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math/0104240 | Let us consider MATH as a pointed simplicial monoid as follows. Given MATH, we let MATH. The constant map with value MATH is the base point. There is a pointed submonoid MATH of MATH consisting of the constant maps. We shall let MATH denote the subring MATH of the pointed monoid ring MATH. From the short exact sequence MATH it follows that we have a weak equivalence MATH. The normalized chain complex of MATH has a generator MATH in degree zero and a generator MATH in degree MATH. The differential takes MATH to MATH. The normalized chain complex MATH of MATH has a generator of the form MATH in degree MATH and a generator of the form MATH in degree MATH. The NAME boundary MATH takes MATH to MATH. It follows that MATH and that the odd homotopy groups of MATH are zero. In order to compute cyclic homology of MATH, we need to evaluate the NAME boundary operator MATH on the chains of the normalized chain complex of MATH. The result is that MATH, and that MATH. It is not easy to compute the higher homology of the bicomplex MATH with MATH and with vertical and horizontal differential induced by MATH and MATH respectively. In degrees up to MATH the horizontal nonzero differentials become isomorphisms after tensoring with MATH. Therefore we have that the homology of the total complex of MATH is a copy of MATH in degree MATH when MATH. We can conclude that if MATH then MATH is a cyclic MATH-group and MATH. To find the order of MATH we can consider the spectral sequence associated to the bicomplex MATH with MATH-term MATH. This spectral sequence is concentrated in even total degrees, and therefore there are no nonzero differentials. We know that in degrees up to MATH all extensions are maximally nontrivial, and we can read off the stated value of MATH. To see that the map MATH is onto when MATH it suffices to check that generators for the group MATH are in the image. This is easy to see from the induced map of MATH-terms of the spectral sequence considered above. |
math/0104240 | The proof goes by induction on MATH. Suppose that MATH for MATH. (By the computation of MATH in CITE this is true for MATH.) We have a cofibration sequence MATH . Applying REF with MATH and MATH and REF we find that MATH . The statement of the lemma can be read off from the long exact sequence associated to the cofibration sequence. |
math/0104240 | It follows from the above lemma that the map MATH is onto and that MATH (see REF and XI). We have that MATH (see for example CITE), and it follows that the map MATH is onto. In CITE NAME and NAME have computed MATH. In the low degrees we are interested in it is MATH in odd degrees and MATH in even strictly positive degrees. It follows that the group MATH is cyclic. Using the cofibration displayed in the proof of the above lemma and the computation of MATH given in CITE we can by induction prove that MATH . The statement of REF now follows from NAME 's REF and from NAME 's computation of MATH in CITE. |
math/0104242 | Immediate from definitions. |
math/0104242 | Immediate from definitions. |
math/0104242 | It is easy to see that for any MATH, the morphism MATH identifies MATH (here we have used REF and functoriality of MATH). Combining this with the definition of MATH and rigidity of MATH we get the statement of the theorem. |
math/0104242 | Define the morphism MATH by the graph shown in REF . Then the composition MATH can be rewritten as shown in REF and thus, MATH which proves that MATH is injective. |
math/0104242 | Using REF, we get MATH . Thus, using MATH (see REF) for MATH we get MATH, which shows that MATH is either simple or zero, and for MATH, we get MATH. |
math/0104242 | Denote this composition by MATH. Since MATH, one has MATH for some MATH. To find MATH, compose both MATH and MATH with the morphism MATH shown in REF . Arguing as in the proof of REF, we see that MATH. |
math/0104242 | Let MATH be as in the proof of REF. Let us calculate MATH. Using REF and functoriality of MATH, we can rewrite MATH as shown in REF . Thus, MATH. On the other hand, it was proved in REF that MATH. |
math/0104242 | Let MATH. Then, by REF, MATH is closed under duality (as usual, we denote by MATH the class of representation MATH). Denote by MATH the multiplicities in the tensor product decomposition: MATH. Then MATH . Indeed: if MATH, then there is an embedding MATH, which gives MATH (by REF), which contradicts MATH. By using MATH, we can rewrite REF as MATH . Let MATH be the full subcategory in MATH generated (as an abelian category) by MATH. Then it follows from REF that MATH is closed under tensor product; it is also closed under duality (by REF) and contains MATH (by REF). By the usual reconstruction theorems, this means that MATH is the category of representations of some group MATH which is a quotient of MATH. But by assumption, the action of MATH on MATH is faithful, which means that the action of MATH on MATH is faithful. Thus, MATH, MATH. |
math/0104242 | Immediate from REF and finite-dimensionality of spaces of morphisms in MATH. |
math/0104242 | It is immediate from REF that MATH lies in MATH and that as an object of MATH, MATH. The structure of MATH-module is determined by the action on the second factor, and the action of MATH by automorphism is defined by the action on the first factor. On the other hand, it is shown in CITE that any algebra in category MATH must be of the form MATH for some subgroup MATH. Combining these statements, we see that MATH. |
math/0104242 | It is immediate from the definition that MATH is the category of MATH-modules with MATH-grading such that MATH . Indeed, the action of MATH and the first component of the grading define MATH as an object of MATH, and the second component of the grading defines the action of MATH: if MATH is homogeneous, then MATH . Define a new MATH-grading MATH by MATH (which implies that MATH if MATH). Then MATH . From this it immediately follows that the functor MATH given by MATH considered with MATH-grading given by the first component of MATH, is an equivalence of categories. The remaining statements of the theorem are straightforward. |
math/0104242 | Let MATH; for now, we consider MATH as a MATH-graded MATH-module, as in the proof of REF. Then explicit calculation shows that MATH . Therefore, MATH where MATH. Comparing it with the usual formula for action of MATH, MATH we see that MATH iff MATH for all MATH. |
math/0104242 | The proof is based on the following easily proved lemma. Let MATH and let MATH be irreducible representations of MATH, MATH respectively. Then there exists MATH - an irreducible representation of MATH such that MATH. Using this lemma with MATH, we see that if MATH, for some MATH, then for all MATH, MATH. Using this lemma again, we see that the set MATH is closed under product. Since MATH, this set is also invariant under conjugation. Thus, MATH is a normal subgroup in MATH. |
math/0104242 | It is well known that MATH is modular; thus, let us prove that if MATH then MATH is not modular. As discussed above, simple objects in MATH are MATH. Let MATH be a formal linear combination of irreducible representations of MATH such that MATH for all MATH; such a MATH always exists if MATH. Then it follows from explicit formulas for MATH (see, for example, CITE) that MATH for all MATH. Thus, MATH is singular. |
math/0104242 | The proof is based on the following lemma. If MATH is simple and MATH is holomorphic, then MATH . Note that MATH is non-zero in any semisimple rigid category in which MATH (see, for example, CITE). In particular, this implies MATH and thus, MATH (see CITE). Let us rewrite the left hand side as shown in REF . Using REF from CITE, we see that the left hand side is the composition MATH where MATH is the projector MATH, and for MATH, MATH is the maximal sub-object of MATH which lies in MATH. But if MATH is holomorphic, then the only simple object in MATH is MATH itself, which is the unit object in MATH. It appears in MATH with multiplicity one, and the right-hand side is exactly the projection of MATH on MATH. For every MATH, define morphism MATH by the following graph: MATH . If MATH is simple, then MATH is an automorphism of MATH as a MATH-algebra. Let us calculate MATH. Using REF we can rewrite the graph defining MATH as shown in REF . (Note that we need to use MATH REF to move the ``ring" through MATH; the last step also uses REF from CITE.) This shows that MATH. In other words, MATH is a morphism of MATH-algebras. But it easily follows from REF that every such morphism is either zero or invertible. Restricting MATH to MATH, we see that MATH is non-zero; thus, MATH is an automorphism. MATH . Using REF and MATH, we can rewrite the left hand side as shown in REF . The statement of the theorem easily follows from these two lemmas. Indeed, combining REF with REF we see that MATH for some MATH. On the other hand, REF gives MATH and thus, MATH. This completes the proof of REF . |
