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math/0011101 | We use induction on MATH. For MATH the result follows since MATH consists of MATH points, MATH is a point, and MATH. For MATH we note that MATH is the complement of a union of complex dimension one subspaces, so that MATH has the homotopy type of the complement of a MATH-torus link and MATH is an n-times punctured two-sphere. Thus MATH. Observe that MATH is formed by attaching MATH one-cells to MATH points. To compute MATH we note that the NAME characteristic MATH is MATH so that MATH . Thus MATH, which is indeed MATH times the first betti number of MATH. The case MATH also follows from the general induction step below. For the general inductive step, we consider the MATH-arrangement MATH with complement MATH. We choose a generic hyperplane MATH, and consider the arrangement MATH as an arrangement in the MATH-dimensional vector space MATH. Since MATH is generic, no intersection of hyperplanes from the original arrangement (a flat in the language of arrangement theory) is contained in MATH (except for ). Thus the lattice of MATH is the same as that of the original arrangement MATH in ranks (= codimension) MATH through MATH. Thus the betti numbers of MATH are the same as those of the complement MATH of the union of the MATH in MATH for rank MATH through MATH CITE. Since the NAME polynomials satisfy MATH, it follows that MATH for MATH. We are attaching MATH-cells to MATH to form MATH. These cells can only change homology groups in degree MATH or MATH . But since MATH all the MATH-cells attached to MATH to form MATH are added with homologically trivial attaching maps, so that the number of MATH-cells added is MATH, as desired. |
math/0011102 | Let MATH be the set consisting of the poles of MATH, MATH and the zeros of MATH. Suppose there exists MATH such that MATH. Then, for every MATH, MATH hence MATH. By induction, for every MATH, we also have MATH, thus MATH. Assume now that all the poles of MATH lie in MATH. Let MATH be a pole of MATH. Let MATH . Suppose MATH and MATH. The first inequality implies MATH. The two inequalities imply MATH for every MATH, that is, MATH that is, MATH . Suppose we have proved that for every integer MATH we have MATH . Then MATH . Thus, MATH that is, MATH . Hence, MATH . Suppose now that MATH, but MATH. Let MATH be a sufficiently negative power of a local parameter at MATH so that MATH. The NAME module MATH is isomorphic to MATH and by CITE MATH. Note that MATH. Then for every MATH we have MATH in particular, MATH. By the argument of the last paragraph we conclude that MATH. If MATH, let MATH be a sufficiently positive power of a local parameter at MATH such that MATH . Once again we take the NAME module MATH which is isomorphic to MATH. By the two last cases and CITE we conclude that MATH. By the non-negativity of the local canonical heights we conclude that MATH. |
math/0011102 | Denote MATH and suppose MATH. Let MATH and MATH the integral part of MATH, then MATH. So we can choose MATH distinct points MATH in MATH. By REF there exist MATH such that MATH for MATH. It follows from the triangle inequality that MATH . Hence, by REF , MATH . |
math/0011102 | Suppose MATH. Then for every MATH, MATH, which contradicts REF . So we take MATH. |
math/0011102 | By REF , MATH. If MATH, then MATH. Thus, since MATH, MATH . The first statement follows from REF . Suppose now that MATH. In this case, since MATH and MATH, we have MATH . The second statement follows from REF . |
math/0011102 | If MATH, then MATH . NAME 's discriminant theorem CITE was first proved in the case of semi-stable elliptic curves. However, this result was extended by NAME and NAME to any elliptic curve CITE. It follows from REF and MATH that MATH . The result now follows from REF and MATH. Suppose now that MATH, then MATH . The result now follows from REF and MATH. |
math/0011104 | Let us prove the first item. Let MATH be the function given by MATH . Since the geodesics in MATH are products of geodesics in MATH and MATH, the function MATH is constant along the orbits of the geodesic flow of MATH. It follows from REF that MATH . If we write MATH, it is easy to check using REF that MATH from which we obtain right away the first equality in the lemma. To prove the second item, let MATH be the set of all horizontal unit vectors. Clearly the geodesic flow of MATH leaves MATH invariant. Let MATH be the restriction to MATH of the differential of the submersion map. Since horizontal geodesics project to geodesics, MATH is a surjective map that intertwines the geodesic flow of MATH restricted to MATH with the geodesic flow of MATH. It follows from REF that MATH. |
math/0011104 | It is shown in CITE (compare also CITE) that if MATH is rationally elliptic then, MATH . Since MATH is simply connected the NAME isomorphism theorem implies that MATH . Since MATH, using REF we obtain MATH. |
math/0011104 | Obviously REF implies REF. Let us prove that REF implies REF. Suppose that MATH is rationally elliptic. By REF , MATH. Since MATH is smooth, the NAME obstruction vanishes. Therefore by NAME 's theory CITE, the homeomorphism type of MATH is completely determined by the intersection form of MATH. It follows that if MATH, MATH is homeomorphic to MATH and if MATH, MATH is homeomorphic to MATH. When MATH, the possible intersection forms are MATH . These forms correspond to MATH, MATH and MATH respectively. On the other hand MATH, MATH and MATH are homogeneous spaces and hence they are elliptic (see REF above). By REF above, MATH and MATH are elliptic. Finally, it is well known that the homology of the loop space with rational coefficients can either grow polynomially or exponentially. |
math/0011104 | It follows from REF and CITE that the homology of the loop space can be computed as MATH where MATH is the differential graded algebra given by the normalized singular cochains with coefficients in MATH. (In fact, NAME and NAME mention in his paper that this special case of their theorem has to be attributed to CITE.) It can be seen that for a manifold MATH satisfying the hypotheses of the theorem there exists a quism between MATH and MATH. This means a morphism of differential graded algebras with the property that induces isomorphisms in homology. Since a quism preserves MATH it follows that MATH . We now make use of the following lemma whose proof will be given after completing the proof of the theorem. The sum of the dimensions of MATH grows exponentially unless MATH. Conversely if MATH then the sum of the dimensions of MATH grow polynomially. A result of CITE (see also the corollary before Lemma F in CITE) using the linking form ensures that the torsion part of MATH always has the form MATH or MATH for some finite abelian group MATH. Hence if MATH is elliptic, the lemma implies that MATH must be zero and when the MATH factor appears the rank of MATH should be zero. Proof of the lemma. Let us set for brevity MATH. Observe that MATH is a (graded) commutative local ring with residue field MATH that satisfies NAME duality. We note that it suffices to prove the lemma ignoring the grading of MATH because MATH for MATH (the first integer indicates the resolution degree and the second the internal grading). Let MATH and let MATH be the maximal ideal of MATH. Given a finitely generated MATH-module MATH, let MATH. MATH is a finite dimensional vector space over MATH. Below we will use the following form of NAME 's lemma: if MATH is a morphism of MATH-modules such that the induced morphism MATH is surjective, then MATH is also surjective. To compute MATH we need to take a projective resolution of MATH regarded as a MATH-module in the obvious way. Since MATH is local a MATH-module is projective if and only if is free. Hence, we will construct a resolution of the form: MATH . The first map MATH is given simply by MATH where MATH and we identify MATH with MATH. Clearly MATH. We will now define a surjective morphism MATH. Let MATH. Clearly MATH generates MATH and hence given any free module MATH, the elements MATH for MATH generate MATH. Hence, to define MATH it suffices to indicate the images of the MATH's. Pick a basis of MATH (which has dimension MATH) and let MATH be determined by a bijection between the generators of MATH and this basis. Note that MATH. Hence MATH is an isomorphism and by NAME 's lemma MATH is surjective. Let MATH be the ideal given by those elements of the form MATH. Note that CASE: MATH; CASE: MATH has dimension MATH. To define MATH, we take MATH and we map the canonical MATH generators of MATH onto a basis of MATH. This gives a surjective morphism as before. By continuing in this fashion we find that at the i-th step of the construction of the resolution we have: CASE: MATH; CASE: MATH has dimension MATH. Therefore MATH. This implies that the growth of sequence MATH is exponential if MATH (with exponent MATH) and at most linear if MATH. Now observe that we have the isomorphism MATH and under this isomorphism the map MATH is zero. Thus the differential of the complex MATH is zero, so the dimensions of MATH over MATH grow exactly as the MATH's. |
math/0011104 | Let MATH be the quotient of MATH by the circle action. A simple argument with the long exact sequence of the fibration shows that MATH is simply connected and MATH. By the NAME theorem MATH. Note that MATH contains a copy of MATH given by the projection to MATH of the subset of MATH given by MATH and hence MATH is not spin. It follows from the NAME classification that the only closed simply connected non spin REF-manifold with MATH is MATH. |
math/0011104 | First consider a metric MATH on MATH which is invariant under the MATH-action. This is obtained as usual by averaging any given Riemannian metric over the orbits. Now consider the manifold MATH and for any MATH the Riemannian metric MATH (where MATH is the Euclidean metric on MATH) on MATH. Define a (free) MATH-action on MATH by MATH . The quotient of MATH by this action is diffeomorphic to MATH and the metric MATH is invariant through the action; therefore it induces a metric MATH on MATH. The projection MATH is a Riemannian submersion, and therefore the entropy of MATH is bounded above by the entropy of MATH (see REF) which is actually equal to the entropy of MATH (see REF). Therefore to prove the theorem it is enough to show that the volume of MATH approaches REF as MATH approaches REF. We will prove this now. First we identify the quotient (of MATH by the MATH-action) with MATH via the diffeomorphism which sends MATH to the class of MATH. Let MATH be the vector tangent to the MATH-action at MATH and let MATH be the tangent to the canonical MATH-action on MATH (which gives the usual trivialization of the tangent space of MATH). Let MATH. The tangent vector to the action on MATH is MATH. If MATH, the MATH-orthogonal subspace to this vector is spanned by MATH and the subspace of vectors of the form MATH where MATH, the subspace of vectors MATH-orthogonal to MATH. It is clear that MATH and MATH coincide on MATH. Moreover, MATH for all MATH. Since MATH we have that MATH . This implies the following equation for the volume elements of the two metrics: MATH . This formula will be enough to show that the volume of MATH approaches REF with MATH. Note first that the formula shows that the volume of any region computed with MATH is always at most the volume of the same region computed with MATH (independently of MATH). Given any MATH, we can find an open neighborhood of the fixed point set of the MATH-action on MATH which has MATH-volume less than MATH. Then the MATH-volume of this neighborhood will also be less than MATH for any MATH. Away from the neighborhood, MATH has a positive lower bound, and the volume formula clearly shows that the MATH-volume of the complement of the neighborhood is of the order of MATH for MATH small. Therefore, for MATH small enough the volume of the complement will also be less than MATH. This completes the proof of the theorem. |
math/0011104 | Pick a point MATH so that MATH lies in only one of the open subsets of the MATH-structure (for this one might need to do some harmless changes in the MATH-structure, like eliminating any open subset which is contained in the union of the others). We can also assume that the torus acting on the open subset containing MATH is of dimension one and that MATH lies on a regular orbit. Now pick a small MATH-ball MATH centered at MATH and transverse to the MATH-action. The union of the orbits through MATH form an embedded solid torus MATH. Repeat the same procedure to obtain an embedded solid torus MATH in MATH containing a point MATH. We will perform the connected sum inside MATH and MATH. First divide MATH into an inner ball and an outer annulus: MATH. We can identify MATH with MATH, where MATH is a small REF-dimensional ball centered at a point in the middle of MATH and transverse to MATH in MATH. The component of the boundary corresponding to the boundary of the deleted MATH is identified with the boundary of MATH. We can now describe the MATH-structure on MATH. On MATH leave the initial MATH-structure. On MATH consider any non-trivial MATH-action on the MATH-factor (here is where we need the hypothesis MATH). The action induced on each component of the boundary glues to the canonical action on the MATH-factor to create a MATH-action (in case MATH is even it will have orbits of dimension REF). Finally on MATH leave the initial MATH-structure. |
