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math/0007062 | A presentation of MATH is MATH . By conjugating the last relation by MATH, we may assume the set of relators is MATH and the MATH with MATH. The latter are precisely the relators obtained from MATH by applying the relation MATH. It follows from CITE that MATH is not finitely presented. NAME better, REF gives MATH. |
math/0007062 | Rephrasing of CITE. |
math/0007062 | Rephrasing of CITE. |
math/0007062 | We follow REF . Consider first the group MATH. Clearly, MATH. Using the computer algebra program Gap, we compute a presentation for MATH, and rewrite its relators as words in MATH, where MATH is a generating set for MATH. We also construct a group homomorphism MATH. Then REF gives a finite MATH-presentation for MATH with generators MATH. We now note that MATH can be extended to a homomorphism MATH, where MATH and MATH have order MATH. The substitution MATH can be used instead of MATH, giving rise to a simpler presentation with generators MATH. Finally, we note that the presentation can be simplified from MATH iterated relators to MATH by introducing two extra substitutions MATH induced by group automorphisms. |
math/0007062 | We start by computing a MATH-presentation for MATH. As above, MATH contains MATH; but MATH and neither MATH nor MATH are branch. First we chose generators of MATH: MATH . Then, we chose generators of MATH: MATH . The fact that MATH is normal can be seen in the following conjugation relations: MATH . Define now the group MATH by its generators MATH and relators MATH for MATH and MATH, where MATH is the word in the above table. Note then that MATH is generated by the words MATH. As in the proof of REF , consider the map MATH corresponding to MATH, and by the NAME method compute a presentation for MATH. Assume MATH. A presentation for MATH is MATH. The image of MATH can be described as MATH. We choose MATH as NAME transversal for this subgroup, and denote the NAME generators MATH . The relators we obtain, in terms of MATH, are MATH . This clearly gives a finite MATH-presentation for MATH - compare with the proof of REF . Now the computation of a presentation for MATH can be finished as in the proof of REF . Finally, a finite MATH-presentation for MATH can be obtained using REF . |
math/0007062 | We follow REF . Consider first the group MATH. Clearly, MATH. Using the computer algebra program Gap, we compute a presentation for MATH, and rewrite its relators as words in MATH, where MATH is a generating set for MATH. We also construct a group homomorphism MATH. Then REF gives a finite MATH-presentation for MATH with generators MATH. We now note that MATH can be extended to a homomorphism MATH, where MATH has order MATH. The substitution MATH can be used instead of MATH, giving rise to a simpler presentation with generators MATH, where MATH is a generating set for MATH. Finally, we note that the presentation can be simplified from MATH iterated relators to MATH by introducing an extra substitution MATH, induced by a group automorphism. |
math/0007064 | Approach: In the rational homology sphere case, we have the following crossing change formula MATH for the NAME invariant from Subsection REF: MATH . Recall that for rational homology spheres, MATH. Using the fact that MATH-signMATHdet-MATH, we may multiply the above equation MATH by MATH to yield the crossing change formula for NAME 's invariant in the case of rational homology spheres, that is, when det-MATH: MATH . Notice that if we divide MATH by sign-MATH that the result is a polynomial in the components of MATH, for MATH such that det-MATH. If we can show that the general crossing change formula for NAME 's invariant divided by sign-MATH must be a polynomial in MATH, then the two polynomials (the one obtained by rescaling the NAME invariant as above and the one directly from NAME 's formula) will agree on the set det-MATH, which is dense in MATH. Therefore these two polynomials will agree everywhere in MATH (that is, their difference is zero). So, the crossing change formula in the rational homology sphere case will extend to arbitrary REF-manifolds. Taking this approach, it now suffices to show that the general crossing change formula for NAME 's invariant divided by sign-MATH is polynomial for every framed link. We will do this via NAME 's formula in terms of the NAME polynomial as presented in Subsection REF The goal is to find a formula for MATH. Recall MATH . If we compute MATH from this formula for NAME 's invariant, because MATH and MATH are in the same link homology class (that is, have all the same linking numbers) the terms involving only linking numbers cancel. This leaves us with MATH . If we consider instead MATH, we see, as desired, that this quantity is a polynomial in the variables MATH just as the other crossing change formula. |
math/0007064 | This result is similar to but slightly different from several standard surgery modifications presented in the literature. A proof is presented here for completeness. Verifying this lemma is merely a matter of carefully tracing through surgery definitions. Let MATH. Consider a ``bulging neighborhood" of the zero-framed component to include the MATH framed component. If we perform the two indicated surgeries within this solid torus, we are returned with a solid torus. The remaining question is how this torus is reattached to the exterior of this neighborhood. In other words, we see that the surgery instructions can be reduced to one local component as indicated on the left, but we have yet to determine the framing of that component. MATH . For any surgeries on this solid torus, the resulting first homology will be generated by the meridians of the two link components, denote them MATH and MATH, as indicated, and the longitude of the entire torus, denoted MATH. To help keep track of curves, let us also denote by MATH the longitude of REF-framed component. Because the MATH-framed component links the other component trivially once, the longitude of the MATH-framed component is isotopic to MATH, the meridian of REF-framed component. Performing REF-surgery reveals the relation MATH in first homology, while performing the MATH-surgery provides MATH. It is apparent from the diagram that MATH. This reduces to MATH. Substituting into the first relation gives us MATH. In fact, MATH and MATH are isotopic to the meridian and longitude of the solid torus, so we recognize this as MATH surgery on the core of that torus, as indicated in the surgery diagram on the left. |
math/0007064 | Start with the presentation on the left. Apply a NAME move of the second kind, sliding the component with MATH framing over the component with MATH framing. This unlinks the MATH component from the MATH component, but links the MATH component with the MATH component. Because the framing of the central component is MATH the framing of the MATH component does not change in the process. |
math/0007064 | First write out the surgery coefficients explicitly so that we may see them clearly. MATH . Then insert a trivial component. MATH . Perform MATH left-handed twists to separate the link. MATH . Next apply REF to the central component, changing the framing of the central component to zero and introducing a new component. MATH . Perform the same maneuver to the left component. MATH . Slide the rightmost component over the leftmost zero-framed component (by REF ) and link it to the now MATH component. MATH . At this point we now collapse the diagram by applying REF in the opposite direction. Eliminate one zero framed component MATH then another. MATH . It is a well-known result (stated in REF ) that if a manifold is presented as a ``chain" with integer surgery coefficients (or a rational coefficient on one end) then the manifold is a lens space. In particular, for MATH and MATH integers, the following manifold MATH is a presentation of MATH where MATH and MATH are given by the continued fraction MATH . By this result, the lens space we have here is given by MATH . Recalling the condition that MATH this gives us MATH . Therefore, MATH. |
math/0007068 | Choose a regular cardinal MATH which is large enough so that the following are true: CASE: MATH is dense in MATH, CASE: MATH-filtered colimits of weak equivalences in MATH are again weak equivalences, CASE: MATH has a cofibrant replacement functor MATH which preserves MATH-filtered colimits, CASE: MATH has a fibrant replacement functor MATH which preserves MATH-filtered colimits, CASE: the right adjoint MATH to MATH preserves MATH-filtered colimits (see CITE). Let MATH be the set consisting of all the natural maps MATH where MATH. The condition that MATH is homotopically surjective shows that the derived functor of MATH takes maps in MATH to weak equivalences, and so MATH descends to a map MATH. It is readily checked that this new map is also homotopically surjective. To check that this is a NAME equivalence one must verify that for every object MATH in MATH, the composite MATH is a weak equivalence in MATH. But any MATH in MATH is a MATH-filtered colimit of objects in MATH by REF , and all the functors in sight commute with such colimits by REF - REF . So the map in question is a MATH-filtered colimit of maps in MATH, which are weak equivalences in MATH. Finally, REF says that MATH-filtered colimits preserve weak equivalences in MATH, and it's easy to check that this property is inherited by any localization of MATH. This completes the proof. |
math/0007068 | Let MATH be the category guaranteed by the above proposition. The model category MATH is left proper and combinatorial, so by REF we can find a set of maps MATH in MATH which become weak equivalences in MATH, and such that MATH is a NAME equivalence. |
