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math/9905032 | We follow the argument in REF . Let MATH be an arbitrary function on MATH such that MATH for all but a finite number of MATH's. Form the probability generating functional: MATH . Then, viewing MATH as a diagonal matrix, we get MATH where the last equality is justified by REF applied to the operator MATH. Now, set MATH, so that MATH for all but finitely many MATH's. Then we can rewrite this relation as follows MATH where the last equality follows by REF . Next, as MATH is in MATH (it is even finite - dimensional), this can be rewritten as MATH . On the other hand, by the very definition of MATH, MATH . This implies MATH, as desired. |
math/9905032 | Set MATH, MATH. By the inclusion-exclusion principle we have MATH . This alternating sum is easily seen to be identical to the expansion of MATH by linearity using MATH . |
math/9905032 | Without loss of generality one can assume that the block MATH is a nonnegative diagonal matrix, MATH. Write the blocks MATH and MATH as matrices, too, and let MATH and MATH be their diagonal entries. Since MATH, we have MATH and therefore MATH . |
math/9905032 | Clearly, REF implies REF . To check the converse claim, write MATH in the block form, MATH, where MATH is of finite size and MATH is small. Write all the MATH's in the block form with respect to the same decomposition of the NAME space, MATH. Since MATH weakly, we have convergence of finite blocks, MATH, which implies MATH. Since MATH, we get MATH, so that all the traces MATH are small together with MATH provided that MATH is large enough. Write MATH and similarly for MATH. Then MATH . In the right-hand side, the first and the third summands are small because of the lemma, while the second summand is small because it is equal to MATH. |
math/9905032 | The fact that MATH is holomorphic in MATH for any trace class operators MATH is proved in CITE. The continuity of the map follows from the inequality MATH which holds for any MATH, see CITE. |
math/9905034 | Let MATH be a local parameter on MATH near MATH, so that the sheaf MATH is locally generated by an element MATH which is well defined up to a MATH-th root of unity. Define the map MATH by MATH. It is easy to check that this definition is independent of local parameter, and hence defines a canonical isomorphism. |
math/9905034 | If MATH has no automorphisms, then each term MATH in the MATH-spin structure MATH is locally free (since MATH is smooth) of rank one and has automorphism group MATH, which acts on MATH by multiplication. However, an automorphism of the MATH-spin structure must also be compatible with the structure maps MATH. In particular, compatibility with the isomorphism MATH shows that any automorphism MATH of MATH must satisfy MATH. Moreover, compatibility with MATH shows that MATH for every MATH dividing MATH and MATH dividing MATH. Thus MATH corresponds to some MATH, and for every MATH dividing MATH, the automorphism MATH is just MATH. This proves the first part of the proposition. For the second part, it is easy to see that any automorphism of the whole MATH-spin curve MATH induces, by restriction, an automorphism of MATH for each MATH, and the map MATH is injective. Moreover, for any edge MATH corresponding to half edges MATH attached to vertex MATH and MATH attached to vertex MATH, an automorphism MATH of MATH will induce automorphisms MATH and MATH; and these automorphisms must agree whenever MATH is NAME (locally free) at the node MATH corresponding to edge MATH. Similarly, if MATH is NAME at MATH, then MATH must equal MATH. However, if the sheaf MATH is NAME at MATH, then MATH and MATH act on distinct vector spaces (the sheaf MATH is not the same as MATH), and so MATH imposes no compatibility condition on MATH and MATH. Since MATH is NAME at the node MATH precisely when MATH divides both MATH and MATH, we have that the condition imposed by compatibility for MATH and MATH is precisely the equality MATH for MATH . It is clear that any MATH which meets this compatibility condition at each edge MATH defines a global automorphism MATH. |
math/9905034 | In the case of MATH, smooth MATH-spin curves of type MATH and index MATH correspond to the torsion points of MATH of exact order MATH. It is well known that the involution MATH acts without fixed points on the points of exact order MATH, unless MATH is MATH or MATH, in which case the involution fixes all MATH-torsion points (including the identity, corresponding to the trivial bundle MATH). It is easy to check (for example, by explicitly writing out the coordinates) that for sheaves MATH of index MATH (or MATH), corresponding to MATH-torsion, there is a canonical choice of isomorphism MATH such that CASE: MATH is trivial, CASE: any other isomorphism MATH differs from MATH by an element MATH, and CASE: MATH commutes with all elements of MATH. Thus MATH has order MATH and is Abelian; and if MATH is even, then every automorphism has order dividing MATH. Thus the proposition follows in genus MATH. The proof in genus MATH is similar, but simpler, since every MATH-spin structure is fixed by the involution. |
math/9905034 | When MATH the bundle MATH is trivial because the sections are disjoint. In the case MATH the result follows from the fact that taking residues gives an isomorphism between MATH and MATH. |
math/9905034 | The map MATH pulls back, via MATH, to give MATH . Since MATH is disjoint from the nodes of MATH, MATH is an isomorphism, even on the boundary strata where MATH fails to be locally free. Consequently, we have MATH . |
math/9905034 | This is a direct consequence of REF and the previous theorem. |
math/9905034 | First, we claim that because MATH is a regular imbedding it has finite Tor dimension; that is, there is an integer MATH such that for every coherent MATH-module MATH, the MATH-modules MATH vanish for MATH. This can be seen as follows. We may assume that MATH is MATH and MATH is MATH for some regular element MATH in a ring MATH. This gives the free resolution MATH of MATH, and shows MATH has finite Tor dimension. Since MATH is flat, the sheaves MATH and MATH are Tor independent over MATH; that is, MATH for all MATH. Let MATH and MATH be the left derived functors of MATH and MATH, respectively. REF states that if MATH has finite Tor dimension, and if MATH and MATH are NAME over MATH, then MATH. However, since MATH is flat over MATH, we have MATH; and since MATH commutes with the NAME character, the lemma holds. |
math/9905034 | The degree of MATH is an integer and is equal to MATH. Thus MATH, and the degree of MATH is strictly negative. Therefore, when MATH is irreducible, MATH has no global sections. When MATH is not irreducible, but MATH is locally free REF at each node, the same argument holds. If MATH is NAME at some nodes, then normalization MATH at the nodes of MATH where MATH is not locally free gives MATH, where MATH is locally free on MATH. Restricting MATH to MATH we obtain a MATH-th root of MATH, where the points MATH are either marked points or inverse images of nodes, and thus the collection MATH still meets the hypotheses of the proposition, but now MATH is locally free on each component, and hence has no global sections. Since MATH is finite, MATH. |
math/9905034 | It is clear from the construction of the classes MATH that they satisfy REF (convexity). REF follows from the fact that since MATH, the top-dimensional NAME class MATH of MATH, is simply the product of the top-dimensional classes of MATH and MATH. Now we will show that REF holds. If MATH is the MATH-th root from the universal MATH-spin structure on MATH, then since MATH, MATH is convex. Repeating the argument in the proof of REF with the NAME character replaced by the top NAME class yields MATH where MATH denotes the edge set of MATH. Also, we can compute the degree of MATH from the diagram The morphism MATH has degree MATH, since MATH is a tree, and therefore MATH also has degree MATH. The morphism MATH has degree MATH, as can be seen from the fact that the coarse moduli space map induced by MATH has degree MATH, and there are MATH automorphisms of a generic MATH-spin structure. Thus the map MATH also has degree MATH, the map MATH has degree MATH, so MATH has degree exactly MATH. Now we can compute MATH using REF (replacing MATH with MATH) to obtain MATH as desired. All that remains to check is REF (vanishing). Let MATH be a point corresponding to a tail marked by MATH. Taking the tensor product of MATH with the exact sequence MATH gives the exact sequence MATH where MATH. Thus MATH corresponds to a root with MATH marked by MATH. Since MATH, and since the residue isomorphism REF MATH shows that MATH is a trivial bundle, we have MATH . By REF the sheaves MATH and MATH are both convex; thus MATH and MATH are both locally free and have the same NAME classes in all dimensions. However, the vector bundle MATH has dimension MATH, and so MATH. This gives REF . |
math/9905034 | Let us show first that the conditions of the theorem are necessary. If MATH, then by REF , MATH must vanish; therefore, we can assume that MATH. In this case the dimension MATH (given in REF ) of the class MATH is equal to zero, and since MATH has two connected components MATH and MATH it will be sufficient to find MATH for two graphs MATH and MATH, such that the intersections MATH and MATH are non-empty. Let MATH be the graph with one genus-zero vertex, MATH tails, and MATH . NAME (that is, all half-edges are decorated with zeroes) loops. In this case, MATH in REF , so by REF (cutting edges) the class MATH pulls back to MATH, where MATH is the graph with one vertex of genus zero and MATH tails. Since the genus is zero, the universal square root MATH of MATH on the universal curve over MATH is convex by REF . Therefore, if MATH is the set of edges of MATH, we have MATH where the first equality follows from REF , the second from REF , and the third from REF . To find MATH consider the graph MATH with a single vertex of genus one, MATH tails, and MATH . NAME loops. REF again shows that MATH pulls back to MATH, where MATH has a single vertex of genus one and MATH tails. Since MATH, REF (forgetting tails) shows that MATH is a pullback from MATH, and MATH by REF . Now let us show that the classes MATH defined above for MATH indeed satisfy REF - REF (convexity) holds when MATH, since in this case the class has dimension MATH, and if MATH is convex (and, therefore, even) MATH and MATH as required. If MATH then MATH is not convex on the universal curve over MATH for any MATH. In particular, consider the degenerate curve of genus MATH which has two irreducible components MATH and MATH joined at a single node, the component MATH of genus zero, containing all MATH marked points, and the component MATH of genus MATH. For degree reasons the node must be NAME, and so MATH corresponds to MATH, for MATH a square root of type MATH of MATH, and for MATH a theta-characteristic on MATH. In general, MATH has non-zero global sections. Thus MATH also has non-zero global sections, and the universal square root is not convex. When MATH and MATH, the sheaf MATH is convex by REF . By REF (and REF ) the class MATH vanishes, and so agrees with our definition of MATH. REF holds because of the simple observation that the parity of a root MATH over a curve with the graph MATH is equal to the sum modulo MATH of the parities of the restrictions of MATH to the components corresponding to MATH and MATH. To