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math/9912059
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Denote by MATH the MATH-functor which corresponds to MATH by NAME. Then MATH for some MATH and some MATH. And necessarily, MATH (where the notation MATH means the smallest element).
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math/9912059
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We have MATH . But MATH is a natural transformation from MATH to MATH. By REF , we get MATH .
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math/9912059
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One has MATH . Therefore if MATH for some MATH-category MATH, then MATH for MATH and we obtain a fillable MATH-shell in the sense of REF .
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math/9912059
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We are going to show the formula by induction on MATH. The case MATH is trivial. If MATH, then MATH. And if MATH, then MATH and MATH . So the labelings MATH and MATH of MATH are the same ones.
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math/9912059
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It is an immediate consequence of REF and of the uniqueness of REF .
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math/9912059
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Set MATH where MATH and set MATH. By hypothesis, the equality MATH holds and let MATH . Take any thick MATH-functor MATH from MATH to MATH. Then MATH . By REF , it is clear that MATH induces a MATH-functor.
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math/9912059
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If MATH, then MATH and by construction of MATH, MATH for any MATH and MATH is MATH-dimensional for any MATH. For MATH equal to MATH, MATH or MATH, the converse is obvious. Suppose the converse proved for MATH and let us prove it by induction for MATH. By hypothesis, as soon as MATH, then MATH is MATH-dimensional. For MATH, one has MATH and MATH therefore MATH and MATH satisfy the induction hypothesis. So MATH and MATH. Since the MATH are thin MATH-cubes for all MATH between MATH and MATH, then MATH and MATH. Therefore MATH and MATH. For MATH, one has MATH therefore an easy calculation shows that MATH.
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math/9912059
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One has MATH. Therefore MATH . So one has MATH . And MATH . Consequently one has MATH . One has MATH. Therefore MATH . So MATH . And MATH therefore MATH . One has MATH. Then MATH . So MATH . One has MATH. Therefore MATH so MATH . Now let us calculate MATH. One has MATH .
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math/9912059
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Equalities REF are obvious. Equalities from REF to REF are immediate consequences of the definitions. With REF , one sees that MATH . For a given MATH, the above equalities are equalities in the free cubical MATH-category generated by MATH. Therefore, they depend only on the relative position of the indices MATH, MATH and MATH with respect to one another. Therefore, we can replace each index MATH by MATH, each index MATH by MATH and each index MATH by MATH to obtain the required formulae. In the same way, it suffices to prove the last two formulae in lower dimension and for MATH. One has MATH and MATH .
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math/9912059
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It is obvious for MATH. We are going to make an induction on MATH. Let MATH and MATH. Then MATH . The equality MATH makes sense if MATH is a MATH-cube. And in this case, MATH is MATH-dimensional and MATH. This equality holds in the free cubical MATH-category generated by MATH, and therefore MATH . Now suppose that MATH. Then MATH .
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math/9912059
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Equalities REF are obvious. For the sequel, one can suppose MATH. In the cubical singular nerve of a MATH-category, two elements MATH and MATH of the same dimension MATH are equal if and only if MATH and for MATH and MATH, one has MATH. Now we want to prove Equality REF. Since MATH, then MATH for any MATH of dimension MATH REF is equivalent to MATH for MATH and MATH. REF implies that MATH for MATH or MATH. For MATH, proving Equality MATH is equivalent to proving it for the case MATH and to replacing each index MATH by MATH, each index MATH in by MATH and each index MATH by MATH. And in the case MATH, the equality is a calculation in the free cubical MATH-category generated by MATH. So we can suppose that MATH is of dimension as low as possible. In our case, this equality makes sense if MATH is MATH-dimensional. Therefore it suffices to verify Equality REF in dimension MATH for MATH. And one has MATH . In the same way, to prove Equality REF, it suffices to prove it for MATH and in the MATH-dimensional case. And one has MATH and MATH . In the same way, one can verify that MATH .
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math/9912059
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It is easy to see that MATH. Now we suppose that MATH. Set MATH. We are going to prove that MATH by verifying that the second member satisfies the characterization of REF . Let MATH. REF implies that for MATH, the dimension of MATH is zero (or equivalently that it belongs to the image of MATH). With REF , one gets MATH for MATH. It remains to prove that for MATH, MATH for any MATH. One has MATH .
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math/9912059
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First let us make the proof for MATH and MATH. Let us consider the following MATH-functor from MATH to MATH : MATH . Then MATH where MATH is a thin element. Therefore MATH and MATH are NAME. We claim that the above construction is sufficient to prove that MATH and MATH are NAME for any MATH and for any MATH. The labeled MATH-cube MATH is actually a certain thin MATH-dimensional element of the cubical MATH-category MATH and it corresponds to the filling of a thin MATH-shell. So MATH where MATH is a function which only uses the operators MATH, MATH, and MATH. In this particular case, MATH could be of course calculated. But it will not be always possible in the sequel to make such a calculation : this is the reason why no explicit formula is used here. And one has MATH, MATH and all other MATH-faces MATH are (necessarily) thin MATH-faces. The equalities MATH and MATH do not depend on the dimension of MATH. Therefore for any MATH and for any MATH, one gets MATH where MATH is a linear combination of thin elements. Now we want to explain that the above construction is also sufficient to prove that MATH and MATH are NAME for any MATH and any MATH and for any MATH. The equalities MATH and MATH do not depend on the absolute values MATH, MATH. But only on the relative values MATH, MATH and MATH. So let us introduce a labeled MATH-cube MATH by replacing in MATH any index MATH in by MATH, any index MATH by MATH and any index MATH by MATH. Then one gets a thin MATH-cube MATH such that MATH and MATH. If the reader does not like this proof and prefers explicit calculations, it suffices to notice that MATH by REF . Set MATH. Then MATH and one completes the proof by an easy induction on the dimension of MATH.
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math/9912059
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It suffices to make the proof for MATH and MATH. And to consider the following thin MATH-cube MATH . Notice that the above MATH-cube is exactly MATH by REF .
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math/9912059
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It suffices to make the proof for MATH and MATH. Set MATH . One has already seen that MATH . It suffices to construct a thin MATH-cube MATH such that MATH and MATH. If the MATH-cube is conventionally represented by REF , the thin <<<<<<< coin.tex labeled MATH-cube of REF with MATH meets the requirement. ======= labeled MATH-cube of REF with MATH meets the requirement. >>>>>>> REF The latter labeled MATH-cube can be defined as the unique thin MATH-cube MATH which fills the MATH-shell defined by MATH .
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math/9912059
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Let us denote by MATH the following property : ``for any MATH and with MATH, for any MATH-morphisms MATH and MATH of any MATH-category MATH such that MATH exists, there exists a thin MATH-cube MATH and a thin MATH-cube MATH which lie in the cubical singular nerve of the free globular MATH-category generated by MATH and MATH, and even in its branching nerve, such that MATH in the normalized branching complex (that is, the equality holds modulo degenerate elements of the branching simplicial nerve) and such that for any MATH-morphisms MATH and MATH, MATH." Since MATH in the normalized chain complex of the branching simplicial nerve, then MATH is a cycle in the branching homology of the free globular MATH-category MATH generated by two MATH-morphisms such that MATH. The MATH-category MATH is of length at most one and non-contracting. Therefore its branching nerve coincides with the simplicial nerve of MATH, the latter being the globular MATH-category freely generated by the composable pasting scheme whose total composition is MATH where MATH and MATH are two MATH-dimensional cells. Therefore this simplicial nerve is contractible. Consequently there exists MATH lying in the cubical singular nerve of MATH (and also in its branching nerve) such that MATH . The MATH-cube MATH is necessarily thin because there is no morphism of dimension MATH in MATH. By setting MATH, we obtain MATH. We are going to prove MATH by induction on MATH. Suppose MATH proved for MATH. Then MATH . Therefore we can set MATH and we have MATH because of the globular equations. So we get a thin MATH-cube MATH such that MATH is a cycle in the normalized chain complex associated to the branching simplicial nerve of MATH. This cycle lies in the branching nerve of the free MATH-category generated by two MATH-morphisms MATH and MATH such that MATH. This MATH-category is of length at most one and non-contracting. Therefore its branching nerve is isomorphic to the simplicial nerve of the globular MATH-category freely generated by the composable pasting scheme whose total composition is MATH where MATH and MATH are two MATH-dimensional cells. Therefore it is contractible. Therefore there exists MATH such that MATH . The cube MATH is necessarily thin because there is no morphism of dimension MATH in the cubical sub-MATH-category generated by MATH and MATH. And MATH is proved.
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math/9912059
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By definition, one has MATH. If MATH was equal to the globular sub-MATH-category generated by MATH then MATH still would exist in the cubical singular nerve. Therefore, MATH can be written as an expression using only the composition laws MATH of MATH and the variables of MATH and moreover, the variables MATH and MATH can appear at most once. By REF , MATH is therefore NAME to MATH, MATH or MATH. Now suppose that MATH. Let MATH. Then MATH, MATH and MATH. Since MATH is a thin element, then all other faces MATH are thin (this can be verified directly by easy calculations). Therefore MATH is NAME to MATH. As illustration, let us notice that for MATH and MATH, MATH is equal to MATH .
