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math/9912088 | The first map is an isomorphism by REF , and the second map is an isomorphism by REF . |
math/9912088 | Without loss of generality, we can assume MATH. We need to show that the following diagram is commutative: MATH . In this diagram the only map which has not been defined is MATH. But since MATH it follows that MATH fixes MATH, so we can replace MATH by MATH in REF , and obtain a map which we also denote by MATH. Since MATH fixes MATH and MATH, we can use REF to replace (to pick an example) MATH by MATH, where MATH. We have to show that the following diagram is commutative: MATH . Now observe that both composites are equal to MATH. |
math/9912088 | We start by showing that the construction of MATH is natural. Let MATH and MATH be two finite MATH complexes, and MATH a MATH-equivariant map between them. We want to define a map of sheaves MATH with the properties that MATH and MATH. Consider two MATH-cell decompositions of MATH and MATH, and let MATH be the collection of all compact subgroups MATH of MATH such that MATH appears in the cell decomposition of either MATH or MATH. Let MATH be a cover adapted to MATH. Since MATH is MATH-equivariant, for each MATH we get by restriction a map MATH. This induces for all MATH a map MATH, which commutes with restrictions. Therefore, we obtain a sheaf map MATH. To get our global map MATH, we only have to check that the maps MATH glue well, that is, that they commute with the gluing maps MATH. This follows easily from the naturality of ordinary equivariant cohomology, and from the naturality in MATH of the isomorphism MATH. (See REF with MATH, MATH and MATH.) Let MATH be a pair of finite MATH complexes, that is, MATH is a closed subspace of MATH and the inclusion map MATH is MATH-equivariant. We then define MATH as the kernel of the map MATH, where MATH is the inclusion map. If MATH is a map of pairs of finite MATH complexes, then MATH is defined as the unique map induced on the corresponding kernels from MATH. The last definition we need is of the coboundary map. If MATH is a pair of finite MATH complexes, we want to define MATH. This is obtained by gluing the maps MATH where MATH is the usual coboundary map. The maps MATH glue well, because MATH is natural. To check the usual axioms of a cohomology theory (naturality, exact sequence of a pair, and excision) for MATH, recall that it was obtained by gluing the sheaves MATH along the maps MATH. Since the sheaves MATH were defined using MATH, the properties of ordinary MATH-equivariant cohomology pass on to MATH, as long as tensoring with MATH over MATH preserves exactness. But this is implied by REF . |
math/9912088 | Notice that if MATH is a point, translation by MATH induces an isomorphism of the corresponding sheaf MATH on MATH to MATH. Via this isomorphism, the gluing map MATH is the usual restriction of holomorphic functions. Therefore MATH. |
math/9912088 | Recall that we defined MATH as the complex algebraic variety MATH. As in the proof of REF , denote by MATH the algebraic structure sheaf of MATH, and by MATH the analytic structure sheaf. Let MATH. We need to show that the natural map MATH is faithfully flat. We first show it is flat: By the local characterization of flatness (for example, REF), it is enough to show that for any MATH, the natural map MATH is flat. But this is true, and the proof goes exactly as in REF . To prove that MATH is in fact faitfully flat, we also need to show that the induced map of spectra of maximal ideals, MATH is surjective (see REF). This is clearly true, since maximal ideals of MATH and MATH correspond to the points MATH. |
math/9912088 | This follows from MATH and from MATH, where the last isomorphism comes from the NAME formula. |
math/9912088 | First assume that MATH is a line bundle. Then from the definition of the NAME character we get MATH. We have already seen that MATH. Since MATH is finite, MATH is a finite module over MATH. Let MATH be a set of generators for MATH. Choose a set of elements MATH so that MATH . Denote by MATH. The element MATH is in MATH, so there exist elements MATH such that MATH . We can also calculate MATH, MATH, etc. Choose some coordinates on MATH such that MATH. We say that a polynomial MATH is dominated by another polynomial MATH if all coefficients of MATH are positive and all the coefficients of MATH are less in absolute value than the coefficients of MATH. Also, if MATH then we can write MATH and MATH, with MATH. We then say that MATH is dominated by MATH if MATH is dominated by MATH for all MATH. Similar definitions of domination can be made for the power series ring MATH and for MATH. Let MATH be a polynomial such that MATH and MATH are dominated by MATH for all MATH. Then one can show by induction that MATH is dominated by MATH, which in turn is dominated by MATH. So MATH is dominated by MATH. But this last element belongs to MATH, so MATH also belongs to MATH. Second, assume MATH is a rank MATH complex MATH-vector bundle. Let MATH be the splitting space of MATH (or the flag variety of MATH - see CITE). By construction MATH comes with an equivariant map MATH. Then the splitting principle says that the pull-back bundle MATH decomposes as a sum of MATH-equivariant line bundles over MATH. Say MATH. Calculation using the NAME - NAME spectral sequence shows that MATH where MATH is the MATH'th symmetric polynomial in the MATH's. The classes MATH are called the NAME roots of MATH. Moreover, we can identify MATH as the subring of MATH generated by the polynomials in MATH which are symmetric in the MATH's. By tensoring with MATH or MATH the same statement is true about MATH and MATH. Now consider MATH. Since MATH is a line bundle and MATH, the first part of the proof implies that MATH for all MATH. Therefore MATH, and since it is symmetric in the MATH's it follows that MATH, which is what we wanted. |
math/9912088 | The proof of this lemma is mostly formal, and just makes intensive use of the identifications and definitions we have made so far. Start by using the exponential map MATH . The element MATH is in the image, so pick MATH such that MATH. Denote by MATH the element corresponding to MATH via the map MATH. Then one can apply MATH to MATH and get a complex number that we denote by MATH. Now it is easy to check the formula MATH. We also know that MATH. So, via the exponential map, what we have to prove becomes MATH with the equality being now regarded in MATH. Let us look more closely at MATH. We saw in the proof of REF that there is an identification MATH, and that the class MATH can be identified to MATH if this is regarded in MATH via MATH. Denote by MATH the polynomial function in MATH corresponding to MATH. Then we have to prove that MATH . But this is obvious, it is just saying that MATH is a linear function. |
math/9912088 | We notice that, by REF , MATH does indeed belong to MATH, which by REF is the stalk of MATH at MATH. Fix MATH a cover of MATH adapted to MATH. Let MATH with MATH and MATH. This implies MATH and also MATH. Denote by MATH and MATH. REF implies that MATH. Now we have to prove that MATH, that is, that MATH. Consider the surjective map MATH induced by the inclusion MATH. If MATH, we have MATH. Therefore it is enough to show that for all MATH we have MATH, where MATH. But this is equivalent to MATH. Denote by MATH. So it is enough to show that, for all MATH and MATH, MATH . Let MATH and MATH. REF applied to equivariant MATH-theory gives a natural isomorphism MATH . Via the identification above, we can think of MATH as a tensor product MATH, with MATH a MATH-bundle and MATH some element in the preimage of MATH via the map MATH. (At least, we know that MATH is generated by such elements.) Via the same identification, translation by MATH becomes MATH where MATH is the image of MATH via the natural map MATH, and the second MATH is regarded in MATH via the usual inclusion MATH. But notice that MATH, because of the exact sequence MATH. Also, MATH becomes MATH. So via the above correspondence we have MATH . Since MATH, it follows that MATH, and it is sufficient for us to show that, for all MATH and MATH, MATH . But this is directly implied by REF , so we are done. |
math/9912088 | Because of the NAME - NAME sequence, it is enough to verify the isomorphism for ``equivariant points" of the form MATH, with MATH a compact subgroup of MATH. Choose an identification MATH such that via this identification MATH . Then MATH. We use now REF , which is also true if we replace MATH by MATH-theory (because it is true for MATH-theory). Since the map MATH commutes with the isomorphisms of REF , it is enough to check that the following maps are isomorphisms: CASE: MATH; CASE: MATH; CASE: MATH. To prove MATH, notice that MATH. Then we have MATH. By REF , MATH. Now notice that, by definition, the map MATH is the identity. For MATH, denote MATH. Then we have MATH. But we know that MATH and MATH. So we deduce MATH. This last ring can be identified with MATH, since the condition MATH makes all power series finite. In conclusion, MATH. Let us now describe the sheaf MATH. Let MATH. If MATH, MATH, so the stalk of MATH at MATH is zero. If MATH, MATH, and the stalk of MATH at MATH is MATH. But MATH, concentrated in degree zero (MATH is MATH-torsion in higher degrees, so the components in higher degree disappear when we tensor with MATH). It follows that MATH is a sheaf concentrated at the elements of MATH, where it has the stalk equal to MATH. Then the global sections of MATH are MATH, MATH copies, one for each element of MATH. The map MATH comes from the ring map MATH. Since MATH generates the domain of MATH, it is enough to see where MATH is sent. Let MATH. Then MATH represents the standard irreducible representation MATH of MATH, where MATH acts on MATH by complex multiplication with MATH, which is regarded as an element of MATH. Notice that MATH corresponds to the element MATH. MATH, because MATH lies in MATH. Then MATH, and the stalk of MATH at MATH is MATH. Therefore MATH sends MATH to MATH. One can easily check that this map is an isomorphism. For MATH, denote MATH. As in MATH, MATH, and MATH, MATH copies. The proof that MATH is an isomorphism is the same as above. |
math/9912088 | By REF the images of the two maps MATH and MATH are the same. But MATH is injective, so we can identify MATH with the image of MATH. This implies that MATH factors through a map MATH, and this yields a splitting MATH. |
