paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0102200 | We use the notation and the results of the decomposition in REF , so we can assume that our SRWL is a convex combination of SRWLs supported on simple paths and a dead-end RW law. For every simple path component MATH, consider the stopped random walk law there, and the flow defined in REF for this walk. Let MATH, as defined above REF . The parameters MATH and MATH equal zero for the dead-end random walk law component. Thus REF follow from linearity and the simple path case (formulas REF ). The rest of the proof concerns the bound REF for the parameter MATH; it is relevant for REF , but not for REF , and should be omitted at first reading. For the dead-end random walk law component, the MATH is bounded above by the expected lifetime of the walker MATH (in fact, this is sharp, achieved when the graph consists of the vertex MATH and a self-loop; if we outlaw self-loops, the sharp bound becomes MATH). From this and the simple path REF we get MATH . Unfortunately, because of the possible self loops, this expression for MATH is messy, making the solution of the optimization problem messy, too. To avoid this, we sacrifice sharpness for simplicity, bounding the MATH and MATH terms by MATH. This yields the bound REF . |
math/0102200 | We have seen in REF that it suffices to bound MATH in terms of MATH. Using the notation and results of REF , we can write MATH . The first inequality follows from NAME 's inequality and the fact that MATH is convex for MATH. The second, from the fact that the function MATH is increasing in MATH for MATH, and that MATH. The third inequality follows from NAME 's inequality and the fact that MATH is convex in MATH for MATH. Solving the above inequality for MATH, and using REF that MATH we get MATH . Note that MATH is the sum of MATH independent copies of MATH. Conditioning on the first step yields MATH, and solving this equation gives the equality in REF , a standard result. Thus the inequality in REF , in terms of MATH, is a comparison of the NAME transforms of MATH and MATH, as required. |
math/0102200 | The expected value inequality follows from differentiating the NAME transforms at REF. For the large deviation inequality, note that MATH for every MATH by NAME 's inequality. Replacing the NAME transform on the right by that of MATH we get MATH . For the last equality, we used the fact that the infimum over MATH is achieved when MATH; this can be checked by direct calculation. Direct computation, or the law of large numbers, implies the expression MATH for MATH, and a standard computation using the NAME transform of MATH yields its large deviation rate function MATH given by REF . |
math/0102200 | Using the notation and results of REF , we can write MATH where MATH . The first inequality follows from NAME 's inequality and the fact that MATH is convex for MATH. The second, from the fact that the function MATH is increasing in MATH for MATH, and that MATH. We will keep the parameter MATH fixed and try to minimize the lower bound (given by REF ) MATH as the parameters MATH and MATH range over the set MATH. We first claim that the infimum is achieved on this set. If MATH converges to MATH, then MATH will converge to MATH, so the lower bound in REF can only converge to a small value if MATH converges to MATH, in which case MATH is a better choice. The same argument can be made with the roles of MATH and MATH reversed, so the infimum must indeed be achieved on this set. REF below, where the hard part of the optimization is done, implies that MATH (respectively, MATH) have to be the same for every MATH, so we may drop the indices MATH. The lower bound REF reduces to MATH and we have MATH. Since MATH is increasing in MATH, it is clear that increasing MATH while keeping MATH fixed will not change MATH but will decrease the numerator on the right hand side of REF . Therefore the minimum is achieved when MATH is maximal, so we may take MATH. After cancellations, the bound REF reduces to MATH . We are left to minimize this while keeping MATH fixed. The solution is MATH . Clearly, MATH, so we have MATH. If we set MATH then this gives exactly the inequality in REF . Therefore, to conclude the proof it suffices to show that MATH is bounded above by the MATH solution of MATH. Equivalently, it suffices to show that MATH. This follows if we combine the simple bound MATH with REF . |
math/0102200 | For MATH the expression MATH is increasing in MATH, and can be bounded below by MATH, so it suffices to prove that this expression is at least MATH. We substitute REF : MATH and replace the last MATH by MATH to get the lower bound MATH . This can be bounded below by MATH times MATH which is easily checked to be at least MATH. |
math/0102200 | CASE: The minimum can only occur for MATH. Otherwise, we may define new values MATH, this will not violate the constraints, and will decrease REF by the convexity of MATH. CASE: Minimum must be achieved in the interior of MATH. By REF and symmetry, the only other case is MATH, for this we may assume MATH (it makes no difference). Let MATH be the value of the left hand side in REF . It is straightforward to check that the unique solution of REF and MATH for which MATH and MATH gives a smaller value for REF . CASE: Minimum is in fact achieved when equality holds in REF , but we do not need to show this. From now on we will only use that REF is also minimal when REF is replaced by the equality constraint REF , where MATH is the actual value of the left hand side of REF . CASE: Since we excluded the case that the minimum occurs on the boundary, it can only occur where REF dimensional gradient of REF is perpendicular to REF dimensional surface determined by REF , or the derivative evaluated at two linearly independent vectors tangent to this surface is REF. This means that the Jacobian of the map given by the function REF and the left hand sides of REF has a two-dimensional nullspace, so any MATH submatrix must be singular. The Jacobian is a MATH matrix; the first and third columns are computed as MATH . We get the other two columns by replacing the index MATH by MATH. Now we set MATH and divide the first two columns by MATH, and the last ones by entries in the second row: MATH . The determinant of the right MATH submatrix has to be REF, and this happens if and only if MATH with MATH . The left MATH submatrix simplifies if we divide the first row by MATH: MATH . After computing the determinant, we get that this matrix is singular iff MATH with MATH . To complete the proof, we have to show that the map MATH is injective. This follows from the fact that if MATH, then MATH is increasing in MATH, MATH and MATH is decreasing in MATH and increasing in MATH. Consider MATH; the two interesting cases are MATH, and MATH. In the first case MATH, in the second, MATH. |
math/0102200 | Set MATH . Note that MATH as well as MATH, and MATH, MATH are bounded by some constants MATH and MATH for all MATH. Also, for all large MATH, we have MATH. For these MATH we apply REF with parameters MATH, MATH, MATH to get MATH . The definition of MATH implies that MATH and therefore MATH . This and REF proves the proposed expected value bound. Clearly MATH. The function MATH and its derivative are continuous on the set MATH, and hence bounded. Therefore MATH . The boundedness of MATH implies that we can ignore the first few MATH for the price of increasing the constant MATH. Thus REF imply the proposed large deviation bound. |
math/0102200 | Let MATH be arbitrary, and let MATH, MATH denote the hitting time and the total weight of the set of edges at distance MATH from MATH, respectively. Then MATH . For all large MATH we have MATH, so by REF each event MATH has probability at most MATH. By the NAME Lemma only finitely many of these events happen. Thus REF is at most MATH, and since MATH was arbitrary, the Corollary follows. |
math/0102200 | We apply REF to the graph in question. The parameters we use are MATH, MATH which is the solution of MATH, and MATH. This will cover the graphs in question, since MATH and this dominates MATH for large MATH. The theorem yields MATH, and using the fact that MATH, the first REF follows. REF also yields the bound REF on MATH which we rewrite as follows: MATH . Substituting the parameters for our case and taking logarithms we get MATH . Multiplying by MATH and substituting the formulas for MATH and MATH we get MATH . Exponentiation yields the bound REF . |
math/0102200 | Without loss of generality we may assume that MATH is even. Let MATH. Then MATH . We now use the binomial formula to get that for MATH even MATH . By NAME 's formula and the fact that MATH we get MATH . The remaining factor can be written as MATH . Using REF for MATH, the fact that MATH and that MATH we get that MATH . The convergence is uniform over all MATH. This and REF imply the claim of the lemma. |
math/0102200 | Let MATH, let MATH, and let MATH . Let MATH an integer, for every MATH, let MATH denote the distance of the farthest vertex visited up to time MATH, and let MATH be the greatest integer so that MATH. Then MATH . Taking lim sup we get MATH . We are taking MATH-th powers to make a sequence of probabilities summable. Now consider the function MATH an upper bound for MATH, so that MATH for all large MATH. For all large MATH is increasing, and we have MATH . The last inequality follows from REF . For large MATH the right hand side is bounded above by MATH for some MATH, so it is summable over the subsequence of MATH-powers if MATH. Therefore by the NAME lemma MATH eventually a.s., so the right hand side of REF is at most REF. Since MATH was arbitrary, the corollary follows. |