math/0011104 | For the proof we will need smooth descriptions of the surfaces: see CITE for details. Every elliptic surface of NAME characteristic REF is obtained by performing logarithmic transforms on a basic elliptic surface. Every elliptic surface is obtained by taking the fiber sum of an elliptic surface of NAME characteristic REF and some rational elliptic surfaces, and then blowing up some points. Basic surfaces are fiber bundles with fibers MATH and structure group in MATH. Hence they admit a polarized MATH-structure whose orbits are the fibers. Now let MATH be a neighborhood of a fiber on a basic surface MATH, where MATH is identified with the unit ball in MATH. Fix a positive integer MATH and integers MATH, MATH such that MATH has order MATH in MATH. Let MATH be given by MATH. MATH generates a group MATH of diffeomorphisms of MATH of order MATH. The quotient of MATH by this group is again diffeomorphic to MATH. Consider also the map MATH given by MATH . MATH generates a group MATH of diffeomorphism of MATH of order MATH which acts freely on MATH. The map MATH, MATH is a diffeomorphism. The logarithmic transform (of order MATH) at the fiber over MATH in MATH is the elliptic surface MATH obtained by gluing MATH and MATH via this diffeomorphism. Clearly the obvious MATH-action on MATH (which fixes MATH) induces a MATH-action on MATH which commutes with the action on the fibers. Hence MATH admits a MATH-structure (with orbits of dimension REF). Rational elliptic surfaces are diffeomorphic to MATH and therefore admit MATH-structures by the previous theorem. Nevertheless we will need to perform fiber sums and so we will give another MATH-structure on it, compatible with the elliptic fibration. To do this we need first to give a description of the surface as an elliptic surface (see CITE). Let MATH and consider the involution MATH of MATH. Let MATH be rotation of MATH around the MATH-axis. The diffeomorphism MATH has REF fixed points. Identify a neighborhood of each of these points with a ball MATH in MATH. Consider MATH. The canonical projection MATH induces an isomorphism away from the preimage of MATH. Construct a surface MATH by replacing the eight copies of MATH with MATH in MATH. The involution MATH extends to an involution MATH on MATH which has REF spheres as the set of fixed points. Then MATH. Let MATH be the projection. Then MATH induces a map MATH which commutes with MATH and so induces a map MATH whose generic fiber is MATH; this map expresses MATH as an elliptic surface. Note that the MATH-action on MATH given by MATH commutes with MATH and induces a MATH-action on MATH. We can extend this action to a MATH-action defined on a neighborhood of the fibers of MATH over the north and south poles. This actions commutes with MATH and so induces an action on a neighborhood of the fibers of MATH over the poles. Away from the fibers over the poles MATH actually is the total space of a fiber bundle with structure group MATH. There is then a polarized MATH-structure defined on this piece, whose orbits are the fibers. On the boundary of the neighborhoods around the fibers over the poles the two actions commute. This defines a MATH-structure on MATH. The fiber sum of two elliptic surfaces is done as follows: pick regular fibers on each surface identifying neighborhoods of them with MATH (MATH is a small REF-ball). Delete the corresponding regular fiber from each surface and then glue both surfaces along MATH. The diffeomorphism class of the resulting surface will depend only on the isotopy class of the diffeomorphism chosen to identify the fibers with MATH. We can therefore take the diffeomorphism to be in MATH and we can see that the MATH-structures we defined on the surfaces of NAME characteristic REF and the rational elliptic surfaces glue well along the fiber sum. Finally blowing up points means, in terms of diffeomorphisms, to take connected sums with MATH's. Such a connected sum admits a MATH-structure by the previous theorem. |
math/0011104 | Let MATH be a tubular neighborhood of the union of the orbits through MATH. Consider the unit MATH-sphere MATH. Pick a non-trivial MATH-action on MATH, for instance complex multiplication in the first REF coordinates. In case MATH is odd we can pick a free MATH-action. Choose a regular orbit of the action and a disc MATH transverse to the orbit. Pick a canonical embedded MATH-sphere MATH and a tubular neighborhood MATH of the union of the orbits through MATH. The manifold MATH is obtained by gluing MATH and MATH along MATH and MATH. But gluing two copies of MATH along the MATH-spheres is the same as taking the product of a MATH-sphere with the connected sum of two copies of MATH. Hence in this glued part we can consider the MATH-structure we defined in the previous theorem, which on each component of the boundary coincides with the structure of MATH and MATH, respectively. This clearly defines a MATH-structure on MATH. This structure is polarized if MATH and MATH are odd. |
math/0011104 | Consider an orthonormal basis MATH of MATH. If MATH and we let MATH then MATH. But MATH is a positive definite symmetric matrix, and therefore it has MATH positive eigenvalues MATH and MATH. Now consider an orthonormal basis MATH of MATH and let MATH. Let MATH. If MATH then MATH. Again, MATH is a positive definite symmetric matrix. Moreover, since MATH for any MATH we have that no eigenvalue of MATH is greater than REF. Therefore the smallest eigenvalue is at least MATH. This means that for all MATH, MATH. Now if MATH we can find the unique vector MATH whose first coordinate is MATH. Then MATH and so MATH . Therefore, if MATH and MATH with MATH and MATH, we have that MATH . But MATH, and therefore either MATH or MATH. In any case, MATH. This means that MATH and so MATH, proving REF . CASE: Given any MATH write MATH (in MATH). The map MATH, MATH, is a monomorphism. Moreover, the image of MATH is included in the MATH-orthogonal complement of MATH, MATH, and therefore MATH gives an isomorphism between MATH and MATH. Pick any MATH. The map MATH is an isomorphism. Therefore there exists a unique MATH such that MATH . Since MATH, we have that MATH . But since MATH, we have that MATH. Therefore MATH . There is a unique MATH such that MATH. Then since MATH and MATH, we have that MATH. Therefore MATH . Hence for any MATH, MATH and REF follows. |
math/0011104 | Let MATH be the open covering corresponding to a MATH-structure on MATH; with corresponding actions MATH by tori MATH on the coverings MATH's. We can construct a regular atlas for the structure as in REF . Namely we construct a new open cover MATH of MATH obtained by considering all non-empty intersections of the MATH's and then removing from each set the ``unnecessary" parts. Each MATH has a finite cover MATH where there is defined an effective torus action. For instance, if one had MATH then one would consider MATH, MATH, MATH so that MATH and both are invariant through the corresponding action (note that on MATH one has defined a MATH action). A Riemannian metric MATH on MATH is called invariant if on each of the open subsets MATH of the MATH-structure the corresponding sheaf of torus acts by isometries. An invariant metric always exists, at least after replacing the open subsets MATH by slightly smaller ones. Such a metric is constructed in REF . Let us then fix a Riemannian metric MATH on MATH invariant through the MATH-structure. Each MATH is, essentially, the intersection of certain number of MATH's. Assume that MATH are ordered in a non-increasing way with respect to the number of the MATH's intersecting. Therefore, if MATH and MATH the torus action on MATH, restricted to MATH, is embedded in the action on MATH. Consider smooth functions MATH, supported in MATH which are constant along the orbits and such that MATH covers MATH. Let MATH and let MATH and MATH (where MATH is the standard Euclidean metric on the MATH-torus). For each open subset MATH consider the following (free) MATH-action on MATH: MATH where MATH. These formulas clearly define a MATH-structure on MATH. But what is more important to us is that it is actually a polarized MATH-structure of positive rank. Note that on the MATH's all the torus actions corresponding to the MATH's which are intersecting glue together to get a free torus action on MATH (where the dimension of the torus acting is the sum of the corresponding MATH's). Pull back the functions MATH to obtain smooth functions MATH on MATH. Note that the functions MATH are invariant through both the torus action (on MATH) coming from MATH and the canonical MATH-action on the MATH-factor of MATH. The same is true for the metric MATH. Now we proceed to collapse the metric MATH along the orbits of the MATH-structure on MATH. This is done in REF . We will describe the procedure, since we need to make some computations on it. Fix a small MATH. For technical reasons it is convenient to first replace MATH by MATH. We construct a metric MATH on MATH by multiplying the metric MATH by MATH in the directions tangent to the orbits of the torus action on MATH (and leaving the same metric in the directions orthogonal to the orbits). Note that the MATH-action on MATH given by the canonical action on the MATH-factor is isometric with respect to MATH. Repeating this procedure MATH-times we get a metric MATH which is invariant under the MATH-action just mentioned. Let MATH be the metric induced on MATH. The projection MATH is a Riemannian submersion. Therefore the entropy of MATH is bounded above by the entropy of MATH (see REF). The entropy of MATH on the other hand is bounded above by MATH, where MATH is an upper bound for the absolute value of the sectional curvature of MATH (see REF). But it is proved in REF , that the sectional curvature of MATH is bounded independently of MATH. Therefore we got that: MATH where MATH is some constant independent of MATH. We will now estimate the volume of MATH. We will do this by comparing the volume element of MATH with that of MATH. Let MATH be a MATH-orthonormal basis of MATH. Then MATH . Since the volume element at a point depends only on the value of the metric at the point, it is the same to work on MATH or on the corresponding finite covering. Therefore from now on we will think that we are working with a MATH-structure to simplify the notation. Fix any point MATH and any point MATH which projects to MATH. We have to check how the volume element changes at each step in the construction of MATH. Of course there is no change in the step MATH if MATH does not belong to MATH. So let us assume for instance that MATH. Moreover, assume that MATH is not a fixed point for the torus action (the set of fixed points has volume REF with respect to any Riemannian metric). We want to compare the volume elements at MATH of MATH and MATH (MATH is of course the quotient of MATH under the MATH-action on MATH). Assume that the orbit through MATH of the torus action has dimension MATH. There is then an orthonormal set of vectors MATH and some linearly independent vectors MATH so that the vectors MATH are tangent to the orbit on MATH, and the directions orthogonal to the MATH's act trivially on MATH at MATH. Let MATH be the subspace of MATH spanned by this MATH vectors (the tangent space to the orbit on MATH), let MATH be the tangent space to the orbit in MATH and let MATH. Let MATH be a MATH-orthonormal basis of the space MATH-orthogonal to the orbit (in MATH). Note that MATH are also MATH-orthogonal to the orbit and MATH. Therefore MATH . Recall that the metric MATH is obtained by multiplying by MATH the values of MATH on MATH. From now on we restrict our attention to MATH, since its orthogonal complement plays no real role in the construction of MATH. We can assume that for any unitary tangent vector to any of the tori (acting on any of the MATH), the derivative of the action in that direction has MATH-norm at most one. Therefore we are under the hypothesis of our Linear Algebra REF . Consider now a MATH-orthonormal basis of MATH; call them MATH. For each MATH write MATH where MATH and MATH (the MATH-orthogonal complement of MATH in MATH). Let MATH. Now, for each MATH, consider the vector MATH . The vector MATH is MATH-orthogonal to the tangent space to the torus factor. Its first coordinate is, of course, MATH. Assume that MATH. Then MATH . Note also that: MATH . Therefore, MATH . Now, in the region where MATH, we have from REF that MATH and, therefore, MATH . Therefore the MATH-volume of the region where MATH and MATH approaches REF as MATH does. The MATH-volume of the region MATH is of the order of MATH. Therefore the MATH-volume of the region where MATH is of the order of MATH (using REF ) and therefore it also approaches REF with MATH. This of course implies that MATH approaches REF with MATH. Finally note that in the passage from MATH to MATH there are two steps: first we multiply by MATH to obtain MATH and then we collapse along the orbits multiplying by MATH. REF , tells us that the second of these steps does not increase volumes (on MATH with the quotient metric). To go from MATH to MATH only the second step is performed. Therefore when passing from MATH to MATH the volume of the region MATH will remain small, while by taking MATH small we can make the volume of the region MATH small. Hence for MATH small enough the MATH-volume of the whole MATH will be as small as desired. Since the entropy of MATH is bounded above independently of MATH, the theorem is proved. |