math/0007068 | If MATH is a combinatorial model category then the Theorem gives us a NAME equivalence MATH for some MATH and MATH. The point is that the universal model category MATH is simplicial and left proper, and these properties are inherited by the localization MATH. We must work a little harder to show that MATH is NAME equivalent to a model category in which every object is cofibrant. Recall that the diagram category MATH has a Heller model structure CITE in which a map MATH is a weak equivalence (respectively, cofibration) if MATH is a weak equivalence (respectively, cofibration) for every MATH. The Heller model structure is related to MATH by a NAME equivalence MATH (where the `H' is for `NAME). This map will still be a NAME equivalence when we localize, so that we get a zig-zag of NAME equivalences MATH . But now the point is that the Heller model structure is simplicial, left proper, and has the property that every object is cofibrant; these properties all pass to the localization. |
math/0007068 | Consider the NAME embedding MATH, together with its canonical cosimplicial resolution induced by the simplicial structure on MATH. In CITE we showed that for any MATH the natural map MATH gives a cofibrant-approximation to MATH. We aim to apply this in the case where MATH is MATH. It's easy to see using adjointness that the overcategory MATH is isomorphic to the overcategory MATH. It's then clear that applying the realization MATH to MATH gives precisely MATH. So we've recovered MATH by starting with MATH, taking a certain cofibrant-approximation in MATH, and then applying the realization MATH. This is precisely what we needed to prove. |
math/0007068 | REF follows directly from the fact that MATH is a NAME pair: the weak equivalence between fibrant objects MATH yields a weak equivalence MATH, and therefore the map MATH is also a weak equivalence. For REF recall that any two cosimplicial resolutions of MATH can be connected by a zig-zag of weak equivalences. So it suffices to prove the result in the case where there is a weak equivalence MATH. We will use MATH for the NAME pair corresponding to MATH. It is easy to see - using the formulas of CITE, for instance - that there are natural transformations MATH and MATH induced by MATH, and these have the properties that MATH and MATH are weak equivalences when MATH is cofibrant and MATH is fibrant. (In the language of CITE, this is a NAME homotopy from MATH to MATH.) If MATH is fibrant we have a weak equivalence MATH, and hence a weak equivalence MATH where MATH denotes any cofibrant-replacement functor in MATH. Consider the square MATH . All the maps in the square are readily seen to be weak equivalences, and so we've shown that MATH and MATH are weakly equivalent via a zig-zag. By the above proposition, this is what we wanted. |
math/0007068 | In light of the above Proposition, this is just a restatement of the definitions. |
math/0007068 | Postponed until REF. |
math/0007068 | The proof of MATH is very easy: Since the underlying category of MATH is locally presentable, for sufficiently large regular cardinals MATH the maps MATH are isomorphisms for all MATH. On the other hand the indexing categories MATH are MATH-filtered, and in combinatorial model categories one has that MATH-filtered colimits are the same as MATH-filtered homotopy colimits for large enough MATH (compare REF ii). So by picking MATH large enough we may ensure both that the map MATH is a weak equivalence and that the map MATH is an isomorphism. This finishes MATH. For MATH we must be more careful. Choose MATH large enough so that MATH is satisfied, but also so that MATH has a cofibrant-replacement functor which maps MATH to itself REF . Let MATH denote this functor, let MATH, and let MATH. Observe that one has maps MATH where MATH is the obvious inclusion, and MATH is the functor sending MATH to MATH. These functors come to us with natural transformations MATH and MATH induced by the natural transformation MATH. Let MATH be the canonical diagram sending an object MATH to MATH. The criteria of REF are readily checked, and so we can conclude that MATH is a weak equivalence. But this is precisely the natural map MATH. By REF the codomain is weakly equivalent to MATH, so we are done. |
math/0007068 | The objects of MATH are pairs MATH where MATH is an object of MATH and MATH is some map in MATH. Since MATH is simplicial, this map has an adjoint MATH. In this way we see that the category MATH is isomorphic to MATH. The map in which we are interested is isomorphic to the map MATH induced by MATH. Now by our choice of MATH we know that MATH is naturally weakly equivalent to MATH, for any MATH. So the above map is weakly equivalent to MATH, which of course is a weak equivalence because MATH was fibrant. |
math/0007068 | By REF we must show that for any fibrant MATH in MATH, the natural map MATH is a weak equivalence. Consider the diagram MATH . The above lemma, together with REF , shows that the horizontal map is a weak equivalence. The diagonal map is a weak equivalence by our choice of MATH REF . Therefore the vertical map is also a weak equivalence, which is what we needed to prove. |
math/0007068 | Let MATH denote the subcategory of MATH consisting of the objects MATH for which MATH is in MATH for all MATH (unlike for MATH, we are not requiring MATH to be a cosimplicial resolution). It is possible to find a functor MATH with the following properties: CASE: Each MATH is a Reedy cofibrant object contained in MATH; CASE: There is a natural weak equivalence MATH; CASE: The object of MATH in dimension MATH is equal to MATH, and the map MATH is the identity. The map MATH is just a certain Reedy cofibrant-replacement functor defined on a subcategory of MATH. In order to construct Reedy cofibrant-replacements, one starts with the MATH-th object and first makes that cofibrant in MATH. For MATH the MATH-th object is already cofibrant, so we can just let it be. Next one moves inductively up the cosimplicial object and factors the latching maps as cofibrations followed by trivial cofibrations (see CITE). By (REF iii) and our choice of MATH, there are factorization functors which will never take us outside the category MATH - this is all that we wanted. If MATH then we will write MATH for the result of applying MATH to the constant cosimplicial object MATH consisting of MATH in every dimension. Consider the maps MATH and MATH induced by MATH and MATH, respectively. These maps have the following behavior: MATH . The composite MATH is the identity by REF of MATH. We will show that the other composite MATH can be connected to the identity by a zig-zag of natural transformations, and then we'll apply REF . The composite MATH sends an object MATH of MATH to the object MATH. Consider the map MATH which maps MATH to MATH. The transformation MATH from REF gives a natural transformation MATH. On the other hand, for any cosimplicial object MATH there is a natural map MATH and therefore a map MATH. This gives a natural transformation MATH. It is easy to check that these transformations satisfy the conditions of REF (see REF as well). So we conclude that MATH is a weak equivalence, and the same for MATH going in the other direction. |
math/0007068 | Let MATH denote the map of simplicial sets which includes MATH as the last vertex of MATH. For any cosimplicial object MATH there is a corresponding map MATH; from this we can define a functor MATH sending the object MATH to MATH. Let MATH denote the functor sending MATH to MATH. The map we are concerned with in the statement of the lemma is MATH. Note that the composite MATH is the identity. We will show that the the other composite MATH can be related to the identity by a zig-zag of natural transformations, and then we'll apply REF . There is a map of simplicial sets MATH such that MATH restricts to the identity map on MATH, and MATH restricts to the map MATH on MATH (left to the reader). Recall that if MATH is a cosimplicial object and MATH is a simplicial set then one gets a new cosimplicial object MATH in a natural way (see the appendix). So MATH induces a map MATH, and by looking at the objects in dimension MATH we get a map MATH which is natural in MATH. Consider the functor MATH defined by MATH . The two inclusions MATH are readily seen to induce natural transformations MATH and MATH. The hypotheses of REF are easily checked to hold, and so we may conclude that MATH, together with MATH going in the other direction, are both weak equivalences. |
math/0007068 | Again, by REF we must show that for any fibrant object MATH in MATH the natural map MATH is a weak equivalence. Consider the commutative diagram MATH . REF says that MATH is a weak equivalence. REF , together with REF , implies the same about MATH. Finally, our assumption on MATH guarantees that MATH is a weak equivalence REF . We therefore conclude that MATH is also a weak equivalence, which is what we wanted. |
math/0007068 | The usual factorizations provided by the small object argument will have the required properties, as long as we use the transfinite version of the small object argument for a sufficiently large ordinal. See CITE. |
math/0007068 | Pick a regular cardinal MATH large enough to satisfy the previous proposition, and also large enough so that MATH is dense in MATH (using locally presentability). The category MATH is small, so applying our given factorizations to maps MATH in MATH only produces a set of new objects. Therefore there exists a regular cardinal MATH such that applying our factorizations to maps between MATH-small objects always produces MATH-small objects. Let MATH be any regular cardinal larger than both MATH and MATH, and let MATH be a map between MATH-small objects. It follows from CITE that we can write MATH as a colimit of maps MATH where MATH, MATH are MATH-small and where the indexing category is both MATH-small and MATH-filtered. Applying our factorization produces maps MATH which are isomorphic to the colimit of the maps MATH. Each MATH is MATH-small (hence MATH-small) by our choice of MATH, and so MATH is a MATH-small colimit of MATH-small objects, hence is itself MATH-small CITE. This completes the proof. |
