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math/0106238 | REF , and REF yield the identity: MATH . Because MATH is a cocycle representing the cohomology class MATH on MATH by REF and because of the relation REF between relative and absolute NAME classes and the definition of MATH given by REF , we see that MATH where MATH is the inclusion map REF. Combining REF for the intersection number, REF , and the definition of MATH in REF then completes the proof of REF . |
math/0106238 | We define a bundle map MATH by the obvious projection so that the following diagram commutes: MATH . By construction, the projection MATH is MATH-equivariant where the circle acts by scalar multiplication on the fibers of MATH and by the action in REF defining MATH on the fibers of the bundle on the left-hand-side of the diagram. Let MATH be the classifying map for the circle bundle defining MATH, covered by the bundle map MATH, so that the following diagram commutes: MATH . By construction of the map MATH, we have MATH. Then, because the maps MATH and MATH are MATH-equivariant, the product MATH defines the map MATH on the circle quotients: MATH . Observe that the intersection of the pre-image under MATH of the zero-section of the bundle REF with MATH is the base space REF: MATH . Hence, the map MATH induces a map on relative cohomology, MATH where MATH denotes the complement of the zero-section. Because the restriction of MATH to MATH defines a bundle map, MATH the NAME class of the normal bundle of MATH in MATH is given by MATH. The relative fundamental class of the manifold with boundary MATH is equal to MATH, as noted in REF. Hence, because the NAME class of the normal bundle of a submanifold is equal to the NAME dual of that submanifold CITE, the fundamental class of MATH is given by MATH where MATH is defined following REF and MATH is the inclusion of pairs. Observe that MATH where MATH is the inclusion REF. Thus, for any MATH, we have MATH which proves REF and thus REF . Because REF commutes, we observe that MATH; this observation and the fact that MATH proves REF . Because REF commutes, we have MATH and this proves REF , completing the proof of the lemma. |
math/0106238 | By the splitting principle we may suppose that MATH, with MATH, and so REF implies that MATH . The negative sign above arises because the circle action is diagonal, as explained in REF . The bundle MATH has total NAME class MATH, and thus MATH . This expression for the formal inverse of MATH yields MATH and REF follows. Because the pushforward MATH is the composition of division by the NAME class MATH and integration over MATH CITE, we have MATH which yields REF , since MATH has dimension MATH. |
math/0106238 | Applying the equality MATH from REF and using the abbreviations MATH, MATH, and MATH yields MATH where the last equality follows from REF . This proves the desired REF . |
math/0106238 | To prove the lemma, we must compare the second bundle in REF with the circle bundle MATH . The circle action in REF is trivial on MATH and given by the action described prior to REF on MATH; observe that the action of the group MATH in the bundle REF is defined in REF, for MATH, MATH, and MATH, by MATH . Let MATH be the unit sphere of the bundle REF, so MATH with the action REF of MATH. Consider the bundle MATH where the action of MATH is given for MATH and MATH by MATH . By REF , the first NAME class of the bundle REF is MATH. To prove REF, it suffices to show that the bundles REF are isomorphic. Because MATH acts anti-diagonally in the definition of MATH, the map MATH is a well-defined bundle isomorphism. The final statement of the lemma follows from the fact that MATH CITE. |
math/0106238 | In CITE, a rank-two, Hermitian vector bundle, MATH, where MATH is a finite-degree cover is constructed with MATH a branched cover of degree negative two. (Similar bundles are constructed in CITE.) The circle action on MATH induced by the action of MATH and the inclusion MATH maps to twice the circle action on MATH. This implies that the map on quotients, MATH, has degree negative one and that the class MATH pulls back to MATH where MATH is the first NAME class of the circle bundle MATH. Pulling back the cohomology classes in REF to MATH and applying the computation of the NAME classes of MATH in CITE then yield the formulas in REF. |
math/0106238 | According to REF , the intersection number on the left-hand side of REF is equal to MATH where MATH has the standard orientation defined in REF. Writing MATH we apply REF to expand the cup product in the pairing REF to give MATH where, for notational simplicity, we omit explicit mention of pullbacks. Denoting MATH using the fact that the classes MATH are pulled back from MATH, and multiplying out, we see that REF becomes MATH . If MATH denotes the rank of the complex vector bundle MATH, then (see REF ) MATH . The identity REF and the fact that MATH when MATH, because the class MATH is pulled back from MATH, imply that REF can be rewritten as MATH where it is understood that binomial coefficients MATH are by definition zero when MATH. By applying REF for division by the NAME dual of MATH, we see that REF yields MATH . We then make the substitution MATH from REF and expand the powers of MATH in REF as binomial sums. For MATH, the equality MATH implies that only one term from each of the binomial expansions of the powers of MATH in REF will not vanish. Therefore, REF yields MATH where we use REF to see that MATH and the constants MATH are defined by MATH . See CITE for identities involving binomial coefficients and extensions of the definition of MATH to the case MATH. Before applying REF to compute each of the pairings with MATH on the right-hand side of REF , we simplify the combinatorial factors appearing in each term in REF using the identities, for MATH, MATH where MATH is the NAME polynomial REF, with constants MATH . Because the constant on the left-hand side of REF is independent of MATH, we can factor it out of all the terms in REF except from the fourth term (the one containing MATH), where the binomial factors can be rewritten as MATH and thus, by replacing MATH by MATH in the definition of MATH and MATH in REF, the combinatorial expression from the fourth term of REF corresponding to REF will give the NAME polynomial MATH. Hence, by substituting REF takes the shape MATH . REF provide formulae for the pairings: MATH . Hence, REF of MATH imply that REF can be written as MATH . We write the power of MATH as MATH, since MATH by REF , with the factor MATH being absorbed into the coefficients MATH, MATH, MATH, whose resulting formulae agree by inspection with those following REF implies that the power of MATH in REF simplifies to MATH which agrees with the power of MATH in REF. Finally, we simplify the combinatorial coefficients MATH in REF. Because MATH we have, for MATH as defined in REF, MATH agreeing with the claimed formula for MATH in REF. Then, using MATH and MATH, we have, for MATH as defined in REF, MATH agreeing with the claimed formula for MATH in REF. This completes the proof of REF . |
math/0106238 | For MATH and MATH it is convenient to define MATH . We then recall the following identities (see CITE): MATH . The preceding identities yield MATH . Let MATH denote right-hand-side of REF and observe that, using REF , we can write MATH in the form MATH . With the substitution MATH, REF becomes MATH where we have set MATH. Applying the NAME convolution identity CITE to the sum over MATH in the right-hand side of REF, we see that MATH . We now express this sum in terms of the hypergeometric function CITE MATH . Using the first identity in REF yields MATH . Then applying the identities (see CITE) MATH to REF gives MATH . Finally, we note that MATH by the second identity in REF. This completes the proof of the lemma. |
math/0106238 | From CITE there is a splitting MATH if and only if MATH, recalling that MATH is a MATH structure with MATH, MATH, and MATH. Hence, MATH is contained in the level MATH if and only if MATH . Substituting REF into REF, we see that MATH . REF follows from the preceding formula relating levels and the relation REF between the dimensions of the NAME moduli spaces. REF follow from the blow-up formula CITE and the definition of NAME type. |
