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math/0011211 | We prove by induction on MATH the inequality above for MATH. The case MATH is trivial. Now let MATH, and consider MATH where MATH is the kernel of MATH. Then for every integer MATH we have the exact sequence MATH . We get MATH where the last inequality follows from the induction hypothesis. Analogously we obtain the inequality for MATH. |
math/0011211 | Let MATH be the minimal graded free resolution of MATH over MATH and MATH be the minimal graded free resolution of MATH over MATH. Then MATH is the minimal bigraded free resolution of MATH over MATH with MATH. Since MATH, the assertion follows. |
math/0011211 | By symmetry it suffices to show the inequality for MATH. Let MATH be the minimal bigraded free resolution of MATH over MATH. Since MATH is an exact functor, we obtain the exact complex of finitely generated MATH-modules MATH . By REF we have MATH . Since MATH one has MATH . We have to compute MATH. Let MATH be the relative NAME modules of MATH and MATH be the relative NAME modules of MATH. That is MATH for MATH and MATH for MATH. Then MATH where MATH mod MATH for MATH and MATH mod MATH for MATH. By CITE the relative NAME modules over a polynomial ring have a linear resolution over the NAME algebra. Hence REF yields MATH . This concludes the proof. |
math/0011216 | Given an object MATH, let MATH be the coproduct MATH indexed by all maps MATH and all objects MATH in MATH. The natural map MATH is clearly a MATH-epimorphism. Moreover, if MATH is in MATH, then this map is split epic, and so MATH is a retract of a coproduct of objects of MATH. These two facts show that MATH is a projective class. |
math/0011216 | MATH : Assume MATH is a strong projective class and let MATH be a complex in MATH such that MATH has trivial homology for each MATH-projective MATH. Then for each MATH and each MATH we have a short exact sequence MATH of abelian groups. Because the projective class is strong, the sequence is split. This implies that the complex MATH is isomorphic to MATH, and in particular that it is contractible. MATH : Let MATH be a map in MATH such that MATH is a quasi-isomorphism for each MATH in MATH, and let MATH be the cofibre of MATH. Then by the long exact sequence, MATH has trivial homology for each MATH in MATH. By REF this complex is contractible. This implies that MATH is a chain homotopy equivalence. MATH : Let MATH be a MATH-epimorphism with kernel MATH. Then the complex MATH has trivial homology after applying MATH, for each MATH-projective MATH. By REF , it is contractible after applying MATH, for each MATH. In particular, MATH is split epic. |
math/0011216 | Some of the properties necessary for a model category are evident from the definitions. It is clear that MATH is bicomplete, since MATH is so. Also, MATH-equivalences have the two out of three property, and MATH-equivalences, MATH-fibrations, and MATH-cofibrations are closed under retracts. Furthermore, MATH-cofibrations have the left lifting property with respect to MATH-trivial fibrations, by definition. It remains to show that MATH-trivial cofibrations have the left lifting property with respect to MATH-fibrations, and that the two factorization axioms hold. The remaining lifting property will be proved in REF , and the two factorization axioms will be proved in REF , assuming that cofibrant replacements exist. Properness will be proved in REF , which also defines the term. That cofibrant replacements exist in cases A and B will be proved in REF, respectively. |
math/0011216 | We can write MATH, where the differential is defined by MATH (we use the element notation for convenience, but it is not strictly necessary), and MATH can be any family of maps such that MATH. Suppose we have a commutative square as below. MATH . In terms of the splitting MATH, we have MATH, where the family MATH satisfies MATH. We are looking for a map MATH making the diagram above commute. In terms of the splitting, this means we are looking for a family of maps MATH such that MATH and MATH. Since MATH is relatively projective, and MATH is MATH-epic, there is a map MATH such that MATH. The difference MATH may not be zero, but at least MATH. Let MATH denote the kernel of MATH. Then there is a map MATH such that MATH. Furthermore, one can check that MATH, so that MATH is a chain map. By hypothesis, MATH is chain homotopic to MATH by a map MATH, so that MATH. Define MATH. Then MATH defines the desired lift, so MATH has the left lifting property with respect to MATH. |
math/0011216 | Suppose first that MATH is MATH-cofibrant. If MATH is a MATH-epimorphism, then the map MATH is a MATH-fibration. It is also a MATH-equivalence, since it is in fact a chain homotopy equivalence. Since MATH is MATH-cofibrant, the map MATH is surjective. But this map is isomorphic to the map MATH, so MATH is relatively projective. If MATH is weakly MATH-contractible, then the natural map MATH is a MATH-trivial fibration. Since MATH is MATH-cofibrant, any map MATH factors through MATH, which means that it is chain homotopic to MATH. The converse follows immediately from REF , since the kernel of a MATH-trivial fibration is weakly MATH-contractible. |
math/0011216 | Suppose first that MATH is a MATH-cofibration with cokernel MATH. Since MATH-cofibrations are closed under pushouts, it is clear that MATH is MATH-cofibrant. The map MATH is a MATH-fibration and a MATH-equivalence. Since MATH is a MATH-cofibration, the map MATH that is the identity in degree MATH extends to a map MATH. In degree MATH, this map defines a splitting of MATH. Conversely, suppose that MATH is a degreewise split monomorphism and the cokernel MATH of MATH is MATH-cofibrant. We need to show that MATH has the left lifting property with respect to any MATH-trivial fibration MATH. But this follows by combining REF . |
math/0011216 | REF follows immediately from REF . CASE: Let MATH be a bounded below complex of relative projectives and write MATH for the truncation of MATH that agrees with MATH in degrees MATH and is MATH elsewhere. Then the map MATH is degreewise split monic and has a MATH-cofibrant cokernel (by REF ), so is a MATH-cofibration by REF . Since MATH for MATH and MATH is MATH-cofibrant, each MATH is MATH-cofibrant. Therefore, so is their colimit MATH. CASE: The proof is just a transfinite version of the proof of REF , combined with the fact that a retract of a cofibrant object is cofibrant. |
math/0011216 | Suppose first that MATH has the left lifting property with respect to MATH-fibrations. Then MATH is a MATH-cofibration, by definition, and the cokernel MATH also has the left lifting property with respect to MATH-fibrations. In particular, since the map MATH is a MATH-fibration, MATH is contractible. Hence MATH is a chain homotopy equivalence, and in particular a MATH-trivial cofibration. Conversely, suppose that MATH is a MATH-trivial cofibration with cokernel MATH. By REF , in order to show that MATH has the left lifting property with respect to MATH-fibrations, it suffices to show that every map from MATH to any complex MATH is chain homotopic to MATH. This is equivalent to showing that MATH is contractible. Since MATH is degreewise split monic, for each relative projective MATH there is a long exact sequence MATH . Since MATH is a MATH-equivalence, MATH for each relative projective MATH and each MATH, and so MATH is a MATH-trivial fibration. Since MATH is MATH-cofibrant, the identity map of MATH factors through MATH, and so MATH is contractible. |
math/0011216 | Suppose MATH is a map in MATH. Let MATH be the cofibre of MATH, so MATH with MATH, and let MATH be the fibre of the composite MATH, so MATH with MATH (the desuspension of the cofibre). Consider the diagram MATH whose rows are triangles in MATH. There is a natural fill-in map MATH defined by MATH, where MATH is the projection. The map MATH makes the left-hand square commute in MATH and the middle square commute in MATH (with the chain homotopy MATH). The map MATH is a MATH-cofibration since it is degreewise split and its cokernel MATH is MATH-cofibrant. Furthermore, since MATH is degreewise MATH-epic and MATH is degreewise split epic, it follows that MATH is degreewise MATH-epic. Applying the functor MATH gives two long exact sequences, and from the five-lemma one sees that MATH is an isomorphism when MATH is a relative projective. Thus MATH is the required factorization. |
