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math/0102129 | The proof is easy. |
math/0102129 | REF is obvious. To prove REF, let MATH be the critical points of MATH and MATH be the critical points of MATH. Notice that MATH and MATH. We saw in the proof of REF that the inner itineraries of MATH and MATH are the same. Thus one gets that MATH iff MATH. If the interval MATH contains a preimage of MATH then MATH. But this occurs iff MATH and so the interval MATH contains a preimage of MATH. Since the points in MATH and MATH have the same inner itinerary, the proof is finished. |
math/0102129 | The first statement it is easy. Let's to prove the second one. It is sufficient to prove that if MATH and MATH are MATH-admissible m.c.d., where MATH is transitive, then any critically finite multimodal map of type MATH with combinatorial type MATH has a restrictive interval of combinatorial type MATH. Let MATH be an extended map for MATH. Consider the representation of MATH given by the definition of MATH-product. Then MATH. In other appropriate representation (given in REF ) MATH-c-MATH. Let MATH be the unique isomorphism with maps the first representation to the second one. For MATH, let MATH be the minimal interval in MATH which contains all points in MATH. Then MATH is a periodic interval for MATH with combinatorial type MATH. |
math/0102129 | The proof is exactly the proof of REF in pg. CASE: We will omit the details. The argument is by contradiction. Assume that MATH . Denote MATH. Using the argument as in CITE, we can prove CASE: For MATH: MATH CASE: We also have, for MATH: MATH . It follows MATH . Apply this inequation recursively to obtain MATH . Select MATH and MATH such that MATH and MATH such that there is a critical point MATH such that MATH. Then MATH . Because MATH and since MATH is a critical point to the extended map MATH, one gets MATH . But MATH . Which is a contradiction with REF . |
math/0102129 | By NAME fixed point theorem (see the apendice), there exists a fixed point to the operator MATH associated to a good family MATH and the essential combinatorial type MATH. |
math/0102129 | The following statement, proved by induction, is sufficient to prove the lemma: Let MATH: if MATH then MATH. If MATH then MATH, where MATH are multi-variable polynomials and MATH. |
math/0102129 | We claim that the connectivity locus is contained in the set MATH. Indeed, take a polynomial MATH in MATH outside this set. Let MATH such that MATH. Consider a critical point MATH such that MATH. We claim that MATH goes to infinity. This is consequence of a simple fact: if MATH is such that MATH then MATH for all MATH. Let MATH any polynomial. Take MATH like above. Then, using the fact above, MATH. In particular, in the connectivity locus the NAME sets are in a fixed neighborhood of zero. Now is easy to see that the connectivity locus is closed. |
math/0102129 | Let MATH be a polynomial of type MATH in MATH. In particular MATH. Then for any MATH and MATH there exists MATH such that MATH for all MATH, MATH. Furthermore there exists a constant MATH such that MATH. Suppose that MATH. For MATH, we obtain MATH . Thus if MATH is small enough, then the map MATH is a strict contraction, and the lemma follows. |
math/0102129 | Take MATH. Let MATH be the distinct critical points, and MATH the critical values, with MATH and MATH. Consider the polynomial MATH. Observe that MATH and MATH, MATH. We conclude that MATH. Since that MATH we obtain MATH. This implies MATH (because MATH and MATH are distinct of zero). Hence MATH. |
math/0102129 | Each polynomial MATH has MATH distinct critical points. Order in an arbitrary way the critical points of MATH. Define, for P in a neighborhood MATH of MATH, MATH, the closest critical point of MATH to MATH. The functions MATH are analytic. Define MATH by MATH. We claim that MATH is a local diffeomorphism. Indeed, if MATH, then the derivatives of MATH are MATH (in this formula, MATH). Because the critical points are distinct, the Jacobian of MATH is not zero. Reduce MATH, if necessary, to assume that MATH a diffeomorphism. Take MATH. Then there exists only one partition of the critical points MATH such that MATH has MATH elements and MATH is atomic. Moreover, if MATH is close to MATH, MATH and MATH respect the same partition. In other words, MATH if and only if MATH. In MATH, consider the afinne space MATH . Observe that MATH is MATH-dimensional and if MATH is close to MATH, MATH iff MATH. Hence MATH is a n-dimensional manifold. |
math/0102129 | First of all, we can assume, using the uniformization NAME mapping, that MATH, for MATH, and MATH. Assume that the diameter of MATH is very big. Hence MATH are polynomial like maps. We will obtain quasiregular extensions MATH of MATH and MATH of MATH whose are compatible-MATH. Choose MATH small and define MATH and MATH. Suppose that we had defined MATH, where MATH is a very fine ring such that the extern boundary of MATH is exactly the boundary of MATH, MATH is the pre-image of MATH by MATH and MATH is a analytic homeomorphism. Define MATH and MATH as an analytic homeomorphism such that the following diagram commute MATH . Let MATH be a quasiconformal map which glues the maps MATH, extending the map to identity outside MATH. Now, we are able to define the quasiregular extensions MATH . Here MATH. It is easy to see that these extensions are compatible. Make the pullback of the trivial NAME field outside MATH by the quasiregular mapping MATH. We obtain an invariant NAME field MATH for MATH (defining the field trivial under K REF ). Define MATH. These NAME fields are trivial in a neighborhood of infinity. Let MATH be the quasiconformal map so that MATH, MATH, MATH, MATH. Define MATH. Then MATH are polynomials. Moreover MATH is hybrid conjugated to MATH. |
math/0102129 | Let MATH and MATH . Denote MATH and define MATH by MATH and MATH. The diagram MATH commutes. Fix an arbitrary MATH and assume, without loss of generality, that the domain of MATH is MATH, for some MATH and MATH is contained in MATH. Consider the fundamental annulus MATH and a compact set MATH such that MATH. Then for any point MATH, there exist MATH such that MATH. Select a homeomorphism MATH such that MATH. The transformation MATH is defined in the open set MATH and with values in MATH, where MATH and MATH. Since MATH, one gets MATH . For some MATH and for points in MATH. Since MATH we can assume, replacing MATH by an appropriate translation of MATH by an integer, that MATH. Apply REF recursively to obtain MATH for MATH and points in MATH. Let MATH . Then for any MATH, select MATH such that MATH, with MATH. If MATH then MATH . The above proof is a variation of the NAME and NAME 's proofCITE when MATH. So MATH, for MATH, because MATH is a local isometry. Hence MATH in the Euclidean topology, when MATH, since MATH near to MATH. |
math/0102129 | Using notations as in the proof of REF , we have that, since the maps MATH are homotopic, that MATH, where MATH does not depend on MATH. Moreover the constant MATH in REF can be select independent of MATH, which is sufficient to prove the lemma. |
math/0102129 | As in the proof of REF, we can construct a isotopy MATH so that MATH, MATH is a conjugation and furthermore MATH is quasiconformal. Consider MATH. For each MATH, MATH is an automorphism of MATH in MATH. It is well know that there is a quasiconformal map MATH which is a conjugacy between MATH in MATH and MATH, where MATH. So MATH satisfies the hypothesis of REF , and furthermore MATH. In particular, MATH is at a finite hyperbolic distance of MATH, independent of MATH. Since MATH is quasiconformal, the same is true for MATH with respect the hyperbolic metric of MATH. In particular we can extend MATH in a continuous way to MATH setting MATH, for MATH. Extend MATH setting MATH. |
math/0102129 | Let MATH be a sequence in MATH. Replacing MATH by a polynomial like map which is affine conjugated to it, we can assume MATH. Consider MATH-quasiconformal maps MATH and polynomial maps MATH as in REF . Since MATH is compact, select a subsequence, if necessary, such that MATH. Since the NAME set of MATH has the diameter away of infinity and zero, and MATH, selecting a subsequence we can assume that MATH converges to a MATH-quasiconformal map. It is not difficult to see that MATH is an analytic map in MATH. |
math/0102129 | Easy. |
math/0102129 | Assume that MATH. Note that, since the boundary of MATH is a quasicircle, the map MATH has a quasiregular extension in a neighborhood of MATH. Indeed, let MATH be a MATH-quasiconformal map which is conformal in MATH and maps MATH in MATH. Then MATH extends to a rational map MATH which is a expansive map of degree MATH in MATH. Hence MATH is a regular map of degree two in a neighborhood of MATH. The same can be done for MATH. Let MATH be a MATH-quasiconformal map which maps MATH in MATH. Since MATH and MATH are contained in the NAME sets of MATH and MATH, these points are at a definitive Euclidean distance of MATH and MATH. Thus, by REF , we can assume that MATH. By induction, suppose that we have constructed MATH-quasiconformal maps MATH, for MATH between MATH and MATH, such that CASE: MATH in MATH. CASE: MATH. CASE: The quasiconformality of MATH is bounded by a constant MATH. Let MATH be a lift of MATH (in other words: MATH), which has the same quasiconformality than MATH. Because the critical values of MATH does not intercept MATH, and the modulus of this annulus is bounded below, MATH is at a bounded distance of MATH in the hyperbolic metric on MATH. Since the same can be said about MATH, by REF , if necessary modify MATH such that MATH, and additionally the new MATH is MATH-quasiconformal. In particular MATH in MATH. We can find a MATH-quasiconformal map MATH such that REF MATH is equal to MATH outside REF MATH is equal to MATH in MATH. Hence MATH is a quasiconformal map which maps MATH in MATH and conjugates MATH and MATH in the boundary of this fundamental annulus. Now, with the usual pullback argument, construct a MATH-quasiconformal conjugacy MATH between MATH and MATH in MATH in MATH such that MATH. For the last step, to obtain a conjugacy which extends to MATH, the result follows of the particular case when MATH is a polynomial and the annulus MATH is invariant by the extern automorphisms of MATH. Select a extern automorphism MATH so that MATH. Thus the conjugacy MATH in MATH glue with the external conjugacy MATH and the new map MATH has the same quasiconformality of MATH outside MATH. |