math/0102201 | By the above remark, it is enough to show that if MATH is a log resolution of MATH, with MATH, then MATH . This follows from the fact that MATH and the invariance of the total discrepancy under proper birational morphisms in the case of divisors (see REF ). |
math/0102201 | We have MATH log canonical if and only if MATH, or equivalently MATH, for all MATH. By replacing MATH with the complement of the union of those images of MATH which do not intersect MATH, we get our result. |
math/0102201 | It is enough to prove the first assertion. This follows from the fact that MATH if and only if MATH is effective or equivalently, all the coefficients in MATH are MATH. |
math/0102201 | The assertion follows from the fact that the functor MATH has a left adjoint, and therefore it commutes with direct products. |
math/0102201 | By restricting to an open neighbourhood of MATH, we may assume that MATH and that MATH is the origin. We have an embedding MATH which induces an embedding MATH such that MATH corresponds to the origin. The action of MATH on MATH extends to an action on MATH induced by MATH. Therefore MATH is a subscheme of a weighted projective space, hence it is projective, and MATH is the cone over it. |
math/0102201 | Consider the projection MATH and let MATH be the inverse image by this morphism of MATH. We have a natural action of MATH on MATH and the quotient by this action is a scheme MATH which is proper over MATH. The properness follows from the fact that it is locally projective, an assertion which is just the globalization of REF and can be proved in the same way. But for every closed point MATH we have MATH (or MATH if MATH) and our assertion follows from the semicontinuity of the dimension of the fibers of a proper morphism. |
math/0102201 | The idea of the proof is similar to that of REF, so that we refer to that paper for some of the details. With the notation in the theorem, note that we have MATH log canonical (NAME log terminal) if and only if MATH, for all MATH. Since the case MATH is trivial (we follow the convention that MATH), we assume from now on MATH nonempty. We fix a function MATH, such that for every MATH, MATH where MATH is a constant such that MATH, for all MATH. We extend this function by defining MATH. For the proof of the ``if" part, we will add later an extra condition of the same type on MATH. We integrate over MATH the function MATH. Computing the integral from the definition, one can see that MATH, where MATH . It is clear that every monomial which appears in the MATH-th term of MATH has degree bounded above by MATH and below by MATH, where MATH, and MATH, for every MATH (recall our convention that MATH). We have precisely one monomial of degree MATH, namely MATH whose coefficient is MATH, the number of irreducible components of MATH of maximal dimension. Similarly, every monomial which appears in the MATH-th term of MATH has degree bounded above by MATH and below by MATH, where MATH . There is exactly one monomial of degree MATH, namely MATH with coefficient MATH. From the above evaluation of the terms in MATH and MATH and REF we see that MATH is integrable. Moreover, it follows from REF that MATH for every MATH. We have also MATH for every MATH, with equality if and only if MATH and MATH. REF implies therefore that we have strict inequality for infinitely many MATH. This shows that in MATH we have monomials of degree MATH for infinitely many values of MATH. We apply now the change of variable formula in REF for the morphism MATH to get MATH . Using the fact that MATH and that MATH has simple normal crossings, one can compute explicitely the integral. For a subset MATH let MATH. With this notation we have MATH . Every monomial in the term of MATH corresponding to MATH has degree bounded above by MATH and below by MATH, where MATH and MATH. Note that MATH. Let us introduce the notation MATH. We see that if MATH, then MATH . Moreover, REF implies that if MATH, then MATH and MATH. This shows that the only monomial of the form MATH which could appear in the part of MATH corresponding to MATH is for MATH. We prove now MATH. Suppose first that MATH is log canonical, so that MATH, for all MATH. Assume that for some MATH, we have MATH. It follows from the above discussion that MATH does not appear in MATH. As we have seen, this implies MATH. In particular, we have MATH. Continuing in this way, we deduce MATH for all MATH, which is impossible. Conversely, suppose that for some fixed MATH such that MATH for all MATH we have MATH, and that MATH, for some MATH. We choose the function MATH such that in addition to MATH it satisfies the following condition. For every MATH, let MATH be the finite set consisting of all the pairs MATH such that MATH. The extra condition we require for MATH is that for every MATH and every MATH, with MATH, we have MATH . This means that MATH. Note that from MATH we can deduce that if MATH, with MATH, then MATH . On the other hand, the top degree monomials which appear in different terms of the sums MATH (for possible different MATH) don't cancel each other, as they have positive coefficients. Let MATH be the highest degree of a monomial which appears in a term corresponding to some MATH. The previous remark shows that the corresponding monomial does not cancel with the other monomials which appear in terms corresponding to MATH. Obviously we have MATH and therefore our assumption that MATH and MATH gives MATH. We also deduce from MATH that MATH. We see from MATH and MATH that this monomial of degree MATH does not cancel with other monomials in terms corresponding to MATH, if MATH. Therefore in the integral MATH we have a monomial of degree MATH, with MATH, a contradiction. The proof of MATH is entirely similar. |
math/0102201 | Note first that the formula is trivial when MATH, since in this case for every MATH, we know that MATH is a smooth variety of dimension MATH. Suppose from now on that MATH. For a fixed MATH, we can find an open neighbourhood MATH of MATH such that MATH and MATH. Therefore, in order to prove the second assertion of the corollary we may assume MATH, in which case the formula follows from REF . In order to prove the first assertion, it remains to be proved that MATH, for every MATH. We reduce again to the case when MATH, when we conclude by applying REF . |
math/0102201 | The statement is a consequence of the formula in REF . |
math/0102201 | Since the cases MATH and MATH are trivial, we may assume that we are in neither of them. We obviously have MATH, for every MATH, so that the formula in REF gives MATH . On the other hand, the formula giving the lower bound for the dimension of the fibers of a morphism implies that MATH . Note that by REF , there is a positive integer MATH such that MATH if MATH. Therefore we deduce MATH for every MATH, such that MATH. This implies that for every MATH, MATH so that in fact MATH which completes the proof of the corollary. |
math/0102201 | Since MATH is a subscheme of MATH, it follows that MATH is a subscheme of MATH, so that MATH for every MATH. The assertion now follows from REF . |
math/0102201 | Using REF , we have MATH . |
math/0102201 | The cases MATH (when MATH) and MATH (when MATH) are trivial, so that we may suppose that we are in neither of them. There is MATH such that MATH. If MATH is the canonical projection, then MATH. Indeed, every local morphism MATH factors through MATH, as MATH. This gives MATH. REF implies that MATH. For the other inequality we use REF . This shows that for every MATH and every MATH, we have MATH. Since MATH for every MATH, the same is true on an open neighbourhood MATH of MATH. We deduce MATH for every MATH, and an easy computation gives MATH. REF implies MATH. |
math/0102201 | REF gives a canonical isomorphism MATH, for every MATH. Therefore we have MATH. We pick MATH such that MATH, MATH, MATH for all MATH, MATH and MATH, where MATH, MATH and MATH are the coefficients appearing in log resolutions of MATH, MATH and MATH. We conclude by applying the second assertion in REF . |
math/0102201 | Let MATH be an open neighbourhood of MATH in MATH such that MATH. Since MATH, by restricting everything to MATH, we may assume that MATH. From the local description of the jet schemes, it follows that since MATH is defined locally in MATH be one equation, we have MATH defined locally in MATH by MATH equations, for every MATH. Moreover, if MATH is the canonical projection, then for every irreducible component MATH of MATH such that MATH, we have MATH. Indeed, it follows from our discussion in REF that if MATH is the ``zero section", then MATH. Moreover, since the ``zero section" is functorial, we have MATH, and we deduce that MATH contains MATH. It follows that MATH, so that MATH . Applying REF , we deduce the inequality in the proposition. |
math/0102201 | If we cover MATH with open subsets MATH, then we obviously have MATH and a similar relation for the restrictions to MATH. Since MATH is smooth, it is locally a complete intersection, so that we can find an open cover of MATH such that on each subset MATH, MATH is an intersection of smooth divisors. It is therefore enough to apply REF inductively on each of these open subsets. |
math/0102201 | Consider the following cartesian diagram: MATH where the horizontal maps are the natural embeddings and MATH is the diagonal embedding. Applying REF to the embedding MATH, the pair MATH, and the closed subset MATH, we deduce MATH . On the other hand, REF implies MATH and these two relations prove the proposition. |
