paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0107199 | REF follows by an elementary calculation. Next, consider the time interval MATH. The variable MATH satisfies the differential equation MATH . Note that MATH is monotonously decreasing and MATH for the times under consideration. The linearization of the drift term MATH at MATH is MATH. It follows that for MATH, MATH and thus MATH . More careful estimates, based on the inequalities MATH show that MATH is also bounded below by a constant times MATH. Now MATH can be computed in the same way, by performing the change of variables MATH, yielding REF. Finally, REF follows easily from the fact that we are again in the stable case for MATH. |
math/0107199 | The proof is almost the same as the proof of REF , the only difference lying in the different behaviour of MATH, given in REF, which requires to distinguish between MATH, MATH, and the remaining MATH up to MATH. |
math/0107201 | To choose MATH first choose any REF-form MATH on MATH with MATH and MATH. If MATH is compact, we then let MATH to be the average MATH of MATH over the group MATH. If MATH is not compact, then due to the existence of slices we may assume that MATH for some representation MATH of a compact NAME group MATH. We average the restriction MATH over MATH and then extend it to all of MATH by MATH-invariance. |
math/0107201 | It's easy to see that MATH is basic and hence descends. For a proof that MATH is contact if MATH is contact see CITE or CITE. Note that the zero level set MATH is the set of points of MATH where MATH-orbits are tangent to the contact distribution MATH. Thus MATH depends only on MATH. It is not hard to see that MATH depends only on MATH as well. |
math/0107201 | Suppose not. Then for some point MATH the orbit MATH is tangent to the contact distribution MATH. Therefore the tangent space MATH is isotropic in the symplectic vector space MATH where MATH and MATH is a MATH-invariant contact form with MATH and MATH. We now argue that this forces the action of MATH not to be effective. More precisely we argue that the slice representation of the connected component of identity MATH of the isotropy group MATH of the point MATH is not effective. The group MATH acts on MATH preserving the symplectic form MATH and preserving MATH. Since MATH is isotropic, MATH as a symplectic representation of MATH. Here MATH denotes the symplectic perpendicular to MATH in MATH. Note that since MATH is a torus, the action of MATH on MATH (and hence on MATH) is trivial. Observe next that the dimension of the symplectic vector space MATH is MATH. On the other hand, since MATH is a compact connected Abelian group acting symplecticly on MATH, its image in the group of symplectic linear transformations MATH lies in a maximal torus MATH of a maximal compact subgroup of MATH. Since the maximal compact subgroup of MATH is the unitary group MATH, MATH, the dimension of MATH is MATH. Therefore the representation of MATH on MATH (and hence of MATH) is not faithful. Since the fiber at MATH of the normal bundle of MATH in MATH is MATH, the slice representation of MATH is not faithful. Consequently the action of MATH in not faithful in a neighborhood of an orbit MATH. Contradiction. |
math/0107201 | Consider the family of MATH-invariant REF-forms MATH, MATH. For all MATH and all MATH we have MATH and MATH is nondegenerate. Therefore the forms MATH are contact in a neighborhood of MATH for all MATH. It is no loss of generality to assume that this neighborhood is all of MATH. Denote the NAME vector field of MATH by MATH. Since the NAME vector field is uniquely defined by MATH, MATH and since MATH is MATH-invariant, MATH is MATH-invariant as well. Define a time dependent vector field MATH tangent to the contact distribution MATH by MATH where MATH is the derivative with respect to MATH. Clearly MATH is MATH-invariant. Note that MATH for all MATH. This is because MATH for MATH. We claim that the NAME derivative of MATH with respect to MATH satisfies MATH . Indeed, since MATH, MATH. By definition of MATH, MATH. On the other hand, MATH. This proves REF . Hence MATH . Denote the isotopy generated by MATH by MATH. Since MATH vanishes on MATH, MATH is defined for all MATH on a neighborhood of MATH. Since MATH is MATH-invariant, the isotopy is MATH-equivariant. Let MATH. Then MATH by REF . Therefore MATH. In particular, MATH where MATH. Note also that since MATH is zero at the points of MATH, MATH fixes MATH pointwise. |
math/0107201 | Suppose MATH is a MATH-equivariant pre-isotropic embedding. Choose a MATH-invariant almost complex structure MATH on MATH compatible with MATH. This gives us a MATH-invariant inner product on the vector bundle MATH. Extend it to a MATH-invariant Riemannian metric MATH on MATH by, say, declaring the NAME vector field MATH of the invariant contact form MATH to be of unit length and orthogonal to MATH (as remarked previously, since MATH is MATH-invariant and the NAME vector field MATH is uniquely determined by MATH, MATH is MATH-invariant). By construction of MATH, MATH and MATH are MATH-perpendicular and MATH is a symplectic subbundle of MATH. Let MATH be the MATH-perpendicular to MATH in MATH. The bundle MATH is also MATH-perpendicular to MATH; MATH is isomorphic to the conformal symplectic normal bundle of the embedding MATH. We therefore have a MATH-equivariant direct sum decomposition MATH . Since MATH is transverse to MATH it follows that MATH . Note that the NAME vector field MATH need not be tangent to MATH, and so MATH need not be MATH-perpendicular to MATH. Never the less, MATH is a topological normal bundle for the embedding MATH. Therefore the restriction of the MATH-exponential map MATH to MATH gives an open MATH-equivariant embedding of a neighborhood of the zero section MATH into MATH; the embedding is identity on MATH. Since MATH is isotropic in MATH, the map MATH defined by MATH is an isomorphism. By composing the inverse of this map with MATH we get a MATH-equivariant map MATH, which has the following properties. The map MATH is the identity on MATH. It is an open embedding on a sufficiently small neighborhood of MATH in MATH. For any point MATH in the zero section of MATH we have MATH and MATH where MATH denotes the canonical symplectic form on MATH and MATH. Now consider two pre-isotropic embeddings MATH, MATH satisfying the hypotheses of the theorem. Since MATH, MATH. Denote this distribution by MATH. Let MATH denote the vector bundle isomorphism with MATH. The map MATH has the property that at the points MATH of the zero section MATH . Consider the two MATH-equivariant maps MATH, MATH given by the construction at the beginning of the proof. We have, for any point MATH, MATH . Also MATH while MATH . We now apply the equivariant relative NAME theorem to MATH, MATH and MATH to obtain a MATH-equivariant diffeomorphism MATH with MATH, MATH for some neighborhoods MATH of the zero section and some function MATH. The theorem follows. |
math/0107201 | Since the contact distribution MATH is of codimension REF, in order to show that the orbit MATH is transverse to MATH, it is enough to prove that there is a vector MATH such that MATH, that is, such that MATH. But the latter is exactly the condition that MATH. Next note that the fiber of the characteristic distribution MATH at MATH is MATH . Let MATH. Since MATH is abelian, MATH is a subalgebra. Consequently MATH is an integrable distribution. Since the leaves of the foliation defined by MATH are tangent to the contact structure, MATH is an isotropic subbundle of the symplectic vector bundle MATH, where as usual MATH. |
math/0107201 | The characteristic distributions and the conformal symplectic normal bundles of the embeddings MATH, MATH, are, respectively, MATH where MATH and MATH are the symplectic slice representations. The lemma follows from the uniqueness of pre-isotropic embeddings REF . |
math/0107201 | By REF it is enough to construct on MATH a MATH-invariant contact form MATH so that the embedding MATH, MATH is pre-isotropic, the symplectic slice representation at MATH is MATH and MATH. We construct MATH as a contact quotient (see REF ). Since MATH is abelian, both right and left trivializations identify the cotangent bundle MATH with MATH. Consider the hypersurface MATH in MATH. Since MATH, MATH is a hypersurface of contact type (the expanding vector field MATH is generated by dilations MATH). Consider the action of MATH on MATH given by MATH and the action of MATH on MATH given by MATH. Both actions preserve MATH, MATH and the tautological REF-form MATH. The action of MATH on MATH preserves REF-form MATH where MATH is the radial vector field on MATH. The diagonal action of MATH on MATH preserves the contact form MATH. The corresponding moment map MATH is given by MATH where MATH is the MATH-moment map. Therefore the reduced space at zero for the action of MATH is MATH and MATH descends to a MATH-invariant contact form MATH on MATH. Note that the moment map for the action of MATH on MATH descends to the MATH-moment map for the induced action of MATH on MATH. Hence it is given by the desired formula: MATH . |
math/0107201 | Since the unitary group is the maximal compact subgroup of the symplectic group, given a symplectic representation of a torus MATH there exists on MATH a MATH-invariant contact structure MATH compatible with the symplectic form MATH. We define the weights of MATH to be the weights of the complex representation MATH. Since any two MATH-invariant complex structures on MATH compatible with MATH are homotopic, the weights do not depend on the choice of MATH, that is, they are well-defined. |
math/0107201 | Arguing as in the proof of REF we may assume that MATH maps MATH into the unitary group MATH. Since the dimension of a maximal torus of MATH is MATH and since MATH is faithful, MATH maps the identity component of MATH onto a maximal torus of MATH. Since the centralizer of a maximal torus of MATH is the torus itself, MATH is connected. Finally, the weights of a maximal torus MATH of MATH form a basis of the weight lattice of MATH. |
math/0107201 | Let MATH denote the characteristic subalgebra of the pre-isotropic embedding MATH. By REF a neighborhood of MATH in MATH is MATH-equivariantly diffeomorphic to a neighborhood of the zero section in MATH. Since MATH is abelian, the action of MATH on MATH is trivial. Since by assumption the action of MATH on MATH is effective, the slice representation of MATH on MATH has to be faithful. By definition of MATH, the dimension of MATH is the dimension of the contact distribution minus twice the dimension of the characteristic distribution, that is, MATH. Now MATH and MATH. Therefore, MATH. By REF MATH is connected and the weights MATH REF of the slice representation MATH form a basis of the weight lattice of MATH. On the other hand, MATH. The rest of the lemma follows. |
math/0107201 | Since the isotropy group MATH is connected, the sequence MATH splits. Hence MATH. |
