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math/0010271 | Applying REF , we get that MATH and MATH commute with each MATH. Since MATH are linearly independent, each MATH is a linear combination of the MATH's and so MATH and MATH commute with each MATH. |
math/0010271 | Applying REF (and using its notation) we get that if MATH, then MATH and so MATH and MATH. Next as MATH for MATH, we get MATH for MATH. On the other hand, since MATH we have MATH so that MATH and therefore MATH . Hence there must be a gap in the spectrum of MATH that contains the interval MATH. Next, if we select any MATH such that MATH, then we have MATH and so MATH. Thus, MATH is an exposed point by REF . |
math/0010271 | Suppose that MATH exposes MATH in MATH so that there is a number MATH such that MATH and MATH if MATH and MATH. Write MATH and suppose MATH is a sequence in MATH such that MATH. As MATH is compact, we may assume that MATH converges to a vector MATH. Since MATH, MATH exposes MATH and MATH, we must have that MATH. Hence, we get that MATH separates MATH and MATH. Now suppose that MATH separates MATH and MATH with separation constants MATH and fix MATH. Since MATH, there is a number MATH with MATH and a vector MATH, such that MATH. Hence, MATH and so MATH exposes MATH. |
math/0010271 | Write MATH for the closure of MATH and observe that since MATH is an isolated point in MATH, MATH. Further, if MATH denotes the closed convex hull of MATH, then MATH. Indeed, if we had MATH, then MATH would be an extreme point of MATH. But in this case we would get that MATH by NAME 's ``converse" to the NAME - NAME theorem CITE. As noted in the paragraph before REF , there is a continuous linear functional MATH that separates MATH and MATH and so, by REF MATH exposes MATH in MATH. Hence the set of linear functionals that expose MATH is not empty. Now suppose that MATH is a linear functional that exposes MATH. Applying REF , we get that there are real numbers MATH such that MATH separates MATH and MATH with separation constants MATH and MATH. Now select MATH so that if MATH and let MATH denote a linear functional with MATH. With this we get MATH . If MATH, then we have MATH and so MATH separates MATH and MATH with separation constants MATH and MATH. Applying REF we get that MATH also exposes MATH and so the set of all such linear functionals is open. To get the final conclusion, observe that since the set of linear functionals that expose MATH is a nonempty open set, it contains a linearly independent set of MATH linear functionals, MATH. Thus there are linearly independent vectors MATH such that MATH is given by the inner product with MATH, and the conclusion follows. |
math/0010271 | CASE: Suppose MATH is an isolated extreme point in MATH. By REF , there is a unique projection MATH such that MATH. Next, applying REF and parts MATH and MATH of REF , we get linearly independent spectral pairs-MATH such that MATH . Since MATH, we have MATH . Since these spectral pairs are linearly independent, a straightforward argument shows that the subspace of MATH spanned by MATH has dimension MATH. Hence, relabeling if necessary, we may assume that MATH are linearly independent. Therefore, by REF MATH is a central projection in MATH. CASE: Suppose that MATH, where MATH is a central projection in MATH and let MATH denote a sequence of extreme points of MATH that converges to MATH. Applying part MATH of REF we get a sequence MATH of unique projections in MATH such that MATH for each MATH. Since MATH is finite dimensional, we may pass to a subsequence and assume that the sequence MATH is norm convergent with limit MATH. Since MATH is continuous, we have MATH and since MATH, we get MATH. Now, since MATH is central and the central projections are isolated in the set of all projections in MATH, we must have that MATH (and hence MATH) for all sufficiently large MATH. Hence MATH is an isolated extreme point. If MATH is an isolated extreme point in MATH, then by REF of this Theorem, there is a central projection MATH in MATH such that MATH . |
math/0010271 | If MATH is abelian and has finite dimension, then each projection in MATH is diagonal. Hence MATH contains a finite number of projections and since every extreme point in MATH is the image under MATH of a projection MATH, MATH is finite. For the converse, if MATH has a finite number of extreme points, then each extreme point in MATH is an isolated extreme point and so by REF , each extreme point of MATH has the form MATH, where MATH is a central projection. On the other hand, if MATH is any spectral pair, then MATH is an extreme point by REF and so MATH is a central projection. Thus, for each fixed MATH the NAME algebra MATH is central and therefore each MATH is central. Hence MATH is abelian. Since the projections of the form MATH generate MATH, this algebra has finite dimension. |
math/0010271 | Suppose that the closure of the isolated points of MATH is not all of MATH. By CITE each open subset of MATH is a locally compact NAME space. Hence, there is a nonempty open set MATH in MATH with no isolated points. Since MATH is countable, MATH is countable and we may write MATH. Set MATH for each MATH. As MATH is NAME, each MATH is open and since MATH is not isolated, each MATH is dense in MATH. Hence by the NAME Category Theorem CITE , MATH is dense in MATH. Since, MATH, this is impossible. Hence, no such MATH exists and so the isolated extreme points in MATH are dense. |
math/0010271 | Write MATH, let MATH denote the map from MATH to MATH defined by the formula MATH and write MATH. It is clear that MATH. Next suppose MATH is countable and fix MATH and MATH. For each extreme point MATH in MATH write MATH so that MATH is a closed face of MATH. Now each MATH is nonempty and so contains extreme points. Since MATH and MATH are disjoint if MATH and MATH are distinct extreme points, the extreme points of MATH must be countable. Thus, if MATH is countable, then MATH is countable for each pair MATH. Hence, if the Theorem is true when MATH, then each MATH is abelian and therefore MATH is abelian. |
math/0010271 | We have that MATH for some spectral pair MATH . By REF . Now suppose that MATH has dimension greater then zero. In this case we must have MATH and so MATH. Hence, MATH is transverse to the isotrace slices of MATH and so MATH is true. The remaining assertions now follow easily from MATH. |
math/0010271 | Fix a closed proper line segment MATH in the boundary of MATH. As MATH is in the boundary of MATH, it is in the boundary of MATH and so is contained in a minimal face MATH in MATH. Since faces in MATH of dimension greater than zero are transverse to the isotrace slices by REF , MATH must be two dimensional. Hence MATH is true. Next, fix an extreme point MATH of MATH. If MATH is not an extreme point of MATH, then it is properly contained in a minimal face MATH of MATH. If MATH were two dimensional, then by its minimality, we would get that MATH lies in the relative interior of MATH. But in this case MATH would be a line segment by REF and MATH would be in its interior, which is impossible since MATH is extreme in MATH. Hence, MATH has dimension less than two. Since MATH is properly contained in MATH, this face must have dimension one. Finally, since MATH is not extreme in MATH, it must be contained in the relative interior of MATH and so MATH is true. Now suppose MATH is an isolated extreme point in MATH. Since MATH has dimension REF, its faces consist of extreme points and line segments. Since MATH is an isolated extreme point, it must be the endpoint of adjacent line segments in the boundary of MATH. Hence MATH lies in two distinct two dimensional faces of MATH by REF of this Lemma. Thus part MATH is true. Finally, assume that the set MATH of extreme points of MATH is countable. Observe that each face in MATH of dimension one is uniquely determined by its end points, which are extreme points in MATH by REF . Hence MATH has a countable number of faces of dimension one. Now fix MATH with MATH. We have that MATH is countable. Each remaining extreme point of MATH is contained in a unique face of MATH with dimension one by REF of this Lemma. Since there are a countable number of one dimensional faces in MATH, the set of extreme points of MATH is countable. |
math/0010271 | As MATH is isolated among the extreme points of MATH, there are distinct two dimensional faces MATH and MATH of MATH such that MATH by REF . Next, since MATH and MATH have maximal dimension in MATH, they must be exposed and so we may write MATH . Further, since the MATH's intersect MATH in adjoining line segments, we must have that MATH. Since neither of the MATH's is zero, these vectors are linearly independent. Hence, by REF , we must have MATH where MATH and MATH are central projections in MATH. Finally, since the MATH's have dimension two, MATH must be a (possibly degenerate) line segment. Hence elements of MATH are the images under MATH of convex combinations of MATH and MATH, as desired. |
math/0010271 | Write MATH and MATH. To show that MATH is abelian, it suffices to show that MATH is central for each spectral pair of the form MATH. Indeed in this case, by a standard argument in measure theory, we get that the spectral projections for MATH and MATH are central and so MATH and MATH commute. So now let MATH denote such a projection, write MATH and let MATH denote the isotrace slice containing MATH. By REF , MATH is an extreme point of MATH and so it is also an extreme point of MATH. Since MATH is countable, the extreme points of MATH are countable by REF and since MATH has dimension REF, these extreme points form a closed set. Hence, the isolated extreme points of MATH are dense in MATH by REF . Now let MATH denote a sequence of isolated extreme point in MATH that converges to MATH. Applying REF , we get sequences MATH of central projections in MATH and a sequence MATH of real numbers in MATH such that MATH . Write MATH for each MATH. As the sequence MATH lies in MATH, it is bounded and so it admits a subnet MATH that converges to an element MATH in MATH in the weak-MATH-topology. Since MATH is normal, we have MATH . Since MATH by REF , we get MATH. Since each MATH is central and the center of MATH is closed in the weak-MATH - topology, we get that MATH is central, as desired. |
math/0010272 | If MATH, where MATH, then MATH . The fact that MATH is indeed the NAME involution follows from the details of the calculation in CITE. |
math/0010272 | Every cusp on MATH can be obtained from MATH by means of MATH and the NAME involution. It remains to recall now the value of MATH at MATH, which is computed by means of the NAME involution. |
math/0010272 | Observe first that the relations among MATH imply that MATH satisfies the differential equations up to first order in MATH. More explicitly, by summing REF up for all triples MATH and using the fact that MATH, we obtain the necessary relation MATH . On the other hand, the differential equations lead to recursive relations on the coefficients MATH of MATH of the form MATH . Observe now that these linear equations define MATH uniquely. Moreover, it is straightforward to show by induction that MATH for some constant MATH. This follows since the inverse of the matrix defining the above recursion is MATH as MATH. Thus the convergence of the series follows, and this concludes the proof. We remark that the constant MATH depends on MATH. |
math/0010272 | Observe that this ring is MATH-graded by the subscripts of MATH. The quadratic relations REF imply that the dimension of each MATH-graded component of MATH is at most MATH. Indeed, a spanning set is constructed inductively by multiplying the previous spanning set by MATH and by adding any monomial of the correct MATH-weight that does not contain MATH. We need to show that modulo MATH any two monomials with the same MATH-weight are equivalent. We use induction on the degree and the number of equal factors. If one monomial is MATH and the other one is MATH, then we may reduce the first monomial to MATH unless MATH. This argument shows that the only difficulty could occur in the case of monomials MATH and MATH with MATH and MATH. However, this is not possible for MATH. |