prove REF , we may assume that MATH, and it is sufficient to check the case that MATH has only one edge. We have by REF and MATH for MATH a tree. Let MATH be defined as in REF . The canonical morphism MATH is actually an isomorphism if MATH is a tree, and so we get MATH since the parity of MATH does not change when restricting to the normalization. In the case of a loop, there are two subcases. First when MATH, MATH is isomorphic to two copies of MATH because of the two choices of gluing data - see REF. Consequently, MATH has degree MATH if MATH. In the second case MATH, and MATH is isomorphic to MATH, so MATH has degree MATH. Also, MATH is MATH if MATH, and MATH if MATH. Thus, when MATH, MATH as desired. When MATH, the two choices of gluing give different parities. Since parity is deformation invariant, this can be seen by degenerating to the special case of the curve MATH, whose partial normalization MATH at one node MATH consists of two irreducible components joined at a single node MATH. One component MATH is of genus MATH and contains the marked points MATH. The other component MATH is of genus zero and contains marked points MATH and MATH. Degree reasons force the node to be NAME, and so MATH is simply a direct sum MATH. Moreover, since MATH is a square root of MATH of type MATH, MATH must be trivial (MATH). Gluing MATH via MATH and MATH yields the trivial bundle MATH and another non-trivial bundle MATH of degree zero, respectively. Consequently, MATH. Since the parities of MATH and MATH are simply the parities of MATH and MATH, respectively, MATH and MATH have different parities. The different parities under the two gluings give MATH, since MATH has a tail marked with MATH. This completes the proof of REF is true because the projection MATH respects the parity of components. REF follows from the definition of the classes MATH. |
math/9905034 | We only need to check the case MATH. Since the class has dimension MATH, REF gives MATH where MATH are the numbers of even/odd theta characteristics on a smooth curve of genus MATH and the last factor of MATH is the local (orbifold) degree of the map MATH near a generic point of MATH. |
math/9905034 | This follows from the definition of the potential, the dimensions of the cohomology classes, and the dimensions of the moduli spaces MATH. It encodes the fact that intersection numbers between cohomology classes vanish if the classes do not have proper the dimension. |
math/9905034 | Recall that on the moduli space of stable curves, MATH, the MATH classes (where MATH) satisfy the equation MATH, where MATH is the image of the MATH-th canonical section and MATH. Let MATH be the forgetful morphism MATH, then since the class MATH on MATH is the pullback via MATH of the MATH class on MATH, one can lift the same formula to MATH to obtain MATH, where this equation is now regarded as being on MATH, MATH is the pullback via MATH of the divisors with the same name on MATH, and MATH is the forgetful morphism MATH. Suppose that MATH. Using the lifting formula and canceling trivial terms, we obtain MATH . Since MATH, we have MATH where the right hand side is understood to vanish if an exponent is negative. Integration over the fiber of MATH yields the desired result. Inclusion of the additional MATH and MATH classes into the correlators does not change the argument, since MATH, and similarly for MATH. Finally, the exceptional cases follow from dimensional considerations and the fact that on MATH, MATH is the identity element in cohomology provided that MATH. The dilaton equation is proved by a similar analysis, where the exceptional case can be computed by using the explicit presentation for MATH on MATH to obtain MATH . Finally, the equation MATH is obtained by combining the dilaton equation and the grading REF . |
math/9905034 | The proof follows from the facts that MATH forms a NAME, and that MATH and MATH are lifts of the analogous classes on the moduli space of stable curves, and from REF . |
math/9905034 | On MATH, the class MATH can be written in terms of boundary classes as MATH . This equation is obtained from lifting the analogous relation on MATH. The classes MATH can be written similarly by applying an element of the permutation group MATH. The recursion relation follows from this presentation and the restriction properties of the MATH to the boundary strata. |
math/9905034 | The proof of the first topological recursion relation arises from the relation on MATH which is obtained from lifting the analogous relation from MATH. The action of MATH yields MATH. This, combined with the restriction properties of the MATH classes, yields the desired result. The second comes from the presentation of MATH on MATH and the restriction properties of MATH. |
math/9905034 | CITE write down a formula (see REF for an explicit proof) for the large phase space potential MATH which in our case is MATH where MATH. The term MATH is equal to the small phase space potential MATH, evaluated at MATH for all MATH, but the small phase potential MATH vanishes. |
math/9905034 | MATH is nonempty if and only if MATH, where MATH and MATH for all MATH. The genus zero correlators are given by MATH where MATH is the (top) NAME class of degree MATH . The class MATH vanishes unless MATH for all MATH by REF . Furthermore, the correlator can only be nonzero if MATH. If MATH then the dimensionality condition becomes MATH, in which case MATH is the identity. This proves the first part of the proposition. If MATH then the dimensionality condition becomes MATH . If this condition is satisfied then MATH. The correlator is MATH . The right hand side can be computed using the relation for the class MATH in REF , which becomes, in genus zero, MATH where MATH and MATH are uniquely determined by the divisor MATH. Let MATH denote the divisor MATH on MATH. Plugging in this formula, one obtains (after doing a little case by case analysis to write MATH and MATH in terms of MATH) MATH where MATH. Since each MATH is NAME dual to the (topological) homology class represented by a point, one has MATH . Similarly, each class MATH can be represented by MATH for some MATH. Therefore, one obtains MATH . The right hand side of this equation can be shown to be equal to MATH, an elementary but not obvious identity. |
math/9905034 | The large phase space, genus zero potential MATH is completely determined by its values on the small phase space by the topological recursion relations. Let MATH denote the small phase space potential, which must satisfy the WDVV equation since MATH yields a NAME. Furthermore, the grading REF shows that the small phase space potential is a polynomial in the variables MATH of degree of at most MATH. One then performs an induction on the degree of the polynomial to show that MATH is uniquely determined by the above data and the WDVV equation. The proof is straightforward. |
math/9905034 | REF shows that when MATH the class given by REF satisfies the axioms of a virtual class, and REF implies that the large phase space potential of the corresponding MATH-spin NAME is equal to the generating function of tautological intersection numbers on MATH (the large space potential of pure topological gravity). By NAME 's theorem CITE, this generating function coincides with the potential function of the NAME hierarchy, which is the same as the MATH hierarchy. |
math/9905034 | In this case, the conjecture means that the genus zero part MATH of the large phase space potential REF of the MATH-spin NAME REF coincides with the potential MATH of the semiclassical limit of the MATH hierarchy. In genus zero the virtual class MATH exists by REF . From REF it follows that the corresponding potential function MATH satisfies the string REF . Because of the uniqueness of the potential function of the MATH hierarchy (and its semiclassical approximation) all that remains is the proof of the following proposition. |
math/9905034 | By REF there is a unique formal power series MATH, of the proper grading, satisfying the equations of REF , WDVV, and the genus-zero topological recursion relations. CITE shows by a straightforward computation that any such power series yields a solution of the semiclassical limit of the MATH hierarchy. |
math/9905034 | The proof follows from REF and the fact that the potential of the NAME structure on the base of the versal deformation of the MATH singularity is equal to the potential MATH of the semiclassical limit of the MATH hierarchy (compare CITE). |
math/9905037 | See CITE, CITE. |
math/9905037 | CASE: Let MATH be associated to MATH, let MATH be the NAME morphism (relative to MATH) on the dual plane of MATH, and suppose that MATH is not defined over MATH. Then, since the envelope is defined over MATH and MATH is MATH-rational, MATH would belong to two different components of the envelope, namely MATH and MATH. This is a contradiction because the point is non-singular. CASE: This follows from REF . |
math/9905037 | That a set of MATH lines no three concurrent satisfies the bound is trivial. Let MATH be the equation of MATH, let MATH be the factorization of MATH in MATH, and let MATH be the curve given by MATH. For simplicity we assume MATH even, say MATH. Setting MATH, MATH and MATH we have MATH. The singular points of MATH arise from the singular points of each component or from the points in MATH, MATH. Recall that an irreducible plane curve of degree MATH has at most MATH singular points, and that MATH, MATH (NAME 's Theorem). So MATH . |
math/9905037 | CASE: Let MATH be the equation of MATH over MATH. By REF , MATH admits a factorization in MATH of type MATH with MATH and MATH. Let MATH be the plane curve given by MATH . Then MATH satisfies the hypothesis of REF and it has even degree by REF . From REF and NAME 's theorem, for each line MATH (in the dual plane) corresponding to a point MATH, we have MATH where MATH, and so at least MATH points corresponding to unisecants of MATH belong to MATH. Since MATH (see REF ) and MATH, then MATH and from REF we have that at least one of the unisecant points in MATH, says MATH, is non-singular. Suppose that MATH goes through MATH. The point MATH is clearly MATH-rational and MATH is not a point of the curve of REF: otherwise MATH (see REF ). Then, MATH and so MATH is the tangent of MATH at MATH. Therefore MATH is not an inflexion point of MATH, and the proof of REF is complete. |
math/9905037 | CASE: This follows from the proof of REF . CASE: This follows from REF in the appendix. |
math/9905037 | Let MATH. Suppose that it is non-singular and an inflexion point of MATH. Then, from REF and the definition of MATH, we have that MATH is not the tangent line of MATH at MATH, that is, we have that MATH. Now suppose that MATH is either singular or a non-inflexion point of MATH. Then from REF we have MATH and the result follows from NAME 's theorem applied to MATH and MATH. |
math/9905037 | CASE: Let MATH be an irreducible envelope of MATH and MATH the degree of MATH. If MATH, then MATH so that MATH and the result follows. So let MATH. Then from CITE we have that MATH is NAME classical and REF follows from REF . |
math/9905037 | CASE: As we mentioned in REF , MATH. Since MATH for MATH, REF follows from REF . CASE: If would exist a NAME classical irreducible envelope of MATH, then from REF we would have MATH so that MATH. |
math/9905037 | We can assume MATH. If MATH is not a power of MATH, by the MATH-adic criterion CITE we have MATH and MATH. |
