paper stringlengths 9 16 | proof stringlengths 0 131k |
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math-ph/9904013 | The proof is similar to the one in REF. Define the two subsets MATH and MATH of the interval MATH, MATH . The intersection of MATH with MATH is by definition empty, and the sets MATH and MATH are open, by continuity of the solution MATH as a function of the initial data MATH. We now show that all MATH sufficiently close to MATH are in MATH and that all MATH sufficiently close to MATH are in MATH . This implies that MATH and MATH is neither in MATH nor in MATH, and therefore the function MATH satisfies MATH for all MATH and therefore MATH since MATH . So let MATH be an initial condition. Then, MATH satisfies the integral equation MATH where MATH and where MATH . MATH is strictly negative for MATH and MATH and therefore we find, like in the proof in REF that any solution with an initial condition MATH sufficiently close to MATH will cross the line MATH . Similarly, for an initial condition MATH close to MATH we can use that MATH and that MATH is strictly negative to show that the corresponding solution will cross the line MATH . This completes the proof of the existence of MATH . To prove uniqueness it is sufficient to use that MATH for MATH in the set MATH (see REF ), and to integrate the difference of two solutions from their respective initial condition to zero, which leads to a contradiction, since both solutions have to be equal to MATH at zero. Finally, that MATH is a continuous function follows from the continuity of MATH as a function of MATH and MATH using the uniqueness of MATH . |
math-ph/9904013 | Define the two subsets MATH and MATH of the interval MATH, MATH . By definition, the intersection of MATH with MATH is empty, and the sets MATH and MATH are open, by continuity of the solution MATH as a function of the initial data MATH. We now show that all MATH sufficiently close to MATH are in MATH and that all MATH sufficiently close to MATH are in MATH . This implies that MATH is neither in MATH nor in MATH, and therefore the function MATH is positive and decreasing for MATH which implies that MATH . So let MATH be an initial condition with MATH with MATH . The proof that such an initial condition is in MATH is rather lengthy and we therefore do not give the details here, but on a heuristic level it is easy to understand why such a solution is in MATH. Namely, near zero the asymptotics of the solution MATH is MATH where MATH and where MATH is such that MATH. Neglecting higher order terms we find that MATH, and we conclude that MATH . Therefore, MATH and MATH if MATH is small enough. Therefore, since MATH for all MATH we find that MATH and therefore MATH for some MATH as claimed. Next, let MATH be an initial condition with MATH with MATH to be chosen below. We now show that such an initial condition is in MATH . For all MATH we have the lower bounds MATH and MATH where MATH and MATH and for MATH we therefore have the lower bound MATH and it follows, using again the integral REF , that MATH diverges at (or before) infinity, that therefore MATH for some MATH which implies that MATH provided MATH . For MATH small enough and for MATH large enough REF can be verified without too much difficulty. With the help of a computer one can show that REF is satisfied for the remaining MATH . For REF is not satisfied, since the above bounds on MATH and MATH are too weak. Sufficiently good bounds can be obtained by dividing the interval MATH in two pieces and by integrating lower bounds on each of the subintervals. We omit the details. Finally, uniqueness of MATH can be proved by integrating the difference of two solutions from MATH to infinity, which, using the positivity of MATH leads to a contradiction with the fact that both of the solutions converge to zero at infinity. |
math-ph/9904020 | It is a well-known consequence of the NAME Vanishing Theorem (see for example, CITE) that we can find MATH such that if MATH and MATH with MATH for MATH, then there is a section MATH with MATH and MATH for MATH. We write MATH. Suppose on the contrary that MATH, and chose a nonzero vector MATH such that MATH. Then recalling REF , we have MATH where MATH. Since the MATH span MATH, it follows that for all MATH, we have MATH. But this contradicts the fact that, choosing MATH with MATH, we can find a section MATH with MATH and MATH for MATH. |
math-ph/9904020 | The MATH-point zero correlation MATH is given by REF with MATH. By the NAME formula CITE, the expectation MATH in REF is a homogeneous polynomial (over MATH) of degree MATH in the coefficients of MATH. By REF , the coefficients of MATH are homogeneous polynomials of degree MATH in the coefficients of MATH. The conclusion then follows from REF - REF . |
math-ph/9904020 | We consider the first NAME projector on the reduced NAME group MATH where MATH . (See the remark at the end of REF.) Its kernel can be written in the form MATH where the MATH form a complete orthonormal basis for MATH. (For example, MATH can be taken to be the set of monomials MATH. In fact, MATH is just a ``weighted NAME kernel" on MATH.) We now mimic the proof of REF , except this time we have an infinite sum over the index MATH; this sum converges uniformly on bounded sets in MATH since the sup norm over a bounded set is dominated by the NAME MATH norm (by the same argument as in the case of the ordinary NAME kernel on a bounded domain). We then obtain a nonzero vector MATH such that MATH for all MATH. But then MATH for all polynomials MATH on MATH, a contradiction. |
math-ph/9904020 | By taking the scaling limit of REF , we obtain MATH . Indeed, since the coefficients of MATH are either of degree REF in the coefficients of MATH or of degree REF in the coefficients of MATH, we see by the proof of REF , using REF - REF , that the leading term of the asymptotic expansion of MATH is MATH times the right side of REF . The bound on the error term follows from REF . Substituting into REF the values of MATH and its horizontal derivatives obtained from REF (with MATH) and REF and canceling out common factors of MATH and MATH, we obtain MATH where MATH is a universal polynomial (homogeneous of degree MATH in each of the variables MATH and with integer coefficients). |
math-ph/9904020 | We use REF , which comes from REF - REF as in the proof of REF . To determine the matrices MATH, we let MATH (instead of the right-invariant vector fields we used above). Recalling REF , we have: MATH . By REF , MATH . Recalling REF , we have MATH . We now apply REF ; note that the NAME formula involves terms that are products of diagonal elements of MATH, and products that contain at least two off-diagonal elements of MATH. The former terms are of the form MATH, and the latter are MATH. Similarly, MATH. The desired estimate then follows from REF . |
math-ph/9904026 | If MATH satisfies the pseudoholomorphicity condition then: MATH . That MATH follows by complex conjugation. |
math-ph/9904026 | We perform all computations in a real anholonomic NAME basis. Consider the compatibility equation between the almost complex structure J, symplectic form MATH and metric MATH, viz. MATH. Taking the covariant derivative of both sides and by use of the compatibility equation we find that: MATH . Computing the covariant derivative in local NAME basis on M: MATH . On comparison of REF one obtains REF . |
math-ph/9904026 | The first three equations are obtained from those of REF by direct substitution from REF . The final equation follows by substitution and by identity REF from REF. |
math-ph/9904028 | Since MATH is diffeomorphic to MATH, we have the NAME cohomology group MATH . The form MATH belongs to its second item which is zero. |
math-ph/9904028 | The proof is based on the relative NAME lemma CITE. |
math-ph/9904028 | Every connection MATH on MATH gives rise to the connection MATH on MATH which is a Hamiltonian connection for the frame Hamiltonian form MATH . Let us consider the decomposition MATH, where MATH is a connection on MATH. The assertion follows from the relation REF . |
math-ph/9904028 | Given locally Hamiltonian forms MATH and MATH, their difference MATH is a REF-form on MATH such that REF-form MATH is closed. By virtue of REF , the form MATH is exact and MATH locally. Put MATH where MATH is a connection on MATH. Then MATH modulo closed forms takes the local form MATH, and coincides with the pull-back of the NAME form MATH on MATH by the local section MATH of the fibre bundle REF . |
math-ph/9904028 | Given a Hamiltonian form MATH, its exterior differential MATH is a presymplectic form of constant rank MATH since the form MATH is nowhere vanishing. It is also seen that MATH. It follows that the kernel of MATH is a REF-dimensional distribution. Then the desired Hamiltonian connection MATH is a unique vector field MATH on MATH such that MATH, MATH. |
math-ph/9904028 | The constrained NAME equations can be written as MATH . They differ from the NAME REF for MATH restricted to MATH which read MATH where MATH is a section of MATH and MATH is an arbitrary vertical vector field on MATH. A solution MATH of REF satisfies obviously the weaker REF . |