math/0010272 | The differential equations for the MATH's imply the following differential equations for the MATH's: MATH . To derive the first set of differential equations, we use the formula for the derivatives of MATH and then collect terms with the common factor MATH (where MATH is the index of summation in REF ). We then split off products of MATH and MATH to get MATH . We observe that the sums that contain MATH would cancel if the summation were over all MATH, since MATH results in MATH, MATH. Thus one may reduce all the terms to MATH modulo MATH. The fact that MATH satisfy REF insures that the coefficient of MATH vanishes. The differential equations for MATH are treated similarly. One can check that the equations on MATH and MATH imply that MATH is zero to the order MATH. The vanishing of other NAME coefficients of MATH is proved by induction on the degree of MATH. It is easy to see that the recursion matrix is invertible starting from the coefficients at MATH, because the diagonal entries dominate the rows. The case of the coefficients by MATH is handled separately as follows. It is easy to see that MATH, where MATH . To show that the coefficient of MATH at MATH is zero, it is enough to show MATH for all MATH such that MATH. To prove this, for every MATH, we use the relations REF for triples MATH, MATH, and MATH multiplied by MATH, MATH, and MATH respectively to get MATH . We then sum the above equations for all MATH. Finally we use MATH and the relation REF for MATH to get MATH which is REF above. To finish the proof, observe now that the differential equations for MATH imply that it vanishes to the order MATH as long as all MATH with MATH vanish. Alternatively, observe that MATH is even, because MATH. |
math/0010272 | By REF we can use MATH to map MATH to MATH. We observe that each MATH-graded component of degree MATH is mapped onto a space of dimension at least MATH, because polynomials of degree MATH in MATH and fixed MATH-grading can have an arbitrary singular part in their NAME expansions. The only exception is the zero graded component, where the coefficient by MATH cannot be chosen freely (in fact, it is always zero), but MATH allows us to freely choose the constant term. This provides a lower bound on the NAME function, which together with REF finishes the proof. |
math/0010272 | Let MATH be a solution to the system REF. Then the results of REF allow us to define a dimension one subscheme MATH of MATH cut out by the ideal MATH. Notice that points MATH with coordinates MATH, where MATH, lie on MATH. Moreover, MATH defines a map from the disjoint union of the neighborhoods of MATH non-singular points in MATH to MATH. As a result, the embedding MATH factors through MATH where MATH is the defining ideal of the reduced irreducible components of MATH that pass through points MATH. Indeed, a map to MATH is determined by a collection of MATH maps to the ring of usual NAME series, which is an integral domain. This shows that MATH. In particular, MATH is reduced and has no isolated closed points. Suppose now that MATH is singular. Let MATH be a singular point on MATH. We will first consider the case MATH, which we can normalize to get MATH. Let MATH be a MATH-point of MATH that reduces to MATH. Set up the differential REF for power series MATH. The argument of REF can be extended to the power series over the ring of dual numbers. Unless the solutions MATH are constant, this allows one to move a MATH-point to a nearby location on MATH, which is impossible since MATH is reduced. Therefore, the solution MATH must be constant, which implies MATH . If the point MATH satisfies REF , then MATH correspond to a cusp of MATH. For each MATH let MATH . Observe that because of relation REF, MATH depends on the sum MATH only. REF then implies MATH for all MATH. As a consequence, we have MATH for all MATH. There are now two cases to consider. CASE: One of the MATH, MATH is zero. If any other MATH is non-zero, then the above equation for a triple MATH implies MATH. Then the above equation for MATH gives MATH and so on. Eventually we get MATH, which is a contradiction. Therefore MATH for all MATH. Relations REF now imply that for all MATH, MATH . Let us denote by MATH. Then relations REF become MATH for all MATH. We may scale MATH so that MATH to get MATH . It is easy to see that for any given choice of signs, either there is no solution, or the solution is unique and rational. The action of MATH allows us to assume that MATH is the biggest of all MATH. We have MATH, therefore MATH. Since MATH, MATH must be positive, so MATH and MATH. Therefore MATH for all MATH. For every MATH we have MATH . If the right hand side is MATH, then we have MATH, which can happen only if MATH, MATH, and MATH. This implies MATH, so MATH. Then analogously MATH, and so eventually we get MATH, a contradiction. Therefore, we have MATH for all MATH. Together with MATH this forces MATH which means that MATH has values corresponding to one of the cusps of MATH. CASE: All MATH are non-zero. Then all MATH are non-zero for MATH. Denote by MATH. Notice that MATH for all MATH, when we set MATH. Therefore, MATH where MATH is a MATH-th root of unity. We can multiply each MATH by MATH where MATH is an appropriately chosen MATH-th root of unity to reduce ourselves to the case MATH. So now we have MATH. For every MATH, MATH consider now REF MATH . They imply MATH, so MATH depends on MATH only. Now the quadratic relations REF imply that MATH depends on MATH only, if MATH. This easily yields that all MATH are the same. If we act on this set of MATH by the NAME involution, see REF, we get MATH for which all MATH are the same. Then the argument of REF finishes the proof of the lemma. Proof of REF continued. Returning to the case of the singular point MATH with MATH, observe that REF imply that the point is one of the points MATH. Similar arguments show then that it has to be nonsingular. Because of the equality MATH and the presence of the MATH action, either MATH is a smooth curve of degree MATH and genus one, or it consists of MATH lines. In the latter case, the NAME function together with the symmetry forces the lines to form a MATH-gon. In particular it has singularities, which implies that MATH corresponds then to a cusp of the modular curve. So MATH is an elliptic curve with a MATH-action which permutes the points MATH. Therefore, these points form a subgroup MATH of order MATH on MATH. Observe now that the embedding of MATH by the MATH's is given by the complete linear system MATH. Because the MATH's are eigenvectors of the MATH-action, we get MATH where MATH denotes the NAME theta function and MATH is a uniformizing parameter on the universal cover of the elliptic curve MATH such that the points MATH lift to MATH. Since all the coefficients of the products MATH in the relations REF are one, all MATH are equal to one. Then the MATH's can be uniquely determined from REF up to a multiplicative constant. More precisely, we rescale the MATH's and MATH's to get MATH. Then for any two solutions MATH and MATH their componentwise difference MATH satisfies MATH whenever MATH. This together with the symmetries of the MATH's yields MATH. It remains to observe now that MATH is a solution to REF. This finishes the proof. |
math/0010272 | We need to show that the scheme MATH cut out by these quadratic relations is smooth. Let MATH be a closed point of MATH that is not a cusp. Let MATH be a MATH-point of MATH that reduces to MATH. NAME though MATH and MATH are defined only up to homothety (multiplication by MATH and MATH, respectively), we shall make a choice and fix a solution as a set of numbers. Then the quadratic relations REF define an elliptic curve MATH embedded into MATH by a complete linear system of degree MATH. A MATH-point defines a deformation of the homogeneous coordinate ring of this embedding. In fact, one can set up the system of differential equations on MATH that will now take values in MATH instead of MATH. Then its solutions will satisfy the quadratic relations and will define a map from MATH to the ring of NAME series with coefficients in the dual numbers. As before one shows that the dimension of MATH over MATH is MATH and that it is a free module over MATH with a monomial basis as in REF . Therefore we have a MATH-equivariant deformation of the embedding of an elliptic normal curve in MATH. Moreover the MATH-torsion points MATH, MATH with coordinates MATH, where MATH are fixed under this deformation. Such MATH-equivariant embedded deformations are parameterized by elements of MATH where MATH is the normal bundle of MATH. An easy calculation shows that this space is REF-dimensional. Therefore all such deformations come either from the scaling of the MATH's and MATH's, or from deformations along the modular curve. This shows that the scheme cut out by the quadratic equations in REF is nonsingular except possibly at the cusps. Thus it remains to show that in the neighborhood of the cusps the quadratic relations REF cut out a smooth curve. The action of MATH and the NAME involution allows one to consider only the case MATH . It suffices to calculate the dimension of the tangent space at this point. Denote the coordinates in the tangent space at this point in MATH by MATH with MATH and MATH. REF yield in this tangent space MATH for MATH with MATH, where MATH. Two solutions are easy to see, namely MATH and MATH where MATH is the NAME delta function. By subtracting a linear combination of these solutions one can reduce any other solution to the case MATH. For every MATH we get MATH which implies MATH and MATH . On the other hand, for every MATH one has MATH which now implies MATH . Because MATH, we deduce that MATH, and then that all MATH and MATH must be zero. Thus the dimension of the tangent space is one, after we mod out by the rescaling factor. |
math/0010272 | This is immediate by using relations REF to eliminate all MATH's in REF. Details are left to the reader. |
math/0010273 | It suffices to prove the expansion for MATH, so we can assume MATH. This is a special case of a result proven in CITE using the NAME transform. NAME give a different proof using methods found in CITE. It also suffices to consider a single face, say MATH. NAME work in a product neighborhood MATH and ignore the cutoff MATH in REF . From the definition we have MATH . Let MATH so that REF becomes the estimate MATH . In particular, since MATH we see that MATH approaches a limit as MATH, call it MATH. This MATH is a function on MATH, and since we also have estimates of the form MATH we have uniform convergence of tangential derivatives MATH as MATH, giving us bounds on MATH. This shows that MATH. The same principle applies to the radial vector field estimates. For example, since MATH the limit of MATH exists as MATH and equals MATH. In this way we get an estimate MATH for any MATH. Repeating this argument with higher derivatives and at all boundary faces of MATH, we conclude that MATH . Integrating REF from MATH to MATH gives MATH . Again using MATH, it is easy to see that MATH and hence MATH. Setting MATH, we thus derive from REF that MATH . Now let MATH so that REF implies MATH . This could be written MATH so assuming MATH we apply the above argument to find MATH such that MATH . Rewrite this as MATH . Repeating the argument yet again, we find MATH such that MATH . Then from MATH and MATH we form MATH and MATH by linear combination so that MATH . This procedure may be continued inductively up to MATH, yielding MATH . To remove the remaining derivatives, consider a function MATH satisfying MATH which is equivalent to MATH . Integrating from MATH to MATH gives MATH . Since we have MATH by the estimates on MATH, we conclude that REF implies MATH. Applying this argument MATH times to REF we obtain MATH where MATH. |
