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math/9903120 | Suppose MATH is a completely prime ideal such that MATH has infinite dimension over MATH. Then MATH so assume that MATH and localize at MATH. If MATH we localize at MATH instead and use a similar argument. Then MATH is a nontrivial ideal of MATH. If MATH we can localize at MATH to obtain a nontrivial ideal in MATH where MATH. However it follows from REF and the assumption that MATH, that MATH has infinite order. Hence by CITE, MATH is a simple ring. This contradiction shows that MATH. Now by REF, MATH where MATH and MATH. Now MATH embeds in MATH which is a domain, so MATH is prime and clearly MATH-invariant. By choosing a polynomial of least degree in MATH with coefficients in MATH we see that MATH and the lemma follows easily. |
math/9903120 | As in CITE we can choose MATH such that the image in MATH is a highest weight vector in MATH and that MATH. Now MATH and either MATH or MATH. Assume that MATH. Then since MATH we get MATH so MATH holds. If MATH, then MATH, since otherwise we would find MATH, MATH and then MATH would contradict MATH. If also MATH then MATH holds. If MATH, then since MATH, MATH holds. |
math/9903120 | Let MATH be such that MATH . Hence MATH, MATH, and by CITE we conclude that MATH. |
math/9903120 | Let MATH be a basis for the NAME module MATH. As MATH, MATH for all MATH and MATH. We have that MATH and MATH. Since MATH, MATH. Hence MATH . Combined with the fact that MATH stabilizes MATH, this implies that we can write MATH, for some MATH. If MATH, then MATH. Thus MATH is either MATH or MATH and a short calculation shows that MATH, a contradiction. Hence MATH . |
math/9903120 | Assume that there are sequences as above. First suppose that MATH. As in CITE choose MATH such that the image in MATH is a highest weight vector and such that MATH. By REF the weights of MATH are MATH with MATH. Thus by REF , MATH and MATH are not weights of MATH. On the other hand MATH, so MATH. Similarly since MATH we have MATH, and MATH implies that MATH. Thus MATH. Now MATH and also MATH. Since MATH, MATH. Since MATH an easy computation shows that the only element of MATH annihilated by MATH and by MATH is the zero element. Hence MATH. Since MATH, we have MATH. Since MATH and MATH are simple, we have that MATH and hence MATH is a highest weight module of weight MATH. As MATH, we conclude that MATH and hence such a nonsplit exact sequence can not occur. Finally suppose that MATH, that is MATH. Since MATH, we have MATH. As in REF , we have MATH. Hence MATH is a module over the field MATH, so the sequence splits, a contradiction. |
math/9903120 | We first show the equivalence of REF and MATH. By CITE any submodule of MATH has the form MATH for some MATH. It follows that MATH has length MATH if and only if every finite dimensional homomorphic image of MATH is simple. Thus MATH implies MATH. Conversely suppose that MATH holds and there is a nonsplit exact sequence MATH with MATH finite dimensional. By REF MATH. Since all NAME modules have composition length MATH, REF cannot occur. Thus MATH is a weight of MATH, that is MATH for some MATH. Similarly by dualizing the above exact sequence and applying REF we get MATH for some MATH. Hence MATH, but this contradicts REF . Next we prove MATH implies MATH by showing that if MATH is a NAME module over MATH of length MATH, then MATH, for some MATH. By the description of the submodules of MATH from CITE cited above we have MATH for some MATH. Thus by REF , MATH and MATH . Forming the difference between these equations we get MATH . As MATH . Thus MATH . Finally, if MATH and MATH is chosen so that MATH it follows that MATH and the NAME module MATH has length MATH. |
math/9903120 | Suppose MATH, MATH and MATH . Again this is an equation in MATH which we can identify with a subfield of MATH. Then MATH. Consideration of the imaginary part of this expression shows that MATH where MATH. If MATH, then since MATH we get MATH, and MATH. Thus MATH, MATH and MATH forces MATH. Hence we can assume that MATH. Let MATH, so that MATH. Then MATH is decreasing on MATH so MATH. Therefore MATH forces MATH. |
math/9903120 | If the result is false then from the description of the submodules of MATH, we can find positive integers MATH such that MATH. As in the proof of REF this means that MATH, MATH, MATH, MATH . As before we identify MATH with a subfield of MATH. By the previous Lemma MATH. Now consider the function MATH. Note that MATH. We obtain a contradiction since MATH has at most one zero. |
math/9903120 | Note that MATH is NAME regular of global dimension REF and NAME by CITE. Using the Lemma in CITE and writing MATH as an iterated NAME extension it follows that MATH also satisfies these properties. From now on let MATH denote either MATH or MATH. Note that MATH is graded and MATH is a homogeneous central element of positive degree in MATH which is not a zero divisor. Thus we can use a graded version of NAME 's lemma and the proof of CITE to show that MATH. The result now follows from CITE. |
math/9903123 | It is well-known that REF are equivalent, and they are also equivalent to the condition that the restriction MATH of MATH is positive definite. Thus REF are equivalent to MATH. This condition is equivalent to REF because MATH. |
math/9903123 | Let MATH be a minimal subset of MATH such that MATH. Write MATH with MATH. Let MATH, and set MATH. By REF we have MATH and hence MATH. If MATH, then we have MATH. This is a contradiction. Thus MATH. By the minimality of MATH we have MATH, and hence we have MATH. |
math/9903123 | Let MATH be the equivalence relation on MATH generated by MATH and let MATH denote the set of equivalence classes with respect to MATH. For MATH set MATH. Then MATH for MATH are all non-zero and mutually orthogonal with respect to the natural positive definite symmetric bilinear form on MATH. Hence MATH is a finite set. Thus it is sufficient to show that MATH is a finite set for each MATH. If MATH, then MATH is positive definite, and hence MATH is a finite subsystem of MATH. Thus MATH is a finite set. Assume that MATH. By REF there exists a finite subset MATH of MATH such that MATH with MATH. Since REF implies MATH for any MATH and MATH. Since MATH is an equivalence class with respect to MATH, we obtain MATH. Therefore, MATH is a finite set. |
math/9903123 | Set MATH. By REF we have MATH. Since MATH is positive definite, MATH is a finite subsystem of MATH containing MATH. Hence we can assume MATH from the beginning. Set MATH. Since MATH, MATH is not identically zero on MATH. Similarly MATH REF is not identically zero on MATH. Since MATH is a countable set, there exists some MATH such that MATH and MATH for any MATH. Then we have MATH. Since MATH, there exist only finitely many MATH such that MATH by REF . Hence there exists some MATH such that MATH for any MATH by CITE. Then we obtain MATH with MATH. Since MATH, we have MATH. |
math/9903123 | CASE: Assume MATH and MATH. Take MATH. By REF there exists some MATH such that MATH. For MATH we have MATH and hence we have MATH for any MATH satisfying MATH. Thus MATH. CASE: Assume MATH. By REF we have MATH. Then we have MATH . |
math/9903123 | Since MATH, REF implies MATH for any MATH. Thus we have MATH. |
math/9903123 | Set MATH. It is sufficient to show that the group MATH is generated by the reflections contained in it. Set MATH . Since MATH contains MATH, REF implies that MATH contains a point MATH such that MATH. Thus replacing MATH with such a MATH, we may assume that MATH and MATH. Then the assertion follows from CITE and CITE. |
math/9903123 | CASE: If MATH, then we have MATH by REF . Hence we have MATH by REF . CASE: Assume MATH. Set MATH . For each MATH there exist only finitely many MATH satisfying MATH by REF . Since MATH, we obtain MATH. Thus we have MATH. On the other hand we have MATH by REF , and hence MATH. Thus we obtain the desired result by REF . REF follows from REF by replacing MATH with MATH. |
math/9903123 | Set MATH . Then MATH contains MATH. CASE: By the definition of MATH we have MATH . In particular, we have MATH by MATH. Thus MATH. Hence it is sufficient to show MATH. Set MATH . Let MATH. REF and the assumption MATH imply MATH. Hence MATH for any MATH. Thus we have MATH if and only if MATH for any MATH. By REF , this condition is equivalent to MATH . Thus we obtain MATH where MATH is a set of positive rational numbers. Then MATH contains MATH, and REF implies that MATH. CASE: This follows immediately from REF . |
math/9903123 | By replacing MATH with MATH if necessary, we may assume MATH from the beginning. Let us first show that there exists some MATH such that MATH, MATH and MATH for any MATH. If MATH, then we have MATH by REF , and REF implies the existence of such a MATH. If MATH, then REF implies the existence of such a MATH. By REF there exist only finitely many MATH such that MATH. Thus there exists some MATH such that MATH for any MATH by CITE. We may assume that MATH. Then we have MATH by CITE. For MATH we have MATH and hence MATH. Thus we obtain MATH. Moreover, we have MATH with MATH. Then MATH is a proper subset of MATH by MATH. |
math/9903123 | It is easily seen that MATH. Hence it is sufficient to show that if MATH appears as a subquotient of MATH, then we have MATH. We may assume that MATH for MATH. The central element MATH of MATH acts on MATH via the multiplication of the scalar MATH for any MATH. For MATH we have MATH by the MATH-invariance of MATH, and hence MATH acts on MATH via the multiplication of MATH. Therefore we have MATH for any MATH. If MATH appears as a subquotient of MATH, then we have MATH, and hence MATH . |