math/9912088 | Let MATH. Any compact MATH-manifold admits a decomposition as a finite MATH complex (see for example CITE). Let MATH be such a cell decomposition. We saw that if MATH is a compact subgroup of MATH, MATH. In particular, this implies that MATH. Let MATH, which is again equivariantly formal. By REF there is a natural identification MATH. By REF the map MATH is flat, so tensoring with MATH over MATH yields a splitting MATH. Now, we observed above that MATH. So we finally get a splitting MATH. This is compatible with the gluing maps of the sheaf MATH, so we get MATH. The upshot of the above discussion is that MATH is injective (since it is injective on stalks), and MATH. The global section functor MATH is left exact, so MATH is injective and MATH. This implies that MATH and MATH have the same image in MATH, namely MATH. (Notice we couldn't have done this without using the splitting, because MATH is not right exact, so it doesn't commute with the image functor.) Now recall that we have a natural isomorphism MATH. Translating the above results via MATH, we obtain that the maps MATH and MATH have the same image. But MATH and by REF the map MATH is faithfully flat. So we obtain that MATH and MATH have the same images in MATH, which is what we wanted. |
math/9912089 | The second and and the sixth maps are isomorphisms because MATH, and therefore MATH. The properties of the tensor product imply that the third and the fifth maps are isomorphisms. The fourth map comes from translation by MATH, so it is also an isomorphism. Finally, the second map MATH is an isomorphism because CASE: If MATH is not special, then MATH, so MATH is the identity. CASE: If MATH is special, then MATH. However, we have MATH. Then we can use the NAME - NAME localization theorem in equivariant cohomology from CITE. This says that MATH is an isomorphism after inverting MATH. So it is enough to show that MATH is invertible in MATH, because this would imply that MATH becomes an isomorphism after tensoring with MATH over MATH. Now, because MATH is special, REF in the definition of an adapted cover says that MATH. But MATH, so MATH, hence MATH. This is equivalent to MATH being invertible in MATH. |
math/9912089 | We modify slightly the notations used in REF to indicate the dependence on MATH: MATH. Recall that MATH is sent to MATH via the algebra map MATH. If MATH is another additive uniformizer, we saw at the beggining of this Section that there exists a nonzero constant MATH in MATH such that MATH. Choose a square root of MATH and denote it by MATH. Define a map MATH by MATH. We have assumed that MATH is homogeneous in MATH, and that MATH is the homogeneous degree of MATH. One can easily check that MATH is a map of sheaves of MATH-algebras. We also have MATH, which means that the maps MATH glue to define a map of sheaves MATH. The equality MATH comes from MATH. |
math/9912089 | For MATH to be a cohomology theory, we need naturality. Let MATH be a MATH-equivariant map of finite MATH complexes. We want to define a map of sheaves MATH with the properties that MATH and MATH. Choose MATH an open cover adapted to MATH, and MATH an additive uniformizer of MATH. Since MATH is MATH-equivariant, for each MATH we get by restriction a map MATH. This induces a map MATH. To get our global map MATH, we only have to check that MATH glue well, that is, that they commute with the gluing maps MATH. This follows easily from the naturality of ordinary equivariant cohomology, and from the naturality in MATH of the isomorphism MATH. Also, we need to define MATH for pairs. Let MATH be a pair of finite MATH complexes, that is, MATH is a closed subspace of MATH, and the inclusion map MATH is MATH-equivariant. We then define MATH as the kernel of the map MATH, where MATH is the inclusion map. If MATH is a map of pairs of finite MATH complexes, then MATH is defined as the unique map induced on the corresponding kernels from MATH. Now we have to define the coboundary map MATH. This is obtained by gluing the maps MATH, where MATH is the usual coboundary map. The maps MATH glue well, because MATH is natural. To check the usual axioms of a cohomology theory: naturality, exact sequence of a pair, and excision for MATH, recall that this sheaf was obtained by gluing the sheaves MATH along the maps MATH. Since MATH were defined using MATH, the properties of ordinary MATH-equivariant cohomology pass on to MATH, as long as tensoring with MATH over MATH preserves exactness. But this is a classical fact: see for example the appendix of CITE. |
math/9912089 | One notices that MATH, because of the third condition in the definition of an adapted cover. If MATH, a global section in MATH is a collection of sections MATH which glue, that is, MATH. So MATH in MATH, which means that we get an element in MATH, since the MATH's cover MATH. So MATH. But clearly MATH for MATH, and the gluing maps are compatible. Therefore MATH. |
math/9912089 | First, we show that the construction of MATH is canonical: Let MATH be another identification of MATH. We then have MATH, and MATH is identified with MATH. Since MATH is also identified with MATH, we get a group map MATH. This implies that we have a continuous group map MATH such that MATH. Any such map must be multiplication by a nonzero constant MATH. Moreover, we know that MATH. This implies MATH, and since MATH takes MATH isomorphically onto MATH, it follows that MATH for some integer MATH. Multiplying this by REF, we get MATH. This, together with MATH, imply that multiplication by MATH descends to a group map MATH. But this precisely means that the construction of MATH is canonical. Notice that MATH can be thought canonically as a point on the ``doubled" ellptic curve MATH. We denote by MATH and MATH the other two points of exact order REF on MATH. Then we form the divisor MATH . Although the choice of MATH and MATH is noncanonical, the divisor MATH is canonical, that is, depends only on MATH. Let MATH be an elliptic function on MATH associated to the divisor MATH. The choice of MATH is well-defined up to a constant which can be fixed if we require that MATH at zero, where MATH is the projection map. |
math/9912089 | The difficult part, namely that MATH is holomorphic, is proved in the Appendix, in REF . Consider the usual cup product, which is a map MATH, and extend it by tensoring with MATH over MATH. We obtain a map MATH. The equivariant elliptic NAME class of MATH is MATH, so we have to show that both these classes are holomorphic. But by REF in the Appendix, MATH. And the ordinary NAME class MATH belongs to MATH, so it also belongs to the larger ring MATH. Now, cup product with MATH gives an isomorphism because MATH is an invertible power series around zero. |
math/9912089 | Recall CITE that the ordinary pushforward is defined as the composition of three maps, two of which are NAME isomorphisms, and the third is a natural one. The existence of the elliptic pushforward follows therefore from the previous corollary. The proof that MATH is a map of MATH-modules is the same as for the ordinary pushforward. The last statement is an easy consequence of the topological NAME - NAME theorem (see again CITE), and of the definition of the equivariant elliptic NAME class. |
math/9912089 | From the hypothesis, we know that MATH is an isomorphism, because it is multiplication by the invertible class MATH. Also, since MATH is invertible, the localization theorem implies that MATH is an isomorphism. Therefore MATH is an isomorphism. Start with a class MATH on MATH. Because MATH is an isomorphism, MATH can be written as MATH, where MATH is a class on MATH. Now look at the two sides of the equation to be proved: CASE: The left hand side = MATH, because MATH multiplication by MATH. CASE: The right hand side = MATH, where the last equality comes from the fact that MATH is a map of MATH-modules. |
math/9912089 | Denote by MATH the normal bundle of the embedding MATH. Let us show that, if MATH, then MATH is invertible in MATH. Denote by MATH the nonequivariant NAME roots of MATH, and by MATH the corresponding rotation numbers of MATH (see REF in the Appendix). Since MATH, MATH. Also, the MATH-equivariant NAME class of MATH is given by MATH . But MATH are nilpotent, so MATH is invertible as long as MATH is invertible. Now MATH translates to MATH, which implies that the image of MATH via the map MATH is indeed invertible. To deduce now that MATH, the elliptic MATH-equivariant NAME class of MATH, is also invertible, recall that MATH and MATH differ by a class defined using the power series MATH, which is invertible for MATH small enough. So MATH exists, and by the previous Lemma, the upper part of our diagram is commutative. The lower part is trivially commutative. |
math/9912089 | To simplify notation, we are going to identify MATH with MATH, where MATH is the ``doubled" lattice described in REF. We want now to think of points in MATH as points in MATH, and of MATH as the pullback of MATH on MATH via MATH. Then we call MATH a torsion point if there is an integer MATH such that MATH (notice that torsion points are defined in terms of MATH, and not MATH). The smallest such MATH is called the exact order of MATH. From REF, we know that if MATH, MATH. Since MATH, define MATH by MATH . Now MATH was obtained by gluing the sheaves MATH along the adapted open cover MATH. So to give a global section MATH of MATH is the same as to give global sections MATH of MATH such that they glue, that is, MATH for any MATH and MATH with MATH. From REF , to give MATH is the same as to give MATH so that MATH, or MATH (MATH the inclusion MATH). Because MATH is supposed to globalize REF, we know that MATH. This implies that MATH for MATH in a small neighborhood of MATH. In fact, we can show that this formula for MATH is valid for all MATH, as long as MATH is not special. This means we have to check that MATH exists in MATH as long as MATH is not special. MATH not special means MATH. Then consider the bundle MATH. We saw in the previous subsection that according to the splitting principle, when pulled back on the flag manifold, MATH decomposes into a direct sum of line bundles MATH, where MATH are the rotation numbers. The complex structure on MATH is such MATH acts on MATH by complex multiplication with MATH. Since MATH is fixed by the MATH action, we can apply REF in the Appendix: Let MATH be the equivariant NAME root of MATH, and MATH its usual (nonequivariant) NAME root. Then MATH, with MATH the generator of MATH. Therefore MATH. So we have MATH . We show that MATH belongs to MATH as long as MATH for all MATH: Since MATH has only nonzero rotation numbers, it has a complex structure. But changing the orientations of a vector bundle only changes the sign of the corresponding NAME class, so in the formula above we can assume that MATH has a complex structure, for example the one for which all MATH. We group the MATH which are equal, that is, for each MATH we define the set of indices MATH. Now we get a decomposition MATH, where MATH is the complex MATH-vector bundle on which MATH acts by multiplication with MATH. Now we have to show that MATH gives an element of MATH. This would follow from REF applied to the power series MATH and the vector bundle MATH, provided that MATH is convergent. But MATH is indeed convergent, since MATH is meromorphic on MATH and does not have a zero at MATH. Now we show that if MATH is nonspecial, MATH for all MATH: Suppose MATH. Then MATH, so MATH is a torsion point, say of exact order MATH. It follows that MATH divides MATH, which implies MATH. But MATH, since MATH has exact order MATH, so MATH that is, MATH is special, contradiction. So we only need to analyze what happens at a special point MATH, say of exact order MATH. We have to find a class MATH such that MATH, that is, MATH. Equivalently, we want a class MATH such that MATH, that is, we want to lift the class MATH from MATH to MATH. If we can do that, we are done, because the class MATH is a global section in MATH, and it extends MATH in the stalk at zero. So it only remains to prove the following lemma, which is a generalization of REF and NAME. |
