paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9902102 | By the same argument as in the minuscule case, the only identity we need to verify is the MATH identity, that is, for all MATH, MATH . As before, we interpret this equation as an equality between endomorphisms of MATH. The left hand side represents the action of the NAME operator (plus some multiple of the identity), t... |
math/9902102 | MATH is nondegenerate because MATH, (CITE), hence it is a hypersurface. We check that MATH implies MATH. Introduce the notation MATH and note that MATH. For all MATH, we have MATH. Differentiating, we obtain MATH and MATH for all MATH. Thus MATH. Now recall the identity MATH which was a consequence of the NAME identity... |
math/9902102 | Let the simple root MATH be the smallest weight of MATH. Then the highest root MATH has coefficient one on MATH, and the weights MATH of MATH are the other positive roots with this property. We write MATH. Note that MATH so that MATH . The coefficients MATH are the NAME numbers, which are negative for MATH. Now MATH fo... |
math/9902103 | One checks directly that the elements MATH, and MATH defined above lie in the kernel of the operators MATH. Since MATH commutes with MATH, the derivatives of these elements also lie in the kernel. But MATH. Hence the algebra generated by these elements lies in MATH. But its character equals the character of MATH, compa... |
math/9902103 | We have to check that NAME components of the differential polynomials MATH, and MATH have the same NAME brackets as the corresponding elements of MATH. But this automatically follows from our construction. |
math/9902103 | We follow the same strategy as in the principal case (see CITE). Introduce the functions MATH and MATH on MATH by the formulas MATH . These functions are invariant with respect to the right action of MATH, and hence descend to MATH. Next, we define the functions MATH on MATH. Define the homomorphism MATH, which sends M... |
math/9902103 | According to the results of the previous subsection, as a MATH - module, MATH, where MATH acts trivially on the second factor. Therefore the cohomology of the complex MATH, which by REF is isomorphic to MATH, equals MATH. By NAME 's lemma (see CITE), MATH. |
math/9902103 | The proof proceeds along the lines of the proof of REF. We introduce a system of coordinates on MATH and then find formulas for the action of MATH, and MATH in these coordinates. Introduce the following regular functions MATH, MATH on MATH: MATH . Straightforward computation analogous to that made in the proof of REF g... |
math/9902109 | To show the MATH-symmetry we consider the modified vertex operators associated with the root lattice MATH with MATH. Let MATH, where MATH is the group algebra generated by MATH. Define MATH where the operators MATH and MATH act on MATH as follows: MATH . The components satisfy the NAME algebra relations MATH where MATH... |
math/9902109 | It follows from definition that MATH . Observe that MATH . As an infinite series in MATH we have MATH with MATH, MATH . Plugging the expansion into the integral and invoking REF . We prove the proposition. |
math/9902109 | This follows from our vertex operator calculus of symmetric functions CITE. MATH where MATH. Taking the coefficient of MATH and using NAME REF we obtain the result. The case of MATH is proved similarly. MATH . Taking the coefficient of MATH we obtain the formula. |
math/9902109 | Let MATH, MATH, and MATH in the following computation. MATH . Note that the integrand divided by MATH is an anti-symmetric function in MATH . It follows from REF that the terms MATH (for which some MATH) make no contribution to the integral. Therefore MATH where MATH and we have used MATH . From the orthogonality of NA... |
math/9902109 | The four formulas are proved similarly. Take MATH. Observe that MATH is of weight zero, thus only MATH contributes to the summation in MATH (see REF ). Since the longest element in MATH has inversion number MATH, the sign of MATH is MATH. |
math/9902114 | This follows immediately from the kernel representation. For instance, we have for MATH thus MATH . The estimate for MATH is similar and the continuity statement is obvious. |
math/9902114 | First we show that we can differentiate under the integral in REF . From the preceding considerations, we conclude that MATH is continuously differentiable, hence we have MATH . To show that MATH it is enough to prove the estimate MATH with MATH locally independent of MATH. Then the differentiability is a consequence o... |
