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math/0008108 | Let MATH. For each MATH let MATH denote the corresponding NAME coordinate on MATH. In terms of NAME coordinates, MATH acts on a point of MATH with coordinates MATH by taking it to the point with coordinates MATH. Let MATH be NAME coordinates of MATH in MATH. By assumption, MATH for every MATH. Let MATH be defined by th... |
math/0008108 | Let MATH . For each MATH let MATH denote the corresponding NAME coordinate on MATH. For each MATH let MATH denote the corresponding NAME coordinate on MATH. Let MATH. Let MATH be NAME coordinates of MATH in MATH and MATH be NAME coordinates of MATH in MATH. By assumption, MATH for every MATH and MATH for every MATH. Le... |
math/0008108 | We take two preliminary steps. CASE: If MATH, then MATH and either MATH or MATH. Indeed, let MATH, and assume that MATH. Let MATH . Then MATH. Now, MATH is not a base point of MATH for any MATH by REF . So MATH is the minimum order of MATH at MATH for each MATH. Analogously, MATH is the minimum order of MATH at MATH fo... |
math/0008108 | By REF , MATH for each MATH and each MATH. In other words, each MATH is homogeneous. Let MATH. Clearly, MATH if and only if MATH. So, by REF , MATH if and only if MATH, MATH . On the other hand, MATH and MATH for each MATH and MATH. So, as MATH runs over MATH, the associated numerical data MATH runs over a finite set. ... |
math/0008108 | To ease notation, let MATH . Let MATH consist of the points of the form MATH for all MATH satisfying MATH where MATH means the smallest non - zero number among the four listed. If REF holds, then MATH for every MATH, and hence MATH. We take three preliminary steps. CASE: For each MATH there are ordered tripartitions MA... |
math/0008108 | Let MATH. By REF , we may assume that MATH. Let MATH be the open neighborhood of MATH given by REF . To ease notation, let MATH . In addition, if MATH let MATH. Let MATH, MATH, MATH and MATH be the invertible sheaves given by REF . As in REF, let MATH denote the blow - up of MATH along MATH. Fix isomorphisms MATH and M... |
math/0008108 | By REF , the covering MATH of MATH is finite. Hence, it follows from REF that MATH is projective. Let's prove that MATH is connected. Let MATH. For each MATH let MATH . By REF , each MATH is an interval, and finitely many MATH suffice to get a covering MATH of MATH. Assume MATH. Then MATH and MATH. By REF , for each MA... |
math/0008108 | Let MATH, and denote by MATH the curve gotten from MATH by splitting the branches of MATH at each MATH and connecting them by a chain of MATH rational smooth curves; see REF Let MATH . Let MATH. By definition, MATH and MATH are the limit canonical aspects with foci on MATH and MATH of a certain regular smoothing MATH o... |
math/0008108 | Assume that MATH. Then we may choose MATH such that MATH. Define MATH by letting MATH for each MATH. Then MATH and MATH. Let MATH . Then MATH and MATH. For each MATH let MATH . Let MATH be the dense open subset of points MATH such that MATH and MATH. Then MATH for each MATH. We claim that there are MATH such that MATH ... |
math/0008108 | If MATH then MATH is simply a point by REF , thus proving REF in this case. We may assume from now on that MATH. (The case where MATH is completely analogous.) Given MATH, the number of MATH such MATH is MATH. Thus the number of MATH such that MATH is MATH. If MATH then MATH if and only if MATH. Indeed, MATH because MA... |
math/0008109 | Recall that MATH is the odd automorphism given by the matrix REF . Consider first MATH, that is, the MATH-invariants in MATH. Letting MATH and MATH, it is clear that this space consists of elements of the form MATH and MATH, and hence is isomorphic to MATH. Therefore, since MATH is a subgroup (since all elements are no... |
math/0008109 | By NAME - NAME duality REF we have MATH as MATH module, where the summation is over strict partitions MATH with length MATH. Therefore combined with REF this gives us MATH . Now by REF and the fact that irreducible MATH-modules are self-contragredient we have MATH where the summation is over all strict partitions of le... |
math/0008109 | We will establish REF together. The argument we will present follows closely the one used in CITE to derive the NAME duality from the MATH-duality. The MATH-duality says that MATH, where the summation is over all strict MATH with MATH. Observe that the space MATH may be identified with the zero-weight space MATH. Putti... |
