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math/0101017 | The argument is the same as in CITE. Also see CITE for more details. Essentially the idea is the same as in CITE: the linearization of the problem of constructing the half spheres has one dimensional kernel and zero dimensional cokernel. This allows perturbing everything away from the flat case: MATH and MATH . We can approximate in adapted coordinates (as in REF ) with this example (see CITE). What makes the a priori estimates work is that very small MATH disks near to the origin have small NAME norm of the (nonlinear) operator MATH while they have NAME norm of roughly fixed magnitude for MATH so that we can move MATH just a little, and effectively push toward MATH. |
math/0101017 | To obtain the uniqueness of NAME disks, we need to understand the intersections of such disks. Suppose that we have two disks with boundary in a surface, so that both boundaries are smooth curves and lie close to an elliptic point. Suppose further that the disks are embedded, and transverse to the real surface along their boundaries. Neither disk can touch the elliptic point itself since that would require tangency with the real surface at that point. Suppose that our disks have an interior intersection. Then we know (by NAME 's arguments as adapted above to our situation) that the intersection survives small perturbation. However, we will need to handle boundary intersections as well. For example, consider the real surface MATH. Ignore for the moment the fact that it has no elliptic point, and think about two complex disks with boundary in this real surface. For example, take MATH and slice it into two disks by cutting out the real points. Then we have two intersecting closed disks, which may cease to intersect if we perturb either one slightly. The delicate part is to make the disks intersect ``on the same side." Let us follow essentially the argument of Ye CITE. Because the NAME theory tells us that the deformation theory of these disks is unobstructed, we can make them both slightly larger, extending them so that they pass across the real surface. Then at an intersection point, we can use adapted coordinates, holomorphic up to leading order terms, to see that the intersection of the two disks looks to leading order just like the complex case. But there is a further innovation we can introduce here: we can adapt our coordinates to the real surface near the intersection point. Since it is totally real near that point, there is no obstruction at any finite order to finding adapted coordinates for which the totally real surface is just the set of points at which the adapted coordinates take on real values. This is easy to check with the NAME - NAME theorem. However, we have to be very careful: there could be an obstruction beyond all orders. But we can pick an order at which to match up the real surface to the set of real points of our adapted coordinates, higher than the order of intersection of our disks at that point, and thereby pretend that the real surface is exactly the set of real points of our coordinates, with no loss of generality. Then we see that, with a little manipulation of the coordinates, one disk looks like MATH with MATH while the other looks like MATH . The integer MATH must be at least one. Under small perturbation through MATH disks we see that the intersection point can at worst break into MATH intersection points, distributed around a circle nearly evenly. As in complex geometry, it is clear then that the intersection persists under perturbation, and that moreover the intersection points are nearly complex conjugates of each other in these coordinates. In particular, intersections cannot disappear off the boundary of the disks, since they will actually ``bounce off" the boundary. Therefore if two MATH disks with boundary in a totally real surface intersect, then they continue to intersect after both of them receive a small perturbation, as long as the intersection was at finite order. However, as our example shows, they might fail to intersect after perturbation if they start off with infinite order intersection. By NAME 's lemma, this can only occur if they can be extended to agree, so that we can glue them together as in the example. If we had two half balls of NAME disks, then either we could glue them together (if they lay on opposite sides of the real surface) or they would be identical (near the elliptic point - if we extend them to be maximal then they would agree) or one disk from one half sphere family has to have finite order intersection with one from the other half sphere family. |
math/0101018 | This follows from the fact that the elements MATH, and MATH belong to MATH. |
math/0101018 | In REF , the theorem is proved when MATH is odd and MATH, using the triangular decomposition REF for MATH. The same proof applies in general as the decomposition REF does not depend on the choice of MATH. |
math/0101018 | The module MATH is a free MATH - module of rank MATH by REF. Since MATH is a NAME subalgebra of MATH, the map MATH is a homomorphism of rings. Since the degrees of the polynomials MATH determine the highest weight of MATH, considered as a MATH - module, we obtain that any MATH - subquotient occurring in MATH has highest weight less than or equal to MATH with respect to our partial order. This proves the surjectivity of MATH and REF . |
math/0101018 | REF describes the so-called diagonal finite-dimensional representations of the non-restricted specialization MATH (although MATH is assumed in CITE to be odd, the proof of REF does not depend on this restriction). It states in particular that these representations are highest weight representations. Since we have a surjective homomorphism MATH, any irreducible MATH - module gives rise to an irreducible MATH - module. By REF, an irreducible MATH - module obtained by pull-back from an irreducible MATH - module is diagonal. Therefore we obtain that every irreducible finite-dimensional MATH - module is a highest weight module. Furthermore, according to REF, the irreducible highest weight representation of MATH with highest weight MATH is finite-dimensional if and only if MATH is a rational function MATH where MATH are two polynomials in MATH of equal degrees with non-zero constant coefficients; MATH is regular at MATH and at MATH, and MATH. In addition, the highest weight of an irreducible MATH - module obtained by pull-back from an irreducible MATH - module of type REF satisfies REF and the condition MATH. But then for each MATH there exists a unique acyclic polynomial MATH, such that REF holds. This completes the proof. |
math/0101018 | This statement is proved in REF in the case when the order of MATH is odd (although the algebra MATH in CITE includes the elements MATH, the same proof works for the algebra without the elements MATH, as in our definition of MATH). For general root of unity MATH, one repeats the proof of CITE replacing MATH by MATH, where appropriate. |
math/0101018 | In the case of MATH the analogous statement follows from REF. REF then follows from REF . |
math/0101018 | The first part follows from REF. The second part follows from REF. |
math/0101018 | This lemma follows from REF and the commutation relations in MATH. |
math/0101018 | We use the fact that the automorphisms MATH, commute with the NAME homomorphism (see REF), and so do the automorphisms MATH of the NAME diagram of MATH (see REF). Hence the automorphisms MATH, also commute with the NAME homomorphism. From this we find: MATH . So, it is zero if MATH is not divisible by MATH or if MATH. If MATH we obtain MATH . If MATH is odd then MATH and MATH have the same parity, and all the signs cancel. If MATH is even, then MATH is also even, and we acquire the sign MATH. |
math/0101018 | We embed the NAME diagram of MATH into the NAME diagram of MATH in such a way that the numbers MATH coincide. By REF , we have the corresponding embeddings MATH and MATH. It follows from REF that the image of MATH under the NAME homomorphism MATH is contained in MATH. Hence we obtain an algebra homomorphism MATH, given by the above formula. The lemma follows. |
math/0101018 | This theorem is proved in REF - REF , in the case when MATH is a root of unity of odd order (that is, MATH and MATH is odd). The proof given in CITE goes through for other MATH (under our assumption that MATH is relatively prime to MATH) with the following changes: REF should be replaced by REF in the case of even MATH should be replaced by the identity MATH (which is proved in the same way as the commutativity of REF ). Finally, the proof uses the description of the NAME polynomials of the NAME pull-backs of irreducible MATH - modules. These NAME polynomials are determined in REF below. |
