paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0010096 | Suppose such an action exists. Recall that the MATH - fold join of MATH is MATH. Hence taking joins we can obtain a free action of MATH on MATH, which contradicts a result due to CITE. |
math/0010096 | The irreducible representations of MATH are described as follows: if MATH let MATH . Then MATH. Evaluating on MATH, MATH respectively yields MATH whence we readily infer that MATH and MATH. Hence we deduce that MATH if and only if MATH; but noting that MATH we conclude that MATH if and only if MATH. |
math/0010096 | Note that MATH, MATH, MATH, hence for odd primes MATH they have cyclic MATH-Sylow subgroup and we need only consider the prime MATH. By our lemma and the existence of the characters mentioned above, the free actions on MATH can be constructed using REF . |
math/0010096 | We first consider MATH, note that it has order MATH and that MATH. Now MATH is an extra-special MATH-group MATH with MATH central. Denote by MATH the conjugacy class of this central element of order REF. From the Atlas CITE we see that MATH has one conjugacy class of involutions MATH and two conjugacy classes of elements of order three (MATH and MATH respectively). There is a complex character MATH which takes the following values: MATH . From these values and the previous lemma, we infer that the associated mod REF NAME class is effective. However notice that the subgroup generated by REF has no fixed-points, hence acts freely on the associated sphere MATH. Therefore in mod REF cohomology the NAME class restricts non-trivially to the cohomology of REF and so must be effective. We deduce that MATH acts on MATH with rank one isotropy and so by our theorem MATH acts freely on a finite complex MATH. Now we consider MATH; in this case MATH. According to the Atlas CITE, MATH has only one conjugacy class of involutions, REFA, and furthermore there is a complex character MATH with the following values MATH . This implies that the mod REF NAME class for MATH is effective, and as before (but now for MATH) we also see that REFA acts without fixed-points, hence the NAME class is mod REF effective too. Hence MATH acts on MATH with periodic isotropy and the rest follows as before. |
math/0010096 | Let MATH denote a group with normal MATH - NAME subgroup MATH. Consider an effective class MATH for MATH arising from a MATH - dimensional representation of MATH as in REF . Now induce this to a representation for MATH; by the double coset formula, the resulting NAME class will restrict in MATH to MATH, which is evidently also an effective NAME class. |
math/0010096 | The first part can be found in CITE, page REF. For the second part we use an argument due to NAME CITE. Note that MATH is the minimal degree of a complex irreducible character for MATH. This is also the minimal degree for a faithful irreducible of the NAME normalizer, which is a semidirect product MATH, where MATH is extra - special of order MATH. The cyclic group MATH acts transitively on the non - trivial linear characters of MATH. The given representation restricts to MATH as the sum of the MATH irreducibles of degree MATH. Restricting further to MATH we just get all non - trivial linear characters, each MATH times. This has value MATH on each element of order MATH in the center, in particular any such element acts without fixed - points on the associated linear sphere. We infer from REF that the corresponding NAME class must be MATH - effective. |
math/0010096 | Clearly an NAME class is effective if and only if its MATH-localization at each prime for which MATH is effective. Moreover if MATH is an NAME class then each of its localizations MATH is a MATH-local NAME class. Now assume that MATH and that for each prime MATH for which MATH one has a fibration MATH with homotopy fiber the MATH-local sphere MATH and MATH-local NAME class MATH. So MATH and the universal cover of MATH is a MATH-CW complex that is homotopy equivalent to MATH. Since the fundamental class survives to MATH in the NAME spectral sequence, the action of MATH on the MATH-local homotopy sphere is homologically trivial and hence homotopically trivial. Let MATH be the finite set of primes for which MATH and let MATH be the set of all other primes. There is a unique homologically trivial action of MATH on the rationalization MATH of an odd dimensional sphere (here all primes are inverted). Give MATH the trivial action of MATH. Then one has a diagram of MATH-CW complexes MATH . Using the homotopy pullback diagram for the classifying spaces described previously, we see that the homotopy classes defining the local homotopy actions define a homotopy action of MATH on the homotopy pull-back of the diagram, which by REF (see CITE, page REF) is homotopy equivalent to MATH. From this we obtain a MATH - CW complex whose associated NAME construction gives the required spherical fibration MATH . |
math/0010102 | The first part of the result follows immediately from REF. For the second part we observe that if a point MATH in MATH is covered by the ball MATH, then MATH; so the number MATH of the balls MATH, that cover MATH, is estimated by the greatest number MATH of points MATH in MATH with MATH and we observe that, by MATH, MATH is estimated by a number MATH depending only on MATH. |
math/0010102 | We have to prove that there is a subsequence of MATH convergent in MATH. Taking into account assumption MATH, the ball MATH can be covered by a finite number of balls MATH, MATH, MATH where MATH depends on MATH, such that every point of MATH belongs at most to MATH balls, where MATH does not depend on MATH. Let MATH and MATH . Then MATH . Since MATH then inequality MATH becomes MATH . Choose MATH and MATH such that MATH. Suppose MATH is weakly convergent in MATH then MATH for MATH. This implies MATH and MATH-is a NAME sequence in the space MATH then MATH-is convergent in MATH. |
math/0010102 | Let MATH. After extraction of a subsequence, we have that MATH is weakly convergent in MATH to MATH. We suppose, without loss of generality that MATH and prove MATH when MATH. Let MATH, MATH such that MATH when MATH. Since MATH when MATH, then MATH such that for MATH . Then for MATH . |
math/0010102 | Let MATH be a sequence in MATH such that MATH in MATH as MATH, where MATH denotes the dual space of MATH. From MATH we obtain that there exists MATH such that for MATH . Then MATH . Since MATH is bounded in MATH and from REF , we know that there exists a subsequence strongly convergent in MATH and weakly to MATH . We apply now REF if MATH or REF if MATH of Ref. CITE to the function MATH and to the sequence MATH and we obtain MATH . From the assumption we have that MATH where MATH as MATH. Then from MATH we have MATH . From MATH and MATH we obtain MATH when MATH. This implies that MATH converges to MATH strongly in MATH. |
math/0010102 | First we prove that for MATH small enough MATH for MATH. Consider the case MATH. As in REF of Ref. CITE we obtain that for every MATH there exists a constant MATH such that MATH where MATH if MATH or MATH if MATH. There exists MATH such that MATH . Choose MATH; then MATH and MATH and the result follows from the last inequality. We consider now the case MATH. From the assumption we obtain that for every MATH there exists a constant MATH such that MATH for MATH. We observe that there exists MATH such that MATH . Choosing MATH, we have MATH; then MATH and MATH . The result follows from the last inequality. Let us prove the existence of MATH such that MATH. Let MATH be the potential of the ball MATH with respect to the ball MATH. Then MATH is in MATH and MATH; we recall that MATH for MATH. Let MATH; we have MATH on MATH, so MATH . Since MATH we have for MATH, MATH suitable, we have MATH. The proof is completed with the application of the Mountain Pass Theorem. |
math/0010115 | The equivalence of REF has been shown for countably generated NAME MATH-modules in REF. The implication MATH can be seen to hold setting MATH for the existing unitary operator MATH and for MATH. The demonstration of the inverse implication requires slightly more work. For the given normalized tight frames MATH and MATH of MATH and MATH, respectively, we define a MATH-linear operator MATH by the rule MATH, MATH. For this operator MATH we obtain the equalities MATH which hold for every MATH. Consequently, MATH for any MATH, and the operator MATH is unitary. |
math/0010115 | We have already pointed out that the set of algebraic generators MATH of MATH is a frame with respect to any MATH-valued inner product on MATH which turns MATH into a NAME MATH-module. That means the inequality MATH is satisfied for two finite positive real constants MATH and any MATH, see REF. What is more, for any frame of MATH there exists another MATH-valued inner product MATH on MATH with respect to which it becomes normalized tight. The latter inner product is unique as shown by REF. So REF of the previous theorem demonstrates the complete assertion. |
math/0010115 | Since both the modular NAME bases are sets of modular generators of MATH we obtain two MATH-valued (rectangular, without loss of generality) matrices MATH and MATH with MATH and MATH such that MATH . Combining these two sets of equalities in both the possible ways we obtain MATH for MATH, MATH. Now, since we deal with sets of coefficients MATH and MATH that are supposed to admit special carrier projections, the coefficients in front of the elements MATH and MATH at the right side can only take very specific values: MATH where MATH is the NAME symbol, MATH is the carrier projection of MATH and MATH is the carrier projection of MATH. So MATH and MATH are positive idempotent diagonal matrices with entries from MATH. The NAME relations MATH, MATH, MATH and MATH turn out to be fulfilled. |