math/9905037 | From the MATH-adic criterion CITE, MATH. Then from REF we have MATH and the result follows. |
math/9905037 | From REF and MATH follow that MATH and so that MATH. From MATH and MATH we have that MATH and it follows the assertion on MATH. The bound on MATH follows from REF . |
math/9905037 | Let MATH; so MATH as MATH. From REF we have that MATH so that MATH. Let MATH. If MATH (so MATH), then from REF and MATH we would have that MATH which is a contradiction for MATH. Let MATH. Then MATH and MATH and MATH. Thus from REF we have MATH, that is, MATH. This is a contradiction since by REF we must have MATH. This eliminates the possibility MATH. The other cases can be handled in an analogous way. |
math/9905037 | If MATH, then from MATH we would have that MATH and so, from REF , that MATH, a contradiction. |
math/9905039 | Indeed, REF shows that the subsheaf MATH of MATH of sections having a norm with moderate growth is a meromorphic bundle (compare CITE). This meromorphic bundle contains MATH because of REF . As both bundles MATH and MATH coincide on MATH, they are equal. If MATH is irreducible, the object MATH is stable: if MATH is a proper subbundle of MATH invariant by MATH, REF shows that CASE: either MATH and the existence of MATH does not contradict stability, CASE: or MATH is finite, and then MATH can be extended as a meromorphic subbundle of MATH; this is impossible as MATH is irreducible. The object MATH is thus stable and has degree MATH, because of REF . We may apply REF to get the harmonic metric satisfying the desired properties. |
math/9905039 | Let MATH be the MATH complex associated with the connection MATH and the harmonic metric MATH given by REF (the complete metric on MATH being fixed as above). This complex is equal to the MATH complex associated with MATH and MATH, as MATH and MATH are mutually bounded, hence is isomorphic to MATH, according to REF . Let MATH be the NAME form associated with the fixed metric on MATH and MATH be the form associated with a MATH metric on MATH. Standard arguments show that the previous results imply that the operator MATH induces an isomorphism MATH . The argument of CITE shows that, after the identification MATH, this morphism corresponds to MATH induced by the cup product with MATH. This proves the corollary. |
math/9905039 | Let MATH be such that MATH has a formal decomposition. Let MATH and MATH. As MATH, we have an isomorphism MATH. Let MATH be an elementary meromorphic connection in the MATH-variable equipped with a formal isomorphism MATH, given by NAME 's theorem. Put MATH . As MATH and MATH are elementary, this isomorphism comes from an isomorphism MATH. Hence, there exists a meromorphic bundle with connection MATH in the MATH-variable such that MATH: it is the invariant part of the meromorphic bundle with connection MATH with respect to the automorphism induced by MATH. Notice that, MATH being compatible with MATH and MATH by definition, it defines a formal isomorphism MATH. |
math/9905039 | Recall that for MATH and MATH we have MATH . We will use the following identities: MATH . The matrix MATH of MATH in the basis MATH is MATH and the matrix of the metric connection MATH is MATH . The computation gives MATH . In the basis MATH, the curvature MATH has matrix MATH with MATH . The matrix of MATH in the orthonormal basis MATH is thus MATH and the ``acceptability" is clear. |
math/9905039 | Denote by MATH (respectively, MATH) the matrix of MATH (respectively, MATH) in the basis MATH. We have MATH . Put MATH and MATH, MATH. The matrix MATH of the operator MATH in the basis MATH is MATH. The pseudo-curvature MATH has matrix MATH . Put MATH and MATH. Then MATH . It is thus enough to prove the lemma when MATH. One shows that in the basis MATH with MATH, the matrix of MATH is zero and the matrix of MATH is equal to MATH which is holomorphic, hence MATH. |
math/9905039 | Working in the basis MATH and denoting by MATH the matrix of MATH in this basis, we see that the matrix of the hermitian form MATH is equal to MATH, if MATH denotes the matrix of MATH. As MATH is orthogonal and commutes with MATH and MATH and as MATH, this matrix is equal to MATH. |
math/9905039 | Let MATH be the rank-one MATH-submodule of MATH which is equal to MATH on MATH and which is generated by the section MATH near MATH, for any MATH. The residue formula and the fact that MATH is an integer give MATH . Indeed, if MATH is trivial, this is the usual residue formula for meromorphic differential forms; if MATH is not trivial, there exists on its dual MATH a logarithmic connection which satisfies the residue formula; the residue formulas for the trivial bundle MATH and for MATH give the formula for MATH. On the other hand, in a punctured neighbourhood of MATH, the MATH-form MATH is identically MATH. The curvature of MATH exists as a MATH-current and the degree of MATH can be computed with this current. Taking into account the NAME currents at each MATH, one gets MATH . The lemma follows. |
math/9905039 | Let MATH with MATH. Let MATH containing MATH. We assume that MATH contains at most one zero of MATH and, if it contains one, this zero belongs to MATH. Let MATH (respectively, MATH) be a one-form (respectively, two-form) on MATH such that MATH and MATH (respectively,). We will show that there exists a function MATH (respectively, a one-form MATH) on MATH such that MATH and MATH (respectively,). This will give the vanishing of MATH (respectively, MATH). By assumption, MATH is monotonic with respect to MATH or to MATH on MATH. Choose MATH with MATH. There exists then (see for example, CITE) a sequence MATH converging to MATH and a sequence of one-forms MATH with coefficients in MATH (respectively, of two-forms MATH) with support in MATH such that MATH . CASE: We assume here that MATH does not vanish on MATH. Let us begin with MATH. Let MATH, with MATH on MATH and MATH outside MATH. Put MATH and define MATH . We will consider the decreasing case for instance. One has a NAME inequality (see for instance CITE) MATH with MATH because MATH is decreasing on MATH. We hence have MATH . We thus deduce the existence of MATH on MATH such that MATH . At this step, we can repeat the proof for MATH and get the vanishing of MATH except maybe at points where MATH vanishes. Up to now we did not use the assumption MATH or MATH. As indicated above, we will now assume that MATH, and leave the regular case MATH to the reader. On the other hand we have MATH . We deduce as above that the sequence MATH has a limit in MATH. Therefore, we have MATH and MATH since MATH. We are hence reduced to the case where MATH with MATH. Put MATH. Assume that MATH does not vanish on MATH. Then there exists, for all MATH, a number MATH such that the function MATH is monotonic, as MATH is, on MATH. Let us keep this lemma for granted and let us end the proof of REF if MATH and MATH do not vanish on MATH. Then MATH and MATH are monotonic, as MATH is, on MATH for a suitable MATH. We may assume in the following that MATH by choosing MATH smaller. Define MATH . We then have, using a NAME inequality as above, MATH with MATH if for instance MATH is decreasing. But we then have MATH hence MATH. The case where MATH is increasing is analogous. Assume now that MATH vanishes at MATH; from the assumption made at the beginning of the proof, we may even suppose that MATH. Consider the case where MATH on MATH (the other case is treated in a similar way). We then have MATH near MATH, so MATH is decreasing with respect to MATH on MATH. Let MATH be small enough and put MATH . It is decreasing with respect to MATH. Moreover, there exists MATH such that MATH . It will be clear from the proof of REF that we may apply it to MATH (using there MATH instead of MATH). We may conclude the proof in this case, defining MATH by the formula MATH . The constant MATH is now bounded by MATH. We have MATH with MATH real analytic on MATH depending only on MATH. Since MATH is bounded from below on MATH, there exists MATH such that, for MATH, the sign of MATH is equal to the sign of MATH. CASE: We assume now that MATH does not vanish on MATH but MATH vanishes at one point of MATH. Choose MATH with MATH on MATH. Consider the same sequence MATH as above and put MATH . NAME us consider the second case for instance. For a fixed MATH we have MATH with MATH . As MATH does not vanish in MATH, we may apply an argument similar to that of REF to find MATH. We conclude that MATH is bounded by a constant independent of MATH, hence MATH and therefore MATH tends to MATH in MATH. At this step, we can repeat the proof for MATH and thus get the proof of REF . Let us end the proof of REF . We have MATH and one gets in the same way the convergence of MATH in MATH. This shows that MATH and MATH are in MATH. We thus have MATH with MATH and we are reduced to the case where MATH. Then MATH if MATH on MATH, since MATH exponentially when MATH, MATH being fixed. We may thus assume that MATH on MATH. Let MATH be such that MATH, so MATH is increasing on MATH and decreasing on MATH. Put then MATH . We have, using once more NAME inequalities, MATH with MATH and analogously MATH . We hence get MATH with MATH independent of MATH. Therefore we have MATH . |
math/9905039 | Let us fix a decomposition MATH lifting the formal decomposition, and let MATH be the first projection. One may write MATH where MATH is asymptotic to MATH, and MATH is asymptotic to MATH. Then MATH is equal to MATH, because MATH is a horizontal section of the meromorphic connection MATH; as MATH has a regular singularity, any horizontal section in a sector which is asymptotically equal to MATH vanishes identically. On the other hand, REF shows that MATH. Consequently, MATH and MATH induce the same morphism MATH. |
math/9905041 | Let MATH be the subset MATH in MATH. NAME 's Theorem gives that MATH where MATH is the inward-pointing unit normal to MATH. But for large MATH we have MATH and MATH on MATH, so that the right-hand side of REF is MATH. Since MATH we see that the right-hand side of REF tends to zero as MATH, and this proves REF . The point about the power MATH is that MATH away from REF in MATH. Using the definitions of radius function and ALE manifold one can show that MATH, as we want. Using NAME 's Theorem again we find that MATH . But for large MATH we have MATH and MATH. Thus, letting MATH gives REF . |
math/9905041 | This is an analogue for MATH of REF . If MATH for MATH and MATH, then by CITE we have MATH where MATH is the volume of the unit sphere MATH in MATH. This is NAME 's representation for MATH. Let MATH be a radius function on MATH. Then MATH, so REF gives MATH . We split this into integrals over the three regions MATH, MATH and MATH in MATH. Estimating the integral on each region separately we prove MATH . In REF , if MATH then MATH for some MATH depending only on MATH and MATH, and so MATH and MATH. One can extend this to show that MATH and MATH for some MATH using REF estimates, as in CITE. The difficulty in doing this is to correctly include the powers of MATH involved in the weighted NAME norm. To do this, for each MATH we consider the ball MATH of radius MATH about MATH in MATH. On this ball we have MATH, MATH for MATH, and MATH. Using the NAME interior estimates on the unit ball in MATH and rescaling distances by a factor MATH, we show that MATH for MATH and MATH on the interior of MATH. Thus MATH and MATH, completing the proof of REF . Next we prove REF . Suppose MATH, MATH and MATH. Then MATH by REF , since MATH and MATH and MATH. Thus, given MATH, there can only exist MATH with MATH if MATH. So suppose that MATH, and define MATH by MATH . Since MATH the term involving MATH in this integral vanishes, so the equation reduces to REF and thus MATH. From REF we see that MATH and estimating as before shows that MATH when MATH. Thus MATH and MATH. The rest of REF follows as above. |
