paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0004117 | Suppose in the above construction of the singular REF-cocycle MATH and the singular REF-cochain MATH that we had been unable to define a set map MATH which was a section of MATH and which satisfied REF . We rewrite REF as MATH where MATH denotes the projection from MATH to MATH (compare with REF ) and where MATH is def... |
math/0004117 | To show that this does indeed define a gerbe we first of all need to show that the assignment MATH defines a stack in groupoids on MATH - that is the gluing laws for objects and arrows are satisfied - and that MATH is locally non-empty and locally connected. Note that by REF is a groupoid. Also MATH defines a presheaf ... |
math/0004117 | Recall that the stable morphism MATH corresponds to the trivialisation MATH of MATH which is obtained by descending the MATH bundle MATH on MATH to MATH. In particular, this implies that the following isomorphism is true on MATH: MATH . The two trivialisations MATH and MATH of MATH differ by the pullback to MATH of a M... |
math/0004117 | Choose an open cover MATH of MATH such that there exist local sections MATH of MATH. Form the pullback bundle gerbes MATH. The stable morphism MATH induces a stable morphism MATH by pullback: MATH, and the transformation MATH induces a transformation MATH by pullback as well: MATH. The transformation MATH satisfies the... |
math/0004117 | We first need to define the bundle gerbe morphism MATH which maps MATH . Define MATH covering the identity on MATH by sending MATH to the path MATH given by MATH . Next, we need to define a MATH equivariant map MATH covering MATH and check that it commutes with the bundle gerbe product. So take pairs MATH and MATH wher... |
math/0004117 | Because MATH commutes with the bundle gerbe products on MATH and MATH, MATH commutes with the bundle gerbe products on MATH and MATH. Since MATH is a bundle gerbe connection, so is MATH and therefore MATH and MATH are bundle gerbe connections on MATH. Since any two connections differ by the pullback of a complex valued... |
math/0004117 | Define a map MATH by sending the point MATH of MATH to the point MATH of MATH. The bundle gerbe product on MATH gives the following isomorphisms of line bundles with connection over MATH. MATH . Here the line bundle MATH is equipped with the connection MATH, while the line bundles MATH and MATH are endowed with the con... |
math/0004117 | The isomorphism MATH gives rise to an isomorphism over MATH: MATH . Also by definition we have an isomorphism of line bundles with connection MATH where each bundle is equipped with the obvious pullback connection. Next, the diagram MATH commutes. This is really the definition of MATH. Therefore we will have MATH . |
math/0004117 | Same as for REF . |
math/0004117 | Suppose we are given a bundle REF-gerbe MATH. Let MATH be a bundle gerbe connection on MATH with curving MATH such that MATH, where MATH is the curvature of MATH with respect to MATH. The three curvature MATH then satisfies MATH. Let us look at what happens over MATH. We have induced connections MATH on the bundle gerb... |
math/0004117 | MATH induces an isomorphism MATH as pictured in the following diagram: MATH . This diagram lives over MATH. Clearly over MATH we get the following diagram, induced from the one above by tensor product and pullback, MATH . Let MATH and MATH denote the sections of MATH and MATH induced by the bundle gerbe products respec... |
math/0004117 | All that one really needs to do is to check that MATH does indeed define a cocycle. This follows from the coherency condition on MATH. One can also check that another choice of the sections MATH, or the line bundles MATH, or the sections MATH changes MATH by a coboundary. |
math/0004117 | We have MATH. But MATH and so MATH. Hence we get MATH. Also we have MATH . Therefore we can construct a section MATH of MATH by taking the section MATH of MATH and choosing sections MATH of MATH and forming the tensor product section MATH. From here it is not too hard to show that the cocycle MATH is cohomologous to MA... |
math/0004117 | We calculate the NAME four class of the bundle REF-gerbe MATH and get it exactly equal to the NAME cocycle defined by NAME and NAME in CITE and CITE. We can then apply REF to conclude that this NAME four class is MATH. We set out to calculate the NAME cocycle MATH. First choose an open cover MATH of MATH relative to wh... |