math/0101018 | Note that if MATH then MATH and MATH. Therefore the rule described in the theorem is unambiguously defined. Recall that MATH as well as MATH and MATH, are ring homomorphisms. Since the ring MATH is generated by the fundamental representation MATH, it is sufficient to prove the lemma when MATH. In this case the NAME polynomial is MATH. Let MATH be the highest weight vector in MATH and MATH. By REF, we obtain: MATH . (Note that MATH in this paper differs from MATH in CITE by the factor MATH.) So, MATH unless MATH is divisible by MATH. If MATH, then for odd MATH we obtain: MATH . We have: MATH, unless MATH, and MATH. In addition, MATH. Hence we find: MATH, unless MATH, and MATH. For even MATH we obtain: MATH . We have: MATH unless MATH, and MATH. Hence we find that MATH, unless MATH, and MATH. It remains to determine the MATH - character of MATH. Consider the case when MATH is odd. Since the NAME polynomial of this module is MATH, we find that the highest weight monomial is MATH. By REF, the other monomial in it must be obtained by specialization of a monomial in the MATH - character of the irreducible MATH - module with the highest weight monomial MATH. Moreover, by REF, the degree of this monomial must be equal to MATH. There is only one monomial of such degree, that is, the lowest weight monomial, and it is equal to MATH. Therefore we obtain that the MATH - character of MATH is equal to MATH. On the other hand, we have: MATH, compare REF, and the second assertion of the lemma follows for odd MATH. If MATH is even, we obtain by the same argument that the MATH - character of MATH is equal to MATH. This proves the second assertion of the lemma for even MATH. |
math/0101018 | Restricting to the subalgebras MATH and using REF , we obtain the statement of the theorem from REF. |
math/0101018 | According to REF , the algebras MATH and MATH are isomorphic, if MATH. Therefore the category of type REF finite-dimensional MATH - modules is isomorphic to the category of weight MATH - modules. The statement of the lemma is obtained by combining this fact with REF and the following result about MATH - characters, which follows from REF : Set MATH. Suppose that MATH is an irreducible MATH - module, such that the highest weight monomial MATH in MATH belongs to MATH. Then all other monomials in MATH belong to MATH. In the case of MATH, we use the inclusion MATH and REF follows from REF . |
math/0101023 | The computation of motivic cohomology of weight MATH shows that MATH . The nontrivial element MATH together with the multiplication morphism MATH defines a morphism MATH. The NAME conjecture implies immediately the following result. The morphism MATH extends to a distinguished triangle in MATH of the form MATH where MATH is the MATH - th cohomology sheaf of the complex MATH. Consider the long sequence of morphisms in the triangulated category of motives from the motive of MATH to the distinguished triangle REF . It starts as MATH . By REF there are isomorphisms MATH . On the other hand, since MATH is a homotopy invariant sheaf with transfers, we have an embedding MATH . The right hand side is isomorphic to MATH. This completes the proof of the proposition. |
math/0101023 | Let us consider morphisms in the triangulated category of motives from the distinguished triangle REF to the object MATH. By definition, MATH is a direct summand of the motive of the smooth projective variety MATH of dimension MATH, therefore, the group MATH is trivial by CITE. Using this fact, we obtain the following exact sequence: MATH . By definition (see CITE) the morphism MATH is given by the composition MATH and the composition of the second arrow with the canonical embedding MATH is the fundamental cycle map MATH which corresponds to the fundamental cycle on MATH under the isomorphism MATH (see CITE). On the other hand by REF the homomorphism MATH defined by the first arrow in REF is an isomorphism. This implies immediately that the exact sequence REF defines an exact sequence of the form MATH . By CITE there is an isomorphism MATH . The NAME resolution for the sheaf MATH (see, for example, CITE) shows that the group MATH can be identified with the cokernel of the map: MATH and the map MATH defined by the fundamental cycle corresponds in this description to the map MATH . This finishes the proof of REF . |
math/0101023 | We have to show that the composition of operations MATH is injective. Let MATH be the simplicial cone of the morphism MATH which we consider as a pointed simplicial scheme. The long exact sequence of cohomology defined by the cofibration sequence MATH together with the fact that MATH for MATH shows that for MATH we have a natural isomorphism MATH compatible with the actions of cohomological operations. Therefore, it is sufficient to prove injectivity of the composition MATH on motivic cohomology groups of the form MATH. To show that MATH is a monomorphism it is sufficient to check that the operation MATH acts monomorphically on the group MATH for all MATH. For any MATH we have MATH by CITE and CITE. Therefore, the kernel of MATH on our group is the image of MATH. On the other hand, the cofibration sequence REF together with REF implies that for MATH we have MATH which proves the lemma. |
math/0101023 | Since our maps are homomorphisms of MATH-modules it is sufficient to verify that the cohomological operation MATH sends MATH to MATH. By REF MATH is injective. Therefore, the element MATH iz nonzero. On the other hand, sequence REF shows that MATH and MATH is a generator of this group. Therefore, MATH. |
math/0101023 | Follows immediately from REF and surjectivity of multiplication by MATH REF . |
math/0101023 | Since all the maps in REF are morphisms of MATH-modules, it is sufficient to check the condition for the generator MATH. And the later follows from REF and the definition of MATH. |
math/0101023 | Let MATH, where MATH are pure symbols corresponding to sequences MATH. Let MATH be the norm quadric corresponding to the symbol MATH. For any MATH denote by MATH the field MATH. It is clear that MATH. Let us fix MATH such that MATH and MATH is a nonzero element. Then MATH belongs to MATH . By REF , the kernel is covered by MATH and is generated by MATH. Thus, we have MATH. |
math/0101023 | It is sufficient to prove the statement for MATH. Let MATH be an invertible element of MATH. Since MATH we have MATH. Therefore MATH is a quotient of two elements of MATH. |
math/0101023 | We may assume that MATH lies in MATH. Then there exists a linear function MATH on MATH such that the map of the residue fields MATH is an isomorphism. Let MATH be a coordinate system starting with MATH. Since the restriction of MATH to MATH is an isomorphism the inverse gives a collection of regular functions MATH on MATH. Each of this functions has a representative MATH in MATH of degree at most MATH. The projective closure of the affine curve given by the equations MATH, MATH satisfies the conditions of the lemma. |
math/0101023 | We already know that MATH is surjective. Let MATH be an element of MATH. By REF there exists a field extension MATH such that MATH is a nonzero pure symbol. By a result of NAME and NAME CITE the map MATH is injective on pure symbols. Hence MATH is a nonzero element of MATH. Since the morphism MATH is compatible with field extensions, the element MATH is also nonzero. Therefore, MATH is injective. |
math/0101023 | Denote by MATH a quadratic form which defines the quadric MATH. Assume that MATH is a nonzero element of MATH. Using REF we can find an extension MATH such that MATH is a nonzero pure symbol of the form MATH. Then, since MATH, the corresponding NAME quadric MATH becomes hyperbolic over MATH. Since MATH is hyperbolic the form MATH is isomorphic to a subform of the NAME form MATH for some coefficient MATH by CITE. In particular, MATH. Therefore, we have MATH. |
math/0101023 | Let MATH be an element of MATH which is represented by a quadratic form MATH. As above we have a sequence of quadrics MATH such that MATH is hyperbolic. This means that MATH goes to MATH under the natural map from MATH to MATH. All quadrics MATH have dimensions MATH. By REF , for any MATH, the kernel MATH is trivial. Therefore, applying the NAME conjecture REF , we conclude that the map MATH is a monomorphism for all MATH. Therefore the map MATH is a monomorphism as well. Therefore, MATH belongs to MATH. |
math/0101030 | NAME is obvious. To see that MATH is perfect, begin with MATH such that MATH. Then MATH, so by hypothesis we may write MATH . For each MATH, choose MATH such that MATH and MATH. Note that each MATH and MATH lies in MATH, and let MATH . Then MATH, which implies that MATH; so by hypothesis we may write MATH . Moreover, since MATH, each MATH lies in MATH . Finally, we write MATH which shows that MATH. |
math/0101030 | Let MATH be a generating set for MATH and let MATH. We will show that MATH is the normal closure of MATH. First note that MATH, for each MATH. Hence MATH and therefore the normal closure of MATH, is contained in MATH. As preparation for obtaining the reverse inclusion, let MATH and observe that MATH . With this identity as the main tool, induction on word length shows that MATH. |
math/0101030 | Let MATH be a set of words so that MATH in MATH. Since MATH is surjective, there exists a set of words MATH so that MATH in MATH. Then NAME transformations give the following isomorphisms: MATH . Now MATH is specified by the last presentation via the correspondence MATH. Hence MATH is the normal closure of the MATH elements of MATH and MATH. |
