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math/9904129 | The proof uses the following property of valuations: MATH. Furthermore, when that minimum is attained in only one MATH, we have equality. Let MATH. Writing MATH one can pass to the valuation by: MATH . Subtracting, one obtains: MATH . This concludes the proof. |
math/9904129 | The NAME diagram of MATH at REF is MATH. (This latest set is convex, since the points lie on the curve MATH and this curve is convex). Therefore, there is a unique root MATH of MATH that minimizes MATH. Since MATH, where the sum ranges over all the roots, we have: MATH . Replacing by the actual values of the coefficients, one gets: MATH . Now, suppose that there is a machine MATH that decides MATH in time polylog(MATH). One can assume without loss of generality that this machine has no constant but MATH and MATH. Let its running time be bounded by MATH. Let us fix MATH. We will derive a contradiction. Let MATH be the polynomial defining the canonical path (recall that MATH is fixed now, so this is the path followed by generic MATH). It can be computed in time MATH, so we have the following bounds: MATH . Since MATH is also a root of MATH, there are coefficients MATH and MATH, MATH, such that: MATH . Thus, MATH. This implies: MATH . Replacing by REF , one obtains MATH, a contradiction. |
math/9904129 | We can assume without loss of generality that the ordering of the MATH satisfies: MATH . For MATH we have: MATH . Hence, using MATH: MATH . By the same argument, for MATH: MATH . Hence, MATH . Set MATH for MATH. We know that the MATH are such that MATH. Hence MATH . Hence: MATH and hence MATH . By REF or by REF, MATH and hence, taking logs: MATH . Taking logs again: MATH and hence: MATH . |
math/9904129 | We see from its NAME diagram that the polynomial MATH has distinct roots MATH with: MATH . So we have MATH, and MATH . Assume that there are MATH such that for each MATH, there is a machine MATH over MATH deciding MATH in time MATH. Its generic path is defined by a polynomial MATH of degree MATH. Let us fix MATH. In particular MATH. We are in the conditions of REF , where MATH. From that Lemma, it follows that MATH . Hence, MATH contradicting our choice of MATH. |
math/9904130 | Let MATH be in MATH. Let MATH be the machine that recognizes MATH in polynomial time, where the input MATH is assumed to be in MATH. Although it is possible that a MATH is not in MATH, it is still possible to recognize MATH in polynomial time. Indeed, MATH is also in MATH. The machine MATH will check MATH and MATH. Now we apply elimination of constants REF , and choose MATH to be constant-free. The nodes of the machine MATH are supposed to be numbered. Given an input MATH, the path followed by input MATH is the list of nodes traversed during the computation of MATH. When the input is restricted to one of the affine varieties MATH's, we can define the canonical path associated to MATH as the path followed by the generic point of MATH. This corresponds to the following procedure: At each decision node, at time MATH, branch depends upon an equality MATH, where MATH is the original input. The polynomial MATH can be computed within the machine running time. In case MATH for all MATH, we follow the NAME and say that this branching is trivial. If not, we follow the no-path and say that this branching is non-trivial. The fact that MATH is a variety is essential here, since it guarantees that only a codimension MATH subset of inputs may eventually follow the NAME at this time. The set of inputs that do NOT follow the canonical path can be described as the zero-set of MATH where the product ranges over the non-trivial branches only. The polynomial MATH can be computed in at most twice the running time of the machine MATH restricted to MATH. By hypothesis, this is polynomial time in the size of MATH. Since we assumed that MATH returns only MATH or MATH, the set of the inputs that follow the canonical path (that is, MATH) is either all in MATH or all in its complementary MATH. There are now two possibilities. First possibility, MATH has measure zero in MATH, and therefore it must be contained in MATH. Second possibility, MATH has non-zero measure, hence it contains the complementary of MATH, and hence MATH is a subset of MATH. |
math/9904130 | REF is trivial, refer to REF. CASE: Let MATH be the problem in REF . Since MATH is generic in MATH, all inputs in MATH should escape the canonical path. Hence, if MATH is the polynomial that defines the canonical path, MATH for MATH. But then it cannot be evaluated in time polylog(MATH), by REF . Hence, under REF , the problem MATH is not in MATH. It does belong to MATH, so MATH. CASE: Using REF implies that MATH. However, since MATH, REF implies MATH. Hence MATH over MATH. |
math/9904130 | Let MATH and assume that MATH. We have to show that MATH. For each MATH, one can embed MATH into some MATH as follows: Let MATH be the deterministic polynomial time machine to recognize MATH, and let MATH be the non-deterministic polynomial time machine to recognize MATH. We can assume without loss of generality that MATH and MATH are constant-free REF . Let MATH be the maximum running time of MATH and MATH when the input is restricted to MATH. Let MATH be the combined Register Equations of machines MATH and MATH for time MATH REF . Thus, MATH is a system of polynomial equations with integer coefficients and indeterminate coefficients MATH. The polynomial system MATH can be constructed in polynomial time from MATH, and the size of MATH is polynomially bounded by the size of MATH. We claim that MATH is contained in some MATH, and that in that case MATH and MATH. Indeed, MATH for some MATH, and MATH for some MATH. Then MATH belongs to MATH if and only if the corresponding MATH has a solution over MATH. We now distinguish two cases: CASE: MATH has measure zero in MATH. Thus MATH for an easy-to-compute polynomial MATH. In that case, since MATH gets mapped into MATH, the composition MATH gives the polynomial associated to MATH. CASE: MATH has measure zero in MATH. Thus MATH for an easy-to-compute polynomial MATH. In that case, since MATH gets mapped into MATH, MATH is the polynomial associated to MATH. |
math/9904131 | This follows from REF and the definition of rigidity REF . |
math/9904131 | The proof follows from the identification of the MATH term in CITE or CITE. Also, note that MATH where MATH . Then MATH, MATH, and MATH. |
math/9904131 | If MATH for MATH, then from REF , the MATH term for the spectral sequence converging to MATH is zero. Thus, MATH for MATH. The result now follows from long exact sequence REF. |
math/9904131 | The theorem follows from REF . |
math/9904131 | We first compute MATH on the level of cochains. Consider the vector space basis MATH of MATH given by MATH . From CITE, the class of MATH in MATH is represented by the cochain MATH . The cochain map MATH inducing the isomorphism MATH satisfies MATH . Also, it is know that MATH generates MATH (as a MATH vector space). Let MATH, MATH, MATH, MATH. From the definition of MATH and MATH, the image of MATH in MATH is REF-form which sends MATH to MATH where MATH . By the definition of the MATH's, the image of MATH is thus MATH . To show that the NAME cohomology class of MATH does not depend on the choice of MATH, consider the one-form MATH where MATH is a MATH function with MATH for all MATH. Letting MATH denote the NAME coboundary, we have from REF MATH . Then MATH . Also, MATH . It follows that MATH . Compare with CITE. Of course, MATH . To show that the cohomology class of MATH does not depend on the choice of MATH, consider the one-forms MATH where MATH is a MATH function (which may have zeroes on MATH). Then MATH . It can be checked that MATH . |
math/9904142 | Suppose that MATH is a cross product algebra. Since by REF is itself an algebra one obtains the first and second identity of REF by unitality of MATH and MATH. Using these relations yields the fourth identity of REF by application of MATH to REF. Again using the left unitality of MATH one concludes that MATH from which one immediately derives the third identity of REF. Now the associativity MATH yields the fifth identity of REF by application of MATH. The sixth relation is obtained by applying MATH. To derive the seventh identity one has to apply MATH. Conversely suppose the conditions of the second item of the proposition are fulfilled. Define MATH according to REF. Then MATH where MATH is the canonical MATH-fold multiplication, and MATH is the abbreviation of the identity of a MATH-fold tensor product of (combined) MATH's and MATH's. On the other hand MATH where the fourth condition of REF has been used twice to obtain the first equation, the sixth relation of REF has been applied in the second equation, and with the help of the seventh relation of REF we obtained the third equation. Hence associativity of MATH has been proven. Using the first four identities of REF one easily proves REF and unitality MATH. |
math/9904142 | If MATH is an algebra isomorphism then define MATH and MATH. Using the particular definition of MATH in REF and the identities REF, it is verified immediately that MATH is an algebra morphism since MATH is an algebra morphism. Similarly one proves that MATH and MATH which is therefore an isomorphism. If on the other hand the conditions of the second statement of REF is fulfilled, then MATH is an isomorphism by assumption. Therefore MATH is canonically an algebra through MATH and MATH. We will show that this defines a cross product algebra structure on MATH by MATH and MATH. Using the explicit expression for MATH, inserting several times MATH and using the above definitions, as well as the fact that MATH is algebra morphism, one obtains the following identities. MATH . Similarly one derives MATH, hence MATH. Therefore MATH. |
math/9904142 | Define MATH. Then by REF it follows MATH, MATH, and MATH. Using consecutively that MATH is an algebra morphism, REF , and REF then yields MATH. Therefore MATH is an algebra morphism obeying the conditions of REF . Suppose that there is another MATH meeting the same stipulations as MATH. Then MATH. In the last equation unital properties of cross product algebra have been used. This proves uniqueness of MATH. |
math/9904142 | The proposition can be proven easily by REF and triviality of MATH. |
math/9904142 | The non-trivial part of the proposition can be verified easily with the help of the identity MATH, the associativity of MATH and MATH, and REF or REF respectively. |
math/9904142 | ``REF. REF.": For MATH we define MATH, MATH to be the morphisms splitting the idempotent MATH as MATH and MATH for some objects MATH. Then with the help of REF it follows MATH and MATH. Hence MATH. Now we define MATH, MATH, MATH, and MATH. Then using REF and the (co- )algebra property of MATH one verifies easily that MATH is an algebra and MATH is a coalgebra. Furthermore MATH is an algebra morphism because MATH, and MATH. In a MATH-symmetric manner one proves that MATH is a coalgebra morphism. ``REF. REF.": Conversely we define MATH for MATH. Since by REF is an algebra morphism it follows MATH and using MATH it holds MATH. In MATH-symmetrical analogy it will be proven that MATH and MATH. From REF we conclude MATH and MATH. Thus MATH and MATH split the idempotent MATH. ``REF. REF.": By REF is an algebra and MATH is an algebra morphism, MATH is a morphism in MATH, and MATH is an isomorphism. Define MATH. Then MATH. Therefore all conditions of REF are fulfilled implying that MATH is algebra isomorphic to a cross product algebra MATH. From the proof of REF one reads off that the isomorphism is given by MATH. In a MATH-symmetric way one shows that MATH is coalgebra isomorphic to a cross product coalgebra MATH on the same tensor product with isomorphism MATH. Thus by REF. Since MATH is a bialgebra it follows then from REF that MATH is cross product bialgebra MATH and MATH is bialgebra isomorphism. Furthermore the identities MATH show that MATH is normalized. ``REF. REF.": Given a bialgebra isomorphism MATH. Then in particular MATH is an algebra isomorphism and MATH is a coalgebra isomorphism. We define MATH, MATH, MATH, and MATH. From REF one derives that MATH is an algebra morphism, MATH is a coalgebra morphism, and MATH. By REF is normalized from which follows MATH and similarly MATH. |
