paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0009085 | The image MATH of MATH are the coefficients of the NAME series MATH . But MATH equals to the resultant of the two polynomials MATH and MATH. (This is a less known form of the resultant MATH which can be obtained by multiplying the NAME by a matrix obtained from the coefficients of the NAME series of MATH see CITE.) On ... |
math/0009085 | This is a straightforward generalization of the pullback property of the ordinary NAME dual. |
math/0009085 | The unfolding map MATH is transversal so we can apply REF . |
math/0009092 | The fact that the sets MATH, MATH and MATH are globally invariant follows immediately from the descriptions REF - REF. The étale algebra MATH corresponding to a finite MATH-set MATH may be defined as the algebra MATH where MATH is the algebra MATH and where MATH acts simultaneously on MATH and MATH. In the generic case... |
math/0009092 | By REF, we have in MATH the relations MATH which proves that the natural projection from MATH to MATH is surjective. Moreover one has the relations MATH which gives a homomorphism of MATH-modules MATH and the exact sequence of the lemma. |
math/0009092 | By REF , we have MATH . Thus it is enough to prove that if MATH is an arbitrary étale algebra over MATH corresponding to a MATH-set MATH and if MATH is a prime such that MATH is not ramified at MATH, then MATH . This well-known assertion follows from the fact that the components of MATH are in bijection with the orbits... |
math/0009092 | By a result of NAME (see CITE), MATH . REF implies that MATH . |
math/0009092 | Let us consider the set of quadruples MATH in MATH such that MATH . If MATH then MATH, MATH and MATH. Therefore, for any MATH as above, MATH and MATH. But for any triple MATH in MATH, there exists exactly one MATH verifying REF if MATH mod MATH and exactly three of them if MATH mod MATH. If MATH and MATH belongs to MAT... |
math/0009092 | To compute MATH we shall use its original definition CITE: MATH where the NAME measure on the affine hyperplane MATH is defined by the MATH-form MATH which is characterized by the relation MATH (where MATH is the linear form defined by MATH on MATH and MATH is the form corresponding to the natural NAME measure on MATH)... |
math/0009097 | We first blow up the singular point (the origin) of the threefold MATH in REF . The resulting threefold, denoted by MATH, is isomorphic to the total space of the restriction of the tautological line bundle on MATH to the surface MATH. The exceptional divisor MATH is isomorphic to the zero section of this line bundle, w... |
math/0009097 | We prove this Lemma by induction on MATH. For MATH, there is nothing to prove. We now assume that the Lemma holds for MATH. Namely, we have found coverings MATH of MATH that satisfy the required property. Following the construction, MATH is a small resolution of MATH. Using the explicit description of MATH, we see that... |
math/0009097 | In case MATH lies over MATH, then there is nothing to prove. Now assume MATH lies over MATH. Let MATH be the integer so that MATH has MATH irreducible components. Then by REF , there is an open neighborhood MATH of MATH and two morphisms MATH so that MATH is equivalent to MATH via an arrow. Thus to prove the Lemma it s... |
math/0009097 | This is straightforward and will be left to the readers. |
math/0009097 | Clearly, any MATH can be expressed as above. We now show that such expression is unique. Let MATH be the MATH-sub-module spanned by all MATH satisfying MATH. Clearly, the quotient MATH-module MATH is a free MATH-module with basis MATH. Let MATH be the quotient homomorphism of MATH-modules. Assume MATH has an expression... |
math/0009097 | Let MATH be the maximal idea and let MATH be the normal form of MATH. Since MATH is non-degenerate, the homomorphism MATH induced by MATH and MATH must be of the form (possibly after exchanging MATH and MATH) MATH for some positive integers MATH and MATH. Hence in REF , all MATH for MATH and MATH for MATH are in the ma... |
math/0009097 | We let MATH, where MATH runs through all positive integers, be indeterminants. We consider the ring MATH, where MATH and MATH mean MATH and MATH respectively. We let MATH be the MATH-adic completion of MATH. We let MATH and its inverse be MATH where MATH and MATH are elements in MATH. We consider the ideal MATH that is... |
math/0009097 | It suffices to prove the case where MATH is reduced and irreducible, which we assume now. Let MATH be induced by MATH. In case MATH is dominant over MATH, then REF implies that MATH as topological spaces and hence the Lemma holds. Now assume MATH factor through MATH. Since MATH is reduced, MATH factor though MATH and t... |
