paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0107218 | of REF CASE: Since MATH, we have MATH for MATH, thus MATH and we may assume MATH for some MATH. Further we may assume that MATH and that MATH acts nondegenerately on some NAME space MATH. Then in particular the support projection of MATH is MATH. (If MATH is the zero map, then the closed subspace MATH of MATH will just be the empty set and there is nothing to show.) REF MATH: It suffices to consider MATH. But then MATH for some elementary set MATH and MATH. Now MATH, and since MATH for MATH, we see that MATH. REF says that MATH for some u.c.p. map MATH, but a glance at the proof tells us that in fact MATH . We obviously have MATH. Note that MATH: Take rank-one projections MATH. Since MATH, we see that MATH but MATH because MATH. Using the universal property of the maximal tensor product and the fact that MATH and MATH commute, it is straightforward to check that MATH and that, under this identification, MATH. Note that MATH is a quotient of MATH, the universal MATH-algebra generated by a positive element of norm MATH; so MATH for some closed MATH, and the isomorphism is given by sending MATH to MATH. CASE: The only thing left to show is that MATH is a MATH-homomorphism, for then MATH and with all these identifications MATH is given by MATH. For any rank-one projection MATH we have MATH because MATH; now since MATH is a projection, so is MATH. But then by REF , MATH; since MATH was arbitrary and MATH is spanned by its minimal projections, MATH is multiplicative on all of MATH. CASE: If MATH is a projection, then so is MATH for each MATH (the MATH are mutually orthogonal and add up to MATH). But then we have for each MATH which implies that MATH and MATH for all MATH for which MATH, and this in turn means that MATH is a MATH-homomorphism. CASE: We will obtain MATH from MATH by setting MATH for a suitable element MATH. Then MATH, where MATH and MATH. Note also that MATH, where MATH is the i-th component of MATH. We may therefore again assume MATH for some MATH. Now suppose MATH (it will soon become clear how small MATH has to be). By REF there exists a projection MATH satisfying MATH. For MATH we have MATH and MATH. Let MATH be another element with MATH, then by uniqueness of the inverse MATH and MATH. Now define MATH as above by MATH. Then MATH and MATH is small, because MATH and because for MATH, we have MATH . Therefore MATH . Thus, if only MATH is small enough, we have MATH. Let MATH be some elementary set, then MATH and MATH, MATH, and for every positive continuous function MATH one has MATH , MATH and MATH. Using REF one obtains MATH so again the MATH are close to (pairwise orthogonal) projections MATH. Set MATH, then MATH, MATH. Again MATH, MATH. Furthermore MATH and setting MATH, one has MATH. But MATH and MATH commute, so MATH and MATH. It follows that MATH is a projection, and that MATH if MATH. The MATH were arbitrary, therefore MATH. Furthermore, MATH is a projection, so MATH is a MATH-homomorphism by REF. |
math/0107218 | CASE: MATH for every finite-dimensional MATH-algebra MATH. By REF MATH for a MATH algebra MATH. CASE: The idea of the proof is simple: We have just seen that a unital c.p. map MATH of strict order zero in fact is a MATH-homomorphism. If it is almost unital, it is close to a MATH-homomorphism. Problems only arise when MATH has no unit. But in this case MATH may be chosen to contain a sufficiently large hereditary subalgebra on which MATH is close to a MATH-homomorphism. Thus we can find enough finite-dimensional subalgebras of MATH; using the local characterization of MATH algebras REF this implies that MATH in fact is MATH. Take MATH a positive approximate unit for MATH with MATH. By making MATH smaller if necessary, we may assume MATH, where we obtain MATH from MATH by REF. If we choose MATH with MATH and MATH, sufficiently small, there is a c.p. approximation MATH, with MATH and the following properties: MATH . For REF we approximate the MATH and MATH sufficiently well by MATH, for REF in addition we use REF ; REF follows from the fact, that MATH. We have MATH, where MATH and MATH is elementary. Set MATH, then MATH and MATH . Here, MATH denotes the unit of MATH. Then MATH so MATH. We now have MATH . Now define MATH; then MATH and by REF there is a MATH-homomorphism MATH such that MATH. Clearly, MATH is a finite-dimensional MATH-subalgebra. It remains to be shown that MATH is close to MATH for MATH. We have MATH . Using MATH and REF we obtain MATH . Therefore MATH . Furthermore MATH is small, because MATH, so MATH is close to MATH. But then by REF , MATH is a MATH algebra. |
math/0107220 | CASE: We will construct MATH using the calculus of clovers with two leaves introduced independently by CITE; see also CITE. Clovers with two leaves is a shorthand notation (on the left) for framed links shown on the right of the following figure: MATH . Since clovers can be thought of as framed links, surgery on clovers makes sense. Two clovers are equivalent (denoted by MATH in the figures) if after surgery, they represent the same REF-manifold. By calculus on clovers (a variant of NAME 's calculus on framed links) we mean a set of moves that result to equivalent clovers. For an example of calculus on clovers, we refer the reader to CITE and also CITE. In figures involving clovers, MATH is constructed as follows: MATH . Notice that at the end of this construction, MATH is a surgery presentation for MATH. CASE: Using the discussion of CITE, it is easy to see that the equivariant linking matrix of (a based representative of) MATH is given as stated. CASE: Finally, we show that every pair MATH arises this way. Indeed, choose a NAME surface MATH for MATH in MATH and a link MATH such that MATH. The link MATH may intersect MATH, and it may have nontrivial linking number with the cores of the bands of MATH. However, by a small isotopy of MATH in MATH (which preserves the condition MATH) we can arrange that MATH be disjoint from MATH and that its linking number with the cores of the bands vanishes. Viewed from MATH (that is, reversing the surgery), this gives rise to MATH as needed. |
math/0107220 | There are several ways to prove this result, including an algebraic one, which is a computation of appropriate NAME groups, and an analytic one, which identifies the invariants with MATH-invariants. None of these proofs appear in the literature. We will give instead a proof using the ideas already developed. Fix a surgery presentation MATH for MATH, with equivariant linking matrix MATH as in REF . Letting MATH, it follows that MATH . Taking signatures for any MATH, MATH, it follows that MATH where the last equality follows from the definition of the MATH-ignature, see CITE and CITE. Thus, MATH. Since MATH is a metabolic matrix, it follows that MATH, from which it follows that MATH. Taking determinants rather than signatures in the above discussion, it follows that MATH. |
math/0107220 | Fix a surgery presentation MATH for MATH, with equivariant linking matrix MATH as in REF . Then the linking matrix MATH and MATH of MATH and MATH are given by MATH and MATH with an appropriate choice of basis. The result follows using REF . |
math/0107220 | Recall that MATH. For the first part, we have: MATH . For the second part, we have MATH . |
math/0107220 | We need to show that the group-like basing relations are preserved. With the notation and conventions of CITE, there are two group-like basing relations MATH and MATH on diagrams. It is easy to see that the MATH basing relation is preserved. The MATH relation (denoted by MATH) is generated in terms of a move of pushing MATH on all legs (of some fixed color MATH) of a diagram. Given a diagram MATH with some MATH-colored legs, let MATH denote the result of pushing MATH on every MATH-colored leg of MATH. In order to show that the MATH relation is preserved, we need to show that MATH. Ignoring the matrix part (that is, setting MATH the empty matrix), we can compute as follows: MATH . The same calculation can be performed when we include the matrix part, to conclude that MATH. |
math/0107220 | Observe that MATH . In CITE, it was shown that the ``determinant" function MATH is multiplicative, in the sense that (for suitable matrices MATH) we have: MATH . Let us define MATH to be the projection MATH. It suffices to show that MATH. We compute this as follows: MATH . Since MATH, the result follows. |
math/0107220 | For MATH and MATH, we have: MATH . |
math/0107220 | It follows from the above lemma, together with the fact that MATH shown in CITE. |
math/0107220 | This is proven in CITE and repeated in REF. |
math/0107220 | We need to show that the NAME Relations CITE are preserved. There are two possibilities: the case that all three edges in a NAME Relation are dashed, and the case that two are part of the skeleton and the remaining is dashed. In the first case, the NAME Relation is preserved because there is an obvious correspondence between lifts that admit an admissible labeling. In the second case, the skeleton looks like (for MATH, with the convention that MATH) MATH and again there is a correspondence between MATH-admissible labelings of lifts of the two sides of the equation. |
math/0107220 | CASE: Let us call an element of MATH special if the beads of its skeleton equal to MATH. Using the NAME Relations, it follows that MATH is spanned by special elements. It is easy to see that MATH maps group-like elements of MATH to group-like elements, and special group-like elements in MATH to group-like elements in MATH. Further, it is easy to show that the diagram MATH commutes when evaluated at special elements of MATH. From this, it follows that the left square diagram of the Lemma commutes. For the right square, we need to show that the MATH-flavored basing relations in MATH are mapped to MATH-flavored basing relations in MATH. There are two kinds of MATH-flavored basing relations, denoted by MATH and MATH in CITE. First we consider MATH. Take two elements MATH such that MATH; we may assume that MATH is obtained from pushing MATH to each of the MATH-colored legs of MATH, for some MATH. Corresponding to a diagram MATH appearing in MATH, there exists a diagram MATH of MATH obtained by pushing MATH onto each of the MATH-colored legs of MATH. For example, MATH . There is a REF correspondence between admissible MATH-colorings of MATH and those of MATH (if we cyclically permute at the same time the labels MATH), shown as follows: MATH . Applying MATH basing relations, the two results agree. In other words, MATH. Now, consider the case of MATH, (in the formulation of CITE). Given MATH, there exists an element MATH with some legs labeled by MATH, such that MATH for some MATH, where MATH is the operation that contracts all MATH legs of a diagram to all MATH legs of it. Now observe that MATH . CASE: Notice first that MATH can be defined when beads are labeled by elements of MATH. There is an isomorphism MATH over MATH which gives rise (after composition with the projection MATH) to a map MATH . Using this map, we can define MATH as before and check that the relations are preserved. |
math/0107220 | This follows easily from the definition of the MATH using the fact that the beads of the diagrams in MATH appear only at the gluing site. |
math/0107220 | Consider a pair MATH where MATH is given by MATH . If we write MATH then, observe that MATH . Recall the map MATH of REF . It follows from the above that for any MATH we have MATH . We wish to determine MATH, which we write as MATH . Since MATH, we can solve for MATH in terms of MATH and obtain that MATH . Observe further the following consequence of the ``state-sum" definition of MATH: for diagrams MATH in MATH, we have that MATH . Now, we can finish the proof of the proposition as follows: MATH . |
math/0107220 | It follows immediately from REF . |
math/0107220 | Using the properties of MATH and MATH it suffices to consider only trivalent graphs MATH with edges decorated by elements in MATH, and in fact only those graphs whose edges are decorated by powers of MATH. Moreover, since both MATH and MATH satisfy the push relations, it suffices to consider graphs whose edges along any forest are labeled by MATH. Fix a trivalent graph MATH with ordered edges MATH decorated by MATH. We begin by giving a description of the algebra MATH in terms of local coordinates as follows. Choose a maximal forest MATH and assume, without loss of generality, that the edges of MATH are MATH where MATH. Each edge MATH corresponds to a REF-cocycle MATH. Since MATH, it follows that MATH is a (free) abelian group with generators MATH and relations MATH for all vertices MATH of MATH and for appropriate local orientation signs MATH. It follows that MATH where MATH. This implies that MATH is a MATH-invariant subalgebra of MATH. Now, MATH is obtained by symmetrizing over MATH-automorphisms of the monomial MATH. We may assume that MATH for all MATH. Thus, MATH . On the other hand, an admissible MATH-coloring of MATH necessarily assigns the same color to each connected component of MATH and then the consistency relations along the edges MATH for MATH show that an admissible coloring exists only if MATH, for MATH, and in that case there is are MATH admissible colorings for each connected component of MATH. Thus, the number of admissible MATH-colorings is MATH. After symmetrization over MATH, the result follows. |
math/0107220 | There are two trivalent graphs of degree MATH, namely MATH and MATH. Label the three oriented edges of MATH for MATH where MATH is the label in the middle (nonloop) edge of MATH. For MATH, let MATH denote the corresponding element. For MATH, MATH, we write MATH and compute MATH . Thus, MATH is spanned by MATH for MATH as above. Applying the above reasoning once again, it follows that MATH is spanned by MATH for MATH as above. Applying the MATH relation MATH it follows that the natural map MATH is onto. It is easy to see that it is also REF, thus a vector space isomorphism. |
math/0107220 | Consider the degree MATH part in the Equation of REF . On the one hand, we have MATH (see CITE) and on the other hand, it follows by definition and REF that MATH. REF which compares liftings ad residues concludes first part of the corollary. For the second part, observe that MATH is a rational function on MATH, which is regular when evaluated at complex roots of unity. Furthermore, by definition of MATH, it follows that MATH is the average of MATH on MATH (evaluated at pairs of complex MATH-th roots of unity) and converges to MATH. This concludes the proof of REF . |
math/0107220 | Using a partial fraction expansion of the denominator of MATH, it suffices to assume that MATH for some MATH. In that case, we have MATH where MATH. |
math/0107220 | Consider the commutative diagram MATH where MATH and MATH are universal covering maps. Since MATH is REF on fundamental groups, it follows that MATH is null homotopic in MATH. Choose representatives MATH of the components MATH of MATH in the unversal cover MATH, for MATH where MATH is the number of components of MATH. Then, MATH . On the other hand, MATH is a choice of representatives of the lifts of MATH to MATH, for MATH, and MATH. Furthermore, if MATH is the equivariant linking matrix of MATH, we have MATH . It follows that if we collect all powers of MATH modulo MATH in NAME polymomials MATH such that MATH (for MATH), then MATH . Writing this in matrix form, gives the result. |
math/0107220 | The first identity follows from REF using the identity MATH of CITE. The second identity follows from the first, after inverting the operators involved. Specifically, REF implies that MATH . Setting MATH, we have that MATH and the above implies that MATH . |
math/0107221 | CASE: For any MATH corresponding to the entry MATH in the first column of MATH. For any MATH corresponding to the two entries MATH in the second column of MATH. For any MATH corresponding to the three entries MATH in the third column of MATH. CASE: Slightly extend MATH to a manifold MATH by pasting to MATH collars homeomorphic to MATH and MATH. We also extend the function MATH and the metric MATH to a function MATH respectively a metric MATH in such a way that MATH has the same critical points as MATH and is regular and its gradient points out on MATH and points inside and is regular on MATH. Then, by standard NAME theory, MATH is simple chain equivalent to MATH. On the other hand we have MATH. CASE: The projection from MATH to MATH is given by MATH . The kernel of this projection is precisely MATH and this complex is therefore simple chain equivalent to MATH. CASE: The chain inclusion MATH is a chain representative of the inclusion MATH. Therefore, MATH (which is the co-kernel of the map above) is simple chain equivalent to MATH. CASE: The chain inclusion MATH is a chain representative of the inclusion MATH. CASE: Combine REF . |
