paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0002075 | We have MATH but MATH . Hence, by type, the second term vanishes as well, and we get REF. A similar, but simpler, computation shows REF Next, using REF, we consider MATH . Again we show MATH. Then REF will follow by type. Clearly MATH showing MATH. Further MATH . |
math/0002075 | CASE: If MATH is a holomorphic curve then, for any MATH, MATH . But then MATH a fortiori for all MATH. It follows MATH . Conversely, MATH for MATH implies MATH and therefore MATH . Again for MATH . This shows MATH . By the preceding, MATH maps into MATH. Its MATH-part is MATH but for MATH . This proves REF. CASE: For M... |
math/0002075 | Define MATH using the conformal structure induced by MATH. Then MATH which implies MATH. Hence MATH is conformal, too, and MATH. For the next computations recall REF , and REF: MATH . Then MATH . With MATH this becomes MATH . Next MATH . Therefore MATH or MATH . We now use REF to compute MATH . Similarly, using REF, MA... |
math/0002075 | Choose MATH such that MATH and MATH are dual bases. Then MATH see REF . |
math/0002075 | Note that MATH. By REF we have MATH . |
math/0002075 | Since MATH we interpret MATH. On a dense open subset of MATH then MATH is injective for any MATH. For MATH we get MATH . The injectivity of MATH then proves the lemma. |
math/0002075 | Motivated by the lemma, we relate MATH to MATH rather than to MATH. We put MATH . Then MATH . The proof will be completed with the following lemma which shows that MATH - like MATH - has values in MATH, while the ``NAME take values in MATH. |
math/0002075 | Recall that MATH is MATH-stable. It is of course also MATH-stable, and therefore MATH . Now MATH is immersive, and therefore MATH. Thus REF will follow if we can show MATH for MATH. But, using REF , MATH . Next, for MATH we have MATH . This proves REF. On the other hand, for MATH, MATH . Together with the previous equa... |
math/0002075 | First we have MATH . Now MATH is MATH-stable, and MATH and MATH imply MATH. Therefore MATH, and on a dense open subset of MATH . |
math/0002075 | First MATH is obviously MATH-stable. It is trivially invariant under MATH and MATH and, therefore, under MATH. Finally, the MATH of MATH is MATH and this vanishes on MATH. The unique characterization of the mean curvature sphere by these three properties implies MATH. |
math/0002075 | Let MATH be a local holomorphic vector field, that is, MATH, see REF , and MATH. Then MATH . Now MATH . Similarly MATH . Therefore MATH . |
math/0002075 | If MATH or MATH, then MATH is a twistor projection by REF . Otherwise we have the line bundle MATH, and similarly a line bundle MATH that coincides with the image of MATH almost everywhere. We have the following holomorphic sections of complex holomorphic line bundles: MATH . We proved the statement about MATH. We give... |
math/0002075 | MATH implies MATH see REF . Since MATH, the assumption MATH implies MATH. Then for MATH . Since MATH, this implies MATH . But MATH is an immersion. Therefore MATH implies MATH. The converse is obvious. |
math/0002075 | Let MATH and MATH. This is an open subset of MATH. Let MATH be a boundary point of MATH, an let MATH be holomorphic sections of MATH on a neighborhood MATH of MATH. By a change of indices we may assume that MATH. But this is a holomorphic section of the holomorphic bundle MATH, and hence has isolated zeros, because MAT... |
math/0002075 | MATH is a holomorphic section by REF . By REF there exists a line bundle MATH such that MATH off a discrete set. Assume now that MATH on an open non-empty set MATH. Then MATH, and MATH on MATH. But then MATH by REF . This is a contradiction, because the zeros of MATH are isolated. |
math/0002075 | In general, if MATH and MATH are two connections, then MATH . We apply this to MATH and use REF : MATH . REF follows from MATH because MATH. |
math/0002075 | The holomorphicity of MATH was shown in REF , and that of MATH can be shown in complete analogy. MATH is a holomorphic complex quaternionic vector bundle, and MATH is a holomorphic subbundle, see REF . Therefore MATH and MATH are holomorphic complex quaternionic line bundles, and the complex line bundle MATH inherits a... |
math/0002083 | The first two points are trivial, while the third has to be checked by an explicit calculation. |
math/0002083 | On a symplectic manifold, there is a canonical isomorphism MATH between vector fields and one forms, given by MATH. On the other hand, for each analytic subspace MATH we have two exact sequences, dual to each other, namely, the conormal and the normal sequence, thus, there is the following diagram: MATH . Now the funda... |
