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math/9909168 | Suppose MATH consists of an infinite number of artinian monomial ideals, which are noncomparable with respect to inclusion. Choose MATH. Since MATH for MATH, each MATH contains some of the finite number of standard monomials of MATH. There are thus an infinite number of ideals in MATH which contain the same set of standard monomials of MATH. Call this collection MATH. Let MATH be the intersection of the ideals in MATH. We will now build a strictly ascending chain of monomial ideals. Suppose MATH and MATH have been chosen. Choose an ideal MATH. We can again find an infinite collection of ideals in MATH which have the same non-trivial intersection with the standard monomials of MATH. Let MATH be this collection, and let MATH be the intersection of the ideals in MATH. We have MATH, since MATH contains some standard monomials of MATH, so in this fashion we get an infinite ascending chain of monomial ideals in MATH, which is impossible. |
math/9909168 | Since MATH is NAME, MATH contains maximal ideals. There are only finitely many maximal ideals by REF , so set MATH to be a maximal ideal in MATH containing an infinite number of ideals of MATH, and repeat, setting MATH. |
math/9909168 | Every associated prime of a monomial ideal is a monomial prime, of which there are only a finite number. We can thus restrict to an infinite collection of MATH all of which have the same set of associated primes, which we will also call MATH. Now for each ideal in this set we find an irredundant primary decomposition, writing the ideal as the intersection of monomial ideals primary to an associated prime in such a way that each associated prime is used only once. Let MATH be the primary component of MATH primary to the monomial prime MATH, where MATH. For a fixed MATH either MATH is finite, so there is an infinite number of MATH with the same MATH, or we can apply REF to the polynomial ring MATH. In either case we get an infinite collection MATH of ideals in MATH such that MATH, where the inclusions need not be proper. Since there are only a finite number of associated primes, by appropriate restrictions we can find a sequence MATH such that MATH for each MATH such that MATH is an associated prime. But since MATH is the intersection of the MATH, where MATH ranges over all over associated primes MATH of MATH, this means that MATH, where the inclusions are proper, since the MATH are all distinct. |
math/9909170 | Clearly, the induced subgraph on the vertex set of any facet is an NAME. For the converse let MATH be an NAME. There is a good acyclic orientation MATH of MATH such that MATH is initial with respect to MATH. Choose a sink MATH among the vertices of MATH. By assumption there is a facet MATH with MATH. Because MATH is good, the vertex MATH is the unique sink of MATH. Moreover, all the vertices of MATH are contained in MATH because MATH is initial with respect to MATH. Comparing MATH with the NAME MATH corresponding to MATH yields MATH due to the minimality of MATH. |
math/9909170 | Consider all possible acyclic orientations of the graph MATH. In view of REF it suffices to exhibit all good acyclic orientations of MATH. Fix an acyclic orientation MATH and a vertex MATH. We want to compute the number MATH of faces in which the vertex MATH is a sink with respect to MATH. This is the number of faces containing MATH built from edges incoming at MATH only. All faces through MATH can be enumerated because the face lattice of the vertex figure is known, compare REF . Filter these faces for the incoming edge condition. Note that for MATH simple the number MATH solely depends on the in-degree of MATH with respect to MATH, which is not true for general polytopes. In particular, we do not have a formula in closed form. Let MATH and let MATH be the number of non-empty faces. We proceed as NAME in his proof. As each non-empty face has at least one sink we have MATH. Further, MATH is good if and only if MATH. Because good acyclic orientations always exist we have that MATH. |
math/9909170 | No cube contains any odd cycle. Assume there is a constructible cubical complex MATH which contains an odd cycle MATH and which is not a cube. Then there are constructible cubical complexes MATH such that MATH. By an induction on the number of construction steps necessary to build up MATH we can assume that MATH is contained in neither MATH nor MATH. That is, MATH passes through the intersection MATH, which is connected, because of our assumption on the dimension. It is conceivable that MATH enters and leaves MATH several times. Pick two vertices MATH on MATH in MATH such that neither half of MATH, obtained from cutting at MATH and MATH, is contained in MATH. If there is only one vertex in the intersection, then either MATH or MATH contains an odd cycle and we are done. Choose a path MATH in MATH between MATH and MATH. Depending on the parity of the length of MATH, combining MATH with either half of MATH yields an odd cycle MATH. Observe that MATH enters and leaves the intersection one times less than MATH. An obvious induction now gives the result. |
math/9909170 | Assume MATH is a ridge, compare REF . Choose an edge MATH in MATH which is not contained in MATH or MATH. Let MATH with MATH and MATH. There is a unique edge MATH through MATH which is contained in MATH, but not contained in MATH. Now choose a linear objective function MATH such that: CASE: the edge MATH is the unique maximal face of MATH with respect to MATH, and CASE: the function MATH attains the same value on MATH and at MATH. Let MATH be the maximal face of MATH with respect to MATH. Clearly, MATH contains the edge MATH. If MATH were strictly greater than MATH, then the intersection MATH would be strictly greater than MATH. This would contradict the maximality of MATH. We have MATH. Denote by MATH the vertex of MATH which is contained in MATH. Let MATH be the maximal face of MATH with respect to MATH. Now MATH contains the points MATH and MATH. As in the argument above, MATH contradicts the maximality of MATH in MATH. We conclude that MATH is an edge. The vertices MATH, MATH, MATH form a triangle in the graph of the cubical polytope MATH. Due to REF we arrive at the final contradiction. |
math/9909170 | Consider the set MATH of all the ridges contained in neighbors of MATH in MATH, but which are disjoint from MATH. By assumption the sublattice of the face lattice of MATH generated by MATH is isomorphic to the face lattice of a MATH-cube. We have to show that the ridges of MATH are the facets of a convex (combinatorial) cube, that is, we have to show that the ridges in MATH are contained in a hyperplane. A capped cubical MATH-polytope can also be seen as a MATH-dimensional (constructible) cubical complex. The facets of this complex are the (combinatorial) cubes which are attached one by one. Each facet of a capped cubical polytope is contained in a unique facet of the complex, that is, a unique MATH-cube. Say, MATH is a facet of the MATH-cube MATH. Then the ridges in MATH are contained in the unique facet of MATH which is opposite to MATH. |
math/9909171 | The fixed points of the hyperelliptic involution are the MATH . NAME points of MATH. The projection MATH exhibits MATH as a double cover of MATH, ramified at the roots of MATH; thus, its genus equals MATH. On the affine variety MATH, we have MATH; thus MATH. Since the functions MATH and MATH have no common zeroes (the polynomial MATH has no multiple roots), we conclude that the differentials MATH are regular on MATH, so long as MATH. Let MATH. On the affine variety MATH, we have MATH. Since MATH the differentials MATH are regular on MATH so long as MATH. We have exhibited MATH linearly independent algebraic one-forms on MATH; since MATH has genus MATH, they form a basis of MATH. |
math/9909171 | Under the action of MATH, MATH transforms into MATH . Expanding the right-hand side in terms of the basis MATH, we recover the action of MATH on MATH. |
math/9909171 | MATH . |
math/9909171 | All of the following calculations are performed in the ring MATH. Using the formulas MATH and MATH, we see that MATH . The third factor of the denominator may be simplified by the factorization MATH . To simplify the factors involving the variable MATH, we use the formulas MATH and, similarly, MATH . Hence MATH . No representation of MATH of the form MATH with MATH odd can have a non-trivial space of invariants; we conclude that we may discard the terms which are odd in MATH before taking the space of invariants under the group MATH. |