math/0004117 | We need to show that MATH is associative, that is MATH. We have MATH where we have denoted composition by bundle gerbe product by juxtaposition and composition via the bundle REF-gerbe product with MATH. We also have MATH . We have to show these two expressions are the same. For this we will need the following lemma. U... |
math/0004117 | Let MATH be a `nice' open cover of MATH. So each non-empty finite intersection MATH is contractible and there exist local sections MATH of MATH. Define maps MATH by MATH. Pullback the bundle gerbe MATH with this map to obtain a new bundle gerbe MATH. So we get a manifold MATH and an induced projection map MATH. If MATH... |
math/0004117 | We first prove that the sheaf of bicategories MATH is locally non-empty. Choose an open cover MATH of MATH such that there exist local sections MATH of MATH. Form the maps MATH sending MATH to MATH. Let MATH. So MATH is a bundle gerbe on MATH. Now pullback the stable morphism MATH using the map MATH, MATH where MATH. L... |
math/0004117 | We have already seen REF that a trivial stable bundle REF-gerbe has zero class in MATH. Conversely, given a stable bundle REF-gerbe with zero class in MATH, the associated REF-gerbe also has zero class in MATH and hence has a global object. This global object trivialises the stable bundle REF-gerbe MATH. |
math/0004119 | REF is obvious (use induction). To prove REF , suppose that MATH are finite metric spaces, MATH, and let MATH be an isometric embedding. Pick an isometric embedding MATH, and use MATH-homogeneity of MATH to find an isometry MATH of MATH such that MATH extends the isometry MATH. Then MATH is an isometric embedding that ... |
math/0004119 | There exists a family MATH of continuous left-invariant pseudometrics on MATH which determines the topology of MATH and has the cardinality MATH. Replacing, if necessary, each MATH by MATH, we may assume that all pseudometrics in MATH are bounded by REF. For every MATH let MATH be the metric space associated with the p... |
math/0004119 | Let MATH be the NAME space of all bounded real functions on MATH which are uniformly continuous with respect to the right uniformity. The natural left action of MATH on MATH, defined by the formula MATH, yields an isomorphic embedding of MATH into MATH. The weight of MATH may exceed the weight of MATH, but it is easy t... |
math/0004119 | It suffices to show that for every MATH the map MATH from MATH to MATH is continuous at the unity. If MATH, then MATH, and we have MATH. Let MATH be given. If MATH is close enough to the unity, then MATH, MATH, and therefore MATH. |
math/0004119 | Since MATH is a NAME function, for every MATH we have MATH and MATH. Hence MATH, and at MATH the equality is attained. |
math/0004119 | We may assume that MATH is a subspace of MATH and that MATH for every MATH. Let MATH be the NAME function defined by MATH for every MATH, where MATH is the metric on MATH. Let MATH. We claim that the extension of MATH over MATH which maps MATH to MATH is an isometric embedding. It suffices to check that MATH for every ... |
math/0004119 | It is clear that every isometry MATH between any two metric spaces can be extended to an isometry MATH. Such an extension is unique, since every point in MATH (or, more generally, in MATH) is uniquely determined by its distances from the points of MATH REF , and similarly for MATH. In particular, every isometry MATH un... |
math/0004119 | For every MATH pick a continuous homomorphism MATH such that MATH extends MATH for every MATH. By transfinite recursion on MATH define a homomorphism MATH such that MATH extends MATH for every MATH and MATH. Let MATH be the identity map of MATH. If MATH, put MATH. If MATH is a limit ordinal, let MATH be the isometry of... |
math/0004119 | Let MATH be a finite metric space of diameter MATH, and let MATH be an isometric embedding. Pick MATH so that MATH. In virtue of REF , there exists an isometric embedding of MATH into MATH which extends MATH. This means that MATH is NAME. The second assertion of the proposition follows from REF . |