math/0106238 | From CITE we see that MATH and that MATH, so the left-hand side of REF is MATH times MATH where the links MATH have the standard orientation. Suppose MATH, for MATH and MATH for MATH. Applying the polarization identity CITE to REF to compute the pairings MATH and noting that MATH by REF , gives (using MATH for brevity) MATH where, using MATH and MATH, and MATH, MATH and where MATH are given by MATH . Note that the factor of MATH in the coefficient MATH and the coefficient MATH in REF are absorbed by the application of the polarization identity CITE. Using MATH and the identities MATH we simplify the terms on the right-hand side of REF (recall that MATH, MATH, and MATH for MATH) to give: MATH and MATH and MATH . Thus, if MATH is even, the terms REF, and REF will cancel when we substitute them into REF and subtract the results in REF, so in this case the difference REF and hence the pairing REF are zero, as claimed. If MATH is odd, the terms REF, and REF will add when we substitute them into REF and subtract the results in REF. Hence, REF follows from REF together with the preceding substitutions and substitution of REF for the coefficient MATH and REF for MATH. Note that the factor of MATH obtained by adding like terms and the coefficient MATH in REF yields the coefficient MATH in REF and that the factors MATH and MATH (mentioned before REF) yield the factor MATH appearing on the right-hand-side of REF. |
math/0106238 | We shall derive REF from the basic cobordism identity REF applied to the blow-up MATH, REF for the NAME invariant, the blow-up REF for level-one NAME link pairings, and the formula from CITE for level-zero NAME link pairings. Since MATH obeys REF , we can choose an integer MATH such that MATH and MATH . According to the last paragraph of CITE, there is a MATH structure MATH over MATH with MATH, MATH, and MATH. Let MATH be the related MATH structure over MATH defined prior to REF . From REF we see that MATH . Therefore, the hypothesis that MATH is equivalent to MATH and hence the moduli space MATH has real codimension greater than or equal to two in MATH. Moreover, MATH is good and so the cobordism REF applies. In general, by REF, the NAME stratum MATH corresponding to a splitting MATH lies in level MATH of the space of ideal MATH monopoles MATH, since MATH. By hypothesis, MATH and MATH for all MATH with MATH (by REF of MATH), so our choice of MATH implies that MATH if MATH and MATH. By hypothesis, MATH is `effective' in the sense of REF and so the only non-zero NAME contributions to MATH arise from moduli spaces MATH with MATH contained in levels MATH or MATH of MATH or, equivalently, from moduli spaces MATH contained in levels MATH or MATH of MATH, again when MATH by REF . Therefore, the cobordism identity REF applied to the moduli space MATH and REF of MATH give MATH . By the remarks in the preceding paragraphs the sum over REF has potentially non-zero terms when CASE: MATH or, equivalently, MATH, and CASE: MATH or, equivalently, MATH. Recall that the links MATH are empty by definition when MATH. Substituting CITE into REF to compute the terms with MATH and substituting the blow-up REF into REF to compute the terms with MATH yields MATH where the coefficients MATH are as given in the statements of REF (since they coincide with those of REF when MATH). From REF for MATH, we see that MATH and so the power of MATH in REF matches that in REF . Finally, via CITE, we see that MATH and so the power of MATH in REF also matches that in REF . Thus, substituting the preceding formulas for powers of MATH and MATH into REF yields REF . Lastly, we simplify the expressions given in REF for the constants MATH and MATH in MATH. From REF and the fact that MATH, we have MATH . Similarly, REF for MATH, the hypothesis that MATH, and REF for MATH yield MATH so REF for MATH then implies that MATH and thus REF gives MATH . This completes the proof of REF . |
math/0106238 | As MATH and MATH has NAME type by hypothesis, REF imply that MATH . Since MATH, this gives MATH . In particular, MATH and thus MATH for all MATH with MATH. We simplify the expression for the power of MATH in REF , noting that we now have MATH whenever MATH since MATH has NAME type. Substituting REF for MATH yields MATH matching the power of MATH in REF . The NAME polynomial constants MATH are equal to MATH when MATH, irrespective of the values of MATH, MATH. The assertions concerning the power of MATH and the coefficients MATH follow immediately from the fact that MATH and MATH. This completes the proof of REF . |
math/0106238 | For REF , the abundance condition implies that there are classes MATH with MATH and MATH. Let MATH. Then the classes MATH and MATH will do. For REF , the class MATH will satisfy the conclusion when MATH, so we can take MATH, while MATH will work when MATH, and we can take MATH. |
math/0106238 | The existence of MATH and MATH with MATH and MATH characteristic follows immediately from REF . Take MATH and set MATH where MATH is characteristic. We shall apply REF and the vanishing result in CITE. By hypothesis, MATH and so MATH (see first paragraph of proof of REF). Since MATH is even, it is convenient to write MATH, where MATH, so REF would then take the form MATH . The constraints MATH and MATH in REF are thus equivalent to (see CITE): MATH . Therefore, when using REF to compute MATH, we can choose MATH no larger than MATH, with MATH when MATH, or MATH when MATH; as we shall see from REF , we have MATH, while REF shows that the term in MATH of degree MATH in MATH is also zero. We first observe that both sides of the second identity in REF vanish in sufficiently low degree. REF implies that, for MATH characteristic, MATH if MATH or (see CITE) if MATH, since the series has parity (using MATH) MATH . By hypothesis of REF , MATH is characteristic and so we may write MATH for some MATH. Since MATH, then REF implies that MATH if MATH or if MATH. In particular, aside from the fact that MATH need not be characteristic, the vanishing result for the NAME series in REF restates REF. According to the hypothesis of REF , we have MATH so MATH and MATH. Because MATH is characteristic we have MATH and as MATH, then MATH. Therefore, REF show that MATH (Note that the alternative solutions to the MATH and MATH constraints yielding MATH, namely those with MATH, would yield MATH and so for that choice of MATH we could choose MATH no larger than MATH.) Hence, from REF , the potentially non-zero terms in the NAME series MATH take the form MATH . Because MATH, REF gives MATH while REF (with MATH) yields MATH . Replacing the terms MATH in REF by their binomial expansions and applying REF gives MATH . REF , and REF then imply that the terms of MATH of degree less than MATH in MATH are zero, yielding the vanishing result for the NAME series stated in REF. From REF the terms of MATH of degree less than MATH in MATH are also zero. Thus, to obtain the second identity in REF, it suffices to prove that MATH and, noting that MATH, MATH . Since MATH, MATH, MATH, and MATH by REF , the coefficients in REF for MATH (with MATH characteristic) become MATH . According to REF , when we expand the terms MATH in REF into a binomial sum of terms of the form MATH and sum over MATH, the sums of terms with MATH or MATH will vanish. Therefore, applying REF to the right-hand side of REF with MATH and MATH, only the terms of degree MATH in MATH will be non-zero. Hence, only the term with coefficient MATH will be non-zero and REF implies that it only contributes MATH . Since MATH and MATH by REF , the power of MATH the right-hand side of REF reduces to MATH. As MATH, and MATH, and MATH (as MATH), the power of MATH in the right-hand side of REF reduces to MATH. Therefore, by these remarks and REF we obtain REF for MATH. Next, applying REF to the right-hand side of REF with MATH and MATH, only the terms of degree MATH in MATH will be non-zero. In particular, REF implies that the term in the right-hand side of REF with coefficient MATH contributes only MATH . Similarly the term on the right in REF with coefficient MATH contributes only MATH . Finally, the term on the right in REF with coefficient MATH contributes only MATH . Hence, as MATH, MATH, and MATH by REF when MATH, noting that MATH and MATH, and combining REF , and REF, we see that REF yields MATH which proves REF for MATH. Thus, by the remark preceding REF , this proves the second identity in REF and completes the proof of REF . |
