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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 MATH we have MATH . But for MATH we have MATH, and hence MATH . By REF has values in MATH, and MATH is its MATH-component, namely the component in the MATH-eigenspace of MATH. In particular, MATH and MATH if and only if MATH. Since MATH determines MATH by linearity, MATH. CASE: For MATH . Comparison with REF shows MATH. |
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, MATH . Comparison yields REF. |
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 equation we obtain MATH, and, for MATH, MATH . But MATH is an immersion, and therefore MATH, proving REF. Finally, for MATH, MATH . But MATH. So MATH, and this is stable under MATH. Therefore MATH. Since MATH, this proves REF. |
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 the (similar) proofs of the others in the appendix. The degree formula then yields MATH . For MATH, that is, MATH, we get MATH, whence MATH. Then MATH, and MATH . Addition yields MATH . It follows that MATH, that is, MATH, and MATH is MATH-stable, hence constant in MATH. From MATH we conclude MATH. Therefore all mean curvature spheres of MATH pass through the fixed point MATH. Choosing affine coordinates with MATH, all mean curvature spheres are affine planes, and MATH corresponds to a minimal surface in MATH. |
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 MATH. We assume that MATH is its only zero within MATH. Moreover, there exist MATH, a holomorphic coordinate MATH centered at MATH, and a holomorphic section MATH such that MATH . Off MATH the section MATH is decomposable, and since the NAME MATH is closed in MATH, it defines a section of MATH, that is, a MATH-dimensional subbundle of MATH extending MATH. The statement about the kernel follows easily using the fact that MATH is the annihilator of MATH. |
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 holomorphic structure. Then, for local holomorphic sections MATH in MATH and MATH in MATH, MATH . By REF we have MATH hence MATH . Then also MATH and therefore MATH. To prove the holomorphicity of MATH, we first note that MATH carries a natural holomorphic structure. The rest follows from the holomorphicity of MATH, and the product rule. Finally we interpret MATH as a section in MATH. Note that the holomorphic structure on MATH is given by MATH. From the holomorphicity of MATH we find, for MATH, MATH . This shows that MATH is MATH-invariant. Moreover, it is obviously invariant under MATH and, as a consequence of MATH, also under MATH. From REF it follows that MATH is invariant under MATH, and that for a local holomorphic vector field MATH and a local holomorphic section MATH of MATH, MATH . Then MATH . |
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 fundamental fact is that this diagram can be completed: the morphism MATH above commutes with MATH, so we have MATH . Note that the image of an element MATH under MATH is just the hamiltonian vector field MATH. The morphisms MATH we are looking for can now be defined as the map induced by MATH, explicitly MATH . To see that MATH is an isomorphism near a smooth point of MATH it will be sufficient to prove this for the map MATH (because at smooth points MATH we have MATH and the map MATH is injective). So assume the sheaves MATH, MATH, and MATH to be defined in a neighborhood of a smooth point which means that they all become locally free. MATH then has to be identified with the conormal bundle. To prove that MATH is an isomorphism, we will construct an inverse. First note that, by the fact that MATH is coisotropic, the morphism MATH actually sends an element of MATH to a form vanishing on all vectors tangent to MATH. So the restriction of MATH to MATH defines a morphism MATH. The situation is as follows: MATH A diagram chase shows that MATH is injective. On the other hand, we have MATH, as MATH is lagrangian. So MATH is an isomorphism and the inverse of MATH. |
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 sequence MATH where MATH and MATH are the kernel respectively, cokernel sheaves of the map MATH. This sequence can be split MATH with MATH. Applying MATH yields MATH . Now we use REF (see CITE): Given a local ring MATH, two MATH-modules MATH and MATH with MATH and MATH, then for all MATH, the modules MATH vanish. It follows that MATH, so we have MATH. Then obviously MATH and by the argument above MATH so the map MATH is an isomorphism. |
