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math/9911019 | We first observe that if MATH and MATH are infinite subsets of MATH with MATH almost contained in MATH, and MATH is a scalar sequence, then MATH . To prove the lemma, it suffices to find MATH, MATH and MATH, so that for every MATH there exists a scalar sequence MATH with MATH and MATH. Once this is accomplished, then MATH and MATH, satisfy the conclusion of the lemma. We now claim that there exists MATH and MATH such that for all MATH there exists a scalar sequence MATH with MATH . If this is not the case we construct a decreasing sequence MATH consisting of infinite subsets of MATH so that MATH, for every scalar sequence MATH with MATH. Let now MATH be almost contained in MATH, for all MATH, and choose a scalar sequence MATH such that MATH . It follows that MATH for all MATH and thus, MATH for all MATH, which is absurd. Therefore our claim holds. We next claim that there exists MATH and MATH so that for every MATH, there exists a scalar sequence MATH such that MATH and MATH. Again, were this claim false, we could choose a decreasing sequence MATH of infinite subsets of MATH such that for all MATH, if MATH, for some scalar sequence MATH, then MATH. Now let MATH be almost contained in each MATH and choose a scalar sequence MATH such that MATH . It follows that MATH for all MATH, and hence MATH for all MATH which is absurd. |
math/9911019 | Let MATH denote the set of those sub-convex combinations of the sequence MATH which are members of a (not necesserilly the same) sub-convex block subsequence of MATH that satisfies the conclusion of REF . We shall apply REF to the sequence MATH and the family MATH in order to obtain MATH so that MATH. REF will then yield that for some MATH, MATH is a MATH spreading model. To this end, it suffices to show that for every MATH, every MATH and MATH, there exists MATH supported by MATH and such that MATH. Suppose this is not the case and choose according to the hypothesis MATH so that MATH is a MATH spreading model with constant MATH. Without loss of generality, by REF , we can assume that MATH is MATH unconditional with constant MATH. Next, choose by REF , MATH such that if MATH, then MATH. Since MATH has no b.c.c. subsequences, there exist, by REF , a sequence of scalars MATH and a sequence MATH of consecutive subsets of MATH with MATH strongly bounded by MATH, such that for all MATH, MATH . However, MATH, for all MATH, and moreover they are supported by MATH. Thus, there exists, for all MATH, MATH . Next choose MATH such that MATH. Choose also MATH with MATH. Then MATH . Therefore, MATH . However, MATH . This contradiction completes the proof of the lemma. |
math/9911019 | Let MATH and MATH be as in the conclusion of REF applied on MATH. MATH . REF yields that MATH is not a limit ordinal and thus MATH, for some countable ordinal MATH. Choose now MATH so that no subsequence of MATH is a MATH spreading model. Since MATH, there exists MATH so that MATH is a MATH spreading model and of course no subsequence of MATH is a MATH spreading model. It follows by REF , that for every MATH, the family MATH is not contained in MATH. Using REF as we did in the proof of REF , we find MATH and MATH, so that MATH for all MATH. It follows now, by REF , that every member of the family MATH can be expressed as the union of at most MATH consecutive MATH sets. We are going to show that MATH and MATH satisfy REF. Indeed, let MATH and choose MATH so that for all MATH, MATH and MATH is a maximal MATH subset of MATH. Set MATH. By REF , there exist a sequence of scalars MATH and a sequence of consecutive subsets of MATH, MATH with MATH strongly bounded by MATH, so that for every MATH . We next choose, for every MATH, a functional MATH such that MATH and let MATH . It follows now, as MATH belongs to MATH, that there exists, for every MATH, MATH . By extending MATH to a maximal MATH subset of MATH, if necessary, we can assume without loss of generality, that MATH itself is a maximal MATH subset of MATH, for all MATH. We also extend the sequence MATH to a scalar sequence MATH in the obvious manner. Concluding, there exist MATH maximal MATH subsets of MATH and a scalar sequence MATH with MATH strongly bounded by MATH, so that for every MATH . Finally, choose MATH such that MATH, for all MATH, and we are done. |
math/9911019 | If MATH, then according to REF there exist a scalar sequence MATH and MATH, MATH, so that MATH is strongly bounded by MATH and MATH, for all MATH. It follows now that if MATH, then for all MATH, MATH and thus MATH is equivalent to the unit vector basis of MATH. Assume now that MATH and let MATH. It follows that no subsequence of MATH is equivalent to the unit vector basis of MATH and therefore, by REF , we obtain MATH such that MATH is semi-boundedly complete. We next choose, according to REF , a scalar sequence MATH and MATH with MATH strongly bounded by MATH so that for all MATH, MATH . Since MATH, there exists MATH such that MATH, for all MATH, MATH. Set MATH and extend the sequence MATH to a sequence MATH in the obvious manner. Evidently, MATH is MATH-faithful for MATH, for all MATH. We next consider the set MATH which is of course closed in MATH and therefore NAME. Our previous argument yields the existence of MATH such that MATH. Clearly MATH satisfies the conclusion of REF . of this lemma. |
math/9911019 | The proof is similar to those of REF . Choose first a sequence MATH of positive scalars such that MATH. Using the same notation and terminology as in REF , let MATH and MATH. The MATH-tuple of elements of MATH and the infinite subset MATH of MATH, MATH, are said to satisfy property MATH provided that MATH, if MATH, and the following statement holds: If MATH and there exists a scalar sequence MATH which is MATH-faithful for MATH, then there exists MATH which is MATH-faithful for MATH and such that MATH . Let us also say that MATH and MATH satisfy property MATH, if they satisfy property MATH, for all MATH. We shall inductively construct an increasing sequence MATH of elements of MATH and a decreasing sequence MATH of infinite subsets of MATH with MATH, if MATH, so that if MATH and MATH, then MATH and MATH satisfy property MATH. The first inductive step is similar to the general one and so we shall not explicit it. Now we assume that MATH and MATH have been constructed with MATH, if MATH, so that if MATH, and MATH then MATH and MATH satisfy property MATH. Let MATH and fix MATH. We define MATH . Clearly, MATH is closed in MATH and therefore NAME. Suppose that for some MATH, MATH, we had that MATH. Let MATH and set MATH . Since MATH, for all MATH, there exist integers MATH as well as scalar sequences MATH, MATH, so that letting MATH we have that for all MATH, MATH and moreover, if MATH is a MATH-faithful scalar sequence for MATH then MATH . Let MATH and observe that any scalar sequence MATH which is MATH-faithful for MATH, is also MATH-faithful for MATH, for all MATH. Next, let MATH and note that MATH . By the induction hypothesis, since MATH and MATH satisfy property MATH, there exists a scalar sequence MATH which is MATH-faithful for MATH and such that MATH . Thus, MATH . It follows now by our previous observation, that MATH is MATH-faithful for MATH, for all MATH and thus, MATH . Hence, MATH, for all MATH. Now choose MATH, MATH. Then, MATH, for all MATH. But MATH and MATH is arbitrary and so MATH is not weakly null in MATH, contradicting our assumption. Concluding, there exists MATH such that MATH. By repeating the previous argument successively over all possible subsets of MATH, we obtain MATH such that MATH, for all MATH. The inductive construction is now complete. Set MATH. Let MATH and MATH. Let also MATH be a scalar sequence which is MATH-faithful for MATH. Suppose that MATH and let MATH, for all MATH. Our construction yields that MATH and MATH, satisfy MATH, for all MATH. We also have, by stability, that for all MATH, MATH where, MATH, and therefore there exists a scalar sequence MATH which is MATH-faithful for MATH and such that MATH . Thus, MATH . Finally let MATH be any cluster point of the sequence MATH in MATH. Evidently, this is the desired sequence. |
math/9911019 | Assume on the contrary, that no subset of MATH containing MATH is a maximal MATH set. We then claim that MATH belongs to MATH. Indeed, suppose that MATH, MATH, and choose MATH maximal with respect to MATH. If MATH, then REF yields that MATH is a maximal MATH set contradicting our assumption. Thus MATH as claimed. There exists now MATH such that MATH belongs to MATH. It follows that we can find MATH and consecutive MATH sets MATH so that MATH . But now, MATH and thus MATH. This contradiction completes the proof of the lemma. |
