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math/0004137
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If MATH is empty so the shape of MATH is the partition MATH, then we obtain MATH as the product MATH where MATH is the integer which MATH lacks compared to MATH. When this product is formed, the only boxes that can be effected are those containing integers strictly larger than MATH or those containing MATH which are to the left of the original box MATH in the construction of MATH. Since this implies that no shared boxes are modified, the shape of MATH is only one box larger than MATH. When MATH is not a partition, the same trick will work if we pretend that the boxes of MATH are actually filled with small integers when the product MATH is formed.
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math/0004137
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REF implies REF , and REF , while REF implies REF , and REF . For example, to prove REF from the corollary, recall that if MATH then there exists a tableau MATH of shape MATH such that MATH is a reverse lattice word with content MATH. Since MATH, MATH must contain a shared box. We can therefore let MATH be the shape of the tableau MATH of REF. Finally, to prove REF we need to show that if MATH is a tableau of shape MATH with at least one shared box such that MATH is a reverse lattice word with content MATH, then there exists a tableau MATH of a shape MATH such that MATH is a reverse lattice word with the same content MATH. Let MATH be the smallest integer contained in a shared box of MATH, and let MATH be the northernmost shared box containing MATH. Then start by removing MATH from this box. If all integers in the row above MATH are larger than or equal to MATH, then we can add a new box containing MATH at the left end of this row. Otherwise let MATH be the rightmost box in the row above MATH which contains an integer strictly less than MATH. Now replace the integer in MATH with MATH and continue in the same way with this integer in the role of MATH and MATH in the role of MATH. Using the induction hypothesis that some box strictly north of MATH contains MATH, it is not hard to check that this process stops before we reach the top row of MATH, and that the result is a tableau MATH with the desired properties.
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math/0004137
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Let MATH be the set in the leftmost shared box which contains MATH and let MATH, MATH, MATH, and MATH be as in the picture. MATH . Let MATH be the tableau obtained by attaching a box containing MATH to the right side of MATH and let MATH be the skew diagram of this box. Then set MATH, and let MATH be the tableau obtained from MATH by replacing MATH with MATH and MATH with MATH. Notice that MATH must be a single integer, since only integers less than or equal to MATH in MATH are affected when forming MATH, and none of these are in shared boxes. Since MATH is a reverse lattice word, so is MATH. But then REF implies that MATH is a reverse lattice word. The lemma follows from this.
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math/0004137
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It is enough to show that for each MATH there is a partition MATH of weight MATH such that MATH and MATH. We will do this by induction on MATH, the case MATH being trivial. By REF there exists a tableau MATH of shape MATH such that MATH is a reverse lattice word with content MATH. Since MATH, this tableau must have at least one shared box. If this box contains an integer which is larger than MATH, then REF gives us a tableau MATH in which the number of MATH's is the same as in MATH. Otherwise some shared box contains an integer smaller than MATH, in which case we use REF to produce MATH. Now if MATH, then MATH and MATH. Since MATH, the required partition MATH exists by induction.
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math/0004137
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Since the map MATH is surjective by REF and since MATH and MATH are free Abelian groups of the same rank, it is enough to show that MATH when MATH. If MATH this follows from REF since MATH has rank MATH. Now if MATH denotes the universal exact sequence on MATH, we get MATH which is zero if MATH.
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math/0004144
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The implication MATH is trivial. For MATH, we will assume that MATH. Recall that there exists MATH, such that MATH with MATH for every MATH. The proposition will be proved by showing the following lemma: For every MATH with MATH, MATH. To see this lemma, fix MATH and let MATH. Since MATH is equi-integrable, there exists MATH such that whenever MATH satisfies MATH, then for every MATH, MATH. Since both MATH and MATH converge to zero in measure, one can choose MATH such that for each MATH, there exists a projection MATH with MATH, MATH and MATH . For MATH, MATH . A similar estimate works for MATH. The lemma is verified. To complete the proof of REF , choose a mutually disjoint family MATH of projections in MATH with MATH for the strong operator topology and MATH for all MATH. Using a similar argument as in CITE, one can get an at most countable subset MATH of MATH such that for each MATH outside of MATH, MATH for every MATH. Let MATH. Replacing MATH by MATH and MATH by its restriction on MATH, we may assume that MATH. Let MATH. It is clear that MATH and MATH for every MATH. Fix MATH and choose MATH such that MATH . We get that MATH and since MATH is arbitrary, the proof is complete.
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math/0004144
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Let MATH be the polar decomposition of MATH. Then MATH. Also MATH and using NAME 's inequality, MATH .
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math/0004144
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We will show the non trivial implication. Fix MATH, a sequence in MATH. We need to show that MATH. We will assume without loss of generality that MATH is a subset of the unit ball of MATH. For every MATH, MATH . Since MATH is MATH-convex, we get: MATH . Using REF on the second term, we have MATH . Let MATH, choose MATH large enough so that MATH. We conclude that MATH . By REF , the first two terms converge to zero so MATH and since MATH is arbitrary, the proof is complete.
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math/0004144
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The lemma can be obtained inductively. Since MATH is countably decomposable, there exists MATH a faithful normal state in MATH. Since MATH is a semifinite projection, there exists a sequence of projections MATH with MATH for every MATH and MATH. One can choose MATH such that MATH. Set MATH . It is easy to verify that MATH satisfies the requirements of the lemma.
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math/0004144
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Let us show that for every MATH, MATH is a MATH-equi-integrable set. Assume that there exists MATH such that MATH is not MATH-equi-integrable. By definition, there exists a decreasing sequence of projections MATH such that MATH, that is MATH . Choose a strictly increasing sequence MATH of MATH such that MATH . Let MATH be a bounded operator such that MATH. We get that MATH . We recall that MATH and since MATH is decreasing, for MATH, MATH and therefore MATH and since MATH, we obtain that for MATH large enough, MATH . Using REF , with MATH and MATH, we get MATH . This implies that MATH . Let MATH be the left support projection of MATH (this is equal to the right support projection of MATH). We have MATH . By the definition of support projection, MATH for every MATH, so MATH is a sequence of finite projections. As in proof of REF , we note that MATH and as before, MATH. Now since MATH, MATH hence MATH converges to zero which implies that MATH. Therefore, MATH. In summary, we get MATH with MATH for each MATH and for some MATH, MATH . Let MATH. For each MATH, MATH so MATH hence MATH. In particular MATH. Using REF it applies since MATH, MATH . Taking the limit as MATH tends to MATH, one gets from REF that MATH. This is a contradiction since MATH. We conclude that for every MATH, the set MATH is a MATH-equi-integrable set. For the case of MATH, it is enough to repeat the argument above for MATH using the definition of MATH (instead of MATH). Details are left to the reader. This ends the proof of the lemma.
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math/0004144
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We note first that if MATH, then MATH and MATH so for fixed MATH, MATH. Similarly, MATH . REF implies that for every MATH, both MATH and MATH are MATH-equi-integrable sets. Therefore, if MATH is not MATH-equi-integrable, there would be a subsequence MATH of MATH and MATH such that MATH . Using REF on MATH and MATH, we obtain: MATH . Taking into account the identities, MATH, MATH and MATH, one can deduce that, MATH . Let MATH be the norm of the identity map from MATH onto MATH, where MATH (respectively, MATH) denotes the MATH-dimensional MATH-space (respectively, MATH-space). We have MATH . We remark that MATH and since MATH is MATH-convex (with constant MATH), the above inequality implies MATH and taking the limit as MATH, we get from REF that MATH . This is a contradiction since MATH, so MATH is a MATH-equi-integrable set. The lemma is proved. To complete the proof of REF , we note that MATH so if we set MATH and MATH. The proof of is complete.
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math/0004144
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For MATH, we set, as in the proof of REF , MATH . One can choose a strictly increasing sequence MATH in MATH such that for each MATH, there exists a sequence MATH with MATH and MATH . For every MATH, set MATH. Since MATH, it is clear that the sequence MATH belongs to MATH and each of the MATH's and MATH's are attained at MATH. One can complete the proof by procceding as in the proof of REF , simmultaneously on the finite set of sequences and the fixed MATH.
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math/0004144
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Let MATH be a bounded sequence in MATH and suppose (by taking a subsequence if necessary), MATH with MATH, the set MATH is MATH-equi-integrable and MATH for all MATH, be the decomposition of MATH as in REF . Let MATH. Since MATH and MATH . Proposition reforder-continuity REF shows that MATH . Choose MATH such that MATH . Inductively, one can construct MATH such that MATH . Since MATH for every MATH, one gets MATH . For every MATH, set MATH . Since MATH is a MATH-equi-integrable set and MATH, it is clear that MATH is MATH-equi-integrable. Also MATH is mutually disjoint. The proof is complete.
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math/0004144
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Assume that MATH has the NAME property (equivalenty MATH does not contain MATH). Since MATH is symmetric, MATH is equivalent to MATH not containing MATH uniformly, and therefore MATH satisfies the MATH-lower estimate for some MATH and one can renorm MATH so that it satisfies the lower MATH-estimate of constant MATH. All of these facts can be found in CITE.
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math/0004145
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The operator MATH has a factorization MATH where MATH, MATH and MATH are as in the above definition. Note that MATH is MATH-summing so MATH is MATH-summing and since MATH is a MATH-space and MATH has the (CRP), MATH (and hence MATH) is compact.