math/0102129 | Here we have a very nice situation: the map f is a polynomial, it has negative NAME derivative and moreover satisfies properties analogous to the standard conditions (see CITE). It is easy verify in the proof of the complex bounds for analytic multimodal maps CITE that MATH and MATH(for this use REF ) can be select independent of MATH. |
math/0102129 | Let MATH be the restrictive interval associate to the MATH-th renormalization of MATH, MATH. Since the period of MATH is bounded by MATH, the length of MATH can not be small, otherwise MATH will contain a periodic point which attracts zero, which is absurd. So we can assume that MATH converges to a periodic interval MATH for MATH, which proves that MATH belongs to MATH. In particular, all periodic points of MATH in MATH are reppeling (because non reppeling periodic points attracts a critical point), so the periodic point in the boundary of MATH converges to a periodic point in MATH. Thus MATH is the unique restrictive interval associate to the MATH-th renormalization of MATH and the MATH-th restrictive interval moves continuously in MATH and so do the postcritical set. |
math/0102129 | Follows of REF . |
math/0102129 | Follows of the connectedness principle by NAME (pg. REF) that MATH is connect. Let MATH be the connect component of MATH which contains MATH. Then MATH is a polynomial like map, and moreover MATH, since MATH is totally invariant by MATH. Hence we obtain REF , if MATH, or REF, otherwise. The last statement of REF follows of REF. |
math/0102129 | The proof of REF is exactly as in the unimodal case (see CITE): we can assume that MATH is a polynomial. Suppose, by contradiction that MATH has interior. Then, by the NAME 's classification of periodic components, the interior of MATH contains an attractor or a NAME disc: the first case is impossible because MATH is a NAME set (and any attractor attracts a critical point) and the second one does not hold because the boundary of a NAME disc must be contained in the postcritical set. The second statement is consequence of the ergodic or attract theoremCITE. REF is consequence of REF follow of the same statements for polynomials REF . The last item is obvious for polynomials, since MATH is commensurable with the length of MATH, which goes exponently fast to zero REF . Now the general case is easy. |
math/0102129 | It is easy too see there is a bound for MATH which depends only on MATH and MATH. So it suffice to proof that there is a point MATH whose hyperbolic distance to MATH is under control. Firstly, assume that MATH is a polynomial in MATH and MATH is the holomorphic moving fundamental annulus MATH selected in REF. We will prove that there exists MATH such that for each MATH there exist points MATH, MATH and a topological disc MATH such that MATH and MATH satisfying: MATH . Indeed, for each MATH with combinatorics bounded by MATH, select a point MATH and a topological disc MATH which contains MATH and a point MATH. Furthermore MATH. since MATH and MATH moves continuously with MATH, for MATH close to MATH we have MATH. Furthermore there is a point MATH of MATH close to MATH. In particular, MATH is under control. Since MATH is compact, the prove is finished. |
math/0102129 | Firstly note that, by REF and a priori bounds, MATH is contained in a larger small NAME set MATH so that the diameters of MATH and MATH are commensurable. Furthermore the diameters of MATH and MATH are also commensurable. The hyperbolic length of MATH is not large because there is fundamental ring for each renormalization with definitive modulus. Since MATH is a MATH-uniform domain, by REF the Euclidean diameter of MATH is commensurable to MATH. It is easy to see that the Euclidean length of MATH is also commensurable to MATH. If the hyperbolic length is small then the Euclidean diameter of MATH will be small relative to MATH, which is a contradiction. The second statement is consequence of the first one and REF . |
math/0102129 | The prove is quite similar to the proof in the unimodal case (see CITE): Consider a decomposition in ramified coverings of degree two MATH, MATH, and the associate extended map MATH defined in MATH by MATH . Define the postcritical set of MATH by MATH. If MATH denote the hyperbolic metric on MATH (extended to MATH on MATH) then MATH , with MATH in the tangent space of MATH, since MATH. Let MATH, with MATH. Then MATH , since MATH is dense in MATH. For MATH, select a vector MATH in its tangent space so that MATH. There are REF cases: CASE: MATH, CASE: MATH and there exists MATH such that MATH and MATH, CASE: MATH and MATH. Here MATH is sufficiently small so that the NAME 's argumentCITE works in the second case. The first and second cases are more easy and we will omit the proof. For details, see CITE. Assume the last situation. In particular MATH, because MATH is a MATH-uniform domain. Consider the piece, defined by closed geodesics, which contains MATH. Let MATH be the exterior geodesic and let MATH be the interior boundary geodesics. Denote by MATH the subset of the postcritical set bounded by MATH. Select MATH minimal so that we can do the univalent pullback of the domain MATH bounded by MATH along the inverse orbit MATH. This means there exists a simply connected domain MATH satisfying CASE: MATH. CASE: The map MATH is univalent in MATH and moreover MATH. CASE: The domain MATH contains a critical value MATH of MATH. Denote by MATH the domain bounded by MATH and let MATH be the corresponding domain in MATH. Let MATH . Then MATH is a proper map of degree two. Since the postcritical set is contained in MATH, the critical value in MATH is contained in some MATH, for some MATH. Choose an arbitrary MATH, MATH MATH and MATH be two paths inside the piece which contains MATH so that: CASE: The initial point of both is MATH. CASE: The end point of both is a point in MATH. CASE: The NAME curve defined by MATH and MATH is not homotopic to a constant curve in MATH. CASE: The hyperbolic diameter of MATH on MATH is bounded. Let MATH and MATH be lifts with respect to MATH of the simply connected sets MATH and MATH so that MATH is an arc whose initial point is MATH. Note that MATH and MATH are disjoint and one of them, say MATH, does not intersect the postcritical set of MATH. So all inverse branches of MATH are well defined on MATH. So let MATH the inverse branch of MATH, defined in MATH so that MATH. Since MATH and MATH has bounded hyperbolic diameter in MATH, we obtain, by REF , MATH for all MATH. By the maximum principle the same distortion control holds in MATH. There exists a small NAME set MATH inside MATH whose diameter is commensurable to MATH. By MATH-Koebe lemma (use that MATH) and the above distortion control, the set MATH has diameter commensurable with MATH. Moreover MATH, which proves the proposition. |
math/0102129 | Suppose, by contradiction, there exists an invariant line field MATH supported on the NAME set MATH on MATH. Select an almost continuity point MATH to MATH. There are polynomial like maps, with definitive modulus, in all scales around MATH, which preserves MATH. After an affine conjugation, we can assume that a subsequence of these polynomial like maps converge to a polynomial like map which preserves a straight line field, which is a contradiction. The second statement is consequence of the lemma in pg. REF. The last statement is an immediate consequence of the first ones. |
math/0102129 | We will define, by induction, homeomorphisms MATH, real in the real line and increasing, such that MATH. Assume that we have defined MATH. We claim that MATH. Indeed, consider MATH and MATH. Since MATH and MATH, we have MATH. Since MATH is increasing, it suffice to show that if MATH is the MATH-th point in MATH, with respect to the order in the real line, then MATH is also the MATH-th point in MATH. But this follows of REF , since MATH and MATH have the same inner itinerary. Now we can find a homeomorphism MATH, real in the real line and increasing such that MATH commutes, which proves the lemma. |
math/0102129 | This lemma is consequence of the bounded geometry (see REF ). For details see CITE or CITE(last chapter). |
math/0102129 | Replacing MATH and MATH by smaller domains, we can assume that the boundary of these domains are quasicircles. Using similar arguments as in REF , we can construct a quasiconformal map MATH which conjugates MATH and MATH in MATH. Since the maps MATH and MATH are real, we can assume that MATH is symmetric with respect to the real line. Let MATH be a symmetric quasiconformal map which conjugates MATH and MATH in the postcritical set. Construct a MATH-quasiconformal map MATH, for some MATH, which is symmetric, increasing in the real line, equal to MATH in MATH and equal to MATH in a neighborhood of MATH. As MATH and MATH have the same combinatorial type, the relative positions of MATH and MATH are the same. Thus we can use REF . Furthermore MATH and MATH has the same inner itinerary. Define inductly MATH, a MATH-quasiconformal map symmetric, increasing and such that MATH. Note that MATH is MATH-quasiconformal conjugacy in the postcritical set and MATH. Moreover MATH in MATH. Since MATH has empty interior, the sequence MATH has an unique limit MATH, which is a conjugacy between MATH and MATH. Indeed MATH is a hybrid conjugacy by REF . |
