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math/9909083 | This Proposition summarizes our previous analysis: the part of the spectrum of MATH which is outside of the disk of radius MATH around MATH is contained in the half plane MATH for MATH small enough, according to REF of MATH. Therefore, the change of stability is equivalent to the change of sign of MATH given by REF implies that MATH vanishes for some value MATH of MATH, whose asymptotic is given by REF. The results on the multiplicity of MATH will be clear provided that we show that MATH does not vanish; but REF together with REF proves that this is the case. This concludes the proof of our proposition. |
math/9909083 | Let MATH . It is equivalent to find a fixed point of MATH for MATH and to find a solution of MATH. Introducing MATH enables us to start from MATH, where we have MATH. The implicit function Theorem in its classical formulation would give us the existence of an interval of MATH on which we can find a solution of MATH, but it would not ensure that we have a solution up to MATH. The purpose of our estimates is to show that we can go as far as MATH. Let us determine MATH such that MATH is a contraction of ratio MATH for MATH and MATH: MATH . Therefore, we have the inequality MATH . Thus, the first condition that we wish to impose is MATH . The second step is to impose a condition on MATH and on MATH such that MATH maps the ball of center MATH and radius MATH into itself, for MATH. We have MATH . Therefore, it is enough to require that MATH . Thanks to the strict contraction Theorem, there exists for MATH and for MATH a unique solution of MATH . Let us denote this solution by MATH. We may estimate MATH as follows: we have the inequality MATH . Therefore, MATH . But we can obtain a much better estimate, since MATH . Therefore, MATH which concludes the proof. |
math/9909089 | It is clear that REF implies REF . If REF is true then MATH and MATH fit together to form a tableau with MATH in the top MATH rows and MATH below. By taking a horizontal cut through this tableau, we see that it must be the product of MATH and MATH. But then it is equal to MATH and REF follows. Finally, suppose REF is true. When the boxes of MATH are column bumped into MATH to form the product MATH, all of these boxes must then stay below the MATH row. This process therefore reconstructs MATH below MATH and REF follows. The statements about vertical cuts are proved similarly. |
math/9909089 | Let MATH be the vertical cut through MATH after column MATH, and put MATH. Then let MATH be the horizontal cut through MATH after row MATH, and put MATH. We claim that the pair MATH fits around MATH. Using REF and that the entries of MATH are smaller than all other entries, it is enough to prove that the MATH entry in the top row of MATH is strictly larger than the MATH entry in the bottom row of MATH. This will follow if the MATH entry in the top row of MATH is larger than or equal to the MATH entry in the top row of MATH. Since MATH and MATH has at most MATH columns, this follows from an easy induction on the number of rows of MATH. It follows from the claim that MATH is the canonical factorization of MATH, and therefore we have MATH as required. |
math/9909089 | The first statement follows from the observation that the bottom rows of MATH can't influence the top part of MATH, which is a consequence of the row bumping algorithm. REF then shows that the factorization MATH is a horizontal cut, so MATH as required. |
math/9909089 | We will first consider the case where the line segment corresponding to MATH goes up. Let MATH be the path under MATH that cuts short this line segment and its successor. MATH . Then by REF is a factor sequence for MATH, which means that MATH is a factor sequence for MATH. In general MATH lies over a path like the one above, and the general case follows from this. |
math/9909089 | Since MATH is a factor sequence for MATH, there exists a factorization MATH such that MATH is a factor sequence for MATH. By REF we have MATH. It is therefore enough to show that if MATH then MATH is a factor sequence for MATH. Let MATH be the number of rows in (the rectangle corresponding to) MATH, and let MATH be the horizontal cut through MATH after the MATH row. We will do the case where a factor of MATH is moved to MATH, the other case is proved using a symmetric argument. We then have a factorization MATH such that MATH and MATH. We can assume that the paths MATH and MATH go down after they meet, and that the original factor sequence for MATH is MATH. MATH . Put MATH. Then MATH is a factor sequence for the path with these labels in the picture. Now let MATH be the rectangular tableau associated to the lower triangle, and let MATH be the canonical factorization of MATH. Since this is a simple factorization we may assume by induction that MATH is a factor sequence. Using REF we deduce that MATH for some tableau MATH, such that MATH. Since MATH is a factor sequence, so is MATH by REF . This means that MATH is a factor sequence, which in turn implies that MATH is a factor sequence for MATH as required. |
math/9909089 | The ``if" implication follows from the definition. If the sequence MATH is a factor sequence, then MATH applications of REF shows that MATH is a factor sequence for the path with these labels. MATH . It follows that MATH and MATH are empty, and MATH is a factor sequence for the bottom MATH rows. This proves ``only if". |
math/9909089 | Let MATH and MATH be the lengths of the top and bottom rows of MATH. The requirement MATH then implies that MATH must have two rows with MATH boxes in the top row and MATH in the bottom row. Now it follows from the NAME formula CITE that the product MATH of the rows in MATH has at most two rows. Furthermore, since MATH contains a violation, the second row of MATH has at most MATH boxes. Using the NAME formula again, this implies that there is exactly one way to factorize MATH into a row of length MATH times another of length MATH. This establishes the existence and uniqueness of MATH. Explicitly, one may use the inverse row bumping algorithm to obtain this factorization of MATH. This is done by bumping out a horizontal strip of MATH boxes which includes all boxes in the second row, working from right to left. Let MATH be the leftmost violation of MATH, where MATH has the form: MATH . Suppose the parts MATH and MATH each contain MATH boxes. Now form the product MATH and let MATH and MATH be the boxes of this product as in the picture: MATH . Since MATH is a violation in MATH, it must be smaller than all boxes in MATH and MATH. Therefore we have MATH . Now since each MATH it follows that if a horizontal strip of length MATH is bumped off this tableau, MATH will remain where it is. In other words we can factor MATH into MATH such that MATH and MATH are rows of lengths MATH and MATH respectively. Since the entries of MATH and MATH are no larger than MATH, the products MATH and MATH are rows of lengths MATH and MATH. But the product of these rows is MATH, so they must be the rows of MATH by the uniqueness. This proves that MATH has the stated properties. |
math/9909089 | Given a pair MATH, let MATH be the finite set of all non-tableau diagrams MATH with weakly increasing rows, such that MATH and MATH, and so that the minimal violation in MATH is in row MATH. The pair MATH is then identified with one of the diagrams in this set. We will describe an involution of the set MATH and another of the complement of MATH in MATH. The restriction of NAME 's involution to MATH is then obtained by applying the involution principle of CITE to these involutions. The involution of MATH simply consists of doing an exchange operation between the rows MATH and MATH of a diagram. This is possible because all diagrams are required to have a violation in row MATH. Now note that a diagram MATH is in the complement of MATH if and only if MATH has a violation outside the MATH row. We take the involution of MATH to be an exchange operation between the row of the minimal violation outside row MATH, and the row above this violation. This is indeed an involution since the minimal violation outside row MATH stays the same. These involutions now combine to give an involution of MATH by the involution principle. To carry it out, start by forming the diagram with MATH in the top MATH rows and MATH below it. Then do an exchange operation between row MATH and row MATH. If all violations in the resulting diagram are in row MATH we are done. MATH is then the top MATH rows of this diagram and MATH is the rest. Otherwise we continue by doing an exchange operation between the row of the minimal violation outside row MATH and the row above it, followed by another exchange operation between row MATH and row MATH. We continue in this way until all violations are in row MATH. Finally, the properties of MATH and MATH follow from the properties of exchange operations. In particular, the requirement MATH follows because we always carry out an odd number of exchange operations. |
math/9909089 | If MATH are diagrams with weakly increasing rows, for example, tableaux, we will write MATH. With this notation we must prove that if MATH is a path of length MATH, then MATH where the sum is over all factor sequences MATH for MATH. Let MATH be a path under MATH as in REF or REF . By induction we can assume that Conjecture REFA is true for MATH, that is, MATH where this sum is over the factor sequences for MATH. We must prove that the right hand side of REF is obtained by replacing each basis element of REF in the way prescribed by the definition of MATH. If we are in REF then this follows from the NAME rule CITE: If MATH is a tableau of shape MATH and MATH and MATH are partitions, then there are MATH ways to factor MATH into a product MATH such that MATH has shape MATH and MATH has shape MATH. Assume we are in REF . By induction we then have MATH where the sum is over all factor sequences MATH for MATH; MATH and MATH are the labels of the two line segments where MATH is lower than MATH. Let MATH be the rectangular tableau of the corresponding triangle, and let MATH be the number of rows in its rectangle. Then by definition we get MATH where the sum is over all factor sequences MATH for MATH such that MATH has at most MATH rows. Now suppose we have a factor sequence MATH such that the diagram MATH is a tableau. Then this tableau must be the product MATH, and so MATH is a factor sequence for MATH. Thus the term MATH matches one of the terms of REF . On the other hand it follows from REF that every term of REF is matched in this way. We conclude from this that the terms in REF is the subset of the terms in REF which come from factor sequences such that MATH fits around MATH. We claim that the sum of the remaining terms in REF is zero. In fact, if MATH is a factor sequence for MATH such that MATH doesn't fit around MATH and MATH, then we may apply NAME 's involution in the way described above to get tableaux MATH and MATH. If Conjecture REF is true, then the sequence MATH is also a factor sequence, and since MATH, the terms of REF given by these two factor sequences cancel each other out. |
