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math/0105246 | By REF and induction, the estimate REF holds for every MATH. Fix MATH and consider the optimal MATH-coupling of (the laws of) MATH and MATH. Then for MATH we have, using the mean value theorem and the NAME - NAME inequality, MATH . By the optimality of the coupling and REF, MATH and by REF for MATH whence the result follows. |
math/0105246 | Since MATH, by NAME 's inequality MATH, and the monotonicity follows. In particular, MATH, and since REF trivially is satisfied, the result follows from REF . |
math/0105246 | Since MATH, REF yields MATH, and the monotonicity follows. The estimate thus follows from REF . |
math/0105246 | Our proof (similar to that of REF) is by induction on MATH in the first case and on MATH, using REF to get started, in the second case. We may therefore assume as our induction hypothesis that MATH is bounded and continuous. It is easily checked that, for each MATH, the inner integral MATH is a density function for the random variable MATH and, using dominated convergence, that MATH is bounded and continuous. Indeed, MATH, and since MATH by symmetry in REF, MATH, uniformly in MATH and MATH. It follows, by dominated convergence again, that MATH is a bounded continuous density for MATH. |
math/0105246 | With a slight abuse of notation, we find, for MATH with any law, MATH coupling MATH optimally and MATH optimally and choosing MATH, MATH, and MATH to be independent. In calculating the MATH-norm value on the right in REF, condition on MATH and then use subadditivity of MATH-norm together with independence to bound that value by MATH. This establishes the first assertion: MATH. Therefore, MATH is nonincreasing, and hence bounded by MATH, which is bounded by MATH if MATH. |
math/0105246 | Using the common optimal coupling for MATH and MATH, this is immediate from the inequality MATH which in turn follows from the fact CITE that MATH is convex in MATH. |
math/0105246 | MATH . |
math/0105246 | First we note that REF can be written MATH. One root of this equation is MATH, and since MATH is convex, with derivative less than MATH at MATH, it follows that the equation has two positive roots, MATH and MATH, and that MATH for MATH, while MATH for MATH. Next we note that REF holds for MATH, with MATH by REF and the inequality MATH, MATH. We then proceed by induction on MATH. For the induction step, suppose that MATH and that MATH has zero mean and satisfies MATH. By the induction hypothesis, there exist constants MATH, MATH, such that MATH . Using our usual coupling of MATH and MATH in terms of the optimal coupling of MATH and MATH, we find easily by REF , for MATH [with MATH, MATH, and MATH independent], MATH . So by induction on MATH it follows that MATH . By the induction REF this yields, for some MATH (depending on MATH), MATH . Let MATH. We break our treatment into three cases: CASE: If MATH, we write the sum in REF as MATH and thus REF shows that REF holds with MATH. CASE: If MATH, we write the sum in REF as MATH and thus REF holds with MATH. CASE: If MATH, the sum in REF equals MATH. Consequently, REF holds with any MATH. It remains to verify that this yields the MATH given in REF. First, if MATH, then the induction hypothesis yields MATH, so REF gives MATH. Similarly, for MATH, MATH and REF shows that any MATH will do. For MATH, we have MATH, so REF yields MATH. The same applies for MATH, since again MATH and we thus may choose MATH so small that MATH. Finally, for MATH, we have MATH since MATH is increasing and equals MATH when MATH. Hence REF applies. |
math/0105246 | If MATH, then, with MATH, MATH, and MATH independent, by REF and a trinomial expansion we have MATH . We apply this with MATH for both MATH and MATH, and note that by induction on MATH, all terms in the sum with MATH and MATH coincide for the two choices of MATH. Hence, MATH for MATH and MATH and the result follows. |
math/0105246 | Let MATH be an optimal coupling of MATH and MATH. If MATH, then MATH as desired. If MATH, we employ the factorization MATH whence MATH where at the second inequality we have employed NAME 's inequality and at the third we have invoked the optimality of the coupling. Another application of NAME 's inequality, this time with conjugate exponents MATH and MATH, yields MATH for MATH, and REF is trivially an equality when MATH or MATH. Combining REF and rearranging, we obtain the desired result. |
math/0105246 | By REF , we have the bound MATH . Thus REF yield the explicit bound MATH . |
math/0105246 | The cases MATH and MATH follow immediately from REF ; see also REF . When MATH, fix MATH. By REF (the latter applied to MATH), for some MATH we have MATH and thus REF implies REF with MATH. By taking MATH sufficiently large, we obtain the result for any MATH. |
math/0105246 | We may assume that MATH and MATH (in particular, MATH), because otherwise MATH with MATH: see REF . By induction then MATH and MATH for every MATH: see REF . As usual, let MATH, MATH, and MATH be independent. If MATH, MATH, and MATH, where MATH, then MATH . Thus, MATH . If further MATH, and thus by NAME 's inequality MATH, this yields MATH . Hence, by induction on MATH and our assumption on the first two moments of MATH, for any MATH, MATH and in particular MATH . Now suppose that MATH. Using an optimal coupling between MATH and MATH, and the fact that MATH has moments of all orders, we find MATH . Combining this with REF, we obtain, for MATH, MATH which implies that MATH for every MATH such that MATH. Consequently, if MATH for every MATH, then MATH for every MATH, and REF applies to yield MATH. |
math/0105246 | Define MATH, where MATH, and similarly MATH. Then MATH and similarly for MATH, while MATH . Together, these yield the first inequality. The second follows by applying the first to MATH and to MATH and summing or subtracting, depending on the parity of MATH. |
math/0105246 | Let MATH. If we apply REF and then use REF , we find, for any MATH, MATH for any MATH. It follows from REF that MATH, for some MATH not depending on MATH. Choosing MATH thus gives, with MATH, MATH and thus MATH with MATH. The result follows, since MATH as MATH. |
math/0105246 | The estimate REF follows from REF , because (whenever MATH exists) MATH. The estimate REF follows from REF using REF and the discussion following it. |
math/0105246 | The estimate REF is immediate from REF . Next, REF for some MATH and REF imply REF for any MATH by the argument used to show REF in the proof of REF . |
math/0105246 | We use the fact that the moment generating functions MATH and MATH are entire analytic functions in the complex plane MATH. Let MATH. There exists a (unique) function MATH which is continuous on MATH and analytic in MATH such that MATH for MATH and MATH for MATH; this function is called harmonic measure and is probabilistically given by the probability that a Brownian motion starting at MATH hits MATH before it hits MATH. Let MATH and MATH. For MATH, MATH which by REF is bounded by some constant MATH (depending on MATH but not on MATH). Let further MATH; we may of course restrict attention to those values of MATH satisfying MATH. Now MATH for MATH and MATH for MATH; thus (since MATH) MATH for every MATH. Since MATH is subharmonic and the right hand side is harmonic in MATH and continuous on its closure, REF holds for every MATH, compare CITE. In particular, setting MATH, we have MATH or MATH . Let MATH be as in REF . By REF and NAME 's estimates CITE, MATH . Since by REF MATH it follows that MATH with MATH. |
math/0105247 | Suppose that MATH is a simple sub-bimodule of MATH, and that MATH has a balanced symplectic form MATH. Then either MATH, or MATH. In the former case MATH is isotropic, so suppose the latter holds, that is, that the restriction of MATH to MATH is non-degenerate, hence a symplectic form on MATH. Since MATH is a product of full matrix algebras, MATH must be isomorphic to a bimodule of the form MATH where MATH and MATH are simple left MATH-modules. The symplectic form gives a bimodule isomorphism MATH, so MATH. It follows that MATH. Since MATH is simple, the homothety MATH is surjective, and since this is a bimodule map, MATH is a quotient of MATH as a bimodule. Let MATH be the image in MATH of the identity element of MATH. Then every element of MATH is of the form MATH for some MATH. Now MATH. Thus MATH, and since this is true for all MATH, non-degeneracy implies that MATH, a contradiction. |
