paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0107175 | If MATH and MATH are connections compatible with MATH, then the difference MATH is a REF-form on MATH with values in MATH, the center of the NAME algebra of MATH. By the property of connection, REF-form on MATH descends to a REF-form on MATH. Conversely, if MATH is a REF-form on MATH with values in MATH, then MATH is a connection on MATH. Clearly this connection is compatible with MATH. |
math/0107175 | For a splitting MATH we put MATH. If splittings MATH and MATH are given, then we have a map MATH defined by MATH. It is easy to see that MATH is trivial on the subspace MATH and satisfies MATH. Hence MATH induces a map MATH and gives rise to a section of MATH. Conversely, a section of the vector bundle produces a new splitting by the translation. |
math/0107175 | Since MATH and a splitting is the identity on the subbundle MATH, we have the result. |
math/0107175 | First note that the values of MATH and MATH are contained in the center of MATH and of MATH respectively as a consequence of the cocycle condition MATH where MATH is the NAME form of MATH. For REF , the following equalities obtained by direct computation show the cocycle condition of MATH: MATH . We can verify that resulting MATH-ech cohomology class is independent of the choice of local section MATH and of the choice of MATH, MATH. The standard argument shows that the definition of the class is independent of the choice of MATH. For REF , if we have a lifting and a compatible connection whose scalar curvature vanishes, then we have MATH and MATH using the transition functions and the connection forms of the lifting. Because MATH is the local description of the scalar curvature, we also have MATH. Hence the MATH-ech cocycle is trivial. Conversely, if the class is the coboundary of a MATH-ech cochain MATH, then we put MATH . These data give a lifting MATH and a compatible connection MATH. The scalar curvature is calculated as MATH on each MATH. |
math/0107175 | We can directly prove these formulas by using the fact that the adjoint action of MATH on the center MATH is trivial. |
math/0107175 | We write MATH for a splitting MATH. For MATH we have MATH . Hence we get a reduced splitting with respect to MATH. Conversely, let MATH be a reduced splitting. We can uniquely decompose any element MATH as MATH, where MATH and MATH is contained in the center. If we put MATH, then MATH is a splitting of MATH. |
math/0107175 | Note that MATH, where MATH is the transition function of MATH. If we take the section as MATH, then we have MATH. |
math/0107175 | Since we have MATH, REF-form MATH takes values in MATH. The basic properties of the NAME form show that REF-form is a connection on the MATH-bundle. We can easily compute the curvature of this connection using the NAME equation. |
math/0107175 | REF are directly proved. REF is proved as follows. A tangent vector at MATH is a pair MATH such that MATH. The push-forward of the vector under MATH can be expressed as MATH, where MATH. Then we have an equality of tangent vectors MATH, where MATH denotes the fundamental vector field generated by MATH. This establishes REF . |
math/0107175 | This theorem is the consequence of the following two formulas: MATH where MATH is the projection induced by the projection MATH and we omit the subscription of MATH and MATH. To prove the first formula we put a tangent vector into both hand sides and compute the values. In the computation we use REF . Note also the formula MATH, where MATH is the group product and MATH for MATH. The second formula is directly checked by REF . |
math/0107175 | Using MATH and REF , we obtain MATH . Then the formula follows from MATH and REF . |
math/0107175 | A direct computation proves REF . For REF , using MATH and the invariance of the bracket under the adjoint action, we can rewrite the left hand side of REF as MATH. Now the right hand side of REF follows from MATH. |
math/0107175 | This is proved by the following formulas MATH . The first formula follows from REF . To prove the second formula we put tangent vectors into both hand sides and compute the values. In the computation we use REF and the property that the value of a curvature is zero for any fundamental vector field. |
math/0107175 | Easily we have MATH. By REF the reduced splittings MATH and MATH induced from MATH satisfy REF . Thus we obtain MATH. These formulas prove the proposition. |
math/0107175 | We show that the MATH-ech cocycles of the NAME cohomology classes coincide exactly under an appropriate choice. We fix sections MATH of MATH with respect to a good cover MATH. First, we take MATH to define MATH. Recall the proof of REF . We can take the section of MATH as MATH. As a consequence MATH holds. Secondly, to define MATH we can take MATH. Using MATH and MATH we obtain MATH. Lastly, since MATH, we have MATH. |
math/0107175 | If MATH is a compatible with MATH, then REF-form MATH takes values in MATH. The basic properties of connection show that MATH is indeed a connection. It is straight to see MATH. Note that we can regard MATH as a REF-form on MATH. Using this formula and the reduced splitting MATH, we have MATH for MATH. Differentiating REF , we have REF . |
math/0107175 | The lifting bundle gerbe of the pull-back is by REF, where MATH is defined in the same way as MATH. If MATH is the bundle map covering MATH, then we have MATH. Using this relation, we obtain a natural isomorphism MATH of bundle gerbes. Because the pull-back of the connection MATH is MATH, we can see that the isomorphism maps the connection on MATH to that on MATH by REF . If the reduced splitting MATH is induced by the splitting MATH, then the reduced splitting induced by the pull-back MATH is MATH. So we obtain MATH by REF . |
math/0107175 | Using REF , we have MATH . Because MATH is simple, we have MATH. This implies that MATH, and the proof is completed. |
math/0107175 | It is clear that MATH is linear if MATH is fixed. So we show that MATH satisfies REF . We have the following relation between MATH-valued REF-forms on MATH. MATH . Since MATH is invariant under the adjoin action of MATH, the proposition is proved. |
math/0107175 | We can check MATH by the NAME formula of the NAME term CITE. Thus the equivalence relation is well-defined. Since the action of MATH on MATH is free, the fiber of MATH is the orbit of the right action. Therefore we obtain a principal MATH-bundle. It is easy to see that MATH is an equivariant map, and we obtain a string structure. |
math/0107175 | Note that there is the following commutative diagram. MATH . Here the lows are string structures and the columns are principal MATH-bundles. Thus, we can investigate the data of the lower string structure by that of the upper string structure. It is obvious that the connection MATH on MATH is compatible with the connection MATH on MATH. By computation we can show that these connections descend. In particular, the latter connection descends to MATH. Hence we obtain the compatible connection MATH on MATH. For the computation of the scalar curvature, we also use the commutative diagram. By REF and the formula MATH derived from NAME 's formula, we obtain the result. |
math/0107175 | If we use the coordinate MATH, then the relation REF is equivalent to the relation REF . |
math/0107178 | The three sublists MATH, MATH and MATH are the same for every MATH. Therefore, MATH is the same as well. |
math/0107178 | Since equivalent wiring diagrams have the same fundamental groups, we may reorder MATH so that all the intersection points on the first wire are consecutive (as explain in the definition of MATH). Then, by definition, the wiring diagram MATH is generated from MATH by breaking the first wire of MATH at the end points of its two rays. We can smoothly rotate them until we achieve the wiring diagram MATH. Keeping the same angle of both rays along the rotation, the diagram is always topologically the same (even in MATH), and the result follows. |
math/0107178 | We will in fact give an explicit isomorphism between the fundamental groups which were described in REF. Let MATH be a list of NAME pairs, and let MATH and MATH be the wiring diagrams associated to MATH and MATH, respectively. Fix MATH such that MATH. Denote by MATH the geometric generators of MATH, and by MATH the generators of MATH. We define MATH by MATH . This definition is motivated by REF , which corresponds to an appropriate NAME move CITE. In order to show that MATH is well defined and is an isomorphism, we need to show that it carries the set of relations of MATH to the relations of MATH. Recall that by REF . We will first show that the relations in MATH associated to the points MATH are equivalent to the relations in MATH associated to the these points. Let MATH. According to the algorithm in REF, the MATH-th skeleton of MATH is obtained by applying on the initial skeleton MATH the halftwists corresponding to MATH, then to MATH, and so on, down to MATH. At the same time, the MATH-th skeleton of MATH is obtained from the same initial skeleton, by applying MATH down to MATH, without applying MATH. Therefore, in order to get the same skeleton in both cases, one has to act on the MATH-th skeleton of w by the inverse to the halftwist corresponds to MATH. The action of the NAME move (and the map MATH) corresponds to this inverse action: it turns the local region in the disk clockwise (in the direction opposite to that of the halftwist's action). Therefore one gets the same skeletons, and hence equivalent relations in the presentation of the fundamental groups. It remains to show that the relation induced by the skeleton corresponding to the first point, MATH, in w, is equivalent to the relation induced by the skeleton corresponds to the last point, MATH, of MATH. According to the algorithm, the skeleton corresponding to MATH for w is simply the straight segment between MATH and MATH. The induced relation is (see CITE): MATH . For MATH, the skeleton is obtained by applying halftwists corresponding to the rest of the pairs, on the initial skeleton MATH. In the braid group, the action of all the MATH halftwists of a list of NAME pairs (with the unique intersection property) is equal to the action of the general halftwist MATH, so the skeleton in this case can be obtained as follows: begin with the initial skeleton MATH, apply MATH and then apply the inverse of the halftwist corresponding to MATH. Applying MATH on MATH yields the skeleton corresponding to the segment MATH. Then, applying the halftwist corresponds to MATH will give the same result. Therefore we get the same skeleton, and hence equivalent sets of relations. This proves that the two affine groups are isomorphic. Finally, compute that MATH, so the relation defining the projective group of MATH transfers to the corresponding relation for MATH. |
math/0107178 | Let MATH and MATH be two lists of NAME pairs, such that MATH. We have to show that MATH. It is obviously enough to assume the only difference between MATH and MATH is one replacement of MATH by MATH, and by applying MATH we may assume that this replacement happens in the first MATH pairs of each list. Then we have MATH and MATH . We will compute the presentations of MATH and MATH using the algorithm given in REF, where MATH are the generators for the first group, and MATH for the second. We will show that the map MATH is an isomorphism between the two groups. We first show that the relations associated to the intersection points MATH for MATH are the same for both diagrams. According to the algorithm in REF, the MATH-th skeleton of MATH is obtained by applying on the initial skeleton, associated to the pair MATH, the composition of the following halftwists: first, those associated to the pairs MATH, and then the MATH halftwists associated to the pairs in MATH. At the same time, the MATH-th skeleton of MATH is obtained by applying on the same initial skeleton the same MATH halftwists, and then the MATH halftwists associated to the pairs in MATH. Since in the braid group the composition of the halftwists associated to the pairs in MATH is equal to the composition of the halftwists associated to the pairs in MATH (they both equal to the general halftwist MATH), we get the same final MATH-th skeleton for MATH and for MATH. In particular, we get the same induced relations in both cases. It remains to show that the relations induced by the first MATH intersection points of MATH are equivalent to the relations induced by the first MATH intersection points of MATH. We will compute the relations in both cases. The relations induced on the generators MATH by the points listed in MATH are MATH . There are three types of skeletons to consider, corresponding to point before the central point, points after the central point, and the central point itself. CASE: First let MATH, so that the point is MATH. We need to apply the halftwists corresponding to the pairs MATH, MATH, on the initial skeleton which corresponds to MATH. The resulting skeleton is given in REF . This skeleton induces the following relation: MATH . Induction on MATH now proves REF for MATH. CASE: We compute the relations for the other points, and then get back to the central point. Let MATH, and consider the