math/0102201 | We make induction on MATH. It is enough to find a locally closed cover of MATH such that the restriction of MATH over each member of the cover has the desired property. Therefore we may assume that MATH is a smooth variety and it is enough to find a nonempty open subset MATH of MATH over which the restriction of MATH has this property. By REF, we can embed MATH as an open subscheme in a scheme MATH proper over MATH. By replacing MATH with MATH and MATH by its closure in MATH, we may assume that MATH is proper. After restricting to an open subset of MATH, we may assume that MATH is also smooth. Let MATH be a log resolution for the pair MATH. It is easy to see that after further restricting over an open subset of MATH we can assume that for every MATH, the restriction to the fiber MATH is a log resolution of MATH, in which case the assertion is obvious. |
math/0102201 | REF show that the set MATH is finite. Moreover, it follows from REF that we can find MATH such that for every MATH and every MATH with MATH, we have MATH . Fix MATH. Since there are only finitely many log canonical thresholds to consider, it is enough to show that for every MATH, there is an open neighbourhood MATH of MATH such that MATH, for every MATH. Since MATH is flat, by restricting to an open neighbourhood of MATH we may assume that MATH is constant on MATH. Therefore it is enough to find MATH with MATH and MATH such that MATH for all MATH. We fix MATH such that MATH and MATH for all MATH. Using REF , we choose an open neighbourhood MATH of MATH such that MATH for all MATH. The following inequalities show that MATH satisfies our requirement: MATH . |
math/0102201 | We will use the formula in REF , so that we estimate first MATH, for every MATH. Note that MATH can be covered by the locally closed subsets MATH, with MATH for all MATH, where MATH is the set of ring homomorphisms MATH, with MATH, for every MATH. We put MATH if MATH. It is clear that if MATH, then MATH. On the other hand, MATH if and only if for every MATH such that MATH, we have MATH. Let MATH be the polar polyhedron of MATH, defined by MATH . We see that MATH if and only if MATH. From the above discussion, we deduce that MATH where the infimum is taken over all MATH such that MATH for all MATH. The formula in REF gives MATH the infimum being taken over all MATH. We clearly have MATH. Note that if MATH and MATH is defined by MATH for all MATH, then MATH. We deduce that MATH. In order to complete the proof it is enough to note that MATH if and only if for every MATH we have MATH this comes from the fact that MATH is the polar polyhedron of MATH. |
math/0102204 | The binomials MATH defined by the two columns of MATH determine a complete intersection MATH of degree MATH in MATH which coincides with MATH over MATH. The irreducible decomposition of MATH consists of the components of MATH - of which there is only one if MATH is free abelian - together with subschemes supported on coordinate flats MATH whose NAME forms are the bracket monomials MATH . The theorem will be proved if we show that the cycle MATH occurs with multiplicity MATH in the complete intersection. Suppose first that MATH. We may assume that MATH . Then, MATH is not contained in MATH, and thus occurs with multiplicity MATH. Suppose now that MATH. We may assume that MATH and MATH . Then, MATH is contained in MATH, and after localizing and changing variable names, we are lead to the following situation: let MATH and MATH in an extension field MATH of MATH, and consider the univariate resultant MATH . We want to show that MATH appears with exponent MATH as a factor of MATH. Indeed, when MATH the condition MATH holds if and only if there exists MATH such that MATH and so MATH occurs in MATH with exponent MATH equal to the intersection multiplicity at the origin of the artinian ideal MATH in MATH. We claim MATH. The given equations are a NAME basis with leading terms MATH and MATH, for the term order defined by MATH and MATH. Hence MATH, that is, there are MATH roots in the affine plane counting multiplicity. Of those, MATH lie in the torus, that is, have both coordinates non-zero. No root of MATH has precisely one zero coordinate. Therefore the multiplicity of MATH at the origin is the difference MATH. |
math/0102204 | This is a reformulation of Pick's formula which states that the area of a lattice polygon equals the number of lattice points in that polygon minus half the number of lattice points in its boundary, minus one. |
math/0102204 | Consider the rational normal scroll of type MATH, a toric surface of degree MATH in a projective space of dimension MATH. Its NAME form has an exact determinantal formula in terms of a NAME matrix. A nice proof of this fact follows from recent results of CITE, since the rational normal scroll is given by the MATH-minors of a matrix of variables. This NAME form is the unmixed, sparse resultant for three polynomials with support MATH . The three polynomials MATH have exactly this support. Our formula is gotten by specializing the NAME matrix for the scroll. |
math/0102204 | The specialization MATH in REF is equivalent to MATH at the level of primal NAME coordinates. The dual NAME form MATH is a MATH-linear combination of bracket terms MATH of degree MATH. If we substitute MATH for MATH in the expansion of such a bracket term MATH then we get MATH . Hence the specialized dual NAME form on right hand side of REF equals the specialization of the primal NAME form MATH under REF , as desired. |
math/0102204 | Suppose that the variable MATH divides MATH. Then every bracket monomial appearing in the dual NAME form MATH contains the letter MATH. Equivalently, every bracket monomial in the primal NAME form MATH contains a bracket MATH with MATH or MATH. In view of CITE, this means that every regular triangulation of MATH contains a simplex for which MATH is not a vertex. But this is false, since MATH lies in every maximal simplex of the reverse lexicographic triangulation of MATH, for MATH smallest; see CITE. |
math/0102204 | It suffices to prove this theorem for the case when all MATH are non-zero. Indeed, if MATH then CITE implies that MATH where MATH is a NAME dual of the configuration MATH. Our assertion for MATH implies that for MATH. We hence assume that MATH for all MATH. Each vertex MATH of MATH corresponds uniquely to the NAME dual of a regular triangulation MATH, and hence to a pair of adjacent linearly independent vectors MATH (indices are understood modulo MATH; recall that MATH). By CITE, the MATH-th coordinate of MATH equals the sum of the normalized volumes of those simplices in MATH which contain the point MATH. By NAME duality, MATH where the sum is over all indices MATH such that MATH and MATH lie in the cone spanned by MATH and MATH. Let MATH be the vertex of MATH between the edges parallel to MATH and MATH. We claim that MATH is mapped to MATH under the affine isomorphism given above. We note that the maximum MATH of the values MATH is attained at the vertex MATH between the edges parallel to two independent vectors MATH such that MATH and MATH (indices modulo MATH). What we are claiming is the identity MATH. Let MATH denote the set of index pairs MATH such that MATH and MATH lie in the cone spanned by MATH and MATH. The set MATH is NAME dual to our regular triangulation, and, hence MATH equals MATH. If we start drawing MATH from the origin, then, MATH and MATH . Our assertion takes the following form: MATH . After erasing equal terms on both sides, the following remains to be proved: MATH . The proof is straightforward by a case distinction involving the relative positions of the vectors MATH and MATH in the plane. |
math/0102204 | We shall first prove the following claim about the full discriminant: MATH in the NAME polynomial ring MATH. Fix any relevant line MATH. Choose an isomorphism in MATH which maps MATH to MATH, and apply this isomorphism to the rows of MATH. Also reorder the rows of MATH so that the multiples of MATH come first. After this transformation, the first column of MATH has the entries MATH. For MATH and MATH only, substitute MATH into the NAME form MATH. Let MATH be the polynomials resulting from MATH in REF under the same substitution. Then MATH, but MATH is divisible by MATH, and this is the highest possible power of MATH with this property (compare REF ). REF implies that the specialized NAME form factors, and one of its factors is MATH . For all subsequent specializations, the NAME form factors accordingly. When we substitute MATH for MATH into MATH then we get the binomial MATH in REF . Clearly, the residual resultant MATH divides the full discriminant MATH. The above claim follows from this. Moreover, our argument shows that MATH is the highest power of MATH which divides MATH. Consider now the factorization formula given by NAME, NAME and NAME in CITE. Under NAME duality, the proper faces of the polytope MATH which are not simplices correspond to relevant lines MATH, and their face discriminants are precisely the binomials MATH. In other words, the full discriminant MATH equals the MATH-discriminant MATH times the product of the expressions MATH where MATH ranges over all relevant lines. We conclude from our claim that MATH divides MATH in the NAME polynomial ring. Since MATH is irreducible, both of our assertions follow. |
