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math/0103106 | We use the following result of Montesinos CITE (see also CITE). CITE Every closed, oriented connected REF-manifold MATH contains a two component link MATH, such that there is a REF-fold covering of MATH branched over MATH which is a surface bundle over MATH. It is well known that every closed, oriented connected REF-manifold MATH has an open book decomposition with connected binding. Let MATH denote the page and MATH denote the monodromy of an open book on MATH. We use MATH to denote the double of MATH, where two copies of MATH are glued along their boundary. Define MATH . So, MATH is a MATH-bundle over MATH with monodromy MATH, where MATH acts on one copy of MATH and MATH acts on the other copy. Let MATH be the involution on MATH given by MATH where MATH denotes the involution on MATH interchanging the two copies of MATH. Then it is easy to see that MATH. Moreover the fixed point set of the action of MATH on MATH is a disjoint union of two circles. Since every REF-manifold MATH is double branched covered by a surface bundle MATH, and every surface bundle over MATH is weakly symplectically fillable as in the proof of REF , we conclude that every closed oriented connected REF-manifold has a weakly symplectically fillable double cover, branched along a REF-component link. |
math/0103112 | CASE: MATH follows from set inclusion and associativity. MATH for all MATH, and right composition with MATH yields: MATH. CASE: MATH follows from associativity and right composition of states MATH that are equivalent under MATH implies MATH for all MATH. So MATH implies MATH. CASE: This monotone rank property follows directly from REF , because range ordering REF implies rank ordering MATH, so MATH, and partition ordering REF implies rank ordering MATH, so MATH. CASE: It follows immediately that if MATH and MATH have rank MATH k, then so does composition MATH. This closure property means that all elements of rank not exceeding k form a subsemigroup MATH of MATH. In fact, composition of any element MATH with any element MATH yields MATH with MATH, sothat MATH. The same holds for MATH. Hence MATH is both left- and right ideal, that is an ideal of MATH with MATH and MATH (see def-REF next section). |
math/0103112 | Let invariants MATH and MATH be ordered MATH, then MATH is identity for MATH, so their ranges are ordered because MATH. Notice that MATH suffices: MATH is right identity for MATH. Conversely, for commuting invariants: MATH implies MATH. This follows from the state transform structure of an invariant MATH means that each state MATH maps to a state MATH which is fixed under MATH. In other words, no state chains of length MATH occur in the state transition diagram of MATH. Range MATH is the set of fixed states of MATH. Now, if MATH then MATH maps each state MATH into a fixed state of MATH for all MATH, so MATH. Since by REF and MATH commute, we have MATH, which means MATH. Clearly, if MATH for commuting invariants MATH and MATH, then MATH and MATH, and hence MATH commuting invariants with the same range are equal. |
math/0103112 | Ordered invariants have different ranks according to the previous lemma. Let MATH be the lowest rank of an ordered pair of invariants. Then, with REF , MATH has a proper ideal consisting of all elements with MATH, which contradicts MATH being simple. If invariants MATH commute: MATH, then their composition MATH is also invariant: MATH since MATH. Moreover: MATH is ordered under MATH, since MATH so MATH, and similarly MATH. It is easily verified [REF, pREF] that MATH is the greatest lower bound or meet of MATH and MATH. So a commuting pair of invariants is either ordered, or their composition is ordered under both, contradicting simple MATH. Hence no pair of invariants commutes. |
math/0103112 | There are three cases of equivalence for invariants MATH left-, right- and indirect equivalence. In the first two cases of "direct" equivalence, rank-REF yields: MATH implies MATH and MATH, sothat MATH; MATH implies MATH and MATH, sothat MATH. Hence left- or right equivalent invariants have the same rank. Transitivity holds in both cases. For instance let MATH and MATH then MATH, since MATH , and similarly MATH. Also right equivalence is transitive. If MATH and MATH, where MATH differs from MATH and MATH, then MATH are not directly left- or right equivalent, yet they are MATH equivalent, denoted MATH. Here MATH is an equivalence relation, easily verified to be reflexive, symmetric and transitive. If MATH and MATH are indirectly equivalent, via other invariants, then they have the same rank by transitivity. CASE: There are several cases: direct and indirect equivalence, with either MATH=REF or MATH. For MATH=REF, in the MATH equivalent case MATH and MATH the elements MATH and MATH are not different from MATH and MATH, forming MATH and MATH respectively. For MATH equivalence of invariants MATH and MATH, and in case MATH=REF the only other intermediate elements are invariants MATH and MATH, with MATH and MATH, seen as follows. Invariants MATH must have equal rank: MATH REF , hence exact equality holds in MATH, so REF and similarly REF . Composing both sides of REF on the right by MATH and applying REF yields MATH. So sequence MATH permutes MATH. Since MATH is invariant, this is the identity permutation, hence MATH for all MATH, meaning MATH. Similarly, invariance of MATH implies MATH. So strings of MATH are equivalent to strings of MATH, which are just MATH, forming a closure of four invariants, with the next equivalences using MATH - MATH since MATH and MATH since MATH and MATH, - MATH since MATH and MATH since MATH and MATH. These relations are depicted in a rectangular form in REF . The four elements MATH form an invariant semigroup with direct product structure MATH. MATH is represented by a two-component code: MATH with MATH and MATH. In other words, the direct product MATH (for MATH=REF) follows from two complementary congruences (preserved partitions), illustrated by REF . Denote MATH and MATH, then MATH with image MATH, and MATH with image MATH. The direct product is implemented by two independent components MATH the first composes as MATH and the second as MATH. The left- and right equivalences can be plotted pairwise in the plane as shown in REF , which also gives the composition tables of MATH and MATH. From this rectangular display follows the term diagonal equivalence for two indirectly equivalent invariants, since this is the only other form of equivalence. It is denoted by MATH where MATH and MATH are obtained by commutation: MATH and MATH for some MATH and MATH, themselves being diagonal equivalent MATH, with MATH and MATH. Diagonal equivalence occurs in pairs: if MATH then MATH, and vice versa. The above analysis for MATH=REF can be generalized simply to MATH for MATH invariants, with each invariant pair forming either MATH or MATH or MATH. If MATH-REF in MATH and MATH, then MATH and MATH are not invariant, generating invariants MATH and MATH in a MATH-REF cycle, with MATH and MATH. The resulting structure is in general a semi-direct product MATH with a group MATH as subgroup of MATH, occurring MATH times, to be derived next. In case MATH is also an image of MATH, then MATH is direct product MATH. Without going into much detail [REF, Vol.I, appx]: each idempotent MATH, interpreted as left- or right- multiplier, yields (principle) subsemigroups MATH and MATH, respectively represented in the composition table of MATH by the rows and columns REF . Each invariant MATH is the identity of a maximal subgroup MATH, the intersection of MATH and MATH, while MATH contains MATH and its invariant MATH as max-subgroup identity. One readily verifies that all max-subgroups are isomorphic. Equivalencing each to one congruence part, with MATH, yields image MATH where MATH and MATH represent the number of max-subgroups in MATH forming left- respectively, right- copy semigroups MATH and MATH as image. Notice that if the product of invariants is not invariant, MATH is not a sub-semigroup of MATH. On the other hand: although MATH occurs MATH as subgroup of MATH, it may also be an image group MATH, with MATH as direct product. If this is not the case, so MATH occurs as subgroups but not as image of MATH, then MATH is said to be a semi-direct product MATH. |
math/0103112 | CASE: We need to show that MATH commutes with MATH, and is left- and right identity for MATH. Both follow directly from MATH and MATH. CASE: If MATH is not of constant rank, then the minimum rank invariants form a proper ideal MATH REF , and there is an ordered and commuting pair of invariants. Consider invariants MATH and MATH, then invariant MATH is also in MATH and has the same (minimal) rank as MATH, so MATH. Hence strict ordering MATH holds. |
math/0103112 | REF : an anti- commutative semigroup MATH is invariant, because any iteration class MATH is a commutative subsemigroup, so MATH for all MATH, so each element of MATH is invariant. Moreover, MATH is of constant rank; otherwise some pair of invariants MATH would be properly ordered REF and thus commute, contradicting MATH being anti-commutative. REF : REF . CASE: MATH for all MATH pairwise MATH-, MATH- or MATH- equivalent REF . CASE: Pairwise equivalence in MATH implies the direct product structure MATH with MATH as follows. If MATH contains only left- equivalent invariants then MATH where MATH and MATH. The other trivial case occurs when MATH contains n right equivalent invariants, and no left equivalence holds: MATH with MATH=REF and MATH. If both left- and right equivalences occur, the MATH rectangular structure REF is seen as follows. Take any invariant MATH and form two subsets: MATH with all elements MATH that are left equivalent MATH to MATH, and MATH containing all MATH with MATH right equivalent to MATH. They intersect only in MATH, because if MATH is left- and right equivalent to MATH, then it cannot differ from MATH. MATH and MATH are left- and right copy subsemigroups of MATH. Let the orders be respectively MATH and MATH. Pairwise equivalence implies n copies of MATH which form a congruence MATH of MATH with image MATH. Similarly, congruence MATH consists of MATH copies of MATH, yielding image MATH. Since no pair of invariants can be both left- and right equivalent, congruences MATH and MATH are orthogonal: MATH. REF : semigroup MATH consists of pairwise equivalent invariants. Then it is anti- commutative which means that no pair commutes. For assume that one pair of distinct invariants MATH commutes: MATH, then they are either ordered MATH or MATH (in case MATH is MATH or MATH), or their product is a third invariant MATH, their meet, that is ordered MATH and MATH. Either case contradicts pairwise equivalence. |
math/0103112 | Consider the successive ranges MATH which, due to range REF , form a reducing inclusion chain of subsets of MATH. Each range is contained properly in the previous one until the cycle is reached at MATH. As soon as two successive ranges are equal, then so are all next ranges: MATH, etc. (compose left and right by MATH). Once the cycle is reached, the minimum rank is obtained: the initial tail ranks decrease strictly, and all periodic elements in thecycle have equal and minimal rank. |
math/0103112 | CASE: Let periodic element MATH generate invariant e with period MATH, so MATH. Then clearly the inverse of MATH with respect to MATH is MATH. Define MATH for consistency in case MATH=REF (MATH), and denote the inverse of MATH by MATH. If MATH is another periodic element generating MATH, with inverse MATH, then MATH has inverse MATH since MATH, and similarly MATH. It follows that MATH generates the same invariant as MATH and MATH, so closure holds. Inverses are unique, because if MATH has two inverses MATH and MATH then MATH. So all periodic elements generating the same invariant form a group. CASE: Let MATH be two right equivalent invariants MATH so MATH and MATH, then right composition of MATH with MATH is a morphism from MATH onto MATH, meaning MATH is an image of MATH, denoted MATH (divisor relation). This follows, because MATH is identity for each MATH in MATH, while for each REF , where we used MATH. In other words: the image of a composition of elements is the composition of their images. We need MATH to show that MATH, in fact MATH generates MATH upon iteration. This is seen by replacing MATH in REF with MATH, then MATH, and in general MATH. Let MATH be the period of MATH, so MATH, then MATH. So if MATH and MATH, hence MATH and MATH are right-copiers for each other, forming right equivalent invariants MATH, then right composition of MATH with MATH yields image MATH. Similarly, right composition of MATH with MATH yields image MATH. Consequently right equivalent invariants MATH have mutually ordered groups MATH and MATH, so they are isomorphic: MATH. Using left composition by MATH and MATH respectively, it follows that also left equivalent invariants have isomorphic groups. And finally, by transitivity, diagonal equivalent invariants have isomorphic groups as well. In that case MATH with REF MATH, and MATH. The diagonal case covers the other two cases of direct equivalence. |
