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math/0007085 | If MATH, then by REF MATH is a two-dimensional hyperplane. If MATH then taking a MATH-dimensional hyperplane MATH through MATH, transversal to MATH we easily obtain from REF that MATH . Since by REF MATH then by REF MATH. But MATH is also an algebraic cone. Hence MATH is either MATH or a finite number of lines. So, by REF , MATH in the first case or is a finite sum of two-dimensional hyperplanes in the second one. |
math/0007085 | Instead of the parametrizations MATH and MATH, we shall use descriptions of MATH and MATH . Define MATH . Obviously, MATH, MATH. From the form of MATH and MATH we see that MATH . Take the hyperplane MATH transversal to MATH . From REF we easily obtain MATH . Since MATH is an analytic cone in MATH of dimension MATH, then from this equality MATH is either MATH or a finite system of lines. So, it suffices to prove that MATH . By definition of MATH we have obviously MATH . Take now any vector MATH. By REF there exists an analytic curve MATH having a parametrization MATH at MATH such that MATH that is, MATH . Since MATH or MATH we may assume that MATH . Changing unessentially the parameter MATH we may assume that MATH . Then MATH . Since MATH, then there exists MATH such that MATH . Denote by MATH the set of MATH for which the above inequality holds-MATH . Since MATH and MATH then from the above inequality we obtain that MATH has the form MATH . Hence MATH for some MATH . We shall show that MATH . Consider the cases: REF. the coefficients MATH vanish for MATH that is, MATH . Then MATH . Hence we have MATH REF. not all the coefficients MATH vanish for MATH . Let MATH be the smallest positive integer such that MATH . Then MATH . Let us first note that for MATH from REF and the fact that MATH we have MATH . Hence and from REF for MATH we have MATH and for MATH from REF we get MATH . Hence MATH . Now, we have MATH . On the other hand, from definition of MATH and REF we have MATH . Then from REF we finally obtain MATH . This ends the proof. |
math/0007085 | It follows from REF by considering parametrizations of MATH and MATH at MATH in the nonsingular case and singular one. |
math/0007085 | Note that the topology in MATH can be described in the following elementary way: if MATH, MATH, then MATH when MATH in MATH if and only if there exist points MATH, MATH, MATH, and their homogeneous coordinates MATH, MATH, MATH, MATH such that MATH and MATH when MATH in MATH-for MATH . Take MATH. Then there exist MATH, MATH, MATH, MATH, such that MATH when MATH. Since MATH are compact sets we may assume that MATH and MATH . Since MATH then MATH. Hence MATH. Of course MATH . From the above description of topology in MATH we easily obtain that MATH . The opposite inclusion MATH is obvious. |
math/0007087 | For REF see CITE or CITE. For REF see CITE or CITE. Finally for REF , see CITE. |
math/0007087 | This follows from CITE. |
math/0007087 | CASE: This is easily proved by induction on MATH, using the fact that MATH is subgroup closed. CASE: This is also easily proved by induction on MATH, using the fact that MATH is quotient group closed. CASE: Clearly MATH, MATH, and MATH is closed under directed unions. Therefore we need to prove that MATH is extension closed. We show by induction on MATH that MATH; the case MATH being obvious. If MATH for some ordinal MATH, then MATH as required. |
math/0007087 | Without loss of generality we may assume that MATH, MATH, and MATH. Then MATH. Suppose that MATH yet MATH. We may write MATH and MATH as a disjoint union of open intervals, which we shall call MATH and MATH respectively. Then a finite number MATH of these intervals will cover MATH; let these intervals be MATH, MATH, MATH, MATH, , MATH, MATH (so MATH is an odd integer) where the MATH are intervals in MATH, the MATH are intervals in MATH, and MATH, MATH. Note that MATH and MATH, and MATH and MATH for all MATH. Also MATH if MATH is odd, and MATH if MATH is even. To see this let us consider the former case. We have MATH, so certainly MATH. Also MATH, so by replacing MATH with MATH if necessary, we may assume that MATH. Then MATH and since MATH, it follows that MATH. Similarly if MATH is even, we can show that MATH. Now choose MATH if MATH is odd (MATH), and MATH REF if MATH is even. Finally set MATH. Set MATH and MATH, and for MATH define MATH . Now set MATH and MATH. Observe that MATH. Indeed if MATH, then by translating by MATH for suitable MATH, we may assume that MATH, and then the result is clear. If MATH is even, then MATH. Since MATH, we see that either MATH for all MATH, or MATH for all MATH; without loss of generality we may assume that MATH. Now MATH, MATH and MATH, hence there exists a positive integer MATH such that MATH and MATH for all MATH. Let MATH be the maximum of the of the MATH (MATH). Since MATH, MATH, we see that MATH for all MATH and for all MATH, and it follows that MATH for all MATH. Similarly there exists a positive integer MATH such that MATH for all MATH. It now follows from NAME 's Table Tennis lemma CITE that MATH is a free group. |
math/0007087 | Let MATH be a finite subset of MATH, and set MATH. The result will follow if we can prove that MATH, because then clearly MATH, and MATH is abelian by CITE. Choose MATH, and then select MATH such that MATH for all MATH. Now choose MATH such that MATH for all MATH. Suppose the sequence MATH is bounded above, and let MATH be the least upper bound of the sequence. Then MATH for all MATH and MATH is the least upper bound of the sequence MATH. We conclude that MATH for all MATH and hence MATH. Therefore we may assume that the sequence MATH is not bounded above, and similarly we may assume that the sequence MATH is not bounded below. Therefore we may assume that MATH. It now follows from REF that MATH for all MATH. Since MATH commutes with all elements of MATH, we see that every element of MATH fixes a point in MATH and we deduce that the set MATH is bounded above by MATH. If MATH is the supremum of the set MATH, then MATH and the result is proven. |
math/0007087 | By replacing MATH and or MATH with their inverses if necessary, we may assume that MATH and MATH. Set MATH and note that MATH. Since MATH, MATH has no fixed points on MATH, and MATH, we see that there exists a positive integer MATH such that MATH. Similarly there exists a positive integer MATH such that MATH. We now show that the subsemigroup generated by MATH and MATH is free on those generators. Set MATH and MATH. Suppose to the contrary that two nontrivial distinct finite products MATH of the form MATH, where the MATH are positive integers, yield the same element of MATH. By cancelling on the left, we may assume without loss of generality that MATH and MATH or REF, where MATH are also products of the form MATH. Since MATH, we see that MATH and MATH or MATH. Thus MATH and we have a contradiction. We deduce that the subsemigroup generated by MATH and MATH is free on those generators and the result follows. |
math/0007087 | For each MATH, we may write MATH as a disjoint union of open intervals, say MATH, where each MATH is an open interval. Then MATH for all MATH, because MATH. Suppose MATH. If MATH and MATH then MATH, so we may choose MATH such that MATH is not contained in any other open interval. Using the fact that MATH is connected, we may now choose MATH so that MATH has nonempty intersection with MATH, and also does not contain MATH. Clearly MATH. Write MATH, and assume without loss of generality that MATH and MATH. The result now follows by applying REF with MATH and MATH. |
math/0007087 | We may view MATH as a subgroup of the piecewise linear orientation preserving homeomorphisms of MATH such that MATH fixes no point in MATH. Define MATH. Then obviously MATH and we see that MATH. Therefore if MATH and MATH are finitely generated subgroups of MATH, there exists MATH such that MATH is the identity map outside MATH for all MATH. Since MATH does not fix any point in MATH, there exists MATH such that MATH. Then MATH fixes all points outside MATH and the result follows. |
math/0007087 | Write MATH and consider MATH as a group of orientation preserving homeomorphisms of MATH. Let MATH denote the complement MATH of a subset MATH of MATH. Since MATH is a finite union of open intervals and MATH, we see that MATH is a finite union of open intervals. Therefore MATH is a finite union of disjoint open intervals, say MATH where MATH. Set MATH, MATH, and MATH for MATH. Then MATH and MATH for all MATH. Since MATH for all MATH, we see that MATH. Now suppose MATH and MATH are finitely generated subgroups of MATH. Then MATH and MATH are finitely generated subgroups of MATH, so by REF there exists MATH such that MATH (the commutator of MATH and MATH) is contained in MATH. Therefore MATH and the result follows. |