math/0106239 | Recall that MATH is the notation used for MATH when MATH is pseudo-proper, and that MATH. The isomorphisms of functors from MATH to MATH: MATH where the second is obvious and the other two are given by CITE, show that it suffices to establish the first isomorphism in REF . Let MATH, MATH and MATH be ideals of definition such that MATH and MATH. The ideals MATH, MATH and MATH are ideals of definition of the formal scheme MATH, possibly different. For each MATH let MATH be the closed immersion determined by MATH (so that MATH is an ordinary noetherian scheme with the same underlying topological space as MATH, but with structure sheaf MATH). The desired isomorphism results from the existence - to be shown - of a family of isomorphisms MATH compatible with the homotopy colimit triangles given by CITE, for MATH: MATH . For, a basic property of triangles is that such a family of isomorphisms extends (not necessarily uniquely!) to an isomorphism between the ``summits" MATH. Though the isomorphisms REF will be canonical, it does not follow that their extension to the summits is. The definition of REF is based on the fact, well-known, though not yet published in full generality, that on the category of separated finite-type maps of arbitrary noetherian schemes, there is a pseudofunctor MATH taking values in MATH, agreeing with MATH when the map MATH is proper and with MATH when MATH is an open immersion, compare CITE. The aforesaid lack of canonicity obstructs immediate extension of the pseudofunctor in question to the category of formal schemes, an extension whose existence would give a stronger canonical version of REF . (Nevertheless such an extension might always exist for reasons as yet unknown to us.) However the MATH-th homology MATH is canonically isomorphic to the direct limit of MATH; and so we will have produced canonical functorial homology isomorphisms MATH . The isomorphisms REF arise from applying MATH to the below isomorphisms REF, that we describe next. Consider the diagram MATH where MATH, MATH, MATH are the closed immersions given by the ideals MATH, MATH and MATH respectively, the maps MATH and MATH are induced by MATH and MATH respectively, the subdiagrams marked by MATH are fiber squares, and MATH and MATH are the closed immersions induced by MATH and MATH, respectively. The outer hexagon is then a diagram of ordinary noetherian schemes with MATH and MATH proper maps, MATH and MATH open immersions and MATH and MATH closed immersions. Use adic flat base change CITE and pseudofunctoriality CITE to obtain the natural composite isomorphism MATH and analogously, MATH . Using the above pseudofunctor on ordinary schemes we write MATH; and since MATH CITE, there results a natural isomorphism MATH . (This by itself is a canonical version of REF for ordinary schemes, for which MATH, MATH, MATH and MATH in REF are identity maps.) We have then the natural functorial isomorphisms MATH . Still to show is that the isomorphisms REF are compatible with the triangles REF. Let MATH be the natural closed immersion, and let MATH be the natural map. By the definitions involved, the desired compatibility amounts to commutativity of the following diagram in the category of functors from MATH to MATH, where MATH is induced by MATH, and MATH will be defined later: MATH . Let us deal first with the top subrectangle of REF. (The bottom one is essentially the same.) To lighten notation, we set MATH. Consider the following expansion of the left side of REF , where all occurrences of ``REF" in a subscript have been hidden, and where all the vertical arrows represent natural closed immersions, so that for each MATH, MATH. MATH . The squares in the middle are fiber squares, to which are associated base-change isomorphisms of the form MATH (with appropriate subscripts attached). With each MATH indicating the use of such a base-change isomorphism, and MATH the natural composition MATH one sees then that the top rectangle in REF expands naturally as MATH with MATH in REF defined to be the composition of the maps in the bottom row. It remains only to check commutativity of each of the subrectangles, which is a straightforward exercise requiring only the simplest formal properties of functoriality and pseudofunctoriality, except for the subrectangle marked MATH, where one uses the transitivity of flat base change CITE. As for the middle subrectangle in REF, after noting that MATH and MATH one can ``factor out" MATH and MATH, and then use the rather simple duality isomorphism for closed immersions, under which the natural map MATH corresponds to the identity map of MATH to reduce the commutativity question to that for a diagram of isomorphisms of functors from MATH to MATH: MATH . In this diagram only maps between ordinary schemes appear, so as before one can identify MATH with MATH and define the vertical arrows via pseudofunctoriality. The horizontal arrows involve flat base change. However, under the identification of MATH with MATH (when MATH is an open immersion of ordinary schemes), a base-change isomorphism like MATH becomes identified with the pseudofunctoriality isomorphism MATH. (This is not trivial, but is contained in the construction of the pseudofunctor MATH.) With this in mind one finds again that pseudofunctoriality yields the desired commutativity. This completes the proof of REF . |
math/0106239 | This is the particular case MATH-identity, MATH, of REF . |
math/0106239 | As in CITE we may assume that MATH is affine, say MATH, and that MATH can be covered by open subsets MATH such that MATH factors as MATH where MATH is a closed immersion, MATH is an open immersion and MATH is the completion of the projective space MATH along some closed subset. Now, for MATH, REF provides an isomorphism MATH, giving a reduction to the two REF MATH and REF MATH a closed immersion, cases dealt with at the end of the proof in CITE |
math/0106242 | NAME 's recognition principle REF is obtained through the following maps: MATH where all maps are MATH-maps between MATH-spaces. When MATH is a MATH-algebra, we want to define MATH-actions on the spaces involved which induce MATH-algebra structures and such that all maps are MATH-maps. The functors MATH, MATH and MATH restrict to functors in the category of MATH-spaces, where, for any MATH-space MATH, we define the action on MATH, MATH and MATH diagonally as in REF . Hence for any MATH-space MATH the MATH-action on MATH is given by MATH where MATH and MATH. This produces a MATH-algebra structure on MATH such that NAME 's map MATH is a MATH-map, and thus a MATH-map. We extend now these actions on the simplicial spaces MATH, MATH and MATH. Recall that the double bar construction MATH is defined simplicially, for a monad MATH, a left MATH-functor MATH and a MATH-algebra MATH by MATH, where MATH, with boundary and degeneracy maps using the left functor, monad and algebra structure maps. The group MATH acts then on MATH through its action on the functors MATH and MATH, which comes to ``rotate everything". For example, the action of MATH on a REF-simplex of MATH is given by MATH. With these actions, all maps above are MATH-maps between MATH-spaces and MATH is equipped with an explicit MATH-action. On the other hand, we have a weak homotopy equivalence CITE MATH for any MATH-connected space MATH. If MATH is a MATH-space, then this composite is a MATH-map with the actions on MATH and MATH defined as above. |
math/0106242 | Let MATH be the translation category of MATH, having MATH as set of objects and MATH. There is then an obvious ``projection" functor MATH which induces a covering map on the nerves. As MATH is contractible, it is in fact the universal cover of MATH. Also, the operad structure of MATH lifts to a natural braid operad structure on MATH, which is a MATH operad. We conclude by REF . |
math/0106242 | Let MATH be a ribbon braided monoidal category and let MATH be the strictification of MATH as a monoidal category. The category MATH then inherits a ribbon braided structure from the one existing on MATH. Its nerve MATH is a MATH-algebra. The space MATH is not necessarily a MATH-algebra, but it admits a MATH-action induced by the twist on MATH, and the equivalence MATH is MATH-equivariant. Now the space MATH is weakly homotopy equivalent to MATH and is a MATH-algebra. The equivalence is obtained through the following diagram of weak equivalences in MATH. MATH . The group completion of MATH is then equivalent to a double loop space MATH, where MATH and the MATH-action on MATH now induces one on MATH, as explained in REF . |
math/0106242 | An element of the free operad on MATH is described by a tree with vertices labelled by MATH CITE. We define the action of MATH on such element by acting on the labels of the vertices, using the comultiplication of MATH. This is well defined as MATH is cocommutative. It induces a MATH-module structure on MATH which induces one on MATH for all MATH by REF . The operad structure maps are then MATH-equivariant by construction. |
math/0106242 | The MATH-equivariance of the algebra map MATH is given by the commutativity of the following diagram: MATH where MATH and MATH give the action of MATH on MATH and MATH respectively and MATH is the interchange. This diagram translates, for the generators of MATH, into REF. The MATH-equivariance of the structure maps MATH for MATH is a consequence of the fact that MATH generates MATH, that the operadic composition is MATH-equivariant and that the structure maps MATH satisfy the associativity axiom. |
