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math/0002030 | We begin by selecting a grading MATH of MATH which preserves MATH, and recalling that by CITE, the group MATH acts simply transitively on the set of all such gradings MATH. Next, to compute how this transitive action changes the associated MATH representations MATH, write MATH relative to MATH, and observe that MATH . Consequently, by MATH, the decomposition MATH of MATH relative to MATH must have MATH, hence MATH . To prove the existence of a grading MATH which satisfies MATH, we proceed by induction and construct a sequence of gradings MATH in MATH terminating at MATH, by the requirement that MATH upon decomposing MATH relative to MATH. To check the validity of this algorithm, let us suppose the gradings MATH have been constructed, and write MATH . Then, by virtue of REF, we can reformulate condition MATH as the requirement that MATH with MATH denoting the decomposition of MATH relative to MATH. Now, by virtue of the fact that MATH (upon decomposing MATH relative to MATH), it is natural to assume that our element MATH is of the form MATH (again, relative to the eigenvalues of MATH). Imposing condition MATH, it follows that we may take MATH to be of this form if and only there exists a solution MATH of the equation: MATH . To find such an element MATH, one simply observes that MATH . To prove the grading MATH so constructed is unique, suppose that MATH is another such grading and write MATH with MATH relative to MATH. Therefore, upon applying MATH to both sides of REF, we see that MATH and hence, we also must have MATH since MATH. In addition, by combining the observation that MATH preserves the eigenspaces of MATH with the fact that MATH fixes MATH, it follows that we must also have MATH . Thus, MATH is a solution to MATH which is of weight MATH relative to the representation MATH. By standard MATH theory, we therefore have MATH, contradicting our our assumption that MATH. |
math/0002030 | By virtue of the previous remarks, it will suffice to check the two sub-cases enumerated above. To verify the assertion for REF , note that MATH is the weight REFp eigenspace of MATH, and recall that by definition we must have MATH . Consequently, the commutativity relation MATH implies that MATH and hence MATH preserves MATH by REF. Regarding the second case, observe that because MATH, MATH must be the trivial filtration MATH in order for the relative weight filtration MATH to exist. Thus, MATH must be the trivial grading on MATH of weight MATH. |
math/0002030 | There are two key steps: CASE: MATH . Show the monodromy weight filtration MATH is self-dual with respect to MATH, that is, MATH. CASE: MATH . Via semisimplicity, assume the pair MATH defines a MATH representation MATH of highest weight MATH. Imposing self-duality, it then follows that MATH unless MATH, hence MATH is an infinitesimal isometry of MATH. |
math/0002030 | The details may be found in REF. However, the idea of the proof is relatively simple: An element MATH belongs to the subspace MATH if and only if MATH . To verify the decomposition MATH, we observe that CASE: MATH . As a vector space, MATH. CASE: MATH . By definition, MATH and hence MATH. |
math/0002030 | To prove that MATH, one simply checks that MATH acts as scalar multiplication by MATH on MATH, and hence MATH. To verify equation MATH, we note that if MATH, MATH denotes the grading of MATH which acts as multiplication by MATH on MATH then MATH since MATH is defined over MATH, and hence MATH . Upon transferring these computations to the induced metrics on MATH, it then follows that MATH . In particular, MATH since MATH, and hence MATH . |
math/0002030 | For simplicity of exposition, we shall first prove the result under the additional assumption that our limiting mixed NAME structure MATH is split over MATH. Having made this assumption, it then follows from the work of REF that: CASE: MATH . The associate gradings MATH and MATH are defined over MATH. CASE: MATH . The endomorphism MATH is an element MATH. CASE: MATH . The filtration MATH is an element of MATH. [This assertion is a consequence of NAME 's MATH . Orbit Theorem, see CITE for details.] Now, as discussed in REF, the fact that MATH is a MATH-morphism of the limiting mixed NAME structure MATH implies that MATH and hence MATH . Consequently, because MATH preserves MATH, MATH . Next, we note that by REF and Lemma MATH and hence MATH . In particular, upon setting MATH and recalling that MATH preserves MATH, it follows that MATH . In addition, because MATH, MATH and hence MATH since MATH is an element of MATH. Therefore, by REF, MATH . Consequently, by the remark which follows Lemma MATH, we have: MATH for some MATH and norm any fixed norm MATH on MATH. To further analyze MATH, decompose MATH according to the eigenvalues of MATH . Then MATH since MATH. Therefore, denoting the maximal eigenvalue of MATH on MATH by MATH, we have MATH because MATH is a holomorphic function of MATH vanishing at zero. In summary, we have proven that whenever the limiting mixed NAME structure MATH of our admissible variation MATH is split over MATH and MATH is sufficiently large, the following distance estimate holds: MATH . If the limiting mixed NAME structure MATH is not split over MATH, we may obtain the same distance estimate by first applying NAME 's MATH splitting: MATH and then proceeding as above. More precisely, let MATH denote the gradings determined by MATH and MATH. Then, REF becomes MATH with MATH and MATH . Thus, applying equation MATH and the remark the follows the proof of REF, we have: MATH for MATH sufficiently large. |
math/0002030 | Let MATH and MATH denote the gradings associated to the split mixed NAME structure MATH via the methods of REF. Then, a quick review of the proof of REF shows that upon setting CASE: MATH. CASE: MATH. we have both MATH and MATH . In particular, since MATH and MATH is a grading of MATH which is defined over MATH: MATH . Moreover, MATH and hence MATH . Therefore, MATH . |
math/0002033 | As it was said in REF, we have MATH in some neighbourhood of MATH where MATH is some MATH-parametric system. If MATH then setting MATH, we get from REF the equality MATH, with MATH, that is, REF is true. Let us apply the induction on MATH. Suppose that the statement is true for MATH. Since MATH implies MATH and REF turns into MATH we can set MATH and get MATH. Indeed, MATH since MATH identically for MATH (the case MATH is obvious). It is clear that MATH. By the supposition of the induction there exist separable NAME spaces MATH, operators MATH, and a holomorphic function MATH with values from MATH such that MATH and MATH in some neighbourhood of MATH. Then REF is true, and the proof is complete. |
math/0002033 | The case MATH is trivial. For MATH we obtain REF from REF by virtue of the uniqueness of NAME 's expansion for MATH. |
math/0002033 | As it was said in REF there exists a conservative scattering system MATH such that MATH in MATH. If MATH then MATH, thus the statement is valid with MATH. If MATH then MATH and REF holds. Let us show that MATH. Indeed, for the linear operator-valued function MATH corresponding to a conservative scattering system MATH we have proved in CITE that MATH, hence for any MATH-tuple MATH of commuting contractions on some separable NAME space MATH we obtain from REF MATH (for a finite sum in REF one can pass to the limit in REF as MATH). Then MATH (here MATH is the orthoprojector onto MATH in MATH), and by virtue of an arbitrariness of MATH and MATH we get MATH. Analogously, MATH, and by REF the statement of REF is valid for MATH also. |
math/0002033 | It follows from REF that MATH is the minimal subspace in MATH containing MATH and reducing MATH for all MATH. By the assumption, MATH. If MATH then by REF the subspace MATH contains MATH and reduces MATH for all MATH, that contradicts to the assumption. Hence, MATH is closely connected. Analogously, MATH is closely connected. |
math/0002033 | For any MATH is a unitary operator, and from REF and the definition of MATH we have MATH, and MATH is an isometry, thus MATH is unitary. Hence, MATH is a conservative scattering system. For any MATH and MATH is an isometry, thus MATH is unitary. Hence, MATH is a conservative scattering system. It is easy to see now that for any MATH has a form REF, that is, MATH. The second assertion of this theorem follows directly from the definition of cascade connection. |
math/0002033 | The part ``if" is clear since in this case by REF there are conservative scattering systems MATH and MATH such that MATH, and by REF we have MATH with functions MATH and MATH belonging to the corresponding classes MATH. For the proof of the part ``only if" let us assume that REF holds with MATH. Let MATH be some conservative realizations, respectively (which exist by CITE), that is, MATH. Then the conservative scattering system MATH has the transfer function MATH, and the subspace MATH in MATH satisfy REF . Define the subspace MATH in MATH by REF, and operators MATH. Then, by REF, MATH is a reducing subspace in MATH for all MATH, and MATH is a closely connected conservative realization of MATH. Define MATH. Then MATH is an invariant subspace in MATH for all MATH. Since MATH we obtain that MATH is an invariant subspace in MATH for all MATH, that is, REF is satisfied for MATH. For all MATH we have the spaces MATH coinciding, and REF is also satisfied for MATH. The proof is complete. |
math/0002035 | Evaluation of sections determines a surjective map MATH of vector bundles on MATH. The corresponding NAME complex takes the form: MATH NAME through by MATH, and applying the hypothesis with MATH as one chases through the resulting complex, one sees first of all that the multiplication map MATH is surjective. Next tensor REF by MATH and apply the vanishing hypothesis with MATH: it follows that MATH maps onto MATH, and hence that MATH is also onto. Continuing, one finds that MATH is surjective for all MATH. But since MATH is very ample, MATH is globally generated for MATH. It then follows from the surjectivity of REF that MATH itself must already be generated by its global sections. |
math/0002035 | In fact, thanks to NAME vanishing, REF 's Lemma applies to MATH as soon as MATH. |
math/0002035 | The first statement follows easily from the definition. For REF , note that MATH for a suitable MATH-divisor MATH numerically equivalent to MATH. This being said, REF is a consequence of the NAME Vanishing theorem whereas REF follows from REF . |
math/0002035 | REF is a slight variation of NAME 's vanishing theorem in its analytic form. If MATH is ample, the result is true with MATH as well as with MATH (the latter case being obtained by replacing MATH with MATH where MATH is an arbitrary smooth metric on MATH; the defect of positivity of MATH can be compensated by the strict positivity of MATH). If MATH is big and nef, we can write MATH with an ample MATH-divisor MATH and an effective MATH-divisor MATH, and MATH can be taken arbitrarily small. We then get vanishing with MATH where MATH is the singular metric of curvature current MATH on MATH. However, if MATH is so small that MATH, MATH, we do have MATH, as follows from an elementary argument using NAME 's inequality. REF follows from REF , NAME and NAME 's REF . Alternatively, one can argue via a straightforward adaptation of the proof given in CITE, based on NAME 's MATH estimates for ideals of holomorphic functions CITE. |
