paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9904111 | It follows from REF that MATH is an orthogonal basis of MATH, and that their quadratic norms are given by MATH . The theorem follows now immediately by setting MATH, MATH and MATH and using the explicit REF of the function MATH and the constant MATH. |
math/9904114 | Let MATH be the projection on the MATH-th factor. Let MATH and MATH be the subbundles which are generated by the image and kernel of MATH, respectively. Then MATH and MATH are contained in MATH, MATH and MATH are contained in MATH, and we have sequences of vector bundles MATH which are generically short exact. Hence, M... |
math/9904114 | The only assertion that requires proof is the final one. Suppose that MATH is poly-stable and that MATH is a proper stable NAME subbundle of MATH with MATH. By semi-stability of MATH the bundle called MATH in the proof of REF must then satisfy MATH and, since MATH, it follows by stability of MATH that MATH. But MATH is... |
math/9904114 | We give the proof in case of MATH-representations, the MATH case being completely analogous. Let MATH be the poly-stable NAME bundle of the form REF corresponding to MATH, as we already noticed MATH. Without loss of generality we can assume that MATH. In this case MATH, as otherwise MATH would be MATH-invariant, and th... |
math/9904114 | Let MATH be a NAME bundle of the form REF which represents a critical point of MATH. MATH comes from the standard representation of MATH on MATH and the infinitesimal gauge transformation MATH which produces the decomposition MATH is fibrewise in MATH. Hence each of the bundles MATH is of the form MATH where MATH for M... |
math/9904114 | This is analogous to the proof of REF. However, in this case the infinitesimal gauge transformation MATH which produces the decomposition MATH of the NAME bundle of the form REF belongs to MATH, that is, it is fibrewise in MATH. Thus there are only two possibilities: either MATH and MATH with MATH, that is, MATH here w... |
math/9904114 | Let MATH be a reducible NAME bundle of the form REF which is a local minimum of MATH. Consider MATH as the space of solutions MATH to NAME 's equations modulo MATH gauge equivalence. First consider the case MATH. Then, by poly-stability, MATH, and MATH and MATH are poly-stable vector bundles. On the other hand, it is c... |
math/9904114 | Immediate from REF . |
math/9904114 | First consider the case MATH. To see that MATH is connected, consider the continuous map MATH where MATH denotes the moduli space of rank MATH poly-stable vector bundles with fixed trivial determinant bundle. From REF we see that this is surjective and, since MATH and MATH are connected, that MATH is connected. Next co... |
math/9904114 | Follows from REF . |
math/9904114 | From REF it follows in particular that MATH is isomorphic to the moduli space of rank MATH, degree MATH poly-stable vector bundles. Since this space is connected, the result is proved. |
math/9904114 | We have to prove that MATH is stable if and only if MATH is. From REF we know that stability of MATH is equivalent to stability of the MATH-bundle MATH. Thus, all we need to prove is that MATH is stable if and only if MATH is. Because stability is unaffected by tensoring with a line bundle, we can equally well prove th... |
math/9904114 | The only thing there is to remark is that any MATH, of the form given above, is stable. But, MATH is the only MATH-invariant subbundle so, this is obvious. |
math/9904114 | Suppose we have a critical point of the type described in MATH of REF, with MATH. Then, MATH is MATH-invariant and therefore, MATH is semi-stable, but not stable. Since we are considering the moduli space of poly-stable NAME bundles, MATH decomposes as a direct sum of rank MATH . NAME bundles of degree MATH. The only s... |
math/9904115 | It is straightforward to translate the proof of CITE to this setting. |
math/9904115 | With MATH as in REF , let MATH be the ideal in MATH generated by MATH . Define MATH and MATH, where MATH is the canonical projection. Obviously MATH is a NAME representation which generates MATH, and it is NAME covariant because MATH. If MATH is another NAME covariant NAME representation, then the homomorphism MATH of ... |
math/9904115 | Let MATH be universal for NAME representations of MATH, and define MATH and MATH. Since MATH is a NAME representation of MATH, we get a homomorphism MATH. To construct the inverse of MATH, let MATH be the universal NAME representation of MATH in MATH, and fix MATH. By CITE, there is a linear map MATH which satisfies MA... |