math/9909172 | The second statement follows immediately from NAME 's characterizations of lattice crosspolytopes (compare REF , MATH), whereas REF . is contained in a more implicitly way in his paper. However, from the work of NAME we know (compare REF , MATH) that in the first case, that is, MATH, the lattice MATH is admissible if MATH as well as all points arising from permutations of the coordinates of the above points are not contained in MATH. Therefore, in order to prove REF. we have to show that the points MATH, MATH, MATH and the corresponding permutations are not contained in MATH. To this end let MATH be the distance function of MATH, that is, MATH. Then MATH and since MATH the function MATH describes a norm on MATH. Since MATH we find MATH . Obviously, the same inequalities hold for the points given by all permutations of the coordinates of the points of the left hand side and thus MATH . |
math/9909172 | For a fixed point MATH let MATH. Then we have MATH and thus MATH is the shadow boundary of MATH in direction MATH. Hence two points of MATH can be connected by a continuous path. Now let MATH . Then, by construction, the set MATH is edge-connected and so it is the union MATH. |
math/9909172 | Let MATH and let MATH be any hyperplane containing MATH and the origin. Then there exists a facet MATH having a nonempty intersection with MATH. Using the edges of MATH we can easily determine all facets MATH, MATH, with this property. Since the polytope has MATH edges this can be carried out in time MATH. For each MATH the boxes MATH, MATH, intersecting MATH can be determined in time MATH by well know methods from computational geometry about range searching (compare CITE). For each possible choice MATH we use REF to verify whether MATH and thus MATH. Now each REF is a feasibility problem of Linear Programming with MATH constraints in dimension MATH and this can be solved with MATH arithmetic operations (compare CITE, pp. REF). |
math/9909172 | By REF the algorithm finds all MATH and at the end we have MATH. Furthermore, at the end of each loop GREF. the set MATH coincides with MATH. To find a first facet MATH we need by REF at most MATH operations. For the estimation of REF . and GREF. we can proceed as in the proof of REF . The boxes in REF . can be found in time MATH and in REF . we have to solve at most MATH feasibility problems (compare REF) with at most MATH constraints in dimension REF which can be done with MATH arithmetic operations. Finally, we observe that for each facet MATH the loop GREF. is executed at most MATH times. |
math/9909172 | We project MATH by the radial projection MATH onto the unit sphere MATH and we carry out all calculations on MATH. It can easily be checked that our asymptotical estimates remain correct. The facets of MATH will be denoted by MATH, MATH, and in the following we shall denote by MATH certain positive constants. First we observe that the uniformity of our sequence implies that for the spherical diameter MATH and the spherical area MATH of the facets holds MATH . Next we note that for a point MATH we have MATH . Now let MATH, MATH be two facets of MATH such that MATH and let MATH. We find by REF that MATH . As the spherical area of the set on the right is bounded by MATH we find by REF that for fixed MATH there are at most MATH faces MATH such that MATH. For two faces MATH and MATH we have MATH. Proceeding as before we see that there can be at most MATH faces of MATH which intersect MATH. Altogether we have found that MATH . Next we ask for the efficiency of REF to determine these sets. Let MATH be a minimal box containing MATH. By the above estimate for the circumradius of two facets MATH we have MATH for each coordinate MATH. Again using the area bound REF we see that there are at most MATH facets MATH with MATH (or MATH) and MATH (or MATH ) (compare REF). Hence the number MATH of REF is of order MATH. Next we note that on account of our assumptions the maximal number MATH of neighbours of a given facet is constant, if MATH is large enough. Moreover, by REF we also see that the number MATH is of order MATH and therefore REF determines the set MATH as well as the sets MATH in time MATH (compare REF ). Since we have already proven that MATH and that for two given facets MATH there are only MATH many facets MATH with MATH, it remains to show that MATH for MATH (compare REF ). Let MATH. Then we have (compare REF) MATH . The radial projection of the latter set is a spherical parallelogram with ``edge length" MATH. Hence its spherical area is MATH and together with REF this shows MATH. |
math/9909172 | Since the vectors MATH correspond to facets defining hyperplanes the vectors MATH are linearly independent. Otherwise we can assume that MATH and we get MATH. Hence the vectors MATH are linearly dependent, because otherwise MATH. In the same way we find that MATH and MATH are linearly dependent. Thus we can find three nontrivial vectors MATH such that MATH where MATH denotes the orthogonal complement of MATH. Now let MATH with MATH, and let us assume that MATH is a local minimum. By the choice of MATH we have MATH for all MATH and therefore let MATH . It is easy to see that this function has a local extremum at MATH if and only if it is constant, that is, MATH. In particular we have MATH. Since MATH are linearly independent this implies that MATH or MATH belongs to MATH, and in the same way we find that MATH or MATH lies in MATH and MATH or MATH belongs to MATH. However, since MATH this yields the contradiction MATH. |
math/9909172 | Let MATH, MATH. Since there exists a path from MATH to MATH in MATH and since the gradient vanishes on MATH we have MATH. For MATH let MATH. The function MATH is an univariate polynomial in MATH of degree at most three with MATH and the derivative vanishes at MATH and MATH. Thus MATH for all MATH. Since MATH is assumed to be a local extremum there exists a MATH such that MATH is a local extremum, too. Hence we have found three points on the line MATH where the gradient of MATH vanishes. Since all partial derivatives are polynomials of total degree at most two we have shown that MATH for all MATH. |
math/9909172 | Obviously, if MATH is factor of MATH then the statement holds. Without loss of generality let MATH and MATH and let MATH be the polynomial whose coefficients are recursively defined by MATH . Multiplication of MATH and MATH yields MATH. Since MATH, MATH we must have MATH. |
math/9909172 | Let MATH be a linear factor of MATH and let us assume without loss of generality that MATH. Let MATH and MATH. By REF we have MATH and for the points MATH we get MATH and hence MATH, MATH. By the choice of the vectors MATH each linear system has an unique solution. Finally, we remark that the MATH-linearly independent points MATH can be found quite easily. If the leading coefficient of MATH with respect to MATH is constant, then we set MATH, MATH, and MATH, where MATH denotes the MATH-th unit vector. Otherwise we perturb these points a bit. |
math/9909172 | Without loss of generality let MATH. By REF is a factor of MATH if and only if there exists a polynomial MATH, MATH, such that MATH. Hence by comparing the coefficients we get MATH . Since MATH the first MATH identities determine uniquely the coefficients MATH, MATH, and the remaining identities can be used as a verification whether MATH is really a factor of MATH, that is, whether MATH. |
math/9909172 | Since MATH is a conic section, the set of all isolated affine subspaces of MATH consists either of one or of two lines or of an isolated root or it is empty. Using well know formula for conic sections one can determine all of them. |
math/9909172 | Let MATH, MATH and MATH. On account of REF we determine all linear factors of MATH and the set MATH containing MATH. Hence it remains to determine the isolated roots, that is, MATH. Since none of the isolated roots lies in a MATH-dimensional subspace of MATH, we can divide MATH by the linear factors corresponding to the elements of MATH. Therefore, in the following we assume that MATH has no linear factors. If MATH is an univariate polynomial in MATH, say, having a real root MATH, then MATH would be a linear factor of MATH. Thus if MATH is an univariate polynomial we set MATH. Hence let MATH. If MATH is an isolated root then MATH and thus we can look for all isolated roots of the set MATH, where MATH denotes the partial derivative of MATH with respect to MATH, that is, MATH with MATH, MATH. Now let MATH be the resultant of these two polynomials. If MATH then we determine MATH and afterwards MATH. Observe, that or any fixed MATH the polynomial MATH can not vanish, because otherwise it contains a linear factor (see REF ). In this case we set MATH. It remains to consider the case MATH and therefore we may assume MATH. By REF we know that MATH and MATH have a common factor MATH which has positive degree in MATH. NAME more, it is not hard to see that MATH implies that MATH is divisible by some MATH, MATH with MATH. Hence, if MATH then MATH can be written as a product of polynomials MATH with MATH which shows that MATH has no isolated roots. So let MATH. Then the common factor MATH of MATH has degree REF or REF with respect to the variable MATH. If MATH than it is a linear factor, because the leading coefficient MATH is a constant. Also, if MATH than MATH is a linear factor of MATH. Hence MATH and it can be written as MATH with MATH, MATH. Thus MATH is a linear factor and we can determine MATH by the following procedure: For each linear factor MATH of MATH let MATH. Then we have to test whether MATH is a factor of MATH. This can be done with the algorithm described in REF . So we can assume that we have found the factor MATH of MATH with MATH. Let MATH. Then the problem is reduced to the determination of the isolated roots of the two polynomials MATH and MATH, which can be solved by REF . Hence we set MATH. |
