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math/9910156 | According to REF below, it is identical to the one of CITE, using the existence of a good NAME relation (compare for instance CITE and the references given there) in order to prove the existence of a good operator as in CITE. |
math/9910156 | Remark first that MATH if and only if its image in MATH belongs to MATH. Moreover, we may fix a representative for MATH and consider MATH to define MATH or MATH. The condition in the proposition is easily seen to be independent of these choices. The result is then a direct consequence of CITE. |
math/9910156 | First, it is easy to see that the morphism MATH induced by the pairing induces commutative diagrams as in the lemma. The compatibility of MATH with MATH and MATH is thus clear. The nondegeneracy of the pairing is then proved by induction on MATH, the case MATH being easy. |
math/9910156 | According to the previous lemma and to REF , the inductive system MATH is isomorphic, via MATH, to MATH of the projective system MATH. The first part of the theorem then follows from the construction of MATH recalled in REF. The proof for MATH and the other properties also follow from the same arguments. |
math/9910156 | According to REF , it is enough to show that MATH is nondegenerate iff all MATH and MATH are so. Now, MATH is an isomorphism in a neighbourhood of MATH if and only if all MATH and MATH are isomorphisms: this follows from the fact that a regular holonomic module MATH is equal to zero near MATH if and only if all its moderate nearby or vanishing cycles vanish on MATH. The result is then a consequence of the definition of MATH and MATH and of REF . |
math/9910156 | REF and the first part of REF follow immediately from the properties of the NAME filtration on holonomic modules. Let us prove the second part of REF . Let MATH with MATH. According to REF, there exists then MATH with MATH such that MATH. Let MATH be the minimal polynomial satisfying MATH with MATH. Then MATH is supported at the origin and satisfies MATH. Therefore, by REF, we have MATH and MATH satisfies MATH. For REF , remark that there exist MATH such that MATH. We may assume that MATH and MATH. There exists polynomials MATH (respectively, MATH) with roots in MATH (respectively, in MATH), such that MATH belongs to MATH. Applying NAME and the condition in REF we conclude that MATH belongs to MATH. Let us now prove REF . We will first need the following lemma. CASE: We have MATH if MATH or MATH. CASE: For all MATH we have MATH . CASE: For MATH, the classes of MATH REF form a basis of the MATH-vector space MATH. CASE: The classes of MATH REF form a basis of MATH. According to REF, the distribution MATH (MATH and MATH) satisfies MATH . It is then in MATH. It follows that, for any MATH, the correspondence MATH induces a surjective MATH-linear morphism MATH . This implies that, for any MATH, we have MATH because a similar property is easily seen to be true for MATH and any morphism of holonomic MATH-modules is strict with respect to the NAME filtration. By the same argument we also get that, for MATH, MATH . As MATH, REF are clear. REF shows that the elements given in REF or REF generate the corresponding bigraded object. If we have, for MATH, a linear relation between the classes of MATH REF in MATH then, by applying a suitable power of MATH and using relation REF, we would have a relation MATH which is clearly impossible by considering the valuation at MATH. Similarly, a linear relation between the classes MATH REF would imply that MATH. Notice now that MATH is bijective: REF shows that it is onto; it is injective because MATH is so, as follows from REF. So, if MATH, we have MATH, hence MATH and similarly MATH, so MATH, which is impossible because MATH acting on MATH is injective. REF follows from REF in the lemma. REF clearly implies that, for all MATH, we have MATH if MATH, for MATH. Then REF follows immediately. |
math/9910156 | REF immediately extend to MATH. Moreover, REF clearly gives MATH . An argument similar to that of REF for MATH gives the last assertion of REF . |
math/9910156 | Let MATH. Then MATH, hence, by REF, there exists MATH such that MATH is supported at MATH, that is, belongs to MATH which is easily seen equal to MATH. |
math/9910156 | Any such MATH can be written as MATH. Then we have MATH since MATH, and MATH so MATH. |
math/9910156 | This follows from the computation in the proof of REF. |
math/9910156 | Denote by MATH the support of MATH and by MATH the order of the distribution MATH on MATH. For MATH, the functions MATH and MATH are MATH on MATH and depend holomorphically on MATH. Consequently, if MATH, the function MATH is holomorphic for MATH. Let MATH be the NAME polynomial for MATH on MATH: there exists MATH such that MATH. By assumption, the roots of MATH are MATH. Denote by MATH the NAME polynomial for MATH; the roots of MATH are MATH. Fix MATH, choose MATH so large that MATH and consider the polynomial MATH . We have MATH. Hence MATH for some MATH. But MATH is holomorphic for MATH, hence for MATH. As the zeros of MATH are MATH, we conclude that the poles of MATH are MATH. A similar argument for MATH shows that the poles are MATH. |
math/9910156 | If MATH and MATH are MATH, the desired assertion follows from REF . In general, one uses REF together with REF to show the assertion for any pair MATH. |
math/9910156 | Denote by MATH the graph inclusion. Put MATH. As MATH (NAME lemma) we have MATH. We then get a sesquilinear pairing MATH . Let us consider first the case of nearby cycles (MATH). By assumption, and using NAME correspondence for nearby cycles, there is a local section MATH of MATH such that MATH, the class of MATH in MATH belongs to MATH and its class in MATH is nonzero at MATH. As the pairing REF is nondegenerate, there exists MATH in MATH such that MATH in MATH. This means that there exists MATH such that, if we put MATH where MATH are holomorphic in a neighbourhood of MATH, the germ MATH in MATH has a nonzero coefficient on MATH. Hence, there exist MATH and MATH such that MATH has a nonzero coefficient on MATH. The result follows from the computation of NAME transform CITE. The assertion for MATH follows from MATH for which we briefly recall the proof. As MATH is a simple MATH-module, MATH is a simple MATH-module, according to NAME 's equivalence theorem. In particular it has neither submodule nor quotient module supported by MATH. This implies (see CITE forgetting the filtration MATH) that MATH is onto and MATH is injective. From CITE we deduce that for any MATH we have MATH and that the induced morphisms MATH are respectively onto and injective. |
math/9910160 | If MATH is continuous and homogeneous, then an affine function of best approximation on the ball can be taken as a linear functional, MATH say, and then MATH so that MATH . Conversely, suppose MATH is continuous and that MATH . Let us note that any MATH we have MATH . This follows from applying NAME 's one-dimensional result to the line-segment MATH, since MATH . We define MATH on MATH by MATH for MATH and MATH . Then MATH is continuous and homogeneous. We will show first that MATH . Suppose MATH are not both zero. Let MATH and choose MATH so that MATH and MATH . Then for MATH, MATH by applying REF . Similarly MATH . From the definition of MATH we have: MATH . Hence MATH . This implies that there exists MATH so that if MATH . We will choose MATH as an affine approximation to MATH . If MATH then, MATH . Now suppose MATH . We write MATH where MATH and MATH . By REF we have: MATH and hence MATH . This completes the proof. |
math/9910160 | By a well-known result of CITE there is a Euclidean norm MATH on MATH so that MATH for MATH . Now suppose that MATH is continuous and MATH . Restricting MATH to MATH we can find a polynomial MATH with MATH for MATH . Fix any MATH . By the definition of the NAME constant and NAME 's theorem CITE there is a polynomial MATH so that MATH for MATH . Hence for MATH we have MATH . According to the NAME inequality (see for example, CITE p. REF) it follows that for MATH . The result now follows easily. |
math/9910160 | We may suppose that MATH and MATH are two norms on MATH so that MATH for MATH . The first estimate is proved just as in REF . The second estimate follows easily from the definition of MATH using REF . |
math/9910160 | First choose for each MATH real numbers MATH so that for any polynomial MATH in one variable of degree at most MATH we have: MATH . In particular we have MATH for MATH . Hence, if MATH then MATH is a MATH-homogeneous polynomial. Using this, let us first prove that MATH . In fact if MATH is continuous and MATH-homogeneous with MATH then MATH and so there exists a polynomial MATH with MATH . Now MATH by the MATH-homogeneity of MATH and REF , and this inequality leads to the estimate MATH for MATH where MATH is a MATH-homogeneous polynomial. Hence REF follows. Conversely let MATH . Suppose MATH with MATH. Then for each MATH with MATH and MATH we define MATH and extend MATH to be MATH-homogeneous. It is easy to see that each MATH is continuous. We also let MATH for all MATH . By the one-dimensional result CITE for each MATH with MATH there is a polynomial MATH on MATH of degree at most MATH so that MATH for MATH . Hence MATH where MATH . Then, for any MATH we have MATH . Using REF for MATH we have the identity MATH and we can deduce MATH . Hence MATH for MATH . We now deduce from REF that MATH for MATH . Indeed let MATH and let MATH be the linear space generated by MATH. By REF and the multivariate NAME type inequality (in dimension REF) CITE we can find a polynomial MATH of degree at most MATH so that MATH for MATH where MATH . But, arguing as before, we can replace MATH by MATH and this allows us to assume that MATH is homogeneous of degree MATH (by similar arguments to those used above.) Hence MATH . Combining with REF we get REF . Then we can conclude that there is a MATH-homogeneous polynomial MATH on MATH so that MATH for MATH . Finally if we set MATH then MATH for MATH and so MATH for a constant MATH depending only on MATH . |