math/0101030 | After passing to a subsequence and relabelling we have a diagram: MATH . The MATH's are clearly finitely generated but may not be finitely presented. We will use this diagram to produce a new sequence with the desired properties. For each MATH, let MATH be a generating set for MATH, and choose MATH so that MATH. NAME superscripts are indices, not powers. Let MATH be the normal closure of the set MATH, define MATH to be MATH, and let MATH be the quotient map. Since MATH we get induced homomorphisms MATH . Define MATH to obtain the commuting diagram: MATH . Since each MATH is generated by MATH and each MATH has a preimage MATH under the map MATH, it follows (from the commutativity of the diagram) that each MATH is surjective. Lastly, each MATH has a finite presentation which may be obtained from a finite presentation for MATH by adding relators corresponding to the elements of MATH. |
math/0101030 | For the forward implication, we begin by observing that MATH is a homotopy collar. First note that MATH is a homotopy equivalence for any finite MATH. A direct limit argument then shows that MATH is a homotopy equivalence. Alternatively, we may observe that MATH and apply the NAME theorem. To see that MATH is a pseudo-collar we apply the same argument to the subsets MATH. For the converse, assume that MATH is a pseudo-collar. Choose a homotopy collar MATH and let MATH. Then MATH, so MATH is a one-sided MATH-cobordism. Next choose a homotopy collar MATH and let MATH. Repeating this procedure gives the desired result. See REF . |
math/0101030 | REF follow easily from the definition of pseudo-collar; while REF is obtained from REF . |
math/0101030 | First we construct a MATH - neighborhood MATH so that MATH is surjective and MATH is an isomorphism. Since MATH is finitely dominated, MATH is finitely generated, so we may choose a finite collection MATH of disjoint properly embedded p.l. arcs in MATH so that MATH is surjective. Choose a collection MATH of disjoint regular neighborhoods of the MATH's in MATH and let MATH . Clearly MATH (and thus MATH) surjects onto MATH; moreover, since disks in MATH may be pushed off the MATH's, then MATH is also injective. Next we modify MATH to be a generalized MATH - neighborhood. Since MATH is finitely dominated, REF implies that MATH is finitely presentable. Hence, by REF , MATH is the normal closure of a finite set of elements. Let MATH be a collection of pairwise disjoint embedded loops in MATH representing these elements, then choose MATH a pairwise disjoint collection of properly embedded REF - disks in MATH with MATH for each MATH. Let MATH be a pairwise disjoint collection of regular neighborhoods of the MATH's in MATH and define MATH . By NAME 's theorem MATH is an isomorphism, and by general position MATH and MATH are isomorphisms. It follows that MATH is a generalized MATH - neighborhood of infinity and MATH is an isomorphism. |
math/0101030 | By the hypothesis and REF , there exists a neat sequence MATH of generalized MATH - neighborhoods of infinity, a subsequence MATH of MATH, and a commutative diagram: MATH . Each MATH is necessarily surjective, so by REF each MATH is the normal closure of a finite set of elements MATH. For each MATH, choose a finite collection MATH-of pairwise disjoint embedded loops in MATH representing the elements of MATH. By the commutativity of the diagram, each MATH contracts in MATH. For each MATH choose an embedded disk MATH with MATH. Arrange that the MATH's are pairwise disjoint, and all intersections between MATH and MATH are transverse. In order to kill the kernels of the MATH's, we would like to add to each MATH regular neighborhoods of the MATH's. This would work if each MATH was contained in MATH; for then we would be attaching a finite collection of MATH - handles to each MATH and each would kill the normal closure of its attaching MATH - sphere MATH in MATH, and no more. Since this ideal situation may not be present, we must first perform some alterations on the MATH's. NAME exists a nested cofinal sequence MATH of MATH - neighborhoods of infinity which satisfy the following properties for all MATH: CASE: MATH, CASE: MATH is an isomorphism, CASE: MATH, and CASE: each MATH-bounds a MATH - disk in MATH. Roughly speaking, a MATH will be constructed by removing regular neighborhoods of the MATH's from MATH; but in order arrange REF and to maintain ``nestedness", some extra care must be taken. We already have that MATH and MATH intersects finitely many MATH REF transversely. In addition, we would like the outermost component of MATH to lie in MATH. If this is not already the case, it can easily be arranged by pushing a small annular neighborhood of MATH into MATH while leaving MATH fixed. Now choose a pairwise disjoint collection MATH of regular neighborhoods of the collection MATH; then for each MATH, choose a smaller regular neighborhood MATH. Between each MATH and MATH there exists a sequence MATH of regular neighborhoods of MATH such that MATH . For each MATH, let MATH . REF are obvious, and since each MATH was obtained from MATH by removing regular neighborhoods of MATH - complexes, REF follows from general position. Now, along each MATH it is possible to attach an ambiently embedded MATH - handle MATH to MATH. For each MATH, let MATH . The naturally induced homomorphisms MATH are now isomorphisms. Lastly, we must apply REF to each MATH to create a sequence MATH of generalized MATH - neighborhoods of infinity with the same fundamental groups. To regain nestedness, we may then have to pass to a subsequence of MATH (and to the corresponding subsequence of MATH) to complete the proof. |
math/0101030 | REF is just REF . To obtain REF , observe that if MATH is pro-equivalent to MATH, then REF implies that MATH is finitely presentable. Hence we may apply REF to obtain the desired sequence. REF follow similarly from REF , with the necessary algebra being found in REF . |
math/0101030 | Fix a triangulation of MATH and for each MATH, let MATH denote the corresponding MATH - skeleton. Let MATH denote the corresponding MATH - skeleton. Note that the inclusion MATH induces a MATH - isomorphism for all MATH. Hence, we have universal covers MATH. Since MATH, the long exact sequence for the triple MATH provides an epimorphism of MATH - modules MATH. Hence, the desired conclusion will follow if we can show that MATH is finitely generated. This will follow immediately from Theorem A of CITE if we can show that MATH for all MATH. Again we employ the exact sequence for MATH: MATH . Clearly the first term listed vanishes for all MATH and, by hypothesis, so does the third term; thus, forcing the middle term to vanish. |
math/0101030 | If MATH, there is nothing to prove; otherwise we assume inductively that MATH and each generalized MATH - neighborhood of infinity MATH contains a generalized MATH - neighborhood of infinity MATH such that MATH. Now let MATH be a generalized MATH - neighborhood of infinity. By the inductive hypothesis, we may assume that MATH is already a generalized MATH - neighborhood of infinity. We will show how to improve MATH to a MATH - neighborhood of infinity. As we noted earlier, MATH for all MATH, MATH for MATH, and MATH. Furthermore, by REF , MATH is finitely generated as a MATH - module. We break the remainder of the proof into overlapping but distinct cases: MATH . Choose a finite collection of disjoint embeddings MATH of MATH - cells representing a generating set for MATH viewed as a MATH - module. Let MATH be a regular neighborhood of MATH in MATH . Notice that MATH and MATH are both isomorphisms. (If MATH this is obvious. If MATH notice that each MATH already contracts in MATH since MATH is a generalized MATH - neighborhood.) Thus, MATH is the universal cover of MATH . Let MATH. Since the MATH's have codimension greater than REF, then MATH is an isomorphism. It remains to show that MATH is a MATH - neighborhood of infinity. To see that MATH is an isomorphism, recall from above that MATH. Then observe that the pair MATH may be obtained from the pair MATH by attaching MATH - handles (the duals of the removed handles), which has no effect on fundamental groups. To see that MATH for MATH, we will show that the corresponding MATH are trivial. By excision, it suffices to show that MATH for MATH. For MATH, the triviality of MATH can be deduced from the following portion of the long exact sequence for the triple MATH: MATH . The first term is trivial because MATH is a generalized MATH - neighborhood, and the last term is trivial when MATH because MATH is homotopy equivalent to a space obtained by attaching MATH - cells to MATH. Thus the middle term vanishes. In dimension MATH, we use a portion of the same long exact sequence: MATH . The last term above is again trivial for the reason cited above. Furthermore, the map MATH is surjective by the construction of MATH; hence MATH is trivial, so MATH vanishes. MATH . The strategy in this case is similar to the above except that when MATH we cannot rely on general position to obtain embedded MATH - disks. Instead we will use the tools of handle theory. By the inductive hypothesis and the fact that MATH is finitely generated as a MATH - module, we may choose a generalized MATH - neighborhood MATH so that, for MATH, the map MATH is surjective. Applying NAME 's theorem to MATH shows that MATH is an isomorphism, and it follows that MATH is also an isomorphism. Hence MATH is the universal cover MATH of MATH. Claim-MATH for MATH . We deduce this claim from the long exact sequence of the triple MATH: MATH . The third term listed above is trivial for MATH, therefore it suffices to show that the first term vanishes when MATH. Let MATH. Since MATH needn't be an