math/0007087 | Write MATH where MATH and MATH. If MATH, then replacing MATH with MATH and MATH with MATH, we may assume that MATH is finitely generated. By REF , there exists a series MATH such that MATH for all MATH with the property that if MATH and MATH are finitely generated subgroups of MATH, then there exists MATH such that MATH and MATH centralize each other. If MATH, then we may take MATH and we are finished. Otherwise we may let MATH be the smallest integer such that MATH and set MATH, a nontrivial normal subgroup of MATH. If MATH and MATH are finitely generated subgroups of MATH, then there exist finitely generated subgroups MATH and MATH of MATH such that MATH and MATH. Then we can find MATH such that MATH and MATH centralize each other. If MATH, then MATH and MATH centralize each other as required. |
math/0007087 | It is easy to see that MATH is closed under taking subgroups, quotient groups, group extensions and directed unions. Since MATH, it will now be sufficient to show that MATH. However if MATH is the free group on two generators, MATH, MATH and MATH, then there is no MATH such that MATH and MATH centralize each other. We deduce from REF that MATH and the proof is complete. |
math/0007087 | Using REF , we can find MATH such that MATH with the property that if MATH are finitely generated subgroups of MATH, then there exists MATH such that MATH and MATH centralize each other. The result is obvious if MATH for all finitely generated subgroups MATH of MATH, so we may assume that there is a finitely generated subgroup MATH of MATH such that MATH. Let MATH be any finitely generated subgroup of MATH. Then there exists MATH such that MATH and MATH centralize each other. Since MATH by REF , we deduce from REF that MATH for every finitely generated subgroup MATH of MATH. We conclude that MATH for every finitely generated subgroup MATH of MATH as required. |
math/0007087 | First suppose MATH has a nontrivial normal abelian subgroup MATH. Here we set MATH. Since MATH, we see from REF that MATH and MATH for all finitely generated subgroups MATH of MATH, so we may assume that MATH. Let MATH and let MATH. Since MATH by REF shows that MATH is a normal subgroup of MATH and that MATH for all finitely generated subgroups MATH of MATH. Therefore we may assume that MATH. Using CITE, we see that MATH and MATH are abelian and it follows that MATH. Thus we may assume that MATH has no nontrivial normal abelian subgroup, so in particular MATH. We now have two cases to consider, namely MATH and MATH and is finitely generated. In the former case the result follows from REF and CITE, while in the latter case the result follows from REF . |
math/0007087 | For each MATH, let MATH denote the smallest left-relatively convex subgroup of MATH containing MATH, and let MATH and MATH. Then MATH is partially ordered by inclusion. Suppose MATH is a nonempty chain in MATH. Then MATH is a left-relatively convex subgroup of MATH by REF , which is not the whole of MATH because MATH is finitely generated and MATH for all MATH, consequently MATH is bounded above by MATH. But MATH because MATH, hence MATH and we may apply NAME 's lemma to deduce that MATH has a maximal element MATH say. Set MATH, which by REF is a left-relatively convex subgroup of MATH, so using the maximality of MATH we see that MATH (thus MATH). If MATH has a self centralizing torsion free abelian normal subgroup MATH such that MATH is torsion free abelian, then we are finished so we assume that this is not the case. By REF , there is an order preserving action of MATH on MATH with kernel MATH such that MATH, and MATH for all MATH. Replacing MATH with MATH and using REF , we may assume that MATH. Let MATH. Then MATH is a nonempty closed subset of MATH, and MATH is naturally a subgroup of MATH. Using the hypotheses of the Lemma, there is a normal solvable subgroup MATH of MATH such that MATH and is finitely generated. By REF , there is a nontrivial normal subgroup MATH of MATH such that MATH whenever MATH is a finitely generated subgroup of MATH. Write MATH where MATH and MATH is finitely generated for all MATH (if MATH is finitely generated, we may choose MATH for all MATH), and MATH. Then MATH for all MATH, and MATH is an ascending chain of left-relatively convex subgroups of MATH with the property that MATH for all MATH. Furthermore MATH is a left-relatively convex subgroup by REF , which cannot be MATH itself because MATH is finitely generated. We deduce that MATH which contradicts the maximality of MATH and finishes the proof. |
math/0007087 | We shall prove the result by transfinite induction on MATH, so by REF choose the least ordinal MATH such that MATH and assume that the result is true whenever MATH and MATH. Now MATH cannot be a limit ordinal, and the result is clearly true if MATH. Therefore we may assume that MATH for some ordinal MATH, and then there exists MATH such that MATH and MATH. Using REF , we may write MATH where MATH and every subgroup of MATH is in MATH for all MATH. For each MATH, let MATH denote the smallest left-relatively convex subgroup of MATH containing MATH. First consider the case MATH is finitely generated. We have an ascending chain of left-relatively convex subgroups MATH, so their union is also a left-relatively convex subgroup by REF which contains MATH. The inductive hypothesis shows that MATH for all MATH and since MATH is finitely generated, we deduce that MATH. But MATH has the solvable normal subgroup MATH such that MATH, so the result follows from an application of REF . Finally we need to consider the case MATH is not finitely generated. Here we write MATH, where MATH and MATH is finitely generated for all MATH. We now have an ascending chain of left relatively convex subgroups MATH, so their union is also a left-relatively convex subgroup by REF which contains MATH. By the case MATH is finitely generated considered in the previous paragraph, we know that MATH for all MATH and since MATH is finitely generated, we deduce that MATH. Another application of REF completes the proof. |
math/0007087 | By CITE, we may lift the action of MATH on MATH to an action of a group MATH on MATH; specifically MATH is a left orderable group with a central subgroup MATH such that MATH and MATH. Note that MATH is finitely generated because MATH is finitely generated. If MATH is finite, then MATH is a torsion free group with an infinite central cyclic subgroup of finite index and it follows that MATH. We deduce that MATH is cyclic. Therefore we may assume that MATH is infinite. By REF MATH has a normal subgroup MATH such that MATH. If MATH, then MATH is finite, consequently MATH is a finite subgroup of MATH and we deduce that MATH. It follows that MATH has an infinite cyclic quotient, so we may assume that MATH. Note that MATH is a finite cyclic group. Since MATH has finite index in the finitely generated group MATH, we see that MATH is finitely generated. Moreover MATH is infinite, so by REF there exists MATH such that MATH. But MATH and the result follows. |
math/0007087 | Define an action MATH of MATH on MATH by MATH for MATH. This action has the required properties. |
math/0007087 | Write MATH, where the MATH are finitely generated subgroups containing MATH. Then MATH, and if each of the MATH is left orderable, then so is MATH by CITE. Therefore we may assume that MATH is finitely generated. Using REF , we can view MATH as a subgroup of MATH. We may write MATH as a countable disjoint union of nonempty open sets, say MATH. On each MATH, either MATH for all MATH, or MATH for all MATH. Using REF , for each MATH there is an action of MATH on MATH by orientation preserving homeomorphisms with the property that either MATH and MATH for all MATH, or MATH and MATH for all MATH. Then by CITE we may assume that MATH on MATH. We have now defined an action of MATH on MATH, and we extend this to an action MATH on the whole of MATH by defining MATH to be the identity on MATH for all MATH. Clearly MATH and MATH. Thus we can define a group homomorphism MATH by MATH for MATH and MATH for MATH, because MATH for MATH. The result now follows from CITE. |
