paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0104019 | The proof does not make use of MATH being f.g. projective. Suppose MATH is a field and MATH is MATH-separable. Let MATH denote the right global dimension of a ring MATH and MATH denote the projective dimension of a module MATH. Then MATH since MATH is finite dimensional semisimple. By NAME 's Theorem for split extensions CITE, MATH whence MATH and MATH is semisimple CITE. Then MATH and MATH are finite dimensional semisimple algebras. It follows from CITE that MATH and MATH are symmetric algebras. Similarly, we arrive at symmetric algebras MATH and MATH via CITE under the assumption that MATH is MATH-projectively MATH-separable with no restriction on MATH. For then MATH is MATH-projective and MATH-separable by transitivity for projectivity and separability. Now we compute using the bimodule isomorphisms MATH and MATH and the hom-tensor adjunction: MATH . Then, since MATH is f.g. projective, MATH is a NAME extension. |
math/0104019 | We note that MATH is a nontrivial idempotent in MATH. Consider MATH. Of course, MATH. We compute with MATH: MATH . |
math/0104019 | We have seen in the lemma that MATH is separable. It is split since we easily check that MATH below is a bimodule projection: MATH is countably generated projective, since MATH defined by MATH and MATH, where MATH, which satisfies the dual bases equation, MATH . By tranposing the elements and maps above, we similarly find a countable projective base for MATH. The rest of the proof is now clear. |
math/0104019 | For MATH to be a NAME extension, we must have MATH finitely generated from the very start. A right projective NAME extension is right f.g. by CITE. |
math/0104019 | In fact this is an easy consequence of CITE. We restrict to giving a description of the isomorphism MATH, the other isomorphisms are similar. For a natural transformation MATH, we define MATH by MATH . Conversely, for MATH, MATH is defined by MATH . The fact that MATH and MATH are each others inverses is proved using REF . |
math/0104019 | Assume that MATH is separable. By NAME 's Theorem, there exists MATH such that MATH for all MATH. Let MATH be the corresponding natural transformation of REF , that is, MATH, and MATH, and the first implication of the Proposition follows. The converse follows trivially from NAME 's Theorem. All the other equivalences can be proved in a similar way. |
math/0104019 | For MATH, we put MATH. By definition, MATH is left MATH-linear. For any MATH, we consider the left MATH-linear map MATH, MATH. The naturality of MATH implies that MATH, so MATH is also right MATH-linear. MATH is given by MATH, with MATH . Using REF , we easily deduce that MATH is right MATH-linear, and that MATH is left MATH-linear and right MATH-linear. Assume that MATH is finitely generated projective, and, as above, assume that MATH is a dual basis. We can then define the inverse MATH of MATH as follows. We view MATH as a map MATH, and we identify MATH and MATH. We then define MATH by MATH . It is clear that MATH is natural and that MATH and MATH are each others inverses. For MATH, we define MATH by MATH for all MATH in MATH. Straightforward computations yield that MATH is well-defined, and is indeed an inverse of MATH. |
math/0104019 | The proof is similar to the previous one, so we restrict to giving the connecting maps. For a natural transformation MATH, we define MATH. Conversely, take MATH. MATH is defined as follows: for all MATH, we put MATH . Now we define MATH. For MATH, we let MATH be given by MATH . Straightforward computations show that MATH, and that MATH is MATH-bilinear. |
math/0104019 | MATH. If MATH is NAME, then MATH is a left adjoint, and therefore right exact and preserving direct limits, so MATH is necessarily finitely generated and projective. Let MATH and MATH be the counit and unit of the adjunction MATH, that is, MATH for all MATH and MATH. Let MATH and MATH. Putting MATH in REF , we find REF . Then take MATH in REF . Making the identification MATH (MATH is finitely generated projective), we find MATH and MATH, so REF follows. MATH. Let MATH and MATH. MATH and MATH satisfy REF , so MATH is NAME. MATH. REF implies that MATH is finitely generated projective. Let MATH and MATH. We easily compute that MATH and MATH are each others inverses. MATH. If MATH and MATH are isomorphic as MATH-bimodules, then there exist MATH and MATH that are each others inverses. Put MATH and MATH. Straightforward computations show that MATH and MATH satisfy REF . |
math/0104019 | Completely similar to the proof of REF . Let us mention that MATH . |
math/0104019 | The equivalence of the first three statements is obtained in exactly the same way as REF . The equivalence of the third and the fourth statement is one of the equivalences in REF . |
math/0104019 | Let MATH be a natural transformation. Then MATH is left MATH-linear, and right MATH-linear since MATH is natural. Given a MATH-linear map MATH, we define a natural transformation MATH as follows: MATH. |
math/0104020 | If MATH, then MATH where MATH is the spectral decomposition of the symmetric positive definite matrix MATH, and MATH. This gives the eigenvalue decomposition of MATH, with eigenvalues the positive entries of MATH and eigenvectors the columns of MATH. Conversely, suppose MATH, where MATH is diagonal with positive diagonal entries. Then we can write MATH. |
math/0104020 | Let this closed subgroup be denoted by MATH, and let MATH be its NAME Algebra. Since MATH is connected, it suffices to show that MATH, or in other words that the tangent space of the manifold MATH at the point MATH coincides with the set of MATH skew - symmetric matrices over the reals. In fact, use of the exponential mapping MATH shows that MATH and MATH share a neighbourhood MATH of MATH, and parallel transport by left trivialisation shows that MATH and MATH share the neighbourhood MATH of any element MATH. Therefore, MATH is both open and closed as a subset of MATH, and since MATH is connected, the result follows. It remains to show that MATH: Let MATH have eigenvalues MATH. Then MATH for all MATH, since the MATH linearly independent eigenvectors of MATH are also eigenvectors of MATH and correspond to the strictly positive eigenvalues MATH, MATH. The NAME - series development of MATH shows that MATH . Upon taking squares on both sides of the ansatz MATH we get MATH, and hence MATH. REF thus yield the identity MATH and this shows that MATH for all MATH. Clearly, we have MATH as expected. Now, for MATH and MATH we get MATH. Hence, MATH. Let MATH be the permutation matrix that permutes the MATH and MATH variables, and let MATH, where MATH is the MATH - th element in the canonical basis of MATH. Then consider MATH . Clearly, MATH and MATH. But since MATH forms a basis of MATH we find that the elements of MATH span this whole space. This shows the claim. |
math/0104020 | The invariance REF is clearly preserved when taking compositions and limits. Hence, REF implies that the symmetric positive definite matrix MATH satisfies the condition MATH for all MATH. It is a trivial matter to prove that this forces MATH to be a scalar, and the result now follows from REF. |
math/0104020 | It follows from REF and the fundamental theorem of differential and integral calculus that MATH is of the form MATH, where MATH, MATH and MATH is a linear form on MATH, that is, there exists a symmetric matrix MATH such that MATH for all MATH. One of the conditions in the definition of a MATH - self - concordant barrier functional MATH for a convex open domain MATH is that the length of the NAME step MATH at MATH, measured in the Riemannian metric MATH defined by MATH be uniformly bounded by MATH, see for example . CITE, that is, MATH . In particular, in the case of MATH this means that MATH for all MATH. But clearly, this implies that MATH. |
math/0104020 | Upon exchanging the roles of MATH and MATH, REF implies that REF is equivalent to MATH . Setting MATH and using MATH in REF, we get MATH for all MATH. Let us show that MATH for all MATH. Applying REF both to MATH and MATH we get MATH and hence we obtain the identity MATH for all MATH, which generalises REF. Set MATH . It follows from the definition of the geometric mean and from the fundamental formula that MATH . Together with REF this implies MATH . Therefore, the set MATH can be written as MATH. By NAME 's REF, MATH generates MATH . This implies that the point MATH is fixed by all NAME automorphisms MATH . Finally, REF (in which the assumption of irreducibility for MATH is essential) says that the group MATH acts transitively on the set of primitive idempotents of MATH. The spectral theorem applied to MATH therefore implies that MATH for some positive real number MATH. Together with REF this implies that MATH for all MATH. |
math/0104020 | Since the Hessian MATH are positive definite cone automorphisms, REF implies that there exists a well - defined function MATH such that MATH . Since MATH is self-scaled, we have MATH for all MATH where MATH denotes the scaling point of MATH and MATH for the self - scaled barrier MATH. The quadratic representation MATH is injective on MATH, see REF. Therefore, the identity above shows that MATH for all MATH . By REF , we have MATH for all MATH. Now, MATH by definition of MATH, and REF shows that we have MATH, which together with REF shows that MATH for all MATH . The proof is now completed by REF . |
math/0104020 | Similar to that of REF . |
math/0104023 | Since MATH is finite dimensional and the graded algebra MATH is generated by MATH the quotient algebra MATH is a finite dimensional MATH-algebra. Define a homomorphism MATH by MATH. Since the elements of MATH act trivially on MATH, we have an induced embedding MATH. This endows MATH with the structure of an algebraic group over MATH. Since MATH consists of those MATH satisfying the polynomial conditions MATH and MATH, we see that MATH is a closed subvariety of MATH. |
math/0104023 | Denote by MATH and MATH the lower central series of MATH and MATH, respectively. Consider the induced map of associated graded algebras MATH where MATH. This algebra is a NAME algebra over MATH with bracket induced by the commutator in MATH. Note that as an algebra MATH is generated by MATH. Consider the commutative diagram MATH . Since MATH is surjective by hypothesis and the vertical arrows are also surjective, we see that MATH is surjective. Now, say that MATH, but MATH. Then since MATH is surjective for all MATH, we must have MATH; that is, MATH. This can occur only if MATH (since MATH is nilpotent). Consider the commutative diagram MATH . This shows that MATH surjects onto MATH. A simple induction then shows that MATH is surjective. |
math/0104023 | For each MATH, define a map MATH by MATH . The map MATH is a homomorphism since MATH . The kernel of MATH is precisely the subgroup MATH. We therefore have an induced injection MATH . We claim that the map MATH is also surjective. To see this, consider the composite MATH (the map MATH is the quotient map MATH). This composite is clearly the canonical isomorphism MATH. It follows that the map MATH is surjective. Now consider the commutative diagram MATH . Since MATH is surjective and MATH is injective, a diagram chase shows that MATH is surjective. |
math/0104023 | Recall the filtration MATH of MATH defined by MATH. By REF , the graded quotients satisfy MATH. Hence, the algebra MATH is generated by MATH. We have the commutative diagram MATH . The top horizontal arrow is surjective since MATH, and the vertical arrows are surjective since the algebras MATH and MATH are generated by MATH. It follows that the map MATH is surjective. We have an exact sequence MATH . Since MATH is surjective, we see that MATH. This implies that MATH for all MATH. But since MATH, we see that MATH for all MATH. In particular, MATH. Thus we have the commutative diagram MATH which shows that the map MATH is an isomorphism. |
