paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0006174 | By the definition of MATH, there is a unique representative up to isomorphism for the NAME class of MATH and it is both flat and regular. Hence the map MATH is an isomorphism. Then the tautological bundle MATH constructed above identifies an analytic neighborhood of MATH with the local semi-universal deformation space ... |
math/0006174 | Choose a component MATH of the fiber product MATH, and let MATH be the induced map. It suffices to show that MATH in the obvious notation. Since both sides are algebraic, it suffices to show the following: let MATH be an irreducible curve such that MATH. Then MATH. Now choose a line bundle MATH of degree zero on MATH s... |
math/0006174 | This follows from CITE, Corollary to REF , p. REF or CITE. |
math/0006174 | The groups MATH and MATH act on MATH by conjugation, and the action of MATH factors through the projection to MATH. Thus, the sheaves MATH and MATH are identified as well. |
math/0006174 | Let MATH and let MATH correspond to the identity in MATH. For a MATH-algebra MATH, if MATH, let MATH. Then there exists a unique homomorphism MATH corresponding to MATH, so that MATH is a MATH-algebra. Now MATH is the image of MATH under MATH, and the element MATH defines a unique homomorphism MATH. Thus MATH defines a... |
math/0006174 | We claim that the functor MATH is representable by an affine space. The proof is by induction on the length of the filtration MATH. If this length is zero, then MATH and there is nothing to prove. Suppose that the claim has been verified for every group and filtration of length less than MATH, and let MATH be a filtrat... |
math/0006186 | The ideal MATH is the sheaf of sections of the sub-bundle MATH. By its construction, this bundle is MATH-invariant and its fiber over MATH is the subspace MATH. This latter subspace is a (minimal) MATH-invariant subspace of MATH where MATH is the stabilizer in MATH of MATH. Since MATH is a symmetric space, it follows t... |
math/0006186 | Fix MATH with MATH and let MATH denote the symmetric group on MATH. Recall from CITE (whose notation I will follow) that one can associate to MATH an irreducible, unitary representation MATH. For example, MATH corresponds to the trivial representation of MATH, while MATH corresponds to the alternating representation, t... |
math/0006186 | Fix any positive definite Hermitian inner product on MATH and define the corresponding invariant forms MATH on MATH. If MATH, then, since it is represented by MATH, which is positive by REF , and since MATH is NAME, the Generalized NAME Inequality of REF implies that MATH must vanish identically. |
math/0006186 | By REF and the positivity of MATH, it follows that, when MATH and MATH, the form MATH vanishes on the NAME variety MATH. As has already been seen, at each smooth point MATH, the tangent plane MATH is of type MATH and so must belong to MATH. Conversely, every subspace MATH of type MATH is tangent to the NAME cell MATH f... |
math/0006186 | It follows from REF , together with the discussion in CITE of NAME 's construction of the of NAME functors (especially REF ), that, when the MATH-form MATH is written locally in the form MATH for some local MATH-forms MATH, these latter forms must be a local basis of the subspace MATH. The statements of the lemma follo... |
math/0006186 | It has already been noted that MATH for some MATH if and only if MATH for all MATH with MATH and MATH. Since, by REF , MATH is positive, the equation MATH holds if and only if MATH vanishes on MATH. In turn, by REF , this holds if and only if MATH is an integral manifold of MATH for all MATH with MATH and MATH, as clai... |
math/0006186 | REF follows from REF together with REF and the fact that MATH. For REF , one direction is easy: If MATH is an integral element of MATH, then, by the first statement, MATH is an integral element of MATH for all MATH with MATH. Conversely, suppose that MATH has dimension MATH and is an integral element of MATH for all MA... |
math/0006186 | Set MATH and suppose that MATH is such that MATH and MATH. By REF , any subspace MATH of type MATH is an integral element of MATH. Since the dimension of such a MATH is MATH, it follows that MATH has integral elements of dimension MATH. It remains to show that MATH has no integral elements of dimension MATH. By the def... |
math/0006186 | Fix MATH and consider any basis MATH of MATH with the property that MATH is spanned by MATH. Let MATH be the dual basis of MATH. The identification MATH can be written in the form MATH where MATH are a basis for the MATH-forms on MATH. In terms of these MATH-forms, the MATH-forms MATH with MATH generate MATH on MATH (s... |
math/0006186 | Apply REF and the complementarity principle REF . |
math/0006186 | Apply REF and the duality principle REF . |