math-ph/9904028 | Let MATH be the universal enveloping algebra of the NAME algebra of the symmetry currents MATH, MATH, REF . Then each non-zero element MATH of its center of order MATH can be written as a polynomial in MATH, and defines the desired Hamiltonian form MATH. |
math-ph/9904028 | The relation REF takes the coordinate form MATH . Substituting REF , we obtain the relation REF . |
math-ph/9904028 | Since MATH exists, the structure group MATH of the vector bundle MATH is reducible to the subgroup MATH of general linear transformations of MATH which keep its MATH-dimensional subspace, and to its subgroup MATH. |
math-ph/9904028 | The map REF is a solution of the algebraic equations MATH . By virtue of REF , there exist the bundle slitting MATH and a (non-holonomic) atlas of this bundle such that transition functions of MATH and MATH are independent. Since MATH is a non-degenerate fibre metric in MATH, there exists an atlas of MATH such that MATH is brought into a diagonal matrix with non-vanishing components MATH. Due to the splitting REF , we have the corresponding bundle splitting MATH . Then the desired map MATH is represented by a direct sum MATH of an arbitrary section MATH of the bundle MATH and the section MATH of the bundle MATH, which has non-vanishing components MATH with respect to the above mentioned atlas of MATH. Moreover, MATH satisfies the particular relations MATH . |
math-ph/9904028 | By the very definitions of MATH and MATH, the Hamiltonian map REF satisfies REF . Then MATH is weakly associated with REF in accordance with REF . Let us write the corresponding NAME REF for a section MATH of the NAME bundle MATH. They are MATH . Due to the surjections MATH and REF , the NAME REF break in two parts MATH . Let MATH be an arbitrary section of MATH, for example, a solution of the NAME equations. There exists a connection REF such that the relation REF holds, namely, MATH where MATH is a connection on MATH which has MATH as an integral section. It is easily seen that, in this case, the Hamiltonian map REF satisfies the relation REF for MATH. Hence, the Hamiltonian forms REF constitute a complete set. |
math-ph/9904028 | Due to the splitting REF , we have the corresponding splitting of the vertical tangent bundle MATH of the bundle MATH. In particular, any vertical vector field MATH on MATH admits the decomposition MATH such that MATH is a vertical vector field on the Lagrangian constraint space MATH. Let us consider the equations MATH where MATH is a section of MATH and MATH is an arbitrary vertical vector field on MATH. They are equivalent to the pair of REF . REF are obviously the NAME REF for MATH. Bearing in mind the relations REF , one can easily show that REF coincide with the NAME REF . The proof is completed by observing that, restricted to the Lagrangian constraint space MATH, REF are exactly the constrained NAME REF . |
math-ph/9904028 | Let MATH be the holonomic bases for MATH with respect to some bundle atlas MATH of MATH with transition functions MATH, that is, MATH. Then BRST functions read MATH where MATH are local functions on MATH, and we omit the symbol of an exterior product of elements MATH. The coordinate transformation law of BRST functions REF is obvious. Due to the canonical splitting MATH, the vertical tangent bundle MATH can be provided with the fibre bases MATH dual of MATH. These are fibre bases for MATH. Then any derivation MATH of MATH on a trivialization domain MATH of MATH reads MATH where MATH are local BRST functions and MATH acts on MATH by the rule MATH . This rule implies the corresponding coordinate transformation law MATH of derivations REF . Let us consider the vector bundle MATH which is locally isomorphic to the vector bundle MATH and has the transition functions MATH of the bundle coordinates MATH, MATH. These transition functions fulfill the cocycle relations. It is readily observed that, for any trivialization domain MATH, the MATH-module MATH with the transition functions REF is isomorphic to the MATH-module of local sections of MATH. One can show that, if MATH are open sets, there is the restriction morphism MATH. It follows that, restricted to an open subset MATH, every derivation MATH of MATH coincides with some local section MATH of MATH, whose collection MATH defines uniquely a global section of MATH, called a BRST vector field on MATH. BRST vector fields constitute a NAME superalgebra with respect to the bracket MATH . |
math-ph/9904028 | The proof is based on inspection of the transformation laws of the holonomic coordinates MATH on MATH and MATH on MATH. |
math-ph/9904036 | From the mass-gap assumption we obtain MATH for all MATH, where MATH. We claim that the vectors MATH exist (in the sense of the functional calculus). Indeed, denoting the spectral measure of MATH by MATH, REF implies MATH . Thus, by the functional calculus, MATH where we used that MATH commutes with MATH. |
math-ph/9904036 | The proof proceeds in three simple steps. For simplicity of notation, we will give the proof only for the case MATH, the proof for MATH is obtained by analogous arguments. In view of translation covariance, it suffices also to consider only the case MATH. CASE: We will first show that for all MATH . To this end, let MATH . One can see from REF that MATH. Moreover, denoting by MATH the projection orthogonal to the vacuum, we deduce, upon using REF (implying MATH) and REF together with the fact that MATH commutes with MATH, MATH, that MATH . Then relation REF follows from weak asymptotic lightlike clustering, REF . Furthermore, by REF it follows that there is a vector MATH so that MATH . MATH . Thus one obtains MATH and the last expression tends to REF in the limit MATH in view of weak asymptotic lightlike clustering, REF . This establishes relation REF . CASE: Relation REF shows, for any MATH, MATH and hence, to prove the theorem, it suffices to demonstrate MATH . To show this, we fix any MATH and use the abbreviation MATH . Now we define two maps with values in MATH, MATH and the two triangle-shaped regions MATH . The region MATH is bounded by the two lightlike line segments MATH and MATH, and by the spacelike line segment MATH. Similarly, MATH is bounded by the line segments MATH, MATH, and MATH. (Compare REF .) REF . Sketch of the regions and bounding line segments described in the text. Now we use MATH and thus, applying NAME law to the region MATH, we convert the integral of MATH paired with the outer normal along MATH into a sum of two integrals of MATH paired with the inner normals along MATH and MATH. Doing the same with respect to the region MATH (with the roles of inner and outer normals interchanged) yields, with the above parametrizations of the various line segments inserted, MATH . In view of REF , we deduce that the first integral on the right hand side of REF equals MATH as soon as MATH and MATH are large enough. This implies that REF , and hence the statement of the theorem, is proved once it is shown that the two remaining integrals on the right hand side of REF vanish in the limit MATH. CASE: The remaining step in the proof is therefore to show MATH . We will demonstrate this only for the MATH case, the reasoning for the MATH case is similar. It holds that MATH for MATH with some sufficiently large MATH. By REF , MATH for MATH, implying that MATH that is, the integral extends for all MATH only over a fixed interval of finite length. Now choose some wedge region MATH in the causal complement of MATH, and let MATH be arbitrary. According to REF , one can find some MATH so that MATH uniformly in MATH. Then MATH for all MATH, and MATH . The absolute value of the first integral on the right hand side of the last equation can be estimated by MATH. Owing to REF , the other integral on the right hand side of the last equation converges to REF for MATH (note that the integrands are bounded uniformly in MATH). Therefore we can find for the given MATH some MATH so that MATH for all MATH. By REF , this establishes the required relation REF , and thus the proof is complete. |
math-ph/9904036 | We consider only the case MATH and MATH, the general case is proved analogously. Then we observe that MATH holds for all MATH as can be seen from REF together with the fact that MATH, MATH. REF defines a quadratic form MATH on MATH which is by assumption positive, that is, MATH, MATH. It follows that there is an essentially selfadjoint, positive operator MATH with domain MATH so that MATH . Using MATH this implies MATH and hence, by REF , MATH for all MATH. The set of vectors MATH, MATH is dense in MATH, therefore, using also covariance REF , one arrives at MATH . Thus MATH for all MATH and MATH, and in view of REF , this entails MATH . Since we have imposed the mass gap REF , we may apply REF to conclude that this is only possible if MATH is parallel to the vacuum vector MATH. As the set of vectors MATH is dense in MATH, this implies MATH, and by the vanishing of the vacuum-expectation value of MATH, finally MATH for all MATH. |
