paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9904104 | Repeating the construction of REF for the tree , we have MATH as defined above. We also see that the MATH- invariant maps are the collection, MATH . |
math/9904104 | REF applied to MATH gives MATH as desired. |
math/9904104 | It is clear that the right hand side is contained in the intersection, so we need only to prove that an arbitrary element of the intersection is contained in the power series on the right. But the only difference between the power series MATH, and MATH, is that in the former, polynomials in the variable MATH can exist as coefficients of the power series in the variable MATH. But this can not occur in MATH, so the intersection is as given. |
math/9904104 | Each of these is an element of MATH which agree when MATH (and when MATH). Because each satisfies MATH they are completely determined by their values at MATH (or MATH), and so are equal. |
math/9904104 | Let MATH and MATH. We shall show that MATH and MATH compose to give an element of MATH. To do so, we select a pullback object MATH as in REF . The total ordering of the internal vertices of the tree MATH provides a total ordering of both MATH and MATH as MATH and MATH. Such a pullback object also consists of a choice of permutation of labels for each MATH, and so we have uniquely determined objects, MATH . Because MATH and MATH are contained in the pullbacks, we can regard them as elements of MATH and MATH respectively. We know that the internal vertices of the composed tree, MATH consist exactly of the internal vertices of MATH, the internal vertices of MATH, and the root of MATH. We shall therefore prove that MATH and MATH compose to give an element of MATH by induction on the number of internal vertices of MATH, MATH. If the composed tree MATH has zero internal vertices, then we are composing with a tree of the form MATH or MATH, and it is clear from the discussions of REF that maps compose as desired. Now, without loss of generality we may assume that in the total ordering MATH, MATH. We need only to prove that there exists a natural map, MATH because on the right hand side, the operator MATH is acting on a collection of singular maps associated to a pair of trees with MATH internal vertices, which we know compose by induction. We know that MATH can be evaluated at MATH (see REF), so the left hand side of REF can be evaluated at MATH, giving the inner collection of singular maps on the right hand side. The transpose of this evaluation map provides us with the desired map. Now that we have seen that the maps MATH and MATH compose naturally into each pullback object, MATH, our proof will be complete if we show that given two pullback objects, MATH and MATH, the composite of MATH and MATH maps through each of them to the same element of MATH the collection over which we are pulling back. Denote this collection MATH. Since both MATH and MATH are elements of a pullback over objects MATH respectively, then if there exists an injective maps from MATH to the composite of MATH and MATH, then we know that the composite of MATH and MATH maps to the same element of MATH, and hence is an element of the pullback. Clearly MATH and MATH compose to give an element of the collection, MATH and because we are working with a vertex group over a field, there is a natural injection from MATH to MATH, so our proof is complete. |
math/9904104 | The idea of the proof is that we get a map between pullbacks by showing that there is a map between the spaces over which we are pulling back. There is also a map between the pullback objects, and these maps form a commuting diagram, inducing a map between pullbacks. For more of the abstract details see CITE. Explicitly, a tree MATH can arise as a refinement of a tree MATH by a sequence of either replacing nodes of MATH with edges, or shrinking internal edges of MATH down to nodes (see REF). These two moves can be interchanged, and can be carried out one move at a time. Thus we shall consider them separately. And because we have defined MATH and MATH as an iteration of operators associated to each internal node, we need only consider refinements involving flat trees. CASE: We consider the case where MATH arises as a refinement of MATH by replacing a node of MATH with an edge. Taking MATH to be the flat MATH leafed tree, we may either replace the root or a leaf to give MATH. If we replace the root, then we are defining a canonical map MATH . Labelling the interior node of the tree on the left hand side, MATH, we see that an arbritrary pullback object in the definition of MATH, is MATH which is manifestly isomorphic to the pullback object, MATH, of the right hand side. The same holds for the space over which we pullback, and so we see that in fact this map is an isomorphism. If we instead replace a leaf, we are defining a canonical map MATH A similar proof shows that this is also an isomorphism. CASE: We next consider the case where MATH arises by shrinking an internal edge of MATH down to a node. In this case we are defining a canonical map MATH where, without loss of generality, we have chosen to consider refinement of the first MATH leaves (MATH). Labelling the internal node of the tree on the right hand side, MATH, we see that an arbritrary pullback object is of the form MATH where MATH and MATH. We need to show that there exists a natural map from MATH to this object, so considering an arbritrary element of MATH, we consider it as an element of the pullback object MATH, where we define a permutation, MATH: MATH where MATH is the placeholder for the internal edge, and where we exclude MATH when MATH is mapped to MATH. Notice that we have shifted the domain of MATH so that it acts on MATH, and that the permutation, MATH, will arise for MATH different choices of MATH. Consider the canonical map we are trying to exhibit: MATH . It will be an identity on the outer terms corresponding to MATH. So our problem reduces to showing that there exists a map MATH where MATH. We immediately notice that there are fewer copies of the singularity MATH on the right hand side. This is because when we expand the singularities, we will have maps of the form: MATH where the second arrow is achieved by multiplication in MATH (for example, REF ). A complicated but routine calculation shows that a succession of maps of this form give the desired map. |
math/9904104 | In order to show that MATH gives a vertex algebra, we check the axioms for the vertex algebra. The vacuum axioms follow naturally from the discussion of REF, together with previous example, REF, giving MATH . Translation covariance is automatically satisfied because our functions, MATH, are MATH-invariant. And, we saw that the locality condition is satisfied in REF. |
math/9904104 | As in the proof for holomorphic vertex algebras REF , the vacuum defines a map MATH for the empty tree. We described in REF how to construct MATH, and REF showed how how to construct the singular function associated to the flat tree with three leaves. Carrying out a similar process leads to the construction of singular functions associated to all flat trees, and the singular functions associated to trees with internal nodes arise from composition of singular functions associated to flat trees. Closure under refinement follows by construction. |
math/9904104 | Since MATH can be evaluated at zero, we know that it has no singularities. By the first vacuum axiom we know that MATH, so the translation covariance axiom says that for any MATH, MATH. Letting MATH, and evaluating this at zero we have MATH and the lemma is proved. |
math/9904104 | Recall that the translation covariance axiom says that MATH. So we have MATH . From the locality axiom we know that for some MATH the following holds MATH . Combining this with the previous lemma, it just says that MATH . Setting MATH equal to zero we have MATH, and since this is an equality of NAME series, we may divide by MATH to give our result. |
math/9904105 | REF is equivalent to MATH where MATH is the refinement of MATH obtained by replacing all components MATH, for MATH, by MATH. Thus the LHS of REF is equal to MATH . Let MATH be a term of this sum, with MATH. This term can only appear in one summand on the Right-hand side of REF , namely MATH with MATH. To show that it does indeed appear, we need to prove that MATH and MATH. Let MATH be the refinement of MATH obtained by replacing the part MATH by MATH if MATH. We have that MATH which implies that MATH, and MATH. If MATH then MATH. However, if MATH then there are two possible cases: either MATH, or MATH and MATH. In the former case MATH, while in the latter, MATH, whence MATH. Conversely, let MATH be a term belonging to a tensor MATH on the Right-hand side of REF . To show that it appears in REF we must prove that MATH. We have that MATH and MATH, which imply that MATH . If MATH then MATH . If MATH then MATH implies that MATH . Therefore, MATH as desired. |
math/9904105 | If all components of a composition MATH, except perhaps the last, are greater than MATH, then the same is true for all compositions MATH and MATH appearing in the Right-hand side of REF . |
math/9904105 | Each difference MATH is in the kernel of MATH as MATH since MATH. In addition, the non-zero differences are linearly independent as they have different leading terms. Letting MATH denote the MATH-th NAME number, there are MATH such differences, and since MATH our result follows. |
math/9904105 | If the first component of a composition MATH is greater than MATH, then the same is true for all compositions MATH and MATH appearing in the Right-hand side of REF . |
math/9904105 | For each MATH, MATH is forbidden in any monomial MATH appearing as a summand of the function MATH. This is equivalent to saying that MATH is a summand of MATH if and only if MATH for all MATH. Therefore at least one of MATH or MATH must be the largest label of a vertex in a connected component in MATH. Now when going from compositions of MATH to subsets of MATH we can do so using our graphs, MATH. All we have to do is list the label of the vertex that is the largest in each connect component, not listing MATH. We call these vertices the end-points. We are now in a position to prove the equivalence of REF for sqs-functions. The powers of REF agree so we need only show that the indices of summation do too. To see this, take any sqs-function MATH and let MATH. Then MATH is a summand in MATH if at least one of MATH or MATH is an end-point in MATH. Therefore MATH or MATH belongs to MATH, and MATH is a summand of MATH. Conversely, if MATH is a summand of MATH, then this implies that for each MATH, we have that MATH or MATH belongs to MATH, so one of MATH or MATH is an end-point in MATH, so MATH is a summand of MATH. |