math/9909172 | We proceed as in the proof of the last lemma. First we determine all common linear factors of these two polynomials and the set MATH containing MATH (compare REF ). Then we divide the polynomials by these common linear factors and it remains to find the isolated roots MATH. To this end we may assume that MATH. Let MATH. If MATH then we determine MATH, MATH and we set MATH. Therefore we can assume MATH. Since MATH we know that if the polynomial MATH is not irreducible then it has a linear factor MATH, say. Using REF we can determine such a linear factor MATH as well as the remainder MATH. Next we split our system in two systems, namely MATH . Since MATH is a linear factor and MATH and MATH are assumed to be free of common linear factors the second system can be solved by substituting one variable in MATH via MATH. Since MATH we can apply recursively the previous argumentation to the system MATH. Thus, if MATH is not irreducible we set MATH. If MATH is irreducible, and since MATH, the polynomial MATH has to be a factor of MATH. Hence the determination of MATH is reduced to the calculation of the isolated roots of MATH. With the algorithm of REF we can find a set MATH containing MATH and we set MATH. |
math/9909172 | Without loss of generality let MATH and let MATH. By MATH we denote the orthogonal projection of MATH onto the plane MATH. Obviously, MATH is an affine subspace and by the definition of resultants we have MATH for all MATH. Since MATH we have MATH. If MATH, then MATH itself corresponds to a common linear factor of MATH and MATH and thus MATH. Therefore let MATH and so MATH is either an isolated MATH-dimensional subspace or an isolated root of MATH. If MATH is contained in a MATH-dimensional subspace of MATH the statement is certainly true. Thus we may assume that MATH for every common linear factor MATH of MATH and MATH and we have to show that MATH is an isolated root of MATH. Suppose the contrary and let MATH such that there exists a path MATH, MATH, connecting MATH and MATH. Then MATH, because otherwise MATH is a MATH-dimensional set containing MATH. Hence we can assume MATH, MATH, and by REF there exist MATH such that MATH, MATH. Since MATH and MATH we get MATH. However this shows that MATH is not an isolated affine subspace of MATH. |
math/9909172 | We set MATH and apply REF . |
math/9909172 | First we note that after some scaling, subtractions, and renumbering we may assume that MATH is given by the following system MATH of polynomial equations (see REF MATH where MATH, MATH are linear polynomials and MATH with MATH. We may further assume that MATH. Depending on the number of non-trivial polynomials and the number of variables in MATH we have to distinguish several cases. Obviously, if all polynomials in REF vanish then we have MATH and we have to do nothing. CASE: MATH consists of one or two or three polynomials in only one variable. Then we can determine MATH by any algorithm computing the roots of an univariate polynomial. CASE: MATH consists of one polynomial in two variables. Without loss of generality let MATH, with MATH. Via the algorithm of REF we can determine MATH and we set MATH. CASE: MATH consists of two polynomials in two variables. Without loss of generality let MATH. Using the Algorithm of REF we can determine a set MATH (compare REF and we set MATH. CASE: MATH consists of three polynomials in two variables. Without loss of generality let MATH. We may assume that both variables occur in all three polynomials and that the polynomials are linearly independent. Otherwise we can reduce this case to one of the previous ones. First, by REF we determine all common linear factors and the set MATH containing MATH and hence we may assume that MATH have no common linear factors and both variables occur in the polynomials. Next we compute the resultant MATH. If MATH then we determine MATH. Observe that for a fixed MATH not all three polynomials MATH can vanish, because otherwise they have a common linear factor. In this case we set MATH. It remains to consider the case MATH. Since MATH, MATH are assumed to be linearly independent and since MATH the common factor (compare REF) has to be a linear polynomial MATH which can be determined via REF . Then we consider the two systems MATH . Both systems are free of common linear factors and since both systems contain linear polynomials we can easily determine MATH and MATH by substitution. We set MATH. CASE: In the remaining cases we first determine the set MATH and therefore we may always assume that the given partial derivatives have no common linear factors. CASE: MATH consists of one polynomial in three variables. a ) Let MATH . Then we may assume MATH and the dimension of any affine subspace of MATH is REF. Therefore we set MATH. b ) Let MATH. Let MATH. For every MATH we have MATH and hence MATH. If MATH then we can always find a neighbourhood MATH of MATH such that MATH for all MATH and MATH belongs to a REF-dimensional connected component of MATH. Hence we may set (compare REF ) MATH . CASE: MATH consists of two polynomials in three variables. Then we may assume without loss of generality that the variable MATH occurs in both polynomials. a Let MATH with MATH, MATH and we can assume that MATH are linearly independent. Let MATH be the resultant of these two polynomials with respect to MATH. Since the polynomials are linearly independent, MATH, and since we have assumed that they have no common linear factors the resultant MATH can not vanish. By REF is a polynomial of total degree at most REF. By REF we can find a set MATH containing MATH and for each MATH we consider the system MATH . Since MATH is a MATH- or MATH-dimensional affine subspace, MATH is a system of at most two polynomials in at most two variables. Hence by the previous cases we can determine a set MATH containing MATH and we set MATH . On account of REF we have MATH. b Let MATH with MATH. Then we can proceed as in the case above. The only difference is that the total degree of MATH is at most REF. CASE: MATH consists of three polynomials in three variables. Then we may assume without loss of generality that the variable MATH occurs in at least two of the polynomials, MATH compare REF and that each of the three polynomials contains at least two variables. Now we compute MATH and MATH. CASE: MATH. By REF we know that MATH and MATH have a common factor with positive degree in MATH. Thus MATH, MATH are either linearly dependent or they have a common linear factor. If they are linearly dependent we can proceed as in case MATH REF. So let MATH be a common linear factor and let MATH, MATH. Observe, MATH are linear polynomials. Next we consider the two systems MATH . Since MATH it suffices to consider MATH. Since both systems contain linear polynomials we can reduce them to systems in at most REF variables which can be handled by one of the methods described in one of the previous cases. CASE: MATH. It is not hard to see that also in this case we can split our system in two systems containing linear polynomials and we can proceed as before. CASE: MATH and MATH. Since MATH and MATH we can use the algorithm of REF in order to determine a set MATH containing MATH. Now, for each MATH let MATH . Since MATH is a MATH- or MATH-dimensional affine subspace MATH is a system of at most three polynomials in at most two variables. Hence by the previous cases we can determine a set MATH containing MATH and we set MATH . On account of REF we have MATH. |
