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 analogous or follow from REF above. |
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 increasing subset size at most MATH if and only if the off-diagonal subset has increasing subset size at most MATH. Since the points were added via a generalized NAME process, the off-diagonal subset itself corresponds to a generalized NAME process; the lemma is then immediate. |
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 fact that MATH is REF when MATH is even and MATH and is REF otherwise. |
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 that for every MATH with MATH, there is a unique MATH with MATH, and for that MATH, MATH. |
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 occur. In either case, we note that MATH so MATH; it follows that MATH. In the second case, MATH, and thus MATH. As these conditions are readily seen to be sufficient, the result follows. |
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 MATH . |
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 block matrix MATH where MATH . Equivalently, via row and column operations, we could take MATH . Upon applying MATH then MATH, we obtain MATH . But then MATH as required. |
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 that MATH can be defined in terms of a sum over self-dual bitableaux. Clearly, it suffices to show that the usual correspondence between self-dual tableaux and pairs of tableaux extends to bitableaux. Define a domino MATH-bitableau of shape MATH to be a tiling of the diagram of MATH with labelled dominos such that the labels increase weakly in each row and column, and such that REF for MATH, no column hits more than one domino labelled MATH, and REF for MATH, no row hits more than one domino labelled MATH. We readily verify (using the fact that deflation preserves the bitableau property) that we have a bijection between self-dual bitableaux and domino bitableaux. To proceed from domino bitableaux to pairs of ordinary bitableaux, it suffices to show that the usual correspondence preserves the bitableau property. But this follows from the fact that the correspondence is valid for column-strict tableaux (and thus for row-strict tableaux by symmetry), as remarked in CITE. Combining these bijections, we obtain the desired correspondence, and thus the theorem for MATH and MATH. |
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 least one basis vector more than once, so will be taken to REF by MATH. |
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 the lemma, it follows that MATH . Now each permutation MATH on the right hand side agrees with MATH outside MATH, but is no longer decreasing on MATH. It follows that each MATH has strictly fewer inversions than MATH (reduce to the case in which MATH differs from MATH by a REF-cycle). It follows that if we iterate this reduction, we will eventually obtain a linear combination of permutations that cannot be reduced. But this is precisely the desired result. It remains only to show that the elements MATH are linearly independent, which we will do via a triangularity argument. If we choose a basis of MATH, we can view MATH as the restriction to MATH of the corresponding action of MATH on the set MATH. Now, to a given permutation MATH, we associate two words MATH and MATH as follows: MATH is equal to the length of the longest decreasing subsequence of MATH starting with MATH; similarly MATH is equal to the length of the longest decreasing subsequence of MATH starting in position MATH. (Since MATH has longest decreasing subsequence of length MATH, these are indeed in MATH.) We easily see that MATH, and thus that MATH has coefficient MATH on the pair MATH. The claim is then that any other permutation taking MATH to MATH has strictly more inversions than MATH. Define the number of inversions MATH of a word MATH to be the number of coordinate positions MATH such that MATH. By standard arguments, we find that if MATH, then MATH; equality holds if for any pair MATH such that MATH, we have MATH and MATH. We immediately deduce that REF MATH and REF if MATH, then MATH. Finally, if MATH with MATH, then MATH has not only the same number of inversions as MATH, but indeed the same set of inversions; it follows that MATH. |
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 positions of MATH corresponding to the MATH-decreasing subsequence. As before, we have MATH . Now, we readily see that for any permutation that appears in the left-hand-side, the corresponding multiset has at most as many inversions as MATH. Indeed, the number of inversions is strictly smaller unless the corresponding multiset is equal to MATH. Thus it remains only to show that the terms corresponding to MATH do not cancel. The only way that the term MATH can correspond to MATH is if we can write MATH where MATH fixes the complement of MATH and MATH fixes the complement of MATH. Now, by the definition of MATH-increasing subsequence, MATH contains at most one element when MATH; similarly MATH contains at most one element when MATH. In other words, we can write MATH . But then MATH as desired. The proof of linear independence is analogous to that in REF . Of the permutations associated to MATH, there is a unique one REF such that each element of MATH and MATH induces a decreasing subsequence and each element of MATH and MATH induces an increasing subsequence. We define MATH, MATH, and observe that any other multiset MATH with a nonzero coefficient at MATH satisfies MATH. |
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 second coordinates before applying the above arguments. |
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 is to require that the increasing subsequence consist entirely of points MATH with MATH. NAME with this restriction, the above proof does not entirely carry over; for instance, both MATH and MATH have the same number of inversions (that is, REF). For an involution MATH, denote by MATH the set of MATH with MATH. Given a multiset MATH, we then define MATH to be the multiset corresponding to MATH where MATH corresponds to MATH. Given two multisets MATH and MATH of the same size on the same totally ordered set, we write MATH to indicate that we can identify elements of MATH and MATH in such a way that each element of MATH is MATH the corresponding element of MATH. Then the theorem follows from the following observation: If MATH is one of the multisets obtained after eliminating an increasing subsequence of MATH, then MATH. If equality occurs, then either MATH or MATH has strictly more inversions than MATH. Linear independence follows as above. |
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 on MATH such that MATH. The word MATH associated to an involution is now defined so that MATH is equal to half the length of the longest symmetric decreasing subsequence starting or ending with MATH; the sign is positive if the sequence ends with MATH, and negative otherwise. Other than this, the arguments are analogous. |
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 result follows by summing over MATH. |
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 asymptotics results REF as MATH follow from the calculations in REF, pp. REF, of CITE. For the proof of REF , define a new matrix function MATH . Then MATH satisfies the jump condition MATH for MATH, and MATH as MATH. Since the jump matrix for MATH is independent of MATH, MATH, the derivative with respect to MATH, satisfies MATH, and MATH as MATH. Hence MATH has no jump across MATH, and satisfies MATH as MATH. If we write MATH as MATH, we have MATH as MATH. Thus MATH is entire and as MATH, MATH. Therefore by NAME 's theorem, we obtain MATH . Recalling that MATH in REF, we have MATH. Changing MATH to MATH from REF is MATH . This is REF when MATH. The proof of REF is very similar to that of REF , and the detail is left to the reader. We only note that in the derivation of REF , we need the identity MATH which can be obtained from REF by setting MATH as MATH. Finally we prove REF . Note that MATH. From the jump condition at MATH, we have MATH . Letting MATH, MATH, in REF , we have MATH, which together with REF implies that MATH. Thus we have MATH for some MATH, MATH. Also the condition MATH for all MATH implies that MATH for all MATH, and hence we have MATH . Now letting MATH, MATH in REF, we obtain MATH . Thus from REF, MATH, which has the solution MATH . From REF with MATH, we have MATH . Therefore, MATH, which is MATH from REF. Now REF gives MATH, proving REF. |