math/0103112 | REF : REF . CASE: Each element MATH of a constant rank semigroup MATH is periodic REF . Hence MATH is a union of as many maximal subgroups as there are invariants, being the subgroup identities REF . The subgroups are disjoint because no element can generate two invariants. Constant rank implies that no two invariants are ordered REF , hence they are pairwise equivalent and form a direct product MATH REF . CASE: Consider an invariant MATH and elements of form MATH. Let the invariant generated by MATH be MATH with period MATH. Since MATH begins and ends with invariant MATH, we have MATH, meaning MATH, and in fact MATH, since no strict ordering occurs in a constant rank semigroup. Hence MATH, in other words MATH generates invariant MATH for each MATH, and is thus in MATH. So for each MATH in constant rank semigroup MATH, MATH is in the max-subgroup containing MATH, denoted as MATH. If MATH are two equivalent invariants, with maximal subgroups MATH and MATH, then the group isomorphism is MATH with MATH, independent of whether it is a left-, a right- or a diagonal equivalence REF , the last case covers the first two. CASE: Constant rank semigroup MATH contains as many disjoint isomorphic groups MATH as there are invariants. These groups form a direct product image MATH under set product REF . If the two congruences MATH for MATH in the same max-subgroup and MATH if MATH for some invariant MATH REF are orthogonal, with images MATH and MATH, then direct product structure MATH follows. And if the product of two invariants is not invariant then MATH is not a subsemigroup, and MATH not an image of MATH, yielding semi-direct product MATH. CASE: The direct product of simple semigroups is also a simple semigroup [REF, pREF ]. Since MATH and MATH are simple, so is their direct product. Although MATH is an image of MATH, it is not necessarily a subsemigroup, in which case MATH is not an image of MATH, with a coupling from REF to MATH, corresponding to a semi-direct product. In either case, the composition of MATH from simple semigroups MATH and MATH yields transforms of equal rank, so MATH has no proper ideal, thus is simple. |
math/0103122 | We may assume that MATH is locally free, and replacing MATH by some covering, that MATH is a NAME cover of MATH with NAME group MATH. Let MATH be the projective bundles. The induced covering MATH is again NAME. By assumption, for some MATH the sheaf MATH is generated by global sections over MATH. Hence for MATH the sheaf MATH is generated over MATH by MATH-invariant sections, hence MATH is globally generated by sections MATH over MATH. By the NAME, there exists some MATH such that MATH is surjective over MATH. |
math/0103122 | Given positive integers MATH for all components MATH of MATH, NAME constructed a finite non-singular covering MATH see REF, with MATH a normal crossing divisor, such that all components of MATH are ramified of order exactly MATH. In REF we choose MATH to be the zero-divisor of a general section of MATH, and we apply NAME 's construction to MATH, where the MATH are one for the components of MATH and where the prescribed ramification index for MATH is MATH. In REF the semi-stable reduction theorem for families over curves (see CITE) allows to choose the MATH such that the family MATH is semi-stable in codimension one. |
math/0103122 | Let MATH be a finite morphism of non-singular curves such that MATH lifts to MATH, and let MATH be the family obtained by desingularizing the main component of the normalization of MATH. By REF in the definition of a mild morphism, and by the choice of MATH, the family MATH has MATH as a mild model. Applying REF, to MATH and MATH, and using REF, we find MATH . Moreover, by REF, and by base change, one obtains a morphism of sheaves MATH which is an isomorphism over some open dense subset. Let MATH denote the rank of those sheaves. It remains to show, that REF induces an injection from MATH into MATH. To this aim, as in REF, let MATH stand for ``taking a desingularization of the MATH-th fibre product". Then MATH is again birational to a mild morphism over MATH. As in CITE, flat base change and the projection formula give isomorphisms MATH the second one outside of a codimension two subscheme. Since both sheaves are reflexive, the latter extends to MATH. The injection REF , applied to MATH and MATH induces an injective morphism MATH . The left hand side contains MATH as direct factor, whereas the righthand contains MATH, and we obtain the injection for the determinant sheaves as well. |
math/0103122 | It remains to verify, that the construction given in REF, for REF can be modified to guaranty REF along with the others. Let MATH be the induced morphism to the moduli scheme. NAME and NAME constructed a finite NAME cover of the moduli space which is induced by a family (see CITE, for example). Hence there exists some manifold MATH, generically finite over the closure of MATH such that the morphism MATH is induced by a family MATH. By CITE, blowing up MATH, if necessary, we find a projective compactification MATH of MATH and a covering MATH, such that MATH is birational to a projective mild morphism over MATH. Replacing MATH by some generically finite cover, we can assume that MATH has such a model already over MATH. Next let MATH be any variety, generically finite over MATH, for which there exists a morphism MATH. By REF, we are allowed to replace MATH by any manifold, generically finite over MATH, without loosing the mild birational model. Doing so, we can assume the fibres of MATH to be connected. Replacing MATH by some blowing up, we may assume that for some non-singular blowing up MATH the morphism MATH factors through a finite morphism MATH. Next choose a blowing up MATH such that the main component of MATH is flat over MATH, and MATH to be a desingularization. Hence changing notations again, and dropping one prime, we can assume that the image of the largest reduced divisor MATH in MATH with MATH maps to a subscheme of MATH of codimension larger that or equal to MATH. This remains true, if we replace MATH and MATH by finite coverings. Applying REF, to MATH, provides us with non-singular covering of MATH such that a desingularization of the pullback of MATH is semi-stable in codimension one. Again, this remains true if we replace MATH by a larger covering, and using REF, a second time, now for MATH, we can assume that this morphism is as well semi-stable in codimension one. Up to now, we succeeded to find the manifolds in REF such that REF hold true. In REF, the projectivity of MATH and the dominance of MATH over MATH follow from the construction. For the divisor MATH in MATH considered above, we replace MATH by MATH and MATH by MATH, and of course MATH, MATH and MATH by the corresponding preimages. Then the non-equidimensional locus of MATH in MATH will be of codimension larger than or equal to two. MATH is generically finite, by construction, hence finite over the complement of a codimension two subscheme of MATH. Replacing again MATH by the complement of codimension two subscheme, we can assume MATH to be equidimensional, hence flat, and MATH to be finite. The morphism MATH has reduced fibres over general points of divisors in MATH, hence it is smooth outside a codimension two subset of MATH, and replacing MATH by the complement of its image, we achieved REF. Since MATH is smooth, the pullback of MATH to MATH is smooth outside of MATH. Moreover the induced morphism to the moduli scheme MATH factors through an open subset of MATH. Since by construction MATH is proper over its image in MATH, the image of MATH lies in MATH and we obtain REF. For REF remark that the pullback MATH of the mild projective morphism MATH to MATH is again mild, and birational to MATH. By flat base change, MATH . Since MATH and MATH are normal with at most rational NAME singularities, we obtain (as in CITE) that the sheaf on the right hand side is MATH whereas the one on the right hand side is MATH. REF coincides with REF, and it has been verified in CITE as a consequence of the existence of a mild model for MATH over MATH. |
math/0103122 | For REF one uses a variant of REF, which has been shown CITE It also follows from the obvious extension of the ampleness criterion in CITE, to the case ``ample with respect to MATH": Under the assumption made in REF, for all MATH sufficiently large and divisible, there exist positive integers MATH, MATH and MATH such that MATH ample with respect to MATH. Since we do not want to distinguish between the two REF , we choose MATH in REF, and we allow MATH in REF . By REF, respectively, in both cases the sheaf MATH is ample with respect to MATH. By REF, or REF, the number MATH is bounded by some constant MATH, for all smooth fibres of MATH. We will choose MATH to be divisible by MATH and larger than MATH. Replacing MATH and MATH by some multiple, we may assume that there exists a very ample sheaf MATH and a morphism MATH which is an isomorphism over MATH, and that MATH is divisible by MATH. By REF, there exists a non-singular covering MATH and an effective divisor MATH with MATH, and such that the discriminant locus MATH does not contain any of the components of MATH. Replacing MATH by a slightly smaller scheme, we can assume that MATH, hence MATH is non-singular and by flat base change MATH for all MATH. The assumptions in REF or REF remain true for MATH, and by CITE, it is sufficient to show that the conclusions in REF hold true on MATH for MATH. Dropping the primes, we will assume in the sequel that MATH has a section whose zero-divisor is MATH for a non-singular divisor MATH. Let MATH denote the rank of MATH. We choose MATH consider the MATH-fold fibre product MATH and a desingularization MATH . Using flat base change, and the natural maps MATH one finds morphisms MATH and both are isomorphism over MATH. We have natural maps MATH where the last morphism is the multiplication map. Hence we obtain MATH . Thereby the sheaf MATH is a subsheaf of MATH. Let MATH be the zero divisor of the corresponding section of MATH . For the sheaf MATH one finds MATH . By REF, and by the choice of MATH we have a morphism MATH surjective over MATH. The sheaf MATH is weakly positive over MATH and there is a morphism MATH surjective over MATH. Since the morphism of sheaves in REF , as well as the first one in REF , split locally over MATH the divisor MATH can not contain a fibre MATH of MATH and by REF, or CITE, for those fibres MATH . Applying REF to MATH one obtains the weak positivity of the sheaf MATH over MATH. By REF one finds morphisms, surjective over MATH . Since the quotient of a weakly positive sheaf is weakly positive, the sheaf on the right hand side is weakly positive over MATH, hence MATH is ample with respect to MATH. For MATH sufficiently large MATH is ample, and one obtains the second part of REF. |
math/0103122 | If one replaces in the tautological sequence REF the divisor MATH by a larger one, the sheaf on the right hand side does not change, hence MATH . Both, MATH and MATH behave well under pullback to MATH (see CITE, for example). To be more precise, there exists an isomorphism MATH and an injection MATH which is an isomorphism over the largest open subscheme MATH, where MATH is an isomorphism. Since MATH is non-singular outside of MATH, and since the singularities of MATH can only appear over singular points of the discriminant MATH, we find MATH. For REF we use again that MATH is non-singular over MATH. So REF induces an isomorphism MATH . The natural map MATH and the projection formula give MATH and REF follows by flat base change. |