math/0007068 | Let MATH be a regular cardinal large enough so that there are functorial factorizations preserving MATH-filtered colimits, and so that the model category has a set of generating cofibrations whose domains and codomains are MATH-small. Let MATH be a MATH-filtered indexing category, and let MATH be two diagrams. We suppose that MATH is a map of diagrams such that MATH is a weak equivalence for every MATH, and we'll show that MATH must also be a weak equivalence. Start by factoring the map MATH into a trivial cofibration followed by a fibration, using our preferred functorial factorization: MATH . These maps are the colimits of the maps obtained by applying the factorization to each spot in the diagram: MATH . Now the maps MATH were assumed to be weak equivalences, and therefore the maps MATH are actually trivial fibrations. If we can show that MATH-filtered colimits of trivial fibrations are again trivial fibrations then we will be done: the map MATH will be a trivial fibration, and so MATH will be a weak equivalence. But we can test if a map is a trivial fibration by checking the lifting property with respect to our generating cofibrations. Since the domains and codomains of these generating cofibrations are MATH-small, they will factor through some stage of the MATH-filtered colimit and we will get our lift. |
math/0007068 | The point is that the homotopy theory in MATH all comes from simplicial sets: the weak equivalences are the objectwise weak equivalences, and the simplicial structure is the objectwise structure. So the lemma is immediately reduced to the corresponding fact for simplicial sets, which is well-known. |
math/0007068 | Consider the diagram MATH given by MATH. The geometric realization MATH is isomorphic to MATH (using the definition of MATH, together with the fact that for bisimplicial sets the realization is isomorphic to the diagonal). Our above discussion therefore gives a commutative triangle MATH where the horizontal arrow is the NAME map, and therefore a weak equivalence. Notice that every object of MATH is cofibrant, being in the image of MATH - therefore MATH actually has the correct homotopy type, and we may drop the `bad-' prefix. Moreover, MATH is known to be cofibrant in this case. Now we apply the realization functor to the above triangle, to get MATH . The horizontal map is still a weak equivalence because we applied MATH to a weak equivalence between cofibrant objects. |
math/0007068 | Our assumption is that the maps MATH are weak equivalences, for every MATH. But note that MATH is precisely MATH (once again, MATH may be identified with the simplicial replacement of the canonical diagram MATH). The above maps are weakly equivalent to the iterated degeneracies MATH in the simplicial object MATH. From the fact that these are assumed to be weak equivalences it readily follows that every map in the simplicial object is a weak equivalence. This implies that the natural map MATH is also a weak equivalence (see CITE, for instance). At this point we look at the triangle from the above proposition, and conclude by the two-out-of-three property that MATH is a weak equivalence. This is what we wanted. |
math/0007068 | Consider the triangle MATH which commutes up to homotopy. The slanted map is the identity, and the vertical map is a weak equivalence because of REF on MATH (which says that MATH is an objectwise weak equivalence). So it follows that the composite across the top row MATH is a weak equivalence as well. Likewise, in the triangle MATH assumption MATH on MATH again shows that the composite across the top is a weak equivalence. The notation is a little confusing because the maps labelled MATH in the two diagrams are not exactly the same, although they are both induced by MATH. But the maps labelled MATH are the same and this is all we need. If MATH are maps in some category such that MATH and MATH are both isomorphisms, then each of MATH, MATH, and MATH is also an isomorphism. Applying this to our situation shows that MATH and the two MATH's are isomorphisms in the homotopy category of MATH, hence they are weak equivalences. |
math/0007069 | Let MATH. We choose a basis MATH of MATH and define the linear map MATH by MATH. Then one obtains a complex MATH whose dual is the head of the NAME - Rim complex resolving MATH. It follows that MATH, and obviously MATH. Since MATH, one has MATH. This shows REF . We quote some well-known facts about the homology of MATH. Let MATH. Then MATH for MATH and MATH. (See REF for the general statement.) Furthermore MATH for MATH by the grade sensitivity of the NAME complex for MATH. REF on MATH for MATH is easily proved from the long exact (co)homology sequence. For REF we modify the complex MATH to the complex MATH by setting REF MATH and REF MATH. The maps to be added are the natural surjection MATH, the zero map MATH, and those induced by the canonical injections MATH. The truncation of MATH to the ``rectangle" MATH, MATH has exact columns. Moreover, row homology for indices MATH can only occur at MATH, namely MATH, MATH, and at MATH, namely MATH. For an inductive argument we let MATH, MATH, be the MATH-th row of MATH and MATH be the image complex of MATH in MATH. Then we have a series of exact sequences MATH . Thus we can use the long exact (co)homology sequence for each MATH. With MATH one therefore obtains the ``southwest" isomorphisms MATH if MATH is even, and MATH if MATH is odd, MATH. In fact, there is an exact sequence MATH and the extreme terms in this sequence are MATH for all MATH under consideration. Let now MATH be even, and MATH. Since the map MATH is injective, the same holds true for its restriction MATH in MATH, and so MATH. In the case in which MATH is odd we can go one further step southwest, and obtain the isomorphism MATH for MATH. Since MATH, we only have an exact sequence MATH but the arguments above can be applied to MATH; it is zero if MATH is even, and isomorphic to MATH if MATH is odd. - |
math/0007069 | The implication MATH is an easy exercise. (See also the considerations at the end of this note.) For the converse observe that MATH for MATH and that MATH is free of rank MATH. So MATH must be cyclic. If MATH were even, then MATH by REF , and MATH. Thus MATH must be odd. In this case MATH by REF where MATH. So if MATH, then MATH must be cyclic, which in turn means MATH. - |
math/0007069 | As in REF , p. REF, we choose isomorphisms MATH, MATH, and define maps MATH, by MATH. Via the natural isomorphism MATH we regard MATH as a map MATH. One has MATH, and it is an easy exercise to show that the diagram MATH is commutative or anticommutative (see for REF , proof of REF ). Consequently the same is true for MATH where MATH and MATH are induced by MATH and MATH. Now let MATH be the composition of MATH and the canonical injection MATH. Then the equation asserted in the proposition obviously holds. In case MATH, MATH. This proves the remaining statements. - |
math/0007069 | Consider the diagram in REF . Since MATH is injective, MATH must be zero if MATH. Next let MATH. Then we obtain MATH, since MATH. Finally if MATH, then in addition MATH, so MATH as stated. The remaining assertions concerning the homology of REF are contained in REF . Instead of MATH we can consider the induced linear form MATH on MATH. REF yields the statement about the existence of such a linear form MATH satisfying MATH. Assume that such a MATH exists. If MATH, then MATH by REF . If MATH, then, under our assumptions, MATH is an ideal in MATH which must be isomorphic to MATH. Using the NAME Theorem, we have MATH with an element MATH. Since MATH, MATH must be a unit. - |
math/0007069 | Let MATH be a linear form on MATH such that MATH. Set MATH. Then REF and the last property in REF are obvious. Since MATH is torsionfree and MATH is reflexive, MATH must be reflexive. Because MATH has grade MATH, it is orientable. MATH being orientable, the orientability of MATH follows from REF . Conversely let MATH be a submodule of MATH which satisfies REF . Then MATH is torsionfree of rank REF and therefore an ideal in MATH which is orientable since MATH and MATH are orientable. Consequently MATH. So for a prime MATH in MATH which contains MATH, the localization MATH cannot be free. On the other hand MATH is impossible since MATH. In view of the last condition in REF , MATH must have grade MATH. - |
math/0007070 | REF allows us to extend MATH to an adjoint pair of categories MATH. To extend this further to MATH we must add a simplicial direction, and figure out what the realization of objects like MATH should be for MATH. This is accomplished by the theory of cosimplicial resolutions, discussed in the next section. The proof will be completed at that time. |
math/0007070 | Both the lemma and proposition are proven in REF. |
math/0007070 | See REF. |
math/0007070 | This is not hard, but requires some machinery. See REF. |
math/0007070 | We have just seen that factoring a functor MATH through MATH is equivalent to giving a cosimplicial resolution on MATH. But it is a standard result in the theory of resolutions that REF any diagram MATH has a cosimplicial resolution, and REF the category MATH is contractible (both are proven in CITE). So REF is just a re-casting of these classical facts. |
math/0007070 | See REF. |
math/0007070 | The point is that the model categories MATH are simplicial and left proper, and these properties are inherited by the localizations MATH. |
math/0007070 | One simply chooses a presentation MATH and then uses REF to lift this map across the NAME equivalences. |
math/0007070 | The NAME embedding MATH will extend to a map MATH (and for convenience we choose the extension induced by the standard cosimplicial resolution, using the fact that MATH is a simplicial model category). The relations we are imposing in MATH clearly hold in MATH, and so this map descends to MATH. It's easy to check that the left adjoint is the sheafification functor and the right adjoint is the inclusion of simplicial sheaves into simplicial presheaves. Perhaps the easiest way to see that this is a NAME equivalence is to factor the map into two pieces. In fact, to start with let's forget about the MATH-homotopy relations; the map we're considering factors as follows: MATH . Here MATH is the model structure constructed in the last section, and MATH is the model structure of CITE mentioned above. The first NAME pair is an equivalence by REF . That the second is a NAME equivalence is essentially CITE. By the above observation these also give NAME equivalences after we localize at the maps MATH. |