math/0104242 | REF is immediate from the definitions. REF follows from the fact that the only simple object in MATH is MATH. To prove REF , we will use the following lemma which easily follows from REF. MATH. This lemma implies that MATH for some MATH. On the other hand, MATH, and MATH. Since, by REF , MATH for MATH and the operators MATH are linearly independent, we see that these two expressions can be equal only if MATH. To prove REF , note that MATH is simple, which immediately implies that MATH is simple. The identity MATH immediately follows from REF . Finally, to prove REF note that REF imply MATH. By REF , this is only possible if MATH. |
math/0104242 | Define MATH by REF (where we used REF to identify MATH). We leave it to the reader to check that so defined MATH satisfies the associativity REF with MATH. |
math/0104242 | Immediately follows from the commutation relations in MATH, REF . |
math/0104242 | The only part which is not obvious is the fact that MATH is commutative (with respect to MATH, not MATH!), that is, that the composition MATH coincides with MATH. To prove it note that explicit REF for MATH shows that this composition, when restricted to MATH, is equal to MATH . Using presentation of MATH given in REF and associativity of MATH, we can rewrite it as shown in REF , which shows that it is equal to MATH. |
math/0104242 | The only one which is not obvious is the commutativity isomorphisms, proof of which is shown in REF , which uses the same conventions as REF . |
math/0104242 | Follows from the previous corollary and REF . |
math/0104245 | Use REF with MATH. This gives MATH . Passing to MATH gives the result. |
math/0104245 | If MATH is cellular, define MATH by MATH and use REF . In the general case use a fine enough triangulation and a simplicial approximation for MATH and proceed as before. |
math/0104245 | The argument is similar to CITE. Assume MATH, then there exists a sequence MATH with MATH for all MATH. Choose disjoint small balls around the critical points of MATH such that whenever a flowline of MATH leaves a ball it takes a positive time MATH to get back into a ball. For a closed orbit MATH let MATH be the number of how often the closed orbit enters (and leaves) such a ball. Because of MATH we get MATH. The sequence MATH is bounded because otherwise MATH. This follows because MATH and MATH outside the balls for some MATH. By passing to a subsequence we can assume that MATH is constant to MATH. Choose points MATH for MATH on the orbit of MATH away from the balls such that there is exactly one ball on the orbit between MATH and MATH and between MATH and MATH. Also denote by MATH the time it takes from MATH to MATH. We can assume that the MATH converge to MATH and the MATH converge to MATH. Notice that MATH. If MATH the continuity of the flow implies the existence of a flow line between MATH and MATH . If MATH there is a ``broken" flow line from MATH to MATH through a critical point. At least one of the MATH has to be MATH because MATH. As a result we get a broken closed orbit of MATH which contradicts MATH. |
math/0104245 | Let MATH. Then MATH . Since MATH is a boundary we get the result. |
math/0104245 | Look at the composition MATH . It is easy to see that the image of MATH under this composition in MATH is REF. Denote the image of MATH in MATH by MATH. Then MATH gets mapped to MATH, compare CITE. So MATH. To see that MATH look at the composition MATH. |
math/0104245 | Look at REF-chain MATH. It is straightforward to check that its boundary gives the result. |
math/0104245 | Since MATH, there exists a rational NAME form MATH such that MATH satisfies condition MATH with respect to MATH. Let MATH be the homomorphism induced by MATH, MATH the infinite cyclic covering space corresponding to MATH and MATH a smooth function with MATH a regular value and MATH. Furthermore choose MATH such that MATH is a splitting of MATH along MATH. We can assume that MATH satisfies condition MATH with respect to this splitting. Hence there is a NAME function MATH and on MATH for all MATH which is ordered in the sense of CITE. Then we get filtrations of MATH and MATH by MATH where MATH is a real number bigger than the image of critical points of index MATH under MATH and smaller than the image of critical points of index MATH under MATH, and MATH . If MATH is a closed orbit of MATH, it lifts to a trajectory MATH such that MATH is bijective. Assume MATH and let MATH be the prime period of MATH, that is, the smallest MATH such that MATH. There is a negative integer MATH such that MATH. Furthermore MATH, where MATH is the generator of the covering transformation group that satisfies MATH. Let MATH be a number such that MATH. It follows from the definition of REF that MATH for all MATH, see CITE. In fact the points are in the interior of the handle. Therefore we get a partition of MATH into sets MATH consisting of closed orbits passing through MATH. We get the chain homotopy equivalence MATH with torsion REF from REF. We show that the MATH-th summand MATH carries the information of MATH, so let us fix MATH and for ease of notation denote the matrix MATH by MATH. To describe the entries MATH of MATH we need the universal cover MATH. For MATH let MATH. The map MATH induces a map MATH. The matrix MATH described in REF comes in fact from a map MATH induced by MATH. Now MATH is a wedge of thickened MATH-spheres and MATH has a natural MATH-action and modulo MATH it consists of as many spheres as MATH has critical points of index MATH. To get the matrix MATH we have to lift the handles of the critical points to MATH. Hence we pick spheres in MATH that we denote by MATH. So if we denote the composition MATH by MATH, where the last map just retracts every sphere other than MATH to the wedge point, then MATH is the degree of MATH. Now we look at MATH. This term contains information about the map MATH. We say MATH are equivalent if they differ only by a rotation and denote by MATH the set of equivalence classes. Then MATH . Fix MATH. The order of MATH divides MATH. Let MATH be the order of MATH and let MATH be so that MATH. If MATH, then MATH. Let us first look at MATH. By REF MATH is homologous to MATH. For MATH, MATH is the degree of the map MATH . Let MATH be the composition of that map with MATH. Then a fixed point other than the basepoint of MATH corresponds to a closed orbit of MATH. Notice that a closed orbit arising this way has multiplicity REF, since it passes through the spheres MATH without a repeating pattern (recall MATH). Furthermore MATH describes the cells through which the closed orbit passes. To continue we now look at the special case where the vector field MATH only has nondegenerate closed orbits. Then the fixed points of MATH are isolated. Notice that the basepoint is a fixed point and its index is REF, since MATH is constant near the basepoint. Now the NAME number of MATH satisfies MATH so if MATH is the subset of MATH consisting of closed orbits following the pattern MATH, then MATH . Recall that MATH is the fixed point index of the NAME map and hence the fixed point index of the corresponding fixed point of MATH. Notice that MATH contributes REF-chain MATH to MATH and MATH. By looking at all combinations MATH we get that MATH where MATH means homologous and MATH is the subset of MATH consisting of closed orbits following a pattern MATH for MATH and some MATH. Now look at MATH. Then MATH and MATH by REF . Let MATH for MATH. Note that MATH but we can have MATH. We also have MATH implies MATH, so we restrict our attention to MATH. Again MATH is the degree of MATH which is defined just as in the case MATH, but this time fixed points can correspond to closed orbits MATH with multiplicity MATH. We say MATH are equivalent, if they differ only by a rotation of MATH elements, where MATH is an integer, for example, for MATH different MATH and MATH are equivalent for MATH, but not for MATH. Denote the set of equivalence classes by MATH. We now have MATH . Note that MATH only depends on MATH, so we denote this by MATH. Fix MATH and let MATH be the order of MATH. Then MATH divides MATH, so let MATH and we get for MATH that MATH, where MATH repeats MATH times. By REF we get MATH . As mentioned above, fixed points of MATH correspond to closed orbits of MATH. Let MATH be a closed orbit coming from a fixed point of MATH which is not the basepoint. The multiplicity of MATH divides MATH, let MATH be the number with MATH. The closed orbit MATH then provides MATH different fixed points of MATH, all with the same fixed point index. Furthermore MATH is represented by MATH which is homologous to MATH