math/0011216 | It is well-known that we can factor any map MATH in MATH into a degreewise split monomorphism that is also a chain homotopy equivalence, followed by a degreewise split epimorphism. Since every degreewise split epimorphism is a MATH-fibration, we may as well assume MATH is a degreewise split monomorphism and a chain homotopy equivalence. In this case, we apply REF to factor MATH, where MATH is a MATH-trivial fibration and MATH is a MATH-cofibration. Since MATH is a chain homotopy equivalence, MATH must be a MATH-trivial cofibration, and so the proof is complete. |
math/0011216 | We construct a factorization MATH of the codiagonal map MATH in the following way. Let MATH be the chain complex that has MATH in degree MATH. We describe the differential by saying that it sends a generalized element MATH in MATH to MATH. Let MATH be the map that sends MATH to MATH and let MATH be the map that sends MATH to MATH. One can check easily that MATH is a chain complex and that MATH and MATH are chain maps whose composite is the codiagonal. The map MATH is degreewise split monic with cokernel MATH, so it is a MATH-cofibration, since we have assumed that MATH is cofibrant. The map MATH is a chain homotopy equivalence with chain homotopy inverse sending MATH to MATH; this implies that it induces a chain homotopy equivalence of generalized elements and is thus a MATH-equivalence. Therefore MATH is a good cylinder object for MATH. It is easy to see that a chain homotopy between two maps MATH is the same as a left homotopy using the good cylinder object MATH. By REF , two maps are homotopic if and only if they are left homotopic using MATH. Thus the model category notion of homotopy is the same as the notion of chain homotopy when the source is MATH-cofibrant. |
math/0011216 | The group MATH may be calculated by choosing a MATH-cofibrant replacement MATH for MATH and computing the homotopy classes of maps from MATH to MATH. (Recall that all objects are MATH-fibrant, so there is no need to take a fibrant replacement for MATH.) A MATH-resolution MATH of MATH serves as a MATH-cofibrant replacement for MATH, and by REF the homotopy relation on MATH is chain homotopy, so it follows that MATH is isomorphic to MATH. |
math/0011216 | REF is an immediate consequence of CITE, since every object is MATH-fibrant. For REF , let MATH be the cokernel of MATH. Since pushouts are computed degreewise, it follows that MATH is a degreewise split monomorphism with cokernel MATH. Thus we have a map of triangles MATH in the homotopy category MATH. The top and bottom maps are MATH-equivalences, so the middle map must be as well, by using the five-lemma and the long exact sequences obtained by applying the functors MATH. |
math/0011216 | There are two conditions that must hold for a model category to be monoidal. One of them is automatically satisfied when the unit is cofibrant. The unit of the monoidal structure on MATH is MATH, where MATH is the unit of MATH, which has been assumed to be MATH-projective. Thus the unit is MATH-cofibrant, by REF , Therefore, the relative model structure is monoidal if and only if whenever MATH and MATH are MATH-cofibrations, then the map MATH is a MATH-cofibration, and is a MATH-trivial cofibration if either MATH or MATH is a MATH-trivial cofibration. It is easy to see that MATH is a degreewise split monomorphism, with cokernel MATH. By taking MATH to be the map MATH and MATH to be the map MATH, we see that if the relative model structure is monoidal, then the tensor product of two MATH-cofibrant complexes is MATH-cofibrant. Conversely, if MATH is MATH-cofibrant whenever MATH and MATH are MATH-cofibrant, the preceding paragraph implies MATH is a MATH-cofibration whenever MATH and MATH are MATH-cofibrations. If either MATH or MATH is a MATH-trivial cofibration, then one of MATH or MATH is MATH-trivially cofibrant, and hence contractible. It follows that MATH is contractible, and so MATH is a MATH-trivial cofibration. |
math/0011216 | REF says that MATH is MATH-cofibrant for any MATH and MATH, where MATH denotes the cofibrant replacement functor constructed in REF. If MATH and MATH are already cofibrant, then lifting implies that MATH is a retract of MATH and MATH is a retract of MATH. Thus MATH is a retract of MATH, so MATH is MATH-cofibrant. REF completes the proof. |
math/0011216 | Let MATH and MATH be MATH-cofibrant. By REF , it suffices to show that MATH is MATH-cofibrant. By assumption, MATH is a retract of a transfinite colimit of a colimit-preserving diagram MATH such that for each MATH, the map MATH is degreewise split monic with cokernel a complex of relative projectives with zero differential. Since MATH preserves retracts, degreewise split monomorphisms and colimits, it is enough to prove that MATH is MATH-cofibrant for each MATH-projective MATH and each MATH. Applying the same filtration argument to MATH, we find that it suffices to show that MATH is MATH-cofibrant for all MATH-projectives MATH and MATH and integers MATH and MATH. But this follows immediately from the fact that MATH is monoidal and REF . |
math/0011216 | Suppose MATH and MATH are MATH-injective cofibrations with cokernels MATH and MATH, respectively. This means that MATH and MATH are degreewise split monomorphisms. Recall the definition of MATH used in the proof of REF . Since MATH is monoidal and preserves pushouts, MATH. One can easily check that MATH is a degreewise split monomorphism, so MATH is a MATH-injective cofibration. If MATH is a MATH-injective trivial cofibration, then the cokernel MATH of MATH has MATH contractible. Let MATH denote the cokernel of MATH. Since the cokernel of MATH is MATH, which is contractible, MATH is a MATH-injective trivial cofibration. Since everything is cofibrant in the injective relative model structure, the other condition in the definition of a monoidal model category is automatically satisfied. |
math/0011216 | Let MATH be the collection of MATH-epimorphisms and let MATH be the collection of all objects MATH such that every map in MATH is MATH-epic. Then MATH contains MATH, and a map is MATH-epic if and only if it is MATH-epic. It follows that the MATH-exact sequences are the same as the MATH-exact sequences. A map MATH is a MATH-equivalence if and only if the cofibre of MATH is MATH-exact. The same holds for the MATH-equivalences, and since the two notions of exactness also agree, the two notions of equivalence agree. Finally, since cofibrations are defined in terms of the fibrations and weak equivalences, the two notions of cofibration agree. That the pair MATH is a projective class follows immediately from the existence of cofibrant replacements. Our work earlier in this section shows that the objects of MATH are precisely those objects MATH that are cofibrant when viewed as complexes concentrated in degree REF. Thus the projective class is the unique projective class giving rise to the same weak equivalences, fibrations and cofibrations. |
math/0011216 | For REF , note that the MATH-projectives are all retracts of MATH for MATH. Hence MATH is a MATH-fibration if and only if MATH is a surjection for all MATH. This is true if and only if MATH is a degreewise split epimorphism. For REF , a similar argument shows that MATH is a MATH-equivalence if and only if MATH is a quasi-isomorphism for all MATH. We claim that this forces MATH to be a chain homotopy equivalence. Indeed, let MATH denote the cofiber of MATH. Then MATH is exact for all MATH. By taking MATH, we find that MATH is exact and that MATH is a split epimorphism. It follows that MATH is contractible, as in the proof of REF . Thus MATH is a chain homotopy equivalence. For REF , let MATH be the cokernel of MATH. Then MATH is degreewise split monic with cokernel MATH. Suppose that MATH is a MATH-trivial fibration with kernel MATH. Then MATH is MATH-trivially fibrant, so REF implies that MATH is contractible. By REF , to show that MATH is a MATH-cofibration, it suffices to show that every chain map MATH is chain homotopic to MATH. By adjointness, it suffices to show that every chain map MATH is chain homotopic to MATH. Since MATH is contractible, this is clear. REF follows from REF , |
