paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/9905130 | It is clear that MATH commutes with NAME derivatives MATH and contractions MATH for all MATH. To show that MATH is a chain map we show that MATH. By definition, the NAME differential splits into two pieces MATH, with MATH. Again we identify MATH by the symbol map. Since MATH and MATH, we have: MATH . Conjugating by MATH, and using MATH, MATH . In the NAME basis, MATH while MATH . Hence, MATH . Next, apply MATH to this expression. This kills the first term since MATH (by MATH-invariance of MATH). Since MATH, the second term is killed as well. Only the last term survives and using REF we find MATH . Together with REF this shows MATH which completes the proof. |
math/9905130 | We have to compute the composition MATH . Since MATH on basic elements, the operator MATH can be replaced with MATH. The latter commutes with MATH, and since MATH it follows that the above composition equals MATH as claimed. |
math/9905130 | We want to apply the operator MATH to MATH . The term MATH lives in MATH and hence commutes with both the operator MATH on MATH and with the operator MATH. Therefore, MATH . To compute MATH, choose MATH with MATH, and express MATH in the NAME basis MATH. Note that since MATH and MATH, the element MATH squares to MATH. Together with MATH, this shows that MATH . Applying the operator MATH, we obtain MATH . Finally, MATH and MATH combine by REF , and we find MATH . |
math/9905130 | The proof is similar to the proof in REF, that MATH induces a ring homomorphism in cohomology. Consider the following two homomorphisms of MATH-differential spaces, MATH, MATH . It suffices to show that MATH are MATH-chain homotopic, that is, there exists a MATH-equivariant, odd linear map MATH such that MATH commutes with NAME derivatives and contractions and such that MATH. To construct MATH we describe all maps in terms of their kernels. The kernels of the maps MATH are MATH . Since MATH are homomorphisms of MATH-differential spaces, they have the property MATH . The kernel MATH is invertible in the algebra MATH, because its form degree MATH part MATH is invertible at all points MATH. The element MATH then has properties MATH so that MATH is a closed element in MATH. Since MATH is equivariantly contractible, the pull-back under the inclusion MATH induces an isomorphism in cohomology. Since both MATH map the identity to the identity, MATH and therefore MATH. Hence, there exists MATH with MATH and letting MATH and using that MATH is closed we find MATH . Consequently, the linear map MATH with kernel MATH is MATH-equivariant and satisfies MATH. Since MATH the map MATH commutes with MATH-contractions. |
math/9905130 | We begin by showing that the pull-backs of MATH and MATH under the inclusion MATH coincide. We compute MATH using REF : MATH . By another application of REF we find MATH . Since MATH has non-vanishing form degree MATH part, there exists a MATH-invariant neighborhood MATH of MATH such that MATH has non-vanishing form degree MATH part, and is therefore invertible, on MATH. We then have, over MATH, MATH while MATH. Since MATH retracts MATH-equivariantly onto MATH, the NAME homotopy operator for the retraction MATH defines an element MATH such that MATH . Letting MATH this gives REF follows from the equation for MATH. |
math/9905130 | Our proof is a simple adaptation of the proof given in CITE, which in turn is based on the original proofs of NAME and NAME. We may assume that MATH is abelian. For each MATH, let MATH be a MATH-invariant tubular neighborhood of MATH. Let MATH be a function such that MATH vanishes in a neighborhood of the fixed point set and MATH in a neighborhood of MATH. Choose an invariant metric MATH on MATH and let MATH be a MATH-invariant REF-form, defined on the complement of the fixed point set, MATH. Then MATH is invertible. Write MATH . Then MATH. The form MATH is supported on MATH and agrees with MATH near MATH. This reduces the proof to the case where MATH is compactly supported in MATH for some MATH. Let MATH be the NAME form CITE representing the NAME class of MATH. It has compact support in fiber direction and its pull-back to MATH represents the equivariant NAME form, MATH. Let MATH. Then the pull-back of MATH to MATH is MATH-cohomologous to MATH. Consequently the quotient MATH is MATH-cohomologous to MATH. Using the main property of the NAME form that its fiber integral is MATH, and replacing MATH by the MATH-cohomologous form MATH we find: MATH . |
math/9905130 | We apply the localization formula to the class MATH. Note that the cocycles MATH and MATH satisfy the condition from REF , which therefore gives MATH . By the discussion preceding REF , MATH may be replaced with MATH . It remains to compute MATH. For this we we use the description of the NAME form in the NAME model. By REF we have MATH . Therefore, the image of MATH under the map MATH is given by MATH and we obtain MATH . |
math/9905136 | Observe that a MATH-Jacobi field which vanishes at some instant on MATH is automatically orthogonal to MATH, so that we really have MATH. If MATH is in the kernel of (the restriction of) MATH, it follows from REF and usual techniques of calculus of variations that MATH is parallel to MATH and that MATH satisfies REF . Since MATH is also orthogonal to MATH, it follows that MATH is a NAME field, except for the case where MATH is lightlike. In the latter case, we get MATH for some function MATH and therefore MATH is a NAME field, where MATH satisfies MATH and MATH. Observe that MATH because MATH. |
math/9905136 | It is a simple computation that uses the NAME equation, REF and the fact that, for MATH-Jacobi fields MATH and MATH, one has MATH. |
math/9905136 | Set MATH. For MATH, choose NAME fields MATH such that the vectors MATH are a basis of MATH and such that MATH. For MATH, choose NAME fields MATH such that MATH and the vectors MATH form a basis of MATH. If MATH is lightlike choose MATH. Then, the MATH's form a basis of the space of MATH-Jacobi fields orthogonal to MATH. Now, we can write MATH, for piecewise smooth functions MATH. For, define MATH for MATH and MATH, MATH, for MATH. The absence of MATH-focal points along MATH and the fact that, under the hypothesis that MATH, MATH, imply that the vectors MATH are a basis for MATH for MATH. Now, we have MATH, where MATH. The desired inequality follows directly from REF (equality implies that all MATH are constant, except for MATH, in the lightlike case). |