point MATH which comes after the central point. To obtain the skeleton, we begin with the initial skeleton, given in REF . We apply the halftwists corresponding to the pairs MATH, MATH, obtaining the skeleton given in REF ; then the halftwist corresponding to the central point MATH, obtaining the skeleton in REF , and finally the halftwists corresponding to the pairs MATH, MATH, obtaining the final skeleton of REF . The relation induced by this skeleton is MATH . Again by induction on MATH, we obtain the relations REF for MATH. Since MATH trivially holds for MATH, this set of relations is proved. CASE: The central skeleton is obtained by applying the halftwists correspond to the pairs MATH, MATH on the initial skeleton corresponding to MATH, which is given in REF . The resulting skeleton is given in REF . The relation induced by this skeleton is MATH which is REF since by REF MATH commutes with all the generators appearing in this relation. Next, we compute the corresponding relations induced by the skeletons correspond to the first MATH intersection points of MATH. The relations induced on the generators MATH by the points listed in MATH are MATH . Again we treat the three classes of points separately. CASE: Let MATH. The MATH-th skeleton of MATH is obtained by applying the halftwists corresponding to the pairs MATH, MATH on the initial skeleton induced by the pair MATH. The resulting skeleton is given in REF . This skeleton induces the relation MATH, and these are the cases MATH of REF . Let MATH. For the MATH-th skeleton we begin with the initial skeleton corresponding to MATH (see REF ), apply the halftwists correspond to the pairs MATH, MATH to get the skeleton of REF , then apply the central halftwist, corresponding to MATH, and get the skeleton of REF , and finally apply the halftwists correspond to the pairs MATH, MATH, and get the final skeleton, presented in REF . This skeleton induces the relation MATH and ranging over the possible values of MATH we obtain the relation of REF for MATH. CASE: The central skeleton is obtained by applying the halftwists correspond to the pairs MATH, MATH to the initial skeleton induced by the pair MATH, which is given in REF . The resulting skeleton is given in REF . The corresponding induced relations are MATH . Using the relations REF we can simplify this to obtain relation REF . Comparing the two propositions we discover the same set of relations, so the affine groups are shown to be isomorphic. Since our isomorphism is MATH, the projective relation MATH goes to MATH, so we have also proved that MATH. |
math/0107178 | MATH inverts the order of the pairs, so that MATH is the identity even on MATH. By the geometric definition, each application of MATH consists of inverting the direction of one pseudoline. MATH is a wiring diagram which is equivalent to MATH rotated by MATH. In particular, MATH is the identity on MATH. It remains to check that MATH. Let MATH be a list of NAME pairs. As in the definition of MATH, we can decompose some MATH into three disjoint consecutive sublists: MATH, where we write: MATH, MATH, and MATH. Now, compute: MATH . |
math/0107178 | Let MATH. By definition, MATH so that MATH. As was shown in the last lemma, the order of MATH is REF. Therefore, in order to show that MATH is dihedral of order MATH, it remains to verify that MATH. Indeed, MATH . |
math/0107179 | In the limit as MATH, the lifted system REF given by a regular triangulation MATH of MATH becomes the disjunction of facial subsystems of REF , one for each facet MATH of MATH. Thus the number of real solutions in the limit is a constructive lower bound for MATH. By REF , the number of odd cells in MATH is a lower bound on the number of real solutions to the facial subsystem given by the facets of MATH. |
math/0107179 | Since a cubic equation in the plane has REF coefficients, the space of cubics is identified with MATH. The condition for a plane cubic to contain a given point is linear in these coefficients. Given REF general points, these linear equations are independent and so there is a pencil REF of cubics containing REF general points in MATH. Two cubics MATH and MATH in this pencil meet transversally in REF points. Since curves in the pencil are given by MATH for MATH, any two curves in the pencil meet transversally in these REF points. Let MATH be MATH blown up at these same REF points. We have a map MATH where MATH is the cubic curve defined by the polynomial MATH. Consider the NAME characteristic MATH of MATH first over MATH and then over MATH. Blowing up a smooth point on a surface replaces it with a MATH and thus increases the NAME characteristic by REF. Since MATH, we see that MATH. The general fiber of MATH is a smooth plane cubic which is homeomorphic to MATH, and so has NAME characteristric REF. Thus only the singular fibers of MATH contribute to the NAME characteristic of MATH. Assume that REF points are in general position so there are only nodal cubics in the pencil. A nodal cubic has NAME characteristic REF. Thus there are REF singular fibers of MATH and hence REF singular cubics meeting REF general points in MATH. Consider now the NAME characteristic of MATH. Blowing up a smooth point on a real surface replaces the point by MATH, and hence decreases the NAME characterstic by REF. Since MATH, we have MATH. A nonsingular real cubic is homeomorphic to either one or two disjoint copies of MATH, and hence has NAME characteristic REF. Again the NAME characteristic of MATH is carried by its singular fibers. There are two nodal real cubics; either the node has two real branches or two complex conjugate branches so that the singular point is isolated. Call these curves real nodal and complex nodal, respectively. They are displayed in REF . The real nodal curve is homoemorphic to a REF and has NAME characteristic MATH, while the complex nodal curve is the union of a MATH with a point and so has NAME characteristic MATH. Among the singular fibers, we have MATH . Thus there are at least REF real nodal curves containing REF general points in MATH. The pencil of cubics containing REF complex nodal cubics of REF has REF real nodal cubics. Thus there are REF real rational cubics containing any REF of REF points in REF . |
math/0107179 | We induct on MATH to construct numbers MATH having the property that, for all MATH with MATH, MATH is transverse (over MATH) and each of its MATH points are real. The case MATH is trivial, as MATH implies that MATH and MATH is a single (real) point in MATH. Suppose we have constructed MATH with the above properties. Let MATH with MATH and consider MATH . Each summand in the second sum is transverse (over MATH) and consists of MATH real points. The intersection on the left will be transverse and consist of MATH real points only if no two summands in the second sum share a point in common. If two summands MATH with MATH share a point, then MATH is non-empty. By REF , MATH is contained in a union of NAME varieties of dimension less than MATH. By REF the condition for a fixed MATH-plane MATH to meet MATH non-trivially is a polynomial of degree at most MATH in MATH. Thus a fixed MATH-plane MATH lies in at most MATH distinct MATH-planes in the family MATH. Hence the intersection REF is empty, and so the summands in the second sum of REF are disjoint. This argument also shows that the points in the summand indexed by MATH lie in MATH. For MATH, let MATH be the set of MATH-planes MATH satisfying the polynomial REF . For MATH, we have MATH. Since MATH has no vanishing NAME coordinates, the constant term of that polynomial is MATH, and so by REF , MATH . Thus MATH meets MATH transversally (over MATH) in MATH real points. We see that there exists MATH such that if MATH, then MATH meets REF transversally (over MATH) in MATH real points. Since the intersection defining MATH is transverse along each MATH for MATH, we may assume that for MATH, MATH is transverse (over MATH) and consists of MATH real points. Let MATH. Then for any MATH with MATH, MATH is transverse (over MATH) and consists of MATH real points. |
math/0107183 | For MATH, the map MATH induces a map MATH which we claim is surjective. In fact, if MATH, then since MATH is strongly noncharacteristic, we can find MATH and MATH so that MATH. Taking the real part shows that MATH, whereupon taking the imaginary part shows that MATH as desired. The kernel of the map MATH is MATH, so there is an induced isomorphism MATH. Hence MATH is independent of MATH, from which the lemma follows. |
math/0107183 | Let MATH be the basis for MATH chosen above. For any solutions MATH, the fiber MATH is the span of MATH. Since also MATH, by replacing the MATH's by a linear combination we can assume that MATH for MATH and MATH for MATH. Define an initial set of coordinates by setting MATH and MATH for MATH, MATH for MATH, and by choosing functions MATH for MATH such that MATH on MATH and MATH. Since MATH is real, it follows that MATH. Recalling that MATH, it follows then that in the coordinates MATH we have MATH, MATH, and MATH for real functions MATH and MATH satisfying MATH. The NAME expansions to order MATH at MATH of the MATH and MATH may be written MATH and MATH, where the MATH and MATH are real polynomials of degree at most MATH in their respective variables. Observe that on MATH we have MATH . The polynomial map MATH given by MATH may be extended to MATH simply by allowing MATH and MATH to be complex. Denote also by MATH this extension. Since MATH, it follows that MATH, so MATH is invertible near MATH as a map from MATH to MATH. Define new solutions MATH near MATH for the hypo-analytic structure by MATH. It then follows from REF that on MATH we have MATH for MATH and MATH for MATH. Define new coordinates MATH on MATH near MATH by MATH, MATH, MATH, and MATH. Then in these coordinates we have MATH for functions MATH satisfying MATH, and also MATH on MATH. Finally, replace MATH by MATH but leave MATH unchanged. In these new coordinates we have MATH and the solutions MATH have the desired form. |
math/0107183 | We first claim that MATH. (This does not use that MATH is CR.) Note that MATH. Using REF , we have MATH, so it must be the case that MATH as claimed. To prove the lemma, we may assume that MATH. Choose a set MATH which is a basis over MATH for MATH, and for which MATH. Next choose MATH such that the cosets MATH form a basis for MATH. (If MATH is CR, it will be the case that MATH, but it is not necessary to argue this separately.) Using that MATH is CR and MATH strongly noncharacteristic, it is easy to see that MATH forms a basis for a maximally real subspace of MATH. We may then take for MATH any submanifold of MATH near MATH with MATH equal to this subspace. |
math/0107183 | Let MATH and let MATH satisfy MATH; we must show that MATH. Construct a new hypo-analytic manifold MATH by introducing new variables as described in REF. The pullback of MATH satisfies at MATH the analogue of the condition MATH, so it suffices to prove the result on MATH. Therefore we may as well assume that our original hypo-analytic structure on MATH has involutive structure which is CR near MATH. Since MATH, there is MATH so that MATH. By REF , we can choose a maximally real submanifold MATH with MATH and MATH. Necessarily we have MATH. Choose a submanifold MATH of MATH such that MATH and MATH; in particular MATH. Since MATH is maximally real, MATH inherits near MATH a hypo-analytic structure of codimension REF from that on MATH. The involutive structure of MATH is CR near MATH, MATH is spanned (over MATH) by MATH, MATH is a maximally real submanifold of MATH, and MATH is spanned (over MATH) by MATH. We may assume that near MATH, MATH has two connected components, the one of which lies on the side of MATH determined by MATH we denote by MATH. Since MATH, it follows that MATH sufficiently near MATH. Now MATH may be regarded as a wedge in MATH with edge MATH. Considering MATH to be the background space, we have that MATH consists of the positive multiples of MATH. The solution on MATH with boundary value MATH restricts to MATH and defines there a solution of the involutive structure of MATH having boundary value MATH on MATH. According to the definition of the wave-front set, the desired conclusion MATH is equivalent to MATH. Since the hypothesis MATH holds just as well when regarding MATH and replacing MATH by MATH, it follows that we can reach this same conclusion if we prove the theorem on MATH. We therefore consider the situation in which MATH is a maximally real submanifold in the hypo-analytic manifold MATH whose structure satisfies MATH at MATH, and MATH is one side of MATH in MATH. We choose local coordinates MATH for MATH near MATH as in REF taking MATH; in our situation we have MATH. Set MATH and MATH and write MATH; we may assume MATH. Renormalizing the coordinates if necessary, we may arrange that MATH. Then MATH is a positive multiple of MATH and the hypothesis MATH says MATH. Upon relabeling, REF can be written as MATH where MATH is independent of MATH. From REF , we know that there is MATH so that MATH on a fixed neighborhood of the origin. Let MATH be a small open ball about the origin in MATH and MATH be such that there is a solution MATH of MATH on MATH with boundary value MATH. We intend to use REF to show that MATH. Introduce the MATH-form MATH on MATH given by MATH. Since MATH is a solution, MATH is closed. Let MATH satisfy MATH for MATH and MATH, where MATH will be chosen later. Setting MATH in REF shows that MATH takes the form REF with MATH. Define MATH as in REF and let MATH to be determined. Then NAME 's Theorem gives for any MATH, MATH . We shall show that if MATH is chosen sufficiently large and MATH sufficiently small, then each of the two integrals on the right hand side of the above equation satisfies an estimate of the form REF . Set MATH and let MATH denote MATH evaluated at MATH and MATH. If we show that there is MATH so that MATH for MATH, then MATH for the same MATH and for MATH near MATH and MATH in a conic neighborhood of MATH, and the desired estimates on MATH and MATH then follow immediately from seminorm estimates for MATH. Now MATH which, using REF , is MATH . First choose MATH so large and MATH so small that MATH and MATH, and then choose MATH so small that MATH. Then for MATH we have MATH, from which we conclude that MATH on MATH as desired. |