math/9904105 | By direct calculation we obtain that MATH, MATH, MATH, and MATH. To obtain our recurrence, we observe that for each sqs-function, MATH where MATH, we can encode MATH as a binary word of length MATH, by placing a MATH in position MATH if MATH is contained in MATH, and MATH otherwise. By this one-to-one correspondence we see that MATH contains no internal peak if its corresponding binary word does not contain MATH as a subword. We therefore count binary words of length MATH that avoid the subword MATH. Appending either MATH or MATH to such a binary word of length MATH gives one of length MATH, provided that we have not created the subword MATH in the last three positions. Let MATH, MATH, MATH, and MATH enumerate those binary words of length MATH that avoid the subword MATH and end in, respectively MATH, MATH, MATH, and MATH. We then obtain the following REF simultaneous recursions. MATH . Clearly the number of MATH-s in MATH with no internal peaks is given by MATH . However by substituting in our recurrences we obtain MATH . |
math/9904105 | By REF we have that the leading terms of MATH determine the other summands that belong to MATH. Hence by REF it follows that the summands of MATH will be the union of the summands of MATH and MATH. However, those summands that appear in both will be duplicated. By definition these will be the summands of MATH, and the result follows. |
math/9904106 | We only explain the first part. REF are obvious and REF follows by REF regarding the structure of cocommutative graded connected NAME algebras. First we arrange that the components of MATH project to immersions in MATH with transverse self-intersections and so that the framing has its first componenet vector field pointing in the MATH direction. We call such a link generic . Given a two-component sublink MATH of the framed oriented link MATH in MATH, let MATH denote its projection on MATH. Then MATH and MATH intersect transversely at double points. Define MATH to be the sum with signs over all points in MATH that MATH overcrosses MATH, according to the convention MATH . If MATH, then we define MATH by counting the self-intersections of MATH in the above manner. Since MATH is a basis of MATH, this defines, by linearity, the desired map MATH. Since MATH is the sum with signs over all points MATH, it follows that MATH for MATH. |
math/9904106 | For the first part of the theorem, we begin by choosing MATH in MATH to be a generic link. Note that MATH can be recovered from its projection MATH together with a knowledge of the signs at each overcrossing. Given MATH, let MATH be an associated MATH-link in MATH such that MATH. Without loss of generality, we can assume that the leaves MATH of MATH form a generic link , and by abuse of notation, we can write that MATH, with the understanding that we have fixed the signs on the overcrossings of MATH. Now, given MATH, let MATH and MATH be the associated MATH-links in MATH and MATH respectively, and let MATH denote the canonical projection. Then we have that MATH where MATH is a unit-framed trivial link in MATH, whose components encircle some crossings of MATH and MATH, so that surgery on MATH brings MATH below MATH. The following lemma implies that the result of changing an overcrossing of MATH over MATH to an undecrossing can be achieved as a difference of the disjoint union of two graphs minus the disjoint union of two graphs with a leg glued. Together with REF and the definition of the multiplication MATH, it implies that MATH which finishes the proof of the first part of REF . The following identity holds in MATH: MATH where the framing of the unknot on the left hand side of the equation is MATH and where we alternate with respect to the MATH-links of the figure. The vertical arcs are arbitrary tubes. This follows from the second equality above in MATH and from MATH see CITE. The second part of REF is obvious from the definition of the NAME bracket. For the third part, notice that the NAME algebra structure on the quotient MATH is given by MATH where the last equality follows from the first part of REF and where MATH is the symplectic form. |
math/9904106 | (of REF ) For a surface MATH of genus MATH with one boundary component, CITE introduced a homomorphism MATH, where, following the conventions of CITE, MATH and MATH is identified with a submodule of the MATH-th tensor power MATH by defining MATH . In subsequent work, NAME showed that modulo MATH-torsion his homomorphism coincides with the abelianization of MATH, thus one gets, over MATH, an onto map of NAME algebras MATH. For the rest of the proof we will work over MATH. In CITE Hain proved that for genus MATH, the above map of NAME algebras has kernel generated by quadratic relations MATH (Hain's notation for MATH is MATH). Combining the proof of REF with REF ], it follows that the relation set MATH is the symplectic submodule of MATH generated by MATH in terms of a standard symplectic basis MATH of MATH. Using the isomorphism MATH given by mapping MATH to the degree MATH graph MATH as in REF , we obtain a map of NAME algebras MATH. Since for every choice of MATH and MATH we have MATH, the first part of REF implies for every basis MATH we have MATH. This implies, by definition, that MATH, thus obtaining the desired map MATH. Since MATH is generated by its elements of degree MATH, the commutativity of the two diagrams follows by their commutativity in degree MATH; the later follows by definition for the first diagram, and by the fact that surgery on a MATH-link of degree MATH with counterclockwise orientation and leaves decorated by MATH is equivalent to cutting, twisting and gluing by an element of the NAME group (of a surface of genus MATH with one boundary component, imbedded in MATH) whose image under the NAME homomorphism is equal to MATH, see CITE. This concludes the proof of REF . We now prove the statements in REF . For a closed surface MATH of genus MATH, CITE gave a version of his homomorphism MATH where MATH is defined to be the cokernel of the homomorphism MATH that sends MATH to MATH, where MATH is the symplectic form of MATH, for a choice of symplectic basis. Working, from now on, over MATH, CITE gave an identification of MATH with MATH, where MATH is given by MATH and MATH denotes the symplectic form. Explicitly, we will think of MATH as the submodule of MATH which is generated by elements of the form MATH for MATH. In CITE Hain proved that for genus MATH, there is an isomorphism of graded NAME algebras MATH which, in degree MATH, is the inverse of the NAME homomorphism, where MATH is the symplectic submodule of MATH generated by the relations MATH . The slight difference of MATH in the relations that Hain gave and the ones mentioned above are due to the difference in the normalization of the MATH-product between NAME and NAME. The restriction of the map MATH to MATH gives a map MATH which sends the relations MATH to zero (this really follows from the calculation of the surface with one boundary component together with the fact that MATH is sent into the ideal MATH of MATH), thus inducing the desired map MATH. We claim that MATH maps MATH to zero. This follows from the identity MATH of NAME twists on the mapping class group of MATH CITE, where MATH refer to the standard meridian, longitude pairs associated with a symplectic basis of MATH and MATH is the boundary curve of MATH. Thus we have the relation MATH on the mapping class group of MATH (where MATH are simple closed curves in MATH with isotopic images in MATH), together with the fact surgery along the MATH-link MATH corresponds to the NAME twist MATH in MATH, CITE. Since MATH is generated by its elements of degree MATH, the commutativity of the two diagrams follows by their commutativity in degree MATH; this is shown in the same way as for a surface with one boundary component. This concludes the proof of REF . |
math/9904106 | For a closed surface MATH of genus at least MATH, REF follows from REF and the following REF , perhaps of independent interest. For a closed surface MATH of genus less than MATH, fix a disk and consider an imbedding of its complement to a surface MATH in MATH of genus at least MATH. Choose a lifting of MATH to a diffeomorphism of the punctured surface that preserves the boundary and extend it trivially to a diffeomorphism of MATH of MATH. Since the NAME homomorphism is stable with respect to increase in genus, and since MATH, the result follows from the previous case. |
math/9904106 | First of all, recall that MATH if any leaf bounds a disk disjoint from the other leaves of MATH. As explained in CITE, an alternative way of writing REF is as follows: MATH for arbitrary disjoint imbeddings of two based oriented knots in MATH. Given a based knot MATH in MATH, let MATH denote the based knot obtained by a push-off of MATH in its normal direction (any will do) followed by reversing the orientation. The above identity implies that MATH in MATH, where MATH is any imbedded graph in MATH with a distinguished leaf the based oriented knot MATH in MATH. Given MATH as in the statement of the lemma, it follows that its distinguished leaf is the connected sum of disjoint based knots of the form MATH; thus it follows from the above discussion that MATH in MATH. |
