paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0009158 | Transfer MATH to MATH and pull back to MATH. Then MATH becomes a holomorphic section of a trivial vector bundle and so each component of MATH is a holomorphic function that decays at MATH. But by the removable singularity theorem of NAME, each component extends uniquely to a holomorphic function on MATH and by the maximum principle must therefore be identically zero. This argument shows that MATH is identically zero on MATH. But then by uniqueness of analytic continuation, MATH is identically zero on MATH. |
math/0009158 | Recall first that MATH . Here MATH is the canonical-bundle of MATH, and the trivial bundle MATH is embedded in MATH by multiplication by MATH. Let us for the moment write MATH so that MATH. This decomposition is preserved by the connection since MATH is parallel, so that MATH decomposes as a pair of operators MATH (taking the component of MATH) and MATH . Now we shall decompose the other factor of MATH. If MATH is a section of MATH, write its components as follows MATH . Now the condition MATH may be written as the two equations MATH and MATH . The vanishing theorem will be proved according to the following scheme: CASE: MATH implies MATH; CASE: MATH implies MATH; CASE: MATH and MATH implies MATH. In other words, relative to MATH, we eliminate first the component in the trivial bundle, next the components in MATH, finally those in MATH. Remark that complex conjugation carries MATH into MATH so it is enough to deal with the components in MATH, MATH and MATH. Proof of REF As in CITE, the real function MATH satisfies NAME 's differential equation; on a compact manifold it follows that MATH is a holomorphic vector field. The argument requires integration by parts but in our situation we have sufficient decay at MATH so that the conclusion holds. In fact, MATH is holomorphic and decays at MATH, so by REF MATH is constant. Finally MATH is MATH so MATH; the proof of REF is complete. Proof of REF . Referring to REF , MATH satisfies the equation MATH. As in the compact case, this implies MATH (compare the proof of the existence of the conformal factor). In particular MATH is harmonic. Because the NAME and NAME-Laplacians agree (up to a factor of MATH) we infer that MATH where MATH is the decomposition of according to components in MATH and MATH. NAME is applicable now to show that MATH is holomorphic. The proof is completed by applying REF to MATH. Finally we consider implication REF . Since MATH and MATH, MATH, say, where MATH is a section of MATH, and REF gives MATH. These operators are the usual NAME operators, coupled to the holomorphic line bundle MATH and MATH lies in the space MATH. For reasons of degree, and using the NAME identity MATH, MATH where MATH is a section of MATH and MATH is a section of MATH. More precisely, MATH . Since MATH (the curvature is of type MATH), we obtain MATH and hence MATH is a holomorphic section of MATH. Since MATH decays at MATH, we have MATH. Two more steps, each involving an application of REF complete the proof. For with MATH, REF says that MATH is a decaying holomorphic section of MATH, hence zero. Since MATH is an isomorphism, it follows that MATH is a decaying, holomorphic section of MATH. This completes the proof of REF and hence the vanishing theorem. |
math/0009159 | By construction, the MATH-graded chain complex MATH has a bounded filtration REF with the associated graded complex given by MATH . Then by the standard technique of CITE, we derive the existence of a spectral sequence MATH with MATH, and MATH . The higher differentials are induced by MATH. The expression of MATH, as discussed in REF , tells us that the higher differentials acting on MATH are defined by MATH. |
math/0009159 | The result is the consequence of the results on the geometric limits in CITE CITE, together with the dimension formulae for various moduli spaces. Suppose given a solution MATH in a one-dimensional component of MATH, with asymptotic values MATH for MATH. The geometric limits as MATH are determined in CITE. Among these geometric limits, there are two parameterized oriented paths MATH on MATH and MATH, respectively, where both paths connect MATH, consistently with the orientations. Here MATH is the circle of flat MATH-connections on MATH modulo gauge equivalence. There is also a holomorphic map MATH from the unit half-disc to MATH with boundary along the paths MATH and MATH. First we show that, if the moduli space MATH is non-empty, then we have MATH . Under this assumption, the geometric limits of a family of solutions MATH in MATH, for MATH, can be deformed by a homotopy in MATH to geometric limits for a monopole in MATH. In particular, this can happen if and only if the path MATH along MATH can be homotopically deformed to a path along the curve MATH, by a homotopy that moves the endpoints MATH to the corresponding points MATH, along MATH. (We use here the same notation as in CITE, CITE, CITE.) Both paths MATH and MATH must be contractible in MATH. In fact, the geometric limits of any solution in MATH define an approximate solution to the monopole equations on MATH, which can be deformed to a solution in the minimal energy component of MATH. Conversely, in this case, the geometric limits of MATH can be spliced together to form a solution to the monopole equations on MATH between the critical points MATH and MATH. Thus the identification of the moduli spaces implies that MATH . Now assume that the moduli space MATH is empty. Notice that oriented paths MATH which are non-contractible in the cylinder MATH correspond to flowlines in the components with higher energy. Here MATH is a covering of the torus MATH of flat MATH-connections on MATH modulo gauge equivalence, obtained by taking the quotient only by those gauge transformations on MATH that extend to MATH, compare CITE. Since we are considering the flowlines in the minimal energy components MATH, the paths MATH and MATH are also contractible in MATH. The analysis of the moduli space on the cobordant manifold MATH with cylindrical ends MATH and MATH in CITE shows that the holomorphic half-disc MATH, from the geometric limits of MATH in the one-dimensional components of MATH, is the degenerate limit of the assembled holomorphic triangles MATH as MATH, where MATH is the parameter in the perturbation on MATH that ``simulates the effect of surgery". In particular, we have the following identification MATH . Here MATH is the corresponding monopole in MATH, under REF , satisfying MATH, and MATH is given by the non-empty REF-dimensional components of MATH. Here we follow the notation of CITE. Then the dimension formulae developed in CITE can be applied to show that MATH . Thus we have MATH. This completes the proof of the statement. |
math/0009159 | For any MATH, let MATH be another MATH-dimensional submanifold in MATH obtained from the construction of the holonomy map and let MATH be a strata transversal section of MATH over MATH. We will prove that the difference between MATH and MATH defines an element in MATH which is homologous to zero. Over MATH, we can choose a strata transverse section MATH of MATH, with MATH and MATH. We also choose a codimension MATH submanifold MATH in MATH, whose intersection with MATH is MATH, and whose intersection with MATH is MATH, respectively. MATH is also compatible with the boundary strata of MATH. From this construction, the zero set of MATH over MATH is a REF-manifold with boundary which consists of three parts: REF the zero set of MATH in MATH; REF the zero set of MATH in MATH; REF the zero set of MATH in the intersection of MATH with MATH, where MATH is the codimension one boundary of MATH. By choosing the base points MATH in MATH away from the cylindrical end for the MATH-fibration MATH, and noticing that MATH is the generator of MATH, we know that the contribution of REF from the codimension one boundary of MATH times MATH only comes from MATH where MATH runs over all the possible critical points with index MATH. Now the transversality condition over the boundary strata over the intersection of MATH with MATH provides a set of finitely many oriented points. Counting these points gives a number, denoted by MATH. Then the difference of the two choices defines a number MATH thus, MATH is homologous to zero. This proves that the two choices define the same homology class in MATH. |
