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math/0009179 | Consider MATH such that MATH is contained in MATH. Since MATH, MATH is inside a MATH-sector of angle MATH which has as its vertex a point in MATH. We claim that this vertex is the point MATH such that MATH, where MATH is the predecessor of MATH at level MATH. Otherwise, MATH-jumps, by the first statement in REF . Moreover, by the second statement in REF , the inverse branch MATH, defined near to MATH, such that MATH must be real. Let MATH. By REF , if MATH is contained in the MATH-sector whose vertex is MATH, then MATH, MATH, is contained in the MATH-sector whose vertex MATH satisfies MATH. Using the previous arguments along all MATH-cycle, the lemma follows. |
math/0009179 | (compare CITE, subsection REF) By the Main Lemma and REF , if MATH then MATH has complex pullbacks along the MATH-cycle. Moreover, if MATH is such that MATH is contained in MATH, then MATH. Hence MATH, where MATH is bounded. Furthermore, for points satisfying MATH, since MATH is large enough, one has MATH . Hence, let MATH be a disc large enough such that REF holds in MATH. Then the pullback of MATH along the MATH-cycle is contained in MATH. Using arguments as in the proof of REF , the complex pullbacks of MATH are inside MATH. Hence, taking a larger disc, if necessary, the pullback of MATH is compactly contained in MATH (clearly the pullback is contained in MATH, but this could not be true for the closure of the pullback). One gets a polynomial-like extension. Now, it is necessary to control the modulus of the polynomial-like extension. By REF, it is sufficient to prove that the ring MATH has the modulus bounded below, where MATH is the filled-in NAME set of MATH. We claim that MATH is contained in a sector supported on MATH. Indeed, take MATH. Let MATH. If MATH, one gets that MATH. Assume that MATH. Denote MATH. If there exist MATH, MATH and MATH satisfying the following properties CASE: MATH is contained in MATH. CASE: MATH is contained in MATH. CASE: MATH is contained in MATH, for some MATH. Moreover MATH. CASE: MATH. CASE: MATH is minimal with the above properties. then we capture MATH in MATH, by REF . Using arguments as in the proof of REF , we obtain that MATH. In the other case, MATH never MATH-jumps for all MATH. Then, by REF , MATH has the itinerary of a point in MATH. Let MATH be the interval of monotonicity of MATH which contains this point. Clearly MATH. Since MATH is contained in MATH and MATH is bounded, MATH goes exponentially fast to zero. Hence By REF , MATH, where MATH, by REF . We captured the NAME set in three NAME neighborhoods: two of these neighborhoods are based on intervals contained in MATH which contain points in MATH. The other one is MATH. Hence the NAME set is contained in MATH, for some angle MATH. Indeed, the filled-in NAME set is inside a sector supported on MATH. We follows the argument in CITE: Consider the interval MATH of monotonicity of MATH which contains the periodic point MATH in MATH. We have MATH. Let MATH be the inverse branch of MATH such that MATH is deeply inside MATH, for large MATH. The map MATH has exactly one fixed point MATH in MATH and MATH, where MATH, MATH does not depend on MATH. We can linearize MATH inside MATH. Do the pullback of MATH along the k-cycle a sufficient number of times (which does not depend on MATH) such that the pullback will be contained in MATH, for some interval MATH such that the component of MATH which contains MATH is contained in MATH, for a small MATH. In particular, since MATH, the NAME set cuts the fundamental ring MATH at most a certain angle. Using that the linearization has small distortion on MATH, if MATH is small, and the invariance of MATH by MATH, we obtain the sector. By invariance of MATH, we obtain the sector in other point of MATH. |
math/0009182 | First we consider MATH. From REF the number of elements of MATH with MATH rational canonical form data MATH is MATH . From REF , MATH can be written as MATH . Thus the number of elements of MATH with MATH rational canonical form data MATH is MATH . Consequently it is sufficient to prove that MATH . Since only partitions corresponding to the polynomial MATH are involved in this last equation, we simplify notation by suppressing the dependence on MATH and by further replacing MATH and MATH by MATH and MATH. From REF one sees that MATH . Clearly if MATH then MATH for MATH and MATH. Elementary combinatorics then shows that this last expression is equal to MATH where the final equality is REF . This completes the proof of REF. REF follows from REF . |
math/0009182 | (First proof) REF on page REF states that MATH . The result now follows by first setting MATH and taking coefficients of MATH on both sides, and then using the fact that MATH. |
math/0009182 | (Second proof) Setting all MATH in the cycle index of MATH equal to REF gives the identity MATH . Since MATH is a probability measure it follows that MATH . Setting all variables in the cycle index for MATH equal to REF and using the fact that MATH is a probability measure implies that MATH (of course this can be proved directly). The lemma follows. |
math/0009182 | Write the NAME expansion MATH. Then observe that MATH. |
math/0009182 | As explained in the second proof of REF , we know that MATH . Multiplying this by the cycle index of MATH gives that MATH . This proves the first assertion of the theorem. For the second assertion use REF . The proofs of the third and fourth assertions are almost identical so are omitted. |