math/9904115 | Let MATH be an approximate identity for MATH. By REF, MATH is an approximate identity for MATH, and the result follows. |
math/9904115 | Let MATH and MATH. If MATH and MATH, then MATH and hence the product MATH belongs to MATH. Letting MATH vary over an approximate identity for MATH, this product converges in norm to MATH, so the set of products MATH has dense linear span in MATH. Hence to see that multiplication induces an isomorphism MATH, it suffices... |
math/9904115 | Since each fiber MATH is essential, any vector MATH can be written in the form MATH with MATH and MATH. But then MATH, and since elements of the form MATH generate MATH as a MATH-algebra, the result follows from the calculations MATH and MATH . |
math/9904115 | Taking MATH in REF, shows that MATH converges strictly to the identity in MATH; that is, MATH is nondegenerate. For REF of a NAME crossed product, suppose MATH is a nondegenerate representation of MATH; we must show that MATH is a NAME covariant representation of MATH. First note that MATH is nondegenerate since MATH a... |
math/9904115 | CASE: The uniqueness of MATH is obvious. By CITE, there is a unital homomorphism MATH determined by MATH . Let MATH be the isometry which satisfies MATH (see the proof of CITE), and define MATH . Then MATH is a homomorphism, and MATH vanishes on MATH. If MATH and MATH, then for any MATH we have MATH giving REF. Since M... |
math/9904115 | If MATH is a set of pairs MATH consisting of maps MATH and MATH, then MATH is a covariant representation of MATH if and only if each MATH is. The main point here is that the value of MATH on an element of MATH of the form MATH is MATH. Suppose MATH is a nondegenerate covariant representation on a separable NAME space M... |
math/9904115 | We follow CITE. Let MATH be a faithful nondegenerate representation MATH of MATH such that MATH is a covariant representation of MATH, and let MATH be a unitary representation of MATH whose integrated form MATH is faithful on MATH. We claim that MATH is a covariant representation of MATH. Most of the verifications are ... |
math/9904115 | Since MATH is a NAME representation, so is MATH. Let MATH. The range of MATH is MATH . Hence for any MATH, the range of MATH is MATH which is MATH if MATH, and is zero otherwise. If MATH is faithful then so is MATH; since MATH is isometric, this implies that MATH is isometric. |
math/9904115 | Fix MATH. Since MATH is NAME covariant and MATH, CITE gives MATH. But MATH is essential, so the reverse inclusion holds as well, and since MATH is precisely the range of MATH, we deduce that MATH is constant in MATH. Since MATH for all MATH, this implies that MATH is NAME covariant. |
math/9904115 | CASE: See CITE. For the continuity assertion, suppose MATH strictly in MATH, MATH, and MATH. There exists MATH and MATH such that MATH, and then MATH . CASE: Both MATH and MATH are the projection onto MATH. CASE: If MATH and MATH, then MATH since both sides of REF are supported on MATH, this implies that MATH. By REF ,... |
math/9904115 | CITE If MATH is NAME covariant, then MATH where the last equality uses REF . Conversely, suppose REF holds for all compact MATH and MATH. If MATH strictly, then MATH where the convergence is in the strong operator topology. Hence MATH for every MATH. Letting MATH strictly shows that MATH is NAME covariant. |
math/9904115 | Suppose MATH, MATH, MATH, and MATH. If MATH is a total order then either MATH or MATH; either way MATH is compact. If the left action of MATH on each fiber MATH is by compact operators, then by CITE, both MATH and MATH are compact. |
math/9904115 | By checking on an operator MATH, one verifies that MATH, and the result follows easily from this. |
math/9904115 | Express MATH with MATH and MATH; similarly, express MATH with MATH and MATH. Since MATH is NAME covariant, MATH is zero if MATH, and otherwise MATH where MATH. Since MATH is compact it can be approximated in norm by a finite sum of operators MATH with MATH, and hence MATH can be approximated by finite sums of the form ... |
math/9904115 | The proof is formally identical to that of CITE. |
math/9904115 | The proof is formally identical to that of CITE, except that in REF one must also note that MATH since MATH and MATH generates MATH. |
math/9904115 | Let MATH be a nondegenerate representation of MATH on a NAME space MATH, and let MATH be the NAME representation of MATH. By REF , MATH is a NAME representation of MATH. Since MATH is nondegenerate, so is MATH; since MATH is nondegenerate, MATH is as well. The previous Proposition thus gives a covariant representation ... |