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 kernel must be the constant sheaf. To prove that MATH, two things have to be checked: As MATH, we must first identify the elements of MATH with the flat lagrangian deformations. Then we have to show that the image of MATH are the trivial deformations. But this is easy, because for MATH, MATH acts as MATH, thus inducing a trivial deformation. Furthermore, by REF , of all deformations coming from vector fields on MATH, only those induced by hamiltonian vector fields are trivial in the lagrangian sense. Now we choose an open set MATH and sections MATH generating MATH. Take an element MATH, which means that MATH for all MATH. Then MATH corresponds to the deformation given by MATH . The ideal MATH is involutive iff for any two elements MATH, we have MATH, which is equivalent to MATH . Consider MATH, which is an element of MATH, so the condition MATH is equivalent to MATH, that is MATH . This means exactly that MATH. In order to interpret the second cohomology group, we define the bilinear mapping MATH . In this way we get a quadratic form MATH. It can be immediately verified that this induces a map MATH. We will now prove the following: Given a lagrangian deformation MATH. Then there is a lift to second order defining an involutive ideal iff MATH. The last condition is equivalent to the existence of MATH with MATH, that is, MATH . But this means that the following ideal is involutive. MATH . |
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 is closed iff the deformation is lagrangian and the subspace of exact one-forms are deformations induced by hamiltonian vector fields (isodrastic deformations, see CITE), these are the trivial ones. MATH is assumed to be a NAME manifold, in this case the first NAME group is exactly MATH. |
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 fibred by the hamiltonian flow of this function. Explicitly, we can make an analytic change of coordinates, such that MATH, MATH for all MATH and MATH. Than the ideal of MATH is of the form MATH for some functions MATH which are independent of the variable MATH (provided that we have chosen the symplectic form to be MATH). |
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 differential on MATH. We choose local NAME coordinates MATH on MATH and MATH on MATH. Suppose that the two ideals are MATH and MATH (if we consider MATH as embedded in MATH). Let MATH be an element of MATH . Then it can be written as a power series in MATH with coefficients in MATH. A direct calculation shows that the differential on MATH is MATH . It is clear that the morphism MATH must be the obvious inclusion MATH . We will now show that the cokernel of this inclusion is acyclic. Then it follows immediately that MATH induces an isomorphism on the cohomology. So let MATH be an element of MATH, that is, MATH where MATH and MATH for all MATH. But then MATH vanishes in the cohomology because it can be written as MATH with MATH . |
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 sense of REF, that is, MATH). The complex of sheaves in the theorem is the complex MATH, which satisfies the first two properties (MATH is MATH-coherent and the differential is MATH-linear). Our task is to verify the third axiom, that is, we have to find a vector field of class MATH such that MATH is transversally constant (see REF), this will be done in REF . Now the proof of the theorem shows that there is a smaller neighborhood MATH of MATH such that MATH. This gives the result by using CITE in the same way as in CITE or CITE. |
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 that we can find a linear combination of theses MATH-fields which is transversal to MATH. The same is true in the complex case, here we have MATH independent hamiltonian fields MATH which are holomorphic. As the holomorphic tangent space at each point is canonically isomorphic (over MATH) to the real one, we get MATH linear independent MATH-fields by applying this isomorphism to MATH. These can be used to find a field transversal to MATH. |
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 necessarily tangent to MATH, so the cohomology sheaves are constant on the local integral curves of MATH. |
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 element than the same holds for MATH. |
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 function MATH depending only on MATH and MATH. Denote the ideal generated by these two functions by MATH and by MATH the stalk of MATH at the point MATH. Then we can identify MATH with MATH, so MATH is free on the two generators MATH and MATH, where MATH while MATH is just MATH, generated by the homomorphism sending MATH to MATH in MATH. The complex MATH at the point MATH then reads: MATH where the pair MATH denotes the homomorphism sending MATH to MATH and MATH to MATH. Now we have to investigate the modules of differential forms on MATH at MATH. In general MATH where MATH is the ring MATH. This leads to MATH where we have used the following abbreviations: MATH can be described as MATH and MATH are the cokernels of the maps MATH and MATH, respectively. So the result is MATH . As MATH is strongly quasi-homogeneous, we have weighted homogeneous local equations for the transversal slice which gives MATH. |
