paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0104059 | Here we essentially ignore NAME 's construction of a symplectic MATH-handle and instead build the MATH-handle from scratch. Let MATH be the standard convex filling of MATH and let MATH be the standard convex filling of MATH, as in REF . We can clearly construct MATH and MATH so that MATH and so that MATH is a MATH-dimensional symplectic MATH-handle. From the standard model it is clear that there exists a contactomorphism from a neighborhood MATH of the binding MATH to a neighborhood MATH of the binding MATH. Furthermore we can arrange that the handle MATH is attached inside MATH. Thus MATH can just as well be attached to a cobordism from MATH to MATH to get a cobordism from MATH to MATH. |
math/0104059 | Let MATH. Use the standard model as described in REF , so that MATH is Legendrian, with MATH. CITE shows that we can attach a symplectic handle along an arbitrarily small neighborhood of MATH with framing MATH, so that we get a cobordism from MATH to a new contact MATH-manifold MATH. As mentioned earlier, we know that MATH, and we need to show that MATH is supported by this open book decomposition. Split MATH into the sets MATH and MATH as above; since the surgery that produces MATH from MATH is localized near MATH, we get a corresponding NAME splitting of MATH into two handlebodies MATH and MATH, with MATH. Since MATH and MATH also agree on a neighborhood of MATH, we still have that the splitting surface MATH is convex with dividing set equal to the binding MATH. Thus, by REF , we need only show that MATH is tight to complete the proof. Let MATH be the contactomorphic embedding of MATH into MATH. We can also attach a symplectic handle along MATH to the standard convex filling of MATH to get a convex filling of a new contact MATH-manifold MATH, which is therefore tight. Since the contact surgery along MATH must be the same as the surgery along MATH, we see that MATH embeds contactomorphically in MATH. Therefore MATH is tight. |
math/0104059 | Let MATH, with binding MATH and fibration MATH. REF asks us to attach a handle along each component MATH of MATH with framing MATH. REF tells us that we get a symplectic cobordism from MATH to MATH. We must show that MATH is supported by the open book decomposition MATH. Because the framing of the surgery is MATH for each component MATH, the fibers of MATH still meet MATH as longitudes, so that we do have an honest open book decomposition of MATH which supports MATH and with pages diffeomorphic to MATH. To compute the monodromy, note that a meridian before the surgery becomes a longitude with framing MATH, so that our original monodromy MATH is now measuring the monodromy via a flow which is longitudinal near MATH. Correcting this flow to be meridinal changes the monodromy by a left twist on each boundary component, so that, properly measured, the monodromy for MATH is MATH. We should reverse the flow to get the monodromy for MATH, which is thus MATH. |
math/0104061 | Suppose that MATH is a MATH-torus knot then MATH hence the first inequality (with equality only in the case of torus knots). For the second, MATH and note that equality occurs precisely when MATH is a MATH-torus knot. |
math/0104061 | Using the notation of the previous proof, MATH with equality if and only if MATH is a MATH torus knot. |
math/0104061 | If MATH is a MATH-torus knot, then a minimal amount of manipulation gives MATH with equality if and only if MATH is a MATH-torus knot. For the right hand bound, firstly, let MATH and MATH be distinct positive integers, then MATH, so MATH and thus MATH, with equality precisely when MATH and MATH differ by one. Now for MATH a MATH-torus knot, MATH by putting MATH in the above paragraph. Note that equality occurs precisely when MATH is a MATH-torus knot. |
math/0104061 | This is easily verified; note that if MATH is a MATH-torus knot then MATH and MATH. |
math/0104061 | Suppose that MATH is a MATH-torus knot with MATH - this just avoids excessive modulus signs in the calculation - then for the left hand bound, MATH and equality occurs precisely when MATH is a MATH-torus knot. For the right hand bound, MATH and claim that this is non-negative and is zero precisely when MATH. To prove the claim, note MATH as MATH, and so also MATH thus, by taking square roots, MATH from which the claim follows on multiplying through by MATH. |
math/0104062 | By the NAME - NAME Vanishing Theorem (compare CITE and CITE) it suffices to show that MATH is big and nef, i. e. we have to show: CASE: MATH, and CASE: MATH for any irreducible curve MATH in MATH. Note that MATH, and thus by REF we have MATH which gives REF . For REF we observe that an irreducible curve MATH on MATH is either the strict transform of an irreducible curve MATH in MATH or is one of the exceptional curves MATH. In the latter case we have MATH . We may, therefore, assume that MATH is the strict transform of an irreducible curve MATH on MATH having multiplicity MATH at MATH, MATH. Then MATH and thus REF is equivalent to CASE: MATH. Since MATH is in very general position REF applies in view of REF . Using the NAME Index Theorem, REF , and the NAME Inequality we get the following sequence of inequalities: MATH where MATH is such that MATH. Since MATH is nef, REF is satisfied as soon as we have MATH . If this is not fulfilled, then MATH for all MATH, and thus MATH . Hence, for the remaining considerations REF may be replaced by the worst case MATH . Note that since the MATH are in very general position and MATH we have that MATH and MATH (compare REF ). If MATH then we are done by the NAME Index Theorem and REF , since MATH is nef: MATH . It remains to consider the case MATH which is covered by REF . |
math/0104062 | We stick to the convention MATH and MATH for MATH, and we set MATH for MATH and MATH. Let MATH be a small disc around MATH with coordinate MATH, and choose coordinates MATH on MATH around MATH such that CASE: MATH for MATH with MATH, CASE: MATH, and CASE: MATH, where MATH locally at MATH (for MATH). We view MATH as a non-trivial deformation of MATH, which implies that the image of MATH under the NAME map is a non-zero section MATH of MATH. MATH is locally at MATH given by MATH. Show that MATH, which are the stalks of MATH at the MATH, and hence MATH is actually a global section of the subsheaf MATH. Set MATH. By assumption for any MATH the multiplicity of MATH at MATH is at least MATH, i. CASE: MATH for every fixed complex number MATH. Hence, MATH for every MATH. On the other hand we have MATH . Since MATH, we have MATH, and hence MATH. For this note that MATH, if MATH. |
math/0104062 | Set MATH. We start with the structure sequence for MATH: MATH NAME with the locally free sheaf MATH and then applying MATH we get a morphism: MATH . Now tensoring by MATH over MATH we have an exact sequence: MATH . And finally taking global sections, we end up with: MATH . Since the sheaves we look at are actually MATH-sheaves and since MATH is a closed subscheme of MATH, the global sections of the sheaves as sheaves on MATH and as sheaves on MATH coincide (compare CITE III. REF - for more details, see REF ). Furthermore, MATH. Thus it suffices to show that MATH. Since MATH is an isomorphism, we have that MATH is finite. Hence, by REF , MATH . Let now MATH be given. We have to show that MATH, i. CASE: MATH for every MATH. If MATH, then MATH. Thus we may assume MATH. As we have shown, MATH where MATH is a local equation of MATH at MATH. Therefore, there exists a MATH such that MATH (note that MATH!). But then MATH is just the residue class of MATH in MATH, and is thus zero. |
math/0104062 | Using the notation of the idea of the proof given on page REF, we have, by REF , a non-zero section MATH. This lifts under the surjection MATH to a section MATH which is not in the kernel of MATH. Again setting MATH, by REF , we have a non-zero section MATH, where by the projection formula the latter is just MATH. Since MATH has a global section and since MATH is irreducible and reduced, we get by REF : MATH . |