math/9905136 | For MATH, let MATH be the bilinear form REF for the restricted geodesic MATH (omitting the term involving MATH); if MATH, then we set MATH to be just the bilinear form REF for the restricted geodesic MATH. For MATH let's write MATH; observe that MATH. The function MATH is non decreasing (if MATH we can regard MATH as a restriction of MATH, by extending vector fields on MATH to MATH defining them to be zero on MATH). We show that MATH is piecewise constant and left-continuous on MATH, and that MATH for all MATH. Let MATH be fixed and choose a normal partition MATH of MATH such that MATH for some MATH (we allow MATH if MATH and we set MATH). Let's denote by MATH and MATH the spaces defined in REF , replacing the interval MATH by MATH (and using the normal partition MATH of MATH). We observe that the direct sum REF (for the interval MATH) is MATH-orthogonal, that is, MATH for all MATH and MATH, which follows directly from REF . Next, we claim that MATH. To check this, just observe that for MATH we have: MATH . The claim now follows from REF , by taking the NAME field MATH. It follows that MATH; Observe that as in REF the space MATH is isomorphic to the space MATH defined by: MATH and we'll call this isomorphism MATH. If MATH is sufficiently close to MATH or, more precisely, if MATH, the arguments above can be repeated by replacing MATH with MATH (observe, in particular, that the space MATH obtained will be the same). We can use the isomorphism MATH between MATH and MATH to define a symmetric bilinear form MATH on MATH corresponding to MATH. Clearly MATH. We have now a one parameter family MATH of symmetric bilinear forms on the (fixed) finite dimensional space MATH and it's not difficult to see that MATH depends continuously (actually, smoothly) on MATH. Let's consider the decomposition MATH, where MATH is positive (respectively, negative) definite on MATH (respectively, MATH) and MATH is the kernel of MATH. We can also assume that this decomposition is MATH-orthogonal (this is just the NAME inertia Theorem). The dimension of MATH is MATH. Since the decomposition MATH is orthogonal with respect to MATH, we know that the kernel of the restriction of MATH to MATH (which corresponds to MATH by the isomorphism MATH) is just the intersection of MATH and the kernel of MATH, the last one being given by REF . Observe that MATH and denote by MATH the subspace of MATH which corresponds to MATH, that is, MATH. In the lightlike Lorentzian case, write also MATH. Observe that MATH is just the set of MATH-tuples of vectors which are parallel to MATH, so that MATH doesn't change if we replace MATH by MATH in its definition, and therefore MATH is also contained in the kernel of MATH. We see now that MATH, except for the lightlike Lorentzian case where MATH. The dimension of MATH is just the multiplicity MATH of MATH as a MATH-focal point. By the continuous dependence of MATH on MATH we see that for MATH sufficiently small and MATH, MATH is negative definite on MATH so that MATH. For MATH we have also MATH so that MATH, that is, MATH is constant on MATH. This finishes the proof that MATH is left continuous. From now on we suppose MATH. The same continuity argument show that for some MATH, we have that MATH is positive definite on MATH for MATH (and positive semi-definite on MATH for MATH lightlike), so that MATH is bounded above by the codimension of MATH (or MATH, respectively). If MATH is not a MATH-focal point this codimension equals MATH so that MATH for MATH. Finally, if MATH is a MATH-focal point, by the above argument we only obtain the inequality MATH. NAME show below that for MATH and for MATH we have MATH, the inequality being strict if MATH (or if MATH is not parallel to MATH, in case MATH is lightlike). But this hypothesis on MATH holds if MATH and MATH, observing that the corresponding vector field MATH on MATH is an unbroken NAME field. We conclude then that MATH for nonzero MATH and hence for all nonzero MATH, which implies that MATH is negative definite on this space and MATH. We are now left with the proof of the inequality MATH. Towards this goal, let MATH and MATH be the vector fields corresponding to MATH, that is, MATH and MATH. Extend MATH to zero on MATH. Then, MATH and MATH. The vector fields MATH and MATH differ at the most in the interval MATH. The restriction of MATH to MATH is the only NAME field such that MATH and MATH, while the restriction of MATH to MATH is the only NAME field such that MATH and MATH. We have: MATH . We now apply REF to the geodesic MATH (with starting and ending points interchanged), for the NAME field MATH, vector field MATH and submanifold equal to the point MATH. For the strict inequality we need the hypothesis that MATH (respectively, MATH not parallel to MATH, in the lightlike Lorentzian case), since this implies that MATH is not NAME in MATH (respectively, does not differ from a NAME field by a multiple of MATH, in the lightlike Lorentzian case). This concludes the proof. |
math/9905136 | The space MATH is given by the subspace of MATH consisting of those vector fields MATH such that MATH; moreover, the restriction of MATH to MATH is precisely MATH. Defining MATH as in REF , let MATH be any subspace of MATH such that MATH. Clearly, MATH, because MATH; moreover, from REF it follows immediately that this decomposition is MATH-orthogonal, that is, MATH for all MATH and all MATH. Since MATH contains MATH, then MATH. Hence, MATH. To conclude the proof, we simply observe that MATH, because MATH. |
math/9905143 | The existence of the asymptotic expansion REF is clear from REF. The actual expansion coefficients then can be determined from REF. |
math/9905143 | By REF one infers MATH . By REF, and the uniformity of the asymptotic expansion REF with respect to MATH as MATH in MATH, which permits its differentiation in MATH, one derives MATH . Thus, MATH proves REF. |
math/9905143 | By hypothesis, MATH for a.e. MATH and all MATH. Thus the trace REF yields REF. |
math/9905143 | By general NAME theory, MATH consists of a countable union of closed intervals on MATH, possibly separated by gaps in between. Moreover, MATH is bounded from below and it has no eigenvalues, MATH . In fact, one can show that MATH is purely absolutely continuous, MATH as is also clear from the analytic continuation of MATH through MATH implied by REF, but we omit the details. The assumptions of uniform (maximal) spectral multiplicity MATH of MATH guarantees the existence of MATH eigenvalues MATH of the monodromy matrix MATH with MATH for MATH for MATH, in particular, MATH is unitary and hence diagonalizable for MATH. Next, suppose that MATH for MATH, with MATH for MATH are MATH linearly independent normalized solutions of REF. We claim that MATH since eigenfunctions of MATH for MATH would necessarily be linear combinations of MATH. However, none of them can lie in MATH since the fundamental matrix MATH, MATH satisfies MATH compare REF with MATH unitary. Since MATH, REF implies the existence of MATH for MATH. In the following we denote by MATH the discrete set of points (that is, countable without finite limit points) where MATH is not diagonalizable. By our hypothesis of uniform spectral multiplicity MATH of MATH, MATH is diagonalizable for all MATH (MATH denoting the boundary of a subset MATH) and we conclude that MATH. By a well-known argument, see, for instance CITE, one infers that all points in MATH are branch points for the eigenvalues MATH of MATH and hence MATH. Thus we obtain