math/9904106 | Fix a graph MATH. Let MATH (respectively, MATH) denote the MATH-links in MATH (with leaves MATH (respectively, MATH)) such that MATH . It follows by definition that MATH and MATH are homologous links in MATH. After choosing a common base point for each pair MATH of components of MATH and MATH, it follows that the connected sum MATH is nullhomologous in MATH and thus bounds a surface MATH in MATH. The surface MATH might intersect the other components of MATH or MATH at finitely many points; however by deleting disks around the points of intersection of MATH with MATH, we can find a nullhomotopic based link MATH and a surface MATH disjoint from MATH with based boundary such that MATH. REF imply that MATH is a sum of terms over MATH-links MATH in MATH which are trees, with at least one component MATH being nullhomotopic. By choosing a sequence of crossing changes (represented by a unit-framed trivial link MATH) that trivialize MATH and using REF , it follows that MATH modulo terms that involve joining some legs of MATH (thus modulo terms that involve graphs with loops), which concludes the proof. |
math/9904106 | This follows easily from CITE. Suppose MATH is some MATH-fold commutator. Then consider the one-relator group MATH and the projection MATH. Consider MATH of length MATH. By induction we can assume that MATH is uniquely defined in MATH and by CITE, it is well-defined in MATH. Moreover, by naturality under MATH, they take the same value on MATH. If MATH this is zero by CITE. Since this holds for all MATH it follows that MATH. If MATH . REF follows directly from the formula in CITE. |
math/9904106 | The ``if" part follows directly from the above proposition. To prove the ``only if" part we proceed by induction on MATH. The inductive step presents us with a map MATH; consider the diagram MATH . The obstruction to lifting this map is the pullback of the characteristic class in MATH of the central extension MATH. But REF implies that MATH is generated by NAME products of length MATH and so the pullback is zero if and only if all NAME products of length MATH vanish in MATH. Thus, we can inductively lift the map to MATH, thus to a map MATH, which is still a MATH-equivalence. On the other hand, since MATH is a MATH-equivalence, it induces an onto map MATH, which is also a MATH-equivalence. Composing with the map MATH, we get an endomorphism of MATH which is a MATH-equivalence. Stalling's theorem CITE implies that this endomorphism of MATH is an isomorphism which implies that the map MATH is one-to-one and thus a MATH-equivalence. |
math/9904106 | Apply NAME five-term exact sequence to the short exact sequence of groups MATH to obtain MATH . Combining this with the map MATH gives the commutative diagram MATH . This diagram yields the exact sequence of the corollary, where MATH is defined as the composition MATH . To prove the formula for MATH first note that, for any MATH we have MATH as follows directly from REF . But now the corollary follows from the definition of MATH and naturality. |
math/9904106 | We will use NAME 's formulation CITE of the NAME products. Choose cocycles MATH representing MATH, for MATH. Since we are assuming all NAME products of length less than MATH are defined and vanish, we can choose cochains MATH, for MATH, with the exception of the three cases MATH so that MATH and MATH. For two of the three exceptional cases the cochains MATH the MATH are cocycles but not necessarily coboundaries. In fact they represent the NAME products MATH and MATH respectively. Thus MATH is represented by the cocycle MATH and MATH is represented by the cocycle MATH. Now consider the cochain MATH . By grouping the terms in one way we see that MATH . Grouping the terms in another way gives MATH . Now subtracting these two formulae for MATH gives MATH since MATH and MATH are cocycles. Since the left side of this equation represents MATH and the right side is a coboundary the proof is complete. |
math/9904106 | The first part is immediate from REF , using the NAME duality isomorphism MATH. In local coordinates, it implies (see REF ) that the NAME product MATH is given by MATH where MATH, the summation is over MATH of length MATH, MATH indicates cup product and MATH indicates cap product with the fundamental homology class of MATH. Let MATH be the NAME algebra bracket, defined by MATH, for MATH. If we regard MATH then MATH can be expressed by the formula MATH . REF implies that MATH and so MATH . Thus REF follows from the third assertion of REF (or its coordinate version, REF ). |
math/9904106 | Given an element MATH we construct maps MATH, where MATH corresponds to the canonical projection MATH under the identification of MATH with MATH, and MATH. Since MATH, we have MATH and so we can combine the two maps to define a map MATH, where MATH is the double of MATH. We would like to extend MATH to a map MATH, for some compact orientable MATH-manifold with MATH the obstruction to the existence of MATH is the element MATH represented by MATH, where MATH are the oriented bordism groups of MATH. Since MATH we must be careful in our choices to assure that MATH. Redo the construction of MATH and MATH but using MATH instead of MATH and using a lift of MATH to an automorphism MATH of MATH instead of MATH. Our restriction on MATH assures that MATH and so we obtain MATH and an obstruction element MATH. Now this element may not be zero, but since the projection map MATH is zero, and clearly MATH maps to MATH, we conclude that MATH. Thus MATH extends to the desired MATH. Let MATH be the obvious diffeomorphisms onto the domains of MATH. It is clear that if MATH were a homology cylinder over MATH, then MATH. But this is not necessarily true and so we will perform surgery on the map MATH, adapting the arguments in CITE to our situation. See also CITE for similar surgery arguments which are used to show that any finite MATH-dimensional NAME complex is homology equivalent to a closed MATH-manifold. Suppose MATH. Then there exists MATH such that MATH and MATH represents MATH. If MATH is any representative of MATH, then MATH. Choose an element MATH so that MATH. Then MATH and MATH represents MATH. Thus for any MATH we can do surgery on a curve representing MATH and extend MATH over the trace of the surgery. The first step in killing MATH will be to kill the torsion-free part. Note that MATH, since MATH is an isomorphism, and so, under the canonical map MATH maps onto MATH. Choose an element MATH which maps to a primitive element of MATH. Now surgery on a simple closed curve MATH representing MATH will produce a new manifold MATH so that, if MATH is the element represented by the meridian of MATH, then MATH (see CITE). Since MATH is primitive in MATH, there is a MATH-cycle MATH in MATH whose intersection number with MATH is MATH. Thus the intersection of MATH with MATH is a MATH-chain whose boundary is MATH. So, by REF , MATH. A sequence of such surgeries will kill the torsion-free part of MATH. But this implies that MATH is torsion by the following simple homology argument. Consider the exact sequence: MATH . Since MATH and MATH imbeds into MATH. But, since MATH, we conclude that MATH. Therefore MATH and so MATH. We now follow the argument in CITE to kill the torsion group MATH. The linking pairing MATH is non-singular since MATH maps isomorphically to MATH. According to CITE if, for MATH, MATH, then we can choose the normal framing to any closed curve MATH representing MATH so that the element MATH is of finite order smaller than the order of MATH. Thus the torsion subgroup of MATH is smaller than MATH. Continuing in this way we reach the point where all the self-linking numbers are MATH. According to CITE this implies that MATH is a direct sum of copies of MATH. Now choose any non-zero element MATH. We will show that surgery on MATH reduces the rank of MATH. Denote by MATH the trace of the surgery and MATH the result of the surgery. Then we have a diagram of homology groups (coefficients in MATH) with exact row: MATH . Now MATH and so has rank one less than MATH. Since MATH is generated by the transverse disk bounded by the meridian curve representing MATH, the dotted arrow can be interpreted as REF intersection number with MATH. By NAME e duality this map is non-zero and so MATH, proving the claim. As in CITE we can assume the normal framing chosen so that MATH has order MATH or MATH. Thus the possibilities for MATH are either MATH or MATH, where MATH. We can then do a surgery to kill the MATH factor, in the first case, or reduce the order of MATH, in the second case. Continuing this way we eventually kill MATH, producing the desired MATH. |
math/9904106 | Let MATH denote the kernel of the natural projection MATH. We first construct a map MATH as follows. If MATH we can write MATH, where MATH. Then, we define MATH, where MATH and MATH is a lift of MATH. Using the isomorphism MATH this defines a map (denoted by the same name) MATH with corresponding description in local coordinates given by MATH . If MATH, as above, then MATH and so MATH . Therefore MATH, which implies that MATH. It is clear that MATH is one-to-one. We now show that it is onto. Suppose we have an element MATH. Lift MATH into MATH (denoted by the same symbols) and define an endomorphism MATH of MATH by MATH . It follows by Stalling's theorem CITE that MATH induces an automorphism of MATH which restricts to the identity automorphism of MATH. We note that MATH . But MATH represents the image of MATH under the NAME bracket MATH, which vanishes since MATH; thus MATH. This shows that MATH and clearly MATH. The fact that the projection MATH is onto follows immediately from REF . It is not hard, however, to give a direct argument; we leave this as an exercise for the reader. Finally, it is clear by the definitions that the diagram in REF commutes, and that the sequence below it is exact. |
math/9904106 | Let MATH and define MATH, where MATH are two copies of the solid handlebody MATH of genus MATH, which are attached to MATH via the diffeomorphisms MATH so that, referring to a basis MATH of MATH corresponding to a symplectic basis of MATH, the MATH are represented by the boundaries of meridian disks in MATH. Thus MATH, the free group generated by MATH (or, more precisely, their images in MATH). MATH is a REF-manifold with boundary MATH, which we can fill-in to obtain a closed REF-manifold MATH. If MATH, then the inclusion MATH induces an isomorphism MATH and we can consider MATH, where MATH. Suppose that MATH. Set MATH, where MATH-is the projection defined by MATH. Then MATH, where MATH are the classes represented by MATH. This assertion is just the obvious generalization of REF and the proof is the same. Now let MATH be an arbitrary element in MATH, where MATH, that is, MATH. We want to construct MATH such that MATH. Consider the endomorphism MATH of MATH defined by MATH . Denote also by MATH the induced automorphism of MATH. To see that MATH, we compute MATH . Therefore, by REF , MATH for some MATH. Since MATH we have MATH and so MATH. By the discussion above, MATH. |