math/0010115 | Let MATH be the set of all linear bounded operators on the NAME space MATH, and recall that MATH as a direct sum of copies of the NAME space MATH itself. Consequently, there are projections MATH such that MATH for any MATH, MATH, and MATH via partial isometries MATH with MATH and MATH. Define MATH, that is, the set of all sequences of elements of MATH for which the series converges weakly in MATH. It can be identified with the set of all bounded MATH-linear maps MATH of the NAME MATH-module MATH into MATH setting MATH, where MATH. Since MATH is a W*-algebra, the set MATH becomes a self-dual NAME MATH-module CITE. In fact, MATH is isometrically isomorphic to MATH as a NAME MATH-module. To see this fix an orthonormal basis MATH of MATH and an orthonormal basis MATH of MATH and define MATH . The freedom of choice for this isomorphism is the careful selection of the orthonormal bases of the self-dual NAME MATH-modules MATH and MATH. If the sequence MATH is given as described above then it is a (possibly non-standard) normalized tight frame of the self-dual NAME MATH-module MATH since we have MATH for every MATH. There exists a frame transform MATH defined by MATH. The image of MATH is a direct summand and self-dual NAME MATH-submodule of MATH. Moreover, MATH is an isometry. Continuing the isometry MATH to an isometry MATH using the isometric isomorphism between MATH and MATH we obtain MATH for any MATH. The structure of direct orthogonal summands of the NAME MATH-module MATH is well-known, they are all generated by multiplying MATH by a specific orthogonal projection from the right. So we can characterize the image of MATH in MATH as MATH for some MATH of MATH. Note that MATH since the module generator MATH is mapped to the module generator MATH. Let MATH be the isometry linking MATH to MATH with MATH, MATH. Now, we establish some information on the adjoint operator of MATH: MATH . Consequently, we get the frame decomposition MATH for any MATH. The frame coefficients MATH may be not the optimal ones. By REF we have the general inequality: MATH for any MATH. By REF we obtain MATH for some elements MATH with MATH and MATH . Of course the root of MATH seems to be selected in an artificial way, the element MATH would do the job as well. However, the following inequality gives some more information on the background of the choice made: MATH . Since the left end equals the right end and MATH holds for every MATH the equality MATH turns out to be valid for every MATH. Consequently, the linking elements MATH can be selected as partial isometries of MATH mapping the left carrier projection of MATH to the left carrier projection of MATH for each MATH because MATH is a NAME algebra and any root of a given positive operator can be described this way. By REF the left carrier projection of MATH has to be lower-equal than the left carrier projection of MATH for any MATH. |
math/0010115 | Denote the frame transform of the tight frame MATH by MATH for MATH. Because the frame transforms MATH and MATH all possess the same coisometries in their respective polar decompositions we obtain the equality MATH. Since the inequality MATH is valid and both MATH and MATH are diagonalizable with a common set of eigenvectors that forms a basis of MATH we obtain MATH . The right expression is minimal if and only if MATH is the arithmetic mean of the lower and the upper spectral bound MATH and MATH of the positive invertible operator MATH. |
math/0010117 | It is enough to show that the anti-commutators form a MATH-Gröbner basis. To do this, we observe that all obstructions resolve. |
math/0010117 | There are only finitely many square-free monomials on MATH, and only the trivial such monomial if MATH. |
math/0010117 | Let MATH be a stable (square-free monomial) ideal, and let MATH be a generator, MATH . Let furthermore MATH . Then the maximal index MATH such that MATH is MATH. Hence stability implies that MATH, and since MATH is an ideal, MATH. This shows that MATH. |
math/0010117 | We start by noting that if MATH then MATH does not divide MATH. This follows since MATH has no elements of degree REF, and since quadratic monomials are lifted to ``strictly ordered" quadratic monomials. Hence MATH is part of a minimal NAME basis, and MATH is part of a minimal generating system of MATH. Mimicking the proof in CITE, we now show that if MATH, then MATH if and only if MATH or MATH for some MATH. So, suppose first that MATH for some MATH. Then MATH, and MATH . , hence MATH. If on the other hand MATH, but MATH (in this case, we say that MATH is square-free and ordered), then MATH. Since MATH is a section to MATH, MATH. Hence, there exist a MATH with MATH. Put MATH, then MATH, hence MATH. It is easy to see that MATH. Therefore, MATH. For the converse, suppose that MATH. If there is some MATH such that MATH, we are done; suppose therefore that MATH. There exist some MATH with MATH. From the fact that MATH we conclude that MATH, hence that MATH. Therefore MATH. To move on with the proof, we observe that MATH contains all non-commutative monomials that fail to be both ordered and square-free, and that the initial monomials of the anti-commutators form a minimal generating set for these monomials. Now suppose that MATH, where MATH is a minimal generator of MATH, with MATH, and where MATH is any square-free monomial. It remains to show the following: MATH is a minimal generator of MATH if and only if MATH. For the ``only if" part, suppose that MATH is a minimal generator of MATH. If MATH, then it belongs to MATH. Suppose therefore that there exists some MATH such that MATH . We first show that MATH. We can assume that MATH is the maximal index such that MATH . If it were the case that MATH, then MATH. Now MATH, which contradicts the fact that MATH is a minimal generator of that ideal. Secondly, we show that MATH. We can here assume that MATH is the minimal index such that MATH . If it were the case that MATH, then MATH. Now MATH, which contradicts the fact that MATH is a minimal generator of that ideal. Hence MATH. So, we know that MATH, and thus that MATH is a square-free monomial in MATH. So MATH, which implies that neither MATH nor MATH is in MATH (that would contradict the fact that MATH is a minimal generator), hence neither MATH nor MATH is in MATH. For the ``if" direction, suppose that MATH. That means that neither MATH nor MATH is in MATH, hence that neither MATH nor MATH is in MATH. Therefore, we can conclude that MATH is a minimal generator of MATH. |
math/0010117 | Follows from REF . We can also see this directly: since the exterior algebra has finite vector space dimension, so has MATH, hence so has MATH. Hence, for some MATH, MATH contains all monomials of total degree MATH. Hence, MATH is generated in degrees MATH and is therefore finitely generated. |
math/0010117 | It is proved in CITE that (regardless of the characteristic of MATH) the generic initial ideal of MATH is strongly stable. Hence, for a generic MATH, MATH is strongly stable, hence stable, hence squeezed. By REF , MATH has a NAME basis which is the ``lift" of a NAME basis of MATH. Since the anti-commutator ideal is invariant under linear changes of variables, we have that MATH. |
math/0010117 | The proof is almost word-for-word identical with the proof in CITE of the corresponding assertion for the symmetrical algebra. To stress the similarity, we temporarily denote MATH by MATH. We start by fixing a total degree MATH and a basis MATH for the MATH-vector space MATH. Then, we let MATH, noting that there is a REF correspondence between MATH-dimensional subspaces of MATH and MATH-dimensional subspaces of MATH. Furthermore, every element in MATH can be written as a finite sum of terms MATH, where MATH and MATH are (non-commutative) monomials of total degree MATH. We can assume that such a term is written such that MATH. Then, we order such expressions ``lexicographically", that is, MATH iff there is a MATH such that MATH for MATH, and MATH (here of course we use the term order MATH for comparison). With this convention, we can define the initial term MATH of MATH. It is easy to see that in fact MATH . We may perform ``Gaussian elimination" on the MATH, so that their linear hull remains unchanged, but their initial terms become distinct. Now let MATH be a MATH matrix of indeterminates, MATH, and let MATH act on MATH by MATH, where on each component it acts as an element of MATH. Now, let MATH be the largest (with respect to the order just defined) term of MATH with non-zero coefficient MATH. Define MATH to be the open subset consisting of all MATH such that MATH. Define MATH to be the MATH-vector subspace of MATH generated by MATH. Then MATH iff MATH. Define MATH. To see that MATH is a two-sided ideal, it suffices to show that MATH. Since both MATH and MATH are open and non-empty, their intersection is also open and non-empty. Pick a MATH. Then MATH and MATH. Since MATH is a two-sided ideal, we have that MATH hence that MATH. Put MATH. Clearly, MATH iff MATH. If MATH happens to be generated in degrees MATH, then we claim that MATH, hence MATH is open. Suppose that MATH, then MATH for all MATH. Since MATH was supposed to be generated in degrees MATH, this implies that MATH. But for all MATH, MATH and MATH have the same dimension as MATH-vector spaces, namely MATH. Hence we conclude that in fact MATH. |
