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math/0007062 | A presentation of MATH is MATH . By conjugating the last relation by MATH, we may assume the set of relators is MATH and the MATH with MATH. The latter are precisely the relators obtained from MATH by applying the relation MATH. It follows from CITE that MATH is not finitely presented. NAME better, REF gives MATH. |
math/0007062 | Rephrasing of CITE. |
math/0007062 | Rephrasing of CITE. |
math/0007062 | We follow REF . Consider first the group MATH. Clearly, MATH. Using the computer algebra program Gap, we compute a presentation for MATH, and rewrite its relators as words in MATH, where MATH is a generating set for MATH. We also construct a group homomorphism MATH. Then REF gives a finite MATH-presentation for MATH wi... |
math/0007062 | We start by computing a MATH-presentation for MATH. As above, MATH contains MATH; but MATH and neither MATH nor MATH are branch. First we chose generators of MATH: MATH . Then, we chose generators of MATH: MATH . The fact that MATH is normal can be seen in the following conjugation relations: MATH . Define now the grou... |
math/0007062 | We follow REF . Consider first the group MATH. Clearly, MATH. Using the computer algebra program Gap, we compute a presentation for MATH, and rewrite its relators as words in MATH, where MATH is a generating set for MATH. We also construct a group homomorphism MATH. Then REF gives a finite MATH-presentation for MATH wi... |
math/0007064 | Approach: In the rational homology sphere case, we have the following crossing change formula MATH for the NAME invariant from Subsection REF: MATH . Recall that for rational homology spheres, MATH. Using the fact that MATH-signMATHdet-MATH, we may multiply the above equation MATH by MATH to yield the crossing change f... |
math/0007064 | This result is similar to but slightly different from several standard surgery modifications presented in the literature. A proof is presented here for completeness. Verifying this lemma is merely a matter of carefully tracing through surgery definitions. Let MATH. Consider a ``bulging neighborhood" of the zero-framed ... |
math/0007064 | Start with the presentation on the left. Apply a NAME move of the second kind, sliding the component with MATH framing over the component with MATH framing. This unlinks the MATH component from the MATH component, but links the MATH component with the MATH component. Because the framing of the central component is MATH... |
math/0007064 | First write out the surgery coefficients explicitly so that we may see them clearly. MATH . Then insert a trivial component. MATH . Perform MATH left-handed twists to separate the link. MATH . Next apply REF to the central component, changing the framing of the central component to zero and introducing a new component.... |
math/0007068 | Choose a regular cardinal MATH which is large enough so that the following are true: CASE: MATH is dense in MATH, CASE: MATH-filtered colimits of weak equivalences in MATH are again weak equivalences, CASE: MATH has a cofibrant replacement functor MATH which preserves MATH-filtered colimits, CASE: MATH has a fibrant re... |
math/0007068 | Let MATH be the category guaranteed by the above proposition. The model category MATH is left proper and combinatorial, so by REF we can find a set of maps MATH in MATH which become weak equivalences in MATH, and such that MATH is a NAME equivalence. |
math/0007068 | If MATH is a combinatorial model category then the Theorem gives us a NAME equivalence MATH for some MATH and MATH. The point is that the universal model category MATH is simplicial and left proper, and these properties are inherited by the localization MATH. We must work a little harder to show that MATH is NAME equiv... |
math/0007068 | Consider the NAME embedding MATH, together with its canonical cosimplicial resolution induced by the simplicial structure on MATH. In CITE we showed that for any MATH the natural map MATH gives a cofibrant-approximation to MATH. We aim to apply this in the case where MATH is MATH. It's easy to see using adjointness tha... |
math/0007068 | REF follows directly from the fact that MATH is a NAME pair: the weak equivalence between fibrant objects MATH yields a weak equivalence MATH, and therefore the map MATH is also a weak equivalence. For REF recall that any two cosimplicial resolutions of MATH can be connected by a zig-zag of weak equivalences. So it suf... |
math/0007068 | In light of the above Proposition, this is just a restatement of the definitions. |
math/0007068 | Postponed until REF. |
math/0007068 | The proof of MATH is very easy: Since the underlying category of MATH is locally presentable, for sufficiently large regular cardinals MATH the maps MATH are isomorphisms for all MATH. On the other hand the indexing categories MATH are MATH-filtered, and in combinatorial model categories one has that MATH-filtered coli... |