math/0102204 | Suppose first that there are no relevant lines. Then, MATH, and the secondary polygon MATH and the NAME polygon MATH are equal up to translation. More precisely, MATH where MATH is the exponent of MATH as a factor of MATH. Using CITE and NAME duality, we find MATH . In light of REF , it suffices to show that MATH . After cancelling terms common to both sides, what remains to be shown is MATH . This identity holds because both sides are equal to the normalized lattice width of the polygon MATH in the direction orthogonal to MATH. We next assume that relevant lines exist. Then MATH generally differs from MATH. The secondary polytope MATH equals MATH plus the NAME sum of the NAME segments of the binomials REF where MATH runs over all relevant lines. Hence, up to lattice translation, MATH . The minimum value of the linear functional MATH over the line segment MATH is MATH when this value is negative and zero otherwise. Therefore REF translates into the identity MATH . The argument for the case of no relevant lines now completes the proof. |
math/0102204 | The condition MATH is equivalent to MATH being a point. This happens if and only if all vectors MATH lie in a relevant line, and MATH for each relevant line MATH. This last condition is equivalent to MATH being centrally symmetric. |
math/0102204 | According to REF , the NAME polygon of the discriminant MATH is essentially the lattice triangle MATH. The number of terms in MATH is at most the number of lattice points in MATH. Using Pick's formula as in REF , we find that the number MATH equals MATH . Using the inequality MATH, we find that the sum of any two of the three last summands is bounded above by MATH. Therefore, MATH . This is the desired inequality. |
math/0102205 | Left invariance for MATH means MATH. We use this to establish bi-invariance for MATH, which means MATH. This follows from MATH. For the second assertion, note that MATH . Using REF we obtain MATH and MATH, which with the above equations yield the desired conclusion. |
math/0102205 | This follows from the fact shown in the previous proof, that MATH . The right side, when regarded as a measure on MATH, describes the location of the MATH-generated walk. But by the right invariance of MATH, the left side is equal to MATH, which when regarded as a measure on MATH, describes the location of the MATH-generated walk. |
math/0102205 | Define, for MATH, MATH where MATH is the image of MATH under the quotient map from MATH to MATH. From REF , we see that MATH, and hence for all MATH, MATH . On the other hand, NAME 's identity on MATH CITE yields MATH where the MATH denotes the conjugate transpose (here only). Notice that MATH may be rewritten as: MATH . Then MATH may be computed as MATH. For MATH, MATH since MATH. For MATH, a trivial computation shows MATH, and thus MATH . REF and the computations from REF , when substituted into REF , and combined with REF , give MATH where MATH may be chosen arbitrarily. Taking only the dominant term REF in the above expression, and letting MATH, we have MATH, MATH, MATH, and MATH. It follows that MATH as was to be shown. The second inequality in the theorem follows from MATH, using the fact that MATH for all MATH. |
math/0102205 | This theorem is embedded in CITE, but obscured by their exotic notation. We briefly indicate how to ``prove" this theorem from results cited in CITE. The set of functions with absolutely convergent NAME series is denoted in CITE by a symbol that resembles MATH, defined in REF . REF shows that any MATH is equal almost everywhere to its NAME series. REF shows that this NAME series converges uniformly to a continuous function that we have denoted MATH. |
math/0102205 | CITE shows the algebra MATH is commutative if and only if the number of times the trivial representation appears in MATH, the the restriction of MATH to MATH, is zero or one. If one, this trivial representation corresponds to a one-dimensional subspace of MATH fixed by MATH, that is, the left MATH-invariant functions on MATH; choose the unique function MATH on MATH normalized so that MATH. This is sometimes called the spherical function of MATH corresponding to the representation MATH. Complete MATH to a basis for MATH so that the matrices of MATH break into irreducible ``blocks". Then for a right MATH-invariant function MATH: MATH where MATH is NAME measure on the subgroup MATH. The second equality is obtained by choosing a coset representative MATH from each coset in MATH and expressing MATH for some MATH and MATH, and noting that NAME measure MATH decomposes as a product measure MATH. A similar argument holds for a right-invariant measure MATH, noting that MATH decomposes as a product measure MATH because of right-invariance. By the orthogonality relations for matrix entries CITE of irreducible representations of MATH, the left-most integral of REF produces a matrix consisting of zeroes except possibly for the MATH-th entry. Thus MATH (respectively, MATH) has zero entries except possibly for the first row. A similar argument using the left-invariance of MATH (respectively, MATH) shows that MATH (respectively, MATH) has zero entries except possibly for the first column. Together, these statements imply that the only entry that could possibly be non-zero is the MATH-th entry. If the trivial representation does not appear in the the restriction of MATH to MATH, the argument above holds by ignoring the role of MATH when choosing a basis for MATH. Orthogonality then shows that the left-most integral of REF yields a zero matrix. |
math/0102205 | To show MATH is continuous at MATH, consider any sequence MATH such that MATH. It must be shown that MATH. Let MATH and MATH. Since MATH is bounded, all the MATH and MATH, being translates of MATH, are uniformly bounded by some constant function. This constant function is in MATH, since MATH is compact. Also, MATH pointwise for all MATH, since MATH is continuous at those points. By the assumption on MATH we have pointwise convergence almost everywhere. By NAME 's dominated convergence theorem, MATH, which is precisely the statement MATH. This completes the proof of REF . |
math/0102205 | We apply REF setting MATH and MATH. By inspection, MATH is bounded by MATH. The lemma implies MATH is continuous everywhere except possibly at MATH. This may be seen by observing that the discontinuity set of MATH is MATH, the boundary of MATH. This is the pre-image of a circle on MATH. On the other hand, MATH regarded as a measure on MATH is uniformly supported on a circle at latitude MATH from the north pole. Any two circles on MATH intersect in at most two points, unless they are identical. Hence MATH unless the support of MATH intersects MATH, which only occurs when MATH. This corresponds to a discontinuity in MATH at MATH when MATH. We now apply REF again to show that MATH is continuous for MATH. The preceding observations show that MATH is continuous almost everywhere (except possibly at the identity which is not in the support of MATH). It is bounded by MATH. Applying the lemma for MATH and MATH shows that MATH is continuous everywhere. Now proceed by induction on MATH. For MATH, let MATH and let MATH, which is continuous. Then REF shows that MATH is continuous. |
math/0102205 | Let MATH. From CITE, MATH for MATH. Substitute MATH, and set MATH. Obtain MATH . To estimate this integral, square both sides and consider the double integral over a square in first quadrant of the plane, and then change to polar coordinates: MATH . For MATH, MATH which by assumption is less than MATH. Now for MATH, the inequality MATH holds, and yields MATH . Integrating the right hand side gives MATH . |
math/0102206 | We establish the lower bound using an inequality of Su CITE for the discrepancy of an arbitrary probability measure MATH on MATH: MATH . Here MATH denotes the MATH-th NAME coefficient of MATH (viewed on MATH mod MATH). Since every term is positive, we can use a dominant term in the sum as a lower bound. The NAME coefficients for the MATH-generator walk are MATH . Since MATH, for any MATH we have MATH . The inequality MATH holds for MATH and MATH. Then MATH as long as MATH. We ensure this by setting MATH (to be specified shortly) and let MATH. Then REF implies there exists an integer MATH such that MATH, which yields MATH, as desired. (Note that MATH was chosen to ensure MATH.) For this MATH we use just the MATH-th term in REF , and since MATH and MATH, we obtain MATH with MATH . Choosing MATH, we recover the lower bound REF. The upper bound is trickier to compute. We start with the NAME inequality CITE: given a probability measure MATH on MATH, for any integer MATH, MATH where MATH represents the NAME transform of MATH. Note that one may choose MATH in the NAME inequality so as to optimize the bound obtained. Since MATH for all MATH, it follows from REF that MATH . In light of the NAME inequality and MATH, we need to estimate a sum of the form MATH . Since MATH may be chosen freely, choose an integer MATH such that MATH . Recall that MATH is an approximability constant for MATH from REF and MATH is the number of steps in the walk. The reason for this choice of MATH will be evident later. Choose an integer MATH such that MATH . The sum in MATH may be grouped into MATH cohorts of integers MATH for MATH. Within each such cohort, the use of REF yields MATH. This says that the points of the sequence MATH (mod REF) in the unit cube in MATH are bounded away from the corners of the unit cube. In fact, they are bounded away from each other, since if MATH, then MATH as well. Therefore if MATH, then any box of side length MATH can contain at most one of the multiples from the sequence MATH (mod REF), MATH, with no such multiples in boxes containing the origin. So divide up the unit MATH-cube into disjoint boxes of side length MATH and sides parallel to the axes. In the worst case, all MATH points are distributed in the boxes in the MATH