paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0101102 | The proof of this Proposition is exactly the same as the one for REF , except that this time the relevant pushouts in the sequences REF are by maps MATH . The map MATH is again compatible with these pushouts by the map induced by MATH where MATH inserts the pointing MATH into the last MATH slots of MATH and MATH inserts the pointing into the first MATH slots. This map is again a weak equivalence since MATH is a MATH-operad, so we are done. |
math/0101106 | By CITE MATH is diffeomorphic to the product MATH where MATH is a compact nilmanifold. Let MATH be a principal MATH-bundle over MATH with the first NAME class equal to MATH. (Recall that MATH gives a bijection between the set MATH of MATH-bundles over a finite cell complex MATH and the group MATH.) Being a principal circle bundle over a compact nilmanifold, MATH is a compact nilmanifold so by CITE MATH where MATH is a simply-connected nilpotent NAME group, and MATH is a discrete subgroup of MATH. Let MATH be the (central) element of MATH corresponding to a fiber of MATH. By CITE, MATH lies in the center of MATH. Let MATH be the (necessarily central) one-parametric subgroup containing MATH. Consider the nilpotent NAME group MATH and think of MATH, MATH as subgroups of MATH. The subgroup MATH of MATH is generated by MATH (else MATH would be a power of some element of MATH which is then projected to a torsion element of the torsion free group MATH). Thus, the group MATH is a discrete subgroup of MATH with factor space MATH. Now MATH gets the structure of a principal MATH-bundle over MATH which is isomorphic to MATH. (Since the bundle projections induce the same maps of fundamental groups, the bundles are fiber homotopy equivalent which implies isomorphism for principal circle bundles.) Now let MATH be a nonzero element of the NAME algebra of MATH tangent to MATH. Use MATH to define the MATH-action MATH as in REF . Thus, there exists a positive MATH such that for all MATH, the manifold MATH has a complete MATH-invariant metric with MATH, so we can consider the quotient of MATH by the isometric action of MATH. Note that the kernel of the MATH-action is MATH, hence the kernel of the MATH-action is MATH. The actions of MATH and MATH on MATH commute, so MATH is the quotient of MATH by the drop of MATH. The drop of MATH can be described as the MATH-action on the MATH-factor, and the same diagonal MATH-action on MATH. Thus, MATH is the total space of a complex vector bundle over MATH. The bundle is the NAME sum of MATH copies of the same line bundle which we denote MATH. Finally, we show that the first NAME class of MATH is MATH. Indeed, since MATH is constructed via the standard MATH-action on MATH, we get MATH. Look at the associated circle bundles MATH and MATH. There is a covering MATH which is a MATH-fold covering on every fiber. Then the classifying map for MATH can be obtained as the classifying map for MATH followed by the selfmap of MATH induced by the MATH-fold selfcover of MATH. By the homotopy sequence of the universal MATH-bundle, this selfmap of MATH induces the multiplication by MATH on the second homotopy group MATH. By NAME it induces the same map on MATH and hence on MATH. Since the first NAME class of the universal bundle generates MATH, we deduce MATH. Since the first NAME class determines (the isomorphism type of) the complex line bundle, MATH is isomorphic to MATH, and the proof is complete. |
math/0101106 | Let MATH be a class with a nontorsion cup-square. Let MATH be a (unique up to isomorphism) complex line bundle over MATH with MATH. By REF for all sufficiently large MATH the NAME sum MATH of MATH copies of MATH has a complete metric with MATH. The total NAME class of MATH can be computed as MATH hence the first NAME class MATH is given by MATH. It remains to show that the total spaces MATH of MATH are pairwise nonhomeomorphic for different values of MATH. Being the quotient of a NAME group by a discrete subgroup, MATH is parallelizable (because the tangent bundle to any NAME group has a left-invariant trivialization which descends to quotients by discrete subgroups). Hence, the tangent bundle MATH has first NAME class MATH. Assume that MATH is a homeomorphism. Since rational NAME classes are topologically invariant, MATH is equal to MATH up to elements of finite order. Thus, the projections of the classes to MATH are mapped to each other by the automorphism MATH of MATH. Since any automorphism of a finitely generated, free abelian group preserves multiplicities of primitive elements, MATH, and we are done. |
math/0101106 | Since MATH, we can find MATH with a nontorsion cup-square (see REF). Furthermore, we can choose MATH so that MATH, say, by taking MATH to be the pullback of a class with a nontorsion cup-square via an obvious projection of MATH onto the MATH-torus. As in proof of REF, we find a vector bundle MATH over MATH such that MATH and MATH. Since MATH, all the higher NAME classes of MATH vanish. Since MATH is a nontorsion class, MATH does not become stably trivial after passing to a finite cover because finite covers induce injective maps on rational cohomology. Now the product MATH has a (product) metric with MATH which by CITE can be assumed of MATH after replacing MATH with MATH (or equivalently, after replacing MATH by its NAME sum with a trivial rank MATH bundle). By CITE the manifold MATH admits no metric with MATH. It remains to show that the manifolds MATH are pairwise nonhomeomorphic for different values of MATH. The tangent bundle to MATH is isomorphic to MATH. Let MATH be the largest number such that the MATH-th integral NAME class MATH is nontorsion. Since MATH is parallelizable, the bundle MATH has first integral NAME class MATH, and zero higher NAME classes. Up to elements of order two, the total NAME classes satisfy the product formula MATH. In particular, MATH is a nontorsion class since it is a product of nontorsion classes. Assume that MATH is a homeomorphism. Since rational NAME classes are topologically invariant, MATH maps MATH to MATH up to elements of finite order. Thus, the projections of the classes to MATH are mapped to each other by the automorphism MATH of MATH. Since any automorphism of a finitely generated, free abelian group preserves multiplicities of primitive elements, MATH, and we are done. |
math/0101106 | Assume REF , and find a complex line bundle MATH over MATH with first NAME class MATH. Let MATH be the NAME sum of MATH copies of MATH. Hence MATH is nontorsion. This proves REF . Now assume REF and take MATH. Arguing by contradiction, assume MATH is a torsion class, that is MATH for some MATH. As above we get MATH, hence all rational NAME classes vanish. In particular, the NAME character MATH of MATH vanishes. Recall that MATH defines an isomorphism of MATH and MATH, thus MATH has finite order in MATH. In particular, for some MATH, the NAME sum of MATH copies of MATH is trivial, as a real bundle. The contradiction proves REF . |
math/0101116 | CASE: Note that MATH is generated by finitely many elements MATH. Let MATH. Then MATH with non-negative rational MATH. Set MATH. Then MATH . The second summand lies in the intersection of MATH with a bounded subset of MATH. Thus there are only finitely many choices for it. These elements together with MATH generate MATH. That MATH is integrally closed in MATH is evident. CASE: Set MATH, and choose a system MATH of generators of MATH. Then every MATH has a representation MATH. Multiplication by a common denominator of MATH shows that MATH. On the other hand, MATH is integrally closed by REF so that MATH. The elements MATH can now be chosen as the vectors MATH appearing in MATH. Their number is finite since they are all integral and contained in a bounded subset of MATH. Together with MATH they certainly generate MATH as a semigroup. |