math/9911019 | Assume that MATH has no subsequence which is b.c.c. Choose MATH, MATH and MATH satisfying the conclusion of REF applied on MATH. If MATH we are done since some subsequence of MATH is equivalent to the unit vector basis of MATH. Assume now that MATH. Choose according to REF , MATH so that the sequence MATH of functionals biorthogonal to MATH is weakly null in MATH. Next choose, according to REF , MATH such that MATH . Let MATH and choose a sequence of positive scalars MATH such that MATH. Choose also MATH. REF now yields MATH such that for every MATH and MATH, if there exists MATH which is MATH-good for MATH, then there exists MATH, MATH-good for MATH and such that MATH . We continue our choice of infinite subsets of MATH by choosing MATH according to REF . Thus, for every MATH and MATH, if there exists a scalar sequence MATH which is MATH-faithful for MATH, then there exists MATH, MATH-faithful for MATH, and such that MATH . Finally, choose MATH according to REF applied for MATH. It follows now, by REF , that for every MATH and MATH, there exists a scalar sequence MATH which is MATH-faithful for MATH and such that MATH . Let now MATH. Let also MATH be the sequence of ordinals associated to MATH. Repeated applications of REF , now yield an increasing sequence of elements of MATH, MATH, and a sequence MATH of infinite subsets of MATH, so that MATH and MATH for all MATH. We thus obtain, by stability, MATH such that MATH, for all MATH, and therefore, MATH . Let MATH be the MATH spreading model constant of MATH. By our choice of MATH, REF yield that MATH and hence MATH is semi-normalized. We now claim that MATH for all MATH. Our claim of course implies that MATH is equivalent to the unit vector basis of MATH. Were our claim false, there would exist MATH, integers MATH, and MATH such that MATH . Since MATH, REF yields MATH such that MATH . Next we set MATH. It follows that MATH . We now fix MATH and let MATH . Because of REF yields the existence of a maximal MATH subset MATH of MATH such that MATH. Let MATH and set MATH . Since MATH, we obtain through REF , a scalar sequence MATH which is MATH-faithful for MATH and such that MATH . We recall here that for every MATH, MATH . Of course, MATH is strongly bounded by MATH and MATH, for all MATH. Therefore, MATH . Our construction yields that MATH . Indeed, MATH . We also observe that MATH . Hence, MATH . It follows now that MATH and thus MATH contradicting the choice of MATH. Hence, our claim holds and MATH is equivalent to the unit vector basis of MATH. Moreover, the equivalence constant MATH, depends only on MATH, MATH and MATH. It is now easily seen that the set MATH is closed in MATH and therefore NAME. Our previous argument yields MATH so that MATH is equivalent to the unit vector basis of MATH, for all MATH. The proof of REF is now complete. |
math/9911019 | The fact that REF. implies REF. is immediate since MATH sequences are easily seen to be semi-boundedly complete, and every convex block subsequence of a MATH sequence is also MATH CITE. Suppose now that REF. holds. If no subsequence of MATH were MATH, then REF yields a convex block subsequence of MATH equivalent to the summing basis. But the summing basis is not semi-boundedly complete. This contradiction shows that REF. must hold. |
math/9911019 | Let MATH. Our assumptions allow us to choose MATH, an infinite subset of MATH, so that MATH, for all MATH. Stability now yields MATH such that MATH, for all MATH. The assertion of the lemma follows from this since MATH is a closed subset of MATH and therefore NAME. |
math/9911019 | REF yields MATH a convex block subsequence of MATH equivalent to the summing basis. We set MATH which is clearly a weakly null sequence. By the results of CITE there exists a countable ordinal MATH such that no subsequence of MATH is a MATH spreading model. It follows now, by REF , that there exists MATH such that MATH, for every MATH. We next choose MATH, a sequence of positive scalars such that MATH. As a consequence of REF we obtain MATH such that MATH, for every MATH and all MATH. A standard perturbation result now yields MATH such that MATH is MATH-equivalent to MATH, for all MATH. Since the summing basis is uniformly equivalent to all of its convex block subsequences, we obtain MATH such that MATH is MATH-equivalent to the summing basis, for all MATH. This completes the proof. |
math/9911019 | Clearly the statements are mutually exclusive since the summing basis is not semi-boundedly complete. Suppose that REF. does not hold. It follows that no subsequence of MATH is MATH. REF now yields that every subsequence of MATH admits a convex block subsequence equivalent to the summing basis. To prove that REF. holds let MATH be a subsequence of MATH. Without loss of generality, by passing to a subsequence according to REF, we can assume that MATH dominates the summing basis. Next choose MATH, a convex block subsequence of MATH equivalent to the summing basis. We set MATH, for all MATH. If MATH is not semi-normalized, then REF. follows. So assuming that MATH is semi-normalized we claim that there exists a subsequence of MATH equivalent to the unit vector basis of MATH. Indeed, if that were not the case, then by NAME 's dichotomy, REF , there would exist a semi-boundedly complete subsequence of MATH. But since every subsequence of MATH dominates the summing basis (and therefore every subsequence of MATH as well), we obtain that MATH has a semi-boundedly complete subsequence which of course contradicts our assumption that REF. does not hold. Hence, our claim must hold and it immediately yields a subsequence of MATH equivalent to the summing basis in view of the following elementary fact: Let MATH and MATH be sequences in a NAME space with MATH equivalent to the summing basis and MATH equivalent to the MATH basis. Then there exists a subsequence of MATH equivalent to the summing basis. |
math/9911021 | Take an arbitrary slice MATH, where MATH. We will produce two functions in MATH having distance nearly MATH. Let MATH. We first pick some MATH such that MATH . The functional MATH can be represented by a regular NAME measure MATH on MATH with MATH, that is, MATH for all MATH. Let MATH be an open set with MATH; such a set exists since MATH is infinite. Fix a strong boundary point MATH. Using the definition of a strong boundary point, inductively construct functions MATH and nonvoid open subsets MATH such that MATH and MATH . Let MATH and define MATH . By construction, MATH on MATH and MATH on MATH. We claim that MATH. In fact, if MATH, then MATH if MATH, MATH and MATH if MATH, and therefore MATH . We now estimate MATH: MATH . Next, we produce a function MATH such that MATH . We then have MATH, and the functions MATH are in the unit ball of MATH. We have MATH and thus MATH . Consequently, MATH so that MATH; but MATH as MATH. Hence MATH. |
math/9911021 | Every nonvoid relatively weakly open subset of the unit ball of a NAME space contains a convex combination of slices, see CITE or CITE. Thus, if MATH is given as above, there are slices MATH and MATH, MATH, such that MATH. Let MATH with MATH and representing measures MATH. We now perform the construction of the proof of REF with MATH, MATH as before and a nonvoid open set MATH such that MATH for all MATH. We obtain functions MATH and MATH (independently of MATH) such that MATH and MATH. Therefore MATH, and MATH. |
math/9911025 | Obviously every NAME semigroup verifies REF. Conversely, let us assume REF and let MATH, be positive integers such that MATH. We have to prove that MATH. If MATH or MATH, then it is clear. Otherwise, if MATH, let MATH, and write MATH . Note that MATH and MATH. Let MATH be defined by MATH thus MATH, with MATH and MATH. If MATH, then REF implies that MATH; otherwise we can repeat the reasoning, obtaining three increasing sequences of integers MATH, such that MATH with MATH. There are two possibilities: if there exists an index MATH such that MATH or MATH, then MATH; otherwise, if MATH for all MATH, then, by construction, the sequence MATH is strictly increasing, so there exists an index MATH such that MATH, and again we get MATH. |
math/9911025 | As seen before, every hyperelliptic semigroup is an NAME semigroup. Conversely, if MATH is NAME and NAME, then MATH, because otherwise we have MATH, and in the same way MATH, contradicting the fact that MATH. Thus two consecutive integers in the interval MATH cannot be both poles. If MATH is symmetric, the same happens for gaps (if MATH are gaps, then MATH are poles). Since REF is always a pole, we get MATH and MATH is hyperelliptic. |
math/9911025 | If MATH is NAME, then for every MATH we have MATH, so MATH. If MATH then there exist MATH such that MATH what is impossible. |
math/9911025 | If MATH, then for MATH, we have MATH. Thus MATH and MATH, hence MATH. If furthermore MATH, then MATH so MATH and MATH. Thus, all the elements in the set MATH are distinct. |
math/9911025 | Since MATH, then MATH and MATH. The conclusion follows from REF and the fact that MATH for MATH. |
math/9911025 | It suffices to show that MATH. Otherwise, if MATH for some MATH, then we have MATH, what leads to MATH. |
math/9911025 | If MATH then MATH, and the result follows from REF . |
math/9911025 | Since MATH, REF follows from REF . To prove REF, let us first note that for MATH, it holds that MATH. Thus, if MATH, we have MATH, hence, according to REF , we have MATH and the proof is complete. |
math/9911025 | Let MATH, MATH, be three poles smaller than the conductor of MATH (hence MATH). There exist poles MATH such that MATH and MATH. Then, since MATH, the result follows from the fact that MATH is NAME. |
math/9911025 | For every MATH we have MATH, hence MATH is even unless there exists a (unique a fortiori) pole MATH such that MATH, that is, MATH. In this case, if MATH, then for every MATH with MATH, then either MATH or MATH, and hence MATH. |
math/9911025 | According to REF , we have MATH for all MATH odd if and only if MATH for all MATH, and this happens if and only if MATH that is, if and only if MATH for all MATH with MATH. This is equivalent to MATH being NAME according to REF . |
math/9911025 | If MATH, then REF implies MATH. Thus the result follows from REF . |
math/9911025 | If MATH is such that MATH, then we have MATH for some MATH. Since MATH, it holds that MATH. On the other hand, since MATH and MATH, we have MATH and MATH. Hence MATH. The converse is clear. |
math/9911025 | If REF holds then MATH for all MATH odd and MATH is an NAME semigroup as we have seen in REF . Conversely, assume MATH is NAME and let MATH be an odd integer with MATH. According to REF we have MATH. Now if we write MATH, then, according to REF , we have MATH . Since MATH, we obtain MATH. Thus, MATH and MATH verifies REF. |
math/9911025 | If MATH, then (see CITE) all the checks MATH in MATH and MATH are independent, so MATH, MATH and MATH. Now, if MATH, then MATH and MATH. Thus MATH. This proves REF. If MATH, then, according to REF , we have MATH. Thus MATH, hence MATH and MATH. |
math/9911029 | Since the maps MATH are flat, since MATH is a fibered product, and since MATH is the pullback of the universal MATH-th root on MATH via MATH, we have MATH . |
math/9911030 | Balanced implies MATH and MATH. Consider the expansion MATH . For this series to be annihilated by REF it is necessary and sufficient that MATH . This identity holds if and only if the circuit MATH is balanced. |