math/9903123 | By the assumption we have MATH and MATH. Assume that there exists some MATH satisfying MATH. We may assume that its length MATH is the smallest among such elements. Set MATH. Since MATH is the shortest element of MATH, CITE implies MATH . Since MATH is the shortest element of MATH, MATH . By MATH there exists some MATH satisfying MATH. Then we have MATH. Hence we have MATH by REF . If MATH, then we have MATH. This contradicts REF . Thus we obtain MATH. Set MATH . By MATH and MATH we have MATH if MATH and MATH if MATH. Now we have MATH . Since MATH and MATH are elements of MATH, we have MATH. By MATH we obtain MATH by the minimality of MATH. Hence we have MATH. It follows that MATH is an extremal weight of MATH. This contradicts MATH, MATH, and MATH or MATH. |
math/9903123 | Take MATH and MATH such that MATH. Let MATH be the set of weights of MATH. Since MATH we have MATH . This implies MATH . Assume that MATH for MATH and MATH. Then we have MATH, and hence MATH by REF . Thus we have MATH . Hence we obtain MATH. In particular, there exists some MATH such that MATH. By REF , MATH is a free MATH-module. Thus the morphism MATH given by MATH is injective. It follows that MATH contains MATH as a submodule. Hence we have MATH. |
math/9903123 | If MATH, then MATH has finite length. Therefore we can reduce the assertion to the case where MATH with MATH. Then the assertion follows from the preceding proposition. Assume now MATH. It is enough to show MATH for any MATH. Let MATH be the set of weights of MATH. We set MATH. Since MATH implies MATH by REF , we may assume MATH ranges over MATH in REF . If MATH for a sufficiently large MATH, then MATH implies that MATH is sufficiently large. Hence the both sides of REF vanish. Fixing such a MATH we shall argue by the descending induction on MATH such that MATH. Let MATH REF be a highest weight of MATH. Then there is an exact sequence MATH where MATH does not contain MATH (MATH). Hence by the induction hypothesis, REF holds for MATH. Arguing by the induction on the cardinality of MATH, REF holds for MATH. Since MATH by the preceding proposition, REF holds for MATH. Then REF holds for MATH because MATH is an exact functor. |
math/9903123 | Since MATH is an exact functor, MATH is a quotient of MATH. By restricting the non-degenerate contravariant form on MATH we obtain a non-degenerate contravariant form on MATH. Thus we have either MATH or MATH. Assume MATH in the case MATH and MATH in the case MATH. Then there exists MATH such that MATH, MATH, and MATH. Set MATH. Then we have MATH and MATH. By REF we have exact sequences MATH . By applying the exact functor MATH, we obtain exact sequences MATH . Since MATH is an isomorphism, we have MATH. Next assume MATH in the case MATH and MATH in the case MATH. Then we have MATH . Let MATH be the maximal proper submodule of MATH. By applying MATH to the exact sequence MATH we obtain an exact sequence MATH . Thus it is sufficient to show MATH. Hence we have only to prove MATH for any MATH satisfying MATH. By REF there exists some MATH such that MATH and MATH. For such a MATH, MATH is a subquotient of MATH. Therefore it is sufficient to show MATH for any MATH such that MATH. Set MATH. Then we have MATH, MATH and MATH. Since MATH according to MATH, MATH, we have MATH. Hence REF implies MATH. Thus we obtain MATH. |
math/9903123 | Note that MATH, MATH and MATH. CASE: MATH. By REF there exist MATH and a proper subset MATH of MATH such that MATH and MATH for MATH (and hence also for MATH). Take MATH such that MATH for MATH and MATH for MATH. Set MATH, MATH. Then we have MATH where MATH. By taking MATH for MATH sufficiently large, we may assume that MATH and MATH. Moreover, we have MATH for any MATH, and hence we have MATH and MATH. Thus REF implies MATH for any MATH and MATH. The assertion then follows from REF . CASE: MATH. The proof is similar to the one for REF . Take MATH and a proper subset MATH of MATH such that MATH and MATH for MATH. Take MATH such that MATH for MATH and MATH for MATH. Set MATH, MATH. By taking MATH for MATH sufficiently large, we have MATH, MATH and MATH for MATH. Thus REF implies MATH for any MATH and MATH. Hence we obtain the desired result by REF . |
math/9903123 | Let us prove first the case where MATH. We first prove the following statement. MATH . By REF there exist MATH and a proper subset MATH of MATH such that MATH and MATH. Take MATH such that MATH for MATH and MATH for MATH. Set MATH. Then we have MATH for any MATH and MATH. Hence by taking MATH for MATH sufficiently large, we obtain REF . Assume that MATH. Let MATH. By REF there exists MATH such that MATH, MATH, MATH for any MATH, and MATH. By REF there exist MATH and a proper subset MATH of MATH such that MATH and MATH. Let MATH be the longest element of MATH. Take MATH such that MATH for any MATH, and set MATH. Then we have MATH . Since MATH, we have MATH when MATH is sufficiently large. For any MATH we have MATH. If MATH, then we have MATH and MATH, and hence MATH. If MATH, then we have MATH when MATH is sufficiently large. Since MATH is a finite set, we have MATH for any MATH for a sufficiently large MATH. By MATH we have MATH when MATH is sufficiently large. Take MATH satisfying REF . Then we have MATH and MATH satisfies REF for MATH. By REF the integers MATH in REF do not depend on the choice of MATH. Hence REF holds for MATH. Since MATH is the longest element of MATH we have MATH, and REF implies MATH. Then REF implies MATH . The desired result follows then from REF . As the assertion in the case MATH is proved similarly, we shall only give a sketch. By REF and an analogue of REF we may assume that MATH for MATH and MATH are sufficiently small. Take MATH and a proper subset MATH of MATH satisfying MATH and MATH. Take MATH such that MATH for any MATH, and set MATH. Then we have MATH and MATH satisfies REF for MATH. Hence we can take MATH as MATH by REF . Then we have MATH by REF . Hence we obtain the desired result by REF . |
math/9903123 | By the definition of the NAME module MATH is obviously a homomorphism of MATH-modules. Let us show that MATH is surjective. It is sufficient to show that MATH for any MATH. Let MATH satisfying MATH. We show by induction on MATH that MATH. The case MATH is trivial. Assume MATH. Since MATH, we have MATH by the hypothesis of induction. By MATH the linear map MATH is bijective. Hence there exists some MATH such that MATH. Then we have MATH . Next let us show that MATH is injective. Assume MATH. By MATH the NAME module MATH is irreducible unless MATH. Thus there exist subspaces MATH of MATH for MATH such that MATH. Hence there exists some MATH such that MATH. Then we have MATH. This is a contradiction. Thus we have MATH. |
math/9903123 | CASE: We have to show that the canonical morphisms MATH and MATH are isomorphisms. By the symmetry we have only to show that MATH is an isomorphism. Let us show that the canonical morphism MATH is bijective for any MATH. By MATH it is sufficient to show that the canonical morphism MATH is bijective for any MATH. This follows from REF for MATH and REF . CASE: We have MATH by REF . By REF MATH is the unique irreducible quotient of MATH. Thus we have MATH. |
math/9903123 | For MATH and MATH we have MATH . Thus for MATH and MATH we have MATH for any MATH if and only if MATH. The last condition is equivalent to MATH for any MATH. Fix MATH and MATH, and consider the function MATH on MATH. We have only to show that MATH is constant on MATH. By a consideration on the contravariant forms on NAME modules we see that MATH is a constructible function on MATH. In particular, it is constant on a non-empty NAME open subset MATH of MATH. Let MATH be the value of MATH on MATH. We have to show MATH for any MATH. Let MATH. By REF MATH is a constant function on MATH for any MATH. Thus we see by the above argument that MATH is constant on MATH. Assume for the moment that MATH . Since MATH, we have MATH for some MATH. Since MATH is a constant function on MATH, we have MATH for any MATH. In particular, we obtain MATH. It remains to show REF . Set MATH . We have MATH and MATH. By the definition of MATH the natural morphism MATH is an isomorphism. Since MATH is a MATH-subspace of MATH we have MATH. Hence MATH is a MATH-lattice of MATH. It follows that MATH is a NAME dense subset of MATH. |
math/9903123 | Set MATH. It is a highest weight module with highest weight MATH. Set MATH. It is sufficient to show MATH for any MATH. Assume that MATH for some MATH. By MATH and REF we have MATH. Hence under the isomorphism MATH we have MATH. It follows that MATH. This contradicts the irreducibility of MATH. |
math/9903129 | We will prove that, given any partial representation MATH, with MATH a complex vector space, then there exists a positive definite MATH-valued inner product MATH in MATH such that: MATH for all MATH and all MATH. In order to prove this, we argue as follows. For all MATH, let MATH denote the element MATH; it is easy to check that the MATH's form a family of commuting idempotents in MATH: MATH and, for all MATH, MATH from which we compute: MATH . Given a subset MATH of MATH, we denote by MATH the element in MATH given by: MATH . Using REF , it is easily seen that MATH. Observe that, since MATH, MATH is zero unless MATH contains the identity MATH of MATH; moreover, if MATH, then MATH, because MATH. Also, the following identity holds: MATH where the sum is taken over all possible subsets of MATH, including the empty subset. In order to prove REF , it suffices to observe that, since the MATH's commute, we have the following combinatorial formula: MATH . Now, let MATH be any MATH-valued positive definite inner product in MATH; we consider the new inner product MATH on MATH, given by: MATH . In order to prove that MATH is positive definite, since MATH is positive definite, it suffices to show that MATH for all MATH implies MATH. This follows easily from REF . We now prove REF . To see this, we start by observing that, for all MATH, MATH and MATH, the following equality holds: MATH . Namely, observe that if MATH, then MATH and both sides of REF vanish. On the other hand, if MATH, then MATH, hence: MATH which proves REF . Similarly, by symmetry one checks that: MATH . For MATH and MATH, using REF we now compute as follows: MATH which concludes the proof. |