math/0007087 | If MATH then the result follows from CITE, so we may assume that MATH is infinite cyclic. We will assume that MATH is a subgroup of MATH and write MATH, where MATH. Let MATH and identify MATH with MATH via the isomorphism MATH for MATH. We need to prove that MATH is left orderable. Write MATH, where the MATH are finitely generated subgroups containing MATH. Then MATH, and if each of the MATH is left orderable, then so is MATH by CITE. Therefore we may assume that MATH is finitely generated. We shall use induction on the NAME length of MATH (so if MATH is a normal series for MATH with MATH infinite cyclic for all MATH, then MATH is the NAME length of MATH). First suppose the NAME length of MATH is REF. This means that MATH is infinite cyclic, say MATH where MATH has infinite order. Then we can view MATH as a subgroup of MATH by letting MATH act on MATH according to the rule MATH for all MATH. Then MATH and the result follows from REF . Therefore we may assume that the NAME length of MATH is at least REF. Let MATH be a nontrivial cyclic central subgroup of MATH such that MATH is torsion free. Then MATH is left orderable because MATH is a torsion free nilpotent group CITE. Suppose MATH. We have an epimorphism MATH. Let MATH be the kernel of this map. Then MATH and MATH, so applying CITE we see that MATH is free and consequently left orderable. By induction MATH is left orderable, and so the result follows from CITE. Finally we need to consider the case MATH. We have an epimorphism MATH. Let MATH be the kernel of this map, and let MATH. With MATH we have an associated standard tree MATH CITE, and MATH acts on this tree. The vertices of MATH are the left cosets MATH and MATH, and the edges are the left cosets MATH, where MATH. A fundamental MATH-transversal CITE for MATH consists of the vertices MATH and the edge MATH. Let MATH be a transversal for MATH in MATH. Then a fundamental MATH-transversal MATH for MATH consists of the vertices MATH and MATH, and the edges MATH, where MATH. The stabilizers of the vertices of MATH are of the form MATH and MATH, and the stabilizers of the edges are of the form MATH for MATH. It follows that MATH is the fundamental group of a graph of groups CITE of the following form where each MATH is of the form MATH for some MATH (where MATH depends on MATH). We can now define an epimorphism MATH by MATH for MATH and MATH for MATH. The kernel of this map is a free group and hence left orderable. Also MATH is left orderable by induction. We now apply CITE twice to first deduce that MATH is left orderable, and then MATH is left orderable, as required. |
math/0007087 | Obviously MATH. Conversely suppose MATH. Then the centralizer of MATH in MATH has finite index in MATH, consequently it contains a normal subgroup MATH of finite index MATH in MATH. Thus MATH for all MATH, so by CITE we see that MATH for all MATH. Therefore MATH and the result is proven. |
math/0007087 | The result is trivial if MATH, so we may assume that MATH. Let MATH. Then CITE shows that MATH, and we now see from CITE that MATH. Also MATH from CITE and we deduce that MATH has finite index in MATH. Therefore MATH is finitely generated and MATH. But MATH by REF and the first part is proven. Finally if MATH, then MATH from CITE. |
math/0007089 | By considering the graded exact sequence MATH in each degree MATH, we see that REF holds if and only if multiplication by MATH, regarded as a linear map MATH from MATH to MATH, is injective when MATH, and surjective when MATH. Write MATH. For MATH, MATH is a basis of MATH, and MATH is a basis of MATH. Thus, we must show that for each MATH, the matrix of MATH in this basis has maximal rank. This matrix has rows indexed by MATH and columns indexed by MATH. The entry at row MATH, column MATH is MATH . If we specialise this matrix, the rank can only decrease, so if we can prove that some specialised matrix has full rank, then we are done. Putting all MATH, we obtain the incidence matrix of MATH-subsets of MATH into MATH-subsets of MATH, that is, the rows are indexed by MATH-subsets and the columns by MATH-subsets, with a MATH at the MATH'th position iff MATH, and REF otherwise. It has been shown by combinatorialists that this matrix has full rank CITE. |
math/0007089 | We put MATH. Suppose that MATH . The matrix of the map REF is a MATH matrix, MATH, where the rows are indexed by MATH-subsets MATH, and the columns by MATH-subsets MATH. The entry at position MATH is MATH . We must prove that this matrix has maximal rank. Clearly, the rank can not increase under specialisation, so if we prove that the matrix obtained by replacing each MATH with REF has maximal rank, then so does MATH. However, the specialised matrix is nothing but the matrix MATH of REF , so it has full rank. |
math/0007089 | This follows from REF , together with REF. |
math/0007089 | Put MATH. If MATH then clearly MATH. Suppose that MATH. Then MATH so the sum is independent of MATH. Furthermore, we can write MATH as a disjoint union MATH, hence the sum can be written MATH . Now, since MATH has cardinality MATH, the set MATH has cardinality MATH, so the permutation which transforms MATH to MATH has cardinality MATH. Hence, by substituting MATH, we get that the sum is equal to MATH which is the desired result. |
math/0007089 | The lemma is trivially true for MATH. If MATH, we note that MATH for MATH, since the permutation transforming MATH to MATH is even. Furthermore, the signs of MATH alternate in sign as MATH goes from MATH to MATH. Thus, for a fixed MATH, there are either as many positive as negative MATH, or REF more positive than negative, depending on the parity of MATH. By summing over all MATH, we conclude that there are always strictly more positive than negative signs. Now suppose that we have shown that MATH for all MATH such that MATH. We want to show that that MATH. We have that MATH and writing MATH as a disjoint union of its first two element, and the remaining elements, this becomes MATH . |
math/0007089 | Denote the row indexed by MATH by MATH, then MATH can be regarded as an element in MATH, the free MATH-vector space on the MATH-subsets of MATH. If we denote the basis element corresponding to a MATH-subset MATH by MATH, then MATH . The number of rows in MATH is MATH, and the number of columns is MATH. There are less rows than columns if MATH, as many rows as columns if MATH, and more rows than columns if MATH. CASE: If MATH, we must prove that the rows are linearly independent. Suppose that there is a linear relation among the MATH's, so that MATH for some numbers MATH. We shall prove that all MATH. Choose a MATH, MATH, and define a linear functional MATH by MATH . Then if MATH we have that MATH . The last step follows from REF . Applying MATH to REF we get that MATH . Since REF tells us that MATH, we conclude that MATH . Now, for any MATH we have, by exclusion-inclusion, that MATH . Fix MATH and put MATH. Since MATH we have, using REF that MATH . Since MATH was arbitrary, all MATH are zero. This shows that the MATH are linearly independent. CASE: If MATH, then MATH is a square matrix. By the previous case, the vectors MATH are linearly independent, but since there are MATH such vectors, they form a basis of MATH; in particular, they span this vector space. CASE: Finally, let us consider the remaining case MATH, so that there are more rows than columns. We must prove that the rows span MATH. We prove this by induction over MATH. The case MATH is already proved, and forms the basis of the induction. We assume MATH fixed, and that the assertion has been proved for all MATH. Let MATH be arbitrary. If we can express MATH as a linear combination of the MATH's, we are done. To this end, put MATH . Since MATH, it follows by induction that there are scalars MATH such that MATH . For MATH, MATH, put MATH. Define MATH . If we write MATH we have that for MATH, MATH, that MATH which implies that MATH . In either case, MATH has coordinate REF in component MATH, unless MATH. Hence, MATH may be regarded as a vector in MATH. By the induction hypothesis, there exist MATH such that MATH . Defining MATH we get that MATH . |
math/0007091 | We are given that MATH is a free basis for MATH. Fix a lifting MATH of MATH. For any countable subset MATH, let MATH be the smallest (necessarily countable) subset of MATH such that MATH. Similarly, for any countable subset MATH, let MATH be the smallest (necessarily countable) subset of MATH such that MATH. Now start with any nonempty countable set MATH such that MATH. We use finite induction to define two sequences MATH of countable sets by MATH . In words, think of MATH as images of MATH. Starting with a countable subset MATH of the basis MATH of MATH, use our lifting of MATH to get an inverse image MATH and take the smallest countable subset MATH of the basis MATH of MATH whose span contains MATH. Now take images of MATH modulo MATH and find the smallest countable subset MATH of the basis MATH which span a group containing all of the elements of MATH. Iterate a countable number of times. We then have for all MATH, MATH and MATH . Set MATH. Clearly MATH is a direct summand of MATH. Moreover, MATH is countably generated since the indexing set is a countable union of countable sets. Equation REF forces MATH . |