math/0010117 | NAME series are preserved by non-singular linear changes of coordinates, and by passing to the initial ideal. |
math/0010117 | There exists a finitely presented algebra with non-rational NAME series CITE. By REF the gin of the defining ideal must have non-rational NAME series, too. Since finitely generated monomial ideals have rational NAME series CITE, the result follows. |
math/0010121 | CASE: This follows directly from the commutative diagram MATH in which the vertical maps are isomorphisms induced by the two projections MATH and MATH and the top row is the homomorphism induced by MATH. CASE: Let MATH denote the standard comultiplication. Write MATH as the composition MATH and MATH as the composition MATH . By hypothesis, we can factor MATH through the product as MATH, for some MATH. It is straightforward to prove that MATH, by checking that their projections onto each summand are homotopic. |
math/0010121 | It is immediate from REF that MATH induces the identity on homology groups. Since MATH is a REF-connected NAME, MATH is a homotopy equivalence. Hence MATH. This establishes REF , and REF follows similarly. |
math/0010121 | Let MATH. Since MATH for MATH, the hypothesis gives that MATH for MATH. Hence, by REF , MATH. Now suppose MATH is any element in MATH. Since MATH and MATH, we have that MATH. Thus, by the properties of the action listed in the introduction, we have MATH . Therefore MATH is a homomorphism. Now suppose MATH is any element in MATH. Then MATH is zero. Since MATH for positive MATH, REF implies that MATH. Thus MATH restricts to MATH as claimed. |
math/0010121 | This can be argued using the long exact homotopy and homology sequences, together with the relative NAME theorem, for the pair MATH. Alternatively, NAME minimal models can be used. We omit the details. |
math/0010121 | We proceed as in the proof of REF. First, we claim that MATH, for all MATH. Since MATH is odd, we have MATH for MATH. Hence we must only check that MATH in degree MATH. By REF, MATH is surjective. Given MATH, write MATH, for some MATH. Then MATH since MATH, and the claim follows. Now a simple modification of the proof of REF yields that MATH, for each MATH. The remainder of the argument follows exactly as in the proof of REF. |
math/0010121 | Suppose MATH is not the identity element in MATH. Since this latter is a MATH-local group, it contains no non-trivial elements of finite order, and hence MATH, for all MATH. By CITE, we have MATH, for each MATH with MATH. From this it follows that for each MATH there is some positive integer MATH and some element MATH such that MATH. Since MATH is of infinite order in MATH, the same must be true of MATH in MATH. |
math/0010121 | Let MATH and MATH be a complex with MATH and dim-MATH. It suffices to show that MATH for every MATH. Consider the last cofibre sequence in a length-MATH restricted spherical cone decomposition of MATH, MATH where MATH is a wedge of spheres and MATH. Now, since MATH, it follows that MATH. Thus, from the properties of the coaction reviewed in REF, there is some MATH such that MATH. Similarly, there is some MATH such that MATH. Note also that MATH since MATH, and MATH is a wedge of spheres of dimension MATH. Now we have MATH . A similar computation yields MATH. Since MATH is abelian, the proof is complete. |
math/0010121 | This follows by REF. |
math/0010121 | Any map MATH induces a corresponding map MATH of MATH'th NAME sections. This gives us a homomorphism MATH which is one-one. Therefore, MATH. But if MATH is the collection of all cyclic groups, then it is a consequence of REF that MATH. |
math/0010121 | If MATH, then the universal coefficient theorem (compare CITE) gives a commutative diagram with exact rows MATH where the middle homomorphism MATH can be either MATH or the identity REF. Thus there is a homomorphism MATH such that MATH. By hypothesis, MATH and so MATH. Therefore, MATH. |
math/0010121 | We shall prove that MATH acts nilpotently on MATH. Then it follows from CITE that MATH is a nilpotent group. Let MATH and consider the diagram with exact rows obtained from the universal coefficient theorem MATH . Then both MATH and MATH are identity maps. Write MATH as the sum of its free and torsion parts and let MATH and MATH be the projections. Since MATH, it follows that MATH. Thus MATH implies that MATH is the identity. In the same way, since MATH, we have that MATH is also the identity. Therefore, MATH can be written as MATH, with MATH a homomorphism that depends on MATH. Hence MATH is generated by elements of the form MATH. On these elements, any MATH satisfies MATH. Hence MATH, and the result follows. |
math/0010123 | Let MATH be the diagonal submonoid of MATH. Since MATH is finitely generated, it is rational. It follows that MATH is a product of rational sets and so is itself rational. |
math/0010123 | Pick an automaton accepting MATH. Reverse the orientation of each edge and invert the edge label. Make every terminal state an initial state and every initial state a terminal state. |
math/0010123 | Suppose MATH is accepted by a finite automaton MATH over MATH. Construct a context - free grammar with one nonterminal MATH for each vertex MATH of MATH. The start symbol is the nonterminal corresponding to the initial vertex. For each edge MATH there is a production MATH, and for each terminal vertex MATH there is another production MATH. It is straightforward to check that this grammar generates MATH. The main step is to use induction on path length and on derivation length to prove that MATH if and only if there is a path in MATH from MATH to MATH with label MATH. For the converse suppose MATH is generated by a context - free grammar MATH as above. Construct an automaton MATH whose vertices are the nonterminals of MATH plus one terminal vertex. The initial vertex is the start symbol. For each production MATH there is an edge MATH, and for each production MATH there is an edge with label MATH from MATH to the terminal vertex. Again it is straightforward to check that MATH if and only if MATH accepts MATH. |
math/0010123 | If MATH is asynchronously automatic, then by CITE MATH is a rational transduction. The converse is CITE except that the automata used there are more restricted than ours. In terms of our notation the vertices of those automata are partitioned into two sets. All edges leaving the first set have labels from MATH, and all edges leaving the second set have labels from MATH. An automaton in our sense can be transformed into one satisfying the definition in CITE. Replace edges by paths if necessary to insure that edge labels are from MATH. If vertex MATH is a source for edges of both types, add a vertex MATH and make all the edges of one type start at MATH instead of MATH. Add edges from MATH to MATH and MATH to MATH with label MATH. There is one more detail. In CITE MATH is defined as MATH instead of MATH, but if one version of MATH is a rational transduction, then the other one is too. |
math/0010123 | Let MATH be a point on MATH. If MATH, then the distance from MATH to MATH is at most MATH, the maximum length of the finitely many combing paths for elements MATH with MATH. Otherwise estimate the distance by constructing a sequence of triangles as in REF . Let MATH be a MATH - triangle whose base is MATH and whose third vertex is a point as close to the middle of MATH as possible. One side, call it MATH, of MATH is distinct from MATH and contains a point MATH with MATH. If MATH subtends a segment of MATH of length greater than MATH, construct triangle MATH with base MATH in the same way MATH was constructed. Continue until reaching a triangle MATH with point MATH on a side subtending a segment of MATH of length at most MATH. To show that the sequence of triangles terminates consider the sequence of numbers defined by MATH and MATH. From the construction above it is clear that the base of triangle MATH subtends a segment of MATH of length at most MATH. It is straightforward to show that MATH whence the sequence of triangles stops at MATH for some MATH. Since the third vertex of each MATH lies on the segment of MATH subtended by the base of MATH, we have MATH. The distance from MATH to MATH is at most MATH for some constant MATH. To verify the last assertion of the lemma assume that MATH is a geodesic and set MATH. The points on MATH a distance greater than MATH from MATH form a union of subpaths of MATH not containing MATH or MATH. Let MATH be any such subpath, and write MATH. Observe that MATH starts at MATH, MATH ends at MATH, and each point of MATH is a distance at most MATH from MATH or MATH. It follows that there are two adjacent points MATH on MATH and points MATH on the paths MATH such that MATH. Since MATH is a geodesic, the distance along MATH from MATH to MATH is MATH. But then any point on MATH is a distance at most MATH from MATH. |
