paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0007007 | To check that MATH is splitting rigid, we need to start with an arbitrary rank MATH vector bundle MATH over MATH that satisfies (MATH), and prove that MATH virtually comes from MATH. Without loss of generality, we can pass to a finite cover to assume that MATH is the total space of a vector bundle MATH, which is the pr... |
math/0007007 | Let MATH be a self-homotopy equivalence of MATH with MATH, so that MATH, and MATH is a derivation of MATH of degree MATH. By REF , it suffices to show that MATH, for all MATH. Let MATH be a self-map of MATH defined by MATH and for MATH, MATH. The fact that MATH is a derivation and MATH implies that MATH is a homomorphi... |
math/0007007 | Let MATH. Since MATH and MATH, MATH is a rational multiple of MATH. Choose a positive integer MATH such that MATH but MATH (which exists since MATH is finite-dimensional). Since MATH is even, we get MATH so that MATH. |
math/0007007 | By REF , it is enough to show that MATH is splitting rigid. Then by REF , it suffices to show that any negative derivation of MATH vanishes on MATH. Since MATH is simply connected, negative derivations of degree MATH automatically vanish on MATH for degree reasons. Since MATH is finite dimensional, REF implies that all... |
math/0007007 | Let MATH be a basis of the vector space MATH. Given a derivation MATH, define linear self-maps MATH of MATH by MATH. It is routine to check that MATH, and the correspondence MATH gives the promised isomorphism. |
math/0007007 | If MATH is even, both the fiber and the base of the MATH-fibration MATH belong to MATH, and hence MATH by CITE. Thus, we are done by REF . Suppose that MATH is odd. Since the NAME class of MATH is trivial, the NAME spectral sequence of the fibration collapses at the MATH-term, so MATH and MATH are isomorphic, as MATH-m... |
math/0007007 | If MATH is even, then MATH so that MATH by CITE, and we are done by REF . If the NAME class of MATH vanishes, then the result follows by REF . Thus, we can assume that MATH is odd and MATH has nonzero NAME class. It follows from the NAME sequence that MATH is onto, so MATH is generated in dimension MATH. Again, we see ... |
math/0007007 | We begin by constructing the canonical model of the reference fibration MATH. Since this fibration is induced by the diagonal map MATH, it follows that MATH is the diagonal embedding MATH. Consider the map MATH. It is well-known that MATH is rationally homotopy equivalent to MATH and the minimal model of MATH is isomor... |
math/0007007 | By REF , we only have to consider the case when MATH or MATH. By REF it is enough to show that MATH is splitting rigid for any MATH and MATH or MATH. Since MATH and according to REF , to insure splitting rigidity it is enough to show that all negative derivations of MATH vanish on MATH. By passing to a finite cover we ... |
math/0007007 | First, we show that MATH is generated in dimension MATH. Consider the NAME model MATH. It admits a natural grading by the wordlength in MATH's given by MATH. Since the differential decreases the wordlength in MATH's by MATH, this grading induces a natural grading in the cohomology MATH; this is the so-called lower grad... |
math/0007007 | First of all, note that all known even dimensional positively curved manifolds belong to MATH. Indeed, it follows from the classification theorem of positively curved homogeneous spaces CITE that any even dimensional homogeneous space belongs to MATH for some MATH. The only known example of a positively curved even-dim... |
math/0007007 | We only give a proof for the NAME manifolds; the NAME manifolds are treated similarly. Let MATH be a NAME manifold. From the homotopy sequence of the fibration MATH one easily sees that MATH has the same rational homotopy groups as MATH. Let MATH be the minimal model of MATH. It is well-known (for example, see CITE) th... |
math/0007007 | Let MATH be a derivation of MATH that induces MATH. The fact MATH is a derivation implies that the map MATH defined by MATH is a NAME. Being a NAME, MATH defines a (unique up to homotopy) map of rationalizations MATH where we can choose MATH so that MATH. Look at the diagram MATH where MATH is the rationalization, and ... |