math/9904142 | The identities MATH and MATH hold since MATH is an algebra morphism and MATH is a coalgebra morphism, whereas the identities MATH and MATH, as well as the equations MATH and MATH have been shown in the proof of REF . From the proofs of REF one can directly read off the structure of the morphisms MATH, MATH, MATH, and MATH given in REF using MATH and MATH. Since MATH is algebra morphism, MATH is coalgebra morphism and MATH for MATH it follows MATH and MATH. Because of REF it follows MATH. Using MATH and MATH from above we obtain in a similar way MATH. Using consecutively REF two times, REF, the identity MATH, again REF , and the identity MATH one obtains the following series of equations. MATH and MATH-symmetrically MATH. Finally one obtains in a similar manner MATH and MATH by MATH-symmetry. |
math/9904142 | If the conditions of REF hold, then by construction (see the proof of the theorem) the injections and projections of MATH and the morphisms MATH, MATH, MATH and MATH are related by MATH, MATH, MATH, and MATH, where MATH is the given bialgebra isomorphism. Thus MATH and MATH follow directly. |
math/9904142 | Obviously statement MATH and MATH are equivalent due to REF . Hence suppose statements MATH and MATH hold. Then by REF there exists a normalized cross product bialgebra MATH which is isomorphic to MATH via the isomorphism MATH. We define MATH . The morphisms MATH, MATH, MATH, MATH, MATH, and MATH are given in terms of the projections and injections by the corresponding relations in REF . With the help of these data we define structure morphisms MATH, MATH, MATH, and MATH analogous to REF. We will show that the morphisms MATH and MATH defined with these structure morphisms according to REF precisely coincide with MATH and MATH respectively, if the assumptions of REF . (or MATH . ) are satisfied. Before we prove this we will provide several auxiliary identities. From REF we obtain MATH . Furthermore it holds MATH since the following equations are satisfied. MATH . Then REF follows from REF . Observe that REF implies the additional REF have not been used to derive REF. Subsequently we will prove MATH graphically. Henceforth we use the notation MATH for MATH. MATH . To derive the first identity of REF we used the specific form of the structure morphisms MATH, MATH, MATH, and MATH. Then REF, and the third identity of REF for MATH yield the second equality of REF. With REF we derive the third equation, whereas REF and again use of REF yield the fourth identity of REF. The fifth equality comes from application of the second ``projection" relation of REF, and for the derivation of the sixth identity we used REF. Finally the definition of MATH, given in the proof of REF , yields the result. In a MATH-symmetric way the identity MATH will be shown. The (co- )unital identities REF can be verified straightforwardly from the definitions, the assumptions and REF . It remains to prove the ``projection" relations REF. Observe that MATH . Taking into account the relation MATH and the relation MATH one obtains with the help of REF MATH . Gluing the identities of REF and MATH according to the left and right hand side of the third equation of REF (for MATH), using REF again, and eventually multiplying both resulting sides with MATH and MATH yields MATH which is the second identity of REF. Analogously the first equation of REF can be derived. This proves that MATH is cocycle cross product bialgebra. Conversely suppose that MATH is an isomorphism of bialgebras. To prove statement MATH it suffices to verify relations REF. Like in the proof ``REF. REF." of REF we define MATH, MATH, MATH, and MATH. Then the identities REF will be proven for MATH in the way described above. And the previously performed gluing of the identities REF yields MATH . Therefore by REF the third identity of REF for MATH follows. Applying MATH-symmetry the third equation of REF for MATH will be derived. Then all conditions are satisfied which have been needed to derive the first four identities in REF. Hence MATH . On the other hand it holds by REF, where we used REF which has been derived under the more general assumption of REF . Multiplying both expressions of MATH with MATH and MATH, and using that MATH is algebra morphism yields MATH from which the second identity of REF can be derived easily with the help of REF . Similarly by MATH-symmetry the first equation of REF will be proven. |
math/9904142 | We prove REF . The remaining statements can be derived in a similar manner or follow directly by MATH-symmetric reasoning. Without loss of generality we may assume MATH and MATH. CASE:: Suppose that MATH is trivial. This means that MATH. Using REF and the unital identities of REF then yields MATH. On the other hand from the unital identities of REF immediately follows MATH. Conversely if MATH holds then analogous calculations as before yield MATH. And then the triviality of MATH follows straightforwardly. CASE:: Suppose that MATH. Then MATH is associative for as MATH. Hence MATH is an algebra. And MATH. Therefore MATH is algebra morphism. Conversely, if MATH is algebra morphism then MATH. Let MATH be algebra morphism. Then from the last identity of REF and the second identity of REF we derive MATH. If on the other hand MATH then we use REF to obtain MATH which shows that MATH is algebra morphism. It is an easy exercise to prove that triviality of MATH and triviality of MATH are equivalent. |
math/9904142 | By definition cocycle cross product bialgebras are cocycle cross product algebras and cycle cross product coalgebras in particular. Hence MATH is an algebra and MATH is a coalgebra. Since cocycle cross product bialgebras are normalized one deduces with REF that MATH is an algebra morphism and MATH is a coalgebra morphism. Then the conditions of REF hold. The first sixteen (co- )unital identities of REF hold by assumption (see REF). Then from REF one easily derives MATH . Since MATH is especially cocycle cross product algebra, the identities REF are satisfied. Composing the fifth equation of REF with MATH and using REF proves that MATH is right MATH-module. Similarly the weak associativity of MATH is shown by composing with MATH. From REF it will be concluded straightforwardly that MATH and similarly MATH. Since MATH is cross product bialgebra, MATH is algebra morphism, and therefore one easily shows that MATH. Eventually all other (co- )unital identities of REF follow then by MATH-symmetry. The weak associativity for MATH follows now from the sixth equation of REF by adjoining MATH on both sides. The module-algebra compatibility will be derived from the fifth equation of REF through composition with MATH. Making use of the bialgebra identity MATH for MATH we prove the algebra-coalgebra compatibility by application of MATH and MATH on both sides of the bialgebra identity. The first equation of the module-coalgebra compatibility will be shown similarly by composing with MATH and MATH. Application of MATH and MATH yields the second equation of the module-coalgebra compatibility. The module-comodule compatibility is derived from the bialgebra identity by composing with MATH and MATH, whereas the cycle-cocycle compatibility comes from composition with MATH and MATH. Application of MATH to the sixth equation of REF and use of the sixth identity of REF yield the cocycle compatibility. All remaining compatibility relations of REF will be proven by application of MATH-symmetry to the former results. |
math/9904142 | REF can be derived straightforwardly from REF and the relations REF. |
math/9904142 | The first identity in REF has been obtained from REF. With the help of REF the second identity will be derived. Application of MATH-symmetry completes the proof. |
math/9904142 | The relative associativity of MATH and MATH will be derived from the respective weak associativity of REF taking into account REF . |
math/9904142 | Using that MATH is a right module and applying REF yields the first identity of REF. |
math/9904142 | We use REF and the second identity of the comodule-algebra compatibility of REF to show that MATH is a left MATH-comodule algebra. In MATH-symmetric manner it will be proven that MATH is a right MATH-module coalgebra. |
math/9904142 | Since MATH is a is a left comodule algebra by REF the first identity in REF follows from the relative asociativity of MATH according to REF . The verification of REF needs a little more calculation. We prove the first identity of REF. The second one can be derived with similar techniques. We start with the second module-coalgebra compatibility in REF . We apply the relativization (with MATH) to the second tensor factor on both sides of the graphic. The left hand side of this relative module-coalgebra compatibility yields the left hand side of the first identity in REF if we consecutively apply modularity of MATH, the second relation of REF and the second relation of REF. To obtain the right hand side of the first identity of REF we transform the right hand side of the relative module-coalgebra compatibility using successively the third relation of REF, modularity of MATH, the first equation of REF, the modularity of MATH, the second identity of REF, and eventually again modularity of MATH. All other identities can be derived similarly, in particular because of MATH-symmetric reasons. |
math/9904142 | For the proof of the first identity of REF we use REF and the fourth equation of REF . To verify the second identity of REF, the module-algebra compatibility for NAME data, REF and the comodularity of MATH have to be applied successively. Using the module-algebra compatibility and the MATH-symmetric version of the fourth identity of REF yields the fifth identity of REF. Simple calculations yield the sixth identity of REF. The remaining relations of the lemma follow by MATH-symmetric reasoning. |
math/9904142 | We use REF for MATH and the relative associativity of MATH REF to obtain the first identity of REF. The first identity of REF will be used explicitely in the proof of REF . Below we will give its detailed derivation. MATH where the first equation comes from the relative bialgebra property of MATH according to REF. The second equation can be verified with the help of the module-algebra compatibility of REF , and using that MATH is a right module. In the third identity we use the relative associativity of MATH proven in REF . Finally the result follows because MATH is a module coalgebra by REF . The left hand side of the first identity of REF will be transformed to the right hand side of the identity by using successively the algebra-coalgebra compatibility for MATH and MATH, the comodularity of MATH, the first relation of REF, the first identity of REF, the relative associativity of MATH according to REF and finally the second relation of REF. The first identity of REF immediately follows from the relative module coalgebra REF . All other relations of the lemma can be derived easily by applying MATH-symmetry. |
