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math/0004003
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We have already proved most of this. The statement that it defines an equivalence of homotopy theories follows from the fact that it carries suspensions of NAME to suspensions of m-coalgebras and loop-spaces to the cobar construction.
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math/0004003
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Suppose MATH and MATH are semi-simplicial sets such that MATH become equivalent when regarded as objects of MATH. Then there exists a homotopy equivalence MATH which induces a homology equivalence MATH where MATH and MATH are the singular complexes of MATH and MATH, respectively. The conclusion follows from the fact that the inclusions MATH are equivalences in MATH - see REF. Given any morphism in MATH we know that there exists a geometric map MATH realizing it. But this implies the existence of a corresponding morphism of singular complexes and a morphism in MATH.
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math/0004003
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In fact MATH will be a strong deformation retract of MATH via a functorial retraction, and the inclusion of MATH will be a homotopy equivalence. We define MATH inductively on simplices via: CASE: If MATH is a MATH-simplex, MATH, where MATH is some point disjoint from MATH and depending on MATH. Note that MATH. CASE: If MATH is a MATH-simplex, define MATH, where MATH is some point disjoint from MATH - and depending on MATH. The following diagram illustrates this construction: where the shaded area REF was in MATH. It is easy to see that this construction is functorial with respect to inclusions of faces so it extends to a functor on the category of simplicial sets. For every simplicial set MATH there exists a canonical inclusion MATH and a set of MATH-simplices MATH indexed by the simplices of MATH. It is also trivial (see REF) to see that the inclusion REF exists and is functorial. We claim that there exists a functorial retraction MATH making MATH a strong deformation retract of MATH. Simply define MATH to be the unique piecewise-linear map fixing MATH and sending MATH to the barycenter of MATH for all MATH. The restriction of this map to MATH coincides with the well-known homeomorphism MATH. It follows that the inclusions REF both induce homology isomorphisms.
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math/0004003
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The simplicial approximation theorem says that there exist integers MATH and a simplicial map MATH such that MATH is homotopic to MATH - where we have identified MATH with MATH and MATH with MATH, respectively, via the well know homeomorphisms between them. To see that this induced map only depends on the homotopy class of MATH, apply this argument to MATH and MATH.
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math/0004003
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This follows directly from REF.
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math/0004003
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This is straightforward: the MATH maps shuffle the factors of MATH into the appropriate positions.
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math/0004003
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This follows from the fact that the term MATH is expressible in terms of the composition operations in MATH, so it pulls back to MATH.
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math/0004003
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This follows from:
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math/0004003
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The second statement is the easiest to prove - bearing in mind the remark following REF. The first statement (that MATH, with these compositions, is an operad) requires us to verify the defining identities of an operad in REF. We use the hypothesis and REF to conclude that there exists a coalgebra, MATH, over MATH with an injective structure map MATH . It is not hard to see that the induced map MATH will also be injective. Since (by REF in the conclusions) we know that the composition-operations of MATH map to those of MATH and since the latter satisfy the operad identities, it follows that the former do as well.
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math/0004003
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Note that MATH since MATH so that the factor MATH in MATH is closed under compositions.
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math/0004003
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As before, the second statement is the easy one to prove - it follows directly from REF. As in REF, the first statement that MATH is an operad follows from REF - which implies that there exists a coalgebra, MATH, over MATH with an injective structure map MATH . It is not hard to see that the induced map MATH will also be injective. This requires a quick check that we have not used any of the specific properties of m-coalgebras distinguishing them from arbitrary coalgebras over an operad (that is, the fact that they are concentrated in positive dimensions). Since (by REF in the conclusions) we know that the composition-operations of MATH map to those of MATH and since the latter satisfy the operad identities, it follows that the former do as well.
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math/0004003
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Simply set all MATH in the formula in REF.
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math/0004009
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Let MATH be a formal metric on MATH, and MATH a nontrivial harmonic MATH-form. Then MATH and MATH are also harmonic and so MATH has constant length. In particular it has no zeros. It follows that MATH and MATH span MATH for all MATH. As MATH is constant for every constant linear combination MATH of MATH and MATH, the NAME formula MATH allows us to compute: MATH . This shows that REF curvature is everywhere nonpositive. But by the NAME theorem this implies that MATH is flat.
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math/0004009
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That the length of any harmonic form is constant follows from REF . The more general statement follows by polarisation.
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math/0004009
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If MATH is harmonic, then MATH is constant by REF . Using that the length of MATH is also constant by REF , we conclude that MATH is constant. The converse is trivial.
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math/0004009
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Fix a formal Riemannian metric on MATH. It follows from the above Lemmas that the number of linearly independent harmonic MATH-forms is at most the rank of the vector bundle MATH. Similarly, when the dimension is MATH, the number of self-dual or anti-self-dual harmonic forms in the middle dimension is bounded by the rank of MATH. Suppose now that MATH are linearly independent harmonic MATH-forms. Then MATH is also a harmonic MATH-form, and is linearly independent of MATH. Thus MATH implies MATH.
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math/0004009
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Fix a formal Riemannian metric MATH on MATH. We consider the NAME or NAME map MATH given by integration of harmonic MATH-forms. As the harmonic MATH-forms have constant lengths and inner products, MATH is a submersion. It induces an isomorphism on MATH, and products of linearly independent harmonic MATH-forms are never zero, but are harmonic because the metric is formal. In the case MATH, we conclude that MATH is a covering of MATH, and is therefore a torus itself. Every formal metric on MATH must be flat because it admits an orthonormal framing by harmonic MATH-forms.
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math/0004009
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Suppose MATH fibers over MATH with fiber MATH and monodromy diffeomorphism MATH. By NAME 's Lemma, we may assume that MATH preserves a volume form MATH on MATH, so that its pullback to MATH descends to MATH as a closed form which is a volume form along the fibers. We can find a Riemannian metric on MATH for which MATH is the closed MATH-form defining the fibration over MATH, and has constant length. Then MATH and MATH generate the harmonic forms in degree MATH and MATH, and their product is harmonic.
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math/0004009
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This follows from the obstruction-theory definition of the NAME class, and the fact that every nontrivial harmonic form has no zeros because of REF.
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math/0004009
|
Suppose MATH is nontrivial and anti-self-dual. We have MATH which is harmonic if and only if the norm of MATH is constant. If it is constant, it must be a non-zero constant, and then the above equation shows that MATH is a symplectic form inducing the opposite (non-complex) orientation on MATH. In particular, MATH must have non-trivial NAME invariants, see CITE. But the KREF surface contains smoothly embedded (-REF)-spheres, which become (+REF)-spheres when the orientation is reversed, showing that all the NAME invariants vanish, see CITE.
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math/0004009
|
It is clear that every metric on every rational homology sphere is formal because there are no nontrivial harmonic forms. Conversely, assume that MATH is a manifold with a non-zero NAME number MATH, for MATH. Let MATH be a Riemannian metric which has positive curvature operator on an open set, say a ball MATH, and assume it is formal. If MATH is a nontrivial MATH-harmonic MATH-form, then MATH shows that MATH has constant length. Therefore, the NAME formula MATH for MATH-forms allows us to compute: MATH . Here the term MATH is positive on MATH, because there the curvature operator is positive. Thus MATH vanishes identically on MATH. As MATH is harmonic, the unique continuation principle implies that MATH vanishes on all of MATH, contradicting the assumption that MATH is nontrivial.
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math/0004010
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REF : notice that every bounded uniformly continuous function extends over the NAME compactification. REF : choose a bounded uniformly continuous pseudometric MATH on MATH subordinated to MATH (that is, MATH) and apply REF to the function from MATH to MATH whose components are distance functions MATH, MATH, with MATH. REF : see REF (which is, in its turn, an adaptation of an argument from REF.)
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math/0004010
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Let MATH be a finite cover of MATH, let MATH, and let MATH be arbitrary. Find an entourage of the diagonal MATH with MATH. At least one element of MATH, denote it by MATH, satisfies the property MATH . By proceeding to a subnet if necessary, we may assume without loss in generality that MATH . In view of the assumed concentration property of the measures MATH, MATH and by force of the second assumption, one has for every MATH . By REF , each of the nets of measures MATH, MATH concentrates, and consequently MATH . Since MATH, one has MATH . It is therefore possible to choose a MATH so large that each of the numbers MATH, MATH, MATH is greater than MATH. It follows that the intersection of all the translates of MATH by elements MATH, MATH is non-empty, and application of REF finishes the proof.
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math/0004010
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The first statement is self-evident. In order to establish the second claim, it is now enough to prove that for every MATH the orbit map MATH is continuous. Let MATH be any. Using the boundedness of the original action, choose a MATH such that for all MATH and MATH, MATH. The set MATH is a neighbourhood of MATH in MATH, and if MATH is arbitrary, then for every MATH apart from a set of measure MATH one has MATH, where MATH for every MATH apart from a set of measure MATH. This means that MATH, establishing the continuity of the orbit map.
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math/0004010
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Make the set-theoretic disjoint union MATH into a weighted graph, by joining a pair MATH with an edge in any of the following cases: CASE: MATH, with weight MATH; CASE: MATH, with weight MATH, CASE: for some MATH, MATH and MATH, with weight MATH. The weighted graph MATH equipped with the path metric clearly contains MATH and MATH as metric subspaces and satisfies the required property.