math-ph/9904038 | The structure of MATH allows to identify the NAME algebras over the different fields. Indeed, transitions MATH may be represented as transitions from the real coordinates in MATH to complex coordinates of the form MATH, where MATH is an additional basis element MATH (volume element). Since MATH is odd, then the volume element MATH in accordance with REF belongs to MATH. Therefore, we can to identify it with imaginary unit MATH if MATH and with a double unit MATH if MATH. The general element of the algebra MATH has a form MATH, where MATH is a general element of the algebras MATH. |
math-ph/9904038 | Indeed, in accordance with REF MATH. Further, from REF we have MATH, and a square of the element MATH is equal to MATH, therefore MATH. Thus, MATH. For the group MATH a square of MATH is equal to MATH, therefore MATH, MATH is a double unit. As expected, MATH. The isomorphisms for the groups MATH and MATH are analogously proved. |
math-ph/9904038 | We start the proof with a more general case of MATH. According to REF the volume element MATH belongs to a center of MATH (MATH), therefore, MATH commutes with all basis elements of this algebra and MATH. Further, recalling that a vector complex space MATH is associated with the algebra MATH, we see that basis vectors MATH generate a subspace MATH. Thus, the algebra MATH in MATH is a subalgebra of MATH and consists of the elements which does not contain the element MATH. A decomposition of the each element MATH may be written in the form MATH where MATH is a set of all elements which contain MATH, and MATH is a set of all elements which does not contain MATH, therefore MATH. If multiply MATH by MATH, then the elements MATH are mutually annihilate, therefore MATH. Denoting MATH and taking into account MATH we obtain MATH where MATH. Consider now an homomorphism MATH, an action of which is defined by the following law MATH . Obviously, at this point the all operations (addition, multiplication, and multiplication by the number) are preserved. Indeed, let MATH then in virtue of MATH and commutativity of MATH with all elements, we have for multiplication MATH that is, the image of product equals to the product of factor images in the same order. In the particular case of MATH we have MATH and MATH, therefore MATH . Thus, a kernel of the homomorphism MATH consists of all elements of the form MATH, which under action of MATH are mapped into zero. It is clear that MATH is a subalgebra of MATH. Moreover, the kernel of MATH is a bilateral ideal of MATH. Therefore, the algebra MATH, which we obtain in the result of the mapping MATH, is a quotient algebra MATH . Further, since the algebra MATH REF is isomorphic to the full matrix algebra MATH, then in virtue of MATH we obtain an homomorphic mapping of MATH onto the matrix algebra MATH. The homomorphism MATH is analogously proved. In this REF quotient algebra has a form MATH or MATH where MATH, since in accordance with REF at MATH we have MATH, therefore MATH. |
math-ph/9904038 | As known, the transformations MATH at the conditions MATH form an abelian group with the following multiplication table Analogously, for the automorphism group MATH in virtue of the commutativity MATH and the conditions MATH a following multiplication table takes place The identity of the multiplication tables proves the isomorphism of the groups MATH and MATH. |
math-ph/9904038 | First of all, since MATH if MATH and MATH if MATH, then in the case of MATH for the matrix of the automorphism MATH we have MATH . Over the field MATH we can always to suppose MATH. Further, let us find now the matrix MATH of the antiautomorphism MATH at any MATH, and elucidate the conditions at which the matrix MATH commutes with MATH, and also define a square of the matrix MATH. Follows to CITE let introduce along with the algebra MATH an auxiliary algebra MATH with basis elements MATH . In so doing, linear operators MATH acting in the space MATH associated with the algebra MATH, are defined by a following rule MATH where MATH is a general element of the auxiliary algebra MATH, MATH and MATH are correspondingly odd and even parts of MATH, MATH are units of the auxiliary algebra, MATH. Analogously, in the case of matrix representations of MATH we have MATH where MATH are arbitrary complex numbers. It is easy to verify that transposition of the matrices of so defined operators gives MATH . Further, for the antiautomorphism MATH, since in this case MATH, it is sufficient to select the matrix MATH so that MATH or taking into account REF MATH . Therefore, if MATH is odd, then the matrix MATH has a form MATH since in this REF product MATH commutes with all elements MATH and anticommutes with all elements MATH. Analogously, if MATH is even, then MATH . As required according to REF in this REF product REF commutes with MATH and anticommutes with MATH. Let us consider now the conditions at which the matrix MATH commutes or anticommutes with MATH. Let MATH, where MATH is odd, since MATH, then MATH where MATH are the functions of the form REF . It is easy to see that in this case the elements MATH and MATH are always anticommute. Indeed, a comparison MATH is equivalent to MATH, and since MATH is odd, then we have always MATH. At MATH is even and MATH it is easy to see that the matrices MATH and MATH are always commute MATH. Further, let MATH be a quantity of the elements MATH of the product REF whose squares equal to MATH, and let MATH be a quantity of the elements MATH of the product REF whose squares equal to MATH. Then a square of the matrix REF at MATH is odd equals to MATH if MATH and respectively MATH if MATH. Analogously, a square of the matrix REF at MATH is even equals to MATH if MATH and respectively MATH if MATH. It is obvious that over the field MATH we can to suppose MATH. Let us find now the matrix MATH of the antiautomorphism MATH: MATH. Since in this case MATH, then it is sufficient to select the matrix MATH so that MATH or taking into account REF MATH where MATH. In comparison with REF it is easy to see that in REF the matrices MATH and MATH are changed by the roles. Therefore, if MATH is odd, then MATH and if MATH is even, then MATH . Permutation conditions of the matrices MATH and MATH are analogous to the permutation conditions of MATH with MATH, that is, the matrix MATH of the form REF always anticommutes with MATH (MATH), and the matrix MATH of the form REF always commutes with MATH (MATH). Correspondingly, a square of the matrix REF equals to MATH if MATH and MATH if MATH. Analogously, a square of the matrix REF equals to MATH if MATH and MATH if MATH. Obviously, over the field MATH we can suppose MATH. Finally, let us find permutation conditions of the matrices MATH and MATH. First of all, at MATH is odd MATH, MATH, alternatively, at MATH is even MATH, MATH. Therefore, MATH that is, the matrices MATH and MATH commute at MATH and anticommute at MATH. Now we have all the necessary conditions for the definition and classification of isomorphisms between finite groups and automorphism groups of NAME algebras. Let MATH and let MATH, therefore, a group MATH is Abelian. A condition MATH is equivalent to MATH, which, clearly, are compatible. In accordance with REF - REF - REF at MATH and MATH. Therefore, MATH for the signature MATH if MATH. The isomorphism MATH for the signatures MATH REF and MATH REF is analogously proved. It is easy to see that for MATH there are only four isomorphisms considered previously. Further, for MATH all the elements of the group MATH anticommute and in this case MATH. The signature MATH is equivalent to REF, here MATH and we have an isomorphism MATH, where MATH, MATH are the quaternion units. It is easy to verify that for the signatures MATH at MATH and MATH at MATH we have an isomorphism MATH, where MATH, MATH are the units of the algebra MATH or MATH. The eight automorphism groups considered previously, each of which is isomorphic to one from the four finite groups MATH, are the only possible over the field MATH. In contrast with this, over the field MATH we can suppose MATH. At MATH we have only one signature MATH and an isomorphism MATH if MATH, since over the field MATH the signatures MATH and MATH are isomorphic to MATH. Correspondingly, at MATH we have an isomorphism MATH for the signature MATH if MATH. It should be noted that the signatures MATH and MATH are non-isomorphic, since there exists no a group MATH with the signature MATH in which all the elements anticommute, and also there exists no a group MATH with MATH in which the all elements commute. Thus, over the field MATH we have only two non-isomorphic automorphism groups: MATH. |
