paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0101193 | The proof is similar to the one for type B. The only difference is that MATH and MATH are semi-integral which gives condition REFii. |
math/0101193 | Let MATH be a skew diagram (respectively, centrally symmetric) and MATH (respectively, MATH), MATH where MATH (respectively, MATH). Let MATH be such that MATH (respectively, MATH), then by the definition of a skew subdiagram and the fact that MATH, we have MATH (respectively, MATH). Now let MATH be integral centrally symmetric skew diagrams such that MATH and MATH. By the proof of REF , there exists a pair MATH, MATH such that MATH and MATH are southwest corners of MATH and MATH respectively. Thus neither MATH or MATH is contained in MATH or MATH for any pair MATH. Now one verifies easily from REF that the subalgebra spanned by the elements MATH or MATH where Type A - MATH, Types B or C - MATH, Type D - MATH and the pair in MATH as given in the proof of REF which was described above, form an abelian subalgebra of MATH. By REF , its dimension is MATH. |
math/0101193 | By the classification REF and REF , this is just a straightforward case by case computation. |
math/0101194 | If the rank of MATH is REF, there is of course nothing to prove. Since MATH is smooth and has dimension REF, MATH is a subbundle. As is well known, MATH stabilizes MATH, for it takes values in MATH. Furthermore, each MATH satisfies the same assumptions as MATH. Let us thus first consider MATH. If MATH, one performs NAME 's construction REF : MATH embedds into MATH, for MATH, with cokernel MATH, such that MATH extends as a logarithmic connection, the residue of which has the new eigenvalues MATH. Furthermore if the local generators in MATH of MATH are MATH with MATH generating MATH and MATH being an eigenvector to MATH, then the new generators are MATH. In this basis, one has MATH. Repeating the procedure MATH times, one reaches a new MATH with residue MATH. Now by CITE again, this implies that the local monodromy underlying MATH is diagonal (actually even a homothety), a contradiction. Thus MATH. Replacing now MATH by MATH, one proceeds inductively. |
math/0101194 | Let MATH be the algebraic regular connection on MATH with underlying MATH. Then MATH, where MATH underlies MATH. Let MATH be a bundle such that MATH has logarithmic poles in MATH. Then MATH is a subbundle, stabilized by MATH. Let us denote by MATH the induced connection. Then its residue is the restriction of the residue of MATH to MATH. Since MATH in all points MATH, a fortiori none of the eigenvalues of MATH can be an eigenvalue of MATH. Consequently, one has MATH . On the other hand, since MATH is torsion free and supported in MATH, one has MATH, thus MATH is a locally free subsheaf, isomorphic to MATH away of MATH, and thus isomorphic to MATH by REF . If now moreover MATH is semistable, then MATH is semistable as well, and one has MATH. This contradicts REF . |
math/0101194 | For MATH irreducible see CITE. Suppose that MATH is reducible. Let MATH be a vector bundle on MATH with logarithmic connection with poles only in MATH which realizes MATH. Let MATH be a subsystem of rank REF. We denote by MATH the inclusion and define MATH . Then MATH is an exact sequence of bundles, and MATH and MATH are equipped with the induced connection. Then MATH will be semistable if MATH. REFst case: For each MATH the two eigenvalues of the local monodromy around MATH coincide. Following NAME, one can choose MATH such that at each MATH the two residue eigenvalues coincide. Thus in particular, MATH, and MATH is semistable. REFnd case: For some MATH the two eigenvalues of the local monodromy around MATH differ. Let MATH be again a NAME extension. Then the space MATH splits into two one-dimensional eigenspaces MATH and MATH of the residue endomorphism MATH. One can apply NAME 's construction CITE REF to either one of these eigenspaces and increase by one either the degree of MATH or that of MATH. Repeating this one can obtain bundles MATH with logarithmic connections such that MATH. Then MATH is semistable. |
math/0101195 | We can rewrite the hypothesis as MATH and the conclusion as MATH . The claim then follows from NAME. |
math/0101195 | Define the sets MATH by MATH and MATH by MATH . The required properties on MATH and MATH are then easily verified. |
math/0101195 | By REF we see that MATH for all cubes MATH of side-length MATH, if MATH is a sufficiently large constant. The claim then follows by covering MATH with such cubes, extracting the top MATH cubes in that collection which maximize MATH, picking two of those cubes MATH, MATH which are not adjacent, and setting MATH for MATH. We leave the verification of the desired properties to the reader. |
math/0101195 | Most of these results are in CITE. For the last result, observe that the discrete function MATH has a MATH norm MATH and is supported in a set of cardinality MATH by the results in CITE. Thus one can find a MATH such that MATH, and the claim follows by setting MATH. |
math/0101195 | The lower bound is clear from REF and the multiplicative version of REF, so it suffices to show the upper bound. Fix MATH, MATH, MATH. By multiplying REF by MATH, which is MATH, we see that MATH . Integrating this over all possible values of MATH and using NAME 's theorem we obtain MATH . On the other hand, from REF and the multiplicative form of REF we have MATH . The claim follows by combining the above two estimates. |
math/0101195 | As before, the lower bound is immediate from the additive and multiplicative versions of REF, so it suffices to show the upper bound. Fix MATH, and let MATH denote the set in REF. Observe that for all MATH we have the telescoping identity MATH where MATH . Indeed, we have the identities MATH . As a consequence of these identities, REF and some algebra we see the MATH is a diffeomorphism on MATH (recall that MATH, MATH, MATH, MATH are fixed). From REF we thus have MATH . Integrating this over all values of MATH and using NAME 's theorem we obtain MATH . The claim then follows from REF . |
math/0101195 | Fix MATH. We may assume that MATH is sufficiently small depending on MATH, and MATH is sufficiently small depending on MATH and MATH, since the claim is trivial otherwise. By reflection symmetry it suffices to prove the first estimate. Suppose for contradiction that MATH . From NAME REF we thus have MATH . Write MATH, MATH. Observe that MATH . This is because for fixed MATH, MATH can only range in a set of measure MATH thanks to REF and the fact that MATH is a MATH set. Subtracting the two inequalities we obtain (if MATH is sufficiently small) MATH . Since MATH, we may thus find MATH such that MATH and MATH . From Refinement REF MATH can be covered by MATH intervals in MATH of length MATH. From this fact, REF, and the geometry of annuli which intersect non-tangentially, we see that the set in REF can be covered by MATH balls of radius MATH. Since MATH is a MATH set, we see from REF that MATH . But this contradicts REF if MATH is sufficiently small. This concludes the proof of the proposition. |
math/0101195 | From REF it suffices to show that MATH . (The constant MATH is non-optimal, but this is irrelevant for our purposes). In order to have MATH one must either have MATH, or that MATH and MATH is within MATH of the line joining MATH. Let us consider the contribution of the former case. Since MATH is a MATH set, we see that each pair MATH contributes a set of measure MATH to REF. From NAME 's theorem we thus see that the contribution of this case to REF is acceptable. Now let us consider the contribution of the latter case. By NAME 's theorem again it suffices to show that MATH for any strip MATH of width MATH. Fix MATH. From the definition of MATH and NAME 's theorem it suffices to show that MATH . From the definition of MATH, we can estimate the left-hand side by MATH . Since MATH is a MATH set, we have MATH. Also, if MATH makes an angle of MATH with MATH we have MATH by REF and elementary geometry, otherwise we may estimate MATH. Inserting these estimates into the previous and using the MATH-separated nature of the MATH, we see that MATH as desired. |
math/0101195 | From elementary geometry we see that MATH whenever MATH, where MATH is the NAME maximal function of MATH. From NAME REF we thus have MATH . From NAME we thus have MATH . If one defines the constants in MATH appropriately, the claim then follows from REF. |
math/0101195 | Fix MATH. From REF it suffices to show that MATH . From the definition of MATH and the fact that MATH, we can find MATH such that MATH and REF holds. From elementary geometry we see MATH stays within MATH of the line MATH . Since MATH is MATH-discretized, we thus have (if MATH is sufficiently large) MATH . The separation conditions MATH are easily imposed by REF, since MATH is a MATH set. |
math/0101195 | Since MATH, it suffices to show that MATH for all MATH, MATH in MATH, MATH respectively. Fix MATH, MATH. By definition of MATH it suffices to show that MATH . Since MATH, it thus suffices to show that MATH for all MATH. Fix MATH. The set in REF is contained in an annular arc of thickness MATH, angular width MATH, and radius MATH centered at MATH. From REF and elementary geometry that the possible values of MATH thus lie in an interval of length MATH. Since MATH is a MATH set, we thus see that the possible values of MATH are contained in the union of MATH intervals of length MATH. From REF and elementary geometry we thus see that the set in REF can be covered by MATH balls of radius MATH. The claim follows. |
math/0101195 | We shall use NAME 's argument, using REF for MATH as a substitute for the direction-separation property. Expand out the left-hand side as MATH . Since MATH, it thus suffices to show that MATH for all MATH. Fix MATH. The quantity MATH can vary from REF to MATH. We need only consider the contribution when MATH, since the remaining contribution is trivial to handle. By dyadic pigeonholing and absorbing the logarithmic factor into the MATH symbol, it suffices to show the distributional estimate MATH for all dyadic MATH. Fix MATH. The set MATH lies within MATH of the perpendicular bisector of MATH and MATH, and within MATH of MATH and MATH, which are themselves separated by MATH. Similarly for MATH. From elementary geometry we thus see that MATH . Since MATH lie within a MATH-separated subset of MATH, which is a MATH set, we see from REF that MATH . The claim follows. |
math/0101195 | For any integer MATH, we have MATH. Since NAME content is sub-additive, we thus have MATH . The claim then follows by letting MATH. |
math/0101195 | We first prove the latter claim. Since MATH is a MATH set we can cover it (if the constants are chosen appropriately) by about MATH balls of radius MATH, so that MATH . The claim then follows from REF . Now we show the former claim. Fix MATH. For every hyper-dyadic number MATH, we can find a collection MATH of balls covering MATH such that MATH and MATH . By reducing the constant MATH slightly we may assume that the MATH are hyper-dyadic. For each hyper-dyadic MATH, let MATH denote the set MATH . Clearly the sets MATH strongly cover MATH as MATH, MATH both vary. Fix MATH, MATH, and let MATH be a collection of hyper-dyadic cubes MATH of side-length at least MATH which cover MATH and which minimize the quantity MATH where MATH denotes the side-length of MATH. Such a minimizer exists since there are only a finite number of hyper-dyadic cubes which are candidates for inclusion in MATH. From REF one can cover MATH by at most MATH cubes of side-length MATH, hence MATH . In particular, we have MATH for all MATH. From the construction of MATH we see that the MATH are all disjoint, and for all hyper-dyadic cubes MATH we have MATH since we could otherwise remove those cubes in MATH from MATH and replace them with MATH, contradicting minimality. For each dyadic MATH, let MATH denote the set MATH . Clearly MATH is a MATH-discretized set. From REF we see that MATH is in fact a MATH set. Now define MATH. From the constraints MATH and MATH we see that there are at most MATH pairs MATH associated to each MATH. Hence MATH is also a MATH set. By construction we see that the MATH strongly cover MATH, and so we are done. |
