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math/0006033
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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 abelian group is an isomorphism, it suffices to show that MATH is surjective. Let MATH. By REF there exists an element MATH such that MATH for some MATH. Next, by REF there exists a MATH such that MATH. Thus MATH.
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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.
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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 additive, we may assume that MATH is a building block. As in the proof of REF we see that there exists finite dimensional MATH-algebras MATH and MATH such that we have a short exact sequence of the form MATH . Apply CITE to this short exact sequence and the one obtained by tensoring with MATH to obtain two long exact sequences for MATH. It is well-known that the canonical maps MATH and MATH are isomorphisms for MATH (compare CITE), so the theorem follows from the five lemma in algebra.
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math/0006033
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Combine REF with CITE.
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math/0006033
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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 follows by REF that MATH and hence by REF we get that MATH.
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math/0006033
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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.
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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 MATH and MATH for every MATH. But this statement is equivalent to REF by the remark above. Assume REF holds. To prove REF , let MATH be a unitary such that MATH has finite order in the group MATH. Then MATH is constant. By REF , and since MATH, MATH, it follows that MATH . In particular, MATH equals MATH at the exceptional points of MATH. On the other hand, MATH and MATH are constant functions on MATH and are hence equal everywhere. We may therefore use REF to conclude that MATH. REF is trivial. Assume REF . By REF , MATH . Hence MATH for MATH, MATH, and we have REF .
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math/0006033
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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 MATH, MATH. Hence we may assume that MATH, MATH. Set MATH. Let MATH be the induced maps and let MATH be the inclusion, MATH. If MATH then MATH is a MATH-equivalence by CITE. Thus MATH in MATH. Let MATH where MATH is a unitary. Let MATH where MATH. By REF we see that MATH mod MATH and thus MATH mod MATH. Hence MATH modulo MATH.
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math/0006033
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Choose MATH such that MATH and such that MATH as unordered MATH-tuples. Let MATH be the integer such that MATH. Set MATH.
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math/0006033
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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.
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math/0006033
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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, and MATH as unordered MATH-tuples. Choose MATH such that MATH for MATH. Choose continuous functions MATH such that for each MATH, MATH and MATH as unordered MATH-tuples. Set for MATH, MATH, MATH . By REF there exists an integer MATH such that for MATH, MATH . Define for MATH, a continuous function MATH by MATH . MATH satisfy the conclusion of the lemma for MATH.
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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 . Let now MATH be given. As above MATH is approximately unitarily equivalent to a *-homomorphism MATH of the form MATH . Note that MATH . By REF it follows that MATH, MATH. And since MATH we see that MATH. By REF choose continuous functions MATH such that for every MATH, MATH and such that MATH as unordered MATH-tuples for each MATH. Again by REF there exist continuous functions MATH such that for every MATH, MATH such that MATH as unordered MATH-tuples for each MATH, and such that MATH . It follows from REF that MATH as unordered MATH-tuples. Since MATH we see by REF that MATH for each MATH. Similarly, as MATH and MATH have the same winding number, MATH, MATH. Let MATH be numbers such that MATH, MATH, are the exceptional points of MATH. By REF there exist a unitary MATH such that MATH for every MATH. Choose a unitary MATH such that MATH, MATH, MATH and MATH. Note that for every MATH, MATH . It follows that we may define a unital *-homomorphism MATH by MATH for MATH, MATH. Then for every MATH, MATH, MATH . Hence MATH and MATH are approximately unitarily equivalent by CITE. Similarly we see that there exists a unitary MATH such that MATH and MATH, and such that MATH defines a *-homomorphism that is approximately unitarily equivalent to MATH. Finally note that by REF we have that MATH for every MATH.
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math/0006033
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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 distinct representatives. In other words, there exists a permutation MATH of MATH such that MATH for MATH.
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Let MATH. If for example, MATH for some MATH then MATH must map the set MATH into MATH. Contradiction.
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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 see that MATH. By REF we have that MATH for some integer MATH. Hence MATH . The reversed inequality is trivial.
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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.
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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. Let MATH be a continuous function such that MATH for every MATH, and such that for each MATH, MATH is constantly equal to MATH on some arc MATH where MATH. Define a unital *-homomorphism MATH by MATH. Set MATH and MATH. Then MATH . MATH and MATH satisfy REF . Let MATH where MATH and MATH is a self-adjoint element with MATH. Note that MATH . Fix some MATH. Let MATH denote the (unital) inclusion. By REF we have that MATH, MATH, MATH. Choose MATH, MATH, such that MATH. By REF we see that for each MATH, MATH for some unitaries MATH and numbers MATH. By changing MATH and MATH we may by REF assume that MATH and MATH . Let MATH be a MATH-arc. By REF , MATH . The last inequality uses REF and that MATH for some MATH-arc MATH. Hence MATH . Therefore by REF , MATH . By REF , if MATH is a MATH-arc then MATH since MATH. Clearly MATH. Furthermore, MATH . By REF it follows that MATH . Let MATH be the continuous function such that MATH, MATH, and such that MATH is linear when restricted to each of the two intervals MATH and MATH. Note that MATH . Finally, define a *-homomorphism MATH by MATH for MATH, MATH. Define a unital *-homomorphism MATH by MATH . Then by REF MATH . Note that MATH and MATH are equivalent representations of MATH on MATH, MATH. In particular, MATH and MATH have the same small remainders, and hence MATH, MATH by REF . Let MATH be the continuous function MATH . For MATH, MATH where MATH is defined by MATH. Note that by REF MATH and MATH . By REF , MATH, MATH, and MATH are approximately unitarily equivalent to MATH, MATH, and MATH, respectively, where MATH are *-homomorphisms of the form MATH for integers MATH with MATH, MATH, unitaries MATH in MATH with MATH, MATH, and continuous functions MATH, MATH, such that for MATH, MATH and such that for each MATH, MATH . Hence MATH . It follows from REF that for each MATH, MATH . Let MATH and let MATH be a MATH-arc. Then by REF MATH . Hence MATH . It follows from REF that for each MATH, MATH . We conclude that MATH . Since MATH and MATH are equivalent representations of MATH on MATH for MATH, it follows that MATH as unordered MATH-tuples. Therefore, as MATH, MATH, we see by REF that MATH . As MATH, MATH, we may thus define a *-homomorphism MATH by MATH for MATH, MATH. Since MATH we get from CITE that MATH and MATH are approximately unitarily equivalent. By REF we have that for every MATH-arc MATH, MATH . As MATH, we conclude from REF that MATH . Hence MATH . Choose unitaries MATH such that MATH . Set MATH. Then for MATH, MATH .
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math/0006033
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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 .
