paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9905083 | Follows by standard results about the behavior of orthogonal polynomials when the weight function is multiplied by a polynomial. |
math/9905083 | The first group of statements are straightforward via appropriate row and column operations on the NAME matrix. For the limits, consider MATH . Here we need to compute the limit of MATH . Now, MATH so for any polynomial MATH, MATH . Plugging in and simplifying gives the desired result. The other cases are either analog... |
math/9905083 | The key observation is that if an increasing subsequence contains a given point off the MATH diagonal, it can always be extended to contain the reflection of that point through that diagonal; moreover, no increasing subsequence can contain more than one point on that diagonal. Thus the point collection at time MATH has... |
math/9905083 | This follows immediately from the symmetric function identities of REF below. (See, in particular, REF to REF .) |
math/9905083 | Consider the first identity. We have MATH . But the integral on the right is REF when MATH and MATH, and is REF otherwise. The result follows immediately. Similarly, the identity for MATH follows from the fact that MATH is REF when MATH is even and MATH, and is REF otherwise; and the identity for MATH follows from the ... |
math/9905083 | This follows from the computations MATH and MATH . |
math/9905083 | By REF , it suffices to evalute MATH . But this integral is just the coefficient of MATH in MATH; that is, MATH . |
math/9905083 | The point is that given any partition MATH, there is a unique partition MATH such that MATH; simply add one to each odd part of MATH, then divide by REF. In particular, then, MATH, and thus MATH . Similarly, MATH . By the argument in CITE, the expression for MATH follows from NAME 's formula MATH REF , and the fact tha... |
math/9905083 | We have MATH since MATH vanishes on MATH. But then we can apply REF to obtain MATH as desired (recall MATH). The remaining formulas are immediate. |
math/9905083 | We recall MATH that is, we sum over all partitions MATH obtained from MATH by adjoining a vertical strip. On the other hand, MATH . In each case, there is at most one way to remove a vertical strip from a generic partition to obtain one with the desired special form. Thus it remains only to determine which partitions o... |
math/9905083 | A partition has trivial REF-core if and only if its diagram can be tiled by dominos. By a classical result, this can happen if and only if the diagram contains as many points MATH with MATH even as with MATH odd. For MATH, MATH, let MATH be the number of points in the diagram of MATH with MATH. Then MATH . But then MAT... |
math/9905083 | It follows from the lemma that for any MATH with trivial REF-core, there exist unique partitions MATH and MATH such that MATH and which satisfy MATH . Consequently, MATH . It thus suffices to show that for any partition MATH, MATH since then we may set MATH in MATH . By the NAME identity, MATH is the determinant of the... |
math/9905083 | Here we use the following analogue of NAME 's formula: MATH . Thus MATH . |
math/9905083 | Dualizing the proof of REF , we find MATH where the sum is over MATH such that MATH with MATH. But, as in the proof of REF , this condition is simply that MATH. The result follows. |
math/9905083 | For MATH even, we have: MATH . For MATH odd, we simply adjoin an extra element MATH to MATH and add a new function MATH which is REF away from MATH, then apply the identity for MATH even. |
math/9905083 | This is essentially shown in CITE; see also CITE. While the references only deal with the case in which every element of MATH is greater than every element of MATH, the proofs carry over directly. |
math/9905083 | This was known for MATH CITE and for MATH CITE; it thus follows in general. |
math/9905083 | We generalize the argument of CITE. Consider, for instance, the case MATH. In this case, the probability that our random multiset MATH is equal to a given fixed multiset MATH is MATH where MATH with MATH of shape MATH. Thus MATH . Similar arguments hold for types MATH and MATH. For MATH and MATH, we also need the fact ... |