math/0004119 | It suffices to show that for every metric space MATH the weight of MATH is equal to the weight of MATH. Let MATH, and let MATH be a dense subset of MATH of cardinality MATH. Let MATH. Then MATH is a family of separable subspaces of MATH, MATH and MATH is dense in MATH (see the proof of REF). Hence MATH has a dense subs... |
math/0004119 | Let MATH be a metric space, MATH be its completion. Every isometry MATH uniquely extends to an isometry MATH. We show that the homomorphism MATH is continuous. Let MATH be the metric on MATH. Fix MATH and MATH. Pick MATH so that MATH. Let MATH. Then MATH is a neighbourhood of unity in MATH. If MATH, then MATH. This imp... |
math/0004119 | Let MATH be a complete metric space containing a dense NAME subspace MATH. We must prove that MATH is NAME. Let MATH be a finite subset of MATH, and let MATH be a NAME function. It suffices to prove that there exists MATH such that MATH for every MATH. Pick a sequence MATH such that: CASE: if MATH and MATH, MATH, then ... |
math/0004119 | Let MATH be a metric space of diameter MATH, and let MATH be the NAME extension of MATH constructed above. Consider the completion MATH of MATH. REF implies that MATH is NAME. REF shows that MATH. Finally, MATH is MATH-embedded in MATH REF and MATH is MATH-embedded in MATH REF , so MATH is MATH-embedded in MATH. Thus M... |
math/0004119 | Let MATH be a metric space. We first construct a MATH-embedding of MATH into a metric space MATH such that MATH and every isometry between finite subsets of MATH extends to an isometry of MATH. For every isometry MATH between finite non-empty subsets of MATH consider the graph MATH of MATH, and let MATH be the set of a... |
math/0004119 | Consider two cases. CASE: MATH is separable. According to REF , there exists a complete separable NAME space MATH such that MATH is a MATH-embedded subspace of MATH. According to REF , MATH is MATH-homogeneous. CASE: MATH is not separable. Let MATH. Applying in turn REF , construct an increasing continuous chain MATH o... |
math/0004119 | Let MATH be a topological group. According to REF , there exists a metric space MATH such that MATH and MATH is isomorphic to a subgroup of MATH. We may assume that MATH has diameter MATH: otherwise replace the metric MATH by MATH. REF implies that there exists a complete MATH-homogeneous NAME metric space MATH such th... |
math/0004119 | The condition MATH (respectively, MATH) holds if and only if the function MATH (respectively, MATH) is non-expanding for every MATH. The condition MATH (respectively, MATH) holds if and only if MATH (respectively, MATH) for all MATH. |
math/0004119 | We claim that every non-empty closed subset of MATH has a maximal element. Indeed, if MATH is a non-empty linearly ordered subset of MATH, then MATH has a least upper bound MATH in MATH, and MATH belongs to the closure of MATH. Thus our claim follows from NAME 's lemma. The set MATH is a closed subsemigroup of MATH. Le... |
math/0004119 | Let MATH be a closed subset of MATH. It is clear that MATH. If MATH, then MATH for every MATH. Thus MATH is an idempotent. The same is obviously true if MATH. Conversely, let MATH be an idempotent in MATH such that MATH. Let MATH. The function MATH, being non-expanding in each argument, is continuous, hence MATH is clo... |
math/0004119 | We have MATH. Taking MATH, we see that the right side is MATH. On the other hand, for every MATH we have MATH, whence the opposite inequality. The continuity of the left action easily follows from the explicit formula that we have just proved. The argument for the right action is similar. |
math/0004119 | Let MATH be invertible. Let MATH be a MATH-triple corresponding to MATH. This means that MATH is a metric space, MATH and MATH are distance-preserving maps from MATH to MATH, MATH and MATH for all MATH. We saw that elements of MATH correspond to triples MATH satisfying the condition MATH. Thus we must verify this condi... |
math/0004119 | The inequality MATH is obvious: if MATH is a pseudometric on MATH extending MATH and MATH and MATH, then at least one of the numbers MATH and MATH must be MATH. To prove the reverse inequality, we construct a pseudometric MATH on MATH extending MATH and MATH such that MATH . The function MATH is defined by these requir... |
math/0004119 | If MATH is a finite subset of MATH and MATH, let MATH. Let MATH be the set of all pairs MATH such that MATH for all MATH. The sets of the form MATH constitute a base of entourages of the uniformity on MATH. If MATH, we say that MATH and MATH are MATH-close. Our proof proceeds in three parts. CASE: We prove that MATH is... |