math/9902114 | From the preceding proposition we infer for MATH hence it suffices to check the formula for MATH and MATH. MATH is just MATH with NAME boundary conditions, MATH thus MATH where MATH denotes the NAME zeta - function. In view of the well - known formulas MATH we find MATH . To prove the result for MATH it is enough to sh... |
math/9902114 | Introducing the first order operator MATH one checks that MATH (compare CITE). Moreover, since MATH is the NAME extension of MATH, the domain of MATH is the completion of MATH with respect to the norm MATH . But since the latter is the square of the graph norm of MATH we find MATH. From CITE we have for MATH . Now let ... |
math/9902114 | CASE: Since MATH is elliptic of order MATH, we have MATH from which we reach the conclusion immediately. CASE: In view of REF we only have to prove the estimate MATH. But this is an easy consequence of REF and the proven first part of this lemma. |
math/9902114 | REF and the formula MATH which holds for MATH large enough, imply the assertion for MATH. If MATH with MATH invertible, then we conclude from MATH that the operator MATH is NAME - NAME, too. |
math/9902114 | Since MATH are independent of MATH, we have the estimate MATH with MATH locally independent of MATH. We would like to apply the formula MATH . However, as the referee pointed out to the author, the operator MATH need not be of trace class. But, in view of REF the operator MATH is trace class and the kernels of this ope... |
math/9902114 | For simplicity, throughout this proof we are going to write MATH instead of MATH. Let MATH and consider MATH. Differentiation of MATH and MATH gives MATH . Since MATH is invertible, we have MATH. Now note that MATH is independent of MATH and MATH . Now consider MATH . We find MATH and again since MATH is invertible, MA... |
math/9902114 | The resolvent expansion REF shows that the estimate REF holds for MATH, too. Then as in the first part of the proof of REF one infers that MATH is smooth and MATH . Now, since MATH we have MATH . To see this we consider MATH defined in REF . We have MATH . Since MATH is also a fundamental system of solutions of the hom... |
math/9902114 | We put MATH . We have MATH. Since MATH, we have MATH and MATH is non - trivial iff MATH (compare CITE), thus MATH iff MATH and MATH. MATH: We put MATH . It is easy to check that MATH is normalized at MATH and MATH is normalized at MATH and MATH . Using REF we reach the conclusion. MATH: We put MATH . Then MATH is norma... |
math/9902114 | By REF MATH is invertible for MATH. Hence, for MATH, the assertion follows from REF. Now, a straightforward calculation shows: MATH . For MATH we infer from CITE and REF that MATH and hence we find MATH . Moreover, from REF we infer MATH, hence MATH and REF implies MATH thus MATH . |
math/9902114 | That MATH has a meromorphic continuation is well - known. A simple way of seeing this is MATH and the right hand side is a meromorphic function in the whole plane. Moreover we have MATH . Since MATH this shows that MATH is regular at MATH. We find MATH and MATH . |
math/9902114 | In view of REF left and right hand side of the equation are entire holomorphic functions and we find MATH . Since the assertion is obviously true for MATH we reach the conclusion. |
math/9902117 | Suppose MATH with some MATH. Since MATH is NAME, there is a maximal MATH such that MATH divides each MATH. Let MATH. No torsion in MATH implies MATH, so MATH in MATH. Thus each MATH, contradicting maximality of MATH. |
math/9902117 | The skein algebra is generated by curves in the surface REF , which, except for the boundary, may be identified with slopes in MATH. These in turn can be organized as the vertices in a tessellation of the upper half space model of MATH by ideal triangles - their sides are geodesics connecting slopes of curves that cros... |
math/9902117 | Embedding MATH into MATH maps MATH onto MATH, forcing MATH. Any relation that is not a multiple of MATH would imply a non-trivial relation among the standard basis elements in MATH. Therefore, we need only express MATH in the generators MATH. This is accomplished by eliminating MATH from REF . |
math/9902117 | The curves MATH generate as in the proof of REF . Commutators and REF are a matter of direct computation. The relations imply that the set MATH spans the module, so we need only show they are linearly independent. As before, it suffices to consider the case MATH. Here, formal identification of variables gives an isomor... |
math/9902117 | The set MATH is seen to be a basis by working over MATH. |