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math/9912059
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We need, only for this proof, the operator MATH introduced in CITE. One has MATH and MATH . Therefore MATH exists and is NAME to MATH by REF . If we work in the MATH-category generated by MATH and MATH, then MATH is a well-defined element of the branching simplicial nerve of MATH. And MATH is the free MATH-category generated by a composable pasting scheme whose total composition is MATH. Since union means composition in such a MATH-category, then necessarily MATH. Since MATH is the identity map on the reduced branching complex, then MATH is NAME to MATH.
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math/9912059
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Obvious.
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math/9912059
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One has in the reduced branching complex MATH and MATH therefore MATH induces a linear map from MATH to MATH. And MATH. Since MATH is the identity map on MATH, then MATH is generated by the MATH where MATH runs over MATH. Therefore the induced morphism of chain complexes is surjective.
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math/9912059
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Since MATH induces the identity map on MATH, then MATH. Therefore MATH .
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math/9912059
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One has MATH and MATH . Now suppose that MATH. Then MATH . So MATH. In the same way, one has MATH .
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math/9912059
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For all MATH, we have seen that MATH induces the identity map on the reduced branching complex. Therefore for all MATH, MATH. The latter equality can be translated into MATH . If the above equality was in MATH, the proof would be complete. Unfortunately, we are working in the reduced branching chain complex, and so there exists MATH and MATH such that, in MATH . Set MATH where MATH are thin elements of MATH. Each MATH corresponds to a thin MATH-cube in the free cubical MATH-category generated by the MATH-cube MATH which will be denoted in the same way (see the last paragraph of REF). One can suppose that each MATH is MATH-dimensional. In the free cubical MATH-category generated by MATH, either MATH is in the cubical MATH-category generated by the MATH for MATH (let us denote this fact by MATH), or MATH is in the cubical MATH-category generated by the MATH for MATH (let us denote this fact by MATH). Therefore one has MATH and MATH . Because of the freeness of MATH, one gets MATH .
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math/9912059
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Consider the case of the reduced branching homology. Let MATH. If MATH then MATH and MATH are two thin elements of MATH. Therefore MATH in the reduced branching complex of MATH. Now suppose that MATH . By hypothesis, there exists MATH such that MATH. Therefore in the reduced branching complex, one has MATH. So MATH is a boundary.
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math/9912059
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Set MATH for any MATH. It is clear that MATH in MATH. Now suppose that MATH for some MATH. We already mentioned that MATH is the free MATH-category generated by a composable pasting scheme in the proof of REF . It turns out that MATH and MATH belong to MATH and it is possible thereby to use the explicit combinatorial description of CITE. Set MATH equipped with the total order MATH. Let MATH (or MATH) be the set of all subsets of MATH of cardinality MATH. Let MATH an arbitrary subset of MATH. There is a lexicographical order on MATH usually defined as follows : if MATH with MATH and MATH with MATH, then MATH means that either MATH, or MATH and MATH, etc. If MATH, a MATH-packet is a set like MATH. If MATH with MATH, then MATH consists of the sets MATH for MATH. We have lexicographically MATH . A total order MATH on MATH will be denoted by MATH for MATH, that is MATH for MATH. A total order is called admissible by NAME and NAME if on each packet it induces either a lexicographical order or the inverse lexicographical order. The set of admissible orders of MATH is denoted by MATH (or MATH). Two total orders MATH and MATH of MATH are called elementary equivalent if they differ by an interchange of two neighbours which do not belong to a common packet. The quotient of MATH by this equivalence relation is denoted by MATH (or MATH). Suppose that for some MATH, the members of the packet MATH form a chain with respect to an admissible order MATH of MATH, that is, any element of MATH lying between two elements of MATH belongs to MATH. Define MATH the admissible order in which this chain is reversed while all the rest elements conserve their positions. Then MATH is still an admissible order and MATH passes to the quotient MATH. The lemma on page REF claims that MATH where MATH. And the poset MATH is described by the following picture : MATH . It turns out that in the picture MATH, the vertices are exactly the MATH-morphisms of MATH and the arrows are exactly the MATH-morphisms of MATH. This explicit description shows therefore that MATH is equal to a composition MATH where the only morphism of dimension MATH contained in MATH is MATH. And the same description shows that MATH is equal to a composition MATH where the only morphism of dimension MATH contained in MATH is MATH. And one has MATH . Since MATH is thin, MATH. Since MATH is a map from MATH to MATH, then MATH in MATH. Since MATH is the free MATH-category generated by the pasting scheme MATH, then for any MATH between MATH and MATH, MATH is a composition of MATH with other MATH of dimension strictly lower than MATH. Suppose that MATH is odd. There exists MATH and MATH such that MATH for some MATH. If MATH, then only one of the MATH or MATH is of dimension MATH therefore MATH or MATH. If MATH, then since MATH then in this case MATH is MATH-dimensional and MATH is of dimension strictly lower than MATH. Therefore in this case MATH. By repeating as many times as necessary the process, the number of cells MATH included in MATH decreases. And we obtain MATH . Now suppose that MATH is even. Since MATH, then necessarily at one step of the process, we have MATH. Take the last MATH such that MATH. Then MATH and MATH. Since MATH, then for this MATH, MATH is of dimension strictly lower than MATH and MATH is of dimension MATH. Therefore MATH. In the same way, MATH if MATH is odd and MATH if MATH is even. So MATH in MATH by REF . Therefore MATH induces a linear map from MATH to MATH still denoted by MATH. Take MATH. Then in MATH, one has MATH .
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math/9912059
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The assertions concerning the formal branching homology are obvious. Since the negative folding operator induces the identity on the reduced branching complex, then MATH is equal to MATH for MATH and is generated by MATH and MATH for MATH. The point is to prove that there is no relations between MATH and MATH for MATH, that is MATH. Suppose that there exists a linear combination of thin MATH-cubes MATH and a linear combination of thin MATH-cubes MATH such that for some integers MATH and MATH, MATH in MATH. Then MATH and so MATH is necessarily a linear combination of thin MATH-cube therefore MATH. Another possible proof of this proposition is to use REF and to use the homotopy equivalence of REF between MATH and MATH.
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math/9912059
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Analogous to the previous proof.
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math/9912059
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We know that MATH . And the differential maps is also completely known. In the formal branching complex, one has MATH where MATH are the MATH's appearing in the word MATH with MATH. It follows that this chain complex can be splitted depending on the position and the number of the MATH's, and that these positions and numbers are not modified by the differential maps. If the number of the MATH signs is MATH, we are reduced to calculating the simplicial homology of the MATH-simplex which is known to vanish in dimension strictly greater than MATH.
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math/9912065
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Of course, an isomorphism between two MATH-framed surfaces must be a MATH-framing of an isomorphism between two ordinary surfaces in the category of ordinary three-dimensional cobordisms, and thus the surfaces must be diffeomorphic. Suppose MATH and MATH are MATH-framed surfaces which are MATH-framings of the same surface MATH (here we have absorbed the ordinary diffeomorphism of the surfaces for notational simplicity). Viewing MATH and MATH as three-manifolds with boundary MATH we form the closed three-manifold MATH and choose some four-manifold MATH which it bounds. Then MATH is a MATH-framed three-manifold with boundary and MATH is a cobordism from MATH to MATH with the obvious MATH . Likewise glue MATH and choose a MATH-framing to get a MATH-framed cobordism MATH . Now MATH represents a MATH-framing on MATH and thus is the same MATH-framed cobordism as MATH if and only if it has the same signature (here by the signature we mean the signature of the four-manifold which forms the MATH-framed manifold underlying the MATH-framed cobordism). Now by NAME 's nonadditivity of the signature result CITE the signature of MATH is not the sum of the signatures of MATH and MATH but the difference depends only on the boundaries, so by connect summing an appropriate number of copies of MATH to MATH we get a new cobordism MATH such that MATH has the same signature as, and thus is the same morphism as, MATH . Therefore MATH and MATH are isomorphic.
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math/9912065
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The functor from MATH to MATH is just the inclusion functor. For the other direction choose for each object of MATH which is connected an isomorphism to the chosen object of the same genus (using REF ). An arbitrary object MATH of MATH is a union of connected pieces, so it is the domain of an isomorphism MATH which is a union of the chosen isomorphisms. Then the functor sends MATH to the isomorphic object in MATH and a morphism MATH to MATH . The composition of functors one way is the identity, and the other way is isomorphic to the identity via the natural isomorphism MATH .
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math/9912065
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Let MATH be the equivalence functor MATH and MATH be the functor MATH . Then of course MATH is a symmetric monoidal functor MATH and hence a TQFT. If MATH is a TQFT whose restriction to MATH is MATH and MATH is the natural isomorphism from MATH to the identity functor on MATH then MATH is an isomorphism from MATH to MATH . It is natural because MATH is natural, and thus MATH and MATH are equivalent.