math/9904142 | The proof follows straightforwardly from the cocycle and cycle compatibilities of REF and the identities REF. |
math/9904142 | Because of MATH-symmetry we will only demonstrate the first identities of REF. Applying to the left hand side of REF the algebra-coalgebra compatibility (for MATH and MATH) and the entwining property of MATH REF yields the result. To obtain the first identity of REF we transform its left hand side consecutively using the relative bialgebra REF of MATH and MATH, the comodularity of MATH, the first relation of REF, the first identity of REF, the relative associativity of MATH, and again the fact that MATH is left comodule. |
math/9904142 | The lemma follows straightforwardly from the (co- )associativity of MATH and MATH and the identities MATH. |
math/9904142 | We only consider the first identities of REF. The second identities in each row are MATH-symmetric analogues. The identities REF follow from REF by application of the mapping MATH. Relations REF are special cases of REF through composition with MATH. In a first step we derive REF for the case i=REF. For REF we obtain MATH where the second equation is an immediate consequence of the relative entwining REF of MATH and the third equation follows from REF and the right module property of MATH. To derive the first identiy of REF for i=REF we use REF, the relative entwining REF of MATH, REF , the coassociativity of MATH and the entwining REF of MATH. Using REF, coassociativity of MATH and the entwining REF of MATH we obtain the relations MATH for MATH. We set MATH. These relations allow us to perform easliy the step from i=REF to i=REF. Finally, the case i=REF will be obtained from the corresponding identities for i=REF by composition with MATH. |
math/9904142 | The morphisms on right hand side of the corresponding compatibility relations can be decomposed with the help of REF. Then the reduction REF will be applied to the particular tensor factors. And finally we use again REF to obtain the result. |
math/9904142 | The first identity of REF has been derived by successive application of the left module-algebra compatibility of REF and the relations REF. The second identity is its MATH-symmetric counterpart. To get the third identity we use REF, the first and the second equation of REF, the relations REF and the module-comodule compatibility. The first identity in REF is obtained from the left module-algebra compatibility and REF. The identities in REF are derived from REF with the help of REF. |
math/9904142 | Both identities follow from the third identity of REF using REF. |
math/9904142 | We prove the first identity of REF graphically. MATH . The first equation in the graphic has been obtained with the help of the cycle-cocycle compatibility REF. In the second equation we use REF. The third equation is REF. Application of REF to the fourth diagram yields the fifth relation. The identities REF follow from REF with the help of the entwining properties of MATH and MATH respectively. |
math/9904142 | The identity follows straighforwardly from REF. |
math/9904142 | The identities REF follow from REF, whereas REF will be derived from REF and the entwining REF of MATH, MATH. |
math/9904142 | According to REF the identities REF have to be verified in order to prove that MATH is a cross product algebra. Similar MATH-symmetric procedures are needed to demonstrate that MATH is a cross product coalgebra. Without difficulties the unital identities of REF can be verified. The fifth relation of REF has been proven in REF in the first identity of REF. The seventh identity of REF will be proven subsequently. MATH where the first identity has been obtained from REF. We use the weak associativity of MATH and the module-algebra compatibility of MATH according to REF to get the second equation in the graphic. Then the entwining property of MATH (see REF ) yields the third relation. In the fourth identity we use REF and the relative coassociativity of MATH corresponding to REF . To derive the fifth identity the module-comodule compatibility of REF has been used. In the sixth equation we apply REF, and in the seventh equation we use REF. Hence the seventh identity of REF has been verified. Finally we will prove the sixth identity of REF. Its left hand side will be transformed according to MATH where we used REF in the first equation, and REF and the right modularity of MATH in the second equation. To get the third identity in the graphic one has to apply REF and again the fact that MATH is right module. In the fourth relation we use REF and the relative comodule property of MATH according to REF . The fifth identity can be verified with the help of REF. The module-comodule compatibility, the associativity of MATH and the entwining REF of MATH yield the final equation in the graphic. For the right hand side of the sixth identity of REF we get MATH . In the first equation of REF we used REF. Then we applied REF to get the second equation. To derive the third identity we make use of the cocycle compatibility and the weak associativity of MATH according to REF . In the forth equation we apply the right comodule-coalgebra compatibility, and the fifth identity follows from REF. The sixth equation in the graphical calculation REF has been obtained from the entwining property of MATH and the left module-algebra compatibility. Eventually, the seventh graphic in REF and in REF coincide. This can be verified by applying REF for MATH and MATH, coassociativity of MATH, the right modularity of MATH and REF . |
math/9904142 | The proof will be splitted into several parts. At first we verify the following diagram. MATH where the identities REF in the first row are definitions and the equalities REF in the second row are special cases of REF. The morphisms in the third row will be obtained by applying MATH, MATH and MATH using the identities REF respectively. A similar diagram can be set up for the MATH. For that we use the following auxilarity definitions MATH . Then MATH . Again the identities REF in the first row are definitions. The remaining identities will be proven in the subsequent lemmas. MATH and hence MATH. The identities are obtained using REF. MATH and therefore MATH. The proof of the identity REF will be given in the following graphical calculation. MATH where the first equation holds by definition and the second identity is an application of REF. In the third equation we use the cycle-cocycle compatibility, REF, the (co-)associativity of MATH and MATH, and REF. To prove the first identity of REF we transform in the next calculation the morphism MATH with the help of REF, the entwining REF of MATH, and REF of MATH. MATH . From the right hand side of this equation we get the right hand side of REF by applying REF and again the entwining REF of MATH. MATH and therefore MATH. The following identities hold. MATH . The first identity of REF is a consequence of REF and the module-comodule compatibility. Using this result we obtain the second relation in REF with the help of REF and the entwining REF of MATH. We denote MATH . Then we obtain MATH . The first equation in REF results from the algebra-coalgebra compatibility, the right comodule-algebra compatibility, the left module-algebra compatibility, REF and the right module algebra property of MATH. The second identity can be obtained by subsequent application of the right comodule-coalgebra compatibility, the left module-algebra compatibility, REF, the right module-coalgebra property of MATH, and the entwining property of MATH. The first relation of REF can now be derived by recursive substitution of the identities REF into the second identity of REF. Eventually we define MATH . In what follows we will tacitly use the various forms of (weak) (co- )associativity of MATH, MATH, MATH, MATH, MATH, MATH, MATH and MATH which are given by REF in particular. The subsequent graphical calculation yields MATH. MATH where we used the module-comodule compatibility and REF in the first equation. In the second identity we use again REF. To get the third equation in the graphic we applied the right module-algebra and the left comodule-coalgebra compatibilities of REF as well as REF. With the help of the module-comodule compatibility the fourth identity can be derived from the definition of MATH in REF and the definition of MATH in REF. Next we will prove the identity MATH. And therefore all relations in both rows of the diagram in REF are satisfied. MATH . The first identity in this graphical calculation has been obtained from REF . In the second equation we use REF . With the help of the left module-algebra and right comodule-coalgebra compatibilities of REF and the relativizations REF we derive the third identity. The fourth equation holds by definition. In order to complete the proof of the proposition the equation MATH needs to be shown. This will be done in the following calculation. MATH where the first equation has been derived with the help of REF. In the second and the third equation we use REF. Finally we apply REF to obtain the fourth identity. Hence REF and therefore REF have been proven. |
math/9904143 | Given any MATH, the sequence MATH is coherent. Conversely, given any coherent sequence MATH, we can define MATH by MATH where MATH. |
math/9904143 | As a vector space, MATH, where MATH consists of all functions supported on monomials of weight MATH. It follows that MATH as MATH vector spaces. Of course, there are exactly MATH monomials of weight MATH. Finally, if MATH then MATH, hence MATH. |
math/9904143 | Since MATH has a MATH-basis is given by all monomials of weight MATH, the two MATH-algebras are isomorphic as MATH-vector spaces. The multiplication in MATH is induced from the multiplication in MATH, with the extra condition that monomials of weight MATH are truncated. This is the same multiplication as in MATH. |
math/9904143 | We must show that if MATH, and MATH . , then MATH for MATH. We have that MATH. |
math/9904143 | The monomial MATH has bi-degree MATH. |
math/9904143 | We have that MATH, MATH for MATH. Hence REF implies that MATH . Putting MATH, MATH for MATH we can write this as MATH from which REF follows by taking logarithms. This implies REF as well. |
math/9904143 | CASE: Obvious. REF It suffices to show that for any subset MATH of cardinality REF or REF, there is a MATH with MATH. If MATH then there is an unique positive integer MATH such that MATH, and MATH is the desired generator. If MATH with MATH then we claim that there is a positive integer MATH such that MATH. Namely, choose MATH such that MATH, then since MATH one has MATH. Hence MATH, so it is a multiple of some minimal generator. By the definition of MATH, this minimal generator must be of the form MATH for some MATH, which establishes the claim. CASE: We must show that the number of solutions in MATH to MATH is precisely MATH. Obviously, any integer MATH fits the bill; there are MATH of those. CASE: The case MATH follows from REF. Hence, it suffices to show that if MATH, MATH, and if MATH has no prime factor MATH, then MATH. The only way this can fail to happen is if MATH, but then MATH is even, and has the prime factor MATH, a contradiction. CASE: For large enough MATH, the only integers MATH with all prime factors MATH are MATH. There is MATH of these, and they are all MATH. |