math/9912089 | We first study the class MATH on each connected component of MATH in MATH. We will see that it lifts naturally to a class on MATH. The problem arises from the fact that we can have two connected components of MATH inside one connected component of MATH, and in that case the two lifts will differ by a sign. We then show that the sign vanishes if MATH has a spin structure. As in the previous subsection, let MATH be a connected component of MATH, and MATH a connected component of MATH which contains MATH. We now calculate MATH, regarded as a class on MATH. From the decomposition REF MATH and from the table, we get the following formula: MATH . Before we analyze each term in the above formula, recall that we defined the number MATH by MATH. CASE: MATH: Here we chose the complex structure MATH such that all MATH. Then, since MATH, we have: MATH. So we get eventually MATH REF MATH, that is, MATH for some MATH. The complex structure on MATH is such that MATH acts by complex multiplication with MATH. Notice that in the previous subsection we defined the complex structure on MATH to come from the decompostion REF . This implies that MATH, and therefore MATH. Consider MATH the equivariant class on MATH corresponding to the complex vector bundle MATH with its chosen complex orientation, and the convergent power series MATH. Then MATH. Define MATH. Using the above formula for MATH with MATH, we obtain MATH REF MATH. The complex structure on MATH is the one for which all MATH. The rotation numbers satisfy MATH, hence MATH. Consider the power series MATH. MATH satisfies MATH, so MATH is either even or odd. According to REF , since MATH is a real oriented even dimensional vector bundle, MATH defines a class MATH, which is a clas on MATH. Now from the table, MATH and MATH differ by the sign MATH, so REF (with MATH) implies that MATH. Finally we obtain MATH . Now, if we put together REF - REF , and define MATH, we have just proved that MATH, or MATH where MATH . Now we want to describe MATH in terms of the correct rotation numbers MATH of MATH. Recall that MATH are the same as MATH up to sign and a permutation. Denote by MATH equality modulo REF. We have the following cases: CASE: MATH. Suppose MATH. Then MATH, which implies MATH. Therefore MATH. CASE: MATH. Let MATH. Suppose MATH. Then MATH, which implies MATH. So modulo REF, the sum MATH differs from MATH by the number of the sign differences MATH. But by definition of rotation numbers, the number of sign differences in two systems of rotation numbers is precisely the sign difference MATH between the two corresponding orientations of MATH. Therefore, MATH. CASE: MATH. Suppose MATH. Then this implies MATH, so by the same reasoning as in REF MATH. We finally get the following formula for MATH . In the next lemma we will show that, for MATH and MATH two different connected components of MATH inside MATH, MATH and MATH are congruent modulo REF, so the class MATH is well-defined, that is, independent of MATH. Now recall that MATH is a connected component of MATH. Therefore MATH, so we can define MATH . This is a well-defined class in MATH, so by REF is finally proved. |
math/9912089 | The proof follows CITE. Denote by MATH the MATH-sphere with the MATH-action which rotates MATH times around the north-south axis as we go once around MATH. Denote by MATH and MATH its NAME and NAME poles, respectively. Consider a path in MATH which connects MATH with MATH, and touches MATH or MATH only at its endpoints. By rotating this path with the MATH-action, we obtain a subspace of MATH which is close to being an embedded MATH. NAME if it is not, we can still map equivariantly MATH onto this rotated path. Now we can pull back the bundles from MATH to MATH (with their correct orientations). The rotation numbers are the same, since the NAME and the South poles are fixed by the MATH-action, as are the endpoints of the path. Therefore we have translated the problem to the case when we have the MATH-sphere MATH and corresponding bundles over it, and we are trying to prove that MATH modulo REF. The only problem would be that we are not using the whole of MATH, but only MATH. However, the difference between these two bundles is MATH, whose rotation numbers are all zero, so they do not influence the result. Now REF says that any even-dimensional oriented real vector bundle MATH over MATH has a complex structure. In particular, the pullbacks of MATH, MATH, and MATH have complex structure, and the rotation numbers can be chosen to be the MATH described above. Say the rotation numbers at the South pole are MATH with the obvious notation conventions. Then REF says that, up to a permutation, MATH, and MATH modulo REF. But this means that MATH modulo REF, that is, MATH modulo REF. |
math/9912089 | By lifting the MATH-action to a double cover of MATH, we can make the MATH-action preserve the spin structure. Then with this action MATH is a spin MATH-manifold. At the beginning of this Section, we say that if MATH is a compact spin MATH-manifold, that is, the map MATH is spin, then we have the NAME pushforward, which is a map of sheaves MATH . The NAME pushforward MATH, if we consider it at the level of stalks at MATH, is nothing but the elliptic pushforward in MATH-theory, as described in REF . So consider the element MATH in the stalk at MATH of the sheaf MATH. From REF , since MATH is spin, REF extends to a global section of MATH. Denote this global section by boldface MATH. Because MATH is a map of sheaves, it follows that MATH is a global section of MATH, that is, a global holomorphic function on the elliptic curve MATH. But any such function has to be constant. This means that MATH, which is the equivariant elliptic genus of MATH, extends to MATH, which is constant. This is precisely equivalent to the elliptic genus being rigid. |
math/9912089 | We know that the map MATH when regarded at the level of stalks at zero is the usual equivariant elliptic pushforward in MATH. Now MATH is the elliptic genus of the family. We have MATH, where MATH is the bundle of tangents along the fiber. Since MATH is spin, REF allows us to extend REF to the NAME section MATH. Since MATH is a map of sheaves, it follows that MATH, which is the elliptic genus of the family, extends to a global section in MATH. So, if MATH is the inclusion of the fixed point submanifold in B, MATH gives a global section in MATH. Now this latter sheaf is free as a sheaf of MATH-modules, so any global section is constant. But MATH is an isomorphism if we invert MATH. |
math/9912089 | From the embedding MATH, we have the following isomorphism of vector bundles: MATH . So, in terms of MATH-equivariant elliptic NAME classes we have MATH. Rewrite this as MATH where MATH is the twisted cocycle from REF . Notice that we can extend REF to virtual bundles as well. In other words, we can define MATH to be MATH twisted by the cocycle MATH. The above formula then becomes MATH which implies that MATH . So the proposition is finished if we can show that for a general vector bundle MATH . Indeed, multiplication by the equivariant elliptic NAME classes on each stalk gives the following commutative diagram, where the rows are isomorphisms: MATH . Notice that MATH is an invertible sheaf, because it is the same as the structure sheaf MATH twisted by the cocycle MATH. In fact, we can identify it by the same method we used in REF . |
math/9912089 | We have MATH, where we write MATH for MATH. Assume MATH has a trivial MATH-action. It is easy to see that MATH. Hence we can write MATH, with MATH, and MATH nilpotent in MATH. We expand MATH in NAME expansion in multiindex notation. We make the following notations: MATH, MATH, and MATH. Now we consider the NAME expansion of MATH in multiindex notation: MATH . This is a finite sum, since MATH's are nilpotent. We want to show that MATH. MATH lies in MATH, since it lies even in MATH. So we only have to show that MATH lies in MATH. But MATH, with one MATH for each connected component of MATH. If we fix one such component MATH, then the corresponding component MATH lies in MATH. According to REF , MATH is holomorphic around MATH, hence so is MATH. Therefore MATH is holomorphic in MATH around MATH, that is, it lies in MATH. Collecting the terms for the different connected components of MATH, we finally get MATH . But MATH, so we are done. If the MATH-action on MATH is not trivial, look at the following exact sequence associated to the pair MATH: MATH where MATH is the torsion submodule of MATH. (The fact that MATH follows from the following arguments: on the one hand, MATH is torsion, because of the localization theorem; on the other hand, MATH is free, hence all torsion in MATH maps to zero via MATH.) Also, since MATH is a direct sum of torsion modules of the form MATH . Now tensor the above exact sequence with MATH and MATH over MATH: MATH . We know MATH. Then MATH was shown previously to be in the image of MATH, that is, MATH. MATH, so MATH, hence MATH. Thus MATH, so there is a MATH such that MATH. MATH might not equal MATH, but MATH, so MATH. Now, MATH, and MATH, which shows that indeed MATH. |
math/9912089 | CASE: If MATH, MATH is a power series in the equivariant NAME classes MATH. But NAME classes are independent of the orientation, so MATH. CASE: If MATH, then MATH, with MATH. Hence MATH. MATH changes sign when orientation changes sign, while MATH is invariant, because of REF. |
math/9912091 | We note that REF define the affine subspace MATH in the linear space MATH. Since there exists an invariant linear connection on MATH, MATH is non-empty. Clearly MATH is a MATH-module and we need to check that MATH contains a MATH-fixed element. But this is evident since MATH is MATH-invariant and MATH is reductive. Namely, let MATH be a linear subspace corresponding to the affine subspace MATH. Then MATH is a MATH-submodule and there exists a complementary MATH-submodule MATH. The intersection MATH is fixed by MATH and represents the desired connection. |