math/0106242 | Let MATH be an algebra over MATH. By REF , MATH is a MATH-module. As an algebra, MATH, where MATH is the fundamental class. This provides MATH with an operator MATH of degree REF satisfying MATH. REF tells us that MATH is an algebra over MATH. The operad MATH, called the NAME operad, was identified by CITE. This operad is quadratic, generated by the operations MATH and MATH, corresponding to the class of a point and the fundamental class under the MATH-equivariant homotopy equivalence MATH. The class MATH induces a graded commutative product on MATH, while MATH induces a NAME bracket of degree REF, that is, a NAME algebra structure on MATH, the suspension of MATH, where MATH. The bracket is defined on MATH by MATH. NAME proved that the product and the bracket satisfy the following NAME relation: MATH . In order to unravel REF , we must understand the effect in homology of the MATH-action on MATH. Clearly MATH because the rotation of the generator in degree REF gives precisely the fundamental REF-cycle. Moreover MATH for dimensional reasons. As MATH is primitive, REF applied respectively to MATH and MATH provides the following relations: MATH . As MATH, REF expresses the bracket in terms of the product and MATH : MATH . If we substitute this expression into the NAME relation REF we get exactly the NAME REF. We can re-wright REF as MATH which says that MATH is a derivation with respect to the bracket. To conclude we must show that REF and the NAME algebra axioms follow from the NAME. This is shown in REF. |
math/0106242 | CASE: MATH is the free exterior algebra generated by the fundamental class MATH, and MATH by dimension. By REF , MATH is a MATH-algebra if and only if MATH is a REF-algebra admitting a MATH-module structure, that is, an operator MATH of degree REF with MATH, such that the following relations hold: MATH where those equations are obtained by setting MATH and MATH in turn in REF . In this case the operator MATH is equal to the bracket and lies in degree REF and MATH. CASE: The evaluation fibration MATH splits as a product. So MATH, with both generators in degree REF. A class MATH comes from the basis, so MATH, and another class MATH comes from the fibre, so MATH. As in the previous case, we know that a MATH-algebra MATH is a REF-algebra with two operators MATH and MATH both in degree REF, satisfying MATH and MATH and relations obtained by settting MATH, MATH, MATH, and MATH in turn in REF . Using the identification MATH, this gives the following equations: MATH . Note that REF correspond precisely to the equations we had for MATH and the bracket in REF . So, by the same calculations, we know that MATH and the product form a NAME algebra of higher degree, that is, MATH and MATH satisfy REF but the operator MATH has now degree REF. There is an additional operator MATH of degree REF. REF says that MATH is a derivation with respect to the product. Using REF , one can rewrite REF in terms of MATH and the product. This shows that REF is redundant. |
math/0106242 | The homology NAME spectral sequence of the principal fibration MATH collapses at the MATH term if MATH is odd; if MATH is even then there is a non-trivial differential MATH CITE. |
math/0106242 | We give a proof of REF . It is known that the quadratic dual of the operad MATH is its own MATH-fold desuspension MATH, with product MATH dual to original bracket MATH and bracket MATH dual to the original product MATH CITE. If MATH, then MATH. So the class MATH gives an operator MATH of degree MATH with MATH and MATH, which thus induces a MATH structure as in REF but this time with product in degree MATH. |
math/0106242 | If we substitute the expression MATH by MATH, then the result follows from REF by the same proof as REF, with the minimal model replacing the deRham complex. Since MATH is the free commutative algebra on MATH, by the NAME theorem, the homology MATH-algebra structure is uniquely determined by MATH on spherical classes. We identify geometrically the MATH-operator on spherical classes. If MATH is a pointed MATH-space and MATH, consider the map MATH induced by the action. Since MATH let MATH be the restriction to the second summand. Clearly MATH is natural with respect to MATH-equivariant maps. The operation MATH represents MATH on spherical classes, that is, the following diagram commutes for any MATH. MATH . The only subtlety is to show that MATH can be replaced by MATH. Consider the universal example MATH. Clearly MATH, and MATH, because MATH and MATH. Moreover MATH because the composite MATH has degree REF, and MATH for the same reason. Recall now that spherical classes are dual to indecomposables in the minimal model. In the operadic homology the NAME MATH on indecomposables in MATH is induced by the NAME on MATH by definition. By naturality it is sufficient to show that MATH and MATH coincide in the universal case MATH. But MATH, because, for homological reasons, the map MATH induced by the MATH-action on MATH gives MATH. |
math/0106255 | By applying REF , MATH . |
math/0106255 | Since MATH, we have that MATH. It follows then that MATH . |
math/0106255 | Define the numbers MATH as the coefficients that arise in the coproduct MATH. From the defining property of the antipode we have the relation MATH as long as MATH is not the empty composition. The notation for the MATH analog of any operator is given in REF . To show that the MATH-analog of MATH satisfies the proposition we must give a definition in terms of NAME algebra operations and then demonstrate that the action reduces dramatically. For any operator, REF may be restated as MATH . At this point it is a direct computation within the NAME algebra of the non-commutative symmetric functions using relation REF and the definition of MATH to arrive at the formula stated in the proposition. Since the computation is detailed and not necessary for the remainder of this exposition, we leave it to the reader as an exercise. |
math/0106255 | Relation REF will follow from REF by setting MATH and MATH. |
math/0106255 | Due to the rule given in (REF , p. REF) we have the following important recurrence for the NAME symmetric functions indexed by a hook. MATH . Also consider the recurrence that we have developed for these non-commutative symmetric functions. MATH . We also have that MATH. Since we have that MATH and MATH, then we see MATH. This implies MATH . By induction on the length of the hook we see that MATH (which is obviously true when either MATH or MATH). |
math/0106255 | MATH. At the same time we have MATH. Both MATH and MATH are compositions of MATH, hence the proposition follows. |
math/0106255 | Let MATH. We will take the scalar product defined in REF of MATH and MATH. This will give the coefficient of MATH in the expression MATH. By expanding MATH in terms of MATH and using that MATH, we see MATH . Break the composition MATH into the composition consisting of the first MATH `cells' of MATH: MATH, and the last MATH `cells:' MATH, so that either MATH or MATH. If MATH or MATH then clearly both sums will be MATH. If MATH then MATH is non empty and contains at least one element MATH. For each MATH in the sums, either MATH contains MATH or it doesn't. The terms with MATH such that MATH are of opposite sign but the same MATH coefficient as those MATH such that MATH. Therefore the two sums will again be MATH. We need only consider the MATH where MATH and MATH. If MATH, then the second sum is clearly MATH and first sum contains only MATH term, MATH (which agrees with the statement of the lemma). If MATH, then both sums have exactly one non-zero term, and the scalar product is MATH . |
math/0106255 | Expanding MATH in terms of MATH and using REF , yields MATH . Now if MATH is in MATH then the inner product is MATH . This agrees with the formula given for MATH. If MATH is not in MATH, the result is MATH times this result. |
math/0106255 | Fix MATH and for MATH let MATH be a composition of MATH. The coefficient of MATH in the right hand side of REF is MATH raised to the power of MATH . This agrees with MATH where MATH is attach and concatenate of the MATH and hence agrees with the MATH coefficient on the left hand side of REF . |
math/0106255 | MATH . If MATH, then MATH is non empty. Take the smallest element MATH of this set (although any will do) and consider the involution MATH on the set of compositions such that the compositions that contain MATH in the descent set are sent to the compositions that do not contain the element MATH (in the most natural manner). For each MATH, the terms corresponding to MATH and MATH have the same weight but opposite sign, hence the sum is MATH. If MATH, then the sum reduces to MATH where the subsets MATH represent the sets MATH. This is clearly equal to the product stated in the proposition. |
math/0106255 | If we wish that MATH, then using REF , MATH where MATH, since MATH . A simple calculation yields the equation stated in the corollary. |