math/0103122 | By CITE, the sheaf MATH is globally generated over some open subset. However, by REF MATH is contained in MATH . Since MATH one obtains REF, as stated. |
math/0103122 | We use the notations from REF , that is, MATH and MATH. As in the proof of REF, in order to show the existence of the morphisms MATH and the commutativity of REF we may enlarge MATH and MATH to include the discriminant loci, hence assume that MATH . By the generalized NAME formula CITE, MATH and by REF , MATH . The tautological sequence MATH defines a filtration on MATH, with subsequent quotients isomorphic to MATH . Induction on MATH allows to deduce that MATH . On the other hand, the inclusion MATH and flat base change gives MATH hence MATH. The second morphism in REF is an isomorphism on the largest open subset where MATH is an isomorphism, in particular on MATH. The way we obtained REF the morphisms are obviously compatible with the different tautological sequences. Since we assumed MATH to contain the discriminant locus, the pull back of REF to MATH is isomorphic to MATH and REF commutes. |
math/0103122 | The properties REF - REF have been verified in CITE in case the general fibre is canonically polarized. So let us just sketch the arguments. By CITE see also REF the sheaf MATH is a direct factor of MATH. The morphism MATH is induced by the natural inclusions MATH tensored with MATH. Such an injection also exist for MATH replaced by MATH. Since the different tautological sequences are compatible with those inclusions one obtains REF. Over MATH the kernel of MATH is a quotient of the sheaf MATH . In particular MATH is injective. The same holds true for all the MATH in case MATH is fibre wise ample, by the NAME vanishing theorem. By REF holds true. For REF, recall that over MATH the morphism MATH is an isomorphism. By the projection formula the morphism MATH is the restriction of the edge morphism of the short exact sequence MATH . The sheaf on the right hand side is MATH and the one on the left hand side is MATH. For MATH, tensoring the exact sequence with MATH and dividing by the kernel of the wedge product MATH on the left hand side, one obtains an exact sequence MATH where MATH is a quotient of MATH. By definition, the restriction to MATH of the morphism considered in REF is the first edge morphism in the long exact sequence, obtained by applying MATH to REF . The wedge product induces a morphism MATH . This morphism factors through MATH. Hence the exact sequence REF is isomorphic to the tautological sequence MATH . The edge morphism MATH of REF is the NAME map. In order to prove REF, we use as in the proof of REF, NAME 's covering construction to find a non-singular finite covering MATH such that for some desingularization MATH of MATH the induced variation of NAME structures has uni-potent monodromy, and such that MATH is semi-stable. From REF, applied to MATH, MATH and MATH instead of MATH, MATH and MATH one obtains a commutative diagram MATH where MATH, where MATH and where MATH is the edge-morphism. In particular the pullback of the kernel of MATH, the sheaf MATH, lies in the kernel MATH of MATH. Leaving out some codimension two subschemes of MATH and MATH, we may assume that MATH is a subbundle of MATH. Choose a smooth extension MATH of MATH such that the closure of MATH is a normal crossing divisor, and let MATH be the NAME bundle, corresponding to the canonical extension of the variation of NAME structures. For some choice of the compactification MATH will extend to a subbundle MATH of MATH. By CITE, the dual MATH is numerically effective, hence weakly positive. Thereby MATH is weakly positive over some open subset, and the compatibility of weak positivity with pullback shows REF. For REF one just has to remark that on page REF it is shown that MATH for MATH . |
math/0103122 | Using the notations from REF, write MATH. By REF, and by the choice of MATH . By REF, there is a section MATH, generating MATH over MATH, and by REF, MATH can not lie in the kernel of MATH. Hence the largest number MATH with MATH satisfies MATH. By the choice of MATH and REF implies that MATH lies in MATH . We obtain morphisms of sheaves MATH . By REF, the sheaf on the left hand side is big, hence its image MATH is big as well. |
math/0103122 | The equality MATH is obvious by definition. Moreover all the sheaves in REF are torsion free, hence it is sufficient to verify the existence of MATH on some open dense subset. So we may replace MATH by an affine subscheme, and REF holds true for MATH. By REF, the sheaves MATH embed in the sheaves MATH, in such a way that MATH restricts to MATH. One obtains REF from REF. |
math/0103122 | Recall that under REF we have considered above the slightly different sheaf MATH for MATH. So MATH is a subsheaf of MATH of full rank. The compatibility of MATH and MATH implies that MATH is a subsheaf of MATH of maximal rank. Hence the induced morphism MATH is an isomorphism over some dense open subset. By REF, the sheaf MATH is a subsheaf of MATH and by REF, the restriction MATH coincides with MATH. Using the notations from REF, one obtains MATH . By REF, the dual sheaf MATH is weakly positive. REF induces morphisms MATH surjective over some dense open subset, and we obtain REF. |
math/0103122 | Let us consider as in REF some finite morphism MATH, a desingularization MATH of MATH, the induced morphisms MATH and MATH, MATH, and MATH. In REF we constructed an injection MATH compatible with the edge morphisms MATH and MATH. Thereby we obtain an injection MATH . In case MATH is non-singular, MATH and hence MATH are isomorphisms. If MATH, REF hold true. As in the proof of REF, there exists a finite covering MATH with MATH non-singular, such that REF holds true for the pullback family MATH. Since a sheaf is ample with respect to some open set, if and only if it has the property on some finite covering, we obtain the bigness of MATH by applying REF to MATH. In general REF allows to assume that MATH fits into REF constructed in REF. Let us write MATH, and MATH for the sheaves corresponding to MATH and MATH on MATH instead of MATH. As we have seen in REF the smoothness of MATH implies that MATH, and MATH . On MATH we are in the situation where the variation is maximal, hence REF holds true and the dual of the kernel MATH is big. So for any ample invertible sheaf MATH we find some MATH and a morphism MATH which is surjective over some open set. Obviously the same holds true for any invertible sheaf MATH, independent of the ampleness. In particular we may choose for any MATH with MATH and for the number MATH given by REF the sheaf MATH . By REF and by REF , applied to MATH, the sheaf MATH is generically generated. By REF the same holds true for some power of MATH . By REF and by the choice of MATH in REF, one finds MATH see REF. |
math/0103122 | By REF, there exists some MATH, a big coherent subsheaf MATH, and an injective map MATH . Its cokernel MATH, as the quotient of a weakly positive sheaf, is weakly positive, hence MATH is the tensor product of the big sheaf MATH with the weakly positive sheaf MATH. |
math/0103122 | In both REF would imply for some MATH that MATH has a subsheaf MATH of positive NAME dimension. But MATH, as a quotient of a weakly positive sheaf, must be weakly positive, contradicting MATH. |
math/0103122 | Let MATH be an invertible sheaf of NAME dimension MATH. Replacing MATH by some power, we may assume that MATH. We have to verify that there is no injection MATH . For MATH such an injection would contradict the vanishing REF shown in CITE. Hence starting with MATH we will show the non-existence of the subsheaf MATH by induction on MATH and on MATH. The exact sequence MATH induces a filtration on MATH with subsequent quotients MATH for MATH. By induction none of those quotients can contain an invertible subsheaf of positive NAME dimension. Hence either the restriction of MATH to MATH is a sheaf with MATH, hence MATH, or the image of MATH in MATH is zero. In both cases MATH contains an invertible subsheaf MATH with at least two linearly independent sections, hence of positive NAME dimension. Now we repeat the same argument a second time: MATH has a filtration with subsequent quotients MATH . MATH is ample, hence by induction none of those quotients can have a non-trivial section. Repeating this argument MATH times, we find an invertible sheaf contained in MATH and of positive NAME dimension, contradicting the induction hypothesis. |
math/0103122 | Let us return to the notations introduced in REF. There we considered an open dense non-singular subvariety MATH of MATH, depending on the construction of REF . In particular MATH embeds to MATH. Let MATH be an ample invertible sheaf on MATH. In order to prove REF we have to find some constant MATH which is an upper bound for MATH, for all morphisms MATH with MATH. For REF we have to show the same, under the additional assumption that MATH. Let us start with the latter. By REF , or by assumption in REF , one finds the sheaf MATH, defined in REF, to be ample with respect to MATH. For REF of REF, that is, if one just assumes that MATH, we may use REF, and choose MATH a bit smaller to guarantee the ampleness of MATH over MATH. Replacing MATH by some multiple and MATH by some tensor power, we may assume that MATH is generated by global sections over MATH. Assume first that MATH. Let MATH be a morphism between projective manifolds, obtained as a compactification of MATH. By REF is smooth over MATH. In REF we have shown, that MATH . On the other hand, upper bounds for the right hand side have been obtained for REF and in general in CITE. Using the notations from CITE, MATH where MATH is the genus of MATH, where MATH, and where MATH is a positive constant, depending on the general fibre of MATH. In fact, if MATH is a general fibre of MATH, the constant MATH can be chosen to be MATH. Since the latter is upper semicontinous in smooth families see CITE or REF there exists some MATH which works for all possible curves. Altogether, we found an upper bound for MATH, whenever the image MATH meets the dense open subset MATH of MATH. In REF, the assumptions made in REF are compatible with restriction to subvarieties of MATH, and we may assume by induction, that we already obtained similar bounds for all curves MATH with MATH. |
math/0103122 | For MATH, the non-singular compactification of MATH, and for MATH, a compactification of MATH is given by MATH with boundary divisor MATH. Then MATH where the sum is taken over all tuples MATH with MATH. If MATH, each of the factors is the pullback of some sheaf on a strictly lower dimensional product of curves, hence for MATH any morphism from a big sheaf MATH to MATH must be trivial. If MATH is a finite covering, the same holds true for MATH. By REF, there exists such a covering, some MATH and a big subsheaf of MATH, hence MATH. |
math/0103122 | Let MATH be the projective bundle. MATH defines sections of MATH, which are not all identically zero on MATH for MATH. Hence there exists a non-singular curve MATH passing through MATH, such that the composite MATH is surjective over a neighborhood of MATH. For a nonsingular curve MATH and MATH consider a morphism MATH, with MATH. Let MATH and MATH be projective non-singular curves, containing MATH and MATH as the complement of divisors MATH and MATH, respectively. On the complement MATH of a codimension two subset of MATH the morphism MATH extends to MATH. Then MATH induces a morphism MATH whose composite with MATH is non-zero for all MATH in an open neighborhood of MATH, hence MATH is surjective over some open dense subset. Since MATH is ample, this is only possible if MATH factors through the second projection MATH. |
math/0103122 | Assume MATH is infinite, and choose an infinite countable subgroup MATH . Let MATH be the open subset of MATH considered in REF, and let MATH be the open subset from REF. We may assume that MATH and write MATH. Since MATH we can find a point MATH whose MATH-orbit is an infinite set contained in MATH. By REF there are rigid smooth curves MATH passing through MATH. Obviously, for all MATH the curve MATH is again rigid, it meets MATH, hence MATH and the set of those curves is infinite, contradicting REF. |