math/9905083 | We have: MATH and similarly for the other cases. |
math/9905083 | That MATH is injective for MATH is straightforward; we simply observe that if MATH, MATH,MATH are linearly independent vectors, then the vectors MATH are linearly independent, as MATH ranges over MATH. Thus, suppose MATH. That MATH follows from the fact that any tensor product of MATH basis vectors must contain at leas... |
math/9905083 | We first need to show that, given any permutation MATH with a long decreasing subsequence, we can express MATH as a linear combination of MATH with MATH ranging over permutations without long decreasing subsequences. Let MATH be such a permutation, and let MATH be a subset of size MATH on which MATH is decreasing. By t... |
math/9905083 | Given a multiset MATH in MATH, we define an inversion of MATH to be a pair of elements MATH, MATH with MATH and MATH (that is, a strictly decreasing subsequence of length REF). Now, suppose MATH has a MATH-decreasing subsequence of length MATH. Choose a permutation MATH corresponding to MATH, and let MATH be the set of... |
math/9905083 | The argument is analogous. In eliminating a given increasing subsequence, we replace both it and its reflection through the diagonal by non-increasing subsequences, so the number of inversions increases. The proof of linear independence is analogous to that of REF . We simply switch the reverse the inequalities on the ... |
math/9905083 | The main difficulty here is that the na\ïve extension of the above algorithm is no longer guaranteed to terminate; for instance, corresponding to the increasing subsequence MATH of MATH, we have the identity MATH . Since MATH is an increasing subsequence of both sides, we could clearly loop indefinitely. The solution i... |
math/9905083 | Analogous. The only issue is that the long decreasing subsequence we eliminate must be symmetric about the diagonal (clearly always possible). For linear independence, we choose a symplectic basis of MATH indexed MATH, MATH. We thus find that the nonzero coefficients of an involution MATH correspond to words of length ... |
math/9905083 | If MATH is the representation of MATH corresponding to the partition MATH, then the dimension of MATH is given by the number of pairs of bitableau of shape MATH with respective content MATH and MATH. In other words, by the generalized NAME correspondence, this is equal to the number of multisets MATH with MATH . The re... |
math/9905084 | The values at MATH follow from REF. For MATH, note from the RHP REF that we have MATH. |
math/9905084 | Let MATH denote the jump matrix of the RHP REF. Since MATH for MATH, MATH also solves the same RHP. By the uniqueness of the solution of the RHP REF, we have MATH . Thus, MATH is real for MATH, thus proving REF . By the symmetry of the jump matrix, MATH, we obtain, by a argument similar to REF , MATH which is REF . The... |
math/9905084 | Write MATH where MATH . The function MATH is strictly decreasing for MATH, and MATH. Hence MATH for MATH. Note that MATH . When MATH, MATH, hence MATH. On the other hand, when MATH, since MATH, we have MATH if MATH. Therefore REF implies that MATH as MATH. Similar calculations give the desired result for MATH. |
math/9905084 | Note that under the stated conditions, we have MATH . |
math/9905084 | We have MATH . Above proposition shows that REF is true for MATH. For MATH, write MATH . Now REF follows from REF. The estimate REF is proved similarly. |
math/9905084 | For MATH and MATH in the above corollary, take MATH such that MATH. Once we fix MATH, then for MATH is large, MATH, and hence by REF above, MATH. Using MATH for MATH, MATH . But from REF, MATH is bounded for MATH. Hence using REF, we obtain the result for REF with new constants MATH, MATH and MATH. For REF , we note th... |
math/9905084 | We first consider MATH. Let MATH be the number of elements in MATH with no increasing subsequence greater than MATH. Consider the map MATH defined as follows: for MATH, set MATH for MATH, MATH, and MATH for MATH. Then it is easy to see that MATH consists of MATH elements, hence MATH. Moreover if MATH has an increasing ... |