math/0106255 | The proof proceeds by induction, for there is a method of calculating the coefficients MATH from ones preceding it in some order. Say that MATH where we assume that MATH and that this family satisfies the three conditions given above. Assume that the coefficients MATH are known and given by the formula stated in the theorem for all MATH such that MATH or for MATH and MATH. To determine MATH we take the scalar product of MATH and MATH since MATH, MATH and all coefficients in MATH have been calculated already. Since MATH, hence we have the expression MATH . Those values of MATH may be calculated from what we have already determined since MATH . Hence we see that MATH . In addition we may calculate MATH since MATH . Although we may use these formulas to calculate the coefficients, the only conclusion that we are going to draw from them is that the coefficients MATH are determined by assuming that the defining conditions are true, hence the family MATH which satisfies these conditions is unique. It remains to show then that MATH satisfies the conditions listed above. Clearly they satisfy REF . It remains to show that the MATH have the correct expansion in terms of MATH. MATH . We also see for MATH is not strictly smaller than MATH then MATH is non-empty and MATH . Since there is some element MATH in MATH, every composition MATH either has MATH or MATH. There is an obvious involution between these two sets of compositions and they have the same weight but opposite sign, hence the sum is MATH in this case. |
math/0106255 | Idea: same as in MATH case. Show that MATH and by a formula (REF p. REF) we have the same recurrence in the commutative case. That is, MATH . By induction this implies that the commutative versions agree on hooks. Consider the product MATH. This is equal to MATH . We also have since MATH. MATH . While at the same time MATH . From here is easily shown that MATH . |
math/0106255 | We calculate the action of MATH on the column vector MATH. This follows by first expressing MATH in terms of MATH using REF , then using the action of MATH on MATH, then reexpressing the answer in terms of MATH using REF . We calculate that MATH . Therefore we see that the action of MATH on MATH is given by the equation MATH . Now translate this tensor product directly to the action on the MATH basis to arrive at the formula stated in REF . |
math/0106255 | With MATH the action of MATH on MATH is given from REF MATH . |
math/0106255 | MATH . |
math/0106256 | Let MATH be the stable normal bundle of MATH and MATH be the normal invariant, MATH. By the definition, MATH where MATH and MATH the NAME map. Let MATH be a bundle map over the homotopy equivalence MATH. Let MATH, where MATH is the NAME map of MATH. The NAME triple MATH together with the MATH-orientation MATH gives a quadratic form MATH, where MATH is a MATH-orientation of MATH. By REF we get that MATH for all MATH. To prove the desired result, it suffices to prove MATH. Note that MATH and MATH are stably equivalent as spherical fibration since MATH is a homotopy equivalence. Thus we can regard MATH and MATH as the the same and so get two orientations for MATH, MATH and MATH. Since MATH preserves the Spin structures/NAME orientations, MATH, where MATH/MATH is the principal fibration as above. This clearly implies that there exists a fibre automorphism MATH over the identity such that MATH . Notice that MATH gives a unique element MATH, where MATH is the space of self homotopy equivalences of MATH. By a formula in CITE, the MATH-dimensional component of MATH is MATH, where MATH is the transgression of MATH. By assumption, it must vanish since the NAME classes vanish. On the other hand, the MATH-dimensional component of MATH is determined by the Spin structures and so it vanishes since MATH preserves the Spin structures. By REF it follows that MATH the quadratic form associated with the NAME triple MATH and the MATH-orientation MATH. Note that in the definitions of MATH and MATH the only different ingradients are the normal invariants, after identifying MATH with MATH. By REF once again MATH. This implies the desired result. |
math/0106256 | Since MATH is a framed manifold, the stable normal bundle is trivial, that is, the classifying map of MATH factors through a point. Choose a MATH-orientation MATH with MATH the bundle map of MATH to the trivial MATH-bundle on a point, then MATH factors through the stable homotopy group MATH. By REF MATH if MATH and the order of elements in MATH is at most MATH if MATH. On the other hand, by REF the definition of MATH does not depend on the choice of the MATH-orientations since MATH is a framed manifold. This completes the proof. |
math/0106256 | It is easy to know that(since we are computing REF-localization) MATH . Assume MATH from now on. If MATH with MATH nontrivial and MATH a nontrivial cyclic group, then MATH and we have by a result in CITE that MATH . An easy calculation shows that MATH if MATH are nontrivial cyclic groups and thus MATH. On the other hand we know groups MATH and MATH by results in CITE,CITE. To complete the proof it remains to calculate MATH for MATH which will be given in the following results. |
math/0106256 | It suffices to prove that MATH is trivial if MATH and nontrivial if MATH. For MATH there is no MATH such that MATH since MATH is detected by the secondary cohomology operation MATH in CITE. Thus there is no MATH such that MATH by the naturality of secondary cohomology operation and the fact that MATH corresponding to mod MATH reduction induces a homomorphism sending MATH to the corresponding element. It follows that MATH . When MATH,there is a map MATH such that MATH since otherwise MATH. By the above fact on map MATH it is easy to see that there is a map MATH such that MATH follows from the lemma above that MATH. |
math/0106256 | Note that we have a commutative diagram MATH . In the above diagram,the two left horizontal maps are injective by REF , the left vertical map is obviously an isomorphism while the fact that the right vertical one is also an isomorphism follows by comparing the NAME exact sequences of MATH and MATH the other hand ,the fact that the right horizontal map on the bottom line is onto follows from the long exact sequence and the known results about MATH for MATH. |
math/0106256 | The relevant commutative diagram in this case is MATH . By the same argument as in the last Proposition , we know the map MATH is onto. If MATH,the two left horizontal maps are trivial by REF , thus MATH. If MATH, what we can get is an exact sequence MATH . To complete the proof, it suffices to prove the last map in the above sequence has a section. To do this we need another diagram MATH where MATH is the natural inclusion. The same argument as above combined with the proof of REF shows that the two MATH's are onto and MATH induces an isomorphism between kernels of two MATH's. Finally we get the following diagram which gives the desired section. MATH . |
math/0106256 | It suffices to prove that the following map MATH induces an isomorphism on MATH-th stable homotopy group. By the calculation in NAME 's book CITE, the first group is generated by the class corresponding to MATH and the second by MATH where MATH are the fundamental classes of the corresponding groups. Now what we want follows from the fact that MATH induces a homomorphism mapping MATH to MATH. |
math/0106256 | Denote the tower in the Proposition by MATH. Denote by MATH a similar tower in which MATH is replaced by MATH. By REF , it is easy to see that there is a map from MATH to MATH which induces an isomorphism on MATH when MATH. On the other hand it is not difficult to see that there is a map from the tower MATH to the tower MATH which induces an isomorphism on MATH. It remains to prove that the natural map MATH can be lifted to MATH and the lifting is compatible to that of the map MATH to MATH. We will give a proof only for MATH , the other cases are similar. Consider the fiber inclusion map MATH, we have the following NAME formula MATH where MATH. By REF and a familiar diagram chase argument as in the proof of REF in Chap. REF(see also CITE) , we have MATH and MATH. It follows easily that MATH. It is not difficult to see from this and a simple computation that MATH and a lifting can be chosen such that MATH lies in its kernel. To complete the proof, note that , as mentioned before , there is a commutative diagram up to homotopy MATH . The lifting from MATH of MATH and the lifting from MATH of MATH can be made compatible by a modification of the lifting from MATH of MATH. The same way the liftings to MATH can also be made compatible. Thus we have the following commutative diagram up to homotopy which completes the proof. MATH . |
math/0106256 | First note that it suffices to show this for the universal spectrum MATH since the map MATH factors through MATH. Notice that MATH for MATH. Thus in the following proof, we may assume that MATH satisfies the same for MATH large. Assuming MATH large, without loss of generality we can assume that MATH is a finite complex. Write MATH for the MATH-dual of MATH and MATH for the MATH-dual of the inclusion MATH. Note that MATH, where MATH is the cohomology fundamental class of the sphere. By the MATH-duality we get a commutative diagram MATH where MATH is the tower in REF and MATH is a lifting of MATH. From the diagram above and REF ,it suffices to show that the homomorphism MATH at the bottom line is injective. From now on we will restrict to the case MATH. The other cases are similar. Let MATH be the fibre of the composite MATH. Note that MATH can be viewed as a fibration over MATH with fibre MATH and MATH-invariant MATH; where MATH is the homomorphism induced by the inclusion MATH and MATH is the basic class of MATH. Consider the following commutative diagrams MATH and MATH where MATH is the homotopy fibre of MATH. MATH in the second diagram above is zero since MATH and thus by duality MATH. Thus the second diagram implies that MATH is a monomorphism. To complete the proof, it suffices to show MATH in the first diagram above. Let MATH, if MATH,then MATH. On the other hand, by duality MATH implies that MATH. Thus MATH . Since MATH is MATH-orientable, that is, MATH. By CITE that MATH. Thus MATH can be lifted to MATH and so MATH. This completes the proof. |
