paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9906111 | Consider a point MATH, in other words a natural transformation MATH for MATH. By putting together the maps MATH we get a function MATH. Next, for any MATH we have projections MATH and we can use the resulting maps to identify MATH with MATH and MATH with MATH and MATH with MATH. There are evident maps MATH and MATH, an... |
math/9906111 | Write MATH as a disjoint union of orbits. |
math/9906111 | Suppose MATH, and define MATH by MATH. By hypothesis, this has the same character as MATH, so there exists an intertwining operator MATH such that MATH for all MATH. As MATH is an irreducible representation of MATH we see that MATH and thus MATH is unique up to multiplication by a scalar matrix. We can thus define a ma... |
math/9906111 | For each irreducible representation MATH of MATH, let MATH denote the sum of the inequivalent MATH-conjugates of MATH. Any MATH-invariant character is a direct sum of copies of the characters of the representations MATH, so it suffices to show that MATH extends to a representation of MATH. Let MATH be the stabiliser of... |
math/9906113 | CASE: By induction on MATH. CASE: Similarly. |
math/9906113 | By induction on MATH. |
math/9906119 | The proof in CITE for MATH extends with minor modifications to the general case. |
math/9906119 | Computed using NAME. |
math/9906119 | It follows from REF that the map REF identifies MATH with the curve class MATH in MATH, so the NAME are integrals over MATH. Localization. Consider the standard action of MATH on MATH. Since the integrands are polynomials in NAME classes of equivariant vector bundles with respect to the induced MATH-action on MATH, the... |
math/9906119 | From REF it follows that MATH. Using REF it is straightforward to check that MATH is the differential operator governing MATH (see for instance CITE). The rest follows from REF . |
math/9906124 | See CITE. See also CITE. |
math/9906124 | This is REF. |
math/9906124 | Sufficiency is due to CITE and necessity to CITE. |
math/9906124 | See CITE. See also CITE. |
math/9906124 | MATH is well defined: Suppose MATH, where MATH. Then MATH is one to one: If MATH, then for some MATH one has MATH . Thus MATH, and MATH. Let MATH. Then MATH, and MATH for some MATH since this subgroup is normal in MATH. Thus MATH. MATH is onto: Let MATH. Since the MATH are onto we have MATH and MATH for some MATH, MATH... |
math/9906124 | Suppose MATH. Let MATH. Then MATH for some MATH, MATH, MATH. So MATH and MATH. Thus MATH . Similarly MATH implies that MATH . Hence MATH. Now suppose that MATH. From the proof that MATH is onto in REF we may assume that MATH for some MATH. Let MATH. Then MATH for some MATH. So MATH and MATH. Hence MATH and MATH. So MAT... |
math/9906124 | First assume that every splitting homomorphism MATH with MATH and MATH has an essential factorization. Let MATH be a closed, orientable REF-manifold with MATH finite. We may assume that MATH is irreducible. Let MATH be a NAME splitting of MATH of minimal genus MATH. Let MATH be the associated splitting homomorphism. By... |
math/9906124 | Suppose the condition on the MATH holds. Let MATH be a closed, orientable, irreducible REF-manifold and MATH a NAME splitting of minimal genus MATH. If MATH, then the NAME Conjecture holds for MATH. So assume MATH. If MATH had an essential factorization, then MATH would be reducible, and, since MATH is irreducible MATH... |
math/9906124 | If MATH is covered by a bundle with fiber a surface MATH, then the image of MATH in MATH is a good subgroup. If MATH contains a MATH subgroup, then a summand is a good subgroup. The converse follows from CITE. For convenience we give the relevant portion of the proof of that result. Let MATH be a good subgroup and MATH... |
math/9906124 | Suppose the extended conjecture is true. If MATH, then by the proof of REF we reduce to the situation in which either MATH has a MATH subgroup or MATH is hyperbolic; in the latter case we apply the Virtual Bundle Conjecture and so conclude in both cases that MATH has a good subgroup. If MATH, then a similar argument sh... |