math/9904115 | Let MATH be a faithful representation of MATH on a NAME space MATH such that MATH is a covariant representation of MATH. By REF , MATH is a NAME representation of MATH, so we can take MATH to be the restriction of MATH to MATH. Suppose MATH is a (nondegenerate) NAME representation of MATH. REF gives us a covariant repr... |
math/9904115 | Let MATH be a faithful nondegenerate representation of MATH on a NAME space MATH, let MATH be the NAME representation of MATH, and let MATH; by REF , MATH is a NAME representation of MATH on MATH. We claim that MATH is faithful. Since MATH is the orthogonal projection of MATH onto MATH (see the proof of REF ), each pro... |
math/9904115 | The proof, based on REF , is identical in form to the proof of CITE. |
math/9904115 | Since MATH is a decomposition of the identity in MATH, and since MATH is a unital representation of MATH, the projections MATH decompose the identity operator. By REF , MATH commutes with each MATH, and thus MATH . Fix an initial segment MATH, and let MATH. By REF , MATH so it suffices to show that MATH . Let MATH . Si... |
math/9904115 | CASE: Since MATH is compactly aligned, the spanning REF holds. Since MATH is continuous and maps onto MATH, we deduce that finite sums MATH in which MATH for all MATH are dense in MATH. It therefore suffices to fix such a MATH and show that MATH. Let MATH be a faithful nondegenerate representation of MATH such that MAT... |
math/9904115 | Denote by MATH the orthogonal projection of MATH onto MATH. Since the MATH's are mutually orthogonal, the formula MATH defines a completely positive projection of norm one which is faithful on positive operators. We claim that MATH . Since MATH is compactly aligned the spanning REF holds, and hence REF follows from REF... |
math/9904115 | Suppose MATH is faithful. By REF , MATH is faithful on positive elements, hence so is MATH; that is, MATH is amenable. Since MATH satisfies REF (see the proof of necessity of REF), the converse follows from REF . |
math/9904115 | Our proof is essentially that of CITE, suitably modified to handle NAME bimodules. The homomorphism MATH induces a coaction MATH of MATH on MATH, and hence a conditional expectation MATH of MATH onto the fixed-point algebra MATH, such that MATH . Since MATH is amenable, MATH is faithful on positive elements. Let MATH b... |
math/9904115 | The group MATH is amenable, and by CITE the canonical map MATH satisfies REF. |
math/9904115 | As in the proof of REF , MATH is nondegenerate. We verify the obvious analogues of REF , and REF . For REF , let MATH be a nondegenerate representation of MATH on a NAME space MATH, let MATH, and let MATH; we must show that MATH is a NAME covariant representation of MATH. Exactly as in the proof of REF , MATH is a NAME... |
math/9904115 | Let MATH be the canonical homomorphism of MATH onto MATH. By CITE, MATH satisfies the hypotheses of REF ; since MATH is compactly aligned, the system MATH is therefore amenable. Identifying MATH with MATH as in the previous Proposition and defining MATH, the initial projection MATH is precisely MATH, and the result fol... |
math/9904115 | Let MATH be a faithful nondegenerate representation of MATH. Then MATH is a partial isometric MATH-representation of MATH which satisfies REF, and applying MATH we see that MATH is as well. Since MATH for every MATH, MATH also satisfies REF, and is hence bicovariant. Since MATH generates MATH linearly and MATH generate... |
math/9904115 | MATH is universal if and only if the representation MATH of MATH is faithful. By REF , this occurs if and only if MATH acts faithfully on the range of MATH whenever MATH, and the result follows from CITE. |
math/9904115 | The first statement is obvious. In the left-invariant partial order on MATH, two elements MATH have a common upper bound if and only if one is an initial word of the other, and then the least upper bound is the longer of the two words. We will show that REF holds if and only if MATH of course a similar statement holds ... |
math/9904115 | Suppose each MATH is nonzero. To see that MATH is universal, we apply REF . If MATH, MATH, and MATH, , MATH, MATH, , MATH, then we can choose MATH such that none of the multi-indices MATH begins with MATH, and none of the multi-indices MATH ends with MATH. Then MATH is nonzero since MATH. Hence MATH is universal. Now d... |
math/9904116 | The Proposition can be proved by modifying the standard argument of, for example, CITE. Briefly, let MATH be a collection of NAME representations on NAME space which are cyclic (that is, generate a MATH-algebra which admits a cyclic vector), and such that every cyclic NAME representation is unitarily equivalent to an e... |