math/0007068 | The objects of MATH are pairs MATH where MATH is an object of MATH and MATH is some map in MATH. Since MATH is simplicial, this map has an adjoint MATH. In this way we see that the category MATH is isomorphic to MATH. The map in which we are interested is isomorphic to the map MATH induced by MATH. Now by our choice of... |
math/0007068 | By REF we must show that for any fibrant MATH in MATH, the natural map MATH is a weak equivalence. Consider the diagram MATH . The above lemma, together with REF , shows that the horizontal map is a weak equivalence. The diagonal map is a weak equivalence by our choice of MATH REF . Therefore the vertical map is also a... |
math/0007068 | Let MATH denote the subcategory of MATH consisting of the objects MATH for which MATH is in MATH for all MATH (unlike for MATH, we are not requiring MATH to be a cosimplicial resolution). It is possible to find a functor MATH with the following properties: CASE: Each MATH is a Reedy cofibrant object contained in MATH; ... |
math/0007068 | Let MATH denote the map of simplicial sets which includes MATH as the last vertex of MATH. For any cosimplicial object MATH there is a corresponding map MATH; from this we can define a functor MATH sending the object MATH to MATH. Let MATH denote the functor sending MATH to MATH. The map we are concerned with in the st... |
math/0007068 | Again, by REF we must show that for any fibrant object MATH in MATH the natural map MATH is a weak equivalence. Consider the commutative diagram MATH . REF says that MATH is a weak equivalence. REF , together with REF , implies the same about MATH. Finally, our assumption on MATH guarantees that MATH is a weak equivale... |
math/0007068 | The usual factorizations provided by the small object argument will have the required properties, as long as we use the transfinite version of the small object argument for a sufficiently large ordinal. See CITE. |
math/0007068 | Pick a regular cardinal MATH large enough to satisfy the previous proposition, and also large enough so that MATH is dense in MATH (using locally presentability). The category MATH is small, so applying our given factorizations to maps MATH in MATH only produces a set of new objects. Therefore there exists a regular ca... |
math/0007068 | Let MATH be a regular cardinal large enough so that there are functorial factorizations preserving MATH-filtered colimits, and so that the model category has a set of generating cofibrations whose domains and codomains are MATH-small. Let MATH be a MATH-filtered indexing category, and let MATH be two diagrams. We suppo... |
math/0007068 | The point is that the homotopy theory in MATH all comes from simplicial sets: the weak equivalences are the objectwise weak equivalences, and the simplicial structure is the objectwise structure. So the lemma is immediately reduced to the corresponding fact for simplicial sets, which is well-known. |
math/0007068 | Consider the diagram MATH given by MATH. The geometric realization MATH is isomorphic to MATH (using the definition of MATH, together with the fact that for bisimplicial sets the realization is isomorphic to the diagonal). Our above discussion therefore gives a commutative triangle MATH where the horizontal arrow is th... |
math/0007068 | Our assumption is that the maps MATH are weak equivalences, for every MATH. But note that MATH is precisely MATH (once again, MATH may be identified with the simplicial replacement of the canonical diagram MATH). The above maps are weakly equivalent to the iterated degeneracies MATH in the simplicial object MATH. From ... |
math/0007068 | Consider the triangle MATH which commutes up to homotopy. The slanted map is the identity, and the vertical map is a weak equivalence because of REF on MATH (which says that MATH is an objectwise weak equivalence). So it follows that the composite across the top row MATH is a weak equivalence as well. Likewise, in the ... |
math/0007069 | Let MATH. We choose a basis MATH of MATH and define the linear map MATH by MATH. Then one obtains a complex MATH whose dual is the head of the NAME - Rim complex resolving MATH. It follows that MATH, and obviously MATH. Since MATH, one has MATH. This shows REF . We quote some well-known facts about the homology of MATH... |
math/0007069 | The implication MATH is an easy exercise. (See also the considerations at the end of this note.) For the converse observe that MATH for MATH and that MATH is free of rank MATH. So MATH must be cyclic. If MATH were even, then MATH by REF , and MATH. Thus MATH must be odd. In this case MATH by REF where MATH. So if MATH,... |