math/0002035 | The strong version of the NAME MATH extension theorem proved by CITE shows that for every singular hermitian line bundle MATH with nonnegative curvature and every smooth complete intersection subvariety MATH (actually, the hypothesis that MATH is a complete intersection could probably be removed), there exists a sufficiently ample line bundle MATH and a surjective restriction morphism MATH with the following additional property: for every section on MATH, there exists an extension satisfying a MATH estimate with a constant depending only on MATH (hence, independent of MATH). We take MATH equal to a smooth zero dimensional scheme obtained as a complete intersection of hyperplane sections of a very ample linear system MATH, and observe that MATH depends only on MATH in that case (hence can be taken independent of the choice of the particular MATH-dimensional scheme). Fix an integer MATH so large that MATH is effective. We apply the extension theorem to the line bundle MATH equipped with the hermitian metric MATH, curv-MATH (and a smooth metric MATH of positive curvature on MATH). Then, for MATH and a prescribed point MATH, we select a zero-dimensional subscheme MATH containing MATH and in this way we get a global section MATH of MATH such that MATH . From this we infer that locally MATH with MATH, hence MATH where MATH is an orthonormal basis of sections of MATH. This implies that MATH contains the ideal MATH. Again, NAME 's inequality shows that this ideal contains MATH for MATH large enough. |
math/0002035 | Since the exceptional set MATH is the divisor where the derivative MATH drops rank, one sees that MATH. Similarly, MATH . Therefore MATH and this has normal crossing support since MATH and MATH do. |
math/0002035 | To lighten notation we will write MATH for the exterior direct sum MATH, so that the formula to be established is MATH . The plan is to compute the multiplier ideal on the left using the log resolution MATH. Specifically: MATH . Note to begin with that MATH thanks to the fact that MATH and MATH have no common components. Furthermore, as MATH and MATH are smooth: MATH . Since MATH, it then follows that MATH . Therefore MATH thanks to the NAME formula. But MATH since MATH and MATH are flat. Finally, MATH by virtue of the fact that MATH is flat for MATH (compare CITE). This completes the proof of the Proposition. |
math/0002035 | We apply REF to the diagonal MATH. Keeping the notation of the previous proof (with MATH, MATH), one has MATH . But it follows from REF that MATH as required. |
math/0002035 | Given MATH, fix MATH plus a general divisor MATH. Then MATH so the assertion follows from the Theorem. |
math/0002035 | This does not follow directly from the statement of REF because the divisor of a general element of MATH is not the sum of divisors of elements in MATH and MATH. However the proof REF goes through to show that MATH and then as above one restricts to the diagonal. |
math/0002035 | Only REF requires a proof, since REF follows again from REF by the restriction principle and the diagonal trick. Let us fix two relatively compact NAME open subsets MATH, MATH. Then MATH is the NAME tensor product of MATH and MATH, and admits MATH as a NAME basis, where MATH and MATH are respective NAME bases. Since MATH is generated as a MATH module by the MATH, we conclude that REF holds true. |
math/0002035 | Clearly, MATH for every effective divisor MATH. We can take MATH so large that MATH is very ample, and we are thus reduced to the case where MATH itself is very ample by replacing MATH with MATH. By definition of MATH, there exists a sequence MATH such that MATH . We now fix an integer MATH (to be chosen precisely later), and put MATH, so that MATH, MATH. Then MATH . Fix next a constant MATH such that MATH is an effective divisor for each MATH. Then MATH is effective, and hence MATH . We select a smooth divisor MATH in the very ample linear system MATH. By looking at global sections associated with the exact sequences of sheaves MATH, we infer inductively that MATH where MATH depends only on MATH and MATH. Hence, by putting MATH, we get MATH and the desired conclusion follows by taking MATH. |
math/0002035 | Note to begin with that it is enough to produce a big and nef divisor MATH satisfying the conclusion of the Theorem. For by NAME 's Lemma one can write MATH where MATH is an effective MATH-divisor, and MATH is an ample MATH-divisor. Then MATH where MATH is ample and the top self intersection number MATH approaches MATH as closely as we want. Fix now a very ample bundle MATH on MATH, set MATH, and for MATH put MATH . We can suppose that MATH is very ample, and we choose a divisor MATH. Then multiplication by MATH determines for every MATH an inclusion MATH of sheaves, and therefore an injection MATH . Given MATH, we use REF to fix MATH such that MATH . We further assume that MATH is sufficiently large so that MATH is big. Having fixed MATH satisfying REF , we will produce an ideal sheaf MATH (depending on MATH) such that MATH . Granting for the time being the existence of MATH, we complete the proof. Let MATH be a log resolution of MATH, so that MATH for some effective divisor MATH on MATH. It follows from REF that MATH is globally generated, and hence nef. Using REF we find: MATH (which shows in particular that MATH is big). This implies that MATH so the Theorem follows upon setting MATH and MATH. Turning to the construction of MATH, set MATH . Since MATH, REF follows from REF applied to MATH. As for REF we first apply REF to MATH, together with the subadditivity property in the form of Variant REF, to conclude: MATH . Now the sheaf homomorphism MATH evidently remains injective for all MATH, and consequently MATH . The required inclusion REF follows by combining REF . This completes the proof of NAME 's Theorem. |
math/0002035 | We start by interpreting MATH geometrically. Let MATH be a log resolution of MATH, with MATH, where MATH is free, and MATH, so that MATH. Then evidently MATH counts the number of intersection points of MATH general divisors in MATH, and consequently MATH . We have MATH for MATH since then MATH is big (and nef), and MATH since MATH embeds in MATH. Hence MATH . On the other hand, an argument in the spirit of REF shows that MATH REF , and so we conclude that MATH for every MATH. For the reverse inequality we use NAME 's theorem. Fix MATH, and consider the decomposition MATH on MATH constructed in REF . Let MATH be any positive integer such that MATH is integral and globally generated. By taking a common resolution we can assume that MATH dominates MATH, and hence we can write MATH with MATH globally generated and MATH . But then MATH gives rise to a free linear subseries of MATH, and consequently MATH . Therefore MATH . But REF holds for any sufficiently large and divisible MATH, and in view of REF the Proposition follows. |
math/0002038 | There is an imprimitivity bimodule isomorphism MATH defined on the generators by MATH since straightforward calculations verify that the above mapping on generators preserves the left action and the right inner product. |
math/0002038 | Straightforward calculations verify that the mapping MATH on the generators preserves the operations, hence extends to a MATH-isomorphism. |
math/0002038 | There is an imprimitivity bimodule isomorphism MATH defined on the generators by MATH since straightforward calculations verify that the above mapping on generators preserves the left action and the right inner product. |
math/0002038 | There is an imprimitivity bimodule isomorphism MATH defined on the generators by MATH since straightforward calculations verify that the above mapping on generators preserves the left action and the right inner product. |
math/0002038 | Straightforward calculations verify that the mapping MATH on the generators preserves the operations, hence extends to a MATH-isomorphism. |
math/0002038 | There is an imprimitivity bimodule isomorphism MATH defined on the generators by MATH since straightforward calculations verify that the above mapping on generators preserves the left action and the right inner product. |
math/0002038 | The desired diagram is the inner rectangle of the diagram MATH (Here and in REF the action and coaction symbols have been omitted for clarity.) We will show how to fill in the bottom arrow so that each of the outer rectangle and the top, bottom, left, and right quadrilaterals commute. Since MATH is surjective, the result will then follow from standard bimodule techniques. Consider the diagram MATH . Since the action MATH of MATH extends to MATH, it follows from standard facts concerning induced actions that MATH extends to an isomorphism MATH. Then an easy check on the generators shows that MATH extends to an isomorphism MATH of MATH imprimitivity bimodules. This shows the left triangle of REF commutes. The inner quadrilateral in REF is the commutative REF . We define the isomorphism MATH at the bottom arrow of REF so that the bottom triangle commutes. On the generators, this isomorphism is given by MATH . Thus the outer rectangle in REF commutes. We noticed in REF that the top quadrilateral in REF commutes. For the right quadrilateral, MATH is functorial in MATH, so the homomorphism MATH yields an imprimitivity bimodule homomorphism MATH with the desired coefficient homomorphisms. By CITE, this implies the quadrilateral commutes. Similarly, the left quadrilateral commutes by functoriality of MATH: the homomorphism MATH yields an imprimitivity bimodule homomorphism MATH with the desired coefficient homomorphisms. Finally, the bottom quadrilateral in REF commutes by a routine computation on the generators. |
math/0002038 | The desired diagram is the inner rectangle of the diagram MATH . Consider the diagram MATH . It follows from CITE that the map MATH extends to an isomorphism MATH, and this serves as the left-hand coefficient map for an isomorphism MATH, hence the left triangle of REF commutes. The inner quadrilateral is the commutative REF . We define the isomorphism MATH at the bottom arrow of REF so that the bottom triangle commutes. On the generators, this isomorphism is given by MATH . Thus the outer rectangle in REF commutes. We noticed in REF that the top quadrilateral in REF commutes. MATH is functorial in MATH, so the homomorphism MATH yields an imprimitivity bimodule homomorphism MATH with the desired coefficient homomorphisms, so the right quadrilateral commutes. Similarly, the left quadrilateral commutes because by functoriality of MATH the homomorphism MATH yields an imprimitivity bimodule homomorphism MATH with the desired coefficient homomorphisms. Finally, the bottom quadrilateral in REF commutes by a routine computation on the generators. |