math/9902117 | We present over MATH only to save the trouble of writing out trivial commutators; in the proof we work MATH-linearly. Given a handle decomposition of MATH, there is a generating set for MATH given in CITE. It is possible to choose the decomposition so that this set is MATH. Since MATH we can replace it with MATH. Relat... |
math/9902117 | For MATH, induct on MATH. The arbitrary case transforms to this one using an element of MATH. |
math/9902117 | Let MATH with each MATH in MATH and each MATH a distinct, non-empty link in MATH. Choose MATH that is not parallel to any MATH. Suppose MATH is a maximal complexity link occurring in the resolution of MATH into the standard basis. The resolution of each MATH lies in a distinct parallelogram, so MATH must be a vertex th... |
math/9902117 | Rotate so that MATH lies in the first coordinate and assume that each MATH. Use both conditions to force the first coordinates of MATH and MATH agree, as well as those of MATH and MATH. Agreement in the other coordinates follows easily. |
math/9902117 | Choose MATH with each MATH distinct and each MATH. Similarly, choose MATH. Suppose that MATH has maximal complexity in the resolution of MATH. Reordering if necessary, we may assume that MATH is one of MATH. If MATH appears in the resolution of any MATH other than MATH, then REF forces a term of either MATH or MATH to ... |
math/9902117 | Assume for now that MATH. Position the link so that near MATH it looks like REF , in which a number next to a strand indicates so many parallel copies. If every crossing in REF is smoothed in the vertical direction the resulting coefficient is MATH. If MATH the link should be isotoped to look like a left-to-right refle... |
math/9902117 | We use the the complexity MATH for MATH. Suppose MATH is a central element. Write it as MATH, where each MATH is distinct but all have the same complexity, and MATH consists of lower complexity terms. REF implies MATH, so MATH. Similar computations with MATH and MATH show that each MATH is a power of the outer boundary... |
math/9902117 | We use the complexity MATH for MATH. Choose MATH with MATH and each MATH distinct. Similarly choose MATH. Reordering if necessary, we may assume MATH is a maximal complexity link in the resolution of MATH. By REF , it occurs in the resolution of MATH with non-zero coefficient. By REF it does not occur in any other MATH... |
math/9902118 | Assume MATH is an embedding off MATH, and let MATH be a line not intersecting MATH. Then the restriction of MATH to MATH is base point free, hence MATH. If MATH, then MATH is a ramified double cover of MATH, contradiction the assumption that MATH is an embedding off MATH. Conversely, choose a length two subscheme MATH;... |
math/9902118 | Let MATH, MATH, be a line not intersecting MATH. By REF , the restriction of the MATH to MATH must satisfy MATH. However, it is easy to check that the only base point free system of quadrics on MATH satisfying MATH is the complete system of quadrics. Hence by REF , MATH is an embedding off MATH. |
math/9902118 | Assume to the contrary that there is a MATH-plane MATH that intersects MATH in a scheme MATH of length MATH. Note that by REF cannot intersect MATH in a scheme of positive dimension. Two conics in MATH intersect in a scheme of length MATH if and only if they have no common component. However, a pair of plane conics can... |
math/9902118 | Take coordinates MATH on MATH and let MATH be the scheme defined by the equations MATH. Clearly, MATH as schemes and it is easy to verify that MATH off of MATH, the exceptional divisor of the blow up. Now, considering a syzygy as a vector of forms, let MATH generate the linear syzygies among the MATH. Let MATH be the s... |
math/9902118 | To prove the first claim, note that because the fibers of MATH are reduced, we need only show that points are separated. Take MATH, and assume MATH. Then by REF , MATH satisfies MATH. There are then two possibilities: CASE: MATH is a MATH-ic hypersurface in MATH, and hence every line in MATH is a MATH-secant line of MA... |
math/9902118 | We construct a morphism MATH whose image is MATH. Let MATH, and push the surjection MATH down to MATH: MATH . The sheaf MATH is locally free of rank MATH, and the map is surjective as MATH maps a fiber of MATH to a linearly embedded MATH. Pulling this surjection back to MATH gives a surjection from a free rank MATH she... |
math/9902118 | Let MATH and form MATH on MATH. Embed: MATH where MATH is the inclusion. Applying MATH to the surjection MATH gives a map: MATH . Recalling MATH, there is a map: MATH where a fiber of the coherent sheaf MATH over a point MATH is isomorphic to MATH. By the above remarks, this map has rank MATH, hence gives a surjection ... |
