paper stringlengths 9 16 | proof stringlengths 0 131k |
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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 MA... |
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 ... |
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 MAT... |
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... |
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 ... |
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 M... |
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... |
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... |
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 c... |
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 ... |
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, MAT... |
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... |
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 ... |
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 cur... |
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 NAM... |
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 proj... |
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... |
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 c... |
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 th... |
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. Therefor... |
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 ... |
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 conne... |
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 res... |
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, MAT... |
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 MA... |
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/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 . L... |
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 assu... |
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 ge... |
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 . ... |
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 correspo... |
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 ... |
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, ... |
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 ... |
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 inter... |
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 im... |
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... |
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... |
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 cons... |
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 compactificat... |
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 impossibl... |
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 n... |
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 poly... |
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 chira... |
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... |
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 ... |
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/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 t... |
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 continuou... |
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 diagr... |
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. ... |
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 pro... |
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 ... |
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 t... |
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 r... |
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 co... |
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 alon... |
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 o... |
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 (c... |
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... |
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 i... |
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... |
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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