paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0002030 | We begin by selecting a grading MATH of MATH which preserves MATH, and recalling that by CITE, the group MATH acts simply transitively on the set of all such gradings MATH. Next, to compute how this transitive action changes the associated MATH representations MATH, write MATH relative to MATH, and observe that MATH . ... |
math/0002030 | By virtue of the previous remarks, it will suffice to check the two sub-cases enumerated above. To verify the assertion for REF , note that MATH is the weight REFp eigenspace of MATH, and recall that by definition we must have MATH . Consequently, the commutativity relation MATH implies that MATH and hence MATH preserv... |
math/0002030 | There are two key steps: CASE: MATH . Show the monodromy weight filtration MATH is self-dual with respect to MATH, that is, MATH. CASE: MATH . Via semisimplicity, assume the pair MATH defines a MATH representation MATH of highest weight MATH. Imposing self-duality, it then follows that MATH unless MATH, hence MATH is a... |
math/0002030 | The details may be found in REF. However, the idea of the proof is relatively simple: An element MATH belongs to the subspace MATH if and only if MATH . To verify the decomposition MATH, we observe that CASE: MATH . As a vector space, MATH. CASE: MATH . By definition, MATH and hence MATH. |
math/0002030 | To prove that MATH, one simply checks that MATH acts as scalar multiplication by MATH on MATH, and hence MATH. To verify equation MATH, we note that if MATH, MATH denotes the grading of MATH which acts as multiplication by MATH on MATH then MATH since MATH is defined over MATH, and hence MATH . Upon transferring these ... |
math/0002030 | For simplicity of exposition, we shall first prove the result under the additional assumption that our limiting mixed NAME structure MATH is split over MATH. Having made this assumption, it then follows from the work of REF that: CASE: MATH . The associate gradings MATH and MATH are defined over MATH. CASE: MATH . The ... |
math/0002030 | Let MATH and MATH denote the gradings associated to the split mixed NAME structure MATH via the methods of REF. Then, a quick review of the proof of REF shows that upon setting CASE: MATH. CASE: MATH. we have both MATH and MATH . In particular, since MATH and MATH is a grading of MATH which is defined over MATH: MATH .... |
math/0002033 | As it was said in REF, we have MATH in some neighbourhood of MATH where MATH is some MATH-parametric system. If MATH then setting MATH, we get from REF the equality MATH, with MATH, that is, REF is true. Let us apply the induction on MATH. Suppose that the statement is true for MATH. Since MATH implies MATH and REF tur... |
math/0002033 | The case MATH is trivial. For MATH we obtain REF from REF by virtue of the uniqueness of NAME 's expansion for MATH. |
math/0002033 | As it was said in REF there exists a conservative scattering system MATH such that MATH in MATH. If MATH then MATH, thus the statement is valid with MATH. If MATH then MATH and REF holds. Let us show that MATH. Indeed, for the linear operator-valued function MATH corresponding to a conservative scattering system MATH w... |
math/0002033 | It follows from REF that MATH is the minimal subspace in MATH containing MATH and reducing MATH for all MATH. By the assumption, MATH. If MATH then by REF the subspace MATH contains MATH and reduces MATH for all MATH, that contradicts to the assumption. Hence, MATH is closely connected. Analogously, MATH is closely con... |
math/0002033 | For any MATH is a unitary operator, and from REF and the definition of MATH we have MATH, and MATH is an isometry, thus MATH is unitary. Hence, MATH is a conservative scattering system. For any MATH and MATH is an isometry, thus MATH is unitary. Hence, MATH is a conservative scattering system. It is easy to see now tha... |
math/0002033 | The part ``if" is clear since in this case by REF there are conservative scattering systems MATH and MATH such that MATH, and by REF we have MATH with functions MATH and MATH belonging to the corresponding classes MATH. For the proof of the part ``only if" let us assume that REF holds with MATH. Let MATH be some conser... |
math/0002035 | Evaluation of sections determines a surjective map MATH of vector bundles on MATH. The corresponding NAME complex takes the form: MATH NAME through by MATH, and applying the hypothesis with MATH as one chases through the resulting complex, one sees first of all that the multiplication map MATH is surjective. Next tenso... |
math/0002035 | In fact, thanks to NAME vanishing, REF 's Lemma applies to MATH as soon as MATH. |
math/0002035 | The first statement follows easily from the definition. For REF , note that MATH for a suitable MATH-divisor MATH numerically equivalent to MATH. This being said, REF is a consequence of the NAME Vanishing theorem whereas REF follows from REF . |