math/0103122 | By REF the existence of such a family implies that for some MATH the sheaf MATH contains a big coherent subsheaf MATH. Replacing MATH by some multiple, one can assume that MATH is ample and invertible, contradicting REF. |
math/0103122 | Write MATH. The surjection MATH defines a morphism MATH. For the ideal MATH of MATH the induced morphism MATH is surjective, as well as the composite MATH where MATH is the blowing up of MATH with exceptional divisor MATH, and where MATH. Let us write MATH for the rank of MATH. For MATH and MATH let MATH be the blowing up of MATH, with exceptional divisor MATH. Then MATH . In particular MATH is equidimensional, hence flat. For MATH and for MATH and MATH . One has an exact sequence MATH and MATH . By flat base change one finds MATH . Moreover MATH is an isomorphism. The inclusion MATH factors through MATH . This map is an isomorphism. We know the surjectivity of MATH so for MATH the morphism from MATH to MATH is surjective as well. By the choice of MATH one has MATH . Starting with MATH, assume by descending induction that MATH . Since MATH one finds MATH as well. |
math/0103126 | The verification is straightforward, using the identities MATH where MATH and MATH is the permutation of the factors. |
math/0103126 | From REF we obtain MATH . Now the formulas for MATH are checked by substituting this for MATH and expanding the commutators of NAME generators. The formulas for other generators are obvious. |
math/0103126 | It is straightforward to check that the formulas above define a representation of MATH on MATH (compare REF). Using REF , we obtain then that the action of MATH is as in REF . |
math/0103126 | We use the triangular decomposition of the universal MATH-matrix of MATH obtained in CITE. We have MATH where MATH is defined in REF. Explicitly we have MATH where MATH is the set of positive roots of MATH, and MATH (see REF for the description of the order of these factors). The explicit formula for MATH is given in REF (see also REF). Therefore we find (compare with REF): MATH and MATH where MATH (note that we have moved the term MATH to the middle factor). |
math/0103126 | The formulas for MATH, MATH, follow from comparing the constant coefficients of the diagonal entries in the above formula for MATH. The formulas for MATH and MATH, MATH, follow from comparing the constant coefficients of the MATH entries in MATH and of the MATH entries in MATH respectively. Finally the formulas for MATH and MATH follow from comparing the MATH coefficient of the MATH entry of MATH and the MATH coefficient of the MATH entry of MATH respectively. After that it is straightforward to check on the generators that this map is compatible with comultiplication. |
math/0103126 | The commutativity of all faces except for the back, lower left and lower right ones follow from the definitions of the maps. The commutativity of the lower left face is checked explicitly on generators. The only nontrivial case is that of MATH. We use the relations MATH which hold in MATH. They can be rewritten in the form MATH . Now, we compute MATH . The computation for MATH is similar. The same argument also proves the commutativity of the lower right face. The commutativity of the back face is proved by standard diagram chasing argument from the commutativity of the other faces. |
math/0103126 | The comultiplication formula MATH implies that the eigenvalues of MATH's satisfy the multiplicative property. Let us extend MATH by adding MATH and MATH. Consider the commutative subalgebra MATH (respectively, MATH) of MATH generated by MATH, MATH, MATH, MATH (respectively, MATH, MATH, MATH, MATH). Then we have an embedding MATH. Using this embedding, it suffices to establish the multiplicative property of MATH and MATH. The former follows from REF, and the latter follows from the comultiplication REF . |
math/0103126 | Let MATH be an irreducible finite-dimensional MATH - module. In order to lift it to a representation of MATH, we need to define an action of MATH and MATH on MATH in such a way that the relations REF and the relations MATH are satisfied. We make MATH act as the identity and define the action of the remaining series MATH, recursively using REF . Likewise, we make MATH act as the identity and define the remaining elements MATH recursively using the formula MATH. The representation of MATH obtained in this way satisfies the following conditions: its restriction to MATH is of type REF, so that MATH; the common generalized eigenvalues of MATH on MATH have the form MATH and MATH. Now observe that MATH has a family of one - dimensional representation MATH (the quantum determinants) on which the subalgebra MATH acts trivially, MATH acts by multiplication by MATH, and MATH acts by multiplication by MATH. By REF, the generalized eigenvalues of MATH and MATH satisfy the multiplicative property. This property implies that the tensor product of MATH with the one-dimensional representations MATH, where MATH runs over the list of the poles of MATH (MATH), has the structure of a MATH - module on MATH, which satisfies the condition of the lemma. |
math/0103126 | The proof proceeds along the lines of the proof of a similar statement in the case of the NAME given in REF (see also CITE; a similar formula may be found in CITE in the NAME context). Since finite-dimensional representations of MATH are MATH - deformations of representations of MATH, the patterns for restrictions of MATH - modules and MATH - modules are the same. In particular, for each NAME scheme MATH (equivalently, a semistandard tableau MATH of shape MATH), there is a vector MATH of MATH, which is contained in the MATH - component of MATH with highest weight MATH. We claim that each of these vectors MATH is a common eigenspace of MATH which contributes the monomial MATH to MATH. In order to show this, we consider an action of the series MATH on MATH. Using REF , we can then recover the action of MATH and MATH. The coefficients of MATH are central elements of MATH and they act by scalar operators on each irreducible MATH module. In particular each MATH is an eigenvector of MATH. Recall that by REF , the evaluation homomorphism commutes with the restrictions. Therefore, if MATH is an irreducible MATH submodule of MATH, then MATH is an irreducible MATH submodule in MATH. To compute the action of MATH on MATH, we consider the lowest weight vector MATH of the corresponding MATH module. In other words, MATH is the lowest weight vector of the the evaluation MATH module containing MATH. Note that by the weight decomposition of MATH, the vector MATH is an eigenvector of each MATH, MATH. Let us compute these eigenvalues. Consider any evaluation module MATH of MATH. It follows from the formulas for the triangular decomposition of MATH that the action of MATH REF on a lowest weight vector MATH of MATH coincides with the action of MATH. Note that MATH by REF , therefore using REF we compute MATH (note that the weight of MATH is MATH). Therefore, using REF we find that the eigenvalue of MATH on MATH is MATH . Using REF and the above rule assigning to MATH a semistandard tableau, we obtain the statement of the lemma. |
math/0103126 | Let MATH be the degree of MATH. The transition matrix from MATH to MATH with entries MATH is inverse to the transition matrix from MATH to MATH with entries MATH. Since MATH is a homogeneous element of the ring MATH of degree MATH, these matrices are unions of blocks corresponding to fixed degrees. These blocks are still of infinite size. However, we can further decompose them into blocks of finite size. Fix MATH, MATH and consider the subring MATH of MATH generated by MATH, where MATH, MATH. We claim that the image of the subring MATH in MATH has a basis of irreducible representations whose MATH - character contains MATH only with MATH. Indeed, let MATH be a monomial in MATH. By REF , the MATH - character of the corresponding tensor product of fundamental representations contains MATH only with MATH. In particular, the MATH - characters of all irreducible representations in the Right-hand side of REF contain MATH only with MATH. On the other hand, let MATH be an irreducible representation whose MATH - character contains MATH only with MATH. Represent MATH as a linear combination of tensor products of the fundamental representations as in REF . Then all MATH's in the Right-hand side belong to MATH. Otherwise, the highest weight monomial of the MATH - character of at least one such MATH's would appear with a nonzero coefficient in the MATH - character of the Right-hand side of REF. This monomial would then contain MATH with some MATH. This contradicts our assumption and hence proves the claim. Finally, observe that the subspace of MATH of degree MATH is finite-dimensional, so our transition matrices are unions of blocks of finite size. The statement of the lemma now follows from REF. |
math/0103126 | We express MATH and MATH as linear combinations of tensor products of the fundamental representations, then multiply them out and express the result as a linear combination of irreducible representations. All of these operation are stable (see REF ). |
math/0103126 | We have already proved the associativity of MATH. NAME of MATH follows from REF and stabilization of MATH. The compatibility of MATH and MATH follows from REF and stabilization of MATH. Thus, we obtain the structure of a bialgebra on MATH. However, we will show in REF below that this bialgebra is isomorphic to the bialgebra of a NAME algebra. Therefore it admits a unique antipode (recall that if an antipode exists, then it is unique). This proves the first assertion of the theorem. It is clear from the definition of multiplication on MATH that MATH is its subalgebra. Furthermore, REF implies that MATH is stable under comultiplication. It is also stable under the antipode, as we will see below. Therefore we obtain the second assertion. |
math/0103126 | We only need to check the compatibility of the maps REF with the comultiplication. This follows immediately by comparison of REF . |
math/0103126 | The verification of the first equality on the basis elements is straightforward using REF. If the second equality is true for MATH and MATH (and for all MATH) then it is also true for MATH. Therefore it suffices to check the second equality in the case when MATH is one of the generators MATH. This is done by a direct computation using REF. The third equality now follows automatically, because if it exists, the antipode is unique. |
math/0103126 | Note that REF is a special case of REF (actually, the two formulas are equivalent, but we prefer to prove REF, since we will need it in REF in a different context). First we prove REF when MATH. If MATH, we obtain MATH . For MATH we have MATH . Now suppose that REF is proved for MATH and MATH. Then for MATH, we obtain MATH . This proves REF in general. |
math/0103126 | We consider the case MATH; all other cases are obtained by a shift of all indices. First, we check the formula MATH . For that, we compute explicitly the action of operators MATH on MATH. Recall that the content of the box MATH is defined to be MATH. From the formula for the MATH - character of MATH given in REF we obtain that if the partition MATH has a removable box of content MATH, then MATH . If the partition MATH has no removable boxes of content MATH, then MATH. Comparing this with the action of MATH in MATH (given by MATH), we obtain REF . This proves the first statement of the proposition. Now we have the diagram of MATH - modules and homomorphisms MATH . To complete the proof, we need to show that this diagram is commutative. Note that the vertical arrows are isomorphisms. Consider the homomorphism MATH, obtained by composing these isomorphisms with MATH. Its dual sends MATH to the vector MATH. Such a homomorphism is unique, and hence coincides with MATH. Therefore the above diagram is commutative. |
math/0103126 | The first statement follows from the uniqueness of the homomorphism MATH sending MATH to MATH, as in the proof of REF. The second statement follows from REF. |
math/0103126 | Since we already know that MATH is a NAME algebra, it remains to show that the MATH - subalgebra of MATH generated by MATH (MATH, MATH), is a NAME subalgebra. Let MATH be an invertible element of some MATH - NAME algebra MATH such that MATH. Then we claim that the MATH - subalgebra of MATH generated by MATH REF is a NAME subalgebra. It is sufficient to check this for some algebra MATH with an element MATH, such that MATH generate MATH. An example of such algebra is MATH with MATH. Then MATH is in MATH (because MATH is a NAME subalgebra) and in MATH. Our claim follows then because the intersection of MATH and MATH in MATH is precisely the MATH - subalgebra generated by MATH and MATH. |
