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math/0012040 | The dimension of the submanifold of such jets in MATH equals REF. Let us estimate the dimension of the orbit at points of the submanifold under the action of the stabilisator of the regular part. The dimension of the orbit under the action of elements from MATH is less than or equals MATH. The dimension of the orbit under the action of elements from MATH is less than or equals MATH . Now one can see that the dimension of the tangent space to the orbit at points of the submanifold is less than or equals MATH. |
math/0012040 | A complete transversal in MATH is trivial for any MATH. In MATH a complete transversal is equivalent to MATH (we write only the second component, the first and the third components are not changed). If MATH we obtain the normal form MATH. If MATH and MATH we can suppose MATH. The tangent space to the MATH-orbit contains the vectors (we write only the singular components): MATH, MATH and MATH. So, using the NAME Lemma we can suppose MATH. If MATH we obtain the normal form MATH. If MATH we obtain the normal form MATH. Now suppose MATH. If MATH and MATH we obtain the normal form MATH. If MATH and MATH we obtain the normal form MATH. If MATH and MATH we obtain the normal form MATH. All these normal forms are MATH-determined. |
math/0012049 | Our first task is to define the NAME of an arbitrary space MATH. The details of this process are in CITE, but the method is wholly due to NAME who makes the construction for complex analytic spaces. Let MATH denote the class of all maps MATH which are compositions of finitely many local blowing up maps. One can define a partial pre-order relation on MATH by calling MATH smaller than MATH, if MATH factors as MATH, for some MATH. We denote this by MATH. The map MATH is necessarily unique and must belong to MATH CITE. If, moreover, the image MATH of MATH is relatively compact (that is to say, the closure of its image is compact), then we denote this by MATH. Any two maps MATH admit a unique minimum or meet MATH with respect to the order MATH CITE, denoted by MATH. This meet MATH is just the strict transform of MATH under MATH (or vice versa). With these definitions, MATH becomes a semi-lattice with smallest element the empty map MATH. A subset MATH of MATH is called a filter, if it does not contain MATH, is closed under meets, and, for any MATH and MATH, with MATH, we have that also MATH. An étoile MATH on MATH is now defined as a filter on the semi-lattice MATH which is maximal among all filters MATH satisfying the extra condition that for any MATH we can find MATH, with MATH. The collection of all étoiles on MATH is called the NAME of MATH and is denoted by MATH. This space is topologised by taking for opens the sets of the form MATH given as the collection of all étoiles on MATH containing MATH, for some MATH. In fact, MATH is isomorphic with MATH via the map MATH, sending MATH to the collection of all MATH for which there exists some MATH such that MATH, see CITE. The NAME is NAME in this topology CITE. Moreover, for any étoile MATH, the intersection of all MATH, where MATH runs through the maps in MATH, is a singleton MATH and any open immersion MATH with MATH, belongs to MATH CITE. We denote the thus defined map MATH by MATH. It is a continuous and surjective map. It is a highly non-trivial result that this map is also proper in the sense that the inverse image of a compact is compact CITE. Next, we will introduce the concept of a flatificator. Let MATH be a map and let MATH. A flatificator of MATH at MATH is a locally closed subspace MATH of MATH containing MATH, such that MATH is flat over it (that is to say, the restriction MATH is flat), and such that, whenever MATH is a second locally closed subspace containing MATH over which MATH is also flat, at least on an open neighbourhood around MATH, then MATH is a subspace of MATH locally around MATH (that is to say, when restricted to some open neighbourhood of MATH). In other words, a flatificator is a largest locally closed subspace over which MATH becomes flat in a neighbourhood of MATH. Such a flatificator is called universal, if it is stable under base change. With this we mean that, if MATH is arbitrary, then MATH is the flatificator of the base change MATH at MATH, for any MATH in the fibre above MATH. In CITE it is shown that any map MATH admits a universal flatificator MATH in each point MATH of MATH. If MATH is moreover reduced, then we can detect flatness via the flatificator: blowing up the flatificator exhibits some non-trivial portion of non-flatness as torsion. More precisely, it is shown in CITE that whenever MATH is not flat in some point of the fibre MATH, then there exists a nowhere dense subspace MATH of MATH, such that the local blowing up MATH with centre MATH renders the fibre above MATH smaller. With this, we mean the following. Let MATH denote the strict transform diagram induced by MATH. Then, for every MATH lying above MATH, we have a non-trivial embedding of closed subspaces MATH where MATH is our universal domain. (Note the extension of scalars is necessary in order to compare these two fibres as subspaces of MATH.) We refer to this result as the Fibre Lemma. Fix an étoile MATH on MATH such that MATH. Put MATH, MATH and put MATH equal to the original map MATH. Finally, to comply with the enumeration below, put MATH and let MATH be the empty map. The Fibre Lemma will enable us to define, by induction on MATH, points MATH and étoiles MATH with MATH, local blowing up maps MATH with nowhere dense centre MATH, maps MATH and non-empty compact subsets MATH of the fibre MATH, at least as long as MATH is not flat in some point of MATH. Each MATH will be the strict transform MATH of the previous map MATH under the local blowing up MATH. Moreover, each MATH will be flat above the centre MATH and have the property on the fibres REF in the point MATH. Let us show how to define from the point MATH a new point MATH and a new étoile MATH on MATH. Apply the Fibre Lemma to the point MATH and the map MATH to obtain a nowhere dense center MATH and a local blowing up MATH rendering the fibre above MATH smaller as explained in REF. Since MATH is nowhere dense and contains MATH, we deduce, from CITE, that MATH, that is, MATH. The isomorphism MATH then yields a uniquely determined étoile MATH on MATH and this in turns uniquely determines the point MATH of MATH. By diagram chasing, one checks that this implies MATH. Finally, we define a compact subset of MATH by MATH . Let MATH denote the coherent ideal of MATH defining the fibre MATH, so that by REF the chain MATH is strictly increasing on the compact set MATH. Therefore this chain must become stationary, say at level MATH, meaning that MATH is flat in each point of MATH. Flatness is an open condition in the source, so that we can find an open MATH of MATH containing MATH, such that MATH is flat. Let MATH, where MATH is the compositum MATH. Note that MATH is compact, since MATH is proper. We claim that after some further local blowing up (in fact, an open immersion will suffice), we may assume that MATH is empty, so that the new strict transform will be flat at each point lying above a point of MATH. To this end, suppose MATH is non-empty and pick some MATH. Since MATH, we can find a compact neighbourhood MATH of MATH in MATH, such that MATH. Hence MATH . The compactness of each MATH means that already a finite number of them, say MATH, for MATH, have empty intersection. Let MATH be the intersection of these finitely many MATH, which is then still a compact neighbourhood of MATH, with the property that MATH . Let MATH be an open of MATH containing MATH and contained in MATH. Then the restriction of MATH above MATH has now the property that it is flat in each point lying above a point of MATH. Summarising, we found for each étoile MATH on MATH with MATH a local blowing up map MATH, such that the strict transform MATH has the property that it is flat in each point lying above a point of MATH. Moreover, there is a canonically defined point MATH on MATH lying above MATH. Let MATH be a relatively compact open neighbourhood of MATH and set MATH . Then, for each MATH, the set MATH is an open neighbourhood of MATH and the union of all the MATH is a neighbourhood of MATH. Let MATH, be the collection of all non-empty relatively compact open neighbourhoods of MATH in MATH. Pick arbitrary MATH with MATH, where, in general, MATH denotes the topological closure of MATH. Consider the open covering of MATH given by the sets MATH . This is indeed a covering, since by the NAME property, we can find for each MATH a MATH for which MATH. Since MATH is proper, we have that MATH is compact and hence there exists a finite subset MATH and a finite subset MATH, such that the collection of all sets of REF with MATH and MATH remains a covering of MATH. Putting MATH equal to the intersection of the MATH, for MATH, this is still a neighbourhood of MATH and MATH . Observing that MATH and using that MATH and MATH are surjective, we deduce that MATH as required. |
math/0012049 | Let MATH and MATH be the corresponding NAME spaces of MATH and MATH respectively and let us continue to write MATH for the corresponding map MATH. Fix an analytic point MATH of MATH (that is to say, a point of MATH), contained in the image of MATH. Let MATH, which is closed in MATH whence compact since MATH is. By REF , we can find a finite collection MATH of maps MATH with MATH affinoid, such that REF - REF hold. For each MATH, let MATH be the corresponding strict transform diagram. By REF we have that the strict transform MATH is flat in each point of MATH. Let us first show that we can modify the data in such way that MATH becomes flat everywhere. Since flatness is open in the source by CITE, we can find an open neighbourhood MATH of MATH in MATH over which MATH is flat. Since MATH and MATH are compact NAME spaces, we can find an open neighbourhood MATH of MATH, such that MATH. Similarly, we can find an open neighbourhood MATH of MATH in MATH, such that MATH. The neighbourhood MATH can be taken inside the union of all the MATH, for all MATH. Set MATH. Note that MATH is the strict transform of the open immersion MATH under MATH. Let MATH be the restriction of MATH to MATH. The strict transform of MATH under MATH is the map MATH which by construction is flat, since MATH . This establishes our claim upon replacing MATH by MATH. Hence we may assume that MATH is flat. Note also that in the above process, we have not violated REF , so that the MATH, for all MATH, form a covering of an affinoid neighbourhood MATH of MATH in MATH. We can translate all these diagrams to the rigid analytic setup and assume that the same diagrams hold with the spaces now rigid analytic varieties (see REF below), where we keep the same names for our spaces and maps, but just replace any blackboard letter, such as MATH, by its corresponding roman equivalent MATH, denoting the corresponding rigid analytic variety. In particular, REF - REF hold and we show how to obtain REF. Let us now vary the analytic point MATH over MATH, so that the MATH cover all analytic points of MATH. Since MATH is compact in the NAME topology so is MATH. Therefore, by CITE, already finitely many of the MATH cover all analytic points of MATH. In particular, there is a finite collection MATH of analytic points, such that the union of all MATH, for all MATH and all MATH, cover MATH. In other words, REF is now verified as well. |
math/0012049 | See REF. |
