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math/0012020 | Let MATH be a finite partition of unity for MATH such that MATH for any MATH. Thus we can write MATH . Now let MATH and choose a chart MATH of MATH with MATH. Let MATH be the corresponding chart of MATH. Set MATH, initially defined as a continuous map MATH. By REF MATH so MATH defines a continuous map MATH which extends to give continuous maps MATH and MATH for any admissible space MATH. Furthermore any two of these extensions agree on functions common to their domains. Using REF it follows that MATH defines a continuous map MATH which has continuous extensions MATH and MATH for any admissible space MATH. Furthermore MATH . From the definition of MATH it clearly follows that the compositions MATH and MATH are both equal to the identity on MATH; that is, MATH is an isomorphism with inverse MATH. On the other hand MATH and MATH are extensions of MATH and MATH so the compositions MATH and MATH must also give the identity map when restricted to MATH. REF then implies MATH is an isomorphism with inverse MATH. Now let MATH be any admissible space. Using REF and the obvious fact that MATH for any MATH, we finally have that MATH is an isomorphism with inverse MATH. |
math/0012020 | Using REF we can write MATH where MATH (the fact that these functions are MATH valued is unimportant in what follows and won't be mentioned explicitly). Now, by assumption, MATH so MATH and MATH. Since MATH and MATH it follows that MATH . On the other hand, for any MATH REF can be rewritten as MATH . The result now follows from the continuity of the dual pairings MATH for MATH (see REF ). |
math/0012020 | From the proof of REF we know that, for MATH, the inverse of MATH is the restriction of the inverse of MATH to MATH. It therefore suffices to prove the result with MATH replaced by MATH. Now REF holds with MATH; see REF for example. It follows that REF is valid for all MATH. REF together with the following claim thus completes the proof. Let MATH. Thus MATH and MATH. By REF we can then find sequences MATH for MATH such that MATH in MATH and MATH in MATH. REF then gives MATH and MATH in MATH. Similarly MATH and MATH in MATH. Setting MATH for each MATH, it follows that MATH is a sequence in MATH which converges to MATH in MATH. |
math/0012020 | It is straightforward to check that the set of uniform scalar operators on MATH can be generated from MATH and the set of first order model operators by taking linear combinations of products. With the help of REF and the fact that scalar differential operators commute to leading order, we can thus write any uniform scalar operator MATH of order MATH on MATH in the form MATH where, for each multi-index MATH with MATH, MATH. However MATH iff MATH for all MATH, whilst MATH for MATH. REF thus complete the result. |
math/0012020 | Let MATH be the uniform operator on MATH given by REF. Using REF we thus have to show that MATH is equivalent to MATH (where MATH denotes the norm on the fibres of the cotangent bundle MATH given by the Riemannian metric on MATH). For each MATH let MATH and MATH denote the principal symbols of the MATH-th entries of MATH and MATH respectively. Using REF and the transformation properties of principal symbols under diffeomorphisms (see REF for example) we have MATH for all MATH and MATH, where MATH and MATH denotes the adjoint of the linear map MATH. By taking determinants of the MATH matrices with entries MATH and MATH we now get MATH for all MATH and MATH, where we have made use of the fact that MATH is homogeneous of degree MATH in MATH. On the other hand, the definition of MATH gives MATH for all MATH and MATH. The equivalence of REF follows from this identity and REF. |
math/0012020 | If MATH is a scalar admissible operator of order MATH on MATH then REF , and REF give MATH for any MATH with MATH (n.b. the operator MATH can be interpreted as acting on either MATH or MATH using REF and the expression MATH). On the other hand, we have MATH . The result follows by applying REF to the individual entries of MATH. |
math/0012020 | Using induction it clearly suffices to prove the result in the case MATH. Now, by REF , we can choose MATH sufficiently large so that the operator MATH is a uniform elliptic operator on MATH. However MATH whilst REF gives us MATH and MATH. REF then implies MATH and MATH . On the other hand MATH is an elliptic operator (on MATH) whilst REF gives us MATH and MATH. REF then implies MATH and MATH (n.b. MATH). The result now follows from REF . |
math/0012020 | Using the hypothesis, REF , we can choose MATH sufficiently large so that we either have MATH and a local inclusion MATH or MATH and a continuous inclusion MATH. REF or REF then gives us a continuous inclusion MATH. The result now follows from REF . |
math/0012020 | From REF we know that for all sufficiently large MATH we have MATH for MATH. Choose MATH for which this is true and set MATH. It follows that we have a NAME series expansion MATH which is convergent in MATH for MATH. Let MATH and set MATH. Then MATH by REF . Now let MATH and suppose MATH . Therefore MATH so REF implies REF also holds with MATH replaced by MATH. The result now follows from REF and induction. |
math/0012020 | Choose MATH so that the conclusion of REF holds with MATH. Now suppose MATH for some MATH. Set MATH and MATH. By REF we have MATH and thus MATH (by the mapping properties of uniform operators on MATH; see also REF ). On the other hand MATH on a neighbourhood of REF whilst MATH by the hypothesis on MATH and REF thus gives MATH and so MATH by REF . Using the fact that MATH is an isomorphism for MATH we also get a norm estimate of the form MATH . REF now completes the result. |
math/0012020 | Choose MATH so that the conclusion to REF holds for MATH. Now suppose we have a sequence satisfying MATH . Let MATH. REF gives us MATH so MATH. Setting MATH we then have MATH. Furthermore, by combining norm estimates given by REF and the fact that MATH is an isomorphism, we get MATH . Also, MATH so MATH and hence MATH . Now define MATH by MATH for all MATH. By the second inequality in REF MATH is a bounded sequence in MATH whilst REF gives us MATH for all MATH. Choose MATH so that MATH. Using REF it follows that MATH is a bounded sequence in MATH. However MATH is a compact map by REF , so we can choose a subsequence MATH which is convergent in MATH. On the other hand, the last part of REF and the first inequality in REF imply MATH in MATH. Since MATH continuously we then get the convergence of MATH in MATH. By combining REF and the last part of REF it follows that MATH is convergent in MATH. Summarising, we have shown that any sequence satisfying REF has a subsequence which is convergent in MATH. A standard argument (see REF or REF) shows this implies MATH has a finite dimensional kernel and a closed range. |
math/0012020 | Choose MATH with MATH. By REF we have a continuous inclusion MATH so MATH. On the other hand MATH so MATH must be finite dimensional by REF . |
math/0012020 | Choose any MATH which lies in the same component of MATH as MATH and MATH, and satisfies MATH. Now MATH and MATH so we also have MATH and MATH by REF then gives MATH and MATH . Applying REF (with MATH and MATH) now gives MATH and MATH, completing the result. |
math/0012020 | We get MATH as a direct consequence of REF . It therefore remains to show MATH. By symmetry, REF and the obvious inclusion MATH for any MATH, it suffices to prove this equality under the assumption that either MATH and we have a local inclusion MATH, or MATH and we have a continuous inclusion MATH. Combining this assumption with REF , we get a continuous inclusion MATH. Suppose MATH is a subspace of MATH with MATH. If MATH then MATH for some MATH. REF then gives MATH and so MATH, completing the claim. If MATH then MATH by an easy application of REF . Now suppose MATH and choose a finite dimensional space MATH which satisfies MATH . REF and the inclusions MATH give MATH . The following claim thus completes the proof. Let MATH and choose MATH so that MATH and MATH lie in the same component of MATH. By REF we can find a sequence MATH such that MATH in MATH. Now REF and the inclusions MATH give MATH . However REF implies MATH is a closed subspace of MATH whilst MATH. Thus MATH is also a closed subspace of MATH, so REF implies MATH. In turn this means we can find MATH and MATH which satisfy MATH. Since MATH . REF then gives MATH. Therefore MATH, completing the claim. |
math/0012020 | Since REF shows that the map MATH is semi-Fredholm we simply have to establish that MATH is finite. Furthermore, REF shows this index to be stable under changes in MATH. It thus suffices to prove the result for a particular admissible space MATH; we take this to be MATH so MATH. By REF we have MATH. Furthermore REF implies MATH. REF then shows the maps MATH and MATH are both semi-Fredholm with finite dimensional kernels. However we have the general identity MATH for semi-Fredholm operators (see REF for example). Therefore MATH which is finite. |
math/0012020 | By REF MATH. Setting MATH it follows that MATH and MATH on a neighbourhood of REF. On the other hand REF gives MATH . Thus REF imply MATH. For MATH set MATH. By REF it follows that MATH for some MATH. Now MATH so, setting MATH, we get MATH by REF . On the other hand MATH (this follows from the definition of MATH) so MATH. Combining this with the definition of MATH and REF we get REF. |
math/0012020 | Since MATH we get MATH. However MATH is closed in MATH with finite codimension (by REF ) so the same must be true for MATH. Now choose a finite dimensional space MATH so that MATH and MATH. The inclusion MATH immediately gives the reverse inclusion. Now let MATH. Put MATH and MATH (n.b. REF ensure that MATH is well defined). With the help of REF we thus have MATH . Hence MATH by REF . On the other hand, by the definition of MATH, MATH for some MATH and MATH. Setting MATH we then have MATH . Therefore MATH, completing the claim. Now MATH so MATH . Combined with the above claim this now completes the result. |
math/0012020 | For MATH set MATH. Since MATH (by REF ; recall that MATH) we immediately get MATH using REF . Now let MATH. Suppose MATH for some MATH and MATH. Thus MATH and so, by REF, there exists MATH and MATH such that MATH . Now MATH so REF implies MATH where MATH. REF then implies MATH. On the other hand MATH so MATH and MATH by REF . However MATH is an isomorphism by REF . Thus MATH. The fact that MATH is a basis for MATH now completes REF . The inclusion MATH is trivial. Now let MATH so MATH and MATH for some MATH. By REF we can thus find MATH and MATH such that MATH . Using the fact that MATH we then get MATH . Hence MATH, completing REF . By REF we have MATH and MATH. Combined with REF it follows that MATH . By combining REF we finally get MATH completing the result. |
math/0012020 | For the sake of convenience we set MATH. The symmetry of MATH about MATH follows directly from REF . Now let MATH be an admissible space and MATH. Using REF and the assumption MATH, we have that the adjoint of the map MATH is MATH where MATH. Using REF together with the fact that the index of a NAME map changes sign when we take its adjoint (see REF for example) we now get MATH which establishes REF. If MATH then we get MATH by setting MATH in REF. Now suppose MATH. Since MATH consists of isolated points (see REF ) it follows that we can find some MATH such that MATH. REF together with REF then imply the existence of MATH such that MATH for all MATH. By REF we also know MATH where MATH and MATH is the algebraic multiplicity of a given MATH. This completes the result. |
math/0012020 | By REF above we know that MATH is a NAME map with index MATH whenever MATH. On the other hand MATH is a discrete subset of MATH (see REF ) whilst REF imply the function MATH is independent of the admissible space MATH, is constant on the components of MATH, and satisfies MATH for all MATH. It follows that MATH is precisely the set of points at which MATH is discontinuous, whilst MATH for any admissible space MATH and MATH. If MATH for some MATH then the MATH-th row of the matrix operator MATH would be REF. As this contradicts the assumption that MATH is elliptic we get MATH. Coupled with REF and the fact that MATH and MATH are vectors of non-negative integers, it is clear that all the terms in the sum defining MATH are equal to REF when MATH. This completes the result. |