math/0009159 | Let MATH denote the boundary embedding maps of MATH into MATH. Then the set of MATH-structures on MATH which agree with MATH over MATH, denoted by MATH, form an affine space over MATH with MATH and MATH. For any MATH, after a generic perturbation, the moduli space MATH is an oriented, compact, smooth manifold with dimension given by MATH . The NAME - NAME invariant is a linear functional MATH where MATH. We summarize this definition briefly. We shall only consider monomials MATH with MATH. For any other degree of MATH, we assign MATH to be zero. For any monomial as above, we can choose smooth REF-dimensional submanifolds representing MATH. Then the holonomy along these loops defines a map: MATH, whose generic fiber is a closed codimension MATH submanifold MATH in MATH. The based monopoles define a MATH-fiber bundle MATH over MATH. Then MATH is defined to be the result of integration of the MATH power of the first NAME class of this MATH bundle over MATH. Equivalently, we can consider the associated rank MATH complex vector bundle MATH over the REFn-dimensional manifold MATH. Then MATH is obtained by counting points with the orientation in the zero set of a generic strata transverse section of MATH over MATH. We can adapt the gluing theory for NAME - NAME monopoles developed in CITE to the setting of CITE and CITE. Thus, when MATH is sufficiently large, we obtain a gluing theorem for NAME - NAME monopoles, which identifies REF-dimensional monopoles on MATH with the following product: MATH . Notice that, for MATH, we write MATH as MATH under the gluing map REF . We have MATH . Now the gluing formulae follow from the definition of the relative invariants MATH and MATH for MATH, with values in MATH and MATH respectively. Note that, in the case of MATH, the choice of REF-cycle MATH fixes the choice of the chamber. In fact, the moduli space MATH, with the perturbation MATH, contains no reducibles. Thus, in particular, since the relative invariants take values in MATH, once MATH is fixed, there are no wall crossing terms for MATH when we change the perturbation and the metric. |
math/0009160 | MATH is MATH-compact. By NAME 's theorem we infer that MATH is compact when endowed with the cartesian topology. We can now identify MATH with a net in MATH to obtain a sub-net MATH of MATH which converges to an element MATH of MATH. Since MATH is MATH-closed in MATH, MATH induces a bounded linear operator from MATH into MATH, which we still denote by MATH. Clearly MATH, for all MATH and thus MATH. Our assumptions yield that MATH, for all MATH and hence MATH is a projection onto MATH. We next demonstrate that MATH is MATH-continuous. By a classical result CITE it suffices to show that for every net MATH in MATH such that MATH, we have that MATH. Note that MATH, for all MATH, and let MATH be any MATH-cluster point of MATH. We will show that MATH. To this end set MATH. Then MATH, as MATH is MATH-continuous. It follows that MATH, for all MATH and MATH, and thus as MATH is MATH-continuous, we obtain that MATH, for all MATH. Hence MATH. Because MATH and MATH is a projection onto MATH, we deduce that MATH, completing the proof of the assertion. |
math/0009160 | Let MATH be a basis for MATH isometrically equivalent to the usual MATH-basis. Since MATH is MATH-continuous and MATH, there exist vectors MATH in MATH such that MATH for all MATH. There also exist scalars MATH, MATH, MATH, MATH, such that MATH for all MATH and MATH. Finally, there exist scalars MATH, MATH, MATH, such that MATH, MATH. Observe that for MATH and MATH we have MATH . Hence, MATH, for all MATH and MATH. Thus MATH for all choices of signs MATH and all MATH. Similarly, MATH for all MATH, MATH, and thus MATH . We first show that there exist vectors MATH in MATH such that MATH for all MATH, MATH, and MATH, for all MATH and all choices of signs MATH. Indeed, let MATH be signs and let MATH be distinct atoms in MATH such that MATH for all MATH. We can assume without loss of generality that the MATH's are pairwise disjoint. We define MATH in MATH as follows: MATH where we regard MATH as an element of MATH. Clearly, MATH and MATH, for all MATH and MATH. Given signs MATH and MATH we have that MATH . We next fix signs MATH, MATH and MATH. We set MATH . We claim that MATH, MATH-almost everywhere in MATH. Indeed, note first that MATH for all MATH, and thus MATH . On the other hand, setting MATH and taking in account that MATH is an atom, we infer that MATH. Indeed, otherwise, MATH and thus MATH, MATH-almost everywhere on MATH. But also, as MATH is an atom, MATH has a constant sign MATH-almost everywhere on MATH and so MATH, contradicting REF. Therefore, MATH and hence MATH, MATH-almost everywhere in MATH, as claimed. We conclude that MATH, for all MATH, MATH, and all choices of signs MATH. Combining with REF we deduce that MATH, for all MATH and all choices of signs MATH. We next set MATH and choose MATH. REF yields a linear operator MATH, MATH, so that MATH and MATH, for all MATH and MATH. Setting MATH, MATH, we obtain that for every choice of signs MATH and all MATH, MATH . Thus MATH, for all MATH. If we define MATH by MATH, we see that MATH is MATH-continuous and MATH. Indeed, for the latter assertion we observe that for every choice of signs MATH, MATH, by the definition of the sequence MATH and the fact that MATH for MATH, and MATH. It follows now, by the choice of MATH, that MATH for every choice of signs MATH, and therefore MATH. We deduce that MATH for all MATH and every MATH. Hence MATH for all MATH. Finally, MATH . The proof of the proposition is now complete. |
math/0009160 | We first choose MATH so that MATH and MATH for all MATH and MATH. We then choose MATH for all MATH, and a sequence MATH of positive scalars such that MATH. REF yields a MATH-continuous linear operator MATH, MATH, such that MATH for MATH, and MATH, for MATH. We next apply REF for MATH, MATH, MATH, MATH and MATH, to obtain a MATH-continuous linear operator MATH, MATH, so that MATH and MATH, for MATH. Continuing in this fashion we construct MATH-continuous linear operators MATH with MATH and such that MATH . Clearly, the sequence of operators MATH converges in norm to a MATH-continuous linear operator MATH such that MATH and MATH, MATH. In addition to that we have MATH and thus MATH. We conclude that MATH . |
math/0009160 | We apply REF for MATH, MATH, MATH and MATH, to obtain a MATH-continuous linear operator MATH, MATH, so that MATH for all MATH. Our assumptions yield that MATH is the desired projection. |
math/0009160 | Assume MATH is infinite and let MATH be an enumeration of MATH. (The argument for finite MATH is implicitly contained in the proof of the infinite case.) It is clear that MATH is isometrically equivalent to the usual MATH-basis. Set MATH, MATH and let MATH be a null sequence of positive scalars. We shall inductively construct MATH-continuous contractive projections MATH such that MATH, whenever MATH. MATH is selected by applying REF for the subspace MATH and the vectors MATH. Suppose MATH have been selected so that MATH whenever MATH. Apply REF for MATH, MATH, MATH, MATH, and the collections of vectors MATH, MATH, in order to obtain a MATH-continuous linear operator MATH, MATH, such that MATH for all MATH, and MATH for all MATH. Clearly, MATH is the required projection onto MATH. This completes the inductive construction. The assertion of the theorem now follows from REF . |
math/0009160 | Clearly, MATH is linearly independent. When MATH is finite the assertion follows immediately from REF as MATH is isometric to MATH. If MATH is infinite let MATH be an enumeration of MATH and set MATH. Of course MATH is isometrically equivalent to the usual MATH-basis, and applying the NAME representation and the NAME theorems, we infer that MATH. A classical result CITE yields that MATH is MATH-closed in MATH. It is not hard to see (compare also REF) that MATH is MATH-isometric to MATH. The result follows from REF . |
math/0009160 | We regard MATH in its MATH-topology and set MATH. Since MATH is separable and MATH is non-separable, MATH is an uncountable MATH-subset of MATH. It follows that MATH is an uncountable Polish space in its relative MATH-topology. We will show that MATH contains a MATH-compact subset MATH homeomorphic to the NAME set MATH, such that MATH. Indeed, let MATH denote negation (MATH). Then MATH is a fixed-point free homeomorphism on the uncountable Polish space MATH and therefore there exists an uncountable relatively open subset MATH of MATH such that MATH. By a classical result MATH contains a compact subset MATH homeomorphic to the NAME set which clearly satisfies MATH. Of course MATH contains homeomorphs of all countable compact metric spaces. REF now yields that MATH contains subspaces isometric to MATH, for every countable compact metric space MATH. Because MATH is separable, a result of NAME REF implies that MATH contains a subspace isometric to MATH. The existence of a contractively complemented subspace of MATH isometric to MATH now follows from a result of CITE. |