math/0010273 | Since MATH is unique up to MATH, independence follows from the invariance of MATH under MATH. To study the MATH dependence, for simplicity let us specialize to the case MATH, that is, a manifold with boundary MATH with boundary defining function MATH. It suffices to consider an index MATH where MATH, and show that MATH. Define the commutator operators MATH . By induction we will show that MATH . Because MATH, for MATH we can then estimate MATH and hence MATH. So the result follows once we establish REF . MATH is multiplicative, MATH and the mapping property MATH is immediate. For the inductive step, assume MATH satisfies REF and consider MATH . By definition we have MATH which takes care of the first term. And the second is a sum of terms of the form MATH and so using REF the desired mapping follows from MATH. The argument is the same in the general case, except that the induction must be done over MATH. |
math/0010273 | The leading terms in the boundary expansion of MATH are MATH where MATH. Observe that for any MATH where MATH and MATH. Provided that MATH we can set MATH so as to have MATH . The remaining terms MATH are obtained by an obvious induction, with the requirement that MATH ensuring that no zeroes occur in denominators. Then using NAME 's lemma we sum the series asymptotically at MATH to get MATH. |
math/0010273 | The first property follows from a standard wave-front set argument. To prove the second, consider MATH and MATH. We can compute MATH by first pull MATH up to MATH through the right, then multiplying by the lift of MATH times the Riemannian density (in the right factor), then pushing forward to MATH through the left. But the lift of MATH vanishes to infinite order at both the right and front faces, and so the push-forward can in fact be written as an integral on MATH: MATH for some MATH. Then MATH is established by moving derivatives under the integral. |
math/0010273 | It is most convenient to argue in the model for MATH given by the upper half-space blown up at the origin. Sticking to our convention that MATH is the boundary defining function, we use coordinates MATH. The model Laplacian is MATH . With radial coordinate MATH, the defining functions for the faces of MATH are MATH and MATH. According to REF, the model resolvent extends meromorphically to a map MATH with poles as indicated above. Let MATH. NAME solve the problem MATH in stages. First a simple computation shows that for MATH, MATH where MATH and MATH. We can use this to solve away the asymptotic expansion in MATH, exactly as in REF , with poles at MATH. The result is MATH such that MATH . Now suppose we want to solve MATH where MATH. By REF we can apply MATH to MATH. Then we have MATH . A straightforward computation shows that MATH and MATH is invariant under MATH. Therefore MATH . The model resolvent can be applied to this expression by REF . The solution to REF is then MATH with poles as indicated. By induction we can extend this trick to higher powers of MATH. By applying this argument to successive terms in the asymptotic expansion of MATH in MATH and asymptotically summing the resulting terms, we can find MATH such that MATH . The solution to the original problem is now given by MATH. |
math/0010273 | Let MATH so that MATH . Assuming that MATH, we can apply the argument from REF to get MATH such that MATH . Now If we define MATH then from REF the relation to MATH is MATH . By solving a linear system for the coefficients MATH in terms of the MATH we can rewrite REF as MATH . Suppose MATH satisfies MATH for MATH. This implies MATH . Integrating from MATH to MATH gives MATH . We note that MATH for MATH, so we conclude that REF implies MATH. Applying this repeatedly to REF gives MATH . Substituting back with the definition of MATH from REF and once again solving a linear system for new coefficients, we get MATH . The matrix relating the MATH's to the MATH's is lower triangular, with diagonal entries of the form MATH. Hence the requirement that MATH ensures the system is non-singular. Now we let MATH be the function in parentheses in REF and simply repeat the argument given above. Assuming MATH, we obtain MATH such that MATH . This construction is possible for MATH, and MATH can be arbitrarily large, so we have full asymptotic expansions. |
math/0010273 | That MATH follows immediately from MATH, so we concentrate on MATH. The asymptotic properties of MATH depend only on the behavior of MATH near MATH and/or MATH. So we can specialize to a particular coordinate neighborhood with coordinates MATH and assume that all functions are compactly supported within this neighborhood. We can rewrite REF as MATH for MATH. On MATH we'll use coordinates MATH, MATH, MATH, and MATH. For convenience, we extend MATH into the interior so as to be independent of MATH within the neighborhood of interest. So MATH does not depend on MATH, MATH, or MATH. Noting that MATH by integration by parts, we can move derivatives under the integral to get MATH . The lift of the vector fields appearing here to MATH is MATH . Writing MATH for MATH, the integrand on the right in REF becomes MATH . Now by the definition of the truncated spaces in REF, MATH . Also MATH and MATH so we conclude that MATH . Applying this argument inductively gives MATH where MATH. For MATH we have MATH. The lift of MATH to MATH through the left lies in MATH, so by passing derivatives under the integral and using the estimates on MATH we have MATH . Hence for MATH we obtain MATH . This has been derived under the assumption that MATH is independent of MATH in the local neighborhood, so the result then follows from REF . |
math/0010273 | Integrating by parts gives MATH . In taking the limit MATH, the boundary term disappears by REF , yielding MATH . Using MATH, we rewrite this as MATH . Thus we can estimate MATH . We take care of the MATH on the right by an estimate of the form MATH and the result follows. |
math/0010273 | Let us single out the operator appearing in the radiation condition by defining MATH . We first note that the only issue is to control MATH at the boundary. Because for MATH, MATH for any MATH, and by REF , we have MATH . So we will work in a product neighborhood of MATH in which the metric has the form REF . Since MATH, the radial component of MATH can be singled out as MATH . We will also let MATH so that MATH . Let MATH be small enough that MATH lies within the product neighborhood. For MATH we choose some MATH such that MATH . We begin by computing a divergence: MATH . In the first term on the right-hand side we have MATH where MATH . Inserting back into REF , we have MATH . Now integrating the divergence on the left-hand side of REF gives MATH and by REF the boundary term is zero. Hence the integral of REF gives MATH . One more integration by parts is needed, to get rid of the second derivative of MATH appearing in the first term of the integrand in REF . To this end, note that MATH . On substituting this into REF , the second derivative of MATH would be contained in the term MATH . This can be integrated by parts to give MATH with a boundary term of MATH (again by REF ). Combining this integration by parts with REF and substituting back into REF we now have MATH . The NAME derivative of MATH may be expressed in terms of MATH: MATH . Since MATH, for some MATH fixed we will have MATH as a tensor, for MATH sufficiently small. Recalling that MATH was to be supported in MATH, we require that MATH be sufficiently small so that REF holds for MATH. Applying REF to the relevant term in REF gives MATH . Using this in REF along with the fact that MATH, we obtain an inequality MATH . Recalling that MATH for MATH, we split the integral on the left-hand side of REF to yield MATH . We can pick MATH, and since MATH, by choosing MATH sufficiently small we can assume MATH. Noting also that MATH, the left-hand side can be bounded below by MATH . Applying this bound together with the estimates MATH we derive from REF the new inequality MATH . By REF the MATH norm of MATH is bounded by that of MATH for MATH. And MATH is estimated as indicated at the start of the proof. With these facts we can reduce REF to MATH . Since MATH, if we set MATH this yields the stated inequality. |
math/0010273 | NAME argued in CITE that the existence of a local parametrix in MATH as in REF shows that there can be no solution of MATH lying in MATH. So the point here is to use the radiation condition to argue that MATH would have to be MATH. Let MATH be a cutoff in MATH. Choose MATH with MATH for MATH, MATH for MATH, and MATH for MATH. Setting MATH, we can interpret MATH as a function on MATH using the product neighborhood on which the metric has the normal form REF . Assume MATH is small enough that MATH is supported in MATH. Assuming that MATH is as stated, we have MATH . Choose MATH small enough that MATH for MATH. Then REF becomes MATH . The commutator is MATH . Substituting this back into REF and taking the imaginary part yields MATH . Let MATH, which lies in MATH by assumption. Then we can write REF as MATH . The first term on the right can be bounded: MATH . By REF and MATH, so MATH is integrable with respect to MATH. Thus this first term can be estimated MATH. For the second term on the right-hand side of REF we have the bound MATH . Since MATH is integrable, this term can be estimated MATH for small MATH. Since MATH is bounded from below in the support of MATH, using these estimates back in REF gives MATH . In fact, since MATH on the interval MATH . We conclude that MATH is finite. By NAME 's uniqueness result, MATH. |
math/0010273 | Let MATH. Suppose that MATH. Then we can define a new sequence MATH so that MATH and MATH . By REF , MATH is a bounded sequence in MATH. Since the inclusion MATH is compact, any subsequence of MATH has a subsequence converging in MATH. The limit MATH of and such subsequence satisfies MATH and by REF it also satisfies MATH. Thus by REF MATH. Since MATH this is not possible. The conclusion is that MATH is a bounded sequence in MATH. By the same reasoning we applied to MATH above, any subsequence of MATH has a subsequence converging in MATH to some MATH such that MATH and MATH. Since such a MATH is unique by REF and therefore independent of the subsequence, we have MATH. |
math/0010273 | Let MATH with locally defined right parametrix MATH as in REF . For notational convenience we'll consider only the global case, since the introduction of cutoff functions for the local case is a simple matter. Assume MATH is as stated. The idea is to apply the transpose of the equation MATH to MATH. To justify this requires integration by parts, and this is where the radiation condition comes in. Consider the real pairing MATH for arbitrary MATH. If the integral is cutoff at MATH, then we can transfer MATH to MATH, picking up a boundary term: MATH . Let MATH denote the boundary correction from the right-hand side of REF . The radiation condition tells us that MATH, and by the structure of MATH and REF we have also MATH. Thus the expression in brackets in MATH lies in MATH, because the leading terms cancel each other. In other words MATH which implies MATH because we can assume MATH. To transfer MATH to MATH, requires only NAME 's theorem, and so we have MATH . Substituting MATH in the left-hand integral then gives MATH . Since these properties hold for any MATH, we conclude that MATH . Note that MATH, so MATH by REF . Also MATH so we can show MATH by moving derivatives under the integral. Hence MATH as claimed. |
math/0010273 | A different choice of boundary defining function could always be written MATH, where MATH and is strictly positive. Thus we are concerned with the difference MATH for MATH. For simplicity we can assume MATH, so that we want to show that MATH . We use an induction similar to the proof of REF . Defining the commutators MATH the result will follow if we can show MATH . The operators MATH and MATH are functions that act by multiplication, and a simple computation shows that MATH and MATH so the mapping result holds for these cases by the characterization in terms of asymptotic expansions. The inductive step follows precisely as in REF , once we note that MATH . |