math/0107183 | First construct a hypo-analytic manifold MATH by introducing new variables as described in REF. If we can show that the result holds on MATH, we can conclude that it also holds on MATH. Therefore we may as well assume that our original hypo-analytic structure on MATH satisfies MATH at MATH. Next use REF to choose a maximally real submanifold MATH such that MATH. We then construct a submanifold MATH and wedge MATH as in the proof of REF . Restriction to MATH gives an injection MATH. If MATH and MATH, then the value of the NAME form MATH agrees when calculated either on MATH or on MATH, since MATH can be extended from MATH to remain tangent to MATH on MATH. Thus all the hypotheses of REF are satisfied on MATH. Since MATH is defined in terms of MATH, if we can prove the theorem for MATH, it follows that it also holds for MATH. Therefore consider the situation in which MATH is a maximally real submanifold in the hypo-analytic manifold MATH whose structure satisfies MATH at MATH, and MATH is one side of MATH in MATH. Choose local coordinates MATH for MATH near MATH as in REF taking MATH, and such that MATH. Then MATH is a positive multiple of MATH. As in the proof of REF , set MATH and MATH. The basic hypo-analytic functions are again given by REF , and we have REF and MATH . The characteristic covector MATH at MATH is in the span of the MATH with MATH; upon making a real linear transformation of these MATH's and corresponding MATH's we may assume that MATH. According to REF , the second order terms in the NAME expansion of MATH take the form MATH with MATH. The hypo-analytic function MATH satisfies MATH and MATH. If we introduce MATH and leave the remaining MATH's, MATH's, and MATH unchanged, then equations of the form REF continue to hold in the new coordinates and REF remain valid, possibly with a different constant MATH. The relations MATH and MATH still hold in the new coordinates and MATH is still a positive multiple of MATH. We may therefore assume that in the coordinates MATH, the basic hypo-analytic functions MATH are of the form REF and in addition to REF , we have for some MATH, MATH on some fixed neighborhood of the origin. It is easily seen that the vector field MATH is a section of MATH extending MATH, where MATH and MATH. An easy calculation then shows that MATH, so MATH. Upon rescaling MATH, MATH, and MATH, we may assume that MATH. Once again we intend to use REF to show that MATH. Let MATH, and as in REF , choose MATH satisfying MATH for MATH and MATH. By REF , the constant MATH defined in REF satisfies MATH. Choose MATH small enough that MATH and take MATH. Just as in the proof of REF , it suffices to show that we can choose MATH and MATH small enough that MATH satisfies MATH for MATH. Now MATH which, using REF , is MATH . If we choose MATH so small that MATH and MATH so small that MATH, then we obtain MATH, which yields MATH on MATH as desired. |
math/0107183 | We first remark that MATH consists of a single ray, so MATH is determined up to a positive multiple. Also, the condition MATH implies MATH, so that the value of MATH is well-defined independent of the extension of MATH from MATH. Choose coordinates MATH for MATH near MATH as in the proof of REF . Thus MATH and MATH, the hypo-analytic functions MATH are given by REF where REF hold, and MATH and MATH is a positive multiple of MATH. We make the same quadratic change of variables to obtain REF ; this time the condition MATH implies that MATH. By REF , the cubic expansion of MATH is of the form MATH where MATH and MATH. A straightforward calculation using the extension REF gives MATH. Therefore our assumption is that MATH . Introduce MATH, where MATH is to be determined. If we set MATH and leave MATH and the other MATH and MATH unchanged, then all of our properties continue to hold in the new coordinates, except that in REF , MATH is replaced by MATH and MATH by MATH. The inequality MATH is precisely the condition that one can choose MATH so that the quadratic form MATH be negative definite. In fact, for MATH, its discriminant is MATH. Hence, after making this change of variables and a subsequent rescaling of MATH, MATH, and MATH, we may assume in REF that MATH for MATH. Next let MATH be small and introduce new variables MATH . Observe first that if MATH is in a fixed neighborhood of the origin and MATH is chosen sufficiently small, then for all MATH with MATH, the corresponding MATH will lie in the neighborhood of the origin in which we have been working. This change of variables has the effect in REF of replacing MATH by new functions MATH given by MATH . We deduce that MATH satisfies REF in the new coordinates with constants independent of MATH for MATH. Also, REF is replaced by an analogous equation for MATH in which the error terms MATH and MATH can both be bounded by MATH for a constant MATH independent of MATH. Combining this observation with REF and removing the tildes from the new coordinates, we see that in addition to REF we can assume that MATH . In order to apply REF , let MATH to be chosen small, let MATH, and choose MATH satisfying MATH and MATH if MATH. By REF , we have MATH. Choose MATH small enough to ensure that MATH and set MATH. Choose MATH and MATH small enough that for all MATH we have MATH and MATH. We shall show that if MATH is chosen small enough, then MATH satisfies MATH for MATH. As in the proof of REF , we have MATH so by REF we obtain MATH . Therefore, if MATH and MATH, then MATH . If MATH and MATH, then MATH . Thus the desired inequality holds with MATH. |
math/0107183 | By introducing extra variables we may assume that MATH is of CR type near MATH. By REF , we can choose a maximally real submanifold MATH so that MATH. According to the definition of the wave-front set, it suffices to show that MATH. The set MATH is also a wedge with edge MATH, and the hypotheses of the Theorem continue to hold if MATH is viewed as the edge. Therefore we may as well prove the theorem with MATH replaced by MATH. If there is MATH such that MATH, then we can choose MATH small and of the appropriate sign so that the vector MATH will satisfy MATH. But for sufficiently small MATH, we have MATH. Therefore in this case the conclusion follows from REF . Hence we may assume that MATH for all MATH. Choose coordinates as in REF with MATH large. Since MATH is of CR type, we have MATH so there are no MATH variables. Now MATH, so we may make a real linear change of coordinates and corresponding change of MATH to arrange that MATH. Similarly we may take MATH. The hypothesis MATH is equivalent to MATH for some MATH; by reordering the coordinates we may assume that MATH. Finally, recalling our assumption that MATH for all MATH, we conclude that MATH for some nonzero MATH. It is straightforward to check using the form of the MATH in REF that one may choose a basis MATH for MATH near MATH of the form MATH in fact, the equations MATH uniquely determine the coefficients MATH and MATH, and one has that the MATH vanish at MATH and MATH. Define a submanifold MATH by MATH, where MATH is to be determined. Define also MATH. We have that MATH, MATH, and since MATH, it follows that MATH sufficiently near the origin. MATH inherits a hypo-analytic structure of codimension REF and CR type in which MATH is a maximally real submanifold. At each point of MATH, the space MATH is spanned by a single complex vector of the form MATH; this normalization ensures that this extended MATH agrees with the MATH we already have at the origin. The MATH are determined by the requirement that MATH should be tangent to MATH. The condition MATH for MATH forces MATH for these MATH. This leaves the one condition MATH, which is satisfied by MATH for MATH. A direct calculation using MATH, MATH, and MATH results in MATH. Now MATH is arbitrary, but by suitable choice of MATH we can ensure that MATH. Therefore the result follows from REF . |
math/0107183 | By introducing extra variables, we may assume that MATH is of CR type near MATH. If MATH, then the conclusion follows from REF . Therefore we may assume that MATH. In this case, the value of MATH is independent of the extension of MATH, so we may as well just consider MATH as an element of MATH. As in the proof of REF , choose a maximally real submanifold MATH such that MATH and a submanifold MATH such that MATH and MATH. Then MATH spans MATH and satisfies MATH and MATH. The conclusion thus follows from REF . |
math/0107183 | It follows from REF that MATH. REF implies that MATH. From these two facts one sees easily that there is an open acute convex cone MATH such that MATH and MATH. By REF , MATH is the boundary value of a holomorphic function on a wedge MATH in MATH with edge MATH with MATH. By REF , this extension agrees with MATH on MATH. |
math/0107183 | It suffices to reason near MATH. Elementary linear algebra (compare the discussion prior to REF ) shows that one can choose coordinates on MATH near MATH of the form MATH for MATH, MATH, MATH, where MATH, so that MATH at MATH and so that: MATH for some MATH and MATH satisfying MATH, MATH, and such that MATH . Write MATH with MATH, MATH, and MATH with MATH, MATH. The corresponding coordinates on MATH are obtained as described above. We consider the case in which MATH corresponds to a line in the normal space MATH which satisfies MATH; the argument in other cases is similar. Thus we introduce coordinates MATH, MATH and obtain coordinates on MATH near MATH and a formula for MATH: MATH . The point MATH corresponds to the direction MATH with MATH . The basis forms in REF can be smoothly extended to sections of MATH near MATH. Their pullbacks under MATH are certainly smooth, so the result follows if we know that these pullbacks remain linearly independent at MATH and have as common kernel the subspace MATH defined above. This is straightforward to check using REF , and REF . |
math/0107185 | The statement follows at once from the formulas given in REF , with standard manipulations involving the notations recalled above. For the first formula, write MATH pushing this expression forward by MATH gives MATH yielding the first formula in the statement. The second formula is proven similarly; it is in fact REF. |
math/0107185 | Let MATH be the irreducible decomposition of MATH; by CITE, each MATH can be identified with the conormal variety of its support. In particular, there is exactly one component MATH over the support MATH of MATH. Following CITE we let MATH, and remark that under our assumption this is a multiple of the characteristic function MATH of MATH, hence of the local NAME obstruction MATH since MATH is nonsingular by hypothesis. By REF, the cycle MATH must equal a constant times MATH; it follows that MATH is the only irreducible component of MATH, and further that MATH . The first assertion in the statement follows, as well as the formula for MATH. The formula for MATH can be obtained similarly from REF, or from the fact that MATH for all MATH (as proved in REF). |
math/0107185 | Since MATH has codimension at least REF in MATH, note that MATH . Therefore with notations as in REF MATH . |
math/0107185 | In view of the first formula in REF , MATH . By REF , MATH, hence MATH . Therefore MATH (using REF). That is, MATH and the formula given in the theorem now follows from the second formula of REF and from MATH. |
math/0107185 | CASE: This is the translation in our notations of the second formula in NAME 's NAME REF: MATH where MATH is the hyperplane class. CASE: From REF , we have MATH therefore, using REF from CITE: MATH with the stated result. |
math/0107185 | Since MATH is assumed to be nonsingular, the weighted NAME class of MATH (compare CITE and REF ) is given by MATH on the other hand, by REF MATH . Therefore MATH . Now arguing as in the proof of REF : MATH and the statement follows from the expression for MATH obtained in the proof of REF . |
