paper stringlengths 9 16 | proof stringlengths 0 131k |
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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 . Moreo... |
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/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 partition... |
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 prove... |
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 ... |
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 ... |
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 fol... |
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... |
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 coeffici... |
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 ... |
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 c... |
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 b... |
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 ge... |
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... |
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... |
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... |
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 p... |
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.... |
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/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... |
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 i... |
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 ... |
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/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 us... |
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 th... |
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 ... |
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 mo... |
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 ... |
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 equivale... |
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 o... |
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 u... |
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... |
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 M... |
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 ... |
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 eas... |
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 MA... |
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... |
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 ... |
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 ... |
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 diviso... |
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 c... |
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 zer... |
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