math/9903129 | If MATH is a unital homomorphism of MATH-algebras, then clearly MATH is a partial representation of MATH in MATH. Conversely, suppose that MATH is a partial representation of MATH. Arguing as in the proof of REF , for all MATH we denote by MATH the element of MATH given by MATH. Recall from REF that the MATH's are commuting idempotents and MATH for all MATH. Similarly MATH . For all elements MATH in MATH, we define: MATH . For MATH and MATH in MATH, we have: MATH . Now, if MATH, that is, if MATH, then, either there exists one element MATH that belongs to MATH but does not belong to MATH, or there is an element MATH that belongs to MATH but not to MATH. In both cases, the product MATH contains the factor MATH and therefore it is null. If MATH, then, since MATH, the product MATH becomes: MATH . This proves that MATH is multiplicative on MATH, and thus its linear extension to MATH gives a homomorphism of MATH in MATH. We now prove that MATH is unital. By definition, since MATH and recalling REF , we have: MATH . Furthermore, we have: MATH . To conclude, we are left with the proof of the uniqueness of the homomorphism MATH satisfying MATH. In order to show this, it suffices to prove that the set MATH generates the whole algebra MATH. To this aim, let MATH be an arbitrary element of MATH, where MATH is a subset of MATH containing the identity. The set of such pairs MATH form a vector space basis of MATH. Let's denote by MATH the subalgebra of MATH generated by MATH. Let MATH be the family of elements of MATH defined by: MATH . We consider the element in MATH given by the product MATH: MATH . Using the rules of multiplication in MATH, it is easily seen that the sum equals: MATH . Hence, for every MATH, MATH contains the element: MATH . Now, suppose that MATH. We have: MATH which means that, for every MATH and every MATH, MATH contains all elements of the form: MATH . Now, MATH hence, for every MATH, MATH contains the element MATH . We repeat the argument, using the fact that: MATH . By induction, we see that MATH contains all elements of the form: MATH which proves that MATH and we are done. |
math/9903129 | The maps MATH, given by MATH, and MATH are partial representations of MATH. By the universal properties of the two algebras MATH and MATH, there exists unital MATH-algebra homomorphisms MATH and MATH such that MATH and MATH. It is easily checked that the composite maps MATH and MATH are the identities on their domains, as they are the identity on a set of generators, hence MATH and MATH are isomorphisms. |
math/9903129 | Let MATH be the units of MATH; for all MATH define MATH by: MATH . Since MATH is connected, then MATH is non empty for all MATH. Namely, if MATH is a sequence in MATH and MATH is a sequence in MATH such that MATH and MATH for all MATH, then the element MATH is in MATH. Clearly, MATH is the (disjoint) union of all the MATH's. Fix a family of elements MATH, with MATH. Then, for every element MATH, MATH and MATH can be written uniquely in the form MATH with MATH, where MATH and MATH. On the other hand, every element of the form MATH, with MATH, belongs to MATH. Observe that, in particular, MATH is constant equal to MATH. The map MATH gives the desired isomorphism between MATH and MATH, proving REF . Finally, let MATH denote the set of matrix units of MATH, that is, MATH is the matrix with null entries, except for the entry in the position MATH which is equal to the identity of MATH. It is easy to check that, for any group MATH, the map MATH extends by linearity to a MATH-algebra isomorphism between MATH and MATH, which gives REF and concludes the proof. |
math/9903129 | Let MATH be any subset of MATH containing the identity of MATH. In the course of the proof, we will identify MATH with the vertex MATH of the graph MATH. We denote by MATH the stabilizer of MATH in MATH, given by: MATH . In the graph MATH, MATH is identified with the set of edges departing and terminating at the vertex MATH. Observe that, since MATH, then MATH. Since MATH acts on the left on MATH, then MATH is the union of right cosets of MATH, say: MATH where MATH . We now prove that the sub-groupoid of MATH corresponding to the connected component of the vertex MATH of the graph MATH is (isomorphic to) the groupoid MATH. First of all, we prove that the connected component of MATH in MATH contains precisely MATH vertices, given by the subsets of MATH of the form MATH, MATH. These sets are all distinct, in fact, if MATH, then MATH and MATH, which implies that MATH. Moreover, if MATH for some MATH, say MATH for MATH, then MATH, which implies that the connected component of MATH contains no other vertex but the MATH's. Observe that the stabilizer of MATH, which is MATH, coincides with the isotropy group of the unit MATH. Hence, by REF the algebra MATH is a direct summand of MATH and MATH is the direct sum of algebras arising from this construction. This proves REF . For the proof of the second part of the thesis, we fix MATH and MATH as in the hypothesis. From the first part of the proof, it is clear that in order to determine the number of occurrences of the algebra MATH in REF we need to count the number of vertices at the level MATH of MATH whose stabilizer is MATH. We define MATH to be the number of such vertices; moreover, we consider the index of MATH in its normalizer MATH: MATH . Since the conjugation is an automorphism of MATH, if MATH and MATH are conjugate in MATH then, by symmetry, MATH for all MATH. Observe that two vertices in the same connected component of MATH have conjugate stabilizers; hence, the number of vertices at the level MATH whose stabilizers are conjugate to MATH is precisely MATH. The connected component of each of these vertices, that gives a contribution of one copy of MATH as a direct summand of MATH, contains MATH vertices. Hence, the number of summands MATH in MATH that arise from this construction is MATH. Finally, by symmetry, the contribution of the single subgroup MATH is: MATH . Every vertex at the MATH-th level of MATH which is fixed by MATH is a subset of MATH (containing MATH) which is union of MATH right cosets of MATH (including MATH); the number of such vertices is MATH. Some of them will have stabilizer which is bigger than MATH, and this number is clearly given by MATH hence we have the following recursive formula for the coefficients MATH: MATH . From REF we obtain immediately REF , which concludes the proof. |
math/9903129 | Under our hypotheses, there exists a bijection between the family of subgroups of MATH and MATH that preserves the order of the subgroups and their inclusions. Hence, REF implies that the coefficients MATH are also preserved. The conclusion follows from REF, considering that the corresponding subgroups have isomorphic group rings. |
math/9903129 | By REF , we have an isomorphism: MATH where the MATH's are (not necessarily distinct) subgroups of MATH. If MATH, then we claim that the summand MATH does not contain a NAME component isomorphic to MATH. Indeed, if it did, then the algebra MATH would contain a direct summand of the form MATH, where MATH and MATH is the dimension of some irreducible MATH-representation of MATH. Let MATH be a prime dividing MATH; in particular, MATH divides MATH. Then, since the characteristic of MATH is zero, MATH divides MATH (see for example, CITE), and consequently it also divides MATH. By hypothesis it follows that MATH must divide MATH. This yields MATH, which is a contradiction and proves our claim. Thus, the summands of REF whose NAME decomposition contains MATH are precisely the ones given by MATH, with MATH. Observe that it must be: MATH for, the equality MATH would imply MATH and MATH, which would give MATH, a contradiction with our hypothesis. Now, each summand MATH in REF , with MATH, contains exactly MATH . NAME components isomorphic to MATH (see for instance CITE). Therefore, in order to compute the multiplicity of the summand MATH in the NAME decomposition of MATH, it suffices to determine the multiplicity of MATH in REF for a fixed subgroup MATH. To this aim, let MATH be a fixed subgroup of MATH and let MATH be a subset of MATH containing the identity, with MATH, and with stabilizer MATH. If MATH, take a prime MATH that divides MATH. Since MATH divides MATH, then MATH must divide MATH. But MATH divides MATH, hence MATH by our hypothesis. Again, this implies that MATH, which shows that MATH. Consequently, every set MATH which is the union of MATH distinct cosets of MATH (including MATH) has stabilizer MATH. The total number of such sets is easily computed as MATH, so the multiplicity of MATH in REF with fixed MATH is: MATH . It follows that the number of algebras MATH contributed into the NAME decomposition of MATH by MATH is given by: MATH . Since the inequality MATH is equivalent to MATH, the result follows. |