math/9911030 | The implication from REF to REF follows from the previous lemma. The equivalence of REF holds because every proper facial subset MATH of MATH is affinely independent. Hence the only non-constant MATH-discriminants arising from facial subsets MATH arise from vertices MATH, in which case MATH. It remains to prove the implication from REF to REF . Suppose that MATH is gkz-rational. Consider a non-Laurent rational MATH-hypergeometric function and expand it as a NAME series with respect to increasing powers of MATH. It follows from the results in CITE that this series is the sum of a NAME polynomial and a canonical MATH-hypergeometric series of the following form: MATH . Here MATH is a suitable integer vector. The series MATH represents a rational function. We may view the series on the right-hand side of REF as defining a rational function of the single variable MATH: MATH . The MATH-discriminant equals MATH where MATH. This implies that the rational function MATH may be written as a quotient MATH where MATH is a polynomial and MATH. It follows from CITE that the coefficients of the series REF must be of the form MATH times a polynomial in MATH. That is, the following expression is a polynomial in MATH: MATH . The rational function MATH satisfies the following general identity CITE for any fixed complex number MATH: MATH . Our rational function MATH has its poles among the points MATH and its zeroes among MATH . We may assume MATH and MATH. Suppose now that MATH. Then, MATH has a pole at a point MATH with MATH and MATH coprime, but since none of the zeroes may be of this form, this contradicts REF . A symmetric argument leads to a contradiction if we assume MATH. This implies that MATH and therefore MATH is also rational function of MATH. Consequently, we can iterate our argument to conclude that, after reordering, MATH for all MATH. |
math/9911030 | REF and the lemma below imply Conjecture REF. The equivalence of Conjectures REF will be shown in REF. |
math/9911030 | We first prove the if-direction. Let MATH be an essential NAME configuration which is a circuit. Each MATH must consist of a pair of vectors in MATH, so that MATH becomes a MATH-matrix. The first MATH rows of MATH show that the kernel of MATH is spanned by a vector MATH. This means that MATH is balanced. Conversely, if MATH is balanced then we can apply left multiplication by an element of MATH to get isomorphically MATH where MATH is a MATH integral matrix of rank MATH. By permuting columns we see that MATH is an essential NAME configuration for MATH. |
math/9911030 | Proceeding by induction, it suffices to consider the case when MATH is obtained from MATH by removing a single point, say, MATH. Since MATH is not contained in any face of MATH, and MATH is a facial subset of itself, the following lemma tells us that the MATH-discriminant MATH divides MATH. Since MATH is not a monomial, this implies that MATH is not a monomial. |
math/9911030 | Let MATH be a generic polynomial with support MATH. By CITE, the principal MATH-determinant is the specialization MATH where MATH denotes the MATH-resultant; see CITE. The irreducible factorization of the principal MATH-determinant ranges over the facial subsets MATH of MATH, MATH where MATH are certain positive integers CITE. Let MATH be the weight vector with MATH and MATH for MATH. The initial form of the principal MATH-determinant with respect to MATH can be factored in two different ways: MATH . Here MATH is the coherent polyhedral subdivision of MATH defined by MATH and the MATH are certain positive integers. The first formula comes from REF and the second formula comes from CITE. Since MATH is a facial subset of MATH, it is also a cell of the subdivision MATH, and hence MATH divides MATH for the facet MATH of MATH. We conclude that MATH divides MATH for some facial subset MATH of MATH. If MATH, this implies that MATH because MATH involves all the variables associated with points in MATH. By our hypothesis, the only facial subset of MATH which contains MATH is MATH itself. Therefore MATH divides MATH. |
math/9911030 | Let MATH be any prime element in MATH. We must show that MATH does not divide MATH. Consider the localization MATH of MATH at the prime ideal MATH. The power series ring MATH is the completion of the local ring MATH with respect to the MATH-adic topology. By assumption, the polynomial MATH lies in the principal ideal generated by MATH in MATH. The basic flatness property of completions, as stated in CITE, implies that MATH lies in the principal ideal generated by MATH in MATH. Since MATH and MATH are relatively prime in MATH, we conclude that MATH is a unit in MATH and so MATH is not divisible by MATH. |
math/9911030 | Suppose MATH is a gkz-rational configuration and let MATH be a rational MATH-hypergeometric function of degree MATH, where MATH are relatively prime, and the MATH-discriminant MATH is not a monomial and divides MATH. We claim that any spanning circuit MATH of MATH is balanced. We shall prove this by induction on MATH. If MATH, then we are done by REF . We may assume that MATH is not a circuit and therefore MATH is a proper subset of MATH. Suppose MATH, and set MATH, MATH, MATH. We may expand the rational MATH-hypergeometric function MATH as MATH where each MATH is a rational MATH-hypergeometric function of degree MATH. Let MATH denote the unique smallest facial subset of MATH which contains the circuit MATH. Then MATH is a spanning circuit in MATH. REF implies that its discriminant MATH is not a monomial. REF implies that MATH divides MATH, the lowest coefficient of MATH with respect to MATH. We now apply REF to the domain MATH, the localization of the NAME polynomial ring at the principal prime ideal MATH. Since MATH is not a unit in MATH, we conclude that some NAME coefficient MATH lies in the field of fractions of MATH but not in MATH itself. This means that MATH divides the denominator of MATH. We have found a rational MATH-hypergeometric function whose denominator contains the non-trivial factor MATH. It follows by induction that the spanning circuit MATH of MATH is balanced. |
math/9911030 | If MATH is a facial subset of MATH then every MATH-hypergeometric function MATH is also MATH-hypergeometric. Indeed, MATH is obviously MATH-homogeneous, but it is also annihilated by the toric operators MATH because the support of MATH lies in MATH if and only if the support of MATH lies in MATH. This proves the if-direction. For the only-if direction, suppose that MATH is weakly gkz-rational and let MATH be a non-Laurent rational hypergeometric function. There exists a facial subset MATH of MATH such that MATH is not a monomial and divides MATH. Our goal is to show that MATH is gkz-rational. We proceed by induction on the cardinality of MATH. There is nothing to show if MATH. Let MATH and form the series expansion as in REF . By applying REF as in the proof of REF , we construct a rational MATH-hypergeometric function whose denominator is a multiple of the MATH-discriminant MATH. This proves our claim, by induction. |
math/9911030 | Let MATH be an interior point of MATH, and let MATH be a minimal size subset of MATH which contains MATH in its relative interior. Then MATH is a circuit of MATH which is spanning and not balanced. |
math/9911030 | Reflexive polytopes possess exactly one interior point. If MATH is bigger than the ambient dimension then MATH times any lattice polytope contains an interior point. |
math/9911030 | It suffices to prove the following two assertions: CASE: MATH . If MATH is a configuration on the line MATH or in MATH-space MATH then MATH is not gkz-rational. CASE: MATH . If MATH is a configuration in the plane MATH then MATH is gkz-rational if and only if the points of MATH lie on two parallel lines with each line containing at least two points from MATH. The case MATH was proved above. We first assume MATH. If the points of MATH lie on two parallel lines then we can write their coordinates as follows: MATH . Thus MATH is the NAME configuration of two one-dimensional configurations. The construction in the next section shows that MATH is gkz-rational for MATH. Conversely, suppose that MATH does not lie on two parallel lines. We may further assume that MATH contains no unbalanced spanning circuit by REF . One example of a configuration satisfying these requirements is MATH . The toric variety MATH is the NAME surface in MATH. Its dual variety is the hypersurface defined by the discriminant of a ternary quadratic form MATH . Suppose there exists a rational MATH-hypergeometric function MATH with MATH a multiple of MATH. Let MATH be the configuration obtained from MATH by removing the fourth and fifth columns. Setting MATH in MATH yields MATH. We can argue as in the proof of REF and construct a rational MATH-hypergeometric function whose denominator contains the binomial factor. REF would imply that the configuration consisting of the first three columns of MATH is gkz-rational, and this is a contradiction to REF . Hence the configuration MATH in REF is not gkz-rational. Another configuration to be considered is MATH where MATH are relatively prime integers. The only spanning circuit of MATH is the balanced circuit MATH. Consider the subset MATH which is an unbalanced circuit on the boundary of MATH. The MATH-discriminant is an irreducible homogeneous polynomial of degree MATH which looks like MATH . Applying the expansion technique with respect to MATH, we get a rational MATH-hypergeometric function whose denominator contains MATH. This contradicts REF . Hence the configuration MATH in REF is not gkz-rational. Our assertion for MATH now follows from the subsequent lemma of combinatorial geometry. Note that four points in the plane, with no three collinear, lie on two parallel lines if and only if they form a balanced circuit. |