math/0007091 | Fix a basis MATH of MATH. Well order MATH. Assume we have MATH for all MATH such that: CASE: Each MATH is generated by a subset of MATH; CASE: MATH is generated by some subset of MATH; and CASE: MATH if MATH. CASE: MATH is countably generated. MATH and MATH are direct summands of MATH since they are generated by subsets of our fixed basis. If MATH, that union cannot map onto MATH. Let MATH be the smallest element of MATH (under the well ordering of MATH) not in MATH. Apply REF to get a countably generated subgroup MATH generated by elements of MATH with MATH and MATH generated by a subset of MATH. Set MATH. Since MATH clearly has the required properties and this process must eventually give all of MATH (at least by the order type of MATH), by transfinite induction we are done. |
math/0007091 | Using the notation of REF , we let MATH where for all MATH, MATH with MATH countably generated. For each MATH in MATH, set MATH, where MATH is the projection of MATH to MATH. If MATH, ignore it and renumber. By assumption, we can lift the direct sum decomposition of the quotient MATH to a direct sum decomposition MATH with units MATH such that MATH. Now set MATH so that MATH lifts MATH. Assume for all MATH, MATH, where MATH is the free group generated by a lifting of the decomposition of MATH generated by the appropriate subset of MATH. Then we have MATH and by the above, MATH. By transfinite induction we get MATH. |
math/0007091 | By REF , it is enough to show that, for a countably generated free abelian group MATH with MATH a basis for MATH, there is a direct decomposition lifting of MATH to the direct decomposition MATH . Form a matrix MATH whose rows are some lifting of MATH. Do infinite Gaussian elimination modulo MATH on MATH. Since the rows of MATH form a basis for MATH and modulo MATH this algorithm agrees with infinite Gaussian elimination, after a finite number of steps, the top MATH rows of MATH will be rows of the identity and all rows of the identity will eventually arise as rows of MATH. Since every entry of MATH which is zero modulo MATH is actually MATH, MATH is row reduced to the identity provided every row at some point stops changing in taking the product MATH. Since all of the entries of row MATH of MATH which are not congruent to MATH mod MATH are contained in a finite number of columns, any row of MATH ceases to change when all the rows of the identity with MATH in those columns have been obtained in the matrix MATH. Hence after an infinite number of steps each row of MATH will have stabilized and the stabilized rows of MATH will form a basis for MATH which lifts the direct sum decomposition. |
math/0007091 | We can take a short projective resolution of MATH over MATH, say MATH is exact with MATH projective and, like MATH, a direct sum of cyclic projectives of the form MATH for some MATH. Then if we let MATH, MATH. This short exact sequence is pure, so tensoring with MATH over MATH gives a short projective resolution of MATH with MATH. Induction on MATH completes the proof. |
math/0007091 | MATH has a projective resolution of the form required in REF . Then REF gives the desired conclusion. |
math/0007094 | Let MATH. For MATH to be real we must have MATH . This implies that either MATH or MATH. If MATH then MATH which is negative or zero if MATH is on or outside of the circle MATH. In the case that MATH put MATH. If MATH has no real root then MATH is always positive. Otherwise MATH . Then MATH and MATH are non-negative, and MATH is not zero in MATH. Therefore MATH will be positive on MATH. |
math/0007094 | We know MATH where MATH varies over eigenvalues of the adjacency operator of the graph. Then MATH is an analytic MATH root for MATH. |
math/0007094 | By REF it is enough to show that MATH has a holomorphic extension on MATH. Let MATH. Here and in the rest of the paper MATH is the principal branch of the logarithm, defined and analytic on MATH. Fix MATH. Then using REF , there exists an open set MATH on which MATH is analytic. Since MATH is self-adjoint and MATH, the spectrum MATH. By the spectral theorem for self-adjoint operators we can write: MATH . Now MATH is well defined, and MATH is a holomorphic function of MATH on MATH. Now for small MATH, MATH . |
math/0007094 | From the remark we need to show that MATH . Let MATH. Let MATH, where MATH is the spectral decomposition of MATH acting on MATH . We now set MATH . Then from CITE, for all MATH, MATH . We know MATH . By REF MATH . If MATH is compact then MATH is bounded uniformly for MATH and MATH in an open interval containing MATH . Now integration by parts shows that indeed MATH as MATH. |
math/0007094 | From CITE we know that MATH . As in the previous example, MATH. Now we have MATH . As MATH, the theorem follows. |
math/0007097 | Compare for example, CITE. |
math/0007097 | To define the meromorphic continuation of MATH in cases where the poles MATH, MATH cross the contour of integration of the integral REF one just needs to deform the contour accordingly. This will obviously always be possible as long as MATH, MATH were initially not separated by the real axis. We will therefore turn to the case that they were initially seperated, and consider without loss of generality the case that MATH was initially in the upper, MATH in the lower half plane. In this case one may deform the contour into a contour that passes above MATH plus a small circle around MATH. The residue contribution from the integral over that small circle is MATH . The Lemma is proven. |
math/0007097 | To verify REF , note that NAME maps MATH, MATH, MATH into the following operators: MATH . The claim follows from the fact that MATH for MATH, MATH. REF is checked by straightforward calculation. |
math/0007097 | It suffices to show that the representation MATH is unitarily equivalent to one of the representations listed in CITE. Consider the operator MATH defined as MATH in terms of the special function MATH (compare REF). MATH is unitary since MATH which follows from REF. Moreover, it follows from the analytic and asymptotic properties of MATH given in the Appendix that MATH maps MATH to the space MATH of entire analytic functions which have a NAME that is meromorphic in MATH with possible poles at MATH . One finally finds from the functional relations of the MATH-functions, REF that MATH . Our representation is thereby easily recognized as the representation denoted by MATH in REF, where MATH. Note that our notation MATH is different from that in CITE and MATH. |
math/0007097 | Any two-variable NAME is contained in MATH. |
math/0007097 | The kernel MATH may be rewritten in terms of the function MATH as follows: MATH . The substitution MATH then leads to the NAME integral REF for the b-hypergeometric function. The rest is straightforward. |
math/0007097 | According to REF one just needs to calculate the residues of MATH for the poles at MATH. We will only need the absolute values of these quantities. The pole at MATH comes from the MATH factor in the expression for MATH. To calculate its residue one needs the following special value of the MATH-function: MATH which follows easily from the fact that the representation REF simplifies to the b-beta integral REF for MATH. We furthermore note that MATH from the reflection property of MATH stated in REF. It thereby follows that MATH . One has MATH, and MATH from the connection between MATH and MATH, as well as the reflection property of MATH (see REF). Therefore MATH. The pole at MATH corresponds to the pole at MATH of MATH. One may determine the singular term for MATH by applying REF to the NAME integral representation REF for the function MATH: MATH . The rest of the calculation proceeds as in the case of MATH and yields MATH. |
math/0007097 | Consider MATH where the NAME of the explicit REF for MATH has been used. The contour of integration for the second term in REF can be deformed into MATH plus contours from MATH to MATH and MATH to MATH. The integral over MATH cancels the first term on the right hand side of REF . Only the contour from MATH to MATH will give nonvanishing contributions in the limit MATH due to the exponential decay of MATH for MATH. In the remaining term one gets in the limit MATH contributions only from the leading terms in the asymptotics of MATH for MATH as quoted in REF . Taking into account that MATH for MATH, it follows that (MATH, MATH) MATH . The expression on the right hand side of REF vanishes by the NAME Lemma for MATH as well as MATH. The remainder is found to be MATH . It follows that MATH by the corresponding well-known property of the kernel MATH, compare for example, CITE. |