math/0006186 | Apply REF and the duality principle REF . |
math/0006186 | I will maintain the basic notation established during the proof of REF , especially the identification MATH, which will be used implicitly throughout the proof. Now, the ideal MATH is generated by the MATH-forms MATH . Note that MATH is skewsymmetric in its upper indices and symmetric in its lower indices. As in the pr... |
math/0006186 | Apply REF and the complementarity principle REF . |
math/0006186 | Combine REF . |
math/0006186 | Again, the notation established in the previous proof-analyses will be maintained. The first difference is that the ideal MATH is generated by MATH-forms of the form MATH where MATH satisfies the relations MATH . (Essentially, MATH is the general element of MATH.) Note that MATH satisfies MATH . As in the proofs of REF... |
math/0006186 | It is immediate that, for any MATH, the submanifold MATH is an integral manifold of MATH. Suppose that MATH satisfies the stated hypotheses and let MATH denote the smooth part of MATH, which is connected since MATH is irreducible CITE. By hypothesis, MATH is an integral manifold of MATH, that is, its tangent planes are... |
math/0006186 | Since MATH, it follows that MATH vanishes on MATH, the smooth part of MATH, for all MATH with MATH and MATH. Consider the positive MATH-form MATH. (This is where the hypothesis MATH is used.) By NAME 's REF MATH . Since MATH, it follows that MATH, so every term on the right hand side of the above equation vanishes on M... |
math/0006186 | Since MATH is generated by its sections, there exists a MATH and a surjective holomorphic bundle map MATH. Let MATH be the rank of the kernel bundle MATH. The mapping MATH defined by MATH then has the property that MATH. Moreover, MATH. Since MATH and MATH is a positive MATH-form, it follows that MATH is a positive MAT... |
math/0006186 | Apply complementarity to the proof of REF . |
math/0006186 | Apply complementarity to the proof of REF . |
math/0006186 | Let MATH be the space of global sections of MATH, a vector space of dimension MATH. Let MATH be the evaluation mapping, which, by assumption, is surjective, so that MATH. The kernel MATH is then a subbundle of rank MATH and can be used to define a mapping MATH that satisfies MATH. Consequently, MATH . Thus, the inequal... |
math/0006186 | Let MATH be an algebraic variety of codimension MATH and degree MATH. It is immediate that MATH is of codimension MATH. Moreover, a simple local calculation shows that, at its smooth points, its tangent spaces are integral elements of MATH (compare REF ). Thus, MATH vanishes on MATH, which implies that MATH for some MA... |
math/0006186 | Let MATH be an algebraic variety of dimension MATH and degree MATH. A simple local calculation verifies that MATH as defined in the proposition has codimension MATH in MATH and that its tangent space at a smooth point is of type MATH. Consequently (compare REF ), it follows that MATH for some MATH. Since MATH if and on... |
math/0006186 | By REF , every integral element of MATH of dimension at least MATH is either an integral element of MATH or of MATH. Moreover, these two ideals have no integral elements of dimension MATH or more in common. Thus, any irreducible integral variety of MATH is either an integral variety of MATH or of MATH. Now apply either... |
math/0006186 | In each case, the homological assumption implies that MATH vanishes on MATH and, hence, that MATH is an integral variety of MATH. Now apply REF . |
math/0006186 | Let MATH be the space of global sections of MATH, a vector space of dimension MATH. Let MATH be the evaluation mapping, which, by assumption, is surjective, so that MATH. The kernel MATH is then a subbundle of rank MATH and can be used to define a mapping MATH that satisfies MATH. Consequently, MATH . Thus, the inequal... |
math/0006186 | Verifying that each of the types listed is indeed an integral manifold of MATH is relatively straightforward. Simple calculations via the moving frame show that the tangent spaces to these subvarieties at their smooth points are integral elements of MATH. Alternatively, the calculations below will provide a direct proo... |
math/0006186 | Combine complementarity and REF . |
math/0006186 | Any variety MATH of pure dimension MATH that satisfies MATH is necessarily an integral variety of MATH. REF implies that any such irreducible variety must, for dimension reasons, fall into the first category listed there. Thus, if MATH is irreducible, then MATH for some MATH. Since the ray generated by MATH is extremal... |
math/0006186 | The proof is very similar to the proofs of REF , so I will only sketch the argument. Such a MATH, is, of course, an integral variety of both MATH and MATH. Conversely, any MATH-dimensional integral variety of both of these ideals is homologous to MATH for some MATH. REF describes the three-dimensional integral elements... |