math/0102129 | Let MATH be the set of primitive, transitive m.c.d with combinatorics bounded by MATH. By REF , for any infinity sequence MATH, with MATH, there exists an infinitely renormalizable real polynomial map of type MATH with this combinatorial type. By the previous theorem, two real polynomial maps of type MATH with same combinatorics are hybrid conjugated and so affine conjugated, since they are polynomials. The point MATH must be a fixed point for this affine map. Thus the conjugacy must be the identity. Let MATH be the application which maps each sequence MATH in the unique real polynomial map MATH of type MATH with this combinatorial type. If MATH is a sequence which converges to MATH, then any accumulation point MATH of the sequence MATH is a real infinitely renormalizable polynomial of type MATH with combinatorics MATH. So MATH. Hence MATH is a homeomorphism between the NAME set MATH and MATH. |
math/0102129 | Since MATH is a covering map MATH and furthermore MATH we obtain MATH . Since the hyperbolic metric in MATH converges to the hyperbolic metric in MATH, we obtain the lemma. |
math/0102129 | We sketch the NAME 's proof: Consider MATH. We have MATH, for some MATH. so MATH . Since MATH, by the inclusion contraction lemma REF : MATH . It follows that MATH. Now it is suffice to observe that MATH. |
math/0102129 | The first statement is an immediate consequence of REF . For the second statement, note that MATH, for same MATH, and MATH. The map MATH is a covering map, with MATH, thus we can apply REF (twice) to obtain REF. |
math/0102129 | Let MATH be a complex number which is not contained in MATH. Then MATH is not in MATH for sufficiently small MATH. Let MATH be maximal such that MATH. For each small MATH let MATH be minimal so that MATH and let MATH be the minimal geodesic between MATH and MATH in MATH. The hyperbolic length of MATH is smaller than the constant MATH in REF , since MATH. By the previous lemma, the length of the lift MATH in the hyperbolic metric of MATH goes exponentially fast to zero, when MATH goes to infinity. |
math/0102129 | Let MATH be an invariant line field to the tower MATH. Because MATH does not support invariant line fields, it is possible select a point MATH where MATH is almost continuous. This means MATH . Here MATH is the NAME measure in MATH. Since MATH is dense and by small NAME sets everywhere theorem, for any MATH there exists a polynomial like map MATH so that: CASE: The map MATH belongs to MATH, for some MATH (indeed, for any MATH small enough); CASE: The map MATH belongs to MATH; CASE: MATH; CASE: MATH. Since MATH is invariant by these maps, normalizing MATH so that MATH, we can select a subsequence which converges to a polynomial like map which preserves a straight line field. This is absurd. |
math/0102129 | Existence: Select a real infinitely renormalizable polynomial MATH of type MATH with combinatorics MATH so that for any MATH-bounded combinatorics there exists a sequence MATH satisfying MATH. Using the complex bounds, select, for renormalizations deep enough, polynomial like representations in MATH. Then the finite tower MATH defined by MATH has a subsequence which converges to a bi-infinite tower in MATH with combinatorics MATH. Unicity: Let MATH and MATH be bi-infinity towers in MATH. Since MATH and MATH have the same combinatorics, there exists one MATH-quasiconformal map MATH which maps MATH in MATH and it is a conjugacy between MATH and MATH in MATH. When MATH we have MATH. Thus MATH admits a convergent subsequence to some quasiconformal map MATH. This map is a conjugacy between the tower MATH and the tower MATH, where MATH is equal to MATH restricts to MATH. Since the NAME field MATH is invariant by the tower MATH, the rigidity of towers implies that MATH is conformal. Thus, up to affine maps, MATH is the identity. |
math/0102129 | Since MATH there exists a MATH-quasiconformal conjugacy between MATH and MATH. Normalizing MATH and MATH so that MATH, we obtain quasiconformal conjugacies MATH so that MATH and MATH. We claim that MATH converges uniformly in compact sets to identity. Indeed, suppose by contradiction there exist sequences of maps MATH and MATH, MATH and MATH , with same combinatorial type so that the corresponding conjugacies MATH does not converge in an uniform way to identity: in other words we can select MATH so that MATH, for some MATH and with MATH. But REF say that a subsequence of MATH converges to a conjugacy between two bi-infinite towers, which do not support invariant line fields, so this conjugacy is a conformal map, hence it is the identity, which is a contradiction. Since MATH, we can select representations MATH which belongs to MATH and furthermore they are restrictions of iterates of MATH. This is possible for MATH. Then MATH, where MATH and MATH, is a representation of MATH. Since MATH, for some MATH, and MATH is close to identity, one gets MATH for MATH. By REF , MATH, which proves the first statement. To proof the second one, note that MATH and MATH, for MATH. By REF MATH is a representation in MATH, where MATH. Since MATH and the diameter of MATH, after the normalization MATH, is bounded yet, the proof is finished. |
math/0102129 | Since the NAME 's proof REF , without modifications, works perfectly well in our situation, we will not repit it here. |
math/0102129 | The proof will be exactly as in REF, with some small modifications to avoid technical definitions: Suppose, by contradiction, there exist sequences MATH, MATH, MATH and MATH so that CASE: MATH, CASE: MATH, CASE: MATH. We can assume that MATH. Then MATH, with MATH, which implies that, for all MATH and MATH, MATH, if MATH is large enough. In particular, by small NAME sets everywhere lemma, for any MATH and MATH there exists a sequence of polynomial-like maps MATH, for MATH large enough, so that CASE: MATH, CASE: MATH, MATH, CASE: MATH, CASE: MATH. Because MATH, with MATH, we can assume that the sequence MATH, up to sums with conformal fields in the NAME sphere, converge uniformly to a quasiconformal vector MATH. Moreover MATH as distributions. In particular MATH is invariant by any limit of the sequence MATH. We claim that MATH. Otherwise, select MATH a point of almost continuity of MATH. Hence MATH is almost a straight NAME field near to MATH, which is impossible since there are polynomial-like maps (which form a compact family after conjugacies by affine maps) in all scales so that MATH is invariant for them. But this is a contradiction, since MATH implies MATH, so MATH. |
math/0102129 | By the dynamic inflexibility theorem CITE, if MATH is a MATH-quasiconformal map between MATH and MATH, MATH, with MATH then MATH is MATH-conformal at MATH. It is not difficult to verify in the proof of dynamic inflexibility theorem that MATH . Here MATH is as in the previous lemma. Note that we can select MATH such that MATH. Hence MATH satisfies MATH . Since MATH for some MATH and MATH, after normalize MATH so that MATH, MATH define a conjugacy MATH satisfying MATH for some MATH. Using arguments as in REF , the proof is finished. |
math/0102129 | Suppose that MATH does not have a fixed point in the interior of MATH. Let MATH be the radial projection of unit ball MATH in MATH. Let MATH. Then MATH by the NAME.-Nagy-Klee Theorem. For a point in the interior of MATH, define the continuous function MATH, where MATH is the unique point where the ray beginning at MATH and containing MATH crosses MATH. We claim that MATH has a continuous extension to MATH such that MATH for any MATH, MATH and MATH, which is absurd, since MATH will be a retraction of MATH in its boundary. Indeed, if MATH are as in REF , we have MATH, MATH and MATH . Hence MATH converges to zero when MATH tends to MATH in an uniform way with respect to MATH. This proves the claim. |
math/0102129 | Note that MATH otherwise there will be a large essential ring in MATH. Denote MATH and fix MATH. Select an arbitrary MATH and define MATH . Let MATH, MATH, be a curve which touch both components of MATH. Then the Euclidean length of MATH is at least MATH and, if MATH is the hyperbolic metric on MATH then MATH which proves the lemma. If the diameter of MATH is large relative to diameter of MATH then MATH crosses many rings MATH, so its hyperbolic length is large, which is absurd. To obtain the lower bound to MATH, notice that MATH for some MATH in MATH and MATH. It is easy to see that MATH ,for any i. Since MATH, the proof is complete. |
math/0102129 | By the NAME distortion theorem we immediately obtain MATH . Since MATH it is suffice to prove that MATH . Since MATH, we gets MATH . Note that MATH. Then MATH and we obtain MATH . The last two equations and the MATH-uniformity of MATH proves the lemma. |
math/0102129 | Let MATH be the minimal geodesic in the hyperbolic domain MATH between MATH and MATH. It is easy to see that there are MATH such that MATH. Here MATH. Thus we can apply the previous proposition (and the NAME distortion lemma) in the inverse branches MATH of MATH in each ball MATH. |
math/0102129 | It follows of REF or the moving lemma at pg. REF. |
math/0102129 | Assume, without loss of generality, that MATH. Consider a MATH- quasiconformal map in MATH such that MATH. After a composition with a NAME transformation which preserves the circle, we can assume that MATH. Furthermore, after translate and rotate MATH, we can assume that MATH and MATH. Since the set of MATH-quasiconformal maps on the plane such that MATH and MATH is compact, there exists MATH such that for MATH, MATH implies MATH. In particular MATH. By the previous lemma, we obtain a MATH-quasiconformal map MATH on the plane which is equal to MATH outside MATH and MATH. |
