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math/0010076 | It suffices to prove an estimate of the type MATH for any choice of MATH . We first prove a weak type MATH estimate for MATH, that is, that MATH . Suppose MATH with MATH . Then if MATH with MATH we can use an appropriate NAME decomposition to find finite sets MATH so that each MATH is an atom of some MATH and MATH if M... |
math/0010076 | This will depend on the following Lemma: Suppose MATH and MATH . Then there is a constant MATH so that if MATH we have MATH . CASE: We may assume MATH . This is a fairly standard application of NAME 's theorem, see CITE. Here we use a version given in CITE. It is simplest to consider the case when MATH is finite with M... |
math/0010076 | Interpolating extra rows of zeros is trivial, so we can assume MATH for all MATH . For the case of columns, we only need to show that MATH . We may suppose that MATH is a finite set with MATH points and that MATH is a finite dyadic filtration of MATH. It is then possible to write MATH where MATH and MATH, and find a dy... |
math/0010076 | This is essentially trivial; we need only to prove that MATH . To do this note that MATH where MATH is obtained from MATH by repeating each column MATH times. The proposition follows then by the triangle law from REF . |
math/0010076 | Suppose MATH . Summation by parts gives MATH thus MATH and the result follows because of the maximal estimate MATH proved in CITE. |
math/0010076 | By combining REF one sees that it is only necessary to show that MATH has a lower MATH-estimate for some MATH . To do this observe that if MATH are disjointly supported sequences, then MATH . Hence MATH . Now suppose MATH and MATH . Then for each MATH, let MATH be the number of MATH so that MATH . Then MATH . This in t... |
math/0010076 | We start by using an argument due to CITE, see also CITE. For any fixed MATH let MATH . Now fix MATH . Then by a result of CITE, MATH if MATH . Then for any MATH we have MATH . Since MATH and since for MATH we have MATH it follows from REF that MATH . In particular if MATH we have MATH . At this point we return to the ... |
math/0010076 | We suppose MATH and that MATH is a matrix satisfying MATH . Consider the operator MATH . The adjoint operator is MATH given by MATH . The dual statement of the result in REF gives that for any sequence of MATH's, MATH we have the estimate MATH where MATH depends only on MATH . Now let MATH be a sequence of independent ... |
math/0010076 | Consider the dilation operator MATH . Then MATH and we have MATH which implies MATH . Likewise we obtain MATH . The corresponding result for the weak type constants follows similarly. |
math/0010076 | By a simple dilation argument it suffices to prove REF when MATH. In this case we have the estimate MATH and also by the self-adjointness of the MATH's and MATH's we have MATH . The required estimate REF (when MATH) will be a consequence of the pair of REF We start by proving REF . We only consider the term MATH since ... |
math/0010076 | Consider the operators MATH defined by MATH . Then MATH . Hence by splitting into REF pieces and using REF we obtain the estimate MATH . Next pick MATH so that MATH and MATH where MATH . Let MATH be a sequence of independent NAME random variables on some probability space MATH . Then for MATH we have MATH . Averaging n... |
math/0010076 | First we show the estimate MATH . Using the atomic characterization of MATH, CITE, we note that it suffices to get an estimate for a function MATH supported in a cube MATH so that MATH for MATH and MATH if MATH . It is then easy to see that if MATH since MATH for each MATH. (Here MATH is the cube with twice the length ... |
math/0010076 | For REF note that MATH . For REF note that MATH . REF is trivial. |
math/0010076 | Suppose that MATH is locally NAME and let MATH be a point of continuity of MATH . Then if we put MATH . REF gives that MATH . Now if MATH it is easy to see that as MATH we have convergence in MATH (and even pointwise) of MATH to MATH . If MATH let MATH be a cube of side MATH centered at MATH in MATH . Let MATH . REF an... |
math/0010076 | In fact REF follows immediately from NAME 's inequality by taking MATH two mutually independent sequences of NAME random variables. To obtain REF , take MATH be a sequence of NAME random variables and for any finite subset MATH write MATH . Now for all MATH, (see also CITE, proof of REF ), MATH by a generalization of N... |