math/0012020 | We have MATH for any MATH and MATH for any MATH. On the other hand, the set of polynomials in MATH variables which are homogeneous of degree MATH can be seen to have dimension MATH. Hence MATH for all MATH. The resulting recurrence relations have the unique solution MATH for all MATH and MATH. A straightforward calculation using REF now shows that if MATH for some MATH then MATH is given explicitly by right hand side of REF or REF when MATH or MATH respectively. The result then follows from REF and the fact that right hand sides of REF are non-increasing functions of MATH which are strictly decreasing outside MATH. |
math/0012021 | Let MATH be a stable model category which admits a MATH-linear equivalence MATH. The image MATH under MATH of the unit object of the smash product is a small weak generator for the homotopy category of MATH. Because the equivalence MATH is MATH-linear, MATH satisfies REF of our main theorem, and so MATH is NAME equivalent to the model category of spectra. Thus the homotopy category of MATH and the category MATH are MATH-linearly equivalent to the ordinary stable homotopy category of spectra. |
math/0012021 | We verify the model category axioms as given in CITE. The category of spectra has all limits and colimits REF , the MATH-equivalences satisfy REF-out-of-REF and the classes of cofibrations, MATH-fibrations and MATH-equivalences are each closed under retracts REF . By definition the MATH-fibrations have the RLP for maps which are both cofibrations and MATH-equivalences. Furthermore a map which is a MATH-equivalence and a MATH-fibration is an acyclic fibration in the stable model structure by REF above, so it has the RLP for cofibrations. This proves the lifting REF . The stable model structure provides factorizations of maps as cofibrations followed by stable acyclic fibrations. Stable acyclic fibrations are in particular MATH-equivalences and MATH-fibrations, so this is also a factorization as a cofibration followed by an acyclic fibration in the MATH-local model structure. REF provide the other factorization REF . |
math/0012021 | The `only if' part holds since the MATH-local acyclic fibrations are level acyclic fibrations, the MATH-fibrant objects are the MATH-spectra with MATH-local homotopy groups (REF above), and MATH-fibrations are in particular level fibrations. For the converse suppose that MATH satisfies REF to REF . We use REF: in order to show that MATH and its adjoint form a NAME adjoint pair it suffices to show that MATH preserves acyclic fibrations and it preserves fibrations between fibrant objects. The MATH-local acyclic fibrations are precisely the level acyclic fibrations, so MATH preserves acyclic fibrations by REF . We claim that every level fibration MATH between MATH-spectra with MATH-local homotopy groups is a MATH-fibration. Given this, MATH preserves fibrations between fibrant objects by REF . To prove the claim we choose a factorization MATH with MATH a cofibration and MATH-equivalence and with MATH a MATH-fibration. Since MATH is MATH-fibrant, so is MATH. Hence MATH is a MATH-equivalence between MATH-spectra with MATH-local homotopy groups, thus a level equivalence. Hence MATH is an acyclic cofibration in the strict model (or level) model structure for spectra of CITE, so that the level fibration MATH has the RLP for MATH. Hence MATH is a retract of the MATH-fibration MATH, and so it is itself a MATH-fibration. |
math/0012021 | Every NAME equivalence between stable model categories induces an equivalence of triangulated homotopy categories. The derived functor of a left NAME functor is also MATH-linear by REF ; if it is an equivalence, then the inverse equivalence is also MATH-linear. So REF implies REF . Now assume REF and let MATH be a cofibrant and fibrant object of MATH which in the homotopy category is isomorphic to the image of the localized sphere spectrum under any MATH-linear equivalence. With this choice, REF holds. Given REF , we may assume that MATH is cofibrant and fibrant and we choose a NAME adjoint pair MATH and MATH as in REF . Since the group MATH is a MATH-module, the functors form a NAME pair with respect to the MATH-local model structure for spectra by REF . By REF the map MATH induced by the left derived functor MATH and the identification MATH is MATH-linear (note that the groups on the left hand side are taken in the MATH-local homotopy category, so that MATH is isomorphic to MATH). Source and target of this map are free MATH-modules, and the generator Id-MATH is taken to the generator Id-MATH. Hence the map MATH is an isomorphism. For a fixed integer MATH, the derived adjunction and the identification MATH provide an isomorphism between MATH and MATH under which MATH corresponds to MATH given by composition with the unit map. For every spectrum MATH the group MATH is naturally isomorphic to MATH, so this shows that the unit map induces an isomorphism of homotopy groups after tensoring with MATH, and REF holds. To conclude the proof we assume REF and show that the NAME functor pair MATH and MATH of REF is a NAME equivalence. Since the group MATH is a MATH-module, the functors form a NAME pair with respect to the MATH-local model structure for spectra by REF . So we show that the adjoint total derived functors MATH and MATH are inverse equivalences of homotopy categories. Note that the right derived functor MATH is taken with respect to the MATH-local model structure on spectra. For a fixed integer MATH, the derived adjunction and the identification MATH provide a natural isomorphism MATH . So the functor MATH reflects isomorphisms because MATH is a weak generator. Hence it suffices to show that for every spectrum MATH the unit of the adjunction of derived functors MATH is an isomorphism in the stable homotopy category. Consider the full subcategory MATH of the MATH-local stable homotopy category with objects those spectra MATH for which MATH is an isomorphism. REF says that the unit map MATH is a MATH-local equivalence, so MATH contains the (localized) sphere spectrum. Since the composite functor MATH commutes with (de-)suspension and preserves distinguished triangles, MATH is a triangulated subcategory of the homotopy category of spectra. As a left adjoint the functor MATH preserves coproducts. By REF above and since MATH is small, the natural map MATH is a MATH-isomorphism of spectra for any family of objects MATH in MATH. Hence the functor MATH also preserves coproducts. So MATH is a triangulated subcategory of the homotopy category of spectra which is also closed under coproducts and contains the localized sphere spectrum. Thus, MATH is the whole MATH-local stable homotopy category, and this finishes the proof. |
math/0012021 | To simplify notation we abbreviate the derived functor MATH to MATH and drop the superscript MATH over the smash product on the homotopy category level. By CITE, the left derived functor MATH is compatible with the actions of the homotopy category of pointed simplicial sets - NAME summarizes this compatibility under the name of `MATH-module functor' CITE. The isomorphism MATH is the special case MATH of a natural isomorphism MATH for a pointed simplicial set MATH which is constructed in the proof of CITE (or rather its pointed analog in CITE). It is important for us that the isomorphism MATH is associative (this is part of being a `MATH-module functor'), that is, that the composite MATH is equal to MATH (as before we suppress the implicit use of associativity isomorphisms such as MATH). In particular the map MATH is equal to the MATH-fold iterate of instances of MATH. Now let MATH be a morphism in the homotopy category of MATH and let MATH represent a stable homotopy element. We have to show that MATH in the group MATH. By the definition of MATH this means proving MATH in the group MATH. Since MATH is an isomorphism we may equivalently show REF after composition with MATH. We note that MATH which is what we had to show. REF use the naturality of MATH. REF uses the defining property of the morphism MATH and the associativity of MATH. |
math/0012021 | By assumption the group MATH is a module over a subring MATH of the ring of rational numbers. Since MATH is a right NAME functor, it satisfies the conditions of REF for MATH. For fibrant MATH, the MATH-th homotopy group of the MATH-spectrum MATH is isomorphic to the group MATH. By the derived adjunction this group is isomorphic to the group MATH, which is a module over the MATH-local endomorphism ring MATH. Hence the homotopy groups of the MATH-spectrum MATH are MATH-local. Thus MATH satisfies the conditions of REF for MATH and it is a right NAME functor for the MATH-local model structure. |
math/0012021 | CASE: For a cosimplicial map MATH the map in MATH is isomorphic to the pushout product MATH CITE of MATH with the inclusion MATH of MATH into MATH. So if MATH is a Reedy cofibration, then MATH is a cofibration in MATH by CITE; hence MATH is a Reedy cofibration. In cosimplicial level MATH, the map MATH is given by the map MATH. If MATH is a Reedy acyclic cofibration, then MATH is an acyclic cofibration in MATH by CITE; hence MATH is also a level equivalence. Suspension then preserves level equivalences between Reedy cofibrant objects by NAME 's lemma CITE. CASE: If MATH is a cosimplicial frame, then MATH is again Reedy cofibrant by REF . A simplicial face map MATH induces an acyclic cofibration MATH by CITE, so MATH is also homotopically constant. CASE: This is the pointed variant of CITE. CASE: If MATH is a Reedy acyclic cofibration, then for every cofibration of pointed simplicial sets MATH the map MATH is an acyclic cofibration in MATH by CITE. By adjointness the induced map MATH is an acyclic fibration of simplicial sets. By NAME 's Lemma CITE, the functor MATH thus takes level equivalences between Reedy cofibrant objects to weak equivalences of simplicial sets. |
math/0012021 | Since MATH is stable there exists a cofibrant object MATH of MATH such that the suspension of MATH in the homotopy category of MATH is isomorphic to the object MATH of REF. By CITE or CITE there exists a cosimplicial frame MATH with MATH. Since MATH is Reedy cofibrant, the map MATH is a cofibration between cofibrant objects in MATH; since MATH is also homotopically constant, these maps express MATH as a cylinder object CITE for MATH. REF-cosimplices of MATH are given by the quotient of the map MATH, hence MATH is a model for the suspension of MATH in the homotopy category of MATH. Since MATH is cofibrant and MATH is fibrant, the isomorphism between them in the homotopy category can be realized by a weak equivalence MATH in MATH. Since MATH is Reedy fibrant and homotopically constant, the map MATH is a Reedy acyclic fibration, where MATH denotes the constant cosimplicial object. Since MATH is Reedy cofibrant, the composite map MATH can be lifted to a map MATH. The lift MATH is a level equivalence since MATH is an equivalence in MATH and both MATH (by REF ) and MATH are homotopically constant. The adjoint MATH of MATH might not be a Reedy fibration, but we can arrange for this by factoring it as a Reedy acyclic cofibration MATH followed by a Reedy fibration MATH, and replacing MATH by the adjoint MATH of the map MATH; by REF the map MATH is a level equivalence, hence so is MATH. Now suppose MATH is a cosimplicial frame and MATH is a cosimplicial map with adjoint MATH. We want to construct a lifting, that is, a map MATH whose composite with MATH is MATH. We choose a cylinder object for MATH, that is, a factorization MATH of the fold map as a Reedy cofibration followed by a level equivalence. The suspension functor preserves Reedy cofibrations and level equivalences between Reedy cofibrant objects by REF , so the suspended sequence MATH yields a cylinder object for MATH. In particular REF-th level of MATH is a cylinder object for MATH in MATH. By CITE the suspension map MATH in the homotopy category of MATH can be constructed as follows. Given a MATH-morphism MATH, one chooses an extension MATH to a cosimplicial map between cosimplicial frames. The map MATH then represents the class MATH. Composition with REF-th level MATH of the level equivalence MATH is a bijection from MATH to MATH. Since MATH is stable, the suspension map is bijective, which means that there exists a cosimplicial map MATH such that the maps MATH and MATH represent the same element in MATH. The map MATH need not be a lift of the original map MATH, but we can find a lift in the homotopy class of MATH as follows. Since MATH is cofibrant and MATH is fibrant, there exists a homotopy MATH from MATH to MATH. Evaluation at cosimplicial level zero is left adjoint to the constant functor, so the homotopy MATH is adjoint to a homotopy MATH of cosimplicial objects. Since MATH is Reedy fibrant and homotopically constant, the map MATH is a Reedy acyclic fibration. So there exists a lifting MATH in the commutative square MATH which is a homotopy from MATH to MATH. Taking adjoints gives a map MATH which is a homotopy from MATH to MATH. Since MATH is a Reedy fibration and the front inclusion MATH is a Reedy acyclic cofibration, we can choose a lifting MATH in the commutative square MATH . The end of the homotopy MATH, that is, the composite map MATH, is then a lift of the original map MATH since MATH. |