math/0107201 | Fix a point MATH. Let MATH denotes the corresponding symplectic slice representation and MATH the characteristic subalgebra. Let MATH and MATH be the embeddings as in REF . By REF , the isotropy group MATH is connected. By REF there exists a MATH-invariant neighborhood MATH of MATH in MATH, a MATH-invariant neighborhood MATH of MATH in MATH, a MATH-invariant contact form MATH on MATH and a MATH-equivariant diffeomorphism MATH such that MATH for some MATH-invariant function MATH. Consequently the MATH- and MATH-moment maps are related by MATH. Recall that the MATH-moment map MATH is given by MATH, where MATH is the moment map for the slice representation. As observed previously the connectedness of MATH implies that MATH is diffeomorphic to MATH. Under this identification MATH. Hence a fiber of MATH is of the form MATH. Since the fibers of MATH are MATH-orbits (see REF ), the fibers of MATH are MATH-orbits. It follows that for any point MATH the set MATH is a MATH-orbit. We next argue that the pair MATH determines MATH and the symplectic slice representation MATH (it obviously determines the characteristic subalgebra). It is no loss of generality to assume that MATH (note that the contact form MATH is not normalized but this won't matter) and that MATH is a neighborhood of MATH in MATH. Since MATH is connected it is determined by its NAME algebra MATH or, equivalently, by its annihilator MATH. Let MATH be the moment cone of MATH. We may assume that MATH is of the form MATH where MATH is a neighborhood of MATH in MATH and MATH is a MATH-invariant neighborhood of REF in MATH. Then MATH. Note that MATH is an open subset of MATH. Hence MATH . Once we determined MATH we have the natural inclusion MATH and the dual projection MATH. Note that MATH. Therefore MATH. By REF , the cone MATH completely determines the representation MATH. |
math/0107201 | See CITE. |
math/0107201 | By REF , the fibers of MATH are connected. By REF the connected components of the fibers of MATH are MATH-orbits. Therefore the fibers of MATH are MATH-orbits and the orbital moment map is injective. Since MATH is compact, the orbital moment map MATH is an embedding. |
math/0107201 | This is an easy consequence of the local normal form theorem, REF . |
math/0107201 | Suppose the action of MATH is not free. Then for some point MATH the isotropy group MATH is not trivial. By REF for any MATH-invariant neighborhood MATH of the orbit MATH there is an open subset MATH of the sphere MATH such that MATH . By the local normal form theorem, REF , we may choose MATH and MATH so that MATH where as in REF MATH denotes the symplectic slice representation, MATH denote the corresponding moment map, MATH is the characteristic subalgebra, and MATH, MATH are the embeddings as in REF . Since MATH is nontrivial, the symplectic slice MATH is not zero. It follows from REF that MATH is a proper cone in MATH. Therefore MATH, that is, MATH is not onto. Contradiction. Therefore the action of MATH is free. By REF , the fibers of MATH are MATH-orbits. Therefore if the action of MATH is free, then MATH is a principal MATH-bundle. |
math/0107201 | Since the action of MATH at MATH is free, for any MATH the induced vector field MATH is nonzero at MATH: MATH. Since the action of MATH preserves MATH, MATH. Therefore, for any MATH . Now MATH . Hence to prove that MATH is onto, it's enough to show that for any MATH with MATH there is MATH with MATH. Since MATH is contact, MATH is nondegenerate. Therefore, for any MATH with MATH there is MATH so that MATH. |
math/0107201 | By REF , the differential MATH is surjective for all MATH. Consequently the image MATH is open in the sphere. On the other hand the image is closed since MATH is compact. Thus the image is the whole sphere and MATH is a submersion. Since MATH is compact it follows that MATH is a fibration. If MATH then, by REF , MATH is a principal MATH-bundle. |
math/0107201 | Denote the orbit map MATH by MATH, MATH. We want to show that for any point MATH there is a MATH-invariant neighborhood MATH and a MATH-equivariant diffeomorphism MATH such that MATH and such that MATH . Pick a point MATH. Then, since MATH, we have MATH. Also, given a MATH-invariant neighborhood MATH of MATH in MATH, let MATH. We have MATH. Therefore by REF if MATH is sufficiently small there exists a MATH-equivariant contactomorphism MATH. Hence, as remarked earlier, MATH and MATH. It remains to show that the map MATH induced by MATH on MATH is MATH. If MATH is sufficiently small, then by REF the maps MATH, MATH, are embeddings. Since MATH, MATH. Therefore MATH. |
math/0107201 | We first consider the special case of MATH. Then the corresponding contact vector field is the NAME vector field MATH. By definition of MATH the NAME derivative MATH satisfies MATH. Hence for any MATH . Since MATH is unique, it is MATH-invariant. Hence MATH for any MATH. And by the previous computation MATH. Therefore MATH . Thus the NAME vector field is tangent to the fibers of the moment map MATH. Since MATH is toric, the connected components of the fibers of MATH are MATH-orbits (see REF ). Therefore the NAME vector field is tangent to MATH-orbits. Hence for any MATH-invariant function MATH, MATH and consequently MATH. That is, the flow MATH of MATH preserves MATH. For any MATH . Since connected components of the fibers of MATH are MATH-orbits, this proves that the contact vector field MATH is tangent to MATH-orbits for any invariant function MATH. Hence the flow of MATH induces the identity map on the orbit space MATH. |
math/0107201 | The argument is standard (compare REF or CITE). Suppose MATH is a contact toric MATH-manifold locally isomorphic to MATH. Fix a homeomorphism MATH. Choose an open cover MATH of MATH such that for each MATH there is a MATH-equivariant contact diffeomorphism MATH inducing MATH on MATH (here MATH is the orbit map). Let MATH it is a NAME REF-cocycle whose cohomology class in MATH is independent of the choices made to define it. Conversely, given an element of MATH we can represent it by a NAME cocycle MATH. We construct the corresponding contact toric MATH-manifold by taking the disjoint union of the manifolds MATH and gluing MATH to MATH along MATH using MATH. The cocycle condition guarantees that the gluing is consistent. |
math/0107201 | Let MATH be an open subset of MATH and let MATH. By REF , the time MATH-flow MATH induces the identity map on MATH and preserves the contact form MATH. Hence for any MATH, MATH is in MATH. We define the map MATH by MATH . We next argue that MATH is onto. Suppose MATH. By REF, there exists a smooth MATH-invariant map MATH such that MATH for all MATH. Moreover, if MATH is contractible, then MATH for some smooth MATH-invariant map MATH. It's not hard to check that MATH is the time-REF flow of the vector field MATH. Note that MATH is a MATH-invariant vector field: for any MATH and MATH where the second equality holds because MATH is MATH-invariant and MATH is abelian. We next prove that the NAME derivative of MATH with respect to MATH is zero: MATH. To do this we recall a few facts about basic forms CITE. Given an action of a compact NAME group MATH on a manifold MATH, a form MATH is basic if it is MATH-invariant and if for any MATH, the contraction MATH is zero (for zero forms we only require invariance). The set of basic forms is a subcomplex of the NAME complex of differential forms, that is, if MATH is basic then so is MATH. Also, if MATH is a MATH-equivariant map inducing the identity on MATH and MATH is basic, then MATH. This is because it is a closed condition that holds on the open dense subset of points of principal orbit type. We claim that MATH is basic for the action of the torus MATH on the manifold MATH. Note that MATH. Since MATH and MATH are MATH-invariant, MATH is MATH-invariant hence basic. Therefore MATH is basic. Also, since MATH and MATH are MATH-invariant, the second term MATH is MATH-invariant. It remains to show that for any MATH, MATH. Now for any MATH because MATH, MATH are MATH-invariant functions and because MATH (since MATH is abelian). Therefore for any MATH . Next let MATH denote the time-MATH flow of the vector field MATH. Clearly MATH is MATH-equivariant and induces the identity map on MATH. Thus , since MATH is basic we have MATH for all MATH. We also know that MATH and that MATH. Therefore MATH . We conclude that MATH is a contact vector field. We define MATH. Then the contact vector field of MATH is MATH and MATH. This concludes the proof that MATH is onto. We define the map MATH by MATH. Thus it remains to show that for any sufficiently small set MATH and any function MATH if MATH for all MATH then MATH for some MATH. In fact it is enough to show that the above equation holds on the open dense subset MATH of MATH consisting of the points where the action of MATH is free. Now the contact vector field MATH of MATH is MATH-invariant and is tangent to MATH-orbits REF . Therefore there exists on MATH a MATH-invariant smooth map MATH such that MATH . Now the time - REF flow of MATH is MATH. Thus if MATH, then MATH for all MATH. Hence MATH and so MATH for all MATH. Therefore, since MATH is discrete and MATH is continuous and since we may take MATH to be connected, MATH for some fixed vector MATH. It follows, since MATH is dense in MATH that MATH for all MATH. Consequently MATH. |
math/0107201 | The sheaf MATH is a fine sheaf, so the long exact sequence in cohomology induced by REF breaks up for MATH into isomorphisms MATH. |
math/0107201 | Since MATH is contact toric, MATH for any MATH REF . Therefore the action of MATH on MATH has no fixed points. By REF all the isotropy groups are connected. Therefore they are either trivial or circles. Suppose the isotropy group MATH at MATH is a circle. Then its NAME algebra MATH equals the characteristic subalgebra MATH. Hence MATH is a multiple of a weight MATH. Also, the dimension of the symplectic slice MATH at MATH is REF. Hence the symplectic slice representation is isomorphic to the standard action of MATH on MATH: MATH. Consequently, by the local normal form theorem, REF , a neighborhood of MATH in MATH is diffeomorphic to MATH, and a neighborhood of MATH in MATH is homeomorphic to MATH. We conclude that MATH is a REF-dimensional MATH manifold with boundary. Since MATH is compact and connected and since MATH it follows that CASE: MATH is homeomorphic to MATH; CASE: there are exactly two orbits MATH, MATH which are diffeomorphic to MATH; CASE: For MATH, MATH where MATH and the character MATH is the map MATH. Since MATH we may identify MATH with MATH and the weight lattice MATH with MATH. Then for any weight MATH of MATH, MATH for some MATH with MATH rational. Since MATH is homeomorphic to MATH, the orbital moment map MATH lifts to a map MATH such that MATH. Note that MATH is an interval with end points being the images of the exceptional orbits MATH, MATH. Hence we may assume that MATH, MATH, and that MATH, MATH are rational numbers. Moreover since MATH is locally an embedding REF , MATH is an embedding. Thus MATH is a homeomorphism. Now suppose MATH is another c.c.c.t. MATH-manifold with MATH. Then MATH is a homeomorphism with MATH. By REF MATH and MATH are locally isomorphic. Since MATH is contractible, MATH. By REF MATH, where MATH is the sheaf in REF . Hence, by REF , MATH and MATH are isomorphic. To prove REF we use an equivariant version of REF: Suppose MATH is a contact manifold, MATH is a manifold with boundary of the same dimension as MATH embedded in MATH. Suppose further that there is a neighborhood MATH in MATH of the boundary MATH and a free MATH action on MATH preserving MATH such that the corresponding moment map MATH satisfies CASE: MATH and CASE: MATH. Let MATH, where, for MATH, MATH if and only if CASE: MATH and CASE: MATH for some MATH, Then MATH is a contact manifold, MATH is a contact submanifold of MATH, and MATH is contactomorphic to MATH. Moreover if there is an action of a NAME group MATH on MATH preserving MATH, MATH and commuting with the action of MATH on MATH, then there is an induced action of MATH on MATH preserving the induced contact structure. Suppose we are given MATH with MATH rational, MATH and MATH. For each MATH there is a weight MATH such that MATH lies on the ray through MATH. Choose MATH sufficiently small so that MATH is non-negative on MATH and MATH is non-negative on MATH and MATH. Consider MATH with the contact form MATH, MATH. Let MATH, MATH. Consider MATH given by MATH for MATH and MATH for MATH. The function MATH is a moment map for a circle action on MATH generated on MATH by MATH and on MATH by MATH. Note that the obvious action of MATH on MATH preserves MATH, MATH, MATH and commutes with the action of MATH defined by MATH. Therefore we may apply REF . The contact manifold MATH so obtained with the induced action of MATH is the desired manifold MATH. |
