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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 wher... |
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 M... |
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... |
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 w... |
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 sequenc... |
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... |
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. W... |
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 pos... |
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... |
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 som... |
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 .... |
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 ... |
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 ther... |
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 ... |
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... |
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/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 a... |
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.... |
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 tha... |
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 larg... |
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... |
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 f... |
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 NA... |
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 ... |
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 comm... |
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,... |
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 M... |
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/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 ... |
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... |
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 s... |
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 MAT... |
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 MA... |
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 con... |
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 MAT... |
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 MAT... |
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-admis... |
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 CI... |
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 resta... |
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 fo... |
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 . T... |
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 un... |
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... |
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... |
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, th... |
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... |
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 ... |
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 app... |
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 ... |
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... |
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 lin... |
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... |
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. ... |
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 M... |
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 a... |
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 de... |
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 pr... |
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... |
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.... |
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 tha... |
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 an... |
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 t... |
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 acyli... |
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 so... |
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, th... |
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 compa... |
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 opera... |
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 . ... |
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/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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