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math/0106193 | Without loss of generality, assume MATH has entries in MATH. Let MATH. We produce a sequence of matrices MATH such that MATH, MATH and MATH; the MATH-adic limit MATH of the MATH will satisfy MATH, proving the proposition. The conditions for MATH are satisfied by the assumption that MATH. Suppose MATH has been defined. Let MATH. For each MATH, let MATH be a solution of the equation MATH. Then MATH and MATH are both congruent to REF modulo MATH. Put MATH; then MATH . Thus the conditions for MATH are satisfied, and the proposition follows. |
math/0106193 | Let MATH be the matrix through which MATH acts on some basis of MATH. By REF , there exists a matrix MATH over a suitable extension of MATH and a matrix MATH over MATH such that MATH. By REF , for some finite extension MATH of MATH, there is a MATH-stable MATH-lattice MATH of full rank within MATH on which the action of MATH is given by a matrix of the form MATH, with MATH invertible over MATH. By applying MATH repeatedly, we may reduce to the case where MATH is separable. By REF , viewed as a MATH-crystal over MATH, MATH has the same special and generic NAME polygons. By CITE, MATH has a subcrystal MATH of dimension MATH with all slopes equal to MATH. Otherwise put, the MATH-module spanned by the eigenvectors of MATH of lowest slope descends to MATH. Since this module admits an action of MATH, MATH admits an action of MATH. We may apply NAME descent to the finite extension MATH of discrete valuation rings to conclude that MATH for some MATH-lattice MATH, which is necessarily MATH-stable. The property of being étale is stable under finite separable extensions, so MATH is also étale, as desired. |
math/0106194 | The proof is an easy direct verification. |
math/0106194 | Denote the operator MATH by MATH. We have MATH where MATH and MATH and MATH will be shown to be cubic in MATH. To eliminate the quadratic terms, we need to set MATH which takes the explicit form: MATH where MATH . Since these coefficients are even in MATH, we will search for even solutions, that is, MATH . REF+REF+REF MATH, and REF+REF MATH lead to MATH where MATH . Therefore we can express MATH in terms of MATH as, MATH . Substituting these expressions into REF, we get MATH where MATH . For MATH, the denominators in REF, and MATH vanish at MATH in a finite subset. For MATH not in this finite subset, MATH is a MATH small perturbation of MATH, and MATH is a MATH small perturbation of MATH. Setting MATH, we have MATH, which leads to the solution given in CITE. To the order MATH, MATH . We will show that the denominator in REF does not vanish except for MATH in a finite subset. Denote the denominator by MATH. As MATH, or MATH, MATH . We also know that MATH where MATH are polynomials in MATH and MATH. For each MATH and MATH, the numerator vanishes at most at four values of MATH. Together with REF , we have that for MATH, MATH does not vanish except for MATH in a finite subset. The denominator in REF has the representation: MATH where MATH . Then MATH . Therefore, for MATH, the denominator in REF, MATH does not vanish except for MATH in a finite subset. REF give the solution to the linear system REF for MATH, where MATH is a finite subset. As in CITE, MATH is also a bounded bilinear map: MATH . We can invert the equation MATH to obtain MATH where MATH is of order MATH, MATH. Thus, terms like MATH and MATH are cubic terms in MATH. |
math/0106194 | For MATH, where MATH is a finite subset, we apply the normal form transform given by REF to REF , then the system REF is transformed into the system REF. By virtue of the estimate REF, the theorem follows from standard argument. For details, see CITE. |
math/0106194 | This theorem follows immediately from the explicit computation in the subsection on Evaluation of NAME Integrals and Second Distance, and the implicit function theorem. |
math/0106194 | We start with the system REF. Let MATH . Let MATH be a time such that MATH for all MATH, where MATH is independent of MATH. From REF, such a MATH exists. The proof will be completed through a continuation argument. For MATH, MATH where MATH is small. Since actually MATH, MATH for any fixed MATH, by REF , MATH whenever MATH, MATH, where MATH and MATH are defined in REF. Thus we only need to estimate MATH. From REF, we have for MATH that MATH . Let MATH. Then MATH . By REF , we have for MATH that MATH . Then MATH . Thus by the continuation argument, for MATH, there is a constant MATH, MATH . |
math/0106194 | From the explicit computation in last subsection and the implicit function theorem, MATH and MATH are zero for the parameter values specified in the theorem . |
math/0106197 | Only the middle bijection requires proof. Recall the following result (NAME REF, III REF): consider a projective system MATH of topologicaly groups acting continuously and compatibly on a topological space MATH; then the canonical map MATH is bijective if CASE: the isotropy group in MATH of each MATH is compact, and CASE: the orbit MATH of each MATH is compact. Apply this with MATH and MATH. Then REF holds because each isotropy group is MATH, and REF holds because MATH is compact. As MATH, we have that MATH is bijective, as required. |
math/0106197 | Let MATH. As MATH stabilizes MATH, MATH for some MATH. This means that there exists a MATH such that MATH as elements of MATH. In particular, MATH. |
math/0106197 | Let MATH. By what we have just proved, there exists a MATH such that MATH and MATH have the same image in MATH. Let MATH. Then MATH, and so MATH in MATH . Therefore, there exists a MATH such that MATH, and so MATH. |
math/0106197 | For any MATH, MATH is a connected component of MATH, and so the lemma shows there exists a MATH in MATH such that MATH. By definition, MATH, and therefore MATH. |
math/0106197 | CASE: The lemma shows that MATH is surjective, and it is obvious that the fibres of the map are the orbits of MATH. CASE: As MATH, it follows from REF that there is an isomorphism MATH . The preceding proposition can be used to show that MATH is an isomorphism. |
math/0106197 | Let MATH; we have to show that MATH-is commensurable with MATH. Note that MATH. Let MATH. The diagram of finite étale maps gives rise to a similar diagram This shows that MATH has finite index in both MATH and MATH. |
math/0106197 | The hypotheses imply that MATH is a pro-algebraic subvariety of MATH and is MATH-invariant. As MATH is dense in MATH, MATH is invariant under MATH, and so MATH is a MATH-invariant subscheme of MATH. Let MATH. Then MATH is MATH-invariant and, as MATH is dense in MATH and MATH acts transitively, this shows that MATH. |
math/0106197 | Let MATH, and let MATH be the set of MATH such that MATH has a component of dimension MATH at MATH. Then MATH is an analytic subset of MATH (NAME REF, p. REF) and is MATH-invariant. Therefore, we can assume that MATH. The image MATH of MATH in MATH (any MATH) is analytic and such that MATH. As MATH, REF (CITE, p. REF) shows that the closure MATH of MATH in MATH is analytic, and NAME 's theorem (CITE, p. REF) shows that MATH is algebraic. Therefore MATH is algebraic, and the proposition applies. |
math/0106197 | There is an isomorphism MATH and we define MATH on MATH to correspond to the map on MATH induced by the MATH-linear map MATH. Clearly, MATH, and MATH is the unique map with this property. The continuity is obvious. If MATH is compact, it is the unique maximal compact subgroup of MATH, and so it is preserved by MATH. For MATH, MATH. |
math/0106197 | NAME REF, VIII REF |
math/0106197 | Let MATH be the volume form associated with MATH. Then MATH for some positive function MATH on MATH. As MATH is MATH-invariant, so also are MATH, MATH, and MATH. Therefore, MATH is constant, and so the NAME tensor of MATH is MATH. |
