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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 product of MATH and a vector bundle over a closed smooth simply-connected manifold MATH. In other words, MATH is the MATH-pullback of a vector bundle over MATH. Fix base points in MATH, MATH, MATH so that the inclusions MATH, MATH, MATH are defined, and let MATH, MATH. We think of MATH and MATH as two vector bundle structures on a fixed manifold MATH, and use the zero sections to identify MATH and MATH with smooth submanifolds of MATH, and identify MATH, MATH with the normal bundles to MATH and MATH. Note that both MATH and MATH are homotopy equivalent to the universal cover of MATH. Let MATH be the homotopy equivalence induced by the zero section of MATH followed by the projection of MATH. Note that MATH is orientable, as the pullback of a bundle over a simply-connected manifold. Fix orientations of MATH and MATH, which defines an orientation on MATH. We orient MATH so that MATH, which defines an orientation on MATH. To simplify notations, assume that MATH; the case of odd MATH is similar. Since MATH has rank MATH, MATH for MATH. As we remarked in CITE, the bundle MATH virtually comes from MATH iff MATH and MATH lie in MATH. Alternatively, since MATH, we see that MATH virtually comes from MATH iff MATH for MATH. By NAME 's theorem, the maps MATH, MATH are homotopy equivalences. Fix their homotopy inverses MATH, MATH. Consider a self-homotopy equivalence MATH of MATH. Note that each of the maps MATH, MATH is homotopic to the identity because, say, MATH is equal to MATH . Now it is routine to check that MATH maps MATH into MATH if and only if MATH maps MATH to itself. By REF maps MATH to itself, so MATH. It was observed in CITE that MATH maps MATH to MATH. Thus MATH, as needed, and it remains to show that MATH also contains MATH for MATH. Since MATH, viewed as a map MATH, is homotopic to the inclusion MATH, we have that MATH. By the NAME sum formula MATH . Since MATH is parallelizable, MATH and MATH. Since MATH is a unit in MATH, we can write MATH for some MATH, and MATH . Also MATH, because MATH is a product of torus and a bundle over MATH. By definition of ``NAME, this means that MATH and MATH lie in MATH for any MATH, and therefore, MATH, MATH for any MATH. The above formula now implies that MATH for MATH, as promised. This completes the proof that MATH virtually comes from MATH. |
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 homomorphism (compare REF above). It is also easy to check that the map MATH is the inverse to MATH and therefore MATH is an automorphism of MATH. Then MATH where MATH are linear self-maps of MATH. Now MATH is a derivation of MATH, so the formulas MATH and MATH for MATH, MATH define an automorphism MATH of MATH. Then MATH where MATH are linear self-maps of MATH, and MATH is a derivation. Continuing in this fashion, we get automorphisms MATH with MATH where MATH is a derivation of MATH, and such that MATH. Thus MATH. Also MATH. Now if MATH, then by REF so that MATH. Thus, MATH as desired. |
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 derivations of degree MATH vanish on MATH as well. |
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-modules. Since MATH, this implies that MATH is an isomorphism. Note that under the above identification of MATH and MATH, MATH corresponds to MATH. By REF , to prove splitting rigidity of MATH for any MATH, it is enough to show that all negative derivations of MATH vanish on MATH. Let MATH be a negative derivation. By REF we can write MATH as MATH where MATH and MATH. Now MATH implies that MATH, and hence MATH, as needed. |
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 from the NAME sequence that MATH for MATH. Hence, REF implies that MATH. So by REF , MATH is splitting rigid for any MATH. |
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 isomorphic to MATH with zero differentials and with MATH. Similarly the minimal model of MATH is isomorphic to its cohomology ring MATH with MATH. Thus MATH can be viewed as a NAME of minimal models of MATH and MATH. Let us construct a NAME model of MATH. Since MATH we compute that MATH for all MATH. Consider the relative NAME algebra MATH where MATH and MATH. Then it is immediate to check that this relative algebra is a NAME model (in fact, a minimal one) of MATH with the quasi-isomorphism MATH given by MATH, MATH, MATH. By the naturality of models of maps CITE, from the fibered square REF , we obtain that a NAME model of the map MATH can be given by the pushout of MATH via the homomorphism MATH; that is, it can be written as MATH where MATH is given by MATH and MATH. In particular, MATH is a model for MATH. Notice that MATH is a free polynomial algebra on a finite number of even-dimensional generators, and thus the model MATH is pure. |
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 can assume that both MATH and MATH are connected. Since MATH is semisimple and MATH is simply connected, the long exact sequence of the fibration MATH implies that MATH is also semisimple. Let MATH be the minimal model of MATH. Since MATH is MATH-connected, we have that MATH. According to REF , the model MATH is a pure. By minimality of MATH, we have that MATH and MATH. By the structure theorem for pure DGAs, CITE, this implies that MATH and MATH for some differential subalgebra MATH such that MATH. Let MATH and MATH. By above MATH. Therefore, by REF , all negative derivations of MATH vanish on MATH. Notice that MATH corresponds to MATH under the isomorphism MATH, and hence applying REF , we conclude that negative derivations of MATH vanish on MATH. |
