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math-ph/0009033 | Let MATH. The first equation follows from MATH and NAME 's formula MATH . To obtain the second equation write MATH instead of REF. |
math-ph/0009033 | We only prove the lemma for the case MATH. The case MATH is similar and easier. The second statement follows from the first one by an argument using the self-adjointness of MATH. To show that MATH is a bounded operator let MATH be an almost analytic extension of MATH with MATH (see REF). Then MATH where MATH. By REF MATH . Insert this into REF . From MATH and the boundedness of MATH it then follows that MATH . This is finite by REF and hence MATH is a bounded operator. |
math-ph/0009033 | We write MATH . The first term on the right-hand side of REF is bounded, because MATH is bounded. To show that also the second term on the right-hand side of REF is bounded, we use the expansion MATH which yields MATH . Here and henceforth we use the notation MATH. The right-hand side of the last equation is clearly a bounded operator, because MATH and MATH are bounded by some constants, while the second factor MATH, which has norm less than MATH, for some constant MATH, ensures absolute convergence of the integral. |
math-ph/0009033 | We use MATH to write MATH . Now we apply the pull - through - formula MATH which can be easily proved by commuting the Hamiltonian MATH with MATH, and we find, from REF MATH . The first term on the right-hand side of REF is bounded by MATH, for MATH sufficiently large. To see this, note that MATH where we used that MATH and that MATH for MATH sufficiently large. From REF , after integration over MATH and sum over MATH, it follows that the first term on the right-hand side of REF is bounded by MATH. We consider now the second term on the right-hand side of REF . We have MATH for all MATH and MATH. Now we expand the commutator in the term on the right-hand side of the last equation, using REF , and we get MATH . After commutation of the exponential MATH in the term on the right-hand side of REF to the right of the factor MATH we find MATH . Inserting this into the right-hand side of REF , the resulting bound into the right-hand side of REF and then in REF we find, after integration over MATH (which gives no problem because of the UV - cutoff), and after sum over MATH, that also the second term on the right-hand side of REF is bounded by MATH, for some finite constant MATH. This proves the lemma. |
math/0009001 | Let MATH be a stable sheaf of MATH. Assume that MATH is not stable. Since the operation MATH preserves MATH-semi-stability, there is a subsheaf MATH such that MATH . Since MATH is general, REF implies that MATH. A simple calculation shows that MATH . This means that MATH is not stable. Hence MATH must be a stable sheaf. REF follows from direct computations. |
math/0009001 | MATH . |
math/0009001 | By CITE, MATH belongs to MATH. By REF , we get MATH . Hence MATH. |
math/0009001 | The first assertion follows from REF . For MATH, we see that MATH . Since MATH, we get REF. |
math/0009001 | CASE: MATH satisfies MATH: Clearly MATH is flat over MATH. Hence we can use base change theorem to show MATH. We first prove the following two claims: CASE: MATH for all MATH. CASE: MATH except for finitely many points MATH. By the stability of MATH, we get REF . Suppose that MATH for distinct points MATH. By CITE, we get a MATH-stable extension sheaf MATH: MATH . Since MATH, we see that MATH. Since MATH, MATH satisfies MATH. Hence MATH must satisfy the inequality MATH. In particular REF holds. Applying base change theorem, we see that MATH and MATH is of dimension MATH. This means that MATH satisfies MATH. In order to prove MATH, it is sufficient to prove MATH. Since MATH, MATH. By using the spectral sequence REF, we conclude that MATH. CASE: MATH is torsion free: Indeed, let MATH be the torsion submodule of MATH. Since MATH is locally free in codimension REF, MATH is of dimension REF. Hence MATH satisfies MATH and MATH is a locally free sheaf of MATH. Since MATH is a quotient of MATH, MATH must be MATH. Hence MATH. Thus MATH is torsion free. CASE: MATH is MATH-stable: If MATH is not MATH-stable, then there is an exact sequence MATH where MATH is a MATH-stable sheaf of MATH. Then we get MATH and an exact sequence MATH . If MATH for any MATH, then by using base change theorem again, we get MATH. This implies that MATH satisfies MATH. By REF, MATH. Hence MATH, which is a contradiction. If MATH for some MATH, then MATH and MATH. Hence MATH and MATH. So we get MATH, which contradicts REF. Therefore MATH is MATH-stable. If MATH, then MATH satisfies the same condition as MATH. Taking into account of REF , we get an isomorphism MATH. |
math/0009001 | CASE: MATH satisfies MATH: We first show that MATH except for finitely many points MATH. Suppose that MATH for distinct points MATH. We shall consider the evaluation map MATH . We assume that MATH, that is, MATH. By the proof of [YREF ], MATH is surjective in codimension REF and MATH is MATH-stable. We set MATH. Then MATH. Since MATH, we get MATH . Since MATH, we get MATH. Therefore MATH except for finitely many points MATH. Base change theorem implies that MATH is a torsion sheaf of dimension REF. Since MATH is a torsion free sheaf on the integral scheme MATH, MATH is torsion free. Hence we get MATH. By the stability of MATH and NAME duality, MATH for all MATH. Hence MATH. Therefore MATH satisfies MATH. CASE: MATH is torsion free: Let MATH be the torsion subsheaf of MATH. By base change theorem, MATH is locally free on the open subscheme MATH. Hence the proof of REF implies that MATH is of dimension REF. Since MATH satisfies MATH and MATH satisfies MATH, MATH must be REF. CASE: MATH is MATH-stable: Assume that MATH is not MATH-stable. Let MATH be the NAME filtration of MATH. We shall choose the integer MATH which satisfies MATH for MATH and MATH for MATH. We shall prove that MATH and MATH. Since MATH, semi-stability of MATH implies that MATH. Hence MATH. On the other hand, we also see that MATH, is of dimension REF. Since MATH is torsion free, MATH. Hence we conclude that MATH. So MATH and MATH satisfy MATH and we get an exact sequence MATH . Since MATH, MATH-stability of MATH implies that MATH and MATH. Thus MATH is of dimension REF. Then MATH satisfies MATH, which is a contradiction. |
math/0009001 | For MATH, MATH satisfies MATH and there is an exact sequence MATH . Since MATH, we get our claim for MATH case. For general cases, we use MATH. |
math/0009001 | We choose a locally free resolution MATH of MATH such that MATH, MATH. Then MATH is quasi-isomorphic to the complex MATH . By our assumption, there is a finite subset MATH of MATH such that MATH is injective for MATH and MATH is surjective for all MATH. Hence we get CASE: MATH, CASE: MATH, CASE: MATH is a vector bundle and CASE: there is an exact sequence MATH . Hence MATH is a vector bundle on MATH. By REF , MATH is torsion free. |
math/0009001 | By the proof of REF , it is sufficient to prove that MATH for some point MATH. We choose a point MATH which is not contained in MATH. Since MATH is of pure dimension REF, we get MATH. |
math/0009001 | By taking account of a NAME filtration, we may assume that MATH is stable. We note that MATH . Since MATH, MATH, obviously REF holds. So we shall prove REF . Let MATH be the decomposition of the scheme-theoretic support of MATH, where MATH consists of all fiber components and MATH consists of the other components. Then we have an exact sequence MATH where MATH is REF-dimensional submodule of MATH. Then MATH is a purely REF-dimensional subsheaf of MATH with MATH. By the stability of MATH, we get MATH . Since MATH is sufficiently large (the condition MATH is sufficient, where MATH satisfies that MATH is ample), we get MATH. Since MATH for all MATH, we shall prove that MATH except finite numbers of points. Proof of the claim: Let MATH be the NAME filtration of MATH with respect to MATH. Then MATH . Since MATH is a subsheaf of MATH, we also have the inequality MATH. If MATH, then MATH and MATH is MATH-equivalent to MATH for some MATH. Hence the choice of MATH is finite. Clearly MATH for MATH. Hence the claim holds. |
math/0009001 | We shall first prove that MATH is MATH-flat. Let MATH be a locally free resolution of MATH on MATH. It is sufficient to prove that MATH is injective for all MATH. We note that MATH and MATH is a torsion sheaf on MATH. Since MATH is torsion free, MATH is injective for all MATH. Thus MATH is a MATH-flat sheaf. Hence we can use base change theorem. Since MATH is relative dimension REF, MATH. Since MATH is semi-stable for general MATH, MATH for a general point MATH of MATH. Thus MATH is a torsion sheaf. By the proof of base change theorem, locally there is a complex of locally free sheaves MATH which is quasi-isomorphic to MATH. Hence MATH, which means that MATH satisfies MATH. Also we get MATH. Hence MATH is of pure dimension REF. |
math/0009001 | We consider the NAME filtration of MATH with respect to MATH, MATH. Applying MATH to this filtration, we get our corollary by REF . |