math/9903129 | We prove more in general that if MATH is a divisor of MATH such that the following two conditions are satisfied: CASE: every prime that divides MATH divides also MATH; CASE: either MATH or MATH is a MATH-group and MATH, then MATH for all MATH. Under our hypotheses, the number MATH satisfies REF above. For the sake of shortness, in the course of this proof, an integer number MATH satisfying REF will be called a small divisor of MATH. We proceed by induction on MATH; the case MATH is clearly trivial. Let MATH be fixed. Observe first that the inequality MATH holds if and only if MATH. Indeed, MATH so MATH obviously implies MATH. Now, if MATH is a MATH-group and MATH, then MATH, and consequently MATH. Thus, MATH . Since both MATH and MATH are powers of MATH, it follows that MATH. Now, if MATH is any group with MATH, then MATH, which gives MATH. It follows MATH and MATH, as claimed. Let now MATH be fixed; observe that, by REF above, since MATH we have MATH. Thus, using REF , we have that the multiplicity of MATH in the NAME decomposition of MATH is: MATH . By the uniqueness of the NAME decomposition, the number in REF is invariant by MATH-algebra isomorphisms. Since the coefficient of MATH in REF is non zero, to conclude the proof it suffices to show that MATH for all MATH. If MATH is a small divisor of MATH, then, by our induction hypothesis, it follows MATH. Suppose now that MATH is not a small divisor of MATH, and let MATH be all the prime divisors of MATH that do not divide MATH. We prove that in this case there exists MATH a small divisor of MATH such that MATH. Since MATH is abelian, every subgroup of MATH of order MATH is a product of the form MATH, where MATH is the (unique) NAME MATH-subgroup of MATH and MATH is some subgroup of MATH such that none of the MATH's divide MATH. Set MATH. It is easily seen that MATH and that MATH is a small divisor of MATH. It follows that MATH, and we are done. |
math/9903129 | Let MATH and MATH be the cyclic factor decompositions of MATH and MATH, and set: MATH . Clearly, MATH for all MATH. We can obviously assume MATH, since the case MATH is trivial. The equality MATH implies that MATH and, in particular, MATH. Now, suppose by contradiction that MATH. For sufficiently large MATH we have MATH and MATH, so we can choose the minimal MATH such that MATH. Clearly, MATH, because MATH. Let MATH (respectively, MATH) the number of the MATH's (respectively, of the MATH's) whose order is equal to MATH. Since MATH, the numbers MATH and MATH must be different, say MATH. Moreover, by the minimality of MATH, it must be MATH for all MATH, because MATH. Therefore, MATH can be obtained from MATH by replacing some direct cyclic factors of order MATH with cyclic factors of order MATH. But, obviously, such a replacement increases the total number of subgroups of order MATH, from which it follows MATH, and, consequently, MATH. Then, by our hypothesis, we must have MATH. On the other hand, MATH has at least one element MATH whose order is greater or equal to MATH, because MATH and MATH. If MATH has only one such MATH, then MATH, MATH, and therefore MATH can be obtained from MATH by replacing a direct cyclic factor of order MATH with a cyclic factor of order greater or equal to MATH. But this means that MATH, which is impossible. Hence, MATH contains at least two elements MATH with order greater or equal to MATH, which implies that MATH, and MATH, a contradiction. |
math/9903129 | We can assume that MATH is algebraically closed. Namely, if MATH is any algebraically closed field that contains MATH, then the isomorphism MATH implies the isomorphism MATH . As we observed in REF , if MATH, then MATH and MATH have the same cardinality. The case MATH is treated separately. The partial group algebra MATH of the cyclic group of order MATH and MATH of the NAME MATH-group are isomorphic to MATH and MATH respectively (see REF). Hence, we have MATH. These are the only (abelian) groups of order MATH, and so we may now assume that MATH. From REF , it follows that MATH for all prime number MATH and all integer MATH such that MATH divides MATH. Let MATH where MATH is the NAME MATH-subgroup of MATH. Then MATH, where MATH is the NAME MATH-subgroup of MATH; clearly MATH for all MATH. Fix a MATH and set MATH, MATH. Since MATH and MATH, REF implies that MATH. As MATH is arbitrary, we finally obtain MATH. |
math/9903129 | It suffices to observe that if MATH, then MATH, where MATH is the field of fractions of MATH. Hence, the conclusion follows from REF . |
math/9903131 | We can easily reduce to the case MATH. If we write MATH then MATH is an isomorphism onto its image, and MATH . We see that MATH kills the first three factors, and is an isomorphism from the fourth factor onto its image; MATH is therefore the sum of the first three factors, which is what we wanted to show. |
math/9903131 | Again, we can assume that MATH. Write MATH where MATH, MATH, and MATH. Then MATH is the direct sum of those MATH's where at least one of MATH or MATH is in the set MATH. Also, MATH is the direct sum of the MATH's where MATH and MATH are both in the set MATH. Thus, their intersection is MATH, as claimed. |
math/9903131 | If MATH then MATH, where we define the group MATH by MATH . Furthermore, up to multiplication by a constant, MATH has the same NAME coefficients as MATH, except that we have to take the MATH-expansion with respect to MATH instead of MATH. Our Theorem, then, is equivalent to the statement that, if MATH satisfies the condition MATH then MATH where MATH. Let MATH; it comes with an action of MATH. If MATH and MATH, define MATH to be MATH. Then MATH: in fact, MATH . The principle of inclusion and exclusion implies that MATH satisfies REF if and only if MATH . Thus, if MATH is an irreducible MATH-representation contained in MATH, it suffices to prove our Theorem for a form in MATH, since the conditions of our Theorem can be expressed in terms of the action of MATH. Let MATH be the prime factorization of MATH. Then MATH, so MATH where MATH is a representation of MATH. Also, MATH acts as the identity on the MATH for MATH. So if we define MATH then MATH and MATH is the space of forms satisfying REF. Thus, REF implies that MATH . Turning now to the question of a form's being in MATH, that is the case if and only if the form is both in MATH and is invariant under the image MATH of MATH in MATH. Also, MATH. Thus, setting MATH to be MATH and MATH to be the space of MATH-invariant elements of MATH, REF implies that an element of MATH is both in MATH and invariant under MATH if and only if it is in MATH . But if MATH is in MATH then it is invariant both under MATH and under projection to the subspace of invariants under the cyclic subgroup generated by MATH; this last condition is equivalent to its being invariant under MATH. Thus, our vector MATH is invariant under MATH and MATH is the set of invariants under MATH that is, the elements of MATH, completing our proof. The cusp form case is similar, replacing MATH by the space of cusp forms. The eigenform case then follows from the facts that the NAME operators are simultaneously diagonalizable and that their action is preserved by the operators MATH. |
math/9903131 | If MATH is an eigenform for MATH with eigenvalue MATH then MATH. Thus, if MATH is a MATH-eigenform then it is determined by its eigenvalues and by MATH. |
math/9903131 | This is part of REF, or of REF. |
math/9903131 | For any positive integer MATH, write MATH for the set of eigenforms in MATH with eigenvalues MATH. By REF , we don't have to worry exactly about which primes are avoided in our set of eigenvalues, so this notation makes sense. Furthermore, let MATH be a minimal level such that MATH is nonzero. By REF , the image of MATH in MATH is one-dimensional. REF shows that any element of the kernel of the map from MATH to MATH is of the form MATH, where MATH. But the minimality of MATH shows that there aren't any such forms; the kernel is therefore zero, so MATH is one-dimensional. To see that MATH is unique, let MATH be the space of adelic cusp forms of weight MATH but of arbitrary level structure; it comes with an action of MATH, and elements of MATH correspond to elements of MATH invariant under the action of a certain subgroup MATH, where MATH varies over the set of all primes and MATH is the highest power of MATH that divides MATH. NAME 's Global Result says that the set MATH of forms in MATH with eigenvalues MATH gives an irreducible representation of MATH; thus, it can be written as a restricted tensor product MATH, and MATH . Since MATH contains MATH, for each MATH it is the case that, if for some power MATH, MATH is nonzero, then there is a minimal such power. Thus, taking MATH to be the product of those minimal powers of MATH, we see that, if for some MATH, MATH is nonzero, then it is nonzero for a unique minimal MATH, namely our MATH. (Alternatively, the uniqueness of the minimal level is part of REF.) Finally, to see that the eigenspace grows as indicated, let MATH be a nonzero element of MATH for MATH minimal. By REF , we can assume that MATH, since our argument above showed that the image of MATH in MATH is nonzero. Fix some multiple MATH of MATH, and assume that we have shown that, for all proper divisors MATH of MATH with MATH, MATH . We then want to show that the same statement holds with MATH in place of MATH. Thus, let MATH be an element of MATH. By REF , the image of MATH in MATH is zero, so by REF , MATH for some forms MATH. Also, MATH unless MATH, since otherwise MATH wouldn't divide MATH, contradicting the unique minimality of MATH. But then REF implies that each MATH, and hence MATH, can be written as a linear combination of the forms MATH for MATH; it is easy to see that such an expression for MATH is unique. |