math/9910160 | Suppose that MATH with MATH . We need to find an extension MATH of MATH with norm majorized by the right-hand side of REF . To do this we follow an extension technique of NAME which is used heavily in CITE. Consider the space MATH that is, MATH algebraically with norm MATH . Then MATH that is, MATH with norm MATH . Since MATH we have MATH . Let MATH . Let MATH and let MATH be the quotient map. Note that MATH maps MATH isometrically onto a subspace MATH of MATH and that by REF there is a projection MATH with MATH . Let MATH be defined by MATH . Then MATH can be regarded as an extension of MATH; more precisely, MATH extends MATH where MATH and MATH is the inverse of MATH on MATH . Then MATH . It therefore remains only to estimate MATH . Fix MATH . Note that MATH is isometric to MATH. Hence by arguments that go back to the paper CITE (see CITE for details) we have the estimate MATH for a suitable increasing function MATH . Now as a direct consequence of NAME 's characterization of MATH-convex spaces CITE we also have that an estimate on the MATH-convexity constant of MATH in terms of MATH. Hence we get an estimate of the form MATH for a suitable increasing MATH . Putting these estimates together we have MATH where MATH . It remains to observe that MATH and we are done. |
math/9910160 | We may choose MATH affine so that MATH . Then clearly MATH dominates the expression on the right of the equation. To prove the converse, we observe (see, for example, CITE, p. REF) that there exist non-empty subsets MATH and MATH of MATH so that MATH and MATH and so that for MATH we have MATH . Let MATH and MATH then MATH and we can find convex combinations so that MATH . Then MATH . |
math/9910160 | We set MATH where MATH is the canonical basis of MATH . Replacing MATH by MATH where MATH is an affine function satisfying MATH for MATH we can obtain an alternate expression for MATH . Suppose then MATH satisfies MATH and MATH for MATH . Choose MATH . Let MATH . Let MATH be a reordering of MATH so that MATH is increasing. Then we may choose signs MATH so that MATH . Then we can write MATH where MATH and MATH . Hence MATH . If MATH then MATH at most MATH times and so MATH . With a similar estimate for MATH we obtain MATH . This leads immediately to the claimed estimate. |
math/9910160 | Suppose MATH . Then MATH . Since MATH, MATH are convex we obtain MATH . This proves REF . To prove REF it suffices to apply REF (compare REF). |
math/9910160 | Let MATH be a bounded function satisfying the condition MATH where MATH is fixed. By NAME 's theorem applied to each line segment we have MATH . Let MATH and apply REF to the function MATH. Thus there is a MATH-dimensional NAME lattice MATH with MATH . To complete the proof we will find a lattice MATH for which MATH . In order to do this we will show the existence of a NAME lattice MATH such that if we put MATH then the spaces MATH and MATH have equivalent norms with the constant of equivalence depending only on MATH . Assuming this fact, let us show how the proof is completed. In this case by REF we have MATH . Using the duality result REF this implies that MATH . Since MATH this establishes REF and combined with REF the theorem is proved. Thus it remains to construct MATH . We will need the following lemma: Suppose MATH is defined by MATH . Then there is a constant MATH depending only on MATH so that for every disjoint family of vectors MATH we have MATH and MATH . Before proving the lemma, let us show how to complete the construction of MATH assuming this lemma. We set MATH . Then MATH. By the lemma both MATH and MATH satisfy upper MATH-estimates with constants depending only on MATH . According to a well-known theorem of NAME and NAME (see, for example, CITE) this implies that MATH and MATH are both MATH-convex with constants depending only MATH . This means that for any MATH we have MATH where MATH depends only on MATH, and a similar inequality holds in MATH . Now by REF there is a lattice MATH so that MATH are MATH-convex with constant one and the MATH-norm is MATH-equivalent to the MATH-norm with MATH depending only on MATH . Finally we use the NAME extrapolation theorem CITE to deduce that there is a NAME lattice MATH so that MATH . |
math/9910160 | Let MATH . Using REF we first estimate MATH where MATH depends only on MATH . Since MATH, REF gives that MATH . Hence MATH where MATH . Now suppose MATH have disjoint supports and that MATH is a convex combination. Then MATH . By duality REF MATH and direct calculation gives us that: MATH . Combining these estimates we have MATH where MATH . Note that we have a precisely similar estimate to REF for MATH in place of MATH using instead the equation MATH . Now suppose MATH have disjoint supports. For any MATH we can write MATH as a convex combination where supp MATH supp MATH and the MATH have disjoint supports. Let us write MATH for MATH and MATH (with MATH as a possible value!). Then by REF MATH where MATH . Now it is a consequence of REF (which is much simpler in our finite-dimensional setting) that this implies MATH . Again the same inequality holds in MATH . Now suppose MATH are any disjoint vectors with MATH . Let MATH . If MATH denotes the cardinality of MATH then MATH and by REF we have MATH . Since MATH this implies an estimate MATH and, combined with the similar estimate for MATH, this establishes the lemma. |
math/9910160 | Suppose first MATH is a bounded continuous function on MATH with MATH . Then there is an affine function MATH defined on MATH with MATH . We can extend MATH to a linear functional on MATH . We also have MATH for MATH . Hence MATH if MATH . Now if MATH we can find MATH and MATH so that MATH . Hence MATH by the one-dimensional NAME result which is essentially the fact that MATH . Since MATH is linear, MATH . It follows that MATH . |
math/9910160 | Let MATH be the quotient map. If MATH is a continuous homogeneous function then there is a linear functional MATH on MATH so that MATH for MATH . For MATH we have MATH and so by the NAME theorem we can find a linear functional MATH with MATH for MATH and MATH . Then there exists MATH with MATH and we have: MATH . REF now follows. For REF suppose MATH satisfies MATH and MATH . Then if MATH is a continuous homogeneous function then MATH . Now there exists MATH so that MATH . Let MATH . Then MATH and the lemma follows. Now suppose MATH . Then for any MATH and MATH there exists MATH so that MATH is MATH-isomorphic to a quotient of MATH. Hence MATH . However, the estimate MATH is proved in CITE (a factor REF was omitted from the argument as pointed out in CITE). Hence MATH for all MATH . Now by REF we have MATH (Note that for MATH we can eliminate a factor of REF and get an estimate of REF.) |
math/9910160 | For this theorem we need the following elementary lemma: Suppose MATH is a NAME space of type MATH where MATH with type MATH constant MATH . Suppose MATH and that MATH . Then there is a subset MATH of MATH with MATH and so that MATH where, as usual, MATH . We prove this by induction on MATH with MATH as the trivial starting point. Suppose MATH is the subset satisfying the conclusions of the lemma for MATH . Then by the definition of the type MATH constant there is a choice of signs MATH with MATH . Without loss of generality we can assume MATH . Let MATH . Then MATH . The induction step now follows easily Returning the proof of REF , we will estimate MATH . Suppose that MATH is any continuous homogeneous function on MATH with MATH . We may pick MATH so that if MATH then MATH . By REF , MATH where MATH is defined in REF . Since MATH is continuous the right-hand side is equal to MATH where MATH . We will show that MATH where MATH . To establish REF choose an integer MATH . By REF MATH and this shows that MATH . In particular, REF holds for all MATH . Suppose now MATH and choose MATH so that MATH . We consider the space MATH; then MATH . If MATH we define elements of MATH by MATH . By REF there is a subset MATH of MATH with MATH so that MATH . In particular, we have if MATH and MATH . Since MATH and MATH is homogeneous, we have MATH . Hence and by REF , MATH by the choice of MATH . We also have from REF MATH . Finally we note that, since MATH . Combining REF gives us MATH and so REF holds. Now REF gives an estimate independent of MATH and so implies that MATH . Since this estimate holds for all such MATH we obtain MATH . Since MATH this gives the required upper estimate in REF . |
math/9910160 | If MATH then MATH (see CITE). It remains to apply REF . |
math/9910160 | The upper estimate is a special case of REF , (or REF ) which we therefore postpone to the next section. For the lower estimate, we use the fact that the space MATH contains a subspace MATH so that every linear projection MATH satisfies MATH where MATH is an absolute constant. This follows from a well-known result of CITE that we may pick MATH with MATH and MATH where MATH is independent of MATH . For convenience let MATH be the space MATH with the norm, REF-equivalent to the MATH-norm, MATH . Then REF holds for every linear projection MATH, with perhaps a different constant. Since MATH is strictly convex, for every MATH there is a unique MATH so that MATH . The map MATH is called the metric projection of MATH onto MATH and the following properties are well-known (see, for example, REF): CASE: MATH is homogeneous and continuous; CASE: MATH is a (nonlinear) projection, MATH for MATH and MATH if MATH . CASE: For MATH . Now let MATH be the standard inner-product on MATH . Let MATH be the orthogonal projection onto MATH and let MATH be the complementary projection onto MATH . Let MATH be the dual norm on MATH . We now define a norm MATH on MATH by the formula: MATH where MATH . Finally let us define the continuous homogeneous function MATH . Now suppose MATH . Let MATH and MATH where MATH and MATH . Then MATH . Now we have MATH by REF . Similarly MATH . Hence by REF have MATH since MATH and MATH . Thus REF implies MATH . Let MATH . Then there is a quadratic form MATH such that MATH for MATH . We can write MATH where MATH is a symmetric MATH matrix or equivalently a symmetric linear operator on MATH . Note for every MATH we have MATH . Hence MATH and MATH . It follows that MATH . We now define MATH . The linear operator MATH is a projection onto MATH; we will use REF and so we estimate MATH . Assume MATH is chosen so that MATH . Then we may pick MATH with MATH and MATH . Now MATH . Note that MATH . Hence MATH by REF . By REF we obtain MATH which implies MATH and hence gives the estimate MATH for suitable MATH . |