math/9909089 | Let MATH be the tableau corresponding to MATH, and suppose MATH is a factor sequence for MATH. Since all tableau on the line going south-west from MATH in the tableau diagram are narrower than MATH, it follows that also MATH has fewer columns than MATH. Similarly MATH has fewer rows than MATH. But this means that MATH fits around MATH and the statement of Conjecture REF is trivially true. |
math/9909089 | It is clear that REF implies REF . For the other implication, put MATH and let MATH be the canonical factorization of MATH. Then it follows from REF that MATH is a factor sequence for MATH. Since MATH by REF , we may assume that MATH. We will handle the case where a factor of the bottom part of MATH is moved to MATH, the other case being symmetric. This means that for some tableau MATH we have MATH and MATH. Since the bottom part of MATH has entries only from MATH, this is also true for MATH. We may assume that MATH and MATH go down outside the displayed angle and that our factor sequence is MATH. MATH . Then by definition there exists a factorization MATH such that MATH and MATH. Since MATH consists of MATH with MATH attached on its right side, and since all entries of MATH are strictly larger than the entries of MATH, it follows that MATH consists of MATH with some tableau MATH attached on the right side. Furthermore MATH by REF . Put MATH. Then we have MATH, MATH, and MATH. It follows that MATH is a factor sequence as required. |
math/9909089 | Let MATH be a factor sequence for MATH which satisfies the conditions in Conjecture REF, and let MATH and MATH be the tableaux obtained from MATH and MATH using NAME 's involution. Since the part of MATH that is wider than MATH has entries only from MATH, the same will be true for MATH by REF . Since MATH has fewer rows than MATH and since MATH is a factor sequence for MATH, it follows from REF that MATH is a factor sequence for MATH. |
math/9909089 | When the rectangle diagram satisfy these properties, then all instances of Conjecture REF follow from either REF or REF. The corollary therefore follows from REF. |
math/9909098 | For each positive even integer MATH, we define an operation MATH on NAME as follows. Let MATH be an NAME. Then MATH where MATH if MATH is even, and MATH otherwise. Then MATH is again an NAME. There exist constants MATH and MATH such that MATH . Let MATH. Then MATH . Now MATH . So MATH . It follows that MATH . Thus MATH where MATH and MATH . We now proceed with the proof of REF . Assume that the congruence ABC conjecture for MATH is true. Then there exists MATH such that MATH for all MATH with MATH. Let MATH. If MATH, REF is trivial. We may therefore assume that MATH is even. Let MATH. Then MATH . Suppose MATH is an odd prime dividing MATH, and let MATH be the largest power of MATH dividing MATH. Then MATH. In particular, MATH. If MATH divides MATH or MATH, then MATH. If MATH divides neither MATH nor MATH, then MATH and MATH are congruent mod MATH; therefore, MATH. Now let MATH be the largest power of MATH dividing MATH. If MATH is even, then so is MATH, and exactly one of MATH and MATH is a multiple of MATH. Thus, one of MATH and MATH is a multiple of MATH. Since, in this case, MATH and MATH, one of MATH and MATH is a multiple of MATH, whence also of MATH. If, on the other hand, MATH is odd, then so is MATH. Then MATH and MATH are both congruent to MATH mod MATH, so MATH. For any MATH, it now follows that MATH for any NAME MATH. Since the right hand side depends only on MATH and MATH, this proves the full ABC conjecture. |
math/9909099 | We begin with the proof that REF are equivalent following CITE and CITE. One computes the first variation of the discrete action MATH with variations that vanish on the set MATH. Thus, MATH where we have used the discrete analogue of integration by parts which simply shifts the sequence MATH where MATH. Since for each MATH, the variations MATH are arbitrary, this establishes the DEL algorithm. We remark that choosing variations which do not vanish at MATH and MATH defines two MATH-forms whose exterior derivative is the unique symplectic MATH-form given in REF . To see that REF is equivalent to REF , notice that since MATH, MATH . Now for REF , we compute MATH and find that MATH where again we have used discrete integration by parts shifting the sequence MATH with MATH, and the fact that MATH. Defining MATH, we obtain the discrete NAME REF for all variations of this form. |
math/9909099 | REF guarantees that the DEL algorithm preserves the symplectic structure MATH on MATH; hence, by REF , the DEL algorithm preserves the NAME structure on MATH. Since the action of MATH on MATH is proper, the general NAME reduction theorem CITE states that the projection MATH is a NAME map. By REF , the projection of the DEL algorithm, MATH is equivalent to the DEP algorithm on G, MATH. Therefore, as the NAME structure on MATH is induced by MATH and as MATH is NAME, we have proven the theorem. |
math/9909099 | The first assertion follows by construction. For the second assertion, using the definition MATH, the DEL algorithm can be reconstructed from DEP algorithm by MATH where MATH is the solution of REF . Indeed, MATH is precisely MATH. Thus, at each increment, one need only compute MATH since MATH is already known. Similarly one shows that in the case of a left MATH action, the reconstruction of the DEP REF is given by MATH . |
math/9909099 | We fix the right action and consider MATH for some MATH. By construction, MATH, whenever MATH, so that the chart is given by MATH. By defintion, both MATH and MATH are always elements of a neighborhood of MATH, so it is clear that they are right invariant. Hence, using the explicit form of the chart MATH together with the right invariance of the Lagrangian MATH, we obtain from REF that MATH . In the case that the group action is on the left, we use MATH as the chart, and proceed with the same argument. |
math/9909099 | Comparing the definitions of MATH and MATH, we see that MATH. Similarly, comparing the definitions of MATH and MATH we conclude that MATH and MATH. Hence, the first equation in REF is precisely the DLP REF . Substituting the second equation of REF into the first results in the following expression: MATH which is precisely the DEP REF when the above identifications are invoked. |
math/9909100 | The proof proceeds in three steps. We begin by computing the first variation using REF . Then we show that the boundary term yields the NAME form. Lastly, we verify the statements related to the interior integral. Choose MATH small enough so that it is contained in a coordinate chart. If in these coordinates MATH, then along MATH, the coordinate expressions for MATH are written as MATH . Using the NAME formula we first compute the second term on the right-hand-side of REF : MATH . Using REF , and the local expression for the vertical vector field MATH, we have that MATH . In the following, we shall use MATH for the formal partial derivative of a function MATH (see REF ), and MATH will denote MATH. Integrating REF by parts, we obtain that MATH . Using the fact MATH, applying the Stoke's formula MATH, and combining the last calculation with REF , we obtain MATH . A form MATH on MATH is contact, if MATH for all MATH. For a smooth Lagrangian density MATH there exist a unique differential form MATH defined by REF such that the boundary integral in REF is equal to MATH . Furthermore, MATH can be written as a sum of MATH and a linear combination of a system of contact forms on MATH with coefficients being functions on MATH. Proof of REF Let MATH be an arbitrary vector field on MATH, and let MATH, a map from MATH to MATH. Then one computes MATH . Using the formula MATH one finds that MATH . Similarly, MATH . Thus if we let MATH, use REF , and recall the local REF for MATH, we obtain that MATH . Next, observe that MATH. Also, MATH . These observations together with the previous identities imply the following important formulas MATH . Substituting these formulas into the boundary integral of the variational principle REF , we obtain that MATH . This proves the existence of a unique differential form MATH and demonstrates how this form naturally arises in the boundary integral of the variational principle. Integration by parts yields the boundary integral with terms that involve partial derivatives of MATH of all orders up to MATH (in our case MATH). REF show that each partial derivative of MATH has an associated MATH-form on MATH, and substitution of these forms yields a unique differential MATH-form as desired. Since MATH and its partial derivatives are functions on MATH, then by REF , MATH is a function on MATH, and therefore MATH is a MATH-form on MATH. It is easy to show that MATH for all integers MATH and for all MATH. Therefore, MATH and MATH are contact forms on MATH. Hence the last statement of the lemma follows. MATH A simple computation then verifies that MATH is the NAME form so that MATH . Next, consider the interior integral of the variational principle REF . Since MATH for all integers MATH, we obtain that MATH where MATH is the variational derivative of MATH in the direction MATH (see REF ). Since MATH is a function of second order by hypothesis, then its variational derivative is a function on MATH. Therefore the form MATH is a MATH-form on MATH. Moreover, the integrand in REF written as MATH defines a unique smooth section MATH as desired in the statement of the theorem. Now we shall prove the following The forms MATH and MATH satisfy the following relationship: MATH for all MATH and all vectors MATH. Furthermore, a necessary condition for MATH to be an extremum of the action functional MATH is that MATH for all vector fields MATH on MATH, which is equivalent to MATH for all vector fields MATH on MATH. Proof of REF . The proof will involve some lengthy computations that we partially present below. To compute MATH, let us write MATH as MATH . Then, for MATH, we obtain MATH . The last step is to pull back MATH by MATH; this eliminates the terms with the contact forms. In addition, using the fact that the pull back commutes with the exterior derivative, and applying formulas such as REF , we obtain that MATH . Some cancellation and further rearrangement yields MATH . Letting MATH, we have that MATH where the right hand side equals MATH by REF . Hence, the relation REF is proved. A necessary condition for MATH to be an extremum of the action functional MATH is that the interior integral in REF vanish for all vectors MATH. From the calculation above, one may readily see that it is equivalent to REF . Now if we let MATH be a vector field on MATH, then MATH . Hence, MATH . Thus, the condition MATH for all vector fields MATH on MATH is equivalent to REF . This completes the proof of the lemma. MATH REF contains two equivalent conditions for MATH to be extremal. Both conditions yield the same coordinate expression of the NAME equations given by MATH which is the final statement of the theorem. |
