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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 an isometric isomorphism mapping MATH onto MATH, hence MATH, being the composition of two isometric isomorphisms, is an isometric isomorphism as we needed to show. |
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 that if MATH for some MATH then MATH. Suppose first that MATH is such that MATH. Since we can rescale MATH and MATH by the same constant we may assume MATH. Let MATH . Then MATH so MATH and MATH . An application of REF completes the proof. |
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 CITE. Let MATH . Then MATH, and we now calculate the NAME transform of the two individual operators making up MATH. First, MATH transforms into MATH, since MATH as claimed. Note that all summation indices range over MATH. We have MATH-convergent summations, that is, relative to the respective MATH-norms, and the exchange of the summations is justified by the norm-isomorphism property of the NAME transform. We now turn to the operator MATH given by MATH, where MATH is the adjoint of MATH. Since MATH is a norm-isomorphism, we have MATH where MATH refers to the respective identity operators. We claim that MATH . The proof is the following calculation, for MATH: MATH which proves the stated REF for MATH. We now calculate MATH where the last step simply represents the NAME series of the final function. The formula MATH is basic in later proofs. Combining the three REF , and REF , we arrive at MATH . This is needed in the calculation of the right-hand side in REF from the lemma as follows: MATH . But if MATH in the summation, then the factor MATH. If MATH, then MATH and MATH. If MATH, then MATH. Continuing the calculation, we get MATH which is the final conclusion of the lemma. |
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 for MATH is trivial to check, but it is not clear which conditions on the MATH's make the right-hand side finite. REF for MATH is useful later since we will be able to check finiteness of the factor MATH. In fact in many cases, we will end up with MATH a constant function on MATH. Checking REF goes as follows: Let MATH, and suppose MATH. Then MATH which is exactly 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 integral is MATH . But the choice of MATH makes it zero: MATH since MATH. We have proved that if MATH, MATH maps MATH onto MATH, and an iteration of the same argument (induction) yields MATH as claimed. We also saw that MATH is isometric if and only if MATH. |
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 holds, we saw in REF that then MATH, and the proof is completed. |
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 the term MATH if MATH, and MATH if MATH. We get MATH which is the identity REF : As in REF , let MATH and MATH be given. Then MATH which is precisely the second identity REF . Note that we obtain the identities in MATH, but since MATH is an isomorphism, MATH, we automatically get the same identities in MATH where the translation to MATH is made via the dictionary of REF . |
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 assertion is this. MATH. We prove the claim by showing that a vector MATH which is orthogonal to all three subspaces MATH, MATH, and MATH, must be zero. The three subspaces are pairwise mutually orthogonal. This is immediate from the definitions except for the last pair. Let MATH and MATH. Then MATH where we used REF , in that order. In the last step, we used MATH. The condition that MATH is orthogonal to all three subspaces amounts to: MATH, MATH, and MATH. Since MATH, we have MATH, so MATH, MATH, MATH. Since MATH, we get MATH. Indeed there is a sequence MATH such that MATH. So if MATH, then MATH since MATH. Hence MATH in the decomposition of MATH: MATH, MATH. But MATH, and we conclude that MATH. MATH is invariant under MATH, MATH. We first show that, if MATH and MATH, then MATH. Using REF , we do this by showing that MATH. But MATH where we used first REF and then REF . Then MATH where we could use REF or, alternatively, REF . This means that MATH has an operator block matrix relative to the orthogonal decomposition MATH of the form MATH with MATH, and the diagonal operators MATH and MATH being endomorphisms of the respective spaces MATH and MATH. Since MATH, the adjoint MATH must be of the same form, and that forces MATH, that is, MATH, and each of the spaces MATH and MATH is then invariant under MATH. In particular, MATH is invariant, which is the claim. Our next assertion merges REF into the following induction: For each MATH, we have the decomposition MATH where the terms in the decomposition are mutually pairwise orthogonal, and further each of the spaces MATH is invariant under MATH, MATH, MATH, where we set MATH. This is a simple induction which is based on REF - REF , and it is left to the reader. Alternatively, we can prove it by using the earlier claims on the operators MATH. We have the decomposition REF with the two REF - REF . For each MATH, let MATH denote the projection onto MATH. Since MATH, this is a decreasing sequence of projections in MATH, By NAME space theory (see, for example, CITE), it has a limit MATH, that is, MATH for all MATH, and MATH is the orthogonal projection onto MATH. Recall that MATH, by definition! In fact, MATH . But REF states that MATH is the projection onto MATH, where we write MATH for the identity operator in MATH. Now MATH is an increasing family of projections, and its limit MATH is the projection onto MATH. Let MATH denote this space. Vectors MATH in MATH are characterized by MATH, or equivalently MATH, and each MATH has the unique representation MATH, MATH, MATH, MATH. Since each of the spaces MATH is invariant under MATH, MATH, by REF - REF , it follows that MATH is also MATH-invariant. Since MATH, it follows that MATH is also MATH-invariant. The proof of REF will be given after the next three corollaries. For each MATH, the space MATH is invariant under MATH, MATH. We showed in REF that MATH and that the right-hand side has the MATH-invariance. Since MATH, we conclude that MATH is also MATH-invariant. Let MATH be as in the statement of REF , that is, MATH is