math/0104062 | W. l. o. CASE: MATH for all MATH. For the convenience of notation we set MATH and MATH. Since MATH is very ample, we may choose smooth curves MATH through MATH and MATH for MATH (compare REF ). Writing MATH for MATH we introduce zero-dimensional schemes MATH for MATH by MATH . MATH. By REF we get MATH and the exact sequence MATH implies with the aid of REF MATH . REF allow us to apply REF in order to obtain MATH . For each MATH there exists a curve MATH with an ordinary singular point of multiplicity MATH at MATH and with MATH for MATH. Consider the exact sequence MATH twisted by MATH and the corresponding long exact cohomology sequence MATH . Thus we may choose the MATH to be given by a section in MATH where the MATH tangent directions at MATH are all different. The base locus of MATH is MATH. Suppose MATH was an additional base point and define the zero-dimensional scheme MATH by MATH . Choosing a generic, and thus smooth, curve MATH through MATH we may deduce as in REF MATH and thus as in REF MATH . But by REF is a base point, and thus MATH which gives us the desired contradiction. MATH with an ordinary singular point of multiplicity MATH at MATH for MATH and no other singular points. Because of REF the generic element in MATH has an ordinary singular point of multiplicity MATH at MATH and is by NAME 's Theorem (compare CITE III. REFEF) smooth outside its base locus. For two generic curves MATH the intersection multiplicity in MATH is MATH. Thus, if REF is fulfilled then MATH and MATH have an additional intersection point outside the base locus of MATH, and NAME 's Theorem (compare REF, NAME REF) implies that the generic curve in MATH is irreducible. MATH, by REF . MATH is NAME at MATH. By REF , we have MATH and thus by REF MATH which proves the claim. |
math/0104062 | Follows from REF . |
math/0104062 | Suppose REF was not satisfied, then MATH . Hence, MATH which implies that REF is sufficient. |
math/0104062 | NAME MATH. Consider the following exact sequence: MATH . Since MATH, the long exact cohomology sequence gives MATH where MATH are local coordinates of MATH. We, therefore, can find a basis MATH of MATH, with MATH, such that CASE: MATH is the curve defined by MATH, CASE: MATH for MATH, MATH, CASE: MATH . Let us now denote the coordinates of MATH w. r. t. this basis by MATH with MATH and MATH. Thus the family MATH parametrises MATH. By the definition of MATH and since MATH, we may choose good representatives MATH for the MATH, MATH. Let MATH and MATH. We should remark here that for any MATH the polynomial MATH is also a good representative, and thus, replacing MATH by MATH, we may assume that the MATH are arbitrarily close to MATH. We are going to glue the good representatives for the MATH into the curve MATH. More precisely, we are constructing a subfamily MATH, MATH, in MATH such that, if MATH denotes the curve defined by MATH, CASE: MATH are the only singular points of the irreducible reduced curve MATH, and they are ordinary singularities of multiplicities MATH, for MATH, and MATH, for MATH respectively, CASE: locally in MATH, MATH, the MATH, for small MATH, can be transformed into members of a fixed MATH-equisingular family, CASE: while for MATH and MATH small MATH has an ordinary singularity of multiplicity MATH in MATH. ``NAME and first reduction for a local investigation. Let us make the following ``NAME: MATH . This gives rise to a family MATH with MATH and MATH where MATH with MATH. Fixing MATH, in local coordinates at MATH the family looks like MATH with MATH . For MATH the transformation MATH is indeed a coordinate transformation, and thus MATH is contact equivalent to MATH . Note that for this new family in MATH we have MATH and hence it gives rise to a deformation of MATH. Reduction to the investigation of the equisingular strata of certain families of polynomials. It is basically our aim to verify the MATH as convergent power series in MATH such that the corresponding family is equisingular. However, since the MATH are power series in MATH and MATH, we cannot right away apply the NAME property of MATH, but we rather have to reduce to polynomials. For this let MATH be the determinacy bound of MATH and define MATH . Thus MATH is a family in MATH, and still MATH . We claim that it suffices to find MATH with MATH, such that the families MATH, MATH, are in the equisingular strata MATH, for MATH. Since then we have, for small MATH, MATH by the MATH-determinacy and since MATH is a coordinate change for MATH, which proves REF . Note that the singular points MATH will move with MATH. It remains to verify REF . Setting MATH, for MATH, we find that MATH is an element inside the linear system MATH, where MATH. Locally at MATH, MATH, MATH induces a deformation of MATH with equations MATH and MATH respectively. Thus any element of MATH has ordinary singularities of multiplicity MATH at MATH for MATH, and since MATH has an ordinary singularity of multiplicity MATH at MATH for MATH, so has a generic element of MATH. Moreover, a generic element of MATH has not more singular points than the special element MATH and has thus singularities precisely in MATH. Replacing the MATH by some suitable multiples, we may assume that the curve defined by MATH is a generic element of MATH, which proves REF . Similarly, we note that MATH in local coordinates at MATH, for MATH, looks like MATH . MATH and thus, for MATH sufficiently small, the singularity of MATH at MATH will be an ordinary singularity of multiplicity MATH, which gives REF . Find MATH with MATH, MATH, MATH, such that the families MATH, MATH, are in the equisingular strata MATH, for MATH. In the sequel we adopt the notation of REF adding indices MATH in the obvious way. Since MATH is NAME at MATH, for MATH, there exist MATH and power series MATH, for MATH, such that the equisingular stratum MATH is given by the MATH equations MATH . Setting MATH we use the notation MATH and, similarly MATH, MATH, MATH, MATH, and MATH. Moreover, setting MATH, we define an analytic map germ MATH by MATH and we consider the system of REF . One easily verifies that MATH . Thus by the Inverse Function Theorem there exist MATH with MATH such that MATH . Now, setting MATH, the families MATH are in the equisingular strata MATH, for MATH. It finally remains to show that MATH, for small MATH, has no other singular points than MATH. Since for any MATH the family MATH, MATH, induces a deformation of the singularity MATH there are, by the Conservation of NAME Numbers (compare REF), (Euclidean) open neighbourhoods MATH and MATH such that for any MATH CASE: MATH, i. e. singular points of MATH come from singular points of MATH, CASE: MATH, MATH. For MATH REF implies MATH and thus MATH must be the only critical point of MATH in MATH, in particular, MATH . Let now MATH. For MATH fixed, we consider the transformation defined by the coordinate change MATH, MATH and the transformed equations MATH . REF then implies, MATH . For MATH very small MATH becomes very large, so that, by shrinking MATH we may suppose that for any MATH and that for any MATH there is an open neighbourhood MATH such that MATH . If we now take into account that MATH has precisely one critical point, MATH, on its zero level, and that the critical points on the zero level of MATH all contribute to the NAME number MATH, then we get the following sequence of inequalities: MATH . Hence all inequalities must have been equalities, and, in particular, MATH which in view of REF finishes the proof. Note that MATH, being a small deformation of the irreducible reduced curve MATH, will again be irreducible and reduced. |
math/0104062 | This follows right away from REF , and REF . |
math/0104062 | See CITE p. CASE: MATH is a locally trivial MATH-bundle, thus MATH is covered by a finite number of open affine subsets MATH with trivialisations MATH which are linear on the fibres. The three disjoint sections on MATH, say MATH, MATH, and MATH, give rise to three sections MATH, MATH, and MATH on MATH. For each point MATH there is a unique linear projectivity on the fibre MATH mapping the three points MATH, MATH, and MATH to the standard basis MATH, MATH, and MATH of MATH. If MATH, MATH, and MATH, the projectivity is given by the matrix MATH whose entries are rational functions in the coordinates of MATH, MATH, and MATH. Inserting for the coordinates local equations of the sections, MATH finally gives rise to an isomorphism of MATH-bundles MATH mapping the sections MATH, MATH, and MATH to the trivial sections. The transition maps MATH with MATH, are linear on the fibres and fix the three trivial sections. Thus they must be the identity maps, which implies that the MATH, MATH, glue together to an isomorphism of ruled surfaces: MATH . |
math/0104062 | pt CASE: The adjunction formula gives MATH . Since MATH covers the whole of MATH and MATH, the two irreducible curves MATH and MATH must intersect properly, that is, MATH is not a fibre of MATH. But then the mapping MATH is a finite surjective morphism of degree MATH. If MATH was a singular curve its normalisation would have to have arithmetical genus MATH and the composition of the normalisation with MATH would give rise to a surjective morphism from MATH to an elliptic curve, contradicting NAME 's formula. Hence, MATH is smooth and MATH. We thus may apply REF to MATH and the degree of the ramification divisor MATH turns out to be MATH . The remaining case is treated analogously. CASE: W. l. o. CASE: MATH. For MATH we have MATH is a fibre of MATH, and since MATH is unramified, MATH. Suppose MATH with MATH. Then MATH, and hence MATH, which contradicts the assumption MATH. CASE: Since MATH, REF imply that MATH. CASE: W. l. o. CASE: MATH. Let MATH. We claim that MATH, and hence MATH. Suppose MATH, then there is a MATH such that MATH, i. CASE: MATH and MATH. Hence, MATH. But, MATH, and thus MATH in contradiction to the choice of MATH. MATH via MATH follows from REF . CASE: By REF we have MATH. CASE: MATH is an isomorphism, and has thus degree one. But MATH. Thus we are done with REF . |