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math/9912065
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If MATH define MATH . For each permutation morphism MATH define MATH to be the corresponding permutation map. Because the permutation groupoid embeds into MATH we have MATH and because of the symmetry axiom we have MATH when MATH is basic. If MATH is not basic because it has an underlying manifold which is not connected, then there exists a permutation MATH such that MATH is a union of basic morphisms. Define MATH . The choice of MATH was not unique, since we could have reordered the factors MATH but it is easy to check that MATH is well-defined, and that MATH on this extended collection of morphisms satisfies all the axioms in the statement of the theorem as well as MATH and MATH . Now if MATH define MATH and MATH where MATH sends MATH to MATH via the identity map and MATH to MATH via the obvious permutation composed with the union of a copy of the MATH map for each MATH . It is easy to check that MATH is the pairing on MATH determined by the pairings MATH on each MATH . Define MATH so that MATH and MATH and define MATH to be the element of MATH dual to the pairing. If MATH and MATH we can construct MATH which is MATH by a sequence of sewings, mendings, and permutations. It then follows from the sewing, mending and symmetry axioms that MATH where MATH is formed by tensoring the identity map with the map associated to MATH . Finally, if MATH is an arbitrary element of MATH define MATH by MATH noting that MATH define MATH . We have only to check that the conditions after the definition of a TQFT are satisfied. CASE: If MATH and MATH then MATH . To see this note MATH and thus MATH . CASE: MATH . By REF . CASE: MATH . This is by definition. CASE: MATH . Suppose that MATH and MATH so that MATH and thus MATH . CASE: MATH . This is by definition. CASE: MATH . This is also by definition.
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math/9912065
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Of course two basic MATH-framed cobordisms are the same if and only if there is they have the same signature and there is a diffeomorphism of the boundaries of the underlying four-manifold intertwining the embeddings of each MATH in CITE NAME proves that given two compact connected three-manifolds with boundary and a diffeomorphism from the boundary of one to the boundary of the other, there is a framed link in the interior of one such that surgery on that link gives a three-manifold with boundary over which the diffeomorphism extends to a diffeomorphism of the entire manifolds. Further, two such links can be connected by a sequence of moves MATH shown in REF together with their mirror images and inverses, where the (respectively) ball, two-handled torus and torus can be embedded anywhere in the manifold. We note first that in the proof of this result, move MATH is used only to make the signatures of the linking matrices of the two links equal, after which every use of it can be a replaced with a use of move MATH and thus the MATH-framed version of NAME 's result would simply drop move MATH . Of course the situation in the present theorem is a special case of NAME 's result, where one of the three-manifolds is MATH with several handlebodies removed, and the parameterizations of the boundary give the diffeomorphism. Thus since it is clear that the MATH-framed NAME moves do not change the MATH-framed three-manifold with embedded handlebodies, we need only prove that moves MATH and MATH can be replaced by a sequence of MATH-framed NAME moves. This relies on the following two observations, essentially due to CITE. The first is that, given a link and embedded handlebodies in MATH and given a choice of curves on the boundary of the handlebodies which generate the homology of the handlebody, a sequence of NAME moves can replace the embedding with one in which all these curves bound disks in MATH minus the embedded handlebodies. The second is that if MATH and MATH are two links plus embeddings in MATH which can be connected by a sequence of MATH-framed NAME moves, and if MATH is a framed link in one of the embedded handlebodies, and MATH are the result of embedding MATH in to MATH via the embedding of the handlebody and then removing that handlebody from the list, then MATH and MATH can be connected by a sequence of MATH-framed NAME moves. For the first observation, notice the embedding of the handlebodies gives an embedding of the boundary curves into MATH and as framed links they admit a projection. It is well-known that for any projection of a framed link, one can, by flipping certain of the crossings, make it a projection of a link in which all components are unlinked zero framed unknots. So apply MATH-framed NAME move I to add sufficiently many MATH framed unknot, and for each crossing that needs to be flipped, apply move II as shown in REF to flip it, noting as indicated by the dotted lines in that figure that the move is meant to be applied to the handlebodies, not just the boundary curves. The result of these moves will have the desired property. For the second observation, notice the result is manifestly true for a single MATH-framed NAME move, and thus for an arbitrary sequence. REF decomposes move MATH into a sequence of MATH-framed NAME moves assuming that the torus is embedded so that the meridian plus the longitude bounds a disk in MATH (here the vertical strands represent any number of link components which might go through the bounded disks). For an arbitrary embedding, use the two observations to show that a sequence of MATH-framed NAME moves will replace any embedding with one whose meridian plus longitude bounds a disk, apply REF , then invert the sequence of NAME moves to return to the original embedding. The two observations again reduce the general problem of move MATH to one where the two-handled torus is embedded as shown in REF , and the moves in the figure prove the result in that case.
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math/9912065
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Of course invariance under the biframed NAME moves guarantees MATH is an invariant of the cobordism. We are given that it satisfies the Symmetry and Nondegeneracy Axioms of REF , and to see it satisfies Sewing and Mending it suffices to check that the embeddings and link shown in REF (shown here only for a genus two MATH-framed surface, represented by its underlying graph) represent the sewing and mending of the basic cobordisms. This follow easily from the definition of the surgery description.
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math/9912066
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Without loss of generality, we may replace MATH with MATH. Let MATH be elements of the MATH-vector space generated by the elements MATH. Now, if MATH is a permutation of MATH, we claim that MATH where MATH is the degree of the left hand side. Since the elements MATH commute and MATH and MATH are at most linear in the variables MATH, it suffices to prove this when MATH is a transposition. By linearity, the assertion is equivalent to the conditions: MATH . We conclude that MATH and that MATH is a commutative MATH-algebra generated by MATH. To see that MATH, it suffices to see that there are no MATH-linear relations between the monomials in MATH. A relation among the monomials in MATH, it would yield a relation among the standard monomials in MATH. However, the standard monomials form a MATH-basis for MATH which completes the proof.
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math/9912066
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Since MATH is positive, MATH is positively graded commutative MATH-algebra and REF implies there are positive integers MATH and MATH such that MATH and MATH. Hence, we have MATH which proves REF follows immediately from REF by applying REF.
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math/9912066
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CASE: Every element MATH can be written uniquely as a finite sum of homogeneous components; MATH where MATH. Let MATH be defined by MATH where MATH. The definition of the MATH-module structure on MATH insures that MATH is a MATH-module homomorphism and the image is a MATH-module. It is clearly surjective. Now, if MATH then MATH. Hence, the element MATH belongs to MATH and we have MATH. It is obvious from the definition of MATH that we have MATH and MATH. CASE: For all MATH, we have isomorphisms MATH . Combining this maps gives the required isomorphism MATH. Moreover, we have MATH and, thus, MATH takes MATH-modules to MATH-modules. Finally, for a graded submodule MATH of MATH, consider the MATH-submodule MATH of MATH generated by the set MATH. To demonstrate that MATH, it suffices to show MATH. Every element of MATH can be written in the form MATH for some MATH and MATH. Applying MATH, we obtain MATH where MATH and MATH. Therefore, we have MATH which completes the proof.
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math/9912066
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CASE: Let MATH, MATH and MATH. By REF , MATH is a subvector space of MATH which implies MATH and MATH. On the other hand, REF also states that the quotient MATH is a MATH-torsion module. Since MATH is a finitely generated module, there exists an integer MATH such that MATH. Thus, we have MATH which implies MATH. Combining the two inequality yields the first part. CASE: The definition of NAME dimension implies MATH . However, a monotonically increasing function MATH and the function MATH are related by the equation MATH and this proves the second assertion. CASE: Applying REF gives MATH and combining this REF yields the third assertion.
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math/9912066
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We begin by observing the following: MATH . Next, notice that MATH. Since MATH is a nonzero-divisor of degree MATH on MATH, we have MATH. Thus, applying the formula MATH for a monotonically increasing function MATH and the function MATH, completes the proof.
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math/9912066
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By definition, irreducible components of MATH correspond to minimal primes in the support of MATH. Hence, if there exists an irreducible component of MATH of dimension MATH, we have a minimal prime MATH in the support of MATH of dimension MATH. Each minimal prime MATH in the support of MATH corresponds to a graded submodule of MATH of the form MATH for some MATH and REF implies the NAME dimension of MATH is equal to its NAME dimension. Thus, we have a graded submodule of MATH with NAME dimension MATH. We complete this proof by showing that the following three conditions are equivalent: CASE: there exist a submodule MATH of MATH with MATH; CASE: there exist a graded submodule MATH of MATH with MATH; CASE: there exist a graded submodule MATH of MATH with MATH; Indeed, we have: CASE: By REF , the graded submodule MATH of MATH has NAME dimension MATH. CASE: Follows immediately from REF . CASE: Follows immediately from REF . CASE: By REF , there exists a graded submodule MATH of MATH such that MATH.