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math/0004010
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We will perform the proof in several simple steps. CASE: By first rescaling the metric MATH on MATH and then replacing it with MATH if necessary, we can assume without loss in generality that the values of MATH are bounded by MATH. CASE: Without loss in generality, we may also assume that MATH supports the structure of an abelian group equipped with a bi-invariant metric, and MATH's are metric-preserving group automorphisms. For instance, one can replace MATH with the free abelian group MATH on MATH, and extend the metric from MATH to a maximal invariant metric on MATH bounded by MATH (the so-called NAME metric, compare CITE); then every isometry of MATH uniquely extends to an isometric automorphism of the metric group MATH. CASE: Let MATH denote a group of isometries of MATH generated by MATH. The semidirect product group MATH is equipped with the bi-invariant metric MATH defined by MATH [The bi-invariance of MATH is established through a direct calculation using the multiplication rule in the semidirect product in question: MATH . As usual, we will identify MATH with a subgroup of the semidirect product under the mapping MATH, and similarly MATH is identified with a normal subgroup of the semidirect product under the mapping MATH. Under such conventions, the automorphism of MATH determined by each MATH is just MATH itself considered as an isometric isomorphism of MATH: MATH . In particular, for every MATH one has MATH . CASE: Let MATH denote the free group on MATH generators denoted by the symbols MATH . Denote by MATH the homomorphism sending each generator MATH to the corresponding element of MATH and each generator MATH to the corresponding element of MATH. Pull the metric MATH back from MATH to MATH by letting MATH . The pseudometric MATH is bi-invariant on MATH, though need not be a metric. By force of the remark at the end of REF , the indexed pseudometric spaces MATH and MATH are isometric (and thus are both metric spaces). CASE: By adding to MATH an arbitrary bi-invariant metric on MATH normalized so as to only slightly change the values of distances between pairs of elements MATH (for instance, let us agree on the discrete metric taking its values in MATH), we can assume without loss in generality that MATH is a bi-invariant metric on MATH, while the indexed metric spaces MATH and MATH are MATH-isometric. CASE: Now replace the metric MATH with the maximal among all bi-invariant metrics on MATH that coincide with MATH on the set MATH of all words of reduced length MATH. To prove the existence of such a metric, say MATH, denote by MATH the family of all bi-invariant metrics on MATH whose restriction to MATH coincides with MATH. Since MATH, the family MATH is non-empty. For any two elements MATH and an arbitrary MATH, the value MATH is bounded from above uniformly in MATH by any sum of the form MATH, where MATH and MATH are two representations having the same length and such that MATH. Now set MATH . The supremum on the right-hand side is finite, and has all the required properties. CASE: Notice that for MATH where the infimum is taken over all possible representations of the above sort MATH, MATH, having the same length and such that MATH. [Proof: the infimum on the right-hand side is a bi-invariant pseudometric, which is greater than or equal to MATH, and whose restriction to MATH coincides with the restriction of MATH. We conclude: this infimum is in fact a metric, and it must coincide with MATH.] CASE: Denote by MATH the smallest positive value of the distance MATH (or, equivalently, MATH) between any two elements of the finite set MATH. It follows from REF that for every word MATH, the value of MATH is at least MATH, where MATH denotes the reduced length of MATH. CASE: Being free, the group MATH is residually finite, that is, admits a separating family of homomorphisms into finite groups. (Compare for example, REF .) Therefore, for every natural MATH there exists a normal subgroup MATH such that the factor-group MATH is finite and the only word of reduced length MATH contained in MATH is identity. Denote by MATH the factor-homomorphism and equip MATH with the factor-distance of MATH by letting MATH . CASE: It is immediate that MATH is a bi-invariant pseudometric. Moreover, for MATH it is a metric: MATH whenever MATH, compare REF . Now assume that MATH. Let MATH and MATH be arbitrary, MATH. Then MATH, because MATH and therefore MATH and REF applies. Therefore, MATH and the restriction of MATH to MATH is an isometry. CASE: Now set MATH and MATH. For every MATH, the inner automorphism of the finite metric group MATH determined by MATH is an isometry, because the metric is bi-invariant. Denote this isometry by MATH. The indexed metric spaces MATH and MATH, MATH are isometric by force of the concluding remark in REF . Taking into account REF , we finally conclude that the indexed metric spaces MATH and MATH are MATH-isometric, as required.
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math/0004010
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One can assume that MATH, where MATH and MATH. Using REF , choose a finite metric space MATH, elements MATH of MATH, and isometries MATH of MATH such that the naturally indexed finite metric spaces MATH and MATH are MATH-isometric. Using REF , isometrically embed MATH and MATH into a finite metric space MATH in such a way that MATH for all MATH. Now extend the embedding MATH to an isometric embedding MATH. According to REF , the (finite) group MATH simultaneously extends to a group of isometries of MATH. Denote the extension of the isometry MATH by MATH. One has for all MATH: MATH and the proof is finished.
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math/0004010
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Let the group MATH act continuously on a compact space MATH. We will show that every finite collection of elements of MATH has a common fixed point in MATH, from which the result follows by an obvious compactness argument. Fix an arbitrary such collection, MATH. Let MATH be an arbitrary element of the unique compatible uniform structure on MATH. Without loss in generality, assume that MATH is closed as a subset of MATH (and consequently compact). Using the boundedness of the action of MATH on MATH, choose a finite MATH and a MATH such that, whenever MATH, one has MATH for all MATH. By REF , there are isometries MATH generating a finite subgroup MATH and such that MATH, MATH. Let MATH be a finite MATH-invariant subset of MATH containing MATH. The iterated NAME extension MATH contains MATH as a subspace made up of all constant functions and is isometric to MATH, and since MATH is finite, an isometry between the two spaces can be chosen so as to extend the canonical embedding of MATH into MATH. Thus we obtain a chain of MATH-embeddings MATH . The group MATH acts on MATH continuously and isometrically REF , and this action canonically extends to a continuous isometric action of the same group on the space MATH. Thus we obtain a continuous group monomorphism MATH with the property that for every MATH one has MATH. Composing MATH with the action MATH, we obtain a continuous action of MATH on MATH. By force of REF , MATH has a common fixed point in MATH, say MATH. In particular, MATH is fixed under the elements MATH, where we identify elements of MATH with constants in MATH. For all MATH and MATH, one has MATH for all MATH and MATH, implying that MATH for MATH. Consequently and by the choice of MATH, MATH for all MATH. Denote by MATH the (non-empty) set of all points MATH with the property MATH for all MATH. Since MATH is closed, so is MATH. If MATH, then MATH. It means that MATH is a centred system of closed subsets of the compact space MATH and therefore has a common point, which is clearly fixed under MATH, as required.
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math/0004010
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Since for every MATH the value of the inner product is uniquely determined by the Euclidean distance between the elements, MATH there is only one way to turn the linear span MATH into a pre-Hilbert space so as to induce the given metric on MATH. The corresponding completion MATH is isometrically isomorphic to the closed linear span of MATH in MATH, that is, MATH. As another consequence of the same observation, every isometry of MATH lifts to a unique orthogonal transformation of MATH. The resulting homomorphism MATH is continuous if the latter group is equipped with the topology of simple convergence on MATH or, which is the same, on MATH. On the groups of isometries of metric spaces the topology of simple convergence on an everywhere dense subset coincides with the topology of simple convergence on the entire space. Consequently, the extended orthogonal representation MATH of MATH in MATH is strongly continuous. It remains to extend MATH to a representation MATH of MATH in MATH. The uniqueness statement is obvious.
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math/0004010
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The claim consists of two parts: first, that all sets of the form MATH are elements of MATH, and second, that each enourage from MATH contains a set of the above type. CASE: Given MATH, MATH, and MATH as above, choose a bounded uniformly continuous pseudometric MATH on MATH such that MATH, and introduce a bounded uniformly continuous function MATH from MATH to the Euclidean space MATH with each component MATH, MATH, defined by MATH . The set MATH is an element of MATH and a subset of MATH. CASE: Let MATH be arbitrary. Choose a bounded uniformly continuous function MATH and a MATH such that MATH. Partition the image MATH into finitely many pieces of diameter MATH each and let MATH be the family of preimages of those pieces under MATH. Define MATH. Clearly, MATH.
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math/0004010
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REF : according to REF , the fixed point on compacta property of a topological group MATH is equivalent to the following: for every bounded right uniformly continuous function MATH on MATH taking values in a finite-dimensional Euclidean space, every finite collection of elements MATH, and every MATH, there is a MATH such that MATH . The mirror image of the above statement applies to left uniformly continuous functions and calls for the existence of a MATH with the property MATH for all MATH. This amounts to the NAME - NAME - NAME property for the pair MATH relative to the left action (with MATH). REF : as the canonical map MATH is uniformly continuous and MATH-equivariant, REF applies. CASE: Let MATH denote the invariant metric on MATH. Fix an arbitrary point MATH. The formula MATH defines a left invariant continuous pseudometric on MATH, and the map MATH factors through to a MATH-equivariant isometric isomorphism between MATH and MATH. CASE: Trivial, as MATH acts on each space MATH continuously and transitively by isometries. CASE: Suppose we are given a finite subset MATH, a finite cover MATH of MATH, and a basic element MATH of the left uniformity MATH, where MATH is a neighbourhood of identity in MATH. Choose a bounded left invariant continuous pseudometric MATH with the property MATH. The sets MATH, MATH, where MATH is the factor-map, form a finite cover of MATH, and by assumption there is a MATH such that MATH is entirely contained in the MATH-neighbourhood of some MATH, MATH. Denote, as before, MATH. The value of MATH is independent on the choice of representatives in left cosets: MATH for all MATH, MATH. Let MATH be any. Since MATH for some MATH, one has MATH, that is, MATH is contained in MATH, and the NAME - NAME - NAME property of MATH is thus verified.
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math/0004010
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REF : assume REF , that is, no finite subspace MATH of MATH containing a copy of MATH is in MATH. Denote by MATH the collection of all finite metric subspaces MATH with MATH. By assumption, MATH. Then for every MATH the set MATH admits a colouring with MATH colours, which we will view as a function MATH, in such a way that the following holds: REF for every isometric embedding MATH and every colour MATH there is an isometric embedding MATH such that the MATH-neighbourhood of MATH in MATH contains no elements of colour MATH. The system MATH is directed by inclusion, and the collection of intervals MATH, where MATH is finite, is a filter on MATH, which we will denote by MATH. Since MATH can be assumed infinite (otherwise there is nothing to prove), MATH extends to a free ultrafilter MATH on MATH. For every MATH, one has MATH, and therefore exactly one of the sets MATH, MATH is in MATH. Consequently, the function MATH determines a colouring of MATH with MATH colours. Now let MATH be an arbitrary isometric embedding, and let MATH be a colour. For every MATH choose, using (MATH), an isometric embedding MATH with no element in the MATH-neighbourhood of MATH, formed in MATH, being of MATH-colour MATH. For every MATH define MATH. (The metric space MATH is finite.) This MATH is an isometric embedding of MATH into MATH with the property that the MATH-neighbourhood of MATH formed in all of MATH contains no elements of colour MATH. Thus, REF is established. REF : evident.