math/0010273 | It is easy to verify that vector fields in MATH lift to vector fields in MATH by considering local coordinates. For example, in the interior of the crossover face the coordinates MATH are valid. The lifts of the vector fields MATH and MATH to MATH are MATH . At the boundary of the regular and crossover faces, we use coordinates MATH, MATH, and the lifts are MATH . The lemma follows immediately from the invariance of MATH under MATH. |
math/0010273 | Extending MATH smoothly into the interior in some arbitrary way, we have MATH because MATH is the indicial root. As in the proof of REF , let us use coordinates MATH, MATH. By the lifts of vector fields computed there we see that MATH where MATH. This shows that the asymptotic expansion of MATH at the regular face may be solved away exactly as in REF , giving MATH as stated. |
math/0010273 | The first part of the construction is the symbolic inversion along the lifted diagonal in MATH. This does not involve MATH and so can be done just as in CITE. We obtain MATH such that MATH . The kernel of MATH, lifted to MATH, lies in MATH. The next step in the construction is to try to solve away this error term at the front face. To this end, we seek an operator MATH such that MATH . Here MATH. For MATH, MATH where MATH is the metric on MATH given by MATH with MATH the constant tensor on MATH defined by the restriction of MATH to MATH. Since MATH is constant, MATH is the pull-back of the standard hyperbolic metric on MATH by a linear change of coordinates. Let MATH denote the resolvent for this metric, which is just given by applying the linear coordinate change to the standard resolvent on MATH. Noting that MATH, we set MATH . For each MATH we have MATH. Although MATH depended smoothly on MATH, MATH does not because of MATH. So this is the first point at which we need the lift from MATH to MATH. Let MATH denote the front regular face (the lift of MATH). Since the dependence of MATH on MATH is analytic, the lift of MATH to MATH has a square root singularity at the crossover boundary coming from that of MATH. Hence we have MATH where the boundaries of MATH are left/right regular and front crossover, and MATH is lifted to MATH through the projection MATH. The front crossover face, denoted MATH, fibers over lift of MATH to the front face MATH in MATH. The lift of MATH from MATH to the front crossover face in MATH will be constant on such fibers. Thus the lift of MATH has no singularity at the front regular boundary of MATH, and MATH . The boundary faces of MATH are left/right/front regular and left/right crossover (see REF ). By REF , this shows that the lift of MATH to MATH has the properties of the restriction of a kernel in MATH to the two front faces. We may therefore choose an extension MATH, having the lift of MATH as leading coefficient at these faces. Furthermore, since MATH was constant on fibers of the front crossover face, we can insist that the leading coefficient of MATH at the left crossover face be constant on the fibers of this face as well. The error term at the next MATH . By construction, the kernel of MATH vanishes at the front face, so MATH . However, there is additional decay on the front face, because MATH is the indicial root. The composition MATH has kernel given by lifting MATH to MATH through the left and applying it to the kernel MATH. Consider local coordinates MATH, where MATH, MATH, MATH, and MATH. This system is valid in a neighborhood of the intersection MATH. The lift of vector fields through the left is easily computed: MATH . Here MATH defines the left regular face, and MATH the left crossover. It is easy to compute that MATH where MATH and MATH. In particular, with MATH we see that MATH vanishes at the left regular face. At the left crossover face we have the same issue as in REF , because the final term on the right in REF is not of lower order in MATH. More specifically, MATH contains terms of the form MATH times smooth coefficients. However, in these coordinates MATH is the fiber variable for the left crossover face. Thus choosing MATH to be constant on the fibers means that in the calculation of MATH the term represented by MATH does not contribute. We conclude that MATH . At the front regular face we may iterate the procedure of solving away terms with the normal operator, exactly as in CITE. We would next apply the model resolvent to MATH (using REF to handle the extra logarithms) and so obtain MATH, from which we determine MATH, and so on. (We cannot do the same at the crossover face, because the derivatives along the fiber appearing in REF cannot be assumed to vanish for any terms beyond MATH. As a result, we could not apply REF to the higher MATH's because of the lack of decay at the left crossover face.) After repeated applications of the normal operator at the front regular face we end up with a sequence MATH which can be summed asymptotically to give MATH such that MATH where MATH. Finally, to get MATH we can appeal to REF repeatedly to solve away the power series at the left regular face, as in REF . This yields MATH such that MATH where MATH is as defined in the statement. Setting MATH gives the result. |
math/0010273 | For convenience let us use subscripts to denote lifts by the MATH's, for example, MATH. Multiplying REF on each side by MATH gives MATH . If MATH is a smooth half-density on MATH, then MATH where MATH is a smooth half-density on MATH. Let MATH be the blow-down. By computing the codimensions of the submanifolds to be blown-up, we can see that MATH so that MATH . Let MATH be the blow down MATH and MATH a smooth density on MATH. Again, by computing codimensions we can see that MATH . Using the liftings REF and combining REF , and REF we then obtain MATH where MATH is the product of all defining functions of the regular faces and MATH the same for the crossover faces. The composition REF then becomes MATH where MATH and MATH. The result follows because MATH . |
math/0010273 | With MATH, we will compute MATH from REF . MATH and MATH are most succinctly described with a chart of index sets akin to REF . Fortunately we can consider the regular and crossover faces separately. At the regular faces we have: MATH and at the crossover faces: MATH . Under MATH the faces which map to the interior are MATH and MATH. From REF we see that the combination MATH vanishes to infinite order at MATH, while at MATH the index set is MATH, so the push-forward is well-defined. Notice that MATH vanishes to infinite order at all of the regular faces of MATH except MATH, where it is polyhomogeneous conormal with index MATH. Thus MATH will be the pushforward of a b-density with index set MATH at MATH and MATH at all the other regular faces. By REF we conclude that MATH has orders MATH at the left, right, and front regular faces. As for the crossover faces, by REF the index sets of MATH are MATH at the left crossover, MATH at the right crossover, and MATH at the front crossover face. It is a trivial observation that for MATH and the result follows. |
math/0010273 | Let us start with the second formula in REF . Suppose that MATH . As in the proof of REF , we examine the behavior of the lifts to MATH. At the regular faces we have: MATH . In order to apply REF we consider MATH, which is the push-forward of a b-density on MATH with index set MATH at MATH, MATH, and MATH, and MATH otherwise. Then pushforward gives indices MATH for MATH at the regular faces. The index chart for the crossover faces: MATH . We conclude that the crossover index sets for MATH are MATH. The first formula in REF is proven in exactly the same way. If MATH is as above, but MATH, then the product MATH will have an interior singularity at the lift of the diagonal through MATH. One can check that MATH annihilates this singularity by standard wave-front set arguments (see CITE for an explicit discussion of this). Then the proof of push-forward to MATH is exactly as above. |
math/0010273 | Let MATH as an element of MATH. From REF we immediately see that MATH . Now consider REF . By construction MATH, which is smooth up to the front face, and since MATH lifts to MATH we conclude that MATH . This means that MATH must annihilate all of the terms in the asymptotic expansion of MATH at the front regular face, since an expansion containing MATH is not allowed in MATH. We can argue term by term using the normal operator to show that these coefficients are zero. For example, at leading order we have MATH for all MATH. From the definition of MATH we see that MATH . Applying REF (in the model case with constant indicial root) to REF , we conclude that MATH. Using this argument inductively, we conclude that MATH vanishes to infinite order at the front face, which puts it in MATH. |
math/0010273 | First note that MATH lifts under MATH to MATH, as may be easily checked in local coordinates. Observe that MATH . Because of the infinite order of vanishing at the two front faces, we deduce that MATH . Let MATH be radial vector fields at the left/right regular and left/right crossover boundary faces of MATH, respectively. We claim the lift by MATH of MATH is a radial vector field MATH at the left regular face, possibly plus a term in MATH. Similarly, MATH lifts to a radial vector field MATH at the left crossover face, possibly plus a term in MATH. The behavior on the right is analogous. These facts are best checked in local coordinates. On MATH we can use coordinates MATH, such that MATH, MATH, MATH, and similarly on the right. For MATH and MATH we can take MATH and MATH. Near MATH and MATH we ignore the blow-up of MATH, since MATH does not intersect these. So we can use the coordinates MATH, where MATH and MATH. Then MATH again, which is indeed a radial vector field at the left regular face MATH. The lift of MATH is MATH which is a radial vector field at the left crossover face MATH. Similarly, with a different coordinate system we could check this in a neighborhood of MATH (then we could ignore the blow-ups of MATH and MATH). Of course, the radial vector fields are not uniquely defined. Another choice of MATH would differ by a field in MATH, where MATH is a defining function for the left regular face of MATH. Since MATH, MATH lifts to MATH, which agrees with the correction term given above. Now we can show MATH by pulling these radial vector fields up to MATH. (Once again the infinite order of vanishing at the front faces is crucial.) For example, MATH entails the estimate MATH and this would be unaffected if we add to MATH a term in MATH. Consequently, it implies the estimate MATH . In a similar way all of the estimates needed to show MATH can be deduced. |
math/0010273 | Let MATH and MATH. As in the proof of REF , we use a chart to describe the boundary behavior of MATH. For the regular faces we have: MATH and at the crossover faces: MATH . The behavior at the middle faces, together with REF , shows that the push-forward is well defined. And the form of MATH is deduced from REF . The second formula is handled in the same way. As in REF the push-forward annihilates the interior singularity. |
math/0010273 | We will prove only the second of the three formulas. The first follows by similar argument because of the cancelation of the interior singularity as in REF . The third formula follows by an analogous but even simpler argument on MATH. Let MATH, where MATH is a section of MATH. As noted in the remarks following REF , there are b-fibrations MATH projecting through the left and right factors. The action of the operator with kernel MATH on MATH, where MATH, is given by MATH . Multiplying both sides by MATH, we can recast this in a form suitable for push-forward: MATH where MATH is shorthand for MATH, MATH, and MATH. The combination MATH has index family MATH, so the push-forward is well-defined. Multiplying MATH by MATH and applying the push-forward result REF tells us that MATH has index family MATH, and the result follows. |