math/0107221 | CASE: It is possible to find a small perturbation of the negative gradient flow of MATH in a neighbourhood of MATH such that the resulting flow MATH points inside MATH on MATH and outside on MATH; it is gradient like and has no stationary points inside MATH. The existence of such a flow implies the claim. To construct this perturbation fix some NAME charts MATH around the critical points of MATH that are situated on MATH. Fix also a collared neighbourhood of MATH that is diffeomorphic to MATH. With respect to this parametrization notice that, for all small enough MATH the projection of MATH on MATH is non-zero for all sufficiently small MATH and MATH. Moreover, for MATH small and MATH we have that MATH is not tangent to MATH and points towards MATH (because of the NAME lemma). This means that by adding to MATH a vector field which is MATH outside MATH (MATH small enough) and is equal to MATH for MATH with MATH decreasing, MATH, MATH the induced flow will satisfy the desired properties with respect to MATH. Of course, a similar construction is possible relative to MATH and produces MATH. CASE: Immediate from REF . CASE: The chain map MATH is the chain homotopy of REF MATH . For any MATH structure on MATH the chain map MATH is chain homotopic to the composite of the simple chain equivalences given by REF MATH . |
math/0107221 | The proof is an immediate consequence of the constructions of NAME cobordisms described in the lemmas below. Let MATH be a NAME function on a closed manifold MATH. Let MATH be such that MATH for all MATH and let MATH be a second metric on MATH. There is a NAME function MATH on MATH such that MATH, MATH, MATH extends the MATH's and MATH for all MATH, MATH. Let MATH be a MATH function such that MATH . Consider the function MATH . We have MATH and this together with the fact that MATH implies immediately the statement (notice that MATH is regular on MATH), except that we also need to remark that there is a metric MATH extending the MATH's such that MATH is NAME with respect to MATH. By using a partition of unity argument it follows that there is a metric MATH that extends the MATH's. Moreover, as MATH is already a NAME function we can slightly modify MATH away from MATH to obtain MATH. Consider a compact cobordism MATH, and suppose given NAME functions MATH. Let MATH be a NAME function with all its critical points in the interior of MATH and which is constant, maximal on MATH, and constant, minimal on MATH. For any neighbourhood MATH of MATH there are suitable constants MATH, MATH and a NAME cobordism MATH between the NAME functions MATH and MATH such that MATH . Inside the neighbourhood MATH of MATH we can find a tubular neighbourhood that we shall identify with MATH. We may assume that MATH and that MATH is constant, regular and equal to MATH on MATH and is constant, regular and equal to MATH on MATH. Pick MATH such that MATH and MATH. Now REF provides a cobordism between MATH and the constant function MATH on MATH. By the same method as in REF we also get an analogous cobordism between MATH and MATH on MATH. Another partition of unity argument shows that these two cobordisms can be pasted together with MATH to provide the function MATH. The metric MATH is obtained by the same argument as in REF. For the cobordism MATH constructed in REF the chain complex MATH of REF coincides with MATH. Let MATH, MATH be NAME functions on a closed manifold MATH such that MATH . A linear cobordism between MATH and MATH is a simple cobordism REF MATH with MATH REF such that for all MATH, MATH is a convex, linear combination of MATH and MATH and MATH for all points MATH. Any two NAME functions MATH, MATH on a closed manifold MATH such that MATH are related by a linear cobordism. If MATH there is a simple cobordism MATH with MATH the product metric. If moreover MATH with MATH constant, then there exists a linear cobordism with MATH the product metric and for any such linear cobordism the chain map MATH is a simple isomorphism. As in the proof of REF we consider a MATH function MATH such that MATH . We define MATH with the immediate consequence that, as above, one can find a metric MATH such that the NAME function MATH satisfies the properties required in the first part of the statement. For the second part, assume that MATH is the product metric MATH. As both MATH, MATH are NAME we can modify MATH outside a neighbourhood of MATH to obtain the desired NAME cobordism. If MATH this modification is not necessary, because the function MATH itself is already NAME with respect to MATH. Moreover, if MATH then for all MATH the flow induced by MATH is tangent to MATH. As MATH, MATH the proof is concluded. The proof of REF follows by applying the construction in REF to the NAME functions MATH and MATH. |
math/0107221 | We first observe that the notion of NAME cobordism introduced in REF can be generalized in an obvious way to the case when MATH and MATH have boundaries and MATH is tangent to MATH (in the sense that MATH for all MATH, MATH). All the previous statements have analogues in this case. In particular, the statements of REF remain true when assuming that the manifold MATH has a non-empty boundary. A general existence result for two-parameter NAME functions follows. Let MATH be a compact manifold without boundary. Let MATH be NAME functions on MATH, MATH and let MATH be simple NAME cobordisms on MATH of MATH and MATH for MATH. Assume MATH and suppose that MATH and MATH are linear cobordisms. There exists a simple NAME cobordism MATH on MATH of MATH and MATH which restricts to MATH on MATH, with MATH . Let MATH with MATH functions with the properties described at the beginning of the proof of REF . Let MATH be another such function. Define MATH . By the same type of argument as those used before it is easy to see that MATH is a NAME function. Moreover, by a partition of unity argument one can construct a metric MATH extending the MATH's. A small perturbation of MATH away from the boundary of MATH leads to a new metric MATH such that the pair MATH satisfies the desired properties. MATH . In the setting of REF above, there is a MATH such that for any MATH and MATH with MATH we have MATH . For MATH, let MATH . As MATH is compact MATH. Let MATH be a small positive constant such that MATH . As a consequence of the fact that MATH is a chain complex we have MATH . The condition imposed to MATH and MATH implies that the two last sums of this expression vanish. Indeed, in the sum before last the only terms that count are those that satisfy MATH. This implies MATH. It follows that MATH. The argument for the vanishing of the last sum is similar. We now use REF to prove the Rigidity REF . We return to the setting of the statement of the Theorem. Thus MATH, MATH, MATH, MATH and MATH are fixed as well as the simple NAME cobordism MATH of MATH and MATH. Let MATH such that MATH. We intend to use REF . We take MATH and MATH a linear NAME cobordism of MATH and MATH with MATH a product metric. As we have MATH for all MATH we note that MATH . Therefore the conditions in REF are verified. We may also take MATH such that this is a linear cobordism with MATH the product metric (see REF ). We use REF to construct MATH. REF can now be applied and by inspecting its proof we see that we may take MATH. We fix a total order on the set MATH such that for MATH the inequality MATH implies MATH. With this total ordering we consider the matrix MATH of the composition MATH as well as the matrix MATH of the composition MATH. Because both MATH and MATH are linear cobordisms with the product metric it immediately follows that MATH. We now want to observe that MATH is an upper triangular matrix with MATH's on the diagonal. Indeed, assume MATH correspond to elements MATH respectively in the fixed order of MATH with MATH. Then MATH and MATH . Therefore, by REF we have MATH and as mentioned above MATH, the NAME symbol. As a consequence we get that MATH has determinant equal to MATH and is therefore an isomorphism. But MATH, and this completes the proof of the Rigidity REF . We now turn to REF . If MATH is sufficiently small, then the number of critical points of MATH equals the number of critical points of MATH. Of course, we may assume MATH sufficiently small such that MATH and this implies that our chain maps MATH and MATH are both isomorphisms. This completes the proof of the Rigidity REF . |
math/0107221 | Immediate from the definitions. |
math/0107221 | CASE: This follows easily from the constructions in REF. We first choose MATH such that MATH contains only regular values of MATH, with MATH and the assumption that MATH is the product metric here. After possibly multiplying MATH with a sufficiently small constant we may assume that the image of MATH belongs to MATH with MATH such that MATH. Let MATH . By using REF for each MATH we construct a NAME function MATH of MATH along MATH such that : MATH agrees with MATH outside MATH; the gradient of MATH is everywhere tangent to MATH and MATH, MATH with MATH small; if MATH then MATH and if MATH, then MATH; on the cobordism MATH the function MATH restricts to a linear cobordism. The construction of the function MATH is done by applying REF to the two cobordisms MATH and MATH and REF to the cobordism MATH. In particular, MATH is a linear combination of the form MATH inside MATH with MATH an appropriate function with MATH and MATH, MATH (see REF); a similar linear combination is valid inside MATH. To understand the pasting between the cobordism MATH and, for example, MATH notice that on MATH we have a cobordism from MATH to MATH. In other words, on both sides of MATH we have cobordisms having MATH as ``high end". This allows the pasting to occur by simply using in the formula above MATH defined on a larger interval, say MATH, and using the same formula to define MATH on both sides of MATH. For a generic choice of MATH the function MATH is already NAME with respect to the metric MATH. In general, to obtain a function MATH such that MATH is NAME we might need to still slightly perturb MATH inside MATH. We may assume that the construction has been made such that MATH and by adjusting the various relevant constants we see that MATH is a splitting of MATH along MATH with the required properties. REF + REF Immediate from the definition and construction of a splitting. |
math/0107221 | CASE: This point is quite immediate. Assume MATH is a MATH-splitting of MATH along MATH as in the statement. By the constructions in REF there is a NAME cobordism MATH between MATH and MATH. If MATH is small enough, then MATH is sufficiently MATH close to MATH such that MATH and we may construct MATH such that it restricts to linear cobordisms with a product metric on MATH and on MATH (we use the notations in REF ). Denote by MATH the chain map induced by this cobordism. Due to the linearity of MATH on the complement of MATH the composite MATH is a simple isomorphism of the form MATH for a chain homotopy MATH . CASE: Clearly, a NAME function MATH is not in general adapted to splitting MATH. We obtain an adapted MATH by starting with an arbitrary MATH and adding pairs of mutually cancelling critical points such that the unstable manifolds of MATH give a MATH decomposition of MATH which is sufficiently fine to ensure that REF is verified. In particular, this means that MATH whenever MATH. After further subdivision we may also assume that for any pair MATH, MATH we either have MATH, or MATH. We then modify MATH such as to perturb the stable manifolds MATH to render them transverse to MATH for all pairs MATH. This modification is performed successively for MATH of increasing indexes and it may be achieved without modifying those unstable manifolds MATH which intersect some unstable manifold of MATH. This happens because if MATH intersects MATH, then MATH and as MATH is NAME we also have that MATH is transversal to MATH (for all MATH). This means that if MATH intersects MATH in a point which belongs also to MATH, then MATH is already transversal to MATH at this point. We also need to modify our function MATH such as to obtain one which also has the property that the second type of intersections in REF are transversal. We perturb successively each unstable manifold MATH, MATH such as to make MATH transversal to the stable manifolds of MATH. Obviously, we need to insure that REF is preserved and therefore the critical points MATH such that MATH intersects some MATH need to be discussed separately. In this case we have MATH. Suppose that MATH, then as MATH is transverse to all stable manifolds MATH (by the NAME condition) it also follows that MATH is transverse to these stable manifolds and no modification of MATH is needed. In case MATH it follows from REF that MATH. We then perturb MATH inside MATH such as to obtain the needed transversality. This will also perturb the higher dimensional MATH's that appear in the union MATH above but can be made such that MATH does not diminish and thus REF continues to be satisfied. It is useful to note that one can obtain in this way an adapted MATH that is arbitrarily close in MATH norm to the function MATH that was given initially. CASE: We now assume that MATH is adapted to splitting MATH. Recall the functions MATH constructed in the proof of REF . We shall work below with functions of this form and with the notations of that proof. In particular, MATH . It is immediate to see that because of the transversality properties of the adapted MATH, the function MATH is NAME with respect to MATH. Therefore MATH is itself a splitting and we will show below that any such splitting with MATH sufficiently small verifies the conclusion. By inspecting the construction in REF, we see that, in each point of MATH, the tangent vector MATH projected onto MATH has the same direction as MATH. We intend to compare the NAME complex of MATH with that of MATH using a special NAME cobordism between MATH and MATH. For this we start with a linear cobordism MATH of NAME functions between MATH and MATH with MATH fixed. We assume MATH and MATH. We denote by MATH the deformation parameter of MATH. We may also assume that in a neighbourhood MATH of MATH the function MATH has the form MATH. We shall identify below MATH and MATH (for example we write MATH etc). We want to show that we can modify MATH only inside MATH such that the resulting function MATH will be a NAME cobordism of MATH and MATH. By a similar method to that used in REF we can define MATH inside MATH by MATH where CASE: MATH is MATH; CASE: MATH is constant equal to MATH if MATH; CASE: MATH; CASE: MATH for all MATH. (Such a function exists, it can be modelled on MATH) and MATH outside of MATH. The function MATH is smooth because MATH outside of MATH and it is clear that it is NAME. Note for each point of MATH the gradient of MATH with respect to the product metric MATH decomposes as an orthogonal sum of a component tangent to MATH which is a multiple of the gradient of MATH and two other components, one in the direction of MATH, and the other into that of MATH. We can change slightly the metric MATH inside a set MATH with MATH thus getting a metric MATH sufficiently close to MATH such that with respect to this new metric MATH remains NAME and its complex is not modified and MATH becomes also NAME. In fact, the set of metrics MATH such that MATH remains NAME with respect to MATH and has the same NAME complex as with respect to MATH is dense and open in some neighborhood of MATH. Therefore, if we now consider the sequence of functions MATH, MATH, MATH, we see that there is a metric MATH on MATH such that MATH equals MATH outside MATH, for all MATH, MATH is NAME with respect to MATH and MATH is also NAME with respect to MATH and has the same NAME complex as it has with respect to MATH (in fact, the set of such metrics, as a countable intersection of open dense sets, is even dense in the mentioned neighborhood of MATH). It will be convenient to use the convention that MATH whenever the difference of indexes between the critical points MATH and MATH of a NAME function MATH is strictly greater than MATH, or if one of MATH is not a critical point of MATH. In the setting above there is some MATH such that in the NAME complex of MATH we have: CASE: MATH if MATH, MATH. CASE: MATH for MATH, with MATH and MATH. CASE: MATH for MATH with MATH, MATH and MATH. Assuming this Lemma, here is the end of the proof of the gluing REF . To simplify notation we denote MATH, MATH with MATH satisfying the conclusion of REF. Denote by MATH respectively the chain maps MATH and MATH. If MATH we let MATH. As MATH is a linear cobordism we have MATH. Consider now the function MATH. It restricts to a linear cobordism on MATH. If MATH, then let MATH. Clearly, MATH with MATH a critical point of MATH. Assume MATH are such that MATH and MATH and MATH. Then MATH . All the counting of flow lines below will be done with respect to the metric MATH which we shall omit from the terminology. As MATH is a chain map and in view of the definition of the NAME complex we have MATH . Now REF implies that: CASE: if MATH, MATH then MATH; CASE: if MATH then MATH except for MATH and then MATH; CASE: if MATH then MATH; CASE: if MATH then MATH. As also MATH, MATH we obtain MATH . Moreover for each MATH, we also have MATH which gives MATH . So we get MATH . We can now define MATH and modulo the obvious identification of MATH with MATH we obtain the desired formula. It remains now to prove the Lemma. Proof of REF . We fix