math/9911030 | We may assume without loss of generality that the origin MATH lies in MATH and is a vertex of the convex hull MATH. Let MATH and MATH be the points of MATH closest to MATH along the edges of MATH adjacent to MATH. Let MATH. Any other point MATH must be of the form MATH, or MATH, or MATH, or MATH, where MATH are positive real numbers and MATH. If MATH, then only the first two cases may occur. Indeed, suppose MATH, then either all the points lie on two parallel lines or there exists a point MATH or MATH in MATH. It is easy to check that in all of these cases, MATH has an interior point. Suppose then that MATH and MATH. We have MATH since the subset MATH lies on two parallel lines. If MATH then the subset MATH contradicts the assumption. Hence, if MATH, either all the points lie on two parallel lines, or MATH which means that MATH is affinely equivalent to REF . On the other hand, if MATH and there exists a point MATH, then either all the points lie on two parallel lines or MATH contains a point of the form MATH or MATH. Since MATH, in all of these cases MATH contains an unbalanced circuit, or MATH is affinely equivalent to the vertex set of a regular pentagon. The only remaining possibility is that all points of MATH be multiples of either MATH or MATH. But if MATH and MATH are in MATH then MATH is unbalanced unless MATH. Hence the only possible configuration not containing the point MATH is affinely equivalent to REF |
math/9911030 | Choose five points from our configuration which are not in a plane. They have the form MATH where the lines MATH and MATH are parallel and MATH lies outside the plane MATH. Suppose that our configuration is not on three parallel lines. There exists a point MATH such that the line MATH is not parallel to the lines MATH and MATH. If, under this hypothesis, the line MATH is still parallel to the plane MATH, then we have created a MATH-dimensional circuit, a contradiction. Therefore the line MATH meets the plane MATH in a point which we call the origin MATH. The origin MATH must be equal to either MATH or MATH; otherwise we would have created a MATH-dimensional circuit. From this requirement we conclude that the configuration MATH is affinely equivalent to MATH. It remains to be seen that MATH is the only point that may be added to the configuration MATH without creating either a MATH-dimensional circuit or an unbalanced MATH-dimensional circuit. A point not a multiple of MATH or MATH obviously creates a MATH-dimensional circuit. A multiple of MATH or MATH creates an unbalanced MATH-dimensional circuit, unless it is the origin. |
math/9911030 | Let MATH be a configuration in affine MATH-space. We shall prove that MATH is not gkz-rational. In view of REF , we may assume that MATH contains no unbalanced spanning circuit. This implies that MATH contains no MATH-dimensional circuit, because such a circuit involves five points and, five being an odd number, that circuit would be unbalanced. Suppose that MATH contains an unbalanced MATH-dimensional circuit MATH. Then MATH lies in a facet of MATH. There must be at least two distinct points MATH and MATH of MATH which do not lie in that facet. Otherwise, MATH is a pyramid and the MATH-discriminant is MATH. If the line MATH is parallel to the plane spanned by MATH then, since MATH is unbalanced, some triangle in MATH has all of its three edges skew to MATH. This triangle together with MATH and MATH forms a MATH-dimensional circuit, a contradiction. Hence the line MATH intersects the plane spanned by MATH. Some triangle in MATH has the property that none of the lines spanned by its edges contains that intersection point. Again, this triangle together with MATH and MATH forms a MATH-dimensional circuit. We conclude that A has no MATH-dimensional circuit and every MATH-dimensional circuit of MATH is balanced. REF tells us what the possibilities are. If MATH lies on three parallel lines, then MATH and thus MATH is not gkz-rational. It remains to examine the special configurations MATH. An affine transformation moves the points MATH and MATH onto the coordinate axes, so that our configuration has the matrix form MATH where MATH and MATH are relatively prime integers, and MATH. The subconfiguration consisting of the last six columns is spanning. We shall prove that it is non-degenerate and not gkz-rational. Our usual deletion technique then implies that the bigger configuration REF is also not gkz-rational. It therefore suffices to consider the following MATH-matrix MATH . We shall distinguish the two cases MATH and MATH. If MATH then MATH represents an octahedron, and if MATH then MATH represents a triangular prism. We shall present a detailed proof for the octahedron case MATH. The proof technique to be employed was shown to us by NAME. We first note that the MATH-discriminant MATH is a homogeneous irreducible polynomial of degree MATH. The NAME polytope of MATH is a simplex with vertices corresponding to the monomials: MATH . Suppose MATH is gkz-rational. There is a rational function MATH, with MATH relatively prime polynomials, such that the MATH-discriminant MATH divides MATH, and MATH is MATH-hypergeometric of some degree MATH. For MATH, the derivative MATH is MATH-hypergeometric of degree MATH and has MATH in its denominator. Replacing MATH by MATH for suitable MATH, we may assume that the MATH-degree of MATH is of the form MATH for some negative integer MATH. We expand MATH around the vertex MATH of the NAME polytope of MATH. This results in a convergent NAME series in the new variables MATH . That hypergeometric series equals, up to a constant, MATH for an appropriate positive integer MATH. The coefficients of this series can be derived directly from the toric operators REF arising from MATH. It is one of the canonical series described for general MATH in CITE. The series MATH represents a rational function in two variables. We denote by MATH the coefficient of MATH in the series REF . Note that MATH. Since MATH is rational, there exist positive integers MATH, and constants MATH, such that MATH and MATH . If we divide MATH by MATH then we get a rational function in MATH and MATH. Hence the following is an identity of rational functions in MATH and MATH: MATH . Let MATH and MATH denote the incremental quotients: MATH . If MATH and we set MATH, MATH, we have MATH . Given now MATH with MATH we have MATH . Note that either MATH or MATH is a factor in the above product. Consider now the point MATH where MATH is an irrational number. We have MATH and therefore both MATH and MATH vanish at MATH. On the other hand, since MATH is irrational, none of the denominators in REF or REF may vanish at MATH. Evaluating the left-hand side of REF at MATH yields MATH which is impossible. We have shown that the matrix MATH in REF is not gkz-rational for MATH. The proof of non-rationality in the triangular prism case MATH, provided to us by NAME, is analogous and will be omitted here. In summary, we conclude that every MATH-dimensional configuration is not gkz-rational. |
math/9911030 | The identity MATH was proved in CITE under the more restrictive hypothesis that the configurations MATH are all full-dimensional; see CITE. The argument given in that proof shows that MATH suffices to imply MATH. On the other hand, the condition of MATH being essential appears in CITE, and CITE shows that it is equivalent to MATH. |
math/9911030 | We must show that a non-degenerate configuration MATH is affinely isomorphic to an essential NAME configuration if and only if its discriminant MATH equals the mixed resultant MATH of some tuple of configurations MATH. The only-if direction is the content of REF . For the converse, suppose MATH. Let MATH be the NAME configuration of MATH. Then MATH. In other words, the toric varieties MATH and MATH in MATH have the same dual variety, namely, the hypersurface defined by MATH. The NAME Theorem CITE shows that MATH, and this implies that MATH and MATH are affinely isomorphic. |
math/9911030 | Let MATH and MATH generic NAME polynomials as in REF . It follows from either the definition or CITE that MATH is a homogeneous function with respect to the grading induced by MATH; that is, it satisfies the equations defined by the operators REF for a suitable parameter vector. It follows from CITE that MATH is also annihilated by the operators REF . Hence MATH is a rational MATH-hypergeometric function. The discriminant associated with MATH equals the resultant MATH, by REF . This resultant is not a monomial, for instance, by CITE. We showed in CITE that MATH is a polynomial. It remains to be seen that the toric residue MATH itself is non-zero for at least one lattice point MATH. Recall from CITE and CITE that the polynomial MATH is supported in MATH and MATH. Here the operator MATH is extended from monomials MATH to the polynomial MATH by linearity. At least one of the residues MATH as MATH runs over MATH, does not vanish and hence is a non-Laurent rational MATH-hypergeometric function. |
math/9911030 | Let MATH be an essential NAME configuration with MATH as in REF . Let MATH be the set of all lattice points in a convex polytope containing all the configurations MATH for MATH. Then, MATH is full-dimensional and MATH. Consider configurations MATH in MATH such that MATH for MATH. The corresponding NAME configuration MATH is still essential, since the NAME sum MATH has affine dimension at least MATH. This property holds for MATH and it does for MATH. We conclude from REF that the NAME configuration MATH is non-degenerate and MATH. We would like to show that, in fact, any such configuration MATH must be gkz-rational. We proceed by induction on the cardinality of MATH. The base case is cardinality zero: if MATH then MATH is gkz-rational by REF . For the induction step we may suppose that MATH is obtained from MATH by removing a point MATH from MATH and assume, inductively, that MATH is a rational MATH-hypergeometric function which contains the discriminant MATH in its denominator. Let us denote by MATH the variable associated with MATH and by MATH the variables associated with MATH. Expand as in REF : MATH where each MATH is a rational MATH-hypergeometric function. We may now argue as in the proof of REF ; since MATH and MATH have affine dimension MATH it follows from REF that the MATH discriminant MATH divides the specialization MATH. Hence, for some MATH, the rational function MATH will lie strictly in the field of fractions of the domain MATH and, consequently, will be a rational MATH-hypergeometric function which contains the discriminant MATH in its denominator. In summary, the configuration MATH inherits the property of being gkz-rational from the configuration MATH. By induction, we conclude that MATH is gkz-rational. |