upon diagonalizing MATH in REF that MATH where MATH denotes the diagonalization of MATH, and MATH, MATH are analytic in MATH with a continuous extension to MATH. Thus each MATH satisfies a quadratic equation and one introduces a canonical set of cuts MATH along MATH (MATH a finite or countably infinite index set), joining the branch points, that is, all points in MATH (as well as MATH and possibly MATH in case MATH is finite) MATH . In this manner, each MATH becomes an analytic function on a (fixed) two-sheeted NAME surface (glued together crosswise along these cuts in a standard manner). In particular, MATH can now be split into pairs MATH such that MATH represents the analytic continuation of MATH (and vice versa), whenever MATH crosses tranversally through one of the cuts. Next, pick a MATH, that is, MATH for some MATH and pick a MATH for MATH in a sufficiently small neighborhood MATH or MATH of MATH. Suppose MATH for MATH along a path in MATH transversally approaching MATH and intersecting MATH at MATH. By analyticity of MATH, the analytic continuation MATH of MATH will satisfy MATH in an appropriate neighborhood MATH (or MATH) of MATH. Hence we may identify MATH with MATH in REF - REF. Thus upon possibly reordering the eigenvalues along the diagonal in REF we may write MATH where MATH such that MATH is the analytic continuation of MATH through the interior of the cuts MATH and hence through MATH. Since the similarity transformations connecting MATH and MATH can be chosen as matrices whose column vectors are the eigenvectors MATH of MATH, and MATH have the same branching behavior as the associated eigenvalues MATH (see, for example, CITE), one infers that MATH is the analytic continuation of MATH through MATH. By REF, MATH is the analytic continuations of MATH through MATH. Consequently, MATH . Since MATH was arbitrary we obtain REF. |
math/9905143 | In the following, let MATH and MATH . By REF , the normal limits MATH and MATH exist for all MATH and REF holds. By REF is equivalent to MATH and hence taking MATH implies REF. By REF, and REF, one infers MATH and MATH . Hence, MATH and MATH exist for all MATH and we may differentiate REF with respect to MATH to obtain MATH . Taking MATH in REF then implies REF and hence we have shown that REF implies REF . Next we prove that REF implies REF . We could immediately invoke REF, but prefer to show a simple argument that permits extensions to non-periodic cases to be discussed elsewhere. By REF is equivalent to MATH and hence to MATH . Thus, REF is equivalent to MATH . In order to exploit REF one computes from REF, MATH . Consequently, REF is equivalent to MATH . A simple manipulation in REF (adding and subtracting MATH) then yields MATH and thus MATH . Taking into account REF results in MATH and hence in MATH since MATH are NAME matrices, implying MATH . Combining REF yields REF and hence REF . Given REF and MATH one computes from REF MATH and hence MATH since MATH and MATH are self-adjoint for all MATH . Thus REF implies REF . |
math/9905144 | Construct inductively NAME sets MATH such that for each MATH: MATH, MATH and MATH. Put MATH if MATH and MATH if MATH. This MATH works. Let MATH enumerate the (clopen) basis of MATH. If MATH is a name for an open set then we have a name MATH for a subset of MATH such that MATH. Let MATH be a NAME set such that MATH and put MATH. Since MATH we get that MATH. Thus we are done for open sets. Next apply easy induction (note that MATH and MATH). |
math/9905144 | Let MATH be a sequence of MATH-REF-generators. Construct inductively infinite maximal antichains MATH such that CASE: for each MATH the set MATH is an infinite maximal antichain above MATH, and CASE: MATH is a maximal antichain above MATH. Use these antichains to define MATH in such a way that MATH. |
math/9905144 | It should be clear. Let MATH and let MATH REF be maximal antichains. We want to find MATH such that MATH and MATH. Take MATH such that MATH and find MATH so that MATH (that is, such that MATH). Choose MATH such that MATH and let MATH be such that MATH. Continuing in this fashion we will define MATH which will work. Now suppose that MATH. To show that the singleton MATH is MATH-small it is enough to prove that the set MATH is dense in MATH. Given MATH. Take MATH such that MATH. There is MATH with MATH and MATH. If MATH then MATH and we are done. So suppose that MATH. Take MATH stronger than both MATH and MATH. Then MATH and MATH are incompatible and consequently MATH. To prove the additivity of MATH-small sets note that if maximal antichains MATH REF witness that sets MATH are MATH-small then any maximal antichain MATH refining both MATH and MATH witnesses that MATH is MATH-small. |
math/9905144 | First note that, in MATH, b encodes a ccc NAME forcing notion and MATH is a basis for it (MATH formulas are downward absolute for all models of MATH). Moreover if MATH``MATH is a maximal antichain in MATH" then MATH is really a maximal antichain of MATH. Notice that MATH and the same concerns MATH, MATH. As MATH is a filter in MATH we have MATH. If MATH, MATH then there is MATH such that MATH and MATH (MATH is a basis for MATH). Consequently MATH and MATH so MATH. As MATH it is enough to show that MATH is a filter. For this it suffices to prove that MATH contains no pair of incompatible elements. Thus suppose that MATH are incompatible. Let MATH be a maximal antichain in MATH such that (in MATH) for each MATH either there is MATH such that MATH and MATH or there is MATH such that MATH and MATH. By the choice of MATH we have that MATH for some MATH. Let MATH be such that MATH and MATH (or MATH). Then MATH and MATH (or MATH). Consequently either MATH or MATH. |
math/9905144 | Since MATH for any MATH (by REF) we easily get that MATH implies MATH. Suppose now that MATH. Let MATH be a real encoding the NAME set MATH. Let MATH be a countable transitive model of MATH such that MATH. Since MATH we find a real MATH such that MATH . By REF we get that MATH is a MATH-generic filter over MATH and MATH. As MATH we find MATH such that MATH (where MATH stands for the NAME set coded by MATH). We claim that MATH . Suppose not and let MATH be a real from the intersection. As earlier we have that MATH is a MATH-generic filter over MATH, MATH. Since MATH we get a contradiction to MATH. For a real MATH let MATH be the sequence of reals coded by MATH. Let MATH. Clearly MATH is the intersection of a MATH-set and a MATH-set. Now MATH is a NAME code for a set from MATH . The first part of the conjunction is MATH, the second part is MATH. Hence the formula is MATH. On the other hand, by REF., MATH is a NAME code for a set not belonging to MATH. Easily the last formula is MATH too. Consequently both formulas are MATH and this fact is provable in ZFC. As MATH formulas are upward absolute (for models of MATH) and MATH formulas are downward absolute (for models of MATH) we are done. |