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math/0006033
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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 by REF , we may assume that REF holds. It is easy to reduce to the case where MATH is a single building block. Since MATH in MATH for MATH, there is by REF a unitary MATH such that MATH for every MATH. Hence we may assume that MATH, MATH. Set MATH. It follows from REF that MATH, MATH. Let MATH be the (normalized) trace of MATH. Let MATH be the induced maps. Note that MATH is a building block by REF . Fix some MATH. Every tracial state on MATH is of the form MATH for some MATH. Therefore MATH and MATH satisfy REF , with MATH replaced by MATH. Note that MATH by REF . Since MATH by REF , we have that MATH by REF , which is REF for MATH and MATH. Similarly we get that MATH, MATH, which is REF . Hence there exists a unitary MATH such that MATH . Set MATH. Then MATH is a unitary and MATH .
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math/0006033
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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 unit, and MATH, because MATH. By REF we have for MATH, MATH, MATH where MATH and MATH are integers such that MATH. Note that MATH. For MATH, choose integers MATH, MATH, and MATH such that MATH and note that MATH for some integers MATH, MATH. Then MATH . Let for each MATH for some integers MATH and MATH and set MATH . Note that for MATH, MATH . Hence MATH . Therefore MATH . Since by REF MATH we see that MATH . By this and REF , MATH . Hence from REF we conclude that MATH . Let MATH denote the exceptional points of MATH and let MATH be those of MATH. Set for MATH, MATH then by REF MATH . Set for MATH, MATH and note that MATH . By REF we see that MATH, MATH, and hence there exists a continuous function MATH such that MATH . Let for MATH and MATH, MATH be the MATH matrix MATH . Since MATH is a block-diagonal matrix with MATH blocks of the form MATH, MATH, MATH blocks of the form MATH, MATH blocks of the form MATH, MATH, and MATH blocks of the form MATH, there exists a unitary MATH such that MATH for every MATH. Set MATH. For each MATH, we have by REF that MATH . Choose for MATH continuous functions MATH such that for each MATH as ordered tuples. Choose a unitary MATH such that MATH for MATH. Define a continuous function MATH such that MATH . Then by REF , and REF we have that MATH for MATH. Define a unital *-homomorphism MATH by MATH . By the remarks following the definition of MATH we see that for MATH, MATH and hence MATH in MATH. Furthermore, as MATH is the identity map on MATH, we have that for MATH, MATH and by REF , for MATH, MATH . Hence MATH in MATH by REF . It follows from REF that MATH in MATH. Finally, for MATH and MATH, MATH . Hence MATH .
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math/0006033
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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 REF there exists a unital *-homomorphism MATH such that MATH are eigenvalue functions for MATH and such that MATH . Since for MATH, MATH we get that MATH .
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math/0006033
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Note that MATH is a building block by REF . Hence by REF there exists a unitary MATH such that MATH . Then MATH by REF .
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math/0006033
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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 settled. Let MATH be the projection and let MATH be the inclusion. As the diagram MATH commutes for MATH, and since MATH for MATH, we get unital *-homomorphisms MATH such that MATH . Define MATH by MATH. Then MATH . So assume MATH. Note that by REF in MATH for MATH. Let MATH. Choose by REF orthogonal non-zero projections MATH, for MATH, with sum MATH such that MATH. Let MATH be the normalized trace of MATH. Note that we have a well-defined map MATH such that MATH for every self-adjoint element MATH. Define MATH by MATH where MATH denotes the inclusion. MATH is a linear positive map, and it preserves the order unit since MATH . By CITE we get that MATH is a MATH-equivalence. Note that MATH . By REF we have that MATH. Choose by REF a unitary MATH such that MATH . Since MATH for MATH, we get by REF a unital *-homomorphism MATH such that MATH . Now define MATH by MATH . MATH is a unital *-homomorphism and MATH . For MATH, MATH, we have that MATH and MATH . It follows that MATH . Finally, for MATH, MATH modulo MATH.
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math/0006033
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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 is an interval building block is similar, and the matrix algebra case is trivial. Choose MATH such that for MATH, MATH . Let MATH where each MATH is an open arc-segment of length less than MATH. Choose for each MATH, a non-zero continuous function MATH with support in MATH such that MATH is zero at every exceptional point of MATH. Set MATH . Let MATH be given such that MATH, MATH. By CITE we may assume that MATH if MATH is a circle building block and MATH if MATH is an interval building block. Here MATH is a unitary, MATH are continuous functions, and MATH are non-negative integers. Since MATH, MATH, it follows that the set MATH intersects non-trivially with every MATH. If MATH is an interval building block, let MATH be the exceptional points of MATH. If MATH is a circle building block, let MATH be numbers such that MATH, MATH, are the exceptional points of MATH. For each MATH, choose a continuous function MATH such that MATH, MATH, such that MATH for MATH, and such that MATH. Define MATH by MATH if MATH is a circle building block, and MATH if MATH is an interval building block. Note that MATH is injective and MATH, MATH.
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math/0006033
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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.
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math/0006033
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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.
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math/0006033
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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 map MATH such that MATH, MATH, and such that MATH, MATH. Let MATH . Define MATH by MATH and let MATH be restriction. Then MATH, MATH.
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math/0006033
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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 building blocks and matrix algebras with unital connecting maps, unital *-homomorphisms MATH, and unital injective *-homomorphisms MATH such that the diagram MATH becomes an approximate intertwining. Furthermore MATH should be injective unless MATH is finite dimensional. This is sufficient since the proposition is trivial if MATH is an NAME. It is easy to construct MATH, MATH and MATH. Assume that MATH, MATH and MATH have been constructed. Let MATH and finite sets MATH and MATH be given. Choose MATH by REF with respect to MATH and MATH. Since MATH is simple we may choose MATH such that MATH for MATH. Choose MATH, MATH and MATH by REF with respect to MATH and MATH. Set MATH. Then MATH . Since MATH, MATH, there exists by REF a unital *-homomorphism MATH such that MATH and such that MATH is injective if MATH is infinite dimensional.
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math/0006033
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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 if MATH and MATH is a unital *-homomorphism, we see that the elements MATH, MATH, are non-zero and mutually orthogonal. Thus MATH.
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math/0006033
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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 to construct a strictly increasing sequence of positive integers MATH, unital *-homomorphisms MATH, and unital injective *-homomorphisms MATH such that the diagram MATH becomes an approximate intertwining. This is done by induction. Set MATH. Assume that MATH has been constructed. Let MATH and a finite set MATH be given. It suffices to construct MATH, a unital *-homomorphism MATH such that MATH and MATH are approximately unitarily equivalent, and a unital injective *-homomorphism MATH such that MATH . Let MATH and let MATH be the minimal non-zero central projections in MATH. Let MATH be the projection, MATH. Choose by REF a finite set MATH of positive non-zero elements with respect to MATH and MATH. Let MATH be the cardinality of MATH. Since MATH is simple there exists a MATH such that MATH . By REF there exists an integer MATH such that MATH . Let MATH and note that MATH . Hence there exists a unital *-homomorphism MATH such that MATH, MATH. Let MATH be the *-homomorphism that agrees with MATH on MATH and with MATH on MATH. The *-homomorphism MATH from MATH to MATH is by CITE approximately unitarily equivalent to the *-homomorphism induced by MATH. Hence MATH and MATH are approximately unitarily equivalent. Let MATH, MATH and let MATH be the unital *-homomorphism induced by MATH. Since MATH there exists by REF a unital injective *-homomorphism MATH such that MATH . Let MATH be the *-homomorphism that agrees with MATH on MATH and with MATH on MATH. Note that MATH is unital and injective and that REF holds.