math/0002038 | The desired diagram is the outer rectangle of MATH . The upper left triangle commutes by REF, so we must show the lower right triangle commutes. We construct the isomorphism MATH as a composition MATH . Here, MATH is the coaction MATH of MATH on MATH, where MATH is the flip isomorphism. It follows from CITE (see also CITE) that there is an isomorphism MATH of MATH onto MATH defined on the generators by MATH . Perpetuating our perverse numbering scheme, we use MATH to define the coaction MATH of MATH on MATH, where MATH is the unitary element of MATH determined by the bounded function MATH from MATH to MATH. It is easy to see that MATH is a MATH-cocycle (more precisely, the obvious analogue for full coactions of the more usual cocycles for reduced coactions - see CITE). It follows from CITE (also see CITE) that there is an isomorphism MATH of MATH onto MATH defined by MATH where MATH is the regular representation of MATH on MATH for MATH (and MATH is faithfully represented on a NAME space MATH). Finally, from CITE we have the isomorphism MATH given on generators by MATH where MATH and MATH are the left and right regular representations of MATH and MATH denotes the characteristic function of the singleton MATH. (MATH is the isomorphism of NAME 's duality theorem CITE, but for maximal coactions rather than reduced ones.) The arguments of CITE, adapted to our context, show that MATH is equivariant for the coactions MATH and MATH of MATH; we define MATH to be the corresponding isomorphism of the crossed products. Careful study of the isomorphisms MATH, MATH, and MATH now shows that the composition MATH is given on the generators by MATH . Using this, straightforward calculations show that there is a MATH imprimitivity bimodule isomorphism MATH defined on the generators by MATH . |
math/0002038 | Since MATH is discrete, the coaction MATH is automatically nondegenerate, so by CITE there is a unique full coaction MATH of MATH on MATH whose reduction coincides with MATH, and then CITE gives an isomorphism MATH; it is easy to see that this isomorphism is equivariant for the dual actions. Then CITE applies, giving a maximal coaction MATH (the ``maximalization" of MATH) and an equivariant surjection MATH whose integrated form MATH is an isomorphism which is equivariant for the dual actions. Then MATH is also equivariant for the restricted coactions MATH and MATH, hence certainly gives a surjection MATH . Since MATH is the normalization of MATH, CITE tells us that, if MATH, then the ideal of MATH induced from MATH via the NAME imprimitivity bimodule MATH coincides with the kernel of the regular representation MATH so that MATH is canonically a MATH - MATH imprimitivity bimodule. But MATH is amenable, so the regular representation of MATH is faithful. Hence we must have MATH, so MATH is actually an isomorphism of MATH onto MATH. It is now clear from the constructions that the identity map on the ordered pairs MATH extends to an isomorphism MATH of the NAME imprimitivity bimodule MATH used in the present paper onto the version of the NAME bimodule associated to the normal coaction MATH in CITE. Thus the diagram MATH commutes. On the other hand, one of the main points of CITE is that the diagram MATH commutes; combining these shows that the top quadrilateral of the diagram MATH commutes. Since the outer rectangle commutes by REF, and the left, right, and bottom quadrilaterals are easily seen to commute, we conclude that the inner rectangle commutes as well. |
math/0002039 | If MATH, then the map MATH is NAME module automorphism of MATH which intertwines the left actions coming from MATH and MATH. For the converse, suppose MATH in MATH, so there exists a linear bijection MATH such that MATH for each MATH. Define MATH by MATH. Then the first two of these properties imply that MATH is an invertible double centralizer of MATH, so there exists an invertible element MATH of MATH such that MATH for all MATH. Since MATH it follows that MATH is unitary, and MATH implies that MATH. |
math/0002039 | We first claim that the composition of morphisms is well-defined. Suppose we have right-Hilbert bimodule isomorphisms MATH and MATH. Then MATH is easily seen to preserve the actions and inner product, so extends to an isometric bimodule map MATH. The map MATH is an inverse for MATH, which is therefore a right-Hilbert MATH - MATH bimodule isomorphism of MATH onto MATH. Next we establish that composition of morphisms in MATH is associative; it suffices to show that MATH and MATH are isomorphic for any right-Hilbert bimodules MATH, MATH, and MATH. But straightforward calculations show that the usual linear isomorphism of MATH onto MATH respects the module actions and right inner products, so extends to the desired isomorphism. Finally, note that the maps MATH and MATH extend to isomorphisms MATH and MATH for any right-Hilbert bimodules MATH and MATH. Hence MATH is an identity morphism from MATH to MATH in MATH. |
math/0002039 | By definition, MATH is the closed span of the rank-one operators MATH CITE, so it is enough to show that MATH. We can verify this by applying both sides to vectors of the form MATH, which densely span MATH. We then have MATH as required. Since MATH is faithful and MATH is a MATH - MATH imprimitivity bimodule, the induced representation MATH is faithful (this follows from the NAME correspondence CITE), and hence so is its extension MATH to MATH. |
math/0002039 | We have to show that if there is a right-Hilbert bimodule MATH such that MATH and MATH as right-Hilbert bimodules, then MATH is a MATH - MATH imprimitivity bimodule (and in fact MATH will then be isomorphic to MATH). We choose a faithful nondegenerate representation MATH of MATH on MATH, and use the inducing construction of REF to give a representation MATH of MATH on MATH with MATH. We shall prove the result by showing that the presence of MATH implies that MATH is an isomorphism of MATH onto the image MATH of the algebra MATH of compact operators. We begin by noting that because MATH acts faithfully on itself and MATH, MATH must act faithfully on MATH: if MATH for all MATH, then MATH for all MATH, whence MATH for all MATH, and MATH. Thus MATH is injective, and the injectivity of MATH implies that MATH is faithful. To show that MATH has the right image, we construct a representation MATH of MATH on MATH. Let MATH be the representation induced from MATH, given as usual by MATH. By assumption, there exists a right-Hilbert bimodule isomorphism MATH, which we use to define a unitary MATH by MATH . Now we define MATH . For MATH, MATH, and MATH we get MATH and we also have MATH for all MATH, since MATH has right coefficient map MATH. Thus MATH is indeed a right-Hilbert bimodule representation of MATH. In particular, we have MATH. It only remains to show that MATH. For this we observe that for MATH, MATH, and MATH we have MATH hence MATH. Using MATH, it now follows that MATH . This completes the proof of REF. |
math/0002039 | Let MATH and let MATH viewed as a MATH - MATH imprimitivity bimodule (see CITE). Since MATH CITE, we can view the canonical nondegenerate homomorphism MATH as a nondegenerate homomorphism MATH. The map MATH extends to an isomorphism of MATH onto MATH. |
math/0002039 | Adding actions to REF , and REF is routine, except possibly in REF . In the proof of REF, we took MATH and MATH, so we need to show that MATH induces an action on MATH. But for each MATH, MATH is an adjointable operator with MATH, and in fact MATH is an automorphism of MATH. A quick calculation shows that MATH so MATH restricts to an automorphism of MATH. Since MATH acts continuously on MATH, MATH gives an action of MATH on MATH, and REF says that MATH is an isomorphism in MATH. Now recall that MATH to see that MATH is MATH - MATH equivariant; it is then easy to check that the isomorphism MATH of MATH onto MATH is MATH - MATH equivariant. |
math/0002039 | We have MATH . |
math/0002039 | We begin by showing that MATH can be completed to give a NAME MATH-module MATH satisfying REF . Straightforward calculations show that REF make MATH into a pre-inner product MATH-module, so we need only verify that the sesquilinear form of REF is positive definite. To do so, fix, for the remainder of the proof, a faithful representation MATH of MATH on a NAME space MATH. Then for each MATH, REF gives MATH which shows positivity. For definiteness, suppose MATH satisfies MATH. Then REF gives MATH for each MATH, so that MATH for each MATH. Thus, MATH for each MATH and MATH, where MATH defines MATH. It follows that MATH for all MATH, whence MATH in MATH. We next show that the MATH-valued inner product on MATH is full. Since MATH has an approximate identity for MATH, since MATH is dense in MATH, and since MATH acts nondegenerately on MATH, functions of the form MATH, where MATH and MATH, span a dense subspace of MATH. Now letting MATH define MATH, a straightforward calculation shows that MATH, so that MATH is dense in MATH. We now claim that REF makes MATH into a right-Hilbert MATH - MATH bimodule. Again, checking the algebraic conditions of REF at the level of MATH-functions is routine; we need to show that MATH for MATH and MATH to see that this extends to an action of MATH on MATH. To show this, we begin by defining actions of MATH and MATH on MATH by MATH . Now we use these actions to define MATH we claim that MATH is bounded on MATH, and hence extends to an operator on MATH. Again using REF, and writing MATH for MATH, we have MATH where the inequality holds because MATH acts boundedly on MATH via the induced representation MATH. It is straightforward to check that each MATH is unitary, and then that MATH is covariant for MATH; MATH is nondegenerate because MATH acts nondegenerately on MATH, so we get a nondegenerate representation MATH of MATH on MATH. Now for MATH, MATH, and MATH, we have MATH so MATH . It follows that MATH. Finally, to see that the action of MATH on MATH is nondegenerate, note that for MATH, the NAME inequality gives MATH so MATH. Now standard arguments show that we can choose MATH to make MATH arbitrarily small: take MATH of the form MATH as MATH runs through an approximate identity for MATH and MATH runs through an approximate identity for MATH. |
math/0002039 | We first show that the map on morphisms is well-defined. Suppose MATH is an isomorphism of right-Hilbert MATH - MATH bimodules which is equivariant for MATH-compatible actions MATH and MATH of MATH. Then MATH is easily seen to give a bijective map MATH which respects the right-Hilbert bimodule structures (REF - REF ) and hence extends to a right-Hilbert MATH - MATH bimodule isomorphism of MATH onto MATH. REF shows that identity morphisms go to identity morphisms. It only remains to see that the assignment MATH respects composition of morphisms; that is, if MATH is MATH-compatible and MATH is MATH-compatible, we need to show that MATH as right-Hilbert MATH - MATH bimodules. The rule MATH defines a linear map from MATH to MATH which preserves the pre-right-Hilbert bimodule structures. In order to see that MATH extends to an isomorphism of the completions, we need only verify that MATH has dense range for the inductive limit topology. For this, let MATH and MATH, and define MATH by MATH. Then for MATH we have MATH . Now, we can approximate MATH by MATH in the inductive limit topology, and taking MATH of the form MATH for MATH and MATH we can thus approximate the function MATH. But such functions have inductive-limit-dense span in MATH. |