math/9912091 | We will prove only the assertion about orbits, leaving the assertion about coverings to the interested reader. Though MATH is not necessarily reductive, the homogeneous space MATH is always reductive, that is, MATH, where MATH. Suppose that MATH is a linear map satisfying REF for MATH, MATH. Then we can extend MATH to the linear map MATH satisfying REF for MATH, MATH. Namely, we take MATH and MATH. Conversely, if we have an appropriate linear map MATH then we can define a linear map MATH by taking a projection of MATH on the subspace of operators preserving MATH and MATH and by restricting MATH on MATH. |
math/9912091 | This is an easy application of REF . |
math/9912091 | Let MATH be an irreducible MATH-submodule in MATH with the highest weight MATH and the highest vector MATH. Let MATH. The operator MATH induces a structure of MATH-module on MATH. The highest weight is MATH and the highest vector is MATH. We notice that if we have MATH for MATH then we have MATH, therefore MATH. |
math/9912095 | In view of the interpretation of MATH as MATH, the proposition follows from the fact that for a meromorphic function MATH on MATH, we have MATH regular at a point MATH if and only if MATH is regular at MATH. |
math/9912095 | Straightforward. |
math/9912095 | Let MATH be a local defining equation for MATH. Write MATH where MATH is the reduced divisor with support MATH. We claim first that the map MATH is a quasiisomorphism if and only if for all MATH the map MATH defined by the commutative diagram MATH given by MATH is a quasi-isomorphism. This follows by considering the cokernel of MATH and filtering by order of pole. The assertion of the proposition follows because MATH has a pole of order strictly smaller than the multiplicity of MATH at every point of MATH. |
math/9912095 | With notation as in REF, we take MATH, where MATH is a local coordinate. Integrability implies MATH with MATH. Multiplying through by MATH, we conclude that MATH is regular on MATH, which is equivalent to the assertion of the lemma. |
math/9912095 | We must show MATH. Using REF and MATH we reduce to showing MATH. Since MATH, we may do the computation formally locally and replace MATH by MATH. The desired vanishing follows because an exact form has no residues. |
math/9912095 | Again the proof is given in detail in CITE and we omit it. |
math/9912095 | Write MATH. Write the universal element in MATH as a matrix MATH. The assertion that MATH vanishes in the fibre at MATH means MATH for some MATH. Note this is an identity of the form MATH . We first claim that in fact this identity holds already in MATH. To see this, write MATH. Note MATH is a free module on generators MATH. Also, MATH is defined by the equations MATH which are of the form MATH that is are linear equations in the MATH with MATH-coefficients. Thus, we have an exact sequence MATH where MATH is generated by MATH-linear combinations of the MATH. We have, therefore, a reduction of structure of the sequence REF from MATH to MATH, and therefore MATH, where MATH is the MATH-span of the MATH. Hence, any MATH-linear identity among the MATH which holds at a point on MATH holds everywhere on MATH. As a consequence, we can integrate to an identity MATH with MATH. If we specialize MATH we find MATH . We conclude from REF that MATH. |
math/9912095 | We first consider the case when MATH has a pole of order MATH at infinity, so the MATH in REF. A basis for MATH is given by MATH . MATH and MATH has basis MATH . To compute the NAME connection, we consider the diagram (here MATH and MATH) MATH . One deduces from this diagram the NAME connection MATH . We may choose MATH, so by REF MATH . In MATH we have the identity MATH . We conclude MATH . In particular, the determinant connection, which is given by MATH, can now be calculated: MATH . We compare this with the conjectured value which is the negative of REF. Define MATH . One has MATH the divisor of zeroes of MATH. We need to compute MATH. We have MATH . Note that the coefficients of MATH do not involve MATH, so the MATH part of the connection dies on MATH and we get MATH . It remains to evaluate the correction terms MATH occurring in REF. In the notation of REF, MATH, and by REF we have MATH. Clearly, the only contribution comes at MATH. Take MATH. At MATH the connection is MATH . We rewrite this in the form MATH as in REF with MATH as in REF and MATH. We find MATH (defining MATH and MATH to be the numerator and denominator, respectively.) Then MATH . Combining REF, and REF we conclude MATH which is the desired formula. We turn now to the case where MATH has a pole of order MATH at infinity. We write MATH . A basis for MATH is given by MATH A basis for the NAME bundle is given by omitting MATH. As in REF, the NAME connection is MATH . To compute the trace, note that in MATH, we have if MATH . If MATH, MATH . It follows that MATH . To compute the right hand side in conjecture REF, we take as trivializing section MATH where MATH . We have MATH, the divisor of zeroes of the polynomial MATH. Again, MATH does not involve MATH, so if MATH, we have MATH . Thus MATH . Finally we have to deal with the correction term MATH. There is no contribution except at MATH. We put MATH as before, and MATH with MATH . Assume first MATH. Then MATH . In the case MATH we find MATH . The theorem follows by comparing REF, and REF. |
math/9912095 | The NAME complex (of global sections on MATH) for MATH is the tensor product of the corresponding complexes for the MATH, so REF follows from REF . For REF we have for example, the complex of global sections MATH. In MATH this gives for all MATH . REF follows easily. |
math/9912095 | Let MATH be the relative connection on MATH with respect to the map MATH, and take MATH to be the fibre coordinate for MATH. Write MATH. Then MATH so in MATH we have MATH . It follows that MATH has rank MATH (generated for example, by MATH and MATH), and the other MATH as claimed. |
math/9912095 | We can now calculate the connection MATH on MATH just as before. We have the NAME diagram (MATH). MATH . Here MATH . We get MATH . We can now substitute MATH getting finally MATH . For convenience define MATH . Representing an element in our rank two bundle as a column vector MATH the matrix for the connection on MATH becomes MATH . The corresponding connection has a regular singular point at MATH and an irregular one at MATH. Extending MATH to MATH on MATH, we can take MATH but the matrix MATH is not invertible at MATH. In order to remedy this, make the base change MATH, and adjoin to MATH the element MATH. Notice the base change modifies the NAME determinant computation. Let us ignore this for a while and continue with the determinant calculation. Define MATH. Make the change of basis MATH . This works out to MATH . Here of course MATH. Note MATH . At MATH the polar part of MATH looks like MATH . Writing MATH, the matrix for MATH with coefficients in MATH is MATH . Also MATH . With respect to the trivialization MATH the matrix MATH at MATH is MATH . Notice that the matrices for MATH are invertible both at MATH and MATH. Writing MATH the contribution MATH is of course vanishing, as well as the contribution at MATH obtained by changing the trivialization MATH to MATH. At MATH we get REF MATH . Now we compare with the conjectural REF MATH . Here MATH can be taken to be the divisor defined by the trivializing section MATH of the sheaf MATH, that is MATH. Thus one has MATH . Further, we have to write MATH where MATH. Since MATH, MATH as well, thus the local contribution at MATH is given by REF. The conjecture gives (writing MATH) MATH . Notice we have adjoined MATH to MATH so we have lost some MATH-torsion. Bearing in mind that MATH which introduces a minus sign in the determinant calculations and comparing with our earlier calculation REF above, we find that what we need to finish is The NAME determinant for NAME cohomology of MATH is twice the corresponding determinant for MATH. Again we use sheaf notation for working with modules. Recall for the pullback we substituted MATH. We can write MATH. We have MATH so with respect to the above decomposition we can write MATH . The second term on the right is the connection obtained by tensoring MATH with MATH. Using the projection formula and the invariance of the latter connection, this is the same as the connection on MATH, that is, it amounts to replacing MATH by MATH. Using REF , this changes the NAME determinant by MATH which is trivial. It follows that the NAME determinant of MATH is twice that of MATH, which is what we want. This concludes the proof of REF . |
math/9912095 | The essential point is the expansion MATH where MATH runs through permutations of MATH. |
math/9912095 | The differential form MATH on MATH is translation invariant. Indeed, to see this we may trivialize the torsor and take the point MATH to be the identity. Introducing a formal variable MATH with MATH, the group structure is then given by MATH with MATH . Since MATH it follows that the MATH are additive, whence MATH is translation invariant. Note that MATH. Define MATH by MATH. Then MATH. In particular, MATH vanishes at the zeroes of MATH. It follows that since MATH, we have MATH as well. The proof that MATH is the unique point where MATH vanishes is given in REF . We shall omit it here. |
math/9912095 | The proof is close to REF . We write (abusively) MATH for the power sum MATH, taken as a function of the elementary symmetric functions MATH. The quadratic monomials MATH all occur with nonzero coefficient in MATH. By construction, MATH with MATH. If we think of MATH as having weight MATH, MATH is pure of weight MATH, so MATH occurs with nonzero coefficient in MATH. Since the weight MATH is maximal, MATH will occur with nonzero coefficient in MATH as well. Thus, we have MATH where MATH is quadratic and contains MATH with nonzero coefficient, and MATH has no terms of degree MATH. Further, MATH has no terms of weight MATH. In particular, the variable MATH does not occur in MATH. If we replace MATH by MATH for a suitable polynomial MATH, we can eliminate MATH from MATH completely, MATH. The weight and degree conditions on MATH are the same as those on MATH, so we conclude that MATH does not involve MATH. Also, this change does not affect the monomials MATH in MATH for MATH. Thus we may write MATH since MATH, we may continue in this fashion, writing MATH, etc. |