math/0010076 | For simplicity we write MATH below. REF follows directly from REF . To prove REF it is enough to consider the case MATH, since the other cases follow trivially by applying REF and the known case MATH . We therefore suppose MATH and establish both REF . An easy calculation gives that for MATH . NAME, the function MATH h... |
math/0010076 | Recalling the definition of MATH from REF we note that the function MATH is compactly supported and is equal to MATH on the support of MATH. For any sequence MATH with MATH we observe that MATH by the NAME multiplier theorem. Let MATH be the bilinear operator with symbol MATH for some fixed MATH. Let MATH and let MATH ... |
math/0010076 | We give the proof in the case MATH; the only real alteration for the other cases would be to replace the appropriate MATH-norm with the MATH norm and use REF . Suppose MATH and consider MATH . We estimate the first term by noticing that for fixed MATH the NAME transform of MATH is contained in the set MATH . Hence if M... |
math/0010076 | As before we write MATH . Let us consider first the case when MATH unless MATH . Let MATH be a MATH-function on MATH supported on MATH and such that MATH on MATH . Fix MATH and consider the symbol MATH . Note that MATH is supported in MATH . Let MATH be bilinear operator with symbol MATH. For any sequences MATH with MA... |
math/0010076 | It is clear from REF that for any MATH we have MATH . Thus it suffices to consider the case MATH and MATH and establish a bound in this case. To do this we consider the symbols MATH where MATH are MATH-functions satisfying MATH for all MATH . Since MATH is bounded by MATH whenever MATH and there is a similar bound for ... |
math/0010076 | This is a stopping time argument. Suppose MATH with MATH . Note that for each MATH the function MATH is MATH-measurable where MATH is the MATH-algebra generated by the dyadic cubes in MATH . Fix MATH. For each MATH let MATH be the collection of cubes MATH so that MATH on MATH and for each MATH we have MATH on MATH . It... |
math/0010076 | The upper bound is proved in REF so we only need to prove the lower bound. It suffices to prove the results for the case when MATH is a MATH-matrix. We start by considering the case MATH when MATH is strictly lower-triangular. In this case let us estimate the norm of the discrete model MATH . In fact MATH where MATH is... |
math/0010076 | Observe that MATH and therefore MATH is the sum of nine terms of the form MATH where MATH. We now use REF , and REF in that order to obtain MATH where MATH . This proves REF . For REF note that if MATH integration by parts gives MATH provided MATH and MATH are nonzero. Now using the fact that MATH is a norm it is easy ... |
math/0010076 | This follows directly from REF . Indeed, we have MATH . If MATH we have MATH . Since MATH this gives the result. |
math/0010076 | Assume REF ; then it follows from REF that for any MATH we have an estimate MATH . Now it is clear from the definition and from REF that we have an estimate MATH . Hence we can deduce easily that MATH for each multi-index MATH . Repeating the same reasoning with the second variable MATH gives REF . Now assume REF . The... |
math/0010076 | This follows immediately from REF which yields the estimate MATH with MATH . |
math/0010076 | It is only necessary to show that MATH . Note first that REF can be used to give the estimate for any infinite matrix: MATH . Now suppose MATH. Then if MATH by REF . Combining with a similar estimate from REF gives the theorem. |
math/0010076 | Let MATH be the cube MATH and consider the bilinear operator MATH . We will show that if MATH is such that MATH, then MATH is bounded and MATH where MATH is a constant depending only on dimension. Suppose that MATH are functions with support contained in MATH and such that MATH . Then MATH with MATH and MATH . Applying... |
math/0010076 | REF is a classical result on paraproducts when MATH and we refer the reader to CITE p. REF for a proof. Note that for a fixed MATH, the map MATH is a NAME singular integral. The extension of REF to MATH for MATH, is consequence of the that if a convolution type singular integral operator maps MATH with bound a multiple... |