math/0107201 | Suppose MATH is a good polyhedral cone defined by some subset MATH of the integral lattice MATH. As a first step we construct a symplectic cone MATH with a symplectic action of MATH commuting with dilations such that the image of the corresponding moment map MATH is MATH. The construction is a slight adaptation of a well-known construction of NAME (compare CITE, CITE). Let MATH denote the standard basis of MATH. Consider the map MATH given by MATH. Since MATH, MATH induces a map MATH. We write MATH for the image of MATH in MATH. Note that the kernel MATH of MATH is MATH . It is a compact abelian subgroup of MATH with NAME algebra MATH. Note that MATH need not be connected. Consider the standard action of MATH on MATH: MATH . The corresponding symplectic moment map MATH is given by MATH where MATH is the basis dual to MATH. We claim that the symplectic quotient MATH of MATH by the induced action of MATH is the desired manifold MATH, that is, that MATH where MATH and MATH is the inclusion. We claim first that MATH. Indeed, since MATH is exact, MATH and hence MATH. Now MATH . Thus MATH. Next we claim that for any MATH the isotropy group MATH is trivial. It would then follow that MATH is a smooth symplectic cone: the action of MATH on MATH is induced by the action of MATH on MATH given by MATH. Now MATH where MATH. On the other hand, MATH if and only if there is a (unique) MATH such that MATH. Hence MATH if and only if MATH. Since MATH is a good cone, for any fixed vector MATH the set MATH is a MATH basis of MATH. Hence MATH implies that MATH for all MATH. Therefore MATH, that is, MATH is trivial. Finally note that the image of MATH under MATH is precisely MATH. Hence the image of the reduced space MATH under the induced MATH moment map MATH is MATH. Since the sphere MATH is a MATH-invariant hypersurface of contact type in MATH, and since the action of MATH on MATH commutes with the action of MATH is a MATH-invariant hypersurface of contact type in the quotient MATH. Moreover MATH is the corresponding contact moment map, and its moment cone is precisely MATH. |
math/0107201 | We first introduce some notation and a simple fact. Let MATH be a cone and MATH be a face of MATH. Let MATH denote the linear subspace of MATH spanned by the vectors in MATH. Let MATH denote the projection. For any point MATH in the interior of MATH there is an open neighborhood MATH of MATH in MATH such that MATH . Note that the cone MATH is isomorphic to MATH. Now suppose that MATH is the moment cone MATH of MATH and that MATH is a point in MATH. By REF for any MATH invariant neighborhood MATH of MATH there is an open subset MATH of the sphere MATH such that MATH. Let MATH be the cone on MATH. Then MATH . By the local normal form theorem, REF , MATH where as usual MATH is the characteristic subalgebra, MATH is the moment map for the slice representation etc. Note that MATH for any sufficiently ``small" open cone MATH. Thus MATH where MATH is the natural projection. It follows that if MATH is the face of MATH containing MATH in its interior, then MATH and MATH is isomorphic to MATH (once we identify MATH with MATH). Moreover, if we represent MATH as MATH for some minimal set MATH and consisting of primitive vectors, then MATH for some subset MATH, and MATH . Hence MATH. Since MATH, MATH is a subset of MATH. For any MATH we have MATH. Therefore MATH . By minimality of MATH the set MATH is the minimal set with this property. On the other hand, by REF , MATH for some basis MATH of MATH. Therefore the set MATH is a basis of the lattice MATH. |
math/0107201 | Let MATH be a contact toric MATH-manifold REF such that MATH is contactomorphic to the co-sphere bundle MATH of the MATH-torus MATH with the standard contact structure. We will argue that MATH is unique as a contact toric manifold. It was shown in CITE that if MATH, then the action of MATH is necessarily free. The argument roughly goes as follows (see CITE for details). Suppose the action of MATH is not free. Consider first the case of MATH. Then MATH is a lens space (compare REF ), hence cannot be MATH. Next consider the case of MATH. Then the moment cone MATH of MATH is a good polyhedral cone determining MATH uniquely (compare REF ). If the maximal linear subspace of MATH has dimension MATH then MATH is isomorphic to the moment cone of MATH, where MATH gets its contact toric structure as a hypersurface of contact type MATH in MATH. Consequently MATH is homeomorphic to MATH. Finally if the dimension of the maximal linear subspace of MATH is zero, that is, if MATH is a proper cone, then by REF and CITE MATH has a locally free circle action so that the quotient MATH is a compact connected symplectic toric orbifold. The real odd-dimensional cohomology of a compact symplectic toric orbifold is zero. Consequently MATH. Hence MATH, MATH. We conclude that if MATH is a contact toric MATH-manifold REF and MATH then the action of MATH is necessarily free. We argue next that it is unique. Suppose MATH and the action of MATH is free. By the classification theorem MATH, MATH. By REF, MATH and MATH are distinct as contact manifolds for MATH. Since MATH is the standard contact structure on MATH, it follows that there is only one contact toric manifold contactomorphic to MATH. In other words there is only one MATH-action on MATH making it a contact toric manifold. Suppose next that MATH and the action of MATH is free. By REF , MATH is a principal MATH-bundle over the sphere MATH, MATH, and each such principal MATH-bundle carries only one MATH-invariant contact structure. Now, principal MATH bundles over MATH are in one-to-one correspondence with elements of MATH which is REF unless MATH, in which case it's MATH. Note however that no nontrivial MATH bundle over MATH is homeomorphic to MATH. We conclude that if MATH is a contact toric MATH-manifold such that the action of MATH is free, MATH and MATH then MATH is a unique contact toric MATH-manifold with such properties. In other words there is only one MATH-action on MATH making it a contact toric manifold. |
math/0107201 | Suppose MATH, MATH are two effective actions of the torus MATH on MATH preserving the standard contact structure. Let MATH be the moment maps for the actions corresponding to a normalized contact form MATH defining the standard contact structure. We will argue that the images MATH are arcs in MATH of length less than MATH (hence MATH are one-to-one). Moreover we'll show that there is an element MATH preserving the weight lattice MATH such that MATH. It would then follow that the action MATH composed with the isomorphism of MATH defined by MATH is MATH. Note that since MATH, the actions MATH are necessarily not free (compare REF and the proof of REF above). Now consider one of the two actions, say MATH. By REF the action is free except at two orbits MATH and MATH. The isotropy groups MATH of MATH are circles, and the images MATH are of the form MATH where MATH are primitive weights with MATH being the NAME algebra of the circle MATH. It follows from the proof of REF that the contact toric manifold MATH can be obtained by cutting MATH using MATH, where MATH (we identify MATH with MATH and MATH with MATH). Hence as a topological space MATH where MATH for all MATH and MATH for all MATH. Note that MATH , MATH, for otherwise MATH and then MATH is MATH. Next observe that since the standard contact structure on MATH is tight, we must have MATH (compare CITE). Indeed, if MATH and MATH, the image of the cylinder MATH in MATH is an overtwisted disk. One can write a similar formula for an overtwisted disk if MATH. We conclude that the image MATH in MATH is an arc of length less than MATH. Consequently the fibers of MATH are connected. We next argue that the weights MATH, MATH which span the edges of the moment cone MATH span a sublattice of the weight lattice MATH of index two. Let MATH and let MATH be two closed subgroups isomorphic to MATH. Let MATH be the topological space MATH where MATH for all MATH, MATH and MATH for all MATH and MATH. In other words MATH is obtained from the manifold with boundary MATH by collapsing circles in the two components of the boundary by the respective actions of two circle subgroups. Let MATH be the two primitive weights determined by MATH and MATH respectively, that is, the kernel of the character defined by MATH is MATH. Then MATH and MATH. Recall that MATH is isomorphic to the weight lattice MATH and that the isomorphism is given as follows. A weight MATH defines a character MATH by MATH; the class MATH is the element in MATH corresponding to MATH. Here MATH is the obvious REF-form on MATH. Consequently if MATH and MATH is a circle subgroup, then MATH is a character and hence the weight MATH defines an element of MATH. Thus if we identify MATH with MATH and MATH with MATH, then the map MATH becomes the map MATH. The sets MATH and MATH are two open subsets of MATH. We have MATH, MATH is homotopy equivalent to MATH, MATH is homotopy equivalent to MATH, MATH is homotopy equivalent to MATH and the inclusion maps MATH, MATH are homotopy equivalent to projections MATH, MATH respectively. Hence under the above identifications of MATH and MATH with MATH, the inclusions MATH, MATH induce the maps MATH, MATH, respectively. We now apply the NAME sequence to compute the integral cohomology of MATH. We start with MATH. Clearly the map MATH is onto. Given the identifications above the map MATH becomes MATH. We therefore have MATH and the lemma follows. Since MATH, MATH. Hence MATH. Consequently, since MATH are primitive, the parallelogram MATH contains exactly five point of MATH: four vertices plus the point MATH in its interior. Hence MATH is a basis of MATH. Of course MATH. By the same argument the image MATH is an arc in MATH of length less than MATH with endpoints MATH, MATH-where MATH are primitive weights. Moreover MATH is a basis of MATH. The linear map MATH defined by MATH, MATH is the desired map. |