math/9905084 | Note that we have a disjoint union MATH . Set MATH, the probability that the length of the longest decreasing subsequence of MATH is less than or equal to MATH. As the first row and the first column of MATH in MATH have the same statistics, we have MATH . Note that MATH . As MATH (see pp. REF), we have MATH and the mai... |
math/9905084 | As in REF, integrating by parts, MATH where MATH. From REF , we have MATH for a fixed MATH where MATH. Noting that MATH for all MATH, from REF , we obtain MATH . Now using convergence in distribution, the dominated convergence theorem gives REF. |
math/9905084 | We have a disjoint union MATH . Hence again MATH . One can check that MATH . Hence, we have MATH where MATH . For fixed MATH, MATH is unimodal in MATH and achieves its maximum when MATH. And MATH is unimodal in MATH and the maximum is attained when MATH. Hence MATH has its maximum when MATH. Consider the disc MATH of r... |
math/9905084 | Integrating by parts, MATH where MATH. Note that when MATH, MATH, and when MATH, MATH. Let MATH fixed. Consider the case when MATH. From REF, MATH where MATH. We apply REF. Note that we are in the region MATH faster than MATH, hence MATH is bounded, say MATH. So we can apply REF. Then we obtain MATH . Since MATH, we ha... |
math/9905084 | Let MATH for MATH. For large MATH, MATH for some MATH. From REF , using REF , it is easy to see that MATH exponentially as MATH. Now REF imply REF. For MATH, let MATH. Similarly, MATH exponentially as MATH, and we obtain REF. |
math/9905084 | Let MATH be the set of points at time MATH, and let MATH be a largest increasing subset of MATH. Then there will exist some number MATH (not unique) such that MATH and such that every other point of MATH has MATH and MATH. For any MATH, we thus have MATH where MATH is the number of points of MATH with MATH and MATH, an... |
math/9905085 | For MATH the condition MATH for all MATH is exactly MATH so MATH. Then MATH normal in MATH follows directly from MATH normal in MATH. It is required to prove that if MATH is such that MATH, if MATH, and if MATH, then MATH. Setting MATH and, since MATH is connected, this is equivalent to MATH for all MATH. But if MATH a... |
math/9905088 | Let MATH be a continuous nest representation of MATH acting on the NAME space MATH. Let MATH be the ideal set in MATH which corresponds to the ideal MATH. Through a series of facts, we will show that MATH is one of the meet irreducible ideal sets listed in REF ; consequently, MATH is meet irreducible. Suppose that MATH... |
math/9905088 | Any projection in MATH has the form MATH for some NAME subset MATH of MATH. Given MATH, if MATH then MATH and if MATH then MATH. This shows that when MATH, it is an atom of MATH. Now suppose that MATH is an atom of MATH. Let MATH be such that MATH. It is evident that there is at most one point MATH such that MATH; we n... |
math/9905088 | Let MATH be a decreasing sequence of projections in MATH with MATH. If MATH, then MATH for large MATH, in which case MATH. Now suppose that MATH and let MATH be such that MATH. With MATH as above, let MATH, so that MATH and MATH. Then MATH for large MATH; hence MATH and, taking strong limits, MATH. It follows that MATH... |
math/9905088 | Let MATH and assume that MATH contains two unit vectors MATH and MATH such that MATH. Let MATH. Since MATH for any projection MATH for which MATH, we may, by a suitable restriction, reduce to considering two cases: when MATH is contained in the diagonal MATH of MATH and when MATH is disjoint from the diagonal. When MAT... |
math/9905088 | Let MATH be a (nonzero) projection in MATH such that MATH and MATH. (One could just take MATH.) Then MATH. Since MATH dominates MATH, MATH . For the second assertion, suppose MATH and MATH. Now by REF , for any MATH, either MATH is zero or else MATH is contained in MATH for some MATH with MATH. But as MATH, MATH. We ha... |
math/9905088 | Let MATH. We shall show that both MATH and MATH are invariant under MATH. Since MATH is a nest representation, this means that one of them must be zero. If MATH is not invariant, then for some MATH and MATH, MATH. Now MATH, so there exists MATH with MATH. By REF , MATH, for some MATH. Since MATH it follows that MATH; h... |