corners of the unit cube nearest the origin. The nearest points in such boxes (under the MATH metric) are at integral multiples of MATH and extend out in layers to distance at most MATH. A crude upper bound for the number of boxes whose nearest point is at MATH-distance MATH from the origin is MATH, and in the Euclidean metric the nearest point in such boxes is at least distance MATH from the origin. Hence we can bound MATH by grouping first by cohorts, then by corners and layers: MATH . The second inequality used the bound MATH and the definition of MATH. The third inequality follows by noting MATH from the definitions of MATH and MATH in REF . Using MATH, the log derivative of the expression of the innermost sum with respect to MATH can be bounded: MATH for all choices of MATH and MATH. Hence the expression in the inner sum decreases geometrically (by at least ratio MATH) as MATH increases, and so the inner sum is bounded by the first term (at MATH) times the constant MATH. Thus MATH . This sum may be bounded by noting that the largest term occurs when MATH. For MATH, the log derivative of the terms with respect to MATH is MATH for MATH. Thus the sum decreases geometrically (with at least ratio MATH) as MATH decreases, so the sum is therefore bounded by twice the final term at MATH: MATH where the second inequality follows from REF , and the final inequality from noting that the numerator is greatest for MATH. Using the NAME inequality we obtain the upper bound MATH for all MATH. An application of REF produces MATH which establishes the constant MATH in the statement of the theorem. |
math/0102209 | If MATH, MATH such that MATH for all MATH. Then, on an interval MATH, MATH, there are MATH intervals where MATH and MATH intervals where MATH, hence MATH which implies MATH. |
math/0102209 | For all MATH such that MATH implies MATH; hence MATH and MATH which implies MATH, MATH. On the other hand MATH, which implies the thesis. |
math/0102209 | (of REF .) If MATH, then MATH by REF , hence the thesis follows by REF . The same argument applies if MATH, therefore we may assume MATH. Because of REF , the claim is proved if we show that, assuming MATH and MATH, MATH is singularly traceable for any MATH. Observations: CASE: MATH implies MATH . CASE: MATH implies MATH . Setting MATH, we get MATH . Now assume MATH, which is equivalent to MATH and MATH. We shall show that, for any such MATH, MATH is a limit point of MATH and of MATH for any MATH (but we only need the case MATH). Indeed MATH . Let now MATH; then MATH . Finally, let MATH; then MATH . |
math/0102209 | By REF , when MATH, MATH is not trace class; moreover MATH and MATH. Therefore we may assume MATH not to be trace class and MATH. Then, let MATH be such that MATH. We have MATH . The thesis follows by REF . |
math/0102209 | Indeed MATH . |
math/0102209 | First we show that MATH . Indeed MATH . Therefore MATH . Now we observe that MATH . The thesis follows from REF . |
math/0102209 | MATH . |
math/0102209 | MATH is in CITE, MATH follows from REF , MATH follows by the NAME Theorem CITE, compare REF , p. REF. |
math/0102209 | The first statement follows from REF. Since MATH if it exists, the second statement follows from the first, however it has been proved directly in CITE. |
math/0102209 | MATH, so that MATH. MATH . Set MATH, and MATH, MATH. As MATH, there follows, for any MATH, MATH, as MATH, that is MATH is NAME in MATH. Therefore there is MATH such that MATH. Let us prove that MATH is independent of MATH. Indeed, if MATH, then MATH, as MATH, so that MATH and MATH have the same limit. |
math/0102209 | Follows by REF . |
math/0102209 | Define MATH . Then MATH is an isometry and MATH. Therefore, if MATH is an exponent of singular traceability for MATH, the corresponding NAME functional is homogeneous of order MATH. This implies that MATH coincides with MATH, namely MATH is the unique exponent of singular traceability, and the NAME functional corresponds to the MATH-dimensional NAME measure. |
math/0102209 | The proof will appear in CITE . |
math/0102209 | The proof will appear in CITE . |
math/0102209 | The proof will appear in CITE . |
math/0102215 | First we prove necessity. Let MATH be an amalgam of MATH and MATH over MATH, with MATH co-central in MATH. Let MATH, MATH, and MATH be as in REF . Since MATH, and MATH, MATH. But in MATH the commutator bracket acts bilinearly on MATH, so MATH . This proves REF , giving necessity. We will prove sufficiency by showing that the hypothesis that MATH is co-central in MATH together with REF imply REF from REF . Note that a co-central subgroup is necessarily normal, so REF applies. To prove REF , let MATH. Then MATH is necessarily in MATH, since MATH lies in MATH; thus MATH commutes with every element of MATH, and when considered inside of MATH, it must commute with everything in the center of MATH. Since MATH is generated by MATH and MATH, it follows that MATH. Thus, MATH. For the other inclusion, note that since MATH is co-central and MATH is nilpotent of class two, it follows that MATH. In particular, MATH . To prove REF , let MATH, MATH, MATH, MATH, and MATH such that MATH and MATH both lie in MATH. We want to show that MATH . Since MATH, note that MATH if and only if MATH, and that MATH in MATH. So we may assume without loss of generality that MATH. Write MATH, with MATH and MATH. Then MATH, so MATH implies that MATH. Then REF applies to MATH and MATH, so we conclude that MATH. Therefore, in MATH, we have: MATH . The third step can be done because both MATH and MATH lie in MATH. The next to last and last steps follow because MATH. On the other hand, in MATH we have: MATH since MATH is central in MATH. Therefore, MATH. Moreover, since MATH, the two commutators lie in MATH. This proves REF , and thus the theorem. |
math/0102217 | In order to prove MATH, we have to show that if MATH, then MATH . It is clear that we must have either MATH or MATH. In the first case we have MATH and the above inclusion follows. The other case is similar. In order to prove the reverse inclusion, we may assume that MATH and MATH are affine and let MATH and MATH. We identify the multiplier ideals with their global sections. We choose bases for MATH and MATH which are compatible with the filtrations induced by the respective multiplier ideals, as follows. Note that the set of multiplier ideals of MATH with coefficient MATH, MATH is finite and MATH if MATH. Therefore we can choose index sets MATH, possibly empty, for MATH, and elements MATH, for MATH and MATH such that for every MATH, a basis over MATH for MATH is given by MATH. We consider an analogous set of elements MATH, with MATH and MATH, satisfying the corresponding property with respect to the multiplier ideals of MATH. A basis in MATH is given by MATH . Therefore a basis in MATH is given by MATH . It is enough to prove that if MATH then MATH. Indeed, the above intersection has a basis given by a subset of MATH, because so has each member of the intersection. For every MATH and MATH such that MATH, we must have either MATH or MATH. Therefore for every MATH, we have either MATH or MATH. This gives MATH and finishes the proof of the lemma. |
math/0102217 | By REF , it is enough to prove that MATH . If there is MATH such that MATH, then MATH and therefore MATH, for every MATH. It follows that the right hand side of REF is MATH and the inclusion is obvious. We may therefore assume that MATH for every MATH, and by symmetry, that MATH, for every MATH. Suppose that for some MATH and MATH, we have MATH but for some MATH, MATH with MATH, we have MATH and MATH. We use NAME 's description for multiplier ideals of monomial ideals in CITE. It says that if MATH is a nonzero monomial ideal and MATH is the convex hull of MATH, then for every MATH, MATH . Here MATH denotes the unit vector MATH. Our hypothesis on MATH and MATH implies that MATH and MATH (where MATH). This means that there are linear maps MATH and MATH such that MATH if MATH and MATH if MATH, but MATH and MATH. Therefore we have MATH. If MATH, since MATH and MATH, we get MATH and MATH. We deduce that MATH. If MATH, then MATH and MATH and we have analogous inequalities if MATH. This shows that the linear map MATH, given by MATH has the property that MATH if MATH . Since we have MATH one more application of NAME 's theorem gives MATH, a contradiction. |
math/0102217 | Let MATH and MATH be log resolutions for all pairs MATH and MATH, and also for MATH and MATH, respectively. Let MATH and MATH. If MATH and MATH are the canonical projections, then we use the notation MATH . If MATH, then MATH. Suppose first that the assertion of the theorem is true for MATH, MATH, MATH and MATH. The change of variable formula for multiplier ideals gives MATH . Using also REF , we deduce MATH . Indeed, the sets MATH and MATH are finite, so that the above intersection has finitely many distinct terms, and therefore commutes with push-forward. Note that if MATH, MATH are quasicoherent subsheaves of MATH - modules and if MATH, then we have MATH. Indeed, this follows by applying MATH to the exact sequence MATH . Note that MATH vanishes. This follows by applying the NAME formula and the Local Vanishing theorem (see CITE) which gives MATH and MATH. Using the fact that MATH and MATH, one more application of the NAME formula and of the change of variable formula for the multiplier ideals gives: MATH . Putting everything together, we get via REF the statement of the theorem. To finish the proof, it is therefore enough to consider the case when all MATH and MATH are ideals defining effective divisors on MATH and MATH, respectively, whose union has simple normal crossings. Since the statement of the theorem is local in MATH and MATH, we may assume that we have étale morphisms MATH and MATH whose images contain the origins in the respective affine spaces, and principal monomial ideals MATH and MATH such that MATH and MATH, for all MATH. Since MATH and MATH are étale, the hypothesis implies MATH and MATH, for all MATH. Moreover, taking multiplier ideals commutes with the pull-back by étale morphisms, so that we can reduce the theorem to the case of monomial ideals, when it follows from REF . |