math/0010077 | By CITE (pp. REF), only the lens spaces MATH with MATH can admit orientation-reversing homeomorphisms. For MATH or MATH, MATH clearly has the required properties. For all cases with MATH, we have MATH . Therefore MATH normalizes MATH and induces an orientation-reversing isometry MATH on MATH. The properties of MATH are... |
math/0010077 | It is clear that MATH normalizes MATH. To determine normality of MATH in MATH, we need check only one other coset representative. We have MATH so MATH normalizes MATH if and only if MATH that is, if and only if MATH. This establishes the first two conclusions, so from now on we assume that MATH is normal. Let MATH be t... |
math/0010077 | The fact that the isometries preserve the fiberings in the nonexceptional cases in REF is immediate from fact that MATH or MATH is contained in MATH, as remarked at the ends of REF. For the exceptional cases in REF , the dimension of the group of isometries is MATH for the cases of MATH and MATH and is MATH for the oth... |
math/0010077 | For MATH this was first proven by NAME (see CITE). For the remaining cases, we will rely on calculations of MATH due to several authors. We utilize the composite homomorphism MATH which takes each isometry to the outer automorphism it induces on MATH. As noted in REF, elementary covering space theory shows that MATH is... |
math/0010079 | As MATH, we may identify MATH and MATH. Under this identification, it is easy to see that MATH and MATH are identified. Since MATH as in REF , this gives an isomorphism MATH, which is a MATH -isomorphism. The MATH -isomorphism MATH is trivial, because the definition of MATH is symmetric in MATH and MATH. It remains to ... |
math/0010079 | Consider the map MATH of REF . Clearly this maps MATH to MATH. As MATH and MATH and the map MATH is injective, we see that the kernel of MATH on MATH is MATH. Similarly, the kernel on MATH is MATH. Thus the kernel of MATH is MATH . But this is contained in MATH. Now MATH, since if MATH lies in MATH then MATH in MATH, s... |
math/0010079 | A short calculation shows that MATH as MATH -modules. Let MATH be a finite-dimensional MATH -module. Then MATH is an injective MATH -morphism, by REF . It is easy to show that MATH is identified with MATH by MATH. Now MATH is a subalgebra MATH of MATH isomorphic to MATH, and clearly MATH is closed under MATH. Choose a ... |
math/0010079 | Regard MATH and MATH as MATH-modules in the obvious way. Define a bilinear map MATH by MATH for MATH, MATH, MATH, MATH and MATH. Recall that MATH and MATH are MATH-submodules of MATH. Define a subspace MATH of MATH by MATH if MATH whenever MATH or MATH. Then MATH is a MATH-submodule of MATH. Now MATH is a MATH-submodul... |
math/0010079 | Let MATH, MATH, MATH and MATH. Then REF shows that MATH, where MATH. Let MATH be the MATH -submodule of MATH generated by the images MATH, where MATH and MATH. Then MATH is the maximal semistable MATH -submodule of MATH. We shall prove the proposition by explicitly constructing MATH elements of MATH, that are linearly ... |
math/0010079 | Let MATH and MATH. REF shows that MATH. As MATH is stable, MATH for nonzero MATH. As MATH is semistable, REF shows that MATH for generic MATH. Thus MATH for generic MATH. Also, REF shows that MATH is semistable. Combining these two facts with REF , we see that MATH, as we have to prove. It remains to show that if MATH ... |
math/0010079 | Let MATH be the NAME of real MATH-planes in MATH. Then MATH, and MATH. The condition for MATH to be a MATH -module is that MATH. A calculation shows that this fails for a subset of MATH of codimension MATH. Thus for generic MATH, MATH is a MATH -module. Suppose MATH is a MATH -module. Let MATH be the maximal semistable... |
math/0010079 | Since MATH is surjective, MATH is injective, and thus MATH is injective. Now MATH induces a map MATH. Suppose that MATH lies in the kernel of this map. Then MATH. By exactness, MATH. As the sequence MATH is also exact, MATH for some MATH. But MATH is injective, so MATH. Thus the map MATH is injective. It follows that t... |
math/0010079 | Let MATH, MATH, MATH, MATH, MATH, MATH, MATH and MATH. Then by REF we have MATH . REF calculates the dimensions of a quaternionic tensor product of stable and semistable MATH -modules. Examining the proof, it is easy to see that these dimensions are actually lower bounds for the dimensions of a quaternionic tensor prod... |