math/9902118 | Let MATH be the exceptional divisor of this blow up. Let MATH be the image of: MATH is flat of degree MATH over MATH. Indeed, by the structure of MATH the fiber over a point in MATH is exactly the corresponding length MATH subscheme of MATH, hence MATH induces the identity morphism MATH. By the universal property of MA... |
math/9902118 | Note that MATH and MATH are isomorphic rank two vector bundles on MATH and that MATH is generated by its global sections. Hence there is a surjection MATH which induces a morphism MATH. This gives the diagrams: where MATH is the natural projection map. MATH makes both triangles commute, and so is a finite (by the secon... |
math/9902118 | The smoothness of MATH is immediate from REF . For the second claim, note that MATH maps MATH to the diagonal in MATH, which is the projectivized tangent bundle to MATH, hence smooth (by the diagonal in MATH, we mean the proper transform on the diagonal under the birational morphism MATH). Therefore, MATH is smooth as ... |
math/9902118 | Denoting tangent sheaves by MATH, it is easy to see that MATH . Let MATH be the universal quotient bundle on the exceptional divisor MATH and MATH the inclusion. Because the intersection of MATH with MATH is a scheme of length two, pulling the exact sequence CITE MATH back to MATH gives MATH ample for MATH. The sequenc... |
math/9902118 | To give a map MATH it is equivalent to give a surjection MATH for some line bundle MATH on MATH. By the above diagram, this is equivalent to a surjection MATH, which we obtain from the natural surjection MATH on MATH. As the fibers of MATH are isomorphic to MATH, it is clear that the induced map is an isomorphism. |
math/9902118 | Push the exact sequence MATH down to MATH. Because MATH is an isomorphism in a neighborhood of MATH, the map MATH is the zero map, hence there is an exact sequence on MATH: MATH where MATH. Since MATH is (very) ample, MATH for all MATH sufficiently large. Therefore there is a section of MATH that does not vanish at MAT... |
math/9902118 | Clearly, the base locus of MATH contains MATH. Note, however, that as MATH is base point free, the base locus of MATH will stabilize for MATH sufficiently large. Hence it suffices to show that if MATH is a point not in MATH, then MATH is free at MATH for all MATH. Now take MATH and MATH in REF , and use REF . |
math/9902118 | Letting MATH be an irreducible curve not contained in MATH, we have MATH for MATH by REF . Letting MATH, MATH is globally generated on MATH for MATH by MATH, hence MATH and MATH is nef. |
math/9902118 | We first show that for MATH sufficiently large MATH restricts to the complete linear system on MATH associated to the invertible sheaf MATH (Compare REF ). For this it suffices to prove MATH . Writing MATH and noting MATH: MATH . Let MATH and rewrite the right side as MATH. For MATH, MATH is a nef MATH-divisor by REF ,... |
math/9902118 | Note first that MATH is normal since it is the image under a morphism of a normal variety with reduced, connected fibers. Let MATH, where MATH. We have the normal bundle sequence: MATH . Because MATH is a fiber of the MATH-bundle MATH over MATH by REF , this sequence becomes: MATH . This sequence clearly splits and we ... |
math/9902123 | The proof is similar to CITE and so details are omitted. |
math/9902123 | First note that MATH. Therefore an easily calculation shows that MATH . Now the proof follows from CITE (see also CITE). |
math/9902123 | We will show more generally that for a link MATH, MATH is divisible by MATH, where MATH is the two-parallel of MATH with respect to the zero-framing and MATH is the number of components in MATH. We will use NAME 's bracket polynomial CITE MATH for a framed link MATH defined by MATH where MATH is the trivial knot with z... |
math/9902123 | We only prove the case where MATH is obtained from an algebraically split, framed link since the general case follows from NAME 's diagonalizing lemma CITE. Now we assume that MATH is given as described before. From the lemmas above, we easily see that MATH belongs to MATH and divisible by MATH (Note that MATH and divi... |
math/9902124 | Without loss of generality, it is sufficient to show that MATH, where MATH and MATH. Since MATH is a nonzerodivisor and MATH is nonsingular, we have MATH. The other isomorphism can be proved analogously. |
math/9902124 | Let MATH, MATH, MATH be matrices over MATH and MATH be a scalar of MATH such that MATH hold. Further, let MATH . In order to prove this lemma it is sufficient to show that the ideals MATH and MATH are equal. Suppose that MATH is an element of MATH. Then there exists a matrix MATH such that MATH. Multiplying MATH on the... |