math/0103126 | The algebra MATH is generated by MATH, MATH and MATH, MATH (MATH, MATH). Moreover, by REF , the evaluation map commutes with the embedding MATH. Therefore we can use REF to compute the evaluation map on MATH. For MATH, MATH, MATH, we have MATH . In particular, we obtain that the image of MATH under MATH contains MATH. In order to find MATH, we use the presentation REF. We then obtain MATH . Therefore, the result is in MATH. Finally we prove that MATH. Note that the evaluation map commutes with the left upper corner restrictions to MATH (MATH). Therefore, we are reduced to the case of MATH. By REF , we obtain MATH . Therefore using REF we find MATH . This completes the proof. |
math/0103126 | This lemma is proved in exactly the same way as the corresponding statement in the MATH case using the triangular decomposition of REF. |
math/0103126 | It follows from the explicit formula for MATH given in REF that the only dominant monomial in the specialized character MATH is the highest weight monomial. |
math/0103126 | The restriction of MATH to MATH coincides with the NAME homomorphism MATH, as defined in REF. The latter is an algebra homomorphism. It is also clear that the restriction of MATH to the subalgebra generated by the elements MATH is a homomorphism. Thus, we need to check that the map MATH respects the relations between MATH and MATH. We have MATH (MATH, MATH, MATH) and these relations are preserved by MATH: MATH where MATH. Thus, it remains to check that MATH preserves the relations between MATH and MATH. In particular, we are reduced to the case MATH. So let MATH. Due to the triangular decomposition of REF, the relations between MATH and MATH are generated by the relations REF. Let MATH be the automorphism of MATH defined by MATH (MATH, MATH, MATH). Suppose MATH is well defined. Then for odd MATH, the automorphism MATH commutes with MATH. For even MATH, we have MATH for MATH of degree MATH (MATH), where the degree is defined by setting MATH . Therefore, applying MATH, it is sufficient to consider only the specializations of the relations REF between MATH and MATH (for MATH the proof is the same): MATH . For MATH, MATH, define MATH inductively by MATH . It is shown in the proof of REF that MATH. Denote by MATH and MATH the images of MATH and MATH in MATH. Then we obtain from the identities REF: MATH . By induction on MATH, we obtain that MATH if MATH or MATH is not divisible by MATH or MATH is not divisible by MATH. For MATH we obtain MATH (at the initial step of our induction, when MATH, we have MATH, so the above formulas obviously hold). This concludes the proof of the lemma. |
math/0103126 | Let MATH be two MATH - modules. By REF , the eigenvalues of MATH on MATH are products of their eigenvalues on MATH and MATH. Since the multiplication on MATH is defined via the MATH - character homomorphism (see REF ), which records the generalized eigenvalues of MATH, the assertion of the lemma may be reformulated as follows: the eigenvalues of MATH on MATH are products of their eigenvalues on MATH and MATH. Now observe that the eigenvalues of MATH REF are completely determined by the eigenvalues of MATH REF and MATH. According to REF, the restriction of the NAME homomorphism to MATH is compatible with the operation of tensor products. Applying REF again, we obtain that the eigenvalues of MATH REF satisfy the desired multiplicative property. Next, we obtain from REF that the pullback of a generalized eigenspace of MATH with eigenvalue MATH is a generalized eigenspace of MATH with eigenvalue MATH. Therefore the eigenvalues of MATH, also satisfy this property. This completes the proof. |
math/0103126 | The proof is similar to that of REF. Namely, we first check the commutativity of this diagram with MATH replaced by MATH. It follows from REF , using the fact that NAME map commutes with the braid group action (see REF). We also use the identities MATH for odd MATH and MATH. Second, we check commutativity of the analogous diagram for the generalized eigenvalues of MATH using REF . Combining these results, we obtain the assertion of the lemma. |
math/0103126 | In the case of MATH the statement of this lemma is proved in REF . This gives us the eigenvalues of MATH REF on MATH. Next we compute the eigenvalues of MATH using the definition of the NAME map. Combining these two results, we obtain the lemma. |
math/0103126 | Associativity, coassociativity and cocommutativity are clear. We only have to check that comultiplication is the algebra homomorphism, MATH. Note that if MATH are MATH - sets, then the element MATH is present in MATH if and only if MATH. Clearly, MATH is present in MATH if and only if MATH is present in MATH. Moreover, the equality of corresponding coefficients follows from the binomial identity MATH . |
math/0103126 | Consider the homomorphism MATH sending the generators MATH to MATH. The verification of the NAME relations is straightforward, so this homomorphism is welldefined. Moreover, it is clearly compatible with comultiplication on the generators of MATH. Therefore it is a homomorphism of NAME algebras. Note that the image of root vectors MATH, MATH (see REF) is a single snake of length MATH with head at MATH and tail at MATH. Also note that if MATH are two MATH - sets such that all heads of MATH are located to the right of all heads of MATH, then MATH. Therefore, the image of an element of the PBW MATH - basis REF of products of divided powers is a MATH - set consisting of snakes corresponding to elements MATH. It follows that the map in the lemma is surjective and injective. |
math/0103126 | Define a partial order on the set of all MATH - sets. Let MATH be MATH - sets with snakes of lengths MATH and MATH. We say MATH if MATH or MATH, MATH or MATH, MATH, MATH, etc. Let the head of the longest snake of MATH be located at REF and MATH. Let we have MATH snakes of length MATH with heads at MATH. Let MATH be the same as MATH except that all MATH longest snakes with heads at MATH are cut into two snakes of length MATH and length REF (the zero length snakes are at REF). Then MATH. We have MATH where the dots stand for the terms smaller than MATH. Therefore, every MATH - set can be reduced to product of MATH - sets which have only snakes of length REF. |
math/0103126 | By REF , it is sufficient to express MATH in terms of MATH and MATH. The set MATH has a natural partial order: MATH if all components of MATH are greater than or equal to the corresponding components of MATH. Note that the homogeneous component of MATH of degree MATH is generated by MATH with MATH. Fix MATH and suppose MATH. Then without loss of generality we may assume that MATH and that MATH for all MATH. Then it is sufficient to show that MATH. In particular our assumption implies that all MATH-sets of degree MATH are in MATH. For MATH, denote MATH the MATH - set of degree MATH which has MATH snakes of length REF with tails at REF and all other snakes are REF. Let MATH be the MATH - set of degree MATH which has MATH snakes of length MATH with head at REF and all other snakes of length REF. We note that by our assumption MATH and compute MATH for some nonnegative integers MATH. Therefore MATH. In particular MATH is in MATH. |
math/0103126 | We need to show that elements of the form MATH, where MATH and MATH, span MATH over MATH. We denote the span of elements of this form by MATH. Because of REF, it suffices to show that all products MATH belong to MATH. The NAME algebra MATH has an automorphism induced by the rotation of the quiver MATH. This automorphism sends MATH to MATH and MATH to itself. Therefore we only need to show that MATH for all MATH. We will show this by induction on MATH. For MATH the statement is trivial. We will prove the induction step by another induction on MATH. Let MATH be the degree of MATH. By our assumptions all MATH-sets of degree smaller then MATH are in MATH. Then following the proof of REF we obtain that MATH. Let MATH, MATH be the MATH-sets of degree MATH which have MATH snakes of length REF with head at REF and all other snakes of lengths REF. Then MATH is a linear combination of MATH and it is enough to prove that MATH. Let MATH be be the MATH-sets of degree MATH with all snakes of length MATH. We have MATH. We compute MATH for some nonnegative integers MATH. We already showed that MATH is in MATH, therefore by induction on MATH all MATH are in MATH. |
math/0103126 | We have already proved that this map is surjective. Hence it is sufficient to compare (graded) ranks of both MATH - modules. For a MATH - set MATH of degree MATH define the total degree by the formula MATH. Set the degree of each MATH to be REF and the degree of MATH to be MATH. Then the map in lemma is degree preserving. Moreover, the homogeneous MATH - modules of a given degree MATH, MATH, MATH and MATH, all have finite rank, so we compute the corresponding formal characters: MATH . The first two formulas are obvious and (the complexification of) the last one is well-known (for example, it follows from REF ). Therefore the lemma follows. |
math/0103126 | The following formula is checked by direct computation: MATH . It follows that MATH . The lemma is obtained by integrating this identity. |
math/0103126 | Denote by MATH the NAME algebra appearing in the right hand side of this formula. It is sufficient to prove that each MATH belongs to MATH. We will do it by induction on MATH. Suppose that MATH for all MATH. Then in particular all MATH - sets of degree MATH such that MATH and this inequality is not an equality for at least one vertex belong to MATH. Let MATH REF be the MATH-set of the same degree as MATH which has MATH snakes of length MATH with tail at MATH with all other snakes of length MATH. In particular, we have MATH. First we claim that for all MATH we have MATH for some integers MATH (here and below the dots denote terms in in MATH). Indeed, let MATH where MATH, be the MATH-set of the same degree as MATH which has the following snakes of positive length: MATH snakes of length MATH and MATH snakes of of length MATH all with tails at MATH. Let MATH be the MATH-set obtained from MATH obtaining by removing all snakes of length REF at vertex MATH. We have MATH if MATH. Now we compute MATH for some nonnegative integers MATH. Therefore we can compute MATH via MATH modulo terms in MATH and the claim follows. Thus, we have MATH for some integer MATH. Consider MATH. In particular, the degree of MATH is the same as the degree of MATH. To complete the proof it is sufficient to show that MATH is in MATH. The difference MATH is a sum of MATH-sets with positive integer coefficients. Each such MATH-set MATH has MATH snakes of length MATH. Moreover, not all of the snakes of MATH have the same tail. Suppose that MATH has exactly MATH snakes with tail at MATH. Then we can write MATH where MATH has exactly MATH snakes of length MATH with tail at REF. In particular MATH. Note that no snake of MATH has a head at MATH. Therefore we have MATH. This completes the proof. |
math/0103126 | Injectivity and compatibility with comultiplication is clear. We only have to show the equality MATH. It is readily seen by induction on the number of snakes in the MATH - set MATH. Indeed, suppose MATH has a snake of length MATH and head MATH. Then the coefficient of some MATH in product MATH can be computed in the following way. Denote by MATH the snake of length MATH with the head MATH. First, we fix the length MATH of the snake in MATH with head MATH, such that its complement of MATH is a snake in MATH. We set MATH, MATH, MATH be such that MATH, MATH and MATH. Then the coefficient of MATH in MATH is MATH, where MATH is the number of snakes in MATH equivalent to MATH. Now consider a term in MATH. There are MATH ways to represent it in the form (a term in MATH)MATH(a term of MATH). Each such MATH - set will give the contribution MATH to the coefficient of MATH in MATH, by the induction hypothesis. |