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math/0004145
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The proof is a refinement of the argument used in REF. We will include most of the details for completeness. Without loss of generality, we can and do assume that MATH is separable. Let MATH. From REF, MATH where MATH is a faithful normal state in MATH. If MATH is completion of the prehilbertian space MATH where MATH then we have the following factorization: MATH where MATH is the inclusion map, MATH for every MATH and MATH. We recall that MATH is a well defined bounded linear map since MATH is dense in MATH and MATH. Let MATH. Claim: MATH is compact. For this, let us consider MATH. Since MATH is separable, it is isometric to a subspace of MATH. Let MATH be the isometric embedding of MATH in MATH and MATH be the natural inclusion of MATH into MATH. Define the following map MATH from MATH into MATH by setting MATH for every MATH. (Here, MATH is the map MATH for MATH with MATH being the conjugate of the complex number MATH). Clearly, MATH is linear and bounded and it can be shown that MATH is MATH-summing and is weak-MATH to weakly continuous. In fact, if MATH is a finite sequence in MATH then MATH so MATH is MATH-summing with MATH. Moreover if MATH is a net that converges to zero weak-MATH in MATH so does the net MATH and since MATH is weak-MATH to weakly continuous, MATH converges to zero weakly in MATH and hence MATH is weakly null which shows that MATH is weak-MATH to weakly continuous. To complete the proof, consider MATH . Since MATH has the NAME extension property and MATH is MATH-summing, MATH is an integral operator. Let MATH such that MATH (such operator exists since MATH is weak-MATH to weakly continuous); MATH is integral CITE and since MATH is a complemented subspace of its bidual MATH (see for instance CITE), MATH is strictly integral and therefore MATH is strictly integral and by REF , MATH (and hence MATH) is compact. Let MATH be a bounded sequence in MATH. There exists a subsequence MATH so that MATH is norm convergent in MATH. Since MATH and MATH are isometries, we get that MATH is norm convergent so MATH. As in CITE, we get MATH which proves that MATH is norm-convergent in MATH. The proof is complete.
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math/0004145
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Let MATH be an operator with MATH. One can choose, by the NAME Factorization Theorem, a probability space MATH such that MATH where MATH is a subspace of MATH, MATH is the closure of MATH in MATH and MATH is the restriction of the natural inclusion MATH. Denote by MATH the closure of MATH in MATH, by MATH the restriction of the natural inclusion and MATH the natural inclusion of MATH into MATH. Claim: MATH is weakly compact. To see this, let MATH be a bounded sequence in MATH. By NAME 's Theorem, there exists a subsequence MATH and a function MATH such that MATH for a.e. MATH. Since MATH, MATH . This shows that MATH and MATH converges to MATH in MATH and the claim follows. Using the factorization of weakly compact operator CITE, MATH factors through a reflexive space and since MATH is MATH-summing, the theorem follows from REF .
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math/0004145
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MATH is trivial. For the converse, let MATH be a weakly null sequence in the unit ball of MATH. It is clear that MATH is also weakly null so without loss of generality, we can assume that MATH is a sequence of self-adjoint operators. For each MATH, set MATH the spectral decomposition of MATH and for every MATH, let MATH . It is clear that for every MATH and MATH, MATH . By the NAME 's characterization of relatively weakly compact subset in MATH (see for instance REF p. REF), we conclude that for any given MATH, there is MATH such that for every MATH, MATH. Moreover MATH is a bounded sequence in MATH and since MATH is compact, there is a compact subset MATH of MATH such that MATH. The proof is complete.
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math/0004146
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For each MATH, let MATH and MATH be the left and right support projection of MATH respectively. Both sequences MATH and MATH are mutually disjoint and for every MATH, MATH. For any finite sequence of scalars MATH, MATH . Note that MATH is disjointly supported by the projections MATH. For each MATH, the semi-finiteness of MATH implies that the family MATH of all projections in MATH of finite trace satisfies MATH. Since MATH is order-continuous, it follows that MATH. For each MATH, choose a projection MATH such that MATH and MATH. Claim: The sequence MATH is equivalent to MATH. Let MATH. For any MATH, we have MATH so the series MATH is convergent whenever MATH does. Conversely, if MATH is a bounded sequence of scalars such that MATH is convergent, then for any subset MATH of MATH, MATH . This shows that the series MATH is convergent. Let MATH and MATH be positive constants so that for any finite sequence of scalars MATH, MATH . If MATH and MATH, set MATH for MATH. The sequence MATH is disjointly supported in MATH and MATH is isometrically isomorphic to MATH. For any finite sequence of scalars MATH, MATH . The proof is complete.
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math/0004146
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Assume that MATH is not strongly embedded into MATH and set MATH the natural inclusion. Since MATH is not strongly embedded into MATH, the restriction MATH is not an isomorphism. There exists a sequence MATH in the unit sphere of MATH which converges to zero in measure. Note that the bounded set MATH cannot be MATH-uniformly integrable. Applying the non-commutative NAME subsequence decomposition to the sequence MATH on MATH (see REF ), there exist a subsequence of MATH (which we will denote again by MATH for simplicity) and a mutually disjoint sequence of projections MATH in MATH such that the set MATH is MATH-uniformly integrable and since MATH converges to zero in measure, we get that MATH. Assume now that MATH has a basis MATH. We will show that the sequence MATH above can be chosen to be a block basis of MATH. In fact since MATH cannot be a neighborhood of zero for the (relative) measure topology on MATH, for every MATH, MATH (where MATH denotes the ball centered at zero and with radius MATH relative to the metric of the measure topology). Denote by MATH the projection MATH onto MATH. Fix MATH and choose MATH so that MATH. The restriction of MATH on MATH cannot be an isomorphism. As above, one can choose MATH and MATH. Inductively, one can construct a sequence MATH in MATH and a strictly increasing sequence of integers MATH such that: CASE: MATH for all MATH; CASE: MATH for all MATH; CASE: MATH for all MATH. Set MATH for all MATH. Clearly MATH is a block basic sequence, MATH for all MATH and MATH converges to zero in measure. The proof is complete.
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math/0004146
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For MATH, we set (as in CITE), MATH and MATH . Assume first that for every MATH, there exists MATH such that MATH. We remark that MATH is supported by the finite projection MATH. There exists a subsequence MATH such that MATH converges to zero in measure. In particular, MATH converge to zero in measure. By the NAME subsequence decomposition on MATH (see REF ), there exists a further subsequence (which we will denote again by MATH) and a disjoint sequence of projections MATH so that the set MATH is MATH-uniformly integrable so by REF , MATH which shows that a subsequence of MATH can be taken to be equivalent to a disjoint sequence of MATH. On the other hand, if MATH for some MATH then MATH . So for every MATH, MATH. Since MATH is of cotype REF CITE, there exists MATH such that MATH . The proof is complete.
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math/0004146
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Inductively, we will construct a sequence MATH in the unit sphere of MATH, strictly increasing sequences of integers MATH and MATH such that: CASE: MATH for all MATH; CASE: MATH for all MATH; CASE: MATH for all MATH. Fix MATH a finitely supported vector in MATH and let MATH so that MATH. Since MATH, there exists MATH such that MATH. Assume that the construction is done for MATH. Let MATH. Since MATH is not isomorphic to MATH, there exist MATH such that MATH. By perturbation, we can assume that MATH is finitely supported. If we fix MATH so that MATH then MATH and the lemma follows.
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math/0004146
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We claim first that for MATH, there exists an absolute constant MATH such that: MATH . Indeed, for MATH, REF follows from non-commutative NAME inequality (see REF-REF-REF). For MATH, we remark that for any given finite sequence MATH in MATH, MATH for some fixed constant MATH. We observe that MATH so MATH which implies that MATH and since MATH is of cotype MATHCITE, there exists a constant MATH so that, MATH . For each MATH, set MATH. Applying the above inequality to the finite sequence MATH, we get that MATH and therefore REF is verified. For the proof of the lemma, we note as above that MATH therefore MATH hence MATH and by REF , MATH . For every MATH, the map MATH is bounded as a linear map from MATH into MATH. Let MATH be an element of MATH and consider MATH. Clearly, MATH. Let MATH be the projection in MATH defined by MATH with MATH and MATH for MATH. We have MATH so MATH and as above, MATH and the sublemma follows. By interpolation, it is also a bounded map from MATH into MATH which shows in particular that there exists an absolute constant MATH such that MATH . By taking adjoints, the other inequality follows. The proof of the lemma is complete.
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math/0004146
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Using the notation above, let MATH, MATH. Clearly, MATH is a semi-finite NAME algebra on the NAME space MATH. Set MATH and MATH. As above, MATH can be identified as a NAME subalgebra of MATH with MATH being the restriction of MATH on MATH. Let MATH be a basic sequence in MATH. Consider the sequence MATH in MATH. Claim: The sequence MATH is right disjointly supported. To verify this claim, recall that elements of MATH are infinite matrices with entries in MATH. For MATH, let MATH with MATH and MATH for MATH. Clearly MATH-is a mutually disjoint sequence of projection in MATH and for each MATH, MATH. For each MATH, let MATH and consider the sequence MATH in MATH defined by MATH. Claim: The sequence MATH is left and right disjointly supported. First note that, as above, the sequence MATH is right disjointly supported so its adjoints MATH is left disjointly supported. To prove that it is right disjointly supported, consider the following sequence MATH in MATH: MATH where MATH and MATH for MATH. It is clear that the MATH's are projections in MATH and since MATH is mutually disjoint in MATH, MATH is mutually disjoint and one can see that for every MATH, MATH. To complete the proof, we use REF , MATH . Applying REF on the NAME algebra MATH, MATH . The proof is complete .
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math/0004146
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Let MATH be a subspace of MATH and assume that MATH contains MATH. Since for MATH, as above, MATH, for some constant MATH (see REF p. REF). There exists an inclusion map from MATH into MATH. If MATH is strongly embedded into MATH, then MATH is isomorphic to a subspace of MATH. In particular MATH embeds into MATH. Contradiction. The converse is a direct consequence of REF .
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math/0004146
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Since MATH is not strongly embedded into MATH, REF implies the existence of a block basic sequence MATH of MATH and a sequence MATH of mutually disjoint projections in MATH such that: MATH . Note that MATH. By taking a subsequence (if necessary), we will assume that for every MATH, MATH . For MATH, set MATH. If MATH is a finite sequence of scalars then MATH where MATH. This shows that MATH . The other inequality can be obtain with similar estimate.
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math/0004148
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Differentiating the equality MATH and REF, we obtain: MATH where MATH denotes evaluation at MATH. Since MATH is an isomorphism, the conclusion follows.
