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math/0011115 | We consider the set MATH of reparametrizations of a fixed map MATH with projectivization MATH . We need only check the action of the group MATH on the space MATH parametrization of maps MATH . But, considering MATH as a (trivial) vector bundle over the affine variety MATH, this is a free action of the group MATH on the vector bundle MATH over a free action on the base space MATH and this group acts as a group of vector bundle isomorphisms. So, by descent theory for coherent sheaves and faithful flatness, the quotient is a vector bundle with fiber isomorphic to MATH. |
math/0011115 | NAME proved CITE that MATH is rational. But by REF MATH is birationally a vector bundle over MATH. Since vector bundles are locally trivial in theZariski topology, we conclude that the product of MATH with the vector-bundle fiber MATH is rational. That is MATH . So MATH is stably rational. Now suppose a component MATH of MATH dominates MATH via the map MATH induced from the projection map MATH . Again by REF , MATH is birationally a (locally trivial) vector bundle over MATH with fiber MATH. So MATH is always stably rational and is rational whenever MATH . |
math/0011116 | Consider the function MATH. Since this function is subharmonic at infinity with respect to the metric induced by MATH, it has no interior maximum point outside of MATH. Suppose that it has a maximum at a boundary point outside MATH, say, at MATH and that MATH . By the strong maximum principle, we have MATH unless MATH is constant, which is not possible if MATH. We note that MATH . On the other hand, we must have MATH at the maximum point MATH. This implies MATH where MATH. Since MATH is Legendrian in MATH with respect to the induced contact structure (in fact, the induced MATH-structure) by the assumption that MATH is transverse to MATH for sufficiently large MATH, MATH is tangent to the contact distribution, which implies MATH . This gives rise to contradiction to REF . |
math/0011116 | It is enough to consider Hamiltonian perturbations of given MATH that are fixed near MATH. The proof of this is standard which we omit. |
math/0011116 | We recall MATH . Since MATH is compact by hypothesis, MATH is compact. Therefore we can choose a non-negative function MATH so that CASE: MATH for MATH near REF or REF. CASE: MATH for MATH such that MATH for some small MATH. We just choose MATH as our new Hamiltonian. |
math/0011116 | Since we extend MATH so that MATH outside MATH which is obviously transverse, it is enough to check the transversality over MATH. In this region, we may consider the function MATH in place of MATH. Recalling MATH it is easy to check that the tangent space of MATH at MATH is spanned by the vectors MATH where MATH is the differential for MATH, MATH and MATH. Applying this vector to MATH, the non-transversal points are characterized by the equation MATH on the collar or on the symplectic cone attached to MATH. Since MATH, MATH on the cone. On the other hand, since the growth of MATH is linear and the growth of MATH is super-quadratic over the radial coordinate, the first equation of REF cannot hold at infinity in MATH. This finishes the proof. |
math/0011116 | Under the hypotheses given in the statement, it follows that MATH . Once we have this, the standard argument in the NAME theory proves the chain REF . To prove the independence of MATH on MATH and MATH, it will be enough to prove that the family of approximations MATH and the change of MATH's satisfying the condition above can be realized by compactly supported Hamiltonian deformations of one another among them. But this follows from the construction of the approximation MATH of MATH. We refer to CITE for the details of this limiting argument. |
math/0011116 | We will be sketchy in the proof because similar gluing arguments have been used many times in the literature by now. Note that MATH is again a Hamiltonian cobordism. We choose a Hamiltonian that is postively admissible to MATH. In fact, by adding a bump function supported in a neighborhood of the hypersurface MATH, we can make the corresponding Hamiltonian MATH so that MATH is ``above" MATH as in REF . We glue each given pair MATH and MATH with the obvious holomorphic strip in the middle. This gluing is possible, as long as MATH is sufficiently small and so MATH is sufficiently large and the strip is sufficiently narrow. This finishes the proof. |
math/0011116 | Recall MATH. We study the equation MATH . Since MATH, MATH and MATH all split, REF splits into MATH and MATH . Noting that REF has the unique solution with index REF (up to translations) and with the asymptotic condition MATH solutions MATH of REF with index REF consist of the pairs MATH such that MATH is a solution MATH of REF with index REF and MATH is the unique solution of REF satisfying REF . In particular, MATH must be constant. Hence we have proven that MATH which in turn implies that the chain map MATH becomes the identity map. This finishes the proof. |
math/0011116 | We compose MATH with MATH where MATH which generates the isotopy MATH. It is immediate to check that the composition MATH is Hamiltonian isotopic to the product cobordism between MATH and MATH via compactly supported Hamiltonian isotopy MATH. Therefore we can apply the standard procedure of using REF to prove the construction of chain isomorphisms between the cases of the identity cobordism and MATH. This proves the theorem. |
math/0011116 | We first prove REF . Let MATH. If MATH, then we should have MATH for some MATH. Since MATH, we must have MATH . Since MATH is transverse to MATH and MATH, REF uniquely determines MATH and so MATH. It remains to show that MATH. To show this, it is enough to prove that MATH because MATH. On the other hand, this follows from REF noting that MATH and that we have extended the NAME function MATH on MATH quadratically to its tubular neighborhood and so MATH. This finishes the proof of REF . Once we prove this, REF immediately follows from transversality of intersections of MATH and MATH. |
math/0011119 | The `if' part of the corollary is well-known and easy to see. It is enough to show the `only if' part. Since MATH bounds a meridian disk MATH of MATH, MATH bounds a REF-chain MATH, where MATH is a disjoint union of MATH copies of a meridian disk MATH of MATH. So we have MATH. Note that MATH is the trivial knot, and hence MATH. Suppose that there is an orientation-preserving homeomorphism MATH. Then MATH and MATH. Therefore, by REF , we have MATH or MATH (mod MATH). This implies that MATH (mod MATH) or MATH (mod MATH). |
math/0011119 | In REF , a tangle MATH and the full-twists part are commutative. Therefore by arguments similar to that in the proof of REF, we obtain MATH . Since MATH (mod MATH), we have the conclusion. |
math/0011119 | We may assume that both MATH and MATH are contained in MATH. We note that the MATH-fold cover MATH is obtained from MATH and MATH by identifying their boundaries in such a way that MATH, where MATH is a solid torus that is the MATH-fold cover of MATH, and MATH and MATH are the meridian and longitude of MATH. If two knots in MATH are homologous, then they are also homotopic. So they differ in a finite number of crossings. Thus it suffices to consider the case in which MATH is obtained from MATH by changing a single crossing MATH. Then MATH and MATH are MATH-periodic in MATH, MATH is obtained from MATH by changing the crossings MATH, and a covering translation MATH of MATH generates the MATH-action on MATH. Set MATH. Then we note that MATH, where MATH and MATH for MATH. This implies that there is a sequence MATH of MATH-periodic links in MATH such that MATH is obtained from MATH by changing the crossings MATH. By REF , we have MATH . This completes the proof. |
math/0011119 | Let MATH be a knot in MATH that represents MATH. We may assume that MATH is homologous to MATH for some MATH (MATH). Then MATH is the MATH-torus knot since MATH is obtained from MATH and MATH by identifying their boundaries such that MATH. It is well-known that MATH where MATH. By REF , MATH where MATH is a divisor of MATH and MATH is a prime integer. By elementary calculations, we obtain MATH or MATH (mod MATH). So we have MATH or MATH (mod MATH). Since MATH, we have the required result. |
math/0011120 | We give MATH degree two, which makes everything homogeneous. Write MATH. We know that MATH so MATH say. Each MATH has strictly negative cohomological degree (because MATH) so it lies in the ideal MATH. Let MATH be the sum of those monomials in MATH that lie in MATH, but do not lie in MATH. It is clear that MATH is divisible by MATH, say MATH. This gives series MATH with MATH, as required. It is easy to see that they are unique. Note that the degree of MATH is the same as the degree of MATH, which is MATH, so MATH has degree MATH. As MATH and MATH for MATH we see that MATH is divisible by MATH. We thus have MATH, say. One checks that MATH, so that MATH. Given that MATH, we see that MATH, and thus MATH. Similarly, the image of MATH modulo MATH is a series in MATH of degree MATH, so it has the form MATH for some MATH. Because MATH, we see that MATH and MATH. |