math/9905144 | Let MATH and MATH. Suppose that MATH is MATH-small. Let MATH be a maximal antichain in MATH such that MATH. Look at MATH (where MATH is the embedding given by productivity of MATH). It is a maximal antichain in MATH. Suppose now that MATH, MATH and MATH. Then MATH and hence MATH. As MATH we conclude MATH. Consequently MATH witnesses that MATH is MATH-small. Now we can easily conclude that MATH implies MATH. Assume now that MATH. Let MATH such that MATH and MATH. Let MATH. Clearly MATH is a MATH subset of MATH and MATH. Thus if we prove that MATH then we will have MATH and the proposition will be proved. Suppose MATH. Since it is a MATH-subset of MATH we find MATH such that MATH. By the first part we get MATH . Let MATH be such that MATH. As MATH we have MATH and hence MATH for some MATH. This means MATH and by the definition of MATH this implies MATH. Thus we have proved that MATH what implies MATH - a contradiction. |
math/9905144 | We want to show the invariance of the basis MATH. For permutations MATH we define MATH (where MATH is defined on MATH, MATH is defined on MATH, MATH and MATH). We claim that MATH, MATH have the properties required in REF . Clearly both are automorphisms of MATH. Moreover, MATH. Hence MATH. Similarly MATH . Thus MATH. The basis MATH is productive. If for MATH and MATH we define MATH (where MATH-c means that we extend the function putting MATH whenever it is not defined) then MATH is a complete embedding of MATH into MATH with the required property. |
math/9905149 | The first statement follows from REF and NAME 's theorem that MATH has MATH unipotent elements. The second statement follows from the first and from elementary manipulations applied to NAME 's principal specialization formula (page REF). Full details appear in CITE. |
math/9905149 | Immediate from REF . |
math/9905149 | NAME (page REF), using NAME symmetric functions, establishes for any partitions MATH, the equation: MATH . Fixing MATH, summing the left hand side over all MATH of size MATH, and applying REF yields MATH . Fixing MATH, summing the right hand side over all MATH of size MATH, and applying REF gives that MATH proving the lemma. |
math/9905149 | For the first assertion, observe that complete flags correspond to cosets MATH where MATH is the subgroup of all invertible upper triangular matrices. Note that MATH fixes the flag MATH exactly when MATH. The unipotent elements of MATH are precisely MATH. Thus the number of complete flags fixed by MATH is MATH. It follows that the sought probability is equal to MATH multiplied by the probability that an element of MATH is unipotent of type MATH. The first assertion then follows from REF . The second part follows from the first part since by page REF, MATH is the number of complete flags of a MATH-dimensional vector space over a field of size MATH which are fixed by a unipotent element of type MATH. The third part follows from the second part and a formula for MATH on page REF. The fourth part follows from the third part and a formula for MATH in CITE. For the fifth assertion, a result on page REF gives that the number of maximal length chains of subgroups in an abelian MATH-group of type MATH is equal to MATH. Observing that for a partition MATH of MATH, MATH, the result follows. |
math/9905149 | For a standard NAME tableau MATH, let MATH be the partition formed by the entries MATH of MATH. It suffices to prove that at step MATH the division algorithm goes from MATH to MATH with probability MATH, because then the probability the NAME 's algorithm gives MATH at step MATH is MATH as desired from REF . The fact that the division algorithm goes from MATH to MATH with probability MATH follows, after algebraic manipulations, from NAME 's principle specialization formula (page REF) MATH . |
math/9905149 | As discussed at the beginning of REF, the matrix MATH may be assumed to be MATH where MATH is the MATH matrix of the form MATH . Let MATH, where MATH is the identity matrix. From this explicit form all eigenvalues of MATH are MATH. Thus if MATH fixes a line, it fixes it pointwise. Hence the number of lines fixed by MATH is one less than the number of points it fixes, all divided by MATH, and we are reduced to studying the action of MATH of non-zero vectors. It is easily proved that MATH has order MATH, where MATH. Hence if MATH with some MATH non-zero, and MATH is the smallest non-negative integer with this property, then MATH is a power of MATH. Thus all orbits of MATH on the lines of MATH have size MATH for MATH. We next claim that MATH fixes a vector MATH if and only if MATH . It suffices to prove this claim when MATH has one part MATH of size MATH. Observe that the MATH-th coordinate of MATH is MATH. If MATH, then MATH fixes all MATH because MATH for MATH. If MATH, then MATH fixes all MATH such that MATH, for the same reason. Finally, if MATH and MATH for some MATH, let MATH be the largest such subscript. Then the MATH-th coordinate of MATH is equal to MATH mod MATH, showing that MATH does not fix such MATH. This explicit description of fixed vectors (hence of fixed lines) of MATH yields the formula of the lemma for MATH, because the number of lines in an orbit of size MATH is the difference between the number of lines fixed by MATH and the number of lines fixed by MATH. The formula for the number of lines in an orbit of size REF follows because there are a total of MATH lines. |
math/9905149 | By REF , MATH . Observe that MATH is the number of subgroups of MATH of type MATH. This is because the total number of elements of order MATH in MATH is MATH, and every subgroup of type MATH has MATH generators. Therefore, using REF , MATH . |
math/9905149 | Let MATH denote the fractional part of a positive number MATH. REF and the writing of MATH in terms of the MATH's imply that MATH . We suppose for simplicity that MATH for some MATH (the case MATH is similar). Continuing, MATH . |
math/9905149 | We proceed by joint induction on MATH and MATH, the base case MATH being clear. Let MATH denote the probability that NAME 's growth algorithm yields the standard NAME tableau MATH after MATH steps. Let MATH be the column number of MATH in MATH. With all sums being over standard NAME tableaux, observe that MATH . |
math/9905152 | Every oriented MATH-dimensional manifold MATH without boundary carries a uniquely defined fundamental class MATH. If MATH is a manifold with boundary MATH then MATH is well-defined in MATH. Every open subset MATH inherits an orientation from MATH so that the natural restriction map MATH gives MATH. Without loss of generality we may assume that MATH . Otherwise, we replace MATH by the open, nonempty subset MATH. Since NAME homology with arbitrary supports is functorial with respect to proper maps of locally compact NAME spaces we redefine the map MATH . By definition of MATH, MATH is proper. The integral class MATH is well-defined. From the exact homology sequence for MATH and the pair MATH, MATH we obtain by assumption the isomorphism MATH. Hence, MATH is well-defined. We consider now an open subset MATH such that MATH and MATH. We carry out the same procedure as before for the proper map MATH and relate it to MATH by the following commutative diagram with respect to the natural restriction homomorphism MATH, MATH . Since MATH it follows that MATH . Let us consider now the bordism MATH. Without loss of generality we can assume that MATH so that we have the proper map MATH . In NAME homology theory, we know that the fundamental class MATH is the image of the fundamental class of the manifold with boundary MATH under the boundary homomorphism MATH in the exact homology sequence of the pair MATH. We obtain the commutative diagram MATH therefore by REF MATH because MATH. Since MATH is injective due to MATH it follows that MATH vanishes. |