math/0009182 | Let MATH denote the the shape obtained by decreasing the size of column MATH of MATH by one and let MATH if MATH is not a partition. To prove the theorem it is sufficient to show that for MATH, MATH . REF shows that MATH where the inner sum is over all paths in the NAME lattice from the empty partition to MATH to MATH and all other notation is as in REF . Simplifying further and using REF again gives that MATH . |
math/0009182 | The MATH probability of choosing a partition with MATH for all MATH is MATH . REF shows that MATH. Thus it is enough to prove that MATH for all MATH. Letting MATH be the MATH probability that MATH it is proved in CITE that MATH. Next one calculates that for MATH . Similarly, observe that MATH . Thus the ratio of these two expressions is MATH as desired. The case MATH must be checked separately but the same ratio results. |
math/0009182 | The survey CITE proved that if one replaced REF in the above algorithm by stopping then one would sample from MATH given that MATH. Let MATH denote MATH with the size of column MATH decreased by one and let MATH be the probability that REF of the Affine NAME Tableau Algorithm adds to column MATH of MATH. The result follows because MATH . |
math/0009182 | We sum over all NAME tableaux MATH with MATH parts the chance that the Affine NAME Tableau Algorithm outputs MATH. Recall from the second remark after REF that one can compute the probability that the Affine NAME Tableau Algorithm outputs any given NAME tableau. The point is that one can in fact compute the probability that the Affine NAME Tableau Algorithm produces a tableau MATH with given values of the MATH's. To do this, consider what happens during REF of the Affine NAME Tableau Algorithm. Suppose that one moves along the NAME lattice from a partition with MATH parts. REF implies that the weight for adding to column MATH is MATH, and that the sum of the weights for adding to some other column is MATH. Thus the chance that REF of the Affine NAME Tableau Algorithm yields a tableau with given values MATH and all other MATH=REF is MATH . In order for the Affine NAME Tableau Algorithm to generate a partition with MATH parts, either REF generates a partition with MATH parts and REF increases the size of column MATH or else REF generates a partition with MATH parts and REF increases the size of column MATH. Thus the probability that the Affine NAME Tableau Algorithm generates a partition with MATH parts is MATH . |
math/0009182 | In the equation of part one of REF set MATH if MATH has k parts and MATH otherwise. Also set MATH for MATH. It follows from REF that MATH . For the second assertion of the theorem use REF to conclude that MATH as desired. |
math/0009182 | In the equation of REF set MATH if MATH has k parts and MATH otherwise. Using REF one sees that the sought number is MATH . |
math/0009182 | Expanding MATH as a geometric series and using unique factorization in MATH, one sees that the coefficient of MATH in the reciprocal of the left hand side is MATH times the number of monic polynomials of degree MATH, hence REF. Comparing with the reciprocal of the right hand side completes the proof. |
math/0009182 | The first equality follows from REF together with the fact that an element MATH of MATH is separable if and only if all MATH have size at most REF. The second equality follows from the cycle index for MATH. |
math/0009182 | REF implies that MATH . Taking coefficients of MATH on both sides and using the fact that MATH one obtains that MATH . For any MATH as above, page REF gives the bound MATH . Thus by the triangle inequality, MATH . Now summing over MATH and using the triangle inequality gives MATH . For MATH observe that taking coefficients of MATH on both sides of MATH gives that MATH . Now use NAME 's bound for MATH to conclude that MATH . |
math/0009182 | The first equality follows from REF together with the fact that an element MATH of MATH is cyclic if and only if all MATH have at most REF part. The second equality follows from the cycle index for MATH. |
math/0009182 | CITE shows that MATH is analytic in MATH except for a simple pole at MATH which has residue MATH. REF implies that MATH is analytic in MATH except for a simple pole at MATH which has residue MATH. Thus MATH is equal to MATH. |
math/0009182 | REF imply that MATH . |
math/0009182 | Taking coefficients of MATH on both sides of the equation in REF gives that MATH . Using the triangle inequality and the fact that MATH, it follows that MATH . |
math/0009182 | An element MATH of MATH is semisimple if and only if all MATH have at most one column. Now use REF . |
math/0009182 | The discussion in CITE shows that MATH is analytic within a circle of radius greater than REF, except for a simple pole at MATH. Since MATH is also analytic within a circle of radius greater than REF, it follows that MATH . Now simply use both NAME identities and the formula for MATH in CITE stated at the beginning of this subsection. |
math/0009182 | The first inequality follows because semisimple matrices are separable. The second inequality follows because a matrix which is not separable is either not cyclic or not semisimple. |