math/0011104 | We will prove that MATH admits a MATH-structure and then apply Theorems A and B. By REF it is enough to show that each of the building blocks of the classification (see REF) admits a MATH-structure. Consider a smoothly embedded REF-sphere MATH representing MATH-times a generator of MATH (MATH). MATH is obtained by performing surgery on MATH. In the same way MATH is obtained by performing surgery on a sphere representing MATH-times the generator of MATH (note that even multiples of the generator have trivial normal bundles). This is easy to check since these manifolds are characterized by their homology groups and whether they are spin or not. Of course, MATH and MATH admit free MATH-actions. MATH also admits a free MATH-action since the NAME action on MATH commutes with the structure group of the bundle. If we consider MATH as the quotient of MATH by the MATH-action MATH then the NAME action is given by complex multiplication in the last two coordinates. But to construct MATH-structures on all the MATH's consider the MATH-action on MATH given by complex multiplication on the last coordinate. This action has fixed points, of course. Call this action MATH, and MATH the free ``NAME. The second homology of MATH is generated by the image (under the projection) of MATH. Call this REF-sphere MATH. Now, given any small MATH consider REF-sphere MATH . MATH is homologous to MATH and they intersect only at MATH. If we set the imaginary part of MATH to be REF, we get the non-trivial MATH-bundle over MATH, MATH, which is diffeomorphic to MATH. We can modify MATH inside MATH to obtain a smoothly embedded sphere MATH representing twice the generator of MATH. The orbits of the MATH-action passing through MATH form an embedded MATH. Its normal bundle is MATH, where MATH is the MATH-bundle over MATH with NAME characteristic REF. MATH can be represented as the quotient of MATH under the MATH-action MATH . There is then a canonical MATH-action MATH on MATH given by complex multiplication in the last coordinate. Define a MATH-structure on MATH by leaving the MATH action on MATH and giving to MATH the MATH action. The zero section of MATH is the embedded MATH and is exactly the fixed point set of MATH. For any MATH, MATH, consider a REF-sphere embedded in MATH representing MATH-times the generator of the second homology group. This sphere represents MATH-times the generator of MATH. Its normal bundle is MATH, which is isomorphic to the trivial bundle MATH. MATH is usually presented as the union of two copies of MATH glued along MATH by the map MATH . Namely, MATH is considered as a map MATH and then we identify MATH with MATH. The identification of MATH with MATH is obtained by an homotopy of the loop MATH in MATH with the constant loop REF. The action MATH can then be viewed in MATH as: MATH for a map MATH. Here MATH means complex multiplication in the first two (real) coordinates. Since MATH the map MATH is null-homotopic. Therefore we can define a MATH-action on MATH which is equal to MATH in an exterior annulus and to MATH in an inner annulus. MATH is obtained by deleting MATH of MATH and gluing MATH along the boundaries. Giving any MATH-action to the MATH-factor of the glued MATH clearly defines a MATH-structure on MATH. These MATH-structures are not polarized. The NAME MATH admits a locally-free MATH-action: simply embed MATH in MATH by sending MATH to the diagonal matrix with MATH, MATH, MATH as the diagonal coefficients and then follow by matrix multiplication. Finally, for any MATH, MATH, MATH is obtained by performing surgery on a sphere MATH representing MATH-times the generator of MATH, which can be represented by a completely transversal sphere for the NAME action on the MATH-factor (in the sense of REF). One then obtains by REF a polarized MATH-structure on MATH. This finishes the first part of the theorem. The last statement follows because the fact that MATH is either non-cobordant to zero or it is cobordant to REF but either MATH or MATH, means that in the factorization of MATH as connected sum of building blocks only appear MATH's, MATH, MATH and MATH and we have put polarized MATH-structures on these manifolds (MATH). |
math/0011105 | The fact that MATH is an isolated singular point of MATH implies that the first partial derivatives of MATH form a MATH-regular sequence by CITE. The rest follows from REF . |
math/0011105 | REF imply that MATH has only top-dimensional cohomology. The same follows for all links of faces in MATH, since these are always joins of order complexes of regular complex polytopes which are medial polytopes of MATH (see CITE or the proof of REF below). |
math/0011105 | The first assertion is CITE. For the second assertion, note that MATH is a subset of the stabilizer of MATH, and hence it suffices to show that they have the same cardinality. By the first assertion, the order of a NAME group MATH is the number of maximal flags in the corresponding polytope MATH, and a group element MATH may be identified with the image MATH of the base flag MATH. In particular, the stabilizer of MATH has the same cardinality as the set of maximal flags in MATH which pass through the partial flag MATH. This cardinality is clearly the product of the numbers of maximal flags in each interval MATH for MATH (where we adopt the convention that MATH, MATH, and MATH). However, each such interval is again the poset of faces in a regular complex polytope MATH, the medial polytope CITE associated with MATH. Since the NAME group associated to MATH may be identified with the subgroup MATH where MATH we conclude that the stabilizer of MATH has cardinality MATH. On the other hand, since MATH, and MATH commute for MATH by the presentation of MATH discussed in REF, we conclude that MATH, as desired. |
math/0011105 | Let MATH be the exterior derivative of MATH. Since MATH is invariant, MATH is invariant, and hence MATH is an anti-invariant MATH-form. By REF , MATH can thus be written as a combination of the MATH with coefficients from MATH: MATH . Since MATH for each MATH, each MATH must have positive degree and thus lie in MATH. By comparing the coefficient of MATH on each side of REF above, we see that each MATH is in MATH. |
math/0011106 | This follows from standard results on the pasting of pullback squares. MATH . |
math/0011106 | MATH and MATH are both REF-element sets, hence isomorphic in a unique way. Suppose inductively that MATH and that we have constructed a commuting REF MATH then MATH for some MATH and MATH; then define MATH to be MATH . It is easy to check that the map MATH thus defined is a bijection and commutes with the MATH's. MATH . |
math/0011106 | CASE: That MATH has a right adjoint is immediate: it is the right NAME extension of the inclusion MATH. However, it will be useful to have the following explicit description of MATH: if MATH then MATH . (When MATH the second line does not make sense, and we instead define MATH as MATH; essentially we are `taking MATH'.) The source and target maps are the obvious ones. As is shown in REF, MATH is naturally a monad opfunctor MATH, whose natural transformation part MATH is an isomorphism. Under the adjunction MATH, the mate of this isomorphism is a natural transformation MATH and this gives MATH the structure of a monad functor MATH. Further checks reveal that the conditions of REF are satisfied, so that there is an induced adjunction between categories of multicategories; moreover, MATH and MATH each preserve terminal objects, so this restricts to an adjunction MATH . MATH has the obvious restriction effect on MATH-operads; in the other direction, if MATH is a MATH-operad, MATH and MATH, then MATH . Next we bring in precontractions. Any precontraction on a MATH-operad MATH evidently gives rise to a precontraction on MATH; conversely, any precontraction on a MATH-operad MATH extends uniquely to a precontraction on MATH. The precontractions produced by these two constructions are preserved by the unit and counit maps of the adjunction REF , so we obtain an adjunction as required. CASE: Any adjunction MATH restricts to an equivalence between MATH and MATH, where MATH is the full subcategory of MATH whose objects are those at which the unit of the adjunction is an isomorphism, and similarly MATH with the counit. In the present case we have MATH, and the counit of the adjunction is the identity transformation. On the other hand, let MATH be a MATH-operad with precontraction and consider the unit map MATH . This is the identity in dimensions less than MATH, and in dimension MATH it consists of the maps MATH (MATH). This is always surjective as MATH carries a precontraction, and is injective precisely when MATH satisfies the condition for precontractions on it to be called contractions. So the unit at MATH is an isomorphism if and only if MATH is an object of MATH. CASE: The proof is just like that of REF . Again it will be useful to have an explicit description of the right adjoint MATH of MATH: it is given by MATH . The source and target maps in dimensions MATH are as in MATH; from dimension MATH to dimension MATH they are first and second projection; and in dimensions above MATH, they are identities. MATH . |
math/0011106 | The functor MATH constructed in REF has a right adjoint, so MATH is initial in MATH. The functor MATH constructed in REF is an equivalence, so MATH is initial in MATH. MATH . |
math/0011106 | First we identify the initial object MATH of MATH; and since MATH, this means examining MATH. A precontraction on a MATH-operad MATH consists of a function MATH for each MATH, such that MATH for all MATH, MATH. A MATH-operad MATH with MATH is merely a plain operad - call it MATH - and a precontraction on MATH consists of a distinguished element of MATH for each MATH. The operad MATH described in REF, together with the element MATH of MATH for each MATH, therefore defines a MATH-operad with precontraction. Using the fact that MATH is the free plain operad on the terminal (free monoid)-graph, it is easy to see that this is the initial object of MATH. By REF , MATH is MATH applied to this initial object: that is, MATH (MATH). In dimension MATH, the MATH-operad structure is as in the plain operad MATH. Given that the source and target functions MATH are first and second projection, the MATH-operad structure in dimension MATH is uniquely determined. This fully describes MATH. An algebra for MATH is, therefore: CASE: a REF-globular set MATH CASE: for each MATH and MATH, a function MATH CASE: for each MATH and MATH, a function MATH satisfying axioms concerning the source and target of MATH in terms of MATH and MATH, together with the axioms for an algebra (which we regard as `all reasonable coherence axioms'). Rephrasing this a little, an algebra for MATH consists of CASE: a set MATH (which is the MATH of the previous paragraph) CASE: for each MATH, a directed graph MATH CASE: for each MATH and MATH, a function MATH CASE: for each MATH, each MATH, and each array of arrows MATH an arrow MATH in MATH, satisfying `all reasonable coherence axioms'. These axioms imply that if MATH then the function MATH is the identity. Now taking MATH and MATH in the fourth item, we have a function which assigns to each string of arrows MATH in MATH an arrow MATH, that is, MATH. This gives the directed graph MATH the structure of a category. By the preliminary comments on functors and natural transformations (page REF), a MATH-algebra therefore consists of CASE: a set MATH CASE: for each MATH, a category MATH CASE: for each MATH and MATH, a functor MATH CASE: for each MATH and MATH, a natural transformation MATH satisfying `all reasonable coherence axioms'. Writing MATH as MATH and MATH as MATH, we see that this is just the description of a (small) unbiased bicategory given by REF and the comments thereafter. MATH . |
math/0011106 | REF is a straightforward induction on the structure of MATH. REF is just the definition of MATH. REF is also straightforward. MATH . |