math/9912095 | Let MATH be the blowup. Then MATH is the strict transform of MATH in MATH. Since MATH is the NAME divisor MATH in MATH, it follows that the strict transform of MATH in MATH or MATH is isomorphic to MATH. We consider the structure of MATH locally around the exceptional divisor. We may assume some MATH is invertible and write MATH. The local defining equation for MATH is MATH. Thinking of MATH as MATH with MATH, we have open sets MATH and MATH. (The tilde indicates we view these as projective coordinates on the NAME.) We have the following coordinates and equations for MATH and MATH: MATH . The strict transform MATH of MATH lies in the locus MATH and so doesn't meet MATH. Both defining equations for MATH are smooth, and MATH is smooth on MATH. Finally, MATH is also smooth. |
math/9912095 | Write MATH for the above integrand. Let MATH be some neighborhood of MATH. Then MATH is rapidly decreasing on MATH near MATH, where the size is defined by some metric on the holomorphic MATH forms on MATH. Since the chains are compact, a chain MATH on MATH will be supported on MATH for a sufficiently small neighborhood MATH of MATH. Thus, integration defines a functional MATH . It remains to show MATH for a MATH chain MATH. Let MATH be an open neighborhood of MATH not meeting MATH. Let MATH be an open neighborhood of MATH in MATH. Write MATH where MATH and MATH. Since MATH is closed, MATH. On the other hand, the volume of MATH can be taken to be bounded independent of MATH. It follows that MATH. |
math/9912095 | Let MATH be a point. We first show MATH . We calculate MATH using the NAME spectral sequence. For MATH, MATH is a smooth, affine quadric of dimension MATH. So MATH away from MATH for MATH, and MATH is a rank MATH local system. The monodromy about MATH and MATH is induced by MATH and MATH, respectively. Both actions give MATH on the fibres. It follows that, writing MATH, we have MATH . It follows that the natural map MATH is surjective and we get a distinguished triangle in the derived category MATH where MATH is a sheaf supported over MATH. In particular, MATH where the vanishing comes by identifying with group cohomology of MATH acting on MATH with the generator acting by MATH. An easy NAME argument yields MATH . It also follows from REF that MATH. The spectral sequence thus gives MATH . To compute these stalks, note the fibre of MATH over MATH is a singular quadric with singular point MATH away from MATH. Thus the fibre of MATH over MATH is the homogeneous affine quadric MATH which is contractible. Further, because MATH meets the fibre of MATH smoothly, one has basechange for the non-proper map MATH, so MATH. At infinity, we have seen again that MATH meets the fibre smoothly, so again one has basechange for MATH. Let MATH. It follows that MATH . Combining REF yields REF. To finish the proof of the lemma, we must show the inclusion MATH is a homotopy equivalence. We can define a homotopy from MATH to MATH by flowing along an outward vector field MATH. For example, if MATH and one has cartesian coordinates MATH, one can take MATH. Since MATH is smooth over MATH, one can lift MATH to a vector field MATH on MATH. Since MATH meets the fibres of MATH smoothly over some larger MATH, we can arrange for MATH to be tangent to MATH along MATH. Let MATH be a smooth function on MATH which is positive on MATH and vanishes on MATH. We view MATH as a vector field on MATH. Flowing along MATH lifts the flow along MATH, carries MATH into MATH and stabilizes MATH over MATH. This is the desired homotopy equivalence. |
math/9912097 | Let MATH be a MATH-point of MATH. We have to prove that the Cartesian product MATH is representable by a scheme proper over MATH. For a weight MATH consider the NAME scheme MATH whose MATH points are pairs: (A map MATH; a subsheaf MATH of MATH on MATH such that MATH is locally free of rank MATH and the quotient MATH is MATH-flat.) It is well-known that MATH exists and is proper over MATH. Now, let MATH be a MATH-point of MATH. To it we can attach a MATH-point of MATH for every MATH, by setting MATH. Moreover, since every collection of MATH's satisfying the NAME relations is uniquely determined by its values on the fundamental weights, that is, by the MATH's, MATH, we obtain that the above morphisms of functors define a closed embedding MATH . Therefore, our Cartesian product is also representable by a scheme proper over MATH. |
math/9912097 | Consider the embeddings MATH, MATH. We take MATH to be the maximal open subset of MATH, such that none of the above MATH's has zeroes on MATH. From the NAME relations, we obtain that on MATH, the other MATH's have no zeroes either. Hence, MATH is the thought-for MATH-bundle on MATH. Let MATH. To each of the points MATH, we can assign an element MATH. Namely, MATH is such that MATH is the order of zero of MATH at MATH. If we put MATH, we obtain a system of maximal embeddings MATH satisfying the NAME relations. Hence MATH is a MATH-bundle on MATH, which on MATH coincides with MATH. |
math/9912097 | The proof will use the same idea as the proof of REF, of which the latter is a particular case, since for MATH, MATH. Let MATH be a rerpesentation of MATH; put MATH, MATH. Let MATH denote the stack of rank MATH vector bundles on MATH and let MATH denote the stack classifying pairs MATH where MATH is a rank MATH vector bundle on MATH and MATH is a rank MATH subsheaf of MATH. We have a natural map of stacks MATH, which sends MATH to the vector bundle MATH. In addition, we have a map of functors MATH, that sends a triple MATH to MATH. It is easy to see that when MATH is large enough so that the image of MATH in MATH generates it as an algebra, the above map MATH is a closed embedding of functors. This proves the proposition, since the map MATH is known to be representable and proper (compare CITE). |
math/9912097 | The proof is a combination of REF and of the following (obvious) statement: Let MATH be an open sub-stack of finite type. Then the set of MATH for which the intersection MATH is non-empty has the form MATH where MATH is some fixed element of MATH. |
math/9912097 | Let MATH and let MATH denote MATH, viewed as a local system just on MATH. Let MATH denote the NAME acting on MATH and on MATH. To construct the sought-for automorphic function it is sufficient to produce an object MATH together with a NAME structure on it, which is a NAME eigen-sheaf with respect to MATH in a MATH-compatible way. In other words, we need that there exists an isomorphism: MATH such that for each MATH the diagram MATH is commutative. Let MATH be a reduction of the MATH-local system MATH over MATH to MATH (such a reduction exists according to our assumtion on MATH). We have now the following assertion: CASE: Let MATH be as above and assume (as in the formulation of REF) that MATH is irreducible. Then the MATH-local system MATH is regular. CASE: There exists an element MATH and its lift MATH such that MATH and such the identification MATH is induced by MATH. We define now the object MATH as MATH. Using REF we construct an isomorphism MATH that corresponds to the above MATH. REF guarantees that MATH possesses all the required properties. |
math/9912097 | A MATH-point MATH of MATH consists by definition of the following data: MATH; a pair of MATH-bundles MATH and MATH on MATH identified with one another outside on MATH; a collection of line bundles MATH, each embedded as subsheaf into the corresponding MATH, such that the NAME relations hold. Recall that the fact that the pair MATH belongs to MATH means, according to REF that for each MATH we have: MATH . Therefore, the embedding MATH gives rise to an embedding MATH and we set by REF to be the object of MATH that corresponds to MATH, MATH and the system of embeddings MATH described above. |
math/9912097 | For MATH let us denote by MATH the map MATH . By definition, for MATH, the sheaf MATH is canonically the same as MATH. We have: CASE: The maps MATH and MATH from MATH to MATH coincide. CASE: For every MATH, the maps MATH and MATH from MATH to MATH coincide. For the proof of REF, observe that for an object MATH, the sheaf MATH on MATH identifies (by base change and the projection formula) with MATH . However, the map MATH decomposes locally in the smooth topology with respect to MATH into a direct product, which implies that MATH . Therefore, MATH and hence MATH . NAME to point REF and the projection formula, the last expression can be rewritten as MATH and finally, by applying REF and the projection formula we obtain that the sheaf MATH identifies with the direct sum over MATH of the expressions MATH . Now, by REF , MATH . Since the maps MATH and MATH from MATH coincide, we obtain that MATH can be identified with MATH as required. |
math/9912097 | It is easy to see that the assertion of the proposition amounts to the following: Let MATH be as MATH-point of MATH, such that for every MATH-module MATH, the sheaf embedding MATH has no zero along the divisor MATH. Suppose that MATH is a MATH-point of MATH such that the a priori meromorphic maps MATH extend to regular maps on MATH, which do not have zeroes along MATH. In this case we have to show that MATH in fact equals MATH, that is, that MATH induces regular maps MATH for every MATH-module MATH. First, let us assume that MATH is isomorphic to MATH for some MATH-module MATH. In this case, the above map MATH must factor as MATH because MATH is maximal at MATH in MATH, by assumption. In particular, this means that MATH is regular. The same argument shows that if MATH is MATH-dimensional, corresponding to a character MATH, the map MATH is an isomorphism. Since the semigroup MATH generates MATH, we obtain that MATH is an isomorphism for all MATH-dimensional representations. To finish the proof, it remains to observe that every MATH-module can be embedded into a tensor product of a MATH-dimensional module and a one of the form MATH. |
math/9912097 | The fact that MATH is contained in MATH follows from the definitions. Therefore, to prove the proposition we have to show that every MATH-point MATH of MATH is contained is some MATH. The latter amounts to showing that there exists a MATH-bundle MATH and an isomorphism MATH, such that the composition MATH extends to a regular map on the entire MATH, which has no zeroes at MATH. The construction of such MATH repeats the proof of REF in its parabolic variant. |
math/9912097 | Point REF follows immediately from REF. Moreover, by the same reason, for a perverse sheaf MATH on MATH, the sheaf MATH is perverse. Therefore, to prove point REF, it is enough to show that whenever MATH is irreducible, MATH is irreducible as well. First of all, since the map MATH is smooth (and has connected fibers), it is clear that MATH is irreducible. Since the situation is NAME self-dual, it is enough to show, therefore, that the MATH-restriction of MATH to MATH lives in the cohomological degrees MATH. Let MATH be the cone of the map MATH. By definition, MATH lives in the cohomological degress MATH. Moreover, REF implies that the sheaf MATH is also ULA with respect to MATH. However, MATH and the required assertion follows from REF. |
