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math/0103011 | Let MATH be a non-contracting MATH-category. The only thing one has to prove is the surjectivity of MATH. There is nothing to prove for MATH and MATH. So let us suppose that MATH. The group MATH is generated by the elements of the form MATH where MATH. The canonical map MATH is generally not surjective on the underlying sets but by REF , any element of MATH is a composite of elements of MATH by only using the composition laws MATH for MATH. Since the branching semi-globular nerve is regular, then MATH is therefore equal to a sum of elements of the form MATH where MATH. Hence the surjectivity. |
math/0103013 | Use REF to obtain a set of generators (over MATH) for the cohomology of MATH. Then use REF to convert these generators into cohomology generators for MATH whose elements are identified with the cochains in the algebraic NAME complex on MATH. |
math/0103013 | Let us sketch a proof of the lemma. By REF it is possible to find a finite dimensional subcomplex of the algebraic NAME complex that captures all the cohomology (namely, just take all the cohomology generators, with zero differential). However, this may not include the given forms MATH. Thus, as a first approximation MATH of the desired complex MATH we take the union of the cohomology generators, the given forms MATH and their boundaries MATH. This is a complex, but the cohomology may be too big. (It is at least as big as the actual NAME cohomology but we may have added extra kernel elements.) We must find forms that reduce the cohomology. By an exhaustion of a MATH-module MATH we mean a sequence of MATH-subspaces MATH such that MATH, MATH and each MATH is finite-dimensional as a MATH-vector space. If MATH is finitely generated over MATH then one may produce an exhaustion for MATH from one for MATH. Input: MATH, the subcomplex of MATH spanned by CASE: the output of the algorithm of REF , CASE: forms MATH with MATH, and their boundaries MATH. Output: A finite dimensional complex MATH containing MATH and all MATH. CASE: Initialization: set MATH. CASE: Let MATH. If MATH, return MATH and exit. CASE: Let MATH be an exhaustion of MATH. For example, let MATH. CASE: Derive an exhaustion MATH of the finitely generated left MATH-module MATH. CASE: For MATH test by trial and error whether there is an element in MATH that maps onto a nonzero element in MATH. As soon as such an element is found, add it to MATH, call the enlarged complex MATH, replace MATH by MATH and move to REF . CASE: If MATH, reenter at REF . CASE: Reenter at REF . End. |
math/0103013 | The inclusions MATH induce a map MATH. To see that this is a quasi-isomorphism consider the induced map between the long exact sequences and recall the five-lemma. |
math/0103014 | Cyclic group MATH of units MATH (MATH for some MATH) has order MATH, namely MATH minus MATH multiples of MATH. Then MATH, the direct product of two relative prime cycles, with corresponding subgroups MATH and MATH, sothat MATH where extension group MATH consists of all MATH residues mod MATH that are REF mod MATH, and core MATH, so MATH contains sub-semigroup MATH. |
math/0103014 | Let MATH be in core MATH, so MATH mod MATH. Then MATH mod MATH, using MATH. Clearly the second core digit, of weight MATH, is not found this way as function of MATH, but requires actual computation (unless MATH as in lemREF). It depends on the MATH produced in computing the MATH-th power of MATH. Recursively, each next core digit can be found by computing the MATH-th power of a core MATH residue with MATH+REF digit precision; here core MATH remains fixed since MATH mod MATH. |
math/0103014 | The period of MATH, which is the smallest MATH with MATH mod MATH, implies MATH (re REF ). So MATH, yielding period MATH. No smaller exponent generates REF mod MATH since MATH has only divisors MATH. MATH consists of all MATH residues which are REF mod MATH. The order of each subgroup MATH must divide MATH, sothat MATH REF and MATH. Then MATH mod MATH, where MATH, sothat MATH mod MATH. Hence no subgroup of MATH sums to REF mod MATH. |
math/0103014 | For even MATH: MATH in MATH implies pairwise zero-sums. In general: MATH for all MATH in MATH, and MATH, so MATH, writing MATH for MATH. Now for any MATH in MATH: MATH sothat MATH=REF implies MATH not in MATH, hence MATH for some MATH in MATH and MATH or MATH. Then: MATH with MATH if MATH. So MATH=REF yields MATH and MATH=REF. |
math/0103014 | If MATH mod REF then MATH implies a core-subgroup MATH of three MATH-th powers: the cubic roots of REF in MATH that sum to REF mod REF . Now MATH, so if MATH then MATH, hence MATH solves REF : a root-pair of inverses, with MATH. MATH in core consists of MATH-th power residues with MATH mod MATH. Write MATH for MATH, then MATH and MATH, sothat MATH mod MATH. Notice the Exponent Distributes over a Sum (MATH), implying inequality MATH for the corresponding REF-extensions MATH of core residues MATH mod MATH. |
math/0103014 | Core function MATH mod MATH REF mod MATH has just MATH distinct residues with MATH mod MATH, and MATH mod MATH (MATH). Including (non-core) MATH makes MATH mod MATH periodic in MATH with period MATH : MATH mod MATH, so MATH suffices for core analysis. Increment MATH, as difference of two functions of period MATH, also has period MATH. CASE: MATH is a polynomial of odd degree with odd symmetry MATH. CASE: Difference polynomial MATH is of even degree MATH with leading term MATH, and residues REF mod MATH in extension group MATH. The even degree of MATH results in even symmetry, because MATH. Denote MATH, then for MATH follows: MATH and MATH, yielding: MATH. By binomial expansion and MATH mod MATH in core MATH: MATH mod MATH. With MATH mod MATH: MATH mod MATH, causing MATH mod MATH. |
math/0103014 | Multiplying by MATH respectively, maps REF to REF if MATH, and REF to REF if MATH, and REF to REF if MATH. All three conditions imply MATH mod MATH. |
math/0103014 | Assume MATH equations MATH form a loop of length MATH (indices mod MATH). Consider function MATH, composed of the three elementary functions: Inverse, Complement and Successor, in that sequence. Let MATH be the identity function, and MATH to prevent division by zero, then under function composition the third iteration MATH, since MATH (repeat substituting MATH for MATH). Since MATH and MATH commute, MATH, REF! = REF permutations of MATH yield only four distinct dual-folded-successor "dfs" functions: MATH. By inspection each of these has MATH, referred to as loop length REF. For a cubic rootpair dfs=E, and REF-loops do not occur since there are no duplets (next note REF). Hence solutions of REF have only dfs function loops of length REF: inverse pair and triplet. |
