paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0006033 | By REF there exist finite abelian groups MATH and MATH such that MATH, MATH. By the universal coefficient theorem, CITE, MATH . By the universal coefficient theorem again, MATH and MATH. Hence MATH . Note that MATH. Thus MATH and MATH are isomorphic groups. Since any surjective endomorphism of a finitely generated abel... |
math/0006033 | Let MATH have standard form MATH . Let MATH denote the NAME product. By assumption we have that MATH in MATH. Thus MATH in MATH. Hence MATH for MATH. This implies that MATH since MATH . Therefore MATH is an isomorphism by REF . By REF there is a unital *-homomorphism MATH such that MATH. Thus MATH. |
math/0006033 | Following CITE we let MATH for every integer MATH. By CITE it suffices to show that the canonical maps MATH and MATH are isomorphisms, where MATH denotes the set of compact operators on a separable infinite dimensional NAME. As noted in CITE it follows from CITE that MATH is a continuous functor. Since it is obviously ... |
math/0006033 | Combine REF with CITE. |
math/0006033 | First note that MATH in MATH by REF . Hence MATH by REF for some self-adjoint element MATH. Since MATH it follows that MATH for some MATH and all MATH. Hence MATH in MATH by REF . By applying MATH we get that MATH, where MATH. Since MATH we see that MATH, MATH. Thus MATH where MATH. But then MATH for some MATH. It foll... |
math/0006033 | Choose a continuous function MATH such that MATH. Define a unitary MATH by MATH. Since MATH we can define a unitary MATH by MATH . Set MATH. |
math/0006033 | For each MATH, MATH, let MATH and MATH be the integers with MATH, and MATH, MATH. By REF there exist MATH and unitaries MATH such that MATH . Since MATH we remark that if REF holds then MATH, MATH. Note that MATH . By REF we see that MATH if and only if for every MATH, MATH . It follows that REF holds if and only if MA... |
math/0006033 | We may assume that MATH is a building block rather than a finite direct sum of building blocks. Let MATH where each MATH is a building block, and let MATH denote the inclusion. Let MATH be the minimal non-zero central projections in MATH. Since MATH in MATH, it follows from REF that there is a unitary MATH such that MA... |
math/0006033 | Choose MATH such that MATH and such that MATH as unordered MATH-tuples. Let MATH be the integer such that MATH. Set MATH. |
math/0006033 | Choose MATH such that MATH and choose MATH such that MATH . Assume that MATH for some integers MATH with MATH, MATH, and an integer MATH. Then MATH . Hence MATH. By assumption it follows that for every MATH, MATH . Thus MATH as unordered MATH-tuples. Therefore MATH . Hence MATH for MATH. From this it follows that MATH. |
math/0006033 | Choose a positive integer MATH such that MATH . We will prove by induction in MATH that there exist continuous functions MATH that satisfy the above for MATH. The case MATH follows from REF . Now assume that for some MATH we have constructed continuous functions MATH such that MATH, and such that for each MATH, MATH, a... |
math/0006033 | By CITE it follows that MATH is approximately unitarily equivalent to a *-homomorphism MATH of the form MATH for MATH, where MATH are integers, MATH, MATH, MATH are continuous functions, and MATH is a unitary. Let MATH denote the winding number of MATH. Let MATH be a unitary MATH matrix such that MATH for all MATH. Set... |
math/0006033 | For MATH, set MATH and MATH . Let MATH be arbitrary. We will show that MATH. Let MATH be the connected components of MATH. Choose for each MATH a MATH-arc MATH such that MATH and MATH. Then MATH . If MATH then MATH for some MATH. Hence MATH . By NAME 's marriage lemma, see for example, CITE, the sets MATH, MATH, have d... |
math/0006033 | Let MATH. If for example, MATH for some MATH then MATH must map the set MATH into MATH. Contradiction. |
math/0006033 | Let MATH. Note that MATH. There exist MATH such that MATH as unordered tuples, and such that MATH . By REF we may assume that MATH and still have that REF hold. Choose an integer MATH, MATH, such that MATH and MATH. Then MATH since MATH by REF . Choose MATH such that MATH and MATH as unordered MATH-tuples. By REF we se... |