math/9905152 | The well-defined homology class MATH follows from combining REF with REF . Suppose now that MATH is a MATH-cycle given by a smooth singular chain. By pairwise identifying and sufficiently smoothing the MATH-dimensional faces of the MATH-simplexes involved in MATH the cycle-property of MATH implies that we obtain a MATH-dimensional manifold MATH, not necessarily compact, with a smooth structure, such that the singular chain gives a map MATH meeting the pseudo-cycle condition, since MATH is covered by the images of the faces of codimension MATH and higher. Two cohomologous singular chains lead to cobordant pseudo-cycles in the sense of REF . |
math/9905152 | Let us consider MATH and MATH with MATH and MATH such that MATH, in particular, MATH. Let MATH and MATH be two relatively compact neighborhoods of MATH. In view of REF we consider the following local coordinates at MATH, respectively for REF MATH, MATH where MATH is the gluing map from REF and MATH is small enough depending on the compact set MATH. Thus, we have to show that CASE: MATH is smooth for MATH and MATH sufficiently small, and that CASE: MATH is smooth at MATH. Let us recall the definition of MATH from CITE. Let MATH be a cut-off function with MATH and MATH. We write MATH . For every MATH and MATH large enough we define MATH by MATH . In particular, MATH. One can find a MATH and a bundle MATH with MATH such that there exists a unique section MATH providing MATH . The bundle MATH can be completed fiberwise in terms of a NAME space yielding a smooth bundle such that MATH is a smooth section. (Details can be found in CITE.) Moreover, there is an exponential estimate for the correction term MATH between MATH and MATH. Namely, there exists a MATH such that MATH for some MATH uniformly for MATH. Moreover, also the covariant derivatives of MATH with respect to MATH and MATH satisfy such an exponential estimate as MATH. This is due to the fact that MATH as MATH and that the gradient flow trajectories MATH and MATH converge exponentially fast towards MATH, MATH . The construction of MATH and MATH in CITE is such that MATH and MATH coincide over MATH. One obtains a unique smooth gluing map MATH extending MATH and MATH and REF follows. Let us consider now the coordinate chart MATH with the exponential estimate for the correction term MATH, MATH where MATH are the covariant derivatives of the section MATH with respect to the variables MATH and MATH. We obtain for the evaluation map MATH, MATH . Thus, the smoothness of MATH follows from REF and the standard identities for the covariant derivatives of MATH at MATH. |
math/9905152 | Consider a point MATH such that MATH . That is, there exists a sequence MATH such that MATH in MATH but MATH contains no convergent subsequence in MATH. We can assume that every MATH corresponds to an element in MATH for some MATH such that MATH for MATH. The compactness result for the space of negative gradient flow trajectories provides a convergent subsequence MATH with MATH. Since MATH does not converge in MATH we obtain MATH. This shows that MATH is covered by the images of the evaluation maps MATH . Thus, MATH is a pseudo-cycle. |
math/9905152 | Consider MATH and MATH such that MATH. Similar to the construction of MATH we now set MATH as in REF . We can obtain the boundary of the manifold MATH as MATH such that by setting MATH we obtain the smooth MATH-dimensional manifold MATH with boundary MATH. Note that MATH is a MATH-dimensional open submanifold with MATH and, analogously to REF , MATH and MATH are covered by the at most MATH-dimensional submanifolds MATH . Altogether, we have MATH and MATH so that REF is applicable. |
math/9905152 | We have to show that the pseudo-cycles MATH and MATH can be related by a suitable pseudo-cycle cobordism such that REF applies. Since asymptotically constant families MATH as above form a convex set we can consider the continuous path MATH . Given MATH, the space MATH is a smooth manifold of dimension MATH, where MATH is the unstable manifold of MATH associated to the pair MATH. Its boundary is the disjoint union of MATH and MATH, that is, the latter with reversed orientation. Analyzing the non-compactness of MATH, we consider a sequence MATH which contains no convergent subsequence. There exists a subsequence MATH such that MATH and either MATH for MATH, or MATH converges in MATH towards a MATH with MATH. Moreover, we can prove a MATH-parameterized version of the gluing result in REF yielding a gluing map MATH for every relatively compact, open subset MATH and MATH. This allows us to build MATH as in REF and to construct a smooth manifold MATH together with a smooth map MATH which extends the given maps MATH and MATH on the boundary. Since MATH for MATH with MATH, we meet the conditions of REF . |
math/9905152 | We have to compare the pseudo-cycles MATH for any NAME cycle MATH. That is, we have to show that the MATH can be extended to a suitable cobordism MATH such that, again, REF applies. Let us consider the space similar to MATH in the proof of REF , MATH for MATH. (If MATH is replaced by a compact interval we are in the situation of REF .) Now we have to deal with additional non-compactness for MATH. Let MATH be such that MATH. Then, there exists a subsequence MATH such that MATH for some MATH. Necessarily, MATH. If both critical points MATH and MATH have equal NAME index then, up to choosing a subsequence, MATH . In that case we denote this weak convergence again by MATH . For the converse, we have a gluing theorem analogous to REF : Let MATH. Given MATH, an open and relatively compact subset, there exists a MATH and a smooth map MATH such that the corresponding REF - REF as in REF hold true. Extending the construction from the proof of REF based on the MATH-parametrized gluing, we now glue in boundary manifolds to MATH such that MATH . Note that it is not necessary to glue in the codimension-MATH manifolds MATH for MATH. Building the quotient manifold MATH analogously as above, we obtain a smooth MATH-dimensional manifold with boundary MATH where MATH is the open subset MATH with complementary strata of codimension at least MATH. The evaluation maps MATH extend from the boundary manifolds to MATH and it is straightforward to verify that the conditions for REF are satisfied. |
