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math/0105173 | Take MATH so that MATH. It is enough to show that MATH for some MATH, MATH if MATH and MATH. By the assumption, there is a nonzero constant MATH such that MATH for all MATH, MATH. By the stability condition, we have MATH if MATH for some MATH. Let MATH be the maximum of such MATH, and suppose MATH. Since MATH, we have MATH . |
math/0105173 | In CITE we have shown that MATH has an partition into locally closed subvarieties MATH with the following properties: CASE: MATH is closed in MATH for each MATH. CASE: each MATH is a vector bundle over a nonsingular projective variety whose homology groups vanish in odd degrees. A partition satisfying REF is called a MATH-partition. (More precisely, it was shown in CITE that MATH is a fiber bundle with an affine space fiber over the base with the above property. The above statement was shown in CITE.) By the long exact sequence in homology groups, we have MATH. On the other hand, by the property of the virtual NAME polynomial, we have MATH. Since MATH satisfies the required properties in the statement, it follows that MATH satisfies the same property. We have a MATH-partition with REF also for MATH, so we have the same assertion. |
math/0105173 | CASE: This can be shown exactly as in REF. CASE: The rank of the vector bundle MATH is equal to MATH (see REF). By the property of virtual NAME polynomials, we get the assertion. CASE: Exactly the same as REF. |
math/0105173 | We only prove the second statement. In fact, it is not difficult to show that the following argument also implies the first statement. We will prove that our MATH satisfies REF in the next section. Therefore, it is enough to check the assertion for MATH given by the geometric REF . By REF, we get the statement for the multiplication. Similarly, for the proof of the statement for the involution, it is enough to show MATH . This follows from MATH . In fact, the same argument as in the proof of REF shows that the right hand side is contained in MATH. Let us prove REF. Since MATH is homotopic to MATH CITE, its usual homology group is isomorphic to that of MATH. Since MATH is compact, the usual homology group is isomorphic to the NAME homology. Therefore, the NAME duality for MATH, which is applicable since MATH is nonsingular, implies MATH . Since MATH (see REF), we get REF. |
math/0105173 | By a generaly theory, it is enough to show that MATH stays in a compact set. Since MATH is proper, it is enough to show that MATH stays in a compact set. By CITE the coordinate ring of MATH is generated by functions of forms MATH where MATH is a linear form on MATH, and MATH, MATH, , MATH. By the MATH-action, this function is multiplied by MATH where we assume MATH and MATH is the integer such that MATH and MATH. Then MATH is nonnegative. Therefore MATH stays in a compact set for any MATH. |
math/0105173 | Fix a representatitve MATH of MATH. Then MATH is a fixed point if and only if there exists MATH such that MATH . Such MATH is unique since the action of MATH is free. In particular, MATH is a group homomorphism. Let MATH be the weight space of MATH with eigenvalue MATH. The above equation means that MATH . Let us define a MATH-graded subspace MATH of MATH by MATH . The above equations imply that MATH is contained in MATH and MATH-invariant. Therefore MATH by the stability condition. This means that MATH is in the image of REF if we set MATH . Conversely, a point in the image is a fixed point. Since MATH is unique, the map REF is injective. Let us consider the differential of REF. The tangent space of MATH at MATH is the middle cohomology groups of the complex MATH . Similarly the tangent space of MATH is the middle cohomology of a complex with MATH, MATH are replaced by MATH, MATH. We have a natural morphism between the complexes so that the induced map between cohomology groups is the differential of REF. It is not difficult to show the injectivity by using the stability condition. |
math/0105173 | Let MATH . Then the condition for MATH is equivalent to the following system: MATH . The equation can be rewritten as MATH . Let MATH be the right hand side. We can solve this system uniquely by induction: If MATH's are given, MATH is uniquely determined by the above equation and MATH, provided MATH. We can check this condition by the induction hypothesis: MATH where we have used MATH for MATH. |
math/0105173 | REF is clear. REF By the fact that MATH is semismall, we have MATH (in REF) for MATH, hence MATH (in REF) is a constant. Then MATH is a polynomial in MATH by REF. Therefore MATH makes sense. We have MATH again by REF. Thus we get the assertion. CASE: By REF , we have MATH . Since MATH is the ordinary character, the assertion is clear. |
math/0105173 | By the characterization of MATH in REF, we have MATH for all MATH. Therefore MATH . Hence we have MATH is MATH plus non l - dominant terms. |
math/0105178 | The proof of the result in the case MATH is straightforward. Let us study the case MATH. We claim that MATH and MATH cannot overlap. Indeed, assume, for instance that there exist a pair of linear words MATH and MATH such that MATH and MATH starts with MATH. Notice that, by definition, MATH. Therefore, MATH. Since MATH is reduced this cannot happen. Finally, if one of the words, for instance, MATH is empty, then since MATH is non-empty, MATH is not reduced, which contradicts our hypothesis. |
math/0105178 | Observe that, by REF , the linear words MATH and MATH are non-empty. Hence in order to prove that the cyclic words MATH and MATH are non-empty, it is enough to prove that MATH and MATH are reduced. Assume that MATH satisfies REF . Now, MATH is not reduced if and only if MATH and MATH is not reduced if and only if MATH. Since MATH, MATH and MATH. If MATH satisfies REF then MATH (respectively, MATH) is not reduced if and only if MATH (respectively, MATH). Since MATH is reduced, none of those equations can be satisfied. Finally, if MATH satisfies REF then MATH is not reduced if and only if MATH and MATH is not reduced if and only if MATH. On the other hand, by REF , MATH and MATH. Hence the proof is complete. |
math/0105178 | Let us prove MATH and leave the proof of MATH to the reader. If MATH satisfies REF , the result holds by definition. Otherwise, observe that if MATH is a reduced cyclic word of three letters then MATH or MATH, which implies the result. |
math/0105178 | We first prove MATH. Let MATH be the number of linked pairs satisfying MATH, MATH and MATH respectively of REF for a pair of cyclic reduced words MATH and MATH. Let MATH be the set of pairs of the form MATH such that MATH is an occurrence of a subword of MATH, and MATH is an occurrence of a subword of MATH. Notice that the cardinality of MATH is MATH. Let MATH be the set of pairs MATH such that MATH and MATH. Clearly, the set of linked pairs satisfying REF MATH is contained in MATH and so any upper bound of the cardinality of MATH is larger than MATH. We claim that each linked pair satisfying REF MATH determines two different elements in MATH, and that for every positive integer MATH, MATH different linked pairs satisfying REF MATH determine MATH such elements. By this claim, since there are MATH pairs satisfying REF MATH, there are MATH pairs in MATH. Then, the cardinality of MATH is at most MATH and so MATH. In order to prove the claim, consider a linked pair satisfying MATH, let MATH be the ``middle" linear word of this pair as in REF and let MATH be the first letter of MATH. Let MATH, MATH by the elements of MATH, such that MATH and MATH start with the first letter of MATH and MATH and MATH end with the first letter of MATH. Since all the linear words MATH are different and, by definition, MATH and MATH are not in MATH, the claim is proved Let MATH be the set of pairs MATH such that MATH and MATH. The set of linked pairs satisfying MATH is contained in MATH. By an argument similar to the one we used to determine the cardinality of MATH, one can show that the cardinality of MATH is MATH. Hence MATH. Adding both inequalities, one gets MATH. Hence MATH, and the proof of MATH is complete. In order to prove MATH we proceed analogously as we did in MATH, but defining MATH as the set of pairs of linear two-letter subwords MATH of MATH such that MATH and MATH do not start at the same letter, that is, MATH and MATH are different occurrences of subwords of MATH. Hence MATH has MATH elements and all the pairs satisfying REF are in MATH. Now, we can complete the proof as above. |
math/0105178 | Let us first consider the case that MATH is a subword of both, MATH and MATH. Assume that MATH. Then we can write MATH where MATH. Since MATH is a subword of MATH, MATH . Since MATH is a subword of MATH, MATH where MATH is the remainder of dividing MATH by MATH. Since MATH is not a power of MATH, MATH. By REF , there exists a linear word MATH and a positive integer MATH such that MATH divides MATH and MATH. So MATH. Since MATH divides MATH and MATH, MATH divides MATH. Thus, MATH is also a power of MATH, contradicting our hypothesis. Therefore, MATH. To prove the result in the other case, one observes that MATH is a subword of a cyclic word MATH if and only if MATH is a subword of MATH. Then replacing MATH by MATH, the result follows. |
math/0105178 | This proof follows the same ideas of the proof of REF , except that in the case MATH, the words MATH or MATH may be empty. Hence it is necessary to see that they cannot both be empty. Observe that if MATH and MATH are empty, then MATH, MATH, for some positive integers MATH, and some linear representative MATH of MATH and MATH of MATH. But in this case, if we write MATH, MATH, as in REF , we have that MATH, MATH. So MATH does not satisfy MATH of REF . |