math/9909100 | We follow REF. Define the MATH-forms MATH and MATH on MATH by MATH and MATH so that by REF , MATH . Furthermore, MATH . Since MATH, we have that MATH . Given vectors MATH we may extend them to vector fields MATH on MATH by fixing vector fields MATH on MATH such that MATH and MATH, and letting MATH and MATH. If MATH covering MATH is the flow of MATH, then MATH is the flow of MATH. Notice that MATH and MATH, hence REF becomes MATH . Recall that for any MATH-form MATH on MATH and vector fields MATH on MATH, MATH . Also recall that for a vector field MATH on MATH and a function MATH on MATH, MATH. We now use the latter and REF on MATH. We have that MATH . Similarly, MATH . Let MATH and MATH be a curve in MATH through MATH such that MATH . Now we restrict MATH to MATH. We shall give a detailed computation of the first term on the right hand side of REF . We have that MATH where the last equality was obtained using NAME 's formula. We have also used the fact that MATH and MATH have the same MATH-th prolongation. Furthermore, using Stoke's theorem, noting that MATH is empty, and applying NAME 's formula once again to MATH, we obtain that MATH . Similarly, MATH . Now, MATH; hence, MATH . Recall that for a differential form MATH on a manifold MATH and for vector fields MATH on MATH, MATH . Therefore, MATH where we have again used Stoke's theorem and NAME 's formula twice. Substituting REF , and REF into REF , we obtain that MATH . We now compute REF . Similar computations as above yield MATH which vanishes for all MATH and MATH. Similarly, MATH for all MATH and MATH. Finally, MATH for all MATH. Therefore, the equation REF vanishes for all MATH and MATH. Using the latter and REF becomes MATH for all MATH and all MATH, as desired. |
math/9909100 | Consider two MATH-vertical vectors MATH and MATH in MATH. Then MATH and MATH are vertical-over-MATH vector fields on MATH whose components we shall denote via MATH or just numerate by MATH. Thinking of the components of the transformation MATH as functions on MATH, we immediately see that MATH . Using REF we express MATH as: MATH . Next, combining with REF we obtain that MATH so that MATH . The integral on the right-hand-side of the above equation leads us to introduce two degenerate skew- symmetric matrices MATH, MATH on MATH: MATH . To each matrix MATH, we associate the MATH-form MATH on MATH given by MATH, where MATH. With the definition of MATH and the use of REF , the multisymplectic form REF becomes, for MATH, MATH . Hence by the Stoke's theorem, MATH . Since MATH is arbitrary, we obtain the desired conservation law REF . In the special case, when the components MATH and MATH, one may verify that MATH so that, letting MATH denote an element MATH, REF takes the special form MATH which is the complete analogue of NAME 's conservation law REF for the wave equation. Next, since the inner product MATH is independent of MATH, the Hamiltonian system of REF on the multi-symplectic structure MATH may be written as MATH which results in MATH . With the choice of the Hamiltonian REF the last four equations yield identities, and the first equation becomes MATH . Using the NAME transformation REF for MATH, the latter equation recovers the NAME REF , hence REF . In other words, the NAME equations on MATH are equivalent to NAME 's equations on the multi- symplectic structure MATH. |
math/9909104 | Since the assertion is claimed for every polynomial, it suffices to prove that the expectation MATH converges. To show convergence of expectation we may let MATH be a monomial. Indeed the monomial MATH will do, since some of the factors may be equal. Expanding the expression MATH using the definition of MATH, it has a term for each function MATH from MATH to MATH: MATH . Let MATH be an individual term in this expansion. Since we are computing the expectation with respect to the product state, we can arrange the factors with respect to MATH rather than MATH: MATH . In this form it is clear that MATH if there is a MATH such that MATH has one element. At the same time, if MATH is the set of those functions MATH whose images have fewer than MATH elements, then MATH because MATH . In other words, there are sufficiently few such functions MATH that they are negligible in the limit. What is left is the set of functions that are exactly REF-to-REF, which only exist when MATH is even. Thus if MATH is the set of perfect matchings of MATH, then MATH where MATH is the odd factorial function. In the limit MATH exactly matches the corresponding expectation MATH of the classical variables MATH. |
math/9909104 | We apply REF . First, we change MATH from a subscript to an argument in certain quantities that depend on it (and implicitly on MATH): MATH where dependence on MATH is implicit in the notation. Let MATH be the set of all centered partitions MATH (the weight lattice). Define a finite difference operator MATH by MATH so that MATH by REF . The two important properties of the operator MATH are first, that it is antisymmetric under the NAME group MATH after translation by MATH, and second, that it has degree MATH. The degree of MATH follows from a factorization that appears in proofs of the NAME dimension formula CITE: MATH where MATH is the set of positive roots of MATH and MATH . Each MATH has degree REF and there are MATH of them. Note that the only antisymmetric polynomial of degree MATH is MATH . When MATH is large, MATH because in the limit MATH becomes an antisymmetric differential operator of degree MATH. When applied to a symmetric Gaussian, it produces an anti-symmetric polynomial factor of degree MATH. The polynomial MATH is the only choice for this factor up to scale. Thus the operator MATH explains one factor of MATH in the statement of the corollary. The other factor is given by MATH, which is also proportional to MATH in the limit as MATH goes to MATH. REF then establishes the stated approximation for MATH, where MATH and MATH each contribute a factor of MATH. |
math/9909108 | MATH is an acyclic complex, because there is a contracting homotopy MATH, given as MATH. The left pentagon in the bow-tie diagram implies that the map MATH is an augmentation. Similarly, the right triangle implies that MATH is an augmentation as well. It is obvious that all the MATH are MATH-bimodule maps. |
math/9909108 | Clearly, MATH is injective. The fact that MATH is the map between cochain complexes follows from the right triangle in the bow-tie diagram. |
math/9909108 | The equivalence of the first two assertions is clear since MATH is the field so that MATH is a free MATH-bimodule. Suppose that REF holds, that is, there is a MATH-bimodule map MATH such that MATH. Define MATH and MATH. Then for all MATH, MATH one has: MATH . Applying MATH to both sides of this equality one immediately obtains that MATH is a MATH-cocycle. The normalisation of MATH follows from the equality MATH applied to MATH and the right triangle in the bow-tie diagram. Now suppose that REF holds. Denote MATH (summation understood), and define MATH, MATH. Then for all MATH, MATH, MATH where we used the left pentagon and the left triangle in the bow-tie diagram together with the normalisation of MATH. Therefore MATH splits MATH. Clearly, MATH is a left MATH-module map. Furthermore for all MATH, MATH we have: MATH where we used the right pentagon to derive the second equality, the left pentagon to derive the third one and finally the fact that MATH is a MATH-cocycle to obtain the fourth equality. This proves that MATH is a MATH-bimodule splitting as required. |
math/9909108 | Throughout the proof of this lemma, MATH is an arbitrary element of MATH. Notice that MATH. CASE: One easily finds that for all MATH, MATH so that the map MATH is well-defined. Consider the map MATH, given by MATH. Clearly, MATH is a left MATH-module map. Since MATH is a REF-cocycle, we have for all MATH, MATH . Therefore MATH is well-defined. For any MATH one easily finds that for all MATH, MATH. Using this fact one obtains MATH as well as MATH . Therefore MATH is the inverse of MATH, MATH. CASE: Suppose MATH for some MATH. Then MATH. The result then follows from the isomorphism MATH given by MATH, where MATH. |
math/9909108 | CASE: Denote the action of the translation map MATH by MATH (summation understood). Notice that for all MATH, MATH, MATH, MATH (compare CITE), and MATH, where we use the NAME notation for the coaction, MATH (summation understood). Consider the map MATH . The map MATH is well-defined because MATH and furthermore, MATH so that MATH is a zero-cocycle, hence belongs to the zero-cohomology group. Let MATH. We claim that the map MATH is the inverse of MATH. Firstly we need to check that MATH is in the centraliser of MATH in MATH. The key observation needed here is that MATH is a MATH-module, that is, for all MATH, MATH. In particular this implies that for all MATH, MATH, and for all MATH, MATH (compare CITE). Since MATH is a zero-cocycle we have for any MATH: MATH so that MATH as claimed. Furthermore, since MATH, we have for all REF-cocycles MATH, MATH. In particular we have MATH . Therefore MATH . CASE: For any MATH consider the map MATH, given by MATH . We will show that MATH is a contracting homotopy, that is, MATH. We use properties of the translation map listed above and the definitions of MATH and MATH to compute MATH . Therefore MATH is a contracting homotopy, so that for all MATH, MATH as claimed. |
math/9909108 | In general MATH . Now, CITE yields the assertion. |
math/9909108 | The contracting homotopy is MATH, MATH, while the augmentations are MATH and MATH. |
math/9909108 | The proof is dual to that of REF . The isomorphism in REF is MATH while the contracting homotopy in REF is MATH, MATH where MATH is the cotranslation map. |
math/9909108 | In general MATH . Now, CITE yields the assertion. |
math/9909108 | This follows from REF below, but it can also be proven by straightforward manipulations with the definition of an entwining structure. In particular, the associativity of MATH follows from the right pentagon in the bow-tie diagram, while to prove the derivation property of MATH one needs to use the definition of MATH and both pentagons in the bow-tie diagram. |
math/9909108 | This lemma follows from REF below too, but it can also be proven by a straightforward application of definitions of MATH and MATH, which uses both pentagons of the bow-tie diagram. |
math/9909108 | We prove this proposition for the operation MATH, the proof for MATH is analogous. Take any MATH, MATH and MATH. To prove the associativity of MATH we need to show that MATH equals to MATH. First using REF one easily finds that MATH, so that only the equality MATH needs to be shown. We have MATH where we used REF to derive the first, second and the fifth equalities, REF to derive the third equality and REF to obtain the fourth one. |
math/9909108 | The first part is a simple consequence of REF and easily verifiable fact that MATH. The second part can be proven by direct computation. We display it for the cup product MATH. Explicitly, for any MATH, MATH one needs to show that MATH . Using definitions of MATH, MATH and MATH this amounts to showing that MATH is equal to MATH . The first term in MATH equals the first term in MATH by the same calculation as in the proof of REF . Using a chain of arguments as in the proof of REF but without the fourth step, one easily shows that the second term in MATH is the same as the second term in MATH. Again, a part of the argument in the proof of REF allows one to transform the first term inside the final brackets in MATH to the form MATH and then REF implies that the last line in MATH vanishes. Now consider the third term in MATH. If MATH then REF imply that MATH so that the part of the sum in the last term in MATH for MATH is the same as the third term in MATH. Similarly, REF implies that the remaining part of this sum is the same as the fourth term in MATH. This completes the proof that MATH is a derivation in the algebra MATH. Similar arguments show that MATH is a derivation in the algebra MATH. |