math/0007069 | As in REF , p. REF, we choose isomorphisms MATH, MATH, and define maps MATH, by MATH. Via the natural isomorphism MATH we regard MATH as a map MATH. One has MATH, and it is an easy exercise to show that the diagram MATH is commutative or anticommutative (see for REF , proof of REF ). Consequently the same is true for M... |
math/0007069 | Consider the diagram in REF . Since MATH is injective, MATH must be zero if MATH. Next let MATH. Then we obtain MATH, since MATH. Finally if MATH, then in addition MATH, so MATH as stated. The remaining assertions concerning the homology of REF are contained in REF . Instead of MATH we can consider the induced linear f... |
math/0007069 | Let MATH be a linear form on MATH such that MATH. Set MATH. Then REF and the last property in REF are obvious. Since MATH is torsionfree and MATH is reflexive, MATH must be reflexive. Because MATH has grade MATH, it is orientable. MATH being orientable, the orientability of MATH follows from REF . Conversely let MATH b... |
math/0007070 | REF allows us to extend MATH to an adjoint pair of categories MATH. To extend this further to MATH we must add a simplicial direction, and figure out what the realization of objects like MATH should be for MATH. This is accomplished by the theory of cosimplicial resolutions, discussed in the next section. The proof wil... |
math/0007070 | Both the lemma and proposition are proven in REF. |
math/0007070 | See REF. |
math/0007070 | This is not hard, but requires some machinery. See REF. |
math/0007070 | We have just seen that factoring a functor MATH through MATH is equivalent to giving a cosimplicial resolution on MATH. But it is a standard result in the theory of resolutions that REF any diagram MATH has a cosimplicial resolution, and REF the category MATH is contractible (both are proven in CITE). So REF is just a ... |
math/0007070 | See REF. |
math/0007070 | The point is that the model categories MATH are simplicial and left proper, and these properties are inherited by the localizations MATH. |
math/0007070 | One simply chooses a presentation MATH and then uses REF to lift this map across the NAME equivalences. |
math/0007070 | The NAME embedding MATH will extend to a map MATH (and for convenience we choose the extension induced by the standard cosimplicial resolution, using the fact that MATH is a simplicial model category). The relations we are imposing in MATH clearly hold in MATH, and so this map descends to MATH. It's easy to check that ... |
math/0007070 | We only give a sketch. The reader can also consult CITE for a similar statement. Consider the subcategory MATH whose unique object is the one-point manifold. This inclusion induces a NAME map MATH. The composition MATH is the usual realization/singular functor pair, and is therefore a NAME equivalence. So the homotopy ... |
math/0007070 | Left to the reader. |
math/0007070 | The fact that MATH is cofibrant implies that MATH is a cofibration, and so the map MATH is also a cofibration. Then MATH is a sequential colimit of cofibrations beginning with MATH, hence cofibrant. |
math/0007070 | First observe that MATH has a free degeneracy decomposition: we take MATH to be the coproduct MATH in which no map MATH is an identity map. Each MATH is a coproduct of representables, hence cofibrant. So MATH is itself cofibrant by the above corollary. We must next show that MATH is a weak equivalence in MATH - that is... |
math/0007070 | One again shows that MATH has a free degeneracy decomposition in which the MATH are coproducts of representables. This takes a little more work than for MATH, but we will leave it to the reader. The fact that MATH is cofibrant follows from REF . To see that MATH is a weak equivalence we consider the bisimplicial object... |
math/0007070 | We explained in REF why MATH was cofibrant, therefore the only thing to prove is that the natural map MATH is a weak equivalence in MATH. So we need to show that for every MATH the map MATH is a weak equivalence of simplicial sets. For brevity let MATH denote the category MATH. The object MATH is the homotopy colimit o... |
math/0007070 | Suppose we have a factorization of MATH through MATH: so we have a NAME pair MATH and a natural weak equivalence MATH. Then for each MATH we get a cosimplicial resolution of MATH by taking MATH to be MATH . This is clearly functorial in MATH, and so gives a resolution of MATH. Conversely, suppose we start with a resolu... |
math/0007070 | Recall that there exists an equivalence between maps of model categories MATH and the following data: CASE: A functor MATH whose image lies in the cofibrant objects, and CASE: A cosimplicial resolution on MATH. Giving a NAME homotopy between two maps MATH corresponds to giving a natural weak equivalence MATH and a lift... |