math/9905088 | If MATH and MATH are not in the same orbit, it follows that MATH and MATH REF . Since MATH and MATH, MATH and MATH are not lineraly ordered, a contradiction. |
math/9905088 | If MATH, REF implies that MATH is contained in the orbit of MATH. If MATH for all MATH, we are done. Suppose then, that MATH for some MATH in the orbit of MATH. Without loss of generality we may assume that MATH. Let MATH be such that MATH. Then MATH and MATH REF . In particular, MATH. Let MATH be a point in the orbit ... |
math/9905088 | The first assertion follows from REF , since all atoms have the form MATH, for some MATH. Since the set MATH is a collection of commuting, rank-one atoms whose ranges span MATH, the NAME algebra which they generate is a masa in MATH. |
math/9905088 | By hypothesis, MATH contains an atom, necessarily of the form MATH, for some MATH. By REF , there is a nonempty interval MATH in an orbit in MATH with the following property: for any MATH-normalizing partial isometry MATH, MATH if, and only, if MATH intersects MATH. In other words, the complement of the spectrum of the... |
math/9905088 | The first assertion is clear. For the second, first note that, since MATH is a masa, MATH. Let MATH, let MATH, and let MATH. Then MATH is the immediate predecessor of MATH in MATH and MATH. Thus, every atom from MATH, and hence MATH itself, is contained in the NAME algebra generated by MATH. The reverse inclusion is ob... |
math/9905088 | Assume that MATH is a nest representation but that MATH is not meet irreducible. Let MATH and MATH be two ideals in MATH such that MATH and MATH differs from both MATH and MATH. By the inductivity of ideals, there exist matrix units MATH and MATH. These matrix units must lie in some MATH; since we may replace the seque... |
math/9905090 | CASE: These are the well known classical NAME relations. For completeness' sake we include a proof. Let MATH and consider the induced linear mapping MATH. Its image, MATH, is contained in each linear subspace MATH of MATH with MATH. Thus MATH is the minimal subspace with this property. MATH is decomposable if and only ... |
math/9905090 | The representation MATH may be realised as those tensors MATH which are symmetric in the pairs MATH for MATH, skew in MATH, and have the property that symmetrising over any three indices gives zero. The corresponding NAME projection of MATH is obtained by skewing over MATH and symmetrising over each of the pairs MATH f... |
math/9905094 | We show the assertion by induction over the number MATH of arguments of the cumulant MATH. To begin with, let us study the case when MATH. Then we have MATH and by the defining relation REF for the free cumulants our assertion reduces to MATH which is true since MATH. Let us now make the induction hypothesis that for a... |
math/9905094 | We only give a sketch of the proof. Applying REF in the form mentioned above in REF , we get MATH where we have to sum over MATH . Because of the assumption ``MATH free" we obtain with REF that all cumulants vanish with the exception of those which have only elements from MATH or only elements from MATH as arguments. T... |
math/9905094 | Applying REF in the particular form of REF yields MATH with MATH . We claim now the following: The partitions MATH which fulfill the condition MATH are exactly those which have the following properties: the block of MATH which contains the element MATH contains also the element MATH, and, for each MATH, the block of MA... |
math/9905094 | We examine a cumulant MATH with MATH for MATH. According to the definition of MATH-diagonality we have to show that this cumulant vanishes in the following two cases: CASE: MATH is odd. CASE: There exists at least one MATH such that MATH. By REF , we have MATH where MATH. The fact that MATH and MATH are MATH-free impli... |
math/9905094 | CASE: In order to show that the joint distributions of MATH and MATH are identical, we have to prove according to REF that MATH for all MATH, MATH and MATH . In the cases when MATH is odd or when with even MATH the elements MATH do not alternate, the cumulant MATH vanishes because of the MATH-diagonality of MATH. By RE... |