math/0107202 | We prove that the subset MATH of MATH defined by MATH is equal to the NAME variety in MATH given by the permutation MATH. This is true when MATH. Let MATH denote the projection. Then it is easy to check that MATH, so the lemma follows from the above remarks about images and inverse images of projections MATH. |
math/0107202 | If MATH then we have MATH for all MATH. The proposition therefore follows from REF . |
math/0107202 | Let MATH be a rational curve of multidegree MATH, let MATH be the kernel of MATH, and let MATH be the span. It suffices to prove that MATH, which in turn is easy to deduce from the identities MATH for all MATH. To prove this, let MATH be the image of a map MATH, and write MATH for MATH. Since MATH and MATH are the kernel and span of the curve MATH in MATH we see that MATH. Write MATH where MATH is a REF-dimensional subspace. Since MATH it is enough to show that MATH. If this is false, we can find a basis MATH for MATH with MATH. Now for suitable coordinates MATH on MATH we have MATH. Since MATH contains both MATH and MATH, we conclude that MATH whenever MATH. In other words the curve MATH is constant which contradicts that its degree is one. |
math/0107202 | The first sum is dictated by the classical Monk's formula CITE. The second sum is equivalent to the following statement. If MATH is a non-zero multidegree and MATH are permutations such that MATH then the NAME invariant MATH is equal to one if MATH for some MATH such that MATH and MATH; otherwise MATH. Suppose MATH and let MATH be a rational curve of multidegree MATH which meets three NAME varieties MATH, MATH, and MATH in general position. Let MATH be the kernel of MATH and set MATH. Then MATH for all MATH. By REF we have MATH. Since the flags are general this implies that MATH. On the other hand the inequalities MATH, MATH, MATH, and MATH imply that MATH. Since this is the maximal possible dimension of MATH we conclude that all inequalities are satisfied with equality. This first implies that MATH. Furthermore, since MATH we deduce that MATH for some MATH. Thus MATH is the variety of partial flags with subspaces of all dimensions other than MATH. Since MATH it follows that MATH. The fact that MATH implies that MATH by the definition of MATH. Similarly we have MATH. Now since MATH and MATH we conclude that MATH and MATH are dual with respect to MATH, that is, MATH or equivalently MATH as required. It remains to be proven that if MATH and MATH for some MATH then there exists a unique rational curve of multidegree MATH which meets the three given NAME varieties. Set MATH and MATH. Since MATH it follows that MATH and similarly MATH. Thus MATH where MATH. Since MATH we conclude that there is a unique partial flag MATH. Similarly, if we set MATH and MATH then there exists a unique partial flag MATH where MATH. In fact, we can say precisely what these partial flags look like. For MATH we set MATH. Since the flags MATH and MATH are general, these spaces have dimension one, and MATH. Now MATH for each MATH and MATH for MATH. Otherwise stated we have MATH for MATH and for MATH. For MATH we have MATH while MATH where MATH. In particular we get MATH and MATH so by REF there is exactly one rational curve of multidegree MATH with kernel MATH and span MATH. This curve consists of all flags MATH where MATH is a one dimensional subspace. When MATH we have MATH, while MATH when MATH. Finally, MATH belongs to MATH if and only if MATH. Now take any non-zero element MATH and let MATH be the MATH-component of MATH in MATH. Taking MATH then gives a point MATH. This completes the proof. |
math/0107205 | Let us take arbitrary MATH. Then MATH . Note that MATH . Continuing the line of equalities, we obtain: MATH . |
math/0107205 | We choose MATH such that MATH is finite. Suppose MATH has compact support. Then MATH is an entire function. Moreover, MATH uniformly for all MATH from some finite interval MATH. It is an immediate consequence of the following equality: MATH . Now using NAME 's theorem and REF we get MATH and the result follows. |
math/0107205 | CASE: This is a part of REF from CITE. CASE: It follows from REF that MATH and MATH. By REF , MATH a.e. and MATH a.e. for some MATH. So, MATH a.e. for every MATH and MATH, where MATH. Now let us fix MATH, MATH, and consider a function MATH, where MATH and MATH have disjoint supports. Then MATH. So, we get the following estimates: MATH . Since the functions MATH are dense in MATH, the proof of REF is finished. CASE: Suppose REF holds for some MATH, MATH. Then, by the transference principle (see, for example, CITE), MATH is a multiplier in MATH for all MATH, where MATH is the unit circle. So, using results from CITE or CITE, we conclude that MATH for all MATH. Thus, MATH and hence MATH is hyperbolic. This completes the proof of REF. CASE: It is easy to see using the resolvent identity, that there exists a MATH such that MATH is a MATH-multiplier for all MATH such that MATH. CASE: Without loss of generality, assume MATH. Denote MATH and fix MATH, MATH, MATH. Let us take a function MATH. By REF we have MATH . So, MATH. This implies that there is a MATH, MATH, such that MATH . Let us apply the functional MATH to the left-hand side of the inequality. Then we have: MATH . By REF the expression under the absolute value sign is equal to MATH . By the triangle inequality, we have MATH which is what we wanted. Now we turn to the second part of the theorem. First, we prove an auxiliary NAME lemma (probably, well-known). If MATH, then the integral MATH converges to MATH a.e. Moreover, MATH . MATH . The inner integral is equal to MATH. One can easily check that MATH is a positive kernel in MATH, that is, MATH tends to MATH a.e. and in MATH as MATH. Suppose MATH. Then by REF we have that MATH. By REF , there is a MATH such that MATH . Let us apply the operator MATH. Then using REF we obtain: MATH . Since MATH and MATH is a bounded linear operator from MATH to MATH, we conclude that the MATH-integral of the second summand converges and equals MATH. This means, in particular, that MATH converges. Let us denote it by MATH. Also let MATH for MATH, MATH, MATH. Now we introduce the following operators: MATH . It is easy to see that MATH. On the other hand, we have just proved that MATH exists for all MATH, MATH. So, by the boundedness principle for bilinear operators, MATH, where MATH does not depend on MATH and MATH. Let MATH, MATH, be a kernel in MATH, that is, MATH as MATH for each MATH. Then MATH and hence, MATH. Let us show that MATH exists for all MATH. This will be enough to prove that MATH exists for all MATH. Fix MATH and notice that MATH . So, MATH . Finally, it is only left to verify that MATH is indeed the NAME 's function. Let us prove the first equality in REF , the second one being analogous. We have MATH where we use the ordinary NAME 's theorem. It follows from the above that MATH is an exponentially decaying function. So, MATH is a bounded operator on MATH. On the other hand, MATH for all MATH. Hence, MATH for all MATH. The proof of REF can be found in CITE. |
math/0107205 | CASE: REF is evident. CASE: Assume for simplicity that MATH. First we claim that MATH maps MATH into MATH. To prove this, let us take an arbitrary function MATH of the form MATH , where MATH and MATH. Then, by REF , MATH . It implies that for every MATH there exists a MATH such that MATH where MATH is the characteristic function of MATH. For any fixed MATH, let us apply the operator MATH to the right-hand side of this inequality. Then we get MATH . Now using equality REF we obtain the following MATH . Thus, MATH. By the choice of MATH and MATH we have the same inequality on the whole real line. Since MATH was chosen arbitrary, the claim is proved. Let us observe that the boundedness of MATH is equivalent to the fact that MATH is a MATH multiplier. Denote by MATH the sun dual to MATH on which the dual semigroup is strongly continuous (see CITE). One can easily check, by duality, that for a test function MATH one has MATH where MATH and MATH is the generator of the sun dual semigroup. In other words, MATH maps MATH into MATH. By what we just proved, MATH is bounded from MATH to MATH, and again by duality, MATH maps MATH into MATH, which proves REF with MATH. The proofs of all other implications are completely analogous to those of REF . Let us now turn to the second part of our theorem. Although its proof is also essentially the same, some comments will be in order. By REF , proved below, REF is equivalent to MATH. So, the operators MATH, introduced in REF, are bounded from MATH to MATH. Uniform boundedness follows from the fact that MATH exists for all MATH and MATH by REF . Consequently, MATH. REF still makes sense for all MATH, because then MATH and all the integrals converge absolutely. So, MATH exists for all MATH, and it is continuous in MATH, MATH. Since MATH, by REF we have that MATH almost everywhere and hence, by the continuity of MATH, for all MATH. Thus, MATH and the proof is finished. |
math/0107205 | Since MATH is finite, there are constants MATH and MATH, MATH such that MATH for all MATH and MATH. Pick MATH large enough to satisfy MATH and such that whenever MATH and MATH , then MATH and MATH does not belong to the sector bounded by the contour MATH. For all such MATH we have MATH . Let us consider the following integral: MATH . By the choice of MATH, the integrand does not have singular points between MATH and MATH. By the NAME Theorem, we have MATH REF implies that the absolute value of the last integral is bounded from above by a constant that does not depend on MATH, whenever MATH, MATH. The analogous estimate from below follows from geometric considerations. Thus, MATH for some positive MATH and MATH. Suppose MATH. Then MATH . Let us notice that MATH for some MATH and all MATH, MATH, MATH, whereas MATH. Consequently, MATH . In combination with REF this gives the following estimates: MATH for all MATH, MATH and MATH, which proves the lemma. |
math/0107205 | If MATH, then our statement is the ordinary inversion formula (see REF ). Otherwise, by the resolvent identity, we have MATH for all MATH, MATH. So, in view of REF , MATH uniformly in MATH. Then, by the NAME Theorem, we get MATH . |
math/0107205 | CASE: Denote by MATH the linear operator that maps MATH to the corresponding solution of REF . By the Closed Graph Theorem, MATH is bounded. We prove that actually MATH. Indeed, by REF , MATH and MATH, for every MATH. On the other hand, a straightforward computation shows that if MATH is a MATH-function with compact support, then MATH and MATH. Thus, MATH. However, if MATH for some MATH, then again by REF , MATH is a solution of REF corresponding to MATH. By the uniqueness, we get MATH . So, MATH on a dense subspace of MATH and boundedness of MATH is proved. CASE: Suppose MATH is bounded from MATH to MATH. For a fixed MATH-function MATH having compact support, we show that MATH solves REF . Indeed, using REF, we get MATH which is precisely REF . Now suppose MATH is an arbitrary function from MATH. Let us approximate MATH by functions MATH of considered type. Then MATH converge to MATH in MATH and, without loss of generality, pointwise on a set MATH with MATH. Thus, REF is true for MATH, MATH and all MATH and MATH from MATH. To get REF for all MATH and MATH, we will modify MATH on the set MATH. To this end, let us take a decreasing sequence MATH such that MATH. Observe that the functions MATH defined for MATH are continuous. Since MATH on MATH, we get MATH everywhere in the half-line MATH. Put MATH to be MATH on MATH. By the above, MATH is a well-defined function on all MATH. Obviously, MATH on MATH. Let us show that MATH satisfies REF . Indeed, for any MATH and MATH we have MATH . |