math/0106197 | This is proved in NAME REF, p. CASE: Alternatively, one can combine the following results: A complex homogeneous manifold with an invariant volume form whose NAME tensor is positive definite is isomorphic to a homogeneous bounded domain (NAME REF, p. REF). A connected unimodular NAME group acting effectively and transitively on a bounded domain is semisimple (Hano REF). A bounded domain admitting a transitive semisimple group of automorphisms is symmetric (NAME REF; NAME REF). |
math/0106197 | Let MATH be the bundle of orthonormal frames over MATH. Every automorphism MATH of MATH defines a compatible automorphism MATH of MATH. Let MATH and let MATH be its image in MATH. The mapping MATH embeds MATH as a closed submanifold of MATH (ibid., p. REF). The projection MATH is clearly proper, and its restriction to MATH is the map MATH. |
math/0106197 | Let MATH, and let MATH be the closure of the orbit MATH for some MATH lifting MATH. Then MATH is a Riemannian manifold, and the closure MATH of MATH in MATH acts transitively on MATH (by REF). Since MATH, MATH, is a proper map, it is an embedding and MATH is a regular closed submanifold of MATH. By REF , MATH has a MATH-invariant complex structure - it is therefore a complex analytic subset of MATH. Since it contains MATH for each MATH, it has dimension MATH, and so REF shows that MATH. Now REF can be applied. |
math/0106197 | Put MATH, MATH. Then MATH. Thus, MATH . |
math/0106197 | If MATH, this is precisely NAME 's lemma (NAME REF). Consider the covering map MATH . Then MATH satisfies REF , and so MATH. This implies that MATH. |
math/0106197 | A stronger statement is proved in REF below. |
math/0106197 | Embed MATH in a compact manifold MATH in such a way that MATH is a divisor with normal crossings. Then there is a finite family of open subsets MATH of MATH such that MATH and, for each MATH, the pair MATH is isomorphic to MATH. For each MATH, choose MATH to correspond to an open subset of MATH satisfying the conditions of REF . Then the complement MATH of MATH in MATH is compact, and it is contained in MATH. Thus MATH . |
math/0106197 | Let MATH map to MATH, and let MATH - it is a Riemannian metric on MATH. Define MATH to be the closure in MATH of the orbit MATH. As in the proof of REF , MATH, where MATH is the closure of MATH in MATH, and it is a complex analytic subset of MATH. As it contains the MATH for each MATH, it also open in MATH, and so MATH. We know MATH on MATH satisfies REF , and we now check that it satisfies REF . Suppose MATH at some point of MATH. Then MATH, and because MATH acts transitively on MATH and MATH is MATH-invariant, we must have MATH holding on all of MATH. Since MATH, this shows that MATH satisfies the estimates REF . Therefore, so also does the volume element induced by MATH on MATH, and so REF implies that MATH. We can now apply REF to MATH, MATH, MATH, and we find that MATH is an arithmetic variety. In particular, it is an open algebraic subvariety of MATH, and so REF implies that MATH. We conclude that MATH, and so it is an arithmetic variety. |
math/0106197 | NAME REF |
math/0106197 | Ibid. REF |
math/0106197 | The NAME metric on MATH is the NAME metric, and so from REF we find that MATH . As MATH for all MATH, we see that MATH from which the lemma is obvious. |
math/0106197 | The fundamental group of MATH is MATH, and so every étale covering of it is of the form MATH . The proof is too messy to write out in detail. Consider for example the case MATH, MATH. Then REF Look at MATH - apply REF . CASE: Each other connected component MATH of MATH is simply connected. Thus MATH . Hence MATH and so one only has to arrange things so that MATH . |
math/0106197 | Embed MATH in a compact manifold MATH in such a way that MATH is a divisor with normal crossings. Then there is a finite family of open subsets MATH of MATH such that MATH and, for each MATH, the pair MATH is isomorphic to MATH. For each MATH, choose a neighbourhood MATH of MATH, as in the sublemma, for MATH. Then the complement of MATH on MATH is compact, and it is contained in MATH. Moreover, for any étale covering MATH, MATH . |
math/0106197 | As the NAME volume form MATH on MATH is MATH-invariant, it induces a form MATH on MATH such that MATH where MATH is the covering map MATH. Clearly, MATH . According to REF , there exists an open subset MATH with compact complement, MATH, such that MATH for any étale covering MATH. Because MATH is compact, there exists a finite set MATH of open subsets of MATH such that CASE: there exists isomorphisms MATH (of complex manifolds); CASE: MATH. Because MATH is a disjoint union of copies of MATH, we have MATH and so MATH. Note that MATH (see REF). We conclude that MATH and that MATH . This proves both REF . Now assume that the sequence MATH does not tend to zero. By passing to a subsequence, we can in fact assume that for some MATH, MATH all MATH. Choose MATH as in REF with MATH, and let MATH; then MATH . Let MATH. We showed above that, on MATH, MATH and so, on MATH, MATH. |
math/0106197 | Suppose MATH; then MATH, which contradicts the assertion above that MATH. Hence the sequence MATH does not tend to zero, and so we can choose a subsequence of MATH's for which MATH. Choose MATH such that MATH. Then MATH. Now take a subsequence of the MATH converging to a limit MATH. Then clearly, MATH. |
math/0106197 | Choose a Riemannian metric MATH on MATH, and let MATH be the metric it induces on MATH. Define MATH . Check that these sets have the right property (compare NAME REF, p. REF). |
math/0106197 | Obvious from REF . (Compare NAME REF, p. REF.). |
math/0106197 | Well known - see NAME REF, p. REF. |
math/0106197 | Let MATH be a smooth variety containing MATH as a dense open subvariety. Then REF shows that MATH, and MATH. The lemma is now obvious. |
math/0106197 | Let MATH be such a subbundle. As MATH is dense in MATH (real approximation theorem) and MATH acts transitively on MATH, it will suffice to show that MATH, some MATH, for a fixed MATH. Let MATH be the isotropy group at MATH. Then MATH is a MATH-stable subspace of MATH. With the usual notations, let MATH . Then MATH and MATH . Note that MATH is a MATH-stable subspace of MATH. Almost by definition of what it means for MATH to be irreducible (NAME REF, VIII. REF, p. REF), the action of MATH on MATH is irreducible. Thus MATH, some MATH. |
math/0106197 | For some compact open subgroup MATH of MATH, MATH. The strong approximation theorem shows that MATH is dense in MATH and so, for any open subset MATH of MATH, there exist elements MATH and MATH such that MATH. Clearly MATH, and MATH; thus MATH is dense in MATH. |
math/0106197 | Let MATH be the closed leaf, and let MATH. Then MATH is a closed submanifold of MATH stable under MATH and all MATH for MATH. If MATH, then REF shows that MATH is dense in MATH, which acts transitively on MATH. Therefore, MATH and MATH, whence MATH. |
math/0106197 | Since both MATH and MATH are MATH-invariant and MATH, we can pass to the quotient and obtain MATH and MATH a subbundle of MATH. As MATH, the result recalled above shows that MATH is an algebraic subbundle of MATH. Let MATH be the inclusion MATH. Regard MATH as a sheaf on MATH, and form MATH - this is a coherent algebraic sheaf on MATH. Let MATH be the subset of MATH where MATH is not locally free. Then MATH is an algebraic subset of MATH and its inverse image on MATH-is MATH-invariant (MATH is the set where MATH is not locally free). Therefore REF shows that MATH is empty, and so MATH is locally free. A similar argument applied to the support of the kernel of MATH shows that MATH is a locally free subsheaf of MATH. The fact recalled above shows that it is algebraic. Similar arguments apply to each variety in the projective system MATH, and so we obtain an algebraic subbundle MATH invariant under MATH, and therefore under MATH. Hence MATH is a subbundle of MATH invariant under MATH, and MATH is MATH-invariant. Now REF shows that MATH for some MATH and, because MATH acts transitively on the set of connected components of MATH, this implies that MATH. |