math-ph/9904038 | Indeed, over the field MATH in accordance with REF from all the fundamental automorphisms at the homomorphic mappings MATH and MATH only the antiautomorphism MATH is transferred into quotient algebras MATH and MATH. Further, according to REF the antiautomorphism MATH corresponds to time reversal MATH. Therefore, groups of the discrete transformations of the spaces MATH and MATH associated with the quotient algebras MATH and MATH are defined by a two - element group MATH, where MATH is an automorphism group of the quotient algebras MATH, MATH is a matrix of the antiautomorphism MATH. Thus, at MATH there are the homomorphic mappings MATH and MATH, where MATH are quotient groups, MATH. At this point, a double covering of MATH is isomorphic to MATH. Analogously, over the field MATH at MATH we have a quotient group MATH. Further, according to REF in the result of the homomorphic mapping MATH the antiautomorphisms MATH are transferred into a quotient algebra MATH at MATH. Therefore, a set of the discrete transformations of the space MATH associated with the quotient algebra MATH is defined by a three-element set MATH, where MATH is a set of the automorphisms of MATH, MATH and MATH are correspondingly the matrices of the antiautomorphisms MATH and MATH. It is easy to see that the set MATH does not form a finite group. Thus, at MATH there is an homomorphism MATH, where MATH is a quotient group, MATH. |
math/9904001 | We observe that the composite map MATH verifies MATH. Since MATH is a linear isomorphism REF, we have MATH. Therefore MATH. |
math/9904001 | REF asserts that MATH is not empty and every irreducible component has dimension at least MATH. Suppose that there is an irreducible component MATH of dimension MATH. Then its inverse image MATH has dimension MATH, hence, since MATH is irreducible, MATH and MATH. The last equality can not happen, since otherwise, using translation by an element of the form MATH, we would have MATH. |
math/9904001 | We are going to define MATH as the residual divisor of the restricted divisor MATH, for a given point MATH and then show that it does not depend on the choice of MATH. We first observe that we have an equality of sets MATH which can be seen as follows: for MATH such that MATH the assumption MATH and the formula MATH imply that MATH . If MATH, then MATH. Again a calculation involving NAME tangent spaces shows that MATH is reduced generically on MATH. Hence we can define MATH by MATH. Now we substitute this expression into REF, which we restrict to MATH . Now we fix MATH and we take the limit when MATH. Since MATH, we see that MATH. So by REF we get the line bundle equality claimed in the lemma and we see that the scheme-structure on MATH does not depend on the point MATH. To prove REF, we compute using REF MATH . Now we restrict to MATH and use the commutativity of REF and the divisorial equality MATH to obtain MATH . Since MATH we can divide this equality by MATH and we are done. |
math/9904001 | The equality follows from the commutativity of the right-hand square of the diagram MATH . The commutativity of the right-hand square follows from that of the outside square because MATH generates MATH. In other words we need to check the assertion of the lemma only for hyperplanes of the form MATH for MATH. This follows immediately from REF. |
math/9904001 | By the above Lemma we have MATH . If MATH is irreducible, then the support of one of the divisors MATH or MATH, say MATH, is contained in the support of MATH. This is impossible because MATH is the inverse image of a divisor in MATH and MATH is the inverse image of the codimension MATH support of MATH. |
math/9904001 | Suppose that MATH is reducible. Then a local computation shows that the hyperplane MATH is everywhere tangent to the branch locus of MATH. It is immediately seen that the branch locus MATH of MATH is the dual hypersurface of the canonical curve. The components of the singular locus MATH of MATH are of two different types which can be described as follows CASE: whose points are hyperplanes tangent to MATH in more than one point. CASE: whose points are hyperplanes osculating to MATH. To prove that MATH, we need to prove that there is a point on MATH which is smooth on MATH because the dual variety of MATH is the closure of the set of hyperplanes tangent to MATH at a smooth point and this is equal to MATH. In other words we need to show that MATH is not contained in MATH. Since MATH has pure codimension MATH, it suffices to show that no codimension MATH component of MATH is contained in a hyperplane. Suppose a codimension MATH component MATH of type MATH (MATH or MATH) is contained in a hyperplane MATH in MATH and let MATH be the corresponding point. Then the set of hyperplanes in MATH through MATH and doubly tangent (respectively, osculating) to MATH has dimension MATH. We have For any MATH the restriction MATH of the projection from MATH to MATH is birational onto its image. If MATH, then MATH is either birational onto its image or of degree two onto an elliptic curve. First note that the degree of the image MATH of MATH by the projection is at least MATH because MATH is a non-degenerate curve in a projective space of dimension MATH. If MATH, then the degree of MATH is equal to MATH. The degree MATH of the restriction of MATH to MATH verifies MATH. Therefore MATH. Or MATH which implies MATH. However, MATH cannot be equal to MATH because MATH is odd. If MATH, then the same argument gives again MATH because MATH. Hence, if MATH is not generically injective, then MATH and MATH. Therefore MATH is either smooth rational or an elliptic curve. Since MATH is not hyperelliptic, we have that MATH is an elliptic curve. First suppose that MATH is birational. If MATH, projecting from MATH, we see that the set of hyperplanes in MATH doubly tangent to MATH has dimension MATH=dimension of the dual variety of MATH which is impossible. If MATH, then the set of hyperplanes in MATH osculating to MATH has dimension MATH which is also impossible. If MATH is of degree MATH, then indeed every hyperplane tangent to MATH pulls back to a hyperplane twice tangent (or osculating if the point of tangency is a branch point of MATH) to MATH and we have a codimension MATH family of type MATH contained in the hyperplane corresponding MATH to MATH. Then MATH could be reducible. |
math/9904001 | For a bi-elliptic curve MATH, the NAME locus MATH has two irreducible components, which are fixed by the reflection in MATH REF . For a smooth plane quintic this NAME locus is irreducible, ruling out REF. For a trigonal curve this NAME locus has two irreducible components, which are interchanged by reflection in MATH, ruling out REF. |
math/9904001 | CASE: Given a bundle MATH and a line bundle MATH which is anti-invariant under MATH, that is, MATH, we have a natural non-degenerate quadratic form with values in the canonical bundle MATH where MATH is a local section of MATH. Note that we have canonical isomorphisms MATH . Therefore we are in a position to apply the NAME lemma CITE to the family of bundles (here MATH is fixed, with MATH) MATH which states that the parity of MATH is constant when MATH varies in MATH. From now on, we suppose MATH, with MATH, then MATH . For the first equality we use the fact that MATH and, by NAME and NAME duality, MATH. First suppose that MATH is general. Then the divisor MATH does not contain the NAME variety MATH (for example, because, for general MATH, MATH), so the restriction of the divisor MATH to MATH is a divisor in the linear system MATH. Moreover, for MATH because MATH. Hence any point MATH is a singular point of MATH, which implies that MATH is an everywhere non-reduced divisor. We have Suppose that MATH is a divisor in MATH. Then there is a divisor MATH such that MATH. A local equation of MATH is given by the pfaffian of a skew-symmetric perfect complex of length one MATH representing the perfect complex MATH where MATH is the NAME line bundle over the product MATH and MATH are the projections on the two factors. The construction of the complex MATH is given in the proof of REF. If MATH is of the form MATH for some MATH, we have MATH. It follows from this equality that MATH is reduced for general MATH. So far we have defined a rational map MATH. It will follow from REF that MATH can be defined away form MATH. CASE: First we consider the composite (rational) map MATH . A straight-forward computation shows that for all MATH such that MATH the divisor MATH equals the translated divisor MATH restricted to MATH. Hence, by CITE, the map MATH is given by the full linear system MATH of invariant elements of MATH. By NAME duality REF MATH and we obtain the commutative diagram in the proposition. Geometrically, MATH is obtained by restricting the projection with center the MATH-eigenspace MATH to the embedded moduli space MATH. Since by CITE MATH we see that MATH is well-defined for MATH even if MATH. |
math/9904001 | It will be enough to show that the canonical divisors MATH and MATH are equal for a general element MATH. In both cases the divisor coincide with the divisor MATH, where MATH is the unique effective divisor in the linear system MATH. The computations are straight-forward and left to the reader. |