math/0002039 | We establish the commutativity of both REF by factoring MATH, as in REF , where MATH is an isomorphism in MATH and MATH is a MATH - MATH equivariant nondegenerate homomorphism. We shall prove the commutativity of REF by showing that we have two commutative diagrams MATH and MATH . Because taking crossed products is a functor, this will then give MATH as required. The commutativity of REF amounts to: Suppose MATH, MATH are objects in MATH and MATH is a nondegenerate homomorphism such that MATH. Then MATH as right-Hilbert MATH - MATH bimodules. The second isomorphism is standard. For the first, define MATH by MATH . The usual change-of-variables arguments show that MATH is right MATH-linear (though strictly speaking it is not necessary to prove this). The map MATH is given by MATH using this identity, some convoluted calculations involving several changes of variables show that MATH for MATH and MATH. Thus MATH converts the inner product on the internal tensor product MATH to the usual one on MATH. To see that MATH has dense range, and hence extends to an isomorphism of right-Hilbert bimodules, it is enough to approximate elements in MATH of the form MATH, where MATH and MATH, with elements in the range of MATH. Moreover, a partition of unity argument shows that functions of the form MATH for MATH and MATH span an inductive-limit dense - and hence norm-dense - subspace of MATH, so we can assume that MATH has this form. But now a routine calculation shows that if MATH, then MATH in MATH; this implies that the range of MATH is dense. Another calculation using the analogue of REF for the homomorphism MATH shows that this isomorphism respects the left action of MATH. To prove the commutativity of REF , we use a device from CITE. Recall that if MATH is an imprimitivity bimodule, then the linking algebra is the collection of MATH matrices MATH with multiplication and involution given by MATH and MATH . This has a unique complete MATH-norm, obtained, for example, by identifying MATH with the MATH-algebra MATH of compact operators on the NAME module direct sum MATH (see CITE). The matrices MATH define full projections in MATH which allow us to identify MATH, MATH and MATH with corners in MATH; we can then use the projections MATH, MATH to break up a NAME MATH-module into modules over MATH and MATH. Our key technical result describes some relations among these submodules. It is a mild generalization of CITE. Suppose MATH and MATH are imprimitivity bimodules, and MATH is a right-Hilbert MATH - MATH bimodule. Then CASE: MATH is a right-Hilbert MATH - MATH bimodule; CASE: MATH is a right-Hilbert MATH - MATH bimodule; CASE: MATH is a right-Hilbert MATH - MATH bimodule; CASE: there is an isomorphism MATH of right-Hilbert MATH - MATH modules such that MATH; CASE: there is an isomorphism MATH of right-Hilbert MATH - MATH modules such that MATH. Together, the last two parts of REF say that the diagram MATH commutes in MATH. If we had such a proposition when MATH and MATH are just right-Hilbert bimodules, and the linking algebras are by REF and MATH, we could avoid having to factor our morphisms. However, we need to know that MATH and MATH are imprimitivity bimodules to identify the corners MATH and MATH with MATH and MATH. Since MATH in particular occurs in the middle of the internal tensor products in REF , it is hard to see how this hypothesis might be avoided. Suppose MATH is a right-Hilbert bimodule and MATH, MATH are full projections. Then MATH is a right-Hilbert MATH - MATH bimodule. Since MATH is certainly a MATH - MATH submodule, and since MATH we only have to check nondegeneracy of the left action and fullness on the right. For nondegeneracy, we use the fullness of MATH to see that MATH is dense in MATH. For fullness, we use the fullness of MATH again to see that MATH is dense in MATH. For REF , apply REF with MATH and MATH, and note that MATH and MATH because MATH and MATH are imprimitivity bimodules. REF follow similiarly. For REF , we first note that because MATH is bilinear, there is a well-defined map MATH on the algebraic tensor product MATH with the required property. We next verify that MATH preserves the inner product: if MATH and MATH, then the inner product MATH is given in MATH by MATH; more formally, MATH . Thus MATH and MATH extends to an isometry of MATH into MATH. To see that MATH has dense range and is therefore onto, note that MATH acts nondegenerately on MATH, so MATH is dense because MATH is full. For REF , note that the MATH-valued inner product is given by the product in MATH, so MATH which is MATH because MATH and MATH belong to MATH. Thus MATH extends to an isometry MATH, as claimed, and MATH is surjective because MATH is full and MATH acts nondegenerately on MATH. NAME to the proof of REF. Consider the action MATH of MATH on MATH; by NAME 's theorem, MATH is then a right-Hilbert MATH - MATH bimodule. To show that REF commutes, we aim to apply REF to MATH, but this requires that we first identify MATH with MATH, and similarly for MATH. Now the dense subalgebra MATH consists of MATH matrices with entries in MATH, MATH, etc., and the diagonal corners have their usual MATH-algebraic structures as subalgebras of MATH and MATH. The norms are the same too: every covariant representation of MATH restricts to a covariant representation of MATH, and the inducing construction of REF shows that every covariant representation of MATH extends to a covariant representation of MATH. So we have embeddings of MATH and MATH as corners in MATH. The bimodule crossed product MATH also embeds: because MATH embeds isometrically, the norm of a function MATH is just MATH. These embeddings combine to give an isomorphism of MATH onto MATH, as desired. (This observation is not new: in CITE, Combes defines MATH to be the corner in MATH.) In exactly the same way, MATH. We next have to identify the corners MATH and MATH with MATH and MATH. But viewing elements of MATH as matrices of functions, and similarly for MATH and MATH, gives the desired identifications. It now follows from REF that MATH as right-Hilbert MATH - MATH bimodules. In other words, REF commutes in the category MATH. We have now established the first part of REF, the natural transformation between the functors MATH and MATH. For the second part, we have to prove that REF commutes. We follow the same procedure as before: factor a given morphism MATH as MATH, and prove that two separate diagrams commute. To see the commutativity of MATH we merely note that the isomorphism of REF respects the left action of MATH, and hence is also an isomorphism of right-Hilbert MATH - MATH bimodules. To see that MATH commutes, we again apply REF to MATH, this time viewed as a MATH - MATH imprimitivity bimodule. To do so, we have to identify MATH with MATH, but there is no essential difference in the argument. This concludes the proof of our main theorem. |
math/0002040 | Denote the meridians of the components of MATH by MATH. In MATH the framings MATH can uniquely be expressed as MATH. This implies for MATH that MATH . In MATH we obtain the following unique expression of MATH REF in terms of the meridians MATH REF of MATH: MATH . This implies the lemma. |
math/0002040 | Choose a diagram of MATH such that MATH and put dots on the components of MATH. Let MATH be the linking matrix of MATH and let MATH be the inverse of the linking matrix of MATH. Then for a series MATH (respectively, MATH) of diagrams in MATH (respectively, in MATH) that contains no struts and has degree-MATH-term MATH, we have MATH . Since MATH, REF follows from REF . |
math/0002040 | Let MATH (respectively, MATH) be a knot in the upper part MATH (respectively, in the lower part MATH) of a tubular neighborhood of MATH representing the MATH-th basis element of MATH. Let the knot MATH have the framing MATH induced by the surface MATH. First consider MATH. Define MATH as the composition of MATH with the projection to the part containing only powers of the strut MATH. REF implies that MATH. Represent MATH by a surgery diagram MATH where the tangle MATH consists of the MATH-th and MATH-th framed strands of MATH and MATH is a MATH-framed tangle consisting of two intervals close to MATH. See REF for an example (compare REF ). In this figure the dotted line separates MATH from MATH and is not a part of the diagram. We have MATH and the explicit description of MATH (see CITE) implies MATH . Observe the following property of MATH: MATH . The last two formulas and REF imply MATH . Using REF for MATH we see that MATH. For MATH . REF implies MATH. We apply REF for MATH as above and obtain MATH. By REF we have MATH which completes the proof. |
math/0002040 | We use the notation from above. For suitable distinguished components of MATH and of MATH the tangle MATH coincides with the part of the framed oriented boundary of MATH that belongs to MATH. Let MATH be the part of the framed oriented boundary of MATH that belongs to MATH. We regard MATH as a non-associative tangle with MATH. The invariant MATH depends only on MATH. Since we know that the NAME matrix MATH is chosen with respect to a basis induced by a standard surface, the definition of the map MATH depends only on MATH (see REF ). We will show below that for knots MATH all terms in MATH that contain an internal vertex do not contribute to MATH. REF then imply MATH . This will show that MATH is determined by the NAME matrix MATH. Obviously, the coefficients of MATH in MATH are polynomials of degree MATH in the entries of MATH. This will prove the keylemma. It remains to consider diagrams MATH in MATH with an internal vertex MATH. In MATH each of the edges incident to MATH is either connected to another internal vertex or appears twice, namely as the difference of the two ways of lifting it to the skeleton MATH. We represent this difference by a box in REF . A neighborhood of the internal vertex MATH looks like in one of the possibilities REF . When we push a lifted vertex in the box along the circle MATH, then it will finally cancel with the second lifted vertex. By the (STU)-relation we can replace a box in REF by a sum of diagrams with an additional internal vertex. More precisely, a part of the diagram looking like in REF , or REF is replaced by a sum of diagrams where a neighborhood of MATH looks like in diagrams that can be reached by following a directed arrow in REF . When we apply this procedure to all boxes, we will finally end up with possibilities REF . By REF all diagrams that have a subdiagram as in REF or REF are sent to MATH by MATH. |
math/0002040 | By REF and NAME REF the coefficients of MATH in the two power series MATH and MATH only depend on a NAME matrix MATH of a knot MATH and are polynomials MATH and MATH in the entries of MATH. By REF we have MATH for all NAME matrices of knots in MATH. REF implies that MATH for all MATH. |