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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 *-homomorphisms of the form MATH for continuous functions MATH, integers MATH and MATH with MATH, MATH for MATH, and unitaries MATH. By REF we have that MATH and MATH are equivalent representations of MATH on MATH, MATH. It follows that MATH, MATH, that MATH, and that MATH as unordered MATH-tuples, MATH. Hence MATH . For every MATH, we have that MATH . As MATH at most contains one of the exceptional points of MATH, we see that MATH . Thus MATH . It follows that MATH. Symmetry allows us to conclude that for all MATH, MATH . By REF we can define a *-homomorphism MATH by MATH . Note that MATH . Since MATH, MATH, MATH, it follows that MATH and MATH are approximately unitarily equivalent by CITE. Hence there exists a unitary MATH such that MATH . Choose a unitary MATH such that MATH . Set MATH. Then for MATH, MATH .
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math/0006033
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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 block or, every MATH is a finite direct sum of at least two circle or interval building blocks. Let us first assume that the latter is the case. Let MATH where each MATH is a circle or an interval building block. For each MATH let MATH denote the coordinate projections, MATH. First we claim that we may assume that all the maps MATH are injective. By NAME 's approximate intertwining argument it suffices to show that given a finite set MATH and MATH there exists an integer MATH and a unital *-homomorphism MATH such that MATH, MATH, and such that MATH is injective, MATH. Choose by REF a finite set MATH of positive non-zero elements with respect to MATH and MATH. As MATH is simple and the connecting maps are injective, we have that MATH, MATH. Thus there exists an integer MATH such that MATH, MATH. Hence MATH, MATH, and the claim follows by MATH applications of REF . Define a unital *-homomorphism MATH by MATH . Since the maps MATH are injective, MATH, and as MATH, it follows from REF that MATH is injective. The theorem therefore follows in this case from the commutativity of the diagram MATH . It remains to prove the theorem in the first case. By passing to a subsequence we may assume that each MATH is an interval building block. Let MATH, let MATH be a positive integer, and let MATH be finite. Again by NAME 's approximative intertwining argument, it suffices to show that there exists an integer MATH and a unital and injective *-homomorphism MATH such that MATH . Choose by REF a finite set MATH of positive elements of norm MATH with respect to MATH and MATH. Since MATH is simple and the connecting maps are injective there exists a MATH such that MATH, MATH. Let MATH. By REF there exists an integer MATH such that MATH and such that MATH . Let MATH. By CITE MATH is approximately unitarily equivalent to a *-homomorphism MATH of the form MATH where MATH is a unitary and MATH are continuous functions. Choose a continuous function MATH such that MATH at the exceptional points of MATH and such that MATH is surjective. Define MATH by MATH . Note that MATH is injective, and that for MATH, MATH . Finally, as MATH, MATH, we see by REF that there exists a unitary MATH such that MATH . Set MATH, MATH.
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math/0006033
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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 MATH be an arbitrary point, MATH. Define linear positive order unit preserving maps MATH and MATH by MATH . Note that MATH . Hence there exist linear positive order unit preserving maps MATH where MATH, such that MATH . Let MATH be the standard basis in MATH. As MATH are positive elements with sum MATH in MATH, there exist a positive integer MATH and positive elements MATH such that MATH and MATH . Define linear positive order unit preserving maps MATH by MATH and MATH by MATH. Since MATH we see that MATH . By continuity of MATH there exist an integer MATH and an element MATH such that MATH. As MATH we see that there exists an integer MATH such that MATH. Set MATH and MATH.
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math/0006033
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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 such that MATH . Since MATH is simple there exists a MATH such that MATH . Choose MATH such that MATH and MATH. By REF there exist a positive integer MATH and a linear positive order unit preserving map MATH such that MATH and an element MATH such that MATH . Since by REF on MATH we see that for MATH, MATH . Hence MATH . Choose MATH such that for MATH, MATH . Define MATH by MATH. Define MATH by MATH. Decompose MATH as a finite direct sum of building blocks and let MATH be the projection, MATH. Identify MATH with MATH. Fix some MATH. Set MATH. MATH is a strictly positive function in MATH, since MATH. Thus for each MATH, we can define MATH by MATH . MATH is positive and linear, and it preserves the order unit since MATH . Let now MATH, MATH, for MATH. Since MATH in MATH we have that MATH . Hence if MATH, MATH, then MATH . Define MATH by MATH . Then MATH and hence MATH . Finally, MATH, MATH. It follows that MATH on MATH.
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math/0006033
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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.
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math/0006033
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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. Note that MATH.
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math/0006033
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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 metric space. It follows that it is constant and hence MATH for every MATH. We conclude that MATH.
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math/0006033
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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 .
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math/0006033
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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 such that for every MATH we have that MATH. To see this choose a positive integer MATH such that MATH. Since MATH is torsion free by REF , we may choose MATH such that MATH, MATH. Let MATH. We may assume that MATH. Choose MATH such that MATH. Then MATH. By REF we get a positive integer MATH and a contractive group homomorphism MATH such that MATH for every MATH. Let MATH be an element of order MATH. There is an integer MATH such that MATH for some element MATH. We claim that MATH. To this end we may assume that MATH is a building block. Then MATH for some MATH. Hence MATH . Thus MATH. Since MATH was arbitrary we conclude that MATH.
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math/0006033
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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 assume that MATH such that MATH is a discrete subgroup of MATH. It follows that MATH for some MATH. By continuity of MATH we have that MATH for some integer MATH and some MATH. Define MATH by MATH . Then MATH . Since MATH has order MATH and MATH it follows that MATH has order MATH as well.
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math/0006033
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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 MATH by REF with respect to MATH and MATH. By REF there exist a positive integer MATH and a linear positive order unit preserving map MATH such that MATH and an element MATH such that MATH . Hence MATH, MATH. Choose MATH such that MATH and such that MATH, MATH. Then MATH and hence MATH. Furthermore MATH. It follows from REF that there exists a unital *-homomorphism MATH such that MATH. Set MATH. This proves the first part of the theorem. To prove the second part of the theorem, let us first note that MATH . Hence if MATH is non-cyclic then MATH by REF . We may therefore assume that MATH is cyclic. By REF we see that MATH is the inductive limit of a sequence MATH where each MATH is a finite direct sum of building blocks and each MATH is unital. By CITE there exist a positive integer MATH and *-homomorphisms MATH and MATH such that MATH is homotopic to MATH and MATH is homotopic to MATH. Note that MATH and MATH are unital. Since MATH by REF , there exists by REF a positive integer MATH such that MATH . By the first part of the theorem there is a unital *-homomorphism MATH such that MATH. Note that MATH . Hence MATH by REF .