math/0008112 | Let MATH, for MATH. Clearly, if MATH, then MATH for each MATH. A simple induction argument shows that it suffices to prove the proposition when MATH, which we shall assume for the remainder of the proof of REF . The Proposition will be a consequence of the following lemma. Let MATH, MATH be as in REF with MATH. If MATH... |
math/0008112 | Let MATH, and hence MATH. Without loss of generality, we may assume that MATH, MATH, are linearly independent over MATH and hence MATH, MATH, are also linearly independent over MATH. We write MATH, with MATH, and for example, MATH for the MATH . Jacobian matrix of MATH with respect to MATH. Consider the MATH-linear sys... |
math/0008112 | We shall prove the proposition by induction on MATH. If MATH, then the proposition holds with either MATH or MATH, depending on whether or not all vector fields in MATH vanish at MATH. Now assume that the proposition holds for MATH; we shall prove it for MATH. If all the MATH vanish at MATH, then again MATH is the only... |
math/0008112 | It suffices to prove the proposition using the special choice of coordinates MATH, MATH made above. Thus, we may assume that MATH, MATH. Since MATH is a formal manifold of codimension MATH and MATH, there are MATH such that MATH is generated by MATH. By using the special form of the MATH (that is, substituting MATH for... |
math/0008112 | By REF , it follows that the iterated NAME mapping MATH is of the form MATH where MATH . In particular, we obtain MATH . The identity REF is an immediate consequence of REF . This completes the proof of REF . |
math/0008112 | By iterating REF , we obtain, for any multi-index MATH, MATH . In particular, applying REF with MATH, we conclude that, for each multi-index MATH, MATH . Since MATH, MATH are tangent to MATH, we deduce that MATH for all MATH and all multi-indices MATH. Hence, MATH, which proves that MATH and MATH. Assume that MATH and ... |
math/0008112 | The implication MATH is clear. The opposite implication follows by an inductive argument based on the facts that MATH is a homomorphism, and MATH and MATH are derivations. For instance, we have, for MATH, MATH . A similar identity holds for MATH. The details are left to the reader. The equivalence MATH follows by simpl... |
math/0008112 | A simple computation (which is left to the reader) shows that REF holds with MATH and MATH . |
math/0008112 | Among MATH and all their repeated commutators, we can pick out, by the assumption in the lemma, MATH such that MATH evaluated at MATH span MATH. By REF , we may assume that there are MATH and MATH, MATH for MATH, such that MATH . Let us write MATH . By assumption the vectors MATH evaluated at MATH span MATH. It is not ... |
math/0008112 | Let MATH be such that MATH form a basis for MATH. By multiplying the MATH by suitable power series (clearing the denominators), we obtain MATH and MATH still form a basis for MATH since MATH is linear over MATH. |
math/0008112 | The main step in the proof is to show that there are formal vector fields MATH on MATH and MATH, MATH for all MATH, such that MATH . To prove this, we write MATH and denote by MATH the homomorphism defined by MATH for MATH. By abuse of notation, we also denote by MATH the mapping given by applying MATH to each componen... |
math/0008112 | We first note that the conclusion of the theorem is independent of the choice of the NAME variety mapping MATH. This can be seen by using the notation of REF and observing that MATH, where MATH is a parametrization of the formal manifold MATH . Hence it suffices to prove REF using the special choice of NAME variety map... |
math/0008114 | CASE: Suppose MATH and MATH. Then MATH as MATH implies MATH as MATH and that there exists a MATH such that MATH. Let MATH be given by MATH. Then it is not difficult to see that MATH is an isomorphism. Therefore MATH is isomorphic to MATH by REF . And by the same argument MATH is isomorphic to the groupoid MATH-algebra ... |
math/0008114 | Suppose that MATH is the edge set of the MATH-th coordinate space of MATH. Then by REF MATH . For MATH, MATH with MATH and the canonical projection to the MATH-th coordinate space MATH, define MATH and MATH by MATH . Then every MATH is a unitary element in MATH, and MATH is generated by MATH. If we denote MATH as MATH,... |