MATH. We have MATH with MATH. Consider MATH. We have MATH if MATH and MATH if MATH. We shall now consider two cases: CASE: MATH. In this case, by choosing MATH sufficiently large, we have MATH and therefore MATH, so that CASE: MATH, CASE: MATH for all MATH. As the number of critical points of MATH is finite this implies that for a sufficiently large MATH the three properties of REF are verified for all critical points MATH such that MATH. CASE: MATH. The same argument as above shows that for all MATH with MATH we have that for sufficiently large MATH . There are two other properties that we still have to verify (and it is only here that we shall need to use the specific form of MATH) : CASE: if MATH, then MATH, CASE: if MATH, then MATH. We first note that for each MATH the boundary of MATH has the property that MATH is an entrance set into MATH for the flow induced by the negative of the gradient of MATH. Similarly, MATH is an exit set. Assume that for arbitrarily large MATH REF above is not satisfied. As the number of critical points of MATH is finite this means that there is a critical point MATH of MATH of index MATH and a sequence MATH such that MATH with MATH. Clearly, as MATH this means that for each MATH there is at least one flow line MATH (for the flow induced by MATH) that joins MATH to MATH and that intersects MATH. Let MATH be the entrance point of this flow line into MATH and let MATH be the exit point of this same flow line. Inside MATH the flow line joining MATH to MATH projects onto MATH as a flow line MATH of MATH joining the projection of MATH denoted by MATH to the projection MATH of MATH. This happens because the gradient of MATH can be decomposed into two components, one orthogonal to MATH and the other having the same direction as the gradient of MATH. We now make MATH. Then, by compactness, we may assume that MATH, MATH. As in the exterior of MATH the function MATH coincides with MATH we obtain that MATH and MATH. Because MATH is a linear cobordism this implies MATH and MATH (see REF ). Of course, we also have MATH and MATH. REF of the function MATH implies that MATH. As MATH and MATH were joined by the flow lines MATH of the flow induced by the negative of the gradient of MATH we obtain that MATH also belongs to MATH. But this means that the intersection of some MATH-unstable manifold of dimension MATH with the MATH-stable manifold of MATH is non trivial. By REF of MATH that means MATH . This implies MATH and leads to a contradiction. We use a similar argument to deal with REF . As above, if for arbitrarily large REF is not satisfied there is a critical point of MATH, MATH such that there exists at least one MATH-flow line MATH joining MATH to MATH and MATH. Obviously, when MATH we may assume that MATH with MATH a critical point of index MATH of MATH. As before, let MATH be the entrance point of MATH inside MATH and let MATH be the projection of MATH onto MATH. The portion of MATH inside of MATH projects to a MATH-flow line MATH joining MATH to MATH. Let MATH . As above we obtain MATH . At the same time we also have MATH. REF of MATH implies MATH which means MATH . It follows that MATH, which contradicts the hypothesis on MATH and completes the proof of the Lemma (we have actually proved more: not only the relevant numbers of flow lines counted with signs are zero but actually the respective sets of flow lines are empty), and also of the gluing REF . |
math/0107221 | The proof of the theorem is based on the idea described in REF and, besides the usual approach to comparing the NAME complexes of two different regular pairs, it makes use of monotone homotopies as introduced in CITE, CITE. The standard way to compare the NAME complexes associated to the two regular pairs MATH and MATH is as follows CITE CITE. Take smooth homotopies MATH and MATH, MATH such that there exists MATH with the property that, for MATH, we have MATH and for MATH, MATH, MATH. Consider the equation: MATH where MATH is the gradient of the function MATH with respect to the Riemannian metric induced by MATH. Let MATH be the space of all finite energy solutions MATH of this equation (where the energy MATH is defined by using MATH in REF instead of MATH). Again this space is compact and it decomposes as the union of the spaces MATH containing those solutions MATH that also satisfy MATH for some with MATH and MATH. Moreover, for generic choices of MATH, the associated linearized operator MATH is surjective for each solution MATH of REF . Whenever MATH are regular pairs for MATH and this additional condition is satisfied we shall say that the pair MATH is a regular homotopy. As in the case of REF a consequence of regularity is that the spaces MATH are manifolds and they also admit coherent orientations. The dimension of such a manifold is given by MATH. Define now MATH by MATH where MATH is the number of elements in MATH (counted with signs). Obviously, the key point is that this application is a chain homomorphism which induces an isomorphism in homology (CITE, CITE). For such a homotopy MATH let MATH . We now consider MATH, MATH and by composition define MATH. We derive MATH with respect to MATH and we obtain (as, for example, in CITE p. REF - except for a change in signs) MATH where MATH is the derivative of the functional MATH at MATH in the direction of the vector field MATH defined along MATH. Now MATH and, because MATH verifies REF , this is always negative. If it happens that MATH for all MATH we may take MATH to be a monotone homotopy in the sense that MATH for all MATH (this is quite analogue to our simple morse cobordisms from REF ). These homotopies have been introduced and used by NAME and NAME in CITE. If MATH is monotone, the formula above shows that MATH which shows that MATH . Suppose now that we have two regular homotopies MATH, MATH relating the regular pairs MATH and MATH. It is then possible to glue the two homotopies together thus getting a new homotopy MATH which is defined by MATH for MATH, MATH for MATH and by analogue formulae for MATH. For MATH sufficiently big this homotopy is regular and it verifies MATH (REF ; REF ). Notice also that if both MATH and MATH are monotone, then so is MATH. We now turn to the statement of the theorem. We fix the regular pair MATH and we let MATH be a regular pair with MATH, MATH. We have MATH, MATH. Let MATH, MATH, MATH. We may find monotone homotopies MATH, MATH relating these pairs such that for all MATH we have MATH. For fixed MATH the pair MATH is regular for a generic choice of MATH inside the set of time- dependent, period one, almost complex structures compatible with MATH, REF (the use of time dependent almost complex structures does not change the form or the behaviour of any of the previous equations). We consider the ``glued" regular homotopy MATH defined as above which is monotone and regular for large MATH. In essence, we have now constructed the two non-trivial faces of the cube in the proof of REF and, to end the proof, we need to show that MATH is an isomorphism. We consider the monotone homotopy MATH such that MATH with MATH some convenient, decreasing function and MATH. Obviously, this homotopy is regular and MATH. Let MATH be a smooth homotopy of homotopies MATH such that MATH verifies MATH, MATH and we have similar identities for MATH. There exists a notion of regularity for such a homotopy of homotopies MATH. If our MATH is regular in this sense it induces a chain homotopy MATH defined by MATH where MATH is the number of elements (counted with signs) of the moduli spaces MATH which, are finite by regularity when MATH and also admit a coherent choice of orientations (see REF). With this definition we have MATH. We index the elements in MATH such that MATH and we denote by MATH the matrix of the linear map MATH in this basis. Because both MATH and MATH are monotone homotopies we may take MATH to be a homotopy of monotone homotopies in the sense that MATH is a monotone homotopy for all MATH. Simultaneously, we also need the pair MATH to be regular. This can be achieved generically (with again MATH time-dependent) as discussed in REF. In this case, it follows from REF that MATH. The coefficients of the NAME differential in MATH verify MATH. By looking to the matrix MATH we obtain that MATH. But MATH, MATH so we get that if MATH, then MATH which means MATH. Therefore, MATH is upper triangular and we conclude that MATH is an isomorphism which ends the proof. |
math/0107221 | Let MATH be the based f.g. free MATH-module generated by the index MATH critical points of MATH. For any MATH and MATH so that MATH . |
math/0107221 | CASE: As for REF . CASE: The NAME complex of MATH is MATH and MATH . |
math/0107221 | Order the critical points of MATH following their critical values. Explicitly, we fix a total order on the finite set MATH such that MATH implies MATH for MATH. The proof of the theorem uses the adaptation to the MATH-valued case of the techniques developed in the first three sections of this paper. These methods will produce certain morphisms relating the various chain complexes involved. Showing that these morphisms are isomorphisms is less immediate than in the MATH-valued case, but it will always follow from the fact that our morphisms satisfy the assumptions of the rather obvious lemma below. Consider the free MATH-module MATH generated by a finite ordered set MATH. A MATH-module endomorphism MATH such that for each MATH is a MATH-simple automorphism. Write MATH, with MATH for MATH. The matrix of MATH has entries in MATH, of the form MATH with MATH, MATH. The augmentation MATH sends MATH to an upper triangular matrix of the form MATH which is clearly invertible. The matrix defined by MATH has entries in MATH. Thus MATH itself is invertible, with inverse MATH defined over MATH. We now return to the proof of REF . The first remark is that REF apply ad literam to the case of MATH-valued maps because the relevant conditions are local in nature. Similarly, the statements in REF remain valid if we replace the relevant NAME complexes by the corresponding NAME complexes with one exception: the argument given for the proof of REF only shows that MATH is a chain map in the MATH-valued case. More care is necessary for the constructions of NAME cobordisms. CASE: Let MATH be two NAME functions with MATH closed. Suppose that MATH and MATH are homotopic. Then there exists MATH and a simple NAME cobordism MATH between MATH and MATH. CASE: If there exists a constant MATH such that MATH then there exists a cobordism MATH between MATH and MATH such that MATH is the identity. The difference between proving this statement and the constructions in REF comes from the fact that the formulas for the homotopies given there are no longer applicable in this case. Let MATH be given by MATH. We also use the following convention : MATH is the standard volume form on MATH and for a MATH-valued function MATH we let MATH. CASE: Let MATH be the homotopy of MATH and MATH. By a simple partition of unity argument we may assume that MATH is flat at the ends of the cobordism in the sense that there are collared neighbourhoods MATH of MATH, MATH, MATH, MATH, MATH small, such that for MATH we have MATH. Because MATH is compact there exists MATH such that MATH for all MATH (where MATH). Let MATH be a NAME function with exactly two critical points, a minimum at MATH and a maximum at MATH that has value MATH and such that for MATH we have MATH. Now define MATH by MATH. As MATH is flat close to MATH, to verify that MATH is a NAME cobordism we only need to notice that MATH has no critical points in MATH. This happens because in this set MATH (which is computed by using polar coordinates MATH on MATH). It is obvious that there are metrics MATH that extend MATH and such that MATH is NAME. CASE: This is the analogue of the construction of linear NAME cobordisms in REF . We consider a NAME function MATH as above (without any derivative restrictions). Let MATH be an increasing MATH-function with MATH, MATH. We let MATH be a smooth homotopy of MATH and MATH that is defined by MATH. Finally, we define MATH as before MATH. We take on MATH the metric MATH. It is easy to verify that MATH is a cobordism whose induced morphism is the identity. Indeed, the lift of this pair to MATH is an obvious linear cobordism in the real valued sense and this implies that MATH is the identity. CASE: Given these two lemmas REF follows exactly by the same argument as REF . More precisely, assume that MATH are homotopic NAME functions such that MATH is MATH close to MATH. Then there exists a homotopy MATH of MATH to MATH which is MATH-close to the constant homotopy MATH. This means that by taking MATH sufficiently MATH close to MATH we may also assume the constant MATH in REF to be as small as desired. We now construct NAME cobordisms relating MATH, MATH and MATH for small constants MATH and MATH that are similar to those in REF and the argument in REF applied to the lifts of MATH, MATH to MATH implies that the resulting chain maps MATH have the property that MATH satisfies the assumptions of REF and is therefore an isomorphism. If MATH the fact that MATH is surjective implies that it is an isomorphism (because any module epimorphism between finitely generated free modules of the same rank is an isomorphism). CASE: Again, the proof here consists of adapting the proof of REF to the MATH-valued case. We shall use the notations introduced in REF . In particular MATH and MATH only differ in the interior of a small tubular neighbourhood MATH and on this set, if we identify the interval MATH with a small arc we have MATH. By taking MATH small we can make the function MATH as MATH close to the function MATH as desired. Therefore, we may find a small constant MATH such that MATH. We then construct by the methods in REF a NAME function MATH which is of the form MATH with MATH, MATH, MATH, MATH and such that MATH if MATH. Because, MATH and MATH coincide outside MATH we notice that on MATH we have MATH. Recalling that actually the image of MATH is in MATH we rewrite MATH as MATH. This shows that MATH extends to a function MATH which is equal to MATH if MATH. With an appropriate metric MATH the function MATH is a NAME cobordism between MATH and MATH. By inspecting its lift MATH we see that the induced morphism MATH has the following property: for each generator MATH we have MATH with MATH and MATH. We consider the composition MATH. The description of MATH now shows that MATH with MATH. REF implies that MATH is an isomorphism. CASE: We intend to prove this item by showing that for a fixed MATH and some special choice of function MATH leads by the construction above to a splitting MATH, a cobordism MATH and a corresponding morphism MATH that has the form MATH such that with the notations in REF we have MATH and MATH . Notice that if such a MATH is constructed, then as MATH vanishes on MATH the composition MATH is an isomorphism of the form required. Therefore, we have reduced the proof to the construction of MATH, MATH such that MATH is as above. This construction may be accomplished in a way similar to the proof of REF . We recall that the metric MATH is the product metric MATH inside a neighbourhood MATH. As in REF , the key point is to work with an adapted function MATH (see REF ). In our setting, consider a NAME function MATH which is adapted to splitting the NAME function MATH along MATH for all MATH (the notations are as in REF ). Such a function exists by the same argument as in the real valued case (there are more conditions to be verified but they are of the same type and finite in number). We now construct the MATH-splitting MATH such that inside the set MATH the function MATH has the particular form of the functions MATH in the proof of REF . There is one minor additional point that needs to be looked at: we need to be sure that MATH satisfies the NAME transversality condition. The needed transversality is certainly satisfied with respect to the unstable and stable manifolds of the critical points in MATH by the transversality part in the ``adapted" condition. The rest of the transversalities needed can be insured by possibly slightly perturbing the stable manifolds of MATH. This can preserve the transversalities already achieved and does not affect REF of the adapted condition (see REF ). We now construct the NAME cobordism MATH in the same general way as for REF , but now we need to control the behaviour of the flow lines of MATH over the set MATH. More precisely, we shall construct a cobordism MATH which satisfies MATH for all MATH and MATH for all MATH. This immediately implies our claim for MATH (as in the proof of REF ). The argument providing this MATH follows. It is quite similar to that in the proof of REF , but with the additional difficulty that all the constructions initially accomplished for real valued functions (on MATH) will need to be adjusted such that they are periodic in order to descend to MATH. We first make the properties of MATH more explicit. We assume that the neighbourhood MATH is like in the definition of splittings (after reparametrization MATH has the form MATH and the metric is the product one here). On MATH we