math/0106197 | The reader is invited to check for himself the proof of NAME (NAME REF, p. REF) - essentially the leaves of the foliation are the equivalence classes for the relation defined in p. CASE: Alternatively, it may be possible to give a proof along the following lines: as we observed above, there is a MATH-equivariant map MATH; if we can pass to the quotient by MATH, we get a map MATH whose fibres are the leaves of the foliation. |
math/0106197 | Easy. |
math/0106197 | Otherwise MATH, and so MATH for all MATH. This implies that MATH, which contradicts an earlier statement. |
math/0106197 | The set MATH is stable under MATH and so is either empty of all of MATH. The last lemma shows that it is not equal to MATH, and so we have that MATH. From REF , and REF , we know that MATH is MATH or MATH (recall that we are assuming MATH is arithmetic). Thus MATH on MATH is MATH or MATH. But (by REF), MATH for some MATH. We conclude that MATH (and MATH) or MATH (and MATH). |
math/0106197 | Let MATH as usual. Then MATH . As MATH is a subalgebra of MATH, the result is now obvious. |
math/0106207 | We prove the first statement with the second following in exactly the same way. Set MATH. Then the closure of MATH is MATH. However, MATH, hence MATH. The element MATH is then an eigenvector of MATH with eigenvalue MATH. There are MATH of these eigenvectors, and the eigenvalues are all distinct by CITE. Since MATH is spanned by MATH elements we can deduce that the elements MATH form a basis for MATH and that the eigenspaces are all MATH-dimensional. |
math/0106207 | We prove the result for MATH, with the result for MATH following in exactly the same way. Throughout this proof, we use a standard notation setting MATH. We first define some elements in MATH represented by tangles as shown in REF . Now applying the skein relation once to MATH we obtain: MATH . Repeated application of the skein relation in this way will clearly yield: MATH . Now observe, similar to a result in CITE, we can find: MATH . Combining equations MATH and MATH, we see that we are only left to show that: MATH for MATH. Let MATH. We must now show that for each MATH, with MATH, there exists a MATH such that: MATH . Now, MATH with MATH. The result follows. |
math/0106207 | We prove the first part of the lemma and note that the second part follows immediately due to the observation that MATH. Given MATH we now show how to recover the NAME diagrams MATH and MATH. From the formula for MATH in REF we see that MATH is a NAME polynomial in MATH and MATH, and must be of the form: MATH . Now consider MATH and MATH individually. It is clear that these are also NAME polynomials, this time only in the variable MATH. We have MATH where MATH is the number of cells in MATH with content MATH, and similarly, MATH is the number of cells in MATH with content MATH. Hence we can uniquely construct MATH and MATH. |
math/0106207 | We prove the result for the MATH with an identical argument proving the result for the MATH. Fix an integer MATH such that MATH and MATH (in other words MATH - the case for MATH is identical). Write MATH as MATH and do induction on MATH. For MATH we have that MATH. Now for MATH and MATH we have that MATH. Moreover, in the proof of REF we saw that the MATH with MATH are eigenvalues of MATH. Now since MATH for all MATH, the MATH are also eigenvalues of MATH. Now assume that for MATH and MATH the MATH are eigenvalues of MATH. Since MATH the MATH are also eigenvalues of MATH. Consider the MATH with MATH and MATH. By the inductive hypothesis these MATH are not eigenvalues of MATH since we have MATH eigenvalues and MATH is spanned by MATH elements and by REF we have that if MATH then MATH and MATH. Define elements MATH with MATH and MATH. Clearly MATH. Now by REF , MATH where MATH. We can find a MATH such that MATH. Now consider MATH. This is clearly non-zero. We find: MATH . Hence such MATH are eigenvalues of MATH. Hence by induction, we have that the MATH, with MATH, MATH and MATH, are eigenvalues of MATH. Moreover, we have found at least MATH eigenvalues for MATH. But MATH is known to be spanned by MATH elements, so MATH has at most MATH different eigenvalues. Hence it has exactly MATH eigenvalues each with multiplicity one. |
math/0106209 | To see this, we first compute MATH, MATH. Thus, MATH . Thus, near MATH, the result is rapidly decreasing since the intercluster interactions MATH, MATH are such, and the same holds for the difference MATH. It is only here that we use MATH; see the comments of the following paragraphs for potentials that do not decay rapidly. Hence, in this region, MATH is a union of maximally extended generalized broken bicharacteristics of MATH. That is, if MATH is in MATH, and MATH, then there exists MATH and a generalized broken bicharacteristic MATH such that MATH and MATH for all MATH. (The constant MATH depends on MATH, namely on how long it takes for bicharacteristics to leave MATH.) But if MATH is in addition in the zero section of MATH or of MATH, then MATH, and MATH for MATH, so MATH (as MATH on MATH) contradicting that MATH for all MATH. Here we use again that MATH is not a threshold, for this assumption ensures that there are no stationary bicharacteristics at MATH. Thus, the zero section of MATH and of MATH, that is, the compressed conormal bundle of the diagonal, is disjoint from MATH. A similar argument shows that for MATH, MATH is disjoint from the compressed conormal bundle of the diagonal in the same region. |
math/0106209 | We change the notation slightly and compute MATH as MATH. As usual, we convert the high energy problem into a semiclassical one by letting MATH, and let MATH so MATH, that is, we are interested in the intersection of the spectrum of MATH with a compact interval (namely MATH). Note that the high energy asymptotics MATH corresponds to the semiclassical problem MATH. Note also that MATH, so semiclassically the potential vanishes two orders higher than the semiclassical Laplacian MATH. Let MATH be an almost analytic extension of MATH (so MATH for all MATH), and let MATH. Via the NAME formula, MATH the study of the functional calculus in a semiclassical setting reduces to that of the behavior of the resolvents away from the real axis, and the uniformity in MATH, up to the real axis in MATH. The main fact that makes the calculations uniform in MATH is that by self-adjointness, the MATH operator norm of MATH is bounded by MATH, for this implies (via the parametrix construction) that all seminorms of MATH in the semiclassical version of MATH (compare REF) are bounded by MATH for MATH in a compact set, MATH bounded from above, and for some MATH and MATH depending on the seminorm. Hence it follows immediately that for MATH, MATH . While the integral is trace class, this may not be apparent since the integrand, which is in MATH, is not such due to the diagonal singularity of the resolvents. However, it is easy to rewrite the integral so that the integrand becomes trace class as well. Namely, let MATH, MATH, MATH, so MATH, and write out the NAME integral representation of MATH using an almost analytic continuation MATH of MATH. Thus, MATH . The integrand is now clearly trace class, in fact it is a semiclassical many-body scattering