math/9909108 | Expand the left hand side of the above equality using definition of MATH in REF . This produces six terms, four of which cancel because of REF . One is left to show that MATH . By REF , MATH therefore the required equality holds once the definition of cup products in REF is taken into account. |
math/9909108 | If MATH and MATH are cocycles one has MATH and the corollary follows. |
math/9909108 | REF is clearly satisfied. REF can be proven by a straightforward calculation which uses the right pentagon in the bow-tie diagram. Next, notice that for any MATH take MATH and compute MATH where we used REF to derive the third equality. This proves REF with MATH. The case MATH is proven in a similar way. The proof of condition REF is again straightforward. |
math/9909108 | The condition for MATH to be in MATH can be explicitly written for all MATH, MATH . If MATH this is precisely the left pentagon in the bow-tie diagram. Therefore MATH. We will show that for all MATH, MATH, MATH. Take any MATH, MATH and compute MATH where the right pentagon in the bow-tie diagram has been used in derivation of the second and fourth equalities, and REF for MATH and MATH in derivation of the third and fifth equalities. |
math/9909108 | Apply MATH to REF and then use the definitions of the canonical entwining structure and the translation map to obtain MATH where MATH. |
math/9909112 | We choose in MATH an open convex cone MATH and MATH affine independent vectors in MATH such that in this basis the components of each vector in MATH have strictly positive imaginary parts. Let MATH converge in MATH to the tempered distribution MATH. This means that we can choose MATH positive numbers MATH such that for MATH and any MATH the relation MATH holds. We set MATH and write this in what follows also as MATH . Since MATH can be chosen in such a way that MATH, we obtain a continuous map MATH from the compactum MATH to MATH. The image of this map is also compact and hence bounded. One can then infer the existence of a seminorm MATH in MATH such that all MATH with MATH are uniformly continuous with respect to this norm, that is, there are is a constant MATH and a positive number MATH such that for each MATH and MATH the inequality MATH is fulfilled. In this way the distributions MATH can be extended to linear functionals on the NAME space whose topology is induced by the seminorm MATH. So the MATH can be considered as elements of MATH. Since MATH is an analytic function in MATH, MATH is also analytic in MATH and by NAME 's integral formula we obtain for MATH the representation MATH where each MATH is a closed path in the strip MATH around MATH respectively. This representation is independent of the chosen paths and we may thus take them as borders of the rectangles MATH. If we now let the rectangles approach the strips in which they are contained, the integrals along the borders MATH disappear and we obtain for MATH the expression MATH where MATH and the MATH are equal to MATH or equal to MATH respectively and the sign in the sum over MATH depends on the orientation of the corresponding borders. We put MATH . The functions MATH as well as their derivatives up to order MATH decrease at infinity faster than any polynomial MATH. On the other hand MATH is a continuous map for MATH into the NAME space MATH and MATH remains in a compact set for MATH and fixed MATH. For MATH we have therefore MATH . We obtain for MATH in MATH the expression MATH where MATH is equal to MATH or zero. As a consequence of the continuity REF it follows for MATH the estimate MATH where we have made use of MATH . Differentiation yields for the sum over MATH so that the estimate REF now reads MATH . The MATH are equal zero or equal MATH and we have MATH. Since MATH and the numbers MATH are arbitrary, we obtain for each compact MATH in MATH an estimate of the form MATH for a positive integer MATH and a constant MATH. |
math/9909112 | It can be shown that the sets MATH are given by MATH where MATH and MATH . Since MATH is given as the set of zeros of the analytic function MATH it is a complex submanifold of MATH and since MATH is open in MATH, so is MATH as well. MATH acts transitively on MATH and the isotropy group of a point MATH is given by MATH, so that MATH . MATH is thus the orbit of a point MATH under complex velocity transforms and we can write each MATH in MATH as MATH where the MATH are complex velocity transforms in MATH - direction with MATH and MATH is a fixed point in MATH. We can thus interpret the functions MATH as functions of MATH and write MATH. Since every complex velocity transform can be obtained from any other by adjoining the latter with a homogeneous NAME transform, the analycity of the MATH follows by REF . |
math/9909112 | The element MATH-in MATH is a tempered distribution MATH which depends on the parameters MATH; for MATH we have thus MATH where MATH. Now MATH is strongly continuous for MATH as well as for corresponding values of MATH; in particular we obtain MATH since strong convergence implies weak convergence we have for all elements MATH of MATH . Hence it follows that MATH and by REF we obtain for MATH in MATH an estimate of the form MATH where MATH is a compact set in MATH. Continuing MATH, analytically to MATH in such a way that MATH also has a bound of the given form with respect to MATH we obtain for MATH in all MATH an estimate of the form REF where MATH is now a compact set in MATH. |
math/9909114 | It follows immediately from REF . |
math/9909114 | It follows immediately from the fact that any involutive basis is a NAME basis REF . |
math/9909114 | Let MATH and MATH be a pair of monomial sets associated with the principal and parametric derivatives of a differential indeterminate in MATH. The complementary set MATH can be written as a disjoint union CITE MATH where MATH is the dimension of monomial ideal MATH, MATH is a finite set, and every MATH is a finite disjoint union MATH with MATH . For every MATH we shall take MATH in REF . Thus, for MATH the decomposition REF MATH holds trivially. If MATH we consider the finite set MATH where monomials MATH generate MATH in accordance with REF . We claim that elements in set REF , and the decompositions REF they determine can be written such that the union in MATH with MATH is disjoint in accordance with REF . To prove the claim we define the degree MATH of set MATH as MATH, and choose all the monomials MATH generating MATH in REF such that MATH. Obviously this can always be done by appropriate choice of MATH. Let now MATH be the set MATH, and let MATH be a finite MATH-autoreduced completion of MATH. The existence of MATH is guaranteed by noetherity of MATH. Now consider the set MATH. Its MATH-involutivity and REF of MATH in REF imply MATH . Thus, we obtain the desired decomposition MATH. Disjointedness of this union follows from that in REF of MATH autoreduction. This proves the claim and the lemma. |
math/9909114 | NAME of MATH with respect to an orderly ranking implies that the associated complementary monomial set contains all the monomials associated with the parametric derivatives. This statement is an immediate consequence of the well-known fact CITE that for a graded monomial ordering the NAME function of a polynomial ideal is defined by the monomial ideal generated by the leading monomials of a NAME basis of the polynomial ideal. Furthermore, by association REF , the decomposition REF yields that every parametric derivative associated with a monomial in MATH is produced by differentiation of the uniquely defined parametric derivative (MATH-generator) with respect to its multipliers. Assigning the fixed values to all these parametric derivatives is obviously equivalent to fixing some function of the multipliers. Therefore, given initial point MATH, in addition to the set of arbitrary constants which associated with elements in MATH, all the parametric arbitrariness is determined by functions corresponding to the MATH-generators and which are arbitrary functions of the multipliers at the fixed values of nonmultipliers from coordinates of the initial point. |
math/9909114 | This is identical to the existence proof in NAME theory CITE (see also CITE). |
math/9909115 | Note that each of the three operations described in REF shrinks MATH (remember REF(MATH)). |
math/9909115 | Straightforward (and the proof of the first part is essentially the same as that of CITE; compare the proof of REF). |
math/9909115 | Straightforward. |
math/9909115 | CASE: Let MATH be equipped with the product topology (of countably many countable discrete spaces). So MATH is a Polish space and it should be clear that MATH, MATH are its NAME subsets. To express MATH we have to say that there is a sequence MATH of elements of MATH such that MATH is obtained from MATH by one of the operations described in REF. Each of these operations corresponds to a NAME subset of MATH, so easily MATH, MATH are MATH subsets of MATH. The main difficulty is to show that the incompatibility relation MATH is a MATH subset of MATH. But this follows from the following observation (note that this is the place where we use the assumption that MATH is regular). Conditions MATH are compatible if and only if there are MATH, MATH and MATH such that CASE: MATH, MATH, CASE: MATH, MATH, MATH, CASE: MATH, MATH, CASE: MATH, CASE: MATH and MATH. (If MATH then MATH is not present; similarly on the MATH side.) First assume that REF are compatible and let MATH be stronger than both MATH and MATH. Passing to a stronger condition we may demand that if MATH are such that MATH then MATH, MATH and that MATH and MATH . Now we may apply the regularity of MATH (see REF) to get MATH such that MATH . Put MATH and check that all demands are satisfied. For the other implication suppose that MATH and MATH are as in the second statement. Choose increasing sequences MATH and MATH such that MATH, MATH and CASE: MATH and MATH, and CASE: MATH, MATH. Apply the cutting property to choose (for each MATH) MATH such that MATH . (If MATH then the MATH is not present.) Next use gluing to choose MATH so that MATH . Since MATH is linked we may choose MATH such that MATH. Now look at MATH. It is a condition in MATH stronger than both MATH and MATH. CASE: MATH and REF MATH . Similarly (and much easier). CASE: Let MATH be a regressive function and let MATH be a countable MATH - closed family which is MATH - directed. Suppose that MATH is a local, MATH - linked and really finitary creating pair (because of the ``local" MATH can be omitted as it is trivial). We are going to show that ``being a (countable) pre-dense subset of MATH" is a NAME property. Let MATH, MATH and MATH be equipped with the natural (product) Polish topologies (note that MATH is a countable set). For MATH, MATH, MATH and MATH we define MATH and MATH . Note that MATH is a NAME subset of MATH and the functions MATH are NAME. Now, each MATH is essentially a finitary tree, so MATH . Consequently, for each MATH and MATH, the set MATH is NAME. Since there are countably many possibilities for MATH and MATH, we easily finish the proof using the following observation. Let MATH. Then MATH is pre-dense in MATH if and only if for each MATH and MATH the tree MATH is well - founded. Suppose that, for some MATH and MATH, the tree MATH has a MATH - branch and let MATH be such a branch. Necessarily MATH (as witnessed by MATH). If follows from the definition of MATH that MATH is finite for each MATH and therefore MATH (remember REF). Now assume that MATH is not pre-dense in MATH and let MATH be a condition incompatible with all MATH. We may demand that for some MATH we have MATH. It should be clear that MATH determines a MATH - branch in the tree MATH (remember that MATH is MATH - linked). Similarly we deal with the respective variant of REF . |