consider a NAME function MATH such that MATH for MATH; MATH is a linear cobordism between MATH and MATH on MATH and is a linear combination of the form MATH inside MATH with MATH an appropriate function with MATH and MATH as in REF ; a similar linear combination is valid inside MATH. The function MATH equals MATH on MATH and is equal to MATH on MATH. Similarly, we may define a standard cobordism MATH between MATH and the function MATH which is defined on MATH and where MATH is a very small constant depending on MATH. This cobordism MATH differs from a linear cobordism only in the interior of MATH. The metric MATH is the product metric MATH. We fix MATH such that if MATH, then the flow line of MATH containing MATH does not intersect MATH. Now let MATH and consider MATH a diffeomorphism given by MATH where MATH is a diffeomorphism such that MATH restricts to the identity on the complement of the square of side MATH and the image of the square of side MATH is the square of side MATH. We also require that MATH restricts to diffeomorphisms when MATH and when MATH. Let MATH. The metric associated to the NAME cobordism MATH is the metric MATH. It is useful to note that MATH coincides with a linear cobordism outside of MATH and that MATH and MATH have the same projection on MATH. Clearly, MATH and MATH differ only outside MATH. For MATH we fix the following notation MATH and MATH. We intend to construct a cobordism MATH between a NAME function MATH and MATH such that there exists a sequence of MATH's with the property that MATH is a linear cobordism on MATH, on MATH it agrees with MATH and it coincides with MATH on each set MATH . Moreover, MATH should verify the obvious analogue of REF . To construct MATH we proceed by induction. We start with the trivial linear cobordism with a product metric. Assume now that we have already constructed MATH and the function MATH and MATH such that MATH for all MATH and MATH is a cobordism between MATH and MATH which is linear on MATH and is such that REF is satisfied by MATH. We now want to construct MATH. We first pick some choice for MATH defined with respect to some MATH and which satisfies the properties required from MATH except possibly REF and such that MATH and MATH in the complement of MATH. As in the NAME case we might need to modify the associated metric on MATH (in the real case it was enough to take MATH but here it is useful to use the constant fixed above) such that MATH is NAME (see above REF ). In fact, it is easy to see that we may assume that the metric MATH associated to MATH satisfies REF with respect to MATH (because it can be chosen to be arbitrarily close to MATH) in the exterior of MATH), it coincides with the standard product metric outside MATH, it coincides with MATH on MATH, MATH, and is arbitrarily close to the product metric MATH. In fact, the set of choices for the metrics MATH is dense and open in some neighborhood of MATH viewed as a metric on MATH. As the difference between MATH and MATH appears around MATH, as in the proof of REF , we assume that there exists a pair of critical points MATH of MATH that do not satisfy REF even if we make MATH. By using the induction hypothesis and proceeding as in the proof of that lemma (see also REF we are lead to the existence of a sequence of points MATH with the following properties: MATH is a critical point of MATH such that MATH, MATH, MATH is related to MATH by a flow line of MATH, MATH is related to MATH by a flow line of MATH, MATH is related to MATH by a flow line of MATH and all the points MATH are critical points of MATH and appear as vertices in a sequence of broken flow lines of MATH (in the proof of REF this sequence only h CASE:elements). Moreover, and this is a key point, all the points MATH for MATH belong to MATH. By transversality this means that MATH. By the transversality properties of MATH we have that MATH . Notice that this implies that MATH which contradicts our assumption that MATH. Indeed, if MATH this has already been shown above and if MATH then MATH is in the unstable manifold of the point MATH whose index is at most MATH. Again by transversality the index of MATH is at most MATH. Therefore, the existence of the MATH has been established and hence the construction of MATH follows by recurrence. In order to complete the proof we need to adjust MATH in such a way that the validity of REF is preserved, but the new functions and metric are periodic and hence induce corresponding notions on MATH. Let MATH be a diffeomorphism which leaves invariant MATH and is the inverse of MATH on each set MATH. Consider the composition MATH. This function still satisfies REF but, of course, with respect to the metric MATH. Notice that MATH in MATH and so it is immediate to see that we can choose MATH as described above on MATH but also such that MATH is periodic. On each of the sets MATH the function MATH satisfies MATH where MATH equals MATH on MATH and equals MATH on MATH (these functions MATH appear when the cobordisms MATH are pasted with the linear cobordisms on MATH). This means that there exists a diffeomorphism MATH which satisfies MATH for MATH and with the property that for MATH, we have MATH (we have identified everywhere MATH and MATH). We now define MATH. By factorization this cobordism provides the desired cobordism MATH. Indeed, by construction, MATH satisfies the required properties relative to REF with respect to the metric MATH. The pair MATH might not be a NAME pair even if the NAME condition is satisfied up to order MATH. Therefore, we obtain the metric MATH by a new (and last) perturbation of MATH away from MATH such that MATH is sufficiently close to MATH for REF to continue to be satisfied (this is possible because this condition concerns only the behavior of flow lines of ``length" MATH). On MATH the metric MATH coincides with MATH. Moreover, MATH is equal to MATH up to a constant and MATH is a splitting of MATH. By construction the morphism induced by MATH satisfies REF , and this concludes the proof. |
math/0107222 | We know MATH. Suppose there exists MATH such that MATH. If MATH, then MATH, so suppose not. Then MATH, so MATH implies that MATH. But MATH, so MATH. Now suppose that there exists MATH. Then by the factorisation property there exist MATH such that MATH and MATH. But then MATH, a contradiction. Therefore MATH, and MATH. |
math/0107222 | The NAME relations REF tell us that the projections MATH are mutually orthogonal. NAME relations REF then give MATH . |
math/0107222 | MATH . |
math/0107222 | Fix MATH with MATH; there is at least one such MATH since MATH. If MATH satisfies MATH and MATH, then by REF MATH, so we have the containment MATH . Now suppose MATH. Since MATH, we must have: CASE: MATH, or CASE: MATH. First suppose REF holds. Then MATH, and hence MATH. Also MATH (since MATH, and hence MATH). Taking MATH and MATH gives MATH with MATH and MATH. Now suppose REF holds. Then we can factorise MATH with MATH, so MATH. We claim that MATH. To see this, suppose that MATH. Then there exists MATH such that MATH and MATH, say MATH. By REF we know that MATH, so MATH. Since MATH is locally convex there is a MATH, but this implies MATH, a contradiction since MATH. Hence MATH with MATH and MATH. |
math/0107222 | REF merely consists of specific cases of NAME relation REF , so suppose REF holds. Define a map MATH by MATH. We prove REF by induction on MATH . The MATH case is trivial since it amounts to the tautology MATH, and the MATH case follows directly from REF since MATH. Suppose REF holds for all MATH such that MATH. Let MATH satisfy MATH and choose MATH such that MATH. Using REF we have MATH . Hence REF holds whenever MATH. |
math/0107222 | First suppose that MATH is not locally convex. Then there exists a vertex MATH and MATH for some MATH such that MATH and MATH for some MATH. Considering the partial isometry MATH, we have MATH but since MATH, no such MATH exists. Thus MATH, and so by the universal property of MATH any NAME MATH-family MATH must have MATH. Now suppose that MATH is locally convex. Let MATH, and for each MATH define MATH by MATH where MATH is the usual basis for MATH. Each MATH because MATH. NAME relations REF follow directly from the definition of the operators MATH; it remains to show that relation REF is fulfilled. Since MATH is locally convex, by REF it suffices to show that for each MATH and MATH, MATH. If MATH, then the relation is trivially true, so suppose MATH. For MATH we have MATH . Taking MATH in REF, we can see that it suffices to show that MATH if and only if there exists MATH such that MATH. If MATH for some MATH, then MATH. If MATH, then MATH because MATH and MATH is a boundary path. |
math/0107222 | Because each MATH is nonzero, and MATH has initial projection MATH, each MATH is nonzero, and hence the representation MATH is nonzero on each MATH. It follows from REF that MATH is faithful on MATH, hence on MATH and on MATH. Since MATH, it follows that MATH is faithful on MATH (see CITE, for example). We can now use the argument of CITE. |
math/0107222 | For MATH and MATH, we write MATH for the unique path such that MATH with MATH. We know from the analysis in REF that MATH is faithful on the fixed-point algebra MATH, and hence the standard argument will work once we know that MATH . Recalling that MATH is dense in MATH, we consider arbitrary MATH where MATH is finite and MATH. Let MATH be the least upper bound of MATH . Then MATH . For MATH, we define MATH and MATH. For MATH, we define MATH and using NAME relation REF we then have MATH the point is that now MATH for all MATH. Since MATH decomposes as a direct sum MATH, so does its image under MATH, and there is a vertex MATH such that MATH . Choose a boundary path MATH such that MATH for all MATH; then for each MATH, there exists MATH such that MATH whenever MATH. Let MATH . Let MATH be the least upper bound of MATH. In particular, MATH when MATH is the second coordinate of an element of MATH, MATH, and MATH. Write MATH for MATH. For each MATH we define MATH . Now we define MATH by MATH . Since the MATH are mutually orthogonal projections, we have MATH . We aim to show that MATH and that MATH; this will give us MATH and the proof will be complete. Write MATH for the matrix algebra spanned by MATH. Notice that MATH. For MATH we have MATH so for MATH unless MATH and MATH unless MATH and hence MATH . Using REF , it follows that MATH is a family of nonzero matrix units, and from this we deduce that the map MATH is a faithful representation of MATH . Since both MATH and MATH are faithful on MATH and since MATH it follows from REF that MATH. To establish that MATH, we show that MATH whenever MATH and MATH this shows that MATH kills those terms of MATH which are the images under MATH of terms of MATH killed by MATH. Notice that if MATH then MATH so for MATH unless MATH . Hence, for MATH with MATH, we have MATH which is nonzero if and only if there exist MATH such that MATH . But MATH : if not, then there exists a MATH such that MATH and MATH . But MATH since MATH is a boundary path. Likewise, MATH, and so REF is equivalent to MATH, which is impossible by REF. This proves REF , and the result follows. |
math/0107222 | We use an inductive construction of MATH like that used by NAME for REF-graphs in CITE. For MATH, we define MATH and write MATH for the set obtained by repeating the process MATH times. Notice that if MATH is hereditary, then MATH. We will show that MATH is hereditary and equal to MATH. We begin by showing that if MATH is hereditary, then MATH is hereditary. To see this, suppose that MATH and that MATH . Then there exists MATH such that MATH and MATH . If MATH, then MATH, so suppose MATH, and factor MATH where MATH. We claim that MATH. To see this, choose MATH such that MATH . If MATH or if MATH, then MATH. So suppose that MATH and MATH. Since MATH is locally convex, MATH, so it suffices to show that MATH implies MATH. Let MATH. Then MATH for some MATH. But now MATH and hence MATH because MATH is hereditary. Thus MATH as claimed. By induction on the length of MATH, it follows that MATH, and hence MATH is hereditary. We now know that MATH for all MATH, and that MATH is hereditary for all MATH; thus MATH is also hereditary. It remains to show that MATH. Because applying MATH can never take us outside of a saturated set, we have MATH, so it is enough to show that MATH is saturated. To see this, suppose that MATH and MATH. Then, since MATH is row finite, we have MATH for some MATH and it follows that MATH. Thus MATH is saturated. |
math/0107222 | The proof of REF is the same as the proof of CITE once we establish that MATH and MATH from REF are locally convex row-finite MATH-graphs. This is easy to check for MATH, and the row-finiteness of MATH follows from that of MATH. We need to check that MATH is a MATH-graph and is locally convex. For convenience, write MATH. To show that the factorisation property holds for MATH, take MATH and suppose MATH. We know there exist unique MATH such that MATH, MATH and MATH. Certainly MATH, and if MATH, then MATH, a contradiction. Hence MATH, and MATH is a sub-MATH-graph. Now we show that MATH is locally convex. Consider an arbitrary vertex MATH which has MATH and MATH for some MATH. We know that MATH, and since MATH is locally convex, we also know there exist MATH and MATH. If MATH, then MATH, and similarly, if MATH, then MATH; in either case saturatedness implies that MATH or MATH, a contradiction. Hence MATH and MATH, so MATH is locally convex. |
math/0107225 | We view MATH and MATH as vector spaces over MATH through MATH, MATH. Let MATH. Then MATH defines a MATH-linear map MATH, where MATH is the MATH-dual of MATH. We want to prove that MATH is surjective. Since MATH is finite-dimensional, it suffices to prove that MATH separates points on MATH. Let MATH, and choose MATH with MATH. After decomposing with respect to the bigrading of MATH, we may assume that MATH for some MATH, which implies MATH. This proves that MATH is surjective. In particular, we can find MATH with MATH. |
math/0107225 | First we must check that MATH is well-defined, that is, that MATH for MATH. This is straight-forward. For instance, to prove the second identity, we assume MATH, MATH. We must prove that MATH or equivalently, by REF, that MATH . This is clear from the fact that MATH. It is easy to check that MATH maps into the subspace MATH of MATH and that MATH is a MATH-prealgebra homomorphism. Let us write out the proof that MATH is multiplicative, that is, that MATH . Assume that MATH, MATH, and write MATH, MATH, where MATH, MATH. The left-hand side of REF is then MATH while the right-hand side is MATH . These expressions are equal since MATH, MATH. To prove that MATH is injective, let MATH. We choose a representative MATH of MATH such that MATH are independent over MATH and MATH for all MATH. Choosing MATH as in REF , we have MATH. Again by REF , there exists MATH with MATH. Thus MATH. This completes the proof. |
math/0107225 | It remains to prove that MATH takes values in MATH. Since MATH preserves the bigrading, it is enough to consider MATH for MATH. We choose a representative MATH of MATH such that MATH, MATH, MATH are linearly independent over MATH and MATH are independent over MATH. Then MATH . Choosing MATH as in REF and applying the above identity to MATH gives MATH . We may assume that MATH with MATH. Define MATH through MATH. It is then easily checked that MATH, which gives MATH so that MATH. By symmetry, MATH for all MATH, and by a similar argument we may conclude that also MATH. This completes the proof. |
math/0107225 | It is easy to check that MATH. To see that MATH preserves MATH, choose MATH with MATH, and check that MATH. Next, check that MATH. REF is then equivalent to MATH, MATH, which is easily verified. To check the first identity in REF, we let MATH and MATH with MATH, MATH, MATH, MATH. Then MATH where we used that MATH. By the definition of MATH and by REF for MATH, this equals MATH . On the other hand, MATH . These two expressions are equal since MATH, MATH. This proves the first identity in REF and the second one may be proved similarly. |
math/0107225 | It follows from REF that the equation MATH defines an intertwiner if and only if MATH . Choosing MATH and MATH in REF and using REF, we see that this indeed holds for MATH. To prove the second statement we apply MATH to both sides of REF. Writing MATH, the left-hand side gives MATH . Computing the right-hand side similarly gives the identity MATH . Replacing MATH and comparing with REF, we see that this is indeed the QDYB equation. |
math/0107225 | Let MATH and MATH denote the left- and right-hand sides of REF, respectively. Assuming the existence of a pairing, we have MATH . Computing MATH similarly one obtains the identity MATH . Replacing MATH and comparing with REF wee see that this is equivalent to REF. Similarly, the identity MATH implies REF. Thus, REF implies REF . The converse is proved similarly: using REF we may extend REF to the algebra generated by MATH, MATH, MATH subject to relations REF. By the above argument, REF guarantees that the resulting form factors through the relations REF. The equivalence of REF is contained in REF . The ``only if"-part of the final statement follows by choosing MATH, MATH in REF. The ``if"-part then follows using REF . |