math/0012049 | For closed immersions, the statement is trivial. If MATH is a rational affinoid subdomain, then MATH, where MATH with MATH having no common zero. Hence any function MATH defined on MATH is MATH-definable on MATH (just replace any occurrence of MATH by MATH). Now, any affinoid subdomain is a finite union of rational subdomains by CITE and hence we proved the proposition for any affinoid open immersion as well. From this, the general locally closed immersion case follows easily. This leaves us with the case of a blowing up. Without loss of generality, we may assume MATH to be affinoid. Let us briefly recall the construction of a blowing up map as described in CITE. Let MATH and let MATH be a closed analytic subvariety of MATH defined by the ideal MATH of MATH. We can represent MATH as a quotient of some MATH, with MATH, so that MATH becomes a closed analytic subvariety of MATH. However, in order to construct the blowing up of MATH with centre MATH, we need a different embedding, given by the surjective algebra morphism MATH for MATH, extending the surjection MATH and where MATH. This gives us a closed immersion MATH and after identifying MATH with its image MATH, we see that MATH. Now, the blowing up MATH is given by a strict transform diagram MATH where MATH denotes the blowing up of MATH with centre the linear space MATH. There is a standard finite admissible affinoid covering MATH of MATH where each MATH has affinoid algebra MATH so that MATH for any point MATH, where the latter is considered as a closed analytic subset of MATH via the above representation of MATH. Moreover, MATH is a closed analytic subvariety of MATH. Therefore, if we set MATH, then MATH is a finite admissible affinoid covering of MATH with the affinoid algebra MATH of each MATH some quotient of the affinoid algebra MATH of MATH, where MATH means all variables MATH save MATH. With this notation, let us return to the proof of the proposition. We are given some MATH-semianalytic set MATH of MATH and we seek to describe the image MATH. Let us focus for the time being at one MATH, where MATH. Since MATH is MATH-semianalytic, we can find a quantifier free MATH-formula MATH, such that MATH belongs to MATH if, and only if, MATH holds. Hence, for MATH, we have that MATH . If MATH holds, then in particular MATH and hence by REF, we have that MATH, for all MATH. Now, a point MATH does not belong to MATH, precisely when one of the MATH does not vanish. Therefore, as MATH ranges through the set MATH and using REF, it is not too hard to see that MATH belongs to MATH if, and only if, MATH and MATH which is indeed a MATH-semianalytic description of MATH. |
math/0012049 | The first two statements follow from the fact that the dimension of a subanalytic set is the maximum of the dimensions of each manifold in any finite subanalytic manifold partitioning (as in REF ). The other statements require more work. See CITE and, in particular, for REF , see CITE. See also CITE for the MATH-adic analogues - the proofs just carry over to our present situation, once one has REF . |
math/0012049 | We have already seen that MATH-semianalytic sets are subanalytic. To prove the converse, let MATH be a subanalytic set of MATH. We will induct on the dimension of MATH and then on the dimension of MATH. The zero-dimensional case follows immediately from REF . Hence fix MATH and MATH. In particular, MATH. It suffices to take MATH closed in the canonical topology. Indeed, assume the theorem proven for all subanalytic sets which are closed in the canonical topology. Let MATH be the closure of MATH with respect to the canonical topology. By REF , also MATH is subanalytic and of dimension equal to the dimension of MATH. Hence by our assumption MATH is even MATH-semianalytic. Let MATH be the boundary MATH, which is again subanalytic by REF . Moreover, by REF , MATH has strictly smaller dimension than MATH. Hence, by our induction hypothesis on the dimension of a subanalytic set, we have that also MATH is MATH-semianalytic. Therefore also MATH, as required. Hence we may assume that MATH is closed in the canonical topology. There exists a semianalytic subset MATH, for some MATH, such that MATH, where MATH is the projection on the first factor. The union of finitely many MATH-semianalytic sets is again such. Therefore, without loss of generality, we may even take MATH to be a basic set, that is to say, of the form MATH where the MATH and MATH are in MATH, with MATH and MATH. Introduce MATH new variables MATH and consider the closed analytic subset MATH of MATH given by the equations MATH, for MATH. Let MATH be the basic subset of MATH given by MATH belongs to MATH whenever MATH, for MATH. Let MATH be the product of all the MATH, for MATH, and we obviously can assume that MATH if MATH is non-empty. If MATH denotes the composition of the closed immersion MATH followed by the projection MATH, then MATH, where MATH is the complement in MATH of the zero-set of MATH. Using CITE, we may, after perhaps modifying some of the equations defining MATH, assume that the closure of MATH in the canonical topology equals the whole of MATH and hence the closure (in the canonical topology) of MATH is MATH. Now MATH and so MATH, since MATH is closed and MATH is continuous. Hence MATH. Before giving the details of the remaining steps, let's pause to give a brief outline of how we will go about it. According to our Flattening REF , we can find finitely many diagrams MATH where each MATH is a finite composition of local blowing up maps with REF - REF and such that MATH is contained in the union of all the MATH. Now, in order to study MATH, we will chase MATH around these REF . There are only finitely many MATH to consider; it will suffice to do this for one such MATH since the analysis for the others is identical. First we take the preimage MATH, which is again a semianalytic set defined by inequalities of the form MATH where the MATH are functions on MATH of supremum norm at most one. Next we take the image of the latter set under MATH. Our extension of NAME 's Theorem REF guarantees that this image is semianalytic. Finally we push this set back to MATH via MATH and denote this set temporarily by MATH. If we had the full version of REF , namely, that a local blowing up map preserves MATH-semianalyticity, then this last set would be indeed MATH-semianalytic. Of course, in chasing MATH around the diagram, we might have lost some points. In other words, it may well be the case that MATH. However, this could happen only for points coming from one of the centres of the local blowing ups that make up MATH (since outside its centre, a blowing up map is an isomorphism). Above each of these centres the strict transform is flat so we account for those missing points using REF once more. Hence the only problem in the this reasoning lies in the application of REF : it is not the whole image that we can account for by means of that proposition, but only for the part outside the centre. Now an induction on dimension allows us to dispose of this part too. Our second induction hypothesis says that any subanalytic set in an affinoid variety of dimension strictly smaller than MATH is MATH-semianalytic. From this we obtain a stronger version of REF . Let MATH be any local blowing up of a quasi-compact rigid analytic variety MATH of dimension at most MATH whose centre MATH is nowhere dense. If MATH is MATH-semianalytic, then MATH is also MATH-semianalytic. The key point is that MATH has dimension strictly smaller than the dimension MATH of MATH, which is also the dimension of MATH. We have an equality MATH . By REF we know that MATH is MATH-semianalytic and by our induction hypothesis on the dimension we also have that MATH is MATH-semianalytic in MATH (take a finite affinoid covering to reduce to the affinoid case). This means that MATH is MATH-semianalytic in MATH. We return to the proof of REF . According to REF , there exists a finite collection MATH of maps MATH, such that each MATH induces a strict transform REF with REF - REF . (The intermediate strict transform diagrams are given by REF below). By REF, if we could show that each MATH is MATH-semianalytic in MATH, then the same would hold for MATH, since there are only finitely many MATH. Therefore, let us concentrate on one such MATH and adopt the notation from REF for this map, so that in particular, REF - REF holds. Let each MATH be the blowing up of the admissible affinoid MATH with nowhere dense centre MATH. The diagram of strict transform is given by MATH . Define inductively MATH as MATH starting from MATH. Note that each MATH is a semianalytic set of MATH defined by several inequalities of the type MATH, where each MATH is of supremum norm at most one. Define also inductively, but this time by downwards induction, the sets MATH where we start with MATH. In particular, we have that MATH. By REF each MATH is MATH-semianalytic in MATH. In order to describe MATH, we will furthermore make use of the sets MATH defined as MATH, for MATH. In particular, note that MATH is nothing other than MATH, which we aim to show is MATH-semianalytic. The next claim shows how two successive members in the chain of commutative REF relate the MATH: for each MATH, we have an equality MATH . Assume we have already established REF, for each MATH. We will prove, by downwards induction on MATH, that each MATH is MATH-semianalytic in MATH, so that in particular MATH would be MATH-semianalytic in MATH, as required. First, since MATH is assumed to be flat, we can apply REF to MATH to conclude that MATH is semianalytic whence MATH-semianalytic in MATH. Assume now that we have already proved that MATH is MATH-semianalytic in MATH and we want to obtain the same conclusion for MATH in MATH. Using REF, it is enough to establish this for both sets in the right hand side of that equality. The first of these, MATH, is MATH-semianalytic since we have now the strong version REF at our disposal. As for the second set, MATH, also this one is MATH-semianalytic, since MATH restricted to MATH is flat and since MATH so that REF applies. Note that we already established that MATH is MATH-semianalytic. Therefore, it only remains to prove REF. To show that MATH is straightforward. We show the reverse inclusion. Let MATH. That means that there exists MATH and MATH such that MATH. If MATH, we are done. Hence assume that MATH so that MATH. However, since MATH we have that MATH. Since MATH is the blowing up of MATH with centre MATH whence an isomorphism outside this centre, we can even find MATH, such that MATH. From MATH it then follows that MATH. Put MATH. Commutativity of the strict transform diagram implies that MATH. Since MATH, the blowing up MATH is injective at that point, so that MATH which therefore belongs to MATH, proving our claim, and hence also our main theorem. |
math/0012049 | This follows from the above discussion in the case where MATH is closed in the canonical topology. The reduction to this case uses an induction argument similar to the one in the proof of the theorem. |
