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math/9912131 | In the proof of REF , we showed that MATH. Using MATH and MATH it follows that MATH, and REF is an immediate consequence. |
math/9912131 | Suppose MATH is a spectral pair, then MATH is a spectral pair by REF, hence MATH determines an isometric isomorphism mapping MATH onto MATH. We must show that MATH determines an isometric isomorphism mapping MATH onto MATH . But this is easy since MATH so MATH. By construction of MATH from MATH it follows that MATH is ... |
math/9912131 | Since REF is trivial, we will show that REF . If MATH, then MATH for any MATH. It follows that MATH for MATH-a.e. MATH. Now MATH so MATH and therefore MATH for any MATH and MATH-a.e. MATH. So, either MATH, or MATH. Consequently, MATH . It follows that MATH . Hence, MATH, for any MATH. By REF it is sufficient to show th... |
math/9912131 | This is a simple consequence of the proof of REF . |
math/9912131 | If MATH, then MATH for MATH-a.e. MATH. |
math/9912131 | If MATH, then MATH hence MATH . Similarly, if MATH, then MATH so MATH . The desired equality is immediate. |
math/9912131 | Let MATH . By density of MATH in MATH, we have MATH. If MATH then MATH is continuous and MATH so MATH, hence MATH. Using MATH, we see that there exist MATH such that MATH for MATH in some neighborhood MATH of MATH and MATH, as desired. |
math/9912131 | Let MATH. Then MATH is continuous and MATH. Let MATH be an open set such that MATH and MATH for any MATH. Then MATH whenever MATH. |
math/9912131 | If MATH, MATH and MATH, then MATH . If furthermore MATH and MATH, then also MATH completing the proof. |
math/9912131 | Let MATH be measurable with MATH and let MATH. Then MATH and MATH by the Uncertainty Principle with MATH. The desired conclusion is immediate. |
math/9912132 | Let the functions be as described in the lemma, MATH, MATH. We first calculate the term MATH from the desired REF . Keep in mind that all integrals and summations are convergent relative to the respective NAME norms (due to the isometries which we described before the statement of the lemma). Details are in CITE and CI... |
math/9912132 | MATH where, in the last step, we used the standard NAME series representation of the respective functions. |
math/9912132 | In the following calculation, convergence is governed by the norm-isometric property of MATH, and this also justifies the exchange of summations and integration: MATH since we already proved the identity MATH (in REF ), the proposition follows. |
math/9912132 | We begin with REF since it is universal. Before the estimate, we may restrict to MATH of compact support, and then, after the fact, this restriction is removed by completion. In the following estimates, we use the NAME - NAME inequality two times, for the respective NAME inner products involved: MATH proving REF . REF ... |
math/9912132 | We compute MATH . If MATH, then MATH and the assertion follows from a standard fact on multiplication operators. The same argument also yields the condition for invertibility of MATH. |
math/9912132 | CASE: Suppose MATH; then MATH which is REF . CASE: Suppose REF holds. Then the previous calculation reverses, and shows that MATH where MATH, and we have REF . Suppose REF , and let MATH; then MATH and we conclude that MATH as claimed. Now let MATH, and consider the following calculation: MATH . If MATH, then this inte... |
math/9912132 | By REF , it is enough to check that every operator MATH which satisfies REF MATH REF must be of the form MATH for some MATH. So let MATH be given, and assume REF . Let MATH. Then, for MATH, we have MATH. We are considering only the case when MATH is bounded. Since MATH is dense in MATH, the result follows. If REF also ... |
math/9912132 | In the previous section, we also elaborated on the operators MATH, MATH, MATH, and the cascade operator MATH for MATH, representing the given low-pass filter. The present proof amounts to a combination of the calculations leading up to REF , and the argument from the proof of that lemma. |
math/9912132 | With the aid of REF , the proof of the two commutation relations REF - REF now amounts to the following computations. They take place in the space MATH, that is, the range of the NAME transform, so it is the right-hand column in REF which is used. CASE: Let MATH and MATH. Then MATH . But the summation is over MATH, and... |
math/9912132 | In the proof, we shall refer to the two REF - REF . If MATH is a linear subspace, the orthogonal complement will be denoted MATH . We clearly have MATH. MATH is invariant under MATH, MATH. Let MATH. To show that MATH, we check that MATH . We used REF in the calculation, noting that MATH by definition. Our next assertio... |
math/9912132 | By REF , MATH . In the first step, REF is used on MATH, then commutativity is used, and in the last step, REF is used on MATH. Since MATH is assumed faithful, MATH and MATH where the quadratic property of MATH was used in the last step. Hence MATH as claimed. |