math/0006186 | The inequality MATH is, of course, immediate from REF . Moreover, by REF , if MATH, then MATH is an integral variety of MATH, where MATH. Now apply REF and interpret each of the possible cases. Only two cases require any comment: One of these is the second category of integral varieties of REF , that is, the case in wh... |
math/0006186 | I will first show that if MATH is not isotropic then one can construct forms in each of MATH that do not vanish on MATH. To begin, suppose that the inner product is nondegenerate on MATH. Then there exists an oriented orthonormal basis MATH of MATH, with dual basis MATH of MATH, so that MATH is spanned by the vectors M... |
math/0006186 | Let MATH be the subgroup of MATH consisting of the matrices MATH that satisfy MATH . Then MATH acts on MATH and induces a transitive action on MATH. Let MATH denote the inclusion and write MATH where MATH are regarded as (holomorphic) mappings. Then MATH is a holomorphic principal fiber bundle over MATH. Moreover, the ... |
math/0006186 | Each of MATH is a positive MATH-form and the above analysis shows that MATH . Since MATH, it follows that MATH and MATH generate MATH. Thus, any MATH-dimensional subvariety MATH that satisfies MATH must be a union of irreducible subvarieties whose homology classes are multiples of MATH. Thus, I may assume that MATH is ... |
math/0006186 | Let MATH be the identity map. For any basis MATH of MATH, write MATH, where MATH are REF-forms on MATH. Note that MATH is a basis for the dual space. Now, MATH occurs as a constituent of MATH, but MATH does not. Consequently, the MATH-forms of the form MATH when MATH, MATH, and MATH are symmetric in their indices, must... |
math/0006186 | The first task (which will be needed in the next proposition as well), is to establish the equations of the moving frame for submanifolds of MATH. Define MATH be the subgroup of MATH consisting of the matrices MATH that satisfy MATH . Also, let MATH denote the set of matrices MATH that satisfy MATH and MATH. Evidently,... |
math/0006186 | Recall the moving frame notation and constructions from the first part of the proof of REF . Suppose now that MATH is an irreducible MATH-dimensional integral variety of of MATH and let MATH be its smooth locus, which is connected since MATH is irreducible. By REF , for every MATH, there exists a MATH so that CASE: MAT... |
math/0006186 | First of all, it follows by either CITE or REF and the general results of NAME mentioned above that MATH. Let MATH be the MATH-invariant NAME form on MATH whose cohomology class is a generator of MATH. By REF, there is a sum of the form MATH where MATH and MATH and MATH and MATH are positive MATH-invariant forms dual t... |
math/0006186 | Using arguments that should, by now, be familiar, one sees that the homology class of a codimension MATH irreducible cycle MATH is some multiple of MATH if and only if the form MATH vanishes on MATH, which is the same as saying that, at any smooth point MATH, the normal space to MATH in MATH is an integral element of M... |
math/0006186 | Using arguments that should, by now, be familiar, one sees that the homology class of a codimension MATH irreducible cycle MATH is some multiple of MATH if and only if the form MATH vanishes on MATH, which is the same as saying that, at any smooth point MATH, the normal space to MATH in MATH is an integral element of M... |
math/0006186 | This proof is very similar in spirit and structure to the proof of REF , so, to save space, I will not go into as much detail here as I did there. Instead, I will limit my discussion to the outline, except where some essentially new or different idea is needed. Let MATH be the identity map. For any basis MATH of MATH, ... |
math/0006186 | The first task (which will be needed in the next proposition as well), is to establish the equations of the moving frame for submanifolds of MATH. Define MATH be the subgroup of MATH consisting of the matrices MATH that satisfy MATH . I will regard MATH as a matrix-valued function and denote its columns as MATH where M... |
math/0006186 | Recall the moving frame notation and constructions from the first part of the proof of REF . Suppose now that MATH is an irreducible MATH-dimensional integral variety of of MATH and let MATH be its smooth locus, which is connected since MATH is irreducible. By REF , for every MATH, there exists a MATH so that CASE: MAT... |
math/0006186 | First of all, it follows by either CITE or REF and the general results of NAME mentioned above that MATH. Let MATH be the MATH-invariant NAME form on MATH whose cohomology class is a generator of MATH. By REF, there is a sum of the form MATH where MATH and MATH and MATH and MATH are positive MATH-invariant forms that a... |