math/0012049 | Let MATH. We use the following corollary to the Embedded Resolution Theorem: given MATH, then there exists a finite collection MATH of maps, such that REF hold, for each MATH in MATH, and furthermore either MATH divides MATH, or vice versa, MATH divides MATH, in the affinoid algebra of MATH. See for instance CITE for a proof. From our Quantifier Elimination REF, we know that MATH is MATH-semianalytic. By a (not too difficult) argument, involving an induction on the number of times the function MATH appears in one of the describing functions of MATH (for details see CITE), we can reduce to the case that there is only one such occurrence. In other words, we may assume that there exist a quantifier free formula MATH in the language MATH and functions MATH, such that MATH if, and only if, MATH . After an appeal to the corollary of Embedded Resolution of Singularities applied to MATH and MATH, and since we only seek to prove our result modulo finite collections of maps for which REF holds, we may already assume that either MATH divides MATH or MATH divides MATH. In the former case, there is some MATH, such that MATH in MATH. Therefore, MATH, unless MATH and MATH, in which case it is equal to MATH. Let MATH be the affinoid subdomain defined by MATH and MATH by MATH, so that MATH is an admissible affinoid covering of MATH. Hence MATH belongs to MATH if, and only if, MATH holds, whereas MATH belongs to MATH if, and only if, MATH holds. Observe that MATH belongs to the affinoid algebra of MATH, since MATH does not vanish on MATH. In other words, MATH is semianalytic on both sets and whence on the whole of MATH. In the remaining case that MATH divides MATH, so that there is some MATH, such that MATH in MATH, we have an even simpler description of MATH, namely MATH, if and only if, MATH holds, again showing that MATH is semianalytic. |
math/0012049 | Let MATH be the prime ideal of MATH associated to MATH. Let MATH, so that it is the maximal ideal of MATH associated to MATH. Finally, let MATH be the maximal ideal of MATH associated to MATH, so that the commutativity of REF translates into MATH . Since MATH is flat, the Going Down Theorem (see for instance CITE) guarantees the existence of a prime ideal MATH of MATH, such that MATH and MATH . Since MATH, also MATH. Among all prime ideals MATH of MATH, with MATH and MATH, let MATH be one of maximal height (recall that MATH has finite combinatorial dimension). Note that MATH is at least one prime ideal satisfying both conditions. We claim that MATH determines a MATH-rational point. Assuming the claim, it follows from MATH that MATH is a lifting of MATH. By Formulae REF, we have inclusions MATH . Since MATH, we cannot have that MATH. Since MATH, the only two prime ideals of MATH containing MATH are MATH and MATH itself. Therefore, we conclude that MATH so that the MATH-rational point MATH is a factorization of MATH, as required. It remains to prove that MATH determines a MATH-rational point. In other words, if we set MATH, then we need to show that MATH. Write MATH as a homomorphic image of some MATH. Since MATH is a domain and since MATH, it follows that MATH has no MATH-torsion and consequently it must be flat over MATH. Put MATH, so that MATH is an affinoid algebra containing MATH (here we used flatness). By NAME Normalization CITE, we can find a finite injective homomorphism MATH, for some variables MATH. Moreover, inspecting the proof of NAME Normalization, one checks that MATH is given by MATH with each MATH a polynomial of NAME norm MATH; see CITE for a discussion. In particular, MATH maps MATH into MATH. On the other hand, let MATH and let MATH be such that MATH in MATH. By [BGR REFEF ], the supremum norm of MATH is equal to the maximum of the MATH. Since MATH its supremum norm is at most MATH. Therefore, all MATH have NAME norm at most MATH, whence belong to MATH. In conclusion, the restriction MATH is integral. Therefore, MATH is a maximal ideal of MATH CITE. Since MATH is algebraically closed, it follows that MATH for some MATH. If MATH, then by the Going Down Theorem CITE for the normal domain MATH, we can find a (non-zero) prime ideal MATH of MATH contained in MATH and such that MATH. By the maximality of MATH, we have that MATH, so that MATH. Consequently, MATH, contradiction. In other words, we must have that MATH, so that MATH is integral over MATH. Since the field of fractions of MATH is then algebraic over MATH, it must be equal to it, as we assumed MATH to be algebraically closed. Therefore MATH. Since MATH is a valuation ring and clearly MATH, we conclude that MATH, as required. |
math/0012049 | So we are given a flat map MATH of affinoid varieties and a special set MATH of MATH. Suppose that MATH where MATH are of supremum norm at most one and MATH is either MATH or MATH. We want to prove that MATH is semianalytic. Using REF below, we may reduce to the following case. There exist flat MATH-algebras MATH and MATH which are topologically of finite type, such that MATH and MATH, and such that MATH, for all MATH, and there exists a flat morphism of MATH-algebras MATH which induces the map MATH (after tensoring with MATH). By our observation above, there exists a locally closed set MATH of MATH such that MATH. Since MATH can be identified with MATH, where MATH, we can view MATH as a locally closed subset of the this space as well. If we also put MATH, then since both rings are finitely generated MATH-algebras, we can invoke NAME 's Theorem to conclude that the image of MATH under the induced map MATH is a constructible set MATH. Identifying MATH with MATH, we may consider MATH as a constructible set of the former space as well and as such it is the image of MATH under the map MATH induced by MATH. Let MATH, so that by our above observation MATH is semianalytic in MATH. Hence, we will have proved our theorem once we show that MATH . The commutative REF expressing the functoriality of MATH, provides the inclusion MATH, so we only need to deal with the opposite inclusion. To this end, let MATH. Let MATH be the corresponding MATH-rational point and let MATH be the reduction MATH of MATH. By assumption, MATH and hence it is the image under MATH of some point MATH. Since MATH is flat, we can apply REF to obtain a MATH-rational point MATH factoring through MATH and lifting MATH. In other words, if MATH denotes the point corresponding to MATH, then this translates into MATH and MATH. Since MATH, the latter implies that MATH, as required. |
math/0012049 | Since the MATH are of norm at most one and using CITE, we can find a flat and topologically of finite type MATH-algebra MATH containing all MATH with MATH, a flat and topologically of finite type MATH-algebra MATH with MATH and a MATH-algebra morphism MATH inducing the map MATH. In general, MATH will not be flat. To remedy this, we use CITE, in order to find admissible coverings as asserted, for which REF hold. Moreover, from its proof it follows that MATH is a homomorphic image of MATH. Therefore, also REF is satisfied. |
math/0012054 | In the first part of the proof we will assume that the point MATH is observable, that is, the sheaf MATH is locally free. Let MATH be a proper subspace of MATH and let the image of MATH under the map MATH be MATH. If REF is satisfied then the rank MATH of the sheaf MATH, which is equal to the dimension of MATH at the generic point is greater than MATH. Now the NAME characteristic MATH. So for MATH large enough MATH . Therefore by REF the point MATH is stable. If MATH is not observable our assumption is that the map MATH satisfies REF . The rank of MATH is the rank of the sub-sheaf MATH of MATH generated by MATH. The sheaf MATH is a sub-sheaf of the sheaf MATH which is the sub-sheaf of MATH generated by MATH, so MATH. By the above calculations MATH satisfies REF so MATH satisfies this condition as well. Thus the point MATH is stable. |
math/0012054 | Suppose the point MATH corresponds to the short exact sequence REF . Then the map MATH is represented by the transpose of the matrix MATH. For each point MATH, MATH determines a subspace of MATH given by the row span of the matrix MATH evaluated at the point MATH. If MATH is nondegenerate then for the generic point MATH and a subspace MATH if MATH and if MATH then MATH. Notice that MATH is the kernel of the map MATH. So if MATH is a nondegenerate point at the generic point MATH, if MATH then MATH and if MATH then MATH. Thus the point MATH satisfies REF and therefore it is stable. |
math/0012054 | The stable orbits (and nondegenerate systems are stable by REF ) form a subset of the semi-stable orbits. |
math/0012054 | Consider the set MATH of proper MATH transfer functions of NAME degree MATH. Let MATH and let MATH be the restriction of the morphism MATH to the NAME open subset MATH. Then MATH describes a quasi-projective variety parameterizing the orbits under the extended full feedback group as described in REF. By REF this variety also parameterizes the set of MATH input, MATH output systems of NAME degree MATH as introduced in REF modulo the full feedback group. |
math/0012056 | Let MATH be a link in MATH in general position with MATH and cutting MATH times, let MATH, that is, MATH . Note MATH. Using the basis elements MATH of MATH given in the previous theorem, MATH can be written as a linear combination of the elements MATH, where MATH is MATH with the reversed orientation. A diagram of MATH is given by the following: MATH . By the inductive definition of MATH, an alternative diagram of MATH is given by: MATH . We will consider the sliding relation given by: MATH . From the above observation, we will be interested in the following relation: MATH . From Relation II, and REF , as MATH, in MATH we have MATH . As MATH is invertible in MATH, we have that MATH. By induction, we can eliminate all elements of MATH which cut the MATH-disk MATH non-trivially. Thus MATH is an epimorphism. |
math/0012056 | First note MATH. We need only show that MATH. Let MATH be the projection map MATH. We will show that MATH in MATH, that is, MATH. We show this by proving now that MATH for any MATH. Recall that MATH is the regular neighborhood of MATH in MATH. Let MATH and MATH. Let MATH and MATH, note MATH and MATH are parallel to MATH. MATH . Let MATH, MATH, where locally MATH . Let MATH be the projection map MATH. Note that MATH and MATH for MATH. Let MATH, then MATH in MATH, since MATH is a MATH-disc and closing a relative link along MATH by MATH and MATH in MATH gives isotopic links. Let MATH denote the map which sends MATH to MATH . In general, let MATH. Now consider the following commutative diagram, MATH . Here MATH and MATH are induced by inclusion maps. Also MATH and MATH are induced by MATH and MATH respectively. By an argument similar to the proof of REF , the composition map MATH is an epimorphism. Take MATH, then MATH. As MATH is an epimorphism, there exists MATH such that MATH. By the commutativity of the diagram, MATH and MATH . Thus MATH. |
math/0012056 | CITE calculated the unframed NAME skein module of a handlebody. It follows from this, the universal coefficient property of skein modules and an argument of NAME in REF that MATH is free. As MATH is free, the map MATH induced by MATH is injective. Let MATH. It is enough to show MATH . Let MATH be the map from MATH given by MATH for MATH. MATH. It also follows from NAME 's basis that the map induced by inclusion MATH is injective. Let MATH be the image of a free basis for the module MATH in MATH . MATH also a basis for injective image of MATH in MATH . Let MATH be the subspace of MATH generated by framed oriented links in MATH which intersect the disk MATH times. Then we have a chain of vector spaces: MATH is a basis for MATH . The vector space MATH is generated by elements of the form MATH in a neighborhood of MATH where MATH is MATH. Let MATH be a basis MATH constructed by taking a maximal linearly independent subset of the above generating set. By the proof of REF , each element of MATH where MATH is an eigenvector for MATH with nonzero eigenvalue. MATH is a basis for MATH . Let MATH. Note MATH. So MATH . It follows that MATH induces a one to one map: MATH. Thus MATH. The result follows. |
