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math/0005060 | Notice first that for MATH, MATH and MATH big enough, then the numbers that appear in the right hand side of the estimates in REF form an estrictly decreasing sequence. That is, MATH . Let us check the inclusion MATH, for example. Suppose first that MATH, then MATH . If MATH, the inclusion is obvious. Otherwise, MATH .... |
math/0005060 | Let us proof REF . The second statement is proved in an analogous way. Let MATH be as in REF . If MATH (in particular, MATH), then MATH (the latter inclusion holds provided MATH). Assume now MATH. If MATH, then MATH and the result is trivial. If MATH, we denote by MATH a cube centered at MATH with side length MATH. The... |
math/0005060 | CASE: Let MATH and MATH. If MATH, there exists some MATH with MATH and MATH. Then MATH and so MATH (as in REF ). CASE: Let MATH and let MATH be such that MATH. We know that MATH . So we are done if we see that MATH. As in REF , we have MATH . Thus MATH . CASE: Let us see the first inequality. If MATH and MATH belongs t... |
math/0005060 | Let us see REF first. So we assume that there exist some cube MATH containing MATH with MATH. Since MATH, we have MATH. In particular, MATH. By REF and the second inequality of REF we get MATH . So REF holds if we take MATH big enough. Consider now REF . By REF have MATH . Thus we can write MATH . Let us estimate the f... |
math/0005060 | In the first two cases MATH for some MATH. In the latter case, by REF and our construction, there exists some MATH such that MATH. So in any case MATH for some MATH. Arguing as in REF , for MATH big enough, it is easily checked that MATH, and so MATH. |
math/0005060 | Assume that MATH and that either MATH, MATH or MATH on MATH, or MATH. By the preceding claim, MATH for some MATH. Then, MATH . |
math/0005060 | First we will prove the first statement. By REF , we know that MATH. Let MATH be such that MATH. We must have MATH. Otherwise, MATH and MATH, which is not possible. Since MATH, we have MATH. We know MATH because REF holds for MATH. By REF (for MATH and MATH) we get MATH . The term MATH is also bounded above by MATH bec... |
math/0005060 | Suppose that MATH is such that either MATH, MATH or MATH on MATH. By REF we have MATH for some MATH. By construction, the center of MATH belongs to some cube MATH with MATH. It is easily seen that MATH. Thus MATH . Since MATH and MATH are contained in a common cube of the generation MATH, by REF we get MATH and so MATH... |
math/0005060 | Consider MATH. We want to see that MATH is very close to MATH on this cube. By REF we have to deal with the cube MATH. Let us see that if MATH is such that MATH, then MATH. Notice that MATH. Now, by the definition of good cubes, there exists some MATH such that MATH, which implies MATH. For MATH big enough, we have MAT... |
math/0005060 | If MATH, we have already seen that MATH. If MATH, then MATH (because MATH and MATH). By construction, we have MATH . If MATH on MATH, then MATH. Now we consider the case MATH such that MATH on MATH. By REF there exists some MATH with MATH. Recall that by REF , if MATH, we have MATH . So if MATH and MATH, we have MATH. ... |
math/0005060 | Take MATH such that MATH. By REF we get MATH for MATH. Therefore, MATH . By REF we have MATH . So we only have to estimate MATH. Take MATH with MATH. Since MATH, by REF we have MATH. Applying REF we get MATH . It is easily checked that MATH. Then we get MATH. |
math/0005060 | Consider first the case MATH. If MATH is such that MATH, then MATH and so MATH. Therefore, MATH . By REF , we get MATH on MATH. Assume now MATH. Let MATH. From the definition of MATH, we have MATH for any MATH, MATH, with MATH. Also, by REF we have MATH . Consider now MATH with MATH. Observe that MATH and MATH. We have... |
math/0005060 | Let MATH be fixed. We are going to estimate the sum MATH . Let MATH be such that MATH. Since MATH is a bad cube, there exists some MATH such that MATH. Then we have MATH. Since MATH and MATH, we get MATH, and so MATH. By the finite overlapping of the cubes MATH in MATH, we have MATH . Now, from the construction of MATH... |