math/0107225 | Iterating REF and recalling the notation REF we have in general MATH (recall that MATH). Using this rule and REF to compute MATH we get two non-zero terms, corresponding to MATH, MATH and MATH, MATH. Thus, MATH . Using that MATH and MATH, we have MATH . Plugging this into the previous expression gives MATH . The expression in brackets equals MATH, which proves the statement for MATH. For MATH, MATH, MATH the proof is similar but simpler, since we only get one non-zero term. Finally we prove the statement for MATH using REF. |
math/0107225 | Iterating REF gives in general MATH . Using this and REF we see that MATH-for any complex polynomial MATH. Thus, for MATH, MATH with MATH. This gives MATH from which the statement readily follows. |
math/0107228 | The key point is to explain how MATH is defined: Take any vector field MATH on MATH (locally defined, if necessary) for which MATH and MATH. Now define a mapping MATH by setting MATH. It is not difficult to show that MATH is constant on the fibers of MATH and therefore that there exists a mapping MATH such that MATH. The mapping MATH is defined by a similar abstract diagram chase. The remainder of the proof is a matter of checking details and can be left to the reader. |
math/0107228 | It suffices to show that the quadratic form on the right hand side of REF is invariant under the flow of MATH. However, by REF. Substituting this into the structure equations allows one to compute the NAME derivative with respect to MATH of the right hand side of REF and see that it is equal to zero. |
math/0107228 | The formulae in REF follow immediately from the definition of NAME derivative, the defining properties of MATH, and the second and third equations of REF. Now compute the NAME derivative with respect to MATH of the second line of REF, using the fact that this operation is a derivation that commutes with exterior derivative, and add the result to the third equation. The result is MATH . (The reader who performs this calculation will note that it uses the fact that MATH is fully symmetric in its indices.) Since MATH while MATH and MATH, it follows that MATH and thus that MATH . In particular, the first equation of REF is verified, which shows that MATH is indeed fully symmetric in all its indices. REF also implies that there must exist (unique) functions MATH that satisfy MATH . Now compute the NAME derivative with respect to MATH of the third line of REF, using the fact that this operation is a derivation that commutes with exterior derivative, and subtract MATH times the second equation from the result. This yields the relation MATH . Of course, this implies both MATH and MATH . However, the symmetry of MATH and MATH and the skewsymmetry MATH now combine to show that MATH . This gives the second equation of REF and, in view of REF and MATH as well. |
math/0107228 | Define complex valued MATH-forms MATH . Using this notation and the condition MATH, one finds that the pullbacks of MATH and MATH can be written in the form MATH . Thus, the metric and MATH-form are algebraically compatible and the MATH-form is closed. The only condition remaining to verify in order to show that this metric is NAME with the given MATH-form as NAME form is whether or not the almost complex structure defined by this pair is integrable. Now, the almost complex structure induced on MATH is the one for which the MATH-pullback of a MATH-form is a linear combination of MATH. By the NAME Theorem, the integrability of the almost complex structure is equivalent to the condition that these forms define a differentially closed ideal. To see this, note that the second and third structure REF (when MATH) can be combined in complex form as MATH . Thus, the complex Pfaffian system spanned by the MATH is NAME, as desired. For the final statement, consider the complex-valued MATH-form defined on MATH by MATH . Let MATH be the canonical bundle of MATH (regarded as a complex manifold), that is, MATH is the top exterior power of the complex cotangent bundle of MATH. Let MATH be the tautological holomorphic MATH-form on MATH and let MATH denote the circle bundle of unit complex volume forms on MATH with respect to the NAME structure constructed in the first part of the proof. Evidently, there is a unique smooth mapping MATH that lifts MATH and satisfies MATH. Because MATH satisfies MATH it follows that MATH is an immersion. |
math/0107228 | Choose a hyperplane MATH that is transverse to the line MATH and set MATH. Of course, MATH is diffeomorphic to MATH. The image MATH is a totally real, real analytic MATH-dimensional submanifold of MATH whose complexified tangent space is transverse to the fibers of MATH. Thus, there exists a unique complex hypersurface MATH that contains MATH. By restricting MATH to a sufficiently small tubular neighborhood of MATH (in some metric on MATH), one can assume that MATH is embedded and everywhere transverse to the fibers of MATH since it is along MATH. Moreover, the hypothesis that MATH is strictly convex towards the origin in MATH implies that MATH is convex (in the sense of REF ) on a neighborhood of MATH, so by shrinking MATH again if necessary, one can assume that MATH is convex everywhere. Consider the corresponding MATH, which is a rectilinear generalized NAME structure on MATH with constant flag curvature MATH. By construction, the fiber MATH is compact and convex. It is now not difficult to see that there must be an open neighborhood MATH of MATH in MATH with the property that, for all MATH, the fiber MATH is also compact and nonempty. This MATH is the desired neighborhood. |
math/0107228 | By hypothesis, the MATH-geodesics of MATH are all closed, so, by REF , the MATH-geodesics of MATH (which are the same) are also closed. Moreover, by REF , the data MATH satisfy MATH where MATH is the NAME curvature of MATH. By the discussion at the beginning of this subsection, there is a canonically constructed coframing MATH on MATH, the unit tangent bundle of MATH over MATH, that satisfies the structure REF of a generalized NAME structure of constant flag curvature MATH and that induces the given data MATH on MATH, its space of geodesics. Because its foliation MATH given by MATH has closed leaves and, in fact, has MATH as its leaf space, REF shows that there is an immersion MATH that realizes MATH as a generalized NAME structure on MATH. The reader can easily check that MATH is, in fact, an embedding and defines a genuine NAME structure on MATH, as desired. |
math/0107228 | The integrability condition is equivalent to the condition that the Pfaffian system on MATH generated by MATH and the components of MATH be NAME. By REF , this condition is satisfied when MATH and is satisfied when MATH if and only if MATH. |
math/0107228 | This is a straightforward calculation: Choose a flag that is non-characteristic with respect to the claimed characteristic variety. One then finds that the characters of this flag are as given in REF. However, by combinatorics, one sees that, not only does one have the identity MATH but also that NAME 's test is satisfied, that is, MATH as was verified in the computation of the second NAME identity for integrable, torsion-free MATH-structures. |
math/0107228 | These results follow from the usual NAME construction of an exterior differential system whose integral manifolds are the the local, integrable, torsion-free MATH-structure plus the algebraic result of REF . The proof is similar in all details to those executed in CITE, to which the reader is referred if more detail is needed. |
math/0107228 | This is a matter of computation and expansion of the definitions. The point is that if one reduces to the locus in MATH where MATH, this clearly defines a MATH-substructure MATH as mentioned above. One can then write MATH and write MATH, just as in the previous section. Then the first equation of REF shows how one can define MATH and MATH in terms of the real and imaginary parts of MATH so that REF hold. Finally, applying REF generates the desired NAME structure. |
math/0107231 | The crux of the matter is to show that the MATH's ``span" MATH. Label the elements of MATH by integers MATH with MATH. For each MATH define a MATH matrix MATH by MATH. The orthogonality condition for the MATH's says that for each MATH . (See REF.) But this says exactly that MATH. Because we are in finite dimensions, this implies that MATH is unitary, so that also MATH. That is, MATH . (See REF.) We apply this for MATH to obtain MATH . |
math/0107231 | We use the notation of REF . Set MATH, so that MATH. Viewing MATH as a MATH-module over itself, we construct a MATH-module map MATH by MATH . We note that MATH preserves the MATH-valued inner products: for any MATH we have MATH . Hence MATH gives a MATH-module injection of MATH into MATH. Thus MATH is projective, and so it has a complementary module, MATH, such that MATH . We can take MATH to be orthogonal to MATH with respect to the inner product described above. To see that this orthogonal complement MATH exists, note that the ``orthogonal" projection of MATH onto MATH is given by MATH, so that MATH . Suppose that a high-pass filter family MATH exists for MATH, and set MATH for each MATH. Then from the equation of REF we see that the MATH's for MATH will be an orthonormal family in MATH. From REF we can deduce that they will actually form an orthonormal basis for MATH, so that MATH must be a free MATH-module. Thus to show that a high-pass filter family exists, we need to show that MATH is a free MATH-module, for then we can obtain an orthonormal basis. (Use, for example, the proof of REF.) We can then multiply by MATH to obtain the desired MATH's. Note that we have MATH . (This says that MATH is ``stably-free".) Thus to show that MATH is free we need to be able to ``cancel" one copy of MATH, so that MATH. We treat first the case in which MATH. This has a simple solution, as is widely seen in the wavelet literature. In this case the group MATH has only two elements. Let MATH denote the non-identity element of MATH. Choose a continuous function MATH on MATH such that MATH and MATH for all MATH. This can be done, for example, by choosing MATH to be a character on MATH such that MATH. Let MATH as above be given. Define MATH by MATH. Then a simple standard calculation shows that the pair MATH, MATH is an orthonormal set in MATH. We can now apply REF to conclude that MATH is a basis for MATH. (We remark that this case is a special case of the fact that on any compact space any stably-free line bundle is free. This is because line-bundles are determined by their first NAME class; see REF. But stably equivalent vector bundles have the same NAME class, by REF.) To treat the other case we use REF which shows that for any compact space MATH the projective modules over MATH correspond to the complex vector bundles over MATH. One direction of this correspondence consists of assigning to a vector bundle its MATH-module of continuous cross-sections. This enables us to use the facts about cancellation of vector bundles which are given in CITE. Our MATH-valued inner products correspond to ``Hermitian metrics" on vector bundles. Suppose now that MATH. Then MATH, where MATH denotes as in CITE the least integer greater than or equal to MATH . Then from REF we can deduce immediately that we can cancel MATH, so that MATH is a free module of rank MATH. We have thus proved the existence of the desired family of continuous high-pass filters associated to MATH . Suppose now that MATH is infinitely differentiable, and that we have obtained a corresponding family MATH of continuous high-pass filters, perhaps by use of the first part of this theorem. We can uniformly approximate the MATH's for MATH arbitrarily closely by infinitely differential functions, say MATH. These functions need not be orthogonal. But we can try to apply to them a ``NAME" process using MATH. The only care needed to make this work is that the approximations must be close enough so that, if MATH denotes the orthogonal projection (defined much as in the early part of the proof of this theorem) of MATH into the orthogonal complement of the span of the new MATH, then MATH must still be close enough to MATH so that MATH exists and is an infinitely differentiable function. For then we can ``normalize" MATH to obtain the new MATH. |
math/0107232 | The Enclosing Property of MATH shows that any component of MATH is canonical. Thus MATH consists of non-parallel canonical surfaces and so is contained in a NAME. Now any embedded essential annulus or torus can be isotoped to lie in MATH. Thus the proof of the proposition reduces to showing that if a component MATH of MATH contains an embedded annulus or torus MATH which is essential in MATH, and is not parallel to a component of MATH, then MATH is not canonical. This means that there is some essential annulus or torus MATH in MATH which has non-zero intersection number with MATH, or equivalently that MATH cannot be properly homotoped apart from MATH. This can be proved by careful consideration of the possibilities using the fact that MATH is a NAME fibre space or a MATH-bundle and that MATH can be isotoped to be vertical in MATH, that is, to be a union of fibres in MATH. However, this consideration has essentially been done already in CITE, so we will use their work to avoid a case by case discussion. In CITE, NAME and NAME defined what they called the MATH-system of MATH. This consisted of a maximal system of disjoint annuli and tori MATH in MATH so that no two of the MATH are parallel, and each MATH is canonical in their sense, which means that any embedded essential annulus or torus in MATH can be isotoped away from each MATH. We would like to remark that the MATH-system is another natural concept that can be generalized to groups. The MATH-system is unique up to isotopy and they showed that the MATH-system contains MATH. By the definition of the MATH-system, any essential annulus or torus disjoint from and not parallel into the MATH-system intersects some other embedded annulus or torus in an essential way and so does not belong to a NAME. The only surfaces which are in the MATH-system but not in MATH are the special annuli described in REF. It is easy to see that in these cases (which essentially arise from NAME bundles over twice punctured discs), there is an immersed annulus intersecting a special annulus in an essential way. It follows that these special annuli also do not belong to the NAME, so that the NAME consists precisely of the MATH-system with these special annuli removed, which is the same as MATH. |
math/0107232 | Perform surgery on the frontier MATH of MATH to make it MATH-injective in MATH. This adds to MATH or subtracts from MATH compact sets, namely REF-handles attached to MATH, so that, by changing MATH by a compact set, we can arrange that MATH has MATH-injective frontier MATH. As MATH is isomorphic to MATH, the components of MATH can only be annuli, discs or spheres. As MATH is irreducible, any sphere bounds a ball, so that any spheres in MATH can be removed by adding this ball to MATH or removing it, as appropriate. As MATH has incompressible boundary, if MATH is a disc properly embedded in MATH with boundary MATH, then MATH bounds a disc MATH in MATH. Thus MATH is a sphere in MATH and so bounds a ball. As before, any discs in MATH can be removed by adding this ball to MATH or removing it, as appropriate. Similarly inessential annuli can be removed from MATH. Thus we eventually arrive at a stage where MATH consists of a finite number of compact essential annuli in MATH. If two boundary circles of MATH lie in a single component MATH of MATH, they will cobound a compact annulus MATH in MATH, and we can alter MATH by adding or removing a regular neighbourhood of MATH. This reduces the number of boundary circles of MATH, so by repeating, we can ensure that distinct boundary circles of MATH lie in distinct boundary components of MATH as required. (This move may produce an inessential annulus component of MATH, in which case we can remove it as above.) This completes the proof of the lemma. |
math/0107232 | Recall that MATH is infinite cyclic. This implies that MATH carries a subgroup of finite index in MATH, so that MATH must separate MATH into two pieces MATH and MATH. Now let MATH denote the four components of MATH which contain the four boundary components of MATH and MATH. Thus each MATH is an open annulus and MATH and MATH cut each MATH into two unbounded annuli. As each MATH contains no other boundary component of any MATH, this means that each of MATH and MATH contains two of these unbounded annuli, and so must be unbounded. This completes the proof of the lemma. |
