multiplying through by MATH. |
math/0105046 | Let MATH, which is a continuous Riemannian metric on MATH. The lemma follows from the easily-verified facts that MATH, MATH, and MATH if and only if MATH. |
math/0105046 | From the definition of MATH, we need to show for any NAME chart MATH that the coefficients of MATH are in MATH, with norm bounded independently of MATH. Since MATH, this follows immediately from the coordinate expression for MATH and the fact that the NAME symbols of MATH in NAME coordinates are uniformly bounded in MATH. |
math/0105046 | First consider REF . By definition, MATH iff MATH for MATH. By the product rule and induction, we can write MATH where MATH is a constant, equal to REF when MATH. Since, by the preceding lemma, MATH is bounded as long as MATH, the result follows easily by induction. For REF , the case MATH follows by inspecting the coordinate expression for MATH in NAME coordinates, recalling that the NAME symbols of MATH are uniformly bounded in MATH in these coordinates. The general case follows as above from REF and the fact that MATH. |
math/0105046 | If MATH is a NAME chart centered at MATH, MATH, which is uniformly bounded above and below by constant multiples of MATH. Thus if MATH, REF yields MATH where MATH is an upper bound on the number of sets MATH that can intersect nontrivially. Conversely, if MATH is finite, then MATH . The argument for the NAME case is similar but simpler, because we do not need the uniform local finiteness in that case. |
math/0105046 | We will prove REF by induction on MATH. Suppose MATH, and consider first the case of scalar functions, so that MATH is the trivial line bundle. By REF there are only two distinct cases: MATH and MATH. For MATH, let MATH be a NAME chart and let MATH be the background coordinates of MATH. We estimate MATH . Since MATH is uniformly bounded on MATH, the result follows. When MATH, we need to show that MATH is uniformly bounded in MATH. The NAME estimate for MATH in background coordinates, together with the fact that MATH vanishes on MATH, shows that MATH near MATH, from which the zero-order estimate MATH follows immediately. The NAME estimate is proved by noting that MATH; since MATH is uniformly bounded together with all derivatives on MATH, it suffices to show that MATH is uniformly bounded in MATH. This is proved as follows: MATH . Now let MATH be a tensor bundle of type MATH (and thus of weight MATH). In any background coordinate domain MATH, basis tensors of the form MATH are easily seen to be uniformly bounded in MATH. Since any MATH can be written locally as a linear combination of such tensors multiplied by functions in MATH, the result follows from the scalar case together with REF. Now suppose MATH and MATH, and assume REF is true for MATH when MATH and MATH. If MATH, then MATH by definition, so it suffices to prove that MATH with norm bounded by MATH. If MATH denotes the difference tensor between the NAME connections of MATH and MATH, a computation shows that MATH is the MATH-tensor whose components are given in any coordinates by MATH . Working in background coordinates, we find that MATH since the coefficients of MATH are in MATH. Since MATH by REF, we use REF to conclude that MATH with norm bounded by a multiple of MATH. Therefore, since MATH is a tensor field of weight MATH, the inductive hypothesis implies that MATH, which implies in turn that MATH as desired. For REF , we begin with the scalar case, and proceed by induction on MATH. Let MATH, and suppose MATH. Given any NAME chart MATH, let MATH be the background coordinates of MATH, and let MATH be the function MATH on MATH, so that MATH . The hypothesis means that MATH, with norm bounded by MATH. For MATH, we estimate MATH . Since MATH is bounded above on MATH, this shows in particular that MATH is bounded by a multiple of MATH. Similarly, if MATH and MATH are in MATH, we have MATH . To complete the MATH case, it suffices to extend REF to any MATH and MATH that lie in the same background chart, for then MATH extends continuously to the boundary as an element of MATH. Note that there is a real number MATH such that whenever MATH, the points MATH and MATH lie in the image of the same NAME chart. We estimate as follows: MATH . For the first term, the case in which MATH is taken care of by REF. On the other hand, if MATH, REF gives MATH . To estimate the second term of