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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 MATH . Let MATH and MATH . Then MATH is supported in MATH and thus MATH . On the other hand MATH and MATH. Hence MATH and so MATH which implies that MATH . Selecting MATH and combining with REF we obtain for MATH where MATH. This gives the weak-type MATH estimate for MATH . Now by the NAME interpolation theorem (applied to the sublinear map MATH) we obtain that MATH as long as MATH . We now prove that MATH when MATH. We consider the dual map MATH defined by MATH where MATH . We have that MATH has norm bounded by MATH as long as MATH that is, MATH . Using this MATH as a starting point, we repeat the argument above to show that MATH has norm bounded by MATH . The NAME interpolation theorem can again be used to show that MATH has norm bounded by MATH for all MATH, and thus in particular when MATH. Therefore we obtain that MATH when MATH. |
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 MATH . Consider the map MATH. For each MATH with MATH, let MATH . For MATH with MATH, MATH, and MATH a sequence of independent NAME random variables on some probability space, we have MATH since MATH has type MATH. It follows from CITE that there is a function MATH, with MATH and MATH a.e such that for any set MATH . Now consider the set MATH of all permutations of MATH which induce permutations of the atoms of each MATH for MATH; there are MATH such permutations MATH . For MATH we have MATH or equivalently MATH . Raising to the power MATH, averaging over MATH, and then raising to the power MATH gives MATH . But this implies MATH which gives the required weak type estimate REF . We now return to the proof of REF . We first observe that we always have MATH since MATH. If MATH, REF gives that MATH and the required conclusion follows from REF . Assume therefore that MATH and that MATH maps MATH with norm MATH. Fix MATH with MATH and use the NAME decomposition of REF , to obtain REF as before, but instead of REF the estimate MATH which implies MATH . Selecting MATH with MATH and combining with REF we obtain MATH . This says that MATH maps MATH into MATH with norm at most MATH, in particular that MATH maps MATH into MATH as long as MATH. REF gives that MATH maps MATH into MATH and also MATH into MATH with norms at most a multiple of MATH. By interpolation it follows that MATH maps MATH into MATH for MATH . We conclude that MATH for MATH but since MATH is comparable to MATH, we finally obtain MATH. Since the converse inequality is always valid when MATH, we apply REF to conclude the proof. |
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 dyadic filtration MATH of MATH and a dyadic filtration MATH of MATH so that MATH for MATH and MATH for MATH . Then for MATH let MATH and note that MATH where MATH . Hence MATH . Integrating over MATH gives MATH . This completes the proof. |
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 turn implies that MATH where MATH is chosen so that MATH . Then we obtain an estimate MATH . This establishes a lower MATH-estimate. |
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 NAME space MATH. Let us define MATH, MATH, and MATH for MATH . Let MATH . It will be convenient to normalize REF so that we have MATH . We also note that REF implies the existence of a constant MATH so that we have MATH for MATH . Now suppose MATH . Suppose that MATH is supported on a measurable set MATH and satisfies MATH everywhere. Then we can define a measurable map MATH from MATH into the set of permutations of MATH so that MATH for all MATH . Thus MATH for all MATH . Let MATH when MATH and MATH otherwise. Now for MATH and MATH, let MATH so that MATH. If MATH we can estimate MATH in view of REF since MATH and MATH . Thus MATH . It follows that MATH . Note that for each MATH . Hence we obtain that if MATH is supported on MATH with MATH everywhere and MATH then MATH for a suitable constant MATH . Let us next refine REF . For MATH let MATH . Then by REF we have if MATH and as MATH we obtain, under the assumptions MATH everywhere and MATH, MATH . If we use a fixed value of MATH and the estimate MATH we find that MATH if MATH. This in turn gives us for every MATH . Now let MATH and fix MATH with MATH. We set MATH and MATH for MATH . Applying REF we obtain MATH which completes the proof under the assumption MATH. The general case follows by scaling. |
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 NAME random variables on some probability space MATH . We use MATH to denote expectations on MATH . Using REF we obtain MATH . This gives MATH which completes the proof by using REF . |
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 the term MATH is similar. Let MATH. Then MATH where MATH lies on the line segment between MATH and MATH. It is now easy to see that the sum of the last three expressions above is bounded by MATH which is clearly controlled by MATH. This estimate is useful when MATH. We now turn to the proof of REF . Since MATH is the difference of two MATH's, it will suffice to prove REF where MATH is replaced by MATH. We only work with the term MATH since the other term can be treated similarly. We have MATH . Since the functions appearing inside the sum in the first term above have supports with bounded overlap we obtain MATH and the crucial observation is that MATH which can be easily checked using the NAME transform. Therefore we obtain MATH and the required conclusion will be proved if we can show that MATH . We prove REF by using a purely size estimate. Let MATH be the center of the dyadic cube MATH. For MATH we have the easy estimate MATH since both MATH. We now control the left hand side of REF by MATH . By taking MATH large enough we obtain REF and thus REF . |
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 now gives MATH . Hence MATH where MATH depends only on MATH and MATH. Similarly MATH . By interpolation we obtain MATH . Finally let us write MATH . Thus by REF , MATH . This shows that MATH . For the converse direction we use MATH and REF . We have MATH and so MATH which leads to the estimate MATH . |
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 and the same center MATH as usually.) This gives the estimate MATH . On the other hand, MATH and combining with the previous estimate we obtain MATH . Complex interpolation gives that MATH when MATH and MATH . Since MATH we deduce the estimate MATH . |
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 and the fact that MATH easily imply that MATH . Since MATH is continuous we have MATH. Taking limits as MATH yields the conclusion. |
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 NAME 's inequality due to CITE and REF . The same estimate is also valid for MATH by our assumptions. These estimates together with REF give REF . |
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 has NAME transform supported in the annulus MATH when MATH. It follows that MATH where as usual MATH is a sequence of independent NAME random variables. (If MATH then MATH.) We need to control the last term in REF . Our hypothesis gives the estimate MATH while we can apply REF to obtain MATH . It remains to estimate MATH where MATH is a second (independent) sequence of independent NAME random variables. Hence using again NAME 's inequality we have MATH in view of REF . Using REF , and REF we obtain MATH which combined with REF gives the first of REF for MATH. The second assertions are derived similarly by symmetry. |