math/0009001 | Assume that MATH is not semi-stable. Then, there is a stable subsheaf MATH of MATH such that MATH and MATH is of pure dimension REF. Applying MATH to the exact sequence MATH we get an exact sequence MATH . By REF , MATH. Since MATH, we also get MATH. Thus MATH and MATH satisfies MATH and MATH is a subsheaf of MATH. We set MATH and MATH. Then MATH. Since MATH is semi-stable with respect to MATH, MATH, CASE: MATH, or CASE: MATH and MATH. On the other hand, MATH . We note that the choice of MATH is finite. In REF , we shall show that the choice of MATH is also finite. Since MATH (see REF ), there is an integer MATH such that for MATH, CASE: MATH implies MATH and CASE: MATH implies MATH. Then MATH implies that MATH, which is a contradiction. Therefore MATH is a semi-stable sheaf. |
math/0009001 | We fix an ample divisor MATH. Since MATH is a subsheaf of MATH, MATH . Since MATH and MATH, we get our claim. |
math/0009001 | By NAME 's inequality, MATH . Since MATH, MATH. Therefore MATH. |
math/0009001 | We note that REF imply that MATH satisfies MATH and MATH is torsion free. Assume that MATH is not semi-stable with respect to MATH, MATH. Then there is a destabilizing subsheaf MATH of MATH such that MATH is torsion free. It is easy to see that MATH is semi-stable for general MATH. Since MATH is sufficiently large, MATH and MATH are semi-stable vector bundles of degree REF for general MATH. Then REF implies that MATH and MATH satisfies MATH and we get an exact sequence MATH . In the same way as in REF , we get a contradiction. Thus MATH is semi-stable. |
math/0009001 | For MATH, the following diagram is commutative CITE. MATH . Hence we see that MATH. Since MATH, by taking determinant of MATH, we get that MATH. Therefore the first equality holds. Applying REF again, we get that MATH. Hence we obtain that MATH. Since MATH and MATH, we get the second equality. |
math/0009001 | We shall only prove the first equality. Applying REF, we see that MATH. Hence we obtain MATH . Since MATH, MATH . By REF, MATH, and hence we get the first equality. |
math/0009001 | The first claim follows from REF and the second claim follows from REF. |
math/0009001 | Let MATH be the multiplication map. By the universal property of MATH, there is a line bundle MATH on MATH such that MATH, where MATH is the first projection. Hence MATH. Since MATH is simple, CITE implies that MATH is stable for all ample line bundles on MATH (if MATH, then MATH is a stable vector bundle on an abelian subvariety of MATH). Since MATH is finite, MATH is also ample. Hence MATH is also stable with respect to MATH, which implies that MATH is stable with respect to MATH. |
math/0009001 | MATH . |
math/0009001 | Since MATH, MATH, REF imlies that MATH, where MATH. Since MATH, the homomorphism MATH sending MATH to MATH is an isomorphism (see REF), and hence we get our claim. |
math/0009001 | By CITE, MATH. Since MATH is injective and MATH, we shall prove that the image of MATH is a primitive submodule of MATH. Let MATH be the morphism such that MATH . We shall consider the composition MATH. Let MATH be the basis of MATH. Then we see that MATH . Hence MATH is a primitive subgroup of MATH. Therefore MATH is primitive. |
math/0009001 | We set MATH and MATH. Then we see that MATH. By REF, MATH is an isometry of NAME structure. Hence REF implies our claim. |
math/0009001 | If MATH and MATH are general, then the results follow from the arguments of CITE (compare CITE). Indeed, let MATH be the moduli space of polarized abelian surfaces of MATH. Then MATH is infinite (countable) union of algebraic subsets (compare CITE). We can also find a suitable polarization for a product of elliptic curves. However, in order to treat the last statement, we also need REF and CITE. Hence we do not use NAME 's arguments here. We first assume that MATH. We note that the stability does not change under the operation MATH. Hence by REF, we may replace MATH by MATH, MATH. Thus we may assume that MATH and MATH are ample. By using REF in Appendix, we may assume that MATH, MATH are general with respect to MATH (that is, MATH in REF holds). By using REF again, we may assume that MATH and MATH are general. Replacing MATH by MATH, MATH, we may assume that MATH is a primitive ample class of MATH (we use REF ). By the connectedness of the moduli space of polarized KREF (or abelian) surfaces and REF , we may assume that REF MATH, REF MATH has an elliptic fibration MATH, REF MATH, MATH are general with respect to MATH and REF MATH, where MATH is a section of MATH, MATH a fiber of MATH and MATH. Since MATH, we see that MATH. Hence if MATH, then REF implies our claims. If MATH, then we take a family of polarized KREF (or abelian) surfaces MATH such that MATH and MATH for MATH. Applying REF again, we can reduce to the case where MATH. Therefore we get our claim. If MATH, then REF implies our assertions. |
math/0009001 | Assume that MATH does not contain a semi-stable sheaf. Let MATH be an open subscheme of a suitable quot scheme MATH such that MATH is a birational quotient of MATH by MATH: There is an open subscheme MATH of MATH and a MATH-invariant morphism MATH such that MATH is birationally equivalent to MATH. For a sequence of NAME vectors MATH, let MATH be the set of MATH such that the NAME filtration of MATH: MATH satisfies MATH. Since MATH does not contain a semi-stable sheaf, there is a sequence of MATH such that MATH is an open dense subscheme of MATH. Let MATH be a morphism sending MATH to MATH, where MATH is the MATH-equivalence class of MATH. By CITE, we see that MATH is dominant. Composing MATH with MATH, we get a morphism MATH. Obviously, this map is MATH-invariant. Hence we get a morphism MATH. By REF , MATH. Since the albanese manifold of MATH is MATH, by the universal property of the albanese map, we get that MATH. Moreover by the second claim of REF , we get that MATH. Hence a general member MATH of MATH fits in an exact sequence MATH where MATH. We set MATH, where MATH are primitive. Then MATH, where MATH. Since MATH is simple, MATH. Hence MATH. We denote the moduli stack of semi-stable sheaves MATH of MATH by MATH. Then we get that MATH. Since MATH (see CITE), we have MATH, which implies that MATH and MATH. Assume that MATH. We set MATH, MATH. If MATH, then we get that MATH. Hence MATH is birationally equivalent to MATH. |
math/0009001 | By REF , MATH. Hence we can apply REF to MATH. |
math/0009001 | By our assumption on MATH and REF , MATH. By using MATH, we have an isomorphism MATH. Let MATH be lines on MATH respectively. For a line bundle MATH of MATH, we set MATH. Let MATH be the NAME 's compactification of the moduli space of MATH-stable vector bundles MATH of NAME vector MATH and MATH. By CITE, MATH is a projective scheme and there is a contraction MATH. By this contraction, each MATH are contracted to points on MATH. Hence each MATH are perpendicular to MATH. Let MATH be the pull-back of an ample line bundle on MATH. Then MATH for any curve which does not contained in MATH. Therefore MATH. Let MATH be the center of the elementary transformation and MATH a line on MATH. Assume that MATH is also NAME. Since MATH, MATH is also of dimension REF. Since MATH is trivial on a neighborhood of MATH, we get MATH, where we denote the extension of MATH to MATH by the same MATH. Therefore we also get that MATH. Since NAME form has positive intersection with effective REF-cycles, we get that MATH . On the other hand, since MATH is the elementary transform of MATH along MATH, for an ample line bundle MATH on MATH, MATH and MATH. By REF, this is impossible. Therefore MATH is not NAME. |
math/0009001 | Let MATH be an element of MATH. We set MATH where MATH is the NAME line bundle on MATH. Then MATH and MATH. By CITE, MATH where MATH is the set of MATH-torsion points of MATH. Since MATH and MATH CITE, we obtain our lemma. |
math/0009001 | We first treat original NAME functor. By REF , MATH is defined over MATH. The second assertion follows from the same computation in REF . Let MATH be a product of two elliptic curves MATH. Let MATH be a universal family on relative jacobian on MATH. By a direct computation, we see that MATH. Hence the first claim holds. The second assertion follows from the same computation in REF . |
math/0009001 | Let MATH be an element of MATH. Then MATH is a MATH-semi-stable vector bundle of MATH. MATH is a MATH-semi-stable vector bundle on MATH with respect to MATH. Since MATH is sufficiently large, MATH is MATH-semi-stable with respect to MATH. Since MATH and MATH is a general ample line bundle, MATH is MATH-stable. By the choice of MATH, MATH is also MATH-stable. Hence MATH is a MATH-stable vector bundle, which implies that MATH consists of MATH-stable sheaves. |
math/0009001 | We assume that MATH. We set MATH, MATH. Let MATH be the elementary transformation of MATH along MATH. Since MATH, we get MATH . Hence we see that MATH. By the choice of MATH, MATH is also MATH-stable (compare CITE). Hence MATH. Since MATH is not rigid (that is, MATH), MATH. Thus MATH. Therefore we get that MATH . In this case, MATH, and hence MATH is a unique stable vector bundle of MATH. It is not difficult to see that the choice of inverse transformations is parametrized by MATH. Therefore MATH is a rational curve. |