math/9903132 | For MATH, we have MATH, where the sum is over all MATH with MATH. Each summand may be written as MATH. Thus, MATH is isomorphic to the direct sum of MATH copies of MATH via the map MATH, MATH. Now consider the boundary map MATH of the complex MATH, induced by left-multiplication by MATH. Let MATH be a generator for MATH. Since MATH for all MATH, we have MATH. Thus, MATH . Write MATH in terms of the basis for MATH specified in REF , where MATH and MATH. Then we have MATH. Checking that MATH in MATH, we have MATH. Thus, with the change in sign, the boundary map MATH respects the direct sum decomposition MATH. |
math/9903132 | Let MATH and MATH be elements of MATH. Write MATH and MATH. From REF, we have either MATH if MATH, or MATH if MATH. It follows from these considerations, and a routine exercise to check the sign, that summands MATH of MATH arise only from MATH-admissible sets MATH. |
math/9903132 | Let MATH be a MATH-admissible set. Associated with MATH, we have the term MATH of MATH from REF . If MATH, it is readily checked that this term contributes nothing to the coefficient MATH of MATH. On the other hand, if MATH, then the above term contributes the summand MATH to the coefficient MATH. |
math/9903132 | From REF, we have MATH and the summand corresponding to MATH is simply MATH. For MATH, write MATH, where the sum is over all MATH. For a fixed MATH, the coefficient of MATH in MATH may then be expressed as MATH where MATH. Note that we have MATH for such MATH. By REF , we have MATH, where the sum is over all MATH-admissible sets MATH with MATH. Thus, MATH and since MATH, we have MATH. |
math/9903132 | The matrix of MATH is MATH, where MATH is the NAME Jacobian. Thus, MATH, where MATH is the map induced by the NAME Jacobian. By the Product REF , we have MATH . The action of MATH on the free group MATH is recorded in REF. Computing NAME derivatives and evaluating at MATH yields the familiar NAME matrix of MATH (see CITE), MATH where MATH. Since MATH, the result follows upon differentiating MATH. |
math/9903132 | The proof is by induction on MATH. If MATH, then MATH is the matrix of MATH, so MATH. Since the derivative of the constant MATH is zero, the entries of MATH are given by REF (with MATH and MATH). In this instance, we have MATH, and a set MATH is MATH-admissible if MATH and MATH. It follows that the case MATH is a restatement of REF . In general, let MATH and write MATH as in REF above. Then we have MATH . By induction, the entries of the matrix MATH are given by MATH where MATH, and for MATH, the second sum is over all MATH-admissible sets MATH for which MATH. By the chain REF , the entries of MATH are given by MATH where MATH. By REF , for MATH, we have MATH . The entries of MATH may be calculated from REF using REF, yielding the formula in the statement of the theorem. We conclude the proof by making several observations which elucidate this calculation. First consider the case MATH. Then MATH and MATH. If MATH, then the first case of REF yields a contribution of MATH to MATH, provided that MATH. Note that this condition implies that the set MATH is MATH-admissible (and that MATH). Note also that in this instance we have MATH, MATH, MATH, and MATH. If MATH and MATH, then the second case of REF contributes MATH to MATH if MATH. In this instance, the set MATH is MATH-admissible, and since MATH and MATH, we have MATH. For general MATH and MATH, suppose that MATH is a summand of MATH for some MATH. Then, by the inductive hypothesis, this summand arises from a MATH-admissible set MATH with MATH. If MATH, then the first case of REF yields a contribution of MATH to MATH, provided that MATH. For such MATH, it is readily checked that the set MATH is MATH-admissible. Also, since MATH, we have MATH. If, as above, MATH is a summand of MATH and MATH, then the second case of REF contributes MATH to MATH provided MATH. In this instance, the set MATH is MATH-admissible, and since MATH, we have MATH, and MATH. Applying these observations to REF above completes the proof. |
math/9903132 | For each MATH, the cohomology of MATH with coefficients in the local system MATH is isomorphic to that of the complex MATH. So MATH if and only if MATH. An exercise in linear algebra shows that MATH . For MATH, we have MATH if MATH. So, as above, MATH . By REF , MATH and MATH for each MATH. Thus, MATH and the result follows. |
math/9903136 | It is easy to verify that MATH and MATH are regular triangulations of MATH. By REF , we know that MATH and MATH are related by not necessarily regular flips. Let MATH be obtained from MATH by a single flip. Then MATH can be transformed into MATH by the sequence of flips and isotopies that is explicitly given in REF . The edges of MATH are drawn bold, and the edges of MATH under flip are dotted. None of these flips introduces a loop. It is possible that some flip for MATH introduces a multiple edge. This happens only if some of the vertices MATH, MATH, MATH and MATH of MATH coincide. We iterate the construction, that is, we replace each flip for MATH by a flip sequence for MATH. Since the four vertices of MATH of each quadrilateral involved in a flip are pairwise distinct, none of these flips introduces a loop or a multiple edge, thus all flips are regular. |
math/9903136 | For any singular triangulation MATH of MATH, we have MATH. Since MATH, we obtain MATH and MATH. It follows easily that MATH for MATH. One obtains MATH from MATH by MATH face subdivisions and some regular flips, see REF for the first barycentric subdivision. The figure shows the neighbourhood of a face, and the edges under flip are dotted. So MATH by the preceding Lemma. |
math/9903137 | Define MATH to be the class of sheaves of the type described in REF . Inductively define MATH to be the class of sheaves MATH such that either MATH is an extension of sheaves in MATH, MATH is a direct summand of a sheaf in MATH, or MATH where MATH is a morphism of projective varieties and MATH. Set MATH. The first four statements clearly hold. REF will be proved by induction. By definition, any sheaf in MATH is of the form MATH with MATH smooth, MATH a morphism, and MATH an ample line bundle on MATH. Given a morphism MATH with MATH projective, choose an ample line bundle MATH on MATH. Then MATH for MATH and MATH by NAME 's vanishing theorem CITE. Therefore by NAME 's vanishing theorem, the NAME spectral sequence and the projection formula, we obtain MATH when MATH and MATH . As the sheaves above can be assumed to be globally generated (again by NAME), we obtain MATH for MATH. Now suppose that GAREF has been established for sheaves in MATH (for all MATH, MATH and MATH). Choose a sheaf MATH and a morphism MATH. If MATH is a summand or an extension of sheaves in in MATH, then clearly MATH for MATH. Therefore we may assume that MATH with MATH with MATH a morphism. By the induction hypothesis, MATH for MATH. Therefore the spectral sequence for the composite collapses to yield isomorphisms MATH . But the sheaves on the right also vanish for MATH by induction. |
math/9903137 | The result is certainly well known, so we will be quite brief. There are two maps: the NAME trace MATH and a map MATH in the opposite direction which corresponds to pullback of forms. The lemma follows from the identity MATH. |
math/9903137 | In the course of the proof, we will construct a large commutative diagram: MATH . We will denote the maps MATH, MATH, MATH and MATH by MATH, MATH, MATH and MATH respectively. Let MATH be a very ample divisor on MATH. If MATH, MATH has a nonvanishing section, and let MATH be the corresponding effective divisor. Let MATH be a birational map of smooth varieties such that the pullback MATH of MATH has normal crossings. Let MATH be a desingularization of MATH such that MATH also has normal crossings CITE. We can choose a positive rational linear combination of exceptional divisors MATH such that MATH is an ample MATH-divisor. Define MATH where MATH is rational, then MATH is an ample MATH-divisor. Therefore MATH . Therefore, after replacing MATH by a large multiple, we can assume that the pullback of the linear system associated to MATH has an integral normal crossing divisor of the form MATH where MATH is smooth, ample and transverse to MATH, and multiplicities of the components of MATH and MATH are less than MATH. Apply NAME 's trick CITE, to obtain a finite cover MATH, branched over MATH such that the pullback MATH of MATH has multiplicity MATH and the multiplicities of the components of the pullback MATH of MATH are unchanged. Let MATH be the fiber product of MATH with MATH. By our assumptions, MATH is smooth and MATH is a normal crossing divisor such that MATH has multiplicity MATH and all other components have smaller multiplicity. Let MATH be a desingularization of the MATH-fold cyclic cover branched along MATH determined by the line bundle MATH (see CITE). Using the formula for the canonical sheaf of a cyclic cover CITE, we see that MATH . By our assumptions about multiplicity, the summand on the right for MATH is just MATH. By REF , we see that MATH is a direct summand of MATH which is in turn a direct summand of MATH. Therefore it is geometrically acyclic. |
math/9903137 | MATH is a direct summand of a tensor power of MATH. |
math/9903137 | Suppose that MATH satisfies the condition of the lemma, but that MATH fails to be nef. Then there exists a curve MATH, with MATH. Since MATH is quotient of MATH where MATH is the projection, MATH . This yields a contradiction when MATH. Conversely suppose that MATH is nef. NAME 's criterion shows that MATH is ample, therefore MATH is ample. There exists a finite branched cover MATH such that MATH possesses a MATH-th root CITE, that is a line bundle MATH such that MATH. Therefore MATH is ample, and this implies that MATH is ample. |