math/0105178 | Clearly, if MATH or MATH hold then MATH has a bigon. We prove now the reverse implication. Let MATH, MATH be the subarcs that bound a bigon, and let MATH be the endpoints of MATH and MATH. Assume that MATH goes from MATH to MATH and MATH goes from MATH to MATH. The proof in the other case is similar. Let MATH, MATH, be such that MATH, MATH, and MATH and MATH are minimal with such a property. Let MATH and MATH. Since the union of MATH and MATH is a null-homotopic loop, and MATH is reduced, MATH. Hence MATH and taking MATH, the integers MATH and MATH satisfy MATH. |
math/0105178 | Consider MATH and MATH thin tubular neighborhoods of the edges MATH and MATH in MATH respectively. The projection of MATH and MATH determine two sides of MATH: the MATH-side, containing MATH and the MATH-side, containing MATH. Now, to each loop MATH such that each small arc of MATH intersects MATH transversely, we associate a cyclic word MATH, which describes the way that MATH crosses MATH: Choose a point MATH in MATH and not in MATH. Let MATH be an ordered sequence of letters of MATH such that, starting at MATH, the first edge crossed by MATH is MATH, from the MATH-side to the MATH-side. The second edge crossed by MATH is MATH from the MATH-side to the MATH-side and, in general, for each MATH, the MATH-th edge crossed by MATH is MATH from the MATH-th side to the MATH-side. Finally, take MATH. Endow MATH with a hyperbolic metric with geodesic boundary such that the arcs MATH are also geodesic. (Such a metric exists because we can assume that MATH is a hyperbolic polygon MATH with geodesic edges and right angles.) Let MATH be a geodesic representative of MATH. Then, MATH does not have bigons, because a bigon cannot be bounded by geodesic segments. Let us prove that MATH is reduced and therefore, MATH. Indeed, suppose that MATH is not reduced. Hence MATH, for some MATH. Then there exists a subarc MATH of MATH, a subarc MATH of MATH such that MATH and MATH bound a disk, which is absurd, being a bigon bounded by geodesics. Now, we claim that there exists an homotopy MATH, MATH such that: CASE: MATH. CASE: For each MATH, MATH does not have bigons. CASE: All the self-intersection points of MATH are double. CASE: All the intersection points of pairs of arcs of MATH are transverse. Indeed, if all the self-intersection points of MATH are double, the claim holds trivially. Otherwise, we remove self-intersection points which are not double in the following way: Let MATH be one such point. Consider a small disk MATH centered at MATH such that each connected component of MATH passes through MATH. First, assume that one of these connected components is included in MATH. In this case, deform MATH as in REF and then as in REF . Observe that in the second step, if MATH is the number of connected components of MATH, then after the homotopy there are exactly MATH transverse double points in MATH. We can assume that in both homotopies, MATH is fixed. Now, assume that none of the connected components of MATH is included in MATH. Deform MATH as we did in the second step of the previous case. Hence we have an homotopy MATH, MATH for some MATH, such that MATH and the number of self-intersection points of MATH which are not double, is strictly smaller than the number of self-intersection points of MATH which are not double. We can assume that all the intersection points of pairs of arcs of MATH are transverse for each MATH. Let us check now that for all MATH, MATH does not have bigons. Indeed, if for some MATH, MATH has a bigon MATH, we can follow MATH in MATH. Since all intersection points of pair of arcs are transverse, there exists a bigon in MATH for every MATH. In particular, there exists a bigon in MATH, a contradiction. (Note that the dotted regions of REF are possible traces of bigons). Now we can extend the homotopy, removing at each step non-double self-intersection points. Thus the claim follows. By MATH, MATH is reduced. Thus, as above, one can prove that MATH. Set MATH. There exists MATH oriented subarcs of MATH, MATH such that MATH and for each MATH, the interior of MATH does not intersect MATH and MATH starts at MATH, runs on the MATH-side and ends at MATH, on the MATH-side. For each MATH there exists a unique arc MATH joining two edges of MATH such that MATH. Thus, MATH satisfies the required properties of a segmented representative without bigons. |
math/0105178 | By REF there exists a segmented representative MATH of MATH which does not have bigons (see REF ). Assume that MATH. Let MATH be an annulus, having MATH as one of the boundary components. Subdivide this annulus in MATH parallel annuli. Let MATH be the boundary components of these annuli. We can assume going along the arc MATH in one of the two possible directions, right after the initial arc of MATH (that is, the arc ``parallel" to MATH ), one finds the initial arc of MATH, then the initial arc of MATH and so on. Each MATH is homotopic to MATH, transversal to MATH and such that for each MATH, where MATH is an arc in MATH from the edge MATH to the edge MATH. Moreover, all the endpoints of the edges MATH are different for every MATH and MATH and if MATH, MATH. For each MATH, let MATH be an arc in MATH joining the initial point of MATH with the last point of MATH. We can assume that for every MATH, if MATH then MATH. Let MATH be an arc in MATH joining the initial point of MATH with the last point of MATH. We may also assume that each of the arcs MATH intersects each of the arcs MATH in at most one point. For each MATH, write MATH where MATH. If MATH, set MATH and if MATH, MATH. By definition, MATH is a loop. Moreover, MATH is a segmented representative. Assume that MATH has a bigon. Let MATH and MATH be as in REF . Write MATH, MATH, with MATH. We claim that MATH, MATH, MATH and MATH are congruent to MATH modulo MATH. Indeed, suppose that REF holds. (If REF holds the proof can be completed with analogous arguments). Since MATH, then either MATH or MATH. If MATH, MATH and MATH. Also, MATH if MATH. Then MATH has a bigon, a contradiction. So MATH. Since MATH and MATH and MATH are congruent modulo MATH, MATH and MATH must be congruent to MATH modulo MATH. By definition, if MATH intersects MATH, then MATH or MATH and the proof is complete. |
math/0105178 | Suppose that MATH. Since MATH is cyclically reduced exactly one of the following holds. CASE: MATH CASE: MATH, and MATH or MATH. CASE: MATH, and MATH or MATH. If MATH holds, setting MATH,taking MATH, MATH and MATH, the result holds. In the cases MATH and MATH one keeps adding segments before and after MATH and MATH, until finding arcs landing in different edges at the beginning and at the end. More precisely, assume that MATH holds. By REF and since MATH, there exist integers MATH, such that MATH and MATH is maximum with the property that MATH . Here, we set MATH. Clearly, REF holds. Thus by taking MATH, MATH, and MATH, MATH, the proof is complete for this case. Finally, assume that MATH holds. Let MATH be non-negative integers such that MATH and MATH is the maximum positive integer such that MATH . Here, we define MATH and MATH. Hence MATH satisfies REF , and so taking MATH, MATH, and MATH and MATH the proof of the lemma is complete. |
math/0105178 | We claim that different points in MATH cannot be assigned to the same pair of arcs by REF . Indeed let MATH, MATH be a semiparallel pair of arcs and assume that MATH are assigned to more than one point. Take two of these points MATH. By REF we can assume that there exist MATH such that MATH, MATH, MATH and there are no points assigned to MATH in the arc MATH. Let MATH, MATH and MATH be as in REF . Let MATH be such that MATH and MATH. The subarc of MATH from MATH to MATH and the subarc of MATH from MATH to MATH bound a disk. Therefore the subarc of MATH from MATH to MATH and the subarc of MATH from MATH to MATH bound the image of a disk. Since the bigons of MATH have endpoints in MATH, MATH, a contradiction. So the claim holds. Now, let MATH, let MATH be the pair of arcs assigned to MATH by REF and let MATH and MATH be the underlying words of MATH and MATH respectively. Let MATH, MATH and MATH be as in REF . By the above claim MATH is the only point in MATH. Therefore MATH is a linked pair (see REF ). Conversely, if MATH is a linked pair of occurrences of subwords of MATH, MATH and MATH are the arcs with underlying words MATH, MATH respectively, then MATH, MATH is a semiparallel pair of arcs. Hence we can apply REF . By the definition of linked pair, one has MATH contains a single point MATH, (see REF ). The pair of arcs assigned to MATH by REF is MATH. Hence the proof of the theorem is complete. |
math/0105178 | First assume that MATH, MATH where MATH and MATH are two different primitive cyclic words, MATH and MATH. Construct two representatives of MATH and MATH and MATH, as in REF , considering two geodesic represented by MATH and MATH, respectively. Observe that MATH does not have bigons. If there are triple intersection points of MATH, deform MATH as in REF to MATH where MATH and MATH are a pair of curves such that MATH does not have bigons, nor triple intersection points. Now, construct MATH and MATH using MATH and MATH respectively as in the proof of REF . Observe that if MATH has a bigon then MATH has a bigon. The minimal number of intersection points of MATH and MATH, equals the number of intersection points of MATH and MATH (where the MATH-th power of a representative means run along the representative MATH times). Using this, it is not hard to see that MATH and MATH have minimal intersection. Now, assume that MATH and MATH are powers (positive or negative) of a word MATH. We can assume that MATH is primitive. Let MATH be a segmented representative of MATH. Consider an annulus MATH in MATH around MATH, so that both boundary components if MATH are also segmented representatives of MATH. Then MATH is the disjoint union of two annuli, MATH and MATH. Let MATH and MATH be arcs representatives of MATH and MATH such that MATH and MATH, and every arc of MATH intersects every arc of MATH in at most one point. Notice that if MATH and MATH have a bigon, then MATH has a bigon. So the proof is complete. |