math/0002083 | Set MATH where MATH. Then it is immediate that MATH is an isomorphism on MATH. To prove that MATH, it suffices to check this in the lowest degrees, that is, we have to show that the diagram MATH commutes. This follows directly from MATH. |
math/0002083 | We have MATH as MATH is torsion free. On the other hand, the kernel is supported on the singular locus of MATH, so it must be a torsion sheaf, hence MATH. |
math/0002083 | We will make use of the following fact: Let MATH be a MATH-module of type MATH, then MATH is reflexive, that is, MATH. The morphism MATH we are looking is obtained by dualizing twice the morphism MATH, this yields MATH as MATH is of type MATH. Clearly, MATH is an isomorphism on the regular locus. We have an exact seque... |
math/0002083 | MATH equals MATH. Take an element MATH of MATH. Then MATH for all MATH. If MATH is not a constant, then the ideal MATH is strictly larger than MATH, not the whole ring and still involutive. This is a contradiction to the fact that MATH is lagrangian, which means that MATH is maximal under all involutive ideals. So the ... |
math/0002083 | The first fact is just the definition of the sheaf MATH. In the second case, note that the space of embedded flat deformations is MATH, where MATH is the normal bundle of MATH in MATH. As MATH is smooth, this happens to be MATH, so each infinitesimal flat deformation corresponds to globally defined one-form on MATH. It... |
math/0002083 | Let MATH be coordinates of MATH centered at MATH. Then the fact that MATH implies that there are coefficients MATH such that the following equation holds in MATH where MATH is an element of MATH vanishing at second order. So we have an element in the ideal describing MATH whose derivative do not vanish. Then MATH is fi... |
math/0002083 | The proof of REF shows that the ideals MATH and MATH describing the two germs differ by exactly one element whose differential do not vanish at the origin. This implies that the conormal sheaves MATH of MATH and MATH of MATH are related by the formula MATH. It follows that MATH . Now we have to describe the differentia... |
math/0002083 | This is obvious since MATH is an exact functor. |
math/0002083 | Choose a representative MATH for the the germ such that MATH iff MATH for all points MATH. We refer the reader to REF. We do not consider a relative situation here, so the map MATH in this theorem is replaced by MATH (Obviously, MATH can be chosen such that this map is a standard representative of the germ MATH in the ... |
math/0002083 | It follows from REF that there exist MATH linear independent hamiltonian vector fields on MATH which respects the stratum MATH. Now we have to distinguish the cases MATH and MATH, in the first one, since MATH is of real dimension MATH and since the intersection of MATH and MATH was transversal, it follows immediately t... |
math/0002083 | Set MATH. Then the last lemma yields a covering MATH of MATH and vector fields MATH defined in a neighborhood of MATH in MATH. Chose a partition of unity subordinate to this covering to obtain a field on MATH which is still transversal to MATH. For each point MATH, which is contained in some stratum MATH, MATH is neces... |
math/0002083 | Let MATH. Then MATH and MATH. But MATH, so MATH for all MATH. |
math/0002083 | By the long exact cohomology sequence, it suffices to prove that the complex MATH is acyclic. This can be done in exactly the same way as for MATH provided that the inner derivative MATH (MATH being the quasi-homogeneous NAME vector field) maps MATH into MATH. But this follows from REF because if MATH is a torsion elem... |
math/0002083 | This is an explicit calculation involving the definition of the complex MATH and the map MATH. It suffices to calculate the rank of MATH and MATH. So suppose that MATH is a decomposable germ. We choose coordinates MATH (with symplectic form MATH) around MATH such that MATH is given as the zero locus of MATH and a funct... |
math/0002090 | For the existence of MATH, we need to check that the elements MATH, MATH REF and MATH in MATH respect the defining relations of the generators MATH, MATH (MATH), MATH in MATH. The MATH-relations for MATH is precisely the content of REF . The MATH-relations for the MATH-tuple MATH follows easily from REF (see also CITE)... |
math/0002090 | We write MATH. Suppose that MATH and write MATH with MATH. We show that MATH. Let MATH such that MATH. If MATH, then MATH. On the other hand, if MATH, then MATH. This proves the assertion for MATH. The case MATH can be obtained by applying the previous case to MATH. |
math/0002090 | Let MATH for MATH be the divided difference-reflection operator defined by MATH . Then for all MATH we have MATH . The proof follows now easily from REF and from REF of MATH. |