a MATH-isometry relative to some NAME operator MATH and representation MATH. We say that a closed subspace MATH is double invariant if MATH is invariant under both MATH and MATH. It is then immediate from the definition of MATH that MATH is double invariant if and only if both MATH and MATH REF are invariant under MATH, that is, MATH, and MATH. Let MATH be as described, and let MATH REF and MATH REF be as in REF . Then both MATH and MATH are double invariant under MATH. From the comment before the statement of the Corollary, it is enough to show that both MATH and MATH are invariant under MATH. Recall from REF that MATH, and MATH. But it is clear from the definition of MATH that MATH. Since arbitrary vectors MATH in MATH may be represented as MATH, MATH, MATH, it follows that MATH. If MATH is encoded with the sequence MATH, MATH, then MATH, that is, MATH and the MATH-invariance for MATH follows from this. By the initial argument we conclude that both MATH and MATH are double invariant. The advantage of the MATH representation of MATH is that MATH takes an especially simple form: If vectors in MATH are represented in the form MATH,MATH, MATH, then the action of MATH on MATH is MATH . The proof follows from the following calculation: MATH, and MATH. Let the representation MATH of MATH in MATH be given as in REF . Let MATH, MATH be as stated at the outset, that is, MATH. On the vectors described in REF , define the NAME norm, and corresponding completion, by MATH and define MATH as in REF . A simple computation, using the corresponding inner product MATH and REF then yields an adjoint operator MATH which turns out to be REF . It is now a simple matter to verify that MATH is the desired sub-isometry. |
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, MATH, we conclude that MATH maps the norm closure of MATH into itself. This norm-closure is MATH by the NAME - NAME theorem. Since MATH leaves MATH invariant, it dualizes to a map on the measures MATH, and we claim that MATH where MATH denotes the NAME point measure at MATH. Indeed, let MATH; then MATH since MATH and MATH. (This is the low-pass property.) It follows that MATH when MATH is the eigenfunction in REF . Suppose we did have a solution MATH to REF as stated. Let MATH, and set MATH. Then MATH. Pick a scaling function MATH for the cascade operator MATH corresponding to MATH, that is, MATH, and MATH. Since MATH is one-dimensional, and MATH by REF , we conclude that MATH. REF then implies that MATH, so MATH. We then have MATH using the argument from the proof of REF . The left-hand side of REF is MATH, which is not convergent when MATH, MATH, MATH, and MATH in MATH. On the right-hand side in REF , some more analysis is needed. Our next claim is that MATH whenever MATH, and the limit is in MATH, or in MATH after a translation of the result via the NAME transform. Using REF , we get MATH . Let MATH. Since MATH, we get for the difference (picking MATH such that MATH) MATH . The MATH-norm is that of MATH; we split up the integral MATH over MATH into two regions, one MATH, and the other the union of the intervals MATH and MATH. We pick MATH such that MATH when MATH. After estimating the two separate contributions, and using REF , the desired result follows. When it is applied to the right-hand side in REF , we get MATH in the MATH-norm. But MATH. Comparing the two results for the limits of the left- and right-hand sides of REF , we arrive at the desired contradiction, and conclude that the eigenvalue REF does not have eigenvectors as stated. |
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 from REF and CITE that MATH . In view of REF , this proves REF . The application of the NAME - NAME theorem (see CITE and line REF above) requires that MATH on MATH, which clearly holds for the present MATH. The argument for REF is similar: MATH . The last step is based on the formula REF MATH . It remains to prove REF . We have MATH and we check that the theorem from CITE applies to that term as well: by the above argument, we have (for MATH) MATH . By a standard summation formula (see, for example, CITE), we have MATH . Hence MATH which is continuous and equals MATH at MATH, so REF also follows from an application of CITE. |
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 sequence MATH (see REF above), and note that MATH . Since MATH, we know from REF and CITE that MATH for MATH in the complement of any neighborhood in MATH of MATH. But the MATH term in REF is MATH . If MATH were continuous, the theorem from CITE would simply give MATH and the equivalence of REF would be clear from this and REF . If REF does not hold, we still have MATH by REF . But MATH is continuous by the conditions on MATH, and MATH in MATH by CITE and REF . Suppose now that REF is given: the argument from above then shows that REF holds in the sense of MATH. |
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 show MATH . We have MATH . This shows REF , and REF is proved. |
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 operator MATH, we get MATH. Now apply this to MATH, and use MATH. Iteration yields MATH . |
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 characteristic polynomial is MATH too. Let us view MATH as the space of sequences MATH. Since MATH for all MATH, we have MATH . But as MATH, MATH is the unique eigenvector of MATH corresponding to eigenvalue MATH, and as the functional MATH is preserved by MATH, it follows from REF that MATH for all finite sequences MATH. Thus, by REF , MATH . But the two assumptions on MATH imply the so-called NAME - Fix condition CITE MATH for all MATH, and hence by the above, MATH . In particular this means that MATH . But by REF , MATH . Since MATH by orthonormality, we conclude that MATH and in particular, MATH for all MATH. Also MATH so, finally, MATH . CASE: We now assume cascade convergence in the sense REF , that is, MATH for the initial vectors MATH which are specified in REF . The object is to derive from this the spectral picture for MATH as specified in REF , and MATH will be identified with its restriction to MATH as mentioned. Of course MATH, and MATH is also, by REF , an operator mapping MATH into itself. Its adjoint