math/0101116 | It is easily seen that the bottom of MATH is the union of finitely many lattice polytopes MATH, all of whose lattice points belong to MATH. We now triangulate each MATH into empty lattice subsimplices. Choose MATH, and consider the line segment MATH. It intersects the bottom of MATH in a point MATH belonging to some simplex MATH appearing in the triangulation of a compact facet MATH of MATH. Let MATH, MATH, be the vertices of MATH. Then we have MATH as in the proof of NAME 's lemma. Set MATH, let MATH be the basic grading of MATH associated with MATH, and MATH for MATH. Then MATH, and at most MATH elements of MATH can appear in a representation of MATH. This shows that MATH. However, this bound can be improved. Set MATH. Then MATH, and it even belongs to the cone generated by MATH. If MATH, one has MATH. If MATH, then MATH is a lattice point of MATH. By the choice of the triangulation this is only possible if MATH for some MATH, a contradiction. Therefore MATH, and so MATH. It follows that MATH. |
math/0101116 | Let MATH be a generating set of MATH and MATH be a finite generating set of MATH as a module over MATH. That MATH is in fact a finitely generated MATH-module, has been stated in REF . Fix representations MATH, MATH, MATH. Then MATH. |
math/0101116 | Consider finite generating sets MATH, MATH. The affine hyperplanes in MATH, spanned by the elements of MATH, cut the cone MATH into subcones which we call elementary cells, i. e. the elementary cells are the maximal dimensional cones in the obtained polyhedral subdivision of MATH. Clearly, the elementary cells are again finite rational cones. So by NAME 's lemma the semigroups MATH are all affine. (The general form of NAME 's lemma used here and below follows from REF and CITE.) MATH is asymptotically covered by the MATH if and only if MATH for every elementary cell MATH, or equivalently MATH where MATH, MATH running through the set elementary cells. We claim that MATH is asymptotically covered if and only if MATH. This clearly proves the first part of the lemma. The ``only if" part of the claim follows easily from the fact that MATH. For the ``if" part we pick elements MATH, MATH REF . Then the assumption MATH implies MATH and we are done because by elementary geometric consideration one has MATH . Now assume the implication MATH of the lemma holds. MATH is contained in finitely many residue classes in MATH modulo MATH. By fixing origins in these classes and taking intersections with the modules MATH, the general case reduces to the situation when MATH. Pick elements MATH. Then we have MATH with the MATH chosen as above. We are done by the following observations: MATH and MATH the latter equality being easily deduced from the condition MATH. Now assume MATH is asymptotically covered by the MATH. Then we have the implication MATH . It only remains to notice that each of these MATH is asymptotically covered by MATH for an arbitrary element MATH, and, similarly, MATH is asymptotically covered by MATH, MATH. |
math/0101116 | REF is a direct consequence of REF follows from the same lemma and the observation that if MATH, then the full rank free sub-semigroups of MATH, generated by elements of MATH, cover MATH asymptotically. This is so because the contribution from degenerate subsets of MATH is ``thin" and cannot affect the asymptotic covering property. REF follows from REF and ASYMPTOTIC. |
math/0101117 | This is a general fact, see CITE. |
math/0101117 | We have to show that if MATH is a standard monomial with MATH and MATH, then MATH for all MATH. For MATH let MATH denote the number of the factors of MATH which are minors of size MATH. By assumption we have that MATH . In view of REF can be rewritten as MATH . We have to show that MATH . Note that MATH . Therefore REF is equivalent to MATH which is true since the left hand side is the sum of non-negative terms. |
math/0101117 | CASE: Let MATH be the valuation ring of MATH. Then evidently MATH and MATH is obviously a NAME domain. All essential valuation overrings of MATH occur in the above intersection. Since, by assumption, MATH is in MATH and in all the MATH with MATH, but not in MATH, we have that MATH is not redundant in the above description of MATH. Then MATH is an essential valuation overring of MATH. The rest is clear by standard results about essential valuations of NAME domains. CASE: Let MATH be an arbitrary element of MATH. Let MATH be a height REF prime ideal of MATH with MATH. We choose an element MATH. Since MATH for all MATH, there is a positive integer MATH such that MATH for all MATH. It follows that MATH. Hence MATH. So we get MATH. To prove MATH, we first note that MATH, where MATH is taken over all height REF prime ideal MATH of MATH. So we only need to show that if MATH, then MATH is a height REF prime ideal of MATH and MATH. That MATH follows from the assumption MATH. To prove MATH, it is sufficient to show that MATH. Let MATH. Then MATH because otherwise there is an element MATH such that MATH, contradicting the facts that MATH, MATH. Thus, MATH is an invertible element in MATH. Hence we can conclude that MATH so that MATH. |
math/0101117 | We apply REF with MATH and MATH. For MATH set MATH where MATH is a MATH-minor, MATH is a MATH-minor and MATH is a large enough integer. The reader may check that these elements have the right MATH-values, that is, MATH for MATH and MATH and MATH. |
math/0101117 | We apply REF with MATH and MATH. To this end it suffices to note that MATH for all MATH. |
math/0101117 | By virtue of REF the equality MATH is simply a re-interpretation of REF. Since we assume that MATH, the primary decomposition of MATH given in REF is irredundant for MATH. It follows that the primary decomposition of MATH is also irredundant. |
math/0101117 | The conclusion follows from REF and from a general result of CITE. Let us mention that one can derive the result also directly from CITE and REF. |
math/0101117 | We apply REF where MATH and MATH. It suffices to take MATH and note that MATH. |
math/0101117 | We apply REF where MATH and MATH. It suffices to prove that MATH has height MATH. To this end take MATH and MATH distinct MATH-minors (they exist by our assumptions). Set MATH. Then MATH and hence MATH. Since the MATH are prime elements, MATH has height MATH. Since MATH is a direct summand of the polynomial ring it follows that MATH has height MATH. |
math/0101117 | It suffices to check that MATH does not contain MATH. Let MATH be a MATH-minor of MATH different from MATH (it exist by our assumptions). Then MATH is in MATH, but not in MATH. The second statement is easy. |
math/0101117 | By CITE and REF there is an exact sequence: MATH . It is well-known that MATH is isomorphic to MATH and that it is spanned by the class of the prime ideal MATH generated by the elements of the form MATH where MATH is a monomial in the MATH of degree MATH. Note that MATH is a torsion free element in MATH. This is because MATH cannot be principal; in fact, it contains all the elements of the form MATH where MATH and MATH is a MATH-minor. Now fix a MATH-minor MATH and set MATH, that is MATH is the ideal generated by all the elements of the form MATH where MATH is a monomial of degree MATH. Evidently MATH is isomorphic to MATH and hence MATH generates MATH. But MATH is MATH and hence MATH is the preimage of MATH with respect to the map MATH. It follows that MATH and MATH generate MATH. By evaluating the function MATH one shows that MATH . Hence MATH. Consequently MATH is generated by MATH, and MATH is torsion free. |
math/0101117 | Let MATH be a monomial, where MATH is a monomial in the variables MATH. Then, for a given MATH, MATH for every MATH if and only if MATH for every MATH. By REF this is equivalent to MATH which in turn is equivalent to MATH. Summing up, MATH for every MATH and MATH if and only if MATH for every MATH. |
math/0101117 | Let MATH be a monomial in the MATH's. We know that MATH if and only if MATH. From equation MATH we may deduce that MATH . Note that MATH is a pure simplicial complex of dimension equal to the dimension of the determinantal ring defined by MATH minus MATH. It follows that MATH for every facet MATH of MATH. In particular, MATH. Since MATH for all monomials MATH and for every MATH, we have MATH. Conversely, let MATH be a facet of MATH such that MATH. Then MATH. Hence MATH, too. |
