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math/9907126 | The partition into macrocells is determined by the choice of one central hexagon for one macrocell. If one chooses this hexagon uniformly at random, the probability that any given corner in MATH is MATH-inner is propertional to the area of the inner size-MATH wall of a macrocell relative to the overall macrocell's area; this probability is therefore MATH. Thus choosing a random macrocell center gives an expected number of MATH-inner members of MATH equal to MATH. The best macrocell center must give at least as many MATH-inner members of MATH as this expectation. |
math/9907126 | Form a planar graph by placing a point at the center of each macrocell, and connecting pairs of points at the centers of adjacent macrocells. Then by assumption this graph is connected, so we can choose a spanning tree. A curve MATH formed by thickening the edges of this tree and passing around the boundary of the thickened tree has two of the three properties we want: it is contained in the union of MATH and passes through all macrocells in MATH REF . Now, suppose that some path MATH in MATH passes consecutively through three pairwise adjacent macrocells MATH, MATH, and MATH, (for example, at points where the spanning tree edges form a MATH angle), and the intersection of MATH with the middle macrocell MATH has more than one component. Then we can simplify MATH by replacing MATH with a curve that passes directly from MATH to MATH REF . This simplification step maintains the two properties that MATH is in the union of MATH and passes through each macrocell. It is possible for such a simplification step to introduce a crossing, but only in the case that more than one path passes through the same triple of macrocells; to avoid this problem we always choose the innermost path when more than one path passes through the same triple of macrocells. Each simplification step reduces the total number of connected components formed by intersecting MATH with macrocells, so the simplification process must terminate. Once this simplification process has terminated, the components of an intersection of MATH with a macrocell (if that intersection has multiple components) must connect non-adjacent pairs of macrocells, so there can be at most three components per macrocell. |
math/9907126 | Draw the wall using regular hexagons, tilted slightly so that no edge is vertical; this gives a monotone embedding. Any vertical line crosses MATH hexagons, and hence MATH edges. |
math/9907126 | According to REF , we can find a curve MATH contained in an embedding of MATH, and passing through each macrocell between one and three times. Find a monotone embedding of MATH, and form a correspondence with the marked corners of MATH (ordered by the positions along MATH where MATH first intersects each macrocell) and the vertices of MATH (ordered according to the monotone embedding). Then we form a MATH routing problem for each component of an intersection of MATH with a macrocell. If the macrocell does not contain a marked vertex, or if the component is not the first intersection of MATH with the macrocell, the routing problem just consists of pairs of boundary vertices of the macrocell, with each pair placed on the two sides of the macrocell crossed by MATH; the number of pairs is chosen to match the number of edges cut by a vertical slice through the corresponding part of the monotone embedding. However, for the first intersection of MATH with a marked macrocell, we instead form a routing problem in which the pattern of connections between the boundary vertices and the marked inner corner matches the pattern of connections in a vertical slice through the corresponding vertex of the monotone embedding. This set of routing problems involves the placement of at most MATH terminals on any side of any macrocell. These vertices can be placed arbitrarily on that side, as long as they can be connected by disjoint paths along the side to the corresponding terminals of the adjacent macrocell. The union of the at most three routing problems within each macrocell is a MATH routing problem and therefore has a solution. Combining these solutions, and contracting the solution paths in each macrocell, forms the desired minor. |
math/9907126 | One direction is easy: we have seen that the apex graphs do not have the diameter-treewidth property, so no family containing all apex graphs can have the property. In the other direction, we wish to show that if MATH does not have the diameter-treewidth property, then it contains all apex graphs. By REF it will suffice to find a graph in MATH formed by connecting some vertex MATH to all the vertices of a wall of size MATH, for any given MATH. If MATH does not have the diameter-treewidth property, there is some MATH such that MATH contains graphs with diameter MATH and with arbitrarily large treewidth. Let MATH be a graph in MATH with diameter MATH and treewidth MATH for some large MATH and for the function MATH shown to exist in REF . Then MATH contains a wall of size MATH. We choose appropriate values MATH and MATH and partition the wall into MATH macrocells of size MATH. Say a macrocell is good if it is not adjacent to the boundary of the wall. Choose any vertex MATH and find a tree of shortest paths from MATH to each vertex. We say that a macrocell is reached at level MATH of the tree if some vertex of the macrocell is included in that level. Since MATH has diameter MATH,the tree will have height MATH. Since all macrocells are reached level MATH, and the number of macrocells reached at level zero is just one, there must be some intermediate level MATH of the tree for which the number MATH of good macrocells reached is larger by a factor of MATH than the number of good macrocells reached in all previous tree levels combined. Let set MATH be a set of corners of the wall formed by taking, in each good macrocell reached at level MATH, a corner nearest to one of the vertices in that level of the tree. By REF , we can find a new partition into macrocells, and a set of MATH corners that are MATH-central for this partition. Each macrocell in this new partition contains MATH of these corners, so by removing corners that appear in the same macrocell we can mark a set MATH of MATH inner corners of macrocells, at most one corner per macrocell. Note that the number of new macrocells reached at level MATH is still MATH, since each old macrocell reached at that level can only contribute vertices to MATH new macrocells. We then contract levels MATH through MATH of the tree to a single vertex MATH. This gives a minor MATH of MATH in which MATH is connected to inner corners of MATH distinct macrocells, and in which MATH other macrocells are ``damaged" by having a vertex included in the contracted portion of the tree. The adjacencies between damaged regions of the wall form a planar graph with MATH vertices and so MATH faces, and there must therefore be a face of this graph containing MATH members of MATH. Let MATH denote this subset of MATH. Now MATH is part of a connected set of undamaged macrocells of size MATH, so by REF we can find a wall of size MATH as a minor of this set of undamaged macrocells. If MATH and MATH, we can find a wall of size MATH. These conditions can both be assured by letting MATH. Combining this wall with the contracted vertex MATH forms the apex graph minor we were seeking. We can carry out this construction for any MATH, and since by REF every apex graph can be found as a minor of graphs of the form of MATH, all apex graphs are minors of graphs in MATH and are therefore themselves graphs of MATH. |
math/9907126 | Let MATH be a minimal connected subgraph of MATH such that all components of MATH are topological disks. Then there must be at most one such component, for multiple components could be merged by removing from MATH an edge along which two adjacent components are connected; any such merger preserves the disk topology of the components and the connectivity of MATH (since any path through the removed edge can be replaced by a path around the boundary of a component). Thus MATH is a graph bounding a single disk face. By NAME 's formula, if MATH has MATH vertices, it has MATH edges. Let MATH be a spanning tree of MATH; then MATH has MATH edges. Note also that MATH has no degree-one vertices, so each leaf of MATH must be an endpoint of an edge in MATH and there are MATH leaves. Any graph formed by adding MATH edges to a tree with MATH leaves must be a subdivision of a graph with MATH edges and vertices. |
math/9907126 | Embed MATH on a minimal-genus surface MATH, so that all its faces are topological disks. Choose a subgraph MATH as in REF , having the minimum number of edges possible among all subgraphs satisfying the conditions of the lemma, and let MATH be a graph with MATH vertices and edges of which MATH is a subdivision (as described in the lemma). Then, each path in MATH corresponding to an edge in MATH has MATH edges. For, if not, one could find a smaller MATH by replacing part of a long path by the shortest path from its midpoint to the rest of MATH. Therefore, MATH has MATH edges and vertices. Now contract MATH forming a minor MATH of MATH. The result is a planar graph, since MATH can remain embedded in its disk, with the vertex contracted from MATH being connected to MATH by edges that cross the boundary of this disk. The contraction can only reduce the diameter of MATH. Therefore, MATH has treewidth MATH, and a tree decomposition of MATH with treewidth MATH can be formed by adjoining MATH to each clique in a tree decomposition of MATH. |
math/9907128 | In fact, the same homomorphism MATH is open even if considered as a mapping from the free abelian topological group MATH to MATH CITE. |