math/9905094 | MATH-diagonality of MATH is clear by REF . So we only have to prove REF . By REF , we get MATH where MATH. Since MATH and MATH are assumed to be free, we also know, by REF , that for a contributing partition MATH each block has to contain components only from MATH or only from MATH. As in the proof of REF one can show ... |
math/9905094 | For notational convenience we deal with the case MATH. General MATH can be treated analogously. The cumulants which we must have a look at are MATH with arguments MATH from MATH. We write MATH with MATH. According to the definition of MATH-diagonality we have to show that for any MATH the cumulant MATH vanishes if (at ... |
math/9905094 | Since we know that both MATH and MATH are MATH-diagonal we only have to see that the respective alternating cumulants coincide. By REF , we have MATH and MATH where in both cases MATH. The only difference between both cases is that in the second case we also have to take care of the freeness between the MATH which impl... |
math/9905096 | Consider MATH with coordinates MATH and canonical basis MATH; let MATH be the curve MATH. We consider the non degenerate symmetric bilinear form MATH on MATH given by MATH for MATH, MATH, and MATH otherwise. The choice of the sign of MATH is done according to whether MATH should be timelike or spacelike, as desired. Le... |
math/9905096 | Let MATH be an open neighborhood of the origin such that the exponential map MATH of MATH maps MATH diffeomorphically onto an open neighborhood of MATH in MATH. Regarding MATH as a coordinate map around MATH, it is well known that the NAME symbols of the NAME - NAME connection vanish at the point MATH. Hence, the covar... |
math/9905096 | If MATH is not lightlike, let MATH be the isomorphism given by the chosen trivialization of the normal bundle to MATH. For a lightlike MATH, let's denote by MATH the isomorphism determined by the choice of the trivialization of the quotient bundle, as described above. By construction, MATH carries MATH (or MATH for MAT... |
math/9905096 | Let MATH be the multiplicity of the focal instant MATH. Let MATH be a basis of MATH such that MATH are a basis for MATH and MATH for MATH. The vectors MATH are a basis of MATH. To prove this, we first observe that they belong to MATH; namely, by REF, if MATH and MATH, we have MATH . To prove the claim, we need to show ... |
math/9905096 | It is an easy consequence of REF . |
math/9905096 | Let MATH be transverse to MATH, and let MATH be the linear operator whose graph in MATH is MATH. Then, MATH is Lagrangian if and only if MATH for all MATH, that is, if and only if MATH . This is just the symmetry of the bilinear form MATH. We now prove that the differential MATH does not depend on the choice of the com... |
math/9905096 | Let MATH be any complementary Lagrangian to MATH, MATH, and let MATH be the corresponding coordinate map around MATH. Recall that the differential MATH at MATH is the isomorphism used to identify MATH with MATH (see REF ). Let MATH and MATH be the projections onto the summands. In the chart MATH, the map MATH is given ... |
math/9905096 | By choosing a symplectic basis for MATH, we reduce the problem to the case MATH and MATH. The group MATH acts smoothly on MATH; we show that the restriction of this action to MATH is transitive on MATH. Let MATH be fixed; we consider bases MATH and MATH of MATH and MATH respectively, which are orthonormal relatively to... |
math/9905096 | By choosing a symplectic basis of MATH, we can reduce to the case MATH, MATH and MATH (see REF ); let MATH be the canonical basis of MATH. Let MATH be any Lagrangian such that MATH; we show that there is an element MATH such that MATH. Let MATH be a linear isometry of MATH such that MATH; now consider the complex linea... |
math/9905096 | By choosing a symplectic basis of MATH, we can reduce to the case MATH, MATH and MATH (see REF ). Let MATH be the canonical basis of MATH and MATH be the subspace generated by MATH, where MATH. Since MATH and MATH are both in MATH, REF gives a symplectomorphism MATH of MATH such that MATH and MATH. Observe that the dia... |
math/9905096 | To prove that MATH is an embedded submanifold of MATH, observe first that, by REF , MATH is an orbit of the action of MATH. It follows that MATH is an immersed submanifold, that is, it does not necessarily have the relative topology. By CITE, an orbit is embedded if and only if it is locally closed, that is, it is the ... |