math/0002035 | REF is a slight variation of NAME 's vanishing theorem in its analytic form. If MATH is ample, the result is true with MATH as well as with MATH (the latter case being obtained by replacing MATH with MATH where MATH is an arbitrary smooth metric on MATH; the defect of positivity of MATH can be compensated by the strict... |
math/0002035 | The strong version of the NAME MATH extension theorem proved by CITE shows that for every singular hermitian line bundle MATH with nonnegative curvature and every smooth complete intersection subvariety MATH (actually, the hypothesis that MATH is a complete intersection could probably be removed), there exists a suffic... |
math/0002035 | Since the exceptional set MATH is the divisor where the derivative MATH drops rank, one sees that MATH. Similarly, MATH . Therefore MATH and this has normal crossing support since MATH and MATH do. |
math/0002035 | To lighten notation we will write MATH for the exterior direct sum MATH, so that the formula to be established is MATH . The plan is to compute the multiplier ideal on the left using the log resolution MATH. Specifically: MATH . Note to begin with that MATH thanks to the fact that MATH and MATH have no common component... |
math/0002035 | We apply REF to the diagonal MATH. Keeping the notation of the previous proof (with MATH, MATH), one has MATH . But it follows from REF that MATH as required. |
math/0002035 | Given MATH, fix MATH plus a general divisor MATH. Then MATH so the assertion follows from the Theorem. |
math/0002035 | This does not follow directly from the statement of REF because the divisor of a general element of MATH is not the sum of divisors of elements in MATH and MATH. However the proof REF goes through to show that MATH and then as above one restricts to the diagonal. |
math/0002035 | Only REF requires a proof, since REF follows again from REF by the restriction principle and the diagonal trick. Let us fix two relatively compact NAME open subsets MATH, MATH. Then MATH is the NAME tensor product of MATH and MATH, and admits MATH as a NAME basis, where MATH and MATH are respective NAME bases. Since MA... |
math/0002035 | Clearly, MATH for every effective divisor MATH. We can take MATH so large that MATH is very ample, and we are thus reduced to the case where MATH itself is very ample by replacing MATH with MATH. By definition of MATH, there exists a sequence MATH such that MATH . We now fix an integer MATH (to be chosen precisely late... |
math/0002035 | Note to begin with that it is enough to produce a big and nef divisor MATH satisfying the conclusion of the Theorem. For by NAME 's Lemma one can write MATH where MATH is an effective MATH-divisor, and MATH is an ample MATH-divisor. Then MATH where MATH is ample and the top self intersection number MATH approaches MATH... |
math/0002035 | We start by interpreting MATH geometrically. Let MATH be a log resolution of MATH, with MATH, where MATH is free, and MATH, so that MATH. Then evidently MATH counts the number of intersection points of MATH general divisors in MATH, and consequently MATH . We have MATH for MATH since then MATH is big (and nef), and MAT... |
math/0002038 | There is an imprimitivity bimodule isomorphism MATH defined on the generators by MATH since straightforward calculations verify that the above mapping on generators preserves the left action and the right inner product. |
math/0002038 | Straightforward calculations verify that the mapping MATH on the generators preserves the operations, hence extends to a MATH-isomorphism. |
math/0002038 | There is an imprimitivity bimodule isomorphism MATH defined on the generators by MATH since straightforward calculations verify that the above mapping on generators preserves the left action and the right inner product. |
math/0002038 | There is an imprimitivity bimodule isomorphism MATH defined on the generators by MATH since straightforward calculations verify that the above mapping on generators preserves the left action and the right inner product. |
math/0002038 | Straightforward calculations verify that the mapping MATH on the generators preserves the operations, hence extends to a MATH-isomorphism. |
math/0002038 | There is an imprimitivity bimodule isomorphism MATH defined on the generators by MATH since straightforward calculations verify that the above mapping on generators preserves the left action and the right inner product. |
math/0002038 | The desired diagram is the inner rectangle of the diagram MATH (Here and in REF the action and coaction symbols have been omitted for clarity.) We will show how to fill in the bottom arrow so that each of the outer rectangle and the top, bottom, left, and right quadrilaterals commute. Since MATH is surjective, the resu... |