math/0009001 | Let MATH be a MATH-semi-stable sheaf of MATH. Assume that MATH is not MATH-stable. Since MATH is general, there is an exact sequence MATH where MATH, MATH are MATH-stable sheaves of MATH. Since MATH and MATH, we see that MATH and MATH. Since MATH, MATH or MATH. Then we get MATH. Since MATH, this is impossible. Therefore MATH is MATH-stable. If MATH is not locally free, then MATH is a MATH-stable locally free sheaf of MATH, MATH. Since MATH, we see that MATH. Hence MATH must be locally free. |
math/0009001 | The proof is similar to CITE. But we repeat the proof since our assumption on MATH is weaker than that in CITE. Let MATH be the relative NAME scheme. We denote the connected component of MATH containing the section of MATH which corresponds to the family MATH by MATH. Since MATH, MATH is a smooth morphism. Let MATH be the moduli scheme parametrizing MATH-equivalence classes of MATH-semi-stable sheaves MATH on MATH with MATH CITE. Let MATH be the closed subset of MATH consisting of properly MATH-semi-stable sheaves on MATH. Since MATH is a proper morphism, MATH is a closed subset of MATH. Since MATH does not contain MATH and MATH is an irreducible curve, MATH is a finite point set. Since our problem is local, we may assume that MATH. Let MATH be a general ample divisor on MATH. Let MATH be the moduli space of simple sheaves MATH on MATH with MATH CITE. Let MATH be the closed subset of MATH consisting of simple sheaves on MATH, MATH which are not stable with respect to MATH and MATH the closure of MATH in MATH. Let MATH be the closed subset of MATH consisting of simple sheaves which are not semi-stable with respect to MATH. In REF , we shall prove that MATH is a subset of MATH. We set MATH. Then MATH is an open subspace of MATH which is of finite type and contains all MATH-stable sheaves on MATH. By using valuative criterion of separatedness and properness, we get that MATH is a proper morphism. Indeed, since MATH is proper, it is sufficient to check these properties near the fibre MATH. The separatedness follows from base change theorem and stability with respect to MATH (compare CITE), and the properness follows from REF below and the projectivity of MATH. Since MATH is a smooth morphism, CITE implies that MATH is a smooth morphism. Thus REF holds. We next prove the second claim. By the proof of CITE, there is a quasi-universal family MATH on MATH. Since MATH is relatively ample, there is a locally free resolution of MATH. Hence we can define NAME character of MATH. Then MATH, MATH, where MATH is the relative NAME class of MATH and MATH is the similitude of MATH. Therefore MATH is a homomorphism between local systems. Assume that MATH is a family of abelian surfaces with a section MATH. Then MATH and we have a flat family of coherent sheaves MATH and MATH on MATH such that MATH, MATH. Moreover there is a universal line bundle on MATH. Hence by using the formal difference MATH, we can define a morphism MATH. |
math/0009001 | We set MATH. Let MATH be the NAME polynomial of MATH with respect to MATH. Let MATH be a locally free sheaf on MATH such that there is a surjective homomorphism MATH, and we shall consider the quot scheme MATH. Then MATH defines a morphism MATH such that MATH, where MATH is the universal quotient. Let MATH be the connected component of MATH which contains the image of MATH. Since MATH is dominant, MATH is dominant, and hence surjective. Since MATH, we get that MATH, MATH, where MATH. Hence MATH as an element of MATH, where MATH is the projection. Therefore we get our lemma. |
math/0009001 | CASE: We first assume that MATH. Since MATH is not semi-stable, there is a quotient sheaf MATH such that CASE: MATH or CASE: MATH . Let MATH be a flat extension of MATH. By REF implies that MATH. Since MATH for any ample divisor MATH on MATH, similar relations to REF hold for MATH. Thus MATH is not semi-stable with respect to any ample divisor on MATH. If MATH, then by using the inequality MATH we can prove our claim. CASE: The proof is very similar to CITE. In CITE, we only consider the case where MATH-semi-stable sheaves are MATH-stable. By using a similar method as in the proof of REF , we can easily modify the arguments in the second paragraph of the proof of CITE. |
math/0009001 | We shall prove our claim by induction on MATH. CASE: Assume that MATH. Then MATH. For MATH and MATH, we get MATH. Hence MATH for general MATH and MATH. CASE: Let MATH be a pair of integers such that MATH and MATH. We set MATH. We may assume that MATH and MATH. We shall choose NAME vectors MATH, MATH such that MATH, MATH. We shall choose MATH, MATH such that MATH. If MATH, then we may assume that MATH is locally free. In this case, we have MATH. Hence MATH. Then there is a non-trivial extension MATH . By CITE, MATH is a MATH-stable sheaf. Therefore our claim holds. If MATH, then MATH, unless MATH. Assume that MATH or MATH. Then we have a non-trivial extension MATH CITE implies that MATH is a MATH-stable sheaf of MATH, where MATH. We choose points MATH such that MATH, where MATH. If we take a suitable surjection MATH, then MATH fits in an exact sequence MATH . Since MATH, MATH is a desired element of MATH. Next we assume that MATH. Then we get MATH. By REF , the claim holds. |
math/0009001 | We choose integers MATH such that MATH and MATH for MATH. We set MATH. By REF , we can choose elements MATH, MATH such that MATH. We set MATH. Then MATH is MATH-semi-stable and MATH. Since MATH, the proof of CITE implies that our proposition holds. |
math/0009001 | By REF , MATH for a general MATH. Since MATH, MATH holds for a general MATH. Then by REF , we get our claim. |
math/0009001 | We first prove REF by induction on MATH. By REF , the assertion holds for MATH. For MATH, we choose a curve MATH of MATH (numerical equiv.). Since MATH is surjective for some MATH because of the claim for MATH, we may assume that MATH does not meet MATH. Then we have an exact sequence MATH . Since MATH, MATH for all MATH. Hence if MATH except finitely many points MATH, then the claim also holds for MATH. For REF , we use REF instead of REF . |
math/0009001 | By our assumption, MATH. We choose an element MATH. By REF , MATH holds for MATH with respect to MATH and MATH belongs to MATH. Since MATH, we may assume that MATH is locally free. We show that MATH satisfies our claim. Let MATH be distinct points of MATH. Let MATH be an element of MATH which fits in an exact sequence MATH . Applying MATH, we get an exact sequence MATH and MATH belongs to MATH. We set MATH. Since MATH, we get our claim. |
math/0009001 | We note that MATH. For general points MATH and a general surjection MATH, MATH is surjective. Hence MATH. Since MATH is MATH-semi-stable, REF implies that there is an element MATH of MATH. |
math/0009001 | We note that MATH for all MATH and MATH. Hence MATH is torsion free. If MATH for some MATH, then MATH. Thus MATH holds for a general MATH. So we shall show that MATH for some MATH. By REF , it is sufficient to show our claim under the assumption MATH. By REF , our claim follows from REF . |
math/0009001 | We note that there is a curve MATH of MATH which does not meet MATH. We first show that MATH for a general MATH. Since MATH, for a general MATH, we have MATH. Hence MATH for a general MATH. Therefore MATH for a general MATH. If MATH, then we have MATH for all MATH. We consider an exact sequence induced by REF: MATH . Let MATH be the image of MATH. Then we have a filtration MATH such that MATH. Since MATH, MATH. Hence REF implies that MATH for a general MATH. Therefore MATH for a general MATH, which implies that MATH for a general MATH. Since MATH for MATH and a general MATH, by using an exact sequence MATH we get that MATH for a general MATH. |
math/0009001 | By REF , we may assume that MATH. By REF , there is a locally free sheaf MATH such that MATH. Therefore we get our claims. |
math/0009001 | Obviously MATH. Since MATH, by REF , we get our claim. |
math/0009001 | We set MATH. Then MATH. Let MATH be a semi-stable sheaf of MATH. Since MATH is semi-homogeneous, MATH. Since MATH, by REF , we get our claim. |
math/0009001 | Let MATH be a quasi-homogeneous vector bundle of MATH. CITE showed that MATH is stable with respect to any ample line bundle on MATH. Let MATH be the multiplication map. Then MATH is a family of quasi-homogeneous vector bundles of NAME vector MATH. Hence we get a morphism MATH. Replacing MATH by MATH, MATH, we may assume that MATH is ample. Then this map is finite. In order to prove the ampleness of MATH, it is sufficient to prove the ampleness of MATH, where MATH is the MATH-th projection. By using NAME theorem, we have MATH. A simple calculation shows that MATH. Since MATH is an abelian surface, MATH and MATH imply that MATH is ample. |
math/0009001 | We set MATH where MATH. Since MATH for MATH, we have MATH. This implies that MATH. Hence we get the relation MATH . Since MATH, MATH. By the definition of MATH, MATH. Hence we get MATH. By the definition of MATH, we get MATH. Replacing MATH by MATH, we may assume that MATH . Since MATH is an isometry, MATH. Hence we get MATH . Since MATH is a universal family of stable sheaves of NAME vector MATH, we get the following relations: MATH where MATH. Since MATH is an isometry, we see that MATH and MATH. Hence we get our lemma. |