math/0007097 | MATH will be entire analytic with respect to MATH by straightforward application of REF , using that MATH is entire analytic in MATH, MATH and the analytic properties of the NAME coefficients summarized in REF. One similarly finds by using REF that the NAME MATH will be meromorphic in MATH with poles at MATH, MATH for any MATH. This establishes the first claim in REF . Note that the analytic continuation of the integral REF that defines MATH can be represented by integrating over a deformed contour MATH. For later use we will present suitable contours for the cases of analytic continuation to MATH and MATH respectively: In the first case one may integrate MATH over the real axis and instead of integrating over MATH one may integrate MATH, compare REF , over a contour consisting of the union of the half axes MATH and MATH, MATH with a half-circle in the upper half plane around MATH of radius MATH. In the second case one may integrate MATH over MATH, and MATH over the contour MATH consisting of the union of the half axes MATH and MATH with a half-circle of radius MATH in the lower half plane around MATH. Now consider the right hand side of REF . The expressions for MATH, MATH contain the shift operators MATH . The shift operator MATH is ``partially integrated" by REF shifting the contour of integration over MATH to the axis MATH, where one will pick up a residue contribution from the pole of the NAME coefficients that lies between these two contours, and REF introducing the new variables of integration MATH. In this way one rewrites the expression for MATH in the form MATH where the MATH denotes the transpose of MATH, and the contours MATH, MATH are just the contours introduced above to represent the analytic continuation with respect to MATH. It is important to notice that due to the fact that only the shift operators REF appear in the expressions for MATH, MATH one does not need to introduce further deformations of the contours in order to treat the poles from the factor in the NAME coefficients that depends on MATH only. It is verified by a straightforward calculation using REF that the NAME coefficients satisfy the finite difference equations MATH . Inserting these relations into REF yields an expression that is easily identified as MATH. |
math/0007097 | This will be a consequence of the following result: MATH satisfies MATH . To see that REF implies the claim, consider the simplified case of a distribution MATH that satisfies MATH, where MATH is a function that vanishes only at MATH and such that MATH if MATH. This distribution has support only at MATH. By REF one has MATH. It is then easy to see that MATH implies MATH for MATH. The generalization to the case at hand is clear. To verify REF one may note that the functions MATH, MATH satisfy eigenvalue equations for the operators MATH and MATH up to an error of order MATH. It follows that MATH . The right hand side of REF will vanish if MATH can be ``partially integrated". To show that this is the case, one needs some information on the form that MATH takes when acting on functions MATH. By straightforward evaluation of its definition one obtains an expression in terms of shift operators MATH . It is convenient to introduce an alternative set of shift operators MATH . The crucial point now is that the expression for MATH when rewritten in terms of MATH, MATH, MATH takes the following form MATH so it contains shifts of MATH, MATH, MATH by positive imaginary amounts up to MATH only. Furthermore note that in REF one may replace MATH by MATH. The analytic properties of the integrand in REF as following from REF now allow to partially integrate MATH by appropriate shifts of the contours of integration over MATH (compare proof of REF ). The verification of the second equation in REF is similar. |
math/0007097 | Introduce MATH . The coefficient of MATH in the expression for MATH coinicides with the sum of the coefficients with which MATH and MATH appear in the asymptotic expansion of the integrand in REF , compare REF identifies the origin of these terms in the asymptotic expansion of MATH, MATH, with the poles in the dependence of MATH, MATH on their variable MATH. It follows that the coefficient of MATH in the expression for MATH is independent of MATH. The result now follows from standard properties of the NAME transformation. |
math/0007097 | Let MATH . The analytic and asymptotic properties of the integrand follow from REF. Let us observe that for MATH one is dealing with absolutely convergent integrals, the integrand being meromorphic both with respect to the integration variables and the parameters. The integral REF therefore does not depend on the order in which the integrations are performed, so we will assume that it is first integrated over MATH. Singular behavior will emerge in the limit MATH. We will call a pole relevant if it has distance of MATH from the real axis, irrelevant otherwise. It then easily follows from REF that the integration over MATH does not introduce any new relevant poles since all the relevant poles in the MATH dependence that have distance of MATH are lying on the same side of the contour. Next one may integrate over MATH. We find from REF that MATH where MATH, MATH are terms that do not lead to relevant poles in the variable MATH after having integrated over MATH. The following abbreviations have been used: MATH . It is then easily found by using REF that the result of the integration over MATH will have poles at the following locations: MATH . The relevant residues can easily be assembled from the expressions given in REF. Moreover, it is straightforward to work out their poles. By again using REF one then finds that all four poles listed in REF will, after doing the MATH integration, produce terms that are singular for MATH, MATH and MATH. The terms that lead to MATH are easily identified by means of MATH . All these terms have as residue an expression proportional to MATH . One just needs to assemble the ingredients to check that REF coincides with what one finds on the right hand side of REF |
math/0007097 | We will start from REF . By using NAME with respect to the variable MATH and REF one may rewrite REF as follows: MATH . Let us introduce sequences of test-functions that tend towards delta-distributions: MATH . Let MATH with MATH. In this case one has MATH . By writing out the definition of MATH and shifting the contours of integration over MATH to MATH, MATH, one reduces the claim to the standard result that MATH for MATH, MATH (Note that MATH is regular for these values of its arguments as follows from REF). We will now consider the sequence with elements MATH . It converges for MATH due to REF . We would like to show that one may exchange the limit MATH with the integration over MATH so that the limit of REF is given by the integral MATH . To this aim it is useful to note that Under the conditions on the variable MATH introduced in REF one finds that the integrand in REF decays exponentially for MATH. The integrand in REF decays at least as fast as the integrand in REF . By a straightforward calculation using the method in the proof of REF one finds that MATH for MATH. The first statement in REF follows. The second statement follows from the first by shifting the contour of integration over MATH in the definition of MATH to MATH. The integrals REF can therefore be transformed into integrals over a compact set, for example, the interval MATH. In order to justify the exchange of limit and integration it therefore suffices to prove the following The convergence of MATH is uniform in MATH. To shorten the exposition, let us consider a slightly simplified situation. Assume that MATH is analytic with respect to both MATH and MATH in open strips that contain the real axis and decays exponentially for either MATH or MATH going to infinity. Let MATH and study the convergence of MATH for MATH. Upon writing MATH, the task reduces to the study of MATH . Convergence for MATH will be uniform in MATH provided that MATH is bounded as function of both MATH and MATH. But this is a consequence of our assumptions: The exponential decay allows us to transform MATH (respectively, MATH) to a function that is analytic on a compact rectangle in MATH, and therefore bounded. The regularity properties of MATH necessary to extend the argument to the present situation follow from REF. We have proved REF provided MATH satisfies the same conditions as MATH in REF follows by analytic continuation. |
math/0007097 | Let us note that MATH if MATH and MATH. A straightforward calculation then shows that MATH . It is useful to also note the commutation relations MATH . We may then calculate in the case MATH . The calculation for the case MATH is identical and the case MATH is trivial. |