math/0002038 | The desired diagram is the inner rectangle of the diagram MATH . Consider the diagram MATH . It follows from CITE that the map MATH extends to an isomorphism MATH, and this serves as the left-hand coefficient map for an isomorphism MATH, hence the left triangle of REF commutes. The inner quadrilateral is the commutativ... |
math/0002038 | The desired diagram is the outer rectangle of MATH . The upper left triangle commutes by REF, so we must show the lower right triangle commutes. We construct the isomorphism MATH as a composition MATH . Here, MATH is the coaction MATH of MATH on MATH, where MATH is the flip isomorphism. It follows from CITE (see also C... |
math/0002038 | Since MATH is discrete, the coaction MATH is automatically nondegenerate, so by CITE there is a unique full coaction MATH of MATH on MATH whose reduction coincides with MATH, and then CITE gives an isomorphism MATH; it is easy to see that this isomorphism is equivariant for the dual actions. Then CITE applies, giving a... |
math/0002039 | If MATH, then the map MATH is NAME module automorphism of MATH which intertwines the left actions coming from MATH and MATH. For the converse, suppose MATH in MATH, so there exists a linear bijection MATH such that MATH for each MATH. Define MATH by MATH. Then the first two of these properties imply that MATH is an inv... |
math/0002039 | We first claim that the composition of morphisms is well-defined. Suppose we have right-Hilbert bimodule isomorphisms MATH and MATH. Then MATH is easily seen to preserve the actions and inner product, so extends to an isometric bimodule map MATH. The map MATH is an inverse for MATH, which is therefore a right-Hilbert M... |
math/0002039 | By definition, MATH is the closed span of the rank-one operators MATH CITE, so it is enough to show that MATH. We can verify this by applying both sides to vectors of the form MATH, which densely span MATH. We then have MATH as required. Since MATH is faithful and MATH is a MATH - MATH imprimitivity bimodule, the induc... |
math/0002039 | We have to show that if there is a right-Hilbert bimodule MATH such that MATH and MATH as right-Hilbert bimodules, then MATH is a MATH - MATH imprimitivity bimodule (and in fact MATH will then be isomorphic to MATH). We choose a faithful nondegenerate representation MATH of MATH on MATH, and use the inducing constructi... |
math/0002039 | Let MATH and let MATH viewed as a MATH - MATH imprimitivity bimodule (see CITE). Since MATH CITE, we can view the canonical nondegenerate homomorphism MATH as a nondegenerate homomorphism MATH. The map MATH extends to an isomorphism of MATH onto MATH. |
math/0002039 | Adding actions to REF , and REF is routine, except possibly in REF . In the proof of REF, we took MATH and MATH, so we need to show that MATH induces an action on MATH. But for each MATH, MATH is an adjointable operator with MATH, and in fact MATH is an automorphism of MATH. A quick calculation shows that MATH so MATH ... |
math/0002039 | We have MATH . |
math/0002039 | We begin by showing that MATH can be completed to give a NAME MATH-module MATH satisfying REF . Straightforward calculations show that REF make MATH into a pre-inner product MATH-module, so we need only verify that the sesquilinear form of REF is positive definite. To do so, fix, for the remainder of the proof, a faith... |
math/0002039 | We first show that the map on morphisms is well-defined. Suppose MATH is an isomorphism of right-Hilbert MATH - MATH bimodules which is equivariant for MATH-compatible actions MATH and MATH of MATH. Then MATH is easily seen to give a bijective map MATH which respects the right-Hilbert bimodule structures (REF - REF ) a... |
math/0002039 | We establish the commutativity of both REF by factoring MATH, as in REF , where MATH is an isomorphism in MATH and MATH is a MATH - MATH equivariant nondegenerate homomorphism. We shall prove the commutativity of REF by showing that we have two commutative diagrams MATH and MATH . Because taking crossed products is a f... |
math/0002040 | Denote the meridians of the components of MATH by MATH. In MATH the framings MATH can uniquely be expressed as MATH. This implies for MATH that MATH . In MATH we obtain the following unique expression of MATH REF in terms of the meridians MATH REF of MATH: MATH . This implies the lemma. |
math/0002040 | Choose a diagram of MATH such that MATH and put dots on the components of MATH. Let MATH be the linking matrix of MATH and let MATH be the inverse of the linking matrix of MATH. Then for a series MATH (respectively, MATH) of diagrams in MATH (respectively, in MATH) that contains no struts and has degree-MATH-term MATH,... |