math/0102134 | First, we suppose that MATH is an operator MATH-bimodule. That REF immediately follows from the operator module over C*-algebra version of REF, but this result relies on several earlier results. For the convenience of the reader, we also include an outline of the proof of this part. MATH : We only need to show minimality of MATH. Let MATH be a tight MATH-injective operator MATH-bimodule such that MATH. Then MATH extends to a completely contractive MATH-bimodule map MATH. By REF , MATH has to be a complete isometry, so that MATH. MATH : The same as the proof of REF. When applying this proof, note that, if MATH is a MATH-projection on MATH (that is, a completely contractive MATH-bimodule map on MATH with MATH and MATH), then Im-MATH is a tight MATH-injective operator MATH-bimodule, so that Im-MATH by minimality of MATH. Hence MATH is a minimal MATH-projection on MATH and MATH is its range. MATH : The same as the proof (Necessity) of REF. MATH : Let MATH be a MATH-injective envelope of MATH. By MATH , MATH is a tight MATH-essential extension of MATH, so that MATH is a tight MATH-essential extension of MATH, hence MATH by maximality. Thus MATH is tight MATH-injective. MATH : Let MATH be a tight MATH-essential extension of MATH. By injectivity of MATH, MATH extends to a completely contractive MATH-bimodule map MATH. But by essentiality of MATH, MATH has to be a complete isometry. Thus MATH. Finally, we show completeness. Let MATH satisfy the equivalent REF - REF and let MATH be its completion. Then MATH is a MATH-bimodule and obviously a tight MATH-essential extension of MATH. By injectivity of MATH, MATH extends to a completely contractive MATH-bimodule map MATH with MATH completely isometric. But by an essentiality of MATH, MATH has to be a complete isometry, so that MATH. |
math/0102134 | Throughout the proof, without loss of generality, we may assume that MATH REF . That MATH was proven in REF. That MATH follows by noting that any MATH-injective extension of MATH which is completely boundedly isomorphic to a minimal MATH-injective extension as left MATH-modules is also minimal and that MATH is a minimal MATH-injective extension by REF. MATH : Without loss of generality we may assume that MATH. Let MATH be any operator left MATH-module and MATH a completely bounded left MATH-module map which is a completely bounded isomorphism on MATH. Let MATH be the inverse of this map and extend MATH to a completely bounded left MATH-module map MATH. Since MATH is the identity on MATH, it is the identity on MATH by rigidity REF . This shows that MATH is completely boundedly isomorphic to MATH. Maximality: Suppose that MATH is a MATH-essential extension of MATH. A completely boundedly isomorphic left MATH-module map MATH extends to a completely bounded left MATH-module map MATH. By essentiality of MATH, MATH has to be a completely bounded isomorphism, so that MATH. MATH : We may assume that MATH for some NAME space MATH by the representation theorem for operator modules (CITE, CITE, CITE). Extend the identity map on MATH to a completely bounded left MATH-module map MATH, then MATH is completely boundedly isomorphic to MATH. The map MATH extends to a completely bounded left MATH-module map MATH . By rigidity REF , MATH is the identity on MATH. Hence we have MATH with MATH completely boundedly isomorphic to MATH. This makes MATH an operator MATH-essential extension of MATH by applying MATH that we have already proved. Thus, by maximality of MATH, MATH so that MATH is completely boundedly isomorphic to MATH as left MATH-modules by MATH. Now we have that MATH , since clearly REF imply REF . Next, we show that MATH . By REF , the identity map on MATH extends to a completely bounded left MATH-module map MATH and the map MATH extends to a completely bounded left MATH-module map MATH. Then MATH is the identity on MATH. But by rigidity REF , MATH is the identity on MATH. MATH : Without loss of generality, we may assume that MATH. Then REF immediately follows from REF. MATH : By injectivity of MATH, the identity map on MATH extends to a completely bounded left MATH-module map MATH. Similarly, by injectivity of MATH, the identity map on MATH extends to a completely contractive MATH-bimodule map MATH. But MATH and MATH by rigidity of MATH REF and MATH, respectively. |
math/0102134 | MATH is clear. We show MATH. Let MATH and MATH a right approximate identity of MATH, then MATH. |
math/0102134 | MATH : Suppose MATH for MATH, then MATH, so that MATH. MATH : Let MATH be a closed right ideal in MATH with MATH, then MATH by REF , so that MATH by REF . MATH : Clear. MATH : Obviously MATH is a closed two-sided ideal in MATH, so we only show essentiality. Let MATH, then MATH is a closed two-sided ideal in MATH. Suppose that MATH and MATH, then MATH for all MATH, so that MATH, hence MATH. REF yields MATH. |
math/0102134 | MATH : Without loss of generality, we may assume that MATH. Let MATH be any operator space, MATH any complete contraction which is completely isometric on MATH. Since MATH is tight MATH-injective, MATH extends to a complete contraction MATH. Then, again, by tight MATH-injectivity of MATH, MATH extends to a complete contraction MATH with MATH. By rigidity REF , MATH, hence MATH has to be a complete isometry. MATH : Clear. MATH : Since MATH is tight MATH-injective, MATH extends to a completely contractive left MATH-module map MATH. By REF , MATH is a complete isometry. Since MATH is tight MATH-injective, MATH extends to a completely contractive left MATH-module map MATH. Then MATH and MATH are completely contractive left MATH-module map, so that MATH and MATH by rigidity REF . Hence MATH and MATH are onto complete isometries, so that MATH is completely isometrically isomorphic to MATH . MATH : Let us define MATH by MATH. And let MATH-Ker-MATH, then MATH is an operator space with a well-defined matrix norm MATH where MATH-Ker-MATH, MATH. Define MATH by MATH-Ker-MATH. Then MATH is a completely contractive liner map which is completely isometric on MATH, so that, MATH is completely isometric on MATH by REF . Hence MATH, MATH. That MATH , that MATH , and that MATH are clear. MATH : Let MATH be any operator right MATH-module and MATH any completely contractive right MATH-module map which is completely isometric on MATH. Then MATH and MATH, which shows that MATH is completely isometric on MATH . Where MATH is defined by MATH . MATH : Clear. MATH : Let MATH, then MATH is a closed two-sided ideal in MATH and MATH. Let MATH be equipped with the quotient norm that makes MATH a C*-algebra, hence an operator MATH-bimodule with induced matrix norms and natural module actions. Let MATH be the quotient map that is a *-homomorphism and also a completely contractive MATH-bimodule map. Since MATH, MATH is one-to-one on MATH, hence *-isometric on MATH. Thus MATH is completely isometric on MATH, especially on MATH. By REF , MATH is one-to-one on MATH, so that MATH, which means MATH is an essential left ideal. MATH : Let MATH be any operator right MATH-module and MATH any completely bounded right MATH-module map which is completely boundedly isomorphic on MATH. Then there exists MATH such that MATH with MATH, MATH. Hence MATH and MATH, Thus, MATH is a completely bounded isomorphism. MATH : Clear. MATH : Just the same as that MATH . MATH : Suppose MATH. Let MATH be the matrix such that MATH and all other entries are MATH. Then, by the assumption, MATH which implies MATH from REF , so that MATH. Hence MATH is an essential left ideal in MATH so that MATH is an essential left ideal in MATH . MATH : Fix MATH . Since MATH is an essential left ideal in MATH, the canonical embedding MATH is one-to-one REF , hence it is a *-isomorphism. MATH : Clear. MATH : Let a MATH . Let us consider the following maps. MATH . Where we regard MATH as a right NAME C*-module over MATH with the inner product MATH defined by MATH . MATH is the set of adjointable maps on MATH . MATH is the set of compact adjointable operators on MATH . Namely, MATH where MATH is defined by MATH for MATH and the closure is taken in MATH . The map MATH is the canonical embedding. The map MATH is defined in the following way: MATH is a well-defined *-homomorphism and extends to an onto *-homomorphism MATH . By REF , MATH is an essential left ideal in MATH thus MATH is one-to-one and hence a *-isomorphism. Therefore, we obtain a *-isomorphism MATH . The map MATH is defined in the following way: MATH where MATH . For a proof that MATH is a *-isomorphism, see CITE, CITE. We can easily see that MATH where MATH . Hence REF follows. MATH : The same as that MATH . |
math/0102134 | MATH : Clear. MATH : Similar to that MATH in REF . That MATH`` REF " is clear. |
math/0102135 | We repeat the argument from CITE, but in a more flexible formulation. Define MATH; our task is to show that MATH. Define the set MATH of vertical line segments by MATH . We introduce the function MATH defined by MATH. Since MATH we see that MATH is determined by MATH and also by MATH. From MATH and REF we also see that MATH is parameterized by MATH. To exploit these observations we apply the iterated popularity argument. Pick any MATH in the refinement MATH of MATH. By construction we have MATH . For each MATH there is at most one MATH which contributes, by REF. Since MATH is determined both by MATH and by MATH, we thus have MATH. On the other hand, since MATH is parameterized by MATH and MATH, we have MATH. Combining these two estimates we obtain MATH. On the other hand, from REF we have MATH, and the claim follows. |
math/0102135 | Since MATH is determined by MATH on MATH, it is determined by MATH, and we thus have MATH . REF then follows from REF. Define MATH. Since MATH is a refinement of MATH, we may find a MATH such that MATH . Fix this MATH, and define MATH by MATH. Since MATH is determined by MATH, we see that MATH so MATH is a refinement of MATH. Now we prove REF. Fix MATH. Since MATH is determined by MATH on MATH, we have some identity of the form MATH for some non-zero scalars MATH and all MATH. In particular, we see that MATH is determined by MATH on MATH. Thus to prove REF it suffices to show that MATH . On the other hand, from the construction of MATH we have MATH for all MATH. The claim follows. |
math/0102135 | By an obvious limiting argument it suffices to show that for any finite set of proper slopes MATH, we can find a finite set MATH of proper slopes such that MATH . Fix MATH. We now pick MATH, MATH, and MATH in such a way that the slopes in REF are proper and distinct. It is clear that this distinctness property holds for generic choices of MATH, MATH, and MATH. We then set MATH to equal to REF. Let MATH, and let MATH obey REF be such that MATH for all MATH. Then we may apply REF to obtain a MATH and MATH obeying the conclusions of that Lemma. We now apply the hypothesis MATH with MATH replaced by the smaller set MATH (note that REF is still true). From REF we have MATH; combining this with REF we obtain after some algebra MATH. Combining this from the bound MATH from REF we obtain the claim. |