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math/0004010
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Combine REF .
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math/0004010
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Denote by MATH the two-element metric space with MATH. Partition the set MATH of all isometric embeddings of MATH into MATH into two disjoint subsets MATH in such a way that whenever an injection MATH is in MATH, the `flip' injection MATH is in MATH, and vice versa. Since the space MATH is MATH-discrete, the MATH-neighbourhood of a subset MATH is MATH itself, and `monochromatic up to MATH' means in this context simply `monochromatic.' One concludes that, with respect to the colouring MATH, no pair of injections of the form MATH is monochromatic up to MATH, and thus the metric space MATH, upon which the group MATH acts transitively and continuously by isometries, fails the NAME - NAME - NAME property.
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math/0004010
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The topological group MATH is isomorphic to the semidirect product MATH of the full orthogonal group MATH equipped with the strong operator topology and the additive group of the NAME space MATH with the usual norm topology, formed with respect to the natural action of MATH on MATH by rotations. (Compare CITE.) Suppose MATH acts continuously on a compact space MATH. Since the group MATH (identified with a subgroup of MATH) is extremely amenable (CITE; compare also Subsection REF), it has a fixed point MATH. The mapping MATH, where MATH is viewed as a closed normal subgroup of MATH, is MATH-equivariant, continuous, and has everywhere dense image in MATH, and thus MATH is an equivariant MATH-compactification of the homogeneous factor-space MATH. Let MATH be an arbitrary continuous function, MATH. Its pull-back, MATH, to MATH is right uniformly continuous. (A standard result in abstract topological dynamics.) If MATH is arbitrary, then for some neighbourhood MATH of identity in MATH one has MATH whenever MATH. Without loss in generality and slightly perturbing the points of MATH if necessary, one can assume that elements of MATH are affinely independent. Let MATH be two arbitrary elements with the property MATH for each MATH. Find an isometric copy of MATH, say MATH, such that MATH is isometric to MATH (or, equivalently, to MATH). There is an isometry MATH of MATH taking MATH to MATH, and an isometry MATH taking MATH to MATH. In particular, MATH, and consequently MATH. Thus, the function MATH is MATH-constant on every affine sphere of codimension MATH having the form MATH. Another way to say it is that, up to MATH, the function MATH only depends on the collection of distances MATH. Now let MATH be an arbitrary collection of isometries. By slightly perturbing them if necessary, one can assume without loss in generality that all the vectors MATH and MATH, MATH, MATH, are affinely independent. Because of infinite-dimensionality of MATH, every element MATH of some affine subspace of MATH of finite codimension has the property that for every MATH and each MATH, one has MATH. Fix any such MATH. Then the values of MATH at the points MATH, MATH differ by less than MATH. Now we can apply REF to conclude that MATH has a fixed point for MATH.
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math/0004010
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Without loss in generality, we can assume that MATH is separable (in fact, even countable). Such a MATH can be MATH-embedded into the NAME space MATH (see CITE, or else REF), and therefore we can further assume that MATH. Choose any element MATH and isometries MATH of MATH with the property MATH, MATH. Denote by MATH the free non-abelian group on generators MATH, MATH. The group MATH acts on MATH by isometries. The formula MATH where MATH denotes the metric on the NAME space, defines a left-invariant pseudometric MATH on the group MATH: MATH . The indexed metric subspace MATH of MATH is isometric to the metric subspace MATH of MATH. Indeed, MATH . Notice also that the latter subspace is contained in the set MATH of all words having reduced length MATH. (The reduced length will always mean that with regard to the generators MATH, MATH.) By adding to MATH a left-invariant metric on MATH taking sufficiently small values on pairs of elements of MATH, we can assume without loss in generality that MATH is a left-invariant metric on MATH. (For instance, add a metric whose only non-zero value is MATH.) Form the NAME graph MATH of MATH with regard to the set of generators MATH. Vertices of MATH are elements of MATH, and MATH are adjacent if and only if MATH. This graph is connected. Now make MATH into a weighted graph by assigning to an edge MATH, MATH the value MATH. Denote by MATH the path metric of the weighted graph MATH. Its value for MATH is given by MATH where the infimum is taken over all natural MATH and all finite sequences MATH, with the property MATH for all MATH. It is easily seen that MATH is a left-invariant metric on the group MATH. Generally, MATH, but restrictions of MATH and MATH to MATH coincide. If one denotes by MATH the minimal value of MATH as MATH and MATH, then MATH, where MATH denotes the word metric with respect to the set of generators MATH. In particular, if a MATH has reduced length MATH, then MATH and accordingly MATH. (As a consequence, the infimum in REF is always achieved.) Let MATH denote the maximal value of the metric MATH between pairs of elements of MATH. Choose a natural number MATH so large that MATH, for instance, set MATH. Every free group is residually finite, that is, admits a separating family of homomorphisms into finite groups. (Compare for example, REF .) Using this fact, choose a normal subgroup MATH of finite index so that MATH. The formula MATH defines a left-invariant pseudometric on the finite factor-group MATH. The triangle inequality follows from the fact that, for all MATH, MATH and the infimum of the right-hand side taken over all MATH equals MATH. NAME of MATH is obvious. Let MATH. Closely approximate the infimum in REF by some value MATH with MATH, then MATH where MATH. The value MATH, MATH, cannot get smaller than MATH. Indeed, unless MATH (in which case MATH), one has MATH and so the word distance from MATH to MATH is at least MATH, and MATH. We conclude: the restriction of the factor-homomorphism MATH to MATH is an isometry. One can now perturb the pseudometric on MATH by adding to it a left-invariant metric taking very small values (for example, taking the only non-zero value MATH) so as to replace MATH with a left-invariant metric, MATH. Take now MATH, MATH, MATH, and let MATH be left translates made by the elements MATH, MATH, in the finite group MATH. The indexed metric space MATH of MATH is MATH-isometric to the metric subspace MATH of MATH. Consequently, the conclusion of the Theorem is verified.
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math/0004013
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: Let MATH be a simply connected REF-manifold with integral cohomology as in REF and linking form equivalent to a standard form. Pick an orientation on MATH and an isomorphism, MATH so that MATH. Let MATH; then MATH. For any manifold with the above cohomology, MATH, the fourth NAME - NAME class of the tangent bundle. An easy calculation using the NAME classes shows that for such manifolds, the total NAME - NAME class is trivial and in particular MATH. Hence, MATH, an even class. So the NAME data for MATH is equivalent to the NAME data for the manifold MATH. From the previous discussion, it follows that when MATH is odd or when MATH is twice an odd number, MATH is MATH-homeomorphic to MATH.
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math/0004015
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By a direct application of REF we have MATH where MATH, MATH. From the assumptions on MATH we see that MATH and the claim follows by some algebra (with MATH).
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math/0004015
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Fix MATH, MATH, MATH, and write MATH for short. We also introduce a dummy tube, setting MATH, MATH. We choose vectors MATH which have the values MATH for MATH, so that MATH and each MATH is a distance MATH from the hyperplane spanned by the remaining MATH vectors MATH. The key observation is that for every MATH, the set MATH is essentially constant in the direction MATH. More precisely, we have the elementary pointwise estimate MATH where MATH is the averaging operator MATH . Summing this in MATH we obtain MATH where MATH . We now claim that MATH. Indeed, the lower bound is trivial, while the upper bound comes from covering MATH by finitely overlapping and essentially parallel MATH tubes. From REF we thus have MATH . To utilize REF we invoke Let MATH, be any MATH linearly independent vectors in MATH, and let MATH be a subset of MATH. Then for any functions MATH on MATH, we have MATH where MATH is an absolute constant, and MATH ranges over all parallelepipeds with edge vectors MATH. We remark that the MATH version of this lemma was proven in CITE. We begin with some reductions. The statement of the lemma is invariant under affine transformations, so we may rescale MATH, where MATH are the standard basis of MATH. It suffices to show that MATH for all unit cubes MATH, since the claim follows by summing over a partition of MATH and using NAME 's inequality. We may assume that MATH is centered at the origin, that MATH, and that MATH are supported on MATH. By another application of NAME 's inequality, it thus suffices to show that MATH for all functions MATH on MATH. In fact we shall show the more general statement MATH for all MATH. We prove REF by induction on MATH. When MATH the claim is clear. Now suppose MATH, and that REF has already been proven for dimension MATH. For all MATH, write MATH, where MATH and MATH is the MATH co-ordinate of MATH. We have the pointwise estimate MATH where MATH . We can then estimate the left-hand side of REF by MATH where MATH is the function on MATH defined by MATH . By NAME in MATH, we can estimate the previous by MATH . By the induction hypothesis, we can estimate this by MATH . By another NAME, we may estimate this by MATH . From NAME 's inequality we have MATH, and the claim follows. Combining this estimate with REF we obtain MATH where MATH ranges over all parallelepipeds with edge vectors MATH. To complete the proof of REF it thus suffices to show that MATH where MATH ranges over all squares parallel to MATH. Let MATH be the hyperplane generated by MATH, and let MATH be a square centered at the origin whose long sides lie on MATH. We can tile MATH by translates of MATH, and estimate MATH by the union of all the translates of MATH which intersect MATH. This will prove REF provided that MATH for some MATH, where MATH is the dilate of MATH by MATH around the center of MATH. To show this, suppose for contradiction that REF failed. By translation we may assume that MATH is centered at the origin. The failure of REF then implies that MATH is not completely contained inside MATH. Since MATH is convex and symmetric around the origin, this implies by duality that MATH is contained in some slab MATH for some MATH outside of MATH, the dual box of MATH. The dual box MATH has dimensions MATH, is centered at the origin, and has its short sides on MATH. Split MATH, where MATH are the orthogonal projections onto MATH and the orthogonal complement of MATH respectively. Since MATH, we either have MATH, or MATH and MATH. In the former case MATH lie within a MATH-neighbourhood of the hyperplane orthogonal to MATH, contradicting the choice of the MATH. In the latter case MATH lie in the MATH-neighbourhood of a MATH-dimensional subspace of MATH, contradicting the non-degeneracy assumption. This completes the proof of REF, and the lemma follows.