math/0012021 | Let MATH be any left NAME functor with an isomorphism MATH, and let MATH be a right adjoint. We construct natural transformations MATH and MATH where MATH and MATH are the NAME pair which were constructed in REF. Furthermore, MATH will be a stable equivalence of spectra for fibrant objects of MATH and MATH will be a weak equivalence in MATH for every cofibrant spectrum. So any two NAME pairs as in REF can be related in this way via the pair MATH and MATH of Construction REF. We denote by MATH the cosimplicial spectrum given by MATH and we denote by MATH the functor between cosimplicial objects obtained by applying the left NAME functor MATH levelwise. The functor MATH is then a left NAME functor with respect to the Reedy model structures on cosimplicial spectra and cosimplicial objects of MATH. We inductively choose compatible maps MATH of cosimplicial objects as follows. Since MATH is a cosimplicial frame, MATH is a cosimplicial frame in MATH. The map MATH is a Reedy acyclic fibration, so the composite map MATH admits a lift MATH which is a level equivalence between cosimplicial frames. The map MATH has the right lifting property for cosimplicial frames, so we can inductively choose a lift MATH of the composite map MATH . We show by induction that MATH is a level equivalence. The map MATH is a weak equivalence in MATH since the other three maps in the commutative square MATH are. The map MATH is a model for the suspension of MATH. Since MATH is stable and MATH is a map between cofibrant objects, MATH is a weak equivalence in MATH. Since MATH and MATH are homotopically constant, the map MATH is a level equivalence. The adjunction provides a natural isomorphism of simplicial sets MATH for every MATH, and we get a natural transformation MATH . By the way the maps MATH were chosen, the maps MATH together constitute a map of spectra MATH, natural in the MATH-object MATH. For fibrant objects MATH, MATH is a level equivalence, hence a stable equivalence, of spectra by REF since MATH is a level equivalence between cosimplicial frames. Now let MATH be a spectrum. If we compose the adjoint MATH of the map MATH with MATH coming from the adjunction unit, we obtain a natural transformation MATH between the left NAME functors. The transformation MATH induces a natural transformation MATH between the total left derived functors. For any MATH in MATH the map MATH is isomorphic to the bijection MATH. Hence MATH is an isomorphism in the homotopy category of MATH and so the map MATH is a weak equivalence in MATH for every cofibrant spectrum MATH. |
math/0012022 | Consider the subcategory of objects MATH in MATH such that MATH is an isomorphism. By the assumptions on MATH, MATH and MATH, both source and target commute with triangles and coproducts. So this subcategory is a localizing subcategory which contains MATH. Since MATH is a small, weak generator it follows that this subcategory is the whole category. This follows from CITE; see also CITE. Now fix any MATH and consider the subcategory of objects MATH in MATH such that MATH is an isomorphism. Again this is a localizing subcategory which contains MATH, and hence is the whole category. Thus, MATH is an isomorphism for any MATH and MATH. |
math/0012022 | Since MATH has an underlying stable monoidal model category, there is an equivalence MATH with all of the properties mentioned in REF . Since the properties of a small, weak generator are determined on the homotopy category level, the image MATH under MATH of the unit object in MATH is a small weak generator of the homotopy category of MATH. Because the equivalence MATH is monoidal and MATH-linear, MATH is isomorphic to the unit and satisfies the hypotheses on the unit in REF . Thus, the homotopy category of MATH, and hence also MATH, is monoidally equivalent to the ordinary stable homotopy category of spectra. |
math/0012022 | The description of the fibrant objects follows from CITE. The description of the fibrations follows from the fact that the positive stable model structure is a localization of the positive level model structure CITE. In a localized model structure the fibrations between fibrant objects are the fibrations in the original model structure. So here they are the positive level fibrations. This statement also follows from the positive variants of CITE or CITE. |
math/0012022 | REF implies REF since the properties of MATH mentioned in REF hold for MATH and are determined by the MATH-linear triangulated homotopy category. REF implies REF because the positive stable model structure is monoidally NAME equivalent to the usual stable model structure on MATH by REF . Since NAME functors induce MATH-linear triangulated functors on the homotopy categories by CITE and CITE and monoidal functors preserve the unit, REF implies REF . Next we show that given REF the simplicial NAME adjoint pair constructed in REF is a NAME equivalence. Since MATH is strong monoidal, MATH. Also, MATH is a weak equivalence, so MATH. The total derived functor MATH is exact by CITE. So MATH where MATH denotes the MATH-th shift of MATH for any integer MATH. This isomorphism and the derived adjunction for MATH and MATH produce the following natural isomorphisms MATH . Since MATH is a weak generator, MATH detects isomorphisms. So to show that this pair is a NAME equivalence we need to show that for any symmetric spectrum MATH the unit of the adjunction MATH is an isomorphism. Consider the full subcategory MATH of such objects. For MATH in homotopy this map is the map MATH induced by MATH. This map of free MATH-modules takes the identity map of MATH to the identity map of MATH. Hence it is also an isomorphism by REF . So MATH is contained in MATH. Since MATH is small, MATH commutes with coproducts by the display above. Hence, since left adjoints commute with coproducts and total derived functors between stable model categories are exact, the composite MATH is an exact functor which commutes with coproducts. So MATH is a localizing subcategory which contains the generator MATH of symmetric spectra. Hence MATH is the whole category. Thus, these derived functors induce an equivalence of homotopy categories. |
math/0012022 | We first construct the functor MATH. Let MATH be a cofibrant desuspension of the unit in MATH with a weak equivalence MATH. Let MATH be the MATH-fold smash product of MATH where MATH. Notice that in general MATH is not cofibrant. For MATH in MATH, define the MATH-th level of MATH to be the simplicial mapping space MATH. The symmetric group on MATH letters acts on MATH by permuting the factors and hence also acts on MATH. The structure map MATH is induced by MATH with MATH defined as MATH . Since the adjoint of MATH is MATH equivariant, this makes MATH into a symmetric spectrum. Here we have used the fact that the simplicial action and the monoidal product commute. MATH is an example of a categorical construction described in CITE. Since MATH is a simplicial model category and MATH is cofibrant for MATH, MATH of a (trivial) fibration is a (trivial) fibration in levels MATH. Since MATH is cofibrant MATH factors as MATH where MATH is the fixed cofibrant replacement of MATH given in the monoidal model structure on MATH. Since MATH is monoidal and MATH is a weak equivalence, MATH is a weak equivalence between cofibrant objects for MATH. Hence MATH takes a fibrant object to a positive MATH-spectrum, which is a fibrant object in the positive stable model structure. Thus MATH takes trivial fibrations to positive trivial fibrations and fibrations to positive fibrations between positive fibrant objects by REF . So MATH is a right NAME adjoint. Next we consider the left adjoint MATH. Using the definition of MATH, MATH is isomorphic to MATH since both corepresent the functor which takes MATH in MATH to the space of MATH-equivariant maps from MATH to MATH. So MATH is isomorphic to the weak equivalence MATH. To evaluate MATH on an arbitrary symmetric spectrum MATH, note that MATH can be built as the coequalizer of the following diagram: MATH . Here one map is induced by the map MATH and the other is induced by smashing MATH with the map MATH which is the adjoint of the identity map on MATH in level one. Since MATH must commute with colimits, MATH is defined as the coequalizer of the diagram: MATH . Again the first map is induced by MATH and the second map uses the fact that the simplicial action and monoidal product in MATH commute to give the isomorphism MATH along with the map MATH. Next we consider the monoidal properties of these adjoint functors. First MATH is lax monoidal; since the simplicial action and monoidal product commute the product of maps induces MATH. These fit together to give a natural map MATH. The unit map MATH is given by sending the non-base point of MATH to the identity map of MATH in simplicial degree zero of MATH. The left adjoint of a lax monoidal functor is automatically lax comonoidal. That is, there are structure maps in the opposite direction of a lax monoidal functor; see the display below. The adjoint of the unit map is an isomorphism MATH. Denote the adjoint pair by MATH and MATH. Then the counit and unit of the adjunction and the lax monoidal structure of MATH give MATH . Here in fact MATH is strong monoidal because this map is an isomorphism. To show this we only need to consider the special case where MATH and MATH for MATH a MATH-space and MATH a MATH-space since the general case follows by using the coequalizer diagrams above. Then MATH . Again commuting the simplicial action and the monoidal product shows this last term is isomorphic via the transformation displayed above to MATH. These monoidal properites also follow from the more general treatment in CITE. Finally, these adjoint functors MATH and MATH are simplicial functors. This follows by various adjunctions from the isomorphism MATH given by the simplicial structure on MATH. |
math/0012022 | If MATH is cofibrant, then MATH also preserves (trivial) fibrations. Hence MATH is a right NAME adjoint functor from MATH to the usual stable model category of symmetric spectra. The statements follow from the same proof as given in REF . |
math/0012022 | Since the (trivial) fibrations in the categories of MATH-modules and MATH-algebras are determined on the underlying category, the restriction of MATH in both cases is still a right NAME adjoint functor to the positive model structure. Since MATH is assumed to be a weak generator by REF , MATH preserves and detects weak equivalences. So by CITE we only need to show that MATH is a weak equivalence for MATH a positive cofibrant object in MATH-modules or MATH-algebras where MATH is a fibrant replacement. Since fibrations are determined on the underlying category, a fibrant replacement as a module or algebra restricts to a fibrant replacement in MATH. Under the given conditions on MATH, if MATH is cofibrant in the positive model category of MATH-modules or MATH-algebras then MATH is cofibrant as a symmetric spectrum. By CITE the identity functor from the positive stable model structure on MATH-modules to the usual stable model structure on MATH-modules is a NAME left adjoint. So if MATH is a positive cofibrant MATH-module then it is a cofibrant MATH-module. Since MATH is assumed to be cofibrant as a symmetric ring spectrum it is cofibrant as a symmetric spectrum by CITE. Hence, by CITE, MATH is cofibrant as a symmetric spectrum. Again by CITE, if MATH is a positive cofibrant MATH-algebra, then it is a cofibrant MATH-algebra. Then by CITE it follows that MATH is cofibrant as a symmetric spectrum. We now show that MATH is a weak equivalence for MATH any cofibrant symmetric spectrum. It then follows that MATH and MATH restrict to NAME equivalences on the positive stable model categories of MATH-modules and MATH-algebras. The proof of REF shows that MATH is a weak equivalence for MATH any positive cofibrant symmetric spectrum. Given a cofibrant symmetric spectrum MATH, choose a positive cofibrant replacement MATH. Since MATH preserves weak equivalences between cofibrant objects and positive cofibrant objects are cofibrant, MATH is a weak equivalence. Then one can choose fibrant replacements and a lift MATH so that MATH. Thus, MATH is a weak equivalence, since MATH preserves weak equivalences between fibrant objects and MATH and MATH are weak equivalences. If MATH is cofibrant then MATH is a right NAME adjoint functor to the usual stable model structures. So the last statement follows similarly. |