math/0005060 | The set MATH is open because MATH is lower semicontinuous. Since for MATH-a.e. MATH there exists a sequence of MATH-doubling cubes centered at MATH with side length tending to zero, it follows that for MATH-a.e. MATH such that MATH there exists some MATH-doubling cube MATH with MATH and so MATH. The existence of the fu... |
math/0005060 | Given MATH, for each integer MATH, we consider the generalized NAME decomposition of MATH given in the preceding lemma, with MATH. We will adopt the convention that all the elements of that decomposition will carry the subscript MATH. Thus we write MATH, as in REF . We know that MATH is bounded and satisfies MATH (beca... |
math/0005063 | If MATH laxly centralizes MATH, let MATH, MATH, MATH, and MATH be given as in the definition of lax centrality. Let MATH be denoted by MATH. Since MATH, we have MATH, which implies that MATH. If MATH, we let MATH be the subalgebra of MATH of pairs related by MATH, and form the pushout square MATH where MATH is defined ... |
math/0005063 | Let MATH, MATH, MATH, and MATH witness the fact that MATH laxly centralizes MATH. Since MATH, there is a natural one-one homomorphism MATH. It is assumed in the definition of lax centrality that MATH, and the same goes for MATH and MATH. Thus, there are MATH, MATH and onto homomorphisms MATH, MATH. Let MATH. Then the d... |
math/0005063 | Let MATH, MATH, MATH, and MATH be given as in the definition of lax centrality, and such that MATH, MATH, and denote MATH by MATH. Let MATH, MATH be tuples of algebras in MATH, with MATH and MATH disjoint, such that MATH and MATH. Since MATH, we have embeddings MATH. Let MATH be a filter on MATH maximal with respect to... |
math/0005064 | By the usual averaging over time modulations argument we may assume that MATH is a free solution to the wave equation. By time reversal symmetry we may assume MATH is a forward solution MATH in which case it suffices to show the scale-invariant estimate MATH . By the usual NAME arguments (for example, CITE) we may assu... |
math/0005064 | By REF the above estimates split into the spatial estimates MATH and the temporal estimates MATH . But these follow from NAME multiplication laws (REF , or CITE; alternatively, use fractional NAME, NAME embedding and NAME) and the hypothesis MATH. (In fact, one has some regularity to spare in all of these estimates, ex... |
math/0005070 | We prove REF . If the assertion does not hold, there exists an index MATH such that MATH. This implies that the self intersection number of each component of MATH is MATH, which is a contradiction (compare CITE). REF follows from REF as MATH are strictly positive. |
math/0005070 | Take a two dimensional cone MATH and assume that MATH is strictly positive. Let MATH be the canonical subdivision of MATH and let MATH be the vertices of MATH on this cone. Write MATH. Assume that MATH and MATH and MATH. Take MATH with MATH and put MATH and MATH. Then MATH and MATH are positive integers and MATH. This ... |
math/0005070 | The first assertion follows by an inductive argument. Write MATH with positive rational numbers MATH. As MATH and MATH, MATH are positive integers. By the definition of MATH, we can write MATH for some MATH. The assertion for MATH holds and MATH. Assume that MATH with MATH. As MATH and MATH is the vertices of the canon... |
math/0005070 | As MATH has an isolated singularity, MATH must contain a monomial of type MATH or MATH. Suppose that MATH. Let MATH be the vertex of MATH adjacent to MATH by an edge. It is clear that the non-compact face MATH which has MATH as a face and is unbounded to the direction of the MATH-axis has covector MATH. One can see tha... |
math/0005070 | Assume that MATH with MATH. CASE: If MATH, we have MATH. As MATH, MATH divides MATH. Thus MATH if and only if MATH and MATH. In this case, MATH and MATH. Assume that MATH. By the definition of MATH, MATH is a non-compact face with dimension REF. In particular, MATH. By REF , we have MATH. CASE: Suppose that MATH. Then ... |
math/0005070 | Denote by MATH. Suppose that MATH and MATH. Then MATH by REF and MATH. By the assumption, we have MATH. As the continuous fraction MATH is given by MATH (MATH copies of REF), we get MATH and the assertion follows immediately. |
math/0005070 | This follows from REF . |