math/0011216 | We first show that MATH is MATH-cofibrant. There is an increasing filtration MATH on MATH, where MATH and MATH. Furthermore, each inclusion MATH is a degreewise split monomorphism. By REF , each quotient MATH is MATH-cofibrant, so each map MATH is a MATH-cofibration. Hence MATH is a MATH-cofibration. Thus MATH is MATH-cofibrant. The map MATH is induced by MATH, adjoint to the identity. The map MATH sends the other summands of MATH to MATH. We leave to the reader the check that this is a chain map. Since MATH is a split epimorphism, MATH is a degreewise split epimorphism, so is a MATH-fibration by REF . To show MATH is a MATH-equivalence, it suffices to show that the fiber MATH is MATH-contractible, or, equivalently, that MATH is contractible REF . The contracting homotopy is given by MATH. Indeed, on the summand MATH, the MATH component of MATH is MATH, and the MATH component is MATH. (This is where we use that MATH commutes with coproducts.) |
math/0011216 | Recall the filtration MATH on MATH used in the proof of REF . Using this filtration, we find that MATH is the colimit of MATH, and each map MATH is a degreewise split monomorphism with cokernel MATH. It therefore suffcies to show that this cokernel is MATH-cofibrant for all MATH. A similar argument using the filtration on MATH shows that it suffices to show that MATH is MATH-cofibrant for all MATH. But this follow immediately from REF . |
math/0011216 | Let MATH be a chain complex. The basic construction is the following ``partial cofibrant replacement": For each MATH choose MATH-epimorphisms MATH and MATH with MATH and MATH-projective. Then form MATH, which has a natural map to MATH. This map is degreewise MATH-epic (because of the MATH's) and is epic under MATH for any MATH-projective MATH (because of the MATH's). And its domain is MATH-cofibrant. We now construct a transfinite sequence MATH with maps to MATH. Define MATH to be a partial cofibrant replacement. Let MATH be the pullback MATH where MATH is the path complex of MATH. Let MATH be a partial cofibrant replacement. Define MATH to be the cofibre of the composite MATH. The map MATH is a degreewise split monomorphism whose cokernel REF is MATH-cofibrant, so it is a MATH-cofibration. Moreover, since the map MATH factors through MATH, there is a canonical null homotopy, and so there is a canonical map MATH. Let MATH be an ordinal with cofinality greater than MATH, and inductively define MATH for MATH. When MATH is a limit ordinal, MATH is defined to be the colimit of the earlier MATH. Let MATH be the colimit of all of the MATH. MATH is MATH-cofibrant and comes with a map to MATH. Since MATH is already degreewise MATH-epic, so is MATH. That is, MATH is a MATH-fibration. Let MATH be a relative projective that is MATH-small relative to the split monomorphisms with relative projective cokernels. We need to show that MATH is sent to an isomorphism by MATH. To simplify notation, we do the case where MATH. That MATH is epic is clear since MATH is already so. So all that remains is to show that MATH is monic. Let MATH be a chain map such that MATH is null. By REF below, the chain map MATH factors as a chain map through some MATH. It is null in MATH, so the composite MATH factors through MATH. Thus we get a chain map to the pullback MATH. And this lifts through MATH. So the composite MATH must be null. In particular, the original map MATH must be null. |
math/0011216 | Suppose MATH is the colimit of a MATH-indexed diagram whose maps MATH have components in MATH, where MATH has cofinality greater than MATH. For simplicity we treat the case MATH. Suppose we have a chain map MATH. The map MATH factors through some MATH, since MATH is small and the maps MATH are in MATH. Then MATH goes to zero in MATH, and so goes to zero in some MATH, using the other half of smallness. So the chain map MATH factors as a chain map through some MATH. This shows that MATH is surjective. That it is injective is equivalent to the fact that MATH is injective. |
math/0011216 | It suffices to show that the cofibrant replacement MATH constructed in REF is MATH-cellular, since if MATH is cofibrant, a lifting argument shows that MATH is a retract of any cofibrant replacement of MATH. The complex MATH is constructed as a colimit of a colimit-preserving functor MATH, indexed by an ordinal MATH. Each map MATH is a degreewise split monomorphism with cokernel that is easily seen to be purely MATH-cellular. By reindexing, one can show that the colimit of such a transfinite diagram is (purely) MATH-cellular. |
math/0011216 | That MATH is proved in Subsection REF. That MATH is CITE. That MATH is proved in Subsection REF. |
math/0011216 | We begin with REF : The map MATH has the right lifting property with respect to the map MATH if and only if each map MATH factors through MATH, that is, if and only if each MATH-element of MATH is in the image of MATH. Now REF : The map MATH has the right lifting property with respect to the map MATH if and only if for each MATH-element MATH of MATH whose boundary is the image of a MATH-cycle MATH of MATH, there is a MATH-element MATH of MATH that hits MATH under MATH and MATH under the differential. (In other words, if and only if the natural map MATH induces a surjection of MATH-elements.) So suppose that MATH has the right lifting property with respect to each map MATH. As a preliminary result, we prove that MATH induces a surjection of MATH-cycles. Suppose we are given a MATH-cycle MATH of MATH. Its boundary is zero and is thus the image of the MATH-cycle MATH of MATH. Therefore MATH is the image of a MATH-cycle MATH. It follows immediately that MATH induces a surjection in MATH-homology. Now we prove that MATH induces a surjection of MATH-elements. Suppose we are given a MATH-element MATH of MATH. By the above argument, its boundary, which is a MATH-cycle, is the image of a MATH-cycle MATH of MATH. Thus, by the characterization of maps MATH having the RLP, we see that there is a MATH-element MATH that hits MATH. A similar argument shows that MATH induces an injection in MATH-homology. We have proved that if MATH has the right lifting property with respect to the maps MATH, then MATH induces an isomorphism in MATH-homology and a surjection of MATH-elements. The proof of the converse goes along the same lines. |
math/0011216 | We check the hypotheses of the Recognition Lemma from the previous subsection. Since MATH is complete and cocomplete, so is MATH; limits and colimits are taken degreewise. The class MATH of MATH-equivalences is easily seen to be closed under retracts and to satisfy the two-out-of-three condition. The zero chain complex is certainly small relative to MATH. It is easy to see that a map is in MATH if and only if it is a degreewise split monomorphism whose cokernel is purely cellular REF . In particular, every map in MATH has components in MATH. We assumed that each MATH in MATH is small relative to MATH, and so it follows from REF that each MATH is small relative to MATH. Since the projectives in MATH are enough to test whether a map is a MATH-fibration or a MATH-equivalence, REF tells us that MATH is the collection of MATH-trivial fibrations and that MATH is the collection of MATH-fibrations. Thus we have an equality MATH, giving us two of the inclusions required by the Recognition Lemma. We now prove that MATH. Since MATH, it is clear that MATH, so in particular MATH. We must prove that MATH, that is, that each map that is a transfinite composite of pushouts of coproducts of maps in MATH is a MATH-equivalence. A map in MATH is of the form MATH for some MATH. Thus a pushout of a coproduct of maps in MATH is of the form MATH, with MATH a contractible complex, and a transfinite composite of such maps is of the same form as well. Thus such a map is in fact a chain homotopy equivalence, so it is certainly a MATH-equivalence. We can now apply the Recognition Lemma and conclude that MATH is a model category with weak equivalences the MATH-equivalences. The fibrations are the maps in MATH, which, as we noted above, are the MATH-fibrations. The cofibrations are the maps in MATH, that is, the maps with the left lifting property with respect to the MATH-trivial fibrations, and this is precisely how the MATH-cofibrations were defined. The recognition lemma tells us that the MATH-cofibrations consist precisely of the retracts of maps in MATH. As discussed above, MATH is the class of degreewise split monomorphisms with purely cellular cokernels. So the MATH-cofibrations are the degreewise split monomorphisms whose cokernels are cellular. |