math/0107205 | CASE: Note that MATH for all MATH. So, MATH . Thus, REF implies REF. Now assume REF. Denote MATH. Then MATH for all MATH, and REF follows. Following CITE we denote by MATH the operator of convolution with the semigroup, that is, MATH . Clearly, MATH is bounded on MATH for all MATH and MATH. Now we define the discrete multiplier operator MATH by the rule MATH where MATH is a trigonometric polynomial. One can check the identity MATH . By the assumption and the spectral mapping theorem for the point spectrum the operator MATH is one-to-one. Suppose REF holds. Then MATH has the left inverse MATH defined on MATH. By the Closed Graph Theorem MATH is bounded as an operator from MATH to MATH. Then REF says that MATH on trigonometric polynomials with values in MATH. So, MATH maps MATH into MATH for all MATH, which is what is stated in REF. If the assertion in REF is true only for some MATH, then as in the proof of REF , MATH maps MATH into MATH. By duality, MATH maps MATH into MATH and hence into MATH. So, MATH is a bounded operator from MATH to MATH, which proves REF with MATH. Assume REF. Then for every MATH there is a MATH such that MATH . Applying MATH in the above inequality and using REF we have: MATH . In particular, for MATH the last inequality yields REF. If REF holds, then taking MATH, with MATH having disjoint supports we get the following estimates MATH . So, we proved REF. Finally, similarly to the proof of convergence of the MATH-integral in REF , we can show that REF follows from MATH-boundedness of MATH. |
math/0107205 | Suppose that MATH maps MATH into MATH for MATH. By the Uniform Boundedness Principle MATH are uniformly bounded for all MATH. Then, by transference, MATH is a multiplier uniformly in MATH and MATH. In view of just proved REF we get MATH for all MATH from some open annulus MATH containing MATH. In order to show uniform boundedness of MATH, let us look at identity REF first. It shows, in particular, that MATH. Just like in the second part of the proof of REF , one can estimate MATH by the multiplier norm of MATH. Rescaling gives the same conclusion for all MATH . Since norms of the corresponding multipliers are uniformly bounded, the desired result is proved. Now let us prove the converse statement. Clearly, the family MATH obeys the resolvent identity on vectors from MATH. Since, in addition, it is bounded, the mapping MATH is strongly continuous on vectors from MATH and, hence, on all MATH. Again by the resolvent identity, MATH is strongly analytic on MATH. Since for any MATH, MATH is the uniform limit of a sequence MATH with MATH, MATH is analytic. It suffices to show that the integral MATH converges for all MATH and there exists a MATH such that MATH. So, let us fix MATH and MATH. Then for any MATH, MATH. Thus MATH . From this we get MATH where MATH is the NAME kernel. Passing to limit inside the integral we get MATH if MATH. And MATH otherwise. In either case, replacing MATH by MATH, if MATH, or by MATH, otherwise, we get the desired exponential decay. |
math/0107205 | Let us prove necessity. If MATH is MATH-hyperbolic, then there is a splitting projection MATH. Suppose MATH. Then by REF applied to the semigroup MATH, MATH-we have MATH. In particular, MATH. On the other hand, if MATH, then by the same reason, MATH, where MATH is defined by MATH . So, MATH. Since MATH is dense in MATH, this shows that MATH continuously extends to all of MATH and equality REF is true. The uniqueness of MATH follows automatically from REF . Since the MATH-norm of MATH is exponentially decreasing and MATH for all MATH, MATH is bounded from MATH to MATH. To show boundedness of MATH, it is enough to notice that if MATH is MATH-hyperbolic, then the scaled semigroup MATH is also MATH-hyperbolic, for small values of MATH. Now we prove sufficiency. Let us introduce the operator MATH. Since REF is valid, and hence REF are true, the norm of MATH on MATH is exponentially decaying. Consequently, by the ordinary inversion formula for NAME transform, we get MATH, for all MATH. This implies MATH on all MATH , in view of the continuity of MATH. So, MATH is a projection. Obviously, MATH. On the other hand, since MATH, we have MATH and REF is proved. To show invertibility of MATH on MATH, we apply REF . It implies that MATH, for MATH in MATH, and hence, MATH is invertible. Another application of REF and the second part of REF proves REF . |
math/0107206 | For every MATH, we choose an element MATH such that MATH. Take MATH. If MATH is a well-ordered subset of MATH such that MATH for all MATH, then we set MATH . Observe that the support of MATH is contained in MATH and thus, it is again well-ordered. Note also that MATH . Indeed, let MATH be the least element in MATH. Then MATH. On the other hand, if MATH and MATH then MATH: if MATH then MATH (by minimality of MATH) and MATH; if MATH then MATH and MATH. Now suppose that MATH is an order preserving embedding such that the image MATH is convex in MATH. We wish to deduce a contradiction. The idea of the proof is the following. Let ON denote the class of ordinal numbers. We shall define an infinite MATH matrix with coefficients in MATH, such that each column MATH is a strictly increasing sequence in MATH. Since MATH is a set, every column of this matrix will provide a contradiction at the end of the construction (compare figure). To get started, we have to define the first row of the matrix. We construct sequences MATH, MATH, and MATH, MATH, in MATH. We take an arbitrary MATH. Having constructed MATH, we choose MATH and MATH as follows. Since MATH has no last element, we can choose MATH such that MATH. Hence, MATH . Let MATH be the least element in MATH for which MATH and MATH the least element in MATH for which MATH . Since MATH is cofinal in MATH, we can choose MATH such that MATH . Further, we set MATH . Then by REF , MATH . Thus, MATH by convexity, and we can set MATH . Now for every MATH we have that MATH, hence every well-ordered set MATH with smallest element MATH has the property that MATH for all MATH; and moreover, MATH . Thus, MATH by convexity. Suppose now that for some ordinal number MATH we have chosen elements MATH, MATH, MATH, such that for every fixed MATH, the sequence MATH is strictly increasing. Then we set MATH for every MATH. If MATH, then MATH and thus, MATH by REF . So for every ordinal number MATH, the sequences MATH can be extended. We obtain strictly increasing sequences of arbitrary length, contradicting the fact that their length is bounded by the cardinality of MATH. |
math/0107206 | For MATH and MATH, set MATH . (here, Im MATH denotes the image of MATH). Now, it is straightforward to check the assertion of the lemma. |
math/0107206 | Let MATH, and set MATH, for MATH. Let MATH be such that MATH. If MATH, then MATH. So assume MATH and set MATH. Suppose that MATH. If MATH, then MATH. On the other hand, MATH (otherwise, MATH). Thus, MATH, a contradiction. Similarly, we argue that if MATH, then MATH, a contradiction. Hence, MATH. Since MATH is a final segment of MATH, this implies that MATH, which proves our assertion. |
math/0107206 | The first assertion follows from REF , applied to the final segment consisting of the single last element of MATH. For the second assertion use REF. |
math/0107206 | Set MATH. Since MATH has a last element, MATH embeds convexly in MATH. Consequently, MATH embeds as a final segment in MATH. By induction on MATH we define MATH, and obtain an embedding of MATH as a final segment in MATH. We set MATH. Since every MATH is a final segment of MATH, every well-ordered subset MATH of MATH is already contained in some MATH (just take MATH such that the first element of MATH lies in MATH). Hence, an element of MATH with support MATH is actually an element of MATH, for some MATH. This fact gives rise to an order isomorphism of MATH onto MATH. To prove the second assertion, we set MATH. Since MATH has a last element by assumption, MATH embeds convexly in MATH, and the same arguments as above work if we define MATH. |
math/0107206 | Assume MATH is not last, and choose some element MATH. This provides us with characteristic functions. If MATH is well-ordered, then let MATH denote the characteristic function on MATH defined by: MATH . Note that these characteristic functions reflect inclusion: if MATH is a proper well-ordered subset of MATH, then MATH. Now assume for a contradiction that MATH, and let MATH. We shall construct a strictly increasing sequence MATH in MATH. Set MATH, and assume by induction that MATH is defined, and strictly increasing in MATH. Then define MATH . It follows that MATH, whenever MATH. Since MATH is order preserving, it follows that MATH as required. |
math/0107206 | The ``if" direction is just the second assertion of REF . So assume now that MATH is a non-empty solution. Assume for a contradiction that MATH has no last element. Then by REF MATH has no last element as well. Thus, the same holds for the solution MATH. This contradicts REF . |
math/0107206 | Recall that MATH embeds as a final segment in the given solution MATH. Thus, MATH embeds as a final segment in MATH as well. In particular, MATH has a last element MATH. Since MATH and MATH, we find that MATH. But MATH, since MATH is a final segment of MATH. |
math/0107206 | Necessarily, MATH is last in MATH (by REF ). Thus, MATH has a last element, so by REF , MATH embeds as a final segment in MATH, and thus in MATH. |
math/0107206 | Since MATH, MATH is last in MATH REF . Therefore, MATH is last in MATH by REF , and so also MATH has a last element MATH. The assumptions imply that MATH. This is equivalent to the assertion that MATH embeds as a final segment in MATH. Now note that MATH is a special solution by REF , that is, MATH embeds as a final segment of MATH. Since MATH is infinite this implies that MATH embeds as a final segment in MATH, as required. |
math/0107206 | REF implies REF by REF implies REF trivially. Finally, REF implies REF by REF . |
math/0107206 | Let MATH be another solution. Then it is a special solution, by REF . So MATH embeds as a final segment in MATH. Since MATH is last in MATH, MATH embeds as a final segment in MATH. By induction, MATH is a final segment of MATH for every MATH. Thus. MATH embeds as a final segment in MATH as well. |