math/0012056 | From the above, we have the following commutative diagram MATH . |
math/0012056 | We proceed by induction on MATH the number of the MATH-handles to be added to MATH to obtain MATH . If MATH we are done by REF . If MATH let MATH be the MATH-manifold obtained from MATH by adding MATH of those MATH-handles added to MATH. Suppose the result is true for MATH, that is, MATH . Suppose that the MATH-th MATH-handle is added along a curve MATH in the boundary of MATH where MATH is disjoint from MATH and the curves where the other MATH-handles are attached. Let MATH and MATH be two points on MATH . By the proof of the Epimorphism REF , MATH . Using CITE, we have the following commutative diagram with exact rows. MATH . The vertical map on the right is an isomorphism by the five-lemma. MATH is obtained from MATH by adding the MATH-th MATH-handle along MATH. Thus MATH and MATH. |
math/0012056 | Since MATH. |
math/0012056 | We consider MATH . Here we start with the string corresponding to the last cell in the last row of MATH, we pull the encircling component successively through the vertical strings, working to the left through columns and upward through the rows. Repeating the above process, for MATH: MATH . The result follows from REF below. |
math/0012056 | CASE: We will borrow the notation of NAME and NAME and use a schematic dot diagram to represent the element in the NAME category MATH, which is between MATH and MATH as shown on the left-hand side. Recall that MATH. Now in the diagram of the left-hand side of REF , introduce a schematic picture MATH as follows: MATH . This indicates that the last MATH strings were pulled out, the MATH-th string marked by MATH starts and finishes at MATH. The arrow on the MATH-th string shows the string orientation when we look at it from above. The MATH-th string encircles the remaining MATH strings in the clockwise direction. Here all strings shown by single dots are going vertical. The left-hand side of REF can be expressed as MATH. We will be working on MATH. Using the NAME skein relations and the inseparability in REF, we have, MATH . Where MATH is given by: MATH . Since MATH, where: MATH . First we have MATH by the property MATH ; secondly, MATH by the property MATH, CITE . It follows that MATH. By a similar argument as in the proof of REF, MATH. Thus MATH, where c is the cell indexed by MATH. CASE: We prove the result with all string orientations reversed. As string reversal defines a skein module isomorphism, this suffices. As MATH, we can use the following schematic picture to denote the left-hand side of REF as MATH, where MATH the MATH-th string is indexed by MATH and circles the remaining strings in the clockwise direction. Again, we have MATH . Where MATH is given by: MATH . Since MATH, where: MATH . We have MATH and MATH by the properties MATH and MATH. We have MATH . It follows that MATH. By the idempotent property, MATH. The result follows. |
math/0012060 | Above we showed that MATH is special Lagrangian if and only if REF - REF hold. We will show that REF is equivalent to REF , and REF - REF are equivalent to REF or REF . Fix MATH, and let MATH. Then MATH. As MATH are oriented conformal coordinates and MATH, one can show that if REF holds then MATH are an oriented orthonormal basis of a special Lagrangian REF-plane in MATH. Let MATH be the unique complex coordinate system on MATH in which we have MATH, MATH and MATH. Then MATH are related to the usual coordinates MATH on MATH by a MATH transformation. Thus, as MATH and MATH are MATH-invariant, by REF we have MATH . In the coordinates MATH, write MATH for MATH. Using REF - REF become MATH cancelling factors of MATH. Setting MATH and substituting MATH and MATH by REF give MATH which is equivalent to MATH . The MATH matrix appearing here has determinant MATH. If this is nonzero then the matrix is invertible, so the column matrix must be zero. If the determinant is zero then MATH. So REF holds if and only if either CASE: MATH, or CASE: MATH. These two possibilities correspond to REF. As the transformation from the MATH coordinates to the MATH coordinates lies in MATH, and MATH is MATH-equivariant, the formula for MATH in the MATH coordinates is the same as REF in the MATH coordinates. Thus, combining REF we see that MATH. Since this holds for all MATH, REF implies REF . Conversely, it is easy to show that REF implies REF . Suppose REF holds, so that MATH and MATH. Then by REF we have MATH . Clearly MATH as MATH. REF give MATH . Thus REF holds at MATH, with MATH. Conversely, if REF holds at MATH then REF holds, so REF - REF hold, and thus REF hold at MATH. Now suppose REF holds, so that MATH. Then REF gives MATH . That is, MATH and MATH are real vectors in the MATH coordinates. It follows that REF holds at MATH. Conversely, if REF holds at MATH then REF - REF hold, and so REF hold at MATH. We have shown that REF is equivalent to REF , and REF hold if and only if REF or REF holds at every point MATH in MATH. These are not exclusive options; both REF may hold at some points. But we need one of REF to hold at every point in MATH, rather than REF at some points and REF at others. Let MATH be a generic point in MATH. Then if REF or REF holds at MATH, it also holds in a small neighbourhood of MATH in MATH. But MATH is connected, MATH are real analytic, and REF are closed conditions, so if one of REF or REF holds in an open set in MATH then it holds in all of MATH. This completes the proof. |
math/0012060 | We shall apply REF . Define MATH to be MATH, with coordinates MATH, and let MATH. Then MATH is a real analytic, nonvanishing section of MATH, as in the theorem. Consider a family of maps MATH depending smoothly on MATH, of the form MATH, where MATH are real analytic maps with MATH and MATH. We do not yet assume that MATH maps to MATH in MATH, but we will prove this later. Then MATH and MATH, so REF becomes MATH by the definition of MATH in REF . Equating linear and constant terms in MATH on each side, we see that REF is equivalent to REF . The only problem in applying REF in this situation is that MATH is not compact. However, as the MATH are defined using maps MATH on MATH, which is compact, this does not matter, and the theorem applies. This can be checked by examining its proof in CITE. Thus, by REF , there exists MATH and unique real analytic maps MATH with MATH and MATH, satisfying REF , and such that MATH is special Lagrangian. It remains only to prove REF , and that MATH maps into MATH. REF follows from the equation MATH, which holds as MATH is Lagrangian. And MATH, by REF and the definition of MATH. Thus MATH is independent of MATH. But MATH as MATH lies in MATH, so MATH, and MATH maps MATH. |
math/0012060 | Suppose MATH is a ruled SL REF-fold in MATH, with r-oriented ruling MATH satisfying REF but not REF . As MATH is real analytic wherever is is nonsingular by CITE, we may take MATH to be real analytic, at least locally. Choose any embedded real analytic curve MATH. Define MATH and MATH, MATH by MATH and MATH. Apply REF . This gives real analytic maps MATH and MATH, and constructs a ruled SL REF-fold MATH in MATH from them, with a ruling MATH that satisfies REF . Now MATH and MATH intersect in the ruled surface MATH in MATH. Therefore, by CITE, MATH and MATH coincide locally. Thus, locally MATH admits a ruling MATH satisfying REF . Since by REF does not satisfy REF , MATH and MATH must be different. It is not difficult to show that different, nearby curves MATH in MATH will yield different rulings MATH of MATH. Thus MATH admits not just two, but infinitely many different rulings. But CITE shows that any SL REF-fold with more than two distinct rulings is planar, that is, locally isomorphic to MATH in MATH. |
math/0012060 | The first part of the theorem follows from the material of REF. If MATH then the second part is trivial. So suppose MATH, so that MATH has only isolated zeros. Let MATH be a point where MATH is nonzero. Then it is easy to show that as MATH is a holomorphic vector field on MATH, there exist oriented conformal coordinates MATH on an open neighbourhood MATH of MATH in MATH such that MATH in MATH. Therefore MATH in MATH. Taking MATH of the first equation of REF , we find MATH as MATH is antisymmetric and MATH. Similarly, taking MATH of the second equation of REF gives MATH as MATH is antisymmetric. Using MATH and MATH, this becomes MATH . Comparing REF with REF , we see by REF that the subset MATH of MATH is a ruled SL REF-fold. Now every point MATH has such a neighbourhood MATH, except for the isolated zeros of MATH. Thus MATH is an r-oriented ruled SL REF-fold, except possibly along a discrete set of lines. But to be special Lagrangian is a closed condition on the nonsingular part of MATH, so MATH is special Lagrangian. |
math/0012060 | Any NAME surface MATH may be written as MATH, where MATH is a lattice in MATH, and the coordinates MATH on MATH are oriented conformal coordinates. Write MATH in this way. Then the holomorphic vector fields MATH on MATH are of the form MATH for MATH. Lift MATH to a MATH-invariant map MATH. For each MATH, define MATH . Then MATH is a ruled special Lagrangian REF-fold in MATH wherever it is nonsingular, by REF , and it clearly has asymptotic cone MATH. It remains to prove that MATH is asymptotic to MATH with order MATH, in the sense of REF . Let MATH, and define MATH by MATH for MATH and MATH. Then MATH is well-defined, and using the expansions MATH for large MATH, one can show that MATH for large MATH, which is the first equation of REF , with MATH. The other equations of REF may be proved in the same way. Therefore MATH is asymptotic to MATH with order MATH. |
math/0012062 | When MATH, the short sequence MATH is a real form of the sequence MATH . This is (a real subspace of) the direct sum of two elliptic sequences, and so is elliptic. Thus we have ellipticity at MATH whenever MATH. This leaves us to consider the case when MATH, giving (a real subspace of) the sequence MATH . This fails to be elliptic. An easy and instructive way to see this is to consider the simplest REF-dimensional example MATH. Let MATH, MATH, MATH and MATH form a basis for MATH, and let MATH denote MATH . Then MATH, so MATH. The map from MATH to MATH is just the exterior differential MATH. Since MATH the symbol map MATH is not injective, so the symbol sequence is not exact at MATH. Consider also MATH. Then MATH, since there is no bundle MATH. But MATH has no MATH-factor, so is not the image under MATH of any form MATH. Thus the symbol sequence fails to be exact at MATH. It is a simple matter to extend these counterexamples to higher dimensions and higher exterior powers. For MATH, the situation is different. It is easy to show that the complex MATH is elliptic everywhere except at MATH. |
math/0012062 | Let MATH be a torsion-free linear connection on MATH preserving the quaternionic structure. Then MATH. As Sp REF -representations, this is MATH . Using the NAME formula we have MATH. Thus the image of MATH under MATH is contained in the MATH and MATH summands of MATH. Since MATH is torsion-free, MATH, so MATH maps (sections of) MATH to the MATH and MATH summands of MATH. Thus MATH. |
math/0012062 | The first equation is equivalent to REF . The rest follows immediately from decomposing the equation MATH. |
math/0012062 | We have MATH, where MATH and MATH. Applying the NAME operator gives MATH . Rearranging these equations gives MATH and MATH. |