math/0103014 | In a triplet for some prime MATH the core increment form REF holds for three distinct values of MATH, where scaling by respective factors MATH in MATH mod MATH returns REF-complement form REF . Consider each triplet equivalence separately, and for a simple notation let MATH be any of the three MATH, with successive core residues MATH mod MATH. Then MATH mod MATH, where MATH mod MATH, has both summands in core, but right hand side MATH mod MATH is not in core, with deviation MATH mod MATH. Hence MATH mod MATH (with MATH mod MATH in the cubic root case), and MATH mod MATH. This equivalence has no finite (equality preserving) REF-extension to integer MATH-th powers since MATH, so the assumed MATH case-MATH solution cannot exist. |
math/0103014 | Let MATH extend the residues MATH, such that MATH mod MATH are not both in core MATH. So the extensions do not extend core precision MATH, and without loss take MATH, due to a scalefactor MATH in REF . Write MATH, then binomial expansion upto quadratic terms yields: MATH mod MATH, and similarly: MATH mod MATH, and: MATH mod MATH, where: MATH mod MATH, and MATH mod MATH, but not so mod MATH: MATH are such that not both MATH are in core MATH, hence core precision MATH is not increased. By REF-extension of MATH (so MATH) does not yield the required equality MATH. To find for which maximum precision equivalence MATH hold, choose MATH sothat: MATH mod MATH . REF . yielding MATH mod MATH. A cubic root solution has also MATH in core MATH, so MATH mod MATH, then MATH with MATH would require MATH mod MATH, readily verified for MATH=REF and any prime MATH. Such extension REF implies inequivalence MATH mod MATH for non-zero extensions MATH. Because MATH together with MATH yields MATH. So any (zero- or nonzero-) extension yields inequivalence mod MATH. |
math/0103014 | A MATH solution is a linear transformed extension of a MATH root in core REF . By REF it has no finite MATH-th power extension, yielding the theorem. |
math/0103014 | In a case-MATH solution MATH divides a lefthand term, MATH or MATH-REF, or the right hand side MATH. NAME the multiple of MATH to the right hand side, for instance if MATH we have MATH, while otherwise MATH. So the sum or difference of two MATH-th powers coprime to MATH must be shown not to yield a MATH-th power MATH for any MATH: CASE: MATH has no solution for integers MATH. Notice that core increment form REF does not apply here. However, by MATH the two lefthand terms, coprime to MATH, are either complementary or equivalent mod MATH, depending on their sum or difference being MATH. Scaling by MATH for some MATH mod MATH, so MATH mod MATH, transforms one lefthand term into a core residue MATH mod MATH, with MATH mod MATH. And translation by adding MATH mod MATH yields the other term MATH or MATH mod MATH respectively. The right hand side then becomes MATH, equivalent to MATH mod MATH. So an assumed equality REF yields, by two equality preserving tansformations, the next equivalence REF , where MATH mod MATH (MATH in core MATH for MATH with MATH mod MATH) and MATH mod MATH: CASE: MATH mod MATH, where MATH , MATH mod MATH. Equivalence REF does not extend to integers, because MATH, and MATH, where MATH are REF-extensions of MATH mod MATH respectively. But this contradicts assumed equalities REF , which consequently must be false. |
math/0103015 | This lemma is obvious because the NAME moves do not change the fundamental group of the complement of the singular graph. |
math/0103015 | Suppose there is a non-trivial knot on a singular arc inside MATH. Let MATH be the complement to the singular set in MATH. Consider decomposition MATH such that MATH contains the knot and MATH has precisely MATH punctures (see REF ). Here MATH could be obtained from the tubular neighborhood of the singular set, so the decomposition is always possible. We now apply NAME theorem to MATH. For MATH: MATH and MATH can not be obtained as a word of MATH, MATH since the knot is non-trivial. By NAME theorem MATH . Here again MATH can not be obtained as a word of the other generators. This means that the rank of MATH is greater then MATH, which contradicts the definition of the common generalized triangle group. |
math/0103015 | Let MATH be the common generalized triangle orbifold, and MATH - the MATH-manifold obtained as the complement to the singular set of MATH in MATH. To prove the lemma it is sufficient to consider the first homology group MATH. The canonical projection MATH maps non of the generators of MATH to the homology cycle that corresponds to the singular cycle in MATH. Since MATH with no any elements of finite order, this immediately follows that MATH which is impossible. Note, that this argument does not work for the orbifold fundamental and homology groups because the homology groups are generated by the elements of the finite order (See example in REF). |
math/0103015 | Consider the axes of the half-turns MATH in MATH. We endow the axes by orientations and join them in couples by common perpendiculars. The result is a hyperbolic right-angled hexagon REF . Vector MATH represents three complex lengths of the sides of our hexagon which are common perpendiculars of the given axes. By theorem from CITE (p. REF) three pairwise non-adjacent sides define right-angled hexagon uniquely up to simultaneous change of orientations of the other three sides. It means that any other hexagon defined by another group with the same parameters can be translated to the given one by a hyperbolic isometry. This isometry obviously defines the required conjugation MATH. It may seem that we have missed the possibility of changing of the orientations of the axes but we have already remarked that the orientations effect only representation of the group in MATH but not isometries themselves. So in this situation we can change the orientations as it is needed. |