pseudo-differential operator of order MATH, and the asymptotics as MATH follows from the construction of MATH, etc. Our first remark is that since MATH is a differential operator (rather than a more general pseudo-differential operator), for all odd MATH, MATH, as in CITE. This follows from the fact that these terms are given by integrals of odd functions of the momentum MATH, hence these integrals vanish. Indeed, due to the additional MATH vanishing in MATH, which implies similar vanishing for the difference of the various resolvents, the leading term is at least two orders higher, that is, MATH (and we have already mentioned that MATH). This is exactly the same order of vanishing as for two-body scattering. In fact, more is true. To see this, we perform local calculations near MATH, that is, we replace the full trace by the trace of MATH times REF where MATH is supported away from the MATH for MATH. Making sure that the MATH form a partition of unity, REF becomes the sum of the local traces, so it suffices to show that the MATH term of each local trace vanishes. Thus, MATH, MATH, MATH, are already trace class. Using MATH, a similar formula for MATH, that MATH, MATH can be replaced by MATH up to an error of MATH, and the various operators can be commuted up to an error with an additional power of MATH, it is easy to see that the MATH term vanishes, where we also take into account that MATH. (In addition, MATH since the odd coefficients vanish.) Changing back to the MATH-independent notation finishes the proof. An alternative, but somewhat formal calculation (in the sense that it does not use the regularization procedure) proceeds as follows. Note first that the resolvents can be rewritten as MATH . Hence, MATH . Thus, the combination of REF and a regularization argument as in REF, though it is rather cumbersome to keep track of the regularizing factors, shows that there is an additional MATH vanishing here, that is, MATH. Note that REF is trace class when MATH, hence for two-dimensional particles, so in this case we do not need the regularization procedure. |
math/0106209 | This statement is empty for MATH, since MATH. We proceed by induction, assuming that we have shown that for all MATH, MATH, MATH is in MATH away from MATH. Suppose that MATH, MATH; in particular, MATH (for then MATH would hold). Then MATH is in MATH near MATH unless MATH, that is, unless MATH, that is, MATH. Now, MATH and we have just seen that each term in the last sum is in MATH near MATH. On the other hand, let MATH be the maximal element with the property MATH and MATH. Note that this maximal element is unique, namely it is given by MATH, that is, MATH, which is a collision plane since MATH is closed under intersections. In particular, MATH since MATH, hence the first sum in REF is MATH, so MATH . Since MATH, it is in MATH (respectively, NAME) near MATH if the potentials MATH are in MATH (respectively, NAME on MATH), hence the local nature of the construction of MATH and MATH (functional calculus and resolvent construction) yields that MATH is in MATH near MATH, hence providing the inductive step. |
math/0106209 | (Sketch, see CITE for complete details.) We give the full commutator construction at the symbol level, and indicate why it gives rise to a microlocally positive commutator. In fact, the commutator will be positive in part of phase space, negative (or not necessarily positive) in another part of phase space. The propagation of singularities estimates, which should be thought of as microlocal energy estimates, work by estimating MATH in the former region in terms of MATH in the latter region and MATH in the union of both regions. Employing an iterative argument, we may assume that for all MATH with MATH, MATH, and we need to show that MATH. (We can start the induction with a MATH such that MATH.) For points MATH in MATH, MATH . For MATH, MATH, so MATH . We define MATH . Due to REF, and due to MATH, we deduce that MATH. Thus, there exists MATH such that for all clusters MATH with MATH, and for all MATH that satisfies MATH, MATH, MATH, we deduce that MATH . Our positive commutator estimates will arise by considering functions MATH where MATH will be fixed later and MATH is arbitrary as long as it is sufficiently small. Note that for all MATH with MATH, MATH under the decomposition MATH, so MATH on MATH, and similarly, possibly by increasing MATH, MATH. Now suppose that MATH . Then we conclude that MATH . Let MATH. For MATH small, REF thus implies that MATH, MATH, MATH, so we deduce from REF that MATH . The positive commutator estimate then arises by considering the following symbol MATH and quantizing it as in REF. Let MATH be equal to MATH on MATH and MATH for MATH. Thus, MATH, MATH, and MATH, MATH. Let MATH be MATH on MATH, MATH on MATH, with MATH and MATH on some small interval MATH, MATH. Furthermore, for MATH large, to be determined, let MATH . Thus, MATH, and on MATH we have MATH which is REF, so MATH is a subset of REF. We also see that as MATH decreases, so does MATH, in fact, if MATH then MATH on MATH. Note that by reducing MATH, we can make MATH supported in an arbitrary small neighborhood of MATH and MATH. Let MATH be identically MATH near MATH and supported close to MATH. We also define MATH . Let MATH be the operator given by REF with MATH in place of MATH. Note that this includes a spectral cutoff in the definition of MATH. The commutator MATH is given to top order by MATH. This is the commutator that gives microlocal positivity on the MATH eigenspace of MATH, see for example, the NAME proof of the NAME estimate CITE. We proceed to esimate MATH directly. Thus, MATH . Then MATH with MATH . Hence, with MATH we deduce that MATH . Moreover, MATH since MATH on MATH, so MATH . On the other hand, MATH is supported where MATH . By our assumption, this region is disjoint from MATH, if we choose MATH sufficiently small. Moreover, by REF, for MATH sufficiently small, we deduce from the inductive hypothesis that MATH (hence MATH) is disjoint from MATH. Let MATH be a quantization of MATH as in REF. Suppose that MATH and MATH. By choosing MATH large, depending on MATH, MATH, (using REF), one can derive a positive commutator estimate from REF using the many-body pseudo-differential calculus, see CITE for details. Apart from technical details it essentially corresponds to using the NAME estimate and the functional calculus microlocally, namely that when localized in phase space in the region of interest, the commutator of a quantization of MATH is positive. We deduce that there exists MATH, such that for MATH is supported in MATH, MATH near MATH, MATH, MATH with MATH in a small neighborhood of MATH, MATH such that MATH . By MATH we mean the support of the function MATH, which is independent of MATH in REF. Here MATH is the error term, it has first order decay, hence it is `negligible'. On the other hand, MATH has the same order as MATH, and it is negative (that is, has the opposite sign of MATH) in part of the phase space. As mentioned above, positive commutator estimates for approximate solutions MATH, that is, MATH microlocally NAME, work by estimating MATH in terms of MATH (plus error terms), that is, MATH is estimated on MATH by its estimate on MATH. One can now use MATH, chosen sufficiently large, to deal with arbitrary weights MATH. A standard commutator and regularization argument then proves that MATH, which in turn finishes the proof. We refer to CITE for details. Instead of following this route, we prove the corresponding resolvent estimate. So suppose that MATH and MATH is disjoint from the region of interest, and it is, say, in MATH, MATH sufficiently small, so that MATH converges to MATH as MATH in sufficiently large weighted NAME spaces. As above, assume that MATH is uniformly bounded in the region of interest in MATH; we want to prove that it is also uniformly bounded in MATH. For MATH supported sufficiently close to MATH, with MATH, MATH, we deduce from REF that MATH, MATH, with similar properties as in REF. Since MATH we conclude that MATH . Since MATH, the second term on the left hand side can be dropped. Since MATH in MATH for MATH, we conclude that for MATH the right hand side stays bounded as MATH, for MATH is uniformly bounded in MATH on MATH and it is uniformly bounded in MATH on MATH. Thus, MATH is uniformly bounded in MATH, and as MATH in MATH, we conclude that MATH. |