math/9905096 | We start by determining the universal covering of the quotient MATH. Towards this goal, we consider the transitive action of MATH on MATH given by MATH, for all MATH and MATH. The isotropy group of MATH is MATH; we have therefore a diffeomorphism MATH given by MATH, for all MATH. Since MATH is discrete, then the map MA... |
math/9905096 | We can clearly assume that MATH, MATH, with MATH the canonical basis of MATH, hence, MATH. We apply REF to MATH, MATH; by REF , we can identify MATH with MATH, and the quotient map MATH is given by MATH. Let MATH be the NAME group of unitary MATH complex matrices having determinant equal to MATH; the universal covering... |
math/9905096 | We define the space: MATH it is easy to see that MATH. We also denote by MATH the complement of MATH in MATH; MATH is given by the union of MATH open connected components MATH, given by MATH . Observe that each MATH is indeed path connected, because, by NAME 's Inertia Theorem, it admits a transitive action of the conn... |
math/9905096 | Let MATH. We start with the case where MATH is positive semidefinite, that is, MATH, and MATH is positive definite, that is, MATH. Let MATH be a subspace of MATH which is complementary to MATH and such that MATH is positive definite on MATH. We need to show that MATH is positive definite on MATH for MATH small enough. ... |
math/9905096 | We consider the case of positive intersections; the other case is then easily obtained by passing to the backwards orientation. By reparameterizing, we can assume that both curves intercept MATH at the same instant MATH. By REF , there exists MATH such that MATH. Let MATH denote the diffeomorphism of MATH given by MATH... |
math/9905096 | We start observing that, by REF , the NAME index MATH depends only on the numbers MATH and MATH. To prove the statement, it suffices to exhibit for each MATH a curve MATH, such that MATH, MATH, and such that the curve MATH has NAME index equal to MATH. Clearly, since we can consider curves reparameterized backwards, it... |
math/9905096 | If there are infinitely many MATH such that MATH, then the right hand side of REF is infinite, and the statement of the Corollary is trivial. Otherwise, let MATH be such that MATH and let MATH be a Lagrangian complementary to both MATH and MATH (see REF ). Set MATH; then, MATH is a curve in MATH defined in a neighborho... |
math/9905096 | Let MATH be such that MATH and let MATH be a Lagrangian complementary to both MATH and MATH (see REF ). Set MATH; then, MATH is a curve in MATH defined in a neighborhood of MATH. It is easily seen that MATH, and it follows from REF that MATH and MATH coincide in MATH. Applying REF around MATH, once to MATH and again to... |
math/9905096 | Using REF implies that the number of MATH-focal instants is finite, and so MATH is well defined (see REF ). Now, using REF , we compute as follows: MATH for all MATH and for all MATH such that the pairs MATH and MATH belong to MATH, that is, for all MATH. By REF , MATH is non degenerate on MATH; moreover, we have: MATH... |
math/9905096 | By REF , since MATH is not MATH-focal, then MATH is not a MATH-focal instant. The conclusion follows at once from REF . |
math/9905096 | For each MATH, define the objects MATH, MATH, MATH and MATH relative to the quadruple MATH as in REF respectively. A simple calculation using the charts described for MATH and for MATH shows that MATH in MATH, and therefore in MATH. Obviously, MATH tends to MATH uniformly on MATH; by standard results about the continuo... |
math/9905096 | We choose a local trivialization of the tangent bundle MATH around MATH by linearly independent smooth vector fields MATH. Since MATH as MATH, we can assume without loss of generality that MATH is in the domain of the MATH's, for all MATH. Now, we trivialize the tangent bundle along each MATH, MATH, by considering the ... |
math/9905096 | Using the NAME - NAME inequality, we compute easily: MATH . REF follows easily. |