math/9911031 | First we claim that there exist subcomplexes MATH and MATH satisfying the following conditions: CASE: MATH and MATH are free abelian groups of the same rank as MATH for all MATH; CASE: MATH and MATH are torsion free for all MATH; CASE: MATH and MATH are isomorphic complexes of abelian groups. CASE: The sequences MATH and MATH are exact for all MATH. This claim can be proved by induction. First since MATH and MATH are bounded complexes of finite generated abelian groups, without loss of generality we suppose MATH and MATH . Consider the subgroup MATH of MATH. Let MATH be the rank of MATH and let MATH be a maximal independent set in MATH. We can enlarge it into a maximal independent set MATH of MATH. Set MATH be the subgroup generated by MATH. Then MATH is finite. Now consider the inverse image of MATH, it is a subgroup of MATH. Moreover, it must have the same rank as MATH. Since MATH is contained in the inverse image of MATH, so is MATH. Find MATH such that MATH. This set is an independent set in the inverse image of MATH and has only trivial intersection with MATH. We select a maximal independent set in MATH, enlarge it to a maximal independent set in MATH, together with MATH, we get a maximal independent set MATH in the inverse image of MATH. Set the free subgroup generated by MATH as MATH. Continuing this setup, we obtain a subcomplex MATH of MATH such that MATH is free, MATH is finite and MATH is torsion free. Similarly for the complex MATH, we can construct a subcomplex MATH of MATH such that MATH is free, MATH is finite and MATH is torsion free. Hence MATH and MATH satisfy REF . But REF easily follow from REF . Hence we proved the above claim. Now choose an isomorphism MATH of complexes. We have MATH . Here we use the facts: CASE: If MATH is a complex of finite abelian group, then REF . If MATH is a complex of MATH-vector spaces, MATH is an automorphism of MATH, then MATH . |
math/9911031 | Consider the complexes MATH and MATH with the restriction map MATH. Note that: CASE: MATH for all MATH. CASE: Since MATH for all MATH, MATH and MATH are both finite and MATH. We have MATH for all MATH. CASE: Now for MATH, consider the map MATH. We have MATH. Then MATH where MATH and MATH . Now applying REF to the case MATH, MATH and MATH, we immediately get REF. |
math/9911031 | First note that the given identity REF is nothing but MATH . Since for the spectral sequence, MATH, we always have MATH . Hence MATH which means that for MATH, MATH . Therefore we have a shorter complex: MATH . Now we set to prove the following fact: MATH . Observe that the set MATH, the only term not finite is MATH. If we substitute it by its torsion, we still get a group of complexes composed of finite abelian groups and with differential MATH. The cohomology groups are MATH(or MATH). By the invariance of NAME characteristic under cohomology, REF is proved. Note that MATH is free and MATH . REF now follows immediately. |
math/9911031 | CASE: See CITE. CASE: Define MATH . It is routine to check that MATH is actually an isomorphism. Since we don't need this fact in the latter context, we omit it here. |
math/9911031 | By straightforward calculation. |
math/9911031 | By direct calculation. |
math/9911031 | First notice that MATH is invariant under MATH. Moreover, let MATH, for any MATH, define MATH then clearly MATH is invariant under MATH. By definition, we have MATH. Put MATH . We can see that MATH is a real vector space with a basis MATH. Furthermore MATH has a natural MATH-module structure. Actually it is a free MATH-module of rank MATH. MATH induces an automorphism in MATH: MATH . We calculate its determinant first. Let MATH . For MATH, define MATH . Note that MATH and MATH where MATH. Then we have MATH . For any MATH, let MATH be the smallest number satisfying MATH. Since the map MATH can be regarded as the left multiplication by the group ring element MATH in MATH, then by REF , we have MATH and MATH . Now by the NAME Principle, we have MATH . Hence MATH . By the NAME inverse formula, MATH . Therefore we have MATH . Now let's look at the right hand side of the above identity. The exponent of MATH is MATH here MATH. Write MATH, then MATH which is exact the right hand side of the identity REF. |
math/9911031 | The proof is similar to the proof of REF . Note that MATH is a real vector space with a basis MATH. On the quotient space MATH, MATH . Now the restriction of MATH on MATH is MATH . We still have MATH where MATH. Similar to the calculation of MATH in REF , we have MATH . We have MATH . |
math/9911031 | First note that MATH. For REF , since MATH it suffices to show that the given collection generates MATH. This can be easily deduced by induction to MATH, with the fact that for any MATH, MATH . For REF , note that the identity MATH holds for any MATH and integer constants MATH, therefore REF follows immediately from REF . |
math/9911031 | For each prime number MATH, we define operators MATH, MATH and MATH on MATH by the rules: MATH where MATH runs through the basis given in REF . It is easy to check: MATH where MATH is the NAME symbol, and MATH . By REF , for any fixed positive integer MATH, we can check that MATH . Now for a fixed positive integer MATH, we define MATH, MATH and MATH on MATH by the rules: MATH here we abuse the notations MATH, MATH and MATH with their restrictions on MATH. It is easy to check the definition MATH here coincides the one we defined in MATH. We can also check that MATH for any elements MATH of the basis given in REF . Now we have MATH . By the above argument, the cochain map MATH is homotopic to the cochain map MATH. But in the negative degree, MATH is nothing but the zero map. Therefore we showed that MATH is acyclic in negative degrees. Now since for the map MATH(respectively, MATH, MATH), MATH(respectively, MATH, MATH), there exists a unique operator MATH(respectively, MATH, MATH) with restriction at MATH the operator MATH (respectively,MATH, MATH). MATH is a cochain map homotopic to the identity map of the cochain complex MATH and vanishes at negative degrees. Therefore MATH is acyclic at negative degrees. Now for MATH, the cohomology groups MATH and MATH easily follow from the definitions of MATH and MATH . Now by a parallel argument to MATH and MATH, we construct MATH, MATH and MATH, MATH respectively. the remaining assertions follow immediately . |
math/9911031 | By the identity REF, the condition in REF is satisfied. For MATH, by REF, then the exponent of MATH in MATH is equal to MATH . The case MATH immediately follows from REF . |
math/9911031 | We consider the following diagram: MATH where MATH and MATH, MATH and MATH . MATH is well defined and all the above maps are isomorphisms of vector spaces. Then we have MATH . Here for the second equality, we use the property that if MATH and MATH are two vector spaces and MATH is an isomorphism from MATH to MATH, then MATH. Now for the three factors at the last line of REF, we have: MATH . We first consider the following diagram which is exact at the rows: MATH where MATH is the natural inclusion map. By REF , if MATH is not a power of MATH, MATH; if MATH is a power of MATH, then MATH. Therefore, MATH . Now let MATH be the map from MATH to MATH such that MATH, then we have MATH. Then on one hand, MATH on the other hand, MATH . But we know MATH, and by the definition of MATH, MATH and MATH. Now the lemma follows from the above results and MATH . Let MATH, then MATH . Note that if we let MATH then MATH is just the left multiplication of MATH on MATH. By REF , we have MATH . MATH . This follows from the abstract index REF . Now let MATH approach MATH, then MATH . If we let MATH, by REF, with the class number formula, MATH and since MATH, then we have MATH . But by REF, MATH is nothing but MATH in CITE. and by REF , we have MATH. This is enough to finish the proof of the theorem. |
math/9911032 | We let MATH be the complex MATH . Hence it suffices to show that MATH is exact. Let MATH be a maximal element in the order ideal MATH. Let MATH be the order ideal whose maximal element set is the maximal element of MATH excluding MATH, then MATH . By REF , we have MATH . Now we prove the Proposition by induction on the cardinality of maximal elements of MATH. If MATH has only one maximal element, this is just REF . In general, both MATH and MATH have less maximal elements than MATH has. Thus the exactness of MATH follows from the exactness of the three complexes MATH, MATH and MATH. |
math/9911032 | Note that MATH then it is easy to see that MATH is nothing but the quotient complex MATH. The short exact sequence MATH induces a long exact sequence MATH . By REF , for MATH and MATH, both MATH and MATH are MATH, so is MATH. Therefore the above long exact sequence is just the exact sequence MATH . Since the map from MATH to MATH is injective, the proposition follows immediately. |
math/9911032 | The complex MATH by definition, is a complex with MATH for MATH and the differential MATH. In general, MATH is exactly the standard tensor product of MATH for all MATH. Write MATH . Since now MATH has differential MATH, MATH. The restriction map is easy to see. This finishes the proof of REF. For the cup product, the diagonal map MATH given above naturally induces a map: MATH which defines the cup product structure. More specifically, the cup product map MATH is induced from MATH. Now the claim follows soon from the explicit expression of MATH. |