by REF . Denote a representing chain of MATH by MATH. Now we can basically proceed as in the case MATH, MATH where MATH consists of closed orbits following a pattern MATH for MATH and the right side is the contribution of those closed orbits to MATH. Summing over all MATH gives MATH . Therefore we get that MATH . Here MATH consists of those MATH that satisfy MATH and MATH for a lift MATH of MATH to MATH. Because of REF summing over all MATH and all MATH gives the desired result for vector fields that only have nondegenerate closed orbits. Now we have to allow degenerate closed orbits for our vector field MATH. A key fact in proving the theorem for vector fields with only nondegenerate closed orbits was the equality REF which still holds in the general case. The fixed point index has to be taken in a more general sense, see CITE. What needs to be shown is that the fixed point index contains the right information for MATH. So fix MATH and look at MATH. The matrix MATH comes from the flow of MATH. Recall the maps MATH which have MATH as degree. By abuse of notation we denote the MATH-handle in MATH corresponding to the thickened MATH-sphere MATH also by MATH. If MATH is a point whose MATH-trajectory leads into a critical point of MATH before it reaches MATH, then MATH, the basepoint of MATH. By the definition of MATH we get that points near MATH will also be mapped to MATH under MATH. Also points on a closed orbit have to be in the interior of the handle by the definition of REF. Points in MATH that do not get mapped to MATH under MATH have trajectories avoiding critical points of MATH between MATH and MATH. A compactness argument gives that the information to define MATH and up to MATH compositions of these maps is contained in a finite piece of the flow. Let MATH be such a piece. The fixed point set of MATH does not consist of finitely many circles in general (aside the fixed points of MATH, MATH and stationary points). But by transversality we can change MATH to a homotopy MATH whose fixed points are finitely many circles. By changing only near the fixed points of MATH we can use the homotopy between MATH and MATH to get a map MATH homotopic to MATH which has finitely many fixed points, each corresponding to a circle of MATH. Now MATH and by CITE MATH can be computed by the transverse intersection invariant MATH, see also CITE for an easy interpretation of MATH in terms of REF. But MATH gives us the right comparison with MATH just as the NAME series did in the case of nondegenerate orbits only. This finishes the proof of REF . |
math/0104245 | We need an argument similar to the proof of REF . Define MATH and MATH. Again we get MATH as MATH. We claim that MATH. So let MATH and assume MATH. Now we proceed as in the proof of REF to get a broken closed orbit MATH of MATH for some MATH. But by continuity we get MATH contradicting the fact that the one parameter family is MATH-controlled. Now let MATH satisfy MATH and let MATH be the flow of MATH. Then MATH is a smooth homotopy of flows and by the argument above there exists a MATH such that MATH contains all closed orbits MATH with MATH. By REF we get MATH where MATH is given by MATH. But MATH, since MATH has no fixed points corresponding to MATH as such fixed points would give a closed orbit of some MATH with period MATH corresponding to MATH. |
math/0104245 | We have MATH for any MATH-gradient, since there are no critical points. Let MATH. By REF we have MATH, where MATH is the zero map. But for any chain homotopy equivalence MATH between acyclic complexes we have MATH. Hence MATH. Now MATH since the one parameter family MATH is MATH-controlled. |
math/0104245 | Since MATH is an almost transverse MATH-gradient, we can rearrange MATH to a NAME function MATH which is self-indexed in the sense of CITE, near the critical points MATH is constant and such that MATH is also a MATH-gradient. By CITE every weak MATH-gradient MATH close enough to MATH is also a weak MATH-gradient. But such a vector field is almost transverse, since for critical points MATH with MATH we have MATH and so MATH. |
math/0104245 | Let MATH. Given MATH we need to find a neighborhood MATH of MATH in MATH such that for all MATH with MATH we have MATH for all MATH. Assume first that MATH is rational. As in REF we have the infinite cyclic covering space MATH and a smooth function MATH with MATH and MATH as a regular value. Let MATH as in REF. For any two regular values MATH is a NAME function on the cobordism MATH. The vector field MATH lifts to a transverse MATH-gradient MATH on MATH whose restriction to MATH is a MATH-gradient. If we choose another MATH-gradient MATH close to MATH, then its lift MATH will also be close to MATH. Since MATH is almost transverse on MATH, a MATH-gradient MATH close enough to MATH will lift to a MATH-gradient MATH such that its restriction to MATH is an almost transverse MATH-gradient by REF . Furthermore the same is true for the weak MATH-gradients MATH for MATH. Now choose a negative integer MATH such that MATH. We set MATH. We claim that any weak MATH-gradient MATH that lifts to an almost transverse weak MATH-gradient satisfies MATH. Assume not, then there exists a broken closed orbit MATH of MATH with MATH. Let MATH be a critical point in the image of MATH. Lift MATH to MATH. The loop MATH lifts to a path MATH in MATH starting at MATH and MATH is a concatenation of trajectories of MATH between critical points and it ends in a translate of MATH. But since MATH, the path MATH actually is in MATH. Now MATH contradicts almost transversality of MATH on MATH. Therefore the family MATH is MATH-controlled and the statement follows by REF . So now assume MATH is irrational. By REF there is a rational form MATH agreeing with MATH in a neighborhood of the critical points such that MATH is a MATH-gradient. We can also choose MATH arbitrary close to MATH. In particular we can choose MATH so close that we have MATH, if MATH is not in a neighborhood of a critical point. By compactness and continuity we can also get this for tangent vectors near MATH. To be more precise, a Riemannian metric on MATH induces a norm MATH on every MATH and we can find a MATH such that MATH for MATH with MATH and all MATH outside the neighborhood of the critical points where MATH and MATH agree. Now the first part of the proof applies to MATH. We choose a neighborhood MATH of MATH such that every MATH satisfies CASE: MATH is a MATH-gradient. CASE: MATH for all MATH. CASE: MATH is MATH-controlled with respect to MATH. It remains to prove that MATH is MATH-controlled with respect to MATH. So let MATH be a broken closed orbit of some MATH. Write MATH for the homomorphism induced by MATH. Then MATH since MATH. Therefore MATH is MATH-controlled as a one parameter family of weak MATH-gradients and the statement follows again by REF . |
math/0104245 | We can use the flow of MATH to define homotopy inverses to the inclusions. For details see CITE . |
math/0104245 | Assume MATH is rational. Let MATH with MATH having MATH as a regular value. Also let MATH be the composition with the universal covering projection. We can assume that the MATH from REF is REF and the liftings of the critical points are chosen in MATH. Define MATH where MATH is the lift of MATH to MATH and the same for MATH. Also let MATH with MATH so small that MATH. Now define MATH and MATH. Both complexes calculate the homology of MATH. The exact case described above gives a chain isomorphism MATH such that the diagram MATH commutes. In fact all the arrows are just induced by inclusion. Passing to the inverse limit gives almost the NAME complex; we only look at MATH. But MATH is a finitely generated free MATH complex, where MATH is the subring of MATH consisting of elements MATH with MATH. Now MATH and similarly for MATH. The chain map MATH is represented by intersection numbers, since the MATH are. In particular we have MATH. Also MATH with MATH and MATH with MATH and MATH is the coefficient of MATH. So we can order the critical points such that the matrix of MATH is of the form MATH, where MATH is nilpotent and MATH satisfies MATH for all entries. The matrix MATH is elementary and MATH where the entries of MATH satisfy MATH. Therefore MATH is an isomorphism of chain complexes and MATH. It remains to prove the proposition for irrational MATH. We can assume that there exists a rational approximation MATH that agrees with MATH near the critical points such that MATH and MATH are also MATH-gradients for we can find a sequence MATH of MATH-gradients such that MATH and MATH have a common rational approximation. Let MATH be the homomorphism induced by MATH. By the rational case above we find a chain isomorphism MATH for the MATH . NAME complexes MATH and MATH. If we can show that the matrix entries of MATH lie in MATH, then MATH induces a chain map MATH by the remarks at the end of REF. Notice that the entries are intersection numbers MATH. So MATH gives a point MATH, a trajectory MATH of MATH from MATH to MATH and a trajectory MATH of MATH from MATH to MATH. Now MATH follows from the next lemma. To see that MATH is an isomorphism with MATH notice that in the irrational case we can choose liftings of the critical points of MATH in MATH, where MATH is an arbitrarily small interval. Then the matrix of MATH in a basis corresponding to these critical points is of the form MATH with MATH for all the entries of MATH, compare CITE . |