math/0105247 | Since bimodules over MATH are semisimple, MATH has a bimodule complement MATH. The fact that MATH is balanced implies that the map MATH induced by MATH is a bimodule map. It is injective, for if MATH were a simple sub-bimodule of the kernel, then MATH would be isotropic by REF , contradicting maximality. Dualizing, the map MATH induced by MATH is surjective. NAME for the category of bimodules then implies that the map MATH induced by MATH factors through MATH by a bimodule map, say MATH. Thus MATH for all MATH. By skew-symmetry, also MATH. Let MATH. Clearly this is a bimodule complement to MATH. Moreover MATH . Now the last term on the right hand side is zero since MATH is isotropic, and the second and third terms each give MATH. Thus the right hand side vanishes. Now MATH and MATH are isotropic, so both have dimension bounded by MATH. But this implies that they have dimension equal to this, and hence both are Lagrangian. Now the map MATH induced by MATH is an isomorphism, and this gives the final assertion. |
math/0105247 | Extend MATH to a maximal isotropic sub-bimodule, which by semisimplicity we may write as MATH, and choose a Lagrangian complement MATH to this. Then MATH is a complement to MATH, and it is coisotropic since MATH. |
math/0105247 | This means that MATH where MATH, MATH and MATH. Here MATH is the symplectic form induced by MATH on MATH and MATH is the vector field on MATH induced by MATH. Each tangent space of MATH can be identified with MATH itself, and MATH can be identified with MATH. The vector field induced by MATH is then the map MATH sending MATH to MATH. The equation thus becomes MATH for MATH and MATH. Now the formula for MATH immediately gives MATH, and using the properties of MATH this becomes MATH, as required. |
math/0105247 | Using the trace pairing MATH to identify MATH with its dual, the map MATH coincides with the map MATH and MATH coincides with MATH as an element of MATH. |
math/0105247 | Let MATH be a quadruple. Choose a maximal isotropic sub-bimodule MATH of MATH and identify MATH with MATH as in REF . The vertex set MATH of MATH is chosen to index the simple left MATH-modules up to isomorphism. The dimension vector MATH is defined by letting MATH be the dimension of the MATH-th simple module. Clearly MATH. If MATH and MATH are simple MATH-modules, then MATH is a simple MATH-bimodule, and all simple modules are obtained in this way. We define the quiver MATH by letting the number of arrows from MATH to MATH be the multiplicity of this simple module in MATH. Finally, MATH is a MATH-invariant element of MATH. Using the trace pairing it corresponds to a MATH-invariant element of MATH. The MATH-invariance means that each of the matrices involved is a scalar multiple of the identity matrix, and combining the scalars gives the element MATH-with MATH. |
math/0105247 | Since MATH is balanced and skew-symmetric, for MATH we have MATH . But this can be rewritten as MATH. |
math/0105247 | The stabilizer MATH of MATH is reductive by NAME 's Theorem, see CITE. Now MATH is the group of units of MATH, and the elements MATH with MATH in the radical of MATH form a closed connected unipotent normal subgroup. Thus the radical must be zero. |
math/0105247 | Let MATH. We show that there is an element MATH with MATH. Since MATH is non-degenerate, MATH, so by dimensions MATH. NAME now implies that there are elements MATH and MATH with MATH. Thus MATH. Now since MATH and MATH we have MATH. The claim follows on writing MATH corresponding to this decomposition, and observing that MATH. The rest is clear. |
math/0105247 | We consider the map MATH sending MATH to MATH. The formula for MATH shows that MATH if and only if MATH, from which the first assertion follows. The conjugation action of MATH on MATH gives a map MATH, and the fact that MATH implies that the differential of this map at MATH is surjective. Let MATH. (See CITE.) By dimensions it follows that the induced map MATH is etale at MATH. Thus Luna's Fundamental Lemma CITE applies to MATH. There is an affine open subset MATH of MATH which contains MATH, is saturated for MATH, the restriction of MATH to MATH is etale, the image MATH is an affine open subset of MATH, saturated for MATH, the morphism MATH is etale, and the morphisms MATH and MATH induce a MATH-isomorphism MATH. Let MATH. Since MATH is a closed subvariety of MATH, one can identify MATH with a closed subvariety of MATH, and clearly MATH where MATH and MATH. Now CITE, applied to the map MATH and the subvariety MATH, implies that the morphism MATH is etale. The fact that MATH is saturated for MATH implies that MATH is saturated for MATH, so the domain of MATH is an open subset of MATH. Similarly, the fact that MATH is saturated for MATH implies that MATH is saturated for MATH, so the codomain of MATH is an open subset of MATH. The claim follows. |
math/0105247 | Since MATH we have MATH. Thus MATH where the last equality holds because MATH. |
math/0105247 | Identifying MATH and MATH with their tangent spaces at the origin, the differential of MATH at the origin is given by MATH so MATH. By the dimension formula MATH must be surjective, so MATH is smooth at REF. |
math/0105247 | MATH, so MATH. Since MATH is coisotropic, this implies that MATH. The left hand side can be rewritten as MATH, and hence MATH. Finally, MATH. |
math/0105247 | Let MATH be the projection coming from the decomposition of REF . Since MATH is isotropic, it follows that MATH. Let MATH be the restriction of MATH. By REF it is etale at REF. We apply CITE to the MATH-morphism MATH. Since the codomain is a vector space it is smooth, so certainly normal. Thus there is an affine open subvariety MATH of MATH containing REF such that the restriction of MATH to MATH is etale. In particular MATH is etale at REF. Now since MATH we have MATH. Thus the pullback of MATH along the closed immersion MATH is a morphism as in the statement of the lemma. The assertion follows. |
math/0105247 | Combine REF . |
math/0105247 | MATH is locally closed since the conditions that MATH and MATH be MATH-module maps and that MATH are closed, while the conditions that MATH be injective and MATH surjective are open. Since MATH is finitely generated as a MATH-algebra, it follows that MATH is a finitely-presented MATH-module, so that MATH is finite dimensional. Choose an exact sequence MATH with the property that the connecting map MATH is surjective. For example one can take MATH to be the universal exact sequence, with MATH where MATH. Let MATH be the set of tuples MATH where MATH is a MATH-module map for the module structure given by MATH, and the diagram MATH commutes. Clearly MATH is a closed subset. We relate the dimensions of MATH and MATH by computing the fibres of the projection MATH. Suppose MATH. By the surjectivity of the connecting map, there is some MATH. Moreover, given homomorphisms MATH and MATH, we have MATH if and only if MATH and MATH, so if and only if MATH and MATH for some MATH. Thus MATH is isomorphic to MATH. It follows that MATH. There is also a projection MATH. To compute the fibre over MATH, we construct the pushout MATH . Any invertible linear map MATH induces a MATH-module structure on MATH such that MATH is a module homomorphism. If MATH is given by invertible linear maps MATH (MATH), this structure corresponds to an element MATH with MATH and MATH. Conversely, if MATH is in MATH, then by the universal property of the pushout we have constructed, there is a map MATH giving a commutative diagram MATH and clearly MATH is an isomorphism. Thus the fibre over MATH is isomorphic to the product over MATH of the set of invertible linear maps MATH. Thus each fibre is irreducible of dimension MATH, so MATH . The assertion follows. |
math/0105247 | Let MATH be the set of quadruples MATH forming an exact sequence as above, with MATH, and with MATH. Clearly MATH is constructible. The first projection MATH has image MATH, so that MATH is constructible. If MATH, since the components of MATH must form a basis of MATH, any other surjective map MATH must be obtained by composing MATH with an automorphism of MATH. It follows that MATH acts freely and transitively on the fibres of the projection, so MATH . The second projection MATH has image contained in MATH, and the fibre over any point MATH is contained in the variety MATH of REF for MATH and MATH. It follows that the fibre has dimension at most MATH. Thus MATH, giving the result. |
math/0105247 | For simplicity of notation, let MATH. First suppose that MATH. Applying MATH to the exact sequence MATH gives an exact sequence MATH . Thus MATH by dimensions. Now suppose that MATH. If there is no MATH with MATH then MATH has no composition factors isomorphic to MATH, so the Hom space is zero. Thus suppose otherwise, and choose MATH maximal with MATH. By the definition of top-type we have MATH. Now applying MATH to the exact sequence MATH and using the fact that MATH has no composition factor isomorphic to MATH, we deduce that MATH embeds in MATH. |