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 grading on MATH. According to CITE, (also compare CITE), MATH for MATH. In our case MATH, and hence MATH for MATH. Next observe that MATH since MATH is odd for any MATH and MATH. Similarly MATH. Therefore, MATH which is equal to the quotient of MATH by MATH. Since MATH is a torus, MATH is generated by MATH-dimensional classes and hence by above the same is true for MATH. By REF this implies that MATH which by the splitting REF means that MATH is splitting rigid for any MATH. |
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-dimensional manifold which is nondiffeomorphic to a positively curved homogeneous space is the space MATH CITE. This space is a MATH-bundle over MATH and, therefore, it lies in MATH. Now we establish splitting rigidity for all known odd-dimensional positively curved manifolds. Those are the standard spheres, the NAME MATH-dimensional homology sphere MATH, the NAME MATH-manifolds MATH CITE all obtained as biquotients of MATH by MATH, and the NAME MATH-manifolds MATH CITE obtained as biquotients of MATH by MATH. By REF, any odd-dimensional rational homology sphere is splitting rigid, so it remains to deal with the NAME and NAME manifolds. By REF below, all NAME manifolds are rationally homotopy equivalent to MATH, and all NAME manifolds are rationally homotopy equivalent to MATH. Since both MATH and MATH belong to MATH, the proof of REF implies that any negative derivation of MATH or MATH vanishes on even cohomology. Now REF implies splitting rigidity of all NAME and NAME manifolds. |
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) that MATH. Therefore, MATH is a free graded algebra on generators MATH with the degrees of the generators given by the subscripts. Obviously, MATH and MATH for some rational MATH. Note that MATH, else we would have MATH which is known not to be the case by CITE. Replacing MATH with MATH, we can assume that MATH. It is clear that MATH must be equal to MATH for some rational MATH. We claim that any such minimal model is isomorphic to the one with MATH. Indeed, let MATH. Consider the map MATH given by MATH. This map is easily seen to be a NAME with the inverse given by MATH. Since MATH is a minimal model of MATH, the proof is complete. |
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 try to find a finite covering MATH, and a map MATH that makes the diagram commute, and satisfies MATH. It follows from an obstruction theory argument as in CITE that such MATH, MATH can be constructed, where the key point is that all obstructions are torsion since the homotopy fiber of MATH has torsion homotopy groups, and all torsion obstruction vanish after precomposing with a suitable finite cover MATH. Furthermore (compare the proof of REF ), one can choose MATH satisfying MATH, MATH, where MATH is the MATH-th power map, so that, for some positive integer MATH, the induced map on MATH satisfies MATH . Finally, by NAME 's theorem, the map MATH is a homotopy equivalence with the desired properties. |
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 the NAME class of some rank MATH vector bundle MATH over MATH. If MATH, then by REF , MATH is proportional to the MATH-th NAME class of some bundle of rank MATH vector bundle MATH over MATH. Say, suppose that MATH is nonzero on the NAME class MATH. By REF , there is a self-homotopy equivalence MATH of MATH such that MATH does not lie in MATH. Now there are two cases to consider. If MATH, then by NAME 's embedding theorem CITE the homotopy equivalence MATH is homotopic to a smooth embedding MATH. The normal bundle MATH has rank MATH. Since MATH and MATH are homotopic, MATH and MATH have equal NAME and NAME classes. This follows from the intersection pairing interpretation of the NAME class, and the NAME sum formula for the total NAME class (see CITE for details). Thus, MATH does not lie in MATH. So MATH does not virtually come from MATH, while MATH is diffeomorphic to MATH, and this means that MATH is not splitting rigid. If MATH for all MATH, then MATH. In other words, MATH, so by REF , there is a diffeomorphism MATH of MATH such that MATH does not lie in MATH. Then the pullback bundle MATH does not virtually come from MATH, while its total space is diffeomorphic to MATH; thus MATH is not splitting rigid. |
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 of REF , the normal bundle to this embedding does not virtually come from MATH and, therefore, MATH is not splitting rigid. If MATH vanishes on MATH, then MATH so MATH preserves MATH. Hence, by REF , replacing MATH with some power of MATH, we can assume that MATH is homotopic to a diffeomorphism. If MATH is nonzero on MATH for some MATH, then MATH is nonzero on the NAME class of some bundle MATH over MATH. Looking at the bundle MATH shows that MATH is not splitting rigid. |
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 equivalently, that the NAME and NAME classes of MATH lie in MATH. We shall borrow notations and arguments from the proof of REF . As in REF, we can assume that MATH is the total space of a vector bundle MATH which is the product of MATH and a vector bundle MATH over a closed smooth simply-connected manifold MATH. Let MATH and MATH by the homotopy equivalence as in REF. Note that MATH. Indeed, as in REF, MATH. So MATH. The only NAME classes of a rank MATH vector bundle that have a chance of being nonzero are MATH, MATH, MATH. Since MATH has zero cohomology in dimensions MATH, MATH, MATH, we get MATH. By the same argument, MATH. We have MATH hence MATH for all MATH, and MATH. Then one easily sees that MATH maps MATH to MATH. Hence, MATH maps MATH to MATH, and therefore, MATH preserves MATH. As in REF, it suffices to show that MATH preserves MATH. We think of MATH as an exterior algebra on generators MATH corresponding to spheres MATH, MATH, MATH, MATH so we need to show that MATH. By rescaling we can assume that MATH. By dimension reasons MATH unless MATH is MATH or MATH. Similarly, MATH unless MATH is MATH, MATH or MATH. Now collecting terms next to MATH's in the identity MATH, we conclude that MATH, and hence MATH for all MATH. Thus, MATH as promised. |