math/0104245 | For every pair of critical points MATH of MATH we can choose a path MATH from MATH to MATH. Then there is a constant MATH such that MATH for MATH and all pairs of critical points. Let MATH be the isotopy between id and MATH. For every MATH we get a path MATH from MATH to MATH. By compactness we can also assume MATH for MATH and all MATH. Since MATH and MATH agree near the critical points there exists a MATH such that MATH again by compactness. Now let MATH be as in the statement. Then MATH which gives the result. |
math/0104245 | Let MATH be induced by the flow of MATH, that is, stop once the boundary is reached. There is a MATH such that MATH. Since MATH is adjusted to MATH there is a MATH such that MATH, where MATH is the MATH-skeleton of the triangulation. Furthermore MATH gives a homotopy between id and MATH. Since MATH is adjusted to MATH we have MATH. We can modify the homotopy away from the endpoints to get a homotopy MATH between id and MATH such that MATH. The modifications are done skeleton by skeleton, compare the proof of CITE and can be done arbitrarily close to the original homotopy. Now define MATH be sending MATH to MATH. Then MATH . Notice that MATH represents MATH and using MATH gives MATH. |
math/0104245 | Assume MATH is rational. We use the notation from the proof of REF . We can assume that MATH contains MATH as a subcomplex. Let us also set MATH. Let MATH be the chain homotopy given by REF . Actually in the nonexact case it comes from a homotopy MATH and it satisfies MATH and MATH. We want to get a chain homotopy MATH based on MATH. First we need MATH. Notice that MATH. So we take the homotopy MATH and flow along MATH for a little bit longer. Call this homotopy MATH. Then MATH, but not necessarily MATH. We need to adjust the homotopy slightly to achieve this. So do this skeleton by skeleton to get a homotopy MATH so close to MATH that passing MATH from MATH to MATH is done within MATH. Then if MATH is induced by MATH as in the proof of REF we get the commutative diagram MATH . Passing to the inverse limit as in the proof of REF gives the result in the rational case. For the irrational case notice that nonzero terms of the chain homotopy give a trajectory of MATH from some MATH to a MATH and a trajectory of MATH from MATH to a critical point. Thus we can use a similar approximation argument as in the proof of REF , we omit the details. |
math/0104245 | For MATH let MATH. The collection MATH forms a neighborhood basis of MATH, so MATH forms a neighborhood basis of MATH. Let MATH. To see that MATH is continuous, we have to find for every MATH a neighborhood MATH of MATH such that MATH satisfies MATH. By REF we have MATH. Assume that MATH is rational. Let MATH be the infinite cyclic covering space corresponding to MATH and MATH such that MATH is the pullback of MATH and MATH a regular value. For simplicity assume that the MATH from REF is REF. Since MATH is transverse, so is its lift MATH to MATH. Choose an integer MATH with MATH. We can find a self-indexing NAME function MATH such that MATH is a MATH-gradient and MATH near the critical points and MATH. CITE gives a neighborhood MATH of MATH in MATH such that every MATH lifts to a MATH-gradient on MATH. Now for MATH with MATH we get MATH, since MATH and MATH. By choosing the isotopy MATH of MATH close enough to the identity we still have MATH. Now we choose liftings of the critical points within MATH to get a basis for the NAME complex. For every MATH the coefficients MATH of MATH then have the property that any MATH with MATH implies MATH, compare the proof of REF . Thus MATH is represented by a matrix MATH where MATH for all entries of MATH. By NAME elimination we see that MATH with MATH, so MATH for all MATH. The irrational case is now derived from the rational case by analogy to the proof of REF , we omit the details. |
math/0104245 | Clearly the homomorphism MATH is continuous. So the statement follows from REF , since MATH is dense in MATH. |
math/0104245 | The composition MATH induces the homomorphism MATH from CITE, compare REF. By CITE, MATH is a power series of a logarithm, so we get MATH, where MATH is induced by the augmentation MATH. By REF we get MATH . |
math/0104245 | Assume that the closed orbits of MATH are nondegenerate. Then for a closed orbit MATH of MATH of multiplicity MATH, we get a summand MATH in MATH. Let MATH be so that MATH represents the conjugacy class MATH. Then MATH, but MATH is homologous to MATH by REF . Therefore MATH (recall the NAME series from REF) and we get the result. The general case now follows by continuity, compare the end of REF. |
math/0104248 | We have MATH where MATH is the graded commutative polynomial ring in two variables of weight MATH (the first isomorphism in REF is just REF , while the second follows immediately from the defining relations REF .) Hence MATH has the properties stated in the lemma. Since MATH is NAME and has global dimension REF, MATH is also NAME and has global dimension REF (see REF respectively). That MATH is NAME follows from REF . According to REF , the NAME parameter MATH is equal to the degree of the inverse of the NAME series MATH. Since MATH is isomorphic (as a graded vector space) to the commutative polynomial ring in three variables of weights MATH, we have MATH and therefore the NAME parameter is MATH. |
math/0104248 | This is proved in CITE in the case MATH; more precisely, it is shown in CITE that MATH can be covered by the two NAME sets consisting of the powers of MATH and of MATH. The proof in general is similar, but we shall work with a ``finer" covering, that is, with larger NAME sets. This covering will be needed in REF. For an element MATH we denote by MATH its homogenization in MATH, that is, we set MATH, where MATH. Now define MATH . We claim that MATH is a covering of MATH. Clearly, MATH and MATH are multiplicatively closed subsets of MATH; and they satisfy the NAME condition (on both sides) because they consist of (homogeneous) locally ad-nilpotent elements in MATH. On the other hand, for any (nonzero) polynomials MATH and MATH, the ideal MATH has finite codimension in MATH and hence coincides with MATH. It follows that MATH for some MATH, and therefore MATH for some MATH. If either MATH or MATH, REF holds trivially. So we may assume that MATH and MATH both have positive degree, say MATH and MATH. Then MATH and MATH, hence MATH and MATH for any MATH. Taking MATH, we find that there is a MATH such that MATH which shows that the pair of NAME sets MATH covers the algebra MATH, as required. |
math/0104248 | If MATH is a graded MATH-module with MATH, it follows at once from REF that the graded quotient MATH is canonically isomorphic to MATH. Applying the (exact) functor MATH to MATH we get the lemma. |
math/0104248 | We have MATH (see REF ). The embedding of MATH in MATH induces an embedding of MATH in MATH as a homogeneous ideal. Now, MATH is just a commutative polynomial algebra in two variables; hence if MATH is the greatest common divisor of the elements of MATH, then MATH is a (homogeneous) ideal of finite codimension in MATH. Denoting by MATH the degree of MATH in MATH, we therefore have an exact sequence of graded MATH-modules MATH with finite-dimensional quotient term. The quotient functor MATH annihilates finite-dimensional modules, so applying MATH to REF , we get the desired isomorphism MATH . The uniqueness of MATH follows from the fact that MATH in MATH only if MATH. Indeed, assuming the contrary, by REF we have MATH for some MATH, and therefore MATH for all MATH. This implies that the sequence of numbers MATH is bounded, which is obviously not the case. |
math/0104248 | As in the proof of the preceding Lemma, we identify MATH with an ideal in the polynomial ring MATH; then MATH is the minimum degree of elements in MATH, and hence MATH. If MATH, then MATH is cyclic (generated by the greatest common divisor MATH above), and hence MATH is also cyclic. |