math/0103126 | The elements MATH are clearly orthogonal to MATH. Now the lemma follows by dimension counting. Namely, we set the degree of MATH to be MATH. The pairing clearly respects degrees, and the dimensions of the corresponding homogeneous components are equal. |
math/0103126 | Recall from REF that MATH is equal to the orthogonal complement of the augmentation ideal of MATH in MATH. This implies that MATH is equal to the subspace of MATH - invariants of MATH, or equivalently, the intersection of kernels of the operators MATH, (MATH). Let us show that each MATH is also annihilated by the MATH's. By REF, the operator MATH acts as the restriction operator MATH. REF immediately imply that MATH for any MATH module MATH. Therefore we find that MATH is contained inside MATH. Comparing the graded characters of the two spaces, we find that MATH . Both algebras are MATH - graded NAME algebras, freely generated by the generators MATH, and MATH, respectively whose degrees are given by the formulas MATH. Therefore MATH is equal to a non-zero multiple of MATH plus a polynomial in MATH, of degree MATH. The comultiplication for the MATH's is given by REF , and we find the comultiplication formula for the MATH's from REF specialized to the case when MATH and the fact that the the map MATH is a homomorphism of NAME algebras. The result is MATH (where we set MATH), which coincides with the formula for MATH. This readily implies that MATH is equal to a non-zero multiple of MATH. It remains to show that this multiple is equal to MATH. In order to do that we find the coefficient of one of the monomials in both MATH and in MATH. We take this monomial to be MATH. It enters with coefficient MATH in the decomposition of MATH into a linear combination of the products of the fundamental representations MATH, because the tensor product MATH has the same highest weight as MATH. On the other hand, recall that MATH corresponds to MATH. In the loop group realization of MATH introduced in REF each element of MATH is represented as a MATH matrix with entries being formal power series in MATH. In terms of the MATH's, the MATH-th entry of this matrix is equal to MATH where MATH, if MATH and MATH, if MATH. The generator MATH is the MATH - coefficient of the determinant of this matrix. Therefore it contains the monomial MATH. This completes the proof. |
math/0103126 | The upper floor is commutative by REF . The middle floor is commutative by REF . The bottom floor is commutative by REF . The upper left face is commutative because the evaluation homomorphism for MATH is by definition induced by the evaluation homomorphism for MATH. The bottom left face is commutative because the homomorphism MATH for MATH is by definition induced by the homomorphism MATH for MATH. The back upper face is commutative by definitions of all participating maps. The back lower face is commutative because all participating maps are maps of MATH modules, and uniqueness of such maps. The front upper face is commutative by REF . The front bottom face is commutative by REF . The upper right face is commutative by definitions of all participating maps. The lower right face is commutative by REF . |
math/0103129 | The identity MATH was proved by CITE. We will prove now that MATH holds for every nonnegative operator MATH. Our method is based on proofs of the formula MATH from articles CITE. If MATH is a nonnegative operator, then the function MATH is a well - defined increasing function with the range MATH, where MATH denotes the projection on the kernel of MATH. It means that the MATH - transform of MATH is a well defined analytical function on MATH. If MATH then MATH is not well defined and we define MATH, therefore MATH holds by definition. Let us consider now the case MATH; we take MATH if this limit exists and MATH otherwise. For every MATH such that MATH we have that MATH so for every MATH we have MATH . Take the limit MATH; since MATH we have MATH and therefore MATH . |
math/0103129 | The first identity was proved in papers CITE. The second identity follows from the first one and REF . |
math/0103129 | The proof follows closely the proof the corresponding scalar - valued Theorem from CITE and thus we will be quite condensed in the following. Proof of the implications MATH . Suppose that MATH is a scalar unitary. Let MATH be a linear map which acts on formal noncommutative polynomials of MATH and MATH with coefficients in MATH and which replaces MATH by MATH: MATH for MATH, MATH, MATH. It is easy to see that every polynomial of MATH, MATH and MATH is a linear combination of an element MATH and expressions of type MATH therefore the property that MATH and MATH are identically MATH - distributed is equivalent to the property that for every MATH as in REF we have MATH. We have that MATH therefore for MATH as in REF there are MATH such that MATH (the factors in brackets might be absent). CASE: In REF we can replace MATH by MATH, where MATH, and distribute it. Since MATH and MATH, for each summand we can directly apply the freeness REF , therefore MATH and REF holds. CASE: Let MATH be a sequence of scalar unitaries identically distributed as MATH, such that the family MATH is free with amalgamation over MATH (it is always possible to extend the probability space MATH in order to find such unitaries). It follows by induction that MATH and MATH have the same MATH - valued distributions, where MATH are scalar unitaries. For each MATH we have that MATH and MATH are free and MATH converges to zero. It follows that without any loss of generality we can assume that MATH. We have that if MATH then MATH and therefore MATH. Similarly if MATH then MATH; therefore if MATH, MATH then MATH. From REF and the freeness REF we have MATH therefore condition MATH holds. Proof of the equivalence MATH REF Let us consider one of the moments MATH where MATH and MATH. It can be written in the form REF , where MATH and MATH. Simple combinatorial arguments show that only special non - crossing partitions contribute in REF , namely noncrossing partitions MATH such that MATH does not depend on MATH and condition MATH holds. CASE: Suppose that the statement holds for every MATH and let us consider MATH, MATH, such that MATH is equal neither to MATH nor to MATH. As above, we can write REF in the form REF , where MATH and MATH. From the inductive assumption we see that the only summand on the right hand side of REF which depends on MATH or MATH is the summand corresponding to MATH; therefore it must be equal to zero and the statement follows. Of course the implication MATH is trivial. |
math/0103129 | Suppose that MATH is MATH - diagonal. It trivially fulfills REF . We use REF to compute a cumulant MATH, where for each MATH we have that either MATH or MATH, where the sequence MATH is balanced; furthermore let each of these two possibilities occur for at least one index MATH. From the following REF it follows that the sum on the right - hand side of REF is equal to zero; therefore REF implies that REF is fulfilled. Suppose now that MATH fulfills REF . We will prove by induction over MATH that MATH for all sequences MATH which are not alternating, that is .are not of the following form: MATH or MATH. Consider first the case MATH. From REF follows that for every MATH, MATH we have MATH therefore MATH. We consider now the case MATH. We define MATH to be the index such that the partial sum MATH takes the biggest value and a sequence MATH given by a cyclic rotation of the sequence MATH: MATH . Since the scalar - valued cumulants MATH have the cyclic property (compare REF ) it is sufficient to show our theorem for the sequence MATH. Partial sums of MATH have the following property: MATH and MATH. It follows that MATH can be written as a concatenation of the following nonempty sequences: MATH where for each MATH we have that either the sequence MATH is given by MATH or MATH is balanced. Every possibility occurs at least once unless the original sequence MATH was not equal to MATH or to MATH. Therefore REF implies that MATH . On the other hand we can use REF to express the left - hand side of REF in terms of cumulants of type MATH. At the first sight REF concerns only MATH - diagonal operators; however the inductive hypothesis assures us that all non - alternating cumulants of less than MATH factors vanish; therefore we have that the only nonvanishing term in REF is given by the full partition MATH: MATH . REF imply that MATH. |
math/0103129 | Suppose that MATH and MATH are free with respect to the state MATH. We will prove by induction over MATH that MATH . Suppose that the statement is true for all MATH. REF gives us MATH . Directly from the definition of the free cumulants we have MATH . The left - hand sides of these equations are equal. From the inductive hypothesis we see that if MATH then corresponding summands on the right - hand sides of REF are equal; therefore also for MATH they must be equal as well: MATH . Since MATH is faithful and the above identity holds for all MATH, it follows MATH . Conversely, if REF holds for all MATH, let us define MATH. We have that MATH . If we take MATH we see that MATH indeed fulfills the definition of a cumulant. REF implies that MATH and MATH are free. |
math/0103129 | From the operator - valued version of REF it follows that for MATH and MATH we have MATH . We apply state MATH to both sides of the equation; REF implies that MATH . This means REF that the joint moments of MATH and MATH are as if MATH and MATH were free. But this means exactly that they are free. |
math/0103129 | We apply REF for algebra MATH generated by MATH. It follows that the assumptions of REF are fulfilled. |
math/0103129 | Let MATH be the unital algebra generated by MATH and let MATH. Furthermore we define MATH. We define MATH . Let sets MATH and MATH be as in REF for MATH, respectively, for MATH (in this case we use MATH instead of MATH). We have that MATH for all MATH, therefore the theorem follows from the following stronger lemma. |
math/0103129 | We shall prove the lemma by induction over MATH. Since MATH, therefore we can assume that for all MATH such that MATH we have either MATH or MATH. Note that if MATH then MATH. (For MATH observe that if MATH, where MATH and MATH then for every MATH we have MATH by freeness. Since MATH is faithful we have that MATH.) Furthermore MATH; therefore if MATH have the additional property that MATH for all MATH such that MATH then REF implies MATH . If the above case does not hold, let MATH the index with the property that MATH and MATH. If MATH then MATH by inductive hypothesis, similar argument is valid for MATH. For MATH let us consider a product MATH . We will show that MATH, therefore the inductive hypothesis can be applied. For MATH the proof is very simple. For MATH we use the assumption that MATH is tracial and that MATH and MATH are free. |
math/0103129 | We can take REF as a definition of the function MATH. It is not difficult to show that the corresponding compound functions are given by MATH for every MATH. Thus we see that the defining relation REF for cumulants is indeed fulfilled. |
math/0103129 | Denote the entries of the matrix MATH by MATH (MATH). We use REF to compute MATH . Since the family MATH is free, the only nonvanishing cumulants are of the type MATH where the sum is taken over such MATH as in REF . This expression will be evaluated in REF below. It follows that the only nonvanishing cumulants of the matrix MATH are MATH . Let the expectation value MATH be as in REF show that MATH is MATH - diagonal with amalgamation over MATH. Furthermore, the above computation and REF show that MATH and MATH are free. It follows from REF that MATH is MATH - diagonal. The distribution of any MATH - diagonal element MATH is uniquely determined by the distribution of MATH and the above calculation shows that the distribution of MATH coincides with a distribution of MATH, where MATH is a free circular NAME element with parameter MATH. |