math/0010273 | Since the index family MATH implies MATH decay at the left face, the composition is represented by a convergent integral. The necessary estimates can be obtained by differentiating under the integral using dominated convergence. |
math/0010273 | Using REF we solve for MATH and asymptotic to MATH near MATH, such that MATH. Then let MATH, which is in MATH by REF . Thus MATH has the stated properties. |
math/0010278 | All claims follow by an easy induction. Using the skein relation REF we express a link as the sum or difference of two links which are either of shorter word length or are ``less linked". |
math/0010278 | Let MATH, MATH be a braid with exponent sum MATH satisfying our conditions. The polynomial MATH can be written as MATH for some polynomials MATH of degree less than MATH in MATH. Now the coefficients of MATH in MATH is a (framing independent) NAME invariant, which depends on MATH and MATH. Since, by our condition, all NAME invariants up to degree MATH vanish, we know that MATH must be the coefficient MATH of MATH in MATH. We know that MATH must have degree less than MATH, but MATH has degree MATH. Thus MATH must be trivial for all choices of MATH, which means, MATH has to be a root of MATH. Hence, by applying the NAME expansion formula to MATH, we conclude that MATH is one of the numbers MATH. |
math/0010278 | That the given dimensions form an upper bound, follows from REF . It remains to construct the following: Let MATH be the degree. We will show that there are braids in MATH such that the dimension of the coefficients of MATH is MATH. For a braid MATH the polynomial MATH satisfies: MATH . Furthermore, we have for arbitrary MATH: MATH . In particular, the coefficient of MATH is MATH. Combining these two formulas, one sees that the coefficient of MATH in MATH is MATH. Replacing successively pairs MATH by MATH, where MATH has the same parity as MATH, yields polynomials with coefficients of MATH equal to MATH. Hence the dimension is equal to the number of MATH with the same parity as MATH. |
math/0010278 | By REF we know that the polynomials MATH and MATH are polynomials in MATH of degree less than or equal to MATH. Let MATH (MATH, respectively) be the exponent sum of the braid MATH (MATH, respectively) on MATH (MATH, respectively) strands. We know that the NAME polynomial MATH - with our reparametrization - is given by a power series in MATH . The coefficient of MATH in MATH is a NAME invariant of order MATH. Since MATH is of degree in MATH less than or equal to MATH, MATH is determined by all NAME invariants up to order MATH. In turn, MATH itself is determined by all terms in MATH of degree less than or equal to MATH. The same argument works with MATH and we are done. |
math/0010282 | To show REF it suffices to show that MATH sends the skein relation MATH to MATH. We have MATH because MATH and MATH. The last statement follows from the fact that if MATH is an oriented framed link and MATH its diagram such that the framing of MATH is the flat framing of MATH then MATH where the sum is taken over all crossings of MATH. MATH and MATH are clearly algebra epimorphisms and MATH and MATH, where MATH are left inverses of MATH and MATH respectively. Thus MATH is an isomorphism on the subspace of MATH generated by trivial links. In fact our proof works for any rational homology sphere, and assuming MATH, for any MATH-manifold. REF follows immediately from REF , because MATH are linearly independent in the ring of polynomials. |
math/0010282 | Part MATH follows from the fact that relations REF follow from the NAME relations, REF. If the trivial link MATH is linearly independent in MATH as is the case for MATH or more generally for MATH (that is, the product of a surface and the interval), and as is conjectured for any MATH-dimensional manifold, then we have a monomorphism MATH from MATH, the submodule generated by MATH to MATH REF, which is left inverse to MATH (MATH). As observed before, for MATH invertible MATH in the fourth skein module, thus MATH of the proposition follows. |
math/0010282 | Any tangle of REF is symmetric with respect to MATH, MATH, and MATH axes. |
math/0010282 | The fact that elements of MATH and MATH are MATH-algebraic follows from the definition. For a MATH-bridge tangle, we start from the MATH-tangle which is equal to MATH for MATH or MATH, and then we add crossings one by one (compare Figure MATH, the figure is drawn to stress the fact that we deal with a REF-bridge tangle. In the algebra of REF we would describe it as MATH). |
math/0010282 | It is sufficient to show that MATH, MATH, and MATH using the fourth skein relations. MATH. MATH. MATH. |
math/0010282 | Let MATH be the number of crossings of MATH. If MATH, then there exists an element MATH of MATH such that MATH is strongly equivalent to MATH. If MATH, then MATH is an alternating expression of MATH or MATH. Then, MATH is dot equivalent to MATH from REF . There is no irreducible element of MATH with more than four crossings. |
math/0010282 | We prove the theorem by an induction on the number of crossings of the element. If MATH, then clearly the statement holds. Assume that the statement holds in the case MATH fewer than MATH. Consider the case MATH. If the element is reducible, then it is generated by elements with fewer crossings than MATH, which can be generated by elements of MATH from the assumption of the induction. If the element is irreducible, then it is generated by an element of MATH and elements with fewer crossings than MATH from REF . This induces that the irreducible element is also generated by elements of MATH. |
math/0010282 | If both elements of the pair are MATH-braids, then the statement follows from REF . The other cases are easy to be checked. For simplicity, we ignore trivial components in the consideration. |
math/0010282 | We need to show only for MATH and MATH. From Figure MATH, we have MATH . |
math/0010284 | We can follow the same argument as we did for REF . Let MATH be the number of isogeny classes of MATH-dimensional abelian varieties over MATH such that MATH modulo MATH. Then our bounds on MATH and MATH give us MATH . On taking the limit, the theorem follows. |
math/0010288 | We choose an open cover MATH for MATH of coordinate balls of radius MATH and assume this is such that on each ball we can choose coordinates MATH with MATH represented by MATH. Further we assume that in fact the balls MATH also cover MATH. In each MATH we look at the disks MATH. We may assume that as MATH varies these disks are all disjoint in MATH. By taking MATH sufficiently small, any given MATH can be isotoped to the identity in a neighbourhood of each MATH. To do this, we note that MATH intersects each MATH transversally so we can project MATH along MATH and assume that MATH maps each MATH to itself. The maps MATH must then be isotopic to the identity and this isotopy can be extended to a small neighbourhood at the expense perhaps of increasing MATH slightly. The idea is now to isotope MATH to the identity between the MATH. Suppose a disk in some MATH is connected by flow-lines of MATH, say of length no more than MATH, to a disk in MATH. Let MATH be the union of the flow-lines. Then there is an isotopy of MATH, which leaves MATH equal to the identity near MATH and MATH and away from MATH, such that the resulting map is the identity on MATH except perhaps for a MATH-small neighbourhood of its boundary. Furthermore, if the original MATH leaves a subset of flow-lines in MATH fixed, parameterized by a subdomain of MATH with finitely many components and a smooth boundary, then the whole isotopy will leave these flow-lines unchanged. This is an application of the following fact. Fact Given a diffeomorphism MATH of the disk, equal to the identity near the boundary, there exists an isotopy of MATH to the identity leaving the boundary fixed. The construction can also be carried out for a family MATH, that is, there exists a corresponding smooth family of isotopies. If the MATH are also equal to the identity on a smooth subdomain with finitely many components, then the family of isotopies can be arranged also to leave these components fixed. Without the condition on leaving a certain subdomain of the disk fixed, this is described exactly in CITE, section MATH. To get the result as stated above one reduces to the original case by first isotoping to the identity in a connected domain enclosing the fixed-point set. The isotopy required by the lemma is now easily constructed. We start with a homotopy between the vector fields MATH and MATH, say MATH where MATH and MATH. This can be arranged so that MATH for all MATH. We cut-off the MATH to remain as MATH near the boundary of MATH. Let MATH be the diffeomorphism of a neighbourhood of MATH defined by leaving MATH fixed and flowing along the vector field MATH to MATH then back along MATH for a similar time. Then MATH and MATH away from the boundary of MATH, see figure MATH. Unfortunately such MATH are not necessarily equal to the identity near MATH, except of course near its boundary. However we can rescale MATH to ensure that the map preserves MATH and then compose with suitable isotopies in the MATH-planes provided by the above fact to correct this. On a subset of MATH where MATH we may assume that MATH for all MATH and the maps MATH can be taken to be the identity all along. MATH . We now apply this lemma repeatedly between different MATH and MATH. Notice that a map is equal to the identity if it is equal to the identity on the flow-lines connecting sets of the form MATH since these regions cover MATH. The order in which we apply the lemma must be chosen carefully however, so as not to disturb regions in which MATH has already been isotoped to the identity. We first isotope MATH to the identity on the union MATH of flow-lines between all MATH and MATH such that the flow-lines are of length less than MATH and the MATH do not intersect any other MATH. The disks MATH referred to here are the same as those MATH defined above, although we are not assuming that MATH and MATH are equal to MATH and MATH for the same MATH. REF divides the region between two disks MATH and MATH into three numbered regions showing the order in which the isotopy provided by REF should be carried out if another disk MATH intersects the flowlines between them. Next we apply the lemma to all remaining pairs of MATH. The point now is that if any MATH happens to sit between MATH and MATH then, away from a MATH-neighbourhood of its boundary, it must lie on a complete set of flow-lines of MATH from MATH to MATH on which MATH has already been isotoped to the identity. Thus the new isotopy will not affect MATH here. After all of the above isotopies then, the resulting map MATH is equal to the identity. MATH . |
math/0010288 | The kernel of any degenerate MATH must be at least MATH-dimensional and so have non-trivial intersection with MATH, which is clearly a contradiction. MATH . |
math/0010288 | For any MATH, MATH and MATH satisfy the conditions of the above lemma on MATH with MATH as MATH is orientation preserving and approximately preserves the NAME vector field. MATH . Now, since MATH and MATH agree near MATH, we can apply NAME 's method to find an isotopy of MATH, fixed near MATH, which generates a symplectomorphism between MATH and MATH. Specifically, the isotopy can be taken to be the time-MATH flow of the time-dependent vector field MATH uniquely defined by MATH. Note that MATH both near MATH and near the zero-section. The composition of this isotopy with our original diffeomorphism, denoted again by MATH, is now the required symplectomorphism between MATH and MATH. We now represent MATH explicitly as a Hamiltonian diffeomorphism. Observe that associated to any MATH-form MATH on MATH is, in the terminology of CITE, a `contracting' vector field MATH defined by MATH. In the case of MATH or MATH, this vector field vanishes only along the zero-section MATH and the associated flow contracts a disk towards each point on MATH. For MATH, these disks are just the cotangent fibers. Now, MATH maps MATH into MATH. The only map doing this which is fixed near MATH is defined as follows. Flow along MATH until we are in the region where MATH, then flow out along MATH for the same time. Let MATH and MATH denote the time-MATH flows of MATH and MATH respectively. Assume that MATH lies in the region where MATH. Note that as MATH near MATH we can extend MATH and MATH smoothly to MATH. Define an isotopy MATH, MATH by MATH. Then MATH and MATH. Now, MATH and similarly MATH so we have MATH and the MATH are all symplectomorphisms. Let MATH, then MATH. Hence the form MATH is closed and the isotopy is Hamiltonian if it is exact. But MATH vanishes near the zero-section and so we can use a parameterized version of the Relative NAME Lemma to construct a smooth family of MATH on MATH such that MATH as required. |