math/0104062 | pt CASE: MATH if and only if MATH or MATH. CASE: Let us first consider the case that MATH is irreducible. If MATH or MATH, then MATH is algebraically equivalent to a multiple of a fibre of one of the projections MATH, MATH. In this situation MATH and thus the irreducible curve MATH does not intersect any of the fibres properly. Hence it must be a union of several fibres, and being irreducible it must be a fibre. That is we are in one of the first two cases. Suppose now that MATH. Thus MATH intersects MATH properly, and MATH and MATH. It now remains to show that the mentioned algebraic systems contain irreducible curves, which is clear for the first two of them. Let therefore MATH and MATH be positive. Then obviously the linear system MATH contains no fixed component, and being ample by REF its general element is irreducible according to REF . CASE: Follows from REF . CASE: By REF is nef if and only if MATH for every irreducible curve MATH. Thus the claim is an immediate consequence of REF . CASE: Since by the NAME ampleness depends only on the numerical class of a divisor, we may assume that MATH. Moreover, by REF MATH is ample if and only if MATH and MATH. If MATH, then MATH and the effective divisor MATH, thus MATH is ample. Conversely, if MATH is ample, then MATH and MATH, thus MATH. CASE: By REF MATH is very ample. If MATH, then the system MATH is basepoint free, which is an immediate consequence of the existence of the translation morphisms MATH, MATH. But then MATH is globally generated and MATH is very ample. Conversely, if MATH, then MATH is a divisor of degree MATH on the elliptic curve MATH and hence not very ample (compare REF ). But then MATH is not very ample. Analogously if MATH. |
math/0104062 | Show first that for MATH is a closed subset of MATH, where MATH. Being the reduction of a connected component of the NAME scheme MATH, MATH is a projective variety endowed with a universal family of curves, giving rise to the following diagram of morphisms MATH where MATH is an effective NAME divisor on MATH with MATH. Let MATH be a global section defining MATH. Then MATH . We may consider a finite open affine covering of MATH of the form MATH, MATH and MATH open, such that MATH is locally on MATH given by one polynomial equation, say MATH . It suffices to show that MATH is closed in MATH for all MATH. However, for MATH we have MATH if and only if MATH . Thus, MATH is a closed subvariety of MATH. MATH is a closed subset of MATH. By REF for MATH the set MATH is a closed subset of MATH. Considering now MATH we find that MATH, being the image of a closed subset under a morphism between projective varieties, is a closed subset of MATH (compare REF ). |
math/0104062 | Fixing some embedding MATH and MATH, MATH is a projective variety and has thus only finitely many connected components. Thus the NAME scheme MATH has only a countable number of connected components, and we have only a countable number of different MATH, where MATH runs through MATH and MATH through MATH. By REF the sets MATH are closed, hence their complements MATH are open. But then MATH is an at most countable intersection of open dense subsets of MATH, and is hence very general. |
math/0104062 | pt CASE: By REF is a connected component of MATH, whose number is at most countable. If in addition MATH, then MATH which proves the claim. CASE: Curves of negative self-intersection are not algebraically equivalent to any other curve (compare CITE p. REF). CASE: Follows from REF . |
math/0104062 | It suffices to find an infinite number of irreducible curves MATH in MATH such that MATH and MATH where MATH and MATH, since the expression in REF is the self intersection of the strict transform MATH of MATH and REF gives its arithmetical genus. In particular MATH cannot contain any singularities, since they would contribute to the arithmetical genus, and, being irreducible anyway, MATH is an exceptional curve of the first kind. We are going to deduce the existence of these curves with the aid of quadratic NAME transformations. If for some MATH and MATH with MATH there is an irreducible curve MATH, then MATH is an irreducible curve, where CASE: MATH are such that MATH, CASE: MATH is the quadratic NAME transformation at MATH, CASE: MATH CASE: MATH CASE: MATH . Note that, MATH, i. e. we may iterate the process since the hypothesis of the claim will be preserved. Since MATH, there must be a triple MATH such that MATH. Let us now consider the following diagram MATH and let us denote the exceptional divisors of MATH by MATH and those of MATH by MATH. Moreover, let MATH be the strict transform of MATH under MATH and let MATH be the strict transform of MATH under MATH. Then of course MATH, and MATH, being the projection MATH of the strict transform MATH of the irreducible curve MATH, is of course an irreducible curve. Note that the condition MATH ensures that MATH is not one of the curves which are contracted. It thus suffices to verify MATH and MATH . Since outside the lines MATH, MATH, and MATH the transformation MATH is an isomorphism and since by hypothesis none of the remaining MATH belongs to one of these lines we clearly have MATH for MATH. Moreover, we have MATH . Analogously for MATH and MATH. Finally we find MATH . This proves the claim. Let us now show by induction that for any MATH there is an irreducible curve MATH of degree MATH satisfying REF . For MATH the line MATH through MATH and MATH gives the induction start. Given some suitable curve of degree MATH the above claim then ensures that through points in very general position there is an irreducible curve of higher degree satisfying REF , since MATH. Thus the induction step is done. |
math/0104062 | Since there is at least one connected reduced fibre MATH, semicontinuity of flat, proper morphisms in the version CITE IV. REFEF implies that there is an open dense subset MATH such that MATH is connected and reduced, hence a single reduced point, MATH. (MATH dense in MATH is due to the fact that MATH is irreducible.) Thus the assumptions are stable under restriction to open subschemes of MATH, and since the claim that we have to show is local on MATH, we may assume that MATH is affine. Moreover, MATH being finite, thus affine, we have MATH is also affine. Since MATH is flat it is open and hence dominates the irreducible affine variety MATH and, therefore, induces an inclusion of rings MATH. It now suffices to show: MATH is an isomorphism. By assumption there exists a MATH such that MATH is a single point with reduced structure. In particular we have for the multiplicity of MATH over MATH which implies that MATH is an isomorphism. Hence by NAME 's Lemma also MATH is an isomorphism, that is, MATH is free of rank REF over MATH. MATH being locally free over MATH, with MATH an integral domain, thus fulfils MATH is an isomorphism for all MATH, and hence the claim follows. |
math/0104062 | Considering points in the intersections of the finite number of irreducible components of MATH we can reduce to the case MATH irreducible. NAME (compare CITE III. REFEF) gives a factorisation of MATH of the form MATH with CASE: MATH connected (i. e. its fibres are connected), CASE: MATH finite, CASE: MATH locally free over MATH, since MATH is flat, and CASE: MATH is connected and reduced, i. e. a single reduced point. Because of REF it suffices to show that MATH is connected, and we claim that they are reduced as well. Since MATH is finite REF is equivalent to saying that MATH is flat. Hence MATH fulfils the assumptions of REF , and we conclude that MATH and the proposition follows from CITE III. REF Alternatively, from CITE IV. REFEF it follows that there is an open dense subset MATH such that MATH is connected for all MATH. Since, moreover, by the same theorem the number of connected components of the fibres is a lower semi-continuous function on MATH the special fibres cannot have more connected components than the general ones, that is, all fibres are connected. |
math/0104062 | Consider the universal family MATH over the connected projective scheme MATH. Since the projection MATH is a flat projective morphism, and since the fibre MATH is connected and reduced, the result follows from REF . |
math/0104062 | Suppose MATH, then the NAME polynomials of MATH and MATH are different in contradiction to MATH. |
math/0104062 | Since MATH is irreducible by REF MATH and MATH do not have a common component. Suppose MATH, then MATH in contradiction to MATH. |