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 conclude that MATH maps MATH to MATH. Next note that all MATH, MATH lie in the subalgebra generated by MATH. Indeed, MATH has the structure group MATH, so MATH is a pullback of a bundle over MATH. Since the cohomology of MATH is generated by MATH, by naturality of the characteristic classes, we see that MATH for any MATH. Since MATH, we conclude that all the characteristic classes of MATH lie in MATH, hence MATH virtually comes from MATH. |
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 for example, CITE) MATH is onto. By naturality of the assembly and since MATH is the classifying space for MATH, MATH factors as the composition of the map MATH induced by the projection MATH, and the universal assembly MATH. The former map is onto, since it has a section induced by a section of MATH, while the latter map is an isomorphism since MATH CITE. Thus, MATH is onto. It is known that the map MATH has a finite cokernel (see for example, CITE), and hence so does the map MATH. By exactness of the smooth surgery exact sequence, the MATH-action on MATH has finite orbits. Since MATH preserves the total NAME class, MATH is tangential for some MATH, so replacing MATH by MATH, we can assume that MATH is tangential. Hence for any integer MATH, MATH is tangential so that the normal invariant of MATH lies in the image of the map MATH induced by the fibration MATH. Since MATH is rationally contractible, MATH is a finite set, so there exists an infinite sequence of positive integers MATH such that the elements MATH have the same normal invariant. By exactness, MATH lie in the same MATH-orbit which is a finite set by above. In particular, for some MATH we have MATH. Thus, for some self-diffeomorphism MATH of MATH, we get that MATH and MATH are homotopic, or MATH is homotopic to MATH, as wanted. |
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 to lift it to MATH. In other words, try to find MATH which would make the following diagram commute up to homotopy: MATH . The obstructions to lifting MATH lie in torsion groups MATH, and are generally nonzero. Each factor MATH of MATH is a MATH-space, so let MATH be the MATH-th power map. It is easy to check that the endomorphism of MATH induced by MATH is the multiplication by MATH. The MATH-algebra MATH is generated in dimension MATH, so we get an obvious generating set for MATH such that the MATH-th power map MATH of MATH act on the generators as the multiplication by MATH. Let MATH be the first nontrivial obstruction to lifting MATH. Since MATH is torsion and MATH is a finite NAME, the naturality of obstructions implies that MATH for some MATH. Repeating this process finitely many times, we find some MATH such that all the obstructions to lifting MATH vanish, and thus there is a map MATH such that MATH is homotopic to MATH. MATH . Now the bundle MATH has the desired properties. |
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 is a positive harmonic majorant for MATH in MATH. In the general case, we use the NAME - NAME multiplier theorem CITE: there exists a function MATH in the NAME class with MATH, MATH. Applying the previous argument, we obtain that MATH and MATH, and hence, MATH, are of bounded type in MATH. Now, let MATH be an entire function of non-zero exponential type belonging to the NAME class. Suppose that MATH is of bounded type in MATH. Then the function MATH has a positive harmonic majorant MATH; without loss of generality we may assume that h is symmetric, MATH. By REF , MATH. Since the function MATH has non-zero exponential type, REF implies that MATH for large MATH. This implies that MATH is an NAME - NAME set. |
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 bounded type in MATH. Take one of them, say MATH, and represent it as the sum of two functions, MATH where MATH, MATH, MATH. Let us verify that MATH is of bounded type in MATH. Indeed, this function is analytic in MATH and satisfies the following properties: MATH and, for MATH, MATH . Thus, the image MATH omits the ray MATH. Applying the NAME theorem, we get that the function MATH is of bounded type in MATH. (Alternatively, we could just consider the functions MATH, MATH, MATH; MATH has absolute values bounded by one in MATH. Hence, MATH is of bounded type there.) The same argument works for MATH, and we conclude that MATH is of bounded type in MATH, MATH, and finally, MATH are of bounded type in MATH. |