math/9904001 | First we need to show that for a general semi-stable bundle MATH with MATH the divisor MATH is smooth at a general point MATH. For this decompose a general NAME divisor into two effective divisors of degree MATH, that is, MATH. Put MATH. Then MATH. If MATH, then MATH, MATH and MATH. At a general point MATH, we see immediately that the tangent space to MATH does not contain the tangent space to MATH, that is, MATH is smooth at MATH. Next we compute the tangent space to the divisor MATH at a smooth point MATH. The smoothness of MATH at MATH implies that MATH. We choose a basis MATH of the MATH-dimensional vector space MATH. Then by REF and the same reasoning as in the proof of REF , we see that the projectivized tangent space MATH to MATH at MATH, which is a hyperplane in MATH is the zero locus of the section in MATH, which is the image of MATH under the exterior product map MATH . Since MATH, we see that MATH. We will now describe the map MATH. Note that, since MATH, we have MATH for MATH general. Hence MATH is also a basis for MATH. Consider the inclusion MATH and decompose MATH, MATH with MATH and MATH. Then the element MATH is the image of MATH under the exterior product map MATH, that is, MATH. Since MATH, we have MATH. So for general MATH, MATH is the unique divisor of the pencil MATH containing MATH. Hence we can conclude that the section MATH considered as a tensor in MATH is MATH. |
math/9904001 | Consider the invertible sheaf MATH and the corresponding embedding MATH . The curve MATH is the curve MATH translated by MATH. A straight-forward computation shows that MATH and by a result of NAME (see CITE page REF) the induced linear map on global sections MATH is surjective. We observe that MATH and that the projectivized tangent line to the curve MATH at the point MATH (respectively, MATH) is the point MATH (respectively, MATH) in MATH (respectively, MATH). Let MATH (respectively, MATH) denote the embedded tangent line in MATH to the curve MATH at the point MATH (respectively, MATH), so that MATH (respectively, MATH) passes through the point MATH (respectively, MATH) with tangent direction MATH (respectively, MATH). Then the lemma will follow if we show that these two tangent lines intersect in a point MATH, that is, MATH . This property follows from a dimension count: since MATH is non-hyperelliptic, we have MATH, so MATH. Since MATH, the tangent lines MATH and MATH are contained in a projective MATH-plane, hence intersect. To get the equivalence stated in the lemma, let MATH denote the hyperplane in MATH corresponding to the divisor MATH. Assume for example, that MATH. This means that MATH contains MATH. Since MATH, it follows from REF that MATH also contains MATH, so MATH. |
math/9904005 | It's clear that one may suppose that MATH is base point free. Taking the NAME of MATH, we can get MATH where MATH has connected fibers. A generic irreducible element of MATH is a smooth projective curve. If MATH is not composed of a pencil of curves, then it's sufficient to verify the birationality of MATH by virtue of NAME 's principle REF . If MATH is composed of a pencil of curves, then MATH is a fibration onto the smooth curve MATH. In this case, MATH is a general fiber of MATH. When MATH is a rational curve, MATH can distinguish different fibers of MATH since MATH. When MATH is irrational, suppose MATH and MATH are two general fibers of MATH. By assumption, we have MATH where MATH is an effective divisor on MATH and MATH is a nef and big MATH-divisor. We consider the system MATH. According to NAME vanishing theorem, we have the surjective map MATH . By REF , we have MATH and MATH. Thus MATH can distinguish MATH and MATH, so can MATH. Therefore it's also sufficient to verify the birationality of MATH. By REF , we can write MATH where MATH is an effective divisor and MATH is a nef and big MATH-divisor. We consider the system MATH. By vanishing theorem, we have the surjective map MATH where MATH is a divisor on MATH with MATH. Thus MATH is an embedding, so is MATH. |
math/9904005 | We may suppose that MATH is basepoint free. Denote by MATH a generic irreducible element of MATH. Then the vanishing theorem gives the exact sequence MATH where MATH is a divisor of positive degree. It is obvious that MATH since MATH is a curve of genus MATH. The proof is completed. |
math/9904005 | First we take a birational modification MATH, according to NAME, such that REF MATH is smooth; REF the movable part of MATH defines a morphism; REF the fractional part of MATH has supports with only normal crossings. Denote by MATH a generic irreducible element of the movable part of MATH. Then MATH is a smooth projective surface of general type by NAME 's theorem. By the vanishing theorem, we have the exact sequence MATH where MATH is a given integer and MATH has the property MATH according to the assumption. If MATH, then MATH by REF. If MATH, we still have the following exact sequence MATH where MATH is a generic irreducible element of the movable part of MATH and MATH is a divisor of positive degree on MATH. Since MATH is a curve of genus MATH, we have MATH . Thus we can see that MATH. The proof is completed. |
math/9904005 | This is obvious according to REF . |
math/9904005 | Taking the same modification MATH as in the proof of REF , we still denote by MATH the general member of the movable part of MATH. Note that both MATH and MATH have the same movable part. For a given integer MATH, we have MATH . It is sufficient to prove the birationality of the rational map defined by MATH . Because MATH is effective by the proof of REF , we have MATH . Thus we only need to prove the birationality of MATH . We have the following exact sequence by the vanishing theorem MATH which means MATH . Noting that MATH where MATH, we want to show that MATH is birational. Because MATH gives a generically finite map, we see from REF that MATH is effective. On the other hand, let MATH be a generic irreducible element of MATH, then MATH. So MATH and thus MATH. By REF , MATH gives a birational map. The proof is completed. |
math/9904005 | First we take the same modification MATH as in the proof of REF . We also suppose that MATH is the movable part of MATH. For a given integer MATH, we obviously have MATH . Thus it is sufficient to verify the birationality of the rational map defined by MATH . By REF , MATH is effective. Thus we only have to prove the birationality of the restriction MATH for the general MATH. The vanishing theorem gives the exact sequence MATH . This means MATH . Suppose MATH is the movable part of MATH. We have to study some property of MATH. Note that MATH is also the movable part of MATH . We have MATH . The vanishing theorem gives the exact sequence MATH where MATH. Denote by MATH the movable part of MATH and by MATH the movable part of MATH . By REF , we have MATH . Noting that MATH is a free pencil, we can suppose MATH is a generic irreducible element of MATH. Now the key step is to show that MATH. In fact, the vanishing theorem gives MATH where MATH is a divisor of positive degree. Because MATH is a curve of genus MATH, MATH gives a finite map. This shows MATH thus MATH. Therefore MATH and so MATH. Noting that MATH we get from REF that MATH is effective. Now REF implies the birationality of the rational map defined by MATH . Because MATH is birational. We have proved the theorem. |
math/9904005 | For all MATH, denote by MATH the movable part of MATH. By NAME 's method (CITE or see CITE), we have MATH. By the vanishing theorem, one has MATH . By REF , we see that MATH . Repeatedly performing this process, one has MATH for all integer MATH. This means that we can write MATH where MATH is an effective MATH-divisor only relating to MATH. Thus MATH . We can write MATH where MATH is the movable part and MATH the fixed one. According to NAMECITE, MATH is composed of a pencil of curves if and only if MATH and MATH. We prove this theorem step by step. CASE: MATH is effective for all MATH. Denote by MATH a positive integer. We have MATH . By the vanishing theorem, one has MATH . Now we consider a sub-system MATH . Because MATH where MATH for big MATH. Thus we have MATH where MATH is divisor on MATH with MATH. Therefore MATH since MATH. CASE: The birationality. Denote by MATH a positive integer. Considering the system MATH, we have MATH . By REF , it's sufficient to verify the birationality of MATH. It's obvious that MATH . Denote by MATH. We have MATH . So MATH is also nef and big for big MATH. Now we have MATH . By REF , one can easily see that the above system defines a birational map onto its image. The theorem follows. |
math/9904005 | Denote by MATH a generic irreducible element of the movable part of MATH. Then it's well-known that MATH is a smooth curve of genus REF. According to Claim in REF, we have MATH. For a positive integer MATH, we have MATH . Since MATH, taking MATH and applying REF , one has MATH. Taking MATH and applying REF once more, one has MATH. This means MATH where MATH is an effective MATH-divisor. Thus we have MATH . Now MATH. It is easy to see that, for MATH, MATH. REF implies that MATH is birational and so is MATH. |
math/9904005 | We still denote by MATH a generic irreducible element of the movable part of MATH. It's well-known that MATH defines a generically finite map and MATH is a smooth curve of genus REF. By a parallel argument as in the proof of REF , we have MATH for any positive integer MATH. This means MATH where MATH is an effective MATH-divisor depending on MATH. Thus we have MATH . Therefore MATH for all MATH. So MATH. We want to verify the birationality of MATH for certain MATH. Because MATH . Fix a big MATH, one can see that MATH for MATH. Thus, by REF , MATH is birational and so is MATH. |