math/0009160 | We first choose an infinite MATH-convergent subsequence MATH of MATH and set MATH. Clearly, MATH. If MATH then MATH is MATH-closed in MATH by REF. We deduce from REF that MATH is the range of a MATH-continuous contractive projection in MATH. It is easy to see that if MATH is an enumeration of MATH then the subspace MATH is the range of a MATH-continuous contractive projection in MATH. Hence by composing the projections previously obtained we see that the assertion of the theorem holds in this case. We shall next deal with the case of MATH. Suppose that MATH and choose a sequence of positive scalars MATH such that MATH. Choose also MATH so that MATH. We shall inductively construct increasing sequences MATH and MATH with MATH, and MATH-continuous contractive projections MATH, where MATH and MATH, MATH, so that the following conditions are fulfilled: MATH . Once this is accomplished, REF will enable us to apply REF and deduce that MATH is the range of a MATH-continuous projection in MATH. Note that MATH is MATH-closed in MATH by REF as MATH is isometrically equivalent to the usual MATH-basis. We first choose MATH in MATH with MATH, and apply REF for MATH, MATH, MATH (MATH), MATH, MATH, and MATH (MATH), MATH, MATH. We obtain a MATH-continuous contractive projection MATH such that MATH for all MATH. Suppose that we have constructed MATH, MATH and MATH so that REF are satisfied. We next choose MATH so that REF is satisfied for MATH. By REF of the induction hypothesis we infer that MATH, for MATH. We can therefore choose MATH in MATH with MATH, such that MATH and MATH for all MATH. Hence REF is satisfied for MATH. We next put MATH and set MATH, for MATH, MATH and MATH. We claim that there exist vectors MATH in MATH so that MATH . Having achieved this and taking in account REF of the induction hypothesis, we employ REF for MATH, MATH, MATH, MATH, MATH (MATH), and the collections of vectors MATH, MATH described above, to find a MATH-continuous linear operator MATH, MATH, such that MATH, for all MATH, and MATH for all MATH. It is easy to verify that MATH is a projection onto MATH so that MATH, MATH and MATH satisfy REF. The collection MATH is explicitly defined in REF. It remains to define MATH. We first choose scalars MATH, where MATH and MATH, such that MATH. Note that MATH for every MATH since MATH. We also define scalars MATH by MATH and set MATH for MATH and MATH. Observe that MATH, by REF of the induction hypothesis. We now set MATH . It follows now by REF of the induction hypothesis that REF is satisfied. To establish REF we have MATH . Finally, we show that REF holds. Indeed, when MATH, this is a consequence of the choice of MATH and MATH. When MATH the assertion follows from REF of the induction hypothesis. When MATH it follows from REF of the induction hypothesis and the fact that MATH. |
math/0009160 | If MATH is non-separable the assertion follows from REF . If MATH is separable, then MATH is countable and MATH is isometric to MATH. Let MATH be a basis for MATH isometrically equivalent to the usual MATH-basis, and choose a MATH-convergent subsequence MATH of MATH according to REF . Let MATH. Then it is easy to see that MATH is MATH-isometric to MATH. The result follows from REF . |
math/0009160 | Let MATH be a basis for MATH isometrically equivalent to the usual MATH-basis. Let MATH, MATH, and let MATH be a sequence of positive scalars such that MATH. The argument in the proof of REF now yields a sequence MATH of MATH-continuous contractive projections in MATH with MATH and such that MATH whenever MATH. Given MATH in MATH, we set MATH. Clearly MATH is a MATH-continuous contractive projection onto MATH. Moreover, our assumptions yield that MATH, for all MATH and thus the sequence of operators MATH converges in norm to a MATH-continuous contractive projection MATH from MATH onto MATH. It is easily seen that MATH, whenever MATH and hence MATH when MATH. We now let MATH. MATH is naturally identified to a subspace of MATH as MATH is MATH-continuous, and of course it is isometric to MATH for all MATH. Since MATH we deduce that MATH for all MATH. It is also easily verified that MATH acts as a contractive projection from MATH onto MATH. NAME 's theorem now yields that MATH, weakly, for all MATH and thus MATH is dense in MATH. |
math/0009161 | REF is just REF in the present notation. CASE: This is immediate from REF since MATH defines MATH outside MATH. Note that the boundary points MATH of MATH correspond to MATH and MATH, respectively. CASE: NAME of MATH near MATH means asymptotic expansions at the interiors of MATH (that is, for MATH with MATH bounded away from REF and MATH, and vice versa), plus uniformity (`joint asymptotics') as both MATH approach zero. By definition, this means that, for all MATH, MATH the space of MATH times differentiable functions (in MATH) vanishing to order MATH at MATH, with MATH. Since MATH, it is easily seen that this is equivalent to REF. CASE: From REF we have MATH . Setting MATH and taking MATH-derivatives, we get MATH so REF is equivalent to MATH . Now polyhomogeneity of MATH near MATH with the index sets MATH implies that MATH for each MATH and each MATH. REF holds precisely when the leading term in the asymptotics MATH of MATH has a positive MATH-power, that is, when MATH. |
math/0009163 | The conditions are obviously necessary. So we only need to prove the sufficiency. Assume that MATH and MATH is not a constant on MATH. Then there exists a curve MATH such that MATH is a smooth point of MATH, and MATH. Let MATH where MATH. Then MATH and MATH that is, MATH . By REF , there exists a MATH such that MATH . Let MATH and MATH . Then for any curve through MATH we have MATH . Let MATH . Then MATH is invertible and the Jacobian MATH is onto. By the Dominant Morphism REF , MATH contains a nonempty NAME open set of MATH, so does MATH. Since the set of MATH's such that MATH is almost onto is a NAME open set, and we just showed that it is nonempty, MATH is almost onto for a generic set of matrices MATH. |
math/0009163 | We will repeatedly use the projective dimension theorem CITE which says that if MATH and MATH are MATH-dimensional and MATH-codimensional projective varieties, respectively, then MATH. In particular, MATH is not empty if MATH. Let MATH . Then MATH must have co-dimension MATH because of the condition MATH. Therefore the linear equation MATH has solutions in MATH for any MATH, and the set of all solutions for each MATH is in the form of MATH where MATH is a particular solution; that is, the solution set is given by MATH where MATH is the unique MATH-codimensional projective subspace through MATH and MATH. Since MATH, we must have MATH and by NAME 's Theorem CITE, there are MATH many points in MATH counted with multiplicities. If MATH is a monic polynomial of degree MATH, then from REF one can see that all the solutions are in MATH. |
math/0009164 | The connected component of a point MATH in MATH is by definition the maximal connected subset of MATH, which contains MATH. The connected set MATH lies in MATH and contains MATH. Therefore, MATH. |
math/0009164 | Let MATH and MATH be any open neighborhood of MATH. Designate the set MATH by MATH. From property MATH follows that there exist MATH, MATH, such that MATH, MATH. But MATH. Therefore, MATH, MATH. |
math/0009164 | Obviously, MATH, MATH. On the other hand, MATH, MATH. So MATH, MATH, since the sets MATH are closed by the definition. |
math/0009164 | This statement is a direct corollary of the previous statement and the following theorem: if the domains MATH and MATH in the plane do not meet, but MATH, then MATH is connected (see REF ). |
math/0009164 | Fix MATH. It is known (see REF ), that for every MATH points MATH and MATH can be connected with the MATH-chain in the set MATH, i. e. final sequence MATH of points could be found in MATH to comply with the condition MATH . For every MATH select MATH-chain in MATH connecting points MATH and MATH. Since MATH, for every MATH in the sequence MATH could be found an element which does not belong to MATH. Therefore, for all MATH indexes MATH are defined to meet the conditions CASE: MATH, MATH; CASE: MATH. The set MATH is compact, being the closed subset of the compact set MATH. Therefore sequence MATH has a limit point MATH. Without loss of generality we shall suppose that MATH (otherwise, it is always possible to pass to a subsequence). Since MATH and MATH then MATH. Therefore, MATH and MATH. On the other hand, for every MATH there exist MATH and MATH, such that MATH and MATH. Then points MATH and MATH could be connected in MATH by the MATH-chain MATH . Consequently, (see REF ) points MATH and MATH belong to one connected component of the set MATH. |