math/0007070 | We only give a sketch. The reader can also consult CITE for a similar statement. Consider the subcategory MATH whose unique object is the one-point manifold. This inclusion induces a NAME map MATH. The composition MATH is the usual realization/singular functor pair, and is therefore a NAME equivalence. So the homotopy theory of topological spaces is a retract of that of MATH, and what we have to show is that MATH doesn't contain anything more. This is where our relations come in, because they are enough to allow us to unravel any manifold into a simplicial set. If MATH is a manifold we may choose a cover MATH whose elements are homeomorphic to open balls in Euclidean space, hence contractible. For each intersection MATH we may do the same, and so on for all the multiple intersections - in this way we build a hypercover for MATH in which all the open sets are contractible. Relation REF allows us to express MATH (the object in MATH) as a homotopy colimit of these contractible pieces, and relation REF allows us to replace each contractible piece by a point, up to weak equivalence. So we find that any representable object in MATH may be expressed as a homotopy colimit of points, which of course is just the data in a simplicial set. In addition we know that every object of MATH is canonically a homotopy colimit of representables, so it follows that every object can be decomposed into just a simplicial set. |
math/0007070 | Left to the reader. |
math/0007070 | The fact that MATH is cofibrant implies that MATH is a cofibration, and so the map MATH is also a cofibration. Then MATH is a sequential colimit of cofibrations beginning with MATH, hence cofibrant. |
math/0007070 | First observe that MATH has a free degeneracy decomposition: we take MATH to be the coproduct MATH in which no map MATH is an identity map. Each MATH is a coproduct of representables, hence cofibrant. So MATH is itself cofibrant by the above corollary. We must next show that MATH is a weak equivalence in MATH - that is, we must show that MATH is a weak equivalence of simplicial sets, for every MATH. Let MATH denote the subcategory of MATH consisting of the same objects but only identity maps. Consider the adjoint pair MATH where MATH is the restriction functor and MATH is its left adjoint. Then MATH is a cotriple, and the cotriple resolution MATH can be seen to exactly coincide with MATH. Now of course if we apply MATH again then we pick up an extra degeneracy, and the map MATH is a weak equivalence in MATH. But applying MATH to a simplicial presheaf gives precisely the collection of all its values, and so we have that MATH is a weak equivalence for every MATH. |
math/0007070 | One again shows that MATH has a free degeneracy decomposition in which the MATH are coproducts of representables. This takes a little more work than for MATH, but we will leave it to the reader. The fact that MATH is cofibrant follows from REF . To see that MATH is a weak equivalence we consider the bisimplicial object MATH whose MATH-th row is MATH, as well as the `constant' bisimplicial object MATH whose MATH-th row is the discrete simplicial presheaf consisting of MATH in every level. The map MATH is the diagonal of a map MATH. But we have already shown that MATH is a weak equivalence for every MATH, which says that MATH is a weak equivalence on each row. It follows that the map yields a weak equivalence on the diagonal as well. |
math/0007070 | We explained in REF why MATH was cofibrant, therefore the only thing to prove is that the natural map MATH is a weak equivalence in MATH. So we need to show that for every MATH the map MATH is a weak equivalence of simplicial sets. For brevity let MATH denote the category MATH. The object MATH is the homotopy colimit of the diagram MATH which sends MATH to MATH. Because the simplicial structure in MATH is the objectwise structure, homotopy colimits are also computed objectwise. This says that MATH is equal to the homotopy colimit of the diagram MATH sending MATH to MATH. This latter object may be identified with MATH, which is MATH - it is a coproduct of copies of MATH, one for each map MATH. Consider the functor MATH which sends MATH to the set MATH. From this functor we may form its NAME construction MATH: this is the category whose objects are pairs MATH where MATH and MATH, and a map MATH is a map MATH in MATH such that MATH. An object of MATH corresponds to the data MATH, so define a functor MATH which sends this object to the simplicial set MATH. NAME has a theorem about homotopy colimits over NAME constructions CITE, and in our situation it gives us a weak equivalence MATH . If MATH corresponds to the data MATH, then the homotopy colimit inside the brackets is just a coproduct of copies of MATH, one for each element of MATH. In this way the double homotopy colimit on the right is readily identified with MATH, and we have already seen that this is MATH. Now consider the category MATH, defined so that the objects consist of the data MATH - this is equal to the category of simplices of the simplicial set MATH (defined in CITE, for instance). We again let MATH denote the diagram which sends MATH to MATH. The colimit of this diagram is just MATH, and the natural map MATH is a weak equivalence of simplicial sets. There is a functor MATH which sends MATH and this induces a map of homotopy colimits MATH. The map of categories has a retraction which is easily checked to be homotopy-cofinal, so it follows that the map of homotopy colimits is a weak equivalence. NAME what we have is the following diagram: MATH . We have shown that every map is a weak equivalence except the vertical one, but then the vertical map must be one as well. This is the statement that MATH is a weak equivalence, which was our goal. |
math/0007070 | Suppose we have a factorization of MATH through MATH: so we have a NAME pair MATH and a natural weak equivalence MATH. Then for each MATH we get a cosimplicial resolution of MATH by taking MATH to be MATH . This is clearly functorial in MATH, and so gives a resolution of MATH. Conversely, suppose we start with a resolution MATH for the functor MATH. Define the functors MATH and MATH by the formulas MATH . REF says that these are an adjoint pair. To see that these are a NAME pair we will check that MATH preserves fibrations and trivial fibrations. For this we need to know that if MATH is a cosimplicial resolution and MATH is a fibration (respectively, trivial fibration) then MATH is a fibration (respectively, trivial fibration) of simplicial sets. But this is CITE. The last thing is to give a natural weak equivalence MATH. But MATH is isomorphic to the object of MATH in degree MATH, and our cosimplicial resolution came with a weak equivalence from this object to MATH. So we're done. Checking the equivalence of categories MATH is fairly routine at this point: we have given the functors in either direction. |
math/0007070 | Recall that there exists an equivalence between maps of model categories MATH and the following data: CASE: A functor MATH whose image lies in the cofibrant objects, and CASE: A cosimplicial resolution on MATH. Giving a NAME homotopy between two maps MATH corresponds to giving a natural weak equivalence MATH and a lifting of this to a natural weak equivalence between the resolutions. Using these facts, proving the lemma is just a matter of getting zig-zags between the resolutions. But this is standard - see CITE. |
math/0007070 | Let MATH be the composite MATH. We will begin by lifting MATH, and this can be accomplished just by lifting MATH. Define MATH by MATH where MATH is the right-adjoint to MATH. We may extend MATH to a map MATH. CASE: The composite MATH is NAME to MATH. To see this, observe that there are natural weak equivalences MATH . Since MATH is an extension of MATH and MATH is an extension of MATH, the claim follows directly from the above lemma. CASE: The map MATH takes elements of MATH to weak equivalences. For this, note that by hypothesis the derived functor of MATH takes elements of MATH to weak equivalences. The same must be true for MATH, since MATH is homotopic to MATH (maps which are NAME will have isomorphic derived functors on the homotopy categories). But MATH was a NAME equivalence, and so the derived functor of MATH must also take elements of MATH to weak equivalences. From REF it follows that MATH descends to a map MATH. The fact that MATH is homotopic to MATH is just a restatement of REF . |
math/0007074 | If we project MATH from this line MATH to MATH, we get a variety MATH of degree MATH and dimension MATH in MATH. Since MATH is non degenerate, MATH. It follows that MATH, hence MATH is of minimal degree. The rational scroll MATH is the image of a smooth rational normal scroll MATH of degree MATH and dimension MATH, by a linear projection. We deduce from the classification of varieties of minimal degree (CITE,CITE) that MATH is a rational normal scroll. The induced rational map MATH from MATH to MATH is thus an elementary transformation (compare CITE) of MATH along MATH. It follows that MATH is the tautological embedding of a projective bundle, elementary transform of MATH; therefore MATH is smooth. |