math/9912132 | CASE: Let MATH. Then MATH and this proves REF . CASE: If REF holds, the calculation shows that the second and the third terms must agree for all MATH, and, by duality, this means that REF must hold a.e. on MATH. |
math/9912132 | Immediate from REF in the previous section. The calculation of the expansion REF follows from checking that the NAME coefficients of MATH are as stated, that is, that MATH . |
math/9912132 | First note that if MATH is assumed continuous, then MATH maps MATH into itself. We see this by first approximating MATH with finite sums MATH. For each such finite sum, the corresponding MATH maps MATH (MATH the ring of finite NAME series) into itself. Since the norm of MATH, as a operator in MATH, is one, that is, MAT... |
math/9912132 | We have the formula MATH directly from a NAME transform of MATH. It follows that MATH if MATH. The lemma now follows from REF . |
math/9912132 | We have MATH . But if MATH, then MATH (from REF of MATH in the Introduction), so MATH. REF then implies that MATH if MATH, which is the desired conclusion. |
math/9912132 | This is the same argument as the one used in REF above, and it is based on MATH . |
math/9912132 | Contained in the previous argument. |
math/9912132 | Note that, when REF - REF are combined with REF , we get MATH, but REF is a better approximation. We first prove REF by checking that CASE: MATH, and CASE: MATH, as MATH. For the first term REF we have MATH where MATH . We take MATH, and recall the formula MATH. Hence MATH. Since this function is continuous, it follows... |
math/9912132 | Since MATH, it follows from REF that MATH, that is, we have both of the families MATH and MATH orthogonal in MATH. For MATH, we also have MATH . Similarly, MATH so, for the norm-difference in REF , we have MATH which means that REF is equivalent to MATH . This last term computes out as follows: first introduce the sequ... |
math/9912132 | From REF , we have MATH and MATH for all MATH, so MATH and the result is immediate from this. |
math/9912132 | In the discussion of REF , we showed that MATH and, as MATH, the limit of that term is MATH. Let MATH, MATH. Then we saw that MATH . This follows from the identity MATH from REF . Hence MATH and it is clear from REF that MATH in MATH if this term is nonzero. |
math/9912132 | We have MATH and the claim follows. |
math/9912133 | If MATH, its NAME transform is MATH, and the NAME operator REF transforms into MATH . This follows from the next computation on MATH . We have MATH . The terms with odd MATH disappear in the last sum, so MATH and REF follows. If MATH, MATH are MATH-functions with compact support, define MATH . To prove REF , we have to... |
math/9912133 | CASE: By NAME 's identity in MATH, MATH . CASE: We have MATH . CASE: Since MATH by REF , we need only calculate MATH, where MATH is the cascade operator MATH given by the low-pass filter MATH. MATH where MATH is the cascade operator corresponding to the product filter MATH. CASE: From the definition of the cascade oper... |
math/9912133 | We have already commented on one inclusion in REF , and the second follows from MATH where MATH is chosen such that MATH. (Equivalently, MATH.) |
math/9912133 | CASE: As an aside, remark that MATH having MATH as a simple eigenvalue means here that the corresponding eigenspace is one-dimensional. But since MATH, and hence MATH is bounded in any equivalent norm MATH on the linear operators on MATH, it follows from NAME 's theorem that the multiplicity of MATH in the characterist... |
math/9912133 | From the definition of MATH we have MATH where we used the reality assumption in the form MATH. |
math/9912139 | The first inequality is essentially REF and the second one follows directly from the definition. The converse is also true. If a measure satisfies the inequalities with MATH then it is locally doubling. |
math/9912139 | Assume that this is not the case. Then there is a subinterval MATH such that MATH and such that the square of side length MATH that it is halved by MATH is inside MATH (see REF ). We can construct a doubling measure MATH in the interval MATH which is the base of the square that contains MATH. The measure of any set MAT... |
math/9912139 | We split the integral into two. In the first we integrate over the region MATH. In this region MATH, therefore the integral is bounded by some constant times MATH. In the second we integrate over the region MATH. We split it into coronas of doubling size and we may estimate it by MATH where MATH is such that MATH. Cons... |
math/9912139 | We start by assuming that MATH, the general case follows if we use the same construction with the measure MATH instead of the measure MATH. We will first find a partition into rectangles MATH in such a way that MATH, MATH and with the ratio between side-lengths bounded. Later on, we will refine this partition in order ... |