math/0105178 | We prove that MATH. Write MATH where MATH is a primitive reduced cyclic word. Let MATH be a segmented representative of MATH and let MATH be the set of self-intersection points of MATH. Recall that MATH is the disjoint union of two subsets, MATH, and MATH where MATH is the set of intersection points of MATH with MATH, for MATH. By REF , MATH is in correspondence with the pairs of linked pairs of MATH of the form MATH. For each MATH, if MATH and MATH are as in REF then MATH . By definition of MATH, MATH . Consider MATH and let MATH be the linked pair assigned to MATH by REF . If MATH then MATH and MATH and if MATH then MATH and MATH. Hence MATH . The proof of MATH is analogous. |
math/0105178 | Linked pairs of MATH and MATH have the form MATH and MATH where MATH and MATH respectively are subwords of MATH, MATH, MATH and MATH (in the first case, REF MATH or MATH holds and in the second case, REF MATH holds). Observe that if MATH (respectively, MATH) is a linked pair then MATH . Hence if MATH are linked pairs such that MATH then either MATH and MATH or MATH and MATH. Assume that MATH, MATH, MATH, where MATH. There exist (possibly empty) linear words MATH, MATH such that MATH, where, if MATH (respectively, MATH) is empty then the last letter of MATH (respectively, MATH) can coincide with the first letter of MATH (respectively, MATH). Hence MATH . If MATH is a cyclic word, denote by MATH the set of subwords of MATH of the form MATH with MATH and MATH. Thus, MATH . Since MATH, we have that MATH . Hence MATH and then, finally, MATH, as desired. In order to prove the above result for the other case, one considers MATH, the set of subwords of MATH of the form MATH with MATH and MATH and proceeds as above. |
math/0105178 | One chooses a non-selfintersecting arc MATH transversal to MATH, with endpoints on boundary components and intersecting MATH exactly once. Then one cuts MATH open along MATH and MATH. Now, one studies the separating and non-separating case. In the separating case, this procedure yields two surfaces. Then one continues cutting open these two surfaces along non-selfintersecting arcs MATH with endpoints in the boundary components of MATH, until obtaining two disks. In the non-separating case one cuts the surface open along non-selfintersecting arcs MATH with endpoints in the boundary components of MATH, until obtaining a disk. After gluing this disk along MATH, one obtains a cylinder. Then one chooses an arc MATH with endpoints in the boundary components of MATH such that cutting the cylinder open along MATH yields a disk. Denote by MATH the disk obtained by cutting MATH along MATH. Then taking the identity map as the homeomorphism MATH completes the proof. |
math/0105178 | We can assume that MATH and MATH are not powers of the same word because if MATH is a power of MATH, then MATH and MATH have disjoint representatives. By REF , MATH and MATH have minimal intersection. Moreover, the number of terms of the brackets is the minimal number of intersection points. Therefore, by REF if the bracket of two words in a certain alphabet has cancelation, then the bracket of the corresponding words in another alphabet also has cancelation. Thus by REF we can assume that MATH has only one letter. Then the result follows from REF . |
math/0105178 | Let MATH be a free homotopy class and let MATH be a geometric representative of MATH in general position. MATH is a sum of free homotopy classes over certain ordered pairs MATH of self-intersection points of MATH. Denote this set of pairs of self-intersection points by MATH. Thus, MATH if and only if the arcs of the circle between the two preimages of MATH contain exactly one preimage of MATH. Hence for each MATH, there are four arcs, MATH and MATH of MATH such that MATH and MATH go from MATH to MATH, MATH and MATH go from MATH to MATH and MATH runs along MATH then MATH, MATH and finally, along MATH. Clearly, MATH if and only if MATH. We claim that for each MATH, the two terms of MATH corresponding to MATH cancel with the two terms corresponding to MATH. The two terms of MATH corresponding to MATH are: MATH where MATH if the orientation given by the tangent vector of MATH at MATH and the tangent vector of MATH at MATH coincides with the orientation of the surface and MATH otherwise. Now, the terms of bracket of the above linear combination corresponding to MATH are MATH where MATH if the orientation given by the tangent vector of MATH at MATH and the tangent vector of MATH at MATH coincides with the orientation of the surface and MATH otherwise. The same computation for MATH gives MATH so the claim is proved and so also the proposition. |
math/0105180 | Let MATH denote the edge length of the regular simplex. Then the form of the cubic equation follows from computing MATH, MATH. Obviously, for MATH the cubic polynomial factors into MATH (compare CITE). For MATH, assume that there exists a factorization of the form MATH with MATH. Since REF does not contain a monomial MATH, we have either MATH or MATH for MATH. If there were more than one vanishing coefficient MATH, say MATH, then the monomials MATH could not be generated. So only two cases have to be investigated. CASE: MATH for MATH. Then MATH for MATH. Furthermore, MATH for MATH and MATH for all MATH. Hence, the coefficient of the monomial MATH is REF, which contradicts REF . CASE: There exists exactly one coefficient MATH, say, MATH. Then MATH, MATH. Further, MATH for MATH and MATH for MATH. Hence, the coefficient of the monomial MATH is REF, which is again a contradiction. |
math/0105180 | The space of lines in MATH is the NAME of REF in MATH. The NAME embedding CITE realizes this as a projective subvariety of MATH of degree MATH . The theorem follows from the refined NAME theorem CITE and from the fact that the condition for a line to be tangent to a quadric hypersurface is a homogeneous quadratic equation in the NAME coordinates for lines CITE. |
math/0105182 | This follows immediately from the NAME theorem. |
math/0105182 | We sketch a proof, adapting the exposition in REF, which treats the case MATH. It is enough to prove the result after extending the base field from MATH to MATH, since base extension of linear maps defined over MATH does not change the dimension of their kernels or images. This principle will be used constantly in this article. We assume without loss of generality that MATH, and begin with the following exact sequence of vector bundles on MATH (using the notation of CITE): MATH . Here MATH is the ``evaluation" map, and MATH. We have used the fact that MATH is base point free, that is, that it is generated by global sections. NAME REF with MATH and taking cohomology reduces our problem to showing that MATH. To this end, choose MATH points MATH on MATH in general position (this is where we really need to work over MATH). Then, by REF and the beginning of the proof of REF, we obtain the following exact sequence (where we have temporarily lapsed back into writing line bundles multiplicatively): MATH NAME this sequence with MATH and taking cohomology reduces our problem to showing, first, that MATH for all MATH (as is clear by using NAME duality and looking at degrees), and, second, that MATH. This last statement is equivalent to requiring MATH, where we write MATH for the canonical bundle on MATH. Since we have assumed that MATH, we see that the line bundle MATH has degree at most MATH. Therefore MATH, because we can embed MATH into the MATH-dimensional space MATH, where MATH is any effective divisor of degree MATH. Since the MATH points MATH impose MATH zeros, we obtain that MATH for generic MATH, as desired. |
math/0105182 | The fact that MATH is a subset of the space defined in REF is immediate. As for the reverse inclusion, it follows from the fact that MATH has no base points. To see why, extend scalars to MATH, and write MATH, and MATH, where only finitely many of the MATH and MATH are nonzero. For any point MATH on MATH, base point freeness means that we can find MATH which (when viewed as a section of MATH) vanishes at MATH to order exactly MATH. Thus, if MATH satisfies the condition in REF , then MATH vanishes at MATH to order at least MATH, from which we conclude that MATH vanishes at MATH to order at least MATH. Since MATH was arbitrary, we obtain that MATH, as desired. |
math/0105182 | The first statement is trivial, as we are requiring sections to vanish at MATH simultaneously to order at least MATH and at least MATH. The condition on the degree of MATH is to ensure that MATH is base point free by REF , so that we can recover MATH. Moreover, MATH in that case. As for MATH, we always have MATH. If we know that MATH, then we can conclude equality by comparing dimensions, since MATH, and an analogous equality holds for the degrees of the corresponding divisors. |
math/0105182 | In order to apply Lemma/REF , we need to observe that MATH is base point free. This holds by REF , since our assumptions imply that MATH. |
math/0105182 | REF gives us the surjectivity of MATH. As for REF, it follows from a modification of the argument in Lemma/REF , since MATH has no base points. Note again that the computation on REF can be done effectively by letting MATH range only over a basis for MATH. |