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 REF , MATH has a positive harmonic majorant MATH in MATH. Using the representation REF for MATH together with REF we obtain REF . Then, applying a standard NAME - NAME argument to the subharmonic functions MATH in the domain MATH, we obtain REF . By REF , MATH . If the last integral is finite, then MATH, and by Theorem C, the polynomials are not dense in MATH. Now we suppose that the polynomials are not dense in MATH, and as a consequence, are not dense in MATH where MATH is to be chosen later on. We apply Theorem D, and get an entire function MATH in the NAME class, of zero exponential type, MATH, with zeros MATH, such that MATH and for every polynomial MATH, for every MATH, MATH . Using relation REF we get MATH . REF implies that the function MATH is of bounded type in MATH, and hence, by REF , MATH . Using this fact and relation REF , we obtain MATH for MATH, where MATH is, as above, the set of polynomials MATH such that MATH. Therefore, to complete the proof of the theorem we need only to verify that MATH . Then, using NAME 's inequality, we conclude that MATH has a harmonic majorant in MATH, and by REF we get that REF does not hold. Let us return to REF . By NAME 's inequality, for every MATH we have MATH . Furthermore, by REF , for MATH we get MATH . Therefore, to prove REF it remains to check that for some MATH, MATH . Since MATH are the zeros of an entire function of zero exponential type, MATH for some MATH. Thus, REF follows from the estimate MATH with MATH independent of MATH, and with some MATH. Our conditions on MATH imply that for some MATH, MATH, and every MATH, there exists MATH, MATH-such that for MATH we have MATH where intervals MATH satisfy the condition MATH, MATH. The following elementary estimate of harmonic measure, MATH shows that for MATH, MATH . Furthermore, MATH . Thus, estimate REF is true for MATH, MATH. |
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 . Finally, since MATH, MATH, we get MATH . CASE: We use an auxiliary estimate of harmonic measure. Let MATH be intervals of length MATH such that MATH, MATH, MATH. Then for some MATH and MATH, independent of MATH, MATH . We postpone the proof of this lemma till the next section. Fix MATH with MATH, and consider the canonical product MATH and the entire function MATH of zero exponential type, MATH. Denote by MATH the zero set of MATH, MATH. It follows from the results of NAME in CITE that the expression MATH tends to a finite non-zero limit for MATH with MATH. Put MATH, MATH. Since MATH, MATH, MATH, we get for some MATH: MATH whence MATH . We define an even weight MATH by the formula MATH . Since MATH . Theorem D implies that the polynomials are not dense in MATH. The even log-convex function MATH is increasing on the positive half-line. Let us verify that MATH . First, without loss of generality, we assume that MATH is MATH-smooth. By convexity of MATH, for every MATH we have MATH. Now we fix MATH. If MATH, then MATH. Furthermore, for some MATH we have MATH. Since MATH, we get MATH, and as a result, MATH. Therefore, MATH; for every MATH, MATH, we have MATH, thus the estimate REF is proved. Relations REF imply that MATH . Choose a sequence of intervals MATH with MATH, MATH, MATH, and set MATH. Apply REF , and add to MATH a union MATH of small intervals such that MATH, and MATH . Then by REF , MATH . |
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. Since MATH is positive, harmonic and symmetric in MATH, is subharmonic in the plane, and has at most order one and mean type there, the subharmonic version of the NAME representation (see CITE) implies that for a finite positive measure MATH on MATH, MATH . Now, REF follows immediately. Using REF and the second estimate in REF , we get MATH . Denote by MATH the union of two horizontal sides of the domain MATH, and by MATH the union of its two vertical sides. An estimate of harmonic measure in MATH gives us that for some positive MATH independent of MATH and MATH, MATH . To obtain this estimate we use an easy generalization of REF (see also CITE). This generalization claims that for every square MATH with horizontal sides MATH and vertical sides MATH, the following inequality holds: MATH . To verify REF we note first that by symmetry, on the diagonals of the square MATH we have MATH. Therefore, applying the maximum principle to the difference of these functions, we get MATH . Finally, MATH . To deduce REF note that the function MATH is positive and continuous in MATH. Therefore, for MATH we have MATH. Hence, for sufficiently big MATH, MATH . For every MATH with MATH consider the square MATH with MATH, and note that MATH, MATH. Therefore, MATH, MATH, MATH, and the estimate REF implies that MATH. Thus, REF is proved. Now, if MATH is sufficiently small, then applying the theorem on two constants to the symmetric harmonic function MATH in the domain MATH, and using REF , and the property MATH we obtain MATH . Thus, MATH, MATH. Applying REF we come to a contradiction. |
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 estimate the function MATH as follows: MATH . The asymptotical relation MATH, MATH, implies now that MATH, MATH, and MATH, MATH. Next, we choose MATH, and apply the theorem on two constants to the function MATH in MATH: MATH . Hence, for MATH, MATH, MATH . Let MATH be the union of two horizontal sides of MATH. REF mentioned in the proof of REF claims that MATH . Therefore, for MATH, MATH . CASE: The NAME function MATH for MATH is positive, bounded and harmonic on MATH. Therefore, applying the NAME formula in the half-planes MATH we get MATH . REF imply that for MATH, MATH . Set MATH. Since MATH is positive and harmonic on MATH, by NAME 's inequality we get MATH . CASE: Finally, we consider an auxiliary function MATH. It is harmonic on MATH, vanishes on MATH, and MATH for MATH. Therefore, for MATH, MATH and MATH . The estimate for MATH is obtained in an analogous way. |
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, we have MATH, MATH. Consequently, MATH, MATH. Once again, by the NAME formula, for some MATH, MATH . For every MATH we denote by MATH the center of MATH. Arguing as in REF of the previous proof, we compare MATH with MATH in MATH, and deduce 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 MATH . It then follows that MATH is independent of MATH. It now suffices to show that MATH . Recall that MATH is the union of the MATH-cells MATH . We see that MATH etc., so that MATH. It follows that MATH . By similar reasoning, MATH and so on, eventually giving MATH as required. This completes the proof. |