math/0004119 | Let MATH, MATH, MATH be neighbourhoods of MATH, MATH and MATH, respectively. We must show that MATH meets the set MATH. We may assume that for some finite set MATH and MATH we have MATH . We saw in the last paragraph of the preceding section that there exist a metric space MATH and isometric embeddings MATH REF such th... |
math/0004119 | Let MATH and MATH be the left and the right uniformity on MATH, respectively. In each of REF - REF the map MATH under consideration is an automorphism of the uniform space MATH. This is obvious for REF . For REF , observe that the uniformities MATH and MATH are invariant under left and right shifts, hence the same is t... |
math/0004119 | According to REF , every idempotent MATH is of the form MATH for some closed MATH. If MATH is invariant under inner automophisms, then MATH and hence MATH for every MATH. Since the action of MATH on MATH is transitive, no proper non-empty subset of MATH is MATH-invariant. Thus either MATH or MATH. Accordingly, either M... |
math/0004119 | Let MATH be a normal subgroup of MATH. We show that MATH is not compact. Fix MATH and MATH such that MATH. Let MATH, and let MATH be the sphere of radius MATH centered at MATH. We claim that the orbit MATH contains MATH. Fix MATH. Since MATH is MATH-homogeneous, there exists an isometry MATH which leaves the point MATH... |
math/0004119 | It suffices to show that for every MATH the map MATH from MATH to MATH is continuous. Fix MATH, MATH and MATH. Let MATH, MATH and MATH. If MATH and MATH, then MATH. |
math/0004119 | Let MATH be the set of all self-maps of MATH, equipped with the product uniformity. The group MATH can be considered as a subset of MATH. The uniformity MATH on MATH induced by the product uniformity on MATH coincides with the left uniformity MATH. Indeed, a basic entourage for MATH has the form MATH, where MATH is the... |
math/0004119 | For every metric space MATH we have MATH. If MATH is homogeneous, then for every MATH the map MATH from MATH to MATH is onto, whence MATH. |
math/0004119 | We saw that MATH is NAME REF . REF shows that MATH is complete, and REF shows that MATH. Let MATH be a continuous onto homomorphism. According to REF , to prove that MATH is minimal and topologically simple, it suffices to prove that either MATH is a homeomorphism or MATH. Since MATH is NAME, so is MATH. Let MATH be th... |
math/0004120 | In the context of NAME operators, REF is proved in CITE; in the corresponding NAME situation REF is proved in CITE. REF then follows by combining REF , taking into account the asymptotic expansions (along any ray with MATH) MATH in the case of NAME operators (compare CITE) and MATH in the case of NAME operators (compar... |
math/0004120 | In order to prove REF, let MATH be the fundamental system of solutions of REF as defined in REF and observe that MATH in the NAME operator case. Thus, any nonnormalized solutions MATH, MATH of MATH for MATH are of the type MATH for some nonsingular MATH matrices MATH. Thus, MATH by REF. In particular, MATH is independe... |
math/0004120 | In order to prove REF one only needs to observe MATH for a.e. MATH, using REF. By continuity of the right-hand side of REF with respect to MATH, REF extends to all MATH. REF then proves REF, and using REF again, REF proves REF. Similarly, to prove REF, one computes MATH for a.e. MATH, using REF. |
math/0004120 | Suppose there is a MATH such that MATH. Then there is a sequence MATH, MATH, MATH, such that MATH . Any NAME function MATH with values in MATH admits the representation MATH where MATH, MATH, MATH is a NAME space, MATH, MATH is a self-adjoint operator in MATH, and MATH with MATH. By REF, MATH where MATH is invertible i... |
math/0004120 | Introducing the map MATH the estimate MATH combined with REF shows that MATH maps MATH into itself. Moreover, using MATH REF implies that MATH is a strict contraction. An application of NAME 's fixed point theorem then proves the existence of a unique fixed point MATH. Moreover, the fixed point MATH can be obtained as ... |
math/0004120 | For MATH the estimate MATH shows that MATH . Analogously, one proves that MATH . Combining REF yields REF. |