math/0105182 | The identity in the first step (which amounts to computing the image of the degree two elements of MATH in the quotient MATH) does not need REF , but turns out to be much simpler. In fact, multiplication by MATH is a bijection between MATH and MATH, the inverse being division by MATH (which does not introduce any poles!). Then the second step above (which is the degree one part of the left hand side of REF, whose computation involves some degree two elements on the right hand side) follows from Lemma/REF , since MATH has no base points. |
math/0105182 | This is an almost verbatim adaptation of the argument in Theorem/REF . The hypothesis on MATH ensures that MATH and MATH have no base points. Note that, as in Theorem/REF , we cannot hope to recover the actual divisor MATH unless MATH. |
math/0105182 | If MATH is of the form MATH, then the resulting MATH is equal to the space MATH in Theorem/REF , where the degree of MATH is MATH. Then MATH has degree MATH, and the result follows. Conversely, given any MATH, define a divisor MATH by the property that for each MATH, MATH is the minimum order to which any nonzero section MATH vanishes at MATH (viewing MATH as a section of MATH). In other words, MATH is the largest possible divisor for which MATH, and MATH is the only possible candidate for a divisor which might give rise to MATH. (One easily checks that MATH is defined over MATH, even if the individual points are only defined over MATH, but we do not actually need this fact for our proof, and give our proof by working entirely over MATH. As usual, there is no problem in extending scalars from MATH to MATH for the proof.) Note that we do not yet know whether MATH, in contrast to our previous notational convention. We do know, however, that MATH, since otherwise, by REF , the codimension of MATH would be at least MATH, contradicting the inclusion MATH. We see in fact that MATH, and that MATH if and only if MATH. Now our choice of MATH gives us in particular a nonzero element of MATH, so the divisor of MATH is of the form MATH, where MATH . The same reasoning as in Lemma/REF then shows us that MATH (the main point is that MATH, just as in Theorem/REF ; on the other hand, for each MATH, we can already find a section MATH that vanishes at MATH precisely to the minimum possible order MATH). Now suppose that MATH were different from MATH. In that case, REF would imply that MATH. Let MATH be the sum of any MATH points of MATH; then the codimension of MATH in MATH would be greater or equal to the codimension of MATH, which we know to be MATH (we are using the fact that MATH, so as to apply REF ). This would contradict our assumption on the codimension of MATH; hence we conclude that MATH after all, and hence by REF we conclude that MATH and that MATH, as desired. |
math/0105182 | The second method to calculate MATH works by the same reasoning as in the proof of Theorem/REF . Now if there exists a nonzero MATH, then its divisor of zeros MATH (viewing MATH as a section of MATH) is an effective divisor of degree MATH which contains MATH. By comparing degrees, we see that MATH. However, MATH and MATH are linearly equivalent (both belong to the class of MATH), from which we obtain MATH. Conversely, if we have MATH, then MATH, and it follows that MATH. |
math/0105182 | Immediate. Note that REF is analogous to drawing the line through two points on a NAME model elliptic curve in the plane, and REF is analogous to finding the third point on the intersection of the elliptic curve with the line. Also note that in REF , our use of Theorem/REF or REF means that we do not need to give special treatment to the case where MATH. In the NAME model, this would mean that our line through a double point is automatically calculated as the appropriate tangent line. It is however simpler in the NAME model to find the line through two distinct points. The analogous statement in our situation is: if we know beforehand that MATH and MATH are disjoint, then we can avoid using the more complicated theorems/algorithms, and instead simply calculate MATH as in Proposition/REF . This will save a lot of time. In practice, the divisors MATH and MATH are likely to be disjoint, so one should first calculate the intersection, and only use the more complicated algorithms for adding divisors in case the codimension of MATH is not MATH. |
math/0105182 | The first method proposed is completely straightforward. As for justifying the second method, we note that the assertion in REF follows from the usual argument, since our requirement that MATH implies that MATH has no base points. REF is analogous to the identity in REF of Theorem/REF . Finally, the assertion in REF follows since MATH also has no base points. |
math/0105182 | The first method is completely straightforward. As for the second, it boils down to the facts that MATH and that MATH. |
math/0105182 | This is entirely analogous to Theorem/REF . The idea is that we hope to have MATH, whence MATH and MATH. We have had to use the higher-degree line bundle MATH in addition to MATH, in order to ensure that the degrees are large enough for us to be able to use Lemma/REF . |
math/0105182 | Immediate. Note the trouble that we went to because MATH is not necessarily base point free, which meant that we had to use REF instead of doing a regular ``flipping" algorithm after REF above. However, we still point out that, generically, MATH is base point free, since its degree is MATH. |
math/0105182 | This is just like Theorem/REF , but we have included the algorithm to show in REF the technique of going from MATH to MATH by virtue of first ``going up," then ``going down." This is because the degree of MATH is too small for us to automatically be able to ``divide" by MATH. If we were originally careful enough to select MATH so that MATH were base point free, we could have begun as in Theorem/REF . However, we cannot in any way guarantee that MATH is base point free, since MATH is arbitrary; hence, concluding with the computation in REF was never an option in the setting of the small model. |
math/0105182 | We know that MATH, and our generous assumptions about degrees ensure that MATH is base point free. Note that we do not assume that the smaller space MATH is base point free; indeed, its degree is MATH, which may be quite small. The rest of the proof is standard by now. |
math/0105190 | This follows from the linearity of the NAME integral, MATH, together with the fact that the exponential map satisfies MATH. |
math/0105190 | The NAME integral is a universal NAME invariant in the sense that if MATH satisfies the hyphotheses then MATH. The result follows immediately from the definition of the total NAME class. |
math/0105190 | This follows immediately from the fact that MATH is multiplicative, and the fact that MATH is multiplicative on the universal enveloping algebra. |
math/0105190 | Let MATH be a knot. By the commutativity of REF above, for the quantum normalizations, MATH. This can be written in the multiplicative normalizations as MATH . But MATH and this is an invertible element in MATH, so MATH as required. |
math/0105190 | This follows from the multiplicativity of the weight system MATH and the fact that both the maps labelled MATH can be expressed in terms, respectively, of the maps MATH and MATH via the same formula which is polynomial in each degree. |
math/0105190 | In the appendix an expression for the NAME integral is given in terms of integer valued knot invariants, MATH, MATH, MATH, MATH, MATH, MATH and MATH. If the transformation to the total NAME class is made, then one finds: MATH . The problem here is with the degree four piece. One can check that the degree four weight system that maps MATH is integer valued on all degree four diagrams. Evaluating this on MATH one obtains the number MATH. From the example of the trefoil in the introduction, it is seen that MATH and hence the total NAME class of the trefoil is not in the integer lattice of MATH. |
math/0105190 | It follows from CITE that the NAME integral has this form where the invariants are rational valued. Stanford CITE has calculated a basis for rational valued additive NAME invariants up to degree six which consists of integer valued invariants. To express the above invariants in terms of Stanford's invariants it suffices to calculate the NAME integral for four suitably chosen knots, compare the values with Stanford's table and then solve the requisite linear equations. I calculated the NAME integral up to degree five of the knots MATH, MATH, MATH, and MATH by using my formulae for torus knots and NAME 's NAME program (see CITE). Denote Stanford's invariants MATH, MATH, MATH, MATH, MATH, MATH and MATH. Define an alternative integral basis in degree five by MATH, MATH and MATH. Then one finds MATH . The left-hand set of equations together with the integrality of Stanford's invariants proves the theorem. |
math/0105190 | These congruences are consequences of the right-hand set of equations in the previous proof together with the fact that all of the invariants are integer valued. The third congruence also requires Stanford's observation that MATH is always even. |
math/0105196 | The proof is a (simplified) replica of the one holding in the case of complex tori CITE. |
math/0105196 | An element of MATH restricted to MATH, with MATH, yields a global section of MATH, which is zero unless MATH. By a density argument we get the result. |
math/0105196 | We compute the cohomology groups MATH. We represent MATH as MATH, with MATH, and MATH, with MATH. Let MATH be the inverse image of MATH in MATH. We work now in a gauge where the factor of automorphy of MATH is MATH, and the operator MATH is the MATH-part of the exterior differential. An element in MATH may be written as MATH, where MATH is a function on MATH satisfying the automorphy condition MATH . If MATH is a coboundary, MATH, one has MATH . The function MATH must satisfy the automorphy condition, which amounts to the following condition on MATH: MATH . If MATH this condition may be solved for MATH, so that MATH. Thus MATH is a skyscraper supported at MATH. If MATH, REF may be solved if and only if MATH so that MATH. This proves the claim. |