math/0012062 | In terms of representations, the situation is of the form MATH where MATH and MATH. For MATH to be in the representation MATH, its components in the representations MATH and MATH must both vanish separately. The component in MATH comes entirely from MATH, so for this to vanish we must have MATH independently of MATH. Similarly, for the component in MATH to vanish, we must have MATH. |
math/0012062 | The first isomorphism is trivial, as is the last (since the hypercomplex structure acts trivially on MATH). The second isomorphism is REF , and the fourth follows in exactly the same way since MATH also. The middle isomorphism follows a similar argument. |
math/0012062 | Replacing MATH with MATH in the formula for MATH obtained in REF , we have MATH as required. |
math/0012062 | We already know that MATH, by definition. Using REF , we see that MATH increases the number of differentials in the MATH-direction by one, so the index MATH increases by one. The only action in the other directions is the MATH-action, which preserves the irreducible decomposition of the contribution from MATH, so the index MATH remains the same. |
math/0012062 | Let MATH. We calculate the dimensions of the spaces MATH for MATH. Recall the notation MATH from REF . It is clear that MATH, since MATH on MATH is sinply MATH on MATH. Thus MATH and MATH. The cases MATH and MATH are easy to work out since they are of the form MATH. For MATH, we have MATH and MATH. For MATH, MATH and MATH. The case MATH is slightly more complicated, as we have to take into account exterior products with the self-dual REF-forms MATH and anti-self-dual REF-forms MATH in MATH. The spaces MATH and MATH receive contributions only from the self-dual part MATH, from which we infer that MATH and MATH. Finally, the space MATH has dimension MATH and the space MATH has dimension MATH, giving MATH a total dimension of MATH. It is now a simple matter to verify that for the top sequence of REF MATH for the middle sequence MATH and for the bottom sequence MATH . |
math/0012062 | Consider first the top sequence MATH . The NAME formula shows that there are no spaces MATH or MATH. Thus MATH on MATH and MATH, so MATH for the top sequence. It is easy to check using the relevant lie in conditions that the map MATH is injective and the map MATH is surjective. To show exactness at MATH, consider MATH. A calculation using REF shows that MATH . Since the MATH have no MATH-components, MATH if and only if all these components vanish. This occurs if and only if MATH, MATH, MATH (since as remarked in REF these equations also guarantee that MATH etc), in which case it is clear that MATH . This shows that the sequence MATH is exact. Restricting to MATH and MATH, we see that exactness holds at these spaces in the middle and bottom sequences respectively of REF . Consider MATH. Then MATH . Since these are linearly independent, MATH if and only if MATH, and MATH is injective. Hence the bottom sequence MATH is exact. Finally, we show that the middle sequence MATH is exact at MATH, which is now sufficient to show that the sequence is exact. Let MATH. Recall the lie in REF that MATH must take the form MATH for some MATH. (The MATH-factor makes no difference here and is useful for cancellations.) Thus a general element of MATH is of the form MATH for MATH. A similar calculation to that of REF shows that MATH . But this is exactly the lie in REF which we need for MATH to be in MATH in which case we have MATH . This demonstrates exactness at MATH and so the middle sequence is exact. |
math/0012063 | For the necessity of the condition we have MATH thus MATH, that is MATH . Since MATH and MATH are relatively prime, the last equation implies that MATH divides MATH and MATH divides MATH. Now, let MATH be such that MATH and MATH, respectively. Then it follows from REF that MATH . Hence, MATH. The proof of the converse is obvious. |
math/0012063 | We have MATH . And, MATH . Thus, MATH . That is, MATH is a NAME polynomial in MATH with MATH . |
math/0012063 | MATH is the fraction field of MATH. Thus, if an element MATH satisfies MATH and MATH then, by REF , MATH with MATH relatively prime NAME polynomials. If the hypothesis is true, this contradicts the fact that MATH and MATH are relatively prime. |
math/0012063 | Suppose that the hypothesis of the proposition is true. According to our notation MATH. Consider the set of subindices MATH, MATH, ordered with the lexicographical order. That is, MATH if and only if the first coordinates MATH and MATH from the left, for MATH above, which are different satisfy MATH. Let MATH be the largest subindex such that MATH occurs in MATH. Let MATH . Then MATH . For MATH we have MATH, thus it does not occur in MATH by the choice of MATH. Also, it does not occur in MATH. Thus the above equation implies that MATH must occur in MATH. Let MATH be its coefficient in MATH and let MATH . Then we have MATH or MATH . But the total degree of MATH is strictly less than the total degree of MATH. This forces MATH . |
math/0012063 | This is a consequence of REF . |
math/0012063 | This is a direct consequence of REF . |
math/0012063 | Let MATH . Then MATH . By REF , for each MATH with MATH the corresponding coefficient of MATH in MATH is MATH . Since MATH, it must be MATH or, equivalently, MATH . This means that for each MATH, the coefficient MATH of MATH in MATH divides the expression MATH . Thus, for each MATH, there is MATH such that MATH . As in the proof of REF , order the triples MATH, MATH, with the lexicographical order. Let MATH be the largest subindex such that MATH occurs in MATH. We have MATH and MATH. Now, for each MATH such that MATH occurs in MATH we have that MATH will occur in MATH but not in MATH or in MATH by the choice of MATH. Therefore, it must occur in MATH. Let MATH then MATH . So MATH occurs in MATH only in MATH . Since MATH occurs in MATH and not in MATH it must occur in MATH. Let MATH be the coefficient of MATH in MATH. Then it has to be MATH . The above equation implies that MATH divides MATH. But this is impossible since the total degree of MATH is strictly less than the total degree of MATH. This contradiction yields the result. |
math/0012063 | We have MATH with MATH . Thus, MATH . By REF , the coefficient of MATH in MATH is MATH . Hence, it must be MATH . From this, MATH . The coefficient of MATH in the above expression is MATH, for MATH. Since this expression for MATH is valid for any index MATH, the ``in particular" part follows immediately. |
math/0012063 | This is an immediate consequence of the ``in particular" part in REF . |
math/0012063 | By REF and since MATH, the coefficient of MATH in MATH is MATH . The MATH are the coefficients of the power products MATH, with MATH, such that MATH contains an expression of the form MATH. By REF , these power products are MATH all of which violate REF for MATH and MATH. Therefore it must be MATH, for all MATH; MATH. But now, substituting back in REF , we see that MATH . Hence, MATH . |
math/0012063 | To prove that MATH for MATH we first show that MATH for MATH. Indeed, for MATH we have MATH, so MATH and MATH . Since MATH has no MATH with MATH, each term in MATH containing such a MATH must be cancelled. In particular we need to cancel the terms containing MATH for MATH above. For that we can only use the derivatives of power products of the form MATH . But these are all strictly greater than MATH (the leading power product of MATH), and they may not occur in MATH. As a consequence, it has to be MATH. Now let MATH be such that MATH for MATH. Then MATH and MATH . Likewise, we need to cancel all the terms in MATH that contain MATH, for MATH. In particular, we need to cancel MATH for MATH. For that we can only use the power products of the form MATH for MATH. But all of them are strictly greater than MATH and cannot occur in MATH. Thus, it has to be MATH. Since this argument is valid for any MATH, it follows that MATH, for MATH. This makes the statement that MATH for MATH, MATH, true for MATH. Now assume that MATH is such that MATH for MATH. So MATH and for MATH occurs in MATH. Thus we need to cancel it. For that we can only use the derivatives of power products of the form MATH with MATH since MATH for all MATH by hypothesis. But all such power products are strictly greater than MATH and therefore they cannot occur in MATH. This forces MATH for MATH. We can repeat this process until MATH and get MATH for all MATH, MATH, MATH, that is, MATH . Now we show that MATH for MATH, MATH, MATH. The process is analogous to what we just did. First we show that MATH for MATH. Indeed, for each MATH we have for MATH that MATH occurs in MATH. So, in order to cancel it, we need to use the derivatives of power products of the form MATH with MATH, all of which are strictly greater than MATH if MATH, and for MATH we cannot simply have one of those since MATH for MATH. Thus such power products cannot occur in MATH and it has to be MATH for MATH. Let MATH be such that MATH for MATH, MATH. We have MATH and for all MATH, MATH, we have that MATH occurs in MATH and in order to cancel it we only have the derivatives of power products of the form MATH with MATH since MATH for MATH, MATH. For MATH, all these power products are strictly greater than MATH and therefore they cannot occur in MATH. For MATH we cannot simply have such power products since for MATH, MATH if MATH. Thus it has to be MATH for MATH. We can repeat this process until MATH and get MATH, MATH, MATH, MATH. This completes the proof of the first part of the lemma. To prove that MATH, for all MATH, suppose that there is MATH such that MATH and let MATH be such that MATH. Then MATH will contain MATH or MATH . As noted above, since MATH does not contain any MATH with MATH, we need to cancel the terms in MATH involving either of the above. But that is impossible since MATH for all MATH and by REF all the power products MATH in MATH must have MATH for MATH. In particular, we cannot have in MATH power products of the form MATH as in REF . |
math/0012063 | Let MATH be the coefficient of MATH in MATH. We have MATH . In order to cancel MATH above, we can only use the derivatives of the power product MATH since for MATH we have MATH. Let MATH be the coefficient of MATH in MATH. Then MATH . On the other hand, in order to cancel MATH above, the only power product that we can use is, again, MATH since MATH for MATH. Thus it must be MATH as well. From REF it follows that, for MATH, MATH . |
math/0012063 | This is a consequence of REF . |
math/0012063 | This is a consequence of REF . |
math/0012063 | This is just a restatement of REF . |
math/0012063 | Let MATH so that, MATH . By REF we can write MATH, with MATH reduced with respect to MATH. Now, MATH . On the other hand, we have MATH . Therefore, it has to be MATH. But MATH is reduced with respect to MATH. It follows, by REF , that MATH. The statement about the form of MATH is just the content of REF . |
math/0012063 | Since MATH is a basis of MATH we have MATH with MATH. Thus, MATH where MATH. Now, MATH is a matrix of change of basis so it is invertible. Also the MATH are contants for MATH, thus the map MATH is a differential bijection. In other words, the differential rings MATH and MATH are isomorphic and therefore we can apply REF to MATH . |