math/0004120 | Since MATH uniquely determines MATH (that is, MATH or MATH) by REF , it suffices to show that MATH and MATH uniquely determine MATH in REF and similarly, MATH, MATH, and MATH, MATH uniquely determine MATH in REF . We start with the NAME REF . By REF, and REF, MATH and hence MATH by REF. Applying REF (with MATH and MATH... |
math/0004120 | In the NAME context one observes from REF , and REF (for MATH) that MATH . Here MATH is a closed and counterclockwise oriented NAME contour which encircles MATH with winding number MATH and MATH with winding number MATH, and whose existence is guaranteed by REF (compare REF). Using the asymptotic expansion (compare CIT... |
math/0004120 | Combining REF, MATH yields REF after straightforward matrix computations. |
math/0004120 | Using REF, and the asymptotic representations MATH one infers MATH . Here REF is explicitly needed in the case of the MATH-sign in REF. By REF implies MATH and hence MATH . Introducing the orthogonal polynomials MATH associated with the normalized spectral measures MATH, MATH, REF implies that MATH and MATH which prove... |
math/0004120 | REF . A direct computation utilizing REF shows that MATH . In addition, the matrix MATH can be read off from the asymptotic expansion of MATH as MATH by a simple NAME series-type argument. Indeed, by REF MATH and hence by REF MATH along any ray with MATH, MATH. Therefore, combining REF, and REF one infers that MATH alo... |
math/0004120 | CASE: Since MATH in REF implies that MATH and we denote the latter by MATH. By REF the MATH . NAME matrices MATH associated with MATH, MATH can be represented as MATH and they admit the estimate MATH . Using the asymptotics MATH the estimate REF combined with REF proves MATH . Since MATH REF yield MATH . REF yield MATH... |
math/0004127 | Without loss of generality we can assume that MATH. Let MATH be the NAME isomorphism witnessing that MATH. Find NAME sets MATH and MATH and the NAME isomorphism MATH extending MATH (exercise in CITE). By removing a countable set from MATH we can arrange that MATH is MATH-dimensional. Let MATH be a meager set such that ... |
math/0004127 | Suppose that MATH. By REF there exists MATH, a NAME set MATH and NAME isomorphism MATH such that CASE: MATH is continuous, CASE: MATH is not meager in MATH. Apply Axiom MATH to to find a compact set MATH such that MATH is not meager in MATH. Set MATH and note that MATH is a homeomorphism between MATH and MATH. Under th... |
math/0004127 | Let MATH be a continuous increasing sequence such that MATH . By properness, MATH, has the required property. |
math/0004127 | For the sake of clarity we will break the proof into three lemmas. The main idea of the proof is already present in CITE. Let MATH be the canonical name for a MATH-generic real and let MATH be the NAME forcing represented as MATH with MATH being the canonical name for the NAME real. For MATH let MATH. Note that MATH is... |
math/0004130 | For the sake of simplicity we will write MATH, instead of MATH, to denote a canonical divisor of MATH. By contradiction, assume that MATH does not impose independent conditions to MATH. Let MATH be a minimal REF-dimensional subscheme of MATH for which this property holds and let MATH. This means that MATH and that MATH... |
math/0004132 | We prove the result by induction on MATH. The cases MATH are nothing but REF respectively. Suppose the statement is true for MATH. Then replacing MATH by MATH and noting MATH, we see that MATH . Then by REF , we conclude that MATH . |
math/0004132 | Recall REF and the fact that MATH for positive integers MATH (compare REF ). We have MATH . Now the result follows by induction, noting that MATH . |
math/0004132 | Since the proofs of REF are almost identical we only prove REF . The special case MATH follows from REF . So we can proceed by induction on MATH and assume that MATH . Then by REF we have MATH as required. |
math/0004132 | If MATH this is REF , so we may assume that MATH . First we deal with the case MATH . Following REF we have MATH . Continuing in this fashion, the result follows. |
math/0004132 | Recall from CITE the vertex operator MATH . It is easy to see that MATH for MATH for all MATH . So it is immediate that MATH . |