math/0011106 | As in the previous proof, REF is by induction on MATH, REF is immediate, and REF is straightforward. MATH . |
math/0011106 | REF is immediate, and REF follow from REF . MATH . |
math/0011106 | REF is by induction on MATH; REF are immediate. MATH . |
math/0011106 | It is enough to prove this when MATH, in which case MATH. The proof is then another easy induction on MATH. MATH . |
math/0011106 | These all follow from the last two propositions. MATH . |
math/0011106 | Essentially we have to check that our data satisfies a large collection of axioms, but our coherence results cover almost all of these checks automatically. Here is the list of the things to be checked and which coherence result each one can be inferred from. CASE: MATH is a functor MATH. This means: CASE: MATH is a bicategory for any MATH: CASE: MATH is a lax morphism for any MATH: CASE: MATH preserves identities: CASE: MATH preserves composition: really we should deduce this from `coherence for a composable pair of unbiased lax morphisms' (which we did not prove), but a direct check is easy. CASE: MATH is a functor MATH. This means: CASE: MATH is an unbiased bicategory for any MATH: CASE: MATH is an unbiased lax functor for any MATH: CASE: MATH preserves identities: CASE: MATH preserves composition: as for MATH above. CASE: MATH. This means: CASE: MATH for any MATH: by construction, MATH and MATH are the same in all respects except perhaps their associativity and unit isomorphisms; and these too are equal by REF CASE: MATH for any MATH: CASE: MATH are natural transformations. This means: CASE: MATH are unbiased lax functors for any MATH: CASE: MATH and MATH are natural in MATH: CASE: MATH and MATH for any MATH: CASE: MATH . |
math/0011106 | Consider the diagram where MATH and MATH are as in the proof of REF , and similarly MATH and MATH. Applying MATH to the isomorphism MATH gives an isomorphism `inside' the upper square, making MATH into a weak map of monads MATH . There is also a natural isomorphism inside the lower square, expressing the fact that the free enriched category construction is natural in a suitable sense, and this gives a weak map of monads MATH . (The checks involved here use the fact that MATH is induced by MATH; again, this is an unnecessarily strong hypothesis, but serves our purpose.) Gluing together these two weak maps of monads gives a third weak map of monads, MATH as required. One can easily check that the diagram in the last sentence of the Proposition commutes. MATH . |
math/0011106 | By definition, a locally finitely presentable category is cocomplete, so MATH has an initial object. The functor MATH has a left adjoint (being monadic), which applied to the initial object of MATH gives an initial object of MATH. MATH . |
math/0011108 | The original NAME proof has never appeared, but a proof of the first claim was given in CITE. The second follows purely formally from the first: the functor MATH is an equivalence, so we have MATH . NAME 's lemma completes the argument. |
math/0011108 | Write MATH. Let MATH be the unique MATH-typical formal group law over MATH with MATH, and let MATH be the formal group scheme over MATH associated to MATH. If MATH then there is a unique lift MATH such that MATH, and there is also an endomorphism MATH of MATH given by MATH. We also have an endomorphism MATH of MATH given by MATH. It is well-known that there is a unique ring map MATH such that MATH and MATH for all MATH, and moreover that MATH is an isomorphism. On the other hand, we can let MATH act on MATH via the NAME automorphism MATH. As the coefficients of MATH lie in MATH, there is a natural identification MATH. Using this, it is easy to identify MATH with the group of pairs MATH, where MATH is an isomorphism of schemes and MATH is an isomorphism of formal groups over MATH. Now put MATH and MATH so that MATH is a formal group over MATH and is the universal deformation of MATH in the sense of CITE (see also CITE for an account in the present language.) Let MATH denote the group of pairs MATH where MATH and MATH. As MATH is the subscheme of MATH defined by the unique maximal ideal in MATH we see that MATH. As MATH, we also see that MATH, so we get a homomorphism MATH sending MATH to MATH; deformation theory tells us that this is an isomorphism. The general theory of NAME ring spectra CITE gives an isomorphism MATH; see CITE for an account in the present language. |
math/0011108 | First, we note that MATH is a MATH-adic analytic group of dimension MATH over MATH. Duality phenomena in the cohomology of profinite groups have been studied for a long time CITE, but the more recent paper CITE is the most convenient reference for the particular points that we need. Write MATH, which is a torsion-free open subgroup of finite index in MATH. It follows from CITE that MATH is a NAME duality group of dimension MATH in the sense used in that paper, which means precisely that MATH has cohomological dimension MATH and MATH and the other cohomology groups are trival. As MATH has finite index in MATH, NAME 's lemma gives an isomorphism MATH; this proves the proposition except for the fact that MATH acts trivially. There is a unique way to let MATH act on MATH such that the subgroup MATH acts by conjugation and the subgroup MATH acts via MATH. On the other hand, MATH acts on MATH by conjugation and thus on the MATH-Lie algebra MATH of MATH. There is an evident MATH-equivariant isomorphism MATH. Using the results of CITE we get a MATH-equivariant isomorphism MATH . (This is just a tiny extension of an argument of NAME, which could be applied directly if MATH were torsion-free.) We write MATH for the determinant of the action of MATH on MATH; it is now enough to check that MATH. Suppose MATH, so MATH acts on MATH by MATH. Let MATH be the subfield of MATH generated over MATH by MATH. Put MATH, so that MATH as left MATH-modules. Using this, we see that the determinant of left multiplication by MATH is just MATH. By a similar argument, the determinant of right multiplication by MATH is MATH. It follows that the conjugation map has determinant one, so that MATH. Moreover, the action of MATH on MATH is the same as the action of MATH by conjugation, which has determinant one by the same argument. |
math/0011108 | Suppose REF holds, so MATH say. Here the coproduct is the MATH-localisation of the ordinary wedge, which means that MATH, where MATH runs over a suitable family of open ideals in MATH. In this second expression, it makes no difference whether the wedge is taken in MATH or in the category of all spectra, so MATH, and it follows easily that MATH is the completion of MATH and MATH. Thus REF . Conversely, if REF holds, choose a topological basis MATH for MATH and use it to construct a map MATH of MATH-modules in the usual way. Then MATH is the completion of MATH and so MATH is an isomorphism, so REF holds. Using this we find that for any MATH-local MATH-module MATH, the group of MATH-module maps MATH is just MATH. If we write MATH for the category of MATH-modules satisfying REF , it is now clear that MATH is an equivalence. We know from CITE that MATH is closed under arbitrary products and retracts, and that any pro-free MATH-module is a retract of a product of copies of MATH. Given this, we can easily deduce that MATH . |
math/0011108 | If MATH then this holds by a well-known compactness argument based on the fact that MATH is finite for all small MATH. As any MATH can be written as a retract of a product of copies of MATH, the claim follows in general. |
math/0011108 | It is easy to reduce to the case MATH. By the argument of CITE it suffices to check that MATH is concentrated in even degrees. We know by standard calculations that MATH (with MATH) and that MATH by NAME exactness, and the claim follows. |
math/0011108 | Write MATH; the reduction map MATH is well-known to be an isomorphism. For any MATH there is a unique way to write MATH with MATH and MATH. This defines functions MATH, and MATH is constant on the cosets of MATH and thus is locally constant. NAME interpolation shows that the evident map from MATH to the ring of all functions MATH is surjective, and thus an isomorphism by dimension count. Our maps MATH give a bijection MATH . Putting these facts together, we see that the ring of functions from MATH to MATH is generated over MATH by the functions MATH subject only to the relations MATH and MATH. The direct limit of these rings as MATH tends to MATH is the ring of all locally constant functions from MATH to MATH, which is thus isomorphic to MATH . Now let MATH be the characteristic function of MATH (for MATH), so the ring of functions from MATH to MATH is just MATH. Recall that MATH; this can be identified with MATH as a set, so the ring of locally constant functions from MATH to MATH is just the tensor product of the rings for MATH and MATH, which we now see is a free MATH-module. The ring of all continuous functions is the completion of the ring of locally constant functions, and thus is pro-free. |
math/0011108 | Consider a map MATH and the resulting map MATH, which can also be thought of (using the evident action of MATH on MATH) as the map MATH. The set MATH is finite by an easy thick subcategory argument CITE so the stabiliser of MATH has finite index in MATH. As the subgroup MATH is a finitely topologically generated pro-MATH group of finite index in MATH, we see from CITE that every finite index subgroup of MATH is open. It now follows easily that the map MATH is continuous when MATH is given the discrete topology. It follows in turn that if we give MATH the profinite topology and MATH the natural topology then the inclusion map is continuous. We also know from CITE that MATH is NAME. A continuous bijection from a compact space to a NAME space is always a homeomorphism, and the lemma follows. |
math/0011108 | Define a map MATH by MATH . Using CITE we see that MATH is continuous, so we have an adjoint map MATH. We claim that this is an isomorphism. As both source and target are pro-free, it suffices to show that the reduction of MATH modulo MATH is an isomorphism. Let MATH be the representing spectrum for the functor MATH; this is a wedeg of finitely many suspended copies of MATH, and it can be made into a MATH-algebra spectrum with MATH. It is not hard to see that MATH and MATH. To analyse MATH, let MATH be the standard MATH-typical coordinate on MATH and let MATH be the resulting formal group law over MATH - see CITE for details and useful formulae. A standard NAME exactness argument shows that MATH is the universal example of a ring MATH equipped with maps MATH and an isomorphism MATH of formal group laws. As MATH has height MATH we see that the same must be true of MATH, so MATH, so we can regard MATH as a map MATH. The coefficients of MATH modulo MATH actually lie in MATH and there is only one map MATH so MATH; we just write MATH for this formal group law. Using the standard form for isomorphisms of MATH-typical FGL's we can write MATH, where MATH is invertible because MATH is an isomorphism. (Readers may be more familiar with the graded case where one gets strict isomorphisms with MATH, but we are working with the degree zero part of two-periodic theories and MATH need not be MATH in this context.) As MATH commutes with MATH we have MATH and thus MATH. In fact, this condition is sufficient for MATH to be a homomorphism of FGL's (see CITE, for example) and we deduce that MATH . We next claim that MATH can be identified with the ring MATH of functions from the NAME group MATH to MATH. Indeed, we can define a map MATH by MATH. The MATH-linear dual of this is the evident map MATH which is injective by NAME 's lemma on the independence of automorphisms, and this bijective by dimension count. Thus MATH is an isomorphism, and we obtain an isomorphism MATH. After some comparison of definitions we see that this is the same as the map MATH, as required. |
math/0011108 | Given MATH, we define MATH . We then define MATH by MATH. Note that if MATH then MATH; this means that the map MATH is MATH-equivariant. The results of CITE show that MATH is continuous, so we can regard MATH as a map MATH. Now let MATH be the dual of a generalised NAME spectrum CITE of type MATH for some ideal MATH so that MATH. We know from REF that MATH and that this is a free module over MATH so that MATH . After some comparison of definitions, we find that MATH is an isomorphism when MATH. Moreover, the construction MATH gives an exact functor from finite discrete Abelian groups to Abelian groups, so the construction MATH gives a cohomology theory on the category of small MATH-local spectra. By a thick subcategory argument, we deduce that MATH is an isomorphism when MATH is small. Now let MATH be a general MATH-local spectrum, and let MATH be the diagram of small spectra over MATH. As MATH is a pro-free MATH-module we see from REF that MATH. We also see that MATH, and it follows that MATH is an isomorphism as claimed. |