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math/9912066
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By definition, MATH is closed under addition, so it is enough to show that it is closed under left multiplication by the elements MATH. Consider the element MATH in MATH. One clearly has MATH where MATH is the MATH-th standard basis vector. Similarly, because MATH belongs to MATH, we obtain MATH where MATH has an initial form which is strictly smaller than MATH. When MATH, we note that MATH is a complete set of representatives for the cosets of MATH. Hence, there exists a bijective set map between MATH and MATH and one easily verifies that the MATH-module structure of MATH and MATH agree under this correspondence.
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math/9912066
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See REF is CITE.
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math/9912066
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REF are analogous to REF. REF follows from the fact that the exponent vectors appearing in MATH for MATH form a subset of the exponent vectors appearing the standard expression of MATH in MATH.
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math/9912066
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See REF.
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math/9912066
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Since MATH, the set MATH generates MATH if and only if MATH generates MATH. Thus, it suffices to study the initial ideals. If MATH then we have MATH. Since MATH is a NAME basis, it follows that MATH for some MATH and MATH. NAME, we obtain MATH which implies MATH as required.
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math/9912066
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See REF.
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math/9912066
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Consider homogenization MATH of MATH with respect to a positive vector MATH. By REF there exists only finitely many distinct reduced NAME basis for MATH with respect to term orders. Let MATH be their union. Fix a term order MATH on MATH and let MATH denote the multiplicative order on MATH obtained from the multiplicative order MATH on MATH. By construction, MATH is a NAME basis with respect to MATH when MATH is positive. Applying REF , we see that MATH is, in fact, a NAME basis with respect to MATH when MATH. If MATH is the dehomogenization of MATH, then REF implies that MATH is a NAME basis of MATH for MATH where MATH. Finally, REF shows that MATH is a universal NAME basis for MATH.
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math/9912066
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See REF.
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math/9912066
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Let MATH be any term order and let MATH be the multiplicative monomial order on MATH defined as follows: MATH . Next, fix a universal NAME basis MATH for MATH and choose MATH small enough so that the following assertions hold: MATH and for all elements MATH in MATH, the standard form of MATH breaks into four pieces MATH such that MATH, MATH, and MATH. In particular, we have MATH . Since MATH is a NAME basis with respect to MATH, REF provides two additional NAME bases in the polynomial ring MATH: CASE: The initial forms MATH for MATH are a NAME basis for the initial ideal MATH with respect to MATH. CASE: The initial forms MATH for MATH are a NAME basis for the initial ideal MATH with respect to MATH. Now, the definition of MATH, REF imply that the polynomials MATH for MATH are a NAME basis for the ideal MATH with respect the vector MATH. Moreover, MATH indicates that the polynomials MATH for MATH generate the ideal MATH and therefore REF completes the proof.
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math/9912066
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See REF.
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math/9912066
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Given the above lemmas and propositions, the proof is now identical to REF.
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math/9912066
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By subsection REFEF, we know that if MATH has a good finite dimensional filtration, then MATH. Moreover, the NAME dimension and NAME dimension of MATH are equal by REF . Since the NAME dimension of a finitely generated module is the NAME dimension of its support, it suffices to consider ideals. In particular, one reduces to proving that, for a MATH-ideal MATH, the initial ideal MATH is independent of the choice of positive vector MATH. We prove this statement by constructing a homotopy or ``NAME walk" between any two initial ideals. Let MATH and MATH be two positive vectors in MATH. We claim that MATH. Moreover, REF implies that each equivalence class MATH is a convex cone and, thus, we have MATH for any positive MATH. Hence, we may, if necessary, replace MATH by a positive scalar multiple, to guarantee that MATH is a positive vector. It follows that MATH is a positive vector and belongs to MATH for all MATH. Define MATH to be the MATH-ideal MATH. Since the line segment from MATH to MATH intersects finitely many distinct walls of the NAME fan, there are real numbers MATH such that the ideal MATH remains unchanged as the parameter MATH ranges inside the open interval MATH; we denote this ideal by MATH. By REF , we have MATH. By applying REF, we see that MATH. Combining these equalities for MATH completes the proof.
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math/9912066
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Fix a positive vector in MATH and let MATH and MATH be associated NAME ring and module. Now, suppose MATH has an irreducible component of dimension MATH. Since MATH this means that the MATH-ideal MATH has a minimal prime MATH of dimension MATH. By REF , we have MATH . Because MATH is a nonzero-divisor on MATH, MATH pulls back to a minimal prime of dimension MATH over MATH. Furthermore, there exists, by REF , a positive vector MATH such that MATH . It follows, by REF , that MATH has a submodule MATH such that MATH. Finally, REF implies MATH is a submodule of MATH with NAME dimension MATH.
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math/9912066
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If the characteristic variety MATH has an irreducible component of dimension strictly less than MATH, then REF implies that there exists a submodule MATH of MATH with NAME dimension strictly less than MATH. However, this contradicts our hypothesis.
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math/9912066
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This is immediate from REF .
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math/9912069
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We use standard facts about composita of NAME extensions, compare CITE or CITE. Since the right-hand sides of the defining equations are linearly independent over MATH, the field MATH is a NAME extension of MATH with NAME group MATH, and the genus of MATH is the sum of the genera of all the degree-MATH subextensions of MATH. The formula for the genus now follows from the easy fact that if MATH and MATH are positive integers coprime to MATH, and MATH has degree less than MATH and is coprime to MATH, then the curve over MATH defined by MATH has genus MATH. The remaining assertions are clear.
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math/9912069
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We prove the lemma by induction on MATH. Throughout the proof, if MATH is an integer, then MATH denotes the unique integer such that MATH and MATH (mod MATH). For MATH we must find positive integers MATH and MATH which are coprime to MATH such that MATH (mod MATH) and MATH. This is easily done: if MATH, then set MATH and MATH; and if MATH, then set MATH and MATH. Now assume, inductively, that we have sequences MATH and MATH of positive integers coprime to MATH such that each MATH and MATH is congruent to MATH (mod MATH). We will find integers MATH and MATH which are coprime to MATH, where MATH and MATH, such that MATH and MATH (mod MATH). This last congruence is equivalent to MATH (mod MATH), where MATH. Now we let MATH where MATH is chosen so that MATH and MATH are both coprime to MATH. Then the desired conditions on MATH and MATH are satisfied, and we have completed the induction.
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math/9912069
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Fix a finite field MATH of characteristic MATH. Let MATH be a monic irreducible polynomial of even degree. For any MATH, consider the system of REF where both MATH and MATH are increasing sequences of positive integers coprime to MATH. This is the fiber product of a degree-MATH cover MATH (as discussed in REF ) with a degree-MATH cover MATH from the curve MATH to MATH. Since the two covers have coprime degrees, the system of equations describes a curve MATH over MATH. Moreover, the induced degree-MATH cover MATH is abelian, since it is the fiber product of abelian covers. Write the degree of the different of MATH as MATH. Then by the NAME formula applied to MATH, the genus MATH of MATH is given by MATH . Since MATH is monic and has even degree, MATH splits completely under the map MATH, so MATH has at least MATH rational points over MATH. Pick any MATH; we now describe choices of the parameters above so that MATH. Let MATH be the largest integer such that MATH. Our assumption implies MATH. REF yields sequences MATH and MATH of positive integers coprime to MATH such that each MATH and MATH is congruent to MATH mod MATH. Define MATH by the equation MATH . Then MATH is an integer, and our bound on MATH implies MATH is positive. Let MATH be a monic irreducible polynomial in MATH of degree MATH; then the degree of the different of the degree-MATH cover from MATH to MATH is MATH. Now MATH, MATH, and MATH satisfy all the conditions of the previous paragraph, and yield a curve MATH of genus MATH such that MATH. Our choice of MATH implies that MATH and MATH, so MATH . We have shown that MATH is positive. To complete the proof, we must show that MATH for all MATH. For this, let MATH be squarefree of degree MATH, and note that the curve MATH has genus MATH and has a MATH-rational point.
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math/9912069
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Let MATH be an increasing sequence of odd positive integers, and consider the system of REF . Just as in the proof of REF , we see that these equations define a curve MATH over MATH such that MATH and the genus of MATH is MATH . For a fixed positive integer MATH, we can find sequences MATH as above yielding curves of any genus greater than MATH: there is a unique choice of MATH, MATH, , MATH for which MATH attains any prescribed integer value in the interval MATH, and for any MATH there is then an odd integer MATH yielding a genus-MATH curve. For a given nonnegative integer MATH, let MATH be the unique positive integer such that MATH. Then the previous paragraph shows that MATH, and for MATH this implies that MATH.