math/0101195 | The proof of trivial if MATH is large, so we will assume that MATH is sufficiently small depending on MATH, MATH. From NAME REF with MATH and NAME 's inequality we have MATH . Thus to show REF it suffices to show that MATH . Suppose for contradiction that REF failed. Let MATH be a maximal MATH-separated subset in the set in REF; we thus have MATH . By construction, for each MATH we can find a line MATH in the direction MATH such that REF holds. Let MATH denote the set MATH. From the construction of MATH we thus have MATH . Let MATH be a collection of squares MATH of side-length MATH which covers MATH and which minimizes the quantity MATH . As in the proof of REF , a minimizer MATH exists and the squares in MATH are disjoint and satisfy MATH for all squares MATH. Also, for all MATH we have MATH since otherwise we could replace MATH by all the MATH-cubes contained in MATH, contradicting the minimality of MATH. Summing this over all MATH we obtain MATH . From REF we thus obtain MATH . We thus have MATH . If we choose MATH sufficiently large, we thus have (for MATH sufficiently small) MATH . In particular, we have MATH . On the other hand, since MATH is bad with respect to MATH, we have MATH . Thus, if we let MATH denote the set MATH then we have MATH . Since MATH is MATH-discretized, we have in particular that MATH . The set MATH is covered by the dilates of those cubes MATH for which MATH. From this and REF we see that MATH is a MATH set but with MATH replaced by MATH. From REF we thus have MATH . On the other hand, from Separation REF we have MATH . Summing this on MATH using REF and noting that each MATH can be in at most MATH of the above sets, we obtain MATH . If MATH is sufficiently small with respect to MATH we obtain the desired contradiction, if MATH is sufficiently small. |
math/0101195 | Let MATH be a constant to be chosen later. We shall show that MATH from which REF follows from a suitable choice of MATH. Partition MATH, where MATH . Let us first deal with the contribution to REF of the case when at least one of MATH is in MATH. By NAME 's theorem and symmetry it suffices to show that MATH . By NAME REF it suffices to show that MATH . Let us first consider the contribution of the case MATH. For each MATH, MATH, the set of MATH which contribute to the above expression has measure MATH by REF, and so this contribution is acceptable by NAME 's theorem. It thus remains to show MATH . By NAME 's theorem again, it suffices to show that MATH for all MATH such that MATH. Fix MATH, MATH. By REF again, it suffices to show that MATH . The set MATH can be covered by MATH intervals of length MATH. Let MATH denote the collection of those intervals MATH in this cover such that MATH. It suffices to show that MATH . For fixed MATH, MATH, the set described above is contained in an annular arc of thickness MATH, radius MATH, and angular width bounded by MATH as one can easily compute using elementary geometry. By the construction of MATH and a simple covering argument, we can thus bound the left-hand side of REF by MATH . To complete the proof of REF it thus suffices to show that MATH for all MATH. But this follows easily by dyadically decomposing the MATH based on MATH and noting from REF that for each MATH, there are MATH intervals MATH for which MATH. Note that any logarithmic factors can be absorbed into the MATH notation. To conclude the proof of REF it remains to show that MATH . From the definition of MATH and REF we see that MATH is contained in a MATH set MATH. From REF and a covering argument we have MATH . The claim then follows from REF. |
math/0101195 | Fix MATH. The space of all strips of width MATH which intersect MATH is a two-dimensional manifold, which we can endow with a smooth metric MATH. Let MATH be a maximal MATH-separated subset of this space of strips; note that MATH. For each MATH, let MATH denote the index MATH which maximizes the quantity MATH and for each MATH, define MATH to be the set MATH . Clearly the sets MATH are contained in MATH and form a finitely overlapping cover of MATH. We shall show the preliminary estimate MATH . It suffices to show MATH . From the bounds on MATH it suffices to show that MATH for all MATH such that MATH. Fix MATH, MATH, and rewrite the above as MATH . By NAME it suffices to show that MATH . By definition of MATH, we have MATH so it suffices to show that MATH . This will obtain if we can show MATH . We first consider the contribution of the case when MATH. In this case we estimate MATH integral by REF and then integrate in the MATH and MATH variables to show that the contribution of this case is acceptable. Similarly we can handle the case MATH. Thus it remains to show that MATH . Suppose MATH is in the above set. Since MATH, we see from elementary geometry that MATH . Thus the desired claim follows from REF, if MATH is sufficiently large. The MATH have most of the properties that we desire for MATH, but need not be covered by a small number of annuli. To remedy this we shall refine MATH slightly. Fix MATH and perform the following algorithm. Initialize MATH to be the empty set. If one has MATH for all MATH, we terminate the algorithm. Otherwise, we choose a MATH for which REF fails, and add the set in REF to MATH. We then repeat this algorithm, continuing to enlarge MATH until REF is finally satisfied for all MATH. Since each iteration of this algorithm adds a set of measure MATH to MATH, this algorithm must terminate after at most MATH steps. Since MATH is a MATH set, we see that each set of the form REF can be covered by annuli MATH with width MATH, radii MATH and MATH-separated, and center at MATH. Thus we have the desired covering REF. It remains to prove REF. From REF and the bounds on MATH it suffices to show that MATH for all MATH. Since MATH it suffices by symmetry to show that MATH for all MATH. But this follows by integrating REF over all MATH. |
math/0101195 | Fix MATH; by symmetry we may assume that MATH. By NAME 's theorem it suffices to show that MATH for all MATH. Fix MATH. Since MATH and MATH are fixed, there are significant constraints on the number of MATH which can contribute to REF. Indeed, if there are two values of MATH, say MATH and MATH, which contribute to REF, then MATH cannot exceed MATH and MATH cannot exceed MATH, due to the presence of REF in the definition of MATH, MATH. Because MATH is constrained to be hyper-dyadic, we thus see that there are at most MATH values of MATH which contribute to REF. Thus it suffices to show REF for a single value of MATH. Since the MATH are finitely overlapping as MATH varies, we see that for each MATH there are at most MATH values of MATH which contribute to REF. Hence it suffices to show that MATH for all MATH. Fix MATH. We may of course assume that REF holds, else REF is vacuously true. By definition of MATH and the fact that MATH it suffices to show that MATH for all MATH. Fix MATH. By REF it suffices to show that MATH for all MATH. Fix MATH. The set MATH is contained in a rectangle of width MATH, hence contained in a rectangle MATH of width MATH by REF. Let MATH denote the distance from MATH to MATH. From elementary geometry we see that MATH is the union of two sets, each of which having diameter at most MATH where MATH is the outer radius of MATH. From REF we thus have MATH . If we arrange the MATH in order of distance from MATH, we have MATH since the MATH are MATH-separated. Since MATH, the claim then follows from REF. |
math/0101201 | Denote by MATH the function MATH. The Airy function is a solution of the Airy differential equation MATH. It follows that MATH . This translates into a first order ODE for MATH. MATH from which it follows that MATH where MATH is some function of MATH and MATH. The obvious equations MATH translate into the equations MATH which imply that MATH is constant. The fact that MATH follows by taking the NAME transform of REF with MATH and comparing it with the known formula for it, see REF. |
math/0101203 | It will be convenient to recast REF so that the solution has zero trace on MATH. For any boundary data MATH, we may choose MATH such that MATH. Let MATH . We rewrite the system REF as an evolution equation in MATH where MATH . We define the vector MATH since MATH-class vector fields form a NAME ring for MATH, we may define the maps MATH and MATH . Thus, the vector MATH . We are using the fact that the projector MATH maps MATH to itself. To see this, we write MATH where MATH solves the NAME problem MATH . Since MATH is in MATH and MATH is in MATH, by elliptic regularity MATH is in MATH so that MATH is in MATH, as desired. One sees that MATH is also in MATH by a similar argument. Next, we define the semigroup MATH . We can now express the system REF as the integral equation MATH . Since MATH and MATH are strongly continuous semigroups, it follows that MATH and that for MATH, MATH; furthermore, we have the usual estimate (see, for example, CITE) MATH . Using the fact that for MATH, MATH is a bounded projection, we obtain that MATH namely, MATH where MATH and MATH depend on MATH, MATH, MATH, MATH, and MATH. Fix MATH and set MATH . We want to choose MATH sufficiently small so that MATH is a contraction. By REF , we can choose MATH so that MATH . If MATH, then by REF we have a bound MATH . Using REF , we have that MATH hence, for MATH, and with MATH, MATH . Choosing MATH small enough so that MATH, we see that by the contraction mapping theorem, MATH has a unique fixed point in MATH, and this proves the theorem. |
math/0101203 | We keep the same notation as in the proof of REF . For MATH, we have that MATH . Thus, it suffices to prove that for MATH, the map MATH is locally NAME. Using REF CITE, we have that for MATH. It follows that MATH, MATH, and MATH. The fact that MATH follows because MATH forms a NAME ring. Hence, MATH is indeed locally NAME, and the remainder of the proof is identical to the one for REF . |
math/0101203 | We compute the pointwise inner-product of REF with MATH, and use the fact that MATH. Hence, MATH, we obtain MATH . Now suppose that MATH occurs at MATH, an element of the parabolic interior; then MATH and MATH. This implies that MATH but MATH which is a contradiction. |
math/0101203 | This again follows from the maximum principle above. |
math/0101203 | Using REF , we rewrite REF as MATH where MATH is any local orthonormal frame. Adding the MATH inner-product of REF with MATH to the MATH inner-product of REF , we obtain the basic energy law MATH . In the case of a flat manifold, such as a bounded domain in MATH, MATH, and REF reduces to the basic energy law REF. From REF , we have that MATH . It follows that MATH where the second and third inequalities follow from NAME 's REF , and the last inequality follows from the NAME inequality for MATH, a positive constant depending on MATH. Taking MATH sufficiently small so that MATH the basic energy law REF on a Riemannian manifold yields the following differential inequalities: MATH where MATH . Using the classical NAME lemma, we obtain MATH . Thus, MATH . |
math/0101203 | When MATH, from the energy law REF , we see that there exists and MATH absorbing set, so that for some MATH all bounded subsets of MATH will enter the MATH ball of radius MATH. For MATH, we take the pointwise inner-product of REF with MATH and integrate over MATH to obtain the differential inequality MATH . Using the interpolation inequality (see CITE for details and further applications) MATH we see that MATH . Using NAME 's trick in the differential REF , we get a uniform bound for MATH which is independent of MATH (even if the constant MATH tends to infinity), and thus we may pass to the limit as MATH. |