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). By REF it follows inductively that MATH for MATH in MATH. In particular, the identity MATH in MATH shows that the generators MATH, MATH REF and MATH satisfy the compatibility condition in MATH. Hence the algebra homomorphism MATH exists. We write MATH for MATH, so that MATH for MATH. Since the MATH REF and the MATH REF generate MATH as an algebra, we see that MATH is surjective. On the other hand, all fundamental relations of MATH as given by CITE can be easily checked for the generators MATH REF and MATH REF of MATH. This implies the injectivity of MATH. |
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 determined by computing the leading coefficient of MATH using the identity MATH and using REF . Suppose now that MATH, then REF imply that MATH is of the form REF for some constant MATH, with MATH given by REF. If MATH, then leading term considerations using REF show that MATH. The expression for MATH when MATH follows now easily by applying MATH on both sides of REF and using the quadratic relation MATH, compare with CITE for the proof in the rank one setting. |
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 such that MATH. Let now MATH and choose a reduced expression MATH. We set MATH . Then MATH for MATH by REF . Iterating REF, we thus obtain MATH . Proof of REF The recurrence REF implies that MATH for all MATH. On the other hand, REF implies that the symmetric NAME polynomial MATH is a non-zero element in MATH for all MATH. Proof of REF Again by REF, we have MATH. Let MATH. Let MATH be a simple reflection in MATH which stabilizes MATH. Then REF implies that MATH. Hence the coefficient of MATH in the expansion of MATH as a linear combination of the MATH (MATH), is MATH. On the other hand, MATH, hence we conclude that MATH. Then REF implies MATH for all MATH, hence MATH. This proves that MATH if MATH. Let now MATH. Then MATH, where MATH is the longest NAME group element. Furthermore, by REF we have MATH with MATH, so MATH is a non-zero element in MATH. Hence MATH if MATH. The triangularity statement follows from REF and from the fact that the coefficient MATH in the above expansion of MATH is non-zero. Proof of REF We have to show that MATH where the MATH are the expansion coefficients of MATH in terms of non-symmetric NAME polynomials MATH (MATH), and where MATH is given by REF. We use again the recurrence REF for the coefficients MATH. We set MATH then the MATH REF are mutually different and MATH see REF . Now observe that MATH for all MATH, so that MATH . Furthermore, observe that MATH by REF . Substituting REF and using REF and the characterization of the roots MATH (see REF), we obtain REF. |
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 in the left-hand side of REF. Combined with REF we then derive that MATH where MATH is given by MATH . Since MATH . , we can apply REF together with the duality REF of the non-symmetric NAME polynomials to obtain the desired REF . |
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 lemma follows now from REF . |
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 invariant under the action of MATH for MATH, where the action of MATH is extended from MATH to (suitably nice) functions MATH in the MATH variables MATH via REF . This is an immediate consequence of the well-known fact that the roots MATH are permuted by the simple reflection MATH. Hence MATH can be rewritten as an integral over MATH with MATH-alternating integrand for all MATH. Now MATH for MATH follows from the fact that the measure MATH is MATH-invariant. The case MATH is more subtle. The behaviour of the measure MATH under the action of MATH is given by MATH which now implies that MATH . For fixed MATH, the integrand in the right-hand side of REF depends analytically on MATH. Indeed, by a direct computation using REF and the second expression of MATH in REF, we see that the MATH-dependent factor of MATH is given by MATH which has the desired analytic behaviour due to the conditions on the parameters MATH and MATH. Thus by NAME 's theorem we conclude that MATH. This completes the proof of the proposition. |