math/9904106 | We apply the constructions and results of CITE. Consider the mapping cone MATH of the natural map MATH of NAME spaces. MATH is constructed from MATH by adjoining MATH-cells MATH along the generators MATH. Then MATH is simply-connected and MATH for MATH. So we have the NAME epimorphism MATH, where MATH denotes the third homotopy group. Suppose MATH. Then the NAME construction gives us a map MATH representing MATH, such that, if MATH is some interior point, then MATH, a zero-framed imbedded circle in MATH, see CITE. Now let MATH be the result of framed surgery on MATH along the MATH. Then MATH induces a map MATH and it is clear that MATH. Finally we note that MATH. |
math/9904106 | It follows from the definitions that we need to establish the commutativity of the following diagram MATH where MATH is defined in REF , MATH is the inclusion map inducing MATH and the NAME homomorphism MATH, and MATH is the NAME map MATH when MATH. Now suppose MATH is represented by a MATH-cycle MATH in MATH . since MATH is homologous to MATH, there exists a MATH-chain MATH in MATH such that MATH. Consider the chain MATH in MATH with boundary MATH. Since MATH is identified with MATH in MATH, the chain MATH is a MATH-cycle in MATH - let MATH denote its homology class in MATH. It follows from the definition of MATH that MATH. On the other hand MATH since the MATH-cycle MATH intersects MATH transversely in the MATH-cycle MATH, if MATH. This establishes the desired commutativity. |
math/9904107 | For MATH and MATH the statement is clearly true and we use induction on MATH. Suppose we know the statement for all positive integers smaller than MATH. Then we distinguish two cases: CASE: If REF and MATH are in the same block of MATH, then the construction of MATH starts by putting the entry MATH in the last slot of MATH, then deleting the element MATH from MATH. This does not alter either the set of minimum elements of the blocks nor the set of descents. Therefore, this case reduces to the general case for MATH, and is settled by the inductive hypothesis. CASE: If the largest element MATH of the block containing REF is smaller than MATH, then as we have seen above, MATH is the concatenation of MATH, and MATH is not empty. Clearly, by the definition of MATH, MATH and the element MATH is the minimum of its block. From this and the inductive hypothesis applied to MATH and MATH, the proof follows. |
math/9904107 | We show that our bijection MATH is an order-reversing map MATH. The conclusion then follows from the self-duality of the lattice of noncrossing partitions. Suppose MATH in MATH. This means MATH is a finer partition than MATH, so every element which is the minimum of its block in MATH is also the minimum of its block in MATH. By REF this implies MATH, so MATH in MATH. |
math/9904107 | The properties of the rank sizes of MATH are immediate consequences of REF and the corresponding properties known to hold for MATH. Moreover, every antichain of MATH is, via the bijection MATH, an antichain of MATH, and the strong NAME property of MATH follows from the strong NAME property of MATH. |
math/9904107 | We use induction on MATH. For MATH the statement is true. Now suppose we know it for all positive integers smaller than MATH. Denote by MATH the smallest element of MATH, and let MATH be a REF-avoiding MATH-permutation whose descent set is MATH. CASE: Suppose that MATH. Then we have MATH and, because MATH avoids the pattern REF, the values of MATH are consecutive integers. So, for given values of MATH and MATH, we have only one choice for MATH. This implies MATH where MATH is the set obtained from MATH by subtracting MATH from each of its elements. On the other hand, we have MATH, meaning that in any permutation MATH counted by MATH the chain of REF holds. To avoid forming a REF-pattern in MATH, we must have MATH. Therefore, MATH where MATH denotes the set obtained from MATH by removing its last MATH elements. Clearly, MATH by the induction hypothesis, so REF imply MATH. CASE: If MATH, but MATH, then let MATH be the smallest index which is not in MATH. Then again, to avoid forming a REF-pattern, the value of MATH must be the smallest positive integer MATH which is larger than MATH and is not equal to any MATH for MATH. So again, we have only one choice for MATH. On the other hand, the largest index in MATH will be MATH. Therefore, in permutations MATH counted by MATH, we must have MATH as REF implies that MATH must be the rightmost left-to-right minimum in such permutations, and that is always the entry REF. In order to use this information to reduce our permutations in size, we define MATH as follows: MATH if and only if either MATH and then, by the definition of MATH, MATH, or MATH and MATH. In other words, we decrease elements larger than MATH by REF; intuitively, we remove MATH from MATH, and translate the interval on its right one notch to the left. If we now take MATH, that will consist of entries MATH so that MATH and MATH. So in other words, we simply remove MATH from MATH (there has been nothing on the right of MATH in MATH to translate). Note that the size of MATH decreases with this operation as MATH. As we have seen in the previous paragraph, we had only one choice for MATH and MATH, so removing them this way does not change the number of permutations with a given descent set. Thus we have MATH, and also MATH. By induction hypothesis, the right hand sides of thes two equations agree, and therefore the left hand sides must agree, too. If MATH and MATH, and so MATH, then MATH, MATH, and indeed, MATH and MATH. CASE: Finally, if MATH, then the statement is trivially true as MATH . So we have seen that MATH in all cases. |
math/9904107 | It is clear that, in MATH, permutations which have the same descent set will cover the same elements and they will be covered by the same elements. The permutations with a prescribed descent set MATH form an orbit of MATH and they can be permuted among themselves arbitrarily by elements of MATH. On the other hand, REF shows that the orbits corresponding to MATH and to its reverse-complement MATH are equinumerous. Hence, a map MATH which establishes a bijection between MATH and MATH for each MATH provides an order-reversing bijection on MATH. |
math/9904107 | For each MATH, let MATH be the set of REF MATH-permutations with excedence set MATH. Let also MATH be the set of REF MATH-permutations with descent set equal to MATH, the reverse-complement of MATH. Thus, in the notation of the previous subsection, the cardinality of MATH is MATH. We construct a bijection MATH (illustrated by REF ). If MATH, then, as seen earlier in the definition of MATH, the entries MATH with MATH form an increasing subsequence. This, and the definition of excedence imply that MATH is a right-to-left minimum (that is, smaller than all entries on its right) if and only if MATH. Now let MATH be the reverse of MATH. Then MATH is a REF-avoiding permutation having a left-to-right minimum at position MATH exactly if MATH. There is exactly one REF-avoiding permutation MATH which has this same set of left-to-right minima at these same positions CITE. Namely, MATH is obtained by keeping the left-to-right minima of MATH fixed, and successively placing in the remaining positions, from left to right, the smallest available element which does not alter the left-to-right minima. We set MATH. REF then tells us that MATH if and only if MATH, in other words, when MATH, and so MATH belongs indeed to MATH. It is easy to see that MATH is invertible. Clearly, MATH can be recovered from MATH as the only REF-avoiding permutation with the same values and positions of its left-to-right minima as MATH. (All entries which are not left-to-right minima are to be written in decreasing order). Then MATH can be recovered as the reverse of MATH. The bijections MATH for all the choices of MATH produce an order-reversing bijection from MATH to MATH. But MATH is self-dual, so the proof is complete. |
math/9904107 | It is immediate from its definition and REF that MATH is a ranked poset (namely, MATH) and has rank-sizes given by MATH, equal to the rank-sizes in MATH. Also, MATH is a self-dual poset: clearly, if MATH is the MATH-avoiding signed permutation which corresponds to the pair MATH, then the mapping MATH is an order-reversing involution on MATH. Toward checking that there is an order-preserving bijection from MATH to MATH, we first recall a fact from CITE: every partition MATH can be encoded by a pair MATH of subsets of MATH whose cardinality is the number of pairs of non-zero blocks of MATH. Informally, these sets consist of the Left and Right delimiters of non-zero blocks when the elements are read in clockwise order (in the circular diagram of MATH). More precisely, if MATH or if MATH has only a zero-block, we set MATH. Otherwise, MATH has some non-zero block consisting of cyclically consecutive elements in its diagram. If such a block consists of MATH in clockwise order, then MATH belongs to MATH and MATH belongs to MATH. By deleting this block and its image under barring, a type-B noncrossing partition of MATH is obtained and the construction of the sets MATH and MATH is completed by repeating this process as long as non-zero blocks arise. For instance, if MATH, then MATH and MATH. Now suppose that MATH in MATH, and that this is a covering relation (that is, MATH). Then there exist MATH and MATH such that MATH and MATH, as a result of the merging of blocks entailed by the covering relation. Thus it is clear that if MATH is mapped to the signed permutation MATH with the property that MATH, then one obtains an order-reversing embedding of MATH into MATH. Combining this with the self-duality of MATH we obtain the desired embedding of MATH into MATH. Finally, the strong NAME property of MATH follows as in type A, from the strong NAME property of MATH (see CITE) and the rank-preserving embedding of MATH into MATH. |