math/0105247 | Let MATH. Let MATH be the subset of MATH consisting of those elements MATH such that MATH has top-type MATH and with MATH. By induction on MATH it is constructible and MATH . By REF , for any element MATH there is an exact sequence MATH with MATH. Thus by REF we have MATH where MATH . Now by CITE we have MATH so MATH. The result follows. |
math/0105247 | Any element of MATH has top-type MATH for suitable MATH and MATH. It thus suffices to prove that MATH for each possible top-type. We group the first sum according to the value of MATH. Fix MATH, and let the values of MATH with MATH be MATH. It clearly suffices to prove that MATH . If MATH, then MATH, and the inequality is trivial. Thus suppose that MATH. Defining MATH for MATH, we have MATH if MATH, and MATH. Thus the inequality becomes MATH or equivalently MATH . Now the first of the two terms on the left is clearly at least MATH, and the second is at least REF, for it is the (positive definite) quadratic form for the NAME graph of type MATH applied to the dimension vector MATH. |
math/0105247 | Suppose that MATH is an irreducible closed subset of MATH. Let MATH be the complement of MATH. By CITE there is a long exact sequence MATH so it suffices to prove that if MATH then MATH and MATH both vanish. Now MATH embeds in MATH by CITE, and this space vanishes since MATH has property MATH and MATH has codimension at least MATH in MATH. Also MATH by excision CITE, and this vanishes since MATH has MATH. |
math/0105247 | MATH is regular in codimension REF since MATH is normal and and its complement has codimension at least REF. Thus we just need MATH to have property MATH, and this is a special case of the theorem. |
math/0105247 | Observe that the preprojective relations involve commutators, and if MATH is a loop and the dimension vector at the corresponding vertex is REF, then MATH automatically. Thus MATH and MATH can be removed from the relations. The result is now clear - in REF MATH is an affine space, and in REF it is the product of a Kleinian singularity and an affine space. |
math/0105247 | We suppose that REF fail, and derive a contradiction. We have MATH. Since REF fails this must be an equality, which since the MATH have non-negative integer components, implies that the MATH must be distinct coordinate vectors. Since MATH is sincere, they are exactly the coordinate vectors in some order, and the multiplicities MATH are the corresponding components of MATH. If MATH for some MATH then we are in the nearly Kleinian case, a contradiction. Thus the decomposition of MATH as a sum over MATH of MATH copies of each MATH has at least two terms, so since MATH we know that MATH. Since REF fails, it follows that MATH, and MATH whenever MATH, that is, whenever there is a loop at the corresponding vertex. Now we have MATH . As observed above, we have MATH, so MATH for some MATH by CITE. Then MATH . Now since REF fails we have MATH for all MATH, that is, that MATH. Equivalently MATH . Let MATH be the quiver obtained from MATH by deleting all loops. Since the MATH are distinct coordinate vectors we have MATH . Now this is zero, for either MATH, or there are loops in MATH at the vertex, in which case MATH. Since MATH is a vector for MATH with positive integer components, connected support, and the bilinear form with each coordinate vector gives REF, it follows that MATH is an extended NAME quiver and MATH is a multiple of the minimal positive imaginary root MATH (see CITE). Now if MATH contains any loops then some MATH, so that MATH. Otherwise MATH is extended NAME itself, and the fact that MATH implies that MATH. Thus we are in the nearly Kleinian case, a contradiction. |
math/0105247 | In view of the etale local structure result, it suffices to prove normality for MATH. We prove it for all quivers and dimension vectors MATH by induction on MATH. Since a product of normal varieties is normal, and a symmetric product of copies of a normal variety is normal, by the main result of CITE we may assume that MATH. Clearly we may suppose that MATH is sincere. Also, we may assume that we are not in the nearly Kleinian case, since that has been dealt with separately. By the lemma, for any representation type MATH, REF or REF hold. By the etale local structure, a point in the stratum of this representation type has a neighbourhood which is isomorphic in the etale topology to a neighbourhood of REF in MATH. By the remark before REF , the points of strata where REF holds are normal by induction. On the other hand, if a stratum satisfies REF , then REF and CITE imply that the inverse image of the stratum in MATH has codimension at least REF. We now apply REF . Let MATH be the normal locus of MATH, an open subset. By the observations above, its complement has inverse image in MATH of codimension at least REF. Thus in fact MATH is normal. |
math/0105247 | It suffices to observe that MATH is normal and that if MATH is chosen so that MATH-semistables are MATH-stable, then since MATH is a symplectic quotient, it has a symplectic form, and hence has trivial canonical class. |
math/0105247 | Let MATH be the inclusion, and let MATH be the composition of the isomorphism MATH and the inclusion MATH. Since MATH is smooth, the codimension condition implies that any locally free sheaf MATH on MATH is MATH-closed CITE, so the map MATH is an isomorphism by CITE. Similarly, if MATH is a locally free sheaf on MATH then MATH is injective by MATH-purity CITE. Let MATH and MATH have dimension MATH. Since MATH has trivial canonical class, there is a nowhere vanishing MATH-form MATH on MATH. Then MATH is such a form on MATH, and since MATH is an isomorphism, there is a MATH-form MATH on MATH with MATH. Now MATH, so the fact that MATH is injective implies that MATH. In particular MATH is a nowhere vanishing MATH-form on MATH. Now if MATH then MATH is the composition MATH where MATH is the induced map on tangent spaces. Since this is nonzero, MATH must be invertible. Thus MATH is etale, and hence it has finite fibres. Then by NAME 's Main Theorem, MATH is bijective. Since it is also birational, it is an isomorphism. |
math/0105247 | Since MATH does not depend on the orientation of MATH, see for example CITE, and since MATH has no loops, we can reorient it so that it has no oriented cycles. Choosing MATH so that any MATH-semistable is MATH-stable, the semisimplification map is a resolution of singularities, and the quotient MATH classifies MATH-orbits of MATH-stable elements of MATH. By the lemma, if MATH is smooth, the semisimplification map is a bijection. Thus, considering the fibre over REF, the set of nilpotent MATH-stable elements of MATH must form a single orbit, so must have dimension MATH (taking into account the fact that MATH has a subgroup isomorphic to MATH which acts trivially). Since MATH has no oriented cycles, by CITE the nilpotent elements of MATH form an equidimensional subvariety of MATH of dimension MATH. Since the nilpotent elements which are MATH-stable form an open subset of this, we must have MATH . In other words MATH, so that MATH is a real root. The equivalence with coordinate vectors follows from the results in CITE. |
math/0105247 | Clearly MATH is constant on orbits of MATH and equivariant for the conjugation action of MATH. The First Fundamental Theorem of Invariant Theory and an induction on MATH shows that MATH induces a closed embedding MATH, see CITE. The image of MATH contains MATH, since if MATH one can identify MATH with the image of MATH, and take MATH and MATH to be the inclusion. In CITE the proof is completed using the NAME Theorem, and perhaps our lemma could be proved in a similar way; here we use the NAME reductions. Clearly it suffices to prove that MATH is irreducible of dimension MATH where MATH. By deframing, MATH for the quiver MATH illustrated in the introduction, MATH and MATH, where MATH is chosen so that MATH. Clearly MATH so by CITE it suffices to show that MATH. Any decomposition of MATH as a sum of positive roots must include one term whose component at vertex MATH is nonzero, say MATH with MATH. All other terms must be of the form MATH with MATH, where MATH . The decompositions involved in the definition of MATH are those in which the summands satisfy MATH, that is, MATH. Let MATH (with MATH and MATH). Clearly the sequence MATH is obtained from MATH by decreasing MATH by one, and increasing MATH by one, for each term of the form MATH in the decomposition of MATH. Since MATH, to show that MATH we need MATH for a nontrivial decomposition, or equivalently MATH. By including additional terms of the form MATH in the decomposition, we may reorder the MATH, and hence assume that MATH whenever MATH. This will reduce the size of MATH, but not change the value of MATH. We now partition the sums according to the value of MATH. If MATH and MATH with MATH, then MATH is the number of NAME blocks of eigenvalue MATH of size MATH, so MATH. It follows that MATH, strict if not all MATH. This is just a numerical condition, but to avoid writing out a proof, observe that it is implicit in the NAME Theorem (see CITE). |