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 that MATH and MATH are bounded for MATH from REF. |
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 subsequence of length greater than MATH, then MATH has an increasing subsequence of length greater than MATH. Thus MATH. But since MATH, we obtain MATH. A similar argument works for the other cases. Note that MATH. |
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 main contribution to the sum comes from MATH. Fix MATH. We split the sum in REF into two pieces : MATH where MATH is the region MATH and MATH is the rest. For MATH, the quantity MATH is unimodal for MATH, and the maximum is achieved when MATH as MATH. Hence MATH . Using NAME 's formula for REF, for any fixed MATH, when MATH, MATH . Hence using REF, we have MATH . But MATH is increasing in MATH, is decreasing in MATH, and MATH. Therefore there are positive constants MATH and MATH such that for large MATH, MATH . On the other hand, REF says that (recall REF) for any fixed real number MATH, there is a constant MATH such that for MATH, MATH for sufficiently large MATH. Since MATH, REF for MATH yields MATH . Let MATH. For MATH, hence for MATH, MATH . Also note that from the asymptotics REF, MATH is bounded for MATH. Hence using REF , we obtain MATH . Therefore we have MATH . Similarly, MATH . But from REF, MATH . Thus using REF, we obtain REF. |
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 radius MATH centered at MATH. We will show that the main contribution to the sum in REF comes from MATH. Set MATH . By NAME 's formula, MATH . Hence from the unimodality discussed above, MATH and by summing up using REF, MATH . Hence we have MATH and the main contribution to the sum in REF comes from MATH. As in REF , we write MATH . From REF, MATH . On the other hand, by the remark to REF , (recall REF) MATH for large MATH. We have a similar inequality of the other direction with MATH, MATH and MATH replaced by MATH, MATH and MATH. Let MATH. In the region MATH, MATH and MATH . Hence as in REF, using REF MATH for a constant MATH which may depend on MATH and constants MATH and MATH which is independent of MATH. Thus for large MATH, MATH with some error MATH such that MATH. Similarly we have an inequality for the other direction. Recalling MATH from REF, we obtain REF. |
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 have MATH thus, MATH . On the other hand, when MATH, similarly we have MATH . Using REF, we obtain MATH . Since MATH, MATH thus MATH . Therefore, using dominated convergence theorem, we obtain REF. |
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, and where MATH is the largest increasing subset of MATH lying entirely in the (part-open) trapezoid with MATH, MATH. Since MATH is binomial with parameters MATH and MATH, Let MATH be sufficiently large and fixed. For all MATH, there exist positive constants MATH, MATH independent of MATH such that for MATH, MATH while for MATH, MATH . For MATH, we have: Let MATH be sufficiently large and fixed. For MATH, there exist positive constants MATH and MATH independent of MATH such that for all MATH, MATH and for all MATH, MATH . We first show the corresponding large-deviation result for the Poissonization. Define MATH to be the length of the longest increasing subsequence when the number of points in the trapezoid is NAME with parameter MATH. Then MATH is bounded between the corresponding processes for the rectangle MATH, MATH and for the triangle MATH. In particular, if MATH deviates significantly from MATH, so must the appropriate bounding process; the result follows immediately from the corresponding results for rectangles and triangles. The corresponding large-deviation result when the number of points is fixed then follows from REF . In our case, the number of points in the trapezoid is binomial with parameters MATH and MATH; the lemma follows via essentially the same argument used to prove REF . As we will see, the value MATH deserves special attention: Let MATH be sufficiently large and fixed. There exist positive constants MATH, MATH such that for MATH, MATH and for MATH, MATH . Moreover, if we define MATH then MATH converges to a standard normal distribution, both in distribution and in moments. That MATH converges as stated follows from the fact that if we write MATH, with MATH then MATH converges in distribution and moments to a standard normal distribution, and MATH converges in distribution and moments to REF. For the large deviation bounds, we note that if MATH, then either MATH or MATH. Thus for any MATH, we have MATH the result follows by balancing the two terms. In the other case, the MATH term always dominates. For any sufficiently small MATH, there exist positive constants MATH, MATH such that MATH and MATH for all sufficiently large MATH. Define a sequence MATH by taking MATH for all MATH. Similarly define a sequence MATH by MATH with MATH for MATH and MATH for MATH. Note that MATH is strictly decreasing and MATH is strictly increasing. For all MATH, MATH . For all MATH, MATH . In the first case, we have MATH . In the second case, it suffices to verify the formula with MATH replaced by MATH. For MATH, MATH . For MATH, MATH . Finally, for MATH, MATH . Let MATH, MATH. Then there exist constants MATH, MATH such that for MATH, MATH . Since MATH is an upper bound on MATH with MATH, it follows that MATH . Similarly, for MATH, MATH and thus MATH . Since there are only MATH such events to consider, the result follows. In particular, with probability MATH, we have MATH . But then using the fact that MATH and MATH are NAME, and using the large deviation behavior of MATH, we find that MATH and MATH . So MATH converges to REF in a fairly strong sense; in particular, they must have the same limiting distribution and limiting moments. |
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 and MATH then by the definition of MATH so differentiation in MATH gives MATH. Then MATH since MATH and MATH and MATH is normal in MATH, so MATH . |
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 has a gap above or, equivalently, that MATH is a projection in MATH not equal to the identity. Then MATH is invariant for MATH and hence MATH. Let MATH; view MATH as a MATH function on MATH. For any MATH, MATH . We use the fact that MATH implies that MATH in MATH. If MATH has a gap above and if there is a point MATH, then MATH and MATH is a non-trivial invariant projection for MATH. If we were not assuming that MATH is normalized to a MATH- representation of MATH, then MATH would be a non-trivial idempotent whose range is an invariant subspace for MATH. Since MATH and MATH is an ideal set, MATH. If MATH, then MATH; hence, MATH. Thus MATH, a contradiction. If MATH has a gap below and if there is a point MATH, then MATH. Essentially the same as for REF : MATH, so MATH. Assume MATH has a gap above, MATH has a gap below, MATH, and MATH. If MATH for all MATH (or, for all MATH in a set with linear span dense in MATH), then MATH is not a nest representation. Since MATH is closed, MATH, for all MATH. As a consequence, we have MATH, for all MATH. Let MATH be the range of MATH and MATH be the norm closure of MATH. Then MATH and MATH are non-zero invariant subspaces for MATH and MATH. So MATH is not totally ordered by inclusion; MATH is not a nest representation. In what follows we use the standard identification of MATH with the diagonal of MATH, that is, with MATH. If MATH is a nest representation, then MATH is an interval in MATH. Suppose not. Then there are three points MATH with MATH in MATH such that MATH, MATH, and MATH. It follows that if MATH with MATH and MATH, then MATH. (Since MATH is open there is a neighborhood of MATH which is contained in