math/0011108 | The axiomatic treatment of NAME resolutions discussed in CITE can be transferred to many other triangulated categories; this will certainly work for unital stable homotopy categories in the sense of CITE, and thus for MATH. The spectra MATH clearly form a MATH-Adams resolution of MATH, so we have a spectral sequence whose MATH page is as described. Let MATH be the fibre of the unit map MATH; it is known that our spectral sequence is associated to the filtration of MATH by the spectra MATH. The map MATH is split by the product map MATH, so MATH. It follows that if MATH lies in the thick subcategory generated by MATH then the map MATH is zero for MATH. However, we know from CITE that MATH lies in this thick subcategory, so MATH for MATH. Using this and the definition of our spectral sequence, we see that it converges strongly to MATH. |
math/0011108 | We use the spectral sequence of REF . To analyse the MATH term, we claim that MATH as left MATH-modules, where we let MATH act in the obvious way on MATH and trivially on MATH. To see this, define maps MATH and MATH by MATH . Clearly MATH and MATH and MATH (because MATH is a ring map). By dualising REF we see that MATH extends to give an isomorphism MATH, and it follows from the above formulae that MATH extends to give an isomorphism MATH with the required equivariance. Next, using the obvious description of MATH in terms of monomials, we see that MATH is isomorphic as a topological MATH-module to a product of copies of MATH, say MATH. It follows that MATH as MATH-modules, and using this that MATH . These identifications can easily be transferred to nonzero degrees, and they respect the cosimplicial structure, so it now follows from REF that in our spectral sequence we have MATH . As the spectral sequence is strongly convergent, we have MATH as claimed. It is not hard to check that this is an isomorphism of MATH-modules, and it follows in the usual way that MATH as MATH-module spectra. One can see from the construction that this is compatible with the action of MATH. |
math/0011108 | First, we have MATH . We can deduce from the above by a thick subcategory argument that MATH is finite in each degree and thus that MATH is dualisable. (We could also have quoted this from CITE; the proof given there is only a slight perturbation of what we've just done.) This implies that MATH for all MATH. In particular, we have MATH so that MATH as claimed. |
math/0011108 | This is an instance of the well-known principle of invariance of residues. Most of the formulae involved are very old, and were originally interpreted in a complex analytic context; starting in REF's they were transferred into algebraic geometry CITE but only in the very recent paper CITE do they appear in the particular technical context that we need. First note that MATH is the smallest closed subring of MATH containing all the MATH'th roots of unity in MATH, so it is preserved by MATH and in particular by MATH. REF of the cited paper gives a canonical map MATH and the short exact sequence MATH gives an isomorphism MATH and MATH gives a canonical map MATH. By putting all these together, we get a map MATH. All the constructions involved are functorial and thus MATH-invariant. By examining the formulae in CITE we see that MATH. |
math/0011108 | This is essentially the local duality theorem (see CITE for example). We first translate the claim using the isomorphism MATH; it now says that MATH. Using the evident isomorphism MATH we can reduce to the case MATH. In that case the construction above gives a MATH-invariant element MATH and thus a MATH equivariant map MATH, defined by MATH. This satisfies MATH . As the map MATH is a perfect pairing, it is easy to conclude that MATH is an isomorphism. |
math/0011108 | See CITE (which relies heavily on CITE). |
math/0011108 | This follows from REF after noting that MATH. |
math/0011109 | Put MATH; we need to show that this is an inner product. Put MATH; it will suffice to check the identities in REF . The symmetry conditions are clear, so we just need the two compatibility conditions for MATH and MATH. One of them is proved as follows: MATH . The first equation is just the definition of MATH and MATH, the second is the interchange axiom, and the third uses the (co)unit properties of MATH and MATH. The other compatibility condition follows because MATH and MATH are symmetric. |
math/0011109 | By REF is an inner product on MATH, and trivially the canonical isomorphism MATH is an inner product on MATH. We can thus define MATH, so MATH is the unique map such that MATH, or in other words the unique map giving the following equality of NAME diagrams: MATH . We claim that MATH is a unit for MATH, or in other words that we have the following equality: MATH . To prove this, we observe that for any two maps MATH we have MATH if and only if MATH . In view of this, the claim follows from the following diagram, in which the first equality comes from the associativity of MATH and the second from the defining property of MATH. MATH . We next equip MATH with the inner product MATH and define MATH. As MATH is a commutative and associative monoid object, it is easy to deduce that MATH is a commutative and associative comonoid object. Thus, to prove that MATH is a NAME object, we need only check the interchange axiom. It follows directly from the definition that MATH is the unique map giving the following equality: MATH . Using the perfectness of MATH, we see that two maps MATH are equal if and only if we have MATH . In view of this, the interchange axiom is equivalent to the following equation: MATH . This equation can be proved as follows: MATH . The first equality uses associativity of MATH, the second uses the defining property of MATH, and the third uses the same two ideas backwards. We still need to check that MATH and MATH are the unique maps giving a NAME structure. For MATH this is easy, because the unit for a commutative and associative product is always unique. For MATH, suppose that MATH is another map giving a NAME structure. We then have the following equations: MATH . The first equality is the interchange axiom, the second is the counit property of MATH, and the third is the associativity of MATH. This shows that MATH has the defining property of MATH, so MATH as required. |
math/0011109 | This is implicit in the proof of the proposition. |
math/0011109 | We saw in the proof of the proposition that MATH is the unique map giving the following equality of NAME diagrams. MATH . The claim follows by working in the opposite category and using REF . |
math/0011109 | The adjunction between the functors MATH and MATH is given by two maps MATH and MATH. It follows from the basic theory of pairings and duality CITE that the following diagrams commute: It follows that the following diagram commutes: On the bottom row, the composite of the first two maps is MATH so the whole composite is just MATH. Thus, MATH. To complete the proof, it is easiest to think in terms of TQFT's. Let MATH be a torus with a small open disc removed. We leave it to the reader to check that MATH is represented by MATH, considered as a cobordism from MATH to MATH. Moreover, the maps MATH, MATH and MATH are all represented by MATH considered as a cobordism from MATH to MATH. The proposition follows. |
math/0011109 | If we make MATH into a symmetric monoidal category using the cartesian product, it is clear that MATH and MATH make MATH into a comonoid object. As MATH and MATH are monoidal functors, the first covariant and the second contravariant, we see that MATH is a monoid object under MATH and MATH, and a comonoid object under MATH and MATH. For the interchange axiom MATH, we note that the following diagram is a transverse pullback and apply the NAME property. Similarly, to prove that MATH, we note that the following square is a transverse pullback: We then apply the NAME property, noting that MATH; this gives the following commutative diagram: We then compose with MATH, noting that MATH and MATH and MATH. We conclude that MATH, so MATH as claimed. |
math/0011109 | Let MATH be a connected groupoid. Choose an object MATH and let MATH be the group MATH. Let MATH be the indiscrete groupoid with MATH, and for each MATH choose a map MATH in MATH. Put MATH, so MATH and MATH for all MATH. Composition is given by multiplication in MATH. Define MATH by MATH on objects, and MATH on morphisms. This is easily seen to be functorial and to be an isomorphism. The generalisation to the disconnected case is immediate. |
math/0011109 | We need to verify the following axioms, numbered as in CITE: CASE: MATH has finite limits and colimits. CASE: If we have functors MATH and two of MATH, MATH and MATH are weak equivalences then so is the third. CASE: Every retract of a weak equivalence is a weak equivalence, and similarly for fibrations and cofibrations. CASE: NAME have the left lifting property for acyclic fibrations, and acyclic cofibrations have the left lifting property for all fibrations. CASE: Any functor MATH has factorisations MATH where MATH and MATH are cofibrations, MATH and MATH are fibrations, and MATH and MATH are equivalences. MCREF: This follows from the fact that MATH is the category of models for a left-exact sketch CITE. More concretely, for limits we just have MATH and MATH. Similarly, for coproducts we have MATH and MATH. NAME are more complicated and best handled by the adjoint functor theorem. MCREF: This is easy. MCREF: Let MATH be an equivalence and let MATH be a retract of MATH. Then MATH is a retract of MATH, so MATH is a bijection and so MATH is essentially surjective. If MATH then MATH is a retract of a map of the form MATH and thus is a bijection, so MATH is full and faithful. Thus MATH is an equivalence as required. It is clear that a retract of a cofibration is a cofibration. For fibrations, let MATH be the terminal groupoid. Let MATH be the groupoid with objects MATH and two non-identity morphisms MATH and MATH. Let MATH be the inclusion of MATH. Then fibrations are precisely the maps with the right lifting property for MATH, and it follows that a retract of a fibration is a fibration. MCREF: Consider a commutative square as follows, in which MATH is a cofibration and MATH is a fibration. Because MATH is a fibration, it is easy to see that the image of MATH is replete: if MATH is isomorphic to MATH then MATH has the form MATH for some MATH. Suppose that MATH is an equivalence; we must construct a functor MATH such that MATH and MATH. As MATH is essentially surjective and the image is replete, we see that MATH is surjective. By REF is a cofibration so MATH is injective. Define a map MATH by putting MATH for MATH and choosing MATH to be any preimage under MATH of MATH if MATH. Clearly MATH and MATH on objects. Given MATH we define MATH to be the composite MATH . One can check that this makes MATH a functor with MATH. Also MATH on morphisms and MATH on objects and MATH is faithful; it follows that MATH on morphisms, as required. Now remove the assumption that MATH is an equivalence, and suppose instead that MATH is an equivalence. We must again define a functor MATH making everything commute. As MATH is injective on objects we can choose MATH with MATH. As MATH is a bijection we find that MATH so we can choose isomorphisms MATH for all MATH. If MATH for some (necessarily unique) object MATH, we have MATH and MATH, and we choose MATH in this case. There is a unique way to make MATH a functor MATH such that MATH is natural: explicitly, the map MATH is the composite MATH . Next, if MATH we define MATH and MATH. If MATH we instead apply the fibration axiom for MATH to the map MATH to get an object MATH and a morphism MATH such that MATH and MATH. Note that these last two equations also hold in the case MATH. There is a unique way to make MATH into a functor such that MATH is natural. Clearly MATH as functors, and MATH on objects. Given MATH in MATH we can apply MATH to the naturality square for MATH and then use the naturality of MATH to deduce that MATH; thus MATH on morphisms, as required. MCREF: Consider a functor MATH. Let MATH be the category whose objects are triples MATH, with MATH and MATH and MATH. The morphisms from MATH to MATH are the pairs MATH where MATH and MATH and the following diagram commutes: We also consider the category MATH with the same objects as MATH, but with MATH, so there is an evident functor MATH. There is also a functor MATH given by MATH and a functor MATH given by MATH; we put MATH and MATH. It is clear that MATH and that MATH and MATH are cofibrations and that MATH is full and faithful. If MATH then MATH so MATH is essentially surjective and thus an equivalence. The functor MATH is clearly full and faithful, and its image is the repletion of the image of MATH. We next claim that MATH and MATH are fibrations. Suppose that MATH and MATH. Then MATH and MATH and MATH. This shows that MATH is a fibration, and the same construction also shows that MATH is a fibration. We now have a factorisation MATH as required by REF . If MATH is essentially surjective then the same is true of MATH and thus MATH is an equivalence and so the factorisation MATH is also as required. If MATH is not essentially surjective then we let MATH be the full subcategory of MATH consisting of objects not in the repletion of the image of MATH and let MATH be the inclusion. We then have an acyclic fibration MATH and a cofibration MATH whose composite is MATH, as required. |