math/0010079 | Let MATH be nonzero, and apply REF with MATH, the semistable MATH -module defined in REF. Let MATH and MATH as in the proof of the proposition. Since equality holds in REF , MATH, so that MATH. But this is the main condition for MATH to be stable. Thus, it remains only to show that MATH is semistable. Let MATH be the m... |
math/0010079 | We shall prove that MATH is stable, by induction on MATH. Firstly, MATH is stable, by definition. Suppose by induction that MATH is stable. The sequence MATH is MATH -exact, and MATH and MATH are stable. Therefore, REF shows that MATH is stable, so all MATH are stable, by induction. By a similar argument involving indu... |
math/0010079 | As MATH is an NAME, MATH is a stable MATH -module, by definition. Let MATH be integers. Putting MATH, REF defines a MATH -morphism MATH. Let MATH be the unique MATH -morphism, such that the restriction of MATH to MATH is MATH. It is elementary to show that because MATH is an NAME, MATH makes MATH into an NAME, and we l... |
math/0010079 | Let MATH, and define MATH, regarding MATH as an endomorphism of MATH. Let MATH. It is easy to verify that MATH. Thus MATH if MATH, and MATH is q-holomorphic whenever MATH is q-holomorphic. This proves the first part. If MATH is holomorphic with respect to MATH, then MATH by the NAME equations. But MATH so that MATH is ... |
math/0010079 | Suppose MATH is a smooth function, such that for each MATH, the function MATH lies in MATH, and for each MATH, the function MATH lies in MATH. The condition for MATH to be q-holomorphic is MATH. But MATH, where MATH involves only derivatives in the MATH directions, and MATH only derivatives in the MATH directions. Let ... |
math/0010079 | We must show that Axiom A is satisfied. REF and MATH are trivial. For part MATH, observe that the permutation map MATH that swaps round the factors, is induced by the map MATH given by MATH. Since the diagonal submanifold is invariant under this, it follows that MATH is invariant under permutation, and so MATH. Let MAT... |
math/0010079 | By REF there is an NAME morphism MATH, and as MATH generates MATH, MATH is surjective. Let MATH be the kernel of MATH. As MATH is a free NAME, each element MATH of MATH defines a unique NAME morphism MATH, that restricts to MATH on MATH. Clearly, if MATH, then MATH if and only if MATH. For each MATH, define a function ... |
math/0010079 | Let MATH, and write MATH. Now MATH is the first derivative at MATH of some element of MATH, which is a q-holomorphic function on MATH. By definition of q-holomorphic function, it follows that MATH, where MATH are the complex structures on MATH. Let MATH be the subspace of elements MATH satisfying MATH. Then MATH has di... |
math/0010079 | As MATH, we deduce that MATH is injective. Together with the surjectivity assumption, this implies that MATH is a diffeomorphism near MATH. Also, MATH for MATH near MATH. The conclusion follows from the lemma. |
math/0010079 | Because of REF , it is sufficient to show that MATH for each MATH. Suppose for a contradiction that MATH and MATH. As MATH is a diffeomorphism, MATH is injective. Let MATH, so that MATH is a proper MATH -submodule of MATH, and MATH. Since MATH is injective and MATH, we have MATH. Now MATH is a MATH-module with MATH, so... |
math/0010079 | Let MATH be a q-holomorphic function on MATH, of polynomial growth of degree MATH. Let MATH. By REF , MATH is isomorphic to MATH, so that the universal cover MATH is isomorphic to MATH. Restricting MATH to MATH and lifting to MATH, the result is a q-holomorphic function on MATH, of polynomial growth of degree MATH. Now... |
math/0010079 | In REF we constructed a MATH -submodule MATH of MATH. By REF , there is a unique NAME morphism MATH. To complete the proof we must show that MATH is an NAME morphism, where the NAME structure on MATH is defined by REF , so we must show that MATH identifies the NAME brackets on MATH and MATH. In REF we remarked that MAT... |
math/0010084 | As for any presentation result, we have to prove two assertions. CASE: The elements MATH satisfy the relations REF and generate the MATH-algebra MATH. CASE: If MATH, MATH and MATH in a MATH-algebra MATH satisfy the relations REF , then there exists a morphism of MATH-algebras MATH sending MATH, MATH and MATH. The proof... |