math/0009164 | Fix MATH and open neighborhood MATH of MATH. Let us show that MATH. The set MATH divides the plane into two connected components. Therefore it consists more than of one point. Take MATH such that MATH and MATH. According to REF there exists MATH. From REF it follows that MATH. Besides, MATH. The set MATH is zero-dimensional. Therefore the space MATH admits a base of open sets MATH such that MATH in space MATH (in relative topology) for all MATH. Let MATH. Then MATH and MATH for some MATH. MATH is an open-closed subset of space MATH. If MATH, the sets MATH and MATH will form a partition of space MATH. Assume that MATH. Since MATH then MATH and MATH. Therefore, the sets MATH and MATH form a partition of the set MATH contrary to it's connectedness. So MATH. The relation MATH is proved similarly. By virtue of arbitrariness in the choice of a point MATH and its neighborhood MATH we conclude that MATH is MATH-set. |
math/0009164 | Let MATH. There exists MATH, such that MATH. According to REF , we can find MATH, MATH, such that MATH, MATH. Designate by MATH the segment with end-points MATH and MATH, MATH. Obviously, MATH, MATH. The set MATH is connected and MATH. We fix a point MATH. Since MATH for every MATH, the set MATH is connected. |
math/0009164 | Fix MATH. For MATH designate by MATH the connected component of MATH in the set MATH. Then in accord with REF the family MATH forms a base of opened neighborhoods of the point MATH in the topology MATH. On the other hand let us denote MATH for every MATH. Then by easy direct verification we receive MATH for every MATH. CASE: E. D. |
math/0009169 | We just have to apply the isomorphisms REF : MATH and similarly MATH . |
math/0009169 | For the first line, remember that MATH if MATH, and zero otherwise. But here we also have that MATH. For the second line, using the properties of the isomorphisms MATH and MATH we obtain that MATH . For the last line we have used the properties of the NAME invariants of MATH REF . |
math/0009169 | Let us write MATH. By applying REF we obtain MATH which proves the lemma. |
math/0009169 | Similar to the proof of REF . |
math/0009170 | We define MATH recursively. Suppose MATH are such that MATH satisfies MATH. Note that since MATH is Hermitian, so is MATH. We need to find MATH so that MATH satisfies MATH up to order MATH. But this happens if and only if MATH. Then MATH is a solution. |
math/0009170 | The MATH-linearity and the injectivity of MATH are obvious since MATH and MATH are deformations of MATH and MATH. To prove surjectivity, let MATH be given. Then MATH whence MATH. Thus defining MATH, we have MATH up to order MATH. Since MATH starts with order MATH, we can repeat the argument to find a MATH such that MATH coincides with MATH up to order MATH. Then a simple induction proves that MATH is onto. |
math/0009170 | For the existence, note that since MATH is f.g.p.m., it follows that we can identify MATH, for some MATH and MATH idempotent. Let MATH be an idempotent deforming MATH and consider the (right) f.g.p.m. over MATH given by MATH. By REF (choosing MATH to be MATH in the upper right corner and zero elsewhere), we can use the isomorphism MATH to pull this MATH-module structure back to MATH, that is, MATH for MATH and MATH. So MATH is a f.g.p. deformation of MATH. Now assume MATH is another deformation of MATH. Let MATH and MATH be the natural projections, which are surjective MATH-module homomorphisms. Then it follows by projectivity of MATH that there exists a MATH-module homomorphism MATH satisfying MATH. Since MATH as MATH-modules, we can write MATH and it is readly seen that MATH. So MATH is an equivalence. |
math/0009170 | As in REF , we choose MATH, a projection deforming MATH, and consider the MATH-module MATH, which we know to define a deformation MATH of MATH. Let MATH be the MATH-valued inner product on MATH obtained from MATH restricted to MATH. A simple computation shows that MATH is a deformation of MATH and hence MATH is a Hermitian deformation of MATH. Let MATH be another Hermitian deformation of MATH. By REF , we may assume that MATH as a MATH-module, with some MATH-valued inner product MATH deforming MATH. Recall that any MATH-valued inner product MATH on the free MATH-module MATH can be written as MATH for some Hermitian element MATH, where MATH, MATH. Since MATH is projective, one can check that the same holds for this submodule. Hence, there is a Hermitian element MATH so that MATH. But since MATH and MATH are deformations of MATH, we can write MATH with MATH. It then follows from REF that we can find MATH so that MATH. It is then clear that MATH is the desired equivalence. |
math/0009170 | Since all deformations of MATH are equivalent we choose a deformation MATH of the projection MATH. Moreover, we choose a unitary MATH with MATH and MATH which is possible due to REF . Now MATH is again a projection with classical limit MATH. Thus it defines a deformation MATH which is equivalent to the first one by an equivalence transformation MATH. Moreover, MATH descends to a unitary MATH-right linear map MATH with lowest order MATH. Then MATH is the desired deformation of MATH. |
math/0009170 | Existence and uniqueness of (Hermitian) deformations follow from REF and the observation before the theorem. Suppose now MATH is local/differential/of NAME type. Choose a deformation MATH of MATH and let us consider the MATH-action on MATH induced by MATH as in REF . Note that if we write MATH, it follows from REF that each MATH is local/differential/differential of order r. Moreover, MATH has the same property. From MATH and MATH, it follows directly that MATH and MATH have the same desired properties. |
math/0009170 | As we have remarked, the algebras MATH and MATH are NAME equivalent, and MATH is a MATH-bimodule defining this equivalence. By symmetry of NAME equivalence, it follows that there exists an idempotent MATH, for some MATH, so that MATH (MATH as row vectors) as a left MATH-module and MATH as a unital algebra. In the Hermitian case, we note that, by CITE, we can actually choose MATH satisfying MATH. Note that since MATH is commutative (and so is MATH), it follows from CITE, CITE that in fact MATH and MATH are isomorphic as MATH-algebras (these algebras are NAME MATH-equivalent in the sense of CITE). Therefore, we can define a map MATH just as we did for MATH. Let MATH and MATH. Let MATH be a deformation of MATH and MATH be the deformation induced by MATH, in such a way that MATH. Let MATH, which is naturally a MATH-bimodule. Note that, by NAME theory, we have MATH. Now pick MATH, a deformation of MATH and let MATH. Then MATH is induced by MATH, and MATH can be identified with MATH. Finally, it is not hard to check (see the existence part of REF ) that MATH and MATH are both deformations of MATH corresponding to MATH. It then follows from REF (for left modules) that these deformations are equivalent and hence so are MATH and MATH. Therefore MATH and a similar argument shows that MATH. This concludes the proof. |
math/0009170 | From REF it easily follows that MATH satisfies the natural NAME rules (see the last two equations in CITE) and this implies the first statement. A simple computation shows that MATH. Note that MATH is equivalent to the first equation in CITE and the last statement follows directly from CITE . |
math/0009170 | Note that if MATH and MATH, we have MATH, where MATH is the Hamiltonian vector field corresponding to MATH. Observe that the curvature tensor corresponding to MATH, MATH, satisfies MATH, for all MATH and MATH. This implies the result. |
math/0009170 | For MATH, write MATH. It is not hard to check that MATH . It is then clear that MATH. The bracket MATH defines an action of the NAME algebra MATH on MATH by derivations and this implies that MATH. Hence the NAME rule for MATH yields MATH, MATH, and this immediately shows that the bracket MATH on MATH coincides with MATH after the identification MATH. |