math/0107187 | The multiplicative structure MATH will be defined using the NAME construction. We therefore begin by recalling some properties of this construction. Let MATH be an embedding of closed manifolds. Let MATH be a tubular neighborhood of MATH, which we identify with the total space of the normal bundle of the embedding. Let MATH be the NAME collapse map to the one point compactification, defined by MATH . If we identify the compactification with the NAME space of the normal bundle, MATH, then in the oriented case we can apply the NAME isomorphism MATH, to get the ``push-forward", or ``umkehr" map, MATH . Recall that in the case of the diagonal embedding of a MATH - dimensional closed manifold, MATH that the normal bundle is isomorphic to the tangent bundle, MATH so that the NAME map is a map MATH. Furthermore, in the oriented case the push - forward map in homology, MATH is simply the intersection product. Now the NAME construction also applies when one has a vector bundle over the ambient manifold of an embedding. That is, if one has an embedding MATH as above, and if one has a vector bundle (or virtual bundle) MATH, then one obtains a NAME map MATH where MATH is the tubular neighborhood of the induced embedding of total spaces MATH. Now MATH is the NAME space MATH, and MATH is the NAME space MATH. So the NAME map is a map MATH . Moreover this construction works when MATH is a virtual bundle over MATH as well. In this case when MATH, where MATH is a MATH - dimensional vector bundle over MATH, then the NAME spectrum MATH is defined as follows. Suppose the bundle MATH is embedded in a MATH dimensional trivial bundle, MATH. Let MATH-be the MATH - dimensional orthogonal complement bundle to this embedding MATH. Then MATH . Notice that when MATH is oriented, the NAME isomorphism is of the form MATH. In particular, applying the NAME construction to the diagonal embedding MATH, using the virtual bundle MATH over MATH, we get a map of NAME spectra, MATH or, MATH . When MATH is oriented, this map still realizes the intersection pairing on MATH after applying the NAME isomorphism. The map MATH defines a ring spectrum structure on MATH that is well known to be the NAME dual of the diagonal map MATH. To construct the ring spectrum pairing MATH, we basically ``pull back" the structure MATH over the loop space. To make this precise, let MATH be the product of the evaluation maps, and define MATH to be the fiber product, or pull back: MATH . Notice that MATH is a codimension MATH submanifold of the infinite dimensional manifold MATH, and can be thought of as MATH . Notice that there is also a natural map from MATH to the loop space MATH defined by first applying MATH and then MATH. That is, MATH where MATH . Notice that when restricted to the product of the based loop spaces, MATH, then MATH is just the MATH - space product on the based loop space, MATH. Remark. This definition needs to be modified slightly since MATH may not be smooth at MATH or MATH. This is dealt with in a standard way by first ``dampening" MATH and MATH by using a smooth bijection MATH with the property that all of its derivatives approach zero at the endpoints MATH and MATH, to reparameterize MATH and MATH. This will allow them to be ``spliced" together by the above formula without losing any smoothness. This is a standard construction, so we leave its details to the reader. Notice that by its REF the embedding MATH has a tubular neighborhood MATH defined to be the inverse image of the tubular neighborhood of the diagonal MATH: MATH . Therefore this neighborhood is homeomorphic to the total space of the MATH - dimensional vector bundle given by pulling back the normal bundle of the embedding MATH, which is the tangent bundle of MATH: MATH . Thus there is a NAME collapse map MATH . As described earlier, we ease the notation by refering to this NAME spectrum as MATH. By the naturality of the NAME construction, we have a commutative diagram of spectra, MATH . Since in the oriented case MATH is the intersection product, then MATH can be viewed (as is done in CITE) as taking a cycle in MATH, and ``intersecting" it with the codimension MATH submanifold MATH. Now observe that the map MATH defined above REF preserves the evaluation map. That is, the following diagram commutes: MATH . Thus MATH induces a map of bundles MATH, and therefore a map of NAME spectra, MATH . Now consider the composition MATH . In the oriented case the homomorphism MATH takes a cycle in MATH, intersects in with the codimension MATH - submanifold MATH, and maps it via MATH to MATH. This is the definition of the NAME product MATH. Now as we did before with the diagonal embedding, we can perform the NAME construction when we pull back the virtual bundle MATH over MATH. That is, we get a map of NAME spectra MATH . But since MATH, we have MATH . Now by the commutativity of REF , MATH induces a map of NAME spectra, MATH and so we can define the ring structure on the NAME spectrum MATH to be the composition MATH . A few properties of this map MATH are now immediately verifiable. First, MATH is associative. This follows from the naturality of the NAME construction, and the fact that the map MATH is associative. (Strictly speaking, REF is MATH - associative as is the usual formula for the product on the based loop space, MATH. However the standard trick of replacing MATH with ``NAME loops" changes the MATH structure to a strictly associative structure. The same technique applies to the map MATH. Otherwise, the spectrum MATH will have the structure of a MATH ring spectrum.) Also, notice tha MATH has a unit, MATH, defined by the composition MATH where MATH is the unit of the ring spectrum structure of MATH, and MATH is the map of NAME spectra induced by the section of the evaluation map MATH defined by viewing points in MATH as constant loops. Notice furthermore that in the oriented case, after applying the NAME isomorphism, MATH induces the same homomorphism as MATH, and so by REF the following diagram commutes: MATH where MATH is the NAME product. This proves REF . Now by the naturality of the NAME construction, REF spectra commutes (compare REF ) MATH . Thus the evaluation map MATH is a map of ring spectra, which proves REF . We now verify REF . Let MATH be a base point, and consider the following pullback diagram: MATH . Thus the embedding MATH is an embedding of a codimension MATH submanifold, and has tubular neighborhood, MATH equal to the inverse image of the tubular neighborhood of the inclusion of the basepoint MATH. This tubular neighborhood is simply a disk MATH, and so MATH where MATH reflects the MATH dimensional trivial bundle over MATH. Thus the NAME construction makes sense, and is a map MATH where the last space is the MATH - fold suspension of MATH with a disjoint basepoint. In homology, the homomorphism MATH denotes the map that is obtained by intersecting a MATH -cycle in MATH with the codimension MATH submanifold MATH. By performing the NAME construction after pulling back the virtual bundle MATH over MATH, we get a map of NAME spectra MATH . But by the commutativity of REF , MATH, which is the trivial, virtual MATH dimensional bundle which we denote MATH. So the NAME map is therefore a map of spectra MATH where MATH denotes the suspension spectrum of the based loop space of MATH with a disjoint basepoint. To complete the proof of REF we need to prove that MATH is a map of ring spectra. Toward this end, consider the following diagram of pull back squares: MATH . This gives NAME maps MATH . Notice that by the naturality of the NAME construction, the above composition is equal to MATH . Now notice that by the formula for the map MATH, the following diagram commutes: MATH where MATH is the usual multiplication on the based loop space. Pulling back the virtual bundle MATH over MATH, and applying the NAME construction, we then get a commutative diagram of spectra, MATH . Now as observed above REF , the top horizontal composition MATH is equal to MATH. Also, MATH is, by definition, the ring structure MATH. Thus the following diagram of spectra commutes: MATH . Thus MATH is a map of ring spectra, which completes the proof of REF . |
math/0107187 | Given a cactus MATH, define MATH to be the mapping space MATH . This space consists of maps from the union MATH. The map from the circle MATH defines a map from MATH to the loop space, MATH . Now MATH can also be viewed as the pullback of an evaluation mapping of the product MATH defined as follows. For each component of the cactus MATH, let MATH denote the vertices of MATH that lie on MATH. MATH is the number of these vertices. Let MATH. We define an evaluation map MATH as follows. Let MATH be the identification of the unit circle with MATH obtained by scaling down the unit circle so as to have radius MATH, and rotating it so the basepoint MATH is mapped to the marked point MATH. Let MATH be the points on the unit circle corresponding to MATH under the map MATH. Define MATH . Now define MATH . Now let MATH be the ordered collection of vertices in the cactus MATH (ordered by the ordering of the components MATH and the ordering of the vertices in each component as described above). For each such vertex MATH, let MATH be the number of components of the cactus on which MATH lies. We think of MATH as the ``multiplicity" of the vertex MATH. Notice that we have the relation MATH . The ``tree" condition on the dual of the cactus also imposes the following relation: MATH . Now consider the diagonal mapping MATH defined by the composition MATH where MATH is the MATH -fold diagonal. Observe that the following is a cartesian pull - back square: MATH where MATH evaluates a map MATH at the MATH vertices MATH. The normal bundle MATH of the diagonal embedding MATH is equal to MATH where MATH denotes the MATH - fold direct sum of MATH with itself as a bundle over MATH. By the tubular neighborhood theorem, we have therefore proven the following. The image of the embedding MATH has an open neighborhood homeomorphic to the total space of the pullback MATH. We now consider these constructions in a parameterized way, by letting MATH vary. Namely, let MATH . We then have a map MATH . We also have a map MATH where MATH is as defined in REF . REF allows us to identify an open neighborhood of the image of MATH, as follows. Let MATH be the MATH dimensional vector bundle over MATH whose fiber over MATH is the sum of tangent spaces, MATH where, as above, MATH are the vertices of MATH, and MATH-is the multiplicity of the vertex MATH. It is immediate that MATH is a well defined vector bundle. REF now yields the following. The image of the embedding MATH has an open neighborhood homeomorphic to the total space of MATH. Notice that this will allow us to perform a NAME collapse map MATH . But in order to prove REF , we will need to twist this map by a virtual bundle MATH over MATH. To define this bundle, first fix an embedding MATH with MATH, having normal bundle MATH of dimension MATH. Define a vector bundle MATH of dimension MATH over MATH whose fiber at MATH is given by MATH where MATH are the points on MATH that map to MATH under the parameterization MATH. (Notice that by the ``tree" condition on the cactus, there is at most one of the MATH on any one circle in MATH.) It is clear that these fibers fit together to give a vector bundle MATH over MATH. REF implies the following result. Consider the embedding of total spaces of bundles MATH then the image of MATH in MATH has an open neighborhood homeomorphic to the total space of the bundle MATH over MATH. This implies we have a NAME collapse map, MATH . Now by definition, MATH is a vector bundle of dimension MATH whose fiber at MATH is the sum of the vector spaces, MATH . But since MATH is canonically trivial, we have an induced trivialization MATH . Therefore the NAME collapse is a map MATH . Let MATH be the virtual bundle over MATH, MATH where MATH refers to the MATH - dimensional trivial bundle. Notice that MATH has virtual dimension MATH, and its NAME spectrum is the MATH - fold desuspension of the NAME spectrum of MATH. We then define MATH to be the composition, MATH . Now recall from CITE that the cactus operad structure on MATH is given by the composition MATH where MATH is the push -forward map. (Note that our notation is different than that used in CITE.) The theorem now follows from the well known fact that the NAME collapse map realizes the push -forward map in homology. |
math/0107187 | Let MATH be the NAME diagonal map. That is, for MATH, then MATH. By the definition of the cosimplicial structure of MATH, to prove the theorem it suffices to prove that the following diagrams of spectra commute: CASE: MATH . We verify the commutativity of these diagrams in several steps. First observe that the maps MATH restrict to MATH to define a map MATH whose image is in MATH making the following diagram commute: MATH where the bottom horizontal map is the diagonal map: MATH . By the naturality of the NAME construction, we therefore have a commutative diagram of spectra REF MATH . Notice further that by the definition of the maps MATH, MATH and the loop composition MATH defined in the last section, the following diagram commutes: MATH . Passing to NAME spectra this yields the following commutative diagram: CASE: MATH . Now consider the following diagram of spectra: MATH . Now the top left square in this diagram clearly commutes. The bottom left diagram is REF above, and so it commutes. The right hand rectangle is REF above, so it commutes. Therefore the outside of this diagram commutes. Now the top horizontal composition is, by definition the map MATH . The bottom horizontal map is seen to be MATH by recalling that the ring multiplication MATH is the NAME map MATH applied to the diagonal embedding MATH. With these identifications, the outside of this diagram is then REF above. As observed earlier, the commutativity of REF proves this theorem, and this completes the proof of REF . |