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math/0006033
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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 MATH and a unital *-homomorphism MATH such that MATH for some MATH. Choose a unital *-homomorphism MATH such that MATH. Since MATH we see that MATH . Hence MATH and MATH. By REF this conclusion also holds if MATH is non-cyclic. Let MATH. Choose an element MATH of finite order such that MATH. Choose a finite direct sum of building blocks MATH and a unital *-homomorphism MATH such that MATH. Choose a unital *-homomorphism MATH such that MATH in MATH. Since MATH we see that MATH. Finally, since MATH is a divisible group there exists by CITE a group homomorphism MATH such that MATH for every MATH of finite order. Set MATH . It is easy to see that the diagram commutes.
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math/0006033
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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 projections in MATH. Since MATH is simple and the connecting maps are injective, there exists a MATH such that MATH, MATH. By REF there exist a positive integer MATH, a linear positive order unit preserving map MATH, and an element MATH such that MATH . Choose an integer MATH by REF with respect to MATH and MATH. Choose a positive integer MATH and unitaries MATH such that MATH . Note that MATH in MATH. Hence MATH . Since MATH we see that for MATH, MATH . Hence there exists an integer MATH such that MATH and such that MATH . It follows that MATH and that MATH . Therefore by REF there exists a unital *-homomorphism MATH such that MATH . It follows that MATH . Note that for MATH, MATH . Hence MATH . By REF , MATH is an isometry when MATH is equipped with the metric MATH. It follows that MATH . Thus MATH . Since MATH and MATH agree on the torsion subgroup of MATH, we see by REF and the definition of MATH that MATH . Hence for MATH, MATH .
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math/0006033
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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 as a MATH-algebra and such that MATH. Choose by uniqueness, REF , a positive integer MATH with respect to MATH and MATH. Since MATH is simple and the connecting maps are injective there exists a positive integer MATH such that MATH . Next, there exists a positive integer MATH such that MATH . Finally choose MATH such that MATH and such that MATH . Choose for each MATH finite sets MATH such that MATH, such that MATH, and such that MATH is dense in MATH. Next, choose finite sets MATH such that MATH for MATH, such that MATH, and such that MATH is dense in MATH. We will construct by induction strictly increasing sequences MATH and MATH and unital *-homomorphisms MATH such that CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH. The integers MATH, MATH, and the *-homomorphism MATH are constructed in REF. The case MATH follows immediately from REF . Assume that MATH, MATH, and MATH have been constructed such that REF hold. Choose MATH such that MATH . Choose by REF a positive integer MATH and a unital *-homomorphism MATH such that MATH . It follows that MATH . Hence there exists an integer MATH such that MATH . By uniqueness, REF , there exists a unitary MATH such that MATH . Set MATH, MATH. It is easily seen that REF are satisfied with MATH in place of MATH. This completes the induction step. By NAME 's approximate intertwining argument, see for example, CITE, there exists a *-homomorphism MATH such that MATH . Clearly, MATH is unital. Let MATH, MATH. The sequence MATH converges to MATH in MATH as MATH. Hence it follows that MATH . On the other hand, from REF it follows that MATH . Hence MATH on MATH and thus MATH on MATH. If MATH then MATH. This is clear in the case that MATH since then MATH, and otherwise it follows since MATH and MATH are isometries and MATH is contractive (with respect to MATH and MATH). Thus MATH is continuous and by arguments similar to those applied above we see that MATH. Let finally MATH be a positive integer. Since MATH is semiprojective there exists by CITE a positive integer MATH such that MATH is homotopic to MATH. Hence MATH in MATH. It follows from CITE that MATH in MATH.
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math/0006033
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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 .
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math/0006033
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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 .
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math/0006033
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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 with unital connecting maps. Let MATH where each MATH is a building block. Let MATH be the set of minimal non-zero central projections in MATH. Let MATH be a finite set and let MATH. It suffices to see that there exists a unitary MATH such that MATH . We may assume that MATH for a positive integer MATH and a finite set MATH. Choose by uniqueness, REF , a positive integer MATH with respect to MATH and MATH. Since MATH is simple and the connecting maps are injective there exists an integer MATH such that MATH . Next choose MATH such that MATH . Finally, choose MATH such that MATH, MATH, and such that MATH . Since MATH by REF is semiprojective there exist by CITE a positive integer MATH and *-homomorphisms MATH such that MATH is homotopic to MATH and MATH is homotopic to MATH, and such that if MATH then MATH . By increasing MATH we may assume that MATH and MATH are unital. Note that MATH and MATH . Choose an integer MATH such that MATH . By REF there exists a unitary MATH such that MATH . If we put MATH we have that MATH .
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math/0006033
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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 .
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math/0006033
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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 REF we have that MATH and MATH as MATH. By CITE there exists an invertible element MATH such that MATH on MATH and MATH on MATH. By REF there exists a group isomorphism MATH such that the diagram MATH commutes and such that MATH for MATH in the torsion subgroup of MATH. By REF there exists a unital *-homomorphism MATH such that MATH on MATH, such that MATH on MATH, and such that MATH in MATH. Note that MATH. It is easy to see that MATH is a bijection with inverse MATH. Hence MATH on MATH. Thus there exists a unital *-homomorphism MATH such that MATH on MATH, such that MATH on MATH, and such that MATH in MATH. By REF the *-homomorphisms MATH and MATH are approximately unitarily equivalent. Similarly MATH and MATH are approximately unitarily equivalent. Thus there are sequences of unitaries MATH and MATH, in MATH and MATH respectively, such that, upon setting MATH and MATH, the diagram MATH becomes an approximate intertwining. Hence by for example, CITE there is a *-isomorphism MATH such that MATH . It follows that MATH on MATH, that MATH on MATH, and that MATH on MATH.
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math/0006033
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This follows easily from REF .
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math/0006033
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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 we will construct a unital projectionless building block MATH such that MATH, and a unital and injective *-homomorphism MATH such that the (constant) functions MATH are eigenvalue functions for MATH, such that MATH on MATH (under the identification MATH) and such that MATH under the identification MATH, where MATH . First choose by REF a unital projectionless building block MATH such that MATH. Assume that MATH has been constructed. We will construct MATH and MATH. Choose MATH by REF with respect to MATH, MATH and the integer MATH. By REF there exists a unital projectionless building block MATH such that MATH and MATH. By REF there exists a unital *-homomorphism MATH such that the identity function on MATH and each of the functions MATH are among the eigenvalue functions for MATH and such that MATH . This completes the construction. Set MATH. MATH is infinite dimensional since the connecting maps are injective, and it is unital projectionless since the connecting maps are unital. By CITE MATH, and hence MATH and MATH are affinely homeomorphic. Clearly MATH. Let MATH be a closed two-sided ideal in MATH, MATH. By (the proof of) CITE, MATH . Choose a positive integer MATH such that MATH. Choose MATH such that MATH. Choose MATH such that MATH. Then MATH for every MATH and MATH. Hence by REF we see that MATH for every MATH. It follows that MATH. Thus MATH is simple.