MATH; use the assumption that all orbits are dense and the ideal set property for MATH.) Choose MATH and MATH so that MATH, MATH, MATH has a gap above, and MATH has a gap below. Then MATH; hence MATH and the range of MATH is in MATH. Also, MATH, so MATH. For any MATH, MATH . Consequently, MATH. REF now implies that MATH is not a nest representation, contradicting the hypothesis. Assume MATH is a nest representation and that MATH is the left endpoint of MATH and MATH is the right endpoint of MATH. Each of MATH and MATH may or may not be elements of MATH. However, if MATH has a gap above, then without loss of generality, we may assume that MATH (simply replace MATH by its immediate successor, if necessary). Similarly, if MATH has a gap below, we may assume that MATH. If MATH and MATH, then MATH. Suppose that MATH, MATH, and MATH. Choose MATH and MATH so that MATH, MATH has a gap above, and MATH has a gap below. [Exceptions: if MATH, choose MATH and note that this element is not in MATH; if MATH, choose MATH and note that this is not in MATH.] It follows that MATH is non-zero and has range in MATH and that MATH. Let MATH be arbitrary. If MATH is in MATH, then MATH and MATH; hence MATH. Thus MATH, for all MATH. REF now implies that MATH is not a nest representation, a contradiction. If MATH is a nest representation and MATH and MATH are the endpoints of the interval MATH, then MATH . The conclusion follows from the following implications for a point MATH: MATH . Given MATH and MATH in MATH, let MATH . If MATH has no gap above, then MATH. If MATH has no gap below, then MATH. This follows immediately from the fact that MATH is an open subset of MATH. If MATH has a gap above, then either MATH or MATH. Assume the contrary. Then MATH and there is MATH such that MATH and MATH. Consider two cases. First, assume that MATH. In this case, since MATH has a gap below, we also have available the assumption that MATH. (See the comments after REF .) Since MATH and MATH, both MATH and MATH are non-zero. Furthermore, for any MATH, MATH. Indeed, if MATH, then MATH and MATH. If either MATH or MATH, the MATH. If MATH and MATH, then MATH; since MATH, we also have MATH. REF implies that MATH is not a nest representation, a contradiction. In the alternative case, MATH and there is MATH such that MATH and MATH has a gap below. Since MATH, MATH. As before, MATH. Let MATH, where MATH is any element of MATH. If MATH then MATH. If MATH then MATH, whence MATH. Once again, REF yields a contradiction If MATH has a gap below, then either MATH or MATH. The idea behind the proof is essentially the same as in the proof of REF . This time, if the conclusion does not hold, then MATH and there is MATH such that MATH and MATH. If MATH then take MATH; otherwise, take MATH so that MATH and MATH has a gap above. Now apply REF to MATH and MATH. Conclusion REF imply that MATH has one of the following two forms: MATH . Note that the latter is a possibility only if MATH and MATH is open. If MATH has a gap above and MATH has a gap below and either MATH with MATH or MATH, then MATH is not a nest representation. Apply REF to MATH and MATH. This effectively ends the proof of REF . If MATH is a nest representation, then MATH is one of the ideal sets listed in REF . Since these are all meet irreducible, we have proven that MATH is meet irreducible. |
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 need to prove the existence of such a point. Since MATH is a NAME set, we can find, for each MATH, MATH disjoint clopen sets MATH, MATH, whose union is MATH, with the further property that any decreasing sequence of these sets has one-point intersection. Since MATH is an atom of MATH, for each MATH there is a unique integer MATH in MATH such that MATH. Let MATH be such that MATH. Clearly, MATH. But MATH is an atom when it is non-zero; hence MATH. The uniqueness of MATH follows immediately form the orthogonality of MATH and MATH when MATH. |
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 and, hence, that MATH and MATH have orthogonal ranges when MATH. If MATH and MATH, then MATH. It is possible that MATH even when MATH and 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 MATH is contained in the diagonal, MATH is a projection in MATH and MATH either dominates MATH or is orthogonal to MATH. In this case, either MATH and MATH or MATH and MATH. When MATH is disjoint form the diagonal of MATH, there is MATH such that MATH and MATH. In this case, MATH and MATH. In particular, MATH and MATH for all MATH. It now follows that MATH and MATH while MATH and MATH. But MATH is a nest representation and MATH and MATH are MATH-invariant; hence one must contain the other. Thus, the rank of MATH is at most MATH. |
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 have seen that MATH, the smallest MATH-invariant subspace containing MATH, is disjoint from MATH; since MATH is a nest representation, it follows that MATH. Thus, for some MATH, MATH. Since the ranges of MATH and MATH are one-dimensional, MATH. For such a MATH we have MATH. By the first paragraph, MATH can be replaced by any MATH with MATH. |
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; hence MATH is an atom in MATH. As MATH majorizes no atoms, MATH. On the other hand, MATH, so that MATH. This contradiction shows that MATH is invariant. If MATH is not invariant, there exists a vector MATH and a MATH-normalizing partial isometry MATH in MATH such that MATH. Since MATH is the sum of the atoms it majorizes, MATH for some MATH. By REF , there is an element MATH with MATH, whence MATH. In particular, MATH, which contradicts the fact that MATH majorizes no atoms. Thus, MATH is also invariant. |
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 of MATH with MATH. Since MATH is strongly maximal, there exist MATH with MATH and MATH. Now MATH; hence MATH. As MATH, it follows that MATH and, hence, MATH. This shows that MATH is an interval. |
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 ideal MATH contains MATH. But the complement is a closed set, so it contains MATH. On the other hand, if MATH is a MATH-normalizing partial isometry with MATH disjoint from MATH, then MATH. Thus, MATH . By CITE, MATH is a meet-irreduclble ideal. |
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 obvious. |
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 sequence MATH by a subsequence, we may assume that MATH and MATH lie in MATH. Since MATH, MATH. By REF , MATH where the sum is taken over all pairs of atoms and is convergent in the strong operator topology. Consequently, there exist points MATH and MATH in MATH such that MATH. For each MATH, let MATH and MATH be the unique diagonal matrix units in MATH such that MATH and MATH. Since MATH, MATH, for all MATH. On the other hand, MATH, so MATH, for all MATH. Let MATH and MATH in MATH and MATH and MATH in MATH be analogously defined for the ideal MATH. This time, MATH and MATH. We shall show (after possibly reversing the roles if MATH and MATH) that for infinitely many MATH, MATH and MATH. This leads to a contradiction with the hypothesis that MATH is a nest representation and so shows that MATH is necessarily meet irreducible. Indeed, from MATH it follows that MATH and hence that MATH and MATH, where MATH and MATH are the smallest MATH-invariant subspaces containing MATH and MATH respectively. But then MATH and MATH are not related by inclusion, a contradiction. Each finite dimensional algebra MATH is a direct sum of MATH's and the matrix units MATH and MATH are in the same summand, as are MATH and MATH. If these two summands differ, then MATH and MATH. Should this occur for infinitely many MATH, then we are done. So we need consider only the case in which, for all MATH, all of MATH, MATH, MATH and MATH are in the same summand in MATH. If MATH and MATH are diagonal matrix units (minimal diagonal projections) in MATH, let MATH be the matrix unit in MATH with initial projection MATH and final projection MATH (if there is such a matrix unit). If MATH, then MATH in the diagonal order on minimal diagonal projections. We shall need the following property of ideals in MATH: if MATH and if MATH is in an ideal, then MATH is also in the ideal. Since MATH and MATH are in the same MATH-summand of MATH, they are related in the diagonal order. By interchanging MATH and MATH and passing to a subsequence, if necessary, we may assume that MATH, for all MATH. The facts concerning the membership of MATH and MATH in MATH and MATH may be rephrased as MATH . As a consequence MATH, for all k. (If MATH for some MATH, then MATH. Since MATH, we have MATH, a contradiction.) But now, MATH . Thus MATH; hence MATH. Also, since MATH, MATH. As pointed out earlier, this implies that MATH is not a nest representation; so, when MATH is a nest representation with an atomic lattice, MATH is meet irreducible. |