math/0012063 | First we need to show that MATH is a NAME extension with differential NAME group MATH. We will use the characterization of REF . We have CASE: MATH, where MATH is the finite dimensional vector space over MATH spanned by the MATH. CASE: The group MATH acts as a group of differential automorphisms MATH with MATH and MATH. This follows from the fact that MATH is the function field of MATH. CASE: MATH has no new constants. This is a consequence of REF . Now, suppose that MATH is a NAME extension of MATH with differential NAME group MATH. By REF , we have that in this situation MATH is isomorphic to MATH (the function field of MATH) as a MATH-module and as a MATH-module. Any MATH equivariant derivation MATH on MATH extends the derivation on MATH in such a way that MATH with MATH. Since MATH is a NAME extension for MATH, then so is MATH, the derivation on MATH being the corresponding restriction of MATH. From this NAME extension one can retrieve MATH by extension of scalars from MATH to MATH. In this way, any NAME extension MATH with differential NAME group MATH can be obtained from MATH via the specialization MATH. This means that MATH is a generic NAME extension of MATH for MATH. |
math/0012063 | (Sufficiency) Suppose that MATH properly contains MATH. Let MATH be the transcendence degree of MATH over MATH. Since MATH is algebraically closed, MATH has to be at least one. We have the tower of fields MATH where the transcendence degree of MATH is MATH and the transcendence degree of MATH is MATH. Since MATH the transcendence degree MATH of MATH has to be MATH and therefore there is an algebraic relation among the MATH over MATH. Let MATH, MATH, be such that MATH . Then MATH . Since the MATH and MATH are polynomials in the MATH with coefficients in MATH, the last equation gives an algebraic relation among the MATH and the MATH over MATH. For the necessity we only need to point out that by construction the set of all the MATH and all the MATH are algebraically independent over MATH. |
math/0012063 | First assume that MATH is a NAME extension. Then the field of constants MATH of MATH coincides with MATH. So we can apply REF and get the result. Conversely, if the set of all the MATH and all the MATH are algebraically independent over MATH, by REF , MATH is a no new constant extension. On the other hand, MATH is obtained from MATH by the extension of scalars: MATH where MATH is the coordinate ring of MATH and MATH acts on MATH fixing MATH. So, MATH. Counting dimensions we get that MATH since MATH, the function field fo MATH. Finally, MATH, where MATH is the finite-dimensional vector space over MATH spanned by the MATH. By REF , MATH is a NAME extension. |
math/0012064 | Here we sketch a simple existence proof. If MATH and MATH are quasi-isomorphic as complexes, then the dg NAME algebras MATH will be quasi-isormorphic as dgLie algebras (one must use the cofreeness of the symmetric product). Since MATH extends to an integrable differential MATH on MATH, so must MATH extend to a differential MATH making the dg NAME algebras MATH quasi-isomorphic. |
math/0012064 | This is a computation. We introduce as a notational aid, the bilinear function MATH defined for MATH and MATH by MATH . While MATH does not satisfy the NAME identity, nor the NAME identity up to homotopy (it is not even skew-symmetric for arbitrary elements of MATH), it does behave well with respect to the differentials - both differentials are derivations of MATH: MATH . In this notation, the lemma asserts that for the special case of MATH and MATH, we have MATH . Let MATH and MATH. Select elements MATH with MATH . We have MATH and MATH identified with MATH. Then MATH can be considered as the cocylce MATH . Now, we compute MATH. We set MATH and find elements MATH satisfying MATH . Then MATH . One finds that MATH . One finds that MATH and at the end MATH as claimed. |
math/0012064 | From REF , the complexes of sheaves MATH and MATH are quasi-ismorphic. Therefore, by taking exterior powers, MATH and MATH are quasi-ismorphic (as complexes of sheaves). Since quasi-isomorphic complexes induces isomorphisms in hypercohomology CITE, MATH and MATH are quasi-isormorphic as complexes. By REF , we have a quasi-isomorphism between MATH and MATH, establishing the theorem. |
math/0012064 | Consider the double complex MATH and the filtration on the single complex MATH . Let MATH denote the spectral sequence associated to this filtered complex. We will see that spectral sequence degenerates at the MATH term: MATH . We compute MATH. To describe the classes in MATH that survive in MATH (see REF ), first note that since the MATH are locally free sheaves on MATH, MATH . CASE: The (homogeneous) elements generating MATH that are MATH closed and not necessarily MATH exact are given by MATH where MATH . A MATH as above is a MATH boundary if and only if MATH for some MATH . So, MATH and MATH . CASE: Each class in MATH is of the form MATH . Now, MATH . Because MATH can be expressed as a sum of terms, each of which has a positive power of a MATH in the numerator, MATH . Therefore, each MATH is a class in MATH of total degree MATH (NAME degree MATH and degree MATH in MATH) and MATH . Note, too, that it is impossible for MATH to kill any of the NAME degree zero classes, hence MATH . Together, we have MATH . |
math/0012066 | We just sketch the proof here. The complete proof will appear somewhere. This proof is based on an unpublished joint paper with NAME. Consider the space MATH. It is a MATH-graded vector space. Consider the MATH formality morphism on it. The polyvector fields MATH is isomorphic to MATH. (This phenomenon can be considered roughly as a kind of the NAME duality). There is an odd vector field MATH on MATH such that MATH (which generates the cochain differential). In coordinates, MATH where MATH are the structure constants of the NAME algebra MATH in some basis MATH and MATH are the odd coordinates on MATH corresponding to MATH. We want know to localize the formality morphism on MATH at the solution to the NAME equation MATH. We claim that the only graphs which appear are unions of the wheels. It follows from REF , LemmaREFEF. REF . Note that here the wheels are not the same wheels as for MATH: here we have one outgoing edge and two incoming edges for each vertex whence for MATH we have two outgoing edges and one incoming. It reflects the fact that MATH is a (quadratic) vector field whence the NAME structure MATH is a (linear) bivector field. It is not straightforward to compute the NAME weights of these wheels corresponding to MATH but it turns out it is possible. The answer is exactly the formula in the Theorem. |
math/0012066 | The natural isomorphism of algebras MATH, MATH is equal to MATH (see REF ). We just apply the map MATH to both sides of REF and use that MATH is a map of algebras. |
math/0012070 | CASE: Assume first that the model is the line MATH with MATH and not the circle. Now, for given MATH, the probability that there is no edge between any MATH and any MATH is MATH . Define a cut to be a vertex with this property. Take MATH such that MATH. So, by the ergodic theorem, if MATH is large enough then, with probability as high as we like, there are at least MATH cuts, and therefore the diameter is, with the same probability, at least MATH. Returning to the cycle, we can divide it into two lines, of length MATH each. Each of these halves is of diameter at least MATH. The same kind of calculation yields that, with high probability, the edges between the two halves of the cycle don't reach the middle third of each of the lines of the vertices, and therefore the diameter stays above MATH. In order to prove that the limit of MATH exists, we do the following: First, consider long-range percolation on MATH. MATH will be the diameter of the long-range percolation restricted to MATH. By the sub-additive ergodic theorem, MATH a.s. for some MATH. In order to prove the convergence for the diameter of the cycle, we divide the cycle MATH into two intervals MATH and MATH of length MATH. The diameter of each half is (with very high probability) approximately MATH. The longest connection between the two halves is of length MATH, so there are cut points MATH and MATH such that MATH and MATH, there are no edges between the arcs MATH and MATH, and, by the strong NAME property, the diameter of each of these arcs is MATH. So, we are done. It is of interest to study the fluctuations of MATH from MATH. CASE: The graph dominates the MATH random graph with edge probability MATH. It is known (see, for example, CITE) that there exists a constant MATH such that MATH . Since the diameter is a decreasing function (with respect to the standard partial order), REF applies also for our model with MATH. Actually, in this case we can even say more: The infinite graph whose vertices are the integers, such that every two vertices are attached with probability MATH has, a.s., a finite diameter, see next section. CASE: Here we use an argument in the spirit of NAME and NAME 's renormalization (see CITE): Again, assume that the model is a line instead of a circle. This assumption creates a measure which is dominated by the original one, and therefore it suffices to prove the result for the line. Take MATH where MATH is such that MATH . Let MATH be a large number, and define MATH to be MATH . Taking MATH small enough and MATH large enough, we can get that MATH . Now, take MATH . We divide the interval of length MATH into MATH intervals of length MATH. Each of these, we divide into MATH intervals of length MATH, and so on. This structure has a lot in common with the one used in CITE for proving the existence of the infinite cluster. We use the following terminology: The MATH intervals of length MATH obtained by this division from MATH, are called components of degree MATH. The components of degree MATH inside such a component are sub-components. two intervals (or component) MATH and MATH are said to be attached to each other, if there exists a bond between a point in MATH and a point in MATH. It is enough to show that for some constant MATH, MATH because MATH for MATH, and if MATH, then we can bound MATH by MATH. This will be enough because for some constant MATH, we have MATH. We now prove REF : Take some MATH. We will show that for MATH large enough, MATH. Define MATH. Then the probability that two intervals of length MATH of distance MATH from each other have an edge between them is at least MATH. Take MATH so large that for every MATH, we have MATH, and so that MATH. For every MATH, consider the line of length MATH. Divide it into MATH components of size MATH. The probability that not all of the MATH components are attached to each other is bounded by MATH . Now, take MATH such that MATH where MATH is such that MATH. Consider the following event, denoted by MATH: for every MATH, and for every component of degree MATH, all of its sub-components of degree MATH are attached to each other. Given MATH, the diameter is no more than MATH for MATH. Therefore, we want to estimate the probability of MATH: Take MATH such that MATH. The probability that there exist a component of degree MATH, such that not all of its sub-components of degree MATH are attached to each other could be bounded by the number of components of degree MATH times the probability for this event at each of them, that is, by MATH by REF for large enough MATH. Therefore, the probability of MATH is at least MATH for MATH large enough. So, MATH for every MATH. CASE: When MATH, the expected value of the number of vertices attached to a certain vertex is finite. Therefore, the graph's growth rate is bounded by the growth rate of a NAME tree. Thus, its growth rate is (bounded by) exponential, and so the diameter cannot be smaller than a logarithm of the number of vertices (MATH, in this case). |