math/9912097 | The assertion of the proposition is in fact an easy corollary of the following fact proven in CITE: Let MATH be a semi-simple group and let MATH and MATH be two MATH-bundles on MATH. Then for any MATH, the restrictions MATH and MATH become isomorphic. Let MATH and MATH be two MATH-points of MATH and let MATH and MATH denote their reductions to MATH, where MATH is the connected component of the identity of MATH. The map MATH is surjective, therefore one can find MATH and a point of MATH which projects to a point MATH, where MATH and MATH have the same reduction to MATH. We have: MATH, where MATH is some MATH-bundle. Now the proof follows from the fact that MATH spans MATH. |
math/9912097 | of REF Let us first prove the "if" part of the proposition. We will fix MATH and let us denote by MATH the fiber product MATH, where MATH is mapped to MATH by means of the projection MATH. Let also MATH (respectively, MATH, MATH) denote the relative version of the stack MATH (respectively, MATH, MATH) introduced in REF. In other words, the fiber of MATH over MATH is MATH and similarly for MATH and MATH. As in REF, there exists a second projection MATH: MATH . Let MATH be the following locally closed substack: MATH (MATH is, according to our conventions, an open substack of MATH.) Thus, MATH classifies the data of MATH where MATH, MATH, MATH, MATH and the data of MATH are compatible in the sense that over MATH, MATH. We have the projections MATH that send a point MATH as above to MATH respectively. If follows from REF that the first of the above projections is smooth, since the map MATH is smooth and that the second projection is an isomorphism, since the map MATH is an isomorphism. Let us denote by MATH and MATH the maps MATH that send MATH as above to MATH and MATH, respectively. Let MATH be the preimage of MATH under MATH. MATH is ULA with respect to MATH. According to REF, MATH is ULA with respect to the map MATH. Now, MATH is the composition MATH and the assertion follows from REF. Consider the map MATH . It is smooth and its composition with MATH equals MATH. Therefore, in order to prove that MATH is ULA with respect to MATH in a neighbourhood of MATH, it is enough to show that the preimage MATH is contained in the image of MATH in MATH under the above map. Thus, let MATH be a MATH-point of MATH. As the map MATH is smooth and, in particular, open, the fact that MATH for some MATH, implies that there exists a triple MATH such that REF MATH. CASE: The embeddings MATH have no zero at MATH. CASE: MATH. But this exactly means that the point MATH belongs to MATH. The "only if" part of REF follows from similar considerations by interchanging right and left. |
math/9912097 | The fact that MATH is representable is evident, since both the source and the target are representable over MATH. Hence, for a MATH-point of MATH, its preimage under MATH is a set, rather than a category, and to prove that MATH is a locally closed embedding, one has to show that this set consists of at most one element. The latter is, however, obvious from the definitions. |
math/9912097 | Let MATH be a MATH-point of MATH. To prove the proposition, it is enough to construct an irreducible stack MATH with a map MATH such that REF MATH is non-empty. CASE: MATH. Let MATH be the set of points where MATH has singularities and let MATH be the corresponding defects. Let us choose elements MATH in such a way that the weight spaces MATH are nonzero. The collection MATH corresponds to a unique MATH and MATH. We define MATH as a fiber product MATH, where MATH is mapped to MATH by means of MATH. The stack MATH splits into connected components (which are in a bijection with the connected components of MATH, and hence of MATH) and we take MATH to be the connected component of MATH corresponding to the coweight MATH. We define the map MATH as in REF. Now, it follows from REF that MATH satisfies REF above. |
math/9912097 | For every collection of non-negative integers MATH, MATH, consider the corresponding map MATH . REF implies that the function on MATH corresponding to the sheaf MATH equals MATH . By summing up along the fibers of the projection MATH, we derive the formula of REF. |
math/9912097 | Assume that MATH. In particular, the corresponding point MATH belongs to MATH. By definition, we must have that for MATH the map MATH is regular on MATH, which implies that MATH. The latter inequality means exactly that MATH. Conversely, let us assume that MATH. Let MATH be a MATH-module of the form MATH, where MATH is a MATH-module. Without restricting the generality, we way assume that the weights of MATH are MATH for some MATH. Then the weights of MATH are MATH. Hence, by definition, MATH means that MATH . However, MATH, by assumption. |
math/9912097 | Let MATH be a point of MATH, where MATH is a MATH - bundle on MATH and MATH is an isomorphism between MATH and MATH defined outside of MATH. We attach to it a point of MATH as follows: The corresponding MATH-bundle is induced from MATH, that is, MATH and the corresponding MATH-bundle is MATH. Now, the data of MATH is obtained as a composition MATH . Conversely, let MATH be an (MATH)-point of MATH. Let MATH be the corresponding point of MATH, in particular, let MATH be the corresponding MATH-bundle. By definition, MATH and MATH are identified outside MATH. By assumption, for each MATH-module MATH, we have the embedding: MATH and the maximal embedding MATH . Hence, the (a priori) meromorphic map map MATH is regular. |
math/9912097 | As we have seen above, the fibers of the map MATH are never empty. Therefore, it is enough to prove that MATH is dense in MATH. Let MATH be a MATH-point of MATH. As in the case of REF, it is enough to construct an irreducible stack MATH with a map MATH, whose image contains MATH and such that MATH. To simplify the notation, we will assume that MATH has a singularity only at one point, call it MATH. Let the defect be MATH. We have: MATH, in the terminology of REF. Let MATH be such that: CASE: MATH for MATH; CASE: MATH. It is easy to see that such MATH indeed exists (compare REF). Note that REF means that MATH lies in the group of cocharacters of MATH. Set MATH, where MATH is mapped to MATH by means of MATH. We define the map MATH as follows: For a point MATH, the resulting MATH-bundle is MATH, the MATH-bundle is MATH and the new MATH are obtained from MATH as in REF. We define MATH as the preimage in MATH of the appropriate connected component of MATH. The fact that MATH satisfies the required properties follows from REF. |
math/9912097 | Set MATH and for a a MATH - bundle, we will denote by a subscript MATH the corresponding induced MATH - bundle. CASE: Let MATH and MATH be two MATH - points of MATH. First we will show that MATH is MATH - equivalent to another MATH - point MATH, such that MATH and the induced isomorphism MATH is regular on the whole of MATH. Pick MATH. As MATH is semi-simple, the MATH-bundles MATH and MATH are isomorphic over MATH (compare REF). Moreover, since MATH is affine, the group of automorphisms of MATH over MATH is dense in MATH. Therefore, since at every MATH the relative position of MATH with respect to MATH is the same as the relative position of MATH with respect to MATH, the above isomorphism MATH can be chosen in such a way that the isomorphism between MATH and MATH is regular on MATH. Consider the corresponding point MATH. Let MATH be such that it projects onto MATH under MATH and consider the corresponding stack MATH. The composition MATH is surjective and let MATH be such that under the above composition it maps to the point MATH . We define MATH as the image of MATH in under the projection MATH. By construction, it satisfies the required condition. CASE: Using the assertion of REF , we can assume that there exists an isomorphism MATH which gives rise to an isomorphism MATH. For a MATH - divisor MATH and a MATH - bundle MATH, let MATH denote the new MATH - bundle MATH. It is clear that we can choose MATH-valued divisors MATH and MATH on MATH, so that there exists an isomorphism MATH. In this case, the identification MATH is automatically regular on the whole of MATH. Therefore, by arguing as above, we can replace our two points MATH and MATH by MATH - equivalent points (we will abuse the notation and denote the latter by the same characters) such that there exists an isomorphism MATH, for which the induced meromorphic map MATH is a global isomorphism. Moreover, by replacing MATH and MATH by MATH and MATH, where MATH is sufficiently large, we can ensure that the cohomology MATH for all irreducible MATH - modules MATH which appear in the NAME - Holder series of the NAME algebra MATH, viewed as a MATH - module via the adjoint action. However, REF implies that any two MATH-bundles, whose reductions modulo MATH are isomorphic to MATH, are necessarily isomorphic. This implies that our two points MATH and MATH are isomorphic and, in particular, MATH - equivalent. |
math/9912097 | Let MATH and MATH be two MATH-points of MATH. Let us denote by MATH (respectively, MATH) the corresponding map. Consider the complexes MATH over MATH. We will normalize (by making a cohomological shift and NAME 's twist) in such a way that their highest (respectively, lowest) cohomology is MATH in degree MATH. To prove the theorem, it is enough to show that for any MATH and MATH as above, MATH . Using REF, we can assume that MATH and let MATH be a point such that MATH, MATH. Let MATH denote the (common) preimage of MATH in MATH under MATH or MATH. Since both maps MATH and MATH are smooth we have: MATH where MATH and MATH are also normalized in the above way. This proves our assertion. |
math/9912097 | It is enough to prove that for each MATH, every irreducible subquotient of MATH is a constant sheaf on MATH. Therefore, it suffices to show that MATH is a constant function on MATH for all MATH. For that purpose, consider two points MATH and for an extension MATH, let MATH and MATH denote the corresponding MATH - points of MATH. As in the above proof of REF, we infer from REF that when MATH is sufficiently large, the sheaves MATH over MATH are isomorphic. Since this is true for infinitely many MATH, we obtain that MATH which is what we had to prove. |
math/9912097 | We have: MATH . As the projection MATH is proper, it follows from REF that it is enough to show that the sheaf MATH on MATH is ULA with repect to the projection MATH. However, since MATH is a fibration (locally trivial in the smooth topology), the needed assertion follows from REF. |