math/9904116 | CASE: We use REF to calculate MATH noting that MATH since MATH is abelian. CASE: Take MATH in REF to see that the linear span of monomials of the form MATH is closed under multiplication, and hence a MATH-algebra. CASE: For each MATH the map MATH is a NAME representation, and hence integrates to an endomorphism MATH of... |
math/9904116 | For each MATH, there is a unique homomorphism MATH such that MATH for all MATH. Let MATH be an orthonormal basis for MATH. Since MATH we deduce that MATH for all MATH, and hence there is a homomorphism MATH such that MATH for every MATH. From REF it is obvious that MATH maps MATH onto MATH. If MATH admits a nontrivial ... |
math/9904116 | Define MATH, MATH, and for each MATH let MATH be an orthonormal basis for MATH. Suppose MATH; we will eventually take MATH. By REF we have MATH and by REF MATH so it suffices to find a unit vector MATH such that MATH . Let MATH be an enumeration of these pairs. We claim that for each MATH there is a unit vector MATH su... |
math/9904116 | Suppose MATH is a MATH-algebra and MATH is a nonzero homomorphism. We will show that MATH is injective, thus establishing the simplicity of MATH. It does not harm to assume that MATH is surjective, so that MATH is unital and MATH. Recall from REF that there is a faithful expectation MATH of MATH onto MATH, given by ave... |
math/9904116 | By REF , MATH admits a nontrivial NAME representation, so one direction follows from REF . For the converse, suppose MATH is not injective. Fix MATH such that MATH and MATH. We claim that MATH generates a proper ideal MATH, so that MATH is not simple. Let MATH be the distinguished NAME representation defined in REF . S... |
math/9904116 | Fix MATH and MATH. Our first goal is to define MATH. In any expression of MATH as an ordered sum MATH of basis elements we have MATH. For each of these finitely-many ordered MATH-tuples MATH, it is easy to see that there is a unique way of factoring MATH as a product MATH. These factorizations can be obtained from one ... |
math/9904116 | Let MATH and MATH. Since MATH, we can define MATH by MATH . It is easy to see that MATH is a NAME representation of MATH, so there is a homomorphism MATH such that MATH. Note that for any MATH and MATH we have MATH since elements of the form MATH generate MATH, MATH is surjective. To see that MATH is injective, we cons... |
math/9904116 | By REF , MATH is the universal MATH-algebra for collections MATH of isometries satisfying MATH that is, MATH. Suppose MATH. If MATH is irrational, then MATH is injective, and MATH is simple by REF . The existence and uniqueness of MATH is elementary when MATH is rational, and repeated applications of REF give MATH. |
math/9904116 | Let MATH be a twisted unit for MATH. By replacing MATH with MATH, we can assume that each MATH is a unit vector. Then MATH determines a multiplier MATH of MATH. As in REF, let MATH be the inductive limit MATH under the embeddings MATH, and let MATH be the canonical embedding of MATH in MATH. Since tensoring on the left... |
math/9904116 | Since MATH is a twisted unit for MATH, REF applies. |
math/9904117 | The theorem follows from the following two facts, applied to MATH and MATH. CASE: Let MATH be a closed sub-manifold and MATH a smooth function. Then there exists a smooth function MATH such that MATH. Moreover, if MATH is bounded from below, MATH can be chosen to be bounded from below too, and if MATH is proper and bou... |
math/9904117 | For any MATH, on every component MATH of MATH, we have MATH . |
math/9904117 | Let MATH be a point in MATH, let MATH be the infinitesimal stabilizer of MATH, and let MATH be the element assigned to the orbit type stratum containing MATH. Let MATH be any element whose projection to MATH is MATH. Pick an open neighborhood MATH of the orbit MATH which equivariantly retracts to the orbit. The constan... |
math/9904117 | The function MATH is well defined, proper, and bounded from below. Since MATH is closed, MATH extends to a function MATH that is proper and bounded from below. (See REF in the proof of REF .) For each MATH, let MATH be an element whose projection to MATH is MATH. We choose MATH to meet the following additional requirem... |
math/9904117 | Let MATH be a minimal stratum and let MATH be the connected component of identity of MATH for MATH. Assume MATH. Then the equivariant NAME class MATH of the normal bundle to MATH is a non-zero torsion element in MATH. In fact, MATH is annihilated by the image of MATH. Alternatively, MATH, because MATH. |