math/0104248 | It is obvious that MATH, so we have only to prove that the Ext term in REF is zero. Let MATH. We show first that MATH . For brevity, set MATH. Applying the functor MATH to MATH, we get the exact sequence MATH . The first term in this sequence is obviously zero; and the NAME property (see REF ) of MATH implies that the last term is also zero, because MATH has finite length. Thus MATH for all MATH. Passing to the limit as MATH, we get REF . Hence REF follows if we show that MATH. Suppose that MATH represents a MATH-orsion element in MATH. This means that for some MATH we have MATH, and hence MATH. Since MATH, we find that MATH, and hence MATH represents zero in MATH. Thus MATH. |
math/0104248 | The map of sheaves MATH induced by multiplication by MATH is clearly injective (for any MATH). Using REF and bearing in mind that MATH we get the short exact sequence MATH . By REF, we have MATH . On the other hand, by REF , we have MATH where MATH is the MATH-th graded component of the commutative polynomial algebra MATH of weight MATH. Therefore the first and last terms of the exact sequence MATH coming from REF are isomorphic to MATH and MATH respectively. Since the grading on MATH is positive, REF follows. The assertion about MATH in REF is immediate in view of REF . To prove the assertion about MATH we observe (again looking at the long cohomology exact sequence of REF ) that the map MATH is an isomorphism for MATH. By the Vanishing REF MATH, MATH is zero for MATH, hence it is zero for all MATH. It remains to prove part MATH of the Theorem. The fact that MATH again follows from REF . From REF (with MATH) we get the exact sequence MATH whence the last statement in the Theorem. |
math/0104248 | The Proposition is trivial if MATH. In general, we calculate: MATH (we used the facts that MATH commutes with MATH and MATH, and that MATH is still well defined on MATH). Similarly, MATH. So MATH . Thus the image of MATH is contained in the kernel of MATH. By REF , this map is surjective with one-dimensional kernel. Therefore MATH has rank MATH, as required. |
math/0104248 | An argument of NAME and NAME (see REF ) shows that MATH is contained in MATH. REF then shows that it has finite codimension (and also that this codimension is independent of MATH). Interchanging the roles of MATH and MATH, we obtain the same result for MATH. Since MATH and MATH are both isomorphic to MATH, the ideals MATH and MATH are isomorphic. An ideal class of MATH has a unique representative of finite codimension, so MATH follows from MATH. Since MATH, we have MATH (the symbol map is multiplicative). So MATH follows from MATH. |
math/0104248 | This is a reformulation of REF MATH (compare the proof of REF above). |
math/0104248 | From REF we easily find (using REF ) that MATH so MATH (where MATH denotes the NAME characteristic). From REF again, we then get MATH . On the other hand, by REF the NAME characteristics of the sheaves MATH satisfy MATH and by REF , MATH for all MATH. By REF , we have MATH, so by induction MATH . Combining REF yields the Lemma. |
math/0104248 | The first statement is obvious, because the coboundary map MATH takes MATH to MATH. The second statement follows from REF . |
math/0104248 | By REF , the two spaces in REF have the same (finite) dimension if MATH, so it is enough to prove that MATH is injective if MATH. In fact, MATH is injective for all MATH. To see that, let MATH and suppose MATH; we have to show that MATH. Equivalently, by REF , we are given MATH and MATH such that MATH and MATH; we have to show that MATH and MATH are in MATH. Clearly, it is enough to show that MATH. We extend MATH to a map from MATH to MATH by setting (as in REF) MATH . It is easy to see that MATH is well defined and respects the MATH-filtration for any MATH. Note that MATH by REF , so MATH. We have MATH, hence MATH; since we are given MATH, we get MATH. But we claim that if MATH then MATH, hence MATH, as required. To prove the last claim, write MATH with MATH. By REF MATH, we may normalize MATH so that MATH. We then have MATH, where MATH; hence MATH. Since MATH, we have MATH, and MATH. Hence MATH, as claimed above. |
math/0104248 | According to REF , it is enough to check that the sequence of sheaves MATH (regarded as MATH-complexes in MATH) is a complete strongly exceptional collection. Here ``complete" means that these objects generate MATH as a triangulated category, while ``strongly exceptional" means that MATH . Property MATH is trivial, since MATH . Similarly, MATH. By REF we have MATH and MATH for all MATH. If MATH then MATH, hence MATH. REF above follows. It remains to show that the collection MATH is complete. Denote by MATH the smallest strictly full triangulated subcategory of MATH containing the objects MATH. We must show that MATH. Since any derived category is generated by its abelian core, it suffices to prove that the MATH-complexes MATH are in MATH for all MATH. We know that MATH has finite global dimension, so every MATH has a finite projective resolution. Moreover, every graded projective MATH-module is a finite direct sum of shifts of MATH (see REF ). Therefore every MATH has a finite resolution by finite direct sums of sheaves MATH. Such a resolution gives a complex isomorphic to MATH in the derived category, and hence (see REF ), we need only to show that MATH belongs to MATH for any MATH. According to CITE (combine REF ), the trivial MATH-module MATH has a graded resolution of the form MATH . By shifting degrees in REF and passing to the quotient category we get (for any integer MATH) a NAME exact sequence in MATH: MATH . Letting MATH above, we observe that MATH is quasi-isomorphic to a complex each term of which is in MATH; therefore, MATH. Similarly, for MATH, it follows that MATH. Arguing in this way by induction (going both in negative and positive directions), we conclude that MATH for all MATH. This finishes the proof of the theorem. |
math/0104248 | Let MATH denote the valuation on MATH corresponding to the given filtration, that is, if MATH then MATH is the smallest integer MATH such that MATH. Let MATH be the valuation on MATH corresponding to the ring filtration we are using. Then MATH . The assumption on MATH is equivalent to MATH . Fix an embedding of MATH in MATH. Then MATH, hence every element MATH of the NAME quotient field MATH can be written in the form MATH with MATH and MATH. Moreover, we can choose MATH so that MATH. Indeed, for any MATH we have MATH, and we cannot have MATH for all MATH because MATH is finite-dimensional. Now define a function MATH as follows: if MATH as above with MATH, let MATH. To show that MATH is well defined, suppose that MATH are two such expressions for MATH. Since MATH is an NAME domain, we have MATH for some MATH. So MATH and MATH. Using REF and the similar fact for the filtration on MATH, we get MATH . Hence MATH, as desired. Now, if MATH then MATH. Thus, setting MATH, we have MATH . Also, for any element MATH, if we choose MATH so that MATH, we have (using REF ) MATH so MATH . Let MATH denote the induced filtration on MATH. By REF , we have MATH that is, MATH . Similarly, by REF , if MATH, we have MATH equivalently, MATH, and hence MATH . The lemma now follows at once from REF . |
math/0104248 | According to CITE (see REF), we have to show that MATH . By NAME Duality (see REF MATH), it is equivalent to show that MATH and MATH vanish for MATH. The vanishing of MATH is part of REF MATH. The statement about MATH follows from REF MATH by a useful argument which would not be available in the commutative case. Because the spaces MATH are finite-dimensional, REF MATH implies that the maps MATH induced by multiplication by MATH are isomorphisms for MATH. If we use these isomorphisms to identify the spaces MATH for MATH, the action of the generators MATH and MATH gives us a finite-dimensional representation of the NAME algebra, which is impossible unless the representation space is zero. |
math/0104248 | The inverse map MATH is constructed as follows. Let MATH be a line bundle over MATH, trivial over MATH, and let MATH be a graded MATH-module with MATH. By (the proof of) REF, MATH can be embedded in a direct sum of sheaves MATH; hence the MATH-module MATH is essentially torsion-free. We set MATH, where the direct limit is taken over the maps MATH. Then MATH is a rank one torsion-free MATH-module, filtered by (the images of) the components MATH. Forgetting this filtration, we obtain a map MATH. Since MATH is trivial over MATH, we have MATH for MATH, and therefore the filtration on MATH coincides (in sufficiently high degrees) with the normalized induced filtration (see REF ). It follows easily that MATH. |