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math/0006033
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Combine REF .
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math/0006039
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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, if MATH, then by REF we get MATH . Notice that, in particular, this implies MATH. We only have to look at the definitions of MATH and MATH again. CASE: Suppose that MATH. In this case we must show that MATH . It is enough to see that MATH. This follows from the inclusion MATH, which holds because MATH and then we can apply REF . On the other hand, since by definition we have MATH we are done.
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math/0006039
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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 . So the second inequality in REF holds if we take MATH big enough. Consider now REF . The first estimate follows easily from REF . Let us see the second inequality of REF . By REF have MATH . Thus we can write MATH . Let us estimate the first integral on the right hand side of REF . Using the first inequality in REF we obtain MATH . Let us consider the last integral in REF (only in the case MATH). By REF we have MATH . From REF we get REF .
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math/0006039
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As mentioned above, MATH and so MATH. On the other hand, we also have MATH, and then MATH and so MATH.
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math/0006039
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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 . Let us estimate the term MATH. Operating as in REF , we obtain MATH . On the other hand, since MATH, MATH, and MATH for all MATH, we get MATH . So MATH verifies REF . Let us consider the term MATH now. By REF we obtain MATH . Since MATH, we have MATH . It is easy to check that MATH. Indeed if MATH, then MATH and so MATH and so MATH. Thus the term MATH also satisfies REF . Let us turn our attention to MATH. In this case we have MATH . Thus we only have to check that MATH. Because MATH, we have MATH and so MATH. Let us see that MATH. If MATH, then MATH . If MATH, then MATH and so MATH, which yields MATH.
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math/0006039
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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 MATH, we have MATH . By REF (taking into account that MATH), MATH . By REF we have MATH for some MATH. Therefore, MATH . Also, we have MATH and MATH. Indeed, if MATH then there exists some MATH, and so MATH and MATH. Then we obtain MATH . In an analogous way, we get MATH . Therefore, by NAME 's Lemma we have MATH for all MATH if MATH. On the other hand, for MATH, operating in a similar way, we also obtain MATH, and if MATH, then we have MATH. Thus REF holds in any case. If there exist stopping cubes, then by REF the kernels of the operators MATH satisfy properties which are similar to the ones stated in REF , and some estimates as the ones above work. If MATH is an initial cube, then MATH, in general. However in the arguments above it is used MATH only to estimate MATH in the case MATH, and notice that MATH for MATH.
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math/0006039
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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 valued NAME operators. These techniques, which are standard in the classical doubling case, have been extended by CITE to the case of non homogeneous spaces. Let us denote by MATH the NAME space of sequences of functions MATH, MATH, such that MATH . Let us consider the operator MATH given by MATH. By REF , MATH is bounded from MATH into MATH. From the results in CITE, it follows that if the kernel MATH of MATH satisfies CASE: MATH for MATH, and CASE: MATH, then MATH is bounded from MATH into MATH, MATH, because MATH is a vector-valued NAME operator. Thus we only have to check that these conditions are satisfied. Let us see that the first one holds. Given MATH, MATH, let MATH be such that MATH. Since MATH, we have MATH . Now we will show that REF is also satisfied. Since MATH, we only have to deal with the term MATH. Let MATH be such that MATH, and suppose that MATH for some MATH. Notice that MATH if MATH. Indeed, we have MATH . If MATH, then MATH, and if MATH, then we have MATH. Assuming MATH, we get MATH since MATH, with MATH. Therefore, MATH . Then, using REF we obtain MATH .
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math/0006039
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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.
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math/0006039
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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. Observe that if MATH, then MATH, and so MATH . Thus MATH. By the symmetry of the assumptions, we also have MATH. By NAME 's Lemma we get MATH, and we are done.
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math/0006039
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For MATH, by REF , we have MATH . Since the matrix MATH originates an operator bounded on MATH, we obtain MATH .
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math/0006039
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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 there exists some constant MATH independent of MATH such that the kernels MATH of the operators MATH satisfy the inequality MATH . This an easy estimate that is left to reader. We take MATH so that MATH, and MATH with MATH. By REF we have MATH where MATH may depend on MATH and MATH and, in particular, we may have MATH. We split MATH as follows MATH . Let us estimate MATH: MATH . Let us consider the term MATH. Since MATH is in MATH uniformly on MATH, we have MATH . From REF we get MATH . Therefore, MATH (for MATH). Now we write MATH . The first integral on the right hand side tends to MATH. The second one tends to MATH, by an application of the dominated convergence theorem, because MATH if MATH .
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math/0006039
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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 and that the constants MATH and MATH in REF tend to MATH as MATH. Recall that MATH, with MATH. First we deal with REF . In REF we have shown that if MATH, then the kernel MATH of MATH satisfies MATH . Moreover, just below REF we have seen that MATH if MATH or MATH. For MATH, MATH, let MATH be the largest integer such that MATH. Since MATH for MATH, we get MATH if MATH. Taking into account that for MATH we have MATH, from REF it easily follows that MATH . An analogous estimate can be obtained for MATH. Thus the kernel MATH of MATH satisfies MATH . Now we have to show that the constant MATH in REF corresponding to the kernel of MATH tends to MATH as MATH. First we will deal with the term MATH, assuming MATH. Let MATH be the largest integer such that MATH. Using REF it is easy to check that MATH . So we only have to estimate the integral MATH. Notice that MATH and MATH, and so MATH. Thus we may assume MATH. Let us consider the case MATH, that is, MATH. By REF we obtain MATH . Assume now MATH, that is MATH (and MATH too). Observe that MATH . It is easily checked that the integrand above is null unless MATH. Since MATH, we have MATH . Also, for MATH, MATH . Therefore, MATH . Using MATH, we obtain MATH . Thus we get MATH . Let us consider the term MATH now. As in the case of the term MATH, we have MATH and we only have to consider the integral MATH. Moreover, it is easily seen that we also have MATH in this case. Thus we may assume MATH again. Operating as above, by REF we obtain MATH . Suppose now that MATH, that is, MATH. We have MATH . Since MATH, we have MATH . Thus MATH . Let us estimate MATH: MATH . We consider MATH now. On the one hand we have MATH . On the other hand, MATH . Thus we have MATH in any case. Therefore, MATH and so MATH . When MATH is negative (MATH), we have analogous estimates. As a consequence, the kernel of MATH satisfies NAME 's condition with constant MATH, and we are done.
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math/0006039
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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 since MATH, then MATH which contradicts the maximality of MATH. Therefore we get MATH . By REF and the finite overlap of the cubes MATH, REF follows. Let us prove REF now. We have MATH . We consider the maximal operator MATH . This operator is bounded on MATH (see REF below). We set MATH. Then we have MATH . By REF we obtain MATH and we are done.
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math/0006039
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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.
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math/0006039
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By REF , for MATH we have MATH .