math/9911032 | Immediately from REF (respectively, REF for MATH), we see that MATH is MATH-acyclic, by spectral sequence argument, it is thus MATH-acyclic. On the other hand, MATH is surjective. Thus MATH is a quasi-isomorphism. Now REF follows directly from the quasi-isomorphism. For REF, just consider MATH, which is also a quasi-isomorphism. |
math/9911032 | Let MATH be the subcomplex of MATH with the same components as MATH except at degree MATH, where MATH . We only need to show that MATH is exact. We show it by double induction to the cardinalities of MATH and MATH. If MATH, we get a trivial complex. If MATH consists of only one element, or if MATH consists only one element, it is also trivial to verify. In general, suppose MATH. Let MATH and MATH. Then we have the following commutative diagram which is exact on the columns: MATH . Here MATH means projection and MATH means inclusion. The differential MATH is induced by the differential MATH of the second row. Notice that the third row is a variation of the chain complex MATH, the first row is the chain complex MATH. By induction, the first row and and the third row are exact, so is the middle one. |
math/9911032 | Let MATH and let MATH . Through the map MATH, we have a commutative diagram MATH where MATH and MATH are given in REF . By this diagram, we identify MATH with MATH. By REF , we have MATH . Then by the proof of REF , MATH where the second and the last identities we use the isomorphisms given in the above diagram. Hence we have MATH . On the other hand, MATH . Since MATH in MATH, the spectral sequence of MATH by the first filtration(that is, by MATH) degenerates at MATH. We have MATH . Since the projective map from MATH to MATH in the commutative diagram is nothing but the restriction of the quotient map MATH at MATH, by the above analysis, we get an isomorphism MATH . Thus the spectral sequence of MATH and MATH are isomorphic at MATH for MATH. In our case, the first filtration is finite, thus strongly convergent, therefore MATH is a quasi-isomorphism(see CITE, Page REF ). The case MATH is similar. In this case, MATH and MATH . Now follow the same analysis as above. |
math/9911032 | We only prove REF. The proof of REF follows the same route. By REF , we know that MATH . Now MATH . REF follows immediately. |
math/9911032 | CASE: By REF , we have induced quasi-isomorphism: MATH . Now since the induced differentials of MATH and MATH in MATH are MATH. Consider the cocycle MATH in MATH, there exists a cocycle MATH(unique modulo boundary) which is the lifting of MATH by the quotient map MATH. Hence MATH is a cocycle in the complex MATH. Let MATH denote the cohomology element in MATH represented by the cocycle MATH. Then MATH is a canonical MATH-basis for the cohomology group MATH. This finishes the proof of REF. CASE: Similar to REF, just consider the map MATH. CASE: For the cup product, there is natural homomorphism MATH therefore MATH and also MATH has a natural MATH-module structure. By the theory of spectral sequences(see, for example CITE, Chap. MATH, MATH), we have the cochain cup product MATH . By using the diagonal map MATH defined in MATH, it is easy to check that: MATH hence we can pass the cup product structure to the quotient and have MATH . Now REF follows immediately from the explicit expression of MATH. This concludes the proof. |
math/9911032 | We only prove REF . The proof of REF is similar. First look the spectral sequence of MATH with the second filtration(that is, the filtration given by MATH), then MATH . Next for the differential MATH induced on MATH, with the same analysis as in computing the MATH terms of MATH(see MATH, REF ), we have MATH . Furthermore, let MATH be the double complex generated by all symbols MATH satisfying MATH and MATH, which can be considered as a quotient complex of MATH. Similar to the proof of Theorem A, the quotient map induces an isomorphism between the cohomology groups of each other. Now let MATH be the canonical lifting of the cocycle MATH in MATH, then MATH is a cocycle in MATH with the leading term MATH and the remainder contained in the direct sum of MATH where MATH and MATH. |
math/9911032 | We only prove the first part. The MATH-terms of the spectral sequence are MATH . Note that MATH in general for any spectral sequence, then MATH . By Theorem A and REF , the left hand side and the right hand side of the above inequality have the same number of elements. hence the inequality is actually an identity. Therefore, the spectral sequence of MATH with filtration given by MATH degenerates at MATH. |
math/9911032 | CASE: For each MATH, there exists some prime MATH dividing MATH such that MATH, and one has MATH whence the result. CASE: The family MATH generates MATH by what we have already proved. The family MATH is of cardinality MATH . Therefore the family MATH is a basis for MATH. CASE: These assertions follow trivially from what we have already proved. |
math/9911032 | Clear. |
math/9911032 | We have only to prove the first statement. By REF and a straightforward calculation that we omit, one has MATH where the direct sum is indexed by pairs MATH with MATH and MATH is the largest positive integer such that MATH. Each of the subcomplexes MATH is an isomorphic copy of MATH, and the latter we have already observed to be acyclic in nonzero degree. |
math/9911037 | Let MATH, and MATH be elements in the unit ball of MATH such that MATH is MATH-separated. Choose MATH such that MATH . Without loss of generality, we may assume that MATH converges pointwise (as a sequence of functions on MATH) to some MATH. It is clear that if MATH and MATH, then MATH. It follows easily that MATH. Let MATH. It may be assumed that MATH converges. As MATH is MATH-separated, so is MATH. We may thus further assume that MATH for all MATH. By going to a subsequence and perturbing the vectors MATH, MATH and MATH by as little as we please, it may be further assumed that REF they all belong to MATH, REF MATH, and REF MATH for all MATH such that MATH, where MATH is the sup norm and MATH. Claim. Let MATH be an admissible set such that MATH. If MATH, and MATH, then there exists an admissble set MATH such that MATH, and MATH. To prove the claim, let MATH be such that MATH, and MATH for all MATH with MATH. Then MATH. Now, for each MATH, MATH for some MATH with MATH. It follows that MATH . Hence, there exists MATH such that MATH. Now let MATH . It is easy to see that the set MATH satisfies the claim. Suppose MATH. Let MATH be normed by a sequence of admissible sets MATH. Denote by MATH, MATH, MATH, and MATH respectively the sequences MATH, MATH, MATH, and MATH. Now MATH . But MATH. Similary, MATH, MATH. By REF , we obtain that MATH, MATH, and MATH are all MATH. Let MATH be the largest integer such that MATH. Note that this implies MATH; hence MATH for all MATH. Thus, MATH for all MATH. Now MATH . Moreover, MATH . Hence MATH . Therefore, MATH . Note that by the first part of the argument above, we also obtain that MATH . Since MATH, we may apply the claim to obtain an admissible set MATH. Using the sequence of admissible sets MATH to norm MATH yields MATH . As the last expression is MATH by REF , we have reached a contradiction. |
math/9911037 | Let MATH be a MATH-separated sequence in the unit ball of MATH. Choose MATH so that MATH . As in the proof of the previous proposition, it may be assumed that there exists a sequence MATH in MATH such that MATH, MATH for all MATH, and MATH whenever MATH and MATH, where MATH . We may also assume that MATH converges. Since MATH is MATH-separated, MATH. The choice of MATH in REF guarantees that MATH. Hence there exist MATH such that MATH where MATH. We may now further assume that MATH for all MATH. Now suppose that MATH for all MATH. Claim. For all MATH in MATH, there exists an acceptable set MATH such that MATH for MATH. First observe that there are acceptable sets MATH such that MATH. Let MATH, MATH and MATH be the sequences MATH for MATH respectively. Then MATH and MATH. Similarly, MATH. It follows from the parallelogram law that MATH. Note also that MATH. Similarly, MATH. Let MATH, respectively, MATH, be the largest MATH such that MATH, respectively, MATH. Since MATH, MATH. Similarly, MATH. Moreover, MATH. Let us show that MATH. For otherwise, MATH. Then MATH . Consider the set MATH. Choose MATH such that MATH and MATH for all MATH with MATH. Note that MATH. Let MATH. If MATH, MATH. Hence there exists MATH such that MATH. It is easy to see that the set MATH is acceptable, and MATH. Hence MATH. Thus MATH . Therefore, MATH. It follows that MATH . However, MATH . Combining REF with the choice of REF yields a contradiction. This shows that MATH. Applying the facts that MATH and MATH, we obtain that MATH . Similarly, MATH . Hence MATH. Thus the set MATH satisfies the requirements of the claim. Taking MATH, MATH, and MATH, MATH respectively, we obtain acceptable sets MATH and MATH from the claim. Since MATH, if MATH, MATH for some MATH such that MATH. This implies that there exists MATH such that MATH. Let MATH and MATH. For MATH, define MATH if there exists MATH such that MATH; otherwise, let MATH. Also, let MATH if there exists MATH; otherwise, let MATH. Finally, let MATH for all MATH. Then MATH for all MATH, MATH, and MATH. Hence MATH. Similarly, MATH. Therefore, MATH. Let MATH be the set of all nodes in MATH that are comparable with some node in MATH. Then MATH . Hence MATH. Now let MATH. Divide MATH into MATH and MATH. Since MATH and MATH are acceptable sets such that MATH, MATH . Thus MATH . Finally, since MATH is acceptable, MATH . This contradicts REF . |