math/0012022 | The first statement is a special case of the second with MATH and MATH. Since the weak equivalences and fibrations are determined on the underlying category the restriction of MATH is a right NAME adjoint functor. Since the equivalent conditions in REF hold, MATH preserves and detects weak equivalences between fibrant objects. By CITE it thus suffices to show that MATH is a weak equivalence for cofibrant commutative MATH-algebras where MATH is the given fibrant replacement functor. Since fibrations are determined on the underlying category a fibrant replacement as a commutative MATH-algebra restricts to a fibrant replacement in MATH. We first consider the case where MATH, the sphere spectrum. Let MATH for a positive cofibrant symmetric spectrum MATH. We claim that to show MATH is a weak equivalence it suffices to show that MATH is a weak equivalence for each symmetric power MATH. MATH does not necessarily commute with coproducts, but it does commute up to weak equivalence with homotopy coproducts because it is naturally isomorphic to the identity on MATH. Here the coproduct is a homotopy coproduct because it is created levelwise and each level of each summand is cofibrant. This is our general strategy; we follow the outline of the proof of CITE but there the composite of the adjoints commutes with colimits on the nose and here it may only commute up to weak equivalence with homotopy colimits. But each of the colimits we must consider is in fact a homotopy colimit of symmetric spectra because such homotopy colimits can be computed levelwise CITE. Now consider each summand. Since MATH is a left NAME adjoint on the spectrum level it commutes with colimits and smash products with spaces, so MATH applied to MATH of MATH in MATH is isomorphic to the weak equivalence MATH of the cofibrant object MATH in MATH. Thus it is sufficient to show that MATH is a weak equivalence for MATH because MATH in MATH and MATH are weak equivalences by CITE and the second hypothesis on MATH. The extended power is the homotopy colimit of the MATH action on MATH. So since MATH is a weak equivalence for the positive cofibrant object MATH and MATH commutes with homotopy colimits, we conclude that MATH is a weak equivalence on the extended power. Following CITE, we proceed by building cofibrant MATH-algebras using modified generating cofibrations. Let MATH denote the simplicial MATH-simplex and MATH its simplicial boundary. Let MATH be the simplicial bar construction which is the bisimplicial set with MATH-simplices MATH. Its geometric realization is isomorphic to MATH. The inclusion MATH induces an inclusion of the horizontally constant bisimplicial set MATH into MATH. Define MATH as the pushout of this inclusion over the map MATH. The geometric realization of the composite gives a map MATH with the geometric realization MATH isomorphic to the unreduced cone MATH. We can use these composite maps with MATH instead of the simplicially homotopic maps MATH to construct generating cofibrations. The model category of commutative symmetric ring spectra is cofibrantly generated with MATH a set of generating cofibrations. So we need to show that MATH is a weak equivalence when MATH is a MATH-cell complex CITE. We have shown MATH is a weak equivalence when MATH is built in one stage. We next consider MATH for MATH constructed from MATH by finitely many pushouts. Assume the result for those MATH-algebras built in MATH stages, and consider MATH with MATH built in MATH stages and MATH a coproduct of maps in MATH. Since MATH commutes with colimits and smash products with spaces, it commutes with geometric realization and the bar construction above. If MATH and MATH, then MATH, the geometric realization of the simplicial symmetric spectrum with MATH-simplices the coproduct of MATH copies of MATH and one copy of MATH. Statements analogous to CITE and CITE show that the category of commutative MATH-algebras is tensored over simplicial sets and the underlying symmetric spectrum of the geometric realization of a simplicial commutative MATH-algebra is isomorphic to the geometric realization of the underlying simplicial symmetric spectrum. Since MATH commutes with colimits and converts smash products with spaces to tensors with spaces, MATH commutes with geometric realizations. Hence, MATH . Since tensors with simplicial sets and colimits in MATH are levelwise, the geometric realization is constructed on each level. But the geometric realization of a bisimplicial set is weakly equivalent to the homotopy colimit by CITE. So the geometric realization MATH is weakly equivalent to the homotopy colimit and MATH commutes with the homotopy colimit. So it is enough to show that MATH is a weak equivalence on each simplicial level of MATH. The MATH-simplices here are given by MATH. These MATH-simplices can be constructed in MATH stages, so by induction MATH is a weak equivalence here, as required. Finally, we must consider a filtered colimit of these commutative MATH-algebras built in finitely many stages. Filtered colimits of commutative MATH-algebras are created on the underlying symmetric spectra which in turn are created on each level. Since the maps in question here are constructed from MATH, they are injections. So the colimit is over level injections between level cofibrant objects and it is weakly equivalent to the homotopy colimit. By induction we have shown that MATH is a weak equivalence at each spot in the colimit and MATH commutes with homotopy colimits, so MATH is a weak equivalence on the colimit as well. So we conclude that MATH is a weak equivalence for any cofibrant commutative symmetric ring spectrum MATH (that is, any retract of a MATH-cell complex). For the second statement we consider cofibrant commutative MATH-algebras MATH for MATH a cofibrant commutative symmetric ring spectrum. Since MATH is cofibrant the unit map MATH is a cofibration of commutative symmetric ring spectra. Since MATH is cofibrant as a commutative MATH-algebra, the unit map MATH and hence the composite MATH is also a cofibration. Hence, by the above, MATH is a weak equivalence. |
math/0012022 | For REF , note that MATH is homotopically constant since the smash product of two maps which are each weak equivalences between cofibrant objects is a weak equivalence. The monoidal product also preserves Reedy cofibrant objects. By CITE, since MATH preserves cofibrations and MATH commutes with colimits, the prolongation MATH also preserves cofibrations. This uses CITE to recognize that the Reedy model category on MATH agrees with the Reedy model category on MATH. For REF , the map MATH is isomorphic to the map MATH. By CITE, if MATH is Reedy cofibrant then this map is a cofibration. So MATH is Reedy cofibrant. Since each map MATH is a trivial cofibration, the coface maps of MATH are trivial cofibrations CITE. Since MATH the codegeneracy maps are also weak equivalences. For REF , the coend defining MATH is a colimit of copies of MATH indexed by the non-base point MATH-simplices of MATH. Use the map MATH in MATH determined by the MATH-simplex of MATH to induce a map MATH. These maps are all compatible and define a map MATH. Here MATH and we have used the fact that MATH commutes with colimits. Since REF show that this is a map between cosimplicial frames, it is a level equivalence if degree zero is a weak equivalence. The map MATH is a level equivalence between Reedy cofibrant objects by REF and the monoidal model structure on MATH. Hence MATH is a weak equivalence by CITE. |
math/0012022 | To define the right adjoint MATH we consider cosimplicial resolutions related to MATH. First, let MATH be the constant cosimplicial object on MATH. Since MATH is not necessarily cofibrant, MATH is not necessarily a cosimplicial resolution. Since MATH has a good desuspension of the unit, one can build a cosimplicial resolution MATH of the cofibrant object MATH. Define MATH and MATH. Define the coface maps as the two inclusions MATH and define the codegeneracy map as the map MATH. Using the factorization properties in MATH one can inductively define the higher levels of MATH, see the proof of CITE. Since MATH is the cofiber of MATH, MATH is the cofiber of MATH, that is, a model for the suspension of MATH. Then the weak equivalence MATH extends to a level equivalence MATH. Define MATH for MATH. Define the right adjoint MATH in level MATH to be MATH. The symmetric group on MATH letters permutes the factors of MATH. The structure maps are induced by the map MATH. REF provides a level equivalence MATH. The isomorphism of MATH with the MATH-fold iterated suspension of MATH induces a MATH-equivariant level equivalence where MATH acts trivially on the target: MATH . Applying MATH to the displayed composition and taking adjoints gives the MATH-equivariant structure map MATH . Let MATH denote the constant cosimplicial object on MATH, the chosen cofibrant replacement of MATH. Then since degree zero of MATH factors through MATH, MATH factors as two level equivalences MATH. Hence MATH for any cosimplicial resolution MATH is a level equivalence by the monoidal model structure on MATH. Since MATH for MATH is a cosimplicial resolution by REF , each map MATH with MATH is a weak equivalence. By the pointed version of CITE, MATH preserves fibrations and trivial fibrations when MATH is a cosimplicial resolution and for MATH fibrant MATH takes level equivalences between cosimplicial resolution to weak equivalences. So MATH takes fibrant objects to positive MATH-spectra and (trivial) fibrations to positive level (trivial) fibrations. Thus MATH is a right NAME functor by CITE since positive stable fibrations between positive MATH-spectra are positive level fibrations by REF . The left adjoint, MATH is formed as in the simplicial case except here the tensors of cosimplicial resolutions with simplicial sets are given by coends. To show that the total left derived functor MATH is strong monoidal we first show that MATH is lax monoidal. For the unit map take the non-base point of MATH to the identity map in simplicial degree zero of MATH. The monoidal product on MATH induces a natural map MATH. Assembling these levels produces a natural map MATH. Hence, MATH is lax monoidal. So its left adjoint MATH is lax comonoidal. Also, MATH because they represent the same functor in MATH. So MATH takes the cofibrant replacement MATH to the weak equivalence MATH. The comonoidal structure on MATH induces a natural transformation MATH where MATH is the total left derived functor of MATH. Since MATH, this map is an isomorphism for MATH and any MATH. For fixed MATH both the source and target are exact functors in MATH which commute with coproducts, so the objects MATH where this transformation is an isomorphism form a localizing subcategory which contains the generator MATH. Hence this transformation is an isomorphism for all MATH and MATH. So MATH is strong monoidal. |
math/0012030 | Assume that the structure map MATH is a MATH-equivalence, where MATH is a nondecreasing sequence with infinite limit. The structure maps of MATH then have the same property. An easy diagram chase shows that the MATH-th structure map of MATH factors as the composite MATH . The first map is a MATH-equivalence and for MATH the second map is a MATH-equivalence, by REF below. The composite is therefore a MATH-equivalence. |
math/0012030 | This follows from the NAME suspension theorem applied to MATH and to MATH, together with the factorization of the cited map as the composite MATH where MATH and MATH are the unit and counit of the MATH-adjunction. |