math/9907128 | Let MATH be topologically generated by the union of a countable sequence of its subgroups MATH, MATH, each of which is topologically isomorphic to the circle group MATH. Fix a translation invariant metric, MATH, generating the topology on MATH. For each MATH choose recursively a number MATH and an element MATH satisfying the following properties for each MATH. CASE: MATH. CASE: The first MATH powers of the product MATH form a MATH-net in the compact subgroup MATH of MATH. CASE: Whenever MATH, the first MATH powers of the element MATH are contained in the MATH-ball of zero having radius MATH. As the base of recursion, choose any element MATH having infinite order and contained in the MATH-neighbourhood of zero formed with respect to the metric MATH. To perform the recursive step, recall the classical NAME Lemma: if MATH are rationally independent real numbers, then the MATH-tuple MATH made up of their images under the quotient homomorphism MATH to the circle group MATH generates an everywhere dense subgroup in the MATH-torus MATH. Now assume that MATH and MATH with REF - REF have been chosen. If the closed subgroup, MATH, generated by the product MATH coincides with all of MATH, we set MATH. Otherwise, MATH forms a proper closed subgroup of the group MATH, and clearly the latter is isomorphic to the topological direct sum MATH. Moreover, the compact abelian NAME group MATH is itself isomorphic to a torus group MATH of a suitable rank MATH, and the image of the topological generator MATH in MATH under such an isomorphism is a MATH-tuple, say MATH, of rationally independent elements of the circle group. Now choose MATH to be an element that is rationally independent of the elements MATH and such that all the powers MATH are contained in the MATH-neighbourhood of zero having radius MATH. It follows from the NAME Lemma that the powers of the product MATH are everywhere dense in the group MATH. Consequently, it is possible to choose MATH as a sufficiently large natural number so that the first MATH powers of MATH form a MATH-net in MATH. The step of recursion is thus accomplished. We claim that the element MATH, which is clearly well-defined since the metric MATH on MATH is complete, forms a topological generator for the group MATH. It is enough to demonstrate that for every number MATH the closure of the cyclic group MATH generated by MATH contains MATH. Let MATH and let MATH by any. Let now MATH be arbitrary. Since MATH, REF implies the existence of a MATH such that the MATH-th power of MATH is at a distance MATH from MATH. The MATH-th power of the remainder of the infinite product, MATH, is at a distance from zero which is less than MATH . (Here we have used REF .) Finally, MATH . Since MATH can be chosen arbitrarily large, the latter inequality means that MATH is the limit of a sequence of suitable powers of MATH, and the proof is finished. |
math/9907128 | Let MATH be an arbitrary pseudometric on MATH as above, and let MATH . It is clearly enough to show that MATH has the required property, that is, if MATH and MATH, then MATH. An arbitrary element, MATH, of MATH is of the form MATH . Let MATH. Then one can assume without loss in generality that in the above expansion all the integer coefficients MATH, and that for different MATH the pairs MATH. Now let MATH be arbitrary. According to our earlier convention, we will identify MATH with the element MATH. One has MATH and the claim follows. |
math/9907128 | Indeed, the image of MATH under the second coordinate projection MATH (which is of course a homomorphism of topological vector spaces) is a discrete subgroup of MATH, formed by all linear combinations of the standard basic elements with integer coefficients. |
math/9907128 | Let MATH be arbitrary. For every MATH, the linear span of MATH contains the element MATH, and for each continuous pseudometric MATH on MATH . Consequently, MATH is in the closed linear span of MATH. Since the set MATH is everywhere dense in MATH, it follows that the closed linear span of MATH contains MATH. Further, each element of the form MATH where MATH, is in the closed linear span of MATH as well. But MATH is an orthonormal basis for MATH. The claim is established. |
math/9907128 | According to REF , the topological group MATH is topologically generated by the union of countably many one-parameter subgroups passing through the elements of the form MATH. Therefore, MATH is topologically generated by the union of images of all such one-parameter subgroups. But those images are tori, and there are countably many of them. |
math/9907128 | Since MATH, the homomorphism MATH is in fact an algebraic isomorphism, and clearly it is continuous. It remains to prove that MATH is open on its image. Let MATH be an arbitrary neighbourhood of zero in MATH. According to REF , there is a neighbourhood of zero MATH contained in MATH with the property MATH, that is, MATH. Consequently, the interior of MATH in MATH is non-empty (it contains the open set MATH in the subspace topology induced from MATH), and the proof is finished. |
math/9907128 | The metrizability of MATH - and therefore of MATH - is quite obvious, and monotheticity of MATH follows from NAME 's Lemma and REF . |
math/9907128 | Denote by MATH the underlying vector space of MATH equipped with the topology MATH. The identity mapping MATH is a continuous linear operator, and as such, it extends over the completions of the two topological vector spaces in a unique way, giving rise to a continuous homomorphism (in general, no longer an algebraic isomorphism) MATH . Choose a countable everywhere dense subset MATH, MATH. Then MATH remains everywhere dense in MATH as well. Now apply our construction to both spaces MATH and MATH. To distinguish between the emerging pairs of objects, we will be using subscripts MATH and MATH, respectively. Thus, MATH, etc. Obviously, the subgroups MATH and MATH coincide as abstract groups, or, more precisely, MATH is a topological group isomorphism. Consequently, the homomorphism MATH factors through MATH to give rise to a continuous homomorphism MATH. Quite evidently, the image of MATH forms an everywhere dense subgroup of MATH. Since the former of the two groups is completely metrizable, NAME 's Lemma coupled with REF imply that MATH is a monothetic group. But the image of a monothetic group under a continuous homomorphism having dense image is again monothetic. |
math/9907128 | Indeed, let MATH, that is, for some monothetic group MATH, one has MATH. Let MATH be a closed subgroup of MATH. Denote by MATH the closure of MATH in MATH; one has MATH. Then it is a standard result, repeatedly used in theory of varieties of topological groups (compare for example, CITE), that the topological factor-group MATH is isomorphic to a topological subgroup of the factor-group MATH in a canonical way. At the same time, the group MATH is clearly monothetic. |
math/9907129 | The relation REF makes it clear that MATH is contained in the center. Conversely, suppose MATH where MATH. Let MATH be maximal with respect to NAME order such that MATH. If MATH then there exists some MATH such that MATH. As MATH for some MATH, the coefficient of MATH is different in MATH and MATH. Hence MATH. But then REF implies MATH is MATH-invariant. |
math/9907129 | As MATH is a free MATH-module of rank MATH, MATH is a free module over MATH of rank MATH. NAME 's version of NAME 's lemma implies that the center of MATH acts by scalars on absolutely irreducible modules, and hence MATH is an irreducible module for a finite dimensional algebra of dimension MATH. |
math/9907129 | Write MATH. Let MATH be an eigenvector for MATH and for MATH. It is enough to show that the eigenvalues of MATH and MATH on MATH differ by a power of MATH. Consider the space spanned by MATH and MATH. This is MATH-stable. If it is two dimensional, then with respect to the basis MATH we have MATH has matrix MATH and MATH has matrix MATH. It follows that MATH has matrix MATH where MATH. By assumption, MATH, so MATH has eigenvalues MATH and we are done. If MATH is a multiple of MATH, then either MATH or MATH. In the first case MATH has eigenvalue MATH, in the second case MATH, and we're done. |
math/9907129 | Let MATH be an exact sequence. Then MATH is a submodule of MATH, if MATH, by NAME reciprocity, and as MATH, MATH is also a submodule of MATH if MATH. As MATH occurs with multiplicity one, either MATH or MATH is zero, and so MATH is irreducible. |
math/9907129 | MATH is the composite of the exact functors of restriction and the functor of generalized eigenspace, which is exact on torsion (and in particular finite dimensional) MATH-modules. |
math/9907129 | If MATH is a MATH-module generated by a MATH-subspace MATH, then MATH is a MATH-module generated by the image of MATH. In particular, if MATH is finitely generated, then MATH is finite dimensional (it is a quotient of the finite dimensional vector space MATH). Now if MATH-mod is generated by a subspace MATH, MATH is generated by MATH and the system MATH are all quotients of a fixed MATH-module MATH, which if MATH is finitely generated is finite dimensional. Hence the inverse system stabilizes in this case. A more careful analysis shows that the system stabilizes for MATH greater than a fixed constant which depends on MATH and not on MATH. This gives the proposition in general, but as we will not use it for non-finitely generated modules, we will omit further details. |
math/9907129 | We prove MATH is left adjoint; the proof for MATH is similar. This is almost, but not quite, a formal result. To see that, observe that if MATH, MATH where the second limit is taken over the system MATH used in the definition of MATH. This equals MATH if MATH is finite dimensional, or more generally MATH-torsion, but not in general. (For example, if MATH and MATH they clearly differ, and indeed MATH is not exact on the category of all MATH-modules.) However, if MATH-mod, then as MATH is finite dimensional, these are equal, and the direct and inverse limits above stabilize after finitely many terms (for MATH, to be crude). Hence if MATH-mod, and MATH-mod, then MATH where we have used, once again, the fact that this limit of Hom's stabilizes after finitely many terms. |
math/9907129 | Let MATH be an indecomposable MATH-module, and MATH a non-zero element in the weight space MATH for some MATH. Then all vectors in MATH are in the generalized eigenspace with eigenvalue MATH, and hence all subquotients of MATH are in MATH. But if MATH, then MATH, and so MATH. (The proof for MATH is similar.) |
math/9907129 | For MATH, MATH acts on MATH with eigenvalues in MATH, so REF is immediate from the definition of MATH. But then REF follows as both MATH and MATH are left adjoint to MATH, and left adjoints are unique when they exist. |
math/9907129 | As MATH are direct summands of the exact functor MATH, they are exact. REF identifies them as the same direct summands. |
math/9907129 | As MATH is semisimple, MATH . |
math/9907129 | We may finish the above proof differently, noticing MATH is still semisimple, to get MATH . It follows that both MATH and this new operator define adjoints to MATH. But the symmetric pairing MATH is non-degenerate, and so these two operators must be equal. |