math/0006033 | By REF there exists an integer MATH such that MATH . Note that MATH . Fix some MATH. Set MATH . Since MATH we see by REF that MATH cannot contain a MATH-arc. Thus MATH. It follows that MATH, MATH. |
math/0006033 | Choose MATH such that for MATH, MATH . Let integers MATH, a building block MATH, and unital *-homomorphisms MATH be given such that REF are satisfied. Choose MATH such that for MATH, MATH . Let MATH denote the exceptional points of MATH and let MATH be those of MATH. Let for each MATH, MATH be the number such that MATH... |
math/0006033 | Let MATH. We may assume that MATH. Let MATH be a self-adjoint element such that MATH modulo MATH and MATH. Define MATH by MATH. Since MATH we have that MATH modulo MATH. Thus MATH . MATH is a building block by REF , and therefore it follows from REF that MATH modulo MATH. Thus MATH . |
math/0006033 | For each MATH, let MATH be the inclusion and let MATH be the projection. Choose by REF a positive integer MATH with respect to the finite set MATH and MATH. Set MATH. Let integers MATH, a finite direct sum of building blocks MATH, and unital *-homomorphisms MATH be given such that the above holds. Since REF implies REF... |
math/0006033 | We may assume that MATH and, by repeating the functions MATH, that MATH. Let MATH be a positive integer such that MATH . Let MATH, MATH, MATH, and MATH be as above. By REF there are integers MATH, MATH, MATH, with MATH for MATH, such that MATH . As in the proof of REF we see that MATH since MATH preserves the order uni... |
math/0006033 | We may assume that MATH, MATH. Choose by REF an integer MATH with respect to MATH and MATH. We may assume that MATH and that MATH. Then choose by REF an integer MATH with respect to MATH and MATH. Now let MATH, MATH, MATH, MATH and MATH be given as above. Choose continuous functions MATH such that in MATH, MATH . By RE... |
math/0006033 | Note that MATH is a building block by REF . Hence by REF there exists a unitary MATH such that MATH . Then MATH by REF . |
math/0006033 | Let MATH be the projection and MATH be the inclusion, MATH. Let MATH denote the minimal non-zero central projections in MATH. Choose by REF an integer MATH with respect to MATH, MATH and MATH. Set MATH. Let MATH, MATH, MATH, and MATH be as above. We may assume that MATH. To see this, assume that the case MATH has been ... |
math/0006033 | By REF (and the corresponding result for interval building blocks) we may assume that MATH is a building block, an interval building block or a matrix algebra rather than a finite direct sum of such algebras. We will carry out the proof in the case that MATH is a circle building block. The proof in the case that MATH i... |
math/0006033 | Note that we may assume that MATH for every positive integer MATH and every minimal non-zero central projection MATH. By REF it follows that MATH for every projection MATH. Let MATH denote the unit. Since MATH is an approximate unit for MATH there exists a positive integer MATH such that MATH for all MATH. Hence MATH, ... |
math/0006033 | Let MATH where MATH. Set MATH. Set MATH. We may assume that MATH unless the interior of MATH intersects non-trivially with MATH. On each MATH let either MATH be the identity map (if MATH) or a continuous map onto MATH that is constant on the set of boundary points of MATH. Set MATH. |
math/0006033 | We may assume that MATH is a quotient of a building block rather than of a finite direct sum of building blocks. Hence by REF MATH where MATH is a closed subset and MATH. Choose MATH such that MATH . Choose by REF a closed subset MATH with finitely many connected components such that MATH, and a continuous surjective m... |
math/0006033 | By REF we have that MATH is the inductive limit of a sequence MATH where each MATH is unital and injective and each MATH is a quotient of a finite direct sum of building blocks. We will construct a strictly increasing sequence of positive integers MATH, a sequence MATH of finite direct sums of building blocks, interval... |
math/0006033 | The lemma is well-known if MATH is an NAME. We may therefore assume that MATH is infinite dimensional for some MATH. Let MATH be a positive integer. Let MATH be positive non-zero mutually orthogonal elements. Since MATH is simple and the connecting maps are injective, there exists an integer MATH such that MATH . Hence... |