math/0010290 | We prove only the invariance of MATH. It follows from REF that we have MATH . |
math/0010290 | We can easily show that MATH . So we have proved the lemma. |
math/0010290 | We can show that MATH . The last equation is realized by using the NAME operators. In fact, we can show that MATH . Here, the NAME operator MATH on the irreducible representation space MATH is defined by MATH and acts as the constant MATH on MATH. Thus we have proved the proposition. |
math/0010291 | This is an immediate consequence of the corresponding property for the law of the pinned sites, given in REF below. |
math/0010291 | REF . is very simple: In the square-well REF is equivalent to MATH which follows from NAME 's inequality. The MATH-pinning case is similar. REF is easy, too: For any local increasing function MATH (of the random set MATH) with support inside MATH, one has MATH . Translation invariance is a simple consequence of this. Indeed, let MATH and MATH. Denoting by MATH (respectively MATH) the biggest (respectively smallest) square box centered at MATH contained in (respectively containing) MATH, we have MATH provided MATH is big enough. Taking the limit MATH and using the fact that MATH, and the corresponding statement for MATH, we get MATH which implies the desired result. |
math/0010291 | We recall that the variance of the Gaussian field can be written MATH where MATH is the law of the random walk in MATH, with transition probabilities MATH, starting at MATH. Inserting this in REF, we get MATH . (Remember that MATH.) Taking the expectation with respect to MATH inside, we get MATH . The corollary then follows from an application of the estimates of REF . |
math/0010291 | We suppose first, for simplicity, that MATH is connected. The changes for the general case are the same as those described in CITE, and we'll indicate their effects on our bounds at the end of the proof. Let MATH, and define MATH as the union of MATH and all its nearest neighboring cells in MATH; let MATH be the largest MATH for which MATH. We then write MATH and MATH . It was proved in CITE, see the proof of REF , that MATH for the class of sets MATH containing exactly one point in each cell of MATH. Therefore, summing only over such MATH's in REF (notice that there are MATH choices for which site is occupied in a given cell), we get, choosing MATH (MATH from the formula above), MATH . From this we easily prove the claim for the one-component case, by summing over MATH. Indeed, we can use the trivial estimate MATH . To treat the case of multiple components, one proceeds as in the proof of REF. The idea is to grow simultaneously all components in a suitable way. This procedure only modifies the value of the constant in the exponent, provided the components are all big enough. In our present situation, this is enforced automatically as soon as MATH is sufficiently small (the cells from which MATH is built are growing when MATH decreases). |
math/0010291 | CASE: If MATH, FKG property of MATH implies MATH for any set MATH, see REF. Therefore MATH which proves the claim since the probabilities are bounded by MATH. CASE: Using the expansion REF, we can write MATH . We therefore have to bound the conditional probability. This can be done as follows: MATH where we used REF to bound the numerator and the bound on the denominator follows from MATH and the local CLT. Therefore the sum over MATH is smaller than MATH, which proves the first claim. To prove the second claim, notice that MATH since MATH (one can restrict the sum over sets MATH not containing MATH, since otherwise the probability of reaching MATH is REF). |
math/0010291 | This follows from a standard subadditivity argument, since MATH where MATH, and the inequality follows from the FKG property. The constant MATH, which depends only on MATH, takes care of the possible discrepancy between MATH and MATH. |
math/0010291 | REF is proved in CITE. REF follows from REF by a standard argument, see CITE. The proof there is for the nearest neighbor random walk only, but it can be easily adapted to cover the more general case considered here. REF follows from REF and PREF REF follows from a standard local limit theorem: MATH for some positive MATH . Under the assumptions of the existence of a third moment, this is a standard NAME type estimate with MATH . We don't know of an exact reference under the assumption of a MATH-moment only. The paper CITE treats the case of a one-dimensional random walk. The method there can easily be adapted to prove REF on the two-dimensional lattice. Finally, REF is proved in REF , for the simple random walk. Again, the proof can easily been adapted to cover the more general case. |
math/0010291 | We use the standard approximation of MATH by tilting the measure and applying a local central limit theorem with error estimate. For MATH in a neighborhood of MATH we consider the tilted measure MATH where MATH . Clearly MATH and MATH . Therefore, the mapping MATH is an analytic diffeomorphism of a neighborhood of MATH to a neighborhood of MATH leaving MATH fixed. Therefore, for any MATH in a neighborhood of MATH in MATH, there exists a unique MATH with MATH . Using this, we see that for MATH small enough, we can write MATH where MATH . Evidently, MATH and a simple computation yields MATH . Furthermore, MATH has now mean exactly MATH and covariance matrix MATH where MATH depends analytically in MATH and satisfies MATH . Applying a local central limit theorem with standard NAME type error estimate we get MATH . |
math/0010291 | Approximate MATH in REF by an appropriate quadratic function. |
math/0010291 | REF is well known. For the convenience of the reader we give a crude proof, sufficient for our purpose. MATH for some MATH by choosing MATH appropriately, for large enough MATH . CASE: Let MATH . Then, by standard large deviation estimates, MATH for MATH large enough, when MATH is chosen appropriately for example, MATH . We consider the independent differences MATH . We have for MATH for any MATH . According to REF, we can choose MATH such that MATH and then, for any MATH we may choose MATH large enough, such that MATH that is, MATH if MATH . This proves the claim. |
math/0010291 | The proposition is an easy consequence of the one-dimensional result. Indeed, write the random walk in (dependent) components MATH where MATH are one-dimension random walks, possessing an exponential moment. The first time MATH when MATH leaves MATH is also the first time where one of the MATH leaves the interval MATH . Assume for instance that at MATH, MATH for the first time leaves the above interval to the right. (There are of course MATH other cases). This is then the first time it surpasses MATH . Furthermore, the number of MATH-cells visited by the MATH-dimensional walk is at least the number of intervals among MATH visited by MATH . For the other MATH cases, similar statements hold, of course. From this observation, REF follows immediately from REF . |
math/0010291 | Remark first that under the conditions of the lemma, MATH for the large enough MATH . This easily follows from irreducibility and aperiodicity. Therefore, MATH is well-defined. We first derive a simple exact expression for this expectation: MATH . This readily follows from a standard ``last exit - first entrance" decomposition. MATH . MATH . MATH . Implementing this into REF and summing over MATH yields MATH . From this, REF follows. We next use this together with the information on MATH in REF to get the desired estimate. MATH the last equation by recurrence of the two-dimensional random walk. From REF , we get MATH and it therefore suffices to estimate the second summand on the right-hand side of REF . If MATH is large enough depending on MATH then MATH whenever MATH . We use REF and obtain for MATH and therefore MATH where MATH means that there is a constant MATH depending on MATH such that MATH . Furthermore, by NAME expansion, we get MATH . Remark that MATH . Combining these observations, we get MATH for large enough MATH . This is much better than required, and therefore proves the proposition. |
math/0010291 | We can take MATH . If MATH then MATH and there is nothing to prove. We write MATH for the law of the random walk MATH conditioned on MATH . (For simplicity, we neglect trivial parity problems.) Let MATH be the number of points visited by MATH on the torus (assuming MATH for simplicity to be even). Then it suffices to prove for MATH . The left hand side of this equals MATH because for MATH we have MATH and for all MATH . We can replace MATH by MATH the number of points visited by a random walk of length MATH on MATH . (We replace MATH by MATH just for notational convenience). MATH by the NAME property. We write MATH for MATH. MATH . MATH . Implementing this into REF , this gives MATH . As MATH this proves the claim. |
math/0010291 | We use the same notations as in the proof of REF : MATH denotes the law of the random walk of length MATH (on the torus), conditioned to start in MATH and to end in MATH . If MATH and MATH then MATH . MATH for MATH . Let MATH which for the region of summation is in MATH, and MATH . Then MATH . Therefore, we get MATH . |
math/0010291 | We rescale the random walk by defining MATH . This random walk depends on MATH through MATH . It takes values in MATH which we regard as a (discrete) subset of the continuous torus MATH with lattice spacing MATH . Remember the setting MATH. The transition probabilities of the MATH-chain are given by MATH . Here MATH is the MATH-th matrix power. By the local central limit theorem (and our aperiodicity assumption) we have that for any MATH there exists MATH such that for MATH and any MATH and MATH . We denote by MATH the set of unions of square MATH with total area at most MATH . In order to prove the lemma, it suffices to prove that for any MATH for small enough MATH uniformly in MATH and MATH . We estimate the above probability in a standard way. For any MATH we have MATH . In order to estimate the right hand side, we use REF . We split the summation on MATH alternatively in intervals of length MATH and MATH the former being called ``short" intervals, the others ``long". We begin with a short interval. Remark that the contribution of all short intervals to the exponent in the expectation on the right-hand side of REF is at most MATH . Therefore, we can leave this part out, replacing the first factor on the right-hand side of REF by MATH . If we choose MATH we have by REF MATH where MATH is the expectation with respect to an uniform starting distribution. We therefore get MATH where MATH is a Brownian motion on MATH with covariance matrix MATH . For MATH we have MATH and we therefore get MATH . We therefore get MATH . Choosing MATH appropriately, this proves the claim . |
