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math/0101206 | We are free to use, metrics near MATH which are Euclidean. In these neighborhoods, then, the non-linear part of the second-order expansion vanishes. But these are the regions where the map depends on the neck-length MATH. |
math/0101206 | Given MATH, we find, for all sufficiently large MATH, a holomorphic MATH by finding MATH with the property that MATH . This can be done, since the map MATH is uniformly quadratically contracting (this is a consequence of REF ), and the error term MATH goes to zero in MATH, according to REF . |
math/0101206 | We can work in an ambient manifold by embedding the algebraic variety MATH into a complex projective space, from which it inherits the NAME form MATH. As in Subsection REF, we must obtain an a priori bound on the MATH-energy of any disk MATH (thought of as a disk in MATH). We imitate the discussion from Subsection REF, only now in families. The product form MATH gives a NAME form on the product MATH. Restricting this to the subvariety MATH, defined by MATH we obtain a quadratic form MATH on the tangent cone of MATH. There is a holomorphic map MATH, under which we can pull back MATH to the subvariety, to obtain another quadratic form MATH on the tangent cone. Note that MATH is nowhere vanishing. Thus, we can form the ratio MATH, to obtain a continuous function on the projectivized tangent cone of MATH. By compactness, this function is bounded above by some constant which we will denote by MATH (since it is the constant MATH appearing in REF ). It follows that if we have a MATH-holomorphic map MATH, then MATH . But we must consider perturbations of these. To control the energy integrand in the region where the complex structures are varying, consider first the (holomorphic) commutative diagram MATH where the vertical maps are induced by the quotients, and the maps all commute with projection to the base MATH. Letting MATH be some open subset of the diagonal in MATH, and MATH be the pre-image of its closure in MATH, then the restriction of MATH to MATH is a covering space. Thus, we can consider the differential form MATH over MATH. Since the maps are all holomorphic local diffeomorphisms, observe that the complex structure MATH tames the form MATH. It follows that if MATH is a sufficiently small perturbation of the constant path MATH (which is the restriction of MATH), then the taming condition is preserved. With respect to such a path MATH, then, the energy of a MATH-holomorphic map MATH from MATH to any given MATH-fiber of MATH can be controlled as in REF by the MATH-integral of its associated branched cover MATH. On the other hand, for a given MATH, this integral is controlled by some multiplicity (depending on the maximal multiplicity in MATH, which is, of course, a topological quantity) of the the MATH-area a fiber of MATH. But the latter area is bounded in the family, since it is obtained from a smooth symplectic form which extends over the family MATH. This gives us the uniform energy bound on the MATH required by NAME 's compactness theorem, and hence gives rise to a limiting cusp-curve which maps into MATH. Indeed, by continuity of the limiting process, it follows that the cusp-curve maps to the MATH fiber, MATH. |
math/0101206 | NAME 's compactness theorem adapted to our situation REF gives a map MATH which is the limit of the MATH, where MATH here is a bubbletree obtained by attaching MATH spheres MATH to the disk MATH . For generic choice of MATH, its image lies in the union of MATH and MATH. Since MATH is contained in the first set, it follows that the main component of the bubble-tree is mapped to the first set. Indeed, this gives rise to the MATH. Moreover, by dimension counts, it follows that the spheres are mapped into the second subset, giving rise to the map MATH. (Note also that the limiting almost-complex structures MATH all agree with the constant complex structure, over this latter subset.) To view the domains of all the MATH as MATH for fixed MATH, we use the diffeomorphisms generated by vector fields MATH (MATH). Specifically, in the proof of the compactness theorem (see, for instance CITE) one rescales around a point where energy is accumulating, to construct the bubbles. These points where the energy concentrates form subsequences of MATH-tuples MATH, which converge to MATH. Now, after rescaling, we obtain a MATH convergent subsequence to spheres whose energy-measures are centered at the origin. This means that there is a sequence of points MATH also converging to MATH, with the property that MATH. We let MATH be the corresponding translation taking MATH to MATH, so that MATH. Thus, the pairs MATH satisfy the extended holomorphic condition on MATH considered earlier. Moreover, the MATH convergence of the rescaled MATH ensures MATH convergence of the restrictions of MATH. The limit is a holomorphic map which is centered (in the sense of REF ). The usual MATH on the main component of the bubble-tree gives us REF and the corresponding part of REF , while the MATH convergence on the spheres gives us REF , and the rest of REF . The MATH convergence at the bubble points is equivalent to REF , in light of the following. Given MATH, MATH, and letting MATH be the wedge-point in MATH, a neighborhood of MATH meets the nearby fiber MATH of the one-parameter family in a subset identified with the product MATH . |
math/0101206 | The point here is that the norm of a holomorphic function MATH on a cylinder is controlled by its behaviour near the boundary, provided that MATH. More precisely, for each integer MATH, we have constants MATH with the property that if MATH is a holomorphic map with MATH, then we have for each sufficiently small MATH, MATH . This follows from looking at the NAME coefficients of MATH. We content ourselves here with the case where MATH. Write MATH. Then, we have MATH . The estimate for larger MATH follows in a similar manner. The corresponding estimate for MATH follows from the uniform, MATH-independent inclusion (provided MATH is bounded below) MATH where MATH, which in turn follows from the fact that the norm of the inclusion MATH is uniformly bounded, provided MATH is bounded below. To apply this, let MATH. Clearly, the MATH have a convergent subsequence. The above argument gives a uniform MATH-bound on MATH. |
math/0101206 | Let MATH denote the natural inclusion. Consider next the difference MATH. If the MATH norm of the difference between MATH and the standard inclusion of the circle is sufficiently small (up to translation), then the two maps are homotopic. Hence, we can lift the difference MATH to a sequence of holomorphic maps MATH where MATH is thought of as the universal covering space of the annulus. We then apply the argument from REF . |
math/0101206 | The strongly MATH-admissible NAME diagrams exist according to REF , which also shows that they can be connected by a sequence of strongly MATH-admissible isotopies, handleslides, and stabilizations. Invariance under those isotopies in the above sense was established in REF . Invariance under handleslides was established in REF for a special choice of isotopy type of the handleslide. Of course, any handleslide (which does not cross the basepoint) can be brought to this form after an isotopy of the attaching circles (which does not cross the basepoint), so general handleslide invariance follows. Finally, stabilization invariance was established in REF (and REF for MATH). When the NAME diagram is only weakly MATH-admissible, REF gives us a strongly MATH-admissible diagram which is isotopic to the given diagram, and REF gives the necessary identifications between MATH and MATH. We then reduce to to strongly MATH-admissible case considered before. |
math/0101208 | See REF. |
math/0101211 | We write MATH, where MATH. Due to the properties of the lexicographic order we deduce that MATH. Therefore the MATH-th block entry of MATH is MATH . This formula can be used in an inductive argument in order to conclude the proof. |
math/0101211 | For sake of completeness we indicate the main ideas of the proof. This will also show that our approach is quite elementary. Assume MATH and let MATH be a factorization of MATH with MATH for some NAME space MATH. From REF we deduce that MATH . In matrix form, MATH . Define MATH and MATH, then we deduce from REF that there exists an unitary operator MATH such that MATH . It follows that there exist NAME spaces MATH, MATH, and an unitary extension MATH of MATH, hence this extension satisfies the relation MATH . Let MATH, MATH, MATH, be the matrix coefficients of MATH. It is convenient to rename some of these coefficients. Thus, we set MATH . From REF we deduce that MATH and MATH . By induction we deduce that MATH where MATH are monomials of length MATH in the variables MATH, MATH,MATH, MATH, MATH, MATH. Since MATH is unitary it follows that all MATH are contractions. We define MATH and for MATH, MATH . Then we define MATH, MATH, MATH, by REF and MATH for MATH. It can be checked that MATH belongs to MATH. Also, since MATH for all MATH, we deduce from REF that MATH. |
math/0101211 | First assume that MATH is positive. Define the operators MATH, MATH, by REF . Let MATH and set MATH, where MATH was defined by REF. Then the unique solution of the displacement equation MATH is MATH, which is positive. By REF , there is MATH such that MATH. We can check that MATH . From MATH we deduce that MATH, MATH. By REF we deduce that MATH, which implies that MATH, MATH. Conversely, assume that there is a MATH such that MATH, MATH. Then , with the previous notation, we deduce that MATH and so, MATH since MATH is a contraction. |