math/0101117 | CASE: Let MATH be a NAME basis of MATH and MATH a NAME basis of MATH. We may assume that these polynomials are monic and homogeneous. Consider the presentation MATH of MATH obtained by mapping MATH to MATH and the presentation MATH of MATH obtained by mapping MATH to MATH. Let MATH a system of binomial generators for the toric ideal MATH, say MATH. Then for each MATH we have expressions MATH with MATH and MATH for every MATH. It is known that the polynomials MATH generate MATH. For each MATH we may take a presentation MATH with MATH and MATH for all MATH. Then, by construction, the preimage of MATH in MATH is generated by the elements MATH and the preimage of MATH in MATH is generated by the MATH. Hence MATH is the quotient of MATH defined by the ideal MATH that is generated by the polynomials REF , and MATH is the quotient of MATH defined by the ideal MATH that is generated by the polynomials MATH and MATH. If we can find a positive weight MATH on MATH such that MATH, then there is MATH-parameter flat family with special fiber MATH and general fiber MATH. This implies that MATH is NAME, provided MATH is. Let us define MATH. First consider a positive weight MATH on MATH such that MATH for every MATH and MATH for every MATH and MATH. That such a MATH exists is a well-known property of monomial orders; for instance, see CITE. Then we define MATH as the ``preimage" of MATH in the sense that we put MATH. It is clear, by construction, that the initial forms of the polynomials REF with respect to MATH are exactly the MATH and the MATH. This proves that MATH contains MATH. But MATH and MATH have the same NAME function by construction, and MATH and MATH have the same NAME function because they have the same initial ideal if we refine MATH to a term order; for instance see CITE. Here we consider NAME functions with respect to the original graded structure induced by the weights MATH. It follows that MATH and we are done. CASE: That MATH is NAME follows from CITE. Since MATH is a NAME positively graded MATH-algebra which is a domain, to prove that MATH is the canonical module of MATH it suffices to show that MATH is a maximal NAME module whose NAME series satisfies the relation MATH for some integer MATH where MATH CITE. The relation MATH holds since by assumption the corresponding relation holds for the initial objects and NAME series do not change by taking initial terms. So it is enough to show that MATH is a maximal NAME module. But MATH is a height MATH ideal since it is the canonical module CITE, and hence also MATH has height MATH. Therefore it suffices to show that MATH is a NAME ring. But this follows from MATH since MATH is NAME (it is even NAME) CITE. |
math/0101117 | By virtue of REF it suffices to show that MATH and MATH are the canonical modules of MATH and MATH respectively. A description of the canonical modules of MATH and MATH has been given in REF. Therefore it is enough to check that MATH is exactly the ideal described in REF and MATH is the ideal described in REF. Note that we may write MATH and MATH . Furthermore, by virtue of REF, MATH and MATH . Since, by the very definition of MATH, we have MATH and MATH, it suffices to show that MATH . But this has (essentially) been proved in CITE. |
math/0101117 | The second formula for MATH follows from the first, since MATH and MATH by REF. We have seen that MATH . We can get rid of MATH and obtain a representation of MATH as a fractional ideal, namely, MATH . It follows that MATH. As MATH, we are done. |
math/0101117 | Note that MATH by definition. Let MATH be a product of minors. We say that MATH has tight shape if its degree is divisible by MATH and it has exactly MATH factors. In other words, MATH has tight shape if MATH and MATH where MATH. Let MATH be a product of minors with tight shape and MATH. Note that MATH . Now fix MATH and set MATH; the product MATH has tight shape and hence MATH. Note that, for obvious reasons, MATH does not contain MATH. It follows that MATH is a prime ideal of height MATH and that MATH. Now take MATH and set MATH; the product MATH has tight shape and hence, as above, we conclude that MATH is prime of height MATH and MATH. Since we know already that MATH, we are done. |
math/0101117 | Assume that MATH. Set MATH. We have seen that MATH . Consequently MATH . Note that MATH can be written as the intersection of MATH and MATH. But the latter is the irrelevant maximal ideal of MATH, whence MATH. Further MATH can be written as an intersection of ideals MATH. Taking into consideration the shape of MATH and REF, we have MATH . Summing up, we get the desired result. |
math/0101118 | It is enough to show that MATH . The proof splits into four cases, corresponding to the four quadrants of the MATH-plane. For example, if MATH, then MATH and the remaining cases are similar. Finally, note that the symbol of MATH is comparable to MATH. |
math/0101118 | By REF , which is equivalent to MATH there exist MATH for MATH such that MATH and MATH. By linearity, MATH so by NAME 's integral inequality, REF and the NAME inequality, MATH . This concludes the proof of REF , and to prove REF we simply write MATH, use the multilinearity of MATH, and apply REF . |
math/0101118 | We prove REF for MATH; the proof for MATH is similar. We have MATH whence MATH . But the last expression equals MATH . As for REF , using the decomposition REF and the bilinearity of MATH, we have MATH so it suffices to prove MATH and MATH. We will only prove the case MATH . The NAME transform of the right hand side, at fixed MATH, is MATH and applying the NAME transform in MATH yields the following expression for the space-time NAME transform at MATH of the right hand side of REF: MATH . But the latter is, by the definition of MATH, equal to the space-time NAME transform of the left hand side of REF. The remaining cases are proved in a similar manner. |
math/0101118 | Assume that MATH is a NAME sequence in MATH. Then MATH is NAME in MATH, so it converges in the latter space to some limit MATH. It remains to prove that MATH as MATH. Fix MATH. There exists MATH such that MATH for all MATH. We claim that MATH for all MATH. To see this, fix MATH such that MATH and MATH, where MATH are conjugate to MATH. Then MATH for all MATH, so it suffices to prove that MATH for fixed MATH. To prove this, we write MATH where MATH. Since MATH converges to MATH in MATH, we conclude that REF holds. |
math/0101118 | Since the norm MATH is compatible with the relation MATH and only depends on the size of the NAME transform, it suffices to prove the inequality when MATH. Thus, we fix MATH and MATH such that MATH and MATH. Let MATH etc. denote the dual exponents. Then MATH where the supremum is over all MATH such that MATH and MATH. But by NAME 's inequality, MATH finishing the proof. |
math/0101118 | Since MATH is an isomorphism of MATH onto MATH, which preserves the NAME class and positivity of the NAME transform, we may take MATH. Since the NAME transform is an isomorphism of both MATH and MATH, it then suffices to prove that the set MATH is dense in MATH. But the standard proof that MATH is dense in MATH shows this to be true. |
math/0101118 | We have MATH where the supremum is over all MATH with MATH. REF follows. For REF , we have MATH where the supremum is over all MATH such that MATH and MATH. |
math/0101118 | We apply REF . By REF , we get MATH . |
math/0101118 | Set MATH. Since MATH, it suffices to show that MATH for all MATH whose inverse NAME transform MATH is nonnegative. The integral on the left hand side is dominated by MATH where MATH . By NAME 's inequality, REF is bounded by MATH times MATH, where MATH, so it suffices to show that MATH . But MATH, where MATH, and since MATH, it follows that MATH . By Theorem A (see also REF ), the right hand side is dominated by MATH. |
math/0101118 | If MATH, the hypotheses of REF are satisfied, so the result follows by REF . |
math/0101118 | By REF , it suffices to prove the estimates REF for MATH. Fix MATH and MATH. Let MATH be any extension of MATH (meaning MATH). Since MATH is local in time, we have MATH on MATH, so by REF, MATH . Let MATH. Since MATH is continuous, REF follows. To prove REF, fix MATH, and let MATH be any two extensions of MATH and MATH. By REF, MATH . Let MATH, and write MATH . We may assume that MATH is increasing. Thus MATH . Let MATH and MATH. Since MATH we conclude that REF holds. |