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math/0004148
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By REF , we have: MATH . Hence, by definition of MATH, we get: MATH .
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math/0004148
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Let MATH be a critical point of MATH. Let MATH be an interval and consider a chart MATH in MATH whose domain contains MATH. Choose an arbitrary MATH with support contained in MATH; by standard computations it follows that: MATH . The fact that the equality above holds for every smooth MATH with support contained in MATH implies that the term MATH is of class MATH; this will follow from the generalized functions calculus developed in Subsection REF (see REF ). Integration by parts in REF and the Fundamental Lemma of Calculus of Variations imply then that REF is satisfied. Observe also that the coordinate representation of the map MATH is given by MATH, so that REF is satisfied. REF follows easily by integrating by REF in intervals of the form MATH and MATH. Conversely, if REF are satisfied, equality REF follows easily, which implies that MATH for all MATH with small support. Since such MATH's span MATH, it follows that MATH is a critical point of MATH.
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math/0004148
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Let MATH be a critical point of MATH; set MATH. Since MATH and MATH are mutually inverse, REF follows. Moreover, by REF , MATH is of class MATH and REF holds. We now prove that the second NAME equation holds, using a chart MATH of MATH. To this aim, we differentiate with respect to MATH the equality: MATH obtaining: MATH where MATH. Using that MATH and MATH are mutually inverse, we get MATH; it follows from REF that: MATH . The second NAME equation now follows from REF and from the NAME - NAME REF . Conversely, suppose that MATH is a solution of the Hamiltonian MATH which admits a Hamiltonian lift MATH satisfying REF. Since MATH and MATH are mutually inverse, from REF it follows that MATH. Finally, equality REF and the second NAME equation imply the NAME - NAME REF , and the conclusion follows from REF .
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math/0004148
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It is a simple application of the Inverse Function Theorem on NAME manifolds (see CITE for details on a similar construction).
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math/0004148
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See CITE.
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math/0004148
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Applying REF to the vector bundle MATH and to a complementary vector bundle of MATH in MATH we obtain a time-dependent referential MATH of MATH such that MATH is a time-dependent referential for MATH; moreover, we may choose MATH so that its domain MATH contains the graph of any prescribed continuous curve in MATH. If MATH is a local chart of MATH which sends MATH to an open subset of MATH then the corresponding chart REF on MATH sends MATH to an open subset of MATH.
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math/0004148
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The proof is a minor adaptation of the proof of CITE where we consider the case that MATH is a point and we use MATH curves instead of MATH curves.
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math/0004148
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Follows from REF observing that a characteristic MATH that vanishes at some MATH is identically zero (see CITE).
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math/0004148
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Consider the NAME spaces MATH, MATH, MATH, and the continuous linear maps MATH, MATH, given by composition of MATH with the quotient map MATH. We know that both MATH and MATH are surjective and have complemented kernel. We have to show that MATH is surjective with complemented kernel if and only if MATH is surjective with complemented kernel. To this aim, observe first that MATH is surjective if and only if MATH and the latter condition is symmetric in MATH and MATH. Finally, to complete the proof we show that, given MATH, then MATH is complemented in MATH if and only if it is complemented in MATH. If MATH is complemented in MATH then by intersecting a closed complement of MATH in MATH with MATH we obtain a closed complement of MATH in MATH. Conversely, if MATH is a closed complement of MATH in MATH and MATH is a closed complement of MATH in MATH then MATH is a closed complement of MATH in MATH because MATH has the product topology of MATH and MATH.
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math/0004148
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Let MATH denote the extension of MATH to MATH which is again defined by MATH. The conclusion will follow by applying REF with MATH, MATH, MATH, MATH, MATH, MATH and MATH equal to the endpoint map MATH. Since MATH is obviously a submersion, we only need to show that MATH is a submersion. Choose a distribution MATH with MATH and let MATH be the time-dependent referential of MATH over MATH which is dual to MATH, that is, MATH for MATH and MATH for MATH. Choose a time-dependent referential MATH of MATH over an open neighborhood of the graph of MATH. The coordinate representation of MATH in the chart REF corresponding to MATH is the natural projection of MATH onto MATH. This shows that MATH is a submersion and concludes the proof.
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math/0004148
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We first consider the case MATH. If MATH, then MATH for all MATH, hence MATH for all MATH with MATH. Choose MATH with MATH; set MATH. It is easily seen that MATH. For the general case, observe that for all MATH, the product MATH has vanishing derivative, and hence it is constant. Since MATH is arbitrary, it follows that MATH is constant.
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math/0004148
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Consider the map MATH given by MATH. It is easily seen that MATH is injective with closed image. It follows that the transpose map MATH is surjective; clearly, the derivative operator for generalized functions is MATH, which proves the first part of the thesis. For the case MATH, let MATH denote the NAME space of absolutely continuous functions MATH having square integrable derivative, and such that MATH. Again, the derivation map MATH is injective and has closed image. Therefore, given MATH, we can find MATH with MATH. It follows that MATH.
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math/0004148
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We prove, for example, the first item. By REF , we can find MATH with MATH. By REF , it follows that MATH is constant, hence MATH. The other items are proven similarly.
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math/0004148
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Consider a partition MATH such that MATH is of class MATH for all MATH. Since the operation of restriction for generalized functions gives the standard operation of restriction for functions, it follows that: MATH for MATH. Hence there exists a MATH map MATH on MATH such that MATH for all MATH. We know that MATH if MATH has support contained in some interval MATH; but such MATH's span a dense subspace of the domain of the linear functional MATH and therefore MATH.
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math/0004148
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If MATH denotes the subspace of MATH consisting of affine maps MATH then obviously: MATH . It is easy to see that we can find MATH such that both sides of REF agree on MATH. Since both sides of REF vanish on MATH, the conclusion follows.
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math/0004148
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The point MATH is critical for MATH if and only if MATH vanishes on MATH. The conclusion follows from elementary functional analysis arguments.
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math/0004148
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We start by choosing an arbitrary complementary distribution MATH to MATH, that is, a smooth distribution of rank MATH in MATH such that MATH for all MATH; moreover, we fix an arbitrary smooth Riemannian structure MATH on the vector bundle MATH. Let MATH and MATH denote the projections and define an extension MATH of MATH by: MATH where MATH . Then MATH is open in MATH and MATH is a Lagrangian on MATH as in REF ; we denote by MATH the corresponding action functional in MATH, defined as in REF. Let MATH, MATH, MATH and MATH be as in the statement of REF (recall also REF). Then, since MATH is regular, the map MATH is a submersion at MATH; moreover, MATH is a critical point of MATH in MATH if and only if it is a critical point of MATH. By REF multipliers REF , this is equivalent to the existence of MATH such that MATH is a critical point of MATH in MATH. We will prove in REF below that the NAME multiplier MATH is of class MATH, that is, that it is given by: MATH for some MATH map MATH. Therefore, MATH is the action functional corresponding to the Lagrangian MATH in MATH defined by: MATH where MATH. We now prove that MATH and MATH are hyper-regular and we compute their NAME transforms. The fiber derivatives MATH and MATH are easily computed as: MATH where MATH. The hyper-regularity is proven by exhibiting explicit inverses: MATH by MATH in the above formula we mean the inverse of MATH seen as a linear map from MATH to MATH. We now compute the NAME transforms MATH and MATH of MATH and MATH respectively. Using REF , we compute easily: MATH and, using REF, we therefore obtain: MATH . We now compute the NAME equations of the Hamiltonian MATH with the help of local coordinates MATH in MATH and of a local MATH-orthonormal referential MATH of MATH. We write: MATH and, using REF, the NAME equations of MATH are given by: MATH . By REF , MATH is a critical point of MATH if and only if it admits a lift MATH satisfying REF with MATH and MATH. Now, it follows easily from REF that MATH is in MATH; since MATH is horizontal, that is, MATH, from the first equation of REF it follows that MATH for all MATH. Setting MATH in REF we obtain the NAME equations of MATH, which concludes the proof.
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math/0004148
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We set MATH we first prove the regularity of the generalized function MATH. To this aim, we localize the problem by considering variational vector fields along MATH having support in the domain of a local chart MATH in MATH. Let MATH be such that MATH is contained in the domain of the local chart; we still denote by MATH the restriction of MATH to MATH. Since MATH is a critical point of MATH, by standard computations it follows that the following equality holds: MATH for every vector field MATH of class MATH along MATH having support in MATH; in the formula above we have regarded the derivative MATH as a MATH-valued bilinear map in MATH. In terms of the local coordinates, the maps MATH, MATH, MATH and MATH evaluated along MATH will be interpreted as follows: CASE: MATH; CASE: MATH; CASE: MATH, where MATH denotes the space of linear maps between two given vector spaces. Using the definition of derivative for generalized functions, from REF we get: MATH for every MATH map MATH having support in MATH, and, by density, for every MATH. It follows: MATH . Let MATH be a referential of MATH along MATH; in terms of the local coordinates the MATH's will be thought as elements of MATH. Moreover, we set MATH where the MATH-tuple MATH is identified with the linear map that takes the MATH-th vector of the canonical basis of MATH to MATH. Composing REF with MATH, we obtain: MATH . Evaluating REF at MATH with MATH and using the horizontality of MATH we get: MATH hence: MATH . Now, considering that MATH is invertible, by REF we can write REF in the form: MATH with MATH and MATH. Applying three times REF , from REF we conclude that MATH belongs to the space MATH; now REF implies that MATH. By REF , there exist MATH such that: MATH . To conclude the proof we show that MATH. Let's show for instance that MATH; the proof of the equality MATH is analogous. Using local charts around MATH, for MATH close to MATH, we consider variational vector fields MATH of class MATH supported in MATH, with MATH. Arguing as in the deduction of REF , we get the following equality: MATH . From REF it follows that MATH is of class MATH, and we can thus use integration by parts in REF to obtain an equality of the form: MATH for some MATH, whenever MATH is chosen with MATH. By considering arbitrary MATH supported in MATH, from REF we obtain that MATH in MATH, so that the integral in REF vanishes for all MATH. Now, we can choose MATH with MATH and MATH arbitrary, so that REF implies that MATH, because MATH is surjective. This concludes the proof.