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 if MATH, and this is the case if and only if MATH for each MATH. But MATH for MATH is the typical element in MATH. CASE: This well known variant of the NAME relations follows by duality (see CITE): MATH . CASE: This is due to CITE. There it is proved using exterior algebra. Apparently, this result is included in REF , page REF. CASE: Another proof using representation theory will be given below. Here we prove it by induction on MATH. Let MATH. Suppose that MATH for all MATH. Then for all MATH we have MATH. Interchange MATH and MATH in the last expression and add it to the original, then we get MATH and in turn MATH for all MATH and MATH, which are the original NAME relations, so MATH is decomposable. Now the induction step. Suppose that MATH and that MATH for all MATH. Then we have MATH for all MATH, so that by induction we may conclude that MATH is decomposable for all MATH, and then by REF MATH is decomposable. CASE: Again this follows by duality. |
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 for MATH. Its vanishing, therefore, is equivalent to the vanishing of MATH for all MATH. According to REF , this means that MATH is simple. Therefore, the theorem is equivalent to REF . |
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 an integer MATH the theorem is true for all MATH. We want to show that it also holds for MATH. This means that for MATH, a sequence MATH, and random variables MATH we have to prove the validity of the following equation: MATH where MATH. The proof is divided into two steps. The first one discusses the case where MATH and the second one treats the case where MATH. CASE: The validity of relation REF for all MATH except the partition MATH is shown as follows: Each such MATH has at least two blocks, so it can be written as MATH with MATH being a non-crossing partition of a MATH-tuple MATH and MATH being a non-crossing partition of a MATH-tuple MATH where MATH and MATH. With these definitions, we have MATH . We will apply now the induction hypothesis on MATH and on MATH. According to the definition of MATH, both MATH and MATH are products with factors from MATH. Put MATH the tuple containing all factors of MATH and MATH the tuple consisting of all factors of MATH; this means MATH (and MATH). We put MATH and MATH, that is, we have MATH. Note that MATH factorizes in the same way as MATH. Then we get with the help of our induction hypothesis: MATH . CASE: It remains to prove that REF is also valid for MATH. With REF , we obtain MATH . First we transform the sum in REF with the result of REF: MATH where we used the fact that MATH is equivalent to MATH. The moment in REF can be written as MATH . Altogether, we get: MATH . |
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. This means that all partitions MATH contributing to the sum must have the form MATH with MATH being in MATH and MATH being in MATH. Obviously, for each such MATH we have MATH . One can now convince oneself, that for each MATH there exists exactly one MATH such that MATH fulfills the condition MATH and that this MATH is nothing but the complement of MATH, that is, we have to sum exactly over all MATH with MATH. This is the assertion. |
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 MATH which contains the element MATH contains also the element MATH . Since the set of those MATH fulfilling the claimed condition is in canonical bijection with MATH and since MATH goes under this bijection to the product appearing in REF , this gives directly the assertion. So it remains to prove the claim. It is clear that a partition which has the claimed property does also fulfill MATH. So we only have to prove the other direction. Let MATH be the block of MATH which contains the element MATH. Since MATH is MATH-diagonal the last element of this block has to be a MATH, that is, an even number, let's say MATH. If this would not be MATH then this block MATH would in MATH not be connected to the block containing MATH, thus MATH would not give MATH. Hence MATH implies that the block containing the first element MATH contains also the last element MATH. MATH . Now fix a MATH and let MATH be the block of MATH containing the element MATH. Assume that MATH does not contain the element MATH. Then there are two possibilities: Either MATH is not the last element in MATH, that is, there exists a next element in MATH, which is necessarily of the form MATH with MATH . MATH or MATH is the last element in MATH. In this case the first element of MATH is of the form MATH with MATH. MATH . In both cases the block MATH gets not connected with MATH in MATH, thus this cannot give MATH. Hence the condition MATH forces MATH and MATH to lie in the same block. This proves our claim and hence the assertion. |
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 implies, by REF , that only such partitions MATH contribute to the sum each of whose blocks contains elements only from MATH or only from MATH. Case CASE : As there is at least one block of MATH containing a different number of elements MATH and MATH, MATH vanishes always. So there are no partitions MATH contributing to the sum in REF which consequently vanishes. Case CASE : We assume that there exists a MATH such that MATH. Since with MATH also MATH is MATH-diagonal, it suffices to consider the case where MATH, that is, MATH and MATH. Let MATH be the block containing MATH. We have to examine two situations: CASE: On the one hand, it might happen that MATH is the first element in the block MATH. This can be sketched in the following way: MATH . In this case the block MATH is not connected with MATH in MATH, thus the latter cannot be equal to MATH. CASE: On the other hand, it can happen that MATH is not the first element of MATH. Because MATH is MATH-diagonal, the preceeding element must be a MATH. MATH . But then MATH will again not be connected to MATH in MATH. Thus again MATH cannot be equal to MATH. As in both cases we do not find any partition contributing to the investigated sum in REF this has to vanish. |
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 REF and the fact that MATH is MATH-diagonal, we get that MATH is MATH-diagonal, too, and therefore MATH also vanishes. Hence we have to consider the case where the arguments MATH alternate (which implies alternating arguments MATH). We inductively show the validity of MATH and MATH for any natural MATH. First, consider MATH. On one hand, the equation MATH holds by definition of MATH. With both cumulants MATH and MATH vanishing because of the MATH-diagonality of MATH the second term of the sum is equal to zero. Since MATH and MATH are assumed to be free, we can write the moment with the help of REF as MATH . So we get MATH. On the other hand, with MATH being MATH-diagonal we obtain MATH . Induction hypothesis: Assume the following to be true for any MATH: MATH . We have to show the validity of these equations for MATH. It suffices to consider the first equation. According to definition of the free cumulants we have MATH . Because of the freeness of MATH and MATH and with the help of REF we get MATH . It follows that MATH . The only partitions MATH contributing in the foregoing sum are those where all blocks are alternating in MATH and MATH. According to our induction hypothesis, we can then replace in all blocks the element MATH by MATH and the element MATH by MATH. So we finally obtain MATH . CASE: We assume that MATH. As, by REF , MATH is MATH-diagonal, MATH is MATH-diagonal, too. |