math/9907129 | By induction on MATH, MATH, and MATH is either MATH or MATH, so MATH is either MATH or MATH. So if MATH, then MATH also, and if MATH, it equals MATH (by the theorem), so MATH, and so MATH. Hence MATH here also. |
math/9907129 | Clearly, REF are equivalent to the existence of a nonzero map of MATH-modules MATH . By NAME reciprocity, this is equivalent to the existence of a nonzero map of MATH-modules MATH . |
math/9907129 | By the NAME formula, MATH admits a filtration whose graded pieces are MATH where MATH runs over representatives for all the cosets MATH except for the coset MATH. Now consider MATH as a module for MATH, and suppose MATH is in its support; that is, MATH. Fix MATH, not the identity double coset MATH. By the shuffle REF , there must be some MATH, MATH, such that the MATH-weight space of MATH on MATH is nonzero, that is, MATH. As MATH, MATH. So we have shown there is some MATH , MATH, with MATH. It follows that MATH, when considered as a module for MATH has support disjoint from MATH. As this is the support of MATH, we see that all subquotients of MATH, considered as a MATH-module, are in different blocks from MATH. The proposition follows. |
math/9907129 | CASE: As MATH is semisimple, we will know it is irreducible if we show MATH equals REF. Note that MATH only if MATH, so MATH is nonzero, and we must show MATH. Write MATH for the kernel MATH . As MATH is exact, we have an exact sequence MATH and by the technical REF MATH. So MATH, where MATH is a quotient of MATH, and MATH is a quotient of MATH. Now, MATH by the technical REF . As MATH is irreducible, either MATH is zero, and MATH, or MATH, and MATH. We have already observed MATH, so MATH is irreducible, nonzero. CASE: As the sequence REF is exact, this also shows that MATH embeds into MATH. Now let MATH be an irreducible subquotient of MATH. Then to show MATH, it suffices to show MATH is zero. But MATH is a subquotient of MATH, and the technical REF gives the result. As MATH it is clear that MATH. CASE: Suppose MATH is irreducible. Then by semisimplicity of MATH and adjunction, we get MATH . So MATH. CASE: To show REF , observe that as we have a surjection MATH we also get a surjection MATH . Given a surjective map MATH, we get an induced surjection on cosocles, MATH, so a surjective map MATH . As MATH is irreducible, this is an isomorphism. |
math/9907129 | It is clearly enough to prove it for MATH-modules, as REF implies the result for cyclotomic NAME algebras. Suppose MATH is an irreducible MATH-module, and write MATH. Then if MATH is semisimple, MATH . So if MATH is a simple summand of MATH, then MATH and REF shows MATH. If MATH is another simple summand of MATH, then MATH, so MATH. It follows that MATH. We have just seen that every irreducible module is of the form MATH, where MATH. It follows that MATH is irreducible, for all irreducible modules MATH. Finally, let MATH be a summand of MATH, where MATH, MATH is irreducible. Then MATH, for some MATH with MATH, as MATH. It follows that MATH, so MATH, whence MATH and MATH. So MATH, as desired. |
math/9907129 | It is enough to prove REF . For, if MATH and MATH are semisimple MATH-modules, then by REF and adjointness MATH and so both MATH and MATH are adjoint to MATH. As the symmetric pairing MATH is non-degenerate, they must be equal. So we prove REF . In fact, we prove a stronger statement. Let MATH be any indecomposable module with cosocle MATH, so MATH. Define MATH, for any MATH. Clearly MATH. We will show that if MATH, MATH is irreducible. Now, the socle filtration of MATH has subquotients isomorphic to MATH. It follows that we can filter MATH with subquotients isomorphic to MATH. Hence MATH is some summand of some copies of MATH (for any filtered module MATH, MATH is a summand of MATH). But MATH . It follows that MATH. Note that this implies if MATH, and MATH, then MATH. Now, if MATH is any simple MATH-module, MATH for some MATH with MATH. Take MATH. Then MATH is irreducible, as desired. |
math/9907129 | We may suppose that MATH, where MATH is an irreducible MATH-module and MATH. By the NAME formula and ``shuffle lemma," if MATH is some module covering MATH we have an exact sequence MATH where MATH has no MATH-eigenspace on MATH, that is, MATH . Hence taking the MATH-eigenspace of MATH, we get MATH and that MATH, where we denote the maximal size of a NAME block of MATH on a module MATH by MATH. We now apply this for MATH. Write, for MATH for the exact sequence defining MATH. We prove REF . Apply MATH to this exact sequence, to get MATH . Now, in MATH, MATH is a multiple of MATH, as this is the unique irreducible module with this central character. Comparing dimensions, we see that MATH, and so in MATH we have MATH . By REF , every subquotient MATH of MATH has MATH, and so the right hand side of the proposition is of the desired form. Finally, as MATH is in MATH, it follows the non-zero terms in this expression are also (after cancelling). We now prove REF . As MATH, REF implies MATH . But by REF and its proof, the MATH-module MATH does occur as a submodule of MATH (in fact, with multiplicity one). It follows that MATH . It remains to show MATH. We show inductively that if MATH in the exact sequence MATH then MATH. in the NAME group of MATH modules. As MATH is the smallest integer MATH for which MATH, and MATH, this will finish us up. But MATH is MATH, which is MATH as it is irreducible and non-zero. Our induction starts with MATH, where it is clear by these remarks. Likewise, if MATH, then if MATH is non-zero it must be a surjection onto MATH, hence the result. We now show that REF imply REF . By REF , the operators of multiplication by MATH are non-zero for MATH, and zero for MATH. Hence they are linearly independent, and MATH. For the reverse inequality, observe that for any module MATH with cosocle MATH as one sees by filtering MATH so it has semisimple quotients and applying REF . Now take MATH. |
math/9907129 | We first show REF are equivalent. Suppose the surjective map MATH is not an isomorphism. We claim that it follows that MATH where we mean equality in the NAME group, and smaller terms means a sum of modules MATH for which MATH. The proof is by induction. We have an exact sequence MATH . By assumption, MATH, so the image of MATH contains a copy of the irreducible module MATH. As MATH, the induction is complete if we show that MATH is an isomorphism implies MATH is also, for MATH. But this is clear - factor MATH, where MATH is the map induced from MATH. But now MATH and so MATH and hence MATH is an isomorphism. Write MATH for the smallest integer MATH for which MATH. We have just shown that for MATH, MATH is a submodule of MATH. It follows that for such MATH, MATH is a submodule of MATH and so MATH. On the other hand, as in the proof of REF, for any module MATH with cosocle MATH, we have MATH . Taking MATH we see REF is equivalent to REF . Now let us show MATH. But this is immediate from NAME reciprocity: any embedding MATH induces a map MATH which when restricted to MATH induces an injection MATH. It follows that MATH. The reverse inequality will be proved in REF. |
math/9907129 | We prove REF ; as REF are similar (or indeed are formal consequences, upon applying the obvious automorphism of MATH). By the classification of MATH-modules, when MATH, MATH is irreducible; hence by transitivity of induction MATH . But the NAME formula implies that MATH occurs with multiplicity one in MATH, hence the useful observation REF shows MATH is irreducible. To prove REF , observe that if MATH is a surjection, then MATH, so as MATH is a quotient of MATH for some MATH with MATH, it is enough to show REF when MATH. We play with the NAME formula again as in REF , so we will be terse. Write MATH. Then the sequence MATH splits, as the shuffle lemma shows the central character of MATH differs from that of MATH. It follows, as in REF, that MATH where MATH is some quotient of MATH, and that MATH the last equality as MATH is irreducible by REF . |
math/9907129 | REF and the first part of REF are immediate from the shuffle lemma. As MATH, where MATH by REF , the last part of REF follows by another application of the shuffle lemma. Finally, to prove REF , it is enough to show MATH. Further, we may again assume MATH, as the cosocle of MATH surjects onto the cosocle of MATH. Now, if MATH is an irreducible quotient as in REF , NAME reciprocity gives a non-zero homomorphism MATH; hence MATH. But MATH by the previous proposition, and so MATH. |
math/9907129 | We have proved everything for MATH; to finish we must only observe that MATH, and so we have the assertions for MATH also. |
math/9907129 | Write MATH . First suppose MATH. Then the surjection MATH induces a surjection, hence isomorphism MATH for all MATH. It follows that MATH. To show MATH, we compute MATH for all MATH and invoke REF . The shuffle lemma shows MATH, hence the definition of MATH and REF shows MATH. Similarly, the shuffle lemma shows that for all MATH, MATH and MATH by REF and the assumption MATH. We clearly have MATH for MATH, and all MATH. Thus another application of REF shows MATH, so MATH. Now suppose MATH. The argument of the last paragraph applies equally well here, so to show MATH we must only show MATH for all MATH. But this is immediate from the shuffle lemma, as in this case MATH is a quotient of MATH. Finally, observe that NAME reciprocity implies that MATH is contained in MATH, so if MATH, then MATH also. It follows that MATH, and so MATH in this case. |
math/9907129 | We merely sketch the necessary modifications in the definitions needed to prove this. Recall that there are REF irreducible representations of MATH with central character MATH. Denote them MATH with MATH, MATH, MATH. The representations MATH and MATH are irreducible, so if we define MATH then the analogue of REF is that each of these operators takes irreducible modules to irreducible modules (or zero). Using the exact sequence MATH, we see that for MATH there is a REF step filtration of MATH where MATH . Arguing as before, one sees that the cosocle of each of these subquotients consists of a sum of terms with fixed MATH: respectively MATH is MATH, MATH, and MATH. Hence MATH, which is a quotient of MATH, is a quotient of precisely one of those three subquotients (for MATH). The rest of the proof is as before. |
math/9907129 | If MATH is an irreducible MATH-module, then MATH CITE MATH is clearly zero unless MATH. So we may assume MATH, where MATH is an irreducible MATH-module, and then MATH . But MATH, by REF . Now suppose that MATH. Then the terms MATH in the above sum have MATH, so none are isomorphic to MATH. It follows that if MATH, MATH unless MATH, and MATH. The lemma is immediate. |