math/9905152 | The main ingredient of this transversality theorem is the following The universal stable manifold MATH for MATH, MATH a NAME function, admits a submersion MATH away from the critical point MATH. It is also a submersion everywhere if MATH. Let us recall some analytic constructions from CITE. The space MATH is the MATH-Sobolev completion of the space of smooth curves MATH with sufficiently fast convergence toward MATH as MATH. It is in fact a NAME manifold. The tangent space to the NAME manifold of smooth Riemannian metrics on MATH is MATH for some fixed Riemannian metric MATH. The function space MATH is a MATH-dense subspace of MATH with a NAME space norm. Let us now consider the smooth map MATH where the right hand side space is a NAME space bundle over the manifold MATH with fiber MATH of MATH-vector fields along the curve MATH. Choosing a Riemannian connection MATH on MATH we obtain the linearization of MATH as MATH . Observe that MATH is an endomorphism of the pull-back bundle MATH. Hence, any variation of MATH as a function of time MATH can be achieved through a variation of MATH over MATH if MATH is injective. Altogether we obtain the tangent space of the universal stable manifold as MATH because MATH is a regular value for MATH, as it will be clear below. Given MATH and MATH we have to show that for each MATH there exist MATH such that MATH, if either CASE: MATH, that is, MATH, or CASE: MATH and MATH. In the latter REF we have MATH with MATH where the Hessian at MATH is positive definite. This implies MATH and hence the submersion property of MATH. In REF let us simplify the operator MATH by using coordinates with respect to an orthonormal parallel frame of MATH. We obtain the operator, MATH where MATH is a smooth path in the space of symmetric MATH-matrices with MATH and MATH with MATH for all MATH. We shall now prove that for all MATH and MATH there exist MATH and MATH such that MATH and MATH. This concludes the proof of REF in view of the fact that each such MATH arises from a MATH since MATH is injective if MATH. Suppose that there exist MATH and MATH such that MATH . This implies that MATH and MATH and therefore MATH if MATH. Moreover, REF implies that MATH for all MATH and we have MATH. If MATH we can find MATH with support in MATH such that MATH contradicting REF . Hence we obtain MATH and by REF MATH for all MATH which implies MATH. Since the cokernel of MATH in MATH is finite-dimensional it follows that MATH is surjective. The proof of REF now follows from the parameter version of the NAME theorem. There exists a residual subset MATH such that for each MATH the map MATH intersects the diagonal MATH transversely. For such a generic MATH, MATH is a smooth manifold of dimension MATH. Note that for MATH intersections MATH for a regular MATH can only occur if MATH. Therefore, it is obvious how the solution space MATH inherits its orientation from an orientation of MATH, MATH and MATH. |
math/9905152 | Consider a sequence MATH. After choosing a suitable subsequence we have MATH . In the case that MATH, we use that MATH can be covered by a map MATH, so that MATH, MATH. Since the intersection of residual sets is residual it follows from REF that for a generic MATH has to be empty by dimensional reasons. Thus MATH can be excluded. Sharpening the convergence result REF we can deduce weak convergence towards a broken trajectory MATH . For such multiply broken trajectories we must have MATH. Hence, if MATH we cannot have MATH and MATH must be compact and hence finite. |
math/9905152 | Computing MATH we have to show that MATH . This follows readily from the MATH-dimensional compactness result for MATH analogous to REF and the corresponding gluing operation. Namely, since MATH, non-compactness of MATH for generic MATH can only occur in terms of simply broken trajectories in the limit. But exactly as for the proof of the fundamental fact MATH, the corresponding gluing result completely analogous to REF shows that the oriented number of boundary components of MATH equals MATH and has to vanish since each component of MATH is diffeomorphic to an interval. This proves REF . Given a pseudo-cycle cobordism MATH in the sense of REF , that is, MATH, we can define the MATH-dimensional manifold MATH for MATH, and generic MATH. The same compactness-gluing argument as before shows MATH . |
math/9905152 | The proof of the compatibility MATH can be carried out exactly analogous to that for REF using the argument from the proofs of REF . Consider now the MATH-dimensional pseudo-cycle MATH associated to a NAME MATH. Let MATH. Then in view of REF for a NAME pair MATH with generic MATH we have intersections MATH only for MATH on the MATH-dimensional strata which are exactly the unstable manifolds MATH in MATH. We therefore have MATH . Using a homotopy operator as before we can show easily that this is homologically equivalent to the definition of the operator MATH, that is, MATH . This proves MATH. |
math/9905171 | Suppose, by contradiction, that there is a sequence MATH, MATH, MATH with the property that for a chosen MATH there is a sequence MATH satisfying MATH. By REF , there are MATH and MATH such that the maps MATH have nonlinearity bounded by MATH for all MATH. By NAME 's Theorem, there is a subsequence MATH which converges in the MATH topology to a map MATH. Hence, the map MATH is contained in the boundary of MATH and is infinitely renormalizable. However, a map contained in the boundary of MATH is not renormalizable, and so we get a contradiction. |
math/9905171 | In the proof of this lemma we will use REF below. Let MATH be maps with MATH norm bounded by some constant MATH, and let MATH be MATH maps. By induction on MATH, and by the Mean Value Theorem, there is MATH such that MATH . Set MATH, where MATH is defined as in REF , and MATH is defined as in REF . Set MATH with MATH. We start by considering the simple REF , where MATH and MATH do not have the same renormalization type, and conclude with the complementary REF . In REF , by REF , there is MATH with the property that MATH . In REF , there is MATH such that MATH, and MATH, where MATH and MATH. By REF , there is a positive constant MATH bounding the nonlinearity of MATH. Since the set of all infinitely renormalizable unimodal maps MATH with nonlinearity bounded by MATH is a compact set with respect to the MATH topology, and since MATH varies continuously with MATH, there is MATH with the property that MATH. Again, by REF , and by REF , there is MATH such that MATH . Thus, MATH . Now, let us consider the cases where REF MATH and REF MATH. In REF , we get MATH . In REF , using that MATH and MATH, we get MATH, and thus, by MATH, we obtain MATH . Hence, again by MATH and MATH, there is MATH with the property that MATH . Therefore, this lemma is satisfied with MATH. |