isomorphism, MATH needn't be the universal cover of MATH. In fact, MATH will be connected if and only if MATH is surjective. In general, MATH has path components MATH (one for each element of MATH) each of which is a covering space for MATH. See REF . Moreover, MATH, and for each MATH we have MATH. Thus each MATH is a ``covering pair" for MATH. It follows that MATH is trivial for all MATH, so by the NAME Theorem, MATH for all MATH and for all MATH. Therefore MATH for MATH, implying (via excision) that MATH vanishes for MATH, thus completing the proof of the claim. We now have a cobordism MATH with MATH and MATH for MATH (where MATH). By REF, there is a handle decomposition of MATH built upon MATH which contains no handles of index MATH and so that the existing handles have been attached in order of increasing index. This give rise to a cellular chain complex for the pair MATH of the form: MATH where each MATH is a finitely generated free MATH - module with one generator for each MATH - handle of MATH. For MATH write MATH where each MATH and each MATH is a MATH - handle of MATH with a preferred base path. Let MATH denote MATH - handles, where MATH is a closed collar on MATH in MATH. We may represent MATH with a single MATH - handle as follows: introduce a trivial cancelling MATH - handle pair MATH to MATH, then do a finite sequence of handle slides of MATH over the other MATH - handles until MATH is homologous to MATH. (Again see CITE.) Now, since MATH, we may apply the NAME Lemma in MATH to move the attaching MATH - sphere of MATH off the belt spheres of all the MATH - handles. NAME the case MATH, the belt spheres of the MATH - handles have codimension MATH in MATH, so a special case of the NAME Lemma (p. REF) is needed. In particular we need to know that the belt spheres are MATH - negligible in MATH, that is, that MATH. Since MATH (attaching the MATH - handles does not kill any MATH), this condition is satisfied. See REF for the dual version of this fact. We may now assume that MATH was attached directly to MATH. By repeating this for each element of a finite generating set for MATH we obtain a finite set MATH of MATH - handles attached to MATH, so that if MATH, then MATH is surjective. The same argument used in REF will now show that MATH is a generalized MATH - neighborhood of infinity. |
math/0101030 | Consider the cobordism MATH. Since MATH it is easy to check that MATH. Hence, MATH for MATH so we may eliminate all MATH - and MATH - handles from a handle decomposition of MATH on MATH. Then the dual handle decomposition of MATH on MATH has handles only of index MATH and, by arguing as in the Claim of REF , we see that MATH for MATH, so we may eliminate all handles of index MATH from this handle decomposition. (In the process we increase the numbers of MATH - and MATH - handles.) Collapsing the remaining handles to their cores gives us MATH. |
math/0101030 | Roughly speaking, the first assertion is obtained by applying REF to each MATH. Since this process is infinite, there are some technicalities to be dealt with. We refer the reader to CITE for details. If MATH is a generalized MATH - neighborhood of infinity then we already know that MATH for MATH. Moreover, our first assertion guarantees that MATH is trivial for MATH. Thus, MATH is trivial for MATH, so by REF (see REF) MATH is a homotopy equivalence-MATH . |
math/0101030 | Finite generation of MATH follows from REF . For projectivity, consider the cellular chain complex for the universal cover MATH of the CW pair MATH provided by REF MATH . Triviality of MATH implies that MATH is surjective, so we have a short exact sequence MATH which splits since MATH is a free MATH - module. Thus MATH, so MATH is a summand of a free module, and is therefore projective. The identity MATH now follows immediately from REF. An alternative argument which relies only on CITE can be found in CITE. |
math/0101030 | The equivalence of REF - REF follows immediately from REF and our earlier discussion of MATH. Since REF is obvious, we need only show how to ``improve" a given MATH with stably free MATH to a generalized MATH - neighborhood MATH with free MATH. This is easily done by carving out finitely many trivial MATH - handles as described below. Fix MATH, and let MATH be a free MATH - module of rank MATH so that MATH is a finitely generated free MATH - module. Let MATH be a closed collar on MATH and let MATH be trivial MATH - handle pairs attached to MATH. Set MATH, and let MATH. NAME 's Theorem and general position show that each of the inclusions: MATH, MATH, MATH and MATH induce MATH - isomorphisms. Thus, MATH is a generalized MATH - neighborhood of infinity, moreover, we have a triple MATH of universal covers. Clearly MATH so for MATH the long exact sequence for triples yields: MATH . Hence, MATH for MATH, and by excision, MATH is a generalized MATH - neighborhood of infinity. In dimension MATH we have: MATH . Since MATH is free this sequence splits, so MATH as desired. |
math/0101030 | If MATH is the matrix of the basis MATH in terms of MATH, then the matrix of MATH in terms of MATH is MATH. Now MATH where MATH is a product of matrices obtained by elementary moves. |
math/0101030 | Let MATH be the universal covering projection. Then MATH is the universal cover of MATH. Also, MATH covers MATH and MATH which is perfect by an application of REF . For a fixed MATH, we will show how to construct MATH. Let MATH be a component of MATH, and let MATH be a small MATH - disk in MATH intersecting MATH transversely in a single point. Since MATH is perfect, then MATH is trivial; so MATH bounds a surface MATH in MATH. Let MATH represent an element of MATH and apply the NAME isomorphism to find a REF - sphere MATH in MATH representing the same element. Since they are invariants of homology class, the MATH - intersection number of MATH with MATH is MATH; while the MATH - intersection number of MATH with any other component of MATH is MATH. Thus, with an appropriately chosen arc to the basepoint, the MATH - intersection numbers of MATH with the MATH's are as desired. If necessary, use general position to ensure that MATH is embedded. |
math/0101035 | The signature function of a knot MATH, MATH, has jumps only at roots of the NAME polynomial, and if these roots are simple the jump is either MATH CITE. The MATH - torus knot has cyclotomic NAME polynomial MATH with MATH simple roots on the upper unit circle in the complex plane. Hence the signature MATH. On the other hand, this MATH signature is easily computed from the standard rank MATH . NAME form for MATH to be exactly MATH, and so all the jumps must be positive REF. The first of these jumps occurs at a primitive MATH - root of unity, so all MATH - signatures must be positive as desired. |
math/0101035 | Suppose that MATH and MATH. Then the MATH - cyclotomic polynomial, MATH would divide MATH. But MATH while MATH. |
math/0101035 | According to CITE the order of the homology of the MATH - fold cyclic branched cover of a knot MATH grows exponentially as a function of MATH if the NAME polynomial has a root that is not a root of unity. Hence, we only need to consider the case that all irreducible factors of the NAME polynomial are cyclotomic polynomials, MATH. Using REF , the result is reduced to the case that that MATH. As in the proof of REF , MATH cannot be a prime power. An elementary argument using the resultant of polynomials (see for instance CITE) gives MATH where the second product is taken over all primitive MATH - roots of unity. Let MATH and let MATH. One has that MATH for some primitive MATH - root of unity and with a bit of care one sees that the product can be rewritten as MATH where now the product is over all primitive MATH - roots of unity and MATH. (Though we don't need it, a close examination shows that if MATH is greater than or equal to the maximal power of MATH in MATH then MATH, otherwise MATH.) If MATH has three distinct prime factors then MATH has at least two distinct prime factors and this product is REF (see for instance CITE). On the other hand, if MATH has two distinct prime factors, then by letting MATH be one of those factors and letting MATH be large, it is arranged that MATH is a prime power and the product yields that prime and in particular is greater than REF. This concludes the proof of the first statement of REF . Finally, suppose that all cyclic branched covers of MATH are homology spheres. By the above discussion we just need to show that no factor of the NAME polynomial is MATH for any MATH. But from REF we see that if MATH divides the NAME polynomial then the MATH - fold cyclic branched cover would have infinite homology. This concludes the proof. |
math/0101036 | This follows quickly from CITE. |
math/0101036 | Considering MATH as an ordinary first-order structure in a language MATH in which it has elimination of quantifiers, let MATH be an expansion of MATH with NAME functions, MATH the language of MATH. Let MATH. Using a standard application of the NAME Theorem there exists in some model MATH of MATH, a sequence MATH indiscernible over MATH in MATH such that for every MATH, there exists MATH in MATH with MATH . Without loss of generality, MATH is the NAME hull of MATH, hence every complete type in MATH in finitely many variables realized in MATH is also realized in MATH. Assuming that MATH, by the MATH-homogeneity of MATH there is MATH, such that MATH . This completes the proof. |