math/0102135 | Fix MATH, MATH, MATH. The first step is to construct the set MATH of slopes. The construction will be quite involved, but is necessary in order to perform the rest of the argument rigorously. Let MATH be distinct proper slopes. We can then form as before the set MATH of corners, and the map MATH on MATH. Given any other slope MATH, recall that there exists a function MATH on MATH such that REF holds. For all but a finite number of exceptional MATH, this MATH is of the type discussed in the previous section, and in particular for any fixed such MATH, and all but a finite number of proper slopes MATH, there exists a proper slope MATH (distinct from MATH) such that MATH is a linear combination of MATH and MATH. It is too unreasonable to expect all of these slopes to lie in MATH. On the other hand, from REF and the hypothesis MATH we have MATH whenever MATH is a fractional linear transformation on MATH which preserves -REF. This gives us much more flexibility since we can always choose such a MATH even if a finite number of exceptional MATH are somehow prohibited. For instance, for fixed MATH, and for MATH avoiding a finite number of exceptional values (depending on MATH), one can find a set MATH of proper slopes such that MATH holds, and such that a dual proper slope MATH distinct from MATH exists for each MATH. Similarly, for fixed MATH, there exists a set MATH of proper slopes MATH avoiding MATH, MATH, MATH, and the exceptional values mentioned earlier, such that MATH holds. Fix the sets MATH and MATH. For any three distinct proper slopes MATH, we define the set of slopes MATH and set MATH of triples of distinct proper slopes by MATH . Let MATH (for instance), and define recursively MATH by MATH for MATH; note that the MATH always consist of triples of three distinct proper slopes. We then set MATH equal to MATH. The idea will be to try to run the heuristic argument using some triples from the set MATH. If at least one of these triples satisfies a certain uniformity property then the argument will run smoothly. If all the triples from MATH fail to be uniform, then we pass to the triples MATH. The point is that the failure of uniformity for MATH will allow us to be less strict about the uniformity required for MATH (we can lose an additional factor of MATH or so). We repeat this process as long as necessary. In the worst case we fall back all the way to MATH, but the uniformity requirement is now trivial (compare CITE, CITE). Henceforth all implicit constants in the MATH notation will be allowed to depend on MATH, MATH, MATH, MATH, while constants denoted by MATH are only allowed to depend on MATH. Let MATH obey REF and set MATH; we may assume MATH. Our task is to show that MATH . Define refinement MATH of MATH, where MATH is an arbitrary enumeration of the finite set MATH. By construction we can choose for each MATH and MATH a set MATH such that MATH. Fix the sets MATH. We define the modified vertical line segment sets MATH for MATH by MATH. Clearly we have MATH . Let MATH, and let MATH be an element of MATH. We say that MATH is MATH-uniform if we have MATH where MATH. For MATH close to MATH, this property asserts that the map MATH maps MATH maps evenly onto MATH. However this property becomes weaker as MATH decreases. For instance, it is clear that the singleton element of MATH is MATH-uniform. With the notation of the previous paragraph, we say that MATH is MATH-chunky if there exists a subset MATH of MATH such that MATH and MATH . Clearly if a triple MATH fails to be MATH-uniform, then it is MATH-chunky (just set MATH equal to the appropriate elements of MATH). Also, every triple in MATH is trivially MATH-chunky. From the above discussion, it is clear that we can find a MATH and a triple MATH which is MATH-uniform, and such that every triple in MATH is MATH-chunky. Henceforth MATH and MATH are fixed. The main issue here is to ensure the powers of MATH one loses in the MATH-uniformity property will be compensated for by the gains in MATH one will obtain from the MATH-chunkiness property. For each MATH, we have MATH, and hence that MATH is MATH-chunky. Hence we have MATH and MATH . We introduce the refinement MATH of MATH and observe MATH . We now run a modified iterated popularity argument. For any map MATH between finite sets, define MATH . Observe the strong refinement property MATH. We now define the refinement MATH of MATH, where MATH is some arbitrary enumeration of the projections MATH and the projections MATH are thought of as mapping to MATH. From the strong refinement property we have (for MATH sufficiently large) MATH; combining this with REF we obtain MATH . Let MATH be a slope in MATH, and let MATH be the associated map on MATH. Introduce the sets MATH. From REF we see that MATH. If we then introduce MATH of MATH then MATH . We shall need a bound on the cardinality of MATH. To do this we shall first repeat the argument in REF . By the construction of MATH we have MATH for all MATH; by REF we thus have MATH . Since MATH is determined by MATH, we thus have MATH . Fix MATH, and define the set MATH by MATH. Since MATH is parameterized by MATH, we see from construction that MATH . Inserting this into the previous we obtain MATH, we thus see from REF that MATH . Applying MATH we thus obtain that MATH. By REF we thus have MATH . Since this holds for all MATH, we thus obtain MATH which simplifies using REF to MATH . Having defined MATH, we now introduce the set MATH of corners. Since MATH is MATH-uniform, we see that MATH . From the construction of MATH we thus have MATH . If we therefore define the set MATH by MATH then we see from REF that MATH . From the construction of MATH we thus have MATH . For any MATH, define the map MATH by MATH where MATH is the map associated to MATH. We once again apply the iteration arguments of REF , defining the refinement MATH of MATH, where MATH is an arbitrary enumeration of the functions MATH. Since MATH is a refinement of MATH, we may fix a MATH such that MATH is a refinement of MATH. Let MATH denote the set MATH. We now obtain upper and lower bounds for the size of MATH. To obtain lower bounds, we argue as in REF . From the construction of MATH we have MATH for any MATH. Using this and the fact that MATH is determined by MATH we obtain MATH. On the other hand, from REF we have MATH . Combining these bounds with REF we obtain MATH . Since MATH is parameterized by MATH, we have MATH . Applying MATH we obtain the lower bound MATH . Now we obtain upper bounds on MATH. We observe as in the Heuristic proof that MATH is determined by MATH and MATH, so we have MATH . By REF we thus have MATH . Since the range of MATH has cardinality at most MATH, we thus have MATH. Combining this with REF we obtain MATH which simplifies using MATH to REF as desired. |
math/0102135 | Fix MATH, MATH, MATH. By raising MATH if necessary we may assume that MATH fails. It will then suffice to prove that MATH, since the claim follows by letting MATH. Since MATH holds and MATH fails, we may find an ambient dimension MATH, a saturated MATH-collection MATH of lines, and a saturated shading MATH such that MATH where MATH. It then suffices to show that MATH . We now perform a reduction similar to the two-ends reduction, which ensures that MATH has dimension MATH in an appropriate sense. Let MATH be a large constant to be chosen later, and let MATH be a dyadic radius. Call a ball MATH heavy if one has MATH . Let MATH denote the set of all points in MATH which lie in at least one heavy ball MATH. We claim that MATH if MATH is sufficiently large. To see this, suppose for contradiction that the above estimate failed. Then the shading MATH is saturated. We may then find a refinement MATH of MATH such that MATH has density MATH on MATH. The set MATH is contained in a union of heavy MATH-balls. By applying the NAME hypothesis at eccentricity MATH, applied to an appropriate blurring of the collection MATH and the shading MATH, we thus see that the number of heavy MATH-balls needed to cover MATH is MATH. From REF we thus have MATH contradicting REF if MATH is sufficiently large. This proves REF. Henceforth the implicit constants may depend on MATH. Set MATH and define the shading MATH on MATH. Since MATH is saturated on MATH, we see from REF that MATH is also saturated on MATH. Since MATH is saturated on MATH, we may find a refinement MATH of MATH such that MATH has density REF on MATH, so in particular MATH . We now apply the ``bilinear reduction" CITE. From REF and pigeonholing we have MATH for some MATH; the diagonal contribution MATH can be discarded since MATH. By another pigeonholing we can find a direction MATH such that MATH . As in CITE we can then rescale MATH to equal REF to obtain MATH where MATH. We refer to elements MATH as angles, and write MATH for MATH, MATH, MATH. MATH . Angles and pivots are the analogue of vertical line segments and values of MATH in REF. Let MATH denote the set of angle-pivot pairs. The following lemma asserts that a large fraction of the points coplanar to MATH and MATH are indeed pivots for MATH. For every MATH we have MATH . The upper bound is trivial, so it suffices to prove the lower bound. Fix MATH, and let MATH denote the set of triples MATH . Since MATH we see that MATH (if the constants are chosen correctly). For each triple MATH, we can find a point MATH which is distance MATH from MATH and MATH, is essentially coplanar with MATH, is essentially collinear with MATH, MATH, and is such that MATH is essentially parallel to MATH. Let us fix this map MATH. Since the MATH are essentially coplanar to MATH and MATH we see that MATH. On the other hand, we