math/0103016 | We prove the lemma by induction on MATH. If MATH then MATH by the previous computation, hence MATH and the initial case is proved. Supposing the lemma holds for MATH, we have MATH where we set MATH . By this last formula, it is clear that the constants involved are universal. Moreover, if MATH is a function MATH then the term MATH vanishes and the same formula says that MATH does not contain MATH without being differentiated. |
math/0103016 | Distributing the derivative in MATH on the terms of the product, we have seen that the derivatives of the metric coefficients depends linearly on MATH, it lasts to check the derivative of MATH. By the last assertion of REF , we have MATH and since MATH we get MATH which proves the first part of the lemma as MATH and MATH are linear in MATH. The second statement clearly follows by the previous computations and the first part of the lemma. |
math/0103016 | Applying NAME inequality to the last term of REF , we get MATH where MATH is given by MATH then MATH. Hence, we can interpolate MATH between a small fraction of MATH and a possibly large multiple of MATH, MATH . Choosing MATH such that MATH and collecting terms we obtain MATH . |
math/0103016 | Suppose first that MATH is embedded and MATH, clearly MATH is bounded by a value depending on the constant MATH. We consider MATH as a subset of MATH via the embedding MATH and MATH as a measure on MATH which is supported on MATH. Then the following result holds REF : let MATH be the ball of radius MATH centered at MATH in MATH, for every MATH we have MATH . Hence, MATH and choosing MATH, for every MATH we get the inequality MATH . Then we need the following formula which is proved in REF, as a consequence of the tangential divergence REF . For every MATH we have MATH where MATH and MATH is any smooth non negative function. Noticing that MATH and using NAME inequality we estimate MATH where in the last passage we set MATH used the previous estimate on MATH. The function MATH is integrable since MATH and we get MATH now sending MATH to zero, on the left side we obtain the value of MATH times MATH which is the volume of the unit ball of MATH, hence MATH . For a general MATH we apply this inequality to the function MATH, thus MATH . Since MATH was arbitrary we conclude that MATH for a constant MATH depending on MATH, MATH and MATH. If MATH is only immersed, we consider the embeddings of MATH in MATH given by the map MATH, where MATH is an embedding of MATH in some Euclidean space. Then, repeating the previous argument (it is possible since the starting inequalities from CITE hold for embeddings in any MATH) we will get the same conclusion with a constant MATH. Finally, as MATH depends only on Vol MATH and MATH, and all the geometric quantities converge uniformly when MATH goes to zero, we conclude that the inequality holds also in the immersed case. Now, given any MATH, we choose MATH, then clearly MATH. By the inequality above we have MATH then using NAME inequality and an interpolation argument as in the proof of REF we get MATH . Applying again NAME inequality, as MATH, we conclude that MATH which gives the thesis since MATH depends only on MATH, MATH and MATH. |
math/0103016 | By REF applied to the tensor MATH we get MATH . Since the MATH are related by MATH and MATH, we have MATH and the first part of the corollary is proved. The second part follows analogously using also REF . |
math/0103016 | The cases MATH and MATH are REF , respectively, the intermediate cases, when MATH, are obtained immediately by the MATH - convexity of MATH in MATH, which is a linear function of MATH, and the fact that the right side is exponential in MATH. If MATH is negative then MATH and MATH hence, the MATH estimate of REF together with REF gives the inequality for every MATH. |
math/0103016 | Keeping in mind the NAME - NAME relations REF and the equations above, we compute MATH . Expanding MATH we continue, MATH . Interchanging repeatedly derivatives in the first term we introduce some extra terms of the form MATH and we get MATH then using REF we conclude MATH . |
math/0103016 | With a reasoning analogous to the one of REF applied to the tensor MATH and by the previous lemma, we have MATH . Interchanging the operator MATH with the NAME in the first term and including the extra terms in MATH, we obtain MATH . |
math/0103016 | By the previous results we have MATH . Interchanging the covariant derivatives in the first term we introduce some extra terms of the form MATH, hence we get MATH where the last integral comes from the time derivative of MATH. Then, carrying the MATH derivatives MATH on MATH by means of the divergence theorem, we finally obtain the claimed result, MATH . The leading coefficient became MATH since we multiplied MATH for MATH while doing the MATH integrations by parts. |
math/0103016 | By REF it follows that MATH, hence MATH and since MATH, we get MATH . Then, by an induction argument we can express MATH as MATH where MATH does not contain derivatives of order higher than MATH. Taking the norm of both sides we get MATH and we conclude the proof computing MATH . |
math/0103019 | Let MATH denote the field of fraction of MATH, and MATH the matrix units in MATH. We first note that MATH, since MATH is a left MATH-isomorphism if we define the left MATH-module structure on MATH by MATH for every MATH. By the NAME isomorphism theorem, MATH as MATH-modules determined by a matrix MATH as MATH. Then the mapping MATH for every MATH determines a MATH-isomorphism MATH. But for every MATH we have MATH whence MATH is a left MATH-module isomorphism as desired. |
math/0103019 | If MATH, then MATH . |
math/0103019 | (MATH.) We compute using the NAME relation: MATH since MATH is an invertible module. (MATH.) Given MATH and MATH, then MATH for every MATH. Then MATH via MATH. If MATH is a finite projective base for MATH, one finds MATH, MATH such that MATH. Setting MATH for each MATH and MATH, we have for every MATH, MATH . We merely reindex to get REF follows from a computation showing MATH for MATH, which is similar to CITE. (MATH.) Suppose MATH. Then MATH is finite projective. Define MATH by MATH for every MATH. Then MATH is epi since for every MATH we have MATH for every MATH. Since MATH is an epimorphism between finite projective modules of the same local rank, (that is, MATH-rank for every prime ideal MATH in MATH), MATH is bijective CITE. A similar argument shows that we may establish REF from REF . |