on the dual space of measures MATH is given by MATH, MATH. The NAME point-measure MATH, given by MATH, is invariant by REF , that is, MATH. Consider the eigenvalue problem: MATH . Then MATH, so MATH if MATH. We assume this, and since MATH, the discussion may be restricted to MATH. We claim that, if MATH, MATH, then MATH, so we cannot have nontrivial peripheral spectrum. By REF , MATH, where MATH is any initial vector with the stated conditions, for example, MATH. Using REF , we also get MATH whenever MATH. For example, take MATH to have this satisfied. Then MATH, that is, a divergent sequence if MATH and MATH, supposing MATH. Using REF , we will show that MATH where the last convergence is in the finite-dimensional subspace of MATH and thus in any norm. This will contradict the divergence of MATH. REF can be verified in two ways: since MATH, MATH and MATH all have support inside a common compact set, the convergence is immediate from the finite sum REF , and REF . Alternatively one can use REF , and MATH. In checking the conditions in REF , we note that MATH, in MATH, so we must verify that MATH is bounded in MATH. But the function inside MATH equals MATH . This contradiction completes the first part of the proof of REF . It remains to show that REF implies that MATH has multiplicity one in the spectrum of MATH, where again MATH is identified with its restriction to MATH. Since MATH, we need only exclude that the multiplicity is MATH or more. But MATH for all MATH by REF and MATH and it follows from REF that MATH. By REF it follows that MATH has multiplicity MATH. |
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 MATH is defined as MATH, where MATH is the set in the square that projects orthogonally onto MATH. Since MATH is locally doubling, then MATH is doubling, therefore it has no atoms. This implies that MATH. |
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. Consider now the ball MATH of center MATH and radius MATH and the ball MATH of center MATH and radius MATH. The constant MATH is chosen in such a way that MATH. This is always possible, since MATH and MATH are equivalent whenever MATH is close to MATH. Therefore MATH, the radius of MATH is smaller than MATH and we may apply REF . We estimate MATH by MATH and the integral is bounded by a constant times MATH . In this last quotient we may again apply REF and compare the quotient of measures by the quotient of radius (if we think of the numerator MATH) and we obtain MATH provided that we choose a MATH large enough such that MATH. |
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 to obtain unitary mass rectangles. Recall that there is some MATH such that MATH for all MATH. Let us partition the plane into parallel strips of width MATH. Then, we slice each strip in rectangles of mass a natural number (the sides of the rectangle have no mass because of REF ). The length of any piece will be between MATH and MATH. Since any square of size MATH has mass at least MATH, it is possible to slice the strip in such a way that the resulting rectangles have a ratio between the sides bounded by MATH. We have no control on the upper bound of the mass of these rectangles; we only know that it is a natural number. In the case of the domain being the disk, one has to replace the strips by annuli centered at the origin of width between MATH and MATH and in such a way that they all have mass which is a natural number. Now we split each annulus in rectangles of integer mass. The length of the sides will be between MATH and MATH, except possibly the last one which closes the circle and which has to been taken of length-side comprised between MATH and MATH. In any case, the resulting rectangles have a ratio between the lengths of the sides bounded by MATH and again without control on the upper bound of the mass. From now on the procedure in the disk and in the plane will be the same. We will break each rectangle in two. All the resulting rectangles will still have integer mass and the ratio of the sides will always remain bounded by MATH. We will proceed to the bisection of each rectangle until the mass is smaller than the doubling constant of the measure. The bisection is done as follows: consider a rectangle centered on the original one with mass one as the filled rectangle in REF . It is important that we build it over the longer of the two sides of the larger rectangle just as in the picture. Its side MATH cannot be larger than one third of the longest side of the original rectangle, because if this was so, the original rectangle would have a mass smaller than the doubling constant, and so we would not need to bisection it. There is a straight line in the filled rectangle (the dashed line in the picture) that splits the original rectangle into two rectangles, each of them of integer mass, and moreover the two resulting rectangles have the ratio of the sides still bounded by MATH, as we claimed. This far, the rectangles are not very deformed and all have a mass between MATH and MATH. In order to obtain rectangles of mass MATH, we split each of them in rectangles of mass one by cutting along the direction of the longest side. The local doubling condition ensures that all of them will be essentially of the same proportion (we use REF ), and since at most we are dividing each rectangle in MATH parts, the resulting rectangles have a bounded ratio of side-lengths as desired. |
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 are the points that will be used in the construction of the multiplier; they will be in fact the zeros of it. Note that all the points MATH appear with a multiplicity MATH since there are MATH points with equal weights. For later use, it is convenient to have an alternative set of zeros MATH at our disposal which are separated from the original ones and still have the same moments. This is easily done. It can be checked immediately that MATH, for any MATH and any polynomial of degree MATH. Thus, we could take as an alternative set MATH, MATH, MATH, where MATH is some number so that all MATH are outside MATH and inside MATH. |