math/9904005 | Denote by MATH a generic irreducible element of the movable part of MATH. Recall that MATH is the contraction onto the minimal model. MATH is the movable part of MATH. It's easy to see that MATH has two types: CASE: MATH, where MATH and MATH is smooth curve of genus REF. CASE: MATH , where MATH is a smooth curve of genus REF. In either cases, we always have MATH. By REF, MATH is basepoint free for MATH. Now NAME 's technique gives MATH and so MATH. By a parallel argument as in the proof of REF , we get MATH for all positive integer MATH. Thus MATH . So MATH. Now by the same argument as in the proof of REF , one can easily obtain the birationality of MATH. |
math/9904005 | We consider the natural map MATH where MATH is the image of MATH. Since MATH, MATH. CASE: MATH. In this case, MATH defines the bicanonical map of MATH. By CITE, MATH is base point free. Thus we see that MATH. We can write MATH, where MATH is an effective MATH-divisor. Denote by MATH a general member of MATH. Now we have MATH and MATH. By REF , MATH is birational and so is MATH. CASE: MATH. Because MATH defines a morphism, the movable part of MATH forms a complete linear pencil. The pencil is rational because MATH. Denote by MATH a generic irreducible element of the movable part of MATH. By REF below, MATH. Because MATH, we can write MATH, where MATH is an effective MATH-divisor. Now MATH and, by REF , MATH. According to REF , MATH is birational and so is MATH. CASE: MATH. In this case, MATH is composed of a pencil of surfaces. One can see that MATH, whence MATH. Applying NAME 's R-R formulaCITE, one has MATH. We can write MATH,where MATH and MATH is an effective divisor. We hope to prove the birationality of MATH. Because MATH, it's sufficient to prove the birationality of MATH. By virtue of NAME 's method, one can see that MATH. Therefore we are reduced to prove the birationality of MATH, where MATH is a general member of MATH. By the vanishing theorem, one has MATH where MATH. REF below is still true when MATH is of type REF , that is, MATH where MATH is an effective MATH-divisor. Thus MATH, where MATH. Because MATH for all MATH, we see that MATH whenever MATH is large. Thus MATH is birational and so is MATH. |
math/9904005 | By CITE, MATH is basepoint free for all MATH. Denote by MATH the movable part of MATH. Then MATH. It's sufficient to prove the birationality of MATH. By NAME 's method, MATH. We only need to verify the birationality of MATH. The vanishing theorem gives MATH . Applying REF , we get MATH. Repeatedly proceeding the above process while replacing "REF" by "REF, MATH", one can obtain MATH and so MATH where MATH is a positive integer. Thus we have MATH . It's easy to see that MATH. Now taking a very big MATH, one has MATH where MATH and MATH . Therefore we have proved that MATH is birational. So MATH is birational. |
math/9904005 | We keep the same notations as in the proof of REF . The proof is almost the same except that we have here MATH. Thus we can prove that MATH is birational by the same argument. This, in turn, proves the birationality of MATH. We conclude the claim. |
math/9904005 | We can suppose MATH is a free pencil. Otherwise, we can blow-up MATH at base points of MATH. Denote MATH. Then MATH. Suppose MATH. Then MATH. Because MATH, we can see that MATH. From MATH, we get MATH, that is, MATH. Thus MATH and MATH. This means MATH by virtue of CITE, which is impossible because MATH. So MATH. |
math/9904006 | For each pair MATH the MATH-vector space MATH is graded. Let MATH be the homogeneous component of degree MATH, and MATH . This is a connected algebraic variety. Since MATH is generated as MATH-algebra by the idempotents and the arrows, and the only relations in MATH are the monomial relations arising from incomposability of paths, it follows that any element MATH extends uniquely to a MATH-algebra automorphism MATH of MATH that fixes the idempotents. Conversely any automorphism MATH restricts to an element MATH of MATH. This bijection MATH is an isomorphism of varieties. Hence MATH is connected. |
math/9904006 | CASE: Since MATH is basic we have MATH. By NAME theory we have MATH. Any auto-equivalence of the category MATH extends to an auto-equivalence of MATH (using projective resolutions), and this induces an isomorphism of groups MATH. The class of auto-equivalences MATH is actually a group here. In fact MATH can be identified with the subgroup of MATH consisting of automorphisms that permute the set of idempotents MATH. Define a homomorphism of groups MATH by MATH if MATH. Thus we get a commutative diagram MATH . For an element MATH we have MATH, and hence MATH. According to REF , MATH is connected. Because MATH is a morphism of varieties we see that MATH is connected. But the index of MATH is finite, so we get MATH. In order to split MATH we choose any splitting of MATH and compose it with the homomorphism MATH. CASE: When MATH is a tree the group MATH is a torus: MATH. In fact MATH consists entirely of inner automorphisms that are conjugations by elements of the form MATH with MATH. Thus MATH. |
math/9904006 | Let MATH be a MATH-linear triangle auto-equivalence of MATH. By REF there exits a two-sided tilting complex MATH with MATH in MATH. Replacing MATH with MATH we may assume that MATH. Hence MATH, and MATH is an equivalence. NAME theory says that MATH for some invertible bimodule MATH. So replacing MATH by MATH we can assume that there is an isomorphism MATH. Now for every object MATH we can choose an isomorphism MATH with MATH (compare REF ). Define MATH to be the composition MATH . According to the proof of REF , for any morphism MATH one has MATH, so MATH is an isomorphism of functors. |
math/9904006 | The group homomorphism MATH is injective, say by REF , and it is surjective by the theorem. |
math/9904006 | CASE: This is implicit in REF and I. REF. In particular REF shows that for any indecomposable object MATH the ring MATH is local. CASE: See REF . According to REF , for each MATH there exists such an NAME triangle. By REF these are all the NAME triangles, up to isomorphism. CASE: Since source and sink morphism depend only on the structure of MATH-linear additive category on MATH (compare REF) we may use REF . |
math/9904006 | This is essentially REF . |
math/9904006 | Consider a sink morphism in MATH ending in MATH, MATH. By REF , it is of the form MATH with MATH (compare Notation REF). From the definition of a sink morphism we see that this is also a sink morphism in the category MATH. According to REF (dual form), both MATH-modules MATH and MATH have the morphisms MATH as basis. And there are no irreducible morphisms MATH for indecomposable objects MATH not isomorphic to one of the MATH, in either category. Thus the lemma is proved for MATH. Let MATH and MATH. If MATH then necessarily MATH. This is clear for MATH, since MATH is a projective module, and an easy calculation shows that for MATH, MATH . In general we can translate by MATH. Now take an arbitrary segment MATH. The paragraph above implies that for MATH and MATH, MATH. Hence MATH. |
math/9904006 | Let MATH be the full subquiver with vertex set MATH. Given a vertex MATH in MATH, denote by MATH the number of its predecessors, that is, the number of vertices MATH such that there is a path MATH in MATH. For any MATH let MATH be the full subquiver with vertex set MATH. MATH is a translation quiver with polarization, and MATH is a full subcategory. By recursion on MATH, we will define a functor MATH satisfying REF and CASE: Let MATH be a pair of vertices and let MATH be the arrows MATH. Then MATH is a basis of MATH. Take MATH. It suffices to define MATH for an arrow MATH in MATH. These arrows fall into three cases, according to their end vertex MATH: CASE: MATH, in which case any arrow MATH ending in MATH is in MATH, and MATH is already defined. CASE: MATH and MATH. Any arrow MATH ending in MATH is in MATH, so we define MATH. By REF holds. CASE: MATH and MATH. In this case MATH is a non-projective vertex in MATH, and we consider the mesh ending at MATH. The vertices with arrows to MATH are MATH, where MATH; MATH and MATH (compare Notation REF). Since MATH the arrows MATH are all in the quiver MATH, and hence MATH are defined. According to REF it follows that there exists an NAME triangle MATH in MATH. Define MATH . Note that the mesh relation MATH in MATH is sent by MATH to MATH, so we indeed have a functor MATH. Also, by REF , for any MATH the set MATH is a basis of MATH. Thus we obtain a functor MATH. By symmetry we construct a functor MATH for negative vertices (that is, MATH), extending MATH. Putting the two together we obtain a functor MATH satisfying REF . Let us prove MATH is fully faithful. For any MATH there is a full subquiver MATH, on the vertex set MATH. Correspondingly there are full subcategories MATH and MATH. It suffices to prove that MATH is fully faithful. By REF the quiver of MATH is MATH, which is pre-projective. So we can use the last two paragraphs in the proof of REF almost verbatim. Finally we shall prove that MATH is unique up to isomorphism. Suppose MATH is another MATH-linear functor satisfying REF. We will show there is an isomorphism MATH that is the identity on MATH. By recursion on MATH we shall exhibit an isomorphism MATH. It suffices to consider REF above, so let MATH be such a vertex. Then, because MATH, we have MATH . Applying MATH to the triangle REF we obtain a morphism MATH such that MATH. Because MATH is faithful we see that MATH, and since MATH it follows that MATH is invertible. Set MATH. This yields the desired isomorphism MATH. By symmetry the isomorphism MATH extends to MATH. |