math/0011120 | For brevity, write MATH and MATH and MATH, so we have a cofibration MATH . We also write MATH . Note that MATH can also be thought of as the sphere bundle MATH in the MATH'th tensor power of MATH, or as the MATH-skeleton of MATH. Note also that MATH . An easy connectivity argument shows that there is a unique element MATH that hits MATH. Indeed, let MATH be the fibre of the map MATH, so that MATH starts in dimension MATH. The bottom cell of MATH is in dimension MATH, so the bottom cell of MATH is in dimension MATH. This is strictly greater than MATH because MATH and MATH. It follows easily that MATH, or in other words MATH, so there is a unique element MATH as described. We next consider the diagram Here MATH and MATH are the zero-section inclusions, and MATH is obtained in the obvious way from the MATH'th power map of total spaces MATH. There is also a corresponding diagram with MATH replaced by MATH. By applying MATH to this, we get a diagram Here MATH and MATH are the NAME classes of MATH and MATH; the corresponding NAME classes are MATH and MATH. It is well-known that MATH and MATH, and it follows easily (because MATH is not a zero-divisor in MATH) that MATH . By mapping the MATH diagram into the MATH diagram, we see that the same fomula holds for the map MATH . In this context, however, we simply have MATH, so MATH . We next apply the octahedral axiom to the maps MATH (which comes down to replacing the maps by cofibrations and using the fact that MATH). As MATH and MATH, we see that the cofibres of MATH and MATH are homeomorphic to MATH and MATH. With these identifications, the map MATH is the MATH'th power map MATH, and thus the induced map MATH is just MATH. We therefore have an octahedral diagram MATH (A circled arrow MATH means a map MATH. The diagram can be made to look more like an octahedron by lifting up the outer three vertices and drawing in an extra arrow to represent the composite MATH.) In particular, we see that the stable fibre of MATH is just MATH. We next claim that the map MATH is an isomorphism, and sends MATH to the NAME class MATH. To see this, let MATH be the restriction of MATH to the basepoint in MATH, so MATH is just a one-dimensional complex vector space. The bottom cell of MATH is just the circle MATH, and the bottom cell of MATH is the one-point compactification of MATH, written MATH. The restriction of MATH to this bottom cell is the standard homeomorphism of MATH with the unreduced suspension of MATH, followed by the projection to the reduced suspension. The claim follows easily from this. We now consider the following diagram. We see from the octahedron that the top line is a cofibration, and the bottom line is a cofibration by construction. The left hand square commutes because MATH. It follows that there exists a map MATH (that is, MATH) making the whole diagram commute. From our discussion of MATH in cohomology and the commutativity of the middle square, we see that the image of MATH in MATH is just MATH. By the uniqueness of MATH, we deduce that MATH. Thus, the commutativity of the right hand square tells us that MATH is the image of MATH in MATH. Under the usual identification of MATH with MATH, the element MATH becomes MATH, and the map MATH is a restriction of the usual map MATH. It follows that MATH as claimed. |
math/0011120 | We may assume that MATH. For any MATH, we can write MATH with MATH and MATH, and then write MATH. Let MATH be the ideal of topological nilpotents, so that MATH. For any MATH-module MATH that is complete with respect to MATH, we can define a map MATH by MATH. Suppose that MATH, and that MATH. Given the form of MATH mod MATH, we see easily by induction that there is a unique sequence of MATH's such that MATH and thus that MATH is an isomorphism. Moreover, if we have a pair of modules MATH such that MATH and MATH are isomorphisms, then so is MATH (by a five-lemma argument). It follows by induction that MATH is iso for all MATH, and thus by taking inverse limits that MATH is iso for all MATH. In particular, MATH is an isomorphism. This means that for any series MATH, there are unique series MATH such that MATH, which proves the proposition. |
math/0011120 | This is a graded version of the equivalence REF, taking into account the equation MATH from REF of that theorem, and the fact that regular local rings are NAME. |
math/0011120 | Suppose that MATH, say MATH with MATH. As MATH has nonzero chromatic depth, we see that MATH and MATH so MATH. This shows that MATH is not a zero-divisor in MATH. Now suppose that MATH, so that MATH is regular on MATH. Consider a series MATH such that MATH is divisible by MATH. We claim that MATH is divisible by MATH. Indeed, suppose that MATH. We then have MATH in MATH. However, MATH is not a zero divisor in MATH, so MATH is not a zero divisor in MATH by the previous paragraph. Thus MATH in MATH, or MATH in MATH say. This means that MATH in MATH, but MATH is not a zero-divisor so MATH as claimed. The conclusion is that MATH is regular on MATH. If MATH then we can replace MATH by MATH and MATH by MATH and MATH by MATH and run the same argument again. It follows by induction that MATH has chromatic depth at least MATH. |
math/0011120 | This follows easily from the proposition, using the obvious fact that MATH has the same chromatic depth as MATH for any algebra MATH over MATH. |
math/0011120 | Write MATH and MATH, so MATH and MATH is a circle bundle over MATH with NAME class MATH. Thus MATH is a circle bundle over MATH with NAME class MATH. This gives a long exact NAME sequence MATH . It follows easily by induction from REF that these sequences are short exact, and that MATH. The case MATH gives the corollary. |
math/0011120 | Suppose that MATH and that MATH hold up to stage MATH (which is trivial for MATH). Define MATH. We first claim that MATH in MATH. Indeed, we have MATH and this divides MATH over MATH. As MATH is a unit multiple of a monic polynomial, it is not a zero divisor in MATH, so we conclude that MATH divides MATH in MATH. We now set MATH to conclude that MATH in MATH as claimed. It is also clear that MATH divides MATH, so it divides MATH over MATH, so MATH is an algebra over MATH. This means that MATH is a quotient of MATH in a natural way. Next, we claim that MATH is a regular local ring in the graded sense. For this, we write MATH. This is clearly a regular local ring, and MATH. By REF it suffices to check that the list MATH is a regular system of parameters for MATH. The length of MATH is MATH which is the same as the NAME dimension of MATH, so it is enough to check that MATH generates the maximal ideal of MATH, or equivalently that MATH, or equivalently that MATH vanish in MATH, or equivalently that MATH is divisible by MATH over MATH. However, we know that MATH is an algebra over MATH so MATH over MATH. Moreover, as MATH vanish in MATH we see from the definition that MATH in MATH. This means that MATH over MATH, so MATH in MATH and MATH. On the other hand, we have MATH and MATH so MATH so MATH as required. This completes the proof that MATH is a regular local ring. It is clear from the above that MATH is a regular system of parameters for MATH, and thus that the NAME dimension of MATH is MATH. As MATH is an integral domain, it follows that any proper quotient of MATH has dimension strictly less than MATH. Thus, if we can show that MATH has dimension MATH, we can deduce that the quotient map MATH is an isomorphism. For the rest of the argument we need to separate the cases MATH and MATH. If MATH we argue as follows. It follows from the definition of MATH that MATH is the largest quotient of MATH over which MATH divides MATH. However, in this context MATH which is automatically divisible by MATH, so MATH. This has the same NAME dimension as MATH, so we must have MATH. In particular, we see that MATH in MATH, and thus that the formal group law MATH becomes MATH in MATH. It follows that REF hold with MATH and MATH. We now consider the case MATH, so that MATH. For this, we use the theory of multiple level structures developed in CITE; we will assume that the reader is familiar with this. Let MATH be the graded ring MATH. Write MATH. It is easy to see that MATH, or MATH if MATH. Moreover, we have MATH. Let MATH be the usual formal group law over MATH, and define MATH, so that MATH is a formal group law over MATH. Let MATH be the associated formal group over MATH, which has height MATH and strict height MATH. This puts us in the context studied in CITE. Now write MATH, and MATH. It is clear from the definitions that MATH is just the ring MATH which classifies MATH-fold level-MATH structures on MATH, and MATH. We know from CITE that MATH is an integral domain and a finitely generated free module over MATH. It is nonzero because MATH. It follows that MATH is a graded domain and a finitely generated free module on homogeneous generators over MATH. We now have a diagram as follows. For MATH we let MATH be the ideal in MATH generated by MATH. This is clearly prime, with MATH and MATH. As MATH is integral over MATH, the going-up theorem CITE gives us a chain MATH of primes in MATH such that MATH. Let MATH be the preimage of MATH under the map MATH. It is clear that the preimage of this under the map MATH is just MATH, so we have a chain of strict inclusions of prime ideals MATH. Let MATH be the maximal ideal in MATH. As MATH is a unit in MATH it is clear that MATH, but MATH so we have a chain of strict inclusions MATH. This shows that MATH has NAME dimension MATH, so MATH as explained earlier. We now define MATH, and for MATH we define MATH . We also write MATH . Clearly, the map MATH defines a MATH-fold level-MATH structure on MATH over MATH. If we compose this with the projection MATH we get an ordinary (MATH-fold) level-MATH structure. It follows from CITE that MATH is a coordinate on a quotient formal group of MATH, and that the kernel of the quotient map is contained in the kernel of MATH. This means that there is a unique formal group law MATH defined over MATH such that MATH is a homomorphism MATH, and that there is a unique power series MATH defined over MATH such that MATH. We define MATH and MATH. It is easy to deduce that MATH is the unique formal group law over MATH such that MATH is a homomorphism MATH, and that MATH is the unique series such that MATH. We next remark that MATH is a NAME domain, so it is not hard to see that the maps MATH are injective. Now write MATH and MATH. We know from REF that MATH is a free module over the subring MATH on generators MATH for MATH. We can thus write MATH for uniquely determined series MATH. Similarly, MATH is a free module over MATH on MATH, so the equation MATH is the unique way to write MATH in terms of the generators MATH. On the other hand, we also have MATH. It follows that MATH for MATH and that MATH, so the series MATH is actually defined over MATH rather than MATH. It is clearly the unique formal group law over MATH for which the series MATH is a homomorphism MATH. Similarly, we see that MATH is defined over MATH, and it is the unique series over MATH for which MATH. By putting MATH we see that MATH is divisible by MATH, so it can be written as MATH. This completes our induction step. |