math/0107189 | We denote by MATH the map MATH with MATH, MATH, and by MATH the map MATH with MATH, MATH. Since MATH, and-MATH, the maps MATH, MATH give a measure-preserving bijection of MATH to itself. We shall show that MATH is a finite composition of maps of the form MATH or MATH. The proof will be accomplished by induction on MATH . Case MATH . In this case, it is sufficient to take MATH as follows: MATH . Induction hypothesis Suppose by induction hypothesis that the proposition is valid for all polynomials MATH of the form REF satisfying MATH, with MATH. Case MATH . Let MATH be a polynomial of the form REF satisfying MATH, MATH, and MATH. By applying the Euclidean algorithm to MATH, MATH, we have that MATH for some MATH, MATH. Because MATH, MATH are coprime, necessarily MATH . We set MATH, that is, MATH, MATH thus MATH with MATH non vanishing identically on MATH, and MATH with MATH . Since MATH and MATH, by applying the induction hypothesis applied to the polynomial MATH in REF , there exists a map MATH, that is a finite composition of maps of the form MATH or MATH, such that MATH, and MATH has all announced properties in the proposition. Therefore, it is sufficient to take MATH. The case MATH is proved in an analogous form. |
math/0107189 | REF implies that MATH . We set MATH and MATH . With the above notation, if MATH, then MATH for every MATH. We put MATH . Then, if MATH, and MATH, with MATH it holds that MATH for every MATH. From the above calculations follow that MATH . By subdividing MATH into equivalence classes modulo MATH, and using REF , and REF , we have that MATH for MATH, where MATH is a polynomial with rational coefficients. Moreover, the polynomial MATH and the constant MATH can be computed effectively. |
math/0107189 | The proof is based on an explicit computation of the MATH-adic absolute value of MATH, when MATH, MATH. In order to accomplish it, we attach to MATH the convex set MATH defined as follows. We associate to each term MATH of MATH (see REF ) an straight line of the form MATH and associate MATH the convex set MATH . Now, we set MATH with MATH. Then MATH (see REF ). The homomorphism MATH sends the topological boundary of MATH into the topological boundary of MATH. Thus the vertices of MATH are MATH where the MATH are the abscissas of the vertices of MATH (compare REF ). If MATH then MATH can be computed from MATH, as follows (we use the notation introduced at REF , and REF ): If MATH, MATH, then MATH . If MATH, then MATH . If MATH, MATH, then MATH with MATH and MATH here MATH is the straight line corresponding to the term MATH . In addition, we note any MATH, MATH satisfies only one of the following conditions: MATH or MATH . Then by using REF , and the previous observation, it follows that MATH with MATH . Since MATH, and MATH do not have singular points on MATH(compare REF ), there exists a constant MATH such that MATH for MATH (compare REF ). Finally, the result follows from REF , by using the algebraic identity MATH . The constant MATH . |
math/0107189 | By taking a simplicial conical subdivision, the computation of MATH is reduced to the computation of integrals of type MATH, MATH-and MATH, with MATH a proper face of MATH (see subsection REF). By REF the real parts of the poles of MATH belong to the set MATH. The computation of the integrals MATH, for MATH a proper face of MATH, involves two cases, according if the semi-quasihomogeneous polynomial MATH with respect to MATH, is geometrically non degenerate or not. If MATH is a one-dimensional cone, and MATH does not have singularities on MATH, then the real parts of the poles of MATH belong to the set MATH where MATH, MATH, is the equation of the supporting line of the facet MATH (compare REF ). If MATH is a two-dimensional cone, MATH is a monomial and then it does not have singularities on the torus MATH, by REF , MATH is a entire function. If MATH is a one-dimensional cone, and MATH has singularities on MATH, then MATH is a semi-quasihomogeneous arithmetically non degenerate polynomial, and thus the real parts of the poles of MATH belong to the set MATH (compare REF ). Therefore the real part of the poles of MATH belong to the set MATH. |
math/0107190 | The construction presented here is a modification of a construction invented by NAME. REF was also proved in CITE using different methods. Let MATH be a family of functions in MATH such that CASE: MATH is strictly increasing for MATH, CASE: MATH. For a perfect tree MATH let MATH denote the set of its branches. Every perfect subset of MATH is a set of branches of a perfect tree. We will build a MATH-tree MATH of perfect subsets of MATH. Let MATH denote the MATH-th level of MATH. We require that CASE: MATH, CASE: MATH, CASE: MATH is countable. Successor step. Suppose that MATH is given. For each MATH choose MATH such that CASE: MATH, CASE: MATH, CASE: MATH for MATH, CASE: MATH. Set MATH to be the successors of MATH on level MATH. Limit step Suppose that MATH is a limit ordinal and MATH are already constructed. For each MATH and MATH we will construct an element MATH belonging to the level MATH as follows. Suppose MATH, MATH and construct sequences MATH, MATH and MATH such that CASE: MATH, CASE: MATH, MATH, CASE: MATH for all MATH, CASE: MATH, CASE: MATH is a perfect tree. The last condition is guaranteed by the careful choice of the sequence MATH. Let MATH be the set obtained by selecting one element out of every tree MATH. The following lemma gives the first part of the theorem. If MATH is an unbounded family in MATH then MATH is meager additive. Proof Suppose that MATH is a meager set. It is well known (see REF ) that there exists MATH and a strictly increasing function MATH such that MATH . Since we work with translations without loss of generality we can assume that MATH for all MATH. As MATH is unbounded there is MATH such that MATH . Fix sequences MATH such that MATH . Fix MATH and let MATH be defined as follows: given MATH let MATH be an enumeration of MATH. Define MATH for MATH. This defines MATH on an infinite subset of MATH, extend it arbitrarily to a total function. Let MATH . We claim that MATH. Note that if MATH then there exists MATH such that MATH . Fix one such MATH and note that for sufficiently large MATH there is MATH such that MATH which implies that MATH. It follows that MATH which means that MATH. Let MATH be the collection of points selected from levels MATH. We have MATH which finishes the proof. To prove the second part of the theorem we will use the following lemma: If MATH is a dominating family then MATH is null additive. Proof The following is well known: Suppose that MATH is a null set. There exists a sequence of clopen sets MATH such that that for all MATH, CASE: MATH, CASE: MATH. Since MATH has measure zero there are open sets MATH covering MATH such that MATH, for MATH. Write each set MATH as a disjoint union of open basic intervals, MATH for MATH, and order these sequences lexicographically in a single sequence MATH. Put MATH and for MATH let MATH. Let MATH . Clearly MATH and, since all basic sets have been accounted for, MATH . Let MATH be a measure zero set and MATH the sequence given by the lemma. Since each clopen set is a union of finitely many basic sets we can find a function MATH such that for every MATH, MATH. Let MATH and MATH be such that MATH . By modifying finitely many MATH's we can assume that MATH. Suppose that MATH. Note that MATH . Since MATH, it follows that MATH . Therefore, MATH, and MATH. The rest of the proof is identical to the proof of the first part. |
math/0107190 | CASE: Let MATH be a perfect set and MATH a countable dense set. By REF without loss of generality we can assume that MATH is a MATH relative to MATH. In other words MATH, where MATH are open sets. Clearly MATH's are dense in MATH. It follows that MATH. CASE: Since MATH there exists a MATH set MATH such that MATH, and there exists a MATH set MATH such that MATH is meager in MATH and MATH. Now the set MATH is the set we are looking for. CASE: Suppose that MATH is countable. By NAME Theorem, CITE, MATH, where MATH is perfect, MATH is countable and open relative to MATH. If MATH then MATH is countable and closed and MATH is a MATH set in MATH. Thus assume that MATH and let MATH be closed nowhere dense sets such that MATH and MATH is closed nowhere dense in MATH for each MATH. Let MATH be closed sets such that MATH. Let MATH and consider sets MATH. Since the family MATH has the finite intersection property we can find a set MATH such that CASE: MATH is finite for all MATH, CASE: MATH is dense in MATH. Let MATH. Note that each set MATH is MATH since it differs from MATH by a finite set. Put MATH . It follows that MATH, which finishes the proof. REF is obvious. |
math/0107190 | CASE: The following argument is a small modification of a proof from CITE. Suppose that MATH. Let MATH be a NAME isomorphism onto a non-meager set. By NAME 's theorem, MATH is continuous on a dense MATH set. It follows that there exists a continuous one-to-one function MATH, where MATH. Let MATH be enumeration of clopen subsets of MATH such that MATH is uncountable. For each MATH, choose a countable dense-in-itself set MATH. We will show that every MATH set which is disjoint from MATH is also disjoint from one of the MATH's. Let MATH be a MATH set containing MATH. For every MATH let MATH be a closed set such that MATH. If for every MATH, MATH is countable then MATH, which is impossible. Thus there exists MATH such that MATH. CASE: Let MATH denote the NAME algebra. The following is a (small) fragment of REF from CITE. For a subset MATH of a perfect Polish space MATH, the following are equivalent: CASE: MATH. CASE: For every MATH-ideal MATH in MATH such that MATH there is a NAME set MATH such that MATH. The implication MATH that is of interest to us is a consequence of NAME 's theorem (CITE): if MATH is a MATH-ideal in MATH such that MATH, then there is NAME automorphism MATH such that MATH . Suppose that MATH are given. Consider the ideal MATH . It is easy to see that MATH (as MATH is atomless and has a countable dense subset). Therefore there exists a set MATH such that MATH. CASE: Let MATH. Let MATH be a MATH set containing MATH such that MATH is meager in each MATH. As in the proof of REF, build by induction a set MATH such that for MATH, CASE: MATH, CASE: MATH is finite. As before MATH is a MATH set and let MATH. The sets MATH for MATH are as required. |
math/0107190 | Suppose that MATH, MATH is countable and MATH is not a MATH relative to MATH. Write MATH, where each MATH is discrete, that is MATH for MATH. It follows that if MATH and MATH then MATH, which finishes the proof. To construct a set MATH as above, recall that a set of reals is called a MATH-set if all of its countable subsets are relative MATH sets. REF implies readily that all MATH-sets, and unions of countable sets with MATH-sets, are universally meager. Let MATH and a countable set MATH be such that CASE: MATH is a MATH-set, CASE: MATH is not a MATH-set, that is MATH is not a MATH set relative to MATH. NAME showed that such sets can be constructed in MATH, REF . |
math/0107192 | The unique intersection property implies that the total number of intersection points is MATH. Moreover, if a wire MATH was below a wire MATH at MATH it will be above it from their intersection point on. Conversely, if every two pseudolines changed order, then we have at least MATH intersection points, and since this is the total number, every two lines intersect exactly once. |
math/0107192 | Let MATH where MATH is any line in MATH. Then: MATH . |
math/0107192 | Let MATH be the point of multiplicity MATH, and consider the remaining MATH lines. If they form a diagram with signature MATH (so that MATH), then in the original diagram the multiplicity of every point cannot be larger by more than MATH. Since we now count the point MATH too, we have that MATH. |
math/0107192 | There are some solutions to MATH except for the ones listed above. For MATH we have MATH, which is forbidden by the case MATH of REF . For MATH, we get MATH and MATH. The first is forbidden by REF , case MATH, and the second by REF , case MATH. For MATH, we get MATH, MATH, MATH, MATH, MATH and MATH. The first five options are forbidden by REF . As for MATH, this is impossible since then a line MATH would meet five other lines, two in each intersection point; but MATH does not divide MATH. |
math/0107192 | Except for the ones listed above, the only solutions to MATH are: MATH, MATH, MATH which are forbidden by the case MATH of REF ; MATH, MATH, MATH, MATH, MATH which are forbidden by the case MATH of this lemma; MATH, MATH, MATH, MATH, MATH, MATH, MATH and MATH. The first seven options are excluded by the case MATH of the lemma, since we must have MATH (note that here MATH counts the points of multiplicity MATH beyond the first one). As for MATH, this is the projective geometry on MATH points, which is well known not to be embedded in the affine plane. |