math/9912139 | The first assertion follows since MATH has bounded excentricity and constant mass. The second one is an immediate consequence of REF . |
math/9912139 | We want that MATH for all polynomials of degree smaller or equal to MATH. We may take any NAME quadrature formula with MATH nodes in MATH that is exact for polynomials of degree MATH. This can be done, eventually taking MATH much larger than MATH (see CITE, for a survey on quadrature formulas with equal weights). These... |
math/9912139 | Take the partition of MATH in rectangles given by REF . We distribute the rectangles in a finite number of families of rectangles MATH with the property that any two rectangles of the same family MATH, MATH are very far apart (that is, MATH, for some large constant MATH). This is possible with the NAME covering lemma. ... |
math/9912139 | We will split MATH in two measures MATH. To describe the measure MATH, let us tile the plane into squares MATH of diameter MATH (dyadic squares in the case of the disk) in such a way that MATH for all MATH. This is feasible because of the hypothesis on the measure. The measure MATH is defined as MATH. The measure MATH ... |
math/9912139 | This is an immediate consequence of NAME 's inequality. |
math/9912139 | The proof of this theorem is the same as in REF when MATH, except that at some points it is easier. For instance, it is trivial to split the real line into intervals all of mass MATH. |
math/9912145 | The standard symplectic form on MATH is MATH. Let MATH and let MATH. This vector field and some level sets of MATH are shown in REF . Notice that MATH is a symplectic dilation and is positively transverse to the level sets of MATH as long as MATH and as long as MATH is defined, but that MATH does not extend across MATH... |
math/9912145 | That each MATH is a positive contact form is a straightforward calculation. To see why the rest is true, note that since MATH, REF-form MATH is closed and therefore exact. Choose MATH so that MATH. Then MATH. |
math/9912145 | Let MATH be the Legendrian knot, MATH a neighborhood of MATH and MATH a framing of MATH. Without loss of generality, by NAME 's theorem for contact structures, we may assume that MATH has the form MATH where MATH, MATH and MATH are coordinates on the disk MATH of radius MATH, and MATH is the MATH-coordinate. This is be... |
math/9912145 | The first result follows from the fact that MATH and MATH are solutions to the same ordinary differential equations with the same initial conditions. For the second result, let MATH, let MATH and let MATH. We will show that there exist unique vector fields MATH on MATH such that MATH. Then the vector fields MATH and MA... |
math/9912145 | Everything follows from the explicit expressions MATH and MATH. |
math/9912145 | First we prove existence of MATH with its dilation-contraction pair MATH. Construct MATH as in REF . Let MATH be an open neighborhood of MATH in MATH such that flow along MATH in MATH starting from MATH is defined for all times MATH with MATH. This gives an embedding MATH such that MATH and MATH. Since both MATH and MA... |
math/9912145 | Let MATH on MATH with MATH, and let MATH . Calculation shows that MATH is a dilation-contraction pair which transversely covers the level sets of MATH as long as MATH. Let MATH and note that MATH (because MATH and MATH). Choose any MATH with MATH. Then the induced contact pair on MATH is given by: MATH with MATH . Forw... |
math/9912145 | As mentioned earlier we may assume that MATH. To avoid too much notation, we will use MATH on MATH to refer to MATH. For a given MATH we will have MATH on MATH and REF gives us conditions for MATH to be transverse to MATH. When MATH is also transverse to MATH we get MATH. Choose a constant MATH with MATH (we have MATH ... |
math/9912145 | Enlarge MATH using the positive symplectification of MATH (see REF ). Attach the subset MATH of this positive symplectification, where MATH comes from REF , using the uniqueness of the symplectic germ MATH. |
math/9912145 | We will first argue that we can enlarge MATH so as to arrange that MATH is in fact partially convex and partially concave, with induced contact pair MATH, and so as to arrange that there are coordinates near each component of MATH, realizing the desired framing, with respect to which MATH is well-behaved satisfying the... |
math/9912148 | Suppose that MATH is obtained from MATH by adding to column MATH. Writing everything out, one sees that MATH . Using the fact that MATH and multiplying terms, the result follows. |
math/9912148 | By REF and the definition of MATH, any path from MATH to MATH yields the equality MATH . In the following equations, paths from MATH to some MATH such that MATH are chosen so as to first go to MATH (in a way independent of MATH) and then go to MATH. Consequently, MATH . Since MATH, the final equality is simply REF on p... |