math/0009001 | By REF, MATH, MATH is a simple vector bundle on MATH. Since MATH and MATH is an isotropic NAME vector, MATH is a stable vector bundle on MATH CITE. Since MATH, MATH is MATH-stable. Hence there is a morphism MATH. By REF, this morphism is injective. Since MATH is a smooth projective surface, MATH is an isomorphism. |
math/0009001 | We set MATH . It is sufficient to prove that MATH. This follows from the following relations which come from REF : MATH . |
math/0009001 | MATH . |
math/0009001 | Since MATH and MATH are semi-stable, it is sufficient to show that MATH. Since MATH, MATH. |
math/0009001 | Considering NAME filtration of MATH with respect to MATH-stability, we may assume that MATH is MATH-stable. If MATH is locally free, then obviously the claim holds. Hence we assume that MATH is not locally free. Under the notation REF, if MATH, then clearly MATH. Hence MATH for all MATH. If MATH, then MATH for MATH. |
math/0009002 | The decomposition MATH gives a partition MATH. By additivity of the NAME characteristic, CITE MATH . If MATH then MATH defines a locally trivial fibration onto MATH. Then MATH . The NAME characteristic of MATH is zero. Hence MATH and MATH. Conversely, if MATH is a primitive polynomial then by NAME formula CITE: MATH . If MATH then MATH, but if MATH then MATH (see CITE), then MATH. |
math/0009002 | Let MATH be small MATH-balls around the affine singularities of MATH and set MATH. Then MATH can be isotoped into MATH and we denote MATH the induced morphism. Then, by CITE or CITE, the invariant cycles for MATH are MATH. Suppose that one of the components of MATH has genus, then MATH has genus and the cycles corresponding to genus induced be MATH are invariant cycles. On the other hand the cycles invariant by all the monodromies associated to elements of MATH are cycles corresponding to the boundary of MATH (see CITE or CITE). Here there is only one monodromy and invariant cycles by all the monodromies are exactly cycles invariant by MATH. It provides a contradiction. |
math/0009002 | REF and CITE asserts firstly, that the cycles of MATH correspond to NAME MATH-blocks MATH for the monodromy MATH and secondly, that MATH does not have any such blocks since MATH is a tree. Now, as MATH, MATH has no cycle. |
math/0009002 | Let us suppose that in the minimal NAME decomposition of MATH there exists two distinct NAME pieces. This decomposition can be obtained, as described in CITE, from the boundary of a neighborhood of the divisor MATH; moreover a NAME twist between two NAME pieces can be calculated (see CITE) and is non-positive. But the decomposition of MATH can also be obtained as the boundary of a neighborhood of MATH (because MATH). Then the same formula proves that the NAME twist is non-negative since the orientation of the second boundary is the opposite of the first; now the NAME twist is non-negative and non-positive, hence equal to zero. That contradicts the fact that the essential pieces were distinct. |
math/0009002 | If MATH is simply connected then the irregular fiber is connected and it is reduced because MATH is a reduced polynomial, hence the generic fiber is also connected, then MATH is a primitive polynomial. Moreover MATH so MATH and by REF , MATH. Let MATH be the tube MATH where MATH is a small disk centered at MATH. Then, as in the proof of REF , by additivity of the NAME characteristic we have MATH. Since the generic fiber is connected then MATH is connected and MATH. The morphism MATH induced by inclusion, is an isomorphism if and only if MATH is a regular value at infinity (see CITE for the case where MATH is connected, and CITE for the general case). In our situation MATH is an isomorphism since MATH hence MATH. Conversely, let suppose now MATH and MATH. As MATH then MATH has the homotopy type of MATH. But, always because there is no critical value at infinity, MATH is just a product MATH (where MATH is a tubular neighborhood in MATH of the link MATH). Then MATH has the same homotopy type as MATH. Now the polynomial MATH is primitive since MATH is reduced and MATH, hence MATH is connected. So MATH is a connected set. As the NAME characteristic of the connected set MATH is MATH, it implies that all the irreducible components of MATH are disks (possibly singular), crossing together, without cycle REF . As a conclusion MATH is a simply connected set. |
math/0009002 | By NAME theorem an equation for MATH is MATH. As in CITE a parameterization of MATH is MATH, with MATH. If MATH does not intersect MATH then MATH is a non-zero constant. If MATH intersects MATH transversally then, as in the proof of NAME theorem by NAME and NAME in CITE, polynomial automorphisms of type MATH enable us to choose MATH as an equation of MATH. |
math/0009002 | If there is no essential singularity, then singularities are ordinary quadratic singularities. As MATH is simply connected then it is the union of smooth disks MATH. Let us suppose that MATH and MATH intersect transversally. Then, by the lemma above, an equation of MATH is MATH, moreover another disk MATH can not intersect MATH and MATH otherwise there is a cycle in MATH or an essential singularity. Then MATH has equation, for instance, MATH. The other disks MATH, MATH are parallel to MATH otherwise there are cycles in MATH, thus MATH has equation MATH. Then MATH is algebraically equivalent to MATH. Let us suppose that there is an essential singularity, then by REF there is only one essential singularity. All the other singularities are ordinary quadratic singularities. Moreover, as MATH and as the tube MATH is a NAME manifold REF , then the link at infinity MATH (MATH is a MATH-dimensional sphere with radius MATH) is a NAME link (that is to say MATH admits a NAME fibration, and the components of the link are fibers for this fibration). By CITE, as MATH and MATH, this link at infinity is the connected sum of the local links of the singularities of MATH, that is to say the link at infinity is the connected sum of the local link of the essential singularity with NAME links. But a NAME link can not have such a structure, then there is only one singularity. So the local link and the link at infinity are isotopic and are a sublink of MATH . Let explain the notations: in the sphere MATH of MATH, MATH, MATH are unknots such that MATH is the NAME link; MATH denotes a torus knot of type MATH (MATH and MATH are relatively prime non-zero natural numbers) such that MATH is isotopic to the link MATH . We now prove that MATH can be written, up to algebraic equivalence, as required. We discuss according to the number of smooth disks in MATH. CASE: We assume that, in MATH, there are two smooth disks with transversal intersection at the essential singularity. By REF , up to algebraic equivalence, an equation of MATH is MATH. We have to prove that MATH. Let us consider the polynomial MATH and let MATH be a polynomial injective parameterization of the curve MATH. The local link for the locally irreducible singularity is a link of type MATH so this parameterization can be written: MATH with MATH and MATH. As MATH is the only point of intersection between MATH and MATH then MATH implies MATH and then MATH. So MATH is monomial: MATH. For similar reasons MATH, and then MATH. CASE: If MATH has only one smooth disk then for some coordinates an equation of MATH is MATH. As before we denote by MATH a parameterization of MATH; we obtain again MATH but with MATH. We can conclude as in CITE: the parameterization MATH is injective so MATH has only one root MATH for all MATH-th root MATH of unity. Hence MATH is of the form MATH and the polynomial change of coordinates MATH, that preserves the axis MATH, gives a parameterization MATH in these new coordinates. So MATH in these coordinates. CASE: If MATH has no smooth disk we can assume that for the decomposition MATH we have MATH by Za\ĭdenberg-Lin theorem for the irreducible component MATH. Let MATH be another component with parameterization MATH. The link at infinity for MATH is an iterated torus knot of type MATH (see CITE), with MATH and MATH. But the link at infinity for MATH is isotopic to local link of the affine singularity of MATH and then is of type MATH. As in CITE either MATH and then MATH so MATH, MATH and the result is proved; or MATH is the unknot and then MATH divides MATH or MATH divides MATH. It implies that MATH. As the components MATH and MATH have only one intersection (at MATH) the one variable polynomial MATH is equal to MATH. For example if we assume that MATH then MATH. But the valuation of MATH is the intersection multiplicity of MATH and MATH at MATH and it is equal to MATH. Thereby MATH and MATH and as before MATH. |