math/0103019 | In the proof of the last theorem we established MATH by showing MATH, for every MATH, is an isomorphism. As we noted, we may equally well establish MATH in this proof by showing that MATH is an isomorphism MATH. Since MATH for each MATH, it follows that there is a unique MATH such that MATH. One defines MATH and easily checks that MATH is an automorphism. |
math/0103019 | We give only the proof of the second equation, the first being similar. By REF , we compute: MATH . |
math/0103019 | Since MATH and MATH freely generate MATH as right MATH-modules, REF is clear with MATH an invertible in MATH. To verify REF , we note that MATH for every MATH. There is an isomorphism MATH given by MATH, for every MATH, since MATH and MATH. REF follows from the injectivity of this mapping and REF . We note that for every MATH the last equation implying that for all MATH, MATH which is equivalent to REF . |
math/0103019 | We compute using the identity MATH for all MATH: MATH and by applying MATH to both sides we obtain MATH . It follows from REF that MATH is a NAME homomorphism with dual bases MATH, MATH, MATH. We compute the NAME automorphism MATH for MATH in terms of MATH and MATH: for all MATH, MATH by applying REF twice. Since MATH freely generates MATH, it follows that MATH, whence MATH . |
math/0103019 | First, MATH is finite projective as a MATH-module. Secondly, MATH since MATH, MATH, MATH and MATH are finite projective MATH-modules. |
math/0103019 | The natural left MATH-module structure on the dual algebra MATH induces a comodule structure mapping MATH, determined by MATH for every MATH. The right MATH-module structure on MATH is given by MATH for every MATH and MATH. A rather long computation shows this compatible with the MATH-comodule structure in the sense of NAME modules. |
math/0103019 | One shows that the map MATH given by MATH is a MATH-linear projection onto MATH. Then the mapping MATH given by MATH has inverse given by the NAME module map MATH defined by MATH. |
math/0103019 | Since MATH and MATH, MATH have the same local ranks, it follows that the finite projective MATH-module MATH has constant rank MATH. Then MATH and MATH is invertible CITE. |
math/0103019 | Assuming that MATH, one then notes that multiplication from the right by MATH on MATH is zero by the existence of the (MATH-module) isomorphism MATH in REF . If MATH is field MATH and it is clear that MATH is then zero. The general case follows from a localization argument. NAME for MATH is apparent if MATH is a field, and the general case follows again from a localization argument. |
math/0103019 | We set MATH, MATH, where we note that the right MATH-module structure is related to the standard left MATH-module structure on MATH via a twist by MATH: for every MATH . Let MATH which is canonically isomorphic to the dual of MATH, and satisfies MATH by REF . Define MATH as the composite of the right MATH-module isomorphisms MATH . It is easy to check that MATH for all MATH and MATH. Now let MATH. MATH is a NAME isomorphism MATH, since MATH is an anti-automorphism of MATH and MATH . |
math/0103019 | We note that the NAME homomorphism MATH satisfies by REF , for every MATH, MATH and MATH since MATH. Since MATH and MATH is finite projective over MATH, we canonically identify MATH, and compute MATH whence REF . |
math/0103019 | We use the fact that the MATH-submodule of integrals of an augmented NAME algebra is free of rank MATH (compare REF , CITE or CITE). Then MATH. It follows from REF that the dual NAME algebra MATH is a NAME algebra. Whence MATH and MATH is an NAME. |
math/0103019 | We compute as in CITE, using REF at first and REF next (for every MATH): MATH . It follows from REF that MATH are dual bases for MATH. |
math/0103019 | Let MATH. It suffices to show the forward implication. Let MATH be such that MATH. Then REF implies that MATH is a MATH-Frobenius system for MATH, where we identify MATH via the obvious isomorphism. |
math/0103019 | MATH is left integral by REF . We recall the isomorphism MATH given by MATH. Given MATH, denote MATH, and note that, for all MATH, MATH . Whence MATH . Thus, MATH generates MATH and the mapping of MATH given by MATH is a MATH-linear projection. If MATH such that MATH, then MATH so MATH freely generates MATH. |
math/0103019 | REF shows that MATH is a NAME homomorphism for the dual NAME algebra MATH. The concepts of left and right norm relative to MATH make sense in the MATH-bimodule MATH. But REF implies that MATH is a left norm for MATH. Similarly, MATH is a NAME homomorphism MATH by applying the anti-automorphism MATH as in REF , and MATH is a right norm. |
math/0103019 | The rightmost equation follows from noting that MATH is a group-like element in MATH, whence MATH and MATH: that is, MATH and MATH fix MATH. The leftmost equation is computed below and follows CITE until REF : for every MATH, MATH by REF , respectively. |
math/0103019 | The dual bases REF follows directly from REF follows from REF since MATH is a coalgebra anti-automorphism. To compute the NAME automorphism we first need to find the inverse of REF : for all MATH, MATH . Next we apply REF where MATH is the anti-automorphism: MATH since MATH and MATH is an algebra and coalgebra automorphism. |
math/0103019 | We first show that MATH is a right integral in the MATH-bimodule MATH. Recall that MATH is canonically identified with MATH . Let MATH, then MATH since MATH. Since MATH is a right norm it follows that there is MATH such that MATH. But comparing REF to the application below of REF : MATH shows that MATH (compare REF ). |
math/0103019 | On the one hand, the NAME automorphism MATH for the NAME homomorphism MATH is by REF given by MATH for every MATH. On the other hand, the NAME automorphism MATH of MATH for the NAME homomorphism MATH is by REF MATH for every MATH. By REF , MATH, so by the comparison theorem MATH for every MATH. Substituting the first two equations in the third yields, MATH which is equivalent to REF since MATH fixes MATH and MATH, and for every group-like MATH, we have MATH. |
math/0103019 | The forward implication is proven by first letting MATH be the separability element for MATH. Next set MATH. Then MATH . The reverse implication is proven by noting that MATH is a separability element for MATH. By hypothesis, MATH where MATH is the multiplication mapping. MATH is in the center MATH of the natural MATH-bimodule MATH as a consequence of REF . |