math/0009097 | The existence of such a parameterization follows immediately from the deformation of nodes CITE. This also follows from the local parameterization constructed in CITE. Now let MATH and MATH be two such parameterizations of MATH along MATH. Because MATH and MATH define the same divisor in MATH corresponding to the subsc... |
math/0009097 | Clearly, should MATH be represented by a subscheme in MATH, then set-theoretically it must be MATH. Hence it suffices to give MATH a closed scheme structure so that it represents the functor MATH. As before, we let MATH be MATH endowed with the reduced scheme structure and MATH be the formal completion of MATH along MA... |
math/0009097 | Let MATH be any integer between MATH and MATH. We let MATH be the group action given in REF . For any closed purely one-dimensional subset MATH of MATH we define the relative automorphism group MATH to be the set of pairs MATH where MATH is an automorphism leaving MATH and MATH fixed and MATH is an element in MATH such... |
math/0009097 | The proof of these two Lemmas are standard and will be omitted. |
math/0009097 | Following the definition of stacks (compare CITE), it suffices to show the following: CASE: For any scheme MATH over MATH and two stable morphisms MATH and MATH, the functor MATH which associates to any morphism MATH the set of isomorphisms in MATH between MATH and MATH is a sheaf in the étale topology; CASE: Let MATH ... |
math/0009097 | We let MATH be represented by MATH. Let MATH and MATH be the standard choices of relative ample line bundles on MATH and on MATH. It follows from the uniqueness of these line bundles that whenever MATH is a morphism and MATH is a pair of isomorphisms shown below that makes the diagram MATH commutative then MATH . Hence... |
math/0009097 | To prove that MATH is unramified over MATH, we only need to check the case where MATH is itself a closed point. Namely MATH, in which case MATH is either empty or is isomorphic to MATH. Since MATH, it is well-known that MATH is unramified over MATH if there are no vector fields MATH of MATH and vector fields MATH of MA... |
math/0009097 | Without lose of generality we can assume that MATH is represented by MATH with MATH the associated morphism so that MATH. We will prove the case where MATH does not factor through MATH. The other case is similar and will be left to readers. We let MATH and MATH be the generic and the closed point of MATH, as stated in ... |
math/0009097 | Let MATH be represented by MATH. We will prove the case where MATH does not factor though MATH and leave the other case to the readers. Since MATH does not factor through MATH, we can assume without loss of generality that MATH is represented by MATH. Then MATH is an ordinary stable morphism over MATH. By the property ... |
math/0009097 | The fact that the moduli stack MATH is separate over MATH follows from REF and that it is proper over MATH follows from REF . It remains to show that it is algebraic. Namely, MATH admits an étale cover by a scheme of finite type. We now show that it admits an étale covering by a quasi-projective scheme. Let MATH be the... |
math/0009097 | The proof is straightforward and will be omitted. |
math/0009097 | Clearly, the composite of MATH with MATH is trivial. Hence by REF MATH which is isomorphic to MATH. This proves the Corollary. |
math/0009097 | The proof is similar to that of REF and will be omitted. |
math/0009097 | The proof is straightforward and will be omitted. |
math/0009097 | The proof is parallel to that of MATH and will be omitted. |
math/0009097 | The proof is straightforward and will be omitted. |
math/0009097 | The proof that the morphism MATH is finite and étale is straightforward, and will be left to readers. We now check that the degree of MATH is as stated. Let MATH be any element. It follows from the discussion preceding to the statement of this Proposition that there is a MATH so that the decomposition of MATH along the... |
math/0009103 | The main point is that induction takes projectives to projectives. All modules in this proof are assumed to be direct sums of simple finite-dimensional MATH-modules. The category of locally finite MATH-modules has enough projectives. This may be checked as follows. In the category of semisimple locally finite MATH-modu... |
math/0009103 | Let MATH be an extension of MATH by MATH, where MATH; then we need to show that the centre of MATH acts semisimply. Let MATH be the action of MATH. Let MATH be the NAME vector field, and consider MATH. The NAME decomposition gives MATH, where MATH (respectively, MATH) is semisimple (respectively, nilpotent). Furthermor... |