math/0007097 | First of all note that one has MATH for any MATH. This follows by shifting the contour of the integration that represents MATH to the line MATH. The fact that MATH is symmetric is then seen by a simple calculation remembering that MATH, MATH. The fact that MATH and MATH are dense in MATH is easily seen by noting that any NAME is contained in these sets. |
math/0007097 | The action of MATH is represented on the NAME transform MATH as multiplication with MATH . The statement on the analyticity properties of MATH is then clear after recalling that the function MATH is entire analytic and of rapid decay being the NAME transform of a MATH function CITE. The statement that MATH is analytic in the strip MATH follows from the asymptotic decay properties of MATH by means of the NAME Theorem. In fact, the rapid decay of MATH ensures convergence of the inverse NAME transformation for any x-derivative of MATH even in the extremal cases MATH and MATH. |
math/0007097 | We will rewrite MATH in a form that allows us to estimate its asymptotics for large MATH. One may write MATH . In the last step we have used that MATH weakly solves the eigenvalue equation, for which one needs to check that MATH: One point of having introduced MATH is that it improves the asymptotic behavior of MATH for MATH by cancelling the poles of its NAME transform in MATH. The regularity theorem for tempered distributions CITE allows us to furthermore write MATH for some positive integer MATH and a polynomially bounded continuous function MATH. The functions MATH may be represented by expressions of the form MATH where MATH are some polynomials in MATH. The functions MATH have main support around MATH, and by choosing MATH large enough one can achieve decay stronger than MATH for any MATH. It is then convenient to split the integral in REF into an integral MATH obtained by integrating over MATH and the remainder MATH. In order to estimate MATH one may use the polynomial boundedness of MATH to estimate its absolute value by some constant times MATH, where MATH can be as small as one likes. The absolute value of MATH can in MATH be estimated by some inverse power of MATH, which is bounded by the chosen value of MATH. It follows that the exist MATH, MATH such that MATH where MATH can be made arbitrarily large by choosing MATH large enough. In the case of MATH one may estimate MATH by some constant times MATH and MATH simply by a constant, which easily gives existence of MATH, MATH such that MATH . This proves the claim about the asymptotics for MATH. In the case of MATH one uses the operator MATH in a completely analogous fashion . |
math/0007097 | To begin with, note that MATH represents the NAME of the distribution MATH of compact support CITE. It follows that MATH is polynomially bounded. Since the convergence in REF is absolute, one concludes that MATH is polynomially bounded as well. In the evaluation of MATH against a test-function MATH one may therefore insert REF and exchange the orders of integration and summation to get MATH where we used that fact that the set MATH represents a partition of unity in the last step. |
math/0007097 | The proof is to a large extend analgous to that of REF , so we will only sketch some necessary modifications. In order to get an estimate of MATH for MATH one may use the eigenvalue equation to rewrite it as MATH . It follows as in the proof of REF that MATH for MATH. In the case of MATH one may use instead MATH which gives MATH for MATH. |
math/0007097 | Consider MATH, where now MATH is chosen proportional to MATH. One has MATH . Now if there were terms with exponential decay weaker than MATH in the asymptotic expansion of MATH for MATH one would find terms terms that grow exponentially with MATH on the right hand side of REF . But polynomial boundedness of MATH excludes the occurrence of such terms on the left hand side of REF . |
math/0007097 | If one introduces MATH via (recall MATH) MATH one may verify by direct calculation using the functional equation of the function MATH that the equation MATH is equivalent to the following equation for MATH: MATH . By using REF and the properties of MATH that are summarized in REF one may deduce the following properties of the NAME transform MATH of MATH from REF : CASE: MATH has a NAME transform MATH that is analytic in MATH, and CASE: MATH has the following asymptotic behavior for MATH: MATH where MATH is a constant, MATH has exponential decay for MATH and MATH has exponential decay stronger than MATH for MATH. REF is equivalent to the following first order difference equation for MATH: MATH . Now there exists a solution to REF , namely MATH that has all the required analytic and asymptotic properties. If there was a second solution MATH of these conditions one could consider the ratio MATH. This ratio must be a solution to MATH. Since MATH has no zeros in the open strip MATH one concludes that MATH is holomorphic in any such strip. The function MATH must furthermore be asymptotic to the constant function for MATH. But this implies that MATH . : The function MATH is holomorphic and regular on the whole NAME sphere, therefore constant. |
math/0007097 | From the relation (recall MATH) MATH which easily follows from the analyticity and asymptotic properties of the MATH-function by means of NAME 's theorem one finds the following functional equation for MATH: MATH . By the MATH self-duality of MATH one also has the same equation with MATH. For irrational values of MATH it follows that REF and its MATH counterpart determine MATH uniquely up to a function of MATH. The expression on the left hand side of course satisfies REF . To fix the remaining ambiguity one may note that the integral defining MATH can be evaluated in the special case of MATH by means of CITE: MATH . REF follows. |
math/0007097 | In order to study the limit MATH it is convenient to split the integral into two integrals MATH and MATH over the intervals MATH and MATH respectively. In the case of MATH one may use the asymptotics of the MATH functions containing MATH for imaginary part of their argument going to MATH, REF , to get MATH where REF was used in the second step. To study the behavior of MATH for MATH it is convenient to change the integration variable in the second integral to MATH. One gets MATH . In this expression one may now use the asymptotics of the MATH functions containing MATH for imaginary part of their argument going to MATH, REF , which yields as previously MATH . The behavior for MATH is studied similarly. |
math/0007097 | Abbreviate the integrand in REF by MATH. A direct calculation shows that it satisfies the equation MATH . The Lemma follows from NAME 's theorem. |
math/0007097 | Direct consequence of analytic and asymptotic properties of the MATH-function given in REF. |
math/0007097 | One needs to calculate MATH . By inserting REF and changing variables MATH, MATH one finds that the integration over MATH produces MATH. MATH is therefore given by the integral MATH . It is then useful to employ the NAME integral representation REF for the b-hypergeometric function that appears in REF of the function MATH. The order of integrals in the resulting double integral may be exchanged, and the MATH integration carried out by means of REF . Up to prefactors that are entire analytic in MATH, MATH one is left with the following integral: MATH where the coefficients are given by MATH . The claim now follows by straightforward application of REF . |
math/0007099 | Multiplication in the ring MATH yields the right action of MATH on MATH: MATH . To see that the induced map is well-defined, observe that, for all elements MATH and MATH, we have MATH . This action is clearly compatible with the left MATH-module structure. It follows that MATH is a MATH bimodule. Since MATH and MATH implies MATH, MATH is bigraded. In other words, if we let MATH denote the opposite algebra, then MATH is a MATH-graded ring and MATH is a graded module over MATH. |
math/0007099 | Since we have MATH it is enough to show that MATH. However, for every MATH, we have MATH which establishes the lemma. |
math/0007099 | It suffices to give MATH-linear maps: MATH . Locally, a section MATH can be identified with an element of MATH, where MATH is a cone in MATH. Moreover, the action of MATH on MATH descends to action on MATH which increases degrees by MATH. Thus, we may define MATH. One verifies that MATH maps into MATH and that these local definitions glue to give the required map. |
math/0007099 | It suffices to show that MATH and MATH have the same action on MATH. On one hand, we have MATH . On the other hand, we also have MATH which establishes the claim. |