math/0105248 | We basically follow REF, making all estimates explicit. We study in this paper only properties of the univariate distributions of MATH. We thus take the liberty of letting MATH denote any random variable with the appropriate distribution CITE. We then may choose MATH defined on the same probability space as MATH and such that MATH . Further, let MATH be an independent copy of MATH and let MATH be independent of everything else. For convenience we write MATH. Observe, by REF, that MATH and recall from REF that MATH . Therefore, MATH . Now MATH say. Given MATH, the random variables MATH and MATH are independent with zero mean, while MATH is a constant. Hence MATH and thus, taking expectations, MATH . By symmetry (replacing MATH by MATH), MATH. We estimate this term by conditioning on MATH, using the independence of MATH and MATH. If MATH, with MATH and MATH, then MATH and MATH; hence NAME 's inequality yields MATH . Consequently, MATH . We postpone the estimation of MATH, and introduce the notation MATH . Combining REF - REF, we obtain our fundamental recursive estimate MATH . We unwrap this recursion partly, by concentrating on the first sum on the right-hand side and regarding the second as known. Thus, writing MATH we define recursively MATH and find by REF and induction MATH . Now, the recursion REF is easily solved (see, for example, CITE), giving MATH . We substitute REF, treating the three terms separately, into REF. The first term in REF yields the sum MATH and the total contribution MATH . The second term in REF yields the sum MATH and the total contribution MATH . Hence we find from REF MATH . We next use the following estimate of MATH, whose proof we postpone. For MATH, MATH . Using this lemma in REF, we find in analogy with REF MATH and thus MATH . We claim that REF implies the sought estimate MATH. Indeed, assume that MATH and that MATH is a number such that MATH for MATH. Then, using MATH, MATH . In particular, for MATH, MATH and thus REF yields (trivially for MATH, too) MATH . Consequently, by REF, MATH . If MATH, which holds for example for MATH, then this yields MATH, and thus REF holds for MATH, too. By induction, REF holds for all MATH, and we have proved the explicit estimate MATH . This is the desired estimate, apart from the value of the constant. To improve the constant, we use numerical calculations by computer. Indeed, for REF, MATH where MATH and MATH is given by REF; so, given any integer MATH, MATH can be computed exactly for MATH. Next, for MATH, an upper bound MATH to MATH can be computed recursively from REF or, equivalently, REF, using the already computed MATH, MATH, to bound MATH in the right-hand side. (We do not know how to compute MATH exactly even for MATH.) For larger MATH, we use the estimates REF . Let MATH . Then for MATH, arguing as in REF, for any MATH such that REF holds for all MATH, MATH and thus by REF MATH . Similarly, with MATH, for MATH, by REF , we have MATH . Consequently, REF yields, using REF again and REF, MATH . In other words, REF holds for MATH, with MATH replaced by MATH . For MATH we find (using NAME or Maple), rounded to four decimal places, MATH and MATH, and thus, taking MATH as in REF, MATH. Moreover, the computer verifies that MATH for MATH; thus REF holds for all MATH with MATH. Using this value in REF we find MATH, and the theorem is proved. |
math/0105248 | Let MATH. Thus MATH form a partition of MATH. We choose a point MATH for each MATH (where the bar here indicates closure) and define MATH that is, MATH when MATH. By NAME 's inequality, MATH . To estimate the second term in REF, note that for MATH, MATH . The NAME - NAME inequality yields MATH and thus (for any choice of MATH), MATH . We have MATH and find MATH and MATH consequently, MATH . Hence REF yields MATH . For the first term in REF, let us first assume that MATH. For MATH we have MATH . For MATH we choose MATH. This yields, using REF, MATH . In the opposite direction, by REF, still for MATH, MATH . Consequently, for MATH, MATH . For MATH we choose instead MATH. The symmetries of MATH and MATH then yield MATH, and since MATH, REF shows that MATH for MATH, too, that is ,REF holds for all MATH. In other words, MATH for all MATH; in particular, MATH for all MATH. This holds trivially for MATH, too, for any choice of MATH, and together with REF yields the result. |
math/0105248 | Since MATH when MATH, it suffices to consider integer MATH. The case MATH is REF (with MATH), so we assume further that MATH. We use induction on MATH and assume that the result holds for smaller positive integer values of MATH. Let MATH be as in REF, and note that for every MATH, MATH by the fact CITE that there is an optimal coupling for MATH that is optimal for every MATH. Using the notation of REF, we have, for MATH, MATH . We use a simple lemma to estimate this. Let MATH, MATH, and MATH be three independent random variables, and let MATH be an integer. Then MATH . By the binomial theorem and independence, MATH . If MATH and MATH we estimate MATH (which holds also for MATH, disregarding the central expression) and similarly MATH and MATH. Hence all terms in the sum, except MATH and MATH, are bounded by the corresponding terms in the trinomial expansion of MATH. Conditional on MATH, the three variables MATH, MATH, and MATH are independent, so the lemma is applicable. Fix MATH and let MATH, so MATH. Then, given MATH, MATH and thus, for any MATH, MATH . Similarly, MATH . Further, given MATH, MATH is a constant, for which we use the simple estimate (from REF) MATH . We first use REF with MATH together with the induction hypothesis MATH, MATH, to obtain (also for MATH) MATH where MATH, like MATH below, denotes some constant depending on MATH only. By similar argument using REF, we obtain MATH . Hence, using REF for MATH, too, REF yields MATH . Taking the average over all MATH we finally find the recursive estimate MATH . The proof is now completed by another induction, this one on MATH. Suppose that MATH for MATH. Then REF yields MATH . Since MATH, we have MATH, and thus, if MATH is sufficiently large, MATH . For such MATH, REF yields MATH, which completes both inductions and the proof. |
math/0105248 | Let MATH, where MATH denotes the fractional part of MATH. For any coupling of MATH and MATH, MATH where MATH, MATH, and thus MATH . We regard MATH as a random variable taking values in MATH, and find that its distribution, MATH say, has NAME coefficients MATH where MATH is the characteristic function of MATH. In particular, MATH. By our hypothesis on MATH and the NAME - NAME lemma, MATH as MATH. Thus, for each fixed MATH, MATH as MATH. This implies that MATH converges weakly (as measures on MATH) to the uniform distribution, that is, MATH where MATH. Consequently, as MATH, MATH which together with REF leads to REF. The proof of the proposition is completed by observing MATH for every MATH, because MATH. |
math/0105248 | Recall that MATH and MATH have mean MATH and that MATH while by REF MATH and thus MATH . Consequently, for the MATH-optimal coupling of MATH and MATH, by NAME 's inequality, MATH . |
math/0105248 | Consider an optimal MATH-coupling of MATH and MATH. Then, for MATH and MATH, denoting the distribution functions of MATH and MATH by MATH and MATH, MATH . Similarly, MATH . Consequently, MATH and thus MATH . |
math/0105248 | By CITE, MATH has a bounded density function, so REF yield, for every fixed MATH, MATH . The result follows by choosing MATH so large that MATH. |
math/0105248 | For any MATH and MATH, MATH . Letting MATH we find, since MATH is continuous, MATH . The result now follows from the following lemma applied to MATH. |
math/0105248 | Let MATH and MATH. By NAME 's inequality, MATH and thus MATH . The interval MATH contains at most MATH integers, and thus it must contain an integer MATH such that MATH . |
math/0105248 | We apply REF with MATH and observe that MATH and that MATH. Indeed, REF is trivial for MATH and easily verified for MATH, while for MATH it holds because then, by REF, MATH . |
math/0105248 | By REF , MATH while MATH . The first estimate follows, and the second is an immediate consequence. |