math/9904107 | Let MATH and MATH be the reverse of MATH. Let MATH be the ``barred complement" of MATH, that is, MATH, and MATH is barred if and only if MATH is not barred. Then it is straightforward to verify that MATH if and only if MATH. Therefore, the reverse complement operation reverses the inclusion of excedence sets for signed permutations. (Thus, the entire hyperoctahedral group MATH ordered by containment of the excedence set is a self-dual poset.) But, clearly, this involution preserves the MATH-avoidance property, and thus MATH is self-dual. |
math/9904109 | The first relation in REF is trivial on MATH, so we only need to show it for MATH since MATH. Note that MATH, therefore REF yields MATH hence MATH . For the second relation we use the fact that MATH for MATH: MATH for all MATH. |
math/9904109 | Let MATH and MATH be isometries. Then MATH and MATH. Now MATH. Inserting this in REF yields the statement. |
math/9904109 | Since MATH must be a subsector of MATH for MATH a conjugate of MATH, there is an isometry MATH. Note that then MATH. Hence by naturality and REF we compute MATH and hence also MATH. We also obtain MATH and MATH by REF . Note that MATH by restriction. REF -REF follow now by naturality, REF . Next, we note that MATH, and hence MATH. Therefore MATH and hence also MATH. Now REF follow from REF . |
math/9904109 | With the normalization convention as in REF , this is just the expansion of the identity in REF , and this certainly holds as well using similar expansions with other intertwiner bases. |
math/9904109 | That each MATH is a minimal central projection and that all minimal central projections arise in this way is obvious from the description of the matrix units. The vertical product MATH is given graphically by the left-hand side of REF . We can use the expansion of REF for the two parallel wires MATH and MATH in the middle. Now note that the horizontal unit is given by MATH. Therefore, by multiplying MATH from the left and from the right, we obtain the diagram on the right-hand side of REF . Reading the diagram from left to right, we observe that intertwiners in MATH and MATH are involved here. Hence we first obtain a factor MATH. Next, we can use the trick of REF to turn around the small arcs at the trivalent vertices involving MATH. This yields a factor MATH. This way we see that the diagram on the right-hand side of REF represents the same element of the MATH as the diagram in. CASE: Now let us look at the part of this picture inside the dotted box. Reading it from the left, this part can be read for fixed MATH and MATH as MATH, and the sum over MATH runs over a full orthonormal basis of isometries MATH in the NAME space MATH since we have the summation over MATH. Next we look at the part inside the dotted box of the diagram in REF . Here, since we introduced the sum over MATH, the part can be similarly read for fixed MATH and MATH as MATH, where the sum over MATH runs over another orthonormal basis of isometries MATH in the NAME space MATH. Since such bases MATH and MATH are related by a unitary matrix transformation (this is essentially ``unitarity of MATH-symbols"), we conclude that the diagrams in REF represent the same element in MATH. We now see that we first obtain a factor MATH. Next we can turn around the small arcs at the outer two trivalent vertices involving MATH and MATH so that we obtain a factor MATH. Then, by ``stretching" the diagram a bit, we can read the diagram for fixed MATH as MATH . Now proceeding with the summations over MATH yields the statement. |
math/9904109 | By REF , we compute MATH . Since the horizontal unit MATH is given by MATH we find that MATH. As MATH sends off-diagonal matrix units to zero and the diagonal ones to strictly positive numbers, this proves that MATH is a faithful state. Obviously also MATH sends off-diagonal matrix units (with respect to MATH) to zero and the diagonal ones to strictly positive numbers, and hence it is a strictly positive functional but it is not normalized. The trace property MATH is clear from the definition of MATH using matrix units for MATH and MATH. |
math/9904109 | We only show the statement for the MATH-sign; the other case is analogous. First we fix MATH and MATH. For each MATH we choose orthonormal bases of isometries MATH, MATH, so that MATH. Using NAME reciprocity, we obtain an orthonormal basis of isometries MATH. Here we chose an isometry MATH such that there is an isometry MATH subject to relations MATH and MATH, as usual. Choosing also orthonormal bases of isometries MATH, MATH, for each MATH (so that MATH) we find that MATH gives an orthonormal basis of isometries of MATH. Finally, using REF , we find that putting MATH defines an orthonormal basis of isometries MATH of MATH. Then we have for any MATH by the elementary relations for the intertwiners MATH the following identity: MATH . The second line yields graphically exactly the diagram in REF where we read the diagram from the left to the right in order to interpret it as an intertwiner in MATH. Now let us take on both sides first the summation over MATH. Then the left-hand side gives exactly the MATH part of MATH (in MATH) as defined in REF . Next we divide by MATH and we proceed with the summation over MATH and MATH. On the left-hand side we obtain the MATH part of MATH this way, and this is exactly the MATH part of MATH. On the right-hand side we now have a summation over the full basis MATH of MATH. Therefore we can use the graphical convention of REF to put a small semi-circle around the wire labelled by MATH at the two trivalent vertices. This gives us a factor MATH so that only a factor MATH remains from the original prefactor in REF . Thus, by repeating the above procedure for all MATH and making finally the summation over MATH, we obtain on the left the full MATH whereas the right-hand side gives graphically the diagram in REF . The diagram on the left-hand side in REF is obtained from REF , up to the factor MATH, by a topological move. |
math/9904109 | From REF we obtain MATH . Hence MATH . Application of the horizontal state MATH of REF and multiplication by MATH yields REF since MATH and MATH decompose into sectors MATH with MATH, and by REF . Now the right-hand side of REF is given graphically by the diagram on the left in REF , and we can slide around the trivalent vertices to obtain the diagram on the right-hand side. Without changing the scalar value we can now open the outer wire labelled by MATH and close it on the other side, as in REF . This way we obtain the picture in REF up to a REF degree rotation, but a rotation is irrelevant for the scalar values. |
math/9904109 | Using the diagram for the matrix elements MATH in REF , the sum MATH can be represented by the diagram on the left-hand side of REF . Using REF and also the trick to turn around the small arcs given in REF , we obtain the right-hand side of REF . We can now slide around the lower trivalent vertex of the wire MATH to obtain the left-hand side of REF . Next, we can use REF to replace the two parallel horizontal wires with labels MATH and MATH by a summation over a thin wire MATH. Similarly, but the other way round, we can then use REF to replace the summation over the wire with label MATH by two straight horizontal wires with labels MATH and MATH. This way we obtain the right-hand side of REF . Now it should be clear how to proceed: We slide around the upper trivalent vertex of the wire MATH counter-clockwise. Then we see that the result gives us the diagram for MATH, rotated by REF degrees. This proves MATH. Next we show commutativity of MATH with MATH. We have to show MATH. Using the graphical expression for the statistics phase MATH on the left-hand side of REF , we can represent MATH by the left-hand side of REF . We now start to rotate the upper oval consisting of the thick wires MATH and MATH in a clockwise direction. This way we obtain the right-hand side of REF . It should now be clear that, if we continue rotating to a full rotation by REF degrees, then we remove the twist from the wire MATH whereas we obtain a twist in the wire MATH which is of the type displayed on the right-hand side of REF , thus representing MATH. Hence MATH. |
math/9904109 | The sum MATH is given graphically by the left-hand side of REF . By using REF for the two parallel vertical wires MATH on the bottom and the IBFE moves we obtain the right-hand side of REF . For the summation over the thin wire MATH we can use REF again to obtain the left-hand side of REF . Now we can slide around the right trivalent vertex of the wire MATH, and this yields the right-hand side of REF . Next we can use the trick of REF to turn around the small arcs from the wire MATH to the wire MATH. This yields a factor MATH. Then we can proceed with the summation over MATH, using REF once more, and this gives us the left-hand side of REF . Now we observe that the summation over MATH provides a killing ring, and hence we obtain a factor MATH. The normalization convention for the small arcs yields another factor MATH, and hence we get exactly the right-hand side of REF . The circular wire MATH cancels the factor MATH, and thus we are left exactly with the global index MATH times a summation over two straight horizontal wires, and the latter is exactly the horizontal unit MATH. The rest is application of the isomorphism MATH. |
math/9904109 | As in particular MATH, we can write MATH with MATH according to the direct sum structure of MATH, MATH. Assume MATH. Then clearly MATH for all MATH. Now the MATH part of MATH is given by MATH, hence MATH. A similar argument applies to MATH, and hence the element MATH is independent of the linear expansions of the MATH's. Therefore REF defines a sesqui-linear map MATH. Now assume MATH. Then in particular MATH for all MATH, and hence MATH, proving strict positivity. That the sesqui-linear form MATH on MATH is non-degenerate follows now from positive definiteness of MATH. It remains to show that MATH. But this is clear since any element of the form in REF can be ``pulled through" the diagram in REF by using the IBFE's. |