math/0007070 | Let MATH be the composite MATH. We will begin by lifting MATH, and this can be accomplished just by lifting MATH. Define MATH by MATH where MATH is the right-adjoint to MATH. We may extend MATH to a map MATH. CASE: The composite MATH is NAME to MATH. To see this, observe that there are natural weak equivalences MATH . ... |
math/0007074 | If we project MATH from this line MATH to MATH, we get a variety MATH of degree MATH and dimension MATH in MATH. Since MATH is non degenerate, MATH. It follows that MATH, hence MATH is of minimal degree. The rational scroll MATH is the image of a smooth rational normal scroll MATH of degree MATH and dimension MATH, by ... |
math/0007074 | Let us denote by MATH the projective bundle MATH in MATH, where MATH. The second projection MATH from the product MATH embeds MATH in MATH as a cone MATH of vertex MATH over the smooth rational normal scroll MATH. Its restriction to MATH is a desingularization of MATH of exceptional locus MATH. A simple NAME class comp... |
math/0007074 | Let MATH be a desingularization of MATH of exceptional locus MATH. We want to determine the linearly normal variety MATH of which MATH is projection. We show by induction on MATH that MATH is a variety of minimal degree MATH. To do so, we need the existence of an hyperplane section of MATH containing MATH that is desin... |
math/0007074 | The MATH-regularity of MATH is equivalent to the MATH-normality of MATH and the vanishing of MATH. Note that by degeneration of the NAME spectral sequence MATH so that we only have to show that MATH . The embedding MATH factors through an embedding MATH of MATH in MATH and the second projection MATH onto MATH, hence co... |
math/0007076 | Let MATH be a collecting of affine MATH-varieties covering MATH. For every MATH the ring MATH is noetherian and integrally closed and therefore a NAME ring. Now MATH equals the intersection of all the MATH considered as subrings of the function field MATH. Hence MATH is a NAME ring. |
math/0007076 | This is immediate, because MATH is the intersection of two NAME rings, namely MATH and MATH where MATH denotes the quotient field of MATH. |
math/0007076 | Let MATH denote the quotient field of MATH. Let MATH and MATH. Since MATH, the prime ideal MATH of MATH does not contain MATH. Hence there is an element MATH. By REF for some natural number MATH. Since MATH, this implies MATH. We will now show the converse. Let MATH be a regular function on MATH, that is, MATH, and let... |
math/0007076 | Let MATH and MATH. Then there is an element MATH such that MATH. Now MATH implies MATH. Conversely, assume MATH is an element in MATH such that MATH for all MATH. Let MATH. Then MATH is finite and MATH for all MATH. This implies MATH for MATH sufficiently large. |
math/0007076 | Let MATH be the quotient field of MATH, MATH, MATH. Choose MATH and define MATH. Now MATH is finite, and MATH being a prime ideal of height one for every MATH implies that MATH is not contained in MATH for any MATH. Hence there is an element MATH with MATH but MATH for all MATH. Then MATH for MATH sufficiently large. |
math/0007076 | Let MATH denote the discrete valuation corresponding to MATH. By MATH there is an element MATH with MATH and a number MATH such that MATH. Thus MATH defines a rational function on MATH whose poles are contained in MATH. Now assume MATH but MATH. This would imply that MATH and MATH vanishes on MATH. Since the poles of M... |
math/0007076 | The equivalence relation MATH defines a subset MATH via MATH . Then MATH is the MATH-sub variety defined by the radical of the ideal of MATH generated by MATH with MATH running through MATH. If MATH are two MATH-sub algebras of MATH, then MATH. Since MATH is noetherien it follows that for any MATH-sub algebra MATH ther... |
math/0007076 | Let MATH, this is an affine MATH-variety. The prime ideals of height one in MATH correspond to the irreducible hypersurfaces in MATH. Let MATH denote the morphism induced by MATH. Let MATH be a prime ideal of height one in MATH. We have seen above that there is an irreducible subvariety MATH such that MATH. From MATH w... |
math/0007076 | This is an immediate consequence of MATH and MATH. |
math/0007076 | Let MATH, MATH the corresponding hypersurface and MATH. We claim that MATH is not NAME in MATH. Indeed, if it were NAME, there would exist an irreducible subvariety MATH with MATH. But this implies MATH. Now let MATH be a prime ideal of height one contained in MATH (such an ideal exists, because MATH is a prime ideal a... |
math/0007076 | Let MATH, MATH, MATH and MATH be as in the theorem. Note that MATH is a finitely generated field extension of MATH, because MATH is finitely generated and MATH. By REF there is a finitely generated MATH-algebra MATH with MATH, MATH and MATH being a NAME ring. We may adjoin finitely many further elements of MATH to MATH... |