math/0011120 | The claim is that the sequence MATH (of length MATH) is regular on MATH. By REF , it is enough to check that the quotient MATH has dimension at most MATH. Let MATH be a graded integral domain which is a quotient of MATH. It is enough to show that every such MATH has dimension at most MATH, and thus enough to show that the maximal ideal MATH of MATH needs at most MATH generators. From now on we work in MATH. Write MATH. This is a subgroup of MATH, of dimension MATH say. Note that MATH acts on MATH in a natural way. After applying a suitable element of this group, we may assume that MATH is the evident copy of MATH in MATH, spanned by the first MATH standard basis vectors, so that MATH in MATH. We write MATH for the space spanned by the remaining standard basis vectors, so that MATH. Define MATH, so that MATH. As MATH is an integral domain, we see that MATH and thus that MATH, where MATH means the largest integer MATH such that MATH divides MATH. As MATH in MATH we see that MATH. On the other hand, we have MATH. Thus MATH. Now consider the list MATH so that MATH has length MATH. It will be enough to show that MATH, or equivalently that MATH in MATH and MATH in MATH. We already have MATH in MATH and the remaining MATH's are in MATH so all MATH's vanish in MATH, as required. We next note that MATH becomes MATH over MATH, and MATH, and MATH so MATH has height at least MATH over MATH. This means that MATH over MATH and the remaining MATH's are in MATH so they vanish in MATH also. Thus MATH, and MATH needs only MATH generators, as required. |
math/0011120 | The proposition is trivial when MATH (so MATH). We may thus assume the statement for MATH and prove the one for MATH. As MATH, we see that the annihilator of MATH is the preimage of the annihilator of MATH in MATH. We write MATH, so that MATH. It follows from REF that the annihilator of MATH in MATH is just the ideal generated by MATH. Thus, we have MATH as required. |
math/0011120 | Take MATH in REF . We know from REF that MATH has chromatic depth at least MATH, and it follows from REF that MATH has chromatic depth at least MATH, so MATH is regular on MATH. It follows easily that MATH is regular on MATH, and thus that the MATH-torsion is contained in the ideal MATH. We know from REF that the annihilator of MATH is the same as the kernel of the map MATH. We also know from REF that MATH. It follows immediately that the MATH-torsion is precisely the ideal generated by MATH and that it is a free module over MATH. It is easy to see from the definitions that MATH. Moreover, modulo MATH we see that MATH is just the product of all linear polynomials of the form MATH for some MATH. In particular, it is nonzero. As MATH is an integral domain, it follows easily that the map MATH is injective. |
math/0011120 | Using column operations and the fact that MATH, we see easily that MATH for all MATH. It is also clear that the element MATH transforms in the same way, and thus that MATH for all MATH and MATH. It is immediate from the definition that MATH lies in MATH and that it is divisible by MATH. It is well-known that MATH is a derivation with MATH and MATH, and from this it follows that MATH is a sum of terms of the form MATH for various permutations MATH. In particular, we see that MATH lies in MATH and that it is divisible by MATH. As MATH and MATH are divisible by MATH and invariant under MATH, we see that they are both divisible by each of the terms MATH in the product formula for MATH. As these terms are inequivalent irreducibles and MATH has unique factorisation, we see that MATH and MATH are divisible by MATH. It is easy to check that MATH and MATH have degree zero, so they lie in MATH. By comparing coefficients for the monomial MATH, we see that MATH. Finally, we can define an (ungraded) ring map MATH by MATH and MATH. We then have MATH, from which it follows easily that MATH, which completes the proof of the proposition. |
math/0011120 | Given REF , all that is left is to prove that MATH. We first show that MATH for MATH. Recall that MATH is the determinant of the matrix with entries MATH for MATH and MATH. As we work modulo MATH we have MATH, so MATH for all MATH. This shows that when we multiply the MATH'th row of our matrix by MATH, it becomes a linear combination of the other rows, so the determinant becomes zero. It follows that MATH annihilates the determinant of the original matrix, in other words MATH. Next, choose elements MATH lifting our elements MATH. We then have natural maps MATH for all MATH. Working in MATH we see that MATH and thus MATH. After applying the forgetful functor to the category of spectra, we conclude that MATH. In view of REF , it is now enough to check that MATH, MATH and MATH have the same image modulo MATH, or equivalently the same image in MATH. One can check from the definitions that these are the same as the elements MATH, MATH and MATH of REF , so they are all the same, as required. |
math/0011120 | We actually prove that the ring MATH admits such a filtration whenever MATH. The case when MATH and MATH gives the proposition. When MATH we have MATH so the claim follows from REF , so we work by induction on MATH. We have MATH, so we may assume that MATH. Note that MATH, so we can apply REF . This gives us a two stage filtration of MATH in which one quotient is MATH and the other is a finitely generated free module over MATH. The proposition follows by induction. |
math/0011122 | REF is proved as REF . In REF , the fact that products exist is CITE; we also give a proof in REF , which is slightly closer in spirit with our other proofs. REF form REF is explained in more detail and proved in REF. |
math/0011122 | Using the cofibration MATH and the fact that MATH, we find that MATH is injective. It is clear that MATH gives zero on the right hand side, so it is zero on the left hand side as claimed. |
math/0011122 | There is a cofibration MATH. The lemma tells us that the first map is zero, so MATH is a split monomorphism, and any splitting is clearly a product. |
math/0011122 | By hypothesis, MATH is the identity on the bottom cell of MATH. We observed earlier that MATH, and it follows that MATH. Similarly MATH. |
math/0011122 | Consider the following diagram: MATH . We now apply the functor MATH and make repeated use of the cofibration MATH . The conclusion is that all maps involving MATH become monomorphisms, all maps involving MATH become epimorphisms, and the bottom row and the middle column become short exact. The first claim follows by diagram chasing. For the second claim, consider the diagram MATH . We apply the same logic as before, using the first claim (with MATH replaced by MATH) to see that the middle column becomes left exact. |
math/0011122 | Let MATH be the set of products. As MATH, it is clear that the above construction gives an action of MATH on MATH. Now suppose that MATH. We need to show that there is a unique MATH such that MATH. Using the unital properties of MATH and MATH given by REF , we see that MATH . Because of REF , we can apply REF to see that MATH for a unique element MATH, as claimed. |
math/0011122 | Let MATH be a product, and write MATH so the claim is that MATH is nullhomotopic. Using the unital properties of MATH we see that MATH . Using REF , we conclude that MATH for a unique element MATH (because MATH is odd). Thus MATH as claimed. |
math/0011122 | Let MATH be the twist map. Clearly, if MATH is a product then so is MATH. Thus, there is a unique element MATH such that MATH . We define MATH. Next, recall that the twist map on MATH is homotopic to MATH, because MATH is odd. It follows by naturality that MATH. Consider a second product MATH. We now see that MATH . Thus MATH as claimed. |
math/0011122 | We choose a product MATH on MATH and define MATH. This is well-defined, by the lemma. If MATH then MATH for all MATH, so there is no commutative product. If MATH then MATH, say, so that MATH is a commutative product. In this case, the commutative products are precisely the products of the form MATH where MATH, so they have a free transitive action of MATH. |
math/0011122 | Write MATH, so the claim is that MATH. It is easy to see that MATH, so by REF we see that MATH factors through a unique map MATH. This is an element of MATH, which is zero because MATH is odd. |
math/0011122 | The statement about the existence and uniqueness of MATH follows immediately from the cofibration MATH, and the fact that MATH. Suppose that MATH exists; it follows easily using the product structure on MATH that MATH is zero. Now let MATH be the given product on MATH, and let MATH be a product on MATH. Consider the map MATH . By the usual argument, we have MATH for a unique map MATH. We define MATH. It is obvious that this vanishes if and only if MATH is a ring map, and that MATH. Now suppose that MATH is commutative, so MATH. On the one hand, using the fact that MATH we see that MATH. On the other hand, from the definition of MATH and the fact that MATH, we see that MATH . Because MATH is a split monomorphism, we conclude that MATH in MATH. |