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math/0004161
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In REF we will construct a parametrix MATH of MATH such that MATH. Moreover, this construction will provide a family MATH that belongs to the class MATH for some MATH, compare REF. Fix now MATH. In MATH we have the estimate MATH . Hence for some MATH we get MATH for all MATH. Therefore, MATH is invertible in MATH for MATH, and MATH. In fact, the inverse of MATH is given by MATH. Furthermore, MATH for some constant MATH. The last inequality is the symbol estimate of MATH as an operator-valued symbol of order MATH. In order to see that for MATH the operator MATH is invertible even for all MATH, let MATH. Then the inverse of MATH can also be written as MATH which belongs to MATH for all MATH and MATH. Observe that the inverse of MATH appearing between MATH and MATH is the inverse with respect to MATH.
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math/0004161
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For simplicity we omit the variable MATH and assume MATH; the general case is completely analogous. We first show the holomorphy of the principal symbol MATH making use of the relation MATH . If the convergence in REF is compact, that is, uniformly on any compact subset of MATH, then MATH is holomorphic there since so is MATH; consequently, MATH is also holomorphic and we can proceed as above to assert the same for MATH. Induction then yields the holomorphy of every MATH. To prove the compact convergence of REF in the topology of MATH let MATH and MATH be given. For MATH and MATH we have MATH . Denoting MATH then MATH because MATH and MATH for MATH. So, MATH . This implies that the convergence in REF is compact for MATH.
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math/0004161
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The variable MATH will be omitted again since it does not play any role along the proof. Let us set MATH . For every MATH the mapping MATH is bounded by definition, thus MATH is uniformly continuous (fundamental theorem of calculus) implying that MATH exists in MATH; note that MATH. In particular, for every MATH and MATH we have the pointwise convergence MATH as MATH, and so MATH . Therefore, MATH since MATH is bounded in MATH. Now, for every MATH a NAME expansion yields MATH . The left side, written in terms of MATH, is exactly the expression in REF. Denoting the integral above by MATH we have for every MATH, MATH . We want to show that MATH uniformly for MATH, but this follows from the estimates MATH taking into account that MATH and so, in particular, the supremum in the latter integral is uniformly bounded.
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math/0004161
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Without loss of generality we omit the variable MATH. It is true that MATH . Let MATH, that is, MATH and MATH. Hence MATH leading to the relation (because MATH is holomorphic in MATH) MATH . To prove the assertion we need MATH for every semi-norm as in REF. Now, setting MATH we have MATH . Because MATH, and since MATH is a symbol of order MATH, then for MATH and MATH the derivatives MATH can be estimated by terms of the form MATH . Using now the inequality MATH for MATH and MATH, we get MATH with a constant MATH depending uniformly on MATH and being independent of the variable MATH. Hence MATH is finite.
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math/0004161
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As it was done before we ignore for a moment the variable MATH. For any MATH the symbol MATH admits an asymptotic expansion MATH with symbols MATH that are homogeneous for MATH, and are holomorphic in MATH for MATH due to REF . In order to prove the claim we have to investigate MATH for any semi-norm in MATH. To this end let MATH be given. For any homogeneous component each semi-norm is bounded due to REF , so we only need to show MATH for some MATH. In fact, choosing MATH we achieve MATH . In the last estimate one uses the relations MATH for MATH, MATH, and MATH for MATH.
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math/0004161
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Due to the inclusion MATH it is enough to prove the statement for MATH. For simplicity, let us drop the variable MATH. REF assures the existence of an expansion REF of MATH for every MATH. As proven in REF the coefficients MATH are indeed given by REF on every line MATH. It only remains to prove that MATH. In other words, we have to verify: CASE: MATH uniformly for MATH in compact intervals, CASE: MATH. To this end let MATH be a compact interval. Since every semi-norm MATH is uniformly bounded for MATH and MATH then REF follows by a pointwise consideration as in the proof of REF . Furthermore, the convergence in REF, taking place in MATH, is uniform for MATH hence compact in MATH. Thus MATH is holomorphic with values in MATH. By using now REF and the fundamental theorem of calculus we obtain for neighboring MATH and MATH the estimates MATH with MATH. That is, the mapping MATH is continuous which implies, in particular, that the NAME integral MATH exists in MATH for every disk MATH. Hence REF holds and the proof is done.
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math/0004161
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Let MATH be fixed. We need to show certain mapping properties according to REF . First of all, observe that (integration by parts) MATH for every MATH. Thus MATH and its formal adjoint MATH are for every MATH bounded operators in MATH and MATH, respectively. Hence they are smoothing operators in the interior, but in order to be in MATH these operators have to improve the weights by some MATH. We prove this by making use of the parametrix MATH of MATH given in REF. Then MATH is a parametrix of MATH and MATH belongs to MATH. Moreover, for MATH we can write MATH where MATH, MATH, MATH, and MATH. Notice that all the contributions involving the smoothing NAME element MATH from REF are now contained in MATH or in MATH since they are either supported in the interior of MATH or multiplied by a NAME element or multiplied by a factor MATH which improves the weight whenever MATH. Clearly, the nonsmoothing parts of MATH improve the weight MATH at least by MATH while the families MATH and MATH has the same gain MATH as the smoothing part of MATH.
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math/0004161
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First of all, we have MATH . Since MATH belongs to MATH, it is for every integer MATH an element of MATH, and its norm is MATH for every MATH. Thus it remains to verify the norm estimate for MATH. For simplicity of notation we will check it explicitly only for MATH. In this case, MATH is just MATH where MATH. From the standard parameter-dependent calculus (compare CITE) we get the norm estimate MATH for MATH. On the other hand, MATH so that MATH with MATH, where MATH and MATH are the constants of growth corresponding to MATH and MATH, respectively (compare REF ). By means of the estimates REF together with REF we get MATH for MATH. Set MATH. For an arbitrary MATH the proof is quite the same once MATH is split as MATH in REF. This can be done by expanding MATH and MATH in REF inside their corresponding pseudodifferential calculi. Notice that MATH in REF is rapidly decreasing.
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math/0004161
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We combine REF to write MATH . Let MATH. In MATH we get MATH after making the change of variables MATH, and applying REF .
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math/0004161
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Let MATH. Split the integral MATH in MATH, and denote the components by MATH and MATH, respectively. Because MATH on MATH we have MATH . With the change MATH and REF we get MATH . Therefore, the change of variables MATH yields MATH . With the same calculations we also get MATH . NAME expansion at MATH yields MATH with MATH . In particular, for MATH we get MATH . Since MATH is for any MATH uniformly bounded, and because MATH decays rapidly as MATH, the last integral is uniformly bounded as desired.
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math/0004161
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Noting that MATH satisfies the relation REF from REF with MATH and MATH (MATH is twisted homogeneous), the assertion follows making the same calculations as in the proof of REF .
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math/0004161
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The proof of this lemma is rather long. For this reason, we wish to outline our strategy in order to make the structure of the proof clearer. First of all, let MATH be given, and recall that MATH and MATH are fixed. Let us simplify the notation by writing MATH . In local coordinates the function MATH is given by MATH where MATH, MATH, MATH, and MATH is a local symbol of MATH. Thus there are symbols MATH, anisotropic homogeneous in MATH of degree MATH for MATH, such that for every MATH, MATH . This expansion induces a decomposition of MATH, say MATH, and a decomposition of MATH, say MATH. Our purpose is to expand in MATH all these components. REF : (Expansion of MATH). First of all, let MATH. Following the proof of REF we can write, exactly as in REF, MATH where MATH. For MATH the kernel MATH decays fast enough as MATH so that the integrals exist, and the last one is uniformly bounded. Thus MATH with MATH depending on MATH. REF : (Splitting of MATH). For MATH let us split (recall that MATH) MATH . This equality split likewise MATH with MATH for MATH, where MATH denotes the corresponding domain of integration. REF : (Expansion of MATH). The change of variables MATH yields MATH due to the homogeneity of MATH. If we denote MATH, and set MATH, then for MATH the integral REF becomes MATH . Finally, REF leads to the expansion MATH with coefficients depending on MATH. In particular, MATH for MATH. REF : (Expansion of MATH). Since MATH depends holomorphically on MATH, we can apply REF with MATH, MATH and MATH in order to expand MATH as MATH for any MATH, with MATH and MATH uniformly for MATH with MATH. For MATH, the degenerate symbol MATH can then be written as MATH and so MATH . Set MATH which is finite as MATH is compact. Now, since MATH, we get MATH due to the change of variables MATH and MATH. Similarly as in the proof of REF we expand MATH. Thus MATH . Since MATH and MATH are uniformly bounded for MATH, MATH and MATH, then the integral at MATH exists (presence of MATH) and is also uniformly bounded. Choosing MATH we finally get MATH with MATH depending on MATH. REF : (Expansion of MATH). According to REF let MATH so that MATH . Recall that in REF each MATH is homogeneous in MATH of degree MATH and consequently MATH is anisotropic homogeneous of degree MATH for MATH. Then MATH . Taking now MATH such that MATH the latter integral can be split in two integrals MATH. We denote the second integral again by MATH, and make the change of variables MATH in the first one. Thus MATH . Now, with the usual change of variables of MATH and MATH, MATH with MATH and MATH. These integrals are of the same type as REF, so REF is of the form MATH with coefficients depending on MATH. In particular, MATH for MATH. In order to complete the expansion of MATH we next treat those terms involving MATH. By means of polar coordinates and the homogeneity in MATH of MATH, MATH which we denote by MATH. Then, for MATH, the corresponding component of MATH becomes MATH . If MATH, then REF equals MATH since MATH for every MATH. Making use of REF this expression can be written as a linear combination of MATH, MATH and MATH. Moreover, the term MATH disappears when MATH. If MATH, then REF becomes MATH since the integral MATH is finite and MATH. Applying REF once more we finally obtain MATH with coefficients depending on MATH. In particular, MATH for MATH. Summary: By virtue of the relation MATH the assertion follows summing up REF.