math/0101118 | For MATH, we only have to note that MATH where MATH for any MATH. This proves REF . Let MATH be the space in REF . Since MATH, REF gives MATH for any MATH. We claim that MATH for all MATH; this finishes the proof of REF , if we choose MATH sufficiently small. To prove the claim, notice that MATH where the supremum is over all MATH such that MATH and MATH. But for such MATH, we have MATH by NAME inversion. Finally, we prove REF . Since the norm only depends on the size of the NAME transform and is compatible with the relation MATH, it suffices, by REF, to prove that MATH for some MATH. Let MATH and choose MATH satisfying MATH. By REF, NAME and NAME, MATH for any MATH. This finishes the proof. |
math/0101118 | Since MATH, we may assume MATH. Let MATH and MATH be defined by MATH and MATH. Since MATH we have MATH . Thus, it suffices to prove MATH for all MATH such that MATH. Since MATH for all MATH, we have MATH and MATH for all MATH, so MATH maps the set MATH into itself. Thus, if MATH, MATH where MATH denote the dual exponents of MATH and the supremum is over all MATH such that MATH and MATH. But by NAME 's inequality and NAME embedding, MATH for any MATH, and we have MATH. |
math/0101118 | The symbol of MATH is MATH. Recall that MATH denotes the NAME inner product on MATH, while MATH always denotes the Euclidean norm. Since MATH we have MATH . Take this to the power MATH, and take the trivial estimate MATH to the power MATH. The product of the left hand sides of the resulting inequalities is then bounded by the product of the right hand sides, and keeping in mind that the symbol of MATH is MATH, we get the desired estimate. |
math/0101118 | Since MATH by the triangle inequality, REF implies MATH . To finish the proof, combine this with MATH . The latter is proved by considering two cases: MATH and MATH, where MATH and MATH are the frequencies of MATH and MATH respectively. |
math/0101118 | REF follows from REF is proved by considering two cases: MATH and MATH, where MATH and MATH are the frequencies of MATH and MATH respectively. |
math/0101118 | By REF , MATH . Since MATH whenever MATH, we conclude that MATH . Since MATH whenever MATH, the second term on the right hand side is subsumed in the first. It is easy to see that MATH provided MATH. Combining this with REF yields the desired estimate. |
math/0101118 | It is readily verified that the symbol MATH of MATH satisfies MATH where MATH, MATH and MATH. We consider three cases, corresponding to MATH, MATH or MATH being the maximum of the three. If MATH is the maximum, then MATH. If MATH is the maximum, then MATH. If MATH is the maximum, we consider two subcases: CASE: MATH; and REF MATH. In REF . In REF . |
math/0101118 | Assume that MATH and MATH (MATH). By NAME 's inequality, MATH . Once we have these estimates, the others follow by interpolation: CASE: Assume MATH (so MATH). Interpolation between REF gives MATH . CASE: By REF , we have MATH . CASE: Assume MATH (so MATH). By REF , MATH . CASE: By REF , MATH . This concludes the proof. |
math/0101118 | We have MATH and a similar computation gives the proof for MATH. |
math/0101118 | Set MATH and write MATH, where MATH is supported in the region MATH and MATH is supported in MATH. By REF , MATH where MATH is the angle between MATH and MATH. Thus, for all MATH, integration in polar coordinates MATH yields MATH and the last integral is finite iff MATH. By symmetry, this implies that MATH is also finite, so we may apply REF . |
math/0101119 | The only point which is not evident is that MATH. This is clearly true when MATH, since then MATH, so MATH is necessarily given by NAME 's formula. The general case then follows by density, since clearly MATH is linear and bounded from MATH into MATH, and the latter space embeds in REF. |
math/0101119 | This is a triviality. To prove REF, simply note that the NAME transform of MATH is MATH. To prove, REF observe that the NAME transform of MATH is MATH and that MATH for MATH and MATH. |
math/0101119 | Observe that MATH for any MATH. Thus, REF follow immediately from REF , respectively. |
math/0101120 | To go from REF to REF, observe that MATH is divergence free by REF; apply MATH to REF to get REF ; apply MATH to REF and use REF to get REF; finally, REF follows from REF using REF. To go the other way, observe that by REF, the right hand side of REF is divergence free; thus MATH, and since the initial data of MATH are divergence free, REF follows. Then, in view of REF are equivalent. Finally, to go from REF to REF, it suffices to check that the right hand side of the latter is divergence free. But this follows from REF. |
math/0101120 | To see that REF is in MATH, first apply REF, then estimate as in REF. That REF is the only MATH solution can be seen by taking the NAME transform of both sides of REF. |
math/0101120 | Recall that MATH, as defined in REF, is a NAME space with inner product MATH (by NAME 's theorem), and that MATH. We denote by MATH the corresponding real NAME space, with inner product MATH. By definition, MATH solves REF in the sense of distributions iff MATH for all MATH. Since MATH is dense in MATH and MATH we conclude that MATH solves REF iff REF holds for all MATH. Taking MATH gives MATH and since MATH for all MATH, we conclude that MATH where MATH. Therefore REF holds, and uniqueness follows. Of course, MATH must be real, since if MATH solves REF, then MATH solves the same equation with MATH, and therefore MATH by what we just proved. To prove existence, observe that the left hand side of REF defines an inner product on MATH, and in view of REF, the corresponding norm is equivalent to the usual norm. Moreover, by REF, the right hand side of REF is a bounded linear functional MATH on MATH. Existence therefore follows from the NAME representation theorem. |
math/0101120 | Subtracting the equations gives MATH . Then by a density argument as in the previous proof, MATH giving the desired conclusion. |
math/0101120 | Proceed as in the proof of REF , but with the right hand side of REF replaced by MATH. Thus REF is replaced by MATH . Existence then follows, and any MATH solution satisfies MATH where MATH is independent of MATH and MATH, and REF follows. Subtracting REF from REF gives MATH and applying REF gives the desired estimate. |
math/0101120 | Set MATH. For every test function MATH in MATH, MATH . The left hand side is a bounded linear functional in MATH. In fact, MATH . Here we used NAME 's inequality and the embedding MATH. But MATH is dense in MATH, so it follows that REF must hold for all MATH. Taking MATH gives MATH . This implies MATH, hence MATH (MATH, as we have defined it, does not contain any nonzero constants). |
math/0101120 | We have MATH whence MATH . The same proof works for MATH. If MATH, then (discarding the time variable) MATH . Thus, by NAME 's integral inequality and the translation invariance of the norm, MATH . This is certainly valid for smooth MATH, and hence in general by using an approximation of the identity. For MATH we write MATH and note that MATH. It follows that MATH. |
math/0101121 | Recall CITE that the natural map MATH is an isomorphism, where MATH is the evident vector bundle over MATH with trivial MATH - action. By applying the ordinary splitting principle to MATH, we are reduced to the case that MATH, where MATH is a complex line bundle over MATH with trivial MATH-action. If MATH is suitable then the NAME class of MATH is MATH . Since MATH is a compact manifold, MATH is nilpotent in MATH, so MATH is a unit of MATH because MATH is a unit of MATH. |
math/0101121 | Let MATH be a complete local MATH algebra, and suppose that MATH is not torsion in MATH. We see from REF that there is an isomorphism MATH which is compatible with the projection and clearly induces an isomorphism MATH . |
math/0101121 | In the exact sequence MATH the first and last terms are zero because MATH acts nilpotently on MATH and as an isomorphism on MATH. |
math/0101121 | Let MATH be a test ring. Suppose that MATH is a lift of MATH. Then MATH is a lift of MATH. According to CITE, there is a unique pair MATH consisting of a map MATH and an isomorphism MATH of lifts of MATH. On the other hand, MATH induces an isomorphism MATH . Assembling these gives an extension MATH which defines an isomorphism MATH . MATH pro-represents the right-hand side by REF . |