math/0005070 | Let MATH be the primitive covectors in MATH inserted by the canonical subdivision from MATH. If MATH, the assertion is immediate from REF , as MATH. We assume that MATH. If MATH, the assertion is obvious. Assume that MATH. By REF , MATH is given by MATH for MATH. Thus MATH. If MATH, MATH is monotone increasing by REF a... |
math/0005070 | The last assertion follows from by REF as MATH. |
math/0005070 | From the congruence MATH modulo MATH, it is clear that MATH. Thus MATH. The equality REF results from MATH . Thus MATH and MATH. The last congruence equation is equivalent to MATH. Assume that MATH. As MATH, there exists positive integer MATH, MATH, such that MATH modulo MATH. Put MATH. We see that MATH. Take MATH whic... |
math/0005070 | The first statement is a conclusion of CITE. REF follows from the NAME criterion and CITE. |
math/0005070 | We first check when the central exceptional divisor MATH is rational by using REF (see also CITE). If this is the case, we compute the number of arms from MATH. If this number is less than REF, we show that MATH. Recall that the number of arms in the resolution graph is the sum of MATH for non-regular cones MATH. CASE:... |
math/0005070 | This is a summary of the following three lemmas. |
math/0005070 | We mainly use REF . Let MATH. The equation is MATH . Hence, we have the following conclusions. CASE: MATH if and only if there exists an integer MATH such that MATH and MATH. This is equivalent to MATH. And in this case MATH. CASE: MATH if and only if there exists an integer MATH such that MATH. And in this case MATH. ... |
math/0005079 | CASE: Let MATH be a MATH-extension of MATH. Since MATH is irreducible, so is MATH. Set MATH and MATH. Note that MATH, or MATH according as MATH is of real, complex, or quaternionic type, respectively. Since MATH, it follows from the NAME reciprocity that MATH where MATH denotes the multiplicity of MATH in MATH. On the ... |
math/0005079 | For MATH and MATH, denote by MATH, MATH, and MATH the set of generators of MATH according to MATH, MATH, and MATH, respectively. Note that the dimension of MATH is twice that of MATH and MATH. Then it is easy to find the generators and relations of MATH according to MATH as in REF . |
math/0005079 | See CITE for the former statement in REF . To see the uniqueness in REF , we note that if MATH and MATH are MATH-extensions of MATH, then MATH. Since MATH acts trivially on MATH and MATH is of odd order, MATH must be a trivial MATH-module. Therefore MATH is also isomorphic to MATH, which means that MATH and MATH are is... |
math/0005079 | The proof of CITE holds in the real category with slight modification. For reader's convenience we shall give the argument when MATH is finite. The case when MATH is infinite is easy since the action of MATH on MATH is transitive, see CITE for details. We first note that if there exists an equivariant isomorphism MATH,... |
math/0005079 | There is an element MATH in MATH with MATH as the character of the fiber MATH-module by REF , and we have the semi-group isomorphisms MATH where the former isomorphism is given by sending MATH to MATH and the latter is given by taking orbit spaces by the MATH-action. In fact, the map sending MATH to MATH is the inverse... |
math/0005079 | CASE: We note that MATH acts on MATH by conjugation. Since the minimal polynomial of any element in MATH has distinct root it is diagonalizable. So two elements in MATH are in the same orbit if and only if they have the same eigenvalues which are MATH because MATH. This implies that MATH has exactly MATH connected comp... |
math/0005079 | It follows from REF that MATH and MATH. Therefore the lemma follows from REF . |
math/0005079 | Let MATH be the set of (not necessarily invariant) complex structures on the fiber MATH. The collection MATH of MATH over MATH forms a MATH-fiber bundle over MATH, the MATH-action on MATH being induced from that on MATH. Then a MATH-invariant complex structure on MATH can be viewed as a continuous MATH-equivariant cros... |
math/0005079 | Since MATH is of real type, the character of MATH is also MATH; so we may view MATH as a complex irreducible character of MATH. We have proved in CITE that the semi-group MATH of isomorphism classes of complex MATH-vector bundles over MATH with multiples of MATH as the character of fiber MATH-modules is generated by fo... |