math/9907129 | We may suppose that projective covers of irreducible modules in MATH are linearly independent for MATH, and that MATH. Suppose we have a relation MATH in MATH with not all the MATH equal to zero. Choose a MATH and a simple module MATH such that MATH, and MATH is maximal among terms in this sum. We may choose MATH so that MATH, as for a module MATH, MATH for all MATH implies that MATH, which is absurd for MATH. Now apply MATH to this sum. By the lemma, we get an equality MATH in MATH, where MATH is a sum of terms of the form MATH, with MATH and MATH. In particular, all the terms in the sum are distinct projective modules in MATH. By our inductive assumption these terms are linearly independent, hence MATH. This contradicts our choice of MATH, and shows the NAME form is non-degenerate. It remains to show it is symmetric. Again, induct on MATH. Clearly the form is symmetric on MATH, and MATH where we have used adjunction twice, and the inductive hypothesis. So the form is symmetric on the image of MATH under the operators MATH and MATH. We must merely show that this is everything. This follows from the following two lemmas. |
math/9907129 | Again, we assume the result for MATH, and suppose MATH. Fix MATH, and MATH, and suppose the result is true for all irreducible MATH with MATH. As there are finitely many irreducible modules in MATH, our induction on MATH starts successfully, somewhere. Let MATH be irreducible, and MATH. Apply the above lemma to MATH, to get MATH which by induction is of the desired form. Arguing in this way for each MATH, we see the result is true except perhaps for modules MATH for which MATH for all MATH. Such a module would have MATH, which is absurd. Hence the result is true for all modules. |
math/9907129 | This reduces to checking the result on irreducible modules for MATH (if MATH), MATH (if MATH, and MATH), and MATH (in the remaining case). To see this, observe that as MATH commutes with MATH, and MATH is injective, it is enough to check that the MATH satisfy the NAME relations on MATH. But, as observed in REF, MATH is just the component MATH of MATH and MATH is coassociative. So it is enough to check the relations involving a word of length MATH in the MATH's on MATH. We must now check the NAME relations for the irreducible MATH-modules, MATH. These modules were listed in REF; and the relations follow by examining each module individually. |
math/9907129 | It is enough to show they satisfy the NAME relations as maps from MATH. But the NAME form of REF is non-degenerate, and MATH is adjoint to MATH with respect to this form. As the MATH satisfy the NAME relations, so do the MATH. |
math/9907129 | As before, to check MATH, it suffices to check on simple MATH-modules. Both left and right hand sides are zero on all simple MATH-modules except MATH, where both are MATH. This implies MATH, by an argument similar to the above one. |
math/9907129 | For MATH we have a surjection MATH . Apply MATH. As MATH is exact, and MATH is right exact, we still get a surjection MATH . But by the NAME formula, we have an exact sequence MATH and hence, as MATH is irreducible, an exact sequence MATH for some MATH, if MATH. Hence MATH where we write MATH if for each irreducible MATH, the multiplicity MATH of MATH in MATH is not less than the multiplicity MATH of MATH in MATH. Sum REF over MATH, we get MATH for some MATH. Next we claim that in MATH . Granting this for the moment, let MATH be an irreducible MATH-module with MATH. Then comparing the multiplicity of MATH in REF and in REF, we see that all inequalities in REF must be equalities, and so the multiplicity of MATH in MATH equals the multiplicity of MATH in MATH; so MATH is a multiple of MATH. Furthermore, if MATH then the multiplicity of MATH in MATH and MATH is zero, as the central characters of MATH and MATH differ. So to prove the proposition it remains to show REF. This is immediate from the NAME formula for MATH. A weak form of this may be immediately deduced from the NAME formula for MATH; we omit further details. |
math/9907129 | As every irreducible module is obtained from MATH by a sequence of raising operators MATH, it is clear that REF are equivalent to REF . REF is immediate from the definition of MATH and the description MATH. So we prove REF . We first observe that as MATH if MATH, we have MATH. Further, as MATH if MATH, for such MATH, in agreement with REF . Hence for MATH, the content of the theorem is the assertion MATH . This is immediate from REF . Likewise, if MATH, REF is equivalent to the assertion of the theorem. |
math/9907129 | CASE: We have MATH, where MATH is an irreducible MATH module with MATH. Set MATH, so that in the NAME group MATH is some number of copies of MATH plus terms with strictly smaller MATH. In particular we have that MATH for all MATH that occur in MATH, and MATH is just some copies of MATH. We will show by decreasing induction on MATH that there is a non-zero map MATH . If MATH, MATH is a quotient of MATH so our induction starts. Now suppose MATH exists, and MATH. Consider MATH. By the NAME formula, this has a three step filtration MATH with successive quotients MATH where MATH is the obvious permutation. As MATH, NAME reciprocity gives a copy of MATH in the image of MATH, and so MATH is non-zero. Suppose that MATH is zero when restricted to the MATH-eigenspace of MATH on MATH. As there is no MATH-eigenspace of MATH on MATH, we must have a non-zero homomorphism from MATH to MATH, that is, a non-zero homomorphism MATH . But MATH, for any constituent MATH of MATH. So this is not possible, and it must be that MATH restricts to a non-zero homomorphism on the MATH-eigenspace of MATH on MATH. As MATH has a filtration with subquotients MATH, there must be a non-zero map MATH. We now take MATH and conclude there is a non-zero homomorphism MATH hence MATH is a subquotient of MATH. But MATH is a multiple of MATH, so we have indeed shown that MATH. CASE: Again write MATH. As multiplication is commutative in the bialgebra MATH, MATH equals MATH in the NAME group. Hence MATH is MATH plus terms MATH with MATH. As MATH surjects onto MATH, if MATH it must be that MATH. |
math/9907129 | Let MATH be such that MATH, and suppose MATH. Then REF tells us that MATH means MATH and that REF shows that MATH. Take MATH, where MATH, and define MATH by setting MATH, and setting MATH to be any integer much greater than MATH, when MATH. It then follows from the previous paragraph applied to MATH that for all MATH and as MATH it follows that MATH that is, that MATH. Applying this to MATH we get REF . To see REF , observe that MATH precisely when MATH and that otherwise MATH. But MATH, and so REF follows if we show that MATH if and only if MATH. But MATH, by REF , and MATH always. Similarly MATH, and MATH always. Finally, REF i shows that MATH if and only if MATH; hence REF . Now REF is immediate from REF . |
math/9907139 | If MATH has finite volume then the vectors MATH span MATH and the NAME matrix is indecomposable CITE. So for each MATH, there is a MATH such that MATH. Successively choose indices MATH such that the MATH-th row contains a non-zero entry in the MATH-st column, for MATH. We can ensure that the MATH are distinct. For, if the only non-zero entries of the MATH-th row are those in the columns with indices MATH, throw away MATH and go back to the MATH-st row to rechoose a different column. Eventually, by discarding and moving backwards, we must be able to rechoose, in the MATH-th row, an index different from all the MATH discarded. Otherwise, MATH are orthogonal to the other basis vectors, contradicting indecomposability. In this way we must arrive at a sequence MATH of length MATH. Hence, for any MATH, MATH for some MATH, and MATH with coefficient non-zero. Thus, the vectors MATH span MATH over MATH and hence MATH is MATH-dimensional over MATH. Now, if MATH is a MATH-basis for MATH and MATH, then the system of REF has a unique solution, since the matrix with MATH-th entry MATH is invertible. But the solutions MATH, since MATH for all MATH. Thus MATH is a MATH-basis for MATH. |
math/9907139 | Suppose MATH so that MATH where MATH and we can take the g.c.d. of the entries of MATH to be REF. From MATH we have MATH . Thus MATH and so MATH. Since MATH, MATH is odd. Suppose MATH is unramified in the extension MATH, then either MATH or MATH with MATH and MATH. So MATH so MATH. This is a contradiction. Now suppose that MATH is ramified. Then MATH where MATH and by REF. Expanding as above, but to three terms, gives MATH . This yields the contradiction MATH. |
math/9907139 | If MATH, then either MATH, in which case MATH, or MATH, in which case MATH, hence MATH. Conversely, if MATH, then MATH for all MATH. If MATH, then MATH since all three terms are MATH. A similar argument deals with the MATH. |
math/9907144 | Let MATH and MATH. Construct MATH by applying the operators MATH for all MATH to a chain. Since the order of applying these operators is arbitrary, we may choose to apply first those for which MATH is not a subset of MATH. At this point for every MATH of rank MATH and MATH of rank MATH with MATH, the interval MATH is isomorphic to a chain of rank MATH. Applying the remaining operators MATH leaves the elements of rank at most MATH or of rank at least MATH unchanged, and has the same effect on MATH as applying the operators MATH to a chain of rank MATH. |
math/9907144 | From the definition of MATH and REF , MATH . By REF , the coefficient in MATH of MATH is MATH which is an empty sum if MATH is not contained in MATH, zero if MATH is properly contained in MATH, and MATH if MATH. This gives the recursion of the lemma. |