math/9910160 | We use the following fact proved in an equivalent form in CITE, p. CASE: There is a universal constant MATH and for each MATH a subspace MATH of MATH, MATH with MATH (where MATH) so that: REF the NAME distance REF there is a projection MATH with MATH . Applying REF to MATH we can find a continuous REF-homogeneous function MATH with MATH and MATH where MATH is a universal constant. Defining MATH we easily have MATH but MATH and this proves the result. |
math/9910160 | By REF we have MATH and by REF we have MATH . So it will suffice to show a similar estimate for MATH . We obtain the result by a linearization technique. We can regard MATH as MATH with an appropriate norm. Now if MATH is a MATH positive-definite matrix, we can define a MATH-valued Gaussian random variable MATH with covariance matrix MATH . Let MATH be the cone of positive-definite matrices. Suppose now that MATH is a MATH-homogeneous continuous function on MATH with MATH . We define a function MATH on MATH by putting MATH . Then MATH is MATH-homogeneous on the cone MATH . Let MATH be the convex hull of the set of matrices MATH where MATH denotes the rank one matrix MATH . We need the estimate: There is a universal constant so that for any MATH we have: MATH . By the main result of CITE there is a constant MATH so that MATH for all MATH-dimensional subspaces. Let MATH . By REF there is a quadratic form MATH on MATH so that MATH for all MATH (where again MATH is a universal constant). This immediately yields the lemma. Returning to the proof of the theorem we note that if MATH and MATH are independent then MATH has the same distribution as MATH . Hence MATH . Now suppose that MATH . Then we can write MATH where MATH for MATH and MATH with MATH . Then MATH has the same distribution as MATH where MATH are independent normalized Gaussian random variables. Hence as is well-known (see, for example, CITE, p. REF) MATH . Using the similar inequality for MATH, we obtain MATH for a universal constant MATH . Hence MATH . Since MATH we can apply REF to MATH to deduce the existence of an affine function MATH on MATH so that MATH where MATH is again a universal constant. In particular MATH is dominated by MATH so we can assume that MATH is linear on the linear span of MATH. Let MATH . Then MATH is a quadratic form. Since MATH where MATH is a normalized Gaussian, we have from REF MATH for all MATH . This gives the desired estimate of MATH and completes the proof of the upper estimate. (The lower estimate.) We establish a lower estimate for MATH we first achieve this for the case of MATH . There is an absolute constant MATH so that for all MATH . Let MATH for MATH . Then, by the Mean Value Theorem MATH for some MATH . Hence MATH . Now define for MATH, MATH . Then for MATH . Hence MATH . Since MATH is even and MATH we can find a quadratic form MATH on MATH so that MATH . As the points MATH for MATH the left-hand side is at least MATH . As MATH for MATH we have MATH . Putting these inequalities together gives MATH . Next we need a lemma using the extension constants from REF . Let MATH be a MATH-dimensional NAME space and let MATH be a linear subspace of MATH . Let MATH and MATH . Then MATH . It will be convenient to regard MATH as MATH with an appropriate norm and let MATH be the usual inner-product on MATH . Suppose MATH is a MATH-homogeneous continuous function on MATH with MATH . Let MATH be the quotient map. Then MATH is continuous and MATH-homogeneous on MATH and MATH . Hence there is a quadratic form MATH such that MATH for MATH . We can assume MATH where MATH is a symmetric matrix. Since MATH we have MATH . Assume MATH . Then MATH and so MATH . Replacing MATH by MATH and MATH by MATH and minimizing the right-hand side gives MATH . This implies that MATH when MATH . From the definition of the extension constant there exists a MATH matrix MATH so that MATH for MATH and MATH has norm at most MATH as an operator from MATH into MATH . Then MATH maps MATH to MATH and hence the transpose MATH maps MATH to MATH . Now MATH and so MATH . Using the extension constant again we can find a MATH matrix MATH which maps MATH into MATH and such that MATH . Let MATH . Then MATH maps MATH to MATH and MATH into MATH . It follows that MATH where MATH is a linear operator from MATH to MATH and we can define a quadratic form MATH on MATH by MATH . Then MATH . Now for given MATH we can choose MATH with MATH and MATH . This implies MATH . We can now complete the proof of the lower estimate in REF . Suppose MATH is a NAME space of dimension MATH . We use the following powerful form of the NAME theorem due to CITE (see REF , where the theorem is formulated in the form required here). There is a subspace MATH of MATH which is MATH-isomorphic to MATH with MATH . We note that the lower estimate in REF is trivial for spaces such that MATH . We therefore will consider only those spaces MATH for which MATH. Then REF gives MATH . Let us put MATH . Since MATH is isometric to MATH and MATH we can apply REF to obtain MATH where, as usual, MATH is an absolute constant. Now we use REF to estimate the constants MATH of REF as follows: MATH where MATH is a suitable increasing function. Since MATH is isometric to MATH we have MATH and so MATH . Together with REF this yields MATH . Combining this with REF we have: MATH . Applying now REF and the inequality MATH we have MATH for an absolute constant MATH . The proof of REF is now complete. |
math/9910160 | This is established by expanding in each variable separately and collecting terms. We omit the details. |
math/9910160 | Suppose first that MATH is continuous and MATH-homogeneous and MATH is defined by REF . Suppose MATH and MATH . Let MATH . Then MATH and so by the NAME type result of CITE there is a constant MATH so that MATH . By REF we also have MATH . Since MATH there is a homogeneous polynomial of degree MATH on MATH so that MATH for MATH . We can express MATH in the form MATH where MATH is a symmetric MATH-linear form. Using the polarization formula from multilinear algebra, we have MATH whenever MATH and MATH . Let MATH . It now follows that MATH and so MATH . We now turn to the converse. Suppose that MATH for MATH . Using REF we have MATH . Hence MATH as required. |
math/9910160 | Let MATH . By the previous lemma, MATH . Hence there is a symmetric MATH-linear form MATH so that if MATH then MATH . Now let us define MATH using REF to be separately homogeneous and for MATH . Note that MATH since MATH unless MATH for all MATH . Hence MATH . It follows, by REF that for MATH . We also have, again using REF MATH and the lemma follows by homogeneity. |
math/9910160 | Using REF with MATH and the NAME inequality MATH (CITE, CITE) we have MATH, but MATH by REF . |
math/9910160 | Since MATH (see for example, CITE) by REF with MATH we have MATH . Conversely by REF we have MATH . |
math/9910160 | Apply REF with MATH and use REF . |
math/9910160 | It is easily checked that the proof of REF goes through with trivial changes for MATH-normed spaces when MATH (see Remark after REF ). Of course the constant MATH in its formulation depends now on MATH . Applying this result to MATH with MATH we therefore have MATH . But MATH and it is essentially proved in CITE (in an equivalent formulation related to the notion of a MATH-space) that MATH with MATH an absolute constant independent of MATH . This proves the Theorem. |
math/9910162 | To see that REF is true, suppose otherwise, and let MATH and MATH in MATH witness its failure. Then for every MATH divides MATH. Subtracting, it follows that MATH divides MATH, and hence MATH satisfies MATH. By REF , this contradicts the hypothesis that MATH contains no isomorphic copy of MATH. REF is immediate because MATH is an embedding, and the elements MATH satisfy the formula MATH in MATH. Let MATH be the logical conjunction of the following formulas: CASE: MATH CASE: MATH CASE: MATH are independent CASE: MATH. Observe that MATH is a MATH-formula. Let MATH be the MATH-sentence MATH. |
math/9910162 | The implication MATH is easy, since MATH, and one can use the embedding MATH to find witnesses in MATH for the existential quantifiers of the conjuncts of MATH. Conversely, suppose that REF holds, and MATH are elements of MATH satisfying MATH. The map MATH given by MATH is an embedding, so to establish REF , it is enough to find an embedding MATH (and then MATH embeds MATH into MATH). For an element MATH, since MATH, there exists MATH such that MATH. By REF , the element MATH is uniquely defined. Claim: the map MATH given by MATH is an embedding. This is easy to show: to see that MATH is a homomorphism, note that for example, for MATH, for every MATH divides MATH, and hence MATH (otherwise REF will reveal a copy of MATH in MATH). A similar argument using the appropriate conjunct of REF in MATH proves that MATH is one-to-one. |
math/9910162 | The implication MATH is analogous to REF . As regards the direction REF , we sketch only the slightly different points, arising from MATH, concerning the formula MATH. By REF , MATH. If for some MATH, then, for every MATH divides MATH, since MATH divides MATH. But also, if MATH, then MATH divides MATH. Hence MATH satisfies MATH, contradicting the hypothesis that MATH does not contain a copy of MATH. Thus MATH is unique such that MATH. The remaining conjuncts of REF hold in MATH because MATH is an embedding. So MATH. |