math/0007007 | Arguing by contradiction, let MATH be a negative derivation of MATH with MATH such that MATH does not vanish on MATH. First, we show that MATH is nonzero on either NAME or NAME class of a rank MATH bundle MATH over MATH. Assume, say that MATH so that MATH. If MATH satisfies MATH, then by REF , MATH is proportional to t... |
math/0007007 | As in the proof of REF , use MATH to construct a self-homotopy equivalence MATH of MATH. If MATH is nonzero on MATH for some MATH, then MATH is nonzero on the MATH-th NAME class of some bundle MATH over MATH. Then for some large MATH, the map MATH is homotopic to a smooth embedding. By the same argument as in the proof... |
math/0007007 | By a surgery argument as in CITE, there is a closed smooth manifold MATH which is homotopy equivalent to MATH and has MATH with MATH. To check that MATH is splitting rigid, we need to start with an arbitrary rank MATH vector bundle MATH over MATH that satisfies (MATH), and prove that MATH virtually comes from MATH, or ... |
math/0007007 | Passing to a finite cover, we can assume that MATH is orientable, and that MATH virtually comes from MATH, where MATH is a diffeomorphism. As before, to show that MATH virtually comes from MATH it is enough to check that its rational characteristic classes lie in MATH. Proceeding exactly as in the proof of REF , we con... |
math/0007007 | Let MATH, MATH, MATH, MATH, MATH so that MATH. Look at the following commutative diagram whose rows are the smooth and the topological surgery exact sequences: MATH . First, note that MATH is onto. Indeed, by the NAME duality with MATH-theory coefficients MATH, so it suffices to show that the homology assembly map (see... |
math/0007007 | We only give a proof for MATH; the even case is similar. Let MATH be the universal MATH-bundle over MATH. We think of MATH as a map MATH. It is well-known that the map MATH is a rational homotopy equivalence, and thus the homotopy groups of its homotopy fiber MATH are torsion. Similarly, consider the map MATH and try t... |
math/0007008 | Let MATH be an NAME - NAME set, and let MATH be in the NAME class, and of exponential type MATH. First we suppose that MATH, MATH. Applying the NAME - NAME principle to the function MATH with MATH, in the upper and in the lower half-planes, we conclude that MATH is non-positive everywhere in MATH, and therefore, MATH i... |
math/0007008 | We are to prove that the function MATH is of bounded type in MATH. Multiplying, if necessary, MATH by a polynomial with real zeros, we obtain MATH . Without loss of generality, we may assume that MATH are real. Furthermore, MATH where MATH, MATH, and MATH . It suffices to verify that each MATH in REF is a function of b... |
math/0007008 | To obtain the result in one direction, we prove that MATH . First, observe that our assumptions on MATH imply that MATH . Indeed, take a sequence of points MATH tending to MATH sufficiently rapidly (for example, MATH suffices), and consider the entire function MATH. Then MATH is in the NAME class, and MATH as MATH. By ... |
math/0007008 | CASE: Consider a set of disjoint intervals MATH, such that MATH, MATH, MATH, and MATH for MATH. We define a weight MATH as follows: MATH, MATH, and MATH elsewhere. By Theorem D, the polynomials are not dense in MATH. Indeed, no entire function MATH of zero exponential type, with real zeros MATH, satisfies the condition... |
math/0007008 | Suppose that for a sequence MATH we have MATH . By NAME 's inequality, for some positive MATH independent of MATH, MATH . For some positive MATH, consider the function MATH harmonic on MATH. Then MATH . Next, we use the following fact (compare REF): MATH . For the sake of completeness, we give an argument from CITE. Si... |
math/0007008 | CASE: For MATH denote by MATH the square MATH. In what follows we use a function MATH; it is positive and harmonic in MATH, vanishes on MATH, and MATH, MATH. The function MATH vanishes on a closed set MATH, MATH, for some MATH independent of MATH. Furthermore, MATH is non-negative on MATH, and harmonic on MATH. We esti... |
math/0007008 | As in the previous proof, let us consider the NAME function MATH for MATH. It is positive, harmonic and bounded in MATH. Denote MATH . Applying the NAME formula in the lower half-plane, we get MATH . By NAME 's inequality, at least on one half of the length of the interval MATH, we have MATH. Therefore, for some MATH, ... |
