paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9903052 | We verify: MATH . |
math/9903052 | Since MATH can be viewed as the kernel of the identity map which obviously commutes with contractions and with the differential, this follows from REF . |
math/9903052 | Under the pairing of MATH with MATH, the projection MATH becomes dual to the projection map MATH induced by the inclusion of the origin. Let MATH be the standard NAME homotopy operator, so that MATH . Let MATH be minus the dual operator to MATH. The Proposition follows by taking the dual of REF. |
math/9903052 | Using REF that conjugation by MATH is the MATH-action on the NAME algebra, and that the element MATH is invariant, we find MATH . |
math/9903052 | The NAME differential is MATH . Since MATH is MATH-invariant, MATH. We have MATH and MATH . Hence MATH . This result combines with the expression for MATH from REF to MATH which is indeed dual to MATH. The calculation for the contractions is simpler: MATH which is dual to MATH. |
math/9903052 | This follows from REF since MATH is the kernel of the identity map. |
math/9903052 | Since MATH is the kernel of the multiplication map on MATH, it suffices to show that MATH . Write MATH and similarly for MATH. We calculate: MATH . For the last equality we used that MATH and for the third equality we used that MATH if MATH are elements of a commutative super-algebra. |
math/9903052 | The properties MATH and MATH is obvious. Let us show that MATH is a chain map. For clarity we denote the differential on MATH by MATH and the differential on MATH by MATH. We want to compare the expression for MATH obtained in REF to MATH . Using the invariance property MATH we compute: MATH . It follows that MATH is g... |
math/9903052 | We have to show that on the subspace of basic elements MATH, MATH . Since basic elements are in particular horizontal, the operator MATH appearing in the definition of MATH can be replaced with MATH, which then commutes with MATH. |
math/9903052 | NAME MATH one reduces the Lemma to the case MATH and MATH. In this case the equation becomes MATH which follows immediately by writing MATH and MATH. |
math/9903052 | Recall that the set of regular elements MATH is the set of all elements whose stabilizer is a maximal torus. Since both sides depend continuously on MATH we may assume MATH. Choose any orientation of MATH. We will apply the Lemma with MATH. Let MATH denote the Pfaffian of an operator on the orthogonal complement of MAT... |
math/9903052 | By continuity it suffices to verify the equation on the open dense subset MATH of regular elements of MATH. We define REF-forms MATH by MATH . At any point MATH we have a decomposition of the tangent space into the spherical part, that is, the tangent space to the orbit MATH, and its orthogonal complement, the radial p... |
math/9903052 | For MATH let MATH be defined as MATH then MATH are integral kernels for the maps MATH. From the fact that these are chain maps intertwining contractions, one obtains MATH . Since MATH is just MATH (pull-back under the addition map MATH), MATH . Notice that MATH is an invertible element of the algebra MATH, and the same... |
math/9903052 | Consider the function MATH. The right-invariant NAME form MATH has the property MATH. Hence the right-invariant vector field satisfies MATH where MATH. Similarly MATH . From this the Lemma follows since MATH . |
math/9903052 | It is well-known (see for example, CITE, p. REF) that MATH with MATH as in the proof of the previous Lemma. Therefore MATH . Since the anti-symmetric part of MATH is equal to MATH the final result is MATH. |
math/9903052 | We begin begin by observing that at any point MATH, the tensor MATH on the left hand side of this equation takes values in MATH. This follows because MATH takes values in MATH, and the same is true for radial derivatives MATH. We calculate the spherical part of MATH in two ways. First, since partial derivatives commute... |
math/9903052 | Since MATH, REF shows that the spherical part of MATH is given by MATH . Using this Equation the proof of REF becomes parallel to that of REF - the only difference being that MATH which accounts for the term MATH on the right hand side. |
math/9903052 | As a consequence of REF , the pull-back to MATH of the radial part of MATH is equal to the radial part of MATH. Hence, combining REF we find that MATH . The equivariance property of MATH shows that MATH . Taking the sum over cyclic permutations of MATH, adding to REF, and using MATH we obtain the identity REF. |
math/9903053 | Since MATH is assumed to be compact we find a smooth partition of unity MATH such that MATH, MATH has compact support, and MATH is equal to MATH in an open neighborhood of MATH. Then one verifies easily that MATH is a well-defined extension having the desired properties which proves the existence. Now let MATH be anoth... |