math/0009103 | The condition on MATH ensures that all the NAME modules involved are MATH-typical. By the remarks above, in MATH there is a projective resolution MATH of MATH, with MATH. The MATH-module MATH is typical, therefore semisimple, so it is a direct sum of NAME modules. We can calculate MATH as the cohomology of the complex ... |
math/0009103 | Consider the cohomology MATH as MATH varies. The complex computing this cohomology is finite-dimensional, and shifting the weights by MATH does not change the dimension of the components. We can therefore view it as a complex with fixed terms with a differential that depends polynomially on MATH. By REF , MATH for gene... |
math/0009103 | Consider a generalised weight module MATH, so that MATH as a vector space (recall that, for MATH, the subspace MATH is simply MATH in this case). We have the following simple fact: if MATH, then MATH. An immediate consequence is that the MATH, defined above, are submodules. |
math/0009103 | First, define a parity MATH by MATH for some MATH; this is well-defined, by our hypothesis. It extends linearly to a function MATH. Now, suppose MATH is a generalised weight module whose support is contained in a single MATH-coset MATH. Shift the parity function to MATH by setting MATH, where MATH is fixed arbitrarily.... |
math/0009103 | We define a relation MATH on the set of highest weights, such that MATH implies the existence of a finite-dimensional indecomposable having both MATH and MATH as subquotients, and, consequently, that MATH and MATH are in the same MATH-block. Finally, we show that if MATH, that is, if MATH, then we can get from MATH to ... |
math/0009103 | Let MATH be a weight such that MATH, MATH is simple, MATH for every root MATH of MATH, and all the MATH are also simple. For example, any sufficiently dominant weight, that is, MATH, with MATH, will do. Then, by REF , there exists a nontrivial extension of MATH by MATH. Therefore, the MATH-quiver of each block contains... |
math/0009103 | Suppose MATH is NAME dense. First of all, MATH and therefore MATH by REF . Next, applying the NAME Theorem, produce a basis MATH of MATH over MATH, and write MATH as MATH, with the MATH. Applying MATH to MATH gives MATH and, since the MATH are linearly independent, each MATH and therefore MATH. This shows that MATH and... |
math/0009103 | For any MATH, we have MATH, therefore MATH acts by some scalar MATH on MATH. It therefore also acts by multiplication by MATH on the finite-dimensional quotient MATH of MATH. Furthermore, if MATH and MATH are in the same block, then MATH. However, by REF , any block MATH is already a NAME dense subset of MATH. |
math/0009104 | As usual we note that the components of the total NAME class are universal polynomials in the NAME classes MATH and we may let MATH be the universal bundle. Then as we may think of the coefficients in the universal polynomials as rational numbers we can note that the NAME classes of degree REF to MATH of MATH vanish if... |
math/0009104 | By twisting MATH by a line bundle on MATH so that it is trivial along the zero section we may assume that it is of order MATH on MATH. Denoting the class of MATH in MATH by MATH we then either have that MATH has support of codimension MATH if MATH or is zero if MATH. Indeed, from the relation MATH and the fact that MAT... |
math/0009104 | This follows directly from the structure of MATH and NAME 's prime number theorem. |
math/0009104 | For any prime MATH larger than MATH we can apply REF on the cover obtained by adding a line bundle everywhere of order MATH. Projecting down to MATH again and using that that cover has degree MATH gives MATH. We then finish by using REF (and noting that the factor MATH causes no trouble as by using several primes we se... |
math/0009104 | The proof is almost the same as that of REF only that now we consider instead the cover given by putting a full level MATH-structure on MATH. This time we therefore get that MATH, where MATH is some irrelevant positive integer. Using again REF we conclude. |
math/0009104 | The first part follows immediately because MATH is a symplectic vector bundle. As for the second part we may assume that the characteristic is MATH as the case of positive characteristic follows from the characteristic MATH case by specialisation. In that case we may further reduce to the case of the base field being t... |
math/0009104 | The MATH-adic part implies the integral cohomology part so we may pick a prime MATH different from the characteristic of the base field and look at MATH in MATH-adic cohomology. What we want to show is that if MATH is odd and MATH and MATH is the largest integer such that MATH, then MATH has order at least MATH and sim... |
math/0009104 | This follows immediately from REF . |