math/9906124 | We may assume that the covering is regular CITE. As pointed out by NAME, the covering translation in MATH corresponding to an element of MATH is by NAME rigidity homotopic to a unique isometry. The lifts of these isometries to the universal cover MATH give a subgroup MATH of MATH. The quotient MATH is then the desired ... |
math/9906128 | Let MATH be fixed, and let MATH satisfy MATH . By MATH, there exist MATH, such that MATH . If MATH is an admissible set then MATH, so this inclusion holds for all MATH . Hence MATH . Applying consecutively REF , the above inclusion and condition MATH, we obtain condition MATH. |
math/9906128 | Condition MATH is trivially satisfied, because the uniform base MATH of MATH consists of only one element MATH. Now we observe that MATH is continuous as a linear map on a finite-dimensional space, so MATH of REF holds for MATH the class of all continuous maps. |
math/9906128 | First, MATH is normal, so MATH is NAME. Next, for any MATH, let MATH . Then the set MATH determines the equivalence relation MATH . To ensure that MATH has a uniformity, we need to check that for any MATH any MATH there are MATH such that MATH which means that the equivalence relation MATH and the uniformity of MATH ar... |
math/9906128 | By REF MATH is NAME and NAME is compact. Hence MATH is closed in MATH . Then, there exists an open neighborhood MATH of MATH and a continuous function MATH that extends MATH and MATH is uniformly continuous. Now if we extend MATH on the whole MATH by putting MATH for MATH then MATH is uniformly u.s.c. (since MATH is op... |
math/9906128 | Let MATH . Then condition MATH reads as follows: there exists MATH such that MATH . Let MATH. By definition of MATH, MATH has a subcover MATH with MATH. But MATH is l.s.c., so MATH consists of open sets. Moreover, if MATH then MATH for some MATH. Hence MATH belongs to MATH, so MATH is an open cover of MATH. By REF , we... |
math/9906128 | Let MATH. Without loss of generality we can assume that MATH . Then according to REF (with MATH), for any nonempty convex-valued l.s.c. map MATH for any MATH, there is a continuous MATH with MATH . We inductively construct a sequence of continuous functions MATH, such that MATH . By REF , there is MATH so that REF hold... |
math/9906128 | In REF , we let MATH and notice that MATH, so REF holds-MATH . |
math/9906128 | In REF we let MATH be discrete and MATH. Then MATH is l.s.c., so REF holds. But since MATH, REF also holds. Finally, we notice that an almost selection with respect to the discrete topology is in fact a selection. |
math/9906128 | By REF with (MATH), for any MATH there exist MATH and a continuous function MATH with MATH finite such that MATH for all MATH . Since the convexity is global, we can define a map MATH by MATH . Since MATH, the multifunction MATH has a fixed point, that is, there exists a MATH. Therefore, there exist such MATH and MATH ... |
math/9906128 | Suppose MATH . By REF , for any MATH, there is a MATH . Now we need to show that MATH for some MATH. By REF , the net MATH has a convergent subnet. Therefore, we can simply assume that MATH. Then for any MATH, there exists a MATH such that MATH for all MATH, or MATH . By REF , MATH for all MATH. Therefore MATH . Since ... |
math/9906128 | In REF we let MATH . Then REF hold because the convexity is regular: MATH. Also, since MATH is compact, we have MATH and MATH, so REF holds. |
math/9906128 | Notice that if MATH is a discrete uniform space, then a multifunction MATH has open fibers if and only if it is l.s.c. Therefore REF of this theorem are exactly REF follows from the fact that MATH is u.s.c. with respect to the discrete topology. Also, since MATH is compact and MATH is discrete, we have MATH and MATH, s... |
math/9906138 | The theorem follows from the equality MATH . Hence the value of MATH on a DD(MATH)-singular NAME having MATH pairs of double points descends to the sum of values of MATH on singular links each of which has MATH double points. |
math/9906138 | The proof of REF is best illustrated by an example shown in REF , where we show a singular link on the left hand side and its corresponding double dating diagram on the right hand side. For the proof of REF we show how to build a singular link for a given DD diagram. Let D be a DD diagram with some full lines and pairs... |