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 be two sequences of mutually independent NAME random variables. Then for MATH we have MATH by our hypothesis, REF , and REF . We now use REF twice to deduce that MATH . This proves the required assertion. |
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 MATH we have MATH . If MATH we obtain the same estimate by noticing that MATH and using the corresponding square-function estimates in MATH. Now we have MATH . If we let MATH be the matrix with entries MATH if MATH and MATH otherwise, then MATH where MATH is the matrix with entries MATH if MATH and MATH otherwise. It is trivial to see that one has the estimate MATH so that MATH . Hence REF give MATH . The same argument shows that the third term in REF is controlled by MATH . The middle term in REF is easy. For MATH we have MATH . Combining we obtain the required upper estimate: MATH . |
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 MATH and MATH we have MATH by considering the supports of the NAME transforms. But then for fixed MATH hence MATH using REF . Now note that if MATH then all MATH vanish. Since MATH, we integrate over MATH to obtain symbols MATH with corresponding bilinear operators MATH satisfying MATH whenever MATH . Note that MATH is supported on MATH . Also if MATH we have that MATH is constant in MATH . Next let MATH be the orthogonal group of MATH and let MATH denote the NAME measure on this group. Define MATH and let MATH be the corresponding bilinear operator. If MATH we can compute that MATH where MATH is a constant depending only on dimension. On the other hand, since MATH, REF gives that MATH whenever MATH . Note that MATH . Let us take MATH and MATH to be residue classes modulo REF. Then if we replace MATH by MATH and MATH by MATH we obtain a bilinear operator whose symbol coincides with MATH on MATH for MATH . Using REF and the multipliers MATH and MATH we obtain that the bilinear operator MATH with symbol MATH satisfies MATH . Summing over REF different pairs of residue classes gives a similar estimate for the symbol MATH . The last step is to remove the factor MATH . But this can be done by using REF since MATH is MATH on MATH . |
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 we have an immediate estimate; MATH . Now let MATH . Then we can compute MATH where MATH . Since the functions MATH for MATH are linearly independent on the support of MATH we can use the above estimate for a linear combination of a finite number of choices of MATH and MATH so that MATH except when MATH and MATH, so that MATH for some fixed constant MATH . By REF we have MATH . This and the similar argument for the case MATH gives the result REF . For REF we observe that the above argument actually also yields a bound on MATH when MATH since MATH also verifies the hypotheses of REF . By choosing a similar linear combination we can then ensure that MATH and obtain the desired result. |
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 is not difficult to see that MATH and this is a disjoint union. Also note the left-hand side has finite measure. For each MATH be MATH be a MATH-measurable function such that MATH everywhere and MATH . Let MATH . Then MATH and MATH . Hence MATH so that we have MATH . This implies that MATH and the result follows from REF . |
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 defined in the proof of REF . Using REF we obtain MATH . (All these quantities are finite since MATH has only finitely many non-zero entries, and so there is a uniform bound on MATH . ) It follows that we have an estimate (for a suitable MATH) MATH . Next we estimate MATH . If MATH it is clear that MATH remains lower-triangular and the invariance properties of MATH imply that MATH . If MATH then it is easy to estimate MATH and MATH . We deduce that MATH for all MATH . Thus we have for a suitable constant MATH . Now we may pick an integer MATH large enough so that MATH . Then we can combine REF to obtain MATH . At this point REF gives the conclusion that MATH . Now suppose MATH is arbitrary. If we let MATH be the bilinear operator with symbol MATH . REF implies that we can use REF to deduce that MATH for some absolute constant MATH . Thus the above argument yields MATH . Similarly MATH and REF is enough to show that MATH . Combining these we have the estimate MATH . The proof is completed by a simple interpolation technique. We will argue first that an estimate of the type MATH for some fixed MATH implies the estimate MATH for every MATH . We only need to consider the first case and MATH (when MATH one repeats the step). Then we may find MATH and MATH so that MATH . The NAME interpolation theorem yields MATH . Since MATH, using REF , and REF we obtain estimate REF as required (recall that we assume MATH is a MATH-matrix so that all these quantities are finite). Repeated use of this argument starting from MATH gives the theorem in the cases MATH. Finally in the case where either MATH or MATH (or both) one can use complex interpolation to deduce MATH where MATH and MATH . This clearly extends the lower estimate to the cases MATH . |
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 to see that by choosing an appropriate MATH or MATH for each pair MATH one obtains the estimate MATH . If MATH the same estimate follows directly from REF . Finally we turn to REF . For fixed MATH with MATH let us define MATH and MATH . Then it follows from REF that for any multi-indices MATH we have MATH . This implies that for fixed MATH and any MATH with MATH we have MATH . We now use either REF if MATH or we refer back to REF or REF according to the values of MATH or MATH, when MATH. For example when MATH and the MATH-th entry of MATH has maximal size MATH, then MATH . Now by REF we can estimate the last expression side above by MATH . Using REF we obtain REF . |
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 . Then for any multi-index MATH one can see easily by differentiation that for any pair of multi-indices MATH we have that REF is satisfied by the symbols MATH and MATH in place of MATH . Applying REF gives REF . |
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 REF we obtain that MATH where MATH is an absolute constant. It follows that MATH extends unambiguously to any MATH with MATH and REF holds. Next fix MATH so that MATH and MATH has support contained in MATH . Now for any MATH let MATH and MATH . Then REF gives MATH . We also note that MATH . Now consider the linear map MATH . Since MATH we have that, if MATH, MATH is bounded with norm controlled by MATH (again using REF .) Hence since MATH . Similarly MATH . Combining these estimates gives MATH . We now use a NAME type argument as earlier in REF . Suppose MATH and MATH satisfy MATH and that MATH . Then if MATH and MATH are two independent sequences of NAME random variables we have MATH . Again by using the result of CITE, we obtain an estimate MATH . Extracting the diagonal gives MATH . We now use CITE as before. There is a weight function MATH with MATH a.e. and MATH so that for any MATH with MATH and any measurable MATH we have MATH . Now suppose MATH are supported in MATH and MATH . Let MATH . Then the above equation yields MATH . On the other hand if we apply REF to MATH where MATH and note that MATH we also obtain that MATH . Raising to the power MATH and averaging gives: MATH . Thus MATH maps MATH into MATH with norm at most MATH . Now let MATH. If we define MATH then we have MATH and we can apply this result to MATH. Notice that MATH where MATH and MATH . This implies that for any MATH we have the estimate MATH for MATH supported in MATH. Letting MATH gives the result. |