math/0012030 | We are given a convergent symmetric spectrum MATH. We first show that both lemmas hold when MATH is replaced by MATH for any fixed integer MATH. We show that MATH is a symmetric MATH-spectrum, and the map MATH is a MATH-isomorphism. The group MATH is the colimit of the sequence MATH . We can calculate the maps MATH induced by the structure maps MATH as maps of colimits arising from the diagrams MATH . Since MATH is convergent, the vertical maps in each such diagram are eventually isomorphisms and so induce isomorphisms of the colimits. Thus each structure map MATH is a weak equivalence. The map on homotopy groups induced by REF is the map MATH obtained by mapping the terms in the source to the terms with MATH in the target. Extending the diagram above vertically and arguing similarly, we see that this map is an isomorphism. We should note here that the subtlety mentioned in Warning REF appears because the vertical maps here differ from the horizontal maps up to isomorphism, see CITE. But this difficulty is avoided here since MATH is convergent. For REF , since MATH is convergent the colimit, MATH, is attained at MATH for MATH. This directly implies the hypotheses required by REF to show that MATH is a weak equivalence. We now deduce that the lemmas hold for MATH. We first show that MATH is a symmetric MATH-spectrum. Using the maps MATH, we define MATH. The compatible maps MATH induce a level weak equivalence of symmetric spectra MATH. Observe that MATH. Similarly, using the induced maps MATH we define MATH. The compatible maps MATH induce a map of symmetric spectra MATH such that the following diagram commutes: MATH . The top horizontal arrow is a MATH-isomorphism since it is obtained by passage to telescopes from the MATH-isomorphisms REF . The map MATH is a level weak equivalence since MATH is a level weak equivalence. Using REF , we see that MATH is a symmetric MATH-spectrum since each MATH is a symmetric MATH-spectrum. Thus it suffices to show that MATH is a level weak equivalence to conclude both that MATH is a symmetric MATH-spectrum and that the bottom horizontal arrow in the diagram is a MATH-isomorphism, giving REF . On passage to telescopes, the weak equivalences MATH given by REF induce a weak equivalence MATH . This is the MATH-th term of a level weak equivalence MATH. It is easily seen that MATH factors as the composite MATH . Since MATH is a level weak equivalence, so is MATH. Therefore MATH is a level weak equivalence. This completes the proof of REF . Finally, to complete the proof of REF , we observe that the level weak equivalences MATH and MATH induce the maps MATH and MATH displayed in the following commutative diagram: MATH . The top horizontal arrow is a weak equivalence because it is the telescope of the weak equivalences REF . The vertical arrows labelled MATH are homeomorphisms obtained by commuting MATH with MATH. The vertical arrows labelled MATH are weak equivalences induced by weak equivalences of REF . The maps MATH and MATH are weak equivalences since MATH and MATH are level weak equivalences. Therefore the bottom horizontal arrow is a weak equivalence, which is the conclusion of REF . |
math/0012030 | We define the symmetric spectrum MATH to be the (levelwise) homotopy fiber of the map MATH. For each MATH in MATH, we define MATH to be the left vertical arrow in the following pullback diagram: MATH . It is clear from the functoriality of our NAME sections that MATH defines a functor on symmetric spectra and that MATH is natural. Since MATH is a MATH-equivalence, MATH is strictly MATH-connected. Since MATH for MATH, MATH induces an isomorphism on homotopy groups in degrees MATH and above. The maps MATH are induced by the maps MATH, and the equation MATH follows from the corresponding equation MATH. |
math/0012030 | Let MATH be the constant functor that takes the value MATH. By inspection of definitions, MATH is homeomorphic to MATH, and MATH is contractible since MATH has an initial object. Therefore the inclusion MATH is a homotopy equivalence. Applying MATH to the unique map from MATH to each object of MATH, we obtain a natural transformation MATH. By the previous lemma, the induced map MATH is a weak equivalence. Clearly MATH restricts to the inclusion on MATH. |
math/0012030 | Let MATH be the category with two objects MATH and MATH and one non-identity morphism MATH. Then MATH. It is standard that MATH determines and is determined by a functor MATH that restricts to MATH on MATH and to MATH on MATH. Let MATH be the projection. Then MATH also determines a natural transformation MATH. For an object MATH of MATH, MATH and MATH. The required natural homotopy MATH is the composite MATH where the first isomorphism is a direct inspection of definitions. |
math/0012030 | Define a functor MATH by concatenation with MATH; explicitly, MATH and MATH for an injection MATH. Let MATH be the unique map, define MATH, and define MATH for MATH. Then MATH is a natural transformation MATH, MATH is a natural transformation MATH, and MATH. Via two applications of REF and a little diagram chasing, we find that the composite MATH is a homotopy inverse to MATH. |
math/0012030 | It suffices to show that MATH is a MATH-equivalence for each MATH. Choose MATH such that MATH. Then for every object MATH in MATH the map MATH is a MATH-equivalence. By REF , this implies that MATH is a MATH-equivalence. The proposition now follows from REF . |
math/0012032 | REF may be found in any book on basic functional analysis. REF is often stated for functionals MATH but the result as stated follows from this by considering MATH. REF is found in fewer books, the proof is quite obvious from REF . For by REF , MATH is clearly w*-closed in MATH, and the restriction of MATH to MATH then takes w*-closed (and thus w*-compact) sets to w*-compact (and thus w*-closed) sets in MATH. Thus the inverse of MATH restricted to the ball is w*-continuous, so MATH is w*-continuous by REF . |
math/0012032 | If MATH is an oplication, consider the associated linear map MATH. Note that MATH for MATH. The last matrix may be regarded as the formal product, via MATH, of the diagonal matrix MATH and the column in MATH with entries MATH and MATH. By the hypothesis, one clearly sees that MATH is contractive, and an analogous argument with matrices shows that it is completely contractive. Thus by REF we see that MATH is a contractive map into MATH. Given MATH, we consider MATH as a map MATH, and argue as above using matrices to see that MATH. Thus it follows that MATH is a complete contraction from MATH. Alternatively the complete contractivity may be deduced from the contractive case, and the relation MATH. It is clear that MATH for all MATH. The last few statements are immediate (see REF). |
math/0012032 | Proceed exactly as in the proof above to obtain a complete contraction MATH such that MATH. One does need to permute the entries of the matrices involved in the calculation as to get them into the form of the hypothesis, but this is a simple matter. The result may then be completed as in our REF-line proof of BRS from REF: for example MATH. |
math/0012032 | We will use NAME. Let MATH be a net converging weak* in MATH to MATH, say. Using REF we will now check that MATH is in MATH. For suppose that MATH. Write MATH as a column MATH. Then MATH by REF. Let MATH. Consider the operator space predual MATH of MATH; any MATH is given by a pair MATH of functionals in MATH. The duality pairing is: MATH . Since MATH in the limit we have that MATH . Thus MATH is contractive. A similar argument, picking MATH in the unit ball of the predual of MATH, and using REF , shows that MATH is completely contractive. Thus MATH, so that MATH is w*-closed. |
math/0012032 | (Of REF :) By symmetry it is enough to show that the multiplication is weak*-continuous in the first variable. One can see this symmetry by considering the `opposite algebra' MATH (compare REF ) which has operator space predual which is MATH with the transposed matrix norm structure. From the lemma it is evident that the map MATH is weak*-continuous with respect to the weak*-topology of the Lemma. For, if MATH w*- in MATH, then MATH w*- in MATH. By REF , MATH is weak*-continuous. But this says exactly that the multiplication on MATH is weak*-continuous in the first variable. |
math/0012032 | Take any operator space MATH which is the dual of a NAME space MATH, such that MATH with the associated w*-topology derived from MATH is not completely isometrically isomorphic via a w*-w*-homeomorphic linear map to a dual operator space (see CITE for such an example, and CITE for an example which has a unique NAME space predual). Build the `canonical MATH unital operator algebra' MATH which has MATH contained completely isometrically as REF-corner and scalars on the diagonal. This is an adaption of an example from CITE where NAME puts zeroes on the diagonal. Suppose that MATH is any isometric w*-homeomorphic linear embedding (such embeddings exist since any NAME space is a quotient of a MATH). Then MATH may be identified isometrically with a subalgebra MATH of MATH via the obvious map MATH . Here MATH is the identity of MATH. That this is an isometry follows for example from CITE REF . It is clear that this subalgebra MATH is w*-closed, so that MATH is a dual NAME space. Moreover the inclusion of MATH into MATH as the corner, is a w*-homeomorphism. That is, MATH is completely isometrically isomorphic to a w*-closed subspace of MATH. Therefore if MATH with its given weak*-topology was w*-homeomorphic and completely isometric to a MATH-weakly closed subspace of some MATH, then so would MATH be, which is false. Finally, suppose that MATH is as in the example at the end of CITE, an operator space which has a unique NAME space predual, but no operator space predual. The MATH constructed above is a dual NAME space. If MATH were a dual operator space, there would exist by REF , a completely isometric unital homomorphism MATH of MATH onto a concrete dual operator algebra MATH. However it is very easy to check that the diagonal idempotents in MATH force the NAME space on which MATH acts, to split as a direct sum MATH, such that MATH may be written as MATH matrices exactly like our original MATH. Moreover, since there are projections onto MATH and MATH, REF-corner of MATH is w*-closed in MATH. Also, MATH restricted to the copy of MATH, maps MATH completely isometrically onto this MATH corner of MATH. But this forces MATH to be a dual operator space. |
math/0012032 | We may assume that MATH for compact MATH, by the usual argument (if MATH for locally compact MATH then NAME implies the existence of extreme points. Any such extreme point, by a simple application of NAME 's lemma has constant absolute value REF, so that MATH is compact). Replace the argument above for REF by an almost identical argument in MATH instead of MATH, and using the very easy REF instead of our REF , to obtain that the multiplication on MATH is separately w*-continuous. Then we may use NAME 's proof of REF from CITE to finish. Alternatively, there is a commutative proof using measure theory which avoids NAME 's argument. We will omit this since this result is quite well known. |
math/0012032 | We know that MATH is a unital operator algebra, and it is easy to check that the natural map MATH is a completely isometric homomorphism. Let MATH. It is easy to see that MATH, and that MATH is a unital subalgebra of MATH. Clearly MATH contains MATH as a left ideal. Therefore by REF, MATH is a complete left NAME in MATH. By REF, MATH is a complete left NAME of MATH. Thus MATH for a projection MATH. Clearly MATH, so that MATH, from which it follows that MATH is a right identity for MATH. |