math/0010300 | By a result of NAME, see CITE, MATH is a symplectic manifold, and the fibers are symplectic submanifolds. As has been observed before CITE, the assumption of relative minimality here implies that MATH is minimal and not ruled, because any pseudo-holomorphic sphere in MATH would have to be contained in a fiber because MATH. Thus NAME 's extension CITE of NAME 's results CITE implies MATH, which we can write as MATH . Every reducible singular fiber contains a curve of negative selfintersection, and all of these are linearly independent in homology and independent of the class of a smooth fiber, which has selfintersection zero. Therefore MATH. Substituting this in REF and using MATH, we obtain MATH . As REF is trivial for MATH, we may assume MATH, and therefore MATH. The NAME characteristic of MATH is MATH . We estimate the signature of MATH using NAME additivity by decomposing the fibration into two pieces. Let MATH be an embedded MATH-disk containing all the critical values of MATH. If MATH, then a result of CITE gives MATH . Let MATH denote the restriction of MATH to MATH. As MATH can be decomposed into MATH pairs of pants, and the signature of MATH over each of them is given by the NAME cocycle CITE and therefore bounded by MATH, we conclude MATH . Combining REF, we obtain MATH . Passing to finite covers of MATH and applying REF to the pulled-back fibrations, we finally have MATH . As MATH, a result of CITE ensures that the canonical class MATH of MATH is represented by a symplectically embedded surface MATH. It may be disconnected, but because MATH is minimal, MATH has no spherical component. In the argument below we will tacitly assume that it is connected. In the general case the same argument works by summing over the components. For the genus of MATH we have the adjunction formula MATH. Using REF we obtain: MATH . The fibration MATH induces a smooth map MATH of degree MATH equal to the algebraic intersection number of MATH with a fiber. This is calculated from the adjunction formula applied to a smooth fiber MATH: MATH . Now NAME 's inequality MATH gives: MATH which together with REF completes the proof of REF. |
math/0010300 | We consider a NAME fibration over the MATH-disk MATH with precisely MATH singular fibers, such that with respect to a basepoint on the boundary of MATH the vanishing cycles of all the singular fibers can be identified with MATH. Then the monodromy of the fibration around the boundary of MATH is MATH. If this can be expressed as a product of MATH commutators, then we can find a smooth surface bundle with fiber MATH over a surface of genus MATH with one boundary component and the same restriction to the boundary. Let MATH be the NAME fibration over the closed surface MATH of genus MATH obtained by gluing together the two fibrations along their common boundary. By construction, no fiber contains a sphere, so MATH is relatively minimal. Thus we can apply REF to conclude MATH as MATH in this case. |
math/0010300 | If MATH, then we have a surjective homomorphism MATH given by collapsing a boundary component to a point. We also have surjective forgetful homomorphisms MATH so it is enough to prove the claim in the case MATH. But this case is immediate from REF . |
math/0010300 | From the presentation of MATH due to CITE, it follows that the NAME of MATH is a finite cyclic group of order MATH if MATH is odd, and of order MATH if MATH is even. Let MATH be a nontrivial separating simple closed curve on the surface of genus MATH which is invariant under the hyperelliptic involution. Then MATH, and using REF we obtain: MATH . Thus the claim follows from REF . |
math/0010300 | For MATH this follows from the result of Mess CITE that MATH is a free group (on infinitely many generators). CITE proved that for MATH the commutator subgroup of the NAME group is: MATH where MATH is the subgroup generated by all squares of elements of MATH, and MATH is the subgroup generated by the NAME twists along separating simple closed curves. Thus, if MATH is a separating NAME twist, then MATH. Using REF, we have MATH and so the claim follows from REF . |
math/0010303 | Since MATH is a two-generated simple group, MATH is an epimorphic image of a free group of rank REF. So it is enough to prove that MATH and MATH. The first equality follows since there is no nontrivial homomorphism of MATH into MATH. On the other hand, there is a canonical embedding MATH, so MATH. |
math/0010304 | Let MATH. We first show that MATH. Choose any MATH. By requirement REF, clearly MATH. So, MATH for some MATH. By REF , we can find two continuously decreasing sequences MATH and MATH in MATH such that MATH, MATH, and whenever MATH and MATH, then MATH and MATH. Standard back-and-forth arguments involving closed unbounded subsets of uncountable cardinals show that MATH is a closed unbounded subset of MATH. Since MATH is stationary, there is MATH with MATH. So, MATH and MATH. Thus MATH as claimed. By a similar argument, it now follows that MATH for each MATH. The final statement is then immediate. |
math/0010304 | Clearly, MATH is doubly homogeneous and dense in MATH. We identify MATH. By REF , each MATH and each MATH REF is invariant under MATH. By homogeneity we get MATH for each MATH. Hence each MATH REF is an orbit of MATH in MATH. By REF , any outer automorphism MATH of MATH determines an automorphism MATH of MATH which permutes the orbits of MATH in MATH and hence by requirement REF, permutes the orbits MATH REF among themselves. By REF , this permutation MATH determines an element MATH of MATH. Conversely, by homogeneity of MATH , any element MATH of MATH can be realized in this way by an automorphism MATH of MATH permuting the sets MATH (MATH), and hence MATH is realized by an outer automorphism MATH of MATH. This correspondence constitutes the required isomorphism. |
math/0010304 | Choose MATH-sequences MATH and MATH such that MATH and MATH for each MATH and MATH, MATH, MATH, MATH. For each MATH, there is an isomorphism MATH permuting the colours as prescribed by MATH. Patching the MATH's together, we obtain the required isomorphism MATH. |
math/0010304 | CASE: Observe that MATH in MATH and MATH in MATH. Now continue by induction through MATH and construction of MATH, MATH. CASE: Put MATH and MATH REF to obtain the result. |
math/0010304 | Choose MATH and MATH, MATH with MATH and MATH. By assumption, there are nice embeddings MATH and MATH with MATH, MATH, MATH, MATH. First, from MATH we obtain two continuously decreasing sequences MATH and MATH such that: CASE: MATH and MATH; CASE: MATH for each MATH; CASE: MATH for each MATH. By REF , for each MATH there is an isomorphism MATH permuting the colours as prescribed by MATH. Patching these isomorphisms MATH REF together, we obtain an isomorphism MATH which permutes the colours as prescribed by MATH. Secondly, let MATH. Then MATH and MATH, and we may assume that MATH and MATH (otherwise consider appropriate upper segments of MATH and MATH subsequently). Observing REF , we see that MATH maps MATH onto MATH permuting the colours in these subsets of MATH as prescribed by MATH. We have MATH and MATH where the two unions are taken over all gaps MATH in MATH. Moreover, for each gap MATH in MATH, we have MATH, and by REF there is an isomorphism MATH permuting the colours as prescribed by MATH. Patching all these isomorphism MATH together with MATH above, we obtain an isomorphism MATH which permutes the colours as prescribed by MATH. Again by REF , there is also such an isomorphism MATH. Now MATH maps MATH isomorphically onto MATH and permutes the colours as prescribed by MATH. |
math/0010304 | We will first consider the case MATH (although the case MATH will be quite similar). We first construct MATH which is MATH-homogeneous for MATH-intervals. Let MATH be the coloured linear ordering defined above. Let MATH be the set of all gaps of MATH. For each gap MATH of MATH, choose a copy MATH of the good MATH-set given by REF . Define MATH . Next we define a linear order on MATH in the unique way so that it extends the given orders of MATH and each MATH and so that MATH for each MATH. By construction of MATH, whenever MATH with MATH, there is MATH with MATH and so there is MATH with MATH. Therefore we regard MATH as a subset of MATH, and we put MATH . Clearly, MATH satisfies REF - REF . By REF , observe that MATH for each gap MATH. Hence, if MATH with MATH, there is a continuously increasing sequence MATH such that MATH and MATH for each MATH. However, note that since MATH satisfies REF , the above is impossible for points MATH with MATH. Since MATH, it follows that MATH also satisfies REF and hence MATH. Moreover, the set MATH with MATH REF is a final (initial) segment of MATH, respectively, and has countable coterminality, so MATH. We wish to obtain the required structure MATH as the union of a tower (indexed by MATH) of structures MATH. For this, we employ the construction in the proof of REF (REF - REF ) of CITE; here our description will be less formal. We let MATH be the set of all quintuples MATH such that MATH, MATH and MATH, and enumerate MATH by a suitable subset MATH. If we deal during the construction with such a quintuple MATH for the first time at step MATH, we employ Basic Construction A to obtain MATH and an isomorphism from MATH onto MATH permuting the colours as prescribed by MATH. Later on, we employ Basic Construction B to extend isomorphisms constructed at previous stages. For limit ordinals MATH, we just put MATH, in the natural way. Since we perform the extension of a constructed isomorphism MATH many times, for each MATH and MATH we finally obtain an isomorphism MATH permuting the colours as prescribed by MATH. This shows that MATH is MATH-homogeneous for MATH-intervals. Also since Constructions A and B are carried out only at points MATH, that is, ``inside" MATH, the set MATH remains unbounded above and below in each MATH REF and in MATH. So, MATH has countable coterminality. We have to ensure that all structures MATH REF and MATH belong to MATH. To see that each element MATH has countable coterminality, observe that this is true in MATH, and that if MATH, then in the construction of MATH new elements only get inserted at points of MATH; so, in particular, each MATH has the same character in MATH as in MATH. Moreover, if MATH and MATH is a nice embedding with MATH, then MATH should remain nice for each MATH. This is the case, if all insertion processes of Basic Construction A and B are only carried out at points MATH. Indeed we have MATH, and we declare all points of MATH ``forbidden points" in the terminology of CITE. This means that we are never allowed, later on, to perform insertion processes at the cuts MATH. This ensures that MATH for each MATH, and MATH is a nice embedding. Furthermore, during the Basic Constructions A and B, to construct MATH we insert copies of the whole intervals of MATH only into particular points MATH with MATH which are not ``forbidden" in MATH. Then we declare in the copy MATH of MATH all elements MATH which correspond to a forbidden point MATH is MATH as forbidden points in MATH (compare CITE, pp. REF). This ensures that our isomorphisms also preserve forbidden points, and if MATH is a nice embedding, then MATH remains nice for each MATH, and so is MATH. Also, MATH is MATH-dense in MATH. Now assume that MATH and MATH with MATH. Suppose that in the construction of MATH the chain MATH gets inserted into MATH at MATH. Then by REF of Basic Construction B, we may assume that MATH has countable coterminality and the elements MATH, MATH of MATH become forbidden points. This ensures that MATH, MATH retain their countable character in each MATH REF and in MATH. We have to show that the forbidden points do not prevent our constructions. By induction, we may assume that if MATH, then there are only MATH many points MATH into which elements have been inserted in order to construct MATH. Hence at stage MATH, for any MATH in MATH there are still MATH many cuts MATH with MATH into which no element got inserted and which are not forbidden in MATH, and these can be used when we deal again with a quintuple in MATH. Moreover, then at stage MATH again only MATH many new forbidden points are created, keeping the above induction hypothesis. Since MATH is regular, also for limit ordinals MATH for any