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math/0004015
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Suppose for contradiction that REF failed. Then we have MATH for some MATH. From REF we thus have MATH . From REF the right-hand side is MATH. From REF we thus have MATH . For every degenerate MATH, let MATH be an affine subspace satisfying REF; one can easily ensure that MATH is a measurable function. Let MATH denote the set MATH . From REF we see that MATH is very large, in fact MATH . On the other hand, we observe that the MATH-projection of MATH does not concentrate in a thin slab. (This kind of observation also appears in CITE, and implicitly in CITE). If MATH, and MATH is a hyperplane in MATH, then MATH . Note that the MATH factor on the right-hand side of REF gives an improvement over the trivial estimate coming from REF. Let MATH denote the set on the left-hand side of REF. From REF we see that MATH for all MATH. Integrating this on MATH, we obtain MATH . From elementary geometry we have MATH . Summing this in MATH, using the MATH-separated nature of the directions MATH, one obtains MATH . The claim follows by combining the above estimates. The idea is to derive a contradiction by interacting REF with REF. Let MATH denote those tubes MATH such that MATH where MATH is the set MATH . If the constant MATH in REF is chosen sufficiently large, then we see from REF that MATH and so by REF again we have MATH . In particular, from REF we have MATH . We now use REF and the ``hairbrush" argument of CITE to show that the tubes in a hairbrush in MATH cannot concentrate in a thin slab. If MATH, MATH is a hyperplane in MATH, and MATH, then MATH . Here MATH is an absolute constant depending only on MATH. As with REF, the key point of REF is that it contains the decay MATH. We first dispose of the portion where MATH, where MATH is some small constant. In this case we note that every MATH which contributes to REF must satisfy MATH and hence REF. In particular, each MATH can contribute at most MATH tubes MATH to REF. Since MATH, the claim then follows from REF and NAME 's theorem. Now consider the contribution when MATH . Let MATH denote all the tubes in MATH which contribute to this portion of REF. By elementary geometry, each MATH contributes a set of measure MATH to REF. Thus it suffices to show that MATH . For each MATH, let MATH denote the set MATH . From REF, and elementary geometry we see that MATH if the constants are chosen appropriately. Thus we have MATH . On the other hand, the function MATH is supported on the set in REF. From NAME we thus have MATH . We now use a NAME argument. We can expand the left-hand side as MATH . We split this sum dyadically based on the angle between MATH and MATH: MATH . Fix MATH. From elementary geometry, a tube MATH can only contribute to the sum if it lies within MATH of REF-dimensional plane generated by MATH and MATH, and even then the contribution is MATH. From the MATH-separated nature of the tubes MATH we thus see that there are only MATH tubes MATH which contribute to the inner sum. Combining these observations we thus have MATH . Inserting this into REF and doing some algebra we obtain REF as desired, if MATH is chosen sufficiently small. We now use REF and the low dimension of the MATH to contradict REF. For each MATH, we have MATH by REF. We may clearly improve this to MATH for appropriate choice of constants. Summing this over all MATH and using REF we obtain MATH . We rewrite this as MATH . From REF and NAME we thus have MATH . We write this out as MATH where MATH . We now claim that For any MATH, we have MATH . From REF it suffices to show that MATH for every MATH, MATH. For fixed MATH, MATH, the set of MATH which can contribute is MATH by REF. So it suffices to show that MATH for all non-degenerate MATH and MATH such that MATH. Fix MATH, MATH. From our assumptions on MATH we see that MATH. The claim then follows from REF (since MATH can of course be embedded in a hyperplane). Combining REF we see that MATH for appropriate choices of constants. We rewrite this as MATH . Using REF and NAME as before we thus have MATH which we write out as MATH . We now find an upper bound for the left-hand side of REF which will achieve the desired contradiction. The key lemma is For each MATH, we have MATH for some absolute constant MATH. Fix MATH, and let MATH be the hyperplane containing MATH and parallel to MATH. (Note that this hyperplane is well defined thanks to the condition MATH). In order for MATH to contribute to REF, we must have MATH . In particular, we have MATH . Since MATH and MATH is parallel to MATH, we thus have MATH . Since MATH, we conclude that MATH . Since MATH and MATH, we thus conclude that MATH . Also, we have MATH . Using that MATH, we see from elementary geometry that for fixed MATH, MATH the set of all possible MATH which contribute is contained in a set of measure MATH. Also, for fixed MATH, MATH, MATH there is at most MATH possible tubes MATH which contribute, thanks to the separation condition MATH. Combining all these observations we can thus estimate the left-hand side of REF by MATH . The claim then follows from REF. In light of REF we may estimate the left-hand side of REF by MATH . In order for MATH to contribute to the above, MATH and thus REF must hold. In particular, there are at most MATH tubes MATH which can contribute for each MATH. We thus have MATH . From REF we have MATH . Combining these two estimates together we obtain a contradiction to REF, if MATH and then MATH is chosen sufficiently small, and the constant MATH used to define MATH was chosen sufficiently large so that MATH.
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math/0004015
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When MATH the claim is trivial with MATH empty if MATH is sufficiently small, since MATH while MATH . Now suppose that MATH. We say that two squares MATH are separated if MATH and MATH. We define MATH to be a maximal pairwise-separated set of squares MATH which satisfy MATH . It is easy to see that REF holds. To show REF, we take advantage of the known bounds for the NAME transform MATH which takes functions on MATH to functions on MATH. (It is also possible to obtain REF by more elementary means). From the construction of MATH we see that MATH . On the other hand, one has the restricted weak-type estimate MATH for all sets MATH (see CITE). In particular we have MATH . On the other hand, since MATH is the union of MATH-balls we have MATH by REF. Combining all these estimates we obtain the result.
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math/0004015
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From REF we have MATH . In particular, we have MATH . Using this, REF, and REF, we find that REF will follow if we can show MATH . By REF , this is equivalent to MATH . On the other hand, from REF we have MATH . Also, from REF, and REF we have MATH . It thus suffices to show that MATH . Consider the right-hand side of REF. For each MATH, the set of MATH which contribute has volume MATH. For each MATH, the set of MATH which contribute has volume MATH by REF. Finally, the total number of pairs MATH which contribute is MATH by REF. So the right-hand side is MATH. Now consider the left-hand side of REF. Using the sets MATH defined in REF, we can write this as MATH . By REF, we can estimate this by MATH . We rewrite this as MATH . The expression inside the norm is supported inside MATH, which has measure MATH by REF. By NAME 's inequality, we may therefore estimate the above as MATH . We now estimate this norm as If MATH holds and MATH is non-degenerate, then we have MATH for some absolute constant MATH. Fix MATH. Raising both sides of REF to the MATH-th power and expanding, it suffices to show that MATH . We first deal with the contribution when the tubes MATH are not coplanar. In this case we use REF to estimate the above by MATH where the supremum is over all squares MATH parallel to MATH. By REF and the finite overlap of the balls MATH we have MATH . The claim then follows from REF. It remains to control the contribution when the tubes MATH are coplanar. In this case we use REF to make the crude estimate MATH . By REF, it thus suffices to show that MATH . For any MATH and any tubes MATH we choose a MATH dimensional space MATH through MATH which is parallel to MATH. This choice of space may not always be unique, but we select it in such a way that MATH is measurable. From elementary geometry we see that if MATH are coplanar, then we must have MATH for some MATH. REF then follows from the non-degeneracy of MATH. By this lemma, we can estimate REF by MATH . Since the integral is clearly bounded by MATH, we can estimate this by MATH as desired by REF, if MATH is sufficiently small.
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math/0004015
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Let MATH be a maximal MATH-separated set of directions, and for each MATH let MATH be a finitely overlapping cover of MATH by MATH-tubes with direction MATH. We can arrange matters so that every MATH obeys MATH for some MATH and MATH. Call a direction MATH sticky if MATH and define MATH . Clearly REF holds. To prove REF it suffices to show that MATH . Since MATH is direction-separated, each non-sticky direction MATH can contribute at most MATH elements to the above set. Hence it suffices to show that MATH where MATH is the set of non-sticky directions. By construction, for each MATH we can find a subset MATH of cardinality MATH such that each MATH contains at least one tube MATH. Let MATH be the union of all these MATH as MATH ranges over MATH. By construction, the MATH have directional multiplicity MATH, and we have MATH and MATH . In particular, from REF we have MATH . On the other hand, from REF we have MATH . From NAME 's inequality we have MATH . However, by REF we have MATH . Combining these two inequalities we obtain REF as desired.
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math/0004015
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We can find unit directions MATH and numbers MATH for all MATH such that MATH and MATH where the dot product is taken in MATH. Fix MATH and MATH. Let MATH be a tube in MATH. We can find MATH with MATH such that MATH so in particular we have MATH for MATH. Since MATH is direction-separated, it thus suffices to show that the set of all possible velocities MATH which obey REF for some MATH can only support MATH-separated values at best. By linearity, we may assume that MATH. We define the rank MATH to be the least integer MATH such that there exist distinct MATH in MATH and co-efficients MATH such that MATH and MATH where MATH is an absolute constant to be chosen later. Since the MATH live in MATH and have magnitude REF, we see that the rank is well-defined and is either REF, or REF. Fix MATH to be the rank, and let MATH be as above. Clearly we may normalize so that MATH . If we multiply REF for MATH by MATH for MATH and add, we obtain MATH . From REF we have MATH whereas by the definition of rank and the fact that MATH we have MATH . Since MATH, we thus see that MATH is constrained to lie in the MATH-neighbourhood of a hyperplane. This means that any MATH-separated set of such MATH can have cardinality at most MATH, as desired.