math/9907144 | Using REF we can show by induction on MATH that for every MATH, MATH is zero unless MATH is the union of some intervals of MATH. In particular, if MATH is an even system of intervals, then MATH whenever MATH is not an even set. The same observation holds for every interval MATH as well, since by REF MATH is isomorphic to MATH for some MATH and some even system of intervals MATH. Therefore the conditions of REF are satisfied by MATH for every MATH, if MATH is an even system of intervals. Now assume MATH is a system of intervals that is not even. First consider the case where MATH contains an interval MATH with MATH even (hence MATH is odd). Let MATH. For MATH nonempty, MATH, so MATH . So MATH for MATH. Fix MATH, and choose MATH and MATH in MATH with MATH, MATH, and MATH. Then by REF , MATH, with MATH odd. So MATH is not half-Eulerian. Now suppose MATH contains only even intervals, but some two intervals have an odd overlap. Let MATH and MATH, where MATH and MATH and MATH are odd, but MATH is even. Then MATH is also even. We show that we may assume no other interval of MATH is in the union MATH. Suppose MATH is another interval of MATH with MATH (and MATH is odd). Since MATH is an antichain, MATH. If MATH is even, then MATH, which is odd, because it is the sum of three odds and two evens. If MATH is odd, then MATH, which is odd because it is the sum of three odds. Thus, if two intervals of MATH have odd intersection and their union contains a third interval of MATH, then two intervals of MATH with smaller union have odd intersection. So we may assume MATH and MATH have odd intersection, and their union MATH contains no other interval of MATH. Let MATH. Then MATH . So MATH . So MATH for MATH. Fix MATH, and choose MATH and MATH in MATH with MATH, MATH, and MATH. Then by REF , MATH, with MATH odd. So MATH is not half-Eulerian. |
math/9907144 | We define a one-to-one correspondence between even interval systems on MATH and sequences MATH satisfying MATH if MATH is even and MATH if MATH is odd. Clearly there are MATH such sequences. For MATH an even interval system, define MATH, where MATH if MATH is an endpoint of an interval of MATH, and MATH otherwise. (Note that for an even interval system, no number can be an endpoint of more than one interval.) For MATH an even interval system, summing MATH over the endpoints of intervals gives REF. So MATH . On the other hand, given a sequence MATH satisfying MATH if MATH is even and MATH if MATH is odd, construct an even interval system as follows. Let MATH be the sequence of indices MATH for which MATH. Then MATH, so MATH. Thus the sequence of MATH's contains the same number of even numbers as odd. Construct an interval system MATH REF recursively as follows. Let MATH and let MATH where MATH is the least index such that MATH and MATH are of opposite parity. Then MATH, where MATH is the interval system associated with MATH with MATH removed. Clearly MATH is of even length. If MATH for some interval MATH of MATH, then MATH, so by the choice of MATH, MATH has the same parity as MATH. Thus MATH is of even length. Furthermore, MATH and MATH are of the same parity, since MATH and MATH are, so again by the choice of MATH, MATH. So the interval MATH is not contained in the interval MATH. The interval system MATH, is even, so by induction MATH is an even interval system. These constructions are inverses, giving the desired bijection. |
math/9907144 | Fix a MATH-word MATH with MATH's and weight MATH. Write the elements of MATH in increasing order as MATH, MATH, MATH, MATH, , MATH, MATH, and let MATH be the interval system MATH. Let MATH. Rewrite the MATH-polynomial MATH as a MATH-polynomial. Recall from REF that MATH, MATH, and MATH, so MATH. Thus, MATH is rewritten as a sum of MATH terms. Each is the result of replacing some subset of the MATH's by MATH, and the rest by MATH; the coefficient is MATH, depending on whether the number of MATH's replaced by MATH is even or odd. Thus MATH where MATH, with MATH if MATH and the remaining MATH's are MATH. By the MATH-vector version of REF , this is precisely the MATH-index of MATH. |
math/9907144 | The idea is that since no two elements of MATH are in the same gap of MATH, elements with ranks in MATH can be inserted independently in chains with rank set MATH. For MATH a MATH-chain (that is, a chain with rank set MATH) and MATH, let MATH be the number of rank MATH elements MATH such that MATH is a chain of MATH. Since every interval of an Eulerian poset is Eulerian, MATH for all MATH and MATH. So MATH . So the flag vector inequality is proved. The second inequality is simply the translation into MATH-vector form. |
math/9907144 | First order the rank MATH elements of MATH in the following way. Choose any order, MATH, MATH, , MATH for the components of the NAME diagram of the rank-selected poset MATH. For each rank MATH element MATH of MATH, identify the component containing MATH by MATH. Order the rank MATH elements of MATH in any way consistent with the ordering of components. That is, choose an order MATH, MATH, , MATH such that MATH implies MATH. A rank MATH element MATH belongs to MATH if MATH is the least index such that MATH in MATH. Write MATH for the number of rank MATH elements belonging to MATH, and MATH for the number of rank MATH elements MATH such that MATH, but MATH does not belong to MATH. Similarly, a rank MATH element MATH belongs to MATH if MATH is the least index such that MATH in MATH. Write MATH for the number of rank MATH elements belonging to MATH, and MATH for the number of rank MATH elements MATH such that MATH, but MATH does not belong to MATH. Note that MATH and MATH, since MATH is Eulerian. A flag MATH belongs to MATH if MATH and MATH is the least index such that either MATH or MATH. Let MATH. Let MATH be the contribution to MATH by elements and flags belonging to MATH. Thus, MATH . If MATH, then MATH. If MATH, then MATH. In all other cases it is easy to check that MATH. Suppose that the rank MATH elements in component MATH are MATH, MATH, , MATH. Then MATH, so MATH. Furthermore, MATH, because any rank MATH element MATH related to MATH must also be related to at least one other rank MATH element, and it is in the same component. That rank MATH element has index less than MATH, so MATH does not belong to MATH. This in turn implies MATH, so MATH. For all MATH, MATH, either MATH or MATH, by the connectivity of the component, so MATH. Thus MATH. This is true for each component MATH, so MATH. |
math/9907144 | We work in the closed cone of MATH-vectors of Eulerian posets. The cone of MATH-vectors of Eulerian posets is contained in the subspace of MATH determined by the equations MATH for MATH not an even set. To prove that the MATH-vector of MATH generates an extreme ray, we show that it lies on linearly independent supporting hyperplanes, one for each nonempty even set MATH in MATH. Fix an even interval system MATH. For each nonempty even set MATH, we find a set MATH such that MATH and MATH satisfy the hypothesis of REF and MATH. CASE: Suppose MATH is the union of some intervals in MATH. Let MATH, MATH, , MATH be all the intervals of MATH contained in MATH. Set MATH. Then for each subset MATH, the corresponding union of intervals contributes MATH to MATH, for MATH. Thus MATH. CASE: If MATH is not the union of some intervals in MATH, let MATH be the union of all those intervals of MATH contained in MATH. Choose MATH, and set MATH. For MATH, MATH unless MATH. But if MATH then MATH cannot be in MATH. So MATH. Now MATH determines a supporting hyperplane of the closed cone of MATH-vectors of Eulerian posets, because the inequality of REF is valid, and the poset MATH lies on the hyperplane. The hyperplane equations each involve a distinct maximal set MATH, which is even, so they are linearly independent on the subspace determined by the equations MATH for MATH not an even set. So the doubled limit poset MATH is on an extreme ray of the cone. |
math/9907144 | The dimension of the cone MATH equals the number of even subsets (a NAME number). So it suffices to show that the vectors MATH are linearly independent. To see this, note that for every set MATH not contained in MATH, MATH. By the disjointness of the intervals in MATH, there is a unique way to write MATH as the union of intervals in MATH. So by REF , MATH. Thus, MATH is the unique maximal set MATH for which MATH. So the MATH-vectors of the posets MATH, as MATH ranges over sets different from MATH, are linearly independent. |
math/9907144 | Apply the Facet REF with MATH. For a nonempty even set MATH, the interval system MATH of MATH is nonempty, so MATH. |
math/9907144 | If MATH, then conditions REF force MATH (or MATH if MATH). The resulting inequality, MATH, gives a facet, as shown in REF . Now assume that MATH. REF is to prove that REF holds for all Eulerian posets. Note that MATH is a nonempty collection of intervals of length two. From each such interval choose one endpoint adjacent to an element of MATH. Let MATH be the set of these chosen elements. The Inequality REF applies to these MATH and MATH because each interval of MATH contains at most one interval of MATH, and hence at most one element of MATH. The resulting inequality is MATH. Now MATH for all MATH if MATH contains an odd interval. So we can restrict the sum to even sets MATH. Since MATH must be contained in MATH, such a MATH must contain the intervals of MATH. Thus, MATH. REF is to prove that if MATH is an interval of cardinality at least REF and MATH contains an element MATH not in MATH, then MATH contains an element adjacent to an interval of MATH. If an interval from MATH ends at MATH, then either MATH or MATH by (MATH) (since MATH). Similarly, if an interval from MATH begins at MATH, then either MATH or MATH. So assume no interval from MATH begins at MATH or ends at MATH. The hypothesis of the theorem states that every interval from MATH has cardinality at most three. Thus the interval MATH belongs to MATH. Hence MATH and MATH. If MATH then MATH, condition REF applied to MATH yields MATH, and MATH is adjacent to MATH. The case when MATH is dealt with similarly. Finally, if MATH and MATH are both endpoints of intervals from MATH, then, since MATH, condition REF applied to MATH and condition REF applied to MATH yield MATH and MATH. Either MATH or MATH belongs to MATH and each of them is adjacent to an element of MATH. Recall that for MATH an even interval system, the vector MATH is in the closed cone of MATH-vectors of half-Eulerian posets. REF is to show that for each even set MATH, there exists an even interval system MATH with MATH such that MATH. Let MATH be an even set not equal to MATH. If MATH, then for every MATH containing MATH, MATH. Now suppose MATH, but MATH. Let MATH be an interval of MATH such that MATH. Then MATH contains an element adjacent to an interval of MATH. Since MATH and MATH is a maximal interval in MATH, MATH. Thus every union of intervals of MATH containing MATH must contain MATH and thus an element not in MATH. So MATH, because all terms are zero. Finally, suppose MATH and MATH. Let MATH be the interval system of MATH consisting only of intervals of length REF. Then every interval of MATH is in MATH. This is because every interval of MATH is of length REF, with at least one of its endpoints adjacent to an element not in MATH. So MATH, since MATH implies MATH. By the Facet REF , the inequality MATH gives a facet of MATH. Now we show that under the added condition MATH for every MATH, the facets obtained are distinct. Note that two MATH pairs can give the same inequality only if they have the same MATH, because MATH is included in the linear form for MATH, and MATH is the minimal (by set inclusion) set for which MATH is in the form. Now for fixed MATH, we show that MATH and MATH give distinct linear inequalities when MATH. Since the sets MATH and MATH are different, there is an interval MATH such that MATH occurs in exactly one of MATH or MATH. Let MATH be a maximal interval with this property. Without loss of generality assume MATH. Then MATH is contained in no interval of MATH. CASE: MATH. Then for every MATH, MATH, the term MATH occurs in the inequality for MATH. At least one of these terms does not occur in the inequality for MATH, because MATH. CASE: MATH. Since MATH and MATH, MATH. By the strengthened hypothesis on MATH, MATH. Then for every MATH, MATH, the term MATH occurs in the inequality for MATH. At least one of these terms does not occur in the inequality for MATH, because MATH. CASE: MATH. The proof is similar to REF . Thus, with the condition MATH for every MATH, the facets given by the theorem are all distinct. |