math/0101211 | This is similar to the proof of REF and we can omit the details. |
math/0101211 | We prove by induction that MATH and with MATH . Similarly, MATH and with MATH . Now, each MATH is a column matrix with MATH entries MATH. From REF it follows that MATH and this implies the required formula. |
math/0101218 | For any MATH, MATH proving that MATH. Using REF with MATH we find MATH proving that MATH is a homomorphism. To prove that MATH is injective note that MATH implies MATH whence, by applying MATH we find MATH (we have denoted by MATH the multiplication map of MATH, MATH). Now, consider a generic element MATH and decompose it in the form MATH with MATH. From REF it immediately follows that MATH showing that MATH, because MATH. In particular, assume MATH. Then MATH since MATH is injective, all factors MATH are linearly independent and therefore MATH whence we conclude that MATH and MATH. |
math/0101218 | The proof of the second statement is immediate. As for the first, if MATH then MATH and, applying the homomorphism MATH, MATH; on the other hand MATH by REF . Hence, MATH commutes with MATH, that is, with MATH, by REF . Similarly if MATH then MATH; on the other hand MATH by REF . Hence MATH commutes with MATH. Therefore MATH. NAME, by REF any MATH can be expressed in the form MATH with MATH. If in particular MATH, then it must be on one hand MATH implying MATH by the linear independence of all factors MATH; on the other hand it must be MATH implying MATH, by the linear independence of all factors MATH and the injectivity of MATH. Therefore MATH. |
math/0101218 | MATH . |
math/0101218 | By REF is a direct consequence of REF . To prove the latter, MATH . |
math/0101218 | As anticipated, the fact that MATH are injective algebra homomorphisms MATH follows from REF . Now, by the NAME decomposition MATH a generic element MATH can be decomposed in the form MATH with MATH; again, the NAME notation at the right-hand side means that a sum of many terms is understood. From REF it immediately follows that MATH showing that MATH. In particular, assume MATH. Then MATH since MATH are injective, all factors MATH are linearly independent and therefore MATH whence we conclude that MATH and MATH belongs to MATH. The proof of the claims with MATH in the inverse order follow in the same way from the NAME decomposition MATH. |
math/0101218 | MATH . |
math/0101218 | The first statement amounts to the preceding proposition applied to MATH. As for the second, the proof is just a small variation: MATH . |
math/0101219 | Since MATH is an abelian category, it suffices to prove that for MATH, MATH and MATH are actually MATH-submodules in MATH respectively. The check is straightforward and is left to the reader. Let MATH. By definition we have: MATH . But since MATH has multiplicity one in MATH, one has MATH for some constant MATH. Thus, MATH. |
math/0101219 | Let MATH. Define MATH where MATH are defined by MATH . This defines MATH as an object of MATH. Define MATH to be MATH or MATH which obviously give the same morphism. One easily sees that this defines a structure of MATH-module on MATH and that so defined tensor product is associative. To check that MATH is the unit object, consider morphisms MATH and MATH. Straightforward check shows that they are well defined, commute with the action of MATH (that is, satisfy REF) and thus define morphisms in MATH and finally, that they are inverse to each other. |
math/0101219 | REF is obvious; for REF , define maps MATH and MATH as shown in REF ; it is easy to deduce from the axioms that these maps are inverse to each other. To prove that MATH is a tensor functor, define functorial morphisms MATH by MATH . It is immediate to check that they are well-defined and inverse to each other. |
math/0101219 | Suffices to take MATH. |
math/0101219 | Let us first show that for MATH, MATH. This follows from the sequence of identities shown in REF . The notation MATH for MATH means that the induced operators MATH are equal, that is, MATH, where MATH is the canonical projection. Next, we need to show that the commutativity isomorphism MATH descends to isomorphism MATH. This is equivalent to showing that MATH, where MATH is the kernel of the canonical projection MATH. To do so, let us rewrite composition MATH as shown in REF . Thus, MATH, or, equivalently, MATH. Abusing the language, we will also use the same notation MATH for the descended morphisms MATH. Then it is immediate from the definition that these morphisms are MATH-morphisms. Finally, since the commutativity isomorphism in MATH satisfies the hexagon axioms, the same must hold for the descended operators; thus, the descended operators MATH define a structure of a braided tensor category on MATH. |
math/0101219 | The identity for MATH immediately follows from commutativity of multiplication. For MATH, it suffices to prove that MATH satisfies the rigidity axioms, that is, both compositions below are equal to identity MATH . To prove the first identity, we represent the composition by a graph and manipulate it as shown in REF . The second identity is proved in the same manner. |
math/0101219 | Recalling the relation between the twists MATH and MATH (see, for example, CITE), we see that the statement of the lemma is equivalent to the following equation: MATH . But it easily follows from symmetry of MATH that the right hand side is the identity morphism. |
math/0101219 | Let MATH. Define the dual object MATH as follows: MATH is the dual of MATH in MATH and MATH is defined by REF . This definition implies the following identities: Define now the maps MATH by REF (we leave it to the reader to check that these formulas indeed define morphisms in MATH). It is easy to check by using identities in REF and isomorphisms MATH defined in the proof of REF that these two maps satisfy the rigidity axioms. |
math/0101219 | To prove REF , we construct linear maps between MATH and MATH as shown in REF ; we leave it to the reader to check that these maps are inverse to each other. To prove REF , note that as object of MATH, MATH, where we used rigidity to identify MATH. Consider the morphism MATH. Again, we leave it to the reader to check that this morphism is actually a morphism of MATH-modules MATH. This shows that MATH is rigid. To prove rigidity of MATH, it suffices to show that for MATH which easily follows from the definition of MATH and the definition of dual object given by REF . |
math/0101219 | REF follows from MATH and MATH; REF immediately follows from REF . To prove REF , it suffices to prove that MATH is a MATH-morphism. We can rewrite MATH in terms of MATH as follows (see CITE): MATH . It now follows from the identities shown in REF (which uses REF and identities from REF ) that MATH is a MATH-morphism. |
math/0101219 | Formula MATH holds in any pivotal category and can be easily deduced from REF. Using definition of rigidity morphisms in MATH shown in REF , we see that MATH is defined by the following identity: MATH . Both sides are MATH-morphisms MATH. Composing them with MATH, we get MATH. Applying this to MATH, we get MATH . |
math/0101219 | Let MATH be rigid; assume MATH is a submodule. By rigidity, MATH. On the other hand, since MATH is a submodule, this implies that MATH. By unit axiom, this implies MATH. Conversely, assume that MATH is simple as MATH-module. Consider MATH and define on it the action of MATH as in REF. Then one easily sees that the morphism MATH where MATH is as in REF is a morphism of MATH-modules. On the other hand, usual arguments show that if MATH is a simple MATH-module, then so is MATH. Thus, such a map is either zero (impossible because of the unit axiom) or an isomorphism. |
math/0101219 | By definition, an object of MATH is a MATH-module MATH with a decomposition MATH such that MATH, and tensor product in MATH is given by MATH. Define functor MATH by MATH and MATH by MATH (note that it follows from definition of the induced module that MATH has a natural decomposition MATH). It is trivial to check that these functors preserve tensor product and are inverse to each other. |
math/0101219 | First, a MATH-algebra is just a commutative associative algebra over MATH on which MATH acts by automorphisms. Next, if MATH is rigid, then MATH is semisimple as a commutative associative algebra over MATH. Indeed, let MATH be the radical of MATH; then MATH is invariant under the action of MATH and thus is an ideal in MATH in the sense of MATH-algebras. By REF , MATH. Thus, MATH is the algebra of functions on a finite set MATH (which can be described as the set of primitive idempotents of MATH) and MATH acts by permutations on MATH. Since MATH appears in decomposition of MATH as MATH-module with multiplicity one, this implies that the action of MATH on MATH is transitive, so MATH. |
math/0101219 | Immediately follows from REF and the fact that for a simple object MATH, MATH. |