math/0009191 | Note that inversions and permutations do not affect MATH, so we need only consider the case where MATH is a NAME twist. Let MATH be a cyclically reduced element of length MATH. Let MATH, where MATH. Consider MATH where MATH denotes the reduced word obtained from MATH. By the Bounded Cancellation Lemma CITE there is a constant MATH such that at most MATH cancellations occur after concatenation of the words MATH and MATH. Hence MATH can decrease by at most MATH (cancellations may occur at the beginning and at the end of MATH). Let MATH. We now have MATH . If we take MATH, our claim is proved for elements of MATH. Using a similar argument, we see that there is a constant MATH such that MATH . |
math/0009191 | We prove our claim by induction on the (minimal) index, MATH, of the filtration element that contains a path MATH. If MATH there is nothing to be proved since MATH contains only one edge MATH which is fixed by MATH. Suppose the claim is true for the subpaths contained in MATH that satisfy out hypothesis, and let MATH be a path in MATH for which MATH. By REF for every MATH, MATH splits into subpaths that are either single edges or exceptional paths. Denote MATH by MATH, so that MATH, where MATH is either a single edge or an exceptional path. Assume there is an exceptional path MATH which is not fixed by MATH. Without loss of generality we may assume that MATH, where MATH, MATH and MATH. Now we have that MATH and MATH . Since MATH is not fixed, MATH and MATH cannot be equal. Therefore MATH . If all exceptional paths in MATH are fixed, there exists an edge MATH which is not fixed by MATH. We know that MATH, where MATH is a closed path contained in MATH. If MATH, our claim is proven since MATH and so MATH . If MATH, there exists MATH such that MATH. We now have MATH . |
math/0009191 | Let MATH be as in REF . Since MATH there is a closed path MATH which is not fixed by MATH. We know that for every MATH, MATH splits into subpaths that are either single edges or exceptional paths. Denote MATH by MATH, so that MATH. If there is an exceptional path MATH in this splitting which is not fixed by MATH, we get MATH as in REF . If all exceptional paths in MATH are fixed, there exists an edge MATH such that MATH, where MATH is a closed path contained in MATH. By REF there exists MATH such that MATH . Hence, in all the above cases, there is MATH such that MATH . |
math/0009195 | Suppose that MATH and MATH and put MATH, MATH with the above constructed MATH. Then MATH for any MATH, which implies the inequality MATH for the corresponding kernels. Denote by MATH the integral operator induced by the kernel MATH. Due to REF the operator MATH is bounded, MATH, and MATH. Moreover, MATH for any MATH by construction. Since MATH strongly as MATH, the norms MATH are bounded uniformly in MATH and the set MATH is weakly compact in MATH. Consequently there exists a sequence MATH and an operator MATH such that MATH weakly as MATH. Passing to the limit MATH in MATH and MATH, we find that MATH and MATH. Since MATH was arbitrary, these inequalities prove that MATH is the weak limit of MATH as MATH and that MATH is the desired absolute value MATH of MATH. |
math/0009195 | It suffices to notice that MATH is an integral operator with the nonnegative kernel MATH and hence belongs to MATH. |
math/0009195 | Since MATH, we have MATH. By REF MATH, whence MATH and MATH, where MATH. It follows that the kernels MATH and MATH of the operators MATH and MATH satisfy the inequality MATH in particular, MATH. Denote by MATH the integral operator induced by the kernel MATH. By the above inequality MATH is a bounded operator belonging to the algebra MATH, MATH, and MATH. Therefore the norms MATH are bounded uniformly in MATH and there exist a bounded operator MATH and a sequence MATH such that MATH converge weakly to MATH as MATH. Passing to the limit MATH in the relations MATH we find that MATH solves the equation MATH and satisfies MATH. The proof is complete. |
math/0009195 | Since MATH by REF and the assumptions of the corollary, REF applies with MATH and MATH instead of MATH and MATH, and the claim follows. |
math/0009195 | We shall seek for MATH of the form MATH where MATH are found recursively from the relations MATH with MATH. In virtue of REF we find successively MATH, MATH such that MATH. Therefore MATH, which shows that the series MATH converges absolutely in the uniform operator topology and its sum MATH satisfies the equality MATH . The theorem is proved. |
math/0009195 | Observe that the solution MATH constructed in the proof of REF satisfies the inequality MATH. Therefore MATH, and it suffices to note that MATH if MATH. |
math/0009195 | It suffices to notice that the assumptions of REF are satisfied for the integral operator MATH with the kernel MATH . |
math/0009195 | We prove first that MATH is bounded in MATH. In fact, MATH whence MATH by the NAME test CITE. Next, put MATH . Then the induced integral operator MATH is a NAME operatorin MATH (see, for example, CITE) and MATH as MATH in view of MATH and the above arguments. Therefore MATH is a NAME operator as well and the lemma is proved. |
math/0009197 | Since MATH, we have MATH . By the preceeding lemma this is equivalent to MATH . Let MATH be any index between MATH and MATH. It follows from MATH that MATH and by induction we have MATH . Thus MATH . |