math/0104062 | By REF MATH is connected. We are going to apply CITE I. REF, and thus we have to verify three conditions. CASE: MATH for all MATH by REF . Thus MATH is an element of the annihilator of MATH with MATH for all MATH. CASE: MATH for all MATH. CASE: Since MATH is connected there is no non-trivial partition MATH of MATH such that MATH for all MATH and MATH. Thus CITE I. REF implies that MATH is positive semi-definite. |
math/0104062 | Suppose MATH and MATH such that MATH and MATH have no common component. We have MATH and thus MATH which implies that MATH where each summand on the left hand side is less than or equal to zero by REF , and the right hand side is greater than or equal to zero, since the curves MATH and MATH have no common component. We thus conclude MATH . But then again REF implies that MATH and MATH, that is, MATH and MATH have no common component. Suppose MATH, then MATH would be a contradiction to MATH. Hence, MATH. |
math/0104062 | MATH is an irreducible projective surface and MATH is surjective. The universal property of MATH implies that MATH is an effective NAME divisor of MATH, and thus in particular projective of dimension at least MATH. Since MATH is projective, its image is closed in MATH and of dimension REF, hence it is the whole of MATH, since MATH is irreducible. By REF the fibres of MATH are all single points, and thus, by REF , MATH is irreducible. Moreover, MATH . MATH . MATH is irreducible. Let MATH be any irreducible component of MATH of dimension one, then we have a universal family over MATH and the analogue of REF for MATH shows that the curves in MATH cover MATH. But then by REF there can be no further curve in MATH, since any further curve would necessarily have a non-empty intersection with one of the curves in MATH. |
math/0104062 | Since MATH and MATH are irreducible and reduced, and since MATH is surjective, we may apply CITE III. REF, and thus there is an open dense subset MATH such that MATH is smooth. Hence, in particular MATH is flat and the fibres are single reduced points. Since MATH is projective and quasi-finite, it is finite (compare REF ), and it follows from REF that MATH is an isomorphism onto its image, i. CASE: MATH is birational. |
math/0104062 | By REF the fibres of MATH over the possible points of indeterminacy of MATH are just points, and thus the result follows from REF . |
math/0104062 | We just have MATH. |
math/0104062 | Let MATH be the normalisation of the irreducible curve MATH. Then MATH is a smooth irreducible curve. Moreover, since MATH is irreducible and smooth, and since MATH is surjective, MATH factorises over MATH, i. e. we have the following commutative diagram MATH . Then MATH is the desired fibration. |
math/0104062 | We write the power series as MATH. MATH for every MATH implies MATH . The identity theorem for power series in MATH then implies that MATH . |
math/0104062 | We denote by MATH and MATH respectively the given embeddings. CASE: By REF we have: MATH . By the projection formula this is just equal to: MATH . CASE: Using REF we get: MATH . |
math/0104062 | Since MATH is an isomorphism, we have for any sheaf MATH of MATH-modules and MATH: MATH . In particular, MATH and MATH . Moreover, the morphism MATH becomes under these identifications just the morphism given by MATH, which is injective. Thus, MATH, and MATH. |
math/0104062 | The multiplication by MATH gives rise to the following exact sequence: MATH . Since MATH and MATH are coherent, so is MATH, and hence MATH is closed in MATH. Now, MATH . But then the complement MATH is open and is contained in MATH (since MATH implies that MATH), and is thus empty since MATH is irreducible and MATH of lower dimension. |
math/0104062 | Considering the embedding into MATH defined by MATH there is a MATH-dimensional family of hyperplane sections going through two fixed points of MATH, which in local coordinates w. l. o. g. is given by the family of REF. Since the local analytic rings of MATH in every point are smooth, hence, in particular complete intersections, they are given as MATH modulo some ideal generated by MATH power series MATH forming a regular sequence. Thus, having MATH free indeterminates in our family MATH of equations, a generic equation MATH will lead to a regular sequence MATH, i. e. the hyperplane section defined by MATH is smooth in each of the two points, and thus everywhere. |
math/0104062 | REF ) implies MATH and thus by NAME (compare REF MATH . Consider now the exact sequence MATH . The result then follows from the corresponding long exact cohomology sequence MATH . |
math/0104062 | As already mentioned, if either MATH or MATH is zero, then we may take MATH. Suppose that MATH. Given an elliptic curve MATH there is a countable union MATH of proper subvarieties of MATH such that for any MATH the NAME number of MATH is two - namely, if MATH and MATH denote the periods as in REF, then we have to require that there exists no invertible integer matrix MATH such that MATH. Compare also CITE p. CASE: We, therefore, may assume that MATH and MATH. The claim then follows from REF , which is due to NAME. |
math/0104062 | We note that a curve MATH with MATH induces a non-trivial morphism MATH, where MATH, MATH, denote the canonical projections. It thus makes sense to study the moduli problem of (non-trivial) maps from curves of genus MATH into MATH. More precisely, let MATH and let MATH be given, where MATH is the NAME variety of divisors of degree MATH on MATH. Following the notation of CITE we denote by MATH the moduli space of pairs MATH, where MATH is a smooth projective curve of genus MATH and MATH a morphism with MATH. We then have the canonical morphism MATH just forgetting the map MATH, and the proposition reduces to the following claim: For no choice of MATH and MATH the morphism MATH is dominant. Let MATH be any morphism with MATH. Then MATH is not a contraction and the image of MATH is a projective curve in the abelian variety MATH. Moreover, we have the following exact sequence of sheaves MATH . Since MATH is a non-zero inclusion, its dual MATH is not zero on global sections, that is MATH is not the zero map. Since MATH we have MATH, and thus MATH has global sections. Therefore, the induced map MATH is not the zero map, which by NAME duality gives that the map MATH from the long exact cohomology sequence of REF is not zero. Hence the coboundary map MATH cannot be surjective. According to CITE p. REF we have MATH . But if the differential of MATH is not surjective, then MATH itself cannot be dominant. |
math/0104065 | We will show that both of the triples MATH and MATH are equivalent to a triple MATH, defined by an orientation-preserving diffeomorphism MATH. We start with the ``half" MATH of REF : handle-sliding of each node-component over the corresponding leaf-component, followed by some isotopies of framed links gives REF , where only part of the link is drawn. Up to a MATH-framing correction, the three depicted components MATH can be normally pushed off at once towards the boundary: we obtain three disjoint curves MATH on MATH. Note that during this push-off, none of the three components MATH is intersected. Then, the components MATH can also be pushed off so that the framing correction is now MATH: the result is a family of three disjoint curves MATH. After a convenient isotopy of the handles, the curves MATH and MATH can be depicted as on REF . We define MATH, where MATH and MATH are the following composites of (commuting) NAME twists: MATH . According to the NAME trick CITE, a MATH-surgery is therefore equivalent to a MATH-surgery. On the other hand, from REF , we deduce that a MATH-surgery is equivalent to a MATH-surgery where MATH. Let MATH denote the meridian of the MATH handle of MATH, for MATH. Then, the equation: MATH holds for both MATH and MATH so that MATH. |
math/0104065 | According to REF , MATH and MATH are diffeomorphic. Let us identify MATH. Recall that MATH was defined as: MATH . Write MATH as MATH where MATH (respectively, MATH) is the sublink containing the leaf(respectively, node)-components. Note that MATH can be isotoped in MATH to some Borromean rings contained in a MATH-ball disjoint from MATH: make MATH leave the handlebody where it was lying, towards the handlebody with the minus sign in REF . So MATH is obtained from MATH by surgery on some Borromean rings contained in a little MATH-ball. The lemma then follows from the fact that the latter is nothing but MATH. |
math/0104065 | For MATH or MATH, let MATH be the principal MATH-bundle derived from MATH and the inclusion of groups: We still denote by MATH the canonical map from MATH to MATH. Then, is an isomorphism. The bundle MATH can be identified with MATH. In particular, there is an inclusion map MATH. Moreover, the diffeomorphism MATH induces a further identification: such that the total space MATH is homeomorphic to the glueing: MATH . We denote by MATH the corresponding inclusion of MATH into MATH. But now, by the NAME sequence, we have: Note that MATH sends each MATH to MATH, while MATH sends each MATH to MATH. Note also that, since MATH is connected, the map MATH is injective. The whole lemma then follows from these two remarks and from the exactness of the NAME sequence. |