math/0012032 | REF is undoubtedly well known, we give a proof since we have no reference for it. Thanks go to NAME for helping to simplify my original argument. That REF implies REF is clear (as indeed is the converse of REF , by the way). Given REF , we see by the Principle of Uniform Boundedness that MATH and therefore MATH is bounded. We may assume without loss of generality that MATH is a contraction. Suppose that MATH is the predual of MATH, and let MATH, a linear subspace of MATH. The canonical map MATH is REF and contractive and MATH . On the other hand, given MATH let MATH for MATH, then MATH. If MATH for a MATH then we'd be done, for in this case its clear that MATH. So suppose, by way of contradiction that MATH. By REF is w*-closed, so by the NAME theorem, there exists MATH such that MATH and MATH for all MATH. The latter condition implies that MATH, whereas the former condition implies the contradictory assertion that MATH. That MATH is w*-continuous with respect to this predual of MATH is now clear. Now we prove REF . We will use the fact (see CITE) that if an operator space MATH is the dual of a NAME space MATH, and if MATH is equipped with its natural matrix norms as a subspace of MATH via the natural inclusion, then MATH is the dual operator space of MATH if and only if the unit ball of MATH is MATH-closed. To check that the latter condition holds, let MATH be a net in MATH, converging to MATH in MATH, the convergence being the MATH-convergence. That is, MATH for all MATH and MATH. Equivalently, MATH for all MATH. By REF , the matrices MATH converge w*- to MATH in MATH. By hypothesis, MATH, so that MATH, and we are done. |
math/0012032 | CASE: Using REF, we need to show that if MATH is a net in MATH converging w*- in MATH to MATH, then MATH. We will use the fact from REF that MATH, and we will test for MATH using REF. So we begin with two matrices MATH (rows indexed by MATH, columns indexed by MATH), and we need to check that MATH where MATH is the canonical completely isometric identification. However we do know that MATH . And of course MATH weak* in MATH for all MATH and MATH, by hypothesis. The first matrix in the last displayed equation is an element of MATH. Now pick a norm REF functional MATH in the predual of the MATH, and apply it to this first matrix in displayed equation. Using REF as in the proof of REF shows that indeed MATH. Thus we have now proved REF . REF follows from REF and the definition of the weak*-topologies concerned (see proof of REF). REF follows from REF . Finally, the assertions of REF are now fairly obvious. If MATH, with MATH in multiplier norm, then MATH in cb-norm. Since the canonical image of MATH in MATH is norm closed, we see that MATH is norm closed. |
math/0012032 | From the above we now know that we can represent MATH as a w*-closed subalgebra MATH of some MATH, with MATH. Then the set of adjoints of operators in MATH is a w*-closed subalgebra of MATH, and MATH. So MATH is w*-closed in MATH, and consequently is a dual space. |
math/0012032 | We may assume that MATH in the MATH-algebra MATH. By basic operator theory, any such MATH is REF-corner of an orthogonal projection MATH. However MATH. Moreover the left adjointable projections are exactly the complete left MATH-projections, which are shown in CITE to be weak* continuous (the argument for this is quite simple and very similar to the classical one). Thus MATH is weak*-continuous. We have MATH, where MATH and MATH are the canonical maps. Since these latter maps are clearly weak*-continuous, so is MATH. |
math/0012032 | First apply REF to the MATH action on MATH (using REF ). This provides the NAME spaces MATH and the map MATH. We also obtain a normal *-homomorphism MATH, with MATH for all MATH. Next apply the oplication theorem, to get a unique unital completely positive MATH such that MATH for all MATH. The last assertion in the statement of our theorem follows also from the oplication theorem. It is elementary to check from the definition of the w*-topology on MATH from REF, the fact that MATH is weak*-continuous in the first variable, and REF of the NAME theorem, that MATH is weak*-continuous. Let MATH . Finally, if MATH, and MATH w*- in MATH, then MATH w*- in MATH. That is, MATH (since MATH is a w*-w*-homeomorphism). Thus MATH is weak*-continuous in the second variable. |
math/0012032 | This follows immediately from the oplication theorem which provides us with a unital MATH such that MATH for all MATH. As in REF MATH is weak*-continuous. A similar argument proves REF . |
math/0012032 | By REF there is a unique w*-continuous unital complete contraction MATH such that MATH. If MATH is left normal then the proof proceeds exactly as in REF, but using REF instead of REF, to obtain the result. So we assume henceforth that MATH is separately weak*-continuous. Let MATH, a closed unital subalgebra of the dual operator algebra MATH. On MATH we consider the relative weak*-topology inherited from MATH. Consider the natural operator module action MATH. Then MATH is separately weak*-continuous. If one follows through the proof of REF carefully one finds that, even without assuming that MATH is a dual space, we have a weak*-homeomorphic completely isometric embedding MATH, and a completely contractive unital homomorphism MATH, such that MATH for all MATH . Moreover for any bounded net MATH weak*- in MATH, we have MATH weak*- in MATH. Let MATH. Then MATH restricted to MATH is w*-continuous, so MATH is w*-continuous. The rest is clear. |
math/0012032 | If MATH is not left normal, then the action of MATH on MATH is only weak*-continuous in the left variable. Conversely, if MATH is left normal, and we are given a left module action of a dual operator algebra on MATH which is weak*-continuous in the left variable, then it is weak*-continuous in the second variable exactly as in REF, but using REF instead of REF. |
math/0012032 | The idea is the same as for the proof of REF. We first observe that the MATH action on MATH (which clearly makes MATH a normal dual MATH-bimodule by REF) has such a representation. To see this use REF that there exist such MATH as above, MATH, such that MATH for all MATH, MATH. Then, since MATH we have that MATH maps into MATH. Now proceed as in REF. We leave the omitted details to the reader. |
math/0012032 | Suppose that MATH is norm dense in MATH. By the result three paragraphs above REF, it follows that MATH is norm dense in MATH. It is easy to see that MATH commutes with MATH inside the multiplier algebra of the MATH-algebra we called MATH in CITE. We refer to that paper, particular REF there, for background on what follows. Since MATH is norm dense in MATH, it follows from the definition of MATH, that the latter will be a commutative MATH-algebra. Note also, that since MATH is also singly generated as a right MATH module, if MATH, then by REF , MATH is strictly positive in MATH. Hence the set MATH is dense in MATH, by NAME. By REF , we may write MATH, for some MATH. Since MATH is dense in MATH (see the beginning of REF), the map MATH from MATH, is an isometric MATH-module map onto a dense subset of MATH. Hence it extends to a completely isometric surjective MATH-module map MATH say, from MATH. Thus MATH and therefore also MATH, are NAME spaces with the MIN structure. In this case we are in the setting of CITE and REF. That is, MATH is a topologically singly generated function module over MATH. We see from REF that MATH is algebraically singly generated iff MATH is e.n.v. as a NAME space, and in this case MATH is actually a commutative unital MATH-algebra. Finally, if MATH is a dual space and MATH is w*-dense, then the above shows that the norm closure of MATH is a MIN space. However the weak*-closure of a MIN space is also a MIN space (we leave this fact as an exercise), so that MATH is a MIN space. |
math/0012034 | : Differentiating REF and using REF we derive: MATH . Setting MATH and using REF we write MATH or since MATH . Therefore, MATH . Finally we write MATH where MATH denote the terms above which are cubic in MATH with coefficients depending only on the structure functions MATH and their derivatives MATH with respect to the frame. This is precisely REF . |
math/0012034 | : We start with REF to which we apply the projection MATH. Therefore, MATH . The proof of REF is an immediate consequence of the following Lemmas We have, MATH where MATH. The term MATH is an acceptable error term. |
math/0012034 | : We start with the equation MATH . We need to find the potential MATH such that MATH is small. Clearly MATH . Thus according to REF and the constancy of the structure and connection coefficients, MATH with MATH a matrix whose entries are quadratic in MATH. Henceforth, MATH . We now estimate, with the help of REF with MATH. MATH . Also, proceeding in the same way, MATH . Therefore all the components of the exterior derivative MATH verify the estimates MATH . Proceeding in precisely the same manner we find that MATH . We define MATH by requiring that the spatial components of MATH verify the equation, MATH . Consider now the divergence -curl system, MATH . By standard elliptic estimates, taking into account the fact that the NAME support of MATH is included in the dyadic region MATH and using REF we infer that, MATH . On the other hand we also have good estimates for MATH. In view of the divergence condition MATH we derive MATH, with MATH the NAME in MATH, MATH. Therefore using standard elliptic estimates and REF we infer that, MATH . We have thus derived the estimate REF. The estimate REF follows in the same manner from REF. We now estimate MATH. We first observe that the divergence equation MATH takes the form MATH. This uniquely defines MATH and we have, MATH which gives REF. To prove REF we write MATH. Therefore, in view of REF, MATH establishing the estimate REF. To end the proof of REF it remains to observe that since each MATH is antisymmetric so is the MATH. We also need to check that the terms MATH generated when we pass from REF to REF are indeed error terms. We have, using REF MATH as desired. |
math/0012034 | : The first part of the proposition is an easy consequence of the identity REF as well as the estimates REF. To prove the crucial second part we write MATH . We estimate MATH using NAME and NAME as follows MATH . Thus, picking MATH sufficiently large, MATH . To end the proof of REF it suffices to prove that for all MATH we have MATH where MATH . Now using the definition MATH, MATH . Therefore, using REF as well as REF MATH as desired. The estimates REF of the proposition are immediate consequences of the estimates REF. REF can be derived immediately from REF. |
math/0012040 | MATH-jet of the second component is MATH which is equivalent to MATH or MATH. Consider the first case. Let MATH be the minimal number such that the MATH-jet is not MATH (if k is infinity then the pair is not simple). Then the multigerm is MATH-determined and is equivalent to MATH. In the second case the multigerm is MATH-determined and we obtain the normal form MATH. |
math/0012040 | Suppose the MATH-jet of the second component is MATH. Let MATH be the minimal number such that the MATH-jet is not MATH. Suppose MATH is odd and is equal to MATH. Then the MATH-jet is equivalent to MATH. If there are no more nontrivial monomials then we obtain the normal form MATH since it is at most MATH-determined. Note that a complete transversal in MATH is trivial for odd MATH. Therefore a nontrivial complete transversal, for some MATH, is MATH . If MATH then we obtain the normal form MATH. If, for all MATH, the MATH-jet equals the MATH-jet then we obtain the normal form MATH, otherwise, for some MATH in MATH, a complete transversal lies in MATH . If MATH then we obtain the normal form MATH. Let MATH. The tangent space to MATH contains the vectors MATH, MATH, MATH and MATH. Therefore we obtain the normal form MATH. Consider the case when MATH is even and is equal to MATH, that is, when the MATH-jet is equivalent to MATH. Let MATH be the minimal order of jet which is different from MATH. In MATH it is equvalent to MATH . If MATH then the second component is equivalent to the normal form MATH. Suppose MATH. Then the MATH-jet is equivalent to MATH. If the higher jets are equal to the MATH-jet then we obtain the normal form MATH. In the other case there exists MATH where a complete transversal is MATH . Note that the tangent space to the MATH-orbit contains the vectors MATH, MATH and therefore the vector MATH. It gives the restriction MATH. If MATH then we obtain the normal form MATH. Otherwise the second component is equivalent to the normal form MATH since the tangent space to the MATH-orbit contains the vectors MATH and MATH, MATH. Suppose that the MATH-jet is MATH. If, for all MATH, the MATH-jet is equal to the MATH-jet then the multigerm is not simple. Otherwise the second component is equvalent to the normal form MATH or MATH. |