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. For MATH this is immediate by REF . As usual, the proof for the affine part of the statement (the case MATH) is more subtle. We begin by rewriting MATH as a (kind of) multiple residue of MATH at MATH, where MATH. This can be done using the MATH-invariance of the weight function MATH. The result is as follows. Let MATH be the component in MATH of the minimal coset representative MATH with respect to the semi-direct product structure MATH, and let MATH. Then we have MATH for all MATH, where the multiple residue at MATH is defined by MATH . In particular, we obtain MATH for all MATH. Now we consider REF with MATH replaced by MATH . We first consider the changes in the multiple residue. By the proof of CITE, we have MATH for all MATH, that is, MATH and MATH. Secondly, MATH by REF . So if we replace the residue at MATH by the residue at MATH in the definition of the multiple residue at MATH, then we obtain minus the multiple residue at MATH. On the other hand, we know by the proof of REF that MATH is invariant under the action of MATH. So the invariance of REF under replacement of MATH by MATH follows from the simple observation that MATH when MATH is a function depending on a single variable MATH, having a simple pole at MATH, and satisfying the invariance condition MATH. |
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 when MATH. We fix MATH, MATH. Since MATH, it follows that MATH. Comparing this outcome with the right-hand side of REF, we obtain MATH which yields the desired result. |
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 MATH . Similarly, since MATH, we derive from REF that MATH can be rewritten as MATH where MATH. Using these alternative descriptions for MATH and MATH together with the orthogonality relations for the symmetric NAME polynomials (see REF ), we obtain MATH for all MATH. On the other hand, by REF , we have MATH for MATH with MATH . Comparing the two different outcomes for MATH, we obtain the desired result in case MATH. The off-diagonal case is covered by the orthogonality relations for the symmetric NAME polynomials, see REF . |
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 duality anti-isomorphism MATH. It follows from REF that MATH with MATH. On the other hand, MATH with MATH by REF . Combined with REF we obtain MATH . In particular, if MATH, then MATH . The ratio MATH can now be evaluated inductively using similar techniques as in the proof of REF . This gives the desired result. |
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 a reduced expression of MATH in MATH, then we obtain from REF and from the previous paragraph that MATH where MATH should be read as MATH when MATH. The lemma follows now easily from REF . |
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 second formula of REF to arrive at MATH . The theorem follows now from REF , combined with the inclusion MATH. This inclusion is a direct consequence of the inequality MATH for all MATH, see REF. |
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 an affine translate of MATH is a generic fiber of MATH. It follows that MATH, so MATH. |
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, MATH, so MATH the last equality by modularity of MATH. CASE: Suppose MATH. Then MATH so MATH . CASE: Suppose MATH. Then MATH for some MATH. Note that MATH, since MATH. It follows that MATH since MATH while MATH. Since MATH we have MATH for MATH, which case is treated above. These calculations verify that MATH can be identified with MATH as described above, with the same rank function. |
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 hyperplanes of MATH. Using this basis, it is clear that elements of MATH are uniquely determined by their linking numbers with the hyperplanes of MATH. By naturality, these linking numbers agree with linking numbers in MATH with the hyperplanes of MATH. Since these linking numbers take values in a discrete space, and vary continuously, they remain locally constant under translation of the fiber, and thus are globally constant under translation around a loop in the base. This proves triviality in degree one. Since MATH is generated by MATH, and the monodromy action respects cup products, it follows that the monodromy acts trivially on MATH. |
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 the rational completion of MATH is aspherical. The assertion then follows from the homotopy sequence of this fibration. |
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 differential of MATH is trivial. |