math/9903053 | Let MATH then also MATH whence MATH due to REF which proves the first part. To avoid trivialities assume MATH. Considering MATH we find MATH whence MATH again due to REF . Thus the second part follows. Finally we observe that for MATH we have MATH by the NAME inequality since MATH. Hence we have MATH and similar we fin... |
math/9903053 | Choose an open neighborhood MATH of MATH such that MATH. Then let MATH be a smooth partition of unity subordinate to the cover MATH, that is, MATH and MATH. Since MATH due to MATH and REF one obtains MATH whence MATH. Setting MATH the first part follows since clearly MATH. Secondly, choose two open sets MATH such that ... |
math/9903053 | Denote by MATH the equivalence class of MATH. Then the Hermitian product of MATH and MATH is given by MATH and the canonical isomorphism to MATH is given by MATH. Since MATH this is clearly well-defined and isometric, hence injective. The surjectivity follows from REF . The second part follows directly from REF . Final... |
math/9903053 | For the first part the inclusion MATH follows from REF hence consider MATH. Assume MATH for some MATH and some MATH. Choosing some MATH with MATH in an open neighborhood of MATH one has MATH and thus by the NAME inequality one finds MATH. Then MATH shows MATH. Thus the first part follows since MATH. Secondly, consider ... |
math/9903053 | We have to determine the support of MATH. Without restriction assume MATH and consider MATH. Then MATH since MATH. Thus MATH follows. Moreover, MATH due to REF and the locality of MATH. Thus REF follows. |
math/9903053 | Assume MATH exists and let MATH be arbitrary where we may assume MATH to avoid trivialities. Choose MATH such that MATH. Then it is sufficient to show MATH since in this case MATH. Hence let MATH then we compute MATH using MATH and MATH and obtain MATH. Now MATH is local and due to REF we even have MATH. With MATH we c... |
math/9903053 | Since MATH we can find two open sets MATH such that MATH and MATH. Now we apply REF to obtain MATH. Then clearly the projectors on MATH, MATH, commute with MATH for all MATH and are clearly MATH-local operators proving the proposition. |
math/9903053 | Let MATH be faithful and let MATH for some non-empty open set MATH. Then MATH and thus MATH. This implies MATH. Now assume MATH, and assume we have found a function MATH such that MATH. Without restriction we can assume that already the lowest order of MATH is non-zero. Hence there is a non-empty open subset MATH and a... |
math/9903053 | The inclusion MATH is in general true due to REF . Hence let MATH and assume MATH to avoid trivialities. Then for all MATH we have MATH whence MATH since MATH is faithful. But this implies MATH and thus the claim follows. |
math/9903053 | Note that MATH is always understood to be extended to MATH. Assume first that MATH is faithful then MATH is represented on MATH by left multiplications which is clearly faithful. On the other hand assume that MATH is not faithful. Then MATH is a non-empty open subset due to REF . Due to REF we have MATH for all MATH. T... |
math/9903053 | Notice that MATH is indeed well-defined and clearly MATH. Then the unitary equivalence is a simple computation. |
math/9903053 | We have to show that if an arbitrary MATH-linear endomorphism MATH of MATH commutes with all MATH then it is a multiple of the identity. To this end we can use the additional canonical ring structure of MATH and in particular the `vacuum vector' MATH. Let MATH be such an endomorphism. Since any left multiplication by e... |
math/9903053 | Let MATH be a local MATH-linear operator commuting with all MATH. Since MATH is MATH-linear it is of the form MATH and clearly all operators MATH are local again. Since MATH commutes with all left multiplications by functions MATH the lowest order MATH commutes with all such left multiplication. On the other hand by NA... |
math/9903053 | For a given local operator MATH we have to construct a sequence of elements MATH such that MATH strongly. First consider a formal series of local operators MATH whose coefficients have support in one common compact set contained in some open subset MATH of MATH. By NAME 's theorem we know, that each coefficient is then... |
math/9903053 | Let MATH be a MATH-strong NAME sequence. Since MATH this implies that MATH is NAME with respect to the strong operator topology, too, and by REF convergent to some MATH. Similarly MATH converges strongly to some MATH. Thus it remains to show that MATH and MATH-strongly. But this is a simple verification using the MATH-... |