math/0009170 | Let MATH. Then a straightforward computation shows the relation MATH . It follows immediately that MATH is a positive algebra element and that MATH is full. The fullness of MATH follows from MATH . We observe that REF can be easily shown as in CITE. In order to prove REF , we recall that this just means REF for the complex-conjugate bimodule MATH. Equivalently, we can consider MATH-representations of MATH on pre-Hilbert spaces from the right. Thus let MATH be such a MATH-representation of MATH from the right and let MATH as well as MATH. Then by REF MATH and this shows the positivity needed for REF . This concludes the proof. |
math/0009173 | The proof is easy. |
math/0009173 | This is a straightforward computation. |
math/0009173 | This is proved in REF and also follows from the proof of the main theorem, REF (see REF for details). |
math/0009173 | Clear. |
math/0009173 | See REF. |
math/0009173 | This is obvious. |
math/0009173 | This follows from nilpotency. It is easy to see that we need only show that MATH for any MATH and MATH. Equivalently, we have to show that the componentwise addition of all MATH is not MATH. First, we generalize MATH and MATH. Clearly, every MATH can be written as MATH for some MATH. We say that MATH for MATH if MATH and MATH for nonnegative integers MATH. In this case, we define MATH. Note that, when MATH, MATH. Now, we set MATH and MATH. Let MATH and let MATH. We see that componentwise addition of two elements of MATH, if in MATH, yields another element of MATH, where the orders are summed. In particular, this means that, when MATH, for MATH, then MATH. Since MATH, it follows that any componentwise sum of elements of MATH, if in MATH, yields a nonzero element of MATH. This is all we need. |
math/0009173 | If MATH, then it is clear from MATH that MATH, negating in each component. But then MATH, which is impossible. |
math/0009173 | In the first case, if MATH reversed orientation, then MATH, thus MATH by nilpotency - this is a contradiction. On the other hand, when MATH reverses orientation, MATH must reverse orientation, and thus by nilpotency cannot be defined on all of MATH. Thus MATH, hence the desired result. The second case follows easily from nilpotency. |
math/0009173 | The fact that the maps are well-defined, inverse to each other, and preserve MATH is easy to see from construction when orientations are preserved (it helps to draw a picture). Also, when orientations are preserved, MATH is trivially preserved. So it remains to consider orientation-reversing cases. Given MATH, if orientation is reversed in MATH or MATH, it can only be reversed in MATH, and MATH by REF . In the case MATH, it follows that MATH, where MATH preserves orientation, MATH reverses orientation, MATH, and MATH. So MATH and MATH are preserved, and MATH. If, instead, MATH, then MATH, where MATH, MATH, and MATH reverses orientation. Again, MATH and MATH are preserved and MATH. Now consider an element MATH in which either MATH or MATH reverses orientation. By REF , only MATH reverses orientation and MATH. So, we get MATH where MATH. In this case, MATH reverses orientation and MATH does not, so MATH and MATH are preserved, and clearly MATH. Finally, suppose MATH reverses orientation. By definition, this means that MATH reverses orientation on MATH. By nilpotency, MATH, and it follows that MATH. We then have MATH and MATH. It follows that MATH reverses orientation, and so MATH and MATH are preserved, and MATH. The decreasing and left cases follow in exactly the same way as the increasing and right cases. |
math/0009173 | This follows directly from the argument above. |
math/0009173 | Fix a choice of sign MATH for this proof. Clearly, whenever MATH, MATH, and MATH, then MATH. In this case, MATH, as MATH also has reversed orientation and the same order as MATH and MATH. It remains only to see that, for any MATH with MATH, there are MATH ways of writing MATH for MATH, and they all are of this form. The formula follows immediately. |
math/0009173 | Take some MATH such that MATH. We show in the following paragraph that either MATH divides MATH, or MATH and MATH. This proves the lemma - all that remains is to see that, in the latter case, MATH. If, instead, MATH, then applying the above result also to MATH, we find that both MATH and MATH are perpendicular to MATH. By space concerns on the diagram, it follows that MATH, but then MATH, which would show that MATH divides MATH, in contradiction to MATH. So, take any MATH such that MATH, and assume that MATH is not a multiple of MATH. Define MATH by MATH if MATH preserves orientation on MATH, and MATH otherwise. Define MATH as follows. For any MATH (MATH is the length of the NAME diagram), let MATH be given by MATH. Then we define MATH. Define MATH. Clearly, MATH is defined so that if MATH is a connected segment of the diagram, then MATH is constant on MATH, for each fixed MATH, MATH. Since MATH for MATH, it follows that MATH is MATH-periodic in the first component. For the same reason, MATH is MATH-periodic in the first component. If MATH, MATH must be MATH-periodic in the first component. This follows since MATH, MATH implies MATH whenever MATH, MATH. By minimality, MATH, which is impossible. Hence, MATH. Furthermore, MATH, because otherwise MATH, which would imply that MATH by the above results applied to MATH, which is clearly contradictory. |
math/0009173 | First, it is clear that MATH divides MATH iff MATH for MATH an integer. In this case, the theorem is satisfied; so suppose not. By applying MATH some number of times to MATH or MATH, it suffices to assume MATH. Now, if MATH is defined on MATH, then MATH together with the Lemma gives the desired result. So assume MATH is not defined on MATH, and hence it is not defined on MATH either. Suppose MATH for some positive integer MATH. Write MATH. Now, by applying MATH some number of times to each MATH, we can obtain MATH for some MATH, showing that MATH is defined on MATH, which is a contradiction. But then MATH is defined The direction-reversed case is the same. |
math/0009173 | Let MATH and MATH for some MATH (which exists by construction). Suppose that MATH. We will analyze all possible cases by considering the value of MATH. Write MATH where MATH. First, I claim that MATH. Suppose instead that MATH. In this case, MATH, which immediately implies from the proof of REF that MATH preserves orientation on MATH. So MATH preserves orientation, and by REF preserves orientation on MATH. Now, if we set MATH, we find that MATH because MATH. Now, MATH shows that MATH. Finally, MATH, because MATH. These facts, however, contradict REF . So, it must be that MATH. We divide into the two cases, REF MATH and REF MATH. First consider the case MATH. Set MATH. Then we have three cases: CASE: MATH, REF MATH, and REF MATH. First consider REF . Now, MATH, so MATH. Conversely, whenever MATH, clearly MATH with MATH (we use that MATH always preserves orientation). In this case, MATH and MATH, as desired. This situation, characterized by MATH, falls into REF and we will call it REF . Next, take REF . In this case, MATH, so MATH and MATH. Conversely, it is clear that MATH iff MATH from the construction of MATH, and in this case, MATH with MATH. Thus, MATH and MATH, as desired. This is a different case of REF , so let us call it REF . Finally, consider REF . In this case, MATH. Set MATH. Then it follows that MATH, so that MATH with MATH. Conversely, if MATH, then MATH with MATH, as desired (this can be checked separately when MATH reverses orientation - here MATH so there is no difficulty.) Hence, MATH and MATH. This is the final case of REF , so let us call it REF . Next, consider the case MATH. Set MATH. Because MATH, it follows that MATH, hence MATH with MATH. Since MATH iff MATH, we have MATH. Conversely, whenever MATH, then MATH with MATH. Hence, MATH and MATH. This accounts for REF . We have proved the first part of the Lemma, because we have considered all possible nonzero values of MATH and MATH, and grouped them into REF . We have shown that each of these is associated with different values of MATH, which justifies the zero values of MATH and MATH in each case. Next, we apply the same analysis used in the first part to show that exactly one of REF holds. Let MATH and MATH. We suppose that MATH and divide into the cases MATH and MATH. First consider MATH. Then it is clear that MATH, with equality iff MATH. Now, MATH is defined on MATH, and MATH, so MATH preserves orientation on MATH. This shows that MATH, so that MATH with MATH. That is, MATH. Conversely, whenever MATH, we know from the fact that MATH preserves orientation that MATH preserves