math/0002040 | The invariants MATH and MATH of MATH differ only by normalization (see REF ). Let MATH. Then we have MATH with MATH. The following four steps show that MATH can be calculated from MATH and vice versa. This will complete the proof. CASE: MATH depends only on the wheel-part MATH of MATH REF . CASE: MATH can be calculated from MATH because MATH where MATH is a formal series of connected wheels (see REF ), MATH, and MATH is injective on connected wheels REF . CASE: MATH contains no struts because MATH is MATH-framed (see REF ). All remaining non-vanishing diagrams in MATH have at least as many internal vertices as univalent vertices. This implies that MATH depends only on MATH. CASE: The map MATH is injective on wheels (see REF , use the MATH-weight system on MATH to see that MATH is injective on connected wheels). Therefore MATH can be calculated from MATH. MATH is invertible. |
math/0002040 | By REF and by REF it is sufficient to show that for a null-homotopic knot MATH in a rational homology sphere each of the invariants MATH and MATH can be computed from the other one. This statement follows from REF . |
math/0002041 | The circle MATH acts on the product manifold MATH by MATH. The function MATH is MATH invariant. The level set MATH is a (set theoretic) disjoint union of two manifolds MATH . The points of the first manifold are regular points of MATH because they are regular points of the function MATH. The points of the second manifold are regular points of MATH by assumption on MATH. Moreover, by assumption on MATH and by construction the circle acts freely on MATH. Therefore the quotient MATH is a smooth manifold. The composition of MATH, MATH with the orbit map MATH descends to a homeomorphism MATH . This gives the cut MATH the structure of a smooth manifold. Moreover, MATH restricted to MATH is an open embedding and MATH is MATH. |
math/0002041 | We may assume that a neighborhood of the boundary MATH in MATH is diffeomorphic to MATH. Then MATH. The circle action on MATH extends trivially to a circle action on MATH making the projection map MATH-invariant. By REF , MATH is a smooth manifold. Moreover, it is easy to see that MATH hence is smooth as well. |
math/0002041 | (compare CITE) Consider the symplectic product MATH. The map MATH is a moment map for an action of MATH on MATH. Arguing as in REF we see that MATH is a regular value of MATH and that the reduced space MATH is the cut MATH. The pullback of the symplectic form MATH by the embedding MATH is MATH. Consequently the induced embedding MATH is symplectic. Similarly one checks that the natural embedding MATH is symplectic as well. |
math/0002041 | By REF MATH is a smooth manifold, MATH is a submanifold and MATH is diffeomorphic to MATH. We now assume for simplicity that MATH is connected. Otherwise we can argue connected component by connected component. By the equivariant coisotropic embedding theorem the product MATH carries and MATH-invariant closed REF-form MATH (MATH acts on MATH by MATH) such that CASE: MATH CASE: the MATH action on MATH is Hamiltonian with a moment map MATH. Moreover there is an open MATH equivariant embedding MATH of a neighborhood MATH of MATH in MATH into MATH such that CASE: MATH for all MATH, CASE: MATH for all MATH and CASE: MATH. In particular MATH is non-degenerate near MATH. We have MATH. By REF MATH is a symplectic manifold, the embedding of MATH into MATH is symplectic and the difference MATH is symplectomorphic to MATH. Therefore MATH is a symplectic manifold with the desired properties. |
math/0002041 | The argument is an adaptation of NAME 's proof of the existence of invariant Riemannian metrics on manifolds with proper group actions CITE. Suppose first that the group MATH is compact. Then there is on MATH a bi-invariant measure MATH normalized so that MATH. We then define MATH to be the average of MATH: MATH for all MATH. Now we drop the compactness assumption. None the less, for every point MATH there exists a slice MATH to the action of MATH. That is, MATH is MATH -invariant embedded submanifold of MATH such that the union MATH of MATH-orbits through the points of MATH is open and such that for every MATH the orbit MATH intersects MATH in a single MATH-orbit (see CITE). Note that the stabilizer MATH of MATH is compact because the action is proper. Since the contact structure MATH is MATH-invariant its annihilator MATH is a MATH-invariant line subbundle of the cotangent bundle. Hence the restriction MATH is completely determined by the restriction MATH. Moreover, any MATH-invariant section MATH is completely determined by its values on MATH: MATH for all MATH, MATH. A contact form MATH is a nowhere vanishing section of MATH. Given a slice MATH we produce a MATH-invariant section MATH of MATH by averaging MATH over MATH. We then extend MATH to the open set MATH by the formula MATH. The form MATH is well-defined on MATH because MATH for any MATH. Next cover MATH by open sets MATH of the form MATH where MATH are slices. In the proof of CITE NAME showed that this cover may be chosen to be locally finite and that there exist a MATH-invariant partition of unity MATH subordinate to the cover. The form MATH is the desired invariant contact form. Here MATH is the average of MATH. |
math/0002041 | The proof is identical to the proof of REF. The main idea of the proof is that a point MATH lies in the zero level set of the moment map if and only if the orbit MATH is tangent to MATH; hence MATH descends to a REF-form MATH on MATH. |
math/0002041 | Consider the contact manifold MATH with the circle action MATH. The map MATH is the corresponding moment map. Arguing as in REF we see that MATH is a regular value of MATH and that the reduced space MATH is the cut MATH. The pullback of the contact form MATH by the embedding MATH is MATH. Consequently the induced embedding MATH is contact. Similarly one checks that the natural embedding MATH is contact as well. |
math/0002041 | By REF MATH is a smooth manifold and MATH is diffeomorphic to MATH. We would like to show that MATH is contact. We have MATH. By REF MATH is a contact manifold, the embedding of MATH into MATH is contact and the difference MATH is contactomorphic to MATH. Therefore MATH is a contact manifold with the desired properties. |
math/0002041 | The vector field MATH descends to a vector field MATH on a neighborhood of MATH. It is not hard to see that the reduced symplectic form MATH satisfies MATH and that MATH is the reduced contact form MATH. |
math/0002041 | The hypersurface MATH in MATH is of contact type. The cut MATH is the reduction of MATH, and the cut MATH is the reduction at zero of MATH. Now apply REF . |
math/0002041 | As in REF let MATH denote the manifold MATH with coordinates MATH, MATH and MATH and a contact form MATH. For each non-negative integer MATH consider an embedding MATH of MATH into MATH: MATH . Let MATH. Consider the action of MATH on the boundary MATH generated by MATH on MATH and by MATH on MATH. Then as in REF , the cut manifold MATH is MATH. By REF each of the contact forms MATH induces a contact from MATH on MATH. Note that by REF the contact from MATH defines the standard tight contact structure on MATH. We now argue that CASE: All contact forms MATH are homotopic as non-vanishing one forms, that is, for all MATH there is a family of REF MATH with MATH, MATH, and MATH for all MATH. Hence the corresponding contact structures are all homotopic as REF-plane fields. CASE: For MATH the contact structures MATH are overtwisted. Since overtwisted contact structures on REF-manifolds are classified by the homotopy type of the corresponding REF-plane fields CITE, it would then follow that all the contact structures MATH are equivalent provided MATH. CASE: For any two distinct integers MATH, there is no contactomorphism MATH which is MATH equivariant. REF together then prove the theorem. To construct a path between MATH and MATH consisting of nowhere vanishing REF-forms, consider first the straight line homotopy between MATH and MATH on MATH. It fixes the end points and so descends to a homotopy between MATH and MATH on MATH. If the homotopy vanishes at a point MATH at a time MATH correct it by pushing the path out in the MATH direction. To see that the contact structures defined by MATH REF are overtwisted, fix MATH and consider the subset MATH given by MATH where MATH is the class of MATH. The set MATH is an overtwisted disk in MATH. To prove REF consider two contact connected manifolds MATH, MATH with an action of a NAME group MATH preserving the contact forms. Let MATH REF denote the corresponding moment maps. If MATH is a MATH-equivariant contactomorphism, then MATH for some nowhere vanishing MATH-invariant function MATH on MATH. Since MATH is connected, MATH is either always positive or always negative. Say MATH. Then for any nonzero vector MATH the preimages MATH of the ray through MATH have the same number of connected components (if MATH then MATH). Now for any MATH the map moment map MATH for the action of MATH is given by MATH . Hence MATH . Therefore REF follows and so does the theorem. |
math/0002041 | The proof is essentially the same as the proof of REF . We start by giving a description of lens spaces which is convenient for our purposes. Consider again MATH. Consider the action of MATH on the boundary of MATH defined on MATH by the vector field MATH and on MATH by the vector field MATH. By REF the cut MATH is a manifold. Note that the standard action of MATH on MATH descends to an action on MATH. One can show that MATH. Next fix MATH and consider for each MATH the embedding MATH where MATH is the unique angle with MATH and MATH. Let MATH. By REF each of the contact forms MATH induces a contact from MATH on MATH. As in the proof of REF one shows that CASE: For all MATH the contact structures MATH and MATH are homotopic as REF-plane fields. CASE: The contact structures MATH are overtwisted for all MATH, hence are all equivalent. Let us denote the contact structure defined by MATH, MATH by MATH. This is the contact structure in the statement of the theorem. Finally by examining the number of connected components of the fibers the appropriate moment maps one sees that for any two distinct integers MATH, there is no contactomorphism MATH which is MATH equivariant. |
math/0002042 | Of course, MATH is regular at each regular point of MATH. Furthermore, given MATH singular point of MATH, we have MATH and MATH, hence MATH a singular point of MATH iff MATH is a twist point of MATH. Now, let MATH the twist points of MATH and MATH their projections by MATH. Then, MATH is regular over MATH and, by compactness, it satisfies REF of NAME fibrations, the regular fiber MATH with MATH being simple covering of MATH branched over the (transversal) intersection with MATH, by the restriction of MATH. On the other hand, since MATH is simple, over each singular value MATH there is only one singular point MATH. In order to verify REF of NAME fibrations, we have to check that the monodromy around each MATH is a NAME twist. Let MATH be local fiber preserving complex coordinates of MATH centered at MATH and making MATH into the surface MATH. We can assume that MATH is orientation preserving on MATH, so that MATH, with MATH sufficiently small, is a counterclokwise parametrization of a simple loop MATH around MATH. Then MATH is the closed braid in MATH, corresponding to a half twist around an arc MATH between two branch points of the restriction of MATH over MATH, whose meridians have the same monodromy. Such a half twist is right-handed (respectively, left-handed) if MATH is a positive (respectively, negative) twist point of MATH and lifts to the right-handed (respectively, left-handed) NAME twist along the unique simple loop MATH contained in MATH (compare REF ), which represents the monodromy of MATH. Finally, assuming MATH, we have that each component of the regular fiber MATH has non-empty boundary, since it is a branched covering of MATH. Similarly, for the loop MATH considered above, we have that each component of MATH has non-empty boundary. Then, we can conclude that MATH is allowable if MATH. |