math/0101036 | REF follows from REF follows from REF by taking MATH. |
math/0101036 | MATH follows from the definition of dividing and MATH is trivial. To prove that REF implies REF let MATH be an infinite MATH-indiscernible sequence with MATH. By REF , there is an ordered set MATH extending MATH with MATH (see MATH at the beginning of the section for the definition of MATH) such that MATH is MATH-indiscernible. Since MATH does not divide over MATH there is MATH in MATH realizing MATH. REF gives the existence of MATH indiscernible over MATH such that for MATH there exists MATH in MATH satisfying MATH . Since MATH for each MATH we've proved REF implies REF . For REF implies REF , let MATH and MATH indiscernible over MATH with the same diagram over MATH as MATH such that MATH, for MATH. By REF there is a sequence MATH indiscernible over MATH indexed by MATH with the same diagram over MATH as MATH. Now MATH and MATH are both indiscernibles indexed by MATH with the same diagram over MATH. Thus, there is an automorphism MATH of MATH fixing MATH and taking MATH to MATH, for MATH. Then MATH realizes MATH and MATH is indiscernible over MATH, proving the lemma. |
math/0101036 | Let MATH be any infinite indiscernible sequence over MATH containing MATH. By the preceding lemma there is MATH in MATH realizing MATH such that MATH is indiscernible over MATH. Let MATH be an automorphism fixing MATH and sending MATH to MATH. Then, MATH does not divide over MATH, hence there is MATH realizing MATH such that MATH is indiscernible over MATH. Since MATH realizes MATH and MATH the proposition is proved. |
math/0101036 | Suppose to the contrary that MATH is not a MATH-Morley sequence over MATH. Let MATH be finite subsets of MATH such that for some finite subsets MATH and MATH, MATH divides over MATH, and MATH is minimal with this property. Let MATH be the largest element of MATH and MATH. Let MATH and MATH. Then, MATH does not divide over MATH by the minimality assumption on MATH. By the hypotheses of the lemma, MATH does not divide over MATH. So, by Pairs Lemma REF , MATH does not divide over MATH. This contradicts our assumption that MATH divides over MATH to prove the lemma. |
math/0101036 | First find an infinite MATH-indiscernible sequence that is MATH-free using REF as follows. Let MATH be an ordinal MATH and suppose sequences MATH, MATH, have been chosen so that MATH realizes MATH and MATH is MATH-free from MATH over MATH. By the Extension Property in the definition of MATH-simple there is MATH realizing MATH which is MATH-free from MATH over MATH. Since MATH satisfies MATH there is MATH such that MATH is MATH-indiscernible and for any MATH there are MATH with MATH. This latter property controls the type diagram of MATH and by REF guarantees that MATH is a MATH-Morley sequence. Given an infinite linear order MATH, MATH, by REF there is a sequence MATH indiscernible over MATH with the same type diagram as MATH. Thus, MATH is also a MATH-Morley sequence in MATH over MATH. Since MATH is MATH-simple, MATH-freeness has finite character. Thus, MATH is also a MATH-Morley sequence, proving the lemma. |
math/0101036 | MATH is simply by the definition of dividing. MATH because such a NAME sequence exists by REF . We now prove MATH . Let MATH be the given NAME sequence and suppose MATH divides over MATH. Notice that MATH is also a NAME sequence in MATH over MATH and satisfies MATH, and MATH divides over MATH. So, replacing MATH by MATH if necessary, we may as well assume MATH. For each MATH and MATH with MATH and MATH, MATH divides over MATH. Simply because MATH realizes MATH, MATH divides over MATH. This fact is witnessed by an infinite MATH-indiscernible sequence MATH containing MATH. Let MATH. Since MATH is MATH-free from MATH over MATH and MATH, MATH does not divide over MATH. Thus, there is a sequence MATH such that MATH. By REF , we can assume that MATH is indiscernible over MATH. Since MATH there is an automorphism MATH fixing MATH taking MATH to MATH. Let MATH. Then, MATH is infinite, indiscernible over MATH and MATH is inconsistent. This witnesses that MATH divides over MATH. To continue with the proof suppose to the contrary that MATH is realized by some MATH. By the Bounded Dividing Property there is MATH of cardinality MATH such that MATH is MATH-free from MATH over MATH. Since MATH satisfies MATH there is MATH, MATH. By the claim, MATH divides over MATH. Since MATH satisfies MATH we contradict that MATH is MATH-free from MATH over MATH. This proves the proposition. |
math/0101036 | If MATH is small then MATH does not divide over MATH REF . Suppose MATH is large and let MATH be an arbitrary MATH-indiscernible sequence in MATH. For any MATH, any realization of MATH also realizes the small type MATH. Using REF and cardinality properties we find a MATH realizing MATH and an infinite sequence MATH indiscernible over MATH with the same type diagram over MATH as MATH, such that MATH realizes MATH. By REF , MATH does not divide over MATH. Using MATH and REF there is no infinite MATH-indiscernible sequence in MATH. Thus, MATH does not divide over MATH. |
math/0101036 | This is simply a counting argument. Let MATH. Since MATH is large there is an infinite MATH-indiscernible sequence MATH in MATH. Since MATH is small for any MATH there cannot be a MATH realizing MATH such that MATH is indiscernible over MATH and MATH, for MATH. By REF , MATH divides over MATH. |
math/0101036 | Let MATH. Let MATH, MATH, be such that MATH is free from MATH over MATH and MATH. Notice that MATH is large since MATH is large and MATH is small. Let MATH be a cardinal MATH and MATH be a NAME sequence in MATH over MATH, where MATH is the reverse order on MATH. For MATH let MATH enumerate the set of realizations of MATH. Since MATH is small we can choose MATH so that MATH is indiscernible over MATH and free from MATH over MATH. Given MATH an arbitrary sequence of length MATH, MATH is free from MATH over MATH since MATH is small REF . By Pairs Lemma REF , MATH is free from MATH over MATH, hence, MATH is free from MATH over MATH. In particular, MATH is free from MATH over MATH. By REF the sequence MATH is disjoint from MATH. We conclude that MATH is inconsistent for MATH. Since MATH is a NAME sequence, MATH divides over MATH REF . |
math/0101036 | If MATH is small then MATH does not divide over MATH REF . Suppose MATH is large. Since MATH does not divide over MATH, MATH is large REF . Thus, there is a NAME sequence MATH in MATH over MATH, where MATH is the reverse order on MATH. Let MATH. Since MATH is a sequence of indiscernibles over MATH in MATH, MATH realizes MATH. By REF , MATH does not divide over MATH, proving the lemma. |
math/0101036 | By Local Symmetry MATH does not divide over MATH and MATH does not divide over MATH. Applying Pairs Lemma and MATH, MATH does not divide over MATH; hence, MATH does not divide over MATH. Thus, MATH does not divide over MATH, again by Local Symmetry. |
math/0101036 | To begin we claim that MATH does not divide over MATH. Suppose, to the contrary, that MATH divides over MATH. Let MATH. Since MATH is free from MATH over MATH, MATH is free from MATH over MATH. Since MATH must be large (or MATH would be free from MATH over MATH) there is an infinite MATH that is a NAME sequence over MATH in MATH and indexed by MATH. Using that MATH is free from MATH over MATH and MATH is indiscernible over MATH, MATH is consistent. Since MATH is a NAME sequence over MATH, this contradicts that MATH divides over MATH and REF . Let MATH be a sequence of length MATH. By symmetry and the choice of MATH, MATH does not divide over MATH. (In fact, the type doesn't divide over MATH, but that is more than we need.) Combining this fact and the claim with Pairs Lemma, MATH does not divide over MATH. By symmetry, MATH is free from MATH over MATH. Thus, MATH is free from MATH over MATH. |
math/0101036 | Let MATH, MATH and MATH be arbitrary sequences of length MATH. Let MATH, MATH, be such that MATH is free from MATH over MATH. Since MATH is free from MATH over MATH, REF implies that MATH does not divide over MATH. By Local Symmetry REF MATH does not divide over MATH. Thus MATH does not divide over MATH and MATH does not divide over MATH (see REF ) Thus, MATH is free from MATH over MATH. This proves the theorem. |
math/0101036 | Let MATH and MATH be arbitrary sequences of length MATH. Let MATH, MATH, be so that MATH is free from MATH over MATH, and let MATH, MATH, be such that MATH is free from MATH over MATH. By Weak Transitivity REF MATH is free from MATH over MATH and MATH is free from MATH over MATH. By Local Transitivity, MATH is free from MATH over MATH. Thus, MATH is free from MATH over MATH, proving the corollary. |
math/0101036 | Let MATH. CASE: By Transitivity, for any MATH, MATH is free from MATH-over MATH. By REF , MATH is a NAME sequence in MATH over MATH. CASE: Without loss of generality, MATH is isomorphic to MATH. Let MATH and MATH be sequences of length MATH, MATH, MATH, MATH and MATH, for MATH, MATH. Let MATH. Since MATH holds for all MATH, MATH does not divide over MATH by REF . By symmetry, MATH does not divide over MATH, for MATH. Thus, MATH is a NAME sequence in MATH over MATH (using REF ). |