math/0007009 | CASE: From the presentation of MATH and REF , there is a unique morphism MATH such that MATH when MATH and MATH are cells. The formula then holds for a general product MATH because it is a composite of cells. CASE: One can check bifunctoriality by considering the values of the appropriate morphisms on generators. |
math/0007009 | Let MATH be the operation underlying MATH in dimension MATH. The operations underlying MATH, MATH and MATH in dimension MATH are then given by MATH . One finds that MATH, from which it follows that MATH and then MATH. One also finds that MATH for any cell MATH in MATH. For a cell MATH in MATH it follows that MATH and M... |
math/0007009 | CASE: These relations are straightforward consequences of the definitions. CASE: Since MATH, we have MATH. From REF , if MATH then MATH . Also from REF , if MATH then MATH . From REF , if MATH then MATH . It now follows that MATH using REF . CASE: From REF , if MATH then MATH . Also from REF , if MATH then MATH . It no... |
math/0007009 | From REF , if MATH then MATH and MATH are in MATH. To complete the proof, suppose that MATH and MATH are in MATH; it suffices to show that MATH. Now, MATH so that the MATH are identities for MATH. It follows that MATH as required. This completes the proof. |
math/0007009 | Since MATH, it follows from REF that MATH . Conversely, suppose that MATH for MATH and MATH; it suffices to show that MATH, and for this it suffices to show that MATH for MATH. But MATH for MATH, so that MATH by REF . This completes the proof. |
math/0007009 | By REF , MATH for some MATH. It follows that MATH so that MATH and MATH as required. |
math/0007009 | We use the characterisation in REF . We first show that the family is closed under MATH. Indeed, if MATH then MATH since MATH. Next we show that the family is closed under MATH. Indeed, if MATH, then MATH trivially, and for MATH we have MATH . It now follows from REF that the family is closed under MATH for all MATH. S... |
math/0007009 | We first show that for a fixed value of MATH the given structure maps MATH and MATH make MATH into a MATH-category. By REF , MATH is closed under the structure maps for MATH, and the same result holds trivially for MATH. From the identities in REF, if MATH then the triple MATH makes MATH into the morphism set of a cate... |
math/0007009 | We have already shown that MATH is a bijection, and it is clearly natural. It remains to show that MATH is a homomorphism. It suffices to show that MATH is a homomorphism for each MATH; in other words, we must show that MATH for MATH and that MATH for MATH a composite in MATH. Suppose that MATH and MATH. Noting that MA... |
math/0007009 | This follows from the identities in REF. |
math/0007009 | Recall the matrix notation used for certain composites: if MATH and MATH are equal by the interchange law, then we will write MATH for the common value. In such a matrix, we write MATH for elements in the image of MATH (which are the identities for MATH), and we write MATH for elements in the image of MATH (which are t... |
math/0007009 | If MATH then MATH since MATH is idempotent by REF . For MATH, it follows from REF that MATH . For MATH, it now follows that MATH as required. |
math/0007009 | From the definitions we get MATH and we similarly get MATH. |
math/0007009 | The first of these results was given in REF . The second result then follows from REF : indeed, we get MATH as required. |
math/0007009 | Note that we have MATH for MATH to be defined. The proof for the cases MATH and MATH consists in evaluating in two ways each of the matrices MATH Note that MATH and MATH are identities for MATH because MATH. The other case follows from the identities in REF. |
math/0007009 | We use relations from REF . For MATH we have MATH or MATH or MATH. For MATH we have MATH or MATH or MATH or MATH. For MATH we have MATH or MATH or MATH. For MATH we have MATH or MATH or MATH. |
math/0007009 | If MATH then MATH with MATH a generalised MATH, and the result is immediate from REF . Now suppose that MATH. Then MATH with MATH a generalised MATH and with MATH a generalised MATH. By REF MATH with MATH a generalised MATH. By REF , this is a composite of factors of the form MATH with MATH and with MATH a generalised ... |