math/0101203 | It follows from REF that MATH so MATH . Therefore, MATH . Now let MATH . Using REF , we have that MATH . In the case that MATH, it follows from a similar argument as in REF that for some constants MATH, MATH . When MATH, we find that for MATH, MATH . The last term is bounded by MATH, so by taking MATH sufficiently small and adjusting the constants as necessary, we see that REF still holds. Thus, using REF and appealing to the uniform NAME lemma (see, for example, CITE), we see that MATH . Because of REF , we may extract a uniform bound for MATH. Hence, we have an a priori uniform bound for MATH in the MATH topology and for MATH in the MATH topology. |
math/0101203 | Taking the MATH inner-product of REF with MATH and adding the MATH inner-product of REF with MATH, we find that MATH . We shall estimate each of the nonlinear terms on the right-hand-side of REF ; as we showed in the proof of REF , the projection MATH acting on the nonlinear terms, maps MATH into itself continuously, so it suffices to estimate MATH and MATH in the third and fourth terms. Using REF , we may interpolate the nonlinear terms between MATH and MATH and MATH and MATH, respectively. We have that MATH where we set MATH so that MATH, and the last equality follows from the fact that MATH, since MATH. For MATH, we use REF to estimate MATH . Using NAME 's inequality, MATH it follows that MATH . For the next term, we have that MATH where MATH. In the case that MATH, REF is bounded by MATH where the first inequality follows from repeated use of REF , and the last inequality follows from MATH, where MATH. One more application of REF shows that REF is bounded by MATH . In the case that MATH, REF is bounded by MATH where the first inequality follows from REF , and the last two inequalities follow from REF . It follows that MATH . We next compute that MATH since for MATH, we have that MATH. We estimate the case MATH first in REF : MATH where the last two inequalities follow from two applications of the NAME 's inequality. For the cases MATH, REF is bounded by MATH, so by REF , we find that for MATH, MATH where we have used NAME 's inequality for the last step. Another application of NAME 's inequality yields the estimate MATH . For the final nonlinear term, we have that MATH . Using the estimate MATH together with REF and NAME 's inequality, we have that MATH . Letting MATH and taking MATH sufficiently small so that MATH the basic REF takes the form MATH . Letting MATH, the classical NAME lemma gives MATH so that MATH . Thus, since the time interval of existence from REF only depends on the initial data, the a priori bound REF together with the continuation property gives the global well-posedness result. Moreover, because of the absorbing sets that exist in MATH by virtue of REF , we obtain using REF, the global attractor that we asserted. |
math/0101206 | If we surger out the MATH, we replace MATH by the two-sphere MATH, with MATH marked points MATH (that is, the pair MATH corresponds to the zero-sphere which replaced the circle MATH in MATH). Now, MATH induces a NAME curve MATH-Â in this two-sphere. If MATH does not separate any of the MATH from the corresponding MATH, then it is easy to see that the original curve MATH had to be null-homologous. On the other hand, if MATH is separated from MATH, then it is easy to see that MATH is obtained by handlesliding MATH across some collection of the MATH for MATH. |
math/0101206 | This is proved by induction on MATH. The case MATH is obvious: if two embedded curves in the torus represent the same generator in homology, they are isotopic. Next, if the two subsets have some element, say MATH, in common, then we can reduce the genus, by surgering out MATH. This gives a new NAME surface MATH of genus MATH with two marked points. Each isotopy of a curve in MATH which crosses one of the marked points corresponds to a handleslide in MATH across MATH. Thus, by the inductive hypothesis, the two subsets are related by isotopies and handleslides. Consider then the case where the two subsets are disjoint, labeled MATH and MATH. Obviously, MATH is not null-homologous, so, according to REF , after renumbering, we can obtain MATH by handlesliding MATH across some collection of the MATH (MATH). Thus, we have reduced to the case where the two subsets are not disjoint. |
math/0101206 | We begin by proving the isomorphism on the level of homology. There is an obvious map MATH induced from the inclusion MATH. To invert this, note that a curve (in general position) in MATH corresponds to a map of a MATH-fold cover of MATH to MATH, giving us a homology class in MATH. This gives a well-defined map MATH, since a cobordism MATH in MATH, which meets the diagonal transversally gives rise to a branched cover MATH which maps to MATH. It is easy to see that these two maps are inverses of each other. To see that MATH is Abelian, consider a null-homologous curve MATH, which misses the diagonal. As above, this corresponds to a map MATH of a MATH-fold cover of the circle into MATH, which is null-homologous; that is, there is a map of a two-manifold-with-boundary MATH into MATH, MATH, with MATH. By increasing the genus of MATH if necessary, we can extend the MATH-fold covering of the circle to a branched MATH-fold covering of the disk MATH. Then, the map sending MATH to the image of MATH under MATH induces the requisite null-homotopy of MATH. |
math/0101206 | The isomorphism MATH is given by the intersection number with the submanifold MATH, for generic MATH. Specifically, if we take a hyperelliptic structure on MATH, the hyperelliptic involution gives rise to a sphere MATH, which we can then use to construct a sphere MATH. Clearly, MATH maps to MATH under this isomorphism. Consider a sphere MATH in the kernel of this map. By moving MATH into general position, we can arrange that MATH meets MATH transversally in finitely many points. By splicing in homotopic translates of MATH (with appropriate signs) at the intersection points, we can find a new sphere MATH homotopic to MATH which misses MATH; that is, we can think of MATH as a sphere in MATH. Note that since this splicing construction makes no reference to a basepoint, this operation is taking place in MATH. We claim that MATH, for MATH. One way to see that MATH is to observe that MATH is homotopy equivalent to the wedge of MATH circles or, equivalently, the complement in MATH of MATH points MATH. Now, MATH can be thought of as the space of monic degree MATH polynomials MATH in one variable, with MATH for MATH. Considering the coefficients of MATH, this is equivalent to considering MATH minus MATH generic hyperplanes. REF states that the homology groups of the universal covering space of this complement are trivial except in dimension zero or MATH. This proves that MATH and hence completes the proof that MATH for MATH. In the case where MATH it is easy to see that MATH is diffeomorphic to the blowup of MATH (indeed, the NAME map gives the map to the torus, and the exceptional sphere is the sphere MATH induced from the hyperelliptic involution on the genus two NAME surface). In this case, the calculation of MATH is straightforward. To verify the second claim, note that the NAME dual of MATH is characterized by the fact that: MATH where the latter equation holds for all MATH. It is easy to see that MATH satisfies these properties, as claimed. Finally, in the case where MATH, we verify that the action of MATH is trivial. Fix maps MATH and MATH . According to REF , we can arrange after a homotopy that MATH where MATH, and according to our calculation of MATH, we can arrange that MATH has the form MATH, where MATH. Now, the map MATH admits an obvious extension MATH . Since the action of MATH on MATH is trivial, the claim now follows immediately. |
math/0101206 | See CITE for the calculation of MATH. The rest follows from this, together with REF . |
math/0101206 | Note that the projection map MATH is a holomorphic local diffeomorphism away from the diagonal subspaces (consisting of those MATH-tuples for which at least two of the points coincide). Since MATH misses the diagonal, the claims about MATH follow immediately from the fact that MATH is a totally real submanifold (for the product complex structure), which follows easily from the definitions. |
math/0101206 | Suppose that MATH. The space MATH is naturally identified with the fundamental group of the space MATH of paths in MATH joining MATH to MATH, based at the constant REF path. Evaluation maps (at the two endpoints of the paths) induce a NAME fibration (with fiber the path-space of MATH): MATH whose associated homotopy long exact sequence gives: MATH . But under the identification MATH, the images of MATH and MATH correspond to MATH and MATH respectively. Hence, after comparing with the cohomology long exact sequence for MATH, we can reinterpret the above as a short exact sequence: MATH . The homomorphism MATH provides a splitting for the sequence. The proposition in the case where MATH follows. The case where MATH follows similarly, only now one must divide by the action of MATH. In the case where MATH and MATH, then MATH is non-empty, so the above reasoning applies. |
math/0101206 | Write MATH (where, by assumption, MATH). If MATH is the domain MATH, then we let MATH denote the coefficient MATH. The surface MATH is constructed as an identification space from MATH where MATH is a diffeomorphic copy of the domain MATH. The MATH-curves are divided up by the MATH-curves into subsets, which we call MATH-arcs; and similarly, the MATH-curves are divided up by the MATH-curves into MATH-arcs. Each MATH or MATH-arc MATH is contained in two (not necessarily distinct) domains, MATH and MATH. We order the domains so that MATH . MATH is obtained from MATH by the following identifications. For each MATH-arc MATH, if MATH, then for MATH, we identify MATH where MATH. Similarly, for each MATH-arc MATH, if MATH, then for MATH, we identify MATH . The map MATH, then, is induced from the natural projection map from MATH to MATH. It is easy to verify that the space MATH is actually a manifold-with-boundary as claimed. |
math/0101206 | For MATH, let MATH be the gradient flow line passing through MATH (connecting the index zero to the index three critical point). Clearly, MATH. Now the difference MATH is a closed loop in MATH, which is clearly homologous to a loop in MATH which meets the attaching disk for MATH in a single transverse point (and is disjoint from the attaching disks for MATH for MATH). The formula then follows. |