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math/9912069
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For any MATH, there is a genus-MATH curve which has at least one MATH-rational point and which satisfies REF - REF : for instance, let MATH be squarefree of degree MATH, and take MATH for even MATH and MATH for odd MATH. Thus, it suffices to show that for sufficiently large MATH there exist curves in toric surfaces whose genus is bounded above by a constant times the cube of the number of rational points. Moreover, it suffices to consider the case where MATH is a prime MATH. We use REF (which follows at once from REF ). We may consider each residue class of MATH mod MATH separately. By dint of REF , we are reduced to showing that for MATH in a given residue class mod MATH, there are increasing nonnegative sequences MATH with MATH bounded, so that the sums MATH take on all sufficiently large positive integers. Consider sequences MATH with MATH and MATH. Since nonnegative linear combinations of MATH and MATH achieve all integer values greater than MATH, there are sequences of this type (for fixed MATH) for which MATH takes on any prescribed value greater than MATH. This proves the lower bound in REF .
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math/9912069
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A toric surface is a two-dimensional normal variety MATH, equipped with the action of the two-dimensional algebraic torus MATH and a dense equivariant embedding MATH. The combinatorial object attached to MATH is a fan, which we denote MATH, consisting of cones in a two-dimensional lattice MATH; the dual lattice MATH (usually denoted MATH) is identified with the lattice MATH in which the NAME polygon sits (so MATH and MATH can be regarded as coordinates on MATH). Cones of the fan correspond to affine coordinate charts on MATH. Each ray (one-dimensional cone) of MATH defines a family of half-planes in MATH, namely MATH defines a half-plane MATH for any MATH. Let us say that any half-plane in this family is associated with MATH. Consider the variety in MATH defined by the equation MATH, with closure MATH in MATH. We show now, by transforming MATH into local coordinate systems, that a complete toric surface MATH has all its fixed points disjoint from MATH if and only if every supporting half-plane which meets MATH in an edge is associated with some ray MATH. Indeed, consider a two-dimensional cone MATH; the corresponding affine chart MATH has coordinate ring MATH, where MATH is the semigroup MATH . Now MATH is disjoint from the origin of MATH if and only if there is an element in MATH, with nonzero constant term, which is equal to MATH for some integers MATH and MATH. Such an element exists if and only if some translate of MATH is contained in MATH and contains the origin, that is, if the corner point of the intersection of the pair of supporting half-planes of MATH associated with MATH and with MATH is a vertex of MATH. The nonsingular toric surfaces are those with MATH for every two-dimensional cone MATH, or equivalently, with every two-dimensional cone generated by vectors which form a MATH-basis for MATH. In this case we say MATH is nonsingular. A toric surface is projective if and only if it is a complete variety, and this is the case if and only if the union of the cones in its fan is equal to MATH: such a fan is called complete. Any finite set of rays in MATH is contained in a nonsingular complete fan MATH. Consequently, given any nonzero polynomial MATH, there exists a nonsingular projective toric surface MATH such that the subvariety MATH of MATH defined by MATH (as above) is disjoint from the fixed points of MATH. Now we consider the long exact sequence of sheaf cohomology groups MATH . The first and last terms in REF vanish because they are invariant under blowing up a rational point on a projective surface, and they vanish for MATH. So the arithmetic genus of MATH is equal to MATH. By NAME duality this equals MATH, where MATH is the canonical bundle of MATH. But MATH is the sum of the toric divisors (closures of one-dimensional torus orbits) of MATH, so one can identify the set of lattice points in the interior of MATH with a basis for MATH.
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math/9912069
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We use the notation of the proof of REF . Fix MATH, and let MATH be a curve of genus MATH which embeds in a toric surface MATH over MATH. Without loss of generality, we may assume MATH is nonsingular projective, and has fixed point set (for the torus action) disjoint from MATH. Since every rational point of MATH is either in the torus MATH or in one of the nontrivial intersections REF , the number of rational points on MATH is at most MATH, where MATH is the number of vertices of the NAME polygon of a defining equation MATH for MATH (in the coordinates of some toric chart). So, it suffices to show that the minimum number MATH of interior lattice points in a convex lattice MATH-gon satisfies MATH whenever MATH, for appropriate positive constants MATH and MATH. CITE showed that any convex lattice MATH-gon has area at least MATH. The desired bound REF follows by combining this, Pick's theorem, and the observation that there is a MATH-minimal MATH-gon with no lattice points on the boundary other than vertices (CITE: removing the triangle bounded by two vertices and one interior edge point of a convex lattice MATH-gon yields a MATH-gon with as many or fewer interior lattice points and strictly smaller area).
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math/9912069
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Fix MATH, and let MATH be the least prime not dividing MATH. For any MATH, define MATH to be the least integer such that MATH . Note that the right side might be only slightly larger than the left, or it might be larger by a factor of as much as (slightly more than) MATH. We will modify the set of primes under consideration in order to find an analogous product which is only slightly smaller than MATH; we do this by replacing two (suitably chosen) primes MATH and MATH by a third prime MATH. We now define MATH, MATH, and MATH; each definition makes sense for MATH sufficiently large. Let MATH be the smallest prime whose removal from the right side of REF would reverse the inequality; that is, MATH is the least prime such that MATH and MATH . (Here, and in the remainder of this proof, any sum or product indexed by MATH is understood to be taken over all primes MATH such that MATH and MATH.) Let MATH be the largest prime such that MATH and MATH. Let MATH be the largest prime such that MATH . We apply REF to the set of primes MATH . (The hypotheses of REF are satisfied when MATH is sufficiently large, since then MATH and also there will be some MATH for which MATH (mod REF).) It follows that MATH. It remains only to determine the asymptotics when MATH. We can rewrite REF as MATH . The Prime Number Theorem implies that the right hand side of REF is asymptotic to MATH as MATH, so MATH. Note that MATH is asymptotic to MATH; this is much larger than MATH, MATH, and MATH, since MATH and (for MATH large) MATH. Since the left and right sides of REF are asymptotic to each other as MATH, it follows that MATH . This completes the proof.
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math/9912069
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Let MATH be the maximal tamely ramified abelian cover of MATH which is unramified outside MATH and in which MATH splits completely. By Class NAME Theory for MATH, the NAME group MATH of MATH fits in a short exact sequence MATH where MATH is the diagonal embedding into the product. Moreover, the inertia group over the place MATH is the image of MATH in MATH. Since MATH divides MATH, the group MATH is a quotient of MATH; let MATH be the corresponding cover. Then MATH is a tame abelian cover with NAME group MATH in which MATH splits completely and in which all places outside MATH are unramified. Moreover, the inertia group over the place MATH is MATH. Hence the genus MATH of MATH satisfies REF . Finally, if the MATH are pairwise coprime then MATH is cyclic.
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math/9912069
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Assume MATH is odd, so MATH. Let MATH be positive integers which will be specified later. Put MATH and MATH for MATH; set MATH for all MATH. Note that MATH (mod MATH) for all MATH. An easy count shows that, for any MATH, there are at least MATH finite places on MATH of degree MATH; since MATH, it follows that we can choose distinct finite places MATH on MATH with degrees MATH. REF yields a tame cyclic cover MATH (over MATH) of degree MATH such that some MATH-rational point of MATH splits completely in MATH and the genus MATH of MATH satisfies MATH . We must show that, for any MATH satisfying REF , we can choose the MATH so that MATH. Pick any MATH satisfying REF . Rewrite the expression for MATH as MATH . Note that, for MATH, we have MATH (mod MATH). For each MATH, let MATH be the unique integer such that MATH and MATH (mod MATH). Then MATH (mod MATH), so REF implies there is a unique integer MATH for which MATH. It remains to show MATH; this follows from REF , since MATH . Finally, we indicate how the argument must be modified to handle the case of even MATH. It suffices to prove the result for MATH. In this case, let MATH satisfy MATH and MATH (mod REF). Define MATH as above for MATH, and let MATH. As above, there is a tame cyclic cover MATH (over MATH) of degree MATH such that some MATH-rational point splits completely in MATH and the genus MATH of MATH satisfies REF . It remains to choose the MATH so that MATH. For MATH, choose MATH as above; for MATH, let MATH satisfy MATH and MATH (mod MATH). Then REF implies there is a unique positive integer MATH such that MATH, which completes the proof of REF , and thus the proof of REF .
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math/9912069
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Fix MATH. For any MATH, put MATH and let MATH be places of MATH of degrees MATH and MATH. Let MATH be the maximal tame abelian cover in which all places besides MATH are unramified and in which MATH splits completely. Then MATH is a cyclic cover of degree MATH in which MATH is totally ramified. Moreover, since MATH splits completely, the cover MATH is defined over MATH and MATH. The genus of MATH is MATH. Finally, MATH so MATH and thus the proof is complete.
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math/9912070
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See for instance CITE REF
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math/9912070
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Let MATH be an indecomposable vector bundle. Then for any base of MATH and MATH, MATH, otherwise MATH for some vector bundle MATH. Let MATH of dimension MATH: if MATH and MATH is such that MATH, then the restriction of MATH in MATH defines a morphism of vector bundles MATH. Let us suppose MATH, then MATH: REF implies that the degeneracy locus of MATH is not empty, which leads to a contradiction. Thus: MATH and in particular, if MATH, it results MATH, that is, MATH is G.I.T. stable.