math/0009002 | Notice that, since MATH the components MATH REF are disjoint, then MATH and one of the component has positive NAME characteristic. But as MATH for all MATH, we can suppose that the components of NAME characteristic MATH are MATH (MATH). We firstly assume that MATH; all the other components verify MATH for MATH, this implies that MATH (MATH). As a conclusion the component MATH is a disk and the others are annuli. Secondly we suppose that MATH. Because of the NAME theorem REF we can assume that these disks MATH are parallel lines with equation MATH. All the other components MATH REF have at least MATH branches at infinity because of the non-intersection with the lines: such a component MATH has MATH branches at infinity whose tangents at infinity are the MATH parallel lines that intersect the line at infinity at one point; if there is one other point at infinity for MATH then there exists one other branch, if there is no other point at infinity then the line at infinity is tangent to MATH, that gives one more branch. Particularly we have MATH for MATH; then MATH . Thus this inequality is an equality; it implies that MATH and MATH, particularly there are exactly MATH branches at infinity. All the components are disks, except the last one which is a MATH-punctured sphere. This completes the proof. |
math/0009002 | The proof is similar to the standard proof, see for example CITE. By abuse, we also denote MATH and MATH where MATH is the union of small open disks around the points of MATH. Let MATH be a triangulation of MATH with ramification points contained in MATH, we denote MATH, MATH,. There exists a triangulation MATH of MATH above MATH such that MATH, MATH and MATH. Then MATH. |
math/0009002 | By NAME theorem we can assume that MATH is the equation for the disk MATH, let MATH be an equation of MATH in these coordinates, there exists MATH such that MATH with MATH, MATH and MATH, MATH. If MATH denotes the curve of REF then the ``blow-up" MATH gives an isomorphism from MATH to MATH, so MATH is homeomorphic to a disk and according to Za\ĭdenberg-Lin theorem the polynomial MATH equal MATH, the new coordinates are given by MATH and MATH, then MATH. Returning to MATH by MATH, and distinguishing the cases MATH and MATH leads to MATH and MATH, with MATH. By triangular automorphisms MATH we can assume that MATH. That ends the proof. |
math/0009002 | We deal with the second case of REF , the disks are given by an equation MATH and the equation of MATH is MATH with MATH, MATH. The projection MATH given by MATH, is of degree MATH and verify the hypothesis of our NAME formula since the points at infinity of MATH correspond to MATH. As MATH then MATH and MATH is a constant. The equation of MATH is now MATH and by some triangular automorphisms MATH we can assume that MATH. |
math/0009002 | Let us recall that from REF we know that there is at most only one essential singularity, and affine non-essential singularities are ordinary quadratic singularities. The non-essential singularities at infinity correspond to a bamboo for the divisor at infinity MATH for the value MATH which intersects the compactification of some smooth disks and another component (possibly singular) of MATH, moreover the multiplicities of MATH equal to MATH on all the components of the bamboo. A typical example is given by NAME polynomial MATH, another example is given in paragraph REF. Let us notice that one of the components of MATH is a disk (possibly singular) because MATH. We firstly suppose that no affine singularity is essential. Then affine singularities are ordinary quadratic singularities and a disk of MATH is smooth because it can not intersect itself as there is no cycle in MATH. In a second time if there exists one essential affine singularity, it is unique and singularities at infinity are non-essential. As MATH such singularities do exist. Then one of the disks associated to a non-essential singularity at infinity is smooth. |
math/0009002 | Let MATH be the disk of REF . Let denote MATH the smooth disks parallel to MATH. According to NAME theorem, we can suppose that equations for these disks are MATH. Let MATH be one of the MATH REF which does not intersect one of the MATH: such a MATH exists otherwise MATH is a connected set and as MATH this is impossible by the remark below REF . After reordering the disks MATH we denote by MATH REF the disks that do not intersect MATH. Then as above REF MATH, and MATH has exactly MATH branches at infinity. Components other than MATH do not contribute to NAME characteristic: that is to say the other disks have intersection with one of the MATH at one point and at exactly one point because MATH has no cycle; components that are not disks are annuli. For MATH we have MATH and the points at infinity correspond to MATH. The NAME formula for the covering MATH defined by MATH proves that MATH is non-branched and that MATH is smooth. Singularities coming from intersections with other components can only be transversal intersections of a smooth disk and another component: the key point is the topology of MATH. First of all, to keep MATH, two components with non-positive NAME characteristic can not intersect; in a second time a disk that intersects the disk MATH is smooth, otherwise it contradicts the configuration for non-essential singularities at infinity; and finally to avoid cycles in MATH, only two directions for disks (for example MATH and MATH) can occur, hence there are no multiple points of order greater than MATH. We have just proved that the affine singularities were ordinary quadratic singularities. We end the classification as in REF . The main difference comes from some lines that give ordinary quadratic singularities. If MATH and MATH is not smooth in MATH then, for NAME characteristic reasons, there can be only one more disk MATH, and MATH intersects transversally MATH at one point. With the algebraic classification of annuli, we see that only one kind of annuli can occur: MATH where MATH is a family of distinct non-zero complex numbers. For similar reasons, if MATH and the disk MATH is smooth in MATH, then only one annulus can occur, but disks MATH REF parallel to MATH can intersect this annulus. Then MATH . The case MATH is treated as in REF with parallel lines added: MATH . This completes the proof. |
math/0009002 | Let MATH and MATH be polynomials with just one critical value MATH and with equivalent colored graphs. Let MATH, MATH come from the resolution of MATH and MATH. One can suppose, after some blowing-ups and absorptions, that their graphs are equal. We set MATH and MATH. By standard arguments (CITE, CITE, CITE), a small neighborhood of MATH is homeomorphic to a small neighborhood of MATH. As all the components of MATH and MATH are rational the monodromies for MATH and MATH induced by a small circle around the value MATH act equivalently: that is to say the following diagram commutes: MATH where MATH and MATH are homeomorphisms and MATH and MATH topological closed disks of MATH with MATH. Let MATH be the part of MATH that corresponds to the irregularity at infinity of the value MATH (MATH is set in the same way). Then MATH defines an homeomorphism between MATH and MATH. Then the homeomorphism MATH from MATH to MATH can be restricted to an homeomorphism MATH that respects the fibration because MATH on the set MATH. We have proved that MATH and MATH are topologically equivalent in a neighborhood of the zero fiber: MATH . We now explain how to continue theses homeomorphisms. As the only critical value for MATH is in MATH the fibration MATH is isomorphic to the fibration MATH. It provides homeomorphisms MATH and MATH (see diagrams). For MATH we obtain homeomorphisms MATH and MATH. But the fibration MATH above MATH is isomorphic to the fibration MATH above MATH, the corresponding homeomorphisms are MATH and MATH. MATH . Then MATH can be continued by MATH and MATH by MATH. |
math/0009002 | We firstly have to prove that the list of polynomials up to algebraic equivalence can be reduced, up to topological equivalence, to the list above. Finally we shall prove that two distinct polynomials of this list are not topologically equivalent. For the cases with MATH, replacing MATH by MATH does not change the polynomial, up to topological equivalence. Moreover the list, for theses cases, is not redundant. Let study what happens for the case MATH. Let MATH be one of the polynomials coming from the algebraic list, and let MATH be the corresponding polynomial with the constant MATH instead of the polynomial MATH or MATH and with MATH instead of MATH. We may find MATH by proving that the graphs MATH and MATH are equivalent. As MATH and MATH have the same behavior at finite distance, we just have to study what happens at infinity. Let MATH be the homogeneous polynomial associated to MATH, MATH and MATH are the two points at infinity of MATH; we denote MATH, MATH the local equations of MATH at the points MATH, MATH. To calculate the part of MATH at infinity, we have two - equivalent - choices. Firstly we can calculate the irregular link at infinity MATH (respectively, MATH), it is a sufficient condition since the (single) irregular link determines the regular links at infinity MATH (MATH), see CITE. Secondly, we can calculate the NAME expansions of the branches of MATH (and MATH) and the intersection multiplicities between the branches of MATH (and between the branches of MATH) by taking into account the line at infinity with local equation MATH. It is a sufficient condition since if we know the topology of MATH then one can