math/0007009 | This follows from REF . |
math/0007009 | By considering the first place where MATH and MATH differ, we see that MATH for some MATH and MATH such that MATH, with MATH or MATH and with MATH a generalised MATH. Since MATH, it follows from REF that MATH since MATH, it follows from REF that MATH is a composite of factors MATH with MATH a generalised MATH. By REF ,... |
math/0007009 | The result for MATH comes from REF by iterated application of REF ; recall from REF that MATH is a composite of MATH and MATH. Now suppose that MATH. By REF MATH . From REF , this is a composite of factors MATH with MATH a generalised MATH and with MATH a MATH-fold product of face operators. By repeated application of ... |
math/0007009 | We first show that the values prescribed for the MATH really define a homomorphism on MATH; in other words, we must show that they respect the relations given in REF . Let MATH be a MATH-dimensional cell in MATH. We must show that MATH; if MATH we must also show that MATH is the appropriate composite of the MATH, where... |
math/0007009 | Consider the composite MATH . By REF , MATH is an isomorphism; it therefore suffices to show that the composite MATH is the identity. This amounts to showing that MATH for MATH. Now, from the definitions of MATH and MATH, we find that MATH since MATH and MATH is idempotent REF , it follows that MATH as required. This c... |
math/0007009 | Let MATH be a MATH-shell. Guided by REF , we let MATH be the MATH-tuple MATH such that MATH . From REF and the identities in REF, it is straightforward to check that MATH is a well-defined function from MATH to itself, and it is easy to see that MATH. |
math/0007009 | This amounts to showing that MATH is a bijection for each MATH in MATH. Since MATH is a composite of operators MATH, it suffices to show that MATH is a bijection for each MATH in MATH. Given MATH, it is straightforward to check that there is a composite MATH and that MATH. We will carry out the proof by showing that MA... |
math/0007009 | By REF , MATH induces isomorphisms from MATH to MATH. Since MATH is the identity operation, MATH induces a bijection from MATH to MATH. By an inductive argument using REF , MATH induces a bijection from MATH to MATH for all MATH. Therefore MATH is an isomorphism. |
math/0007009 | Suppose that MATH is a composite of elements of the forms MATH and MATH. Then MATH is in the image of MATH by REF , so x is thin by REF . Conversely, suppose that MATH is thin. It follows from the proof of REF that MATH is a composite of MATH with elements of the forms MATH and MATH. By REF , MATH is in the image of MA... |
math/0007009 | CASE: In dimension MATH there is by adjointness a natural bijection MATH . These bijections combine to form the natural isomorphism REF of cubical MATH-categories because, on both sides, the cubical MATH-category structures are induced by the corresponding operators MATH, etc. in MATH. CASE: This isomorphism may be pro... |
math/0007009 | The functor MATH is left adjoint to MATH, and this is what the proposition says in dimension REF. In dimension MATH we have a natural bijection MATH and these bijections are compatible with the cubical operators. |
math/0007009 | For any cubical MATH-category MATH, there are natural isomorphisms of cubical sets MATH . |
math/0007009 | CASE: There are natural bijections MATH . CASE: This is a special case of REF . CASE: It follows from REF that MATH has left adjoint MATH. But the obvious isomorphism MATH induces an isomorphism MATH, so MATH is naturally isomorphic to MATH. The result now follows from REF . |
math/0007010 | For MATH providing REF . Multiplying by MATH we get MATH . When we take convex combinations, the first two summands cancel out, so the lemma follows from the inequality MATH for positive invertible operators MATH and MATH. MATH . |
math/0007010 | Let MATH be finite dimensional and MATH. By hypothesis and REF there exists MATH such that MATH. Thus by REF MATH . MATH . |
math/0007010 | The inequality MATH follows by monotonicity. To prove the opposite inequality we consider MATH in its NAME with respect to MATH, so we may assume MATH and MATH are NAME algebras and MATH and MATH normal. Let MATH be as before. Then MATH, hence by REF MATH. Thus by REF MATH . It follows from REF that MATH. But this ineq... |