math/9904143 | The first and the last assertions are obvious. The second one follows from the proof of REF in the previous lemma. |
math/9904144 | By CITE, the line bundle on MATH associated with the divisor MATH is MATH. Let MATH be the canonical section of this line bundle; then MATH is an eigenvector of MATH, because its divisor MATH is MATH-stable. The closure in MATH of MATH is the divisor of a regular function on MATH, eigenvector of MATH of weight MATH, and unique up to scalar multiplication. Thus, the weight of MATH is MATH. As the MATH generate the NAME group of MATH, the existence of MATH and the formula for its divisor follow immediately. Finally, uniqueness of MATH up to scalar is a consequence of the fact that MATH has a dense orbit in MATH. |
math/9904144 | Let MATH be the NAME module for MATH; it is a simple, self-dual MATH-module CITE. On the other hand, the line bundle MATH is MATH-linearized by construction of MATH, and the MATH-module MATH contains an eigenvector of MATH of weight MATH, unique up to scalar, by REF . Further, the image of this eigenvector under restriction to MATH is non-zero, since no MATH contains MATH. Using NAME reciprocity CITE and self-duality of MATH, we obtain a MATH-homomorphism MATH such that the composition MATH is non-zero. Since the MATH-module MATH is isomorphic to MATH, hence simple, it follows that MATH is an isomorphism. We thus obtain a MATH-homomorphism MATH . Moreover, the composition MATH is surjective, because the product map MATH is CITE. Now, by CITE, there is a natural MATH-isomorphism MATH and there is a unique MATH-homomorphism (up to a constant) MATH . Further, for MATH and MATH in MATH, the map MATH is a splitting of MATH (up to a constant) if and only if MATH where MATH is the MATH-invariant bilinear form on MATH. Finally, if MATH and MATH for sections MATH, MATH of MATH, then the zero subschemes MATH, MATH in MATH are compatibly MATH-split. Because MATH is a surjective MATH-homomorphism, we can identify it with MATH. Let MATH (respectively, MATH) be a highest (respectively, lowest) weight vector in MATH. Set MATH, MATH, MATH and MATH. Then MATH, MATH are in MATH and they satisfy MATH. Thus, MATH splits MATH compatibly with MATH and MATH. Set MATH and consider MATH a global section of MATH. Recall from CITE that the canonical sheaf of MATH is MATH . Thus, MATH. By CITE, MATH splits MATH compatibly with MATH. Set MATH and MATH. Then MATH and MATH are in MATH, and MATH (respectively, MATH) is an eigenvector of MATH (respectively, MATH) of weight MATH (respectively, MATH). By REF , we have MATH where MATH is the sum of the classes of the MATH over all simple reflections MATH, and MATH. Thus, MATH splits MATH compatibly with MATH and MATH. This implies the theorem, as in the proof of CITE. Namely, one uses CITE and the fact that each MATH is an irreducible component of an iterated intersection of irreducible components of MATH. |
math/9904144 | Let us prove REF . Since the divisor MATH is ample, this is a consequence of CITE when MATH. Moreover, since MATH, MATH are defined over MATH, it follows from the construction of MATH CITE that MATH, the boundary divisors MATH and the large NAME varieties MATH are all defined and flat over some open subset of MATH (in fact, they are defined over MATH by CITE.) Therefore, by the semicontinuity theorem, REF holds in characteristic zero as well. Moreover, by the proof of REF follows (in arbitrary characteristic) from REF . |
math/9904144 | By REF , the scheme-theoretic intersection MATH is reduced; this is equivalent to REF . For REF , we have to check that the image of each MATH in MATH is not a zero divisor. But the scheme-theoretic intersection MATH is reduced. Further, by CITE, each irreducible component of MATH has codimension MATH in MATH, and is not contained in MATH. Thus, the restriction of MATH to MATH does not vanish on any such component. It follows that MATH is not a zero divisor in MATH. For REF , observe that MATH is smooth in codimension one, as the preimage in MATH of a NAME variety in MATH (this goes back to CITE.) Further, the intersection MATH is reduced for MATH, so that each irreducible component of this intersection contains smooth points of MATH. Thus, MATH is smooth in codimension one (this also follows from CITE.) By NAME 's criterion, it is enough to prove that MATH has depth at least two. Because MATH acts on MATH with finitely many orbits and a unique fixed point MATH (the base point of MATH), it suffices to prove that MATH has depth at least two at MATH. This is clear if MATH, because the local equations of MATH at MATH form a regular sequence in the local ring MATH. On the other hand, if MATH then each MATH is smooth. We have indeed MATH, MATH, and MATH is the projectivization of the space of MATH matrices where MATH acts by left and right multiplication. So MATH is either MATH or the projectivization of the subspace of upper triangular matrices. |
math/9904144 | We will use the duality between line bundles and curves: each closed curve MATH in MATH defines an additive map MATH where MATH is the degree of the restriction of MATH to MATH. In fact, MATH only depends on the classes of MATH and MATH up to rational equivalence. Further, MATH is rationally equivalent to a positive integral combination of closed irreducible MATH-stable curves CITE. Examples of such curves are the ``NAME curves" MATH and MATH in MATH. Note that MATH for all MATH. We first show the following The closed irreducible MATH-stable curves in MATH are the MATH and MATH for MATH. They are contained in MATH. Further, each MATH is rationally equivalent in MATH to MATH. The first assertion follows from the description of all MATH-orbits in MATH given in CITE. And as MATH contains both MATH and MATH, it contains the MATH and MATH. For the latter assertion, we begin by the case where MATH. Then we saw that MATH and MATH. Further, MATH is a smooth quadric in MATH, and both MATH, MATH are embedded lines. Thus, they are rationally equivalent in MATH. The general case reduces to the previous one, as follows. Set MATH, then MATH is the closure of a unique MATH-orbit MATH in MATH. Let MATH be the parabolic subgroup generated by MATH and MATH; let MATH be the opposite parabolic subgroup containing MATH, and let MATH be their common NAME subgroup. Then the MATH-variety MATH fibers equivariantly over MATH, with fiber the canonical completion of the adjoint group MATH (this follows for example, from CITE). This group is isomorphic to MATH. Set MATH, the parabolic subgroup generated by MATH and MATH. Now MATH fibers equivariantly over MATH and the fiber over the base point is a closed MATH-stable subvariety MATH of MATH, isomorphic to MATH. Restricting this fibration to MATH, we obtain the canonical map MATH. Thus, MATH contains both MATH and MATH. Further, MATH has a unique closed orbit MATH in MATH; and MATH is contained in MATH (because MATH meets all MATH-orbits). Thus, the closure of MATH in MATH is isomorphic to MATH, and contains both MATH and MATH as embedded lines. We return to the proof of REF . For injectivity, let MATH be a weight such that the restriction of MATH to MATH is trivial. Then the restriction of MATH to each MATH is trivial. It follows that MATH for MATH, and that MATH. For surjectivity, we first prove that the abelian group MATH is free of finite rank. For this, we identify MATH to the group of all NAME divisors on MATH up to rational equivalence (this holds because MATH is normal.) Let MATH be the MATH-fixed point of MATH. Let MATH be the set of all MATH such that the orbit closure MATH contains MATH. Then MATH is an open affine MATH-stable subset of MATH, containing MATH as its unique closed MATH-orbit (it is the image under MATH of the affine chart MATH defined in CITE.) Because MATH, the intersection MATH is a non-empty open affine MATH-stable subset of MATH, containing MATH as its unique closed MATH-orbit. It follows that the NAME group of MATH is trivial (because MATH is normal), and also that any regular invertible function on MATH is constant. Therefore, any NAME divisor on MATH is rationally equivalent to a unique NAME divisor with support in the complement MATH. Now the abelian group of NAME divisors with support in MATH is free of finite rank. We now prove that any NAME divisor MATH on MATH which is numerically equivalent to zero (that is, MATH for each closed curve MATH in MATH) is rationally equivalent to zero. Indeed, by CITE, there exists a positive integer MATH such that MATH is algebraically equivalent to zero. But algebraic and rational equivalence coincide for NAME divisors on MATH, by freeness of MATH and CITE. Thus, the class of MATH in MATH is zero, and we conclude by freeness of MATH again. For a line bundle MATH on MATH, define a weight MATH by MATH so that MATH for MATH. By REF , it follows that MATH for all closed curves MATH in MATH. By the previous step, MATH is isomorphic to MATH. This proves that MATH. For the remaining assertions of REF , let MATH be a line bundle on MATH such that MATH is generated by its global sections (respectively, ample). Then MATH (respectively, MATH) for any closed curve MATH in MATH. Applying this to the curves MATH, one obtains that MATH is dominant (respectively, dominant regular). Conversely, for dominant MATH, the line bundle MATH admits a global section MATH which does not vanish identically on MATH, by REF . Thus, the MATH-translates of MATH generate MATH. If moreover MATH is regular, then MATH is ample by CITE. |
math/9904144 | F rom the exact sequence of sheaves on MATH: MATH we see that MATH injects into MATH. The latter is equal to MATH . We have indeed MATH because MATH is generated by the regular sequence MATH by REF . We now need the following For a weight MATH, the following conditions are equivalent: CASE: MATH is dominant. CASE: MATH is nonzero. CASE: If MATH is dominant, then the restriction to MATH of MATH is generated by its global sections. CASE: Recall that MATH and that the restriction of MATH to MATH is equal to MATH. Thus, there exists MATH such that both MATH and MATH are non-zero. But MATH implies that MATH for each MATH such that MATH; this follows from CITE, see also CITE. Similarly, MATH implies that MATH for each MATH such that MATH. Since MATH permutes the simple roots, the latter is equivalent to MATH for each MATH such that MATH. Now, for each MATH, we have either MATH or MATH, because MATH. Returning to the proof of REF , let MATH such that the space MATH is nonzero. Then MATH is dominant by REF . By REF , the restriction MATH is surjective; therefore, the restriction MATH is surjective. It follows that, firstly, MATH where the sum is taken over all MATH such that MATH and that MATH is dominant, and, secondly, that MATH is equal to MATH . Since MATH for MATH, this implies our statements. |
math/9904144 | The canonical filtrations of the MATH fit together into a filtration MATH of MATH. REF implies that MATH (the MATH-th power of the ideal generated by MATH) and that MATH . Thus, the associated graded ring is isomorphic to the polynomial ring MATH. By CITE MATH form a regular sequence in MATH. Further, by CITE, the graded ring MATH is a quotient of MATH, and the latter ring is generated by its subspaces MATH. So MATH is generated by MATH and the MATH (which lift the MATH.) |