math/0104023 | Denote by MATH the NAME closure of the image of MATH in MATH. Since the composite MATH is an isomorphism REF , the map MATH is surjective. Note that if MATH is a unipotent group over MATH, then MATH is an abelian MATH-group and hence MATH. REF implies that the map MATH is surjective; that is, MATH. |
math/0104023 | Note that MATH is a prounipotent group over MATH and that the image of MATH in MATH is NAME dense. Denote by MATH the unipotent MATH-completion of MATH. By the universal mapping property, there is a unique map MATH making the diagram MATH commute. We show that MATH is an isomorphism. Since the map MATH has NAME dense image, so does the map MATH. It follows that MATH is surjective since MATH is a closed subgroup of MATH CITE, p. CASE: To see that MATH is injective, suppose that MATH is a unipotent group over MATH. Then MATH is a subgroup of some MATH for some MATH (here, MATH is the subgroup of upper triangular unipotent matrices). A representation MATH induces a ring homomorphism MATH defined by MATH. The image of MATH under MATH lies in the subalgebra MATH of nilpotent upper triangular matrices. Thus, the kernel of MATH contains MATH and hence MATH induces a map MATH . Since MATH, we see that MATH is contained in MATH. If the image of MATH is NAME dense in MATH, then MATH. Thus, the diagram MATH commutes. Now, MATH, with each MATH unipotent over MATH. Applying the above construction to the compositions MATH, we see that MATH is injective as follows. If MATH, then MATH maps to MATH in each MATH. But the diagram MATH shows that MATH for each MATH; that is, MATH. |
math/0104023 | The group MATH is the set of grouplike elements in MATH. On the other hand, the group MATH is the set of grouplike elements of MATH. |
math/0104023 | Let MATH be a MATH-vector space with basis MATH. Then we have a bijection MATH . On the other hand, if MATH is a vector in MATH, the map MATH identifies MATH as a unipotent group over MATH. The universal mapping property of unipotent completion then provides a bijection MATH that is, MATH and MATH both represent maps of MATH into MATH. It follows that the natural map MATH is an isomorphism. |
math/0104023 | This is essentially the same as REF . Each MATH is a unipotent MATH-group by hypothesis and since MATH surjects onto each MATH, the image of MATH in MATH is NAME dense. Since MATH is finite dimensional, the unipotent MATH-completion of MATH is the group MATH. The universal mapping property provides a unique map MATH of prounipotent MATH-groups; it is surjective since MATH is a closed NAME dense subgroup of MATH. The injectivity of MATH follows since the composition MATH must factor through some MATH. |
math/0104023 | CITE The group MATH acts on the completion MATH by conjugation. This action preserves the filtration by powers of MATH, so MATH acts on the associated graded algebra MATH . If MATH is finite dimensional, then each algebra MATH is finite dimensional. Thus, each of the groups MATH is an algebraic group. Since MATH is generated by MATH, it follows that MATH, the group of augmentation preserving algebra automorphisms of MATH, is a proalgebraic group which is an extension of a subgroup of MATH by a prounipotent group MATH: MATH . If the action of MATH on MATH factors through a rational representation MATH, then we can form a proalgebraic group extension MATH of MATH by the prounipotent group MATH by pulling back the extension REF along MATH. Since the map MATH factors through MATH, we can lift the map MATH to a map MATH whose composition with MATH is MATH. By the universal mapping property of relative completion, this induces a homomorphism MATH. Since the composite MATH is the action of MATH on MATH by inner automorphisms, we see that the kernel of this map is the center of MATH. It follows that the kernel of MATH is central in MATH. |
math/0104023 | We first show that the composition MATH is surjective. Let MATH be the cokernel of this map; it is a rational representation of MATH. Pushing out the extension MATH along the map MATH, we obtain an extension of algebraic groups MATH . This splits; say MATH is a splitting. Since the image of MATH in MATH is NAME dense, the image of MATH in MATH is NAME dense. But since the diagram MATH commutes, we see that the image of MATH is NAME dense in MATH. This forces MATH to be trivial. We have shown that the map MATH is surjective. In particular, both of these vector spaces are finite dimensional (MATH is finite dimensional by REF ). Also, both MATH and MATH are their own unipotent MATH-completions. Thus we have isomorphisms CITE MATH and MATH, where MATH is the set of grouplike elements of MATH. Now, MATH and MATH, where MATH is the augmentation ideal in MATH. Since MATH is surjective, we see that the map MATH is surjective. But then the induced map MATH is surjective (CITE, p. REF). This implies that MATH is surjective since MATH maps grouplike elements to grouplike elements. |
math/0104023 | Let MATH. This class corresponds to a central extension MATH in the category of proalgebraic MATH-groups. Observe that MATH is prounipotent. Suppose that MATH restricts to MATH in MATH. Then we have a commutative diagram of extensions MATH . Since MATH maps to MATH in MATH, the bottom extension splits. We then have the composite map MATH. By the universal mapping property of MATH, we get a unique map MATH making the diagram MATH commute; that is, the top extension splits. Thus, MATH. If MATH is finite dimensional, then the unipotent MATH-completion of MATH is the group MATH. By REF , the map MATH is an isomorphism for each MATH. It follows that the map MATH is an isomorphism. |
math/0104023 | We have the commutative diagram MATH . The vertical arrow is an isomorphism by REF . |
math/0104023 | Note that MATH is the inverse limit of the groups MATH with each MATH an extension MATH . For each MATH, we have a NAME - NAME spectral sequence MATH . Taking the direct limit of these sequences, we get a spectral sequence MATH satisfying MATH that is, MATH . Since MATH is an isomorphism we have MATH REF . Since the extension MATH splits, we have MATH for MATH; that is, we have an isomorphism MATH where MATH is the NAME - NAME spectral sequence for the above extension. Moreover, the differential MATH vanishes for both spectral sequences since the extensions are split. By REF , there is an inclusion of MATH-modules MATH and hence MATH injects into MATH. These facts together imply the following: CASE: MATH; CASE: MATH; CASE: MATH. It follows that MATH injects into MATH. |
math/0104023 | We have an injection MATH . If MATH is a number field then by CITE, the group MATH coincides with MATH. We then have the chain of equalities MATH . Since MATH is a torsion group when MATH is a number field, we see that MATH. If MATH is a finite field, then we use the following chain of equalities MATH where the last equality is the fundamental theorem of algebraic MATH-theory. Since MATH for finite fields, we are done. |
math/0104024 | Assume there is no such MATH. Then we can find a sequence of normalized MATH with MATH . The additional condition MATH allows us to integrate by parts in the NAME formula to obtain: MATH . Since MATH is a bounded tensor of MATH we obtain a uniform bound on MATH . By replacing MATH with a subsequence we can assume that MATH converge weakly in MATH norm to some MATH such that MATH . Furthermore, since the restriction map from MATH to MATH is compact, we can also assume that MATH in MATH . Hence MATH on MATH (NAME boundary value condition) and MATH is harmonic on MATH so MATH . This contradicts the strong MATH convergence MATH and the fact that MATH . |
math/0104024 | We start by deriving a pointwise bound on the spinorial part. Consider a trajectory of finite type MATH . Let MATH be the supremum of the pointwise norm of MATH over MATH . If MATH for some MATH since MATH we have MATH at that point. Here MATH is the four-dimensional Laplacian on MATH . By the standard compactness argument for the NAME - NAME equations CITE, we obtain an upper bound for MATH which depends only on the metric on MATH . If the supremum is not attained, we can find a sequence MATH with MATH . Without loss of generality, by passing to a subsequence we can assume that MATH and MATH (hence MATH). Via a reparametrization, the restriction of MATH to each interval MATH can be interpreted as a solution MATH of the NAME - NAME equations on MATH . The finite type hypothesis and REF give uniform bounds on MATH and MATH . Here we identify connections with REF - forms by comparing them to the standard product flat connection. We can modify MATH by a gauge transformation on MATH so that we obtain MATH on MATH and MATH on MATH . Using REF we get a uniform bound on MATH . After this point the NAME - NAME equations MATH provide bounds on all the NAME norms of MATH and MATH by elliptic bootstrapping. Here MATH could be any compact subset in the interior of MATH for example MATH . Thus, after to passing to a subsequence we can assume that MATH converges in MATH to some MATH up to some gauge transformations. Note that the energies on MATH are positive, while the series MATH is convergent because MATH is bounded. It follows that MATH as MATH so MATH . In temporal gauge on MATH, MATH must be of the form MATH where MATH and MATH are constant in MATH giving a solution of the NAME - NAME equations on MATH . By REF , there is an upper bound for MATH which depends only on MATH . Now MATH converges to MATH up to some gauge transformation, hence the upper bound also applies to MATH . Therefore, in all cases we have a uniform bound MATH for all MATH and for all trajectories. The next step is to deduce a similar bound for the absolute value of MATH . Observe that MATH for all MATH sufficiently large. As before, we interpret the restriction of MATH to each interval MATH as a solution of the NAME - NAME equations on MATH . Then we find that a subsequence of these solutions restricted to MATH converges to some MATH in MATH . Also, MATH must be constant in temporal gauge. We deduce that a subsequence of MATH converges to MATH, where MATH is a solution of the NAME - NAME equations on MATH . Using REF , we get a lower bound for MATH . An upper bound can be obtained similarly. Now let us concentrate on a specific MATH . By a linear reparametrization, we can assume MATH . Let MATH . Then MATH satisfies REF - dimensional NAME - NAME equations. REF and the bounds on MATH and MATH imply a uniform bound on MATH . Via a gauge transformation on MATH we can assume that MATH on MATH and MATH on MATH . By REF we obtain a bound on MATH and then, by elliptic bootstrapping, on all NAME norms of MATH and MATH . The desired MATH bounds follow. |