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 coordinates. By definition, the total space of the pullback of MATH along the projection MATH is the set of pairs MATH such that MATH, and MATH in MATH. But this means that the last MATH components of MATH match the first MATH components of MATH. Then each such MATH corresponds to a unique point of MATH which, by the first two conditions, lies in MATH. Under this identification, the projection MATH coincides with MATH. This identifies MATH with the pullback of MATH, as claimed. |
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 lines MATH for MATH . The coned fiber arrangement MATH pictured in REF is a MATH arrangement. Indeed MATH is precisely REF. Alternatively, MATH is a factored arrangement of rank three, hence MATH by CITE. We conclude by REF that MATH is MATH. |
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 remaining two conditions follow from the facts that MATH and MATH are right MATH-module maps respectively. |
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-linear, it suffices to take an element of MATH of the form MATH and compute MATH . We used REF and the fact that MATH is a unit in MATH to derive second and third equalities, and then definition of the right action of MATH on MATH to obtain the fourth one. To prove that MATH is a right MATH-module map it is enough to consider any MATH of the above form, use the same equations as for the preceding calculation, and compute for all MATH, MATH . The coassociativity of MATH follows from the coassociativity of the coproduct MATH. Now, it is clear from the definition that MATH is a left MATH-module map. To prove that it is a right MATH-module map as well take MATH as above and compute for all MATH, MATH where the first of REF was used to obtain the second equality. Again, directly from the construction of MATH it follows that MATH. On the other hand using the definitions of MATH and MATH, the second of REF we obtain for any MATH of the above form MATH . This completes the proof that MATH is a MATH-coring. CASE: If MATH with the right MATH-coaction MATH, then MATH with the coaction MATH since MATH. Conversely, if MATH then the coaction MATH provides MATH also with the weak entwined module structure. |
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 and MATH are the required adjunctions. |
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 particular, MATH. On the other hand, MATH is a right MATH-module morphism, thus we have, MATH, so that MATH is MATH-central as required. ` REF . Suppose there exists MATH such that MATH. For any MATH define MATH by MATH. Since MATH is MATH-central we have for all MATH, MATH: MATH hence MATH is a right MATH-module map. Furthermore MATH, that is, MATH is split by MATH. Finally for any MATH we have MATH, which means that MATH is natural in MATH. |
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 consider MATH given by MATH. The naturality of MATH implies for all MATH, MATH, that is, MATH. Therefore MATH is a left MATH-module map, and, consequently, MATH is a left MATH-module morphism as a composition of two such morphisms. Since MATH splits MATH we have MATH . Now, for all MATH consider the morphism MATH, MATH, and also the morphism MATH. By the naturality of MATH we have MATH . From the first of equations REF we have for all MATH, MATH. Combining this expression with the second of equations REF we obtain MATH . Applying MATH to this equality one obtains MATH . On the other hand, MATH is a right MATH-comodule map, so that MATH. Applying MATH to this formula we obtain MATH . Thus MATH can be expressed in terms of MATH in two different ways. Comparison gives REF as required. ` REF . Suppose there exists MATH as in the theorem. Then for all MATH define an additive map MATH, MATH. Clearly MATH is a right MATH-module map since MATH is such a map. Furthermore, for any MATH, MATH we have MATH where we used REF . This means that MATH is a morphism in MATH. Furthermore, take any MATH. Then for all MATH, MATH we have MATH where we used the fact that MATH is a right MATH-comodule map. Finally we take any MATH and compute MATH . This shows that MATH is the required splitting of MATH. |
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, MATH. Applying MATH to the first equality in REF and MATH to the second equality in REF one deduces REF . By REF , the forgetful functor is separable as required. |
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. The first of the above conditions for MATH is equivalent to the normalisation of MATH, MATH, while the second condition gives for all MATH, MATH. By the faithfully flat descent the latter is equivalent to MATH. |
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 morphism in MATH such that MATH, MATH. Taking MATH and using above equality we obtain MATH. Let MATH. Since MATH is a right MATH-comodule map we have MATH. Applying MATH to this equality we obtain for all MATH, MATH, that is, MATH is a finitely generated left MATH-module. |