math/9904144 | Set MATH. For MATH, let MATH be the sum of MATH-weight spaces in MATH (for the right MATH-action) over all weights in the coset MATH. Then each MATH is a MATH-submodule of MATH, and we have MATH . Further, the MATH-module MATH is freely generated by any MATH-eigenvector. Choose MATH and a dominant weight MATH in the coset MATH. By REF , MATH contains a submodule isomorphic to MATH, for some MATH. Thus, MATH contains a right MATH-eigenvector MATH of weight MATH, and one deduces that MATH . Now, let us filter MATH by the order of pole along the boundary of MATH. Specifically, consider the section MATH of MATH, where MATH. Then MATH is invariant by MATH, and its zero set is the boundary MATH. Therefore, the MATH-module MATH is the increasing union of its finite dimensional submodules MATH for MATH. The associated graded of this filtration satisfies MATH . Let MATH be the multihomogeneous coordinate ring on MATH, then MATH and MATH . Consider the decreasing filtration of MATH, image of the filtration of MATH by the ideals MATH. As MATH form a homogeneous regular sequence in MATH, the associated graded of MATH satisfies MATH where MATH is the multihomogeneous coordinate ring of MATH. Taking homogeneous components of degree MATH, we see that each MATH has a finite decreasing filtration with associated graded MATH . Reordering the indices, we obtain a canonical increasing filtration of MATH satisfying the requirements of REF . For REF , recall that the irreducible components of MATH are exactly the MATH for MATH. We first construct an increasing filtration of MATH by partial unions of these components, as follows. Choose an indexing MATH which is compatible with the NAME order, that is, MATH if MATH. In particular, MATH and MATH. Set MATH . Then we have the following MATH. Let MATH, MATH in MATH such that MATH, that is, MATH. If moreover MATH, then MATH for some MATH. It follows that MATH, so that MATH . For the opposite inclusion, let MATH such that MATH, that is, MATH. Then MATH for some MATH. Thus, MATH is contained in MATH. Returning to the proof of REF , let MATH be the ideal sheaf of MATH in MATH. Then MATH identifies to the ideal sheaf of MATH in MATH. By definition, we have an exact sequence of sheaves of MATH-modules: MATH. Thus, the sequence MATH is exact. Further, MATH as MATH is a union of NAME varieties in MATH, and MATH is dominant. So we obtain an exact sequence MATH . Now REF implies that MATH . By induction on MATH, we thus obtain a filtration of MATH with associated graded MATH . Further, we have MATH by CITE. And MATH if MATH is the element of minimal length in its coset MATH (where MATH is the isotropy group of MATH in MATH.) Otherwise, we claim that MATH. Indeed, by CITE, the restriction map MATH is an isomorphism. It follows that the MATH-module MATH equals MATH if MATH has maximal length in its coset MATH, and equals MATH otherwise. Thus, the MATH-module MATH has a filtration with associated graded MATH, sum over all MATH such that MATH has maximal length in its MATH-coset. This implies REF . |
math/9904144 | With notation as in REF , we have for any MATH and any index MATH: MATH by REF and the argument thereafter. Since MATH, it follows that MATH for all MATH. Recall now that each NAME variety MATH is NAME. Denote by MATH its canonical sheaf, a MATH-linearized sheaf. By CITE, we have an isomorphism of MATH-linearized sheaves MATH where MATH denotes the shift by the character MATH in the MATH-linearization. Thus, we obtain by using NAME duality on each MATH to pass from the first to the second line: MATH . On the other hand, set MATH and define similarly MATH, MATH. Then we obtain as in REF that MATH. As above, it follows that MATH for any MATH. Thus, MATH and the first identity is proved. In particular, MATH which completes the proof of the second identity. |
math/9904144 | We begin by proving that MATH is NAME. For this, we use the notation introduced in the proof of REF . We check by decreasing induction on MATH that each MATH is NAME. If MATH then MATH, a non-singular variety. For arbitrary MATH, we have an exact sequence MATH . Further, we know that MATH is NAME; and, by the induction hypothesis, the same holds for MATH. On the other hand, MATH by REF . The canonical sheaf of MATH is the tensor product of the ideal sheaf of MATH with the invertible sheaf MATH. By CITE, it follows that MATH is NAME, of depth MATH. Thus, the depth of MATH is MATH. Together with the exact sequence above, this implies easily that MATH is NAME, see CITE. We now prove that the ring MATH is NAME. For this, let MATH be the corresponding affine scheme. Then MATH is the multicone over MATH in the sense of CITE; we now recall some constructions from that paper. Let MATH be the total space of the vector bundle over MATH, equal to the direct sum of the line bundles MATH; let MATH be the projection. Then MATH . In particular, MATH so that we have a morphism MATH. The torus MATH acts on MATH and on MATH (because MATH is graded by the character group of MATH), compatibly with the action of MATH. Clearly, MATH and MATH are equivariant for the action of MATH. As the line bundles MATH are generated by their global sections, MATH is proper. Further, we have MATH as MATH is affine and MATH. In particular, MATH is surjective. Let MATH be the total space of MATH minus the union of all sub-bundles MATH for MATH; let MATH and MATH be the restrictions of MATH and MATH. Then MATH is an isomorphism onto an open subset MATH of MATH, and MATH is a principal MATH-bundle over MATH. As a consequence, the restriction of MATH to each irreducible component of MATH that is, to each MATH is birational. Thus, MATH is equidimensional of dimension MATH. We claim that MATH for MATH. Indeed, as MATH and MATH are affine, this amounts to: MATH which follows from REF . Because MATH is NAME, the same holds for MATH, and we have MATH as a MATH-linearized sheaf; here MATH denotes the shift of the MATH-linearization by the character MATH. We claim that MATH for MATH, that is, MATH for MATH. Indeed, we have MATH as MATH is equidimensional of dimension MATH. For MATH, the line bundle MATH is ample. Because MATH is NAME, we have therefore MATH for MATH and large MATH. But MATH, being an union of NAME varieties in MATH, is NAME split. Thus, MATH for MATH by CITE. This proves our claim. We now recall a version of a result of NAME, see for example, CITE. Let MATH be a proper morphism of algebraic schemes. Assume that MATH is NAME, MATH is equidimensional of the same dimension as MATH, MATH and MATH for MATH. Then MATH is NAME with dualizing sheaf MATH. The statement is local in MATH, so that we may assume that MATH is a closed subscheme of a smooth affine scheme MATH. Denote by MATH the inclusion and set MATH. Then MATH and MATH for MATH. Applying the duality theorem to the proper morphism MATH and the sheaves MATH and MATH, we obtain MATH that is, MATH for MATH, and MATH . This means that MATH is NAME with canonical sheaf MATH. REF implies that the graded ring MATH is NAME with canonical module MATH . Further, MATH for MATH. The module MATH is MATH-graded and each homogeneous component is a finite-dimensional MATH-module. Thus, we can consider the NAME series MATH where MATH, MATH are in MATH, and the MATH are the canonical basis of the group algebra MATH. Now we have MATH . Together with REF , it follows that MATH . Using CITE, we have therefore MATH . Now a result of CITE would imply that MATH is NAME if MATH were a domain. This is not the case, but MATH is the quotient of the domain MATH by the ideal generated by the regular sequence MATH. It follows that MATH is NAME with NAME series MATH because each MATH is the restriction of a MATH-invariant section. As a consequence, we obtain MATH . Thus, by the result of NAME quoted above, MATH is NAME and its canonical module is generated by a homogeneous element of degree MATH, eigenvector of MATH of weight MATH. It follows that MATH is NAME as well and that its canonical module is generated in degree MATH and weight MATH. It remains to prove that MATH and MATH are NAME and to determine their canonical sheaves. For this, consider the isomorphism MATH where MATH is an open subset of MATH, and the principal MATH-bundle MATH. Then MATH as a MATH-linearized sheaf, because the same holds for MATH. Further, MATH so that MATH . Taking invariants of MATH, we obtain MATH that is, MATH is NAME with canonical sheaf MATH. The argument for MATH is similar. |
math/9904144 | By CITE, MATH is equidimensional. Further, MATH is NAME, as MATH is. Because MATH is smooth, it follows that MATH is flat. For MATH, the scheme-theoretic fiber MATH identifies to MATH . Set MATH. Then MATH is stable under right multiplication by MATH, and left multiplication by MATH (the isotropy group of MATH in MATH). The quotient of MATH by the right MATH-action is MATH, whereas the quotient by the left MATH-action is isomorphic to the scheme-theoretic intersection of MATH with the orbit MATH. This intersection is reduced by REF ; thus, MATH and MATH are reduced, too. The remaining asssertions are direct consequences of these facts. |
math/9904144 | As the map MATH is flat and MATH-invariant, and MATH is isomorphic to affine line, the fibers MATH and MATH have the same class in MATH. Thus, the direct images of these fibers under MATH are equal in MATH. This proves the first equality. For the second one, we use the notation introduced in the proof of REF . Set MATH . Then MATH, MATH, and we obtain as in REF that: MATH . As a consequence, we have an exact sequence of sheaves MATH where MATH fits into an exact sequence MATH . It follows that MATH . By decreasing induction on MATH, we thus have MATH. For the third equality, it suffices to prove that MATH in MATH. But the definition of MATH implies that MATH. Further, the NAME function of the partially ordered set MATH is given by MATH if MATH and MATH otherwise, see CITE. |
math/9904144 | Observe that MATH where the second equality follows from REF . In particular, we have for MATH: MATH . As the MATH-linear forms MATH are linearly independent, we obtain MATH . |
math/9904144 | Let MATH be the class of the MATH-linearized line bundle MATH in MATH. As the restriction of this line bundle to the diagonal is MATH, we have MATH . By REF , the latter is equal to MATH . To complete the proof of the first equality, it suffices to check that MATH . For this, using NAME duality as in the proof of REF , we obtain MATH . Further, the NAME character formula implies that MATH for all weights MATH. It follows that MATH by NAME duality once more. Now the second equality follows from REF in the proof of REF . For MATH, the third equality follows from the vanishing of the MATH CITE, and the fourth one from NAME 's dimension formula. |