math/0104248 | Note that we are not allowed deduce this result from REF since the proof of that theorem depends on REF ! Below we will construct a (left) inverse to REF which is independent of MATH. This means that in principle we have to prove the lemma only for one particular MATH. If MATH then the lemma can be deduced from the results in CITE, CITE, CITE although the point of view in these papers is slightly different. We will give a proof which works equally well for all MATH. Perhaps the method has some independent interest. Let MATH. We need to prove two things: CASE: MATH, that is, MATH has its only non-vanishing cohomology in degree -REF. This has the effect that MATH is in the image of some object in MATH. CASE: MATH is actually in the image of MATH, that is, MATH. Now MATH has the pleasant property that if MATH then MATH. From this we easily deduce that REF.,REF. above are actually equivalent to the following single statement: CASE: MATH. Let MATH and MATH. Then MATH is the path algebra of the quiver MATH . Observing that MATH defines an equivalence between MATH and MATH we want to understand the composition MATH . Checking on projectives, and then on complexes of projectives we find that on an object MATH in MATH the composition REF is given by a length two complex concentrated in degrees MATH of the following form MATH . It is now clear that if MATH then the image of MATH under the composition REF is equal to MATH where MATH is the simple object in MATH defined by MATH. Since MATH corresponds to MATH we are done. |
math/0104250 | Using the orthonormal frame MATH on MATH, the components of the NAME - NAME - connection can be obtained via the formulas MATH resulting from the NAME - formula. A direct computation of the commutators yields MATH . Now, a short calculation gives MATH by which one further calculates MATH which shows that the first coefficient of MATH vanishes. Similarly, it can be seen that the second coefficient is also zero, since MATH and by using REF again one sees that the commutator MATH vanishes completely. In REF therefore only the terms MATH are non - trivial, and for the forms MATH this gives the stated expressions. |
math/0104250 | We calculate the components of MATH by using the structure REF of the hypersurface MATH. By REF - forms MATH and MATH vanish and MATH . Further, the differentials MATH of the connection forms are given by MATH and one obtains the stated formulas for the components MATH of the curvature tensor by using REF. Notice that MATH implying that MATH vanishes. |
math/0104250 | One computes MATH obtaining thus, with the previous proposition, for the components MATH of the NAME tensor that MATH the remaining coefficients being equal to zero. In the same way as the components of the Riemannian curvature tensor turn out to be bounded when MATH, the components of MATH and thus MATH stay bounded, too. Explicitly one has MATH showing that the divergent terms cancel out and the assertion follows. |
math/0104250 | By using the symmetries of the NAME symbols MATH one has MATH . The first component of MATH reads MATH since by the relations REF one has MATH . The second component is given by MATH as can be verified by an analogous calculation. As far as the third component is concerned, using also the relations REF one computes MATH and MATH so that MATH . In the same way one verifies for the fourth component that MATH . The stated expression for MATH then follows by noting that the equalities MATH hold and that the derivatives MATH of the components of the normal vector with respect to MATH are given by the components MATH according to MATH thus finishing the proof. We remark that, since MATH, one has that MATH for all vector fields MATH, and a computation indeed shows that the normal part of MATH vanishes. |
math/0104250 | One computes MATH . Once again we calculate the components of MATH separately. By using the symmetries of the MATH and REF one obtains MATH and, moreover, MATH . A similar calculation gives for the second component the expression MATH . One calculates further MATH as well as MATH thus obtaining for the third component of MATH that MATH . Finally, by an analogous calculation one finds that the fourth component reads MATH . Since MATH the desired statement follows by noting that MATH and MATH . |
math/0104250 | Again, MATH . By the symmetries of the MATH one has for MATH odd that MATH and one obtains for the first and third component MATH . In an analogous way one has for MATH even MATH the second and fourth component being given by MATH and the assertion follows. Again, one verifies that MATH. |
math/0104250 | By REF one has MATH where we made use of the relations REF as well as MATH . Since further MATH, MATH and MATH one obtains for the expression above that MATH . Here we made use of the relation MATH and MATH as well as MATH . Because of MATH one finally obtains MATH since MATH. By REF , MATH and using the equalities MATH in addition to the above relations, one also sees that MATH vanishes, since by REF , MATH . Analogously, MATH . Finally, MATH and the remaining components are determined by the symmetry of MATH. |
math/0104250 | One easily sees that MATH as well as MATH . So MATH. We further remark that MATH and MATH . Let us now compute the geodesic curvature of MATH regarded as a curve in MATH using the stereographic projection. With respect to the coordinates MATH the induced metric on MATH reads MATH . Writing MATH for MATH the geodesic curvature of MATH is then given by (see for example, CITE) MATH the NAME symbols MATH being obtained from REF , where now the MATH denote the components of MATH and the MATH should be replaced by the corresponding MATH. Note that MATH. We put MATH, MATH and obtain MATH a direct calculation then yields MATH and by noting that MATH and MATH one finally has (up to a sign) MATH and thus the assertion. |
math/0104250 | As computed in the proof of the previous corollary, the components of MATH with respect to the orthonormal frame REF are given by MATH . The roots of the characteristic polynomial-MATH are then MATH, MATH. |
math/0104250 | Let MATH be a geodesic of positive energy MATH from the curve MATH to the point MATH with coordinates MATH. For MATH the geodesic MATH is precisely the geodesic line REF already described. If MATH were not equal zero, at least MATH would be different from zero almost everywhere; then REF would imply that there exists a critical value MATH for which MATH . For smaller values of MATH the right - hand side of REF would become negative, implying that MATH must hold for all MATH. This means that for MATH the geodesic MATH can never reach the curve MATH. Assume therefore MATH, MATH being arbitrary. By REF we have MATH . In case that MATH, this expression becomes negative for small MATH so that MATH can never reach the set MATH. However, for MATH we have that MATH is non - negative for all MATH, as well as MATH so there are infinitely many geodesic lines MATH reaching the set MATH in MATH in a spiral motion. In this case, REF implies for MATH the relation MATH that is, MATH as well as MATH are strictly monotone increasing as functions in MATH and the point MATH is reached earliest, that is, for smallest MATH, in case that MATH is also zero. Since the length of a geodesic is given by MATH the distance of the point MATH to the set MATH must be given by the length of the geodesic MATH. The integral MATH cannot be represented by elementary functions and one has MATH where MATH is the hypergeometric function introduced above. For MATH the defining series converges even in MATH; the hypergeometric function has an analytic continuation for MATH and under the assumption that MATH it can be written for all MATH as the integral MATH where MATH denotes the Gamma function and MATH is assumed in order to make the integrand uniquely defined. If MATH is real, differentiation under the integral with respect to MATH gives the stated equality REF if one takes the relation MATH, MATH, MATH arbitrary, into account additionally. For MATH we finally deduce from REF MATH and thus MATH finishing the proof. |
math/0104250 | Let MATH and MATH be given as in the previous proposition and let MATH be a harmonic function on MATH. By REF one has MATH for all MATH. Now, by the NAME - NAME uniqueness theorem for second order differential operators MATH with elliptic metric principal symbol CITE MATH cannot vanish identically on MATH unless it vanishes everywhere. Therefore, for MATH not being trivial, the set MATH is not empty, and by REF there exist constants MATH and MATH such that MATH for all MATH. In particular MATH for all MATH. |
math/0104250 | Assume that a WK spinor with WK number MATH is given on MATH. Then, by REF MATH must hold, where MATH has been computed in REF Using the relation MATH one computes with respect to the trivialization REF MATH thus obtaining for the Laplacian of MATH that MATH compare REF . One computes further that MATH . Since MATH, MATH, one obtains MATH for the left - hand side of REF and MATH for the right - hand side, so that REF reads MATH and one sees that in case MATH, it cannot be satisfied for any choice of the curve MATH. Note that since the integrability REF is purely geometric, the assertion of the proposition holds for any spin structure. |