math/0012070 | Divide the circle into two arcs, MATH and MATH, of length MATH each. The expected value of the number of edges connecting the two halfs is: MATH . For some constant MATH. So, the expected value of the size of the boundary of MATH divided by the size of MATH is bounded by MATH for some constant MATH, and this, using the NAME inequality, gives the desired result for every MATH. |
math/0012071 | Suppose there would be such a MATH. Then necessarily MATH and MATH as well as MATH would have to start in order MATH. But this is incompatible with MATH. |
math/0012071 | It is straightforward to check that MATH is well-defined, unitary, and an intertwiner between MATH and MATH. |
math/0012071 | This follows immediately from the fact that the classical limit functor preserves orthogonal sums. |
math/0012073 | The proof is essentially the same as in the NAME algebra setting. For example, to show REF , fix MATH and consider the algebra MATH with convolution product MATH and unit element MATH. Using REF , and REF, one easily checks that MATH and MATH. Hence we can conclude that MATH. |
math/0012073 | Set MATH. Firstly MATH (since MATH) and so MATH. Then let MATH. Using REF, MATH. Therefore MATH and so MATH. Finally, let MATH. By REF , MATH. Thus MATH and hence MATH. |
math/0012073 | Suppose that MATH is a right MATH-comodule over MATH. Firstly, for any MATH, MATH, by REF. Secondly, for any MATH, MATH, and MATH, MATH . Moreover, by construction, MATH for any MATH. Hence MATH is a MATH-graded left MATH-module. Conversely, suppose that MATH is a left MATH-graded MATH-module. Since, for all MATH and MATH, MATH, REF is verified. To show that REF is satisfied, let MATH and MATH. Set MATH . Suppose that MATH. Then there exists MATH such that MATH. Now MATH is dense in the linear topological space MATH endowed with the MATH-topology (see CITE). Thus MATH, where MATH. Then there exists MATH such that MATH. Now, for all MATH, MATH, and MATH, MATH that is, MATH. Therefore MATH, which is a contradiction. We conclude that MATH and then MATH. Hence MATH is a right MATH-comodule over MATH. |
math/0012073 | Let MATH be the map defined as in REF . It is easy to verify that MATH. Thus MATH is a MATH-graded left MATH-module and hence, by REF , MATH is a right MATH-comodule over MATH. |
math/0012073 | REF follows directly from REF . To show REF , let MATH be a graded submodule of a rational MATH-graded left MATH-module MATH. Let MATH defined by MATH. Suppose that there exist MATH and MATH such that MATH. Since MATH is rational, we can write MATH. Without loss of generality, we can assume that the MATH are MATH-linearly independent and MATH. Let MATH such that MATH and MATH for MATH. Now MATH, contradicting the fact that MATH is a graded submodule of MATH. Thus MATH for all MATH. Hence MATH is rational. Let us show REF . Denote by MATH the left action of MATH on MATH. Set MATH and MATH given by MATH. Recall MATH can be viewed as a subspace of MATH via the embedding MATH given by MATH. Define MATH for any MATH, and set MATH. Fix MATH, MATH, and MATH. Let MATH such that MATH. We can write MATH. Now, for any MATH, MATH. Then MATH and so MATH. Hence MATH. Therefore MATH is a graded submodule of MATH. Moreover one easily checks at this point that MATH for any MATH in MATH. Thus MATH is rational. Suppose now that MATH is another rational graded submodule of MATH and denote by MATH its corresponding MATH-comodule structure maps (see REF ). Let MATH and MATH such that MATH. By the definition of MATH and MATH and of the embedding MATH, it follows that MATH. Thus MATH, and so MATH. This holds for all MATH such that MATH. Thus MATH for any MATH. Hence MATH. Therefore MATH is the unique maximal rational graded submodule of MATH and is the sum of all rational graded submodules of MATH. |
math/0012073 | We will denote by MATH (respectively,MATH) the right action of MATH on MATH (respectively,on MATH) and by MATH (respectively,MATH) the MATH-comodule structure maps of MATH (respectively,of MATH). For any MATH, define MATH by MATH. Remark first that, for any MATH, MATH is a coinvariant of MATH on MATH. Indeed, if MATH, MATH . For any MATH, define MATH by MATH. Then MATH is MATH-linear since MATH for all MATH and MATH. Moreover MATH for all MATH. Indeed let MATH and MATH. By the definition of MATH, there exists a coinvariant MATH of MATH on MATH such that MATH. In particular MATH. Thus MATH . Then MATH is a NAME MATH-comodule morphism. To show that MATH is an isomorphism, we construct its inverse. For any MATH, define MATH by MATH. The map MATH is well-defined since MATH is a coinvariant of MATH on MATH for all MATH, and is MATH-linear since, for any MATH and MATH, MATH . Moreover MATH for all MATH. Indeed, for any MATH, MATH and so, since MATH is a MATH-coinvariant of MATH on MATH, MATH . Thus MATH is a NAME MATH-comodule morphism. It remains now to verify that MATH and MATH for any MATH. Let MATH and MATH. By the definition of MATH, there exists a coinvariant MATH of MATH on MATH such that MATH. In particular, MATH and MATH. Then MATH . Finally, for all MATH, MATH . Hence MATH and MATH and MATH are NAME MATH-comodule isomorphisms. |
math/0012073 | Let MATH be a left MATH-integral for MATH such that MATH for some MATH and let MATH such that MATH. Then MATH (by REF ) and so MATH. Using REF, we have that MATH. Hence MATH. The right case can be done similarly. |
math/0012073 | Suppose that MATH is a left MATH-integral for MATH. Fix MATH. Let MATH such that MATH. We have that MATH since MATH for all MATH. Therefore MATH, see REF . Hence, since MATH for all MATH, MATH is a coinvariant of MATH on MATH. Conversely, suppose that MATH is a coinvariant of MATH on MATH. Let MATH. Then MATH, that is, MATH for all MATH. Hence MATH is a left MATH-integral for MATH. |
math/0012073 | Let us first show that for any MATH, MATH, MATH, and MATH, MATH where MATH denotes the natural pairing between MATH and MATH. Remark first that MATH . Then, for all MATH, MATH and so MATH where MATH is the left MATH-action on MATH defined by MATH for any MATH and MATH. Then MATH and hence REF is proved. Recall that the MATH-comodule structure map MATH of MATH is, via the natural embedding MATH, the restriction onto MATH of the map MATH defined by MATH. Let MATH. By REF, we have that, for any MATH such that MATH, MATH, and MATH, MATH . Therefore, by REF , MATH. Hence the action of MATH on MATH is well-defined. This is a right action because MATH is unitary and anti-multiplicative (see REF ). Finally, REF is satisfied since REF says that MATH for any MATH, MATH, and MATH. Thus MATH is a right NAME MATH-comodule over MATH. |
math/0012073 | Suppose that MATH is a non-zero left MATH-integral for MATH. Let MATH. If MATH, then the result is obvious. Let us suppose that MATH. Then MATH by REF and so MATH (by REF ). Let MATH such that MATH. By REF , MATH. Now MATH (since MATH). Thus MATH (since MATH is an isomorphism) and so MATH (since MATH). |
math/0012073 | For any MATH, since MATH is finite dimensional and MATH, we have that MATH. Therefore the natural embedding MATH is an isomorphism. Thus MATH is a rational MATH-graded MATH-module (see REF) and so MATH for all MATH. Now MATH since MATH, MATH, and MATH (by REF ). Hence there exists a MATH-coinvariant MATH of MATH on MATH such that MATH. Set MATH for any MATH. By REF , MATH is then a left MATH-integral for MATH. Moreover MATH and so MATH is non-zero. Suppose now that MATH is another left MATH-integral for MATH. Let MATH such that MATH. By REF , MATH and MATH are injective (since there exists a non-zero left integral for MATH) and so MATH. Therefore MATH since MATH and MATH. Now MATH by REF and MATH (by REF ). Hence there exists MATH such that MATH. If MATH is such that MATH, then MATH and thus MATH (since MATH and MATH). If MATH is such that MATH, then MATH and so MATH. Hence we can conclude that MATH is a scalar multiple of MATH. To show the existence and the uniqueness of right MATH-integrals for MATH, it suffices to consider the coopposite NAME MATH-coalgebra MATH to MATH (see REF). Indeed MATH is a right MATH-integral for MATH if and only if MATH is a left MATH-integral for MATH. This completes the proof of the theorem. |
math/0012073 | To show REF , let MATH. By REF , MATH and MATH are injective. Thus MATH and so MATH is bijective. To show REF , let MATH be a non-zero left MATH-integral for MATH and fix MATH. If MATH, then the result is obvious. Let us suppose that MATH. Recall that MATH and MATH defined by MATH is an isomorphism. Since MATH, MATH, and MATH is bijective, the map MATH defined by MATH is an isomorphism. Thus MATH is a free left MATH-module of rank REF with vector basis MATH. Using MATH (see REF), one easily deduces the right case. |
math/0012073 | Let MATH be a non-zero right MATH-integral for MATH. Let MATH. For any MATH, MATH is a right MATH-integral for MATH and thus, by REF , there exists a unique MATH such that MATH for all MATH. Now MATH (MATH). Thus there exists a unique MATH such that MATH for any MATH and MATH. Then MATH for any MATH and MATH and hence MATH for all MATH. Let MATH. If MATH, then either MATH or MATH (by REF ) and so MATH. If MATH, then, for any MATH and MATH, MATH and thus MATH (since MATH by REF ), that is MATH. Moreover MATH and so MATH (since MATH by REF ). Then MATH is a MATH-grouplike element of MATH. Since all the right MATH-integrals for MATH are scalar multiple of MATH, the ``existence" part of the lemma is demonstrated. Let us now show the uniqueness of MATH. Suppose that MATH is another such MATH-grouplike element of MATH. Let MATH be a non-zero right MATH-integral for MATH. Fix MATH. If MATH, then MATH. If MATH, then MATH (by REF ) and so there exists MATH such that MATH. Therefore MATH. This completes the proof of the lemma. |
math/0012073 | To show REF , let MATH. Define MATH by MATH for any MATH. If MATH denotes the product in the convolution algebra MATH (see REF), then, for any MATH, MATH . Therefore, since MATH is invertible in MATH with inverse MATH, we have that MATH, that is MATH for all MATH. REF can be deduced from REF using the NAME MATH-coalgebra MATH (see REF). |
math/0012073 | We use the same arguments as in the proof of CITE, even if we cannot use the duality (since the notion a NAME MATH-coalgebra is not self dual). We can assume that MATH is a non-zero right MATH-integral (otherwise the result is obvious). To show REF , let MATH and MATH. Since MATH is a non-zero right integral for the NAME algebra MATH, there exists a left integral MATH for MATH such that MATH (compare CITE). By REF for MATH, we have that MATH . It is easy to verify that MATH is a right MATH-integral for MATH and MATH is a right integral for MATH such that MATH. Thus REF for MATH gives that MATH . Hence, comparing with REF , we obtain MATH . Now MATH is a right MATH-integral for MATH and MATH is a left integral for MATH such that MATH. Thus MATH (by applying REF for MATH), that is MATH . Finally evaluating REF with MATH and using REF gives that MATH. To show REF , let MATH and MATH. For any MATH, let us define MATH by MATH for all MATH. Using REF , one easily checks that MATH is a right MATH-integral for MATH. Let us denote by MATH the multiplication of MATH and by MATH the right action of MATH on MATH defined by MATH. Then, since MATH is the distinguished grouplike element of MATH, REF with MATH and MATH gives that MATH, that is MATH. Let us show REF . For any MATH, define MATH by MATH for all MATH. Since MATH and MATH are left MATH-integrals for MATH which are non-zero (because MATH is non-zero, MATH is invertible and MATH is bijective), there exists MATH such that MATH for all MATH (by REF ). As above, let MATH be a left integral for MATH such that MATH. Recall that MATH. Thus MATH. Hence MATH for all MATH, that is MATH for all MATH and MATH. This completes the proof of the theorem. |