math/9912097 | Let MATH be equal to MATH. Consider the non-symmetrized NAME stack MATH, defined as MATH, where MATH. Let MATH, MATH and MATH denote the projections from MATH to MATH and MATH, respectively. The map MATH is an étale covering. Therefore, it is enough to show that the complex MATH lives in non-positive cohomological degrees and is ULA with respect to the projection MATH. Observe that the sheaf MATH on the affine NAME is an extention of sheaves of the form MATH, MATH, MATH. Hence, the complex MATH is an extention of complexes of the form MATH for finitely many values of MATH and MATH. This proves the required assertion in view of the definition of ``good"' perverse sheaves and the above REF. |
math/9912097 | It is clear that if MATH is a ``good" perverse sheaf on MATH, then so is MATH. Therefore, to prove the theorem we have to show that the restriction of MATH to MATH lives in negative cohomological degrees if MATH is ``good". In other words, we must show that for every MATH, the sheaf MATH lives in negative cohomological degrees. For MATH as above, let MATH be the corresponding element of MATH. The map MATH is the long vertical map in the diagram MATH where the map MATH is a finite (symmetrization) map. Moreover, the long horizontal map in this diagram is nothing but MATH. Let MATH be an irreducible subquotient of MATH for some MATH (of course, MATH). REF implies that every such MATH is a pull - back of MATH, where MATH is a perverse sheaf on MATH and MATH is the dimension of the corresponding connected component of MATH. By applying the projection formula, we obtain that MATH is an extention of sheaves of the form MATH where MATH is as in REF above and MATH. Therefore, it suffices to show that MATH lives in non-positive cohomological degrees. However, this follows immediately from REF. |
math/9912097 | Consider the following general set-up: Let MATH be a map of algebraic stacks with MATH smooth. Let MATH be an open substack such that the map MATH is smooth as well. Assume, in addition, that the sheaves MATH and MATH are ULA with respect to the map MATH. Now let MATH be another algebraic stack mapping to MATH. Let MATH be the Cartesian product: MATH . In the above situation MATH . The proof of this lemma follows immediately from the definition of the ULA property (compare proof of REF). We apply this lemma in the situation when MATH, MATH, MATH and the assertion of our theorem follows from REF. |
math/9912097 | A point of the stack MATH is by definition a triple of bundles MATH plus a collection of embeddings MATH which satisfy the NAME relations. We set: MATH, where for MATH, the map MATH is the composition: MATH . |
math/9912097 | By definition, for MATH, MATH which, according to the projection formula, can be rewritten as MATH . However, the maps MATH and MATH from MATH to MATH coincide and so do the maps MATH and MATH from MATH to MATH. Hence, REF can be identified, using once again the projection formula, with MATH which, by REF is the same as MATH . |
math/9912097 | Consider the fiber product MATH . It classifies the data of MATH, with MATH, MATH and MATH. As in REF, there are two projections MATH and MATH from this stack to MATH that ``forget" MATH and MATH, respectively. By applying the projection formula, we obtain that for MATH, MATH and MATH, MATH can be written as: MATH where MATH is the pull-back of MATH under the map MATH. However, the maps MATH and MATH from MATH to MATH coincide. Hence, REF can be rewritten using the projection formula as MATH . Moreover, it is easy to see that for any sheaf MATH on MATH and therefore, REF can be identified with MATH which is what we had to prove. |
math/9912097 | Since MATH is the only fundamental weight, MATH is a evidently a closed substack in the stack classifying triples MATH as above. To show that this closed embedding is an isomorphism, one has to show that any map MATH satisfies the NAME relations, which is evident. To show that MATH is smooth we argue as follows: consider the algebraic stack MATH which classifies coherent sheaves on MATH of generic rank MATH. It is known (compare CITE) that MATH is smooth. We have a natural map MATH, which sends a triple MATH as above to the quotient MATH. It is easy to see that this map is smooth, hence MATH is smooth too. |
math/9912097 | By construction, we have a natural map MATH, which corresponds to the open embedding MATH. Its cone is by REF. However,the latter complex is the same as the pull-back under MATH of MATH (compare REF). This implies our assertion. |
math/9912097 | Indeed, let us consider the fiber product MATH. We have the evaluatuion map MATH and let us denote by MATH the pull-back of the NAME sheaf on MATH under this map, tensored by MATH. To construct the isomorphism MATH, it is enough to show that both MATH and MATH can be canonically identified with the MATH - direct image of MATH under the projection MATH. However, the !-direct image of MATH under the projection MATH is evidently isomorphic to MATH and hence is further push-forward onto MATH can be identified with MATH. The assertion for MATH follows by symmetry. |
math/9912097 | Consider the fiber product MATH, where MATH is the map MATH. By definition, this stack classifies quintuples MATH, where MATH are rank MATH-bundles, MATH is an embedding of coherent sheaves such that MATH is a skyscraper sheaf at MATH, MATH is a line bundle and MATH is an arbitrary map. We have a natural map MATH, which sends a quintuple MATH as above to MATH, where MATH and MATH is the composition MATH . Let MATH we the closed substack of MATH, which corresponds to triples MATH such that MATH. To prove point REF, it is enough to show that MATH . Remark. If we knew that MATH is MATH up to cohomological shift and NAME 's twist, the last asserion would be obvious from the decomposition theorem (compare below). Since the latter fact is not in general true, we have to proceed differently. Take MATH and let MATH be as above. Let MATH be as above as well, and let MATH be the constant sheaf on the MATH-section of MATH. Consider MATH; as above, we have a natural map MATH. Let MATH be the closed substack that classifies triples MATH, for which MATH. It is clear that over the open subset MATH, the map MATH is an isomorphism and over MATH this is a fibration with the typical fiber MATH. Since MATH is smooth, from the decomposition theorem we obtain that MATH . By construction, MATH. Moreover, since MATH, MATH does not "distinguish" between MATH and MATH and we obatin: MATH . Therefore, MATH which is what we had to prove. The finishes the proof of point REF. Point REF is a tedious but straightforward verification, which we omit. |
math/9912097 | Assume that REF does not hold, that is, that there exists a root MATH of MATH such that MATH. Let MATH denote the corresponding map. Then it is easy to see that MATH commutes with the image of MATH. Hence MATH contains MATH and therefore it is not commutative. To prove REF let us note that MATH. Hence, if MATH is commutative, it follows that MATH. On the other hand, MATH clearly normalizes MATH. Hence MATH. |
math/9912097 | (of REF .) Assume that MATH is not commutative. Let MATH denote its NAME algebra. We claim that there exists a nilpotent element MATH, such that MATH for a scalar MATH. Indeed, MATH defines an automorphism of MATH (it is obvious that MATH and hence also MATH is normalized by MATH). Therefore, the above lemma shows that there exists a NAME subgroup MATH, which is normalized by MATH. Let MATH be its NAME algebra and let MATH be the nilpotent radical of this NAME algebra. Then MATH is also normalized by MATH. We choose MATH to be an eigenvector of MATH. Thus, let MATH be as above. It is well-known that to every nilpotent element MATH in MATH one can canonically associate a parabolic subgroup MATH in MATH. The subgroup MATH can be uniquely characterized in the following way: Let MATH be the NAME algebra of MATH. Fix a homomorphism MATH in such a way that MATH . Let MATH be the weight decomposition of MATH with respect to the adjoint action of MATH. Then MATH. On the one hand, the fact that MATH implies that MATH. On the other hand, since MATH is an eigenvector of MATH and the assignment MATH is canonical, the group MATH is also normalized by MATH. But as MATH is a parabolic subgroup in MATH, it follows that MATH, which is a contradiction. |
math/9912100 | We may assume both, MATH and MATH to be semi-stable. The natural inclusion MATH induces an inclusion MATH, for all MATH. Hence if MATH is not birationally isotrivial, REF is satisfied. |
math/9912100 | It is easy to find a finite cover MATH and an isomorphism MATH of polarized manifolds. In fact, there exists a coarse moduli space MATH of polarized manifolds, and NAME and NAME constructed a finite cover of MATH which carries a universal family (see CITE, p. REF). Of course one may assume MATH to be NAME with group MATH. In different terms, one has a lifting of the NAME action on MATH to MATH, giving MATH as a quotient. Let MATH be the ramification group of a point MATH. Then MATH acts trivially on the fibre MATH. On the other hand, the automorphism group of a polarized manifold of non-negative NAME dimension is finite, hence the action of MATH on MATH must locally be the pullback under MATH of the action on MATH. Necessarily the same holds true globally and MATH . |
math/9912100 | Let us write again MATH. Then MATH. If MATH is not étale, there exists some MATH such that MATH contains a multiple point. This remains true, if we replace MATH by any finite cover MATH. In particular, in order to prove the first part of REF, we may assume that MATH, for MATH the image of a section of MATH. The same can be assumed for the second part. In fact, if MATH is étale but some fibre of MATH non reduced, then the same remains true after replacing MATH by an étale covering. Consider for some MATH a point MATH which lies exactly on MATH of the components MATH of MATH, say MATH . Let MATH denote the reduced fibre of MATH or MATH over MATH, let MATH and let MATH be the multiplicity of MATH in MATH. Finally let MATH be the multiplicity of MATH as a fibre of MATH and MATH the intersection cycle, a positive multiple of MATH. For all MATH the natural map MATH induces an isomorphism MATH where MATH denotes the fractional part of a real number MATH. Since a similar equation holds true for MATH, one obtains MATH . Choosing for MATH the lowest common multiple MATH the left hand side of REF is zero, hence MATH divides MATH. Choosing MATH, one finds that each MATH divides MATH, hence MATH. For MATH one has MATH and MATH . Therefore REF implies that MATH. This is only possible for MATH and if MATH is reduced. |