math/0002090 | The proof is based on the following consequence of NAME 's REF : let MATH and MATH, then MATH with MATH given by REF. Suppose now first that MATH, that is, that MATH. Using REF , we see that MATH, so we have to show that MATH. Now REF imply that MATH is a constant multiple of MATH. The constant multiple can be determin... |
math/0002090 | We first introduce some notations and deduce some preliminary results which we will need for the proof. Let MATH and let MATH be an element in MATH. By REF we have for MATH and MATH that MATH with MATH . It follows now from the relations MATH that the coefficients MATH satisfy the recurrence relations MATH for MATH and... |
math/0002090 | This follows directly from the presentation of MATH as given by CITE. |
math/0002090 | By REF we have MATH for all MATH. So it suffices to prove that MATH for all MATH and all MATH. REF is easy when MATH is stabilized by MATH since then we have MATH where the last equality follows from (the proof of) REF . So we assume for the remainder of the proof that MATH . We can use now REF to commute MATH and MATH... |
math/0002090 | The first statement follows from CITE with the indeterminates in CITE specialized to MATH for MATH, MATH for MATH and MATH for MATH. The identity REF follows then from REF and the invariance of the measure MATH under the action of MATH. |
math/0002090 | By REF of MATH we have MATH . We consider a term MATH in this sum with MATH. Then there exists a simple root MATH REF which is orthogonal to MATH, and which is mapped to a negative root MATH by MATH. Now REF implies that the factor MATH of MATH is zero. Hence the contribution in the sum REF is zero unless MATH. The lem... |
math/0002090 | The proposition is obviously correct for MATH (MATH), so it suffices to prove it for MATH (MATH). Let MATH. It follows by direct computations that MATH for MATH, with MATH and with the action of MATH as defined in REF. Now observe that MATH is MATH-alternating, that is, MATH for MATH. On the other hand, MATH is invaria... |
math/0002090 | The intertwining REF can be proved by checking it for the algebraic generators MATH, MATH and MATH REF of MATH using REF . The fact that REF defines an action of MATH on MATH follows then from REF since MATH is an algebra isomorphism and MATH is bijective. |
math/0002090 | The proof for MATH with MATH is immediate. Hence it suffices to check the intertwining property for MATH (MATH). Let MATH. By REF we have for MATH, MATH with MATH given by MATH . Since MATH for MATH and MATH, it thus suffices to prove that MATH is invariant under replacement of MATH by MATH for all MATH and all MATH. F... |
math/0002090 | By REF we have MATH . Furthermore, it follows from REF that MATH for MATH. REF reduces to MATH when MATH, with the constant MATH as given in the statement of the theorem. Combined with REF it follows that MATH. Since MATH is bijective, we then also have MATH. It remains to prove REF. By REF , we only have to prove REF ... |
math/0002090 | First of all, observe that MATH for all MATH and that MATH, where MATH (respectively MATH, MATH) is the idempotent corresponding to the trivial representation of the underlying finite NAME algebra of type MATH. Combined with REF we can rewrite MATH as MATH . In particular, the symmetric NAME transform MATH maps into MA... |
math/0002090 | We prove the lemma using the non-affine intertwiners MATH (MATH). For the moment, we fix MATH and MATH such that MATH. We first give some additional properties of MATH which we will need for the proof. The intertwiner MATH is self-adjoint, that is, MATH, where MATH. This can be checked most easily in the image of the d... |
math/0002090 | Let MATH. It follows from REF that MATH is the unique element of minimal length in MATH which maps MATH to MATH under the dot-action. In particular, any element MATH which is smaller than MATH with respect to the NAME order, maps MATH to a non-zero element in MATH. Since MATH, we have MATH, hence MATH . Let now MATH be... |
math/0002090 | First of all, observe that MATH for all MATH. Indeed, the first formula is trivial, while the second formula is a direct consequence of REF , the definition of the non-symmetric NAME transform MATH and its intertwining properties given in REF . We use now successively the first formula of REF, then REF and finally the ... |
math/0002090 | This follows from the evaluation of MATH (MATH) (see REF ), REF . |
math/0002094 | Dualizing the description of complete principal truncation to hyperplane arrangements, we see that MATH is the matroid of the arrangement MATH obtained by choosing a generic subspace MATH of codimension MATH containing MATH, and restricting MATH to MATH. Then MATH has dimension MATH, MATH is a hyperplane of MATH, and a... |