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 that the requirement MATH is equivalent to the following properties of MATH: The block containing REF must also contain MATH and, for each MATH, the block containing MATH must also contain MATH. (This couples always MATH with MATH and MATH with MATH, so it is compatible with the MATH-freeness between MATH and MATH.) The set of partitions in MATH fulfilling these properties is in canonical bijection with MATH. Furthermore we have to take care of the fact that each block of MATH contains either only elements from MATH or only elements from MATH. For the image of MATH in MATH this means that it splits into blocks living on the odd numbers and blocks living on the even numbers. Furthermore, under these identifications the quantity MATH goes over to the expression as appearing in our REF . |
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 least) one of the following things happens: CASE: There exists a MATH with MATH. CASE: MATH is odd. REF yields MATH where MATH. The MATH-diagonality of MATH implies that a partition MATH gives a non-vanishing contribution to the sum only if its blocks link the arguments alternatingly in MATH and MATH. Case CASE : Without loss of generality, we consider the cumulant MATH with MATH for some MATH with MATH. This means that we have to look at MATH. REF yields in this case MATH where MATH. In order to find out which partitions MATH contribute to the sum we look at the structure of the block containing the element MATH; in the following we will call this block MATH. There are two situations which can occur. The first possibility is that MATH is the first component of MATH; in this case the last component of MATH must be a MATH and, since each block has to contain the same number of MATH and MATH, this MATH has to be the third MATH of an argument MATH. But then the block MATH gets in MATH not connected with the block containing MATH and hence the requirement MATH cannot be fulfilled in such a situation. MATH . The second situation that might happen is that MATH is not the first component of MATH. Then the preceeding element in this block must be a MATH and again it must be the third MATH of an argument MATH. But then the block containing MATH is again not connected with MATH in MATH. This possibility can be illustrated as follows: MATH . Thus, in any case there exists no MATH which fulfills the requirement MATH and hence MATH vanishes in this case. Case CASE : In the case MATH odd, the cumulant MATH has a different number of MATH and MATH as arguments and hence at least one of the blocks of MATH cannot be alternating in MATH and MATH. Thus MATH vanishes by the MATH-diagonality of MATH. As in both cases we do not find any partition giving a non-vanishing contribution, the sum vanishes and so do the cumulants MATH. |
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 implies that only such MATH contribute which do not connect different MATH. But the MATH-diagonality of MATH implies that also in the first case only such MATH give a non-vanishing contribution, that is, the freeness in the second case does not really give an extra condition. Thus both formulas give the same and the two distributions coincide. |
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. Let MATH be endowed with the conformally flat metric MATH, where MATH is a smooth function in MATH that vanishes together with its partial derivatives on the MATH-axis. The factor MATH will be chosen so that the corresponding metric MATH will satisfy the required properties. To this goal, we recall some formulas about the covariant derivative and the geodesic equation in general conformal metrics. Let MATH and MATH denote the covariant derivative or the gradient operators in the metrics MATH and MATH respectively; note that the covariant derivative MATH is the usual directional derivative in MATH, although the gradient MATH is not the usual gradient in MATH. For smooth vector fields MATH in MATH, we have: MATH moreover, the geodesic equation in MATH is: MATH . Since MATH on MATH, then MATH is a geodesic in MATH; moreover, by REF, the parallel vector fields along MATH in MATH are just the constant vector fields. Hence, we trivialize the normal bundle along MATH in MATH by choosing the first MATH vectors of the canonical basis as a parallel moving frame. To compute the NAME equation along MATH in MATH, we linearize the geodesic REF , obtaining: MATH where MATH is the ordinary second derivative in MATH and MATH is the MATH-symmetric linear operator given by MATH. In the deduction of REF we have used the fact that MATH and MATH vanish on MATH. Observing that the covariant derivative along MATH in MATH equals ordinary derivative in MATH and comparing REF with the general NAME REF we see that the curvature tensor MATH of MATH along MATH is given by: MATH . It is easily checked that: MATH if we set: MATH and consider an arbitrary smooth extension of MATH on MATH, then the assignment MATH gives the required function. To conclude the proof, we now need to exhibit a submanifold MATH of MATH, passing through MATH with tangent space MATH, and whose second fundamental form in the normal direction MATH equals MATH. This will follow immediately from the next Lemma, in which we prove something slightly more general. The last assertion in the statement of the proposition is totally obvious. |
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 covariant derivative at this point coincide with the usual directional derivative in MATH. If MATH is a submanifold of MATH passing through MATH and MATH, then, since MATH is the identity map, the tangent space MATH is MATH; moreover, by the above observation about the covariant derivative, the second fundamental form of MATH at MATH equals the second fundamental form of MATH at MATH in the flat space MATH. We define MATH to be the smooth submanifold of MATH given by the graph of the map MATH in the decomposition MATH, namely: MATH . The conclusion follows from an elementary calculation of the second fundamental form of MATH. |
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 MATH lightlike) to MATH. For MATH non lightlike, it is easily checked that MATH carries MATH onto MATH by observing the correspondence between MATH-Jacobi fields orthogonal to MATH and MATH-solutions of REF. Similarly, if MATH is lightlike, MATH carries the quotient MATH onto MATH (see REF ). For MATH non lightlike, taking the orthogonal complements of MATH in MATH and of MATH in MATH, using REF we conclude that MATH induces the desired isomorphism between MATH and MATH. If MATH is lightlike, we take the orthogonal complements of MATH in MATH and of MATH in MATH. Again, using REF we get that MATH induces an isomorphism between the image of MATH in the quotient space MATH and MATH. To conclude the proof, we observe that REF implies that MATH does not belong to MATH, which implies that it maps isomorphically into MATH. |
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 that the vectors MATH are linearly independent, because MATH, by REF. To see this, observe that the fields MATH are linearly independent in MATH, hence the pairs MATH are linearly independent in MATH. The conclusion follows from the fact that MATH. We now define a family of continuous vector fields MATH along MATH, by setting: MATH and MATH . The vectors MATH are now a basis for MATH. Namely, the first MATH vectors MATH are a basis for MATH, and the remaining MATH vectors MATH are a basis for MATH; moreover, MATH is non degenerate on MATH, which implies that MATH. By continuity, the vectors MATH are a basis for MATH for MATH sufficiently close to MATH. But that implies that, for MATH sufficiently close to MATH and MATH the vectors MATH are a basis for MATH, which implies that there are no MATH-focal instants around MATH. The case MATH is treated similarly, observing that MATH and considering that MATH is non degenerate on MATH. |