REF, let MATH be a positive integer such that MATH lies in the interval MATH. Then, since MATH and MATH lie in the image of a single NAME chart, as do MATH and MATH, REF gives MATH where the last inequality follows from the fact that MATH when MATH. Since both terms in parentheses above are bounded by MATH and thus by MATH, this completes the argument for the MATH case. (I am indebted to NAME for pointing out a gap in an earlier version of this proof, and providing helpful suggestions for fixing it.) For MATH, to show that MATH, it suffices to show that MATH is bounded and MATH whenever MATH is a smooth vector field on MATH. It is straightforward to check that any such vector field maps MATH into MATH. Thus if MATH, then MATH by induction, with norm bounded by MATH. Since MATH is obvious uniformly bounded by MATH, it follows that MATH. Finally, let MATH be a MATH tensor bundle. If MATH, to show that MATH we have to show that the components of MATH in any background coordinate domain MATH are in MATH. These components are given by complete contractions of tensor products of MATH with tensors of the form MATH, each of which is in MATH. By REF, these tensor products are in MATH. Since complete contraction clearly maps this space into MATH, the result for tensor fields follows from the scalar case. |
math/0105046 | Working directly with the definition of MATH, for any NAME coordinate chart MATH, we have to show that MATH is uniformly bounded in MATH. Since MATH on MATH, MATH will be identically equal to REF or REF on MATH unless MATH. Under these restrictions, it is easy to verify that MATH is uniformly bounded in MATH. |
math/0105046 | Suppose MATH. First we show that MATH can be approximated in the MATH norm by compactly supported elements of MATH. Let MATH be as in the preceding lemma. We will show that MATH in MATH as MATH, which by REF is the same as MATH in MATH whenever MATH. By the product rule, MATH . Since MATH is bounded and supported where MATH, we have MATH . By REF , MATH is integrable, so the integral on the right-hand side above goes to zero as MATH by the dominated convergence theorem. Next we must check that if MATH is compactly supported in MATH, it can be approximated in the MATH norm by smooth, compactly supported tensor fields. The classical argument involving convolution with an approximate identity shows that MATH can be approximated in the standard NAME MATH norm on a slightly larger compact set by tensor fields in MATH. However, on any fixed compact subset of MATH, it is easy to see that the MATH norm is equivalent to the MATH norm, thus completing the proof. |
math/0105046 | As noted above, for any section MATH of MATH, MATH can be written as a sum of contractions of terms like REF. If MATH is covariant of degree MATH and contravariant of degree MATH, then this tensor product has MATH upper indices and MATH lower indices. (The MATH lower indices are the undifferentiated indices of the MATH copies of MATH.) Because we are assuming that MATH is the same type of tensor as MATH, MATH of the upper indices must be contracted against MATH of the lower indices, so in particular we must have MATH . It is obvious that tensoring with MATH, MATH, and MATH are all uniformly degenerate operators. Using REF for the components of the difference tensor MATH, we see that the components of MATH in background coordinates are MATH up to the boundary. It follows that MATH is a uniformly degenerate operator for MATH. Since MATH is also uniformly degenerate, it follows by induction that MATH is uniformly degenerate as well. A straightforward computation (compare CITE) shows that the components of MATH are given by MATH where MATH and MATH are universal polynomials, so MATH. Moreover, an easy induction shows that for MATH, MATH. It follows that tensoring with MATH is a uniformly degenerate operator. By virtue of REF, therefore, we can rewrite REF in the following manifestly uniformly degenerate form: MATH . Since contraction of a lower index against an upper one is also uniformly degenerate, the result follows. |