math/9911037 | Assume that MATH is the first non-negative integer where the proposition fails. Let MATH be the nodes obtained by applying the proposition for the case MATH. (If MATH, begin the argument with any node MATH.) For each MATH, MATH, let MATH and MATH be a pair of incomparable nodes in MATH. (If MATH, let MATH be any node in MATH.) Since the proposition fails for the nodes MATH, there are distinct branches MATH, MATH passing through MATH, MATH, and a number MATH so that MATH have separated at level MATH, but MATH for any sequence of nodes MATH. However, since the proposition holds for the nodes MATH, we obtain a sequence of nodes MATH such that MATH . (Note that the preceding statement holds trivially if MATH.) For each MATH, choose a node MATH in MATH such that MATH. Then MATH and MATH is a pair of incomparable nodes in MATH, and the argument may be repeated. (If MATH, repeat the argument using the node MATH.) Inductively, we thus obtain sequences of branches MATH, a sequence of numbers MATH, and sequences of nodes MATH such that CASE: the branches MATH have separated at level MATH, MATH, CASE: MATH for any sequence of nodes MATH CASE: MATH, and MATH CASE: MATH whenever MATH, and MATH. It follows that if MATH, then MATH . Let MATH, MATH. By REF , MATH. Moreover, because of REF , if MATH, then MATH . Thus MATH is MATH-separated in the norm MATH. By the choice of MATH, there are MATH such that MATH. Therefore, MATH. But this contradicts REF . |
math/9911037 | In the notation of the statement of REF , we obtain for each MATH, nodes MATH such that MATH, and MATH. Hence MATH cannot be an equivalent norm on MATH. |
math/9911037 | The case MATH is trivial. Suppose the proposition holds for some MATH, MATH. Let MATH, MATH, and let MATH. Divide MATH into disjoint subsets MATH and MATH such that MATH. By the inductive hypothesis, there exist acceptable sets MATH and MATH, MATH, such that MATH, MATH, MATH, and MATH; and also MATH, MATH, MATH, MATH, and MATH for all MATH. It is easily verified that the sequence MATH is MATH-separated and has norm bounded by MATH with respect to MATH. It follows that there exists MATH such that MATH. The induction is completed by taking MATH to be MATH. |
math/9911043 | CASE: For MATH, let MATH be the local expansion of MATH at MATH. As there exists MATH such that MATH (for example, MATH satisfying MATH), we can consider the following hyperplane in MATH: MATH . Then, thanks to REF follows, once we have shown that MATH. Taking into consideration REF , MATH yielding the desired relation MATH. CASE: By the Fundamental Equivalence REF , MATH belongs to the osculating hyperplane at MATH for every MATH. Then from REF we infer for all but a finitely many points MATH that MATH and REF follows. CASE: This is clear because once the projective coordinates are fixed, then the osculating hyperplane at any point is uniquely determined modulo a non-zero element of MATH. |
math/9911043 | Since MATH is MATH-rational we can take it to the point MATH by means of a MATH-linear transformation. The new coordinates still satisfy REF . In addition, we can assume that MATH so that MATH for MATH. Now, the set up and the results of the computation involving local expansion in the proof of REF together with REF allow us to limit ourselves to check that MATH for every point MATH chosen such that MATH. As we have already noted, MATH for MATH. Then, taking also into consideration REF , we only need to see that MATH. As a matter of fact, this follows from REF , and hence the proof of the lemma is complete. |
math/9911043 | Choose a point MATH. Here, MATH for some MATH with MATH. In fact, if MATH for MATH, then MATH would belong to the hyperplane osculating at general points of MATH and so MATH would be in the above MATH which is impossible as we have shown before. Now consider the hyperplane MATH of REF which can be regarded as a hyperplane of MATH. Let MATH such that MATH. MATH. We have that MATH so that MATH. This shows that the osculating hyperplane to MATH at MATH passes through MATH REF . Since MATH, this is only possible when MATH. Thus we have proved that MATH contains no point different from MATH. We want to show next that the divisor MATH of MATH is MATH. To do this we have to show that MATH where MATH denotes the valuation at MATH, MATH, MATH is a local parameter at MATH and MATH. (Recall that MATH.) After a MATH-linear transformation of MATH we may assume that MATH and that MATH where MATH and MATH is the MATH-order sequence of MATH. Then we have to show that MATH. From REF we deduce that MATH . On the other hand, we claim that MATH. By REF MATH . From the definition of MATH it follows that MATH for almost one index MATH. Since MATH and MATH the only possibility is MATH, and MATH with MATH. The latter relation proves the claim. Now, this together with Eq. MATH yield that MATH. Hence, MATH of MATH is MATH from which the first part of REF follows. The second part follows from the Fundamental Equivalence REF . |
math/9911043 | Without loss of generality we may suppose that MATH. For MATH, let MATH with MATH. Note that MATH for MATH and that the matrix MATH has rank MATH. We prove that MATH is actually a Hermitian matrix over MATH. To do this, we re-write REF in the following manner: MATH . Taking into account the uniqueness of the MATH-tuple MATH proved in REF , comparison with REF gives MATH . Since MATH are linearly independent over MATH, this yields MATH for every MATH. This proves that MATH is Hermitian. After a MATH-linear transformation of MATH we may assume that the matrix MATH is the diagonal matrix with MATH units. Then REF becomes MATH and hence MATH lies on the Hermitian variety of REF . |
math/9911043 | By a way of contradiction, assume that MATH lies not only on MATH but also on the non-degenerate Hermitian variety MATH which is assumed to be the image of MATH by a non-trivial MATH-linear collineation fixing MATH. Choose any point MATH. Then the MATH and MATH have the same tangent hyperplane at MATH, as each of these tangent hyperplanes coincides with the osculating hyperplane to MATH at MATH. To express this geometric condition in algebraic terms, set MATH, and write the equations of MATH and MATH explicitly: MATH; MATH where MATH, and MATH is a non-singular non-identity unitary matrix of rank MATH. Then the above geometric condition in algebraic terms is that the homogeneous MATH-tuples MATH and MATH are equal up to a non-zero factor. Another meaning of the latter relation is that the non-trivial MATH-linear collineation associated to the matrix MATH fixes MATH pointwise. But this is impossible as MATH is not contained in a hyperplane of MATH; a contradiction which proves the theorem. |
math/9911043 | We show that MATH cuts out on MATH the divisor MATH. From REF , MATH . Writing the lower order terms in MATH, we have MATH . Hence MATH and equality holds if and only if MATH. We show that if MATH, then MATH. From REF , MATH . Thus, MATH. Since MATH for MATH, the claim follows. Since MATH is birational and MATH, we obtain MATH for every MATH, which shows the lemma for every MATH. For the case MATH, we also need to check that MATH. This inclusion occurs when MATH. Since the latter relation is a consequence of REF , the claim follows. Hence, MATH because MATH is birational and MATH, |
math/9911043 | Multiplying the last column by MATH and adding to it MATH times the first column plus MATH times the second column etc. plus MATH times the penultimate column gives MATH . Each element but the last one in the last column is actually MATH. In fact, this follows from the relation REF by derivation. Furthermore, the MATH-th NAME derivative of the same relation gives MATH and this completes the proof. |
math/9911043 | By CITE, MATH where MATH. Actually, MATH. In fact, MATH together with MATH would imply that the point MATH lies on MATH but this contradicts REF . Since MATH is a local parameter at MATH, we also have MATH, and the claim follows. |
math/9911043 | From the proof of REF we obtain the following result. For any point MATH, CASE: MATH if and only if MATH CASE: MATH if and only if MATH but MATH . On the other hand, MATH . Hence CASE: MATH if and only if MATH CASE: MATH if and only if MATH but MATH . Now, comparison with the previous result proves REF . |
math/9911043 | Take a MATH-rational point MATH. Since the number of MATH-rational points of MATH is MATH, there are at most MATH chords through MATH and another MATH-rational point of MATH. But, since MATH, the number of MATH-rational lines through MATH is at least MATH and hence one of these lines is neither a line contained in MATH, nor a tangent line to MATH at MATH, nor a chord through MATH and another MATH-rational point of MATH. Now, any MATH-rational point MATH outside MATH is a good choice for MATH. |
math/9911043 | Assume on the contrary that MATH meets MATH in a non MATH-rational point MATH. Then MATH is the line joining MATH and MATH. This implies that MATH is contained in the osculating hyperplane of MATH at MATH. Hence the common points of MATH with MATH are only two, namely MATH and MATH. But this contradicts the hypothesis that MATH. |