math/9905041 | The theory of weighted NAME spaces on AE manifolds and the Laplacian is developed by CITE, who restrict their attention to the case MATH. In particular, they prove REF for the case MATH, MATH and MATH, CITE. Their proof uses a result equivalent to REF in the case MATH and MATH. By using REF together with the methods of CITE one can show that REF applies not only to MATH with its Euclidean metric, but also to any ALE manifold MATH asymptotic to MATH. This proves REF of of the Theorem immediately. For REF , let MATH, and define MATH by REF . Then by REF we have MATH. Also MATH by REF , and so MATH lies in MATH and has integral zero on MATH. Since MATH we have MATH for MATH, as we have to prove. Applying REF for MATH to MATH, we see that there is a unique MATH with MATH, which satisfies MATH . Defining MATH gives MATH as we want. Clearly MATH, and the inequality MATH then follows from REF and the estimate on MATH above. |
math/9905041 | Let MATH be the NAME form of MATH. Then if MATH is a smooth function on MATH we have MATH . Also, if MATH is a real MATH -form on MATH and MATH it can be shown that MATH where MATH is the NAME star and MATH the volume form of MATH. REF hold on any NAME manifold of dimension MATH. Define a function MATH on MATH by MATH. Since MATH, it follows that MATH. Now suppose for simplicity that MATH. Then by REF there exists a unique function MATH with MATH. Set MATH, which is an exact REF-form in MATH. As MATH we can use the last part of REF to prove that MATH, for some MATH. By REF we have MATH, so REF gives MATH . Let MATH be a radius function on MATH, and define MATH for MATH. Integrating REF over MATH and using NAME 's Theorem gives that MATH . But for large MATH we have MATH, MATH and MATH on MATH, and MATH. Thus the right-hand side of REF is MATH. As MATH, taking the limit as MATH shows that MATH, and so MATH on MATH. Thus MATH, as we have to prove. We have proved the theorem assuming that MATH, but we wish to prove it for all MATH. If MATH and MATH then MATH for any MATH with MATH, and so from above we have MATH for some unique MATH in MATH. However, if MATH and MATH, one can show that MATH as we want. This is because MATH is a stronger derivative of MATH than MATH is, and contains more information. |
math/9905041 | Let MATH be an ALE NAME metric on MATH, with NAME form MATH. By REF there exists a closed, compactly-supported, real MATH -form MATH on MATH with MATH in MATH. Define MATH. Then MATH is an exact real MATH -form on MATH. Now the NAME form of MATH on MATH is MATH. So from the definition of ALE NAME metric we see that MATH, and therefore MATH as MATH has compact support. Thus by REF there is a unique real function MATH with MATH, and we have MATH. Let MATH be a smooth function with MATH for MATH and MATH for MATH. For each MATH define a closed MATH -form MATH by MATH . Then MATH wherever MATH, and MATH wherever MATH and outside the support of MATH. It is easy to show that MATH is a positive MATH -form for large MATH, which therefore defines a NAME metric MATH on MATH. Define MATH to be MATH for some MATH sufficiently large that MATH is positive, MATH on the support of MATH and MATH. Then MATH is an ALE NAME metric in the NAME class of MATH, and where MATH we have MATH, since the NAME form of MATH is MATH, the NAME form of MATH is MATH, and MATH as MATH. |
math/9905041 | Let MATH in MATH. Then there are coordinates MATH on MATH in which MATH acts by MATH . As MATH acts freely on MATH we can take MATH for MATH. Since MATH we know that MATH preserves a complex symplectic form on MATH, and we can choose MATH so that this form is MATH. Thus REF gives MATH for MATH. But as MATH this implies that MATH for MATH. Therefore MATH for all MATH in MATH. So by CITE it follows that MATH is a terminal singularity, as defined in CITE. Terminal singularities are essentially singularities which have no crepant partial resolutions. To be more precise, a crepant resolution of a terminal singularity has no exceptional divisors. Thus, if MATH is a crepant resolution of MATH then MATH. By NAME duality for manifolds with boundary we see that MATH, which is a contradiction, as MATH must contain a complex curve. So MATH has no crepant resolutions. |
math/9905043 | If MATH is a generic point of MATH then MATH is locally isomorphic to MATH near MATH, and MATH is locally isomorphic to MATH near MATH, as MATH is a local product resolution. But if MATH then MATH, and MATH is a generic point of MATH, so that MATH is also locally isomorphic to MATH near MATH, and MATH is locally isomorphic to MATH near MATH. Hence MATH and MATH are locally isomorphic. In fact they are globally isomorphic, and this gives an isomorphism MATH, which makes REF commutative. |
math/9905043 | Let MATH be fixed throughout the proof. Above we defined a lot of notation such as MATH and so on, associated to MATH. The corresponding data associated to MATH will be written MATH, etc., in the obvious way. We will express the data for MATH in terms of that for MATH. Define the indexing set MATH by MATH, and for each MATH set MATH. Then MATH is a subgroup of MATH is a finite set of subspaces of MATH. The two special elements of MATH are MATH and MATH. Also MATH, MATH and MATH are the same as MATH, MATH and MATH for each MATH. Let MATH, MATH and MATH be as in REF . It can be shown that there is a unique map MATH, such that the product MATH of MATH with the identity on MATH makes the following picture into a commutative diagram: MATH . It easily follows that MATH is a local product resolution of MATH. |
math/9905043 | We shall use the notation defined in the proof of REF . In addition, let MATH be the Euclidean metric on MATH. Let MATH, so that MATH and MATH. Writing the Euclidean metric MATH on MATH as MATH, REF with MATH replaced by MATH becomes MATH on MATH. Using MATH to pull REF from MATH back to MATH, and substituting MATH since MATH by REF , gives MATH on MATH. Hence, subtracting REF gives MATH . By restricting to MATH for MATH and taking MATH to be large, we show that MATH on MATH, for all MATH. Thus MATH is a QALE NAME metric on MATH. |
math/9905043 | Let MATH be smooth with MATH for MATH and MATH for MATH, where MATH is the constant in REF . For each MATH, define a smooth function MATH on MATH by MATH . The idea here is that MATH at distance at least MATH from the pull-back of MATH in MATH, that MATH at distance no more than MATH from the pull-back of MATH, and that between distances MATH and MATH we join the two possibilities MATH and MATH smoothly together using a partition of unity. Note that MATH in MATH. For MATH, let MATH be integers satisfying MATH . It can be shown that these equations have a unique solution MATH. Now define MATH . As MATH in MATH, we see that MATH is smooth. It turns out that this MATH is of NAME potential type on MATH and asymptotic to MATH for all MATH, so that it satisfies the conditions of the theorem. The proof of this is rather complicated, and we will not give it in full. Instead, we will explain the important points under the simplifying assumption that MATH for all MATH. In this case MATH, so MATH. Thus we may rewrite REF as MATH . Now MATH acts on MATH and MATH, and MATH is MATH-invariant by REF . Thus MATH is also MATH-invariant, and pushes down to a function on MATH. From REF, MATH induces an isomorphism between MATH and MATH. In fact MATH is the push-forward of MATH under this isomorphism. The factor MATH in REF ensures this, because each point of MATH pulls back to an orbit of MATH in MATH, consisting of MATH points. Thus, abusing notation a little, we can write MATH, because the factor MATH in REF compensates for the fact that each generic point in MATH pulls back to MATH points in MATH. But MATH away from MATH, and so away from MATH in MATH we have MATH, again by an abuse of notation. We must prove that MATH satisfies REF . One way to interpret this is to say that MATH is divided roughly into overlapping regions corresponding to MATH, where in the MATH region MATH is locally isomorphic to MATH and MATH. Now on the MATH region we have MATH if MATH and MATH if MATH, by REF . Therefore on the MATH region we have MATH by REF , as we want. This idea can be used to show that REF holds away from MATH in MATH. It remains to consider the parts of MATH near MATH. Within distance MATH of MATH in MATH, the functions MATH in REF are not all equal to REF, and so we do not have MATH for all MATH. The reason for introducing the MATH is that the push-forward of MATH to MATH may not extend smoothly to MATH. So we modify MATH to get MATH, whose push-forward is zero near MATH and does extend smoothly to MATH. But we then have to make sure that the `errors' due to the MATH are within the bounds allowed by REF . One can show using REF that this is always so, and the proof is complete. |
math/9905043 | Let MATH be as in the proof of REF . Then MATH is positive outside a compact set in MATH, because at large distances from MATH we have MATH for some MATH, and so MATH is positive because MATH is positive. Let MATH, and choose a smooth function MATH such that MATH at distance less than MATH from MATH in MATH, and MATH at distance more than MATH from MATH in MATH, and MATH, MATH. For large MATH it turns out that MATH satisfies the conditions of the proposition. |
math/9905043 | Let MATH, and suppose for simplicity that MATH. Then MATH, so by REF there exists a unique MATH such that MATH and MATH as MATH in MATH. Since MATH, we see that MATH on MATH. Suppose that MATH at some point of MATH. Then MATH is non-constant and has a maximum in MATH, since MATH and MATH as MATH in MATH. But this contradicts the maximum principle, as MATH. Therefore MATH, and MATH. Similarly we show that MATH, and so MATH. To complete the proof, it is enough to show that MATH and MATH for some MATH independent of MATH. We shall do this by applying REF to balls of radius MATH about MATH, for small MATH. Let MATH be the balls of radius REF about REF in MATH, where MATH. Fix MATH, and choose a vector space isometry MATH. Let MATH be small, and define a map MATH by MATH, where MATH is the exponential map, and we have identified MATH and MATH. Using the definition of QALE metric, we can show that if MATH is small, then CASE: MATH is a diffeomorphism between MATH and MATH, CASE: MATH is a diffeomorphism between MATH and MATH, and CASE: the metric MATH on MATH is close to the Euclidean metric MATH on MATH in MATH, and MATH is bounded independently of MATH. The idea here is that MATH is the approximate length-scale at which the metric on MATH near MATH differs from a Euclidean metric. Thus, balls of radius MATH and MATH should resemble Euclidean balls of the same radius provided MATH is sufficiently small. Define an operator MATH and functions MATH on MATH by MATH . Then MATH is the Laplacian of MATH on MATH, which is close to MATH in MATH, and therefore MATH is close in MATH to the Euclidean Laplacian MATH on MATH. Also MATH, as MATH. On MATH we use the Euclidean metric MATH, and on MATH we use the metric MATH. Since MATH, it follows that MATH . Thus, as MATH on MATH we have MATH on MATH, and as MATH on MATH we have MATH on MATH. Therefore REF shows that MATH on MATH for MATH, together with the appropriate NAME estimate. So MATH on MATH for MATH, together with the appropriate NAME estimate. We have shown that MATH is bounded in MATH, which completes the proof. |