math/0105046 | It is an immediate consequence of the definition of the indicial map that if MATH and MATH are uniformly degenerate operators with sufficiently smooth coefficients, then MATH . Since a sum or composition of two MATH bundle maps is again of class MATH, it suffices to display MATH as a sum of compositions of uniformly degenerate operators with MATH indicial maps. Writing MATH as a sum of contractions of terms of the form REF, we see that it suffices to show that each of the following uniformly degenerate operators has MATH indicial map: CASE: MATH; CASE: MATH. CASE: MATH; CASE: MATH; CASE: MATH. The last three operators above are themselves MATH bundle maps over MATH, whose indicial maps are just their restrictions to MATH, so there is nothing to prove in those cases. For REF, observe that MATH can be written as a sum of compositions of the operators MATH and MATH. Since the commutator MATH is a smooth bundle map and therefore has smooth indicial map, we need only consider the operator MATH. Let MATH be the difference tensor as in the preceding proof, and observe that for any MATH, MATH . As noted above, MATH is a smooth bundle map; and it follows from REF that MATH is a MATH bundle map. Therefore the indicial map of MATH is MATH as claimed. To analyze the remaining operator, MATH, let MATH be the MATH-tensor field on MATH whose components are MATH and let MATH be MATH . Because the assumption that MATH is asymptotically hyperbolic means that MATH on MATH, we can write MATH, where MATH. Using REF, therefore, we can write MATH where MATH. It follows that MATH is just tensoring with MATH, which is a MATH bundle map. On the other hand, REF shows that MATH . Because MATH is parallel, MATH. In particular, this implies that MATH so MATH has vanishing indicial map for MATH. |
math/0105046 | Let MATH be arbitrary, and let MATH be any background coordinates defined on a neighborhood of MATH in MATH. By an affine change of background coordinates, we may arrange that MATH and MATH at MATH. Let MATH be the NAME chart given by MATH in terms of these background coordinates, defined on some neighborhood MATH of MATH in MATH. Let MATH be the Riemannian metric MATH on MATH, let MATH, and let MATH be the indicial map of MATH. Then MATH, so the characteristic exponents and multiplicities of MATH at MATH are the same as those of MATH at MATH. Thus it suffices to show that MATH at MATH. For convenience, here is a summary of the several metrics being considered in this proof: MATH . Arguing as in the proof of the preceding lemma, it suffices to show that each of the following operators has vanishing indicial map at MATH: CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH. If we let MATH be the difference tensor between the two connections, a computation based on REF yields MATH . Because MATH at MATH, it follows that MATH is a zero-order operator that vanishes at MATH. The next three operators, tensoring with REF - REF, are obviously zero-order operators vanishing at MATH, so their indicial maps have the same property. To complete the proof, we need to show that the indicial map of REF vanishes at MATH. This is obvious from the argument in the preceding proof when MATH, because both MATH and MATH individually have vanishing indicial maps. On the other hand, the preceding proof showed that the restriction of MATH to MATH is equal to MATH, which in turn is equal to MATH at MATH. This completes the proof. |
math/0105046 | Choose any point MATH, and let MATH be background coordinates in a neighborhood MATH of MATH in MATH. Then we can write MATH locally as MATH where each coefficient MATH is a matrix-valued MATH function (the coordinate representation of a bundle map from MATH to itself). Suppose now that MATH is a section of MATH that has a MATH extension to MATH. Then MATH if MATH, so MATH is given locally by MATH . To compute MATH, we first compute the formal adjoint of a monomial of the form MATH as follows. Let MATH, which is a MATH metric on MATH. The inner products on MATH defined by MATH and MATH are related by MATH, and the volume elements by MATH. If MATH are smooth sections of MATH compactly supported in MATH, then MATH where MATH is a constant-coefficient polynomial in the components of MATH, MATH, and MATH. From this, it follows easily that MATH where MATH is MATH up to MATH. Therefore, MATH and we conclude that MATH which was to be proved. |
math/0105046 | The preceding proposition shows that MATH. Thus if MATH is a characteristic exponent of MATH, then MATH, and hence also MATH, is singular, which means that MATH is also a characteristic exponent. Since MATH and MATH, the result follows. |
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