math/0011122 | This will follow immediately from the NAME sequence if we can show that MATH. For this, it suffices to show that the map MATH is surjective for all MATH. This follows from the cofibration MATH and the fact that MATH acts trivially on MATH. |
math/0011122 | Because MATH admits a product, we know that MATH acts trivially on MATH. Because MATH has the form MATH, we see that MATH acts trivially on MATH. Thus MATH acts trivially on MATH, and REF assures us that MATH. Let MATH be the product on MATH. By the above, there is a unique map MATH which is compatible with the maps MATH. It is easy to check that this is an associative and unital product, and that it is the only one for which the MATH are commuting ring maps. It is also easy to check that MATH is commutative if and only if each of the maps MATH commutes with itself, if and only if each MATH commutes with itself. Now let MATH be any ring in MATH. We may assume that each MATH maps to zero in MATH, for otherwise the claimed bijection is between empty sets. As MATH is a ring, this means that each MATH acts trivially on MATH, so that MATH. We see from REF that ring maps from MATH to MATH biject with systems of ring maps MATH for MATH such that MATH commutes with MATH for MATH. The claimed description of ring maps MATH follows easily. |
math/0011122 | We can use REF to reduce to the case where MATH where MATH is invertible in MATH and MATH is generated by a regular sequence MATH. We know from REF that there is a unique commutative product MATH on MATH. If MATH and MATH in MATH then in the notation of REF we have MATH and thus MATH, so the unique unital map MATH is a ring map. It follows that MATH is a strong realisation of MATH, and thus that MATH is a strong realisation of MATH. Using REF , we get a ring MATH which is a strong realisation of MATH. |
math/0011122 | After using REF , we may assume that MATH. Choose a regular sequence MATH generating MATH. As MATH, we can choose a product MATH on MATH such that MATH. We let MATH be the ``infinite smash product" of the MATH, as in REF , so that MATH. Because MATH maps to zero in MATH, we see easily that the map MATH commutes with itself. By REF , we conclude that MATH is commutative. Let MATH be an even commutative ring, and that MATH has no MATH-torsion. The claim is that MATH. The right hand side has at most one element, and if it is empty, then the left hand side is also. Thus, we may assume that there is a map MATH of MATH-algebras, and we need to show that there is a unique ring map MATH. By REF , we know that ring maps MATH biject with systems of ring maps MATH (which automatically commute as MATH is commutative). There is a unique unital map MATH, and REF tells us that the obstruction to MATH being a homomorphism satisfies MATH. Because MATH has no MATH-torsion, we have MATH, so there is a unique ring map MATH, and thus a unique ring map MATH as required. |
math/0011122 | First, observe that if MATH and MATH are MATH-modules, there is a natural map MATH which is an isomorphism if MATH is a wedge of suspensions of MATH (in other words, a free MATH-module). Choose a homogeneous basis MATH for MATH over MATH, where MATH has degree MATH. Define MATH, so that MATH is a free MATH-module with a given isomorphism MATH of MATH-modules. Define MATH and MATH and MATH . The product map MATH gives rise to evident maps MATH which in turn give isomorphisms MATH of MATH-modules. The multiplication map MATH corresponds under the isomorphism MATH to a map MATH. After composing this with MATH, we get a product map MATH. A similar procedure gives a unit map MATH. We next prove that this product is associative. Each of the two associated products MATH factors as MATH followed by a map MATH, corresponding to a map MATH. The two maps MATH in question are just the two possible associated products, which are the same because MATH is associative. It follows that MATH is associative. Similar arguments show that MATH is commutative and unital. Now consider an object MATH equipped with a map MATH (and thus a map MATH). As MATH is a strong realisation of MATH, there is a unique map MATH compatible with the map MATH. This makes MATH into a MATH-module, and thus gives an isomorphism MATH. There is thus a unique MATH-module map MATH inducing the given map MATH. It follows easily that MATH is a strong realisation of MATH. |
math/0011122 | We may as usual assume that MATH, and write MATH. Let MATH be the infinite smash product of the MATH's, so that MATH. It will be enough to show that there is a unital isomorphism MATH. Moreover, any unital map MATH is automatically an isomorphism, just by looking at the homotopy groups. There is a unique unital map MATH. Write MATH, and let MATH be the map MATH where the second maps is the product. Because MATH is a ring and each MATH goes to zero in MATH, we can apply REF to get a unital map MATH as required. |
math/0011122 | It is not hard to see that MATH, with a sign coming from an implicit permutation of suspension coordinates. We also have MATH and thus MATH. Given any finite subset MATH of the positive integers, we define MATH where MATH. The claim is that one can make sense of homogeneous infinite sums of the form MATH with MATH, and that any graded map MATH of MATH-modules is uniquely of that form. Write MATH, and let MATH be the evident map. It is easy to check that MATH if MATH, and a simple induction shows that MATH is a free module over MATH generated by the maps MATH for which MATH. Moreover, REF implies that MATH. The claim follows easily. |
math/0011122 | Choose a regular sequence MATH generating MATH. Write MATH, and let MATH be the map MATH . It is easy to see that MATH is the homotopy colimit of the objects MATH (although there may not be a ring structure on MATH for which MATH is a homomorphism). We also write MATH for the smash product of the MATH for which MATH and MATH, and MATH for the evident map MATH. Consider a derivation MATH, and write MATH. Because MATH is a derivation, we see that MATH is a sum of MATH terms, of which the MATH'th is MATH times the composite MATH . Now consider an element MATH of MATH. It is easy to see that there is a unique unital map MATH, and that MATH. Now consider the following diagram. MATH . The left hand square commutes because the terms MATH for MATH become zero in MATH. It follows that there exists a map MATH making the whole diagram commute. However, MATH is the unique map making the middle square commute, so the whole diagram commutes as drawn. Thus MATH (thinking of MATH as an element of MATH). As MATH, we conclude that MATH. Thus MATH. This shows that MATH is actually a homomorphism MATH. It is easy to check that the whole construction gives a homomorphism MATH. If MATH then all the elements MATH are zero, so MATH. As MATH is the homotopy colimit of the objects MATH, we conclude from REF that MATH. Thus, MATH is a monomorphism. |
math/0011122 | It is clear that MATH is generated by the elements MATH. Suppose that we have a relation MATH in MATH (not MATH). We claim that MATH for all MATH. Indeed, it is clear that MATH so by regularity we have MATH say; in particular, MATH. Moreover, MATH, so by induction we have MATH for MATH, and thus MATH as required. Now suppose that we have a relation MATH, say MATH. We then have MATH, so by the previous claim we have MATH, so MATH. This shows that the elements MATH generate MATH freely. |
math/0011122 | It is easy to see that MATH is a derivation and that MATH (NAME 's delta). This shows that MATH is surjective, and the rest follows. |
math/0011122 | Let MATH be the universal FGL over MATH and put MATH. Let MATH be the unit map, so that MATH. Using the universality of MATH, we see that MATH is an IPO if and only if we have MATH . The left hand side is of course MATH. There is an evident map MATH sending MATH to MATH and MATH to MATH, and one can check that this is injective. Thus, MATH is an IPO if and only if MATH . The right hand side here is MATH and MATH so the proposition will follow once we prove that MATH. To do this, we use the usual isomorphism MATH, so that MATH, MATH and MATH are NAME classes, so MATH and MATH and MATH. As MATH is a natural multiplicative operation we also have MATH, which gives the desired equation. |
math/0011122 | The FGL over MATH coming from MATH is just the multiplicative FGL MATH, so MATH. If we put MATH then MATH. We thus need only verify that MATH. This is a straightforward calculation; some steps are as follows. MATH . |
math/0011122 | Working rationally and modulo MATH, we have MATH so MATH . Note that MATH is integral and its constant term is MATH, so the above equation is between integral terms and we can sensibly reduce it modulo MATH. We next recall the formula for MATH given in CITE. We consider sequences MATH with MATH and MATH for each MATH. We write MATH and MATH. We also write MATH where MATH . The formula is MATH . The only terms which contribute to MATH modulo MATH are those for which MATH, so MATH for all MATH. If MATH has this form and MATH then MATH. Thus MATH . As remarked in REF , we have MATH, so MATH as claimed. |
math/0011122 | The first statement is clear, just because MATH. For the second statement, write MATH and MATH and MATH. Let MATH be the coefficient of MATH in MATH. Because MATH we have MATH, and MATH . This expression is to be interpreted in MATH, so we need to interpret MATH in MATH. Thus REF tells us that MATH and MATH and all other MATH's are zero. Thus MATH as claimed. For the last statement, REF gives MATH . By the previous paragraph, this can be written as MATH. |