math/0101123 | To make the paper as self-contained as possible, we will give a direct proof of this for regular weights MATH, since the general case relies on an unpublished result of NAME, CITE. It is enough to prove this in the ungraded case. Let MATH be all distinct (up to isomorphism) projective indecomposable MATH-modules. The algebra MATH has a simple socle CITE. Therefore if Ann-MATH is non-zero it must contain the socle of MATH. The equality MATH, implies that MATH acts faithfully on at least one projective indecomposable, say MATH. By REF and a result of CITE the dimension of MATH, so MATH generates MATH. Now assume MATH is regular. It follows from CITE that we can find a translation functor MATH such that MATH and, by CITE, that MATH is a self-equivalence of MATH. Since MATH can be identified with the endomorphism ring of the identity functor on MATH, conjugation by MATH induces a ring automorphism of MATH, say MATH. It follows that MATH is generated by MATH, as claimed. If MATH is not regular the above argument fails since it need no longer be true that we can find a translation functor MATH which is a self-equivalence and sends MATH to MATH. In this situation we use the following fact, CITE: if the highest weight of MATH (in the graded category) belongs to MATH and does not lie on the affine wall, then the canonical map MATH is an isomorphism. Hence it is enough to show that we can find a representative of the isomorphism class of MATH whose highest weight belongs to MATH and does not lie on the affine wall. A straightforward calculation shows that this follows from CITE. |
math/0101123 | It is enough to prove this for ungraded MATH-modules. Suppose that MATH is a composition factor of MATH. Thus MATH appears as a direct summand of MATH for some MATH. Hence we have a commutative diagram MATH . Thanks to REF there exists MATH such that the above endomorphism of MATH is multiplication by MATH. Then MATH since, by hypothesis, the composition factor MATH does not lie in MATH. Since MATH is local it follows that the endomorphism is an isomorphism and so MATH lies in the head of MATH as required. |
math/0101123 | Let MATH be a finite dimensional algebra with simple modules MATH. The NAME quiver of MATH is the directed graph with vertices labelled from MATH to MATH and MATH edges from MATH to MATH. By CITE MATH is NAME equivalent to the the path algebra of its NAME quiver factored by some admissible ideal, that is an ideal generated by linear combinations of paths of length at least two. Let MATH be such that MATH, or, equivalently, MATH. Set MATH for MATH and set MATH. Since MATH appears only once as a composition factor of MATH, MATH acts by scalars on MATH, making MATH a MATH-module. By CITE, for MATH modulo MATH and MATH . Thus the NAME quiver of MATH is of the form REF, possibly with loops added at the vertices. Let MATH be the quotient of this quiver which is NAME equivalent to MATH. We will show MATH is isomorphic to MATH. The projective covers of the simple MATH-modules are spanned by the paths ending in a fixed vertex. Hence, REF shows that there can be no loops at vertices and further, that MATH is therefore a quotient of MATH. In particular, its dimension is at most MATH. Let MATH be the kernel of the sum of two projections MATH. By REF MATH . Since MATH is a quotient module of MATH, the dimension of MATH can be estimated by MATH . We deduce that MATH, proving the first statement of the theorem. This also proves that MATH is the projective cover of MATH in MATH. Let MATH be regular, so that MATH, and let MATH. Thanks to REF the algebra MATH has a MATH-grading. By CITE MATH and MATH are equivalent MATH-categories if MATH as a MATH-graded algebra. But, up to a choice of scalars, MATH corresponds to a MATH-module homomorphism sending MATH to MATH, and MATH corresponds to the a MATH-module homomorphism sending MATH to MATH. This proves the second claim. |
math/0101123 | It is straightforward to check that MATH, MATH and MATH are central elements. By construction MATH is a free MATH-module with basis MATH. It follows from the relations in MATH that the degrees of the homogeneous components of any non-zero central elements must be a multiple of MATH. Since MATH and MATH are central we must find which polynomials in MATH are central. Let MATH be such a polynomial. Since MATH we deduce that the roots of MATH are invariant under integer addition. It follows that MATH is a polynomial in MATH as required. Using the defining relations once more we have MATH . Since MATH has degree MATH it follows that MATH. Hence, the centre of MATH is a quotient of the ring of functions of a Kleinian singularity of type MATH. Any proper quotient of the ring of functions on a Kleinian singularity has dimension REF or REF. Thus since MATH is finitely generated over its centre and has NAME dimension REF, the centre must be the entire ring of functions. |
math/0101123 | Let us obtain all relations in REF recursively. For MATH, MATH . Similarly, for MATH, MATH . Now we need to show that all remaining compositions are trivial. The highest terms of defining relations are MATH, MATH, MATH, MATH, MATH, and MATH. Let us make certain that all compositions are trivial. Firstly, MATH . Similarly, MATH . Then MATH . NAME, the compositions MATH and MATH are trivial. Then MATH . Another possible composition to consider is MATH . The remaining compositions MATH, MATH, MATH, and MATH are trivial by a similar argument. |
math/0101123 | The direct sum decomposition REF follows at once from the description of the NAME basis of MATH. Adding dimensions of summands, we arrive at the dimension of MATH, that is MATH. |
math/0101123 | If MATH then the roots of MATH are MATH. If MATH then MATH are already roots of MATH. Thus, the dimension of the first summand in REF is MATH where the number of summands in the parenthesis is MATH. This sum equals MATH. Similarly, the second summand has dimension MATH, so that the total dimension is MATH. |
math/0101123 | CASE: For the generator MATH one observes that MATH . Thus if MATH is not a root of MATH then MATH and MATH. Let MATH be a root of MATH. Let MATH be the NAME basis of MATH constructed in REF . The module MATH is determined by the pair MATH and this turns out to be a NAME pair. Indeed, there are three elements in MATH whose leading monomials end with MATH: MATH . Elements of MATH whose leading monomials end with MATH fall into two types: MATH . Direct computation now yields the formulas for the action. CASE: The proof is analogous. |
math/0101123 | Set MATH and MATH. Note that if MATH then MATH. Since every MATH-module has a MATH-fixed point we see that any simple MATH-module is a quotient of MATH for some MATH. By REF MATH is isomorphic to MATH as a MATH-module. Since MATH is a local algebra it follows that MATH has a simple head. In MATH the element MATH is annihilated by MATH and belongs to the MATH component, so there is a graded MATH homomorphism MATH . The cokernel of MATH, say MATH, has dimension MATH and is simple if MATH since it is generated by any basis vector MATH it contains. If MATH it follows from REF that MATH has a unique MATH-fixed point, namely MATH. Hence if MATH and MATH are non-zero MATH and MATH are isomorphic if and only if MATH. This proves REF . Since MATH is simple-headed, MATH and MATH the chain of homomorphisms MATH proves REF . The proof of REF is similar. REF is clear. |
math/0101123 | Let MATH be such that MATH. Let MATH be the projective cover of MATH in MATH. Recall the general formula, CITE MATH . Let MATH be the kernel of the sum of two projections MATH. Then MATH has head isomorphic to MATH so is a quotient of MATH. By REF MATH. Using REF we find MATH proving that MATH. Let MATH. The basic algebra of MATH is MATH. Let MATH (respectively MATH) be the homomorphism MATH (respectively MATH) associated to the composition factor MATH of MATH (respectively MATH) lying in the second NAME layer of MATH (respectively MATH). It is straightforward to check that MATH and MATH, together with the idempotents arising from the projections in MATH, generate the basic algebra and satisfy the relations of the no-cycle algebra. Since MATH the first statement of the proposition follows. The second statement is proved in the same manner as REF . |