recover the topology of the family MATH (see CITE) as MATH and MATH are the only critical values for this family. We will use the second method: MATH if and only if MATH and MATH (and MATH and MATH) have equivalent NAME expansions and the same intersection multiplicities. We will detail the calculus for MATH with MATH, MATH and MATH, the calculus are similar for the other polynomials. Then MATH and MATH. The local equation of MATH at MATH is MATH . A similar formula holds for MATH. The branches of MATH and MATH are smooth and intersect the line at infinity MATH transversally. Moreover the intersection multiplicities for the branches of MATH are independent of the coefficients MATH, of MATH, and of the MATH: let MATH then MATH; for MATH, MATH (see how to calculate intersection multiplicities below), so MATH and MATH have equivalent NAME expansions and the same intersection multiplicities. The following lemma allows us to calculate intersection multiplicities; the first point is well-known (see CITE or CITE), the second point is a consequence of the first. Let MATH be irreducible plane curve germs at MATH. CASE: Let MATH be the local links of MATH and MATH. Then the intersection multiplicity verify MATH with MATH is the linking number, MATH is the valuation and MATH is a NAME parameterization for the curve MATH (which is supposed not to contain MATH). CASE: Let MATH be complex numbers with MATH and MATH. Then MATH . For the second point at infinity MATH the local equation of MATH is MATH . All the branches intersect transversally the line at infinity, and the topology of each branch is given by one of the NAME expansions MATH, MATH and MATH and is independent of MATH, of MATH and of the MATH. Moreover intersection multiplicities are also independent of the coefficients: let MATH then MATH, MATH, MATH, and for MATH, MATH. As a conclusion MATH, MATH and MATH, MATH have the same branches and the branches have the same tangency, so MATH and MATH are topologically equivalent. Finally, we shall prove that the list is non-redundant. As before, we detail the calculus for the polynomial MATH with MATH; for the other polynomials the method is the same. Let suppose that another polynomial, MATH, of the topological list verify MATH. Then MATH has the same type as MATH, that is to say that MATH with MATH. As MATH, the localizations MATH and MATH (respectively, MATH and MATH) at MATH (respectively, MATH) have equivalent NAME expansions and the same intersection multiplicities. We deduce from the calculus of intersection multiplicities at MATH that MATH and at MATH that MATH and MATH. It implies that MATH, MATH, MATH and then MATH. |
math/0009007 | Using REF, we obtain that the category of MATH-modules supported at MATH, is equivalent to the category of continuous modules over MATH, on which MATH acts as the identity. However, these are the same as modules over MATH. |
math/0009007 | Consider the NAME MATH. We claim that it has a natural NAME* algebra structure, whose components are as follows: MATH . Consider the universal enveloping chiral algebra MATH. It contains as a chiral subalgebra MATH, where MATH is regarded as a NAME* algebra with a trivial bracket. However, since MATH is already a chiral algebra, we have a homomorphism MATH . Let MATH denote the ideal inside MATH generated by the kernel of this map. We set MATH and we claim that it satisfies all the requirements. We define the filtration on MATH by declaring that MATH is the image under the chiral bracket of MATH, where MATH is the corresponding term of the filtration of the universal enveloping chiral algebra, compare REF. From this definition it is easy to see that the embedding MATH induces an isomorphism MATH. REF of CADO holds in view of REF , since MATH is commutative. The chiral bracket induces a map MATH . When we compose it with the projection MATH, we obtain a map, which factors as MATH . However, MATH, hence we obtain a map MATH, which is easily seen to satisfy the three conditions of point REF. The map MATH is a surjection, as follows from REF. Since the LHS is MATH-flat, to prove that this map is an isomorphism, it is enough to do so on the level of fibers. Let MATH denote the ring of functions on the group-scheme MATH. It follows from REF that MATH where MATH is a MATH-module via the action by left-invariant vector fields. (In the above formula the induction is MATH, where MATH is the quotient of the universal enveloping algebra by the standard relation that MATH equals the identity.) The filtration induced on MATH coincides with the standard filtration on the induced module. Therefore, as in REF, MATH . Therefore, the map MATH is an isomorphism. |
math/0009007 | It suffices to show that MATH is supported scheme-theoretically on the diagonal and verify that its NAME projection under MATH equals MATH. The latter fact is obvious from the anti-symmetry property of the chiral bracket. Let us multiply MATH by a function on MATH of the form MATH. We obtain MATH . However, MATH, which implies our assertion. |
math/0009007 | Let us multiply both sides by a function on MATH of the form MATH. In both cases we get MATH. Therefore, the expression MATH is killed by the equation of the diagonal. However, since MATH we obtain that REF vanishes identically. |
math/0009007 | For MATH, MATH and the NAME* bracket on it comes from the usual NAME algebra structure on MATH. The map MATH comes from the action map MATH. To prove the theorem, it suffices to construct a NAME* algebra map MATH such that the induced map from MATH to MATH is MATH. let MATH and let MATH. Then the image of MATH in MATH is the value at MATH of the element MATH . The required map MATH is defined by a formula similar to the one of REF. Namely it sends a section MATH to the MATH projection of MATH . The fact that MATH is MATH follows immediately from REF. To show that it commutes with the NAME* algebra bracket, we proceed as in the proof of REF. Namely, by the same argument as in REF, it suffices to show that the difference MATH is killed by a function of the form MATH. By applying the (super-) NAME identity, we obtain that the second term in REF is the projection under MATH of the sum over MATH and MATH of four terms MATH . Let us analyze the above expression term-by-term. First of all, it is easy to see that the second and the third terms vanish. Secondly, the first term is killed by multiplication by MATH. The fourth term yields that: MATH . However, MATH, which is what we had to prove. |
math/0009007 | We will use the following statement, valid for an arbitrary locally free NAME* algebra: Let MATH be a module over MATH, which is induced from a MATH-module, that is, MATH where induction is understood in the restricted sense, that is, MATH acts as identity. Under the above circumstances there is a canonical isomorphism MATH . We will apply this lemma in the case MATH. It is easy to see, as in REF, that the embedding MATH induces an isomorphism of MATH-modules MATH . Moreover, the restriction of the MATH-action on the LHS to MATH, that comes from the embedding MATH coincides with the action that comes from the natural MATH-action on MATH by left-invariant vector fields. This follows from the fact that the embeddings MATH and MATH *-commute with one another. Hence, we obtain an isomorphism MATH and it respects the MATH-action on both sides. Since the kernel of the evaluation map MATH is pro-unipotent, we have an isomorphism of MATH-modules MATH where the MATH-action on the Right-hand side is trivial. To prove the proposition, it remains to show that the MATH-action on MATH is trivial as well. Consider the group MATH of automorphisms of the formal disc. This group acts on the whole picture. In particular, our homomorphism MATH is MATH-equivariant. However, it is easy to see that any such homomorphism, which is, moreover, trivial on MATH is zero. |
math/0009007 | By construction, MATH is a quotient of the universal enveloping chiral algebra of MATH. Therefore, MATH, if it exists, is uniquely described by a NAME* algebra map MATH . On the second and the third summands MATH is determined by REF , respectively. The restriction of MATH to MATH is the identity map onto MATH in the target and the restriction to MATH is determined by the following condition: The adjoint action gives rise to a map MATH. From it we obtain a NAME map MATH . (On the level of fibers, this map is the co-action MATH.) We need that the diagram MATH commutes. The fact that the resulting map of NAME commutes with the NAME* bracket is straightforward. Thus, one obtains a map of chiral algebras MATH and one easily checks that it factors through the natural surjection MATH . The fact that REF is satisfied follows from the formula for the embedding MATH, compare REF. The fact that MATH is an isomorphism is easy, since one can construct its inverse by a similar procedure. |