math/0104024 | First let us keep the metric on MATH fixed and prove that MATH are naturally isomorphic for different MATH . In fact we just need to do this for MATH because MATH does not depend on MATH and MATH . It is not hard to see that for any MATH and MATH the finite energy trajectories of MATH are contained in MATH . Let MATH . Then MATH is an isolating neighborhood for all MATH . By REF of the NAME index, for MATH the flow of MATH on MATH . Let MATH so that MATH is the orthogonal complement of MATH in the MATH metric, a span of eigenspaces of MATH . Another isolating neighborhood of MATH is then MATH where MATH is a small closed ball in MATH centered at the origin. The flow MATH is then homotopic to the product of MATH and a flow MATH on MATH which is generated by a vector field that is identical to MATH on MATH . From the definition of the NAME index it is easy to see that MATH . Here MATH is the linear flow generated by MATH on MATH and the corresponding equivariant NAME index can be computed using REF of the NAME index: it equals MATH . By the same REF we obtain MATH . This implies that MATH and MATH are canonically isomorphic. Next we study what happens when we vary the metric on MATH . We start by exhibiting an isomorphism between the objects MATH constructed for two metrics MATH on MATH sufficiently close to each other. Consider a smooth homotopy MATH between the two metrics, which is constant near MATH . We will use the subscript MATH to describe that the metric in which each object is constructed is MATH . Assuming all the MATH are very close to each other, we can arrange so that: - there exist MATH large enough and independent of MATH so that REF is true for all metrics MATH and for all values MATH - there exist some MATH and MATH such that neither MATH nor MATH is an eigenvalue for any MATH . Hence the spaces MATH have the same dimension for all MATH so they make up a vector bundle over MATH . Via a linear isomorphism that varies continuously in MATH we can identify all MATH as being the same space MATH - for any MATH we have MATH . Here we already think of the balls as subsets of the same space MATH . Then MATH is a compact isolating neighborhood for MATH in any metric MATH with the flow MATH on MATH . Note that MATH varies continuously in MATH . By REF of the NAME index, MATH . The difference MATH is the number of eigenvalue lines of MATH that cross the MATH line, counted with sign, that is, the spectral flow MATH as defined in CITE. NAME, NAME and NAME prove that it equals the index of the operator MATH on MATH with the metric MATH on the slice MATH and with the vector MATH always of unit length. Choose a REF - manifold MATH as in the previous section, with a neighborhood of the boundary isometric to MATH . We can glue MATH to the end of MATH to obtain a manifold MATH diffeomorphic to MATH . Then MATH by excision. From REF and using the fact that MATH and MATH do not depend on the metric we get MATH . It follows that MATH because the MATH operator has no spectral flow (for any metric its kernel is zero since MATH). The orientation class of this isomorphism is canonical, because complex vector spaces carry canonical orientations. Thus we have constructed an isomorphism between the objects MATH for two different metrics close to each other. Since the space of metrics MATH is path connected (in fact contractible), we can compose such isomorphisms and reach any metric from any other one. In order to have an object in MATH well-defined up to canonical isomorphism, we need to make sure that the isomorphisms obtained by going from one metric to another along different paths are identical. Because MATH is contractible, this reduces to proving that when we go around a small loop in MATH the construction above induces the identity morphism on MATH. Such a small loop bounds a disc MATH in MATH and we can find MATH and MATH so that they are not in the spectrum of MATH for any metric in MATH . Then the vector spaces MATH form a vector bundle over MATH which implies that they can all be identified with one vector space, on which the NAME indices for different metrics are the same up to canonical isomorphism. The vector spaces MATH are also related to each other by canonical isomorphisms in the homotopy category. Hence going around the loop must give back the identity morphism in MATH . A similar homotopy argument proves independence of the choice of MATH in REF . Thus MATH must depend only on MATH and on its MATH structure, up to canonical isomorphism in the category MATH . |
math/0104024 | REF states that there is a NAME set of forms MATH for which all the critical points of MATH are nondegenerate. Nondegenerate critical points are isolated. Since their moduli space is compact, we deduce that it is finite, so there exists MATH as required in REF . Furthermore, the condition MATH is equivalent to the fact that the reducible MATH is nondegenerate. |
math/0104024 | We choose an isolating neighborhood for MATH to be MATH . Here MATH the constant in REF , is chosen to be large enough so that MATH contains the image under MATH of the ball of radius MATH in MATH . By virtue of REF , all we need to show is that MATH and MATH satisfy REF in its hypothesis. REF that there exist sequences MATH and a subsequence of MATH (denoted still MATH for simplicity) such that the corresponding MATH do not satisfy REF for any MATH . Then we can find MATH and MATH such that MATH with MATH . We distinguish two cases: when MATH and when MATH has a convergent subsequence. In the first case, let MATH be the trajectory of MATH such that MATH . Then, because of our hypotheses, MATH and MATH for all MATH . Since MATH by REF we have that MATH for MATH sufficiently large. This is a contradiction. In the second case, by passing to a subsequence we can assume that MATH . We use a different normalization: MATH is the trajectory of MATH such that MATH . Then MATH and MATH for all MATH . By the NAME - NAME Theorem we know that MATH converges to some MATH in MATH norm, uniformly on compact sets of MATH . This MATH must be the NAME projection of a NAME - NAME trajectory. Let MATH . From REF we know that the convergence MATH can be taken to be in MATH but only over compact subsets of MATH . However, we can get something stronger than MATH for MATH as well. Since MATH is self-adjoint, there is a well-defined compact operator MATH . We have the estimate: MATH . But since MATH and MATH are trajectories of the respective flows, if we denote MATH we have MATH so that MATH . Fix MATH . We break each of the two integrals on the right hand side of REF into MATH . Recall that MATH live in MATH . This must also be true for MATH because of the weak convergence MATH in MATH . Since MATH and MATH are continuous maps from MATH to MATH there is a bound: MATH where MATH is a constant independent of MATH . On the other hand, on the interval MATH we have MATH and MATH is a compact map from MATH to MATH . We get MATH . In addition, MATH live inside a compact set of MATH and we know that MATH uniformly on such sets. Therefore, MATH . Similarly, using the fact that MATH in MATH uniformly in MATH for MATH we get: MATH . Putting REF , and REF together and letting MATH we obtain: MATH . Since MATH in MATH the same must be true for MATH . Recall that MATH is the boundary value of an approximate NAME - NAME solution on MATH with MATH . Equivalently, MATH . Since MATH are uniformly bounded in MATH norm, after passing to a subsequence we can assume that they converge to some MATH weakly in MATH . Changing everything by a gauge, we can assume without loss of generality that MATH . Now REF says that: MATH . We already know that the last term on the right hand side goes to MATH as MATH . Let us discuss the first term. First, it is worth seeing that MATH . Let MATH be the decomposition of MATH into its linear and compact parts; MATH and MATH are direct summands of MATH and MATH respectively. We have MATH in MATH (because MATH by construction), and MATH . Using the fact that MATH weakly in MATH we get that each term on the right hand side converges to MATH weakly in MATH . Hence MATH . Now the first term on the right hand side of REF is MATH . It is easy to see that this converges to MATH in MATH norm. We are using here the fact that MATH uniformly on compact sets. Similarly one can show that the second term in REF converges to MATH . We already know that MATH converges to MATH . This was proved starting from the boundedness of the MATH on the cylinder on the right. In the same way, using the boundedness of the MATH on the manifold MATH on the left (which has a cylindrical end), it follows that MATH in MATH . Thus, MATH in MATH . Let MATH with MATH . We know that MATH converges. Also, MATH weakly in MATH hence strongly in MATH . This implies that MATH in MATH and MATH in MATH . Since MATH they must converge in MATH just like the MATH . We are using the NAME multiplication MATH . Putting all of these together, we conclude that the expression in REF converges to MATH . Thus MATH in MATH . We also know that MATH . In addition, since MATH in MATH and using MATH we get that MATH in MATH . This implies that MATH . Now it is easy to reach a contradiction: by a gauge transformation MATH of MATH on MATH we can obtain a solution of the NAME - NAME equations on MATH with MATH . Recall that MATH was the starting point of MATH the NAME projection of a NAME - NAME half-trajectory of finite type. By gluing this half-trajectory to MATH we get a MATH monopole on the complete manifold MATH . From REF we know that there are ``universal" bounds on the MATH norms of the monopole (in some gauge) restricted to any compact set, for any MATH . These bounds are ``universal" in the sense that they depend only on the metric on MATH . In particular, since NAME projection is continuous, we obtain such a bound MATH on the MATH norm of MATH for all MATH . Recall that MATH because MATH and that MATH . When we chose the constant MATH we were free to choose it as large as we wanted. Provided that MATH we get the desired contradiction. REF proof is somewhat similar to that in REF . Assume that there exist sequences MATH and a subsequence of MATH (denoted still MATH for simplicity) such that the corresponding MATH do not satisfy REF for any MATH . Then we can find MATH such that MATH with MATH . Let MATH be the half-trajectory of MATH starting at MATH . Repeating the argument in REF , after passing to a subsequence we can assume that MATH converges to some MATH in MATH uniformly over compact sets of MATH . Also, this convergence can be taken to be in MATH for MATH while for MATH we get that MATH in MATH . Observe that MATH is the NAME projection of a NAME - NAME half-trajectory of finite type, which we denote by MATH . We can assume that MATH . Then, just as in REF , we deduce that MATH converges in MATH to MATH a solution of the NAME - NAME equations on MATH with MATH . By gluing MATH to MATH we obtain a MATH monopole on MATH . By REF , this monopole must be smooth in some gauge, and when restricted to compact sets its MATH norms must be bounded above by some constant which depend only on the metric on MATH . Since the four-dimensional NAME projection from MATH to MATH is continuous, we get a bound MATH on the MATH norm of MATH . But MATH in MATH and MATH . Provided that we have chosen the constant MATH to be larger than MATH we obtain a contradiction. |
math/0104024 | Let MATH be the complexification of the map MATH . Note that MATH where MATH is the representation MATH . Using the equivariant NAME mapping degree, tom NAME proves in CITE the formula: MATH where MATH is the usual mapping degree, and MATH is the MATH - theoretic NAME class of MATH in our case its character evaluated at MATH equals MATH . Since MATH the limit only exists in the case MATH . |
math/0104024 | A characteristic element MATH is one that satisfies MATH mod REF for all MATH-Torsion. Given such a MATH there is a MATH structure MATH on MATH with MATH . Let MATH . In REF we constructed an element: MATH . The restriction to the fixed point set of one of the maps MATH which represents MATH is linear near MATH and has degree MATH because MATH . Hence MATH . Together with REF , this completes the proof. |
math/0104024 | If we apply REF for MATH we get MATH for all characteristic vectors MATH . By REF from CITE, the only unimodular forms with this property are the diagonal ones. |
math/0104024 | Since MATH is even, the vector MATH whose first MATH coordinates are MATH and the rest are MATH is characteristic. We have MATH and we have shown that MATH . Rather than applying REF , we use the bound MATH directly. This gives that MATH . But the only even, negative definite form of rank at most REF is MATH . |
math/0104024 | Since MATH and MATH is compact, it suffices to show that for any MATH with MATH we have MATH . Let MATH be such a sequence, MATH so that MATH . Since MATH is compact, by passing to a subsequence we can assume that MATH . If MATH have a convergent subsequence as well, say MATH then by continuity MATH and MATH . Thus MATH as desired. If MATH has no convergent subsequences, then MATH . Given any MATH for MATH sufficiently large MATH so MATH . Letting MATH and using the comapctness of MATH we obtain MATH . Since this is true for all MATH we have MATH . This takes care of the case MATH since we obtain a contradiction. If MATH we reason differently: MATH is equivalent to MATH letting MATH we get MATH so MATH . Thus MATH as desired. |