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math/9912070
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Let us suppose that REF holds, then MATH and MATH have the same degeneracy locus and this implies: MATH . Since MATH, MATH, MATH and MATH are irreducible, it results MATH and MATH or MATH and MATH, thus REF holds. NAME let us suppose MATH and MATH and let MATH be the respective base change matrices. Then MATH . Otherwise if MATH and MATH then if MATH are the respective base change matrices, then MATH . Thus REF holds.
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math/9912070
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Let MATH. Then MATH and thus if we consider REF-dimensional parameter subgroup defined by the weights MATH and MATH, it results: MATH . Thus the points of MATH are in one-one correspondence with the orbits of the matrices MATH by the action of MATH. The previous lemma implies the isomorphism in REF .
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math/9912070
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By the proof of REF we have that for any MATH, MATH, where MATH is the image of MATH as a subspace of MATH. Thus we have, up to a base change, MATH where MATH and MATH are linearly independent. It is easily checked that MATH and since MATH, it must be MATH: this implies that MATH. Moreover MATH for any MATH otherwise MATH cannot be stable.
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math/9912070
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See CITE and CITE.
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math/9912070
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It is just a direct computation.
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math/9912070
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By REF , it follows that MATH for any MATH defined as in REF . Thus, by REF , the points represented by the matrices MATH, do not give any contribute to MATH. Moreover it is not difficult to see that, if MATH, the only matrix MATH, as in REF , such that MATH is given by MATH and MATH, where MATH.
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math/9912074
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To prove the proposition, it is enough to show that over an algebraic closure MATH of MATH, the category MATH is semi - simple and every irreducible object in it is MATH - equivariant. Indeed, this would imply that every object of MATH has a form MATH, where MATH is absolutely irreducible (and hence MATH - equivariant) and MATH is a ``perverse sheaf over MATH", that is, a vector space acted on by the NAME. The two facts mentioned above are well-known and we include the proof for completeness. First of all, every irreducible MATH - equivariant perverse sheaf on MATH is an intersection cohomology sheaf on a closure of a MATH - orbit and each such orbit is MATH-stable. To prove the semi-simplicity assertion, we must show that if MATH and MATH are two MATH - orbits, then MATH. First, let us assume that MATH. The natural map MATH is an injection. However, the right-hand side is nothing but MATH, and it vanishes, since any MATH - orbit MATH is a isomorphic to a principal bundle over a (partial) flag variety of the group MATH with a unipotent structure group. Thus, let MATH and without restricting the generality, we can assume that MATH. (It is easy to see that when neither MATH nor MATH, the above MATH is automaticaly zero). It is enough to show that the MATH-restriction of MATH to MATH lives in the perverse cohomological degrees MATH. It is known due to CITE, that the stalks of MATH have the parity vanishing property. Since the dimensions of MATH and MATH have the same parity, MATH has perverse cohomologies only in the even degrees. In particular, its MATH-st perverse cohomology sheaf is MATH, which is what we had to prove.
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math/9912074
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By taking NAME expansions of the map MATH, we obtain a map MATH. Consider the map MATH obtained from the first map MATH and MATH . Our two MATH - points of MATH are the compositions of the above map followed by the projection MATH in the first case, and by the action map MATH in the second one. Hence, the proposition follows from the MATH - equivariance of MATH.
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math/9912074
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Let MATH be a MATH - point of MATH with MATH. Then the data of MATH trivializes MATH. Therefore, MATH is a reduction to MATH of the trivial MATH - bundle on MATH, that is, a map MATH. This defines a map MATH and it is straightforward to see that it is an isomorphism. The fact that MATH follows immediately from REF.
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math/9912074
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Let MATH be the support of MATH in MATH. Choose an integer MATH so that MATH acts on MATH through the quotient MATH. Thus, we have to show that MATH is MATH - equivariant. Let MATH be the sheaf of groups (on the category of all schemes with NAME topology), which attaches to a scheme MATH the group of maps from the localization of MATH around MATH to MATH, with the condition that MATH maps to MATH. By taking NAME expansions at MATH, we obtain a map of sheaves MATH (the underline means ``the sheaf represented by"). The composition MATH is a surjection of sheaves of groups. The proof will be given below. From this lemma we infer that it suffices to prove the following: Let MATH be a smooth scheme and let MATH be an open subset in MATH containing MATH. Let MATH be a section of MATH. We have two maps from MATH: one is the projection on the second factor and the other one is obtained by composing MATH and the action of MATH on MATH. We must show that the pull-backs of MATH under these two maps from MATH to MATH are isomorphic. From MATH we obtain two maps MATH: the map MATH is just the projection on the second factor and the map MATH is described as follows: Let MATH be another scheme and let us consider a MATH-point of MATH. This amounts to having a MATH-point MATH of MATH and a map MATH such that the map MATH factors through MATH. Let MATH denote the preimage of MATH under the map MATH. Using MATH, we obtain a map MATH. We need to produce another MATH-point MATH of MATH. By definition, MATH is the trivial bundle on MATH and is identified with MATH over MATH. We have: MATH. Therefore, to define MATH we need to define a gluing isomorphism MATH. The latter is obtained by composing MATH and MATH. The data of MATH is by definition induced by MATH, via the identification MATH, since MATH contains MATH. The data of MATH follows from the construction. Both MATH and MATH are maps of schemes over MATH. Note that their values over MATH are the two maps MATH described above. In addition, MATH defines a map MATH, and over MATH the maps MATH and MATH factor by means of MATH through the action of MATH on MATH, as in REF. Both MATH and MATH are smooth and it is well-known that the functor of nearby cycles commutes with pull-backs under smooth morphisms. Therefore, it suffices to show that MATH . However, as in REF we obtain that MATH.
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math/9912074
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We will prove that if MATH is affine, the map MATH is a surjection. The proof goes by induction. The assertion is obvious for MATH, since MATH. Therefore, it suffices to show that any map MATH whose restriction to MATH is trivial, can be lifted to a map from the localization MATH to MATH. However, the group of maps MATH as above is abelian and is isomorphic to the group of functions on MATH with values in the NAME algebra of MATH. In addition, we know that MATH is spanned by MATH-dimensional subspaces that correspond to subgroups of MATH isomorphic to either MATH or MATH. Hence, it suffices to analyze separately these two cases. For MATH the assertion is obvious: we can assume that MATH is affine and then any function on MATH can be extended to a function on MATH. Similarly for MATH: Let MATH be a coordinate on MATH around MATH. Our map MATH is a MATH-jet of a regular function on MATH of the form MATH, where MATH is a function on MATH. This MATH is the required map MATH.
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math/9912074
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For two integers MATH we introduce a (relative) NAME scheme MATH, where for a scheme MATH, MATH consists of a map MATH and a coherent subsheaf MATH of MATH over MATH, such that the quotient MATH is MATH - flat. For MATH, there is a natural closed embedding MATH. It is easy to see that for MATH, MATH identifies naturally with the inductive limit MATH. For general MATH, we choose a faithful representation MATH and show as in REF that MATH is a closed subfunctor inside MATH.
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math/9912074
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The fact that MATH and MATH follows immediately from REF. Hence, we must analyze the situation over MATH. We will prove the assertion for MATH, since the proof for MATH is the same. We will construct canonical morphisms in both directions between the corresponding functors. CASE: Let MATH be as above with MATH. First, we define new MATH - bundles MATH and MATH as follows: MATH (respectively, MATH) is by definition trivial over MATH (respectively, MATH) and is identified with MATH over MATH (respectively, over MATH). Since MATH, in order to have well-defined MATH and MATH over MATH, we must define a gluing data over the intersection MATH. However, the corresponding gluing data for both MATH and MATH are provided by the isomorphism MATH. By construction, we have the trivializations MATH . Since MATH, the data of MATH gives rise to a reduction MATH of MATH to MATH. Thus, to MATH above we attach the point MATH. CASE: Let MATH be a MATH - point of MATH. We attach to it a point of MATH as follows: The MATH - bundle MATH is by definition identified with MATH over MATH and with MATH over MATH. The gluing data for MATH over MATH is given by the composition: MATH . Thus, we obtain a well-defined MATH - bundle over MATH, which is trivialized, by construction, over MATH. Finally, the data of MATH for MATH defines a data of MATH for MATH, as MATH. Thus, we have constructed maps MATH and it is easy to see that they are inverses of one another.