math/0005079 | The map MATH is surjective by REF and injective except for Cases A and B by REF . In both Cases A and B the target of MATH is a semi-group generated by one element by REF while the domain of MATH is generated by two elements with the relation as in REF . This implies that MATH is two to one. Finally, we note that tenso... |
math/0005079 | REF - REF follow from REF , and REF follows from REF . We now prove the last statement in the theorem. After setting MATH and MATH, it is obvious that the inverse images of the pairs MATH in the remark after REF by the semi-group homomorphism MATH in REF are real MATH-line bundles with trivial fiber MATH-module. Moreov... |
math/0005079 | The proof is elementary and left to the reader. |
math/0005079 | CASE: Since MATH acts freely on MATH, every real MATH-line bundle over MATH with trivial fiber MATH-module is the pull-back of a real line bundle over MATH by the quotient map MATH. Suppose MATH is even. Then MATH is trivial, so pullback line bundles by MATH have trivial first NAME classes, which means that the underly... |
math/0005079 | Recall from the last statement in REF that all generators are related through tensor product with the real MATH-line bundles MATH and MATH according as MATH and MATH, respectively. These line bundles are all trivial if MATH is even by REF . So the existence of one trivial generator implies triviality of the other gener... |
math/0005082 | By the first resolvent identity one has MATH . Therefore MATH and, by duality (here MATH is considered as an element of MATH), MATH . This ends the proof. |
math/0005082 | By REF one has MATH and so MATH solves REF ; by linearity MATH is also a solution. As regard REF let us at first note that MATH and MATH . Therefore one has MATH which immediately implies that MATH satisfies REF . |
math/0005082 | By REF MATH necessarily differs from (the restriction to MATH of) MATH by a MATH-independent, densely defined operator MATH. Being densely defined, MATH has an adjoint and MATH. Therefore, being MATH injective, MATH is closable and so, being MATH bounded, MATH is closable. Denoting by MATH the closure of MATH, the clos... |
math/0005082 | We have already proven that, under our hypotheses, MATH is a pseudo-resolvent, that is, MATH . We proceed now as in the proof of [REF ]. By [REF] MATH, being a pseudo-resolvent, is the resolvent of a closed operator if and only if it is injective. Since MATH would imply MATH by REF we have MATH (see REF ) and so MATH. ... |
math/0005082 | Writing MATH by REF one has MATH . Thus by REF there follows MATH and so, since MATH implies MATH, one has MATH . Injectivity of MATH and MATH for any MATH then follows by injectivity of MATH (see REF ), REF , injectivity of MATH, and the definitions of MATH. Being MATH densely defined, one has MATH and so injectivity ... |
math/0005082 | By our hypotheses MATH necessarily differs from (the restriction to MATH of) the bounded sesquilinear form associated to MATH by a MATH-independent NAME form MATH. Therefore MATH is a semi-bounded, densely defined, closable NAME form. If MATH denotes the unique semi-bounded self-adjoint operator corresponding to the cl... |
math/0005083 | Suppose the conditions hold. The cone MATH yields a toric open embedding MATH, hence MATH is quasiaffine. To see that the map MATH is surjective, consider an affine chart MATH, where MATH is a maximal cone. Since MATH induces a bijection of maximal cones, there is a MATH such that MATH. Moreover, MATH was assumed to ha... |
math/0005083 | Suppose MATH is a quotient presentation. Given an invariant affine open subset MATH, the preimage MATH is affine as well. There is an effective invariant principal divisor MATH with support MATH, because MATH is quasiaffine. So MATH is an effective NAME divisor with support MATH. By construction, MATH lies in the image... |
math/0005083 | Suppose MATH is a quotient presentation and consider two invariant affine charts MATH, MATH of MATH. Since MATH is an affine toric morphism, the preimages MATH are invariant affine charts of MATH. The restriction of MATH defines a quotient presentation MATH. Since MATH is separated, the intersection MATH is even affine... |