math/0010123 | By the Pumping Lemma for context - free languages there exists a constant MATH such that for all MATH each MATH of length MATH contains a subword MATH of length at most MATH which can be replaced by a shorter word MATH to obtain another element of MATH. Note that MATH as all words in MATH represent the same element of MATH. By REF MATH is a rational transduction. Hence so is MATH, the union of MATH over all pairs MATH which occur above for nonterminals of rank REF. Define MATH. MATH consists of all words in MATH which cannot be reduced to other words in MATH by a substitution of the form MATH. Since these reductions are length reducing, MATH contains all words in MATH which are of minimal length among words in MATH representing the same element of MATH. Consequently MATH is still a combing. MATH is regular as it is a difference of regular languages. Let MATH, and consider a derivation MATH. Each nonterminal of rank zero appearing in this derivation derives a subword MATH of MATH,MATH, or MATH. If MATH, then one of MATH can be shortened by a substitution of MATH for MATH contrary to our choice of MATH. |
math/0010128 | To see that REF implies REF , suppose (as we may) that MATH is MATH-equivalent to MATH. Take MATH and let MATH be a MATH-dominated perturbation of MATH. Then MATH is norm bounded; and so, since MATH is equivalent to MATH, we may define a bounded linear operator MATH by setting MATH . Moreover, for MATH, we have, setting MATH, that MATH . Since MATH, by a well-known result, MATH is an isomorphism from MATH onto MATH satisfying MATH for each MATH, as desired. It is obvious that REF implies REF . To see that REF implies the first (and non-trivial) part of REF , first assume that MATH has a weakly NAME subsequence MATH and let MATH be an arbitrary normalized sequence in MATH. Let MATH be the MATH-dominated perturbation of MATH defined by MATH . Then, by hypothesis, MATH is equivalent to MATH. From this, it follows that the linear map MATH given by MATH for each MATH extends to a well-defined linear isomorphism MATH. In particular, the map MATH is weak-weak continuous. Thus, since MATH is weakly NAME, so is MATH. But then we arrive at the contradiction that the arbitrary normalized sequence MATH is weakly NAME. It must be that MATH has no weakly NAME subsequence, whence the result follows by NAME 's MATH-theorem CITE. Towards showing that REF implies REF , let MATH be a normalized sequence in MATH equivalent to MATH. Let MATH be the MATH-dominated perturbation of MATH defined by MATH . Then, by hypothesis, MATH is MATH-equivalent to MATH for some pair of positive numbers MATH. Thus, for any sequence MATH of scalars we have MATH whence, MATH . REF now follows since MATH is equivalent to MATH. |
math/0010128 | First note that if MATH is a semi-normalized sequence in MATH such that MATH for some MATH, then we necessarily have MATH for each MATH. Indeed, this is obvious for MATH, while for MATH and for MATH we have: MATH . Thus, MATH is semi-normalized if both MATH and MATH are, whence the result follows. In particular, REF holds for MATH. Using the triangle inequality and REF , we have, for any sequence of scalars MATH, MATH as desired. |
math/0010128 | We need to show that MATH for any sequence of scalars MATH. The right hand inequality follows from the triangle inequality. Towards proving the left hand inequality, let MATH be given and write MATH where MATH denotes the standard unit vector basis of MATH. Then for each MATH, we have MATH; whence, MATH as desired. |
math/0010128 | To construct the sequence heralded by REF , it suffices to find, for each integer MATH, a basis MATH of MATH such that both of the following statements hold: CASE: MATH is MATH-equivalent to the standard unit vector basis of MATH. CASE: MATH. Towards this end, let MATH and let MATH denote the standard unit vector basis of MATH. Define MATH (we drop the superscripts for notational clarity). Then we have REF , and MATH is a linearly independent sequence with coefficient functionals MATH given by: MATH . Using REF , it is easily verified that MATH is MATH-equivalent to MATH. |
math/0010139 | It follows from CITE that the length is at least REF. Let now MATH denote a bounded unital homomorphism of MATH into MATH then by CITE there exists a *-representation MATH of MATH on MATH and an invertible MATH in MATH such that MATH. Let MATH then it follows that MATH also is a *-representation and we can - and will - replace MATH by MATH and assume that MATH and MATH is a positive and invertible contraction. The weak closure, or the bi-commutant MATH of the unital algebra MATH splits into a sum of an infinite NAME algebra - say MATH - and a finite one - say MATH. If one of the summands is missing then the following computations will all be simplified, so we assume that both MATH and MATH are non trivial. The NAME space MATH splits accordingly via a central projection - say MATH - in the weak closure of MATH such that MATH acts on MATH and MATH acts on MATH. Finally the representation MATH splits via MATH into the sum of an infinite representation - say MATH and a finite one - say MATH. The latter representation is automatically ultra weakly continuous since any ultra weakly continuous functional on MATH factors through the trace on MATH. On the other hand this trace induces a trace on MATH and since this algebra is a factor it has only got a single ( normalized ) trace. Hence the composition of the representation MATH and an ultra weakly continuous functional on MATH has a density with respect to the trace on MATH and therefore is ultra weakly continuous on MATH. In particular this means that MATH is a normal isomorphism of MATH onto MATH. We will define MATH by MATH, then for any unitary MATH in MATH we have MATH. Since MATH multiplication of this inequality from the left and from the right with MATH yields, as in CITE the following inequality MATH . The unitaries in M form a group so we can extend the inequality above to the formally stronger inequality MATH . Since the NAME Density Theorem makes it possible to approximate unitaries in the weak closure MATH of MATH strongly with unitaries from MATH the validity of REF can be extended to MATH . It is hopefully clear that REF above also imply that the norm of MATH is at most MATH and therefore in order to prove that MATH has length REF it is by REF sufficient to prove that for any MATH and any unitary MATH the inequality just below - named (Goal) - holds. MATH . In the rest of the proof we will fix MATH and the unitary MATH. Further we will use the convention that for any operator MATH we will let MATH be given as MATH. It is clearly enough to prove the inequality (Goal) for any vector state on MATH so we will also fix a unit vector MATH, a positive real MATH and verify the inequality (Goal) in the state MATH up to MATH. We will have to divide the computations according to the two representations MATH and MATH. Here we are faced with the problem that MATH does not commute with the central projection MATH, so we will have to replace MATH by one which does. In order to do so we first split MATH as the sum MATH by MATH and MATH. Since MATH is assumed to have property MATH we can by NAME 's result CITE find a set MATH of pairwise orthogonal and equivalent projections in MATH with sum MATH such that all the norms MATH are small. Here the norm MATH is the one induced by the pre NAME space structure on MATH coming from the unique trace state. It is a well known fact that for any ( uniformly ) bounded subset of MATH the ultra strong topology is the same as the one coming from this norm. In particular this means that given the vector MATH in MATH and the fact that the representation MATH is normal we can find the set of projections MATH such that MATH . Having the projections MATH, we will replace h by a positive contraction commuting with a finite dimensional subfactor of MATH which contains MATH in its main diagonal algebra. In order to do so we find a set of matrix units MATH in MATH such that MATH and the set of matrix units generates a subfactor - say MATH - of MATH isomorphic to MATH. In the infinite algebra MATH we find a pair of unital and commuting subfactors MATH and MATH of MATH such that, MATH is isomorphic to MATH and MATH is isomorphic to the NAME algebra tensor product MATH. Finally we will let MATH denote the NAME subalgebra of MATH acting on K given as the sum MATH. This is a NAME algebra of type I and hence injective. Further since MATH is infinite and MATH is finite MATH is also a central projection in this algebra. We can then average over the unitary translates MATH of h with unitaries from MATH and we can find a positive contraction MATH in the commutant MATH of MATH such that MATH . First we remark that since MATH is a central projection in MATH we must have that MATH commutes with MATH and secondly we see from the construction of MATH and REF, that this inequality must hold for MATH too. MATH . The next observations with respect to MATH have to be performed according to the decomposition of the NAME space MATH via MATH and MATH, so we will split REF. Let us define MATH and MATH then we get from REF MATH . MATH . With respect to REF it is quite easy to see that this one extends to unitaries in any matrix algebras over MATH with the same constant MATH because MATH is in the commutant of MATH and MATH is an infinite tensor factor of MATH. Hence we get immediately the following inequality with respect to MATH. MATH . We will now show that we can obtain a similar inequality with respect to MATH. Here we will use an argument which is based on the one NAME uses in the proof of CITE where it is proved that an operator in MATH which is supported on a projection of trace MATH ( with respect to the normalized trace ) can be factored in the way described in REF . We then define partial isometries MATH in MATH. Let MATH denote a set of matrix units for the MATH part of the product MATH, then we can write the the partial isometries MATH as sums of tensors as below and we get MATH . From REF we get since MATH and MATH commutes with MATH and MATH is a contraction that MATH . Since MATH commutes with MATH we can use the equality MATH to write MATH and then transform REF into a set of MATH inequalities regarding operators on MATH by identifying MATH with the subspace of MATH corresponding to the projection MATH. We will now look at each of the terms in the sum above. Hence we identify define vectors MATH in MATH and operators MATH in MATH by MATH . By construction, each operator MATH is a contraction and hence since MATH is a finite NAME algebra there exist unitary operators - say - MATH in MATH such that MATH. REF then yields MATH . The last equality is due to the fact that MATH commutes with all the MATH. Going back to REF we get MATH and MATH so REF show that MATH and consequently MATH is completely bounded and satisfies MATH . Hence the similarity degree is REF and the length is too. |