math/9907148 | Proceeding according to the genus, suppose MATH has signature MATH with MATH. Map MATH onto MATH, free of rank two, by sending MATH, MATH, and all the other generators to the identity. Since MATH is MATH-generated for MATH (see CITE), we are done. A group of genus one with MATH can be surjected onto MATH for MATH, by comment REF above. The map MATH sending all generators to the permutation MATH has kernel isomorphic to MATH by comment REF above, hence the result holds for groups of genus one with MATH. For groups of genus one with no periods, hence signature MATH for MATH, we may surject onto MATH. But this is easily seen to be free of rank two, so the result holds here also. A group of genus zero with no periods must, by REF , have at least three parabolic generators, and hence surject MATH. But this is free of rank two also. With a single period we have MATH, and the group surjects MATH, the free product of MATH and MATH. This surjects MATH, which in turn surjects any Fuchsian triangle group of the form MATH. With two periods and one parabolic, we have MATH, where MATH, so we can surject any Fuchsian triangle group like MATH. A group with more parabolics, MATH for MATH, surjects MATH done above. Finally, MATH, MATH, surjects either MATH or MATH for MATH. NAME the former onto a Fuchsian MATH. The latter has already been handled. |
math/9907148 | The hyperbolic triangle group MATH surjects MATH for MATH and MATH some prime divisors of MATH and MATH. If MATH is Fuchsian, we have by REF that MATH. If MATH and MATH are distinct, we have a group listed in REF. If MATH, the map MATH that sends the generators of orders MATH and MATH to the permutation MATH and the generator of order MATH to the identity has kernel MATH. We have MATH, hence MATH, and the theorem holds for MATH if it holds for MATH, a group listed in REF. If MATH isn't Fuchsian, it must be, after a possible reordering, one of MATH for MATH, MATH or MATH. The first gives that MATH must have the form MATH, for MATH and MATH. If MATH or MATH then MATH, as MATH is spherical and MATH . Euclidean, so the group surjects MATH or MATH, both of which are listed in the lemma. For MATH, MATH surjects MATH. This in turn surjects MATH, MATH prime, and we have a group listed in REF unless MATH or MATH. In the first case, MATH, so MATH surjects MATH. In the second, MATH, and the group surjects MATH when MATH, or MATH otherwise. The cases MATH or MATH are entirely similar. This accounts for the MATH . Fuchsian groups, and the case of a general triangle groups is much the same. Similarly for the groups with four or five elliptic generators - either they can be surjected directly onto triangle groups or eliminated from consideration using comment REF at the beginning of the section. The only exceptions are those listed in the lemma. Finally, a group with six or more elliptic generators can always be surjected directly onto a Fuchsian group with five. No doubt the reader can fill in the details. |
math/9907148 | Suppose on the contrary that MATH is imprimitive with block system MATH. For MATH, let MATH be the permutation induced by MATH on MATH, and MATH the group generated by the MATH. The map MATH is an epimorphism from MATH onto MATH, and MATH acts transitively on MATH. All blocks MATH thus have the same size, say MATH. If MATH is in MATH, the support of MATH, then MATH and its image under MATH are distinct blocks, and so MATH is contained in MATH. Taking the union of all the blocks in MATH thus gives MATH . Now MATH has order MATH a prime, and MATH is a homomorphic image of MATH. Thus, if MATH, then MATH has order MATH, and so MATH. Since MATH is non-trivial, we have MATH, and hence by REF , MATH. This contradicts the fact that MATH is a MATH-cycle, so we must have MATH. This means that MATH fixes every block, or equivalently, any point and its image under MATH lie in the same block. But MATH is a single cycle, so there is a block MATH with MATH. By the condition stated in the Lemma, MATH and its image under MATH intersect for all MATH, so are equal. Since the MATH generate MATH, the whole group must fix MATH, and by transitivity, MATH, so there is just one block. This is the desired contradiction. |
math/9907148 | The underlying graph of MATH is clearly a MATH-graph, so it remains to show that all faces have boundary labels of the required form. If the boundary of a face does not contain a MATH-edge with initial vertex one of the MATH or MATH, then all edges are contained in a single MATH-diagram MATH, and we are done. Otherwise, we obtain the boundary label for the face by starting at a MATH or MATH and traversing a path with label some power of MATH or some power of MATH, until it closes (which it does by repeating an arc). The path obtained by traversing just MATH-edges passes through the vertices MATH or MATH, before closing with label MATH, so such faces are as they should be. Observe that before composition, the path starting at MATH with label some power of MATH arrived at vertex MATH after MATH directed edges, proceeded to traverse the MATH-loop at MATH and then a MATH-edge. After composition, the path from MATH with such a label arrives instead at MATH after MATH directed edges, traverses the new MATH-edge to MATH, and is then identical with the path before composition. So the boundary label behaves as if the composition never happened, and is thus of the required form. The number of vertices is obvious. |
math/9907148 | Only cycles in MATH that pass through handle points are affected by the composition. If MATH and MATH lie in such a cycle, then in MATH the cycle is identical, except that MATH is replaced by MATH. |
math/9907148 | Let MATH be distinct primes not dividing MATH and MATH. For MATH and MATH arbitrary non-negative integers we construct a sequence of REF as follows: for the MATH-th step, if either MATH, take MATH, otherwise, MATH. At step MATH, MATH, take MATH identical copies of MATH and let MATH be the composition, MATH . In particular, the two handles on each MATH allow us to perform the composition, which is a MATH-diagram by REF . Observe that MATH has at least two handles. At step MATH, MATH, take MATH identical copies of MATH and let MATH be a composite diagram of the form REF but with MATH copies of MATH instead of MATH copies of MATH. Finally, at step MATH, if MATH, let MATH be as in the previous step. Otherwise, take a diagram of the form REF but replace one of the MATH's by a MATH (using its sole handle). A quick sketch may help the reader to see what is going on. Now MATH, and since MATH and MATH are relatively prime, so too are MATH and MATH. By choosing MATH and MATH suitably, MATH can thus be made to equal any integer greater than MATH. So, if MATH is the homomorphism arising from MATH, we have permutation representations of MATH for all but finitely many degrees. By REF the permutation MATH contains the MATH-cycle MATH and no other cycles of length divisible by MATH, so some power of MATH is just MATH. NAME, REF give MATH or MATH, but the generators of MATH have odd order, so in fact MATH. |
math/9907149 | Let MATH and MATH be isometries. Then we have MATH and MATH . Since MATH is an isometry we can compute MATH which gives the desired REF . Here we have used that MATH by CITE. Since MATH we can compute by virtue of naturality (compare CITE) MATH which gives the desired REF . |
math/9907149 | Put MATH for MATH and MATH. This defines square matrices MATH and we have MATH where the MATH's are the fusion matrices of MATH within MATH. With these, the matrices MATH therefore share the simultaneous eigenvector MATH, defined by entries MATH, MATH, with respective eigenvalues MATH. Note that the sum matrix MATH is irreducible since each MATH with MATH is a subsector of some MATH by definition. Now define another vector MATH with entries MATH, MATH. Note that all entries are positive. We now compute MATH that is, MATH. Hence MATH is another eigenvector of MATH with the same eigenvalue MATH. By uniqueness of the NAME eigenvector it follows MATH for all MATH with some number MATH. We can determine this number in two different ways. We first find that now MATH. By computing the dimension we obtain MATH, establishing REF . On the other hand we can compare the norms: For MATH we have MATH. For MATH we compute MATH hence MATH whereas MATH. But note that MATH . We have found MATH and this proves the proposition. |
math/9907149 | The implication MATH follows again from MATH arising from MATH. We next show the implication MATH . : From REF we here obtain MATH. Because we have in general MATH we here find MATH. Consequently, MATH in the MATH fusion rule algebra. Assume for contradiction that some MATH is reducible. Then MATH if MATH is a subsector. But MATH appears on the left hand side with a coefficient larger or equal MATH whereas with coefficient MATH on the right hand side which cannot be true. Hence all MATH's are irreducible and as MATH, they must also be distinct. Therefore MATH, and consequently MATH must be a permutation matrix: MATH with MATH as MATH. Moreover, by virtue of the homomorphism property of MATH-induction we have two isomorphisms MATH from the MATH into the MATH fusion rule algebra and consequently MATH defines an automorphism of the MATH fusion rules. Finally, the implication MATH is trivial. |