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 complementary Lagrangian MATH; observe that, by REF , we can always find complementary NAME to MATH. To prove the claim, let MATH and MATH be two complementary NAME to MATH; the two charts MATH and MATH map MATH to the zero bilinear map. We have to prove that the differential of the transition map from MATH to MATH at MATH is the identity of MATH. The transition map is given by: MATH where MATH is the restriction to MATH of the projection MATH and MATH is the identity on MATH. The differential of REF at MATH is easily computed to be the identity. It remains to prove the commutativity of REF. Let MATH and MATH be the domains of the charts MATH and MATH respectively. Then, it is easy to check the commutativity of the diagram: MATH . The conclusion follows by differentiating REF. |
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 by: MATH where MATH and MATH. REF holds for MATH in a neighborhood of MATH, where MATH is invertible. The differential of REF is then easily computed as: MATH where MATH. The conclusion follows at once from the definition of MATH. |
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 the Euclidean inner product of MATH. Since the imaginary part of the Hermitian product is MATH, and MATH vanishes on both MATH and MATH, then MATH and MATH are orthonormal basis of MATH with respect to the Hermitian product. Hence, there exists an element of MATH that carries MATH to MATH, and MATH acts transitively on MATH. Obviously, the isotropy group of MATH is MATH, which concludes the proof. |
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 linear extension of MATH to MATH. Such a map has the required property. Let MATH be the subspace generated by MATH. Then, MATH is Lagrangian, and MATH. It remains to prove that, given a Lagrangian MATH with MATH, there exists an element MATH such that MATH. To prove this claim, we define the following spaces. Let MATH be the space generated by MATH; MATH be generated by MATH and MATH be generated by MATH. Observe that MATH is the orthogonal complement of MATH with respect to MATH; also, MATH, and MATH restricts to the canonical symplectic forms of MATH and of MATH, that will be still denoted by MATH. Let MATH be the restriction to MATH of the projection MATH. It is easy to check that MATH for all MATH. Since MATH is Lagrangian, we have MATH; moreover, it is easily seen that MATH is Lagrangian in MATH. Since MATH, we have that MATH is complementary to MATH in MATH. By REF , there exists a symplectomorphism MATH of MATH that is the identity on MATH and carries MATH into MATH. Finally, the required element MATH is given by: MATH . Indeed, MATH, because MATH and MATH are both subspaces of MATH containing MATH that have the same image under MATH. This concludes the proof. |
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 diagonal MATH is a Lagrangian subspace of MATH which is complementary to both MATH and MATH; the desired Lagrangian MATH is, for instance, MATH. |
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 intersection of an open and a closed set. Now, recall REF and simply observe that MATH. We now compute the codimension of MATH in MATH. Let MATH be any Lagrangian complementary to MATH; the NAME group MATH is diffeomorphic to MATH. Namely, we have a diffeomorphism: MATH that associates to each pair MATH the symplectomorphism MATH of MATH whose restriction to MATH is MATH and whose restriction to MATH is equal to: MATH where MATH denotes the transpose map of MATH, and MATH is seen as a linear map MATH. It follows that the dimension of MATH is equal to MATH. The group MATH is the image under MATH of the product MATH, where MATH is the group of orientation preserving isomorphisms of MATH. It follows that MATH and hence MATH is connected. Now, we choose an element MATH and we calculate the dimension of its isotropy group in MATH. To this aim, let MATH be any MATH-dimensional subspace and let MATH be the image under MATH of the annihilator of MATH in MATH. Then, MATH is a Lagrangian in MATH and MATH. The isotropy group of MATH is the image under MATH of the set of pairs MATH such that MATH and MATH vanishes on MATH. It follows that the dimension of this isotropy group is MATH. Hence, using REF , the codimension of MATH in MATH is computed as MATH. We now compute the tangent space MATH at any point MATH. Such a space is given by the image of MATH under the differential MATH, defined in REF : MATH . The elements of MATH vanish on MATH. A simple dimension counting shows that MATH consists precisely of those elements. This completes the proof of the first part of the statement. We now consider the submanifold MATH; by the formula computed above, its codimension in MATH is equal to MATH. The transverse orientation is well defined in the statement of the Proposition, and the naturality follows easily from the commutative REF . |
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 MATH given by MATH is a covering map. Considering the composition of the two maps above, we obtain a covering map MATH given by MATH, MATH. Since MATH is simply connected and MATH is connected, the quotient MATH is simply connected, and so MATH is the universal covering map of MATH. We now determine the group of covering automorphisms of MATH, which is isomorphic to MATH. We recall that an automorphism MATH of MATH is a homeomorphism of MATH such that MATH. For all MATH, the map MATH is an automorphism of MATH which is trivial if MATH. Thus, we have an action of MATH in MATH by automorphisms of MATH, which is transitive and simple (that is, without fixed points) on the fibers of MATH. It follows that MATH is the group of covering automorphisms of MATH, concluding the proof of the first part of the statement. To construct an explicit isomorphism between MATH and MATH, one uses the standard procedure of taking homotopy classes of loops obtained as the images under MATH of curves in MATH that connect the point MATH and a generic point in the fiber MATH. |
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 group of MATH is MATH, with covering map MATH. The group MATH is easily computed as: MATH where by MATH we mean MATH if MATH is even and its complement MATH if MATH is odd. The connected component MATH equals MATH, and MATH is isomorphic to MATH. Such an isomorphism is given by mapping each term of the union in REF into the integer MATH. As a generator for MATH, we choose the term in REF corresponding to MATH; such element is of the form MATH, where MATH is chosen to be the pair MATH, with MATH the diagonal matrix: MATH . Observe that MATH is a traceless Hermitian matrix, so that MATH belongs to the NAME algebra MATH of MATH. In order to determine a generator for MATH, we choose the curve MATH given by MATH, connecting the neutral element of MATH with MATH. The curve MATH in MATH is now easily computed as given in the statement of the Corollary. |
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 connected group MATH. If we set MATH and MATH, then we can find a continuous curve MATH (and MATH) in MATH (in MATH) from MATH to MATH (from MATH to MATH). Define MATH, MATH. Then, since MATH and MATH do not intersect MATH, the concatenation MATH is homologous to MATH in MATH. Let MATH; then, MATH. Since MATH and MATH have the same endpoints, they admit a fixed endpoint homotopy in the vector space MATH. The composition of such homotopy with MATH gives a homotopy between MATH and MATH through curves with endpoints outside MATH. Hence, MATH and MATH are homologous in MATH, and we are done. |