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 now follows that MATH . |
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. Similarly, MATH is closed under MATH. It remains to show that MATH when MATH and MATH are in MATH and the composite exists. Suppose that MATH and MATH. If MATH then MATH if MATH then MATH is MATH or MATH, so MATH is certainly in MATH; and if MATH then MATH (note that MATH is an identity for MATH because it lies in the image of MATH). This completes the proof. |
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 category (with MATH and MATH the left and right identities of MATH and with MATH as composition), and these structures commute with one another. Trivially the triples MATH for MATH provide further commuting category structures. To show that these structures make MATH into a MATH-category, it now suffices to show that an identity for MATH is also an identity for MATH if MATH; in other words, it suffices to show that MATH for MATH and MATH. For MATH, this is trivial; we may therefore assume that MATH. But REF gives us MATH as required. We have now shown that the MATH are MATH-categories. We know from REF that MATH maps MATH into MATH, and it remains to show that this function is a homomorphism. That is to say, we must show that MATH for MATH, and we must show that MATH for MATH a composite in MATH. But if MATH then MATH and MATH by identities in REF; if MATH we get MATH and MATH and if MATH then MATH and MATH trivially. This completes the proof. |
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 MATH and using REF , we find that MATH . Suppose that MATH and MATH. Then MATH . Suppose that MATH is a composite in MATH with MATH. Let MATH be the morphism such that MATH and MATH; then MATH . Let MATH be the inclusion and let MATH be MATH with its codomain restricted to MATH, so that MATH. Since MATH and MATH are in MATH, we have MATH and MATH; we therefore get MATH . Now let MATH be the MATH-category with one generator MATH and with relations MATH. We see that there is a factorisation MATH through the obvious push-out of MATH . We also see that MATH . It now follows that MATH therefore MATH . Finally, suppose that MATH is a composite in MATH with MATH. We must have MATH, and we get MATH . This completes the proof. |
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 the identities for MATH). We first compute MATH. It is straightforward to check that MATH . Similarly, MATH is as a composite MATH . We now evaluate the rows of the matrix for MATH. The first and last rows yield MATH and MATH. The composite of the non-identity elements in the third row is MATH, which is an identity for MATH, so the third row can be omitted. Similarly, the fifth row can be omitted. The second, fourth and sixth rows have the same values as the rows of the matrix for MATH. It follows that MATH also. Therefore MATH. This completes the proof. |
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. Since MATH, it follows from the case already covered that the factors can be written as MATH with MATH a generalised MATH. The factors now have the form MATH with MATH a generalised MATH, as required. |
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 is then a composite of factors of the form MATH with MATH a generalised MATH and with MATH a generalised MATH. These factors have the form MATH with MATH a generalised MATH, as required. |
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 REF , there is a further decomposition into factors MATH with MATH a MATH-fold product of face operators. This completes the proof. |
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. The first of these equations, MATH, is an immediate consequence of REF . For the second equation, let MATH be a cell of positive dimension MATH. By REF , MATH which may be identified with MATH. By REF , this is the appropriate composite of the MATH, as required. We have now constructed functions MATH, and we must show that these functions form a homomorphism of cubical MATH-categories. We must therefore show that MATH, that MATH, that MATH, and that MATH. First we consider MATH, where MATH. Let MATH be a cell in MATH of dimension MATH, and let MATH. We then have MATH, so MATH and MATH. It follows that MATH and MATH . Therefore MATH as required. Next we consider MATH, where MATH. Let MATH be a cell in MATH of dimension MATH. From REF , we see that MATH has the form MATH or MATH. Let MATH, so that MATH in the first case and MATH in the second case. Let MATH be MATH or MATH as the case may be, and let MATH be the underlying MATH-category homomorphism. We now see that MATH and MATH with MATH. It follows that MATH . Using REF , we also see that MATH. We now get MATH (recall that MATH is to be identified with MATH) and MATH so that MATH as required. Next we consider MATH, where MATH is a composite in MATH. Let MATH be a cell in MATH of dimension MATH. From REF , MATH is equal to MATH or MATH or to MATH for some MATH. In any case, using REF if necessary, we see that MATH is a composite of factors MATH such that MATH is the corresponding composite of the MATH, where MATH are the functions expressing MATH as a push-out. Let MATH be the function such that MATH we see that MATH as required. Finally we consider MATH, where MATH. Let MATH be a cell in MATH of dimension MATH. From REF , MATH has the form MATH or MATH or MATH. We can now use the same argument as for MATH, noting that MATH and that MATH by REF . This completes the proof. |