orientation on MATH, and hence that MATH with MATH. It is then clear that MATH. This accounts for REF . Now, suppose that MATH with MATH. We divide into the cases MATH and MATH. First suppose MATH. In particular, this implies that MATH is defined on MATH, and by nilpotency, it must preserve orientation. Also, MATH must preserve orientation on MATH. So, we see that MATH, thus MATH with MATH. Conversely, if MATH and MATH, we see from MATH and REF that MATH. It follows that MATH with MATH, and MATH, as desired. This accounts for one situation of REF ; call this REF . Next, suppose MATH and MATH. We further divide into the three REF MATH, REF MATH, and REF MATH. In REF , we use that MATH preserves orientation to see that MATH, so that MATH with MATH. Conversely, whenever MATH, it follows that MATH using REF : otherwise we would have MATH with MATH while MATH. Hence, MATH with MATH. Call this situation REF . In REF , we have that MATH. Conversely, whenever MATH, we must have MATH by REF , and then MATH and MATH, as desired. If MATH, then it also follows from REF considering MATH that MATH, and then MATH. Call this situation REF . Finally, we consider REF . Now, MATH shows that MATH. By nilpotency, MATH must preserve orientation on MATH, and it follows that MATH with MATH. Conversely, if MATH with MATH, then it follows that MATH, MATH. Hence, MATH with MATH, and MATH. This has the same passing properties as REF , so call this situation REF . We have finished the second half of the Lemma, since we have accounted for all possible values of MATH, and MATH in REF . Each of these are associated with distinct values of MATH with rexspect to the MATH, and MATH, once MATH is fixed. To obtain the result for decreasing quadruples, simply reverse all directions and permute the components of all MATH-quadruples (not pairs!) in this proof. |
math/0009173 | Clearly MATH iff MATH. Suppose MATH. Hence MATH. Suppose for a contradiction that MATH. Then, MATH. However, MATH and REF imply that MATH, a contradiction. So, the identity MATH easily follows (since MATH). The rest of the proof is almost exactly the same as the proof of the first part of REF , getting rid of MATH. Again, the results follow with simple modifications in the decreasing case. |
math/0009173 | The proof is the same for both REF , does not mention MATH, and is given in the following paragraphs. First, we note that MATH iff MATH and MATH reverse orientation, which is true iff MATH and MATH reverse orientation, which is true iff MATH. Now, we show that MATH implies MATH. To reach a contradiction, suppose that MATH and MATH. By nilpotency, MATH and MATH preserve orientation. Thus MATH and MATH also preserve orientation. Write MATH. First suppose MATH. Then MATH, so that MATH is defined on MATH and therefore on MATH. However, this implies that MATH, while MATH. This contradicts REF . So MATH. Then MATH and MATH. Since MATH, it must be that MATH, which implies that MATH. So MATH. Now, write MATH, for MATH. Since MATH, it follows that MATH REF or MATH (MATH). By REF , it follows that MATH divides both MATH and MATH. But then MATH divides MATH and MATH, which shows that MATH, contrary to assumption. Next, suppose MATH. We show MATH. Find MATH and MATH such that MATH and MATH. In this case, MATH. Hence, MATH. By assumption, MATH. Since MATH, this in particular implies that MATH, as desired. |
math/0009173 | Again, REF have nearly the same proof, which follows. Suppose, on the contrary, that MATH. Clearly MATH, else MATH would have reversed orientation, which is not possible by REF . So MATH. Now, MATH is defined on MATH since, in REF is defined on MATH and MATH, which follows from MATH, and in REF is defined on MATH. In particular, MATH is defined on MATH. But now, MATH but MATH, contradicting REF . |
math/0009173 | First we note that, if MATH, then MATH, for the following reason. Suppose MATH and write MATH or MATH, depending on whether MATH is bad or good. Then MATH, so REF applies and shows that MATH, as desired. By symmetry, MATH implies that MATH. So, it suffices to show that MATH if MATH and MATH if MATH. By the symmetry of the situation, we need only prove the first. Assume, for sake of contradiction, that MATH and MATH. First suppose that MATH. Then we note that MATH is defined on MATH because it is defined on MATH and MATH. Next we note that MATH because MATH. Also, MATH, and MATH. This contradicts REF . On the other hand, it is impossible that MATH. If this were true, then it would follow that MATH shifted MATH to the right by MATH, while MATH. This would contradict REF . |
math/0009173 | Clearly MATH would imply that MATH and MATH reverse orientation, which is not possible since MATH. Suppose instead that MATH. We have that MATH, and hence MATH, is defined on MATH, and hence MATH. Since MATH, we find that MATH, contradicting REF . So MATH. |
math/0009173 | First, note that MATH is defined on MATH since it is defined on MATH, and the former is a subset of the latter on the diagram. If MATH, then set MATH. If MATH and MATH, set MATH. In either case, MATH is defined on MATH. Indeed, MATH is defined on MATH, and MATH. Note that, by REF , MATH and MATH must preserve orientation. Hence, MATH and MATH must preserve orientation on MATH. Set MATH. We have MATH, contradicting REF . |
math/0009173 | This follows easily from REF . |
math/0009173 | This easily follows from REF . |
math/0009173 | Clear. |
math/0009173 | CASE: This follows immediately from REF in the cases that MATH and MATH. (Note that it is impossible to have MATH by REF .) In the case that MATH and MATH or in the case that MATH and MATH, it is clear that MATH. CASE: It follows that only one of the MATH can be in MATH from REF . If MATH, then no MATH is outer by REF . By REF , if MATH, there is nothing to prove. If MATH and MATH, then REF gives the desired result. For MATH the result is easy. CASE: The two statements follow from REF , respectively. |
math/0009173 | That MATH, and MATH are disjoint from MATH follows from REF , and the definition, respectively. The other facts regarding disjointedness are obvious. To check that the union is all of MATH, we apply REF .ii, which shows that MATH. It is clear, though, that MATH. This proves the desired result. |
math/0009173 | CASE: This follows immediately from REF . CASE: This follows immediately from REF .ii, REF, and REF. CASE: This follows immediately from REF . CASE: This follows immediately from the definition of MATH. CASE: This follows from REF , since we know MATH is disjoint from MATH. |
math/0009173 | For every MATH, we see from the Lemma, REF , that MATH. Furthermore, MATH and MATH where these are defined (Lemma, REF). Now take any MATH. First note that either MATH or MATH because MATH. Note that MATH because MATH and MATH are disjoint from MATH REF . Now, if MATH, then using REF , MATH (of type a or b, depending on whether MATH or MATH, respectively). If MATH, then clearly MATH. In this case, MATH by the Lemma, REF , and REF , using REF . Furthermore, in this case MATH, again from the Lemma, REF . Then, MATH. Thus, we see that all elements MATH fit into at least one of the sets REF or REF . To see that all elements in MATH fit into at most one, note that the way in which MATH appears is uniquely determined by which of the following holds: CASE: MATH, MATH; CASE: MATH, MATH; CASE: MATH, MATH. Now, it remains to consider those elements of MATH. For any MATH, we know that MATH by REF . The former two are subsets of MATH, so we know that MATH for any MATH, and clearly MATH is unique. The negative case is almost identical to this one. |
math/0009173 | The terms in the expansion of MATH are MATH for each MATH, as well as the terms MATH for MATH and MATH for each MATH. |