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math/0004018
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First, we choose coordinate systems on each space. Let MATH and MATH, so that the discrete quotient map REF is given by MATH, and the continuous quotient map by MATH. Recall that the fiber derivative MATH in these coordinates has the following form (see, for example, CITE) MATH . Then the above diagram is given by: MATH where MATH stands for MATH. To close this diagram and to verify the arrow determined by MATH compute the derivative of MATH using the chain rule MATH where we have used that, according to the definition of MATH, the partial derivative MATH is given by the linear operator MATH. REF is precisely the NAME transformation MATH for a right invariant system (see the previous subsection).
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math/0004018
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The proof follows from the results of the previous subsection, in particular, REF relates the DLP dynamics on MATH with the DEP dynamics on MATH which, in turn, is related to the DEL dynamics on MATH via the reconstruction REF .
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math/0004018
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The proof is based on the commutativity of REF and the MATH invariance of the unreduced symplectic forms. Notice that MATH and MATH in REF are NAME manifolds, each being foliated by symplectic leaves, which we denote MATH and MATH for MATH and MATH, respectively. Denote by MATH and MATH the corresponding symplectic forms on these leaves. Below we shall prove the compatibility of these structures under REF . Repeating this proof leaf by leaf establishes then the equivalence of the NAME structures and proves the theorem. Recall that the NAME MATH-form MATH on MATH derived from the variational principle coincides with the pull-back of the canonical MATH-form MATH on MATH (see, for example, CITE). Recall also that for a right-invariant system, reduction of MATH to MATH is given by right translation to the identity MATH, that is, any MATH is mapped to MATH. Thus, for any MATH, where MATH, MATH so that MATH pulls back MATH to MATH. Henceforth, MATH shall denote the inverse map of MATH restricted to MATH. Let us write down using the above notations how the symplectic forms are being mapped under the transformations in REF ; we see that MATH . Then, using the coordinate notations of REF , for any MATH and MATH, MATH where MATH. Continuing this equation using REF , we have that MATH where MATH and MATH denotes MATH. Using REF , it follows that MATH and, hence, for the tangent maps, we have that MATH . So, if MATH in REF are images of some MATH invariant vector fields MATH on MATH, that is, MATH, then from REF it follows that MATH where MATH and MATH. The last equation precisely means that MATH is the discretely reduced symplectic form, that is, the image of MATH under the quotient map MATH.
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math/0004021
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We will use an induction on the number MATH of normal generators. If MATH the relations imply that all MATH commute in MATH. Since by assumption these elements generate MATH it follows that all commutators vanish in MATH. Now assume the statement holds for groups with MATH normal generators and let MATH be normally generated by MATH. Define MATH to be the normal closure of the element MATH. Since all the conjugates of this element commute, MATH is abelian. Moreover, the intersection MATH of all MATH lies obviously in the center of MATH. Now consider a commutator MATH with MATH. Since all quotients MATH are NAME groups with MATH generators it follows by induction that MATH and thus MATH is central that is, MATH in MATH. This shows that MATH.
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math/0004021
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The statement for the generators follows from the standard rewriting process in nilpotent groups: If a nilpotent group MATH is normally generated by MATH then it is also generated by these elements. One uses an induction on the nilpotency class of MATH based on the fact that MATH mod MATH implies MATH mod MATH for all MATH. Moreover, the fact that MATH is generated by MATH - fold commutators MATH if the MATH generate MATH shows that MATH is finitely generated if MATH is. An induction on the nilpotency class together with the fact that a (central) extension of finitely presented groups is finitely presented implies that a finitely generated nilpotent group is also finitely presented.
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math/0004021
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For MATH ordinary homology, the theorem comes from CITE. To get the general statement, one uses an eventual NAME Theorem as in CITE. It is therefore necessary to reduce to the case of simply-connected spaces. This can be easily done in our context by picking maps MATH, which represent the generators MATH of MATH, and attach REF - cells to get simply-connected complexes MATH. Then MATH for all MATH, so the again by CITE the maps MATH are eventually zero. By the eventual NAME Theorem for simply-connected spaces it follows that the maps MATH are eventually null homotopic and thus the induced maps MATH are eventually zero. Moreover, there is an exact sequence MATH . By assumption and excision we have MATH which proves that MATH is a monomorphism. The same exact argument works for MATH replaced by MATH which implies our claim.
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math/0004021
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The NAME - NAME sequence MATH implies the first statement. The NAME tori represent trivial elements in MATH, since they bound solid tori there.
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math/0004021
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To simplify notation we will assume MATH. The general case can be worked out analogously. Let MATH be a parameterization of a neighborhood of MATH in MATH such that MATH and MATH. We may assume that MATH where MATH . This can in fact be taken as a definition of a finger move. Let MATH be the compactification of MATH. We have MATH . Note that MATH, where MATH is a meridian to MATH and MATH is an element corresponding to the handle. MATH is the complement of two intersecting slice disks (MATH and MATH) in the MATH - ball and so by NAME duality MATH. (Actually MATH since both slice disks are unknotted.) The corresponding map MATH sends meridians MATH and MATH of MATH and MATH to the generators. MATH is a trivial two component link and thus MATH. We may assume that MATH sends the meridians to MATH and MATH to MATH and MATH. NAME 's Theorem gives now a map MATH . The inclusion MATH sends MATH to MATH and MATH to MATH (up to some power of MATH) and so the inclusions MATH induce the desired map.
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math/0004021
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Consider MATH. Here MATH is the spin structure on MATH constructed above and MATH is the map that gives MATH on the fundamental group. Since MATH is a spin boundary we have MATH. The factorization of MATH implies that MATH is in the image of the composition MATH for every MATH. Using MATH, REF implies now MATH and hence MATH bounds a spin manifold MATH such that MATH commutes. It is now a standard procedure to do spin structure preserving surgeries on circles in the interior of MATH to obtain MATH.
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math/0004021
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Recall that MATH is bounded by MATH by REF . Let MATH be MATH copies of MATH, the manifold from REF . We may write MATH. Now we glue MATH to MATH along parts of their boundary to obtain a manifold MATH with boundary MATH. MATH . Using REF and MATH we can find a map MATH making the triangle MATH commute. By the NAME - NAME sequence MATH . All manifolds on the right hand side are spin and have therefore vanishing second NAME - NAME class. Hence MATH is also spin. Again we can do spin structure preserving surgeries in the interior of MATH to obtain MATH.
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math/0004021
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Let MATH be the spin manifold obtained in REF . Recall that MATH . Now let MATH be the disjoint union of MATH copies of MATH. Then MATH and MATH both contain the disjoint union of MATH copies of MATH which we will denote by MATH. Now glue MATH to MATH along MATH to obtain MATH . Then MATH and MATH bounds disjointly immersed slice disks in MATH. We will modify MATH to the MATH - ball without destroying the immersed slice disks. In order to do so it is enough to make MATH REF - connected (using NAME duality and the h - cobordism Theorem). Again by NAME 's Theorem MATH and this group is normally generated by the meridians MATH (the MATH are identified with products of meridians). But the meridians bound MATH - disks in MATH and so MATH. To kill MATH we will do surgery on MATH - spheres in MATH (and hence away from the immersed slice disks). The horizontal line of the diagram below is part of a NAME - NAME sequence. By REF MATH, and this explains that MATH below is trivial. MATH . Now according to REF MATH maps onto MATH and the NAME tori represent elements in the kernel of this map. But the NAME tori also generate MATH CITE, and so MATH maps also onto MATH. The exact sequence MATH proves now that MATH maps onto MATH. It is a classical result of NAME and NAME in CITE that MATH can be changed to the MATH - ball by a sequence of surgeries on classes in MATH. (Their original formulation considers framed manifolds, but here a spin structure is enough to guarantee trivial normal bundle on MATH - spheres.) We just saw that MATH maps onto MATH, so we can represent these classes by MATH - spheres in MATH (and by general position embedding these spheres comes for free). So we can do surgery to MATH and change MATH to the MATH - ball.
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math/0004021
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Let MATH (respectively, MATH) be the attaching maps into MATH that guide the construction of MATH (respectively, MATH) upwards (respectively,downwards) from MATH. Now index MATH index MATH implies MATH . By general position we may assume that MATH and MATH are disjoint in the level MATH. But this means that MATH and MATH are independent in the following sense: MATH can be pushed down to an embedding MATH into MATH that does not intersect the attaching map MATH for MATH into MATH. Now first MATH is constructed from MATH and then MATH from MATH.
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math/0004021
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Using NAME 's jet transversality theorem CITE we can assume that MATH is a generic immersion except for a finite number of cross caps CITE. As in NAME 's original immersion argument, these cross caps can be pushed off the, say, lower boundary MATH. This changes the lower boundary MATH by cusp homotopies which we may assume are small enough such that they don't change the link homotopy class of MATH. Let MATH be a NAME immersion satisfying the conclusion from REF . If all the MATH are of types MATH and MATH then MATH is in fact a link homotopy and we are done. Let MATH be the components of MATH. We know that MATH. Consequently, all critical points of index MATH of MATH on MATH can be canceled (by NAME cancellation) by critical points of index MATH. By applying REF several times to move other critical points up respectively down, we may assume that the corresponding elementary NAME immersions for the canceling REF - and REF - handles of MATH are consecutive. Let's say they are MATH and let MATH. Then MATH and MATH is a product on MATH. Hence the observation from above applies and we can replace MATH by elementary NAME immersions of type MATH and MATH. More precisely, we replace MATH by MATH which doesn't change anything away from MATH and removes the critical points on MATH. Finally, we can make this map generic, producing a NAME immersion with singularities of types MATH and MATH only. We can apply the same procedure to all components and all MATH of type MATH and (by symmetry) MATH. Hence we may now assume that all the MATH are of type MATH,MATH,MATH and MATH. Again by REF we can order the types of the MATH, such that MATH are consecutive elementary NAME immersions that induce critical points on MATH. We still know MATH and the same argument as before allows us to replace MATH by elementary NAME immersions of type MATH and MATH. Repeating this procedure on MATH finishes the proof.
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math/0004022
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Let MATH be a local chart of MATH. Then for any MATH and MATH in the domain of this local chart we have: MATH then, using the local coordinates REF , one gets easily all the results of the proposition.