math/0009197 | The cycle decomposition of MATH contains a cycle of the form MATH, where MATH. To show that MATH it suffices to show that MATH . MATH . By REF we have MATH since MATH . By REF , MATH and by REF we have MATH since MATH . Thus MATH . It follows trivially from MATH that the remaining two conditions for MATH are satisfied. Now we show the uniqueness of this index MATH . Pick any MATH . The decomposition of MATH may contain a cycle of the form MATH then MATH . Since this is a MATH-cycle, MATH . Hence MATH . (Note that this argument holds when such a MATH does not appear in MATH . In this case delete all the MATH-s.) Finally, we prove the second statement in the lemma. Pick any MATH . In MATH there may be a cycle of the form MATH with MATH . For the second statement in the lemma it suffices to show that MATH . In cycle notation, MATH . This is a MATH-cycle, so MATH . Thus MATH . |
math/0009197 | First pick any MATH . There are no cycles in MATH that involve the index MATH but there may be a cycle of the form MATH where MATH . In cycle notation MATH . Since this is a MATH-cycle, MATH . It follows that MATH . Now pick any MATH . In MATH there may be a cycle of the form MATH . In this case MATH . Since this is a MATH-cycle, MATH and it follows that MATH . |
math/0009197 | The decomposition of MATH contains a cycle of the form MATH . Let MATH . By REF we have MATH and hence MATH . The uniqueness of this index MATH and the second statement in the lemma can be proved with arguments similar to those in the proof of REF . |
math/0009197 | Suppose MATH with MATH. Since MATH but MATH the cycle decomposition of MATH must contain a cycle involving MATH or MATH . Assume both are present and write MATH . Since MATH we have MATH and it follows that MATH . Let MATH be the least index, MATH such that MATH . Then MATH . Let MATH and MATH . Since MATH we have MATH . Also, MATH since MATH . Hence MATH and it follows that MATH . The converse is proved similarly. |
math/0009197 | Let MATH denote the set of indices for MATH and let MATH denote the set of indices for MATH . Since MATH all objects in the definition of MATH coincide with those in the definition of MATH . |
math/0009197 | CASE: Let MATH denote the set of indices for MATH . Since MATH . It follows by definition that MATH and MATH . We observe that the simple reflections MATH occuring in the reduced words for MATH all have MATH . Hence MATH . CASE: From the definition we see that MATH . For MATH, MATH and MATH have no simple reflections in common, so MATH . By REF , MATH . |
math/0009197 | Let MATH and let MATH be the objects involved in the construction of this element. Let MATH and let MATH be the objects involved in the construction of this element. Since MATH there is a MATH such that MATH if MATH, MATH and MATH if MATH . If MATH then MATH so we assume MATH . Let MATH . CASE: First observe that MATH and MATH . It remains to show that MATH and this requires a small trick. We recall that MATH and write MATH . From MATH and MATH it follows that MATH . Also, MATH so MATH . Hence MATH . CASE: Now we show by direct computation that MATH for any MATH . It will then follow by REF that MATH . A rather laborious computation using the relationship between MATH and MATH yields MATH and MATH . This is a reduced decomposition, so MATH . CASE: Finally, we observe that MATH is a reduced word. Hence MATH and by REF MATH . |
math/0009197 | The set MATH of indices for MATH is also the set of indices for MATH . MATH . |
math/0009197 | By REF , MATH . By REF from REF, MATH . |
math/0009197 | Applying emma REF we get MATH . The term in braces may be rewritten as MATH and we apply REF to obtain MATH . Thus MATH and gathering together the coefficients of MATH, we arrive at the conclusion of the lemma. |
math/0009198 | See CITE. |
math/0009198 | REF follows from REF applied to the vector space MATH with the filtration induced by the MATH-grading, and the set of operators MATH . |
math/0009198 | The statement is clear for MATH, that is, for NAME triples. The general case follows from REF. |
math/0009198 | By REF, the numbers computing the cardinality of MATH satisfy the same recursion relation as the NAME numbers MATH. |
math/0009198 | For MATH we have to check that MATH, where MATH are given by REF. First, MATH, because MATH. Among REF for MATH we have new cases only for the triples MATH and MATH. They follow from MATH . (The last inequality follows from REF with MATH.) Among REF for MATH we have new cases only for MATH. The case MATH is weaker than MATH. The case MATH (respectively, MATH) follows from MATH (respectively, MATH) and REF for MATH with MATH. Next, we have to check that the images of the maps MATH are disjoint. Suppose they are not, and MATH and MATH for some MATH,MATH,MATH. Then, the first equality implies MATH. Therefore, we have MATH. It follows that the positive part terms, that is, MATH, in the second equality are equal and after its cancellation MATH. Therefore, we have MATH and MATH. Finally, let MATH. Note that MATH where MATH . Moreover, the triple MATH is admissible, because MATH by REF for MATH. Therefore, it suffices to show that MATH. This is also straightforward. The combinatorial path MATH looks like MATH and the combinatorial path MATH looks like MATH . The relation MATH follows from MATH. REF are clearly preserved. REF for MATH is also clear in all places except when MATH; moreover the case MATH obviously follows from the case MATH. If MATH then we have MATH and REF for MATH with MATH follows from REF for MATH with MATH. If MATH then we have MATH and REF for MATH with MATH follows from REF for MATH with MATH. |