math/0011216 | Let MATH be a set of weak generators for MATH and, without loss of generality, assume that each MATH in MATH is MATH-cofibrant. Let MATH. Then each MATH in MATH is MATH-projective. Suppose MATH in MATH is MATH-epic. We must show that MATH is MATH-epic. Consider the chain complex MATH, with zeroes elsewhere. Because the map MATH is MATH-epic, it is easy to check that every map from a generator MATH to the complex MATH is null homotopic. But since MATH is a set of weak generators, this implies that this complex is MATH-equivalent to the zero complex. That is, the complex MATH is exact for each MATH in MATH. In particular, the map MATH is MATH-epic. |
math/0011216 | We use the NAME invariants, as described in CITE. Recall that for each prime MATH, each abelian group MATH and each ordinal MATH, the NAME invariant MATH is a cardinal number. From the definition it is clear that MATH for MATH larger than the cardinality of MATH. Furthermore, MATH takes direct sums of abelian groups to sums of cardinals CITE. Hence we only need to find an abelian group MATH with MATH. Such a (MATH-torsion) group exists for every MATH by CITE. |
math/0011216 | For any set MATH of abelian groups, we must exhibit a map MATH that is MATH-epic but not split epic. So fix a set MATH and let MATH be a cardinal larger than the cardinality of each group in MATH. Let MATH be an abelian group that is not a retract of any direct sum of abelian groups of cardinality less than MATH, using REF . Let MATH denote the direct sum of the MATH in MATH, with one copy of MATH for each homomorphism MATH. Then there is an obvious map MATH, and this map is not split epic since MATH is not a retract of MATH. However, if MATH is an abelian group in MATH, any homomorphism MATH obviously factors through MATH. |
math/0011216 | Fix a cardinal MATH. Since MATH is difficult, there is a map MATH of MATH-modules that is not split epic such that MATH is epic whenever MATH has cardinality less than MATH. Let MATH be the map MATH of MATH-modules. Since MATH splits off of MATH, MATH is the sum of MATH and another map as a map of MATH-modules. Thus MATH is not split epic as a MATH-module map and hence as a MATH-module map. Since MATH is right adjoint to the forgetful functor, we have MATH for any MATH-module MATH. In particular, MATH is surjective for all MATH-modules MATH of cardinality less than MATH. Since any set of MATH-modules will have sizes bounded above by some cardinal, we have shown that the ring MATH is difficult. Next we show that the relative projective class MATH on MATH is also not determined by a set. Recall that the MATH-projectives are the retracts of extended modules MATH and the MATH-epimorphisms are the maps of MATH-modules that are split epic as MATH-module maps. So for each MATH we must exhibit a map MATH of MATH-modules that is not split epic as a MATH-module map such that, for every extended module MATH of cardinality less than MATH, the map MATH is epic. Note that neither requirement uses the MATH-module structure on MATH. We choose MATH as in the previous paragraph, where we already noted that it is not split epic as a MATH-module map. Since MATH is surjective for all MATH-modules MATH of cardinality less than MATH, of course MATH is surjective for all extended modules MATH of cardinality less than MATH. |
math/0011219 | REF through REF are obvious. REF follows from the fact that the growth of the NAME metric is logarithmic in this case CITE. |
math/0011219 | We shall prove that the filtration MATH is determined by the local monodromies of the cohomology sheaf MATH around the branches of MATH which are known to be quasi-unipotent, where MATH. We take an open neighborhood MATH of MATH in the classical topology which is isomorphic to a polydisk with coordinates MATH such that MATH is defined by MATH. To simplify the notation, we write MATH instead of MATH. There exists a finite surjective and NAME morphism MATH from a smooth variety which is etale over MATH such that, for the induced morphism MATH from a desingularization MATH of the fiber product MATH, the local system MATH has unipotent local monodromies around the branches of MATH, where we set MATH, MATH and MATH. Let MATH be the induced morphism. We may assume that MATH is isomorphic to a polydisk centered at a point MATH with coordinates MATH, and the morphism MATH is given by MATH for some positive integers MATH, where MATH for MATH. The NAME group MATH is isomorphic to MATH. Let MATH be generators of MATH such that MATH for some roots of unity MATH of order MATH. The group MATH acts on the sheaves MATH and MATH equivariantly such that the invariant part MATH is isomorphic to MATH, because MATH and MATH for MATH. The vector space MATH is decomposed into simultaneous eigenspaces with respect to the action of MATH. Let MATH be a simultaneous eigenvector such that MATH for some MATH with MATH. Let MATH be a section of MATH which extends MATH. Then the section MATH satisfies that MATH and MATH. On the other hand, MATH is a generating section of MATH. Therefore, MATH descends to a section MATH of MATH. If the MATH varies among a basis of MATH, then the corresponding sections MATH make a basis of the locally free sheaf MATH. We have MATH, since the NAME metric on the sheaf MATH has logarithmic growth along MATH. Therefore, the sections MATH form a basis of a locally free sheaf MATH. |
math/0011219 | CASE: Since the MATH are derived from the basis in the case of unipotent monodromies, we obtain our assertion by REF . CASE: We can check the assertion locally. We write MATH for some nonnegative integers MATH. Then the left hand side is generated by the sections MATH for MATH, while each component of the right hand side is by MATH . Since MATH, we should compare MATH and MATH . We have MATH hence MATH . On the other hand, if we set MATH for MATH, then MATH . Therefore, they are equal. Since the minimum is attained at a value of MATH which does not depend on MATH but only on MATH, we obtain the equality. |
math/0011219 | By CITE, there exists a normal crossing divisor MATH such that MATH which satisfies the following: there exists a finite surjective and NAME morphism MATH from a smooth projective variety which is etale over MATH and such that MATH has integral coefficients and all the local monodromies of MATH are unipotent under the notation of REF . We replace MATH by MATH and let MATH be the NAME group of MATH. By REF , we have MATH for MATH. Since MATH, we have by REF MATH . We want to calculate the MATH-invariant part of the locally free sheaf MATH . For this purpose, let MATH be the largest divisor on MATH such that MATH . This is equivalent to the condition MATH . Hence we obtain MATH . Since MATH our assertion is proved. |
math/0011220 | Take NAME 's MATH-Dougall sum CITE MATH and let MATH, MATH, MATH and MATH. After some simplifications this gives REF. |
math/0011220 | Suppressing their MATH-dependence we denote the left sides of REF by MATH and MATH, respectively. By application of the recurrence REF MATH . Since the second term on the right changes sign after the variable change MATH it vanishes. The first term is again split using REF leading to MATH . Next we let MATH and MATH denote the right sides of REF. A single application of REF show that REF again holds. Since REF trivializes to MATH for MATH this settles the lemma. |