math/9909172 | For simplification we assume that MATH, MATH. With MATH, MATH, we may write MATH for suitable matrices MATH. Now for MATH and for MATH let MATH. If MATH we set MATH. According to REF we have to find a MATH such that MATH with MATH in case I., MATH in case II., and MATH in the cases III. and IV. Furthermore in the cases I. and II. we have the additional restrictions MATH . Let us denote by MATH, MATH, the vectors of the test set MATH. This means we have MATH if MATH, and MATH otherwise. Observe, by construction we know that all vectors MATH lie in supporting hyperplanes of the polytope. Therefore, if MATH we can verify REF by just checking the facet defining inequalities of the polytope for the corresponding points MATH, etc. Thus in the following we assume MATH. Then we define a set MATH by (compare notation at the beginning of this section) MATH . Using standard methods from Linear Programming we can easily decide whether MATH or we can find a point MATH. If MATH then the affine subspace does not contain an admissible lattice. So let MATH. In the cases III. and IV. or if MATH also satisfies REF in the cases I. and II., we are done and we have found an admissible lattice. So let us assume that we are in case I. or II., MATH and MATH violates REF. The most simple way to decide whether there exists a MATH satisfying REF is the following: In case I. we consider for MATH the sets MATH . In REF we set for MATH . Obviously, in the case I. (case II.) there exists an admissible lattice in the affine subspace MATH if and only if there exist MATH REF such that MATH (MATH). Again, using tools from Linear Programming we can either find a MATH in one of sets MATH (MATH), and thus an admissible lattice MATH, or we know that MATH contains no admissible lattices. |
math/9909172 | REF . |
math/9909176 | In the basis MATH, the commutation relations of MATH have the form, MATH . The computation of the three terms entering the NAME bracket yields MATH whence MATH. |
math/9909176 | The proof of the first two equalities is by computation, using the formula MATH and the derivation property of the algebraic NAME bracket. To prove the third equality, we use REF to obtain MATH . The first term vanishes by the graded NAME identity and the second term vanishes because MATH is MATH-invariant. |
math/9909176 | Since MATH, and since left- and right-invariant vector fields commute with each other, we obtain MATH . Using REF and the fact that MATH is MATH-invariant, the right-hand side is MATH. Since MATH, we arrive at REF . |
math/9909176 | Here and below MATH is identified with MATH. By REF , MATH . To conclude we use the fact that MATH and, again, the fact that left- and right-invariant vector fields commute. The proof for right-invariant vector fields is similar. |
math/9909176 | Maximal isotropic subspaces in MATH form a NAME which we denote by MATH. Since MATH is a connected NAME group, the subspaces MATH and MATH belong to the same connected component of MATH. Let MATH be an isotropic complement of MATH in MATH. The NAME MATH being an algebraic variety, the set of isotropic complements MATH of MATH in the connected component of MATH is a NAME open set. Since MATH is in the same connected component of MATH as MATH, the set of isotropic complements to MATH in the connected component of MATH is also a NAME open set. An intersection of two nonempty NAME open sets being nonempty, one can always find a MATH which is an isotropic complement of both MATH and MATH. Any such subspace is admissible at the point MATH. |
math/9909176 | We choose an isotropic complement MATH of MATH in MATH, admissible in an open neighborhood MATH of MATH. Let MATH be an element in MATH. We apply the map MATH, where MATH, to the MATH-form MATH. By definition, we obtain, MATH which proves REF . The complement MATH being admissible, the MATH-forms MATH form a basis of the cotangent space to MATH at each point in MATH. Hence, REF gives a characterization of the bivector MATH on MATH. |
math/9909176 | The proof is identical to that in the case of NAME actions. See CITE CITE. |
math/9909176 | First, we show that the bracket of MATH-invariant functions, MATH, is MATH-invariant. Indeed, MATH for any MATH, because MATH and, hence, MATH annihilates MATH. Next, we observe that the bracket defined by MATH on invariant functions is a NAME bracket, MATH because MATH and MATH and MATH are MATH-invariant. Finally, if one modifies the complement MATH to MATH, the bivector on MATH is modified by a twist, MATH, and the NAME bracket of invariant functions is unchanged, MATH because MATH annihilates invariant functions. |
math/9909176 | To prove the proposition, we use both REF and the above definition of the moment map and its equivariance. Thus, at point MATH such that MATH, MATH . Here we have used the equivariance property of the moment map and REF , which imply that, for any MATH in MATH, MATH when MATH, and hence MATH . The proposition is therefore proved. |
math/9909176 | At each point MATH, let us apply the construction of REF , with MATH, to obtain an admissible complement by means of which we formulate the definition of the moment map. The bivector on MATH, after the twist MATH, is MATH . If MATH is in the orthogonal complement of the kernel of MATH, then MATH, and MATH. Taking into account REF, we see that REF is then equivalent to the definition MATH . For MATH, the moment map conditions are MATH . We replace MATH by their values found in REF . Using the equivariance of the moment map, and the fact that MATH, for MATH, we find, for all MATH, MATH . Thus the moment map condition, for MATH, reduces to MATH and therefore it coincides with REF. |
math/9909176 | Let MATH for an admissible MATH. Let MATH and MATH be arbitrary functions on MATH, and let MATH denote the vector field on MATH such that MATH for any MATH-form MATH on MATH. Then MATH . Since MATH, and since the tangent space to the MATH-orbit at each point is contained in the distribution MATH, we conclude that MATH. |
math/9909176 | By the definition of the moment map, for all MATH, MATH while, by the characteristic property of MATH, MATH. Thus, we see that MATH follows from the equivariance condition, MATH, and from the fact that the MATH-forms MATH span the cotangent space to MATH. |
math/9909176 | We assume that MATH is nondegenerate and we let MATH be a MATH-form in the kernel of MATH. Then, MATH . Since MATH is admissible, there exists MATH, such that MATH. (If MATH, we can set MATH.) By the nondegeneracy of MATH and the definition of the moment map, MATH. Applying REF, we obtain MATH, where MATH. The equivariance of the moment map implies that MATH, which in turn implies that MATH. If both complements MATH and MATH are admissible at MATH, the operator MATH is invertible, and therefore from MATH, we obtain MATH. Since MATH, the vector field MATH vanishes. Therefore MATH, and MATH is nondegenerate. |
math/9909176 | The proof is analogous to that in the case of NAME actions. See, for example, CITE . Let MATH be as stated and let MATH be its projection in MATH. Let MATH be the symplectic leaf passing through MATH in the NAME manifold MATH. We choose any complement MATH admissible at MATH. We need to prove that the projection of the tangent space to the level submanifold, MATH, coincides with the tangent space to the symplectic leaf, MATH. We denote the NAME bivector which is locally defined on MATH near MATH by MATH. Let MATH be a vector in MATH. By definition, MATH, where MATH is a MATH-form in MATH, and we can also represent MATH as MATH. Vectors MATH that are tangent to the level submanifold MATH are characterized by the property MATH for any MATH. The bivector MATH being nondegenerate, any vector MATH tangent to MATH at MATH can be represented as MATH for some MATH. Then MATH is tangent to MATH if and only if MATH, for all MATH, in other words, the MATH-form MATH is the inverse image of a MATH-form MATH, MATH. We conclude that MATH which proves the theorem. |
math/9909180 | REF tells us that MATH . From REF of MATH, we can write this as MATH . If we define MATH and use REF of MATH, REF becomes MATH where we have used REF to estimate the final sum in REF . It therefore suffices to show that MATH . The inner sum in REF can be evaluated by REF , yielding MATH . Using the asymptotic formula given by REF for the main term in this last expression and the estimate in REF to bound the error term, we see that MATH which establishes REF and hence the theorem. |
math/9909180 | We remark that it suffices to show that REF holds when MATH is sufficiently large, by adjusting the constant implicit in the MATH-notation if necessary. Recall that the content of a polynomial MATH is the greatest common divisor of all of its coefficients (so that a polynomial is primitive precisely when its content equals REF). Since MATH is integer-valued, the coefficients of MATH are all rational, so we can choose a positive integer MATH such that MATH has integer coefficients. Write MATH where MATH is the content of MATH, the MATH and MATH are positive integers, and the MATH and MATH are distinct primitive, irreducible polynomials with positive leading coefficients satisfying MATH for MATH and MATH for MATH. Set MATH, so that MATH is a balanced polynomial with integer coefficients. If MATH is an integer such that MATH, then it is clear from REF that the largest prime factor MATH of MATH is the same as the largest prime factor of MATH provided that MATH exceeds all prime divisors of MATH. In particular, as long as MATH then MATH is MATH-smooth precisely when MATH is MATH-smooth. But MATH and each MATH, so we see that REF always holds when MATH is sufficiently large (in terms of MATH and MATH), since MATH. Therefore MATH the error arising from values of MATH for which MATH, of which there can be at most MATH. Now we choose a positive real number MATH such that for each MATH, we have MATH and MATH for all MATH. If we set MATH, then MATH is again a balanced polynomial with integer coefficients, and moreover MATH is effective by our choice of MATH. Finally, we see that MATH . This together with REF establishes the lemma. |