math/9906138 | The proof of this theorem is basically identical to that of REF. The only difference occurs when we reduce the number of axons in MATH so that it would be the same as for MATH. as already mentioned there, we cancel a pair of axons by the use of double crossing flips. In general these crossings do not have to be close. ... |
math/9906138 | The invariants of type MATH are the constants. There is a unique NAME invariant of type MATH for knots, which is MATH (the second coefficient of the NAME invariant). Therefore for a link with MATH component, we have MATH each one for each component of the link. The extra invariant in MATH is the NAME MATH. If MATH then... |
math/9906138 | The proof of the theorem is done by using REF-T, REF-T and REF-T which give sequences of equalities between double dating diagrams. Those equalities are given in REF. |
math/9906138 | In order to prove the theorem we introduce another space MATH having MATH as a subspace. Further we will define a map MATH from MATH into MATH which extends MATH and we will show that MATH is mapped into the subspace of strutless diagrams. The objects of MATH are diagrams having trivalent vertices some of which externa... |
math/9906147 | Without any loss of generality one can suppose that the curve is reduced. Therefore one can choose the infinite line in the complex projective plane MATH so that it intersects the curve at MATH different points. Let the affine part of the curve MATH be given by the equation MATH (where MATH is a polynomial of degree MA... |
math/9906147 | Let MATH, MATH, and let an (affine plane) curve MATH be given by the equation MATH where MATH. The degree MATH of the curve MATH is equal to MATH. Let MATH (MATH and MATH are local coordinates on the plane near the origin). Then MATH . Since MATH terms of higher degree, one has MATH terms of higher degree. Substituting... |
math/9906153 | Let MATH, where MATH. For MATH define MATH. It is clear that MATH. Define MATH if MATH and MATH otherwise. Define MATH to be the sum of all letters MATH such that MATH. Then form the following system of equations: MATH . There are MATH right linear equations in MATH unknowns satisfying the conditions of REF . Therefore... |
math/9906153 | The semigroup defined is the set of equivalence classes of MATH with respect to the second two relations (that is, the NAME extension rules MATH and MATH) with a zero adjoined and multiplication of any two classes of MATH defined to be zero. |
math/9906153 | Define an automaton MATH where MATH, MATH and MATH is defined as follows: MATH . It is clear from the definitions that the extended state transition MATH is such that MATH if and only if MATH. Hence MATH. Therefore MATH is regular over MATH. |
math/9906153 | Recall the (`target') functions MATH and MATH. We use the following definition to restrict sets to those elements whose `target' is MATH. MATH . Then define MATH be the set of irreducible forms of the terms MATH with respect to MATH. For each object MATH we define an incomplete non-deterministic automaton MATH with inp... |
math/9906159 | We use additive notation for the group operation in MATH. Let MATH be a prime factorization. Then MATH, and MATH restricts to an automorphism of each of the factors. If MATH is odd, MATH is cyclic, so each order two automorphism of MATH sends a generator to its inverse. However, MATH, so if MATH, MATH has three automor... |
math/9906159 | Recall that MATH, MATH, and MATH. A group MATH is said to be complete if MATH is trivial and every automorphism of MATH is inner. For MATH, MATH is complete (see CITE). Alternating groups are never complete: for MATH, MATH, and conjugation by the transposition MATH is an automorphism of MATH which is not inner in MATH.... |
math/9906159 | We use additive notation for the group operation in MATH, but multiplicative notation in the (possibly nonabelian) group MATH. Let MATH generate MATH. MATH is generated by MATH, together with MATH and MATH, where MATH and MATH. We assume MATH and MATH are normalized so that MATH and MATH lie in the NAME REF-subgroup of... |