math/0106211 | Let MATH be a twisted curve over a scheme MATH, with moduli space MATH. If MATH is a balanced twisted MATH-cover, we have seen in REF that MATH is an algebraic space. We claim that the composition of MATH, with MATH the morphism to the moduli space, is an admissible MATH-cover. The fact that MATH is a principal bundle follows from the definition, since MATH and MATH is a principal bundle. The fact that MATH is an admissible cover, as well as REF above, are identical to the argument of REF . Conversely, given an admissible MATH-cover MATH over a scheme MATH, consider the stack quotient MATH. Now MATH is a principal MATH-bundle, the morphism MATH is representable since MATH is, MATH is nodal being the quotient of a nodal curve, MATH is the moduli space of MATH, and the action of MATH on MATH is balanced, showing that MATH is balanced. It is easy to see that these correspondences are functorial, and that they are inverse to each other in the usual sense. |
math/0106211 | The statement is local in the smooth topology on MATH, so we may assume that MATH is a scheme; in this case the result is standard. |
math/0106211 | Consider the natural sequence MATH . The group MATH is naturally isomorphic to MATH since MATH is connected and MATH is a principal bundle REF . The term on the right is MATH by REF . |
math/0106214 | By the proof of coherence theorem in appendix, together with the observation discussed above, the problem is reduced to checking the compatibility of triangulated vector spaces of labeled pentagons which have at most one vertex of interaction. Consider the pentagon of type MATH with MATH, MATH, MATH, MATH, MATH. If one of MATH, MATH, MATH and MATH is equal to the unit element MATH, then the pentagonal relation is reduced to the trivial identity of associativity transformations such as MATH . So we may assume that MATH, MATH, MATH and MATH belong to MATH. If they have no interaction point, the pentagonal relation is trivially satisfied. Otherwise, there are three cases according to the position of the interacting point MATH. Suppose, for example, that the point MATH is shared by MATH and MATH. Write MATH and MATH. Then we have MATH and similarly for others. Thus the coherence is reduced to that for the labeling of lower level MATH and we can apply the induction hypothesis. |
math/0106214 | If MATH and MATH are planar diagrams in MATH and MATH respectively with the composed diagram MATH out of MATH, then the image of MATH in MATH is factored into the form MATH, where MATH and MATH with MATH a strict subsequence of MATH. Then, by an easy induction, we see that MATH (the highest spin part MATH in MATH does not appear in MATH for MATH), whence MATH is in the kernel of the linear map MATH. |
math/0106214 | We prove the assertion by an induction on MATH. Assume that the statement is true for MATH. Consider the MATH-module MATH with MATH and let us begin with looking at how it is decomposed into simple components when restricted to MATH. Let MATH with MATH and write MATH. By the obvious isomorphism MATH together with the NAME reciprocity MATH we need to decompose MATH: If MATH, MATH and MATH is forced to be MATH. Then we have the isomorphism MATH and the MATH-module MATH is equivalent to the MATH-module MATH, which is irreducible by the induction hypothesis. Thus the MATH-module MATH is irreducible as well. Next we consider the case that MATH with MATH (MATH). From MATH, we have two simple components MATH and MATH, which are mapped into MATH by MATH with MATH or MATH and MATH in MATH or MATH respectively. The MATH-module MATH is then decomposed into a direct sum of two inequivalent MATH-modules MATH and MATH by the induction hypothesis. To see the irreducibility of the MATH-module MATH, we choose a planar diagram MATH in MATH and let MATH be its image in MATH, which corresponds to MATH in MATH. Let MATH be the diagram obtained by taking the contraction of MATH for two through strings colored by MATH (and MATH) at the right end and set MATH REF . Then we have MATH with MATH and MATH REF . By NAME reciprocity, MATH has the expression MATH and its quantum trace is calculated by MATH if MATH and MATH, which does not vanish. Thus MATH is a non-zero constant multiple of MATH, which belongs to the image of MATH in MATH; MATH and MATH are mixed up by the action MATH on MATH. This, together with the fact MATH, shows that MATH, proving the irreducibility of the MATH-module MATH. Since the image of MATH in MATH is clearly non-trivial, the irreducibility particularly shows the surjectivity of MATH. On the other hand, we know the exact sequence of MATH-modules MATH (see the proof of CITE) and the induction hypothesis implies MATH proving the bijectivity of the map MATH. The inequivalence of MATH-modules MATH for different MATH is now obvious because MATH can be recovered by checking the irreducible decomposition of the restriction to MATH. |
math/0106214 | By the previous lemma, we see that MATH is surjective because MATH is semisimple and inequivalent irreducible representations of MATH gives rise to inequivalent irreducible representations of MATH. (Use the double commutant theorem for the semisimple MATH-module MATH.) From the analysis in CITE, we know that MATH which is checked for generic evaluation parameters but the formula itself clearly holds without restrictions. Since MATH is surjective, the equality of dimensions shows that it is in fact bijective. |
math/0106214 | With the choice MATH, the condition MATH for any MATH is equivalent to require that MATH does not belong to the set specified above. Conversely assume that the NAME algebra MATH is semisimple for any MATH and look at the coloring MATH (MATH). In the planar algebra MATH with MATH an odd integer, taking all the possible pairings of the form MATH or MATH (multiplied by MATH or MATH respectively), we obtain an idempotent MATH of middle pattern MATH. By the choice, the reduced algebra MATH is isomorphic to the NAME algebra, which is semisimple by our assumption. It is well-known CITE that the NAME algebra of MATH strings CITE is semisimple for all odd MATH (if and) only if the parameter MATH is out of the range of numbers in question. |
math/0106214 | We prove by a (reverse) induction on the length of MATH. If MATH, MATH is adjacent to the unique triangulation of length MATH and the problem is reduced to that for MATH. For MATH, MATH contains a diagonal line MATH passing through the vertex MATH. Let MATH and MATH be subpolygons of MATH separated by MATH with the induced triangulations MATH and MATH respectively (MATH). We may assume that MATH contains the bottom edge MATH without loss of generality (see REF ). Since any short-cut to MATH does not change the line MATH, associativity transformations in a short-cut MATH REF are operations on MATH or MATH. Thus, gathering these into two groups, we obtain short-cuts MATH and MATH for MATH and MATH respectively. By associativity of tensor products of linear maps, the isomorphism MATH specified by the short-cut MATH is equal to the one given by the short-cut MATH. In this way, the problem is reduced to those for MATH and MATH, which obviously have smaller lengths than MATH. |