math/0006033 | By REF we have that MATH is the inductive limit of a sequence MATH where each MATH is unital and injective and each MATH is of the form MATH for a finite (possibly trivial) direct sum of building blocks MATH and a finite dimensional MATH-algebra MATH. Set MATH and let MATH be the canonical *-homomorphism. It suffices t... |
math/0006033 | We may assume that MATH for MATH. Let MATH be the exceptional points of MATH. Choose a positive integer MATH such that MATH . For MATH, define a continuous function MATH by MATH . Set MATH . Let MATH be unital *-homomorphisms that satisfy REF . By CITE we see that MATH and MATH are approximately unitarily equivalent to... |
math/0006033 | By REF we see that MATH is the inductive limit of a sequence MATH where each MATH is a finite direct sum of circle and interval building blocks and each MATH is a unital and injective *-homomorphism. By passing to a subsequence, if necessary, we may assume that either, every MATH is a circle or an interval building blo... |
math/0006033 | We may assume that MATH, MATH. Decompose MATH as a finite direct sum of building blocks and let MATH denote the projection, MATH. For every MATH, identify MATH and MATH. Choose open sets MATH such that MATH and such that MATH . Let MATH be a continuous partition of unity in MATH subordinate to the cover MATH and let MA... |
math/0006033 | We may assume that MATH, MATH. Decompose MATH as a finite direct sum of building blocks. Let MATH be projections such that MATH generate MATH. By factoring MATH through the MATH-algebra obtained from MATH by erasing those direct summands MATH for which MATH, we may assume that MATH, MATH. There exist positive integers ... |
math/0006033 | The image of the canonical map MATH is dense by CITE, since MATH is a simple countable dimension group. By REF is the composition of this map with the linear bounded map MATH induced by MATH. It follows that MATH is dense in some subspace of MATH. |
math/0006033 | Since MATH in MATH there is an integer MATH such that MATH. By REF we see that MATH for some positive integer MATH and an element MATH. Since MATH is discrete and since MATH we see that MATH for some positive integer MATH and an element MATH. Since MATH is injective we may choose an integer MATH such that MATH in MATH.... |
math/0006033 | Let MATH be a unitary such that MATH. Let MATH be a homotopy connecting MATH to MATH. We may assume that MATH for MATH. Thus MATH where MATH is a continuous path of self-adjoint elements in MATH. Since MATH we see that MATH has finite order in MATH. Thus MATH is a continuous path in a totally disconnected subset of a m... |
math/0006033 | By CITE and REF there exist a positive integer MATH and *-homomorphisms MATH such that MATH is homotopic to MATH and MATH is homotopic to MATH. By increasing MATH we may assume that MATH and MATH are unital. There exists an integer MATH such that MATH in MATH. Thus MATH by REF . Hence MATH by REF . |
math/0006033 | By REF we may assume that MATH is the inductive limit of a sequence MATH of finite direct sums of building blocks with unital connecting maps. Similarly MATH is the inductive limit of a sequence of finite direct sums of building blocks MATH with unital connecting maps. Let MATH. There exists a positive integer MATH suc... |
math/0006033 | By continuity of MATH there exist a positive integer MATH and an element MATH in MATH such that MATH in MATH. Since the short exact sequence of REF splits we may assume that MATH. Note that MATH and hence MATH for some MATH with MATH in the group MATH. If MATH is non-cyclic then we see that MATH by REF . Thus we may as... |
math/0006033 | Let a finite direct sum of building blocks MATH and a unital *-homomorphism MATH be given. Let MATH be a continuous affine map such that MATH . Let MATH be the minimal non-zero central projections in MATH. As in the proof of REF we see that we may assume that MATH, MATH. Choose MATH such that MATH. Choose an integer MA... |