math/9904144 | Because MATH and MATH are NAME, the same holds for MATH. And because MATH is birational and MATH is normal, we have MATH. We now show that MATH for MATH. For this, it suffices, by REF (see for example, CITE), to show that MATH for MATH and for any regular dominant weight MATH. Consider the line bundle MATH and its higher direct images MATH for MATH. Then MATH is the MATH-linearized sheaf on MATH associated with the MATH-module MATH, and MATH is the restriction to MATH of this MATH-linearized sheaf. As MATH for all MATH by REF , we have MATH for MATH. For a MATH-module MATH, denote by MATH the corresponding homogeneous vector bundle on MATH. Then we obtain from the NAME spectral sequence for MATH that MATH . By REF , the left MATH-module MATH has a filtration with associated graded a direct sum of MATH's for certain dominant weights MATH. Further, we have for MATH: MATH as follows from CITE or CITE. Thus, MATH and, therefore, MATH for MATH. We now determine the canonical sheaf MATH; we begin with the relative canonical sheaf MATH of MATH. Observe that the relative canonical sheaf of MATH equals MATH as a MATH-linearized sheaf, where MATH denotes the shift by MATH of the MATH-linearization. Indeed, MATH is the MATH-linearized sheaf on MATH associated with the MATH-linearized sheaf MATH on MATH. On the other hand, the sheaf MATH is MATH-linearized, and the associated MATH-linearized sheaf on MATH is MATH by REF . As MATH is the pull-back of MATH under the inclusion MATH, it follows that MATH. In particular, MATH is invertible. Thus, MATH . We now claim that MATH . For this, observe that MATH contains the preimage under MATH of MATH, a NAME divisor. Further, the complement MATH of that divisor is equal to MATH. As the line bundle associated with MATH is MATH, it follows that MATH which proves the claim. We conclude that MATH . Further, since MATH and MATH, then MATH and, therefore, MATH . We now prove that MATH for MATH. Using NAME 's lemma, again, it suffices to prove that MATH for MATH and for MATH big enough (we consider here MATH). We argue by induction over MATH, the case where MATH being obvious. In the general case, choose a decomposition MATH where MATH is a simple reflection, and MATH. Let MATH be the parabolic subgroup generated by MATH and MATH. This defines the variety MATH together with the map MATH. Let MATH with projections MATH and MATH. Let MATH be the composition of MATH and MATH, then MATH is an isomorphism above MATH. Further, MATH is a locally trivial fibration with fiber MATH. The MATH-action on MATH lifts to MATH, where MATH acts trivially on MATH and MATH. We claim that MATH, MATH and MATH for MATH. This follows from CITE. In more detail, consider a reduced expression for MATH and let MATH denote the NAME resolution associated to the corresponding reduced expression of MATH. Observe that MATH factors through MATH, say MATH. Since MATH, MATH, MATH are proper and birational and MATH, MATH are normal, then MATH, MATH and MATH. Let MATH denote the complement of the open MATH-orbit in MATH, then MATH and MATH, so that MATH. Further, by CITE, one has MATH and MATH for MATH. Since MATH is proper with fibres being points or projective lines, then MATH for MATH and, therefore, one obtains, by using the projection formula, that MATH . This proves the claim. Since MATH is the pull-back of MATH under the locally trivial fibration MATH then, using again the projection formula, it follows that MATH . This yields MATH and it suffices to prove that the right-hand side vanishes for MATH and for MATH big enough. Embed MATH into MATH, as a NAME divisor; then MATH embeds into MATH. Observe that MATH whereas MATH. Thus, we have an exact sequence MATH . Together with the induction hypothesis, it yields an exact sequence MATH and isomorphisms for MATH: MATH . Consider the projection MATH . Then the higher direct image sheaf MATH is the homogeneous vector bundle on MATH associated with the MATH-module MATH . The latter vanishes for MATH and large MATH, by the induction hypothesis. As MATH is a projective line, it follows that MATH for MATH. And setting MATH then MATH gives an exact sequence MATH . To complete the proof, it remains to show that MATH. For this, it is enough to check that MATH is generated by its global sections, that is, that MATH is the quotient of a MATH-module. Now, using MATH and MATH, observe that MATH . Further, MATH (indeed, MATH is obviously contained in MATH; and MATH is not contained in MATH, because MATH is not stable by MATH), and this intersection is reduced as large NAME varieties are compatibly split in MATH. Therefore, MATH, and the restriction map MATH is surjective for MATH big enough, by NAME 's theorem. Thus, MATH is a quotient of a MATH-module. This completes the proof of REF . Now the previous arguments and REF imply that MATH is NAME with canonical sheaf MATH, which proves REF . Then REF follows by arguing as in the proof of REF . |
math/9904144 | Since MATH is the complete intersection in MATH of the NAME divisors MATH, by REF , it follows that MATH is NAME. Similarly, MATH is NAME and its canonical sheaf is the restriction to MATH of MATH . The latter is equal to MATH, as we saw in the proof of REF . This proves REF . The multiplication by MATH defines exact sequences MATH and MATH . By REF , it follows that MATH and MATH for MATH. Iterating this argument, we obtain MATH and MATH for MATH. The vanishing of MATH and the equality MATH follow similarly from the exact sequences MATH and MATH together with REF . This proves REF . It also follows, using REF , that MATH . But MATH as MATH is a locally trivial fibration, and MATH as MATH. This completes the proof of REF . Finally, REF is checked as in the proof of REF . |
math/9904144 | Recall that MATH. By REF , the latter is isomorphic to MATH . Using REF , we obtain similarly that MATH. We prove that MATH by descending induction on MATH. If MATH then MATH and MATH. In this case, the assertion follows from CITE. In the general case, let MATH be a simple reflection such that MATH; let MATH be the parabolic subgroup of MATH generated by MATH and MATH. Then, using CITE, we obtain that MATH . Further, the natural map MATH is surjective by REF . Thus, MATH embeds into MATH . The latter equals MATH by CITE, and we conclude by the induction hypothesis. |
math/9904148 | This lemma is well known, but in view of the difficulty we have to locate an explicit reference, we present a proof here. For simplicity, we only present the proof for the case MATH. The arguments in general case are exactly the same, but the notation is more complicated. We also identify the one dimensional torus with the unit circle MATH in MATH. Fix MATH and consider the free action of MATH on the product of spheres MATH given by MATH . Let us denote by MATH the quotient of MATH by the action of MATH. Since the reduced cohomology of MATH vanishes up to dimension MATH it follows that there is a natural MATH equivariant isomorphism between the cohomology MATH and MATH for any MATH less than MATH (this may be shown by the same arguments as in the proof of REF) Clearly, MATH acts on MATH as complex conjugation on the first factor and multiplication by MATH on the second factor. Hence, the induced action on the cohomology MATH is multiplication by MATH. This proves the lemma. |
math/9904148 | It follows from REF , that there exists a formal power series MATH with non-negative coefficients, such that MATH . Our goal is to show that MATH. It follows from REF, that both the equivariant NAME counting series and the equivariant NAME series are additive with respect to MATH. More precisely, if MATH denotes the direct sum of two representations then MATH . Hence, it suffices to prove the lemma for the irreducible representations of MATH. Moreover, it follows from REF that it is enough to prove that MATH when MATH is a reducible representation, which contains any of the irreducible representation as a subrepresentation. In particular, it is enough to prove that MATH when MATH is the regular representation. However, REF implies that, if MATH is the regular representation, then REF reduces to the MATH-equivariant NAME inequalities with trivial coefficients. It was shown by CITE that the later inequalities are exact, that is, MATH. |
math/9904159 | The isomorphism on the left-hand side has been discussed in REF . The one in the middle is the special case MATH, MATH of the isomorphism of graded MATH-algebras MATH induced from the NAME class homomorphism MATH that associates to a character MATH the first NAME class of the line bundle MATH. Here MATH denotes the one-dimensional MATH-module with weight MATH. For a cone MATH and the corresponding subtorus MATH of MATH, the restriction mapping MATH and the natural mapping MATH induce a commutative diagram MATH since we have MATH (the restriction of MATH to MATH is just the line bundle associated to the character MATH) and NAME classes are functorial, thus proving the naturality of the isomorphism in the middle. We now discuss the isomorphism MATH on the right hand side: The affine orbit splitting MATH of REF and the MATH-equivariant contraction MATH onto the distinguished point induce isomorphisms MATH . The whole construction is natural with respect to some face relation MATH. This is easily seen from any splitting of the torus in the form MATH, since the choices of MATH and MATH do not play any role. |
math/9904159 | We proceed by induction on the number of cones in the fan MATH. For MATH, the assertion is obvious. For the induction step, we choose a maximal cone MATH and consider the NAME sequences associated to MATH and MATH, both for MATH and for MATH. It suffices to prove that in the commutative diagram MATH obtained from REF , the rows are exact, the vertical arrows are isomorphisms, and MATH vanishes. Applying the induction hypothesis to the fans MATH and MATH, and REF to MATH, we see that the assertion holds for the second and the third arrow, and we obtain the leading MATH in the upper row. Furthermore, since the fan MATH is simplicial, it is not difficult to see that the sheaf MATH is flabby; hence, the map MATH, and thus also MATH, is surjective. This proves MATH, as, by induction hypothesis, we know that MATH vanishes. By the Five Lemma, the first vertical arrow is an isomorphism as well, thus proving our claim. |
math/9904159 | For a cone MATH, let MATH be the character which coincides with MATH on MATH. Since the map MATH is an isomorphism of sheaves, it suffices to show that the ``local" equivariant NAME class MATH is mapped onto MATH. Observing that the inclusion MATH of the fixed point MATH induces an isomorphism MATH, we may further restrict our attention to the fibre MATH of MATH over MATH. As a MATH-module, this fibre is nothing but MATH, and the character MATH is the NAME class of that bundle. This completes the proof. |
math/9904159 | Let MATH be a simplicial refinement of the defining fan MATH for MATH, and denote with MATH the corresponding equivariant MATH-resolution of singularities. By REF , the assertion holds for MATH, since then MATH. By the equivariant version of REF, NAME, NAME, and NAME as stated in CITE (see also CITE or CITE), we may interpret MATH as a subspace of MATH. That proves the assertion. |
math/9904159 | Since there are only finitely many open subsets in MATH, it suffices to verify the sheaf axioms for two open subsets MATH. We thus have to prove the exactness of the sequence MATH . That follows from REF : The exactness is obvious if MATH is odd; for even MATH, the vector space MATH vanishes and thus, the sequence is part of the exact NAME sequence for MATH. As a consequence, MATH is a sheaf of MATH-modules, since MATH is the associated sheaf to the presheaf MATH, and each MATH is a MATH-module. |
math/9904159 | We define the sheaf MATH inductively on the MATH-skeleton subfans MATH starting with MATH for MATH. Suppose that for some MATH, the sheaf MATH has already been constructed on MATH, so in particular, for each MATH-dimensional cone MATH, the module MATH is given. It thus suffices to define MATH together with a restriction homomorphism MATH inducing an isomorphism on the quotients modulo MATH. This is achieved by setting MATH with the restriction map being induced by some MATH-linear section MATH of the residue class map MATH. The unicity is proved similarly in an inductive manner; we refer to our companion article CITE for details. |