math/9909115 | It follows from REF that we may find an infinite set MATH such that for each MATH we have CASE: MATH. Let MATH be such that MATH. We claim that the function MATH dominates MATH, that is, MATH . If not, then we may choose an increasing sequence MATH of integers such that MATH and CASE: MATH, CASE: MATH. Define MATH by MATH (for MATH). It follows from REF that MATH for each MATH, so we may apply REF to conclude that MATH. But look at the clauses MATH and REF above. Whenever MATH, there is MATH such that MATH, MATH and MATH so easily we get a contradiction. |
math/9909115 | In all cases we will use REF for functions MATH defined by MATH, MATH for MATH (for MATH) and a suitably chosen name MATH for a function in MATH. CASE: We consider the case when MATH is a local creating pair only. Let MATH. Using the normality of MATH, choose an increasing sequence MATH and a sequence MATH such that for each MATH and MATH: CASE: MATH, MATH, CASE: MATH, MATH. Let MATH be the name for MATH - generic real, that is, MATH (see CITE). Let MATH be a MATH - name for a function in MATH defined by MATH (if MATH then MATH). CASE: MATH, CASE: Assume that MATH is such that MATH for all MATH. Then MATH`` MATH ". MATH . Let MATH and let MATH, MATH. Take MATH so that for some MATH we have MATH and MATH. Since MATH is normal, we may find MATH such that MATH (remember MATH). Clearly the condition MATH forces that MATH. MATH . Let MATH be such that MATH for all MATH and let MATH. Let MATH be such that MATH (so MATH for each MATH). We may assume that if MATH then MATH and that MATH. For each MATH choose MATH as follows: CASE: if MATH, then MATH, CASE: if MATH, MATH, then MATH is such that MATH (remember that MATH is linked). Since MATH we easily see that MATH, and clearly it is a condition stronger than MATH. As MATH, the claim follows. Now, the first clause of the theorem is an immediate consequence of REF. CASE: The proof is similar to the one above. Let MATH. Choose fronts MATH of MATH (for MATH) such that for each MATH: CASE: MATH, CASE: MATH (clearly possible; see CITE). For each MATH and MATH choose a sequence MATH such that MATH . Let MATH be the name for the MATH - generic real and let MATH be a MATH - name for a real in MATH such that (the condition MATH forces that) if MATH (for some MATH) and MATH, then MATH. CASE: MATH, CASE: If MATH is such that MATH for all MATH, then MATH`` MATH ". MATH . Like REF MATH. MATH . Let MATH. We may assume that for some MATH we have: MATH, MATH for all MATH, and MATH for all MATH. We build inductively a tree MATH and a system MATH as follows. We declare that MATH. Suppose we have declared that MATH. If MATH, then we let MATH and we declare MATH. If MATH for some MATH, then we choose MATH such that MATH, and we declare MATH. Finally, we let MATH and we notice that MATH. CASE: Let MATH. By ``gluing", we may assume that MATH for each MATH. Using semi - normality we may choose an increasing sequence MATH and a sequence MATH such that CASE: MATH, MATH, MATH, MATH (for MATH), CASE: if MATH, MATH, MATH, and MATH, MATH, then MATH . Now we define the name MATH like before, so MATH . CASE: MATH, CASE: If MATH for all MATH, then MATH`` MATH ". MATH . Suppose MATH, and MATH. Passing to a stronger condition (using ``gluing and cutting") we may assume that CASE: MATH and CASE: MATH, MATH for some MATH. Choose MATH such that MATH and look at the condition MATH. MATH . Let MATH. Passing to a stronger condition if necessary, we may assume that for some increasing sequence MATH we have: CASE: MATH, MATH, MATH for all MATH, CASE: if MATH then MATH. Using ``super - gluing" choose creatures MATH (for MATH) such that CASE: MATH, MATH, and CASE: MATH, and CASE: MATH. Apply ``linked" to choose creatures MATH such that MATH . Then MATH is a condition in MATH, MATH and it forces that MATH. |
math/9909115 | CASE: Let MATH be a creating pair for MATH and let sets MATH (for MATH) witness that it is NAME - producing. Let MATH be a MATH - name for a real in MATH defined by MATH (where MATH is the name for the MATH - generic real). Suppose that a condition MATH is such that MATH for all MATH. Let MATH, MATH. It should be clear that there is MATH such that MATH . Hence easily MATH is a name for a NAME real. CASE: Let MATH, MATH and let MATH . Suppose that MATH is a maximal antichain such that, for each MATH and MATH, we have MATH. There is MATH such that MATH . Assume not. Then we may choose a sequence MATH such that for each MATH the conditions MATH and MATH are incompatible. We inductively build a tree MATH and a system MATH together with a sequence MATH so that CASE: MATH, MATH, CASE: MATH, CASE: MATH, MATH. Fix a bijection MATH such that MATH implies MATH. We declare that MATH, MATH, MATH. Suppose we have arrived to the MATH stage of the construction and MATH have been already defined so that the clauses MATH above are satisfied. Let MATH be such that MATH . Let MATH consist of these MATH that MATH (remember MATH). Apply MATH of REF to the sequence MATH to choose an infinite set MATH such that, letting MATH and MATH we have MATH . Finally, we let MATH. This finishes the description of the inductive step. After the construction is carried out we let MATH. It should be clear that MATH (and even MATH). Consequently we find MATH such that the conditions MATH and MATH are compatible. Suppose that MATH. Then necessarily MATH. It follows from our construction (remember clause MATH) that we may find MATH such that MATH. But then, using the assumption that MATH is MATH - linked and MATH is MATH - closed and MATH - directed, we immediately get that the conditions MATH are compatible, contradicting the choice of the MATH. Similarly, if MATH then taking any MATH we get that the conditions MATH are compatible, again a contradiction. Since the conditions of the form used above are dense in MATH one easily concludes that the forcing notion MATH is nice. CASE: Similarly. |
math/9909115 | CASE: It follows from NAME 's Theorem (see CITE) and the definitions of the norms. REF - REF Straightforward (compare CITE). |
math/9909115 | Let MATH consist of all creatures MATH such that CASE: MATH for some MATH such that, if MATH, MATH, CASE: MATH, CASE: if MATH then MATH, otherwise MATH. (Note that MATH implies MATH.) For MATH such that MATH (for MATH) let MATH . It should be clear that MATH is a composition operation on MATH, MATH is countable and forgetful, and if each MATH is finite then MATH is really finitary. For a creature MATH we define MATH as follows. It consists of all finite sets MATH (a suitable enumeration) such that CASE: MATH, and CASE: MATH. It is clear that MATH is a decomposition operation on MATH, so MATH is a MATH - creating triple. It follows from REF that MATH is linked, and using REF one easily shows that it is super - gluing. Similarly, MATH has the cutting property by REF. Note that CASE: if MATH, then MATH. Hence, using REF, we may easily conclude that MATH is reducible. However, it is not normal - one can build MATH such that MATH but MATH (which is in some sense paradoxical, and this is why we modify this example in REF). MATH is semi - normal. Let MATH, MATH be such that MATH, MATH. We may assume that MATH (remember MATH). We choose inductively a sequence MATH such that CASE: MATH, MATH, MATH, CASE: MATH, CASE: MATH (for MATH), CASE: MATH. There are no problems for MATH (note that there are practically no restrictions on MATH, and MATH is determined). So suppose that we have arrived to a stage MATH of the construction. It follows from REF that MATH. Let MATH . Necessarily MATH. Now, using MATH and REF, we may choose MATH such that MATH. Next, we pick MATH satisfying MATH and MATH. This finishes the construction. For MATH let MATH be such that MATH. It should be clear that CASE: MATH, MATH, MATH (for MATH). We claim that additionally CASE: if MATH, MATH, MATH and MATH, MATH, then MATH (what will finish the proof of the claim). So suppose MATH, MATH, MATH and MATH, MATH. Let MATH. If MATH, then we may take MATH such that MATH and MATH, so clearly MATH. Assume now that MATH, so MATH. Since MATH, there is MATH such that MATH. Clearly MATH. Finally note that the forcing notion MATH is not trivial. |
math/9909115 | It is similar to REF, but instead of MATH we use MATH. So MATH consists of MATH such that CASE: MATH for some MATH such that, if MATH, MATH, CASE: MATH, CASE: if MATH then MATH, otherwise MATH. For MATH such that MATH (for MATH) let MATH . For MATH, MATH consists of all finite sets MATH such that CASE: MATH, and CASE: MATH. Clearly MATH is a forgetful super - gluing MATH - creating triple. It is linked, and it is really finitary provided MATH is finite for each MATH. Suppose MATH are such that MATH. Then MATH . We may assume that MATH, MATH (otherwise trivial). It follows from the assumptions (and the definition of MATH) that MATH. If MATH, then the conclusion is immediate, so assume MATH. Thus MATH and MATH. Choose MATH such that MATH and MATH. Note that, by REF, we have MATH. Consequently, we may choose MATH such that elements of MATH have pairwise disjoint domains, MATH and MATH. Now, for some MATH we have MATH. By the choice of MATH we may build MATH such that MATH and MATH. Since MATH we are done. Using REF we see that MATH is almost normal in the following sense: the reducibility demand (see REF) holds for those MATH for which MATH. However, this is enough to carry out, for example, the proof of REF with almost no changes. MATH is regular and has the cutting property. First we show the regularity. So suppose that MATH are such that MATH, MATH, MATH, and MATH, where MATH is such that MATH. It follows from the definition of MATH (see REF) that MATH . Let MATH be such that MATH, MATH, MATH and MATH . Then MATH and clearly MATH. The only thing left is to show that the norm of MATH is at least REF. If MATH, then it is clearly true as MATH. So suppose that MATH and we have to argue that MATH. But this is clear as MATH. Now let us show that MATH has the cutting property. Let MATH, MATH, MATH. Choose MATH such that elements of MATH have pairwise disjoint domains, MATH and MATH (like in the proof of REF). Put MATH . Let MATH be such that MATH, MATH, MATH and MATH . Now check. |