math/0103019 | We make use of the dual bases MATH given by REF . If MATH is MATH-separable, then by the proposition above there is MATH such that MATH . Applying MATH we obtain MATH whence MATH is NAME. Conversely, if MATH is NAME with inverse MATH such that MATH, then we let MATH. Note that MATH whence MATH is MATH-separable by REF . |
math/0103019 | Suppose MATH satisfies MATH. From this and REF , we easily see that MATH is a NAME separability element. |
math/0103019 | If MATH is the MATH-Frobenius system for MATH given by REF , we note here that MATH, so that the MATH-element of REF , MATH is NAME by REF . |
math/0103019 | Since MATH in MATH for every MATH, it follows that the NAME isomorphisms MATH and MATH induce the first and second isomorphisms above (between MATH and MATH). |
math/0103019 | Since MATH is assumed finite projective, we need only show that MATH. We compute using the hom-tensor adjointness relation and two applications of REF : MATH . |
math/0103019 | The natural module MATH is finite projective as a corollary of the NAME Freeness theorem CITE. Furthermore, the NAME automorphism MATH for every MATH by REF , whence MATH. Thus the hypotheses of REF are satisfied. |
math/0103019 | It is enough to show that MATH. As an algebra, MATH, the tensor product algebra of MATH and the opposite algebra of MATH. Now MATH is MATH-Frobenius algebra if and only if MATH is, since they have the same NAME system with a change of order in the dual base. By REF , MATH is a MATH-Frobenius algebra. It follows from REF that MATH is a NAME algebra, since MATH. Now the MATH-space of integrals of an augmented NAME algebra is free of rank one, which proves our theorem. |
math/0103019 | Let MATH be the localization homomorphism MATH. Then it follows easily that MATH. |
math/0103019 | We need the following result proved in CITE: Suppose MATH is some linearly ordered set and for each ordered MATH, MATH is a commutative ring and there is an epimorphism MATH such that the restriction to the group of units MATH is a surjection. If MATH then the induced map MATH is injective. The hypotheses of this proposition are fulfilled by the mappings MATH whence MATH is injective. |
math/0103023 | Let MATH and MATH be Legendrian knots whose topological types are MATH and MATH, respectively and MATH and MATH. We also regard MATH and MATH as fronts. Further we can assume that MATH and MATH (respectively, MATH) lies in the left (respectively, right) region of MATH-plane, that is, MATH (respectively, MATH) as in REF . Then we connect MATH and MATH by joining a right cusp of MATH and a left cusp of MATH as in REF . This procedure produces a Legendrian knot whose topological type is MATH and NAME - NAME invariant is MATH. This completes the proof. |
math/0103024 | Since MATH by the MATH case of REF, we have MATH . |
math/0103024 | Since MATH by the MATH, MATH, MATH, MATH, case of REF , we have MATH . We arrived at a sum of MATH infinite multisums. In the MATH-th sum, for MATH, due to the factor MATH in the numerator of the summand, we shift the index MATH. We then obtain MATH . Next, we simplify all the MATH series according to REF , and obtain MATH . Now we apply the MATH, MATH, MATH, and MATH, MATH, case of REF , which can be rewritten as MATH to simplify the expression obtained in REF to MATH . Finally, using MATH we can easily transform the expression in REF into MATH which is the right side of REF, as desired. |
math/0103024 | To establish REF, we apply NAME 's argument successively to the parameters MATH using REF . Both sides of the multiple series identity in REF are analytic in each of the parameters MATH in a domain around the origin. Now, the identity is true for MATH and MATH, by the MATH summation in REF (see below for the details). This holds for all MATH. Since MATH is an interior point in the domain of analyticity of MATH, by the identity theorem of analytic functions, we obtain an identity for MATH. By iterating this argument for MATH, and analytic continuation, we establish REF for general MATH where MATH. The details are displayed as follows. Setting MATH, for MATH, the left side of REF becomes MATH . We shift the summation indices in REF by MATH, for MATH, and obtain MATH . Next, using the identities MATH and MATH together with the MATH, MATH, and MATH, MATH, case of CITE, specifically MATH and further the MATH, MATH, MATH, MATH, and MATH, MATH, case of the multidimensional summation formula in REF, we simplify the expression in REF to MATH . Now, this can easily be further transformed into MATH which is exactly the MATH, MATH, case of the right side of REF. |
math/0103025 | For the proof in NAME case see CITE. The statement for a reductive MATH easily follows from the statement for the factor of MATH by its center. |
math/0103025 | The ``if" part of REF follows from the definition of a nilpotent MATH-module. The ``only if" part of REF is proved in CITE. REF is proved in CITE. Recall that the NAME graph is assumed to be of finite (ADE) type. |
math/0103025 | The smoothness part of REF in the case of MATH are proven by CITE. The fact that MATH is connected is proven by CITE). The proof of REF is analogous to the one of REF, or one can deduce the former from the latter by transposing MATH. Either argument also proves the part of REF concerning MATH. REF (and the corresponding part of REF follows from the definition of MATH (MATH), REF is proven by CITE (see also CITE). |
math/0103025 | REF are proven in CITE (see the proof of REF for a similar argument). REF follows from REF, and REF. Let MATH denote the variety of all MATH-graded subspaces of graded dimension MATH in MATH. It is a MATH-homogeneous variety with connected stabilizer of a point, and MATH. The map MATH given by MATH is a locally trivial fibration over MATH with the fiber over MATH equal to MATH. Now REF follows from REF follows from REF. |