math/0104024 | Choose MATH a small compact neighborhood of MATH such that MATH . We claim that if we choose MATH sufficiently small, we have MATH . Indeed, if there were no such MATH we could find MATH with MATH . Let MATH for some MATH such that MATH . By passing to a subsequence we can assume MATH . If MATH has a subsequence converging to some MATH then by taking the limit MATH and MATH which contradicts MATH . If MATH has no such subsequence, then MATH . Since MATH by taking the limit we get MATH . Thus MATH . On the other hand MATH which contradicts the fact that MATH . Let MATH be as above and let MATH be an open neighborhood of MATH such that MATH . Since MATH and MATH is compact, by making MATH sufficiently small we can assume that MATH . Let us show that there exists MATH such that MATH for any MATH . If not, we could find MATH with MATH . Since MATH is compact, there is a subsequence of MATH which converges to some MATH such that MATH or, equivalently, MATH . This contradicts the fact that MATH and MATH are disjoint. Let MATH be as above. For each MATH either MATH or there is MATH so that MATH and MATH . In the first case we choose MATH a compact neighborhood of MATH such that MATH . In the second case we choose MATH to be a compact neighborhood of MATH with MATH and MATH . Since MATH is compact, it is covered by a finite collection of the sets MATH . Let MATH be their union and let MATH . Then MATH is compact, and we can assume that it contains a neighborhood of MATH . We choose the index pair to be MATH . Clearly MATH and MATH . It remains to show that MATH is an index pair. First, since MATH is compact and disjoint from MATH by REF above MATH is compact. Since MATH is compact as well. We need to check the three conditions in the definition of an index pair. REF is equivalent to MATH . We have MATH because MATH and MATH contains a neighborhood of MATH . We have MATH because if MATH is of the form MATH for MATH such that MATH then MATH which contradicts MATH . REF can be easily checked from the definitions: MATH is positively invariant in MATH by construction, and this implies that it is positively invariant in MATH as well. REF requires more work. Let us first prove that MATH . We have MATH and we already know that MATH . For MATH there exists MATH such that MATH and MATH . Recall that we chose MATH so that MATH for any MATH . If MATH this implies MATH . If MATH then, because MATH is in some MATH for MATH the fact that MATH implies again MATH . Therefore MATH so MATH . To prove that MATH is an exit set for MATH pick MATH and let MATH . It suffices to show that MATH . Assume this is false; then MATH . Note that MATH . Also MATH so MATH is contained in MATH . It follows that for MATH sufficiently small, MATH . Since MATH is positively invariant in MATH and MATH we get MATH . This contradicts the definition of MATH . Therefore, MATH . We conclude that MATH is a genuine index pair, with MATH and MATH . |
math/0104026 | Using equation MATH we get MATH which means that MATH . So similarly to proof of REF , this theorem holds. |
math/0104026 | By REF we get for all positive integer MATH which means that MATH hence the proposition holds. |
math/0104026 | By REF the first equality holds. Now let us prove the second equality. By REF we get MATH therefore MATH . Hence, since MATH and MATH, the second equality holds. |
math/0104027 | We have MATH, MATH. Multiplication by MATH is an invertible linear isometry MATH, that is, MATH; it intertwines MATH with MATH. NAME prove that it sends MATH to MATH and MATH to MATH. First, MATH for a trivial reason: for all MATH the function MATH belongs to MATH, since MATH. In order to prove that MATH take polynomials MATH such that MATH for MATH, and MATH, whenever MATH; say, we may take MATH choosing MATH such that MATH. We have (for every MATH) MATH pointwise, and MATH. The majorant belongs to MATH; polynomials MATH belong to MATH, therefore to MATH, and to MATH. So, MATH. Now we apply the equality MATH to measures MATH, MATH (symmetric to MATH, MATH); these are related by MATH; thus, MATH. The isomorphism MATH between MATH and MATH (as well as MATH and MATH) transforms MATH to MATH, MATH to MATH, and the function MATH into the function MATH. So, MATH . However, MATH for MATH (namely, MATH); therefore MATH . So, MATH, that is, MATH. |
math/0104027 | A moderate measure MATH is related to a finite measure MATH by MATH; thus MATH where MATH. REF for MATH is equivalent to the condition MATH for MATH. The latter holds if and only if MATH has a density MATH of the form MATH, where MATH and MATH. It means that MATH has the density MATH; note that MATH. |
math/0104027 | It suffices to prove the latter, MATH, since MATH, and MATH, and MATH. In order to prove that MATH implies MATH, consider MATH for some MATH; we have to prove that MATH, where MATH is a linear combination of functions MATH, MATH. Such MATH is NAME transform of a linear combination of functions MATH for MATH (otherwise MATH), except for the case MATH; in that case MATH is constant, and we need NAME transform of a measure (concentrated at the origin) rather than a function of MATH. The same difficulty appears for MATH, when MATH; in that case MATH does not belong to MATH, since MATH jumps at the origin, and MATH is a finite measure rather than a function of MATH. However, a smoothing, say, MATH, does the job for MATH. For MATH the function MATH belongs to MATH. So, MATH. In order to prove that MATH implies MATH, consider MATH where MATH, MATH for MATH. The function MATH on the closed half-plane MATH is continuous, and tends to MATH for MATH. Therefore the function MATH on the closed disk MATH is continuous (and vanishes at MATH). Take polynomials MATH such that MATH uniformly on the disk; then functions MATH belong to MATH and converge to MATH in MATH. So, MATH. |
math/0104027 | REF is equivalent to REF. By REF is equivalent to MATH in the FHS sense. The latter is REF for MATH. By REF it is equivalent to REF for MATH. It remains to note that MATH has no poles except for MATH. |
math/0104027 | First, integration in REF may be restricted from MATH to MATH. Indeed, using the property MATH we get the kernel MATH equivalent to MATH. Second, MATH which is just a change of variable, MATH. Let MATH satisfy REF; we have to check REF. Consider MATH. The triangle inequality gives MATH, since MATH. Also, MATH for almost all MATH due to REF. Taking MATH such that MATH and MATH we get MATH, which is REF. Let MATH satisfy REF; we need to check REF, or equivalently, MATH. We have MATH for all MATH; similarly, MATH for all MATH. So, MATH. |
math/0104029 | Set MATH and let MATH be the dual flag bundle of MATH with tautological flag MATH. Define MATH by MATH, and set MATH for MATH. Then MATH is a full flag of subbundles in MATH, and REF applies to give a formula for the structure sheaf of the NAME variety MATH. Set MATH for MATH. Then there is a unique section MATH such that the dual flag MATH is the pullback of the tautological flag MATH on MATH, and furthermore we have MATH as subschemes of MATH. Since the loci MATH and MATH have the same codimensions and are NAME, this implies that MATH which completes the proof. |
math/0104029 | By the splitting principle we may assume that MATH and MATH come equipped with full flags MATH and MATH. Let MATH be the NAME permutation for MATH with descent at position MATH. Then MATH is a permutation in MATH where MATH. Set MATH for MATH and MATH for MATH, and let MATH be the map MATH, that is, the map MATH is extended by zeros on the trivial parts of MATH and MATH. It is now easy to check that MATH as subschemes of MATH, so by REF we get MATH where MATH and MATH. Notice that MATH. This means that the formula does not change when we shift the permutation MATH, so in fact we have MATH . This finishes the proof. |
math/0104029 | To cut down on the notation, we shall prove this in the case where MATH and MATH are empty and MATH. The proof of the general case is exactly the same. For convenience we will also write MATH for MATH. Using the rule MATH repeatedly we get MATH . The lemma follows from this since MATH . |
math/0104029 | Define a ``potential" function MATH where MATH. Then MATH for all sequences MATH. We proceed by induction on MATH. If MATH then MATH must be weakly decreasing. In fact it must be a partition because the assumption MATH implies that MATH. Therefore MATH already has the desired form. If MATH then for some MATH we must have MATH. We can now apply REF with MATH and MATH to write MATH as a linear combination of other determinants MATH, and it is easy to check that these satisfy MATH and MATH for all MATH. Each of these new determinants is therefore a linear combination of the polynomials MATH by induction, which proves the claim for the sequence MATH. The fact that the coefficients MATH are independent of MATH follows because the formula of REF is independent of MATH. |
math/0104029 | It follows from REF that the asserted identity is true if we sum over all CMYDs relative to MATH. We will prove that the terms for which MATH has special boxes cancel each other out in the right hand side. Notice that each column of a CMYD can have at most one special box. We will group each CMYD MATH for which the leftmost special box is of type A with two other CMYDs whose leftmost special boxes are of type B, such that the contributions from these three diagrams cancel. Similarly a diagram with a leftmost special box of type C will be grouped with two diagrams with leftmost special boxes of type D. Notice that if MATH is a CMYD relative to MATH such that MATH and MATH contains a special box, then MATH has at least MATH gray boxes, so the top row of MATH contains an unmarked gray box which is outside MATH. Notice also that since MATH whenever MATH, we have MATH for any diagram MATH such that MATH. Now let MATH be a CMYD whose leftmost special box is of type A. The conditions for a type A box then make it possible to change this box into a white box or a marked white box, while the diagram continues to be a CMYD. Here it is important that the top row of MATH contains at least one unmarked gray boxes outside MATH, since this ensures that the modified diagram satisfies REF . MATH . The signs of the contributions from the two new diagrams are the opposite of the sign of the contribution from MATH. Since MATH, the contributions from all three diagrams therefore add to zero by REF . Notice that any special box of type B can be changed to a gray box. Therefore all diagrams with a leftmost special box of type B get canceled in this way. Now suppose the leftmost special box in MATH is of type C. In this case the box can be changed to a marked gray box while the box above is either marked or unmarked white. MATH . Again the new diagrams give contributions of opposite sign from that of MATH, and MATH, so the contributions of all three diagrams cancel by REF . Finally all diagrams with a leftmost special box of type D are taken care of in this way, since any diagram with a type D box can be changed so the special box turns into type C. |