math/9909180 | We shall strive first for admissibility and then for exclusiveness (as if we were climbing the social ladder). If MATH is a prime greater than MATH, then the primitivization MATH of MATH is not the zero polynomial (mod MATH), since any primitive polynomial has at least one nonzero coefficient modulo every prime; and in fact MATH has degree at most MATH. Therefore MATH has at most MATH zeros (mod MATH), and so takes at least one nonzero value (mod MATH). If MATH is a prime not exceeding MATH, then from the identity REF applied to MATH, we see that the content of MATH is MATH . Thus for any prime MATH, MATH where all of the terms of the form MATH are nonnegative since the quantities involved are all integers. By general properties of the discriminant of a polynomial, we know that if MATH divides both MATH and MATH, then MATH divides MATH. Put another way, MATH where the last inequality is one of our hypotheses on MATH. In particular, one of the first two terms on the right-hand side of REF is less than MATH, and hence REF can be simplified to MATH . If MATH, then MATH for some integer MATH that is not divisible by MATH, and so MATH . On the other hand, if MATH, then MATH for some integer MATH that is not divisible by MATH, and so MATH . In either case we see that MATH takes a nonzero value (mod MATH), and so MATH is admissible. To show that MATH is exclusive we want to show, given two distinct irreducible factors MATH and MATH of MATH, that MATH and MATH have no common zeros modulo any prime. Note that any irreducible factor MATH of MATH is the primitivization of an irreducible factor MATH of MATH, by NAME 's lemma on the contents of polynomials with integer coefficients; and any irreducible factor MATH of MATH has the form MATH for some irreducible factor MATH of MATH. Similarly, there is an irreducible factor MATH of MATH such that MATH is the primitivization of MATH. If MATH is a prime not dividing MATH, then MATH and MATH have no common zeros (mod MATH) by general properties of the resultant of two polynomials, whence the same is clearly true for MATH and MATH and thus for MATH and MATH; so it suffices to consider the case where MATH divides MATH. Again, by general properties of the resultant of two polynomials, we know that if MATH divides both MATH and MATH, then MATH divides MATH. Put another way, MATH . By exchanging the MATH-s and MATH-s if necessary, we can assume without loss of generality that MATH where the last inequality is another of our hypotheses on MATH. From the equation analogous to REF for MATH, we see that REF implies that MATH. Therefore MATH for some integer MATH that is not divisible by MATH, and so MATH independent of MATH, so that MATH has no zeros (mod MATH) whatsoever. Hence certainly MATH and MATH have no common zeros (mod MATH), which shows that MATH is exclusive. |
math/9909180 | Let MATH be an integer-valued polynomial, let MATH be the number of distinct irreducible factors of MATH, and let MATH, , MATH be the degrees of these factors. Let MATH, and let MATH be the number of distinct irreducible factors of MATH whose degree equals MATH. Let MATH and MATH be real numbers satisfying MATH; let MATH be a real number and set MATH. We are trying to show that the asymptotic REF holds. If MATH then MATH, in which case MATH. Thus we are trying to prove that MATH holds uniformly for MATH and MATH, where here and throughout this proof, all constants implicit in MATH-notation may depend on MATH and MATH. Notice that MATH where MATH. Therefore, by REF we can find an effective polynomial MATH with integer coefficients that is the product of MATH distinct irreducible polynomials of degree MATH, such that MATH . Let MATH be the integer given by REF , so that MATH satisfies the hypotheses of REF . For each fixed integer MATH, define the polynomial MATH, and let MATH be the primitivization of MATH. Each of these polynomials MATH has integer coefficients, and is balanced and effective because MATH is. Moreover, by REF each MATH is admissible and exclusive as well. By the hypothesis of the proposition to be proved, we know that REF holds for each MATH. Therefore (assuming Hypothesis UH) we have MATH since each MATH, like MATH, is the product of MATH distinct irreducible polynomials of degree MATH. On the other hand, every integer MATH is congruent (mod MATH) to some integer MATH, and so every value MATH for MATH corresponds to a value MATH for some MATH. Furthermore, the corresponding value MATH simply equals MATH, a difference which does not affect whether the value is MATH-smooth as soon as MATH exceeds MATH. Therefore, when MATH is sufficiently large in terms of MATH, we have MATH where MATH . The function MATH satisfies MATH for all MATH, and so MATH. Using the asymptotic REF , we conclude that MATH . Together with REF , this shows that the asymptotic REF holds for the original polynomial MATH, which establishes the proposition. |
math/9909180 | This is simply inclusion - exclusion on the factors of MATH that, for a given argument MATH, have large prime divisors. For any integer MATH define MATH . Then for any nonempty subset MATH of MATH, REF of MATH for MATH is equivalent to MATH . But REF also implies that MATH since MATH is MATH-smooth if and only if each MATH is MATH-smooth. We therefore find that MATH which is equivalent to the statement of the proposition. |
math/9909180 | We recall REF of MATH: MATH . Since MATH for each MATH when MATH is sufficiently large, there is a unique MATH-tuple MATH of primes for each argument MATH that is counted by MATH. If we write the values MATH as MATH for suitable integers MATH, we see that MATH . Because MATH is exclusive, no prime can divide two different values MATH at the same argument MATH; therefore, we may insert the condition of summation MATH without changing the sum. Furthermore, if MATH divides MATH for each MATH, then MATH is congruent to some element MATH of MATH by its REF , and conversely. Therefore MATH . Making the change of variables MATH, we see that MATH by REF of the MATH. The polynomial MATH is effective and hence strictly increasing for MATH, so if we define MATH to be the smallest real number MATH such that MATH for each MATH, then we see that MATH . As for the size of MATH, we see from REF of MATH that MATH since each MATH has degree MATH, and so MATH. This establishes the proposition. |
math/9909180 | By definition, MATH . Since MATH does not divide MATH, we may multiply both sides of the latter congruence by MATH. Then making the bijective change of variables MATH, we see that MATH as claimed. |
math/9909180 | Write MATH for some integer MATH. Making the bijective change of variables MATH in the latter congruence in REF , we see that MATH as claimed. |
math/9909180 | Write MATH for some integer MATH. Since MATH does not divide MATH, we may multiply both sides of the latter congruence in REF by MATH. Then making the bijective change of variables MATH, we see that MATH as claimed. |
math/9909180 | By definition, MATH . Now every root of MATH is certainly a root of MATH, and so it follows that every MATH is congruent to some MATH. Therefore MATH from the last line of REF , which establishes the lemma. |
math/9909180 | Factor MATH into powers of distinct primes. We first note that by REF , MATH . For every MATH-tuple MATH such that each MATH, the NAME Remainder Theorem gives us a MATH such that MATH (mod MATH) for each MATH, and this correspondence is bijective. Therefore MATH since MATH only depends on MATH (mod MATH) by REF . This last sum now factors: MATH by the definition of MATH and by REF with MATH and MATH. Since the last expression of REF is just MATH, the lemma is established. |
math/9909180 | From REF of MATH, we see that MATH by the change of variables MATH. We may change the lower limit of integration from MATH to REF, incurring an error that is MATH. Therefore MATH as claimed. |
math/9909180 | By the definition of MATH, and using the remark following REF to factor the polynomial MATH, we have MATH . Since the MATH are pairwise coprime, MATH does not divide any of the MATH other than MATH. Also, since MATH is a root of MATH and MATH is exclusive, MATH cannot be a root of any other MATH. Therefore, since MATH, we see that MATH for all MATH. We may therefore divide the congruence in REF by MATH for each MATH, obtaining MATH which establishes the lemma. |