math/0010084 | We prove that MATH satisfy the relations REF . The formulae MATH and REF are true, REF is equivalent to the fact that MATH is a MATH-form and REF says that MATH for any MATH in MATH, that is, that MATH is a morphism of MATH bimodules. Let MATH be an orthonormal basis of MATH and let MATH be an orthonormal basis of MATH... |
math/0010084 | The unital NAME MATH-algebra MATH being presented by the relations corresponding to MATH, MATH and MATH, its tensor MATH-category of corepresentations has to be completion of the tensor MATH-category MATH generated by MATH, MATH and MATH. (This is a direct consequence of tannakian duality CITE, compare REF.) On the oth... |
math/0010084 | The canonical trace MATH is a MATH-form, with MATH. Its restriction to MATH is the canonical trace of MATH, so it is a MATH-form, with MATH. Also MATH being a trace, the projection MATH has to be a MATH bimodule morphism. Thus MATH is a MATH-form, with MATH . Thus REF applies to MATH. In terms of quantum groups, we get... |
math/0010087 | It is not difficult to show that MATH is affine linear in each connected component of MATH and that the gradient image MATH is equal to MATH minus some of its boundary points. This readily implies the statement. For details we refer to CITE. |
math/0010087 | It is not difficult to show that for MATH symmetric, positive definite matrices MATH the inequality MATH holds, with equality precisely if MATH and MATH are real multiples of each other. Applying this to the sum REF and using the fact that it contains at least two terms for all MATH, the first statement follows. Combin... |
math/0010087 | Let MATH be compact convex subsets of MATH. From the monotonicity properties of mixed volumes it follows that MATH with strict inequality holding unless one of MATH is a point or MATH and MATH are two parallel segments. Assume now that we have a non-trivial factorization MATH and let MATH denote the NAME polytopes and ... |
math/0010087 | In REF it is shown that MATH is proportional to the average number of solutions in MATH to the system of REF as MATH ranges over the real torus MATH. Note that the set of critical values of the mapping MATH is a semialgebraic set. Thus it is contained in a real-algebraic curve MATH. Consider the product space MATH with... |
math/0010087 | The Lemma follows from the fact that MATH maps conjugate points to the same point, MATH. |
math/0010087 | The curve MATH is non-singular and therefore MATH is a smooth surface with the boundary MATH. |
math/0010087 | We claim that the singular points of MATH may only arise as the intersection points of two non-singular branches of MATH. Consider the map MATH. Over MATH each of the two branches of MATH must be non-singular. Indeed, it maps REF to MATH and, therefore, the link of each point of this branch is an unknot. By a similar r... |
math/0010092 | To prove REF , note that the inclusion MATH follows from the definition of intersecters. We are left with proving the inclusion MATH . Assume that it does not hold, and consider such an intersecter MATH for MATH that MATH. In this case, the inclusion MATH holds not for all sets MATH from the family MATH, hence there ex... |
math/0010092 | There is nothing to prove for MATH or MATH. So suppose that MATH and MATH. Let MATH be an intersecter for MATH. According to REF , for all MATH, MATH, we have MATH, and in view of REF , for every atom MATH we have the inclusion MATH the left-hand part of which is nonempty. Hence, for all MATH the inclusion MATH implies... |
math/0010092 | The atom set MATH of the poset MATH is MATH, therefore, by REF , the subposet of intersecters for MATH in MATH is MATH and the statement follows. |
math/0010092 | Since the atom set MATH of the poset MATH is MATH, we have, according to equality REF , MATH and the statement follows. |
math/0010092 | There is nothing to prove for the trivial blockers MATH and MATH. So suppose that MATH is nontrivial. Choose an arbitrary antichain MATH. With regard to reciprocity property for intersecters, every element of MATH is an intersecter for the antichain MATH. In other words, for each element MATH we have the inclusion MATH... |
math/0010092 | There is nothing to prove for a trivial blocker MATH, so suppose that MATH is nontrivial. Choose two antichains MATH. According to REF , we have the following equalities in the lattice MATH: MATH . Therefore MATH . Hence MATH. |