math/0104029 | We start by observing that MATH has no marked white boxes. If MATH has such a box, then since it is not special, there must be a gray box below it. But this gray box must then be special of type C or D, a contradiction. Notice also that no unmarked gray boxes can be contained in MATH, since these would necessarily be special of type A. Now suppose MATH contains a marked gray box, and consider the northernmost such box. Since this box is not special (and not in the top row of MATH), the white box above it is not in the right vertical strip of MATH. Now consider the row of boxes in MATH to the right of this white box. If this row contains a box in the right vertical strip of MATH, then this would necessarily be a special box of type B. We conclude that if MATH contains a marked gray box then some box northeast of this box is contained in MATH but not in MATH. Now assume that MATH is not contained in MATH and consider the northernmost row where MATH is missing boxes from MATH. Since MATH contains at least MATH gray boxes, this can't be the top row, and the row above must contain a box in the right vertical strip of MATH which has no box below it. Since this box can't be marked gray by the argument above, it must be special of type MATH, again a contradiction. We conclude that MATH is contained in MATH and that all boxes from MATH are white. To prevent these white boxes from being special, there must furthermore be a gray box in each column of MATH. This proves the result. |
math/0104029 | If MATH, it follows from CITE that MATH. Alternatively this can be deduced from REF , see for example, CITE. Notice in particular that MATH. For MATH the lemma therefore follows from the identity MATH. When MATH the lemma is true because MATH is the inverse power series to MATH. |
math/0104029 | Suppose at first that MATH. If MATH is a CMYD relative to MATH with no special boxes such that its coefficient MATH is non-zero, then since MATH we conclude by REF that MATH where MATH. But then we have MATH and MATH, so MATH. The theorem therefore follows from REF in all cases where MATH. For the general case it is enough to show that MATH where MATH; this is sufficient because any partition MATH such that MATH occurs in either side of the claimed identity must have length at most MATH, and the stable NAME polynomials for partitions of such lengths are linearly independent when applied to MATH variables. For the rest of this proof we will let MATH denote the MATH variables MATH. Let MATH be the cofactor obtained by removing the first row and the MATH'st column of the determinant defining MATH. Notice that this does not depend on MATH, and we have MATH . Now using REF we obtain MATH . Since REF is equal to REF for all MATH, the theorem follows from the following lemma. |
math/0104029 | Since the form of each fixed degree in REF must be zero, we can assume that each MATH is a polynomial and that MATH for MATH for some MATH. Assume at first that MATH and let REF be true whenever MATH. By assumption we then have MATH . Since the determinant of the matrix is the NAME polynomial MATH, we conclude that each MATH. Now assume MATH. If MATH then since MATH we get MATH . So if we put MATH, the left hand side of REF is equal to MATH . Since this is equal to zero for all large MATH, we conclude it is zero for all MATH by induction on MATH. |
math/0104029 | This is clearly true if MATH. For MATH we write MATH, and the lemma follows because MATH for all MATH and MATH. Finally, if MATH we set MATH and form the bundle MATH. Then construct the fiber square: MATH . We will suppress pullback notation for vector bundles. By CITE the locus MATH in MATH is mapped birationally onto MATH. Using REF and the fact that determinantal varieties have rational singularities CITE we therefore get MATH . Since pullback along the vertical maps are isomorphisms which are compatible with the horizontal pushforward maps, the lemma follows from this. |
math/0104029 | Notice that for any MATH we have MATH and MATH for all MATH. Since MATH for MATH this means that MATH if either MATH, or MATH and MATH. If MATH then MATH for MATH and MATH for MATH, so MATH as required. For MATH we have MATH, so the left hand side in the lemma is MATH as required. |
math/0104029 | Let MATH be the variety of partial flags MATH such that MATH has rank MATH and MATH has rank MATH. Then form the commutative diagram from CITE: MATH . The formula can now be proved by calculating the pushforward to MATH of the class MATH in two different ways, using descending induction on MATH. Here MATH is the sequence MATH. We are therefore reduced to the case MATH where MATH is a projective bundle. Notice that since MATH is now a line bundle we have MATH for MATH. Using this we get the following identities in MATH. MATH . The step replacing MATH with its power series expansion is valid since MATH is zero for MATH. Now using REF , and REF we get MATH which is what we want to prove. |
math/0104029 | In CITE it is shown that the linear map MATH defined by MATH is a ring homomorphism. Using this we get MATH. On the other hand we have MATH, so MATH. The corollary follows from this. |
math/0104042 | For the proof of REF , let MATH be a MATH - homology sphere. Pick a spin manifold MATH with boundary MATH. Now suppose we are given another spin manifold MATH such that MATH. Let MATH. As MATH is a MATH - homology sphere, MATH is spin and MATH. Hence MATH where we used NAME 's Theorem CITE in the third line. Since MATH does not depend on MATH we can conclude that MATH is finite and bounded from above by MATH. Since this is true for every MATH we also obtain that MATH and the finiteness of MATH. Now suppose we are given a MATH - homology sphere MATH such that MATH. Then we can find spin REF - manifolds MATH and MATH with MATH such that MATH . Now consider the spin manifold MATH. Since MATH and MATH, the above equality implies that MATH. By NAME 's Theorem this is only possible if MATH, and we obtain that MATH. Hence we have MATH and MATH as claimed. Now let us proof REF . Pick a spin manifold MATH with boundary MATH such that MATH. Then MATH bounds MATH, and we obtain that MATH . Hence we have MATH. If we choose a spin manifold MATH with boundary MATH such that MATH, then MATH bounds MATH and we have MATH . These two inequalities imply the desired result. As to REF first note that - thanks to REF - it suffices to prove this for MATH. Choose spin manifolds MATH and MATH with boundaries MATH and MATH such that MATH. Then MATH is spin with boundary MATH and MATH. To prove REF note that again we only have to prove the required property for MATH. The condition MATH implies that there exists a spin REF - manifold MATH with boundary MATH such that MATH. Pick manifolds MATH such that MATH and MATH. Consider the spin manifolds MATH . Then MATH, MATH and MATH. Similarly MATH, MATH and MATH. So we obtain MATH and the claim follows. To prove the second part of the assertion we only have to prove that MATH. Since MATH we have MATH and MATH. By REF , we obtain MATH and therefore MATH. The first part of REF is an immediate consequence of REF. As to the second part, the assumption MATH implies that there exists a spin manifold MATH with boundary MATH such that MATH. Now suppose there is some MATH such that MATH is the boundary of a MATH - acyclic manifold MATH. Consider the spin manifold MATH . Then MATH which, by NAME 's Theorem, implies that MATH. But MATH, hence we obtain that MATH, in contradiction to MATH. To prove REF , note that given a surface MATH in MATH with boundary MATH, there is a double covering MATH branched along MATH with boundary MATH. The manifold MATH is spin, and it is well known that MATH whereas MATH (see CITE for a proof). This implies the claimed inequalities. |
math/0104042 | Let MATH denote the knot MATH. The double covering MATH of MATH branched along this knot is MATH, hence MATH by REF . It is not hard to verify that MATH. By REF we obtain that MATH so MATH. Now the knot MATH is build up from the rational tangles MATH and MATH. The rational tangle MATH can be changed to the trivial tangle by MATH crossing changes if MATH is even and by MATH crossing changes if MATH is odd, so we obtain that MATH. Of course MATH, and therefore we can conclude that MATH. |
math/0104042 | The trace of a homotopy given by the crossing changes is a union of immersed annuli MATH in MATH with MATH positive and MATH negative self - intersection points connecting MATH and MATH (see CITE). Pick a connected surface MATH with boundary MATH such that MATH. By gluing this surface with MATH we obtain a connected immersed surface MATH having boundary MATH whose genus is MATH and which has MATH positive and MATH negative self intersection points. Now let us assume that MATH. Since we can join two self - intersection points of opposite signs by a handle we can construct a surface bounding MATH which has genus MATH and MATH positive self - intersection points. Replacing the remaining self - intersections points by MATH handles, we end up with an embedded surface with boundary MATH which has genus MATH and the claim follows. In the case that MATH a similar argument applies. |
math/0104042 | Suppose that we have an index MATH such that MATH. If MATH is odd, we can deform the knot MATH into the knot defined by the sequence MATH by performing MATH negative crossing changes. If MATH is even, we can do MATH negative crossing changes to obtain MATH. Repeating this for every index MATH for which MATH is positive, we eventually obtain a knot MATH for which MATH if MATH and MATH otherwise after having performed MATH negative crossing changes. A similar reduction can be done if MATH. In this case we can do MATH respectively MATH positive crossing changes, depending on whether MATH is even or odd. So we see that after performing MATH additional positive crossing changes, we end up with a link MATH where MATH and MATH if and only if MATH is even. Observe that the knot MATH is the boundary of an obvious NAME surface which has genus MATH. Now assume that MATH, that is, MATH. By REF we can conclude that MATH . If we have MATH, we can use the same argument to obtain the lower bound MATH . Since of course either MATH or MATH the claimed inequality follows. |
math/0104042 | It is well known that the lens space MATH is a double covering of MATH, branched along the two - bridge knot MATH, see for instance CITE. By REF , there exists a surface MATH with boundary MATH which has genus MATH . Using the expression for the signature of MATH derived in REF now immediately yields the desired result. |
math/0104042 | Let MATH denote the simply connected REF - manifold which is obtained from MATH by adding a handle along MATH with framing MATH. Then MATH is a MATH - homology sphere, in fact MATH. Furthermore MATH. As the boundary of MATH is a MATH - homology sphere, the MATH - intersection form is non - degenerate and hence the fact that MATH implies that MATH is not spin, that is, MATH is the non - zero element of MATH. Pick a NAME surface MATH for MATH and let MATH denote the surface which is obtained by gluing MATH with the core of REF - handle. Then MATH, and therefore MATH. Let MATH denote the homology class of MATH and denote the NAME duality map MATH by MATH. Then the fact that the self - intersection of MATH is odd implies that the image of MATH under the restriction MATH is the non - zero element, that is, we have MATH . Now pick a spin manifold MATH with boundary MATH and consider the closed manifold MATH. The fact that MATH is a MATH - homology sphere implies that MATH. As MATH is spin the surface MATH represents MATH, that is, MATH is characteristic. Therefore we can conclude that MATH. By NAME additivity, MATH. By definition of the NAME invariant we also have that MATH, and therefore we obtain that MATH. By CITE, we obtain MATH for the NAME invariant of the surface MATH. However it is also known CITE that MATH, and the proof of the proposition is complete. |
math/0104042 | Suppose that MATH is obtained by surgery on a knot MATH with framing MATH. First let us consider the case that MATH is positive (note that MATH must be odd). Then MATH, and by REF , we have that MATH. If MATH is negative, we obtain MATH, and multiplying this by minus one gives the desired result. |