math/0011122 | To show that MATH exists, it is enough to show that the formal group law on MATH obtained from the map MATH is MATH-typical. Let MATH be an odd prime, so the associated cyclotomic polynomial is MATH. We need to show that MATH . (This is just the definition of MATH-typicality for formal groups over rings which may have torsion.) Consider the ring MATH. As MATH we have MATH, and by looking at the coefficient of MATH we find that MATH. Now write MATH and MATH, so that MATH. We find that MATH . This gives us a ring map MATH; we claim that this is injective. Indeed, it is easy to see that MATH is a basis for MATH, and that MATH is a permutation of this basis. Suppose that we have MATH . Using the evident map MATH, we see that MATH for all MATH. As MATH, we see that MATH . As the elements MATH are a permutation of the elements MATH, we see that MATH for all MATH. We may thus regard MATH as a subring of MATH. Next, we know that MATH because MATH is MATH-typical over MATH. By REF , we also know that MATH where MATH. It is easy to see that MATH, so that MATH. We also have MATH for all MATH. This means that MATH . By combining REF to REF , we see that MATH as required. |
math/0011122 | Using NAME 's formulae as in the proof of REF , we have MATH . When MATH we have MATH, with equality only when MATH or MATH. It follows easily that MATH . By inverting this, we find that MATH and thus that MATH. |
math/0011122 | Because MATH in MATH, it follows immediately from the lemma that MATH. Using MATH, we see that MATH, and the first claim follows. If we now put MATH then MATH, and the second claim follows. |
math/0011122 | Let MATH denote the image of MATH in MATH and write MATH. Recall that the NAME generators MATH are characterised by the formula MATH . By applying the ring map MATH and putting MATH we obtain MATH . The first term can be evaluated using REF . For the remaining terms, we have MATH . We can use REF to rewrite this as MATH . After using the formula MATH to collect terms, we find that MATH . On the other hand, we know that MATH . The first term is zero because MATH is divisible by MATH. For the remaining terms, REF gives MATH . Thus, we have MATH . By comparing this with REF and equating coefficients of MATH, we find that MATH . After some rearrangement and reindexing, this becomes MATH . In particular, we have MATH. We now define MATH . The claim of the proposition is just that MATH for all MATH. Using the recurrence relation given in REF , one can check that for all MATH we have MATH . In particular, we have MATH, and it follows inductively that MATH for all MATH. We also have MATH so MATH. |
math/0011122 | First consider the case MATH, and write MATH. By inspecting REF , we see that MATH for all MATH, and thus REF tells us that MATH. We may thus assume that MATH. Write MATH, thought of as a subring of MATH. We will recursively define a sequence of elements MATH for MATH such that CASE: MATH CASE: MATH if MATH. It is clear that we can then take MATH. We start by putting MATH. Suppose that we have defined MATH with the stated properties. There is an evident map MATH which is an isomorphism in degree MATH. Let MATH be the image of MATH in MATH, and write MATH. We can lift this to get an element MATH of MATH such that MATH and every coefficient in MATH is MATH or MATH. It is easy to see that REF is satisfied, and that MATH. However, we still need to show that MATH is divisible by MATH. By assumption we have MATH for some MATH. Recall from REF that MATH. It follows after a small calculation that MATH also. Moreover, we have MATH, so MATH. It follows easily that MATH, as required. |
math/0011122 | If MATH we have MATH and MATH, so MATH, as required. Thus, we may assume that MATH, and it follows easily from the formulae for MATH and MATH that MATH induces a ring map MATH. Note that MATH over MATH. Write MATH. Arguing in the usual way, we see that MATH . It follows easily that we must have MATH . It follows that MATH for MATH, and that MATH, as required. |
math/0011122 | The claims involving MATH and MATH follow from REF , and those for MATH follow from REF . The claim MATH follows from REF , as the condition MATH is trivially satisfied for dimensional reasons. The claim for MATH can be proved in the same way as REF after noting that all the obstruction groups are trivial. |
math/0011122 | Choose an ideal MATH as in REF and set MATH. Everything then follows from REF . |
math/0011122 | This follows easily from the fact that MATH, given by REF . |
math/0011122 | We treat only the case of MATH; the other cases are essentially identical. Any ring map MATH commutes with the given map MATH, because the latter is central. It follows that maps MATH of MATH-algebras biject with maps MATH of rings, which biject with systems of commuting ring maps MATH for MATH. For MATH we have MATH, so REF tells us that the unique unital map MATH is a ring map. This remains the case if we replace the product MATH on MATH by MATH, or in other words replace MATH by MATH. There is an obstruction MATH which may prevent MATH from being a ring map. If it is nonzero, we have MATH . This shows that MATH is a ring homomorphism. After replacing MATH by MATH if necessary, we may thus assume that all the MATH are ring maps. The obstruction to MATH commuting with MATH lies in MATH. If MATH and MATH are different then at least one is strictly less than MATH; it follows that MATH and thus that the obstruction group is zero. Thus MATH commutes with MATH when MATH, and we get a unique induced map MATH, as required. |
math/0011122 | The cofibration MATH gives a fibration MATH of spaces. The usual theory of NAME sequences and fibrations tells us that the image of MATH is the union of those components in MATH that map to zero in MATH, so MATH is surjective. In particular, we find that MATH is nonempty. Similar considerations then show that the MATH-space MATH acts on MATH, and that for any MATH the action map MATH gives a weak equivalence MATH. This shows that MATH acts freely and transitively on MATH. This is easily seen to be compatible with our free and transitive action of MATH on MATH REF , and the claim follows. |
math/0011122 | Consider the following picture of MATH. MATH . The axes are set up so that MATH and MATH, so MATH . We write MATH and MATH for the upper and lower hemispheres and MATH for the unit disc in the plane MATH. Thus MATH and MATH, so MATH can be identified with the cofibre of the inclusion MATH. Note also that MATH is MATH-equivariantly homeomorphic to MATH (by radial projection from the MATH-fixed point MATH, say). Let MATH be the closed disc of radius MATH centred at MATH and let MATH be the closure of MATH. Define MATH by MATH on MATH and MATH on MATH. We see by obstruction theory that MATH can be extended MATH-equivariantly over the whole of MATH. Moreover, if we identify MATH with MATH as before then the restriction of MATH to MATH represents the same homotopy class MATH as considered in REF , as one sees directly from the definition. Next, note that MATH retracts MATH-equivariantly onto MATH, so we can extend our map MATH over MATH equivariantly. As MATH is MATH-connected, we can extend it further over the whole of MATH, except that we have no equivariance on MATH. Now define MATH by MATH. We claim that MATH. Away from MATH this follows easily from the equivariance of MATH and MATH, and on MATH it holds because both sides are zero. Using this and our identification of MATH with the cofibre of MATH we see that the restriction of MATH to MATH represents the class MATH in REF . Now observe that MATH is MATH-dimensional and MATH is MATH-connected, so our map MATH is nonequivariantly homotopic to the constant map with value MATH. This implies that MATH is homotopic to MATH, so MATH as claimed. |
math/0011123 | We temporarily write MATH for the ideal called MATH above. Put MATH and let MATH be the inclusion. Then MATH factors as MATH, where MATH is surjective. For any ideal MATH we see that MATH is surjective mod MATH, so MATH is zero mod MATH iff MATH is zero mod MATH. This condition depends only on the map MATH, so we can legitimately define MATH. Now suppose we have another presentation MATH, where MATH has rank MATH. Define MATH by MATH. It will suffice to prove that MATH and by symmetry we need only check the first of these. By projectivity we can choose a map MATH with MATH, and define MATH by MATH. It is easy to check that this gives another presentation MATH . If MATH then MATH is certainly nonzero, because the composite MATH is the identity, and MATH. If MATH and MATH then (by restricting to MATH) we see that MATH. For the converse, notice that MATH is a graded ring for any module MATH, and that MATH is a ring map for any homomorphism MATH of MATH-modules. One can check that MATH is contained in the ideal in MATH generated by MATH. It follows that if MATH then MATH. This shows that MATH, and thus that MATH, as required. |
math/0011123 | The presentation MATH shows that MATH . If this is actually an equality we must have MATH or in other words MATH, so MATH, so MATH. The remaining claims follow by symmetry. |
math/0011123 | Clearly MATH is a quotient of the ring MATH, and it is free of rank MATH, so MATH. |