math/0103025 | The fiber of the map MATH over a point MATH in MATH consists of all linear maps MATH (where MATH, MATH) subject to the following linear equations MATH for any MATH. Now the proposition follows from the fact that the linear map MATH given by MATH is surjective. To prove this let MATH be orthogonal to the image of the above map with respect to the MATH pairing, that is MATH for any MATH, MATH, MATH, and any MATH. It follows that MATH for any MATH, MATH. In particular the kernel of MATH contains the image of MATH and is MATH-invariant. Since MATH is stable it follows that MATH and hence the map REF is surjective. The proposition is proven. |
math/0103025 | REF follows by induction in MATH using REF . The base of the induction is provided by REF follows from REF using the fact that MATH is a fibration with fibers isomorphic to MATH over the variety of graded MATH-step partial flags in MATH with dimensions of the subfactors given by MATH and MATH. The dimension of this flag variety is equal to MATH . REF follows from the fact that MATH is a union of a finite number of locally closed subsets MATH (for different MATH) having identical dimensions (compare REF). REF follows from the fact that MATH is open in MATH and from REF. To prove REF (respectively, REF) note that MATH (respectively, MATH) and the action of the group MATH on MATH (respectively, MATH) is free. Now REF (respectively, REF) follows from REF (respectively, REF). |
math/0103025 | The proof is analogous to the proofs of REF . Namely given MATH the fiber of MATH over this point is vector bundle MATH over an affine space consisting of all linear maps MATH subject to the equations MATH . Now REF follows from the condition that MATH. The proof is similar to REF (compare CITE), or one can deduce REF from REF and its dual using the fact that MATH if and only if MATH and MATH . The fiber of MATH over a point MATH is an affine space of all linear maps MATH subject to the equations MATH . One has to check that the systems of linear equations used in the proof are not overdetermined, which follows from the fact that MATH, and MATH (compare the proof of REF ). |
math/0103025 | REF follows from REF follows directly from definitions and the fact that MATH (a connected smooth variety) is the only element of MATH which is killed by MATH for all MATH. To prove this let MATH be an irreducible component of MATH, and MATH be a generic point of MATH. Let MATH be as in REF and MATH. If MATH then there exists MATH such that MATH (compare CITE). It follows that MATH, and hence MATH. Therefore if MATH for all MATH then MATH, which means MATH. To prove REF note that as sets both MATH and MATH are in natural bijections (induced by the vector bundles MATH - compare REF) with the set of irreducible components of the variety MATH, where MATH, and MATH. In this way one obtains a bijection between MATH and MATH, and it follows from definitions that this bijection is a crystal isomorphism. |
math/0103025 | Because of commutativity of REF it is enough to consider the case MATH, and because the definition of the crystal structure uses NAME restriction which commutes with tensor product (compare REF) it is enough to consider the case MATH. Now the theorem follows from REF . |
math/0103025 | The proposition follows from REF , and REF . |
math/0103025 | Proofs of REF, and REF are analogous to the proofs of REF , and REF , respectively, or one can deduce the former statements from the latter. To prove REF note that an element of the set MATH is an irreducible component of the variety MATH (compare REF), where MATH, MATH. A point of this variety is a stable triple MATH. Since it is stable MATH. Therefore MATH, which implies REF. |
math/0103026 | For the proof in NAME case see CITE. The statement for a reductive MATH easily follows from the statement for the factor of MATH by its center. |
math/0103026 | Fix MATH, MATH. The operator MATH defines a flag MATH . The variety MATH is a variety of all subspaces MATH such that MATH, MATH, MATH, MATH. The idea of the proof of the Proposition is to construct a subspace MATH satisfying the above conditions in several steps. On each step the choices are parametrized by a certain variety. Start with MATH. It is an arbitrary subspace of dimension MATH in MATH. So one has a NAME MATH of such subspaces. Having fixed MATH choose a subspace MATH inside of it (the choices are parametrized by the NAME MATH). Next step is to select MATH. The intersection MATH is already fixed. Thus one proceeds in two steps. First choose a subspace MATH inside MATH (which gives MATH). Then each choice of MATH corresponds to an element of the affine space MATH. At this point MATH and MATH are fixed (the latter is determined by MATH). Thus the remaining choices for MATH correspond to elements of MATH. A rigorous way to spell out the above argument is to say that one has the following chain of locally trivial fibrations MATH with fibers isomorphic to MATH, MATH, MATH, MATH, and MATH respectively. The proposition follows. |
math/0103026 | REF follow from definitions. For REF write MATH as a composition of fibrations (compare the proof of REF ): MATH . Here the notation is as follows: MATH is a variety of tuples MATH, where MATH and MATH is a subspace of MATH of dimension MATH; MATH is a variety of tuples MATH, where MATH and MATH (respectively, MATH) is an isomorphism MATH (respectively, . MATH); MATH is a variety of tuples MATH, where MATH and MATH is a subspace of MATH such that MATH, MATH; MATH, MATH, MATH are natural projections. The fiber of MATH is a NAME of dimension MATH, the fiber of MATH is isomorphic to MATH, the fiber of MATH over MATH is isomorphic to the affine space MATH, and one can describe the fiber of MATH over MATH as follows. Choose a subspace MATH (respectively, MATH, MATH) complimentary to MATH in MATH (respectively, to MATH in MATH, to MATH in MATH), and consider MATH (respectively, MATH) as a map MATH (respectively, MATH). Then MATH for some matrices MATH, MATH, and the fiber of MATH consists of operators MATH that have the following block form MATH with respect to the presentation of MATH as a direct sum MATH . In REF MATH is an arbitrary MATH matrix. REF follows. Proof of REF is analogues. The crucial step is to describe the set of matrices of the form MATH such that MATH. REF follows from the fact that matrices of rank equal to MATH form an open dense subset in the set of matrices of the form REF follows from REF follows from REF or REF together with REF. |
math/0103026 | Let MATH, MATH. Choose a basis MATH in MATH such that MATH and consider a subspace MATH defined as the linear span of the set MATH where it is assumed that MATH. Then the flag MATH belongs to MATH. |