math/0010092 | We have to prove that for all MATH, it holds MATH. There is nothing to prove when one of the blockers MATH is trivial. Suppose that both MATH and MATH are nontrivial. With the help of REF , we write MATH . According to REF , we have the following equalities in MATH: MATH . |
math/0010092 | Let MATH. If MATH then MATH, and we see that MATH, by REF . Conversely, the relation MATH implies the relation MATH, in view of REF . Because the restriction map MATH is bijective, we see that it is an antiautomorphism of MATH. |
math/0010092 | The only missing step is to prove REF , but the equality MATH immediately follows from the self-duality of the lattice MATH, in view of the existence of its anti-automorphism MATH. With the help of equality REF , we obtain REF . |
math/0010093 | We prove the second relation. The other ones are derived similarly or by observing that the four subalgebras generated by MATH, MATH, MATH and MATH are all isomorphic. It is clear from the defining relations that MATH for some MATH. Multiplying with MATH from the left gives MATH leading to the recursion relation MATH w... |
math/0010093 | We will first prove the case MATH, which may be written as MATH . It is clear from the defining relations that MATH for some coefficients MATH. To find a recursion formula for MATH we write MATH where we used REF in the last step. This leads to the recursion MATH for MATH, where MATH. We must check that this holds for ... |
math/0010093 | Let MATH denote the set of invertible elements in MATH. First we observe that the element MATH of REF is a polynomial in MATH of degree MATH with the leading coefficient in MATH. By REF , MATH is a polynomial in MATH of degree MATH, with the leading coefficient in MATH. Applying NAME elimination, we can expand MATH, ag... |
math/0010093 | That MATH is a MATH-subspace means precisely that MATH for some MATH. Then the invariance of MATH means that MATH for MATH, MATH. By REF , this is impossible unless MATH or MATH. |
math/0010093 | First one checks that the kernel and the image of an intertwiner are always invariant MATH-subspaces. It follows that an intertwiner between irreducible corepresentations is either zero or bijective. Counting dimensions (over MATH) then gives the first statement. For the second statement, we assume that MATH. Since MAT... |
math/0010093 | By definition, MATH which gives MATH where we used REF in the last step. Inserting MATH and simplifying gives the desired expression. |
math/0010093 | We first check that MATH preserves the bigrading and the moment maps, and consequently the commutation relations REF. Then it suffices to check the first four defining relations of REF , two relations from REF and the determinant relation MATH. This is straight-forward; for instance, the relation MATH is equivalent to ... |
math/0010093 | We use the expressions for MATH given in REF . First suppose that MATH, MATH, so that MATH . By REF , the element MATH acts on MATH by multiplying with an element of MATH which is obtained from MATH by replacing MATH, MATH, MATH. Then MATH acts by replacing MATH, multiplying with MATH and shifting MATH. In conclusion, ... |
math/0010093 | There are two things to be checked: first, that MATH maps into the subspace of MATH which maps onto MATH, second, that MATH factors through the defining relations of MATH. For the first part, note that the left hand side of REF splits into a sum of quotients of spaces of the form MATH. This component is mapped to MATH ... |
math/0010093 | We sketch two proofs. For the first one, we make the NAME MATH where MATH is the dynamical determinant of REF . Writing MATH and using the commutation relations of REF , one derives the recursion relations MATH for MATH, where MATH. This leads to an identity equivalent to REF. Alternatively, one may plug the expression... |
math/0010093 | For the first statement, we apply MATH to REF. Using the unitarizability we obtain MATH which means precisely that MATH is intertwining. For the second statement, we apply MATH to REF. This gives indeed MATH where we used that MATH. |
math/0010093 | Let MATH be a left-invariant integral on MATH. Then MATH for some MATH. The left-invariance of MATH means that MATH . By REF , this implies that MATH unless MATH, in which case the normalizing condition shows that MATH is given by REF. For the right-invariance, the proof is similar. |