math/0011220 | As a first step we add zero in the form MATH to the left side of REF. By the recurrence REF we are then to prove MATH . Suppressing the MATH-dependence in the first, we denote the left sides of REF by MATH and MATH, respectively. Using REF we then get MATH . Next we let MATH and MATH denote the right sides of REF. One application of REF shows that the same recurrence again holds. Since REF is true for MATH we are done. |
math/0011225 | As the algebra is finite dimensional, it follows from the weight space decomposition that there exists at least one weight MATH which is "extreme" in the following sense : MATH . Then for any vector MATH we have MATH, where MATH denotes the adjoint operator in MATH. Thus the vector is central, and the corresponding vertex of MATH is isolated. |
math/0011225 | As we have imposed MATH for any weight MATH, we can reorder the weights of MATH in such manner that a relations MATH corresponds to a sum MATH, where MATH. Thus the maximal number of sums of weights equals the number of possibilities MATH with MATH and MATH. It is easily seen that this number is precisely MATH. |
math/0011225 | For MATH the assertion is obvious. Now suppose it holds for MATH. Then MATH . |
math/0011225 | Suppose that MATH is bipartite. As it has a point of degree MATH, say MATH, the partition of the point set MATH must be MATH. This implies that there are no lines in the weight graph joining points of MATH, which implies that the subgraph whose point set is MATH is totally disconnected. Taking the complementary, it follows that MATH has complete subgraph of degree MATH, isomorphic to MATH. Now this graph has MATH lines, but MATH has at most MATH lines. The contradiction follows from the preceding lemma. |
math/0011225 | The only case to be considered is when the weight graph MATH is associated to a nilpotent NAME algebra whose center is one dimensional. For any higher dimensional centers, let MATH be two vertices corresponding to central weight spaces, and MATH another arbitrary point. In the complementary MATH these points are totally disconnected, thus they induce a triangle in the weight graph. So we can assume that the graph MATH has a unique isolated point. If the weight graph does not contain a triangle, then MATH contains a complete subgraph isomorphic to MATH, which is not possible in view of REF . |
math/0011225 | Let MATH and MATH be two nonequivalent weight systems associated to nilpotent NAME algbras of rank MATH, where MATH . Suppose that both their associated weight graphs MATH and MATH and its fundamental subgraphs MATH and MATH are isomorphic. Then we can find a permutation MATH such that MATH for the complementary of the fundamental subgraphs. Define the map MATH . Then, whenever the weight MATH is joined with MATH, its image MATH is joined with MATH, as it corresponds to a line of the subgraph MATH. It follows at once that MATH and MATH are the same weight system, which contradicts the hipothesis. |
math/0011225 | Let MATH be a basis as in MATH. As the semidirect product MATH is REF-step solvable, whenever there are vectors MATH such that MATH, we have MATH with MATH or MATH with MATH. Thus for any vector MATH with MATH we have MATH . This implies, in view of the construction of the graph MATH, that the complementary to the fundamental subgraph MATH is totally disconnected, so that MATH is isomorphic to the graph MATH, where MATH. The converse follows at once. |
math/0011225 | Let MATH be complete and fix a weight, MATH for example. Then the removal of the set MATH from the weight graph gives the fundamental subgraph MATH, as the intersections of the sets MATH are empty. Thus MATH . By recurrence it follows that MATH . |
math/0011225 | If the semidirect product were REF-step solvable, then its weight graph would be complete. Then its complementary graph MATH-is totally disconnected, which implies that there are no nontrivial brackets in MATH . |
math/0011225 | If the algebra MATH is rigid, then the torus MATH must be a maximal torus of derivations for MATH, and by REF , the algebra MATH is at least three step solvable, which contradicts the assumption. |
math/0011227 | The NAME duality in MATH combined with the exact cohomology sequence of a pair MATH gives MATH where MATH is the inclusion map. If MATH is oriented and MATH is connected, then the NAME Theorem yields MATH, and thus MATH is cyclic with a generator presented by a loop around MATH. The same property holds for the fundamental groups of MATH and MATH, since they are abelian by the assumption of REF . |
math/0011227 | A solid torus MATH can be viewed as the complement MATH of an open tubular neighborhood MATH of an unknot, so that MATH represent meridians of this unknot. Taking a fiber sum of MATH with MATH along MATH is equivalent to knotting MATH in MATH via MATH. So, performing MATH-fiber sum twice, along MATH and MATH, we obtain the same result as after taking fiber sum along MATH once, via MATH. |
math/0011227 | A pair of MATH-disc membranes, MATH, MATH, on each of MATH is constructed like in the proof of REF . Namely, MATH consists of MATH disks which yield the disks MATH, that are glued along MATH. Furthermore, MATH splits also into MATH disks, MATH. Let us choose their orientations induced from a fixed orientation of MATH and cyclically order in accord with the ordering of MATH, then the unions MATH provide the required discs MATH, which are glued along MATH. More precisely, MATH are the parts of the components of MATH bounded by the intersections of the components with the tori MATH, whereas MATH are obtained from MATH by a small shift making them membranes on MATH. |
math/0011227 | It is enough to show that MATH, since it implies that MATH and thus the inclusion map MATH is monomorphic. The first inclusion map in the composition MATH that we analyze, is just MATH, and has kernel MATH, as stated in the Proposition. Now note that MATH is a deformational retract (spine) of MATH, so it is enough to check the triviality of MATH. This triviality follows from that MATH is MATH, with a generator represented by a loop around MATH (say, by the computation in CITE reproduced in the Appendix), and thus MATH. |
math/0011231 | Suppose that MATH. Take MATH so that MATH and MATH for each MATH and let MATH. Since MATH and MATH, by REF we have a homomorphism MATH such that MATH for each MATH. Then, MATH for each MATH, which contradicts REF . To show the second proposition by contradiction, suppose that MATH. Then, we have MATH such that MATH but MATH. Let MATH. Since MATH and MATH, we apply REF and have a homomorphism MATH such that MATH for each MATH. Then, we have a contradiction similarly as the above. |
math/0011231 | We construct MATH by induction as follows. Suppose that we have constructed MATH. If MATH satisfies the required properties of MATH, we have finished the proof. Otherwise, there exist MATH such that MATH and MATH. We claim that this process finishes in a finite step. Suppose that the process does not stop in a finite step. Then, we have MATH's and so let MATH. Then, MATH by REF . Since MATH, MATH for each MATH. Now, MATH and and hence MATH by REF , which is a contradiction. |
math/0011234 | Consider the diagram MATH associated to the poset MATH. Note that inclusion is reversed such that MATH becomes MATH. The inclusion maps MATH and MATH clearly are closed cofibrations. By the Projection Lemma CITE we have that MATH. The inclusion maps being homotopic to the constant map, it follows from the Wedge Lemma CITE that MATH. |
math/0011234 | The poset of faces of a path is obtained by a subdividing of each edge of the path, when constructing NAME matchings, the acyclic property is trivially satisfied. Therefore, the pure NAME complex of the path with MATH edges is simply the complex of partial matchings that are extendible to perfect matchings for the path with MATH edges. We will call this the pure matching complex for the path with MATH edges. Assuming that the MATH edges are labeled MATH, it is clear that there is exactly one perfect matching which contains MATH , namely MATH and every other perfect matching contains the edge MATH. It is easy to see that MATH can be collapsed onto the face MATH which is contained in the facet MATH. As a result, we are left with a cone with apex MATH, which is obviously collapsible. Note that a path with exactly two nodes, that is, an interval has a discrete NAME complex isomorphic to MATH. |