math/9906159 | We use the commutative diagram MATH where the vertical maps are induced by the coefficient reduction. The right-hand map is an isomorphism for all the subgroups except for MATH, where MATH. The fact that MATH and MATH are squares of one-dimensional classes makes calculation easy for every column except MATH. In that ca... |
math/9906159 | The main claim that needs to be established is the exactness of the following sequence: MATH . Since a rotation in MATH is homotopic to the identity, it is clear that any MATH is in the kernel of MATH. For the reverse inclusion, suppose MATH. Since MATH acts trivially on homology, it has a fixed point MATH. A fairly st... |
math/9906159 | Since MATH acts trivially on homology, we must have MATH . Let MATH. Since each of MATH and MATH preserves orientation on MATH, each has a fixed point on MATH. By the assumption of pseudofreeness, each fixed set must be REF-dimensional unless MATH. Claim: Let MATH and MATH denote the two obvious projections from MATH. ... |
math/9906159 | We begin with some observations about the geometry of MATH. Suppose MATH and MATH are nontrivial elements of MATH. Each has a well defined axis of rotation - we denote them respectively by MATH and MATH. Observe that MATH for some MATH if and only if MATH leaves MATH invariant. If MATH fixes MATH, then MATH, and MATH. ... |
math/9906159 | As a special case of NAME 's theorem CITE, an involution MATH on MATH with MATH on MATH has a fixed point set MATH with MATH. If the sequence MATH splits, then MATH contains such an involution. The second statement follows from the observation that for MATH, Fix-MATH . Fix-MATH. |
math/9906159 | This follows immediately from REF . |
math/9906159 | We know from REF that if MATH splits, then MATH cannot act pseudofreely. It follows from REF that MATH must be of type REF or REF (and that MATH is even). Recall from the beginning of REF that the structure of MATH is determined by the data MATH and MATH. By REF , a hypothetical MATH has the form MATH, where MATH, and ... |
math/9906159 | We know it is necessary that the sequence not split. To prove the converse, we assume it does not split and we construct a pseudofree MATH-action. Let the data MATH and MATH be given. As in the previous proof, MATH, and we will choose MATH and MATH appropriately. In this case, choose any MATH and MATH so that MATH, and... |
math/9906159 | Let MATH and MATH, and suppose an extension MATH exists, with MATH acting pseudofreely. Let MATH, and MATH. By REF , we know that MATH must be cyclic of even order, with MATH or MATH, and that the sequence MATH must not split. REF describes those groups MATH which satisfy these criteria. If MATH, the geometry of MATH p... |
math/9906159 | We have already seen that REF is necessary and sufficient in the homologically trivial case, and that REF is necessary and sufficient in the MATH case. REF is necessary by REF (NAME 's theorem). On the other hand, this condition implies that MATH does not split, and REF show this to be sufficient in the MATH case. Fina... |
math/9906159 | It is shown in CITE that, without the pseudofree assumption, such an isotropy group MATH must be abelian of rank MATH or MATH. But a rank MATH group cannot act freely on the linking sphere to MATH, and thus cannot act pseudofreely on MATH. |
math/9906159 | Since MATH for some MATH, MATH. Observe that for any MATH, if MATH, then MATH. In other words, MATH acts on Fix-MATH. Two points of Fix-MATH are in the same MATH-orbit if and only if they are in the same MATH-orbit, for if MATH, MATH, and MATH, then MATH. But each MATH-orbit of the action on Fix-MATH has cardinality MA... |
math/9906159 | Since the groups occur in pairs, MATH is even. Let MATH, and rearrange the list so that MATH, MATH, and so forth. Since MATH, it follows that MATH. As in the MATH case, the possible MATH are MATH, and MATH. In each case, these numbers represent the sizes of maximal cyclic subgroups of MATH, and each conjugacy class of ... |
math/9906159 | Say MATH is nice if for each singular point MATH, MATH meets REF or REF MATH-orbits. If MATH is nice, then its isotropy groups are repeated in pairs. We assume inductively that every proper subgroup of MATH is nice, but that MATH is not. Then by REF , there is some MATH so that MATH; that is, MATH acts transitively on ... |