math/0105248 | It is well known that the number MATH of NAME comparisons has the same distribution as the internal path length of a random binary search tree (under the random permutation model) with MATH internal nodes - see, for example, CITE. Moreover, it was shown by CITE that when MATH is reinterpreted as the internal path length of an evolving random binary search tree after MATH keys have been inserted, the process MATH is a martingale, which is MATH-bounded and thus converges a.s. and in MATH to some limit MATH. It follows that also MATH a.s., and thus in distribution; hence this random variable MATH is (a realization of) the same MATH as above. The martingale property can be written MATH, for the appropriate MATH-field MATH. Since MATH is convex, it now follows by NAME 's inequality for conditional expectations that MATH; and thus, taking expectations, MATH. By the same argument, MATH for each MATH, which together with NAME 's lemma yields MATH as MATH. |
math/0105248 | MATH with MATH. |
math/0105248 | By NAME 's inequality, MATH . The result follows from REF , since MATH by REF. |
math/0105248 | Take (for MATH) MATH in REF . |
math/0105248 | Consider a MATH-optimal coupling of MATH and MATH. Then, using the mean value theorem, the NAME - NAME inequality, REF , and REF, MATH . The result follws by REF . |
math/0105254 | Notice that MATH intersects properly the diagonal, MATH in MATH, along MATH. Thus, a sequence of local equations of MATH in MATH at any point of MATH restrict to a regular sequence in the local ring of MATH at that point. Since MATH is NAME, it follows that the same holds for MATH, and that MATH for any MATH. Let MATH (respectively, MATH) denote the codimension of MATH (respectively, MATH) in MATH. Then we may choose a locally free resolution MATH (respectively, MATH) of the sheaf of MATH-modules MATH respectively, MATH of length MATH (respectively, MATH). Now MATH is a locally free resolution of the sheaf of MATH-modules MATH. By the preceding step, it follows that MATH is a locally free resolution of MATH. Thus, MATH for all MATH. Since MATH is NAME, the dual complex MATH is a locally free resolution of the sheaf MATH . Likewise, MATH (respectively, MATH) is a locally free resolution of MATH (respectively, of MATH). It follows that MATH and that MATH for all MATH. |
math/0105254 | The first assertion follows from NAME 's transversality theorem ; we recall the proof, since we shall repeatedly use its ingredients. Let MATH be the inclusion and let MATH be the ``multiplication" map. Notice that MATH acts on MATH via left multiplication on MATH, and that MATH is a MATH-equivariant morphism to MATH. Thus, MATH is a locally trivial fibration with fiber MATH. The latter is isomorphic to MATH, the pullback in MATH of a NAME variety. Therefore, the fiber of MATH has rational singularities. Now consider the cartesian product MATH with projections MATH to MATH, and MATH to MATH. Let MATH be the projection and let MATH as displayed in the following commutative diagram: MATH . By definition, the square on the right is cartesian, so that MATH is also a locally trivial fibration with fiber MATH. As a consequence, MATH is a variety, and we have MATH . Furthermore, the morphism MATH is proper, and its fiber at each MATH identifies with MATH. Thus, for general MATH, the intersection MATH is either empty or equidimensional of dimension MATH. This proves the first assertion. If in addition MATH is NAME, then by REF , the intersection MATH is NAME whenever it is proper. If in addition MATH is normal, then MATH is nonsingular in codimension MATH, for general MATH (since MATH is nonsingular). Therefore, MATH is normal, by NAME 's criterion. Assume now that MATH has rational singularities ; then MATH has rational singularities as well. Now the following easy result completes the proof of REF . Let MATH and MATH be varieties and let MATH be a morphism. If MATH has rational singularities, then the same holds for the general fibers of MATH. We may reduce to the situation where MATH is affine, MATH is nonsingular and MATH is flat with connected fibers. Let MATH be a general fiber of MATH ; then MATH is NAME, since MATH is. Choose a desingularization MATH and let MATH. Then MATH is a general fiber of MATH, and hence is connected. By generic smoothness, MATH is nonsingular, so that MATH restricts to a desingularization MATH. Since MATH has rational singularities, the map MATH is an isomorphism ; since MATH is the restriction of MATH to MATH, and similarly for MATH, it follows that the map MATH is an isomorphism as well. |
math/0105254 | If MATH then MATH is locally isomorphic to MATH, so that the ideal sheaf of MATH in MATH is NAME. Since MATH is also NAME and NAME has pure codimension MATH, it follows easily that MATH is NAME as well. In the general case where MATH, notice that the natural map MATH restricts to a proper surjective morphism MATH which maps isomorphically MATH to MATH (since MATH). Thus, MATH (as sets). By REF , we have MATH . It follows that MATH. We claim that MATH . To see this, choose a reduced decomposition of MATH and let MATH be the corresponding standard desingularization (see CITE). Then MATH acts in MATH, and MATH is MATH-equivariant. Furthermore, MATH contains a dense MATH-orbit, mapped isomorphically to MATH by MATH ; the complement MATH of this orbit is a union of nonsingular irreducible divisors MATH intersecting transversally. By REF , we have MATH . Let MATH this is a desingularization of MATH. Since MATH is ample, we have MATH for MATH, by the NAME vanishing theorem (see CITE) ; furthermore, MATH vanishes for MATH as well, and MATH . Now the NAME spectral sequence for MATH implies the claimed vanishing. Using that claim and duality for the morphism MATH, we obtain MATH . Thus, to prove that the sheaf MATH is NAME, it suffices to check the vanishing of MATH for MATH. We deduce this from the NAME vanishing theorem (see for example, CITE) as follows. Notice that MATH is the support of a very ample divisor of MATH (to see this, consider a MATH-linearized very ample invertible sheaf MATH on MATH and a MATH-semi-invariant section MATH of MATH that vanishes on MATH ; then the zero set of MATH is exactly MATH). Thus, we may choose positive integers MATH such that the divisor MATH is very ample. Choose also a positive integer MATH and let MATH. Finally, let MATH and MATH. Then the invertible sheaf MATH is very ample. By REF , it follows that MATH for MATH, that is, MATH. As above, the NAME spectral sequence for MATH yields the vanishing of MATH for MATH ; thus, MATH, and hence MATH, is NAME. By REF , the same holds for MATH for general MATH. Since MATH has pure codimension MATH in MATH, and the latter is NAME, the ideal sheaf MATH is NAME as well. This implies the vanishing of MATH for MATH. If in addition MATH is normal, then so is MATH by REF . Thus, the sheaf MATH is reflexive of rank MATH ; this implies the latter assertion. |
math/0105254 | We begin with the case where MATH ; we then set MATH. We show how to obtain MATH by base change from a degeneration of MATH constructed in REF. Let MATH be the adjoint group of MATH and let MATH be its wonderful completion ; this is a nonsingular projective variety where MATH acts with a dense orbit isomorphic to MATH, and a unique closed orbit isomorphic to MATH. Let MATH and let MATH be its closure in MATH. Since MATH is invariant under the action of MATH, we may form the associated fiber bundle MATH . On the other hand, the map MATH, MATH factors through a map MATH which is clearly surjective and MATH-equivariant. Furthermore, the product map MATH is a closed immersion, with image the ``incidence variety" MATH . By REF, MATH is flat, with reduced NAME fibers ; these identify with closed subschemes of MATH via MATH. The fiber at the identity element of MATH (respectively, at the unique MATH-fixed point MATH) identifies with MATH (respectively, MATH). Furthermore, the closure MATH of the torus MATH in MATH is a nonsingular MATH-equivariant completion of that torus, containing MATH as a fixed point. Let MATH be the unique MATH-invariant open affine neighborhood of MATH in MATH. Then MATH is equivariantly isomorphic to affine MATH-space where MATH acts linearly with weights MATH. Thus, for the action of MATH by left multiplication, MATH is isomorphic to MATH. We claim that the subvariety MATH of MATH equals MATH. To see this, note that MATH since MATH. It follows that MATH contains MATH as an irreducible component. Furthermore, MATH restricts to a flat morphism from the complement MATH, to MATH. If this complement is not empty, then its image meets the open subset MATH of MATH, a contradiction. This proves the claim, and hence all assertions of REF in the case where MATH. In the