math/9910162 | By REF , together with REF , and REF , the following sentence MATH is as required: MATH . REF follow immediately from REF . |
math/9910162 | MATH is slender CITE. |
math/9910163 | First note that by REF is equivalent to MATH , whenever MATH with MATH . To prove REF implies REF we note that if MATH is a MATH-weight so that MATH we can find an outer function MATH so that MATH a.e. Then REF gives MATH for MATH with MATH . This in turn implies that MATH for all MATH with MATH . By the NAME Theorem this implies there exists MATH so that MATH or MATH where MATH . For the reverse direction just note that if MATH with MATH then MATH so that REF follows from the NAME inequality. |
math/9910163 | In this case MATH a.e. and MATH a.e. Furthermore: MATH a.e., so that we obain the result from REF . |
math/9910163 | First suppose MATH and MATH where MATH . Then there exists MATH with MATH a.e. In particular, MATH a.e. and so MATH maps MATH into the same sector. It follows that we can define MATH for all MATH . Choose MATH so that MATH and let MATH . Then MATH and MATH so that MATH . Now by REF we have that MATH is a MATH-weight. However MATH so that MATH . We deduce that MATH . For the converse direction assume that MATH is a MATH-weight. Then there exists MATH so that MATH where MATH . Arguing as above we have MATH and MATH is a MATH-weight with MATH . Note that MATH, and this establishes the other direction. |
math/9910163 | It is enough to show MATH is bounded and MATH since MATH is also a monotone multiplier. To see this note that if MATH is finitely nonzero and MATH then MATH so that MATH . |
math/9910163 | Note that if MATH then MATH whence a calculation as in REF gives MATH . It follows that MATH . It remains therefore only to select MATH so that MATH and MATH . For convenience we write MATH where MATH and MATH . For any MATH pick MATH to be the greatest integer so that MATH . Then MATH and MATH . Now MATH and MATH . This yields the desired result. |
math/9910163 | In fact MATH so it is clear that MATH . For the other direction suppose MATH is a trigonometric polynomial in MATH . Then for large enough MATH we have MATH and then MATH. This quickly yields MATH . |
math/9910163 | We shall prove that if MATH then the existence of an invertible MATH so that MATH is equivalent to the existence of a weight MATH equivalent to MATH so that MATH . Once this is done, the result follows from REF . In one direction this is easy. Assume MATH equivalent to MATH and MATH . This means that there is an equivalent inner-product norm on MATH in which the basis constant of MATH bounded by MATH . It follows from REF that in this equivalent norm we have MATH . Hence MATH is similar to an operator MATH such that MATH . We now consider the converse. Let MATH be the operator MATH . Suppose MATH is an invertible operator such that MATH . We will define a new inner-product on MATH by MATH where MATH denotes any NAME limit (see for example, CITE p. REF). Since MATH is an isometry on MATH and MATH is invertible this defines an equivalent inner-product MATH norm on MATH . Now for any MATH and fixed MATH we have MATH where MATH is given in REF . Hence MATH . Now MATH . Thus with respect to the new norm MATH the basis constant is at most MATH . Now let MATH for MATH and let MATH when MATH . Then it follows easily that MATH for all MATH and that for all finitely nonzero sequences MATH of complex numbers we have that MATH . This implies (see CITE p. REF) that there is a finite positive measure MATH on MATH so that MATH . Thus MATH . However this norm is equivalent to the original norm so that MATH is absolutely continuous with respect to NAME measure and of the form MATH where MATH . It follows that in MATH the basis constant of the exponential basis is at most MATH and so by REF we have MATH and the proof is complete. |
math/9910163 | That REF implies REF is a consequence of NAME 's inequality (see CITE). Similarly REF implies REF is trivial. It therefore remains to prove that REF implies REF . We shall treat the case when the MATH are distinct; small modifications are necessary in the other cases. We shall also suppose the measure MATH is a probability measure so that MATH for all MATH . First note that if MATH then for any MATH then MATH is well-defined where MATH and if MATH is polynomially bounded we have an estimate MATH or equivalently MATH whenever MATH is finitely non-zero. Letting MATH we obtain MATH . Recall that by NAME 's theorem CITE the sequence MATH is interpolating (compare CITE p. REF) so that there is a constant MATH such that for any sequence MATH there exists MATH with MATH and MATH for all MATH . Hence MATH for all finitely non-zero sequences MATH . Hence by the parallelogram law we have MATH from which it follows that MATH . |
math/9910163 | The equivalence of REF is proved in REF . The equivalence of REF is due to CITE or REF and its proof REF . |
math/9910166 | We make the blowing-up procedure explicit, in terms of open affine coverings. For each MATH we define a finite index set MATH, consisting of all pairs MATH with the property that MATH and MATH for MATH and that MATH for some MATH, if and only if MATH. Observe that for each MATH there is a surjection MATH, which maps the triple MATH to the element MATH of MATH. (Here we have used the convention that MATH for any permutation MATH). Furthermore, this surjection is in fact a bijection in the case of MATH. Let MATH and chose an element MATH in its preimage under the surjection MATH. We define subrings MATH of MATH together with ideals MATH, MATH, distinguishing three cases. CASE: MATH CASE: MATH CASE: MATH . Observe that the objects MATH, MATH, MATH thus defined, depend indeed only on MATH and not on the chosen element MATH. By induction on MATH one shows that MATH is covered by open affine pieces MATH (MATH), such that MATH (equality as subrings of the function field MATH), and such that the ideals MATH and MATH are the defining ideals of the closed subschemes MATH and MATH respectively. |
math/9910166 | This is immediate from the local description given in REF. |
math/9910166 | Let MATH be a scheme and MATH a tupel defined over MATH, which has the properties stated in the lemma. Let us first consider the case, where all the sheaves MATH, MATH are trivial and where there exists a MATH, such that MATH and MATH is nowhere vanishing for MATH and MATH respectively. Observe that under theses conditions there exists a unique set of trivializations MATH, MATH, MATH such that MATH for MATH, MATH for MATH, and such that the diagrams MATH commute for MATH. Let MATH be defined by MATH for MATH and MATH for MATH, and let MATH be the morphism defined by MATH. It is straightforward to check that the induced morphism MATH does not depend on the chosen number MATH and that it is unique with the property that the pull-back under MATH of the universal tupel is equivalent to the given one on MATH. Returning to the general case, observe that there is an open covering MATH, such that for each MATH there exists a MATH with the property that over MATH all the MATH, MATH are trivial and that MATH and MATH is nowhere vanishing over MATH for MATH and MATH. The above construction shows that there exists a unique morphism MATH such that for each MATH the restriction to MATH of the pull-back under MATH of the universal tupel is equivalent to the restriction to MATH of the given one. Thus it remains only to show that the isomorphisms defining the equivalences over the MATH glue together to give a global equivalence of the pull-back of the universal tupel with the given one. However, this is clear, since it is easy to see that there exists at most one set of isomorphisms MATH, MATH establishing an equvalence between two tuples MATH and MATH. |
math/9910166 | Let MATH be the preimage of MATH under the morphism MATH. By definition of MATH, we have for all MATH: MATH . Let MATH be the Torus in MATH. We have an isomorphism MATH defined by MATH, MATH,MATH, and a commutative quadrangle MATH where the vertical arrows are the natural inclusions. Furthermore, the map MATH induces an isomorphism MATH for MATH. Using the fact that MATH is separated and that MATH dense in MATH, the lemma now follows easily. |
math/9910166 | It suffices to show that for each MATH the restriction of MATH to the open set MATH is a generalized isomorphism from MATH to itself. Let MATH REF be the upper (lower) triangular MATH matrix with MATH on the diagonal and entries MATH REF over (under) the diagonal MATH. For MATH we define MATH . Here we interprete the matrices MATH and MATH as automorphisms of MATH. Accordingly we view the sheaves MATH and MATH as subsheaves of MATH. We have to show that the tupel MATH is a generalized isomorphism from MATH to itself. We have for MATH the following equality of subsheaves of MATH: MATH . This is easily checked by restricting both sides of the equations to the open subsets MATH, MATH of MATH and using REF. Observe that the morphisms MATH are described by the matrices MATH and the morphisms MATH by the matrices MATH respectively, where we have abbreviated MATH by MATH, and MATH by MATH. Furthermore the matrix-decomposition on page REF (for MATH) shows that MATH is the diagonal matrix with entries MATH. With this information at hand, it is easy to see that MATH is indeed a generalized isomorphism from MATH to itself. |
math/9910166 | The the morphism MATH defining the action is given on MATH-valued points by MATH where MATH is a generalized isomorphism as in REF from MATH to MATH and MATH is the generalized isomorphism where for MATH the bf-morphisms from MATH to MATH, the ones from MATH to MATH and the isomorphism MATH are the same as in the tupel MATH, and where the bf-morphisms MATH in the tupel MATH are replaced by the bf-morphisms MATH respectively. The invariance of the divisors MATH and MATH is clear, since they are defined by the vanishing of MATH and MATH respectively. |