math/9903053 | Since we allowed for finitely many negative powers of MATH we notice from REF that any differential operator on MATH can be expressed as MATH with a suitably chosen MATH. Thus the claim easily follows from REF . |
math/9903053 | With the above definitions the proof is a simple computation. |
math/9903053 | It is straightforward to see that MATH is complete in the finite topology and since clearly MATH we only have to construct, for a give MATH, a sequence MATH of differential operators converging to MATH. But this is essentially NAME 's theorem: let MATH be an approximate identity then by NAME 's theorem MATH is a differ... |
math/9903053 | We shall first show the following lemma which is a particular case of the proposition: MATH is complete in the strong operator topology. Let MATH be a NAME sequence. Since MATH is complete in the MATH-adic topology we observe that for any MATH the sequence MATH is NAME and thus convergent in MATH. Thus MATH clearly def... |
math/9903054 | For a homogeneous polynomial MATH of degree MATH, the NAME identity gives MATH . Invariance of MATH yields MATH . Using an auxiliary variable MATH, MATH . By orthogonality of MATH, MATH . Equating expressions for MATH reveals equivariance: MATH . |
math/9903054 | For the change of variable MATH and MATH, the chain rule yields MATH . Since MATH, MATH . |
math/9903054 | Noting that MATH, MATH . |
math/9903056 | The first statement is immediate from the fact that the degree of a map MATH is zero if and only if the map is homotopic to a constant (NAME 's Theorem). REF can be verified using the geometric description of the action of MATH and MATH on MATH, and REF follows from the fact that the NAME characteristic is equal to the... |
math/9903056 | The action of any MATH on MATH is local, only changing MATH in a MATH-ball in MATH, and so MATH, where MATH is the restriction to MATH of the unique framing of MATH. But MATH, where MATH is the image of MATH in MATH. Evidently MATH or MATH for MATH or MATH, repectively, and REF follows. The forward implication in REF i... |
math/9903056 | REF are immediate from the analogous properties for MATH in REF . REF is proved in REF . The last property follows from REF d and the fact that MATH. |
math/9903056 | Choose any simply-connected spin MATH-manifold MATH with spin boundary MATH, for example constructed by attaching MATH-handles to MATH along an even framed link CITE. To compute MATH, we use the restriction MATH to MATH of the unique framing of MATH. First observe that MATH . Indeed MATH, where MATH can be expressed as... |
math/9903056 | Let MATH denote the oriented MATH-plane bundle over MATH which is orthogonal to MATH. Clearly MATH extends to a framing of MATH if and only if MATH has a nonvanishing section. The obstruction to finding such a section is given by the NAME class MATH. Since the projection MATH is covered by a bundle map MATH, it follows... |
math/9903056 | The existence of MATH was shown above, and the uniqueness (up to not necessarily fiber-preserving homotopy) is clear. Calculating the relative NAME number for the disk bundle MATH over MATH bounded by MATH, as in the example, we have MATH . The NAME defect is then gotten by subracting MATH. |
math/9903056 | After a homotopy, we can assume that MATH and MATH are identified under MATH. Extend each MATH to a framing of a collar neighborhood of MATH inside MATH. The result of gluing MATH and MATH together using MATH is a priori a MATH-manifold with corners, which we denote by MATH, and MATH and MATH clearly combine to give a ... |
math/9903056 | If MATH is regular, then some multiple of MATH bounds, MATH. It follows that MATH. Dividing by MATH gives MATH. For irregular MATH there exist regular covers MATH and MATH of degrees MATH and MATH such that MATH, as discussed above. Then by REF (since MATH and MATH are regular). Dividing by MATH gives MATH. |
math/9903058 | The spectral sequence in REF is derived from a double complex for computing the hyper-cohomology MATH. On the other hand, since MATH is affine, MATH for MATH for any coherent MATH. Thus the MATH consists of two adjacent non-zero rows. The result follows from standard arguments. Notice that MATH by normality. |
math/9903058 | The first statement is proven in CITE and the second one follows from the same argument, so we repeat it here. We may assume MATH and consider the trivial relation MATH. Thus all MATH and MATH. Also - MATH since MATH and MATH is a domain. So some MATH and some MATH. |
math/9903058 | Assume MATH and let MATH represent MATH. After changing MATH to MATH we may assume MATH for MATH. (Remember that the MATH are independent.) This contradicts REF , since for all relations MATH. |