math/0006033 | It will be convenient to set MATH, MATH, and MATH, MATH. Note that MATH and MATH by REF . Hence MATH can be identified with a matrix of the form MATH where MATH is a group homomorphism, MATH. Let MATH. If MATH is cyclic then MATH for some positive integer MATH and MATH. Choose a finite direct sum of building blocks MAT... |
math/0006033 | Let MATH where each MATH is a building block. By REF there are for each MATH an element MATH in MATH, integers MATH, and an element MATH in the torsion subgroup of MATH such that MATH . Choose MATH such that MATH. Set MATH. Choose MATH such that MATH and such that MATH . Let MATH denote the minimal non-zero central pro... |
math/0006033 | We may assume that MATH is infinite dimensional. Hence by REF we may assume that each MATH is unital and injective. Let MATH where each MATH is a building block and let MATH be the set of minimal non-zero central projections in MATH. For each positive integer MATH, choose a finite set MATH such that MATH generates MATH... |
math/0006033 | By REF we have that MATH is torsion free such that MATH is defined. It follows by REF that MATH for MATH in the torsion subgroup of MATH. Apply REF . |
math/0006033 | Choose an element MATH such that MATH on MATH and such that MATH on MATH. By REF there exists a group homomorphism MATH such that MATH and MATH agree on the torsion subgroup of MATH and such that the diagram MATH commutes. The conclusion follows from REF . |
math/0006033 | We may assume that MATH is infinite dimensional, and hence by REF we see that MATH is the inductive limit of a sequence MATH of finite direct sums of building blocks with unital and injective connecting maps. By REF we have that MATH is the inductive limit of a sequence MATH of finite direct sums of building blocks wit... |
math/0006033 | As above, but with the following changes. Instead of REF we get by REF that MATH . By REF we may now replace REF by MATH . Finally, REF follows again by REF . |
math/0006033 | By REF we may assume that MATH is the inductive limit of a sequence MATH of finite direct sums of building blocks with unital and injective connecting maps. Similarly we may assume that MATH is the inductive limit of a sequence MATH of finite direct sums of building blocks with unital and injective connecting maps. By ... |
math/0006033 | This follows easily from REF . |
math/0006033 | By CITE MATH is isomorphic to an inductive limit in the category of order unit spaces of a sequence MATH . It is easy to see that this implies that MATH is isomorphic to an inductive limit of a sequence of the form MATH . Choose a dense sequence MATH in MATH and a dense sequence MATH in MATH. For every positive integer... |
math/0006033 | Combine REF . |
math/0006039 | CASE: Let MATH and MATH. If MATH, then MATH. Then MATH and so MATH (as in REF ). CASE: Let MATH. We know that MATH . So we are done if we see that MATH. As in REF , we have MATH . Thus MATH . CASE: The first inequality follows from the definition of MATH and MATH. The second statement is also straightforward. Indeed, i... |
math/0006039 | Let us see REF first. So we assume that there exist some transit cube MATH of the MATH-th generation containing MATH. Since MATH, we have MATH. In particular, MATH. So the inequality MATH is a direct consequence of REF . We consider now the second inequality in REF . By REF and the second equality of REF we get MATH . ... |
math/0006039 | As mentioned above, MATH and so MATH. On the other hand, we also have MATH, and then MATH and so MATH. |
math/0006039 | From REF we see that if MATH, then there exists some MATH such that MATH and MATH. Thus MATH, and from REF we get MATH. CASE: By REF we have MATH . Since MATH, we get MATH . Similarly it can shown that MATH. So it only remains to see that MATH. Recall that MATH and MATH. Then we have MATH . CASE: Using REF we get MATH ... |
math/0006039 | For simplicity we assume that all the cubes MATH, MATH, are transit cubes. In the final part of the proof we will give some hints for the general case. Moreover, we only have to prove REF . The others follow from REF by the NAME Lemma, as in CITE. Assume MATH. The kernel of the operator MATH is given by MATH . Since MA... |