math/0011109 | Consider a pullback square as follows, in which MATH is a weak equivalence and MATH is a fibration. Suppose that MATH and put MATH and MATH. By the construction of pullbacks in MATH, we see that the following square is a pullback square of sets: As MATH is a weak equivalence, the map MATH is a bijection, and it follows that the same is true of MATH. This means that MATH is full and faithful. Next suppose we have MATH, so MATH. As MATH is essentially surjective there exists MATH and MATH in MATH. As MATH is a fibration there is a map MATH in MATH with MATH and MATH. By the pullback property there is a unique MATH with MATH and MATH. Thus MATH, proving that MATH is essentially surjective and thus an equivalence. |
math/0011109 | If there exists such a map MATH then the square is visibly equivalent in MATH to a homotopy pullback square, and thus is homotopy cartesian. For the converse, suppose that the square is homotopy Cartesian. We can then find a diagram as follows which commutes in MATH, such that the outer square is a homotopy pullback, and the diagonal functors are equivalences. There is a ``tautological" natural isomorphism MATH, and we write MATH so that MATH. As the top and left-hand regions of the diagram commute in MATH, we have natural maps MATH and MATH, which we can use to form a natural map MATH . Using the remark in the preceeding definition, we see that MATH. As MATH is an equivalence, we see that the same is true of MATH. Next, we note that the functors MATH are joined by the natural map MATH where the first and third maps come from the commutativity of the right-hand and bottom regions of the diagram. This gives a functor MATH; we leave it to the reader to check directly that this is an equivalence. Next, consider the composite MATH . As MATH is full and faithful, this composite has the form MATH for a unique natural map MATH, which gives rise to MATH. One checks directly that MATH, and both MATH and MATH are equivalences, so MATH is an equivalence, as required. |
math/0011109 | Suppose for the moment that MATH is indiscrete and MATH is connected. Then for MATH we have MATH and MATH is injective but the codomain has only one element so the same is true of MATH. Thus MATH is full and faithful. It is also a fibration and MATH is connected so it is surjective on objects. If MATH then the unique map MATH in MATH must become an identity map in MATH but MATH reflects identities so MATH. We now see that MATH is an isomorphism so MATH is a homeomorphism and thus certainly a covering. If MATH is indiscrete and MATH is disconnected, we can still show that MATH is a covering by looking at one component at a time. Now suppose merely that MATH is connected. We can then split MATH as MATH, where MATH is indiscrete and MATH is a group, as in REF . Let MATH be the indiscrete category with object set MATH, and define MATH by sending the unique morphism MATH to MATH. One checks that MATH and that MATH is the usual covering map. Thus, MATH is indiscrete and MATH is a covering with the property that MATH is also a covering. Now form a pullback square as follows: Note that MATH is a covering. As MATH is indiscrete we know that MATH is a covering by the first paragraph. Thus, the pullback of MATH along the surjective covering map MATH is a covering, and it follows easily that MATH is a covering. Finally, if MATH is disconnected we just look at one component at a time. |
math/0011109 | This is a simple translation of NAME 's analysis of coverings of MATH. Suppose we start with a functor MATH. We then define a category MATH whose objects are pairs MATH with MATH and MATH; the morphisms MATH are the maps MATH in MATH such that MATH sends MATH to MATH. There is an evident forgetful functor MATH sending MATH to MATH; one checks that this is a covering. Conversely, suppose we start with a covering MATH. For each MATH, we define MATH. Given a morphism MATH in MATH and an element MATH, the definition of a covering gives a unique morphism MATH in MATH with MATH; we define a map MATH by MATH. We leave it to the reader to check that these constructions give the claimed equivalence. |
math/0011109 | CASE: This is clear. CASE: Let MATH be a functor. We define a new groupoid MATH as follows. The objects are equivalence classes of triples MATH, where MATH and MATH and MATH; the equivalence relation identifies MATH with MATH if and only if MATH and there is a map MATH such that MATH. The maps from MATH to MATH are the maps MATH in MATH such that there exists a map MATH in MATH with MATH. Equivalently, MATH gives a map MATH if and only if MATH. There is an evident functor MATH defined by MATH. Given MATH we find that MATH can be thought of as a map MATH in MATH, so MATH is essentially surjective. Moreover, we find that MATH is just the image of MATH in MATH, and thus that MATH is full. Thus we have MATH. There is also an evident functor MATH defined by MATH. It is easy to check that MATH is a covering and MATH as required. In terms of REF , the covering MATH corresponds to the functor MATH defined by MATH. CASE: Suppose we have a square as in the statement of the proposition. We first define a map MATH as follows. Suppose that MATH. As MATH is essentially surjective, we can choose MATH and MATH in MATH. We apply MATH to get MATH. As MATH is a covering, there is a unique pair MATH with MATH and MATH such that MATH and MATH. We would like to define MATH. To check that this is well-defined, consider another MATH and another MATH, giving rise to a unique pair MATH. As MATH is full there exists MATH such that MATH and one checks that MATH has the defining property of MATH. Thus MATH as required. This means that we have a well-defined map MATH with MATH on objects. It is easy to check that MATH on objects as well. Now suppose we have a map MATH in MATH. We can choose maps MATH and MATH with MATH. By the definition of MATH on objects we have maps MATH and MATH such that MATH and MATH. As MATH is full we can choose MATH such that MATH. One then checks that the map MATH has MATH. As MATH is faithful, there is at most one map MATH with this property, so MATH is independent of the choices made. We can thus define MATH on morphisms by MATH, so that MATH. Using the faithfulness of MATH, we check easily that MATH is a functor and that MATH. Thus MATH fills in the diagram as required. Finally suppose that MATH is another functor making the diagram commute. We must check that MATH. As MATH is faithful it is enough to check this on objects. Given MATH we choose MATH as before and write MATH and MATH. We then have MATH, so the definition of MATH gives MATH as required. |
math/0011109 | See CITE. Of course, in our case, many of these things are immediate from the definitions. |
math/0011109 | As equivalences and coverings are faithful, we see that quasi-coverings are faithful. For the converse, let MATH be faithful. We can factor MATH as MATH where MATH is a covering and MATH is full and essentially surjective, as in REF . As MATH is faithful we see that MATH is faithful and thus an equivalence, as required. |
math/0011109 | We can factor MATH as MATH with MATH a covering and MATH full and essentially surjective. Now consider the following diagram: The square is defined to be the pullback of MATH and MATH, and MATH is the unique functor such that MATH and MATH. By REF we know that MATH is an equivalence. It will thus be enough to show that MATH is an isomorphism in MATH. As MATH is full and essentially surjective, and MATH is an equivalence, we see that MATH is full and essentially surjective. We next show that MATH is surjective on objects. Suppose MATH, and put MATH and MATH so that MATH say. As MATH is essentially surjective, we can choose MATH and MATH in MATH. Thus MATH in MATH. As MATH is an equivalence, there is a unique MATH such that MATH. As MATH is a covering, there is a unique MATH and MATH such that MATH and MATH. Thus MATH satisfies MATH. As MATH is a covering, it reflects identity maps, so MATH and MATH. Thus MATH is surjective on objects, as claimed. Now consider the following diagram: It follows from REF that there is a unique map MATH making everything commute. In particular, we have MATH. It follows that MATH and MATH is full and surjective on objects so MATH. Thus MATH is an isomorphism, as required. |
math/0011109 | Using REF , it is not hard to reduce to the case in which MATH is the standard homotopy pullback of MATH and MATH. As MATH is a quasicovering we can factor it as MATH where MATH is a covering and MATH is an equivalence. We can then define MATH, MATH and MATH so that the bottom square is cartesian, which implies that MATH is a covering. Our next task is to define the functor MATH. An object MATH is a triple MATH. As MATH is a covering and MATH, we see that there is a unique morphism MATH in MATH such that MATH and MATH. Thus MATH and we can define MATH on objects by MATH. Note that MATH and MATH so we can define MATH . Next, consider a morphism MATH in MATH, where MATH for MATH. We define MATH and MATH as above, and define MATH so that the following diagram commutes. We now define MATH on morphisms by putting MATH. It is easy to check that this makes MATH into a functor, and that MATH is a natural map. Thus, the top square in our diagram commutes up to homotopy. It is also clear that MATH. Thus, all that is left is to check that MATH is an equivalence. Let MATH and MATH be as above, and suppose given MATH. As MATH is faithful and MATH is an equivalence, we see that there is at most one map MATH making the upper trapezium of the above diagram commute, and at most one map MATH making the lower trapezium commute. Moreover, MATH exists if and only if MATH does, and they determine each other by MATH and MATH. Note also that MATH is the set of pairs MATH such that the top trapezium commutes, and MATH is the set of pairs MATH making the bottom trapezium commute. It follows easily that MATH is full and faithful. Now consider an object MATH, so MATH. As MATH is essentially surjective we can choose an object MATH and a map MATH in MATH. We thus have an object MATH of MATH. Clearly MATH so MATH is surjective on objects, and thus an equivalence as claimed. |
math/0011109 | By well-known results of NAME, the space MATH also has the homotopy type of a CW complex. Evaluation at the basepoint of MATH gives a surjective NAME fibration MATH whose fibres have the form MATH. As MATH we know that these fibres are acyclic and so MATH is a weak equivalence, and thus a homotopy equivalence. By a standard result (the dual of CITE, for example) we deduce that MATH is fibre-homotopy equivalent to MATH. One can also see that for any MATH we have MATH and our fibre-homotopy equivalence shows that this is contractible. The result follows. |
math/0011109 | It follows from the lemma that MATH. Recall that MATH is an equivalence. Thus, for any MATH we have MATH . As MATH is an equivalence we conclude that MATH for all MATH, and it follows by NAME 's lemma that MATH as claimed. |