math/9910166 | Restricting to a neighbourhood of MATH, we may assume that the sheaves MATH, MATH, MATH are free. Let MATH and MATH be global frames for MATH and MATH respectively. After permutation of their elements, and restricting to a possibly smaller neighbourhood of MATH, we may further assume that the morphisms MATH induced by MATH and MATH respectively, are isomorphisms. Let MATH . Then we have direct-sum decompositions MATH which are respected by MATH and MATH. Let MATH be a nowhere vanishing section of MATH. The frames MATH, MATH of MATH, MATH, defined by setting MATH, MATH for MATH and MATH, MATH for MATH, have the desired property. |
math/9910166 | CASE: An easy calculation shows that the morphism given by the prescription MATH does not depend on the chosen local frames MATH, MATH. Therefore using REF, we may define MATH by this local prescription. CASE: This can be proven along the same lines as REF. Alternatively, it follows by applying REF. to the bf-morphism MATH obtained from MATH by setting MATH and MATH. |
math/9910166 | This follows immediately from the local description of MATH and MATH respectively. |
math/9910166 | From the proof of REF it follows readily that the restriction of MATH to MATH is MATH as an element of MATH . On the other hand, REF tells us that MATH is a generator of CITE MATH over MATH. |
math/9910166 | Let MATH be a generalized isomorphism from MATH to MATH. For MATH denote bf-morphisms as follows: MATH and let MATH the isomorphism MATH. CASE: We may assume that MATH. We show by induction on MATH that admissibility of MATH for MATH implies the diagonalizability of MATH with respect to MATH. The case MATH is trivial, so assume MATH. By assumption, the morphism MATH is an isomorphism. Let MATH and MATH . Then we have natural direct sum decompositions MATH . Since the bf-morphisms MATH and MATH respect these decompositions, we can write MATH and MATH where MATH (respectively MATH, MATH, MATH) is a bf-morphism from MATH to MATH (respectively from MATH to MATH, from MATH to MATH, from MATH to MATH) for MATH. By the same reason, we can write MATH, where MATH and MATH. Observe that MATH and MATH for MATH and that MATH . Now we define MATH where we identify MATH with MATH via the isomorphism MATH and MATH with MATH via the isomorphism MATH . Let MATH where MATH is the composition MATH and where the other morphisms are the ones from the MATH and the MATH. It is easy to see that MATH is a generalized isomorphism from MATH to MATH. Furthermore, it follows from REF that MATH where MATH and MATH are the canonical global frames of MATH and MATH respectively. Therefore we have MATH . Since, by assumption, the sections MATH are nowhere vanishing, the above equation implies that the same is true also for the sections MATH. In other words, MATH is admissible for MATH. By induction-hypothesis, we conclude that there exists a diagonalization MATH of MATH with respect to MATH. Let MATH and MATH . Observe that there are natural isomorphisms MATH . We set MATH , MATH for MATH, and MATH for MATH. Finally, we let MATH be the isomorphism induced by MATH. It is now easy to see that the tupel MATH is a diagonalization of MATH with respect to MATH. Conversely, assume that there exists a diagonalization MATH of MATH with respect to MATH. Observe that the diagram MATH where MATH, is commutative for MATH. Therefore we may assume that MATH. But then MATH is the section induced by the isomorphism MATH for MATH. In particular the MATH are nowhere vanishing on MATH, which is precisely what is required for the admissibility of MATH for MATH. CASE: By REF, it suffices to show that in the case MATH (MATH a field) there exists a pair MATH which is admissible for MATH. We apply induction on MATH, the case MATH being trivial. It is an easy exercise in linear algebra, to show that the morphism MATH has at least rank one. Consequently there exist indices MATH, such that the composition MATH is an isomorphism. Let the sheaves MATH, MATH be defined as on page REF, with MATH replaced by MATH, and using these sheaves, let MATH be defined as on page REF. This is a generalized isomophism from MATH to MATH. By induction-hypothesis there exists a pair MATH, which is admissible for MATH. Let MATH be defined by MATH and let MATH be defined analogously. As on page REF we have MATH for MATH, that is, the pair MATH is admissible for MATH. CASE: This follows from the proof of REF, since it is clear that the construction of the diagonalization there is unique. |
math/9910166 | CASE: Let MATH . By REF there exists a diagonalization MATH of MATH with respect to MATH. Let MATH (respectively MATH) REF be the nontrivial entries of the lower (respectively upper) triangular matrix MATH (respectively MATH). Let MATH and MATH be the morphisms defined by MATH and MATH respectively. Furthermore, let MATH be the morphism induced by the tupel MATH (compare REF ). Thus we have a morphism MATH where the right isomorphism is the inverse of the one in REF . It is clear that MATH . Denote by MATH the pull-back under MATH of the diagonalization of MATH, which exists by REF. By the uniquness of diagonalizations (compare REF), we have MATH, MATH and MATH. Therefore the isomorphisms MATH induce an equivalence between MATH and MATH. This proves the existence part of the proposition. For uniqueness, assume that MATH is a further morphism from MATH to MATH, such that MATH is equivalent to MATH. Let MATH, MATH, MATH be an equivalence. Note that by REF. Let MATH be the pull-back under MATH of the diagonalization with respect to MATH of MATH. Then MATH is a diagonalization of MATH with respect to MATH. By REF we conclude that MATH, MATH and MATH. But this implies that the composite morphism MATH equals MATH and thus that MATH. CASE: Denote for a moment by MATH the open subset of MATH, defined by the nonvanishing of MATH for MATH. We have already seen in REF that MATH is contained in MATH. Let MATH. Since MATH is dense in MATH, there exists a generalization MATH of MATH. Then there exists a morphism MATH, where MATH is the Spec of a valuation ring, such that the special point of MATH is mapped to MATH and its generic point to MATH. By definition of MATH, the pair MATH is admissible for the generalized isomorphism MATH. Therefore REF tells us that there exists a morphism MATH, which coincides with MATH at the generic point of MATH. Since MATH is separabel, it follows that MATH and thus that MATH. CASE: This follows immediatelly from REF. |
math/9910166 | . Let MATH be a scheme and MATH a generalized isomorphism from MATH to itself. By REF , there is a covering of MATH by open sets MATH, and for every index MATH a pair MATH, which is admissible for MATH. REF now tells us that there exists for each MATH a unique morhpism MATH with the property that there is an equivalence, say MATH, from MATH to MATH. By REF , the MATH glue together, to give a morphism MATH. It remains to show that also the MATH glue together, to give an overall equivalence from MATH to MATH. For this, it suffices to show that for two generalized isomorphisms MATH and MATH from MATH to itself there exists at most one equivalence from MATH to MATH. The question being local, we may assume by REF that MATH is diagonalizable with respect to some pair MATH. Composing the diagonalization of MATH with any equivalence from MATH to MATH gives a diagonalization of MATH with respect to MATH. Since different equivalences from MATH to MATH would yield different diagonalizations of MATH, REF tells us that there exists at most one equivalence. |
math/9910166 | Observe that MATH is naturally isomorphic to the closed subscheme MATH of MATH. The restriction of MATH to MATH induces in an obvious way a complete collineation MATH of MATH to itself on MATH. The corollary now follows from REF . |
math/9910166 | In the case of MATH and MATH, the theorem is a consequence of REF, where the representing objects are MATH, MATH and MATH respectively. Let MATH an open covering such that there exist trivializations MATH . Let MATH and MATH the projection. By REF , over the intersections MATH the pairs MATH induce isomorphisms MATH. These provide the data for the pieces MATH to glue together to define MATH. Using REF it is easy to check that MATH has the required universal property. This proves the existence of MATH. The existence of MATH and of MATH is proved analogously. |