math/9907149 | By CITE we have MATH hence MATH . Application of MATH now yields the claim since MATH by CITE . |
math/9907149 | Using chiral locality, it was proven in CITE that MATH whenever MATH is a subsector of some MATH. Hence MATH for MATH. |
math/9907149 | (Similar to the proof of CITE.) We only show it for MATH; the proof for MATH is analogous. 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. |
math/9907149 | (Similar to the proof of CITE.) It is clear that we obtain a scalar which is zero unless MATH. To compute the scalar, we put MATH and then we can close the wire MATH on the left hand side, what has to be compensated by a factor MATH. We can now open the wire MATH on the right and close it on the left, and this way we can pull out the wire MATH, yielding a closed loop. Hence the summation over MATH gives the global index, and the resulting picture is regularly isotopic to REF . |
math/9907149 | We compute the sum MATH graphically. The proof for MATH is analogous. This sum is given by the left hand side of REF . Using the expansion of the identity (compare CITE) for the parallel wires MATH on the top and MATH on the bottom we obtain the right hand side of REF . Using such an expansion now the other way round for the summation over MATH we arrive at the left hand side of REF . The crucial point is now the observation that left and right part of this wire diagram are only connected by wires MATH and MATH. Let us start again with the original picture, namely the left hand side of REF , and make an expansion for the open ending wires MATH and MATH on the left side with a summation over wires MATH. Then it follows that only the wires with MATH contribute because MATH is always zero unless MATH. This establishes equality with the right hand side of REF . The wire MATH can now be pulled in and application of the naturality move of REF for the relative braiding yields the left hand side of REF . Then, using the graphical identity of REF gives us the right hand side of REF , as only the wire MATH survives in the sum of the chiral horizontal projector. The two ``bulbs" yield just a scalar factor MATH, but due to the summation over the fusion channels MATH it appears with multiplicity MATH. Hence the total prefactor is calculated as MATH and this is the prefactor of MATH. |
math/9907149 | Starting with REF we can slide around the, say, left trivalent vertex of the wire MATH to obtain the left hand side of REF . Using now the relative braiding naturality move of REF and turning around the small arcs, giving a factor MATH, yields the right hand side of REF . We now see that the summation over the wire MATH is just an expansion of the identity which can be replaced by parallel wires MATH and MATH (compare CITE). Hence we obtain a closed loop MATH which is just another factor MATH, and we are left with the original diagram for MATH, together with a prefactor MATH. |
math/9907149 | Using REF we compute MATH . On the other hand we obtain by expanding the vectors MATH in basis vectors MATH hence MATH. |
math/9907149 | We show MATH; the MATH case is analogous. The dimensions MATH are counted as MATH . Using now the graphical representation of the scalar product in REF , then we obtain with the normalization convention for the small semicircular wires exactly the wire diagram for MATH, compare REF . |
math/9907149 | We prove MATH for MATH. The proof of MATH for MATH is analogous. First we can turn around the small arcs at the trivalent vertices of the wire MATH which gives us a factor MATH. The we use the expansion of the identity (compare CITE) for the parallel wires MATH and MATH. This we way we obtain the left hand side of REF . Now let us look at the part of the 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. Next we look at the part above the dotted line on the right hand side of REF . In the same NAME annulus, this can be similarly read as MATH where the sum runs over another orthonormal basis of isometries MATH. Since such bases are related by a unitary matrix (``unitarity of MATH-symbols"), we conclude that both diagrams represent the same vector in MATH. Now turning around the small arcs at the trivalent vertices of the wire MATH and using the expansion the identity in the reverse way leads us to the left hand side of REF . Then we look at the part of the picture inside the dotted box. In a suitable NAME annulus, this can be read as an intertwiner in MATH. Since any element in this space can be expanded in the basis basis given in the dotted box on the right hand side of REF , we conclude that MATH is in fact a linear combination of MATH's, hence it is in MATH. |
math/9907149 | We prove the representation property of MATH; the proof for MATH is analogous. For MATH, the vector MATH is given graphically by the left hand side of REF . Next we use the expansion of the identity (compare CITE) for the parallel wires MATH and MATH on, say, the left hand side of the crossings with the wire MATH. Note that only MATH can contribute because MATH otherwise. Application of the braiding fusion relation for the relative braiding, REF , yields the right hand side of REF . Using expansions of the identity also for the parallel pieces of the wires MATH and MATH on the left and on the right, we obtain a picture where the bottom part coincides with the wire diagram in CITE, up to the crossing with the wire MATH. In fact we can use the same argument (``unitarity of MATH-symbols") as in the proof of CITE to obtain the desired result MATH . As the prefactors coincide with those in the decomposition of the vertical product MATH into MATH's, the claim is proven. |
math/9907149 | From Lemma CITE we obtain MATH. The left hand side is displayed graphically by the left hand side of REF . We can ``pull in" the wire MATH since it admits relative braiding with both MATH and MATH, and this way we obtain the right hand side of REF . We can use the expansion of the identity for the parallel wires MATH on the top and bottom (compare CITE) to obtain the left hand side of REF . Here only ambichiral morphisms MATH contribute in the corresponding sums over MATH since they appear between MATH and MATH. Application of the naturality moves of REF for the relative braiding yields the right hand side of REF . Now we see that intertwiners in MATH appear so that we first obtain a factor MATH. Then we take the scalar part of the loop separately to obtain REF , where we need a compensating factor MATH. By using the (MATH and MATH version of the) graphical identity of REF we obtain the left hand side of REF . Here we used the fact that only the wire MATH survives the summations over MATH of the chiral horizontal projectors. The ``bulbs" give just inner products of basis isometries. Due to the summation over internal fusion channels we obtain therefore a multiplicity MATH with a closed wire MATH, evaluated as MATH. Thus we are left with the right hand side of REF . Note that MATH. Now the MATH part of the right hand side of REF must be equal to the MATH part of MATH. Sandwiching this with basis (co-) isometries yields the identity displayed in REF . This is the orthogonality relation showing that the braiding on the ambichiral system is non-degenerate (compare CITE). Consequently the number MATH must be MATH, the ambichiral global index. |
math/9907149 | Let MATH denote the MATH part of MATH and similarly MATH the MATH part of MATH. Then the MATH part MATH of MATH can be written as MATH . Thus each component of MATH is obviously linear in the components of the vectors in MATH and MATH, proving bi-linearity. It remains to be shown that MATH it is in MATH. But this is clear since any element of the form given in CITE can be horizontally ``pulled through". As such elements span the whole double triangle algebra, the claim is proven. |
math/9907149 | Using the expansion of the identity (compare CITE) for the parallel wires MATH and MATH on the top yields the left hand side of REF . We then slide around the trivalent vertices of the wire MATH along the wire MATH so that they almost meet at the bottom of the picture. Turning around their small arcs yields a factor MATH, and we can then see that the summation over MATH is just the expansion of the identity (compare CITE) which gives us to parallel wires MATH and MATH. This way we arrive at the right hand side of REF . Then we apply the expansion of the identity four times: First twice for the parallel wires MATH and MATH on the bottom, yielding expansions over MATH and MATH. Next we expand the parallel wires MATH and MATH in the middle lower part of the picture, resulting in a summation over a wire MATH. Finally we expand the parallel wires MATH and MATH in the center of the picture, yielding a summation over a wire MATH. This gives us REF . Now we can pull the circle MATH around the middle expansion MATH, just by virtue of the IBFE moves as well as the NAME relation for thin wires. Due to the prefactor MATH, the summation over MATH yields exactly the orthogonality relation for a non-degenerate braiding (compare CITE), the ``killing ring". Therefore we obtain zero unless MATH, and our picture becomes disconnected yielding two intertwiners in MATH and MATH. Hence we obtain a factor MATH, and the whole diagram represents a scalar. To compute the scalar, we can proceed exactly as in the proof of CITE: We go back to the original picture on the left hand side of REF and put now MATH and MATH. Then we close the wires MATH and MATH on the right which has to be compensated by a factor MATH. Next we open the wire MATH on the left and close it also on the right. Then the MATH loop can be pulled out and the summation over MATH gives the global index MATH; we are left with the right hand side of REF . |
math/9907149 | We show MATH. The proof of MATH is analogous. It follows from REF that MATH. Therefore it suffices to show that MATH. Such an element is given graphically in REF . If we multiply horizontally with some MATH either from the left or from the right, then the resulting picture contains a part which corresponds to an intertwiner in MATH or MATH, respectively. Hence this is zero unless MATH. But MATH is spanned by elements MATH, MATH, and MATH is the subspace spanned by those with MATH. As the MATH's are horizontal projections, the claim follows. |
math/9907149 | We only show the first relation; the proof for the second one is analogous. It suffices to show the relation for vectors MATH. Then the vertical product MATH is given graphically by the left hand side of REF . Since MATH admits relative braiding with MATH, we can slide around the right trivalent vertex of the wire MATH and apply the naturality move of REF for the relative braiding to obtain the right hand side of REF . In the lower left corner we now recognize the vector MATH of REF , hence the whole diagram represents the vector MATH yielding the statement. |