math/0101206 | Given MATH, let MATH denote the MATH trajectories for MATH connecting the index one to the index two critical points which contains the MATH-tuple MATH; similarly, given MATH, let MATH denote the corresponding trajectory from the index zero to the index three critical point. Thus, if MATH, MATH is a closed loop in MATH. A representative for MATH is obtained by modifying the vector field MATH in a neighborhood of MATH. It follows then that MATH can be represented by a cohomology class which is compactly supported in a neighborhood of MATH (we can use the same vector field to represent both MATH structures outside this neighborhood). It follows that the difference MATH is some multiple of the NAME dual of MATH (at least if the curve is connected; though the following argument is easily seen to apply in the disconnected case as well). To find out which multiple, we fix a disk MATH transverse to MATH; to find such a disk take some MATH so that MATH (if no such MATH can be found, then MATH, and REF is trivial), and let MATH be a small neighborhood of MATH in MATH. Our representative MATH of MATH can be chosen to agree with MATH near MATH; and the representative MATH for MATH can be chosen to agree with MATH over MATH. With respect to any fixed trivialization of MATH, the two maps from MATH to MATH corresponding to MATH and MATH agree on MATH. It makes sense, then, to compare the difference between the degrees MATH and MATH (maps from the disk to the sphere, relative to their boundary). Indeed, MATH . To calculate this difference, take another disk MATH with the same boundary as MATH, so that MATH bounds a three-ball in MATH containing the index one critical point corresponding to MATH (and no other critical point); thus we can assume that MATH over MATH. Now, since MATH does not vanish inside this three-ball, we have: MATH . Thus, MATH since MATH vanishes with winding number MATH around the index MATH critical points of MATH. It follows from this calculation that MATH. Letting MATH be a collection of arcs with MATH, and MATH be such a collection with MATH, we know that MATH represents MATH. On the other hand, if MATH is one of the arcs which connects MATH to MATH, then it is easy to see that MATH is homotopic relative to its boundary to the segment in MATH formed by joining the two gradient trajectories connecting MATH and MATH to the index one critical point. It follows from this (and the analogous statement in MATH) that MATH is homologous to MATH. REF follows. REF follows from similar considerations. Note first that MATH agrees with MATH away from MATH. Letting now MATH be a disk which meets MATH transversally in a single positive point (and is disjoint from MATH), and MATH be a disk with the same boundary as MATH so that MATH contains the index zero critical point, we have that MATH (note now that MATH vanishes with winding number MATH around the index zero critical point). It follows that MATH. Now, MATH is easily seen to be NAME dual to MATH. |
math/0101206 | In a neighborhood of MATH, we are using an integrable complex structure, so the disk MATH must either be contained in the subvariety (which is excluded by the boundary conditions) or it must meet it non-negatively. |
math/0101206 | It follows from the excision principle for the index that attaching a topological sphere MATH to a disk changes the NAME index by MATH (see CITE, CITE). On the other hand for the positive generator we have MATH according to REF . |
math/0101206 | Given MATH we consider the lift MATH obtained by pulling back the branched covering space MATH. (That is to say, MATH is defined to be the covering space of the image of MATH away from the diagonal MATH, and in a neighborhood of MATH, MATH is defined as a subvariety of MATH - it is here that we are using the fact that each of the MATH agree with the standard complex structure near MATH.) We break the energy integral into two regions: MATH . To estimate the integral on MATH, we use the fact that each MATH tames MATH, from which it follows that there is a constant MATH for which MATH where MATH. To estimate the other integrand, choose a NAME form MATH over MATH. Over MATH all the MATH agree with MATH, so MATH is MATH-holomorphic in that region, so there is some constant MATH with the property that MATH (the constant MATH depends on the Riemannian metric used over MATH and the choice of NAME form MATH). Moreover, the right hand side can be calculated using MATH according to the following formula: MATH . Now, fix any two-form MATH over MATH. Then there is a constant MATH with the following property. Let MATH be any map which is MATH-holomorphic on MATH. Then we have the inequality MATH . This holds for the constant with the property that for each tangent vector MATH to MATH and MATH where MATH. Such a constant can be found since MATH is compact and MATH determines a non-degenerate quadratic form on each tangent space MATH. Applying REF for the form MATH, and combining with REF , we find a constant MATH with the property that MATH . Moreover, with respect to the symplectic form MATH, the preimage under MATH of MATH and MATH are both Lagrangian. This gives a topological interpretation to the right-hand-side of REF : MATH which makes sense since MATH defines a relative cohomology class in MATH. Note that the correspondence MATH induces a right inverse, up to a multiplicative constant, to the map on homology MATH thus, the homology class MATH depends only on the relative homology class of MATH, thought of as a class in MATH - in particular, it depends only on the equivalence class MATH of MATH. Thus, given a class MATH, this gives us an a priori bound on the MATH-energy of the (branched) lift of any holomorphic disk MATH, combining REF , and REF , we get that MATH (for some constant MATH independent of the class MATH). |
math/0101206 | Given a holomorphic map MATH which does not lie in the diagonal, we can find a branched MATH-fold cover MATH pulling back the canonical MATH-fold cover MATH, that is, making the following diagram commutative: MATH . Indeed, MATH inherits an action by the symmetric group on MATH letters MATH, and MATH is equivariant for the action (and its quotient is MATH). Let MATH denote projection onto the first factor. Then, the composite map MATH is invariant under the action of MATH consisting of permutations which fix the first letter. Then, we let MATH, and MATH be the induced map from MATH to MATH. It is easy to verify that MATH has the desired properties. |
math/0101206 | The hypotheses ensure that for all MATH, we have that MATH where MATH for any MATH. This is true because the MATH-injectivity hypothesis ensures that the corresponding map MATH, coming from REF , is injective (with injective linearization, by elementary complex analysis) on the region mapping to the MATH-curves MATH. Thus, following CITE, by varying the MATH in a neighborhood of the MATH, one can see that the map MATH is a smooth point in a parameterized moduli space (parameterized now by variations in the curves). Thus, according to the NAME theorem, for generic small variations in the MATH, the corresponding moduli spaces are smooth. |
math/0101206 | The spheres in MATH for the complex structure MATH are all contained in the set of critical points for the NAME map MATH which is a degree one holomorphic map. Thus, the set of spheres is contained in a subset of real codimension two. |
math/0101206 | To investigate compactness, note first that a sequence of elements in MATH has a subsequence which either bubbles off spheres, or additional disks with boundaries lying in MATH. However, it is impossible for a sequence to bubble off a null-homotopic disk with boundary lying in MATH, since such disks must be constant, as they have no energy (according to the proof of REF , see REF ). Moreover, sequences in MATH cannot bubble off homotopically non-trivial disks, because then one of the components in the decomposition would have negative MATH-integral, and such homotopy classes have no holomorphic representatives. This argument also rules out bubbling off spheres, except in the special case where the subsequence converges to a single sphere (more precisely, the constant disk mapping to MATH, attached to some sphere). But this is ruled out by our hypothesis on MATH, which ensures that for any MATH sufficiently close to MATH, the MATH-holomorphic spheres are disjoint from MATH. To prove smoothness, note first that any holomorphic disk in MATH for MATH has a dense set of injective points. To see this, fix any point MATH. The intersection number of MATH with MATH is MATH, and both are varieties; it follows that there is a single point of intersection, that is, there is only one MATH for which MATH. Thus, MATH is injective in a neighborhood of MATH. It follows that for any MATH sufficiently close to MATH, all the MATH-holomorphic degenerate disks are injective in a neighborhood of MATH. Thus, according to the usual proof of transversality, these pairs are all smooth points in the parameterized moduli space. Thus, the result follows from the NAME theorem. |
math/0101206 | Fix a genus one NAME surface MATH. Let MATH denote the complex structure on MATH, thought of as the connected sum of MATH copies of MATH, connected along cylinders isometric to MATH. As MATH, the NAME surface degenerates to the wedge product of MATH copies of MATH, MATH. If for each MATH, the moduli space were non-empty, we could take the NAME limit of a sequence MATH in MATH to obtain a holomorphic map MATH into MATH (a linear chain of MATH tori meeting in MATH nodes). (In this argument, we have a one-parameter family of symmetric products, which we can embed into a fixed NAME manifold, where we can apply the usual NAME compactness theorem, see also REF.) The latter symmetric product decomposes into irreducible components MATH . These components meet along loci containing the connected sum points for the various MATH. Moreover, the torus MATH can be viewed as a subset of the irreducible component MATH (corresponding to all MATH). The NAME limit MATH then consists of a holomorphic disk MATH with boundary mapping into MATH, and a possible collection of spheres bubbling off into the other irreducible components. But MATH, so it follows that MATH is constant, mapping to MATH (which is disjoint from the connected sum points). Since MATH misses the other components of the symmetric product, it cannot meet any of the spheres, so MATH is the NAME limit of the MATH. But, MATH, while we have assumed that MATH. |
math/0101206 | We connect MATH and MATH by a generic path MATH in MATH. As in the proof of REF , this gives rise to the required compact cobordism. Note that the possibility of bubbling off a sphere is ruled out, choosing MATH small enough to ensure that MATH is disjoint from all MATH-holomorphic spheres with MATH. |
math/0101206 | The first part follows directly from REF . Compactness follows from the usual compactification theorem for holomorphic curves (see CITE and also CITE, CITE, CITE), which holds thanks to the energy bound REF . Specifically, the compactness theorem says that a sequence of points in the moduli spaces converges to an ideal disk, with possible broken flowlines, boundary degenerations, and bubblings of spheres. Broken flowlines are excluded by the additivity of the NAME index, and the transversality result REF . Spheres and boundary degenerations both carry NAME index at least two, so these kinds of degenerations are excluded as well. |
math/0101206 | This follows in the usual manner from the compactifications of the one-dimensional moduli spaces MATH with MATH (together with the gluing descriptions of the neighborhoods of the ends). Note that if MATH, then MATH, so there are no flows connecting MATH to MATH. We note also that there are no spheres in MATH or degenerate holomorphic disks (whose boundary lies entirely in MATH or MATH), so the only boundary components in the compactification consist of broken flow-lines. |