math/0006039 | The right inequality follows from the left one (with MATH instead of MATH). Indeed, by an argument similar to the one used for MATH in REF , it follows that MATH . In REF below we will show that MATH is bounded and invertible in MATH, and so MATH. The left inequality in REF will be proved using techniques of vector val... |
math/0006039 | For MATH big enough and MATH, MATH if MATH. Thus MATH . Now we only have to check that MATH for MATH. We set MATH . For each MATH, we denote by MATH the least integer such that MATH. We have MATH . Since MATH, we get MATH and so MATH . For MATH, since MATH, we have MATH. Thus MATH and then REF holds. |
math/0006039 | The kernel of MATH is given by MATH. We will apply NAME 's Lemma, using the estimates of the preceding lemma, interchanging MATH and MATH when necessary. We have MATH . By REF we get MATH . If MATH, then MATH. Thus MATH . We estimate MATH now. By REF we obtain MATH . We have MATH . Finally we turn our attention to MATH... |
math/0006039 | For MATH, by REF , we have MATH . Since the matrix MATH originates an operator bounded on MATH, we obtain MATH . |
math/0006039 | We know that MATH and MATH converge respectively to MATH and MATH in MATH. Since we are assuming that the kernel MATH of MATH is bounded, we have MATH, for all MATH and MATH. As a consequence, MATH converges to MATH uniformly on MATH as MATH. Therefore, for any compact set MATH, we have MATH . It can be checked that th... |
math/0006039 | Notice that we only have to show that MATH is a HCZO such that MATH as MATH, taking into account REF and the fact that HCZO's are bounded on MATH (see CITE, and also CITE for a different proof). We have already seen in REF that MATH as MATH in the operator norm of MATH. Thus it only remains to see that MATH is a HCZO a... |
math/0006039 | Let us see that REF holds. We may assume that MATH is bounded. For each MATH we choose the biggest cube MATH (that is, with MATH minimal). By NAME 's covering theorem, there exists a family cubes MATH with finite overlap such that MATH. Therefore we have MATH . Observe also that if MATH, then MATH. Otherwise MATH, and ... |
math/0006039 | By REF and the subsequent remark in REF , since MATH, we have MATH for any doubling cube MATH of the MATH-th generation. Therefore, the lemma follows from REF in the preceding lemma taking MATH. |
math/0006039 | By REF , for MATH we have MATH . |
math/0006039 | Let MATH be the kernel of MATH. Observe that MATH where MATH. Since MATH, it follows that MATH, with MATH (the details are left to the reader). Let us see that MATH satisfies REF . Take MATH and let MATH be such that MATH. We have MATH . Let us remark that the constant MATH above equals MATH. It is not difficult to che... |
math/0006039 | We only have to prove that the operators MATH are bounded uniformly on MATH. For a fixed MATH, we denote MATH and MATH. Since MATH and MATH is weakly bounded, from REF it follows MATH. It is straightforward to check that MATH. We also have MATH, MATH. Indeed, MATH . Because of REF , the operator MATH converges strongly... |
math/0006045 | Consider a graph MATH and a surface MATH as above. MATH can be thought of as an embedded disk with bands. We can assume that MATH is disjoint from the (interiors of the bands) of MATH and thus MATH intersects the (interior of) MATH only in the embedded disk. Cut each band along arcs (in the normal direction to the core... |
math/0006045 | It follows from two applications of the MATH relation that MATH . |
math/0006045 | First we construct the map MATH. Let MATH be a MATH-spanning link in MATH. Given a graph MATH with colored legs choose an arbitrary embedding of it in MATH. For every coloring MATH of each of its univalent vertices (where MATH are integers), push MATH disjoint copies of MATH (using the framing of MATH), orient them the... |
math/0006045 | The first statement follows immediately from the fact that if MATH is a basis then no nontrivial linear combination is nullhomologous, thus the MATH relation is vacuous. For the second statement, since we are using MATH coefficients, we may assume that the link MATH is a basis for MATH, and choose a link MATH to span t... |
math/0006045 | Let MATH be a MATH-spanning link and MATH be a MATH-basis. Then, we have over MATH which concludes the proof of the corollary. |