math/9905043 | From REF we see that MATH lies in MATH, and hence in MATH. So there exists a unique MATH with MATH. Clearly MATH is smooth, MATH and MATH for some MATH. Also, using REF we can show that MATH for some small MATH, and thus by the proof of the previous theorem we have MATH. Thus MATH satisfies REF , as we want. |
math/9905043 | Suppose for a contradiction that MATH are distinct NAME QALE NAME metrics on MATH in the same NAME class, and let MATH be of the smallest dimension in which this can happen. Clearly MATH, since otherwise we can replace MATH by MATH, which has smaller dimension. Let MATH be asymptotic to metrics MATH on MATH for MATH, as in REF . Then MATH are NAME, and in the same NAME class. Thus MATH for all MATH in MATH, since MATH when MATH. Let MATH be the NAME forms of MATH. Then by REF we have MATH, where MATH is a function of NAME potential type on MATH. Since MATH for MATH, the functions MATH of REF are zero for MATH, and MATH satisfies REF . But MATH have the same NAME curvature, so that MATH. Thus MATH and MATH satisfy the second version of the NAME conjecture for QALE manifolds. By uniqueness in the conjecture we see that MATH, so that MATH, a contradiction. Thus MATH is unique in its NAME class. |
math/9905043 | Let MATH have NAME forms MATH. Then REF shows that MATH, where MATH is a unique function of NAME potential type on MATH. As MATH is the only NAME QALE metric in its NAME class by REF , we see that MATH, and MATH is asymptotic to MATH when MATH and MATH. Using these facts one can show that the MATH satisfy REF . Therefore we may apply REF to find a function MATH of NAME potential type on MATH such that MATH is the NAME form of a QALE NAME metric MATH on MATH, which is asymptotic to the NAME metrics MATH on MATH. Also, as MATH, by construction MATH satisfies REF outside a compact subset MATH of MATH. Let MATH be the NAME form of the Euclidean metric MATH on MATH, and let MATH be the holomorphic volume form on MATH, which is well-defined as MATH, since MATH is a crepant resolution. Calculation shows that MATH on MATH, where MATH. Let MATH. Then MATH is a nonsingular holomorphic volume form on MATH. Define a smooth real function MATH on MATH by MATH. So MATH has NAME form MATH, as we want. We must show that MATH satisfies REF . For simplicity we will restrict our attention to the case MATH for all MATH. Then REF gives MATH . Let MATH satisfy MATH. Then putting MATH in MATH on MATH shows that MATH near MATH. Thus MATH and MATH, and so MATH near MATH. If MATH in MATH and MATH with MATH, then MATH near MATH. Also the function MATH defined in REF satisfies MATH near MATH, since MATH near MATH for all MATH. But MATH, and so MATH, where MATH is defined by REF . Therefore the metric MATH on MATH has NAME form MATH near MATH. But MATH is NAME. So MATH is the NAME form of a NAME metric near MATH. It is then not difficult to show that MATH near MATH. Thus we have MATH wherever MATH on MATH, for all MATH. Multiplying out MATH and rearranging gives MATH where `' are terms of order at least REF in MATH. Substituting REF into the equation MATH, rearranging and using REF shows that MATH where `' are terms of order at least REF in the MATH or at least REF in MATH, and the equation holds for MATH with MATH. Thus MATH is roughly quadratic in the MATH, to highest order. Using REF and the special properties of the MATH and MATH, we can show that REF holds. The proof of this is complicated, and we will not give it. The basic idea is that MATH is made by combining the NAME metrics MATH. The dominant terms in MATH result from interference between MATH and MATH for MATH, and contribute MATH to MATH on MATH, and corresponding terms to MATH. However, if MATH with MATH then MATH is NAME and asymptotic to both MATH and MATH. So we introduce no extra NAME curvature by combining MATH and MATH in this case, which is why the sum in REF is restricted to MATH with MATH. More details of this argument are given in CITE. |
math/9905046 | CASE: The first part is proven in CITE. As for the second part, note the following equivalences, which hold for each MATH using REF MATH . CASE: Using MATH in REF we obtain especially that a MATH-submodule of the form MATH is a behavior. Thus, write MATH with some appropriate MATH. Then MATH is a behavior (see also CITE) and moreover MATH . REF follow now from REF with REF . |
math/9905046 | follows from REF , and REF . |
math/9905046 | CASE: By REF we have MATH. Without loss of generality we may assume that MATH is column-reduced, that is, MATH is the sum of the column degrees of MATH. From CITE we obtain matrices MATH such that MATH. Indeed, the parameter ord-MATH in CITE is equal to the degree, compare CITE. Setting MATH and using REF we obtain the desired representation. Furthermore, CITE shows that the triple MATH is minimal with respect to row and column size of the matrix MATH (or MATH). Hence, use of CITE leads to REF - REF . REF follows from CITE and REF . |
math/9905047 | The proof of this theorem follows from the definitions of mean curvature and NAME map. A complete proof can be found in CITE, and the original proof in CITE. |
math/9905047 | Because of the above observations we see that the condition MATH implies the existence of a MATH function MATH such that MATH and which makes the second variation of area REF negative. Namely MATH is the initial point of a REF-parameter family of minimal surfaces with boundary equal to MATH, such that each element in this family has area smaller than the area of MATH. Hence MATH cannot be locally area-minimizing (and therefore cannot be stable). |
math/9905047 | The proof of the three assertions in the claim follows easily from the fact that the surfaces being considered are embedded and stable, and that MATH is the homology boundary of these surfaces. |
math/9905047 | First notice that the theorem certainly holds if MATH consists of the union of the two planar regions bounded by MATH and MATH. Hence we need to prove the theorem for connected MATH. The proof will be divided in five steps. CASE: For MATH sufficiently close to MATH, the components of MATH away from the crossing points are graphs over MATH. For each MATH, let MATH be a disk in MATH with center MATH and radius MATH, chosen so that the following two conditions are satisfied: CASE: REF the disks MATH are mutually disjoint; CASE: REF for each MATH, MATH is topologically a MATH. Let MATH, and let MATH be the vertical cylinder over MATH. Moreover, let MATH. Let us notice that along the curves MATH and MATH, the normal vector to MATH must become arbitrarily vertical as MATH approaches MATH. To see this, for example along MATH, suppose MATH is vertical at a boundary point MATH of MATH. Consider, above and away from MATH, a piece of a half strictly unstable vertical catenoid with boundary curves a larger circle in the plane MATH and a smaller circle in the plane MATH (this catenoid is a graph above the plane MATH, except along the smaller boundary circle). Translate vertically down this catenoid piece until it makes first contact with MATH at some point MATH; the maximum principle implies first that MATH is a boundary point both of the catenoid piece and of MATH, and then that the two minimal surfaces MATH and the catenoid piece coincide in a small neighborhood of MATH. This contradiction shows that MATH cannot become vertical along its boundary, away from the crossing points. Moreover we have that locally the surface MATH is situated on the same side with respect to the vertical cylinder MATH, otherwise one could translate a catenoid in such a way that the first point of contact would be an interior point, which contradicts the maximum principle. With the same notation as above, let us notice that there exists the upper lower bound of the angle MATH formed by MATH with an arbitrary catenoid piece intersecting MATH only in MATH; in fact this infimum is given by the angle between the tangent plane to MATH at MATH and the horizontal plane MATH. Moreover, for an arbitrary MATH in MATH, the upper lower bound of MATH approaches zero as MATH gets closer to MATH. Let MATH . The compactness of MATH guarantees that MATH is well defined. Furthermore, what has been said above implies that MATH . By NAME 's estimate CITE there exists a universal constant MATH such that MATH, where MATH is the Gaussian curvature at a point of a stable minimal surface and MATH is the distance between that point and the boundary of the surface. Now let us choose MATH sufficiently close to MATH so that MATH. Let MATH be a point in MATH, and let MATH be the orthogonal projection of MATH on the plane MATH. Let MATH be the horizontal distance between MATH and MATH. Let MATH be the part of the vertical cylinder MATH with height MATH and containing MATH in its interior. Let MATH be the part of the vertical cylinder MATH with height MATH which is contained in MATH. Let us suppose that our assertion in REF is not true. It will be therefore possible to find a point MATH in MATH whose normal vector MATH is horizontal, namely MATH. There follows the existence of a geodesic MATH in MATH with unimodular velocity, and such that MATH, MATH, and the curvature MATH of MATH in MATH is bounded above by MATH because of NAME 's estimate. Notice that, since MATH (and since MATH implies MATH), MATH lies in MATH for MATH. Considering the estimate for the curvature of MATH, the estimate MATH for MATH, and the integral MATH we can conclude that the length of the curve MATH in the unit sphere MATH is less than MATH, for MATH. Since MATH, we have that MATH and MATH are both less than MATH, and that MATH is larger than MATH, for MATH. Hence it is MATH . Likewise we have: MATH and MATH because of the condition MATH. But this implies that MATH intersects one of the two horizontal disks of MATH for some value of MATH between MATH and MATH, which is impossible, given the way MATH was constructed. Hence MATH is never horizontal on MATH, which therefore is union of graphs over MATH. Moreover, notice that at most two sheets can lie over MATH, otherwise MATH would not be embedded (at least along the boundary). If the surface is area-minimizing, then at most one sheet lies above each point, but if no assumption of area minimality is made, then there could be points above which there are two sheets. Let us remark that for MATH the above graphs are ``almost horizontal", since it is clear the above argument can be strengthened to show that, for any MATH, it is possible to make MATH for each MATH, for MATH sufficiently close to MATH. CASE: MATH is the union of two disjoint graphs in the neighborhood of each MATH in MATH, where MATH and MATH are components of MATH with multiplicities MATH and MATH respectively. To see this, consider a small vertical cylinder MATH whose vertical axis contains MATH and whose height is MATH. Moreover, let MATH. Now homotetically expand MATH, with coefficient of homothety MATH, and center of the homothety in MATH. The image of MATH under the homothety converges, as MATH, to a minimal surface having a simply connected component MATH which is a stable minimal surface