math/9907149 | All we have to show is the completeness relation REF ; the rest is clear since then each MATH, MATH can be expanded in the chiral matrix units. We have MATH and this is given graphically by the left hand side of REF . Looking at the middle part we observe that we obtain a factor MATH, and therefore we only have a summation over MATH. Then the middle bulb gives just the inner product of basis isometries, so that only one summation over internal fusion channels remains and we are left with the right hand side of REF . But now we obtain a factor MATH and this yields exactly MATH by virtue of the non-degeneracy of the ambichiral braiding, REF . |
math/9907149 | It suffices to show the relation using elements given in REF instead of matrix units MATH. The product MATH is given graphically by the left hand side of REF . Here we have used the expansion of the identity to replace parallel wires MATH and MATH by summations over wires MATH and MATH. By virtue of the unitarity of braiding operators, the IBFE symmetries and the NAME relation for thin wires, the wire MATH can now be pulled over the trivalent vertices and crossings to obtain the right hand side of REF . Here we have already resolved the summations over MATH back to parallel wires MATH and MATH, respectively. Then we slide the trivalent vertices of the wire MATH along the wire MATH so that we obtain the left hand side of REF . Next we turn around the small arcs at the trivalent vertices of the wire MATH, yielding a factor MATH, so that the summation over MATH is just identified as another expansion of the identity. Thus we arrive at the right hand side of REF . The circle MATH around the wire MATH is evaluated as the statistics character MATH (compare CITE). Therefore the resulting diagram represents MATH . The proof for MATH is analogous. Finally we consider MATH for MATH. We proceed graphically as in the proof of REF . But now we can slide around the trivalent vertices of the wire MATH and apply the naturality moves of REF on both sides as MATH is ambichiral. Therefore we obtain REF . Then the small arcs of the trivalent vertices of the wire MATH can again be turned around so that we obtain a factor MATH and that the summation over MATH yields just the expansion of the identity leaving us with parallel wires MATH and MATH. We conclude that the resulting diagram represents MATH completing the proof. |
math/9907149 | From the decomposition of the chiral centers in REF it follows that the (left) regular representations MATH of MATH decompose into irreducibles as MATH. It follows similarly from CITE that the (left) regular representation MATH of MATH decomposes into irreducibles as MATH. Representations of the corresponding fusion rule algebras of MATH sectors are obtained by composition with the isomorphisms MATH mapping the MATH fusion rule algebra to MATH. It was established in CITE that the representation MATH of the full MATH fusion rule algebra obtained by left action multiplication on the MATH sectors decomposes into irreducibles as MATH. The claim follows now since MATH fulfills MATH by definition (compare CITE) and MATH by the identification theorem CITE. |
math/9907149 | First note that the subfactor MATH arising from the loop group construction for MATH in CITE is isomorphic to MATH, where MATH is a hyperfinite MATH factor, MATH is the NAME subfactor CITE with principal graph MATH, and MATH is an injective MATH factor, by CITE. This shows that the subfactor MATH for MATH is isomorphic to MATH, where MATH is the NAME tower of MATH and MATH is a sum of three minimal projections in MATH corresponding to MATH. It is thus enough to prove that the subfactor MATH is a basic construction of some subfactor. We recall a construction in CITE. Let MATH be one of the NAME diagrams of type A, D, E. Let MATH be an abelian NAME algebra MATH and MATH be a finite dimensional NAME algebra containing MATH such that the NAME diagram for MATH is MATH. Using the unique normalized NAME trace on MATH, we repeat basic constructions to get a tower MATH with the NAME projections MATH. Let MATH be the NAME of MATH with respect to the trace and MATH its NAME subalgebra generated by MATH. We have MATH by NAME 's lemma. For a projection MATH, we have a subfactor MATH, which is called a NAME (GHJ) subfactor. Let MATH be MATH and MATH be the projection corresponding to the vertex of MATH with minimum NAME eigenvector entry. We study the subfactor MATH in this setting. Set MATH, MATH. The sequence MATH is a periodic sequence of commuting squares of period REF in the sense of CITE. For a sufficiently large MATH, we can make a basic construction MATH so that MATH is also a basic construction. We can extend the definition of MATH to small MATH so that the sequences MATH is a periodic sequence of commuting squares of period REF. For a sufficiently large MATH, the graph of the NAME diagram for MATH stays the same and the graph for MATH is its reflection. This graph can be computed as in REF in an elementary way (see for example, CITE, CITE), so we also have the graph for MATH, and we see that MATH is MATH and the three minimal projections in MATH correspond to the REFth, REFth, and REFth vertices of MATH. (The graph in REF is actually the principal graph of MATH by CITE, but this is not important here.) Then we see that the NAME diagram for the sequence MATH starts with these three vertices and we have the graph MATH or a part of it as the NAME diagram at each step, as in REF . Each algebra MATH is generated by the NAME projections of the sequence MATH. Similarly, if we choose MATH as MATH and MATH be the projection corresponding to the first vertex of MATH, we get a periodic sequence MATH of commuting squares. (Note that we start the numbering of the vertices of MATH with REF.) It is well-known that the resulting subfactor MATH is the NAME subfactor CITE with principal graph MATH. We make basic constructions of MATH for REF times in the same way as above and get a periodic sequence MATH of commuting squares. Let MATH be a sum of three minimal projections corresponding to the REFth, REFth, and REFth vertices of MATH in MATH. Setting MATH and MATH, we get a periodic sequence of commuting squares MATH such that the resulting subfactor MATH is isomorphic to MATH defined in the first paragraph. Now we see that the NAME diagram of the sequence MATH is the same as the one for MATH as in REF and each algebra MATH is generated by the NAME projections for the sequence MATH. This shows that the two periodic sequences of commuting squares MATH and MATH are isomorphic. Thus the resulting subfactors MATH and MATH are also isomorphic. Since the subfactor MATH is a basic construction of MATH, we conclude that the subfactor MATH is also a basic construction of some subfactor, as desired. |
math/9907149 | By CITE, it is enough to prove that the two subfactors have the same higher relative commutants. Let MATH be the subfactor arising from the conformal inclusion and MATH the inclusion map MATH. We label MATH morphisms as MATH, where MATH or MATH. We set the finite dimensional MATH-algebras MATH, MATH, to be as follows. (For MATH, MATH starts at REF.) MATH . We then naturally have inclusions MATH, and similarly embeddings MATH as well as MATH. With these, we have a double sequence of commuting squares. Note that the sequence MATH is a usual double sequence of string algebras as in CITE (compare CITE) and we now have an extra sequence MATH here. Set MATH to be the NAME of MATH with respect to the trace. Then we have the NAME tower as MATH . The NAME diagram of MATH is given by reflections of the NAME diagram of type MATH or MATH, so the algebra MATH is generated by the NAME projections. The NAME diagram of MATH is given by reflections of the NAME diagram of type MATH or MATH since we know the fusion graph of MATH on the MATH sectors, so the subfactor MATH is isomorphic to the NAME. Then we next show that the higher relative commutants of this subfactor are given as MATH which are also the higher relative commutants of MATH from the above definition, so the proof will be complete. The definition of MATH shows that MATH and MATH commute. Then NAME 's compactness argument CITE (compare CITE) or NAME 's dimension estimate CITE gives MATH. We similarly have MATH. In general, we have MATH so that we can compute MATH which shows equality MATH. We then have MATH which completes the proof. |
math/9907152 | This follows from REF below. |
math/9907152 | Since the NAME diagram of MATH is symmetric, we have MATH if and only if MATH. Elements MATH where MATH, MATH have the same sign thus come in pairs; the first part of the definition of MATH just takes the one with positive MATH and MATH. The remaining elements of MATH must be of the form MATH for MATH. These form a sequence of nested intervals MATH . Since each of the sets MATH and MATH must be a union of pairs in MATH, we see that MATH is even for all MATH. The second part of the definition above thus puts MATH in MATH for all MATH. Since any MATH appears in exactly one pair in MATH, the first statement follows. The fact that MATH is a tree now follows easily from the fact that MATH is a tree. |
math/9907152 | This follows from REF below. |
math/9907152 | Using the quiver relations REF (or REFs and REFs), we see that if MATH, and MATH is the parent of MATH we have MATH on MATH. The product above is thus a telescoping product, since MATH for all sufficiently negative MATH. The telescope starts with MATH, where the sign is determined by MATH, that is, by whether MATH appears first or second in a MATH-pair. |
math/9907152 | First note that MATH is unipotent if and only if MATH is, since they are of the form MATH and MATH, respectively. We proceed by downward induction in MATH. If MATH is the maximal element in MATH (or MATH), then the quiver relation REF implies MATH, since all MATH act as the identity on MATH. Now suppose the proposition holds for all MATH, and take some MATH. REF and the quiver REF can be used to express MATH as a product of powers (positive and negative) of MATH for MATH; these are unipotent by the inductive assumption, and they commute by quiver relation REF. The unipotence of MATH and MATH now follow. |
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