math/0101219 | The proof is based on the following lemma. If MATH is rigid, then every MATH is a direct summand in MATH for some MATH. Consider the map MATH. It is surjective and is a morphism of MATH-modules. Moreover, both MATH and MATH have canonical structures of MATH-bimodules, and MATH is a morphism of MATH-bimodules (we leave the definition of MATH-bimodule as an exercise to the reader). This map has one-sided inverse: if we define MATH by MATH then MATH is a morphism of MATH-bimodules and it immediately follows from REF that MATH. Thus, MATH splits: we can write MATH for some MATH-bimodule MATH so that under this isomorphism, MATH is the projection on the first summand. Therefore, MATH is a direct summand of MATH. From this lemma, the proof is easy. Indeed, it easily follows from exactness of MATH and adjointness of MATH and MATH (see REF) that for every MATH, MATH is a projective object in MATH. Since a direct summand of a projective object is projective, the lemma implies that every MATH is projective and thus, MATH for every MATH. |
math/0101219 | Recalling the definition of rigidity isomorphisms in MATH and isomorphisms MATH, we see that MATH is given by MATH . Restricting both sides to MATH, we get MATH which is easily seen to be equivalent to the statement of the lemma. |
math/0101219 | If MATH, then the statement immediately follows from REF . Thus, let us assume that MATH and prove that in this case, MATH. First, note that the composition MATH can be rewritten as shown in REF . From this presentation one easily sees that MATH is a morphism of MATH-modules; since MATH is simple, this implies MATH for some MATH. Next, let us calculate MATH: MATH . Thus, MATH is a projector. On the other hand, it follows from REF that MATH. Combining these two results, we get MATH. If we assume that MATH, then this implies that MATH; by REF, this is impossible if MATH. Thus, MATH. |
math/0101219 | Since both sides are linear in MATH it suffices to prove this formula when MATH is simple. If MATH, the statement immediately follows from REF . Thus, we only need to prove that if MATH is simple, MATH, then the right-hand side is zero. To prove this, let MATH be defined by MATH . On one hand, it easily follows from REF that MATH . On the other hand, we can deform the figure defining MATH as shown below MATH . By REF , this implies MATH. |
math/0101219 | Proof for MATH is obvious from the definition. As for MATH, it suffices to prove that MATH for any MATH. Using adjointness of MATH and MATH, this reduces to MATH . (note that MATH, but in general, not in MATH). Using REF and definition of MATH, this can be rewritten as the following identity of figures: MATH which can be proved by rewriting the graph in left hand side as shown in REF and using REF . Similarly, the identity involving MATH is equivalent to MATH which is also equivalent to REF. This completes the proof of REF. |
math/0101219 | The proof is based on the following lemma. Let MATH be a semisimple rigid braided balanced category over MATH, with finitely many isomorphism classes of simple objects. Then MATH is modular iff the matrix MATH, defined by REF , satisfies MATH for some MATH. This lemma is not new; however, for the sake of completeness, we include its proof below. Thus, to prove that MATH is modular, it suffices to prove MATH for some MATH. But by REF, MATH commutes with MATH up to a constant; thus, MATH . Thus, MATH is modular; all other statements of the theorem immediately follow from REF. If MATH is modular, the statement is well known and in fact MATH (see, for example, CITE). Thus, let us assume that REF holds and deduce from it non-degeneracy of MATH. First, note that MATH, where MATH are simple objects in MATH and MATH. Thus, REF implies MATH which can be rewritten as MATH . Let us now choose some MATH and let MATH . On one hand, it is easy to show using the definition of MATH that MATH (see, for example, CITE) and thus, MATH . On the other hand, decomposing MATH in a direct sum of irreducibles and using REF, we get MATH which is a non-singular matrix. Therefore, MATH is non-singular and thus MATH is modular. This completes the proof of the lemma and thus of REF. |
math/0101219 | We give a sketch of the proof; details will appear in the forthcoming paper CITE. If MATH is a VOA, then for every MATH we have the vertex operator MATH. Restricting it to MATH, we get a structure of a MATH-module on MATH. It is immediate from the definitions that the map MATH is an intertwining operator of the type MATH and thus gives a morphism of MATH-modules MATH, where MATH is the ``fusion" tensor product. It follows from the usual commutativity and associativity axioms for a VOA (see CITE) that MATH defines a structure of a commutative and associative algebra on MATH. Existence and uniqueness of unit follow from existence and uniqueness of the vacuum vector in a VOA (see REF above). Condition MATH follows from the fact that eigenvalues of MATH on MATH are integer. A straightforward check shows that the arguments above can be reversed and that the category of representations of MATH as a VOA coincides with MATH. |
math/0101219 | Let MATH be a rigid MATH-algebra with MATH. In this case, MATH is a monoidal category and, by REF , a module category over MATH. This implies that the NAME ring MATH is a module over MATH. By REF, MATH is semisimple, so MATH has a distinguished basis (classes MATH of simple objects) so that in this basis, multiplication by any MATH, has coefficients from MATH. In addition, this module has the following properties: CASE: The module MATH is indecomposable: it is impossible to split the set of simple objects MATH as MATH so that MATH are MATH-submodules in MATH. Indeed, every simple module MATH appears with non-zero multiplicity in MATH for some MATH. This follows from MATH. CASE: There exists a map MATH such that MATH and MATH. Indeed, it suffices to let MATH and use REF. CASE: There exists a symmetric bilinear form MATH on MATH such that MATH for any MATH. Indeed, we can let MATH and use rigidity and MATH (not canonically). All modules MATH over MATH which have REF - REF above were classified in CITE, where it is shown that they correspond to finite NAME diagrams with loops with NAME number equal to MATH. Under this correspondence, vertices of the NAME diagram correspond to the elements of distinguished basis of MATH, and the matrix of multiplication by MATH is MATH, where MATH is the NAME matrix of the NAME diagram. (NAME diagrams with loops, in addition to the usual NAME diagrams, include ``tadpole" diagrams MATH shown in REF ; in CITE, this diagram is denoted by MATH. By definition, the NAME matrix for such a diagram is the same as for MATH but with MATH, and the NAME number for MATH is equal to MATH). In an interesting note CITE, it was shown that the dimension vector MATH can also be obtained from so-called ``semi-affine" NAME diagrams, which give MATH both for finite and affine NAME diagrams. Now we have to check which of these modules can actually appear as NAME ring MATH for some rigid MATH-algebra MATH. First, note that if MATH is indeed the NAME ring of a rigid MATH-algebra MATH, then not only we have a distinguished basis MATH and an inner product MATH but in fact, the distinguished basis is orthonormal with respect to MATH. This implies that the matrix of tensor product with MATH is symmetric in this basis. Thus, only simply-laced NAME diagrams can possibly come from MATH. This leaves us with the ADET type diagrams. Next, we need to determine which vertex of the NAME diagram corresponds to the unit object, that is, to MATH itself. If MATH is a rigid MATH-algebra, then MATH corresponds to the end of the longest leg of the corresponding NAME diagram. By an ``end" we mean a vertex which is connected to exactly one vertex; in particular, the vertex with a loop in the diagram of type MATH is not considered an end vertex. Let MATH be the object corresponding to one of the ends of legs of the NAME diagram. Then MATH is simple. Since in a rigid category, tensor product of non-zero objects is always non-zero, this implies that MATH is simple. Thus, MATH is connected to exactly one vertex, which means that MATH itself is an end of one of the legs. To prove that MATH is the end of the longest leg, note that if MATH is an end of the leg of length MATH (that is, consisting of MATH edges), then MATH are simple but MATH is not. This implies that MATH are simple, which means that the leg containing MATH has length at least MATH. This determines the vertex corresponding to MATH uniquely up to an automorphism of the NAME diagram. Once we know the vertex corresponding to MATH, we know the class of MATH in MATH; since MATH is a tensor functor and MATH generates MATH, this uniquely determines the map MATH at the level of NAME rings, and thus, the adjoint map MATH. In other words, we can write for each vertex of the NAME diagram the structure of the corresponding object MATH as an object of MATH. In particular, this gives decomposition of MATH itself as an object of MATH. Doing this explicitly for diagrams MATH gives the answer shown in REF (no, it was not found using a computer - it is done easily by hand), which agrees with the one given in NAME classification. Next step is to find which of the possible MATH given in this table do have a structure of a MATH-algebra. Type MATH: in this case, MATH obviously has a unique structure of commutative associative algebra, and MATH. Type MATH. Let us introduce the notation MATH . It easily follows from explicit formulas that MATH and MATH; in particular, MATH. The object MATH in MATH has a structure of a rigid MATH-algebra iff MATH. In this case, the structure of an algebra is unique up to isomorphism, and this algebra