math/9910166 | The isomorphism MATH is given on MATH-valued points by the bijectiv correspondence MATH where MATH is a generalized isomorphism from MATH to MATH with MATH for MATH and MATH, MATH are flags of type MATH and MATH in MATH and MATH respectively, MATH is a complete collineation from MATH to MATH for MATH, MATH is a complete collineation from MATH to MATH for MATH and MATH is a generalized isomorphism from MATH to MATH. The mapping MATH is defined as follows: Let MATH as above be given. For convenience we set MATH, MATH and we let MATH and MATH be the isomorphism MATH and its inverse respectively, whereas we let MATH and MATH both be the zero morphism. For what follows, the picture below may help to keep track of the indices: MATH . Let MATH, MATH, MATH and MATH . It is then clear from the definition of generalized isomorphisms that MATH are flags of type MATH and MATH in MATH and MATH respectively. Let MATH. We set MATH and MATH where MATH. Observe that the sheaves MATH thus defined are locally free of rank MATH. Indeed, this is clear for MATH. For MATH it suffices to show that MATH is generated by the two subsheaves MATH and MATH. For this in turn, it suffices to show that the image of MATH by the morphism MATH is MATH. But this is clear, since by the definition of generalized isomorphisms we have MATH . Since MATH we have a natural isomorphism MATH. We define MATH to be the zero morphism. Thus we have a bf-morphism MATH of rank zero. For MATH let MATH be the bf-morphism induced by the bf-morphism MATH. Observe that MATH and that the morphism MATH maps MATH injectively into MATH. Therefore we have a natural isomorphism MATH by which we identify these two sheaves. It is not difficult to see that MATH is a complete collineation in the sense of REF from MATH to CITE MATH. In a completely symmetric way the generalized isomorphism MATH induces also complete collineations MATH from MATH to MATH for MATH. It remains to construct the generalized isomorphism MATH. We set MATH for MATH and MATH for MATH. It is then clear that the MATH and MATH are locally free of rank MATH. It follows from REF . that the MATH and MATH are nowhere vanishing for MATH and MATH. Therefore we may identify MATH with MATH and MATH with MATH for MATH and MATH respectively. This implies in particular that we have MATH and MATH. Let MATH be the bf-morphisms induced by the bf-morphisms MATH respectively. We have MATH and therefore MATH . By the same argument: MATH . Thus the isomorphism MATH induces an isomorphism MATH. Again it is not difficult to check that MATH is a generalized isomorphism from MATH to MATH. This completes the construction of the mapping MATH . We proceed by constructing the inverse of this mapping. Let data MATH be given. Let MATH and MATH for MATH and MATH respectively. Now let MATH and MATH. Let MATH and MATH be defined by the cartesian diagrams MATH respectively. For a moment let MATH. We have a commutative diagram MATH where the left vertical arrow is induced by MATH and the upper horizontal arrow is the composition MATH . The diagram MATH induces a morphism MATH. Analogously, we have a morphism MATH, where we have employed the abbreviation MATH. Let MATH and MATH be defined by the cocartesian diagrams MATH respectively. We define MATH by the cartesian diagram MATH . Observe that the composed morphism MATH maps the submodule MATH of MATH isomorphically onto the submodule MATH of MATH. Therefore we have canonical injections MATH for MATH. Analogously, we have canonical injections MATH for MATH. Now let MATH and MATH. We want to define MATH and MATH in this case. Let first MATH and MATH be defined by the cocartesian diagrams MATH where we have set MATH and MATH. Let furthermore MATH and MATH be defined by the cocartesian diagrams MATH . Now we define MATH and MATH by the cocartesian diagrams MATH respectively. For MATH and MATH this gives formally a new definition of MATH, but it is clear that we have a canonical isomorphism between the two MATH's. A similar remark applies to MATH. We define the invertible sheaves MATH together with their respective sections MATH as follows: MATH . Let the MATH and MATH be defined symmetrically (that is, by replacing in the above definition the letter MATH with MATH, MATH with MATH, MATH with MATH, MATH with MATH and MATH with MATH). It remains to define the bf-morphisms MATH for MATH and MATH. Again we restrict ourselves to the left hand side, since the right hand side is obtained by the symmetric construction. For MATH (respectively for MATH) the bf-morphism MATH is induced in an obvious way by the bf-morphism MATH (respectively by the bf-morphism MATH ). For the definition of the bf-morphism MATH consider the two canonical exact sequences MATH where MATH. Observe that we have canonical isomorphisms MATH . The isomorphism MATH follows from the observation we made after the definition of MATH, namely that the composed morphism MATH maps MATH isomorphically to MATH. The isomorphism MATH comes from the fact that for MATH the left vertical arrow in the defining diagram for MATH vanishes, and the isomorphism MATH follows since for MATH the left vertical arrow in the defining diagram for MATH is an isomorphism. Thus we have morphisms MATH which make up the bf-morphism MATH. For MATH the bf-morphism MATH is constructed similarly from the exact sequences MATH and the canonical isomorphisms MATH where MATH. This completes the construction of MATH . It is not difficult to see that MATH is a generalized isomorphism from MATH to MATH and that the mapping MATH is inverse to the mapping MATH constructed before. We leave the details to the reader. |
math/9910166 | The MATH-valued points of MATH are the generalized isomorphisms MATH where MATH for MATH and MATH and where MATH, MATH are nowhere vanishing for MATH, MATH. It is clear that MATH is invariant under the operation of MATH. From the proof of REF it follows that we have the following isomorphism MATH where MATH . There is a left MATH-operation on the right-hand side of this isomorphism, given on MATH-valued points by MATH where MATH is an isomorphism (up to multiplication by an invertible section of MATH) from MATH to MATH for MATH, MATH an isomorphism (up to multiplication by an invertible section of MATH) from MATH to MATH for MATH and MATH is an isomorphism from MATH to MATH. It is easy to see that this operation is transitiv and that the isomorphism MATH is MATH-equivariant. |
math/9910166 | Let MATH be a MATH-scheme. A MATH-valued point of the fibre product MATH is given by a generalized isomorphism MATH-from MATH to MATH such that the induced morphism MATH identifies MATH with the subbundle MATH. Let MATH and MATH be the ranks of MATH and MATH respectively. Observe that MATH. We restrict ourselves to the case, where MATH and MATH are both strictly smaller than MATH. (The cases where one or both of MATH are equal to MATH are proved analogously). Then the sections MATH and MATH are invertible and MATH. From the proof of REF it follows that such a MATH may be given by a tupel MATH where MATH are the filtrations given by MATH is a complete collineation from MATH to MATH, MATH is a complete collineation from MATH to MATH, and MATH is the isomorphism MATH . We see in particular that the tupel MATH is already determined by the subbundle MATH (that is, by the morphism MATH) and the pair MATH. |
math/9910171 | Trivially MATH so we need only prove the reverse inclusion. If MATH then we must show that MATH . Let MATH and, for MATH, inductively define MATH so each MATH is non-empty, compact and MATH . Then define MATH . Since MATH is a nested sequence of non-empty compact subsets of MATH, it follows that MATH is a non-empty compact subset of MATH. So MATH and for MATH, MATH . Thus every point of MATH has a MATH pre-image in MATH. Define MATH to be any point of MATH (so MATH) and inductively define MATH to be a point in MATH such that MATH . If we also define MATH for MATH then MATH, given by MATH, is a solution through MATH . It follows that MATH . |
math/9910171 | If MATH and the sequence MATH converges to MATH then for any MATH the sequence MATH is well defined for MATH sufficiently large and converges to a point MATH. By continuity of MATH, if MATH then MATH and if MATH then MATH. It follows that MATH . |
math/9910171 | We may assume MATH is open. From REF it follows that there is a MATH with MATH . Our proof is by contradiction. Suppose there is no MATH with MATH. Then every MATH there is a MATH that is, there is a MATH-chain MATH with MATH and MATH . Truncating these when MATH or extending them to be orbit segments if MATH we produce for each MATH a MATH-chain MATH with MATH . Choosing a sequence MATH converging to MATH as MATH, we may assume that each of the sequences MATH has a limit, say MATH. Clearly MATH and MATH . Also it is clear by continuity of MATH that MATH for MATH . Hence MATH which is a contradiction. |
math/9910171 | Let MATH . If MATH is not an isolating block for MATH then there is a point MATH such that MATH and MATH are all in MATH, but MATH . By continuity of MATH there exists MATH such that if MATH then MATH. Since MATH and MATH there are MATH-chains MATH with MATH and MATH, and MATH with MATH and MATH. Then the sequence MATH is a MATH-chain from MATH to MATH containing MATH. This implies MATH which is a contradiction. |
math/9910171 | To check REF it suffices to show MATH . Clearly MATH is in the complement of MATH and hence MATH . Thus if MATH is a sufficiently small neighborhood of MATH, we have MATH . Since MATH is invariant MATH could only be non-empty if MATH were. Hence MATH and MATH . REF is trivially satisfied. To check REF note first that since MATH we have MATH . Also MATH where the superscript MATH indicates complement. It follows that MATH because MATH . Since MATH and MATH are disjoint it is clear that if MATH is a sufficiently small neighborhood of MATH then MATH and MATH are also disjoint. Thus REF is satisfied. Finally we check the fact that MATH is a filtration pair for maps which are MATH close to MATH. First we note that MATH is an isolating block for nearby maps. If this were not the case then there would exist a sequence MATH converging to MATH and a sequence MATH with MATH . Choosing a subsequence we can assume MATH converges to MATH which would contradict the fact that MATH is an isolating block for MATH. Similarly if MATH and MATH then a subsequence of MATH will converge to a point MATH contradicting the hypothesis that MATH is a neighborhood of MATH . Thus MATH is a filtration pair for any map sufficiently MATH close to MATH. |
math/9910171 | We may assume the given neighborhood MATH is open in MATH and MATH is a compact subset of MATH. Let MATH be an isolating block for MATH inside MATH. Let MATH be a smooth function which vanishes exactly on MATH . For any regular value MATH is a compact manifold with boundary. If we choose MATH small enough MATH will be an isolating block for MATH. Now, let MATH be a smooth map which vanishes exactly on MATH. For any regular value MATH the space MATH is a manifold with boundary. If MATH is small enough and MATH, then MATH is a small neighborhood of MATH. Modifying MATH slightly we may assume that the boundary of MATH and the boundary of MATH intersect transversely. If we define MATH and MATH was chosen sufficiently small then it follows from REF that MATH is a filtration pair. Triangulating MATH with MATH as a subcomplex shows that MATH is homeomorphic to a finite simplicial pair. |