math/0005083 | First, suppose that the quotient presentation MATH is defined by an inclusion of rings MATH. The weight module MATH is given by the homogeneous component MATH of degree MATH. Let MATH and MATH be the primitive lattice vectors generating the rays in MATH and MATH, respectively. Choose MATH representing MATH. Note that t... |
math/0005083 | The problem is local, so we can assume that MATH is given by an inclusion of rings MATH. By REF , the ring of invariants MATH is nothing but MATH. |
math/0005083 | Suppose that MATH maps to MATH. According to REF , the homogeneous components in MATH are invertible. NAME easily check that the multiplication maps MATH are bijective. So by REF , the quotient presentation MATH is a principal homogeneous MATH-space. Hence the condition is sufficient. Reversing the arguments, you see t... |
math/0005083 | First, we check sufficiency. Let MATH be the preimage of the subgroup MATH. The group scheme MATH is finite, so its action on MATH is automatically closed. Consequently, the quotient MATH is a geometric quotient. NAME directly see that MATH is quasiaffine. Consider the induced toric morphism MATH. The strict transforms... |
math/0005083 | By definition, we have MATH. Since MATH is an open embedding, MATH holds for each quasicoherent MATH-module MATH. This gives MATH. Because MATH, we have MATH. Consequently, MATH. |
math/0005083 | Choose MATH such that the homogeneous elements MATH define effective NAME divisors with support MATH. Then MATH has the same radical as the irrelevant ideal MATH. Suppose that MATH. Then the restrictions MATH are zero as well. Note that the preimage MATH is affine, with global section ring MATH. The NAME subring MATH d... |
math/0005083 | First, we claim that MATH factors through the open subset MATH. For MATH choose MATH such that MATH is a unit in MATH. Then MATH is invertible on a MATH-saturated neighbourhood of MATH. Clearly, this neighbourhood is mapped into MATH. According to REF , the projection MATH is a categorical quotient for the MATH-action ... |
math/0005083 | Let MATH. Then MATH is invertible on a neighbourhood of the fibre of MATH over MATH. Looking at the commutative REF , we see that MATH is invertible at some point of the fibre of MATH over MATH. Since MATH is saturated, this means MATH. The reverse inclusion is clear by definition. |
math/0005083 | Suppose that MATH are two base-point-free MATH-algebras, which define two morphisms MATH, with MATH. First, assume that MATH. Let MATH be saturated. Using REF , we infer MATH. To check the second condition for equivalence, note that MATH are NAME polynomial algebras. So the map MATH induces the desired isomorphism. Con... |
math/0005083 | In REF , we already saw that the assignment is functorial in MATH. By REF , it is well-defined on equivalence classes and gives an injection from the set of equivalence classes to the set of morphisms. It remains to check that the identity morphism MATH arises from a base-point-free MATH-algebra. Indeed: you easily che... |
math/0005083 | Let MATH be a base-point-free MATH-algebra on the scheme MATH defining a morphism MATH. Set MATH. The map MATH defines a homomorphism MATH. Clearly, it suffices to show that this map is bijective. The problem is local, so we may assume that MATH is affine, hence each weight module MATH is trivial and MATH holds. Accord... |
math/0005083 | By assumption, MATH has invertible homogeneous components. By the preceding Proposition, each base-point-free MATH-algebra MATH is isomorphic to the preimage MATH. |
math/0005091 | It will be shown that the bijection between MATH and MATH given by MATH induces an isomorphism of posets MATH. To establish this, it suffices to show that to each codimension MATH flat MATH there corresponds a codimension MATH flat MATH. Write MATH, where MATH, and let MATH. The flat MATH may be realized as the set of ... |
math/0005091 | For MATH, define MATH by MATH where MATH and MATH are in MATH. It is then readily checked that MATH is the identity map. Furthermore, if MATH is another hyperplane of MATH, the composition MATH is given by MATH so is null-homotopic. Consequently, the classes MATH form a basis. Finally, using stratified NAME theory, one... |