math/0012040 | In MATH a complete transversal is trivial. In MATH it is equivalent to MATH . We can eliminate the monomials in the other co-ordinates by simple permutation of co-ordinates and left equivalences. If MATH we obtain the MATH-jet MATH. This jet is MATH-determined. A complete transversal in MATH is trivial. If MATH a complete transversal in MATH is equivalent to MATH . If MATH we obtain the normal form MATH. If MATH and MATH we obtain MATH if MATH or MATH REF if MATH. In the last case, if MATH we obtain MATH. The tangent space to the MATH-orbit contains the vectors MATH and MATH, so, using the NAME Lemma, we obtain the normal form MATH. If MATH and MATH our jet is equivalent to MATH. We have to move to higher jets. A complete transversal in MATH, where MATH, is equivalent to MATH since the tangent space to MATH-orbit contains the vectors MATH and MATH. If MATH we obtain the normal form MATH. If MATH and MATH we obtain MATH. If MATH for every MATH such that MATH we obtain MATH. If in MATH, for every MATH such that MATH we obtain MATH. Similar reasonings with the MATH-jet MATH produce the normal forms MATH - MATH and MATH - MATH. Let's now suppose that MATH and MATH in MATH. We obtain the MATH-jet MATH. This jet is MATH-determined. A complete transversal in MATH is trivial. If MATH then a complete transversal in MATH is equivalent to MATH . If MATH we obtain the normal form MATH. If MATH and MATH we obtain the MATH-jet MATH. If MATH we have the MATH-jet MATH. Since the tangent space to the MATH-orbit contains the vectors MATH and MATH (here we write both components), we obtain MATH, that is, the normal form MATH. If MATH we move to the higher jets. Consider a complete transversal in MATH, where MATH. Since the tangent space to the MATH-orbit contains the vectors MATH, MATH and MATH, a complete transversal is equivalent to MATH . If MATH we obtain the normal form MATH. If MATH, for any MATH such that MATH we obtain MATH. If in MATH, for any MATH such that MATH we obtain MATH. Similar reasonings with the MATH-jet MATH produce the normal forms MATH - MATH. |
math/0012040 | Let MATH be the MATH-dimensional space MATH . Let MATH be a point of MATH with nonzero co-ordinates: MATH. For a group MATH acting on the MATH denote by MATH the intersection MATH, where MATH is the tangent space to MATH and MATH is the tangent space to the MATH-orbit of the point MATH. Let's evaluate the dimension of MATH. The group MATH is the product of MATH and MATH. If MATH, where MATH is a MATH-change and MATH is a point of MATH then, for any MATH-change MATH, the point MATH lies out of X. Moreover if MATH is a point of MATH then the points MATH and MATH lie in MATH too. It means that a vector MATH from MATH is the sum of some vectors MATH and MATH. Therefore MATH is the sum of MATH and MATH. MATH is the MATH-dimensional space and includes the vectors MATH and MATH, since these vectors are the linear combinations of MATH and MATH. The preimage of MATH with respect to the tangent mapping MATH is MATH . Therefore the intersection MATH is at least MATH-dimensional (it contains the vectors MATH, MATH, MATH) and the dimension of MATH is not more than MATH. This implies that any open neighbourhood of a point MATH intersects with infinite number of different orbits. |
math/0012040 | First note that the second component is at most MATH-determined and a complete transversal in MATH and MATH is trivial. A complete transversal in MATH is MATH. If MATH then the MATH-jet is equivalent to MATH. Therefore we obtain the normal forms MATH and MATH. Suppose MATH. Then the MATH-jet is equivalent to MATH. In MATH a complete transversal is MATH. If MATH then we obtain the normal form MATH. Suppose MATH and MATH then the second component is equivalent to the normal forms MATH or MATH. If MATH and MATH then there are two possibilities: the normal form MATH or MATH. |
math/0012040 | A complete transversal in MATH is MATH . Consider MATH. Then the multigerm is REF-determined and we obtain the normal forms MATH - MATH . Suppose MATH and MATH. A complete transversal in MATH is MATH which is equivalent to MATH since the tangent space to the MATH-orbit contains the vectors MATH, MATH, MATH and MATH. If MATH then the multigerm is REF-determined and we obtain the normal forms MATH and MATH. Suppose MATH. A complete transversal is not trivial only in MATH and MATH. Since MATH is equivalent to MATH (the tangent space to MATH includes the vectors MATH and MATH) we obtain the normal forms MATH, MATH and MATH. Consider the case MATH and MATH. As above a complete transversal in MATH is equivalent to MATH. If MATH then we obtain the forms MATH and MATH. Suppose MATH, then the MATH-jet is equvalent to MATH. Note that a complete transversal is not trivial only in MATH and MATH. In MATH it is equivalent to MATH. If MATH we obtain the normal form MATH. Consider MATH. There are two cases: MATH and MATH. In the first REF complete transversal in MATH is equivalent to MATH and we obtain two normal forms: MATH and MATH. In the second case we obtain the MATH-jet MATH which produces the normal forms MATH and MATH. |
math/0012040 | As in REF consider the MATH-dimensional space MATH: MATH . In a point MATH with nonzero co-ordinates MATH the dimension of MATH is MATH and the dimension of MATH is MATH. Therefore the dimension of MATH is not more than MATH. From this it follows that, for any neibourhood MATH of MATH, the number of orbits which intersects MATH is infinite. |
math/0012040 | In MATH a complete transversal is MATH . Let MATH. A complete transversal in MATH is MATH . If MATH, then from REF it follows that there are two simple components MATH and MATH. Consider MATH. Then the MATH-jet is equivalent to MATH. It produces the forms MATH, MATH and MATH. Let MATH and MATH. Then the MATH-jet is equivalent to MATH and a complete transversal in MATH lies in MATH . Note that if MATH, then the pair is adjoint to the pair of REF and therefore is not simple. So the MATH-jet is equvalent to MATH. From REF we obtain that a simple component with such a MATH-jet is equivalent to MATH. Let MATH. A complete transversal in MATH lies in MATH. Suppose MATH, then the component is adjoint to the component of REF . So MATH and the MATH-jet is equvalent to MATH. From REF it follows that the MATH-jet is equivalent to MATH, what produces the last three forms. |
math/0012040 | The dimension of the space MATH: MATH is MATH. At the point with nonzero co-ordinates the dimensions of MATH is not more than MATH and that of MATH is MATH. Therefore the dimension of MATH is less than MATH and the pair is not simple. |
math/0012040 | From REF it follows that the MATH-jet of the second component is equivalent to MATH. A complete transversal in MATH lies in MATH . If MATH then we obtain the first three normal forms. Otherwise the MATH-jet is equivalent to MATH. REF implies that the MATH-jet is equivalent to MATH and we obtain the normal forms MATH and MATH. |
math/0012040 | Let 's consider the tangent space to the MATH-orbit. It is generated by the following MATH vectors: MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH. Note, that MATH . So, these vectors are linearly dependent and we can remove MATH from the list above. Hence it is obvious that the tangent space to the MATH-orbit does not contain the vector MATH from the tangent space to our REF-dimensional submanifold. But if the tangent space to the MATH-orbit does not contain the tangent space to the submanifold, for each point of it, the multigerm fails to be simple. |
math/0012040 | We shall not calculate tangent spaces in obvious cases, but in other cases we have to do it. In MATH a complete transversal is trivial for any MATH. In MATH it is equivalent to MATH . There exists MATH such that MATH (we suppose that the multiplicity of the first component is less than or equals the second one). If MATH then we obtain MATH. It is clear, that this multigerm is MATH-determined, that is, we obtain MATH, MATH. Now suppose MATH. If MATH then using multiplications of co-ordinates in the target and parametrs in the sources by constants we obtain MATH . If MATH we obtain MATH . We shall discuss REF later. Now consider REF . In MATH a complete transversal is trivial for any MATH. In MATH a complete transversal is given by MATH . If MATH then we obtain MATH. If MATH and MATH we obtain the MATH-jet MATH . Starting from this moment, it is not important for our consideration that MATH, so we can also consider REF . A complete transversal in MATH is trivial. Let 's consider it in MATH. We can simply obtain the following vectors: MATH and MATH. Note, that the tangent space contains the vector MATH, the power of the last monomial is greater than or equals MATH. So, the tangent space contains MATH. Now, we want to obtain the vector MATH. It is obvious, if we note that the tangent space contains MATH, MATH and MATH, the power of the last monomial is greater than MATH. After it we can conclude that a complete transversal is equivalent to MATH . Now we shall prove, for every MATH that this MATH-jet is MATH -determined. If, using left equivalence, we eliminate a monomial with even degree in one component, then we obtain a monomial of higher degree in another component, so we have to eliminate monomials of odd degree. Let 's consider the MATH-jet MATH of our multigerm. We shall eliminate the monomial MATH in the second component. Our construction does not depend on MATH in REF . We can obtain this monomial as MATH. In the first component we obtain MATH. Now we have two possibilities: MATH or MATH. In the first case we consider our multigerm in MATH and our monomial in the first component is MATH. It can be obtained as MATH. So, in the second component we obtain a monomial of degree MATH. In the second case we consider our multigerm in MATH. The monomial in the first component is MATH, it can be obtained as MATH. So, in second component we obtain a monomial of degree MATH. In this construction we did not use that MATH, so it is suitable for eleminating the monomials in the first component as well. Therefore, in REF , if MATH we obtain the normal form MATH, if MATH we obtain the normal form MATH. |
math/0012040 | In MATH a complete transversal is equivalent to MATH . If MATH then we obtain the normal form MATH which is MATH-determined. If MATH and MATH we have MATH . One can obtain the vector MATH as a linear combination of MATH, MATH and MATH. Then we can use the NAME Lemma and obtain the normal form MATH, which is MATH-determined. If MATH we obtain the normal form MATH which is also MATH-determined. |
math/0012040 | Using REF , we have only one possibility for REF-jet MATH. Any REF-jet of this curve is equivalent to MATH . The tangent space to MATH-orbit contains the vectors MATH (so we can suppose MATH), MATH, MATH, MATH, MATH (so we can suppose MATH), MATH, MATH and MATH (so we can suppose MATH). Hence, if MATH we obtain the first curve, if MATH we obtain the second curve. Note, that both this REF-jets are REF-determined. |
math/0012040 | Simple calculations in MATH show that the tangent space to the MATH-orbit does not contain the vector MATH, so MATH is a module in our family. |
math/0012040 | A complete transversal in MATH is equivalent to MATH . Let 's suppose that MATH. Then our REF-jet is equivalent to MATH. This jet is REF-determined. The tangent space to its MATH-orbit contains the vectors MATH, MATH and MATH, so we obtain the curves MATH - MATH. Now let us consider the case MATH. If MATH this REF-jet is equivalent to MATH . If MATH this REF-jet is equivalent to MATH . If MATH we obtain MATH . Now we have to consider REF-jet. Using REF we see that in the second component the fourth co-ordinate equals MATH. In REF we obtain the normal forms MATH and MATH, which are REF-determined. In REF we see that a complete transversal in MATH is trivial and this curve is REF-determined. So we obtain the normal forms MATH, MATH, MATH and MATH. |
math/0012040 | A non-regular curve is ajacent to the curve MATH. That is why the first curve is adjacent to MATH, the second one to MATH and the third one to MATH. Note that the third curve in the triple is ajacent to MATH therefore the triple is ajacent to MATH . By changing indexes of axis and curves we can obtain the triple MATH . The second and the third curves are ajacent to curves with REF-jet MATH. So it is remaines to prove that a triple with the MATH-jet MATH is not simple. NAME we have to prove the following statement. |