math/9903137 | Let MATH be a very ample line bundle such that MATH is also ample. Then MATH for all MATH and MATH. In other words, MATH is MATH-regular CITE. This implies that MATH is a quotient of a sum of MATH's, and therefore ample. Thus MATH is nef by the previous lemma. |
math/9903137 | Denote the projection MATH by MATH. Let MATH be an ample line bundle on MATH which is necessarily of the form MATH where MATH is a line bundle on MATH. As MATH is big, it is enough to check that MATH has a nonzero section for some MATH (see the comments preceding REF ). Let MATH be a general point of MATH; it lies over a general point MATH. MATH will denote the residue field at MATH. Choose a trivialization of MATH at MATH. Then a section of MATH can be identified with a section of MATH, and this determines a hypersurface in the fiber MATH. Choose MATH so that all elements of MATH extend to global sections, and choose a section MATH, so that the corresponding hypersurface avoids MATH. Then MATH lifts to a global section of MATH which can be identified with a global section MATH of MATH. Pulling MATH back to MATH yields a nonvanishing section of MATH. |
math/9903137 | Let MATH be a locally free quotient of a big vector bundle MATH. Suppose MATH is very ample, choose MATH so that MATH is generically generated by its global sections. Then the restriction of these sections also generates MATH generically. |
math/9903137 | The proof will be reduced to a series of observations. Fix a very ample line bundle MATH. The tensor product of two or more generically globally generated coherent sheaves has the same property. Therefore the set MATH of integers MATH for which MATH is generically globally generated forms a semigroup. Given a coherent sheaf MATH, choose an integer such that MATH is globally generated. Then MATH is generically globally generated for any MATH. Apply the result of the previous paragraph to obtain MATH such that MATH is generically globally generated. This implies that MATH. A semigroup containing two relatively prime integers contains all but finitely many positive integers, and this concludes the proof. |
math/9903137 | Let MATH and MATH be two big vector bundles and MATH a very ample line bundle. Then for each MATH, there is an integer MATH such that for both MATH and all MATH, MATH is generically globally generated. Choose MATH so that MATH is globally generated for all MATH. Then choose MATH. Then one sees, after grouping terms appropriately, that MATH is generically globally generated for any MATH. Therefore the same holds for MATH. |
math/9903137 | Suppose that MATH is a big vector bundle, and MATH an ample line bundle. For any partition MATH, there exists integers MATH such that MATH as the weight MATH and such that MATH is a direct summand of MATH by CITE. This implies that MATH is generically globally generated for MATH. Therefore MATH is generically globally generated for arbitrary MATH and MATH, since the first factor can be decomposed into a sum of NAME powers of large weight. If MATH and MATH are both big, then MATH must be big since it is a direct summand of MATH. |
math/9903137 | See CITE . |
math/9903137 | Let MATH and let MATH denote the NAME embedding. The restriction of MATH to MATH is MATH. If MATH is nef, then so is MATH, and hence also MATH. This argument requires only that all MATH. Suppose that MATH is nef and big, then MATH is also nef and big by the preceding lemmas. Let MATH be the NAME embedding. Then MATH is nef and big by REF . This together with the previous paragraph implies that MATH is again nef and big. |
math/9903137 | Let MATH be the list of indices MATH for which MATH, and let MATH. Also let MATH and MATH. Let MATH and let MATH . Then MATH is nef or nef and big according to whether MATH is. Let us suppose that MATH is nef and prove that MATH whenever MATH by induction. Suppose that MATH where MATH is smooth and MATH is an ample line bundle on MATH. There is no loss of generality in assuming that MATH is surjective, since otherwise we can replace MATH by its image. Consider the Cartesian diagram: MATH where MATH. By REF , MATH . Therefore MATH is nef and also big because the restriction of MATH to any fiber is ample so that MATH . Therefore MATH is geometrically acyclic by REF . Suppose that we have proved the result for all sheaves in MATH. Then clearly it holds for a direct summand or an extension of sheaves in MATH. If MATH with MATH, then MATH. Now suppose that MATH is both nef and big. Let MATH be a surjective morphism from a smooth projective variety MATH. Construct a Cartesian square as above. By REF MATH is nef and big, therefore MATH is geometrically acyclic. |
math/9903137 | Let MATH and MATH, MATH is positive by the definition of MATH. Then MATH. As MATH is nef, the corollary is immediate. |
math/9903137 | By the assumption on the singularities, we can replace MATH by a desingularization without affecting MATH. Then by CITE, there exists a morphism from a smooth projective variety MATH such that MATH is a direct summand of MATH. |
math/9903137 | From the NAME spectral sequence, it suffices to kill MATH for all MATH. In fact, these groups vanish for all MATH by the previous lemma. And they vanish for MATH, by CITE . |
math/9903137 | Since the section is regular, we have the NAME resolution: MATH . This can be broken up into a series of short exact sequences. MATH is geometrically positive because it is a summand of MATH. Therefore the lemma follows by tensoring these sequences by MATH, and using the vanishing of MATH for MATH . |
math/9903137 | MATH. |
math/9903137 | Let MATH be the map such that MATH is the pullback of the universal quotient bundle MATH. Set MATH, and let MATH denote the NAME embedding. By REF the NAME spectral sequence collapses to yield isomorphisms MATH for MATH. Furthermore, the vanishing theorem implies that MATH is MATH-regular CITE. Therefore by [CITE, page REF], MATH is surjective for MATH. A simple diagram chase using the maps MATH and the isomorphism REF finishes the proof. |
math/9903137 | Since these conditions are stable by pullback under a generically finite map, there is no harm in assuming that the base variety MATH is smooth and projective. Let MATH then there exists a smooth variety MATH with a line bundle MATH and a finite map MATH such that MATH. The existence of MATH follows from CITE or CITE. Therefore MATH is nef, where MATH. The NAME embedding MATH shows that MATH is also nef. Consequently, so is MATH. The theorem implies that MATH is geometrically semipositive. Since MATH is a direct summand of MATH, the corollary follows. |
math/9903137 | We will only sketch the proof since the lemma is not used here. Proceed as above to construct a map MATH of smooth projective varieties such that MATH has a MATH-th root MATH. Then MATH is nef, hence also is its restriction to any complete intersection curve MATH corresponding to a multiple of MATH. The same goes for MATH, so it has nonnegative degree. This last condition is equivalent to the above inequality. |
math/9903137 | By REF and the comments preceding it, a vector bundle of the form MATH, with MATH unitary flat and MATH a nef line bundle, is both nef and strongly semistable and therefore geometrically semistable. Moreover a vector bundle is clearly nef and strongly semistable if and only if its pullback under a finite map is. This along with the fact that the class of geometrically semipositive vector bundles is closed under extensions and direct summands implies the corollary. |
math/9903137 | By our assumptions, there exists a partition MATH, such that MATH . Consequently, MATH. The right hand side is multiplicative and a deformation invariant by NAME. Furthermore, this equality persists under étale covers because the pullback of a nef (and big) vector bundle is nef (and big), and it persists under small deformations by upper semicontinuity of cohomology. |
math/9903137 | Let MATH be the rank of MATH at the generic point. Let MATH be the blow up of the MATH-st NAME ideal of MATH. Let MATH be the blow up of the MATH-nd NAME ideal of MATH, and so on. By CITE, MATH frees all of the MATH. To finish the construction, choose a desingularization MATH. |
math/9903137 | We first make a few preliminary observations. REF follows from REF by duality CITE. Next suppose that MATH is a birational map with MATH smooth, then the equality MATH along with the projection formula shows that it is enough to prove REF result for MATH. The heart of the argument involves globalizing. Let MATH be a projective variety with a point MATH such that MATH. Choose presentations MATH for each MATH. The presentation matrices can be extended to matrices of regular functions in a neighbourhood of MATH. Hence after blowing up MATH (away from MATH) these matrices extend to maps MATH where MATH is an effective NAME divisor containing all the poles of the matrix entries and such that MATH. We can assume that MATH is ample, since otherwise we can replace it with MATH with MATH, where MATH is a very ample divisor avoiding MATH. Set MATH. Then by construction MATH. Let MATH be a resolution of singularities which frees the sheaves MATH. By blowing up further, we can assume that MATH dominates MATH. The key point is that we arranged that each MATH is a quotient of a direct sum of MATH; as MATH is ample, this implies that MATH is nef and big. Therefore by REF , MATH . We obtain REF for MATH since cohomology commutes with flat base change. |