math/9912148 | The second expression for MATH implies that MATH for all MATH. The fact that MATH follows from the hypotheses on MATH and the MATH's, together with the skew-expansion rule (REF ' on page REF) for NAME polynomials. Thus MATH for all MATH. From the definition of MATH it is a probability measure. For larger MATH this foll... |
math/9912148 | In general the transition probabilities from MATH to MATH to sample from a coherent family MATH is MATH. These sum to MATH by the definition of coherence, and sample from MATH because MATH . This principle together with the formula for MATH inside the proof of REF , imply the proposition. |
math/9912148 | This follows by comparison with the formula in REF. |
math/9912148 | This follows easily from the following five ingredients: REF , homogeneity of MATH (which implies that MATH), NAME 's principal specialization formula (page REF), and a piece of paper. |
math/9912148 | We use the ergodic method REF . The map MATH is the same as for the NAME lattice and the NAME kernel is defined as MATH. To see that the first condition of a boundary is met, recall that the function MATH is harmonic if it exists (here MATH is a sequence of vertices of a path with each MATH). In fact it is true that MA... |
math/9912152 | Evident. |
math/9912152 | Assertions a and b follow from routine verification. To prove assertion c, first recall that the connected component of the group completion MATH of a topological monoid MATH is homotopy equivalent to the colimit MATH, where the MATH's are connected components of MATH, and MATH is a collection contained countably infin... |
math/9912152 | REF follows from a simple routine verification. See REF below. The filtration preserving property follows from the construction of the maps. It follows from REF that pull-back under MATH gives an isomorphism between MATH and the invariants MATH under the action of MATH. On the other hand, it is well-known that MATH als... |
math/9912152 | It is clear that MATH induces an isomorphism MATH, for all MATH. Since an abelian topological monoid is simple, the result follows. |
math/9912152 | Let MATH denote the natural map, defined as the composition MATH; compare REF . It suffices to show that MATH for all MATH. Since MATH induces an injection in rational cohomology, we will then show that MATH, for all MATH. Let MATH be the generator of the cohomology ring of MATH, and let MATH, MATH, be the fundamental ... |
math/9912152 | Just observe that the construction of MATH implies that MATH, and hence MATH. The result now follows from REF . |
math/9912152 | The argument is standard. The projection MATH gives a morphism MATH which in turn it induces a ``transfer map" MATH which is easily seen to be a homotopy equivalence. The same applies one one replaces MATH by MATH in the construction. The observation preceding the proposition completes the argument. |
math/9912152 | One just needs to observe that these maps are induced by morphisms of inverse systems whose components are maps of the type MATH where MATH are generalized flag varieties and MATH is a finite group. Furthermore, all spaces are MATH-local. The result follows. |
math/9912152 | We know that MATH is a monoid morphism, hence MATH. Therefore, MATH as maps. In other words, MATH. This is equivalent to say that MATH in the language of REF. We now use REF with MATH replaced by MATH, compare REF , to conclude that MATH is homotopic to MATH. This together with REF and the definition of MATH gives the ... |
math/9912152 | Denote MATH, MATH and MATH. The multiplicative system MATH of the NAME ring MATH is sent by MATH to the multiplicative subgroup MATH of the units of MATH. Recall that MATH is isomorphic to the localization MATH; compare CITE. Therefore, there is a unique ring homomorphism MATH satisfying MATH. Since both MATH and MATH ... |
math/9912152 | The result follows from the naturality of the constructions and the case MATH. |
math/9912152 | This is just a careful diagram chase using the definitions. |
math/9912152 | Assertion a is proven in CITE, and the arguments in the proof are outlined in the proof of REF . Assertions b and c follow from NAME 's complex suspension theorem CITE and the splittings of CITE. Assertions d and e follow from CITE and a mere inspection of the definitions of the filtrations. The last assertion is prove... |
math/9912152 | Follows directly from the definitions. |
math/9912152 | The directed system can be visualized with the aid of the following diagram. MATH . Note that the bottom horizontal arrow MATH is homotopic to MATH; compare REF . A minor modification of the arguments in REF ends the proof. |
math/9912152 | In the first diagram, the left vertical face commutes by REF , and the right vertical face commutes by REF . The top and bottom faces commute because MATH can be seen as an additive endomorphism of MATH which sends MATH to MATH. The commutativity of last diagram follows from an inspection of the definitions. |