math/9906159 | Let MATH act homologically trivially, so that MATH is exact. Claim: MATH acts freely on MATH. Let MATH, so that MATH generates MATH. If, for some MATH, MATH lies in the same orbit as MATH, then for some MATH, MATH. But MATH has NAME number MATH, so if it has a fixed point, its fixed point set must be REF-dimensional. F... |
math/9906159 | Given the necessity of REF , and REF , it suffices to rule out the possibility that MATH, where MATH. In the case of the linear models REF , the argument divided into two parts: If MATH, then MATH fixes a torus, contradicting pseudofreeness. And if MATH or MATH, then the fact that MATH exchanges factors of MATH means t... |
math/9906160 | Suppose the sequence of constituent lengths starts with MATH where MATH is even. We will prove that this is followed by another short constituent. We have to prove MATH . Since MATH is even, we can consider the integer MATH . We will obtain REF from the expansion of the following expression with the generalized NAME id... |
math/9906160 | Let the intermediate constituent have length MATH. Suppose we have already proved that this intermediate constituent is followed by MATH short ones ending in MATH, with MATH, so that there is a segment MATH in the sequence of constituent lengths of MATH. We will show that the next constituent is also short and ends in ... |
math/9906160 | Deflating MATH times, according to REF , we may assume MATH. Write MATH, where MATH is a power of MATH (possibly MATH), and MATH. By REF , we know that the first intermediate constituent is followed by at least MATH short ones ending in MATH. Let MATH . If MATH, we have MATH. Therefore we obtain MATH . We will prove MA... |
math/9906160 | By REF and induction, we may assume that we have the subsequence MATH in the sequence of two-step centralizers of MATH, for some MATH. Let MATH, so that MATH, and let MATH be a homogeneous element at the end of the MATH-th two-step centralizer before the long one, so that MATH. We have to prove MATH . We have MATH, fro... |
math/9906161 | Suppose first that MATH and MATH, MATH are as above. By shrinking MATH if necessary, we can make sure that MATH implies MATH for every MATH . By shrinking MATH further if necessary, we can arrange that the exponential map of MATH at MATH identifies a neighborhood MATH of the origin in MATH and MATH. Moreover, the eleme... |
math/9906161 | We claim that for any MATH REF-tuple MATH, MATH, is a normal collection in the sense of NAME. Indeed, this is clear if MATH is disjoint from MATH; otherwise MATH by our assumption. So assume that MATH. By the normality of MATH, near any point MATH in MATH there are local coordinates MATH, MATH, on MATH such that MATH a... |
math/9906161 | Since any total order compatible with inclusion can be obtained from any other one by repeatedly interchanging the order of adjacent elements, but keeping the order compatible with inclusion, it suffices to show that MATH are naturally isomorphic if both of these total orders respect inclusion. Now, either MATH, in whi... |
math/9906161 | It suffices to show that MATH lifts to be an element of MATH where MATH is minimal with respect to inclusion and MATH . Taking into account that for any MATH, MATH is a normal collection of p-submanifolds of MATH, this claim follows from CITE, or it can be checked directly by using projective coordinates on MATH. |
math/9906161 | By the first part of the statement, for any MATH, MATH, the commutator MATH is in MATH. |
math/9906161 | It is convenient to use the explicit description of MATH in terms of localization and quantization REF. Thus, we may assume that MATH. Note that the pull-back of MATH to MATH is a polyhomogeneous symbol of order MATH which we denote by MATH. Then the kernel of MATH is MATH where MATH is the kernel of MATH. But by the f... |
math/9906161 | This result essentially reduces to the fact that MATH is an algebra. Indeed, write MATH, and note that MATH where MATH denotes the composite of MATH with the multiplication operator MATH by MATH. Since the latter is in MATH, hence in MATH, we conclude that MATH. Thus, we only have to analyze MATH. Again, we can reduce ... |