math/0012040 | Consider the MATH-dimensonal subspace of the space of REF: MATH . Using MATH-changes each such triple can be reduced to the same form with MATH . Note, that if one can reduce REF-jet to the form, where MATH, then one can do it using only the following changes: MATH, MATH, MATH and MATH. Hence we have the following equations for MATH: MATH . Therefore we have MATH. Then MATH and MATH. And it implies that MATH. So REF-jet under consideration has a continious invariant MATH, that is, it is not simple. |
math/0012040 | Consider REF-jet of our multigerm. Suppose that first MATH curves can be (and are) reduced to the form MATH and REF-jets of all other curves have zero co-ordinates with number greater than MATH (otherwise one can reduce MATH curves to the form MATH). If the total number of curves is equal to MATH, we obtain the normal form REF Suppose that MATH. Then our multigerm can not contain more than MATH components. The reason is that MATH-lines in MATH is not a simple multigerm. For MATH they have a continious invarian - double ratio. For MATH consider the space of REF of MATH lines in MATH. It is MATH-dimensional. The dimension of the tangent space to the MATH-orbit is less than or equals MATH. The dimension of the tangent space to the MATH-orbit is less than or equals MATH. For our multigerm to be simple it is necessary to have MATH that is, MATH. Hence MATH. So, we have MATH and a non-singular curve with zero co-ordinates with numbers greater than MATH. By MATH-transformations and permutations of curves and co-ordinates we can reduce REF-jet to MATH and MATH, where MATH. If MATH we obtain the normal form MATH (since MATH equals MATH for the last curve and zero for first MATH curves, MATH). If MATH we move to higher jets. A complete transversal in MATH is equivalent to MATH . If MATH we obtain the normal form MATH. If MATH and there exists MATH such that MATH we obtain the normal form MATH (since it is MATH-determined). Now suppose MATH. There are at least REF components in our multigerm. If the multigerm has at least REF components, then it fails to be simple, since its REF-jet is adjacent to REF lines in MATH. So, the multigerm contains REF components. Note, that by REF the family MATH is not simple. So we have only one possibility for REF-jet: MATH jet is REF-determined and we obtain the normal form MATH. |
math/0012040 | A complete transversal in MATH is trivial for any MATH. A complete transversal in MATH is equivalent to MATH . If MATH we obtain the normal form MATH. If MATH but there exists MATH such that MATH we obtain the normal form MATH (MATH depends on the number of coefficients MATH different from zero). |
math/0012040 | The normal form MATH is REF-determined since MATH. Consider REF-jet MATH. It is REF-determined so we have to consider REF-jet. A complete transversal in MATH is equivalent to MATH . If MATH we obtain the normal form MATH which is REF-determined. If MATH we can eliminate MATH in the first and MATH-st co-ordinates, since the tangent space to the MATH-orbit contains the vectors: MATH, MATH and MATH. Hence we obtain the normal form MATH (MATH depends on the number of coefficients MATH different from zero). |
math/0012040 | The dimension of the submanifold of such jets (in MATH)equals MATH. Now we shall estimate the dimension of the stabilisator of the regular part that is, the subgroup in MATH which preserves the regular part. The dimension of the tangent space to the MATH-orbit is less than or equals MATH, since it is generated by the images of the vectors MATH with MATH. The dimension of the tangent space to the MATH-orbit is less than or equals MATH, since it is generated by the images of two vectors: MATH and MATH. Hence the dimension of the tangent space to the MATH-orbit is less than or equals MATH. If the dimension of the tangent space is less than the dimension of the submanifold then this REF-jet fails to be simple. Hence MATH that is, MATH. |
math/0012040 | We shall describe all simple multigerms with REF-jet MATH. A complete transversal in MATH is trivial for any MATH. In MATH it is equivalent to MATH since the tangent space to the MATH-orbit contains the vector MATH. If MATH we obtain the normal form MATH since it is MATH-determined. If MATH and MATH we obtain the MATH-jet MATH which is MATH-determined, since MATH is equal to MATH for the singular component and to zero for both regular components. Hence we have to move to the MATH-jet. A complete transversal in MATH is equivalent to MATH . This is so since the tangent space to the MATH-orbit contains the vectors MATH, MATH, MATH and MATH. MATH is equal to MATH for the singular component and to zero for both regular components. If MATH we obtain the normal form MATH, if MATH we obtain the normal form MATH. |
math/0012040 | Consider a complete transversal in MATH. It is equivalent to MATH . Suppose MATH. We obtain the MATH-jet MATH. This jet is MATH-determined since MATH is equal to MATH for the singular curve and to zero for the regular ones. This jet is not enough, so we have to move to higher jets. A complete transversal in MATH is trivial for any MATH. Consider it in MATH. It is equivalent to MATH . If MATH we obtain the normal form MATH. If MATH and MATH we obtain the normal form MATH if MATH. If MATH we can eliminate MATH in the third co-ordinate, since the tangent space to the MATH-orbit contains the vectors MATH and MATH. Hence we obtain the normal form MATH. If MATH and MATH we also obtain the normal form MATH. Now suppose MATH and MATH. We obtain the MATH-jet MATH, where MATH. We have to consider a complete transversal in MATH. It is equivalent to MATH since the tangent space to the MATH-orbit contains MATH and MATH. If MATH we obtain the normal form MATH. If MATH and MATH we obtain MATH. If MATH and MATH, for every MATH such that MATH, we obtain MATH. If MATH in REF we obtain the normal form MATH. Now suppose MATH and MATH in REF . We have the MATH-jet MATH. It is MATH-determined, since MATH is equal to MATH for the singular curve and to zero for regular ones. A complete transversal in MATH is trivial for any MATH. In MATH, where MATH, it is equivalent to MATH . If MATH we obtain the normal form MATH. If MATH and MATH we move to higher jets. We suppose MATH, since if MATH the tangent space to the MATH-orbit contains the vectors: MATH and MATH. A complete transversal in MATH, where MATH, is equivalent to MATH. If MATH we obtain the normal form MATH. If MATH for every MATH such that MATH, we obtain the normal form MATH. If in REF MATH for every MATH such that MATH, we obtain the normal form MATH. Now we shall consider a complete transversal in MATH. It is equivalent to MATH . Suppose MATH or MATH. This jet is not finite determined, so we need to move to higher jets. In MATH a complete transversal is trivial for any MATH. First, suppose MATH. Then this MATH-jet is equivalent to MATH. In MATH a complete transversal is equivalent to MATH . If MATH we obtain the normal form MATH, which is MATH-determined. Now suppose MATH. If MATH and MATH we obtain the normal form MATH, where MATH. If MATH and MATH we obtain the normal form MATH. These both forms are MATH-determined. Now suppose MATH and MATH. Then we have the MATH-jet MATH. This jet is MATH-determined, since MATH is equal to MATH for the singular component and to zero for the regular ones. A complete transversal in MATH is trivial for any MATH. In MATH it is equivalent to MATH since the tangent space to the MATH-orbit contains the vectors MATH and MATH. If MATH we obtain the normal form MATH. If MATH and MATH we obtain the normal form MATH. If MATH for every MATH such that MATH, we obtain the normal form MATH. Now, suppose MATH and MATH in REF . We obtain the MATH-jet MATH. A complete transversal in MATH is trivial for any MATH. In MATH it is equivalent to MATH . Suppose MATH. We obtain the normal form MATH which is MATH-determined. If MATH and MATH we have the MATH-jet MATH which is MATH-determined since MATH is equal to MATH for the singular component and to zero for the regular ones. A complete transversal in MATH REF is equivalent to MATH . If MATH we obtain the normal form MATH. If MATH for each MATH such that MATH, we obtain the normal form MATH. |
math/0012040 | The dimension of the submanifold of such jets in MATH is equal to MATH. Let us estimate the dimension of the orbit at points of the submanifold under the action of the stabilisator of the regular part. It is generated by the images of the following vectors from MATH: MATH and MATH. Hence the dimension of the orbit under the action of elements from MATH is less than or equals MATH. Only MATH, MATH and MATH can give rise to non-zero vectors. Hence the dimension of the tangent space to the MATH-orbit is less than or equals REF. Now one can see that the dimension of the tangent space to the MATH-orbit at points of the submanifold is less than or equals MATH. By similar resoning as in REF we conclude that if the multigerm is simple then MATH, that is, MATH. We have a contradiction with the condition of the Lemma. |
math/0012040 | A complete transversal in MATH is equivalent to MATH . If MATH we have REF-jet MATH. A complete transversal in MATH is equivalent to MATH . For any MATH and MATH this jet is REF-determined. If MATH we obtain the normal form MATH. If MATH we obtain the normal form MATH. If MATH REF-jet is equivalent to MATH. Using REF we note that there is only one possibility for REF-jet: MATH . If MATH this jet is REF-determined and we obtain the normal form MATH. If MATH we move to higher jets. A complete transversal in MATH is trivial. A complete transversal in MATH is equivalent to MATH . If MATH this REF-jet is REF-determined and we obtain the normal form MATH. If MATH and there exists MATH such that MATH we obtain the normal form MATH. If MATH, and for each MATH we obtain the normal form MATH with MATH which is REF-determined. |
math/0012040 | Using REF we conclude that REF-jet is equivalent to MATH. If MATH and MATH then we obtain the normal form MATH which is REF-determined. Let MATH and MATH. We have REF-jet MATH. This jet is REF-determined, so we have to consider a complete transversal in MATH. It is equivalent to MATH since the tangent space to the MATH-orbit contains the vectors: MATH, MATH, MATH and MATH. If MATH in REF we obtain the normal form MATH. If MATH and MATH we obtain the normal form MATH. If MATH we obtain the normal form MATH. |
math/0012040 | The dimension of the submanifold of such jets in MATH is equal to MATH. Let us estimate the dimension of the orbit at points of the submanifold under the action of the stabilisator of the regular part. It is generated by the images of the vectors: MATH, MATH, MATH; MATH, MATH. Hence the dimension of the orbit under the action of elements from MATH is less than or equals MATH. Only MATH, MATH, MATH and MATH can give rise to non-zero vectors. Hence the dimension of the tangent space to the orbit under the action of elements from MATH is less than or equals REF. Now one can see that the dimension of the tangent space to the orbit at points of the submanifold is less than or equals MATH. Since MATH, the multigerm is not simple. |
math/0012040 | A complete transversal in MATH is equivalent to MATH . Since the tangent space to the MATH-orbit contains the vectors MATH, MATH and MATH, we can suppose MATH. If MATH we have REF-jet MATH. This jet is REF-determined, so we move to MATH. A complete transversal is equivalent to MATH. If MATH we obtain the normal form MATH. If MATH but MATH and MATH we obtain the normal form MATH. If MATH and MATH or MATH and MATH we obtain the normal form MATH (these both cases are equivalent since we can permutate co-ordinates and regular components). If MATH we obtain the normal form MATH. If MATH in REF , we have (by similar reasoning) three possibilities for REF-jet: MATH, MATH and MATH. Using REF we see that there is only one multigerm for each of these three REF-jets - the normal forms MATH - MATH. All of them are REF-determined. |
math/0012040 | REF-jet REF is not finite determined. A complete transversal in MATH is trivial for any MATH. In MATH it is equivalent to MATH . If MATH we obtain the normal form MATH, which is MATH-determined. If MATH and MATH then the jet is equivalent to MATH . Since the tangent space to the MATH-orbit contains the vectors (we write only non-zero (the second) component): MATH, MATH, MATH, MATH and MATH then the tangent space also contains MATH. Hence by the NAME Lemma we conclude that the MATH-jet is equivalent to the normal form MATH. If MATH and MATH we obtain the normal form MATH. |
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