math/0101036 | The hypotheses of the lemma and REF yield a NAME sequence MATH in MATH over MATH, where MATH is the reverse order on MATH. A fortiori, MATH is a NAME sequence over MATH. Let MATH be a type over MATH for some MATH, MATH, satisfied by some MATH, MATH. Since MATH is indiscernible in MATH, MATH holds for all MATH. Since MATH is a NAME sequence over MATH, MATH does not divide over MATH by REF . Thus, MATH is free from MATH over MATH. |
math/0101036 | Let MATH. The equivalence relation defined by MATH is MATH-invariant with a bounded number of classes (since MATH is small). |
math/0101036 | By REF and the MATH-invariance of MATH, MATH can be extended to any length MATH. If MATH then MATH for any MATH in MATH. Hence, there are unboundedly many equivalence classes. |
math/0101036 | REF implies REF follows from the previous lemma and transitivity of equivalence. For REF implies REF , call MATH the equivalence relation defined by REF . Notice that MATH is MATH-invariant. Suppose MATH had unboundedly many equivalence classes and let MATH be inequivalent elements for some suitably large MATH. By REF , there exists MATH indiscernible over MATH, such that MATH for some MATH. By the definition of MATH, we have MATH. Hence, MATH, by MATH-invariance, contradicting the choice of MATH. Thus, MATH. Therefore, if MATH, then MATH holds so that REF holds. |
math/0101036 | REF implies REF follows immediately from the definition. We show REF implies REF . Define MATH if there exists MATH with MATH. This is clearly an equivalence relation since MATH is a group. Notice that it is also MATH-invariant since MATH is normal in MATH. Hence, it is enough to show that MATH. Suppose not and let MATH be a large set of MATH-inequivalent elements. Let MATH be a bounded set extending MATH containing a representative of every Lascar strong type over MATH. By the pigeonhole principle, there exists MATH such that MATH. Let MATH such that MATH. But MATH, a contradiction. |
math/0101036 | This is immediate by REF . |
math/0101036 | By REF , there exists MATH such that MATH. Let MATH. Another application of the that lemma shows the conclusion. |
math/0101036 | Since MATH is large there is an infinite NAME sequence MATH over MATH with MATH. Using REF there is an infinite sequence of indiscernibles MATH such that MATH is indiscernible over MATH and MATH is indiscernible over MATH. Then MATH is a NAME sequence over MATH and MATH. By REF , given MATH, MATH is a NAME sequence in MATH over MATH. In particular, MATH is free from MATH over MATH. |
math/0101036 | Without loss of generality, MATH where MATH is the reverse order on MATH. Let MATH be a suborder of MATH such that MATH is coinitial in MATH, MATH is coinitial in MATH and MATH is isomorphic to MATH. Let MATH be an injective function from MATH into MATH such that MATH, for MATH. Let MATH. It is routine to show that MATH is a NAME sequence in MATH, for MATH. For MATH, let MATH. Since MATH does not divide over MATH there is a MATH realizing MATH. Thus, MATH realizes MATH. By REF , MATH does not divide over MATH, proving the lemma. |
math/0101036 | Let MATH be the order obtained by adding a single element MATH to the end of MATH. Let MATH be a sequence in MATH such that MATH is MATH-indiscernible. By the MATH-simplicity of MATH there is MATH, MATH, such that MATH is MATH-free from MATH over MATH, MATH. Let MATH and MATH. If MATH, then MATH and MATH have the same type over MATH. Thus, MATH is MATH-free from MATH over MATH. By REF and MATH-simplicity, MATH is a NAME sequence over MATH. This proves the lemma. |
math/0101036 | Without loss of generality, MATH is indexed by MATH. By REF , there is MATH so that MATH is a NAME sequence over MATH, for MATH. For MATH, MATH does not divide over MATH, hence it does not divide over MATH. By the Extension Property there is a MATH realizing MATH such that MATH does not divide over MATH. Since MATH is a NAME sequence over MATH, for MATH, MATH does not divide over MATH. Thus, there is MATH realizing MATH such that MATH does not divide over MATH. Since MATH realizes MATH, MATH does not divide over MATH. By Transitivity REF , MATH does not divide over MATH. A fortiori, MATH does not divide over MATH. |
math/0101036 | By REF , let MATH, MATH be a sequence such that there exists MATH-indiscernible sequences containing MATH and MATH, for MATH. By the extension property, we may find MATH realizing MATH such that MATH does not divide over MATH. Hence, by using an automorphism fixing MATH and MATH, we may assume that MATH does not divide over MATH. Hence, MATH does not divide over MATH, so it is enough to show the conclusion when MATH, MATH belong to the same indiscernible sequence MATH (since then MATH does not divide over MATH, and so MATH does not divide over MATH, etc.). There is a MATH-indiscernible sequence MATH such that MATH, for each MATH. Let MATH be a cardinal, MATH, and MATH the order with MATH reversed with the elements of MATH denoted MATH, for MATH. Since MATH is MATH-indiscernible, we may in fact assume that it is of the form MATH, MATH and MATH. Notice that MATH is also indiscernible over MATH. Since MATH does not divide over MATH, REF says we can choose MATH so that MATH is indiscernible over MATH. Notice that MATH and MATH, for each MATH. By the indiscernibility of MATH and the homogeneity of MATH, for MATH there is a MATH such that MATH for MATH and MATH for MATH. By REF , there is a MATH-indiscernible sequence MATH such that MATH for some MATH. In particular, MATH, completing the proof of the claim. Since MATH does not divide over MATH, MATH does not divide over MATH, by REF . In particular, MATH does not divide over MATH. Since, MATH the lemma is proved. |
math/0101036 | Suppose that MATH does not divide over MATH. There is a subsequence MATH of MATH of length MATH such that MATH does not divide over MATH so we may as well assume MATH. For MATH a Lascar strong type over MATH let MATH be the set of MATH, MATH such that there is MATH realizing MATH and MATH does not divide over MATH. If MATH, for some Lascar strong type MATH over MATH, then MATH. Given MATH in MATH, for MATH, let MATH in MATH such that MATH does not divide over MATH. By Type Amalgamation there are, for MATH, MATH realizing MATH such that MATH does not divide over MATH. Thus, MATH, for MATH, hence MATH, proving the claim. Since the class of Lascar strong types over MATH is bounded, MATH contains a bounded number of parallelism classes. |
math/0101036 | In the special case where MATH is MATH-indiscernible, this follows basically from REF . In general we will use this fact and that MATH is an amalgamation base. Recall REF about independence of NAME sequences. There is a sequence MATH of length MATH such that MATH is MATH-indiscernible, MATH is independent over MATH and MATH is free from MATH over MATH. To see this, first let MATH be a sequence indexed by MATH such that MATH is indiscernible over MATH. Consider MATH as an indiscernible sequence in the reverse order (indexed by MATH) and apply REF to obtain a sequence MATH of length MATH such that, letting MATH, MATH is a NAME sequence over MATH (under the reverse order). Thus, MATH is independent over MATH. Since MATH is indiscernible over MATH, MATH is independent over MATH. Now suppose MATH is a finite subset of MATH and suppose, towards a contradiction, that MATH divides over MATH. There is a MATH, MATH, such that MATH is free from MATH over MATH. By the indiscernibility of MATH over MATH we can assume MATH is disjoint from MATH, hence free from MATH over MATH. Since MATH is free from MATH over MATH transitivity of independence implies that MATH is free from MATH over MATH. This contradiction proves the claim. Arguing as in the claim there is also a (nonempty) MATH, MATH, such that MATH is free from MATH over MATH . Let MATH. Let MATH be any element of MATH and MATH. There is MATH free from MATH over MATH such that MATH and MATH hold. We first show there is MATH realizing MATH and MATH. Since MATH is an amalgamation base, MATH. So, there is MATH, a strong automorphism over MATH taking MATH to MATH. Then MATH and MATH holds. Since MATH is free from MATH over MATH and MATH is free from MATH over MATH . Type Amalgamation yields MATH free from MATH over MATH such that MATH and MATH hold. The following chain of arguments shows that MATH. MATH (transitivity) MATH (symmetry) MATH (transitivity and the independence of MATH and MATH over MATH) MATH (transitivity) MATH (transitivity and the fact that MATH holds) MATH (transitivity) MATH (symmetry) MATH (transitivity and the fact that MATH holds) MATH (a fortiori). Finally, to show that MATH, it suffices to prove that MATH by transitivity and the independence of MATH and MATH over MATH. It is enough to show that MATH, for any MATH with MATH. The argument above can be repeated with MATH replacing MATH to prove that MATH. An application of transitivity then completes the proof. |