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 completes the proof. |
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 MATH for MATH and that MATH for MATH. Let MATH be a member of MATH. Then MATH . The first and third rows are in the image of MATH by REF , so they are identities for MATH and can therefore be omitted. This leaves the second row in which MATH and MATH are identities for MATH. It follows that MATH. Now let MATH be a member of MATH. By REF , MATH, so MATH . By REF the first and third columns are in the image of MATH, so they are identities for MATH and can be omitted. This leaves the second column so that MATH. This completes the proof. |
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 MATH, so MATH is itself a composite of elements of the forms MATH and MATH. |
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 proved directly, or, as is well known, be deduced from the axioms for a monoidal closed category. |
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 inequality must hold for MATH, MATH, as well. Hence MATH and MATH. MATH . |
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 dense in MATH, MATH. MATH . |
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 circle algebra MATH in a sequence as above, then by REF MATH, hence MATH, which proves that the circle algebra C*MATH cannot be MATH in a sequence MATH as above. MATH . |
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, proving the proposition. 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 converse inequality follows from easy estimates using REF . MATH . |
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 temperature MATH if MATH . |
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, and consequently so is their sum MATH. The first inequality in REF then follows from REF . For the second, set MATH and note that one has a surjective map MATH of vector bundles on MATH. By definition of MATH, the bundle on the left becomes nef when twisted by the MATH-divisor MATH. Since quotients of nef bundles are nef, this implies that MATH is nef, and the required inequality follows. For REF , we use the fact (compare CITE, p. REF) that MATH and MATH have the same normalization MATH, which sits in a commutative diagram: MATH . Moreover, the exceptional divisors MATH and MATH of MATH and MATH pull back to the same divisor MATH on MATH. Invoking again REF one has MATH as required. |
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 (possibly redundant) primary decomposition MATH. In particular every associated prime of MATH must occur as the radical of one of the MATH, that is, as one of the distinguished subvarieties. |
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 to MATH then MATH is globally generated. Therefore MATH for every MATH, and in particular MATH. We next show that MATH. To this end, fix any MATH. Then we can choose large positive integers MATH such that MATH so that in particular MATH is globally generated. Writing as before MATH for the blow-up of MATH, with exceptional divisor MATH, it follows that MATH is globally generated and hence nef. Therefore MATH, as required. It remains to prove that MATH. To this end we use REF to the effect that NAME Vanishing remains valid even after twisting by arbitrary nef divisors. Specifically, consider an irreducible projective variety MATH, and fix an ample divisor MATH plus a coherent sheaf MATH on MATH. NAME shows that there is an integer MATH such that for any nef divisor MATH: MATH (The important point here is that MATH is independent of MATH.) We propose to apply REF on the blowing-up MATH of MATH. Given MATH, choose large integers MATH such that MATH . Then MATH is ample, so there exists an integer MATH such that if MATH then for any nef divisor MATH on MATH, the bundles associated to the divisors MATH have vanishing higher cohomology. Now fix any integer MATH, and write MATH . Then MATH is nef (in fact ample), and consequently we have the vanishing of the higher cohomology of the line bundle MATH . It now follows from REF below - and this is the crucial point - that MATH provided that MATH is sufficiently large. Therefore MATH is MATH-regular for MATH, and consequently MATH . By taking MATH (and hence also MATH) to be large enough, we can arrange that the second term on the right is MATH, so that MATH for MATH. Therefore MATH, and we are done. |
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 MATH, which in turn determines an embedding MATH in such a way that MATH. Write MATH for the projection. NAME vanishing for MATH, applied to the ideal sheaf MATH, shows that if MATH then the natural homomorphism MATH is surjective. On the other hand, recalling that MATH for every MATH, one sees that the image of REF is exactly MATH. It follows that MATH for MATH, as asserted. |
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 governs the leading term of the NAME polynomial of the sheaf in question. We leave details to the reader. |