math/0010139 | Let MATH denote a bounded projection from MATH onto MATH. We will first prove that we may assume that MATH is generated - as a NAME algebra - by a countable set of operators. Suppose that the theorem has been proven for such NAME algebras. Then for a general finite factor as MATH and any NAME subalgebra say MATH of MATH there exists a conditional expectation of norm one MATH of MATH onto MATH coming from the trace by MATH for MATH and MATH, so in particular for each subfactor MATH of MATH with a countable set of generators we have the bounded projection MATH of MATH onto MATH. Moreover it follows nearly immediately from the construction of MATH that the property MATH of MATH is inherited by MATH. Then by assumption there exists a projection of norm one from MATH onto MATH and since the unit ball in MATH is ultra weakly compact we can perform a limit over norm one projections from MATH onto larger and larger subalgebras MATH of MATH and by compactness obtain a net of projections of norm one from MATH onto subalgebras of MATH which converges pointwise ultra weakly to a projection of norm one from MATH onto MATH. Let us now assume that M is generated by a countable set of operators. Then there exists a countable set MATH which is ultra weakly dense in the unit ball of MATH. The result of CITE then implies MATH . The choice of the projections MATH and the density of the sequence MATH imply that we get the following result MATH . In order to be able to use REF we have to make a modification of MATH corresponding to each of the sets MATH. Hence we choose for each MATH a set of matrix units - say MATH - for a subfactor - say MATH - of MATH such that MATH is isomorphic to MATH and MATH . Let MATH denote the group of unitaries in MATH, then this is a compact group and it consequently has a NAME probability measure say MATH. we can now define a projection MATH of MATH onto MATH which is modular with respect to elements from MATH by MATH . We can not prove that these projections are completely bounded but we can construct a sequence MATH of mappings from MATH into MATH which has the property that MATH and MATH ultra strongly for MATH. We define MATH by MATH and prove that MATH. In order to do so we fix a set of matrix units MATH for MATH and define partial isomteries MATH by MATH. Then MATH and MATH . In the following computations we will identify the algebra MATH with MATH in the natural way and hence we get for MATH . Hence, since we are working with the operator norm; MATH . Now for each k we use the properties of MATH and the MATH modularity of MATH to see that MATH . The last sum inside MATH is obtained in MATH via the identification mentioned above and the norm of the sum is dominated by MATH since the sum is nothing but MATH. A combination of this and the results REF give that MATH . The sequence of uniformly bounded mappings MATH of MATH into MATH has a subnet which converges pointwise ultra weakly to a linear mapping say MATH of MATH into MATH. By REF we get that MATH, so MATH is completely bounded. Further we get from REF and the fact that the MATH all are projections onto MATH that MATH. We have then proved that MATH is a completely bounded projection from MATH onto MATH. By CITE it then follows that there exists a completely positive projection of norm one from MATH onto MATH and the proof is completed. |
math/0010139 | As in the case above where the algebra is a complemented subspace of some larger C*-algebra we will like to show first that it is sufficient to prove the result for a NAME factor which is countably generated. So suppose that the result has been established in this case and let MATH denote a continuous REF-cocycle on MATH. Then for any subalgebra MATH of MATH we have - as demonstrated in the proof of REF - a completely positive projection MATH of norm one from MATH onto MATH. This projection is also a MATH-bi-module mapping, so a simple algebraic manipulation shows that the composed bilinear map given by MATH is a continuous REF cocycle on MATH. In the proof below we will show that when MATH is countably generated then this cocycle is the coboundary of a continuous linear mapping MATH which satisfies MATH. If one defines MATH one gets a bounded net of mappings of MATH into MATH indexed by the set of countably generated subalgebras of MATH. We may then find a subnet which converges pointwise ultra weakly to a continuous REF-cochain MATH on MATH and in turn get that MATH is the coboundary of MATH . Let us now suppose that MATH is countably generated and that M acts standardly on MATH with MATH a cyclic and separating unit trace vector for M. The involution induced by MATH is denoted MATH. The start of the proof follows the proof of CITE where the set. The pre dual of MATH is now separable and by NAME 's result CITE there exists an injective subfactor MATH in MATH such that MATH . By CITE we may assume that MATH is multimodular with respect to MATH, separately ultra weakly continuous and vanishes whenever any of the arguments is in MATH. Further by CITE there is a an ultra weakly continuous MATH-bimodular mapping MATH such that the coboundary MATH equals MATH. Further by CITE we can choose MATH such that MATH . Since we are only dealing with REF-cocycles some direct estimates can be made to show that much less than a factor of REF will do as well. By CITE we get a norm estimate for the action of MATH, but in order to understand the inequality below we must say that we use the term MATH on any finite factor to mean REF-norm with respect to the trace state - or the normalized trace on the algebra. Then we can quote CITE as: MATH . We are now in the position to use the MATH-property in a similar way as above. Let MATH be a sequence in the unit ball of MATH which is dense with respect to the MATH topology. For each MATH we choose using CITE a set of pairwise orthogonal and equivalent projections MATH in MATH with sum MATH such that MATH . For each MATH we will modify MATH by a coboundary say MATH which is related to the set of projections MATH in such a way that we can prove that MATH is cohomologous to a REF-cocycle which is completely bounded in the first variable. By CITE this is sufficient in order to see that MATH is a coboundary too. MATH . This MATH is clearly an ultra weakly continuous linear map of MATH into MATH such that MATH, consequently for the coboundary - say MATH - we have MATH. In order to clarify the following computations we introduce some bilinear operators from MATH to MATH by MATH . We can then examine MATH. The fact that MATH is a REF-cocycle is used from the second to the third line just below. MATH . We know that MATH so we can see that MATH is expressed by MATH . This decomposition shows that for a fixed y in MATH and a fixed natural number MATH we get using REF that for MATH and any MATH . It is clear from the construction of MATH in REF that this sequence of bilinear operators on MATH is uniformly bounded and converges pointwise ultra strongly towards MATH. A combination of this with the estimates from REF shows that if we take a subnet of the sequence MATH which converges pointwise ultra weakly to a continuous REF-cochain - say MATH on MATH then REF-cocycle - say MATH on MATH given by MATH is completely bounded in the left variable, and by the methods from CITE we can show that a REF-cocycle which is completely bounded in the left variable is the coboundary of a continuous REF-cochain - say MATH - on MATH which is in the pointwise ultra weakly closed convex hull of the set of continuous REF-cochains of the form below; MATH . We do have MATH and by a combination of REF we get that MATH . We have now proved that if MATH is countably generated any continuous REF-cocycle MATH is inner and we have moreover obtained a universal bound on the cochains implementing MATH so we may conclude that the theorem is proved for a general continuous NAME factor with property MATH. |
math/0010141 | Suppose MATH and MATH immerses in MATH. Then MATH has a thickening (an immersed in MATH regular neighborhood) which is an aspherical MATH-manifold with fundamental group MATH. But then the universal covering action violates REF . |