math/0104250 | With respect to the realization of the previous section one has for MATH that MATH that is, MATH for every MATH. Then MATH implies that the NAME operator on MATH is given by MATH since MATH . By taking MATH into account one obtains on MATH, for the NAME operator, the system of partial differential equations MATH where MATH . Let now MATH and MATH be of the form MATH. Clearly one has then MATH as well as MATH . Equating these expressions yields the relation MATH for MATH, so that by integration MATH . Putting MATH one sees that all spinors of the form MATH are harmonic on MATH, where MATH are constants. Since, further MATH for an open region MATH and MATH, the harmonic spinors MATH are in MATH. |
math/0104250 | To begin with, note that MATH converges pointwise to MATH as MATH, and we therefore introduce the function MATH replacing in MATH the parameter MATH of the NAME potential by the new parameter MATH. One computes MATH as well as MATH since MATH. Each other function in MATH and MATH of the functional dependence MATH is also harmonic with respect to MATH and MATH. We put MATH . For MATH one has then MATH . As remarked, MATH, so that MATH . Then one computes, since MATH, MATH where MATH, as well as MATH i. e., the MATH are MATH - approximations of MATH - harmonic spinors, MATH being in MATH, too. |
math/0104250 | Let MATH be as in REF . One has MATH therefore MATH is strictly increasing and tends to MATH as MATH, so that it seems natural to estimate MATH from below according to MATH . Here MATH is a cutting point to be determined in a convenient manner, such that the resulting lower bound for MATH is as great as possible. A possible choice would be the turning point MATH of MATH, which can be calculated by the condition MATH by solving the equation of third degree in MATH . Since this turns out to be a little bit involved and does not necessarily lead to optimal estimates, we look for a condition for MATH instead such that MATH i. e., MATH . This is fulfilled if MATH where we assumed MATH. For small MATH and MATH this does not represent a much stronger condition. Again this is assured if MATH and we put MATH . Then one calculates MATH the function MATH being given by MATH . We remark that, as MATH, the functions MATH and MATH tend to a finite value that is independent of MATH, namely MATH the cutting point MATH also remaining finite. We finally obtain an estimate for MATH of the form MATH . Note that MATH tends to MATH if, additionally, MATH so that the value of MATH at the point MATH becomes arbitrarily close to MATH. This can always be achieved by choosing MATH small enough, though for big MATH the cutting point MATH becomes big, too. Nevertheless, we will see that this is of no relevance for later arguments. For small MATH we do not lose too much by the above estimate, since then MATH is also small. We turn now to estimating MATH. First, one has MATH and we set MATH which yields MATH. The function MATH vanishes only for MATH. The zeros of MATH are MATH and the solutions of the equation of fifth degree in MATH, MATH . Now MATH becomes zero for MATH and is strictly increasing; MATH is equal to MATH for MATH and strictly decreasing. REF has therefore exactly one real solution; it is positive and will be denoted in the following by MATH. Note that MATH is greater than REF and bounded from above by MATH. Since MATH is non - negative and MATH, the numbers MATH and MATH are the only absolute minima of MATH. The absolute value of MATH can then be estimated according to MATH . The relation MATH as well as MATH imply the estimate MATH . In a similar way one sees by MATH that MATH has a maximum at MATH with MATH and we obtain the estimate MATH . Now MATH tends asymptotically to MATH as MATH and one computes MATH so that for MATH one sees that MATH has a maximum at MATH otherwise it is strictly increasing. Inserting MATH in MATH we obtain MATH where MATH, and thus, for MATH, the estimate MATH . As MATH, the function MATH tends to MATH. Summarizing we find that, under the assumption that MATH, MATH can be estimated from above according to MATH where MATH; finally we obtain for MATH, assuming MATH to be small, that MATH . Under the assumption that MATH we obtain the estimate MATH for the NAME quotient. The expression MATH tends to zero as MATH, so that the NAME quotient itself becomes arbitrarily small for MATH. Since for closed curves MATH the hypersurfaces MATH are complete, both MATH and MATH are self - adjoint, and by the MATH - MATH principle, see e. CASE: CITE, one has MATH since MATH is bounded from below. The domain of definition of the closure MATH of the NAME operator is given by MATH and in case MATH, one has MATH . The first assertion of the theorem then follows by noting that the inequalities MATH, MATH and MATH imply that MATH lies in MATH and MATH, respectively, since MATH is assumed to be complete. To see this, let MATH be fixed and MATH be the function defined in REF. Following CITE we put MATH . Then MATH on MATH and MATH. Further one sees that MATH is NAME - continuous and, hence, almost everywhere differentiable with MATH, where MATH is a constant. Since MATH is complete, the closed envelopes of the geodesic balls MATH are compact in MATH and therefore MATH . Since MATH, one has MATH in MATH. In the same way MATH implies with the relation MATH that MATH in MATH. Consequently, one obtains MATH, and in a similar way MATHly, by setting MATH, we obtain a sequence of elements in MATH of unit length for which MATH as MATH, which implies that MATH. |
math/0104250 | Let MATH and MATH. By our previous considerations MATH satisfies the differential REF and we consider its continuation MATH to the whole complex domain. For MATH both MATH and MATH, MATH, are meromorphic functions with poles of first and second order at zero, respectively. The differential REF is therefore of NAME type and zero is a regular singular point. Let MATH, MATH form a fundamental system of solutions of REF; they can be expanded around the origin into the uniformly convergent series MATH where MATH, MATH are constants and MATH, MATH are the roots of the equation MATH with MATH see e. CASE: CITE. One obtains MATH, MATH, which yields in our case that MATH, and hence MATH, MATH. Evidently, analogous considerations hold for MATH, too. Now, MATH . In order that the above integral remains bounded it is necessary that MATH and MATH decrease with order greater than one for MATH, since MATH therefore MATH, MATH, must hold for large MATH. As, moreover, MATH is smooth, there exists a MATH such that MATH. However, by REF one has that MATH is monotone increasing so that MATH must hold. Consequently, MATH is monotone decreasing and strictly monotone decreasing for MATH. Let us now assume that MATH without loss of generality. If MATH is not identically zero, it follows that, in a neighbourhood of the origin, its components MATH and MATH must have the developments MATH where MATH are constants; otherwise one would have MATH. Let now MATH be sufficiently small so that MATH and MATH can be developed as above and, in particular, MATH . Then MATH and hence MATH. |
math/0104250 | Let MATH be a circle in MATH with center around the origin and radius MATH. Without loss of generality we can assume that MATH. For MATH and by the previous considerations MATH are harmonic MATH - spinors on MATH, where MATH, MATH are constants and MATH. By REF , apart from the trivial representation no other representations of the MATH - action MATH can occur in the MATH - kernel of the NAME operator and we obtain REF in case that MATH. The general statement then follows from the fact that MATH and MATH are isometric for MATH. If, further, MATH is a harmonic spinor with respect to MATH, then the regularity theorem for solutions of elliptic differential equations implies that MATH. However, since all MATH - harmonic spinors have to be linear combinations of the MATH, which, nevertheless, are not regular at MATH, the MATH - kernel of the NAME operator and its closure turn out to be trivial. Since, by REF , zero belongs to the spectrum of MATH, it follows that MATH. |
math/0104250 | By REF , MATH. Further, since MATH is strictly increasing one has that MATH so that MATH, where MATH. The assertion then follows from the proposition above. |
math/0104250 | By REF , MATH is MATH - integrable over MATH for MATH. One computes further that MATH the derivatives MATH and MATH being zero so that MATH . For MATH, and assuming MATH, the monotony of the integral implies MATH . Similarly, under the assumption that MATH one computes MATH showing that MATH. Summing up we have MATH using REF one then obtains the stated bound from above for the essential spectrum of the Laplacian since, by REF , zero can be no MATH - eigenvalue of MATH and, hence, of MATH. |