math/0107192 | There are REF solutions to the equation MATH except for the above listed ones. REF excludes three cases with MATH, nine cases with MATH and fifteen cases with MATH. Out of the nine possibilities with a point with multiplicity REF, the lemma (with MATH) excludes eight, the remaining one is MATH. Suppose there are at least two points in which four lines meet. If they do not share a common line, the signature must be MATH. Otherwise, there is one remaining line, which can meet at most REF otherwise-simple intersection points, so we cannot have a third point of multiplicity MATH. The last two cases are MATH and MATH. Assume signature of the form MATH. Every line meets seven other lines, so the number of multiple points on a line is no more than MATH. Together, there are no more than MATH multiple points (since each point is counted three times). This argument rules out the signature MATH. It also shows that in a diagram with signature MATH, on each line there are exactly one simple point and three multiple points. We could not find a geometric argument to rule this case out (see REF below), but our algorithm REF found no wiring diagrams for this signature. |
math/0107193 | See CITE or REF or REF. |
math/0107193 | The existence is proved above. The uniqueness up to isotopy is proved as follows: Each regular neighborhood fibers over a MATH-orbifold with fibers connected MATH-orbifolds in the orbifold sense. A regular neighborhood can be isotoped into any other regular neighborhood by contracting in the fiber directions. We can modify the proof of REF to be adopted to an annulus with a finite group acting on it and an imbedded circle. |
math/0107193 | Form regular neighborhoods of the involved boundary components of the split orbifold and those of the original orbifold. They have zero NAME characteristic. Since their boundary MATH-orbifolds have zero NAME characteristic, the lemma follows by the additivity REF . |
math/0107193 | First, we show that the image of MATH is in MATH. By REF,the holonomy group a MATH-orbifold of negative NAME characteristic fixes no line, since a finite-index subgroup is a fundamental group of a closed surface of negative NAME characteristic. Since MATH is finitely presented by REF , the holonomy map is a local homeomorphism follows from CITE. |
math/0107193 | It is shown in CITE using NAME 's result CITE that MATH is an open subset of MATH if MATH is a surfaces. This is done by taking product MATH and finding an affine structure on it corresponding to a MATH-structure on MATH. An affine structure on MATH is convex if and only if the MATH-structure on MATH is convex. Since all discussions in CITE apply to differentiable orbifolds as well, we can replace MATH by MATH. |
math/0107193 | Since MATH admits a hyperbolic structure, MATH is isomorphic to a discrete cocompact subgroup of MATH. There is a torsion-free finite-index normal subgroup MATH of MATH by NAME 's lemma CITE. Let MATH be MATH in MATH. There is a finite covering surface MATH of MATH corresponding to MATH and MATH of MATH corresponding to MATH. The finite group MATH maps injectively into MATH. Similarly, MATH maps into MATH. Thus, the commutative diagram MATH holds where MATH is an induced isomorphism. By the NAME realization theorem of CITE, MATH acts on MATH and MATH on MATH. A homeomorphism MATH realizes MATH and for each MATH, MATH is homotopic to MATH for MATH. Give MATH and MATH arbitrary hyperbolic metrics which are MATH- and MATH-invariant respectively (that is, using NAME 's orbifold hyperbolization of MATH and MATH). Then choose a unique harmonic diffeomorphism MATH in the homotopy class of MATH as obtained by CITE. Since MATH by uniqueness, MATH induces a desired orbifold diffeomorphism MATH. |
math/0107193 | Choose a sequence of representations in MATH so that MATH for a representation MATH; that is, MATH converges to MATH algebraically. We show that MATH is in the image proving the closedness. Let MATH be the orbifold MATH with a MATH-structure corresponding to MATH, and let MATH be the universal cover of MATH. Let MATH be the universal cover MATH with the induced MATH-structure from MATH, and MATH a developing map of MATH associated with MATH. Then MATH maps onto a strictly convex domain MATH in an affine patch of MATH. (This follows since MATH is a universal cover of a convex MATH-surface finitely covering MATH. See CITE.) MATH induces a MATH-diffeomorphism MATH. There is a unique MATH-structure on the sphere MATH such that the covering projection MATH is a projective map. The nontrivial deck transformation is represented by the antipodal map of the sphere. Its collineation group MATH is isomorphic to MATH, generated by MATH and the antipodal map, We can show that MATH always lifts to an embedding MATH and MATH lifts to a homomorphism MATH (see CITE): we can lift first, and for a deck-transformation MATH of MATH, MATH is another developing map, and hence it must equal MATH for MATH. Defining MATH, we see that MATH is a lift of MATH. The image MATH of MATH is a convex open subset of an open hemisphere in MATH with a standard geodesic structure. By choosing a subsequence if necessary, the sequence of the closures MATH converges to a compact convex subset of MATH in a closed hemisphere CITE. We claim that the limit MATH is neither a point, a line segment, a lune, nor a closed hemisphere. Otherwise, (taking a finite subcover MATH of MATH if necessary), all MATH are images of a sequence of developing images of convex MATH-structures on a closed surface MATH. We showed that such a degeneration cannot happen in CITE. (See also CITE.) Thus, MATH is a compact convex subset of an open hemisphere in MATH. By choosing a subsequence, MATH converges to a representation MATH lifting MATH. As in CITE, MATH acts on MATH. Since MATH is a map to MATH, MATH is discrete and faithful by REF. (MATH has a finite index subgroup which is torsion-free. Apply REF here and the finite index extension argument is trivial.) Therefore, MATH acts on an open disk MATH with quotient orbifold MATH. By REF , there is a diffeomorphism MATH inducing MATH. Since MATH is an embedding onto a MATH-invariant convex open domain MATH, we see that MATH is realized also as the quotient space MATH. Thus, MATH is realized as a holonomy homomorphism of a convex MATH-structure on MATH and lies in the image of MATH. |
math/0107193 | Let MATH be a closed MATH-orbifold with MATH. Let MATH the connected subset of MATH consisting of hyperbolic MATH-structures on MATH. Since by Theorem A, MATH is connected, and MATH sends MATH into MATH, it follows that MATH sends MATH into MATH. By REF , MATH is an open subset of MATH. Since MATH is an open map, MATH is an open subset of MATH. By REF , the image is a closed subset of MATH. Hence, the image equals MATH. The holonomy group of a convex MATH-orbifold is discrete since it acts on an open domain discontinuously. Thus, MATH consists of discrete embeddings. Recall also that since MATH acts properly on MATH, MATH is a subset of MATH. To complete the proof of Theorem B, we show that MATH is injective: That is, given two holonomy representations MATH and MATH for MATH and MATH for convex MATH-orbifolds MATH and MATH. Then, MATH for holonomy homomorphisms of MATH and MATH of MATH and MATH respectively. We show that if MATH, then MATH and MATH are isotopic equivariantly with respect to MATH. (According to the definition of deformation spaces in CITE, this will prove the injectivity of MATH.) Let MATH be a closed surface finitely covering MATH. Let MATH be the image of MATH composed with MATH and MATH that of MATH composed with MATH. Since MATH and MATH restricted to MATH are the same, REF shows MATH. The images of MATH and MATH are the same, and they are both equivariant under the homomorphism MATH. The map MATH is so that MATH and is equivariant under the homomorphism MATH . Let MATH have a hyperbolic metric MATH. Then MATH and MATH induce metrics on MATH and MATH respectively. We now show that MATH and MATH are MATH-equivariantly isotopic. MATH has induced hyperbolic metrics MATH and MATH induced from MATH and MATH respectively. There is a path of Riemannian metrics MATH for MATH from MATH to MATH. By the equivariance, they induce metrics on MATH to be denoted by same letters. Recall that MATH is the closed surface covering MATH. Let MATH denote the corresponding closed surface covering MATH. Let MATH denote the Riemannian metrics of MATH corresponding to MATH. By REF of Tromba CITE, there exists a smooth one-parameter family of harmonic diffeomorphisms MATH. Since these harmonic diffeomorphisms are unique in their homotopy classes, they should be equivariant under automorphisms of MATH and MATH, and MATH descend to a one-parameter family of diffeomorphisms MATH. One can lift the inverse map MATH of MATH to a smooth one-parameter family of diffeomorphisms MATH using analytic continuations. By uniqueness of harmonic diffeomorphisms, we see that the inverse map MATH lifts to MATH, and by analytic continuation MATH lifts to MATH for some deck transformation MATH of MATH, where MATH is isotopic to MATH equivariant with respect to MATH. MATH must equal MATH. Since MATH, REF implies that MATH equals the identity since the center of MATH is trivial. Therefore MATH equals MATH. Thus MATH and MATH are MATH-equivariantly isotopic. Applying MATH to MATH and MATH and MATH to MATH, implies that MATH and MATH are equivalent. Therefore, MATH is a homeomorphism, inducing one MATH . |
math/0107193 | A path in the subsurface MATH is homotopic to a geodesic in MATH. Since there are no bigons, the geodesic itself is in MATH. A closed geodesic in MATH is always principal. Thus, MATH is convex (see CITE). |
math/0107193 | Pass to a finite covering surface MATH with MATH. Let MATH be a development pair and MATH the universal cover of MATH. Since MATH is convex, its holonomy representation is faithful. Let MATH and MATH be two deck-transformations corresponding to simple closed curves not homotopic to each other. Then these curves are homotopic to principal closed geodesics and MATH and MATH act on geodesic lines MATH and MATH in the universal cover MATH corresponding to the closed geodesics respectively. The endpoints of MATH are the attracting and repelling fixed points of MATH by the principal conditions; and the endpoints of MATH those of MATH. No two of these points coincide. Otherwise, MATH can be sent arbitrarily close to MATH by MATH. Since MATH map to a simple closed curve and MATH is an embedding, this cannot happen. No three of these points are collinear since MATH acts on MATH freely, and by convexity. Suppose that MATH for a collineation MATH. Then MATH for MATH, and MATH acts on each of the two pairs of four noncollinear points. Since the images of four points, no three of which are collinear, determine the collineations, MATH is the identity map or a unique reflection determined by the four points. Therefore MATH is constant. |
math/0107193 | Since the silvered MATH has empty boundary, MATH is a local homeomorphism. The one-to-one correspondences discussed above obviously makes a commutative diagram: MATH . The desired conclusion follows. |
math/0107193 | By REF , MATH is an open subset of MATH where MATH is obtained from MATH by silvering boundary components. By the above one-to-one correspondence between MATH and MATH and that MATH between MATH, we obtain the result. |
math/0107193 | We have the following commutative diagram: MATH . Since the image of MATH is closed, our result follows. |
math/0107193 | REF show that the image under MATH is an open set. The proof follows from REF and the fact that MATH is connected by REF . |