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 of MATH, then it also maps MATH into itself with bound a multiple of this constant. REF follows from a similar observation while REF is a dual statement to REF . To prove REF set MATH. We have that MATH and the NAME transform of MATH is supported in the annulus MATH. It follows that MATH where MATH is the NAME maximal operator which is certainly bounded on MATH. To prove REF we freeze MATH and look at the linear operator MATH whose kernel is MATH. It is easy to see that MATH . This estimate together with the fact that the linear operator MATH maps MATH gives that MATH maps MATH using the NAME decomposition. This proves REF . To obtain REF we use REF (with MATH) and we apply to the NAME decomposition to the operator MATH for fixed MATH. Finally REF is a consequence of REF . |
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 immediate from the facts that MATH and MATH. |
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 the subgroup of MATH generated by MATH, and let MATH be the preimage of MATH in MATH, that is, the subgroup generated by MATH and MATH. Of course, MATH. We have MATH, so when MATH is odd, MATH is cyclic of order MATH, and MATH with the MATH factor generated by MATH. Put MATH and MATH. When MATH is odd, MATH, and since MATH, MATH is either MATH or MATH, and MATH is either MATH or MATH. In particular, when MATH is odd, MATH. Suppose first that MATH is odd. We may assume that MATH is odd. For suppose that MATH is even. Consider the element MATH, which generates MATH. We have MATH . Under the automorphism of MATH which interchanges the two coordinates, MATH is carried to MATH. Consequently, the quotients of MATH by the subgroups generated by these two elements are isomorphic. Following the notation of CITE, for a subgroup MATH we will denote by MATH the projections of MATH to MATH for MATH, and put MATH and MATH. Then, MATH. Since MATH is odd, MATH where MATH is cyclic of order MATH generated by MATH and MATH is generated by MATH. The order of MATH is MATH and the order of MATH is MATH, so the order of MATH is MATH, which shows that MATH. Since MATH and MATH are odd, the quotient of MATH by MATH is isomorphic to MATH. The image of MATH in the quotient is MATH, so MATH, as in REF . Now suppose that MATH is even. Since MATH is odd, MATH has order MATH and does not contain MATH. Since MATH must contain an involution, MATH contains the order MATH subgroup generated by MATH and MATH. The quotient of MATH by this order MATH subgroup is MATH, into which MATH descends to a subgroup MATH of order MATH generated by MATH. The order of MATH is MATH and the order of MATH is MATH. If MATH, then we must have MATH, and MATH, giving REF . So assume that MATH, so that the index of MATH in MATH is MATH. The quotient of MATH by MATH is isomorphic to MATH. The image of MATH must be the subgroup generated by MATH, since otherwise MATH would have equaled MATH, and taking the quotient by this subgroup gives MATH as in REF . |
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 other cases. Since the isometries act as a group of orbifold diffeomorphisms on the quotient orbifold, the dimension of a group of fiber-preserving isometries is at most MATH more than the degree of symmetry of the quotient orbifold, the extra dimension coming from the isometries that move fibers vertically. For MATH and MATH this sum is MATH, and for the other cases in REF it is MATH, so MATH cannot preserve the NAME fibering. For MATH and MATH, there are other fiberings, but none can be preserved by a group of dimension greater than MATH, since the maximum degree of symmetry of a quotient orbifold is MATH (achieved by the NAME fiberings). For the other three cases in REF , the fiberings are unique up to isomorphism and hence all fiberings have the same quotient orbifold, whose degree of symmetry is MATH. Since MATH has dimension MATH, it cannot preserve any fibering. For the additional quaternionic manifold in REF , some of the isometries MATH with MATH do not preserve the NAME fibering. In fact, MATH cannot preserve any NAME fibering isomorphic to this one. For the fibering is nonsingular, with homotopy exact sequence MATH, where MATH is the subgroup of MATH represented by the fiber. Since no cyclic subgroup of order MATH of MATH is invariant under all automorphisms induced by elements of MATH, no such fibering can be invariant. The quotient orbifold for the other NAME fibering on this manifold is MATH. Some isometries from each path component do preserve this fibering, but since MATH, these other fiberings cannot be preserved by all of MATH. REF is due to the fact that the NAME fiberings on the MATH all have nonzero NAME class. Any fiber-preserving diffeomorphism must preserve the NAME class, but any orientation-reversing diffeomorphism must multiply it by MATH, so there are no fiber-preserving orientation-reversing isometries. |
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 the outer automorphism induced on MATH by conjugation by MATH. In all cases except the tetrahedral and icosahedral manifolds, there are explicit published calculations of MATH which show it to be abstractly isomorphic to MATH. In these cases, it is sufficient to check that MATH is injective. For lens spaces other than MATH, CITE gave a complete calculation of MATH, obtaining groups isomorphic to the MATH obtained in REF (these were also calculated in CITE). In particular, MATH, and since MATH contains orientation-reversing elements, MATH is an isomorphism. In all cases with MATH, MATH is injective. For if we regard our standard generator MATH of MATH as MATH, we can check that MATH sends MATH to MATH, MATH sends MATH to MATH, and MATH sends MATH to MATH, an automorphism of order MATH in the cases when MATH. In all cases MATH is injective. For the MATH that contain one-sided NAME bottles, the mapping class groups were calculated in CITE and CITE. These include all quaternionic and prism manifolds, and the lens spaces MATH. Again, the groups are isomorphic to MATH. For the quaternionic manifolds, the abelianization MATH is isomorphic to MATH. The MATH factors in the isometry groups act as MATH on the MATH factor. When MATH, the isometry MATH acts as multiplication by MATH on the MATH factor. Since MATH is odd, we have MATH, so this is a nontrivial automorphism on MATH. It follows that MATH is injective for these cases. For the prism manifolds, note first that MATH sends MATH to MATH, which is the nontrivial outer automorphism of MATH. For the cases of MATH, MATH is odd, and MATH induces multiplication by MATH on the MATH factor, so MATH is injective for these cases as well. When MATH is an index MATH diagonal subgroup in MATH, the automorphism induced by the nontrivial element of MATH multiplies by MATH in the MATH factor, so is not inner. The mapping class groups of the octahedral manifolds were calculated in CITE, obtaining groups isomorphic to MATH. The only nontrivial element is MATH, for the cases MATH. Since MATH, this induces a nontrivial outer automorphism. For the remaining cases, we do not know an explicit calculation of MATH. But