math/9904159 | Since the fan space MATH is covered by finitely many affine fans, it suffices to prove that for each cone MATH, the restriction map MATH is surjective. Using the results on graded modules recalled in the ``NAME, that is a consequence of condition (LME). - The vanishing of MATH follows immediately from the same condition, since MATH and thus MATH ``live" only in even degrees. |
math/9904159 | The Normalization property is obviously satisfied, since we have MATH . The Pointwise Freeness condition will be verified in REF , and the Local Minimal Extension requirement, in REF . |
math/9904159 | Since REF says that for intersection cohomology, MATH behaves like the product MATH, the implication ``i REF" is obvious, and ``ii REF" follows immediately from the Vanishing REF . For the implication ``iii REF", we observe that the assumption implies the degeneration of the intersection cohomology NAME spectral sequence MATH associated to the fibering MATH at the MATH-level: Since MATH vanishes for odd MATH, the spectral terms MATH vanish for odd total degrees MATH, and consequently, the differentials MATH are trivial. By induction on MATH, that holds also for every MATH. |
math/9904159 | Let MATH denote a toric MATH-resolution as in the proof of the Vanishing REF . By the ``classical" (that is, non-equivariant) Decomposition Theorem of NAME, NAME, NAME and NAME, we know that MATH is isomorphic to a direct summand of MATH, and, according to REF and NAME, that module vanishes in odd degrees. |
math/9904159 | By the ``relative affine orbit splitting" MATH and REF as well as its analogue MATH, we see that it is sufficient to consider the case of a MATH-dimensional cone MATH. We use the same notations as in the previous section; in particular, we write MATH and MATH. Furthermore, by the arguments of the preceding section, we may replace MATH, MATH, and MATH by MATH, MATH, and MATH, respectively, where MATH is a suitable finite subgroup, without changing the base ring MATH and the above homomorphism. Hence we may assume that MATH is a principal MATH-bundle. First we collect in a big commutative diagram all the objects we have to consider: MATH . Here, MATH again denotes the homomorphism given by the cup product with the NAME class MATH of the line bundle MATH as in the previous section. Once having established the diagram, the proof is achieved as soon as we have identified the quotient MATH - this is just the image of the ``edge homomorphism" MATH relating the equivariant and the non-equivariant theory - with MATH. To that end, we consider the right hand side of the diagram, carefully keeping track of the different module structures in the top row: Whereas MATH and MATH are both modules over MATH, we have to look at MATH as a module over the ring MATH, where MATH. In particular, the ``overlined" modules in the second row are quotients modulo the maximal ideals of the respective base rings: For MATH and MATH, this is the ideal MATH, whereas for MATH, we have to consider MATH (with MATH). As the projective MATH-toric variety MATH is equivariantly formal, we may identify this graded MATH-module with MATH. We may now consider MATH as a subring of MATH since MATH is canonically isomorphic to a submodule of MATH. Hence, writing the horizontal arrow MATH as MATH we see that it is an epimorphism, being induced by the horizontal isomorphism MATH in the top row. Thus, using the isomorphism in the bottom row, we are done if we can show that the kernel of MATH equals the image of the ``hard NAME homomorphism" MATH. Before we do that let us give some further remarks on the big diagram: The isomorphism MATH in the top row is obtained as follows: The bundle projection MATH induces a compatible family of projections MATH between the finite-dimensional approximations of MATH and MATH. As these projections are placid maps, there is a (unique) induced homomorphism MATH (for a discussion, see CITE or CITE). We now note that the bundle is locally trivial and that there is a finite open affine covering of MATH by toric subvarieties MATH such that the restricted bundle MATH actually is trivial. By REF , we thus have isomorphisms MATH; gluing these by a NAME argument yields the result. - We note that, by construction, the isomorphism REF is a morphism of MATH-modules; it thus induces an isomorphism of the quotients modulo the homogeneous maximal ideal MATH, while MATH refers to the bigger base ring MATH and thus is a quotient of MATH. We recall why we obtain the other isomorphisms occuring in the diagram: By the attachment condition, the restriction mapping MATH factors through the inclusion of MATH, inducing the oblique isomorphism. The two vertical isomorphisms follow from the fact that both MATH and MATH are equivariantly formal. The lower horizontal isomorphism has been obtained in REF in the previous section. We now continue with the proof of REF . First of all, we lift the ``hard NAME map" MATH to a map in equivariant intersection cohomology: We make the line bundle MATH a MATH-toric line bundle by choosing some sublattice MATH complementary to MATH, thus providing a direct sum decomposition MATH. From REF, we recall that the MATH-equivariant NAME class MATH of MATH is a lifting of the ``usual" NAME class MATH. It follows that the cup product with MATH yields a homomorphism MATH that is a lifting of the mapping MATH given by the cup product with MATH. We further recall that MATH is a sheaf on the defining fan MATH for MATH, and it is a module over the ``structure sheaf" MATH corresponding to the fan MATH. Using the canonical isomorphism MATH, we may identify the equivariant NAME class MATH with the MATH-piecewise linear function MATH (see REF , and REF ). Now the quotient projection MATH induces an isomorphism MATH and the image of MATH coincides with the restriction to MATH of the ``global" linear form MATH, the projection MATH mapping MATH to MATH and having MATH as kernel, compare REF. On the other hand, we have an isomorphism of polynomial rings MATH and thus, the equality MATH for the homogeneous maximal ideals. The proof of the assertion that the restriction homomorphism MATH induces an isomorphism on the quotients with respect to the submodules generated by MATH is now obtained as follows: Under the inverse of the isomorphism REF , the submodule MATH is mapped onto MATH. We thus have an isomorphism of quotients MATH . As explained above, the mapping MATH given as multiplication by MATH lifts the ``Hard NAME homomorphism" MATH to the equivariant theory. Since MATH, as a projective MATH-toric variety, is equivariantly formal, we may eventually rewrite the right hand side of the above isomorphism REF as follows: MATH . The proof is now achieved using the isomorphisms REF , and the fact that the contractible affine toric variety MATH is MATH-formal. |
math/9904159 | REF ` REF ': Since MATH is a sheaf, we see that we have a natural inclusion MATH of MATH-modules. By REF , each MATH is a free MATH-module. For MATH, we have MATH, so the right hand side is a free MATH-module. Moreover, every submodule of a torsion-free module is again torsion-free. ` REF ': If MATH is a maximal cone of dimension MATH, let MATH. The product of some non-zero polynomial function MATH vanishing on MATH (such a function can be obtained as a product MATH of non-zero linear functions MATH with MATH) and of a non-zero section in MATH yields a non-zero section MATH (recall that MATH is a free MATH-module!) that vanishes on MATH. We extend it trivially outside of MATH and thus get a non-trivial torsion element, since it is ``killed" by every non-zero global linear function in MATH vanishing on MATH. REF: ` REF ': This is clear by REF. ` REF ': We only sketch the argument, leaving details for future exposition: Consider the intersection cohomology version of the NAME spectral sequence (see, for example, CITE) that computes the intersection cohomology of the pull back of a bundle. Here we look at the bundle MATH and take as map the inclusion of a one point set MATH into MATH. Since MATH is simply connected, the spectral sequence converges: We have MATH with MATH . Here MATH, MATH, and MATH, respectively, are considered as differential graded algebras respectively, modules with trivial differential, and we can compare with the classical Tor functors of commutative algebra: If MATH, MATH are graded MATH-modules, the corresponding NAME are again in a natural way graded MATH-modules: MATH such that MATH . Since MATH is a free MATH-module, we obtain that MATH and MATH for MATH. So in particular, we have MATH for MATH odd, and hence also MATH in that case. CASE: This follows immediately from REF, since free modules are torsion-free. CASE: According to REF, NAME, NAME, and NAME, we know that MATH is a direct summand of MATH, thus inheriting the property that the ``usual" intersection cohomology vanishes in odd degrees. |
math/9904159 | We intend to prove that the intersection cohomology NAME number MATH is non-zero. With MATH and MATH, we consider the following part of the exact NAME sequence: MATH . The zeroes at both ends are due to the fact that the toric varieties MATH and MATH are equivariantly formal. The ``affine orbit splitting" REF provides an isomorphism MATH with MATH, where MATH, as a contractible affine MATH-toric variety, is known to be equivariantly formal. By the NAME formula, we have MATH; in particular, we get MATH and MATH. By the results of CITE, the NAME number MATH of an arbitrary MATH-dimensional toric variety given by a non-degenerate fan MATH (that is, such that MATH spans MATH) is determined by the number MATH of rays, namely, we have MATH. Denoting with MATH and MATH the respective number of rays, we clearly have MATH. We thus obtain MATH, MATH, and MATH. Since the NAME characteristic of an exact sequence vanishes, we obtain MATH. |
math/9904159 | It is known (see, for example, CITE) that the natural homomorphism MATH is surjective (and even an isomorphism, though we do not need this stronger result). To investigate the target, let MATH be a completion of the fan. The set of cones MATH defines a closed invariant subvariety MATH of the compact toric variety MATH that has the same number MATH of connected components as MATH. Combining that with the isomorphism MATH and with the exact sequence MATH eventually yields the following chain of REF that proves the assertion. |
math/9904159 | We have to prove that the toric variety MATH has vanishing intersection cohomology in each odd degree MATH if MATH has. To that end, we look at the exact sequence MATH where the final term vanishes since MATH is odd. It clearly suffices to prove that MATH vanishes. We may identify this relative group with MATH by excision. We thus consider the analogous exact sequence MATH . As MATH is simplicial, there is an isomorphism MATH, where MATH is a finite subgroup of MATH acting diagonally on MATH. By the NAME formula, we thus have isomorphisms MATH, so MATH vanishes for each MATH. For MATH, these facts immediately yield MATH. The remaining case MATH follows from the fact that the restriction mapping MATH is an isomorphism. |