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 sequence MATH is exact for every MATH. This sequence shows that MATH for every MATH, where by abuse of notation we are writing MATH for the class of the restriction MATH to MATH. Consequently MATH. Similarly, REF shows that MATH for fixed MATH provided that MATH. As we are working over an uncountable ground field, we can assume by taking MATH to be very general that this holds simultaneously for all MATH. Since of course also MATH, if MATH it therefore suffices to prove the Proposition for MATH. So by induction on MATH we can assume that MATH. Supposing then that MATH, we need to bound as a function of MATH the length of the (finitely supported) subsheaf MATH of sections having zero-dimensional support. Equivalently, we need to bound for MATH the dimension MATH. To this end, observe first of all that for every integer MATH there is an inclusion MATH . The plan is to estimate the dimension of the group on the right for a suitable integer MATH. Fix MATH plus large integers MATH such that MATH where MATH. Then MATH, and so the exact sequence MATH together with REF shows that MATH . But NAME implies that as a function of MATH, MATH . It follows from REF that by taking MATH (and hence MATH) sufficiently large, and MATH sufficiently small, we can arrange that MATH where MATH is a constant. The result then follows from REF . |
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 computation we get the following conditional probabilities MATH . Using REF we have MATH. On the other hand, direct computation using conditional probabilities gives MATH. The resulting equation has four roots when solved for MATH: the double root MATH and two roots MATH. Solution MATH corresponds to the independent sequence with MATH. Since MATH, therefore the only non-trivial solution is MATH, which gives MATH and MATH. REF in this case is verified by direct computation with conditional probabilities. |
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 that since MATH, the recurrence has unique solution MATH. |
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 only need to show that the left-hand side of REF is a linear function of the appropriate variables. Indeed, in the non-degenerate case the coefficients MATH in a linear regression are uniquely determined from the covariances; the covariance matrices are non-degenerate since MATH and MATH. Using routine properties of conditional expectations, the case of general index MATH reduces to two values MATH. By symmetry, it suffices to give the proof when MATH. Conditioning on additional variable MATH we get MATH . Now adding MATH to the condition we get. MATH . This gives the system of two linear equations for MATH, which has the unique solution which is a linear function of MATH when MATH. It remains to notice that if MATH then MATH. Indeed, the latter is equivalent to MATH and holds true because MATH, MATH, and MATH. Therefore the regression MATH is linear, and REF holds for MATH. This proves REF by induction. Passing to the limit as MATH in REF with MATH we get REF . |
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 both sides of REF , with MATH and substituting REF , we get MATH . Replacing MATH by the right hand side of REF we get REF . |
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. CASE: If MATH has only two values then by REF MATH and MATH is a non-random constant, ending the proof. CASE: If MATH has at least three values, then MATH are linearly independent. Therefore REF implies MATH . Since REF implies that MATH, the only solutions of REF are MATH or MATH. Since MATH, both solutions imply MATH. Clearly, MATH implies REF . On the other hand if MATH and MATH, then MATH. Thus REF hold true. |
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 hand side of REF is zero. Since MATH has more than two values, this implies that MATH and MATH. Therefore REF implies REF . Now consider the case when MATH. From REF we have MATH where MATH. This shows that MATH is quadratic. By REF we have MATH; since MATH this implies that MATH. We also know that either REF holds true, which is equivalent to MATH, or REF holds true, which is equivalent to MATH. We now compare these two expressions with REF : since MATH and MATH is non-constant, the coefficients at MATH must match. That is, either MATH or MATH. By a simple algebra the former implies REF and the latter implies REF . |
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 constant conditional variance and REF , we write MATH in two different ways as MATH and as MATH . Combining these two representations and using REF , and MATH we get after simple algebra MATH . Similarly, we rewrite MATH in two different ways as MATH and, using REF , as MATH . Using REF , after some algebra we get MATH . Solving the system of REF for MATH we get MATH . Substituting REF , and denoting MATH we have MATH . Therefore by REF and a simple calculation we have MATH . Since MATH this implies that either MATH or MATH. |
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. Let MATH denote the probability measure which orthogonalizes MATH (see for example, NAME REF ), and for fixed MATH define MATH where MATH are normalized orthogonal polynomials MATH. By NAME REF , for MATH REF defines a NAME transition function with invariant measure MATH. For MATH, let MATH be a stationary NAME chain with the initial distribution MATH and transition probability MATH. It is known that MATH is either gaussian or of bounded support, see NAME REF , and hence the joint distribution of MATH is uniquely determined by mixed moments MATH. We will show by induction with respect to MATH that MATH for all MATH and all non-negative integers MATH. By NAME REF marginal distributions are equal, MATH; this shows that equality REF holds true for all integer MATH when MATH. Suppose REF holds for all MATH. Fix integer MATH. Expand polynomial MATH into orthogonal expansion, MATH. Then MATH . Repeating the reasoning that lead to NAME REF , we have MATH. Therefore MATH is expressed as a linear combination of moments that involve only MATH. Since the same reasoning applies to MATH, we have MATH, and REF follows. |