as the authors of CITE note (and carry out for the case of MATH, for which MATH is the NAME sphere), the following result from their paper leads easily to one. Let MATH be a tetrahedral or icosahedral manifold. Then any diffeomorphism of MATH is isotopic to a fiber-preserving diffeomorphism. Consequently, any diffeomorphism which is homotopic to the identity is isotopic to the identity map. To utilize this theorem, we use the following lemma. Let MATH be an elliptic MATH-manifold MATH with MATH. Suppose that MATH is a diffeomorphism which induces the identity outer automorphism on MATH. Then MATH is homotopic to the identity map. Triangulate MATH. We may assume that MATH is the identity on the MATH-skeleton of MATH (first make it preserve the basepoint and induce the identity automorphism, then make it fix a maximal tree in the MATH-skeleton MATH, then since MATH induces the identity automorphism we may deform it to be the identity on the remaining edges of MATH). Since MATH, we may further make MATH the identity on MATH. There is a subcomplex MATH of MATH such that MATH is obtained by attaching a single MATH-cell MATH to MATH along MATH. The characteristic map of MATH and its composition with MATH agree on MATH, so taking them as the maps on the top and bottom hemispheres defines a map MATH. Via the isomorphism MATH, the degree MATH equals MATH. Since MATH factors through the universal covering MATH, MATH for some MATH. Since MATH and MATH, it follows that MATH has degree MATH and hence MATH is homotopic to the identity map. By the lemma, REF shows that MATH is injective. For the remaining cases, note that if MATH is relatively prime to the order of the finite group MATH, then MATH. In each of these cases, we will show that MATH is injective and that all outer automorphisms of MATH that are realizable by diffeomorphisms lie in the image of MATH, from which it follows using the injectivity of MATH that MATH is an isomorphism. Suppose first that MATH. The abelianization MATH is MATH, on which MATH induces the nontrivial automorphism. From CITE, MATH is MATH, so MATH is an isomorphism. Suppose that MATH with MATH (and hence MATH since MATH must be odd). By REF , every diffeomorphism is isotopic to one that preserves the fiber, so must induce multiplication by MATH or MATH on the MATH factor which is represented by the fiber. So the image of MATH must be contained in MATH, and MATH carries MATH isomorphically to this subgroup. Finally, suppose that MATH is an index MATH diagonal subgroup in MATH. As before, any diffeomorphism must induce multiplication by MATH in the first factor. Multiplication by MATH only preserves the diagonal subgroup if we also apply the nontrivial outer automorphism to the elements of the second factor. So the automorphisms realizable by diffeomorphisms lie in a MATH subgroup of MATH, and MATH carries MATH isomorphically to this subgroup. For MATH, we have from CITE that MATH, but that the nontrivial element is not induced by any diffeomorphism or even homotopy equivalence of MATH. So (as calculated explicitly in CITE), MATH, and we have seen that MATH as well. For MATH, the nontrivial element of MATH cannot be in the image of MATH, for if it could be realized by a diffeomorphism, then this would lift to a diffeomorphism of the covering corresponding to the subgroup MATH (this subgroup is characteristic since MATH). The lift would induce the nontrivial outer automorphism, and we have already seen that this is impossible. So the only realizable outer automorphism is multiplication by MATH on MATH, and MATH carries MATH isomorphically to this subgroup. |
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 show that MATH. The maps MATH and MATH compose to give a linear map MATH, defined in the obvious way. Define a linear map MATH by MATH, for each MATH. Suppose that MATH, and MATH. Then MATH for each MATH. It can be shown that this implies that MATH. Thus MATH is injective. Now from REF and the definitions, it is easy to show that MATH interpreting this equation as we did REF . As MATH is injective, MATH is isomorphic to the right-hand side of REF . By the same argument, MATH is also isomorphic to the right-hand side of REF . This gives a canonical isomorphism MATH. It turns out to be a MATH -isomorphism, and the lemma is complete. |
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, so MATH as MATH is injective. Thus MATH, and MATH is injective. |
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 basis MATH of MATH over the field MATH. Suppose that MATH in MATH, for MATH. Let MATH be nonzero, and such that MATH. Then MATH splits as MATH. Using this splitting, write MATH, with MATH. If MATH, then MATH, so MATH implies that MATH. As this hold for all MATH, we have MATH and MATH. But MATH is a basis over MATH and MATH. Thus MATH, and MATH. We have proved that MATH are linearly independent over MATH. It is now easily shown that MATH, and that MATH, as we have to prove. Now let MATH be semistable, with MATH and MATH. From above, MATH. But we have MATH . Since MATH, MATH, and so REF shows that MATH, with equality if and only if MATH. Thus MATH for all nonzero MATH, as we have to prove. To complete the proposition, it is enough to show that MATH for generic MATH. As MATH is semistable, it is generated by the images MATH. So suppose MATH is generated by MATH for MATH, where MATH. Let MATH, and suppose that MATH for MATH. This is true for generic MATH. Clearly MATH, as MATH. Thus MATH, as MATH. We deduce that MATH is contained in MATH. But MATH is generated by the spaces MATH, so MATH. This completes the proof. |
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-submodule of MATH, and clearly MATH . But MATH, and thus MATH so that MATH, where MATH. Therefore MATH if and only if MATH. Suppose that MATH is a MATH -module, and MATH is a MATH -morphism. Then MATH, so that MATH. We have MATH and MATH. It is easy to show that MATH. Let MATH, and put MATH, and MATH. In this case MATH. The argument above shows that MATH if and only if MATH. But by this holds by REF , as MATH is stable. Thus MATH, and MATH as MATH. It follows that MATH, so MATH. Now MATH is semistable. Therefore MATH is generated by submodules MATH of the above type, and the intersection of the subspaces MATH for all nonzero MATH, must be zero. So MATH, giving MATH, and MATH from above, which completes the proof. |