math/9904117 | By using a partition of unity on MATH, the corollary can be reduced to the case where MATH is a linear space and MATH. This case follows immediately from REF when MATH is replaced by MATH with the trivial MATH-action on the second factor. |
math/9904117 | Let us first examine the case where the action is locally free near MATH. Fix a basis MATH in MATH. Then the vector fields MATH form a basis in the tangent space to the orbit at every point of a MATH-invariant neighborhood MATH of the orbit through MATH. By setting MATH we thus obtain a form defined along the orbits in... |
math/9904117 | By REF , there exists an open covering of MATH by invariant sets MATH, and on each MATH there exists an invariant one-form MATH such that MATH for all MATH. Let MATH be a partition of unity subordinate to this covering, with MATH supported in MATH for each MATH. Define MATH. Then MATH on MATH for all MATH. |
math/9904117 | Consider the new abstract moment map MATH. By REF , there exists an invariant neighborhood MATH of MATH in MATH and a MATH-invariant one-form MATH on MATH such that MATH on MATH. Let MATH; then MATH on MATH. |
math/9904117 | Pick an element MATH whose restriction to MATH is equal to MATH, and apply REF to the abstract moment map MATH. |
math/9904117 | Take a tubular neighborhood of MATH which retracts to MATH, and apply REF to each of its connected components. |
math/9904117 | It is clear by definition that MATH if MATH is Hamiltonian. Conversely, assume that MATH. Then there exists a MATH-equivariant equivariantly closed two-form MATH such that the assignment of MATH is also MATH. The difference MATH is an abstract moment map with the zero assignment. By REF , MATH is exact and therefore Ha... |
math/9904117 | The exactness of the left column is a particular case of a more general fact, that the cohomology of the basic NAME complex of MATH is equal to MATH, if MATH is compact or, more generally, if the MATH-action is proper, even when the action is not free. This result, due to CITE, is similar to the NAME theorem and can be... |
math/9904117 | First note that by adding, if necessary, an additional copy of MATH (with the trivial MATH-action) to MATH, we can always make MATH even. There exists a MATH-invariant complex structure on the vector space MATH; fix one. We obtain a representation MATH of MATH on MATH with weights MATH. The infinitesimal action of the ... |
math/9904117 | By REF it suffices to prove REF for the cohomology of the complex MATH. The first part of the theorem follows from the fact that if MATH then MATH. Therefore, the longest possible tuple MATH of distinct strata has MATH. Thus MATH for MATH. When MATH, the maximal stratum MATH is the open dense stratum in MATH, on which ... |
math/9904117 | Denote by MATH the tuple MATH in which each MATH occurs MATH times and the strata MATH are ordered and distinct: MATH. Denote by MATH the number of MATH's such that MATH; call this number the fatness of the tuple. As is easy to check, the fatness of a MATH-tuple is no greater than MATH. For each integer MATH let MATH b... |
math/9904117 | The theorem is an immediate consequence of the fact that the sequence of complexes MATH is exact. To prove the exactness of REF, note that the space MATH is the kernel of the restriction map MATH by its definition. To see that the restriction map is onto, note that every cochain MATH in MATH can be extended to a cochai... |
math/9904117 | The short exact sequence REF naturally induces a short exact sequence of complexes, MATH and hence the long exact sequence REF in cohomology. |
math/9904118 | Suppose that MATH is a local biholomorphic change of coordinates with MATH. Since REF is independent of the choice of defining function we can choose MATH as a defining function for MATH. By the chain rule we have that MATH . Now the only crucial point is that the matrix on the right hand side has holomorphic entries. ... |
math/9904124 | If MATH, this is CITE Hence we will assume that MATH, and fix some MATH for which REF on MATH hold true. Let MATH be a blowing up, such that MATH is non-singular, MATH a normal crossing divisor and such that, for MATH, the image of MATH is an invertible sheaf, isomorphic to MATH over MATH. So MATH, for some divisor MAT... |
math/9904124 | By REF it is sufficient to show, that for some MATH is generated by global sections, or that MATH is ample. By definition of nef, this hold true, if the sheaf MATH is nef. To this aim see REF we can replace MATH by a covering, unramified in MATH, and MATH by the pullback family. Thereby we may assume that MATH for an i... |