math/0008114 | We have the following NAME exact sequence. MATH . We consider MATH as the automorphism MATH given by the NAME isomorphism of NAME. Define MATH by MATH for MATH. Then the induced automorphism MATH is the required isomorphism. For MATH, let MATH be the induced unitary element as in the proof of REF . Then MATH is homotop... |
math/0008127 | Let MATH denote the unitary group of MATH, endowed with NAME measure MATH. Let MATH . For each MATH, we get MATH so MATH for all MATH. Furthermore, MATH . Lastly, for MATH, set MATH . Assume now that MATH, that is, MATH for all MATH. Then MATH . Since whenever MATH, MATH and any MATH is a linear combination of unitarie... |
math/0008127 | Let MATH be as above. Consider the NAME - NAME MATH - algebra MATH, generated by the operators MATH and MATH. Then MATH. The automorphism MATH is identified with the NAME automorphism MATH: indeed we have MATH, MATH and MATH. Choose MATH as in REF . Let MATH. Then MATH. Let MATH. It follows that MATH. It is easily seen... |
math/0008127 | Fix MATH. By the lemma in CITE we can find a MATH such that for every pair of elements MATH in the unit ball of MATH such that MATH we have the implication MATH . Let MATH. So assume that MATH is a finite set such that each MATH, MATH and MATH, MATH and MATH is an element of MATH such that MATH for MATH. By our choice ... |
math/0008127 | By CITE it suffices to show the inequality MATH. Let MATH be given and MATH be any finite set containing the unit of MATH and such that MATH, MATH. Let MATH be an approximate unit such that MATH, for all MATH. Since we can manufacture a quasicentral approximate unit out of the convex hull of MATH CITE, we may further a... |
math/0008127 | Denote by MATH the range of MATH. Define recursively MATH . Let MATH be the cardinality of MATH. Then MATH has dimension at most MATH times the dimension of MATH, so that MATH. Let MATH be the orthogonal projection onto MATH; then MATH are clearly an increasing sequence, and MATH for all MATH. Let MATH . Then MATH for ... |
math/0008127 | We have a short exact sequence MATH with splitting MATH given by MATH. Let MATH, MATH and MATH be self-adjoint elements, each of norm at most MATH, so that each MATH has finite rank. Let MATH be the sum of the ranks of MATH. Let MATH . Fixing MATH and a positive integer MATH, let MATH . Then the sum of the ranks of MAT... |
math/0008127 | REF follows from REF and the results of CITE. Hence it is sufficient to prove REF ; for that one only needs to prove that MATH, in view of the behavior of entropy with respect to inductive limits. We now proceed by induction on MATH. Since MATH, the statement is true for MATH. Applying REF to MATH gives MATH, which is ... |
math/0008127 | Let MATH be the GNS NAME space associated to MATH, and let MATH be the cyclic vector associated to MATH. Let MATH be the NAME algebra on two generators CITE. Without loss of generality, by replacing MATH with MATH and MATH with MATH we may assume that the GNS representation MATH satisfies MATH. Let MATH be the unitary ... |
math/0008127 | Since MATH is free from MATH and commutes with MATH, we see that for all MATH, MATH. Now let MATH be such that MATH. Then MATH because MATH and MATH are free with respect to MATH. But then MATH and MATH are free with respect to MATH, since for all MATH, MATH, and any element in MATH has the form MATH, for some MATH. Th... |
math/0008127 | We will denote by MATH the defining map from MATH onto a dense subspace of MATH, and similarly for the other MATH spaces. We will also need the isomorphism MATH given by MATH. CASE: Let MATH. Choose MATH so that MATH, that is, MATH. There exists and element MATH such that MATH that is, MATH . Thus MATH. CASE: Given MAT... |
math/0008127 | Consider in MATH the unitary MATH. Consider the subalgebras MATH . Since MATH is free from MATH with amalgamation over MATH, it is easily seen from the definition of freeness and REF that the algebras MATH and MATH are also free with amalgamation over MATH. Furthermore, the restriction of MATH to MATH is MATH, and henc... |
math/0008127 | Since MATH has trivial MATH - theory, it follows from NAME 's exact sequence for free products (see REF) that MATH is zero. Once MATH is known to be simple and purely infinite, it will follow from NAME 's fundamental results CITE, that there exists a partial isometry MATH so that MATH and MATH. Set MATH. Then MATH, in ... |