math/0107232 | REF tells us that we can arrange that MATH consists only of embedded essential annuli, and that distinct boundary circles of MATH lie in distinct boundary components of MATH. First we will show that MATH is connected. Otherwise, pick two components of MATH. There is a compact incompressible annulus MATH embedded in MATH and joining these two components. It is possible that MATH meets other components of MATH in its interior, but we can isotop MATH so that it meets each component of MATH in essential circles only. Thus MATH has a sub-annulus which joins distinct components MATH and MATH of MATH and whose interior does not meet MATH. We will replace MATH by this sub-annulus and will continue to call this annulus MATH. Without loss of generality we can suppose that MATH lies in a component MATH of MATH, and cuts MATH into pieces MATH and MATH. REF shows that MATH and MATH must each be unbounded. Let MATH and MATH denote the components of MATH which contain MATH and MATH respectively and let MATH denote the submanifold MATH of MATH. Then all four of the intersection sets of MATH and MATH with MATH and MATH are unbounded, so we have found an ending submanifold of MATH with non-zero intersection number with the ending submanifold MATH, contradicting our hypothesis that MATH-is associated to a canonical splitting of MATH over MATH. This contradiction shows that MATH must be connected. Next we show that MATH must carry MATH. Otherwise, MATH carries a proper subgroup MATH of finite index in MATH. Let MATH denote the pre-image of MATH in MATH, so that MATH is the pre-image of MATH in MATH and consists of at least two embedded essential annuli. Note that distinct boundary circles of MATH must lie in distinct boundary components of MATH. Now we can apply the argument of the preceding paragraph with MATH in place of MATH and MATH in place of MATH. This will yield an ending submanifold of MATH with non-zero intersection number with the ending submanifold MATH and hence with MATH. This contradicts our hypothesis that MATH-is associated to a canonical splitting of MATH over MATH, completing the proof of the lemma. |
math/0107232 | REF tells us that if we start with such a splitting of MATH over MATH, the associated ending submanifold MATH of MATH can be chosen to have frontier consisting of a single essential annulus MATH which carries MATH. We will show that MATH can be chosen so that its projection into MATH is an embedding whose image we again denote by MATH. Thus MATH induces the given splitting of MATH over MATH. As pointed out earlier, it is trivial that MATH must be topologically canonical. In order to prove that we can choose MATH so as to embed in MATH, it will be convenient to choose MATH to be least area in its proper homotopy class. It is shown in CITE that any such least area annulus will also be embedded in MATH. (There are two options here. One can choose a Riemannian metric on MATH and then choose MATH to minimise smooth area, see CITE or CITE, or one can follow the ideas of CITE and choose a hyperbolic metric on the MATH-skeleton of some fixed triangulation of MATH and choose MATH to minimise MATH-area. It does not matter which choice is made.) The fact that MATH has zero intersection number with any ending submanifold of any MATH, with MATH-isomorphic to MATH, implies, in particular, that the pre-image MATH of MATH in MATH does not cross any of its translates. As the frontier MATH of MATH is an infinite strip covering MATH and so is connected, it follows that MATH crosses none of its translates. This means that, for all MATH, at least one of MATH and MATH has bounded image in MATH. In turn, this implies that the self-intersection number of MATH is zero so that MATH must cover an embedded surface in MATH, by CITE. If MATH properly covers an embedded annulus, then the stabiliser of MATH will strictly contain MATH. But as MATH is associated to a splitting of MATH over MATH, we know that the stabiliser of MATH is exactly MATH. It follows that either MATH embeds in MATH as required, or that it double covers an embedded MATH-sided NAME band MATH in MATH. In the second case, we can isotop MATH slightly so that it embeds in MATH as the boundary of a regular neighbourhood of MATH. Thus in all cases, we can choose MATH so that its projection into MATH is an embedding as required. This completes the proof that the given algebraically canonical splitting of MATH comes from a topologically canonical annulus. |
math/0107232 | Let MATH be an ending submanifold of MATH. As in the proof of REF , we can arrange that MATH consists of essential annuli and tori, and that distinct boundary circles of MATH lie in distinct boundary components of MATH. We can also arrange that MATH has at most one torus component, as any two incompressible tori in MATH cobound a product region in MATH, which can be added to MATH or subtracted from MATH, as appropriate. This completes the proof of the lemma. |
math/0107232 | We can assume that MATH is chosen as in the immediately preceding REF . First we show that MATH cannot have an annulus component which is mixed. For suppose that MATH has an annulus component joining a left annulus component MATH and a right annulus component MATH of MATH. As MATH has left and right annulus components, it does not have a torus component so there is an essential torus in MATH. This torus is the frontier of an ending submanifold MATH of MATH, and we consider the four intersections MATH. None of these can be compact as each contains a non-compact piece of MATH or of MATH in its boundary. This uses the fact that distinct boundary circles of MATH lie in distinct components of MATH. It follows that MATH and MATH have non-zero intersection number contradicting our hypothesis that MATH is associated to a canonical splitting of MATH over MATH. This contradiction shows that MATH does not have a mixed annulus component. Note that this implies that each component of MATH separates MATH. Second we will show that MATH cannot have any components which are left or right annuli. The only remaining possibility will then be that MATH consists of a single torus as claimed. Note that in the following proof, we will only use the fact that the given splitting is MATH-canonical, but not the fact that it is algebraically canonical. This will allow us to use the same argument in the last part of the proof of REF . Suppose that MATH has a left annulus component MATH which carries the subgroup MATH of MATH. Then the pre-image of MATH in MATH consists of infinitely many copies of MATH. Let MATH denote the full pre-image of MATH in MATH. Of course, MATH is not an ending submanifold of MATH as its frontier MATH is not compact. The components of MATH may be compact annuli, infinite strips or an open annulus if MATH has a torus component. A compact annulus component of MATH will be called left or right depending on whether it projects to a left or right annulus component of MATH. Recall that distinct boundary circles of MATH lie in distinct boundary components of MATH. It follows that distinct boundary circles of MATH lie in distinct boundary components of MATH. As MATH contains more than one left annulus component, there is an incompressible annulus MATH embedded in MATH and joining two such components. It is possible that MATH meets other components of MATH in its interior, but we can isotop MATH so that it meets each component of MATH in essential circles only. Hence, if MATH meets a component MATH of MATH, then MATH must be an annulus, compact or open, and must separate MATH. It follows that if MATH meets MATH in more than one essential circle, we can alter MATH so as to remove two circles of intersection with MATH, essentially by replacing a sub-annulus of MATH by a sub-annulus of MATH. As the boundary circles of MATH lie in left annulus components of MATH, we can arrange, by repeating this argument, that MATH meets only left annulus components of MATH. Now MATH must have a sub-annulus which joins distinct left annulus components MATH and MATH of MATH and whose interior does not meet MATH. We will replace MATH by this sub-annulus and will continue to call this annulus MATH. Without loss of generality, we can assume that MATH lies in a component MATH of MATH, and we let MATH and MATH denote the components of MATH which contain MATH and MATH respectively. This annulus MATH separates MATH into two pieces MATH and MATH, and the argument of REF shows that neither MATH nor MATH can be compact. Recall that if MATH is any left annulus component of MATH in MATH, then MATH carries the group MATH and cuts MATH into two pieces one of which also carries MATH. It follows that any left annulus component of MATH cuts MATH into two pieces one of which projects into MATH by a homeomorphism. As MATH and MATH are such annuli, it follows that at least one of MATH and MATH projects into MATH by a homeomorphism. In particular, at least one of MATH and MATH is an ending submanifold of MATH. Assume that MATH is an ending submanifold. Cut MATH along MATH, and let MATH denote the piece so obtained which contains MATH. Thus MATH is also an ending submanifold. If MATH contains no translate of MATH other than MATH itself, we remove one such translate from MATH to obtain a new ending submanifold MATH such that MATH contains at least two translates of MATH. Otherwise we let MATH equal MATH. Now we let MATH and note that MATH is an ending submanifold of MATH and that MATH contains a translate of MATH. This implies that all four of the sets MATH are unbounded. If MATH and MATH denote the pre-images of MATH and MATH respectively in MATH, it follows that each of the four sets MATH has projection into MATH which is unbounded, so that MATH and MATH have non-zero intersection number. This contradicts the assumption that MATH-is associated to a canonical splitting of MATH over MATH, so we conclude that MATH cannot have a component which is a left annulus. Similarly, MATH cannot have a component which is a right annulus. This completes the proof of the lemma. |
math/0107232 | REF tells us that if we start with such a splitting of MATH over MATH, the associated ending submanifold MATH of MATH can be chosen to have frontier consisting of a single essential torus MATH. It is automatic that MATH carries MATH. We will show that MATH can be chosen so that its projection into MATH is an embedding whose image we again denote by MATH. Now MATH induces the given splitting of MATH over MATH. As pointed out earlier, it is trivial that MATH must be topologically canonical. In order to prove that we can choose MATH so as to embed in MATH, it will again be convenient to choose MATH to be least area. See CITE or CITE for the existence results in the smooth and MATH cases respectively. Again CITE shows that a least area torus is embedded. The fact that MATH has zero intersection number with any ending submanifold of any MATH, with MATH-isomorphic to MATH, implies in particular that the pre-image MATH of MATH in MATH does not cross any of its translates. Hence the frontier MATH of MATH, which is a plane above MATH, crosses none of its translates. This means that, for all MATH, at least one of MATH and MATH has bounded image in MATH. In turn, this implies that the self-intersection number of MATH is zero so that MATH must cover an embedded surface in MATH, by CITE. If MATH covers an embedded torus, then the stabiliser of MATH will contain MATH. But as MATH is associated to a splitting of MATH over MATH, we know that the stabiliser of MATH is exactly MATH. It follows that either MATH embeds in MATH as required, or that it double covers an embedded MATH-sided NAME bottle in MATH. In this case, we can isotop MATH slightly so that it embeds in MATH as the boundary of a regular neighbourhood of the NAME bottle. Thus in all cases, we can choose MATH so that its projection into MATH is an embedding as required. This completes the proof that the given algebraically canonical splitting of MATH comes from a topologically canonical torus. |
math/0107232 | Let MATH denote the group carried by MATH, so that we have a splitting of MATH over MATH. Lift MATH into MATH, and let MATH denote one side of MATH in MATH. Let MATH denote the pre-image in MATH of MATH. We consider the annulus boundary components of MATH. There are two which contain the boundary circles of MATH. The others lie on the left or right of MATH. As all these annuli carry exactly the same group, namely MATH, we cannot have such annuli on both sides of MATH, as this would immediately yield a compact annulus in MATH crossing MATH in an essential way, contradicting our hypothesis that MATH is topologically canonical. Hence we can suppose that MATH contains no annulus component of MATH. Let MATH be a subgroup of MATH isomorphic to MATH, and let MATH be an ending submanifold of MATH with pre-image MATH in MATH. We will show that MATH and MATH do not cross. If this is known for every such MATH-and MATH, it will follow that MATH has intersection number zero with every such MATH, as required. REF tells us that we can choose MATH to be a submanifold of MATH with frontier consisting only of essential annuli, and all these annuli carry the same group. Thus MATH has frontier consisting of finitely many infinite strips, which are ``parallel" in the sense that each lies within a uniformly bounded distance of each other. We consider how these strips project into MATH. If one (and hence every) such strip has compact image in MATH, that image will be a (possibly singular) annulus properly mapped into MATH. As MATH contains no annulus component of MATH, it follows that each of these annuli can be properly homotoped to lie in MATH. After this homotopy, one of MATH or MATH would be empty. It follows that, before the homotopy, one of these two sets projected to a bounded subset of MATH, so that MATH and MATH do not cross, as required. Otherwise, the image of each of these strips in MATH is non-compact and the map into MATH is proper. Thus each end of each strip is mapped to the left or to the right side of MATH. If one of these strips has its two ends on opposite sides of MATH, the corresponding annulus component of MATH has non-zero intersection number with MATH, contradicting our hypothesis that MATH is topologically canonical. It follows that each of these strips has both of its ends on the same side of MATH, and hence that all the ends of all the strips are on one side of MATH. Hence they can be properly homotoped so as to be disjoint from MATH, which again shows that MATH and MATH do not cross as required. Now let MATH be a subgroup of MATH isomorphic to MATH, and let MATH be an ending submanifold of MATH with pre-image MATH in MATH. We will show that MATH and MATH do not cross. This will show that MATH-has intersection number zero with every such MATH. REF tells us that we can choose MATH so that MATH consists of essential annuli and possibly a torus in MATH. Thus MATH has frontier MATH consisting of infinite strips and possibly a plane. In any case, if MATH denotes an incompressible torus in MATH, it yields a plane MATH in MATH such that MATH lies in a bounded neighbourhood of MATH. Now we consider how MATH projects into MATH. There are two cases. Either MATH projects into MATH to yield a properly immersed plane, or MATH projects to a properly immersed annulus. In the first case, as a plane has only one end, and as a properly immersed plane in MATH can only meet MATH in a compact set, it follows that the intersection of MATH with MATH or MATH is compact. As MATH lies in a bounded neighbourhood of MATH, it follows that MATH meets MATH or MATH in a compact set. This implies that one of the four intersection sets MATH is bounded, so that MATH and MATH do not cross as required. In the second case, there are two subcases depending on whether the two ends of the annulus image of MATH are mapped to the same or opposite sides of MATH. If they are mapped to the same side, then much as above, it follows that one of the four intersection sets MATH has bounded image in MATH, so that MATH and MATH do not cross as required. If they are mapped to opposite sides of MATH, it follows that the torus MATH in MATH has non-zero intersection number with MATH, which contradicts our hypothesis that MATH is topologically canonical. This concludes the proof that a topologically canonical annulus in MATH determines an algebraically canonical splitting of MATH. |