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. Now for each family MATH we can construct a multiplier MATH in such a way that it has no zeros in any of the rectangles of the family MATH and not even in their immediate neighbors. The way to proceed to construct MATH is the following: For any rectangle MATH that is neither from the family MATH nor one of its immediate neighbors we take the set of points MATH given by REF . For the rectangles MATH from the family or its adjacent rectangles we use the alternative set of points MATH also defined in REF . We build as before a multiplier MATH with zeros at the selected points. It has the right growth and the additional property that it has no zeros in the rectangles from the family MATH and its adjacents. This is clear because we can choose a constant MATH in REF in such a way that the points MATH are neither in MATH nor in its immediate neighbors. Moreover they are not so far apart from MATH that they reach another rectangle from the family (this can be prevented by choosing a very large MATH in the splitting of the rectangles into families). Thus the exceptional set for MATH does not include any rectangle from the family MATH. |
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 is the rest. It follows from the definition that MATH, therefore MATH is a locally doubling measure. It is also true that MATH is locally doubling because MATH does not change abruptly in neighboring squares and moreover MATH. We will regularize the measure MATH by taking the convolution (the invariant convolution when MATH is a disk) of it with the normalized characteristic function of a very large disk: MATH. The measure MATH in the plane satisfies MATH when MATH, it satisfies MATH. It is clear from their definition that MATH in MATH and MATH in the disk. The same is true for MATH. We take integral operators MATH and MATH that solve the NAME equation MATH. The operator may be defined as MATH . In the case of MATH we choose MATH . This makes the integrals defining MATH and MATH convergent. In the case of the disk MATH . CITE and CITE estimated this kernel by: MATH therefore the integrals defining MATH and MATH are convergent. We take as MATH. The Laplacian of MATH is MATH which has the desired properties. Moreover MATH. This difference is bounded by MATH . This integral is bounded by a constant times MATH, whenever MATH a locally doubling measure. This is REF. The disk MATH is covered by a bounded number of cubes MATH, therefore the difference between MATH and MATH is bounded as claimed. |
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. Choose constants MATH and MATH with MATH. The handle MATH will be a subset of MATH, with MATH, MATH and MATH. First we calculate the contact forms induced by MATH on MATH and MATH. Both are restrictions of the form MATH. On MATH, natural polar coordinates consistent with our orientation convention are MATH. With respect to these coordinates we get MATH. Natural polar coordinates on MATH are MATH and with respect to these coordinates we get MATH. Now, given any small positive MATH, we will show how to modify the contact structure MATH inside a neighborhood MATH of MATH to get a new contact structure MATH which extends across MATH and still satisfies the property that MATH is nondegenerate. Let MATH where MATH and MATH goes smoothly to MATH as MATH goes to MATH, has positive derivative and is equal to MATH on MATH. This agrees with MATH outside MATH because MATH, is contact because MATH and satisfies the nondegeneracy condition because MATH. Intuitively, MATH twists too far as we move in towards MATH to extend across MATH, so we back off to MATH and then twist more slowly so that MATH does extend. Forward flow along MATH for time MATH starting at a point MATH gives a map MATH from a subset of MATH into MATH defined by the following equations: MATH . Letting MATH and MATH, we see that forward flow for time MATH defines a diffeomorphism MATH. Note that MATH and that MATH. Given any three radii MATH we can construct a symplectic handle MATH as follows: Choose a smooth function MATH which is equal to MATH on MATH, is decreasing on MATH and is equal to MATH on MATH. Then let MATH be the union of all the forward flow lines starting at points MATH flowing for time less than or equal to MATH, together with MATH. The attaching boundary MATH is MATH while the free boundary MATH is the image under MATH of the graph of MATH in MATH together with MATH. This construction is illustrated in REF . In the figure MATH, MATH and MATH are rather far apart for the sake of clarity, but in general one would carry out this construction with these radii only slightly larger than MATH. A few forward flow lines for MATH are shown, starting on MATH. The construction of MATH on MATH gives a positive contact structure MATH on MATH by letting MATH on MATH and letting MATH elsewhere. The nondegeneracy condition on MATH and the fact that MATH is positively transverse to MATH where defined gives us that MATH is nondegenerate. Furthermore MATH agrees with MATH on MATH, so that if we succeed in attaching MATH to MATH via a contactomorphic embedding of MATH into MATH, the contact structures will patch together to give a contact structure MATH on the boundary MATH of the new symplectic manifold MATH with MATH nondegenerate. Now we show how to find this embedding and attach MATH. Given the transverse link MATH which is fat with respect to a framing MATH, consider a component MATH of MATH with neighborhood MATH with almost normal coordinates MATH which go out as far as MATH. Notice that we can choose these coordinates so that MATH. This means that the contact structure MATH near MATH is spanned by MATH and MATH and that MATH. Choose some MATH such that MATH. Now consider the coordinates MATH on MATH; these are almost normal coordinates and we could explicitly reparametrize MATH to make them normal. However it is sufficient to notice that MATH is spanned by MATH and MATH and that this implies that, after reparametrizing MATH, the coordinates would go out as far as the handle framing MATH. This is because we construct MATH with MATH. This means that if we construct MATH with MATH small enough we can guarantee a contactomorphism from MATH to MATH, taking the handle framing to 