math/0008127 | Consider the free product conditional expectation MATH. The GNS representation associated to MATH is faithful, by definition. Let MATH. By REF , MATH also gives rise to a faithful GNS representation. By REF, MATH and MATH are free with respect to MATH. It follows from the assumptions and the embedding result (see REF) ... |
math/0008127 | Using the embedding result (see REF), we can reduce to the case that MATH and MATH, by replacing MATH with the algebra generated by MATH and MATH. We have the following sequence of covariant inclusions, justified below: MATH . Inclusion REF is implied by REF , together with the embedding result (see REF). Isomorphism R... |
math/0008127 | Because of the embedding result (see REF) and the behavior of entropy with respect to inductive limits, it suffices to prove the statement when MATH is a set with two elements. By monotonicity of MATH, the inequality MATH in REF is clear. Let MATH be the NAME algebra generated by the unilateral shift MATH and denote by... |
math/0008127 | Consider the reduced group MATH - algebra MATH of the nonabelian free group on MATH generators, where MATH is the cardinality of MATH. Let MATH be the unitary generators of MATH corresponding to the free generators of MATH. Let MATH be the automorphism such that MATH. Then from CITE and CITE, MATH, Let MATH where MATH ... |
math/0008127 | If MATH has no cycles then MATH is (conjugate to) a free permutation so MATH by REF . In general, by making a cycle decomposition of MATH and using REF , we see that in order to prove REF we may without loss of generality assume that MATH itself is a cyclic permutation, MATH, of a finite set MATH. However, then MATH, w... |
math/0008127 | Assume without loss of generality that MATH. Given MATH we can find a MATH - invariant state MATH such that MATH. Let MATH be the canonical conditional expectation and define a state MATH. Then one checks that MATH and MATH. But under these conditions, CNT - entropy is also monotone (compare CITE) and so MATH. Since MA... |
math/0008127 | Since MATH is faithful, by REF the GNS representation of MATH is faithful. Consider the MATH - algebra MATH, where MATH is the free shift and let MATH be the implementing unitary. It is fairly easy to see (compare the proof of CITE) that there exists a MATH - isomorphism MATH such that MATH. Hence MATH. But REF above a... |
math/0008127 | Replacing MATH with MATH, if necessary, we may assume that there exists a NAME unitary in the centralizer of MATH. In this setting, CITE ensures that MATH is a simple MATH - algebra, where MATH is the NAME algebra and MATH is the vacuum state. Moreover, this free product is nuclear since it is isomorphic to a NAME - NA... |
math/0008127 | By adding a unit to MATH and replacing MATH by its crossed product by MATH, we may assume that MATH is unital and MATH is inner. We may also assume that there exists a MATH - invariant state MATH with faithful GNS representation. Indeed, it follows from REF that the covariant embedding MATH, where we regard MATH and MA... |
math/0008130 | Let MATH and MATH be as in the statement and MATH. Because MATH is surjective, it follows that MATH are linearly independent at MATH. Each function MATH is the defining function of a hyperface because MATH must map faces of codimension MATH to faces of codimension MATH. |
math/0008130 | We may assume that MATH is dense in MATH. Fix MATH and we let MATH if MATH, for some MATH and MATH. We need to show that MATH is well-defined and bounded. Thus, we need to prove that MATH, if MATH and MATH, for some MATH and MATH. We will show that, for each MATH, there exists a constant MATH such that MATH . Let MATH,... |
math/0008130 | This is just the summary of the above discussion. |
math/0008130 | For brevity, let MATH be the space on the right-hand side in REF. Also, let MATH. Then we get for all MATH that is, MATH, and MATH. On the other hand, for MATH, there exists MATH such that for all MATH hence, MATH which gives the second equality in REF. By CITE, it remains to show MATH . Because of MATH, MATH is ellipt... |
math/0008130 | Let MATH, MATH. We shall prove first that, for small MATH, there exists MATH such that MATH and MATH, for all non-degenerate representations MATH on MATH. Because the family MATH, MATH, extends to a first-order differential operator on MATH, the adiabatic groupoid of MATH, we obtain that MATH induces an element in MATH... |
math/0008130 | The proof is the same as that of the boundedness of operators of order zero, using NAME 's trick CITE. Let us briefly recall the details. It suffices to show MATH, for some constant MATH independent of MATH. Choose MATH with MATH. This is possible because MATH is defined and continuous on the sphere bundle MATH of MATH... |