math/0002040 | Let MATH (respectively, MATH) be a knot in the upper part MATH (respectively, in the lower part MATH) of a tubular neighborhood of MATH representing the MATH-th basis element of MATH. Let the knot MATH have the framing MATH induced by the surface MATH. First consider MATH. Define MATH as the composition of MATH with th... |
math/0002040 | We use the notation from above. For suitable distinguished components of MATH and of MATH the tangle MATH coincides with the part of the framed oriented boundary of MATH that belongs to MATH. Let MATH be the part of the framed oriented boundary of MATH that belongs to MATH. We regard MATH as a non-associative tangle wi... |
math/0002040 | By REF and NAME REF the coefficients of MATH in the two power series MATH and MATH only depend on a NAME matrix MATH of a knot MATH and are polynomials MATH and MATH in the entries of MATH. By REF we have MATH for all NAME matrices of knots in MATH. REF implies that MATH for all MATH. |
math/0002040 | The invariants MATH and MATH of MATH differ only by normalization (see REF ). Let MATH. Then we have MATH with MATH. The following four steps show that MATH can be calculated from MATH and vice versa. This will complete the proof. CASE: MATH depends only on the wheel-part MATH of MATH REF . CASE: MATH can be calculated... |
math/0002040 | By REF and by REF it is sufficient to show that for a null-homotopic knot MATH in a rational homology sphere each of the invariants MATH and MATH can be computed from the other one. This statement follows from REF . |
math/0002041 | The circle MATH acts on the product manifold MATH by MATH. The function MATH is MATH invariant. The level set MATH is a (set theoretic) disjoint union of two manifolds MATH . The points of the first manifold are regular points of MATH because they are regular points of the function MATH. The points of the second manifo... |
math/0002041 | We may assume that a neighborhood of the boundary MATH in MATH is diffeomorphic to MATH. Then MATH. The circle action on MATH extends trivially to a circle action on MATH making the projection map MATH-invariant. By REF , MATH is a smooth manifold. Moreover, it is easy to see that MATH hence is smooth as well. |
math/0002041 | (compare CITE) Consider the symplectic product MATH. The map MATH is a moment map for an action of MATH on MATH. Arguing as in REF we see that MATH is a regular value of MATH and that the reduced space MATH is the cut MATH. The pullback of the symplectic form MATH by the embedding MATH is MATH. Consequently the induced... |
math/0002041 | By REF MATH is a smooth manifold, MATH is a submanifold and MATH is diffeomorphic to MATH. We now assume for simplicity that MATH is connected. Otherwise we can argue connected component by connected component. By the equivariant coisotropic embedding theorem the product MATH carries and MATH-invariant closed REF-form ... |
math/0002041 | The argument is an adaptation of NAME 's proof of the existence of invariant Riemannian metrics on manifolds with proper group actions CITE. Suppose first that the group MATH is compact. Then there is on MATH a bi-invariant measure MATH normalized so that MATH. We then define MATH to be the average of MATH: MATH for al... |
math/0002041 | The proof is identical to the proof of REF. The main idea of the proof is that a point MATH lies in the zero level set of the moment map if and only if the orbit MATH is tangent to MATH; hence MATH descends to a REF-form MATH on MATH. |
math/0002041 | Consider the contact manifold MATH with the circle action MATH. The map MATH is the corresponding moment map. Arguing as in REF we see that MATH is a regular value of MATH and that the reduced space MATH is the cut MATH. The pullback of the contact form MATH by the embedding MATH is MATH. Consequently the induced embed... |
math/0002041 | By REF MATH is a smooth manifold and MATH is diffeomorphic to MATH. We would like to show that MATH is contact. We have MATH. By REF MATH is a contact manifold, the embedding of MATH into MATH is contact and the difference MATH is contactomorphic to MATH. Therefore MATH is a contact manifold with the desired properties... |
math/0002041 | The vector field MATH descends to a vector field MATH on a neighborhood of MATH. It is not hard to see that the reduced symplectic form MATH satisfies MATH and that MATH is the reduced contact form MATH. |
math/0002041 | The hypersurface MATH in MATH is of contact type. The cut MATH is the reduction of MATH, and the cut MATH is the reduction at zero of MATH. Now apply REF . |
math/0002041 | As in REF let MATH denote the manifold MATH with coordinates MATH, MATH and MATH and a contact form MATH. For each non-negative integer MATH consider an embedding MATH of MATH into MATH: MATH . Let MATH. Consider the action of MATH on the boundary MATH generated by MATH on MATH and by MATH on MATH. Then as in REF , the... |