math/0002094 | As in the proof of REF , the arrangement MATH can be identified with the restriction MATH of MATH to the linear subspace MATH of codimension MATH spanned by MATH and MATH. Then MATH . There are three cases. CASE: Suppose MATH satisfies MATH. By modularity of MATH. Then there exists MATH such that MATH. Then MATH. Since... |
math/0002094 | The fiber of MATH over MATH is the complement of the arrangement MATH in MATH. Since the base MATH is path-connected, REF implies that the arrangements MATH are lattice-isotopic. Then the assertion follow from CITE. |
math/0002094 | The complement MATH coincides with the open stratum MATH of MATH. So REF implies that the restriction of MATH to MATH is a fiber bundle projection, by the NAME Isotopy Lemma CITE. |
math/0002094 | This follows immediately from the long exact homotopy sequence of the fibration MATH. |
math/0002094 | Since the fiber MATH is the complement of an arrangement, the cohomology of MATH is free abelian, and is generated by MATH. First of all we argue that the monodromy action on MATH is trivial, by the same reasoning as in the corank-one case CITE. The group MATH has a free basis consisting of elements dual to the hyperpl... |
math/0002094 | The argument is the same as in the corank-one case CITE. Because the monodromy action is trivial, hence nilpotent, on the cohomology of the fiber, the map MATH induces a fibration of the rational completion of MATH over that of MATH, with fiber the rational completion of MATH. Since MATH, the hypothesis implies that th... |
math/0002094 | The first assertion follows from the triviality of the monodromy action established in REF . The second is a consequence of the factorization identity among the NAME series. Indeed, according to CITE, the formula of REF holds for a general spectral sequence MATH, with a correction term that vanishes precisely when the ... |
math/0002094 | For these special realizations, the bundle map MATH is the restriction of the projection MATH onto the last MATH coordinates. Similarly, the map MATH is the restriction of the projection MATH onto the last MATH coordinates. The map MATH can be identified with the restriction of the projection MATH onto the first MATH c... |
math/0002094 | In case MATH is a point, then MATH is modular in MATH, and the modular fibration MATH is a trivial bundle with fiber MATH, by CITE. The pullback of a trivial bundle is trivial. |
math/0002094 | Let MATH. Let MATH denote the vertex-induced subgraph of MATH on vertices MATH. Then MATH is a modular coline of MATH by REF , and MATH is supersolvable, hence MATH, by REF . Let MATH, and let MATH . Then MATH for MATH, and MATH for MATH. The fiber arrangement MATH is the affine arrangement in MATH consisting of the li... |
math/0002105 | One can easily check that given an entwining structure MATH one has the coring MATH as described in the proposition. Conversely, let MATH be a MATH-coring with structure maps given in the proposition. The properties of the right MATH-action imply REF and the first of REF required for the entwining map MATH. The remaini... |
math/0002105 | For any MATH, using the definition of MATH and REF we have MATH . Therefore MATH is a projection as claimed. CASE: Clearly MATH is a left MATH-module. The fact that it is a right MATH-module can be checked directly using REF . CASE: First we need to show that MATH. Since MATH is MATH-linear and, evidently, left MATH-li... |
math/0002105 | For any MATH define, MATH, MATH, while for any MATH define MATH, MATH. Clearly both MATH and MATH are natural in MATH and MATH respectively. Furthermore, because MATH is a coaction MATH, for all MATH. It remains to be shown that for all right MATH-modules MATH, MATH. Take any MATH and MATH and compute MATH . Thus MATH ... |
math/0002105 | ` REF . Suppose that MATH is separable, and let MATH be split by MATH, the latter defined in REF . Since MATH is a right MATH-module we can take MATH and define MATH. Since MATH is split by MATH, we have MATH. Now for any MATH define the morphism MATH by MATH. Since MATH is natural we have: MATH . This implies, in part... |
math/0002105 | Recall the a ring extension MATH is separable iff there exists an invariant MATH such that MATH (compare CITE). Since in the canonical coring MATH the counit is given by the multiplication projected to MATH the separability of MATH is equivalent to the separability of MATH by the preceding theorem. |
math/0002105 | ` REF . Suppose MATH is separable. Let MATH be the adjunction defined in REF and let MATH be the splitting of MATH. Since MATH is a right MATH-comodule via MATH there is a corresponding MATH and we can define MATH. The map MATH is a right MATH-module morphism as a composition of two such morphisms. Next for all MATH co... |