math/0101123 | Set MATH and let MATH be the number of non-zero MATH's. Let MATH be such that MATH. Let MATH (respectively MATH) be the simple MATH-modules (respectively graded MATH-modules) appearing in REF (respectively REF ) and let MATH (respectively MATH) be their projective covers. We have MATH and MATH . Thanks to our construction of MATH in REF and MATH in REF we have a graded isomorphism MATH proving the first statement of the theorem, together with an equivalence. The equivalence is a MATH-equivalence by CITE. |
math/0101123 | Let MATH be the basic algebra of MATH. There exists a MATH-bimodule MATH, flat over MATH, such that the functor MATH induces an equivalence between the categories of finite dimensional MATH-modules and finite dimensional MATH-modules. Let MATH, a flat family of MATH-modules over MATH. Given a primitive idempotent MATH, it suffices to show that the dimension of MATH is independent of MATH, a NAME open neighbourhood of a point. Without loss of generality we can trivialise the family MATH. Then MATH defines an algebraic family of projection operators on the finite dimensional vector space MATH, MATH . Since the dimension of MATH is equal to the rank of MATH, and the latter is obviously constant, the lemma is proved. |
math/0101123 | (Alternative) Without loss of generality we can assume that MATH is irreducible. Let MATH. For any MATH let MATH be the characteristic polynomial corresponding to the action of MATH on MATH. As this a polynomial of degree MATH we obtain a regular map MATH sending MATH to the coefficients of MATH (coefficients written in order MATH). Given a simple MATH-module, MATH, let MATH. Thanks to CITE we have MATH . For any MATH let MATH be the closed set defined by the vanishing of the first MATH co-ordinates. Then MATH is closed and consists of the values MATH for which MATH. Hence, for MATH, there exists an open set, MATH such that for all MATH the value MATH is minimal. Let MATH be this minimum. The set MATH has a minimum, denoted MATH, which is achieved at MATH say. Then for all MATH we have MATH. We deduce that the set MATH is dense in MATH since it contains MATH. Repeating the above for all simples we see there is a dense set in MATH where all the functions MATH take minimal values. But MATH is constant and therefore they must take minimal values everywhere, proving the lemma. |
math/0101123 | It follows from REF that if MATH is not gradable it is necessarily a band module and hence indecomposable. Thus, by REF , it suffices to show that MATH does not admit a MATH-grading. Suppose for a contradiction that MATH admits a MATH-grading. Let MATH be the MATH-subalgebra generated by MATH and MATH. Any NAME subalgebra belonging to MATH is uniquely determined by its intersection with MATH. Let MATH. A straightforward calculation shows that the restriction of MATH to MATH has a direct summand isomorphic to the baby NAME module for MATH induced from MATH with highest weight MATH. For MATH the element MATH is a highest weight vector for the NAME subalgebra MATH, yielding an isomorphism MATH. Since MATH is regular MATH, and so, by CITE, the baby NAME modules for MATH induced from different NAME subalgebras of MATH with highest weight MATH are not isomorphic. Hence, by the last paragraph, the isomorphism MATH forces MATH to have infinitely many non-isomorphic direct summands, a contradiction. |
math/0101123 | Let MATH. We have the following natural isomorphisms MATH where the last isomorphism is the projection formula. The same equations hold for MATH, so MATH is a MATH-functor. Using the same formalism, one checks the inverse is also a MATH-functor. |
math/0101123 | Let MATH be the projection map. Since there is a natural isomorphism MATH, we have MATH . |
math/0101123 | In the non-equivariant setting the complex REF was studied in CITE. It is shown in CITE that its homology vanishes in degrees REF and is MATH in degree REF if MATH. Thus we have a quasi-isomorphism MATH . For MATH the bundle MATH restricted to MATH is MATH CITE. If MATH a similar argument using CITE, shows that MATH. It remains to prove that MATH. For MATH let MATH be the subscheme of MATH consisting of the components MATH. Associated to this subscheme we have the maps of coherent sheaves arising from the inclusions MATH and MATH . This is an exact sequence; so we have MATH. Induction yields MATH . By CITE we have MATH and MATH, which yields MATH . By CITE we have MATH and MATH. The result follows. |
math/0101123 | We first calculate MATH. If MATH then, using REF , we see that MATH equals the complex MATH . It is straightforward to check that this complex only has homology at its end term, equal to MATH. Therefore MATH. If MATH (the case MATH is analogous) we find MATH is the complex MATH . Again, it is straightforward to check that this complex only has homology at its end term. A basis for this homology group is MATH, showing that this is the unique bigraded MATH-module with head MATH and socle MATH (multiplication by MATH sends the head MATH to the socle MATH). Call this MATH. If MATH, then MATH since the relevant NAME vanishes, REF . The description of MATH follows immediately, noting that MATH thanks to REF. |
math/0101123 | Given any object MATH and MATH evaluation gives a natural isomorphism MATH in MATH. Hence, by CITE, MATH . By construction MATH so we deduce that MATH. The same statement with MATH follows since MATH . Finally, since MATH and MATH we find MATH, as required. |
math/0101123 | REF follows immediately from the variance properties of MATH and MATH. By REF we have MATH and MATH proving REF . Finally, we have MATH . In MATH the identity provides an identification MATH, which, together with above equation, proves REF . |
math/0101123 | By construction MATH for MATH and MATH. The claim about duality then follows from the sentence following the proof of REF . The calculation of NAME in REF shows that MATH . Calculating the image of this in MATH under MATH yields REF . Thus the pairings agree since REF correspond to REF |
math/0101124 | With the notations MATH and by computing derivatives of inverse functions, we obtain from REF MATH . The positivity of the left-hand side is assured in case MATH, and is equivalent to the condition MATH by positivity of MATH. The function MATH is analytic in MATH is strictly positive, hence the left-hand side of REF is bounded on the interval MATH for each MATH. Due to the unbounded behavior of MATH on any interval containing zero, there exist MATH for which REF and hence convexity of MATH is satisfied. |
math/0101124 | Stationarity means MATH after some changes of variables, by straightforward computations we obtain from REF MATH where MATH . We eliminate the expressions MATH for all MATH with the use of REF MATH to obtain MATH . For MATH fixed, let us consider bounded cylinder functions MATH, which depend on the variables MATH. By stationarity of MATH, REF is satisfied for all of them. Hence it is necessary that MATH does not depend on the variables MATH and its mean is zero according to MATH. Only MATH, and the second term in MATH are the terms which can contain product of functions of different variables MATH. Each of them is positive by monotonicity of MATH. Thus it follows that each of these three terms must not depend on more than one variable. This implies MATH due to the form of the second term in MATH. The value MATH of this constant is a consequence of REF. Thus we conclude that MATH is necessarily exponential, the rates are that of the EBL model REF. MATH and MATH can also contain at most one variable, hence we obtain MATH for MATH and MATH for MATH. This means that we have at most two kinds of marginals of MATH, one on the left-hand side of the defect tracer and an other one on its right-hand side. In REF, computing the expectation value of MATH times the terms of MATH, summed up for MATH and for MATH, gives zero. The reason for this is that the variables in these terms are independent of the variables which MATH depends on. For the rest of the indices, note that we have a telescopic sum for MATH. Due to this and using the rates of the EBL model, we can simplify our expressions to MATH . It is easy to check that simply choosing MATH does not eliminate the variables MATH from MATH. Hence this can not be a solution for MATH to make REF be satisfied for all MATH. This means that the marginals on the left-hand side of the defect tracer are different from those on the right-hand side. When taking expectation value in REF, this leads to having constant times MATH from the terms containing MATH and MATH in the first part of the expression of MATH. In order to make REF be satisfied for all MATH, it is necessary that we obtain other constants to have zero together with. They can only come from MATH and MATH. Thus we conclude MATH with the use of the form REF of MATH. In view of REF, we have the measures MATH with MATH. We know that the normalization MATH in the right-hand side of REF is periodic in the parameter MATH with period one, which tells us MATH . Using this result together with REF and with the property MATH, we see that REF is satisfied, which completes the proof. |