math/0101206 | As is usual in NAME 's theory, one considers the ends of the moduli spaces MATH, where MATH satisfies MATH. This space has a priori three kinds of ends: CASE: those corresponding to ``broken flow-lines", that is, a pair MATH and MATH with MATH CASE: those which correspond to a sphere bubbling off, that is, another MATH and a holomorphic sphere MATH which meets MATH CASE: those which correspond to ``boundary bubbling", that is, we have a MATH, and a holomorphic map MATH from the disk, whose boundary is mapped into MATH or MATH, which meet in a point on the boundary. (In principle, several of the above degenerations could happen at once - multiple broken flows, spheres, and boundary degenerations, but these multiple degenerations are easily ruled out by dimension counts and the transversality theorem, REF .) In REF , we argue that MATH (note that a disk whose boundary lies entirely inside MATH or MATH also has a corresponding domain MATH, which, in this case, must be a multiple of MATH; if MATH is pseudo-holomorphic, then MATH for MATH according to REF , and if MATH, the disk must be constant). Thus it follows from REF that MATH. From transversality REF , it follows that MATH and MATH must be constant; so that, in particular, MATH. Now for generic MATH, we know that the holomorphic spheres miss the intersection points MATH; hence, the case of spheres bubbling off is excluded. Thus, when MATH, boundary bubbles are excluded, so, counting the ends of the moduli space MATH, we get that MATH . When MATH, there are additional terms, corresponding to the boundary bubbles (and the gluing descriptions of the ends), giving a relation MATH see for example CITE. But the terms MATH and MATH both vanish, according to REF . From the additivity of MATH under juxtapositions of flow-lines, it follows that the double sums considered above are coefficients of MATH. |
math/0101206 | The fact that MATH is a subcomplex is an easy consequence of REF . |
math/0101206 | We begin with the claim about MATH. It is clear that the kernel of MATH consists of those homology classes MATH for which MATH for sufficiently large integers MATH. To see that MATH can be chosen independent of MATH, observe that MATH is a finitely generated chain complex over the ring MATH, and hence its homology is finitely generated over MATH, as well. Thus, the sequence of submodules of MATH must stabilize. Now, we claim that for MATH large enough that MATH, it is also the case that MATH. First, since MATH is an automorphism of MATH, it is clear that MATH. Conversely, if MATH, then writing MATH , we have that MATH (since MATH); thus MATH. |
math/0101206 | The isomorphism follows readily from REF and the long exact sequence. Since MATH is a finitely-generated MATH-module which is annihilated by MATH, it follows immediately that this module is also finitely generated over MATH. |
math/0101206 | The existence of such a volume form obviously implies weak admissibility, since each non-trivial domain has positive area. Assume, conversely, that each non-trivial periodic domain has both positive and negative coefficients. By changing the volume form, we are free to make each domain in MATH have arbitrary positive area. Thus, the claim now reduces to some linear algebra. We say that a vector subspace MATH is balanced if each of its non-zero vectors has both positive and negative components. The claim, then, follows form the fact that a vector subspace of MATH which is balanced admits an orthogonal vector each of whose coefficients is positive. This fact is true by induction on the dimension of the ambient vector space (and it is vacuously true for MATH). Now, suppose MATH is a balanced subspace of MATH, and let MATH denote the projection map MATH. Either MATH is also balanced, or MATH contains a vector MATH whose MATH component is MATH, all other components are non-positive, and at least one of them is negative. In this latter case, we construct the required positive orthogonal vector as follows. Apply the induction hypothesis to find a vector MATH with MATH for MATH, which is orthogonal to MATH. The required vector, then, is MATH. If, on the other hand, all MATH of the vector spaces MATH are balanced, then by induction we can find vectors MATH and MATH with MATH for MATH, and MATH for MATH. Then, MATH is our required vector. |
math/0101206 | Fix some initial MATH with MATH. Then, in view of REF , any other MATH with MATH has the form MATH where MATH is some periodic class, MATH its associated periodic domain, and MATH is the positive generator of MATH. If MATH, this forces MATH for some periodic domain whose associated homology class is annihilated by MATH. If MATH, then MATH. Thus, the lemma follows from the observation that for any fixed MATH, there are only finitely many periodic domains MATH in the set MATH . We see this as follows. Let MATH denote the total number of domains (components in MATH). We can think of MATH as lattice points in the MATH-dimensional vector space generated by the domains MATH. Given MATH, written as MATH, we let MATH denote its naturally induced Euclidean norm MATH . If MATH had infinitely many elements, we could find a sequence of MATH with MATH. In particular, the sequence MATH has a subsequence which converges to a unit vector in the vector space of periodic domains with real coefficients which annihilate MATH. We write the vector as MATH. Since the coefficients of MATH are bounded below, but the lengths of the MATH diverge, it follows that all the coefficients of MATH are non-negative. Of course, if the polytope the subspace of MATH annihilated by MATH, corresponding to periodic domains with only non-negative multiplicities has a non-trivial real vector, then it must also have a non-trivial rational vector. After clearing denominators, we obtain a non-zero periodic domain (with integer coefficients) annihilating MATH, with only non-negative coefficients. This contradicts the hypothesis of weak admissibility. |
math/0101206 | Fix a reference MATH with MATH. Then, as in the previous lemma, any other class MATH with MATH can be written as MATH . Thus, each class MATH with MATH corresponds to a periodic domain MATH with MATH . The lemma follows from the fact that (for fixed MATH) there are only finitely many periodic domains satisfying this inequality. This follows as in the proof of REF : an infinite number of such periodic domains would give rise to a a real periodic domain MATH for which MATH from which it is easy to see that there must be an integral periodic domain with the same property. But such a periodic domain would violate the strong admissibility hypothesis. |
math/0101206 | When MATH is strongly MATH-admissible, the key point is to observe that the boundary operators REF is actually a finite sum. This follows from REF . Similarly, when the diagram is only weakly admissible, REF ensures that the differentials for MATH and MATH are finite sums. With these remarks, the proof proceeds exactly as in the proof of REF . |
math/0101206 | Let MATH and MATH be a pair of isomorphic coherent orientation systems, Fix a reference point MATH. Given any other MATH and path MATH, there is a sign MATH with the property that for MATH. As the notation suggests, if MATH and MATH are isomorphic orientation systems, the number MATH is independent of the choice of MATH, so we can define a map MATH by MATH. It is straightforward to verify that MATH induces an isomorphism of chain complexes. |
math/0101206 | This is a variant on the usual proof that MATH. Suppose that MATH satisfies MATH, and let MATH. Then, since MATH (since MATH is a cocycle), we get that MATH . (Note that boundary degenerations do not contribute to the above sum, as in the proof that MATH.) Summing over all MATH with MATH and MATH, we get the MATH-coefficient of MATH. |
math/0101206 | If MATH is a coboundary, then there is a zero-cochain MATH (a possibly discontinuous map from MATH to MATH) with the property that if MATH is an arc in MATH (a one-simplex), then MATH. Let MATH where the evaluation of MATH on MATH is performed by viewing the latter as a constant path from MATH to MATH. Then, it follows from the definitions that MATH . |
math/0101206 | Begin with any NAME diagram MATH for MATH and let MATH be a collection of pairwise disjoint curves which are dual to the MATH, in the sense that for all MATH and MATH, MATH (the right hand side is NAME delta, and the left hand side denotes both the geometric and algebraic intersection numbers of the curves). By isotoping the MATH if necessary, we can arrange that MATH. Choose a basepoint MATH distinct from MATH, MATH, and MATH (indeed, choose MATH to be disjoint from the neighborhood of the MATH where the winding is performed). Let MATH, labelled so that MATH for MATH. Each time we wind MATH along MATH. we create a new pair of intersection points near MATH between MATH and the new copy of MATH. Winding along each MATH times, then, we can label these intersection points MATH (ordered in decreasing order of their distance to MATH, and with sign distinguishing which side of MATH - in its tubular neighborhood - they lie in). Thus, we have induced intersection points MATH labeled by MATH. Note that with our conventions, the short arc in MATH connecting MATH to MATH, followed by the short arc in MATH with the same endpoints, is homologous to MATH in MATH. No matter how many times we wind MATH along MATH, the MATH structure of the farthest intersection point MATH remains fixed (this is clear from the definition of MATH: the winding isotopy induces an isotopy between the induced non-vanishing vector fields induced over MATH). Moreover, by the definition of the difference map MATH introduced in Subsection REF, together with REF , we have that MATH . Thus, we can find NAME diagrams which realize the MATH structures which differ from some fixed MATH structure MATH by non-positive multiples of the MATH,.,MATH. Moreover, if we choose parallel copies MATH of the MATH, only with the opposite orientations, and wind along those in addition, we can realize all MATH structures which differ from MATH by arbitrary multiples the MATH,.,MATH. Now, it is easy to see that the group MATH is generated by the NAME duals of the MATH. Hence, we can realize all MATH structures. |
math/0101206 | In view of REF , we can start with a MATH-realized NAME diagram. We will show that after winding the MATH sufficiently many times along curves MATH as in the proof of the previous lemma, we obtain a pointed NAME diagram for which each renormalized MATH-periodic domain has both positive and negative coefficients. Such a NAME diagram is strongly MATH-admissible. We find it convenient to use rational coefficients for our periodic domains. Write MATH, and choose a basis MATH for the MATH-vector space of renormalized periodic domains. Note that a renormalized periodic domain MATH is uniquely determined by the part of its boundary which is spanned by MATH; that is, the map which associates to a periodic domain the MATH-coefficients of its boundary gives an injection of vector spaces. After a change of basis of the MATH and reordering the MATH, we can assume that for all MATH, MATH where MATH. For each MATH choose points MATH which are not contained in any of the MATH or MATH. Let MATH and then choose some integer MATH so that MATH . Choose parallel copies MATH of the MATH for MATH, and let MATH be the new periodic domains, obtained after winding the curves MATH times along the MATH and MATH times in the opposite direction along the MATH. Note that MATH . In a similar manner, we see that MATH . It is a straightforward matter, then, to verify that for any linear combination of the MATH, one can find some point MATH for which MATH is positive, and another MATH for which MATH is negative. |
math/0101206 | First, note that if MATH is strongly MATH-admissible, then if we choose curves along which to wind the MATH (disjoint from the basepoint MATH), then the winding gives an isotopy through strongly MATH-admissible pointed NAME diagrams. The reason for this is that, in the complement of a small neighborhood of the winding region, the various renormalized periodic domains remain unchanged; thus, if some renormalized periodic domain has positive coefficients, then it retains this property as it undergoes winding. Thus, it suffices to show that if two NAME diagrams are isotopic (via an isotopy which we can assume without loss of generality takes place only among the MATH - taking MATH to MATH), then if we wind their MATH-curves simultaneously along some collection of MATH to obtain MATH, then the pointed NAME diagrams MATH and MATH are isotopic through strongly MATH-admissible NAME diagrams. To see this, we choose MATH curves and their translates MATH as in the proof of REF . Now, we choose constants MATH where we think of MATH as the parameter in some isotopy taking MATH to MATH, and MATH is the corresponding one-parameter family of renormalized periodic domains. (Strictly speaking, the point MATH generically lies on the translates of the MATH for finitely many MATH, so that for those values of MATH, the multiplicity MATH does not make sense as we have defined it; for those values of MATH, we use a small perturbation MATH of the basepoint MATH.) Using these constants MATH as in the proof of REF , the present lemma follows. |