math/0104065 | In that case, MATH is a handlebody and so MATH is zero. Therefore, the restriction map MATH is injective. |
math/0104065 | The equality MATH is obvious. By construction, both of MATH and MATH are extensions of MATH. The lemma then follows from the fact that the restriction map: is injective since the relative cohomology group MATH is zero. |
math/0104065 | According to REF , we can think of the MATH-torus as: MATH . The surgered manifold MATH will be thought of concretely as in REF . Pick a spin MATH-manifold MATH spin-bounded by MATH, and a spin MATH-manifold MATH spin-bounded by the MATH-torus with MATH as a spin structure. Glue the ``generalized" handle MATH to MATH along the first handlebody of the MATH-torus in decomposition REF , using MATH as glue. We obtain a MATH-manifold MATH. Orient MATH coherently with MATH and MATH, and then give to MATH the spin structure obtained by glueing those of MATH and MATH (see REF ). It follows from definitions that the spin-boundary of MATH is MATH. According to NAME theorem on non-additivity of the signature (see CITE), we have: MATH . The involved correcting term is the signature of a real bilinear symmetric form explicitely described by NAME. It is defined by means of the intersection form in MATH, with domain: MATH where MATH,MATH,MATH are subspaces of MATH defined to be respectively the kernels of: No matter who is MATH, since the Borromean diffeomorphism MATH lies in the NAME group, we certainly have MATH. The space MATH then vanishes and so does the correcting term. The announced equality then follows by taking REF modulo MATH. |
math/0104065 | Last statement is clear from REF and from the fact that the NAME function of the MATH-torus takes values in MATH. Let us show that the NAME invariant is at most of degree MATH. Take a closed spin MATH-manifold MATH and two disjoint MATH-graphs MATH and MATH in MATH. According to REF , in order to show that: MATH it suffices to show that: MATH where the left MATH is defined by MATH and the right MATH is defined by MATH. But this follows from definition of the maps MATH and from the fact that MATH extends MATH. It remains to show that the NAME invariant is not of degree MATH (and so it will be ``exactly" of degree MATH). For instance, all of the spin structures of the MATH-torus are related one to another by MATH-surgeries (Compare REF below) and so are not distinguished one to another by degree MATH invariants. But NAME distinguishes one of them from the others. |
math/0104065 | In the following, all (co)homology groups are assumed to be with coefficients in MATH. We use the above fixed notations. Let us consider the map: where MATH is the obstruction to extend any MATH to the whole of MATH. Let also MATH denote the connecting homomorphism for the pair MATH. Note that the following equation holds: MATH . Since MATH is injective, it follows that MATH is injective. The same map MATH can be defined for MATH relatively to MATH, and for MATH and MATH relatively to MATH. We have thus the following commutative diagram: where the letter MATH stands for a NAME duality isomomorphism, the vertical arrows are induced by inclusions and the map MATH is defined by planar commutativity. From intersection theory, we deduce that: MATH . Let now MATH be a spin structure of MATH such that the corresponding characteristic solution MATH of MATH satisfies REF. We define MATH. By REF , there exists a unique spin structure MATH of MATH with MATH as associated characteristic solution of MATH. We want to show that MATH. Diagram chasing shows that proving MATH should suffice. This follows from REF and from the fact that MATH and MATH. |
math/0104065 | This follows from the definitions and from REF : note that MATH is nul-homologous in MATH, and that here MATH is merely a handlebody. |
math/0104065 | Replace in the lhs of REF , this simple MATH-graph by REF-component link of REF such that MATH links the MATH component of MATH and use REF to obtain the intermediate link of REF . Perform then some spin NAME moves to obtain the right-hand side of REF . |
math/0104065 | The two spin structures of MATH are equivalent, so are the eight ones of MATH. Furthermore, MATH can be obtained from MATH by surgery along a trivial MATH-framed three-component link. Surgery on the MATH-framed Borromean rings gives rise to the MATH-torus MATH, and this link can be obtained from the trivial link by a simple MATH-surgery. |
math/0104065 | Since MATH, we have: MATH . |
math/0104065 | Let MATH be a quadratic form on a finite MATH-group MATH, going with a linking pairing MATH. Let also MATH be a special linking pairing, giving rise to MATH by NAME construction MATH. We want to compare the invariants MATH and MATH of MATH and MATH, in order to quantify how much MATH determines MATH, and so MATH, up to isomorphism. Recall that MATH. Denote by MATH the canonical projection. The map MATH induces a morphism from MATH onto MATH with kernel MATH. So, MATH induces a natural isomorphism: In particular, the MATH-vector spaces MATH and MATH have the same rank. So, MATH . Besides, since MATH is special, we have: MATH . The isomorphism MATH makes MATH and MATH commute because of REF . As a consequence, MATH sends MATH to MATH. Furthermore, when these (simultaneously) vanish, the natural isomorphism between MATH and MATH induced by MATH, make MATH and MATH commute (because of REF ). As a consequence, these two quadratic forms will have the same NAME invariant. So, to sum up, MATH . Since MATH, MATH vanishes. It remains to be noticed that MATH is nothing but MATH. Thus, MATH . Now, from REF and NAME theorem, we see that MATH together with MATH determine MATH up to isomorphism. What has been remaining to be proved for REF , then follows. |
math/0104065 | Take some quadratic forms MATH and MATH over MATH, and let MATH be such that MATH (see REF ). By the hypothesis on MATH, there exits some MATH such that MATH, and so, by homogeneity of MATH, we have: MATH . But since MATH is then of order at most MATH, MATH has to be of order at most MATH (see CITE). It follows that MATH and so, by REF , we obtain that MATH. REF allows us to conclude. |
math/0104065 | Let MATH denote the boundary of a regular neighbourhood of MATH in MATH. The normal bundle of MATH in MATH is naturally trivialized, so MATH inherits from MATH a spin structure. Let MATH be the quadratic form associated to the spin smooth surface MATH as defined by NAME in CITE. The following identity then holds for each parallel MATH: MATH when MATH is thought of as a curve on MATH. Since MATH is quadratic with respect to the modulo MATH intersection form MATH on MATH, we have: MATH . |
math/0104065 | Consider the MATH-manifold MATH obtained from MATH by attaching a MATH-handle to MATH along MATH. Identify MATH with MATH. Since MATH, MATH extends in a unique way to a spin structure MATH of MATH. MATH is then a spin cobordism between MATH and MATH, where MATH is the closed oriented MATH-manifold obtained from MATH by the corresponding surgery, and where MATH is the restriction of MATH to MATH. Note also that the core of the MATH-handle is a MATH-disc MATH in MATH with boundary MATH in MATH, and whose normal bundle can be trivialized in accordance with the trivialization of the normal bundle of MATH in MATH given by MATH. The framed knot MATH will briefly be said to have property MATH in MATH. According to NAME Theorem (see CITE), the spin MATH-manifold MATH admits an even surgery presentation in MATH (that is, the linking matrix is even and its characteristic solution corresponding to MATH is the trivial one). Denote by MATH the trace of the surgery and by MATH the unique extension of MATH to the whole of MATH. By glueing MATH to MATH along MATH, we obtain a spin MATH-manifold MATH with boundary MATH. The MATH-handle from MATH to MATH can be reversed. After a rearrangement, the MATH-manifold MATH appears as MATH to which have been simultaneously attached some MATH-handles (one more than MATH), with boundary MATH, and to which MATH can be extended. So, MATH is the trace of an even surgery presentation. As a summary, we have found so for an even surgery presentation MATH for MATH such that MATH has property REF in MATH. Let us work with this surgery presentation of MATH. Notations of REF will be used: MATH, MATH stands for the intersection form of MATH and so on. Let the MATH-disc MATH give an element of MATH. The latter is identified with an element MATH of MATH. Recall from the definition of MATH that, in this even case, MATH where MATH is the rational extension of MATH and where MATH is such that MATH. Let MATH such that MATH. Then, there exists MATH such that MATH. So MATH works as a MATH. REF can be rewritten as: MATH . When MATH is seen as belonging to MATH, the integer MATH is equal to MATH, where MATH and MATH are MATH-cycles representatives for MATH in transverse position in MATH. By means of a ``collar" trick appearing in CITE, we will be able to give examples of such MATH and MATH. We add a collar MATH to MATH such that MATH is identified with MATH. Let MATH be a NAME surface for MATH in MATH in transverse position with MATH, and MATH be a NAME surface for MATH in MATH. Because of the property MATH, MATH can be pushed off to a disc MATH in such a way that MATH and MATH. REF is a good summary. We define MATH and MATH. Then: MATH . The last MATH in REF is intersection in MATH. The lemma follows from REF and the definition of a framing number. |