math/0007076 | Given a MATH-algebra homomorphism MATH, choose a finitely generated MATH-sub algebra MATH such that the quotient fields of MATH and MATH coincide. The restriction of MATH yields a MATH-morphism MATH from MATH to MATH and the inclusion MATH yields a birational morphism MATH. Now MATH is the desired rational map. |
math/0007076 | We apply REF with MATH and set MATH. There is an inclusion MATH. By the preceding lemma this induces a rational map MATH with MATH. Since MATH, it follows that MATH. Finally note that for every affine MATH-variety MATH and every MATH-invariant morphism MATH we obtain an inclusion MATH which implies that there exists a ... |
math/0007076 | The implication MATH follows from REF . MATH is trivial and MATH follows from the proposition below. Finally, MATH for the case of characteristic zero is implied by the well-known correspondence between locally nilpotent derivations and MATH-actions on affine varieties in characteristic zero. |
math/0007076 | Let MATH be an open embedding in a normal affine MATH-variety MATH, and let MATH. Let MATH denote the union of codimension MATH-components of MATH and choose a regular function MATH on MATH such that MATH is contained in the zero set of MATH. Then choose a regular function MATH on MATH such that MATH vanishes on MATH, ... |
math/0007076 | Let MATH and MATH. Each element in MATH is a birational self-map of MATH such that the induced field automorphism of MATH stabilizes MATH. In particular MATH. Due to REF there exists an object MATH and a MATH-morphism MATH such that MATH. Consider now an object MATH with a MATH-invariant MATH-morphism MATH. Then MATH i... |
math/0007082 | The first term in the formula is clearly determined by the others. By induction on MATH, the mixed invariants of degree MATH are therefore determined by the one-point invariants of degree MATH or less. |
math/0007082 | We apply the main theorem in the case MATH. Only the double sum appears, in this case as the single sum MATH . The first and last of the terms are: MATH and MATH . Multiply REF through by MATH, and the main theorem tells us we obtain an element of MATH when we integrate against MATH. For example, by the projection form... |
math/0007082 | The existence of MATH follows from the functoriality of NAME spaces. The two projections are just the two maps: MATH and MATH which clearly commute with the action of MATH. Over the open subset of MATH consisting of smooth curves, MATH is an isomorphism. The last parametrization is of the same component as the first, a... |
math/0007082 | Given a stable map MATH, represent MATH by a tree with vertices and edges corresponding to the nodes and components of MATH, respectively. For a MATH to map to MATH under the forgetful map, each MATH must parametrize a different curve (edge of the tree) mapping with degree MATH to MATH, and each MATH must be a node (ve... |
math/0007082 | The two sides of the formula represent the contribution of MATH to localization formulas for MATH which, by uniqueness, must coincide. |
math/0007082 | Let MATH. Then applying correspondence of residues to the map MATH of smooth NAME stacks (here we use MATH) of REF , and using the enumeration of fixed loci in REF , we get: MATH . (The computation MATH is easily made.) Now, the equivariant NAME classes MATH appearing in the denominators depend entirely on the nodes of... |
math/0007082 | It suffices by induction to prove the lemma for the case MATH. In that case, we will consider a (non-commuting!) diagram of stacks: MATH where the horizontal maps are the ``cross-ratio" maps defined as follows. The universal curve MATH over MATH maps to MATH via the forgetful map. Together with the evaluation map to MA... |
math/0007082 | We may assume that MATH is very ample. Indeed, suppose the polynomiality condition holds for the expression MATH for some MATH. Only the MATH's divisible by MATH will produce non-zero terms, because the degree of every curve (measured against MATH) is a multiple of MATH. But replacing MATH by MATH in the twisted tensor... |
math/0007082 | As in the proof of the rank one version, we may assume that MATH,,MATH are not just eventually free, but free, by replacing them with positive multiples (which can be taken to be the same multiple). The MATH define a morphism MATH and the theorem results from applying the correspondence of residues to the class MATH, w... |
math/0007082 | The only term in the main theorem involving a MATH-function of MATH variables and curves of (multi) degree MATH is MATH . The product MATH is a polynomial in MATH, expanding as MATH for some MATH depending on MATH. It follows by downward induction on the power of MATH and the main theorem that every term in the expansi... |