math/0010141 | The case of MATH was discussed above. For the general case, assume on the contrary that MATH violates the proposition. Define MATH by MATH. The following lemma then implies that MATH classifies into MATH, a contradiction. |
math/0010141 | We may assume that MATH since otherwise MATH is homeomorphic to MATH. Let MATH be a (small) open set homeomorphic to MATH. Consider the diagram MATH . Note that MATH fibers over MATH with fiber MATH; thus MATH and similarly MATH; moreover, inclusion MATH is a homotopy equivalence. Since for MATH the two spaces in the first row of the above diagram are simply-connected, it follows that MATH induces an isomorphism in homotopy groups, and is therefore a homotopy equivalence. |
math/0010141 | Again we first consider the case MATH. Assuming the contrary, consider the homotopy MATH defined by declaring that MATH is the class of parallel rays containing the ray from MATH through MATH. Then MATH covers a classifying map and therefore has degree REF. Let MATH be a Euclidean ball centered at the origin containing MATH and assume that MATH is chosen so that MATH. There is a homotopy MATH of MATH defined by setting MATH be the equivalence class of rays containing the ray from MATH through MATH. Now MATH visibly factors through the projection MATH and therefore by REF in the definition of obstructor complexes the degree of MATH, and hence of MATH, is REF. Contradiction. For the case of a general MATH, replace the definition of MATH by MATH, and replace the ball MATH by a compact set in which MATH can be homotoped to a point. |
math/0010141 | Let MATH be the cycle for MATH. We define MATH to have twice as many elements: for every MATH put MATH and MATH into MATH. It is straightforward to check REF . To verify REF , choose a general position map MATH, and let MATH be a perturbation to a general position map. Put the cone point high above the hyperplane MATH and let MATH be the natural extension of MATH. Then MATH and the claim follows. |
math/0010141 | Let MATH be the cycle for MATH, MATH. We define MATH to have MATH elements: for each MATH and MATH we put the following two pairs in MATH: MATH and MATH. REF is vacuous for MATH as there are no MATH-simplices in MATH. To verify that MATH is a cycle, suppose first that MATH and MATH are disjoint simplices of MATH (with MATH and MATH simplices of MATH and MATH simplices of MATH) and that the sum of their dimensions is MATH. If MATH or if MATH then the corresponding MATH-cell is not a face of any MATH-cells in MATH. So without loss of generality we may assume that MATH and MATH. Since MATH represents a MATH-cell, REF for MATH implies that there is an even number MATH-cells MATH and MATH in MATH that contain MATH. Since MATH and MATH are precisely the MATH-cells in MATH that contain MATH the verification of REF in this case is finished. Now suppose that MATH and MATH are disjoint simplices of MATH and that the sum of their dimensions is MATH. The number of ways of enlarging this cell to a MATH-cell in MATH is either REF (if MATH) or it equals the number of vertices MATH such that MATH, which is even by REF for MATH. Thus REF is verified for MATH. It remains to verify REF . Let MATH be general position maps and MATH the total number of intersection points of MATH-images of unordered pairs of simplices in MATH. View MATH as MATH and MATH as MATH. Perturb MATH to a general position map MATH and let MATH be the linear join of MATH and MATH. The number of intersection points of MATH-images of unordered pairs of simplices in MATH is MATH. The details are left to the reader. |
math/0010141 | The special cases when MATH are clear. Let MATH be a MATH-obstructor complex and MATH a proper, NAME, expanding map. Let MATH be a contractible manifold with a proper metric and MATH a uniformly proper NAME map satisfying the contractibility function requirement. Consider the composition MATH. Now extend MATH inductively over the skeleta of MATH to get a map MATH. Using the contractibility function, we can arrange that the diameter of the image of each simplex of MATH is uniformly bounded. It follows that MATH is a proper expanding map, and therefore MATH by the Linking REF . |
math/0010141 | The latter two statements are obvious, while the first one follows from the Join REF . The product MATH can naturally be viewed as MATH and the product map into MATH satisfies the requirements. (If one of the two groups is REF-ended, use the Cone Lemma instead.) |
math/0010141 | Let MATH be as in the definition. Since MATH is a MATH-set in MATH, there is a homotopy MATH with MATH, MATH and MATH. Restricting to MATH and reparametrizing yields an expanding map MATH. |
math/0010141 | Functions MATH are constructed by induction on MATH, with the case MATH being the definition. The inductive step consists of defining MATH on the boundary of MATH so that the last item above holds and then extending to the interior by coning off from the first vertex. More precisely, if MATH is a linear map with MATH and MATH belongs to the face with MATH then MATH where MATH is the image of MATH under the (partially defined) MATH. Checking the properties listed above is straightforward (by construction, the restriction of the function to a face containing MATH is already the cone on the opposite face). |
math/0010141 | If MATH (or MATH) is finite, then MATH is quasi-isometric to MATH (or MATH) and equality holds. If MATH (or MATH) is REF-ended, we can use MATH (or MATH) in the proof below and appeal to the Cone Lemma. If both MATH and MATH are REF-ended, then MATH is virtually MATH and thus MATH, MATH, so equality again holds. Let MATH and MATH be proper NAME expanding maps defined on the vertices of a fine triangulation of the cones on obstructor complexes MATH and MATH. Define MATH by MATH . CASE: MATH is a proper map. Indeed, let MATH be a sequence in MATH leaving every finite set. If the sequence MATH leaves every finite set, the same is true for MATH. Otherwise, after passing to a subsequence, we may assume that the sequence MATH stays in a finite set MATH. Then MATH stays in the finite set MATH. Since MATH is a proper map, the sequence MATH stays in a finite set, and thus the sequence MATH leaves every finite set. Since MATH is a proper map, we see that the sequence MATH leaves every finite set. CASE: If MATH and MATH are disjoint simplices of MATH, then MATH and MATH diverge. Indeed, let MATH and MATH be sequences in MATH and MATH respectively, leaving every finite set. Note that MATH is a NAME map, so if one of two sequences MATH and MATH leaves every finite set in MATH, then MATH (since MATH and MATH diverge) and consequently MATH . Now assume that both sequences MATH and MATH are contained in a fixed finite set MATH. Then we have MATH . Since MATH and MATH stay in a finite set and MATH it follows that MATH and MATH and the claim is proved. The remaining problem is that MATH is not NAME. CASE: The restriction of MATH to MATH is NAME with the NAME constant independent of MATH. Indeed, let MATH be two adjacent vertices in MATH. MATH and the claim follows from the assumption that MATH is NAME. For every MATH let MATH denote the restriction of MATH to the slice MATH. Recall that MATH is the left translation by MATH and it induces an isometry between MATH (with the MATH-metric) and MATH. CASE: MATH is NAME with respect to the word-metric on MATH (but the NAME constant depends on MATH). In particular, MATH is NAME. Indeed, MATH which is NAME. We next order all vertices of MATH and then extend (simplex-by-simplex) MATH for each MATH to the map MATH using the weak convexity of MATH and REF . Then define MATH . Let MATH be defined as MATH on each MATH. We now note that for MATH and for MATH from REF we have that: CASE: For each MATH there is MATH so that sets of diameter MATH in a simplex of MATH are sent by MATH to sets of diameter MATH. CASE: If MATH are adjacent vertices in MATH then MATH and MATH are MATH-close for any MATH, where MATH is a NAME constant for MATH. For each MATH choose a positive integer MATH so that the simplices of the MATH barycentric subdivision of MATH have diameter MATH for all vertices MATH at distance MATH from MATH. We now define a triangulation of MATH. Start with a decomposition into cells of the form MATH where MATH is a simplex of MATH and MATH is a simplex of the MATH barycentric subdivision of a simplex of MATH with MATH. Now triangulate each such cell inductively on the dimension so that the vertex set of the triangulation is precisely MATH . The restriction of MATH to the vertex set is now a NAME function. CASE: This restriction is still proper and expanding. The proof closely follows proofs of REF . If the sequence MATH (respectively, one of the sequences MATH or MATH) leaves every finite set, the proof is exactly the same as in REF . Otherwise, without loss of generality, the sequence MATH (respectively, sequences MATH and MATH) belong to finitely many slices of the form MATH and therefore lie a bounded distance away from sequences considered in REF coming from the vertices of the original triangulation. The proof follows. |