see from elementary geometry that MATH for all MATH, and thus MATH . Since all the elements MATH in this set are pivots for MATH, the claim follows. From the preceding Lemma and REF we thus have MATH . Let MATH denote the map MATH. We have MATH. This will be a variant of NAME 's argument. We first observe that MATH essentially determines MATH (up to a multiplicity of MATH), since MATH is parallel to MATH, which has magnitude MATH, and the MATH are direction-separated. (This is the analogue of MATH being determined by MATH in REF). Thus we may find a refinement MATH of MATH such that MATH is determined by MATH on MATH. By REF and a pigeonholing it suffices to show that MATH for each MATH. Fix MATH, and consider an element MATH. If MATH contributes to the above set, then MATH and MATH, MATH, and MATH are essentially coplanar, so the number of possible MATH is MATH since MATH is separated. For each such MATH, the number of MATH which can contribute is MATH. Since MATH is essentially parallel to MATH, the number of MATH which can contribute is MATH. The claim follows. We use this lemma to fix a refinement MATH of MATH such that MATH is parameterized by MATH. We now lift our NAME set to the space MATH (this is the analogue of the iteration argument in REF). If MATH, we define MATH to be a (discretized) line which contains the points MATH and MATH, and whose MATH variables ranges over the region MATH. For each MATH, define MATH. We also define MATH to be those elements MATH of MATH such that MATH appears in the left-hand side of REF. By construction, MATH is a shading with density REF on MATH. From elementary geometry we see that MATH for all MATH. MATH is a MATH-collection of lines for each MATH. Fix MATH. Since the direction of MATH is affinely determined by MATH, it suffices to show MATH for all balls MATH. On the other hand, from the construction of MATH we have MATH. Since MATH is parameterized by MATH, the claim follows. From this lemma and the hypothesis MATH we can now obtain a multiplicity bound on MATH. More precisely: Fix MATH. Then there exists a set of points MATH such that the shading MATH is saturated on MATH, and one has MATH on MATH. We set MATH where MATH is to be chosen later. Now suppose for contradiction that MATH is not saturated. Then the shading MATH must be saturated on MATH. There must therefore be a refinement MATH of MATH such that MATH has density REF on MATH. By the hypothesis MATH we thus have MATH . By construction, MATH is outside MATH. From the definition of MATH we thus have MATH . On the other hand, the left-hand side is clearly MATH. Thus we have a contradiction if MATH is sufficiently large. Fix MATH, and let MATH be as in the previous lemma. The counting function MATH has a MATH norm of MATH and a MATH norm of MATH. From NAME and a summation MATH we thus have the lower bound MATH. To prove REF, it thus suffices by REF to obtain the upper bound MATH . We expand the left-hand side of REF as MATH . Suppose MATH, MATH, MATH are such that MATH, MATH, and MATH are essentially collinear. Then there exists a MATH such that MATH. In light of this, the previous sum can be bounded from below by MATH where MATH . The previous sum can be simplified to MATH . By the NAME inequality, this can be bounded from below by MATH . From the definition of MATH it is clear that MATH . Also, for fixed MATH we see from the definition of pivot that MATH . Thus the left-hand side of REF is bounded from below by MATH and the claim follows from REF. |
math/0102139 | Let MATH (respectively, MATH) be the involutions defining MATH (respectively, MATH). Obviously, MATH, MATH and MATH. Moreover, we have MATH. |
math/0102139 | Statements a' and a" follow from REF . As regards REF , from the same lemmas we get: CASE: if MATH is odd and MATH, then MATH; CASE: if MATH is even and MATH, then MATH. It is easy to check that b'MATH-b" is equivalent to REF. |
math/0102139 | As explained above, the MATH-residues of the graph MATH are exactly MATH and they are all of length MATH. When MATH, the same property holds for the MATH-residues (and possibly for the MATH-residues) of MATH if and only if MATH and MATH, while it does not hold for the other types of residues. This proves the statement. |
math/0102139 | The ``if" part follows from REF . The ``only if" part will be proved in the Appendix. |
math/0102139 | Since MATH, MATH and MATH is odd, the condition MATH cannot be satisfied and the condition MATH is equivalent to MATH. This proves the statement. |
math/0102139 | Apply REF (including the note REF of page REF) and REF (see tables of page REF). |
math/0102139 | CASE: See REF. CASE: See CITE. |
math/0102139 | of REF Let MATH (respectively, MATH) be the involutions defining MATH (respectively, MATH). Moreover, let MATH and MATH be the maps MATH . Since MATH, the map MATH is a bijection. The pair MATH is an isomorphism between the graphs MATH and MATH. In fact, we get: CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH. |
math/0102139 | We have MATH. |
math/0102139 | of REF Let MATH (respectively, MATH) be the involutions defining MATH (respectively, MATH). Moreover, let MATH and MATH be the maps MATH . Since MATH sends the MATH-residues of MATH injectively onto the MATH-residues of MATH, then it is a bijection. We claim that MATH is an isomorphism between MATH and MATH. From the definition of MATH, we immediately get MATH and MATH. A direct computation shows that MATH if MATH is even and MATH if MATH is odd, where MATH and MATH only depend on MATH. Since MATH, MATH, MATH, MATH, MATH, MATH, MATH and MATH, for every MATH, from REF we get: MATH if and only if MATH and MATH if and only if MATH. Now we show, by induction, that REF MATH, for every MATH. By REF we only need to prove REF for MATH. First of all, we have MATH and MATH. Therefore, REF holds in MATH, for all MATH. Let us suppose that REF holds in MATH, with MATH. From REF we get, if MATH is odd, MATH and, if MATH is even, MATH. Therefore, REF holds in MATH, for all MATH. Then we show, by induction, that REF MATH, for every MATH. By REF we only need to prove REF for MATH. First of all, we have MATH and MATH. Therefore, REF holds in MATH, for all MATH. Let us suppose that REF holds in MATH with MATH. From REF we get, if MATH is odd, MATH and, if MATH is even, MATH. Therefore REF holds in MATH, for all MATH. b' Let MATH (respectively, MATH) be the involutions defining MATH (respectively, MATH). Moreover, let MATH and MATH be the maps MATH . Since MATH sends the MATH-residues of MATH injectively onto the MATH-residues of MATH, then it is a bijection. We claim that MATH is an isomorphism between MATH and MATH. From the definition of MATH, we immediately get MATH and MATH. A direct computation shows that MATH if MATH is even and MATH if MATH is odd, where MATH and MATH only depend on MATH. Since MATH, MATH, MATH, MATH, MATH, MATH, MATH and MATH, for every MATH, from REF we get: MATH if and only if MATH, MATH if and only if MATH, MATH if and only if MATH and MATH if and only if MATH. Now we show, by induction, that REF MATH, for every MATH. By REF we only need to prove REF for MATH. First of all, we have MATH and MATH. Therefore, REF holds in MATH, for all MATH. Let us suppose that REF holds in MATH with MATH. From REF we get, if MATH is odd, MATH and, if MATH is even, MATH. Therefore, REF holds in MATH, for all MATH. Then we show, by induction, that REF MATH, for every MATH. By REF we only need to prove REF for MATH. First of all, we have MATH and MATH. Therefore, REF holds in MATH, for all MATH. Let us suppose that REF holds in MATH with MATH. From REF we get, if MATH is odd, MATH and, if MATH is even, MATH. Therefore, REF holds in MATH, for all MATH. b" Follows directly from point b' and REF . |
math/0102139 | of REF Let MATH (respectively, MATH) be the involutions defining MATH (respectively, MATH). Moreover, let MATH and MATH be the maps MATH . It is easy to check that the map MATH, defined by MATH, is the inverse map of MATH; therefore, MATH is a bijection. The pair MATH is an isomorphism between MATH and MATH. In fact we get: CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH. |
math/0102139 | Since MATH, both MATH and MATH are bijections. Then we have: CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH. CASE: MATH, CASE: MATH; CASE: MATH; CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH. |
math/0102139 | First of all, observe that the vertex MATH of MATH lies in a MATH-residue of length MATH. All the MATH-residues (respectively, the MATH-residues) of MATH are mapped by any isomorphism either to the MATH-residues or to the MATH-residues of MATH. a' Each of the two MATH-residues of length MATH is mapped to a MATH-residue (where each vertex has second coordinate MATH) and each of the two MATH-residues of length MATH is mapped to a MATH-residue. a" Each of the two MATH-residues (respectively, the two MATH-residues) of length MATH is mapped either to a MATH-residue (where each vertex has second coordinates MATH) or to a MATH-residue (where each vertex has second coordinates MATH). CASE: The MATH-residue, the MATH-residue, the MATH-residue and the MATH-residue of length MATH are mapped to either the MATH-residue or the MATH-residue or the MATH-residue or the MATH-residue. Since each vertex of these four residues has second coordinate MATH, the statement holds. |