math/0106217 | . CASE: We may assume MATH. The real numbers MATH satisfy MATH. If MATH, then MATH and MATH is MATH-ordered. Otherwise, let MATH be the smallest integer such that MATH. Then MATH . If MATH, one gets MATH and clearly MATH is MATH-ordered. If MATH, then MATH implies REF There are two cases. If MATH, then MATH. From MATH, MATH, it follows that MATH. Let MATH be the integer such that MATH. Then MATH . If MATH, then MATH and consequently MATH. |
math/0106217 | . The condition MATH implies that MATH is MATH-ordered. By the translation rule, we get that MATH is MATH-ordered. The insertion rule shows that MATH is MATH-ordered and, again by the insertion rule, one gets that MATH is MATH-ordered. |
math/0106217 | . The discussion before REF shows that MATH and MATH intersect if and only if MATH or MATH are MATH-ordered. From REF , it follows that MATH or MATH is MATH-ordered. Moreover, MATH and MATH since otherwise MATH is MATH-ordered and the intervals are disjoint by REF . If MATH (or equivalently if MATH), then MATH. Thus, we may assume that the numbers MATH are distinct. Assume the first ordering holds. The formula for the intersection is straightforward if MATH. If MATH, then MATH and MATH. The two other cases are proved in the same way. |
math/0106217 | . Arguing by contradiction, suppose that MATH. Since MATH, the interval MATH is not empty, therefore by REF MATH or MATH is MATH-ordered (or the sequence obtained by exchanging MATH and MATH). Consider the first case, the second is the same by exchanging MATH and MATH. Since MATH is MATH-ordered, the translation rule shows that MATH is MATH-ordered which gives, applying twice the insertion rule, that MATH is MATH-ordered. From this, we get that MATH is MATH-ordered. From REF , we know that MATH, and this is then disjoint from MATH. |
math/0106217 | . This is a direct consequence of the preceding discussion. |
math/0106217 | . Suppose that MATH with MATH. The other case is symmetric. We first prove that MATH . Set MATH. Since MATH, the interval MATH is not empty. By REF there are two cases: either MATH is MATH-ordered, or MATH is MATH-ordered. We show that this second case cannot happen. Indeed in this case, MATH . Moreover for each MATH, the sequence MATH is MATH-ordered. By translation and insertion, the sequence MATH is MATH-ordered. This shows that MATH, and consequently MATH for each MATH in MATH, contradicting the assumption that MATH . Thus, MATH is MATH-ordered. This implies that MATH . If MATH then MATH is MATH-ordered. By translation, MATH is MATH-ordered. By insertion of MATH into MATH one gets MATH is MATH-ordered. Again by insertion of MATH, the sequence MATH is MATH-ordered. Thus MATH . The second part of the proof deals with MATH . In this intersection the index MATH runs through the set MATH. The set MATH is partitioned into three possibly empty subsets as follows: MATH iff MATH, MATH iff MATH and finally MATH iff MATH . Of course, MATH . If one of the sets MATH is empty it does not contribute to the intersection. Clearly MATH. Next MATH. If MATH is not empty then MATH is in MATH and MATH. Finally, MATH. If MATH is not empty then MATH is in MATH and MATH. To finish the proof, we just have to verify that in each case, MATH is equal to MATH. If MATH then MATH and the sequence MATH is MATH-ordered (case MATH). If MATH then the sequence MATH is MATH-ordered (case MATH). If MATH then MATH and the sequence MATH is MATH-ordered (case MATH). If MATH then the sequence MATH is MATH-ordered (case MATH). CASE: If MATH and MATH are nonempty then MATH. As the sequences MATH and MATH are MATH-ordered, by the insertion rule either the sequence MATH or MATH is MATH-ordered. The first case is impossible because MATH is not empty. The second case implies that MATH. CASE: If MATH and MATH then MATH and the sequences MATH, MATH are MATH-ordered. By insertion the sequence MATH is MATH-ordered. Thus MATH. Case MATH is symmetric to case MATH. CASE: If MATH and MATH then MATH and the sequences MATH, MATH are MATH-ordered. By insertion rule either the sequence MATH, or MATH is MATH-ordered. The first case is impossible because MATH is non empty. The second case implies that MATH. |
math/0106217 | If MATH or MATH then MATH . As MATH by translation rule we have MATH. Furthermore, MATH . By the preceding remark and by identification, the only possibility is MATH and MATH . It follows that MATH . If MATH then MATH . By hypothesis we have MATH . But MATH implies that MATH . If MATH then MATH should be empty in contradiction with the hypothesis. Thus MATH. The interval MATH is equal to MATH . That is MATH and MATH . By identification, we have MATH . If MATH then MATH . As MATH implies MATH, by contraposition MATH implies MATH for all MATH. Thus MATH is MATH-ordered for all MATH. As MATH and MATH is not in MATH the sequence MATH is MATH-ordered. Thus MATH is MATH-ordered. Consequently MATH is equal to MATH . As MATH by identification we find MATH. |
math/0106226 | Using the above notation, let MATH be a minimal free resolution of MATH with MATH for some MATH. First we show that MATH. By REF, there exist an element MATH such that MATH. If MATH, then in the complex MATH, the element MATH satisfies MATH since MATH. However, MATH, and hence MATH is a MATH-cycle which is not a boundary. Therefore MATH, which contradicts the assumption. Hence MATH, and the projective dimension of MATH, MATH, is finite. Observe that REF implies that the depth of MATH is zero. Hence by the NAME formula, MATH, MATH is projective. |
math/0106226 | Let MATH be a minimal free resolution of MATH. Then in the complex MATH, all the differentials are zero and therefore, MATH. |
math/0106226 | Let MATH be a minimal free resolution of MATH with MATH. Then MATH. On the other hand, in the complex MATH, all the differentials are zero. Hence MATH. Thus, MATH which is finite since MATH is NAME. |
math/0106226 | First note that MATH is a regular sequence on MATH since MATH is again a MATH-regular sequence. If MATH then from the long exact sequence MATH associated to the short exact sequence MATH we conclude that MATH. Now consider the long exact sequence MATH associated to the short exact sequence MATH . From the argument above we get that if MATH then MATH, and hence by REF , MATH. Inductively, if MATH for MATH then MATH. Let MATH be a minimal free resolution of MATH. Then MATH where MATH and MATH. By assumption, MATH and hence, by REF , MATH. Thus, by the inductive argument, we cannot have MATH for all MATH. |
math/0106227 | We have to show that MATH is not bounded from below on MATH whenever MATH, MATH, MATH. By the above, MATH is not bounded from below on MATH; hence, given MATH, for some MATH we have MATH. Now pick MATH such that MATH; then there is some MATH such that MATH that is, MATH and MATH. |
math/0106227 | The implications REF are evident. CASE: Assume there is some closed absolutely convex subset MATH with the property from REF that does not coincide with the whole space MATH. By the NAME theorem there is a functional MATH and a number MATH such that MATH for every MATH. If MATH and MATH are arbitrary, pick MATH; this intersection is nonempty by assumption on MATH. It follows that MATH, hence REF fails. CASE: Suppose MATH and put MATH. By the definition of a strong NAME operator this MATH satisfies REF . So MATH and hence MATH. |