math/0104042 | Suppose for a moment that MATH can be obtained as the result of integral surgery on some knot. The trace of this surgery is a simply connected REF - manifold MATH with second homology MATH. If MATH denotes the framing of the surgery the intersection form on MATH is MATH times the standard form on MATH. Using this one easily sees that the linking form on MATH maps the generator of MATH to MATH, see for instance CITE for a proof. Note that MATH. Now it has been pointed out in CITE that - as a consequence of the fact that the unknotting number is one - the linking form on MATH takes precisely the values MATH in MATH for some sign MATH, in particular MATH is in the image. Hence there exist integers MATH such that MATH . Multiplying this by MATH leads to MATH in contradiction to our assumption. |
math/0104042 | Clearly we can restrict ourselves to the case that the self - intersection number of MATH is positive, the case of negative self - intersection number then follows by reversing the orientation. Let us first consider the case that the NAME invariant MATH is zero. Let MATH denote the manifold which is obtained from MATH by attaching MATH copies of MATH and consider the surface MATH given by gluing MATH with the exceptional divisors. Then MATH, and clearly MATH. As explained in CITE we can now construct a sphere MATH, where MATH is obtained from MATH by attaching MATH copies of MATH such that MATH and such that MATH is still characteristic. Blowing down this sphere gives a spin REF - manifold MATH as required. In the case that MATH, consider MATH and the surface MATH obtained by gluing MATH with a torus representing REF times the generator of MATH. Then MATH and the claim follows from what we just proved. |
math/0104042 | By assumption, MATH is obtained by surgery on a knot MATH. The trace of this surgery is a REF - dimensional handlebody MATH with boundary MATH which is the result of attaching a REF - handle along MATH to MATH. Consequently MATH and MATH. By REF , MATH, and clearly MATH, that is, MATH. Now pick a surface MATH with boundary MATH such that MATH and let MATH denote a NAME surface for MATH. Then the surfaces MATH and MATH formed by gluing the core of REF - handle with MATH respectively MATH clearly have the same homology class MATH. As in the proof of REF , this homology class is characteristic, and MATH. It is known that the NAME invariant of the surface MATH obtained by gluing MATH and the core of REF - handle is MATH, and as MATH, the same is true for MATH . Now we can apply REF to MATH to obtain a spin REF - manifold MATH with boundary MATH and the claimed estimates follow from the definitions of the invariants MATH and MATH. |
math/0104042 | Let MATH denote the framing of the knot. Of course MATH. Once we can show that MATH must be positive the claimed inequality follows from REF . So let us assume that MATH. Then MATH, and by REF , we have MATH . However combining this with the assumption MATH shows that MATH. But this is only possible if MATH, which contradicts our assumptions. |
math/0104047 | By using CITE and the fact that MATH for MATH, it is enough to show that the set of syzygies MATH forms a homogeneous basis for the set of all syzygies MATH among the elements of MATH. To simplify the calculations, we are going to make all the polynomials of MATH monic by dividing each one by its leading coefficient; let MATH. So without loss of generality, we will assume that all MATH are monic. It is easy to see that MATH is a homogeneous syzygy. In fact, MATH . Claim: For all MATH and MATH, MATH is generated by elements of MATH. In fact MATH . Let MATH. The proof will proceed by induction on MATH. For MATH, MATH. Now assume the claim is true for MATH, that is, MATH is a combination of elements of MATH. We need to show REF for MATH. Hence, we need to show that MATH can be written as a combination of elements of MATH. For MATH, MATH . And for MATH, MATH . Thus, the set MATH is a homogeneous basis of the set of all syzygies on MATH. Therefore, by CITE and the way that we constructed MATH, MATH is a NAME basis for the ideal MATH. |
math/0104047 | The proof follows from the fact that MATH is a NAME basis for the ideal MATH with respect to the graded reverse lexicographic order. Hence the initial ideal of MATH with respect to this order is MATH . |
math/0104047 | Let MATH and let MATH such that MATH and MATH. We need to show that MATH. Since MATH, there exist MATH and a MATH such that MATH. Thus MATH. Since MATH, there exists a MATH such that MATH. But MATH, so MATH . Therefore, MATH is divisible by MATH and hence MATH. |
math/0104048 | By REF MATH is either the surface of REF or the surface of REF . In the former case the bicanonical map has degree REF by REF. In the latter case, the degree is REF by REF. |
math/0104048 | By REF , MATH is a disjoint union MATH, where MATH. The sets MATH are open since MATH is a topological invariant, thus we only have to show that MATH and MATH are irreducible of dimension MATH and MATH respectively. The points of MATH are in one-to-one correspondence with the isomorphism classes of curves of genus REF, thus the result is well known in this case. If MATH, then MATH is the only irrational pencil of genus REF of MATH. This can be seen in several ways, for instance by observing that a base-point free pencil is determined uniquely by the span of its class in MATH and that MATH, being of dimension REF, has only two isotropic lines, spanned by the classes of the general fibre of MATH and MATH. It follows that the double cover MATH is also determined uniquely, since it is the étale cover of MATH that kills the monodromy of the pencil MATH. So MATH is determined by the choice, up to isomorphism, of a curve MATH of genus MATH with an elliptic involution MATH and of a curve MATH of genus MATH with a free involution MATH. The pair MATH is determined by the quotient curve MATH, by the line bundle MATH of degree REF on MATH such that MATH, where MATH is the quotient map, and by the branch divisor MATH of MATH. So, taking into account the action of the automorphism group of MATH, we see that MATH depends on REF parametres. In addition, it is not difficult to write down an irreducible global family containing all the isomorphism classes of double covers of genus REF of elliptic curves. Thus the isomorphism classes of the pairs MATH form an irreducible family of dimension REF. Analogously, the isomorphism class of MATH is determined by the genus REF curve MATH and by the choice of a MATH-torsion line bundle MATH of MATH. The pairs MATH form a MATH-dimensional irreducible family (compare for instance REF). |
math/0104048 | REF follows from NAME - NAME inequality MATH and from the inequality MATH of CITE. REF correspond to REF, respectively. |
math/0104048 | REF is REF. The fact that the components of MATH are translates of abelian subvarieties follows from REF and the fact that the translation is by a torsion point follows from REF. REF follows from REF . REF follows by combining REF. REF is a consequence of REF. |
math/0104048 | We remark first of all that, in view of the assumption, we have MATH by REF. Assume now that there is a MATH-dimensional component MATH of MATH. By REF, MATH, where MATH is an abelian subvariety of MATH and MATH is a torsion point. Notice that MATH, since by REF MATH is an isolated point of MATH. Fix a torsion point MATH, denote by MATH its order and assume that MATH is minimal. This is the same as saying that if MATH then MATH. Let MATH be a nonzero vector tangent to MATH and let MATH be the conjugate of MATH. The vector MATH lies in the tangent cone to MATH at MATH and therefore the map MATH is not injective by REF. Denote by MATH a nonzero element in the kernel of this map. Let MATH be the connected étale cyclic cover associated to the finite subgroup MATH. The group MATH can be naturally identified with the dual group of the NAME group MATH of MATH, and MATH is the eigenspace of MATH on which MATH acts via the character corresponding to MATH. Since MATH, by the NAME - NAME Theorem there exists a fibration MATH onto a curve of genus at least MATH, such that MATH and MATH are pull-backs from MATH. The fibres of MATH are integral curves of MATH. Since MATH is MATH-invariant, it follows that MATH permutes the fibres of MATH and thus induces a pencil MATH. More precisely, we have a commutative diagram: MATH . Let MATH be the MATH-form on MATH such that MATH. By commutativity of the diagram, MATH is MATH-invariant and thus it is a pull-back from MATH. This shows that MATH is not rational. In view of the assumptions, we conclude that MATH has genus MATH. Since MATH is a pull-back from MATH, comparing the tangent spaces at the origin one sees that MATH. Now let MATH be the MATH-form on MATH such that MATH. By commutativity of the diagram, MATH is an eigenvector for MATH with character corresponding to MATH. Since this character generates the dual group MATH, the action of MATH on MATH is effective, that is, MATH is a MATH-cover. We claim that MATH is totally ramified, namely it does not factor through an étale cover of MATH. Assume otherwise, and denote by MATH the subgroup generated by the elements that do not act freely on MATH. Set MATH, MATH. Then we have a commutative diagram: MATH where the maps are the obvious ones. In particular, MATH is étale by construction. If we denote by MATH the order of MATH, then the cover MATH corresponds to the subgroup of MATH generated by MATH. On the other hand, MATH is obtained from MATH by taking base change with MATH, so the subgroup of MATH corresponding to MATH is actually a subgroup of MATH, that is, MATH. This contradicts the minimality assumption on MATH, and we conclude that MATH is totally ramified. The MATH-action on MATH gives a decomposition MATH, where the MATH are line bundles and MATH is the eigenspace of MATH on which MATH acts via the character MATH. Saying that MATH is totally ramified is the same as saying that MATH for every MATH. The standard formulas for abelian covers (compare REF ), give MATH. Denote by MATH the genus of the general fibre of MATH. The Corollary on page REF gives: MATH. Thus we either have MATH or MATH, MATH. Recall that MATH is also the genus of the general fibre of MATH. Then MATH is impossible: indeed by the Lemma on page REF the equality MATH would imply that MATH is the product of MATH with a curve of genus MATH, contradicting the fact that MATH is of general type. So we are left with the case MATH, MATH. Notice that in this case we also have MATH, so that MATH. So, again by the Corollary of page REF, the fibration MATH is isotrivial with smooth fibres, and the only singular fibres of MATH are two double fibres with smooth support, occurring at the two branch points of MATH. Hence by REF , there exist a curve MATH of genus MATH, a curve MATH of genus MATH, and a finite group MATH acting faithfully both on MATH and on MATH such that: CASE: MATH is isomorphic to MATH; REF the diagonal action on MATH defined by MATH is free; CASE: MATH is isomorphic to the quotient MATH, and the fibration MATH corresponds to the map MATH induced by the second projection MATH. In addition, the following hold: CASE: MATH has genus MATH (since MATH by REF); CASE: MATH (since MATH). Applying the NAME formula to the quotient map MATH we see that MATH has order MATH. REF now implies MATH, hence MATH is the surface of REF and MATH is the pencil MATH. Thus MATH has also an irrational pencil of genus MATH, contrary to the assumptions. |
math/0104048 | Let MATH. By REF , MATH is a component of MATH. It follows from REF , v) that for all MATH the map MATH is injective. Using NAME theory with twisted coefficients it follows that the map induced by cup product MATH is non zero on (non zero) simple tensors. Therefore, by a result of NAME (see CITE pg. CASE: MATH. Hence MATH . This is the required contradiction. |
math/0104048 | Compare REF . |