math/0002041 | The proof is essentially the same as the proof of REF . We start by giving a description of lens spaces which is convenient for our purposes. Consider again MATH. Consider the action of MATH on the boundary of MATH defined on MATH by the vector field MATH and on MATH by the vector field MATH. By REF the cut MATH is a m... |
math/0002042 | Of course, MATH is regular at each regular point of MATH. Furthermore, given MATH singular point of MATH, we have MATH and MATH, hence MATH a singular point of MATH iff MATH is a twist point of MATH. Now, let MATH the twist points of MATH and MATH their projections by MATH. Then, MATH is regular over MATH and, by compa... |
math/0002042 | First of all, since MATH and MATH are connected, there exists a REF-fold simple branched covering MATH, with MATH as in the statement, such that any NAME twist of MATH along a non-separating simple loop can be realized, up to isotopy, as the lifting of a half twist around an arc in MATH between two branch points of MAT... |
math/0002042 | The connection of MATH follows immediately from the connection of MATH, since the monodromy of MATH is generated by NAME twists, so it preserves the components of MATH. We also observe that, for the same reason, the monodromy of MATH fixes the boundary of MATH. Now, if MATH or MATH is already connected, we can set MATH... |
math/0002042 | CASE: Given an analytic branched covering MATH, we have that MATH is a NAME surface without boundary, since the restriction of MATH to MATH is a finite holomorphic map (see CITE, p. REF). Let MATH be a proper strictly plurisubharmonic function and MATH be the plurisubharmonic function defined by MATH. By the transversa... |
math/0002042 | Let MATH the branch points of MATH and MATH meridian loops around them, such that MATH in MATH. Putting MATH, we have that the restriction MATH is a locally trivial bundle with fibre MATH and monodromy MATH, where MATH is the homomorphism indiced by the inclusion of MATH into the complement of the branch points MATH. T... |
math/0002042 | Given a open book MATH with page MATH, we can write MATH, with MATH right-handed NAME twist along MATH and MATH. Then, fixed MATH and MATH meridian loops around them, such that MATH in MATH, we consider the NAME fibration MATH determined by the branch points MATH and the monodromies MATH for MATH(compare REF). By REF ,... |
math/0002042 | Looking at the double branched covering MATH shown in REF , we see that MATH covers twice the loop MATH encircling all the MATH branch points, where MATH denotes the genus of MATH. Then MATH is the lifting of a double right-handed twist along MATH. By expressing the corresponding braid in terms of the standard generato... |
math/0002042 | By REF , the oriented boundary of any compact NAME surface if orientation preserving diffeomorphic to a positive open book. NAME, given a positive open book MATH, we can assume, up to the plumbing operation REF introduced in CITE (compare proof of REF above), that the binding of MATH is connected. Then, by REF , MATH i... |
math/0002042 | Let MATH be an open book with page MATH and binding MATH, such that MATH is orientation preserving diffeomorphic to MATH. Since MATH is generated by NAME twists along non-separating simple loops, we can express MATH as a product of such twists. Now, by REF , any left-handed twist along a non-separating loop can be obta... |
math/0002043 | The MATH-bundles over a finite cell complex MATH are classified by homotopy class of maps from MATH to MATH. Such a class determines a conjugacy class of homomorphisms MATH . In our case MATH is a surface with non-empty boundary, so MATH has cohomological dimension REF (being free). Thus MATH-bundles over a surface wit... |
math/0002043 | The proof is immediately obtained from REF and the fact that the disjoint union MATH is cobordant to MATH with MATH by a toric cobordism with base the sphere MATH with MATH holes (it can be constructed similarly to the proof of REF ). |
math/0002043 | By MATH we denote the subgroup of MATH generated by the squares of matrices with negative determinant MATH . It is evident that MATH and is normal in it. We show that MATH and this implies the claim. We use the following presentations of MATH and MATH; see (CITE). For MATH , MATH and MATH , MATH . The commutator subgro... |
math/0002043 | It follows from REF that for MATH there exists an oriented toric cobordism between the torus bundles MATH and MATH if and only if MATH . Thus MATH . The generator of MATH is the conjugacy class of the element MATH . The subgroup MATH lies in the subgroup MATH. Thus, if there exists an unoriented toric cobordism with an... |