math/0011109 | We can easily reduce to the case where MATH is a group rather than a groupoid. It was observed in the proof of of CITE that a certain map MATH (arising from the NAME theory of generalised NAME spectra) is an isomorphism. It is thus enough to show that MATH. We will need some notation. Firstly, we will need to consider various unlocalised spectra, so in this proof only we write MATH for the ordinary, unlocalised sphere spectrum, and MATH for MATH. Similarly, we write MATH for the unlocalised smash product and MATH. Next, we will work partially in the equivariant categories of MATH-spectra and MATH-spectra, indexed over complete universes CITE. We write MATH and MATH for the corresponding MATH-sphere objects. Also, we can regard MATH as a naive MATH-spectrum with trivial action and then extend the universe to obtain a genuine MATH-spectrum, which we denote by MATH. We define a genuine MATH-spectrum MATH in the analogous way. We next recall the definition of MATH. It is obtained from a certain map MATH by observing that MATH is MATH-local and that any map from MATH to a MATH-local spectrum factors uniquely through MATH. It will be enough to check that MATH is adjoint to MATH, where MATH is the composite MATH . We thus need to show that two elements of the group MATH are equal. REF (applied to MATH) gives a natural isomorphism MATH . Let MATH and MATH be the collapse maps. NAME Construction II. REF gives a pretransfer map MATH of genuine MATH-spectra. By smashing this with MATH and passing to orbits we get the transfer map MATH. Using this and the proof of CITE we find that MATH corresponds to the composite MATH in MATH. We now return to the definition of MATH. We have a map MATH of MATH-spectra. We next apply the fixed point functor, noting that MATH and that CITE gives an equivalence MATH. The resulting map MATH is MATH (see CITE). To understand this better, we need to follow through the construction of MATH. We use the notation of CITE, noting that in our case we have MATH. The construction uses the group MATH, the semidirect product of MATH with MATH using the action by conjugation. There are two natural maps MATH given by MATH and MATH. In our case we find that the resulting map MATH is an isomorphism, so we can replace MATH by MATH everywhere. The subgroup MATH becomes MATH, the standard embedded copy MATH of MATH becomes MATH, and the maps MATH and MATH become the projections MATH. The relevant spectrum MATH is MATH, so MATH and MATH. The map MATH is obtained from MATH by shrinking the universe, passing to orbits and adjointing as described in CITE. It follows that MATH is obtained from the composite MATH by a similar procedure. We can identify MATH with MATH, and we find that the adjoint of MATH is obtained by applying another similar procedure to the map MATH . This procedure amounts to just composing with MATH and using our isomorphism MATH. It follows that the adjoint of MATH is MATH, as required. |
math/0011109 | Let MATH be the diagonal map. We first claim that the following square is homotopy-cartesian: To see this, let MATH be the homotopy pullback of the functors MATH and MATH. The square is clearly cartesian, which means that MATH embeds as a full subcategory of MATH; we need only check that the inclusion is essentially surjective. The objects of MATH are MATH-tuples MATH where MATH and MATH and MATH and MATH. The maps from MATH to MATH are triples MATH making the following diagram commute: The canonical functor MATH is given by MATH. We define MATH by MATH. Then MATH, and we have a natural map MATH given by MATH. This proves that MATH is an equivalence, and if we compose it with the projections MATH and MATH we get the functors MATH and MATH. This proves that our original square is homotopy-cartesian, so the NAME property tells us that MATH . We now use the fact that MATH and compose with the projection MATH to get MATH . We next note that MATH and MATH-localise to conclude that MATH, as required. |
math/0011109 | We may assume that the square is actually a pullback square of fibrations (see REF ), so in particular it commutes on the nose. As MATH is a perfect pairing, it suffices to check that MATH. By transposition, this is equivalent to MATH . To verify this, we consider the following diagram: We claim that this is homotopy-cartesian. It is clearly cartesian, so it suffices (as in the previous proof) to show that the obvious functor from MATH to the homotopy pullback is essentially surjective. Suppose we are given an object of the homotopy pullback, in other words a MATH-tuple MATH where MATH, MATH, MATH and MATH. As MATH is a fibration we can choose MATH and MATH such that MATH and MATH. Thus MATH is an object of MATH and the following diagram gives an isomorphism MATH: This shows that our square is homotopy-cartesian. The vertical functors are faithful and thus are quasicoverings, so the NAME property tells us that MATH . We next compose with the map MATH, noting that MATH and that MATH. We conclude that MATH, as required. |
math/0011109 | This is formally identical to the proof of REF ; we need only check that the following square is homotopy-cartesian, and that is easy. Alternatively, the result can be deduced from the proof of REF . |
math/0011109 | The key point is that the following square is homotopy-cartesian: To see this, let MATH be the homotopy pullback of MATH and MATH. The objects of MATH are tuples MATH where MATH and MATH. The morphisms from MATH to MATH are pairs MATH where MATH and MATH and MATH and MATH. We can define a functor MATH by MATH and a functor MATH in the opposite direction by MATH. We find that these are equivalences and that either projection MATH composed with MATH is just MATH; it follows that the square is homotopy-cartesian, as claimed. We conclude that MATH. We also know from REF that MATH and MATH. Everything now follows from the evident fact that MATH. |
math/0011109 | We have a commutative diagram as follows, which is easily seen to be both cartesian and homotopy-cartesian: The vertical functors are faithful and thus are quasicoverings. The NAME property now tells us that MATH as claimed. Next, consider the groupoid MATH. It is easy to see that this is just a disjoint union of MATH copies of MATH, and that the functor MATH just sends each copy isomorphically to MATH. The remaining claims now follow easily from REF . |
math/0011109 | For any MATH-good group MATH, we can define MATH . We see from Scholium REF that MATH is the unique map such that MATH. Now take MATH, so MATH and MATH. Write MATH; we know from CITE that MATH (a simpler proof appears in CITE). Put MATH; it follows from REF that MATH. Now consider the form MATH so that MATH. In view of the above, it will suffice to check that MATH. For this, we note that MATH so MATH so MATH so MATH. Now, MATH for some power series MATH with MATH and this differential is invariant under translation, which implies that MATH also. Thus MATH as required. |
math/0011109 | For the integral two-periodic version of MATH we have MATH. When MATH it follows easily that MATH. We also have MATH so MATH and the claim follows easily. In the case MATH we have MATH and the stated formula follows in the same way. |
math/0011109 | We may assume that the square is actually a pullback square of fibrations (see REF ), so in particular it commutes on the nose. Fix MATH, so MATH. We need to check that MATH. Because MATH is a fibration, any isomorphism class in MATH that maps to MATH in MATH has a representative MATH such that MATH. Using this, we find that MATH . We also know that MATH is a fibration, so every isomorphism class in MATH that maps to MATH contains a representative MATH with MATH, in other words MATH has the form MATH for some MATH with MATH. It follows that MATH . Fix MATH with MATH. The coefficient of MATH in MATH is then MATH. For MATH we need to be more careful, because there will typically be objects MATH with MATH but MATH. Put MATH where MATH iff there exist MATH and MATH such that MATH. It is easy to see that MATH iff MATH in MATH, and it follows that the coefficient of MATH in MATH is MATH . To analyse this further, we introduce the set MATH where MATH iff MATH. Using the fact that MATH is a fibration, one checks that the map MATH gives a bijection MATH, so that MATH. On the other hand, there is an evident projection MATH sending MATH to MATH. If MATH then we can choose MATH and MATH such that MATH, and then observe that there is a unique MATH such that MATH. One checks that the coset MATH depends only on the equivalence class MATH and that this construction gives a bijection MATH . Note also that MATH because MATH. It follows that MATH . This is the same as the coefficient of MATH in MATH, as required. |
math/0011109 | We first construct a map MATH of MATH-algebras. By juggling various adjunctions we see that it suffices to construct, for each functor MATH, a map MATH of MATH-algebras, such that MATH when MATH is isomorphic to MATH. We know from our previous remarks that MATH must factor through MATH for some MATH. We thus get a map MATH, and we know from CITE that MATH, and MATH is a quotient of this ring, so we get the required map MATH as the composite MATH . One checks easily that this is independent of the choice of MATH. Isomorphic functors MATH give homotopic maps MATH and thus MATH as required. The resulting map MATH is easily seen to be natural for functors of groupoids and to convert equivalences to isomorphisms. Both source and target of MATH convert disjoint unions to products. Any finite groupoid is equivalent to a finite disjoint union of finite groups, so it suffices to check that MATH is an isomorphism when MATH is a group. This is just CITE. To say that this isomorphism respects the construction MATH is just to say that MATH is a natural map, which is clear. We also need to check that for any functor MATH, the following diagram commutes: We first make this more explicit. The functor MATH induces MATH. By applying MATH and noting that MATH we get a map MATH. After tensoring with MATH we obtain the left hand vertical map in the above diagram. On the other hand, MATH also induces a functor MATH and thus a map MATH. By dualising and tensoring with MATH we obtain the right hand vertical map. We first prove that the diagram commutes when MATH is a quasi-covering. This reduces easily to the case where MATH is a group and MATH is connected. It is not hard to see that in this case MATH is equivalent to the inclusion of a subgroup MATH and MATH is just the MATH-localisation of the transfer map MATH. It follows from CITE that MATH, where the sum runs over cosets MATH such that MATH maps MATH into MATH. The right hand side can be rewritten as MATH, where the sum now runs over elements rather than conjugacy classes. Fix a homomorphism MATH that becomes conjugate to MATH in MATH. Then the number of MATH's for which MATH is the order of the group MATH, so MATH . If we want to index this sum using conjugacy classes of MATH's rather than the MATH's themselves, we need an extra factor of MATH, the number of conjugates of MATH. This gives MATH . On the other hand, MATH is just the automorphism group of MATH in the category MATH, so the map MATH is given by MATH . The claim follows easily by comparing these formulae. We have an inner product on MATH obtained from the inner product MATH on MATH, and an inner product on MATH obtained from the standard inner product on MATH for any MATH. By taking MATH to be the diagonal functor MATH in the above discussion, we see that our isomorphism MATH converts the former inner product to the latter one. Thus MATH is compatible with taking adjoints and with the construction MATH, so it is compatible with the construction MATH as well. |
math/0011109 | The isomorphism MATH can be proved either by reducing to the case of a group and quoting CITE, or by taking the fixed-points of both sides in REF under the action of MATH. From the latter point of view, the term in MATH indexed by MATH corresponds to the terms in MATH coming from homomorphisms MATH with image MATH, so MATH becomes the function MATH that sends MATH to MATH. REF identifies this with MATH, as required. |
math/0011110 | We prove by induction on MATH that there exists a relation among the entries of any MATH cells, namely MATH . The above relation is certainly true for MATH. We evaluate, for MATH, MATH . Using the boundary REF , we obtain, for MATH, MATH . Thus, MATH . Therefore, we obtain the system of sequences MATH . From REF we get MATH, which replaced in REF gives the recurrence MATH . Similarly, MATH . The initial conditions are MATH. |
math/0011110 | By induction on MATH we prove that MATH which will imply the first assertion. Assume MATH. We need to show MATH, which will be proved by showing that the left hand side expression satisfies the NAME recurrence with the initial conditions of MATH. Denote by MATH the left hand side expression. First, MATH. Now, MATH. Assume MATH. Then MATH where MATH . Using MATH in the previous recurrence we get MATH relation satisfied by the NAME sequence. Since MATH has the same initial conditions as MATH we have MATH. The first step of induction is proven. Now, MATH . We shall prove that the matrix MATH acts as an index-translation on the NAME sequence, namely MATH . If this is so, then by taking MATH, the step of induction will be done. Let MATH (MATH is assumed fixed). First, MATH. Then, MATH. Now, for MATH, MATH where MATH . Using MATH in the previous recurrence, we get MATH. Therefore, MATH, since MATH. Using REF , with MATH, we get the recurrence between the entries of the higher power of MATH, namely MATH . The fact that MATH is the unique matrix with the given properties follows easily observing that such a matrix could be defined inductively as follows: let MATH. Assume MATH and construct MATH by bordering MATH with the first column and the last row (left and bottom). The first column is MATH and the last row is given by: MATH and MATH. |
math/0011110 | Using REF , with MATH, we obtain the first two identities. Now, with the help of REF and the trivial identity MATH, we get MATH . Since MATH, we obtain the third identity. Using MATH and NAME 's identity (see CITE, p. REF) (usually given for the NAME numbers, but certainly true for the sequence MATH, as well, as the reader can check easily, by induction) MATH we get the fourth identity. |