math/0009007 | First, let MATH be a commutative chiral algebra. Consider the projective limit MATH, where MATH runs over the set of all chiral subalgebras MATH with MATH. Then MATH is a commutative algebra, which carries a natural topology. Moreover, a structure of a chiral MATH-module on a vector space MATH it amounts to a continuous MATH-action on MATH. When MATH is of the form MATH, for a q.c. sheaf of MATH-algebras MATH, MATH represents the functor on the category of MATH-algebras given by MATH. In particular, MATH. Let us suppose now that MATH is a NAME* module over a NAME* algebra MATH. Then we obtain a natural continuous action of MATH on MATH. It is easy to see that in the above example of MATH and the action MATH of REF, the resulting MATH-action on MATH coincides with the natural action by left-invariant vector fields. That said, the assertion of the proposition becomes a direct corollary of the construction of MATH as a quotient of the universal enveloping algebra of MATH combined with REF. |
math/0009007 | Let MATH be the NAME* algebra map corresponding to the embedding MATH by means of right-invariant vector fields. Let is consider the corresponding NAME* bracket MATH . As in the proof of REF above, we need to show that if we compose the above NAME* bracket with the natural surjection MATH, we obtain a commutative diagram MATH . But this follows from the construction of the map MATH and REF . |
math/0009007 | Let us view MATH as a functor from MATH to the category of MATH-modules supported at MATH. The difference of the two actions of MATH for a given MATH is a map MATH which commutes with both the left MATH- and the MATH-actions on MATH. In other words, MATH is a map from MATH to the endomorphism ring of the functor MATH. Our goal is to prove that MATH. Let us first consider the case when MATH is the MATH-function NAME MATH on MATH, that is, the direct image of MATH under MATH corresponding to the coset of MATH. It is easy to see that the corresponding MATH-module identifies with MATH with the natural MATH- and MATH-actions, compare REF. In particular, it is generated by a canonical element MATH. Therefore, the map MATH vanishes for MATH. Indeed it is enough to show that MATH, but this follows immediately from the construction. Observe now that the action of MATH by right translations induces endo-functors of MATH and MATH-mod, which commute with the functor MATH in the natural sense. By construction, for every MATH, MATH commutes with this MATH-action. In particular, since every MATH for MATH, can be obtained from MATH as a MATH-translate, we conclude that MATH vanishes for all MATH. The above observations imply what we need: First, every subscheme of MATH admits a NAME cover over which the morphism MATH admits a section. Thus, let MATH be a locally closed subscheme of MATH and MATH be a map MATH. Then, any NAME MATH on MATH is isomorphic as a MATH-module to MATH. In particular, MATH defines a map from MATH to the endomorphism ring of the the forgetful functor MATH, and it is known that MATH. However, since MATH ``kills" all the NAME of the form MATH, the map MATH vanishes identically. |
math/0009009 | Since MATH is NAME, for every MATH there is an open set MATH such that its closure MATH does not contain MATH. By NAME property for separable metric space MATH, there is a countable subcover MATH of MATH. A compact NAME space MATH is normal. So there are continuous functions MATH such that MATH and MATH. To end the proof take MATH. |
math/0009009 | Let MATH be defined by REF . Thus MATH which implies MATH. To end the proof we need therefore to establish the converse inequality. Fix a bounded continuous function MATH and MATH. Let MATH. Clearly MATH. By REF again, for every MATH, there is MATH such that MATH. Therefore MATH . This means that the sets MATH form an open covering of MATH. Using compactness of MATH, we choose a finite covering MATH. Then, writing MATH we have MATH for all MATH. Using REF , and REF we have MATH . Since REF implies MATH this shows that MATH. Therefore MATH, proving REF . |
math/0009009 | Let MATH be the NAME compactification of MATH. Since the inclusion MATH is continuous, we define MATH by MATH. It is clear that MATH is a maximal NAME Functional, so by REF there is MATH such that MATH. Using MATH-continuity REF it is easy to check that MATH for all MATH. Indeed, given MATH by REF there are MATH such that MATH on MATH, but MATH. Then from REF we get MATH. Since MATH is arbitrary, MATH. This shows that MATH for all MATH. It remains to observe that since MATH is a NAME compactification, every function MATH is a restriction to MATH of some MATH, see CITE. Therefore REF holds true for all MATH. To prove that the rate function is tight, suppose that there is MATH such that MATH is not compact. Then there is MATH and a sequence MATH such that MATH for all MATH. Since Polish spaces have NAME property, there is a countable number of open balls of radius MATH which cover MATH. For MATH, denote by MATH one of the balls that contain MATH, and let MATH be a bounded continuous function such that MATH and MATH on the complement of MATH. Then MATH pointwise. On the other hand REF implies MATH, contradicting REF . |
math/0009009 | Let MATH . Clearly, MATH. By positivity REF this implies MATH. |
math/0009009 | Let MATH be defined by MATH and fix MATH. Recall that throughout this proof we assume MATH. By the definition of MATH, we need to show that MATH where the supremum is taken over all MATH and the infimum is taken over all MATH. Moreover, since REF implies that MATH for all MATH, therefore MATH. Hence to prove REF , it remains to show that there is MATH such that MATH (also, for this MATH, the supremum in REF will be attained) To find MATH, consider the following sets. Let MATH and let MATH be a set of all finite convex combinations of functions MATH of the form MATH, where MATH. It is easily seen from the definitions that MATH and MATH are convex; also MATH is non-empty since MATH, and open since MATH is closed. Furthermore, MATH and MATH are disjoint. Indeed, take arbitrary MATH . Then MATH where the first inequality follows from the convexity of MATH and the second one follows from REF applied to MATH and MATH. Therefore MATH and MATH can be separated, i. e. there is a non-zero linear functional MATH such that for some MATH see e. CASE: CITE Claim: MATH is non-negative. Indeed, it is easily seen that MATH belongs to MATH, and, as a limit of MATH as MATH, MATH is also in the closure of MATH. Therefore by REF we have MATH. To end the proof take arbitrary MATH with MATH. Then MATH and by REF MATH . This ends the proof of the claim. Without loosing generality, we may assume MATH; then it is well known, see e. CASE: CITE, that MATH for some MATH; for regularity of MATH consult CITE. It remains to check that MATH satisfies REF . To this end observe that since MATH, by REF we have MATH for all MATH. This ends the proof of REF . |
math/0009009 | REF gives the variational representation REF with the supremum taken over a too large set. To end the proof we will show that MATH on measures MATH that fail to be countably-additive. Suppose that MATH is additive but not countably additive. Then NAME theorem implies that there is MATH and a sequence MATH of bounded continuous functions such that MATH for all MATH. By REF and MATH-continuity MATH. Since MATH is arbitrary, therefore MATH for all MATH that are additive but not countably-additive. Thus REF implies REF . |
math/0009010 | It is not difficult to see that the order of vanishing of the NAME form along MATH is independent of the choice of basis MATH of the CR vector fields, and characteristic form MATH near MATH. Thus, it suffices to prove REF using the special choices introduced above. Since MATH is assumed to be of MATH-infinite type along MATH, we may write MATH where MATH is a homogeneous polynomial of some degree MATH with MATH. In particular, the MATH matrix MATH is not identically REF. A straightforward calculation shows that MATH which completes the proof of REF . |
math/0009010 | The conclusion follows immediately from REF . |