math/0101206 | It is easy to see that if we take a weakly MATH-admissible NAME diagram, then REF provides an isotopy to a strongly MATH-admissible NAME diagram, and that the given isotopy is an isotopy through weakly MATH-admissible NAME diagrams. |
math/0101206 | Let MATH be a codimension one constraint in MATH representing the class MATH, chosen to miss all the constant paths (corresponding to the intersection points MATH). Consider the map MATH defined by MATH . We claim that MATH . This follows by considering the ends of the one-dimensional moduli spaces MATH where MATH. The ends where MATH correspond to the commutator of MATH and MATH, while the ends where the maps MATH bubble off correspond to the commutator of MATH with the corresponding boundary maps. REF , of course, says that MATH commutes with MATH, on the level of homology. |
math/0101206 | Given a sequence of NAME moves, we can clearly introduce isotopies as needed to arrange that no handleslides, only isotopies (of the MATH or the MATH), cross the basepoint MATH. Since the roles of MATH, MATH are symmetric, it suffices to consider the case where the isotopy of MATH, say MATH, crosses MATH. We denote the new isotopic curve by MATH. We claim that MATH can be moved by a series of handle-slides and isotopies to MATH all of which are supported in MATH. This can be seen by first surgering out MATH to get a torus MATH with MATH marked points. Clearly, in MATH the curves induced by MATH and MATH are isotopic. We can follow the isotopy by moves MATH where isotopies across the marked points MATH are replaced by handle-slides in MATH. See REF for the MATH case. |
math/0101206 | Existence is ensured by REF . Given two strongly MATH-admissible NAME diagrams, REF gives a sequence of pointed NAME moves which connect them. Now, by introducing additional isotopies as in REF , we can arrange for all the isotopies to go through strongly MATH-isotopic NAME diagrams. |
math/0101206 | Let MATH be points contained in the interiors of the domains before the pair-creation, and MATH be a point in the new domain. Let MATH be the torus MATH. As before, if MATH, we let MATH denote the space of homotopy classes of NAME disks for MATH, MATH; if MATH we let MATH denote the homotopy classes with moving boundary conditions defined above, and we let MATH denote the homotopy classes of NAME disks for the pair MATH and MATH, now thinking of MATH and MATH as intersections between those tori. Fix MATH, and MATH. It is easy to see that each homotopy class MATH has a representative MATH which is constant for MATH. As such, MATH can be thought of as representing a class MATH. Indeed, this induces a one-to-one correspondence MATH. In a similar manner, if MATH, we have identifications MATH, which preserve all the local multiplicities MATH for all MATH. Let MATH be a sequence of homotopy classes in MATH which support holomorphic representatives, and have a fixed NAME index. Next, fix MATH. Since the MATH all support holomorphic representatives, the local multiplicities at the MATH for MATH are non-negative; it follows that for MATH, MATH. But MATH is a homotopy class connecting MATH with MATH, which are intersection points which existed before the pair creation, so we can consider the corresponding element of MATH. From the above observations, the multiplicities at all MATH for MATH are are bounded below, and the NAME index is fixed, so there can be only finitely many such homotopy classes, according to REF . The lemma now follows. |
math/0101206 | According to REF , we can find a weakly MATH-isotopic, strongly MATH-admissible NAME diagram. (Indeed, it is easy give exact Hamiltonian isotopies of MATH which are suitable for the purposes of REF .) Note that the analogue of REF also holds in the weakly admissible context (where now both MATH and MATH are fixed), so we can construct chain homotopy equivalences MATH and MATH by modifying REF (for example, in the definition of MATH we drop terms involving MATH with MATH) showing that the corresponding groups are isomorphic. |
math/0101206 | First, note that the boundary homomorphism MATH is injective, and its image is MATH. The first isomorphism then follows from the long exact sequence in homology for the pair MATH, bearing in mind that MATH (by excision), and that the map MATH is trivial: the NAME surface is obviously null-homologous in MATH. The second isomorphism follows from the fact that under the natural identification MATH, the image of MATH is identified with MATH. The final isomorphism follows from the fact that MATH . |
math/0101206 | The first statement is clearly true. Let MATH denote the space of NAME triangles connecting MATH. Then, evaluation along the boundary gives a fibration MATH whose fiber is homotopy equivalent to the space of pointed maps from the sphere to MATH (the base space here is a product of path spaces). In the case where MATH, this gives us an exact sequence MATH . By definition, MATH. The evaluation MATH provides a splitting for the first inclusion, so that MATH . That image, in turn, is clearly identified with the kernel of the natural map MATH (we are using here the fact that MATH is Abelian). The proposition when MATH then follows from REF . The case where MATH follows as above, after dividing by the action of MATH. |
math/0101206 | Recall that MATH, MATH, and MATH determine the two-plane field on the boundary minus finitely many three-balls. Fix MATH and choose extensions over the three-balls (compare REF). This data specifies the two-plane field over MATH. The above discussion shows that the MATH structure extends over this region, and, indeed, since the deleted region is topologically a MATH, it follows from a cohomology long exact sequence that the extension is unique. It is easy to see also that the induced MATH structure does not depend on the extension of the two-plane fields to the three-balls in the boundary. Changing MATH by a homotopy moves MATH by an isotopy, so it is easy to see that the induced MATH structure depends only on the homotopy class of MATH. Indeed, it is easy to see from the above discussion that the restriction MATH determines the MATH structure. It follows that adding spheres to the homotopy class in MATH does not change the induced MATH structure. |
math/0101206 | Since addition of NAME disks to a given homotopy class MATH changes the induced two-plane field in a neighborhood of the boundary, it follows that the above constructions descend to give a map MATH what remains to verify that we have identified the image, and also that MATH is one-to-one. To see that we have characterized the image. Recall that for an arbitrary four-manifold-with-boundary MATH there is a canonical map MATH which is defined as follows. Choose a MATH structure MATH over MATH, and let MATH where MATH is the coboundary map. It is easy to see that MATH is independent of the choice of MATH, and that it vanishes if and only if MATH extends over MATH. Next, we argue that MATH. To see this, isotope MATH, MATH, and MATH so that there are intersection points MATH, MATH, and MATH for which MATH, so that there is a triangle connecting them. We have explicitly constructed the corresponding MATH structure, thus MATH, as well. It is easy to see that MATH . Similarly, MATH . It follows that MATH: the obstructions to extending a MATH structure are the same as the obstruction to finding a NAME triangle. Suppose that MATH and MATH are a pair of triangles in MATH with MATH, so that their difference from REF can be interpreted as the triply-periodic domain MATH. We claim that this triply-periodic domain corresponds to a relative cohomology class in MATH whose image in MATH is the difference MATH, which in turn verifies that the map MATH is injective as claimed. But this now is a local calculation since, as is easy to verify, the restriction map MATH is injective, and each of the latter groups is generated by the NAME duals to the cylinders MATH (where MATH, MATH, or MATH, and MATH). On the one hand, the evaluation of a triply-periodic domain on, say, MATH is easily seen to be simply the multiplicity of MATH in the boundary of the triply-periodic domain. On the other hand, the pair of two-plane fields representing MATH and MATH differ over MATH only at those points where one of MATH or MATH contains MATH. The fact that the constant appearing here is one could be determined by calculating a model case (see CITE). |
math/0101206 | Let MATH be a fixed homotopy class representing MATH. Fix an arbitrary orientation MATH. Next, we construct MATH, where MATH are periodic classes. To this end, let MATH denote the subgroup of MATH consisting of elements MATH which satisfy the property that MATH for some periodic classes MATH and MATH in MATH and MATH respectively. We claim that REF given MATH, the corresponding MATH and MATH are uniquely determined, and also that REF the quotient MATH of MATH is a free Abelian group. According to REF , we obtain a splitting MATH . Now, for MATH, we define MATH so that MATH . According to REF , this is a consistent definition. We then define MATH arbitrarily on a basis of generators for MATH, and allow that to induce the orientation on all MATH. Both REF follow from the following diagram: MATH where MATH. Interpreting the second homology as periodic domains, we see that the quotient group MATH is the image of MATH. Now, by excision, MATH (where here MATH is a handlebody and MATH is an interval); in particular, MATH is injective, establishing REF , and also the image of MATH is non-torsion, establishing REF . We have thus consistently oriented all triangles in MATH. As a final step, we choose a complete set of paths MATH for MATH over which we choose our orientations (for MATH) arbitrarily, and use them to define the orientation for all the remaining MATH in the given MATH-equivalence class. |
math/0101206 | Given MATH with MATH and MATH, the difference MATH is a triply-periodic domain which, in view of REF , can be written as a sum of doubly-periodic domains. Given this, finiteness follows as in the proof of REF . |
math/0101206 | Suppose that MATH satisfy MATH, and MATH. Then we can write MATH; so by the additivity of the index, it follows that MATH (which is identified with the first NAME class evaluation). The proof then follows from the proof of REF . |
math/0101206 | This follows as in REF : we wind transverse to all of the MATH, MATH, and MATH simultaneously. |
math/0101206 | The fact that MATH is a chain map follows by counting ends of one-dimensional moduli spaces of holomorphic triangles (compare CITE). Fix MATH, MATH, MATH, and consider moduli spaces of holomorphic triangles MATH where MATH, MATH, and MATH. The ends of this moduli space are modeled on: MATH . In the above expression, the pairs of homotopy classes MATH and MATH range over MATH and MATH with MATH, MATH, MATH (with analogous conditions for the MATH and MATH). Counted with signs, the first two unions give the MATH-coefficient of MATH (using the natural differential on the tensor product), while the last gives the MATH-coefficient of MATH. Recall that if MATH has a holomorphic representative, then MATH. Thus, MATH maps the subcomplex MATH into MATH. Similarly, MATH as above also gives a chain map. The MATH-equivariance MATH (and indeed for the other induced maps, where stated) follows immediately from the definitions. |