math/0007074 | Let us denote by MATH the projective bundle MATH in MATH, where MATH. The second projection MATH from the product MATH embeds MATH in MATH as a cone MATH of vertex MATH over the smooth rational normal scroll MATH. Its restriction to MATH is a desingularization of MATH of exceptional locus MATH. A simple NAME class computation shows that MATH, with MATH for all MATH. Let us denote by MATH the smooth rational normal scroll MATH in MATH. Since MATH, the map MATH defines a morphism MATH from MATH to MATH, which image MATH is a non degenerated rational subscroll of MATH in MATH. Moreover since MATH is linearly normal and MATH non degenerated, MATH is the restriction to MATH of a linear projection. Since the restriction MATH of MATH to MATH has to be an injection of MATH-modules, it defines an elementary transformation between MATH and MATH, induced by a linear projection from MATH to MATH, which factors through MATH and the projection of MATH from the line MATH. Therefore MATH is smooth of dimension MATH, so that MATH induces an isomorphism of rational scrolls between MATH and MATH and MATH. We denote by MATH the strict transform of MATH in MATH. Let MATH denote the embedding map MATH. Notice that the diagonal MATH in MATH is isomophic to MATH by MATH, so that the schemes MATH and MATH are isomorphic by MATH. For each MATH, the injections induced by MATH, MATH, define in MATH over MATH a rational curve MATH. Let MATH denote the cokernel of MATH . Remark that MATH. Since for each MATH, the line MATH meets MATH, the scheme MATH is supported on MATH in MATH. If MATH, the smoothness of each curve MATH at each point of MATH, shows that MATH is on MATH for both curve MATH and MATH. The cotangent space of MATH at MATH is then the direct sum of the cotangent spaces of MATH at this point. Therefore the multiplicity of MATH at MATH is MATH. From this, we deduce that the length of MATH is MATH. |
math/0007074 | Let MATH be a desingularization of MATH of exceptional locus MATH. We want to determine the linearly normal variety MATH of which MATH is projection. We show by induction on MATH that MATH is a variety of minimal degree MATH. To do so, we need the existence of an hyperplane section of MATH containing MATH that is desingularisezd by MATH. CASE: The variety MATH is smooth in a neighborhood of MATH. CASE: The variety MATH satisfies MATH. This is true for MATH since then MATH is a smooth rational curve CITE. Let us assume it is true for MATH. Consider the linear system MATH of hyperplanes MATH containing MATH. The generic element of the restriction to MATH of MATH satisfies our induction hypothesis so that any MATH in MATH is a smooth point of MATH for MATH generic. Since any smooth point MATH on a NAME divisor MATH of a variety MATH is a smooth point of MATH, MATH is a smooth point of MATH. Since MATH is a birational invariant for smooth varieties, we deduce that MATH for any desingularization MATH of the variety MATH. Moreover since MATH is a variety of minimal degree, hence a cone over a smooth rational variety, we have MATH. Let us now prove that MATH is the image by a regular projection of a variety of minimal degree MATH. Let MATH be a generic hyperplane section of MATH. Recall that MATH is a desingularization of MATH. CASE: A generic element of the linear system MATH is smooth and the dimension of this system is MATH. CASE: The rational map MATH defined by this system is in fact regular and maps MATH onto a variety MATH of dimension MATH and degree MATH. For MATH, CITE, the curve MATH is a smooth rational curve of degree MATH, hence is a regular projection of some rational normal curve in MATH. For MATH assume that for any k-dimensional varieties MATH having an extremal secant line, the total transform MATH of the linear system of hyperplane sections of MATH, gives a map from MATH to MATH which image MATH is a MATH-dimensional variety of degree MATH, that projects onto MATH. Let us consider on MATH the total transform MATH of the system cut out on MATH by the system of hyperplanes containing MATH. Let MATH be a generic hyperplane section of MATH containing MATH; a generic element of MATH is irreducible and MATH of MATH is smooth. Indeed MATH is smooth away from its base locus by NAME 's theorem on singularities of a generic member of a linear system. The variety MATH is smooth at MATH, hence the base locus of MATH consists of the MATH points MATH. The divisor MATH is an irreducible hypersurface of MATH by NAME 's irreducibility theorem (MATH); it has an extremal secant line by construction, hence MATH is smooth in a neighborhood of MATH by REF. The NAME divisor MATH is then smooth at MATH. This also implies that a generic element of MATH is smooth. We can thus find an irreductible and non degenerated hyperplane section of MATH, MATH such that MATH is a smooth element of MATH. We have the following exact sequence: MATH . By REF, MATH. The restriction of MATH to the closure of MATH in MATH is a desingularization of MATH so that we can apply the induction hypothesis to a generic hyperplane section MATH of MATH to get: MATH . Since a generic element of MATH doesn't meet MATH, MATH . Since MATH is base point free, the degree of MATH is MATH and the rational map MATH that MATH defines is regular. The dimension of MATH is at least MATH. Since MATH, is the union of the points in the pencil formed by MATH independent hyperplane sections MATH and MATH of MATH containing MATH, MATH is cut out by the pencil generated by the images of the strict transforms of MATH and MATH in MATH and therefore has dimension MATH. The linear system MATH generically separates points, so that the map MATH is birational onto its image. Its inverse MATH is then a rational map from MATH onto MATH that extends to a linear projection MATH from MATH onto MATH, since MATH is linearly normal. The map MATH is regular since MATH and MATH have the same dimension and the same degree. We can now use the NAME theorem to conclude. Assume that MATH, then MATH is a cone over the NAME surface MATH in MATH and MATH is the linear projection of MATH by a point. Since the extremal secant line MATH meets MATH at smooth points, it must be the image of a MATH-secant MATH-plane of MATH, hence the center of projection lies in the linear span of MATH and MATH is a cone over the NAME surface in MATH. If MATH the variety MATH can be neither a cone over the NAME surface nor the NAME surface itself, unless MATH. Indeed, if it were so, MATH would be a cone MATH over the generic projection of the NAME to MATH or over the NAME surface in MATH, with MATH. So clearly, we would have MATH. If MATH, we cannot have any projection of the NAME surface either. Hence MATH must be a rational scroll unless MATH was already a variety of minimal degree. Since the projection MATH is regular, if MATH is smooth, MATH is a smooth rational scroll. If MATH is a cone MATH over a smooth scroll of minimal degree MATH, MATH must be the cone MATH, hence is a cone over a smooth rational scroll. Any extremal secant line moreover has to lie in the linear space generated by MATH, for it has to meet MATH at smooth points. |
math/0007074 | The MATH-regularity of MATH is equivalent to the MATH-normality of MATH and the vanishing of MATH. Note that by degeneration of the NAME spectral sequence MATH so that we only have to show that MATH . The embedding MATH factors through an embedding MATH of MATH in MATH and the second projection MATH onto MATH, hence corresponds to the data of an exact sequence of MATH-modules MATH such that MATH . Let us denote by MATH the first projection from the product MATH. From the NAME exact sequence on MATH, twisted by MATH and pulled back by MATH to MATH and from the pull back by MATH of the morphism MATH we get a morphism MATH of MATH-modules which vanishing locus is scheme-theoretically MATH. The NAME complex associated to MATH gives a locally free resolution of MATH: MATH . Indeed this complex is exact since MATH is generically surjective and its cokernel MATH is of expected codimension REF in the locally NAME scheme MATH (CITE p. REF). Since the article of NAME, NAME and NAME, it is now a standard trick , to twist this resolution by the pull back by MATH of a suitable line bundle MATH on MATH, in order to be able to push down to MATH in MATH the pieces of information of this resolution of MATH. There is a line bundle MATH on MATH which satisfies MATH . For such a line bundle we have then MATH and MATH. The proof of this is similar to CITE REF . First we have a strictly decreasing filtration by vector bundles MATH such that MATH is a line bundle of negative degree MATH . In order to get MATH and MATH, it is enough to choose MATH such that MATH and MATH. Since a generic line bundle of degree MATH is non special, it is enough to take MATH generic, so that MATH, that is to say MATH. Indeed we have MATH. Since MATH the smallest possible degree for MATH is MATH, so that we also have MATH,. Note that MATH, so that by NAME 's theorem MATH. Since MATH and MATH we have MATH. Applying NAME 's formula to the push forward by MATH of the NAME resolution of MATH twisted by MATH we get the following complex MATH . Let MATH denote the MATH-projective bundle structure map of MATH. For the choice of line bundle MATH as in the previous lemma, by the same argument as in REF , this complex is exact. We get MATH by the projection formula. The degeneracy locus of the morphism MATH that we obtain this way is set-theoretically MATH. The cokernel of MATH is isomorphic to MATH by the projection formula applied to MATH and MATH. A generic section MATH of MATH induces an exact sequence MATH whith MATH and where the MATH's are distinct points of MATH. The push forward by MATH of this exact sequence is MATH where MATH denotes MATH. Moreover since MATH is a birational morphism and MATH is smooth, MATH. On each fiber MATH, we have MATH. Therefore we get MATH, so that MATH where MATH is the fiber of the scroll MATH over the point MATH. As in CITE (proof of REF p. REF) the existence of the commutative diagram MATH which allows us to compare the regularity of MATH with the regularity of MATH, follows from the minimality of MATH . In CITE the morphism corresponding to MATH is constructed from a minimal free resolution; the local minimality we seek here is taken care of by the following lemma. (A minimality criterion) Let MATH be the NAME complex associated to some map MATH where MATH is a local ring of maximal ideal MATH. Let us assume that there is a local morphism MATH turning MATH into a finite free MATH-module of rank MATH. Any MATH-module MATH has via this morphism a MATH-module structure that we denote by MATH. Consider the map of MATH-modules MATH, obtained by composing MATH with MATH, where MATH is the wedge product of the first MATH elements of a base of the free MATH-module MATH. The NAME complex MATH of MATH-modules associated to MATH surjects onto the complex of free modules MATH. In particular this last complex is minimal if MATH is minimal. Moreover if MATH, where MATH is the maximal ideal of MATH, then the complex