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math/0006039
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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 check that MATH. Thus we have MATH . Now we will show that REF is also satisfied. First we will deal with the term MATH. We have MATH . We only have to estimate MATH for MATH, because otherwise MATH . Since MATH and MATH (we are assuming MATH), we have MATH . Therefore, MATH . We derive MATH . The estimates for the term MATH are similar.
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math/0006039
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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 to MATH in MATH as MATH. With estimates analogous to the ones in REF , it can be seen that MATH is a HCZO, and taking into account that MATH, it follows that this operator is bounded on MATH uniformly on MATH. Arguing as in CITE, it can be shown that for any function MATH bounded with compact support and any MATH, MATH . As in CITE, this implies MATH . Now it only remains to apply the version of the MATH theorem stated in REF to the operator MATH. Recall that that lemma applies to CZO's with bounded kernel. However, as explained in the remark above, it is enough that the operator fulfil REF , instead of the usual gradient condition demanded from the kernels of CZO's. Moreover, the additional hypothesis in REF about the MATH-boundedness of the kernel was useful in the preceding section to deal with the convergence of some integrals and also for the proof REF . Although the kernels of MATH and MATH are not MATH-bounded, we already know that these operators are bounded on MATH and that they are weak limits of operators MATH, MATH, with `nice' kernels. This allows the extension of the arguments for proving REF to the present situation. We omit the detailed arguments.
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math/0006045
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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 of the band) using the Cutting and Sliding REF as shown MATH (where MATH is a surface of genus MATH, and the solid arcs represent arbitrary tubes in REF-manifold). The above calculation reduces to the case of a surface MATH hawith no bands, that is, a disk. Using the Cutting and Sliding REF once again, we may assume that the leaf MATH of MATH is zero-framed and that the disk MATH intersects geometrically once a leaf of MATH and is otherwise disjoint from MATH. The following equality MATH which follows by REF , concludes our proof.
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math/0006045
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It follows from two applications of the MATH relation that MATH .
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math/0006045
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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 same (respectively, opposite) way from MATH if MATH (respectively, MATH), and finally take an arbitrary band sum of them. We can arrange the resulting knots, one for each univalent vertex of the embedding of MATH, to be disjoint from each other, and together with the embedding of MATH to form an embedded graph with leaves in MATH. Although the isotopy class of the embedded graph with leaves is not unique, the image of MATH is well-defined. This follows from REF . We need to show that the relations MATH, MATH, MATH, MATH and MATH are mapped to zero, which will define our map MATH. For the MATH and MATH relations, see for example CITE. The MATH relation follows from REF . The MATH and MATH relations follow from REF . We now show that MATH is onto, over MATH. Note first that MATH is generated by MATH for all simple graphs of degree MATH, where a simple graph is a disjoint union of graphs of degree MATH. Each of the leaves of MATH are isotopic to some connected sum of (possibly orientation reversed) components of MATH and contractible knots. Using the Cutting REF , we may assume that each leaf is isotopic to one of the components of MATH (with possibly reversed orientation) or is contractible in MATH. From this point on, the proof is analogous to the case of MATH. Let MATH be the link consisting of all contractible leaves of MATH. There exists a trivial, unit-framed link MATH in MATH with the properties that MATH each component of MATH bounds a disk that intersects MATH at at most two points. MATH . Under the diffeomorphism of MATH with MATH, MATH becomes a zero-framed unlink bounding a disjoint collection of disks MATH. Such a link MATH was called MATH-untying in CITE. REF imply that we can assume each of the disks MATH are disjoint from MATH and intersect MATH in at most two points MATH. REF imply that MATH is onto, over MATH. In order to show that MATH is independent of MATH, up to isomorphism, we need the following: Every two MATH-spanning links MATH and MATH in MATH are equivalent by a sequence of moves CASE: Add one component (after possibly changing its orientation) of MATH to another. CASE: Change the framing of a component of MATH. CASE: Insert or delete a null-homologous zero-framed component of MATH. It suffices to show that under these moves MATH is equivalent to MATH. Consider a component MATH of MATH. Since MATH is a basis of MATH, we can add a multiple of components of MATH (after perhaps changing their orientation) so that MATH is nullhomologous, in which case we can change its framing to zero, and erase it. The lemma now follows by induction on the number of components of MATH. If MATH is obtained from MATH by applying one of the three moves above, we will now define MATH (abbreviated by MATH in what follows) and show that REF holds. For the first move, if MATH and MATH (where MATH is an arbitrary oriented band sum of MATH with MATH) then MATH sends a MATH colored vertex of an abstract graph MATH to a MATH colored vertex of MATH. It is easy to see that this defines a map MATH whose inverse sends a MATH colored vertex of MATH to a MATH colored vertex of MATH. Similarly, one can define a map MATH where MATH. REF follows from REF . For the second move, let MATH and MATH where MATH is a knot whose framing differs from that of MATH by MATH. For graph MATH with MATH legs colored by MATH we define MATH where the summation is over all functions MATH such that the cardinality MATH of MATH is even and MATH is the result of gluing the MATH colored legs MATH of MATH for which MATH pairwise and recoloring the remaining MATH colored legs with MATH colored legs. It is easy to see that MATH is well-defined (that is, that it respects the relations in MATH) and that its inverse is given by MATH . Let MATH denotes a MATH-framed unknot in MATH which bounds a disk that geometrically intersects MATH in one point and intersects no other components of MATH. Then MATH is diffeomorphic to MATH under a diffeomorphism that sends the image of MATH in MATH to MATH in MATH. Since MATH and MATH, REF (or rather, its equivalent form MATH) follows from the following: For a graph MATH of degree MATH as above, we have in MATH: MATH . Using the Cutting REF each MATH-colored leaf MATH of MATH can be split along an arc in two leaves; one that bounds a disk MATH intersecting MATH once and disjoint from MATH, and another that is isotopic to MATH but disjoint from MATH. For MATH, let MATH denote the graph MATH. REF implies that MATH. Let MATH denote the graph in MATH that corresponds to MATH under the diffeomorphism MATH; we obviously have MATH. Note that MATH has a collection of MATH leaves each of which is unknotted bounding a disk with linking number MATH with every other leaf of this collection. An application of REF MATH times together with REF implies that MATH (respectively, MATH) for even (respectively, odd) MATH. For the third move, let MATH and MATH where MATH is a null-homologous zero-framed knot, and consider the natural map MATH. Choose a surface MATH that MATH bounds. The MATH relation in MATH for MATH colored vertices defines a map MATH; this map is independent of MATH since the difference between two choices of MATH equals to a choice of a closed surface and the resulting difference vanishes due to the MATH relation on MATH. It is easy to see that MATH is inverse to MATH. REF follows essentially by definition. This completes the proof of REF .
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math/0006045
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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 the torsion part of MATH. Then, we have that MATH. There are integers MATH and surfaces MATH such that MATH for all components of MATH. The MATH relation for MATH colored legs gives a map MATH which is independent of the choices of MATH and is inverse to the map MATH. Thus, MATH. Since MATH for every MATH-spanning link MATH, the result follows. The third statement follows immediately from the fact that if the intersection form on MATH vanishes, then the MATH relation is vacuous. The forth and fifth statements are immediate consequences of those above.