math/0011123 | In the proof it will be convenient to write MATH and MATH instead of MATH and MATH, to display the dependence on MATH. First consider the case where MATH is a point and MATH. Then MATH is a homeomorphism and MATH. It is a standard fact that MATH can be identified with the inclusion MATH, and thus that MATH. In the general case, note that we have a map MATH of spaces over MATH. The projection MATH is a homeomorphism which we regard as the identity. If we let MATH be the projection, we have a splitting MATH. The inclusion MATH gives an inclusion MATH and thus an inclusion MATH, or equivalently a map MATH covering MATH. As MATH and MATH acts trivially on MATH we see that MATH. Next, we note that tensoring with MATH gives an isomorphism MATH and thus an isomorphism MATH with MATH. One can check that the following diagram commutes: It follows that MATH, and the previous discussion identifies this with MATH. It follows that MATH, as claimed. Now let MATH be the partial derivative of MATH with respect to MATH evaluated at MATH. This is characterised by the equation MATH; it is clear that MATH, and by applying MATH we see that MATH in the torsion-free case. As MATH we see that the NAME class of MATH is MATH. Thus, we have MATH . The MATH'th NAME class of MATH is the coefficient of MATH in this series, which is MATH as claimed. |
math/0011123 | Put MATH for MATH. Given a sequence MATH with MATH, put MATH. We first claim that the elements MATH form a basis for MATH over MATH. This is very well-known in the case where MATH is a point (so MATH) and MATH represents ordinary cohomology; it can proved using the NAME spectral sequence of the fibration MATH. For a more general theory MATH we still have an NAME spectral sequence MATH. It follows easily that the elements MATH form a basis whenever MATH is a point. A standard argument now shows that they form a basis for any MATH. Indeed, it follows easily from the above that they form a basis whenever MATH is trivialisable. We can give MATH a cell structure such that MATH is trivialisable over each cell, and then use NAME sequences to check that the elements MATH form a basis whenever MATH is a finite complex. Finally, we can use the NAME exact sequence to show that the elements MATH form a basis for all MATH. The ring MATH is graded-commutative so we certainly have MATH and in particular MATH for all MATH. Suppose we can show that MATH. Then MATH extends to give a map MATH of NAME algebras, and from the previous paragraph we see that this is an isomorphism. Combining this with the isomorphism of REF gives the required isomorphism MATH. All that is left is to check that MATH. For this we consider the case of the tautological bundle MATH over MATH, and take MATH. (We use this MATH-periodic version of MATH simply to comply with our standing assumptions on MATH; we could equally well use MATH itself.) Here it is standard that MATH is a formal power series algebra over MATH and thus is torsion-free. The ring MATH is a free module over MATH and thus is also torsion-free. As MATH we must have MATH as required. More generally, for an arbitrary bundle MATH over a space MATH we have a classifying map MATH giving rise to a map MATH. Moreover, for any MATH we can choose an orientation in degree zero and thus a ring map MATH. Together these give a ring map MATH, which carries MATH to MATH. As MATH in MATH, the same must hold in MATH. |
math/0011123 | For notational convenience, we will give the proof for a vector space; it can clearly be done fibrewise for vector bundles. We regard MATH as the set of unit complex numbers and define MATH as follows: MATH . Thus MATH is just the usual pinch map MATH. Note that if MATH and MATH then the eigenvalues of MATH lie in MATH so we can interpret MATH as an endomorphism of MATH as in REF. As MATH we see that MATH for all MATH and thus that MATH, so MATH gives a map from MATH to itself. We now define MATH by MATH. It is clear that the filtration of MATH is the number of eigenvalues of MATH (counted with multiplicity) lying in the open upper half-circle, and the filtration of MATH is the number in the open lower half-circle. Thus, the filtration of MATH is the number of eigenvalues not equal to MATH, which is less than or equal to the filtration of MATH. On the other hand, each map MATH has degree MATH and thus is homotopic to the identity, so MATH is homotopic to MATH. |
math/0011123 | For brevity we write MATH. We also write MATH and MATH and MATH. Because the filtration of MATH is stably split, the restriction map MATH is a split surjection, with kernel MATH say. Note that MATH and MATH are both projective over MATH. We need to show that MATH. First, we have MATH and it follows easily that MATH. We next claim that MATH for all MATH. Indeed, MATH is the image in cohomology of the map MATH, and so MATH is contained in the image in cohomology of the map MATH . Note that MATH is homotopic to the map MATH, which sends MATH into MATH. It follows that the restriction of MATH to MATH is null, and thus that MATH as claimed. It follows inductively that MATH for all MATH. This gives us a natural surjective map MATH. We previously gave a natural basis MATH for MATH, and it is clear that the subset MATH is a basis for MATH. It will be enough to prove that the images of these form a basis for MATH. The argument of REF allows us to reduce to the case where MATH is a point, MATH, and MATH represents ordinary cohomology. A proof in this case has been given by CITE (and possibly by others) but we will sketch an alternate proof for completeness. As the map MATH is surjective, it will suffice to show that the source and target have the same rank as free Abelian groups. For this, it will suffice to show that MATH has the same rank as MATH for MATH. As MATH has rank MATH, it is clear that MATH has rank MATH. On the other hand MATH is the NAME space MATH. Note that although MATH is not a complex bundle, it is necessarily orientable because MATH is simply connected. Thus, the NAME isomorphism theorem tells us that the rank of MATH is the same as that of MATH. By counting NAME cells we see that this is again MATH, as required. (This will also follow from REF .) |
math/0011123 | Suppose we have isometric linear embeddings MATH such that MATH for all MATH. We must show that MATH. Put MATH and MATH. Recall that MATH and that MATH. As each MATH embeds in MATH we see that MATH divides MATH and there is a natural surjection MATH. By combining these maps we get a map MATH, whose cokernel is MATH. From the definition of the NAME rank, we must prove that MATH. For this, we first note that an isometric embedding MATH of vector spaces gives rise to a homomorphism MATH by MATH . The alternative description MATH makes it clear that MATH depends continuously on MATH and MATH. We now extend this definition fibrewise, and define MATH by MATH. We have MATH and MATH . Using the fact that MATH is primitive in MATH, we find that MATH. Next, observe that if MATH for MATH we have MATH and MATH so MATH. Thus MATH factors through MATH, and it follows that MATH is mapped to zero by MATH, as required. |
math/0011123 | If MATH then MATH, so MATH is an isometry. Conversely, if MATH is an isometry then it preserves inner products so MATH for all MATH which means that MATH. NAME if MATH is not an isometry we have MATH which implies that MATH is injective. It is thus a strictly positive self-adjoint operator on MATH, so we can define MATH by functional calculus (as in REF). We then define MATH. This is the composite of MATH with an automorphism of MATH, so it has the same image as MATH. It also satisfies MATH, so it is an isometric embedding. |
math/0011123 | It follows immediately from REF that REF MATH REF MATH(MATH). CASE: Define MATH, MATH and MATH by the following pushout square: Equivalently, we can write MATH for the orthogonal complement of MATH in MATH and then MATH. CASE: Put MATH. By hypothesis, for each MATH we can choose an orthonormal sequence MATH in MATH. We can then choose elements MATH and MATH such that MATH. We find that MATH. This implies that the elements MATH are linearly independent, so MATH as required. CASE: Note that MATH is a nonnegative self-adjoint operator with the same kernel as MATH, and thus the same rank as MATH. Similarly, MATH is a nonnegative self-adjoint operator on MATH with the same rank as MATH. More basic facts about these operators are recorded in REF . As in REF we let MATH be the MATH'th eigenvalue of MATH (listed in descending order and repeated according to multiplicity). We see from REF that MATH is a continuous function of MATH. Moreover, as MATH has rank at least MATH we see that MATH. Now define MATH by MATH, and define MATH and MATH. (Here we are using functional calculus as in REF again.) One checks that MATH and MATH. We now have maps MATH which we combine to get a map MATH . Similarly, we define MATH . It is easy to check that MATH and MATH, so MATH and MATH are isometric embeddings. Now choose an orthonormal sequence MATH of eigenvectors of MATH, with eigenvalues MATH. Put MATH; these vectors form an orthonormal sequence of eigenvectors of MATH, with the same eigenvalues. For MATH we have MATH so MATH so MATH and MATH so MATH. This is the same as MATH, so it lies in MATH. Thus, this intersection has dimension at least MATH, as required. |
math/0011123 | We divide the sphere bundle MATH into two pieces, which are preserved by the evident action of MATH: MATH . The inclusions MATH give inclusions MATH which are easily seen to be homotopy equivalences. It follows that MATH. We also have MATH . Given a point in this space we have a map MATH sending MATH to MATH. This has norm MATH and is unchanged if we multiply MATH by an element of MATH. Using this we see that MATH. Of course, we also have MATH. We therefore have a homotopy pushout square as shown on the left below, giving rise to a commutative square of formal schemes as shown on the right. This evidently gives us a map MATH. To be more precise, we use the NAME sequence associated to our pushout square. This gives a short exact sequence MATH where MATH . We have seen that MATH, and the map MATH just corresponds to our map MATH . This map will thus be an isomorphism if MATH is injective, as claimed. |