contained in a half-space and bounded by a straight line, MATH. The image MATH of MATH via the NAME map is either a single point or a great-circle on MATH containing the north and south pole of MATH, and near the boundary MATH the image of MATH via the NAME map is entirely contained in one of the two hemispheres determined by MATH, because of the way the barrier to get MATH was constructed (theorem MATH). Hence the image under the NAME map of MATH is entirely contained in such half hemisphere, by the hypothesis of stability. By reflection with respect to the line MATH one obtains a complete minimal surface containing a line and having total curvature between MATH and MATH. The two only complete minimal surfaces with total curvature MATH are the catenoid and NAME surface CITE. Since the catenoid does not contain a line of reflective symmetry, and NAME surface's NAME map does not satisfy our conditions, then MATH must be a half plane, and its image via the NAME map must ba a point. Finally, notice that the points contained in the other simply connected component of the homothetic expansion of MATH correspond to points which are contained in the interior of MATH, and hence NAME 's curvature estimate applies to them. These observations complete the proof. CASE: MATH is the union of two disjoint graphs in the neighborhood of each crossing point having multiplicity MATH. This follows from the previous two steps, and from the analysis of the possible liftings to MATH of small circles in MATH around the point under consideration. Our next aim is to study MATH around a crossing point whose multiplicity is MATH. CASE: MATH is topologically a disk in the neighborhood of each crossing point with multiplicity MATH. We will show that in a closed cylindrical neighborhood MATH of each crossing point with multiplicity MATH, the compact surface MATH has genus zero for MATH sufficiently close to MATH. In order to do this, let us homothetically expand the spherical neighborhood MATH with center in a point MATH, with expansion coefficient MATH, where MATH is a point of maximum Gaussian curvature inside MATH. Let us denote by MATH the expanded neighborhood, and notice that the expansion transforms the planes MATH and MATH to two new planes which have distance equal to MATH from each other. Moreover the homothety takes the arcs MATH and MATH to arcs MATH and MATH in MATH which are segments of an almost straight line. Let MATH, and MATH, with the convention that the orthogonal projections of MATH and MATH on MATH lie in the boundary of the same multiplicity one component of the interior of MATH, and the orthogonal projections of MATH and MATH lie in a different multiplicity one component. Now let us join MATH to MATH and MATH to MATH by two geodesic arcs contained in the homothetic expansion MATH of MATH. Because MATH is a graph over MATH away from the crossing points with multiplicity MATH, and because for each MATH we have MATH away from these crossing points, for MATH (namely the normal vector to MATH in MATH is almost vertical if MATH is away from the crossing points with multiplicity MATH), we may assume that the two geodesic arcs defined above project orthogonally on two different multiplicity one components of the interior of MATH, and hence do not intersect each other. We will now show that the piece MATH contained in MATH and bounded by the loop union of the four curves MATH, MATH, and the two geodesic arcs previously defined, is a disk. Since MATH is contained inside the boundary of the convex hull of MATH, MATH separates its convex hull (which is simply connected) into two distinct types of regions, one associated with the MATH sign and the other with the MATH sign. Hence MATH is orientable, and consequently also MATH is so. This implies that the NAME map, from the oriented MATH to the unit sphere MATH, is well defined. The Gaussian image MATH of the boundary curve MATH is a curve that lies in a small neighborhood of the spherical region bounded by the union of two great semicircles joining the north and south poles of MATH. Since MATH is stable, MATH is also stable. Because the Gaussian image of a stable minimal surface cannot contain a hemisphere, the image of MATH under the NAME map can only contain one of the two regions in the complement of this neighborhood in MATH, and must be disjoint from the other region. Since the winding number of the NAME map around MATH is one, the NAME map can only cover this region once. This implies that for MATH sufficiently close to MATH it is: MATH . Now, the geodesic curvature MATH is zero along the two almost straight arcs contained in MATH, and along MATH and MATH the geodesic curvature is approximately equal to zero. The sum of the exterior angles where these smooth arcs intersect is between MATH and MATH. Therefore by the NAME theorem, one has that the NAME characteristic of MATH is either zero or one. Since MATH consists exactly of one curve, we can conclude that the NAME characteristic is MATH, and that MATH is topologically a disk. Hence, in small neighborhoods of the crossing points with multiplicity MATH, MATH is topologically a disk. CASE: MATH is approximately helicoidal around crossing points having multiplicity MATH. Normalize MATH by a homothety with center a point MATH of maximum Gaussian curvature, in such a way that MATH on the normalized surface, which we shall denote by MATH. Modulo a translation, we can suppose that the point MATH is the origin. Let us notice that MATH on MATH as MATH approaches MATH, since the NAME map MATH is almost vertical on MATH, except near the crossing points where MATH changes very quickly, and hence the modulus of MATH must be large near the crossing points. Therefore normalizing MATH involves a dilation factor which becomes arbitrarily large as MATH approaches MATH, and hence the two planar curves in the boundary of MATH become arbitrarily straight as MATH approaches MATH, around the crossing points. A result of CITE states that for each sequence of surfaces MATH, MATH, it is possible to extract a convergent subsequence (in the MATH norm) MATH in the compact spherical neighborhood MATH with radius MATH in MATH. This sequence of surfaces can be chosen in such a way that MATH is a subsequence of MATH if MATH. The sequence MATH is the NAME diagonalization, and converges in the MATH norm in arbitrary compact regions to a surface MATH having one or two boundary curves, which must be straight lines. Moreover NAME 's compactness theorem CITE implies that MATH is embedded. The limit surface MATH is simply connected in any compact spherical neighborhood, and therefore is simply connected in MATH. If the boundary of MATH consists of two straight lines, then by NAME reflection principle MATH can be extended to a simply connected minimal surface in MATH properly embedded, without boundary and with infinite symmetry group. By virtue of REF and CITE, such extended minimal surface is a plane or a helicoid. However this surface contains two straight lines which do not intersect and are not parallel to each other, and hence MATH is a piece of a helicoid, with one or two boundary lines. If MATH had only one boundary line, then it could be extended via a rotation of angle MATH around the straight boundary line, producing a properly embedded minimal surface. Because the convergence of the above subsequence to MATH is with respect to the MATH norm, the normal vectors are converging as well, and the extended surface has finite total curvature. By a result of NAME and NAME such a surface must be a plane or a catenoid; however since it is simply connected, it must be a plane. Hence MATH is a half-space. Since the convergence to MATH is of class MATH, we know that the normal vectors along MATH are not constant as MATH, which implies that MATH cannot be a plane. Therefore MATH must be a piece of helicoid with two boundary lines. Since MATH is a graph over MATH and because we are considering a point with multiplicity MATH, MATH is a graph over the components of MATH having the MATH sign. Hence MATH is totally determined. If there was some subsequence which would not be eventually contained in a given MATH-neighborhood of MATH in a given sphere MATH in MATH, then it would be possible to find a subsequence MATH converging to a point not belonging to MATH, which is a contradiction. This allows us to conclude that any sequence MATH such that MATH eventually lies in a predetermined arbitrarily small MATH-neighborhood of MATH inside each compact region of MATH. For an arbitrarily given MATH, let us choose MATH big enough in such a way that the surface MATH be contained in a MATH-neighborhood of MATH. For each pair of points MATH and MATH, with MATH in MATH with normal vector MATH to MATH and MATH in MATH with normal vector MATH to MATH such that MATH, the estimate on the function MATH implies that MATH is bounded away from zero. In fact by choosing MATH sufficiently close to zero and MATH sufficiently large, we will be able to achieve MATH to be arbitrarily close to MATH. It follows that MATH is union of graphs on MATH for MATH sufficiently large, and hence MATH is a one-sheeted graph on MATH around a crossing point with multiplicity MATH, which is in accordance to the end of the proof of REF. This allows us to conclude that MATH converges to MATH in the MATH-norm as one-sheeted graphs, and that the normal vectors are convergent as well; hence we have in addition that MATH converges to MATH in the MATH-norm in any compact sphere, and that MATH is approximately helicoidal in a neighborhood of each crossing point with multiplicity MATH, for MATH sufficiently close to MATH. |
math/9905047 | This follows easily from the description of MATH, and from the observation that a cell decomposition of MATH can be computed via a cell decomposition of the unique varifold MATH determined by MATH, and corresponding to MATH. Such a cell decomposition is equivalent to one that has: CASE: - number of vertices equal to MATH. CASE: - number of edges equal to MATH. CASE: - number of faces equal to MATH. The unique area-minimizing varifold gives rise to a minimal surface having largest genus, since for this varifold MATH. Hence the proof is finished. |
math/9905047 | Since it is not ambiguous, after an appropriate choice of MATH, we will drop the MATH indicating dependence on MATH, in this proof. Let us consider the foliation MATH, and consider the surface MATH which is obtained by deforming MATH via the NAME vector field MATH, namely the set of all the points MATH, as MATH varies in MATH. Let us consider the ``restrictions" of the foliation and of the surface MATH to the subset MATH of MATH, and let us denote by MATH and MATH the transformed images of MATH via the foliation and the NAME vector field, respectively. Because all the surfaces we are considering are compact, there exists the maximum value MATH for which the surface MATH intersects the foliation MATH (after such value the translated surfaces MATH are situated ``above MATH"). By the maximum principle at a boundary point, such last point of contact must be a point MATH contained in the boundary curve MATH. Let MATH be the point where MATH attains its maximum value; since MATH is an almost horizontal graph, there is no loss of generality in the assumption that the maximum value of MATH is attained at MATH (in fact this can always be achieved by slightly deforming the curve MATH). Let now MATH be a very thin strip which is a MATH-neighborhood of MATH with respect to the metric of MATH, MATH, and MATH a point contained in MATH corresponding to the last intersection point of MATH with MATH. Let us notice that: MATH and that certainly, if MATH denotes the value of MATH corresponding to MATH, then we have MATH, which implies that MATH where MATH denotes the ordinary MATH-coordinate of a point MATH in MATH. Now, since the unit normal vector to MATH approaches MATH in a continuous fashion away from the crossing points, the above says actually that the difference between MATH and MATH is at most MATH. Finally, by perturbing MATH slightly in all directions around MATH and applying the above argument to these perturbed curves, the proof of the lemma is complete. |