math/0010093 | Applying MATH to REF gives MATH where we inserted REF for the NAME - NAME coefficients. Next we observe that MATH . This follows easily from REF , and can also be proved without using explicit expressions for the matrix elements, similarly to the proof of REF . Combining REF completes the proof. |
math/0010093 | We write out the proof for MATH, the case of MATH being slightly easier. It suffices to evaluate both sides on MATH. Let us write MATH, MATH. One then has MATH . On the other hand, MATH using REF in the last step. By the counit axioms this implies MATH using REF and the fact that MATH unless MATH, MATH. Applying MATH t... |
math/0010096 | For any fibration sequence MATH there is a natural map MATH and the action of MATH on MATH is determined by the composition MATH. The functor MATH induces a map MATH; in the proposition, MATH, MATH and MATH. So the functor MATH induces an isomorphism MATH (indeed, for MATH, we have isomorphisms MATH by REF sections). T... |
math/0010096 | The lemma is proved by induction on MATH. For MATH, let MATH be a map that represents the cohomology class MATH and let MATH be the fibration defined by the pullback square MATH where MATH is the path fibration. The fiber of MATH is MATH which is MATH. By construction the generator of MATH transgresses to MATH. Hence e... |
math/0010096 | Let MATH; then for each integer MATH we will construct by induction a fibration sequence MATH and a map MATH over MATH. The induction begins at MATH with the fibration sequence MATH that one has by hypothesis. The inductive step is next. Suppose that one has a fibration sequence MATH. The MATH-invariant MATH extends to... |
math/0010096 | Let MATH and MATH be orientable spherical fibrations with fibers that are odd dimensional spheres of dimension greater than one. Then there is a pushout square MATH . The projections MATH and MATH are spherical fibrations with odd dimensional fibers. If MATH and MATH are finitely dominated then the fiber product MATH a... |
math/0010096 | First we observe that by taking fiber joins if necessary (that is, MATH), we can assume that MATH is an orientable spherical fibration. Let MATH be a CW - complex. Recall that MATH denotes the topological monoid of self homotopy equivalences of MATH and that MATH is the connected component of the identity in MATH. By C... |
math/0010096 | By taking fiber joins if necessary, we can assume that the spherical fibration is orientable. Now if MATH is an oriented spherical fibration and the total space is homotopy equivalent to a finite dimensional complex then the NAME sequence with local coefficients shows that MATH has periodic cohomology. Conversely, assu... |
math/0010096 | As before we can assume orientability for our fibration by taking fiber joins if necessary. Now if MATH is an orientable spherical fibration and MATH is homotopy equivalent to a finite complex then the NAME sequence shows that MATH has periodic cohomology. Conversely, suppose MATH has periodic cohomology. By REF there ... |
math/0010096 | Assume that all the isotropy subgroups have periodic cohomology; then we can choose MATH such that for any isotropy subgroup MATH, MATH is a periodicity generator (see CITE, CITE). If MATH is described by the extension MATH where MATH denotes the MATH - fold dimension shift of the trivial MATH - module MATH, then this ... |
math/0010096 | Let MATH be an oriented spherical fibration having MATH as its NAME class. Since MATH is finite, MATH and hence MATH are of finite type. Moreover, since MATH is a projective MATH-module we infer from the NAME sequence that the map MATH is an isomorphism for every coefficient MATH and all integers MATH which are larger ... |
math/0010096 | We apply the main result in CITE, namely given a complex as above, there exists a real representation MATH of MATH such that MATH for all subgroups MATH. Applying REF completes the proof. |
math/0010096 | Let MATH denote the one-dimensional representation of MATH defined by MATH, and MATH for MATH. Let MATH. Using the double coset formula and the fact that MATH is central, we readily see that MATH and MATH. |
math/0010096 | Take MATH as before, constructed from the representations induced from the center. If any MATH fixed a point in MATH, then MATH. Therefore the subgroup generated by MATH and MATH would have MATH - rank at least as large as MATH, a contradiction. |
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