math/0106256 | For MATH, let MATH. For MATH large, MATH is the following composition of maps MATH where MATH for the basic class MATH. Identifying MATH with MATH . It is readily to see that MATH, here MATH is the composition MATH where MATH is the NAME constuction of MATH. Now the cofibration MATH is also a fibration at least in the stable range. It follows immediately that MATH is surjective. On the other hand, it is easy to know that the generator MATH satisfies MATH. Thus, for the inclusion map MATH, the composition MATH induces a nontrivial homomorphism on the MATH-th homology and thus MATH is an isomorphism. Moreover, the generator MATH satisfies that MATH. Thus the composition MATH is null homotopy if and only if MATH. By REF , the proof now follows by the commutative diagram MATH . |
math/0106256 | Let MATH denote the fiber multiplication. Since MATH, MATH is the composition MATH where MATH is the second difference of MATH and MATH, that is, the secondary obstruction to deform MATH to MATH. Consider the commutative diagram: MATH where MATH is a lifting of MATH, MATH, MATH, and MATH, MATH the inclusion of the fibre. Write MATH, here MATH and MATH are the factors of the wedge. Note that MATH. We are going to show MATH. As MATH factors through the map MATH, it suffices to prove that MATH is zero. Note the homomorphism MATH is an isomorphism as it induces an ismomorphism on the MATH-th homology groups. The composition MATH is null homotopy. Thus MATH. This completes the proof. |
math/0106256 | Identify the reduced framed bordism group MATH with the stable homotopy group MATH. Recall that MATH. By CITE it is easy to see that the homomorphism MATH is identified with the following geometrically defined homomorphism MATH . By REF follows since MATH is an isomorphism. To prove REF , note that there is an exact sequence by REF MATH . This completes the proof. |
math/0106256 | For each MATH, consider the reduced bordism class MATH. It is easy to see that MATH is a bordism invariant. One verifies the following map defines a homomorphism MATH . By REF the reduction homomorphism MATH is surjective if MATH is even. If MATH, let MATH be a generator. By REF it follows that MATH lies in the image of the inclusion map MATH. By the remark above this implies that the sphere bundle MATH has a section. Therefore MATH for all MATH. On the other hand, one can verify that MATH is a generator if MATH is nonzero. This proves REF . If MATH, by CITE there is an element MATH so that MATH is nonzero. This proves REF . If MATH is odd, by REF the homomorphism MATH is an isomorphism. Thus there is a MATH so that MATH for all MATH. In particular, if MATH can be lifted to the MATH-coefficient class with order MATH, MATH and so MATH. This completes the proof. |
math/0106256 | By REF the data of invariants are homotopy invariants of the manifolds. Thus the homotopy and homeomorphism classification of such manifolds are the same. There is an isomorphism MATH . Therefore from REF there is a reduced framed bordism class MATH corresponding to the given algebraic data MATH (respectively, MATH ) if MATH is even (respectively, odd). This together with REF implies this is an REF correspondence. Add some MATH to MATH if necessary so that MATH is surjective. By surgery on MATH we may assume further that MATH is an isomorphism and MATH. Therefore the data can be realized by a MATH-connected MATH-dimensional MATH-manifold, MATH, so that MATH and MATH respectively, MATH. Now it suffices to prove that the map MATH is injective. Suppose that MATH, MATH, are two framed smooth manifolds with the same data (for TOP manifold, the similar argument works identically). Note that the NAME invariants of MATH must vanish since MATH. By the assumption there are maps MATH, so that MATH and MATH are reduced framed bordant, where MATH induces an isomorphism on the MATH-th homology groups. Since both MATH framed cobordant to some homtotopy spheres, there is a framed homotopy sphere, MATH, so that MATH and MATH are framed bordant. By CITE or CITE it follows that MATH and MATH are diffeomorphic since MATH. Therefore MATH and MATH are almost diffeomorphic. The same argument as above applies to show that MATH and MATH are homeomorphic to each other. This completes the proof. |
math/0106257 | It follows immediately from REF that MATH and MATH have the same homotopy type and thus MATH is an NAME for each n. |
math/0106257 | If MATH is an NAME , then MATH is an NAME and NAME. Since MATH is an NAME by REF , it follows by REF that MATH which is in contradiction to the condition. |
math/0106257 | By REF , the rationalization MATH which is an NAME induces an isomorphism of groups MATH . It follows from REF that it suffices to prove that there is no essential NAME from MATH to MATH. On the other hand the map MATH which represents a generator of MATH induces an isomorphism of groups MATH . Again REF implies that it suffices to prove that there is no essential NAME from MATH to MATH which is well known to be equivalent to the injectivity of the following homomorphism MATH where MATH is defined as follows: Let MATH be the join . There is a well defined map MATH by k[x,t,y]= MATH . It is well known that MATH is homotopy equivalent to MATH. If MATH is an NAME with multiplication MATH , then MATH is the composite map MATH where the last map is the map MATH and MATH, MATH are the projection of MATH to the first and second factors respectively . Consider the following commutative diagram where the horizontal maps which are isomorphisms come from the universal coefficient Theorem MATH . It follows that it suffices to prove that the map MATH or equivalently the map MATH is injective . On the other hand, it is well known that the map MATH is dual to the reduced coproduct which is an isomorphism in this case and thus completes the proof. |
math/0106257 | That MATH and MATH have the same Mislin genus follows from the condition that MATH is a special phantom map. On the other hand REF apply here. Thus the Theorem above follows from the following Proposition. |
math/0106257 | Let MATH be any essential map . If MATH we will prove that this leads to a contradiction which concludes the proof. Since MATH is a phantom map , MATH is also a phantom map. Thus we have the following commutative diagram up to homotopy. MATH . If MATH , then MATH is the composite MATH where MATH is the homotopy fiber of the rationalization MATH. On the other hand we claim that Any map MATH factors through a map MATH where MATH. Assuming this , note that MATH admits a right inverse, we have that MATH is the composite MATH. It is easy to know that MATH is MATH-connected and thus MATH is MATH-connected. It follows that MATH is the composite MATH. By REF , MATH which contradicts REF . |
math/0106257 | NAME and NAME proved in CITE that , if MATH have the same Mislin genus, then MATH. So we have a map MATH which is a composite MATH where MATH is the projection. To prove MATH induces a homotopy equivalence it suffices to prove that MATH induces a homotopy equivalence MATH which follows directly from the fact that MATH is weakly contractible and the fact MATH which can be found in CITE , for a stronger result , see CITE. |
math/0106259 | The first part can be found in REF. The second part can be proved by an argument similar to that of REF. |
math/0106259 | First part of the theorem follows from REF as noted by CITE. The second and the fourth parts follows from long homotopy exact sequence. For the proof of third part, it follows from the following exact sequence that it suffices to prove the third part for each MATH . The above sequence is exact since an easy argument shows that there is a cofibration for each MATH and MATH is the direct limit of MATH. An induction argument using above cofibration reduces the proof of third part to that of a nonequivariant analogue of it which is of course true. |