math/9909115 | Let MATH consist of creatures MATH such that CASE: MATH for some MATH and MATH such that, if MATH, MATH, CASE: MATH, CASE: if MATH, then MATH, otherwise MATH. The operation MATH gives non-empty results for singletons only and MATH . Clearly, MATH is a really finitary creating pair which is local, forgetful and linked (remember REF). MATH is normal and NAME - producing. First note that MATH is reducible (remember MATH of the construction for REF; note that here there are no problems caused by MATH). Next note that (the proof of) REF applies here too. To show that MATH is NAME - producing fix (for MATH) a MATH and let MATH . Suppose that MATH, MATH, MATH and let MATH. If MATH, then we easily choose MATH extending MATH and such that MATH, MATH (remember MATH). So suppose that MATH and thus MATH. Then we may find MATH such that MATH, elements of MATH have pairwise disjoint domains all of size MATH. Now we easily build MATH, both extending MATH and such that MATH, MATH. |
math/9909115 | Let MATH be the collection of all MATH such that CASE: MATH for some MATH and MATH such that MATH, CASE: MATH, CASE: if MATH then MATH; otherwise MATH. The operation MATH is natural: it gives non-empty results for singletons only and MATH . It should be clear that MATH is a local forgetful creating pair for MATH. To show that it is MATH - linked suppose that MATH, MATH, MATH and MATH. Then MATH and thus MATH. Let MATH be such that MATH, MATH. Clearly MATH. If MATH then clearly MATH, so suppose MATH. Necessarily MATH, so MATH and therefore MATH. Hence MATH . Let us show now that MATH is NAME - producing. For each MATH choose a set MATH such that MATH if MATH is finite, and MATH is infinite co-infinite if MATH is infinite. Suppose that MATH, MATH. Then MATH, MATH, so we may choose MATH and MATH, and we easily finish. Finally, let us argue that MATH is of the BCB - type, To this end suppose that MATH, MATH, MATH. Then MATH for each MATH. If MATH is finite, then the demand in MATH of REF is trivially satisfied (just take MATH). So suppose that MATH is infinite and to simplify notation let MATH. Let MATH be the increasing enumeration; MATH. We may find an infinite set MATH and MATH such that CASE: MATH for each MATH, CASE: MATH is constant for each MATH, CASE: MATH is strictly increasing for each MATH. Suppose MATH for some (equivalently: all) MATH. Then, for sufficiently large MATH, for every MATH we have MATH. Consequently MATH, so we may easily finish. |
math/9909115 | The only thing that should be shown is that being a maximal antichain is a NAME relation (remember REF and so REF ). Put MATH, MATH, and MATH and MATH. Then MATH are Polish spaces (each equipped with the respective product topology), MATH and MATH are MATH and MATH subsets of MATH, respectively, and for our conclusion it is enough to show that MATH is a NAME subset of MATH. Plainly we may assume that MATH (as we may modify suitable the family MATH without changing the forcing notion). Now, for MATH we define MATH . Clearly each MATH is a compact subset of MATH. Suppose that MATH, where MATH are such that MATH (for all MATH). Then the following are equivalent: CASE: There is a condition MATH incompatible with every MATH (for MATH). CASE: There are MATH and MATH and MATH such that CASE: for every MATH, if MATH then MATH, and CASE: for every MATH, if MATH, MATH and MATH and MATH and MATH, then MATH. Assume REF MATH and pick MATH and MATH such that MATH and MATH is incompatible with all MATH (for MATH). For MATH, MATH, let MATH be an enumeration of MATH. Note that, as MATH, if MATH and MATH, then for some MATH and MATH we have MATH. Letting MATH we may now easily define MATH so that MATH witness REF MATH. Suppose now that MATH and MATH witness REF MATH. Let MATH and for MATH, MATH and MATH let MATH be such that CASE: if MATH, MATH and MATH and MATH, then MATH. (The choice of the MATH's is possible by MATH .) Now, for each MATH pick MATH so that MATH . Notice that for sufficiently large MATH we have MATH, so for those MATH we will also have MATH. Hence MATH and easily it is a condition incompatible with all MATH's. For MATH and MATH, let MATH consist of all MATH such that CASE: MATH, MATH, where MATH are such that MATH for all MATH, and CASE: MATH, and CASE: for all MATH we have MATH . It should be clear that MATH is a closed subset of MATH and hence (as MATH is compact) the set MATH is a closed subset of MATH, and MATH is NAME. Now, pick a NAME function MATH such that if MATH and MATH, then CASE: for each MATH, MATH and MATH, CASE: for each MATH, for some MATH and MATH we have MATH CASE: if MATH, MATH, MATH, and MATH, then MATH. Clearly, MATH is pre-dense if and only if so is MATH. Hence, for MATH we have MATH is pre-dense if and only if there are MATH and MATH such that MATH (remember REF). Since both MATH and MATH are countable, the proof of the proposition is completed. |
math/9909115 | For MATH and a creature MATH let a creature MATH be such that CASE: MATH, CASE: MATH, CASE: MATH. [The creature MATH is defined only if MATH.] Let MATH be the collection of all (correctly defined) MATH (for MATH, MATH). For MATH (defined as above for MATH) we let MATH . |
math/9909115 | The family MATH consists of tree - creatures MATH such that CASE: MATH such that MATH, MATH and MATH, MATH, CASE: MATH, CASE: MATH. The tree composition MATH is natural: it gives non empty results for singletons only and then MATH . Now check. |
math/9909115 | Let MATH. For each MATH such that MATH choose a tree MATH satisfying MATH . Let MATH be such that MATH (defined if MATH, MATH). Our aim is to show that for some MATH, MATH is defined and belongs to MATH (that is, MATH). So suppose that for each MATH, either MATH or MATH. Let MATH and let MATH be such that MATH and MATH (clearly MATH). Choose a front MATH of MATH such that for each MATH, if MATH is defined and MATH, then MATH, and if MATH then MATH. For each MATH we have CASE: MATH, where MATH (for MATH), and if MATH and MATH is not defined or MATH then we stipulate MATH (and similarly MATH if MATH). We show this by downward induction on MATH. First suppose that MATH. Let MATH. Then MATH and MATH . Suppose now that MATH has been shown for all MATH, MATH. Let MATH and MATH. It follows from the inductive hypothesis and REF that MATH . [Note that though REF guarantees the additivity of MATH only when MATH, MATH, we can first prove that MATH . Next, we may reduce the needed additivity to the one postulated in REF by dividing all terms by suitably large MATH.] Now, by REF , MATH and hence MATH . Since MATH we may easily finish. Now we apply REF to MATH. We get then MATH a contradiction. |
math/9909115 | Straightforward. |
math/9909115 | Like REF. |
math/9909115 | CASE: Let MATH. Choose a front MATH of MATH such that MATH (so necessarily MATH). Take MATH such that MATH . Clearly MATH, MATH is a front of MATH, MATH and MATH. Hence, MATH and MATH. Now, using REF , we may conclude that MATH, finishing the proof. CASE: Straightforward. |
math/9909115 | Let MATH. For each MATH define MATH by: MATH (with the convention that MATH). It follows from REF that each MATH is a MATH - measure on MATH. MATH. First note that, by a suitable modification of REF, we have MATH. So suppose that MATH, and let MATH be such that MATH. Let MATH and let MATH be a tree such that MATH and MATH, MATH imply MATH. Clearly MATH, so we may choose MATH so that MATH. If MATH, then (MATH and) we may choose a condition MATH stronger than MATH which decides the value of MATH. By REF we find MATH such that MATH, what contradicts MATH. Therefore MATH, and like in REF we may build a condition MATH and a front MATH of MATH such that CASE: MATH, MATH, CASE: MATH, CASE: MATH for each MATH, and CASE: MATH. Now, we may easily conclude that MATH, getting a contradiction. The conclusion of the proposition follows immediately from REF and the definition of MATH's (so we take MATH and suitable MATH's for MATH). |
math/9909115 | Let MATH . It follows from REF that, if MATH is a condition stronger than MATH, then MATH. If MATH, then we are clearly done, so suppose MATH. Note that if MATH then MATH. Thus MATH is a tree with MATH, MATH, MATH. This MATH determines a condition MATH, MATH. It follows from the previous remark that MATH. Take a front MATH of MATH such that MATH and MATH. For each MATH fix a condition MATH such that MATH, MATH, and MATH decides MATH on a front. Let MATH be such that MATH . It should be clear that MATH is a condition as required . |
math/9909115 | Straightforward. |
math/9909115 | The proof follows the lines of the appropriate proof for the NAME forcing notion (see for example, CITE). Let MATH witness that MATH satisfies MATH of REF, MATH. For simplicity, we assume that MATH (for MATH). For a condition MATH and MATH let MATH . Choose inductively conditions MATH such that for each MATH: CASE: MATH, MATH, MATH, and MATH for MATH, CASE: if MATH and there is MATH such that MATH and MATH decides MATH, then MATH decides MATH. (Note that ``strongly linked" implies that if MATH, then MATH are compatible; also remember that MATH is full.) Since MATH is NAME, we may choose an increasing sequence MATH such that MATH . For MATH let MATH . It should be clear that MATH is a condition stronger than MATH and for every MATH, MATH we have MATH . Hence easily MATH approximates MATH at all points of the form MATH (for MATH), and by the additional demand on MATH (in ``strongly-MATH") we conclude that MATH approximates MATH at all MATH. |
math/9909115 | Let MATH be the name for MATH - generic real and let MATH be a name for a subset of MATH such that MATH . (Clearly MATH is infinite.) Let MATH be given by MATH if and only if MATH. It should be clear that MATH is a name for a NAME real over MATH. |
math/9909115 | Straightforward. |
math/9909115 | Straightforward. |
math/9909115 | It is easy to check that the relation MATH of MATH is transitive (so MATH is a forcing notion). Let us argue that it is MATH - centered when MATH is countable (the case of MATH - directed MATH can be handled similarly). For MATH, MATH let MATH . Each MATH is a directed subset of MATH. Let MATH (for MATH). (Thus MATH, MATH.) Let MATH be the increasing function given by REF(MATH). Pick a sequence MATH such that MATH and (for MATH and MATH) MATH (possible by the definition of the forcing MATH and REF(MATH)). Now build inductively a system MATH as follows. We declare that MATH, and if MATH, MATH, then MATH and MATH. Suppose we have defined MATH up to the level MATH, so we know MATH for MATH. Let MATH be the tree of height MATH built from these MATH (so it is the respective ``initial part" of our future MATH), and assume that MATH (for MATH). Apply REF to get MATH such that MATH and MATH, MATH and MATH (for MATH). We declare that MATH up to the level MATH is MATH (and the respective MATH are as chosen above). Plainly, MATH is a condition stronger than both MATH and MATH. The rest should be clear. |