satisfies MATH. Let MATH be the multiplication map MATH. All components of such a map are uniquely determined by the unit axiom, except for MATH. Since MATH, such a map is unique up to a constant. Rigidity implies that MATH. This proves uniqueness. To check existence, fix some non-zero MATH. Then associativity and commutativity are equivalent to MATH . To check the second equation, we use the following lemma For generic values of MATH, let MATH be a nonzero homomorphism. Then MATH where MATH is the universal twist and MATH. To prove this lemma, note that it immediately follows from balancing axiom in MATH that MATH, which gives the formula above up to a sign. To find the sign, it suffices to let MATH. Since this formula works for generic values of MATH, it should also be valid for MATH being a root of unity. In particular, applying this lemma to MATH and MATH, we get MATH . We have MATH. Thus, MATH is equal to one iff MATH is divisible by REF. Therefore, the map MATH is commutative iff MATH. To check associativity, note that both sides are equal up to a constant (since MATH); to find the constant, take composition of both sides with MATH and use MATH. The category of representations of this algebra is described in detail in REF. It follows from the analysis there that the structure of MATH as MATH-module is described by the diagram MATH. Type MATH. The diagram MATH can not appear as MATH for a commutative associative algebra MATH. Indeed, in this case MATH must be isomorphic to MATH, but it was proved in REF that there is at most one structure of a rigid MATH-algebra on this object, and if it exists, MATH is described by MATH, not MATH. Type MATH. This diagram can not appear as MATH for a commutative associative algebra MATH. Indeed, in this case the table gives MATH. Obviously, MATH is a subalgebra in MATH, and multiplication on MATH defines a structure of MATH-module on MATH and morphism of MATH-modules MATH. By rigidity, this morphism is non-zero, which also implies that the restriction of MATH to MATH is non-zero. But it immediately follows from REF that such a morphism can not be symmetric. Type MATH. In this case, there is a unique up to isomorphism MATH-algebra structure on MATH. Existence follows from the discussion of the previous section and existence of a conformal embedding of affine NAME algebras MATH (see REF). To prove uniqueness, note that the only non-trivial components of the multiplication map MATH are MATH, MATH. Both of them are unique up to a constant factor. We can fix some non-zero morphisms MATH . Then MATH for some MATH. It follows from rigidity that MATH. Using isomorphism of MATH-algebras MATH given by MATH, we see that without loss of generality we can assume MATH, so MATH. Condition that MATH be associative gives the following quadratic equation on MATH: MATH where MATH are morphisms MATH given by MATH . It is easy to see that MATH, so the equation MATH is non-trivial. Thus, such an equation may either have no solutions at all or have exactly two solutions differing by sign: MATH. These two solutions actually would give isomorphic algebras: the map MATH given by MATH gives the isomorphism. Type MATH. In this case, there again exists a unique structure of a rigid MATH-algebra on MATH. Existence follows from existence of conformal embedding MATH (see REF). To prove uniqueness, let MATH be the subalgebra generated (as a MATH-algebra) by MATH. Let MATH be the vertex operator algebra corresponding to MATH; by results of REF, it is an extension of the VOA MATH. From the definition, MATH is generated as a VOA by MATH and MATH. Since MATH is an irreducible MATH module, it is generated (as MATH module) by its lowest degree component (degree stands for homogeneous degree, that is, eigenvalue of MATH). This lowest degree is equal to MATH. Since it is well known that MATH is generated as a VOA by its degree one component MATH, we see that MATH is generated as a VOA by MATH. It is also easy to check that conformal dimensions (that is, lowest eigenvalues of MATH) for MATH and MATH are greater than one, so MATH, where MATH is an irreducible MATH-module with highest weight REF. By CITE, if MATH for MATH and MATH is generated as VOA by MATH, then MATH is a NAME algebra with an invariant bilinear form, and MATH is naturally a module over MATH; moreover, MATH is a quotient of the NAME module MATH over MATH for some MATH. Thus, we see that embedding MATH defines an embedding MATH. Rigidity of MATH also implies that the multiplication map MATH is non-zero, which implies that the restriction of the commutator in MATH to MATH is non-zero. Now we can use the following lemma. Let MATH be a finite-dimensional NAME algebra which contains a subalgebra isomorphic to MATH and as a MATH-module, MATH . If, in addition, restriction of the commutator map MATH is non-zero, then MATH. It is easy to see that in such a situation, MATH must be simple (indeed, the only possible ideals are MATH and MATH, and none of them is an ideal). But the only REF-dimensional simple NAME algebra is MATH. Therefore, embedding MATH gives rise to an embedding MATH. Since the NAME central charge is the same for MATH, this embedding extends to a conformal embedding MATH. But it is well known (see, for example, CITE) that such a conformal embedding uinique, namely MATH. |
math/0101219 | By definition, MATH. As an object of MATH, MATH we need to check which of the modules MATH are in the image of MATH. To do so, we use the following lemma. Let MATH be even, MATH and let MATH be defined by the compositions MATH where MATH, MATH and MATH are arbitrary non-zero morphisms. Then MATH. To prove the lemma, it suffices to consider the identity shown in REF and then apply REF to both sides. This proves the lemma. This lemma implies that for MATH, MATH consists of those MATH with MATH even and MATH, while for MATH, MATH consists of those MATH with MATH even and MATH. This determines the decomposition of MATH as on object of MATH. By REF, this determines this tensor product as a representation of MATH uniquely except for ambiguity in the choice of the action of MATH on MATH; in other words, we do not know if MATH or MATH appears in decomposition of MATH. To answer this, note that we already know enough to deduce that for MATH, MATH. Thus, using rigidity we find MATH since we already know decomposition of MATH. Similar arguments show that for MATH, MATH. This completes the proof of the theorem. |
math/0101220 | We only have to note that the group MATH has a classifying MATH-space MATH whose fundamental crossed complex MATH is homotopy equivalent to MATH. |
math/0101220 | Certainly MATH is homotopic to MATH for some MATH since the set of pointed homotopy classes MATH is bijective with the morphisms of groups MATH. The result follows from CITE (`if MATH is MATH-realisable, then each element in the homotopy class of MATH is MATH-realisable'). |
math/0101220 | The proof is modelled on that given by NAME in CITE. If MATH is a complete graph the result is clear, since finite products of aspherical spaces are aspherical. We now work by induction on the number of vertices of MATH. Suppose MATH is not complete. Then there are vertices MATH of MATH which do not form an edge of MATH. Let MATH be the vertex set of MATH, and let MATH be the full subgraphs of MATH on the complements in MATH of MATH respectively. Let MATH be the corresponding graph products. By the inductive assumption, these are aspherical. It is clear that MATH . But the maps on fundamental groups induced by the inclusions MATH are injective. REF now implies that MATH is aspherical. |
math/0101220 | This is immediate from previous results on the fundamental crossed complex of a product of MATH-spaces, and the Generalised NAME Theorem. |
math/0101221 | We suppose that MATH is an odd prime. The case MATH is similar. Let MATH denote the mod MATH . NAME algebra. The degree of an element MATH is denoted MATH. Recall from CITE, that the NAME spectral sequence is a strongly convergent second quadrant cohomological spectral sequence of MATH-modules MATH . More precisely, there exists a convergent filtration of MATH-modules on MATH: MATH such that MATH. Here MATH denotes the MATH-th desuspension of a MATH-module. Let MATH such that the class MATH is non zero. We want to prove that MATH. As a MATH-module, MATH is the MATH-th homology group of a complex of MATH-modules, namely the NAME construction, whose MATH-th term is MATH. The element MATH is represented by a cycle of the form MATH, where MATH, MATH and MATH. So MATH. CASE: MATH. Then MATH. Therefore, by a degree argument, the element MATH of MATH is zero. CASE: MATH. Since the NAME formula applies, MATH is represented by the element of the NAME construction, MATH . So MATH is zero for degree reasons. Therefore MATH belongs to MATH which is concentrated in degrees MATH, thus MATH. |
math/0101221 | The NAME spectral sequence for the previous fiber product satisfies MATH . Here MATH denotes the NAME homology. As a MATH-module, MATH is the MATH-th homology group of a complex of MATH-modules, namely the NAME complex, whose MATH-th term is MATH. The same arguments as in the proof of Theorem A allow us to conclude except in REF for MATH. If MATH, we can only affirm that MATH. The evaluation map MATH admits a section MATH. So MATH admits MATH as retract. The edge homomorphism MATH correspond to MATH. Since MATH, MATH. For degree reason, all MATH-th powers are zero in MATH. So MATH. |