math/0002042 | First of all, since MATH and MATH are connected, there exists a REF-fold simple branched covering MATH, with MATH as in the statement, such that any NAME twist of MATH along a non-separating simple loop can be realized, up to isotopy, as the lifting of a half twist around an arc in MATH between two branch points of MATH, whose meridians have the same monodromy (see CITE and remember that all the non-separating simple loops in MATH are equivalent). Then, any element of MATH can be represented by the lifting of a diffeomorphism of MATH onto itself isotopic to the identity, since NAME twists along non-separating simple loops generate MATH. Let MATH be the branch points of MATH and MATH be disjoint disks such that MATH and MATH is an arc in MATH for every MATH. Then, the restriction of MATH over MATH is a locally trivial fiber bundle. Given a band presentation MATH with bands REF MATH, we construct a branched covering MATH as follows: start with the covering MATH; cut each MATH along MATH and each MATH along MATH, where MATH is a transversal arc for the band MATH; glue them back respectively by MATH and MATH, where MATH is the monodromy of a simple loop MATH which goes once through MATH and MATH is a homeomorphism isotopic to the identity which lifts to MATH by means of MATH. In order to extend MATH to a branched covering MATH, we consider a branched covering MATH whose branch set is a surface braided over MATH with only one positive twist point over MATH and whose restriction over MATH coincides with MATH. As we have seen in the proof of REF , the composition MATH is a NAME fibration branched over MATH with regular fiber MATH, such that the monodromy of a counterclockwise meridian loop around MATH is a right-handed NAME twist along a non-separating simple loop MATH. Now, for any MATH, let MATH, where MATH is a counterclockwise meridian loop around MATH, MATH is the right-handed NAME twist along MATH and-MATH. Since MATH is allowable, MATH cannot separate MATH, so there exist diffeomorphisms MATH and MATH such that: MATH preserve or invert the orientation according to MATH; MATH is orientation preserving and lifts to MATH with respect to MATH; MATH. Then, assuming that the arcs MATH do not meet REF-handles MATH, we can glue MATH copies of MATH to MATH, by means of the diffeomorphisms MATH and MATH. Calling MATH the branched covering of MATH obtained in this way, we have that MATH is a NAME fibration whose branch points a monodromy coincide with that ones of MATH, by REF and its proof. So, up to orientation preserving diffeomorphisms, MATH and in particular the total space of MATH is MATH. |
math/0002042 | The connection of MATH follows immediately from the connection of MATH, since the monodromy of MATH is generated by NAME twists, so it preserves the components of MATH. We also observe that, for the same reason, the monodromy of MATH fixes the boundary of MATH. Now, if MATH or MATH is already connected, we can set MATH. NAME, in order to connect the boundary of MATH, we consider the following plumbing operation for NAME fibrations with connected bounded fiber, which is analogous to the operation REF introduced by NAME in CITE for open-book decomposition. Let MATH the surface obtained by gluing an oriented band MATH to MATH (we are assuming MATH) and MATH be a simple loop which goes once through MATH (we are also assuming MATH connected). Then, we consider the new NAME fibration MATH with regular fiber MATH, branch points MATH and respective monodromies MATH, where MATH are the branch points of MATH, MATH are the respective monodromies for MATH thought as NAME twists of MATH and MATH is the right-handed NAME twist along MATH. By the definition of MATH, we get MATH, in fact MATH can be obtained by adding to MATH a cancelling pair of handles: one REF-handle MATH glued to MATH (remember that the monodromy of MATH fixes MATH), due to the change of the fiber, and one REF-handle attached along MATH with MATH, due to the new branch point MATH (compare CITE and CITE). On the other hand, if MATH is not connected and the band MATH joins two different components of MATH, then MATH has one component less than MATH and MATH is non-separating in MATH. Then we can get the required NAME fibration MATH from MATH, by iterating the plumbing operation, until the boundary of the fiber becomes connected. |
math/0002042 | CASE: Given an analytic branched covering MATH, we have that MATH is a NAME surface without boundary, since the restriction of MATH to MATH is a finite holomorphic map (see CITE, p. REF). Let MATH be a proper strictly plurisubharmonic function and MATH be the plurisubharmonic function defined by MATH. By the transversality of the branch set of MATH with respect to MATH, we have MATH for MATH (regular value) sufficiently large. Now, the function MATH is proper and strictly plurisubharmonic on MATH, for every MATH. By choosing MATH sufficiently small, we have also MATH, hence MATH is a NAME surface with boundary. CASE: Let MATH a covering branched over a positive braided surface MATH. By REF , MATH is analitically branched (see CITE for the definition). Then, by REF and CITE (compare CITE), MATH is a true analytic covering of MATH. CASE: This implication follows immediately from REF . CASE: Let MATH be a NAME surface with boundary. By REF , it is enough to find a simple branched covering MATH, whose branch set is a positive braided surface. We start with a handle decomposition MATH,where MATH consists of MATH- and MATH-handles and the MATH's are Legendrian REF-handles attached to MATH. In order to make the proof easier to read, we consider first the special case of one REF-handle attached to MATH. This allows us to explain the crucial ideas of the proof, avoiding many technical details. Then, we show how to deal simultaneously with different REF-handles and how to work the presence of REF no REF-handles and one REF-handle. In this case, we have MATH, for a Legendrian REF-handle MATH. Let MATH the Legendrian attaching knot of MATH. Then, MATH can be represented by a front projection diagram MATH as described above. An example of such a diagram is depicted in REF ; all the diagrams in the following REF have to be considered as successive modifications of this one. First of all, we smooth all the cusps and add a negative kink at each right one. In this way, we get a new diagram MATH of MATH (in fact of a transversal knot parallel to MATH, compare CITE) whose blackboard framing represent the Legendrian framing of MATH (see REF ). Then, we redraw MATH as a polygonal diagram with smoothed corners and edges of slope MATH or MATH, paying attention to not introduce local minima or maxima for the abscissa other than the ones coming from cusps, and rotate everything of MATH radians. The resulting diagram MATH (see REF ) has the following properties: all the edges of MATH are horizontal or vertical; at each crossing the vertical edge crosses in front; any vertical edge belongs to one of the three types shown in REF , depending on the local structure of MATH in a neighborhood of it. Finally, we apply to MATH the moves described in REF , in order to get a new diagram MATH, satisfying the same properties of MATH, with all the vertical edges of types REF respectively in the left-most and the right-most positions. Of course, also MATH is a diagram of MATH (up to smooth equivalence) whose blackboard framing represent the Legendrian framing of MATH. The vertical edges of the types REF come respectively from the left cusps and the right cusps of the diagram MATH. Hence, putting MATH, we have exactly MATH vertical edges of type REF and MATH vertical edges of type REF. Let MATH be all such edges, numbered starting from the uppermost one of type REF and following the orientation of the diagram which induces on it the up-down orientation. We can assume that MATH has been constructed in such a way that, going from left to right, we have in the order MATH on the left side of MATH and MATH on the right side of MATH (see REF ). Now, we consider the simple branched covering MATH with MATH sheets labelled from MATH to MATH, whose branch set consists of disks MATH parallel to the second factor and whose monodromy around MATH is MATH, for every MATH. We think MATH as parallel disks in MATH-with interiors pushed inside MATH and represent their boundaries as vertical lines MATH in the diagram. Furthermore, we assume that: MATH and MATH for MATH; MATH lies immediately on the right (respectively, left) of MATH for MATH odd (respectively, even); MATH crosses in front of MATH at all the crossings except the upper (respectively, lower) one near to MATH for MATH odd (respectively, even), as shown in REF . Let MATH be new vertical edges with the following properties: MATH is collinear with MATH for any MATH; all the MATH's lies above all the MATH's; the projections of the edges MATH on MATH have disjoint interiors and their union coincides with MATH; the bottom end of MATH and the top end of MATH have the same ordinate for any MATH. Then, we join the MATH's by horizontal edges, in order to get a trivial knot diagram linked with the MATH's as shown in REF , where the horizontal edges crosses behind MATH at all the crossings except the lower one near MATH and the lowermost one too if MATH is odd, for any MATH. Finally, we connect this diagram with MATH by means of a vertical band as show in REF , in such a way that the resulting diagram MATH is again a diagram of MATH intersecting MATH along an arc and the corresponding blackboard framing still represent the Legendrian framing of MATH. Let MATH be the arc represented by MATH. Then MATH is the disjoint union of MATH arcs and a knot MATH equivalent to MATH by an ambient isotopy of MATH, which makes the lifting of the blackboard framing along MATH into the Legendrian framing of MATH with one left-twist added. In fact, by unfolding the sheets of MATH we get a diagram of MATH, which is the connected sum of a copy of MATH in the sheet MATH with a trivial loop going forth and back in the other sheets. Moreover, the unfolding process, applied to the lifting of the blackboard framing along MATH, gives us a framing which coincides with the blackboard one except for a right (respectively, left) half-twist for each vertical segment MATH or MATH with MATH odd (respectively, even). The knot MATH obtained starting from REF , together with the lifting of the blackboard framing, is represented in REF . At this point, the method