math/9909180 | By REF we can write MATH . Thus to establish the lemma, it suffices to show that MATH . If MATH is a prime that does not divide MATH, then we know that MATH by REF . Thus MATH where we have used REF to change MATH to the MATH in the last equality. All that remains to establish REF , and hence the proposition, is to show that MATH . For every MATH-tuple MATH such that each MATH, the NAME remainder theorem gives us a MATH such that MATH (mod MATH) for each MATH, and this correspondence is bijective. Therefore MATH since for a given prime MATH the quantity MATH depends only on MATH by REF . This last sum now factors as MATH by REF . This establishes REF and hence the proposition. |
math/9909180 | Let MATH be the discriminant of MATH, which is nonzero since MATH is squarefree, and write MATH where all but finitely many of the MATH are zero. CITE gives a bound for MATH that implies MATH for any squarefree polynomial MATH and any prime power MATH (this estimate is improved by CITE, though it will suffice for our purposes as stated). In particular we see that MATH for all prime powers MATH, which establishes the lemma. |
math/9909180 | From general facts about polynomials over finite fields, for a given prime MATH either MATH or MATH. However, the fact that MATH is admissible means MATH, and so MATH for all primes MATH. This implies that for all primes MATH, MATH . This shows that MATH for all prime powers MATH (by adjusting the implicit constant if necessary). Similarly, MATH by REF , showing that MATH. |
math/9909180 | It is well-known that the number MATH of distinct prime divisors of MATH satisfies MATH, which implies that MATH for any positive constants MATH and MATH. Therefore, any multiplicative function MATH satisfying MATH for all prime powers MATH automatically satisfies MATH. By REF , respectively, both MATH and MATH have this property for some constant MATH depending on the polynomial MATH, and so the lemma is established. |
math/9909180 | If we include in the sum those (nonnegative) terms for which the MATH are not necessarily pairwise coprime, and use the trivial inequality MATH, we see that MATH . Given MATH, we see from REF that MATH . Since MATH by the upper bound REF on MATH, we can choose MATH so small (depending on MATH and MATH) that the right-hand side is MATH, which establishes the proposition. |
math/9909180 | The constants implicit in the MATH and MATH-notations in this proof may depend on the polynomial MATH and thus on MATH as well. By REF we know that MATH using REF , and so MATH by the asymptotic REF and the fact that the last sum is convergent. Again by REF , we see that MATH, and so MATH as before. This establishes the lemma. |
math/9909180 | We want to apply REF to establish the first claim of the lemma. The function MATH is nonnegative and multiplicative, and by REF we know that MATH, say. Furthermore, REF verifies REF with MATH. Therefore we may apply REF to see that MATH . This establishes the first claim of the lemma; the second claim follows easily from the first by a simple partial summation argument. |
math/9909180 | All constants implicit in the MATH and MATH-notations in this proof may depend on the polynomial MATH and the parameter MATH. We recall REF of MATH: MATH . Notice that MATH for MATH; in particular, when MATH satisfies the bound REF , the exponent MATH is strictly less than REF. Consequently the coefficients of the polynomial MATH are certainly MATH in size, while we have the lower bound MATH . We may therefore apply Hypothesis UH with MATH to the error terms MATH in REF ; this yields MATH using Hypothesis UH. We see from the lower bound REF that the logarithmic term in the denominator can be replaced by MATH, yielding MATH and by REF and the definition of MATH, this is the same as MATH . Since the MATH are pairwise coprime, we may factor the multiplicative function MATH into MATH in the first sum on the right-hand side of this last equation. By then deleting the restrictions MATH from the conditions of summation in both sums, and noting that MATH and MATH, we see that MATH by REF (applied with MATH in the second sum). Clearly each MATH, while MATH again since MATH by the upper bound REF on MATH. Therefore the estimate REF becomes MATH, which establishes the proposition. |
math/9909180 | If MATH does not divide MATH, then MATH by REF . On the other hand, if MATH exactly divides MATH, then MATH by REF ; moreover, REF applied with MATH and MATH certainly implies that MATH. In either case, we have MATH for some positive integer MATH, and REF tells us that MATH. This establishes the lemma. |
math/9909180 | If MATH is any prime power, then putting MATH and MATH in REF , we see that MATH. This implies that MATH . Since each MATH, it follows that for any prime MATH dividing MATH, we have MATH and hence MATH . Therefore to establish the lemma, it suffices to establish the upper bound REF for primes. But MATH is a nonnegative integer, and it cannot be REF since we are assuming that MATH is admissible. Therefore MATH, so we can write MATH by REF . This establishes the lemma. |
math/9909180 | If the polynomial MATH is not admissible then MATH (see the remarks following the statement of Hypothesis UH on page REF). Clearly MATH from its definition, and so the upper bound REF holds easily in this case. Thus for the remainder of the proof, we may assume that MATH is in fact admissible. We first bound MATH by noting that MATH since MATH for each MATH. This casts the problem of estimating MATH in terms of a sieving problem. The version of NAME 's sieve that we now employ can be found in NAME - CITE. Given a set MATH of integers, assume that a real number MATH and a multiplicative function MATH can be chosen such that MATH . Assume that MATH satisfies the two conditions MATH . Then we have the upper bound MATH uniformly for MATH, where the implicit constant depends only on the implicit constants in REF on MATH. For our application, we take MATH and MATH; with these choices, the left-hand side of the bound REF is exactly MATH. We also set MATH and MATH. Of course REF is satisfied, since the polynomial MATH has MATH zeros (mod MATH) in every block of MATH consecutive integers. On the other hand, since we are assuming that MATH is admissible, REF show that REF are satisfied. The bound REF thus implies that MATH using REF . The first product converges to MATH as MATH tends to infinity, while the factors MATH in the second product can be deleted for the purpose of finding an upper bound. Therefore MATH . In light of REF , this establishes the lemma. |
math/9909180 | REF is easy to see by noting that MATH . As for the second assertion, the hypothesis MATH implies that MATH . Since the exponent MATH of MATH is positive by the upper bound REF on MATH, this establishes the lemma. |
math/9909180 | All of the constants implicit in the MATH-notation in this proof may depend on the polynomial MATH. We use the bound given by REF on each term in the inner sum, yielding MATH . By the second assertion of REF , we may replace the denominator MATH by MATH. In addition, since the MATH are pairwise coprime we may factor the term MATH; and the final sum in REF is precisely MATH by the definition of MATH. Therefore MATH where we have deleted the condition MATH. By the first assertion of REF , this estimate becomes MATH . For the term with MATH, for instance, we may use the second claim of REF with MATH to bound the sum over MATH, and the first claim of REF to bound the remaining sums over MATH: MATH since MATH. The other terms are similar, and we see from the estimate REF that MATH which establishes the proposition. |
math/9909180 | All of the constants implicit in the MATH-notation in this proof may depend on the polynomial MATH (and thus on MATH and its irreducible factors MATH as well). For each MATH, the function MATH is a nonnegative multiplicative function satisfying MATH by REF (note that each factor MATH of the admissible polynomial MATH is itself admissible). Moreover, by REF the asymptotic REF holds for each MATH with MATH. We may therefore conclude from REF that MATH . We have MATH since each MATH, and so to establish the lemma it suffices to show that MATH . Since each MATH, we see from REF that MATH . Rewriting MATH by the definition of MATH, we note that the inner sum on the right-hand side is a telescoping series by the definition of MATH: MATH . Thus REF becomes MATH . Since MATH is an exclusive polynomial, for any prime MATH the number of roots of MATH is equal to the sum of the numbers of roots of each MATH; in other words, MATH. Therefore MATH which establishes REF and hence the lemma. |