math/0101221 | The proof follows the lines of CITE. Since MATH, MATH survives till MATH. Therefore by REF and CITE (See also CITE), MATH survives till MATH. Since MATH, MATH . |
math/0101222 | First, MATH is generated by MATH, the image of MATH in MATH. Since MATH is generated by MATH, in particular MATH is generated by the finitely many MATH-fold products of elements of MATH; this proves the first point. Actually, far fewer generators are required for MATH; in the extremal case when MATH is a free group, a basis of MATH is given in terms of ``standard monomials" of degree MATH; see Subsection REF or CITE. For the second claim, assume more generally that MATH for all MATH, so that MATH is a MATH-vector space. We use the identity MATH, due to NAME. Let MATH be a generator of MATH, with MATH and MATH. Then MATH by induction, so MATH and MATH is a MATH-vector space. |
math/0101222 | Let MATH be branch over MATH of finite index, and set MATH, the core of MATH. Then obviously MATH; and since MATH for all MATH, we have, writing MATH, MATH and MATH, so MATH. |
math/0101222 | The subgroup MATH of MATH is isomorphic to MATH through MATH, and its MATH conjugates under powers of MATH commute, since they act on disjoint subtrees. |
math/0101222 | The proof follows by induction, and we may suppose MATH without loss of generality. Multiplying by terms in MATH, we may assume MATH by some element acting only on the last MATH subtrees below the root vertex. Then MATH . |
math/0101222 | We interpret MATH in the NAME graph as MATH in MATH. The generator MATH is then MATH. By REF , the adjoint operators MATH correspond to the arrows labeled MATH. The arrows connect elements whose degree differ by MATH, so the degree of the element MATH is MATH as claimed. The power maps MATH are all trivial on the elements MATH, so the NAME algebra and restricted NAME algebra coincide. The elements MATH for MATH belong to MATH, and hence are trivial in MATH. |
math/0101222 | Direct computation, using the decompositions MATH etc. and linearizing. |
math/0101222 | The proof proceeds by induction on length of words, or, what amounts to the same, on depth in the lower central series. First, the assertion is checked ``manually" up to degree MATH. The details of the computations are the same as in CITE. We claim that for all words MATH with MATH we have MATH, and similarly MATH. The claim is verified by induction on MATH. We then claim that for any non-empty word MATH, either MATH (if MATH starts by MATH) or MATH for MATH (if MATH starts by MATH). Again this holds by induction. We then prove that the arrows are as described above; this follow from REF . For instance, MATH . Finally we check that the degrees of all basis elements are as claimed. For that purpose, we first check that the degree of an arrow's destination is always one more than the degree of its source. Then fix a word MATH, and consider the largest MATH such that MATH belongs to MATH. There is then an expression of MATH as a product of MATH-place commutators on elements of MATH, and therefore in the NAME graph there is a family of paths starting at some element of MATH and following MATH arrows to reach MATH. This implies that the degree of MATH is MATH, as required. The modification giving the NAME graph of MATH is justified by the fact that in MATH we always have MATH, so the element MATH appears always last as the image of MATH through the square map. The degrees are modified accordingly. Now MATH, and MATH, with equality only when MATH. This gives an additional square map from MATH to MATH, and requires no adjustment of the degrees. |
math/0101222 | Consider the sequence of coefficients of MATH. They are, in condensed form, MATH . The MATH-th coefficient is MATH if there are MATH and MATH of degree MATH in MATH, and is MATH if there is only MATH. All conclusions follow from this remark. |
math/0101222 | We perform the computations in the completion of MATH, still written MATH. With REF in mind, MATH is the subgroup generated by all MATH and MATH, for MATH. We claim inductively that if MATH at all positions MATH, then MATH, and similarly for MATH. Therefore some terms may be neglected in the computations of brackets. Now we compute MATH for MATH. Here MATH means some terms of greater degree have been neglected: MATH . All asserted arrows follow from these equations. Finally, we prove that the degrees of MATH and MATH are as claimed, by remarking that MATH and MATH, that MATH for MATH and all words MATH (so the claimed degrees smaller of equal to their actual value), and that each word of claimed degree MATH appears only as MATH for words MATH of degree at most MATH (so the claimed degrees are greater or equal to their actual value). The last point to check concerns the cube map; we skip the details. |
math/0101222 | Define polynomials MATH . Then one checks directly that the polynomials MATH satisfy the same initial values and recurrence relation as MATH, hence are equal. All convergence properties also follow from the definition of MATH. The words of degree MATH are MATH, MATH, and all the words that can be obtained from these by iterating the substitutions MATH, MATH, MATH, MATH along with MATH and MATH at the beginning of the word. This gives MATH words in total, half of the form MATH and half MATH. There is a unique word of degree MATH, and that is MATH. Note that these last two claims have a simple interpretation: there are MATH ways of writing MATH in base MATH using only the digits MATH; there is a unique way of writing MATH in base MATH using these digits. |
math/0101222 | Apply REF to the series MATH, which is comparable to the NAME series of MATH. |
math/0101222 | Clearly true for MATH; then a direct consequence of MATH (obtained by NAME in CITE) and MATH for MATH. |
math/0101222 | The proof is similar to that of REF , but a bit more tricky. Again we perform the computations in the completion of MATH, still written MATH. Again MATH is the subgroup generated by all MATH and MATH, for MATH. We claim inductively that if MATH at all positions MATH, then MATH, and similarly for MATH. Therefore some terms may be neglected in the computations of brackets. Now we compute MATH for MATH. Here MATH means some terms of greater degree have been neglected: MATH . Note that in the last line the ``negligible" term MATH has been kept; this is necessary since sometimes the MATH term cancels out. Now we check each of the asserted arrows against the relations described above. First the MATH arrows are clearly as described, and so are the MATH arrows on MATH; for instance, MATH which holds by induction on the length of MATH. Next, the MATH arrows on MATH agree; for instance, MATH . All other cases are similar. Note how the calculation for MATH explains the definition of MATH: both MATH and MATH have degree smaller than MATH in MATH, but they are linearly dependent in MATH. Finally, we prove that the degrees of MATH and MATH are as claimed, by remarking that MATH and MATH, that MATH for MATH and all words MATH (so the claimed degrees smaller of equal to their actual value), and that each word of claimed degree MATH appears only as MATH for words MATH of degree at most MATH (so the claimed degrees are greater or equal to their actual value). |
math/0101222 | Consider the sequence of coefficients of MATH. They are, in condensed form, MATH . The MATH-th coefficient is MATH if there are MATH and MATH of degree MATH in MATH, and is MATH if there is only MATH. All conclusions follow from this remark. |
math/0101222 | It is a general fact for a MATH-generated group MATH that MATH. Since MATH and MATH (modulo MATH), we have MATH and therefore MATH. Next, MATH, so MATH, and the claimed formula holds for all MATH by induction. Finally MATH for all MATH. |
math/0101222 | Assume MATH acts on a MATH-regular tree, and write as before MATH. The proof relies on an identification of the NAME action on group elements and the natural action on tree levels. We first claim that for any MATH and MATH where MATH is the length of a minimal word moving MATH to MATH in the tree MATH. Therefore the growth of MATH and MATH may be compared just by considering the degrees of elements of the form MATH for some fixed MATH; indeed the other MATH will contribute a smaller growth to the NAME growth series than the corresponding vertices to the parabolic growth series, and the MATH finitely many values MATH may take in a branch portrait description will be taken care of by the constant MATH. Now there is a constant MATH such that MATH has greater degree than MATH for all MATH. Indeed there exists MATH and MATH such that MATH, and then MATH, proving the claim. We may now take MATH. The NAME growth series is the sum over all MATH and coset representatives MATH of the power series counting the growth of MATH over words MATH of length MATH. There are MATH choices for MATH, and for given MATH at most MATH of these power series overlap. |