math/9902124 | We first prove REF . Let MATH and MATH be matrices over MATH with MATH. Then we have MATH. The other relation MATH can be proved in a similar way. Next we prove REF . Suppose that MATH is a unit of MATH. We prove the following relations in order: CASE: MATH, REF MATH, REF MATH, REF MATH. CASE: The proof of REF follows ... |
math/9902124 | (Only If) Suppose that a plant MATH is stabilizable. Let MATH be a stabilizing controller of MATH. First, MATH is in MATH, since MATH and MATH. From REF in the proof of REF , if MATH, then MATH. (If) Because MATH, the ``NAME part can be proved analogously. |
math/9902124 | This proof mainly follows that of REF. If the matrix MATH itself is MATH-nonsingular, then we can select the zero matrix as MATH. Hence we assume in the following that MATH is MATH-singular. Since REF is MATH-nonsingular, there exists a full-size minor of MATH in MATH by NAME 's expansion of REF . Let MATH be such a MA... |
math/9902124 | Let MATH be the matrices over MATH as in REF . Recall that MATH denotes the MATH-module generated by rows of the matrix MATH. By REF , there exists a matrix MATH over MATH such that MATH holds for some positive integer MATH. Then we have a factorization of the matrix MATH over MATH as MATH, where all entries of the mat... |
math/9902124 | This lemma is an analogy of the result given in the proof of REF. See the proof of REF. |
math/9902124 | We prove the following relations in order: REF `` REF ," REF `` REF ," and REF `` REF ." REF `` REF ": Suppose that MATH is a stabilizing controller of the plant MATH. Then, the MATH-module MATH is obviously free. By the relation MATH in REF , we have that the MATH-module MATH is projective. By using REF , the projecti... |
math/9902124 | In the construction of the stabilizing controller in REF of the proof of REF , the denominator matrix of REF is MATH-nonsingular. Suppose that the obtained stabilizing controller is expressed as MATH with the matrices MATH and MATH over MATH such that MATH is MATH-nonsingular. Then since the relation MATH holds, every ... |
math/9902124 | Suppose that the plant MATH is stabilizable and strictly causal. Suppose further that MATH is a stabilizing controller of MATH. We employ the notation from REF of the proof of REF . Thus, MATH and MATH with MATH from REF . Let MATH. Then this MATH is again a principal ideal of MATH. Observe here that REF and its proof ... |
math/9902137 | Suppose that MATH. Then MATH, but we also have that MATH, hence MATH. This is impossible since MATH is cancellative and MATH. |
math/9902137 | If MATH is convergent, then so is all its sub-products. By REF , this means that no element can occur infinitely many times in the product. |
math/9902137 | Note that the subspace MATH is a sub-monoid of MATH, since the closure of a sub-monoid is a sub-monoid. The only non-trivial inclusion is MATH. Let, as in REF, MATH be a convergent product, with MATH any set; let MATH be the directed set of all finite subsets of MATH, let for MATH, MATH, and let MATH be the correspondi... |
math/9902137 | Since MATH is reduced, MATH is irreducible if and only if it can not be written as a finite product of non-units, all different from MATH. Thus if MATH is topologically irreducible, it is irreducible. For the converse, we need to use that MATH allows finite decimation. Suppose that MATH is irreducible, and that MATH. L... |
math/9902137 | It suffices to show that an irreducible element is topologically prime, so suppose that MATH is irreducible. Since MATH is topologically prime atomic, MATH may be written as a convergent product of topologically prime elements. We claim that this product must have only one factor. Hence, MATH is topologically prime. To... |
math/9902137 | Suppose that MATH is an atom in MATH, and that MATH . , with MATH. Then there exists MATH with MATH. Since MATH is topologically factorial, we can uniquely factor MATH into atoms: MATH . We can assume that MATH are pair-wise disjoint. Applying CITE we have that MATH . Since MATH is topologically factorial, factorisatio... |
math/9902137 | Suppose that MATH is an atom in MATH. By the previous proposition, MATH is prime. Suppose that MATH . , with MATH, and write each MATH as a convergent product MATH. MATH but on the other hand, MATH for some MATH. Write MATH, then MATH . Since MATH allows dissociation, we get that MATH and since MATH is topologically fa... |