MATH is minimal. The natural surjection of MATH-modules MATH gives a commutative diagram MATH that induces a surjective map of complexes of free MATH-modules between MATH and MATH. Indeed by construction the map of MATH-modules MATH is surjective hence MATH is also surjective. It thus induces an automorphism MATH of MATH. It is then clear that MATH. The surjection between the complexes of MATH-modules induced by MATH gives a surjection MATH as claimed. If MATH after wedging by MATH, we have MATH so that MATH is minimal when MATH is minimal. By NAME formula MATH, so that MATH is a finite locally free MATH-module of rank MATH. We are exactly in the situation of the criterion since locally, the complex MATH is simply the NAME complex associated to MATH. Since the map MATH induces a surjection on global sections, none of the sections defining MATH vanishes. This shows then that MATH is locally minimal. We have MATH for all MATH. This is equivalent to show that MATH for all MATH . Indeed since MATH is projective and only has fibers of dimension MATH, MATH vanishes on any coherent sheaf for MATH, so that MATH is surjective. By non speciality of MATH and use of the NAME formula we get MATH. Consider now the NAME resolution of MATH twisted by MATH . Let us denote by MATH the cokernel of MATH . We have MATH for MATH. Since the fibers of MATH are of dimension MATH and MATH is projective we know that MATH vanishes on any coherent sheaf, so that MATH is right exact. Then MATH for all MATH, since MATH and MATH. Therefore MATH for all MATH. Repeating this argument, we deduce that MATH for all MATH when MATH and MATH . Moreover MATH, since MATH is coherent (MATH is projective) and MATH. Therefore MATH. To show that MATH is MATH-regular, it suffices to show that MATH for MATH and MATH. We have MATH, for MATH and MATH. We deduce from the local minimality of MATH that the following complex extracted from the previous commutative diagram MATH is also locally minimal. Now the sheafification of a minimal resolution of MATH is of the form MATH . Since the exact sequence MATH comes from the sheafification of a locally free minimal resolution of MATH for some MATH, we can construct the following commutative diagram MATH . Since MATH has a resolution of type MATH we have MATH for MATH ( by REF). Therefore to show that MATH for MATH, it suffices to show that MATH is MATH-regular, where MATH is defined by the following trianglular diagram MATH . This holds by the now standard trick due to NAME, NAME and NAME, consisting in applying REF to the NAME complex associated to MATH twisted by MATH. |
math/0007076 | Let MATH be a collecting of affine MATH-varieties covering MATH. For every MATH the ring MATH is noetherian and integrally closed and therefore a NAME ring. Now MATH equals the intersection of all the MATH considered as subrings of the function field MATH. Hence MATH is a NAME ring. |
math/0007076 | This is immediate, because MATH is the intersection of two NAME rings, namely MATH and MATH where MATH denotes the quotient field of MATH. |
math/0007076 | Let MATH denote the quotient field of MATH. Let MATH and MATH. Since MATH, the prime ideal MATH of MATH does not contain MATH. Hence there is an element MATH. By REF for some natural number MATH. Since MATH, this implies MATH. We will now show the converse. Let MATH be a regular function on MATH, that is, MATH, and let MATH. For MATH consider the associate prime ideal MATH. Now MATH implies that MATH, hence MATH. Thus MATH for all MATH and MATH. Since MATH is finite, it follows that MATH for MATH sufficiently large, that is, MATH. |
math/0007076 | Let MATH and MATH. Then there is an element MATH such that MATH. Now MATH implies MATH. Conversely, assume MATH is an element in MATH such that MATH for all MATH. Let MATH. Then MATH is finite and MATH for all MATH. This implies MATH for MATH sufficiently large. |
math/0007076 | Let MATH be the quotient field of MATH, MATH, MATH. Choose MATH and define MATH. Now MATH is finite, and MATH being a prime ideal of height one for every MATH implies that MATH is not contained in MATH for any MATH. Hence there is an element MATH with MATH but MATH for all MATH. Then MATH for MATH sufficiently large. |
math/0007076 | Let MATH denote the discrete valuation corresponding to MATH. By MATH there is an element MATH with MATH and a number MATH such that MATH. Thus MATH defines a rational function on MATH whose poles are contained in MATH. Now assume MATH but MATH. This would imply that MATH and MATH vanishes on MATH. Since the poles of MATH are contained in MATH, it follows that MATH for MATH large enough. On the other hand MATH implies MATH. Thus the assumption MATH yields a contradiction to the requirement that MATH. For the final statement, let MATH denote the irreducible components of MATH. Then MATH. Since both MATH and all of the MATH are prime ideals, it follows that there exists a MATH such that MATH. |
math/0007076 | The equivalence relation MATH defines a subset MATH via MATH . Then MATH is the MATH-sub variety defined by the radical of the ideal of MATH generated by MATH with MATH running through MATH. If MATH are two MATH-sub algebras of MATH, then MATH. Since MATH is noetherien it follows that for any MATH-sub algebra MATH there exists a finitely generated MATH-sub algebra MATH such that MATH. Finally, if MATH is a NAME ring, then MATH is integrally closed in its quotient field. Hence the integral closure of MATH is again contained in MATH. Furthermore the integral closure of MATH is again finitely generated as a MATH-algebra. Thus we may assume that MATH is integrally closed. But integrally closed finitely generated MATH-algebras are NAME rings. |
math/0007076 | Let MATH, this is an affine MATH-variety. The prime ideals of height one in MATH correspond to the irreducible hypersurfaces in MATH. Let MATH denote the morphism induced by MATH. Let MATH be a prime ideal of height one in MATH. We have seen above that there is an irreducible subvariety MATH such that MATH. From MATH we infer that there exists an irreducible subvariety MATH such that MATH is generically quasi-finite and MATH is the smallest MATH-saturated subvariety containing MATH. Now let MATH be an irreducible subvariety of MATH such that MATH and MATH. Let MATH. Then MATH. On the other hand MATH and MATH. Thus MATH implies MATH. It follows that MATH must be NAME in MATH. Since MATH it now follows from MATH being generically quasi-finite that MATH either is a hypersurface or equals MATH. The latter is excluded since MATH and MATH. Thus MATH has to be a hypersurface implying that MATH. |
math/0007076 | This is an immediate consequence of MATH and MATH. |
math/0007076 | Let MATH, MATH the corresponding hypersurface and MATH. We claim that MATH is not NAME in MATH. Indeed, if it were NAME, there would exist an irreducible subvariety MATH with MATH. But this implies MATH. Now let MATH be a prime ideal of height one contained in MATH (such an ideal exists, because MATH is a prime ideal and MATH is a NAME ring). Then MATH is a prime ideal of height one. Since MATH is of height one, it follows that MATH contrary to our assumption MATH. Thus MATH is a hypersurface in MATH with MATH not being dense in MATH. Since MATH is dominant, there are only finitely many such hypersurfaces in MATH. |
math/0007076 | Let MATH, MATH, MATH and MATH be as in the theorem. Note that MATH is a finitely generated field extension of MATH, because MATH is finitely generated and MATH. By REF there is a finitely generated MATH-algebra MATH with MATH, MATH and MATH being a NAME ring. We may adjoin finitely many further elements of MATH to MATH and thereby assume that the quotient fields of MATH and MATH coincide. Then we choose MATH as the integral closure of MATH in MATH. Since MATH is integrally closed, we have MATH. Furthermore MATH and MATH is again finitely generated, because it is the integral closure of a finitely generated MATH-algebra. Thus MATH fulfills the ``NAME Assumptions". Statement MATH of the theorem is clear, and statement MATH follows from REF . Next we define MATH and MATH as in the corollary above. By REF the difference set MATH is finite. Hence we may define an ideal of MATH by MATH with MATH. Since MATH is of height one for every MATH, it is clear that MATH for MATH. Therefore MATH by REF . Finally, REF follows with the aid of REF . |
math/0007076 | Given a MATH-algebra homomorphism MATH, choose a finitely generated MATH-sub algebra MATH such that the quotient fields of MATH and MATH coincide. The restriction of MATH yields a MATH-morphism MATH from MATH to MATH and the inclusion MATH yields a birational morphism MATH. Now MATH is the desired rational map. |
math/0007076 | We apply REF with MATH and set MATH. There is an inclusion MATH. By the preceding lemma this induces a rational map MATH with MATH. Since MATH, it follows that MATH. Finally note that for every affine MATH-variety MATH and every MATH-invariant morphism MATH we obtain an inclusion MATH which implies that there exists a morphism MATH such that MATH. |
math/0007076 | The implication MATH follows from REF . MATH is trivial and MATH follows from the proposition below. Finally, MATH for the case of characteristic zero is implied by the well-known correspondence between locally nilpotent derivations and MATH-actions on affine varieties in characteristic zero. |
math/0007076 | Let MATH be an open embedding in a normal affine MATH-variety MATH, and let MATH. Let MATH denote the union of codimension MATH-components of MATH and choose a regular function MATH on MATH such that MATH is contained in the zero set of MATH. Then choose a regular function MATH on MATH such that MATH vanishes on MATH, but does not vanish on any irreducible component of MATH. If MATH, we replace MATH by MATH for a sufficiently large MATH. In this way we may assume that both functions MATH are defined over a finite NAME extension MATH with NAME group MATH. Now we may replace MATH by MATH. Therefore we may assume that both MATH are defined over MATH. We obtain a MATH-morphism MATH. By construction MATH is the union of codimension MATH-components of MATH. Since regular functions extend through subvarieties of codimension at least REF on normal varieties, it follows that MATH for MATH. Next we consider MATH and the natural projection MATH given by MATH. This realizes MATH as the quotient of MATH by the MATH-action given by MATH . Now the fiber product MATH is an affine variety and MATH, because the image of MATH in MATH is contained in MATH and MATH. The MATH-action on MATH induces a MATH-action on the fibered product MATH and evidently MATH. |