math/0002046 | Assume that MATH holds. Since MATH and MATH are MATH-Cartier, so is their difference MATH. Consider the decomposition MATH into positive and negative parts. Since MATH and MATH are not contained in MATH, the surface MATH is MATH-factorial near MATH. So MATH is MATH-Cartier. Hence MATH is also MATH-Cartier . As MATH, th... |
math/0002053 | It is a direct consequence of the commutativity of the diagram MATH since the top map MATH is an isomorphism by REF and both vertical maps are the epimorphisms. |
math/0002053 | Because of REF and NAME duality, MATH. |
math/0002053 | It follows from REF, since the homomorphism MATH passes through the trivial group MATH. |
math/0002053 | CASE: This claim follows, because the rank of a matrix is equal to the order of the largest non-zero minor. CASE: This claim follows from REF . CASE: By REF , the set MATH is an algebraic subset of MATH. So, it suffices to prove that MATH. But, by REF , MATH for MATH small enough, and so MATH. |
math/0002053 | The existence of MATH follows from REF . The flexibility of MATH follows, since MATH is a symplectic form for MATH small enough. Indeed, if we set MATH, then MATH. |
math/0002053 | Take a closed REF-form MATH which represents MATH. Then MATH is a symplectic form for MATH small enough. Furthermore, by REF , there exists arbitrary small MATH such that MATH. Now the result follows from REF. |
math/0002053 | CASE: Trivial. CASE: Suppose that there exists MATH with MATH. Take MATH in the interior of MATH. Then, in view of REF, there exists an arbitrary small MATH such that MATH, that is, MATH does not belong to the interior of MATH. This is a contradiction. CASE: This is true because of REF . |
math/0002053 | Let MATH be the usual non-singular pairing given by MATH for MATH and MATH. Define a skew-symmetric bilinear form MATH via the formula MATH for MATH. It is easy to see that the rank of MATH, which must be an even number MATH with MATH, is equal to MATH. The inequality MATH follows from REF. The last claim holds since, ... |
math/0002053 | Because of the NAME duality, MATH. So, MATH, and hence MATH is determined uniquely up to non-zero multiplicative constant. |
math/0002053 | Here we use some ideas from CITE. It follows from REF that MATH where MATH. Indeed, if MATH then there exists MATH such that MATH. Therefore, MATH, and MATH . Because of REF , every class in MATH is symplectically harmonic. So, it suffices to prove that any cohomology class MATH is symplectically harmonic. Let MATH wit... |
math/0002053 | This follows from REF because of REF. |
math/0002053 | It follows from REF since, by REF, MATH. |
math/0002053 | Since MATH, we conclude that the linear form MATH is non-degenerate for some point MATH. So, MATH is non-degenerate since it is homogeneous. Thus, MATH is non-degenerate. |
math/0002053 | According to our assumption about the type of MATH, there exists a basis MATH of homogeneous MATH-forms on MATH such that MATH . Since the NAME cohomology of the nilmanifold is isomorphic to the NAME cohomology of the NAME algebra, we conclude that MATH . In particular, by REF, a REF-form MATH on MATH is symplectic if ... |
math/0002053 | According to our assumption about the type of MATH, there exists a basis MATH of homogeneous MATH-forms on MATH such that MATH . The cohomology groups of degrees REF are: MATH . In particular, if MATH is a symplectic form on MATH then MATH where MATH. The following claim can be proved similarly to REF. Let MATH be a sy... |
math/0002053 | CASE: MATH and REF-form MATH on MATH is symplectic if and only if MATH where MATH. Furthermore, if MATH then MATH; otherwise, MATH. Finally, MATH for every MATH. CASE: MATH and REF-form MATH on MATH is symplectic if and only if MATH where MATH . Furthermore, if MATH, then MATH; otherwise, MATH. Finally, MATH for every ... |
math/0002053 | Because of the NAME isomorphism MATH, we conclude that MATH and the first equality follows from REF. Similarly, MATH . Now, in view of REF, the computation of dimensions completes the proof. |
math/0002053 | If MATH varies then the result follows from the first equality of REF. If MATH varies but MATH does not vary then the result follows from the second equality of REF. |
math/0002064 | Suppose that REF holds, so that MATH, where MATH is a directed set parametrizing elements MATH (and MATH). For any MATH we have natural isomorphisms MATH . For MATH, let MATH be the imbedding; we have a commutative diagram MATH where MATH is such that MATH, MATH is multiplication by MATH and MATH is the projection indu... |
math/0002064 | Let MATH be a countable generating family of MATH. Then MATH generate MATH and MATH for all MATH. For every MATH, we can write MATH, for certain MATH. Let MATH be the MATH-module defined by generators MATH, subject to the relations: MATH . We get an epimorphism MATH by MATH. The relations imply that, if MATH, then MATH... |
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