math/0008130 | The statement follows from the boundedness of MATH, if MATH, so assume that MATH. Then the proof is the same as that of the previous lemma if in the proof of that lemma we replace MATH with MATH and take MATH. |
math/0008130 | The first statement follows from the previous lemma. Fix MATH such that MATH. REF gives MATH for some MATH and all MATH . Consequently, there is MATH with MATH for all MATH. Since MATH is essentially self-adjoint by REF and MATH, its range MATH is dense by CITE. By REF, we obtain for MATH which completes the proof. |
math/0008130 | By the previous corollary, both MATH and MATH extend to bounded operators. |
math/0008130 | We notice that if MATH satisfies the assumptions of the lemma, then MATH satisfies them as well. We can assume then that MATH. We shall check only that MATH. The relation MATH can be proved in the same way or follows from the first one by taking adjoints. Let MATH be with MATH, MATH be a sequence converging to MATH, an... |
math/0008130 | From REF we know that MATH is in MATH. By REF , applied to MATH and MATH, we have MATH, thus, MATH. Taking square roots completes the proof. |
math/0008130 | Assume first that MATH. Then the theorem follows from REF . For arbitrary MATH, MATH, and hence we get MATH by using functional calculus with continuous functions. A look at REF completes the proof. |
math/0008130 | If we change the metric on the compact space MATH, we obtain a new NAME operator, and MATH will be replaced by a different operator MATH. However, by REF , MATH and MATH are bounded for all even integer MATH. By interpolation, they are bounded for all MATH. This proves the independence of the NAME space on the choice o... |
math/0008130 | We have MATH, by REF , because MATH and MATH. The identity for the principal symbol follows from the corresponding one in REF . |
math/0008130 | By construction, MATH is, up to similarity, the restriction of MATH to one of the fibers MATH, with MATH. |
math/0008130 | Using the deformation retract MATH, we define (up to isomorphism) MATH. Then MATH, the isomorphism being uniquely determined up to homotopy. Moreover, we have a (non-canonical) isomorphism MATH of vector bundles, which allows us to define a MATH-module structure on MATH. By replacing MATH with an isomorphic bundle, we ... |
math/0008130 | Let MATH, as above. CASE: Choose MATH such that MATH and define MATH. Then MATH. Similarly, we find a right inverse for MATH up to compact operators. Thus, MATH is NAME. CASE: The operator MATH is the product of the bounded operator MATH and of the compact operator MATH. CASE: For MATH in a MATH-algebra MATH and MATH a... |
math/0008130 | An injective representation MATH of MATH-algebras preserves the spectrum, and in particular, MATH is invertible if, and only if, MATH is invertible. Denote by MATH the algebra of bounded operators on MATH. The morphism MATH induced by MATH is also injective. Fix MATH. Then MATH is invertible if, and only if, MATH is in... |
math/0008130 | Again, REF follow from REF . The assumption MATH implies MATH. Because the groupoid obtained by reducing MATH to MATH is amenable, the representation MATH, MATH is injective on MATH. This gives MATH, MATH, for all MATH. |
math/0008130 | The representation MATH given by the restrictions MATH and the homogeneous principal symbol is injective. This gives REF . For MATH note that we have MATH for the NAME transform MATH of MATH, and MATH. To obtain REF from REF as above, it is enough to observe that the operator MATH, (with MATH elliptic of order MATH, fi... |
math/0008130 | It is enough to prove that the representation MATH is injective on MATH, because we can recover the principal symbol of a pseudodifferential operator from its action on functions, as explained in the previous section. The groupoid MATH is amenable because the composition series of REF are associated to the groupoids MA... |
math/0008130 | We are going to apply REF , with MATH ranging through the set of hyperfaces of MATH; this is possible because of REF . Furthermore, note that by the definition of the groupoid structure on MATH in REF , the boundary hyperfaces MATH of MATH are closed, invariant submanifolds with MATH. For each boundary hyperface MATH o... |