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math/9912074
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We will apply REF to the map MATH. We have: MATH . However, MATH. Hence, it remains to show that MATH, that is, that the vanishing cycles functor MATH applied to MATH yields zero. This follows almost immediately from the fact that MATH was obtained by the twisted external product construction: Let MATH be a closed MATH - invariant subscheme of MATH which contains the support of MATH . Let MATH be a finite dimensional quotient of MATH such that the action of the latter on MATH factors through MATH and let MATH be the corresponding MATH - torsor over MATH. Then MATH is a closed subscheme of MATH that contains the support of MATH. Hence, it is enough to calculate MATH. The map MATH is smooth and it is well-known that the functors of nearby and vanishing cycles commute with pull-backs under smooth morphisms. Therefore, it is enough to check that MATH applied to the pull-back of MATH to MATH is MATH. However, the above pull-back is a direct product MATH. Since the projection MATH is smooth, this implies the required vanishing.
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math/9912074
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First, by the construction of the functor MATH, we have a MATH - action on MATH for every MATH. Moreover, REF implies that the isomorphisms MATH and MATH (for MATH) are compatible with this MATH - action. Therefore, for MATH we have the isomorphisms MATH . In addition, REF applied to the map MATH implies that MATH. Let us study the function corresponding to MATH on MATH. (By enlarging the finite field we may assume that MATH is split.) The fact that MATH and MATH are isomorphic as perverse sheaves defined over MATH for any MATH, implies that the corresponding element of MATH is central. At the same time, it vanishes under the map MATH. Hence, this function is zero, as MATH induces an isomorphism MATH, by NAME 's theorem. Now, the same fact is true not only for MATH, but also for all finite field extensions MATH, which implies that MATH.
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math/9912074
|
MATH REF . Let us first prove the assertion when MATH. In this case for an affine MATH, MATH consists of pairs MATH, where MATH is a projective rank MATH module over MATH and MATH is an isomorphism MATH, where MATH is the standard MATH-dimensional vector space. For an integer MATH consider the MATH - dimensional vector space MATH and consider the functor that associates to a scheme MATH the set of all MATH - flat and MATH - stable submodules of MATH. This functor is representable by a closed subscheme, call it MATH, of the NAME of MATH. In particular, MATH is proper. We have a natural closed embedding MATH, since MATH is canonically a MATH - invariant subquotient of MATH. Hence, we obtain a directed family MATH, and it is clear that the functor MATH is isomorphic to MATH. CASE: To treat the case of an arbitrary MATH, it suffices to show that if MATH is an embedding of reductive groups, then the natural map MATH realizes MATH as a closed subfunctor of MATH. For the proof, it will be more convenient to use the realization of the affine NAME via a curve MATH, instead of the formal disc (compare REF). Consider the quotient MATH. It is a basic fact that under the above circumstances, MATH is an affine variety; let MATH denote the point corresponding to MATH. Thus, let MATH be a scheme and let MATH be a MATH-bundle on MATH, trivialized by means of MATH over MATH. In particular, we obtain a map MATH . Since MATH is affine over MATH, there exists a closed subscheme MATH of MATH, such that for any MATH such that MATH extends to a map MATH, MATH factors as MATH. It is easy to see that this MATH represents the fiber product MATH.
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math/9912083
|
Let MATH be a tubular neighborhood of MATH in MATH, and MATH its preimage in MATH. Then MATH has the structure of MATH. Let us denote by MATH and by MATH the two boundary components of MATH. Let MATH be a function on MATH which is constant and equal to MATH in a neighborhood of MATH and is constant and equal to MATH in a neighborhood of MATH. Moreover, assume that MATH is even under the action of MATH. Let MATH be the natural projection. Then consider the form MATH. Since MATH is a global angular form, MATH is a representative of the NAME class of the normal bundle of MATH. Therefore, if we extend MATH by zero on the whole of MATH, we have that MATH is the pullback of a representative of the NAME dual of the diagonal. This might not be our choice MATH. Anyhow, MATH, and it is not difficult to check that one can choose MATH. So we set MATH, and it is an immediate check that REF hold.
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math/9912083
|
We can pull back the integrand form in REF to MATH and integrate it over there. Using the fact that the pullback of MATH is exact, by NAME 's theorem we can rewrite REF as MATH . The codimension-one faces in MATH are labeled by subsets of MATH. Denote by MATH any of these subsets. Assume now that the cardinality of MATH is MATH. Since points in MATH label vertices in the graph and the graph is trivalent, we have the relation MATH where MATH denotes the number of edges connecting points in MATH, and MATH denotes the number of edges with exactly one end-point in MATH. Now we have four cases, according to MATH . The cardinality MATH of MATH is then MATH, MATH, MATH and MATH respectively. The boundary face labeled by MATH is a bundle over MATH with projection MATH and fiber MATH. So the fiber dimension is MATH, MATH, MATH and MATH respectively. We now write the integrand form MATH restricted to this boundary as MATH, where MATH is the product of the pullbacks of MATH corresponding to edges with at least one end-point not in MATH, times MATH in REF . In REF , the term MATH contributes to MATH only if both MATH and MATH are in MATH. In REF , also terms with either MATH or MATH in MATH contribute. Moreover, MATH contributes to MATH only in REF . As a consequence, the degree of MATH will be: CASE: MATH, REF MATH, REF MATH, REF MATH. By using all the above results, we see that the degree of MATH in REF is MATH, MATH, MATH and MATH respectively. Since MATH must be a constant zero-form, we see that the contribution of the face MATH vanishes unless we are in REF with MATH, in REF with MATH, or in REF with MATH. Notice, moreover, that we can replace MATH by MATH in MATH. Thus, in the last case above we conclude that the contribution vanishes by REF . The first case is taken care of by the fact that MATH is a cocycle. We are then left with REF and MATH. Since MATH is connected, there are only two possibilities: REF only point MATH has collapsed at MATH, REF all points have collapsed at MATH. In REF and the fiber is MATH. After this trivial integration we get MATH. REF yields MATH. To prove that MATH vanishes if MATH is odd, consider the involution MATH, MATH. All the pullbacks of MATH change signs. Since the number of edges is MATH, the integrand form gets the sign MATH. On the other hand, since MATH is MATH with some submanifolds blown up, under the involution the orientation gets the sign MATH.
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math/9912083
|
We work as in the proof of REF . The only difference is that we must distinguish between the cases when the collapse is at a point on MATH or otherwise. Notice that, since MATH does not belong to the image of MATH, there is no such term as MATH. By the same reason, when we consider a collapse at a point on MATH, we only have points in MATH. If MATH is involved, the term vanishes since MATH is basic and is a REF-form. If MATH is not involved, reasoning as in the proof of REF and applying REF shows that the term vanishes unless MATH. But this is taken care of by the fact that MATH is a cocycle.
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math/9912086
|
Note that MATH is a topologically trivial MATH-bundle over MATH. Hence by NAME formula we have MATH . Since the higher direct image sheaves MATH, vanish, it follows that MATH. We deduce that the NAME group MATH is generated by MATH and the subgroup of MATH generated by the irreducible components of MATH. Thus for MATH there exists a divisor MATH with MATH such that MATH; the assertion follows.
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math/9912086
|
CASE: Assume the contrary. Recall that MATH is a compact complex torus with universal covering MATH. We may regard the NAME class MATH as bilinear form on the vector space MATH. Suppose that MATH is not semi-positive definite. Let MATH with MATH and let MATH denote the orthogonal complement of MATH (that is, MATH). Let MATH be a semi-positive skew-Hermitian form on MATH such that MATH and MATH. Now consider the MATH-form MATH on MATH given by MATH . By construction we have MATH. Let MATH so that MATH is effective and has no component of MATH, and MATH. By the NAME duality, MATH . Since MATH, we have MATH . On the other hand, MATH . Note that MATH, because MATH is effective and MATH, and that MATH, because MATH, and MATH vanishes on MATH by construction. Thus we deduced a contradiction. CASE: When MATH, the assertion is well known REF . Assume MATH. Let MATH act on MATH as the MATH-th factor of MATH. Since MATH, one infers that there is an orbit whose closure intersects MATH transversally. Hence, MATH in the form described in REF with MATH. Now let us consider MATH as in the above REF . By REF we know that MATH. Assume that there is a vector MATH with MATH. Then we choose MATH and MATH as in REF . Because of the definition we have MATH . By the flat connection of the bundle MATH, the vector MATH is identified as a vector field on MATH. Observe that MATH. The construction of MATH and REF imply that MATH for all MATH. It follows that the one-parameter subgroup corresponding to MATH must stabilize MATH; this is a contradiction. Thus MATH if MATH. Since all MATH in REF and MATH, it follows that MATH is positive. Thus MATH is ample on MATH.
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math/9912086
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Assume the contrary. Then there exists a stratum MATH such that MATH. Clearly the closure MATH of MATH is likewise contained in MATH. But the closure of any stratum contains a minimal stratum, that is, contains a stratum of MATH. However, this is in contradiction to REF .