math/0101036 | By the definition of parallelism there are MATH, MATH, such that MATH. The proof is by induction on MATH, so we assume there is MATH conjugate to MATH, MATH, and there is MATH realizing MATH such that MATH does not divide over MATH and does not divide over MATH. Since MATH there is MATH realizing MATH such that MATH and MATH. Without loss of generality, MATH. To prove that MATH a NAME sequence in MATH must be introduced and the preceding lemma applied. Since MATH does not divide over MATH there is a sequence MATH of length MATH which is a NAME sequence in MATH, is free from MATH over MATH, and free from MATH over MATH. Without loss of generality, MATH is free from MATH over MATH. Thus, MATH is free from MATH over MATH (by symmetry and transitivity), and MATH is free from MATH over MATH. MATH is free from MATH over MATH by REF . Also by transitivity, MATH is free from MATH over MATH, hence MATH is free from MATH over MATH. Since MATH is free from MATH over MATH, MATH is free from MATH over MATH, hence MATH is free from MATH over MATH. In particular, MATH is free from MATH over MATH, proving the lemma. |
math/0101036 | CASE: Since any element of MATH is the image of a tuple from MATH under a function interpreting a symbol in MATH, this is straightforward. CASE: By REF the elements of MATH are invariant under automorphisms of MATH. The other conditions follow quickly from the construction of MATH and the fact that MATH is a logical structure. CASE: By the construction of MATH, for any finite tuple MATH from MATH there is MATH such that MATH holds if and only if MATH. For the given elements MATH suppose MATH and MATH. Let MATH, MATH and MATH define MATH. The relation MATH expressing MATH is in MATH, hence in MATH. Thus, there are MATH, MATH, MATH and MATH. There is an automorphism MATH of MATH (and MATH) such that MATH, hence MATH and MATH, proving the lemma. |
math/0101036 | Left to the reader with the hint in the preceding paragraph. |
math/0101036 | Given MATH so that MATH there is a MATH such that MATH and MATH is large. Let MATH be a MATH-indiscernible sequence in MATH, whose existence is guaranteed by the MATH-homogeneity of MATH. Then, MATH is a MATH-indiscernible sequence in MATH. |
math/0101036 | Let MATH be a base for MATH. Suppose MATH and let MATH. If MATH is sufficiently large there is an infinite MATH-indiscernible sequence MATH such that for all MATH, there are MATH with MATH. From here the proof is as in REF . |
math/0101036 | MATH . First observe that MATH is large, since MATH divides over MATH. Let MATH be an infinite sequence ultra-indiscernible over MATH in MATH, where MATH is indiscernible over the base MATH for MATH and MATH is inconsistent in MATH. To reach a contradiction, suppose there are MATH such that MATH, MATH and MATH does not divide over MATH. Since MATH, for each MATH there is MATH realizing MATH such that MATH. By standard arguments about indiscernibles there are MATH, MATH, such that MATH is MATH-indiscernible in MATH and MATH. Since MATH does not divide over MATH there is MATH realizing MATH. Then, MATH realizes MATH. This contradiction completes the proof of this direction. MATH . Now suppose that MATH does not divide over MATH, MATH is any base for MATH and MATH. We need to find a MATH such that MATH and MATH does not divide over MATH. If MATH is small then for any MATH such that MATH, MATH does not divide over MATH. So we can assume MATH is large and let MATH be a NAME sequence in MATH where MATH satisfies MATH in REF . Then, MATH is ultra-indiscernible over MATH in MATH. Since MATH does not divide over MATH there is MATH realizing MATH. Let MATH. MATH can be chosen arbitrarily large so there is an arbitrarily large MATH also satisfying MATH in REF such that MATH, for MATH. Thus, for MATH, MATH, REF shows that MATH does not divide over MATH. Since there is a MATH realizing MATH such that MATH, the proof is complete. |
math/0101036 | Let MATH be the equivalence relation on MATH such that MATH holds if and only if MATH is parallel to MATH. Let MATH be MATH. Let MATH be the unique hyperimaginary element obtained by adjoining all hyperimaginaries in MATH. Clearly, MATH is the unique canonical base for MATH. |
math/0101036 | This is immediate by REF . |
math/0101036 | CASE: We show the contrapositive. Suppose that MATH does not divide over MATH and let MATH be indiscernible over MATH where MATH are in the parameters of MATH. Let MATH be such that MATH. We must show that MATH. Consider MATH. By MATH-homogeneity, we may assume that MATH. Notice that MATH is indiscernible over MATH. Then MATH is realized by some MATH, since MATH does not divide over MATH. Hence MATH holds for all MATH, and by stability, MATH holds for all but fewer than MATH elements MATH. This implies that MATH holds for some MATH. Hence MATH, so in particular MATH, that is, MATH. The proof that MATH implies that MATH is similar. CASE: Follows from REF by definition of freeness (freeness has finite character). |
math/0101036 | Assume that MATH is simple. REF follows from REF or REF by counting types just like in the first-order case (we use the fact that each type is free over a subset of size less than MATH of its parameters). To see that REF implies REF , use the preceding lemma and REF, which says that if MATH and MATH do not split strongly over MATH and have nonsplitting extensions over any set, and if in addition MATH then MATH. REF implies REF is also as in the first-order case: Let MATH and MATH containing MATH be given. Assume MATH is stable. We show that REF holds with MATH and MATH. By the preceding lemma, it is enough to show that the number of extensions in MATH which do not split strongly over MATH is at most MATH. Let MATH contain a realization for each MATH for MATH a finite sequence in MATH. Then, MATH has size at most MATH: By the stability spectrum theorem, MATH is stable in MATH, so any set of size MATH contains an infinite MATH-indiscernible subset, and thus contains different realizations of the same Lascar strong type over MATH. Notice that if MATH, MATH do not split strongly over MATH and MATH, then MATH. To see this, suppose MATH. Let MATH realize MATH. Then there exist MATH and MATH for MATH such that MATH, MATH and MATH belong to an infinite MATH-indiscernible sequence. Since MATH does not split strongly over MATH, we have MATH, for MATH. Hence MATH, so MATH, and so MATH by the same argument applied to MATH. Hence, the number of free extensions over MATH is bounded by MATH which is at most MATH, since MATH is stable in MATH. |
math/0101036 | We are assuming MATH, hence MATH. Suppose MATH does not divide over MATH in MATH. If MATH is small, then MATH does not divide over MATH. So, suppose MATH is large and let MATH be any indiscernible sequence in MATH of cardinality MATH. For MATH there is MATH such that MATH. There is MATH, MATH, such that MATH is independent from MATH over MATH. By the stationarity of strong types in a stable theory, MATH realizes MATH. Since indiscernible sequences in a stable theory are indiscernible sets, MATH and MATH have the same isomorphism type over MATH. Thus, there is MATH realizing MATH. This shows that MATH does not divide over MATH. |
math/0101036 | Let MATH be a homogeneous universal model of the universal theory MATH (see CITE or CITE for an earlier reference) that contains MATH and has arbitrarily large cardinality. Let MATH be the quantifier-free definable relations on MATH. Let MATH be the equivalence relation on MATH that holds if MATH, for all MATH. Thus, MATH is MATH-type definable. Let MATH be the logical structure on the sort MATH in MATH. By REF , MATH is homogeneous. The model MATH consists of MATH interpreting MATH and MATH interpreting MATH. The reader should verify that MATH satisfies the axioms for a pre-Hilbert space relative to the field MATH under the operations and relations inherited from MATH. The real numbers are a subring of MATH since each real is a constant symbol of the language. Notice that if MATH, and MATH for some MATH, then MATH. Let MATH. MATH is a NAME space homogeneous with respect to quantifier-free relations. Let MATH. In MATH, MATH, hence MATH. Thus, MATH is a submodel of MATH. The crucial axioms above are universal, so MATH is also a pre-Hilbert space. Since MATH is an uncountable homogeneous universal model of MATH it realizes any consistent quantifier-free type over a countable subset. It follows that MATH defines a complete metric space on MATH. Moreover, it is easy to verify that MATH is homogeneous with respect to the quantifier-free relations. This proves the claim and the lemma. |
math/0101036 | Let MATH be a MATH-indiscernible sequence in MATH. The construction of MATH at the beginning of this section, and the fact that MATH is orthogonal over MATH guarantees the existence of a MATH realizing MATH such that MATH, for all MATH. Checking the possible relations on MATH, this implies that MATH, for all MATH. This witnesses that MATH does not divide over MATH. Continuing, for each MATH, there is MATH such that MATH. Since MATH, MATH has a unique extension over MATH. Thus, for each MATH, MATH. We conclude that MATH does not divide over MATH. |
math/0101036 | Let MATH be a complete orthogonal subset of MATH, chosen so that MATH is a complete orthogonal subset of MATH. Let MATH, which is countable by NAME 's Inequality. Let MATH and MATH. Let MATH be a minimal subset of MATH such that MATH. Since MATH is countable, MATH is countable by REF , and we may assume that MATH. By enlarging MATH and MATH if necessary we can assume that MATH implies that MATH. By REF , MATH does not divide over MATH, proving the lemma. |