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math/0004161
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Since MATH on MATH, MATH .
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math/0004161
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We can write MATH with MATH . Since MATH, the function MATH can be split, inducing a corresponding decomposition MATH with MATH after the change of variables MATH, MATH, and MATH. Now, MATH for some MATH, and so MATH which is MATH since the integral is uniformly bounded.
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math/0004161
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Actually, we only need to put together all the single expansions obtained in this section. The asymptotic summation here means: For a given MATH there exists MATH and MATH such that MATH for every MATH and MATH, where MATH is some expression of 'degree' MATH from the right-hand side above, that is, MATH is a linear combination of terms like MATH with MATH, and MATH. First of all, we choose MATH large enough such that MATH is of trace class. Hence the family MATH from REF is of trace class too, and REF together with REF imply: for a given MATH there are MATH and MATH such that MATH for MATH. Recall that MATH with MATH and MATH as in REF. By definition, every MATH splits into a component supported in the interior of MATH and another one supported near MATH. It is known (compare CITE) that the trace of the interior part is of the desired form due to the homogeneity of the local symbols. On the other hand, all the components supported near the boundary are either operator-valued NAME symbols or parameter-dependent NAME operators with degenerate symbols. By using REF for the integrals of NAME type and REF for the rest, we obtain the desired expansion integrating over the manifold MATH the local expansions obtained there.
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math/0004166
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Fix MATH and find MATH so that MATH . Fix MATH and a finitely supported MATH. It suffices to show that MATH . Find MATH so that MATH and let MATH . Fix MATH. Assume that MATH . Then MATH . Thus by REF , with MATH, MATH and so MATH . But REF is equivalent to MATH which contradicts REF. Thus MATH and so REF holds.
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math/0004166
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Fix MATH. Keeping with the notation in REF , find MATH so that MATH . Fix MATH with MATH . It suffices to show that MATH . Fix MATH. Find MATH so that MATH and let MATH . Fix MATH. Assume that MATH . Then MATH . Thus by REF MATH and so MATH . But MATH and so MATH . A contradiction, thus MATH . Since MATH was arbitrary, REF holds.
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math/0004166
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Fix MATH and let MATH. Let MATH be given by MATH and MATH be a function satisfying the hypothesis in REF . Find MATH so that MATH . Next find MATH and MATH so that MATH . Fix MATH. It suffices to show that MATH . Let MATH . If MATH, then by the proof of REF holds. So let MATH. Find MATH so that MATH . Let MATH be the functional given by MATH where the inner product in the natural inner product on MATH. Let MATH and fix MATH. Assume that MATH . It suffices to find a contradiction to REF. Towards this, let MATH . It suffices to show (keeping with the same notation but with MATH) that REF through REF hold; for then REF holds and so by REF MATH . REF follows from REF since MATH . REF follows from REF since MATH . Towards REF , note that by REF MATH and so MATH . Towards REF , note that by REF MATH . Thus by REF MATH . Thus REF holds provided MATH or equivalently MATH . But by REF and that MATH . Thus REF holds. REF follows from the fact that MATH. Towards REF , since MATH, the vectors MATH and MATH are orthogonal in MATH and so MATH but MATH and so by REF MATH . Thus REF .
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math/0004166
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Let MATH fail the PCP and MATH. By a standard argument (for example, see CITE), there is a closed subset MATH of MATH of diameter one such that each (nonempty) relatively weakly open subset of MATH has diameter larger than MATH. Without loss of generality MATH (just consider a translate of MATH). Let MATH . Note that MATH. If MATH is (nonempty) relatively weakly open subset of MATH, then MATH is a relatively weakly open subset of MATH and so MATH . Thus MATH does the job.
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math/0004166
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Let MATH be a NAME space without the PCP. Fix MATH and MATH. It suffices to show that MATH. Find a subset MATH of MATH which satisfies the conditions of REF with MATH and find MATH so that MATH . Let MATH. It suffices to show that MATH . By REF there exists MATH so that MATH and MATH is almost in MATH; thus, by a standard perturbation argument (for example, see CITE) there exists MATH so that MATH . Thus MATH . Thus MATH.
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math/0004166
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Let MATH and MATH be a NAME space. That REF though REF are equivalent and that REF implies REF follows easily from the below known facts about a NAME space MATH. CASE: MATH is uniformly convex if and only if MATH is CITE. CASE: MATH is uniformly convexifiable if and only if MATH admits an equivalent UKK norm CITE. CASE: MATH is uniformly convexifiable if and only if MATH is uniformly smoothable (compare CITE). Towards showing that REF implies REF , let MATH be asymptotically uniformly smoothable and MATH be a separable subspace of MATH. It suffices to show that MATH is uniformly convexifiable (compare CITE). It follows from CITE that if MATH is separable, then MATH is asymptotically uniformly smooth if and only if MATH has the MATH property. Thus MATH admits an equivalent MATH norm. But MATH cannot embed into MATH since MATH is asymptotically uniformly smoothable and so MATH is asymptotically weak-MATH uniformly convexifiable and so is also asymptotically uniformly convexifiable. Thus MATH is asymptotically uniformly convexifiable where MATH. From REF implies REF it follows that MATH is uniformly convexifiable and so so is MATH.
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math/0004167
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Suppose, to obtain a contradiction, that there are two distinct nondegenerate cones MATH. By REF , the semigroup structure on MATH induces monoid structures on MATH and MATH, and by REF , both have zeroes, say MATH and MATH. The relation MATH, which holds for all MATH (because MATH), must therefore hold for all MATH. Thus, MATH is a zero of MATH. Similarly, MATH is also a zero of MATH. But a semigroup cannot have two distinct zeroes, so MATH. Consequently, MATH and in particular MATH has a zero. But MATH are distinct cones, whence MATH is a proper face of MATH, and is therefore degenerate. This contradicts REF .
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math/0004167
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The direction MATH is clear. For MATH, combine REF .
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math/0004167
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See REF .
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math/0004167
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By REF , there exists a finite cover of MATH by invariant open affine subsets. But every open subset of MATH intersects MATH, since MATH is dense in MATH, and an invariant subset of MATH intersecting MATH must contain MATH, because MATH itself is an orbit under the action of MATH.
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math/0004168
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Let MATH be a p.i. MATH-sequence with respect to MATH. Then MATH for each finite sequence MATH of scalars. Define MATH . Then MATH and MATH satisfy REF ; thus, MATH is an asymptotically isometric copy of MATH. Conversely, let MATH and MATH satisfy REF . Then MATH is a p.i. sequence. To see this, let MATH and MATH . For each MATH, define MATH by MATH; for indeed, MATH for each finite sequence MATH of scalars. For each MATH, let MATH be a norm-preserving NAME extension of MATH. Let MATH endowed with the usual sup norm, and consider the isometric embedding MATH given by MATH . Let MATH be the `truncation' of MATH; specifically, MATH . For each MATH, REF gives that MATH and so by REF MATH . Since for each MATH and MATH it follows that MATH for each finite sequence MATH of scalars. Also, MATH for each MATH. Thus MATH is isometrically equivalent to the unit vector basis of MATH.
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math/0004168
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Let MATH be a NAME space satisfying REF . We shall inductively construct a sequence MATH of countably infinite subsets of MATH and a sequence MATH of separable subspaces of MATH which satisfy, for each MATH, CASE: MATH, CASE: MATH norms MATH, that is, if MATH then MATH CASE: MATH is contained in the MATH-cluster points of MATH, that is, if MATH and MATH are from MATH and MATH then MATH . For the first step of the induction choose a countably infinite subset MATH of MATH and find a separable subspace MATH of MATH which satisfies REF . Suppose that we have chosen MATH and MATH satisfying the three conditions. Since MATH is separable and elements of MATH are of norm one, there is a countable subset MATH of MATH satisfying REF . Next we find a separable subspace MATH of MATH which satisfies REF . This completes the inductive step. Now let MATH and MATH . REF gives that MATH is isometrically equivalent to the usual basis of MATH. REF , along with the fact that MATH, give that MATH is dense-in-itself in the weak-star topology on MATH.