math/0101126 | Let MATH, and let MATH denote the surjective homomorphisms. We consider the fiber bundle MATH . We let MATH whence MATH . Suppose MATH; that is, MATH is a MATH matrix of rank MATH. We consider MATH where MATH. Clearly, MATH and thus MATH . It is easily seen that MATH . Therefore, MATH . |
math/0101127 | First we assume that MATH is a rational homology REF-sphere. When stretching the neck in MATH, as MATH, we get two manifolds, each endowed with an infinite cylindrical end, MATH . The NAME monopole moduli space of MATH, for sufficiently large MATH, can be described in terms of the moduli spaces of MATH and MATH as analyzed in CITE. Notice that the moduli spaces of flat connections on MATH, modulo the subgroups of the gauge transformations on MATH which can be extended to the whole manifolds MATH or MATH, define the following character varieties: MATH . Thus the character variety MATH is MATH . The covering maps between these character varieties are illustrated as follows MATH where the maps MATH are the covering maps with fibers as indicated. Based on the analysis of the space MATH of irreducible monopoles on a REF-manifold with a cylindrical end modelled on MATH, as in CITE, we see that the asymptotic limit map defines a continuous map: MATH . The reducibles on MATH embed in the character variety MATH. Then, for a sufficiently large MATH, the gluing theorem gives a diffeomorphism: MATH here MATH is a small neighbourhood of the ``bad points" MATH in MATH where the twisted NAME operator on MATH has non-trivial kernel, and MATH are the reducible lines in MATH. The above fibred product is obtained (compare CITE) by taking the pullbacks of the images of the boundary value maps under the projections MATH . Let MATH be the coordinates on MATH corresponding to the holonomy around the longitude MATH and the meridian MATH respectively. In the gluing map MATH above, MATH corresponds to the lines MATH. For each line MATH, the image of the gluing map gives a diffeomorphism onto MATH. For each MATH structure MATH, there is a unique reducible monopole on MATH, which is given by the intersection of MATH, the flat connections on MATH, with the line MATH in MATH: MATH . Here MATH is the holonomy of the flat connections in MATH around the longitude MATH. Similarly, we have gluing maps for monopoles on MATH and MATH, respectively. In the gluing map MATH for MATH, MATH is identified with MATH. Similarly, we have the gluing map MATH for MATH, MATH where MATH is given by MATH. For each MATH, the reducible monopole for MATH, consists of the unique point MATH . For MATH with a non-trivial MATH structure MATH (MATH), the set of reducibles is empty for any generic perturbation, and for MATH, it consists of one circle of reducibles MATH in the cylinder MATH, which can be perturbed away by introducing a small perturbation as in REF. Now we apply the perturbation to simulate the effect of NAME surgery. This amounts to a careful choice of perturbation as in REF, which we now briefly describe. Choose a compactly supported REF-form MATH representing the generator of MATH, defined as in REF, such that we have MATH for any point on MATH. Under the isomorphism MATH, given by NAME duality, this form corresponds to the generator MATH. The class of MATH in MATH is trivial, and we can write MATH, where MATH is a REF-form satisfying MATH, that is, MATH. Choose on MATH a metric as in REF. Fix a MATH-connection MATH representing the trivial connection on MATH, For any MATH-connection MATH, define MATH to be MATH . For any given MATH, we can choose a function MATH with the following properties. CASE: MATH is continuously differentiable on MATH and satisfies the periodicity REF the derivative MATH has range MATH for all MATH, and satisfies MATH for MATH. REF the following estimate holds: MATH . Now, for the MATH structure MATH (MATH), consider the function MATH and define a perturbation of the NAME equations on MATH in the following way: MATH . With respect to the chosen metric on MATH, with sufficiently large positive scalar curvature on the support of MATH as specified in REF, the only solutions of the perturbed monopole equations are reducibles MATH, that satisfy MATH . In addition to this surgery perturbation, we consider another perturbation of the NAME - NAME equations on the tubular neighbourhood MATH in MATH, MATH, and MATH. This perturbation has the effect of producing a global shift in the character variety to avoid the bad points on MATH when we deform the unperturbed geometric triangles in MATH to the perturbed geometric triangles in MATH. Let MATH be a compactly supported REF-form in MATH satisfying REF . Let MATH be some small real parameter. Consider an additional perturbation MATH of the curvature equation on MATH inside MATH and inside MATH. This perturbation has the effect of shifting the asymptotic values by an amount MATH. We choose the sign so that the line of reducibles on MATH for MATH becomes MATH, the line of reducibles on MATH for MATH remains the same MATH, and the line of reducibles on MATH for MATH becomes MATH. On MATH inside MATH for MATH we shall consider the perturbed curvature equation MATH . Therefore, we can partition the moduli spaces for MATH REF into the union of the moduli spaces for MATH REF and MATH, as in REF. This completes the proof of the theorem for the case of rational homology REF-sphere MATH with a knot MATH representing a torsion element of order n in MATH. For a general REF-manifold MATH with a smoothly embedded knot MATH representing a torsion element of order n in MATH, the proof is essentially the same as the case of rational homology spheres discussed above, and we omit the details here. |
math/0101127 | In the proof of REF , we know that the perturbed NAME monopoles on MATH and MATH are given by the following gluing models (here we assume that MATH is a rational homology REF-sphere): MATH . Note that the additional perturbation REF of the curvature equation on MATH inside MATH and inside MATH introduce a shift of coordinates MATH to MATH. We can introduce these new coordinates, still denoted by MATH. Then the reducible line for MATH is given by MATH in MATH. We will show that there exists a further surgery perturbation on MATH that suits the purpose of identifying monopoles on MATH and MATH as stated in the Theorem. Without loss of the generality, after possible coordinates change, we can assume that MATH. Fix MATH. We will construct a function MATH, which depends on a small MATH, such that, as MATH, the curve MATH approaches the union of lines MATH where MATH . We identify MATH with the fundamental domain MATH in MATH. The asymptotic values MATH can be lifted to MATH periodically. Using this MATH - periodicity, we only need to construct a function MATH, which depends on MATH, such that, for any given MATH, we have MATH . Such function can be easily constructed as in the proof of REF . Then over MATH, MATH is defined to be MATH. See REF where the perturbation is illustrated in the cases with MATH and with MATH, respectively. The general case can be proved by a similar method. |
math/0101127 | The first claim follows from the gluing models of the monopoles in MATH and MATH. Then, using the results in CITE, we know that, in the case of the rational homology REF-sphere, the spectral flow of the twisted NAME operator along the path of reducibles gives the index shift on MATH and MATH according to the wall-crossing formulae derived in CITE. Again by the results of CITE, for MATH with MATH, the induced grading from the parametrized spectral flow is same as the original relative index on MATH. |