math/0006186 | The analysis is very similar to that for the proof of REF , so I will leave the details to the reader. Only one aspect of the proof requires comment: Since there are two types of integral elements of MATH, there are, correspondingly, two types of integral elements of MATH. This forces a subdivision into two cases like ... |
math/0006186 | Follow the pattern of the proof of REF . |
math/0006191 | A splitting of Type REF is given as follows. Note that MATH is the mapping cylinder of a (single) NAME twist on the torus. Thus, if we begin with the rational elliptic surface MATH let MATH denote the image of a fishtail fiber, and let MATH be a disk around MATH containing no other singular points for MATH, then MATH s... |
math/0006191 | First, we reduce to the case where MATH is absent (that is, zero-dimensional). This is done in two steps, first establishing an inclusion MATH where both are thought of as subsets of MATH, and then seeing that the map MATH is surjective. To see Inclusion REF we describe the geometric representatives for the generators ... |
math/0006192 | Fix an initial metric MATH which is product-like along all MATH, and which agrees with MATH away from a tubular neighborhood of MATH. REF gives us a constant MATH so that for all MATH-tuples MATH with MATH, MATH is allowable. We can view the metric MATH as the result of inserting tubes with parameters MATH into a one-p... |
math/0006192 | According to NAME 's theorem, for any generic MATH, the fiber MATH is a compact, canonically oriented, zero-dimensional manifold. We investigate the conditions necessary to show that the fiber misses the boundary of the domain of MATH. The set of points MATH for which the fiber MATH does not contain boundary points of ... |
math/0006192 | To see this, we make use of the NAME duality map. NAME duality gives rise to a natural involution on MATH, with the property that MATH for any MATH and MATH. For any given metric MATH, this involution preserves the theta divisor (by NAME duality), and fixes the points associated to spin structures. The involution on MA... |
math/0006192 | If MATH is not a torsion class, then MATH and MATH are disjoint, so the proof of REF applies. For the torsion MATH structure MATH (that is, the one corresponding to the spin structure), there are a priori two invariants, depending on the sign MATH for the perturbation MATH. But REF guarantees that MATH and since MATH c... |
math/0006192 | Note that the cylindrical-end metric MATH on MATH is conformal equivalent to the metric MATH on MATH inherited from MATH. Indeed, we can write MATH where MATH is a real-valued function, which agrees with the real coordinate projection on the cylindrical ends. The spinor bundles of the two manifolds can be (metrically) ... |
math/0006192 | Since the operator MATH is translationally invariant in the cylinder MATH, there is a single constant MATH which works for all the manifolds MATH, so that for any form MATH, MATH (see for instance CITE); thus, MATH . This together with the NAME lemmas shows that the forms MATH converge to zero in MATH over any compact ... |
math/0006192 | Fix a genus MATH . NAME decomposition of MATH. There is a ``stabilized" genus MATH . NAME decomposition of MATH, corresponding to the natural decomposition MATH that is, let MATH and consider the NAME decomposition MATH . We would like to show that the invariant MATH associated to the NAME decomposition MATH agrees wit... |
math/0006192 | Our hypothesis on MATH gives us a compact set MATH with the property that MATH. Similarly, our hypothesis on MATH gives a compact set MATH with the property that MATH. We let MATH, MATH be any pair of compact sets whose interior contains MATH and MATH. Consider pairs MATH over MATH which correspond to the intersection ... |
math/0006192 | We would like to apply a version of REF , with more than one neck (note that the proof works in this context as well). Let MATH be the theta divisor of MATH. It contains none of the connect sum points, of course, because it has degree zero. Moreover, the set MATH misses the theta divisor for MATH for MATH (the theta di... |
math/0006192 | The proof will rely on the fact that MATH has bounded variation along any one-parameter family of metrics. Specifically, let MATH be a one-parameter family of metrics, and fix a norm on MATH. Then, there is a constant MATH with the property that for any MATH, MATH, MATH . This follows immediately from the compactness o... |
math/0006192 | Take a weak limit of connections which lie in MATH, as all the MATH curves are stretched. Under this limit, the surface degenerates into a collection of genus zero surfaces (with cylindrical ends), whose ends correspond to attaching circles for MATH or separating curves for MATH. Thus, the weak limit of connections in ... |