math/9909180 | By adjusting the constant implicit in the MATH-notation if necessary, it suffices to establish the asymptotic REF when MATH is sufficiently large (in terms of MATH and MATH). Notice that MATH for each MATH, from the fact that MATH is effective. Notice also that each MATH, where the exponent MATH is strictly less than MATH. Therefore for each MATH, the expression MATH tends to infinity with MATH. We assume that MATH is so large that MATH . By the monotonicity of the MATH and the hypothesis bounding the MATH, this certainly implies that MATH for any MATH. We define MATH, the integrand in REF of MATH. Splitting that integral at the point MATH yields MATH . We estimate the first integral trivially by noting that MATH when MATH lies in the range of integration. In the second integral, the condition MATH holds since we are considering only values of MATH that are so large that REF is satisfied. Therefore MATH and thus to establish the lemma it suffices to show that MATH . We accomplish this by integrating by parts: MATH . For any MATH, we note that MATH for each MATH (see the remarks preceding REF ), and so we have MATH for MATH in this range. Also, by logarithmic differentiation we see that MATH from the estimate REF and the fact that MATH for any polynomial MATH. Given these estimates for MATH and MATH, we see that REF implies REF and hence the lemma. |
math/9909180 | Set MATH . The key to the proof is to notice that MATH can be expressed in terms of MATH by partial summation: MATH upon integration by parts. We proceed by induction on MATH; the base case MATH follows immediately from REF , since MATH for MATH sufficiently large. For the inductive step, suppose that the lemma holds for the case MATH, so that we know the asymptotic formula for MATH; we can then insert this asymptotic formula into REF to obtain MATH . Since MATH for all MATH, this error term is MATH. Moreover, making the change of variables MATH yields MATH . Therefore REF becomes MATH which is the desired asymptotic formula for the case MATH. This establishes the lemma. |
math/9909180 | By REF , MATH . The main term can be evaluated by REF with MATH, while the error term is MATH by MATH applications of REF . We obtain MATH . Since MATH for each MATH, we see that MATH, whence the right-hand side of REF becomes MATH. This establishes the proposition. |
math/9909180 | We could use partial summation to asymptotically evaluate the sum in REF in terms of elementary functions and the dilogarithm function, and then verify that the resulting expression satisfies the differential-difference equation characterizing the function MATH. However, if we take it as known that MATH uniformly for MATH and MATH, then we can use the following simpler argument. For MATH in this range we can write MATH (where the final sum on each line is empty if MATH). Using NAME 's formula this becomes MATH . Comparing this to the known asymptotic REF establishes the lemma. |
math/9909180 | Define the multiplicative function MATH . For the particular polynomial MATH, it is an easy exercise to compute from REF of MATH that MATH where MATH is the twin primes constant MATH . If we assume that MATH is even, this gives MATH . Note that MATH for any positive MATH (see the proof of REF ), and that MATH since the sum MATH converges. We can thus apply REF with MATH to see that MATH where MATH is as defined in REF . In light of REF , therefore, to establish the lemma (for even MATH) it suffices to show that MATH. But by REF , MATH as desired. This establishes the lemma when MATH is even, and the same argument slightly modified holds when MATH is odd. |
math/9909180 | Using the same techniques as in the proofs of REF , we can see that MATH and so MATH . The sum in the error term is MATH by REF , whence the error term is MATH in its entirety. Using partial summation (see the proof of REF ), we can show that REF implies MATH . This establishes the first claim of the lemma. Similarly, MATH . Again the error term can be shown to be MATH, while the inner sum in the main term can be evaluated by a similar partial summation argument: MATH . This establishes the second part of the lemma. |
math/9909180 | All of the constants implicit in the MATH- and MATH symbols in this proof may depend on the multiplicative function MATH, and thus on MATH and MATH as well. We begin by examining an analogue of MATH weighted by a logarithmic factor. We have MATH say. If we define the function MATH by MATH then MATH becomes MATH . Since MATH for MATH and MATH by partial summation, we can rewrite REF using REF as MATH where we have defined MATH . We integrate both sides of REF against MATH, obtaining MATH . Some cancellation can be obtained on the left-hand side by switching the order of integration in the double integral and evaluating the new inner integral; REF becomes simply MATH . We can substitute this into REF , divide by MATH, and rearrange terms to get MATH . An upper bound for MATH is now needed. Since MATH is bounded from its REF and the asymptotic REF , we have MATH . We also have MATH . Because the sum MATH converges by REF , the last product in REF is bounded as MATH tends to infinity. Therefore REF implies that MATH . The terms MATH and MATH can be estimated by MATH and MATH and so both MATH and MATH are MATH by the estimate REF . Therefore, by REF of MATH, we see that MATH . In particular, since MATH, we have MATH and so REF and the bound REF give us the asymptotic formula MATH for MATH, where MATH . To complete the proof of the proposition, we need to show that MATH can be written in the form given by REF ; we accomplish this indirectly, using the asymptotic REF . Consider the zeta-function MATH formed from MATH, defined by MATH . From the estimate REF and partial summation, we see that MATH converges absolutely for MATH (we shall only need to consider real values of MATH), and thus has an NAME product representation MATH for MATH. We can also use partial summation to write MATH for MATH. Since MATH for MATH, it is certainly true that MATH in that range; using this together with the asymptotic REF becomes MATH valid uniformly for MATH. Making the change of variables MATH in all three integrals and multiplying through by MATH yields MATH as MATH, where the exponent MATH is positive and at most REF (since MATH). Because the NAME MATH-function satisfies MATH as MATH, REF implies MATH . On the other hand, from REF we certainly have the NAME product representation MATH for MATH, and one can show that in fact this NAME product converges uniformly for MATH. The important contribution comes from the sum MATH, and we see from REF and partial summation that MATH uniformly for MATH and MATH. The remaining contributions can be controlled using REF . Consequently, taking the limit of both sides of REF as MATH gives us MATH (where we have just shown that the product on the left-hand side converges), which is equivalent to REF . This establishes the proposition. |
math/9909180 | We would like to apply REF to the multiplicative function MATH defined by MATH . Certainly MATH, and so the estimates REF for MATH follow from the same estimates for MATH. We also have MATH from the assumption that MATH satisfies REF . Therefore MATH satisfies REF as well, with the error term being MATH uniformly in MATH. If we keep this dependence on MATH explicit throughout the proof of REF , the only modification necessary is to include a factor of MATH on the right-hand sides of the estimates REF , and REF and in the error term in REF . Therefore, the application of REF to MATH yields MATH where the implicit constant is independent of MATH. Because MATH the proposition is established. |
math/9909180 | We have MATH since MATH. Hence REF is satisfied. Also, since MATH is a nonnegative function, we have MATH by the asymptotic REF , which shows that REF is also satisfied with MATH. Therefore the two parts of this proposition are just special cases of REF , respectively. The form given in REF for MATH is equivalent to the form given in REF , since the assumption that MATH is nonnegative implies that the sum MATH is nonzero. |
math/9909180 | All of the constants implicit in the MATH- and MATH symbols in this proof may depend on the multiplicative functions MATH, and thus on MATH, MATH, and the MATH as well. Define MATH . We establish the desired asymptotic REF for MATH by induction on MATH. The base case MATH of the induction is exactly the statement of REF . Supposing now that we know that the asymptotic REF holds for sums of the form MATH, we wish to show that it holds for MATH. We rewrite MATH as MATH and we can use REF to obtain an asymptotic formula for this inner sum. For any index MATH, let MATH be the multiplicative function defined on prime powers by MATH which satisfies MATH by the hypothesized estimate on the size of MATH, and set MATH . Then by REF becomes MATH where MATH is defined in REF , and where we have defined MATH (using the multiplicativity of MATH and the fact that the MATH are pairwise coprime) and MATH . First we obtain an asymptotic formula for MATH. The sum MATH is precisely of the form MATH with each MATH replaced by MATH. By REF , for each MATH the multiplicative function MATH satisfies MATH since MATH. We may therefore invoke the induction hypothesis to show that the asymptotic formula MATH holds for MATH. Next we obtain an estimate for MATH. By the definition of MATH we have MATH where we have defined MATH . We can rewrite MATH, for example, as MATH . In this innermost sum of nonnegative terms, we can delete the restriction that MATH be coprime to MATH and extend the range of summation from MATH to MATH; this yields (after renaming MATH to MATH) the upper bound MATH . The same analysis shows that MATH is an upper bound for MATH for each MATH. With this upper bound, REF becomes MATH . By the induction hypothesis for MATH, we see that MATH. Furthermore, both MATH and each MATH are MATH (the latter by REF ). Therefore MATH since MATH. Using this estimate for MATH and the asymptotic REF for MATH, we see that REF becomes MATH . This would establish the lemma if only we had MATH in place of the product MATH. However, the MATH-factors of these two expressions are certainly equal by inspection. For each prime MATH, moreover, the power of MATH in the infinite products of the two expressions equals MATH in both cases, and we also have MATH . Therefore the local factors in the infinite products of MATH and MATH are also equal, and so the asymptotic REF is equivalent to the statement of the lemma. |