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. First, since MATH is positive definite on MATH, there is MATH such that MATH is positive definite on MATH for MATH (the set of positive definite symmetric bilinear forms is open). Let MATH be an arbitrary norm on MATH and define: MATH . It is easy to see that, for all MATH small enough, we have MATH so that MATH is positive definite on both MATH and MATH for MATH small enough. We want to show that, if MATH is small enough, then for all MATH and MATH, MATH is positive definite on the two dimensional subspace of MATH generated by MATH and MATH. By the positivity on MATH and MATH, it suffices to prove that, for MATH small enough, the following inequality holds: MATH for all MATH, MATH. Obviously, we can assume MATH. As MATH vanishes on MATH, there exists MATH such that, for all MATH small enough, we have: MATH for all MATH, MATH with MATH. By REF, for all MATH small enough we get: MATH for all MATH, MATH with MATH. This yields REF and concludes the first part of the proof. For the general case, we consider decompositions MATH and MATH, where MATH is positive definite on MATH, negative definite on MATH, and MATH is positive definite in MATH and negative definite in MATH. We then apply the result proven in the first part of the proof to the restriction of MATH to MATH once, and again to the restriction of MATH to MATH. The conclusion follows by observing that MATH is positive definite on MATH and negative definite on MATH, which implies that MATH and MATH for MATH small enough. Clearly, this also implies that MATH is non degenerate. |
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; we deduce from REF that the curve MATH has a unique intersection with MATH, which is transverse and positive. Moreover, by REF , MATH and MATH are homologous in MATH. Without loss of generality, we can therefore assume that MATH. Let MATH be a Lagrangian which is complementary to both MATH and MATH (see REF ). Since MATH, MATH, has a unique intersection with MATH at MATH, then the restriction of MATH to any closed subinterval containing MATH in its interior is homologous to MATH in MATH. Thus, we can assume that the images of MATH and MATH are contained in the domain MATH of the chart MATH. Let MATH, MATH; one checks easily that MATH and, since the intersection of MATH with MATH is unique, MATH is non degenerate for all MATH. It follows that MATH is constant for MATH and for MATH. By the positivity of the intersection, the restriction of MATH to the one dimensional subspace MATH is positive definite (see REF ). Moreover, by REF , it follows that MATH is positive definite on MATH. The fact that MATH is homologous to MATH in MATH will follow from REF once we prove that MATH and that MATH. Applying twice REF around MATH, we obtain the following equalities for each MATH: MATH for the second equality we have applied REF to the curve MATH reparameterized backwards. The conclusion follows from the fact that MATH. Using the first part of the Proposition, to conclude the proof we need to exhibit a smooth curve MATH whose homology class is a generator of MATH, and that intersects MATH exactly once, with such intersection transverse. To this aim, let MATH be a symplectic basis of MATH such that MATH is the Lagrangian generated by MATH (see REF ). Consider the curve MATH introduced in the statement of REF . It intersects MATH only at the instant MATH and MATH, because MATH. To check the transversality, we make computations using the chart MATH (rather than MATH, see REF ), where we choose MATH to be the Lagrangian generated by the vectors MATH, MATH, which is complementary to both MATH and MATH. We set MATH, and we obtain for each MATH a symmetric bilinear form in MATH which, in the basis MATH is given by the diagonal matrix: MATH with MATH . Since MATH, then MATH intersects MATH transversally (with negative intersection), and the proof is complete. |
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 suffices to consider the case MATH. For MATH, a constant curve with positive type number equal to MATH would do the job. It is indeed sufficient to exhibit curves MATH as above for all MATH. To prove this claim, observe in first place that, if such a curve MATH is found and MATH is any other curve having the same positive type numbers at the endpoints, then the corresponding curves in MATH have the same NAME index. Now, if the curves MATH are chosen in such a way that the endpoint of MATH coincides with the initial point of MATH, then the concatenation MATH has the desired properties. To complete the proof, we now show how to construct the curves MATH as above. Choose any basis of MATH and define a curve MATH such that MATH is given in the chosen basis by the diagonal MATH matrix having diagonal vector MATH. Let MATH; we need to show that MATH. It is easy to see that every MATH intersects MATH only once at MATH, and that MATH. Since MATH is positive definite on the one dimensional space MATH, the intersection of MATH with MATH is transverse and positive (see REF ). By definition, the NAME index of MATH is equal to MATH, and we are done. |
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 neighborhood of MATH. It is easily seen that MATH. By elementary arguments, we have that, for MATH sufficiently close to MATH, the following inequality holds: MATH . Hence, for MATH small enough, we get: MATH . The conclusion follows easily from REF . |
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 a backwards reparameterization of MATH, we conclude that if MATH is small enough, then MATH is non degenerate for MATH, and that: MATH . Subtracting REF from REF, we get MATH . We have proven that the intersection of MATH with MATH at MATH is isolated, and, using REF , it follows from REF that: MATH . The conclusion follows from the additivity of the NAME index with respect to concatenation. |
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 . The conclusion follows from REF . |
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 continuous dependence on the data for ordinary differential equations, from REF we get that MATH tends to MATH uniformly (actually, in the MATH-topology) as MATH. By the continuity of the action of MATH in MATH, it follows that MATH tends to MATH in the compact-open topology. Since MATH and MATH is open in MATH, we have that MATH, that is, MATH is not MATH-focal for MATH for MATH sufficiently large. It is not hard to prove (see REF below) that there exists a MATH such that there are no MATH-focal instants on the interval MATH relatively to the quadruple MATH, for all MATH. Hence, the curve MATH does not intercept MATH in the interval MATH, for all MATH. The NAME index MATH is by definition equal to the NAME index MATH of the restriction of MATH to MATH. Since MATH and MATH are locally path connected, and since MATH is locally simply connected, the convergence of MATH to MATH (over the interval MATH) in the compact-open topology implies that, for MATH sufficiently large, MATH is homotopic to MATH through curves with endpoints outside MATH. Therefore, MATH for MATH large enough, and we are done. |
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 parallel transport of the vectors MATH along MATH. Associated to these trivializations, we produce quadruples MATH admissible for the differential problem in MATH, MATH. We emphasize that we are considering trivializations of the entire tangent bundle along the geodesics MATH; recall REF for a discussion about this issue. We also observe that the condition that MATH implies in particular that MATH is non degenerate on MATH. Clearly, under our hypothesis, MATH tends to MATH as MATH in the sense of REF . The conclusion follows now easily from REF . |
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. Using integration by parts, for all MATH we have the following: MATH . Moreover, for all MATH, we have: MATH . Let MATH be such that MATH; we apply to such a function REF , obtaining: MATH . Using REF, we obtain that the right side of REF is bounded from below for MATH with MATH, that is, there exists MATH such that: MATH . Let now MATH be any eigenvalue of MATH in MATH and MATH be a corresponding eigenvector with MATH. From REF we compute easily: MATH where MATH is the operator norm of MATH in MATH. This concludes the proof. |
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 where in the first equality we have used REF and in the second one we have used REF. Hence, the derivative of the pull-back REF is given by: MATH where in the first and in the third equality we have used the fact that MATH is a symplectomorphism and in the second one we have used REF. Observe that, by REF, MATH is simply given by: MATH . Integration of REF on the interval MATH using REF gives: MATH because MATH (see REF ). Finally, let MATH be elements of MATH; evaluating REF in the pairs MATH and MATH in MATH, we get: MATH which was obtained by using REF . This concludes the proof. |