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math/9912086
|
First, if MATH is invariant under one of the MATH direct factors of MATH in REF , we take the corresponding quotient. Thus we may assume that MATH does not contain anyone of the MATH coordinate factors of MATH. Assume that MATH. Let MATH be a subgroup of MATH isomorphic to MATH. Then there are integers MATH such that MATH . By re-arranging indices and coordinate changes of type, MATH, we may assume that there is a natural number MATH such that MATH for MATH and MATH for MATH. Let MATH with MATH . Then MATH. Consider MATH. If MATH, then MATH would be MATH-invariant and in particular would be invariant under the coordinate factor groups contained in MATH. Since this was ruled out, we have MATH. Now observe that for every MATH we have MATH . Hence it follows from MATH and MATH that MATH . This violates the boundary REF because of REF . Thus MATH is zero-dimensional, and hence finite. As a consequence, MATH is compact. After a finite covering, MATH maps injectively in MATH and therefore is regular.
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math/9912086
|
Let MATH (respectively, MATH) be the universal covering of MATH (respectively, MATH), and MATH the preimage of MATH. We may assume that MATH . Let MATH be MATH-linear independent real numbers with MATH . Let MATH or MATH be an arbitrary integer as assumed in the proposition, and set MATH where MATH is a linear map such that the image MATH in MATH is NAME dense. Moreover, by a generic choice of MATH and MATH we have that MATH is NAME dense in MATH. Let MATH be a finite collection of relatively compact holomorphically convex open subsets of MATH such that there are sections MATH and such that the MATH cover MATH. Set MATH. For every MATH the restricted divisor MATH is defined by a homogeneous polynomial MATH of multidegree MATH, where the coefficients are holomorphic functions on MATH. Let MATH denote the associated inhomogeneous polynomial. Then MATH is a polynomial of multidegree MATH. Due to MATH, MATH does not carry the highest degree monomial, MATH. Recall MATH where MATH is a lattice in MATH and acts on MATH via MATH where MATH is a group homomorphism into the product of MATH. Together with REF this implies that there is a constant MATH such that MATH for all MATH and MATH with MATH and MATH. Note that for every MATH there exists an element MATH and an index MATH such that MATH. Then there is a constant MATH such that MATH for all MATH, MATH with MATH. From REF it follows that for MATH . Hence, MATH for all MATH with MATH. Therefore, MATH . On the other hand one deduces easily from REF that MATH. Hence, MATH .
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math/9912088
|
It is easy to see that MATH, the group algebra of MATH. We define a map MATH by noting that MATH extends to a non-zero MATH-algebra map MATH. We then take MATH, which is a maximal ideal of MATH. Since the domain and codomain of MATH both take products of groups to products of varieties, it suffices to check that MATH is an isomorphism when MATH or MATH, which we leave to the reader. For the second statement, notice that when MATH is a torus, there is a natural isomorphism MATH.
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math/9912088
|
Define MATH, and MATH the set of all connected components of the elements in MATH. Put a metric on MATH which yields its usual topology. Denote this metric by MATH. Let MATH. If MATH for all MATH (this is possible only when MATH is connected), then choose MATH such that MATH, and MATH is small. If, on the contrary, there exists a connected component MATH such that MATH, then take MATH a ball of center MATH and radius MATH, with MATH and such that MATH is small. We show that MATH is adapted: REF is trivially satifsfied. To prove REF , let MATH and MATH be such that MATH. Suppose we have neither MATH, nor MATH. Then by the definition of MATH there exist two compact subgroups MATH and MATH of MATH such that MATH and MATH. But from the definition of MATH it follows that MATH is a ball of center MATH and radius MATH. Similarly, MATH is a ball of center MATH and radius less than MATH, so MATH and MATH cannot possibly intersect, contradiction. REF is obviously satisfied, by construction. Finally, to show REF , let MATH be such that MATH. Suppose MATH and MATH belong to different connected components of MATH. Then by the same type of reasoning as above, it follows that the radii of MATH and MATH are smaller than MATH, so MATH and MATH cannot possibly intersect, which again leads to a contradiction.
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math/9912088
|
MATH is the group of irreducible characters of MATH, so for MATH consider the complex vector bundle MATH over the classifying space MATH, where the map MATH is given by MATH. Then the first NAME class of MATH gives a natural isomorphism MATH. If MATH is a torus, we saw in REF that MATH can be identified with the dual of the integer lattice MATH, so by tensoring the map MATH with MATH, we get the natural isomorphism MATH. Taking symmetric products, we get the desired isomorphism. If MATH is a general (non-connected) compact abelian NAME group, the isomorphism still holds, since both domain and codomain depend only on the connected component of MATH containing the identity, which is a torus.
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math/9912088
|
All the results we mention in this proof are from CITE. Identify the complex NAME algebra MATH with MATH. We denote by MATH the algebraic structure sheaf of MATH, and by MATH the analytic structure sheaf. Let MATH be the completion of the local ring MATH with respect to its maximal ideal. It is sufficient to show that the natural maps MATH and MATH are flat, and that for any MATH open in MATH, the map MATH is flat. We start by noticing that MATH is the completion of MATH with respect to its maximal ideal at zero, so by REF , we know that MATH is flat over MATH. The completion of MATH with respect to its maximal ideal is also MATH, so MATH is flat. It is in fact faithfully flat, because it is local: see REF . Now one can check directly by the definition of flatness that having MATH flat and MATH faithfully flat implies that MATH is flat. Notice also that the same proof can be used to show that MATH is flat. Next let MATH be an open set. By the local characterization of flatness, REF , in order to show that MATH is flat, we have to show that for any MATH the natural map MATH is flat. But we have already shown this when MATH, and the proof for general MATH is the same. Now in order to prove the proposition, just transfer the results we have proved via the exponential map, MATH. This is where we need MATH small.
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math/9912088
|
First we show that MATH is a sheaf of MATH-modules. If MATH is an open cover of a topological space MATH, denote by MATH, etc. Then a presheaf MATH is a sheaf if and only if for any MATH and any finite cover MATH the following sequence is exact MATH where MATH, MATH, etc. Denote this sequence by MATH. Since MATH is a sheaf of MATH-modules on MATH, the sequence MATH is exact. Now, by REF , each MATH-module in the sequence MATH is flat. Therefore, MATH can be thought as a flat resolution of the last term to the right, MATH. Using the Flat Resolution Lemma (see REF ), one can easily show that by tensoring MATH with any MATH-module, we also get an exact sequence. In particular, by tensoring with MATH we obtain MATH, which therefore is exact. But that means that MATH is a sheaf (of MATH-modules). To show that MATH is a sheaf of MATH-modules, if MATH, we need an action of MATH on MATH. The translation map MATH, which takes MATH to MATH gives a map MATH, which sends MATH to MATH. Then we take the result of the action of MATH on MATH to be MATH. It is an easy exercise to show that MATH is coherent: one uses that MATH is a finitely generated MATH-module.
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math/9912088
|
The case when MATH is trivial, so we assume the contrary. Then there exists a compact subgroup MATH of MATH such that MATH and MATH. Then REF from REF implies that MATH. But MATH, so we deduce that MATH. This means that for every MATH we have MATH. Now we will prove that whenever MATH, MATH becomes an isomorphism if tensored with MATH. This is enough for us, because we can regard the statement to be proved as a statement about sheaves, and we are proving the isomorphism stalkwise. Now notice that REF implies that MATH is built from MATH by adding cells of the form MATH, with MATH and MATH. So it is enough to show that, when tensored with MATH over MATH, the ring MATH becomes zero. Or, equivalently, it is enough to have MATH become zero after tensoring with MATH. We therefore proceed to show that MATH is the zero algebra. Via the correspondence MATH, it is enough to show that if MATH is the complexification of the NAME algebra of MATH, and MATH, then MATH. Denote by MATH the kernel of the surjective map of algebras MATH. It is clear that MATH kills MATH as a MATH-module. The vector space MATH is the vanishing set of MATH, so since MATH, there exists a polynomial MATH in MATH such that MATH. This means that MATH is invertible in MATH, hence MATH is invertible in MATH. By the balancing property of the tensor product, we know that MATH, because MATH goes to zero via the map MATH. So we found an element in MATH which is invertible and zero at the same time. This can only happen if MATH is the zero algebra.
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math/9912088
|
Since MATH, it follows that MATH. Let MATH be the connected component of MATH containing the identity. Because both MATH and MATH belong to MATH, REF from REF implies that MATH and MATH belong to the same connected component of MATH, or equivalently that MATH. Let MATH be the complexification of the NAME algebra of MATH. The exponential map MATH is a surjective group homomorphism, because MATH is a torus. Consider an element in the coset MATH. Denote it also by MATH. Let MATH be the global sections functor. Using the exponential map, it is enough to show that MATH is a map of MATH-modules. This is equivalent to showing that MATH for all MATH. But this is easy, since we can identify MATH with the ring of polynomial functions MATH in MATH such that MATH for any MATH in the complexification of the NAME algebra of MATH.
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math/9912088
|
MATH is fixed by MATH, so by REF we know that MATH, with MATH. Therefore it is enough to define a map MATH . But MATH was shown in REF to be a map of MATH-algebras, so we can just take MATH. More formally, we define MATH as the composition of the following isomorphisms: MATH.
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