math/9902137 | For any topologically prime element MATH, we have that MATH iff MATH for some MATH. To see this, first note that if MATH then MATH. The right hand side is a product of MATH and a convergent since MATH allows finite decimation. Conversely, if MATH then by definition of topologically primeness there is a MATH such that M... |
math/9902137 | This is a direct consequence of our assumption that MATH allows arbitrary decimation. |
math/9902137 | To show that MATH is a algebraic monoid, we must show that if MATH then MATH is in fact in MATH. So, we must show that MATH is convergent. Let MATH be a neighbourhood of MATH. Since multiplication in MATH is continuous, there is a neighbourhood MATH of MATH and a neighbourhood MATH of MATH such that MATH. Since MATH an... |
math/9902137 | Since MATH is reduced and cancellative for all MATH, we have that MATH is a submonoid of a reduced, cancellative monoid, and hence it is reduced and cancellative. Let MATH be a convergent sum in MATH, and let MATH. We want to show that MATH is convergent, that is, that the following two conditions hold: CASE: MATH conv... |
math/9902137 | The fact that MATH is discrete means that MATH consists precisely of the finitely supported maps MATH. Suppose that MATH. We put MATH, and note that since MATH, and since MATH is discrete, there is a MATH such that MATH for all MATH. We have that MATH is supported on a finite set MATH, and for each MATH, there is a MAT... |
math/9902137 | If MATH then MATH. Since MATH is irreducible in MATH, it is irreducible in MATH. If MATH, then MATH, so if MATH for MATH, then MATH is not topologically irreducible, hence REF not irreducible. We have thus shown that MATH. By REF we have that all MATH are topologically prime. A topologically prime element is prime, hen... |
math/9902137 | The underlying algebraic monoid of MATH is isomorphic to MATH, and MATH is an order ideal. |
math/9902137 | Each element MATH can be written uniquely as MATH; this sum is convergent with respect to the topology of point-wise convergence, and by construction, its image under MATH is also convergent. |
math/9902137 | We define MATH where we have used the fact that MATH REF . This map is obviously an injective homomorphism of algebraic monoids. Since MATH if and only if MATH if and only if MATH if and only if MATH we get that the image of MATH is exactly MATH, thus that MATH as algebraic monoids. Henceforth, we regard MATH as a map ... |
math/9902142 | Let MATH denote the adiabatic limit of MATH with similar notation for MATH. Fix MATH and let MATH. Since MATH is isomorphic to the kernel of MATH on MATH, which in turn is isomorphic to MATH, REF implies that MATH . Thus the dimension of the intersection of MATH with MATH is independent of MATH. This implies that MATH.... |
math/9902142 | The difference MATH is a smoothing operator and using REF the difference MATH can be represented as MATH . All entries in the formula presented above are smoothing operators due to the fact that MATH has a smooth kernel. |
math/9902147 | The equalities involving MATH in REF follow from REF since MATH . The other equalities in REF are obtained by taking adjoints, and REF is a direct consequence of REF . |
math/9902147 | Set MATH. Then, by REF and the proof of REF, the operators MATH on MATH define bounded operators on MATH. This easily yields REF . Now REF follows from REF since MATH and MATH are of order zero, and MATH and MATH vanish on MATH and preserve each MATH. |
math/9902147 | For MATH we have MATH . Then MATH by REF, and the result follows. |
math/9902147 | Take any MATH. For MATH, there exists some MATH such that MATH. So MATH is defined in MATH. But MATH because MATH. |
math/9902147 | For MATH, we have MATH . |
math/9902147 | Since the orthogonal projection MATH preserves smoothness on MATH, the result follows by REF. |
math/9902147 | Take any MATH, and fix some MATH with MATH, where MATH. We know that MATH by REF . On the other hand, if MATH and MATH denote the extensions of MATH and MATH to continuous maps MATH, we have MATH where MATH, and MATH . Hence MATH by REF . Therefore the result follows by REF . |
math/9902147 | The inclusion MATH of the first equality is obvious, and the inclusion MATH of the second equality follows from REF -REF . To prove the inclusion MATH of the first equality, by REF it is enough to prove that MATH for all MATH. This obviously holds if we prove MATH for every such a MATH since the inclusion MATH of the s... |
math/9902147 | This follows from REF . |
math/9902147 | This follows from REF -REF . |
math/9902147 | This follows from REF -REF . |
math/9902147 | The result follows from REF with easy arguments (see REF). |
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