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 because MATH is Legendrian. Note that in this model MATH is the ``zero-framing" coming from the product structure on MATH. We will measure framings relative to this product framing, so that MATH. On MATH, we have MATH, where MATH and MATH. The MATH-form MATH is positive on MATH. We will construct a symplectomorphism MATH from the disk MATH of radius MATH in MATH with the standard symplectic form MATH onto a region MATH with the symplectic form MATH. Then on MATH let MATH and note that MATH. Thus REF gives a contactomorphism MATH from MATH onto MATH taking the zero framing to the zero framing. Finally note that MATH is fat with respect to the framing MATH. We construct MATH directly. Choose two positive constants MATH and MATH, define MATH by: MATH and verify that MATH. The map is only defined when MATH, but as long as we choose MATH, MATH will be defined on MATH. By choosing MATH small enough we can guarantee that MATH. |
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 MATH, on MATH and MATH respectively, are given by the following formulae: CASE: MATH CASE: MATH . We first show the existence and uniqueness of MATH, then show that MATH as described in these formulae satisfy the conditions of the lemma, and then show uniqueness of MATH. There exists a unique MATH such that MATH because contraction with MATH gives a linear isomorphism from MATH to MATH (this depends on working in dimension MATH), and MATH is constructed to be in this latter subspace. But MATH is also in MATH because MATH, and thus MATH. On MATH, letting MATH and MATH, we need to show that MATH . First note that MATH. REF is quick: MATH. To show REF , note that MATH and that MATH . Next we will prove that MATH is the unique vector field on MATH satisfying these equations. Suppose MATH and MATH are two solutions. Let MATH; we will prove that MATH and thus conclude that MATH. REF implies that MATH, REF implies that MATH and REF implies that MATH everywhere. Thus MATH is invariant in the MATH direction and vanishes when MATH, so MATH everywhere. On MATH the argument is a mirror image of the argument for MATH. |
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 MATH induce the same contact forms on MATH and are both symplectic dilations, we can conclude that MATH. By the uniqueness in REF we also know that MATH, so that MATH preserves all the relevant structure. Let MATH and MATH. Now choose two functions MATH such that MATH but MATH everywhere, let MATH and let MATH. Finally let MATH. If we choose MATH small enough we can guarantee that MATH is NAME. Since MATH and MATH, we know that the symplectic forms and the dilation-contraction pairs patch together to define a symplectic form MATH on MATH with a dilation-contraction pair MATH transversely covering MATH inducing MATH. (See REF .) For the uniqueness result, we need to construct a bundle isomorphism MATH covering the identity and preserving the symplectic forms, and then we can apply NAME 's theorem. To do this we construct a pair of vector fields MATH along MATH in MATH with open domains MATH covering MATH, both positively transverse to MATH, such that MATH and such that MATH. Then there exists a unique MATH sending MATH to MATH by REF . To see the existence of the pair MATH, first extend MATH to a maximal rank REF-form on MATH to get MATH along MATH such that MATH. Then by REF there exists a unique MATH along MATH such that MATH and MATH. Let MATH, a maximal rank REF-form on MATH which extends MATH. Then there exists a maximal rank extension MATH of MATH to MATH which agrees with MATH outside a closed set MATH inside MATH containing MATH in its interior. Let MATH be the corresponding vector field such that MATH and let MATH. |
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 . Forward flow along MATH gives a map MATH from some subset of MATH into MATH defined by: MATH . Letting MATH and MATH (which is positive because MATH), we see that forward flow for time MATH defines a diffeomorphism MATH. We can make MATH arbitrarily small by choosing MATH small enough. Now for any choice of radii MATH we can build MATH exactly as in REF, using a function MATH which is MATH on MATH, decreasing on MATH and MATH on MATH. By construction MATH is transverse to MATH. The only part of MATH which could fail to be transverse to MATH is MATH. Using REF we can state conditions for MATH to be transverse to MATH. Using the notation from the proof of REF we get: MATH . Thus MATH needs to satisfy MATH for all MATH. Since we will have MATH, MATH and MATH, this will work as long as MATH . Therefore if we build MATH with MATH we can guarantee that MATH is transverse to both MATH and MATH. We see that MATH, MATH and MATH can be chosen arbitrarily small, as long as we choose MATH small enough, so that we can arrange for MATH to be an arbitrarily small neighborhood of MATH. |
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 because MATH and MATH). Note that we then have MATH (because MATH and MATH). We will construct MATH so that MATH on MATH for some positive MATH; this together with MATH gives a contact pair on MATH which is prepared for surgery on MATH. Notice that this in fact determines MATH on MATH because we must have MATH for MATH. We should check that MATH so defined is in fact positive: MATH as long as MATH, which will hold for small enough MATH as long as MATH, which is how we chose MATH. In other words, if we choose MATH small enough we can guarantee that MATH on MATH. Next we check that, with MATH thus defined on MATH, MATH is transverse to MATH. Calculating and applying REF we see that transversality will hold if MATH which, for our given MATH, becomes: MATH . This holds for MATH because MATH and MATH. Now we should check that we can extend MATH to MATH to have support inside MATH for some MATH, in such a way that MATH remains transverse to MATH. On MATH the transversality REF above will be satisfied because MATH will be identically MATH and MATH. For MATH, if we choose MATH small enough we can replace REF by the following simpler condition: MATH . Using the facts that MATH and MATH and MATH are positive, it is easy to extend MATH to MATH