math/9903137 | Consider the spectral sequence MATH . If we plot the MATH terms in the MATH-plane, then our assumptions imply that the nonzero terms are concentrated along the MATH and MATH axes. Since the abutment vanishes for MATH, we can deduce equalities MATH for MATH, and isomorphisms MATH for MATH. The vanishing of the first two local cohomologies implies that MATH is reflexive. Since it coincides with the reflexive sheaf MATH on the punctured spectrum, it coincides everywhere. |
math/9903137 | Choose a birational map MATH which frees MATH and with MATH smooth. Then MATH is nef or nef and big in accordance with our assumptions about MATH. Therefore MATH, and hence also MATH, is geometrically acyclic. There is an injection MATH and a surjection MATH which yield the inclusions after applying MATH. |
math/9903138 | We begin the proof of REF with a definition. Let MATH be a generating set for MATH. A MATH-deformation of MATH with respect to this generating set is a representation MATH of MATH into MATH such that MATH for MATH. Here, for an element MATH of MATH, we take MATH to be the matrix norm of a matrix representative of MATH. By the quasiconformal stability theorem of CITE, there exists a constant MATH so that a MATH-deformation MATH of MATH for MATH is induced by a quasiconformal homeomorphism of MATH. In particular, MATH is an isomorphism from MATH to MATH, and MATH is convex co-compact. Since MATH converges to the identity representation of MATH into MATH, we have that MATH converges to MATH for each MATH. In particular, for MATH, there exists MATH so that the restriction MATH of MATH to MATH is a MATH-deformation of MATH for MATH, and we are done. |
math/9903138 | Let MATH be a subgroup of MATH which is locally free but not free, and let MATH. We first show that MATH contains a subgroup which is locally free but not free, namely the intersection MATH. Every finitely generated subgroup of MATH is a finitely generated subgroup of MATH, and so is free. In particular, this shows that MATH is locally free. To see that MATH is not free, we first note that since MATH has finite index in MATH, MATH has finite index in MATH. If MATH is free, then MATH is a torsion-free group containing a free subgroup of finite index, and so by REF and CITE, we have that MATH is free, a contradiction. Hence, MATH contains the subgroup MATH which is locally free but not free. Since MATH is a subgroup of MATH, we see that MATH contains a subgroup which is locally free but not free. |
math/9903138 | We begin the proof of REF with a definition. For a Kleinian group MATH, define the commensurability subgroup MATH of MATH in MATH to be MATH . Note that MATH. Moreover, if MATH and MATH are commensurable, then MATH. Suppose there exist infinitely many manifolds MATH, MATH, in MATH which are commensurable. Since MATH and MATH are commensurable, we have that MATH for all MATH, MATH. Set MATH. Though we do not give a precise definition here, we note that there exists a special class of Kleinian groups, the arithmetic Kleinian groups, which roughly are Kleinian groups defined by number theory. For a detailed discussion of arithmeticity, we refer the interested Reader to the forthcoming book of CITE. For the purposes of this note, it suffices to make use of a major result of CITE, see also the discussion in CITE, which states that a finite co-volume Kleinian group MATH is arithmetic if and only if MATH has infinite index in MATH. Note that arithmeticity of MATH implies that MATH is not discrete, and hence is dense in MATH. We now apply a result of CITE, which states that given a constant MATH, there exist only finitely many arithmetic Kleinian groups MATH with co-volume MATH at most MATH. In particular, only finitely many of the MATH can be arithmetic, and so there must exist some MATH for which MATH is not arithmetic. In particular, MATH is a subgroup of finite index in MATH, and so MATH is discrete. We now make use of a fact about the set of volumes of hyperbolic MATH-manifolds, namely that there is a minimum volume hyperbolic MATH-manifold, and so there exists some MATH so that MATH for all MATH. In particular, the index of MATH in MATH is bounded above by MATH for all MATH. In particular, infinitely many of the MATH have the same index in MATH. Since MATH is finitely generated, it has only finitely many subgroups of a given finite index, and so infinitely many of the MATH are equal, a contradiction. This contradiction completes the proof. |
math/9903138 | This Corollary follows immediately from REF and the fact, due to NAME, that the volumes of the manifolds in MATH are bounded by MATH; for a discussion of this fact, see for instance CITE or CITE. |
math/9903138 | Suppose there exist infinitely many elements MATH, MATH of MATH which are pairwise commensurable. Write MATH, so that MATH has finite index in both MATH and MATH. Let MATH be the commensurability subgroup of MATH. Since MATH and MATH are commensurable for each MATH, we have that MATH for MATH. Set MATH. We first show that MATH. Since MATH is a subgroup of MATH, we immediately have that MATH. For the opposite inclusion, let MATH be any element. Since MATH and MATH are commensurable, we have that MATH. Since MATH, we have that MATH, and so MATH is a non-empty closed subset of MATH invariant under MATH. Since MATH contains MATH, no smaller non-empty subset of MATH can be invariant under MATH, and so MATH. There are now two cases. In the case that MATH is not a circle in MATH, it is a standard fact, see for instance CITE, that MATH is a discrete subgroup of MATH with MATH. Since each MATH is non-elementary, there is a canonical metric of constant curvature MATH on MATH which is invariant under MATH, and so descends to the hyperbolic surfaces MATH, MATH, and MATH. By the NAME finiteness theorem CITE, MATH has finite area. Since MATH covers both MATH and MATH, both of these hyperbolic surfaces have finite area. Since there is a lower bound on the area of a hyperbolic surface, this implies that each MATH has finite index in MATH, and so MATH is finitely generated and discrete. The index MATH is bounded by the quotient of the areas MATH, and so as in the proof of REF infinitely many of the MATH must be equal. In the case that MATH is a circle in MATH, the stabilizer of MATH in MATH is conjugate to MATH, and so we are done by REF . |
math/9903138 | We begin by recalling REF from REF of a convex co-compact Kleinian group MATH which contains a subgroup which is locally free but not free. The boundary of the compact MATH-manifold MATH corresponding to MATH is a connected surface of genus two which is incompressible in MATH. Let MATH be a compact hyperbolizable acylindrical MATH-manifold whose boundary MATH is the union of a torus MATH and a surface MATH of genus two. Such manifolds are common. To construct one, begin with a compact hyperbolizable acylindrical MATH-manifold MATH whose boundary is a closed surface of genus two. Such manifolds are known to exist by work of CITE, see also CITE. Here, we are using the fact that the condition that MATH be acylindrical is equivalent to the existence of a hyperbolic structure with totally geodesic boundary on MATH. It is a result of CITE that for each element MATH in MATH, there exists a simple closed curve MATH in MATH representing MATH so that MATH is hyperbolizable, where MATH is a regular neighborhood of MATH in the interior MATH of MATH. We now consider how to choose MATH so that MATH is acylindrical. Let MATH be a meridian-longitude system on the torus MATH, as described in REF. Choose MATH so that MATH is not homotopic into MATH. Suppose there exists an essential annulus MATH in MATH so that one component of MATH lies in MATH and the other component of MATH lies in MATH. Write MATH, where MATH lies in MATH and MATH lies in MATH. Express MATH in the chosen meridian-longitude system, so that it is a MATH curve on MATH, where MATH, MATH are relatively prime. If MATH is a MATH curve on MATH, then MATH bounds a disc MATH in MATH. Then, the union MATH is a properly embedded disc in MATH whose boundary MATH is a homotopically non-trivial curve on MATH, which contradicts the incompressibility of MATH. If MATH is a MATH curve on MATH for MATH, then MATH is homotopic to MATH in MATH. However, the acylindricality of MATH implies that if MATH is homotopic into MATH, then MATH is homotopic into MATH, a contradition. To see this, we use a result of CITE, which states that if MATH is a finitely generated Kleinian group and if MATH and MATH are two components of MATH, then MATH . So, if MATH is homotopic into MATH, then MATH stabilizes a component MATH of MATH, where MATH is a realization of MATH as a Kleinian group. In particular, the fixed points of MATH lie in MATH. Since MATH is acylindrical, the components of MATH have disjoint closures, see for instance CITE. So, it cannot be that MATH, as then the two components MATH and MATH of MATH would have disjoint components with intersecting closures. So, MATH, and so MATH and hence MATH are homotopic into MATH. The acylindricality of MATH implies that there is no essential annulus in MATH with both boundary components in MATH, as then would then give an essential annulus in MATH. Also, there can be no essential annulus in MATH with both boundary components in MATH. Hence, there are no essential annuli in MATH. Let MATH be an orientation-reversing homeomorphism, and consider the MATH-manifold MATH . We now make use of NAME 's hperbolization theorem for MATH-manifolds, see CITE, which states that a compact, orientable, irreducible, atoroidal MATH-manifold with non-empty boundary is hyperbolizable. As mentioned in the Introduction, this part of the argument is standard and is included for the sake of completeness. Note that MATH is compact, as both MATH and