math/9905047 | Let us consider MATH . Then for each MATH, if MATH is a NAME vector field, one has, as observed in paragraph REF, MATH . Let us choose MATH. Because of our hypotheses, MATH induces a MATH such that REF above is satisfied, namely MATH which implies MATH . Moreover REF above implies MATH which yields the conclusion: MATH . If MATH can be chosen arbitrarily small at the beginning, when MATH one has MATH namely MATH. Clearly in our case we can exchange the roles of MATH and MATH, since MATH is contained in MATH, and have that as MATH, MATH, and vice-versa. In particular, we have that if MATH is unstable, so is MATH, and vice-versa. |
math/9905047 | Let MATH be the NAME vector field on MATH, and without loss of generality let us suppose that the maximum of MATH is attained at some point belonging to MATH (otherwise the proof of the lemma is still valid, as one can easily see: this hypothesis takes care of the ``worst possible case"). Moreover let MATH be a bump function of class MATH on a thin strip MATH containing MATH and having area less than MATH such that MATH is constantly equal to MATH on MATH, constantly equal to zero on MATH (MATH is a strip containing MATH and contained in MATH), and such that MATH. Define now MATH. We get: CASE: MATH, because of the property shown in REF. CASE: MATH, namely MATH . CASE: MATH, since MATH . The assertion hence follows by choosing MATH. |
math/9905047 | Suppose that the assertion stated in the theorem is false. Then there would exist a sequence MATH, corresponding to which there would be a sequence of unstable minimal surfaces MATH with boundary MATH, and described according to REF. Hence there would exist a sequence of NAME vector fields MATH defined on MATH, all having the property proven in REF. Therefore, by induction and by REF, MATH, MATH, it would be possible to fix a positive MATH such that in a MATH-neighborhood (strip) of MATH one could define a bump function having bounded gradient which, when multiplied by MATH would yield a new function MATH defined on the stable part given by MATH (the fact that this would be possible for each MATH follows from REF). But the NAME quotient associated to such a function can be made (dependently on MATH) arbitrarily close to a number which is strictly less than MATH, thus producing a contradiction. For the sake of completeness, let us notice here that REF is of fundamental importance in this proof, because it ensures that none of the functions MATH is the function identically equal to zero. |
math/9905047 | The theorem will be proven by contradiction. Let MATH and MATH be two distinct families (for MATH sufficiently close to MATH) of minimal surfaces with the same boundary MATH, existing for all MATH, and having the same limit varifold MATH. Let MATH be the unbounded connected component of the region MATH of space given by the collection of points in the slab which are ``outside" of MATH, MATH be the union of the bounded connected regions given by the points ``in between" MATH and MATH, and MATH be the closure of MATH. Notice that MATH contains the truncated cylinder above MATH, and that MATH is contained in a small neighborhood of the truncated cylinder above MATH. Notice that MATH strictly contains MATH. Since MATH and MATH are stable by REF, applying the existence theorem by NAME and NAME stated in section MATH to MATH, we obtain the existence of a stable minimal surface MATH ``above" MATH. By REF and NAME, applied to the region MATH, we obtain the existence of a stable minimal surface MATH ``below" MATH. By construction, MATH and MATH are disjoint from each other. Moreover, for MATH sufficiently close to MATH, the two stable minimal surfaces MATH and MATH obtained in this way can be described as stated in REF, namely as approximately helicoidal around the crossing points with multiplicity MATH, and as almost horizontal graphs away from the crossing points with multiplicity MATH, by virtue of REF. Hence MATH and MATH are homeomorphic to the MATH, and furthermore MATH and MATH are normal graphs above MATH, and hence over each other, namely there exists a function MATH such that: MATH that is, for each MATH there exists a unique MATH such that MATH where MATH is the normal vector to MATH in MATH. Hence MATH and MATH converge to the same limiting varifold, as MATH. It also follows from MATH that MATH bounds a product region, say MATH. Since MATH and MATH are normal graphs over each other, we know that the angle between the two normal vectors to MATH and MATH at a boundary point is strictly between MATH and MATH. Then by the minimax theorems due to CITE, and generalized by CITE to the case of nonempty boundary, MATH is also the boundary of an unstable embedded minimal surface MATH, contained in MATH, for all MATH sufficiently small. We now wish to show that MATH is homeomorphic to MATH and MATH, and that MATH actually has the same geometric description as MATH and MATH, which will imply, for example, that MATH. First notice that the connected components of the complement in MATH of the union of small neighborhoods of the crossing points, is the union of almost horizontal graphs, just like MATH and MATH. In fact, the area of each such connected component is almost equal to the area of the corresponding components of MATH and MATH. The proof of this fact follows from comparing the area of MATH with the maximum area of the family MATH, indexed by MATH. The definition of minimax implies that the area of MATH cannot be larger than the maximum area of the family; but since every surface in the family is a graph over MATH, we can estimate the area of MATH by that of MATH and MATH, and the proof of the claim follows as in REF of the proof of REF, since the area estimates just proven allow us to apply NAME 's curvature estimates. So we know that MATH given by very flat graphs away from the crossing points. In the neighborhood of a crossing point with multiplicity MATH, consider the part of MATH bounded by four geodesic arcs, constructed as in REF of the proof of REF. Such quadrilateral region has sum of the external angles strictly less than MATH. By NAME 's MATH-theorem stated in section MATH, applied to an analytic smoothing having total curvature less than MATH of the quadrilateral defined by the geodesic arcs, we know that this quadrilateral region must bound a stable minimal surface which is topologically a disk. This means that MATH can be described as in REF, for all MATH sufficiently small, and that MATH is a graph over MATH. But then, for MATH sufficiently close to MATH, MATH must be stable, by REF. This produces a contradiction and finishes the proof of the uniqueness theorem. |
math/9905049 | The inverse images of MATH and MATH in MATH are represented by algebraic spaces (respectively, schemes), because MATH and MATH are representable (respectively, strongly representable). But these inverse images are open substacks which cover MATH. |
math/9905049 | Since MATH is finitely presented over the ground scheme, we may assume that MATH is obtained by base change from a stack of finite type over MATH. Hence to obtain a cover we may assume that MATH is of finite type over MATH. Also, since the morphism MATH is finite and surjective we can assume MATH is reduced. By working with each irreducible component separately we can assume MATH is integral. Finally by normalizing we can assume that MATH is normal. Suppose that MATH has an open cover MATH such that MATH has a finite cover by a scheme MATH. The composite morphism MATH is quasi-finite. Thus, by NAME 's Main Theorem CITE the morphism MATH factors as an open immersion followed by a finite representable map MATH. Since MATH is assumed to be irreducible, the finite representable morphism MATH has dense image so it must be surjective. Set MATH. The induced map MATH is finite, representable and has dense image, so it is surjective. Since any finite representable morphism is strongly representable, we can, by applying the Lemma, conclude that MATH is a scheme. Thus, to prove the theorem it suffices to prove that MATH has a cover by open substacks which admit finite covers by schemes. By CITE, MATH has a quasi-finite flat cover by a scheme MATH. Let MATH be an irreducible component of MATH. Once again applying NAME 's Main Theorem, the quasi-finite morphism MATH factors as MATH, where the first map is an open immersion and the second map is finite (and by density surjective). Replacing MATH by MATH we may therefore assume that MATH is generically a scheme. In particular, we can assume that MATH has a generic point MATH. Let MATH be a smooth presentation for MATH. Since we are working locally we can assume that MATH is a normal variety. By CITE, the smooth cover can be refined to a quasi-finite flat cover by a scheme MATH and the morphism MATH is the composition of a closed immersion and an étale morphism. Again since we are working locally we may assume that MATH is irreducible. In particular we may also assume that MATH is normal. Since the morphism MATH is quasi-finite, it is open. Replacing MATH by an open substack, we may assume that MATH is surjective. Now we construct a finite cover of MATH by a scheme. The map MATH is generically finite, so MATH is a finite extension of MATH (recall that MATH is the generic point of MATH). Let MATH be a normal extension of MATH containing MATH. Then MATH is NAME over a field MATH which is a purely inseparable extension of MATH. Let MATH be the normalization of MATH in MATH. Let MATH be the pre-image of MATH in MATH, and for MATH let MATH be the translate of MATH under the action of MATH. Each MATH is a scheme. Since normalization commutes with smooth pullback CITE, we may invoke CITE to deduce that MATH acts transitively on the fibers of MATH. Hence the MATH cover MATH, so MATH is a scheme which is a finite cover of MATH. |
math/9905049 | Let MATH be a point in MATH. Replacing MATH by an étale cover if necessary, we may assume the points of MATH all have residue field equal to the residue field of MATH. Then, for any MATH, MATH, the induced maps MATH are isomorphisms. Thus the composite MATH gives an automorphism of the completed local ring MATH. By assumption on MATH, MATH in a neighborhood of MATH so the automorphism is nontrivial. Thus, MATH must act nontrivially on the vector space MATH for some MATH. Then, there exists MATH such that the stabilizer group MATH acts faithfully on the space of MATH-jets at MATH. By NAME induction on MATH, there is a MATH for which the stabilizer action on MATH-jets is faithful at all points of MATH. |