math/0104048 | By REF, MATH has an irrational pencil of genus MATH iff MATH is the surface of REF . If MATH has no irrational pencil of genus MATH, then MATH by REF . Thus we can apply REF to the NAME map MATH. It follows that MATH is pricipally polarized and MATH is a theta divisor. It is well known (compare REF ,b) that an abelian threefold with an irreducible principal polarization is the Jacobian of a curve MATH of genus MATH. As already explained in REF , the theta divisor is the canonical model of the symmetric product of MATH. Thus MATH is an ample NAME divisor with MATH. To finish the proof, it suffices to show that the map MATH is birational. If we denote by MATH the divisorial part of the ramification locus of MATH, the adjunction formula gives MATH. If we denote by MATH the degree of MATH, then we have: MATH where the first inequality follows from the fact that MATH is nef and the second one from the fact that MATH is nef. Since MATH by REF, it follows that MATH and MATH and MATH are isomorphic. |
math/0104056 | The first conclusion follows from the relation between the standard deformation complex REF and the complex REF involving MATH. Since MATH and MATH, we may write MATH, from which it follows that MATH on MATH. That MATH is due to two simple facts: first, that MATH and, second, that MATH is an isomorphism. |
math/0104056 | We assume the conclusion is false. That is, for each integer MATH, we assume that there is an element MATH satisfying MATH. Rescaling these MATH if necessary, we may assume that MATH and therefore MATH. By our estimate REF (extended by continuity to MATH), we have MATH . The sequence MATH is thus bounded with respect to MATH, the NAME MATH-norm. Any such set is precompact with respect to MATH; this means there is a subsequence MATH that converges weakly in MATH and strongly in the NAME MATH-norm. Let MATH be its limit. On the one hand, MATH as each element MATH is. On the other hand, the closedness of the differential operator MATH implies that MATH and MATH. Thus MATH, so MATH. But MATH, so this is a contradiction. |
math/0104056 | By REF , the quadratic form MATH defines a norm that is equivalent to MATH. We endow MATH with this norm, and let MATH denote the associated symmetric bilinear form. Note that if MATH and MATH are smooth, then MATH. By REF , the linear functional MATH is bounded on MATH for any MATH. The NAME representation theorem then implies that there is a unique MATH such that MATH for all MATH. Thus we have solved REF for MATH. The NAME operator is given by MATH, the solution MATH to MATH in the above sense. This makes sense for MATH, so under the orthogonal decomposition MATH we can extend MATH to all of MATH by declaring that it is identically zero on MATH. We define the harmonic projector MATH as orthogonal projection onto MATH under this decomposition. The operators MATH and MATH project onto orthogonal spaces, so MATH. On the other hand, the decompositions MATH follow immediately from the construction of MATH and MATH. To see that MATH takes a bit more work. From MATH it follows directly that MATH, so we need only show that, say, MATH. This follows easily by considering separately MATH (on which MATH and MATH are separately zero) and MATH, in which case MATH is a straightforward computation based on the formulas MATH, MATH, and MATH. Finally, the isomorphism MATH follows as usual from the existence of the NAME operator, since the harmonic projector MATH restricts to a map MATH whose kernel is exactly MATH by the arguments above. |
math/0104056 | In our local frame, we may write MATH. (Since MATH, there are no MATH terms.) In this case MATH is (see REF ) MATH where we have discarded all the terms without a derivative of a component of MATH. This proves the second claim; the first claim follows from applying the one-form MATH to both sides of REF: MATH where we have simplified using MATH. Finally, we prove REF . To compute this adjoint, we take the inner product of MATH with an element MATH of MATH, and integrate by parts: MATH . If we write MATH for MATH (again, there is no MATH term as MATH), then we can compute MATH. The inside term is not difficult to compute, and we get MATH, so MATH. Undoing the integration by parts above gives REF . |
math/0104056 | We begin by computing MATH. From REF, we have MATH . We expand this to get MATH . Since MATH by REF and MATH for all MATH and MATH, one of the cross terms simplifies: MATH. Four of the other cross terms combine and REF simplifies to MATH . We will deal with the remaining cross terms by adding MATH. By REF , MATH . Since MATH, we have MATH . Moreover, MATH as MATH. Hence MATH . To cancel the cross terms in REF, we will make use of the fact that MATH commutes with MATH and MATH modulo lower-weight terms, and therefore integrating by parts yields MATH . A similar argument shows that three of the cross terms on the right-hand side of REF cancel all the cross terms of REF: MATH . We now have more cross terms, this time involving MATH. We will deal with some of these cross terms using integration by parts. The adjoint of MATH is MATH, and so (using MATH and other commutation relations), we have MATH . Similarly, MATH . Thus MATH . Now the three parts grouped in parentheses can be removed by the NAME inequality. This gives us MATH which is REF . This concludes the proof of the NAME Estimate. |
math/0104056 | What we will show, in fact, is that for each MATH and each MATH there is a constant MATH such that MATH . The constant MATH can be chosen to be dominated by all the different constants MATH, so that the sum of the various individual estimates REF yields the subelliptic estimate REF. We prove this lemma in stages: we produce the estimate REF for each of the components MATH, MATH, MATH, and MATH in turn. The MATH case: We begin by noting that we have the estimate REF for MATH and MATH by the NAME Estimate, REF . Now consider the part of REF that we discarded in the last step of the proof of the NAME Estimate: MATH . Notice that, since MATH and MATH, MATH for any MATH. As we have already estimated MATH, this allows us to obtain an estimate MATH for some MATH. Similarly, we can obtain an estimate MATH for some MATH. From these estimates and REF , we obtain estimates for MATH, MATH, and MATH. We again return to a term we discarded at the end of the proof of the NAME Estimate: we have MATH . We may rewrite part of this as MATH . In the same way as above, we can control the inner product on the right. Since we have an estimate already for MATH, we get estimates for MATH and MATH. We can integrate by parts to write MATH . The previous estimates for the terms on the the right-hand side of this inequality then establish estimates for MATH. We use the fact that MATH to get MATH which gives us estimates for MATH and MATH. Using integration by parts, we get an equality MATH . We thus obtain an estimate on MATH from the estimates on MATH and MATH. Using this same trick, we have MATH and we get an estimate on MATH. Using REF for the local expression of MATH, we get MATH . On the other hand, MATH . Since we've already estimated MATH, this gives us an estimate on MATH and MATH. Similarly, we may use MATH and MATH to obtain estimates on MATH. This completes the proof of the MATH case of REF . The MATH case: Recall that MATH by REF , so this case follows from the MATH case. The MATH case: We begin by recalling that we have our estimate for MATH and MATH by the NAME Estimate, REF . We also remark that we have estimated MATH in the proof the MATH case of REF . In the proof of the NAME Estimate, REF , we did not use the fact that MATH . From this fact we have that MATH . Using the same method as in the proof of the MATH case, and noting that we have estimates for all of the MATH terms, we obtain estimates for MATH and MATH. Using our integration by parts trick, we see that MATH . We have estimates for both terms on the right-hand side, so this gives us estimates for MATH and MATH. Now we produce an estimate for MATH. We can write MATH for MATH, so integration by parts yields MATH . This gives us an estimate on MATH. Since MATH, we get MATH and an estimate on MATH. Similarly, MATH and we may estimate MATH. Finally, integration by parts gives us the equalities MATH which allow us to estimate MATH, MATH, and MATH. This is the last of the required MATH estimates, and so this completes the proof of the MATH case of REF . The MATH case: It is simplest to notice the symmetry between the MATH case and the MATH case. For example, the NAME Estimate gives us an estimate on MATH and MATH as well as MATH and MATH. Making the appropriate changes in the proof of the MATH case will then give us a proof in this case as well. As this is the final case, we have now completed the proof of REF . |
math/0104056 | We will show that MATH whenever MATH. Because MATH is subelliptic, MATH is smooth whenever MATH is smooth, so the required estimate follows by approximating with smooth sections. The proof of REF is by induction on MATH. By using a partition of unity we may assume that MATH is supported in the domain of a frame satisfying REF. Observe that REF and the NAME inequality imply that MATH. As usual, we will let MATH and MATH denote equality and inequality modulo lower-weight terms, which can be absorbed by using standard interpolation inequalities. We begin by considering derivatives in the MATH direction. By REF , MATH . Because MATH commutes with MATH and MATH modulo terms of weight MATH, it follows that MATH is an operator of weight at most MATH. Therefore, after integrating by parts, the second term above can be absorbed to yield MATH . Now we can prove REF for the case MATH. Observe that the commutation relations for MATH and MATH imply that MATH is equal to a constant multiple of MATH modulo lower-weight terms. Therefore, using REF again, we get MATH where MATH is some operator of weight MATH. Integrating by parts and using REF, we find MATH so MATH . Choosing MATH small enough, we can absorb the MATH term and obtain REF when MATH. Now assume that REF holds for some MATH. By induction, we have MATH . If MATH denotes any of the vector fields MATH or MATH, then MATH, where MATH and MATH are operators of weight MATH and MATH, respectively. Thus MATH . Since MATH is a sum of terms of the form and MATH and MATH, this completes the induction. |
math/0104056 | The proof of this proposition is simply the fact that MATH, MATH, and MATH take derivatives only in the MATH directions; thus MATH can be written in a local frame for MATH as a homogeneous quadratic polynomial in the coefficients of MATH and their derivatives, in which each monomial has a total of no more than four MATH derivatives. The assumption that MATH and the NAME embedding theorem yield the result. |
math/0104056 | We will solve this problem first in a NAME space: complete MATH with respect to the norm MATH for some integer MATH to obtain a NAME space, which we denote by MATH. Consider the NAME analytic map from MATH to itself given by MATH . REF implies that MATH is actually mapped to another element of MATH. This is clearly an analytic local isomorphism. The NAME inverse mapping theorem then gives us an analytic inverse map; that is, an analytic function MATH from MATH to itself such that MATH . Our family REF is locally (near the origin MATH) parametrized by the analytic set MATH defined in REF. To see this precisely, notice that REF implies that for MATH, MATH (as MATH on MATH). Combining this with the definition of MATH, we see that MATH . Since MATH depends complex analytically on MATH, our MATH is a complex analytic subset of MATH. |