math/0002105 | Given a map MATH as in REF one defines MATH, MATH. By the ` REF part of the proof of REF , MATH and thus it is a right MATH-comodule splitting of MATH. Using the fact that MATH is a MATH-bimodule map one easily verifies that MATH is a left MATH-comodule map. Conversely, given MATH define MATH. Since MATH splits MATH, M... |
math/0002105 | Recall that an extension MATH is split iff there exists a MATH-bimodule map MATH such that MATH (compare CITE). In the case of the canonical MATH-coring MATH the conditions required for the map MATH read MATH and MATH for all MATH. Since MATH the maps MATH are in one-to-one correspondence with the maps MATH via MATH. T... |
math/0002105 | The proof of this lemma is based on the proofs of CITE. Let for all MATH, MATH, MATH be the natural isomorphism and let MATH. Since MATH is natural and MATH, MATH we have for all MATH, MATH. Evaluating this equality at MATH and the resulting equality at MATH we obtain MATH. Now, for any MATH, let MATH be the unique mor... |
math/0002105 | Given MATH one can view it as a right MATH-module via MATH, for all MATH, MATH. Indeed, it is immediate that MATH. Furthermore, using that the right coaction of MATH on MATH is a right MATH-module map we have for all MATH, MATH, MATH . Let MATH, MATH, MATH be a dual basis of MATH as a left MATH-module. Notice that for ... |
math/0002105 | CASE: By REF implies that MATH is a finitely generated projective left MATH-module. REF shows that the forgetful functor is the restriction of scalars functor MATH. By REF , this functor has the right adjoint MATH. By CITE the restriction of scalars functor has the same left and right adjoint if and only if the extensi... |
math/0002105 | If MATH define MATH. Then MATH . Furthermore MATH, so that MATH is a grouplike as required. Conversely, if MATH is a grouplike, define MATH, MATH. Clearly MATH is a right MATH-module map. The fact that MATH is a right MATH-module map implies for all MATH, MATH. Finally, since MATH is a right MATH-module map we have MAT... |
math/0002105 | For any MATH define an additive map MATH, MATH. Clearly MATH is a right MATH-module map. Furthermore for any MATH and MATH we have MATH as well as MATH . Therefore MATH is a MATH-comodule map. One easily checks that MATH is natural in MATH. Next for any MATH define an additive map MATH, MATH. Notice that MATH so that M... |
math/0002105 | If MATH corresponds to a MATH-Galois extension MATH then MATH as MATH-corings via the canonical map MATH, and hence MATH is NAME. Conversely, if MATH is NAME then MATH is a right MATH-comodule, and by the correspondence in REF it is a MATH-module. The corresponding grouplike in MATH is MATH. Furthermore MATH, since MAT... |
math/0002105 | It suffices to show that MATH, then MATH will provide the required isomorphism of MATH-corings. Notice that from the definition of MATH in REF it follows that MATH. Since a typical element of MATH is of the form MATH and MATH is a left MATH-module map we have MATH. Therefore MATH. On the other hand, since MATH is a wea... |
math/0002105 | Assume that MATH is NAME and NAME is a faithfully flat left MATH-module. First notice that MATH. For all MATH consider the following commuting diagram of right MATH-comodule maps MATH . The maps in the top row are the obvious inclusion and MATH, while MATH is the coaction equalising map. The top row is exact since it i... |
math/0002106 | Suppose without loss of generality that MATH. Consider a basis vector MATH of MATH, so that MATH with each MATH. This vector is fixed by MATH if and only if MATH, MATH and so on. Since MATH equals the number of basis vectors fixed by MATH, the lemma follows. |
math/0002106 | Consider the table of the characters MATH; we are interested in the dimension of the row-span of this table. Since the dimension of the row-span of a matrix is equal to the dimension of its column-span, we can equally well study the dimension of the space spanned by the columns of the table. By the preceeding lemma, th... |
math/0002106 | We proceed by induction on MATH. Equality certainly holds for MATH. For larger MATH, the inductive hypothesis shows that MATH when MATH, and so MATH . |
math/0002106 | The proof is by induction on MATH. It can be checked with a computer that REF is true for MATH. Now assume that MATH and that REF holds for MATH. Let MATH be the unique positive integer satisfying MATH . Thus MATH is just the integer MATH of REF . Explicitly we have MATH . By the definition of MATH we have MATH . It ca... |