math/0011123 | Put MATH. Let MATH be obtained from MATH by removing the MATH'th row and MATH'th column. The matrix MATH has the same general form as MATH so by induction we have MATH. If we expand MATH along the top row then the MATH'th term is MATH. As MATH we have MATH and so MATH only involves variables MATH with MATH. The remaining terms in the row expansion of MATH have the form MATH for MATH, and MATH is either zero (if MATH) or a variable strictly higher than all those appearing in MATH. The lemma follows easily. |
math/0011123 | Let MATH be the ideal in MATH generated by MATH and MATH, so MATH. We also put MATH. As MATH is topologically nilpotent, it will suffice to prove the result modulo MATH. We will thus work modulo MATH throughout the proof, so that MATH, and we must show that MATH is generated over MATH by MATH. Let MATH be defined by MATH, and let MATH be the matrix of MATH with respect to the obvious bases. It is then easy to see that MATH, where MATH is generated by the minors of MATH of size MATH. The entries in MATH are MATH. We next claim that all the generators MATH are nilpotent mod MATH, or equivalently that MATH in the ring MATH for all MATH. By downward induction we may assume that MATH in MATH for MATH. We consider the submatrix MATH of MATH given by MATH for MATH. By the definition of MATH we have MATH and thus MATH in MATH. On the other hand, we have MATH for MATH so MATH is lower triangular so MATH. Thus MATH is nilpotent in MATH but clearly MATH so MATH in MATH as required. It follows that MATH is a quotient of the polynomial ring MATH. Now let MATH be the submodule of MATH spanned over MATH by MATH; we must prove that this is all of MATH. As MATH, it will suffice to show that MATH is an ideal. In the light of the previous paragraph, it will suffice to show that MATH is closed under multiplication by the elements MATH, or equivalently that MATH contains all monomials of weight MATH. We thus let MATH be a multiindex of weight MATH. There is then a unique sequence MATH with MATH and MATH. Put MATH, so that MATH. Let MATH be the submatrix of MATH consisting of the first MATH columns of the rows of indices MATH, so the MATH'th entry of MATH is MATH. Note that the elements MATH lie in MATH. REF tells us that the lowest term in MATH is MATH. It is clear that the weight of the remaining terms is at most the size of MATH, which is MATH. By an evident induction, we may assume that their images in MATH lie in MATH. As MATH we deduce that MATH as well. |
math/0011123 | The previous lemma is the universal case. |
math/0011123 | There is an evident projection MATH, which identifies MATH with the divisor MATH on MATH over MATH. This divisor has degree MATH, so MATH is free of rank MATH over MATH. It follows by an evident induction that MATH is free over MATH, with rank MATH. |
math/0011123 | Put MATH; by elementary combinatorics we find that MATH. Put MATH. Using MATH we obtain a MATH-linear map MATH, which is surjective by REF ; we must prove that it is actually an isomorphism. Now consider the scheme MATH; REF tells us that the ring MATH is a free module over MATH of rank MATH. On the other hand, MATH can be identified with the scheme of tuples MATH where MATH and MATH. In other words, if we change base to MATH we can regard MATH as MATH, and now REF tells us that MATH is free of rank MATH over MATH. Now choose a basis MATH for MATH over MATH. We can combine this with MATH to get a map MATH. This is a direct sum of copies of MATH, so it is surjective. Both source and target of MATH are free of rank MATH over MATH. Any epimorphism between free modules of the same finite rank is an isomorphism (choose a splitting and then take determinants). Thus MATH is an isomorphism, and it follows that MATH is an isomorphism as required. |
math/0011123 | This is the universal case of the lemma. |
math/0011123 | Note that MATH. For MATH we have MATH where MATH may be zero, but MATH is nonzero. It follows that our ring MATH can also be described as MATH, or equivalently as MATH. The claim now follows easily from the previous corollary. |
math/0011123 | We have diagrams as shown below, in which the rows are easily seen to be exact: We see by induction that the maps MATH are all isomorphisms, and the claim follows by taking inverse limits. |
math/0011123 | For any monic polynomial MATH of degree MATH we write MATH. If MATH then MATH. Note that MATH, and that MATH. As we work mod MATH we have MATH. As we work mod MATH we have MATH. We also have MATH and MATH, so MATH and MATH. It follows easily that MATH, so MATH for MATH. We now have to distinguish between the case MATH and the case MATH. First suppose that MATH. Then for MATH we have MATH, and also MATH, and MATH for MATH by the previous paragraph. This implies that MATH. We also have MATH, and by comparing coefficients we deduce that MATH for MATH. The case MATH gives MATH, so MATH for MATH, so the lemma is true with MATH. Now suppose instead that MATH. As MATH for MATH we have MATH . We now multiply this by MATH and use the fact that MATH. By comparing coefficients of MATH we find that MATH. In view of this, our equation reads MATH . The right hand side has the form MATH with MATH, and by comparing coefficients we see that MATH as claimed. |
math/0011123 | REF tells us that MATH is MATH plus terms involving lower powers of MATH. It follows easily that if we filter the source and target of MATH by powers of MATH, then the resulting map of associated graded modules is a isomorphism. It follows that MATH is an isomorphism, and thus that MATH is an isomorphism. It follows that the map MATH is a split monomorphism of modules over MATH (and thus certainly of modules over MATH). We have seen that this map factors as MATH where MATH is surjective by REF . It follows that MATH is an isomorphism and that MATH is a split monomorphism, as required. |
math/0011123 | For each MATH we have a line bundle over MATH whose fibre over the point MATH is MATH. This is classified by a map MATH, which gives rise to a map MATH. The direct sum of these line bundles corresponds to the divisor MATH. This direct sum is a subbundle of MATH, so MATH. This construction therefore gives us a map MATH. In the case MATH we have MATH and MATH so the claim is that MATH, which is true by definition. In general, suppose we know that MATH. We can regard MATH as the projective space of the bundle over MATH whose fibre over a point MATH is the space MATH. It follows that MATH is just the divisor MATH over MATH, which is easily identified with MATH. The proposition follows by induction. |
math/0011123 | Let MATH denote the tautological bundle over MATH. This is a rank MATH subbundle of the pullback of MATH so we have a degree MATH subdivisor MATH of the pullback of MATH over MATH. This gives rise to a map MATH. Next, consider the space MATH. There is a map MATH given by MATH. This lifts in an evident way to give a homeomorphism MATH. Of course, this is exactly parallel to the proof of REF . Over MATH we have points MATH of MATH with coordinate values MATH say. Let MATH be the set of monomials MATH with MATH for MATH. From our earlier analysis of MATH and MATH we see that MATH is a basis for MATH over MATH. We also see from REF (applied to the bundle MATH) that MATH is a basis for MATH over MATH. This means that our isomorphism MATH is a direct sum (indexed by MATH) of copies of our map MATH. It follows that MATH must also be an isomorphism. |
math/0011123 | First, we can pull back the bundles MATH from MATH to MATH (without change of notation). We also have a bundle over MATH whose fibre over a point MATH is MATH; we denote this bundle by MATH, and note that there are natural inclusions MATH. We then have divisors MATH and MATH on MATH over MATH with MATH and MATH, so the triple MATH is classified by a map MATH. We next consider the universal case. As our model of MATH we use the space of orthonormal MATH-frames in MATH, so MATH is just the NAME of MATH-planes in MATH. Given a point MATH we construct a point MATH as follows: CASE: MATH is the span of MATH CASE: MATH is the span of MATH CASE: MATH is the span of MATH CASE: MATH is the span of MATH CASE: MATH is the map MATH that sends MATH to MATH. This gives a map MATH. Next, the group MATH has a subgroup MATH, inside which we have the smaller subgroup MATH consisting of elements of the form MATH. It is not hard to see that MATH, and that MATH gives a homeomorphism MATH. Moreover, MATH is contractible and MATH acts freely so MATH, so MATH as claimed. In the general case we can choose maps MATH classifying MATH, and this gives rise to a pullback square as follows: The vertical maps are fibre bundle projections so this is actually a homotopy pullback square. This give an NAME spectral sequence as in the statement of the theorem. On the edge we have MATH which is the same as MATH . We can now identify this as the tensor product of MATH with MATH over MATH. The part in degree zero is easily seen to be MATH as claimed. |
math/0011123 | A point of MATH is a tuple MATH where MATH, MATH, MATH, MATH and MATH. We can define a map MATH by MATH. It is not hard to see that this is a fibre bundle projection, and that the fibre over a point MATH is the space of linear isometric embeddings from MATH to MATH. This space is homeomorphic to the space of linear isometric embeddings of MATH in MATH, which is well-known to be contractible. Thus MATH is a fibration with contractible fibres and thus is a weak equivalence. |
math/0011123 | Recall that MATH is the scheme of pairs MATH where MATH is a divisor of degree MATH, MATH is a divisor of degree MATH and MATH. There is an evident isomorphism MATH sending MATH to MATH. The proposition tells us that MATH. We already know that MATH, and REF tells us that MATH. We therefore have an isomorphism MATH. (This involves an implicit NAME isomorphism, which is valid because MATH has only even-dimensional cells.) We leave it to the reader to check that this isomorphism is the same as the map considered previously. |