math/0007010 | As pointed out in REF MATH is conjugate to the infinite tensor product of the diagonals MATH considered in REF. Thus there exists an increasing sequence MATH of full matrix algebras with union dense in MATH such that MATH, where MATH, MATH. Then MATH is of type I and contains MATH. By REF MATH. Since MATH is weakly den... |
math/0007010 | Let MATH, where MATH, and let MATH be REF-shift on MATH, and MATH the unique tracial state, see REF . Then MATH. Let MATH (see REF for the detailed definition), and let MATH be the unitary operator in MATH which implements MATH. By CITE MATH is a MATH algebra, and by monotonicity MATH. However, if MATH belongs to a cir... |
math/0007010 | It suffices to show REF . Let MATH. By REF there exists MATH such that if MATH satisfies MATH then MATH. By hypothesis on MATH there exists therefore MATH such that if MATH then MATH. This also implies that if MATH then MATH. Put MATH. Then there exists MATH with MATH and MATH for MATH. Hence by REF, MATH so that MATH,... |
math/0007010 | Let MATH denote the C*-subalgebra of MATH generated by the symmetries MATH, MATH. Then MATH is abelian, and MATH vanishes on each MATH. Thus the C*-dynamical system MATH is isomorphic to REF-shift, hence has entropy MATH, hence by monotonicity MATH. Thus by REF , and the inequality preceeding the lemma MATH . The conve... |
math/0007010 | Use the affinity of the function MATH. Let MATH be asymptotically abelian with locality as before and MATH a self-adjoint local operator. Put MATH . Then MATH is a derivation on MATH and defines a one-parameter group MATH on MATH. Let MATH. We say a MATH-invariant state MATH is an equilibrium state at MATH at inverse t... |
math/0007023 | In fact, supppose that MATH is globally generated. Then MATH is likewise globally generated and hence nef. Therefore MATH. The second assertion is proven similarly. |
math/0007023 | We keep the notation introduced in REF . It is evident that MATH and it follows from the conormal bundle sequence MATH that MATH . The assertion is then a consequence of REF . |
math/0007023 | We will apply REF . Thus for REF , let MATH be a surjective mapping from an irreducible variety MATH which dominates the blowings-up of MATH along MATH, MATH and MATH. Thus MATH carries effective NAME divisors MATH and MATH characterized by MATH . Note that then MATH . Write MATH and MATH. Then MATH and MATH are nef on... |
math/0007023 | One checks right away as in REF , that MATH for every MATH. The stated inclusions then follow from the NAME - NAME theorem (compare CITE). |
math/0007023 | Consider the classes MATH in the vector space of numerical equivalence classes on MATH with real coefficients. Thus MATH is a nef class - so in particular MATH and MATH for all MATH and MATH - and MATH. Arguing as in the proof of REF, one then finds that MATH as required. |
math/0007023 | It is enough to show that every associated subvariety is distinguished. To this end, let MATH be the sheaf of all functions on MATH whose pull-backs to MATH vanish to order MATH along the NAME divisor MATH. Then MATH is a primary ideal, and one has MATH . Since MATH is integrally closed this means that we have the (pos... |
math/0007023 | Set MATH and MATH. Note to begin with that MATH for all MATH, from which it follows that the limit MATH exists. Call this limit MATH. We will prove the theorem by establishing (from right to left) the inequalities MATH . Starting with the right-most inequality in REF , recall that if MATH is MATH-regular with respect t... |
math/0007023 | Since MATH is ample for MATH, it follows from NAME vanishing that MATH . The isomorphism on global cohomology groups is then a consequence of REF thanks to the NAME spectral sequence. As for REF , the assertion is local on MATH, so we may assume that MATH is affine. Choosing generators MATH gives rise to a surjection M... |
math/0007023 | The essential point is to show that if MATH is an equidimensional MATH-module without embedded components, then the restriction MATH of MATH to MATH is also equidimensional without embedded components (see CITE for an argument in a similar setting). Once one knows this, one can deduce the lemma from the fact REF , that... |