math/0007099 | Suppose otherwise: for some MATH, we have MATH. It follows that MATH and hence MATH . We deduce that MATH for all MATH and thus MATH. However, the MATH in any cone MATH are linearly independent. |
math/0007099 | The fact that MATH is a ring homomorphism follows directly from the definition. Thus, the assertion reduces to showing that MATH is surjective and to describe its kernel. To achieve this, we determine the NAME closures of MATH and MATH. Since the NAME closure of the set of integer points inside a rational polyhedral cone is the linear space spanned by that cone, it is easy to check that MATH where MATH. We claim that MATH. Indeed, REF implies that MATH, so it is enough to check that MATH is a radical ideal. However, MATH is the principal ideal generated by MATH . From REF , we see that MATH is reduced and hence MATH. Applying REF , it follows that MATH and MATH are surjective. To prove that MATH is injective, recall that MATH. Thus, it is enough to show that, for every MATH, we have MATH. To see this, observe that MATH implies MATH. Therefore, it suffices to notice that MATH, which follows from REF . |
math/0007099 | We must show that, for every MATH, the homomorphism of graded left MATH-modules MATH is an isomorphism. The first step is to prove that MATH induces an isomorphism of the associated sheaves. Fix MATH and choose MATH, mapping to MATH in MATH, such that MATH is an invertible element in MATH. Since the restriction of MATH to MATH is trivial, one may always find such a MATH. We then have a commutative diagram: MATH where MATH is the morphism in REF and MATH is the analogous morphism induced by MATH. Now, REF implies that MATH is an isomorphism and the vertical arrows are clearly isomorphisms. It follows that MATH is an isomorphism and therefore MATH induces an isomorphism of the associated sheaves. If MATH is a graded MATH-module, we write MATH . For every such MATH, there is an exact sequence MATH where MATH is the irrelevant ideal; see CITE. Hence, if MATH and MATH both vanish, then MATH is an isomorphism. We relegated these vanishing results to REF below. |
math/0007099 | The first assertion implies the second, so it suffices to show that MATH is not a zero divisor. Every MATH can be written uniquely as MATH where MATH and MATH are non-negative and have disjoint support. Consider the action of the torus MATH on the NAME algebra MATH; the corresponding eigenspace decomposition is MATH, where MATH. Let MATH be the ideal generated by MATH for all MATH. Since MATH is generated by linear forms, it is a prime ideal. Let MATH denote the left MATH-ideal MATH. With this notation, we have MATH. For MATH to be a non-zero divisor on MATH, it suffices to prove that, for MATH and MATH, the relation MATH implies MATH. To accomplish this, we first note MATH . Hence, MATH implies MATH. Suppose MATH. Since MATH is a prime ideal, there exists an index MATH such that MATH and MATH. Expressing MATH in terms of the linear generators of MATH, it follows that there is MATH such that MATH. However, for every MATH, REF implies MATH from which we deduce that MATH and MATH giving a contradiction. |
math/0007099 | Since MATH is not a zero divisor on MATH, there is a short exact sequence of MATH-modules MATH and the long exact sequence of local cohomology gives MATH . Because every element in MATH is annihilated by a power of MATH, the injectivity of MATH in REF implies that MATH. Thus, it suffices to prove that MATH . Let MATH be the left MATH-ideal satisfying MATH. To prove that MATH, we must show that if MATH satisfies MATH for some MATH and all MATH, then MATH. Using the notation from the proof of REF , we have MATH. From the decomposition of MATH, we have MATH and we deduce MATH, where MATH is defined to be MATH. Thus, it is enough to consider elements MATH of the form MATH with MATH and prove that MATH. Moreover, we may assume that MATH for MATH. By induction on MATH, we see that MATH for MATH. Hence, for MATH, we have MATH and, for MATH, we obtain MATH where MATH. We deduce MATH where MATH. For each MATH, we define an automorphism MATH given by MATH and we define MATH to be the composition MATH (the map MATH is defined in the paragraph before REF ). It is clear that MATH is surjective and MATH. Applying MATH to REF gives MATH . REF obviously extends to MATH and implies MATH . Since this relation holds for every MATH, a second application of REF shows that MATH and therefore MATH. |
math/0007099 | We present the left MATH-modules case here - the proof for right MATH-modules is completely analogous. For the first assertion, it is enough as well to construct MATH-linear maps MATH for all MATH, satisfying the obvious axioms. To accomplish this, we consider a local version of the left multiplication map REF when MATH. More explicitly, for each MATH, this morphism is MATH given by MATH. We claim that MATH is a morphism of right MATH-modules. To see this, recall that sections MATH and MATH can be identified with elements in MATH and MATH respectively. In particular, we have MATH for all MATH. It follows that, by tensoring the map MATH on the right with MATH over MATH, we obtain a MATH-linear map MATH . These maps glue together to give MATH which makes MATH into a graded left MATH-module. |
math/0007099 | Again, we give the proof only for left modules. For the first part, we consider an object MATH in MATH. By definition, we have MATH, where MATH is a left MATH-module and MATH. In light of REF , we must show that, for every MATH, we have MATH. Now, if MATH, then MATH where MATH and the hypotheses on MATH imply MATH . Therefore MATH has a structure of left module over MATH for every MATH. It is straightforward to verify that these structures glue together to give a MATH-module structure on MATH. By construction, we see that MATH is quasi-coherent sheaf over MATH. Conversely, let MATH be an object of MATH. Applying REF , we know that MATH is a graded left MATH-module, so it is enough to prove that MATH satisfies REF. Fixing MATH and MATH, it suffices to show that MATH for every MATH and all sections MATH. As explained in REF , we have MATH so that we may identify MATH with a linear combination of elements of the form MATH. By definition, we have MATH and we claim that MATH is zero. Indeed, for every MATH, we have MATH. Therefore, if MATH, then MATH and MATH . Finally, the exact sequence REF provides the first natural transformation, once we observe that MATH is MATH-torsion. It follows from CITE that the sheaf associated to MATH is isomorphic to MATH. |
math/0007099 | Follows immediately from REF . |
math/0007099 | Once again, we present the proof only for left modules. Suppose that MATH belongs to MATH. To establish that MATH, we have to check that, for every MATH, MATH is finitely generated over MATH. Thus, it suffices to show that, for every element MATH, there is an invertible element MATH such that MATH has degree zero. Consider MATH in MATH. Since MATH is smooth, there is a divisor corresponding to MATH supported outside MATH and whose class is MATH. It follows that MATH is an invertible element in MATH and MATH has degree zero. For the second assertion, we must show that given a coherent left MATH-module MATH there exists a finitely generated MATH-submodule MATH of MATH such that MATH. Since MATH is coherent, MATH is finitely generated over MATH, for every MATH. Choose, for each MATH, a finite set of homogeneous elements in MATH which are the numerators for a corresponding set of generators of MATH. Setting MATH to be the MATH-submodule of MATH generated by the union of these sets, we have MATH and MATH is finitely generated over MATH. |
math/0007099 | REF provides MATH and MATH. |
math/0007099 | This follows from REF . |
math/0007099 | This condition is clearly necessary. To see the other direction, consider the surjective graded morphism defined by the given generators: MATH. By hypothesis, this factors to an epimorphism MATH and REF implies that MATH. |
math/0007099 | REF implies that there exists MATH such that MATH. Now, REF gives an epimorphism MATH. Taking the corresponding morphism of sheaves and applying REF establishes the claim. |
math/0007099 | We only need to show that the graded components of MATH and MATH are annihilated by suitable NAME operators. However, this follows from the fact that MATH, where MATH. |