math/9905096 | Since MATH is non degenerate on MATH, then MATH and MATH are complementary subspaces in MATH; let MATH be any positive definite inner product on MATH which makes MATH and MATH orthogonal, and denote also by MATH its extension to a Hermitian product in MATH. We denote by MATH the corresponding Hermitian product in MATH.... |
math/9905096 | The map MATH is clearly injective on MATH, and it is onto by REF. We compute the derivative MATH; recalling REF , we have: MATH the pull-back of MATH by MATH is a symmetric bilinear form on MATH given by: MATH . We want to calculate the derivative of the pull-back REF with respect to MATH. First, we differentiate MATH:... |
math/9905096 | It follows immediately from REF . |
math/9905096 | Let MATH be chosen so that MATH for all MATH and all MATH. By REF , to such purpose it suffices to take MATH large enough. By REF , we can find MATH small enough, so that MATH for all MATH and for all MATH. We will consider the restriction of MATH to the rectangle MATH. Now, the NAME index MATH of the curve MATH, MATH ... |
math/9905096 | Let MATH and MATH be the order of zeroes of MATH at MATH and MATH respectively, MATH. Define the following constant: MATH and let MATH be such that MATH and MATH . Finally, let MATH be the infimum of MATH on the interval MATH. The desired neighborhood of MATH is defined by requiring that MATH if and only if: MATH . To ... |
math/9905108 | We may further excise in the relative homology MATH from REF and get an isomorphism: MATH which is induced by inclusion of pairs of spaces. This also shows that each inclusion MATH induces an injection in homology MATH. All the points REF follow from this. Note that REF also follows by excision. |
math/9905108 | REF . A general NAME duality result (see for example, CITE) says that, since we work with triangulable spaces, we have: MATH . Next, the cohomology group splits, through excision, into local contributions, by our localization result REF : MATH where the second equality holds because MATH is contractible, for small enou... |
math/9905108 | CASE: Let MATH denote the projectivised relative conormal of MATH. The key argument we shall use here is the independence of MATH from the multiplicative unit MATH, which was proved by CITE. Since MATH can be identified with MATH, where MATH denotes the space of hyperplanes through MATH in MATH, we may consider the pro... |
math/9905108 | We take back the notations of REF . Since MATH is a finite set of points on MATH, the variation of topology of the fibres of MATH is localisable at those points (compare REF ). Let MATH. For a point MATH, it follows by the classical result of NAME for holomorphic functions with isolated singularity CITE that the pair M... |
math/9905108 | The hypothesis implies that the germ of MATH at MATH is just the point MATH. For any MATH small enough, the germ MATH is locally defined by the function: MATH . We have that, locally at MATH, the singular locus MATH is equal to MATH, in particular included into MATH. Consider the map MATH. Note that the polar locus MAT... |
math/9905130 | To show that MATH are NAME coefficients for a well-defined distribution MATH, we have to show that they are polynomially bounded as a function of MATH. Clearly, this only involves the semi-simple part of MATH (indeed, if MATH is abelian, there is nothing to prove since MATH in that case and the map MATH is the identity... |
math/9905130 | Let MATH and MATH be given. Then MATH while on the other hand MATH . Hence, MATH. Similarly, MATH, proving REF. REF are obtained from REF as follows: MATH . |
math/9905130 | Using the NAME basis for MATH, the only non-vanishing entries in MATH are MATH. Up to permutation of indices, the only non-vanishing structure constants are of the form MATH. One verifies: MATH . REF follows from the calculation, using REF, MATH . The last REF is just the infinitesimal version of the MATH-invariance of... |
math/9905130 | We use the symbol map MATH to identify MATH with the differential REF on MATH. Since MATH, we find MATH . In this expression, the terms cubic in contractions MATH cancel thanks to the NAME REF for MATH. The remaining terms combine, using REF, to MATH . Composing this operator with MATH kills the first term, and also th... |
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