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 independent over MATH. This will imply that MATH. Since MATH and MATH, we see that MATH, so MATH is semistable. Here is some new notation. Let MATH, and define MATH for all MATH. Similarly define MATH. It can be shown that MATH, and similarly for MATH. Thus MATH. Since MATH is semistable, we can choose nonzero elements MATH and MATH, such that MATH and MATH is a basis for MATH over MATH. Since MATH is stable, MATH for each MATH. Therefore for each MATH we may choose elements MATH of MATH, such that MATH and MATH are linearly independent over MATH in MATH. As MATH is stable, it is not difficult to see that there are nonzero elements MATH and MATH, such that MATH, and for each MATH, the set MATH is linearly independent over MATH. This is just a matter of picking generic elements MATH and MATH, and showing that generically, linear independence holds. We shall also need another property. Define MATH by MATH for MATH. Now as MATH for each MATH, the codimension of MATH in MATH is at most MATH. Since MATH, this gives MATH. The second property we need is that for each MATH, if MATH and MATH for all MATH, then MATH. Here MATH means the linear span over MATH. Again, it can be shown that for generic choice of MATH, this property holds. Now for MATH and MATH, MATH and MATH. REF gives an element MATH in MATH. Moreover MATH. Thus MATH. Therefore, we have made MATH elements MATH of MATH. Similarly, for MATH and MATH, the elements MATH exist in MATH, which gives a further MATH elements of MATH. Since MATH, we have constructed MATH explicit elements MATH and MATH of MATH. Suppose that MATH in MATH, where MATH for MATH, MATH and MATH for MATH and MATH. We shall show that MATH. Now this equation implies that MATH for all MATH and MATH, by REF . Let MATH. Then MATH, so MATH. As this holds for all MATH and the MATH are linearly independent over MATH, it follows that MATH for all MATH and all MATH. By the property of MATH assumed above, this implies that MATH for all MATH. It is now easy to show that MATH for all MATH. Thus the MATH elements MATH and MATH of MATH are linearly independent over MATH, so MATH. But MATH by REF . So MATH, and MATH is semistable. This finishes the proof. |
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 is stable, then MATH is stable. Suppose MATH is stable. Then MATH for all nonzero MATH, so MATH for all nonzero MATH. As MATH is semistable, it is stable, by REF . This completes the proof. |
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 -submodule in MATH. Then MATH, for some MATH with MATH. A calculation shows that for given MATH, the subset of MATH with MATH is of codimension MATH. Thus for generic MATH, MATH, and MATH is semistable. Suppose MATH is semistable. Let MATH. Then MATH if and only if MATH. A computation shows that this fails for a subset of MATH of codimension MATH. Thus the condition MATH for all nonzero MATH fails for a subset of MATH of codimension at most MATH, since this subset is the union of a REF-dimensional family of MATH-codimensional subsets, REF-dimensional family being MATH, the unit sphere in MATH. Therefore for generic MATH, MATH for all nonzero MATH, and MATH is stable. |
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 the map MATH is surjective. Since MATH are finite-dimensional, it follows that MATH and MATH. Under these isomorphisms MATH is identified with MATH. Thus MATH is surjective. Using exactness we see that MATH and MATH. By subtraction we find that MATH. But we have already shown that MATH is injective, and MATH is surjective. Using these facts, we see that the sequence REF is exact, as we have to prove. Now MATH is injective, and clearly MATH is injective. Thus REF shows that MATH is injective. So the sequence REF is MATH -exact at MATH, as we have to prove. Suppose that MATH. It is easy to show, using exactness, injectivity, and the definition of MATH, that MATH. Thus MATH, and the sequence REF is exact at MATH, as we have to prove. Recall from REF that as MATH are finite-dimensional, MATH is surjective. Suppose that MATH. Then MATH for some MATH. The map MATH is surjective from above, and thus MATH for some MATH. Thus MATH. But it is easy to see that MATH, as maps MATH. Thus if MATH, then MATH, for MATH. Therefore MATH is surjective. Let MATH, and suppose MATH. As REF is exact, MATH, and as MATH is injective, MATH is unique. We shall show that MATH. It is enough to show that for each MATH, MATH. Since MATH is surjective, MATH for MATH. Then MATH. But MATH and MATH, so MATH. Thus MATH, and MATH. It follows that the sequence MATH is exact at MATH. Therefore REF is MATH -exact at MATH, as we have to prove. The example above shows that the sequence need not be MATH -exact at MATH, and the proposition is finished. |
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 product of general MATH -modules. Therefore MATH . By REF , the sequence REF is MATH -exact at MATH and MATH. The only way MATH -exactness at MATH can fail is for MATH or MATH not to be surjective. We deduce that MATH . But MATH is MATH -exact, so that MATH and MATH. Combining REF , we see that equality holds in REF , because the inequalities go opposite ways. Counting dimensions, MATH and MATH must be surjective. Thus by definition, REF is MATH -exact at MATH, and the proposition is proved. |
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 maximal semistable MATH -submodule of MATH. As MATH is semistable, MATH. Also, from above the map MATH is surjective. As MATH is semistable, we deduce that MATH. But MATH is MATH -exact, so MATH and MATH imply that MATH. Therefore MATH is semistable, so MATH is stable. |
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 induction on MATH, MATH is stable. Now let MATH be as given, and let MATH be the natural projection, for MATH. Because MATH and MATH are MATH -exact sequences of stable MATH -modules, two applications of REF show that the sequence MATH is MATH -exact at the middle term. Now MATH maps MATH to MATH. By identifying the kernels of MATH and MATH, it can be seen that there exists a linear map MATH as in the lemma, such that MATH as maps MATH. Using the MATH -exactness of REF and that MATH is a MATH -morphism, we deduce that MATH is unique and is a MATH -morphism. |
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 leave this to the reader. As MATH satisfies Axiom AG, MATH is graded, so MATH is an NAME, and the proposition is complete. |
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 q-holomorphic. This completes the lemma. |
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. Then the function MATH is equal to some MATH. Thus MATH. But the functions in MATH are q-holomorphic, so MATH. Therefore MATH, and similarly MATH. So MATH, and MATH is q-holomorphic, as we have to prove. It is clear that the space MATH of such functions MATH is closed under addition and multiplication by MATH. Thus MATH is a MATH-submodule of MATH, so MATH is a MATH -submodule of MATH. Now let MATH. Then MATH, so MATH defines a linear map MATH. Define a map MATH by MATH. For MATH, define MATH by MATH. Since MATH, MATH, so MATH, regarding MATH as an element of MATH. Thus MATH. Similarly, defining MATH for MATH, we find MATH for each MATH. To show that MATH, we only need to show that MATH is smooth. In general, if MATH is a function on MATH, such that for each MATH, the function MATH is smooth, and for each MATH, the function MATH is smooth, it does not follow that MATH is smooth. However, because real tensor products involve only finite sums as in REF, MATH implies that MATH lies in some finite-dimensional subspace of MATH for all MATH, and similarly MATH lies in a finite-dimensional subspace of MATH for all MATH. These imply that MATH is smooth. Thus MATH. Define MATH. In this way we define a map MATH. It is easy to show that MATH is a MATH -morphism. Also, if MATH it is easily seen that MATH, so MATH is injective. Thus MATH is an injective MATH -morphism, as we have to prove. Suppose MATH are finite-dimensional, and let MATH. We must find MATH in MATH such that MATH. Choose bases of the form MATH for MATH and MATH for MATH. It can be shown that such bases exist. Let MATH be the unique linear map satisfying MATH for MATH, MATH. One may prove that MATH, and MATH. Thus MATH is an injective and surjective MATH -morphism, and clearly is a MATH -isomorphism. This completes the proof. |