introduced by Montesinos in CITE (see also CITE) allows us to construct a MATH-fold simple branched covering MATH, whose branch set and monodromy coincide with the ones of MATH, except for the attachment to MATH of a ribbon band MATH, which represent the blackboard framing along MATH (see REF for the branch set arising from the diagram of REF ). Then, denoting by MATH the ribbon annulus resulting from this surgery on MATH, the branch set of MATH is the regularly embedded surface MATH. To conclude this part of the proof, we see that the branch set of MATH is isotopically equivalent to a positive braided surface (over the second factor). In fact, MATH is already braided (without any twist point) and MATH can be made into a braided surface by adapting the NAME 's braiding process (see CITE) in such a way that all the MATH's are left fixed. Moreover, due to the special form of MATH, all the twist points arising in the process turn out to be positive. Namely, we deform the parts of the band MATH corresponding to vertical edges of MATH of types REF (including the MATH's with MATH odd), one by one from left to right, to new disks parallel to the MATH's, successively putted in front of the previous ones, as shown in REF . After all these deformations have been performed, we are left with a certain number of parallel disks and bands between them (in particular, some of such bands correspond to the edges MATH with MATH even). All such bands have the form depicted in the left part of REF (up to conjugation), each one being linked to an arbitrary number (possibly none) of vertical lines. The right part of REF , shows how such a band can isotoped to a braided one with a positive twist point (compare CITE). REF no REF-handles. This time we have MATH, for some Legendrian REF-handles MATH. Let MATH be a front projection diagram of the Legendrian link MATH, where MATH is the attanching knot of MATH. New diagrams MATH, MATH and MATH of MATH can be obtained starting from MATH as in REF ; we use the subscript MATH for the part of a diagram corresponding to MATH. Then, putting MATH and MATH, we denote by MATH the vertical edges of types REF of MATH. We assume the MATH's and the MATH's numbered in such a way that: MATH belong to MATH and are ordered in REF (starting from the uppermost of type REF), for any MATH; the first edges of the MATH's have increasing indices from bottom to top, that is we have in the order MATH. We also assume the MATH's placed so that, going from left to right, we have in the order MATH on the left side of MATH and MATH on the right side of MATH. Then, we consider the simple branched covering MATH with MATH sheets labelled from REF to MATH, whose branch set consists of disks MATH parallel to the second factor and whose monodromy around MATH is MATH if MATH and MATH otherwise. As above, we think the MATH's as parallel disks in MATH with the interiors pushed inside MATH and represent their boundaries as vertical lines MATH in the diagram. Furthermore, we assume that: MATH for any MATH; MATH for all the other MATH's; the positions of the MATH's and the crossings of MATH with them are as in REF . Finally, we change each MATH into a new diagram MATH, by the same costruction we have performed in the previous case on the entire diagram MATH for obtaining MATH. Thanks to the choices made above about the position of the MATH's, we can do that without creating any extra crossing. In other words, the new parts of the diagram, representing the unknots and the bands connecting them with the MATH's, do not cross each other nor the remaining part of the old diagram MATH. Moreover, we let the unknot diagram arising from MATH cross in front of all the MATH's with MATH. In this way, we get a new diagram MATH of the link MATH, such that each MATH meets MATH along an arc in MATH and it is a diagram of MATH whose blackboard framing represent the Legendrian framing of MATH (see REF for the diagram MATH obtained starting with the diagram MATH of REF ). Let MATH, where MATH is the arc represented by MATH. Then, MATH is the disjoint union of some arcs and a link MATH equivalent to MATH by an ambient isotopy of MATH, which makes the lifting of the blackboard framing along each MATH into the Legendrian framing of MATH with one left-twist added. We can prove this fact as in REF , after observing that, as in that case, MATH is essentially contained in the sheet MATH, being the component MATH of MATH over MATH contained in the sheets MATH, so that different MATH's interact only in the sheet MATH. In order to get a MATH-fold simple branched covering MATH, we modify MATH by attaching to each disk MATH a ribbon band MATH, which represets the blackboard framing along MATH and is disjoint from the other MATH's. Then, the branch set of MATH is a regularly embedded surface in MATH, consisting of MATH disks and MATH annuli, that can be made into a positive braided surface, by the same method used in REF . General case. Let MATH, where MATH is obtained attaching MATH REF-handles to MATH and the MATH's are Legendrian REF-handles. We represent such handle decomposition by a diagram MATH as in REF and we get diagrams MATH and MATH of MATH as in the previous cases, expanding the dotted circles behind the diagram and representing them by dotted vertical lines. So, MATH crosses in front of these vertical lines at all the crossings, except the ones corresponding to passages of the link MATH through REF-handles, as shown in REF . Then, we push away from MATH all the vertical edges of type REF (including the ones needed to realize the arcs which go through REF-handles), by using the moves of REF . In this way, we get a diagram MATH as in the previous REF . We also assume such vertical edges MATH, as well as the subdiagrams MATH, numbered and placed as in that case. Now, let MATH the MATH-simple branched covering constructed as in REF , starting from the actual diagram MATH, without taking into account the dotted components. In order to make MATH into a simple branched covering MATH, we add to it MATH sheets labelled from MATH to MATH and MATH branch disks MATH parallel to the previous ones, whose meridians have monodromies MATH. Assuming also these new disks as parallel disks in MATH with the interiors pushed inside MATH, we can represent their boundaries in the diagram by MATH vertical lines MATH. We think the MATH-REF of MATH, being realized by the MATH-th sheet together with the pair of branch disks MATH, MATH (compare CITE). Then, we draw the lines MATH and MATH in correspondence of the MATH-th dotted vertical line from the left in REF , letting a horizontal edge of MATH cross in front of them iff it crosses in front of such dotted line (see REF ). At this point, we construct another diagram MATH of MATH, by modifying MATH as in REF and letting all the new horizontal edges introduced in the construction cross in front of the vertical lines MATH. Finally, we define the disjoint union of arcs MATH as above and see, in the same way, that MATH is the disjoint union of some arcs and a link MATH equivalent to MATH and that the blackboard framing along each MATH lifts to the right framing of MATH. Hence, by attaching to each disk MATH a ribbon band MATH as above, we change MATH into a MATH-fold simple branched covering MATH. The branch set of MATH is a regularly embedded surface in MATH, consisting of MATH disks and MATH annuli, that can be made into a positive braided surface, again by the same method used in REF . |
math/0002042 | Let MATH the branch points of MATH and MATH meridian loops around them, such that MATH in MATH. Putting MATH, we have that the restriction MATH is a locally trivial bundle with fibre MATH and monodromy MATH, where MATH is the homomorphism indiced by the inclusion of MATH into the complement of the branch points MATH. Then, MATH is orientation preserving diffeomorphic to the mapping torus MATH of the mapping MATH. On the other hand, MATH, since the restriction MATH is a (locally) trivial bundle with fiber MATH. So, we conclude that MATH. |
math/0002042 | Given a open book MATH with page MATH, we can write MATH, with MATH right-handed NAME twist along MATH and MATH. Then, fixed MATH and MATH meridian loops around them, such that MATH in MATH, we consider the NAME fibration MATH determined by the branch points MATH and the monodromies MATH for MATH(compare REF). By REF , we have MATH. For the second part of the proposition, observe that we can choose the MATH's non-separating if MATH connected and the MATH's positive if MATH is a positive open book. The following REF guarantees that such choices can be made simultaneously. |
math/0002042 | Looking at the double branched covering MATH shown in REF , we see that MATH covers twice the loop MATH encircling all the MATH branch points, where MATH denotes the genus of MATH. Then MATH is the lifting of a double right-handed twist along MATH. By expressing the corresponding braid in terms of the standard generators, it can be easily realized that MATH, where MATH and MATH are the right-handed NAME twists along the loops MATH and MATH depicted in the figure. |
math/0002042 | By REF , the oriented boundary of any compact NAME surface if orientation preserving diffeomorphic to a positive open book. NAME, given a positive open book MATH, we can assume, up to the plumbing operation REF introduced in CITE (compare proof of REF above), that the binding of MATH is connected. Then, by REF , MATH is the oriented boundary of a compact NAME surface. |
math/0002042 | Let MATH be an open book with page MATH and binding MATH, such that MATH is orientation preserving diffeomorphic to MATH. Since MATH is generated by NAME twists along non-separating simple loops, we can express MATH as a product of such twists. Now, by REF , any left-handed twist along a non-separating loop can be obtained as a product of some right-handed twists and of MATH. In fact, using the notations of REF , this is true for the loop MATH, hence the same holds for any non-separating simple loop in MATH, being all such loops equivalent. Then, we have MATH, with MATH a product of right-handed NAME twists and MATH, because MATH is a central element of MATH. So, we can surgery MATH along MATH in order to get a new REF-manifold MATH, orientation preserving diffeomorphic to the positive open book MATH, which is NAME fillable by REF . |
math/0002043 | The MATH-bundles over a finite cell complex MATH are classified by homotopy class of maps from MATH to MATH. Such a class determines a conjugacy class of homomorphisms MATH . In our case MATH is a surface with non-empty boundary, so MATH has cohomological dimension REF (being free). Thus MATH-bundles over a surface with non-empty boundary are in bijection with the homomorphisms MATH. The manifold MATH we are looking for exists if and only if there is a commutative diagram MATH where MATH is induced by inclusion of the boundary, that is, MATH, hence we have shown the first part of the claim. To see when MATH is orientable, use the standard construction of an oriented surface from the disk MATH, by identifying some MATH-disks MATH on its boundary. Then we can explicitly construct MATH starting from MATH by gluing MATH's on its boundary. As MATH is oriented and the gluing must create no orientation-reversing loop, it is not hard to see that the condition MATH assures the orientability of MATH. |