MATH are compact, and is orientable, as MATH is orientation-reversing. Let MATH be the image of MATH in MATH. MATH is irreducible: Let MATH be an embedded MATH in MATH. First, isotope MATH so that MATH is the finite union of disjoint simple closed curves. Let MATH be an innermost curve on MATH, meaning that one of the components of MATH contains no curve in MATH. Note that MATH bounds a closed disc MATH in MATH, namely the closure in MATH of the component of MATH which contains no curve in MATH. Since the interior of MATH is disjoint from MATH, we have that MATH is contained in MATH, where either MATH or MATH. Since MATH is incompressible in MATH by assumption, MATH is a homotopically trivial curve in MATH. So, we can isotope MATH into MATH and thereby get rid of MATH. Repeating this argument for each curve in MATH in turn, working from the innermost out, we can isotope MATH into either MATH or MATH. Since both MATH for MATH and MATH are irreducible, MATH then bounds a MATH-ball in MATH and hence in MATH, and so MATH is irreducible as well. MATH is atoroidal: This argument is very similar to the argument that MATH is irreducible. Let MATH be an incompressible torus in MATH, and isotope MATH so that MATH is the finite union of disjoint simple closed curves. Again performing an innermost disc argument, we can isotope away all the curves in MATH which bound a disc in MATH. Hence, we can assume that all the curves in MATH are homotopically non-trivial curves on the torus MATH. Since both MATH and MATH are embedded surfaces in MATH, the curves in MATH are parallel on MATH. Moreover, since MATH separates MATH, there must be an even number of curves in MATH. If MATH is empty, then MATH is contained in MATH for either MATH or MATH. Since both MATH and MATH are hyperbolizable, MATH is then homotopic in MATH into MATH. If MATH is non-empty, then consider the closure MATH of a component of MATH contained in MATH. Since the curves in MATH are homotopically non-trivial on MATH, MATH is an annulus. Since MATH is incompressible in MATH, the boundary curves are homotopically non-trivial curves in MATH, and so MATH is an incompressible annulus in MATH. Since MATH is acylindrical, we can homotope MATH into MATH, and hence into MATH. Doing this for each component of MATH which lies in MATH, we can homotope all of MATH into MATH. Since MATH is atoroidal, we can homotope MATH into a toroidal component of MATH, which is also a toroidal component of MATH. Hence, MATH is atoroidal. Hence, we have a hyperbolizable MATH-manifold, namely MATH, whose boundary is a torus. The Kleinian group MATH uniformizing MATH is necessarily minimally parabolic and geometrically finite. Also, since MATH contains a quasiconformal conjugate of MATH, we see that MATH contains a convex co-compact subgroup which in turn contains a subgroup which is locally free but not free. If we now perform NAME surgery along MATH, we may combine REF to see that there exist infinitely many commensurability classes of co-closed Kleinian groups which contain a subgroup which is locally free but not free. |
math/9903139 | Let MATH be a bounded operator commuting with MATH. We begin by considering the sets MATH. Assume that MATH. First we will verify that MATH leaves invariant the set MATH with MATH, that is, the set MATH . To do this, assume by way of contradiction that MATH does not leave MATH invariant. This means that there exists some function MATH with MATH and such that the measurable set MATH has positive measure. Pick some MATH such that MATH has positive measure. The commutativity property MATH easily implies MATH for each MATH. Let MATH denote the norm on MATH. We shall reach a contradiction by computing the norm of the function in MATH in two different ways. On one hand, the hypothesis MATH and the fact that MATH on MATH imply that MATH, and so MATH . On the other hand, for the element MATH we have MATH whence MATH for each MATH, contradicting the fact that MATH. Hence, MATH leaves MATH invariant. Let us verify now that MATH leaves invariant each MATH with MATH. Consider MATH. Obviously the multiplication operator MATH also commutes with MATH and MATH. By the previous part MATH leaves MATH invariant. Since MATH we see that MATH leaves MATH invariant as well. Since MATH the set MATH whenever MATH. Thus, for MATH we have proved that MATH leaves any set MATH invariant. The assumption made at the beginning of the proof that the multiplier MATH is nonnegative can be easily disposed of. Indeed, pick any MATH such that the function MATH is positive. Obviously MATH commutes with MATH (since MATH does) and MATH. By the preceding part MATH leaves MATH, that is MATH, invariant. Finally notice that MATH. This shows that the case of the sets MATH follows immediately from the case of the sets MATH consided above. |
math/9903139 | CASE: Let MATH be a compact positive operator dominating MATH, that is, MATH holds for each MATH. Since MATH is a compact operator and MATH is a norm bounded sequence, we can extract from MATH a convergent subsequence. Without loss of generality we can assume that the sequence MATH itself converges in MATH, that is, there exists MATH such that MATH. By passing to another subsequence if necessary, we can also assume without loss of generality that MATH holds for each MATH. Letting MATH we see that MATH and clearly MATH, whence MATH for each MATH. It remains to note that MATH for each MATH. CASE: Assume that MATH is a disjoint sequence and let MATH be a subsequence of MATH. By REF , there exists a subsequence MATH of MATH (and hence of MATH) such that the pairwise disjoint sequence MATH is order bounded. Since MATH has order continuous norm, it follows that MATH in MATH; see CITE. Thus, we have shown that every subsequence of MATH has a subsequence convergent to zero, and consequently MATH in MATH. |
math/9903139 | Let MATH. If MATH is constant on a non-empty open subset of MATH, then MATH commutes with a non-zero positive rank-one operator (see CITE), and so MATH is compact-friendly. For the converse, assume that MATH is compact-friendly, and consequently there exist non-zero bounded operators MATH with MATH positive, MATH compact and such that MATH . Taking adjoints, we see that MATH . The following three properties follow in a rather straightforward way. CASE: For each MATH the support of the measure MATH is contained in the set MATH, where MATH denotes the unit mass at MATH. This claim is immediate from consideration of the identity MATH . CASE: Since MATH, it follows immediately from MATH that for each MATH the measure MATH is also supported by MATH. CASE: Pick MATH with MATH and MATH. Next, choose a non-empty open set MATH on which MATH for some MATH. Then for each MATH we have MATH. Indeed, to see this, notice that MATH . To complete the proof, assume by way of contradiction that the set MATH has an empty interior for each MATH. Then the non-empty open set MATH, chosen in REF must meet infinitely many sets MATH. Pick a sequence MATH in MATH with MATH if MATH, and let MATH for each MATH. Then MATH for each MATH. Furthermore, since each MATH is supported by the set MATH and the sequence MATH is pairwise disjoint, the sequence MATH is also disjoint. However, by REF (which is applicable since the norm in MATH is order continuous) we should have MATH, a contradiction. This completes the proof of the theorem. |
math/9903139 | It was shown in CITE that if MATH has a flat, then MATH commutes with a positive rank-one operator - and hence MATH is compact-friendly. In the converse direction, assume that MATH is compact-friendly and that, contrary to our claim, MATH is not constant on any set of positive measure. Pick three non-zero bounded operators MATH such that MATH and MATH are positive, MATH is compact and MATH . To obtain a contradiction, it will suffice (in view of REF ) to construct a sequence MATH in MATH satisfying the following properties: CASE: MATH for each MATH, CASE: MATH is a disjoint sequence, and CASE: MATH for each MATH and for some MATH. The construction of such a sequence is quite involved and will be presented in a series of lemmas below. |
math/9903139 | Consider the function MATH defined by MATH . Clearly, MATH, and the function MATH is continuous by virtue of the ``no flats" assumption about MATH. Therefore, there exists some MATH such that MATH. Since MATH, and since the sets MATH are essentially disjoint, the MATH-additivity of the norm in MATH implies that MATH . Consequently, MATH that is, MATH, as required. |
math/9903139 | Since MATH and MATH, the MATH-additivity of the norm yields MATH whence MATH. From MATH we have MATH. So, taking into account that (in view of REF ) MATH, we see that MATH . For the last inequality, note that MATH and the proof of the lemma is finished. |
math/9903139 | By REF we have MATH. This and the right inequality in REF imply that MATH. The equalities MATH and MATH imply MATH . Finally, we use the identity MATH and again the above estimate MATH to get: MATH . This implies MATH, as desired. |
math/9903139 | Since, by their definition, the vectors MATH are pairwise disjoint and have the sets MATH (which are disjoint and invariant under our operator MATH) as their support sets, we see that the vectors MATH, MATH, are also pairwise disjoint. Now recalling that MATH and using the right inequality in REF we can easily estimate MATH: MATH . This completes the proof. |
math/9903147 | We have MATH . The first term of the right-hand side is calculated as follows, MATH while the second term is calculated by REF. |
math/9903147 | The proof follows from rearranging REF: MATH . |
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