math/0104065 | Find an even surgery presentation of the spin MATH-manifold MATH. So we are led to apply REF with MATH. |
math/0104065 | Implication MATH is obvious since MATH and MATH are both unimodular. Now suppose that REF is satisfied. For each MATH, there exists a nondegenerate symmetric bilinear form MATH such that: MATH where MATH (note that it is free), MATH, and MATH is the zero form. The form MATH is still even and MATH and MATH are still isomorphic. Consequently, without loss of generality, we can assume both MATH to be nondegenerate. But this case was treated by NAME in CITE. |
math/0104065 | This follows directly from REF . |
math/0104066 | We fix a background metric MATH and write MATH for its volume form. Define MATH by MATH. Let MATH be an eigenfunction branch corresponding to MATH. NAME subscripts, we have MATH . Each object in REF is real-analytic in MATH. By NAME expanding, collecting first order terms, integrating by parts, and using the eigenequation, we find that MATH . Here the symbol MATH denotes the first derivative with respect to MATH evaluated at MATH. By definition, we have MATH for each fixed vector field MATH and function MATH on MATH. Differentiating in MATH yields MATH. Using this identity, REF reduces to MATH . From MATH, we have MATH . By interpreting MATH as the determinant of the matrix representation of MATH with respect to an orthonormal basis of MATH, one finds that the supremum of MATH is bounded by MATH. The claim follows by applying this bound and REF to REF . |
math/0104066 | Straightforward computation gives MATH . From REF and MATH we find that MATH . Integrating by parts over MATH gives MATH . Also note that from REF we have MATH . The claim follows. |
math/0104066 | Apply REF . |
math/0104066 | Let MATH denote the dilated interval MATH, and let MATH belong to MATH with MATH on MATH. By multiplying both sides of REF by MATH and integrating over MATH one obtains MATH . On the other hand, integration by parts gives MATH . We have MATH where MATH in a neighborhood of the origin. Note that MATH adds MATH to the homogeneity of any function. Hence MATH, and thus, since MATH, we have MATH. Therefore, by REF MATH where MATH. To complete the proof it suffices to show that MATH . Since MATH and MATH is bounded on the support of MATH, there exists MATH such that MATH . Let MATH. Integrating by parts gives MATH . Note that the support of MATH belongs to MATH. Integration by parts in MATH gives MATH . Estimate REF then follows from the fact that MATH has support in MATH and equals MATH on MATH. |
math/0104066 | Note that by homogeneity and positivity, MATH for some MATH, and hence MATH. Therefore, the first claim will follow from the second. By REF , we may assume that the function MATH has a uniques minimum at MATH, and thus MATH has a unique maximum there. To prove the claim, it will suffice to show the same for MATH. In other words, it is enough to show that MATH is positive for MATH and negative for MATH. Since MATH is strictly convex, MATH is strictly increasing. Let MATH. The derivative MATH is homogeneous of degree REF, and, therefore, for MATH . Hence MATH is decreasing for MATH. That is, MATH is negative for MATH as desired. An analogous argument shows that MATH for MATH. The claim follows. |
math/0104066 | Our starting point is REF : MATH . Recall that by (global) hypothesis, the supremum of MATH over the unit tangent bundle of MATH is finite. By applying the argument that immediately follows REF to the restriction of MATH to MATH, we obtain MATH for some positive constant MATH. By REF , the restriction of MATH to MATH equals MATH. Thus, by combining REF we obtain MATH . We claim that MATH . To see this, first note that from REF we compute MATH . The function MATH is constant on each fibre MATH, and hence MATH. Therefore, since MATH and MATH, we find that MATH . The operator MATH preseves the decomposition MATH. In particular, MATH and MATH is constant on each fibre. Therefore, MATH, and it follows that MATH . The claimed REF follows. From REF we also have that MATH and hence combined with REF we have MATH . Substitution into REF then yields MATH . Since MATH is homogeneous of degree MATH, REF applies to give MATH as well as the analogous estimate with MATH replaced by MATH. By combining these estimates with REF and absorbing constants, we obtain the claim. |
math/0104066 | We apply REF . Since MATH and MATH are positive and MATH, the first term on the right hand side of REF is nonnegative. Therefore, by REF we have MATH for some constants MATH and MATH. Since MATH, division of both sides by MATH gives MATH . Since MATH, the left hand side of REF is integrable, and, moreover, the negative variation of MATH over MATH is MATH. Thus, since MATH, the function MATH has a limit as MATH tends to MATH. Thus, the claim follows via exponentiation. |
math/0104066 | Let MATH. Note that because MATH, we have MATH, and hence MATH for MATH small. Thus, by using REF , we obtain MATH . If MATH, then the right hand side of REF is positive for MATH small. The claim follows. |
math/0104066 | By REF , we have MATH for MATH small. Thus, it follows from REF , that there exist positive constants MATH, MATH such that MATH . Since MATH, we have upon letting MATH . Dividing by MATH gives MATH . Since MATH, we obtain MATH . It follows that the negative variation of MATH over the interval MATH is MATH. It follows that MATH is either finite, in which case MATH is finite, or MATH in which case MATH. In either case the limit exists. |
math/0104066 | Since MATH is homogeneous of degree zero, it is bounded. By REF , we have MATH, and hence MATH. Therefore, via REF we find that MATH . Hence by REF MATH for some positive MATH and MATH small. Dividing by MATH gives MATH . Since MATH, the right hand side is integrable, and, in particular, the negative variation of MATH is MATH. Therefore MATH exists, and it follows that MATH has a limit. |
math/0104066 | We claim that if MATH, then for sufficiently small MATH . Indeed, from REF , we have MATH if and only if MATH . If MATH, then for sufficiently small MATH, we have MATH, and hence for MATH . Therefore MATH for MATH and sufficiently small MATH. The claimed REF then follows from homogeneity and integration. By applying REF with MATH, MATH, and MATH, we obtain the complementary estimate. Indeed, since MATH is compact, the function MATH - defined by REF - is orthogonal to the zero eigenspace of MATH. Hence for each MATH . Thus, REF applies to give MATH . Combining REF gives us the desired REF for all MATH. |
math/0104066 | By REF we have MATH and hence it follows from REF that MATH. Therefore MATH . Here the last inequality follows from REF . We wish to apply integration by parts to the last integral in REF . Towards this end, let MATH be a smooth function supported in MATH with MATH on MATH and MATH. Since MATH is positive, convex, and homogeneous of degree REF, the set MATH is a closed interval MATH that contains MATH. For each MATH define MATH . Integration by parts shows that the operator MATH is symmetric on MATH with NAME boundary conditions. Thus, MATH . By REF , we have MATH for MATH. Thus, since MATH, estimate REF combine to give MATH . Therefore, to verify REF , it will suffice to show that MATH . By homogeneity, the supremum of MATH over MATH is MATH for any constant MATH. We compute MATH where MATH. The operator MATH adds MATH to the degree of a homogeneous function, and hence MATH . Since MATH, we have MATH, and hence MATH. To estimate the remaining two terms in REF , we need to estimate MATH and MATH. To this end, consider MATH, the inner radius of MATH. By REF , there exists MATH such that MATH for all MATH small. Therefore MATH and, similarly, MATH. The function MATH appearing in REF is homogeneous of degree MATH. Hence since MATH is homogeneous of degree zero, MATH . A similar argument shows that MATH . The desired estimate REF then follows from REF , and REF . |