math/0009198 | First, we prove that for any combinatorial path MATH and for any level MATH . NAME triple MATH, we have a decomposition of MATH, MATH where MATH, MATH and MATH; and a decomposition of the triple MATH, MATH where MATH, MATH, are level MATH triples, such that we have MATH . We prove this statement by induction on MATH using the induction hypothesis that any level MATH combinatorial path of length MATH can be represented as a sum of MATH combinatorial paths of level MATH. It is obvious that the decomposition REF for MATH implies the above hypothesis for MATH. The decomposition of MATH is not unique. We can choose any decomposition MATH. Among the pairs MATH, we have MATH pairs MATH, MATH pairs MATH and MATH pairs MATH. We call these boundary pairs. They are independent of the choice of the decomposition. The decomposition of MATH is unique. We have MATH triples MATH, MATH triples MATH, MATH triples MATH and MATH triples MATH, where MATH are given by REF. We want to label these triple as MATH so that the equality REF holds. Recall REF. The key property in the proof is that for level MATH we have MATH and MATH in all other cases. Now we match up the level MATH triples with the boundary pairs. The triples MATH and MATH must match up with the pairs MATH because of the consistency of MATH. Their numbers are the same, and we match up arbitrarily among them. If MATH we match up all MATH with MATH, and we match up MATH with MATH or MATH. If MATH we match up all MATH with MATH, and we match up MATH with MATH or MATH. It is easy to see REF by using the above remark on MATH for level MATH. Now, we prove the statement of the proposition using the result obtained above. One can group MATH as MATH where MATH so that MATH. We have the corresponding grouping MATH. Then, setting MATH, we have MATH . The statement follows from this. |
math/0009198 | By the construction MATH intertwines the maps MATH and MATH. Now the statement follows from REF by induction on MATH. |
math/0009198 | This follows from REF . |
math/0009198 | Follows from REF. |
math/0009198 | Follows from REF. |
math/0009198 | The assertion follows from REF . Indeed, MATH. Also, we always have MATH. Therefore, for energy MATH and spin MATH we only have to compensate the contribution from MATH and MATH. To prove the equality for MATH we note that MATH . |
math/0009198 | Let us denote the induced actions by MATH, MATH, MATH. We will show MATH. (The proof of MATH is similar, and the rest of the relations are straightforward.) We have MATH. The mapping MATH sends MATH to MATH. For MATH, let MATH be the corresponding element in MATH. Then we have MATH . |
math/0009198 | We give the proof for the first map. The second one is similar. We need to show the validity of the relations REF . The relation MATH follows from MATH. The relation MATH follows from a variant of REF . Finally, the relation MATH follows from the integrability REF , and MATH from a variant of REF . |
math/0009198 | REF follows from REF . |
math/0009198 | The relation MATH for MATH follows from MATH. The relation MATH for MATH follows from MATH by a variant of REF . |
math/0009198 | It is enough to show that the vector MATH satisfies MATH . First, note that the quotient module MATH is generated by MATH where MATH. Note also that that MATH and MATH. If MATH, then MATH. Therefore, we have MATH because MATH . If MATH, we have MATH because MATH . We have MATH because of integrability REF . Consider the case MATH. We have MATH. From REF we have MATH . From REF we have MATH . Finally, consider the case MATH. We have MATH, and by using REF , we obtain MATH . |
math/0009198 | Take an element of MATH of the form MATH where MATH is the highest weight vector of MATH. These vectors form a spanning set. Suppose that MATH, that is, MATH. Let us prove that in MATH we have MATH where the vector MATH is the highest weight vector of MATH. Note that we have MATH . This is the only property of MATH we need for the proof. We use induction on MATH. If MATH then MATH. Using MATH and MATH, we have MATH . Therefore, we have MATH . If MATH, we have MATH . Note that MATH. Changing MATH to MATH, MATH to MATH and MATH to MATH, we apply the induction hypothesis. Thus we obtain MATH . From REF we have MATH . |
math/0009198 | We consider the first sequence. Consider the following commutative diagram. MATH . The columns and the first and second rows are exact. One can check that the third row is exact by the standard diagram chasing. REF follows from the exactness of the third row and the surjectivity of REF. It is clear from REF that the modification of the spaces REF does not break the exactness. |