general case, we consider MATH (scheme-theoretical intersection in MATH), with projection MATH. Notice that MATH is contained in MATH, and that MATH . Thus, MATH is an irreducible component of MATH ; the latter is invariant under the action of MATH in MATH. We claim that MATH intersects properly MATH in MATH, that is, every irreducible component MATH of MATH has dimension equal to MATH . In fact, it suffices to check that MATH. Since MATH is MATH-invariant and MATH acts attractively in MATH with fixed point MATH, it suffices in turn to show that MATH. But MATH . And the latter is equidimensional of dimension MATH, since MATH intersects properly all MATH. This proves our claim. Since MATH and MATH are NAME subvarieties of MATH intersecting properly, then MATH is equidimensional and NAME, by REF . Furthermore, the morphism MATH is equidimensional by the proof of the preceding claim ; therefore, MATH is flat. As in the first step of the proof, it follows that MATH equals MATH as sets. Furthermore, MATH is clearly reduced, so that MATH is generically reduced. Since it is NAME, it is reduced, and MATH equals MATH as subschemes. Likewise, the fiber MATH equals MATH as sets. Furthermore, this fiber is generically reduced (since each MATH is), and NAME (since MATH is flat and MATH is NAME). Thus, MATH is reduced. By semicontinuity, it follows that all fibers are reduced. |
math/0105254 | CASE: We adapt the argument of REF to this setting. We may index the finite poset MATH so that MATH whenever MATH. Let MATH for MATH. Then MATH and MATH. We claim that MATH . To see this, we may assume that MATH. Then MATH is a union of products MATH for certain MATH in MATH. We must have MATH (since MATH) and MATH (since MATH). Thus, MATH is contained in MATH. Conversely, if MATH, then MATH for some MATH such that MATH, whence MATH ; this yields the opposite inclusion. The claim is proved. Now consider the exact sequence MATH where MATH denotes the ideal sheaf of MATH in MATH. Then MATH identifies with the ideal sheaf of MATH in MATH ; by the claim, this is the ideal sheaf of MATH in MATH. This yields the ascending filtration of MATH. With obvious notation, we obtain likewise MATH which yields the descending filtration. CASE: By REF , the sheaf MATH is NAME of depth MATH. Now a descending induction on MATH shows that each MATH is a NAME variety of dimension MATH. Furthermore, we obtain exact sequences MATH that is, MATH . Since MATH and MATH are NAME by REF , we obtain MATH . The latter is isomorphic to MATH, by REF again. This yields a descending filtration of MATH, with associated graded as claimed. The ascending filtration is obtained by replacing MATH with MATH. |
math/0105254 | Since MATH and MATH are two fibers of the flat family MATH over the affine space MATH, we have MATH in MATH. By REF , it follows that MATH . Now let MATH be the projection to the first factor, and denote MATH the corresponding pushforward map. Then MATH, whereas MATH for all coherent sheaves MATH, MATH on MATH. This yields our formulae for MATH. To obtain the formulae for MATH, notice that the sheaf MATH is flat over MATH, since MATH is NAME and flat over MATH. Furthermore, the restriction to MATH to the fiber at MATH (respectively, MATH) is isomorphic to MATH (respectively, MATH). Thus, MATH in MATH. Now both formulae follow from REF by the preceding argument. Alternatively, these formulae may be derived from those for MATH by applying the involution MATH and duality in MATH. For, using REF , we obtain isomorphisms MATH whence MATH . |
math/0105254 | We first consider the case where MATH is the full flag variety ; furthermore, we replace MATH by MATH for simplicity. Recall that each intersection MATH is the fiber MATH, with notation displayed by the commutative diagram MATH where the square on the right is cartesian. Recall also that MATH has rational singularities. Let MATH a subvariety of codimension MATH in MATH. For general MATH, we have MATH . Thus, our REF is a consequence of the following assertion: MATH . We shall deduce REF from the NAME vanishing theorem, like in the proof of REF . Since that theorem concerns nonsingular varieties, we first construct a desingularization of MATH. Let MATH be a desingularization. On the other hand, let MATH be a standard desingularization as in the proof of REF . Composing MATH with the multiplication map MATH defines MATH. Define likewise MATH, MATH and consider the commutative diagram MATH where the square on the right is cartesian. Since MATH is a MATH-equivariant morphism from the nonsingular variety MATH to MATH, it is a locally trivial fibration with nonsingular fiber. Thus, the same holds for MATH, so that MATH is nonsingular as well. The map MATH is a desingularization ; it restricts to a proper morphism MATH which is clearly birational. Thus, MATH is a desingularization of MATH. The subset MATH is a union of nonsingular irreducible divisors intersecting transversally in MATH ; clearly, MATH. We claim that MATH . To verify this, notice that MATH since MATH is a locally trivial fibration. Furthermore, MATH since MATH and MATH. Therefore, MATH which implies the claim. We next obtain the analogue of REF for MATH, that is, MATH . This follows from a relative version of the NAME vanishing theorem. Specifically, recall that MATH is the support of a very ample divisor MATH of MATH, with the notation of the proof of REF . Let MATH for MATH ; then the nonempty MATH are the irreducible components of MATH. Define positive integers MATH as in the proof of REF and let MATH and MATH. Then the invertible sheaf MATH is the pullback under MATH of a very ample invertible sheaf on MATH. Since MATH is generically injective, it follows that MATH is MATH-numerically effective and MATH-big. Therefore, MATH for MATH, by REF . This proves REF . Likewise, MATH is MATH-numerically effective, so that MATH . Finally, we claim that MATH . Together with REF and the NAME spectral sequence for MATH, this will imply REF . To check REF , we factor MATH into MATH, with notation displayed in the commutative diagram MATH where the square on the right is cartesian. Notice that MATH has rational singularities, and that MATH is a desingularization ; furthermore, we obtain MATH by the preceding arguments for determining MATH, applied to the regular locus of MATH. Thus, MATH . It follows that MATH where the latter equality holds since MATH is flat. By the projection formula and rationality of singularities of MATH, this yields MATH . And one may check as above that the latter equals MATH. This completes the proof of REF and hence of REF , in the case where MATH. Finally, in the case where MATH, one argues by reducing to MATH as in the proof of REF ; we skip the details. |
math/0105254 | We apply REF to the normal, NAME variety MATH. This yields MATH in MATH. Multiplying both sides by MATH (where MATH are the projections) and then applying MATH, we obtain MATH for any weight MATH. But since MATH is an isomorphism, we have MATH . This proves our first formula. Recalling that MATH as seen in REF, we obtain MATH . By REF and duality in the variety MATH of dimension MATH, it follows that MATH . Since MATH and MATH, this implies our second formula. If in addition MATH is dominant, then MATH for every MATH, as follows from REF . This yields our third formula. |
math/0105254 | The invertible sheaf MATH has a section with zero subscheme the NAME variety MATH. This yields an exact sequence MATH whence MATH in MATH. Multiplying this equality by MATH yields MATH which implies our first formula. The second formula follows by duality. |
math/0105256 | This follows from NAME 's REF applied to MATH. |
math/0105256 | The finite powerset functor MATH is nontrivial, preserves weak pullbacks and is bounded CITE, hence MATH is prodominical. Moreover there exist two nonempty finite coalgebras MATH and MATH for the finite powerset functor such that MATH CITE. Let MATH be the coproduct injection in MATH, and let MATH be the identity on MATH. By previous remarks, MATH and therefore there exist nonzero MATH with MATH. |
math/0105256 | A coalgebra for the identity functor is given by a self map of a set. Let MATH be a set with MATH, and define MATH by MATH and MATH. Then MATH is a proper subcoalgebra of MATH that is open and dense in the coalgebra topology on MATH. It follows from REF that the partial map MATH of REF is weakly total but not total. |
math/0105256 | Let MATH be the forgetful functor. If MATH is nonempty, so is the diagonal MATH, which is a bisimulation on MATH CITE. REF states that if MATH is a product in MATH with projections MATH, then MATH is the largest bisimulation between MATH and MATH; taking MATH, we have that MATH so that MATH cannot be empty. It follows that MATH for any MATH. |