math/9905054 | Assume without loss of generality that MATH is connected. Since an open surface is a MATH - space, the inclusion MATH is homotopic to a point if and only if the homomorphism MATH is trivial CITE. But MATH is also an open surface. Representing MATH as the union of an increasing chain of compact surfaces with boundary we see that there exists a countable system of embedded circles which generates MATH. Thus we get the proposition. |
math/9905054 | It suffices to show that there exist pairwise disjoint closed embedded discs MATH such that MATH is a component of MATH and MATH. Since MATH is compact it has only finitely many connected components which we denote by MATH; let MATH denote the boundary components of MATH. Now fix some MATH. Since MATH is open and MATH contractible, at least one of the discs MATH intersects the interior of MATH. Denote this disc by MATH. We claim that MATH contains MATH. Indeed, pick any point MATH, and assume on the contrary that there exists a point MATH. Since MATH is connected, there exists a path MATH in the interior of MATH which joins MATH and MATH. But, since MATH lies inside MATH and MATH outside MATH, MATH must intersect MATH which is impossible. This contradiction proves the claim. Notice that for MATH either MATH and MATH are disjoint, or one contains the other. So choose from the set MATH those discs which are maximal with respect to inclusion. This family of discs clearly satisfies all the requirements above. |
math/9905054 | Assume on the contrary that MATH is a regular value of MATH. Then there exists a segment MATH which consists of regular values of MATH and such that MATH. By definition of MATH, the set MATH contains a non - contractible circle. Since the gradient flow of MATH takes MATH into MATH we conclude that MATH contains a non - contractible circle, too. Hence MATH, in contradiction to the choice of MATH. The proof for MATH is analogous. |
math/9905054 | It suffices to show that MATH contains a non - contractible curve from MATH. Write MATH, where MATH are compact surfaces with boundary such that supp-MATH and MATH for all MATH. If some boundary component of some MATH is non - contractible in MATH we are done. Assume therefore that all of them are contractible. Then REF implies that all MATH are contractible in MATH. We conclude that MATH, that is, MATH, in contradiction to our standing assumption. |
math/9905054 | We are going to apply REF with MATH. If MATH is not contractible in MATH then it contains a curve from MATH. This curve lies either in MATH, which implies MATH, or in MATH, in which case MATH. Suppose now that MATH is contractible in MATH. Then MATH cannot contain a curve from MATH. This means that every curve from MATH intersects the set MATH, so MATH and MATH. But, as we have seen in REF , MATH is non - negative and MATH is non - positive. Therefore MATH. |
math/9905054 | When MATH this is proved in CITE. The case of a general open surface of infinite area can be reduced to this one as follows. Assume without loss of generality that MATH consists of just one disc of area MATH. Let MATH be the closed standard disc of area MATH. Since MATH has infinite area, it is an easy consequence of the NAME - NAME theorem (CITE, see also CITE) that there exists a symplectic embedding MATH such that MATH. Clearly, MATH induces the natural homomorphism MATH . It is important to notice that MATH does not increase the corresponding NAME distances. Our flow MATH lies in the image of MATH, that is, MATH where MATH is a one - parameter subgroup of Ham-MATH whose Hamiltonian is supported in int(MATH). Thus, the desired inequality follows from NAME 's original theorem since MATH. |
math/9905054 | Let us decompose the flow MATH into two commuting flows as follows. Fix any MATH, and choose a smooth function MATH satisfying the following properties: CASE: MATH if MATH CASE: MATH if MATH CASE: MATH if MATH CASE: MATH if MATH or MATH . Define the new Hamiltonians MATH and MATH, and denote their flows by MATH and MATH, respectively. Then MATH . Observe that supp REF is contained in the set MATH . Pick any regular value MATH of MATH. Then MATH is a compact REF - dimensional submanifold with contractible boundary. Denote by MATH the hull of MATH. REF implies that MATH is a finite union of pairwise disjoint closed discs. Moreover, MATH is a subset of supp(MATH), so MATH is contained in the hull of supp(MATH). Recall that this hull is compact. Combining this with REF above, we conclude that there is a constant MATH, depending only on supp REF but not on MATH, such that MATH for every MATH. On the other hand, MATH is generated by MATH with MATH, hence MATH . Now, the relation REF implies that MATH . Therefore MATH for every MATH, and the inequality in REF follows. Moreover, if MATH we get that MATH for all MATH, and REF is proven. |
math/9905054 | The first equalities are proved in CITE, and the second ones in CITE. |
math/9905054 | Since MATH, it suffices to show that MATH . Fix an arbitrary MATH. Choose MATH to be a non - contractible circle on MATH such that MATH. REF imply that MATH. Analogously, MATH. Thus MATH, for every MATH. Thus we get the desired inequality. |
math/9905054 | The proof goes along the lines of CITE, and we only give a sketch here. The argument is devided into three steps. CASE: Choose a compact connected submanifold with boundary MATH whose interior contains both MATH and MATH. Let us perform the following surgery on MATH. We remove the complement to MATH and attach to each boundary component of MATH a cylindrical end of infinite area. Note that the loop of Hamiltonian diffeomorphisms MATH extends to this new surface. Therefore we can assume from the very beginning that MATH has a finite number of ends and each end has infinite area. Such a surface MATH is geometrically bounded (or tame) in the sense of NAME 's theory of pseudo - holomorphic curves (see CITE). This will enable us to apply NAME theory in REF below. CASE: We claim that MATH . This fact is well known to experts, however, as far as we know, no reference is available. Here is a sketch of the argument. Denote by Diff-MATH, respectively Symp-MATH, the identity component of the group of compactly supported diffeomorphisms, respectively symplectomorphisms, of MATH. Consider the sequence MATH . The left arrow is a monomorphism (see CITE adjusted to the non-compact case along the lines mentioned in the book). The right arrow is an isomorphism; this follows from NAME 's deformation argument with parameters (compare CITE). But it is shown in CITE that MATH. This completes the proof sketch of the claim. CASE: Finally, recall the so - called Lagrangian suspension construction for Lagrangian submanifolds MATH in a symplectic manifold MATH. Given MATH with MATH, we consider the embedding MATH . If we equip MATH with the split symplectic form MATH then the above map is a Lagrangian embedding. In our case MATH is a circle, and the image of the embedding is a Lagrangian torus which we denote by MATH. In view of REF we know that the loop MATH, is homotopic to the constant loop at the identity. Hence the Lagrangian torus MATH is exact Lagrangian isotopic to MATH. Moreover, since MATH is non - contractible, MATH. Then NAME theory CITE guarantees the existence of an intersection point in MATH, that is, there are MATH and MATH such that MATH . This completes the proof of the lemma and finishes the proof of REF . |
math/9905058 | Let MATH be open and MATH vanish on MATH. We have to show that MATH . Let MATH be any point in MATH. As well-known, there exists a neighborhood MATH of MATH and vector fields MATH, MATH on MATH such that MATH and MATH where MATH depends only on the dimension of MATH. One has, using the fact that MATH is a REF-cocycle MATH . |
math/9905058 | Any differential operator on MATH is indeed determined by its values on the subspace MATH. |
math/9905058 | : straightforward. |
math/9905058 | Check that for MATH, REF coincides with REF , the result follows then by induction. |
math/9905058 | This follows immediately from REF and the fact that the sequence REF is split when restricted to MATH, see CITE. Let us also give an elementary proof. Recall that a REF-cocycle on MATH with values in MATH is a coboundary if it is of the form MATH for some MATH. Moreover, REF-cocycle vanishes on MATH if and only if MATH is MATH-equivariant. (compare CITE). If MATH, there is no MATH-equivariant operators MATH different from zero. In virtue of REF , the property of MATH-equivariance implies, in particular, that MATH has to be proportional to MATH. Now, the commutation relation REF shows that this operator can never be MATH-equivariant. REF follows. |
math/9905058 | Since the sequence REF is split, one has MATH. If in addition MATH, then by REF MATH and the sequence REF is split. |
math/9905061 | The proof is direct and is left to the reader. |
math/9905061 | Fix MATH as in the hypothesis of the theorem, and let MATH be the set of branches associated with MATH. Let MATH. Clearly MATH is a theory in MATH that satisfies REF 's Compactness Theorem. It follows from REF that there exists a MATH-saturated structure MATH such that: MATH . By the definition of MATH in MATH it follows that MATH. It remains to prove that MATH. We claim: For every formula MATH, MATH iff MATH. proof: By induction in the formulas of MATH. The proofs for MATH and for the countable (or finite) conjunction steps are direct, and are left to the reader. Negation. CASE: Assume that MATH. Assume also, in order to get a contradiction, that MATH. By induction hypothesis it follows that MATH which implies that there exists a branch MATH such that MATH . However, since MATH, it follows from the definition of the approximate truth for the negation REF that there exists a function MATH with the following "weak" surjectivity property: MATH and such that MATH. From these two properties of MATH it follows that there exists a MATH such that MATH, which implies from the properties of the weak approximate negation REF that MATH, but this contradicts REF . CASE: Assume that MATH. By induction hypothesis, we get that MATH. Hence, for every branch MATH, there exists an integer MATH such that MATH. We invoke now REF to obtain that there exists an integer MATH such that MATH. Consider now that the collection of all formulas of the form MATH that hold in MATH. From REF it follows that those formulas are finitary and belong to MATH so they are at most countable. This implies that we can construct a function MATH with the "weak" surjectivity property (that is, MATH) and such that MATH. We get then from the definition of approximate formulas for MATH REF that MATH. This completes the proof of the negation step. Existential . There is only one interesting direction. Assume that MATH. Then there exists MATH such that MATH . Since the above formula is a countable conjunction of finitary formulas in MATH and MATH is MATH-saturated, it follows that the conjunction MATH is approximately realized in MATH for some MATH. This implies that MATH, and hence, by induction hypothesis, MATH. This completes the proof of the existential step and of the claim. From the above claim it follows that MATH. This completes the proof of the theorem. |
math/9905061 | Assume, in order to get a contradiction, that there exists MATH such that for every integer MATH there is a normed structure MATH satisfying: MATH . We can now invoke REF to obtain that there exists a normed structure MATH such that MATH and MATH, but this is a contradiction with the hypothesis. |
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