math/0107193 | Consider first the case MATH. The completions of components of MATH have negative NAME characteristic by REF since they are obtained by sewing the components of MATH. Let MATH be a finite cover of MATH which is an orientable surface with principal closed geodesic boundary. Let MATH be the group of automorphisms of MATH so that MATH is projectively diffeomorphic to MATH in the orbifold sense. (We showed the existence of such a cover in REF.) Let MATH, MATH, be the components of the inverse image of MATH. Since MATH is convex, orientable, and MATH, each MATH is homotopic to a simple closed geodesic (see CITE). Let MATH, MATH, be the union of all simple closed geodesics in MATH homotopic to some MATH. Then clearly MATH acts on MATH . Let us take a component MATH, and let MATH be the subgroup stabilizing MATH. Let MATH be the curves homotopic to MATH in MATH, on the union of which MATH acts on. MATH must act transitively on the set MATH as MATH is the unique image in MATH. If MATH has a unique element, then MATH also act on the unique element, and the image of MATH is the unique principal closed geodesic or principal MATH-orbifold MATH isotopic to MATH. (A MATH-equivariant isotopy in MATH can be constructed by perturbations of MATH and MATH-s and using innermost-bigons which occur MATH-equivariantly.) If MATH has more than one element, then let MATH be the unique maximal annulus bounded by elements in MATH. Then it is easy to see that MATH acts on MATH and MATH permutes MATH-s by projective diffeomorphisms. MATH covers MATH embedded in MATH, and MATH is a suborbifold of MATH bounded by MATH. The NAME characteristic MATH since MATH. Since the completions of components of MATH have negative NAME characteristic and the embedded suborbifolds MATH are such completions, we obtain a contradiction. Hence, the element of MATH is unique. We obtain MATH in a similar manner. If MATH and MATH are principal closed geodesics or geodesic full MATH-orbifolds, subsegments of MATH and MATH do not bound a bigon. (To prove this, simply lift to the universal cover which may be considered a convex domain in an affine patch.) Thus, MATH and MATH meet in the least number of points or MATH. If MATH, then a power of MATH is homotopic to that of MATH for some MATH. Since they are disjoint simple closed curves, they bound an annulus MATH in the closed surface MATH. If there are MATH or MATH for some MATH and MATH in the interior of MATH, they are essential in MATH. We find an annulus MATH bounded by them and with interior disjoint from such curves. (Possibly MATH.) Now, MATH covers a suborbifold in MATH with zero NAME characteristic, which is not possible. We conclude that MATH and MATH are disjoint. By induction, there exist principal closed geodesics or principal MATH-orbifolds MATH disjoint from each other so that MATH is isotopic to MATH for each MATH. |
math/0107193 | We essentially follow the sketch of the proof of REF. We remark that we don't really need to have silvered elementary MATH-orbifolds in the conclusion since we can define clarifying as ``decomposing" also; however, we don't need this fact in this paper. In this proof, we won't distinguish between silvered elementary annuli and elementary annuli mainly in indicating their types. Suppose first that MATH has no corner-reflectors or boundary full MATH-orbifolds; that is, singular points of MATH are cone-points or in a closed geodesic of mirror points in the boundary of MATH, and the boundary components of MATH are principal closed geodesics. Let MATH be the cone-points of MATH. Assume for the moment that the number of cone-points MATH is greater than or equal to MATH, and MATH. Let MATH be a disk in MATH containing all MATH. Then MATH has a structure of a MATH-orbifold of negative NAME characteristic (by REF ). If at least one MATH has an order greater than two, then we find a simple closed curve bounding a disk MATH containing MATH and another cone-point, say MATH. We find simple closed curves that bounds a disk MATH including MATH and another cone-point. We find a disk MATH including MATH and another cone-point and so on. Therefore, MATH is divided into a disk MATH and annuli MATH containing unique cone-points. Each of MATH and MATH-s has negative NAME characteristic. MATH contains no singular points and has negative NAME characteristic. Decompose MATH into pairs-of-pants. By REF , we can find a collection of principal closed geodesics in MATH decomposing MATH into convex MATH-suborbifolds of negative NAME characteristic of type REF , and REF . If all cone-points have order two, then we choose a disk MATH in MATH containing three cone-points. Split along a cone-type MATH-orbifold connecting the two cone-points in MATH. Now find MATH as above to obtain the decomposition into elementary MATH-orbifolds of type REF or REF . If MATH, then MATH is homeomorphic to a disk or MATH. In the former case, MATH decomposes as above. In the latter case, the decomposition along a one-sided simple closed curve produces a disk with cone-points In particular the resulting orbifold has negative NAME characteristic. Now apply REF . Suppose that MATH. When MATH, MATH is an elementary MATH-orbifold of type REF . Next suppose MATH and at least two cone-points of order greater than MATH. Then there exists a simple closed curve in MATH bounding two disks containing at least two cone-points, each containing a cone-point of order greater than MATH, and with negative NAME characteristic. Divide the complement of the union of these disks to annuli as above. Next suppose MATH and a single cone-point of order greater than two. Then there exists a cone-type MATH-orbifold connecting two cone-points of order two. Cutting along the MATH-orbifold produces a disk with principal geodesic boundary. Decompose the disk into union of a disk containing the single cone-point of order greater than two, and annuli with single cone-points. If all cone-points are of order two, then MATH. Then decompose along two disjoint cone-type MATH-orbifolds, obtaining an annulus with cone-points of order two. Now proceed as above. Suppose that MATH or MATH. If MATH, then we can introduce one or two annuli or NAME bands containing the one singular point each and we obtain a decomposition into REF type orbifolds. If MATH, then an essential simple closed curve cuts MATH into an annulus. The decomposed orbifold is connected and of negative NAME characteristic. If MATH and MATH is homeomorphic to MATH or MATH, then MATH. If MATH and MATH, then MATH is homeomorphic to MATH. Split MATH along a simple closed curve into a disk with two cone-points, leading to the next case. Finally if MATH is homeomorphic to a disk, then MATH is an elementary MATH-orbifold of type REF and MATH. (This settles REF below.) We now suppose that MATH has corner-reflectors and/or boundary full MATH-orbifolds. Let MATH be the boundary components of MATH containing corner-reflectors or boundary MATH-orbifolds, and MATH denote the remaining components. Let MATH be a simple closed curve in MATH homotopic to MATH for MATH. The component of MATH containing MATH has negative NAME characteristic by REF . Suppose that MATH split along the union of MATH-s has a component MATH with nonnegative NAME characteristic. Then the underlying space of this component is homeomorphic to either a disk, an annulus, or a NAME band. If MATH is homeomorphic to an annulus or NAME band, no singular point of MATH lies on the component. When MATH is an annulus, cut along the center closed curve obtaining MATH decomposed into two annuli with singular points. They can be pasted with other components of MATH without changing their NAME characteristics, obtaining a different decomposition of MATH. When MATH is a NAME band, cut along a one-sided closed curve obtaining an annulus with singular points. The annulus is then pasted with the adjacent component of MATH without changing the NAME characteristic, obtaining a new decomposition. If MATH is homeomorphic to a disk, then there are either two cone-points of order two or a single cone-point of arbitrary order by the NAME characteristic condition or no cone-points, and MATH is homeomorphic to a disk. If there are two cone-points of order two, we can split MATH along a cone-type MATH-orbifold in MATH to obtain an annulus with corner-reflectors. If there is a single cone-point, then MATH must have been a disk with one cone-point. If there is no cone-point in MATH, then MATH is a disk. These facts and REF imply that MATH decomposes into convex MATH-suborbifolds listed below: CASE: A disk with corner-reflectors and boundary full MATH-orbifolds and no cone-points. CASE: An annulus with corner-reflectors and boundary full MATH-orbifolds in one boundary component of its underlying space and no cone-points. CASE: A disk with corner-reflectors, boundary MATH-orbifolds, and one cone-point of arbitrary order. CASE: A MATH-orbifold without corner-reflectors or boundary full MATH-orbifolds. When there are boundary full MATH-orbifolds, we silver them. This does not change the NAME characteristic of any MATH-suborbifold containing it by REF . We mark the full MATH-orbifold to be of boundary type, and we will never cut them apart. After our decomposition, we will clarify them. Let MATH be one of the above REF . Let MATH denote the number of corner-reflectors. In REF , assume MATH. If MATH, the orbifold MATH is an elementary one of type REF . Suppose that MATH. From a side MATH, we choose one MATH-orbifold, say MATH, ending at MATH and each of the other sides except the sides adjacent to MATH. We choose these to be disjoint. If MATH ends at a side was a boundary full MATH-orbifold, we delete MATH. If any component of MATH split along the remaining MATH-orbifolds MATH has NAME characteristic greater than or equal to zero, then we delete one of the MATH-orbifold adjacent to the component. Continuing in this manner, we obtain a collection MATH so that if MATH is split along them, each of the component is an orbifold of type REF , or REF . Finally, clarifying the boundary types yields the desired decomposition. In REF , the underlying space of MATH is a disk minus an open disk in its interior. As before let MATH denote the corner-reflectors and MATH edges. If MATH, then MATH is an elementary MATH-orbifold of type REF . Suppose MATH. From an edge MATH which is not boundary type, we choose MATH-orbifolds MATH ending at the sides MATH so that the two endpoints of MATH are in MATH and other MATH has endpoints in distinct edges. We eliminate MATH-s that ends in the edges adjacent to MATH and ones of boundary type. If we split along MATH-s and obtain a component of NAME characteristic greater than zero, then we eliminate one of the adjacent MATH-s. As in REF , by continuing in this manner, we decompose MATH into an elementary orbifold of type REF and ones of type REF , or REF . In REF , suppose that the cone-point has order greater than two. If the number of corner-reflectors is one, then MATH is an elementary MATH-orbifold of type REF . As above, we find MATH-orbifolds MATH and do similar constructions. Suppose that the cone-point has order two. Draw a mixed-type MATH-orbifold MATH from the cone-point to a mirror point in the nonboundary type full MATH-orbifold. The decomposition results in a disk. NAME the MATH-orbifold which folds to MATH. Decompose the disk and clarify it back. |
math/0107193 | For a proof, see CITE or CITE. |
math/0107193 | We see that the edge lengths or angle measures of MATH completely determine the unique disks. The edges labeled by Greek letters can be made arbitrarily large or small after the angles labeled by Greek letters are assigned as the reader can easily verify. By the above formulas, MATH is a homeomorphisms. |
math/0107193 | In REF , find the shortest segment MATH from one boundary component to the other. Cutting along it, we obtain a hexagon, where the boundary is cut into three alternating sides of the hexagon MATH. Let MATH and MATH be from the boundary full MATH-orbifold. Let the length of MATH and MATH be equal. By symmetry, the lengths of two sides corresponding to MATH become equal. Hence, we can glue back such a hexagon to obtain an elementary MATH-orbifold of type REF always. Thus, we see that given MATH and MATH, we can obtain an elementary MATH-orbifold. Since MATH and MATH are lengths of the boundary components, we have shown that MATH is a homeomorphism. In REF , we silver the boundary component temporarily, find a shortest segment MATH from the mirror edge to the boundary component. Then cutting along MATH, we obtain a pentagon. Let MATH and MATH be the boundary full MATH-orbifolds of the pentagon. Letting the lengths of MATH and MATH equal, we can always glue back to obtain an elementary MATH-orbifold of type REF . Since we can change the length of MATH arbitrarily by changing the common length of MATH and MATH, we see that MATH is a homeomorphism. In REF , we draw a shortest segment MATH from the cone-point to the boundary MATH-orbifold. We obtain a pentagon where MATH corresponds to edges labeled MATH and MATH above. If the lengths of MATH and MATH are equal, we can glue back to obtain an elementary MATH-orbifold of type REF . Here, MATH and MATH are from the boundary MATH-orbifold by cutting, and their lengths are the same. If the lengths of MATH and MATH are the same, then those of MATH and MATH are equal, and we can glue back. Since MATH is the length of the boundary MATH-orbifold, we see that MATH is a homeomorphism. In REF , we draw a shortest segment MATH from the cone-point to the mirror edge. Then we obtain a quadrilateral with angles MATH, MATH, MATH, and MATH where MATH is the order of the corner-reflector and MATH is that of the cone-point. Such a quadrilateral is unique, and we can glue back to obtain an elementary MATH-orbifold of type REF always. Thus the NAME space is a single point. |
math/0107193 | The universal cover of the orbifold is a convex domain and contains no bigons or monogons with geodesic boundary. Any nonsimple arc which is homotopic to a simple arc must have some bigon or monogon in the universal cover. Thus MATH is simple. |