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 all MATH, MATH. On the other hand MATH . This implies that MATH . Using this equality one easily finds that given MATH, MATH is a right MATH-comodule with the coaction MATH. Also, one easily checks that the maps described provide the required isomorphism of categories. |
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 extension MATH is NAME. CASE: Since MATH is a finitely generated projective left MATH-module the map MATH, given by MATH is bijective by CITE. Clearly MATH is a MATH-bimodule map. Thus we have MATH as MATH-bimodules. The extension MATH is NAME iff MATH, that is, iff MATH as MATH-bimodules. CASE: This follows from the bijective correspondence MATH, MATH, MATH. |
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 MATH. On the other hand MATH. Put together this implies that MATH is a right MATH-coaction. |
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 MATH is well-defined. One easily checks that MATH is a right MATH-module map natural in MATH. Finally, with MATH, MATH as before, take any MATH and compute MATH. Then take any MATH, MATH and compute MATH. This proves that MATH and MATH are the required adjunctions. |
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 MATH. For the same reason the MATH-coring isomorphism MATH explicitly reads MATH and thus coincides with the canonical map MATH. This proves that MATH is a MATH-Galois extension and by the uniqueness of the canonical entwining structure, MATH must be the canonical entwining structure associated to MATH. |
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 weak entwined module we have for all MATH, MATH. In the view of the fact that MATH this implies that MATH. |
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 is a defining sequence of MATH tensored with MATH, and MATH is exact. The bottom row is exact too. Since MATH is a bijection, so are MATH and MATH. Therefore MATH is an isomorphism in MATH. For all MATH consider the following commutative diagram of right MATH-module maps MATH . The maps in the top row are: MATH and MATH, and the top row is exact by the faithfully flat descent. The bottom row is the defining sequence of MATH and hence is exact. This implies that MATH is an isomorphism in MATH and completes the proof of the fact that MATH and MATH are inverse equivalences. Conversely, assume that MATH and MATH are inverse equivalences. Notice that MATH, MATH is a MATH-bimodule isomorphism. Since MATH is a right MATH-comodule via the coproduct, there is a corresponding adjunction MATH, and it is bijective. Define MATH, MATH. Explicitly MATH. Clearly MATH is a MATH-bimodule map such that MATH. Furthermore since MATH is a grouplike and MATH is a MATH-bimodule map we have MATH, for all MATH. On the other hand MATH . Therefore MATH is a MATH-coring isomorphism and hence MATH is NAME. If MATH is a flat left MATH-module then both kernels and cokernels of any morphism in MATH are right MATH-comodules (that is, MATH is an Abelian category). Therefore any sequence of MATH-comodule maps is exact if and only if it is exact as a sequence of additive maps. Since MATH is an equivalence, it preserves and reflects exact sequences. By the above observation it does so even as viewed as a functor from MATH to the category of MATH-modules. Therefore MATH is a faithfully flat left MATH-module. |
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, the MATH entry of the column corresponding to the MATH-cycles is equal to the number of nonnegative integer solutions to the equation MATH. Consequently, one easily verifies that MATH is the generating function for the entries of the column corresponding to the MATH-cycles. The dimension of the column-span of our table is therefore equal to MATH, and the proposition is proved. |
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 can be checked that the maximum value of MATH for MATH is MATH. Set MATH. Since MATH, by the induction hypothesis we have MATH . It is routine to check that when MATH the right hand side is less than MATH, and the proof follows. |
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 sufficiently large) shows MATH from which the result follows without difficulty. |