math/0102217 | Consider the diagonal embedding MATH. If MATH and MATH are the projections on the first and, respectively, the second component, let MATH . Note that we have MATH. We clearly have MATH, so that by the Restriction theorem (see CITE) we deduce MATH . On the other hand, REF gives MATH . The above inclusions imply the statement of the corollary. |
math/0102217 | Using the fact that MATH and similar equalities for MATH and MATH, we reduce immediately to the case MATH. By definition, we have to prove that for every MATH, we have MATH . If MATH is a positive integer such that for every MATH we have MATH, then we can apply REF to get MATH . On the other hand we have by REF and a similar inclusion for MATH. This proves the statement of the corollary. |
math/0102217 | This is precisely the statement of REF in the case MATH. |
math/0102217 | The proof of REF applies word by word in this case if we know that we have equality for monomial ideals. Therefore we may assume that MATH, MATH and that MATH and MATH are monomial ideals. We have to prove that if MATH, then MATH . If MATH or MATH, then the right hand side of the above inclusion is MATH, and the statement is trivial. Suppose therefore that we are in none of these cases. Moreover, if MATH, then it is easy to see from the definition of multiplier ideals that MATH (note that REF and its extension to the case of ideals give a more general statement). In this case we get the above inclusion since MATH. Therefore we may assume that MATH, and by symmetry, also that MATH. We use again the description in CITE for multiplier ideals of monomial ideals. First, this description shows that these ideals are generated by monomials. Suppose that we have MATH. Let MATH be a linear map such that MATH if MATH and MATH if MATH. By CITE, it is enough to prove that for every such MATH we have MATH. On the other hand, since MATH we get MATH. Similarly, since MATH we get MATH. This implies that MATH. |
math/0102217 | We apply REF to the sheaves of ideals MATH and MATH and to MATH. Note that the multiplier ideals of MATH are given by MATH for every MATH. We therefore obtain MATH . The statement of the proposition now follows from this and the formula for the multiplier ideals of MATH. |
math/0102217 | We consider first the case when MATH is defined by the vanishing of the last MATH coordinates. Let MATH and MATH be the canonical projections. If MATH is the ideal defining the origin in MATH, then MATH. Moreover, we have MATH. We know that MATH is equal to MATH, if MATH, and it is equal with MATH, otherwise. We may assume that MATH because otherwise the statement of the proposition is trivial. REF gives MATH . Since we have MATH if MATH, we deduce that MATH which finishes the proof of this case. We show now that if MATH is a zero dimensional ideal, then we can reduce the statement to the above case. Since the statement is local, we may assume that MATH, for some point MATH and it is enough to check the equality in the proposition in an open neighbourhood of MATH. Therefore we may assume that there is an étale morphism MATH with MATH such that MATH. Here we view MATH embedded in MATH as before. Note that MATH induces an isomorphism between the completions of the local rings of MATH and MATH at MATH and MATH, respectively. But MATH, so that there is an ideal MATH, such that MATH. Since construction of multiplier ideals commutes with pull-back by étale morphisms, we deduce the proposition in the case of zero-dimensional ideals from the case we have already proved. To finish the proof of the proposition, we show how to deduce the general case from that of zero dimensional ideals. Obviously it is enough to prove that for every MATH we have equality after localizing at MATH: MATH, where MATH and MATH. We fix MATH such that MATH and MATH. Let MATH be the ideal of the point MATH in MATH. Using REF and the fact that we know the statement for the zero-dimensional ideal MATH, for every MATH, we get: MATH . Since this is true for every MATH, NAME 's Intersection theorem gives the inclusion MATH. The reverse inclusion follows similarly: MATH for every MATH, and we apply again NAME 's Intersection theorem. This completes the proof of the proposition. |
math/0102217 | If for example MATH, with MATH, by replacing MATH with MATH and applying REF , we reduce ourselves to the case when MATH and in this case it is enough to prove that for every MATH, we get an induced isomorphism: MATH . As in the proof of REF , we first prove the case MATH. To simplify the notation, whenever there is no danger of confusion, we will identify MATH with MATH via MATH and denote it by MATH. We may clearly assume that the support of MATH consists of only one point MATH. Let MATH. We pick a regular system of parameters MATH for MATH around MATH such that MATH are in the ideal of MATH. After restricting to a suitable open neighbourhood of MATH, this induces an étale morphism MATH such that MATH, where MATH is defined by the vanishing of the last MATH coordinates. Moreover, since MATH, there is a subscheme MATH such that MATH and MATH induces an isomorphism MATH. We get a similar morphism MATH with analogous properties. Using the fact that construction of multiplier ideals commutes with pull-back by étale morphisms we reduce the equality in REF to the case when MATH and MATH are both affine spaces. Moreover, using REF , we see that we may assume that MATH. In this case the isomorphism MATH can be lifted to a local ring homomorphism MATH . Since MATH for MATH, MATH, it follows that MATH induces an isomorphism of the corresponding completion rings that is, it is étale. By restricting further to neighbourhoods of MATH in MATH and MATH, we may assume that MATH is induced by an étale scheme morphism MATH. Using one more time the invariance of multiplier ideals under pull-back for étale morphisms, we deduce REF in the zero-dimensional case. Suppose now that MATH has arbitrary dimension. It is enough to prove that for every MATH, the analogue of REF holds for the images of those two ideals in MATH. Let us denote by MATH and MATH the ideals defining MATH in MATH and MATH respectively. It is clear that we get induced isomophisms MATH, for every MATH. We apply the statement in the case of the zero dimensional schemes MATH and MATH. If MATH is such that MATH, then applying also REF , we get MATH . This shows that the image of MATH by the isomorphism induced by MATH is contained in MATH . NAME 's Intersection theorem gives now the inclusion MATH in REF . The reverse inclusion follows by symmetry. |
math/0102217 | The argument has the same flavor as the one used in REF , so that we just sketch it briefly. Again. it is clear that we may assume that MATH. The main point is to use REF to show that MATH is not a jumping number for a sheaf of ideals MATH on a variety MATH if and only if there is MATH such that for every MATH and every MATH, MATH, we have MATH . Indeed, suppose first that MATH is not a jumping number. Then there is MATH such that MATH . REF gives MATH which gives the inclusion in REF . Conversely, if we have REF , then REF gives MATH for all MATH. NAME 's Intersection theorem implies that MATH . Since the left hand side of the above inclusion contains MATH, one more application of NAME 's Intersection theorem shows that MATH is not a jumping number. Since REF is a statement about zero dimensional subschemes, the proof can be concluded with an argument which paralels the one in REF . |
math/0102219 | The first claim is straightforward. The second claim is a lengthy but straightforward computation. |
math/0102219 | Straightforward computation gives MATH . By hypothesis, we have MATH and hence by REF MATH . Integrating by parts over MATH gives MATH . Also note that from REF we have MATH . The claim follows. |
math/0102219 | Since MATH, the function MATH is orthogonal to the constants. Since the MATH-eigenspace consists of the constants, by the minimax principle MATH . The claim then follows from REF . |
math/0102219 | By REF, and hence MATH is Riemannian complete REF. Therefore MATH depends only on the geometry of the ends. (See, for example, REF ). In particular, let MATH be the NAME extension of MATH restricted to smooth functions supported in the MATH end with respect to the MATH-norm induced by MATH. Then MATH . Hence the claim reduces to the consideration of a single cusp. From REF we have that MATH . Hence the claim is a consequence of REF below. |
math/0102219 | Let MATH be as in REF where MATH. Since MATH is unitary, we have MATH. Moreover, since MATH, MATH and MATH. Thus, with MATH and MATH, identity REF specializes to MATH . The first claim follows. If MATH, then REF becomes MATH where MATH. From REF we have MATH as MATH. Thus, MATH as MATH tends to MATH. Since MATH belongs to MATH, the second claim follows from standard results on the essential spectrum of NAME operators. See, for example, REF. |