math/0004132 | We prove this by induction on MATH . If MATH the assertion is clear. If MATH the assertion follows from the fact that MATH on MATH . Now we assume that MATH . Using REF gives MATH . By induction, both MATH and MATH converge to holomorphic functions in MATH whence so does MATH . Indeed, MATH . It is easy to express MATH... |
math/0004132 | First we take MATH for MATH with MATH . Then from REF we have MATH for some scalars MATH and quasi-modular forms MATH . Now the result follows from REF . For arbitrary MATH we can assume that MATH is a monomial in MATH for MATH and MATH . Use the result for MATH and REF again to see that MATH is a linear combination of... |
math/0004132 | After REF it suffices to show that if MATH in MATH satisfies MATH then the function MATH lies in MATH . This is essentially established in the paper CITE. In the notation of CITE, the present function MATH is denoted MATH where MATH is the integral quadratic form which corresponds to MATH . If MATH is odd then the func... |
math/0004132 | It is enough to prove sufficiency. Let MATH of weight MATH be expressed in the form MATH with each MATH a form in MATH of weight MATH . As MATH is also a generalized modular form of weight MATH each of its MATH-transforms REF has a MATH-expansion. In particular, the MATH-transform of MATH yields an equality of the shap... |
math/0004132 | Recall that MATH is an orthonormal basis of MATH . Note that MATH and both MATH for MATH and MATH annihilate MATH. Since all summands of MATH contain one of these as a factor, we immediately see that MATH. |
math/0004132 | It suffices to consider the condition MATH. Since MATH and MATH annihilates MATH if MATH we see that MATH is equivalent to MATH. Now since MATH, MATH acts as MATH on MATH. |
math/0004137 | It is enough to show that MATH is the sum of the signed monomials MATH for tableaux MATH with no integers larger than MATH. Now number the diagonals of MATH from south-west to north-east as described in the previous section and let MATH be the free Abelian group with one basis element for each partition containing MATH... |
math/0004137 | Since MATH is super-symmetric, it is enough to prove that MATH for any number of variables MATH. Now MATH has a MATH-linear automorphism which sends MATH to MATH for each MATH. Since this automorphism takes MATH to MATH and MATH to MATH, the lemma follows from REF. |
math/0004137 | Suppose MATH was bumped out of box number MATH in MATH, counted from the top. Then since MATH is a tableau, MATH is less than or equal to the set in box number MATH in MATH. This implies that the MATH'th box of MATH is equal to the MATH'th box of MATH for all MATH. For MATH, the MATH'th box of MATH can contain elements... |
math/0004137 | Notice at first that MATH, which follows directly from REF . Since all of the integers in MATH come from MATH, it suffices to show that all integers from MATH which are MATH are also strictly greater than MATH. Let MATH be the number of the box in MATH (counted from the top) which contains MATH. There are then two poss... |
math/0004137 | Suppose the southernmost box of MATH occurs in column MATH. Then let MATH be the tableau consisting of the leftmost MATH columns of MATH and let MATH be the rest of MATH. We will write MATH to indicate this. Similarly we let MATH and MATH, that is, MATH and MATH are the leftmost MATH columns of these products. Finally ... |
math/0004137 | Suppose at first MATH is a single integer, and let MATH. Then by the definition of MATH, the minimal element of MATH will bump out MATH from MATH, after which the remaining elements of MATH will be added to the same box as the minimal element went into. This recovers the tableau MATH. If MATH has more than one element,... |
math/0004137 | Notice at first that if MATH has shape MATH then MATH must be defined by one of REF or REF . Let MATH be the largest subset of the form MATH such that none of REF - REF are used in the definition of MATH, and let MATH. The lemma is easy to prove if MATH is empty, so we will assume MATH. Let MATH and MATH. Then MATH whe... |
math/0004137 | It follows from REF that the maps MATH define an inverse to the map of REF when MATH. If MATH and MATH is any element, there are unique rook strips MATH which split the vertical strip MATH up into disjoint intervals from north to south, such that MATH contains the columns of the extra boxes in MATH. Then set MATH and M... |