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math/0004168
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We shall assume that MATH is a complex NAME space as the proof in the real case is easier. The equivalence of REF follows directly from REF . To see that REF implies REF , let MATH and MATH be sequences satisfying REF . Partition MATH into infinite sets MATH and let MATH be the bounded linear operator that maps MATH to the MATH unit vector of MATH when MATH. Then MATH is an isometric quotient mapping. To see that REF implies REF , let MATH be an isometry from a quotient space of a subspace MATH of MATH onto MATH. Fix a null sequence MATH of positive numbers. Find a sequence MATH in MATH such that MATH is the MATH unit vector of MATH and MATH . Then MATH and MATH satisfy REF . To see that REF implies REF , let MATH and MATH satisfy REF . We shall define the bounded linear operators in the (commutative) diagram below MATH as follows. Let MATH be dense in the unit sphere of MATH. Define MATH by MATH. REF gives that MATH is a surjective norm-one bounded linear operator. Furthermore, MATH is an isometric quotient mapping for if MATH then there is a subsequence MATH converging in norm to MATH and so MATH . Let MATH be the natural embedding and let MATH be the canonical isometric embedding given by point evaluation. Since MATH has the NAME Extension Property, MATH admits a norm-preserving extension MATH. Dualizing gives the commutative diagram MATH where MATH is the canonical isometric embedding given by point evaluation. To see that MATH is the desired isometric embedding, let MATH. Then, since MATH is an isometric embedding and MATH is the identity mapping, MATH . Clearly, REF implies REF . To see that REF implies REF , let MATH be an isometric embedding and let MATH be a null sequence of positive numbers. Then MATH is a weak-star continuous isometric quotient mapping. By NAME 's Theorem, MATH is weak-star dense in MATH. For each MATH, let MATH be a MATH - net for MATH. Let MATH be the tree MATH where MATH, the MATH-level of MATH, is MATH . If MATH then MATH thus, for each MATH . We will define inductively, for each MATH, a collection MATH of disjoint sets of positive REF measure and a function MATH such that, for each MATH and MATH, MATH and, for each MATH, MATH . To start the induction, let MATH . For the inductive step, let MATH and suppose that we have constructed disjoint sets MATH of positive measure. For each MATH, partition MATH into sets MATH of positive measure. Consider the function MATH defined by MATH . Since MATH is weak-star dense in MATH there exists MATH approximating MATH closely enough to ensure that the sets MATH all have positive measure. This completes the proof of the inductive step. For each MATH, select MATH such that MATH. To see that MATH is an asymptotically isometric copy of MATH, let MATH be a finite complex sequence. Define MATH from MATH by MATH thus, MATH. For each MATH, find MATH so that MATH . Then MATH and so by REF MATH for each MATH. Thus MATH . So MATH and MATH do indeed satisfy REF . To show that REF implies REF , let REF hold. Then we have the situation depicted in REF. For MATH, let MATH denote the point mass measure at MATH. Since MATH is a weak-star continuous isometric embedding MATH equipped with the weak-star topology of MATH, is homeomorphic to MATH and is isometrically equivalent to the usual basis of MATH. By the NAME Theorem MATH and so MATH is weak-star compact and satisfies MATH. By REF there exists MATH such that MATH and such that MATH is dense-in-itself in the weak-star topology on MATH. Since for any finite set MATH of scalars MATH the set MATH is isometrically equivalent to the usual basis of MATH. To see that REF implies REF , let MATH satisfy REF . Then by REF , there is a separable subspace MATH of MATH which satisfies REF . From REF and the fact that MATH is linearly isometric to MATH, it follows that MATH satisfies REF . The equivalence of REF gives that MATH also satisfies REF . Thus REF through REF are equivalent. The fact that MATH is linearly isometric to MATH gives that REF implies REF . That REF implies REF when MATH is separable is due to NAME: REF .
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math/0004168
|
We exhibit MATH as a quotient space of MATH with its usual norm MATH . Let MATH . Since each element of MATH has a representative of the form MATH, MATH . Thus MATH is isometrically isomorphic to MATH. Observe that MATH with its usual norm MATH and MATH . It follows that MATH is isometrically isomorphic to MATH where MATH and the mapping MATH given by MATH is an isometry. Since the norm MATH is strictly convex, the space MATH does not contain an isometric copy of MATH. Thus by REF the space MATH does not contain asymptotically isometric copies of MATH. From REF it follows that MATH for each MATH and so REF holds by duality.
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math/0004172
|
Fix MATH and let MATH be a system of generators of MATH. Let MATH be the traces of MATH in the left regular representation of MATH. By hypothesis there exist unitaries MATH in MATH such that MATH if MATH in MATH and MATH if MATH in MATH. Let MATH. Then MATH belongs to MATH and hence so does the limit MATH . Thus MATH for all MATH. Hence MATH.
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math/0004172
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We divide the proof into several steps. We construct first an approximate embedding of the relation MATH into MATH. We then take the free amalgamated product (over MATH) by a unitary that commutes with MATH and perturb MATH with this unitary. This gives an approximate embedding of MATH into the nonscalar unitaries in some free product algebras. Since these algebras are embeddable into MATH CITE CITE, the result will follow, by REF. Construction of an approximate embedding. There exist unitaries MATH, MATH (of zero trace) in MATH with the following properties: CASE: MATH . CASE: Denote by MATH the abelian algebra generated by MATH and let MATH be the corresponding conditional expectation. Let MATH. Then for all MATH, MATH for a universal constant MATH. Here MATH is the normalized NAME trace on matrices. CASE: MATH . We describe first the construction of the unitaries MATH, MATH. Let MATH be the diagonal algebra of MATH, and for convenience we think of MATH as being identified with MATH, MATH. Let MATH, for MATH, be the projection MATH and let MATH . Let MATH be the unitary defined by MATH . Let MATH be a unitary such that MATH, MATH, and such that MATH for all MATH, MATH. (See REF .) For MATH, we define MATH by the requirement MATH . Observe that the definition of MATH, MATH implies, for a universal constant MATH, that MATH . Moreover, MATH, and MATH . In the remaining case, we have that MATH . We now proceed to the proof of REF. Since MATH, MATH, MATH, and MATH commute with MATH, it follows that MATH . But this quantity is less than MATH by REF . This completes the proof of REF . To prove REF we need to describe first MATH. But it is obvious that MATH . Consequently MATH . We use the above formula for MATH and use REF . We take MATH. Then MATH has trace MATH. Since MATH and MATH the above formula shows that MATH . Hence MATH . The computations for MATH, MATH are similar, eventually the factor MATH being replaced by MATH or MATH. This completes the proof of REF . It is obvious that MATH, MATH, MATH, and MATH are vanishing. CASE: In this step we consider the amalgamated free product of the algebra MATH described above and MATH. The amalgamated free product is considered over MATH (the NAME algebra generated by MATH). Let MATH have the canonical generator MATH, a NAME unitary. MATH is endowed with the standard trace. Consider the algebra MATH with the canonical amalgamated free product trace (see section on definitions, CITE, and CITE). By CITE (see also CITE, CITE), we have that MATH is a free group factor. Using the ultrafilter construction (CITE, CITE), we construct algebras MATH, MATH and MATH consisting of bounded sequences of elements in the algebras MATH, MATH and MATH respectively. It is obvious that for MATH in MATH we have MATH . Let MATH be the unitary element MATH and let MATH be the (abelian) NAME algebra generated by MATH. Let MATH be the unitary MATH. We identify MATH with constant sequences with elements in MATH. Since, by CITE, CITE, any type MATH free group factor embeds into MATH, we obtain that the algebra MATH is embedded into MATH and hence MATH . The trace on MATH is the amalgamated free product trace and coincides with the restriction of the trace on MATH. Let MATH be the identity minus the conditional expectation MATH from MATH (or MATH) onto MATH. The following properties hold true: CASE: MATH are NAME unitaries, MATH. CASE: MATH, MATH; MATH, MATH, MATH. To prove REF , note that MATH because of the corresponding property for MATH. Since MATH it follows also that MATH. Let MATH be the standard generator of MATH and let MATH, MATH. CASE: Let MATH be the unitaries defined in REF. Clearly MATH, as MATH commutes with MATH. Let MATH be a word in MATH such that MATH, MATH. Consider the following assumption on the sequence of the indices MATH. Assumption on MATH. One of the following possibilities occurs (about consecutive indices): CASE: Either MATH, MATH are both positive or negative except for the case when MATH. CASE: If MATH, MATH, then MATH. CASE: If MATH, MATH, then MATH. If the word MATH is subject to REF , then MATH is not a multiple of a scalar (and hence MATH). We use the following property of an amalgamated free product MATH where MATH, MATH are finite algebras with faithful traces MATH, MATH whose restrictions coincide on the common unital subalgebra MATH. Assume MATH is a word in MATH, MATH, MATH, such that MATH, MATH, MATH, MATH. Then MATH is not a scalar multiple of the identity. This follows for example from the construction in CITE. Then for the word MATH, we use the fact that MATH, MATH. Since MATH are always zero for all MATH, the only instances in the product in MATH where we could have elements with MATH nonzero are in subsequences of the form MATH or MATH . But in these cases MATH applied to the elements MATH, MATH, and MATH is nonzero. The remaining two cases correspond to subsequences involving MATH with MATH and MATH, MATH both strictly positive or both strictly negative. The case MATH corresponds to a subsequence of the form MATH or MATH. In either case we use the fact that MATH, MATH, for MATH. The case MATH is similar. Hence the property of the amalgamated free product applies, and MATH is non-scalar. CASE: Any word (except the identity) in the NAME group MATH, of total degree zero in MATH, is equal to one of the words MATH for which all REF - REF on consecutive indices, described in REF, hold. Note that by REF MATH we can reduce the proof of the theorem to words of total degree MATH in MATH. To prove the claim of REF, the following two lemmas, dealing with easier situations, will be used. Let MATH and MATH. Then MATH is equal to a product of the form MATH for some strictly positive numbers MATH, MATH, and MATH is nonzero, MATH, MATH. For example MATH, where the product involves MATH occurrences of the letter MATH (not counting powers). We start with MATH, MATH odd. Then MATH. We then split this product as: MATH . We repeat this procedure with MATH and will obtain MATH . By repeating this procedure and stopping when running out of powers of MATH, we get the required result. Similarly one proves: Let MATH, MATH. Then MATH is equal to a product of the form MATH where MATH, MATH. Moreover, MATH and MATH for MATH. We start with one arbitrary word MATH where no obvious cancellations are possible. By moving from the left to the right we look at the first change in sign in the sequence MATH. Say this occurs when MATH. At that point, if MATH when MATH, MATH, or if MATH when MATH, MATH, we apply one of the two preceding lemmas to replace MATH by one of the sequences described in the lemmas. More precisely, if, for example, MATH, MATH, we apply REF for MATH . By replacing MATH in the word by the form given in REF , the structure of the word up to the next power of MATH following MATH would fulfill the requirements of the claim. The only case, when in doing this replacement, a change of structure could occur in the structure of the word, before MATH, is when MATH. But in this case MATH so MATH won't cancel the MATH appearing at the beginning of the word from REF . Here we reiterate the procedure. A similar argument works for MATH, MATH. By induction, this completes the proof of REF. By REF and IV we conclude the proof of our theorem.
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math/0004172
|
Fix MATH in MATH and let MATH be any projection less than MATH. The fact that MATH is a maximum value for MATH on MATH, implies that MATH . Thus for any projection MATH less than MATH we have that MATH . But this gives exactly that MATH . Similarly for MATH.
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math/0004172
|
This follows by writing down explicitly that MATH .