math/9903058 | The first statement is a special case of a well known property of blow-ups, see for example, CITE. We prove the second statement for lack of reference. Let MATH be the relation module for the MATH, and set MATH to be the MATH module generated by MATH. Clearly MATH. Choose some MATH. We may find a MATH such that MATH fo... |
math/9903058 | Consider an affine chart MATH of MATH with MATH as above. The isomorphism in REF is given by the map MATH which is locally MATH. This is easily checked to be injective, and if MATH then it comes from a derivation MATH determined by MATH and MATH. Let MATH be the ideal sheaf of MATH in MATH and MATH the exceptional divi... |
math/9903058 | From the quotient map MATH and REF we get the following commutative diagram with surjective horizontal maps; MATH . Now MATH. For a rational singularity with MATH the ``relations among relations" are generated by independent linear ones REF . We may argue as in REF to show that the images of all MATH are in MATH. So MA... |
math/9903058 | Let MATH be the kernel of the multiplication map MATH and MATH the kernel of multiplication by MATH on MATH. Clearly MATH, but we have assumed that MATH, so MATH. Thus MATH which equals MATH by REF . |
math/9903058 | We may assume by genericity of MATH that a global section vanishes if and only if it vanishes in the chart MATH. (We know for example that MATH and MATH has support at points.) Thus we need to show that the local maps MATH map an element of the kernel of multiplication by MATH to zero if MATH. If MATH is in this kernel... |
math/9903058 | Since MATH is generic, the cokernel of MATH has support at points, so MATH. The result now follows from REF . |
math/9903058 | This follows from REF . |
math/9903058 | There is a well known exact sequence on the resolution of a normal singularity MATH (see CITE). After tensoring this sequence with MATH and applying MATH we get a commutative diagram MATH where all the maps are injective. The sheaves MATH and MATH on MATH have negative degree, so the horizontal maps are also surjective... |
math/9903058 | We have MATH by NAME - NAME. We compute MATH in a standard manner (see for example, CITE for another example). Since MATH is rational we may construct a ``computation sequence" MATH where MATH with the property MATH. If we tensor the exact sequence MATH with MATH, we may compute recursively if we know MATH. To compute ... |
math/9903058 | To prove REF it is enough to show that MATH has support at points. We claim that the isomorphism in REF induces locally a surjection MATH. Indeed if MATH is in the kernel MATH of MATH, then MATH by REF. Thus MATH in MATH and MATH is contained in the kernel of the multiplication map MATH. Consider the commutative diagra... |
math/9903058 | If the fundamental divisor is reduced, then MATH is reduced CITE. In this case MATH has support at points so MATH. From the diagram we get an exact sequence MATH, and MATH by REF . This proves the last two statements. |
math/9903058 | In this case MATH, MATH is the fundamental divisor and MATH. It follows from CITE that MATH where MATH. We claim that MATH. In CITE NAME proves that MATH. The proof actually shows that MATH for any cycle MATH with MATH for all irreducible components MATH of MATH. In particular MATH, and we may use CITE to conclude that... |
math/9903060 | CASE: If MATH is not log canonical in a neighborhood of MATH, there exists a prime divisor MATH on some extraction of MATH such that MATH and MATH. Let MATH be a closed point. Suffices to show that MATH, since MATH. Let MATH be a log resolution such that MATH and MATH are divisors on MATH. Since MATH, there exists a co... |
math/9903060 | Let MATH such that MATH. According to the observation above, MATH for every closed subvariety of MATH passing through MATH. Therefore there exists an open neighborhood MATH of MATH such that MATH is log canonical. In particular, MATH for every MATH. Therefore MATH is open and MATH has log canonical singularities on MAT... |
math/9903060 | Let MATH a nonsingular point and let MATH be the exceptional divisor on the blow-up of MATH. Then MATH, since MATH is effective. If MATH, then MATH according to REF MATH. |
math/9903060 | Since MATH is MATH-nef, the induced map MATH has connected fibers from the Connectedness Lemma. The only candidates for components of MATH are MATH and components of MATH where MATH is attained. Therefore there exists a component MATH of MATH such that MATH and MATH. Finally, MATH, hence the desired inequality. |
math/9903060 | CASE: Assume that MATH is realized as a divisor on the log resolution MATH, with MATH. MATH . Note first the following equivalences: CASE: MATH is NAME log terminal over a neighborhood of MATH in MATH iff MATH. CASE: MATH is the minimal lc center at MATH for MATH iff MATH. Therefore the implication ` REF ' is clear. Fo... |