math/0002106 | From the definition of MATH and REF (and the fact that the right-hand side of REF is increasing), along with the inquality MATH for MATH, it follows that MATH for MATH. For MATH sufficiently large, we can evidently choose MATH such that MATH, so MATH. Since MATH, an application of the quadratic formula (again for MATH ... |
math/0002108 | Let MATH be a normalized basic sequence in MATH with biorthogonal functionals MATH (that is, MATH if MATH, and MATH otherwise) satisfying MATH (one can always take such sequences, see CITE, p. REF, or CITE, p. REF). For MATH set MATH. Let MATH be a MATH function with the following properties: CASE: MATH whenever MATH o... |
math/0002108 | Let MATH be the NAME functional of MATH. We may assume that MATH. By REF there is a closed subset MATH of MATH and a MATH diffeomorphism MATH which is the identity outside MATH. It can be assumed that the origin belongs to MATH. Then the function MATH defined by MATH is MATH smooth on MATH, restricts to the gauge MATH ... |
math/0002108 | Let MATH be the diffeomorphism from REF . We may assume that the origin belongs to the deleted set MATH and that MATH, so that MATH restricts to the identity outside MATH. Then the formula MATH where MATH is the NAME functional of MATH, defines a MATH smooth retraction from MATH onto the boundary MATH. This proves REF ... |
math/0002108 | We can write MATH, where MATH and MATH is finite. Pick a MATH smooth REF bump function MATH such that MATH does not satisfy NAME 's theorem, and let MATH be a MATH smooth NAME bump function on MATH so that MATH whenever MATH. Then the function MATH defined by MATH is a MATH smooth REF bump which satisfies MATH. Indeed,... |
math/0002108 | The proof is similar to that of the preceding corollary. Pick MATH and MATH smooth REF bump functions on MATH and MATH, respectively, so that MATH and MATH do not satisfy NAME 's theorem. Then the function MATH defined by MATH is a smooth REF bump which satisfies MATH. |
math/0002108 | To save notation, let us just write MATH when referring to the usual norm in MATH. We will make use of the following restatement of a striking result due to NAME (see CITE). There is a MATH diffeomorphism MATH from MATH onto MATH such that all the derivatives MATH are uniformly continuous on MATH, and MATH for MATH. Le... |
math/0002114 | The operator MATH may be written MATH . Thus the left-reduced symbol as above is MATH . Hence the sub-principal symbol is MATH which vanishes when MATH. |
math/0002114 | We must show that when the set MATH is blown up inside MATH, the closure of MATH is a smooth manifold with corners which meets the front face of the blowup transversally. Let us restrict attention to a neighbourhood of MATH; the case of MATH is similar. Consider the vector field MATH. By REF, in MATH this is given by M... |
math/0002114 | It follows inductively using REF that the coefficient of order MATH vanishes, for MATH. Thus, actually MATH. Then REF shows that the next coefficient also vanishes. |
math/0002114 | Apply the above argument to the symbol at MATH. |
math/0002114 | To proceed, we give the proof for short range MATH; the proof for long range MATH requires only minor modifications. Let MATH be the subspace of MATH given by the image of MATH on the sum of MATH and the range of MATH applied to MATH. If MATH is not dense, then there is a function MATH orthogonal to MATH. Since for MAT... |
math/0002117 | As MATH is a generalized flag variety, the sheaf cohomology MATH vanishes and it follows that MATH is an isomorphism - see CITE. Hence MATH is a noncommutative model of MATH. This implies the result for MATH. |
math/0002117 | The first statement follows as MATH is prime and the second by REF . |
math/0002117 | Let MATH be any NAME open affine in MATH. Then MATH is generated by the multiplication operators MATH and the order MATH operators MATH where MATH. We obtain an algebra anti-isomorphism MATH by assigning MATH and MATH. This follows by checking the relations among our generators of MATH; see CITE. In particular, MATH. N... |
math/0002117 | Using the geometry of the big cell we get the coordinate expressions for the vector fields MATH and then we work out the twisting correction REF by choosing MATH. We use the fact that MATH is MATH-semi-invariant of weight MATH. See CITE,CITE; there are minor variations in the final answers owing to different normalizat... |
math/0002117 | By REF , MATH as MATH-representations with MATH. Then MATH is multiplicity-free by REF . It follows that MATH contains a unique copy of MATH; let MATH be a highest weight vector in that copy so that MATH for all MATH. Then MATH and MATH are highest weight vectors in MATH of the same weight, and so they are equal up to ... |
math/0002117 | Immediate from REF since MATH for MATH. |
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