math/0107207 | Let MATH be a function from MATH to MATH, that is, MATH is a colouring, and we shall find MATH as required for it. Let MATH list MATH such that MATH . For MATH let MATH : MATH . Clearly MATH by REF . For each MATH we define MATH by: MATH where for MATH, we let MATH be defined by MATH for MATH, recalling that MATH divides MATH as MATH. So MATH is a subset of MATH of cardinality MATH hence we can choose MATH . For MATH let MATH be MATH . If for some MATH the function MATH serve as a weak diamond sequence for MATH, we are done so assume that for each MATH there are MATH and MATH such that: CASE: MATH is a club of MATH. CASE: MATH CASE: if MATH then MATH. Now define MATH by MATH for MATH . Let MATH is divisible by MATH and MATH and MATH . Clearly MATH is a club of MATH hence we can find MATH. So by the definition of MATH we have MATH and for MATH we have MATH is equal to MATH Why? note that MATH as MATH and see the definition of MATH and of MATH, so : MATH . Hence MATH is well defined and by the choice of MATH we have MATH so by its definition, MATH for each MATH satisfies MATH . Now by the choice of MATH we have MATH so MATH but MATH whereas we have chosen MATH such that-MATH a contradiction. REF . |
math/0107207 | Let MATH be a REF-to-REF function from MATH onto MATH, for simplicity, and without loss of generality MATH and let the function MATH for MATH be such that MATH. Let MATH for some MATH of cardinality MATH, we have MATH so by REF clearly MATH, so let us list MATH as MATH with no repetitions, and let MATH. For MATH let MATH a function from MATH to MATH such that for every MATH we have MATH . By clause REF necessarily MATH has cardinality MATH. For each MATH and MATH we define MATH by MATH for MATH. Now for MATH, clearly MATH belongs to MATH. Clearly MATH is a subset of MATH of cardinality MATH which as said above is MATH. But MATH by clause REF, so we can find MATH . For each MATH we can consider the sequence-MATH as a candidate for being a MATH -Wd-sequence. If one of then is, we are done. So assume toward contradiction that for each MATH there is MATH which exemplify its failure, so there is a club MATH of MATH such that CASE: MATH and without loss of generality CASE: MATH . But MATH and so MATH hence we have gotten CASE: MATH . Define MATH by MATH, now as MATH is regular uncountable clearly MATH for every MATH we have MATH and if MATH then MATH is a club of MATH (see the choice of MATH recall that by REF the sequence MATH is good, see REF ). Clearly MATH is a club of MATH. Now for each MATH, clearly MATH; just check the definitions of MATH and MATH. Now recall MATH is the function from MATH to MATH defined by MATH . But by our choice of MATH clearly MATH, so MATH . Hence MATH, however as MATH clearly MATH, together MATH. As MATH clearly MATH, moreover for each MATH we have MATH, see its definition above, is equal to MATH which by the previous sentence is equal to MATH. As this holds for every MATH and MATH are members of MATH, clearly they are equal. But MATH so MATH whereas MATH has been chosen outside this set, contradiction. |
math/0107207 | CASE: By the definition of MATH CASE: By CITE. |
math/0107207 | CASE: The same proof. CASE: Let MATH be increasing continuous with unbounded range and let MATH be stationary, such that-MATH and MATH is a good MATH-Wd- parameter,let MATH and repeat the proof using MATH instead MATH. Except that in the choice of the club MATH we should use MATH: for every MATH . Rang MATH we have MATH and MATH is a limit ordinal and MATH. CASE: Similarly. |
math/0107207 | The point is that CASE: if MATH is a normal filter on MATH is MATH -increasing in MATH and MATH then there is a normal filter MATH on MATH extending MATH such that MATH CASE: if MATH is MATH- increasing MATH, and MATH and MATH then there are functions MATH for MATH such that MATH, and MATH . |
math/0107207 | CASE: Follows by REF CASE: By CITE for some MATH, for every regular MATH we have: MATH. Let MATH . By REF it is enough to prove MATH if MATH then MATH proof of MATH . If not then for every MATH there is MATH and define MATH . This is a MATH-complete filter on MATH see CITE. So for some MATH the set MATH is unbounded in MATH. By CITE (alternatively use CITE on normal filters) REF for MATH from MATH and MATH is a MATH-complete filter on MATH. But as MATH letting MATH which is MATH, so MATH, there is a family MATH such that for every MATH, for some MATH and MATH for MATH we have MATH hence for every MATH, for some MATH we have MATH . So for some MATH and unbounded MATH we have MATH and moreover for some MATH we have MATH and moreover MATH. But this contradict REF . CASE: We can find MATH such that: CASE: MATH is MATH-increasing continuous such that MATH and MATH. CASE: if MATH and MATH then MATH. Let MATH be MATH, so it is well known that MATH is the MATH-th function, in particular MATH where MATH is the club filter on MATH; in fact MATH . Choose MATH a club of MATH of order type MATH, and let MATH enumerate MATH, that is, MATH is the MATH-th member of MATH . For MATH let MATH be constantly MATH, and let MATH be defined by MATH CASE: MATH [why? by an assumption] For MATH we define MATH by: MATH . Note that MATH is a function from MATH to MATH. Now CASE: MATH for MATH [Why? as MATH, so by a hypothesis MATH CASE: for MATH is a club of MATH [Why? Obvious] CASE: for MATH we have MATH [Why? the first statement by the definition of MATH and of MATH. The second by the first MATH.] CASE: if MATH then MATH hence MATH [Why? the first as MATH hence for some MATH, we have MATH . Let MATH, so MATH . This of course suffice by REF. |
math/0107208 | The proof can be found in CITE and is based on the fact that for countable MATH and any pure subgroup MATH of finite rank, MATH implies MATH (MATH a torsion group). |
math/0107208 | Let MATH . By REF MATH. Now it is easy to see that MATH stationary implies that the relative MATH-invariant MATH. Since we are assuming MATH the weak diamond MATH holds (see CITE). Thus REF or REF imply MATH by CITE. Conversely, assume that MATH but REF do not hold. Then, the relative MATH-invariant MATH and hence CITE shows that MATH - a contradiction. |
math/0107208 | Since we are assuming MATH the weak diamond MATH holds. Let MATH be a stationary subset of MATH such that MATH holds. Since MATH is a cub in MATH we may assume without loss of generality that MATH, that is, MATH consists of all limit ordinals of MATH. Choose a tree-like ladder system MATH such that MATH is a successor ordinal for all MATH and MATH. We enumerate the sets MATH by MATH without repetitions, for example, MATH. Now let MATH be the free group generated by the elements MATH. Let MATH and for MATH where MATH. Let MATH be the subgroup of MATH generated by the elements MATH. Then the only relations between the generators of MATH are MATH for MATH and MATH. Now, for MATH let MATH be the pure closure in MATH of MATH. Then the sequence MATH forms a MATH-filtration of MATH. Moreover, for MATH we have MATH where MATH is the free group on the generator MATH and MATH. Finally, MATH is MATH-free by NAME 's criterion. Indeed, if MATH is a finite subset of MATH, then the pure closure of MATH is contained in the pure closure of a finite subset MATH of MATH. By enlarging MATH we may assume that there exists MATH such that for all MATH, MATH if and only if MATH. Then REF show that the pure closure of MATH is free (compare CITE). Finally, let MATH be an infinite subset of MATH. Since MATH is MATH-free there exists no finite rank pure subgroup MATH of MATH such that MATH, hence REF shows that MATH if and only if the set MATH is stationary in MATH. Since for MATH we have MATH it is now easy to see that MATH is stationary if and only if MATH is stationary. Note that MATH is a cub in MATH. |
math/0107208 | Let MATH be given. If MATH, then we choose MATH to be free of cardinality MATH and we are done. Therefore, assume that MATH. Choose a continuous increasing sequence of boolean subalgebras MATH such that each MATH is countable and contains all finite subsets of MATH. Note that this is possible since we are assuming MATH. Let MATH and put MATH and MATH where we assume that each MATH is repeated infinitely many times. Choose for MATH and MATH . Note, that this is possible since MATH is infinite and MATH. Let MATH and let MATH be the group from REF for MATH. Then MATH is MATH-free and of cardinality MATH and by REF it suffices to prove that for a subset MATH we have MATH if and only if there exists MATH such that for all MATH, MATH is finite. Thus let MATH and assume that MATH. Then there exists MATH such that for all MATH, MATH. Fix MATH, then MATH for some MATH. Hence, for all MATH we obtain MATH. Thus, MATH which is finite. Conversely, assume that MATH. Then there exists MATH such that for all MATH, MATH. Fix MATH, then MATH for infinitely many MATH by the choice of the MATH's. But, if MATH, then MATH and hence MATH is infinite. This finishes the proof. |
math/0107208 | Let us first show that REF to REF hold for MATH for any torsion-free group MATH of cardinality less than or equal to MATH. Clearly, REF are true. Moreover, if MATH is countable, then CITE shows that REF hold for MATH. Thus assume that MATH is of cardinality MATH and let MATH be a MATH-filtration of MATH. Let MATH be a prime and assume that MATH. Moreover, assume that MATH is a MATH-group and MATH. By REF there exists either a finite rank pure subgroup MATH of MATH such that MATH or the set MATH is stationary in MATH. If MATH exists, then CITE shows that MATH contradicting the fact that MATH. Thus assume that MATH is stationary in MATH. Again by CITE it follows that for MATH also MATH since all MATH's are countable. Thus MATH and REF shows that MATH - a contradiction. Thus REF holds since the converse implication is trivial. It is straightforward to see that also REF holds for MATH using similar arguments as above. Finally, assume that MATH satisfies REF to REF . We identify MATH with MATH by a bijection MATH. Let MATH. Then it is easy to see that MATH is an ideal on MATH containing all finite subsets of MATH. Thus by REF there exists a MATH-free group MATH of cardinality MATH such that for every subset MATH, MATH if and only if MATH. Since we have already shown that MATH satisfies REF to REF it is now obvious that MATH. |
math/0107208 | Let MATH be the given infinite set of primes. It was shown by NAME in CITE that there exists a strictly decreasing chain of subsets MATH REF such that CASE: MATH is infinite; CASE: MATH implies MATH is almost contained in MATH; CASE: MATH implies MATH is infinite; CASE: MATH is finite. Let MATH be the ultrafilter generated by MATH and let MATH be the group from REF for MATH. If MATH is an infinite subset of MATH, then divide MATH into two disjoint infinite subsets, for example, MATH. Since MATH is an ultrafilter it follows that without loss of generality MATH. Hence, there exists MATH such that MATH is finite. Thus, for every MATH we obtain MATH is finite. Therefore, the set MATH is not stationary in MATH and REF implies that MATH. Finally, MATH for any completely decomposable group MATH since MATH violates CITE which is our REF . |
math/0107210 | CASE: Note that MATH . Now the result follows from the fact that MATH where MATH compare Sec. CASE: The proof of REF is identical. |
math/0107210 | CASE: As before, let MATH be a generator of MATH and let MATH be given by MATH and by MATH respectively. MATH is equal to MATH or MATH depending if MATH is odd or even. Similarly, MATH is equal to MATH or MATH depending if MATH is odd or even, where MATH are given by MATH and by MATH respectively. Now REF follows from the fact that MATH and MATH . CASE: Let MATH . Since MATH is a MATH-group, MATH (see CITE) and by the universal coefficient theorem, MATH . Therefore, it is sufficient to show that MATH and MATH are isomorphic as MATH-modules, and for that it is enough to consider indecomposable MATH-modules MATH only. Such modules are classified in REF . We leave it to the reader to check that in each of the three possible cases we get MATH as needed. |