math/0007076 | Let MATH and MATH. Each element in MATH is a birational self-map of MATH such that the induced field automorphism of MATH stabilizes MATH. In particular MATH. Due to REF there exists an object MATH and a MATH-morphism MATH such that MATH. Consider now an object MATH with a MATH-invariant MATH-morphism MATH. Then MATH is dominant rational map with MATH. Thus we obtain a MATH-algebra homomorphism MATH, which by REF induces a MATH-morphism from MATH to MATH. Therefore MATH. |
math/0007082 | The first term in the formula is clearly determined by the others. By induction on MATH, the mixed invariants of degree MATH are therefore determined by the one-point invariants of degree MATH or less. |
math/0007082 | We apply the main theorem in the case MATH. Only the double sum appears, in this case as the single sum MATH . The first and last of the terms are: MATH and MATH . Multiply REF through by MATH, and the main theorem tells us we obtain an element of MATH when we integrate against MATH. For example, by the projection formula, the first term gives MATH and similarly for the other the terms. When we consider only the terms that are constant in MATH, we obtain the following formula: MATH where MATH are a basis, with respect to which MATH is the inverse of the intersection matrix. This much holds for any ample MATH generating MATH. The fact that MATH is a complete intersection tells us that all the one-point invariants of the form MATH vanish when MATH is a primitive cohomology class. This can either be seen using NAME 's formulas, or by a monodromy argument. Since a basis for the cohomology may be chosen consisting of powers of MATH and (orthogonal) primitive classes, this tells us that we may replace the basis MATH by the smaller set MATH of powers of MATH, resulting in REF . |
math/0007082 | The existence of MATH follows from the functoriality of NAME spaces. The two projections are just the two maps: MATH and MATH which clearly commute with the action of MATH. Over the open subset of MATH consisting of smooth curves, MATH is an isomorphism. The last parametrization is of the same component as the first, and is given by the correspondence MATH induced by the curve in MATH. |
math/0007082 | Given a stable map MATH, represent MATH by a tree with vertices and edges corresponding to the nodes and components of MATH, respectively. For a MATH to map to MATH under the forgetful map, each MATH must parametrize a different curve (edge of the tree) mapping with degree MATH to MATH, and each MATH must be a node (vertex) for MATH,,MATH. Also, the shortest path between two such vertices of the tree cannot contain any such edges, and if one of those edges is removed, the tree either stays connected or it has two components, one of which is an edge corresponding to MATH mapping with degree MATH to MATH. In order for MATH to map to MATH under the other forgetful map to MATH, MATH and MATH must represent different edges, MATH and MATH must represent vertices (possibly the same one) and the shortest path from MATH to MATH may not contain either of the two edges. The only fixed points under the torus action which satisfy both conditions are those of types REF. This proves the first part of the lemma. Under the map to MATH, the parametrization of MATH is forgotten. This may result in an unparametrized component with MATH or MATH nodes, which is then collapsed. Moreover, in the MATH node case, the resulting marked point must also be forgotten. This gives the second part of the lemma. |
math/0007082 | The two sides of the formula represent the contribution of MATH to localization formulas for MATH which, by uniqueness, must coincide. |
math/0007082 | Let MATH. Then applying correspondence of residues to the map MATH of smooth NAME stacks (here we use MATH) of REF , and using the enumeration of fixed loci in REF , we get: MATH . (The computation MATH is easily made.) Now, the equivariant NAME classes MATH appearing in the denominators depend entirely on the nodes of the domain of a general representative MATH. Essentially, there are two types of nodes: those of type I, where at the point MATH, the MATH-th parametrized component meets a component mapping in positive degree MATH to MATH, and those of type II, where at the point MATH, the MATH-th parametrized component meets the point MATH of the MATH-th parametrized component. See REF . Any type I node is a codimension MATH condition - one for the node and one for specifying MATH - while a type II node is a codimension MATH condition - one for the node and two more for specifying MATH and MATH. Set MATH . Then the type I node contributes the factor MATH to MATH, while the type II node contributes MATH . (For this type of computation, see CITE.) Thus, if we let MATH the following computations are valid on any MATH: MATH . Finally, let MATH and consider MATH, the equivariant hyperplane class on the linear space MATH. There is an equivariant morphism MATH which is a composition of the forgetful map remembering only MATH and the last parametrization, and the ``map to the linear sigma model." (The geometry of this second (birational) map was used in CITE to give a proof of the mirror theorem.) This MATH pulls back to the fixed loci as follows: MATH . To see this, note that under the morphism MATH, the fixed loci map to various copies of MATH sitting as the fixed loci in MATH. More specifically, set MATH which are all fixed under the MATH action. One easily computes that MATH restricts to MATH as MATH. Finally, fixed loci of types REFa, and REFc map under MATH to MATH, the fixed loci of type REFb map to MATH, and the loci of type REF map to MATH for appropriate MATH. Substitute in the summands of REF for the equivariant NAME classes and for the choice MATH and push forward under the total evaluation map MATH. Then by the projection formula and the computations above we obtain the following: MATH . We get the theorem by multiplying both sides of REF by MATH, collecting types REFa, and REFb under the double sum, and noting that MATH because MATH is polynomial in MATH,,MATH and the inverse to MATH belongs to MATH. |
math/0007082 | It suffices by induction to prove the lemma for the case MATH. In that case, we will consider a (non-commuting!) diagram of stacks: MATH where the horizontal maps are the ``cross-ratio" maps defined as follows. The universal curve MATH over MATH maps to MATH via the forgetful map. Together with the evaluation map to MATH, this defines MATH. If MATH is a stable map with MATH marked points MATH, MATH, MATH, then the map MATH defined by MATH may be taken to be the unique map with the property that MATH and MATH. This is the cross-ratio if MATH, MATH, MATH belong to the same component of MATH, but is well-defined even if they lie on different components. For MATH, we apply the prestabilization map MATH (see CITE) to the universal curve over MATH followed by the stabilization map MATH. This map has the same pointwise description as MATH. The diagram doesn't commute because MATH stabilizes unstable maps to MATH, while MATH does not. On the other hand, there is a ``good" open substack MATH with the following properties: CASE: MATH and MATH are both isomorphisms over MATH. CASE: The diagram above is Cartesian when restricted to MATH. CASE: Translates of MATH by elements MATH cover MATH, If MATH, then MATH, MATH, and MATH are the preimages of MATH, MATH, and MATH, so MATH is invertible at MATH. If MATH lies over MATH, then MATH, MATH, and MATH all belong to same component MATH of the curve associated to MATH, and MATH imposes the unique parametrization on MATH taking MATH, MATH, and MATH to MATH, MATH, and MATH. Clearly, then, MATH and MATH are isomorphisms over MATH and the diagram is Cartesian over MATH. Since every prestable map MATH of degree one is generically an isomorphism over MATH, it follows that the translates of MATH cover MATH. We finish the proof now by comparing MATH with MATH. Suppose MATH lies over MATH. Then MATH is an isomorphism at MATH, so since MATH is flat everywhere (it is a composition of the flat forgetful maps), it follows that MATH is flat at MATH. But an arbitrary MATH lies over some translate MATH, over which the composition of MATH with translation by MATH is an isomorphism, and we similarly conclude that MATH is flat at an arbitrary MATH. This gives us REF . Thus MATH is flat, and we may use the flat pull-back to define MATH. NAME showed that the relative intrinsic normal cone MATH pulls back under MATH to MATH and the same trick we employed in the previous paragraph shows that it pulls back under MATH to MATH. The flatness of MATH also tells us that MATH pulls back to the corresponding element of the derived category of sheaves on MATH, and we get REF . The last sentence of REF is a consequence of NAME 's work, since the induced maps MATH are always gluing maps of NAME spaces. |
math/0007082 | We may assume that MATH is very ample. Indeed, suppose the polynomiality condition holds for the expression MATH for some MATH. Only the MATH's divisible by MATH will produce non-zero terms, because the degree of every curve (measured against MATH) is a multiple of MATH. But replacing MATH by MATH in the twisted tensor products simply multiplies the expression by MATH. If we now replace the subscript of each MATH by the degree of the curve against MATH (instead of against MATH) we get the desired result for MATH. The embedding MATH defined by MATH allows us to define a morphism MATH and an equivariant NAME class MATH as in the MATH case. Applying REF to the map MATH, we see that MATH is a local complete intersection (l.c.i.) morphism, since it factors through the graph followed by a flat morphism: MATH . Then by REF , MATH . It follows by the projection formula that the correspondence of residues holds for MATH (and any equivariant NAME class MATH) with each MATH replaced by MATH, and MATH replaced by MATH (again, using REF ). The proof of the MATH case now carries over to prove the general rank one case. |