math/0102219 | Without loss of generality, the homogeneity constant MATH equals MATH. Note that the operator MATH is a NAME extension associated to the form MATH . In particular, to prove the claim it will suffice to show that the intersection of the MATH-unit ball with MATH is compact in MATH. (See, for example, REF). We claim that for any MATH and any (smooth) function MATH with MATH and MATH, we have MATH where MATH is the smallest nonzero eigenvalue of MATH. Indeed, REF holds and hence MATH . The claim then follows from inspecting REF . Now let MATH be a sequence of functions such that MATH. Since MATH, given MATH small, there exists MATH such that MATH. By NAME 's theorem, the restriction of MATH restricted to MATH is a MATH . NAME sequence. In particular, there exists MATH such that if MATH, then MATH . Applying REF to MATH then gives the claim. |
math/0102219 | Since MATH, by REF we have that for any MATH, there exists a MATH such that for all MATH . For MATH defined as in REF , let MATH. Note that we have MATH . It follows that REF holds iff MATH . Since MATH and MATH is a unitary transformation, we have MATH. Moreover, since MATH and MATH, from REF we have MATH where MATH. Thus, it follows from REF that MATH is convex and that MATH as MATH. Note that if MATH is a linear, then since MATH, and MATH, the function MATH is also convex. Thus, since MATH is conxex, it follows from REF that MATH is convex. Moreover, from the definition of MATH, we find that MATH satisfies REF with MATH. |
math/0102219 | Since MATH, there exists MATH such that for MATH, we have MATH. It then follows from REF , and REF , that for MATH . Multiply both sides by MATH and integrate over MATH to obtain MATH . Thus, it suffices to show that the integral on the right hand side remains bounded as MATH tends to zero. To verify this, we integrate by parts and obtain MATH where MATH, MATH, MATH, MATH, and MATH are constants that depend on MATH. By REF , MATH is convex and satisfies REF . It follows that both MATH and MATH are bounded for sufficently small MATH. The claim follows. |
math/0102219 | For MATH, define MATH where MATH is the constant mode of MATH. Now define MATH where MATH is the MATH norm. We will show that this renormalized sequence satisfies the claim. Let MATH be nonnegative, have support in MATH, and satisfy MATH on MATH. Using the NAME inequality (for both MATH and MATH) and the fact that MATH, we find that MATH . Using integration by parts and the fact that MATH, one finds a constant MATH depending only on MATH such that MATH is less than MATH times the integral of MATH over MATH. By REF , this latter integral is less than REF, and hence MATH is a bounded sequence in the space of MATH functions that have support outside of MATH. By NAME 's theorem we may pass to a subsequence such that MATH converges in this space to a function MATH. Hence the restriction of MATH to MATH converges in MATH. It follows that the restriction of the constant mode, MATH, to MATH converges to MATH in MATH norm. Thus, by NAME 's embedding theorem (in dimension REF) MATH converges to MATH in MATH, and hence the boundary conditions of REF on MATH converge in MATH as MATH for each nontrivial interval MATH. Since the metrics converge, the coefficients of REF converge in MATH. It follows that MATH extends uniquely to MATH, and that, for each MATH, the sequence MATH converges to MATH in MATH. Since MATH is obtained from MATH by adding on the constant mode MATH, we find that for every interval MATH, the restriction of MATH to MATH is uniformly bounded in MATH-norm. Hence, for each MATH, the argument above can be applied to give a further subsequence MATH whose restriction converges in MATH. Diagonalization yields a further subsequence that converges to a function MATH for every MATH. Since MATH is bounded, we may take a further subsequence such that MATH converges to some MATH. For each test function MATH supported away from MATH, we have that MATH. It follows that MATH is a weak - and hence by elliptic regularity, a strong - solution to MATH. It remains to show that MATH does not vanish identically. Since MATH and MATH, there exists an interval MATH symmetric about MATH such that for all MATH and MATH sufficiently small MATH and MATH . Indeed, the operator MATH adds MATH to the degree of homogeneity. Let MATH with MATH on MATH. From REF , and the self-adjointness of MATH, we obtain: MATH . Note that MATH equals MATH plus a smooth function supported on MATH bounded by MATH. Thus, it follows from REF that MATH . Thus, from the normalization of the MATH-norm in REF , we find that MATH . Therefore, since MATH restricted to MATH converges in MATH to MATH, the function MATH is nontrivial. |
math/0102219 | The operator MATH is defined via the (sesquilinear) form MATH restricted to MATH. For MATH, the domain of MATH is MATH. For any power MATH, the function MATH is uniformly continuous in MATH over closed intervals that do not contain MATH. It follows from REF that MATH is continuous in the `generalized sense' for MATH. Thus, by REF, any finite system of eigenvalues varies continuously for MATH. It follows that each MATH is continuous for MATH. We are left with showing the continuity at MATH. Let MATH be a span of eigenfunctions associated to the first MATH eigenvalues: MATH. Then by the minimax principle, MATH is the minimum value of the functional MATH over MATH where MATH denotes the MATH-norm. Let MATH be an eigenfunction of MATH with eigenvalue MATH. By REF , the function MATH belongs to the domain of MATH for small MATH. From the continuity of MATH, we find that MATH . Hence, by the minimax principle, MATH. For each MATH, let MATH be an eigenfunction of MATH with eigenvalue MATH. By applying REF to a subsequence whose eigenvalues limit to MATH, we obtain an eigenfunction MATH of MATH with eigenvalue MATH less than or equal to MATH. But since MATH is the smallest eigenvalue, MATH. It follows that MATH. The continuity at MATH of MATH for general MATH follows from a straightforward inductive argument involving MATH. (Note that no claim is made about the continuity of the family MATH.) |
math/0102219 | By the NAME monotonicity, we have that MATH is bounded below by the number, MATH, of eigenvalues of the NAME problem on MATH. By NAME monotonicity, MATH is bounded above by the number, MATH, of eigenvalues of the NAME problem on MATH plus the number of eigenvalues of the NAME problem on MATH. The latter is MATH since MATH converges uniformly on MATH. In sum, we have MATH . By REF the NAME (respectively, NAME) spectrum decomposes into the NAME (respectively, NAME) eigenvalues of MATH acting on MATH and those of MATH acting on MATH. By REF , the number of NAME (respectively, NAME) eigenvalues of MATH is MATH. Hence, the claim reduces to REF below. |
math/0102219 | First note that since MATH and MATH is homogeneous of degree REF, the eigenvalue REF on MATH is equivalent to the problem MATH on the dilated interval MATH. By conjuagting both sides of REF by MATH and applying REF , we obtain the equivalent eigenvalue problem MATH on the interval MATH. Let MATH be the endpoints of the interval MATH. Let MATH (respectively, MATH) be the number of eigenvalues MATH of REF on the interval MATH (respectively, MATH). Then by NAME bracketing (see, for example, CITE p REF), we have MATH . We claim that it suffices to show that MATH as MATH tends to zero, and that MATH where MATH is MATH for MATH large. Indeed, then REF would be equivalent to MATH . By REF below, we would then have MATH and the claim would follow from REF . To verify REF , we use homogeneity. We have MATH, and hence MATH as MATH. Estimate REF follows. Moreover, MATH . To prove REF , we use the following fact: If MATH is smooth, homogeneous of degree MATH, and MATH, then MATH as MATH tends to infinity. (Indeed, we have MATH and NAME expand in the first coordinate.) By applying this fact to MATH and MATH, we find that MATH and MATH . REF then follows from REF . |
math/0102219 | Let MATH denote the (necessarily simple) NAME (respectively, NAME) eigenvalues of REF for the interval MATH. By the standard theory, for each MATH, MATH is a decreasing function of MATH with MATH. It follows that MATH . See REF . Let MATH be a nonzero solution to REF with MATH and MATH (respectively, MATH). (The function MATH need not satisfy the NAME REF boundary condition at MATH.). Since MATH is integrable, there exist (see, for example, CITE p. REF) bounded functions MATH, MATH, such that MATH . Note that MATH if and only if MATH (respectively, MATH). Thus by REF , a NAME eigenvalue MATH if and only if MATH (respectively, MATH). In sum, for the NAME problem, MATH . Thus, letting MATH we have MATH and the claim follows for the NAME case. To prove the NAME case, one can use a similar analysis involving differentiating REF . Or one can derive the NAME case from the NAME case via NAME bracketing for NAME problems: the MATH-nd NAME eigenvalue is at least as great as the MATH-th NAME eigenvalue CITE. |
quant-ph/0102014 | Let us prove that task REF can be solved efficiently by a quantum algorithm. In fact, we can reduce the test to an instance of the Abelian hidden subgroup problem as follows. First, we compute the orders of the underlying elements (see CITE for example). Let the orders of MATH and MATH be MATH and MATH, respectively. Then for a tuple MATH from MATH, set MATH. Clearly MATH is a homomorphism from MATH into MATH, therefore this is an instance of the Abelian hidden subgroup problem, and its kernel can be found in polynomial time by a quantum algorithm. The kernel contains an element the last coordinate of which is relatively prime to MATH if and only if MATH is representable as a product of powers of MATH's. Also, from such an element an expression for MATH in the desired form can be constructed efficiently. |