math/9906161 | We prove this statement by finding MATH explicitly in terms of local coordinates. To simplify the notation we assume that MATH. We identify MATH with MATH locally so that MATH is given by MATH, MATH. In the interior of MATH we can use the same coordinates as at the front face of MATH, that is, the ones given in REF. So... |
math/9906161 | Choose MATH with principal symbol MATH and indicial operators MATH. Note that these match up under the projections MATH since those of MATH match up. Thus, they indeed specify the restriction of some MATH to the boundary of MATH. Quantizing MATH gives an operator MATH with the required indicial operators and principal ... |
math/9906161 | Since MATH is bounded, we can replace MATH by a function MATH such that MATH on the spectrum of MATH. Now MATH is invertible for MATH, so MATH for MATH. Again, MATH can be analyzed uniformly up to the real axis, and then the NAME integral representation of MATH now proves the proposition. |
math/9906161 | Suppose that MATH and MATH. With MATH as above, let MATH, so MATH as MATH. Then MATH and MATH. With MATH as above, let MATH. Then MATH and MATH . But MATH by assumption, so by REF MATH . Hence, there exist MATH (in place of MATH), MATH, etc., as in REF , MATH, and the indicial operator of MATH at MATH is just the compo... |
math/9906161 | As MATH in MATH, we have for all MATH that MATH . Thus, with MATH, the previous proposition implies that for any MATH, MATH is differentiable from both the left and the right at MATH, and both of these derivatives are equal to MATH. (We remark that this is proved directly in the Appendix as a first step to the proof of... |
math/9906161 | The previous corollary and the above remarks show that for all MATH, MATH. Let MATH. Thus, MATH vanishes at MATH for all MATH. Since the base variables MATH and MATH are MATH-invariant, we conclude that MATH vanishes identically, hence MATH is constant, and similarly for MATH, proving that MATH for all MATH. The last s... |
math/9906161 | Let MATH . REF implies REF immediately. Let MATH be a MATH-invariant function. Let MATH here we slightly abuse the notation and write MATH. Then MATH is MATH-invariant, so REF applies and gives MATH. Since MATH, the chain rule and a short calculation of MATH gives REF . The first equation in REF follows since along MAT... |
math/9906161 | Our strategy consists of constructing a MATH-invariant function MATH with MATH in a neighborhood of MATH. Thus, by REF , MATH for MATH, MATH sufficiently small, so MATH is increasing there. This will allow us to draw the desired conclusion for the correct choice of MATH. We remark that this MATH will reappear in the pr... |
math/9906161 | Let MATH be as above and introduce local coordinates centered at MATH. We may assume that MATH is given by MATH for a suitable splitting MATH. Thus, MATH is of the form MATH, and as MATH, MATH. By REF , taking into account that MATH is MATH-invariant, MATH . Since MATH, there exist MATH, MATH, such that MATH for MATH, ... |
math/9906161 | If MATH then MATH for MATH near MATH by REF , hence near MATH, MATH is a (MATH-projected) bicharacteristic of MATH (as MATH vanishes at MATH). If MATH then REF applies and proves the result. If MATH, then with MATH as in REF , MATH is a generalized broken bicharacteristic, broken at MATH, with MATH (prime taken with re... |
math/9906161 | Let MATH . Thus, MATH, so MATH. Since the spectrum of MATH is a subset of MATH and MATH, we have MATH where MATH and MATH if MATH is in the spectrum of MATH. By REF , MATH . Let MATH be identically MATH near MATH and vanish near MATH. Then MATH . Now let MATH be identically MATH near MATH and vanish near MATH. Let MATH... |
math/9906161 | We apply a parameter dependent version of the previous lemma to the indicial operators to conclude that for each MATH there exists MATH with MATH . It follows from the NAME integral formula construction of the square root in the calculus and the explicit formulae REF that the indicial operators MATH match up so that th... |
math/9906161 | We define MATH if MATH, otherwise we define MATH as in the previous proposition. The only additional ingredient is the analysis of MATH near MATH with MATH. To do this analysis, we follow the construction of MATH in detail. So let MATH and let MATH . Thus, MATH. Let MATH . By our assumption, there exists MATH such that... |