math/0008130 | Let MATH be a minimal face of MATH (that is, not containing any other face of MATH). Then MATH is a compact manifold without corners and hence the NAME operator on MATH contains MATH in its spectrum. The above theorem then shows that MATH. On the other hand, MATH is positive, and hence MATH. This completes the proof. |
math/0008131 | We shall write MATH, for simplicity, where the filtration is as defined in REF . The filtration of the complex computing the NAME homology of MATH then gives rise to a spectral sequence with MATH, by standard homological algebra. By the definition of the NAME complex of MATH, MATH . This completes the proof for NAME ho... |
math/0008131 | We shall use MATH . Denote MATH for simplicity. The projective limits give rise to a MATH exact sequence MATH for every fixed MATH (see REF from the Appendix). From the assumption that MATH for MATH and MATH, we know that MATH becomes stationary for MATH and MATH. This shows that the MATH term above vanishes for MATH, ... |
math/0008131 | Each of the algebras MATH is a topologically filtered algebra in its own if we let MATH (so the filtration of MATH depends only on MATH and MATH). For the algebras MATH, the NAME complex is defined as for any topologically filtered algebra, except that there is no need to take an additional direct limit with respect to... |
math/0008131 | The assumption that MATH is independent of MATH has as a consequence that we need not take inductive limits with respect to MATH in the definition of the NAME complex of MATH. Denote MATH for simplicity. The NAME complex of MATH is complete, in the sense that MATH . The projective limits give rise to a MATH exact seque... |
math/0008131 | Each of the algebras MATH is a topologically filtered algebra in its own, if we let MATH (so the filtration really depends only on MATH and MATH). We shall write MATH instead of MATH, and hence MATH. The NAME complex of MATH is then given by MATH . Because taking homology is compatible with inductive limits, we obtain ... |
math/0008131 | This is completely analogous to the previous results, so we will be sketchy. The natural filtration on the complex MATH induces a spectral sequence whose MATH term is the MATH-homology of MATH. This spectral sequence is proved to be convergent as in any of the above two theorems. |
math/0008131 | This statement is local because we are dealing with germs. So we can assume that MATH and MATH. The complex whose cohomology we have to compute is then the projective tensor product of the corresponding complexes for MATH and MATH an appropriate number of times. By the NAME Lemma, the complex corresponding to MATH has ... |
math/0008131 | Consider for each face MATH of MATH the usual NAME complex MATH whose homology is, by (a variant of) NAME 's theorem, the absolute cohomology of MATH. Denote by MATH the hyperfaces containing MATH and by MATH their defining functions. Fix a local product structure in a neighborhood of the face MATH and choose a smooth ... |
math/0008131 | That the cohomology of the first complex is isomorphic to MATH is of course well known. The second isomorphism follows from the NAME spectral sequence applied to the map MATH and the above proposition. |
math/0008131 | We shall prove this lemma for MATH, the other case being completely similar, and actually even simpler. The subspaces MATH are defined in the following way. Fix an increasing, countable exhaustion MATH by compact submanifolds. For any vector bundle MATH, we denote by MATH the space of classical symbols supported above ... |
math/0008131 | We know from REF that MATH. Using again the NAME, we obtain that MATH is isomorphic to MATH and the operator MATH is the projection which sends MATH to REF, is the inclusion MATH and the identity on the other factors. The above computation of MATH then gives the result. |
math/0008131 | We shall use REF . For MATH, REF gives MATH . Moreover, the assumptions of REF are satisfied, by REF , and this gives MATH . This gives the result. |
math/0008131 | Whenever the periodicity operator MATH of the exact sequence REF is surjective, we have MATH the projective limit being taken with respect to the operator MATH. The conclusion then follows from REF . |
math/0008131 | The proof is essentially the same as for the corresponding result for the algebra MATH, so we will be brief. The filtration on MATH is induced from the filtration on MATH. This and the specific form of the MATH-terms then give MATH . Moreover, the differential MATH is the same as that for MATH if MATH, but is trivial f... |