math/0105196 | A choice of flat coordinates MATH on MATH fixes an isomorphism MATH. The NAME sheaf MATH on MATH is the product of the NAME sheaves MATH on the MATH factors of MATH, as one can check for instance by describing the NAME bundles by their factors of automorphy. Let MATH be of the form MATH where each MATH lies in a factor of MATH. If MATH, a word-by-word translation of the NAME theorem for NAME cohomology (compare for example, CITE) gives a decomposition MATH whence we have, by REF , MATH . Induction on MATH then yields, for every MATH, MATH . This proves both claims. |
math/0105196 | The proof of this Proposition in given in REF. |
math/0105196 | The restriction MATH is defined as MATH (here MATH and MATH are the natural inclusions). The result is proved by applying the topological base change CITE to the diagram MATH . |
math/0105196 | Both claims follows from REF and the absolute case. |
math/0105196 | It follows from REF . |
math/0105196 | The first claim follows from the fact that the horizontal part of the tangent space to MATH has constant dimension; the second from the Lagrangian condition which implies that the local equations of MATH in MATH are linear in the action coordinates. |
math/0105196 | A proof is given in REF. |
math/0105196 | This will use the proof of REF given in REF. We know that MATH induces a connection on the NAME transform if the curvature MATH of the NAME bundle on MATH vanishes on MATH, where MATH and MATH are the supports of MATH and MATH, respectively. In view of the form of MATH, this condition is met if for each MATH the intersections of MATH and MATH with the fibres MATH, MATH yield subtori of MATH, MATH that are normal to each other. But looking at the equations of the supports, REF , and comparing with the absolute REF , we see that this condition is fulfilled. |
math/0105196 | The connection REF-form of the connection MATH can be written in an appropriate gauge as MATH with the quantities MATH constant. From the proof of REF given in REF we know that the transformed connection MATH is given in local action-angle coordinates by REF-form MATH . Rewriting this in terms of MATH we obtain MATH where the coefficients MATH are constant. Since MATH because of the flatness of MATH, it follows that the curvature of MATH is given by MATH . Since MATH, where the functions MATH are those of REF , the condition MATH can be written as MATH . But this is the system of REF , therefore when MATH is Lagrangian, this condition is automatically satisfied. |
math/0105196 | The WIT condition follows immediately from REF . To show the remaining part of the claim we write local equations for MATH as MATH . Performing a fibrewise transform we obtain the following equations for the support MATH of the transform MATH: MATH where MATH. It remains to show that MATH is Lagrangian and that the family MATH is parallelly transported by the NAME connection. The latter point follows from the complex structure of MATH (compare REF ): the NAME equations for MATH imply that the coefficients MATH and MATH are constant. As far as the Lagrangian property of MATH is concerned, the holomorphicity of MATH and MATH imply REF in the proof of REF (in REF). Therefore MATH is Lagrangian. Observe that the transformed connection MATH has a REF-form given by MATH whence we can immediatly deduce its flatness. |
math/0105202 | This follows from the natural long exact sequence: MATH induced from the short exact sequence of chain complexes MATH . Now, observe that MATH is an isomorphism on MATH if and only if the latter group is trivial, since each element of MATH is annihilated by a sufficiently large power of MATH. |
math/0105202 | Consider a NAME diagram which is weakly MATH-admissible for all MATH structures (that is, a diagram which is MATH-admissible NAME diagram, where MATH is any torsion MATH structure, compare REF and, of course, REF). This diagram can be used to calculate MATH and MATH for all MATH-structures simultaneously. But the tori MATH and MATH have only finitely many intersection points, so there are only finitely many MATH structures for which the chain complexes MATH and MATH are non-zero. |
math/0105202 | Let MATH be a strongly MATH-admissible pointed NAME diagram for MATH. If we switch the roles of MATH and MATH, and reverse the orientation of MATH, then this leaves the orientation of MATH unchanged. Of course, the set of intersection points MATH is unchanged, and indeed to each pair of intersection points MATH, for each MATH, the moduli spaces of holomorphic disks connecting MATH and MATH are identical for both sets of data. However, switching the roles of the MATH and MATH changes the map from intersection points to MATH structures. If MATH is a NAME function compatible with the original data MATH, then MATH is compatible with the new data MATH; thus, if MATH is the MATH structure associated to an intersection point MATH with respect to the original data, then MATH is the MATH structure associated to the new data. (Note also that the new NAME diagram is strongly MATH-admissible.) This proves the result. |
math/0105202 | Changing the orientation of MATH is equivalent to reversing the orientation of MATH. Thus, for each MATH, and each class MATH, there is a natural identification MATH where MATH is the class with MATH, obtained by pre-composing each holomorphic map by complex conjugation. This induces the stated isomorphisms in an obvious manner. |
math/0105202 | Consider the genus one NAME splitting of MATH. Here we can arrange for MATH to meet MATH in precisely MATH points. Each intersection point corresponds to a different MATH structure, and, of course, all boundary maps are trivial. |
math/0105202 | From the picture, it is clear that (for some appropriate orientation of MATH and MATH) we have: MATH . Thus, if MATH is a standard symplectic basis for MATH, then MATH in MATH. It follows that MATH is generated by MATH. We can calculate, for example, MATH as follows. We find a closed loop in MATH which is composed of one arc MATH, and another in MATH both of which connect MATH and MATH. We then calculate the intersection number MATH, MATH. It follows that MATH in MATH. So, MATH. Proceeding in a similar manner, we calculate: MATH for MATH. Finally, MATH, as these intersections can be connected by a square. It follows from this that the equivalence class containing MATH contains three intersection points: MATH,MATH, and MATH. Finally, note that MATH, for some fixed MATH, according to REF, and MATH generates MATH, according to the intersection numbers between the MATH and MATH calculated above. |
math/0105202 | We draw the domains MATH and MATH belonging to MATH and MATH in REF respectively, where the coefficients are equal to REF in the shaded regions and REF otherwise. Let MATH, MATH denote the part of MATH, MATH that lies in the shaded region of MATH. Once again, we consider the constant almost-complex structure structure MATH. Suppose that MATH is a holomorphic representative of MATH, that is, MATH, and let MATH denote the corresponding REF-fold branched covering of the disk (see REF). Also let MATH denote the corresponding holomorphic map to MATH. Since MATH has only REF coefficients, it follows that MATH is holomorphically identified with its image, which is topologically an annulus. This annulus is obtained by first choosing MATH or MATH and then cutting the shaded region along an interval MATH starting at MATH. Let MATH denote the length of this cut. Note that by uniformization, we can identify the interior of MATH with a standard open annulus MATH for some MATH (where, of course, MATH depends on the cut-length MATH and direction MATH or MATH). In fact, given any MATH and MATH, we can consider the annular region MATH obtained by cutting the region corresponding to MATH in the direction MATH with length MATH. Once again, we have a conformal identification MATH of the region MATH with some standard annulus MATH, whose inverse extends to the boundary to give a map MATH. For a given MATH and MATH let MATH, MATH, MATH, MATH denote the arcs in the boundary of the annulus which map to MATH, MATH, MATH, MATH respectively, and let MATH, MATH denote angle spanned by these arcs in the standard annulus MATH. A branched covering over MATH as above corresponds to an involution MATH which permutes the arcs: MATH, MATH. Such an involution exists if and only if MATH in which case it is unique (see REF). According to the generic perturbation theorem, if the curves are in generic position then these solutions are transversally cut out. It follows that MATH. We argue that for MATH and MATH the angles converge to MATH, MATH. To see this, consider a map MATH, which induces a conformal identification between the interior of the disk and the contractible region in MATH corresponding to MATH and MATH. One can see that the continuous extension of the composite MATH, as a map from the disk to itself converges to a constant map, for some constant on the boundary. (It is easy to verify that the limit map carries the unit circle into the unit circle, and has winding number zero about the origin, so it must be constant.) Thus, as MATH, both curves MATH and MATH converge to a point on the boundary of the disk, proving the above claim. In a similar way, for MATH and MATH the angles converge to MATH, MATH. Now suppose that for MATH we have MATH. Then the signed sum of solutions with MATH cuts is equal to zero, and the signed sum of solutions with MATH cuts is equal to MATH. Similarly if for MATH we have MATH, then the signed sum of solutions with MATH cuts is equal to MATH, and the signed sum of solutions with MATH cuts is equal to zero. This finishes the proof for MATH, and the case of MATH is completely analogous. Although the domains MATH and MATH do not satisfy the boundary-injectivity hypothesis in REF, transversality can still be achieved by the same argument as in that proposition. For example, consider MATH, and suppose we cut along MATH, so that the map MATH induced by some holomorphic disk MATH is two-to-one along part of its boundary mapping to MATH. Then, it must map injectively to the MATH-curves so, for generic position of those curves, the holomorphic map MATH is cut out transversally. Arguing similarly for the MATH cut, we can arrange that the moduli space MATH is smooth. The same considerations ensure transversality for MATH. Note also that we have counted points in MATH and MATH, for the family MATH, but it follows easily that the same point-counts must hold for small perturbations of this constant family. |