maintaining this condition if MATH. If MATH then, after perhaps making MATH smaller still, we extend MATH to MATH, with MATH near MATH, in such a way that MATH for all MATH and that MATH for all MATH, for some small MATH. This is enough to conclude that conditon REF is met fir 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 conditions of REF . Recall the notation in the definition of ``nicely fibered". The transverse contact vector field is MATH and the fibration is MATH. With MATH as given, we have some induced contact form MATH on MATH with MATH such that MATH satisfy the definition of ``nicely fibered". Consider a new contact form MATH defined by MATH and MATH. Notice that we then have MATH. The new contact form MATH has the same kernel as MATH, so MATH for some function MATH. By replacing MATH with MATH if necessary we can arrange that MATH. Also note that we can replace MATH with MATH for any constant MATH without changing the ``nicely fibered" condition, and thus we can arrange that MATH (using the compactness of MATH). This means that we can enlarge MATH so as to arrange that the induced contact form on MATH is actually MATH, using the symplectification MATH. Now choose a constant MATH so that MATH on MATH (this depends on the compactness of MATH and on the fact that MATH and MATH are invariant on a neighborhood of MATH so that MATH is constant near MATH). Let MATH and let MATH. Then MATH is a contact pair on MATH with MATH, so using REF we may now regard MATH as partially concave and partially convex with induced contact pair MATH. Of course MATH is still defined on all of MATH while MATH is only defined on MATH, and so for now MATH contains no new information. However, when we show that MATH is well-behaved near MATH we will be able to attach the handles from the previous section, after which MATH will extend across the new boundary and we will be able to forget about MATH to conclude that the new boundary is concave. Thus MATH contains the seed of the concavity which we will achieve after attaching handles. We see that MATH is well-behaved near MATH using the ``nicely fibered" condition again. Near each component we know that there are normal coordinates MATH such that MATH, MATH and such that MATH and MATH are invariant under the flows of MATH, MATH and MATH. This immediately establishes that MATH for some constants MATH and MATH, and that MATH for some constants MATH and MATH. Now we need to arrange that MATH and MATH are positive; this will follow from the orientation condition on the characteristic foliation on the fibers. Looking at our orientation convention, the fact that the foliation points radially inwards means that MATH. Recall that we can take MATH as our model, so we get that MATH for small MATH. Thus either MATH and MATH are both positive, or if they are both negative we can replace the coordinate system with MATH to get MATH and MATH both positive. Now if we arrange that each coordinate system MATH for each component of MATH realizes the desired framing MATH of MATH, we see that the condition that MATH is positive with respect to the fibration MATH means exactly that, near each component, MATH and since MATH this means that MATH. Now we still may not have that MATH. To arrange this we may need to again replace MATH by MATH for some constant MATH (again using compactness of MATH), which will replace MATH and MATH with MATH and MATH. Now REF shows how to enlarge MATH and attach handles. After enlarging MATH, we have MATH partially convex and partially concave with induced contact pair MATH, with MATH and MATH. After attaching the handles MATH is partially convex and partially concave with induced contact pair MATH with domains MATH, with MATH and MATH (where MATH is the union of the ascending spheres). This means that we can ignore MATH and realize that in fact MATH is concave with induced negative contact form MATH, and the characterization of MATH in the addendum follows. |
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 page REF with MATH (a NAME rule). |
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 follows from induction and the equation MATH . |
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 MATH where MATH denotes dimension in the NAME lattice and MATH is the number of paths in the NAME lattice from MATH to MATH. This follows from the observation that the product of the mulitplicities MATH along a path in the NAME diagram of this example depends only on the endpoints of the path. The fact that the second condition of a boundary is met follows from a generating function argument showing that the MATH separate points of the boundary and the NAME theorem. The third condition of a boundary amounts to exactly the same condition as for the NAME lattice, and thus holds. |
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 infinitely many copies of each element in MATH; compare CITE. This argument implies the following lemma. Let MATH be an abelian topological monoid, equipped with a continuous monoid augmentation MATH onto the additive monoid of the non-negative integers, and let MATH be its group completion; compare REF. Define MATH and chose MATH. Then the colimit MATH of the system MATH given by translation by MATH is also an abelian topological monoid. Furthermore, MATH induces a continuous monoid morphism MATH and MATH is homotopy equivalent to the group-completion MATH. In our case MATH, and assertion c follows from the above considerations. To prove assertion d, we shall use the following result. Let MATH be a finite group acting on a smooth, connected projective variety MATH, and let MATH be the quotient map. Then MATH induces an isomorphism MATH, where MATH denote the invariants of the cohomology of MATH under the action of MATH. Furthermore, if MATH is simply connected and the fixed point set MATH is non-empty, then MATH is simply connected. The first part of the theorem is well-known and follows from standard transfer arguments. Consider a fixed point MATH and denote MATH. It follows from CITE that one can find an equivariant triangulation of MATH in which MATH is a vertex, and such the quotient MATH becomes a simplicial map for the quotient