math/0009198 | First, we consider the spaces of coinvariants with respect to MATH without the cutoff MATH. We use the surjection MATH . Noting that MATH, we can induce a surjection from REF . MATH . We continue to another surjection MATH . Therefore, we obtain the surjection MATH . Finally, we show that the cutoff REF does not break the surjectivity. There are two cases we must check: the case MATH and MATH. In the first case, the weight spaces that are cut off in REF are mapped to zero. The proof is similar to the proof of REF . In the second case the proof is more elaborate. If MATH, the weight spaces that are cut off in the MATH-coinvariants, are mapped to the weight spaces that are also cut off. However, if MATH (and therefore MATH), this is no longer true. In this case we show that the weight spaces that are cut off are mapped to zero. Let MATH, and MATH. For a subset MATH of MATH, we denote MATH . We set MATH. We claim that if MATH, then MATH . First consider the case when MATH. Since MATH in MATH, it is enough to show the equality MATH in an arbitrary MATH-module assuming that MATH and MATH. Let MATH be the set of all subsets of MATH with the cardinality MATH, and denote by MATH the complement of MATH. We have MATH . Define the matrix MATH indexed by MATH: MATH . Note that MATH is equivalent to MATH for some MATH because MATH. Therefore, MATH is the incidence matrix of the term MATH in MATH. It is easy to show that the matrix MATH is non-degenerate (for example, by calculating its inverse). Therefore, REF follows. If MATH, we start with showing the equality MATH . If MATH, REF follows from MATH in MATH. If MATH, we have MATH. Since REF follows by induction on MATH. Now we show REF by using REF . For MATH we define MATH by MATH. The left hand side of REF is a positive linear combination of MATH. We will show that MATH for any MATH. Then, our REF follows from REF . The symmetry REF is true for the transpositions MATH and MATH. Therefore, it is enough to show the symmetry with respect to the transposition MATH. We use the notations MATH . For MATH satisfying MATH, MATH and MATH, we set MATH . Here MATH is a subset MATH such that MATH, and MATH is its complement in MATH. We have MATH. Note also that MATH. We will prove that MATH is in fact independent of MATH and MATH . In particular, we have MATH and therefore MATH is symmetric with respect to the transposition MATH. The proof is by induction on the pair of integers MATH in the lexicographic ordering. For the induction steps we use the identities MATH . Using REF with MATH we obtain REF with MATH. Because of the induction hypothesis, we can rewrite REF with MATH as MATH . Here MATH denotes the right hand side of REF and MATH is an arbitrary subset of MATH such that MATH. From this follows that MATH is independent of MATH and it is given by REF . |
math/0009198 | We consider the first map. We need to show that the MATH-action on MATH satisfies MATH, MATH and MATH. The first two relations follow from the corresponding relations for MATH. The last one is proved in a similar manner as the proof of MATH in REF , by using the relation MATH in MATH. |
math/0009198 | REF follows from REF . |
math/0009198 | REF follows from REF . |
math/0009198 | The MATH-module MATH is generated by the vector MATH. The set of vectors of the form MATH is a spanning set of MATH. In fact, the operators MATH can be eliminated either by using the highest weight condition MATH or by taking the quotient with respect to the subspace MATH. The statement of the lemma follows. |
math/0009198 | We show the first line of inequalities, the second one is proved similarly. Using REF we have MATH . From REF we have MATH . This lemma follows from these inequalities along with REF . |
math/0009198 | This corollary follows from REF . |
math/0009198 | Since MATH, MATH and MATH are equivalent to MATH and MATH, respectively. Therefore, the admissibility of the triple MATH reduces to MATH . Namely, the sum in REF is taken over MATH and MATH such that MATH . This is equal to the sum over MATH . Now consider the surjection MATH . The sum in the left hand side is equivalent to the sum over MATH . Set MATH . Then, we have MATH . The sum REF is equal to REF by this identification. |
math/0009198 | We will prove the above equality by induction on MATH in three kinds steps. The first steps are MATH, the second MATH, and the third MATH. In the first and the third steps, we assume that MATH and prove that MATH . In the second steps, we assume that MATH and prove that MATH . In each step, we also prove the equality of the form REF . The base of the induction is MATH, where REF is valid because of REF . Note also that REF follows from REF with MATH replaced with MATH, which is proved in the first steps. Now, we show the first and the third induction steps at the same time. Using REF , we have the chain of inequalities: MATH . Using REF we obtain MATH . Summing up these inequalities for MATH and finally using REF we obtain MATH . In particular, we obtain REF . In the second steps we proceed with MATH. We only append one formula to REF : MATH . The rest of proof is similar. |
math/0009198 | The corollary follows from the proof of REF . |
math/0009198 | This is proved in the proof of REF . |
math/0009198 | The corollary follows from the proof of REF . |
math/0009198 | By REF , all we have to show is the injectivity of the mappings MATH . The injectivity follows from the proof of REF . |
math/0009198 | This follows from REF by letting MATH. |
math/0009198 | The recursion REF gives us a recursive way of constructing monomial basis of the space of coinvariants MATH. As shown in REF , for each MATH satisfying MATH, the mapping REF gives a bijection between the set of MATH satisfying MATH and the set of MATH satisfying MATH and such that MATH is admissible. Recall also that under this identification, MATH given by REF satisfies REF with MATH. Suppose that we have a set of monomials MATH for each MATH such that for each MATH the union MATH forms a basis of MATH. The inverse map of the bijection REF (with MATH replaced by MATH) maps the basis MATH to the basis MATH. By REF we have the basis MATH of MATH. By REF we then have the basis MATH of MATH. In this way we obtain the basis of MATH from the basis of MATH. Noting that MATH and comparing these formulas with REF we see that this recursion gives exactly the basis REF . |