math/9911046 | The homogeneous flow on MATH is topologically transitive, that is, there exists a MATH whose orbit MATH is dense in MATH. We define a function MATH on this orbit by the formula MATH . We need to prove that MATH satisfies a NAME condition on MATH and hence can be extended to MATH as a NAME function. Given MATH, let MATH be such that MATH . We will show that MATH . As it is customary, MATH means that the expression is MATH for some constant MATH. We will use this notation to avoid an accumulation of constants. Since the homogeneous flow is not NAME, we cannot hope to be able to approximate any MATH - close piece of the dense orbit by a closed one as for the geodesic flow. Nevertheless, the close relation with the geodesic flow allows us to prove the following Using the notations above, for a piece of a dense orbit of REF , there is a lift of a closed orbit of MATH to MATH, MATH with MATH, such that for MATH . By REF the projection of MATH onto MATH is a dense orbit MATH of the geodesic flow MATH with MATH, and it follows that MATH . Let MATH and MATH be the corresponding pieces of orbits. Let us denote MATH, MATH and MATH. Since the geodesic flow MATH is NAME, by NAME Closing Lemma there exists a MATH such that MATH and its orbit MATH is closed, that is, MATH with the period MATH satisfying MATH. The point MATH can be chosen such that the leaf of the unstable foliation containing MATH, MATH, and a leaf of the stable foliation containing MATH, MATH intersect, and since they are transversal, they intersect in a point MATH, and MATH. Lift MATH to MATH. Since MATH its lift MATH intersects MATH, so let MATH, and similarly, MATH intersects MATH, MATH, and MATH (see REF which represents the picture in the direction transversal to the orbits). Since the distance between MATH and MATH decreases exponentially for MATH, and MATH, we have MATH and since all leaves are transversal we have the same estimate on MATH; MATH . Similarly, since the distance between MATH and MATH decreases exponentially for MATH, and MATH, we have MATH and the same estimate on MATH: MATH . Therefore, we have simultaneous estimates MATH and MATH . MATH. It follows from REF that MATH and since by the hypothesis MATH we obtain the required estimate REF . This proves the claim. Thus MATH can be extended from the dense orbit to a NAME function in MATH. Since MATH on the dense orbit, it follows that MATH is differentiable in the direction of the homogeneous flow and MATH in MATH. A similar argument shows that the function MATH is constant on MATH - cosets. For, let MATH. There are MATH and MATH on the dense orbit of MATH - close to MATH and MATH respectively. Then the projection to MATH is a MATH - close orbit of MATH. Find a closed orbit of MATH - close to the projection, and lift it back to MATH in the same manner we described before. It follows from the exponential estimates that MATH and as MATH, we obtain MATH. |
math/9911046 | Let MATH . Since the series REF converges absolutely, we can interchange summation and integration and, using standard ``NAME - NAME method" (see CITE p. REF), obtain MATH where MATH is a fundamental domain for MATH. We now make the change of variables MATH, where MATH is as in REF , mapping the ``standard geodesic" MATH into the axis of MATH. The following lemma is checked easily. MATH, where MATH. According to REF we have MATH . Let us denote MATH. Then MATH . Further calculations can be divided onto three parts: CASE: Show that MATH . CASE: Show that MATH . CASE: Show that MATH . CASE: MATH leaves invariant MATH. The fundamental domain MATH for MATH over which we take the integral can be described as MATH with the disc MATH of radius MATH ``attached" at every point MATH (this disc has only one common point with MATH). Change coordinates on the disc MATH: MATH . Then on the disc MATH, MATH . We have MATH so MATH where MATH is a holomorphic function of MATH. By the NAME integral formula MATH hence this integral does not depend on the value of MATH and, taking into account that MATH, we get MATH . CASE: Denote MATH. MATH leaves invariant MATH for any MATH. Repeat REF MATH times more (that is, totally we perform REF MATH times). It is proved by induction that MATH and the step MATH essentially repeats the argument in REF . For MATH we obtain MATH . CASE: The last integral is over MATH, the fundamental domain for a ``standard hyperbolic element" MATH of MATH acting on the unit disc in MATH. The change of coordinates MATH where MATH maps the unit disc to the upper half - plane with polar coordinates MATH, so that MATH is mapped onto the upper half - annulus MATH . Using formulas MATH we obtain MATH . But MATH hence MATH and we have MATH . The last integral does not depend on MATH as the integral of a MATH-invariant holomorphic function of MATH. For the same reason it does not depend the choice of the point MATH and the path from MATH to MATH in MATH. Hence we may take MATH, and since on MATH, and the last integral can be rewritten as MATH where the integration is over a segment of MATH. Using REF again, we go back to the integral over the axis of MATH, and obtain MATH . Finally we obtain MATH with MATH . |
math/9911046 | We make a change of variables MATH used in REF which maps the ``horizontal" geodesic to the axis of MATH in such a way that MATH. According to REF all lifts of the segment of the geodesic MATH are given by MATH, and all lifts of the corresponding segment of the ``horizontal" geodesic, by MATH. Using local coordinates REF on MATH we obtain MATH . The last two integrals are in MATH over the lifts MATH and MATH, and MATH, MATH and MATH, MATH are evaluated at the corresponding value of the parameter MATH: MATH, MATH, MATH. We used the cocycle identity to obtain the last equality: MATH . The required formula now follows from REF . Notice that since MATH is MATH - invariant, the integral is in MATH. |
math/9911046 | The uniform boundedness of MATH on MATH of finite volume would imply the required result. If MATH is compact, it follows from the NAME condition. If MATH is not compact, it is sufficient to show that MATH is bounded at each cusp MATH. The proof is similar to the proof for Fuchsian groups in CITE. It is based on an application of a partial NAME transform MATH described above. Let MATH be the geodesic in MATH given by MATH, MATH. Then MATH is the geodesic in MATH going to the cusp MATH. On MATH we have MATH, where MATH. Then on MATH, considered as an orbit of MATH, MATH may be estimated using REF for any MATH by the integral MATH over MATH, which is finite. The finiteness of the volume implies that for any MATH there exists a neighborhood of the cusp MATH such that MATH, and the uniform boundedness of MATH now follows from the NAME condition. |
math/9911046 | It follows immediately from the fact that has been, apparently, first pointed out in CITE (see also REF and CITE p. REF) that for any MATH . Here MATH is the linear combination of the partial derivatives MATH where MATH are coordinates on the bounded domain MATH and MATH. For MATH this can be easily checked by a direct differentiation along the orbit of the frame flow MATH using the decomposition MATH corresponding to the local coordinates REF with coefficients of MATH independent on MATH. Then MATH . Since MATH is holomorphic we have MATH . |
math/9911047 | Assume that MATH is Lagrangian in MATH for all MATH and choose a MATH-family of symplectomorphisms MATH such that MATH for all MATH. Define MATH and MATH; then MATH is the coefficient matrix of a symplectic system that corresponds to REF by the change of variables MATH. |
math/9911047 | See CITE. |
math/9911047 | For all MATH, the projection onto the second coordinate gives an identification between MATH and the space MATH (recall REF ). From REF, it follows easily that this identification carries the restriction of MATH to the restriction of MATH. The conclusion follows then from REF , observing that, by REF , MATH is non degenerate on MATH. |
math/9911047 | Since MATH can be uniformly approximated by smooth curves, there is no loss of generality in assuming that MATH is smooth. The first step is to prove that MATH can be approximated in the MATH-topology by smooth curves of NAME starting at MATH having intersections with MATH only in MATH and transversally. Let MATH be such that MATH does not intersect MATH. It follows from CITE that we can find a sequence MATH of smooth curves that are transverse to MATH for all MATH and such that MATH converges to MATH in the MATH-topology. Since MATH has codimension greater than one for MATH, transversality to MATH in facts implies that MATH does not intercept MATH. Hence, MATH has only transverse intersections with MATH. Clearly, each MATH can be extended to a smooth curve on MATH such that MATH in MATH and such that MATH converges to MATH on MATH in the MATH-topology. For MATH sufficiently large, it will also follow that MATH does not intercept MATH in MATH. It's now possible to construct a sequence MATH of smooth curves converging to MATH in the MATH-topology such that MATH for all MATH and such that each MATH projects into MATH by evaluation at MATH. We now set, for each MATH, MATH. Obviously MATH is a smooth curve in MATH and MATH tends uniformly to MATH. Moreover, denoting by MATH the upper-right MATH block of MATH, we see that MATH is invertible for MATH sufficiently large, since MATH tends uniformly to MATH and MATH is invertible. Similarly, since MATH is nondegenerate on MATH (recall REF ), MATH is nondegenerate on MATH for MATH sufficiently large. Therefore MATH is a set of data for the symplectic differential problem, and obviously MATH is its associated curve of NAME. It follows that MATH is nondegenerate. Finally, if MATH is not MATH-focal then MATH. Since MATH is open and MATH tends to MATH we see that MATH is not MATH-focal for MATH sufficiently large. |
math/9911047 | First, if MATH is a MATH-solution and MATH, then, using REF, we compute: MATH . Let MATH and let MATH be any MATH-solution such that MATH, so that MATH on MATH for some MATH. It follows from REF that MATH is negative definite on the infinite dimensional space of maps MATH, where MATH is in MATH and has support in MATH. |
math/9911047 | Consider the bilinear form MATH; recalling REF, from REF it follows easily that MATH is represented by a compact operator on MATH. It remains to prove that MATH is represented by a compact perturbation of a positive isomorphism of MATH. For each MATH, we define a positive definite inner product MATH on MATH by setting MATH where MATH is the orthogonal projection with respect to MATH. We have: MATH and it follows: MATH . Clearly, the integral of the first term above gives a NAME space inner product in MATH, and it is therefore represented by a positive isomorphism of MATH. Using again REF , to conclude the proof it suffices to show that the linear map MATH is continuous in the MATH-topology of MATH. To this aim, it clearly suffices to show that the maps MATH are MATH-continuous for all MATH. This follows from the fact that, for MATH the quantity MATH defined by: MATH is constant (see REF ). Using REF and integration by parts, we see that the integral in REF is continuous in MATH with respect to the MATH-topology. Integrating REF on MATH, we see that the functional MATH is MATH-continuous, which concludes the proof. |
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