math/9909115 | Let MATH consist of all tree creatures MATH such that CASE: MATH for some MATH, MATH and MATH such that MATH CASE: MATH, CASE: MATH. The operation MATH is natural: MATH . For MATH let MATH be such that MATH and for MATH let MATH . Let MATH. Check that MATH is a ccc - complete mixing triple for MATH. |
math/9909115 | For notational convenience, let MATH. Note that we may assume that MATH (as we may work with the mapping MATH for some increasing MATH instead). Let MATH be the increasing enumeration, and let MATH be defined by: MATH and MATH (for MATH). [Here we keep the convention that if MATH, then MATH; or just assume that MATH.] Let MATH be a MATH - name for an element of MATH, and let MATH be a condition in MATH. We may assume that MATH, and just for simplicity let MATH. We define inductively a tree MATH, mappings MATH, MATH, a function MATH, and a system MATH of conditions from MATH. We declare MATH. Using REF we choose an increasing sequence of REF and values for MATH (thus defining MATH) such that CASE: MATH, MATH and MATH, CASE: MATH. Since MATH is a NAME ultrafilter we may choose a set MATH (the increasing enumeration) such that MATH . We declare that MATH for all MATH and we let MATH . Next we choose pairwise disjoint sets MATH such that MATH and MATH . Suppose now that we have already defined MATH, MATH, together with MATH, MATH, MATH (for MATH, MATH) and MATH so that the following conditions are satisfied: CASE: if MATH, and MATH are defined, then MATH, CASE: if MATH, MATH, then there are no repetitions in the sequence MATH, CASE: if MATH, MATH, then MATH CASE: the (defined) MATH's are pairwise disjoint, CASE: for each MATH we have: MATH and MATH CASE: if MATH is a condition such that MATH, MATH and MATH for all MATH, then MATH. [Check that these conditions are satisfied at the first stage of the construction.] Note that it follows from clauses MATH and MATH that for each MATH we have MATH and hence (by clause MATH) CASE: MATH. Fix MATH (and note that MATH). Choose an increasing sequence MATH of conditions in MATH and values for MATH such that CASE: MATH, MATH and MATH, CASE: if MATH, MATH, then MATH, CASE: the sequence MATH is constant (and let MATH be the enumeration of MATH; necessarily MATH), CASE: MATH for each MATH (note that the limit exists as the sequence is non-increasing). [Why possible? Use REF, remember our additional assumption on MATH.] Choose MATH (for MATH and MATH) so that: CASE: MATH, CASE: MATH and MATH (remember that each MATH is a NAME ultrafilter). Now we declare that MATH for all MATH and MATH and we let MATH . This finishes the definitions of MATH and of MATH for MATH. It should be clear that (the respective variants of) clauses MATH, MATH, MATH and MATH are satisfied. Using MATH we may easily choose sets MATH (for MATH) so that the demands MATH hold. The construction is finished. The tree MATH is perfect and it determines a condition MATH. MATH (so MATH). First note that the clause MATH is not enough to show this, as there are fronts in MATH which are not included in any MATH. However, we may use the specific way the construction was carried out to build a semi - MATH - measure MATH such that MATH (what is enough by REF). So, if MATH, MATH, then we let MATH; if MATH, MATH then MATH. Now check. Thus MATH is a condition stronger than MATH and it forces that MATH. Since MATH satisfies the ccc, we may easily finish. |
math/9909115 | The family MATH consists of creatures MATH such that CASE: MATH such that MATH and MATH, CASE: MATH, CASE: MATH. The operation MATH is natural: MATH . Now check. |
math/9909115 | Let MATH and let MATH consists of tree creatures MATH for MATH such that CASE: MATH for some MATH, MATH and MATH, CASE: MATH, CASE: MATH. The operation MATH is natural, so MATH if and only if MATH and MATH. Let MATH, MATH. MATH consists of quadruples MATH such that CASE: MATH, CASE: MATH, CASE: MATH, MATH, MATH, CASE: if MATH, MATH then for some MATH we have MATH, MATH and MATH. Easily MATH is a universality parameter and MATH is the Universal Meager forcing notion, |
math/9909115 | For a nonempty set MATH, MATH, we define MATH as the value of MATH for MATH such that MATH and MATH. (Note that under our assumptions on MATH there is exactly one such MATH.) Let MATH consists of all quadruples MATH satisfying the demands of REF and such that for some sequence MATH we have CASE: MATH, MATH, CASE: MATH. Now check. |
math/9909115 | Let MATH, MATH and MATH be as defined in the construction for REF. Let MATH consists of all quadruples MATH satisfying the demands of REF and such that MATH . Let MATH. Note that the forcing notion MATH is equivalent to MATH defined as follows. A condition in MATH is a pair MATH such that MATH and MATH is a tree such that MATH is a measure zero subset of MATH; the order of MATH is the natural one: MATH if and only if MATH, MATH and MATH. |
math/9909115 | REF Straightforward. CASE: See (the proof of) CITE. CASE: Follows from REF and the definition of MATH (remember that ``being a maximal antichain of MATH" is absolute; see CITE). |
math/9909115 | Straightforward (or see CITE). |
math/9909115 | Suppose that MATH is a generic filter over MATH and MATH is MATH - generic over MATH. It follows from the third assumption that there is a MATH - name MATH for MATH such that MATH for every MATH - name MATH for an element of MATH. Consequently, we find a complete embedding MATH such that MATH and MATH. Now continue like in the proof of CITE; see also the original NAME 's article CITE. |
math/9909115 | Plainly, MATH is closed under finite intersections. Since the equivalence classes of each MATH are directed and any two compatible members of MATH have a least upper bound, we may conclude that the elements of MATH are directed and downward closed. Before we verify the demand REF , let us first note that if MATH, MATH, MATH, then for some MATH we have MATH. [Why? Use REF to choose MATH such that MATH . Clearly this MATH is as required.] Now suppose that a sequence MATH converges to MATH in the topology generated by MATH, and MATH for all MATH. Take MATH such that MATH. Choose an increasing sequence MATH such that MATH . Next pick conditions MATH such that MATH (remember that each MATH is directed). It follows from REF that we may find a condition MATH such that MATH and MATH. Then MATH . Since MATH is directed and MATH, the conditions MATH have an upper bound in MATH - let MATH be such an upper bound. |
math/9909115 | Straightforward. |
math/9909115 | Let MATH be MATH - names such that MATH (note that MATH is a model of iterable sweetness " and also MATH for all MATH ", remember REF ). Let MATH be a countable basis of the topology MATH and let MATH . For MATH, MATH and MATH and MATH REF we put MATH . Let MATH be the collection of all sets of the form MATH (for suitable parameters MATH). CASE: The family MATH forms a countable basis of a topology on MATH; we will denote this topology by MATH. MATH is an isolated point in MATH. CASE: If MATH, MATH and MATH, then there is MATH such that MATH and MATH. CASE: Should be clear. CASE: Let MATH and let MATH (for MATH) be such that MATH and MATH. Choose MATH such that MATH and every two members of MATH have a common upper bound in MATH (possible by REF). Next pick MATH such that MATH and CASE: every member of MATH has an upper bound in each of the sets MATH, CASE: every MATH elements of MATH have a common upper bound in MATH. Let MATH be such that MATH and each element of MATH has an upper bound in MATH, and let MATH be such that MATH. Put MATH . First we show that MATH. By our choices, MATH and MATH. For MATH, MATH is a condition stronger than MATH and MATH, so MATH (remember REF ). Next, for MATH, we have MATH and MATH, so MATH. Now, suppose that MATH. Then MATH and we have conditions MATH (for MATH) and conditions MATH (for MATH) such that CASE: MATH and MATH, CASE: MATH and MATH. Pick a condition MATH stronger than all MATH (for MATH). We claim that MATH. Clearly MATH and MATH (for MATH). Fix MATH. By the choice of MATH, we find MATH stronger than MATH. By the choice of MATH, we find a condition MATH stronger than both MATH and MATH. Then MATH and we are done. Suppose that a sequence MATH is MATH - converging to MATH, and MATH. Then there are an infinite set MATH and conditions MATH (for MATH) such that for each MATH and MATH: CASE: MATH, MATH, CASE: MATH. Pick MATH, MATH such that MATH (for MATH). Fix sequences MATH (for MATH) such that CASE: MATH, CASE: MATH forms a basis of neighbourhoods of MATH. Clearly MATH for every MATH, so we may pick an increasing sequence MATH such that MATH . Let MATH be such that MATH, MATH (for MATH, MATH). Each sequence MATH - converges to MATH so we may find an infinite set MATH and conditions MATH (for MATH) such that MATH . Let MATH. Suppose MATH is MATH - converging to MATH. Then there is MATH such that: CASE: if MATH, then for some MATH and MATH (for MATH) we have: CASE: MATH, MATH, CASE: MATH. Let MATH enumerate all sets MATH to which MATH belongs. Apply REF to choose inductively a decreasing sequence MATH such that for each MATH: if MATH, then there are conditions MATH (for MATH) satisfying MATH . Next pick an infinite set MATH almost included in all MATH's. Suppose that MATH - converges to MATH. Then there is a condition MATH stronger than MATH and such that MATH . Let MATH, and let MATH (for MATH) be such that MATH and MATH. Pick MATH such that CASE: MATH, MATH, CASE: any REF elements of MATH have a common upper bound in MATH, CASE: every element of MATH has an upper bound in each of MATH. Apply REF to choose MATH such that the condition MATH of REF holds. Note that then MATH . [Why? If not, then we may pick a condition MATH and an integer MATH such that MATH . Pick MATH such that MATH and any REF members of MATH are compatible, and apply MATH of REF to MATH. We get a condition MATH such that MATH . Since MATH are compatible, we get a contradiction.] Since MATH - converges to MATH, we may find an infinite MATH and a condition MATH such that MATH . Next use MATH of REF to pick conditions MATH and MATH such that MATH . By the choice of MATH we get conditions MATH such that MATH, MATH, MATH (for MATH). Now we are going to define a MATH - name MATH for a condition in MATH. Let MATH be a maximal antichain of MATH such that for each MATH and MATH: CASE: either MATH or MATH are incompatible, and CASE: either MATH or MATH are incompatible. Fix MATH. If MATH are incompatible, then let MATH be MATH. Assume MATH and let MATH . The condition MATH forces that the sequence MATH converges to MATH, and MATH, and MATH (for all MATH). Applying REF(i+iii) we find a MATH - name MATH for an element of MATH such that MATH . Now, let MATH be a MATH - name such that MATH (for MATH). Look at the condition MATH. Clearly MATH (for all MATH) and MATH, MATH (for MATH), so MATH. Moreover, if MATH, then MATH and MATH, so MATH and we are done. If MATH, then there is MATH such that MATH and any two conditions MATH have a common upper bound in MATH. Let MATH and let MATH (for MATH) be such that MATH and MATH. Pick MATH such that MATH and every three members of MATH have a common upper bound in MATH. Also choose MATH such that MATH, and each member of MATH has an upper bound in every MATH (for MATH) and every two members of MATH have a common upper bound in MATH. Put MATH . Clearly MATH. Suppose now that MATH and let MATH be such that MATH and MATH (for MATH and MATH). Also let MATH be stronger than both MATH and MATH, and for each MATH let MATH be stronger than both MATH and MATH and MATH. Now, like in the proof of REF, choose a MATH - name MATH for a condition in MATH such that MATH`` MATH ", and MATH`` MATH " for MATH, and MATH`` MATH " for MATH. (Remember that the sets MATH are forced to be directed by REF(i+ii).) Then MATH is a condition stronger than both MATH and MATH. Now we may put together REF to conclude that the topology MATH satisfies the demand REF , finishing the proof of the theorem. Note that MATH for each MATH and the mapping MATH is a homeomorphic embedding of MATH into MATH, so we may think that MATH is the restriction of MATH to MATH. Moreover, under this interpretation, MATH is an open subset of MATH. |