math/0102139 | (``only if" part:) From REF we get MATH and MATH. Let MATH (respectively, MATH) be the involutions defining either MATH or MATH (respectively, either MATH or MATH) and let MATH be the bijections defined in REF respectively. Moreover, observe that the pairs MATH and MATH defined in REF are automorphisms of both MATH and MATH. a' If MATH is an isomorphism between MATH and MATH then, by REF , MATH and, up to the action of MATH and MATH, we can suppose MATH with MATH. Then we have the following four cases: CASE: If MATH and MATH, then MATH and MATH is an automorphism of MATH. CASE: If MATH and MATH, then MATH and MATH is an isomorphism between MATH and MATH. CASE: If MATH and MATH, then MATH and MATH is an isomorphism between MATH and MATH. CASE: If MATH and MATH, then MATH and MATH is an isomorphism between MATH and MATH. a" If MATH is an isomorphism between MATH and MATH then, by REF , MATH and, up to the action of MATH and MATH, we can suppose MATH, with MATH if MATH and with MATH if MATH. Then we have the four cases of the previous point and the following four cases: CASE: MATH and MATH; CASE: MATH and MATH; CASE: MATH and MATH and REF MATH and MATH. In REF , if MATH is even (respectively, if MATH is odd) we get MATH (respectively, MATH). Since MATH and MATH, we get MATH, MATH, and therefore MATH. REF is a composition of REF is a composition of REF is a composition of REF . Therefore, we have: CASE: If MATH and MATH, then MATH and MATH is an isomorphism between MATH and MATH. CASE: If MATH and MATH, then MATH and MATH is an isomorphism between MATH and MATH. CASE: If MATH and MATH, then MATH is the map MATH and MATH is an isomorphism between MATH and MATH. CASE: If MATH and MATH, then MATH is the map MATH and MATH is an isomorphism between MATH and MATH. CASE: The pair MATH is an automorphism of MATH. Therefore, if MATH is an isomorphism between MATH and MATH then, up to the action of MATH, MATH and MATH, we can suppose that MATH, with MATH. By this assumption, MATH sends the MATH-residue of MATH onto the MATH-residue of MATH and therefore either MATH or MATH. We have four cases: CASE: If MATH and MATH, then MATH and MATH is an automorphism of MATH. CASE: If MATH and MATH, then MATH and MATH is an isomorphism between MATH and MATH (respectively, between MATH and MATH) if MATH is odd (respectively, even). CASE: If MATH and MATH, then MATH and MATH is an isomorphism between MATH and MATH. CASE: If MATH and MATH, then MATH and MATH is an isomorphism between MATH and MATH (respectively, between MATH and MATH) if MATH is odd (respectively, even). Therefore, if MATH is isomorphic to MATH, then REF MATH when MATH is odd and REF MATH when MATH is even. Since REF MATH REF is equivalent to MATH mod MATH, our proof is completed. |
math/0102144 | We are going to prove identities REF ; the other two identities are proved in a similar manner (and are much easier). Obviously, we have MATH . Therefore, MATH . Since MATH is self-adjoint, we have MATH . Using the fact that MATH is skew-adjoint, the left-hand side of REF can be rewritten as MATH so MATH (notice that MATH is real, see REF). This proves REF. Since REF implies MATH this also proves REF. |
math/0102144 | Let us sum REF over MATH. Then we have MATH . NAME 's equality implies that the left-hand side of REF is not greater than MATH. This proves REF. |
math/0102144 | Let us multiply REF by MATH and sum the result over all MATH. We will get: MATH . The left-hand side of REF can be estimated as follows: MATH . (The last equality uses the fact that the expression under MATH is skew-symmetric with respect to MATH.) Now REF imply REF. |
math/0102144 | Acting as in the proof of REF we get MATH . Also, MATH and, similarly, MATH . Therefore, REF can be re-written as MATH . Finally, using REF , we have MATH and MATH . The Theorem now follows by combining REF and using REF. |
math/0102144 | Subtracting identity REF from REF, taking the absolute value, and using the triangle inequality and REF, we get MATH . The left-hand side of this inequality is estimated from above by MATH which implies REF. The estimates REF are obtained by applying exactly the same procedure to REF separately. |
math/0102144 | The equalities REF - REF are proved in CITE; it remains only to prove REF . Using the definitions of MATH, MATH, and integrating by parts, we have MATH . |
math/0102146 | We may assume MATH. Let MATH be any finite subset of MATH. By CITE, there exist vectors MATH and MATH of norm at most MATH in a NAME space MATH such that MATH for every MATH. If we define MATH and MATH by MATH and MATH, then MATH and MATH have norm at most MATH, and the proposition follows from the factorization MATH for every MATH with support in MATH. |
math/0102146 | For every MATH, consider the multiplication operator MATH on MATH defined by MATH. MATH is diagonal for the canonical basis MATH, so that MATH. In the basis MATH, its matrix representation is MATH, so that MATH and MATH. The norm of MATH on MATH is therefore the norm of MATH on the subspace of NAME matrices in MATH. Furthermore, the c.b. norm of MATH on MATH is the norm of the NAME multiplier MATH on MATH CITE. For all MATH and MATH in MATH, we have MATH by matrix unconditionality REF , so that MATH . It remains to note that MATH. |
math/0102146 | If MATH is positive, then MATH is a positive operator on MATH. Henceforth its norm on MATH is attained at MATH, and MATH. This quantity is furthermore a lower bound on all MATH. |
math/0102146 | MATH is trivial. MATH. The c.b. norm of a NAME multiplier MATH on MATH is the supremum of the c.b. norm of its restrictions MATH to finite rectangle sets MATH. Furthermore, the c.b. norm of an elementary NAME multiplier MATH on MATH equals MATH. MATH. If MATH is not a union of pairwise row and column disjoint rectangle sets, then there are MATH, MATH such that MATH and MATH. Let MATH, MATH, be the operator from MATH to MATH with matrix MATH. Its eigenvalues are MATH so that MATH and therefore MATH for some MATH if MATH. MATH. Suppose MATH and let MATH intersect MATH. Then there are pairwise disjoint sets MATH and pairwise disjoint sets MATH such that MATH and MATH which is an average of elementary tensors of norm MATH, so that its projective tensor norm is bounded by MATH, and actually is equal to MATH. |
math/0102146 | MATH. If MATH holds, every sequence of signs MATH is a NAME multiplier on MATH. By a convexity argument, this implies that every bounded sequence is a NAME multiplier on MATH, which may be extended to a NAME multiplier on MATH with the same norm by CITE. MATH holds by NAME 's inequality (see CITE) and an approximation argument. MATH is but the formulation dual to MATH (see CITE). MATH. A computation yields MATH . MATH is CITE. (The proof can be found in CITE and in CITE.) MATH, MATH can be found in CITE. MATH. In fact, a forest is a bisection in the terminology of CITE: it is the union of a row section MATH with a column section MATH. It suffices to prove this for a tree: let the vertices of its edges be indexed by words as described in the Terminology; then let MATH be the set of its elements of the form MATH with MATH of odd length and MATH a letter, and let MATH be the set of its elements of the form MATH with MATH of even length and MATH a letter. MATH is CITE. Note that row sections and column sections form MATH-unconditional basic sequences in MATH and are MATH-complemented in MATH by REF . MATH follows from the open mapping theorem. |
math/0102146 | Let MATH. If MATH, the empty closed walk suits. Suppose MATH; we have to find a closed walk of length MATH that is mapped on MATH. Consider a walk MATH in MATH such that MATH and MATH for every MATH, and furthermore MATH is maximal. We claim MATH that MATH and MATH that MATH. Let MATH. MATH. If MATH, then MATH. Thus there is MATH such that MATH. But then MATH and there is MATH such that MATH: MATH is not maximal. MATH. Suppose MATH. Then MATH and MATH. By hypothesis, they are not row and column disjoint: there are MATH such that MATH or MATH such that MATH. By interchanging MATH and MATH, by relabeling the vertices, we may suppose without loss of generality that for MATH there is MATH such that MATH. Then there is MATH such that MATH. By the argument used in REF, there is a closed walk MATH such that MATH and MATH (with MATH). Then the closed walk MATH shows that MATH is not maximal. |
math/0102146 | This follows from REF . |
math/0102146 | MATH. If MATH for MATH, we may suppose MATH: consider then MATH and MATH. If MATH for MATH, we may suppose MATH: consider then MATH and MATH. MATH. Use MATH in a maximality argument. MATH. Note that the closed walks MATH in MATH are either cycles or of length MATH; in the latter case MATH for some MATH. |
math/0102146 | MATH and MATH follow from the matrix unconditionality of NAME norms (see REF ). MATH follows from REF . MATH. Let MATH if MATH is even and MATH if MATH is odd. Then MATH and MATH and this yields a lower bound on the norm of MATH on MATH. The c.b. norm of MATH on MATH is equal to its norm and thus to the maximum of MATH for MATH and MATH. As MATH if MATH and MATH, we may suppose that MATH. We may also restrict our study to MATH and further to MATH, as MATH if MATH. In that case MATH. It turns out that MATH is maximal if MATH and, as MATH is increasing for MATH and decreasing for MATH, this maximum is MATH. MATH. If MATH is not an even integer and MATH, then MATH is not an isometry on MATH: otherwise the functions MATH and MATH would have the same distribution by the NAME Theorem CITE. If MATH, then MATH contains no cycle of length MATH, so that by REF MATH, every closed walk MATH satisfies MATH and the function MATH of REF is constant in MATH. If MATH, the closed walk relation MATH satisfies MATH by REF . Then the coefficient of MATH in MATH equals MATH and must equal the same quantity with MATH replaced by MATH if MATH defines an isometry on MATH. |