math/0006192 | Fix a metric MATH on MATH as in REF . Let MATH denote the metric which is stretched by MATH along the MATH and MATH along the separating curves. The lemma guarantees that for all MATH, MATH is allowable. Now, for all sufficiently large MATH, REF gives the isotopy of MATH with the subset MATH, where MATH. |
math/0006192 | Let MATH be MATH-allowable metrics for MATH. Given MATH, we construct a natural element MATH closely related to the MATH defined in the beginning of this section. The MATH will be a translate of the following analogue of the MATH, which is assigned to a pair of lifts MATH and MATH (to MATH) of the manifolds MATH and MA... |
math/0006192 | We adopt the notation and most of the argument for the proof of REF . The difference arises when one wishes to prove that the moduli spaces MATH and MATH are cobordant, that is, when one wishes to exclude the possible MATH boundary components. To do this, it is no longer possible to argue that MATH . Rather, to exclude... |
math/0006192 | This is a direct consequence of REF . |
math/0006192 | REF shows that MATH is determined by MATH and the classes MATH and MATH. After a series of handleslides, we can arrange that all MATH and MATH are equal to one, fixed generator of MATH. It follows from REF that all MATH equal MATH up to translation. Since the NAME polynomial MATH (modulo multiplication by MATH) is the ... |
math/0006192 | Since MATH, there is a unique MATH structure over MATH. Moreover, there is a complex-antilinear involution MATH of the spinor bundle which commutes with NAME multiplication (actually, this involution exists in much more general contexts, and can be thought of as the basis for the conjugation action on the set of MATH s... |
math/0006193 | Morally this means that MATH-equivariant sheaves descend naturally to the quotient MATH. To check that the cohomology of complexes MATH and MATH are canonically isomorphic it is sufficient to consider infinitesimal gauge transformation MATH, MATH. Then the map MATH gives the needed chain isomorphism. |
math/0006193 | It is straightforward to check that MATH, MATH. In our case MATH-module is the complex REF with MATH-action MATH, MATH. The operators MATH for MATH define the lifting of this action to MATH-action. |
math/0006193 | MATH . |
math/0006193 | We need to check first that MATH for MATH. It is enough to prove this for MATH. Applying the standard commutation rules MATH one gets: MATH . Assume that MATH, or equivalently MATH . Applying the operator REF to MATH we get MATH on the other hand MATH . Let MATH be a solution to NAME equation describing a deformation o... |
math/0006193 | Let MATH describes a MATH-deformation MATH. Let MATH describes a family of elements of the varying semi-infinite subspaces. Then for any vector field MATH: MATH . |
math/0006193 | It follows from REF that MATH is transversal to MATH. One can show that MATH-module MATH is free (it follows, for example, from REF ). Therefore MATH. |
math/0006193 | It follows from REF that MATH . Therefore there exist MATH such that MATH . On the other hand, MATH . Therefore MATH . The coordinates MATH were chosen so that MATH is linear in MATH. Therefore in these coordinates MATH and MATH. |
math/0006194 | The orientation and partitioning into MATH structures are described in REF. When MATH, the genericity statement follows from REF , which is proved in REF of the present paper. Note that solutions with MATH correspond to points in MATH which map to MATH, which is a discrete set of points (since MATH is a homology three-... |
math/0006194 | Let MATH be a generic allowable homotopy connecting MATH and MATH. Consider the moduli space MATH . According to the generic metrics result, REF , this space is an oriented one-manifold with boundary. Since MATH is an allowable homotopy, the only boundary components are MATH, MATH and MATH. Thus, counting boundaries, w... |
math/0006194 | The vanishing of the kernel follows from the vanishing of the kernel over the three pieces: MATH, MATH, and MATH respectively. Over the handlebodies the kernel vanishes for generic choices of MATH and MATH (this will be proved in REF ). Over the cylinder, it vanishes for generic paths MATH, according to REF . Moreover,... |
math/0006194 | The allowability of MATH and diffeomorphism statement for MATH were proved in REF; note that the hypothesis that MATH was not used in the proof of this fact. One can arrange for MATH and MATH to be generic simultaneously by varying the family MATH in a region MATH which does not contain the connected sum point. This st... |
math/0006194 | The connection MATH naturally induces a connection on the spinor bundle of the boundary. More precisely, under the splitting MATH into the MATH-eigenspaces of NAME multiplication by MATH times the volume form of MATH, the connection MATH naturally induces a connection MATH on MATH. If MATH is compactly supported in MAT... |