math/9905171 | Let MATH and MATH be as defined in REF , and let MATH and MATH be as defined in REF . Take MATH such that MATH. Set MATH such that MATH and MATH. Then, for every MATH, the values MATH are less than MATH. By REF , if MATH with MATH, then the map MATH is contained in MATH. Thus, MATH is once renormalizable, and MATH. By induction on MATH, let us suppose that MATH is MATH times renormalizable, and MATH for every MATH. By REF , we get that MATH. Hence, again by REF , the map MATH is once renormalizable, and MATH. |
math/9905171 | The proof follows from REF . |
math/9905171 | Let us define MATH, and, by induction on MATH, we define MATH, where MATH is the NAME transform of MATH given by the NAME Principal Value of MATH . By REF in page REF, we get MATH, where MATH is given by MATH . By the NAME inequality (see page REF), for every MATH, the NAME operator MATH is bounded, and its norm MATH varies continuously with MATH. An elementary integration also shows that MATH (see page REF). Therefore, given that MATH, there is MATH with the property that MATH . Since MATH, it follows from NAME 's inequality (see page REF) that there is a positive constant MATH such that MATH . By a simple computation, we get MATH . Thus, by REF , and REF , there is a positive constant MATH with the property that MATH . |
math/9905171 | The main point in this proof is to combine the hybrid conjugacy between MATH and MATH given by NAME, with NAME 's pull-back argument, and with NAME 's rigidity theorem for real quadratic maps. Using NAME 's pull-back argument and the hybrid conjugacy between MATH and MATH, we construct a MATH quasiconformal homeomorphism MATH which restricts to a conjugacy between MATH and MATH. Moreover, MATH satisfies REF of this lemma, and the restriction of MATH to the filled in NAME set of MATH extends to a quasi conformal map that is a hybrid conjugacy between MATH and MATH. By NAME 's glueing lemma (see REF) it follows that MATH also satisfies REF of this lemma. Now, we give the details of the proof: let us consider the set of all quadratic-like maps MATH contained in MATH. Using the NAME Distortion Lemma (see page REF), we can slightly shrink MATH for some MATH to obtain an open set MATH with the following properties: CASE: MATH is symmetric with respect to the real axis; CASE: the restriction of MATH to MATH is a quadratic-like map; CASE: the annulus MATH has conformal modulus between MATH and MATH; CASE: the boundaries of MATH are analytic MATH quasi-circles for some MATH, i. e., they are images of an Euclidean circle by MATH quasiconformal maps defined on MATH. Let MATH be the set of all quadratic-like maps MATH contained in MATH for which MATH satisfies REF , MATH, REF of last paragraph. Since for every MATH the boundaries of MATH are analytic MATH quasi-circles, any convergent sequence MATH, with limit MATH, in the NAME topology has the property that the sets MATH converge to MATH in the NAME topology (see REF in pages REF). Therefore, the set MATH is closed with respect to the NAME topology, and hence is compact. Furthermore, by compactness of MATH, and using the NAME Distortion Lemma, there is an Euclidean disk MATH which contains MATH for every MATH. Now, let us construct MATH such that REF , MATH, REF of this lemma are satisfied. Since MATH is symmetric with respect to the real axis, there is a unique NAME Mapping MATH satisfying MATH, and such that MATH. Since the boundaries of MATH are analytic MATH quasi-circles, using the NAME Theorem (see REF in page REF) the map MATH has a MATH quasiconformal homeomorphic extension MATH which also is symmetric MATH. Let MATH be the unique continuous lift of MATH satisfying MATH, and such that MATH. Since MATH is a MATH quasiconformal homeomorphism, so is MATH. Using the NAME Theorem, we construct a MATH quasi-conformal homeomorphism MATH interpolating MATH and MATH with the following properties: CASE: MATH for every MATH; CASE: MATH for every MATH; CASE: MATH. Then the map MATH conjugates MATH on MATH with MATH on MATH, and is holomorphic over MATH. By REF, there is a MATH quasiconformal hybrid conjugacy MATH between MATH and MATH, where MATH is a neigbourhood of MATH. Using the NAME Theorem, we construct a MATH quasiconformal homeomorphism MATH interpolating MATH and MATH such that CASE: MATH for every MATH; CASE: MATH for every MATH; CASE: MATH. Then the map MATH conjugates MATH on MATH with MATH on MATH, and satisfies REF as stated in this lemma. Furthermore, MATH for every MATH, MATH for a.e. MATH, and MATH for a.e. MATH. For every MATH, let us inductively define the MATH quasiconformal homeomorphism MATH as follows: CASE: MATH for every MATH; CASE: MATH for every MATH. By compactness of the set of all MATH quasiconformal homeomorphisms on MATH fixing three points (MATH, MATH and MATH), there is a subsequence MATH which converges to a MATH quasiconformal homeomorphism MATH. Then MATH satisfies REF as stated in this lemma. The restriction of MATH to the set MATH has the property of being a quasiconformal conjugacy between MATH and MATH. Furthermore, the NAME differential MATH has the following properties: CASE: MATH for every MATH; CASE: MATH for a.e. MATH; CASE: MATH for a.e. MATH. Therefore, by NAME 's glueing lemma, MATH is a MATH quasiconformal homeomorphism, and MATH restricted to the set MATH is a hybrid conjugacy between MATH and MATH. |
math/9905171 | The main step of this proof consists of constructing the real quadratic-like maps MATH satisfying MATH, and such that the maps MATH defined by MATH form a holomorphic motion MATH, and have the property of being holomorphic on the complement of a disk centered at the origin. Using REF , we prove that there is a positive constant MATH with the property that MATH. Finally, we show that this implies REF above. Now, we give the details of the proof: let us choose a small MATH, and a small open set MATH of MATH containing the interval MATH such that, for every MATH, the quadratic map MATH has a quadratic-like restriction to MATH, and MATH. Let MATH be a MATH function with the following properties: CASE: MATH for every MATH; CASE: MATH for every MATH; CASE: MATH for every MATH. There is a unique continuous lift MATH of the identity map such that CASE: MATH; CASE: MATH; CASE: MATH varies continuously with MATH. Then the maps MATH are holomorphic injections, and, for every MATH, MATH varies holomorphically with MATH. Let MATH be the interpolation between the identity map and MATH defined by MATH. We choose MATH small enough such that, for every MATH, and MATH, MATH is a diffeomorphism. Then MATH is a holomorphic motion over MATH with the following properties: CASE: the map MATH is a conjugacy between MATH on MATH and MATH on MATH; CASE: the restriction of MATH to the set MATH is the identity map; CASE: if MATH is real then MATH. By REF , MATH extends to a holomorphic motion MATH over MATH, and, by taking MATH, the map MATH is MATH quasiconformal for every MATH. By REF , there is a MATH quasiconformal homeomorphism MATH, and an open set MATH such that REF MATH restricted to MATH is a hybrid conjugacy between MATH and