paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0004137 | 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... |
math/0004137 | 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... |
math/0004137 | 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 obtain... |
math/0004137 | 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 ... |
math/0004137 | 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. |
math/0004144 | 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 exist... |
math/0004144 | Let MATH be the polar decomposition of MATH. Then MATH. Also MATH and using NAME 's inequality, MATH . |
math/0004144 | 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... |
math/0004144 | 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 M... |
math/0004144 | 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... |
math/0004144 | 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 ... |
math/0004144 | 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 ... |
math/0004144 | 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... |
math/0004144 | 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. Al... |
math/0004145 | 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. |
math/0004145 | 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 t... |
math/0004145 | 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 restricti... |
math/0004145 | 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 MAT... |
math/0004146 | 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 ... |
math/0004146 | 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-uni... |
math/0004146 | 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 sub... |
math/0004146 | 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 s... |
math/0004146 | 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 ... |
math/0004146 | 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:... |
math/0004146 | 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. Contr... |
math/0004146 | 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... |
math/0004148 | Differentiating the equality MATH and REF, we obtain: MATH where MATH denotes evaluation at MATH. Since MATH is an isomorphism, the conclusion follows. |
math/0004148 | By REF , we have: MATH . Hence, by definition of MATH, we get: MATH . |
math/0004148 | 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 MAT... |
math/0004148 | 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: MA... |
math/0004148 | It is a simple application of the Inverse Function Theorem on NAME manifolds (see CITE for details on a similar construction). |
math/0004148 | See CITE. |
math/0004148 | 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/0004148 | 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. |
math/0004148 | Follows from REF observing that a characteristic MATH that vanishes at some MATH is identically zero (see CITE). |
math/0004148 | 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 w... |
math/0004148 | 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/0004148 | 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 tha... |
math/0004148 | 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 absol... |
math/0004148 | 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. |
math/0004148 | 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 MAT... |
math/0004148 | 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. |
math/0004148 | The point MATH is critical for MATH if and only if MATH vanishes on MATH. The conclusion follows from elementary functional analysis arguments. |
math/0004148 | 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 M... |
math/0004148 | 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 MA... |
math/0004161 | 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 M... |
math/0004161 | 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 ... |
math/0004161 | 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 MA... |
math/0004161 | 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 ... |
math/0004161 | 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 e... |
math/0004161 | 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 wor... |
math/0004161 | 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 or... |
math/0004161 | 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 ... |
math/0004161 | We combine REF to write MATH . Let MATH. In MATH we get MATH after making the change of variables MATH, and applying REF . |
math/0004161 | 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/0004161 | 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 . |
math/0004161 | 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 w... |
math/0004161 | Since MATH on MATH, MATH . |
math/0004161 | 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. |
math/0004161 | 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 com... |
math/0004166 | 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. |
math/0004166 | 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,... |
math/0004166 | 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 f... |
math/0004166 | 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 M... |
math/0004166 | 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 stand... |
math/0004166 | 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/0004167 | 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. T... |
math/0004167 | The direction MATH is clear. For MATH, combine REF . |
math/0004167 | See REF . |
math/0004167 | 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. |
math/0004168 | 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... |
math/0004168 | 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... |
math/0004168 | 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... |
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 M... |
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 f... |
math/0004172 | 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 s... |
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. |
math/0004172 | This follows by writing down explicitly that MATH . |
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 par... |
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 hen... |
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 an... |
math/0004173 | 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/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... |
math/0004176 | 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 limi... |
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 a... |
math/0004178 | 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 MA... |
math/0004178 | 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 tw... |
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. |
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/0004179 | If we fix MATH and differentiate MATH, we have MATH. |
math/0004181 | 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 ... |
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. |
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 . |
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. |
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 t... |
math/0004181 | The 'if' part is easy and the 'only if' part follows from REF in the same way as REF follows from REF . |
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 a... |
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 t... |
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. |
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... |
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