math/0010142 | We inductively find indices MATH, MATH, as above, open sets MATH and points MATH with MATH and MATH for some index MATH, so that MATH . This will prove the Lemma, for if MATH and MATH for some MATH then, since MATH it will follow that MATH. Since MATH, there is MATH with MATH. Let MATH. Using MATH and the form of the MATH, it follows that there is MATH so that MATH. Now MATH and so there is MATH, an open neighbourhood of MATH, so that MATH. Since MATH, this establishes the base step. For the inductive step, assume we have chosen indices MATH, open subsets of MATH, MATH and points MATH, with MATH and MATH, so that, for MATH, we have MATH . Since MATH, there is MATH so that MATH. Notice that MATH (as in REF ) is in MATH. It follows that there exists MATH such that MATH and so MATH. Hence there exists an open neighbourhood MATH of MATH, contained in MATH, so that MATH . It remains only to show that MATH for all MATH. An element MATH in MATH is of the form MATH for some MATH. Assuming MATH for some MATH, we have MATH completing the induction. |
math/0010142 | Assume that MATH for some MATH-recurrent point MATH. We will find MATH such that MATH has nonzero spectral radius. We may scale MATH so that there exists a relatively compact open neighbourhood MATH of MATH such that MATH for all MATH. Since MATH when MATH, we may also assume that MATH. Since MATH is MATH-recurrent, there exists a sequence MATH which is strictly increasing in the directions of MATH such that MATH. Deleting some initial segment, we may assume that MATH for all MATH. If MATH has all entries going to infinity, then we may apply REF with MATH, to find a strictly increasing sequence MATH such that MATH for all MATH and points MATH such that MATH for all MATH in MATH . If not, (enlarging MATH and passing to a subsequence if necessary) we may assume that the restriction of MATH to MATH takes only finitely many values. Passing to another subsequence, we may further assume that this restriction is constant. Applying REF , we may find a strictly increasing sequence MATH in MATH with MATH and points MATH such that MATH for all MATH in MATH . We may suppose that MATH for all MATH. Thus MATH is an admissible term in the formal power series of an element of MATH. Fix a nonnegative function MATH such that MATH for all MATH and consider MATH . This is an element of MATH since the series converges absolutely. To complete the proof, it suffices to show that the spectral radius of MATH is strictly positive. Note that MATH where MATH is MATH, a nonnegative function satisfying MATH for all MATH. Thus each NAME coefficient MATH of MATH is a finite sum of nonnegative functions, and hence its norm dominates the (supremum) norm of each summand. Since MATH, it suffices to find MATH such that for each MATH there exists MATH such that the norm of some summand of MATH exceeds MATH. If we let MATH, then trivially MATH is a term in MATH. In the next product, MATH, we have the term MATH . Generally, one term in the expansion of MATH is MATH . Claim. If MATH and MATH, then MATH where MATH is as in the definition of MATH and the product is over all MATH in MATH. Proof of Claim. For MATH, the claim holds trivially as MATH. Assuming the claim is true for some MATH, we have MATH for MATH in MATH, proving the claim. Recall that for each MATH there exists MATH such that MATH for all MATH. Since MATH, we have MATH where MATH ranges over MATH and hence MATH. From the claim, it follows that MATH and so, by the earlier remarks, MATH . Thus the proof will be complete if we show that MATH or equivalently MATH for all MATH. Setting MATH, the recurrence relation for MATH becomes MATH and MATH, which has solution MATH. |
math/0010142 | Let MATH be arbitrary and MATH. Then MATH which is zero since MATH is supported on MATH and MATH is supported on the disjoint set MATH. This shows that all NAME coefficients of MATH will vanish, and hence MATH. It follows that all products MATH vanish and hence MATH. On the other hand, choosing functions MATH equal to MATH on MATH and MATH equal to MATH on MATH, we find MATH, so the ideal MATH is nonzero. |
math/0010142 | Let MATH be a relatively compact open set. We wish to find a MATH-recurrent point in MATH. Since MATH is not MATH-wandering, there exists MATH such that MATH. Hence there is a nonempty, relatively compact, open set MATH with MATH such that MATH. Since MATH contains no MATH-wandering subsets, a similar argument shows that there exists MATH such that MATH and the MATH-th entry of MATH is greater than that of MATH for every MATH. Inductively one obtains a sequence of open sets MATH and MATH strictly increasing in the directions of MATH with MATH and MATH all contained in the compact metrisable space MATH. It follows from NAME 's theorem that the intersection MATH is a singleton, say MATH. Since MATH we have MATH for all MATH and so MATH; hence MATH. |
math/0010142 | If the strongly recurrent points are dense in MATH, then by REF there are no nonzero monomials in the NAME radical of MATH. But we have already observed that an element MATH is in the NAME radical if and only if each monomial MATH is. Thus MATH is semisimple and hence semiprime. Suppose that MATH is semiprime. Then REF shows that there are no nonempty MATH-wandering open sets for MATH . Thus, by REF , the strongly recurrent points are dense. |
math/0010142 | CASE: To see that MATH, note that if MATH then for every neighbourhood MATH of MATH (in MATH) the set MATH is a neighbourhood of MATH in the relative topology of MATH, so there exists MATH such that MATH. Thus MATH showing that MATH. On the other hand if MATH then for each relative neighbourhood MATH of MATH, since MATH is a neighbourhood of MATH in MATH there exists MATH such that MATH. Since MATH and MATH is invariant, MATH establishing REF . CASE: Given a closed invariant set MATH, if MATH has no MATH-wandering points, then MATH is dense in MATH by REF , and hence MATH. On the other hand, MATH clearly has no MATH-wandering open sets. |
math/0010142 | As a MATH-recurrent point cannot be MATH-wandering, MATH. If MATH for some MATH, then by REF the set MATH of MATH-recurrent points of the subsystem MATH equals MATH, so MATH; but MATH, and so MATH. Finally, if MATH is a limit ordinal and we assume that MATH for all MATH then MATH. This shows that MATH and so MATH since the sets MATH are closed. But on the other hand, if MATH is the depth of the strong MATH-centre we have MATH, a closed invariant set. Since MATH, the dynamical system MATH can have no MATH-wandering points. Thus it follows from REF that MATH and hence equality holds. |
math/0010142 | If MATH is the support of MATH then MATH; in other words the compact set MATH consists of MATH-wandering points for MATH. This means that each MATH has an open neighbourhood MATH so that the (relatively open) set MATH is MATH-wandering for MATH. Each MATH has an open neighbourhood MATH such that MATH is empty (and so MATH-wandering). The family MATH is an open cover for MATH. Thus, there is a partition of unity for MATH, that is, a finite subcover, MATH, and functions MATH, MATH, with MATH a compact subset of MATH, so that MATH. |
math/0010142 | Since MATH is dense in MATH, it suffices to prove that any MATH is contained in MATH. Suppose MATH . By REF we may write MATH as a finite sum MATH where each MATH is supported on a compact set which is MATH-wandering. Since MATH for some MATH as observed above, by REF we have MATH and so MATH. Thus MATH. Suppose the result has been proved for all ordinals less than some MATH. Let MATH be a limit ordinal. If MATH, we have MATH, hence MATH can be covered by finitely many of the MATH, hence (since they are decreasing) by one of them. Thus MATH has compact support contained in some MATH REF and so MATH. Therefore MATH by the induction hypothesis. Now suppose that MATH is a successor, MATH. By REF , we may write MATH where the support of MATH is compact and contained in an open set MATH such that MATH is MATH-wandering for MATH, that is, MATH when MATH. This can easily be seen to imply MATH. Let MATH be arbitrary. Writing MATH as above, it follows as in the proof of REF that for each MATH all NAME coefficients of MATH are supported in MATH (for some MATH) which is contained in MATH by the previous paragraph. Thus MATH. By the induction hypothesis, MATH must be contained in MATH. Thus MATH is quasinilpotent, hence so is MATH (by the spectral mapping theorem). Since MATH is arbitrary, it follows that MATH for each MATH, so that MATH. Finally, we suppose that MATH is closed. Then the argument above can be repeated exactly up to the previous paragraph, changing MATH to MATH. The previous paragraph can be replaced by the following argument. Thus MATH. By the induction hypothesis, MATH must be contained in MATH. Thus all products MATH are in MATH and so the (possibly non-closed) ideal MATH generated by MATH satisfies MATH. For every prime ideal MATH, we have MATH and so MATH. Hence MATH, and therefore MATH for each MATH, so that MATH. |
math/0010142 | Let MATH be a monomial contained in MATH and let MATH be the support of MATH. Then REF shows that MATH must vanish on MATH. On the other hand, let MATH be as in the statement of the Theorem, so that MATH vanishes on MATH (where MATH). We will show that MATH is in MATH. It is enough to suppose that the support MATH of MATH is compact. Since MATH is contained in MATH, it is contained in finitely many, hence one, MATH. It follows by REF that MATH. In the final statement of the theorem, one direction is obvious. For the other, suppose MATH is closed. Then by the final statement of REF , we have MATH. |