math/9904006 | The fact that MATH is in the center of MATH is trivial. As for MATH, this follows immediately from REF (or by REF , since MATH is the NAME functor of MATH). |
math/9904006 | Given an auto-equivalence MATH of MATH, the formula MATH iff MATH defines a permutation MATH of MATH that preserves arrow-multiplicities. Hence it restricts to a permutation of MATH. By REF , MATH commutes with MATH and MATH. |
math/9904006 | Let MATH. By REF we know that MATH. Hence by REF , MATH iff MATH acts trivially on the set MATH. In particular we see that MATH. Now use REF . |
math/9904006 | According to REF, the group MATH is abelian in all cases except MATH. But a direct calculation in this case (compare REF ) gives MATH. |
math/9904006 | Since MATH we get a permutation MATH. Let's prove that MATH commutes with MATH in MATH. Consider a vertex MATH. In the Notation REF, there are vertices MATH and irreducible morphisms MATH and MATH that form bases of MATH and MATH respectively. Since we have MATH this must be a multiple of a mesh relation. Hence MATH. Finally to define MATH we have to split MATH consistently with MATH. It suffices to order the set of arrows MATH for every pair of vertices MATH consistently with MATH. We only have to worry about this when MATH has infinite representation type. For any MATH choose some ordering of the set MATH. Using MATH and MATH this ordering can be transported to all of MATH. By the isomorphism MATH of REF the ordering is copied to MATH. |
math/9904006 | Choose an equivalence MATH as in REF . If MATH has infinite representation type then the isomorphism MATH we have chosen (as in REF ) tells us how to extend MATH to an equivalence MATH that commutes with MATH (compare REF ). Let MATH be a triangle auto-equivalence of MATH. Then MATH induces a permutation MATH of the set MATH that commutes with MATH. For every vertex MATH choose an isomorphism MATH in MATH. Given an arrow MATH in MATH, define the morphism MATH by the condition that the diagram MATH commutes. Then MATH. If MATH is another choice of isomorphisms MATH then MATH is an isomorphism of functors MATH, so the map MATH is independent of these choices. It is easy to check that MATH respects composition of equivalences. |
math/9904006 | The proof has three parts. CASE: We show that the homomorphism MATH of REF is injective. Let MATH be a two-sided tilting complex such that MATH. Then the permutation MATH fixes the vertices of MATH. Using the fact that MATH we see that MATH in MATH. Replacing MATH with MATH we may assume MATH is a single bimodule. According to REF , we see that MATH is actually an invertible bimodule. Since MATH is full we get MATH. Hence by NAME theory we have MATH as bimodules. CASE: Assume MATH has finite representation type, so that MATH. We prove that MATH is surjective. Consider a MATH-linear auto-equivalence MATH of MATH. Let MATH as in the proof of REF . According to REF , MATH commutes with MATH. Define MATH . Then for any MATH and integers MATH the equivalence MATH of REF produces isomorphisms MATH . Therefore MATH . Also for any MATH there is some integer MATH and MATH such that MATH . Since any object MATH is a direct sum of indecomposables MATH, this implies that MATH if MATH. By REF and the proof of `` REF " of REF there exists a two-sided tilting complex MATH with MATH in MATH (compare REF). Replacing MATH with MATH, where MATH, we can assume that MATH is trivial. Now that MATH is trivial, MATH restricts to an auto-equivalence of MATH, and by REF we have MATH. Then REF tells us MATH. CASE: Assume MATH has infinite representation type. Then the quiver isomorphism MATH of REF induces a group isomorphism MATH and MATH. We prove that MATH is surjective. Take an auto-equivalence MATH of MATH, and write MATH. After replacing MATH with MATH for suitable MATH, we can assume that MATH for all MATH. Because MATH is the preprojective component of MATH (compare CITE), we get MATH . As in REF above, MATH. Since MATH is a complete slice, REF says that MATH is a tilting module. So MATH is a two-sided tilting complex over MATH. Replacing MATH by MATH we can assume MATH is trivial. Let MATH be an invertible bimodule such that MATH. Replacing MATH with MATH we get MATH. Then by REF we get MATH. |
math/9904006 | CASE: By REF the homomorphism MATH is surjective. REF identifies MATH. CASE: If MATH has finite representation type then MATH is a tree, so MATH by REF . By REF we get MATH . CASE: If MATH has infinite representation type then MATH by REF . We know that MATH is in the center of MATH. |
math/9904006 | The isomorphisms are by REF . The data in the third column of REF was calculated in REF, except for the shift MATH which did not appear in that paper. So we have to do a few calculations involving MATH. Below are the calculations for types MATH and MATH; the rest are similar and are left to the reader as an exercise. Type MATH: Choose the orientation in REF . The quiver MATH looks like REF . Therefore MATH where MATH and MATH. Now by REF and VIII. REF, the quiver MATH is the full subquiver on the vertices in the triangle MATH. The projective vertices are MATH and the injective vertices are MATH, where MATH. We see that MATH, and the quiver MATH is the full subquiver on the vertices in the triangle MATH. Hence MATH and MATH. The relation MATH is easily verified. Type MATH: The quiver MATH is in REF , and MATH is a full subquiver. From the shape of MATH we know that MATH should have MATH indecomposable projective modules, MATH having length MATH and one of them simple. From the shape of the opposite quiver MATH we also know that MATH should have MATH indecomposable injective modules, MATH of them simple and one of length MATH. Counting dimensions using NAME sequences we conclude that MATH is the full subquiver on the vertices MATH. The projective vertices are MATH, the injective vertices are MATH, and the simple vertices are MATH, where MATH. For MATH let MATH, MATH and MATH, be the projective, simple and injective modules respectively, indexed such that MATH, and with MATH. So MATH and MATH for MATH. By the symmetry of the quiver it follows that there is a nonzero morphism MATH in MATH for MATH, and hence MATH . The rule for connecting MATH with MATH (see REF) implies that MATH. Therefore MATH for MATH. Now for each such MATH there is an NAME triangle MATH. When this triangle is turned it gives an exact sequence MATH, and hence MATH. The conclusion is that MATH for all MATH, so MATH. |
math/9904006 | The isomorphisms follow from REF . The structure of MATH is quite easy to check in all cases. In type MATH, MATH odd, the automorphism MATH is MATH . |
math/9904006 | As in the proof of REF , the group of auto-equivalences of the path category is MATH. Hence MATH. Given MATH let MATH be its matrix with respect to to the basis MATH, and let MATH. Define an auto-equivalence MATH with MATH and MATH, MATH. Then MATH preserves all mesh relations, and by a linear algebra argument we see that up to scalars at each vertex, the only elements of MATH are of the form MATH. Let MATH be MATH and MATH, with the obvious action on arrows to make it commute with the polarization MATH. Then MATH is generated by MATH and MATH, so MATH. The formula for MATH above shows that MATH for MATH. Finally use REF . |
math/9904006 | CASE: This is because MATH. CASE: Here the group of auto-equivalences of MATH is, in the notation of the proof of REF , MATH, and the group of isomorphisms is MATH. Therefore MATH is isomorphic to MATH as varieties, and as matrix group MATH. The auto-equivalence associated to MATH is MATH and MATH. The quiver MATH has no multiple arrows. Let MATH be the symmetry MATH for MATH, and MATH. Then MATH generates MATH, and we can use REF . The action of MATH on MATH is MATH. CASE: Here MATH, and the subgroup of isomorphisms is MATH. The symmetry MATH of order MATH acts on MATH by MATH. Let MATH be the symmetry MATH, MATH if MATH, and MATH if MATH. Then MATH and MATH commute, and they generate MATH. The action of MATH on MATH is trivial. CASE: Similar to REF . |
math/9904006 | There is an NAME sequence MATH in MATH. Applying the functor MATH to this sequence, and using REF , we get a triangle MATH in MATH. Hence MATH . On the other hand for MATH we have MATH. This proves that MATH; but MATH. |