math/9904109 | Using REF we can replace the left-hand side of REF by the left-hand side of REF . Next we can slide one of the trivalent vertices of the wire MATH around the wire MATH. Using the identity of REF , we obtain a factor MATH, and we can now proceed with the summation over MATH, again using REF . Using also REF for the parallel wires MATH, MATH as well as MATH and MATH, we obtain the right-hand side of REF . Using now REF once again for the wires MATH, MATH, we can pull the wire MATH over the middle expansion. The summation over MATH yields a killing ring which disconnects the picture into two halves, one is an intertwiner in MATH and the other in MATH. Hence we obtain a factor MATH, and we conclude that the left-hand side in REF represents a scalar intertwiner MATH, MATH. To compute that scalar, we can start again on the left-hand side of REF , now putting MATH and MATH. The diagram on the left-hand side of REF clearly represents an intertwiner of the same scalar value MATH. We can now use the move of REF which does not change the scalar value: We open the wire MATH on the left and close it on the right. The resulting diagram is regularly isotopic to the diagram on the right-hand side of REF . Thus we are left with exactly the diagram for MATH. This proves the lemma. |
math/9904109 | Using REF and also the trick of REF , we can draw the diagram on the left-hand side in REF for MATH. Now let us look at the part of this picture above the dotted line. In a suitable NAME annulus, this part can be read for fixed MATH and MATH as MATH, and the sum runs over a full orthonormal basis of isometries MATH in the NAME space MATH since we have the summation over MATH. Next we look at the part above the dotted line on the right-hand side of REF . This can be similarly read for fixed MATH and MATH as MATH, where the sum runs over another full orthonormal basis of isometries MATH. Since such bases MATH and MATH are related by a unitary matrix transformation (this is again just ``unitarity of MATH-symbols"), the left- and right-hand side represent the same vector in MATH. Then, using again REF and also the trick of REF , we conclude that the vector MATH can be represented by the diagram on the left-hand side of REF . Now let us look at the part of the diagram inside the dotted box. In a suitable NAME annulus, this can be interpreted as an intertwiner in MATH. But any element in this space can be written as a linear combination of elements constructed from basis isometries MATH, MATH, as indicated in the dotted box on the right-hand side of REF . The coefficients in its linear expansion depend only on MATH for fixed MATH, but certainly not on MATH. This shows that MATH is a linear combination of MATH's, thus MATH. |
math/9904109 | It follows from REF that MATH unless MATH and MATH. We now show that MATH. (We denote MATH.) The sum is given graphically by the left-hand side in REF . A twofold application of REF yields the right-hand side in REF . Applying REF twice again, we obtain the left-hand side of REF . We can now slide the upper trivalent vertex of the wire MATH around to obtain the right-hand side of REF . Next we can use the trick of REF to turn around the small arcs at the trivalent vertices of the wire MATH, yielding a factor MATH. This gives the right-hand side of REF . Since we have a summation over MATH, we can again use REF , and this gives us the left-hand side of REF . As we have a prefactor MATH, the summation over MATH provides a killing ring, and only MATH survives it: We obtain a factor MATH. Now our picture starts to collapse. The factor MATH yields, with the normalization convention as in REF , a factor MATH. Since our picture is now disconnected into two parts which represent intertwiners in MATH, they are scalars and we obtain a factor MATH. This gives us the right-hand side of REF . Therefore we are now left with a sum over scalars times two straight vertical wires labelled by MATH, representing a scalar intertwiner in MATH. The scalar value of each connected part of the picture is MATH, therefore we can compute the prefactor as MATH . Thus we are left with a sum over two vertical straight wires with label MATH and prefactor MATH. This is MATH. Next, we can expand each vector MATH, in an orthonormal basis as MATH . Inserting this in REF yields MATH . Now using MATH and REF we compute MATH hence MATH . Thus MATH is a projection and we also have MATH. Hence for any MATH we find MATH by REF . Thus each MATH can be expanded in our matrix units, and since MATH is spanned by the MATH's we conclude that MATH is a complete system of matrix units. It follows that the non-zero vertical projectors are minimal central projections in MATH, and that the simple summand MATH is a full MATH matrix algebra. It remains to show MATH. The dimension of MATH can be counted as MATH . Now MATH is given graphically in REF . By the IBFE's we can pull out the circle with label MATH which gives us another factor MATH. We can therefore proceed with the summation over MATH, and this yields a factor MATH, the global index, and then we are left exactly with the picture in REF . |
math/9904109 | The vector MATH is given graphically by the left-hand side of REF . Now note that the upper part of the diagram represents an intertwiner in MATH. Therefore it vanishes unless MATH and then it must be a scalar multiple of MATH. Hence we can insert a term MATH which corresponds graphically to the disconnection of the wires as on the right-hand side in REF and multiplication by MATH. Then the factor MATH disappears because of the normalization convention for trivalent vertices with small arcs, and we are left exactly with the right-hand side of REF . It follows in particular that MATH unless MATH. The claim follows now since the vertical projectors sum up to MATH and MATH is the identity on MATH. |
math/9904109 | For MATH and isometries MATH and MATH we define a vector MATH by the diagram in REF . Using again intertwiner bases, we also put MATH with some multi-index MATH. It follows from the right-hand side in REF that MATH. Conversely, we obtain by REF that MATH unless MATH, hence MATH. With MATH, closing the wires on the bottom and on the top on both sides of REF yields MATH . Hence linear extension of MATH defines a unitary operator MATH. Note that MATH means multiplication by MATH from the right in each block MATH and this corresponds graphically to closing the open ends of the wires MATH in REF and multiplying by MATH. Therefore we find MATH where MATH. Thus MATH. |
math/9904110 | The assertion MATH is proved in REF in the form MATH and MATH for MATH. Let us prove MATH. We write MATH for MATH, and MATH for MATH for convenience. By NAME vanishing theorem (see CITE) MATH for MATH. The sequence REF remains exact after shifting by MATH . Note, that MATH is a free MATH-module of rank MATH and MATH is a free MATH-module of rank MATH when MATH. This observation shows that MATH . Assume for the moment that MATH for MATH. Then REF gives the exact sequence of global sections MATH . Since MATH (see REF ), we have MATH . It remains to show that MATH for MATH. Moreover, it is sufficient to prove the vanishing of MATH. As observed in REF , the restriction homomorphism MATH is surjective. Hence, the sheaf MATH is generated by its global sections since MATH is ample, and all cohomologies MATH for MATH vanish by the NAME vanishing theorem. |
math/9904110 | The statement follows immediately from REF by taking the restriction of the sheaf MATH to MATH. |
math/9904110 | The formula MATH is the assertion of REF. We prove this statement independently. Making use of the exact sequence REF , we get MATH . Now the requested formula follows from REF and the vanishing of MATH for MATH. Let us prove assertion MATH. The short exact sequence REF remains exact after tensoring by MATH. As in the case above, we have MATH and from REF and the NAME vanishing theorem, MATH . We prove the second assertion of the theorem by using the description of the space of global sections of the sheaf MATH. We have a decomposition into a direct sum of MATH-homogeneous components MATH where MATH is the MATH-subspace in MATH generated by the smallest face of MATH containing MATH. Since the support polytope MATH admits the natural partition MATH, where MATH is the relative interior of the face MATH, we have MATH . Thus we have given a different proof for the second formula. |
math/9904110 | We need a special form of the NAME duality (see REF) MATH for any locally free sheaf of the MATH-modules MATH with the dual MATH on a complete simplicial toric variety MATH. Take MATH, MATH. From the NAME duality we have the isomorphism MATH . Hence, using the vanishing of MATH for MATH, we get the equality MATH which is equivalent to MATH and this completes the proof. |
math/9904110 | We have the chain of equalities MATH where we have used the identity MATH which is proved in the Appendix. |
math/9904110 | Using the equality of REF , we get MATH as required. |
math/9904110 | Indeed, from the long exact sequence of REF , we have MATH . Since MATH is ample, MATH applying the generalized NAME formula given in REF , we prove the statement. |
math/9904110 | It follows from REF that MATH and MATH . Each MATH can be uniquely written as MATH, where MATH, MATH. The support polytope MATH is defined by the inequalities MATH in MATH. Let MATH. Then the inequalities above are equivalent to MATH . The same arguments as in REF show that MATH, where MATH, and MATH follows immediately. |
math/9904110 | From the MATH-th exterior power of the sequence REF , we have MATH . Since MATH is ample, for all MATH, we obtain the exact sequence of global sections MATH REF implies the isomorphisms MATH and consequently, the isomorphisms MATH . Hence we obtain the relations MATH . By induction we get the requested formula. |
math/9904110 | Since MATH is a simple, there are precisely MATH-faces of MATH containing a given MATH-face MATH of MATH. Hence, the requested formula is equivalent to the combinatorial identity MATH or MATH . We give a proof of this identity based on the method of integral representations. Rewrite the sum MATH as MATH where the cycle MATH is MATH. One can choose the numbers MATH and MATH small enough, say MATH, MATH, so that the geometric series MATH converges on the contour of integration MATH, and it is possible to reverse the order of integration and summation. Therefore, summing up a geometric sequence, MATH . By the residue theorem, the second integral is MATH where MATH, since MATH for MATH. Finally, MATH which concludes the proof. |