math/0105256 | By a result of CITE, the preimage of a subcoalgebra is a subcoalgebra if and only if MATH weakly preserves pullbacks along monomorphisms; hence pullbacks of subcoalgebras are subcoalgebras in MATH. Paraphrasing Worrel CITE, since the forgetful functor MATH creates colimits and since MATH weakly preserves pullbacks if MATH does, coproducts are disjoint and stable under pullback. Fix coalgebras MATH and MATH; it follows by standard arguments CITE that the pullback operation MATH distributes over coproducts; taking MATH to be the terminal object, it follows that products distribute over coproducts. |
math/0105256 | The map MATH is defined by sending the sub-coalgebra MATH of MATH to the domain MATH, where MATH. The inverse map is defined as follows. Given a domain MATH, we may write MATH, where MATH is a monomorphism of the coalgebras, and where MATH is a morphism in MATH. By definition of composition in MATH one obtains the following diagram in which the squares are pullbacks. We adopt the convention of not showing the structure maps and induced morphisms of the coalgebras occurring in commutative diagrams of coalgebras. MATH . We use this diagram to define the coalgebra MATH corresponding to the domain MATH as the image of MATH under the left vertical column of the diagram. Showing that the meets are preserved involves certain large diagrams such as those occurring in CITE. |
math/0105256 | Let MATH in MATH be given by MATH, where MATH and MATH are in MATH, with MATH mono, and for MATH let MATH in MATH be given by MATH, where MATH and MATH are in MATH, with MATH mono. Consider the following diagram. MATH . We claim that the bottom parallelogram commutes in MATH. Observe that the top parallelogram is a pullback in MATH. It follows that the composite MATH in MATH is represented by the pair MATH . On the other hand, the composite MATH in MATH is represented by the pair MATH which equals REF since the (back) rectangle commutes, as dist is natural in MATH. |
math/0105256 | We first show that if MATH is an epimorphism in MATH, then so is MATH. The following diagram is a pullback in MATH. MATH . Apply the forgetful functor MATH to this and take the pullback in MATH. REF states that if MATH preserves weak pullbacks, then the forgetful functor MATH preserves weak pullbacks; hence the square in the following diagram is a weak pullback, and therefore there exists an induced map as indicated. MATH . Since MATH preserves epimorphisms, MATH is surjective. Since it is possible that MATH, in that trivial case commutativity of the square forces MATH, so that MATH. Otherwise, a diagram chase shows that MATH is surjective and, since MATH reflects epimorphisms, MATH is an epimorphism in MATH. Therefore, if MATH are epimorphisms in MATH, so is MATH. Using the first pullback diagram alone, one has that if MATH is a monomorphism, then so is MATH, and therefore a product of monomorphisms in MATH is a monomorphism. It follows that the epi-mono factorization of a product MATH is the product of the epi-mono factorizations of MATH and of MATH. REF and CITE states that if MATH is a subcoalgebra of MATH for MATH, and if the product MATH exists, then so does MATH, which is a subcoalgebra of MATH. If MATH and MATH are coalgebra morphisms, then by previous remarks, one has two canonically isomorphic epi-mono factorizations of MATH as indicated in the following commutative diagram. MATH . Since the morphisms into MATH are inclusions, the induced map is the identity. |
math/0105256 | Let MATH be a partial monomorphism in MATH. Then MATH, where MATH are monomorphisms in MATH, and where MATH is a subcoalgebra of MATH. The category MATH has REF factorizations, hence one has the following diagram in which MATH has the epi-mono factorization MATH. MATH A morphism in MATH is epi if and only if it is surjective in MATH. The morphism MATH is a monomorphism since MATH and MATH are. Therefore MATH is invertible in MATH with inverse MATH which, since the image of MATH is contained in the image of MATH, is a morphism in MATH by a diagram lemma (for example, CITE). The partial morphism MATH defined by MATH gives the required section. The required properties MATH and MATH can be verified in the following commutative diagram, in which the squares are pullbacks in MATH. MATH . |
math/0105256 | Following the procedure in CITE for obtaining the smallest MATH category containing a given set of morphisms, we think of the elements of MATH as the objects of an as yet unspecified MATH-category MATH, and construct the set MATH of morphisms accordingly. For each MATH, we adjoin an identity morphism MATH to MATH, subject to the functoriality relations MATH, and so on. Next, adjoin the values of the fourteen natural transformations above with arguments in MATH to MATH, subject to the relations REF through REF above and subject to the relations that hold in a MATH-category . Finally, we close the set of morphisms that results under MATH and composition. This produces the category MATH. The homomorphism MATH is extended to MATH in the only way possible, following the three step construction of MATH. Identities in MATH must be preserved by MATH. The image in MATH under MATH of a value of one of the fourteen natural transformations is completely determined by functoriality and by definition of MATH. For example, the value of MATH on MATH is MATH. Finally, we require that MATH commute with MATH and composition; for example, MATH. |
math/0105256 | Since MATH is a MATH-prodominical category with ranges that satisfies the weak axiom of choice, it is an iteration category. |
math/0105256 | The proof we give in MATH presupposes that the endofunctor MATH preserves products, since we assume that products are constructed as they are in MATH; the general case follows from CITE. Let MATH be the free semigroup generated by MATH and define the set MATH . The set MATH is a sub-coalgebra of MATH since, translating from Heller CITE into this context, MATH where MATH, MATH, and where the dot MATH denotes concatenation of sequences in MATH; that is, MATH. Next we apply the diagram lemma in MATH. There exists a unique induced map in the following diagram MATH . We may suppose that MATH. If MATH then MATH, and as both of the indicated sequences are in MATH, both MATH and MATH are in MATH, and therefore MATH. We take MATH . The remaining assertions are immediate. |
math/0105256 | Under the hypothesis on MATH it follows from REF that for each MATH-coalgebra MATH and for each MATH, the subcoalgebra cogenerated by MATH exists; hence the MATH-cogenerated subcoalgebras of MATH form a base for the coalgebra topology on MATH. Each MATH-cogenerated subcoalgebra is connected in the coalgebra topology. |
math/0105257 | The case MATH is trivial. Take MATH, MATH. Let MATH be a NAME surface for MATH. Construct a NAME surface MATH from MATH parallel copies of MATH, each adjacent pair of copies joined, in order, by a MATH-handle with a half-turn of sign MATH. If MATH is a fiber surface, then so is MATH (see CITE); in any case MATH. The proof is finished by contemplating an appropriate figure (see REF ). |
math/0105257 | CITE drew attention to the connected sum of iterated torus knots MATH. (Their motivation was a vague question CITE about relations among the concordance classes of the knots associated to complex plane curve singularities, and they gave an answer to one form of the question by observing that MATH is algebraically slice.) CITE showed that MATH is not ribbon. By the proposition, and the indifference of connected sums (of knots) to the location of the summation, MATH (where all the NAME sums are MATH-gonal and along fiber surfaces). The connected sum of a knot and its mirror image is ribbon, so MATH, MATH, and MATH are ribbon. If MATH is ribbon, let MATH, MATH; if not, let MATH, MATH. In either case, MATH and MATH are ribbon but MATH is not ribbon. On the other hand, since (as mentioned in CITE) the NAME sum in the proposition is obviously ``direct" (that is, the NAME linking between any cycle on MATH and any cycle on MATH, taken in either order, is MATH), MATH is algebraically slice, for clearly a NAME direct sum is algebraically concordant to the connected sum of the same summands. |
math/0105258 | The proof follows by observing that MATH where MATH stands for the vector with the integral parts of MATH as coordinates. Thus by the variational definition of the effective diffusivity MATH and REF follows by observing that the effective diffusivity is invariant under a translation of the medium: MATH. |