math/9904006 | For an orientation MATH let MATH be the quiver of REF . As usual MATH denotes the set of vertices of MATH. Let MATH be the groupoid with object set MATH, and morphism sets MATH for MATH. The groupoid MATH acts faithfully on the family of sets MATH. According to REF there is an injective map of groupoids MATH. Let us first assume MATH is a NAME graph. Then there is a canonical isomorphism of sets MATH. The action of MATH on MATH is MATH. By REF , the action of MATH on MATH is MATH if MATH, and MATH. Since MATH commutes with MATH we have MATH for any MATH and MATH. If MATH is not NAME then MATH, MATH, etc., and the proof is the same after these modifications. |
math/9904006 | We will only treat the NAME case; the general case is proved similarly with modifications like in the previous proof. Let MATH. From the proof above we see that MATH for some MATH. A quiver map MATH with MATH must have MATH for all MATH, since MATH is a tree. Therefore MATH. |
math/9904008 | See REF or REF. |
math/9904008 | See REF . |
math/9904008 | For MATH, let MATH be the set of MATH such that MATH and MATH. By definition, MATH . As MATH is smooth of pure dimension MATH over MATH, NAME 's lemma implies that MATH . Consequently, MATH . As MATH, we have MATH . Now, MATH therefore MATH . Finally, MATH, hence MATH . |
math/9904008 | By definition, MATH . We compute these sums separately. The integral over MATH is equal to MATH. Then MATH . Concerning the integrals over MATH, we have MATH . Finally, MATH . Adding all these terms gives MATH . |
math/9904008 | Indeed, we have MATH . The integral of a non-trivial character over a compact group is MATH, hence this integral equals MATH if MATH, equals MATH if MATH and equals MATH if MATH. This proves the lemma. |
math/9904008 | The inequality is trivially true for MATH. We prove it for any MATH by induction: to lift a point in MATH to a point in MATH, one needs to solve two equations in MATH: MATH . A point in MATH which reduces to a point in MATH modulo MATH has MATH lifts in MATH. On the other hand, a point reducing to a point in MATH has MATH or MATH lifts according to the two linear equations being compatible or not. This implies the lemma. |
math/9904008 | The set MATH is defined by the two equations MATH. Fix the coordinates MATH so that MATH is the first vector. Up to a constant, one may write MATH for some homogeneous polynomials MATH of degree MATH. Then, denoting MATH, MATH is defined by the equations MATH . On MATH, MATH and on MATH, MATH. As MATH is smooth, MATH doesn't vanish on MATH which must therefore be either empty or of dimension MATH. Its degree cannot exceed MATH. The bound on the number of MATH-rational points are a consequence of the following (certainly well-known) easy lemma. |
math/9904008 | We prove this by induction on MATH. If MATH, the result is clear. Then, one can assume that MATH is reduced, irreducible and not contained in any hyperplane. For any hyperplane MATH which is rational over MATH, MATH is a closed subscheme of MATH of dimension MATH and of degree MATH. By induction, we have MATH . Finally, any point of MATH is contained in exactly MATH rational hyperplanes in MATH, so that MATH . As MATH, this implies MATH . |
math/9904008 | Indeed, we see from REF that for MATH, MATH the MATH being uniform in MATH. Consequently, MATH converges to a holomorphic bounded function on MATH. As the finite number of remaining factors converge uniformly in MATH, the existence of MATH is proven. The growth of MATH in vertical strips follows from NAME 's estimates for the NAME zeta function. |
math/9904008 | Recall REF : MATH the right hand side of which we have to estimate all terms. The first one is MATH. Then, as MATH is bounded, the second one is MATH . For the last term MATH, we use REF so that, denoting MATH, MATH . Moreover, MATH so that MATH . The lemma is proved. |
math/9904008 | Write MATH . The convergence of the first infinite product to a bounded holomorphic function follows from the preceding lemma. As in REF, there exists a constant MATH such that MATH . Using the rapidly decreasing behaviour of MATH as a function of MATH established in REF, the proposition is proved. |
math/9904014 | We use an algebraized version of arguments in CITE. CASE: Consider the double filtration for MATH. Since MATH and MATH, the sum in the definition is actually direct. Obviously, MATH. It follows from the definition of double filtration that MATH. Therefore MATH . Now, an easy summation proves that MATH. Thus, MATH . Since MATH is a NAME subalgebra, MATH is even. Hence MATH must be a NAME subalgebra. CASE: Assume that MATH is a nonzero element in MATH. Then MATH lie in the NAME subalgebra MATH. By a result of CITE, MATH is irreducible and the pairs of semisimple elements are dense in MATH. Therefore MATH . Associated with MATH, there is a decomposition MATH, where MATH. It follows that MATH and MATH. Obviously, the first summand is positive and we obtain MATH. This contradiction proves REF . |
math/9904014 | It is already proved that the inclusion ``MATH" holds. Since MATH is NAME and the pair MATH is not principal, it follows from CITE that MATH. Then the assertion follows for dimension reason. |
math/9904014 | By symmetry, it suffices to prove the first half of each item. The proof applies to both pn- and almost pn-pairs. CASE: This part is essentially the same as in CITE. Consider the MATH-limit: MATH, which lies in MATH. Since different summands have different weights relative to MATH, the sum is direct and therefore MATH. The space MATH possesses the MATH-filtration and MATH. For similar reason, MATH and hence MATH. As in the proof of REF , one may conclude by making use of the parity argument: MATH lies in the reductive NAME algebra MATH and therefore MATH must have the same parity as MATH. CASE: Applying the formula in REF with MATH gives MATH . Obviously, dimension of the left-hand side is MATH. Since MATH, we are done. CASE: Since MATH, we have MATH. By either REF or REF , MATH is a regular nilpotent element in MATH. Therefore MATH, where MATH consists of nilpotent elements. Finally, MATH and therefore MATH consists of semisimple elements. Whence MATH. |
math/9904014 | CASE: For CASE : Suppose that MATH, that is, all the eigenvalues of MATH in MATH are integral. We need to prove here that the case MATH is impossible. Assume not and MATH, MATH. A standard calculation with the Killing form on MATH shows that MATH if and only if MATH. By definition, put MATH. For each MATH, consider the finite set MATH, with the lexicographic ordering. This means MATH or MATH and MATH. Denote by MATH the unique maximal element in MATH. Let MATH be an element such that MATH for all MATH. Then MATH is a nonzero element in MATH. By REF , there is MATH such that MATH. Then MATH. Since MATH, we have MATH is nonzero and belongs to MATH. However, MATH. Therefore MATH, which contradicts the choice of MATH. Thus, the case MATH is impossible. For (non-MATH): Suppose MATH. Consider the set MATH. Because MATH is the unique ``non-integral" homogeneous subspace of MATH, MATH lies in the single coset space MATH and has a unique ``north-east" corner. Obviously, MATH is this corner. Since MATH for all MATH, this corner must lie in the positive quadrant. The condition MATH implies MATH. It remains to demonstrate that both MATH must be fractional. Assume not, and MATH, while MATH is fractional. Consider a ``path inside of MATH" connecting the points MATH and MATH: Starting from a nonzero element in MATH, we may always apply either MATH or MATH until we arrive at MATH (MATH). Since MATH is integral, we must intersect somewhere the vertical axis. This means MATH has a fractional eigenvalue in MATH. It then follows from nilpotency of MATH that MATH has a fractional eigenvalue in MATH as well. However, this contradicts to REF . CASE: The pairs MATH such that MATH are said to be bi-weights of MATH. In either case, the bi-weights lie in an open half-plane of MATH, hence the assertion. In the non-MATH case, MATH is the unique nonintegral bi-weight. Since MATH is not a bi-weight (see REF ), this implies MATH. It is also easily seen that MATH is Abelian. |
math/9904014 | Define MATH by MATH. It is an inner automorphism of MATH. Then MATH, MATH, and MATH. As MATH contains no semisimple elements, MATH is semisimple. |
math/9904014 | CASE: Let MATH be the minimal element in MATH with respect to the lexicographic ordering. Then MATH, while MATH (MATH). It is not hard to prove that MATH, but we do not need this. CASE: Now the eigenvalues of MATH are integral and the bi-weights of MATH lie in the upper half-plane. The same argument as in CITE shows that MATH is injective for all MATH and MATH. (Otherwise we would find an element MATH with MATH, MATH.) Then, by duality, MATH is surjective for MATH. In particular, MATH. On the other hand, MATH is not injective. Hence MATH is not surjective, that is, MATH. |
math/9904014 | In view of REF can be restated as: MATH and MATH. Assume now that MATH is special, that is, MATH for some MATH. Then MATH and MATH. A contradiction! |
math/9904014 | CASE: By REF , any almost pn-pair of non-MATH-type yields an inner involution MATH such that MATH is semisimple. But MATH has no such involutions. CASE: Since all nilpotent orbits in MATH are NAME and hence special, there are no almost pn-pairs of MATH-type as well. |
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