math/0104066 | By definition, MATH satisfies MATH . Thus, by homogeneity MATH . By REF , MATH tends to MATH as MATH tends to zero. If MATH, then since the right hand side of REF is positive, MATH must tend to infinity as MATH tends to zero. By homogeneity, REF implies that MATH. Thus if MATH, then by REF we have MATH . By REF and the condition MATH, the function MATH assumes a unique minimum at MATH. Therefore, MATH is strictly bounded away from zero. The claim follows. |
math/0104066 | Since MATH, by REF we have MATH . Note that from REF we obtain MATH. Hence, MATH . By hypothesis we have MATH and hence MATH. Applying NAME 's principle for MATH acting on MATH, we obtain MATH as well as MATH . Thus, it follows from REF that there exists MATH such that MATH . Since MATH, combining REF with REF yields MATH such that MATH for some MATH. By substituting REF into REF and applying REF to the MATH-term in REF , one obtains MATH for some constants MATH and MATH. Thus, since MATH, there exists a positive constant MATH such that MATH . Division by MATH gives MATH and hence, by arguing as in the proof of REF , one finds that MATH converges as MATH tends to zero. Therefore the claim follows from REF . |
math/0104066 | If MATH is less than the smallest positive eigenvalue of MATH, then MATH and the claim follows. Otherwise, let MATH be the largest MATH-eigenvalue that is less than MATH. From the definition of MATH we have MATH . Since, by hypothesis, MATH, there exists MATH such that for sufficiently small MATH . Hence, using the (global) hypothesis that MATH for all MATH, we have MATH . Combining this with REF gives MATH . Therefore, the claim follows from REF . |
math/0104066 | By REF we have MATH . Using REF , one obtains REF with MATH replaced by MATH. From this and REF it follows that the left hand side of REF is bounded above by MATH . Thus it will suffice to show that MATH for some MATH. Towards verifying REF , we will apply REF . Namely, let MATH to be the smallest of all eigenvalues that are greater than MATH and let MATH. Note that from the definition of MATH, for each MATH . Note also that REF follows in this case from REF . Therefore REF provides MATH such that MATH . By REF , there exists MATH such that MATH. Define MATH . It follows from REF that MATH. By homogeneity, we have MATH for all MATH. In particular, estimate REF holds for all MATH. Thus, by integrating this estimate over MATH and combining with REF we obtain MATH . Therefore, since MATH, the claimed REF is proven. |
math/0104066 | We assume that the hypothesis of REF is not satisfied and derive a contradiction. Namely, we assume that MATH is a positive MATH-eigenvalue and obtain a contradiction in the form of two conflicting estimates. In particular, it is enough to show that there exist constants MATH such that for small MATH . This is impossible since MATH and MATH and hence MATH. The respective sides of REF are given below as REF . |
math/0104066 | We first claim that it suffices to show that there exists MATH such that for small MATH . Indeed, since MATH, there exists MATH such that MATH for small MATH. Hence we would have MATH . Note that since MATH . Thus, since MATH it would follow from REF that there exists a constant MATH such that MATH . Note that since MATH and MATH, we have MATH. Thus, since MATH, we could then integrate REF over MATH and would find that MATH . Since MATH, we would then obtain REF by dividing both sides of REF by MATH. Recall from REF that MATH denotes the fibrewise projection of MATH onto the MATH eigenspace of MATH. Letting MATH, we have MATH. We claim that to verify REF it suffices to show that MATH where MATH and MATH. To see this, note that by REF , there exists MATH such that MATH for MATH. It follows that MATH . Since MATH, we have MATH, and hence since MATH, for any MATH, we have MATH for all sufficiently small MATH. Thus, by combining REF , we would obtain MATH for MATH small. Applying the argument in REF with MATH replaced by MATH, we would have MATH (In this and what follows, MATH and MATH represent generic constants.) Recall that in REF we had MATH, and hence we have REF with MATH replaced by MATH: MATH . As pointed out above, MATH for small MATH. Hence by applying NAME 's principle as in REF to MATH, we could combine REF to find that MATH . Combining this with REF would yield MATH such that MATH where MATH denotes a (generic) positive constant. By substituting REF into REF and applying REF to the MATH-term in REF and using REF , one would obtain MATH . Note that MATH for small MATH. Hence from REF one would have MATH for small MATH. Since MATH and MATH for small MATH. Thus by choosing MATH and recalling REF we would obtain REF from REF . And hence REF follows from REF . As a first step toward the verification of REF , we rescale in MATH. In particular, let MATH, and MATH, and for each MATH and MATH, define MATH . Then using homogeneity, we find that MATH and MATH . Since MATH and since MATH belongs to the MATH eigenspace of MATH, we have MATH from REF . It follows from REF and homogeneity that MATH satifies the following ordinary differential equation MATH where MATH and MATH is a bounded smooth function. Hence, by REF , to prove REF it will suffice to prove that there exists MATH such that for any solution MATH to REF we have MATH . Towards verification of REF we apply REF to REF . Indeed, by REF is bounded, and hence REF applies to give a constant MATH such that MATH . It follows that MATH . By REF and the strict convexity of MATH, there exists MATH such that MATH for MATH. It follows that MATH . By substituting REF and choosing MATH we obtain REF . The proof is complete. |
math/0104066 | It will suffice to show that MATH . Indeed, we may suppose that MATH, for otherwise we are done. Hence REF implies MATH . By REF , we have MATH for small MATH, and thus division would give MATH . By integrating over MATH with MATH small we would find that MATH . Exponentiation would then give REF with MATH. To verify REF we will estimate MATH using the methods of REF. From REF we have MATH . By substituting REF into REF one obtains that MATH . Thus, by integrating REF and using REF we find that MATH . We may estimate the righthand side of REF by splitting the integral over the sum MATH where MATH. To be precise, apply NAME 's principle - as in REF - to find that MATH as well as MATH . It follows that if we let MATH denote the right hand side of REF , then MATH . By REF and hence MATH. It follows that MATH where MATH . To estimate MATH, apply REF `in reverse' to find that MATH . Thus by using the assumption MATH, REF , and REF , we obtain MATH such that MATH . By combining REF , and REF , we find (generic) constants MATH such that MATH . Hence by substituting this into REF and applying REF to the middle term in REF , we obtain MATH for possibly different (generic) constants. By using REF , we find that MATH . We claim that there exists MATH such that for all MATH for all small MATH. To see this, first note that we cannot have MATH for all small MATH. For then by REF we would have that MATH for all small MATH. Using the argument of REF , we would then find that MATH is bounded and hence MATH would converge by REF . This would contradict the assumption that MATH. Hence there exists MATH such that REF is true for MATH. A calculation shows that REF holds if and only if MATH . Thus to prove REF for MATH, it suffices to show that MATH is increasing for small MATH. To see that this is true, note that by homogeneity MATH, and hence MATH . By REF , MATH for small MATH. Thus from REF we have MATH, and hence MATH as desired. Therefore, since REF holds true, REF yields MATH . Moreover, for small MATH we have MATH, and hence by REF we have MATH. it follows from REF that MATH . Into this substitute REF , and use both REF and the definition of MATH in REF to obtain REF as desired. |
math/0104066 | Let MATH. Then REF is equivalent to MATH . From REF , we see that MATH satisfies MATH . If MATH is large and positive, then MATH is uniformly convex on MATH, and REF follows. If MATH is large and negative, then MATH oscillates rapidly. In particular, by REF, MATH differs from a solution to MATH in MATH-norm by order MATH. A straightforward calculation shows that REF holds uniformly for MATH and hence MATH for MATH positive and sufficiently large. For MATH bounded, the claim follows from continuous dependence on parameters, the linearity of the equation, and the fact that the MATH-norm cannot vanish on any nontrivial interval. |
math/0104080 | Because MATH and MATH is MATH invariant, MATH if and only if MATH. This establishes the first point. The second one follows in the same manner, using the fact that MATH. |
math/0104080 | The proof of REF is derived from the corresponding proof of the symplectic cross section theorem of NAME and NAME CITE. Indeed, extend the MATH-action to the symplectization of MATH and let MATH be the corresponding symplectic moment map. The symplectic cross section theorem gives MATH as a symplectic submanifold of the symplectization. By hypothesis, MATH is invariant under dilations by MATH and hence REF implies that MATH. It follows that the contact cross section is a hypersurface of contact type in the symplectic cross section, which gives the result. |