math/0007082 | As in the proof of the rank one version, we may assume that MATH,,MATH are not just eventually free, but free, by replacing them with positive multiples (which can be taken to be the same multiple). The MATH define a morphism MATH and the theorem results from applying the correspondence of residues to the class MATH, where the MATH are the equivariant hyperplane classes pulled back from MATH. |
math/0007082 | The only term in the main theorem involving a MATH-function of MATH variables and curves of (multi) degree MATH is MATH . The product MATH is a polynomial in MATH, expanding as MATH for some MATH depending on MATH. It follows by downward induction on the power of MATH and the main theorem that every term in the expansion of MATH in MATH is determined inductively by MATH-functions involving fewer variables and/or lower degrees. Note that by stopping the induction at the MATH term, we determine the constant term, about which the main theorem tells us nothing. This argument only proves the reconstruction theorem when all cohomology is generated by the MATH since it (inductively) requires knowledge of the classes MATH where MATH is an arbitrary cohomology class. This argument does, however, capture the main idea of the proof. We now prove the following by induction on MATH: CASE: For all MATH and MATH, MATH CASE: For all MATH, the invariants MATH are determined by the one-point invariants and the intersection matrix on MATH. In terms of MATH-functions (using the symmetry), this claim is equivalent to CASE: If MATH and MATH then MATH . CASE: For all MATH, MATH is determined by one-point invariants and the intersection matrix on MATH. To start our induction, note that the claim holds for MATH by assumption. Also, the claim holds for MATH: MATH by orthogonality, and MATH and hence is determined by the intersection matrix on MATH. Using the argument at the beginning of this proof, the vanishing in REF will follow by induction (on the power of MATH), once we establish vanishing for all expressions of the form MATH . But these expressions may be rewritten: MATH . To rewrite MATH, choose an orthogonal basis MATH such that MATH with intersection matrix MATH and MATH with intersection matrix MATH. Then MATH . Now suppose REF holds for all MATH such that either MATH or MATH and MATH. Then MATH (taking MATH), and MATH since the first factors in the first double sum and the second factors in the second sum vanish. This proves REF by induction. Similarly, assuming REF , we prove REF by induction, noting that in this case, the second double sum in MATH (but not the first) vanishes. The first double sum and the MATH terms are explicitly determined by the intersection matrix MATH and NAME invariants for lower MATH. |
math/0007084 | The space MATH is the image of MATH by the algebraic map MATH where * denotes the action of MATH on MATH. It follows that MATH is irreducible and constructible. If MATH is the law of MATH, then its inverse image is algebraically isomorphic to the first projection MATH . The algebraic mapping which associates MATH to MATH in the NAME MATH by the image of the set of central ideals of MATH whose factor is isomorphic to MATH. If MATH and MATH give the same image MATH then MATH belongs to the stabilizer of MATH, mapping MATH on MATH. We deduce that the minimum of dimensions for MATH equals MATH. |
math/0007084 | We only observe that the MATH-abelianity is given by the differential form MATH. |
math/0007084 | We analyze the elements of the set MATH which have the prescribed nilindex. For MATH it is immediate to see that the nonexistence of such an extension. For MATH the cocycles MATH defining the extension have to satisfy the relation MATH . The only ones which verify this requirement are the cocycles MATH . Moreover, if MATH is the dual basis of MATH we have MATH . Observe that the space MATH gives the presentation MATH and the underlying NAME algebra is obviously isomorphic to MATH for any MATH. |
math/0007084 | The listed endomorphisms are clearly derivations of MATH. Their linear independence is also easy to verify. As MATH is naturally graded, a derivation MATH decomposes as MATH where MATH and MATH for MATH. From this decomposition it is not difficult to prove that MATH where MATH for MATH. |
math/0007084 | Let MATH be a cocycle. Then, for MATH, MATH we have MATH . The cocycle condition MATH implies the relations MATH for MATH, and MATH . Thus the cocycle MATH can be rewritten as : MATH . So the cocycles MATH constitute a generator system for MATH. Clearly they are linearly independent, so they form a basis. This proves the first assertion. The second follows immediately. |
math/0007084 | From the basis obtained in the previous proposition it follows easily that MATH. Observe in particular that most coboundaries have been previously excluded from the cohomology space MATH by REF. |
math/0007084 | Follows from the previous result, as for MATH . |
math/0007084 | We have seen that the rank of the algebra MATH is MATH for any MATH. From the previous proposition we know that MATH for MATH. Without loss of generality we can take MATH. Let MATH such that MATH . The cocycle MATH implies in particular MATH . Taking the relation MATH, we obtain MATH . This proves that the deformation MATH has rank one. Now let MATH be the solvable NAME algebra whose NAME equations on the basis MATH-are MATH . It is immediate to verify that this NAME algebra is solvable and complete. If MATH or MATH-the algebra is moreover ( see CITE for a standard proof ). As we have MATH, its nilradical is clearly isomorphic to MATH. |
math/0007084 | As the sill cocycle is itself linearly expandable, we know from previous results that the algebra MATH has rank one. If MATH, it is easy to verify that the entries of the matrix MATH of MATH satisfy the relations MATH . Thus the sill cocycle implies MATH . Now let MATH the summand of MATH that satisfies MATH. Then there exists a MATH such that MATH . From this equation we obtain MATH so that MATH and as MATH we have MATH. It is not difficult to see that the other summands of MATH do not affect these relations. It follows that the matrix of MATH is strictly upper triangular, thus the derivation MATH is nilpotent and MATH is characteristically nilpotent. |
math/0007084 | The only manner in which the differential form MATH is closed is that the cocycle MATH only affects the differential form MATH adding the exterior product MATH or changing the differential forms MATH by adding, respectively, the exterior products MATH and MATH. This corresponds to the indexes MATH and MATH. |
math/0007085 | By the NAME theorem it suffices to prove that MATH is an analytic subset of MATH. We will apply the elementary NAME method REF , used there in the case MATH altghough one can also use the method of blowing-ups. Define the holomorphic functions MATH where MATH and MATH. The all functions MATH vanish if and only if MATH or MATH is a multiple of MATH. Set MATH . This is an analytic subset of MATH and hence so is MATH . The set MATH is also analytic. So, MATH is an analytic set in MATH. Then MATH is analytic in MATH. Since MATH if and only if MATH, then MATH is an analytic subset of MATH. |
math/0007085 | Since MATH depends only on the germs of MATH and MATH at MATH, we may assume that MATH and MATH. Consider the analytic set MATH, defined in the proof of the previous Proposition. If we denote by MATH the projection MATH, then MATH and over each point MATH we have MATH and hence MATH . Since MATH then MATH . By the same equality REF no irreducible component of MATH is contained in MATH and in particular in MATH. Hence MATH . |
math/0007085 | Consider the analytic set MATH defined in the proof of REF . We have MATH . Since this point lies in the closure of MATH then there exists an analytic curve MATH passing through MATH such that MATH. Take a parametrization MATH at MATH of one irreducible component of MATH . We have MATH . Since for any MATH, MATH and MATH are linearly dependent and MATH when MATH then MATH . |
math/0007085 | Let MATH, MATH. Since MATH is a cone then MATH. Take analytic curves MATH and MATH having parametrizations MATH and MATH at MATH, MATH such that MATH and MATH. Since MATH and MATH for sufficiently small MATH and MATH then MATH. |
math/0007085 | It suffices to prove MATH . Take MATH. We may assume that MATH. By REF there exists an analytic curve MATH having a parametrization MATH at MATH such that MATH . Since MATH and MATH then MATH . Let MATH . Since MATH then MATH-and MATH are linearly independent. Hence and from REF MATH . So, MATH . |
math/0007085 | The proof follows from REF and the fact that the mapping MATH,MATH is an analytic cover. |
math/0007085 | We may assume that the germs MATH are irreducible. It suffices to prove that MATH . Since MATH are analytic curves and MATH are irreducible at MATH we will consider two possible cases: MATH . MATH. Then by REF MATH. Hence we get REF . MATH . MATH. After a linear change of coordinates in MATH we may assume that MATH, where MATH . Put MATH . Let MATH and MATH be parametrizations of MATH and MATH at MATH, respectively-MATH . Since MATH, we may assume that MATH . Consider descriptions of MATH and MATH where MATH is a sufficiently small positive number. Take now MATH and MATH. From REF there is an analytic curve MATH having a parametrization MATH at MATH such that MATH . Define MATH . Then MATH . Since MATH is a parametrization of a curve we have that MATH or MATH is not identically zero. Without loss of generality, we may assume that MATH and MATH. Put MATH. Hence MATH. Changing unessentially MATH we may assume that MATH. We define MATH . We claim that MATH . In fact, for the first coordinate we have MATH and for the next coordinates MATH . |
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