math/9904111 | Observe that MATH and MATH are well defined as functions on the MATH-interval MATH since MATH. The formula for MATH follows by observing that MATH is of the form REF with MATH and MATH given by MATH . By inserting the explicit functions MATH and MATH, we see that the corresponding MATH and MATH coincide with the ones given by REF. The formula for MATH in the point MATH follows now immediately from REF of MATH. |
math/9904111 | Since MATH and MATH (respectively MATH and MATH) are real valued on the MATH-interval MATH, we may restrict the proof to real valued functions MATH and MATH. Then we derive by a direct computation using REF that MATH . The lemma is now an easy consequence of this formula since the finite sums become telescoping. |
math/9904111 | Using the formula MATH it follows that the behaviour of the weights of MATH at infinity is given by MATH where MATH is the positive constant MATH . This implies that MATH for all MATH. The proof follows now from the definition of the Wronskian REF and the fact that MATH . |
math/9904111 | REF follows immediately from REF . For the proof of REF , we observe first that the weights of MATH around zero behave like MATH . Define now the function MATH by MATH if MATH, MATH if MATH and MATH if MATH. Then MATH follows from REF, and MATH since MATH if MATH. Furthermore, MATH and MATH, hence MATH. By construction we have MATH and MATH, MATH for MATH. For the proof of REF we define the function MATH by MATH if MATH, MATH if MATH, and MATH if MATH. Then MATH. Furthermore, MATH and MATH. So it remains to show that MATH. Using the explicit REF for the MATH-difference operator MATH, we see that MATH, MATH as MATH. Combined with REF, it follows that MATH. |
math/9904111 | We have to show that MATH . By REF and the fact that MATH for MATH, it suffices to show that MATH for all MATH. Since MATH, we have for all MATH, MATH as desired. |
math/9904111 | Since MATH is symmetric, it suffices to prove the inclusion MATH. Let MATH. Then by REF, MATH. For any MATH we have, using REF , that MATH . Now using REF , one derives from the existence of the limits MATH for all MATH that MATH and MATH. It follows that MATH, as desired. |
math/9904111 | We start with computing MATH. The coefficient MATH in REF of MATH is non-zero for MATH since MATH. It follows that solutions of MATH on MATH are in one to one correspondence with solutions MATH of a recurrence relation of the form MATH with MATH by definition. The correspondence is obtained by associating the sequence MATH REF to MATH. The coefficients of the recurrence relation are then given by MATH, MATH and MATH. Any solution of such a recurrence relation is uniquely determined by MATH, hence MATH. Then MATH follows from the fact that functions MATH are in one to one correspondence with solutions MATH of the recurrence relation MATH . Finally, in order to show that MATH, observe that solutions of MATH on MATH are in one to one correspondence with solutions MATH of a double infinite recurrence relation of the form MATH since MATH and MATH REF of MATH are non-zero for MATH. This is a two-dimensional space. CASE: Let MATH. Using REF , the product rule for the MATH-derivative MATH and the second equality of REF, we have for all MATH, MATH . If MATH, then it follows from REF that MATH for all MATH since MATH, hence MATH is constant on MATH. A similar argument shows that MATH is constant on MATH if MATH. CASE: Let MATH. The continuously differentiability of the MATH's at the origin yields that MATH. Indeed, both sides are equal to MATH (compare with the proof of REF ). The result follows now from REF . |
math/9904111 | Suppose that MATH, then we claim that the map MATH is injective. Suppose that the map is not injective. Then there exist two linearly independent functions MATH such that MATH are linearly dependent. The Wronskian MATH is not identically zero as function on MATH by the linear independence of MATH and MATH. Since MATH is constant on MATH by REF , it follows that MATH is not identically zero on MATH, which contradicts the linear dependence of MATH and MATH. So if MATH, then it follows from REF and from the injectivity of the map MATH that MATH. This proves that MATH for all MATH, and it proves the implication MATH. The implication MATH is proved in a similar manner, while the implications MATH and MATH are immediate consequences of REF . |
math/9904111 | The proof for MATH is a direct consequence of REF and the explicit REF for MATH. The proof for MATH follows then from REF. |
math/9904111 | First of all, observe that if MATH and MATH, then MATH, hence MATH is well defined. Let MATH and let MATH, MATH be such that MATH. Using the asymptotics REF for the weights of the inner product MATH . , it follows from REF that MATH because MATH. It remains to show that MATH. For the proof we use some well known results from the theory of the classical moment problem, see for instance CITE or CITE. If MATH satisfies MATH on MATH, then by setting MATH we see that MATH satisfies the recurrence relation MATH with MATH and MATH. It follows from CITE and from the fact that the sequence MATH is bounded that the NAME moment problem corresponding to the recurrence relation REF for MATH is determined. By CITE this implies that MATH, as desired. |
math/9904111 | The explicit expression for the Wronskian follows by computing the limit MATH using the first expression of REF . Since the Wronskian is non-zero, it follows that MATH and MATH are linear independent, hence they form a basis of MATH by REF . |
math/9904111 | We first prove the connection coefficient formula for MATH. Observe that MATH is well defined for MATH since MATH. We fix MATH with MATH such that MATH. Furthermore, we assume that MATH and we fix MATH such that MATH and MATH. By the assumptions on MATH and MATH, we may apply the three term recurrence relation CITE with MATH, MATH, MATH, MATH and MATH. We arrive at MATH . The MATH on the left hand side is the MATH in REF for MATH and the second MATH on the right hand side is the MATH in REF for MATH. Again by the assumptions on MATH and MATH, we may rewrite the first MATH on the right hand side using CITE with MATH, MATH, MATH, MATH and MATH. This gives MATH . The MATH on the right hand side is the MATH in REF for MATH. So substituting this formula in the three term recurrence relation, and simplifying the formulas using in particular the functional relation MATH for the NAME theta function, we arrive at the desired result for restricted choices of MATH, MATH and MATH. The extension to all MATH is made using the fact that the left hand side and the right hand side of the MATH-function expansion are solutions of the eigenvalue equation MATH on MATH. Finally, the restrictions on MATH and MATH can be removed by analytic continuation. The proof of the connection coefficient formula for MATH follows by analytic continuation from the connection coefficient formula for MATH using REF. |
math/9904111 | Using the minimal solution CITE of the three term recurrence relation CITE, it follows that MATH is a solution of MATH on MATH, where MATH is the second order MATH-difference operator defined by REF. In particular, we have MATH since MATH. It follows that MATH. We have MATH by REF and MATH. It follows that MATH for a unique constant MATH. The explicit expression for MATH can be found using MATH and by evaluating MATH using the MATH-Gauss sum CITE. |
math/9904111 | We first assume that MATH. Using REF we then have MATH where MATH and MATH are given by REF, respectively. By REF , the right hand side of REF is equal to MATH where MATH is the positive constant defined by REF. It follows by direct computation that MATH . Now we can apply the MATH-product identity MATH see CITE, with parameter values MATH to obtain MATH . The proposition now follows for MATH by substitution of REF and using MATH. By continuity in MATH, it follows that REF holds for all MATH. |
math/9904111 | We fix MATH satisfying REF. Then the Wronskian MATH is non-zero since MATH, see REF . Hence MATH is a linear basis of MATH by REF . For the proof of REF , we observe first that MATH satisfies the three properties as stated in REF , see REF. If MATH is another function satisfying the same three properties, then MATH since MATH and MATH by REF . Hence MATH on MATH by REF . Finally, for the proof of REF , observe that MATH and MATH are linearly independent by REF and by REF . Since MATH by REF and MATH by REF , it follows that MATH. So MATH, as desired. |
math/9904111 | This is well known, see for instance CITE and CITE. The connection between MATH and MATH with the MATH in the last equality of REF follows from CITE. |
math/9904111 | We first prove REF for MATH. By REF , there exist unique MATH such that REF holds for all MATH. These coefficients can be expressed in terms of NAME by MATH . By REF we have MATH where MATH is the positive constant defined by REF now follows by substituting REF, the explicit REF for the Wronskian MATH, and the explicit REF for MATH and MATH in REF, and using the theta function identities MATH and REF. It follows now by continuity in MATH that REF is valid for MATH. |
math/9904111 | Immediate from REF . |
math/9904111 | The proof for MATH is trivial since MATH, MATH, MATH and MATH are regular at MATH. For MATH (MATH), observe that MATH and MATH have simple poles at MATH and that MATH and MATH are regular at MATH. It follows from REF and the first equality of REF that the singularity of MATH at MATH is removable for MATH if the (at most simple) singularity of MATH at MATH is removable. This can be checked by direct computions using the theta function identities MATH and REF. It follows that MATH exists and that REF holds. The extension REF of MATH lies in MATH because MATH and the derivatives of MATH and MATH at MATH are continuously differentiable at the origin, compare REF . |