math/0107193 | The universal covering of MATH is a convex domain with a projectively invariant NAME metric. This metric induces a NAME metric on MATH with a corresponding notion of arclength. Define MATH for each MATH to be the compact subset of MATH obtained by removing from MATH the union of convex open neighborhoods of the boundary components and the cone-points so that MATH is homeomorphic to a pair-of-pants. We assume that as MATH, MATH is strictly increasing and contains any interior nonsingular point of MATH eventually. First, there exists a topological lamination MATH with this property. We can approximate it by a finite-length three-leaf topological lamination MATH with six (distinct) endpoints either in the boundary components of MATH or the cone-points. We assume that for each MATH, as MATH, MATH approximates MATH very closely in MATH-sense. Moreover, we assume that the endpoints of MATH winds around MATH along the chosen orientation infinitely as MATH. Since MATH is convex, there exists a geodesic lamination MATH homotopic to MATH by a homotopy fixing the endpoints. We may choose MATH for each MATH so that MATH meets it always in a union of six connected arcs near the six end points of MATH. Thus MATH is a finite-length finite lamination with endpoints in MATH by REF . Since MATH has six endpoints, it has three leaves. The length of MATH is bounded for fixed MATH as MATH. Otherwise the NAME limit set is a lamination in MATH containing an infinite leaf. This contradicts REF . Thus for each MATH, we can choose a subsequence so that MATH converges to a three-leaf finite-length finite lamination MATH. By using a diagonal subsequence argument, we obtain a sequence MATH so that MATH converges to such a lamination for each MATH. The direction of the winding is the same as the topological lamination by REF . The limit is now a geodesic lamination with three leaves. The intersection of the geodesic lamination with the compact set MATH has fixed topological type. Therefore the complement in MATH of the union of the boundary, the lamination, and the cone-points is precisely the union of two disjoint convex open disks, each of whose boundary are leaves of the lamination. Lifting to the universal covering convex domain, such disks develop to convex open triangles. |
math/0107193 | An end of any embedded infinite arc in MATH which lifts to a properly embedded arc in a universal cover winds around a simple closed curve parallel to a boundary component or a simple closed curve around a cone-point or end at a cone-point. (That is, its limit points comprise one of these.) Thus, MATH winds around a closed geodesic parallel to a boundary component. Since MATH is convex and with principal boundary and MATH can be covered by a convex surface with principal boundary, the closed curve is a boundary component (see CITE). |
math/0107193 | Since we repeat the proof of REF , we give a sketchy argument. Let a MATH-orbifold MATH, each component of which has negative NAME characteristic, be in a class MATH if the following hold: CASE: The deformation space of hyperbolic MATH-structures MATH is diffeomorphic to a cell of dimension MATH where MATH is the number of cone-points, MATH the number of corner-reflectors, MATH is the number of boundary full MATH-orbifolds. CASE: There exists a principal fibration MATH with the action by a cell of dimension MATH. Let MATH be a MATH-orbifold whose components are orbifolds of negative NAME characteristic, and it splits into an orbifold MATH in MATH. We suppose that REF hold for MATH, and show that REF hold for MATH. Since MATH eventually decomposes into a union of elementary MATH-orbifolds where REF hold, we would have completed the proof by REF . Since we need to match lengths and find gluing maps preserving lengths, the arguments are similar to the rest of the proof of REF . We need the result of REF for hyperbolic cases: CASE: Let the MATH-orbifold MATH be obtained from pasting along two closed curves MATH in a MATH-orbifold MATH. The map resulting from splitting MATH is a principal MATH-fibration, where MATH is the subset of MATH where MATH and MATH have equal invariants. CASE: Let MATH be obtained from MATH by cross-capping. The resulting map MATH is a diffeomorphism. CASE: Let MATH be obtained from MATH by silvering. The clarifying map MATH is a diffeomorphism. CASE: Let MATH be obtained from MATH by folding a boundary closed curve MATH. The unfolding map MATH is a principal MATH-fibration. CASE: Let MATH be obtained by pasting along two full MATH-orbifolds MATH and MATH in MATH. The splitting map MATH is a diffeomorphism where MATH is a subset of MATH where the invariants of MATH and MATH are equal. CASE: Let MATH be obtained by silvering or folding a full MATH-orbifold. The clarifying or unfolding map MATH is a diffeomorphism. Again, using REF adopted to each cases, we prove the theorem. |
math/0107199 | The case MATH is the one considered in CITE, and the other cases can be proved in exactly the same way. |
math/0107199 | First note that we are working in slow time. The estimate is classical for MATH, and starting from there, REF can be obtained by considering a partition of the interval MATH with spacing proportional to MATH. |
math/0107199 | We will only prove the case MATH, as the general case follows along the same lines. Let MATH for some constant MATH to be chosen later and set MATH. Note that, under our condition on MATH, we may assume MATH. Let MATH. We write the expectation of MATH as MATH, where MATH . The first term MATH can be estimated trivially, namely by MATH. To estimate the second term MATH, we employ integration by parts, thereby obtaining MATH . We now split the integral at MATH, MATH and MATH and estimate the resulting terms separately. By REF, MATH . Estimating the remaining part of the integral with the help of REF, MATH follows. Since the expectation of MATH is bounded above by MATH, REF follows from the fact that we can choose MATH large enough to bound all three terms by some constant times MATH. |
math/0107199 | Let us again focus on MATH. Estimate REF is obtained in the same way as REF, the only difference lying in a more elaborate bound on MATH. We use the integral representation MATH of the SDE REF (for MATH), thereby obtaining MATH . The second term on the right-hand side is bounded above by MATH. The first one can be estimated by bounding MATH uniformly in MATH, with the help of the estimate MATH, valid for all MATH, compare REF. The remaining integral behaves like MATH. Thus we obtain MATH while MATH can be estimated as before. Again choosing MATH for MATH large, yields Estimate REF. |
math/0107199 | Since REF are an immediate consequence of the fact that MATH is Gaussian with variance REF, we only need to prove REF. We restrict our attention to the case MATH as the case MATH even follows by an obvious adaptation and the case MATH odd is obtained from the case MATH even by an application of NAME 's inequality. First note that MATH . Estimating the expectation of the product by NAME 's inequality and REF follows. |
math/0107199 | Consider first the case MATH small. By REF, we have for any MATH . The first term on the right-hand side immediately yields the first term in REF due to the Gaussian nature of MATH. We denote MATH by MATH. For MATH, the second term can be bounded by MATH . We choose MATH by MATH and estimate both summands in REF by REF. Note that the first summand dominates the second one by our choice of MATH, since we assumed MATH. Thus we obtain the bound REF. For MATH large, we employ the trivial bound MATH together with Estimate REF, thereby obtaining REF. |
math/0107199 | First note that REF establishes the bound REF on the probability of a sample path crossing the potential barrier. From REF and the preceding lemma, one easily obtains MATH in the stable case, while MATH is trivial. Now the preceding results imply the stated estimates. |
math/0107199 | The first part is a direct consequence of REF with MATH, where MATH is chosen sufficiently small that the relation MATH for all MATH implies that MATH for these MATH. The second part is an application of REF (with MATH). Note that the theorem naturally extends to the case MATH. In fact, the integrand in REF should be the curvature of the potential at the deterministic solution tracking the saddle MATH, but the curvature behaves like MATH, compare CITE. |
math/0107199 | The fact that the drift term MATH has a negative second derivative with respect to MATH for all MATH implies that MATH is unlikely to exit the strip of width MATH through its upper boundary, as was proved for negative MATH in CITE. We also know by REF that MATH is unlikely to exit the strip through its lower boundary if MATH. The stopping time MATH has been defined in such a way that MATH cannot leave a strip of larger width before time MATH. |
math/0107199 | We decompose MATH where MATH is defined in REF. The first term on the right-hand side can be estimated as in REF , as there is no need to distinguish positive and negative deviations for this term. However, REF allows us to obtain bounds valid on a larger domain of MATH. Note that REF remains valid when MATH is replaced by MATH. This is a consequence of the MATH being monotone functions of MATH and a slightly more elaborate estimate showing that MATH obeys the same bound as was used for MATH in REF. Thus we obtain the estimate MATH valid for MATH. An application of REF shows that MATH, MATH, and we already know that MATH. Choosing MATH provides an estimate of the form REF. The second term on the right-hand side of REF can be estimated, using the monotonicity of MATH in MATH, by the relation MATH . The first term on the right-hand side can be estimated by REF , provided MATH is larger than some constant of order MATH. It decreases like MATH. Using the fact that MATH for MATH not too close to MATH (note that the contribution of MATH close to MATH is even smaller) and REF , we obtain that MATH . This last term is easily seen to dominate all others, so that REF is proved. To estimate deviations in the positive direction, we split terms as in REF. The first term can also be bounded by REF. Using the fact (compare CITE) that MATH it only remains to estimate MATH . The expectation of this term also decreases like MATH by REF . Taking REF into account, we have proved REF. |
math/0107199 | The proof being similar to the one of the previous proposition, we only outline the main steps. Introduce a stopping time MATH for some small MATH, compare the comment on the definition of MATH in the beginning of the subsection. We first need to control the behaviour of MATH . Observe that since MATH for MATH, the first term on the right-hand side is positive, while the second one is negative or zero. First note that if MATH is bounded above by MATH, then MATH cannot exceed a value of order MATH. Deviations of MATH in the positive direction can be bounded using a decomposition similar to REF and applying REF for MATH instead of MATH. We find MATH valid for MATH. Deviations of MATH in the negative direction can only be caused by the second term on the right-hand side of REF. However, there is no small lower bound for that term. The reason is that transitions to MATH are only probable in the window MATH. If this opportunity is missed, which happens with a probability of order MATH, then MATH keeps tracking MATH and MATH may reach negative values of order MATH. To complete the proof, we need to show that on MATH . Let MATH be the deterministic solution starting in MATH at time MATH. This solution is attractive, and thus REF holds with MATH instead of MATH as a consequence of REF . But the distance between MATH and MATH decreases exponentially in MATH, which implies that the area between them is at most of order MATH. Thus REF holds for MATH. But for smaller MATH, it is trivially satisfied. |
math/0107199 | By partial integration, MATH . The first integral be evaluated by splitting it at MATH, MATH and MATH, and the second one at MATH and MATH. Each time, the integral over the first interval dominates. The estimate REF is obtained similarly. |
math/0107199 | First note that REF is a direct consequence of REF as we are in the stable case. In order to prove REF, we consider again the stochastic process MATH, satisfying the SDE MATH where REF implies MATH for MATH. Note furthermore that MATH for MATH. By NAME 's inequality, it follows that MATH for MATH, where MATH is defined as the solution of the time-homogeneous SDE MATH . For any MATH, we can write MATH . Since REF is an autonomous SDE, it is easy to see that the first term on the right-hand side can be bounded by MATH which can be made as small as we like by taking MATH sufficiently large. In order to estimate the second term, we introduce MATH, where MATH will be chosen later, and define MATH . Using time homogeneity and the NAME property, we can write MATH . The result is thus proved if we manage to bound MATH by a term exponentially small in MATH. In order to estimate MATH, it is convenient to introduce the process MATH, which obeys the SDE MATH . Let MATH and MATH. Using again NAME property and time-homogeneity shows that MATH, where MATH . Since MATH, each term can be easily estimated by comparison with an appropriate linear or MATH-independent equation. Consider for instance MATH. We know that MATH lies above the solution MATH of the linear SDE MATH the solution of which at time MATH is a Gaussian random variable with mean MATH and variance MATH. We can thus estimate MATH provided MATH, i. e., MATH. Now, MATH and MATH allow for similar bounds, and the result thus follows from REF, taking MATH and MATH sufficiently large. |
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