triangulation of MATH. In particular, given any MATH-simplex MATH, there is a simplex MATH such that the restricion of MATH to MATH is a homeomorphism onto MATH. Standard arguments show that any loop in MATH based on MATH is homotopic to a simplicial loop MATH. Therefore, there is a partition MATH of MATH such that MATH maps MATH onto a MATH-simplex MATH of the triangulation in such a way that MATH becomes a homeomorphism from MATH onto the open simplex MATH. For each MATH let MATH be a lift of MATH, and let MATH be a reparametrization of its characteristic map by the interval MATH. Since MATH one can find MATH such that MATH, and hence the path MATH defined as MATH is a lifting of MATH. One then proceeds inductively to produce a lifting MATH of MATH. Since MATH and MATH because MATH is a fixed point of the MATH-action, it follows that MATH is a loop based on MATH. The result follows. Since the action of MATH on MATH is induced by the natural representation MATH of the general linear group on MATH, one concludes that MATH acts trivially on the cohomology of MATH. The previous lemma then implies that MATH induces an isomorphism between the rational cohomology of two simply-connected spaces. Hence, MATH is a rational homotopy equivalence. It follows that MATH is also a rational homotopy equivalence. Since MATH is compatible with the direct sum operation on MATH, one concludes that it is a map of infinite loop spaces, once we give MATH the infinite loop space structure coming from the abelian topological monoid structure. We leave the details to the reader. |
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 also induces an isomorphism MATH, whenever MATH; compare CITE. Using the isomorphisms MATH exhibited in the proof of REF , together with MATH, one concludes that MATH induces an isomorphism in the MATH-th rational cohomology groups, for MATH. To conclude the proof, first note that a simple inverse limit argument shows that MATH induces an isomorphism in rational cohomology. Then, REF implies that both MATH and MATH are simply-connected, and hence that MATH is a rational homotopy equivalence. |
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 class of a coordinate MATH-plane, the NAME dual to MATH. Define MATH as MATH, where MATH denotes the MATH-th projection. Given a partition MATH where MATH, let MATH be the associated generator of MATH, dual to MATH. Given integers MATH, one has a commutative diagram: MATH where MATH following the notation of REF , and where MATH denotes the natural inclusion, and MATH is the projection. Consider a partition MATH. If some MATH is strictly less than MATH, then MATH, where MATH. In this case, if MATH is the canonical class of MATH and MATH denotes the NAME pairing, then MATH where the last equality follows from the fact that MATH. On the other hand, it follows from the construction of the splittings MATH that MATH and hence MATH . By definition, MATH . It follows from REF that MATH where MATH is the universal quotient MATH-plane bundle over MATH. Combining REF , one gets MATH . Write MATH . Chasing REF one obtains MATH . This concludes the proof. |
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 equalities MATH of elements in MATH. The result now follows from REF . |
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 are MATH-local abelian topological monoids, they are a product of rational NAME spaces and the homomorphism MATH determines a unique homotopy class of maps satisfying the desired property. |
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 proven in CITE. |
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 structure of a MATH - operad space. Since this is a MATH operad in the sense of CITE, this implies that the group completion, MATH has the structure of an infinite loop space. |
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 the holomorphic automorphism group, MATH. The space of holomorphic embeddings of MATH in an infinite dimensional trivial bundle is easily seen to be contractible CITE, and the action of MATH is clearly free, with local sections. Again, see CITE for details. The lemma follows. |
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 algebraic bundles, MATH. Thus MATH is the set of path equivalence classes of such pairs; that is, the set of path equivalence classes of embedded algebraic bundles of rank MATH. We may therefore conclude that MATH is the NAME group completion of the monoid of path equivalence classes of embeddable algebraic bundles. |
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 in the same path component of MATH. Since MATH is a disjoint union of constructible subsets of the NAME monoid MATH, then MATH and MATH lie in the same connected component of a constructible subset in some projective space. Using the fact that any two points in an irreducible algebraic variety MATH lie in some irreducible algebraic curve MATH (see CITE), one concludes that any two points in a connected constructible subset of projective space lie in a connected constructible curve. Let MATH be a connected constructible curve contained in MATH and containing MATH and MATH. Under the canonical identification MATH, one identifies the inclusion MATH with an algebraic morphism MATH . This map classifies the desired embedded bundle MATH over MATH. The converse is clear. |
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 are holomorphic maps from MATH to some NAME. By the above observations MATH . |
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 means that the map MATH is given by the composition MATH . But since MATH is an isomorphism this means MATH is injective. But we already saw in REF that MATH is surjective. Thus MATH, and therefore MATH, are isomorphisms. Clearly from their descriptions, MATH and MATH preserve tensor products, and hence are ring isomorphisms. |