math/0006045 | The proof is a simple application of the locality property of the NAME integral, as explained leisurely in CITE, and a simple counting argument. We now give the details. We need to show that CASE: The part of the LMO=NAME integral MATH of degree at most MATH is an invariant of type MATH. CASE: For a trivalent graph MAT... |
math/0006045 | If MATH is as in the statement of the corollary, colored by a sublink MATH of MATH which is not MATH-spanning , then we can find a MATH, and a closed surface MATH such that MATH for all components MATH of MATH. Cut MATH along an edge, and color the two new leaves MATH and MATH to obtain a graph MATH. By definition, we ... |
math/0006045 | We will first show the result for MATH. Let MATH be as in the statement of the corollary and let MATH be the graph with two more leaves than MATH, colored by MATH and MATH respectively as shown: MATH . The MATH relation implies that MATH . Now, we will show the result for all MATH. Let MATH be the same graph as MATH wi... |
math/0006047 | Let MATH. We first compute the NAME derivatives of T in the directions of the vector fields given by REF . One gets MATH where MATH. The NAME operator intertwines MATH with itself as a module over MATH. It is in particular invariant with respect to the constant vector fields and has therefore constant coefficients. To ... |
math/0006047 | Just evaluate MATH on MATH. |
math/0006047 | Let first MATH be such that MATH. Since MATH is not resonant and MATH is injective, MATH. Assume now that MATH is an eigenvector of MATH of eigenvalue MATH and that MATH has total order MATH. Let us rewrite the condition MATH according to the decomposition MATH with MATH and MATH. We get MATH . The first equation shows... |
math/0006047 | Let MATH. It suffices to prove that MATH for any MATH. Since MATH also belongs to MATH, both members of this equality are eigenvectors of eigenvalue MATH of MATH. They have the same principal symbol and are thus equal. |
math/0006047 | Taking account of the preceding lemmas, we just have to show that the map MATH is such that MATH is the unique equivariant symbol map and that it is bijective. On the one hand, any analogue of MATH prolongs an eigenvector of MATH into an eigenvector of MATH with the same eigenvalue and principal symbols, and hence equa... |
math/0006047 | Notice first that MATH . The second and third terms are respectively multiples of MATH and MATH, and the first equals MATH where MATH. Therefore, MATH . Let MATH. As MATH we observe that MATH also belongs to MATH. The result then follows from the definitions of MATH and MATH. |
math/0006047 | The first three assertions are evident since MATH . To prove the stated inequality, observe first that MATH if MATH. Indeed, one has then MATH . Now, if MATH then MATH . Else, suppose that MATH and notice that MATH. One can write MATH . Moreover, MATH . Denote by MATH the sum of the homogeneous monomials of degree MATH... |
math/0006047 | Assume that MATH, MATH, is the polynomial form of MATH. As it can be seen from REF , the polynomial form of MATH differs from MATH by MATH . This quantity vanishes if the degree of MATH does not exceed one. We have shown in the proof of REF that both MATH and MATH belong to MATH for any vector field MATH. When MATH, th... |
math/0006047 | Proceed as when MATH is not resonant, noticing that both members of the equality MATH belong to MATH provided that MATH. |
math/0006047 | A slight adaptation of the proof of REF shows that such a map is local. Being invariant under the action of constant vector fields, it is thus a differential operator with constant coefficients. The invariance of MATH with respect to the linear vector fields means that the polynomial form of MATH maps MATH onto a tenso... |
math/0006047 | As it was done in the case MATH, it can be checked that one of the weights MATH and MATH must vanish and the other be chosen in MATH. A direct computation then allows to prove that the map which associates to MATH the bidifferential operator MATH is projectively equivariant for any value of the parameter MATH. One defi... |