math/0106259 | To begin the proof , consider the following commutative diagram following CITE MATH . The right hand square is the pull-back by the MATH-Arithmetic Square Theorem as in CITE and horizontal sequences are easy to see to be fibration sequences. Now REF implies that MATH is a MATH-phantom and thus the composite MATH is MATH-phantom . Then REF shows that MATH and thus MATH has a right MATH-homotopy inverse. It follows that MATH is an equivariant NAME. As in nonequivariant case MATH is rational. The proof completes if we proves that the equivariant MATH-invaraints of MATH are all trivial. To prove it , note first that the above result is also true for MATH where MATH is the equivariant MATH-th NAME section of MATH-space MATH CITE. Thus MATH has a right MATH-homotopy inverse. let's consider the following commutative diagram MATH where MATH , MATH and MATH. The fact that MATH is completion and the above commutative diagram imply that MATH is trivial. Since MATH is rational , it follows that MATH is trivial. On the other hand MATH has a right MATH-homotopy inverse . It follows that MATH is trivial which completes the proof. |
math/0106259 | The proof follows easily from REF as in CITE. |
math/0106260 | REF is essentially the same as that of REF. To prove REF ,it suffices to prove that MATH by REF where MATH. The given conditions and REF imply that MATH. The result follows from an induction by NAME tower of map MATH , see the proof of REF for details of a dual proof. |
math/0106260 | It is REF . |
math/0106260 | It is REF. |
math/0106260 | By REF , for each pair MATH in the Theorem, there exist mappings MATH so that MATH. It is easy to know that MATH . REF implies that MATH since MATH is an NAME. The result follows by taking MATH where MATH. |
math/0106260 | The proof is similar to that of REF using REF. Although we modified the notion of maximal map, we have also modified the integer MATH to ensure that it remains valid. |
math/0106260 | Take MATH-equivalences MATH as granted by REF so that MATH. We will try to define the other maps in the above diagram so that the Theorem is true. Without loss of generality, we can assume that there is only one MATH such that MATH if MATH. Choose rational NAME structures on MATH such that MATH are NAME. It is always possible to choose bases-MATH for MATH and MATH for MATH so that MATH is diagonal with respect to these bases and MATH and MATH are bases for MATH and MATH. Let MATH be defined by MATH and MATH where MATH is a basis for MATH and MATH is a basis for MATH. Then MATH is also of the diagonal form under the bases. Choose elements MATH and MATH such that they projects to bases of MATH and MATH respectively. Define the maps MATH similarly. Since MATH are MATH-equivalences and rational NAME, we can write MATH where MATH are prime relative to MATH and MATH are torsions. Certainly we can find MATH and MATH so that MATH . Let MATH be the map which sends each MATH to MATH and MATH be the map which sends each MATH to MATH and set MATH to be the maps which send MATH and MATH to MATH and MATH respectively. Obviously what we have to verify is that MATH. It is easy to know from diagram chasing argument that MATH. Now MATH which are torsion of order dividing MATH for all MATH. MATH for all MATH iff MATH for all MATH iff MATH since MATH is prime relative to MATH. Now it is clear that the front and back squares are homotopy pullback. |
math/0106260 | Without loss of generality, we can assume that there is an integer MATH such that MATH if MATH. By REF it follows that there exists a surjection MATH where MATH . In other words, there is a surjection MATH where MATH where MATH is the set of MATH matrices and MATH represents MATH respectively. Given MATH such that MATH (mod MATH) for MATH, then an elementary calculation shows that MATH, MATH, MATH and MATH. REF implies that we can assume MATH. It follows from REF that MATH. Thus MATH factors through the map MATH. The rest of the proof is the same as that of CITE. |
math/0106260 | Assume, without loss of generality, that the MATH matrix MATH has entries zero unless those on diagonal. Then MATH is equivalent to the equations MATH where MATH, MATH, are the nontrivial terms in matrix MATH. The proof of the existence of the matrices MATH with the prescribed properties will be given by mathematical induction and it is obvious for MATH. Assume it is also true for MATH, we will prove it for MATH. If MATH, using elementary matrix we can find MATH such that MATH and we have the following modulo MATH equation MATH . Similarly we can find MATH if MATH with similar properties. It follows that we can assume MATH. In this case MATH is equivalent to the equations MATH . The equations above imply that MATH and MATH are determined by one matrix. Thus the result follows from the corresponding result in the one matrix case. |
math/0106260 | As mentioned before, the maximal map MATH as in the Theorem exists. Let MATH. It suffices to prove that MATH by REF . The given condition and REF imply that MATH. Now consider the NAME decomposition of the map MATH. A typical term in the decomposition fits into the following MATH where MATH and map MATH is the mapping cone of a map MATH and MATH is the type MATH . NAME space : MATH . By the condition on MATH , there is an integer MATH such that MATH. Let MATH. We will prove that MATH where MATH is the map induced by MATH which will completes the proof since for MATH, MATH and MATH. We will proceed by induction. It is obvious for the case MATH . If the statement is true for MATH , we want to prove it for MATH. From the cofibration MATH the following exact sequence is exact: MATH . Now MATH . Since MATH is a co-H-map, it follows by REF that MATH belongs to image of the group MATH which has exponent dividing MATH. Therefore it follows that MATH which completes the induction. |
math/0106260 | The proof is exactly the dual of that of REF. |
math/0106260 | The proof is dual to that of REF. |
math/0106260 | An exactly dual proof can be given for REF , except that, instead of using REF , we appeal to REF below which is dual to REF. |
math/0106260 | Let MATH be the co-action given in REF with respect to maximal map MATH. Then MATH is the composite MATH where MATH and MATH. It is easy to verify that MATH is what we want. |
math/0106260 | The proof is similar to that of REF using REF. |
math/0106260 | Take MATH-equivalences MATH as granted by REF such that MATH. We will try to define the other maps in the above diagram so that the Theorem is true. As in the dual case , we can assume there is only one MATH such that MATH if MATH. Choose rational co-H-space structures on MATH such that MATH are co-H-maps. As in the dual case we can choose elements MATH, MATH , MATH and MATH such that they project to bases of MATH , MATH , MATH and MATH respectively. As in the dual case MATH is of the diagonal form with respect to these bases and MATH where MATH are prime relative to MATH and MATH are torsions. Certainly we can find MATH and MATH so that MATH . Let MATH be a base for MATH and MATH be a basis for MATH. Define MATH by MATH, MATH respectively and define MATH so that its matrix with respect to MATH, MATH is the same as that of MATH . Let MATH be the map which sends each MATH to MATH and MATH be the map which sends each MATH to MATH and set MATH to be the maps which send MATH and MATH to MATH and MATH , respectively. Obviously the only thing we have to verify is that MATH. It is easy to know from diagram chasing argument that MATH. Now MATH which are torsions of order dividing MATH for all MATH. MATH for all MATH iff MATH for all MATH iff MATH since MATH is prime relative to MATH. |
math/0106260 | Without loss of generality, we can assume that there is an integer MATH such that MATH if MATH. By REF it follows that there exists a surjection MATH where MATH . In other words, there is a surjection MATH where MATH and MATH is the set of MATH matrices and MATH represent MATH respectively. Given MATH such that MATH (mod MATH) for MATH, then an elementary calculation shows that MATH, MATH , MATH and MATH. REF implies that we can assume MATH. It follows from REF that MATH. Thus MATH factors through the map MATH. The rest of the proof is dual to that of CITE. |
math/0106260 | It follows from the equation MATH which can be verified directly. |
math/0106260 | By REF , we have MATH . On the other hand , MATH. Thus MATH is a co-H-map if MATH. Now MATH . The universal coefficient formula for homotopy group with coefficient and REF imply that MATH induces a monomorphism which gives the desired equation MATH. |
math/0106260 | Now the proof of REF follows from an induction by the homology decomposition of the co-H-map MATH together with REF and the fact that cofiber of a co-H-map is a co-H-space and the inclusion of the cofiber is a co-H-map. |
math/0106261 | The first statement follows from the fact that MATH for a reduced group since otherwise there will be nontrivial divisible subgroup in MATH. To prove the second statement , note that the induced map MATH is trivial since MATH is reduced group and MATH is rational thus divisible by the arithmetic square Theorem CITE. It follows that MATH is injective and an easy induction argument shows what we want for MATH. For MATH, it can be proved by an argument similar to that of REF. |