math/0002043 | The proof is immediately obtained from REF and the fact that the disjoint union MATH is cobordant to MATH with MATH by a toric cobordism with base the sphere MATH with MATH holes (it can be constructed similarly to the proof of REF ). |
math/0002043 | By MATH we denote the subgroup of MATH generated by the squares of matrices with negative determinant MATH . It is evident that MATH and is normal in it. We show that MATH and this implies the claim. We use the following presentations of MATH and MATH; see (CITE). For MATH , MATH and MATH , MATH . The commutator subgroup MATH of MATH is a free group of rank REF generated by MATH and MATH . Thus MATH, and as MATH, we have MATH. By using the relations MATH and MATH, each element MATH can be written in the normal form MATH where MATH. If MATH, the element MATH can be written in the form MATH. Now MATH . Thus, MATH. |
math/0002043 | It follows from REF that for MATH there exists an oriented toric cobordism between the torus bundles MATH and MATH if and only if MATH . Thus MATH . The generator of MATH is the conjugacy class of the element MATH . The subgroup MATH lies in the subgroup MATH. Thus, if there exists an unoriented toric cobordism with an orientable base between MATH and MATH, then there also exists an unoriented toric cobordism with a non-orientable base between them. Hence, REF lets us calculate the third group of unoriented toric cobordims, MATH, as well. Thus, MATH . The generators here are the conjugacy classes of MATH and MATH . |
math/0002046 | Assume that MATH holds. Since MATH and MATH are MATH-Cartier, so is their difference MATH. Consider the decomposition MATH into positive and negative parts. Since MATH and MATH are not contained in MATH, the surface MATH is MATH-factorial near MATH. So MATH is MATH-Cartier. Hence MATH is also MATH-Cartier . As MATH, there must be at least one horizontal NAME divisor MATH. Consider the invertible sheaf MATH. Obviously, MATH is MATH-generated on MATH. Since MATH is finite, we can apply REF (over affine open subsets MATH), and deduce that MATH is MATH-generated for some MATH. Consequently, the homogeneous spectrum MATH is a projective MATH-scheme, hence projective. But MATH is ample on the generic fibre of MATH, contradicting the universal property of the projective reduction. |
math/0002053 | It is a direct consequence of the commutativity of the diagram MATH since the top map MATH is an isomorphism by REF and both vertical maps are the epimorphisms. |
math/0002053 | Because of REF and NAME duality, MATH. |
math/0002053 | It follows from REF, since the homomorphism MATH passes through the trivial group MATH. |
math/0002053 | CASE: This claim follows, because the rank of a matrix is equal to the order of the largest non-zero minor. CASE: This claim follows from REF . CASE: By REF , the set MATH is an algebraic subset of MATH. So, it suffices to prove that MATH. But, by REF , MATH for MATH small enough, and so MATH. |
math/0002053 | The existence of MATH follows from REF . The flexibility of MATH follows, since MATH is a symplectic form for MATH small enough. Indeed, if we set MATH, then MATH. |
math/0002053 | Take a closed REF-form MATH which represents MATH. Then MATH is a symplectic form for MATH small enough. Furthermore, by REF , there exists arbitrary small MATH such that MATH. Now the result follows from REF. |
math/0002053 | CASE: Trivial. CASE: Suppose that there exists MATH with MATH. Take MATH in the interior of MATH. Then, in view of REF, there exists an arbitrary small MATH such that MATH, that is, MATH does not belong to the interior of MATH. This is a contradiction. CASE: This is true because of REF . |
math/0002053 | Let MATH be the usual non-singular pairing given by MATH for MATH and MATH. Define a skew-symmetric bilinear form MATH via the formula MATH for MATH. It is easy to see that the rank of MATH, which must be an even number MATH with MATH, is equal to MATH. The inequality MATH follows from REF. The last claim holds since, by the NAME duality, MATH. |
math/0002053 | Because of the NAME duality, MATH. So, MATH, and hence MATH is determined uniquely up to non-zero multiplicative constant. |
math/0002053 | Here we use some ideas from CITE. It follows from REF that MATH where MATH. Indeed, if MATH then there exists MATH such that MATH. Therefore, MATH, and MATH . Because of REF , every class in MATH is symplectically harmonic. So, it suffices to prove that any cohomology class MATH is symplectically harmonic. Let MATH with MATH closed. Since MATH, there exists MATH such that MATH. It is known CITE that MATH is surjective. So, there exists MATH with MATH, and hence MATH. So, if we take MATH, then MATH and MATH. But the equality MATH implies that MATH in view of REF . Thus, MATH is symplectically harmonic by REF . |
math/0002053 | This follows from REF because of REF. |
math/0002053 | It follows from REF since, by REF, MATH. |
math/0002053 | Since MATH, we conclude that the linear form MATH is non-degenerate for some point MATH. So, MATH is non-degenerate since it is homogeneous. Thus, MATH is non-degenerate. |
math/0002053 | According to our assumption about the type of MATH, there exists a basis MATH of homogeneous MATH-forms on MATH such that MATH . Since the NAME cohomology of the nilmanifold is isomorphic to the NAME cohomology of the NAME algebra, we conclude that MATH . In particular, by REF, a REF-form MATH on MATH is symplectic if and only if MATH where MATH, MATH. Let MATH be a symplectic form on MATH. Then the following holds: REF if MATH, then MATH; REF if MATH and MATH, then MATH; REF if MATH and MATH, then MATH. Furthermore, MATH for every symplectic form MATH on MATH. It follows from the NAME duality that MATH. Hence, by REF MATH since the four cohomology classes from above are linearly independent. Furthermore, for every MATH the image of the mapping MATH is MATH which has dimension MATH for MATH, dimension MATH for MATH and MATH, and dimension MATH for MATH and MATH. The result follows from REF . Now, the proof of the proposition follows from REF. |
math/0002053 | According to our assumption about the type of MATH, there exists a basis MATH of homogeneous MATH-forms on MATH such that MATH . The cohomology groups of degrees REF are: MATH . In particular, if MATH is a symplectic form on MATH then MATH where MATH. The following claim can be proved similarly to REF. Let MATH be a symplectic form on MATH. Then the following holds: REF if MATH, then MATH, MATH; REF if MATH, then MATH, MATH. MATH . Now, by REF, MATH is flexible. |
math/0002053 | CASE: MATH and REF-form MATH on MATH is symplectic if and only if MATH where MATH. Furthermore, if MATH then MATH; otherwise, MATH. Finally, MATH for every MATH. CASE: MATH and REF-form MATH on MATH is symplectic if and only if MATH where MATH . Furthermore, if MATH, then MATH; otherwise, MATH. Finally, MATH for every MATH. CASE: The second NAME cohomology group is given by MATH . REF-form MATH on MATH is symplectic if and only if MATH where MATH. Furthermore, if MATH, then MATH; otherwise, MATH. Finally, MATH for every MATH. Notice that in this last case the nilpotent NAME algebra (so the compact nilmanifold) is reducible of type MATH. |
math/0002053 | Because of the NAME isomorphism MATH, we conclude that MATH and the first equality follows from REF. Similarly, MATH . Now, in view of REF, the computation of dimensions completes the proof. |
math/0002053 | If MATH varies then the result follows from the first equality of REF. If MATH varies but MATH does not vary then the result follows from the second equality of REF. |
math/0002064 | Suppose that REF holds, so that MATH, where MATH is a directed set parametrizing elements MATH (and MATH). For any MATH we have natural isomorphisms MATH . For MATH, let MATH be the imbedding; we have a commutative diagram MATH where MATH is such that MATH, MATH is multiplication by MATH and MATH is the projection induced by REF (and similarly for MATH). Since MATH, for all MATH we can find MATH such that MATH is a multiple of MATH. Say MATH; then we can take MATH, so MATH is a multiple of MATH and in the above diagram MATH. Hence one can define a map MATH such that MATH and MATH. It now follows that for every MATH-module MATH, the induced morphism MATH is the zero map. Taking the colimit we derive that MATH is flat. This shows REF . In order to show REF we consider, for any prime number MATH, the following condition REF MATH is generated (as a MATH-module) by the MATH-th powers of its elements. Clearly REF implies REF for all MATH. In fact we have : REF holds if and only if REF holds for every prime MATH. Suppose that REF holds for every prime MATH. The polarization identity MATH shows that if MATH then MATH. To prove that MATH it then suffices to show that for every prime MATH dividing MATH we have MATH. Let MATH be the NAME (MATH); we can denote by MATH the ring MATH seen as a MATH-algebra via the homomorphism MATH. Also set MATH for a MATH-module MATH. Then the map MATH (defined by raising to MATH-th power) is surjective by (MATH). Hence so is MATH for every MATH, which says that MATH when MATH, hence for every MATH. Next recall (see CITE Exp. XVII REFEF) that, if MATH is a MATH-module, the module of symmetric tensors MATH is defined as MATH, the invariants under the natural action of the symmetric group MATH on MATH. We have a natural map MATH that is an isomorphism when MATH is flat (see CITEEF. REF; here MATH denotes the MATH-th graded piece of the divided power algebra). The group MATH acts trivially on MATH and the map MATH REF is an isomorphism. The first statement is reduced to the case of transpositions and to MATH. There we can compute : MATH. For the second statement note that the imbedding MATH is an almost isomorphism, and apply REF . Suppose now that MATH is flat and pick a prime MATH. Then MATH acts trivially on MATH. Hence MATH . But MATH is spanned as a MATH-module by the products MATH (where MATH and MATH). Under the isomorphism REF these elements map to MATH; but such an element vanishes in MATH unless MATH for some MATH. Therefore MATH is generated by MATH-th powers, so the same is true for MATH, and by the above, REF holds, which shows REF . |
math/0002064 | Let MATH be a countable generating family of MATH. Then MATH generate MATH and MATH for all MATH. For every MATH, we can write MATH, for certain MATH. Let MATH be the MATH-module defined by generators MATH, subject to the relations: MATH . We get an epimorphism MATH by MATH. The relations imply that, if MATH, then MATH, so MATH. Whence MATH and MATH is an isomorphism. We consider the diagram MATH where MATH and MATH are induced by scalar multiplication. We already know that MATH is an isomorphism, and since MATH, we see that MATH is an epimorphism, so MATH is an isomorphism, which shows REF . Now REF follows from REF , since it is well-known that a flat countably presented module is of homological dimension MATH (see CITE (Ch.I, REF ) and the discussion in CITE pp. REF). |