math/0011234 | Observe that MATH. We claim that MATH. To show this assume that MATH is a maximal face of MATH. Then both MATH and MATH are facets of MATH and MATH, respectively. It follows that MATH consists of MATH elements and it cannot possibly contain MATH, MATH or MATH and therefore these elements must come from MATH. From the matching property that is required it follows that there exactly one possibility is, that is, MATH. Also, to any proper subset of MATH , say MATH, one can always add the edges MATH and MATH to obtain a NAME matching, which is obviously an element of MATH. This completes the proof of the claim. Now the star MATH is clearly contractible, and as shown above the boundary of the face MATH is entirely contained in MATH. We infer that MATH is homotopic to MATH. Similarly, MATH. As the stars MATH and MATH are contractible, we can apply REF to derive that MATH is homotopic to the suspension of the intersection MATH. Thus MATH. Similarly, MATH. Now we consider MATH, which we claim is equal to MATH, where MATH is the pure matching complex of the path with edges MATH. To show this, assume MATH is a facet of the intersection. It is easy to see from the matching requirement that only way that MATH can have MATH elements is if MATH and MATH, that is MATH is a facet of MATH. It is also evident that in this instance, the set MATH is a facet of the pure matching complex of the path on the edges MATH . Now the facets of MATH are subsets of MATH. It is clear that any such facet MATH is also a facet of the pure matching complex of the path on the edges MATH. Conversely, to any facet MATH of the pure matching complex of this path we can add either MATH or MATH to MATH to get a facet of MATH and we can add either MATH or MATH to get a facet of MATH so that MATH will be facet of MATH. Thus we have MATH is the union of the two contractible complexes MATH and MATH. Note that if we have a facet of the pure matching complex of the path on MATH edges, consecutively labeled starting with an odd number, then every facet consists of a string of odd edges (possibly empty) followed by an even string of vertices (possibly empty). On the other hand, if the labeling starts with an even number, then the facets consist of an even string followed by an odd string. This observation is useful in determining MATH, which following the above arguments, is the intersection of the pure matching complex on the path MATH with that of the pure matching complex on the path MATH. It follows that any face in this intersection consists entirely of odd vertices or entirely of even vertices. Now the two faces MATH and MATH are in the intersection and obviously, they are the unique maximal even and odd sets, respectively, in the intersection. Therefore, we have shown intersection MATH is a disjoint union of two non-empty simplices. That is, it is homotopic to MATH. Due to REF we have that MATH. Finally, the claim follows from REF . |
math/0011234 | A line segment can be oriented in two different ways. Hence the result for MATH. Now consider the MATH-simplex with vertices MATH, MATH, MATH; label the edges of the NAME diagram from MATH to MATH as in REF . For any subset MATH let MATH be the discrete NAME complex of the subcomplex generated by MATH. Moreover, MATH. Clearly, each maximal NAME matching contains either MATH or MATH or MATH, that is, MATH. Now MATH, MATH, and MATH. Moreover, the intersection MATH equals MATH which consists of two isolated points. Thus MATH. Observe that MATH has four isolated points, that is, it is equal to MATH. We infer that MATH. |
math/0011234 | On the contrary assume that MATH. Hence, for some MATH the restrictions MATH and MATH differ. Abbreviate MATH for MATH and MATH for MATH, respectively. Let MATH be the subgraph of the NAME diagram of MATH which is induced on the vertex set MATH. Both, MATH and MATH induce perfect matchings of the bipartite graph MATH. So their symmetric difference is a union of cycles. We arrive at a contradiction because of the acyclicity condition on NAME matchings. |
math/0011234 | The numbers of perfect matchings of the graph of the MATH-cubes for MATH are easy to determine. All these matchings are acyclic and thus are NAME matchings of the respective MATH-simplex. We say that an edge of MATH is in direction MATH if its vertices differ in the MATH-th coordinate. In the following we construct perfect acyclic matchings of cubes which contain edges of all but one direction. Observe that all perfect matchings of MATH, MATH and MATH are of this kind. Choose MATH. Suppose we have two such matchings MATH, MATH in MATH and MATH, respectively. Then MATH is a perfect acyclic matching of MATH. Now MATH contains edges of either MATH or MATH directions. Fix MATH, and let MATH be the unique direction which MATH does not contain an edge of. Now MATH contains edges from MATH directions if and only if MATH contains edges of direction MATH. If there are MATH perfect acyclic matchings of the MATH-cube containing edges of all but one direction, then MATH of them contain edges of a given direction. This gives MATH different perfect acyclic matchings of the MATH, which contain edges from all but direction MATH. Now there are MATH choices for MATH, and all of them yield different matchings. |
math/0011234 | We will prove the result by induction on MATH. The initial case MATH is clear. Further, MATH . |
math/0011235 | The bijections MATH, MATH, and MATH give the equidistribution part of the result. Calculations show that these three distributions differ pairwise on MATH. |
math/0011235 | Recall that the NAME numbers satisfy MATH, and MATH . We show that MATH satisfy the same recursion. Clearly, MATH. For MATH, let MATH, and let MATH be a MATH element subset of MATH. For each MATH-avoiding permutation MATH of MATH we construct a unique MATH-avoiding permutation MATH of MATH. Let MATH be the word obtained by writing the elements of MATH in decreasing order. Define MATH. Conversely, if MATH is a given MATH-avoiding permutation of MATH, where MATH, then the letters of MATH are in decreasing order, and MATH is a MATH-avoiding permutation of the MATH element set MATH. |
math/0011235 | Given a partition MATH of MATH, we introduce a standard representation of MATH by requiring that: CASE: Each block is written with its least element first, and the rest of the elements of that block are written in decreasing order. CASE: The blocks are written in decreasing order of their least element, and with dashes separating the blocks. Define MATH to be the permutation we obtain from MATH by writing it in standard form and erasing the dashes. We now argue that MATH avoids MATH. If MATH, then MATH and MATH are the first and the second element of some block. By the construction of MATH, MATH is a left-to-right minimum, hence there is no MATH such that MATH. Conversely, MATH can be recovered uniquely from MATH by inserting a dash in MATH preceding each left-to-right minimum, apart from the first letter in MATH. Thus MATH gives the desired bijection. |
math/0011235 | This result follows readily from the second proof of REF . We here give a different proof, which is based on the fact that the NAME numbers of the second kind satisfy MATH . Let MATH be the number of permutations in MATH with MATH left-to-right minima. We show that the MATH satisfy the same recursion as the MATH. Let MATH be a MATH-avoiding permutation of MATH. To insert MATH in MATH, preserving MATH-avoidance, we can put MATH in front of MATH or we can insert MATH immediately after each left-to-right minimum. Putting MATH in front of MATH creates a new left-to-right minimum, while inserting MATH immediately after a left-to-right minimum does not. |
math/0011235 | Let MATH be a partition of MATH. We introduce a standard representation of MATH by requiring that: CASE: The elements of a block are written in increasing order. CASE: The blocks are written in decreasing order of their least element, and with dashes separating the blocks. Notice that this standard representation is different from the one given in the second proof of REF . Define MATH to be the permutation we obtain from MATH by writing it in standard form and erasing the dashes. It easy to see that MATH avoids MATH. Conversely, MATH can be recovered uniquely from MATH by inserting a dash in between each descent in MATH. |