math/0106261 | That MATH follows from the fact that MATH for ratinal group MATH and MATH . Then the equation MATH implies the first statement. The equation about cohomology in second statement is true since MATH is rational while the proof of another equation is similar to that as in the first statement. |
math/0106261 | The first and the last statements follow from the last Proposition and the well known NAME Lemma. The second and third statements follow from a MATH argument for MATH where in the second statement, MATH and MATH is the MATH term in the NAME tower of the map MATH while in the third statement , MATH and MATH is the MATH term in the NAME tower of the map MATH. In both case the sequence MATH is a sequence consisting of isomorphisms and thus the MATH is trivial and the wanted isomorphisms follows immediately. |
math/0106261 | MATH is the MATH of a sequence of compact groups and continuous homomorphisms which is will known to be trivial. |
math/0106261 | This is an easy consequence of the Proposition above. |
math/0106261 | The equivalence between the last two statements follows directly from the following commutative diagram where the bottom horizontal homomorphism and the right side vertical homomorphism are isomorphisms by REF . MATH . Now assume the first statement, then we have MATH for all MATH. It follows from REF that MATH. The proof of another direction is similar to that in CITE using REF instead of NAME 's origional result which is stated only for space of finite type. |
math/0106261 | The proof is the same as that in CITE . |
math/0106261 | As first noted by NAME and NAME , similar proof as in that of Theorem B of CITE leads to the following homotopy fibration MATH where the union is over phantom maps MATH. On the other hand , different components of MATH are homotopy equivalent since MATH is the homotopy fiber of a map between two connected spaces. It follows that MATH . |
math/0106261 | The proof is the same as that of the corresponding result in CITE using results of CITE or REF. |
math/0106262 | Let MATH. By REF , to prove REF , it suffices to prove that , for map MATH such that MATH homotopic to MATH, it induces a homomorphism in cohomology given by MATH. Given such MATH, for any MATH, MATH where the first sum is taken over MATH from MATH to MATH and the second sum is taken over all MATH such that MATH. Thus we get a sequence of maps MATH where MATH and MATH. To prove that MATH , it suffices to prove that MATH where MATH and MATH while for the proof of later we need to study the behaviour of these maps with respect to the cup product of cohomology. If MATH , then MATH . On the other hand MATH. Using the formula for MATH and comparing the terms associated with MATH, we find the following equations MATH where MATH with MATH and the sum is taken over all partitions of MATH into MATH and MATH. Let MATH. Then then above formula implies that MATH is a derivation of degree MATH which is trivial by the condition of REF . The above formula in case MATH implies that MATH is a derivation of degree MATH modulo products of derivations of degree MATH which are trivial. It follows that MATH is a derivation of degree MATH and thus is trivial by the condition of REF . Inductively we can prove that all MATH are trivial which completes the proof of REF . |
math/0106264 | Clearly MATH if and only if MATH, from which it follows that MATH. Since the support of MATH is MATH, we obtain REF , from which REF follows on considering adjoints and replacing MATH by MATH. |
math/0106264 | Let MATH; from the lemma we know that MATH where MATH is a positive integer. Hence, up to positive constants, the products MATH are the characteristic functions of all the double cosets, so their linear span is MATH. From this it already follows that MATH is generated by isometries and that the universal C*-enveloping algebra for MATH exists and is a quotient of the universal C*-algebra of isometric representations of MATH. The key here is of course that isometries have norm MATH in any nontrivial NAME space representation, so the basic conditions for the existence of the universal C*-algebra (see for example, CITE) are satisfied. In order to conclude that the spanning set is a linear basis of MATH we have to prove that MATH whenever MATH, with MATH and MATH in MATH. For this it suffices to show that the expression MATH depends only on the left quotient MATH. Since MATH is directed, that is, MATH, it is enough to show that for each MATH the value of the above expression does not change when we replace MATH and MATH by MATH and MATH, respectively, for which it suffices to show that MATH because MATH is multiplicative on MATH. In order to prove this we first compute MATH by evaluating the convolution product at MATH using REF : MATH . The claim then follows on applying the basic fact MATH, valid for every group inclusion MATH, to the inclusions MATH and observing that MATH and that MATH . |
math/0106264 | The assumption MATH says that the action of MATH leaves MATH invariant. This is also equivalent to saying that MATH, because MATH implies MATH. Assume first that MATH is a NAME subgroup of MATH. Then it is clearly also a NAME subgroup of MATH, and for each MATH one has MATH. Assume now that MATH is a NAME subgroup of MATH and that MATH for every MATH. Let MATH and write MATH with MATH and MATH, (an easy argument shows this is always possible because MATH). In order to show that MATH is finite, we write: MATH which is a union of at most MATH right cosets. This proves that MATH is a NAME pair, and hence that MATH, as shown above. Finally, the operations defined on MATH restrict to those on MATH, so the identity map gives a unital embedding of *-algebras MATH. |
math/0106264 | The first assertion is from REF . A somewhat tedious but straightforward computation of convolution products in MATH shows that the MATH and the MATH satisfy the relations (MATH, MATH, MATH). Notice that relation REF simply says that MATH is an isometric representation of MATH (see REF) and relation REF is just a restatement of the embedding of MATH in MATH. In order to see that MATH is a linear basis of MATH, notice first that the support of MATH is the set MATH, and that since MATH, all double cosets arise as supports. Thus the products MATH linearly span MATH, and we only have to prove that they are linearly independent. Different supports are disjoint, so to prove linear independence it suffices to show that if two such products have the same support then they coincide, which we do in the following lemma. Let MATH and MATH, and assume that MATH. Then MATH. First take quotients modulo MATH in MATH to see that MATH. Since MATH is directed there exist MATH and MATH in MATH such that MATH and MATH. Since MATH, REF implies that the function MATH is invariant under multiplication by elements of MATH on the left, so MATH . Since MATH is multiplicative on MATH and MATH, we also have MATH. From the assumption MATH it follows at once that MATH at which point it becomes convenient to let MATH and MATH, and to write MATH and MATH in terms of MATH and MATH. Multiplication of REF on the left by MATH and on the right by MATH yields MATH . Using the multiplication rules in MATH we see that CASE: MATH is the support of MATH, CASE: MATH is the support of MATH, CASE: MATH is the support of MATH, and CASE: MATH is the support of MATH. It follows that MATH because the supports match and since MATH the coefficients also coincide. Now simply multiply by MATH on the left, by MATH on the right, and simplify MATH, and other similar products, to see that MATH. This finishes the proof of the lemma and thus of REF . It is possible to give a direct computational proof of REF along the lines of CITE, but it is more efficient to use the NAME algebra as in REF . To this effect notice that, because of relation REF with MATH, the projection MATH is contained in MATH for every MATH. Since MATH is a semigroup of isometries by relation (MATH), the left hand side of REF defines a semigroup of injective corner endomorphisms: the image of MATH is the corner MATH, on which MATH has an inverse defined by MATH. A standard reinterpretation of the relations (MATH, MATH, MATH) shows that they are in fact a presentation of the *-algebraic semigroup crossed product MATH. Since the MATH's and the MATH's in the NAME algebra satisfy the relations, there is a canonical homomorphism of MATH to MATH, which we must show is an isomorphism. In order to avoid confusion we temporarily denote the generators of the universal *-algebra with the given presentation by MATH and MATH while maintaining MATH and MATH for the generators of the NAME algebra. The linear span of products of the form MATH is multiplicatively closed, and so must be all of MATH, see for example REF. But the image of this spanning set is the linear basis MATH of the NAME algebra obtained in REF . Since a spanning set is mapped to a linear basis, the former is also a linear basis and the map is an isomorphism. This finishes the proof of REF and of the theorem. |
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