math/0007023 | Let MATH be a general divisor linearly equivalent to MATH, and consider the restriction MATH of MATH to MATH. According to REF there are only finitely many prime ideals which appear as associated primes for any of the ideals MATH for MATH. So we may assume that MATH does not contain any of these primes, so that the seq... |
math/0007026 | First notice that MATH, so the values and probabilities in REF are well defined. Indeed, if MATH then we have MATH and hence MATH. A simple computation using REF shows that the one-step correlation coefficient is MATH, and the two step correlation is MATH. Since by REF, this implies that MATH and MATH. By routine compu... |
math/0007026 | Indeed, multiplying REF by MATH we get MATH. In particular, if MATH then MATH and MATH for all MATH. On the other hand, if MATH, then MATH and the correlation coefficients MATH satisfy the recurrence MATH . From this we infer that MATH as MATH. Indeed, since MATH, MATH is finite, and satisfies MATH. Is is easy to see t... |
math/0007026 | By REF , we have MATH, and MATH. We first show by induction that for all MATH where MATH, MATH . For MATH, REF follows from REF when MATH. Clearly, REF trivially holds true when MATH or MATH for all MATH. Suppose that REF holds true for a given value of MATH and all MATH. We will prove that it holds true for MATH. We o... |
math/0007026 | By REF we have MATH. Since MATH, from REF we get MATH . We now give another expression for the left hand side of REF . Substituting MATH into REF we get MATH. By REF this implies MATH. Since MATH, combining the latter with REF we have MATH . We now substitute REF as follows. Taking the conditional expectation MATH of b... |
math/0007026 | If MATH is quadratic then there are constants MATH such that MATH . Since MATH is a homogeneous NAME chain and REF holds true MATH . On the other hand, REF implies, see REF MATH . Combining this with REF we get MATH . Since MATH and MATH therefore MATH must have at least two values. We consider separately two cases. CA... |
math/0007026 | We first consider the two-valued case. If MATH is a non-random constant, then MATH and thus MATH is non-unique; one can take MATH to satisfy REF , or one can take MATH to satisfy REF . Suppose now that MATH has more than two values. We first verify that that the collusion REF holds true when MATH. In this case the left... |
math/0007026 | Conditioning in two different directions in MATH we get MATH. Therefore MATH. Since MATH we have MATH, which ends the proof. |
math/0007026 | Since the conclusion is trivially true when MATH, throughout the proof we assume that MATH. In this REF implies MATH. Since the assumptions are symmetric, and MATH by REF and stationarity we have MATH. Notice that REF implies MATH. Thus MATH . We now compute conditional moments using the approach of NAME REF . Using co... |
math/0007026 | For MATH let MATH . The range of values of MATH implies that MATH. We give the proof for the case MATH. The only change needed for the case MATH, is to use the symmetric two-valued NAME chain defined in REF instead of the NAME chain MATH defined below. Define orthogonal polynomials MATH by the recurrence MATH with MATH... |
math/0007027 | The smoothness of the flow map follows by considering the Lagrangian version of REF given by MATH where MATH denotes the tangent map of MATH (which in local coordinates is given by the REFxREF matrix of partial derivatives MATH), and where MATH is the covariant derivative along the curve MATH (which in Euclidean space ... |
math/0007027 | The ordinary differential REF can be written as MATH (see page REF). Remarkably, MATH is a MATH vector field, and CITE provides the existence of a unique short-time solution to REF in MATH which depends smoothly on MATH, and where MATH only depends on MATH. Thus, it suffices to prove that the solution curve MATH does n... |
math/0007029 | By REF it suffices to show that this is a MATH - representation of MATH. REF - REF are straightforward to verify and REF follows by observing that for fixed MATH and MATH the map MATH given by MATH is a bijection. If MATH is surjective, then it is clear that every generator MATH of MATH is in the range of MATH. |