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 steps. REF k = q. In this case it suffices to show that MATH and MATH lie in the same path component of the morphism space MATH, where MATH is the rank of MATH. Now using the notation of REF we see that these two elements both lie in MATH, which, as proved in REF there is path connected. CASE: General Case: Suppose without loss of generality that MATH. Then clearly the classes MATH and MATH represent the same element of MATH. But this latter class is MATH which we know by REF represents the same MATH - theory class as MATH. Thus MATH is a well defined class in MATH. Clearly the above arguments also verify that MATH only depends on the isomorphism type of MATH. |
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 MATH and MATH are algebraically equivalent, then MATH. So let MATH be an algebraic equivalence. Since MATH is a curve in projective space, we can find a projective embedding of the product, MATH . Now for sufficiently large MATH, MATH is embeddable, and given an embedding MATH, the pair MATH defines an algebraic equivalence between the embedded bundles MATH and MATH, where the MATH are the appropriate restrictions of the embedding MATH. Thus MATH . Thus MATH . But these classes are MATH and MATH. Thus MATH is well defined. Notice also that MATH is surjective. This is because, as was seen in the proof of the last theorem, if MATH is and embedded holomorphic bundle, then MATH. The essential point here being that the choice of the embedding MATH does not affect the holomorphic MATH - theory class, since the space of such choices is connected. Finally notice that MATH is injective. This is follows from two the two facts: CASE: The classes MATH are units in the ring structure of MATH, and CASE: If bundles of the form MATH and MATH are algebraically equivalent then the bundles MATH and MATH are algebraically equivalent. |
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 appropriate category of equivariant spectra. There is a natural MATH equivariant homotopy equivalence of suspension spectra MATH . As a corollary of this splitting theorem we get the following splitting of topological MATH - theory spaces. There is a MATH -equivariant homotopy equivalence of topological MATH - theory spaces, MATH . Given to spectra MATH and MATH, let MATH be the spectrum consisting of spectrum maps from MATH to MATH. We again refer the reader to CITE for a discussion of the appropriate category of spectra. If MATH is the zero space functor from spectra to infinite loop spaces, then MATH where MATH refers to the space of infinite loop maps. Let MATH denote the connective topological MATH - theory spectrum, whose zero space is MATH. Now REF yields a MATH equivariant homotopy equivalence of the mapping spectra, MATH-and therefore of infinite loop mapping spaces, MATH . But since MATH is, in an appropriate sense, the free infinite loop space generated by a space MATH, then given any other infinite loop space MATH, the space of infinite loop maps, MATH is equal to the space of (ordinary) maps MATH. Thus we have a MATH - equivariant homotopy equivalence of mapping spaces, MATH . Notice that REF is the holomorphic version of REF . In order to prove this result, we need to develop a holomorphic version of the arguments used in proving REF . For this we consider the notion of ``holomorphic stable homotopy equivalence", as follows. Suppose that MATH is a smooth projective variety (or union of varieties) and MATH is a spectrum whose zero space is a smooth projective variety (or a union of such), define MATH to be the subspace of MATH consisting of those infinite loop maps MATH so that the composition MATH is holomorphic. Notice, for example, that MATH. Now suppose MATH and MATH are both smooth projective varieties, (or unions of such). A map of suspension spectra, MATH is called a holomorphic stable homotopy equivalence, if the following two conditions are satisfied. CASE: MATH is a homotopy equivalence of spectra. CASE: If MATH is any spectrum whose zero space is a smooth projective variety (or a union of such), then the induced map on mapping spectra, MATH restricts to a map MATH which is a homotopy equivalence. With this notion we can complete the proof of REF . This requires a proof of REF that will respect holomorphic stable homotopy equivalences. The version of this theorem given in CITE will do this. We now recall that proof and refer to CITE for details. Let MATH be a connected space with basepoint MATH. Let MATH denote MATH with a disjoint basepoint, and let MATH denote the wedge of MATH with the two point space MATH. Topologically MATH and MATH are the same spaces, but their basepoints are in different connected components. However their suspension spectra MATH and MATH are stably homotopy equivalent spectra with units (that is, via a stable homotopy equivalence MATH that respects the obvious unit maps MATH and MATH.) Moreover it is clear that if MATH is a smooth projective variety then MATH and MATH are holomorphically stably homotopy equivalent in the above sense. Now by taking smash products MATH -times of this equivalence, we get a MATH - equivariant holomorphic stable homotopy equivalence, MATH . Now notice that the MATH - fold smash product MATH is naturally (and MATH equivariantly) homeomorphic to the cartesian product MATH. Notice also that the MATH fold iterated smash product of MATH is MATH - equivariantly homeomorphic to the wedge of the smash products, MATH . Thus MATH gives a MATH - equivariant stable homotopy equivalence, MATH which gives a proof of REF . Moreover when MATH is a smooth projective variety (or a union of such) this equivariant stable homotopy equivalence is a holomorphic one. In particular, given any such MATH, this implies there is a MATH equivariant homotopy equivalence MATH where MATH refers to those maps of spectra that preserve the units. If we remove the units from each of these mapping spectra we conclude that we have a MATH equivariant homotopy equivalence MATH . But these spaces are precisely MATH and MATH respectively. REF now follows. |
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 on MATH is given by permuting the factors according to the action of MATH on MATH, this implies that the right hand vertical map in this diagram, MATH is a MATH -equivariant homotopy equivalence. Hence the left hand vertical map MATH is also a MATH - equivariant homotopy equivalence. This is the statement of REF . |
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