math/0101222 | The first two assertions are checked directly as follows. Let MATH be the set of finite-index subgroups of MATH not in MATH. Consider the finite quotient MATH, and the preimage MATH of MATH defined as MATH . Clearly the image of MATH in MATH is at most as large as MATH, and the preimage of MATH in MATH is at least as large as MATH. Now we use the algorithms in Gap CITE computing the top of the lattice of normal subgroups for finite groups REF and finitely presented groups (MATH). The number of subgroups not contained in MATH agree in MATH and MATH, so give the structure of the lattice not below MATH in MATH. Let now MATH be a normal subgroup of MATH, contained in MATH. If MATH is non-trivial, then it has finite index CITE. It is easy to see that MATH contains MATH and MATH for some words MATH, using for instance the congruence property CITE; therefore the generators of MATH may be chosen as MATH with MATH and MATH for all MATH. Taking the commutators of these generators with the appropriate generator among MATH, we shift the ranks of the MATH-terms up by MATH, and multiplying a generator by another we may get rid of all generators except MATH and the one with MATH of smallest rank. We therefore consider all subgroups MATH, and seek conditions on MATH, MATH and MATH so that to each normal subgroup in MATH there corresponds a unique expression of the form MATH. Let first MATH be minimal such that MATH; then take MATH minimal such that for some MATH we have MATH . Take also MATH minimal such that MATH for some MATH. Define the functions MATH as follows (MATH stands for ``monomial" and MATH stands for ``squares"): Consider MATH as an element of MATH, truncated at degree MATH. Successive commutations with generators MATH, according the the rules of REF , give rise to other elements of MATH. We stress that we use the complete computations of commutators, and not just those in the filtered NAME algebra. Define MATH as the minimal word MATH such that MATH that arises in this process; if no such word occurs, MATH. Define MATH as the minimal MATH such that MATH occurs in this process; if no such product occurs, MATH. Now, since MATH, we necessarily have MATH. Also, all MATH of degree at least MATH can be replaced by terms of lower degree MATH. This proves the claimed inequalities. Conversely, if there existed another description MATH for another choice of MATH's, then by dividing we would obtain a product of MATH in MATH, contradicting MATH. The data MATH subjected to the Theorem's constraints therefore bijectively correspond to MATH's. The index of MATH can be computed in MATH. Seeing elements of MATH as inside MATH, a vector-space complement of MATH is spanned by all MATH of rank less than MATH, and all MATH of rank less than MATH. We consider finally three cases: first assume MATH and MATH. Then MATH gives MATH by commutation with MATH, which itself gives MATH by commutation with MATH, so we may suppose MATH. Various MATH's can be added, giving the description MATH. Now assume MATH. Then since MATH would produce MATH by commutation with an appropriate conjugate of MATH, we must have MATH so that the same commutation vanishes, giving the description MATH. Finally assume we have MATH and MATH. Then necessarily MATH; and taking appropriate commutations we see that the normal subgroup under consideration contains MATH. We may then replace the generator MATH by MATH, and obtain the description MATH. |
math/0101222 | The automorphism group of MATH is determined in CITE: it also acts on the binary tree, and is MATH . It then follows that MATH is a strict subgroup of MATH; and hence MATH for any normal subgroup that is generated by expressions in MATH and MATH for words MATH. The theorem asserts that all normal subgroups of MATH below MATH have this form; it then suffices to check, for instance using the algorithms in Gap CITE, that the finitely many normal subgroups of MATH not in MATH are characteristic. |
math/0101222 | The number of subgroups of index MATH behaves in a somewhat erratic way, but is greater when MATH is of the form MATH, so that there is a maximal number of choices for MATH and MATH, and is smaller when MATH is of the form MATH. We compute the numbers MATH and MATH of normal subgroups of MATH contained in MATH of index MATH, with respectively MATH and MATH, yielding the upper and lower bounds. The computations are simplified by the fact that for these two values of MATH there are only subgroups of type MATH. Let us start by the upper bound, when MATH. First, for MATH, the subgroups of index MATH are MATH, MATH and MATH, giving MATH. Then, for MATH, the subgroups can of index MATH can be described as follows: CASE: MATH for all MATH counted in MATH, except when MATH, when no subgroup appears in MATH, and when MATH, when MATH should be replaced by MATH; CASE: MATH for all MATH counted in MATH, except when MATH, when no subgroup appears in MATH, and when MATH, when MATH should be replaced by MATH; CASE: MATH, with the same qualifications as above; CASE: MATH. It then follows that MATH, so MATH for all MATH. For the lower bound, we have MATH; and for MATH, when MATH, the subgroups can of index MATH can be described as follows: CASE: MATH for all MATH counted in MATH; CASE: MATH for all MATH counted in MATH; CASE: MATH, with the same qualifications as above; CASE: MATH and MATH. It then follows that MATH, so MATH for all MATH. In summary, the number of normal subgroups of index MATH oscillates between MATH and MATH for MATH (when all normal subgroups of MATH are contained in MATH). These bounds give respectively MATH and MATH. |
math/0101222 | The proof follows from the description of REF . Assume MATH. To determine the parity of the number of subgroups of index MATH, it suffices to consider which MATH expressions have no choices for MATH. These are precisely the MATH with MATH, the MATH with MATH and the MATH with MATH. Now these last two families yield a subgroup for precisely the same values of MATH, namely those satisfying MATH, and therefore contribute nothing modulo MATH. The first family contributes a subgroup for all MATH. |
math/0101226 | First we show the existence of cosingular vectors. Let MATH be the restricted dual of MATH and MATH be the NAME involution: MATH . We define the contragradient representation of MATH: MATH as MATH . Throughout this proof MATH denote arbitrary vectors of MATH. Then there is a one to one correspondence between singular (cosingular) vectors in MATH and cosingular (singular) vectors in MATH such that vectors in a pair have the same weight. Hence we study the singular vectors in MATH. The space MATH can be regarded as a NAME space as MATH . Let MATH be the automorphism which changes MATH index of the NAME generators of MATH: MATH . Set MATH, then MATH is another NAME representation with the highest weight MATH. Then we define an isomorphism MATH by MATH . The last equation comes from MATH . With this MATH a singular vector in MATH can be mapped to a singular vector of the same weight in MATH. Singular vectors in MATH can be constructed explicitly with screening charges. Denoting the highest weight vector of MATH by MATH, MATH is a singular vector in MATH. To verify that the vector REF is non-vanishing, we checked that the matrix element MATH is non-zero by using a formula in REF. Note that the quantity MATH makes the subscript MATH be odd. By the same argument, MATH is a singular vector in MATH. In MATH vectors REF are singular vectors of the weight MATH and this proves existence of MATH and REF. To conclude all cosingular vectors in MATH are given by the argument above, we study the determinant MATH. The matrix MATH connects the elements of MATH and monomials of MATH at the degree MATH CITE, for example, MATH . If we look at MATH as a polynomial of MATH, its zero of the first degree for the smallest MATH corresponds to a cosingular vector of MATH of the degree MATH. Let MATH denote the number of monomials of the degree MATH in MATH, then MATH . The cosingular vectors MATH of MATH give Right-hand side as a divisor of MATH. The order of MATH is the number of MATH (MATH:odd) in the LHS of REF, which is given by MATH where MATH is defined by the polynomial of indetermimates MATH as MATH . And this order coincides with the order of Right-hand side from the following identity: MATH . From this lemma, we can see that MATH has zeros of the first degree only at MATH and MATH respectively. Since these zeros correspond to MATH, there is no cosingular vector other than MATH . This ends the proof for cosingular vectors MATH. Using similar argument, we can also find all the cosingular vectors in MATH and this proves that all singular vectors are given by MATH and REF. |
math/0101226 | In the case of the homogeneous gradation CITE, to prove the same theorem the explicit form of NAME construction is used only to show the existence of singular and cosingular vectors in REF . Since their existence is proved in REF in our case, the theorem can be proved using the same argument. |
math/0101226 | This is proved by the same argument given in REF. Note that as operators neither MATH nor MATH are zero but MATH and MATH hold on the NAME modules. For REF , let MATH be the canonical homomorphism and modules MATH be submodules generated from MATH in REF. Properties of the map MATH are known in the proof of REF with the NAME filtration and we can show MATH . |
math/0101230 | This is obvious (see CITE). |