math/0101206 | Fix first the complex structure MATH over MATH. Consider a one-parameter variation family of maps MATH from MATH into the space of almost-complex structures over MATH, where MATH is a real parameter MATH (which are perturbations of the symmetrized complex structure MATH over MATH). We write down the case of MATH; the other homology theories work the same way, with only notational changes. Consider the map MATH defined by MATH . Now, the ends of MATH count MATH that is, MATH and MATH are chain homotopic. The above argument shows that if we fix a map MATH into the space of almost-complex structures over MATH, with specified behaviour near the vertices of the triangle, then the induced map is uniquely specified up to chain homotopy. Now, we investigate invariance as one changes the one-parameter families of paths used at the corners. To this end, we claim that the following diagram MATH commutes up to homotopy. Here, the maps maps MATH are the homomorphisms induced by variations of these paths (which we encountered in the proof of REF ; note that we have now dropped the variation of path from the notation, and replaced it with the an indication of the NAME diagram we are working with). The homotopy now is obtained by counting triangles with in a formally MATH-dimensional space, but which are MATH-holomorphic with respect to any MATH for MATH in a one-parameter family. We do not spell out the details here: a very similar argument is given in the next proposition. Similar remarks hold as we vary the boundary behaviour at the other vertices. As a final remark, since the induced maps are invariant under small perturbations of the family MATH, it follows also that the induced map is independent under variations of the complex structure MATH over MATH. |
math/0101206 | We begin with isotopies of the MATH. As in the proof of isotopy invariance of NAME homologies, we let MATH be an isotopy (induced from an exact Hamiltonian isotopy of the MATH in MATH), and we consider moduli spaces with dynamic boundary conditions. Specifically, let MATH be a parameterization of the edge MATH, with MATH . Consider moduli spaces indexed by a real parameter MATH: MATH and divide them into homotopy classes MATH, with MATH, MATH, MATH. Note that if MATH, then MATH is generically a compact zero-dimensional manifold, so we can define MATH . Fix, now, any homotopy class MATH with MATH and MATH, and consider the one-manifold MATH . This has ends as MATH, which are modeled on MATH where the first union is over all MATH with MATH, MATH in the sense of Subsection REF, MATH, and MATH (with analogous conditions on the second union). There are also ends of the form MATH where the first union is over all MATH in the same equivalence class as MATH, MATH in the sense of Subsection REF, MATH, and MATH and MATH while MATH (with analogous conditions over the other two unions). Counting ends with sign, we get that MATH where MATH are the chain maps induced by the isotopy MATH, as constructed in Subsection REF note that here we have suppressed the isotopy MATH from the notation. Isotopies of the MATH work the same way; we now set up isotopies of the MATH. Consider moduli spaces indexed by a real MATH . These moduli spaces partition according to homotopy classes MATH (with MATH, MATH, MATH). Note that for MATH, this is the usual moduli space for holomorphic triangles for MATH, MATH, and MATH. Again, when MATH, the union MATH is generically a compact, zero-dimensional manifold, and we can define Note that if MATH, then MATH is generically a compact zero-dimensional manifold, so we can define MATH . Fix a homotopy class MATH with MATH and MATH, and consider the ends of MATH . The ends as MATH are modeled on MATH where the union is over all MATH, MATH with MATH, MATH, MATH, and MATH with MATH. Counting these ends with sign, we get a contribution of MATH while the end as MATH corresponds simply to MATH. There are other ends as before, whose contribution is MATH . Thus, we have exhibited a chain homotopy from MATH with MATH. |
math/0101206 | The given data already allow us to orient all squares representing MATH. Specifically, any such square can be decomposed as a sum of NAME triangles MATH, where here the MATH denote homotopy classes of NAME triangles for the subtriples indicated in the notation, and which represent the MATH-equivalence class induced from MATH. We then define the orientation for the given square by MATH (note that we have suppressed the subscripts from MATH indicating the diagram used for the polygon, as our notation for the homotopy classes already carries this information). Since the induced orientations over MATH are compatible, this definition is independent of the decomposition. In particular, if it follows that for any four NAME disks MATH, for the NAME pairs on the boundary (MATH, MATH, MATH, and MATH respectively), we already have that MATH . Next, fix a pair of triangles MATH, MATH for the corresponding NAME triples, and fix an arbitrary orientation MATH, and then orient MATH so that MATH . Given this, we can now orient all MATH: given any such MATH, in view of REF we can write MATH where MATH, MATH, MATH, and MATH. We now define MATH to be compatible with this decomposition, and observe that, according to REF , the answer is independent of the choices of MATH. We extend these orientations on MATH to an orientation system for MATH in the usual manner. (Note that the isomorphism class of this orientation system is uniquely determined.) This orientation system of MATH and the fixed orientation MATH on the initial triangle MATH suffices to orient the all the remaining triangles for the triples MATH in the MATH-equivalence class corresponding to the restriction of MATH: any other such triangle MATH can be written in the form MATH and can be oriented correspondingly. The orientation obtained in this manner is well-defined, owing to REF . The same remarks hold for the remaining subtriple MATH, and since the induced orientation systems on the MATH agree, it is straightforward now to verify that the induced orientation system on the MATH and MATH are consistent with the orientation system on the squares in MATH, completing the construction. |
math/0101206 | We define a map MATH by MATH . Note that above map is a finite sum by the strong admissibility requirement on the NAME quadruple; indeed, it also implies that the sums appearing in the statement of the theorem are finite sums. Counting ends of one-dimensional moduli spaces MATH with MATH, we see that MATH induces a chain homotopy between the maps MATH and MATH . Again, the other cases are established in the same manner. |
math/0101206 | There are altogether MATH intersection points between MATH. Fix some MATH, and let MATH and MATH be intersection points with MATH, MATH and for all MATH we have MATH. Note that MATH, MATH bound two pairs of disks, which we denote by MATH and MATH, as in REF . Let MATH be homotopy classes given by MATH and MATH respectively. It is easy to see that MATH. By juxtaposing the disks we get a periodic class MATH in MATH, with MATH. It easy to see that MATH and similarly MATH are generated by MATH classes, all of which can be realized in this way, so it follows that all MATH intersection points represent the trivial MATH structure. Moreover, since each non-trivial combination of these periodic classes has both positive and negative coefficients, it follows that MATH is admissible for MATH (since MATH, the strong and weak admissibility notions coincide). It also follows that if MATH denotes the number of positive coordinates for MATH, then the relative grading is given by MATH. We still have to show that all boundary maps are zero. For the purpose of these calculations, we will use the path MATH, where MATH is some fixed complex structure over MATH. In these calculations, transversality for the flow-lines considered is either immediate or it is achieved by moving the curves, as in REF . First let MATH and MATH be as above. Then for all MATH, with MATH and MATH, the moduli space MATH is empty, since MATH has some negative coefficients. Also MATH contains a unique solution that maps the trivial MATH-fold cover of the disk to MATH, so that one of the sheets is mapped to MATH, and other sheets map to MATH for MATH respectively. Similarly MATH also contains a unique smooth solution. Since the two domains differ by a periodic domain, we are free to choose an orientation system for which the signs corresponding to these solutions are different. For this system, the the component of MATH in MATH is zero. Now suppose that MATH and MATH differ in at least two coordinates, and MATH. Then it is easy to see that for all MATH, with MATH the moduli space MATH is empty, since MATH has some negative coefficients. This shows that MATH, and consequently MATH. In fact, since the two disks MATH and MATH separately have a unique holomorphic representative, it follows that the MATH-module structure is given by the identification MATH. Let MATH be the intersection point representing a top-dimensional homology class in MATH. We have shown that for all MATH, MATH. It follows that for all MATH with MATH we have MATH. By REF , when MATH, the moduli space MATH is empty. The case MATH corresponds to the boundary map in MATH which was just shown to be trivial. It follows that MATH. The algebra action on MATH gives then the map MATH which is easily seen to be an isomorphism by properties of the short exact sequence MATH . |
math/0101206 | We can think of MATH as the set MATH. By the NAME reflection principle, we can extend any involution of MATH to the NAME sphere, so that it switches MATH and MATH. Thus, this involution has the form MATH for some MATH. Since the involution takes the set of complex numbers with modulus MATH to those with modulus MATH, it follows that MATH, and the lemma follows. |
math/0101206 | We work out the MATH case, for notational convenience (the general case follows easily). Let MATH for MATH denote the connected components of MATH, see REF . Let MATH, and MATH. Then there are exactly three classes, MATH, with MATH and MATH, and these are given by MATH, MATH, and MATH. As before, MATH. We claim that MATH. Since these cases are symmetric we deal with MATH. According to REF , a holomorphic disk MATH representing MATH gives rise to a branched double cover MATH and a map MATH. In our case MATH has degree REF in MATH and MATH and has degree REF on the other regions. Here, MATH is an annulus, and the image of MATH lies in the union of MATH, MATH, MATH, MATH. In fact the part of the image that lies in MATH is an arc starting at MATH, see REF . More generally, for each MATH, we can consider the subset MATH, which is the interior of the region in MATH obtained by removing a length MATH subarc of MATH starting at MATH, with arc-length normalized so that MATH corresponds to the endpoint MATH. The region MATH is topologically an (open) annulus, and hence can be identified conformally with a standard (open) annulus MATH, where, MATH is a non-zero real number depending on MATH. The identification is given by a map MATH which extends to a continuous map from the closure MATH of MATH to the closure of MATH, which maps the boundary of MATH into the union of MATH, MATH, MATH, and MATH, so as to map onto the length MATH sub-arc of MATH. Let MATH and MATH denote the subsets of MATH which map to MATH and MATH respectively, and let MATH and MATH denote the angles swept out by MATH and MATH. When the map MATH is induced from a holomorphic disk in the second symmetric product, the corresponding branched double-cover (of the disk, by MATH) induces an involution on MATH which switches MATH to MATH. These assertions follow from