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math/0004172
|
Assume first that MATH, and let MATH be any partial isometry mapping MATH onto MATH. We will show that MATH belongs to MATH. Indeed MATH can be identified with MATH in such a way that MATH . But then we take MATH . If the trace of MATH is different from MATH, we may then assume that MATH. By the above argument, any partial isometry MATH mapping MATH into a projection under MATH, determines an element MATH in MATH. Thus MATH contains MATH for any partial isometry MATH, such that MATH, MATH. Let MATH be any unitary in MATH and let MATH be any unitary in MATH. The same argument shows that MATH belongs to MATH. Since any element in MATH and MATH is a linear combination of unitaries, this shows that MATH is always in the linear span of MATH for all MATH in MATH and MATH in MATH. This set is obviously weakly dense in MATH.
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math/0004172
|
Indeed if MATH is such a maximum point for the functional MATH on MATH, then for all MATH in MATH we have that MATH . But then this will give that MATH for all MATH, MATH. Thus for all MATH in MATH we have MATH and hence MATH for all MATH selfadjoint in MATH. Since MATH is also selfadjoint, it follows that MATH and hence that MATH commutes with MATH.
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math/0004173
|
Let MATH and let MATH. We need to show that the modular symbol MATH depends only on the coset MATH. First we assume MATH is a splitting. By the remarks at the end of REF, if MATH and MATH is the tuple obtained by left translation, then MATH. From this it follows that if MATH, then MATH. Indeed, MATH, where MATH, and any element of MATH preserves MATH. Now assume that MATH isn't a splitting. There are two possibilities: CASE: MATH for some MATH; CASE: MATH for all MATH and MATH for some MATH. In the first case, we have MATH for all MATH, so the NAME series is identically MATH. In the second case, we have MATH for all MATH, since left translation of the tuple MATH by MATH preserves the incidence conditions satisfied by the MATH. This completes the proof.
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math/0004173
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First, the twisted series on the right are well-defined, since if MATH and MATH are compatible then so are MATH and MATH for each MATH. We have the following basic relation among modular symbols for the minimal parabolic subgroup, from CITE: MATH . Note that the relations in CITE imply that this equality holds true in MATH for any collection of MATH-dimensional rational subspaces MATH, even with the possibility that some MATH aren't splittings. The result follows immediately from REF and the fact that if MATH with MATH, then MATH .
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math/0004173
|
We begin by recalling some facts from the theory of modular symbols associated to the minimal parabolic subgroup. These facts are equivalent to results in CITE, and are just reformulated in terms of tuples and splittings. Let MATH be the set of all full tuples of MATH-dimensional subspaces. We define a function MATH as follows. From each MATH-dimensional subspace MATH, we choose and fix a primitive vector MATH. Then we set MATH . Let MATH be the subset of tuples for which MATH. The set MATH is finite, where MATH acts by left translations. One can show that any modular symbol MATH can be written as a sum MATH where MATH is a finite subset of MATH (depending on MATH). Moreover, the cardinality of MATH is bounded by MATH, where MATH is a polynomial depending only on MATH CITE. Let MATH and consider the modular symbol MATH. Since MATH is compatible with MATH, the space MATH is the span of the first basis element of MATH. Let us assume for the moment that for MATH, MATH is the span of the MATH-th standard basis element of MATH. This implies that MATH is the absolute value of the determinant of a fixed MATH minor of MATH. Hence MATH where the implied constant depends only on MATH. Let MATH be the right hand side of REF. It follows that there is a polynomial MATH, depending only on MATH, such that MATH . Now consider the value MATH. Since MATH is finite, there is a maximum value MATH that MATH attains on this set. Writing MATH, we have MATH where the implied constant depends on MATH and MATH. The right of REF has the same convergence properties as the usual NAME series, and so the proof is complete under our assumption on MATH. Now assume MATH is a general MATH-dimensional subspace of MATH for MATH. Let MATH be the MATH-th coordinate of MATH, and let MATH . Then MATH where the implied constant depends on MATH and MATH. The rest of the proof proceeds as above.
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math/0004176
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For the first statement, we need a convergent sequence MATH of elements of MATH, each of which represents the oriented matroid MATH, and whose limit is in MATH and represents MATH. This sequence is defined by closing up the angle MATH, leaving the points MATH all at height REF in MATH and in the right order in the limit. Meanwhile, the realizations of the remaining points are obtained by letting each MATH-coordinate be MATH while maintaining the colinearity and intersection properties determined by MATH. The second statement is proven using two facts from elementary projective geometry: CASE: For every two MATH-tuples MATH and MATH of points in general position in affine MATH-dimensional space, there exists a unique affine automorphism taking each MATH to MATH. CASE: If MATH are four points on a line in the affine plane (in the order given), their cross-ratio MATH is invariant under affine automorphism. Now assume by way of contradiction there exists an affine arrangement MATH for some MATH. We will compute the cross-ratio of the points MATH in MATH in two different ways (a MATH-way and a MATH-way) and come up with a contradiction. Construct a sequence MATH of arrangements in MATH such that CASE: the sequence converges to MATH, CASE: the elements MATH all have MATH-coordinate REF in each MATH, and CASE: the subarrangements MATH in each MATH are all projectively equivalent. (That is, for every MATH there is an affine automorphism of the plane taking the subarrangement in MATH to the subarrangement in MATH.) Fix a small MATH. Define MATH to be the unique realization of MATH with CASE: MATH, MATH, MATH, and MATH in the same positions as in MATH, CASE: MATH the point at distance MATH above the position of MATH in MATH, CASE: MATH determined by requiring REF-dimensional affine arrangement MATH in MATH to be projectively equivalent to the corresponding arrangement in MATH, and CASE: All other points determined by the colinearity and convexity conditions of MATH, together with the condition that the MATH-coordinates of MATH and MATH are REF and the MATH-coordinates of all remaining points are MATH. That all elements of the arrangements MATH except MATH and MATH converge to the corresponding elements of MATH is clear. We get convergence of MATH and MATH by noting that there exists some sequence MATH in MATH converging to MATH. As MATH increases, the elements MATH in MATH converge to the corresponding elements of MATH. Since the positions of MATH and MATH are determined by the positions of MATH, MATH and MATH converge as well. Note that the subarrangements MATH-in the MATH are all projectively equivalent (by the affine automorphism fixing MATH and MATH and mapping the corresponding MATH to each other). Thus the cross-ratio MATH of MATH is the same in all the MATH, and so MATH is the cross-ratio of MATH in MATH. Similarly, we get a sequence MATH in MATH and calculate the cross-ratio of MATH in MATH to be MATH. Thus MATH. On the other hand, consider the affine automorphism of the plane fixing the points MATH and MATH in MATH and taking the point MATH in MATH to the point MATH in MATH. This sends the subarrangement MATH in MATH to the corresponding subarrangement of MATH, hence sends the point MATH in MATH to the point MATH in MATH. But it does not send the point MATH in MATH to the point MATH in MATH, since MATH in MATH. Hence MATH, a contradiction.
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math/0004176
|
We assume that there is a tame triangulation of MATH and reach a contradiction. By REF , MATH and MATH for every MATH and MATH. Choose a sequence of simplices MATH so that MATH . Then there exists a sequence of simplices MATH so that MATH and MATH is a face of MATH. Then MATH so, by REF of the above lemma, the MATH's are distinct. Thus there are an infinite number of simplices in a compact set MATH, which is a contradiction.
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math/0004178
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We may take MATH, MATH and MATH, where MATH and MATH. Also choose MATH with MATH for MATH. Then, let MATH be the cycle that starts at MATH, goes to MATH, then runs on MATH counter-clockwise and goes back to MATH(see REF ). We choose a numbering of MATH, where MATH is as in REF , such that the monodromy of MATH along MATH is MATH. Similarly, for MATH, we give a numbering so that the monodromy along the circle MATH, counter-clockwise, is MATH. Let MATH be given. Then, we can talk of monodromy MATH of MATH along MATH. Then MATH is a transposition, since it is the monodromy of the simple ramification over MATH. Also, let MATH be the permutation defined by the path from MATH to MATH and numbering of MATH and MATH. Then, we have MATH. Conversely, such MATH defines MATH.
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math/0004178
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We use notations in REF . If MATH and MATH is the equivalence class of MATH, then MATH if and only if MATH and MATH intersect and MATH if and only if MATH and MATH intersect. There are either one or two cyclic components of MATH that intersect MATH, since MATH is a transposition. If there is only one, say MATH, then two components of MATH intersect MATH, that is, the two components of MATH. In this case, we have MATH and MATH. If there are two, MATH and MATH, then MATH is the sole component of MATH that intersects MATH, and we have MATH, MATH. The other properties are easy to see.
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math/0004178
|
In this case, we have MATH and MATH connects MATH to MATH for some MATH. By symmetry, we may assume MATH. Then, MATH satisfies MATH and the associated graph is isomorphic to MATH if and only if MATH and MATH for MATH. When MATH, the number of such MATH is MATH. Now MATH proves the claim.