math/0009198 | Consider the exact sequence obtained by the same argument as in the proof of REF . MATH . Here MATH is the canonical surjection and MATH is the submodule of MATH generated by MATH. The subspace MATH appears in the canonical filtration of the first kind as MATH. Therefore, by REF . we know the exact decomposition of the adjoint graded spaces for MATH and MATH. In particular, we have MATH and MATH . Therefore, we have MATH . From this the statement of the proposition for MATH follows. Next, consider the exact sequence MATH . The subspace MATH appears in the canonical filtration of the second kind as MATH. Applying the same argument as before, we obtain the statement of the proposition for general MATH. |
math/0009198 | We have to show the validity of the relations REF , or REF . We will give the proofs for the non-trivial ones. The relations MATH for REF follow from REF with MATH. The relations MATH for REF follow from REF with MATH. |
math/0009198 | The proof goes similarly as in REF except for the difference in the formulas of MATH between the first term and the second and third terms. The difference of the ranges of MATH for MATH and MATH is given by MATH . In this range, we have MATH . Therefore, we have MATH . On the other hand, the range of MATH for MATH is MATH . Noting that MATH we have MATH . Therefore, we have MATH . By the correspondence MATH these two ranges of MATH coincide. |
math/0009198 | The proof goes similarly as in the previous proposition. Noting that MATH, we have MATH for MATH and MATH. The range of MATH for MATH is MATH . The subset corresponding to MATH is exactly the range for MATH. Similarly, we have MATH for MATH where MATH . By the correspondence MATH the range of REF overlaps the range of REF exactly when MATH . |
math/0009198 | The exactness of triple REF follows from REF . The other two cases are obtained from REF by applying the automorphisms MATH and MATH. |
math/0009198 | Suppose that an element MATH can be expressed in two different ways: MATH where MATH. We must show that MATH for each MATH. Consider a pair of vectors spaces and their subspaces: MATH. Then, we have MATH. Using this one can prove that MATH . The welldefinedness of REF follows from this, and the rest of the lemma is easy. |
math/0009198 | It is enough to prove the case MATH and MATH. There are four cases: MATH, MATH, MATH, MATH. We use induction on MATH. If MATH, the assertion is clear from REF . In the induction, we use the recurrence relations for the space MATH given by REF . We have the following bijections: MATH . Let us prove the injectivity of MATH. We use the following abbreviated notations. MATH, MATH, MATH. We denote the highest weight vectors in MATH, respectively, by MATH. Note that MATH, and hence MATH . Consider the following mapping diagram. MATH . The first and the third horizontal arrows are the coproduct maps. The downward vertical arrows are canonical surjections, and the upward vertical arrows are the canonical bijections of REF . The second horizontal arrow is induced from either of the other horizontal arrows. Noting that the shift automorphism MATH and the coproduct MATH are commutative, one can check that the induced map is unique, and thereby we have a commutative diagram. By the induction hypothesis, the third horizontal arrow is injective. Therefore, the second horizontal arrow is also injective. From this follows that the kernel of the first horizontal arrow is included in a smaller subset in the filtration. Repeating this argument, that is, using induction on MATH, one can show the injectivity of the coproduct map MATH. In other three cases, the relation REF is modified. Let us consider REF . Using similar abbreviated notations, we have MATH . This induces a mapping of the form MATH . Here we use MATH. If MATH, we consider the following mapping diagram. MATH . The first horizontal arrow is MATH. The downward vertical arrow in the right is assured by REF . The rest of proof is similar. If MATH the second term in the right hand side of REF is absent. In this case, we use MATH . The rest of the proof is similar. In REF , the proof will go with MATH using the bijection of the form MATH . |
math/0009198 | We assume that MATH. Consider the following commutative diagram. MATH . The horizontal arrows are coproducts, the vertical up-arrows are compositions of canonical injections and surjections, and the vertical down-arrows are canonical surjections. The only non-trivial one is MATH which follows from REF . We are to prove that the first horizontal arrow is injective. We know that the third horizontal arrow is injective (REF up to the automorphisms MATH given by REF ). Note also that because of REF we can apply REF . Suppose MATH belongs to the kernel of MATH. If MATH, we take the smallest values MATH such that MATH belongs to the image of MATH. Take a preimage MATH of MATH. We have MATH because MATH. We show a contradiction by finding MATH satisfying MATH. For simplicity set MATH and MATH. Since MATH, MATH belongs to MATH. We will show that MATH . Prepare a set of vectors MATH for each MATH such that MATH forms a basis of MATH. Because of REF MATH forms a basis of MATH. Suppose that MATH where MATH. We write MATH . We have MATH . Note that MATH . From this we conclude that MATH belongs to MATH . Repeating the argument for MATH and so on, we obtain MATH . If MATH, thus we have MATH for all MATH, and thereby MATH. We return to MATH which satisfies MATH. Using REF and the injectivity of the third horizontal arrow, we conclude that MATH. This implies MATH . Therefore, we can replace MATH by MATH keeping the property MATH. |