math/0107232 | Let MATH denote the group carried by MATH, so that we have a splitting of MATH over MATH. Lift MATH into MATH, and let MATH denote one side of MATH in MATH. Let MATH denote the pre-image in MATH of MATH. As MATH cannot have a torus component, it must consist of a (possibly empty) collection of left annuli and right annuli and planes. If there are any left annulus components, it follows that the canonical component of MATH containing the left side of MATH is a peripheral NAME fibre space-MATH . Similarly for right annuli. As we are assuming that MATH is not special, it follows that MATH cannot have both left annulus and right annulus components. Without loss of generality, suppose that MATH has no left annulus components. Let MATH be a subgroup of MATH isomorphic to MATH, and let MATH be an ending submanifold of MATH with pre-image MATH in MATH. We will show that MATH and MATH do not cross. As before, this will show that MATH-has intersection number zero with every such MATH. REF tells us that we can choose MATH to be a submanifold of MATH with frontier consisting only of essential annuli, and all these annuli carry the same group. Thus the frontier MATH of MATH consists of finitely many `parallel' infinite strips. As before there are two cases depending on whether these strips project properly into MATH or project to compact annuli in MATH. Suppose that MATH projects to compact annuli in MATH. The boundary lines of MATH then project to circles in MATH which must all lie on the right side of MATH, as MATH has no left annulus components. Thus we can homotop MATH to arrange that MATH lies entirely on one side of MATH. It follows that before the homotopy, one of the four intersection sets MATH has bounded image in MATH, so that MATH and MATH do not cross as required. Next suppose that MATH projects to infinite strips in MATH. Each boundary line of each such strip must lie in a boundary component of MATH. As the intersection of a strip with MATH is compact, the two boundary lines of a single strip must lie on the same side of MATH, so that the strip can be homotoped to be disjoint from MATH. As the infinite strips forming MATH are parallel, it follows that they all lie on the same side of MATH, and again MATH and MATH do not cross. Now let MATH be a subgroup of MATH isomorphic to MATH, and let MATH be an ending submanifold of MATH with pre-image MATH in MATH. We will show that MATH and MATH do not cross. As before, this will show that MATH-has intersection number zero with every such MATH. REF tells us that we can assume that MATH consists of essential annuli and possibly one torus. Thus MATH consists of infinite strips and possibly a plane. As in the proof of REF , we consider the projection of MATH into MATH. An infinite strip component of MATH must project to a properly immersed infinite strip or to an annulus. As above, in either case, the image of any one strip component of MATH can be homotoped to lie on one side of MATH. If the image of a strip is a singular annulus, its boundary curves can only lie on the right side of MATH. Let MATH denote an incompressible torus in MATH, and let MATH denote the pre-image plane in MATH. Thus MATH lies in a bounded neighbourhood of MATH. There are three cases for the projection of MATH into MATH. We can obtain a proper map of a plane, torus or annulus. If MATH is mapped properly into MATH, the fact that MATH has only one end implies that all except a compact subset of MATH maps to one side of MATH. Also the infinite strips in MATH must all project to infinite strips each within a bounded neighbourhood of the image of MATH. It follows that we can homotop MATH to lie on one side of MATH, so that MATH and MATH do not cross. If the image of MATH in MATH is a singular torus, it follows that every infinite strip of MATH has image a singular annulus and so has boundary lying on the right side of MATH. Thus MATH-can be homotoped into the right side of MATH, and again MATH and MATH do not cross. If the image of MATH in MATH is a singular annulus, it will be convenient to let MATH and MATH denote the canonical pieces of MATH which meet the left and right sides of MATH respectively. Of course, it is possible that MATH. Now the image of MATH is a singular annulus whose two ends must lie on the same side of MATH, as otherwise the tori MATH and MATH would intersect essentially, contradicting our hypothesis that MATH is a canonical torus. If an infinite strip in MATH has image an infinite strip in MATH, that strip must again have its boundary on the same side of MATH as the ends of the singular annulus which is the image of MATH. If an infinite strip in MATH has image a singular annulus, we know that its boundary must lie on the right side of MATH and so can be homotoped to lie entirely on the right side of MATH. Thus either all of MATH can be homotoped into one side of MATH, so that MATH and MATH do not cross, or MATH has some infinite strips which project to singular annuli whose boundary curves lie on the right of MATH, but the ends of the image of MATH lie on the left of MATH. We will show that this last case cannot occur. Suppose it does occur. Then MATH projects to a torus in MATH which is homotopic into MATH. As the image of MATH in MATH is an annulus, MATH and MATH cannot be homotopic, so it follows that MATH is a NAME fibre space. Also there is a singular compact annulus MATH in MATH which has one boundary component on MATH, meets MATH transversely in a single circle, and has its second boundary component in an annulus boundary component of MATH. The existence of MATH implies that MATH is a NAME fibre space. Further the existence of MATH shows that MATH and MATH can be fibred to induce the same fibration on MATH. This contradicts the fact that MATH is a canonical torus in MATH, so we deduce that this last possibility cannot occur. This completes the proof that a topologically canonical torus in MATH, which is not special, determines an algebraically canonical splitting of MATH. |
math/0107232 | Let MATH be a special canonical torus in MATH, and let MATH denote the group carried by MATH, so that we have a splitting of MATH over MATH. Lift MATH into MATH, and let MATH denote one side of MATH in MATH. Let MATH denote the pre-image in MATH of MATH. The hypotheses imply that MATH has at least one annulus component on each side of MATH. As before we will refer to the two sides of MATH-as the left and right. Suppose first that MATH has at least two annulus components on each side of MATH. Pick a compact left-annulus MATH in MATH joining two distinct left annulus components of MATH and let MATH be a similarly defined right annulus. Let MATH denote the component of MATH which does not contain MATH, let MATH denote the component of MATH which does not contain MATH, and let MATH. Then clearly all four of the intersections MATH are unbounded, so that MATH-and MATH have non-zero intersection number and so MATH is not algebraically canonical in this case. If MATH has only one annulus component on one side of MATH, or on both sides of MATH, we simply make the preceding argument in a double cover of MATH chosen so that each annulus component of MATH has two components above it in the double cover. |
math/0107232 | CASE: Lift MATH into MATH, and let MATH denote one side of MATH in MATH. Let MATH denote the pre-image in MATH of MATH. As MATH is special, the canonical pieces of MATH on each side of MATH are NAME fibre spaces which meet MATH either in tori or in vertical annuli. Hence MATH consists of a non-empty collection of left annuli and right annuli and of planes, with the left annuli and right annuli carrying incommensurable subgroups of MATH. Let MATH be a subgroup of MATH isomorphic to MATH, and let MATH be an ending submanifold of MATH with pre-image MATH in MATH. We will show that MATH and MATH do not cross. As before, this will show that MATH-has intersection number zero with every such MATH. REF tells us that we can choose MATH to be a submanifold of MATH with frontier consisting only of essential annuli, and all these annuli carry the same group. Thus the frontier MATH of MATH consists of finitely many `parallel' infinite strips, that is, any strip lies in a bounded neighbourhood of any other strip. As before there are two cases depending on whether these strips project properly into MATH or project to compact annuli in MATH. Suppose that MATH projects to compact annuli in MATH. The boundary lines of MATH then project to circles in MATH which must all lie on the same side of MATH, as the left annuli and right annuli of MATH carry incommensurable subgroups of MATH. Thus we can homotop MATH to arrange that MATH lies entirely on one side of MATH. It follows that before the homotopy, one of the four intersection sets MATH has bounded image in MATH, so that MATH and MATH do not cross as required. Next suppose that MATH projects to infinite strips in MATH. Each boundary line of each such strip must lie in a boundary component of MATH. As the intersection of a strip with MATH is compact, the two boundary lines of a single strip must lie on the same side of MATH, so that the strip can be homotoped to be disjoint from MATH. As the infinite strips forming MATH are parallel, it follows that they all lie on the same side of MATH, and again MATH and MATH do not cross. This completes the proof that MATH is MATH-canonical. Let MATH and MATH denote the NAME fibre spaces on each side of MATH, remembering that it is possible that MATH equals MATH. If MATH has a canonical annulus of MATH in its frontier, this annulus will be vertical and so must carry the fibre group of MATH. Thus if each of MATH and MATH are distinct and each has a canonical annulus of MATH in its frontier, it is immediate that MATH has splittings over two incommensurable infinite cyclic subgroups of MATH. If MATH equals MATH, and has a canonical annulus of MATH in its frontier, this yields a splitting of MATH over an infinite cyclic subgroup of MATH and a suitable conjugate of this splitting will be over an incommensurable infinite cyclic subgroup of MATH. Thus in this case also, MATH has splittings over two incommensurable infinite cyclic subgroups of MATH. If MATH has no canonical annulus in its frontier, it must meet MATH in a torus boundary component MATH. In this case, MATH cannot be homeomorphic to MATH, as this would imply that MATH was inessential, so it follows that MATH contains an essential vertical annulus with boundary in MATH. As before, whether or not MATH and MATH are distinct, this implies that MATH has splittings over two incommensurable infinite cyclic subgroups of MATH, as required. This completes the proof of REF of REF Let MATH be a MATH-canonical splitting of MATH over a subgroup MATH which is isomorphic to MATH, such that MATH has splittings over two incommensurable infinite cyclic subgroups of MATH. As usual, we consider the cover MATH of MATH, whose boundary must consist of at most one torus, some planes and some left annuli and right annuli. If MATH admits a splitting over an infinite cyclic subgroup MATH, it follows that there is an embedded essential annulus MATH in MATH such that MATH carries a subgroup of MATH. Thus the fact that MATH admits splittings over two incommensurable infinite cyclic subgroups of MATH, implies that MATH contains essential annuli MATH and MATH which also carry incommensurable subgroups of MATH. In particular, they must lift to annuli in MATH. If MATH has a torus boundary component MATH, then any loop in MATH is homotopic into MATH. It follows that MATH contains essential annuli MATH and MATH (which may not be lifts of MATH and MATH which carry incommensurable subgroups of MATH and each have one boundary component in MATH. But this implies that MATH is homeomorphic to MATH, and hence that MATH-has finite index in MATH, which contradicts the hypothesis that MATH splits over MATH. This contradiction shows that MATH does not have a torus boundary component. It follows that MATH has both left annuli and right annuli, and the left and right annuli of MATH carry incommensurable subgroups of MATH. Let MATH denote an ending submanifold for the given MATH-canonical splitting MATH of MATH, where MATH is chosen as in REF . As the left and right annulus components of MATH carry incommensurable subgroups of MATH, it follows that no component of MATH can be a mixed annulus. Now we use the same argument as in the second part of the proof of REF to show that as MATH is MATH-canonical, MATH cannot have a component which is a left annulus nor a right annulus. It follows that MATH consists of a single essential torus MATH in MATH, and the argument of REF shows that MATH can be chosen to project to an embedding in MATH. Finally, the fact that MATH has both left annuli and right annuli implies that MATH is a special canonical torus as required. This completes the proof of REF ) of REF . |
math/0107232 | Let MATH denote MATH. REF tells us that there is a natural bijection between the collection of isotopy classes of all canonical annuli and tori in MATH and a certain collection of conjugacy classes of splittings of MATH, some canonical and some MATH-canonical. As this collection of splittings of MATH is defined purely algebraically, it follows that the conjugacy classes of splittings of MATH obtained from MATH and from MATH are the same. Now REF implies that they yield isomorphic graphs of groups structures for MATH, as required. |
math/0107232 | We start by applying REF . Let MATH-denote the NAME of MATH. It follows immediately from REF that we can homotop MATH to a map MATH such that MATH maps MATH to MATH by a homeomorphism and MATH. Further, it is now automatic that MATH induces a bijection between the components of MATH and of MATH, and that if MATH is the closure of a component of MATH, and if MATH denotes the closure of the component of MATH which contains MATH, then MATH is a homotopy equivalence. If MATH does not meet MATH, then MATH is proper, that is, MATH, and so NAME 's Homeomorphism Theorem implies that it is homotopic to a homeomorphism by a homotopy fixed on MATH. Note that this is true whether or not MATH is fibred, so that the restriction of MATH to interior components of MATH must also be a homeomorphism. In order to complete the proof of the Deformation Theorem, it remains to handle those components of MATH which are not fibred and meet MATH. Let MATH be the closure of such a component, and let MATH denote the closure of the component of MATH which contains MATH. Let MATH denote the induced map MATH and recall that MATH restricts to a homeomorphism of MATH with MATH. It was shown in CITE that any simple pair MATH such that MATH contains an embedded MATH-injective annulus not parallel into MATH is fibred. In particular, it follows that no component of MATH is an annulus. Now REF , which we prove below, tells us that MATH can be homotoped to a homeomorphism while keeping MATH mapped into MATH during the homotopy. Note that the final homeomorphism may flip the MATH-factor of some of the annuli in MATH. |
math/0107232 | Let MATH be the cover of MATH corresponding to the image of MATH in MATH and MATH the cover of MATH corresponding to the image of MATH. We have MATH, a lift of MATH. We denote by MATH, a lift of MATH at the base point of MATH, so that MATH is homeomorphic to MATH, and observe that it is enough to deform MATH into MATH. We denote by MATH the inverse image of MATH under MATH. We next consider the map induced by inclusion on homology groups with integer coefficients: MATH . We claim that MATH is the zero map. If MATH is empty, the above assertion is equivalent to showing that MATH, the case that was considered in CITE. In the general case, the proof proceeds in a similar fashion. Consider a component MATH of MATH other than MATH. If MATH is non-trivial, we have a homotopy annulus joining any loop in MATH to a loop in MATH. By our assumption any such annulus can be homotoped relative to its boundary into MATH. It follows that any loop in MATH is peripheral in MATH and thus MATH is either MATH or MATH with a closed subset of MATH removed. Moreover, there is an annulus in MATH joining any non-trivial loop in MATH to a boundary component of MATH. We next observe that MATH is generated by properly embedded, two sided surfaces in MATH. If MATH is such a surface, consider a component MATH of MATH which is not in MATH. Let MATH denote the component of MATH which contains MATH. If MATH is contractible in MATH, it must bound a MATH-disc in MATH which we add to MATH and then push off MATH. If MATH is essential in MATH, we add to MATH an annulus MATH in MATH which joins MATH to a peripheral loop MATH in MATH and push MATH off MATH. Thus, in either case, we can modify MATH without changing its image in MATH so as to remove the component MATH. By repeating this process, we see that the image of MATH in MATH is generated by surfaces MATH whose boundary is in MATH. Now any such surface is homologous to an incompressible surface, and we observe that any incompressible surface in MATH with boundary in MATH is parallel into MATH. Thus the image of MATH in MATH is zero. Using NAME duality for cohomology with compact supports, it follows that the induced map in cohomology from MATH to MATH is zero. As MATH is a homotopy equivalence, the induced map MATH is a proper homotopy equivalence, so that it induces an isomorphism MATH. Further the hypotheses imply that MATH induces a homeomorphism on MATH, so that it also induces an isomorphism MATH. Applying NAME duality in MATH, we see that the induced map in homology (with integer coefficients): MATH is also the zero map. Next consider MATH. Since MATH induces an isomorphism MATH and is already an embedding on MATH, we can homotop MATH relative to MATH to become an embedding. One method to prove this is to choose a metric on MATH which blows up away from MATH, and homotop MATH to be of least area. A more classical argument involves taking a regular neighborhood MATH of MATH, and compressing its frontier MATH by adding MATH-handles or by cutting MATH along a compressing disc for MATH. One can homotop MATH so as to avoid these compressing discs, so that one obtains a submanifold MATH of MATH which contains MATH and has incompressible frontier. We let MATH denote a component of MATH. As MATH induces an isomorphism MATH, it follows that there is a homotopy inverse map MATH. This map will induce a proper map MATH which induces an injection of fundamental groups and an injection of boundaries. It follows that this map is homotopic to a homeomorphism rel MATH. Hence MATH can be homotoped to an embedding with image MATH as required. Since MATH, which is now embedded, represents the zero element in MATH, we see that MATH is parallel into MATH. Thus MATH can be deformed into MATH relative to MATH and hence MATH can be deformed into MATH relative to MATH. This completes the proof of REF . |