math/0011235 | From the proof of REF we see that MATH has MATH blocks precisely when MATH has MATH descents. |
math/0011235 | We give a combinatorial proof using a bijection that is essentially identical to the one given in the second proof of REF . Let MATH be an involution. Recall that MATH is an involution if and only if each cycle of MATH is of length one or two. We now introduce a standard form for writing MATH in cycle notation by requiring that: CASE: Each cycle is written with its least element first. CASE: The cycles are written in decreasing order of their least element. Define MATH to be the permutation obtained from MATH by writing it in standard form and erasing the parentheses separating the cycles. Observe that MATH avoids MATH: Assume that MATH, that is MATH is a cycle in MATH, then MATH is a left-to-right minimum in MATH. This is guaranteed by the construction of MATH. Thus there is no MATH such that MATH. The permutation MATH also avoids MATH: Assume that MATH, then MATH must be the smallest element of some cycle. Then MATH is a left-to-right minimum in MATH. Conversely, if MATH is a MATH-avoiding permutation then the involution MATH is given by: MATH is a cycle in MATH if and only if MATH. |
math/0011235 | CITE observed that the dashes in the patterns MATH and MATH are immaterial for the proof of REF . The result may, however, also be proved directly. For an example of such a proof see the proof of REF . |
math/0011235 | Under the bijection MATH in the proof of REF , a cycle of length two in MATH corresponds to an occurrence of MATH in MATH. Hence, if MATH has MATH fixed points, then MATH has MATH descents. Substituting MATH for MATH we get the desired result. |
math/0011235 | Let MATH denote the number of involutions in MATH with MATH fixed points. Then Porism REF is equivalently stated as MATH . In CITE NAME and NAME showed that the NAME polynomials are given by MATH . To prove REF , multiply REF by MATH and sum over MATH. MATH . We now multiply REF by MATH and sum over MATH. Tedious but straightforward calculations then yield REF from REF . Finally, we obtain REF from REF by identifying coefficients in REF . |
math/0011235 | Given MATH in MATH, let MATH be the result of writing MATH in the standard form given in the second proof of REF , and let MATH. By the construction of MATH the fist letter in each MATH is a left-to-right minimum. Furthermore, since MATH is monotone the second letter in each non-singleton MATH is a right-to-left maximum. Therefore, if MATH is a MATH-subword of MATH, then MATH is left-to-right minimum and MATH is a right-to-left maximum. Thus MATH avoids both MATH and MATH. Conversely, given MATH in MATH, let MATH be the result of inserting a dash in MATH preceding each left-to-right minimum, apart from the first letter in MATH. Since MATH is MATH-avoiding, the second letter in each non-singleton MATH is a right-to-left maximum. The second letter in MATH is the maximal element of MATH when MATH is viewed as a set. Thus MATH is monotone. |
math/0011235 | If MATH is the minimal element of a non-singleton block, then call MATH the first element of that block. Similarly, If MATH is the maximal element of a non-singleton block, then call MATH the last element of that block. An element of a non-singleton that is not a first or last element is called an intermediate element. Let us introduce an ordering of the blocks of a partition. A block MATH is smaller than a block MATH if MATH. We define a map MATH that to each non-overlapping partition MATH of MATH gives a unique monotone partition MATH of MATH. Let the integer MATH range from MATH to MATH. CASE: If MATH is the first element of a block of MATH, then open a new block in MATH by letting MATH be its first element. (A block MATH is open if MATH.) CASE: If MATH is the last element of a block of MATH, then close the smallest open block of MATH by letting MATH be its last element. CASE: If MATH is an intermediate element of some block MATH of MATH, and MATH is the MATH:th largest open block of MATH, then let MATH belong to the MATH-th largest open block of MATH. CASE: If MATH is a singleton block of MATH, then let MATH be a singleton block of MATH. Observe that MATH is monotone. Indeed, it is only in REF that we close a block of MATH, and we always close the smallest open block of MATH. Conversely, we give a map MATH that to each monotone partition MATH be a of MATH gives a unique non-overlapping partition MATH of MATH. Define MATH the same way as MATH is defined, except for REF , where we instead of closing the smallest open block close the largest open block. It is easy to see that MATH and MATH are each others inverses and hence they are bijections. |
math/0011235 | Follows immediately from REF together with REF . |
math/0011235 | Under the bijection MATH in the proof of REF , the number of blocks in MATH determines the number of left-to-right minima of MATH, and vice versa. The number of blocks is not changed by the bijection MATH in the proof of REF . |
math/0011235 | Let MATH be a permutation of MATH such that MATH. Then MATH is MATH-avoiding if and only if MATH, where MATH is a MATH-avoiding permutation of MATH, and MATH is a MATH-avoiding permutation of MATH. We define recursively a mapping MATH from MATH onto the set of NAME paths of length MATH. If MATH is the empty word, then so is the NAME path determined by MATH, that is, MATH. If MATH, then we can use the factorisation MATH from above, and define MATH. It is easy to see that MATH may be inverted, and hence is a bijection. |
math/0011235 | The sufficiency part of the proposition is trivial. The necessity part is not difficult either. Assume that MATH contains a MATH-subword. Then there exist MATH such that MATH is a subword of MATH, and MATH is a segment of MATH. If MATH, then MATH form a MATH-subword in MATH. Assume that MATH. Indeed, to avoid forming a MATH-subword we will have to assume that MATH for all MATH, but then MATH is a MATH-subword. Accordingly we conclude that there exists at least one MATH-subword in MATH. |
math/0011235 | Follows immediately from REF . |
math/0011235 | A return step in a NAME path MATH is a MATH such that MATH, for some NAME paths MATH, MATH, and MATH. A useful observation is that every non-empty NAME path MATH can be uniquely decomposed as MATH, where MATH and MATH are NAME paths. This is the so-called first return decomposition of MATH. Let MATH denote the number of return steps in MATH. In CITE NAME showed that the distribution of MATH over all NAME paths of length MATH is the distribution we claim that MATH has over MATH. Let MATH be a NAME path of length MATH, and let MATH be its first return decomposition. Then MATH. Let MATH, and let MATH be the decomposition given in the proof of REF . Then MATH. The result now follows by induction. |
math/0011235 | We mimic the proof of REF . Let MATH. Since MATH avoids MATH it also avoids MATH by REF via MATH. Thus we may write MATH, where MATH, MATH is a MATH-avoiding permutation of MATH, and MATH is a MATH-avoiding permutation of MATH. If MATH then MATH where MATH, or else a MATH-subword would be formed with MATH as the MATH in MATH. Define a map MATH from MATH to the set of NAME paths by MATH and MATH . Its routine to find the inverse of MATH. |
math/0011237 | Note, first, that MATH on MATH. Thus, since MATH anti-commutes with MATH, we have MATH. Hence, REF holds, when restricted to MATH. We now consider the restriction of MATH to the cylinder MATH. Recall that the function MATH was defined in Subsection REF. Clearly, MATH . Since the operators MATH and MATH anti-commute, we obtain MATH . Since MATH, it follows, that REF holds with MATH. |
math/0011237 | It is well known, compare , for example, CITE, that the Lemma is equivalent to the following statement: For any MATH there exists a compact set MATH, such that if MATH is a smooth compactly supported section of MATH, then MATH . Here, MATH is the Riemannian volume element on MATH, and MATH denotes the Hermitian scalar product on the fibers of MATH. Set MATH. To prove REF note that, since MATH is a bounded, there exists a compact set MATH, such that MATH on MATH. Note, also, that the first summand in REF is a non-negative operator. Hence, we have MATH . |