math/0106227 | It is enough to prove the implications REF for the following assertions about a fixed number MATH: CASE: There exists a closed absolutely convex set MATH not containing MATH that intersects all MATH. CASE: There exists a functional MATH such that for all MATH there exists MATH satisfying MATH. CASE: There exists a closed absolutely convex set MATH not containing MATH for any MATH that intersects all MATH. To see that REF implies REF , pick MATH, MATH. By the NAME theorem we can separate MATH from MATH by means of a functional MATH; then we shall have for some number MATH that MATH for all MATH and MATH. On the other hand, MATH; hence REF holds for MATH. If we assume REF , we define MATH to be the closed absolutely convex hull of the elements MATH, MATH, appearing in REF . Obviously MATH intersects each MATH. If MATH for some MATH, then since MATH on MATH, we must have MATH on MATH, that is, MATH. Therefore, MATH works in REF . |
math/0106227 | REF implies that MATH for every unit basis vector MATH if MATH. [Actually, the theorem quoted is formulated for operators on MATH for compact MATH, but the theorem works likewise on MATH with MATH locally compact.] Hence MATH is the only strong NAME operator on MATH. (Another way to see this is to apply REF .) To show that MATH is not NAME we shall exhibit a closed absolutely convex set MATH intersecting each MATH, yet containing no ball. Let MATH, that is, MATH which is closed in MATH. Fix MATH and MATH. If MATH, say MATH, pick MATH such that MATH. Then MATH for every MATH. Obviously, MATH does not contain a multiple of MATH. CASE: We claim that MATH defines a strong NAME functional on MATH whenever MATH is a sequence of signs, that is, if MATH for all MATH. Indeed, let MATH, MATH and MATH. Pick MATH such that MATH and define MATH by MATH for MATH and MATH for MATH. Then MATH and MATH; hence MATH and MATH. |
math/0106227 | Let MATH be a NAME and NAME be a closed absolutely convex subset which intersects all the elements of MATH. In particular for every fixed MATH the set MATH intersects all the sets MATH, MATH. By definition of a NAME this means that all the points of the form MATH, MATH, belong to MATH, that is, MATH. But MATH is also a NAME, so MATH, and by convexity of MATH, MATH. |
math/0106227 | This follows from the previous proposition; that the NAME is MATH is a consequence of MATH for all MATH; see the above proof. |
math/0106227 | Assume to the contrary that there is a finite-dimensional space MATH that is not NAME. By REF we can find a sequence of functionals MATH such that MATH for each MATH. By compactness of the ball we can pass to the limit and obtain a functional MATH with the property that MATH for each MATH. Denote MATH; this is a norm-compact convex set. Let MATH be an arbitrary point. If we apply the above property to MATH for all MATH, we obtain, again by compactness, some MATH such that MATH, MATH and MATH. We have MATH, so MATH. Therefore MATH hence MATH and MATH is a diametral point of MATH, meaning MATH . But any compact convex set of positive diameter contains a nondiametral point CITE; thus we have reached a contradiction. |
math/0106227 | Denote MATH, MATH. Then MATH but MATH. So MATH. |
math/0106227 | For every MATH select MATH. Then MATH, MATH. Denote by MATH a supporting functional of MATH. By the previous lemma MATH tends to REF when MATH tends to infinity. So by the definition of an exposing functional, MATH tends to MATH. By the same lemma MATH tends to MATH, so MATH also tends to MATH. Hence every weak limit point of the sequence MATH belongs to the intersection of MATH and MATH, so this intersection is nonempty. |
math/0106227 | CASE: Let MATH be a closed absolutely convex set which intersects all the sets MATH, but does not contain MATH. By the NAME theorem there exists a functional MATH such that MATH for every MATH. We fix MATH with MATH. Let MATH be a strongly exposed point of MATH. As before, we denote an exposing evaluation functional by MATH. Now MATH for all MATH. By REF and the evident equality MATH this implies that the set MATH intersects MATH. If MATH is an element of this intersection, we see that MATH. Likewise, since MATH, we find some MATH; hence MATH. Therefore, MATH. CASE: The argument is the same as in REF . |
math/0106227 | Assume that MATH and that its NAME is MATH; then this parameter is strictly smaller than some MATH. Choose MATH as in REF so that MATH for every strongly exposed functional MATH. We now claim that in any MATH-neighbourhood of MATH there is some MATH which can be represented as a convex combination of MATH strongly exposed functionals. First of all, the convex hull of the set MATH of strongly exposed functionals is norm-dense in MATH; in fact, this is true of any bounded closed convex set in a separable dual space CITE. Hence for some MATH, MATH with MATH and MATH . Let MATH and let MATH be the point of intersection of the segment MATH with the relative boundary of MATH, that is, MATH with MATH. Let MATH be the face of MATH generated by MATH; then MATH is a convex set of dimension MATH. Therefore an appeal to NAME 's theorem shows that MATH can be represented as a convex combination of no more than MATH extreme points of MATH. But MATH, and our claim is established. We apply the claim with some MATH to obtain some convex combination MATH of MATH strongly exposed functionals such that MATH. One of the coefficients must be MATH, say MATH. Now if MATH, MATH . By REF we have MATH which contradicts our choice of MATH. |
math/0106227 | The argument of REF implies in the setting of MATH rather than MATH that the NAME of MATH is MATH, and the converse estimate follows from REF . |
math/0106227 | By the definition of an interpolating pair, for an arbitrary MATH there exists an element MATH, MATH, such that MATH on MATH and MATH on MATH. Then the element MATH belongs to MATH, so MATH which completes the proof. |
math/0106227 | Before we enter the proof proper, we formulate a number of technical assertions that are easy to verify and will be needed later. If MATH is a strong NAME operator on a NAME space MATH, if MATH and MATH, then there is some MATH such that MATH . Choose MATH and MATH such that MATH, MATH and pick by REF MATH such that MATH; this MATH clearly works. If MATH, MATH and MATH in a normed space, then MATH whenever MATH. Should MATH for some MATH, then, since MATH for MATH, MATH . (The case MATH is analogous.) If MATH and MATH in a normed space, then MATH. Should MATH, then MATH . To start the actual proof we may assume that MATH. Fix MATH and MATH such that MATH; let MATH so that MATH. Put MATH, MATH and pick MATH such that MATH . We are going to construct functions MATH by induction so as to satisfy CASE: MATH, MATH, CASE: MATH on MATH and MATH on MATH, MATH, CASE: MATH, MATH, MATH. Suppose that these functions have already been constructed for the indices MATH. We then define MATH as in REF . Since by induction hypothesis MATH and MATH, we clearly have MATH. From MATH we conclude using NAME REF (with MATH) that MATH. Thus REF is achieved. To achieve REF it is enough to use that MATH is interpolating along with the induction hypothesis that MATH. Finally REF follows from NAME REF with MATH. Next we argue that MATH . This follows from NAME REF and our choice of MATH. But for MATH we can estimate MATH therefore, letting MATH, MATH . Furthermore we have the estimates MATH and for MATH . By REF and the above we see that MATH. Hence it is left to replace MATH by an element MATH, MATH, to finish the proof. |