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math/0004022
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Let us first assume REF . Then, using the results of REF and the fact that MATH and MATH are smoothing, one proves easily that MATH is an isomorphism whose inverse is given by: MATH . Now let us prove REF . Following CITE page REF, we recall that the NAME kernel of MATH is the finite sum of a smooth function and of oscillatory integrals (supported in small coordinates charts) of the following type: MATH where MATH vanishes for MATH, MATH is an homogeneous phase function parametrizing locally the graph MATH of MATH which satisfies MATH so that locally we have: MATH and MATH. Notice moreover that MATH and MATH are local diffeomorphisms. With these notations, the NAME kernel of MATH is the finite sum of a smooth function and of oscillatory integrals (supported in small coordinates charts) of the following type: MATH . Let MATH which is identically zero in a neighborhood of the zero section, in order to analyze MATH it is enough to study the operator MATH where MATH denotes the operator whose NAME kernel is given by REF . The NAME kernel of MATH is given by: MATH . In this integral we replace MATH by its NAME expansion: MATH . Using the following two identities MATH and integrating by parts we see that MATH is the sum of a smooth function and of MATH . Now for MATH we set MATH and we consider MATH . If we replace MATH by its NAME expansion MATH then, using integrations by parts as above, it follows easily that MATH is the sum of a smooth function and of MATH . We observe that if we apply the NAME rule for the term MATH in the previous integral then the following differential operators will appear MATH . It is clear from REF that, expressed in the coordinates MATH, these differential operators REF are MATH-differential operators. Therefore we have just proved that MATH is the sum of a smooth function and of: MATH where the MATH are MATH-differential operators. Now we recall that the NAME kernel of MATH is the finite sum of a smooth function and of terms of the type REF . So in order to analyze MATH it is enough to study the operator MATH whose NAME kernel is the finite sum of a smooth function and of integrals of the type: MATH . Moreover we can write MATH where MATH and we can assume (at the expense of shrinking the local coordinates charts) that MATH is a local diffeomorphism whose inverse is denoted MATH. With these notations, we set: MATH . Then a change of variable formula allows us to see that the oscillatory integral REF is equal to MATH . We observe that, expressed in the coordinates MATH, the vector fields MATH are MATH-differential operators. Therefore one proves easily the assertion of REF of the Theorem by replacing MATH by its NAME expansion MATH and using, as before, integration by parts.
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math/0004022
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REF are left to the reader. REF is an easy consequence of REF and of REF.
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math/0004022
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For MATH or MATH we set: MATH where MATH denotes the set of smooth functions which vanish of infinite order at MATH. Then we have the following exact sequence: MATH of MATH-algebras. Here MATH denotes the induced formal MATH-deformation of the sphere at infinity. A direct construction of this deformation may be described as follows. Let MATH denote the space of pseudodifferential operators on, say, MATH of order MATH modulo the smoothing operators. Then the space of doubly infinite sequences MATH is a flat module over MATH, where the multiplication by MATH acts as the right translation. If we endow it with the product MATH it is easily seen to be isomorphic to MATH. Any trace MATH on MATH is given by a sequence of MATH-linear, MATH-valued functionals MATH on MATH such that MATH . The MATH-linearity of MATH implies that MATH and the trace condition on MATH implies that each MATH is a trace on the algebra of pseudodifferential operators modulo the smoothing operators. Recall that, on this latter algebra, the NAME residue MATH is the unique trace up to multiplicative constant. Thus MATH is, up to multiplicative constant, uniquely determined by MATH, and hence the space of traces on MATH is one-dimensional. We recall that MATH is NAME (in the sense of NAME, see CITE), so we have the following long exact sequence in cyclic cohomology: MATH . From REF we recall that the space of MATH-linear traces (with values in MATH) on MATH is one dimensional and generated by MATH. By above, MATH is one-dimensional. The connecting map: MATH is given by taking a trace on MATH, extending it to a linear functional on MATH and taking its NAME boundary. In particular, it is not zero (this is equivalent to existence of a pseudodifferential operator with nonzero index!). This implies that MATH is either one or two dimensional. Since, with the notations of the Proposition, MATH are two linearly independent elements of the vector space of traces on MATH, the rest of the statement of above Proposition follows.
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math/0004022
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CASE: A standard NAME sequence argument shows that MATH is indeed two dimensional. The fact that MATH defines a basis is left to the reader. CASE: This is an easy consequence from REF and of the properties (see CITE, CITE ) of the trace density map MATH.
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math/0004022
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One obtain this formula by first applying REF and then by letting MATH. As we will see below, the involved characteristic classes of vector bundles on MATH are in fact standard NAME cohomology classes and hence the regularized integral coincides with the orientation class of MATH.
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math/0004023
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As a first step, note that if MATH is a smooth point, MATH an automorphism and MATH a birational morphism, then it is clear that MATH if and only if MATH. To conclude, it suffices to note that - up to composition with automorphisms - there are only finitely many birational morphisms MATH. Recall that for a given bundle MATH, we have that MATH for all but finitely many smooth points MATH.
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math/0004023
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Set MATH where MATH is the ideal sheaf of MATH. It follows immediately that MATH is a coherent sheaf of MATH-algebras. Define MATH. The existence of MATH, MATH and MATH follows by construction. In order to see that fibers of MATH are of the desired type, let MATH be an arbitrary closed point. We are finished if we show that the fiber MATH has a single singularity which is a simple node or cusp. After replacing the base MATH by an affine neighborhood of MATH and performing a base change, if necessary, we may assume that there exists a relatively ample divisor MATH which intersects every MATH-fiber in a single smooth point. We may furthermore assume that MATH. Write MATH, write MATH and note that both MATH and MATH are affine. By choosing a bundle coordinate MATH on MATH, we may write MATH, MATH and MATH. Because MATH we can decompose MATH. Since MATH by construction, we have the equation of rings MATH . If MATH is chosen properly, tensoring with MATH yields MATH for a number MATH. An elementary calculation shows that this ring is generated by the elements MATH, MATH and the constants MATH. Therefore MATH which defines a cusp if MATH and a node otherwise.
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math/0004023
|
The strategy of this proof is to find a sequence of base changes which modify MATH and MATH so that REF can be applied. First, after finite base change, we may assume that the normalization MATH is a MATH-bundle over MATH. Recall that smooth morphisms are stable under base change. Thus, even after further base changes, the normalization of the pull-back of MATH will still be a MATH-bundle over the base. Now let MATH be the singular locus of MATH-fibers and let MATH be an irreducible component which maps surjectively onto MATH. Since MATH is assumed to be a smooth curve, MATH is actually finite over MATH. If MATH is not isomorphic, perform a base change. Thus, we assume that MATH. As a next step, let MATH be the normalization and consider the relative NAME MATH of zero-dimensional subschemes of length REF in MATH over MATH. Recall the fact that taking the NAME commutes with base change (see for example, CITE). In our setup this means that if MATH is a morphism, then MATH . This has two important consequences. First, by choice of MATH, if MATH is a general point, then MATH is zero-dimensional of length at least two, so that MATH is not empty. It follows that MATH is surjective. Second, let MATH be a subvariety which is finite over MATH. Let MATH be the normalization of MATH and perform another base change. Since taking MATH commutes with base change, MATH can be seen as a subscheme of MATH and therefore defines a subscheme MATH. Thus, all the prerequisites of REF are fulfilled, and we can apply that lemma in order to obtain MATH. Because our construction involves only finite base change, it is clear that MATH satisfies the requirements of REF if and only if MATH does.
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math/0004023
|
Write MATH, MATH where MATH denotes numerical equivalence and MATH is a general fiber of MATH. Since MATH and MATH are effective, MATH and thus MATH. Therefore MATH.
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math/0004023
|
As a first step, we need to find an estimate for the dimension of the subfamily of non-immersed curves. We let MATH be the closed subfamily of non-immersed curves and claim that MATH . Indeed, if MATH, then let MATH be a singular curve which corresponds to a general point of a component of MATH which is of maximal dimension. By REF , we find a smooth point MATH which is general with respect to MATH. We remark that MATH is not an isolated point point of MATH and conclude by REF that MATH cannot be proper. But then MATH cannot be proper, contrary to our assumption. This shows REF . Next, we show that MATH. We will argue by contradiction and assume that MATH. Now, if MATH is any curve which parameterizes nodal curves in MATH, we can apply REF to the family MATH to see that MATH is not proper. It follows that the closure MATH intersects MATH and therefore MATH . Hence our claim follows from REF . In order to show that MATH is at most finite, perform a dimension count. It is clear that MATH which in turn implies that general fibers MATH of the natural projection MATH are of dimension MATH. Now it suffices to note that the natural map MATH is finite, that is, that MATH to obtain that MATH which yields the finiteness result. With REF , the same dimension count, using the family MATH, immediately shows that MATH which means that all curves associated with MATH are immersed. This ends the proof of REF . To prove REF , we argue as above. Let MATH be a singular curve corresponding to a general point of MATH and MATH general with respect to MATH. By REF , the family MATH cannot be proper, contradiction.
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math/0004023
|
If MATH is a morphism which is birational onto its image, if MATH and the image MATH is a curve which is associated with MATH, then it follows from REF that MATH is smooth in a neighborhood of MATH. Now, to conclude that MATH is a morphism, it suffices to realize that MATH can be written as a composition MATH where MATH is the parameter space of morphisms, and MATH sends a morphism MATH to the image of its tangent map: MATH. The existence of MATH and the existence of the quotient space follows from universal properties - see CITE. To show that MATH is finite, we argue by contradiction. If MATH was not finite, we could find a curve MATH such that MATH is a single point MATH. After changing base, we can assume that MATH is smooth, and consider the diagram: MATH where MATH is the normalization of the universal family and hence a MATH-bundle over MATH. There exists a section MATH which contracts to the point MATH, and the restriction of the tangent morphism MATH yields a morphism: MATH where MATH is the normal bundle of MATH and MATH a line in MATH. But since the normal bundle is not trivial, this map has to have a zero! Thus, there exist curves associated with MATH which have non-immersed singularities at MATH. This contradicts REF , and we are done.
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math/0004023
|
Following an argument of NAME, MATH is generically one-to-one if MATH, and the latter follows from REF . See CITE for a proof of NAME 's result.