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math/0004178
|
Let MATH be an element of MATH. CASE: MATH. By symmetry, we may assume MATH. There are two edges MATH such that MATH. Let MATH be the graph obtained by removing MATH and connecting MATH to MATH and MATH to MATH. Write MATH. If MATH belongs to MATH and MATH, then there exist MATH with MATH for MATH and positive integers MATH with MATH such that MATH and that MATH belongs to MATH, as is easily seen from definition. Let MATH be positive integers with MATH. Then, let MATH be the set of MATH such that there exists MATH with the property that MATH(MATH), MATH and that the graph associated to MATH is isomorphic to MATH. If MATH or MATH, MATH maps MATH to the cyclic component of MATH connected to MATH, and MATH to the one connected to MATH. If MATH, MATH maps MATH and MATH to cyclic components of lengths MATH and MATH, respectively. Thus there are MATH choices for MATH in these cases and we have MATH. If MATH and MATH hold, then it is MATH. On the other hand, for each MATH, there are MATH transpositions MATH for which there exists MATH with MATH for MATH such that MATH belongs to MATH. The number of such MATH is MATH, and we have MATH . Here the sum is taken over different MATH's with MATH. Noting that MATH and MATH are the same for MATH if and only if MATH, we see MATH if MATH and MATH if MATH. Finally, since MATH holds, MATH proves the assertion. CASE: MATH. By symmetry, we may assume that MATH and MATH appear as MATH. Let MATH be the edge such that MATH. Let MATH be the graph obtained by removing MATH and connecting MATH to MATH. As in REF , If MATH belongs to MATH and MATH, then there exists MATH such that MATH, where MATH, and that MATH belongs to MATH. For positive integers MATH with MATH, let MATH bet the set of MATH such that there exists MATH with the property that MATH(MATH), MATH and that the graph associated to MATH is isomorphic to MATH. Then there are MATH such MATH, and MATH is MATH. On the other hand, for MATH and MATH with MATH, the number of pairs MATH with MATH REF such that MATH belongs to MATH is MATH times MATH: let MATH be the cyclic permutation corresponding to MATH. Choose MATH and let MATH be the residue of MATH modulo MATH. We take MATH to be MATH, and then MATH has cyclic components of lengths MATH and MATH, containing MATH and MATH respectively. We can choose MATH so that the two components equal MATH and MATH, respectively. Thus we have MATH . Since MATH holds in this case, MATH proves the assertion.
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math/0004179
|
If we fix MATH and differentiate MATH, we have MATH.
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math/0004181
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The proof is based on an idea of NAME, compare CITE. Let MATH be a continuous, self-adjoint and homogeneous lift of MATH such that MATH for all MATH. Such MATH exists by the NAME selection theorem, compare CITE. Define MATH such that MATH. Choose continuous functions MATH, such that CASE: MATH for all MATH, CASE: MATH for all MATH, CASE: for each MATH, there is a MATH such that MATH for all MATH, and all MATH, CASE: MATH. Let MATH be an increasing sequence of finite subsets with dense union in MATH. Write MATH where MATH are compact subsets of MATH. For each MATH, choose MATH as in REF. We may assume that MATH. By REF we can choose elements MATH in MATH such that MATH for all MATH and MATH for MATH, and CASE: MATH for all MATH, CASE: MATH for all MATH, and all MATH, CASE: MATH for all MATH and all MATH, CASE: MATH, for all MATH, CASE: MATH, for all MATH, where MATH and MATH are the compact sets MATH and MATH . Since we choose the MATH's recursively we can arrange that MATH for all MATH and all MATH. By connecting first MATH to MATH via the straight line between them, then MATH to MATH via a straight line, then MATH to MATH etc., we obtain norm-continuous pathes, MATH, in MATH such MATH for all MATH and CASE: MATH, for all MATH, CASE: MATH for all MATH and all MATH, CASE: MATH for all MATH, all MATH and all MATH, CASE: MATH, for all MATH, CASE: MATH, for all MATH. In addition, MATH. Let MATH denote the NAME MATH-module of sequences MATH in MATH such that MATH converges in norm. Writing an element MATH as the sum MATH we define a representation MATH of MATH on MATH such that MATH. Then MATH acts by automorphisms on MATH ( = the adjoinable operators on MATH) such that MATH. Set MATH . Then MATH is a projection in MATH since MATH clearly is. Note that MATH is tri-diagonal because of REF above, and that the entries of MATH are all in MATH, with the notable exception of the MATH-entry which is equal to MATH modulo MATH. We define MATH by MATH . Set MATH for MATH and MATH. We assert that MATH is an asymptotic homomorphism. By using the continuity of MATH and that MATH is a compact set for fixed MATH, it follows readily that the family of maps MATH, is an equicontinuous family. Since each MATH is self-adjoint and homogeneous, it suffices therefore to take a MATH and elements MATH, MATH, and check that MATH and MATH . The first two limits are zero by REF, the third by REF. For each MATH, MATH modulo MATH ( = the ideal of 'compact' operators on MATH). Since MATH is weakly stable there is an equivariant isomorphism MATH of NAME MATH-modules which leaves the first coordinate invariant. We can therefore transfer MATH to an asymptotic homomorphism MATH with the stated property.
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math/0004181
|
Since MATH is strongly homotopic to MATH there is an equivariant MATH-homomorphism MATH, where MATH is a separable MATH-algebra containing MATH, and an equivariant MATH-homomorphism MATH such that MATH. Apply REF to MATH.
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math/0004181
|
Let MATH be the automorphism of MATH given by MATH. It is wellknown that MATH is strongly homotopic to MATH. Hence MATH is asymptotically split by REF .
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math/0004181
|
It follows from REF that MATH and MATH are both asymptotically split. Since MATH and MATH are unitarily equivalent, the conclusion follows because infinite direct sums are well-defined for asymptotically split extensions.
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math/0004181
|
Let MATH be an asymptotic homomorphism such that MATH for all MATH. We may assume that both MATH and MATH are equicontinuous, compare REF. Let MATH be a sequence of finite subsets with dense union in MATH. For each MATH there is MATH with the property that MATH when MATH. Choose then a sequence of functions MATH such that MATH for all MATH and such that MATH for all but finitely many MATH's for all MATH. Set MATH for all MATH. Note that MATH, for all MATH and MATH. Then MATH and MATH define asymptotic homomorphisms MATH. By connecting appropriate permutation unitaries, acting on MATH by permutations of MATH-coordinates, we get a norm-continuous path of MATH-invariant unitaries MATH such that MATH for all MATH. Then MATH for all MATH. Since MATH is weakly stable there is an isomorphism MATH of NAME MATH-algebras which fixes the first coordinate. Applying this isomorphism in the obvious way and remembering the identifications MATH and MATH gives the result.
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math/0004181
|
The 'if' part is easy and the 'only if' part follows from REF in the same way as REF follows from REF .
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math/0004181
|
Since MATH is weakly stable we can write MATH with MATH acting trivially on the tensor-factor MATH. We embed MATH into MATH via MATH. Let MATH be a dense sequence in MATH. For each MATH there is a function MATH such that MATH. Let MATH be the MATH-algebra generated by MATH. Then MATH. Consider a positive element MATH and a MATH. Set MATH. We can then find a sequence MATH of projections in MATH such that MATH . Let MATH be a partition of unity in MATH subordinate to the cover MATH and set MATH. Then MATH, MATH. For each MATH we choose a partial isometry MATH such that MATH, MATH and MATH. Set MATH. Then MATH and MATH. It follows that we can find a sequence MATH of separable MATH-subalgebras of MATH and for each MATH have a dense sequence MATH in the positive part of MATH and elements MATH in MATH such that MATH and MATH for all MATH. It follows then from REF that MATH is a separable stable MATH-subalgebra of MATH such that MATH. Note that MATH contains a sequence MATH with the property that MATH for all MATH since MATH does. Set MATH. By repeating the above argument with MATH substituted by the MATH-algebra MATH generated by MATH, we get a stable MATH-subalgebra MATH which contains a sequence MATH such that MATH for all MATH. It is clear from the construction that we can arrange that MATH. We can therefore continue this procedure to obtain sequences of separable MATH-algebras, MATH in MATH, and MATH in MATH such that each MATH is stable and contains a sequence MATH such that MATH, and MATH for all MATH. Set MATH and MATH. It follows from REF that MATH is stable. By construction MATH for all MATH and MATH. The last property ensures that MATH is an ideal in the MATH-algebra MATH generated by MATH and MATH. There is therefore a MATH-homomorphism MATH. By construction an approximate unit for MATH is also an approximate unit for MATH so MATH extends to a MATH-homomorphism MATH which is strictly continuous on the unit ball of MATH. Since MATH is stable there is a sequence MATH, of orthogonal and NAME equivalent projections in MATH which sum to MATH in the strict topology. Then MATH, is a sequence of orthogonal and NAME equivalent projections in MATH which sum to MATH in the strict topology. Since MATH consists entirely of MATH-invariant elements it follows that all the MATH's are MATH-invariant. Consequently MATH as MATH-algebras, proving that MATH is weakly stable.
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math/0004181
|
The equivalence MATH follows from REF and the implication MATH is trivial, so we need only prove that REF. To this end, let MATH denote the set of equi-homotopy classes of asymptotic homomorphisms MATH. Choose MATH-invariant isometries MATH such that MATH and define a composition in MATH by MATH . It follows from REF that MATH is a group. It suffices therefore to show that the natural map MATH has trivial kernel. If MATH is an asymptotic homomorphism representing an element in the kernel we conclude from REF that there is a norm-continuous path MATH, of MATH-invariant unitaries in MATH and an asymptotic homomorphism MATH such that MATH for all MATH. By a standard rotation argument we can remove the unitaries MATH via an equi-homotopy and we see in this way that MATH in MATH. Hence MATH in MATH.
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math/0004181
|
Let MATH. Then MATH proving that MATH converges in MATH. And MATH proving that also MATH converges in MATH. It follows that MATH exists as a strict limit in MATH. It it then straightforward to check that MATH.
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math/0004181
|
Let MATH be another unit sequence satisfying REF-REF. There is then a unit sequence MATH in MATH such that MATH for all MATH. Connect MATH to MATH by a straight line, then MATH to MATH by a straight line, etc. This gives a path MATH of unit sequences. For each MATH we get then a map MATH such that MATH and MATH in MATH. Let MATH be the map obtained from MATH as MATH was, but by using MATH instead of MATH. Then MATH in the strict topology for all MATH, and MATH in norm for all MATH. Hence MATH in MATH. The same argument with the unit sequence MATH replaced by MATH shows that the class of MATH in MATH is independent of the choice of unit sequence. Once this is established it is clear that a homotopy of asymptotic homomorphisms MATH gives rise, by an appropriate choice of unit sequence, to a homotopy which shows that MATH only depends on the homotopy class of MATH. That MATH is also independent of the discretization and only depends on the homotopy class of MATH follows in the same way as in REF.
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