math/0004137 | Let MATH be any positive integer. It is enough to show that for any partition MATH there exists a polynomial MATH such that MATH is a linear combination of NAME polynomials MATH for partitions MATH of length MATH. Define a partial order on partitions by writing MATH if MATH, or MATH and MATH. Notice that given a partit... |
math/0004137 | If MATH is a reverse lattice word, then the rectification of MATH in the plactic monoid is the semistandard NAME tableau MATH of shape MATH in which all boxes in row MATH contain the integer MATH CITE. This implies that the identity MATH holds even with the weaker relations of the plactic algebra. On the other hand, if... |
math/0004137 | Start by noticing that the only tableau of shape MATH whose word is a reverse lattice word is the tableau MATH. It follows from this that the coefficient of MATH in the right hand side of REF is MATH. On the other hand, the left hand side is equal to MATH. If MATH is a tableau on MATH, then MATH is equal to MATH exactl... |
math/0004137 | By replacing each non-commutative variable MATH with MATH in REF , we obtain the identity MATH for single stable NAME polynomials. Since MATH is super symmetric, the same equation must hold for the double polynomials as well. |
math/0004137 | By REF the left hand side is the signed sum of monomials MATH for all tableaux MATH of shape MATH such that all contained integers are between MATH and MATH. If MATH is such a tableaux, let MATH be the subtableau obtained by removing all integers strictly larger than MATH (as well as all boxes that become empty), and l... |
math/0004137 | Number the rows of MATH such that the top box in the leftmost column of MATH is in row number one, and so that the numbers increase from top to bottom. Then suppose MATH and MATH are tableaux of shapes MATH and MATH for which all contained integers are less than or equal to MATH. Then all integers in row MATH of MATH w... |
math/0004137 | In this proof we will write MATH for the variables MATH and MATH for MATH. Then by REF we have MATH where the sum is over all partitions MATH such that MATH is a rook strip. Notice that when MATH has more than MATH columns, then MATH by REF . Therefore we only need to include terms for which MATH in the sum. For such t... |
math/0004137 | Let MATH be the first MATH columns of MATH and let MATH be rest like in REF. Then since MATH we get MATH. The statement therefore follows from the proposition. |
math/0004137 | Let MATH denote MATH and let MATH denote MATH. Then REF implies that MATH. Now using REF we obtain MATH. |
math/0004137 | It is enough to show this for finitely many variables MATH and MATH, as long as MATH and MATH can be arbitrarily large. Let MATH be a rectangle with MATH rows and MATH columns. If MATH is a partition such that MATH occurs with non-zero coefficient in MATH then first of all MATH. Furthermore, REF shows that MATH is non-... |
math/0004137 | These statements are clear from REF , and REF . |
math/0004137 | We will start by comparing the coefficients MATH and MATH. Suppose MATH is a tableau of shape MATH such that MATH is a partial reverse lattice word with respect to the intervals MATH and MATH and with content MATH. Then all integers in MATH which come from the interval MATH must be located in the upper-left corner in M... |
math/0004137 | It follows from REF that the single polynomial MATH is a possibly infinite linear combination of polynomials MATH for partitions MATH with at most MATH columns: MATH . Using that MATH is super symmetric, this implies that REF is true with the same coefficients. Let MATH be the longest permutation of MATH. Using REF we ... |
math/0004137 | It is enough to show that if MATH is not a rectangle and MATH then the absolute value of some coefficient MATH is at least two. The assumptions on MATH and MATH imply that we can find a partition MATH containing MATH such that MATH is a vertical strip with MATH boxes and at least three columns. Then it follows from NAM... |
math/0004137 | Start by locating the leftmost shared box MATH of MATH which contains MATH. To construct MATH we start by removing MATH from this box. Then look for the nearest box MATH below or to the left of MATH which contains MATH, such that the box above MATH does not contain MATH and the box to the left of MATH does not contain ... |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.