math/0103129 | If MATH then the statement of the lemma follows trivially since one of the factors is equal to zero. If MATH then the statement follows easily from REF . For the other cases we will use the induction with respect to MATH. We define MATH, MATH and MATH. For every noncrossing partition MATH of any subset of the set MATH (where MATH for every value of MATH) we define MATH . Due to the freeness every partition MATH which contributes nontrivially in the sum REF can be written as MATH, where MATH and MATH; therefore the left - hand side of REF is equal to MATH where the first sum is taken over MATH such that MATH and the second sum is taken over MATH such that MATH. For every MATH we have that MATH is equal to a product of MATH expressions of type MATH which is always equal to MATH (we use here that MATH, so MATH and MATH is an element from above of the diagonal of the matrix). Therefore we showed that the left hand side of REF is equal to MATH and the inductive hypothesis can be applied. |
math/0103129 | We consider first the case MATH. For MATH where MATH, the equation MATH reads MATH where MATH. We solve this system of equations and find MATH, MATH (since MATH is strictly positive, MATH is invertible). The equation MATH has a solution since the operator on the right hand side is strictly positive; for every MATH, MATH we have MATH . The general case follows from this special case MATH by induction as follows. We apply the lemma to the operator MATH and two projections MATH and MATH and thus get an operator MATH such that MATH and MATH . By induction hypothesis, we can now apply the lemma for the operator MATH and the projections MATH and thus get a MATH such that MATH and MATH is an upper triangular matrix with respect to MATH. The element MATH gives the wanted triangular square root with respect to MATH. |
math/0103129 | Let MATH be a MATH - diagonal operator. Our goal is to construct a scalar probability space MATH and elements MATH such that MATH are orthogonal projections in respect to which MATH is in the upper triangular form and such that MATH has the same distribution as MATH. If the condition on spectra REF holds then it follows that also the original element MATH has the upper triangular property (compare CITE). First of all note that, for every MATH, the subspace MATH defined in REF fulfills MATH therefore MATH, hence MATH and MATH are free REF . The projections MATH that we want to construct will turn out at the end of the proof to be equal to MATH; these are heuristical motivations why we will choose MATH to be free from MATH. Construction of MATH. Let us first restrict to the case where MATH is strictly positive. For given MATH, MATH let us consider an enlargement MATH of the probability space MATH. This enlargement should contain orthogonal projections MATH such that MATH, MATH and such that MATH and MATH are free (this enlargement is given by the NAME - NAME free product of algebras CITE). REF gives us an element MATH such that MATH and such that MATH has the upper triangular form REF . We define the operator - valued structure MATH on MATH as in REF. The free product of algebras with amalgamation over MATH CITE gives us an enlargement MATH of the space MATH and an operator MATH which is a scalar NAME unitary such that MATH and MATH are free with amalgamation over MATH. As in REF we recover the scalar - valued structure MATH on MATH. Note that MATH is an enlargement of MATH. REF gives us that MATH is MATH - diagonal with amalgamation over MATH; we know that MATH and MATH are free, therefore REF assures us that MATH is MATH - diagonal. The distribution of any MATH - diagonal element MATH is uniquely determined by the distribution of MATH; therefore the elements MATH and MATH are identically distributed. Since MATH is a diagonal matrix, the operator MATH is the wanted upper triangular matrix. Estimates for spectra. The set MATH, defined to be the spectrum of MATH is contained in the spectrum of MATH, where MATH. We can furnish MATH with the operator - valued structure MATH in the standard way (with respect to orthogonal projections MATH); note that MATH and MATH for every MATH. For MATH we have MATH and therefore MATH. This observation implies that the (restriction of) cumulant function MATH in the probability space MATH fulfills the definition of the cumulant function MATH in the space MATH so MATH . We apply now REF and see that MATH and MATH are free in the scalar probability space MATH. We also see that MATH is MATH - diagonal with amalgamation over MATH. REF assures us that MATH is MATH - diagonal. REF implies that MATH where MATH . Due to the upper triangular form of MATH we have MATH and REF implies that MATH . Similarly as above, we see that the spectrum of MATH is a subset of the spectrum of MATH, where MATH. REF implies that MATH where MATH where MATH denotes the inverse of MATH. Let us introduce the notation MATH therefore MATH and MATH and therefore MATH is well - defined, where MATH denotes the inverse of MATH. As before show that MATH is MATH - diagonal and then REF imply that MATH . Lastly, REF shows that MATH. REF from CITE shows that MATH, where MATH is the projection on the space REF . The general case. In the general case where MATH might have a kernel, for every MATH we construct as above a MATH - diagonal element MATH in the triangular form with respect to projections MATH, such that MATH and MATH are identically distributed. There exists a sequence MATH of positive numbers which converges to zero such that the sequence MATH converges for every polynomial MATH of MATH noncommuting variables. We define MATH to be the free algebra generated by elements MATH and a state MATH on MATH by MATH . We have that MATH is in the upper triangular form and MATH and MATH are identically distributed. The spectra estimates can be done similarly. |
math/0103129 | It is well-known CITE that MATH is MATH - diagonal (this follows for example, by REF ). We will show that MATH is also MATH - diagonal. Firstly, note that MATH is MATH - diagonal with amalgamation over MATH (see REF ). It remains to show that MATH is free from MATH. We can extend the algebra MATH in such a way that MATH and sets MATH, MATH, MATH are free. First of all notice that it follows from REF that MATH and MATH are free. Therefore for MATH we have that MATH is uniquely determined by moments MATH and MATH for some MATH. But again REF gives us that MATH and MATH are free; therefore REF is uniquely determined by moments MATH and moments MATH. If we repeat the above discussion for the product MATH, where MATH, we see that MATH . The above result implies that MATH and MATH are free and that MATH and MATH have the same distribution. It follows that MATH and MATH have the same distribution. |
math/0103132 | Choose a point MATH far away from the graph. Since the graph is a tree and no vertices are surrounded by edges, there are MATH arcs that join MATH to each vertex and are disjoint each other and are disjoint from the graph as in REF . Choose a new horizontal axis disjoint from the graph and move each vertex along each arc by a homeomorphism so that it lies on the new axis as REF . Then the result is a subgraph of the inner complete graph. tree . |
math/0103132 | A maximal tree of MATH is equivalent to a connected subgraph of the inner-complete graph from which we can obtain the inner complete graph by adding missing edges that are conjugates of an existing edge by an adjacent edge. |
math/0103132 | Induction on MATH. It is trivial when MATH. Suppose a linear spanned tree MATH with MATH vertices generates MATH with MATH positive relations. We add a new vertex and a new edge MATH to MATH. For each edge MATH of MATH, we will have a new positive relation so that MATH new positive relations will be added. In the following we use NAME transformations CITE that add and delete a generator REF denoted by MATH or MATH to utilize triangular relations. CASE: If MATH and MATH have no intersection, add the positive relation MATH . CASE: For each set of edges MATH simultaneously adjacent to MATH at a vertex of valency MATH as in REF , add MATH positive relations MATH CASE: If MATH and MATH intersect and form a pseudo face as in REF , add the positive relation MATH that is derived from MATH where MATH inductREF CASE: If MATH and MATH intersect and form a pseudo face as in REF , add the positive relation MATH that is derived from MATH where MATH CASE: If MATH and MATH intersect and form a pseudo face as in REF , add the positive relation MATH that is derived from MATH where MATH inductREF |
math/0103132 | Choose a spanning tree MATH of MATH. Then a linearly spanned tree MATH generates MATH with MATH positive relations as in REF . When each edge MATH in MATH are added, a circuit is formed and the circuit give a new positive relation as follows: CASE: If MATH forms a circuit with no intersection with other edges as in REF , add the positive relation MATH where we regard that the circuit bounds a polygonal disk by cutting open all edges inside the circuit as in REF . cutopen CASE: If MATH forms a circuit with some intersections with other edges as in REF , add the positive relation MATH face . |
math/0103132 | Let MATH and MATH be the set of generators given by MATH and MATH, respectively and let MATH be a finite set of positive relations on MATH such that MATH has the embedding property. Let MATH be the set of relations on MATH which is a ``full" subset of MATH in the sense that MATH contains all relations in MATH written on MATH. We will prove by contradiction that MATH has the embedding property. Suppose that there exists a pair of positive words MATH on MATH such that MATH are equivalent in the braid group but are not positively equivalent under MATH. Since MATH has the embedding property, MATH under MATH. Thus there is a sequence of positive words MATH over MATH such that MATH under MATH and each positive equivalence is obtained by one direct application of relations in MATH. The sequence must contain at least one positive word, say MATH, that is not written solely on MATH, otherwise MATH under MATH. In the view of REF , it is impossible that MATH contains any relation MATH such that MATH is a positive word over MATH and MATH is a positive word over the edges that are not incident to any vertex of MATH. Furthermore MATH is a full subgraph of MATH. Thus MATH must contain an edge that joins a vertex MATH in MATH to a vertex MATH not in MATH. Let MATH be such an occurrence that comes last in MATH. Since MATH is connected, the vertex MATH is joined to another vertex MATH in MATH by an edge MATH. As an automorphism of the punctured plane, MATH does not change the circle MATH because the edges for MATH never touch MATH. On the other hand we will show MATH must change MATH and this is a contradiction because MATH and MATH are isotopic as automorphisms of the punctured disk. The automorphism MATH is the composition of the counterclockwise half twists along edges in MATH. Then the counterclockwise half twist along MATH creates an intersection MATH of MATH with the edge MATH. The intersection MATH can disappear only via a clockwise half twist along an edge incident at either MATH or MATH as in REF . But all of half twists in MATH are counterclockwise because MATH is a positive word over edges, circle . |
math/0103132 | An intersection between two adjacent edges always creates a pseudo face that must contain a vertex and so there is no intersection between adjacent edges in a linearly spanned graph. Suppose MATH has more than one intersection. REF shows a typical situation with one intersection and another intersection MATH. The edge MATH must join vertices MATH and MATH, otherwise MATH is an intersection between two adjacent edges. One can easily check that all possibilities of completing MATH create a pseudo face containing at least a vertex. oneinter . |
math/0103132 | Let MATH be any positive finite presentation of the braid group. REF implies that any shorter positive relation than MATH itself can not make MATH positively equivalent to MATH. Choose a large MATH such that MATH is longer than any relation in MATH. Then MATH is not positively equivalent to MATH over MATH. |