math/0009173 | CASE: Let MATH. The negative case is similar (see comments at the end of the proof of this part). Let MATH and let MATH be the part of the formula for MATH which will not change upon applying MATH, MATH, and MATH (where applicable). Let MATH if MATH and MATH otherwise. Set MATH. First we show that MATH . First, by REF , MATH . Hence, MATH . If MATH, then MATH and MATH. In this case, REF follows from the definition of MATH. If MATH, then MATH, and REF implies MATH. Then REF shows that MATH so that the left-hand side of REF is MATH. By REF, this is the same as the right-hand side. Next, we show that MATH . Naturally, we may assume that MATH reverses orientation. Since, in this case, MATH also reverses orientation, it must be that MATH. Suppose now that MATH and MATH. Now, for every MATH such that MATH, set MATH, MATH, and MATH. Then MATH are exactly those reversed chains such that MATH. Now, we consider the possible passing properties of the MATH-pairs involved. Note first that, since MATH, MATH for all MATH. For the same reason, MATH for all MATH. Next, note that MATH for all MATH - otherwise, applying MATH to MATH would contradict nilpotency. For the same reason, MATH for all MATH. Next, note that MATH for all MATH. This follows from the fact that MATH for all MATH, making use of REF . Now, REF follows readily. Let MATH when MATH and MATH otherwise. Finally, we show that MATH . The proof is similar to the proof two paragraphs back. If MATH, then MATH. By REF , MATH. Thus MATH, proving REF in this case (note that MATH and recall the definition of MATH.) If MATH, then MATH and by REF (or the equation above), MATH and MATH. If we set MATH, then it follows from REF that neither MATH nor MATH is passed, so MATH. Hence, MATH, so the left-hand side of REF is MATH, which is equal to the right-hand side by REF and the above analysis. Now, putting REF, and REF together, we get MATH as desired. The negative case is almost the same as the above, except that the ``error term" MATH over-corrects, but this is counteracted by the fact that now MATH is defined on MATH as well as MATH. (The details are omitted.) REF This is almost the same as the proof above. First we note that the result is clear when MATH, as MATH and MATH gives the desired formula (bearing in mind the definition of MATH). Suppose MATH. Let MATH and let MATH if MATH and MATH, and MATH otherwise. Let MATH. Set MATH. First we show that MATH . REF shows that MATH and MATH . First suppose MATH. In this case, REF implies REF easily. If MATH, then MATH, and MATH. Then, MATH. Furthermore, MATH. From this REF follows. Next, we claim that MATH . This follows by exactly the same arguments as used in the second paragraph of the proof of REF . But, MATH, so, putting REF together, the result easily follows. The negative/MATH case is almost the same as this, bearing in mind the final comments in the proof of the previous part. |
math/0009175 | The first assertion can be easily checked. The second part follows from the computation in CITE. For completeness sake we give the argument here: Let MATH be the NAME of MATH along MATH. Then MATH has the presentation MATH . Obviously, we have a epimorphism MATH mapping MATH to MATH, MATH to MATH, and MATH to MATH. It only remains to show that every relation in the given presentation of MATH follows from the relations of MATH. Observe first in MATH that by conjugation with MATH, MATH implies MATH. Moreover, commutativity is commutative, that is, MATH implies MATH. Hence, it remains to prove MATH in MATH for MATH. We will do this by induction on MATH. Assume therefore MATH commutes with MATH for MATH. Conjugate the relation MATH with MATH. We obtain MATH . Now observe that by induction MATH commutes with MATH and with MATH. This second relation also implies (by conjugation with MATH) that moreover, MATH commutes with MATH. Therefore, we can simplify the commutator in REF to the desired MATH . By induction we therefore see that MATH. Using the presentation, we next check that the abelianization of MATH is isomorphic to MATH, and MATH are mapped to two free generators, whereas MATH is mapped to zero. Therefore, MATH is equal to the normal subgroup generated by MATH, which is generated by MATH, MATH, MATH. All these elements are of order MATH, and by conjugation with sufficiently high powers of MATH we see that they all commute. Therefore, MATH is a vector space over MATH with countably many generators, and therefore isomorphic to a countable direct sum of copies of MATH. Observe, however, that MATH is quite different from the base of the NAME MATH. The element MATH is a typical example which is not contained in MATH but in MATH. |
math/0009175 | Observe that if MATH is induced from MATH, that is, MATH (so we can view MATH also as an operator on MATH), then essentially MATH and we deduce that MATH (compare CITE). Therefore, it will be sufficient to find MATH such that MATH. Take MATH of REF . Choosing MATH and MATH, we see that REF is in the spectrum of MATH, and that MATH. |
math/0009175 | We perform a standard construction where one attaching map will be given by the MATH of REF , compare for example, CITE. Let MATH be a finite MATH-dimensional NAME with MATH, for example, the MATH-complex of the finite presentation given above. Let MATH be the wedge product of MATH and MATH. The corresponding map MATH generates a free copy of MATH inside MATH. Define now MATH, where MATH is given by MATH of REF , and where MATH is given using MATH. Choosing an appropriate basis of cells, it follows that on the cellular MATH-chain complex MATH of the universal covering MATH of MATH, the differential MATH is given by the matrix MATH, where MATH denotes transpose and MATH is the number of MATH-cells in MATH. Since there are no MATH-cells, MATH is zero. Consequently, MATH . |
math/0009175 | Choose a finite MATH-dimensional simplicial complex MATH homotopy equivalent to the NAME MATH of REF . Then embed MATH into MATH CITE and thicken MATH to a homotopy equivalent MATH-dimensional compact smooth manifold MATH with boundary MATH, such that moreover the inclusion of MATH into MATH is a homotopy equivalence CITE. Recall that a map MATH between two NAME is called a MATH-equivalence (MATH), if MATH is an isomorphism for MATH, and an epimorphism for MATH. By transversality CITE, every map of MATH or MATH to MATH with MATH is homotopic to a map into MATH. It follows that the inclusion MATH is a MATH-equivalence. Consequently, by CITE, MATH and MATH. If we choose a smooth Riemannian metric MATH on MATH, then by the MATH-Hodge NAME theorem CITE we also obtain MATH. |
math/0009179 | If the lemmas does not holds then there exists critical points in MATH. Let MATH be the closest one of MATH. MATH, otherwise there will be another critical point between MATH and MATH. But this implies MATH, where MATH, which is absurd. |
math/0009179 | If the intersection number is at least REF, the interior of MATH will contain at least two intervals MATH and MATH in the family MATH, for some MATH. Since MATH, for all MATH, the intervals MATH belong to MATH, which is absurd, since MATH satisfies MATH. |
math/0009179 | Let MATH be the maximal interval such that MATH, MATH and MATH is monotone in MATH. We claim that MATH. Otherwise there exists a critical point MATH to MATH in the boundary of MATH such that MATH is in the interior of MATH. Then, by REF , the interior of MATH contains MATH, MATH, in the left or right side of MATH. This proves the first statement. The second statement is obvious, since MATH restricted to MATH is monotone and MATH, for all MATH. |
math/0009179 | Similar to previous lemma. |
math/0009179 | Immediate consequence of the previous lemmas. |
math/0009179 | This follows of the non existence of wandering intervals. |
math/0009179 | Use that MATH and MATH are bi-Lipchitz in a neighborhood of zero. |
math/0009179 | We will use the smallest interval trick. Let MATH be the smallest interval in MATH. Observe that MATH, where MATH depends on MATH. Taking MATH, the interval MATH satisfies MATH. Consider the maximal interval MATH which contains MATH such that MATH is monotone in MATH and MATH. We claim that MATH. Otherwise, there exists MATH such that MATH is a critical point for MATH, MATH. By REF , MATH contains an interval in MATH. This is a contradiction. By the real NAME lemma and the quasi symmetry at the critical points, there exists MATH such that MATH satisfies MATH. Use REF many times and the real NAME lemma to conclude that there is MATH such that MATH satisfies MATH. In particular, MATH is a MATH-neighborhood of MATH. By REF , real NAME lemma and the quasi symmetry at the critical points, one gets control of distortion in the monotone parts of the k-cycle and that the pullback of MATH along the k-cycle is a MATH-neighborhood of MATH. Moreover, this neighborhood is contained in MATH, by REF . Hence MATH. Since MATH goes to zero, by REF , one gets the second statement. In particular, if MATH is small with respect to MATH then MATH contains an attractor which attracts MATH, because the combinatorics is bounded and by the previous lemma. But this is absurd, since MATH is constant for large MATH. |
math/0009179 | Let MATH. Near of MATH, we have MATH. By the real bounds MATH . We learn this argument in CITE. |