math/9910171 | Let MATH be the quotient map and define MATH and MATH otherwise, where we have identified MATH with MATH . By REF MATH is disjoint from MATH. Hence if MATH is a sufficiently small neighborhood of MATH in MATH, then MATH is disjoint from MATH. Hence MATH for all MATH . It is immediate that MATH is continuous on MATH since it is the composition of continuous functions there. To check continuity at MATH note that if MATH is a sequence in MATH converging to MATH then MATH is eventually in MATH so MATH is a sequence which is eventually constant and equal to MATH . |
math/9910171 | Let MATH . There is a continuous function MATH given by MATH and MATH otherwise, where we have identified MATH with MATH and MATH is the quotient map. Clearly MATH . Since MATH contains the exit set of MATH and MATH is disjoint from MATH, there is a MATH such that for every MATH we have MATH for some MATH . Define a map MATH by MATH and MATH otherwise, where we have identified MATH with MATH and MATH is the quotient map. Clearly MATH . It is immediate that MATH is continuous on MATH since it is the composition of continuous functions there. To check continuity at MATH note that there is a neighborhood MATH of MATH in MATH such that MATH and hence for any point MATH, we have MATH for some MATH. Thus for any point MATH, we have MATH . So MATH is constant on a neighborhood of MATH in MATH and hence continuous at MATH. It is also clear from the definitions that MATH and MATH . |
math/9910171 | There is a continuous function MATH given by MATH and MATH otherwise, where we have identified MATH with MATH and MATH is the quotient map. Clearly MATH . It follows from REF that there is a MATH such that MATH . Define MATH by MATH and MATH otherwise, where we have identified MATH with MATH and MATH is the quotient map. Clearly MATH . It is immediate that MATH is continuous on MATH since it is the composition of continuous functions there. To check continuity at MATH note that if MATH is a sequence in MATH converging to MATH then MATH is a sequence which is eventually constant and equal to MATH. |
math/9910171 | By REF we may choose MATH sufficiently small that MATH. Denote MATH by MATH. We will show that if MATH is a sufficiently small compact neighborhood in MATH of the exit set MATH then MATH is a filtration pair with the property that MATH is shift equivalent to both MATH and MATH. The argument is the same in both cases so we will give the proof for MATH. As before we will identify MATH with MATH and hence we may consider MATH as a subset of MATH and consider the map to be MATH. Recall that MATH is the set of MATH such that MATH . Thus from the definition of MATH it is clear that MATH implies there is no MATH-chain from MATH to MATH. Consequently by REF the MATH-limit set of such a MATH can only be the point MATH. From this it follows that there is a MATH such that for any MATH we have MATH . If MATH is a sufficiently small neighborhood of MATH in MATH, the same is true for points of MATH . Define MATH to be MATH . Then MATH because MATH is a neighborhood of MATH and any sequence converging to a point of MATH will have an image under MATH converging to MATH. So any such sequence will eventually be in MATH . We also note that MATH. Thus MATH and MATH are filtration pairs for MATH. We complete the proof by showing MATH is shift equivalent to both MATH and MATH . By REF MATH is shift equivalent to MATH and by REF MATH is shift equivalent to MATH. Thus MATH is shift equivalent to MATH . To see that MATH is shift equivalent to MATH we observe that if MATH is the quotient map then we can identify the associated pointed spaces (and maps) of the filtration pairs MATH and MATH. By REF MATH is shift equivalent to the associated pointed space map for the pair MATH which we have identified with MATH. |
math/9910171 | It is immediate from the proofs of REF that this result holds in the two special cases covered by those lemmas. But in the general case proved in REF the semi-conjugacies constructed were all compositions of semi-conjugacies arising in these two special cases. It is clear from the definition that the composition of semi-conjugacies of standard shift equivalences is again standard, so the result follows. |
math/9910171 | Let MATH. If MATH then clearly MATH . On the other hand, if MATH then MATH since this would imply MATH . Likewise we have MATH. Hence it follows that MATH . From this and the fact that MATH and MATH are semi-conjugacies, it is immediate that MATH . The proof that MATH is similar. |
math/9910171 | By REF we know that MATH and MATH are shift equivalent via a standard shift equivalence. Let MATH be the minimal lag of all standard shift equivalences. Take MATH. Defining MATH in such a way it is clear that REF hold. Let MATH. We construct MATH as follows. Let MATH and MATH be a standard shift equivalence realizing the minimal lag MATH and having semi-lags MATH and MATH respectively. Define MATH . First, notice that since MATH . REF implies that MATH is independent of MATH and MATH. Also, if MATH then MATH and taking MATH we see that REF is satisfied. Let MATH be given. Let MATH and MATH form a standard shift equivalence with semi-lags MATH and MATH respectively achieving minimal lag MATH. Observe that MATH and MATH form a standard shift equivalence between MATH and MATH with semi-lags MATH and MATH respectively. Since MATH it follows that MATH, the lag of the shift equivalence given by MATH and MATH. Similarly, MATH the lag of MATH and MATH. Thus, by REF we have MATH . Thus REF holds. |
math/9910171 | This result is an easy consequence of REF . |
math/9910171 | We use the fact that we may calculate MATH using a filtration pair homeomorphic to a finite simplicial complex. This implies that for any MATH sufficiently close to MATH the maps MATH and MATH are homotopic by a homotopy through maps close to MATH. |
math/9910171 | This result follows immediately from REF . |
math/9910171 | The first of these properties is immediate from the definitions. The second follows from the fact that if MATH is an attracting neighborhood for MATH and MATH then there exists no MATH-chain from a point of MATH to a point of MATH. The only non-trivial part of the third property is that MATH is an isolated invariant set. This holds because whenever MATH is an attracting neighborhood for MATH, then MATH is an isolating neighborhood for MATH. |
math/9910171 | These results are immediate except for the fact that MATH is an attracting neighborhood. To see this observe that since MATH . |
math/9910171 | It suffices to prove the result for an attractor MATH consisting of a single point since we can always consider the map induced by MATH on the quotient space MATH which has the single point MATH as attractor. Thus we assume MATH consists of one point. Let MATH be an attracting neighborhood for MATH, so MATH with MATH is a filtration pair for the invariant set MATH . The associated pointed space MATH consists of MATH together with a disjoint single point MATH . The map MATH has precisely two fixed points, MATH and MATH. Given any neighborhood MATH of MATH it follows from REF that there is filtration pair MATH for MATH inside MATH. Let MATH be one of the semi- conjugacies from the shift equivalence between MATH and MATH . Then MATH is an open and closed subset of MATH which is forward invariant, contains MATH, and is disjoint from MATH . It follows that MATH so if MATH is the quotient map then MATH is an attracting neighborhood for MATH which is contained in MATH. |
math/9910171 | An attracting interval MATH in MATH for the order MATH is also an attracting interval for a total ordering of the set MATH which is compatible with MATH. Thus, without loss of generality we may assume that MATH is a total order. We will give the proof by induction on the number of elements MATH in MATH. Clearly if MATH then MATH so we may take MATH . Hence we may assume the result for MATH and need to prove it for MATH. Let MATH be the maximal element of MATH and define MATH to be the attracting interval MATH. Let MATH and choose a compact neighborhood MATH of MATH which is disjoint from MATH . We wish to show that for some smaller neighborhood MATH of MATH we have the property that MATH for all MATH . We assume this is not the case and show that leads to a contradiction. If no such MATH exists then there is a convergent sequence MATH such that MATH and MATH is not a subset of MATH . From this it follows that we can choose MATH with the properties CASE: MATH, CASE: MATH for MATH and CASE: MATH for some MATH . By choosing a subsequence we may assume that the sequence MATH converges, say to MATH . We claim that MATH. This follows because MATH for all MATH and for any MATH the sequence MATH is well defined for sufficiently large MATH and has a subsequence which converges to a point of MATH . Thus we have shown MATH for all MATH so by REF we have MATH . Also, for every MATH since MATH . Thus MATH is disjoint from MATH and the definition of NAME decomposition implies that MATH. Let MATH be a solution for MATH. Then MATH for some MATH, so MATH . But this leads to a contradiction since MATH is maximal, so MATH is impossible, but MATH would imply that MATH which is false. Thus we have shown that there is a compact neighborhood MATH of MATH with the property that MATH for all MATH . To each MATH we may associate MATH defined to be the smallest non-negative integer such that MATH . The function MATH is upper semicontinuous. Since MATH is compact MATH is finite. Let MATH. If MATH then MATH for some MATH so MATH . Hence MATH . Since MATH and MATH it follows that MATH . The fact that MATH follows from REF . If MATH then we're done. If MATH then MATH is a subinterval of MATH . Note that MATH is a NAME decomposition for the map MATH with attractor MATH. Since MATH has one fewer elements than MATH our induction hypothesis implies there is an attracting neighborhood MATH for MATH . Finally, we remark that by REF we may assume that MATH. |