math/0101230 | We first verify that the NAME multiplication by elements of the space MATH spanned by the generators of MATH is orthogonal. Recall that it suffices to verify that each element MATH acts on MATH as a skew-symmetric endomorphism. We abuse the notation and write MATH for an endomorphism associated with MATH. Clearly, MATH since NAME multiplications on MATH and MATH are orthogonal. Since every element of MATH is a linear combination of such products, our assertion follows. To verify properties of the integral structure, it suffices to show that elements of the basis REF are permuted up to sign by the double products of generators. There are three cases to consider MATH, MATH, and MATH. In the first two cases the action is as desired since the multiplication by MATH's permutes MATH up to sign and the multiplication by MATH's acts the same way on MATH. To treat the third case, note that MATH and MATH. Thus, up to signs, MATH since double products of generators of MATH and MATH permute up to sign the distinguished bases of MATH and MATH respectively. Similarly, MATH . This proves the lemma. |
math/0101230 | All assertions except irreducibility are contained in the lemma above. By CITE MATH is equal to the dimension of an irreducible, MATH-graded NAME module over MATH. It follows immediately that MATH is irreducible. |
math/0101231 | Indeed, the right hand side is generated by all the oriented cycles in MATH starting and ending at MATH and is therefore generated by the MATH and the MATH where MATH is an arrow in MATH starting in MATH and ending in MATH. If we have an algebra morphism MATH then we have an associated algebra morphism MATH defined by sending MATH to the MATH-entry of the MATH matrix MATH and MATH to the MATH-entry of MATH. The defining relations among the MATH and MATH introduced before imply that MATH is indeed an algebra morphism. |
math/0101240 | As mentioned before, it is shown in CITE that MATH gives a lamination near MATH. NAME MATH by a finite number of balls MATH. Let MATH be local parameterizations of MATH near MATH such that, for all MATH, MATH and MATH. Such parameterizations exist if the balls MATH are taken small enough. Write MATH. Let MATH be the image of MATH under MATH , and set MATH. Finally, let MATH be a partition of unity subordinate to the covering MATH of MATH. Then the map MATH whose inverse is given by the formula MATH has the desired properties. |
math/0101240 | For each MATH, one has MATH . Letting MATH, we see that MATH. The proposition now follows (with MATH) from this and the openness of the maps MATH. |
math/0101240 | Let MATH be a function to be specified shortly. It is easily shown (say, using linear coordinates on MATH) that there is a constant MATH, depending only on the rank of MATH, such that for all MATH, MATH, and MATH, MATH where MATH. Now MATH . Thus MATH does the trick provided MATH is bounded from below by a sufficiently large constant. This completes the proof. |
math/0101240 | The statement of the proposition does not change if we replace the metric on MATH by the metric given by REF, so let us work with that metric. By compactness of MATH it suffices to show the estimate REF for large MATH. Let MATH as in the definition of pseudo-linear, and let MATH be the associated exponent of contraction. Thus we have MATH on MATH for all MATH. Pick MATH so small that MATH. For MATH let MATH be the decomposition of MATH relative to the splitting MATH. Then MATH so MATH. For MATH we have MATH where the MATH's are polynomials with no constant or linear terms. The slowly varying nature of the MATH's implies the existence of constants MATH and MATH such that MATH . We now inductively show the following estimate, which clearly implies the statement of the proposition (with MATH replaced by MATH): there exists MATH such that if MATH, MATH and MATH then MATH . This estimate clearly holds for MATH. Suppose it holds for MATH and let us show that it then hold for MATH after possibly increasing MATH. Indeed, if MATH then REF imply that MATH if MATH is large enough. Thus REF holds, which completes the proof. |
math/0101240 | Write MATH and recall our notation MATH. We construct a pair of sequences MATH and MATH of slowly varying polynomial bundle maps such that: CASE: MATH and MATH; CASE: MATH; CASE: there exists MATH such that MATH for all MATH; CASE: MATH is a pseudo-linear map of the flag associated to the splitting of MATH; CASE: MATH converges to an analytic map MATH. The construction is inductive, and proceeds as follows. Set MATH and MATH. Suppose that we have constructed MATH and MATH. Let MATH with MATH and MATH to be determined. Then a simple calculation shows that MATH . Thus, writing MATH, we see that REF holds for MATH if we can find homogeneous solutions MATH and MATH for the equation MATH . The next lemma shows that solutions of REF exist so that REF hold as well. If MATH is a homogeneous, slowly varying, polynomial bundle mapping of degree MATH, then there exist homogeneous, slowly varying polynomial bundle mappings MATH and MATH, also of degree MATH, such that MATH . Moreover MATH can be chosen as follows. If MATH is sufficiently large, then one can take MATH. Otherwise, MATH is a pseudo-linear map of the flag associated to MATH with no linear part, that is, MATH has the form MATH . Before proving REF , we need to develop a few ideas. First, the homogeneous polynomial bundle mapping MATH can be decomposed with respect to the splitting MATH into MATH-homogeneous parts MATH as in REF. By linearity we only need to solve REF for each summand MATH. Recall that if MATH is slowly varying, then so are these summands. In order to solve REF we will make crucial use of the fact that MATH is regular with respect to the splitting MATH. Let MATH be the NAME data associated to MATH and notice that MATH since MATH is contracting. If MATH is MATH-homogeneous, then it follows that MATH . This estimate, which is the key to the analysis in this section, will be made more precise below. Here we only note that the quantity MATH plays an obvious role in REF and leads to the following definition. We say that the pair MATH is resonant if MATH . It is called non-resonant otherwise, and more specifically, super-resonant or sub-resonant if we have MATH or MATH, respectively. There exists MATH such that all pairs MATH with MATH are sub-resonant. Furthermore, if MATH is resonant then MATH. For the first statement pick MATH. The second statement is easily verified. For the rest of this section we work with a fixed but small MATH. Specifically we require that MATH where the supremum is taken over all sub-resonant pairs MATH (it is easy to check that this is possible), and MATH where the minimum is taken over all sub-resonant pairs MATH; the latter form a finite set by REF . With notation as in REF , if MATH is non-resonant and MATH is MATH-homogeneous, then one can find a slowly varying MATH so that MATH . Further, there exists a constant MATH, depending only on on the NAME data MATH, such that MATH where MATH is the operator norm associated to the adapted metric REF. The operator MATH has the following two formal inverses: MATH . We proceed in two cases. CASE: NAME case: let MATH. CASE: NAME case: let MATH. We have to show that this makes sense. Let us consider the sub-resonant REF . We have MATH . It follows from REF that MATH and that MATH does not depend on MATH, MATH or MATH. Thus REF holds, and this implies that MATH is slowly varying. Notice that the construction of MATH does not depend on the choice of MATH. The super-resonant case is treated similarly, using REF. This completes the proof. Simply decompose MATH as a sum of resonant, sub-resonant and super-resonant MATH-homogeneous polynomial bundle mappings MATH. For the non-resonant terms we get MATH from REF and set MATH. For the resonant terms we set MATH and MATH. Then set MATH and MATH. From REF we get that MATH and MATH are slowly varying. The remaining statements follow from REF . Returning to the proof of REF we have constructed our sequences of maps MATH and MATH, and have shown that they satisfy REF - REF above. It remains to show that MATH is analytic and slowly varying. For this it suffices to show that MATH for some MATH-slowly varying MATH and some concave function MATH. Recall that the degree of the inverse of a pseudo-linear map is bounded in terms of the degree of the map itself. In particular, the degree MATH of MATH is finite, so we may write MATH where MATH and MATH is homogeneous of degree MATH, and all the maps are slowly varying. Finally, write MATH, so that MATH is analytic and slowly varying. We denote the homogeneous expansion of MATH by MATH observing that MATH. Keeping MATH fixed we now pick a MATH-slowly varying MATH with the following properties: MATH . Here MATH and MATH will be chosen later. As before we write MATH. That REF - REF are possible follows from the fact that each MATH and each MATH are slowly varying, as are MATH and MATH, while MATH is contracting. We claim that if MATH is small enough and MATH is large enough, then MATH . This is clearly true for MATH by our choice of MATH. Suppose that REF holds for some MATH. To establish it for MATH it suffices, in view of REF , to prove that MATH where MATH is given by REF. To this end, let us take a look at the quantity MATH, keeping in mind that we are only interested in the terms of degree MATH. Thus the terms coming from MATH automatically disappear, and we have MATH where the sum is over all MATH and MATH such that MATH, MATH, MATH and MATH. (The constraint on the index MATH comes from the fact that REF MATH, and REF we need only consider the terms of degree MATH.) The sum REF consists of considerably fewer than MATH terms. We now estimate MATH in all possible cases, using REF and the induction hypothesis, and assuming that MATH is small while MATH is large. CASE: then MATH and MATH . CASE: then MATH and MATH . CASE: then MATH and MATH. Further, a straightforward calculation readily shows that MATH, so MATH . Putting all this together, we have: MATH . Thus the induction step is complete and REF holds for all MATH. REF now follows from the definition of a slowly varying bundle map. |