math/0101036 | Consider the NAME spaces MATH. Let MATH be a complete orthogonal subset of MATH and MATH a complete orthogonal subset of MATH. Notice that each MATH is orthogonal to MATH. Since MATH, MATH has a unique extension over MATH, namely MATH, which also does not divide over MATH. There is a MATH realizing MATH such that each MATH is orthogonal to MATH, and among such MATH there is a unique type over MATH. There is a unique extension of MATH over MATH, and by REF MATH does not divide over MATH. This proves the lemma. |
math/0101037 | Assuming for the moment, that MATH, we get that MATH. Then, if MATH, we conclude that MATH. By definition of MATH, we then have MATH so MATH, hence MATH. Since MATH is a linear operator, MATH holds as well, so MATH. But MATH is a subspace of MATH, so it follows that MATH. This means that MATH only if MATH so that MATH if MATH. This implies our monotonically decreasing upper bound for the dimensions of the MATH. To see that MATH: we use induction. We have MATH, so MATH. If MATH, then MATH. |
math/0101037 | We first observe that MATH or equivalently that MATH,MATH. A little bit of thought is enough to convince oneself that MATH. Therefore we get that MATH.(Here we have used the invertibility of MATH to conclude that MATH). Let us approach the problem a little more generally. We shall use the fact that MATH to show that the set MATH of all MATH in MATH such that MATH is open and of full measure. Here MATH and MATH are linear subspaces of MATH. Define column vectors MATH for MATH that are orthogonal to each other and span MATH. Likewise let MATH for MATH be column vectors orthogonally spanning MATH. Define the MATH by MATH matrices MATH and MATH as follows: MATH and MATH . Then MATH and MATH so that we get MATH . Now we note that MATH . To show that MATH is open in MATH, observe that MATH and therefore MATH . Note that MATH is a continuous function of MATH, that is, MATH is continuous (actually smooth). Assume without the loss of generality, that MATH. Let MATH be the MATH dimensional measure of regions in MATH applied to the parallelepiped with edges equal to the columns of MATH. Then MATH is precisely equal to MATH. Since both MATH and MATH are continuous, we have that MATH is open. To show that MATH has zero MATH-dimensional NAME measure, we first introduce a change of coordinates. Define MATH to be an orthogonal matrix obtained by filling in the zero columns of MATH appropriately. Obtain MATH from MATH analogously. Then MATH where MATH is the identity matrix of dimension MATH and we have set MATH. So MATH . Since MATH and MATH are orthogonal, we have that the MATH dimensional NAME measure of MATH and MATH are equal. We now prove that MATH has measure zero. Any MATH can be written as the block matrix with dimension of MATH being MATH. MATH . We can write MATH out in terms of elements as MATH . Now assume that MATH. Identify MATH with the elements of MATH of a MATH matrix with the elements MATH removed. Next, define a mapping of MATH by MATH which has, as its image precisely those matrices in which MATH has column MATH that is the linear combination of the first MATH columns. If we redefine our mapping to get a series of completely analogous mappings, each of which has a column of MATH being dependent on the other columns of MATH, then we end up with MATH such maps. Since each of these maps are smooth, and singular (the rank of the derivative is not equal to the dimension of the image space) NAME 's theorem CITE tells us that the MATH-dimensional measure of the image of each map is zero. Therefore the union of the images also has measure zero. But this union is exactly MATH. Since the case of MATH is completely analogous, we have now shown that MATH is open and of full measure in MATH. To complete the proof we let the operator change at each step so that MATH , MATH , and so on. Now we have an extended operator MATH. We will show that the set MATH is open and dense in MATH. Define MATH to be the open subset of full measure in MATH whose members, MATH, satisfy MATH. Choose a countable subset, MATH, which is dense in MATH. Now for each element MATH of MATH, define MATH to be the open full measure subset of MATH such that MATH implies that MATH, or equivalently, MATH. Define MATH. Since MATH is dense in MATH we can pick a countable MATH that is also dense in MATH. We have that MATH and MATH implies that MATH and MATH. Continuing this process we obtain MATH for MATH such that MATH implies that all the intersections are transverse, that is, that MATH for MATH. We have therefore found a subset of MATH which is dense in MATH. Now we show that MATH is open. The requirement that each of the intersections are transverse is equivalent to the requirement that MATH which in turn is equivalent to a requirement involving orthogonal complements, specifically that MATH all have full rank where MATH is the matrix with rows equal to independent n-dimensional vectors spanning the linear subspace MATH. This last set of expressions follows from the fact that if MATH and MATH are linear subspaces of MATH then MATH so that the matrices immediately above have rank MATH iff the the previous intersections have dimensions MATH respectively. But this last expression can be seen to be exactly those matrices which satisfy the equations MATH where MATH measures the MATH-dimensional volume of the the parallelepiped spanned by the MATH in MATH. But since the inverse operation is continuous on the set of invertible matrices and these volume functions are smooth, we have that the set of MATH having the full rank property is open. Thus, MATH is open in MATH. |
math/0101037 | As was seen in the proof of the previous theorem, the transversality requirement and the fact that we are considering the case of MATH for all MATH, reduces to MATH all having full rank. But this is equivalent to another full rank condition as follows. Let MATH be the transpose of MATH. In other words, while MATH are the orthogonal row vectors that span compliment of the null space, MATH are the column vectors that span the ``same" space. This condition is equivalent to the following. For a dense and open set of MATH: MATH . That is, for an open and dense set of MATH the sequence of subspaces MATH generated by the iterates MATH, are of maximal dimension. The proof of this fact is well known. For lack of a reference I give a proof here. Without loss of generality let MATH be the k left most columns of the MATH identity matrix. Then the dimension of MATH is the rank of the matrix formed by taking the k left columns of I, followed by the k left columns of L, followed by the k left columns of MATH, and so on. Pick the upper left matrix minor and compute it's determinate this will give a polynomial in MATH which we want to show is nonzero except on an open and dense set. As long as the polynomial is nonzero at one point then we are done since this implies that the set of zeros occupies a submanifold of MATH that is at most, MATH dimensional. (So we even have more . the set of ``good" MATH has full measure and is open!) To show that the determinant of the upper left matrix minor is nonzero at a point, we consider MATH = permutation matrix that shifts everything to the left k clicks. This gives us the identity matrix in the upper left matrix minor and so the determinant in question evaluates to REF! |
math/0101044 | Let MATH, for an appropriate sequence MATH. Recall that MATH. Then combining REF with REF have that some subsequence of the MATH converges to a MATH-invariant measure MATH supported on MATH. Note that in CITE, the notation MATH refers to a subset of a fundamental set of roots corresponding to the face of a NAME chamber containing MATH in its boundary. If MATH converges then both MATH and MATH exist (note the definition of MATH in CITE). Again in the notation of CITE, MATH is the conjugate subgroup MATH in MATH. Moreover, MATH is the orbit MATH. By REF any other convergent subsequence of the MATH produces the same measure in the limit, and therefore the sequence MATH itself converges to MATH uniquely. |
math/0101044 | It suffices to show that given a geodesic segment MATH between two points MATH, there exists some MATH such that function MATH is strictly convex in MATH, and hence on an open positive MATH-measure set around MATH. We know it is convex by the comment preceding the statement of the proposition. If MATH is constant on some geodesic subsegment of MATH for some MATH, then MATH must lie in some flat MATH such that the geodesic between MATH and MATH (which meets MATH at a right angle) also lies in MATH. On the other hand, MATH is in the direction of the algebraic centroid in a NAME chamber, and MATH is perpendicular to this direction. By the properties of the roots, MATH is a regular geodesic (that is, MATH is not contained in the boundary of a NAME chamber). In particular, MATH is contained in a unique flat MATH. Furthermore, MATH is a finite set (an orbit of the NAME group). As a result, for almost every MATH is strictly convex in MATH. For fixed MATH, by the last property listed in REF , we see that MATH tends to MATH as MATH tends to any boundary point MATH. Then for fixed MATH and MATH, MATH increases to MATH as MATH tends to any boundary point MATH. Hence it has a local minimum in MATH, which by strict convexity must be unique. |
math/0101044 | From its definitions, MATH is continuous in MATH and MATH. Observe that for fixed MATH, MATH. If follows that MATH. This implies the proposition. |
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