math/0101127 | Using the gluing REF , we can describe the moduli space MATH for two monopoles MATH of relative index MATH. With our convention on the choice of components for these moduli spaces, we know that MATH is zero-dimensional, if non-empty, and obtained as the gluing of the admissible geometric limits on MATH and the corresponding holomorphic triangles. As we let MATH, it is easy to see that the holomorphic triangles degenerate to certain holomorphic discs, and the admissible geometric limits on MATH for MATH are identified with the admissible geometric limits on MATH for MATH. Since the relative index of MATH is zero, MATH, being a zero-dimensional moduli space of minimal energy, is empty unless MATH, in which case MATH consists of a unique solution. This proves that MATH . Similarly, we obtain MATH . |
math/0101127 | REF Using the diagram chasing, REF , we see that we have MATH . By comparing this expression with MATH we obtain MATH . This completes the proof. |
math/0101127 | The argument is exactly the same as in the proof of REF, therefore we shall omit the proof here. |
math/0101127 | By the gluing theorem for REF-dimensional monopoles as in CITE, we have MATH where MATH is the MATH-curve on MATH. Thus, we obtain MATH . Notice that the set MATH consists of the unique reducible monopole for each MATH. Similarly, we have MATH . Here the reducible set consists of MATH . In order to avoid the circle of reducibles on MATH, we need to introduce a small perturbation such that MATH on MATH is a small parallel shifting of MATH such that the bad point is not contained in the narrow strip bounded by these two parallel curves. We denote this small shift of MATH by MATH, where MATH is a sufficiently small positive number. This can be achieved by a perturbation of the equations as in CITE. Then we have MATH . In order to compare the three countings in REF - REF , we need to choose an oriented REF-chain MATH in MATH whose boundary REF-chain is given by MATH and such that MATH does not contain the bad point in MATH. Then, counting the boundary points of MATH, as a REF-chain, we obtain MATH . As MATH does not contain the bad points, we know that the possible points of MATH all lie on the curve MATH, away from the points MATH. It is easy to see that MATH covers the intervals of MATH between two consecutive points MATH with different multiplicities: the multiplicities are MATH, for the intervals MATH respectively. By the identity REF , we know that MATH . Combining all the identities in REF , we obtain the proof of the proposition. |
math/0101127 | REF follows from the wall-crossing formulae in CITE and the NAME index theorem. The proof of REF is analogous to the proof of REF. We adapt their arguments to our situation. We write the standard surgery cobordism between MATH and the Lens space MATH as MATH . Then the surgery cobordism between MATH and MATH can be identified as MATH . We fix a metric on MATH which respects the product structure MATH and MATH, and agrees with MATH and MATH on the boundaries MATH and MATH, respectively. For a MATH structure MATH in MATH on MATH, whose reducible monopole corresponds to MATH (with MATH and MATH), we consider the MATH structure MATH on MATH whose reducible monopole is MATH. Then we claim that MATH is independent of MATH and depends only on MATH and on MATH . To prove this claim, we choose a MATH structure MATH on MATH whose restriction to MATH and MATH is given by MATH and MATH, respectively, and such that MATH. On MATH, we choose a connection MATH, whose restriction to MATH is the path of reducibles connecting MATH to MATH along the curve MATH. Then we have MATH where the third equality follows from the splitting principle for the index, as the NAME operator has no kernel on the various boundaries and corners CITE. Notice that we have MATH and the connection MATH extends to connection MATH on MATH by a flat connection, whose index on MATH satisfies MATH . In fact, we can choose the metric on MATH with a positive scalar curvature metric on MATH. Therefore, we have MATH which depends only on MATH and MATH, and so does the quantity MATH . When summing the identity REF over MATH and MATH, notice that the term MATH is independent of MATH, hence we obtain the proof of REF by using REF . |
math/0101127 | Following from REF , we know that MATH depends only on MATH and MATH. We denote this term by MATH. By subtracting REF from the surgery formula for the NAME invariants in REF , we obtain the proof of this theorem. |
math/0101127 | We first derive the surgery formula for the invariant MATH from REF and the NAME invariant for MATH (a rational homology MATH, that is, MATH) (see CITE CITE): MATH where MATH and MATH is the coefficient of the symmetrized NAME polynomial of MATH, MATH normalized such that MATH . Set MATH as the normalized NAME invariant. Then the surgery formula in CITE for MATH can be expressed as (compare CITE): MATH . Here MATH is the NAME sum of relatively prime integers MATH and MATH (compare CITE). Comparing REF , we only need to show that MATH . Since MATH is independent of the manifold MATH, we can choose some examples that can be computed explicitly, and use them to identify the coefficient MATH. The Lens space MATH can be obtained by a MATH-surgery on an unknot in MATH. The calculation of CITE for MATH gives us that MATH . This implies that REF holds for MATH. Now we can prove REF by induction on MATH. This is exactly the same argument as in the proof of REF on the equivalence of their theta invariant and the NAME invariant. The example is the NAME manifold MATH, obtained by MATH surgery on a knot of order MATH in MATH. By NAME calculus it is possible to show that MATH can be obtained as MATH-surgery on a knot in the Lens space MATH, and can be obtained as a sequence of surgeries on knots of order less than MATH, see the proof of REF for details. |
math/0101128 | Consider a SFT MATH . Let MATH be the cardinality of the alphabet of MATH and MATH the length of the forbidden blocks. Consider a horseshoe map MATH with MATH - branches. We consider the standard NAME partition for MATH. Define MATH to be the (finite) union of NAME rectangles (for MATH) which correspond the forbidden MATH - blocks which define MATH . NAME rectangles are by definition closed, however since a horseshoe is a NAME set we can find an open hole MATH containing MATH such that the intersection of MATH with the horseshoe is exactly MATH. This yields MATH as an exclusion subshift. |
math/0101128 | Suppose we have a hole which is a union of intervals and yields MATH as an exclusion system. Consider the point MATH. The whole orbit of MATH, except MATH itself is in MATH. Thus MATH must lie in the hole. Let MATH. Clearly MATH and MATH. Thus the hole must consist of an infinite number of intervals: MATH can not be an IES. |
math/0101128 | Let MATH (here the decimal point marks the position between the MATH-st and MATH-th elements of the sequence). The point MATH is not in the even shift, it must fall into the hole under some iteration of MATH. We treat several cases, first of all suppose that MATH falls into the hole at the boundary of MATH. Then at the instant that MATH falls into the hole all the MATH's are to the right (or all are to the left) of the decimal point (that is, MATH with MATH or MATH). Note that the intersection of the rectangular holes with the boundary consists of a finite union of intervals. Thus we can apply the argumentation of the previous example to conclude that it is impossible to have an infinite number of the MATH fall into the hole when they are on the boundary of MATH. In other words all but finitely many MATH which fall into the hole away from the boundary of MATH. For such a MATH consider the code MATH at the instance of falling into the hole. It has the form MATH where MATH. Consider the sequence MATH with MATH and the sequence MATH with MATH. These two sequences get arbitrary near from the left and the bottom to MATH and all their elements are in the even shift. Thus MATH must be a corner of the hole. This contradicts to the assumption that the number of corners of the hole is finite. |
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