math/0009179 | Because MATH, MATH will return to MATH after a bounded number of iterations of monotone and folding parts of the MATH-th renormalization. In particular, by real bounds, the derivative of map MATH is bounded away from infinity. This proves the lemma. |
math/0009179 | Assume by contradiction that MATH is very close to a point MATH with MATH. By the real bounds for the partial monotone parts of the MATH-th renormalization, MATH is very close to MATH with respect to level MATH. But MATH returns to MATH after a bounded number of iterates of MATH. Since MATH is deeply inside MATH and by the control of derivatives of MATH in MATH, this is a contradiction. |
math/0009179 | Let MATH. Let MATH be the closest point of MATH in MATH such that MATH. Note that if MATH is close to MATH, using REF and that MATH, there will be an attractor in the interval MATH which attracts a critical point. This is an absurd. The point MATH cannot be close to the periodic point in MATH, otherwise, by the real bounds, MATH would be close to the boundary of MATH, which is impossible. |
math/0009179 | Note that MATH is commensurable with MATH. By the real bounds, the map MATH has bounded derivative. This proves the lemma. |
math/0009179 | Let MATH be as in REF . In the proof of REF we proved that MATH contains a MATH-neighborhood of MATH. Let MATH. Let MATH be the pullback of MATH along the k-cycle. By REF , MATH. Let MATH be a connect component of MATH. There are two cases: if MATH is a connect component of MATH then MATH clearly contains an interval in the family MATH or a mirror image of one. Otherwise MATH is a connect component of MATH and it contains an interval in the family MATH, by the maximality of MATH. Note that this interval contained in MATH is commensurable with MATH REF and it contains a critical point which cuts it in commensurable parts REF . Hence MATH, since MATH does not contains critical points for MATH. Replacing MATH, if necessary, by a smaller interval between MATH and MATH, we obtain the second statement of ii. |
math/0009179 | Note that MATH is the connect component of MATH which contains MATH. Hence MATH is commensurable with MATH and MATH, for some MATH, by REF . Let MATH be the boundary point of MATH inside MATH. Surely MATH. It is sufficient to prove that the interval MATH is commensurable with MATH. Indeed, if MATH is very small, by the real bounds and REF there will be an attractor which contains a critical point in its immediate basin. This is absurd, since MATH is constant for large MATH. |
math/0009179 | Observe that MATH is not small with respect to MATH. Otherwise, by the real bounds and bounded combinatorics, there will be an attractor in MATH which attracts the critical point MATH. Suppose that a point in the boundary of MATH is close to the boundary of MATH. By the control on the derivatives in the monotone and folding parts of the MATH-th renormalization, such point will return to MATH after a large number of iterations of the monotone and folding parts of the MATH-th renormalization, which is absurd, because the combinatorics is bounded. To prove the second statement, let MATH and MATH be such that MATH. Let MATH be the minimal times such that MATH and MATH. By REF , the real NAME lemma and the quasi symmetry at the critical points, for every MATH, MATH, there is in each side of MATH an interval MATH commensurable with MATH such that MATH is monotone in MATH. Here, MATH is the time of first entry of MATH to MATH. Hence if the space between MATH and MATH is small and MATH, then MATH, MATH are very close distinct points in MATH. Moreover MATH, MATH are distinct points in the orbit of the periodic point contained in MATH. By the control of derivatives of MATH-renormalization and bounded combinatorics, the interval MATH is mapped monotonically into itself by the map MATH, which is absurd, since the negative NAME derivative of MATH in MATH implies that an attractor in MATH contains a critical point in its immediate basin. |
math/0009179 | If the statement does not holds then there exists MATH, for some MATH, very close to MATH with respect to level MATH, hence MATH. After a bounded number of iterations of MATH, MATH is mapped inside MATH. But this is impossible, because MATH is very close to a point in MATH, which never returns to MATH. |
math/0009179 | Since the maps MATH are compositions of univalents maps which map the real line to the real line and the maps MATH belong to the NAME class, we can use REF (see also REF) in pg. REF. |
math/0009179 | Use the control on the first and second derivatives given by complex NAME lemma. |
math/0009179 | If MATH is close to MATH, use REF ; otherwise, capture MATH in a NAME neighborhood whose diameter is commensurable with MATH and use REF . For details, see CITE. |
math/0009179 | (compare with the proof of REF) The pullback of MATH along the MATH-cycle exists. Hence let MATH, MATH, be as in REF . The complex pullback of MATH, MATH, until MATH by MATH is inside MATH, with MATH, by REF . By REF , MATH cuts MATH in commensurable parts, where MATH is the predecessor of MATH at level MATH. Hence , by REF , the complex pullback of MATH by MATH along the orbit of MATH is contained in MATH, where MATH does not depend on MATH. By induction, assume that the complex pullback of MATH along the MATH-cycle until MATH is inside MATH. Again the complex pullback of MATH until MATH by MATH is inside MATH, MATH, by REF . The complex pullback of MATH by MATH along the orbit of MATH is contained in MATH, where MATH is the predecessor of MATH at level MATH, by REF . Hence the complex pullback of MATH along all MATH-cycle is inside MATH, for some MATH. Observe that if MATH, for some MATH, then MATH and MATH are inside MATH, which has diameter commensurable with MATH (because MATH). This proves a. Assume MATH and suppose that MATH is not inside MATH. Every point MATH satisfies MATH, for all interval MATH inside MATH, for a definitive MATH, since MATH is MATH-deeply contained in MATH. This proves b. To prove c, suppose that MATH does not MATH-jump in returns to level MATH, MATH. If MATH then MATH. By REF .ii and REF , MATH does not MATH-jump for each MATH, with MATH. |
math/0009179 | Let MATH be the smallest time satisfying the above properties. We have MATH, for some MATH. Then MATH, by REF (use REF successively, and finally REF). Since MATH is bounded, by REF there exists MATH such that MATH. Hence MATH . Since MATH-jumps and MATH, by real bounds, MATH is contained in MATH. By REF , there exists an interval of monotonicity MATH for MATH which contains MATH and such that MATH contains MATH, for some MATH. Moreover MATH, where MATH depends only on MATH. By REF MATH . |
math/0009179 | (compare with the proof of REF) Let MATH be the maximal interval of monotonicity of MATH which contains MATH. Since MATH is contained in MATH, the interval MATH is contained in MATH. In particular, MATH is very small. Assume that REF does not hold. By REF , MATH never MATH-jumps for MATH. Hence, by REF , MATH, with MATH. Since MATH is MATH-deeply contained in MATH, then MATH, if MATH is small enough. |
math/0009179 | (compare CITE, subsection REF) Let MATH, with MATH, be such that MATH. In other words, MATH is in the 'scale' of level MATH. If MATH, we are done by REF . Otherwise MATH. Note that MATH, where MATH and MATH. If MATH is zero, using arguments as in the proof of REF , MATH is well defined and it is contained in MATH, where MATH and MATH are as in the proof of REF (replacing MATH by MATH). In particular MATH. Since MATH, by REF MATH. By REF (recall REF ), we obtain the lemma. Otherwise, suppose that MATH-jumps for some MATH, where MATH is a return to level MATH. Then we are done, by REF . If MATH-jumps in a return to level MATH does not happen then MATH, by REF . Let MATH be the deepest level such that MATH, with MATH distinct of zero, MATH and MATH. Note that MATH is bounded. Suppose, by induction, that MATH, MATH, and MATH does not MATH-jump, for MATH. By REF , either there is a return to level MATH between MATH and MATH, which MATH-jumps, or MATH and moreover MATH does not MATH-jump for MATH, by REF . If there are not good returns between the times MATH and MATH then MATH. Then either there are good returns to level MATH between MATH and MATH, and we can use REF , or MATH. Using arguments as in the proof of REF, MATH is well defined and it is inside MATH, where MATH and MATH are as in the proof of REF (replacing MATH by MATH). In particular MATH. Since MATH, one gets MATH, by REF . By REF (recall REF ), we obtain the lemma. |
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