MATH in MATH we have MATH for some MATH, and it follows that the set MATH has size MATH. Next we consider the characters of elements of MATH. As noted before, each inserted good MATH-set MATH satisfies MATH. Also, for MATH we have (inside MATH) MATH. So, MATH. We want to ensure that our construction neither destroys these characters nor adds new ones. The first part is clear, since we insert sets with countable coterminality only at points MATH with MATH and add a forbidden point to the left and to the right of the inserted set, thereby preventing further insertions of sets at these points. To ensure the second part, we may proceed as in CITE. That is, suppose MATH is a countable sequence of ordinals MATH and MATH with MATH for each MATH. Let MATH. Then in MATH we have MATH and MATH in MATH. In order to prevent sets getting inserted at later stages at MATH, we declare MATH (and any point arising in this way) as a forbidden point of MATH. This ensures that also MATH in MATH. Therefore MATH. In particular, since MATH, we have MATH but MATH, so MATH is not anti-isomorphic to itself. By similar remarks as above about MATH, we also obtain that each MATH REF and MATH satisfy REF . Now MATH is MATH-homogeneous for MATH-intervals and hence, by REF , also MATH-homogeneous for MATH-intervals. Since MATH, a patching argument similar to the one used for REF shows that MATH is doubly homogeneous in MATH-colours. It remains to consider the case MATH and so MATH. The above construction would also work to give MATH with REF and MATH, but now this does not imply that MATH is not anti-isomorphic to itself. To remedy this, we can proceed as follows. Let MATH be the chain by inserting in the ordinal MATH between each element and its successor a copy of MATH, the converse of MATH, and also adding a copy of MATH to the right of MATH. Then MATH, and there is a unique element MATH with MATH. Now put MATH to the right of MATH and proceed, with MATH replaced by MATH, as above, inserting copies of good MATH-sets into each gap of the chain MATH to obtain MATH. The construction above now produces a chain MATH with MATH. In particular, MATH also satisfies MATH in MATH. By construction, there is a continuously increasing sequence MATH such that MATH and MATH for each MATH. There is also a continuously decreasing sequence MATH such that MATH and MATH for each MATH. But there is no element MATH with MATH and these two asymmetric ascending respectively descending approximation properties interchanged, so MATH is not anti-isomorphic to itself. Since MATH keeps these properties in MATH, but the construction produces no MATH with the interchanged properties, it follows that MATH is not anti-isomorphic to itself. |
math/0010305 | CASE: Should be clear REF So we are given the sequence MATH. We choose the increasing sequence MATH of natural numbers by letting MATH, MATH be the first MATH such that CASE: for every MATH and MATH we have MATH and MATH, and MATH], and MATH mentions only MATH with MATH. Note that MATH is well defined as the sequence MATH converges to MATH, so for each MATH for every large enough MATH we have MATH. We shall prove that MATH is as required in REF. So let a sequence MATH of group words obeying MATH be given (see MATH above). For each MATH we define the sequence MATH of members of MATH as follows. For MATH we let MATH be MATH and now we define MATH by downward induction on MATH letting CASE: MATH. Now shall work on proving CASE: for each MATH the sequence MATH is eventually constant. Why does MATH hold? By the definition of obeying we can find MATH such that CASE: MATH, MATH, MATH, and MATH is trivial for MATH, and MATH . For MATH let MATH . Now let MATH; we claim that: CASE: if MATH and MATH, then MATH restricted to the interval MATH is the identity. [Why? If MATH, this holds by the choice of the MATH as the identity everywhere. Now we prove MATH by downward induction on MATH (but of course MATH). But by the definition of composition of permutations it suffices to show CASE: every permutation mentioned in the word MATH maps every MATH to itself. Let us check this criterion. The MATH for MATH satisfies this as the indexes are MATH and MATH; now apply the choice of MATH. The MATH satisfy this by the induction hypothesis on MATH. So the demands in MATH holds, hence we complete the downward induction on MATH. So MATH holds.] CASE: If MATH and MATH, then MATH is the identity on the interval MATH. [Why? We prove this by downward induction; for MATH this holds by MATH, if it holds for MATH, recall that MATH is trivial, so MATH, so this follows.] CASE: For every MATH we have: for every MATH, the functions MATH agree on the interval MATH, and also MATH agree on this interval. [Why? For MATH by MATH; for MATH by MATH.] CASE: For any MATH and MATH such that MATH we have: MATH . CASE: MATH is MATH. [Why? This holds by MATH.] We prove this by downward induction on MATH (for all MATH and MATH as there). CASE: Proving for MATH, assuming we have it for all relevant MATH (and MATH of course). Let MATH and we concentrate on proving MATH as the proof of MATH is the same. So MATH . So let us write this group expression as the product MATH, where each MATH is one of MATH, or is an inverse of one of them. For MATH let MATH, so MATH is the identity permutation for MATH and is MATH for MATH. Hence it suffices to prove the following CASE: if MATH and MATH, then MATH. [Why does MATH hold? We do it by induction on MATH; now for MATH the permutation is the identity so trivial. For MATH just note that because each MATH can map any MATH only to numbers MATH and that MATH has been proved for MATH when MATH is appropriate.] So we have proved MATH, and hence MATH and thus also MATH. Lastly CASE: for each MATH and MATH the sequence MATH is eventually constant. Why? Same as the proof of MATH. Together, we can defined for any MATH the natural number MATH as the eventual value of MATH. So MATH is a well defined function from the natural numbers to themselves (by MATH), in fact it is one-to-one (as each MATH is) and is onto (by MATH), so it is a permutation of MATH. Clearly the sequence MATH converges to MATH as a permutation, the metric is actually defined on the group of permutations of the family of members of MATH, and MATH is a closed subgroup; so MATH actually is an automorphism of MATH. Similarly the required equations MATH hold. |
math/0010305 | So let MATH be a basis of MATH and as MATH is a separable metric space there is a sequence MATH of (pairwise distinct) members of MATH with MATH. Let MATH, so MATH converges to MATH and MATH. Assume MATH is as in the conclusion of REF , and we shall eventually get a contradiction. Let MATH be a subgroup of MATH generated by some countable MATH and including MATH and let MATH list the members of MATH. Now CASE: MATH satisfies the condition also in MATH. [Why? As there is a projection from MATH onto MATH and MATH.] For each MATH let MATH, where MATH, so this is a sequence of words as mentioned in REF . CASE: The set of MATH which obey MATH is co-meagre. [Why? Easy; for each MATH the set of MATH's which fail the demand for MATH is nowhere dense (and closed), hence the set of those failing it is the union of countably many nowhere dense sets, hence is meagre.] CASE: For each MATH the family of MATH such that there is solution MATH for MATH in MATH satisfying MATH is nowhere dense. [Why? Given a finite sequence MATH of natural numbers note that for any sequence MATH of which MATH is an initial segment and solution MATH satisfying MATH, we can show by induction on MATH that MATH is uniquely determined, call it MATH. Now, if MATH, which is a member of MATH, is not MATH, then for some MATH it has no MATH-th root and we let MATH and we are done. If not, letting MATH, also MATH is well defined and equal to MATH, hence is not MATH. Hence for some MATH has no MATH-th root, so MATH is as required.] Now we can finish the proof of REF : just by MATH . NAME Theorem, for some MATH, the sequence MATH of group words obeying MATH, there is no solution in MATH, hence no solution in MATH . |
math/0010305 | Follows by REF . |
math/0010306 | Let MATH be a model of MATH with the strict order property. So, there are MATH-definable partial order MATH on MATH-tuples in MATH for some MATH and a sequence MATH of MATH-tuples in MATH such that MATH for MATH. By NAME 's Theorem, we can assume that MATH is a MATH-indiscernible sequence in MATH. Also, we can assume that there is a MATH-automorphism MATH of MATH such that MATH. So, MATH is a model of MATH. Now by way of contradiction, suppose that MATH has a model companion, say MATH. Extend MATH to a model MATH of MATH. MATH is a MATH-elementary extension of MATH since MATH is model complete. We can assume that MATH is sufficiently saturated. In the rest of the proof, we work in MATH. Consider the partial type MATH and let MATH. Claim. In MATH, CASE: MATH, and CASE: if MATH is a finite subset of MATH then MATH. If this claim holds, then it contradicts the saturation of MATH. We first show REF . Let MATH be such that MATH. Then MATH satisfies MATH. Suppose MATH satisfies MATH. Let MATH be such that MATH. By MATH, we have MATH. By MATH, we have MATH . By transitivity, we get MATH, which is a contradiction. Now we turn to a proof of REF . Suppose MATH. Let MATH be such that MATH, MATH, and MATH is a MATH-elementary substructure of MATH. For each MATH, let MATH be the set of MATH-formulas MATH satisfied in MATH by some tuple MATH such that MATH and MATH. Here, MATH-formulas are the formulas in MATH with parameters in MATH. Note that if MATH and MATH, then MATH, and by compactness, if MATH then there is MATH such that MATH and MATH. Since MATH is sufficiently saturated, there is MATH such that if MATH and MATH then MATH. We can also assume that MATH. Since the sets MATH and MATH are invariant under MATH, MATH is also invariant under MATH, which means, for any MATH-formula MATH and a tuple MATH, MATH if and only if MATH. Now choose MATH such that MATH and consider MATH. Then MATH. Let MATH be the set of REF such that MATH where MATH is a formula in MATH and MATH. Since MATH is invariant under MATH, we have MATH. By the choice of MATH, MATH and thus MATH. By the definition of MATH and by compactness, there is MATH such that MATH and MATH realizes MATH. Since MATH is a complete MATH-type over MATH, there are MATH-substructure MATH of MATH and a MATH-automorphism MATH of MATH such that MATH, MATH and MATH. Now we have, MATH . Since MATH is a model of MATH, it is an existentially closed model of MATH. Note that the partial order MATH is definable by an existential MATH-formula modulo MATH. So, the formula MATH has a solution in MATH. Hence, we have MATH. This proves REF and we are done. |
math/0010309 | Let MATH. Then MATH, where MATH and MATH denotes NAME multiplication by the one-form MATH. An easy computation establishes that MATH and that MATH. Then MATH, where MATH denotes the projective MATH-action. |
math/0010309 | We need to show that MATH defined in REF is independent of MATH. By REF , it suffices to show that MATH is a norm continuous family of projections in the MATH-algebra MATH. By NAME 's principle, one has the following identity whenever MATH, MATH . It follows that there is a constant MATH independent of MATH such that MATH . Using NAME 's principle repeatedly, one sees that the family of projections MATH is continuous in the norm topology. |
math/0010309 | By hypothesis, the twisted NAME map is an isomorphism. Therefore to compute the range of the trace map on MATH, it suffices to compute the range of the trace map on elements of the form MATH . Here MATH is an even parity MATH-cycle over MATH. By the twisted analogue of the MATH index theorem of CITE and CITE for elliptic operators on a covering space that are invariant under the projective action of the fundamental group defined by MATH, and which was stated and used by CITE (compare Appendix), one has MATH as desired. |
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