math/0012073 | We can suppose that MATH. Let MATH. Remark that it suffices to show that, for all MATH, MATH . Fix MATH. Let MATH be a non-zero right MATH-integral for MATH (see REF ). By multiplying MATH by some (non-zero) scalar, we can assume that MATH. By REF , there exists MATH such that MATH for all MATH. By REF , MATH. Thus MATH . Since MATH is a right MATH-integral for MATH and MATH is a right integral for MATH such that MATH, REF applied to MATH gives that MATH . Then MATH . Now, since MATH is left integral for MATH, MATH . Therefore MATH and so, using REF and then REF , MATH . Now, since MATH for any MATH, MATH, and MATH is an algebra morphism, MATH . Thus MATH. Finally, comparing with REF , we get REF. |
math/0012073 | Let MATH and MATH. If MATH, then the result is obvious. Let us suppose that MATH. Let MATH be a non-zero right MATH-integral for MATH. Then MATH . Now, by REF , MATH is a free right MATH-module of rank REF for the action defined by MATH for any MATH and MATH. Moreover MATH is a basis vector of MATH. Thus, since the above computation says that MATH we conclude that MATH. |
math/0012073 | To show REF , let MATH of finite order MATH. Consider the distinguished MATH-grouplike element MATH of MATH and the distinguished grouplike element MATH of MATH. Using REF , one easily shows by induction that MATH for all MATH and MATH. Recall that the order of a grouplike element of a finite dimensional NAME algebra MATH is finite and divides MATH (see CITE). Therefore MATH has a finite order which divides MATH and MATH has a finite order which divides MATH. Now MATH . Then, for all MATH, by REF, MATH . Hence MATH. REF is REF for MATH, since in this case MATH is an endomorphism of MATH. |
math/0012073 | We have to show that if MATH is semisimple then MATH is semisimple. Suppose that MATH is semisimple and fix MATH. Since MATH is a finite dimensional algebra, it suffices to show that all left MATH-modules are completely reducible. Thus let MATH be a left MATH-module and MATH be a submodule of MATH. Since MATH is a finite dimensional semisimple NAME algebra, there exists a left integral MATH for MATH such that MATH (compare CITE). Let MATH be any MATH-linear projection which is the identity on MATH. Let MATH be the MATH-linear map defined by MATH for any MATH, where MATH denotes the action of MATH on MATH. The map MATH is the identity on MATH since, for any MATH, MATH . Let MATH. Using REF and the fact that MATH is a left integral for MATH, we have MATH and so MATH . Therefore, for all MATH and MATH, MATH . Hence MATH is MATH-linear and MATH is a MATH-supplement of MATH in MATH. |
math/0012073 | Let us show REF . Suppose that MATH is a sum of simple MATH-submodules. Let MATH be a maximal subset of MATH such that MATH is direct. Let us show that this sum is in fact equal to MATH. It suffices to prove that each MATH REF is contained in this sum. The intersection of our sum with MATH is a MATH-subcomodule of MATH, thus equal to REF or MATH. If it is equal to MATH, then MATH is not maximal since we can adjoin MATH to it. Hence MATH is contained in the sum. To show REF , suppose that MATH and let MATH be a MATH-subcomodule of MATH. Let MATH be a maximal subset of MATH such that the sum MATH is direct. The same reasoning as before shows this sum is equal to MATH. Let us show REF . Let MATH be the MATH-subcomodule of MATH defined as the sum of all simple MATH-subcomodules of MATH. Suppose that MATH. Then MATH where MATH is a non-zero MATH-subcomodule of MATH. Let us show that there exists a simple MATH-subcomodule of MATH, contradicting the definition of MATH. By REF , MATH (where MATH) is a rational MATH-graded left MATH-module which is non-zero. Let MATH, MATH. The kernel of the morphism of MATH-graded left MATH-modules MATH is a MATH-graded left ideal MATH. Therefore MATH is contained in a maximal MATH-graded left ideal MATH (by NAME 's lemma). Then MATH is a maximal MATH-graded left MATH-submodule of MATH (not equal to MATH), and hence MATH is a maximal MATH-graded MATH-submodule of MATH, not equal to MATH (corresponding to MATH under the MATH-graded isomorphism MATH). Moreover it is rational since it is a submodule of the rational module MATH (see REF ). So we can consider the MATH-subcomodule MATH of MATH (see REF ). Write then MATH where MATH is MATH-subcomodule of MATH. Therefore MATH and so MATH. Now, since MATH is a maximal MATH-graded MATH-submodule of MATH (not equal to MATH), we have that MATH is a non-zero MATH-graded MATH-submodule of MATH which does not contain any MATH-graded submodule other than MATH and itself. Moreover MATH is rational since it is a MATH-graded MATH-submodule of the rational MATH-graded MATH-module MATH (see REF ). Finally MATH is a simple MATH-subcomodule of MATH. |
math/0012073 | Let MATH be a MATH-subcomodule of a cosemisimple right MATH-comodule MATH. Let MATH be the sum of all simple MATH-subcomodules of MATH and write MATH. Therefore MATH. If MATH, it contains a simple MATH-subcomodule (see the demonstration of REF ). Thus MATH and MATH, which is cosemisimple. Now write MATH. MATH is a sum of simple MATH-subcomodules (it is a MATH-subcomodule of MATH and thus cosemisimple) and the canonical projection MATH induces a MATH-comodule isomorphism between MATH onto MATH. Hence MATH is cosemisimple. |
math/0012073 | REF implies trivially REF . Moreover REF is equivalent to REF . Indeed REF implies REF since MATH (by REF ). Conversely, suppose that MATH is such that MATH. Let MATH such that MATH. Then MATH. Now MATH by REF . Hence MATH. Let us show that REF implies REF . Consider MATH as a a right MATH-comodule over itself (with comultiplication as structure maps). For any MATH, set MATH. Since the comultiplication is unitary, MATH is a MATH-subcomodule of MATH. Therefore MATH is a direct summand of MATH (since MATH is cosemisimple), that is there exists a MATH-comodule morphism MATH such that MATH for all MATH. For any MATH, since MATH, there exists a (unique) MATH-form MATH such that MATH for all MATH. Let us verify that MATH is a right MATH-integral for MATH. Let MATH. Since MATH is a MATH-comodule morphism, we have that MATH . If MATH, then either MATH or MATH (by REF ) and so MATH. If MATH, then there exists MATH such that MATH and, by applying MATH to both sides of REF, we get that MATH. Therefore MATH is a right MATH-integral for MATH. Finally, let MATH such that MATH. Then MATH (since MATH) and so MATH (since MATH). To show that REF implies REF , let MATH be a reduced right MATH-comodule over MATH with structure maps by MATH and MATH be a MATH-subcomodule of MATH. We have to show that MATH is a direct summand of MATH (see REF ). Define MATH by MATH for all MATH. We first prove that, for any MATH, MATH . Indeed, for any MATH and MATH, MATH . Let MATH be any MATH-linear projection and define, for all MATH, MATH . For any MATH, using REF, we have MATH . Thus MATH is a MATH-comodule morphism between MATH and MATH. Let MATH and MATH. If MATH, then MATH (since MATH and thus MATH is reduced) and so MATH. If MATH, then MATH . Therefore MATH is a MATH-comodule projection of MATH onto MATH and consequently MATH is a direct summand of MATH (namely MATH). This finishes the proof of the theorem. |
math/0012073 | To show REF , suppose that MATH is cosemisimple. By REF , there exists a right MATH-integral MATH for MATH such that MATH. Since MATH is a right integral for MATH such that MATH, MATH is cosemisimple (by CITE). Let us show REF . Suppose that MATH is finite dimensional and MATH is cosemisimple. By CITE, there exists a right integral MATH for MATH such that MATH. By REF , there exists a non-zero right MATH-integral MATH for MATH. In particular, MATH is a non-zero right integral for MATH. Therefore, since MATH is finite dimensional, there exists MATH such that MATH (by CITE). Thus MATH is a right MATH-integral for MATH such that MATH. Hence MATH is cosemisimple by REF . This completes the proof of the corollary. |
math/0012073 | By REF , MATH is semisimple if and only if MATH is semisimple, and by REF , MATH is cosemisimple if and only if MATH is cosemisimple. It is then easy to conclude using the fact that, in characteristic REF, a finite dimensional NAME algebra is semisimple if and only if it is cosemisimple (see CITE). |
math/0012073 | Let MATH. If MATH, then MATH. Suppose that MATH. By REF , there exists a right MATH-integral MATH for MATH such that MATH and MATH. Then MATH. By REF , the spaces of left and right MATH-integrals for MATH coincide. |
math/0012073 | REF follow directly from the axioms of a crossing. To show REF , let MATH. Using the axioms, it is easy to verify that MATH in the convolution algebra MATH (see REF). Thus, since MATH is the inverse of MATH in MATH, we have that MATH and so MATH. |
math/0012073 | Let MATH be a non-zero left MATH-integral for MATH. For any MATH, since MATH is a non-zero left MATH-integral for MATH (see REF ) and by the uniqueness (within scalar multiple) of a left MATH-integral in the finite dimensional case (see REF ), there exists a unique MATH such that MATH for all MATH. Using REF , one verifies that MATH is a group homomorphism. Since any left MATH-integral for MATH is a scalar multiple of MATH, the result holds for any left MATH-integral. Finally, let MATH be a right MATH-integral for MATH. Since the antipode is bijective (MATH is finite dimensional), and using REF and the fact that MATH is a left MATH-integral for MATH, we have that, for all MATH, MATH. |
math/0012073 | Let us show REF . Let MATH be a left integral for MATH. We can assume that MATH (if MATH, then the result is obvious). By REF , MATH for any MATH. Thus MATH is a left integral for MATH. Therefore, since MATH is finite dimensional and MATH, there exists MATH such that MATH. Let MATH be a non-zero right MATH-integral for MATH. We have that MATH. Now MATH (because MATH is a non-zero left integral for MATH and MATH is a non-zero right integral for MATH). Hence MATH and so MATH. It can be shown similarly that the result holds if MATH is a right integral for MATH. Let us show REF . If MATH is a left integral for MATH, then, for all MATH, MATH (since MATH is a left integral for MATH). Thus, by the uniqueness of the distinguished grouplike element of the NAME algebra MATH, we have that MATH. To show REF , let MATH be a right MATH-integral for MATH. By REF , MATH is also a right MATH-integral for MATH. Then, for any MATH, using REF , MATH . Hence, by the uniqueness of the distinguished MATH-grouplike element (see REF ), we have that MATH and so MATH for all MATH. |
math/0012073 | Let us show REF . We have MATH . Thus MATH (since MATH is invertible). By applying MATH on both sides, we get the first equality of REF . The second one can be obtained similarly. To show the first equality of REF , set MATH . Let us compute MATH in two different ways. On the first hand, MATH . On the second hand, MATH . Comparing these two calculations and since MATH is invertible, we get the first equality of REF . The second one can be proved similarly by computing the expression MATH. REF is a direct consequence of REF . Finally, REF follows from REF : MATH . This completes the proof of the lemma. |
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