math/0107210 | Let MATH denote the one point space and let MATH be a zero cell in MATH . The MATH-equivariant maps: MATH and MATH induce maps MATH whose composition is the identity. MATH is either MATH or MATH . Hence, if MATH is not MATH for some MATH then MATH and MATH . Hence, MATH is even and, hence, MATH . Therefore MATH cannot be the identity. |
math/0107210 | Since the MATH-action on MATH is trivial, the total complex of MATH is isomorphic to MATH where MATH is the cochain complex MATH . Hence, by NAME formula, MATH is equal to MATH . Since MATH is either MATH or MATH depending if MATH even or odd, the proof follows. |
math/0107210 | MATH is equal as a MATH-module to MATH where the product is taken over all places MATH in MATH and MATH . Since MATH is a simple group, a place MATH in MATH either splits completely in MATH or it is inert, or it is totally ramified. In the first case MATH acts on MATH by freely permuting its components and therefore MATH . In the second and third cases MATH is either MATH or MATH and the cohomology groups of MATH with these coefficients have been calculated in REF: MATH if MATH is inert, and MATH for ramified MATH . (Note that for archimedean valuations, the calculation reduces to calculating MATH where MATH acts on MATH by conjugation). Hence MATH depending if MATH or MATH . |
math/0107210 | Since the functor MATH on MATH-modules is left exact, the bottom horizontal row of REF yields the exact sequence MATH . By considering a MATH diagram for MATH analogous to REF we get the exact sequence MATH which together with the sequence above forms a commutative diagram MATH . By REF, the central vertical arrow is an isomorphism. Hence, by Snake Lemma (compare CITE), MATH . Consider the long exact sequence of cohomology groups of MATH induced by the short exact sequence of coefficient groups MATH . Since for any MATH-module MATH and MATH this long exact sequence gives rise to an exact sequence MATH where MATH coincides with the map MATH of REF . Hence, by REF , MATH . |
math/0107214 | First we prove that MATH defines a bijection. To this end it needs to be shown that MATH is well-defined and that the algorithm is invertible. Adding the letter MATH to the MATH-th column of a tableau in MATH adds a box in row MATH to the MATH-th rigged partition for all MATH. This operation leaves all vacancy numbers unchanged, except MATH which changes by MATH. Since MATH there hence exists a singular box in row MATH in MATH for all MATH. This shows that MATH is well-defined. It is easy to see that the algorithm for MATH is invertible. Given MATH one constructs MATH with MATH by successively determining MATH for MATH and MATH. Set MATH to be MATH with inverted quantum numbers and let MATH be the rigged configuration after determining MATH. To determine MATH, one finds the smallest index MATH such that there is a singular string in row MATH of MATH for all MATH. Set MATH. The rigged configuration MATH is obtained from MATH by removing the selected singular strings in row MATH and making the new boxes in row MATH singular. Then it is clear that MATH (where MATH if MATH) so that MATH will indeed be a MATH-multipartition of shape MATH and content MATH. It remains to prove that MATH preserves the statistic. Since we are dealing with MATH-multitableaux of single rows the inversion statistic of REF simplifies as follows. The pair MATH is an inversion of MATH if MATH and either MATH and MATH or MATH and MATH. Furthermore, the following relation between a multitableau MATH and the corresponding configuration MATH will be needed MATH . Building MATH up successively, the addition of the letter MATH in box MATH changes the inversion statistic by MATH . Denote by MATH the configuration corresponding to MATH built up to letter MATH in the algorithm for MATH. Then, setting MATH and inserting REF yields MATH . On the other hand, the change of the rigged configuration statistic MATH, taking into account that we have to invert the quantum numbers, is given by MATH where the first line comes from the change in MATH and the second line is induced by the change of the vacancy number. Since REF agree at each step in the algorithm and MATH, MATH is statistic preserving. |
math/0107214 | We begin by proving that MATH defines a bijection. Adding the letter MATH to the MATH-th row of a tableau in MATH adds a box in row MATH to the MATH-th rigged partition for all MATH. This operation leaves all vacancy numbers unchanged, except MATH which changes by MATH. Since MATH there exists a singular box in row MATH in MATH for all MATH. This shows that MATH is well-defined. It is easy to see that the algorithm for MATH is invertible. It remains to show that the statistic is preserved. In the single column case MATH is an inversion if MATH, MATH and MATH. Furthermore, the analog of REF becomes MATH . Adding the letter MATH to MATH in box MATH changes the inversion number by MATH where the last line subtracts all contributions which do not satisfy MATH. Denote by MATH the configuration corresponding to MATH built up to letter MATH in the algorithm MATH. Then, setting MATH and inserting REF yields MATH . On the other hand, taking into account that the quantum numbers need to be inverted for the correct statistic, we have MATH from the change in vacancy numbers. Since MATH and MATH agree at each step of the algorithm, MATH is statistic preserving. |
math/0107217 | Suppose MATH is completely isometric and MATH is a boundary representation. Set MATH, and note that it is completely contractive. Suppose MATH dilates MATH. The goal is to show that MATH reduces MATH. To this end, define a map MATH by MATH, MATH. This map is completely contractive, and so by the NAME extension theorem extends to a completely positive unital map MATH with MATH. Observe that the map which takes MATH, MATH is completely positive, and by definition, MATH for all MATH. We have assumed that MATH is a boundary representation, so in fact MATH for all MATH. From this we have for all MATH, MATH where the first inequality is the NAME inequality for completely positive maps CITE. From this we see that MATH. An identical argument gives MATH, proving that MATH reduces MATH. The converse is a straightforward exercise and is left to the reader. |
math/0107217 | The proof is by contradiction. Accordingly, suppose MATH does not lift to a weakly extremal representation relative to MATH. Let MATH be the cardinality of the the set of points in the unit sphere of MATH, MATH the cardinality of the set of elements in the unit ball of MATH. Set MATH. Let MATH be the smallest ordinal greater than or equal to MATH. Note that there is a MATH so that for MATH, MATH for all MATH and MATH. Construct a chain of liftings in MATH by transfinite recursion on the ordinal MATH as follows: if MATH, and MATH has a predecessor, let MATH denote a strong (with respect to MATH) nontrivial lifting of MATH. Such an lifting exists by the assumption that MATH does not lift to a weak extremal. If MATH is a limit ordinal, set MATH to the spanning representation of MATH. For any MATH in the unit sphere of MATH and MATH in the unit ball of MATH, there are at most countably many MATH's with predecessors where MATH. Since the cardinality of the set of ordinal numbers less than or equal to MATH and having a predecessor is MATH, there must be an ordinal MATH with predecessor where MATH for all MATH in the unit sphere of MATH and MATH in the unit ball of MATH, so that MATH is a lifting of MATH which is weakly trivial with respect to MATH; a contradiction, ending the proof. |
math/0107217 | We use REF to prove REF . Let MATH denote a given representation. Lift MATH to a representation MATH which is weakly extremal relative to MATH. Lift MATH to a representation MATH which is weakly extremal relative to MATH. Continuing in this manner, constructs a chain MATH, MATH, with respect to the partial order on liftings with the property that MATH is weakly extremal relative to MATH. The resultant spanning representation MATH lifts MATH and it is easily checked to be extremal, since it is weakly extremal relative to MATH for all MATH. |
math/0107217 | The proof closely follows that of the existence of weak extremals, and is by contradiction. Hence we suppose MATH does not lift to a weak MATH-representation relative to MATH. We define the ordinal MATH as in the proof of REF . Construct a chain of dilations in MATH where each of the representations by transfinite recursion on the ordinal MATH as in REF : if MATH and MATH is a limit ordinal, set MATH to the spanning representation of MATH and if MATH has a predecessor, let MATH be a dilation to a strong (with respect to MATH) nontrivial dilation of MATH, which exists by the assumption that MATH does not lift to a weak MATH-representation. Then for any MATH in the unit sphere of MATH and MATH in the unit ball of MATH, there are at most countably many MATH's with predecessors where MATH or MATH. The same reasoning then gives a representation MATH in our chain dilating MATH which is weakly trivial with respect to MATH, a contradiction. |
math/0107217 | This now follows the proof of REF . Construct a countably infinite chain of representations MATH into the bounded operators on NAME spaces MATH, where MATH is a weak MATH-representation with respect to MATH for each MATH. Let MATH denote the spanning representation on the NAME space MATH. Since a dilation of a weak MATH-representation with respect to a representation MATH is also a weak MATH-representation with respect to MATH, MATH is a weak MATH-representation with respect to MATH for all MATH. It easily follows that MATH is a MATH-representation. |
math/0107217 | A proof follows directly from REF . Viewing MATH as a subspace of MATH, the inclusion mapping MATH is a completely isometric representation and thus, according to this proposition, it dilates to a completely isometric representation MATH which is a MATH-representation. To see that MATH is the MATH-envelope, suppose MATH is also completely isometric. In this case MATH given by MATH is completely contractive (and thus well-defined). By REF, there exists a NAME space MATH containing MATH and a representation MATH such that MATH REF . Since MATH is a representation of MATH and MATH is a MATH-representation, MATH reduces MATH. Thus, MATH extends to an onto representation MATH. |
math/0107218 | This is a variation of REF . We may assume MATH and MATH to be unital and MATH. By NAME 's theorem we have that MATH, where MATH is a MATH-representation and MATH is the projection onto MATH, that is, MATH. Then we get MATH hence MATH. Finally MATH . |
math/0107218 | MATH. |
math/0107218 | This is an argument used frequently in MATH-theory, compare for example, REF . If MATH, then MATH, so MATH, where MATH. But then MATH is continuous on MATH and MATH is a projection satisfying MATH (see REF for the definition of MATH). Finally, MATH, where MATH and one has MATH . |