math/0106227 | The implication MATH follows from REF , as follows. Let us apply REF to MATH, MATH, MATH, where MATH is a positive scalar function vanishing on MATH, and MATH. Then for MATH which we get from REF let us find a point MATH such that MATH. Because MATH we have MATH, that is, MATH. Now select an open neighbourhood MATH of MATH such that MATH for all MATH and put MATH. To prove the implication MATH let us fix a positive MATH, MATH and MATH. Now apply inductively REF to obtain elements MATH, MATH, MATH, MATH, open subsets MATH, closed subsets MATH, MATH and functions MATH with the following properties: CASE: MATH, CASE: MATH, MATH, CASE: MATH, MATH for all MATH, MATH, and MATH. By an argument similar to the one in REF , we have for a proper choice of MATH . Let us put MATH. Then the last inequality and REF of our construction yield that MATH, MATH and MATH. The only thing left to do now is to estimate MATH from above. If MATH, then MATH. If MATH for some MATH then MATH . In this sum all the summands except for the last one satisfy the inequality MATH and the last summand MATH is bounded by MATH. So MATH . The same estimate holds for MATH. To prove the implication MATH fix MATH and MATH. Pick a point MATH with MATH and a neighbourhood MATH of MATH such that MATH . Denote MATH and MATH and apply REF to obtain a function MATH such that MATH, MATH and MATH. For this MATH we have MATH and MATH, so MATH. Let us now pass to the case of a perfect compact MATH. The implication MATH is evident. The proof of the remaining implication MATH is similar to that of REF . Namely, let MATH, MATH and let MATH be small. We have to show that there is an element MATH such that MATH and MATH . To this end let us pick a closed subset MATH (whose complement MATH we denote by MATH) and a point MATH in such a way that MATH, MATH and for every MATH . Denote MATH, MATH and apply REF to obtain a function MATH such that MATH, MATH and MATH. For this REF follows from REF follows from REF . |
math/0106227 | REF immediately provides the proof. |
math/0106227 | The key feature of MATH is that MATH is not a MATH-semigroup; for we have shown in REF that MATH and MATH define strong NAME functionals on MATH, but MATH is not in MATH and hence MATH is not, either. Now if MATH fails to be a MATH-semigroup, pick MATH with MATH. Put MATH for a perfect compact NAME space MATH; then by REF MATH, but MATH. |
math/0106227 | Let us fix MATH, an open set MATH in MATH and MATH. By REF we find a function MATH as described there corresponding to MATH, MATH and MATH. Put MATH and MATH. As above, there is a function MATH corresponding to MATH, MATH and MATH. We denote MATH and continue the process. In the MATH step we get the set MATH and apply REF to obtain a function MATH corresponding to MATH. Choose MATH so that MATH and put MATH. Now using the NAME Lemma we find a continuous function MATH satisfying MATH for all MATH, MATH, MATH and vanishing outside MATH. We claim that MATH. Indeed, by our construction, if MATH, then MATH, and if MATH (with the understanding that MATH stands for MATH), then MATH . Moreover, MATH . Thus MATH, and since MATH was chosen arbitrarily, we are done. |
math/0106227 | CASE: Let MATH be MATH-narrow. We will use REF . Let MATH be a closed subset, MATH, MATH and MATH. According to REF there exists a function MATH vanishing on MATH of the form MATH, where MATH, MATH and MATH is nonnegative, such that MATH. Evidently this MATH satisfies all the demands of REF . CASE: Let MATH be a strong NAME operator, and suppose MATH is separable. Let MATH be a nonvoid open subset. Given MATH and MATH we define MATH . This is an open subset of MATH, and by REF it is dense in MATH. Now pick a countable dense subset MATH of MATH and a null sequence MATH. Then by NAME 's theorem, MATH is nonempty. Let MATH, and fix MATH. We denote by MATH the closure of MATH this is an absolutely convex set. We claim that MATH intersects each set MATH. Indeed, if MATH, MATH, MATH and MATH, then for a function MATH as appearing in the definition of MATH we have MATH. Since MATH is NAME, say with parameter MATH, the set MATH contains MATH. This implies that MATH satisfies the definition of a MATH-narrow operator with constant MATH. CASE: Let MATH; then by REF MATH is a strong NAME operator on MATH. But MATH hence MATH is not MATH-narrow unless MATH. |
math/0106227 | Let MATH, and pick functions MATH as in REF . Then by the NAME Dominated Convergence Theorem, for any given MATH we have MATH . Conversely, let MATH be a sequence of open sets in MATH such that MATH and MATH. By the NAME Lemma there exist functions MATH having the following properties: MATH for all MATH, MATH, and MATH if MATH. Clearly, MATH pointwise and MATH whenever MATH. This means that the sequence MATH is weakly null. Applying the NAME Theorem we finally obtain a sequence of convex combinations of the functions MATH which satisfies all the conditions of REF . This completes the proof. |
math/0106227 | We first prove the more general ``moreover" part. Put MATH. By the definition of a MATH-narrow operator and REF there is a function MATH with MATH, MATH and MATH. Obviously, MATH. Again applying the definition we find MATH with MATH, MATH and MATH. As above MATH. In view of the MATH-narrowness of MATH there exists a function MATH with MATH, MATH and MATH. In the next step we construct MATH such that MATH and MATH. Proceeding in the same way, in the MATH step we find a set of functions MATH and nonempty open sets MATH in MATH such that MATH, MATH and MATH, if MATH. Then we put MATH to start the next step. It remains to show that the set MATH is as desired. Indeed, MATH is clearly a nonempty closed MATH-set and MATH for every MATH. It is easily seen that the sequences MATH and MATH meet the conditions of REF for the operator MATH. So, MATH for every MATH. To prove the converse, let MATH be any open set in MATH and let MATH. By assumption on MATH we can find a closed MATH-set MATH, MATH. Consider the open sets MATH and functions MATH provided by REF . For sufficiently large MATH we have MATH and MATH so that MATH may serve as a function as required in REF . This finishes the proof. |
math/0106228 | REF is an immediate consequence of REF . If MATH, then the coherent restriction of MATH to the preimage of MATH is trivial. Hence the same is true for its coherent restriction to MATH or MATH, which proves REF . It is clear that MATH. Now REF follows also: MATH . |
math/0106228 | The first assertion is clear. The strict transform of MATH in the blowing ups of members of MATH contained in MATH is MATH. The strict transform of MATH meets the latter in MATH. Then MATH must be the closure of the preimage of MATH in MATH. If we next blow up the members of MATH strictly between MATH and MATH, the strict transform of MATH is MATH. Blowing up MATH yields MATH, Finally, blowing up the members of MATH strictly containing MATH gives MATH. The lemma now follows easily. |
math/0106228 | By REF contains the union in the statement. The opposite inclusion also holds: MATH and MATH are related if and only if MATH and both have the same image in MATH. |
math/0106228 | Let MATH be generated by its sections and separate the points of MATH. Then the invertible sheaf MATH has the same property. Let MATH be the morphism defined by a basis of its sections. A possibly higher power of MATH will have a normal variety as its image and so it suffices to prove that MATH separates the points of every stratum of MATH. By hypothesis this is the case for MATH. Given MATH of codimension MATH, then it follows from REF that the pull-back of MATH to MATH is isomorphic to a line bundle on MATH tensorized over MATH with MATH. Any set of generating sections of MATH must therefore determine a set of generators of the vector space MATH. Such a generating set separates the points of MATH. The same is true for a set of generating sections of a power of MATH. So for some positive MATH, MATH separates the points of every stratum of MATH. |
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