math/0103132 | The graphs in REF give the NAME presentation and the band-generator presentation. So they have the embedding property. In order to show that all other linearly spanned graphs with REF vertices do not have the embedding property, we appeal to REF . To check REF, we use the fact that the band-generator presentation has the embedding property so that the positive equivalence in the presentation is the same as the equivalence in the braid group. We also utilize the left and right cancellation theorem and the left and right canonical forms in the band-generator presentation in CITE. In the view of REF , the graphs being considered are divided into two types: planar graphs and graphs with one intersection among edges. CASE: Planar graphs with REF vertices can be divided further into three types: graphs containing neither a triangle nor a rectangle, graphs containing at least a rectangle, and graphs containing at least a triangle but no rectangle, where a triangle (or a rectangle) is non-degenerate, that is, must have REF (or REF, respectively) vertices and must contain no other vertices inside. CASE: This type is further divided into two types REF . CASE: starREF We may have two possible graphs as in REF up to equivalence. For each of these two graphs, let MATH and MATH. We will show that MATH and MATH satisfy REF . Add the edges MATH so that these edges together with MATH form an inner-complete graph. Then MATH and so MATH. Suppose MATH for some positive word MATH in the band-generator presentation. Then MATH by the left cancellation in the band-generator presentation. But MATH may only start with MATH or MATH which commute. Thus MATH can not be written over MATH. Similarly MATH. Thus MATH satisfies REF . Due to the triangular relation MATH, MATH can not be equivalent to a positive word over MATH that contains MATH since MATH is not positively equivalent to a positive word in the band-generator presentation that contains the subword MATH. Thus MATH also satisfy the hypothesis of REF over MATH. CASE: All possible graphs except the NAME graph have multiple edges as in REF . except CASE: We may have two possible graphs as in REF up to equivalence. For each of these five graphs, let MATH and MATH. cycle Add the edges MATH so that these edges together with MATH form an inner-complete graph. Then MATH and so MATH. Suppose MATH for some positive word MATH in the band-generator presentation. Then MATH by the left cancellation in the band-generator presentation. Thus MATH can not be written over MATH. Similarly MATH. Thus MATH satisfies REF . Due to the triangular relation MATH, MATH can not be equivalent to a positive word over MATH that contains MATH since MATH is not positively equivalent to a positive word in the band-generator presentation that contains the subword MATH. Thus MATH also satisfy the hypothesis of REF over MATH. CASE: In this case, graphs are divided two types by the number of triangles. CASE: We have one possible graph as in REF up to equivalence. Let MATH and MATH. outerREF Add the edges MATH so that these edges together with MATH form an inner-complete graph. Then MATH and so MATH. Let MATH. Then MATH. The right canonical form of MATH is MATH. So MATH may finish with MATH over the band-generators and so MATH or MATH must appear in MATH. Similarly MATH. CASE: In this case triangles must share at least REF edge since we consider only REF vertices. However if two triangles share REF edges then it must contain a multiple edge. Thus we have two possible graphs as in REF . For these two graphs, let MATH and MATH. trianREF Add the edges MATH so that these edges together with MATH form an inner-complete graph. Then MATH and so MATH. The left canonical form of MATH is MATH and so it can not start with MATH over the band-generators. Thus MATH. The right canonical form of MATH is MATH and so it can not finish with MATH. Thus MATH. Due to the triangular relation MATH, MATH can not be equivalent to a positive word over MATH that contains MATH since MATH is not positively equivalent to a positive word in the band-generator presentation that contains the subword MATH. Thus MATH also satisfy the hypothesis of REF over MATH. CASE: The edges MATH and MATH that intersect each other are transformed to the diagonals of a rectangle. Then REF more edges are needed to make an inner-complete graph. We have four distinct types of graphs, depending on how many edges are missing from the inner-complete graph. CASE: REF is the only possible graph in this case. Let MATH and MATH. nonplane Add the edges MATH so that these edges together with MATH form an inner-complete graph. Then MATH and so MATH. Let MATH. Then MATH. To remove MATH, MATH must start with MATH. But the left canonical form of MATH is MATH and so it may start only with MATH. Thus MATH. Similarly MATH. CASE: In this case, there are two possible graphs as in REF . For the graph in REF , let MATH and MATH. nonREF Add the edges MATH so that these edges together with MATH form an inner-complete graph. Then MATH and so MATH. The left canonical form of MATH is MATH and so it can not start with MATH over the band-generators. Thus MATH. The right canonical form of MATH is MATH and so it can not finish with MATH. Thus MATH. For the graph in REF , let MATH and MATH. Then MATH satisfy REF by using a similar argument as for REF CASE: The only possible graph is REF . Let MATH and MATH nonREF MATH and so MATH . The left canonical form of MATH is itself and so it can not start with MATH over the band-generators. Thus MATH. Let MATH then MATH and MATH commute, so MATH can not be removed. Thus MATH. CASE: If we add any edge to the inner-complete graph, multiple edges are created. Thus only the inner-complete graph is eligible. |
math/0103132 | We have already discussed about linearly spanned graphs with REF vertices. Let MATH be a linearly spanned graph with more than REF vertices that has the embedding property. First we choose a connect full subgraph MATH with REF vertices from MATH so that there is a separating circle satisfying the hypothesis of REF . Choose REF vertices that form a connected subtree in a spanning tree of MATH. Take the full subgraph with these REF vertices. If there is no other vertices in faces of this full subgraph, then this is a desired full subgraph. If there is other vertices in the faces of this full subgraph and none of them is adjacent to the chosen REF vertices, then there is an edge-path starting at a vertex MATH on a face and ending at one of REF vertices such that the edge-path intersect the full subgraph of REF vertices and so MATH is contained in a pseudo face. Since MATH is linearly spanned, this can not happen. Thus at least one of vertices on faces, say MATH, is adjacent to one of REF vertices. Then we have the less number of unwanted vertices contained in faces of a new connected full subgraph that is obtained by replacing one of REF vertices by MATH. By repeating this process, we eventually obtain a connect full subgraph MATH with REF vertices MATH that can be separated by a circle from other vertices of MATH. REF say that MATH must be either the NAME graph or the inner-complete graph with REF vertices. Choose a vertex MATH in MATH that is adjacent to one of MATH. If MATH is an inner-complete graph, then each full subgraph with REF vertices consisted of MATH and any three vertices from MATH must be an inner-complete graph by REF . Thus the full subgraph with REF vertices MATH is an inner-complete graph. By repeating this process, MATH itself eventually becomes an inner-complete graph. We now suppose that MATH is the NAME graph with REF vertices MATH where MATH are of valency REF and the other vertices are of valency REF. Choose a vertex MATH in MATH that is adjacent to any one of MATH. By REF , MATH can be adjacent to either one or both of vertices MATH. If MATH is adjacent to both vertices, we stop the process. If MATH is adjacent to one vertex, say MATH, then MATH form an NAME graph and we repeat this process. Eventually we see that MATH must contain one of the following two types of graphs as a full subgraph separated by a circle as in REF unless MATH itself is an NAME graph. The proof will be completed when we show that both types of graphs do not have the embedding property. CASE: The graph of this type is depicted in REF . Let MATH and MATH. Then one can show that MATH satisfy the hypothesis of REF by a similar but longer argument as for REF . Thus MATH does not have the embedding property kcycle CASE: The graph of this type is depicted in REF . Let MATH and MATH nonplREF Add supplementary edges MATH as in REF to form a subgraph of the inner-complete graph with MATH vertices. Then MATH and so MATH. We now show that MATH . Suppose that MATH . Then MATH where MATH. Due to the triangular relation MATH, the positive word MATH must end with MATH in order for MATH to disappear. We will show that MATH can not end with MATH by an induction on MATH. Clearly MATH can not end with MATH in the band-generator presentation. MATH . It is easy to check that MATH ends with MATH if and only if MATH ends with MATH if and only if MATH ends with MATH in the band-generator presentation. By the induction hypothesis, MATH can not end with MATH. Similarly we have that MATH . Consequently MATH satisfy the hypothesis of REF . |
math/0103134 | Let MATH. One has MATH . Therefore, MATH and similarly, MATH, are homomorphism. They are smooth because the action of MATH is smooth. If MATH is another choice instead of MATH, there exists MATH such that MATH and one has MATH whence MATH. This proves that the class of MATH in MATH does not depend on the choice of MATH and MATH. Now, if MATH is a MATH-equivariant diffeomorphism over the identity of MATH, then, by choosing MATH and MATH, one has MATH. The proof of REF is then complete. |
math/0103134 | If MATH is an horizontal lifting , so are MATH and MATH for MATH. As MATH, one has MATH for all MATH. |
math/0103134 | Suppose that MATH. This means that there exist MATH, with MATH in MATH, such that the following equality MATH holds in MATH. This implies that MATH with MATH and MATH (the centralizers of the images of MATH and MATH). Therefore MATH, which implies MATH. To prove the converse, observe that CASE: MATH for MATH. CASE: MATH for MATH and MATH. CASE: MATH for MATH in the identity component of MATH. Suppose that MATH. This means that there are MATH, MATH and MATH in the identity component of MATH auch that MATH (MATH can be put in the middle since the identity component of MATH is a normal subgroup of MATH). One then has MATH . |
math/0103134 | Let MATH, MATH and MATH as in the statement of Proposition C. Let MATH. Then, MATH and, by REF, one has MATH in MATH. By REF, MATH. Therefore, MATH, the last equality coming from REF and that MATH can be represented by the transition function CITE. |
math/0103134 | Since MATH is trivial, MATH which is suppose to be connected. Therefore, REF is a particular case of REF of Theorem B. Let MATH parametrizing the meridian arc, as in REF . Let MATH be two homomorphisms representing MATH. One can find MATH and MATH so that MATH and MATH are the isotropy representations associated to MATH and MATH. As MATH is connected, the submanifold MATH of MATH is connected and there is a smooth lifting MATH of MATH such that MATH and MATH. As MATH is trivial, MATH is isotropic. REF of Theorem A then follows from Proposition C. |
math/0103134 | Recall that any element of MATH can be represented by a homomorphism (a geodesic in a maximal compact subgroup MATH of MATH, with a MATH-bi-invariant Riemannian metric, being a MATH-parameter subgroup CITE). Therefore, if MATH is a MATH-bundle over MATH, there exists a homomorphism MATH such that MATH. For MATH, let MATH given by MATH. If MATH is not trivial, the classes MATH are all distinct in MATH. Indeed, the set MATH is in bijection with lattice points in a NAME chamber of the NAME algebra of a maximal torus of MATH and the point representing MATH is MATH times those representing MATH. Suppose first that MATH is not trivial. Hence, MATH is not trivial and MATH are all different classes in MATH with MATH. The result then follows from Propositions C and REF. When MATH is trivial, one takes any non-trivial homomorphism MATH. The classes MATH in MATH represent infinitely many distinct MATH-equivariant MATH-bundles MATH with trivial MATH. |
math/0103134 | If MATH, the group MATH is semi-simple and the set MATH is finite. The latter follows from the following known results: CASE: a homomorphism is determined by its tangent map at the identity (as a homomorphism of NAME algebras). CASE: the NAME algebra of MATH contains only finitely many semi-simple NAME subalgebras, up to inner automorphism CITE. CASE: there are only finitely many homomorphisms between two semisimple NAME algebras. Also, if MATH is compact, the group MATH is compact and then MATH is finite. Proposition E then follows from Theorem B. |
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