math/9912100 | We may assume, that MATH is a normal crossing divisor and that the MATH-invariant defines a morphism in a neighborhood of MATH. We choose MATH to be ramified over MATH of order divisible by the multiplicities of the components of MATH, and such that MATH has a stable reduction MATH. REF allow to assume that REF holds true. Choosing for MATH the normalization of MATH, the fibres of MATH are reduced and MATH has at most rational NAME singularities. Obviously MATH factors through MATH. MATH is a desingularization of MATH, such that REF holds true, and such that the flat relative minimal model, described in REF, exists over MATH. If we take for MATH any desingularization of this minimal model, MATH is invertible, and REF holds true. Up to now, we obtained the first seven properties, and we remark, that to this aim, we can replace MATH by any larger covering. The last three properties will follow from the first ones. Let MATH be an irreducible component of MATH. Since MATH is semistable, MATH must be a reduced normal crossing divisor. In particular, the proper transform of MATH in MATH can neither belong to the multiple locus, nor to the discriminant locus, except perhaps to the part corresponding to NAME polygons. In particular MATH will not be a component of MATH. Let MATH be the reflexive hull. If MATH divides MATH then in a neighborhood of a singular point MATH of MATH the sheaf MATH is isomorphic to MATH where MATH is the MATH-invariant. In fact, MATH can not lie on transversal components of the multiple or discriminant locus, and as remarked above, all others are part of MATH. From REF we obtain an injection MATH hence MATH is invertible for all multiples MATH of MATH. Moreover, since the parts of the multiple locus or of the discriminant locus, which are missing in REF , are all exceptional components for MATH, we obtain REF, as well. REF follows from REF. For the equality in REF one just has to remark that for the fibre MATH of MATH over MATH one has MATH . Let MATH be a local section of MATH in a neighborhood of a point of MATH, which is blown up in MATH. By REF the MATH-th power of this section is the direct image of a section of MATH, hence of MATH with MATH exceptional. Since MATH contains the inverse image of an invertible sheaf on MATH, MATH must be the direct image of a section of MATH and we obtain the reflexivity in REF for all MATH. |
math/9912100 | By the canonical bundle formula CITE, MATH is a subsheaf of MATH. Hence MATH is an invertible subsheaf of MATH, and since both coincide outside of a finite number of points, they are the same. The second equality follows from REF. |
math/9912100 | For a Jacobian fibration MATH write MATH. By REF we can find a covering MATH, étale over MATH, such that the conditions in REF are satisfied for suitable models of both, MATH and MATH. We will drop the MATH and assume MATH. The family MATH is birational to the semistable family MATH. If MATH is not birationally isotrivial, REF implies that MATH is ample, and MATH will be semi-ample with respect to MATH. REF implies that, MATH is ample with respect to MATH. By REF, the same holds true for MATH. Choose the effective divisor MATH, such that MATH . The last condition in REF implies that, for all MATH, MATH hence MATH satisfies REF. MATH is smooth, and choosing MATH as the closure of the zero-section of MATH the assumptions in REF hold true (with MATH). By REF MATH contradicting the choice of MATH in REF . |
math/9912100 | REF follows from REF and from the well-known uniqueness of the decomposition of MATH (see for example REF ). For REF let MATH be a representative of a class MATH, with MATH. Since MATH is numerically equivalent to MATH, for some MATH, one obtains from the NAME formula MATH . Therefore MATH or MATH. Since MATH in MATH, replacing MATH by MATH we may assume that MATH is represented by an effective divisor MATH. Since MATH one finds the MATH to be irreducible components of the fibres of MATH, and one may assume that those are components of reducible fibres. One obtains REF from the classification of the singular fibres (see CITE, for example). |
math/9912100 | In what follows we will work locally in MATH, but by abuse of notations we will write MATH and MATH. Consider first the case where the reduced general fibre of MATH is a smooth elliptic curve. Let us assume for a moment that MATH has ramification order MATH, the multiplicity of MATH. Then the normalization MATH is étale over MATH, hence smooth over MATH. However MATH, for MATH might be singular. The fibres of MATH are smooth elliptic curves or reduced NAME. Therefore MATH has at most rational NAME singularities. Let MATH be one of those singularities, MATH. We choose local parameters MATH on MATH in MATH, such that MATH is the zero set of MATH and such that MATH is the pullback of a local parameter on MATH in MATH. So the branched cover MATH is locally given by MATH and MATH, for parameters MATH and MATH on MATH. Considering the projection given by MATH and the morphism induced in a neighborhood of MATH in MATH, we obtain a family of surfaces with a smooth general fibre and with an isolated rational NAME singularity in the special fibre MATH. By CITE such a singularity allows a simultaneous resolution, after taking a further branched covering, totally ramified over MATH. Hence, replacing MATH by MATH for some MATH depending on MATH, we may assume that the normalization MATH of MATH has a small resolution MATH. In particular, MATH is smooth and the fibres of MATH are reduced curves with at most ordinary double points as singularities, at least over a neighborhood of the given point. To do this simultaneously for all points over MATH, we have to choose MATH to be divisible by MATH and by MATH for all singular points in MATH. For any multiple MATH of MATH let MATH be the corresponding covering. Locally, for the point MATH considered above, MATH can be chosen to be the covering of MATH, totally ramified along MATH of order MATH. Since MATH is non-singular, MATH is non-singular, and REF holds true. REF follow from the construction of MATH. If the general fibre of MATH is a NAME polygons of length MATH (hence of type MATH), the construction of MATH is quite similar. All fibres of MATH are of type MATH, for MATH. Again we start with MATH, totally ramified of order MATH and with the normalization MATH. The non-smooth locus of MATH consists of MATH disjoint sections, say MATH and of a finite number of points in MATH. For the latter, the argument given above works. In fact, if MATH is one of the isolated points, MATH has again a rational NAME singularity in MATH, and choosing a larger covering there exists a simultaneous resolution. Along MATH, the morphism MATH is equi-singular. Hence for MATH totally ramified over MATH, we can simultaneously resolve the singularities in the normalization of MATH which are lying over MATH. |
math/9912100 | Both sheaves are subsheaves of MATH, hence in order to show that the inclusion MATH factors through MATH, we can argue locally in a neighborhood of MATH. Using the notation from REF, one has (locally in a neighborhood of MATH) a diagram MATH and REF holds true. |
math/9912101 | Let MATH and MATH be three vector fields on MATH. Then: MATH so that MATH is compatible with MATH. From the definition of MATH: MATH and this shows that: MATH and the result follows. |
math/9912101 | Let MATH be the orthonormal basis of MATH for MATH which diagonalizes MATH, and let MATH be the associated eigenvalues. We need to prove the upper bound above with MATH replaced by MATH, because MATH is skew-adjoint, and MATH is an orthonormal basis of MATH for MATH. According to the previous proposition and to REF , it is enough to prove that, under our curvature assumptions, for any MATH and for any orthonormal basis MATH of MATH: MATH where MATH is the sectional curvature of MATH on REF-plane generated by MATH and MATH. Let MATH the curvature operator, and MATH the metric on MATH coming from the metric on MATH. We need to prove that, for any MATH, if, when MATH are orthogonal and have unit norm, MATH, then, with the same hypothesis on MATH and MATH, we have: MATH . Let MATH, and let MATH be a REF-plane. Denote by MATH the restriction of MATH to MATH followed by the orthonormal projection on MATH, MATH its eigenvectors, and MATH its eigenvalues. If MATH are orthogonal with unit norm, they can be written as MATH and MATH, so that: MATH . If now MATH, we find that: MATH . This is maximal when: MATH (which is in MATH and corresponds to a possible value of MATH). Replacing MATH by this value in REF shows that: MATH . Since the right side is maximal for MATH, we find the upper bound we need for MATH and the result for MATH follows. |
math/9912101 | We only have to prove the second assertion, concerning the curvature. Let MATH and MATH be the area elements associated to the metrics MATH and MATH on MATH. By the NAME REF : MATH . Let MATH be an orthonormal moving frame on MATH, and let MATH be its connection REF-form, that is: MATH . Then: MATH . But MATH is an orthonormal moving frame on MATH, and its connection REF-form MATH is: MATH . Therefore: MATH . Those equations give the relation we need between MATH, MATH and MATH. The inequalities on MATH are direct consequences of this formula, because: MATH . Now the function: MATH has as derivative: MATH, so its increasing for MATH and decreasing for MATH; for MATH we find the upper bound on MATH is obtained: CASE: if MATH, when MATH, and it is MATH ; CASE: if MATH, when MATH, and it is MATH. The same argument gives the lower bound for MATH, with MATH remplaced by MATH. |
math/9912101 | A direct computation shows that, for MATH and MATH : MATH because MATH is torsion-free. |
math/9912101 | By definition: MATH . |
math/9912101 | From REF : MATH so, if MATH and MATH: MATH so that: MATH . But MATH and MATH, and it follows that: MATH . Take the scalar product (for MATH) with MATH and then with MATH, and use the symmetry of MATH with respect to MATH to obtain the result. |
math/9912101 | According to the previous proposition: MATH . Let MATH, call MATH the restriction of MATH to the NAME of REF in MATH. A simple compactness argument shows that there exists a constant MATH (which does not depend on MATH) such that: MATH where MATH is the maximum of the sectional curvatures of MATH at MATH. Therefore, isolating in MATH a part coming from the derivative of MATH from another coming from the rotation of the tangent plane during a displacement in the direction of MATH shows that: MATH because the norm of the rotation of MATH during displacements along MATH is measured by MATH. But MATH and MATH, so: MATH and, if MATH is the maximal possible value of MATH, that is, MATH: MATH whence the first result. The same computation with MATH and MATH interchanged gives the same bound for MATH. |
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