math/0011110 | Multiplying the recurrence REF by MATH and summing for MATH, we get MATH . Thus, MATH . Solving for MATH, we get MATH . We need to find MATH and MATH. We prove MATH . There is no difficulty to show the relations for MATH. Assume MATH. First we deal with the elements in the first row, MATH . Now we prove the result for the elements in the first column. MATH . Using REF, we get MATH . Using REF and the fact that MATH and MATH, we deduce the result. |
math/0011110 | The MATH entry in MATH is MATH which is MATH, unless MATH, in which case it is MATH. |
math/0011110 | Using REF, if MATH, then MATH . Since MATH divides neither MATH nor MATH (otherwise it would divide MATH), we get MATH . Therefore MATH . Using NAME 's identity MATH, for MATH, we get, if MATH, MATH since MATH. If MATH, then MATH . The previous two congruences replaced in MATH, proves the first claim. REF implies MATH . The residue MATH in the previous relation is just MATH. Thus, if MATH even, then MATH, so MATH for any MATH. The remaining cases are similar. |
math/0011110 | Straightforward, using REF or REF . |
math/0011111 | Multiplying REF by MATH and summing for MATH, we get MATH . Therefore, MATH which implies MATH . The generating function for the first row of MATH is MATH . Thus, MATH and the first claim is proved. It is trivial to find the generating function of the columns of MATH. Now, for MATH, multiplying REF by MATH and summing for MATH, we get MATH . If MATH, the generating function for the first column of MATH is MATH . It is not difficult to obtain that a recurrence of the form MATH has the solution MATH . Using REF in the recurrence for MATH, we get MATH . |
math/0011111 | We need to prove MATH which is a well-known relation CITE. |
math/0011111 | First, we prove that MATH and MATH. In fact a more general result is true; using REF we know that if MATH, then MATH and MATH. In our case, MATH. Thus, the minimal polynomial of MATH modulo REF equals MATH for MATH even and MATH for MATH odd. Since MATH, we deduce that MATH has distinct roots. Thus, MATH is diagonalizable (over a degree REF extension of MATH). First, MATH. Thus, the minimal polynomial of MATH modulo REF equals MATH, which has the solutions MATH (where MATH) in the quadratic extension of MATH, namely MATH. Let MATH be the multiplicities of MATH, respectively. We want to show that MATH. Since MATH has dimension MATH, MATH . Now, using REF , we get MATH . It is known (see CITE) that MATH. Thus, MATH is divisible by MATH, so MATH. Therefore, MATH and MATH, so MATH. The eigenvalues of MATH correspond to eigenvalues of MATH and since, MATH (modulo REF), we get that MATH corresponds to MATH, which corresponds to MATH, and similarly, MATH. We deduce MATH and so, MATH. In a similar manner we can show the other cases. |
math/0011111 | First, we show that MATH is a permutation matrix away from MATH, namely MATH . It is a trivial matter to prove MATH, which will give the equivalence. It suffices to show the second identity, which follows easily, since an entry in MATH is MATH . Now, denote by MATH, the matrix obtained by taking absolute values of entries of MATH. We show that MATH has the same characteristic polynomial (eigenvalues) as MATH, namely we prove their similarity, MATH . Since MATH (by REF), to show REF it suffices to prove that MATH. Therefore, we need MATH which is certainly true. We use a result of CITE to show that MATH in turn is similar to MATH, the diagonal matrix whose diagonal entries are the elements in the eigenvalues set listed in decreasing order according to size of the absolute value. For instance, for MATH, MATH . If MATH is similar to MATH, the theorem will be proved. We sketch here the argument of CITE for the convenience of the reader. Define an array MATH by MATH . Except for the sign, MATH. Let MATH, where MATH . We observe that, in fact, MATH is the companion matrix of the polynomial with coefficients MATH. Let MATH be the matrix with entries MATH. It turns out that the eigenvector matrix MATH of MATH, with columns vectors listed in decreasing order of absolute value of the corresponding eigenvalues, normalized so that the last row is made up of all REF's, satisfies MATH where MATH is the NAME matrix, which is the eigenvector matrix of MATH with eigenvectors listed in decreasing order of the absolute values of the corresponding eigenvalues. Also, MATH . Our theorem follows. |
math/0011111 | We showed that MATH . It follows that MATH, which together with MATH, proves the corollary. |
math/0011112 | The statement is equivalent to the analogous statement for convergence of the integrals of positive functions. Given the convergence of MATH we want to say the same is true for MATH replaced by MATH. By changing the order of integration we can make the following estimate: MATH which proves the statement. |
math/0011112 | First, let us note that each of the differentials MATH is surjective. Moreover, MATH is also surjective when restricted to the MATH for any MATH. This is equivalent to saying that MATH. So the cohomology of the complex MATH can be computed from the cohomology of the MATH-tuple complex with differentials MATH and terms given by the space of MATH-periodic holomorphic functions. Every such function has the NAME expansion which is convenient to write in the form MATH . The form of the NAME coefficients is motivated by the following simple form of the differential MATH . The next observation is that MATH, for MATH are surjective (and remain such on MATH for MATH). Explicitly, if MATH is given by REF , then MATH is also convergent by REF and defines an entire function, and MATH. Here MATH is the coefficient of MATH written in the basis MATH, that is, MATH. So the calculation now reduces to the MATH-complex with the differentials MATH and terms in the form MATH . In this complex the differentials MATH are injective. This easily follows from the fact that vanishing of MATH for MATH means that MATH are constant into the MATH direction. This forces summing over negative definite sublattice. Hence the NAME series for MATH diverges. Thus we see that the cohomology of MATH reduces to a cocycle concentrated in MATH, and given there by a function in the form MATH with finite number of MATH non zero. Finally, since MATH the cohomology MATH can be represented a multiple of the cocycle MATH . |
math/0011112 | This is quite standard (see, for example, the NAME 's book CITE with some minor modifications). Let us, for instance, show that the exponential factor in the transformation of MATH behaves well with respect to the group multiplication. Let MATH, then MATH. The exponential factor in MATH is given by MATH . The exponential factor in MATH consists of two ingredients: MATH and MATH . Combining these together and using the identity MATH in various forms, matching of the two exponential factors reduces to showing the following: MATH . The last identity follows at once using MATH and MATH . |
math/0011112 | It is easy to see from linear algebra that the passage from MATH to MATH can be factored into a sequence of transformations of types REF such that on each intermediate step the basis remains MATH-split. More precisely, any positive primitive sublattice MATH can be transformed to MATH in at most MATH steps by applying the transformations of type REF (with possible conjugation by types REF ). This, in turn, is a consequence of the following assertion: given a positive primitive sublattice MATH and a positive vector MATH, there is a positive primitive sublattice MATH which contains MATH and MATH. Since the primitive integral sublattices form a dense subset in the real NAME of positive MATH-subspaces, it is enough to show the last statement over MATH. Then we can take MATH, the subspace spanned by MATH and the MATH-dimensional part of MATH orthogonal to MATH. Obviously, the quadratic form restricted to MATH is positive definite. According to our previous calculations, the transformations of types REF have no visible impact neither on the cocycle MATH nor on the complex MATH. Hence, we are reduced to show the statement of the proposition for the transformation of type REF . Recall that for MATH of the type REF we have MATH for all MATH, except for MATH. Let MATH denote the positive sublattice of the basis MATH. Next we consider the following entire function MATH . Though each of the two sums are divergent, their difference is the convergent sum over the lattice points in between MATH and MATH. On the picture below, which shows the MATH slice of the lattice, those are represented by bold dots. The hollow dots represent terms taken with negative sign and MATH is negative in the shaded area. The cochain represented by this function in MATH is a coboundary between MATH and MATH. Indeed, all differential vanish when applied to MATH except for MATH and MATH: MATH . From the calculation of REF we recall that MATH has components only in MATH and MATH both given by MATH. This completes the proof. |
math/0011112 | Since MATH defines an action, we need to check the triviality for the set of generators only. CASE: MATH. The action is defined exactly modulo the transformations of this type. Hence in this case the statement holds trivially. CASE: MATH. Here we have MATH, hence MATH. Also the functions representing the cocycles MATH and MATH are identical: MATH. The matrix for the resolution identification is MATH . Using the calculation for the transformation of type II we conclude that MATH already on the level of cocycles. CASE: MATH. This seems to be the most interesting case. As for the classical theta functions the proof involves application of NAME transform and NAME summation formula. We will demonstrate the basic idea on the simple example of an REF-dimensional torus MATH, with MATH (MATH is not an elliptic curve since the polarization is negative). According to the matching map on resolutions of type III we have to show that two cocycles of the double complex MATH are homologous in MATH. To see this we define the entire function MATH by MATH . Then MATH and MATH . The minus sign appearing here reflects the choice of orientation on the line MATH. This ambiguity is suppressed by the choice of REF root of unity for the modular transform. With all that, we see that the difference of the above cocycles is the coboundary of the function MATH. To deal with the higher dimensional case, first, we need to make some comments on the choice of the original split basis MATH. Namely, using the transformation of REF above, we may assume without loss of generality that the reference basis is MATH-split, that is MATH can be chosen to be the reference basis. In such a basis the matrix MATH has the form MATH. So we can apply calculation of type III to conclude that MATH is represented by MATH in MATH. Next, we need to choose a totally real vector subspace in MATH such that the imaginary part of MATH restricted to MATH is positive definite. By induction we may also assume that MATH has a codimension one filtration MATH, such that MATH, MATH. The convenient property of this filtration is that MATH can be cut out by MATH equations MATH. Following the analogy with the example above we define the NAME transform of the function MATH by using the partial Vick rotation: MATH . Then one can use the usual theory of NAME transform for positive definite MATH. In particular, the operators of translation and multiplication by the corresponding character interchange. Also, using the real coordinates on MATH one has MATH . The same identity holds if the integration cycle MATH is shifted by any vector in MATH. We define a cochain MATH in MATH by the functions MATH, MATH: MATH . We want to show that the total differential applied to this cochain gives the difference between MATH and MATH in MATH: MATH . Since MATH, the differentials MATH vanish trivially for MATH. Also, MATH is clearly periodic with respect to MATH, MATH. For the rest we have: MATH . But MATH means that MATH for MATH. Thus the only non-trivial MATH-differential is given by MATH . In particular for MATH we have: MATH . Similarly, the differentials MATH vanish for trivial reasons for MATH. On the other hand MATH . For MATH the domain of integration remains the same, hence MATH. Also, for MATH, we have MATH, where MATH is a fixed complex number with non-zero imaginary part, and MATH. So the integrant is a holomorphic function in some open neighborhood of the subspace MATH in MATH. Hence the only non-trivial MATH-differential is given by MATH . The last equality is given by the NAME residue map evaluated on the corresponding cycles. Thus we see that MATH. Finally, using NAME summation formula for the function MATH of MATH real variables and the lattice MATH, we deduce that MATH . This completes the proof of the theorem. |
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