math/0009104 | The top NAME class of MATH is MATH. |
math/0009104 | The class MATH vanishes up to torsion on MATH; for dimension reasons MATH is then represented by a multiple of the fundamental class of MATH. |
math/0009104 | Consider (for MATH) the space MATH of products of a principally polarized abelian variety of dimension MATH and an elliptic curve. It is the image of MATH in MATH under a morphism to MATH which can be extended to a morphism MATH. Since MATH is the affine MATH-line we find a rational equivalence between the cycle class ... |
math/0009112 | The main formula we need is MATH which follows easily from the properties stated of the BGG operators. Since MATH, we have MATH this last because by the MATH assumption, MATH annihilates MATH. If MATH, then MATH annihilates MATH too, and the Right-hand side is zero. Combining that with the formula for MATH gives the fi... |
math/0009112 | We reduce to the well-studied case (see CITE) of a single flag manifold. Let MATH; since MATH it suffices to consider the map MATH. And MATH, so (omitting the MATH) we're studying the fibers of the composite MATH, as already done in CITE. |
math/0009112 | Let MATH be the set of flags in relative position MATH to MATH, MATH to MATH, and MATH to MATH where MATH are three flags in generic relative position. Then by codimension count (and the usual appeal to NAME 's transversality theorem) the set MATH is finite. However, since none of MATH care what MATH is (since by assum... |
math/0009112 | Let MATH be the set of flags in relative position MATH to MATH, MATH to MATH, and MATH to MATH where MATH are three flags in generic relative position. Then as in the previous proof, the set MATH is a union of MATH's, reflecting the ambiguity in MATH. If we change one of MATH to have a descent at MATH, each of these MA... |
math/0009112 | Since the second argument has no descents, any place MATH that MATH has a descent and MATH does not gives us an opportunity to cycle a descent from the first argument to the third, replacing MATH. This modification keeps the sum of the lengths MATH and neither causes nor breaks the condition MATH. So we can reduce to t... |
math/0009112 | Let MATH be MATH with the numbers in the MATH-th and MATH-th positions switched, decreasing the number of inversions by exactly one (and so that MATH). In particular every number in MATH physically between the MATH-th and MATH-th positions is not numerically between MATH and MATH. We want to show that MATH unless MATH,... |
math/0009115 | There are obvious isomorphisms MATH and MATH, compatible with the first NAME class map, so it suffices to consider the case MATH is a quotient stack, and we are reduced to CITE. |
math/0009115 | A finite cover by a scheme MATH exists by CITE (or in the case of primary interest - MATH a NAME stack - by CITE). We may suppose MATH normal, hence finite flat over MATH away from a closed substack MATH of MATH which is empty or of codimension MATH. This implies (compare CITE) that MATH is a quotient stack, and REF ap... |
math/0009115 | Clearly, REF implies REF , and REF implies REF . As MATH is injective, REF implies MATH lies in the image of the boundary homomorphism MATH of the NAME sequence. If MATH, then MATH is isomorphic to a MATH-quotient of the principal bundle on MATH associated to MATH. So, REF implies REF . Given a line bundle as in REF , ... |
math/0009115 | We follow the program of NAME Intersection Theory CITE: MATH is a MATH-quotient of MATH minus the zero section, so scheme approximations to MATH are complements of zero sections of line bundles over MATH: MATH for MATH sufficiently large. The result now follows from the excision sequence MATH . |
math/0009115 | We compactify MATH by setting MATH, the (complex) projectivization of the NAME sum of MATH and a trivial complex line bundle. Now we have MATH and MATH, complex line bundles over MATH, with MATH. The NAME sequence in cohomology gives MATH . Let MATH denote the tautological complex line bundle on MATH; then the projecti... |
math/0009115 | Let MATH denote the MATH twist of the tautological line bundle on MATH. The complement of the zero section in MATH is homotopy equivalent to MATH, so by standard arguments, if MATH is a complex line bundle on MATH with MATH, then the complement, over MATH, of the zero section of MATH is homotopy equivalent to MATH. Now... |
math/0009115 | We may assume MATH is reduced. Consider the components of the regular locus MATH. For each MATH, if we let MATH denote the restriction of MATH over MATH, then we claim the image of MATH contains MATH. Indeed, this follows by REF when MATH is NAME locally trivial; otherwise by REF combined with REF , MATH induces an iso... |