math/9910171 | The sets MATH with MATH can immediately be seen to satisfy the properties of the definition of NAME decomposition REF with one exception. If MATH with MATH and MATH is a solution for MATH then it is not immediate that MATH for some MATH . However, the proof of REF did not make use of this property so the conclusion of this proposition is valid for MATH. In particular if MATH is an attracting interval then MATH is an attractor. We now consider MATH with MATH and let MATH be a solution for MATH. Since MATH and MATH has a NAME decomposition given by MATH it must be the case that either MATH is a subset of a single MATH or MATH contains points of MATH and MATH with MATH . But if the attracting interval MATH is chosen to contain one of MATH and MATH and not the other then the existence of the attracting neighborhood MATH makes it impossible for MATH to contain points of both MATH and MATH. Thus MATH for some MATH. |
math/9910171 | Without loss of generality we may assume that the hypotheses of REF hold, for if not, then by REF we may simply consider the pointed space for MATH and the associated NAME decomposition there. Moreover, since any interval is also an interval for a total ordering we may assume MATH is a total order. In this order there are attracting intervals MATH and MATH such that MATH . Consider the attracting neighborhood MATH for MATH given by REF and the map MATH . NAME decomposition is given by MATH. Another application of REF implies that MATH is an attractor with attracting neighborhood MATH . Clearly, with respect to MATH the set MATH is the dual repeller to MATH and hence is closed. The set MATH is an isolating neighborhood of MATH . |
math/9910171 | From REF above it is clear that MATH whenever MATH and MATH are attracting intervals with MATH. It is then straightforward to check that MATH satisfies the requirements of REF . |
math/9910171 | We shall build the MATH inductively. The induction hypothesis is as follows. Suppose MATH is an attracting interval and that for each attracting interval MATH there is an attracting neighborhood MATH satisfying the conclusions of the lemma. Moreover, if MATH and MATH with MATH then MATH. For MATH take MATH. Suppose the induction holds for an attracting interval MATH and MATH is minimal in MATH. We will define MATH for all MATH in the attracting interval MATH. Without loss of generality we may assume that for each MATH we have MATH. For if not then we may replace each MATH by MATH for some MATH sufficiently large. By the induction hypothesis and the above remark, there exists an open neighborhood MATH of MATH such that MATH is disjoint from MATH for all MATH, MATH and that MATH is disjoint from MATH for all MATH. By REF there is an attracting neighborhood MATH for MATH. Define MATH and more generally for any attracting interval MATH . It is straightforward to check that this construction completes the inductive step. |
math/9910171 | Let MATH be a filtration pair for MATH and MATH. Let MATH denote the associated NAME decomposition of MATH. By REF there exists a collection MATH with the property that MATH and MATH. To obtain a NAME set filtration for MATH we simply take MATH where MATH corresponds to the NAME set MATH and MATH is the quotient map. It is clear that this collection forms a NAME set filtration. |
math/9910171 | Suppose MATH and MATH are shift equivalent with MATH and MATH. We claim that MATH is an isomorphism with inverse MATH. Because the diagram MATH commutes we see that MATH. Similarly, MATH. Thus, MATH and MATH are inverse isomorphisms. Now suppose that MATH and MATH are isomorphic objects in NAME 's category. In particular, suppose MATH and MATH are inverse isomorphisms. We have MATH and MATH. Thus, we have MATH, MATH and MATH such that the following diagrams commute. CASE: MATH . Let MATH, MATH, MATH and MATH. So defined, we have MATH, MATH, MATH and MATH. Thus MATH and MATH are shift equivalent. |
math/9910174 | We must show that MATH. This almost, but not quite, follows from CITE. There, it was shown that the polynomial MATH which would imply that MATH. Unfortunately, there was a small error in that calculation; namely, two terms were omitted from the calculation REF of the NAME polynomial of MATH in case MATH is of type MATH, associated to the NAME subdiagrams MATH and MATH of MATH. Once they are taken into account (the first contributes MATH, the second MATH), we see that MATH where MATH and MATH equal MATH or MATH. This proves the theorem. |
math/9910177 | Let MATH be a unit map. The main reason why the above homotopy equivalences hold is that the elements MATH are in the image of the homomorphism MATH. Indeed, consider first the spectrum MATH. Let MATH be a map representing MATH. We obtain the cofibration: MATH . Then the composition MATH represents MATH. Let MATH be the map MATH where MATH is a multiplication. Note that the diagram MATH commutes since the map MATH represents a unit of the ring spectrum MATH. We obtain a commutative diagram of cofibrations: MATH where MATH gives a homotopy equivalence by REF - lemma. The proof for the spectrum MATH is similar. Consider the spectrum MATH. First we note that the bordism theory MATH coincides with the theory MATH, where the order of singularities is switched. In particular, the spectrum MATH is a cofiber in the following cofibration: MATH . Here the map MATH is defined as follows. Let MATH be a map of degree REF. Then the composition MATH represents MATH. The spectrum MATH is a module (say, left) spectrum over MATH, that is, there is a map MATH so that the diagram MATH commutes. Then the map MATH is defined as composition: MATH . Note that the diagram MATH commutes since MATH represents a unit, and MATH is a left module over the ring spectrum MATH. We obtain the commutative diagram of cofibrations: MATH . The map MATH gives a desired homotopy equivalence. Thus we have MATH. |
math/9910177 | Recall that for each singularity manifold MATH there is an obstruction manifold MATH with singularity. In the cases of interest, we have: MATH, which is non-trivial; and the obstruction MATH, and MATH. Thus CITE implies that there is no admissible product structure in the cobordism theory MATH, so the spectrum MATH does not admit an admissible product structure. The obstruction element MATH, and since MATH is even, the obstruction manifold MATH is, in fact, a manifold without any singularities (see CITE), so the element MATH is in the image MATH. However, the elements of MATH are divisible by MATH, so they are zero in the group MATH, and, consequently, in MATH. The result of CITE implies that the spectra MATH, MATH and MATH have admissible product structures MATH and MATH respectively. It is also well-known CITE that the element MATH is an obstruction to the commutativity of the product structure MATH. An obstruction to the commutativity for the product structure MATH lives in the group MATH. The obstructions to associativity are REF - torsion elements, (see CITE) so they all are zero. |
math/9910177 | In the case MATH, MATH, the splitting shows that MATH. The map MATH (for any partitions MATH) may be lifted to the MATH - connective cover of MATH, as it is shown in CITE. Let MATH be a unit map, and MATH be a left inverse of MATH. The composition MATH is null-homotopic for MATH. Let MATH be a generator. It is well-known that the image of the map MATH on positive dimensional homotopy groups is MATH. It implies that MATH and MATH for all partitions MATH. Since the unit map MATH is a map of ring spectra, we have MATH, so the elements MATH are even for all partitions MATH. In particular, MATH is even for all MATH. Let MATH be the NAME class corresponding to a partition MATH. NAME, NAME and NAME show that the NAME character MATH, for MATH. It implies that MATH are even elements for all MATH. The NAME classes MATH and MATH determine the oriented cobordism ring MATH in dimension REF, so the NAME element goes to an even element in MATH under the natural map MATH. Thus the composition MATH takes the NAME element MATH to zero. |
math/9910177 | Recall that the NAME map MATH induces a multiplication by the NAME element in MATH and connected covers MATH and MATH. Let, as above, MATH. We apply the NAME REF MATH to obtain the formula: MATH . Let MATH be a space, MATH, MATH be the NAME element. Then MATH. The MATH - characteristic numbers MATH commutes with a multiplication by the NAME element, that is, MATH. The NAME REF gives: MATH . Here MATH, MATH, and MATH by REF . We note that MATH and MATH since the cobordism theory MATH has an admissible product structure by REF . Let MATH be a partition, and MATH. The map MATH lifts to connective cover: MATH. Let MATH be a map representing the NAME element MATH. We denote by MATH the composition MATH . Note that the diagram MATH commutes since MATH is a module over MATH. Let MATH be a partition, so that MATH. The following diagrams commute: MATH A commutativity of the first diagram follows from REF . Recall that a projection of the NAME element into the homotopy group of MATH is zero. Let MATH be a finite spectrum. The map MATH in homotopy coincides with the homomorphism in mod REF homology groups MATH and is trivial for any space MATH since MATH has the NAME filtration REF. It implies that MATH is a trivial map. A commutativity of REF now follows. To complete the proof of REF we notice that REF give the commutative diagram MATH where the map MATH exists since the both rows are cofibrations. The five-lemma implies that MATH is a homotopy equivalence. |
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