math/0006194 | Fix some constant MATH, and fix a smooth, non-decreasing function MATH with MATH . Then, the hypersurface MATH inherits a metric from MATH. The region where MATH and MATH is diffeomorphic to half of the three-ball (its two boundaries, are the loci where MATH and MATH vanish respectively). Since the function MATH is con... |
math/0006194 | Both statements are proved in CITE: the NAME paramatrix is constructed in the proof of REF , and the exponential decay estimate is NAME REF of that reference. |
math/0006194 | We can introduce coordinates on MATH with respect to which MATH takes the form MATH. For any real numbers MATH, MATH, let MATH. Clearly, MATH gives a one-parameter family of complex structures whose differential at MATH acts by MATH . The lemma follows. |
math/0006194 | We calculate the (connection form) difference between the NAME connection and MATH over MATH. Suppose for the moment that MATH. Let MATH be a (time dependent) moving coframe for MATH. There are functions MATH over MATH for which MATH where MATH is the connection matrix for the metric MATH over MATH. Thus, by the NAME f... |
math/0006194 | By REF , the spectral flow localizes to the region where MATH meets the MATH-theta divisor. By homotopy invariance of both quantities, then, it follows that the spectral flow must be some multiple of the above intersection number. The factor is then calculated in a model case, as in REF (note that in that proposition, ... |
math/0006194 | We recall from CITE that MATH, where the MATH structure MATH corresponds to the integer MATH under the identification MATH which sends the spin structure to MATH. |
math/0006194 | Every integral homology three-sphere can be obtained from MATH by a sequence of MATH surgeries. So, it follows that MATH is uniquely determined by its surgery formula and its value on MATH. It is easy to see that MATH: fix a genus one NAME decomposition of MATH, and let MATH be a constant family of metrics on the torus... |
math/0006194 | Smoothness follows from the generic metrics statement REF . There is no MATH boundary since the metric MATH is MATH-allowable. The MATH boundary lies in MATH, which maps near MATH, as above. The MATH boundary corresponds to the intersection of the theta divisor with MATH, which in turn maps to MATH. Indeed, this latter... |
math/0006194 | Fix a real number MATH with MATH. Fix curves in MATH: MATH, MATH, and MATH. Note that the restriction of MATH and MATH to the torus gives the spin structures corresponding to MATH and MATH respectively. Since the moduli space misses these spin structures, it follows that MATH and MATH are well-defined, and indeed by th... |
math/0006194 | Recall that there is a natural cobordism MATH from MATH to MATH. This cobordism is obtained from MATH, and then attaching a two-handle with MATH framing. We have a natural inclusion MATH which maps MATH to the boundary of MATH (that is, taking MATH to the knot complement, as a subset of MATH, and MATH to the correspond... |
math/0006194 | This follows from the surgery formula, and the calculation of MATH from CITE, according to which MATH . |
math/0006194 | Since any rational homology three-sphere MATH can be obtained from MATH by a sequence of surgeries on rational homology three-spheres, REF and the constants MATH determine the sum MATH (see REF ). A suitably normalized version of the NAME invariant satisfies a formula of the same shape, with constants MATH. We spell th... |
math/0006194 | The MATH boundaries correspond to the intersection of the theta divisor with the MATH. Using the identification between the Jacobian and the MATH, MATH corresponds to those MATH representations of MATH which extend to representations of MATH. Since MATH bounds in MATH, these representations must take MATH to a MATH-tor... |
math/0006194 | This follows immediately from the splitting principle for spectral flow, applied to the zero-surgery MATH. |
math/0006194 | Note first of all that MATH is homologous to MATH. Thus, we can consider the oriented MATH-chain MATH which does not contain MATH, and whose boundary is MATH. By elementary differential topology, MATH . These are to be viewed MATH-chains; that is, we have that MATH where MATH denotes the multiplicity of MATH at MATH ti... |
math/0006194 | Since MATH can be thought of as the MATH index of the NAME operator for MATH, with ends attached, the above statement is an easy application of the splitting principle for the index, bearing in mind that the difference MATH is the index of the NAME operator on MATH, and topological terms which depend only on MATH, MATH... |
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