math/0105202 | This is the same argument as in the MATH case, together with the observation that if MATH, and MATH or MATH, and MATH, then the domain MATH contains regions with negative coefficients (so the moduli space is empty). Moreover, since MATH, it follows that MATH. |
math/0105202 | First note that MATH is the boundary of REF-manifold described by the plumbing diagram in REF , where the number of MATH spheres in the right chain is MATH. This gives a description of MATH as the total space of an orbifold circle bundle over the sphere with MATH marked points with multiplicities MATH respectively, where MATH. The circle bundle MATH has NAME data MATH and the canonical bundle is MATH. Now we can apply CITE to compute the irreducible solutions, relative gradings and the boundary maps. Let us recall that for the unperturbed moduli space there is a REF to REF map from the set of irreducible solutions to the set of orbifold divisors MATH with MATH and MATH where the preimage consists of a holomorphic and an anti-holomorphic solution, that we denote by MATH and MATH respectively. Note that MATH, MATH lie in the MATH structures determined by the line-bundles MATH, MATH respectively. In order to simplify the computation we will use a certain perturbation of the NAME equation. Using the notation of CITE this perturbation depends on a real parameter MATH, and corresponds to adding a two-form MATH to the curvature equation, where MATH is the connection form for MATH over the orbifold. Now holomorphic solutions MATH correspond to effective divisors with MATH and anti-holomorphic solutions MATH correspond to effective divisors with MATH . According to CITE the expected dimension of the moduli space between the reducible solution MATH and MATH is computed by MATH where MATH denotes the holomorphic NAME characteristic of the bundle MATH, and MATH is given by the inequalities MATH . Returning to our examples let MATH denote the divisor MATH. It is easy to see that MATH and MATH are in the same MATH structure. Also MATH and MATH are in the same MATH structure. From now on let MATH denote the MATH structure given by the line bundle MATH, and MATH corresponds to the line-bundle MATH. Clearly MATH, since MATH. Since MATH for all MATH with MATH the unperturbed moduli space (with MATH) have no irreducible solutions. It follows that MATH and MATH are generated by MATH and we have the corresponding isomorphisms with MATH, MATH respectively. Clearly the MATH action maps MATH to MATH, so in the light of the MATH symmetry in NAME theory, it is enough to compute the equivariant NAME homologies for MATH. For these MATH structures let us fix a perturbation with parameter MATH satisfying MATH where MATH is sufficiently small. This perturbation eliminates all the holomorphic solutions. It still remains to compute the anti-holomorphic solutions. First let MATH. Since MATH the irreducible solutions in MATH are MATH for MATH. It is easy to see from CITE, see also CITE, that the irreducible solutions and MATH are all transversally cut out by the equations. Computing the holomorphic NAME characteristic we get MATH for MATH, MATH for MATH, and MATH for all other MATH, where MATH. The dimension formula then gives MATH . As a corollary we see that MATH is zero, since all these moduli spaces have negative formal dimensions, and relative gradings between the irreducible generators are even. In MATH the relative gradings between all the generators are even, so MATH is trivial as well. Now the isomorphism between MATH and MATH corresponds to mapping MATH to MATH, and MATH to MATH. Similarly the isomorphism between MATH and MATH corresponds to mapping MATH to MATH, and MATH to MATH. Furthermore MATH is freely generated by MATH and the map MATH gives the isomorphism with MATH. Now suppose that MATH. Then there are no irreducible solutions for the perturbed equation. So MATH and MATH are generated by MATH and we have the corresponding isomorphisms with MATH, MATH respectively. For MATH we get the analogous results by replacing MATH with MATH. |
math/0105202 | Both cases follow from the observation that MATH is independent of the MATH structure MATH. To see this, note that for any MATH, we can wind normal to the MATH so that MATH and MATH are both weakly MATH-admissible, where MATH and MATH are two choices of basepoint which can be connected by an arc which meets only MATH. Now, both MATH and MATH are calculated by the same equivalence class of intersection points, using the basepoint MATH in the first case and MATH in the second. This changes only the boundary map, but leaves the (finitely generated) chain groups unchanged, hence leaving the NAME characteristic unchanged. The result for MATH then follows from this observation, together with REF . For the case where MATH, recall that the NAME decomposition gives MATH a chain complex with MATH one-dimensional generators corresponding to the MATH (each of which is a cycle), and MATH two-dimensional generators corresponding to the MATH. On the one hand, the determinant of the boundary map is the order of the finite group MATH (which, in turn, is the number of distinct MATH structures over MATH); on the other hand, this determinant is easily seen to agree with the intersection number MATH. The result follows from this, together with MATH-independence of MATH. |
math/0105202 | The intersection points between MATH and MATH which are not induced from MATH correspond to the intersection points between the original MATH and MATH. So, suppose that MATH is an intersection point between MATH and MATH (there are, of course, finitely many such intersection points), and let MATH be some basepoint outside the winding region. As we wind MATH times, and place the new basepoint MATH inside the winding region as above (so as not to cross any additional MATH-curves), we see that MATH where we think of MATH a one-dimension homology class in MATH. The lemma then follows. |
math/0105202 | Assume MATH is odd, and let MATH be the class with MATH, and whose MATH boundary lies entirely inside the tubular neighborhood of MATH. We claim that MATH is obtained from MATH by winding only its MATH-boundary (and hence leaving the domain unchanged outside the winding region). This follows from the fact that the NAME index is unchanged under totally real isotopies of the boundary. It follows then that the multiplicities of MATH inside a neighborhood of MATH grow like MATH. Recall that the multiplicities of MATH inside grow like MATH, while outside they grow like MATH. Now, the set of all MATH homotopic classes connecting MATH to MATH is given by MATH . If this class is to have non-negative multiplicities, we must have that MATH or MATH. This proves the assertion concerning classes from MATH to MATH, letting MATH. Considering classes from MATH to MATH, note that all MATH classes have the form MATH . When MATH, these classes have negative multiplicities outside MATH. When MATH, these have negative multiplicities inside the neighborhood of MATH. |
math/0105202 | This follows immediately from the previous lemma. |
math/0105202 | Consider a pair of generators MATH and MATH, for which the coefficient of MATH is non-zero, that is, that gives a homotopy class MATH for which MATH and MATH. Thus, by REF , there are two possible cases, where MATH or MATH (for MATH and MATH). Note also that MATH. The case where MATH, has two subcases, according to whether or not MATH. If MATH, MATH, and it follows easily that MATH. Since the periodic domains have both positive and negative coefficients, the MATH coefficient of MATH must vanish. If MATH, then the domain of MATH must include some region outside the neighborhood of MATH. Moreover, since MATH we have that MATH; but since the support of the twisting region is sufficiently small, it follows that MATH that is, MATH. When MATH, it is easy to see that MATH . It follows that MATH. Moreover, MATH so MATH, by our hypothesis on MATH, so that MATH. |
math/0105202 | This is an algebraic consequence of REF . We can define a right inverse to MATH, MATH where MATH is the disk connecting MATH to MATH. Then, we define a map MATH . Note that the right-hand-side makes sense, since the map MATH decreases the ordering (which is bounded below), so for any fixed MATH, there is some MATH for which MATH . It is easy to verify that MATH is a right inverse for MATH. The map sending MATH induces a map from MATH to MATH, which is injective, since for any MATH, we have that MATH . Similarly, the map MATH supplies an injection MATH. It follows that MATH. |
math/0105202 | According to REF we have the short exact sequence MATH which we compare with the short exact sequence MATH . The result then follows by comparing the associated long exact sequences, and observing that the connecting homomorphism for the second sequence agrees with the map on homology induced by MATH. |