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 is the usual partial time derivative). Since MATH where MATH is the NAME curvature of MATH, and since MATH is MATH and MATH forms a multiplicative algebra whenever MATH, we see that the linear operator MATH maps MATH back into MATH. Denote by MATH the vector field MATH . Then, MATH where MATH for all MATH, MATH and MATH for all MATH, MATH, and MATH. It follows from REF , and REF that MATH is a MATH vector field. Thus REF is an ordinary differential equation on the tangent bundle MATH governed by a MATH vector field on MATH; it immediately follows from the fundamental theorem of ordinary differential equations on NAME manifolds, that REF has a unique MATH solution on finite time intervals which depends smoothly on the initial velocity field MATH, that is, there exists a unique solution MATH with MATH dependence on initial data MATH, where MATH depends only on MATH. When MATH, this interval can be extended globally to MATH by virtue of MATH remaining in MATH. Unfortunately, the global existence and uniquess of a MATH flow map MATH does not follow for initial data MATH for MATH, so we provide a simple argument to fill this gap. We must show that MATH can be continued in MATH. It suffices to prove that MATH and MATH are both bounded in MATH. This is easily achieved using energy estimates. We have that MATH and MATH . Computing the MATH norm of MATH and MATH, respectively, we obtain MATH and MATH . It is easy to estimate MATH where the first inequality is due to NAME and NAME 's inequalities and the second is the NAME embedding theorem. Similarly, MATH . Since the solution MATH to REF is in MATH for all MATH, we have that MATH is bounded for all MATH. Because the vorticity MATH is in MATH, we have by REF that MATH; hence MATH is bounded for MATH. It then follows that MATH and MATH are in MATH for all time. |
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 not leave MATH. Following the proof of REF , and using the fact that the solution MATH to REF remains in MATH for all time CITE, it suffices to prove that MATH is bounded in MATH. Letting MATH denote the potential vorticity, and computing the curl of REF , we obtain the REFD vorticity form as MATH . It follows that for all MATH, MATH is bounded in MATH (conserved when MATH) and therefore by standard elliptic estimates MATH is bounded in MATH, and hence in MATH. |
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 MATH for infinitely many MATH (recall MATH). Choose a subsequence MATH such that MATH. Since MATH is cofinal, there is a unique MATH such that MATH for MATH; then MATH and hence MATH is compact. |
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 proof are similar to those given in CITE. |
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 the inverse image of MATH is MATH and hence MATH is continuous. Now suppose that the image of MATH is cofinal, then the procedure defined above gives a continuous inverse for MATH. Given MATH, then since MATH is cofinal, the intersection MATH contains a single point MATH. Note that MATH depends on MATH continuously. |
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 whenever MATH. To see this we apply REF to give MATH where MATH. Hence the MATH - algebras MATH, form a directed system as required. |
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 to show that it is faithful on MATH, for all MATH. Fix MATH and MATH with MATH we need only show that MATH. Since MATH we have MATH . Hence MATH and MATH is faithful on MATH. Let MATH be a nonzero positive element; then since MATH is faithful MATH and as MATH is faithful on MATH we have MATH hence, MATH and MATH is faithful on MATH as required. |
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 have MATH, since MATH and MATH is injective. NAME follows from the fact that MATH together with the observation that MATH. The same argument shows that MATH and so MATH. For REF we note that there is a surjective MATH-homomorphism MATH such that MATH for MATH (see REF ) which is clearly equivariant for the respective MATH - actions. Moreover by CITE we have MATH for all MATH and so the result follows For REF note that the injection MATH extends naturally to a homomorphism MATH which in turn induces a map MATH characterised by MATH for MATH. Let MATH be the fixed point algebra of the gauge action of MATH on MATH restricted to the kernel of MATH. The gauge action restricted to MATH descends to an action of MATH on MATH which we denote MATH. Observe that for MATH and MATH we have MATH hence MATH (if MATH then MATH). By the same formula we see that MATH and the result now follows by REF. The last assertion follows from REF together with the fact that MATH (see REF). For REF define a map MATH given by MATH; this is surjective as these elements generate MATH. We note that MATH carries a MATH action MATH defined for MATH and MATH by MATH. Injectivity then follows by REF, since MATH is equivariant and for MATH we have MATH. |
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 aperiodic points for every MATH. Conversely, suppose that MATH satisfies the aperiodicity condition, then for every MATH there is MATH which is aperiodic. Then MATH is aperiodic. Hence the aperiodic points are dense in MATH. |
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 follows that MATH and hence MATH; in which case MATH, a contradiction. |
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, suppose that MATH is minimal and that MATH and MATH then there is MATH such that MATH are in the same orbit. Hence there exist MATH such that MATH; then it is easy to check that MATH and MATH have the desired properties. |
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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