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 or MATH; CASE: MATH for MATH; CASE: MATH for MATH; CASE: MATH for MATH; CASE: MATH. For MATH let us define MATH by MATH. It is clear that the functions MATH are all MATH smooth and have NAME constant less than or equal to MATH, MATH on MATH, MATH on MATH, and MATH for all MATH. Our path MATH of linear isomorphisms is going to be of the form MATH where each MATH takes the vector MATH into MATH and for every MATH the mapping MATH is still a linear isomorphism and, moreover, the families of isomorphisms MATH and MATH are uniformly bounded. Let us define the isomorphisms MATH. They are going to be of the form MATH where MATH satisfies MATH, and MATH (the exact definition of MATH will be given later). Their inverses MATH will be MATH . We want the linear mappings MATH to be linear isomorphisms. We have MATH from which MATH and we need to write MATH and MATH as linear functions of MATH. If we apply the functionals MATH and MATH successively to REF , we denote MATH, MATH, MATH, MATH, and we take into account that MATH, then we obtain the system MATH which we want to have a unique solution for MATH. The determinant of this system is MATH and we want MATH to be bounded below by a strictly positive number, and this bound has to be uniform in MATH. For MATH this can easily be done by setting MATH (so that MATH, MATH, MATH, and therefore MATH for all MATH). For MATH, put MATH then MATH, MATH, MATH, MATH, MATH, MATH, and everything is fine (indeed, MATH and MATH for all MATH). Therefore, with these definitions, the linear system REF has a unique solution for MATH, which can be easily calculated and estimated by NAME 's rule, of the form MATH . The linear forms MATH, MATH satisfy that MATH for all MATH, as is easily checked. Now, by substituting MATH and MATH in REF we get the expression for the inverse of MATH, that is, MATH . By taking into account that MATH, MATH and MATH for all MATH, one can estimate that MATH and MATH for all MATH. So let us define MATH by MATH . This path is well defined because the sum is locally finite; in fact, from the definition of MATH it is clear that, for a given MATH there exist some MATH and MATH such that MATH for all MATH, that is, MATH is locally of the form MATH, where MATH. This implies that the MATH are really linear isomorphisms and that the path is MATH smooth. On the other hand, the path MATH is MATH smooth as well, because it is the composition of our path MATH with the mapping MATH, MATH, which is MATH smooth and whose derivative is given by MATH for every MATH (see REF ). This proves REF. Next, by bearing in mind the local expression of MATH and the above estimations for MATH and MATH, we deduce that MATH for all MATH, where MATH will be fixed later. This shows REF . Now, if MATH and we write MATH for MATH as above, then it is clear that MATH is locally of the form MATH and therefore MATH from which we get MATH. Moreover, we have MATH and therefore MATH from which MATH and REF is satisfied as well provided we fix MATH. Now let us define the path MATH by MATH . It is clear that MATH is a MATH smooth path in MATH, and MATH for all MATH (from which it follows that MATH is NAME). Let us see that MATH is bounded. For a given MATH there exists MATH so that MATH and therefore, taking into account the definition of MATH and the fact that MATH for all MATH, we have that MATH . This shows that the image of MATH is contained in the ball MATH and MATH is bounded. Let us also remark that MATH. Finally, let us check that MATH satisfies the separation REF . Let MATH and take MATH so that MATH; then we have MATH . By observing that MATH for all MATH and taking into account the definition of the MATH, it is not difficult to see that MATH where MATH. Then, by applying either MATH or MATH to the expression for MATH above, depending on which the maximum in REF is, and bearing in mind that MATH and MATH for all MATH, we get that MATH and it follows that MATH. This shows that if MATH is large enough then MATH for all MATH. In order to get paths MATH and MATH and a vector MATH with REF - REF and such that MATH is contained in the unit ball, it is enough to multiply them all by MATH. |
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 outside MATH, and has the remarkable property that MATH for all MATH (indeed, MATH is non-zero everywhere because MATH whenever MATH, MATH, and MATH is a linear isomorphism at each point MATH). Now, take a MATH real function MATH such that MATH for MATH, MATH outside MATH, MATH, MATH, and MATH for all MATH. Then, if we define MATH by MATH it is immediately checked that MATH is a MATH smooth bump on MATH which does not satisfy NAME 's theorem and whose support is precisely the body 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 . Once we have such a retraction it is easy to prove REF : the formula MATH defines a MATH smooth self-mapping of MATH without fixed points. On the other hand, if we pick a non-decreasing MATH function MATH so that MATH for MATH and MATH for MATH, then the formula MATH for MATH, MATH, defines a MATH homotopy joining the identity to a constant on MATH, that is, MATH contracts the pseudosphere MATH to a point. |