math/0009198 | Since MATH, by using MATH and MATH, we obtain the first equation of the lemma. The proof of second one is similar. |
math/0009198 | First, we show MATH by induction on MATH. The case MATH is the assumption of the Lemma. Suppose REF is true for MATH. Then MATH and REF holds for MATH. Next, we have MATH . Therefore, the lemma follows by induction on MATH. |
math/0009198 | We use induction on MATH. The case MATH follows from REF . We have MATH . Using the induction hypothesis, we have MATH . The assertion follows from this. |
math/0009198 | First, we prove MATH by induction on MATH. In fact, MATH . Next, we proceed as follows: MATH . By repeating this argument, the assertion reduces to showing that MATH . This follows from REF with MATH. |
math/0009200 | CASE: Using the exact form REF we can describe MATH. Then we deduce the assertion. CASE: The transformation MATH in REF define a group action of the symplectic group on MATH (see CITE). We can check that the equality for every member of the generator system MATH of MATH. |
math/0009200 | We apply the transformation REF for MATH. For it we proceed the preparatory calculations. At first, get the explicit form of MATH by using REF . So we obtain MATH . Using the explicit form of MATH in REF , we get MATH for all MATH by a computer and calculation. By REF , we may put MATH for a certain MATH. Returning to the explicit form of MATH we should get MATH. We check that MATH by a computer aided calculation. Hence we have MATH for all MATH. This implies our assertion since MATH is an REF-th root of REF. |
math/0009200 | We have an explicit form of MATH in REF - REF . We use it and obtain MATH . According to REF , MATH . So the assertion follows. |
math/0009200 | The divisor of the holomorphic REF-form MATH is MATH. Hence MATH is a half integer characteristic (see REF ). For MATH and MATH, applying REF we have MATH since MATH. By REF , we have MATH . Hence it holds MATH by REF . Namely, putting MATH we have MATH . Let us recall that MATH is the symplectic representation matrix of MATH with respect to the basis MATH of MATH. And we have MATH so MATH is the representation matrix of MATH with respect to the basis MATH of MATH. Hence it holds MATH . Recalling REF , this implies that MATH is the NAME constant, that is MATH. Hence we have MATH since MATH is the unique MATH - invariant half integer characteristic. |
math/0009200 | Since MATH is a cycle, we see that MATH mod MATH. And we have MATH . By the same calculation, we see that MATH . Calculating intersection numbers, we have the following equality MATH as homology classes. Hence it holds MATH . By the same way, we obtain the results for MATH, MATH and MATH. |
math/0009200 | Let us consider a curve MATH and its period MATH. We assume that MATH. According to REF , there exist points MATH such that MATH . On the other hand, by REF , we have MATH . Hence it holds MATH . By NAME 's theorem, the divisor MATH is linearly equivalent to the divisor MATH, and we have MATH . For the effective divisor MATH, we have MATH by the NAME. We claim that MATH. In fact, the basis MATH is written as MATH and we have following vanishing orders; MATH . Because any holomorphic REF-form is written in the form MATH we see that there is no holomorphic REF-form MATH such that MATH . Hence we have MATH, that is, MATH contains only constant functions. This contradicts to REF since MATH and MATH is not effective. |
math/0009200 | This follows from REF . |
math/0009200 | Let MATH be a characteristic in MATH. Let MATH denote the theta constant MATH. Using this notation, we have MATH (see REF ). So our assertion is reduced to the inequality MATH, since MATH is a constant multiple of MATH. Set MATH and set MATH . By definition, MATH. For simplicity, we denote MATH by MATH. By elementary calculations, we see that MATH . In case MATH, we have MATH . In case MATH, we have MATH . Consequently, MATH for any MATH. Set MATH and consider the summations MATH . Using a computer, we can evaluate MATH and MATH. We have a approximate value MATH by NAME. On the other hand, we have MATH . The last term is very small. For example, MATH and the same calculations shows MATH. This implies MATH. |
math/0009200 | Set MATH and set MATH with MATH. By the computation same as the one in the proof of REF , we have MATH . Hence it holds MATH . Therefore MATH vanishes on the mirror of MATH provided MATH. This implies REF . REF follows by the same argument with MATH and MATH. |
math/0009200 | By REF , MATH vanish on MATH, and MATH vanish on MATH. By REF , MATH are not identically zero on MATH and on MATH, since MATH. The result is obtained by applying the transformation REF for above theta constants and MATH. For example, we have MATH . Since MATH (see REF ) and MATH, we see that MATH is not identically zero on MATH. |
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