math/0102146 | MATH is trivial. MATH. Suppose that MATH contains a cycle MATH with MATH. CASE: MATH shows that MATH is not a real MATH-unconditional basic sequence in MATH. MATH. A tree on MATH vertices has exactly MATH edges, so that a forest MATH satisfies MATH. Conversely, a cycle of length MATH is a graph with MATH row vertices, MATH column vertices and MATH edges. MATH. Let MATH be a tree and index the vertices of its edges by words MATH as described in the Terminology, so that MATH is a set of form MATH . Define inductively MATH and MATH: let MATH; suppose that MATH and MATH have been defined for all MATH such that the length of MATH is at most MATH. We then set for all words MATH of length MATH and all letters MATH such that MATH: MATH . If MATH is a union of pairwise disjoint trees, we may define MATH and MATH on each tree separately; we may finally extend MATH to MATH and MATH to MATH in an arbitrary manner. MATH may be proved as MATH. MATH. If MATH holds, then every NAME multiplier by signs MATH is simple in the sense that MATH with MATH (vs. MATH) the diagonal operator on MATH ( vs. MATH) of multiplication by MATH (vs. MATH), so that the c.b. norm of MATH on any MATH is MATH. MATH. If MATH holds, every MATH may be extended to an elementary tensor MATH of norm MATH. MATH follows because every element of MATH with norm MATH is the half sum of two elements of MATH: note that MATH. MATH. It suffices to check Equality REF for MATH with support contained in a finite rectangle set MATH. As MATH is MATH-complemented in MATH, REF yields REF . MATH because they are dual statements. MATH. Use Equality REF. MATH may be deduced by the argument of REF MATH. MATH. Taking sign sequences MATH in MATH shows that all relative NAME multipliers by signs on MATH define isometries. Apply REF . |
math/0102146 | MATH follows from REF MATH and REF MATH. MATH. MATH holds if and only if MATH and therefore if and only if MATH holds. MATH. Suppose MATH and let MATH. Let MATH. If MATH, let us define MATH and MATH. Otherwise, if MATH is a singleton MATH, let us define MATH and MATH by MATH if MATH and MATH otherwise. Otherwise, MATH is a singleton MATH and we define MATH and MATH by MATH if MATH and MATH otherwise. Note that the MATH have pairwise disjoint support and are null sequences, as well as the MATH. Then MATH is an average of elementary tensors in MATH of norm MATH, so that this average is also bounded by this norm, which obviously is a lower bound. |
math/0102146 | MATH, MATH are trivial. MATH. Suppose that MATH contains a cycle MATH of length MATH: the corresponding set of couples is MATH. Let MATH be as in MATH and let MATH for some arbitrary MATH. Then MATH. Consider MATH as a function on the group MATH. Then the NAME coefficient of MATH at the NAME character MATH is, by REF MATH, MATH . (Note that MATH.) As this last sum has only positive terms and contains at least the term corresponding to MATH, MATH cannot be constant. MATH. Let MATH, MATH and MATH be as in the proof of the implication MATH. Let MATH be as in MATH. Consider MATH as a function on the group MATH. Then the NAME coefficient MATH of MATH at the NAME character MATH is, by REF MATH, MATH . As this last sum has only positive terms and contains at least the term corresponding to MATH, MATH cannot be constant. MATH. Apply REF MATH. MATH. If MATH contains a cycle MATH, then MATH contains two distinct paths MATH, MATH of length MATH from MATH to MATH. If MATH contains two distinct paths MATH, MATH with MATH, MATH and MATH, let MATH be minimal such that MATH, let MATH be minimal such that MATH and let MATH be minimal such that MATH. Then MATH is a cycle in MATH of length MATH. MATH holds by REF MATH: If each MATH satisfies MATH, then REF shows that MATH as defined in REF is constant in MATH. |
math/0102146 | Let MATH be such that MATH. As MATH has finite rank, there is a finite MATH such that the projection MATH of MATH onto MATH satisfies MATH. Let MATH. Then MATH for all MATH. For each MATH (vs. MATH), let MATH ( vs. MATH) be the diagonal operator on MATH ( vs. MATH) of multiplication by MATH ( vs. MATH): MATH and MATH are unitary, so that MATH is an isometric operation on MATH. Let MATH be the operator defined by MATH . Then, as MATH uniformly on compact sets, MATH . A computation shows that MATH is the operator associated to the NAME multiplier MATH whose support is contained in MATH. |
math/0102146 | MATH is but a reformulation of the definition MATH and implies MATH. MATH is a consequence of MATH and MATH. MATH. If MATH, then there is MATH such that there are infinitely many paths of length MATH from MATH to MATH: there is a path MATH that can be completed with edges outside any given finite set to a cycle of length MATH. |
math/0102146 | Suppose that MATH enjoys MATH in MATH and fails MATH. Then, for some MATH, MATH contains a sequence of cycles MATH with the property that MATH for all MATH with support in MATH and all MATH with support in MATH. With the notation of REF, this amounts to stating that the multiplier on MATH given by MATH if MATH and MATH otherwise actually is an isometry on MATH. As MATH, this implies by REF MATH that MATH. Suppose that MATH enjoys MATH. We claim that for every finite MATH there is a finite MATH such that every closed walk MATH of length MATH in MATH satisfies MATH. That is, given a closed walk MATH and MATH such that MATH and MATH, MATH . Suppose that this is not true: then there is a MATH, there are MATH and there are cycles MATH such that the MATH are injective sequences of vertices and MATH is even for at least one index MATH: let us suppose so for MATH. If MATH, consider the path MATH of odd length MATH; if MATH, consider the path MATH of odd length MATH. Then MATH can be completed with vertices outside any given finite set to a cycle of length at most MATH because MATH is a path of length MATH in MATH for every MATH. This proves that MATH fails MATH. The claim shows that MATH enjoys MATH in MATH for MATH. In fact, if MATH is defined by MATH for MATH and MATH for MATH, then, with the notation of REF , MATH does not depend on MATH, so that MATH if MATH and MATH, and MATH has complex MATH by REF MATH. |
math/0102146 | MATH. If MATH is MATH-independent, consider a closed walk MATH in MATH with MATH; as MATH, MATH is a permutation of MATH. But then, by the hypothesis on MATH and MATH, MATH is a permutation of MATH, so that MATH: MATH is a complex completely MATH-unconditional basic sequence by REF MATH. MATH. If MATH is not MATH-independent, then there are two MATH-uples MATH, MATH that are not a permutation of each other such that MATH, so that if we set MATH and MATH for each MATH, then MATH. But, by the hypothesis on MATH and MATH, MATH is a permutation of MATH and MATH is a permutation of MATH, so that MATH and MATH. Then MATH is a sum of row and column disjoint closed walk relations MATH such that MATH for at least one MATH: MATH is not a real MATH-unconditional basic sequence in MATH by REF MATH. |
math/0102146 | MATH. MATH is MATH-independent if and only if MATH with MATH has only trivial solutions. If this is the case, consider a closed walk MATH in MATH. As MATH, we have MATH or MATH, so that MATH is not a cycle. Conversely, if MATH contains a solution to MATH with MATH, then MATH contains the cycle MATH. MATH. We may suppose that MATH. Consider the cycle MATH. |
math/0102146 | By REF MATH, MATH is a MATH-unconditional basic sequence in MATH, with MATH an integer, if and only if MATH is a graph of girth MATH in the sense of CITE. Therefore MATH and MATH are shown in CITE. REF is CITE and the sufficient condition for equality follows from CITE. |
math/0102148 | From the maximum principle we deduce that MATH . Hence for MATH, MATH and again from the maximum principle, MATH . We conclude that the MATH are monotone in MATH. Their pointwise limit is convex and thus continuous. So they converge locally uniformly according to NAME 's theorem. |
math/0102148 | Since MATH is strictly convex, MATH attains its maximum at the boundary. Tangential derivatives are uniformly bounded there in view of the NAME boundary conditions. The normal derivatives are estimated as follows. Let MATH and MATH the outer unit normal to MATH at MATH. We choose MATH independent of MATH such that the line segment MATH is contained in MATH. Using the convexity of MATH as well as the fact that MATH lies below MATH and MATH, MATH (MATH denotes the directional derivative). For MATH and MATH, we consider similarly the line segment MATH and obtain MATH . We finally use the uniform MATH bounds of REF , in particular that MATH. |
math/0102148 | Since MATH is convex and lies above MATH with MATH, MATH and thus MATH . The MATH estimates of REF imply that MATH, and so the left hand side is MATH. According to REF , MATH, and thus MATH . In order to derive REF , we consider MATH and the barrier functions along the line segment MATH parametrized by MATH, MATH. The boundary values of MATH are MATH. Thus using that MATH lies above MATH and is convex, we obtain the estimate MATH and thus MATH . Using REF , MATH where we have set MATH. These estimates imply REF provided that MATH is sufficiently close to the outer boundary, more precisely if MATH lies in the strip MATH where MATH is some constant independent of MATH. Now suppose that MATH lies outside this strip, MATH. We construct straight lines through MATH which are tangential to the hyperbola MATH. A short explicit calculation shows that there are exactly two such lines, and that they go through the points MATH with MATH . Using that MATH is bounded uniformly in MATH, one sees that MATH, and so we can by increasing MATH arrange that MATH . Thus MATH, meaning that the tangential points are inside the ball MATH. Since MATH is convex and lies below MATH, the line segments joining the points MATH and MATH, respectively, both lie above MATH. Moreover, these line segments are by construction tangential to the hyperbola MATH at MATH. Hence MATH . Using REF , we obtain that MATH and REF gives REF . |
math/0102148 | We may assume that MATH. Then MATH is represented locally as MATH, where MATH . According to the NAME boundary conditions, MATH . We differentiate twice with respect to MATH, MATH, MATH and obtain that at MATH, MATH . According to the decay conditions at infinity REF , MATH and furthermore MATH . Thus the result follows in view of REF . |
math/0102148 | Note that MATH, MATH and MATH. The first inequality follows directly from the MATH estimates of REF , whereas for the second inequality we use furthermore that MATH on MATH, the decay properties of MATH, and the fact that MATH vanishes on MATH. To prove the last inequality, we apply REF , the relation MATH, and the MATH-estimates to obtain that MATH . Now we use REF . |
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