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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 ... |
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, the... |
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/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 c... |
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, MA... |
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 MAT... |
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 sh... |
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, ... |
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 out... |
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 interva... |
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 MAT... |
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 ... |
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 ... |
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 fi... |
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 i... |
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 ther... |
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 in... |
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 none... |
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 finite... |
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 MAT... |
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 sho... |
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 w... |
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 ... |
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 negat... |
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 r... |
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 ... |
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 subset... |
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 ... |
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 row... |
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 MAT... |
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, a... |
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... |
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 . ... |
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 ... |
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 ... |
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 foll... |
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... |
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 ... |
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/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 dependen... |
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... |
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 contou... |
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 a... |
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 ... |
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 fo... |
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 in... |
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 ... |
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 exclud... |
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... |
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 ... |
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 wa... |
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 M... |
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 im... |
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 lo... |
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 co... |
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... |
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, w... |
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 b... |
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 MAT... |
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 . Th... |
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. Cons... |
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 suf... |
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. Bec... |
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 MAT... |
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, t... |
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 lef... |
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 ha... |
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 const... |
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... |
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