quant-ph/0102014 | The proof is similar to the one of REF , where MATH is taken. |
quant-ph/0102014 | We use the presentation of MATH obtained by the algorithm of REF to find generators for MATH. Let MATH be the generating set from the presentation. If MATH generates MATH then it is easy to find generators for MATH. Let MATH denote the set of elements obtained by substituting the generators in MATH into the relators, and let MATH stand for the normal closure (the smallest normal subgroup containing) of MATH. Then MATH since MATH and MATH by definition of MATH and MATH. Still some care has to be taken since it is possible that MATH generates MATH only modulo MATH, that is it might generate a proper subgroup of MATH . Therefore some additional elements should be added to MATH. Let MATH be the generating set for MATH. Using the constructive membership test for MATH we express the original generators from MATH modulo MATH with straight line programs in terms of the elements of MATH. For each element MATH we form the quotient MATH where MATH is the element obtained by substituting the generators from MATH into the straight line program for MATH modulo MATH. Let MATH be the set of all the quotients formed this way. Note that MATH and MATH generate together MATH. Then one can verify that the normal closure of MATH in MATH is MATH. Thus, from MATH and MATH we can find generators for MATH in time polynomial in the MATH using the normal closure algorithm of CITE. We obtained the desired result. |
quant-ph/0102014 | First we extend naturally MATH to MATH: on a coset of MATH, it takes the value MATH for an arbitrary member MATH of the coset. The algorithm is the standard quantum algorithm for the Abelian hidden subgroup problem. We repeat several times the following steps to find a set of generators for MATH. CASE: Prepare the initial superposition: MATH. CASE: Apply the Abelian quantum NAME transform in MATH on the first register: MATH. CASE: Call MATH: MATH. CASE: Apply again the NAME transform in MATH: MATH. CASE: Observe the first register. By hypothesis, the states MATH are orthogonal for distinct MATH, therefore an observation of the first register will give a uniform probability distribution on MATH. After sufficient number of iterations, this will give a set of generators for MATH, which leads then to a set of generators for MATH. Note that in the above steps it is sufficient to compute only the approximate quantum NAME transform on MATH which can be done in polynomial time. |
quant-ph/0102014 | For applying REF , one has to verify that we can perform tasks REF - REF. If MATH is of polynomial size, it is trivial. Therefore we suppose that MATH is solvable. We will closely follow the approach indicated by NAME in CITE for dealing with factor groups. First, let MATH. To compute the order of MATH in MATH, we compute the period of the quantum function MATH, where MATH for some multiple MATH of the order. This function can be computed efficiently since one can prepare the superposition MATH by REF , and for example we can take MATH as the order of MATH in MATH. Therefore by REF one can find this period. Second, let MATH and let MATH be pairwise commuting elements modulo MATH. generating some Abelian subgroup MATH. We compute the orders of the underlying elements on MATH using the previous method. Let the orders of MATH and MATH be MATH and MATH, respectively. Then for a tuple MATH from MATH, set MATH. Then MATH is a homomorphism from MATH into MATH. From REF , the kernel of MATH can be computed in polynomial time by a quantum algorithm. Moreover it contains an element the last coordinate of which is relatively prime to MATH if and only if MATH is representable as a product of powers of MATH-s. Also, from such an element an expression for MATH in the desired form can be constructed efficiently using elementary number theory. |
quant-ph/0102014 | Let MATH be a hidden subgroup of MATH defined by the function MATH. We start with the following observation. If MATH is a normal subgroup of MATH and MATH is such that MATH and MATH, then by the isomorphism theorem, MATH which implies MATH. We will generate such a subgroup MATH for MATH. As the commutator subgroup MATH of MATH consists of products conjugates of commutators of the generators of MATH we can enumerate MATH, and therefore also MATH, in time polynomial in the MATH. We consider the function MATH which can be computed by querying MATH times the function MATH. The function MATH hides the subgroup MATH. Note that MATH is normal since MATH is Abelian. Thus by REF , we can find generators for MATH by a quantum algorithm in time polynomial in the size of the input + MATH since MATH, because MATH is Abelian. For each generator MATH of MATH, we enumerate all the elements of coset MATH and select an element of MATH. The cost of this step is again polynomial in the MATH. We take for MATH the subgroup of MATH generated by the selected elements and MATH. We get MATH, and by the definition of the selected elements MATH. |
quant-ph/0102014 | Let MATH be a subgroup of MATH hidden by the function MATH. The main line of the proof is like in REF : we will generate MATH which satisfies MATH and MATH (or equivalently MATH). Again we start the generation of MATH with MATH which can be computed in polynomial time in the input size by REF since MATH is Abelian. The additional generators of MATH will be obtained from a set MATH which, for every subgroup MATH (in particular, for MATH), contains some generator set for MATH. For each MATH, we will verify if MATH (equivalently MATH or also MATH), and in the positive case we will find some MATH such that MATH. Both of these tasks will be reduced to the Abelian hidden subgroup problem, and the elements of the form MATH will be the additional generators of MATH. If MATH is cyclic, we use REF to find generators for the NAME subgroups of MATH (note that MATH). Each NAME will be cyclic (and unique), therefore a random element of the NAME MATH-subgroup will be a generator with probability MATH. Note that one can check if the choosen element is really a generator by using the order finding procedure of REF . Then, for each MATH we choose a generator MATH for the NAME MATH-subgroup after iterating the previous random choice. The MATH-subgroups of MATH are MATH, where MATH is the order of the NAME MATH-subgroup of MATH. Let MATH stand for the union of the sets MATH over all primes MATH dividing MATH. Note that MATH, and the cost of constucting MATH is polynomial in the input size. MATH contains a generating set for an arbitray subgroup MATH of MATH because for each MATH, it contains a generator for the NAME MATH-subgroup of MATH (namely MATH where MATH is the smallest positive integer MATH such that MATH). In the general case, let MATH be a complete set of coset representatives of MATH. MATH can be constructed by the following standard method. We start with the set MATH. In each round we adjoin to MATH a representative MATH of a new coset, for each MATH and each generator MATH of MATH, if MATH, for all MATH. This membership test can be achieved using a quantum algorithm for testing membership of MATH in the commutative group MATH. The procedure stops if no new element can be added. Then, for each MATH, we consider the function defined on MATH as follows. For every MATH, let MATH and let MATH. Obviously, for MATH and MATH, MATH if and only if MATH, while MATH if and only if MATH. We claim that MATH is either empty or a coset of MATH in MATH. Indeed, if MATH contains MATH for some MATH, then MATH, and conversely for all MATH such that MATH, we have MATH. It follows that in the group MATH, MATH hides either MATH or MATH for some MATH depending on whether MATH is empty or not. Note that this set is indeed a subgroup because MATH is an elementary Abelian MATH-group. We remark that MATH is determined only modulo MATH. As MATH is Abelian, we can find generators for this hidden subgroup in quantum polynomial time. From any generator of type MATH we obtain an element MATH. Repeating this, we collect elements in MATH for each of MATH such that MATH. Let MATH be the subgroup of MATH generated by the collected elements and by MATH. Then by construction MATH is a subgroup of MATH which satisfies the claimed properties. |
quant-ph/0102054 | The element of matrix MATH in MATH-th row and MATH-th column is MATH. The element of matrix MATH in the same row and column is MATH. As in the each row and column of matrices MATH there is a finite number of nonzero elements, it is also finite in the given series. Therefore the elements of the series can be rearranged, and MATH. |
quant-ph/0102054 | Let us consider the matrix MATH. The element of this matrix MATH, where MATH is MATH-th row of the matrix MATH. Let us consider the matrix MATH. The diagonal element of this matrix is MATH . On the other hand, taking into account REF , we get that MATH . Therefore MATH. It means that MATH . This implies that every element of series REF does not exceed MATH. Hence MATH. The last inequality implies that MATH. Therefore MATH. |
quant-ph/0102054 | Let us assume that the rows of the matrix MATH are orthogonal. Let us consider REF from the proof of REF , that is, MATH. As the rows of the matrix MATH are orthogonal, MATH. Hence MATH, that is, MATH or MATH. Therefore MATH or MATH. Let as assume that every row of the matrix has norm MATH or MATH. Then MATH and in compliance with REF , MATH. This implies that MATH. Hence the rows of the matrix are orthogonal. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.