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 MATH be the `diagonal' submanifold in MATH, and let MATH be the `diagonal' submanifold in MATH. We interpret part MATH as follows. MATH is a space of q-holomorphic functions on MATH. The maps MATH and MATH are the maps restricting to MATH and MATH respectively. Thus MATH is the result of first restricting to MATH and then to MATH, and MATH is the result of first restricting to MATH and then to MATH. Clearly MATH, proving part MATH. Interestingly, the proof of REF does not use the associativity of quaternion multiplication. This raises the possibility of generalizing the definition to give `associative algebras over a nonassociative field'. Part MATH is given in REF . Finally, REF follows easily from the fact that MATH is the identity in MATH. Thus all of Axiom A of REF applies, and MATH is an NAME. |
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 by MATH for MATH. It is easy to see that if MATH, then MATH is a MATH-valued polynomial on MATH of degree at most MATH. But MATH if and only if MATH for each MATH. Thus MATH is the zeros of a collection of polynomials on MATH. By NAME 's CASE: Theorem, we can choose a finite number of polynomials, which define MATH. |
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 dimension MATH, and MATH, so MATH, as we have to prove. If MATH, then MATH. But MATH determines MATH and MATH. (For instance, MATH if and only if MATH, so MATH determines MATH.) Thus if MATH, then MATH determines the hypercomplex structure on MATH. Now MATH is a MATH-submodule of MATH, and it can be shown that if MATH is a MATH-submodule of MATH such that MATH, MATH, and MATH is close to MATH, then MATH determines a different hypercomplex structure. It is easy to see that if MATH, then we may choose such a MATH with MATH. Thus MATH is consistent with two different hypercomplex structures. Therefore MATH determines the hypercomplex structure on MATH if and only if MATH. |
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. But MATH, so MATH. By REF , using MATH, we see that the virtual dimension of MATH is greater or equal to that of MATH. Using the stability of MATH, we then prove that for each nonzero MATH, the image of MATH lies in MATH, and as MATH is a proper MATH -submodule, this contradicts the semistability of MATH. Thus MATH for all MATH, and the proposition is complete. |
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 all such functions are in fact polynomials of degree MATH on MATH, by a classical result about harmonic functions of polynomial growth on MATH. Therefore, we may decompose MATH on MATH into a sum of homogeneous polynomials on each MATH. Clearly, these homogeneous pieces lie in MATH for MATH, so MATH. We have shown that if MATH, then MATH, and MATH. But the inclusion MATH is immediate, so MATH. Clearly, the filtrations on MATH and MATH agree, and the lemma is complete. |
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 MATH on MATH. Thus MATH identifies the NAME brackets on this subspace. Because MATH is an NAME morphism and MATH generates MATH, we can deduce from REF that if MATH and the pullback of MATH agree on MATH, they must agree on the whole of MATH. Thus MATH identifies the NAME brackets of MATH and MATH, and MATH is an NAME morphism. |
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 will be based on the results in the paper of CITE, plus some graphic computations for REF and some purely algebraic computations for REF . CASE: First, the relations REF are easily verified by drawing pictures. Let us show that the MATH-subalgebra MATH of MATH is equal to MATH. First, MATH contains the infinite sequence of black and white NAME projections MATH as well as the infinite sequence of bicolored NAME projections MATH which by the result of NAME and NAME CITE generate the diagonal MATH-algebra MATH. (If MATH is a MATH-algebra, its diagonal MATH is defined by MATH and MATH if MATH.) Thus we have inclusions MATH so we can use the following standard argument. First, we have MATH. Second, the existence of semicircles shows that the objects of MATH and MATH are selfdual and by NAME reciprocity we get MATH for MATH even. By tensoring with MATH and MATH we get embeddings MATH and this shows that the dimension equalities hold for any MATH and MATH. Together with MATH, this shows that MATH. CASE: Assume that MATH, MATH and MATH in a MATH-algebra MATH satisfy the relations REF . We have to construct a morphism MATH sending MATH . This will be done in two steps. First, we restrict attention to diagonals: we would like to construct a morphism MATH sending MATH . By constructing the corresponding NAME projections MATH and MATH, we must send MATH . The presentation result for MATH of NAME and NAME CITE reduces this to an algebraic computation. More precisely, it is proved in CITE that the following relations REF MATH, MATH if MATH and REF MATH and REF MATH and MATH if REF MATH and REF MATH and MATH are a presentation of MATH. So it remains to verify that MATH where MATH, MATH are MATH are abstract objects and we are no longer allowed to draw pictures. First, by using MATH and MATH these relations reduce to: CASE: MATH for MATH, MATH and MATH. CASE: MATH for MATH and MATH REF MATH and MATH for MATH REF MATH and MATH REF MATH REF MATH REF MATH . With MATH, MATH, MATH and MATH one can see that most of them are trivial. What is left can be reformulated in the following way. MATH . By multiplying the relation REF by MATH and by MATH to the right we get the following useful formula, to be used many times in what follows. MATH . Let us verify REF . First, we have MATH and by replacing MATH with MATH we get MATH, so REF is true. We have MATH and by replacing MATH with MATH we get MATH, so REF is true. We have MATH and by replacing MATH with MATH, then MATH with MATH we get MATH, so REF is true. The first two commutators are zero, because MATH and MATH are selfadjoint. Same for the others, because of the formulae MATH . The conclusion is that we constructed a certain MATH-algebra morphism MATH that we have to extend now to a morphism MATH sending MATH and MATH. We will use a standard argument (see CITE). For MATH bigger than MATH and MATH we define MATH where MATH and where the convention MATH is no longer used. We define MATH and MATH in MATH by similar formulae. We have MATH. We define a map MATH by MATH . As MATH doesn't depend on the choice of MATH, these MATH's are the components of a map MATH. This map MATH extends MATH and sends MATH and MATH. It remains to prove that MATH is a morphism. We have MATH as well as a similar description of MATH, so MATH sends MATH to MATH. On the other hand we have MATH, so MATH on MATH. Thus MATH commutes, so MATH is multiplicative: MATH . It remains to prove that MATH. We have MATH for certain MATH and MATH, so it is enough to prove it for pairs MATH of the form MATH or MATH. For MATH this is clear, so it remains to prove that the set MATH is equal MATH. First, MATH being a MATH-algebra morphism, we have MATH. On the other hand, computation gives MATH. Also, MATH being involutive and multiplicative, MATH is stable by involution and multiplication. We conclude that MATH contains the compositions of elements of MATH with MATH's and MATH's. But any MATH in MATH is equal to MATH, so it is of this form and we are done. |