math/0105258 | Let MATH. Write MATH , MATH the solutions of MATH and MATH with NAME conditions on MATH. Observe that MATH and MATH are the unique minimizers of MATH and MATH with MATH and MATH. Observe that since MATH and the minimum of the right member in REF is reached at MATH. It follows that MATH, which proves the lemma. |
math/0105260 | We rely on the approximation process described by NAME in CITE. Introduce the following NAME space MATH . Set MATH . One checks that MATH for any orthonormal basis MATH of MATH, and that MATH is the logarithm of a real analytic function. The following result connects the singularities of MATH and MATH. For any point MATH and all MATH, one has CASE: MATH; CASE: MATH. We then conclude the proof of the theorem as follows. As MATH does not charge MATH there exists a point MATH such that MATH (by NAME 's theorem). In particular by REF , MATH does not charge MATH either, hence there exists a holomorphic function MATH which does not vanish identically on MATH. For such a function, we apply ojasiewic's inequality CITE and get MATH with MATH and for some constants MATH. We infer for any MATH and any MATH, MATH . From this, it is easy to see that MATH; hence MATH . In particular, we get for any MATH, MATH which implies the result. |
math/0105260 | This lemma is standard for NAME numbers (see CITE). We emphasize that REF is not obvious and relies on the NAME extension theorem. The generalization is straightforward for NAME numbers. We nevertheless give the arguments for REF for completeness. Let MATH normalized by MATH. The mean value property inequality for subharmonic functions implies MATH . Hence for all MATH, we have MATH . Write MATH. By definition of NAME number we have MATH hence MATH with MATH. We hence get MATH as desired. |
math/0105260 | The multiplicity MATH is always smaller than the degree of the critical set of MATH, which is MATH. Applying REF to MATH and letting MATH we get REF. |
math/0105260 | Since the (global) topological degree of MATH is MATH we have MATH for all MATH. Let us show that MATH is a finite set. We follow the proof of REF. Take a point MATH is the regular locus of MATH such that MATH belongs to the regular locus of MATH, and MATH is a regular point for MATH. One can find local coordinates so that MATH with MATH. On the other hand, we have in terms of currents MATH and MATH. As MATH, we get MATH. Hence outside the finite set MATH we have MATH. We conclude the proof noting that the cardinality of MATH can be bounded only in terms of MATH. If MATH, then the orbit of MATH must visit the finite set MATH infinitely many times and is therefore preperiodic to a periodic orbit in that set. So we may write MATH with MATH. But then MATH, so we must have MATH for MATH. Thus MATH for all MATH, and that implies that MATH for some MATH. We conclude that MATH is periodic and that the orbit of MATH is totally invariant and contained in MATH. |
math/0105260 | This is a local result so we may assume MATH. Write MATH and MATH. Let MATH where MATH is a homogeneous polynomial of degree MATH. The Jacobian determinant of MATH is a homogeneous polynomial of degree MATH or vanishes identically. Thus MATH and MATH. |
math/0105260 | By pre- and post-composing by projective linear maps of MATH we may assume that MATH. Write MATH where MATH, MATH and MATH are polynomials of degree MATH and MATH, MATH. Then clearly MATH can only vanish up to order MATH and so MATH. Further, MATH if and only if MATH and MATH are homogeneous polynomials of degree MATH, which means precisely that MATH is a homogeneous point. |
math/0105260 | REF are consequences of REF, and the last assertion follows from REF are local so we may assume MATH and for sake of simplicity we write MATH, MATH, MATH. Because of REF we only have to show the inequalities MATH and MATH. To prove MATH we use the fact that MATH (see for example, CITE) and (by definition) MATH for some constants MATH. For any ball MATH of radius MATH, one gets MATH for some constants MATH. By letting MATH we get MATH. We now give an analytic proof of MATH. We first note that MATH. It follows from REF that MATH . This concludes the proof. |
math/0105260 | First assume that MATH is a totally invariant curve. We want to show MATH. Given MATH and MATH outside a finite subset (depending on MATH) we may pick local coordinates at MATH and at MATH so that MATH is given by MATH. It follows that MATH on MATH outside a finite set, and thus MATH on all of MATH by upper semicontinuity. This implies MATH on MATH hence MATH. Conversely pick an irreducible component MATH. We will show that MATH belongs to a totally invariant curve. We claim that MATH . This implies that MATH, hence MATH as MATH is finite. Let us show the claim. Pick a point MATH satisfying MATH. By REF , either MATH is preperiodic or MATH for some MATH and for some fixed curve MATH with MATH. In the latter case, we have MATH for some maximal MATH. For a generic point MATH, we have MATH, and for any MATH, MATH. Hence MATH, and we infer that MATH is totally invariant, whence MATH. If MATH is preperiodic, then MATH for some MATH and MATH. Clearly MATH and MATH. Apply REF to MATH and find non-negative integers MATH such that MATH. As MATH is of degree MATH, we have equality. In particular the union MATH of the critical components passing through MATH with MATH is a totally invariant set for MATH. But then the curve MATH is totally invariant for MATH, hence MATH by the argument above. Since MATH we have in fact MATH. This concludes the proof of the claim. Finally, REF shows that MATH is the closure of the set MATH. The remaining statement of the theorem follows from the classification of totally invariant curves REF . |
math/0105260 | Pick holomorphic maps MATH so that MATH, and set MATH. Then MATH are the irreducible factors of MATH. There exist integers MATH so that MATH for some holomorphic MATH with MATH. By ojasiewic's inequality CITE there exist constants MATH such that MATH in a neighborhood of the origin. For fixed MATH write MATH where MATH is a homogeneous polynomial of degree MATH. Similarly, set MATH for a non-degenerate homogeneous polynomial MATH of degree MATH. We know that MATH is an increasing supermultiplicative sequence such that MATH, and that MATH is an increasing sequence such that MATH (see REF). By REF, so for any fixed MATH, MATH with MATH we have MATH for MATH and some constants MATH. We infer that for a generic MATH and for any MATH for some MATH. On the other hand for all MATH and all MATH we have MATH . Let MATH be the MATH by MATH matrix MATH and let MATH be the spectral radius of MATH. Denote MATH and fix MATH so that MATH . For MATH large enough, so that MATH, and for generic MATH, we can apply REF, and get MATH . Hence MATH where we let MATH. By induction and using REF for MATH we have MATH . On the other hand REF shows that MATH . As MATH was generic, we can let it tend to zero and let MATH tend to infinity. We infer MATH, and therefore MATH as MATH was chosen arbitrary. The NAME theorem now implies the existence of an eigenvector MATH for MATH with non-negative coefficients associated to the eigenvalue MATH. We have MATH, which completes the proof. |
math/0105260 | REF shows that MATH is finite and totally invariant. A point MATH, MATH, is superattracting as the NAME series of MATH at MATH vanishes to order MATH. Let us now prove the characterization of MATH. First consider a point MATH. If MATH, then MATH and so MATH by REF . Conversely, if MATH, then MATH by REF so MATH. Next consider MATH. We want to show that MATH if and only if MATH is periodic and the orbit of MATH is totally invariant. Both of these properties are preserved under replacing MATH by an iterate, so we may assume that the line MATH is totally invariant for MATH and that MATH is on this line. Notice that the restriction MATH is a rational map of degree MATH and that MATH. Assume first MATH. Then MATH hence MATH belongs to a totally invariant orbit. Conversely, suppose that the orbit of MATH is totally invariant. After replacing MATH by MATH we may assume that MATH is a totally invariant fixed point for MATH. Thus we may write MATH in local coordinates MATH, for holomorphic MATH, MATH with MATH. One checks that MATH, thus MATH. This completes the proof. |
math/0105260 | Write MATH. Let MATH be a NAME set. We are looking for a lower bound for MATH in terms of MATH. To this end we apply the NAME estimate REF to the function MATH in each chart of a given atlas of MATH. Notice that MATH. We conclude that there exists a constant MATH such that MATH for all MATH. We pick a ``stopping time" MATH defined by MATH, and deduce the following sequence of inequalities: MATH which complete the proof. |
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