math/0104080 | Denote the NAME vector field by MATH and its flow by MATH. As noted earlier, by uniqueness MATH is invariant and therefore MATH is equivariant. Therefore, for any MATH and MATH, it follows that MATH. Thus MATH . This establishes REF . Since MATH, NAME 's formula gives MATH. Therefore, MATH which establishes REF . The proof of REF is an application of REF . For any MATH, the second item implies that for any MATH, MATH . Thus, MATH, whence MATH. The final point follows immediately from the second point. Note that MATH. |
math/0104080 | Choose MATH. Then MATH, where MATH is the differential of the co-adjoint representation. Thus, for any MATH, we have that MATH . Therefore, MATH is a NAME ideal in MATH. |
math/0104080 | If MATH is transverse to MATH, then MATH is a submanifold of MATH. Choose MATH. The transversality condition, MATH is equivalent to the condition MATH . Let MATH be the isotropy subgroup of MATH in MATH and MATH its NAME algebra. Choose MATH and denote the NAME vector field of MATH by MATH. Then MATH since MATH. Because MATH fixes MATH, MATH. Thus, MATH . Therefore, MATH and MATH acts locally freely on MATH. To establish the second statement, suppose that MATH for some MATH and let MATH. Then, by Proposition, REF MATH for some MATH, where MATH is the NAME vector field. Therefore, MATH . Hence, MATH and MATH. Since MATH and MATH, it follows that MATH. Since MATH acts locally freely on MATH, MATH and hence MATH is transverse to MATH. |
math/0104080 | Choose MATH and MATH. Write MATH where MATH and MATH. Then MATH. Hence MATH. |
math/0104080 | Fix MATH. Since MATH is isotropic, MATH. For any MATH, MATH by definition. Hence MATH. Conversely, if MATH, then by unravelling the various definitions, one sees that MATH. |
math/0104080 | Since MATH is transverse to MATH, MATH is a submanifold of MATH and REF implies that MATH acts locally freely MATH. Hence, MATH is an orbifold. Fix MATH. For any MATH, we have that MATH . Hence, MATH descends to MATH on MATH, which is contact if and only if MATH . Let MATH be the contact moment map associated to the action of MATH on MATH. Note that MATH where MATH is the natural projection. Observe that MATH and hence that MATH is an isotropic subspace of the symplectic vector space MATH by REF . Let MATH denote the symplectic perpendicular of MATH in MATH. A vector MATH is in MATH if and only if for all MATH . That is, MATH if and only if MATH. Hence, MATH where MATH. Note that MATH is a submanifold of MATH by the transversality condition. Indeed, MATH. Since MATH is transverse to MATH, MATH acts locally freely on MATH by REF and thus on a neighborhood of MATH and hence on MATH (or, at least on a neighborhood of MATH in MATH). Therefore, also by REF , MATH is a submanifold of MATH. REF implies that MATH. We first show that MATH. If MATH and MATH, then MATH . Therefore, MATH . Note that this implies that MATH. The reverse inclusion is slightly more delicate. Since MATH, we can choose a splitting MATH where MATH. Let MATH. The proof is completed by showing that MATH and MATH are complementary subspaces of MATH which are prependicular with respect to MATH. REF implies the reverse inclusion. We first show that MATH. Choose any MATH and let MATH. Then for some MATH, MATH since MATH. Hence, MATH. Additionally, MATH . Thus, MATH. By equivariance, MATH. Hence, MATH maps MATH to MATH. In fact, one can show that this map is an isomorphism. By definition, MATH. The assumption MATH is equivalent to MATH . Hence, MATH. Finally, if MATH, then MATH. Hence, MATH is a subspace of dimension MATH. A dimension count implies that MATH and MATH are complementary subspaces of MATH, which completes the proof. |
math/0104080 | See CITE for a complete proof. |
math/0104080 | By REF the restriction of MATH to MATH is a contact form and MATH is MATH invariant. The contact moment map for the MATH action on MATH is given by MATH. Note that MATH . The proof will be completed by showing that MATH. From this it follows that MATH. Since MATH, the conclusion follows. Since MATH is MATH-invariant and contains MATH, MATH. Recall that MATH. If MATH, then MATH is tangent to MATH at MATH. On the other hand, since MATH and MATH is convex, it follows that MATH is also tangent to MATH at MATH. Hence MATH . However, because MATH is a slice, MATH. Thus, MATH and MATH. |
math/0104080 | By REF , we may assume that MATH is fixed by the co-adjoint action of MATH on MATH. Let MATH be the moment map for the action of MATH. Because MATH, where MATH is the natural projection, we have MATH. Therefore MATH is an open subset of MATH, which is stratified by REF . Open subsets of stratified spaces are naturally stratified. |
math/0104080 | MATH is a submanifold of MATH and for all MATH, MATH where MATH is the set of MATH-fixed vectors in MATH (see, for example, REF). The NAME vector field, MATH, of MATH, is MATH-invariant and, since the MATH-action preserves the contact structure, MATH . Because MATH is a symplectic subspace of MATH, the restriction of MATH to MATH is a contact form on MATH. It what follows it is useful to cite REF , pg REF, of CITE, which identifies MATH, the dual of the NAME algebra of MATH, with MATH. For any MATH and MATH since MATH. Hence, the image of MATH in MATH under MATH is contained in MATH. Because the moment map is equivariant, MATH. The action of MATH on MATH is defined by MATH, where MATH is the coset containing MATH. This is free by definition. The moment map, MATH, for the MATH action on MATH is given by the restriction of MATH to MATH. Therefore, MATH . The natural inclusion, MATH descends to a map, MATH defined by MATH, where MATH and MATH denote the orbit through MATH under the MATH and MATH actions respectively. If MATH, then the stablilizer of MATH in MATH is conjugate to MATH. This implies that some element of the MATH-orbit through MATH has stabilizer equal to MATH, whence MATH is surjective. To show that MATH is injective, suppose that MATH and that MATH. Because MATH and MATH have stabilizer equal to MATH, it follows that MATH and therefore that MATH is injective. |
math/0104080 | The trivialization MATH defined by MATH is equivariant (where MATH acts trivially on MATH) since MATH is. |
math/0104080 | Set MATH. Then MATH is a submanifold of MATH and the natural inclusion, MATH gives rise to a projection, MATH. By REF , the NAME vector field is tangent to MATH. Split the tangent bundle of MATH at MATH as MATH . We obtain an induced splitting, MATH . If MATH, then MATH. Therefore, MATH embeds MATH into MATH. Moreover, this embedding is equivariant. Denote the symplectic moment map for the lifted MATH action on MATH by MATH. Then MATH . For any MATH, and MATH, we have MATH since MATH implies MATH for all MATH. Hence, MATH embeds MATH into MATH. Let MATH be the induced embedding. The co-tangent bundle reduction theorem of NAME and NAME CITE asserts that there is a symplectomorphsim MATH. Set MATH. Recall that the reduced contact form on MATH is defined to be the unique REF-form, MATH, on MATH such that MATH, where MATH is the orbit map. It follows, by definition, that MATH, giving MATH. |
math/0104080 | Set MATH and MATH . By REF , MATH is a line bundle over MATH. Let MATH be the normalizer of MATH in MATH and MATH. Set MATH . Then MATH is a contact submanifold of MATH and MATH acts freely on MATH, preserving MATH. Denote the corresponding moment map by MATH. Recall that we can identify MATH with MATH and that there is a diffeomorphism MATH. It follows that, without being too fussy about equivalences versus equalities, MATH . Now apply the free case to obtain the result. |
math/0104084 | We first prove MATH is self-dual for the NAME pairing on MATH. Equivalently, we will prove, for all non-zero MATH, there exists a element MATH such that MATH . Let MATH pull-back to MATH: MATH . If MATH then there exists MATH satisfying: MATH as the NAME pairing on MATH is nondegenerate. Let MATH. Then, MATH . If MATH, then we will apply the Hard NAME Theorem (HLT) to prove MATH. Let MATH be the complex dimension of MATH. By HLT applied to MATH, we may assume MATH for MATH. Thus, MATH. By HLT applied to MATH, there exists an element MATH satisfying: MATH . By the NAME Hyperplane Theorem applied to MATH, MATH . Let MATH satisfy MATH. We find, MATH . As MATH and MATH, we find: MATH . By HLT applied to MATH, we conclude MATH. The vanishing MATH then follows. Next, we prove the vanishing of NAME - NAME invariants required for REF . Let MATH and MATH. MATH . Since MATH (see CITE) and MATH is a cohomology class pulled-back from MATH, the above integral may be rewritten as: MATH . |
math/0104084 | When MATH, MATH. Then, REF easily implies the Proposition. |
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