math/9912152 | Consider the following diagram. MATH . It follows from definitions that MATH and that MATH . Hence, one has MATH. Therefore, MATH where the last equality comes from REF . |
math/9912152 | This follows from the universal case MATH and the fact that the forgetful functor from MATH to MATH is a homotopy equivalence. |
math/9912152 | This is similar to the case MATH . The theorem above gives an equivalence MATH . Composing both sides with the projections MATH and using the definitions of the rational NAME classes and the NAME character, concludes the proof. |
math/9912152 | The first assertion is evident. If MATH is a path between MATH and MATH, then MATH provides the natural homotopy between MATH and MATH. |
math/9912152 | It follows from the definitions that, given MATH, one has MATH and MATH . By hypothesis one has that MATH is homotopic to MATH, and the result now follows from REF . |
math/9912153 | Let MATH be the linear isometries operad. That is, MATH is the space of linear (complex) isometric embeddings of MATH into MATH. These spaces are contractible, and the usual operad action MATH give holomorphic maps for each MATH. It is then simple to verify that this endows the holomorphic mapping space MATH with the s... |
math/9912153 | As was described in CITE, elements in MATH are in bijective correspondence to isomorphism classes of rank MATH embedded holomorphic bundles, MATH. By modifying the embedding MATH via an isomorphism between MATH and MATH, we see that MATH is homeomorphic to the space of holomorphic embeddings MATH, modulo the action of ... |
math/9912153 | Recall that MATH . But the set of path components of the NAME group completion of a topological MATH space is the NAME group completion of the discrete monoid of path components of the original MATH - space. Now as observed above the morphism space MATH is given by configurations of isomorphism classes of embedded alge... |
math/9912153 | Let MATH be the (continuous) classifying map for the topological bundle MATH over MATH, which gives the path equivalence between MATH and MATH, and denote MATH and MATH the restrictions of MATH to MATH and MATH. Since MATH is compact, the image of MATH is contained in some NAME MATH. It follows that MATH and MATH lie i... |
math/9912153 | As observed above, the set of path components of the NAME group completion of a topological monoid MATH is the NAME - group completion of the discrete monoid of path components: MATH . Therefore we have that MATH is the NAME group completion of MATH. Thus every element MATH can be written as MATH where MATH and MATH ar... |
math/9912153 | This follows from REF and the fact that short exact sequences split in MATH. |
math/9912153 | Let MATH be a flag - like smooth projective variety. Since every embeddable holomorphic bundle is represented by a holomorphic map MATH, for some NAME, then REF implies that MATH is generated by classes MATH, where MATH is such a holomorphic map. But then MATH clearly is the class represented by MATH in MATH. But this ... |
math/9912153 | We first verify that given any holomorphic bundle MATH, that MATH is a well defined element of MATH. That is, we need to show that this class is independent of the choices made in its definition. More specifically, we need to show that MATH for any appropriate choices of MATH, MATH, MATH, and MATH. We do this in two st... |
math/9912153 | Let MATH be a smooth projective variety and let MATH denote the NAME group of monoid of algebraic equivalence classes of holomorphic bundles over MATH. We show that the correspondence MATH described in the above theorem induces an isomorphism MATH . We first show that MATH is well defined. That is, we need to know if M... |
math/9912153 | By REF , MATH is a MATH - equivariant homotopy equivalence. Therefore it induces a homotopy equivalence on the fixed point sets, MATH . But these fixed point sets are MATH and MATH respectively. |
math/9912153 | Let MATH in the above corollary. MATH. |
math/9912153 | Let MATH be the wreath product MATH viewed as a subgroup of the symmetric group MATH. The obtain an identification of orbit spaces MATH . Finally, apply the above corollary when MATH. |
math/9912153 | In order to prove this theorem we begin by recalling the equivariant stable splitting theorem of a product proved in CITE. An alternate proof of this can be found in CITE. Given a space MATH with a basepoint MATH, let MATH denote the suspension spectrum of MATH. We refer the reader to CITE for a discussion of the appro... |
math/9912153 | By REF we have the following homotopy commutative diagram: MATH . Notice that all the maps in this diagram are MATH equivariant, and by the results of REF the horizontal maps are MATH -equivariant homotopy equivalences. Furthermore, by REF the maps MATH are homotopy equivalences. Now since the MATH action on MATH and o... |
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