math/9903060 | Since MATH for any log resolution MATH, we can assume that we are in the situation and notation of REF MATH. Then MATH is equivalent to MATH . Since MATH, there exists an open subset MATH such that MATH and MATH is effective on MATH. Moreover, the induced inclusion MATH factors as: MATH . In particular, the natural inc... |
math/9903060 | To check the equality MATH we choose an arbitrary prime divisor MATH, with MATH and MATH. From Example MATH, we have to show that MATH. If MATH, then MATH is log canonical over MATH, hence MATH is log canonical over MATH. But MATH, so MATH is log canonical. Hence MATH. In particular, MATH. Conversely, let MATH. After p... |
math/9903060 | There exists a fiber space MATH induced by a birational base change MATH with the following properties: CASE: MATH is nonsingular, MATH is a divisor and MATH is included in a snc divisor MATH; CASE: MATH is nonsingular and MATH is included in a snc divisor MATH; CASE: there exists an index MATH such that MATH, MATH for... |
math/9903060 | We have MATH. On the other hand, MATH. Since MATH is a contraction, we infer that MATH . Therefore MATH. Clearly MATH is MATH-exceptional. Moreover, since MATH is MATH-nef, the negativity of the birational contraction MATH implies that MATH is effective. |
math/9903060 | CASE: We have MATH. CASE: Since MATH is an irreducible component of MATH, there exists an open subset MATH such that MATH is the only lc center for MATH. Since MATH is an exceptional center, MATH, so the the Connectivity Lemma implies that MATH has connected fibers over MATH. The induced morphism MATH has thus connecte... |
math/9903060 | (compare CITE) The first two properties are formal consequences of the definition of the residue and REF MATH. As for the last statement, let MATH be the induced adjunction setting. By the NAME vanishing theorem MATH, hence the morphism MATH is surjective. Let MATH be a prime divisor on MATH and assume by contradiction... |
math/9903060 | It is a formal consequence of the Base Change Conjecture applied to MATH under the birational base change MATH. |
math/9903060 | Since MATH, MATH is an exceptional lc center and the induced morphism MATH is an extraction. By REF MATH, MATH is MATH-Cartier and MATH is a crepant extraction. Therefore MATH is also crepant. The rest is formal. |
math/9903060 | Denote by MATH the unique lc place over MATH. There exists an extraction MATH such that MATH factors through MATH and MATH satisfies the property MATH. Therefore the first two properties hold. From REF MATH and REF MATH, the last part holds for MATH. Finally, let MATH be a MATH-Cartier divisor on MATH whose support doe... |
math/9903060 | CITE We assume MATH, otherwise there is nothing to prove. Then MATH and MATH is log canonical in a neighborhood of MATH according to MATH. Moreover, MATH is normal from the NAME Lemma. There exists an effective MATH-divisor MATH passing through MATH such that MATH is an exceptional minimal lc center of MATH for MATH. F... |
math/9903060 | Let MATH. The cohomological interpretation of ampleness gives the exactness of the following sequence MATH for MATH large enough. Note that MATH by the projection formula. There exist polynomials MATH and MATH, of degrees MATH and MATH respectively, such that MATH for MATH large enough. Note that MATH since MATH is a g... |
math/9903060 | We use the notation from the previous lemma. Denote also MATH and MATH. We have to prove that the natural map of vector spaces MATH has a nontrivial kernel for MATH sufficiently large and divisible. Since MATH is nef and big on MATH, we have MATH for MATH sufficiently large and divisible. From the previous lemma, we de... |
math/9903060 | The above statement is stronger than CITE, but with the same proof, presented here for completeness. CASE: Assume first that MATH is nonsingular. REF gives a divisor MATH such that MATH does not contain MATH in its support and MATH. Then MATH satisfies the required properties at MATH. Indeed, if MATH is MATH-Cartier, M... |
math/9903060 | (compare CITE) The assumptions are invariant under blow-ups on MATH, so we can assume that MATH has normal crossing. By contradiction, there exists MATH such that MATH for every prime divisor MATH on MATH such that MATH. Let MATH be an effective exceptional divisor on MATH with coefficients less than MATH, such that MA... |
math/9903060 | Let MATH be a resolution as in REF , and let MATH be the induced map to the normalization of MATH. Since MATH is normal at MATH, we can harmlessly say MATH. Let MATH be the divisor obtained by applying REF to MATH. From REF applied to MATH and MATH, we deduce that MATH, that is MATH is not klt near MATH. Let MATH. CASE... |