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math/0006045
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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.
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math/0006045
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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 MATH of degree MATH in a rational homology REF-sphere MATH, we have that MATH . For the first claim, recall that a degree MATH clover MATH in a manifold MATH is the image of an embedding MATH of a neighborhood MATH of the standard (framed) graph MATH of MATH, and that surgery of MATH along MATH can be described as the result of NAME surgery on the six component link MATH in MATH shown below MATH is partitioned in three blocks MATH of two component links each. We call each block an arm of MATH. Alternating a rational homology REF-sphere MATH with respect to surgery on MATH equals to alternating MATH with respect to all nine subsets of the set of arms of MATH. Recall also that the NAME integral of a framed link MATH in a REF-manifold MATH (defined by NAME for links in MATH and extended by NAME for links in arbitrary REF-manifolds CITE) takes values in linear combinations of MATH-colored uni-trivalent graphs. Recall also that the LMO=NAME integral of a rational homology REF-sphere MATH (obtained by surgery on a framed link MATH in a rational homology REF-sphere MATH) is obtained by considering the NAME integral MATH, splitting it in a quadratic MATH and trivalent (a better name would be ``other") part MATH, and gluing the MATH-colored legs of MATH using the inverse linking matrix of MATH. Given a clover MATH in a rational homology REF-sphere MATH, (where MATH are of degree MATH), let MATH denote the link that consists of the MATH arms of MATH. When we compute MATH, we need to concentrate on all MATH-colored uni-trivalent graphs that have at least one univalent vertex on each block of MATH. Such graphs will have at least MATH univalent vertices. Since at most three univalent vertices can share a trivalent vertex, it follows that the above considered graphs will have at least MATH trivalent vertices; in other words it follows that MATH. The second claim is best shown by example. Recall that surgery on the (generic trivalent graph) MATH shown below corresponds to surgery on two clovers MATH and MATH, each with arms MATH for MATH and MATH. The linking matrix of the MATH component link MATH and its inverse are given by MATH where MATH is the identity MATH matrix. The relevant trivalent part MATH is shown schematically in four cases here, where the graphs on the left terms of each case come from MATH and the graphs on the right terms of each case come from MATH and the dashed lines correspond to gluings of the univalent vertices: MATH . However, the last three cases all contribute zero, since MATH is a REF-component unlink whose coefficient in MATH is a multiple of the triple NAME invariant and thus vanishes. Thus, we are only left to glue terms in the first case, and this is summarized in the following figure MATH which concludes the proof.
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math/0006045
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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 have MATH, thus the result follows from the MATH relation.
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math/0006045
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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 with MATH leaves colored by MATH, for MATH. Since MATH is primitive and linearly independent from MATH, the MATH case for MATH shown above implies that MATH for all MATH. Since MATH, the result follows.
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math/0006047
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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 compute the NAME operator, we thus only need to collect and sum terms with constant coefficients in REF . Applying the formulas for NAME derivatives once again, we get MATH . Summing these terms gives the announced expression.
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math/0006047
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Just evaluate MATH on MATH.
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math/0006047
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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 that there exists one and only one MATH such that MATH, and that MATH is an eigenvector of MATH. The second, that MATH is uniquely determined by MATH. We may thus define MATH. The value of the map MATH is clearly well-defined on any eigenvector of MATH, hence the proof.
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math/0006047
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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.
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math/0006047
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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 equals MATH. Now, if MATH, the principal symbol of MATH, and therefore MATH, equal MATH. If MATH admits MATH as its principal symbol, then MATH has an order less than that of MATH and an evident induction allows to conclude that MATH is surjective.
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math/0006047
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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.
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math/0006047
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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 in the latter expression. It is clear that MATH. Furthermore, if MATH, MATH . Since MATH, it suffices to compute explicitly the different values of MATH for MATH to prove that MATH for any odd natural number MATH. One proceeds in the same way if MATH and MATH.
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math/0006047
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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, the last two terms equal MATH and thus also belong to MATH.
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math/0006047
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Proceed as when MATH is not resonant, noticing that both members of the equality MATH belong to MATH provided that MATH.
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math/0006047
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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 tensor field with homogeneous component in MATH given by MATH where MATH symbolizes the derivatives of the argument of MATH. It means that the polynomial symbolization of MATH is a linear combination of powers of the operators MATH and MATH. But the last two stabilize the space MATH, because of the definition of this space. Furthermore, MATH and MATH vanish or intertwine irreducible MATH-submodules of the spaces MATH. Hence the conclusion.
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math/0006047
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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 defines similarly an operator associated to MATH. This equivariance also holds for the map that associates to MATH the operator MATH whatever the value of MATH. Moreover, any eigenvector of MATH not belonging to MATH is the principal symbol of a unique eigenvector of MATH.
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math/0006047
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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.
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math/0006049
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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.
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math/0006049
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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 number of critical pints of MATH lying in the interior of MATH is at least the category of MATH. Since MATH (because of REF ), the number of billiard trajectories inside MATH, which start at MATH, end at MATH and make MATH reflections is at least MATH .