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math/0004023
|
Let MATH be the normalization of the universal family MATH and consider the diagram MATH . It follows directly from REF that MATH is birational and it follows immediately that MATH is birational as well. But then MATH is a finite birational morphism between normal spaces, and therefore isomorphic. In particular, MATH and MATH are smooth. The induced map MATH is an isomorphism away from the section MATH, which is contracted: otherwise, NAME 's main theorem asserts that there exists a point MATH and a positive dimensional subfamily of curves passing through both MATH and MATH. But NAME 's bend-and-break argument says that this cannot happen if MATH is unsplit. In particular, since MATH is smooth it follows that MATH is smooth. In this setting, REF yields the claim. See CITE.
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math/0004024
|
Suppose MATH is a countable set of functions. Let MATH denote the set MATH for every MATH. Every MATH is at most countable because MATH is separable. So choose a MATH in MATH, then we know that for every MATH the set MATH is nowhere dense in MATH. For such a MATH the set MATH is countable. Because the set MATH is also countable the NAME category theorem tells us that the set MATH is nonempty, thus showing that MATH is not transitive.
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math/0004024
|
The proof of the first inequality is easy. We simply have to observe that the set of all constant functions on the reals is a transitive set of functions. Now for the second inequality. Striving for a contradiction suppose that MATH. Let MATH be a transitive set of functions such that MATH. We define a set map MATH on the reals by MATH for every MATH. Because MATH, this set map MATH is of order MATH, which is less than MATH. According to the free set lemma there exists a set MATH such that MATH and for every MATH we have MATH. This is a contradiction, because every two reals in MATH provide a counter example of MATH being a transitive set.
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math/0004024
|
We will show by transfinite induction that MATH is a transitive set in MATH for all MATH. For MATH this is obvious. Suppose the theorem is true for all MATH. Let MATH and MATH be reals in MATH. If MATH is a successor ordinal, MATH, then we use REF in the case that at least one of MATH and MATH is not in MATH to show that there exist a continuous function MATH defined in MATH (so MATH) such that in MATH we have MATH or MATH. Since we are forcing with countable support and because reals are countable objects, there are no new reals added by MATH for MATH. So if MATH is a limit ordinal we only have to consider the case where MATH and at least one of MATH is not in MATH. Then we use REF to show the existence of an continuous function MATH defined in MATH such that in MATH or MATH holds.
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math/0004024
|
We will construct a fusion sequence MATH such that each MATH will know all the first MATH splitting nodes of every branch of the perfect tree MATH and MATH for every MATH. Because MATH forces that MATH is a new real, there exists an element MATH with maximal length MATH, such that MATH and MATH does not decide MATH. There exist MATH such that MATH for MATH. Without loss of generality the stems of MATH and MATH are incompatible. Let MATH and let MATH denote the element MATH. Now assume we have MATH. Consider MATH, we have an element MATH of maximal length MATH such that MATH. There exist MATH, MATH such that MATH for MATH. Again without loss of generality the stems of MATH and MATH are incompatible. Let MATH denote the integer MATH and MATH. We let MATH denote the element MATH. Now the induction step is completed, because MATH knows all the first MATH splitting nodes of every branch in MATH and MATH for every MATH. We define the function MATH by MATH . As MATH is a finite approximation of the added NAME real MATH, we have by the construction of our MATH for MATH and the function MATH that MATH for every MATH. And so the fusion MATH of the sequence MATH forces that in the extension MATH the equality MATH holds. This MATH, being a continuous bijection between two NAME sets, is (of course) a homeomorphism.
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math/0004024
|
Suppose MATH and MATH are two reals of MATH. We consider two cases. CASE: MATH is a real in MATH. The constant function MATH is a continuous function defined in MATH, thus a member of MATH, and in MATH it maps MATH onto MATH. REF both MATH and MATH are reals not in MATH. Let MATH be a witness of this, so MATH. According to REF there exists a MATH and a homeomorphism MATH defined in MATH such that MATH, where MATH denotes the added NAME real. If we apply the lemma again we get a MATH and a homeomorphism MATH defined in MATH such that MATH. But now we have that MATH and we see that MATH is the element of MATH we are looking for.
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math/0004024
|
This is an immediate consequence of REF .
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math/0004024
|
Define the element MATH as follows for MATH: MATH . In this way we strengthen the tree MATH above MATH keeping the rest of the perfect tree intact (according to MATH anyway).
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math/0004024
|
Let MATH denote the set of all MATH extending MATH. Because MATH is a perfect tree there exists a MATH-name MATH such that for every MATH we have MATH . According to REF there exists a MATH such that if MATH is consistent with MATH we have a MATH such that MATH. Put MATH. We have MATH for every MATH. Enumerate MATH as MATH. Let MATH be such that MATH, where MATH is such that MATH for every MATH. Use REF to find a MATH such that MATH. We continue this procedure with all the MATH. So if MATH is consistent with MATH we find a MATH such that MATH, and also that MATH for every MATH. And we use REF to define MATH such that MATH. If MATH is not consistent with MATH we choose MATH. We now have for every MATH consistent with MATH a finite tree MATH extending the tree MATH such that every branch in MATH has (at least) MATH different extensions in MATH and MATH. As MATH forces that, for each MATH the size of the set MATH is at least MATH we can find for MATH consistent with MATH a sub tree MATH of MATH such that MATH and whenever MATH and MATH are distinct and consistent with MATH we have MATH. Define MATH such that MATH and choose MATH such that for every consistent MATH we have MATH. If we let MATH be equal to MATH the proof is complete.
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math/0004024
|
By REF we know that there is a MATH and there exist MATH names MATH for a map on the finite sub trees of MATH and MATH for a perfect tree such that MATH. Without loss of generality we assume that MATH. Let us construct a fusion sequence MATH. Let MATH, MATH, MATH and choose MATH in such a way that we are building a fusion sequence. Suppose we have constructed the sequence up to MATH, let us construct the next element of the fusion sequence. We let MATH denote all MATH consistent with MATH. If we choose in REF MATH, MATH and MATH we get a MATH such that for every MATH extending MATH, consistent with MATH, we have a finite sub tree MATH (MATH follows from REF ) of MATH such that CASE: MATH is an extension of MATH, CASE: for every branch MATH in MATH there exist at least two different branches of length MATH in MATH extending MATH, CASE: if MATH and MATH are two distinct members of MATH consistent with MATH we have MATH. We choose MATH with REF such that MATH and MATH. We iteratively consider all the MATH. In the general case if MATH is consistent with MATH then REF gives us a MATH and a MATH such that MATH. We choose MATH in the same way as above, using REF such that MATH and MATH. If MATH is inconsistent with MATH then we choose MATH and MATH. After considering all the MATH's we define MATH and MATH. This ends the construction of the next element of the fusion sequence. For every MATH if MATH is consistent with MATH and extends MATH then MATH . Considering our function MATH, let us denote the finite tree MATH by MATH. We have MATH . When we are building the fusion sequence we can of course make sure that the fusion determines MATH as well. Suppose we have that MATH, MATH of length MATH. With REF we can choose MATH strong enough such that for every MATH consistent with MATH we have a MATH of length MATH such that MATH. So assume we have made sure this is the case and let us define the function MATH in MATH by MATH for every maximal branch MATH for every MATH consistent with MATH for some MATH. The function MATH is well-defined by REF and we have for every MATH and MATH consistent with MATH that MATH and thus MATH.
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math/0004024
|
By applying REF twice we have a MATH in MATH and MATH names MATH, MATH and MATH, MATH for maps and perfect trees respectively such that MATH. Without loss of generality we can assume that MATH. During the construction of possible finite sub trees MATH for MATH, when constructing the fusion sequence in the proof of REF we could of course at the same time also have constructed a similar sequence of finite sub trees MATH for MATH. Without loss of generality we could also have made sure that in the proof of REF is replaced by CASE: for every maximal branch MATH in MATH there are exactly two different branches of length MATH in MATH extending MATH, Following the proof of REF we have for every MATH consistent with MATH finite sub trees MATH and MATH such that MATH . We are ready to define the homeomorphism MATH in MATH that maps MATH onto MATH in the extension. Suppose MATH and MATH such that MATH extends MATH. Every maximal branch in MATH corresponds to exactly one maximal branch in MATH. Let MATH map the splitting point in MATH above any maximal branch in MATH onto the splitting point in MATH above the corresponding maximal branch in MATH. The function MATH thus defined will be a continuous and one-to-one mapping between two NAME sets, so a homeomorphism. Furthermore the fusion MATH forces that MATH maps MATH onto MATH in the extension.
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math/0004024
|
Choose a MATH such that MATH. Let MATH be consistent with MATH and let MATH denote all the MATH extending functions MATH. Because MATH forces that MATH, there is an antichain below MATH of size MATH such that all these elements force different interpretations of MATH in the extension. In other words there exist a sequence MATH of MATH names for elements of MATH and a sequence MATH of MATH names for elements of MATH such that for all MATH we have MATH and if MATH and MATH are distinct then MATH . Repeatedly using REF we see that there exist a MATH and sequences MATH, MATH for some integer MATH such that for every MATH we have MATH . Now let MATH denote the element of MATH such that MATH, and MATH for every MATH consistent with MATH. This completes the proof.