math/0011123 | The basic idea is to refine the proof of REF . We will take MATH as our model for MATH. We write MATH and MATH for brevity. We will need various isometries between infinite-dimensional vector spaces. We define MATH by MATH, and we define MATH by MATH. Next, it is well-known that the space of linear isometric embeddings of MATH in itself is contractible, so we can choose a continuous family of isometries MATH with MATH and MATH. Similarly, we can choose continuous families of isometric embeddings MATH with MATH and MATH and MATH. We now define a map MATH by MATH, where MATH is the orthogonal projection map from MATH to MATH. This acts as the identity on MATH and thus has rank at least MATH. If we choose MATH large enough that MATH and let MATH be the inclusions, then MATH. Next, we need to define a map MATH. Given MATH we can construct maps MATH as in the proof of the implication MATH , so MATH. We can thus define MATH by MATH. Suppose we start with MATH, define MATH to be the orthogonal projection, and then define MATH as above so that MATH. Observe that MATH decreases distances, and acts as the identity on MATH. If we let MATH be the eigenvalues of MATH (listed in the usual way) we deduce that MATH and that MATH for all MATH. It follows from this that MATH and MATH are the respective identity maps, so MATH . In particular, we have MATH for MATH, so MATH. Next, for MATH we define MATH by MATH . One can check that this is an isometric embedding, with MATH and MATH and MATH for all MATH. Similarly, if we put MATH we find that this is an isometric embedding of MATH in MATH with MATH and MATH and MATH for all MATH. It follows that MATH for all MATH, and this gives a path from MATH to MATH. Recall that we chose a path MATH from MATH to MATH. The pairs MATH now give a path from MATH to MATH in MATH. Both of the paths considered above are easily seen to depend continuously on the point MATH that we started with, so we have constructed a homotopy MATH. Now suppose instead that we start with a point MATH; we need a path from MATH to MATH. We have MATH, so MATH, where MATH is the orthogonal projection. One can check that this is characterised by MATH. Next, for MATH we define MATH by MATH . This is an isometric embedding with MATH and MATH. Similarly, we define MATH by MATH and we define MATH by MATH so MATH. The points MATH give a path from MATH to MATH in MATH. Next, we define MATH by MATH. The points MATH give a path from MATH to MATH in MATH. Using REF one can check that MATH . The map MATH is a strictly positive self-adjoint automorphism of MATH, so the same is true of MATH for MATH. The points MATH form a path from MATH to MATH. All the paths considered depend continuously on the point MATH that we started with, so we have defined a homotopy MATH. |
math/0011123 | We first replace MATH by the homotopy-equivalent space MATH. We write MATH and MATH for brevity, and similarly for MATH and MATH. We first claim that there is a commutative diagram as follows. Indeed, the isomorphism MATH comes from REF , and the map MATH comes from REF . It was proved in REF that the top horizontal map is a split monomorphism of MATH-modules, and it follows that the same is true of the map MATH. We now specialise to the case where MATH is MATH, the two-periodic version of the integer NAME spectrum. We then have MATH for all spaces MATH. This splits each of the rings on the bottom row of our diagram as a product of homogeneous pieces, and it is not hard to check that there is a unique compatible way to split the rings on the top row. We know that MATH is a split monomorphism; if we can show that the source and target have the same NAME series, it will follow that MATH is an isomorphism. If MATH then MATH so the claim is certainly true. To work downwards from here by induction, it will suffice to show that MATH for all MATH. To evaluate the left hand side, we consider the space MATH . Let MATH be the space of triples MATH of mutually orthogonal subspaces of MATH such that MATH and MATH. This is well-known to be a model of MATH and thus homotopy-equivalent to MATH; the argument uses frames much as in the proof of REF . Let MATH be the bundle over MATH whose fibre over MATH is MATH. If MATH and we put MATH and MATH then MATH and so MATH. It is not hard to see that this construction gives a homeomorphism of the total space of MATH with MATH. This in turn gives a homeomorphism of the NAME space MATH with the quotient space MATH. By induction we may assume that MATH is concentrated in even degrees, and it is clear from the NAME isomorphism theorem that the same is true of MATH. This implies that MATH is in even degrees and that MATH. As MATH has dimension MATH, we see that MATH. We also know that MATH. The conclusion is that MATH . We next evaluate MATH. Put MATH . We know from REF that MATH is freely generated over MATH by the monomials MATH for which MATH. It follows that the monomials MATH for which MATH form a basis for MATH over MATH. Similarly, those for which MATH form a basis for MATH over MATH. Thus, if we let MATH be the module generated over MATH by the monomials with MATH, we find that MATH. It is not hard to check that the monomials for which MATH form a basis for MATH over MATH. Next, let MATH be generated over MATH by the monomials MATH for which MATH; note that this involves the variables MATH rather than the variables MATH used in MATH. Because MATH we have MATH . Using this, we see that MATH. However, REF essentially says that MATH as graded Abelian groups, so MATH, so MATH . As explained previously, this implies that MATH is an isomorphism in the case MATH. We next consider the case MATH. Let MATH be the kernel of the usual map MATH. Because MATH is free of finite type and concentrated in even degrees, we see that the NAME spectral sequence collapses and that the associated graded ring MATH is isomorphic to MATH. Using this it is not hard to check that MATH is an isomorphism in the case MATH also. Finally, given an arbitrary even periodic ring spectrum MATH we can choose a complex orientation in MATH and thus a ring map MATH. Using this, we deduce that MATH is an isomorphism for all MATH. |
math/0011123 | This is another NAME spectral sequence. |
math/0011123 | The polynomial MATH is evidently divisible by MATH, say MATH. If we put MATH we find that MATH is a polynomial of degree at most MATH and that MATH, so we must have MATH and MATH. Thus MATH is the coefficient of MATH in MATH. As MATH has degree less than MATH and MATH is monic of degree MATH we deduce that MATH. |
math/0011123 | Put MATH. If we work modulo MATH then it is immediate from the lemma that MATH; this shows that MATH. Conversely, if we work modulo MATH then MATH is divisible by MATH and hence by MATH for all MATH, so MATH. It follows from the lemma that MATH for all MATH, and MATH is monic so MATH. Thus MATH. |
math/0011123 | Define MATH by MATH . Next, define MATH by MATH. We identify MATH with MATH using the basis MATH. It is easy to see that MATH and MATH. Moreover, the matrix of MATH has the form MATH so MATH. It follows immediately that MATH. It is easy to see that none of the polynomials MATH (where MATH) are zero-divisors, so MATH is not a zero-divisor. |
math/0011123 | It is enough to prove the corresponding result in the universal case, where MATH is the tautological divisor over MATH. As the map MATH is faithfully flat, it is enough to prove the result after pulling back along this map. In other words, we need only consider the divisor over the ring MATH with equation MATH. Let MATH be the discriminant of this polynomial, so MATH. Put MATH, and let MATH denote the ring of functions from MATH to MATH, with pointwise operations. We can define MATH by MATH, and the NAME Remainder Theorem tells us that the resulting map MATH is an isomorphism, and it follows that MATH. We also have MATH; the element MATH corresponds to the function MATH. Now put MATH and MATH. Let MATH be the remainder when the polynomial MATH is divided by MATH. This corresponds to the function MATH, where MATH is the remainder of MATH modulo MATH. As the discriminant is invertible in MATH we see that MATH iff MATH, and otherwise some coefficient of MATH is invertible. Using this, we deduce that MATH and MATH. If we let MATH be the evident basis of MATH over MATH, this means that MATH is spanned over MATH by the elements MATH for which MATH for some MATH, and these elements satisfy MATH so the map MATH is zero. As MATH is not a zero-divisor we deduce that the map MATH is zero, as claimed. |
math/0011123 | The main point to check is that MATH. Consider an element MATH. Let MATH denote the element MATH, so that MATH and MATH. We are interested in the component of this in MATH, which is easily seen to be MATH. Moreover, one checks that MATH so the relevant component of MATH is MATH, as claimed. Let MATH be the set of multiindices MATH with MATH for all MATH, and let MATH be the subset of those for which MATH. Put MATH. Then MATH is a basis for MATH, and MATH is a basis for MATH. Moreover, if MATH and we write MATH we see that MATH and MATH if MATH and MATH. It follows that MATH is surjective and MATH is a split injection of MATH-modules, as indicated in the diagram. REF tells us that MATH factor as MATH for some MATH, and a diagram chase shows that MATH is surjective. This gives the right hand triangle of the diagram. We simply define MATH to get the left hand triangle. As MATH is MATH-equivariant we have MATH . As MATH is surjective, this proves that MATH, so the top triangle commutes. |
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