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, we have MATH and, since MATH whenever MATH, it is clear that MATH unless MATH. |
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. Let us consider, for MATH, the difeomorphism MATH, MATH, and the function MATH defined by MATH. Then MATH satisfies the following properties: CASE: MATH is MATH smooth. CASE: MATH and MATH, for every MATH. CASE: MATH, for every MATH, MATH. CASE: MATH for every MATH. CASE: MATH is NAME in bounded sets and MATH is NAME. Now, we define the functions MATH by MATH, whenever MATH. We will identify MATH with the infinite sum MATH, where an element MATH belongs to MATH if and only if every MATH is in MATH and MATH, being MATH. Then, we define the function MATH by MATH . First, note that MATH is well-defined, since REF implies that, whenever MATH, MATH . On the one hand, note that, if MATH has NAME constant less than or equal to MATH then MATH is also NAME with constant less than or equal to MATH, since for MATH and MATH in MATH we have MATH . This implies that, if MATH, the functionals MATH satisfy that MATH. Indeed, we have MATH, and therefore MATH. Also, MATH, and then we get that MATH also belongs to MATH. Let us now prove that MATH is MATH smooth. For every MATH and MATH in MATH, we can estimate MATH . Therefore MATH is NAME differentiable and MATH. Moreover, MATH is NAME since MATH. This implies, in particular, that MATH is NAME on bounded sets. Let us check that MATH uniformly approximates MATH. Indeed, from REF on MATH and REF , we have that, for every MATH, MATH and then, MATH . In order to obtain functions which approximate the norm uniformly in MATH let us consider MATH. According to REF we have that MATH for any MATH. Let us check that MATH is bounded. By REF we have, for any MATH, MATH . Consequently, MATH is NAME with NAME constant, say N. In a similar way, we obtain that MATH is NAME, since for any MATH, MATH in MATH, MATH . Finally, note that the set MATH is contained in MATH, which has empty interior in MATH. This concludes the proof of REF . In order to prove REF , we consider a MATH function MATH, MATH for MATH, and supp-MATH. Then, we can define a required bump function as the composition MATH. Indeed, on the one hand, MATH and therefore MATH. On the other hand, MATH, whenever MATH, and hence MATH for MATH. The bump function MATH is clearly NAME with NAME derivative since MATH, MATH and MATH are NAME and NAME is NAME on bounded sets. |
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 MATH . This is equal to MATH times the b-normal vector field MATH plus a sum of vector fields which have the form MATH, where MATH vanishes at MATH and MATH is tangent to lb and MATH (all considerations taking place inside MATH). Thus, under blowup of MATH, MATH lifts to a vector field of the form MATH where MATH is smooth and tangent to the boundary of MATH, and so dividing by MATH yields a nonvanishing normal vector field plus a smooth tangent vector field. As above, such a vector field has a continuation across the boundary to the double of MATH (across the front face) as a smooth nonvanishing vector field. This holds true smoothly up to the corner with lb, so MATH is a smooth manifold with corners. |
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 MATH implies MATH, MATH satisfies MATH where we used that MATH is symmetric on MATH. On the other hand, MATH maps MATH, and for any MATH, MATH, hence MATH. In addition, MATH, with kernel MATH, maps MATH, so we have MATH . If MATH, then MATH has the form MATH, that is, it is incoming. A standard argument then implies that MATH. Let MATH, where MATH. NAME 's formula yields MATH so MATH. It then follows iteratively from REF that the expansion of MATH at the boundary of MATH vanishes identically, that is, that MATH. Finally a unique continuation theorem, see for example, CITE, shows MATH identically. This means that MATH is indeed dense in MATH. |
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. Now it is easy to see that the maps MATH patch together to define an anti-isomorphism MATH of sheaves of algebras. Then MATH evaluated on global sections gives MATH and we have MATH. Finally, MATH is unique since the vector fields MATH generate MATH. |
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 normalizations of MATH. |
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 scaling. |
math/0002117 | Immediate from REF since MATH for MATH. |