math/9904159 | As in the proof of REF , we have to show that MATH vanishes in odd degrees. There is an isomorphism MATH where MATH is a finite subgroup of MATH acting diagonally on MATH, such that MATH. As passing to the quotient by MATH does not influence the rational (intersection) cohomology (see CITE), we obtain MATH thus proving the assertion. |
math/9904159 | We choose a vector MATH such that either MATH lies in the interior of the support REF , or that MATH lies in the interior of the complement REF , and complete the given fan MATH by adding the new ray MATH together with all cones of the form MATH, where MATH is a cone in the boundary of the support. The ``old" fan then is the complement of the star of the ``new" ray. |
math/9904163 | Choose a set MATH such that CASE: there is MATH with MATH and MATH, but CASE: for each MATH and MATH, if MATH then MATH. Clearly it is possible; necessarily MATH (remember that MATH is typical). Fix MATH. Applying bigness to MATH we get MATH such that MATH and MATH. On the other hand, by the choice of the set MATH (and bigness) we find MATH such that MATH and MATH. |
math/9904163 | Let MATH and let MATH witness that MATH is densely representable by MATH. Let MATH (for MATH). Then the sets MATH are dense in MATH by REF . We claim that they witness the failure of MATH. So suppose that MATH is a directed set which meets each MATH. For every MATH choose MATH such that MATH. Look at the set MATH - it is an uncountable REF - selector from MATH, a contradiction. |
math/9904163 | Let MATH, MATH, MATH a basis of a topology on MATH, MATH. Assume that some condition in MATH forces that MATH is not a MATH - family. Then we find MATH, MATH, MATH and MATH - names MATH for elements of MATH (for MATH) such that MATH . For each MATH choose a sequence MATH and a condition MATH such that MATH". (So necessarily MATH.) Let MATH be the following forcing notion: a condition in MATH is a finite set MATH such that MATH the order is the inclusion (that is, MATH if and only if MATH). Note that if MATH, MATH then the conditions MATH are incompatible (we will use it in REF). Assume MATH, MATH (the increasing enumeration; MATH). Then there is MATH such that CASE: for each MATH, if MATH are incompatible in MATH then there are MATH and MATH such that MATH. For each MATH, MATH and distinct MATH choose MATH such that if MATH then MATH, otherwise MATH are such that CASE: MATH, MATH, MATH, and for all MATH, MATH . (Why possible? Remember the definition of MATH and REF .) Each MATH is countable, so there are MATH and an uncountable set MATH such that CASE: MATH. Suppose that MATH and the conditions MATH are incompatible. It means that there are MATH and MATH such that MATH. We are going to show that we may demand MATH (what will finish the proof of the claim). So suppose that MATH. If MATH, then (by MATH) MATH and (by MATH) MATH. But applying the last part of MATH we may conclude now that MATH, a contradiction. So MATH. But then MATH and thus MATH. MATH is a ccc forcing notion. Suppose that MATH is an antichain in MATH. We may assume that, for some MATH, MATH for all MATH. Let MATH be the increasing enumeration. Using REF we may find an uncountable MATH such that for each distinct MATH there are MATH and MATH with MATH. For each MATH let MATH be a sequence of length MATH such that MATH . Look at the set MATH. It should be clear that it is an uncountable MATH - selector for MATH (the clause MATH of REF for MATH is witnessed by MATH). A contradiction. MATH `` MATH is a MATH - family of good graphs ". The only bad thing that could happen after forcing with MATH is that an uncountable MATH - selector was added (for some MATH). If so, then we have MATH, MATH - names MATH (for MATH) and a condition MATH such that MATH . Clearly we may require that for some MATH, for each MATH the condition MATH forces that MATH (for all MATH). For each MATH pick a sequence MATH and a condition MATH such that MATH. Next, choose an uncountable set MATH and MATH such that for MATH, MATH (the increasing enumeration) and MATH of REF holds (possible by REF). Let MATH (for MATH) be sequences of length MATH defined by MATH . Note that for distinct MATH we have: CASE: if MATH are compatible in MATH, MATH, then for some MATH we have MATH; CASE: if MATH are incompatible in MATH then there are MATH and MATH so that MATH. (Why? MATH follows from the choice of MATH, MATH, MATH is a consequence of MATH.) Hence, MATH is an uncountable MATH - selector from MATH, a contradiction. For some MATH we have MATH . As we stated before, if MATH and MATH are distinct then the conditions MATH are incompatible in MATH. Since, by REF, the forcing notion MATH is ccc, there is a condition MATH such that MATH . It should be clear now that the condition MATH forces (in MATH) that MATH fails the ccc. Let MATH be as guaranteed by REF and let MATH be the MATH restricted to elements stronger than that MATH. It should be clear that MATH is as required in the proposition. |
math/9904163 | In REF use REF; in other cases use directly the assumption that MATH is typical. |
math/9904163 | For a condition MATH we may find a stronger condition MATH with the following property CASE: for each MATH, MATH there are MATH, MATH, MATH, and MATH, MATH, MATH (objects, not names) such that the condition MATH forces the following: CASE: MATH, MATH, MATH, CASE: MATH, MATH for each MATH, CASE: for each MATH such that MATH is defined: MATH are in MATH, the conditions MATH are stronger than MATH and MATH, and if MATH, MATH, and MATH is the MATH - name for the MATH generic real (that is, MATH) then MATH CASE: for each MATH and MATH, MATH we have MATH . [Why? Just apply REF (and remember that supports are finite).] From now on we will restrict ourselves to REF with the property MATH (what is allowed as they are dense in MATH). So we will assume that for each condition MATH under considerations and MATH, MATH, the objects (not names!) MATH, MATH, MATH, MATH, MATH, MATH are defined and have the respective properties. Note that, in MATH, if MATH, MATH, MATH and MATH then MATH and MATH (remember REF). Therefore, we may label elements of MATH by pairs from MATH and allow ourselves small abuse of notation identifying MATH with the respective MATH. Next let MATH. Now, suppose that some condition MATH forces that MATH . Then we may find a condition MATH, an integer MATH, MATH (not necessarily distinct) and MATH - names MATH (for MATH) of sequences of length MATH such that MATH . For each MATH choose a condition MATH (satisfying MATH and) stronger than MATH and a sequence MATH such that MATH and MATH . Now we consider two cases. CASE: MATH. Then for some MATH, for uncountably many MATH, MATH. Let MATH be a generic over MATH and work in MATH. Because of the ccc of MATH, the set MATH is uncountable, so we get an uncountable MATH - selector from MATH (in MATH), contradicting the assumption MATH. CASE: MATH. If for some MATH the set MATH is uncountable then we may repeat the arguments of Case A. So assume that MATH is countable for each MATH. Applying ``standard cleaning procedure" and passing to an uncountable subsequence (and possibly increasing our conditions) we may assume that MATH for each MATH and, letting MATH be the increasing enumeration of MATH: CASE: MATH forms a MATH - system with heart MATH, CASE: for some MATH and MATH, we have MATH and MATH (so necessarily MATH), CASE: MATH, CASE: MATH, CASE: if MATH, MATH, MATH and MATH then MATH, MATH, MATH, MATH (see REF), and CASE: if MATH is the order preserving bijection then MATH is the identity on MATH and for each MATH . Let MATH be the set of these MATH that for some (equivalently: all) MATH we have MATH. There are MATH and MATH such that MATH is stronger than both MATH and MATH and MATH. Let MATH be a generic filter over MATH. Work in MATH. By the ccc of MATH, the set MATH is uncountable. Look at the sequence MATH. By REF, it cannot be a MATH - selector, so there are MATH such that MATH . Now, going back to MATH, we easily find a condition MATH such that MATH are as required. Let MATH be as guaranteed by REF. For MATH such that MATH, MATH, let MATH be the order preserving bijection (see clause REF above). We define a condition MATH as follows: MATH and CASE: MATH, CASE: if MATH, MATH, MATH then MATH CASE: if MATH, MATH then MATH, MATH. It should be clear that MATH is a condition stronger than both MATH and MATH. If MATH and MATH then MATH . If MATH, MATH, MATH and MATH, MATH are the names for MATH (MATH respectively) generic reals added by MATH (MATH, respectively,) then MATH . If MATH then look at the choice of MATH (see REF). Putting everything together we conclude that MATH a contradiction . |
math/9904163 | Like REF. |
math/9904164 | All REF follow easily from the properties of quasi-Hopf algebras as stated in REF. We will give a rather detailed proof such that the unexperienced reader may get used to the techniques used when handeling formulae involving iterated non coassociative coproducts. We will in the following denote MATH. Equality REF follows directly from the antipode property MATH. To show REF one uses REF to compute MATH by REF. Equality REF is obtained similarly: MATH . REF follows immediately from REF . To show REF we use REF: MATH . REF follows by using REF for MATH, REF and then REF MATH . Finally, we prove REF by using REF and the left identity in REF to compute MATH . |
math/9904164 | The definitions in REF imply MATH, which by REF further implies MATH . Thus we are left to show the identity MATH (denoting MATH): MATH where the last equality follows from REF. |
math/9904164 | Clearly, REF implies MATH. Conversely, if MATH then REF implies MATH, hence MATH. |
math/9904164 | Using REF and MATH we compute MATH . Conversely MATH by REF . Thus MATH is indeed an isomorphism of vector spaces. We are left to show that MATH also respects the quasi-Hopf MATH-bimodule structures. By definition we have MATH and therefore MATH . Here we have used REF in the second and third line, respectively. Thus MATH is a MATH-bimodule map. Finally, we show that MATH (and therefore MATH) are also MATH-comodule maps. MATH . Here we have used REF in the first line and REF together with the quasi-coassociativity REF in the second line. |
math/9904164 | If MATH then by REF MATH, whence MATH. The inverse implication follows from MATH and MATH. |
math/9904164 | Denote MATH and MATH the projections onto the corresponding coinvariants. To prove that MATH is bijective we claim MATH where MATH is given by MATH . By REF MATH is well defined and we have MATH thus proving REF. To prove REF let MATH and MATH. Then REF MATH where in the second line we have used REF , in the third line REF and in the fourth line the identity MATH which follows easily from REF. Thus, by REF we conclude for general MATH . This proves that MATH is bijective. To prove that it is MATH-linear we compute for MATH and MATH . MATH-linearity of MATH follows by applying MATH to both sides and using REF . |
math/9904164 | If MATH is a morphism of quasi-Hopf MATH-bimodules, then MATH, MATH is MATH-linear and MATH . Thus, MATH is uniquely determined by its restriction MATH, and by REF MATH provides an equivalence of categories with reverse functor given by MATH and MATH. By REF these functors preserve the monoidal structures provided we show that for any three objects MATH in MATH the following diagram commutes MATH . To this end let MATH, then MATH where we have used REF in the first line, MATH-linearity of MATH in the second line, REF in the third line and again MATH-linearity of MATH in the last line. |
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