math/0009115 | Denote by MATH the NAME number of MATH; then MATH is strictly less than the second NAME number MATH of MATH. Let MATH be the NAME group of MATH. Without loss of generality, we may suppose the image of MATH in MATH is a primitive lattice element. Then we can write MATH where MATH has rank MATH. Recall that the hypothese... |
math/0009119 | REF follows from REF : MATH . Now we prove REF : MATH as claimed. Here we have used REF , the definitions and REF . For the proof of REF , we first observe that, if MATH, then MATH . Using that MATH is a coalgebra map, REF , we conclude that MATH . The proof of REF has no difference with the proof of the analogous stat... |
math/0009119 | We have MATH where we used REF. |
math/0009119 | It follows from results of CITE, CITE, CITE, CITE and CITE that MATH is the positive part of the so-called NAME kernel corresponding to the NAME matrix MATH. See CITE for details. The presentation by generators and relations follows from the considerations in the last paragraph of p. REF and the first paragraph of p. R... |
math/0009119 | CASE: Let us first assume that the braiding is symmetric, that is MATH for all MATH. By CITE we can assume moreover that the NAME matrix MATH is connected. From our assumptions on the orders of the MATH we then conclude that the braiding has the form MATH for all MATH where MATH is a root of unity of order MATH. See CI... |
math/0009119 | CASE: As in the proof of REF we first assume that the braiding is symmetric. If MATH, then MATH and hence the corresponding NAME relation REF says that MATH. Thus, we can easily reduce to the connected case. In such case, MATH as before and the Theorem is shown in CITE. CASE: In the general case, we change the group as... |
math/0009119 | If MATH then MATH (otherwise MATH) and MATH (otherwise MATH). If MATH then MATH (otherwise MATH) and MATH (otherwise MATH). Assume that MATH. Then MATH . Then MATH divides MATH and analogously, MATH divides MATH. So that MATH by the assumptions on the order of MATH and MATH; by symmetry, MATH. Assume that a vertex MATH... |
math/0009119 | . Clearly, MATH is an algebra map if and only if REF holds for all MATH, MATH. It follows also easily that REF holds when MATH, or MATH, or MATH and MATH. Next, let MATH and MATH be arbitrary elements; one can then check that REF holds for MATH and MATH if it holds for all the possibilities MATH and MATH; MATH and MATH... |
math/0009119 | By induction on the number of connected components. Here is the first step: REF is true if the NAME diagram corresponding to MATH is connected. Let MATH be a NAME module over MATH with MATH and pick MATH. By REF and the formulas for the biproduct, there exists a unique algebra map MATH such that MATH, MATH. Also, by RE... |
math/0009119 | CASE: Assume that MATH, MATH for some MATH. Substituting MATH and MATH in MATH and using MATH we conclude that MATH . Changing the rôles of MATH and MATH we obtain in the same way MATH . First assume that MATH. In particular, MATH and MATH or MATH. If MATH, respectively, MATH, then we get from REF, respectively, REF, t... |
math/0009119 | It is known that MATH, see for instance CITE. REF then follows from REF . To prove REF , let us assume that MATH. By REF again, MATH, for some MATH. We can choose MATH when MATH or else if MATH. That is, MATH is a linking datum for MATH, MATH and MATH; and REF hold. |
math/0009119 | Let us fix MATH. Let MATH be the algebra presented by generators MATH, MATH, MATH and relations REF; it is a NAME algebra via REF. Let MATH and let MATH be the subalgebra of MATH generated by MATH, MATH, and MATH, MATH. By REF , we know that MATH is a NAME subalgebra of MATH. Note that MATH is a graded NAME algebra wit... |
math/0009119 | By REF , there exists a linking datum MATH and a surjective NAME algebra map MATH, where MATH. But MATH by REF ; hence MATH is an isomorphism. |
math/0009119 | By CITE we can realize the braiding over a suitable finite abelian group MATH and twist with a REF-cocycle MATH such that the resulting braiding MATH is symmetric with elements of odd order, has the same diagonal elements and is of NAME type with the same NAME matrix MATH as MATH. We can now conclude from CITE that MAT... |
math/0009119 | Define MATH, MATH. In both cases we have to show MATH. We assume that MATH is not REF. Let MATH, MATH, with MATH for all MATH. Then action and coaction on MATH are given by MATH, MATH and MATH, MATH for all MATH. Hence MATH, MATH are linearly independent since MATH or MATH. (If both equalities would hold, then MATH, an... |
math/0009119 | We fix a connected component MATH. Let MATH be the NAME submodule of MATH with basis MATH, and MATH the quotient of MATH modulo the NAME relations of all elements MATH with MATH in I. Let MATH. The map MATH factorizes over MATH, since the NAME relations hold in MATH. By REF the subalgebra MATH of MATH generated by the ... |