math/0008131 | We first prove the analogue of REF in our new settings: namely, the MATH-term of the cyclic spectral sequence associated to the topologically filtered algebra MATH by REF is given by MATH if MATH, and otherwise by MATH . Moreover, the periodicity morphism MATH vanishes if MATH and is the natural projection if MATH. Ind... |
math/0008131 | The MATH-unitality follows from REF . For the second part, we proceed essentially as in REF . Choose an anti-symmetric tensor in the last MATH-variables MATH with MATH. We denote by MATH the quantization of MATH. Let MATH be the total degree. Because MATH where the dots represent terms of order at most MATH, the quanti... |
math/0008131 | The MATH-unitality follows from REF , as above. To prove the rest of this proposition, we can either use the same method as the one used to prove REF , or we can argue that this proposition actually follows from REF . |
math/0008131 | The computation of MATH proceeds exactly as in the case of manifolds with boundary. From REF we know that the MATH term of the spectral sequence associated with the order filtration on the NAME complex is given by MATH . In our case, MATH which is the main reason why this result is true, and explains the thinking behin... |
math/0008131 | By a standard argument using the 'SBI'-exact sequence, it is enough to prove this statement for NAME homology. Consider then the spectral sequences associated to the two algebras and the order filtration on their NAME complexes by REF . The induced map is then an isomorphism at the MATH-term, by REF , because MATH . |
math/0008131 | Only the last statement is not similar to some other proofs in the previous sections. Choose an open neighborhood MATH of MATH in MATH such that MATH is a deformation retract of MATH. Also, let MATH be a smooth function supported in MATH which is one in some smaller neighborhood of MATH. This data gives rise to a lifti... |
math/0008131 | Let MATH be the dimension of MATH, as before. The dual of REF holds true for NAME cohomology. This implies that the inclusion MATH is isomorphic (in the sense of the above Proposition) to the dual of the map MATH. But this last map is checked to be an isomorphism. |
math/0008131 | We will proceed by induction on MATH to show that the morphism MATH is zero for any MATH. Let MATH denote the dimension of the manifold MATH, as above. The statement of the theorem is obviously true for MATH. The proof of REF , show that the groups MATH are generated by elements of order MATH with respect to the degree... |
math/0008133 | Consider the complex REF, which computes the continuous cohomology of MATH, and let MATH be the map MATH . As in REF , the map MATH descends to the quotient to induce an isomorphism MATH of complexes, that is MATH, which establishes the isomorphism MATH, as desired. |
math/0008133 | CASE: First we check that MATH and MATH are injective. Indeed, assume that MATH, for some MATH. Then, if MATH, we have MATH and hence MATH. Consequently, we have MATH, with MATH, as desired. The same argument shows that if MATH and MATH have a point in common, then the standard subgroups MATH and MATH are conjugated in... |
math/0008133 | There exists a (not natural) isomorphism MATH of vector spaces. By the above lemma, the inclusion of MATH of MATH-modules induces natural isomorphisms MATH because the functor MATH is compatible with inductive limits and with direct sums. The naturality of these isomorphisms and the Five Lemma show that MATH . This is ... |
math/0008133 | It follows from the definition of MATH that, if MATH, then MATH vanishes in a neighborhood of MATH. Conversely, if MATH is in MATH, then we can find some polynomial MATH, with MATH, such that MATH is bounded from below on the support of MATH by, say, MATH, then MATH. The second isomorphism follows from the first isomor... |
math/0008133 | Let MATH be a standard subgroup of MATH. Recall first that MATH is a finite group that acts freely on MATH, which gives a MATH - equivariant isomorphism MATH . Let MATH be a smooth MATH - module. The NAME - NAME spectral sequence applied to the module MATH and the normal subgroup MATH gives natural isomorphisms MATH . ... |
math/0008133 | The result follows from Luna's Lemma. For MATH-adic groups, Luna's Lemma is proved in CITE, page REF, Properties ``C" and ``D." |
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