math/9904124 | If MATH, this holds true with MATH replaced by the smaller sheaf MATH, since MATH . If MATH, that is, if MATH is semi-stable, then MATH. By CITE, one has an inclusion MATH, hence MATH. |
math/9904124 | Choose a non-singular projective compactification MATH of MATH such that MATH extends to MATH. REF hold true for MATH. By REF the degree of MATH is smaller than a constant depending on MATH and MATH. There exists a finite covering MATH and MATH which induces MATH . We may assume, in addition, that MATH has a semi-stabl... |
math/9904127 | Assume first that MATH is invariant under gauge automorphisms MATH. Since the representation MATH of the gauge group on MATH, given by MATH, is unitary, the MATH also form an orthonormal basis in MATH. Since the endomorphism associated with a NAME space of isometries as in REF is independent of the choice of an orthono... |
math/9904127 | Since MATH commutes with MATH, the implementer MATH is even: MATH . This implies that MATH (see REF) so that, by REF, MATH . |
math/9904127 | If MATH is an antisymmetric operator from MATH to MATH of finite rank, then one readily verifies that MATH (here the expressions of the form MATH are defined by expanding MATH with MATH, and by setting MATH). Approximating MATH by such finite rank operators in the NAME - NAME norm (compare CITE), one finds that MATH be... |
math/9904127 | The subspace MATH is invariant under MATH because MATH commutes with the components of MATH (compare the proof of REF ). Since MATH is an orthonormal basis in MATH, it follows from the canonical anticommutation relations that MATH . Similarly, MATH commutes with the basis projection MATH (compare REF) and leaves MATH i... |
math/9904127 | Assume that MATH is localized in MATH. Let MATH be the standard basis in MATH, let MATH be a component of the complement of MATH, and let MATH with MATH. Then MATH is gauge invariant, and MATH is an observable localized in MATH. Since MATH is localized in MATH, one has MATH. Since the MATH are linearly independent, it ... |
math/9904127 | The rather lengthy estimates of the NAME - NAME norms of these commutators, which are essentially due to NAME, can be found in CITE. |
math/9904127 | Let MATH be the orthogonal projection onto MATH, and let MATH as in REF. Then MATH and MATH commute with MATH because MATH and MATH do so. Therefore the positive part MATH of MATH and the operator MATH defined in REF also commute with MATH. It follows that MATH and MATH commute with MATH as well. Arguing as in the proo... |
math/9904127 | MATH is invariant under MATH because MATH is invariant and because MATH commutes with MATH (see the proof of REF ). The assertion hence follows from REF . |
math/9904128 | MATH where MATH ranges over the MATH sub-matrices of MATH of the form MATH for some MATH. Hence MATH . |
math/9904128 | Apply REF to the polynomial MATH from REF . |
math/9904128 | Let MATH and let MATH where MATH is the MATH matrix obtained by deleting the MATH-th row and the MATH-th column of MATH. Then, by multi-linearity of the determinant, MATH hence MATH . Therefore, MATH . |
math/9904128 | Assume without loss of generality that the first MATH lines of MATH are independent. Let MATH, , MATH, , MATH be the sub-matrices obtained from MATH by deleting the last line and the MATH-th column. Then we can scale MATH in such a way that MATH . We have MATH. By reordering rows and columns, we obtain for each MATH th... |
math/9904128 | REF implies MATH . We claim that MATH. Indeed, MATH and MATH can be obtained by solving systems of linear equations with coefficients in MATH, thus MATH. Also, MATH so MATH as claimed. By REF. Hence, by REF , MATH . |
math/9904128 | According to REF , MATH . Also, MATH and according to REF MATH and hence MATH. On the other hand, MATH . Hence MATH . |
math/9904128 | Let MATH be the sub-matrix of MATH obtained by deleting the MATH-th row and the MATH-th column. By NAME 's rule, MATH. Therefore we should define the degree MATH morphism: MATH . Then by REF , MATH . |
math/9904128 | We apply REF to the system: MATH with MATH to obtain MATH . We can bound MATH and MATH. We can apply REF to the map MATH to get MATH. REF implies MATH, hence: MATH and by REF , MATH . Thus, we can estimate that MATH . |
math/9904128 | By definition of the norm, MATH. By REF , we have: MATH . Knowing that MATH, we can use REF to deduce that MATH . According to REF , MATH where MATH is a universal constant. Thus, MATH where MATH is a universal constant. |
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