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. We denote by MATH the orthonormal basis MATH of MATH. We have MATH so MATH is a MATH-form if and only if MATH. On the other hand, we get MATH so the formula MATH in REF is equivalent to the fact that MATH is a MATH-form on MATH. It remains to check the following three formulae, with MATH. MATH . By using the fact that MATH is a bimodule morphism we get succesively that MATH where MATH is such that MATH. By using the above formula for MATH we get MATH . This allows us to prove the first MATH formula, because we have MATH for any MATH. The second MATH formula follows from MATH with MATH, for any MATH. For the third MATH formula, we have MATH and this gives MATH with MATH given by MATH where MATH is the multiplication of MATH, which was computed in a similar way. Thus MATH satisfy the relations REF , so REF applies and gives a certain MATH-algebra surjective morphism MATH. It remains to prove that MATH is faithful. For, consider the maps MATH where MATH. These make the following diagram commutative MATH and by gluing such diagrams we get a factorisation by MATH of the composition on the left of conditional expectations, which is the NAME trace. By positivity MATH is faithful on MATH, then by NAME reciprocity faithfulness has to hold on the whole 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 other hand, the linear form MATH being a MATH-form, REF applies and gives an isomorphism MATH. |
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 that the fundamental representation MATH of MATH satisfies MATH for any MATH and MATH. With MATH and MATH we get MATH . Together with the canonical isomorphism MATH, this shows that the action of MATH on MATH is ergodic. In particular MATH, so by the above, if MATH is a MATH-factor with minimal action of MATH (the existence of such a MATH is shown by NAME in CITE) then MATH is a subfactor and we are done. |
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. Combining this with REF yields the second part. |
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 the amoebas of MATH and MATH respectively. From REF it follows that MATH. On the other hand, since MATH and MATH, it follows from REF that MATH . This is a contradiction. |
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 the two projections MATH and MATH onto MATH and MATH defined by MATH and MATH. Let MATH. Since the map MATH has discrete fibers, it follows that MATH is a real curve. Hence MATH is a null set in MATH. Since REF has no solutions in MATH for MATH outside MATH, it follows that MATH as required. |
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 reason branches of MATH cannot have singular points over MATH. Indeed, the links of such points are unknots since neighborhoods of those points map REF to small half-disks from MATH. By REF the image of each branch of MATH under MATH has a convex complement. Therefore the images of branches of MATH cannot intersect (that would produce points of MATH with at least REF inverse images under MATH). Thus the only singularities of MATH are intersection points MATH of a pair of conjugate non-singular imaginary branches. If these branches are not transverse then they have a real tangent line MATH. The points of MATH close to MATH will be covered at least twice by each of the two branches of MATH which leads to a contradiction. We conclude that the only singularities of MATH are MATH-singularities. |
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 exists such an element MATH that MATH. Therefore MATH is not an intersecter for MATH, but this contradicts our choice of MATH. Hence, REF holds. |
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 that MATH . This means that MATH and completes the proof. |
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. Taking this inclusion into account, we assign to the antichain MATH the antichain MATH which is the blocker of MATH, by REF . Then MATH, and the theorem follows. |
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 while multiplying with MATH from the right similarly leads to MATH . The coefficients MATH are determined by these two recursion relations together with the initial conditions MATH, MATH. It is easy to check that the solution is indeed given by MATH. |
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 the constant functions MATH. After simplifications one arrives at MATH which is equivalent to the two different NAME 's triangle identities for the MATH-binomial coefficients CITE: MATH . The case MATH of the proposition, which may be written as MATH may be proved in the same way. Multiplying this expression and REF gives MATH by REF . Putting MATH and simplifying completes the proof. |
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, again with the leading coefficient MATH. Combining these facts and using REF , we can for MATH write MATH with MATH. Again by NAME elimination, each element MATH is in the MATH-span of MATH. Similarly, MATH is in the MATH-span of MATH. Thus, by REF , the matrix elements span MATH. A dimension count completes the proof. |
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 MATH preserves the grading we have MATH with MATH. By REF, the intertwining property means that MATH for all MATH, MATH. By REF , this implies that MATH is independent of MATH and MATH, which completes the proof. |
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 . |
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, where MATH . Writing the NAME - NAME polynomial as a MATH and inverting the order of summation, we obtain the desired expression. In the remaining three cases of REF , the lemma can be proved similarly. Alternatively, one can use REF , which leads to a more involved computation, but allows one to treat all four cases simultaneously. |
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 by MATH, and is then projected to MATH. Here we may indeed replace MATH by MATH. For the second part, we write down the relations valid on both sides of REF explicitly. Those on the left-hand side may be written as MATH where, for instance, the first identity is an abbreviation for MATH and those on the right-hand side as MATH . We must show that the second group of relations implies the first group. This is clear except for the relation MATH. However, by REF we have also that MATH, and thus indeed 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 for MATH given in REF into the right-hand side of REF, commute MATH across this expression (using REF ) and then again apply REF to the factor MATH. This results in an identity involving only commuting variables, which turns out to be the terminating MATH summation formula CITE: MATH with MATH a formal quantity satisfying MATH. |
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 to this identity and using that MATH gives MATH . Comparing with REF we see that MATH. |
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). The action of MATH on MATH determines the homomorphism MATH; if the action is trivial then the homomorphism is trivial and the action of MATH on MATH is trivial. |