math/9904111 | REF follow from REF . CASE: Let MATH with MATH, then MATH, hence MATH is well defined. Then MATH follows from REF and from REF . CASE: We first prove the connection coefficient formula for MATH. Then the connection coefficient formula is valid for all MATH, see REF . Since MATH, it follows by the uniqueness property of extensions of eigenfunctions, see REF , that the connection coefficient formula is valid for all MATH. The connection coefficient formula holds then for all MATH by continuity. CASE: This follows from REF, and REF . |
math/9904111 | We first prove that MATH . For the proof we need to consider the two cases MATH and MATH seperately. CASE: MATH. Observe that the product REF for the MATH-derivative MATH and REF imply the following product rule for MATH: MATH . Combined with the easily verified formulas MATH and REF of MATH, we obtain MATH for MATH. By writing out the MATH-terms in this formula, we see that the right hand side reduces to MATH, which in turn is equal to MATH by REF . This completes the proof of REF for MATH. CASE: MATH. Observe that the NAME function in the end-point MATH reduces to MATH since MATH. Using the product REF , we then have that MATH . Combined with the expression of MATH in the end-point MATH in terms of the MATH-derivative and the functions MATH and MATH, see REF , and using the fact that MATH, we obtain MATH . Hence REF is valid for the end-point MATH since MATH by REF . It remains to show how REF leads to a proof of the proposition. Let MATH be the set of functions MATH with finite support. Then MATH as dense subspaces. Let MATH, then MATH is continuously differentiable at the origin since MATH. Furthermore, MATH is a constant multiple of MATH for MATH, hence MATH. By REF it follows that MATH, hence MATH. Combined with REF, this proves that MATH for all MATH. By continuity, REF is valid for all MATH. This completes the proof of the proposition. |
math/9904111 | Recall that MATH is a generalized eigenfunction of MATH with generalized eigenvalue MATH if MATH, MATH and MATH in MATH. This can be checked for all MATH by an elementary computation using REF . Let MATH. Then MATH is part of the spectrum MATH of MATH since there exists a generalized eigenfunction of MATH with generalized eigenvalue MATH. Then MATH is in the continuous spectrum MATH of MATH or in the point spectrum MATH of MATH since a self-adjoint operator does not have residual spectrum, see CITE. It remains to show that MATH. We have to distinguish between the cases MATH and MATH. CASE: MATH, that is, MATH with MATH. It is a straightforward consequence of the asymptotic behaviour of MATH to infinity, see REF, that any non-zero linear combination of the basis elements MATH of MATH does not lie in MATH (see REF for the definition of MATH), hence MATH. CASE: MATH. Observe that MATH since MATH. Using that the asymptotics to infinity of MATH is given by MATH we obtain that MATH compare with the proof of REF . Combined with REF , it follows that MATH is a basis of MATH. It is easy to see, using asymptotics to infinity, that any non-zero linear combination of these basis elements does not lie in MATH, hence MATH. |
math/9904111 | For MATH we write MATH for the unique complex number with modulus less than MATH such that MATH. Fix MATH. By a straightforward computation using REF , we have for MATH and MATH that MATH where MATH is the NAME. Let MATH and let MATH such that MATH and MATH for MATH. Then we have MATH . Using the connection coefficients formula given in REF and using the fact that MATH and MATH REF are regular at MATH, we obtain MATH . It follows now by symmetrization of the double MATH-Jackson integral that MATH . The proposition follows now for all MATH using REF and changing the integration variable to MATH using the map MATH, see REF. The result now also holds when MATH or MATH since the spectral measure MATH is countably additive. |
math/9904111 | We assume first that MATH. In view of REF applied to the special case MATH and MATH, it suffices to observe that MATH, which is a consequence of REF and CITE. Finally, observe that the inequality MATH for MATH and MATH, where MATH is the norm of MATH, holds for all MATH by continuity. Hence MATH can be uniquely extended to a continuous linear map MATH for all parameters MATH. |
math/9904111 | For MATH we have that MATH on MATH. The proof follows now from REF since MATH is real valued for MATH. |
math/9904111 | Observe that there exists a constant MATH such that MATH for all MATH and for all MATH with MATH. The first equality follows then from NAME 's dominated convergence theorem. The second equality follows from REF , using that MATH since MATH is continuously differentiable at the origin, see REF . |
math/9904111 | We only sketch the proof, since it is similar to the little MATH-Jacobi case, see CITE and CITE. Let MATH be the algebra of complex valued, continuous, MATH-invariant functions on MATH. We fix MATH such that MATH is not in the support of MATH and MATH. It suffices to give a proof of REF for such functions MATH and MATH. We start with the second expression of MATH in REF , and replace MATH by its MATH-function expansion, see REF . Using the estimate MATH for MATH, where MATH (compare REF), which can easily be proved using the mean value theorem, we may replace in the expression of MATH the function MATH by its asymptotic value MATH at MATH. Combined with REF it follows now from the bounded convergence theorem that MATH provided that the limit in the right hand side of the equality exists, with MATH . Since MATH is continuous and non-zero on MATH, the NAME lemma implies that the contributions of the sums of MATH with MATH tend to zero in the limit, so MATH may be replaced by MATH . Applying the NAME lemma again and using MATH we arrive at MATH provided that the limit in the right hand side of the equality exists, where MATH is the NAME kernel, MATH . The result follows now from the well known MATH-properties of the NAME kernel. |
math/9904111 | It follows from REF that MATH for all MATH. Hence MATH is the identity on MATH and MATH. Consequently, we have MATH. CASE: Assume that MATH. Let MATH. Observe that MATH by the previous paragraph and by REF , where MATH is the norm of MATH. By REF , this implies that MATH. Let MATH. We have seen that MATH, hence there exists a function MATH such that MATH. Now MATH by the previous paragraph, and MATH is injective by REF , hence MATH, as desired. |
math/9904111 | Let MATH. Then MATH by REF . Furthermore, if MATH, then MATH where the first equality follows from REF . Since MATH and MATH are regular at MATH, it follows that REF is valid for all MATH. Hence MATH and MATH on MATH for all MATH, as desired. |
math/9904111 | Throughout the proof, we use the notations introduced in the proof of REF . CASE: Let MATH and fix MATH satisfying the properties as stated in REF . It suffices to prove that MATH. Observe that REF is still valid in the present setting, but that the analogue of REF is now given by MATH where MATH and where MATH is the unique element satisfying MATH. By the condition MATH, we have for MATH satisfying MATH that MATH . By the bounded convergence theorem we may interchange the limit MATH and the integration in REF. Combined with REF, this gives MATH, as desired. CASE: Let MATH and MATH. Choose arbitrary MATH such that MATH and such that MATH. Then by the first part of the proposition, we have MATH. We compute now MATH using REF. We substitute REF and change the integration parameter MATH in REF to the MATH-parameter using REF. Observe that for MATH and MATH, we have that MATH (respectively MATH) lies in the lower (respectively upper) half plane of MATH. Since MATH has a simple pole in MATH, it follows by NAME 's Theorem that MATH where the factor MATH arises from changing the integration variable in REF to MATH using the map MATH, and from the fact that one has to change sign in order to get a positive oriented curve around MATH. The proof is now completed using the explicit expression of MATH, using REF and by symmetrizing the double MATH-Jackson integral. |
math/9904111 | It follows from REF that MATH. Observe now that MATH by REF and MATH by REF . It follows that MATH and MATH, since MATH is the disjoint union of MATH and MATH, see CITE. |
math/9904111 | Orthogonality is clear since the functions MATH REF are eigenfunctions of the self-adjoint operator MATH with mutually different eigenvalues MATH (MATH), see REF . It remains to derive the explicit expression for the quadratic norm MATH. We first assume that MATH. Observe that MATH for MATH by REF . Since MATH and MATH is a projection, it follows that MATH for MATH, as desired. The result now follows for MATH by continuity. |
math/9904111 | We write MATH for the inner product of MATH. Suppose that MATH and let MATH. Applying REF , we obtain MATH. In order to extend this result to parameters MATH in MATH, we show that MATH and MATH depend continuously on MATH. This is clear for MATH, while for MATH it is clear except for the term MATH where we use the convention that MATH if MATH for the right hand side. The continuity of this term follows by NAME 's dominated convergence theorem, using the asymptotics MATH where MATH, which hold uniformly for MATH in compacta of MATH. To compute the asymptotics REF for MATH with MATH we used REF, and for MATH we used the formula MATH see REF (observe that the right hand side of REF is well defined for MATH with MATH and can be uniquely extended by continuity to MATH). It follows that MATH for MATH and MATH, hence MATH uniquely extends to an isometric isomorphism MATH onto its image for all MATH. Let MATH and MATH be MATH-invariant continuous functions on MATH with compact support. We have MATH for MATH with MATH given by REF and MATH which is a finite sum by the assumptions on MATH. Assume that MATH, then it follows from REF that MATH. Furthermore, it follows from the proof of REF that MATH, where MATH. Hence MATH and MATH. It follows from REF that MATH . This implies that MATH. It follows from REF that REF is valid for all MATH. Furthermore, by continuity arguments, we have MATH for all MATH. Hence, MATH uniquely extends to an isometric isomorphism MATH onto its image for all MATH. A direct computation now shows that MATH for MATH, compare the proof of REF . This implies that MATH, MATH and MATH, which completes the proof of the theorem. |
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