math/0104056 | We must construct MATH and MATH. Suppose that we are given a family of deformations of a neighborhood MATH of MATH, MATH. Let MATH be a covering of MATH by coordinate domains, indexed by some finite set. Let MATH be local holomorphic coordinates on MATH, and let MATH be transition functions: MATH . For brevity, we will write this as MATH . We can extend this to a local coordinate covering MATH for MATH with transition functions MATH defined on MATH, holomorphic in MATH and smooth in MATH. We use a similar abbreviation as above: MATH with the requirement that MATH. For simplicity, we use local complex coordinates MATH depending complex analytically on the parameter MATH. That is, each function MATH is a smooth function on MATH and complex analytic on MATH, and the corresponding complex structure on MATH (as an element of MATH ) is determined by MATH . Similarly, the induced CR structure defined in REF is also determined locally by MATH where MATH. This equality also means that the map MATH is a CR embedding from MATH to MATH, with the complex structure MATH. We have to construct MATH, locally expressed by MATH on MATH, which depends complex analytically on MATH, and a holomorphic map MATH from MATH to MATH, satisfying MATH for all MATH (where, if necessary, we may shrink MATH to a smaller neighborhood of MATH). The proof of the existence of such functions is a standard formal power series argument. Consider the power series expansions MATH . We are using multi-index notation, so if MATH and MATH, then MATH and MATH. In general, if MATH is any vector-bundle-valued function of MATH, we will use the notation MATH to mean the part of the power series for MATH about MATH that is homogeneous of order MATH in MATH. For such homogeneous polynomials, we will use a subscript MATH to indicate the degree in MATH. Similarly, a superscript MATH will indicate a (not usually homogeneous) polynomial of degree MATH in MATH. First we formally construct these power series, then prove convergence. Let MATH and MATH be the MATH-th partial sums in the above power series expansions: MATH . We construct MATH and MATH formally by induction on MATH. At any step MATH, we wish to have MATH and MATH satisfy MATH for MATH near MATH. At our initial step (that is, at MATH), we define MATH and MATH. These obviously satisfy our REF . Now we assume that we have already constructed MATH and MATH satisfying REF. To begin our construction of MATH and MATH, we define a polynomial MATH on MATH, homogeneous of degree MATH in MATH, such that MATH . (In this way, MATH is a rough first approximation of MATH, the homogeneous part of MATH in degree MATH.) To do this, we construct vector-valued polynomials MATH on MATH, again homogeneous of degree MATH in MATH, by the relation MATH . This definition of MATH makes sense as the induction REF implies that the right-hand side of REF has only terms of order MATH and higher in MATH. We use these MATH and a partition of unity MATH subordinate to the covering MATH to define MATH . We will show that such MATH satisfy REF. To do this, we need to know how MATH (or MATH) transform over different coordinate charts. We have the following lemma (compare to CITE). On MATH, MATH . By the definition of MATH, MATH . We replace MATH with MATH to get MATH . We expand the first term on the right-hand side in a power series about the point MATH; this implies MATH . (In the last line we have used the inductive REF and NAME 's theorem applied to MATH. Any error term involving MATH multiplied by itself or by MATH can be absorbed into MATH.) The first and third terms simplify to MATH modulo MATH, and so REF reduces to REF . This proves the lemma. With MATH defined by REF, MATH transforms as in REF . From the definition of MATH and REF, MATH . Thus MATH . By NAME 's theorem, this is equivalent to REF. To define the next term in our formal power series, we will write locally MATH where MATH is the local expression for a homogeneous polynomial MATH of degree MATH in MATH, with values in MATH, and MATH is a homogeneous polynomial of degree MATH with values in MATH. Since the transformation law for sections of MATH is MATH it follows that our prospective MATH transforms the correct way: MATH . We still must construct MATH and MATH so that MATH and MATH, defined as in REF , satisfy the inductive REF . Note first that, by REF , the CR structure defined by MATH must satisfy MATH . From this it follows that MATH . On the other hand, from the definition, it is clear that (see REF ) the map MATH linearizes to the identity, so MATH . Finding solutions to the second equation in REF is thus reduced to the following theorem. There are vector-valued polynomials MATH and MATH, homogeneous in MATH of degree MATH, solving MATH where MATH takes values in MATH, and MATH takes values in the finite-dimensional harmonic space MATH. The proof of this theorem will follow from several lemmas and propositions. There is a homogeneous polynomial MATH of degree MATH in MATH, with values in MATH, such that MATH where we have written MATH for MATH. Since our CR structure is strictly pseudoconvex, the map MATH is an isomorphism. Hence there is a MATH-valued polynomial MATH which such that MATH is a polynomial that takes values in MATH. By the inductive hypothesis, for each MATH, the polynomial MATH already takes values in MATH; thus we may assume MATH. Writing MATH for MATH and MATH for MATH, we thus have MATH . To prove the proposition, it suffices to show MATH . In order to show this, we first prove the next lemma. MATH and MATH hold. In particular, MATH as MATH. For MATH, MATH REF are the parts of the deformation equation that are order MATH in MATH. (Of course, each MATH includes first derivatives of MATH.) The expressions for MATH are given in REF . Since MATH is quadratic, and we may replace each MATH with MATH in turn. On the one hand, MATH by the induction REF . On the other hand, MATH itself satisfies MATH. Together, these facts imply that MATH. The proof for MATH is similar. Continuing the proof of REF , we remark that MATH is, for each MATH, an integrable complex structure. Since MATH is a CR embedding for each MATH, modulo terms of order MATH and higher, the CR structure induced by MATH is also integrable: MATH . Obviously, we may remove the terms of order MATH and higher to see that MATH . From the previous lemma, MATH, and so MATH . A similar computation shows that MATH (and the zero follows from REF ). The integrability condition is thus MATH . Because MATH takes its values in MATH, we have MATH. Hence MATH . This is equivalent to REF , so this proves REF . MATH . We recall that MATH . (The map defined on each MATH by MATH makes sense globally modulo MATH.) Thus MATH . We apply the operator MATH to this equality. By REF , the left-hand side is the image of an element of MATH under MATH, so this makes sense. The decomposition of REF implies that MATH, and from this REF follows easily. MATH . The first term on the left-hand side satisfies MATH as we have seen in the proof of REF . By the construction of MATH REF , we have MATH . Taking the difference of the last two equations implies MATH . The proposition then follows from REF . We wish to solve REF , which can be written as MATH . We begin by solving MATH for MATH and MATH. By REF , the left-hand side of this equation is in the kernel of MATH, modulo MATH. The decomposition of REF implies that MATH and MATH, defined as follows, satisfy REF : MATH . Since MATH, and MATH for elements of MATH, we may define MATH locally by MATH . To solve REF , and thus REF , we simply set MATH. This MATH and MATH solve REF , so we have proved REF . Continuing the proof of REF , we turn to the proof of convergence of the formal series. This part of the proof uses REF and NAME (see CITE). We define a NAME MATH norm on a power series by setting MATH . Consider the power series MATH this converges for any positive MATH. Moreover, for positive MATH, we have MATH, where MATH means every coefficient of the left-hand side is less than the corresponding coefficient of the right-hand side. This implies that MATH for all integers MATH. By choosing suitable MATH and MATH (see CITE or CITE), we wish to show, for any integer MATH, that MATH . (The reason for subtracting MATH in REF is because MATH has no MATH term.) By the NAME embedding theorem, this would give us all the convergence and regularity claimed in REF . Proof of the convergence REF is done by induction on the partial sums. That is, we assume that we have MATH then establish the same inequality for MATH. The special properties of MATH are used here: we bound the MATH-st degree terms with lower degree terms that we have previously bounded. As MATH and MATH are chosen properly, we can bound sums of powers of MATH by MATH itself. The MATH term is well-behaved: for any MATH there is a constant MATH such that the harmonic projector MATH satisfies the estimate MATH for any MATH. However, we may have to correct MATH to ensure convergence, because our construction of of MATH involved first derivatives of MATH. Recall from REF that MATH is a solution to REF , which can be viewed as a linear MATH equation for the standard deformation complex REF. Because MATH is a holomorphic vector bundle, by the results of CITE there is a NAME operator MATH satisfying MATH for all MATH, where MATH and MATH is the projection onto MATH. Arguing as in CITE, we let MATH and MATH . It is true that there is a first derivative of MATH in MATH, but only in the MATH direction. In fact, we recall that MATH is defined on MATH by MATH . (The CR structure defined on MATH by MATH makes sense globally, modulo MATH.) Thus MATH . By the inductive hypothesis, we have MATH while MATH takes its values in MATH. Since the composition MATH of the adjoint operator and the standard NAME operator gains REF in this direction, there is no problem in the convergence of our formal solution. This finishes the proof of REF . |
math/0104059 | By attaching MATH-handles, we can get from MATH to MATH where MATH has one boundary component. By REF , MATH is equal to a product of right-handed and left-handed NAME twists. First note that each homologically trivial right or left twist is equal to a product of homologically nontrivial right or left (respectively) twists, due to the chain relation REF . To see this, observe that any homologically trivial curve MATH is the boundary of a compact subsurface of MATH. Thus we may assume that all the twists are homologically nontrivial. Let MATH be the genus of MATH and let MATH be a right twist about MATH. Now suppose we express MATH as MATH where MATH is a right twist about the homologically nontrival curve MATH and MATH and MATH are arbitrary elements of MATH. Then we may find a chain MATH in MATH, with MATH, such that: MATH . Thus by inserting homologically nontrivial right twists we can change MATH to MATH (since boundary twists commute with interior twists). Repeating this process for every right twist in MATH, we can change MATH to MATH where MATH is a product of homologically nontrivial left twists and MATH is some (possibly large) positive integer. Our task is to reduce MATH to MATH. Note that, by inserting more right twists if necessary, we may assume that MATH is odd. We write MATH where MATH is the corresponding product of right twists. Add more MATH-handles to MATH to get MATH, with MATH connected and with the genus MATH of MATH equal to MATH, so that MATH. Fix a particular chain MATH in MATH such that MATH is a regular neighborhood of MATH and MATH is a regular neighborhood of MATH. Then we have: MATH . Thus, by inserting homologically nontrivial right twists we can change this expression to: MATH . Finally this shows that MATH may be changed to MATH by inserting right twists. |
math/0104059 | Given a contact MATH-manifold MATH, REF tells us that MATH is supported by an open book decomposition, with page MATH and monodromy MATH. Apply REF to get a new pair MATH with MATH as in the lemma. Since we only added MATH-handles and inserted right twists, REF tell us that MATH. Consider the pair MATH as in REF and notice that MATH, a composition of homologically nontrivial right twists. Let MATH be a disk and MATH the identity in MATH. Then we can get from the pair MATH to MATH by attaching MATH-handles and inserting homologically nontrivial right twists, so that MATH (see REF ). Thus we have the following cobordism: MATH . Finally, apply REF to conclude that MATH and put the two cobordisms together to get a concave filling of MATH. |
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