math/0007084 | The space MATH is the image of MATH by the algebraic map MATH where * denotes the action of MATH on MATH. It follows that MATH is irreducible and constructible. If MATH is the law of MATH, then its inverse image is algebraically isomorphic to the first projection MATH . The algebraic mapping which associates MATH to MA... |
math/0007084 | We only observe that the MATH-abelianity is given by the differential form MATH. |
math/0007084 | We analyze the elements of the set MATH which have the prescribed nilindex. For MATH it is immediate to see that the nonexistence of such an extension. For MATH the cocycles MATH defining the extension have to satisfy the relation MATH . The only ones which verify this requirement are the cocycles MATH . Moreover, if M... |
math/0007084 | The listed endomorphisms are clearly derivations of MATH. Their linear independence is also easy to verify. As MATH is naturally graded, a derivation MATH decomposes as MATH where MATH and MATH for MATH. From this decomposition it is not difficult to prove that MATH where MATH for MATH. |
math/0007084 | Let MATH be a cocycle. Then, for MATH, MATH we have MATH . The cocycle condition MATH implies the relations MATH for MATH, and MATH . Thus the cocycle MATH can be rewritten as : MATH . So the cocycles MATH constitute a generator system for MATH. Clearly they are linearly independent, so they form a basis. This proves t... |
math/0007084 | From the basis obtained in the previous proposition it follows easily that MATH. Observe in particular that most coboundaries have been previously excluded from the cohomology space MATH by REF. |
math/0007084 | Follows from the previous result, as for MATH . |
math/0007084 | We have seen that the rank of the algebra MATH is MATH for any MATH. From the previous proposition we know that MATH for MATH. Without loss of generality we can take MATH. Let MATH such that MATH . The cocycle MATH implies in particular MATH . Taking the relation MATH, we obtain MATH . This proves that the deformation ... |
math/0007084 | As the sill cocycle is itself linearly expandable, we know from previous results that the algebra MATH has rank one. If MATH, it is easy to verify that the entries of the matrix MATH of MATH satisfy the relations MATH . Thus the sill cocycle implies MATH . Now let MATH the summand of MATH that satisfies MATH. Then ther... |
math/0007084 | The only manner in which the differential form MATH is closed is that the cocycle MATH only affects the differential form MATH adding the exterior product MATH or changing the differential forms MATH by adding, respectively, the exterior products MATH and MATH. This corresponds to the indexes MATH and MATH. |
math/0007085 | By the NAME theorem it suffices to prove that MATH is an analytic subset of MATH. We will apply the elementary NAME method REF , used there in the case MATH altghough one can also use the method of blowing-ups. Define the holomorphic functions MATH where MATH and MATH. The all functions MATH vanish if and only if MATH ... |
math/0007085 | Since MATH depends only on the germs of MATH and MATH at MATH, we may assume that MATH and MATH. Consider the analytic set MATH, defined in the proof of the previous Proposition. If we denote by MATH the projection MATH, then MATH and over each point MATH we have MATH and hence MATH . Since MATH then MATH . By the same... |
math/0007085 | Consider the analytic set MATH defined in the proof of REF . We have MATH . Since this point lies in the closure of MATH then there exists an analytic curve MATH passing through MATH such that MATH. Take a parametrization MATH at MATH of one irreducible component of MATH . We have MATH . Since for any MATH, MATH and MA... |
math/0007085 | Let MATH, MATH. Since MATH is a cone then MATH. Take analytic curves MATH and MATH having parametrizations MATH and MATH at MATH, MATH such that MATH and MATH. Since MATH and MATH for sufficiently small MATH and MATH then MATH. |
math/0007085 | It suffices to prove MATH . Take MATH. We may assume that MATH. By REF there exists an analytic curve MATH having a parametrization MATH at MATH such that MATH . Since MATH and MATH then MATH . Let MATH . Since MATH then MATH-and MATH are linearly independent. Hence and from REF MATH . So, MATH . |
math/0007085 | The proof follows from REF and the fact that the mapping MATH,MATH is an analytic cover. |
math/0007085 | We may assume that the germs MATH are irreducible. It suffices to prove that MATH . Since MATH are analytic curves and MATH are irreducible at MATH we will consider two possible cases: MATH . MATH. Then by REF MATH. Hence we get REF . MATH . MATH. After a linear change of coordinates in MATH we may assume that MATH, wh... |
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