math/0107232 | We start by considering a component MATH of MATH, and observing that either REF applies, or we are in the special REF . First suppose that the special case of REF does not arise, so that any MATH-injective annulus in MATH is properly homotopic into MATH. Then REF applies to every component of MATH, so we can homotop MATH rel MATH to a proper map. We will continue to call this map MATH. Next we claim that, for each component MATH of MATH, we can deform MATH to an embedding so that during the homotopy each component of MATH either stays fixed or is moved across an annulus component of MATH. The argument here uses the fact that MATH is MATH-injective, but this is not enough. It is also necessary to use the fact that any MATH-injective annulus in MATH is properly homotopic into MATH. This is another omission in NAME 's proof. We let MATH denote the component of MATH which contains MATH. If MATH is closed, then MATH is homotopic to a covering map of MATH of some degree MATH. REF tells us that, as MATH-and MATH are orientable, MATH must equal MATH, so that we have deformed MATH to a homeomorphism onto its image, as required. Next we consider the case when MATH has non-empty boundary. We already know that MATH embeds MATH in MATH. Now we consider the lift MATH of MATH into the cover MATH of MATH whose fundamental group equals MATH. We can homotop MATH rel MATH to an embedding. This homotopy induces a homotopy of MATH. We will now assume that this homotopy has been done and use the same notation for the new maps. Thus MATH is an embedding in MATH, and we denote its image by MATH. Let MATH denote the universal cover of MATH, and let MATH denote the pre-image in MATH of MATH. The full pre-image in MATH of MATH consists of MATH and all its translates by MATH. Suppose that MATH is not an embedding. Then some translate MATH of MATH must meet MATH but not equal MATH. Let MATH denote a component of this intersection, and let MATH denote the image of MATH in MATH. This will be a compact subsurface of MATH. Note that the boundary of MATH consists of circles which project to MATH. Consider an essential (possibly singular) loop MATH in MATH and let MATH denote a line in MATH above MATH. Let MATH denote a generator of the stabiliser of MATH. Then MATH lies in the stabiliser MATH of MATH and in the stabiliser MATH of MATH. Thus there is MATH in MATH such that MATH. As MATH is not equal to MATH, we know that MATH does not lie in MATH. Thus we obtain a MATH-injective (possibly singular) annulus MATH in MATH with both ends on MATH and one end at MATH, such that MATH is not properly homotopic into MATH. But any MATH-injective annulus in MATH is properly homotopic into MATH. It follows that some component MATH of the closure of MATH is an annulus and that MATH can be homotoped to cover MATH, keeping MATH in MATH during the homotopy. As MATH is a union of components of MATH and MATH, and as no component of MATH is an annulus, it follows that MATH is a component of MATH. We let MATH denote the component MATH of MATH. A particular consequence of the preceding argument is that MATH must be homotopic in MATH into a component of MATH. Hence every essential loop on MATH is homotopic into a component of MATH, so that MATH must be an annulus parallel to a component of MATH. As the boundary of MATH consists of circles which project to MATH, it follows that MATH must project to MATH. In this case, we homotop the two parallel components of MATH across MATH. Note that this reduces the number of components of MATH by MATH. Repeating this as needed yields the required homotopy of MATH to an embedding such that during the homotopy each component of MATH either stays fixed or is moved across an annulus component of MATH. After doing this for all components of MATH, if MATH fails to be a homeomorphism, we must have two components MATH and MATH of MATH whose images intersect or even coincide, or we must have components MATH of MATH and MATH of MATH whose images coincide. In the first case, let MATH denote a component of the intersection of MATH and MATH. (Note that MATH and MATH may both be closed, in which case so is MATH.) Then MATH is bounded by some components of MATH, so that MATH must be a union of certain components MATH of MATH and MATH. As in the previous paragraph, any essential loop MATH in MATH yields an annulus in MATH, joining MATH and MATH, which then implies that MATH is not empty and that MATH is homotopic into MATH. This implies that each MATH must be an annulus. As no component of MATH is an annulus, it follows that MATH must be an annulus component of MATH. Thus we can change MATH by flip homotopies on these annuli and obtain a homeomorphism from MATH to MATH. Finally NAME 's Homeomorphism Theorem shows that we can properly homotop MATH to a homeomorphism, completing the proof in this case. In the second case, we have components MATH of MATH and MATH of MATH whose images coincide. Hence MATH and MATH are homotopic in MATH. If MATH and MATH are closed, this implies that MATH is homeomorphic to MATH. Otherwise, there is a submanifold MATH of MATH homeomorphic to MATH, such that MATH corresponds to the union of MATH and MATH. If MATH denotes any component of MATH and MATH denotes the annulus in MATH joining MATH to a component of MATH, then MATH must be parallel into MATH. It follows that MATH is homeomorphic to MATH in this case also. In either case, it is easy to homotop MATH to a homeomorphism keeping MATH mapped into MATH during the homotopy. This completes the proof of REF , so long as the special REF never occurs. Now suppose that for some component MATH of MATH, we are in special REF . As no component of MATH is an annulus, we see that in this special case, MATH is a MATH-bundle, whose base surface is the twice punctured disc or the once punctured NAME band. As MATH is free of rank MATH, so is MATH, so that each of MATH and MATH is a handlebody of genus MATH. MATH is MATH where MATH denotes the twice punctured disc. Thus MATH consists of three annuli. As MATH induces a homeomorphism MATH, it follows that MATH also consists of three annuli. Choosing one boundary component from each of these three annuli in MATH gives us three disjoint simple closed curves in MATH. Each must be essential in MATH, and no two can be homotopic in MATH, for that would yield an annulus in MATH joining boundary circles of distinct components of MATH. As MATH is the maximum number of disjoint essential simple closed curves one can have on the closed orientable surface of genus MATH, it follows that MATH must consist of two pairs of pants, or equivalently, twice punctured discs. Now let MATH denote one of the two components of MATH. The given map MATH has a homotopy inverse MATH, and we consider the composite map MATH, where the last map is simply projection. Then MATH is proper and MATH-injective and so is properly homotopic to a covering map, as MATH is not an annulus. As MATH and MATH each have NAME number equal to MATH, it follows that MATH is homotopic to a homeomorphism rel MATH. Thus we can homotop MATH to a homeomorphism while keeping MATH mapped into MATH during the homotopy, as required. MATH is MATH where MATH denotes the once punctured NAME band. In this case, MATH is a connected double cover of MATH, and MATH consists of two annuli. As before, we let MATH denote a component of MATH. Recall that MATH-is not an annulus. Now let MATH denote a homotopy inverse to MATH, and consider the composite map MATH, where the last map is simply projection. Then MATH is proper and MATH-injective and so is properly homotopic to a covering map, as MATH is not an annulus. The degree of this covering must be even, as MATH is orientable and MATH is not. As MATH is an incompressible subsurface of MATH, we have MATH. As MATH, it follows that the covering has degree MATH. It follows that each of MATH and MATH is the orientable double cover of MATH. Hence we can homotop MATH to a homeomorphism while keeping MATH mapped into MATH during the homotopy, as required. |
math/0107232 | As in the proof of REF , we can homotop MATH to be proper while keeping MATH mapped into MATH during the homotopy. Now we apply REF . We conclude that MATH is properly homotopic to a map MATH which is a homeomorphism unless REF MATH is a MATH-bundle over a closed surface and MATH, or REF MATH (and hence also MATH) is a solid torus and MATH is a branched covering with branch set a circle, or REF MATH is a handlebody and MATH. If MATH is a homeomorphism, we can alter MATH and the homotopy so as to arrange that MATH is mapped into MATH during the homotopy, which completes the proof of the theorem in this case. REF is excluded in the hypotheses of our theorem, so it remains to eliminate the two REF . Suppose that we have REF , so that MATH is a MATH-bundle over a closed surface MATH and MATH lies in a component MATH of MATH. As MATH is a homotopy equivalence, the inclusion of MATH in MATH must also be a homotopy equivalence, so that MATH is homeomorphic to MATH. If MATH is the trivial MATH-bundle over MATH, then MATH must map each component of MATH to MATH with degree MATH. Otherwise, the bundle is non-trivial and MATH maps MATH to MATH with degree MATH. In either case, MATH contains embedded annuli which are not properly homotopic into MATH, so MATH must be non-empty. Let MATH be a component of MATH in MATH and MATH be a component of MATH which maps to MATH by a homeomorphism. Let MATH denote a component of MATH, and let MATH denote the component of MATH mapped to MATH. Whether MATH is a trivial or non-trivial MATH-bundle, there is an annulus MATH embedded in MATH with one end equal to MATH which cannot be properly homotoped into MATH. Let MATH denote the other end. Suppose that MATH meets a component MATH of MATH. As MATH maps MATH to MATH by a homeomorphism, MATH is disjoint from MATH. As MATH is homotopic to MATH, we see that MATH can be homotoped out of MATH, and hence that MATH can be homotoped out of MATH. By repeating this argument for any other components of MATH which meet MATH, we can arrange that MATH lies in MATH. But this immediately contradicts the condition that any embedded incompressible annulus in MATH is parallel into MATH or MATH. This contradiction shows that REF cannot occur. Finally suppose that we have REF , so that MATH, and hence MATH, is a handlebody. If MATH denotes a component of MATH, the argument in the proof of REF shows that either MATH can be modified rel MATH to be an embedding, or that the image of MATH contains an annulus MATH which is a union of components of MATH and MATH. If we regard MATH as an annulus in MATH, the assumptions of our theorem imply that MATH is parallel into MATH or MATH. The first case is only possible if MATH is an annulus, and the second is only possible if MATH is a component of MATH. In the second case, we can further homotop MATH to an embedding by moving two components of MATH across MATH. Thus either, we can homotop MATH to an embedding or MATH is an annulus and the boundary of MATH is a torus consisting of the image of MATH. As MATH is a handlebody, it must be a solid torus, contradicting our assumptions. Thus MATH can be homotoped to an embedding relative to its boundary, for each component MATH of MATH, and we complete the argument as in REF. |
nlin/0107026 | If MATH, then MATH. Hence, MATH. Next, if MATH and MATH then MATH by REF and MATH. Hence, MATH. |
nlin/0107026 | Define the sets MATH and MATH by REF . If MATH, then MATH for some MATH, and therefore, by REF , MATH for all MATH. Hence, the limit MATH exists and equals MATH. If MATH, then MATH for some MATH. By REF (and its proof) MATH . Hence, the limit MATH exists and equals MATH. In particular, MATH. Consequently we obtain REF . Now REF follows if we set MATH . |
nlin/0107026 | The condition MATH on MATH implies MATH. Suppose the contrary. Then we would find two fixed points MATH and MATH and a non fixed point MATH in between MATH and MATH. Let MATH. Since MATH is continuous, MATH. Hence the intermediate value theorem implies that there exist a point MATH in MATH such that MATH. Then we would have MATH, which contradicts MATH. |
nlin/0107026 | Let MATH be a connected component of the set MATH. Then, MATH by the assumption on MATH. Thus, if MATH then the sequence MATH is monotone nondecreasing and bounded and so it has a limit MATH, which is necessarily a fixed point of MATH. In particular, the subsequence MATH converges to a fixed point of MATH. Similarly MATH converges to a fixed point of MATH for any MATH is a connected components of MATH . |
nlin/0107026 | By definition, one can take a semi-open interval MATH which is either of the form MATH or MATH with MATH, MATH as MATH and MATH is one-to-one on MATH. Then, MATH where MATH. Similarly, for each MATH, MATH. Hence, we get MATH and on the other hand MATH . Consequently, MATH . The assertion for MATH follows by induction. |
nlin/0107026 | If MATH for some MATH, MATH. Suppose MATH for all MATH, then MATH . By REF and MATH, MATH . As a result, there exist MATH such that MATH for large MATH and we have MATH. This implies MATH. |
nlin/0107026 | In the proof of REF , a key inequality is MATH for MATH. Now MATH from assumption. So MATH by induction. The proof is similar. |
nlin/0107026 | Assume MATH and MATH. Then, MATH . Hence, MATH implies MATH and REF follows by induction on MATH. |
nlin/0107026 | If we show REF , then REF will be obvious from REF . Assume MATH, then MATH. By REF , MATH. Hence MATH. |
nlin/0107026 | By REF and the definition of MATH, MATH . Now, MATH and MATH. By REF , MATH. Thus MATH. Hence REF follows by induction. |
nlin/0107065 | In the case of MATH, MATH, it is clear that MATH for MATH. Iteration of the evolution equation of MATH gives rise to higher order equations of the form, MATH for MATH, and MATH being a polynomial in MATH. Combining these equations with REF, we obtain the bilinear equations, MATH which can be cast into a single equation, MATH . Next we use REF to obtain MATH for MATH, MATH being a polynomial in MATH. This yields MATH and we have MATH . Similar discussion with the differential equations for MATH gives the desirous result. |
nlin/0107065 | From REF, we obtain MATH for any non-negative integer MATH. Differentiating REF with respect to MATH and applying REF, we have MATH . There exist polynomials MATH and MATH such that MATH where the degree of MATH is at most MATH. In view of REF, we obtain MATH. REF implies MATH and that MATH satisfies REF. Differentiating REF with respect to MATH and applying REF, we have MATH . We can rewrite the second term in the left hand side as follows: MATH . Thus we obtain MATH . Since the expression in parentheses is an polynomial in MATH, we can apply exactly the same argument above to get the unique polynomial MATH and show that MATH satisfies REF. |
nlin/0107065 | Taking the complex conjugation of REF, we have MATH . Then we find that two monic polynomials MATH and MATH are characterized by the same data MATH. This proves the results. |
nlin/0107065 | The translational invariance MATH reads MATH . Since the right-hand-side is analytic on MATH, we have MATH where the contour is taken as the unit circle with the center at MATH. On the other hand, the function MATH satisfies the differential equations MATH for MATH, which entails MATH . From the analyticity requirements, it follows that the left-hand-side is a polynomial of degree at most MATH, which we denote MATH. It is straightforward to shows that MATH which are nothing but the evolution REF . With these equations, we can apply the same argument as the proof of REF . The resulting equation coincides with REF. |
nlin/0107065 | By the OPE REF and the property MATH, we obtain MATH . Comparing the last line to REF, we have the desirous result. |
nlin/0107065 | Since the representation of MATH under consideration is constructed from REF , it is enough to show MATH for MATH and MATH. From REF, we have MATH . These equations and the relation MATH give the commutativity above. |
nlin/0107065 | This is the direct consequence of REF . |
nlin/0107065 | Using MATH instead of REF, we can derive the desirous result. |
nlin/0107065 | This can be proved in the same fashion as REF ; Use MATH instead of REF. |
nlin/0107067 | Using the operator product expansion for MATH, MATH and the property MATH, it is straightforward to show that MATH above satisfies REF. The remaining relations can be checked by direct calculations. |
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