math/0104250 | This is a consequence of REF . |
math/0104251 | Clearly, MATH and MATH. Assume that MATH. Then MATH . Since MATH, we have MATH. Thus MATH and MATH, a contradiction. |
math/0104251 | If MATH for all MATH, there is nothing to prove. Assume that MATH and MATH is not lc. Replacing MATH with MATH, MATH, we may assume that MATH is lc but not klt (and MATH). Let MATH be an inductive blowup of the pair MATH (see CITE) and let MATH be the exceptional divisor. By definition, MATH is irreducible, MATH and MATH is plt. Write MATH where MATH is the proper transform of MATH. Clearly, MATH with MATH. CASE: By CITE, MATH. Pick a point MATH. Then MATH is the only component of MATH, passing through MATH (see CITE). Moreover, MATH is lc at MATH CITE. Hence MATH is plt at MATH and the coefficient MATH of MATH at MATH satisfies the inequality MATH. Therefore, MATH satisfies conditions of REF . This gives us MATH, a contradiction. |
math/0104251 | For some MATH, MATH the pair MATH is lc but not plt at MATH. By REF , we have MATH, that is, MATH is lc. If MATH has only one component, there is nothing to prove. So, we may assume that MATH has exactly two components CITE. Then near MATH we have MATH where MATH and MATH. Take MATH so that MATH. As in the proof of REF, considering the weighted blow up with weights MATH we get MATH. |
math/0104251 | Apply REF . We obtain MATH and MATH. The inverse implication follows by CITE. |
math/0104251 | Assume that there is a horizontal component MATH with MATH. Let MATH be the general fiber. Then MATH and by Adjunction we have equality REF: MATH . By our assumption, MATH. Thus MATH, a contradiction. |
math/0104251 | Denote MATH and run MATH-MMP. Since MATH, each time we contract an extremal ray MATH such that MATH. Hence MATH is not contracted. By REF , at the end we obtain a model with MATH. |
math/0104251 | Since MATH, one can apply CITE to MATH. This gives us that the family MATH of all such MATH is bounded. That is there is a family MATH such that every MATH is a fiber of MATH. Therefore there is a polarization MATH on MATH giving us an embedding MATH over MATH. This induces a very ample divisor MATH on each MATH. For all coefficients of MATH we have MATH. Then MATH . Hence the family of all MATH is represented by a closed subscheme of MATH over MATH. This shows that the pair MATH is bounded. From the equality MATH we obtain the following linear equation in MATH: MATH where MATH . This gives us a finite number of possibilities for the MATH, MATH and MATH. |
math/0104251 | Let MATH be the blowup of all divisors with discrepancies MATH (see CITE) and let MATH be the crepant pullback of MATH: MATH . Then MATH satisfies conditions of REF . Moreover, MATH . Clearly, MATH (see CITE). Replace MATH with MATH. Up to permutations of the MATH we may assume that MATH . Now it is sufficient to show that MATH for all MATH. Consider the boundary MATH with MATH as in REF. Then MATH. For a sufficiently small positive rational MATH, the MATH-divisor MATH is a boundary. It is clear that MATH cannot be nef. By REF the pair MATH is lc. Run MATH-MMP. On each step we contract an extremal ray MATH such that MATH . CASE: We claim that none of the components of MATH is contracted. Indeed, assume that MATH contracts MATH. Take MATH so that MATH and MATH. Since MATH and MATH, there is a component, say MATH, of MATH meeting MATH. Further, take MATH so that MATH. Then MATH. It is easy to see that MATH and MATH. Note that MATH is lc (because so is MATH). As in REF, we can apply REF to MATH to derive a contradiction. This proves our claim. CASE: By REF the lc property of MATH is preserved on each step. At the end of the MMP we get a birational model MATH with nonbirational extremal MATH-positive contraction MATH, where MATH is either a curve or a point. CASE: Then MATH. Let MATH be the general fiber of MATH. Then MATH satisfy conditions of REF . This yields a contradiction. CASE: Then MATH and every two components of MATH intersects each other. By REF , MATH . Similarly, if MATH and the image of MATH on MATH is not a point, then MATH . But if MATH is contracted to a point on MATH, then MATH. In this case, MATH. This proves our lemma. |
math/0104253 | The NAME and dimensional hypotheses guarantee that the derived direct images are defined for complexes not bounded below. There are natural "adjunction" maps MATH giving MATH . Notice that MATH . Now the adjunctions MATH give a natural map MATH . Composing gives us a natural transformation MATH . Whether this is an isomorphism is a local question; hence we may assume that MATH and MATH for MATH. Suppose MATH is quasi-isomorphic to a complex of MATH-flat sheaves; replace MATH with this flat resolution. Then MATH. For MATH let MATH be a finite affine open cover of MATH. Let MATH denote the open affine cover MATH of MATH. Notice that in all these covers arbitrary intersections of the covering sets are affine. Let MATH denote the simple complex associated to the NAME double complex of MATH with respect to MATH. Similarly, let MATH be the NAME complex with respect to MATH. Now MATH is quasi-isomorphic to MATH, and hence MATH is quasi-isomorphic to MATH. But the sheaves of these complexes are MATH-flat by construction, whence MATH . Similarly MATH . Hence we are reduced to showing that the complex MATH is quasi-isomorphic to MATH. But this is showed in the proof of REF of EGA III CITE. |
math/0104253 | Apply the KÃREF⁄REFnneth formula with MATH, MATH, MATH, MATH, MATH, MATH and MATH. |
math/0104253 | We have the commutative diagram MATH . Notice that because both MATH and MATH are flat morphisms, we have MATH . Using this and the projection formula, we have MATH . But MATH fits in a Cartesian square MATH . As MATH is separated, MATH is a closed immersion, and consequently so is MATH. In particular, MATH is an exact functor and therefore equal to MATH. Hence MATH as claimed. |
math/0104253 | See CITE. The proof is a generalisation of NAME 's original proof of this result for the absolute transform MATH in CITE. |
math/0104253 | Consider the commutative diagram MATH . It is immediate that all squares are Cartesian. If MATH is flat, then so are MATH, MATH and MATH; this proves the claim about replacing derived pull-backs. Since in any case MATH is flat, MATH is also flat. So by REF we can do a base change around the leftmost square. We get MATH . |
math/0104253 | We simply use the composition property of derived functors and the projection formula: MATH . |
math/0104253 | This follows from EGA III REF. However, that part of EGA can be somewhat hard to read; one could also follow the simpler proof of REF , making the fairly minor and obvious adjustments for hypercohomology. |
math/0104253 | Our schemes are NAME, and so it suffices to restrict our attention to closed points. Since MATH is flat, MATH is quasi-isomorphic to a complex of sheaves flat over MATH. Moreover, MATH is proper over MATH, and so MATH is a proper morphism. We are then in position to use REF. Let MATH be a closed point. Now MATH on MATH. Hence by hypothesis the natural map MATH is trivially surjective, and by the base change theorem in fact isomorphic, for all MATH. As the hyper direct images of a complex of coherent sheaves are coherent for a proper map, we have MATH for MATH by NAME 's lemma. This proves the first part of the proposition. Now in particular MATH. Thus, by the second part of the base change theorem, MATH is an isomorphism. But as MATH is also surjective and thus isomorphic, MATH is free in a neighbourhood of MATH, again by the second part of REF. |
math/0104253 | By the assumptions and REF, MATH is a locally free sheaf shifted MATH places to the right. Hence REF gives MATH . But this shows that MATH is also a locally free sheaf shifted MATH places to the right. Both statements of the proposition are now immediate. |
math/0104253 | Consider the diagram MATH where the right-hand square is the fibre-product diagram and MATH. It is clear that the left-hand square is also commutative, and that the composition of the two top arrows is just the canonical projection MATH. But this means that the big rectangle is Cartesian, and hence so is the left-hand square too. By definition, MATH where MATH is the pull-back of the NAME sheaf onto MATH. Clearly MATH. Now by the projection formula MATH . Because MATH is flat as a base extension of a flat morphism, we can do a base change REF around the left-hand square to get MATH . But MATH is a closed immersion and thus MATH. Putting these observations together, we get MATH . |
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