math/0101242 | We have MATH . |
math/0101242 | From REF we obtain that MATH where MATH and MATH. (Note that MATH.) Since MATH we may rewrite REF as MATH . The elements MATH are linearly independent over MATH, hence MATH. From REF we have MATH, which implies that MATH, as well as MATH. Therefore, MATH . But MATH which is nonzero unless MATH. |
math/0101242 | We may assume that MATH. The minimal polynomial for MATH (over MATH as well as over MATH) is given by MATH . Hence, by letting MATH be a representative of MATH modulo MATH, we can rewrite the left hand side of REF as MATH with no further relations among the MATH's, and thus MATH if and only if MATH for all MATH not congruent to zero modulo MATH. |
math/0101242 | We first assume that MATH. Let MATH be the factorization of the principal ideal MATH. Since MATH does not ramify in MATH, we have MATH, and hence MATH where MATH is the ramification index of MATH in MATH. Since MATH is congruent to MATH modulo MATH for all MATH, we find that MATH for all MATH. Now, since MATH, we have MATH and by REF , MATH. Thus MATH for all MATH, that is, MATH for all MATH. But if MATH then MATH for all MATH, and since complex conjugation permutes the set of primes of MATH that lies above MATH, and MATH we get that MATH for all MATH, contradicting that the ramification index is MATH. For the general case, the previous argument carries through by noting that MATH is a unit (and thus multiplication of MATH by MATH does not change the ideal factorization) and that MATH if and only if MATH. |
math/0101242 | Collecting terms in REF according to the values of MATH, we obtain MATH . Clearly MATH and MATH for at most MATH values of MATH. Letting MATH we may, by REF , write REF as MATH . Since MATH we get that MATH and hence MATH for all MATH. Since MATH for at most MATH values of MATH, MATH implies that MATH for some MATH, and thus MATH. Since MATH we find that MATH. On the other hand, MATH. Thus MATH, and it follows that MATH for MATH. In other words, MATH for MATH, and since there are at most MATH nonzero values among the MATH's, the remaining ones must all be equal to zero. |
math/0101245 | MATH . |
math/0101245 | CASE: Creating a three iteration partial NAME basis with the relations MATH using the NCGB command NAME, yields a set of polynomials, which includes relations MATH and relations REF . See REF pages REF. The order used is given on page REF, order REF . Since polynomials created through the NAME basis algorithm are in the polynomial ideal generated by the original relations, the validity of relations REF is a consequence of relations REF . For us to write the relations REF in the form REF , we require the invertibility of MATH and MATH. These invertibility relations are provided by the NAME lemma given above since the outer matrix REF , consisting of MATH,MATH,MATH, and MATH (all knowns), is assumed to be invertible. With this, we can solve for MATH explicitly in equations REF and write the relations REF defining MATH. Furthermore, we may use these definitions of MATH to write the relations REF as REF . CASE: The converse is again approached using a NAME basis method. As above, the NAME complement formulas give the invertibility of MATH and MATH, which shows that relations REF follow from REF . The question then becomes whether or not relations REF are in the ideal generated by polynomials REF . We create a seven iteration partial NAME basis MATH from the polynomials REF with the NCGB command NAME, under the graded (length) lexicographic monomial order. One can verify that the original REF reduce to MATH with respect to MATH. This shows that the relations MATH and MATH are elements of the noncommutative polynomial ideal generated by the relations REF and the invertibility of the knowns. The result follows. |
math/0101248 | Consider the cylindrical model of MATH described above. MATH corresponds to a MATH-plane in MATH which is disjoint from MATH. There are two hyperplanes containing MATH which are tangent to MATH, and one of them is tangent to MATH along a line; so there is a unique hyperplane MATH which contains MATH, is transverse to MATH, and is tangent to MATH. MATH intersects MATH along a MATH-dimensional manifold which, by construction, is a totally geodesic hyperplane in MATH. |
math/0101248 | Let MATH; call MATH the vector in MATH corresponding to the variation of the dual point to the horosphere tangent to MATH at a point which moves in the direction of MATH on MATH. MATH is the sum of a term MATH corresponding to the displacement of MATH (with a parallel transport of the tangent hyperplane) and a term MATH corresponding to the variation of the tangent hyperplane, while MATH doesnt move. Using the cylindrical model, one checks that MATH (with both terms seen as in MATH) while MATH. |
math/0101248 | This follows again from REF , and from the correspondence between vectors on MATH and on MATH. |
math/0101248 | Let MATH. For MATH, call MATH the geodesic sphere of radius MATH centered at MATH, with the normal oriented towards the exterior for MATH and towards the interior for MATH. Define a map MATH from MATH to MATH sending MATH to the horosphere tangent to MATH at the point MATH, where MATH is considered as a unit vector in MATH. MATH can then be extended by continuity to a map from MATH to MATH. MATH is the inverse of MATH. By REF , the metric induced on MATH is: MATH . Now, using for example, the cylindrical model described above, with MATH as the ``north pole" in MATH, shows that the surfaces MATH are the intersections of MATH (seen as a cylinder in MATH) with the horizontal hyperplanes, that is, the hyperplanes in MATH which are parallel to the hyperplane containing MATH. Therefore the lines MATH are in the kernel of MATH, and moreover they correspond to the vertical lines. Finally, by definition of their parametrization (by the distance between equidistant horospheres) it is the same as the one they have in MATH. |
math/0101248 | The first point is a direct consequence of REF above. For the second point remark that, if MATH is the point in MATH which is the intersection of the horospheres in MATH, then the map sending a horosphere MATH to its point at infinity is by construction an isometry between MATH with its induced metric and MATH with the visual metric at MATH. It is therefore a conformal map. |
math/0101248 | Isometries correspond by defintion to isometries of MATH, which act conformally on MATH, and thus on MATH by REF ; point REF follows. Conversely, any conformal transformation of MATH defines by REF a conformal transformation of MATH, and therefore an isometry of MATH, and also an isometry of MATH. This proves point REF . For point REF , let MATH be the dual hyperplane of MATH, that is, the totally geodesic hyperplane in MATH such that MATH corresponds to the set of horospheres tangent to MATH. Let MATH be an isometry of MATH. By construction, MATH is isometric to MATH, so that MATH defines an isometry MATH of MATH. Since MATH is an hyperplane in MATH, MATH has two extensions as an isometry of MATH, one of which preserves orientation. We call this orientation preserving extension MATH again. MATH defines a unique isometry MATH of MATH, which leaves MATH stable by construction. The same works for the other extension of MATH. |
math/0101248 | The first point is obviously a consequence of the other. For the second point note that, in the neighborhood of MATH, the extended vector field MATH is of the form: MATH with MATH tangent to MATH. Therefore the definition of MATH shows that: MATH where MATH is the NAME connection of the induced metric MATH on MATH, and the result follows since MATH at MATH. For the third point note that, by REF , MATH, so that the NAME connection MATH of MATH is given by: MATH where vector fields on MATH and MATH are identified through the projection along the vertical lines. Therefore (by the usual conformal transformation formulas, see for example, REF): MATH so that MATH at MATH since MATH at MATH. |
math/0101248 | This follows again from REF . |
math/0101248 | We call also MATH and MATH the projections of the vector fields on MATH, and MATH its metric, which has constant curvature MATH. The metric on MATH is then the pull-back of MATH under the projection of MATH to MATH along the vertical lines. Therefore, the NAME connection MATH of MATH is (see for example, REF): MATH where MATH is the NAME connection of MATH. Thus, using the fact that MATH at MATH, we find that, still at MATH: MATH and the result follows. |
math/0101248 | For the first part of the lemma, we want to show that the connection (again called MATH) defined by: MATH is torsion-free and compatible with MATH. But it is torsion-free because: MATH the last step using the NAME equation on the dual hypersurface. Therefore MATH, and MATH is torsion-free. To check that MATH is compatible with MATH is even more direct. If MATH are vector fields on MATH, then: MATH . The second point of the lemma is easy to prove using the first; if MATH and MATH are vector fields on MATH, then: MATH . |
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