math/9903060 | Set MATH. By REF , there exists an effective MATH-Cartier divisor MATH such that CASE: MATH, MATH for some small enough MATH CASE: MATH is log canonical at MATH with normalized and minimal lc center MATH at MATH CASE: MATH . We repeat the previous step for MATH and so forth, only that we apply REF for MATH instead of M... |
math/9903060 | We just need to show that Conjecture REF implies Conjecture REF. We use induction on MATH. Fix MATH and let MATH be a divisor given by Conjecture REF. We may assume that MATH, the minimal lc center of MATH at MATH, is normalized. In particular, MATH is a normal point of MATH. If MATH, then MATH. We hence assume that MA... |
math/9903060 | We first show that there exists an effective MATH-Cartier divisor MATH such that MATH where the infimum is taken after all prime divisors MATH on birational extractions MATH with MATH. Indeed, since MATH, REF gives the required divisor if MATH is a nonsingular point. If MATH is a base point free ample NAME divisor, the... |
math/9903060 | Let MATH with MATH and MATH. According to REF , we may assume that MATH is a normalized lc center, that is MATH for some neighborhood MATH of MATH. Since MATH and MATH is ample, the extension of NAME vanishing gives the surjection MATH . But MATH, where MATH is a closed subscheme of MATH. In particular, MATH . The lift... |
math/9903069 | Since MATH is surjective by definition and MATH is an isomorphism, for any MATH there exists MATH with MATH. As we described before MATH is spanned by connected sums of web diagrams whose dashed part are wheels with even spokes or uni-trivalent graphs with negative NAME characteristics. Therefore it is sufficient to sh... |
math/9903069 | Suppose for a contradiction that there exists a non-zero web diagram MATH we have MATH. Now we define a weight system MATH by MATH and MATH if MATH (and extend it linearly to MATH). This is well-defined since the set of wheels (with even spokes) are linearly independent with respect to the AS, IHX, STU and FI relations... |
math/9903075 | By definition, MATH if and only if the domain of discontinuity MATH of MATH is empty. If MATH is connected and non-empty, then MATH is a function group, which is a finitely generated Kleinian group whose domain of discontinuity contains a component invariant under the action of the group. Soma CITE shows that MATH is t... |
math/9903075 | For each point MATH of MATH, there exists a hyperplane MATH in MATH containing MATH so that the circle at infinity MATH of MATH bounds a closed disk contained entirely in a component MATH of MATH. Thus, MATH, which implies that MATH. Therefore, MATH, which in turn implies that MATH. For each point MATH of MATH, there e... |
math/9903075 | Since the interior of MATH is an open submanifold (or possibly an open sub-orbifold, in the case that MATH contains torsion) of MATH, it suffices to show that MATH is injective on the interior of MATH. As MATH is covered by MATH, it suffices to show that if MATH, then MATH is empty. Let MATH be any element of MATH. Sin... |
math/9903075 | We argue much as in the proof of REF to show that MATH is injective on MATH. Let MATH be any element of MATH. Since MATH is nicely NAME in MATH, there exists a component MATH of MATH which is a NAME domain, so that MATH and MATH is non-empty. Thus MATH is a proper subset of some component MATH of MATH. In particular, M... |
math/9903076 | The proof divides naturally into three steps. In Step MATH, we show that REF holds in the case that MATH is either a generalized web group, a degenerate group without accidental parabolics or an elementary group. In Step MATH, we show that REF holds in the case that MATH has connected limit set, using the result of REF... |
math/9903076 | NAME 's lemma guarantees that there exists a finite index, torsion-free subgroup MATH of MATH. REF implies that MATH converges strongly to MATH and that MATH converges to MATH. Since MATH for all MATH and MATH, we see that MATH converges to MATH. If MATH is not virtually abelian, then REF implies that MATH converges st... |
math/9903077 | By REF, MATH, but by NAME, antisymmetry, and REF, MATH . This shows that the lemma holds if MATH and MATH do not commute. Otherwise, MATH. |
math/9903077 | Let MATH. Then MATH . Similarly, interchanging MATH and MATH, MATH . Furthermore, MATH . Hence MATH . From this the two equations follow. |
math/9903077 | REF is the first identity of REF with MATH replacing MATH. Identity REF follows from two applications of REF and NAME: MATH . |
math/9903077 | Immediate from Identity REF . |
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