math/0103026 | The fact that MATH is a normal crystal follows immediately from definitions. To prove that it is highest weight it is enough to show that it contains unique element MATH such that MATH . Fix MATH such that MATH, and let MATH be a generic point of a component MATH of MATH satisfying REF (``generic" here means ``not lying in any other component"). By definition of the operators MATH it implies that MATH . From this it follows by induction on MATH that MATH and then using inverse induction on MATH one obtains MATH . Hence the component MATH has one generic point MATH. It means that there is unique highest weight component MATH of MATH, MATH is the point, and MATH. The proposition follows. |
math/0103026 | REF follows from definitions. To prove REF write MATH as a composition of fibrations (compare the proof of REF): MATH . Here the notation is as follows: MATH is a variety of tuples MATH, where MATH and MATH is a subspace of MATH of dimension MATH; MATH is a variety of tuples MATH, where MATH and MATH (respectively, MATH) is an isomorphism MATH (respectively, . MATH); MATH is a variety of tuples MATH, where MATH and MATH is a MATH-step flag in MATH such that MATH, MATH; MATH, MATH, MATH are natural projections. The fiber of MATH is a NAME of dimension MATH, the fiber of MATH is isomorphic to MATH, the fiber of MATH is isomorphic to the affine space MATH and the fiber of MATH is isomorphic to the affine space MATH . REF follows. |
math/0103026 | The proposition follows from REF , and REF . |
math/0103027 | The following lemma will be of higher importance in the following. For each subgraph MATH of MATH, let us denote by MATH the partition of MATH in connected component. Then, if MATH, we have for MATH almost MATH, we have MATH where MATH . Let us define MATH by MATH and MATH. It is easy to see that MATH . We have MATH, and the equality holds if and only if MATH. The quantity residing in connected components intersecting the boundary of MATH can be controlled using well-known results about the distribution of the size of finite clusters. In both subcritical case and supercritical case, we can found MATH and MATH such that MATH . (We can take MATH when MATH and MATH if MATH. See for example the reference book of CITE for a detailed historical bibliography. ) It follows from a standard NAME argument that for MATH almost MATH, there exists a (random) MATH such that MATH . If follows that for each MATH, MATH is completely inside MATH, and therefore MATH . Then, MATH . By the ergodic theorem, we have MATH almost surely MATH . Since MATH , the result follows. REF we forget technical controls, the key point of this proof is the identity REF . It is interesting to note that CITE used an analogous trick to prove that MATH almots surely, when MATH is the number of open clusters in MATH. Let MATH be a sequence such that for each MATH, there exists exactly one MATH connected to MATH in MATH. Then MATH where MATH is the infinite connected component (MATH if there is none). Since MATH it follows from the ergodic theorem that MATH for MATH almost MATH. From now on, we will suppose that MATH is a such a graph, and that, moreover, it is such that the conclusions of REF hold - MATH almost all graph is such that. Now, our goal is to apply a light improvement of the well-known proof of CITE for the law of large number. Let us state our result. Let MATH be a sequence of pairwise independent and identically distributed variables. We suppose that MATH is integrable and note MATH. Let also be MATH be a doubly indexed sequence of non-negative numbers such that CASE: MATH . CASE: MATH . CASE: MATH . CASE: MATH . CASE: MATH . Then, almost surely MATH . By linearity, it is sufficient to prove the theorem for nonnegative random variables. Assume then that MATH. Moreover, we can assume without loss that MATH for MATH. This can be done by permuting columns of the matrix MATH. Let MATH . We define MATH and MATH. We also consider the truncated variables MATH and associated sums and quotients: MATH and MATH. MATH . It follows that there exists MATH such that MATH . Now, fix MATH and define MATH to be the integer which is the closest to MATH. Then MATH . But since MATH, it is easy to prove that MATH . Then, there exists MATH such that MATH . Now, MATH . It follows from NAME 's inequality and the first NAME lemma that MATH . By monotone convergence, MATH . Let MATH be such that MATH for MATH. Then, for MATH . Then, MATH, and since MATH is arbitrary, MATH. It follows that MATH . Now, we go back to not truncated variables. Since MATH, we have MATH . It follows that for almost all MATH, there exists MATH such that MATH for MATH. Then, for MATH . It follows that MATH . If MATH, then since MATH is non-decreasing, we have MATH . Since MATH, it follows that MATH . Since this is true for each MATH, we have proved that MATH . Our goal is to apply this result to the sequence MATH defined by MATH. Since the sequences MATH are non-decreasing, so are the sequences MATH. Indeed, we have MATH. Moreover, MATH . Since the MATH's are natural numbers, it follows that MATH is finite. In fact, MATH is the number of finite components which intersect MATH. Together with the ergodic theorem, REF gives MATH . As already mentioned, NAME has proved that MATH . Then, we have MATH . We must now prove that MATH is bounded. But MATH . Using the conclusions of REF , we get MATH which completes the checking of the assumptions. It follows that MATH . Since MATH, it comes from REF that MATH . |
math/0103027 | The first point is a consequence of the formula given in REF and the second point is a consequence of the first one, because MATH is non constant as soon as MATH is not a NAME measure. |
math/0103027 | Let MATH. We have MATH since MATH for MATH almost MATH. |
math/0103027 | MATH . Then MATH with MATH . By REF , we have for MATH almost MATH. Now, we have just to prove MATH . Therefore, we will prove that for MATH almost MATH, the sequence MATH satisfies the NAME condition. For each MATH, we have MATH with MATH. Then, the NAME condition is fulfilled if MATH. But we have already seen that MATH, whereas MATH. This concludes the proof. |