math/0010148 | We re-write MATH. We shall find the values MATH such that MATH. Since MATH, we assume MATH. Applying NAME 's Theorem repeatedly for the base MATH representations (MATH), MATH and MATH, we obtain MATH . For MATH, we get MATH . Let MATH. Since for any MATH, MATH, if MATH, then MATH. Thus, the numbers MATH with MATH are those positive integers MATH with MATH or MATH. From MATH, MATH, and MATH, we obtain MATH. Thus, MATH. Therefore, if MATH is prime, MATH for any number MATH. Using NAME 's Theorem we easily see that for MATH, the least residue of MATH modulo MATH is MATH. |
math/0010148 | We use the identity MATH. Using REF we get that the power of MATH dividing MATH is MATH . Let MATH. Thus, REF becomes MATH since MATH. |
math/0010148 | As before MATH. A number MATH, which does not satisfy the divisibility, must satisfy (see REF ) MATH which implies MATH. Assume first that MATH. Therefore, MATH, therefore MATH must be even, say MATH, so MATH . Assume now that MATH, and MATH. It follows that MATH, MATH. Therefore, MATH and since MATH, we get that MATH. Then, MATH, MATH. It follows that MATH so MATH must be divisible by MATH. Since MATH, we see that MATH must be an empty sum. Therefore, all MATH. We obtain MATH, and the first claim is proved. Let MATH. Consider MATH. It follows that MATH and MATH attaches to this string the block MATH to the right, so it is of the same form. Since MATH and MATH, except for MATH, we get, using NAME 's theorem, MATH . Now, MATH, MATH and MATH, implies MATH and MATH. Thus, REF becomes MATH, since MATH. Consider MATH, MATH and MATH. Observe that MATH. It follows that MATH . But MATH attaches the block MATH to the right of the base MATH representation of MATH, and since there is a carry in this case, we get MATH. Also, MATH, MATH, MATH except for MATH and MATH, for MATH. Now, applying NAME 's theorem we get MATH since MATH. We prove that the last expression is the multinomial coefficient (this was observed by one of our referees, whom we thank). First, assume MATH. Since MATH in this case, we get MATH and the claim is trivially satisfied. Let MATH. We observe that MATH since in the last sum MATH for MATH. Therefore, MATH . Taking the product MATH, since MATH, which replaced in REF produces the claim. The theorem is proved. |
math/0010148 | Straightforward. |
math/0010148 | By REF , if MATH, then MATH so MATH . If MATH, then MATH, for some MATH. Working modulo MATH implies MATH. Thus, MATH. Assume MATH . We obtain MATH. Modulo MATH, this transforms into MATH which will imply MATH, for some MATH. Since MATH, we get MATH. We obtain, for MATH, MATH, MATH with the condition MATH. |
math/0010148 | First we prove that MATH. By squaring we get MATH which is equivalent to MATH . The last inequality is equivalent to MATH, which is certainly true if MATH. Now, let MATH. We evaluate MATH . Therefore, MATH . Taking MATH, in REF, we get MATH . If MATH, then REF implies our claim that REF is true for MATH sufficiently large, since, by REF, the left side is MATH and the right side is MATH. |
math/0010148 | If MATH, REF , valid for MATH changes into MATH which is certainly true for MATH. REF imply that the exceptions (if they exist) are of the form MATH. Therefore, since the number of pairs MATH, giving different numbers of the above form, is less than MATH, we get the theorem. |
math/0010148 | If MATH, REF , valid for MATH, changes into MATH which is true for MATH. As in the previous proof, we get that the exceptions (if they exist) are of the form MATH. Therefore, since the number of triples MATH, giving different numbers of the above form, is less than MATH, we get the result. |
math/0010149 | We evaluate MATH . If MATH odd, then associating MATH, we get MATH . If MATH even, then then associating MATH, except for the middle term, we get MATH . |
math/0010149 | We use REF . The first two identities are straightforward. Now, MATH since MATH and MATH. |
math/0010149 | We evaluate MATH . Assume MATH odd. Then, associating MATH, we get MATH . Assume MATH even. Then, as before, associating MATH, except for the middle term, we get MATH . |
math/0010149 | Let MATH . If MATH, then MATH, and MATH . |
math/0010149 | Straightforward using the NAME formula for MATH and MATH. |
math/0010149 | We use REF . Associating MATH, except for the middle term in MATH, we obtain MATH . We did not insert the middle term, since it is equal to MATH . Assume first that MATH is odd. Using REF into REF, and observing that MATH, we get MATH . Assume MATH even. As before, MATH . In the same way, associating MATH, except for the middle term, MATH . |
math/0010149 | The second, third and fifth identities follow from the previous theorem. Now, using REF , with MATH, we get MATH . Furthermore, the fourth identity follows from MATH since MATH. |
math/0010149 | We use MATH in REF . Associating MATH in MATH, we obtain MATH since MATH and MATH, by REF . In the same way, associating MATH, with the middle term zero, MATH since MATH and MATH, by REF . Therefore, for MATH even, MATH and for MATH odd, MATH . |
math/0010149 | The first identity is simple application of REF . The identities for even powers are consequences of REF . Now, using REF , we get MATH since MATH. |
math/0010153 | As we mentioned the simplicial relations are already held and it remains to check the following extra relations: MATH . To prove REF , we first compute MATH : MATH . By a similar argument, we can deduce MATH . Continuing, MATH and eventually, MATH . We leave it to the reader to check the remaining equations. To prove the converse it suffices to use just MATH. |
math/0010153 | We should verify that MATH commutes with MATH . For MATH it is obvious that MATH commutes with MATH . For MATH we have MATH where we have made use of the MATH-trace property of MATH. We leave it to the reader to check MATH. Finally, we show that MATH commutes with MATH: MATH and MATH . |
math/0010153 | We must show that MATH commutes with MATH and MATH. In this proof we show MATH and leave the other cases to the reader. At first, let MATH, so MATH. Therefore MATH. Now, MATH so that MATH . Next, we have MATH . On the other hand, MATH . |
math/0010153 | It can be shown that MATH. |
math/0010153 | As always it is needed to verify the relations REF,.,REF We only check REF and leave the others to the reader. MATH . By a similar argument we get, MATH and finally we have MATH . |
math/0010153 | We leave it to the reader the first part of proof and just prove the second part. We must verify that the following diagram is commutative. MATH . We have MATH and MATH . |
math/0010153 | By the above remark we have MATH where MATH denotes cyclic double complex MATH. Since MATH is contractible and MATH is a Homotopy equivalence the double complex MATH is a resolution for MATH, where MATH is MATH . On the other hand, MATH . So, to complete the proof it suffices to compute MATH . But by finding a NAME resolution for MATH, we have MATH . |
math/0010153 | If MATH then it is obvious that MATH is a simplicial module by the following faces and degeneracies: MATH . Now, MATH can be computed by the above complex. Let MATH . We leave to the reader to show MATH is a simplicial map and in fact, MATH is a simplicial isomorphism by the following inverse map MATH . |
math/0010155 | Let MATH and MATH be two mutually independent sequences of NAME variables (that is, each is complex-valued and uniformly distributed on the unit circle). By applying the unconditionality of the NAME and the NAME inequality it is sufficient to show the existence of a constant MATH so that for any MATH we have: MATH . To see this we define MATH for MATH by MATH and MATH . Let MATH . Then MATH is an analytic martingale and so for a suitable constant MATH depending only on MATH we have: MATH . This yields the desired inequality. |
math/0010155 | CASE: We use the remark that it is enough to establish REF for distinct operators MATH . If MATH and MATH then MATH and hence MATH . This proves REF and indeed a rather stronger result. CASE: Let MATH where MATH and MATH is a finitely nonzero collection of complex numbers with MATH . Suppose MATH . We first note that MATH . We will also use the fact REF that there is a constant MATH depending only on MATH so that for MATH we have from REF, MATH . Hence MATH . This proves REF . CASE: Let MATH where MATH is a finitely nonzero matrix with MATH . In this case if MATH and MATH we note that: CASE: We use the proof of REF . This time we again use the fact it suffices to consider the operators without repetition. So we consider MATH and repeat the proof of REF with MATH if MATH and MATH otherwise and replace each MATH by the identity. Using REF in place of REF gives the desired conclusion. |
math/0010155 | We give the proof in the NAME case, the others being similar. We first make the observation that it suffices to consider the case when MATH as one can make the transformation MATH . In this case we have the formula MATH . We write MATH and MATH so that MATH . Note that for a suitable constant MATH we have an estimate MATH whenever MATH . Now suppose MATH . Suppose MATH . Let us suppose that MATH are chosen so that MATH . We have: MATH for a suitable constant MATH . A similar argument can be done for MATH . |
math/0010155 | We can assume MATH . For a suitable constants MATH and MATH we have MATH and the last quantity is finite. For the last part observe that for any bounded sequence MATH and MATH the series MATH must converge to MATH where MATH . |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.