math/0007099 | Since MATH and MATH, the second assertion is a consequence of the first. Because MATH as MATH-modules and MATH as MATH-modules, there is a natural isomorphism of MATH-modules MATH. Thus, it suffices to prove that MATH is compatible with the right MATH-module structures. By taking a presentation MATH, we see that it suffices to establish the claim for MATH. In this case, the restriction map MATH is injective for open subsets MATH. Thus, the claim reduces to showing that MATH is compatible with the right MATH-module structure on MATH. Over MATH, the map MATH is given by MATH, where MATH. Now, it is enough to check that MATH is compatible with right multiplication with a vector field MATH over MATH. Using the notation from REF , we may assume that MATH, for some MATH. We first compute MATH . By definition, we have MATH from which we deduce MATH. Identifying the action of MATH on MATH with the action of MATH, we obtain MATH . On the other hand, we have MATH and we conclude that MATH. In the second part, we consider MATH. For MATH, we construct the natural map MATH, by composing the morphisms: MATH . Since MATH is MATH-saturated, it follows that MATH is also MATH-saturated and hence REF is an isomorphism. Moreover, MATH is a functor and MATH is an isomorphism which implies that REF is also an isomorphism. |
math/0007099 | Consider a point MATH in MATH and MATH satisfying MATH. Writing MATH, we have MATH and we deduce that MATH, for all MATH such that MATH. Because there is MATH such that MATH, we conclude that MATH for every MATH with MATH. On the other hand, MATH belongs to MATH so we have MATH for every MATH. It follows that MATH. Because this holds for each MATH and the MATH form part of a basis of MATH, we also conclude that MATH when MATH and therefore MATH. |
math/0007099 | The first step is to construct the morphism MATH as a categorical quotient - this is a local problem. For every MATH, let MATH be the open subset defined by the non-vanishing of MATH. In other words, we have MATH, which is clearly MATH-invariant. Thus, the categorical quotient is locally MATH. Since MATH, for every MATH and MATH, MATH, we have MATH. Now, if MATH is a face of MATH such that MATH for MATH, then we set MATH. It follows that MATH which provides an open immersion MATH. Thus, we obtain morphisms MATH which glue together to give the categorical quotient MATH. In the second step, we establish that MATH is in fact a geometric quotient. By Amplification REF, it suffices to show that every MATH-orbit in MATH is closed. Consider MATH. Since MATH is a geometric quotient, the projection MATH induces a morphism MATH. By REF , the morphism MATH given by MATH is bijective. Because the characteristic of the ground field MATH is zero and both MATH and MATH are smooth, the map MATH is an isomorphism. Let MATH be defined by MATH. Hence, the map MATH is the identity on MATH. It follows that MATH is the identity map on MATH and we conclude that MATH. Since the projection from MATH onto MATH is MATH-equivariant, the second assertion follows from REF . |
math/0007099 | See REF. |
math/0007099 | By choosing a basis MATH for MATH, we can write MATH. For each MATH, we pick a representative MATH for MATH such that MATH are linearly independent. We then enlarge this collection to obtain a basis MATH for MATH. Setting MATH, for all MATH, it follows that the ideal MATH equals MATH, which has dimension MATH. Hence, the MATH and MATH form a regular sequence and we deduce MATH. To prove that MATH is normal, we apply NAME 's criterion. Because being a complete intersection implies the MATH condition, it suffices to show that MATH satisfies condition MATH, which we check by using the Jacobian criterion. The Jacobian matrix MATH of MATH is given by MATH . Observe that, for MATH, the restriction MATH is surjective. Indeed, if MATH is the cone generated by MATH, then every element in MATH can be represented by a divisor whose support does not intersect MATH. We deduce that if the rank of MATH is strictly less than MATH, then at least two of the pairs of coordinates MATH are zero. By cutting with MATH extra quadrics MATH, we see that the codimension of the singular locus of MATH is at least two. Therefore, the variety MATH is normal. Moreover MATH is a cone which implies it is connected and hence integral. Recall that, for MATH, we have MATH, and for the order filtration the initial term (or principal symbol) of the above elements is MATH. Since the MATH for MATH form a regular sequence in MATH, it follows that MATH. On the other hand, we have already seen that MATH is reduced, so we have MATH. |
math/0007099 | By REF , MATH is a quotient of MATH, for some MATH and some MATH. It follows that MATH. |
math/0007099 | Since MATH is naturally isomorphic to MATH, we see that MATH is isomorphic to MATH over MATH. On the other hand, from the local description of MATH, we know that the inverse image of MATH is MATH and therefore MATH . By REF , we have an isomorphism of filtered rings: MATH . Notice that the graded ring associated to left hand side is MATH. Indeed, following the proof of REF , the initial terms of MATH are equal to MATH and form a regular sequence in MATH. Therefore, by passing to the associated graded rings, MATH induces the required isomorphism. Because the MATH are compatible with restriction, these local isomorphisms glue together to give the required isomorphism . |
math/0007099 | Since MATH, we may choose a finite set MATH of homogeneous generators for MATH. By using these homogeneous elements to define a good filtration of MATH, it follows that MATH is a graded finitely generated MATH-module. Therefore, both MATH and its radical MATH are graded ideals of MATH. Recall that for every MATH, we have MATH. We deduce that every subscheme defined by a graded ideal is MATH-invariant; in particular, MATH is MATH-invariant. To prove the second assertion, we argue locally and use the identification in REF . Over the open subset MATH, the ideal defining MATH is MATH. On the other hand, the characteristic variety of MATH over MATH can be computed as follows: If MATH has the good filtration induced by the images of MATH, then we obtain MATH. To see that the annihilator of MATH in MATH is MATH, it is enough to observe that MATH can be generated by elements of degree zero. Since MATH, we may identify MATH with MATH. Because these identifications are compatible with restriction, they glue together to give the required isomorphism. |
math/0007099 | Follows immediately from REF . |
math/0007099 | This claim follows immediately from standard results about NAME dimension; see REFEF. |
math/0007099 | Suppose otherwise; then we have MATH for all MATH. Let MATH be the maximum submodule of MATH of dimension strictly less than MATH. In other words, MATH is the submodule consisting of all MATH such that MATH has dimension strictly less than MATH. Since MATH is a submodule, there is a short exact sequence MATH . By construction, MATH has no nonzero submodules of dimension strictly less than MATH and, hence, the irreducible components of MATH have dimension at least MATH; see CITE. By hypothesis, the irreducible components of MATH of dimension MATH are contained inside MATH where MATH. Since MATH, it follows that the characteristic variety of MATH is contained inside MATH. Moreover, the support of a MATH-module equals the projection of its characteristic variety (see CITE) which implies that MATH is supported on MATH. Taking the long exact sequence in local cohomology, we have MATH . By REF has no MATH-torsion, so we have MATH. Because MATH is supported on MATH, we have MATH. Choosing a set of generators MATH, the long exact sequence induces MATH . Hence, MATH is a finitely generated MATH-subquotient of MATH. REF then implies that MATH which is a contradiction. |
math/0007099 | Applying REF , we see that MATH is the maximum of the local dimensions of MATH over MATH. Hence, the claim follows from REF . |
math/0007099 | Follows immediately from REF . |
math/0007106 | CASE: All of the ingredients of the definition are unchanged when MATH is replaced by MATH. The MATH's and all of the MATH's except for MATH are unchanged when MATH not beginning with MATH is multiplied on the left by MATH. The same is true for right multiplication by MATH provided MATH does not end in the inverse of a generator. If MATH and MATH does not end in MATH, then right multiplication of MATH by MATH increases MATH and MATH by REF, and leaves all other MATH's and MATH's unchanged. Thus the exponent of MATH in the formula for MATH increases by REF, and all other exponents in the formula are unchanged. CASE: The properties in REF imply that MATH and furthermore permit calculation of MATH on words of a given length from its values on words of smaller length. CASE: If MATH does not end in the inverse of a generator, then MATH . Otherwise, MATH for some MATH not ending in MATH and we have MATH where we have used MATH and MATH. |
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