math/0105202 | The map MATH depends on a base-point and an equivalence class of intersection point. However, according to REF , MATH depends on this data only through the underlying MATH structure MATH (when the latter is negative). Let MATH denote the NAME characteristic MATH. We fix a basepoint MATH as before. We have a map MATH defined as follows. Given MATH, we have MATH where MATH is the canonical homotopy class connecting MATH and MATH, and MATH. (In fact, it is easy to see that the above assignment is actually independent of the number of times we twist MATH about MATH.) There is a naturally induced function (depending on the basepoint) MATH by MATH where MATH is the local intersection number of MATH at MATH. It is clear that MATH . It follows that MATH . We investigate the dependence of MATH on the basepoint MATH. Note first that there must be some curve MATH which meets MATH whose induced cohomology class MATH is not a torsion element in MATH: indeed, any MATH appearing in the expression MATH with non-zero multiplicity has this property. Suppose that MATH and MATH are a pair of possible base-points which can be connected by a path MATH disjoint from all the attaching circles except MATH, which it crosses transversally once, with MATH. We have a corresponding intersection point MATH. We orient MATH so that this intersection number is negative (so that MATH points in the same direction as MATH). Now, we have two classes of intersection points MATH: those which contain MATH (each of these have the form MATH), and those which do not. If MATH lies in the first set, then MATH if MATH lies in the second set, then MATH . Note that there is an assignment: MATH obtained by restricting MATH to MATH, and hence a corresponding map MATH . We have the relation that MATH . It follows from REF that MATH . (note that MATH has finite support). It is easy to see directly from the construction that MATH and the term MATH from REF can differ at most by a sign and a translation with MATH, where MATH and MATH are universal constants. Since MATH and MATH are three-manifold invariants, by varying MATH, it follows that MATH. A simple calculation in MATH shows that MATH, too. It follows that MATH must agree with MATH. |
math/0105202 | This is an immediate application of REF and the NAME formula for chain complexes over the principal ideal domain MATH. Specifically, we have that MATH where MATH denotes the MATH-complex, that is, MATH . It is easy to see then that for any pair of non-negative integers MATH and MATH, MATH while for any MATH-module MATH, MATH and MATH. To see the NAME characteristic statement, we proceed as follows. First, observe that to calculate the NAME characteristic of the graded MATH-module MATH is the same as the NAME characteristic of the MATH-vector space MATH. From above, we have that MATH is freely generated over MATH by MATH with MATH where MATH (observe that all generators of the form MATH inject into MATH) and also generators MATH for MATH and MATH. Observe in particular that when MATH are both non-zero, MATH has a corresponding element MATH whose degree differs by one, so these cancel in the NAME characteristic. The only remaining elements are those of the form MATH with MATH and MATH, and also MATH with MATH and MATH. These contribute MATH and MATH to the NAME characteristic MATH respectively. |
math/0105202 | We consider first MATH structures on MATH of the form MATH. Let MATH be a strongly MATH-admissible pointed NAME diagram for MATH. Consider the NAME diagram for MATH discussed in REF, given by MATH, where MATH is a genus one surface and MATH and MATH are a pair of exact Hamiltonian isotopic curves meeting in a pair MATH and MATH of intersection points. Choose the reference point MATH so that the exact Hamiltonian isotopy connecting the two attaching circles does not cross MATH. Recall that there is a pair of homotopy classes MATH which contain holomorphic representatives, indeed both containing a unique smooth, holomorphic representative (for any constant complex structure on MATH). We can form the connected sum diagram MATH, where we form the connected sum along the two distingushed points, and let the new reference point MATH lie in the connected sum region. This is easily seen to be strongly MATH-admissible. Of course MATH; thus MATH is generated by MATH, where MATH, and MATH, that is, MATH (where the second factor is shifted in grading by one). We claim that when the neck is sufficiently long, the differential respects this splitting. Fix MATH. First, we claim that for sufficiently long neck lengths, the only homotopy classes MATH with non-trivial holomorphic representatives are the ones which are constant on MATH. This follows from the following weak limit argument. Suppose there is a homotopy class MATH with MATH for which the moduli space is non-empty for arbitrarily large connected sum neck-length. Then, there is a limiting holomorphic disk in MATH. On the MATH factor, the disk must be constant, since MATH (here we are in the first symmetric product of the genus one surface), and all non-constant homotopy classes have domains with positive and negative coefficients. Thus, the limiting flow has the form MATH for some MATH (in MATH). REF applies then to give an identification MATH. Indeed, we have the same statement with MATH replacing MATH. Next, we claim that (for generic choices) if MATH is any homotopy class with MATH, which contains a holomorphic representative for arbitrarily long neck-lengths, then it must be the case that MATH, and MATH or MATH. Again, this follows from weak limits. If it were not the case, we would be able to extract a sequence which converges to a holomorphic disk in MATH, which has the form MATH or MATH. Now, it is easy to see that MATH for MATH or MATH (by, say, looking at domains); hence, MATH. It follows that as a flow in MATH, MATH. Thus, there are generically no non-trivial holomorphic representatives, unless MATH is constant. Observe, of course, that MATH, and also MATH. With the appropriate orientation system, these flows cancel in the differential. Putting these facts together, we have established that MATH where MATH is the differential on MATH, and MATH is the differential on MATH. Indeed, it is easy to see the action of the one-dimensional homology generator coming from MATH annihilates MATH, and sends MATH to MATH. When the first NAME class of the MATH structure evaluates non-trivially on the MATH factor, we can make MATH and MATH disjoint, and have a NAME diagram which is still weakly admissible for this MATH structure. Since there are no intersection points, it follows that MATH in this case is trivial. |
math/0105202 | First observe that if MATH is an embedded sphere in MATH, then for each MATH for which MATH, we have that MATH. This is a direct consequence of REF : attach a handle to MATH to get a homologous torus MATH and apply the theorem. Now, let MATH be a representative of MATH whose MATH is minimal, labeled so that MATH for MATH are the components with genus zero. Then, MATH . |
math/0105202 | Using an appropriate trivialization of the tangent bundle MATH, we can view the normalized gradient vector field MATH as constant over MATH. Let MATH be the boundary, which is divided into two hemispheres MATH, so that the sphere MATH contains the index one critical point and MATH contains the index two critical point. We can replace MATH by another vector field MATH which agrees with the normalized gradient over MATH, and vanishes nowhere in MATH (and hence can be viewed as a unit vector field). With respect to the trivialization of MATH, we can think of the vector field as a map to the two-sphere; indeed the restriction MATH, is constant along the boundary circle, so it has a well-defined degree, which in the present case is one, since MATH and MATH . The line bundle we are considering, MATH, then, is the pull-back of the tangent bundle to MATH, whose first NAME number is the NAME characteristic for the sphere. |
math/0105202 | The proof is the same as the proof of REF, only keeping track now of the homotopy classes of the corresponding triangles. |
math/0105202 | Consider a NAME function on MATH with one index zero critical point, MATH index one critical points and MATH index two critical points. Let MATH be the MATH-level of this function, MATH be the curves where MATH meets the ascending manifolds of the index one critical points in MATH, and let MATH be the curves where MATH meets the descending manifolds of the index two critical points. By gluing in the solid torus in three possible ways, we get the manifolds MATH, MATH, MATH. Extending the given NAME function to the glued in solid tori, (by introducing an additional index two and index three critical point), we obtain NAME decompositions for the manifolds MATH, MATH, and MATH. We let MATH and MATH be small perturbations of MATH for MATH. In this manner, we have satisfied REF. To satisfy REF , we wind to achieve weak admissibility for all MATH structures for the NAME subdiagram MATH: in fact, we can use a volume form over MATH for which all such doubly-periodic domains have zero signed area (compare REF). Then, for the MATH and MATH, we use small Hamiltonian translates of the MATH (ensuring that the corresponding new periodic domains each have zero energy). There is a triply-periodic domain which forms the homology between MATH, MATH, and MATH in a torus summand of MATH containing no other MATH or MATH (for MATH). By adjusting the areas of the two triangles with non-zero area, we can arrange for the signed area of the triply-periodic domain to vanish. |
math/0105202 | Note that the three-manifolds described here are MATH-fold connected sums of MATH, so the result follows from REF (or, alternatively, see REF). |