math/9909115 | See CITE. |
math/9909115 | CASE: Let MATH be a MATH - creating triple for MATH. Suppose that MATH is linked, gluing and has the cutting property. For MATH and MATH let MATH . Let MATH be the topology generated by the sets MATH (for MATH and MATH) and MATH. It is straightforward to check that MATH is a model of topological sweetness. Other instances of REF can be handled similarly. CASE: We consider the case when MATH is countable only (if MATH is MATH - MATH - directed the proof is similar). We put MATH and we define relations MATH (for MATH) on MATH as follows: MATH if and only if MATH, MATH and MATH . We claim that MATH is a model of sweetness. Plainly, each MATH is an equivalence relation with countably many equivalence classes, MATH. Similarly as in REF one can show that the equivalence classes of each MATH are directed and that the clause REF is satisfied. Let us show that the demand REF holds. So suppose MATH (for MATH, MATH). Thus MATH, MATH (for MATH). We build inductively a system MATH as follows. First we let MATH and we declare MATH . Suppose that we have defined MATH already. Pick MATH such that for some MATH we have MATH and MATH . Next, choose MATH and MATH (for MATH) such that MATH, MATH-REF, MATH and MATH . Let MATH be such that MATH, MATH, and MATH when MATH and MATH when MATH. Apply repeatedly REF to get MATH such that MATH and MATH, and MATH . Note that then MATH for all MATH. After the construction is carried out one easily checks that MATH is a condition stronger than all MATH's (for MATH) and MATH. |
math/9909115 | For a finite candidate MATH and sequences MATH and MATH we let MATH . Plainly the sets MATH (for suitable MATH) and MATH constitute a basis of a topology MATH on MATH. it is not difficult to check that MATH is a model of topological sweetness. |
math/9909115 | CASE: Suppose that MATH are universes of ZFC, MATH, MATH, MATH is generic over MATH and MATH. Assume MATH. CASE: If MATH and MATH, then for some disjoint NAME sets MATH we have MATH and MATH. Consequently, if MATH and MATH, then MATH. CASE: If MATH are incompatible in MATH, then for some disjoint NAME sets MATH we have (for MATH): MATH . Straightforward if you remember REF. Now, let MATH witness that MATH is explicitly sour to MATH (in MATH). By REF we know that, in MATH, MATH . For some MATH the set MATH is uncountable. Let MATH. It follows from REF that there are distinct MATH such that MATH. By REF we know MATH, so by REF these two conditions are incompatible in MATH. But then, using REF, we find disjoint NAME sets MATH such that MATH and MATH, a contradiction to MATH. CASE: Let MATH contain MATH and MATH. Working in MATH, pick a sequence MATH of pairwise distinct members of MATH. Put: CASE: MATH, MATH, CASE: MATH if and only if MATH. Now check. |
math/9909115 | Since MATH are as in REF, we have a nice description of MATH - names for reals, see REF. So we have NAME functions MATH such that MATH. It follows from the assumed properties of MATH that CASE: for every NAME set MATH, if MATH then MATH. (Note that MATH is coded by the real MATH as well.) Since MATH is not inaccessible in MATH, for some real MATH we have MATH. Let MATH . (Note that if MATH is MATH - generic over MATH, then it determines a MATH - generic filter MATH over MATH and MATH; remember REF and the choice of MATH.) CASE: MATH is a MATH subset of MATH. [Why? Note that, by REF, the formula ``MATH is a NAME code for a set MATH and MATH" is (equivalent in ZFC to) a MATH formula. Hence easily the formula ``MATH is MATH - generic over MATH" is MATH.] CASE: For every NAME set MATH, if MATH then MATH. [Immediate by MATH and REF .] CASE: MATH. [It follows from the sourness and the assumption that MATH.] Finally note that MATH implies that both sets MATH do not have the MATH - NAME property. |
math/9909115 | CASE: Let us think about the NAME forcing MATH as the set of partial functions from MATH to REF with the relation of extension; MATH for MATH (so we interpret MATH as MATH), and MATH is the natural name for the MATH - generic real in MATH. Let MATH be the name for MATH - generic real, that is, MATH. Let MATH be standard MATH - names for functions from MATH to REF such that CASE: MATH if and only if MATH, CASE: MATH. MATH moreover MATH for any non-meager NAME set MATH. Hence MATH are as in REF. Let MATH and let MATH be such that MATH . Pick MATH such that MATH are pairwise distinct and CASE: MATH, and CASE: MATH, where MATH is as in the assumption MATH. Suppose MATH and MATH. Build a condition MATH such that CASE: MATH, MATH if MATH, and MATH whenever MATH, CASE: if MATH, MATH, then for some MATH we have MATH and MATH, where MATH, CASE: if MATH then MATH, CASE: if MATH is not any of the MATH's from clause REF above (for MATH), MATH, MATH, then MATH, where MATH (remember: MATH is linked and REF + REF ), CASE: if MATH, MATH and MATH, then MATH is such that MATH (remember assumption MATH and clause REF ), CASE: if MATH, MATH, then MATH. It should be clear that we can build MATH satisfying the demands REF - REF and that then MATH and MATH. Now we easily conclude that MATH is NAME over MATH; the rest should be clear too. For MATH and MATH let MATH be such that MATH, if MATH, MATH, then MATH, and if either MATH or MATH then MATH. Note that the function MATH is NAME. CASE: For each MATH the set MATH is predense in MATH. CASE: If MATH, MATH, MATH, MATH and MATH, then MATH . CASE: Straightforward. CASE: Note that if MATH, MATH then for each MATH, MATH implies (MATH and) MATH (and, of course, MATH, MATH). Also, if MATH, MATH, then (MATH and) MATH. Now one easily shows that MATH is very explicitly sour to MATH. REF, CASE: Similarly. CASE: Here we think about the NAME forcing MATH as the set of finite partial functions from MATH to REF ordered by the extension; MATH, MATH are interpreted suitably. For each MATH pick MATH. Let MATH be the name for MATH - generic real, and take standard MATH - names MATH for functions from MATH from REF such that CASE: MATH if and only if MATH, CASE: MATH. MATH moreover MATH for any non-meager NAME set MATH. Hence MATH are as in REF. Quite similar to REF. Now, for MATH and MATH let MATH be such that MATH, and if MATH, MATH, MATH, then MATH, and MATH is all other cases. CASE: For each MATH and MATH, the set MATH is predense in MATH. CASE: If MATH, MATH, MATH, MATH and MATH, then MATH . Straightforward. Now we easily finish. CASE: Similarly. |
math/9909115 | Here we interpret the NAME forcing notion MATH in the standard way, that is, it is MATH (and MATH, MATH are natural). For MATH let MATH (remember REF MATH). Let MATH be the canonical name for the MATH - generic real and let MATH be standard MATH - names for reals in MATH such that CASE: MATH if and only if there are MATH such that MATH, MATH and MATH; CASE: MATH. MATH are as in REF. We will show that MATH is (a name for) a NAME real over MATH; then the rest should be clear. So suppose MATH and we may assume that MATH, MATH. For MATH let MATH and let MATH be such that MATH remember REF . Let MATH . Clearly, for all MATH, MATH, and thus we may choose an infinite set MATH such that MATH forms a MATH - system with the heart, say, MATH. Take MATH such that CASE: MATH remember REF and suppose that MATH, MATH. We are going to build a condition MATH such that MATH . To this end, pick MATH such that CASE: MATH, and CASE: for some MATH we have MATH where MATH. (Possible by the choice of MATH and the assumption that MATH is MATH - closed.) Since MATH, we may find MATH such that MATH. For MATH fix MATH such that MATH, MATH. Let MATH enumerate MATH. By induction on MATH define MATH as follows: if there is MATH such that MATH, then MATH is the first such number, otherwise MATH is MATH. Now we choose MATH so that: CASE: MATH, MATH; CASE: if MATH, MATH, MATH, then MATH; CASE: if MATH, then MATH; CASE: if MATH, then MATH CASE: MATH is such that MATH and MATH (where MATH is given by MATH). [Why is the choice possible? Demands REF - REF are easy; REF can be satisfied by MATH, remember REF MATH; REF is possible by REF and MATH.] One easily checks that the demands REF - REF imply MATH is a condition in MATH stronger than MATH. Also, by REF + REF + REF , MATH (remember REF(MATH); thus MATH in clause REF ). Hence MATH. Now we argue that MATH. If not, then for some MATH we find MATH such that MATH . Necessarily, MATH. Moreover, if MATH, then MATH (as if MATH then MATH and MATH). Since MATH, we may conclude that MATH (and thus both are in MATH). However, then we get MATH, what contradicts MATH. For MATH and MATH let MATH be such that MATH, MATH for MATH, and MATH for MATH. Let MATH . It is straightforward to check that MATH witness MATH is very explicitly sour to MATH (note that if MATH, MATH, MATH, MATH and MATH, then MATH`` MATH "). |
math/9909115 | CASE: We are going to apply REF. First note that even though MATH as defined in REF is not complete we can easily make it so, or restrict our attention to the forcing notion below some condition (the problems with completeness come from the technical requirement in the definition of MATH that MATH). We are going to build a sourness system MATH for MATH such that the demand REF holds. Let MATH, MATH. For MATH pick MATH such that CASE: if MATH, then MATH, and MATH, CASE: if MATH, MATH, then there are no repetitions in the sequence MATH. We claim that, letting MATH and MATH, MATH is as required (we identify MATH with MATH, of course). Clauses REF(MATH) - REF are clear. Suppose that MATH, MATH, MATH, MATH. Then, for large enough MATH, MATH, so let MATH. Assume that MATH (for MATH) are pairwise distinct. By MATH, for each MATH there is MATH such that MATH. Hence for some MATH . Hence we easily conclude that REF holds. The demands REF are even easier. For MATH we proceed similarly, but we choose MATH so that MATH. |
math/9909116 | If MATH is not elliptic then for some MATH, MATH belongs to MATH. Now consider the operator MATH on MATH (with respect to the trivial connection on MATH). If MATH denotes the line subbundle of MATH corresponding to the span of MATH then any section of MATH which is independent of MATH belongs the kernel of MATH. Hence the kernel of MATH is infinite dimensional on MATH. |
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