math/0005091 | Denote points in MATH by MATH, where MATH and MATH satisfies MATH for each MATH. Similarly, denote points in MATH by MATH, where MATH and MATH. In this notation, we have MATH and MATH. Let MATH be the total space of the pullback of MATH along the map MATH. It is then readily checked that the map MATH defined by MATH is... |
math/0005091 | By the previous result, the bundle MATH is equivalent to the pullback of MATH along the map MATH of REF. An analogous result for the complements of the subspace arrangements MATH and MATH is established next. For MATH, view MATH as MATH and MATH as MATH. Denote points in the configuration space MATH by MATH, where MATH... |
math/0005091 | For each hyperplane MATH of MATH, let MATH denote the composition of MATH and the natural inclusion MATH. It will be shown that the composition MATH induces the identity in homology if MATH, and induces the trivial homomorphism if MATH, thereby establishing the result. For MATH, write MATH as in REF. Then MATH is given... |
math/0005091 | The proof is by induction on MATH. In the case MATH, MATH is an arrangement of MATH points in MATH, the fundamental group of the complement is MATH, the free group on MATH generators, and it is well known that MATH is isomorphic to the free NAME algebra MATH, see for instance CITE. In general, assume that the fiber-typ... |
math/0005091 | From the exact sequence of NAME algebras REF noted above, it follows that MATH is isomorphic to the semidirect product of MATH by MATH determined by the NAME homomorphism MATH given by MATH for MATH. It suffices to show that the homomorphism MATH factors as asserted. From REF , and the results of NAME and NAME stated i... |
math/0005091 | By REF , one has MATH, where the sum is over all MATH and MATH for which MATH. The result follows. |
math/0005091 | The proof is by induction on MATH. In the case MATH, there is nothing to show since MATH is a free group on MATH generators, MATH is isomorphic to the free NAME algebra MATH, and there are no codimension two flats in MATH. In general, assume that MATH is strictly linearly fibered over MATH and that MATH as before. Then... |
math/0005091 | Given a fibration MATH with a section MATH, there is a homotopy equivalence MATH given by the composite: MATH where MATH is the loop space multiplication and such that the inclusions of MATH and MATH into MATH are maps of MATH-spaces. NAME, if the spaces involved have torsion free homology then MATH. By REF and NAME, o... |
math/0005091 | The realization of the bundle MATH as the pullback of the bundle of configuration spaces MATH along the map MATH from REF yields REF algebras MATH with exact rows, and, on the level of primitives, REF algebras MATH where MATH is induced by MATH. Since the underlying bundles admit cross-sections, the rows in the above d... |
math/0005091 | By REF , one has MATH, where the sum is over all MATH and MATH for which MATH. Since the homology suspension MATH is an isomorphism and MATH is the map in loop space homology induced by MATH, one has MATH, where the sum is over all MATH and MATH for which MATH. The result follows. |
math/0005091 | REF follows from the fact that the homology of MATH is abelian while the homology of MATH is generated by NAME brackets of weight at least MATH in homological degrees greater than MATH. REF follows from the remarks at the beginning of this section. REF follows at once from the fact that the result holds in case MATH, w... |
math/0005094 | The above REF is the statement for MATH. We proceed by induction on MATH. Assume REF for some MATH. Then MATH . Since MATH the above integral can be expressed as a sum of non-negative terms. |
math/0005094 | According to CITE MATH . We want to write the divisor in the form MATH for MATH. This gives MATH . Now MATH . We use the restriction property of MATH with respect to MATH. On MATH the class MATH is invariant under the action of MATH exchanging the punctures. So under the natural map MATH it descends to the restriction ... |