math/0007029 | Fix MATH and MATH with MATH. The sequence MATH defined by MATH and MATH for MATH is cofinal. Set MATH and MATH for MATH and let MATH be defined by the method given in REF. Then MATH has the desired properties. |
math/0007029 | By REF it suffices to show that MATH is compact for all MATH. Fix MATH and let MATH be a sequence in MATH. For every MATH, MATH may take only finitely many values (by REF ). Hence there is a MATH such that MATH for infinitely many MATH. We may therefore inductively construct a sequence MATH in MATH such that MATH and M... |
math/0007029 | One may check that the sets MATH form a basis for a topology on MATH. To see that multiplication is continuous, suppose that MATH. Since MATH are composable in MATH there are MATH and MATH such that MATH, MATH and MATH. Hence MATH and MATH and the product maps the open set MATH into MATH. The remaining parts of the pro... |
math/0007029 | Given MATH choose a sequence MATH such that MATH is cofinal in MATH. Set MATH and let MATH be defined by the condition that MATH. We must show that there is a MATH such that MATH. It suffices to show that the the intersection MATH. But this follows by the finite intersection property. One checks that MATH. Furthermore ... |
math/0007029 | Since MATH is surjective, the map MATH is a homeomorphism (see REF). The map MATH extends to a surjective morphism MATH. Let MATH be a section for MATH and let MATH be an identification of MATH with MATH. Then we get a groupoid isomorphism by the map MATH where MATH. |
math/0007029 | Let MATH, then for MATH define MATH by MATH where MATH is the canonical basis for MATH. Notice that MATH is nonzero since MATH; one then checks that the family MATH satisfies REF - REF . |
math/0007029 | Applying REF to MATH with MATH, to MATH with MATH and using REF we get MATH . By REF if MATH but MATH, then the range projections MATH, MATH are orthogonal and hence one has MATH. If MATH then MATH where MATH and so MATH; REF then follows from REF . The rest of the proof is now routine. |
math/0007029 | Fix MATH and let MATH, MATH, MATH, MATH be such that MATH and MATH, then by REF we have MATH so that the map which sends MATH to the matrix unit MATH for all MATH, MATH with MATH extends to an isomorphism. The second isomorphism also follows from REF (since MATH implies MATH). We claim that MATH is contained in MATH wh... |
math/0007029 | If MATH for some MATH then clearly MATH is not faithful. Conversely, suppose that MATH is equivariant and that MATH for all MATH; we first show that MATH is faithful on MATH. For any ideal MATH in MATH, we have MATH (see CITE, CITE). Thus it is enough to prove that MATH is faithful on each MATH. But by REF it suffices ... |
math/0007029 | For REF we note that MATH for MATH is a MATH-representation of MATH; hence there is a MATH-homomorphism MATH such that MATH for MATH (see REF ). Let MATH denote the MATH-action on MATH induced by the MATH-valued MATH - cocycle defined on MATH by MATH (see CITE); one checks that MATH for all MATH. Clearly for MATH we ha... |
math/0007029 | Observe that if MATH is aperiodic then MATH implies that MATH and hence MATH has trivial isotropy, and conversely. Hence MATH is essentially free if and only if aperiodic points are dense in MATH. If aperiodic points are dense in MATH then MATH clearly satisfies the aperiodicity condition, for MATH must then contain ap... |
math/0007029 | If MATH for some MATH then clearly MATH is not faithful. Conversely, suppose MATH for all MATH; then by REF we have MATH and hence from CITE it suffices to show that MATH is faithful on MATH. If the kernel of the restriction of MATH to MATH is nonzero, it must contain the characteristic function MATH for some MATH. It ... |
math/0007029 | By REF MATH; since MATH is essentially free, MATH is simple if and only if MATH is minimal. Suppose that MATH is cofinal and fix MATH and MATH; then by cofinality there is a MATH and MATH so that MATH and MATH. Then MATH and MATH is in the same orbit as MATH; hence all orbits are dense and MATH is minimal. Conversely, ... |
math/0007029 | Arguing as in CITE one shows that MATH is locally contracting. The aperiodicity condition guarantees that MATH is essentially free, hence by CITE (see also CITE) we have MATH is purely infinite. |
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