math/0105202 | The proof is obtained by suitably modifying REF. Suppose that MATH and MATH are holomorphic representatives of MATH and MATH respectively. We obtain a nodal pseudo-holomorphic disk MATH in the singular space MATH specified as follows: CASE: At the stratum MATH, MATH is the product map MATH. CASE: At the stratum MATH, MATH is given by MATH pseudo-holomorphic spheres which are constant on the first factor. More precisely, for each MATH for which MATH (where the MATH are arbitrary), there is a component of MATH mapping into MATH, consisting of the product of the constant map MATH with the sphere in MATH which passes through MATH. CASE: The map MATH misses all other strata of MATH. As in REF, we can splice to obtain an approximately holomorphic disk MATH (a triangle) in MATH. When the connected sum tube is sufficiently long, the the inverse function theorem can be used to find the nearby pseudo-holomorphic triangle. The domain belonging to MATH is clearly given by MATH described above. Conversely, by NAME 's compactness (see also REF), any sequence of pseudo-holomorphic representatives MATH for arbitrarily long connected sum neck must limit to a pseudo-holomorphic representative for MATH, where MATH for some MATH. However, since MATH, it follows that MATH. Thus, the gluing map covers the moduli space. |
math/0105202 | First observe that the space of MATH structures over MATH extending a given one on the boundary is identified with MATH. In particular, modulo doubly-periodic domains for the three boundary three-manifolds, every triangle MATH can uniquely be written as MATH for some pair of integers MATH and MATH, where MATH is the generator of the space of triply-periodic domains: in fact, the integer MATH is determined by the intersection number MATH, and MATH can be determined by the signed number of times the arc in MATH obtained by restricting MATH to its boundary crosses some fixed MATH. For the triangles MATH this signed count can be any arbitrary integer, so these triangles represent all possible MATH-equivalence classes of triangles. The other claims are straightforward in the case where MATH. In this case, the curves MATH, MATH, MATH lie in a surface of genus one, so the holomorphic triangle can be lifted to the complex plane. Hence, by standard complex analysis, it is smoothly cut out, and unique. The fact that MATH for higher genus follows from induction, and the gluing result, REF . Specifically, if the result is known for genus MATH, then we can add a new torus MATH to MATH which contains three curves MATH, MATH, MATH which are small Hamiltonian translates of one another (and the basepoint is chosen outside the support of the isotopy). The torus MATH contains a standard small triangle MATH, for which it is clear that MATH. Gluing this triangle to the MATH in MATH, we obtain corresponding triangles in MATH satisfying all the above hypotheses. The fact that MATH for MATH follows similarly, with the observation that the other moduli spaces of triangles on the torus are empty. |
math/0105202 | For any system of coherent orientations, associativity, together with REF , can be interpreted as saying that MATH (up to an overall sign), as a formal sum. Of course, if we are using only MATH coefficients, the proof is complete. More generally, the orientation system for MATH is chosen so that MATH is a cycle. But this leaves the orientation system over MATH unconstrained, and any choice of such orientation system determines the choice over MATH (up to an overall sign depending on the MATH structure used over MATH). Now, the relative sign appearing above corresponds to the orientation of the triangles MATH vs. the triangles MATH over MATH, and each such pair of triangles belongs to different MATH-orbits for the square MATH. Thus, we can modify the relative sign at will. We choose it so that the terms pairwise cancel. |
math/0105202 | It is important to observe that the area filtration defined above is indeed well-defined. The reason for this is that if MATH are a pair of homotopy classes in MATH with MATH, then MATH is a triply-periodic domain. It follows from above that it must have total area zero. Suppose that we have a pair of generators MATH and MATH which are connected by a flow MATH. If MATH is a class with MATH, then, of course, MATH is a class with MATH; thus, MATH; but MATH, as all of its coefficients are non-negative (and at least one is positive). |
math/0105202 | There is a natural filtration on MATH, defined by MATH, where MATH is the class with MATH. This is a filtration, in view of the usual positivity of holomorphic disks (see REF ); indeed, the filtration decreases by at least MATH along flows. The filtration induced by MATH and the map MATH, defined by MATH very nearly agrees with this natural filtration, for sufficiently small MATH. To see this, note that there is a unique ``small" triangle MATH which has non-negative coefficients and is supported inside the support of MATH. Clearly, MATH, and MATH. Now, if MATH is the class with MATH the juxtaposition of MATH can be used to calculate the MATH filtration of MATH; thus MATH. In particular, since MATH decreases by at least MATH along flowlines, MATH, too, must decrease along flows. For MATH, there is another filtration, this one induced by squares. Given MATH, consider MATH with MATH, and let MATH . Indeed, if MATH is the minimum area of any domain in MATH, then MATH decreases by at least MATH along each flowline. Note that MATH. Now, we claim that MATH nearly agrees with the filtration MATH induced by MATH and the right inverse MATH: MATH. Again, if we let MATH denote the point in MATH closest to MATH, there is a unique small triangle MATH. If MATH is a triangle with MATH (that is, used to calculate MATH), then, the juxtaposition MATH is a square which can be used to calculate MATH. But MATH, so since MATH decreases by at least MATH for non-trivial flows, it follows that MATH, too, must decrease along flows. |
math/0105202 | The map MATH counts the number of holomorphic triangles in homotopy classes with MATH, with MATH and MATH. One of these triangles, of course is the canonical small triangle MATH. One can calculate that MATH. This gives the MATH component of MATH. Now, no other homotopy class MATH with MATH has its domain MATH contained inside the support of MATH; thus, if MATH is non-empty, then MATH. Moreover, in the proof of REF , we saw that if MATH is the homotopy class with MATH, then MATH . But now MATH can be used to calculate the filtration MATH. Thus, MATH . Next, we consider MATH. As before, if MATH, we let MATH denote the intersection point closest to MATH. Suppose that MATH has a non-zero component in MATH with MATH; thus, we have a MATH with MATH, which supports a holomorphic triangle. Again, MATH cannot be supported inside the support of MATH, so MATH. Fix MATH (for MATH) with MATH, and MATH with MATH. Clearly, the juxtaposition MATH is a square whose area must agree with the square MATH, where MATH is the canonical small triangle, so MATH and hence MATH. |
math/0105202 | REF provides the null-homotopy MATH: the MATH coefficient of MATH counts holomorphic squares MATH with MATH. Our aim here is to prove that if the MATH component of MATH is non-zero then MATH. Now, the filtration difference between MATH and MATH is calculated (to within MATH) by MATH, where MATH has MATH. Adding the smallest triangle in MATH (and hence changing the area by no more than MATH), we obtain another square MATH with MATH, whose area must agree with the area of MATH. Now if MATH is sufficiently small (MATH), it follows that the filtration difference between MATH and MATH is positive. |
math/0105202 | The proof follows along the lines of REF . In this case, letting MATH be the generating periodic domain in the torus, we have that MATH . We must choose MATH so that it is the MATH corner point for the domain containing the basepoint MATH. Note that MATH meets the reference point MATH with multiplicity MATH. This proves the MATH independence of the intersection number MATH of the choice of MATH. (See REF .) |
math/0105202 | This now follows directly from the picture in the torus. In particular, in the present case, there is no need for REF . |
math/0105202 | This time, the curve MATH is isotopic to the juxtaposition of the MATH-fold juxtaposition of MATH with the MATH. Now, we have MATH different intersection points between MATH and MATH. We choose one (so that the analogue of REF holds, for our given choice of basepoint), and label it MATH. We will have no need for the remaining MATH intersection points. Let MATH denote the intersection point between MATH and MATH, and let MATH, MATH, and MATH be as before. We have a corresponding MATH structure MATH corresponding to MATH. If MATH, there is a unique MATH structure MATH with the property that there is a MATH structure MATH on MATH with MATH, MATH, and MATH. We let MATH. Fix a MATH structure MATH over MATH. We consider the chain map MATH defined by MATH . We define MATH as follows. Consider MATH . This gives us maps: MATH . It follows once again from associativity, together with the analogue of REF , that MATH. We homotope the MATH-curve to the juxtaposition of the MATH-fold multiple of MATH with MATH. This gives a short exact sequence of graded groups MATH . The inclusion follows as before: each intersection point MATH of MATH corresponds a unique intersection point between MATH, which can be canonically connected by a small triangle. To see surjection, note that each intersection point of MATH gives rise to MATH different intersection points between MATH, which we label MATH. Note, however, that MATH. Now, MATH is a generator, so there will always be a unique induced intersection point representing the MATH structure MATH over MATH. The rest follows as before. |
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