math/0107218 | Set MATH, then MATH . Furthermore we have MATH and by REF there exists a projection MATH satisfying MATH, thus MATH. Obviously MATH. |
math/0107218 | This basically is REF . If necessary, we adjoin a unit to MATH. Define symmetries MATH and MATH, then there is MATH with MATH such that MATH. Also, one readily checks that MATH. Set MATH, then it is not hard to see that MATH as well. As in REF , one checks that MATH, thus MATH and MATH. Set MATH, then MATH, MATH and MATH . |
math/0107218 | CASE: If MATH is not invertible, then there is some MATH such that (in MATH notation), MATH . But then MATH; in particular MATH is orthogonal to all the MATH. We would thus have MATH linearly independent vectors in MATH, a contradiction. REF follows from REF , for, since MATH is open, one can choose unitaries MATH in MATH such that MATH are linearly independent (where MATH denotes a unit vector corresponding to MATH). |
math/0107218 | CASE: REF is immediate. We show REF: Take a finite open covering of MATH. Because MATH, there is an open refinement MATH of order not exceeding MATH. By a standard argument (compare REF ) we may assume MATH to be finite and set MATH. Take a partition of unity MATH subordinate to MATH and set MATH then MATH also is a refinement of order not exceeding MATH. It suffices to find an open refinement MATH of MATH of strict order not exceeding MATH. View MATH as map MATH, where MATH is the standard simplex in MATH and MATH is the minimal subcomplex of MATH containing MATH (s. REF or CITE for an introduction to simplicial complexes). Let MATH be the vertices of MATH and MATH be the open stars around MATH. Then MATH; as MATH has order less than or equal to MATH, we have MATH (note that MATH, where MATH is the (combinatorial) dimension of the (abstract) simplicial complex MATH, the geometric realization of which is MATH). Let now MATH be the barycentric subdivision of MATH with vertices MATH and open stars MATH compare REF. Set MATH, then MATH is a refinement of MATH is of order less than or equal to MATH, as MATH. Note that MATH is of strict order not exceeding MATH if the following condition on MATH holds: CASE: for any distinct vertices MATH of MATH there exist indices MATH such that MATH and MATH do not sit in one and the same face of MATH, that is, MATH and MATH are not connected by an edge of MATH. The vertices MATH of MATH are exactly the barycenters of simplexes MATH of MATH. But then MATH span a simplex in MATH, if MATH is a face of MATH in MATH for MATH, and every simplex in MATH is of this form compare REF. Let now MATH be distinct vertices of MATH, such that MATH is connected to MATH by an edge in MATH if MATH. But then MATH, because MATH is a proper face of MATH or vice versa. As MATH, we have MATH. Thus REF holds and we are done. |
math/0107218 | Let MATH and MATH and MATH be given. Choose a compact subset MATH such that MATH for MATH and MATH. Then there is an open covering MATH of MATH with MATH and such that MATH (here we used that MATH is compact). Because MATH, there is a refinement MATH with the following properties: CASE: MATH for MATH, REF for every MATH there exists MATH such that MATH, REF MATH has strict order no greater that MATH. Take a partition of unity MATH subordinate to MATH. Because MATH is normal, there is a function MATH such that MATH and MATH for MATH. Set MATH, and note that MATH and that MATH has strict order less than or equal to MATH. Also, MATH if MATH. We are now prepared to define MATH . For MATH we obtain for every MATH: MATH because MATH, if MATH and MATH, if MATH. Therefore, MATH and the proof is complete. |
math/0107218 | It obviously suffices to approximate elements of the form MATH for MATH, MATH, MATH. By REF there is a c.p. approximation MATH within MATH for MATH with MATH and MATH for some MATH. Then MATH is a c.p. approximation for MATH. We show MATH: We have MATH. Now let MATH be elementary, then each MATH lives in a summand MATH, because it is minimal. This means MATH, where MATH is a generator of MATH and MATH is a minimal projection in MATH. If there are MATH such that MATH, then MATH because MATH, but then also MATH. If the MATH are pairwise orthogonal, then, because MATH, there must be MATH such that MATH, thus MATH. Therefore MATH and we are done. |
math/0107218 | By the NAME - NAME lifting REF , MATH has a completely positive contractive lift MATH. Let MATH and MATH be given. Choose a c.p. approximation MATH for MATH within MATH with MATH. Then MATH is a c.p. approximation of MATH within MATH. MATH, because MATH is a MATH-homomorphism, thus preserves orthogonality. |
math/0107218 | Given MATH, MATH, choose c.p. approximations MATH and MATH within MATH for MATH and for MATH, respectively, such that MATH and MATH. As a consequence, MATH is a c.p. approximation within MATH for MATH. Since MATH, we have MATH. Equality follows from REF . |
math/0107218 | Let MATH be the induced maps, then MATH. Given MATH, MATH, choose MATH such that there are MATH with MATH, MATH, and such that MATH. By REF there is a c.p. approximation MATH (of MATH) within MATH for MATH with MATH. After extending MATH to all of MATH by the NAME extension theorem, MATH is a c.p. approximation within MATH for MATH. |
math/0107218 | Let MATH be the map from MATH to MATH. It follows from REF that MATH. But now MATH, where the MATH are nested, so by REF we have MATH and the assertion follows. |
math/0107218 | If MATH, then MATH and there is nothing to show, so we assume MATH. Let MATH and suppose MATH. We (inductively) construct an elementary set MATH, such that MATH for MATH. This will imply MATH; we obviously have MATH, so MATH if MATH. There are orthogonal minimal projections MATH such that MATH; set MATH. Next suppose that for some MATH we have constructed an elementary set MATH such that MATH . Choose a minimal projection MATH orthogonal to MATH; this is possible since MATH. Take MATH and identify MATH with MATH. There is an open neighborhood MATH of MATH in MATH such that MATH . Note that MATH is elementary for all MATH. Set MATH for MATH and MATH, then the MATH and MATH are closed in MATH. If MATH was empty, this would mean MATH but then by REF we would have MATH, a contradiction to MATH. So let MATH; just as above we obtain that MATH must be nonempty. Inductively we get that MATH. MATH is open in MATH and we have MATH . Choose some MATH and set MATH. It might still happen that MATH. To fix this, identify MATH with MATH. There is an open neighborhood MATH of MATH such that MATH, MATH, MATH and MATH are all nonzero for MATH and MATH. Now there must be a nonempty open subset MATH such that MATH for otherwise again by REF und REF we would have MATH, contradicting our construction. Choose some MATH and define MATH then MATH is elementary by construction, MATH and we have MATH . Induction then yields an elementary set MATH with the desired properties. |
math/0107218 | MATH, so MATH. |
math/0107218 | For any MATH and for MATH we have MATH therefore MATH . Now inductively construct pairwise orthogonal projections MATH and MATH, such that MATH . First define MATH. Next suppose MATH, MATH already are constructed. We then have MATH . REF now yields a projection MATH with MATH . Obviously MATH for MATH, if MATH has been chosen small enough. |
math/0107218 | MATH by REF , so we have to show MATH. Unfortunately, the proof is a little technical, so we briefly sketch the strategy: Given an open covering MATH of MATH, choose a refinement MATH, where the MATH are suffiently small. Take a partition of unity MATH subordinate to MATH and a c.p. approximation MATH with MATH for the MATH. By REF , MATH must be a direct sum of matrices of rank no larger than MATH. On suitable subsets of MATH one now can introduce equivalence relations to combine the MATH to a new open covering MATH, which still refines MATH. One can use REF and the special properties of MATH to produce a refinement MATH of MATH, the order of which is less than or equal to MATH. Then MATH will be the desired refinement of MATH. MATH is second countable, thus MATH for compact subsets MATH. By REF we have MATH. Now if we can show that MATH we are done, for the countable sum REF then yields MATH. Therefore we may assume MATH to be compact (and second countable); in particular we may choose a metric MATH on MATH. Set MATH. Choose a partition of unity MATH subordinate to the given (finite) open covering MATH of MATH. Set MATH, then there exists a MATH, such that MATH and MATH with MATH. Again because MATH is compact, there exists a finite open covering MATH with MATH let MATH be a partition of unity subordinate to MATH. We have to define some constants: First set MATH. Then choose MATH as in REF (with MATH) and define MATH. Note that MATH. Take a c.p. approximation MATH, within MATH for MATH. Then MATH, and by REF we have MATH. Let MATH be the unit of MATH and MATH be the MATH'th component of MATH. Now define MATH and MATH . Denote by MATH the equivalence relation on MATH generated by the relation MATH . Let MATH be the equivalence classes with respect to MATH, then MATH. Set MATH; then MATH and MATH, if MATH. Setting MATH, we have MATH and MATH for MATH. It follows that MATH . Define MATH (compare REF); MATH is a projection and for MATH we obtain: MATH . We have MATH so by REF (and by the choice of MATH) there are pairwise orthogonal projections MATH with MATH. We are now prepared to define MATH we will show that MATH is an open covering that refines MATH. Clearly, the MATH are open. By REF , the order of MATH is less than or equal to MATH, because MATH. MATH covers MATH: Let MATH be in MATH. MATH can be decomposed into sums of orthogonal projections: MATH since MATH, for at most MATH projections of the form MATH or MATH, MATH or MATH can be nonzero (again we used REF ). Now for all MATH we have MATH: this is clear if MATH; if MATH, then MATH for some MATH, but then REF yields MATH . We also have MATH, thus for at least one MATH we obtain MATH that is, MATH and MATH is a covering. It remains to check that MATH is a refinement of MATH. We first show MATH: Be MATH. Then MATH, thus MATH for some MATH. If MATH, then again REF gives MATH a contradiction, so MATH and MATH. Next we show MATH for MATH: Suppose there are MATH with MATH. Recall that MATH and that MATH is an equivalence class with respect to MATH. This means there are MATH such that MATH and MATH . But MATH, thus for MATH one can choose MATH, such that MATH . Setting MATH, one obtains MATH with MATH if MATH. Set MATH then MATH if MATH, because MATH and MATH, so MATH. Also, MATH and MATH, thus MATH . But MATH, therefore MATH; altogether we obtain: MATH . We thus have MATH, furthermore MATH . But MATH, and MATH, so MATH. Now we can apply REF to obtain MATH. On the other hand, MATH, so in particular MATH. Then by REF we have MATH, a contradiction. Therefore MATH. Finally it turns out that MATH for some MATH: Take MATH, then there is MATH such that MATH. Then for any MATH we have MATH, therefore MATH, and MATH, so MATH. But this means MATH. Thus we have shown that MATH covers MATH, is of order less than or equal to MATH and refines MATH. Therefore, MATH. |
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