math/0103027 | MATH where MATH is the translation operator defined by MATH and MATH. Moreover MATH is an increasing function and MATH satisfies the FKG inequalities. Then, MATH is a stationary random field of square integrable satisfying to the FKG inequalities. Therefore, according to CITE, the Central Limit Theorem is true if we prove that the quantity MATH is finite. But MATH with MATH. Now, MATH . Hence MATH because MATH and MATH are independent as soon as MATH. Since MATH and MATH, we have MATH . Then, MATH and MATH . Since CITE have proved the existence of MATH such that MATH it follows that the series converges. Of course, a so sharp estimate is not necessary for our purpose. Estimates derived from CITE, and from CITE would have been sufficient. |
math/0103027 | Rearranging the terms of the sum, we easily obtain MATH . We will now put MATH and define MATH and MATH . As usually, it means that MATH. It is also important to emphasize that the following properties are fulfilled under MATH: CASE: MATH is independent from MATH. CASE: MATH is independent from MATH. Therefore, we have MATH . Conditioning by MATH and using the fact that MATH is MATH-measurable, we get MATH, with MATH and MATH . By REF we have for each MATH and MATH almost MATH: MATH . Then, by dominated convergence MATH . Then MATH where the last equality follows from REF . We have just proved that MATH . Since MATH we get MATH . By REF , it follows that MATH . |
math/0103027 | Using the characteristic function, it is easy to see that MATH is Gaussian if and only if MATH does. Let us define, for MATH: MATH . We have MATH . By definition of MATH, MATH is a symmetric measure. So if MATH if Gaussian, it is centered and we have MATH . Then, we have MATH . If we denote by MATH the image of MATH by MATH, we have MATH . Then, we have MATH . It follow that for MATH almost MATH, MATH: MATH is a NAME measure. Therefore, MATH. Since MATH, we necessary have MATH and then MATH. |
math/0103028 | Take MATH and MATH. Then MATH is absolutely continuous, and thus MATH . Moreover, using NAME 's inequality again, MATH and thus MATH . |
math/0103028 | Let MATH. Then REF yields MATH and the result follows by REF . |
math/0103028 | We argue as in the proofs of REF , using REF . |
math/0103036 | List the elements of MATH as MATH. (Note that MATH may be either finite or countably infinite.) For each MATH let MATH be the MATH matrix MATH with a MATH in the MATH position and REF's elsewhere. Also let MATH be the MATH matrix MATH with MATH's along the diagonal and MATH's above the diagonal. Now for each MATH let MATH be the vertices of the tail which is added to MATH to form MATH. Then, by the way that desingularization is defined, we see that with respect to the decomposition MATH the matrix MATH will have the form MATH where the MATH's and MATH's are row-finite. If we let MATH, then MATH. Also MATH. Let us define a map MATH by MATH . We shall show that MATH induces a map from MATH to MATH. Let MATH . Then MATH . Thus MATH induces a map MATH. We shall show that MATH is an isomorphism. To see that MATH is injective suppose that MATH . Then there exists MATH such that MATH. For each MATH let MATH where MATH denotes the MATH entry of the vector MATH. Then, for each MATH define MATH recursively by MATH where MATH denotes the MATH entry of the vector MATH and MATH denotes the MATH entry of the vector MATH. Now for each MATH define MATH by MATH. Then MATH and thus MATH and MATH is injective. Furthermore, since MATH is surjective it follows that MATH is surjective. Thus MATH. Next we shall examine MATH. Note that with respect to the decomposition mentioned earlier MATH will have the form MATH where the MATH's and MATH's are column-finite matrices. If we let MATH, then MATH. Also MATH. Let us define a map MATH by MATH . Note that if MATH, then MATH so MATH restricts to a map MATH. We shall show that this map is surjective. Suppose that MATH . Then for each MATH we must have that MATH. If MATH, then for all MATH we must have MATH . Since MATH we know that MATH is eventually zero. Thus the above equations imply that MATH. Since this holds for all MATH we have that MATH and MATH is surjective. Furthermore, since MATH is clearly injective, MATH is an isomorphism and MATH. Next we shall define a map MATH by MATH . We shall show that MATH induces a map from MATH to MATH. Suppose that MATH. Then there exists an element MATH such that MATH. Hence MATH . Thus MATH maps MATH into MATH and hence induces a map MATH. We shall show that this map is injective. Suppose that MATH equals zero in MATH. Then MATH . But then as before we must have that MATH and the above equation implies that MATH so MATH is injective. We shall now show that MATH is surjective. Let MATH. It suffices to show that there exists MATH such that MATH . For each MATH write MATH and define MATH . Note that since MATH is in the direct sum, all of the above sums are finite, and since MATH we have that eventually MATH and hence MATH . If we then take MATH and MATH which are finite sums since MATH is in the direct sum, we have that-MATH . Thus MATH is surjective. Hence MATH is an isomorphism and MATH. |
math/0103036 | Let MATH be a desingularization of MATH. Since MATH is row-finite and has no sinks it follows from CITE that MATH and MATH. By CITE MATH is NAME equivalent to MATH. Because MATH-theory is stable, we have that MATH and MATH. The result then follows from REF . Furthermore, if MATH satisfies REF , then it follows from CITE that MATH also satisfies REF . Hence by CITE we have that MATH. Since MATH is stable, the result again follows from REF . |
math/0103036 | MATH, so we have MATH, and the result then follows from REF . |
math/0103039 | REF implies that MATH is hereditary, and REF that MATH is saturated. Because MATH is the smallest saturated set containing MATH, it now suffices to prove that MATH. Suppose that MATH. Then either there is a path MATH from MATH to a sink MATH, or there is an infinite path which begins at MATH. In the first case, MATH cannot be in MATH because MATH and MATH is hereditary. In the second case, MATH cannot be in MATH because otherwise we would have an infinite path going round the finite set MATH, and there would have to be a loop in MATH. Either way, therefore, MATH, and we have proved MATH. |
math/0103039 | From REF and CITE, we see that MATH, and the result follows from CITE. |
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