math/0006047 | REF shows that the prolongation of any eigenvector MATH, MATH, of MATH will be impossible only if MATH for some MATH such that MATH and MATH define a critical value of the shift. But then, in view of REF , MATH, hence a contradiction. |
math/0006049 | An easy calculation shows that a configuration MATH is a critical point of MATH if and only if for any MATH the vector MATH is orthogonal to the tangent space MATH. The last condition is clearly equivalent to the requirement that the normal to MATH at MATH bisects the angle between MATH and MATH. |
math/0006049 | Choose MATH small enough such that the conclusions of REF hold. Since at the points of the boundary MATH the gradient of MATH has the outward direction, the critical point theory for manifolds with boundary CITE applies; the conclusion is that the critical points of the restriction MATH should be ignored, and the numbe... |
math/0006049 | Consider the inclusion MATH and the NAME spectral sequence CITE of the continuous map MATH where MATH is the sheaf on MATH associated with the presheaf MATH . To describe the sheaves MATH, consider partitions of the set MATH into intervals, that is, subsets of the form MATH. For any such partition MATH we denote by MAT... |
math/0006049 | If we replace MATH by a field MATH, the result follows directly from REF . In particular, we see that the dimension of the cohomology of MATH do not depend on the field of coefficients. We conclude that the integral cohomology of MATH has no torsion and is nonzero only in dimensions divisible by MATH. Consider the cycl... |
math/0006049 | Using relations REF we see that the additive basis of MATH is given by monomials of the form MATH, where MATH are disjoint multi-indices. Hence it is clear that for MATH the differential algebra MATH can be embedded into MATH; in fact MATH may be identified with the subalgebra generated by MATH and MATH. The factor MAT... |
math/0006049 | We will use induction on MATH. The statement is trivial when MATH. Let's assume that it is true for MATH. Consider the homomorphism MATH . It is clear that MATH is injective and increases the total degree by MATH. Using relation REF one finds MATH . Hence we obtain a short exact sequence MATH and a long homological seq... |
math/0006051 | According to CITE, the function MATH can be computed as MATH where MATH is the measure on MATH defined by MATH . Since for MATH takes integral values, this shows the first statement. Reducing modulo MATH we may replace the function MATH by the function MATH if MATH, which is congruent to it modulo MATH on MATH. This im... |
math/0006051 | Since MATH the order of MATH is prime to MATH and we have MATH for some MATH. By using the definition of MATH repeatedly we find MATH . Since MATH we may move the last term to the left hand side of the equation and obtain MATH and dividing by MATH and reducing modulo MATH we obtain using the proposition MATH . |
math/0006051 | Let MATH. Then one finds MATH with MATH. Write MATH . It is easy to verify that MATH with MATH. To find the coefficients MATH modulo MATH for MATH one can simply reduce the above equations modulo MATH and one easily finds that MATH . It thus remains to show that also for MATH the coefficient MATH is divisible by MATH. ... |
math/0006051 | Using the fact that MATH is a derivation and that MATH and MATH we see that MATH . Here we understand that MATH. Every MATH can be written as MATH with MATH a root of unity in MATH and MATH. We have MATH . If we assume that MATH we now find from the last computation and from REF , MATH . It follows that to make the red... |
math/0006054 | Let MATH be the inclusion map. This fits into a commuting diagram of maps introduced in REF : MATH . Then MATH . But MATH (see REF ), so MATH. Hence, we have a natural map MATH. Since the fibres of MATH and MATH are naturally isomorphic, this must be an isomorphism. Hence, MATH as required. |
math/0006054 | If we define the functor MATH as follows: MATH where MATH and MATH are the projection maps in REF , then MATH. The functor MATH is NAME 's original NAME transform introduced in CITE. Using this notation, it is enough to show that MATH. To see this it suffices to show that MATH over MATH. But MATH is supported at the in... |
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