math/0005098 | Let MATH be a common log resolution of MATH and MATH, with MATH . REF implies that MATH, and hence MATH (that is, the difference MATH is effective). Therefore MATH and the statement follows. |
math/0005098 | Since the relative canonical bundle MATH is effective, it follows from the definition that MATH for any ideal MATH. Then using REF we find that MATH . Taking MATH, this gives REF . For REF , fix MATH and use the subadditivity relation REF to deduce: MATH as asserted. |
math/0005098 | It suffices by REF to show that MATH . But membership in the ideal on the right is tested locally at a general point of each irreducible component of MATH. So we can assume after shrinking MATH that MATH is smooth and irreducible, of codimension MATH, and in this REF is clear. For then MATH for all MATH, and MATH is re... |
math/0005098 | We can assume without loss of generality that MATH is MATH-exceptional and that MATH, so that MATH. Applying REF to the graded family MATH REF , it suffices to prove that MATH. We suppose to this end that we've fixed a large integer MATH such that the multiplier ideal MATH in question is computed on a log resolution MA... |
math/0005098 | The symbolic powers MATH again form a graded family of ideals, so REF will apply as soon as we establish that MATH. Referring to the primary decomposition REF , it is enough to show that MATH . For a given index MATH, inclusion in MATH is tested at a generic point of MATH. So having fixed MATH we are free to replace MA... |
math/0005100 | The hereditary orders in MATH containing MATH form a partially ordered set which we will denote by MATH. If MATH lies minimally over MATH then one proves exactly as in CITE that MATH. Let MATH be a maximal order lying over MATH. We deduce that MATH, where MATH is the length of a maximal chain in MATH, starting in MATH ... |
math/0005100 | By the previous proposition it suffices to prove this for MATH. Let MATH be the regular projective curve associated to the function field of MATH CITE. Then MATH is a regular compactification of MATH. In particular MATH is a finite number of points, whence by the localization sequence MATH is finitely generated if and ... |
math/0005100 | It is clear that the categories in REF. and REF. have finitely generated (free abelian) NAME groups, and that MATH is not finitely generated if MATH is derived equivalent to a hereditary category MATH with all objects of finite length and an infinite number of nonisomorphic simple objects. In view of REF the proof is c... |
math/0005100 | For the convenience of the reader we repeat the proof. We have to show that MATH is generated by MATH and MATH. Let MATH be the full subcategory of MATH consisting of objects isomorphic to finite direct sums of the form MATH. Then MATH is a strict triangulated subcategory of MATH. This can be deduced from the fact that... |
math/0005100 | The inclusions MATH define a map MATH. An inverse to this map is given by sending MATH to MATH (see REF for notations). |
math/0005100 | Using the same method as in REF we find inductively using REF that MATH. Now it is easy to see that sending MATH to MATH defines an isomorphism MATH. This proves what we want. |
math/0005100 | Let MATH and assume that MATH is maximal such that MATH. Then there is a triangle MATH . Applying MATH yields injections MATH. Since for MATH the non-trivial homology of MATH occurs in degrees MATH and since the projective dimension of MATH is less than or equal to REF it follows that MATH for MATH. In particular MATH ... |
math/0005100 | If MATH is a map in MATH then one has to show that MATH, MATH. Since the complex represented by MATH clearly lies in MATH, this follows from the previous lemma. |
math/0005100 | By REF one has MATH . |
math/0005100 | By REF we have to show MATH. By REF it follows that MATH is equal to MATH. So it is sufficient to show that MATH. By REF one has MATH. Since MATH is an abelian subcategory of MATH, it also has finite NAME dimension. In particular if MATH there is a quotient category MATH of MATH which has finite length. Selecting a sim... |
math/0005100 | Since one has MATH it follows that MATH . It is now easy to see that MATH which yields the result. |
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