simple modifications of the NAME mapping theorem, see for example CITE. Indeed, it follows from these classical considerations that if the curves are smoothly embedded, then the objects MATH and MATH also depend smoothly on the parameter MATH. In light of REF points in the moduli space MATH are in one-to-one correspondence with MATH with MATH. For certain (generic) choices of the curve MATH, we can arrange that the graph of MATH and MATH have non-empty, transversal intersection; that is, that there is a finite, non-empty collection of holomorphic disks. By a slight perturbation of the curves, it then follows that the formal dimension of MATH is REF, so MATH. Now let MATH, and let MATH be given by MATH. MATH. Clearly MATH and MATH generate the periodic classes of MATH, and both of them has zero NAME index. Clearly any non-trivial linear combination has positive and negative coefficients as well, so our diagram MATH is MATH-admissible. In order to compute the boundary maps, note that MATH, where the unique solution is again a product of a holomorphic disk and a constant map. We also claim that MATH . This computation follows easily from the previous discussion and some complex analysis. Let MATH denote the annulus given by the domain MATH in MATH, and let MATH and MATH denote the conformal angles of the parts of MATH that correspond to MATH and MATH respectively. By general position we can assume that MATH. Let MATH, MATH be defined as above. Clearly MATH, so for small enough MATH we have MATH. We claim that MATH, MATH, MATH. This follows readily from NAME 's compactness (although in this case, it could also be proved using classical conformal analysis, see for example CITE). It follows that if MATH, then MATH, and if MATH, then MATH. Thus, we have calculated MATH. The computation for MATH is similar. Again we have an annulus with a cut of length MATH. Let MATH denote the conformal angle corresponding to MATH, MATH respectively. In this case MATH, so for small enough MATH we have MATH. As above, as MATH, the conformal angle MATH converges to MATH. It follows that if MATH, then we have MATH, and if MATH then MATH. This proves MATH. Again, we are free to choose an orientation system so that the sign of the flow obtained here cancels the sign in MATH (since the domains differ by a periodic domain), so that the MATH component of MATH vanishes. The same argument shows also that the MATH-component of MATH, where MATH. The remaining components of MATH can be shown to vanish, using the arguments from REF . This shows that MATH. The corresponding statement about MATH also follows as before. The case where MATH follows from the MATH case (where we have the same holomorphic maps multiplied with MATH constant maps). |
math/0101206 | This follows from associativity of the triangle construction. First, we begin with some remarks on admissibility for the pointed NAME quadruple MATH . Note that MATH . To see this, note that all quadruply-periodic domains for our NAME quadruple can be written as sums of doubly-periodic domains for the four bounding three-manifolds (this is an easy consequence of the fact that the spans in MATH of the three MATH-tuples MATH, MATH, and MATH all coincide). Since the map from doubly-periodic domains to quadruply-periodic domains models the map MATH to MATH, it follows that the map MATH is injective. Since the restriction of MATH to the boundary is trivial, it follows that the subgroup itself is trivial. Thus, the orbits MATH needed for admissibility (in the sense of Subsection REF) are trivial. Moreover, it is now easy to see that if we choose MATH, MATH and MATH all sufficiently close to one another, so the six NAME diagrams are strongly admissible, then the quadruple MATH is strongly admissible. The result is now a direct application of REF . |
math/0101206 | The admissibility is easy to check. Now let MATH denote the triangular region in MATH with vertices MATH, compare REF . Let MATH be given by MATH. Then all MATH, with MATH has some negative coefficients so MATH. To study MATH, note that for each MATH there is a unique map from the two-simplex to MATH that satisfies the boundary conditions. The corresponding MATH-tuple of maps to MATH gives the unique solution in MATH, which is easily seen to be smooth. It follows that the coefficient of MATH in MATH is MATH. For all MATH we have MATH. Thus, MATH. More precisely, choosing orientation conventions on MATH and MATH as in REF , that is, for which MATH and MATH are cycles, then for the induced orientation convention on MATH, MATH is a cycle so there, too, we are using the orientation convention of REF . The corresponding result for MATH follows similarly. |
math/0101206 | Straightforward. |
math/0101206 | It is easy to see that the intersection points correspond: MATH. We choose a basepoint MATH in the same path-component of MATH as MATH, the point along which we perform the connected sum to obtain MATH. Let MATH denote the corresponding basepoint in MATH. If MATH, and MATH is the corresponding point MATH, then the induced MATH structures agree MATH, since the corresponding vector fields agree away from the three-ball where the stabilization occurs. Let MATH, and MATH be the class with coefficient MATH. Let MATH, MATH, and MATH be the class with MATH. Then, we argue that for certain special paths of almost-complex structures, the moduli space MATH is identified with MATH (together with its deformation theory, including the determinant line bundles). Hence, the chain complexes are identical. To set up the complex structure, we let MATH and MATH denote the connected sum points for MATH and MATH respectively. Recall that there is a holomorphic map MATH which is a diffeomorphism onto its image. Suppose that MATH is a (generic) family of almost-complex structures over MATH which agrees with MATH in a tubular neighborhood of MATH, a neighborhood which is holomorphically identified with MATH. Using the above product map, we can transfer MATH to an open subset of MATH, and extend it to all of MATH. It follows from our choice of MATH that if MATH, then any MATH-holomorphic representative for MATH must have its image in MATH, and hence its product with the constant map is MATH-holomorphic in MATH. Conversely, a MATH-holomorphic curve in MATH with MATH must be contained in MATH. The identification of deformation theories is straightforward. Since the map we describe identifies individual moduli spaces it is clear (compare REF ) that the map is equivariant under the action of MATH. |
math/0101206 | The map from MATH to the base MATH is the map sending the pair MATH to the sum MATH. The fiber over a base point MATH is represented by pairs of the form MATH with MATH. Two pairs MATH and MATH are equivalent if MATH. Now, the map MATH is an involution whose quotient is a projective line. |
math/0101206 | Choose holomorphic coordinates MATH centered at MATH with respect to which MATH corresponds to the intersection of the coordinate patch with the subvariety MATH. (These are MATH-holomorphic coordinates, which are actually holomorphic for all the MATH, in the specified region.) With respect to these coordinates, we can write MATH as MATH. Next, fix a local coordinate function MATH around the point MATH, so that MATH. We can factor MATH, where MATH and MATH. Thus, in cylindrical coordinates, we can write this as the map sending MATH to MATH . Now MATH vanishes at the origin, so that there is a constant MATH with the property that MATH. Applying the above cut-off construction forces us to consider the new function MATH . Now, MATH of this is given by MATH so we have a constant MATH with the property that MATH pointwise; moreover, this tensor is supported in the strip where MATH is constrained to lie in the interval MATH. Thus there is a MATH with MATH . To handle the other components MATH, we proceed in a similar manner. Noting that MATH, the cutting off construction gives MATH. Applying MATH to this, we obtain MATH . We have a constant MATH with MATH, so that MATH pointwise. Once again, since the tensor is supported in a bounded strip, we have that MATH . Together, REF give REF , with respect to a cylindrical-end metrics on the range MATH (which, over the subset MATH, can be made to be isometric to the corresponding subset of MATH). The second part of the lemma is straightforward. |
math/0101206 | This is true because MATH is a ruled surface (see REF ), the generator corresponds to holomorphic spheres representing the fiber class (according to the proof of that lemma), and the map MATH given by MATH is a section. Thus, the composite MATH is a holomorphic map from the sphere to the torus, which must be constant. Hence, MATH maps to a fiber. Indeed, from our assumption on the homotopy class, it follows that MATH must map isomorphically to a fiber, and each fiber is determined by its intersection with the section. |
math/0101206 | This follows immediately from REF , and the corresponding estimate for MATH (which follows in the same manner). Note that the image of MATH is contained in the region where the divisor on the MATH side is bounded away from MATH, so that the almost-complex structures MATH split as MATH. |
math/0101206 | Consider the operator MATH . Here, MATH is viewed as a map between MATH and MATH, endowed with cylindrical metrics. By the removable singularities theorem, the kernel and cokernel are identified with kernel and cokernel of the flow-line in MATH, MATH . (See the corresponding discussion in CITE.) By assumption, the cokernel vanishes and its kernel is one-dimensional, since we assumed that MATH was a generic flow-line. Moreover, the operators MATH converge to MATH as MATH. The lemma then follows. |
math/0101206 | This follows as in the proof of REF . Note that (by a removable singularities theorem) the kernel and cokernel of MATH can be identified with the corresponding spaces for the operator MATH where MATH and MATH . In turn, these spaces are given by the deformation theory of (parameterized) holomorphic spheres, up to reparameterizations. The fact that the cokernel vanishes follows from the fact that the fiber class of a ruled surface form a smooth moduli space, which in turn follows from an easy NAME spectral sequence argument, which we defer to the next paragraph. For each sphere, then, there is a complex three-dimensional family of holomorphic vector fields which fix the image curve. These generate a three-dimensional space of kernel elements (for each sphere). However, if we restrict the domain to MATH, we are considering spheres which are specified to lie in the subvariety MATH at a fixed pair of points: hence, there is only a one-dimension space MATH of automorphisms left. This kernel is easily seen to be captured by MATH. To see that the MATH-fold fiber class of MATH over MATH considered above has a smooth deformation theory, note that each MATH-tuple of fibers is given as the zero set of a section MATH of a bundle MATH, which is the pull-back of a NAME class MATH line bundle MATH over the base MATH. The cokernel of the deformation complex is identified with the one-dimensional cohomology of the quotient sheaf of MATH by MATH, MATH . First, we prove the vanishing MATH . The NAME spectral sequence for the fibration has a MATH term with MATH where MATH is the derived functor of the push-forward map MATH. It suffices to prove that both MATH and MATH. The first vanishing follows from NAME duality, since MATH, a positive line bundle over an elliptic curve. The second vanishing statement follows from the projection formula MATH (see for example REF). This proves the vanishing in REF . The vanishing of MATH, then, follows from REF , the long exact sequence in cohomology associated to the short exact sequence of sheaves: MATH and the fact that MATH. |
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