math/0011237 | From REF, we see that MATH is a bounded operator, depending continuously on MATH. The lemma follows now from the stability of the index of a NAME operator, compare , for example, CITE. |
math/0011237 | By REF , it is enough to prove the proposition for one particular value of MATH. But it follows from REF that, if MATH is a negative number such that MATH, then MATH, so that MATH. |
math/0011237 | The same calculations as in the proof of REF , show that MATH . Both summands in the right hand side of REF are non-negative. Hence, the kernel of MATH is given by the tensor product of the kernels of these operators. The space MATH is one dimensional and is spanned by the function MATH. Similarly, MATH is one dimensional and is spanned by the one-form MATH, where we denote by MATH the generator of MATH. It follows that MATH . |
math/0011237 | Using the equality MATH we can write MATH . Similarly, MATH . Summing these identities and dividing by REF, we come to REF. |
math/0011237 | Note that MATH for any MATH in the support of MATH. Hence, if MATH, we have MATH. Set MATH and let MATH. Using REF , we obtain MATH . |
math/0011237 | From REF we obtain MATH . |
math/0011238 | The quotient map MATH is a fibration, and since MATH is contractible there is a continuous section MATH. Let MATH and MATH be the given maps. Define MATH by MATH . We have the commutative diagram MATH . Claim. MATH is a proper map. Indeed, let MATH be a sequence in MATH leaving every compact set. If the sequence MATH leaves every compact set, the same is true for MATH. Otherwise, after passing to a subsequence, we may assume that the sequence MATH stays in a compact set MATH. Then MATH stays in the compact set MATH. Since MATH is a proper map, the sequence MATH stays in a compact set, and thus the sequence MATH leaves every compact set. Since MATH is a proper map, we see that the sequence MATH leaves every compact set. Claim. If MATH and MATH are disjoint simplices of MATH, then MATH and MATH diverge. Indeed, let MATH and MATH be sequences in MATH and MATH respectively, leaving every compact set. Since MATH is a NAME map, if one of two sequences MATH and MATH leaves every compact set in MATH, then MATH (since MATH and MATH diverge) and consequently MATH. Now assume that both sequences MATH and MATH are contained in a fixed compact set MATH. Then we have MATH . Since MATH and MATH stay in a compact set and MATH it follows that MATH and MATH . |
math/0011238 | Let MATH be the center of MATH. By CITE the intersection MATH is a lattice in MATH. All claims now follow by induction on MATH from the exact sequence MATH and REF . |
math/0011238 | Let MATH be a maximal compact subgroup in MATH. Now apply REF to the preimage MATH of MATH in MATH. |
math/0011238 | We can assume that the component of the identity of MATH has no compact factors. Further, using the Product Lemma CITE, we may assume that MATH is an irreducible lattice in MATH. If the real rank of MATH is MATH then REF says that MATH is an arithmetic lattice (with respect to some MATH-structure on MATH) and the theorem follows from REF . Now suppose that the real rank of MATH is REF. Then the symmetric space MATH is real, complex, or quaternionic hyperbolic space or the NAME plane (see CITE). If MATH is a uniform lattice acting cocompactly on MATH, then the statement follows from REF , so we may assume that MATH is a nonuniform lattice. If MATH is the real hyperbolic space MATH let MATH be a maximal parabolic subgroup of MATH. Then MATH is commensurable to MATH and there is a proper expanding map of the cone on MATH into a horosphere stabilized by MATH. Adding a ray that diverges from this horosphere but stays within a bounded distance from a MATH-orbit produces a proper expanding map from the cone on MATH and shows that MATH. If MATH is the complex hyperbolic space of complex dimension MATH, then a maximal parabolic subgroup MATH of MATH is a lattice in the NAME group MATH that is a central extension MATH. By REF there is a proper expanding map of the cone on MATH into a horosphere stabilized by MATH. The argument now follows as in the real case. If MATH is quaternionic hyperbolic space or the NAME plane, then MATH is an arithmetic lattice CITE and the claim again follows from REF . |
math/0011238 | First suppose that the center MATH of MATH is finite. If necessary, replace MATH by MATH so that the center is trivial. Then MATH has the structure of (the identity component of) a linear real algebraic group (see for example, CITE) and the proof is reduced to REF . Now suppose that the center MATH of MATH is infinite. Again we may assume that MATH has no compact factors. Then by CITE MATH is a discrete subgroup of MATH, so CITE (with MATH) implies that MATH has finite index in MATH. It now follows from REF and the centerless case applied to the lattice MATH that MATH where MATH is a maximal compact subgroup of MATH. It remains to show that MATH . After passing to finite covers, we may assume that MATH where MATH is a torus and MATH is a simply connected compact group (see for example, CITE). The preimage of MATH in MATH decomposes as MATH where MATH is a torus and MATH. Then MATH can be identified with MATH, so we have MATH . |
math/0011238 | Assume that the adjoint representation has precompact image. Then the NAME algebra MATH of MATH admits a MATH-invariant inner product and hence breaks up as the direct sum of simple and abelian NAME algebras. The simple summands have compact type, so the integral subgroup MATH corresponding to the sum of all simple summands is compact, as well as normal in MATH. The quotient MATH is a noncompact connected abelian group and it therefore maps onto MATH. |
math/0011238 | The isometry group MATH of MATH is a NAME group. Let MATH be the given action, and denote by MATH the closure of MATH. Then MATH is also a NAME group. If MATH has infinitely many components, then the NAME theorem CITE implies that MATH has finite kernel and the image of MATH is closed. It follows that the action of MATH on MATH is properly discontinuous, contradicting the assumption MATH and REF . Now suppose that MATH has only finitely many components. After passing to a subgroup of MATH of finite index, we may assume that MATH is connected. If MATH is compact, then the MATH-orbits, and hence MATH-orbits, are bounded. Thus assume that MATH is noncompact. Then MATH admits a representation MATH whose image has noncompact closure (see REF ). We can arrange in addition that MATH is trivial on the finite central subgroup MATH (by applying REF to MATH). By the NAME 's super-rigidity CITE (applied to an algebraic MATH-group that has MATH as the identity component CITE and to the lattice MATH) there is a continuous extension MATH. The kernel MATH is a normal subgroup of MATH and is therefore contained in the center. It follows from CITE (see also CITE) that MATH is a closed subgroup of MATH, thus the image of MATH is closed and discrete and MATH cannot be connected, contradiction. The last sentence follows from the NAME fixed point theorem CITE. |
math/0011238 | Follow the proof above verbatim until the extension MATH is considered. Now the kernel is a closed normal subgroup and its NAME subalgebra can be assumed, after reordering the factors, to be equal to the NAME subalgebra of MATH for some MATH (if the kernel is discrete the argument concludes the same way). It follows that the product MATH of projections and of MATH has finite kernel. Now CITE again implies that the image of this map is closed, and hence the diagonal action of MATH on MATH is properly discontinuous. REF now implies that MATH. |
math/0011238 | The sphere MATH is the full preimage of the MATH-simplex MATH of MATH and we add the vertex MATH to form MATH. It is straightforward to check that the subcomplex of MATH spanned by the described vertices is precisely a copy of MATH. |
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