math/0009119 | Let MATH be the dual NAME algebra of MATH in the braided sense (see for example CITE). MATH is a graded braided NAME algebra in MATH with MATH, for all MATH. By assumption there are MATH, MATH, with MATH for all MATH, and a basis MATH of MATH with MATH for all i. Let MATH in MATH be the dual basis of MATH. Then MATH wi... |
math/0009119 | Let MATH be a finite-dimensional pointed NAME algebra with coradical MATH and let MATH be the diagram of MATH. Then MATH is a NAME module of finite NAME type by CITE. Hence the claim follows from REF . |
math/0009119 | Let MATH be a finite-dimensional pointed NAME algebra with coradical MATH and diagram MATH and MATH as defined above. Since MATH and MATH are braided NAME algebras over MATH of the same dimension, and MATH, we can iterate this process and after finitely many steps we obtain a graded braided NAME algebra MATH over MATH ... |
math/0009120 | The proof of the Nodal Domain Theorem is based upon deriving a contradiction from Hypothesis W: MATH has MATH weak nodal domains, and Hypothesis S: MATH has MATH strong nodal domains, respectively. We call the domains MATH and define MATH for MATH. None of the functions MATH is identically zero. Since they have disjoin... |
math/0009123 | This is REF. |
math/0009123 | This is REF. |
math/0009123 | Every nontrivial representation of MATH in characteristic MATH has dimension MATH. Indeed, first assume that MATH. Then in the ``generic" case" of MATH such a representation must have dimension MATH, thanks to the already cited REF. If MATH then MATH and the smallest dimension is MATH, according to the Tables in CITE. ... |
math/0009124 | The first and the second properties of MATH comprise the definition of this vector field. The third property is equivalent to that MATH is a MATH-module; the fourth property is clear. Finally, the last property follows from the first (see, for example, the proof of CITE). We leave out the details of the proof for the s... |
math/0009124 | The first assertion is clear by REF . If MATH is homogeneous, the representations MATH are equivalent to each other for all MATH. Furthermore, by applying diffeomorphisms fixing MATH, we conclude from REF that the operators MATH, MATH, are conjugate to each other for all MATH and, hence, have the same eigenvalues. In p... |
math/0009124 | REF is a particular case of REF . Let us outline a direct proof, leaving entirely aside the question of homogeneity. Denote by MATH the natural projection. Fixing the direct product structure on MATH, we embed MATH into MATH as a NAME subalgebra. (Here we use the fact that the symplectic structure on MATH is independen... |
math/0009124 | The argument is similar to the proof of REF with NAME 's theorem used to introduce a local (in MATH) direct product structure on MATH. |
math/0009124 | In what follows we assume that the vector bundle MATH is trivial. The proof of the general case requires only superficial changes. In the calculations below we use the sign convention of CITE. Fix a trivialization MATH and thus identify sections of MATH with functions MATH. Let MATH be a NAME vector field on MATH. Sinc... |
math/0009124 | The argument is by ``pushing the non-linearity of MATH to infinity". Pick a trivialization of MATH so as to identify MATH with MATH as ordinary, not NAME, vector bundles. With the trivialization fixed, the structure of a NAME vector bundle on MATH is given by the MATH-valued vector field MATH on MATH satisfying the NAM... |
math/0009124 | The algebroid MATH is isomorphic to the action algebroid of the coadjoint action on MATH. The latter integrates to the action algebroid MATH. By REF , there is a one-to-one correspondence between NAME and NAME-equivariant vector bundles over MATH. As is shown in CITE (or in CITE for closed balls centered at MATH), ever... |
math/0009124 | The argument is similar to the proof of NAME 's theorem on the formal normal form of a vector field and to NAME 's proof of the formal linearization theorem for NAME structures, CITE. Fix a (formal) trivialization of MATH. We write MATH as a formal power series on MATH and assume that this formal power series contains ... |
math/0009124 | Let us set MATH and MATH. By REF, to establish REF, it suffices to show that MATH and MATH are anti-isomorphic. The required anti-isomorphism MATH is MATH where MATH is the symplectic structure on MATH. Combined with REF, this proves REF. Moreover, it is clear that the bijection REF induces a bijection on the level of ... |
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