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math/0006049
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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 MATH the subset of MATH, consisting of all configurations MATH, satisfying the conditions: MATH . Given two interval partitions MATH and MATH, we say that MATH refines MATH and write MATH if the intervals of MATH are unions of the intervals of MATH. We denote by MATH the number of intervals in the partition MATH. Note that MATH implies MATH and MATH. For the partition MATH with MATH holds MATH (since we assume that MATH). If MATH then MATH is a single point. For MATH the space MATH is homeomorphic to the Cartesian power MATH. As in CITE, we will denote by MATH the subset of MATH satisfying the conditions MATH for MATH. The configuration space MATH is homotopy equivalent to the product of spheres MATH. A homotopy equivalence MATH is given by the map MATH . Fixing an orientation of the sphere MATH, determines a canonical top-dimensional class in MATH, which is the pull-back of the product MATH under REF . If MATH two points, we denote by MATH the subspace of MATH consisting of configurations MATH with MATH; similarly we denote by MATH the subspace of configurations with MATH. Let MATH be a partition of MATH on intervals of lengths MATH, and let MATH . We claim that the stalk of the sheaf MATH at MATH equals MATH . Indeed, by definition, this stalk is MATH where MATH is a small open ball around MATH. If MATH then we may choose points MATH, one for each interval of MATH, so that MATH if MATH belongs to the MATH-th interval. Let MATH be a small open neighborhood of MATH, so that each MATH is diffeomorphic to MATH and the sets MATH and MATH are disjoint when the points MATH and MATH are distinct. Then we may take MATH, and our claim follows. We see that MATH vanishes unless MATH is a multiple of MATH and MATH . For an interval partition MATH of MATH with MATH denote by MATH the constant sheaf with stalk MATH and support MATH. We claim: for any MATH, the sheaf MATH is isomorphic to the direct sum of sheaves MATH the sum taken over all interval partitions MATH with MATH. To prove the claim, let MATH be an interval partition of MATH into intervals of length MATH, where MATH. Then for any interval partition MATH into intervals of length MATH, such that MATH, we have the canonical inclusion MATH . The target space of map MATH has a canonical nonzero MATH-dimensional cohomology class (compare above). The induced map MATH on MATH-dimensional cohomology with MATH coefficients is a monomorphism. Let MATH denote the image of the top-dimensional canonical class under the induced map MATH. Then (similarly to REF) for a fixed MATH, the classes MATH form a linear basis of the cohomology MATH, where MATH runs over all partitions with MATH and MATH. Indeed, using the map into product of spheres REF we see that a linear basis of the cohomology MATH form monomials MATH with MATH, such that for any MATH the indices MATH and MATH belong to the same interval of partition MATH. Let MATH be the partition determined by the equivalence relation on MATH, where MATH. Then MATH and the above monomial coincides with the class MATH. Given a partition MATH of MATH on intervals of length MATH, where MATH, consider the commutative diagram MATH formed by the natural inclusions. Define sheaf MATH over MATH. We want to show that MATH is isomorphic to MATH, that is, it is the constant sheaf with stalk MATH and support MATH. First, MATH vanishes outside MATH (since we are considering the cohomology of the top dimension). Let MATH be a small open neighborhood of a point MATH, such that MATH, where all MATH are small open disks and MATH if MATH and MATH lie in the same interval of MATH. Then MATH (where MATH) has a canonical element (compare above). This gives a continuous section of MATH over MATH, and hence MATH. The commutative diagram above gives a map of sheaves MATH, and summing, we obtain a map of sheaves MATH which, as we have seen above, is an isomorphism on stalks; hence it is an isomorphism, and REF follows. We arrive at the following description of the term MATH of the NAME spectral sequence MATH where MATH runs over all partitions of MATH with MATH. In order to identify this description with the one given in the statement of the theorem, assign to a monomial MATH with MATH the equivalence relation on the set of indices MATH generated by MATH . This equivalence relation defines a partition MATH of the set MATH on MATH intervals. In view of the relations MATH the term MATH is isomorphic to MATH. The monomial MATH corresponds to the partition MATH, which we should ignore since MATH; this explains the relation MATH. Now we prove REF concerning the differentials of the spectral sequence. The first nontrivial differential is MATH. To find MATH it is enough to find the cohomology classes MATH, where MATH. We will use functoriality of the NAME spectral sequence and the following well-known property. Let MATH be a manifold and let MATH be a submanifold of codimension MATH with oriented normal bundle. Consider the NAME spectral sequence MATH of the inclusion MATH. The sheaf MATH is the constant sheaf with support MATH and stalk MATH for MATH and it vanishes for all other values MATH. The only nonzero differential MATH acts as follows: the class MATH is mapped into MATH, the class dual to MATH, where the same orientation of the normal bundle to MATH is used in order to trivialize the sheaf MATH and to define the dual class MATH. In order to show the first relation MATH, consider the diagram MATH and apply the previous remark to the bottom row with MATH and MATH. The sign MATH appears as the degree of the antipodal map MATH: the framing of the normal bundle to MATH, which we use to define the fundamental class MATH, is antipodal to the framing determined by REF , which we use to trivialize the derived sheaf. To obtain relations MATH with MATH we use the commutative diagram MATH and apply the remark above with MATH and MATH. The last relation MATH follows similarly.
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math/0006049
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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 cyclic configuration space MATH (compare CITE) and the fibration MATH which has MATH as the fiber. The nonzero rows of the NAME spectral sequence have numbers divisible by MATH; also, the spectral sequence has only two columns MATH and MATH. We obtain that all differentials of the spectral sequence vanish and the cohomology of the fiber MATH is isomorphic to the factor of the ring MATH with respect to the ideal generated by class MATH (the pull-back of the fundamental class of the base). Comparing the above information with the structure of the ring MATH, described in REF, proves REF .
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math/0006049
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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 MATH has a very simple structure. Each element MATH has a unique representation in the form MATH, where MATH. From REF we obtain that the differential of MATH acts as follows MATH. Hence MATH is equivalent to MATH, which implies that MATH. Thus we obtain that each factor MATH is acyclic. The statement of the Lemma now follows by induction.
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math/0006049
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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 sequence MATH . We will show that the connecting homomorphism MATH is an isomorphism for all MATH and its is an epimorphism with one-dimensional kernel for MATH. This clearly implies the statement of the Lemma. Any element MATH has a unique representation in the form MATH . Then MATH equals MATH and hence we obtain MATH where the first summand corresponds to the class of MATH and the second summand corresponds to the class of MATH. Suppose that MATH is a cycle of the relative complex MATH. In order to calculate MATH, the image under the connecting homomorphism, we have to view MATH as a chain in MATH and compute MATH. We obtain MATH, which shows that MATH is always an epimorphism and it is an isomorphism if and only if MATH; by our induction hypothesis it holds if MATH. This completes the proof.
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math/0006051
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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 implies that MATH is congruent modulo MATH to MATH .
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math/0006051
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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 .
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math/0006051
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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. For this it is easier to consider the function MATH which satisfies MATH, MATH with MATH and MATH. In the situation above we have MATH, where MATH is the MATH-adic valuation and MATH is the MATH-th coefficient in the power series expansion with respect to MATH of any of the functions MATH. Let MATH be the MATH-th coefficient of MATH for any power series MATH. We have MATH . This implies that MATH . The lemma is clearly true for MATH. Suppose it is true for MATH. Then MATH . Since MATH, to finish the proof we have to check that for every MATH we have MATH. The well known estimate MATH and the assumption MATH imply that it is sufficient to require MATH, and this is satisfied for MATH. The case MATH is also OK according to this analysis unless MATH.
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math/0006051
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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 reduction of MATH independent of MATH for all MATH it is necessary and sufficient that for MATH we have MATH . If this is satisfied then the reduction of MATH is MATH so we should also require MATH. Let MATH. Then the relations REF can be written as MATH where MATH is a constant. This implies that MATH and after solving the resulting differential equation that the MATH first equations in REF are equivalent to MATH for some other constant MATH. We have MATH. Now, in MATH the coefficient of MATH is MATH . It is easy to see that for REF with MATH to be satisfied, this coefficient must be MATH and hence MATH. This gives the choice of the coefficients in the theorem and shows that they are the unique choice in MATH. Now if we have coefficients in MATH satisfying the theorem, then we may clear denominators not dividing MATH and using only independence of MATH we obtain that these must be a rational multiple of our MATH. But since the reduction of MATH is non-trivial the multiplier must be MATH.
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math/0006054
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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.
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math/0006054
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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 intersection of translates of MATH and MATH. Hence, MATH is MATH-MATH and MATH is a flat line bundle. On the other hand, the properties of MATH and MATH imply that translating MATH to MATH twists MATH by MATH and so by normalizing the NAME bundles appropriately, we have MATH.
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