math/9906161 | First, MATH follows from REF. Moreover, the infinite order vanishing of MATH at MATH implies that for every MATH, MATH and MATH, MATH . Thus, NAME 's rule shows that for MATH acting in MATH, MATH acting in MATH, MATH and MATH . But this means precisely that MATH and it vanishes to infinite order at the boundary in the ... |
math/9906161 | This can be proved directly from the definition of the indicial operators, that is, by computing MATH where MATH and MATH, similarly to CITE. Since this is equal to MATH, and MATH, we can assume that MATH, the calculation being very similar in the general case. To compute the commutator, it suffices to commute both MAT... |
math/9906161 | Note that the estimate REF is trivial if MATH (with MATH, MATH arbitrary) since then both sides vanish as MATH denoting the subsystem Hamiltonian as in REF, and MATH by the assumption on the absence of bound states of all subsystem Hamiltonians (including MATH with MATH). We prove REF by induction on MATH. First, REF i... |
math/9906161 | Let MATH be MATH-invariant, MATH, satisfy estimates REF, and such that MATH and MATH. (Here MATH can be regarded as a function on MATH.) Let MATH identically MATH near MATH, and let MATH be such that MATH and MATH. For example, MATH can be constructed as in REF . The indicial operators of MATH are MATH since MATH. Thus... |
math/9906161 | The main step in the proof is the construction of an operator which has a microlocally positive commutator with MATH near MATH. In fact, we construct the symbol of this operator. This symbol will not be a scattering symbol, that is, it will not be in MATH, only due to its behavior as MATH corresponding to its MATH-inva... |
math/9906161 | We again employ an iterative argument, so we assume that MATH and we need to show that MATH. We first construct a MATH function MATH of MATH, MATH, MATH, MATH and MATH which measures the distance of bicharacteristics of MATH in MATH from MATH. Thus, MATH will be small along these bicharacteristics. We will take MATH of... |
math/9906161 | First note that there is nothing to prove if MATH, so from now on we assume that MATH. The proof is very similar to the previous one and the positive commutator construction is exactly the same as in three-body scattering CITE, based on the MATH-invariant function MATH used here in the proof of REF . Thus, we take loca... |
math/9906161 | The proof is very similar to the previous ones and now the positive commutator construction follows that of CITE in three-body scattering. Thus, we take local coordinates as above, that is, of the form MATH with MATH defined by linear equations in MATH. Then we construct MATH (defined near MATH) to measure the squared ... |
math/9906161 | We only need to prove that for every MATH, if MATH and MATH then there exists a generalized broken bicharacteristic MATH, MATH, with MATH and such that MATH for MATH. In fact, if this statement holds for all MATH with MATH, let MATH and put the natural partial order on MATH, so MATH if the domains satisfy MATH and MATH... |
math/9906161 | This result is a weak form of the limiting absorption principle and can be proved by a NAME estimate. In the Euclidean setting, it is a combination of the NAME estimate, proved by CITE, and its microlocalized version obtained by CITE. In the geometric setting, the NAME estimate describes the commutator of MATH with a s... |
math/9906161 | As mentioned above, the first part follows from the self-adjointness of MATH, so that for MATH, MATH, MATH, we have MATH; recall that the distributional pairing is the real pairing, not the complex (that is, MATH) one. The wave front statement of REF and the assumption on MATH show the existence of the limit MATH in MA... |
math/9906161 | Just as in the proof of REF , the positive commutator estimate of REF (but now applied with a regularizing factor in MATH) shows that MATH, and then REF shows that MATH . We remark that although we need a regularizing factor here which requires some changes in the proof, for example, see the argument of the paragraph b... |
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