math/0009010 | By the NAME - NAME propagation theorem (see CITE), it suffices to show that MATH extends holomorphically to a full neighborhood of some particular point MATH. We shall first choose a point MATH so that REF is applicable. First, since MATH is of MATH-infinite type along MATH, MATH is in fact of MATH-infinite type MATH outside a proper real-analytic variety of MATH in view of REF . Also, since MATH is finite and MATH, MATH is a connected complex hypersurface contained in MATH. By assumption, MATH is MATH-essential at some point of MATH, for some integer MATH. (It is not difficult to see that, in fact, MATH has to be one.) It follows, exactly as in the finite type case (see for example . REF; compare also the arguments in CITE) that MATH is MATH-infinite MATH-nondegenerate outside a proper real-analytic subvariety of MATH. Since MATH is finite, we can find MATH such that MATH is a local biholomorphism, MATH is MATH-infinite MATH-nondegenerate at MATH, and, in addition, MATH is of MATH-infinite type MATH at MATH. Hence, we may apply REF to the mapping MATH at MATH with MATH and MATH. Let us choose local coordinates MATH, vanishing at MATH, on MATH and MATH, vanishing at MATH, on MATH as described in REF . We shall denote the components MATH by MATH in order not to confuse them with the components of the mapping into MATH. We shall write, for each MATH, each multi-index MATH and each non-negative integer MATH, MATH and also MATH for the vector MATH, where MATH and MATH. Hence, by REF , we have MATH for each MATH and each multi-index MATH and non-negative integer MATH such that MATH. If we add the contact equations MATH for MATH, then we obtain the system MATH where MATH is a real-analytic vector valued function of MATH and MATH, depending only on MATH near MATH and MATH near MATH (and the possibly the value of MATH). Let us fix MATH near MATH and consider REF as a system of ordinary differential equations for MATH. This system has a singularity of so-called NAME - Bouquet type at MATH, and its properties are well understood (see for example, REF; or CITE and further references in these books). We shall use the following result, which is undoubtedly known. However, the author has not found a satisfactory reference for it and, hence, we shall provide a proof. Let MATH, with MATH and MATH, be a MATH-smooth MATH-valued function near MATH such that MATH and MATH where the MATH are analytic functions near MATH. Then, MATH is real-analytic near MATH. A classical result due to NAME (CITE; compare also REF) states that the general solution MATH of REF , with MATH and MATH, is given for MATH by a convergent series of the form MATH where the MATH are integers, and MATH. A similar convergent series represents MATH for MATH. If we assume that the solution MATH is MATH-smooth near MATH, then no fractional powers or logarithmic terms can appear and the series for MATH must match that for MATH. We conclude that MATH is given by a convergent power series in MATH and, hence, MATH is real-analytic near MATH. This completes the proof of REF in the special case MATH. (A similar result for arbitrary MATH is implicit in the literature, but the author has been unable to find an explicit reference for such a result.) To treat the case of a general MATH, we shall use a result by NAME (building on an idea of NAME) reducing the system REF to a special form. After an invertible linear transformation of the MATH if necessary, we may assume that MATH is of the form MATH where MATH, MATH is a MATH-matrix in NAME normal form, and MATH as usual denotes terms of at least order two in all the variables. We shall denote the eigenvalues of MATH, repeated with multiplicity, by MATH. The solution curve MATH satisfies the differential system MATH . The characteristic roots of this system are MATH. By REF CITE there are an integer MATH (determined by the location of the roots MATH in the complex plane), with MATH, and analytic functions MATH, MATH, MATH, MATH, with MATH and MATH such that MATH and the system REF , provided that MATH, pulled back to the MATH-space, becomes MATH where MATH and MATH are polynomials which do not involve the variables MATH. (If MATH (which corresponds to all eigenvalues MATH being located on the negative real line MATH), then there are no variables MATH in REF . This already implies that the MATH, MATH are real-analytic. Thus, in what follows we may assume that MATH.) Observe that the MATH first equations in REF can be solved for MATH, MATH. Hence, by substituting the smooth functions MATH in this identity, we conclude that MATH is a MATH-smooth MATH-valued function that satisfies the system MATH where MATH are as in REF . By applying REF cited above to the equation for MATH (which is REF type for MATH), we conclude that MATH is real-analytic near MATH. By substituting the real-analytic function MATH into the equation for MATH, using the "triangular" form of REF , and again applying NAME 's theorem, we conclude that MATH is real-analytic. By repeating this procedure inductively, we conclude that all the MATH are real-analytic near MATH. The real-analyticity of MATH now follows from REF . This completes the proof of REF . By applying REF for fixed MATH (possibly after subtracting the value MATH from MATH), we conclude that MATH is given by a convergent series MATH . Moreover, the smoothness of MATH also implies that the coefficients MATH are smooth functions of MATH. A simple NAME category argument shows that there are MATH near MATH, MATH, and MATH such that the series REF converge uniformly in MATH, with MATH, for MATH. It is well known (see for example, REF ) that this implies that the CR mapping MATH extends holomorphically to a full neighborhood of MATH in MATH. This completes the proof of REF , in view of the remark at the beginning of the proof. |
math/0009010 | The proof of REF will follow from REF if we can show that MATH is necessarily weakly essential at some point of MATH. Exactly as in the finite type case, one can easily show that in MATH being MATH-essential, for some integer MATH, at some point is equivalent to being of MATH-infinite type MATH for some integer MATH (compare for example, CITE). Also, as mentioned in the proof of REF above, a real-analytic hypersurface MATH is of MATH-infinite type MATH on a dense set of MATH if it is of MATH-infinite type along MATH. Hence, to prove REF it suffices to show that MATH is of MATH-infinite type along MATH for some integer MATH; that is, we must show that MATH is not NAME flat (MATH). But this follows easily from the fact that MATH is not NAME flat REF (see for example, REF ). This completes the proof of REF . |
math/0009011 | Let MATH be an embedding of MATH into an algebraic closure MATH of MATH containing MATH. It is enough to show that for each MATH, MATH. However, MATH and therefore MATH. But MATH, since MATH is NAME, so the lemma follows. |
math/0009011 | First we discuss the possible NAME of NAME MATH with MATH. If MATH, MATH is quadratically closed, so MATH and MATH. If MATH, then MATH or MATH; by CITE if MATH, then MATH is euclidean, and in particular MATH is not a square in MATH, so this case is impossible for a NAME of level MATH. Thus we see that if MATH is a NAME of level MATH and MATH, then MATH, accordingly. Now let us suppose that MATH, where MATH. Then from CITE we have that the NAME of MATH is MATH. So we must prove that when MATH, MATH. As we noted in the first paragraph of this proof, if MATH, this is trivial, so (although it is not logically necessary) we can eliminate the need to keep track of trivial cases by assuming that MATH. To show that MATH we will show show for each MATH that MATH is NAME, so that MATH, and hence MATH. Our method of demonstrating that MATH is NAME will be to prove something a bit stronger, namely that for each MATH there is a MATH such that MATH in MATH. From this it follows that MATH, which is a NAME extension of MATH since MATH. So, let us take MATH and seek an element MATH such that MATH in MATH. Note first that MATH for some elements MATH of MATH, and that we may take MATH to be linearly independent in MATH. Writing MATH and using the proof of CITE, it follows that we may complete our partial basis of MATH to a basis MATH of MATH with the property that MATH is a basis of MATH. (Here MATH is an ordinal number, used as an indexing set, and MATH.) Then in MATH we have MATH where MATH and all but finitely many terms in the product are MATH, so that MATH in MATH, and hence also in MATH since MATH. This completes the proof that MATH for NAME of level MATH. |
math/0009011 | We follow the same plan: we show that for each MATH there is a MATH such that MATH in MATH. It follows that for any MATH that MATH is NAME, so that MATH. The key point is again that there is a basis of MATH of the form MATH such that the set MATH is a basis of MATH. The failure to list MATH as a basis element of MATH is not a mistake. If MATH is a field of characteristic not equal to REF, then MATH and MATH, so that MATH, that is, MATH is a square in MATH. The proof that MATH is then completed my mimicking the details at the end of the proof of REF . |
math/0009011 | As in the proof of REF , it is enough to show that for each MATH that MATH is NAME. Since in the case at hand, MATH is a NAME of level MATH, by CITE, we have a basis for MATH of the form MATH, where MATH and such that MATH is a basis of MATH. We claim that in order to show that for every MATH, MATH is NAME, it suffices to prove this fact for MATH in MATH. Suppose MATH, where MATH. To show that MATH is NAME, by REF it is enough to show that MATH is a square in MATH for each MATH. But if this is true for each MATH, the calculation MATH shows that it is also true for MATH. So we must only show that for each MATH that MATH is NAME. If MATH, this follows from the fact that MATH, while if we take MATH, we may use REF , and note that MATH is either MATH, which is obviously a square, or MATH, which is an element of MATH and therefore a square in MATH. |
math/0009011 | By NAME 's theorem, the inflation map from MATH to MATH is surjective, so MATH, but this is just the kernel of MATH. |
math/0009011 | This follows by considering the following commutative diagram and applying the previous lemma. MATH . |
math/0009011 | We have MATH and MATH. Since MATH and MATH, the result follows from considering the long exact sequence in cohomology associated to the short exact sequence of coefficients MATH, since MATH is just the map MATH. |
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