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 every MATH is a MATH - partial NAME class. The inductive step is next. Assuming that MATH is a MATH - partial NAME class, we show that some cup power MATH is a MATH - partial NAME class. Let MATH be a fibration sequence for which a generator of MATH transgresses to MATH. By REF , the action of MATH on MATH is trivial. So, the MATH term of the NAME spectral sequence for cohomology with coefficients in MATH is MATH . Let MATH be the MATH-invariant and let MATH denote the cohomology class represented by MATH. The class MATH lies in MATH and survives to MATH. The differential MATH is a class in MATH; if MATH is sufficiently large, this group is isomorphic to MATH and MATH. If MATH then MATH is transgressive and MATH. Consider the diagram MATH having as its first row the MATH-fold fiberwise join of the fibration MATH and as its second row the fibration obtained by the fiberwise application of the NAME section functor MATH to the first row. Applying the NAME section functor to the natural map MATH gives a weak equivalence MATH . In dimension MATH one has MATH . The induced map in homology is the MATH-fold sum of maps MATH. By symmetry MATH for all MATH. To understand MATH consider the composite MATH . The adjoint is the map MATH which is determined on the bottom cell. It is obtained by applying the functor MATH to the inclusion MATH. It follows that MATH is the suspension homomorphism and therefore that the map MATH is the fold map followed by the suspension homomorphism. Let MATH denote the MATH-invariant MATH and let MATH be the cohomology class that it represents. By naturality of differentials it follows that MATH is the image of MATH under the map MATH that is induced by the suspension homomorphism on the coefficients. Now the image of the suspension homomorphism is a finite group and if MATH is the exponent of the image then MATH. On the other hand, if MATH then MATH transgresses to a class MATH which is well defined modulo the ideal generated by MATH and MATH transgresses to MATH modulo the ideal generated by MATH where MATH is the image of MATH under the map MATH induced by the suspension homomorphism. Again, if MATH is the exponent of the image of the suspension homomorphism then MATH transgresses to zero modulo the ideal generated by MATH. Let us now choose MATH to be the square of the exponent of the image of the suspension homomorphism. Then by construction, MATH survives to MATH in the NAME spectral sequence. This implies that the map MATH extends to a map on MATH and there is a diagram MATH in which each row and each column is a fibration sequence. Taking the fibration sequence in the first row completes the inductive step. |
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 a map on MATH if and only if the cohomology class represented by MATH survives to MATH in the NAME spectral sequence for cohomology with coefficients in MATH. There are two differentials which might be non-trivial on the class represented by MATH. These are MATH, taking values in the kernel of multiplication by MATH; and MATH, taking values in MATH. But by hypothesis both obstruction groups are zero and one can construct the diagram MATH in which each row and each column is a fibration sequence. Let MATH. The homotopy fiber of the map MATH is weakly equivalent to MATH. By construction the integral cohomology of MATH vanishes above dimension MATH. If in addition one has the condition on cohomology with local coefficients that is given in the hypotheses, then it follows from the NAME spectral sequence with local coefficients that MATH has finite cohomological dimension. So by a result of CITE, MATH is homotopy equivalent to a finite dimensional CW complex. From this we obtain a spherical fibration over MATH with total space MATH homotopy equivalent to a complex of dimension less than MATH. |
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 and the fiber join MATH are finitely dominated. It follows that one has an equation in MATH: MATH . Since MATH is a spherical fibration with a finitely dominated base and an odd dimensional sphere as the fiber, we conclude that MATH (see CITE). Repeated application allows us to prove the lemma. |
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 CITE, MATH classifies fibrations MATH for which the map MATH is trivial. In particular, MATH classifies orientable spherical fibrations. The lemma will be proved by induction on MATH; the case MATH being trivial. Now, assume a trivialization over MATH; the obstructions to extending this to a trivialization over MATH are classes MATH, where MATH is the fiber and MATH ranges over the MATH - cells in MATH. Taking the MATH - fold fiber join replaces this obstruction with MATH, where MATH is the natural stabilization map defined by MATH. Now we know that for a fixed MATH, MATH provided MATH is sufficiently large. Note that these stable homotopy groups of spheres are all finite. Hence given MATH as above, we can choose a large integer MATH such that MATH, ensuring the vanishing of all the obstructions and from there the extension of the trivialization. |
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, assume that MATH has periodic cohomology. Let MATH be an integral cohomology class and MATH be an integer such that MATH is an isomorphism for MATH. By REF , there is an integer MATH and a fibration sequence MATH such that a generator of MATH transgresses to MATH. By REF , since MATH there is an orientable spherical fibration with NAME class MATH, and the total space has the homotopy type of a finite dimensional complex. |
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 is an orientable spherical fibration MATH such that MATH is homotopy equivalent to a finite dimensional CW - complex. By forming the fiber join of the map MATH with itself if necessary we may assume that the fiber of the map MATH is a homotopy sphere of odd dimension MATH and hence that MATH. Since MATH has finite type, MATH has finite type. Hence MATH is finitely dominated and its finiteness obstruction is an element MATH. By REF has finite order MATH. The total space of the MATH-fold fiber join MATH is also finitely dominated. By REF , the finiteness obstruction of MATH is MATH. Hence MATH is homotopy equivalent to a finite complex. |
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 is equivalent to the projectivity of the modules MATH (see REF). Now let MATH denote the cellular co-chains on MATH, and consider the exact sequence MATH . By construction, MATH is MATH-projective, hence cohomologically trivial - this yields a cohomology isomorphism with any coefficients MATH for all MATH which by construction is induced by multiplying with MATH, where MATH is the usual bundle map. For the converse observe that periodicity implies that for any prime MATH, MATH will have NAME dimension one or zero, hence by a theorem due to CITE, all the isotropy must be of rank one that is, with periodic cohomology. |
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 than a fixed integer MATH. By REF there is a spherical fibration MATH such that MATH is homotopy equivalent to a finite CW - complex and the fundamental class of the fiber transgresses to a cup power of MATH. The MATH cover of MATH is homotopy equivalent to a sphere MATH and so the map MATH factors up to homotopy through the MATH-skeleton of MATH. By REF there is an integer MATH such that the spherical fibration MATH is a product fibration when restricted to the MATH-skeleton of MATH. Hence the map MATH in the cartesian square MATH is a product fibration and MATH is a MATH-equivariant spherical fibration such that the fundamental class of the fiber transgresses to a cup power of MATH. So the complex MATH has the required properties. |
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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