math/9909180 | It is easily seen by integration by parts that MATH . First we consider the case where MATH. In this case, we have MATH by REF . Also by REF , MATH . But since MATH we have MATH, and so we see that the error term is MATH, which establishes the lemma in this case. In the remaining case, where MATH, we note that MATH by integration by parts. The integral in the error term is over an interval of length MATH, and the integrand never exceeds MATH, and so the error term is MATH. As for the main term, the fact that MATH implies that MATH, and so we can write MATH where we have defined the function MATH. One can check that this function MATH is bounded on the interval MATH, and so this error term is simply MATH, which establishes the lemma. |
math/9909181 | Either apply REF , using the fact that any MATH-invariant metric on MATH with non-negative scalar curvature is isometric to a compact surface of revolution in MATH (see for example, CITE and the references therein), or see at the end of REF why the proof of REF also proves this corollary (without using the above fact). |
math/9909181 | From MATH and integrating by parts we obtain that MATH . In the case where MATH, we get MATH. Since for MATH we have equality, it follows that MATH. When MATH, we have that MATH. The solution of the corresponding NAME - NAME equations which is given in the Appendix suggests that we now consider the sequence of functions defined by MATH to get MATH and MATH . From this it follows that MATH and so MATH. |
math/9909181 | Assume that MATH has zero average. Then MATH where the inequality follows from REF . We thus have that MATH . |
math/9909181 | The curvature MATH of the surface of revolution is given in the arclength coordinate MATH by CITE MATH while in the moment map coordinate MATH we have that (see REF) MATH . This means that MATH . Defining a function MATH by MATH, we get from REF the following differential equation relating MATH and MATH: MATH . Given MATH, one easily checks that the unique solution MATH of REF satisfying MATH is MATH as required. |
math/9909181 | From MATH we have that MATH . Hence, using REF , we get MATH . |
math/9909191 | The necessity of REF is obvious. As for the sufficiency, suppose MATH is a small geodesic triangle in MATH and that MATH denotes its image in MATH. If MATH intersects the branch set MATH in a vertex or an edge or if it is contained in MATH, then REF insure that the distance between two points of MATH is the same as that of the corresponding points of MATH. If the intersection of MATH and MATH is one or more points in the interiors of the edges, then we can subdivide MATH and apply a standard argument (compare REF , page REF) to again conclude that the CAT REF - inequality holds. |
math/9909191 | From the above discussion, the statement holds when MATH is MATH - compact. Since the set of MATH - compact MATH - flats is dense in the set of all MATH - flats in MATH, and any sequence of MATH - flats in MATH which all meet a compact set will have a convergent subsequence, the general case follows. |
math/9909191 | Since there are MATH - flats in MATH passing through any two given points, we may choose MATH passing through MATH and MATH. Let MATH be an image flat of MATH, MATH. Since MATH by the convexity of the metric, we have MATH, so by passing to a subsequence, we may assume that MATH and MATH is an image flat of MATH, the limit of the MATH. We have MATH is bounded for all MATH; hence, MATH is constant, by the convexity of the metric. |
math/9909191 | Fix a geodesic MATH in MATH. By REF , there exists a flat MATH and an image flat MATH such that MATH has constant distance from MATH. Let MATH be a sequence of MATH - compact flats which converges to MATH. For each n, let MATH be an image flat of MATH, and denote by MATH the corresponding lattice on MATH or MATH. Passing to subsequence if necessary, we may assume that MATH, which is also an image flat of MATH, hence MATH has constant distance to MATH. Let MATH be the projection of MATH onto MATH. We can choose a sequence of MATH - compact geodesics MATH such that MATH. That is, there exists a sequence MATH such that for each MATH, MATH and MATH when MATH. By taking a diagonal subsequence in MATH, we know that there exists sequence MATH such that MATH and MATH as MATH. |
math/9909191 | With the duality condition established as above, the proof of REF can be easily modified to cover the singular case. All we have to do is to estimate MATH the same way for a point MATH on MATH and allow MATH to be an arbitrary but fixed point in MATH, then MATH for any MATH would imply MATH. Here MATH is any accumulation point of MATH. |
math/9909191 | For REF , the proof given in the appendix of CITE works. The only place where care needs to be taken is with regard to the paragraph on page REF, where the angles are used. In a general NAME space, the angle function MATH, while it is still continuous in MATH and MATH, is only upper semicontinuous with respect to the vertex MATH. Also, two different geodesics from MATH could form a zero angle (in fact they could even share a segment). Note that given two distinct points MATH, MATH in MATH, there always exists a point MATH such that MATH. Also, MATH is continuous in MATH or MATH on MATH, and satisfies the triangle inequality MATH . With these properties, the paragraph on page REF can easily be modified and the proof is valid. |
math/9909191 | REF is obvious. REF follows by taking closures from the fact that, for MATH, the point MATH of REF belongs to MATH, together with REF . |
math/9909191 | In the singular case, it is still true that if two geodesic rays meet at MATH with angle MATH, then they form a geodesic. Also, two parallel geodesics bound a flat strip (the `Flat Strip Lemma' or `Sandwich Lemma', see for example CITE or CITE). By the definition of NAME function (see CITE) and the NAME of Cosine in a comparison triangle, it is not hard to see that if MATH, then MATH. Also, if MATH, then MATH. From this it follows that MATH. |
math/9909191 | This is essentially the proof of REF, but since we do not assume that MATH is a proper subset of MATH, we get a weaker result. |
math/9909191 | The proof of REF is valid in the singular case as well. |
math/9909191 | The proof of NAME REF of CITE holds true using REF , instead of REF and NAME REF of CITE. |
math/9909191 | Again, the proof of REF works using REF instead of REF and the fact that the limit of a convergent sequence of (half) flats is still a (half) flat. |
math/9909191 | The proof of REF goes through without change, except for the last paragraph of the proof on page REF, where angles are again used. What is needed to complete the proof is the following lemma. Consider a geodesic triangle with vertices MATH, MATH, and MATH. Let MATH, MATH, be the geodesic segment from MATH to MATH, with MATH. If there is a sequence MATH of small positive numbers that converges to MATH, such that MATH for each MATH, then MATH. Let MATH be the geodesic segment from MATH to MATH, with MATH, MATH, where MATH. Denote by MATH the distance between MATH and MATH. Then the function MATH is monotonically decreasing in MATH and MATH, and it's limit when MATH, MATH is the cosine of the angle MATH. Now assume that MATH. Then there exist MATH and MATH such that MATH. By the monotonicity of MATH in MATH, we have MATH for any MATH. So for MATH sufficiently small, we have MATH. Hence for MATH sufficiently small, MATH which contradicts our assumption. To complete the proof of REF , we apply the lemma to the triangle with vertices MATH, MATH and MATH. We have that two of the internal angles are at least MATH. So the sum of the internal angles is at least MATH, which implies that it must be a triangle in MATH, with two right internal angles. This is clearly impossible. |
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