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math/0004024
|
For the first part of the theorem suppose that we have MATH such that MATH. We will construct a fusion sequence below MATH and define a continuous function MATH in MATH such that the fusion of the sequence forces that MATH holds in MATH. Let MATH, MATH, MATH, and choose MATH in such a way that we are building a fusion sequence. Suppose we have constructed the sequence up to MATH, we will construct the next element of the fusion sequence. Let MATH denote an enumeration of all maps from MATH into MATH consistent with MATH. According to REF there exists a MATH such that for every MATH consistent with MATH we have distinct MATH's in MATH (where MATH follows from REF ), such that MATH. Now use REF to construct MATH such that MATH and MATH. We now iteratively consider all the MATH. In the general case if MATH is not consistent with MATH then we make sure that MATH and MATH. If MATH is consistent with MATH we find by REF a MATH such that for every MATH consistent with MATH we have distinct MATH's in MATH such that MATH. Now use REF to construct MATH such that MATH and MATH. After considering all MATH we define MATH and MATH. If we take a closer look at REF we can also let the fusion sequence that we just constructed determine MATH. Because if we have MATH, following the proof of REF we can make sure that (by some strengthening of MATH or the MATH's, if necessary) there exist MATH's in MATH, not necessarily distinct, extending MATH such that for MATH consistent with MATH we also have MATH. So assume we have done this. We have for every MATH consistent with MATH . Now we are ready to define our function MATH which will map MATH in MATH continuously onto MATH. Let MATH for all MATH and all MATH. Then MATH for MATH consistent with MATH and MATH. It follows that the fusion MATH forces that in MATH we have MATH. Moreover MATH is a continuous function, this follows from REF . For the second part of the theorem suppose that MATH. Just as in REF we can choose not only the MATH's in REF distinct but also the MATH's for MATH and MATH for some MATH. With this, the constructed continuous function MATH is actually a homeomorphism.
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math/0004024
|
Suppose that MATH is a MATH name and MATH an element of MATH such that MATH. So there exists a MATH name MATH and a MATH name MATH such that MATH. Aiming for a contradiction assume MATH is a name for an object not in MATH. There exists a MATH such that MATH does not decide MATH. Now we have MATH does not decide MATH, and MATH does not decide MATH. So we can find in MATH a MATH such that MATH and in MATH a MATH such that MATH. This gives the contradiction we are looking for because MATH and MATH. So MATH must be a name of an element in MATH.
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math/0004024
|
For every MATH we have that there exists a function MATH mapping MATH onto MATH or vice versa. This function MATH is not a member of MATH for the obvious reason that assuming that MATH maps MATH onto MATH we get MATH, which, of course, is false. Using REF and the fact that MATH we see that the size of MATH is at least MATH, because MATH for every MATH. By REF we are done.
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math/0004030
|
CASE: First we must show that MATH is well defined. Observe first that MATH that is, MATH for every MATH. Let now MATH be arbitrary. Then MATH and MATH for every MATH, hence MATH is well defined. CASE: Now we show that MATH is dense in MATH. First the elements with only finitely many entries not MATH are dense in MATH. Further every such element is a linear combination of elements of the type MATH, again all but a finite number equal MATH. Since MATH is cyclic for MATH, we can approximate these elements by elements of the form MATH with MATH, hence MATH is dense in MATH. CASE: In this step we want to show that MATH is closable. Let MATH, MATH (MATH), where MATH is defined analogously to MATH, hence it is a densely defined operator, too (All MATH are closed operators affiliated with MATH and MATH, compare CITE, and MATH). Now MATH . This shows MATH, MATH, and MATH, hence MATH is closable. This shows also, that MATH. CASE: To show that MATH is affiliated with MATH, let MATH be a unitary. Then MATH. Let now MATH. Then MATH . This shows MATH for every unitary MATH, hence, since MATH is a core for MATH, MATH . CASE: In the last step we calculate MATH .
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math/0004030
|
CASE: Let MATH the polar decomposition of MATH, and MATH the spectral resolution of MATH. Then MATH for every unitary MATH and every MATH, that is, MATH for every MATH. Now MATH, and, since MATH, also MATH. CASE: REF shows that MATH, further MATH is dense in MATH. Now MATH is invariant under the unitary group MATH, that is, MATH is a core for MATH and also for MATH. The assertion for MATH follows analogous. CASE: This follows from REF and CITE. CASE: Now for every MATH there exists exactly one MATH such that MATH (MATH, where MATH is the conjugation with respect to MATH). Now define MATH where MATH. Then MATH and MATH . Hence MATH converges and therefore MATH converges in the graph norm of MATH to MATH, that is, also MATH is a core, since MATH is it.
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math/0004030
|
Set MATH (MATH). Then matrix calculation shows that MATH is a family of orthogonal, equivalent, finite projections, such that MATH and MATH. Now MATH defines a selfadjoint system of matrix units MATH such that MATH for every MATH. Now define operators MATH and MATH . Since MATH is dense in MATH (compare proof of REF) and a core both for MATH and for MATH they are well defined and densely defined. Let now MATH and MATH. Then MATH . This means that MATH and MATH, that is, MATH, hence MATH is closable since MATH is densely defined. Let now MATH be the closure of MATH. Then MATH is a core for MATH. Since MATH and MATH for every unitary MATH and every element MATH, it follows that MATH and MATH is affiliated with MATH. Further MATH hence MATH is affiliated with MATH. Now set MATH, where MATH is the canonical partial isometry from MATH to MATH. Now MATH since MATH, and with MATH such that MATH. This means that we can construct the operator MATH according to REF. Now MATH and MATH coincide on the core MATH, and hence they are equal.
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math/0004030
|
REF shows that MATH for every MATH and MATH (compare REF). Now MATH .
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math/0004030
|
Let MATH and MATH be bounding sequences for MATH and MATH, respectively, (compare CITE). Then: MATH since MATH and MATH (compare CITE).
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math/0004030
|
CASE: Since MATH for all MATH , where MATH is the polar decomposition of MATH, we can write the trace, defined in REF, as follows (MATH is the spectral measure of MATH): MATH . Since the MATH are mutually orthogonal, we have MATH . Further MATH . CASE: Let MATH be cyclic. Then there are MATH with MATH for every MATH, where MATH. This means, using REF , MATH that is, since MATH and MATH is dense in MATH, MATH has dense range, that is, MATH is injective. Let now MATH be injective and MATH with MATH. Now MATH and MATH where MATH (MATH is closed), hence MATH, and, since MATH is injective, MATH for all MATH. Because MATH is cyclic for MATH hence separating for MATH, MATH for all MATH, such that MATH. CASE: Let MATH be separating. This means that MATH is cyclic for MATH. Then there are MATH and MATH for every MATH, where MATH (MATH). This means MATH . Since MATH is dense in MATH we have proven that MATH has dense range. For the converse read the argument backwards. CASE: This follows from REF.
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math/0004030
|
Set MATH or MATH. Then MATH fulfills the conditions of REF and is invertible, such that from REF follows that MATH is cyclic and separating.
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math/0004030
|
Since MATH is infinite but semifinite there is a countable orthogonal family of finite equivalent projections MATH in MATH, such that MATH. Now there is a selfadjoint system of matrix units MATH with MATH (compare CITE). This shows that MATH is isomorphic to MATH where MATH and MATH is isomorphic to every MATH (MATH). Since the projections MATH are finite also MATH is a finite factor. Since MATH possesses the separating vector MATH we can represent the algebras MATH by the GNS representation for the faithful state MATH induced by the separating vector MATH on a NAME space MATH, such that the vector MATH implementing the state MATH is a cyclic and separating vector for MATH. Since all the MATH are isomorphic and they possess in this representation a cyclic and separating vector, they are all unitarily equivalent. This means that we can choose as MATH one of the MATH acting on the representation space MATH. Since the factor MATH possesses a cyclic and separating vector if MATH does (see REF ) and it is isomorphic to MATH it is unitarily equivalent to MATH.
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math/0004030
|
The finite case is just REF. In the infinite case the existence and the asserted properties follow from REF infinite case, the uniqueness from REF .
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math/0004030
|
Again the finite case is just REF. In the infinite case the necessarity of the trace condition follows from REF and the sufficiency from REF, respectively,
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math/0004030
|
CASE: First we observe that MATH for every MATH. For let MATH and MATH, MATH, then MATH . Further MATH such that MATH is an (algebraic) conjugation for MATH. CASE: Let now MATH be bounded ( MATH all the MATH and MATH, respectively, are bounded). Then we show that the NAME operator MATH defined by MATH can be written as MATH . For this let MATH and MATH. Then MATH and MATH . Now MATH which proves REF. Now MATH and MATH . Further MATH and all the assertions are proven in the bounded case. CASE: In the last step we approximate the (unbounded) operator MATH by bounded operators MATH in exactly the same way as in the proof of REF and show the assertions like there also in the unbounded case.
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math/0004030
|
Let MATH now be a type MATH or MATH factor and MATH the operator corresponding to the cyclic and separating vector MATH. Let further MATH the spectral resolution of MATH. Then we can define a positive measure MATH on the MATH-algebra of NAME sets in MATH, such that MATH where MATH for all NAME sets MATH. Now MATH. Assume without loss of generality MATH. Then MATH that is, MATH . Since MATH is infinite MATH, that is, MATH hence MATH . Suppose now that also MATH, then MATH which is a contradiction.
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math/0004030
|
We construct the MATH inductively: Since the range of MATH is all of MATH (compare CITE) there is a projection in MATH, such that MATH. Suppose now that for MATH the MATH are pairwise orthogonal with MATH (MATH). Setting MATH the restricted algebra MATH is again a type MATH factor, finite, if MATH is finite, and infinite, if MATH is infinite (compare CITE) with the dimension function MATH if MATH is finite, and MATH else, where MATH . With the same argument as above there is again a projection MATH, such that MATH, if MATH is finite, and MATH else. In both cases MATH and MATH (MATH).
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math/0004036
|
Put MATH so that MATH. Since MATH is decreasing when MATH or MATH and increasing when MATH. Thus (roughly speaking) MATH attains its maximum at MATH. Since there are MATH terms in the summation formula of MATH, we have MATH . Taking MATH and dividing by MATH we have MATH which turns out to be MATH . Since MATH, we have MATH . On the other hand from CITE we have MATH . Therefore MATH and MATH . We finally have MATH completing the proof.
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math/0004047
|
(This idea, and in fact the whole proof, is due to CITE.) Assume that the polynomials MATH are pairwise incomparable, where MATH, and MATH abbreviates MATH. For each MATH there is a natural number MATH such that MATH can be written as MATH, where MATH is a MATH-ary term, and MATH (the entries of the vector MATH are the ``coefficients" of the polynomial MATH). Since there only countably many terms, our assumption MATH implies we can thin out the set MATH to a set MATH of the same cardinality such that all the MATH (for MATH) are equal, say MATH. Now it is easy to check that the ``coefficients" MATH form an antichain in MATH.
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