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 is by definition the NAME index of the quadruple MATH; the NAME index MATH of the curve MATH, MATH is equal to the spectral index, by REF . Finally, the image by MATH of the remaining two sides of the rectangle MATH is disjoint from MATH by our choice of MATH and MATH. The conclusion follows from the homotopy invariance of the NAME index. |
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 check that this choice of MATH works, let MATH be chosen so that MATH has a zero of order MATH at MATH and a zero of order MATH at MATH. Using the MATH-th order NAME polynomial of MATH around MATH, we get: MATH where MATH satisfies: MATH . By our choice of MATH, it follows that MATH has no zeroes in MATH; similarly, MATH has no zeroes in MATH. From MATH, it follows that MATH has no zeroes in MATH, which concludes the proof. |
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 enough ball MATH. CASE: The same NAME duality result may be applied locally to yield: MATH . Note that the decomposition MATH also follows from the localization REF . |
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 projections MATH and MATH. Then MATH is the germ of MATH at MATH, where MATH denotes the hyperplane MATH and is identified to a point of MATH. Now MATH and therefore MATH is either void or of dimension at least MATH. On the other hand, by REF condition we get MATH and since MATH is an isolated point, it follows that MATH has dimension at most MATH. More precisely, the polar locus at MATH is not void (hence a curve) if and only if MATH and this, if and only if MATH. Since MATH, this last condition does not depend on MATH. Our claim is proved. CASE: Suppose MATH has dimension MATH (since if void, the multiplicity in cause is zero). Consider a small enough ball MATH centered at MATH, to fit in the NAME fibration CITE of the function MATH at MATH: MATH where MATH is centered at MATH. The notation MATH will stay for the representative in MATH of the germ MATH. We may choose MATH so small that, for all MATH, those intersection points MATH which tend to MATH when MATH, are inside MATH. This is possible because MATH is a curve which cuts MATH at MATH. We shall compute the homology MATH of the NAME fibre of the fibration REF . Inside MATH, the restriction of the function MATH to MATH has a finite number of isolated singularities, which are precisely the points of intersection MATH. We start with the claim that the space MATH is contractible, for small enough disc MATH centered at MATH. We need the following: Let MATH have isolated MATH-singularities at MATH. Let MATH be a small enough ball at MATH such that the sphere MATH cuts transversely all those finitely many strata of MATH which have MATH in their closure and does not intersect other strata. Then, there exist small enough discs MATH and MATH such that MATH is transverse to MATH, for all MATH. By absurd, if the statement is not true, then there exists a sequence of points MATH tending to a point MATH, such that the intersection of tangent spaces MATH is contained in MATH. Assuming, without loss of generality, that the following limits exist, we get: MATH . Let MATH be the stratum containing MATH. Remark that MATH, since MATH and MATH. This implies that MATH. We have, by the definition of the stratification MATH, that MATH and obviously MATH. On the other hand, MATH, since MATH. In conclusion, the intersection in REF contains MATH. But, since MATH, the limit MATH cannot contain MATH and this gives a contradiction. Let MATH be so small that MATH. By REF above and by choosing apropriate MATH and MATH, the map MATH is a locally trivial fibration. Therefore MATH is homotopy equivalent, by retraction, to the central fibre MATH. This proves our claim. We now remark that the central fibre MATH is just the complex link at MATH of the space MATH. The space MATH is a product MATH at MATH, along the projection axis MATH, hence its complex link is contractible. Hence so is MATH. Pursuing the proof of REF , we observe that MATH is homotopy equivalent to MATH, for MATH and MATH like in REF and, in addition, the radius of MATH much smaller than the radius of MATH. This supplementary condition is meant to insure that MATH. Now, the total space MATH is built by attaching to the space MATH, which is contractible, a finite number of cells of dimension MATH, which correspond to the NAME numbers of the isolated singularities of the function MATH on MATH. The sum of these numbers is, by definition, the intersection multiplicity MATH. We have proven that: MATH . When replacing all over in our proof the function MATH by MATH, we get the same relation REF , with MATH instead of MATH. This concludes our proof of REF. |
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 MATH is (MATH)-connected, where MATH. For MATH, a similar statement turns out to be true. We may invoke the following lemma, which is an extended version of a result by CITE: CITE The pair MATH is MATH-connected, where MATH. We conclude that the space MATH is built up starting from a fibre MATH, then moving it within a fibration with a finite number of isolated singularities. By the above connectivity results and by Switzer's result CITE, at each singular point one has to attach a number of MATH-cells, equal to the local NAME number, respectively, the polar NAME number. The total number of cells is the sum of all these NAME numbers. |
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 MATH is a curve or it is void, since MATH is an isolated MATH-singularity. Following CITE, see also CITE, there is a fundamental system of privileged polydisc neighbourhoods of MATH in MATH, of the form MATH, where MATH is a disc at MATH and MATH is a polydisc at MATH such that the map MATH is a locally trivial fibration over MATH. We chose MATH and MATH such that MATH. Let MATH. Observe that MATH is contractible, since it is the NAME fibre of the linear function MATH on a smooth space. This is obtained, up to homotopy type, by attaching to MATH a certain number MATH of MATH-cells, equal to the sum of the NAME numbers of the function MATH. Since we have the homotopy equivalence MATH, we get, by REF , that MATH. Now MATH is homotopy equivalent to the NAME fibre of the germ MATH, which has NAME number MATH. The space MATH is obtained from MATH by attaching exactly MATH cells of dimension MATH (comming from the polar intersections) and of a number of MATH-cells comming from the intersections with MATH. This number of cells is, by definition, MATH. We get the equality: MATH . Lastly, since MATH, our proof is done. |
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 map). Assume therefore that MATH is simply connected. Then MATH is spanned by fundamental weights MATH of MATH. Let MATH . By REF, MATH using multi-index notation MATH, hence MATH. The image of MATH is contained in the unit polydisk MATH. For any distribution MATH supported on the polydisk MATH, there exist constants MATH and MATH such that the value of MATH on any test function MATH is bounded by derivatives of MATH up to order MATH (compare REF ) MATH . Applying this to MATH and MATH immediately gives the required estimates. To show that MATH is continuous, consider the semi-norms MATH for MATH. Since the coefficients MATH in the NAME decomposition MATH are rapidly decreasing, and since MATH, the function MATH on MATH is well-defined and continuous. Let MATH be the NAME operator on MATH. On MATH it acts as a scalar MATH. Hence MATH is still continuous. By the NAME Lemma, this shows that MATH is smooth. Since MATH it follows that MATH is continuous. |
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. |
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 the second since MATH where we have used REF . We conclude MATH . |
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