MATH; CASE: MATH is holomorphic over MATH; and REF MATH. Let MATH be defined by MATH. Then, for every MATH, MATH is a MATH quasiconformal homeomorphism which conjugates MATH on MATH with MATH on MATH. We define the NAME differential MATH as follows: CASE: MATH if MATH; CASE: MATH if MATH; CASE: MATH if MATH and MATH. Then REF the NAME differential MATH varies holomorphically with MATH; CASE: MATH for every MATH; and REF if MATH is real then MATH for almost every MATH. By the NAME Theorem (see CITE), for every MATH there is a normalized MATH quasiconformal homeomorphism MATH with MATH, MATH, and MATH such that MATH, and MATH varies holomorphically with MATH. Thus, the restriction of MATH to MATH is a holomorphic map, and if MATH is real then MATH for every MATH. The map MATH defined by MATH is MATH quasiconformal, and thus a holomorphic map. Furthermore, the map MATH is hybrid conjugated to MATH, and so MATH is a quadratic-like map whose straightening MATH is MATH. Since the conformal modulus of the annulus MATH depends only on MATH, we obtain that there is a positive constant MATH such that the conformal modulus of MATH is greater than or equal to MATH. If MATH is real then MATH, which implies that MATH is a real quadratic-like map. For the parameter MATH, the map MATH is MATH quasiconformal and fixes three points (MATH, MATH and MATH). Therefore, MATH is the identity map, and since the map MATH conjugates MATH with MATH, we get MATH. Now, let us prove that the quadratic-like map MATH satisfies REF . By compactness of the set of all MATH quasiconformal homeomorphisms MATH on MATH fixing three points (MATH, MATH and MATH), there are positive constants MATH for every MATH with the property that MATH . Thus, there is MATH with the property that MATH is holomorphic in MATH for every MATH, and MATH. Let MATH be the circle centered at the origin and with radius MATH. By REF , we obtain that MATH is at a uniform distance from MATH and MATH for every MATH, and MATH. Hence, by the NAME Integral Formula, and since MATH is a holomorphic motion over MATH, the value MATH varies holomorphically with MATH, and there is a constant MATH with the property that MATH . Thus, REF the map MATH is holomorphic in MATH; CASE: MATH is less than or equal to MATH; and REF MATH. Hence, by REF , there is a positive constant MATH such that, for every MATH, and for every MATH, we get MATH . Since MATH is a holomorphic motion over MATH, and by REF , we get MATH . By REF , and REF there is a positive constant MATH such that, for every MATH, and for every MATH, we obtain MATH . This implies that MATH . Since MATH is a MATH quasiconformal homeomorphism, and fixes three points, we obtain from REF in page REF that there are positive constants MATH and MATH with the property that MATH. Then by REF there is a positive constant MATH such that, for every MATH, and for every MATH, we have MATH . Finally, by increasing the constant MATH if necessary, we obtain that the last inequality is also satisfied for every MATH and MATH contained in MATH. |
math/9905172 | Assume that MATH has more than one dealternator. Let MATH be one of the dealternators. Then, MATH is adjacent to other four crossings (some of them may be the same). Let MATH be another dealternator. Since the crossing change at MATH makes MATH alternating, each of those four crossings must coinsides MATH. Then, MATH is a diagram obtained from a NAME link diagram with two crossings by changing one of the crossings. |
math/9905172 | Suppose that there exists a connected almost alternating diagram MATH of the trivial link MATH with MATH REF components. Here we assume that MATH has the minimal crossings among such diagrams, and then MATH is prime. Since MATH is also splittable, MATH is not reduced from REF . If MATH is REF, we obtain another almost alternating link diagram of MATH with less crossings than those of MATH by applying reducing move II. This contradicts the minimality of the number of crossings of MATH. Therefore, MATH is REF. Then apply reducing move I to MATH. If we obtain a connected diagram, then it is a connected alternating link diagram, which is non-splittable. Thus, we have a disconnected alternating diagram consisting of two connected components. Since MATH has more than two components, at least one of them has more than one component, which is a connected alternating link diagram and thus non-splittable. Thus, MATH is non-trivial, which contradicts our assumption. |
math/9905172 | We may assume that the complexity of MATH is finite. It is sufficient to show just that MATH passes the dealternator, since we have the third condition of REF . Suppose that MATH does not pass the dealternator. Here, we may assume that the dealternator is outside of MATH, that is, in the region with MATH of the two regions devided by MATH. Note that the length of every curve is more than one and even. From the almost alternating property, MATH must contain at least one curve inside of it, say MATH. Then, MATH is also does not pass the dealternator, because it is inside of MATH. Since MATH also must contain at least one curve inside of it, we can inductively find an infinitely many curves inside of MATH. This contradicts the finiteness of the complexity of MATH. |
math/9905172 | Let MATH, MATH, and MATH be the numbers of the vertices, edges, and faces of MATH, respectively. From the NAME 's formula, we have MATH. Since we also have MATH, MATH and MATH, MATH. |
math/9905172 | We show only the case for REF, which has three tangle areas MATH, MATH, and MATH and ten regions MATH, MATH, MATH, MATH. Some of these regions might be the same. Then we have six possibilities which regions are the same from the reducedness and the primeness of MATH (for instance, we have a nonprime diagram if MATH and we have a nonreduced diagram if MATH and MATH). Here, we show the case that all ten regions are mutually different. Other cases can be shown similarly. In addition, let us assume that regions MATH and MATH do not share an arc inside of MATH (in this case, we consider diagram MATH of MATH and diagram MATH of another splittable link instead of MATH and MATH, see REF ). If link MATH has diagram MATH, then MATH has diagram MATH with one less crossings than those of MATH as well. Then, note that all nine regions MATH, MATH, MATH, MATH are mutually different. (Connectedness) Assume that MATH is not connected. Then, we have a simple closed surve MATH in a region of MATH such that each of the two regions of MATH contains a component of MATH (here we call such a curve a splitting curve). If MATH is entirely contained in a tangle area, then it is easy to see that MATH is not connected as well. Therefore, MATH and MATH is in region MATH, MATH, MATH, or MATH. We may assume that MATH has minimal intersection with MATH. Take a look at one of outermost intersections of MATH and MATH. Then the intersection is one of the following REF . In the cases of REF , we can have another splitting curve MATH which is entirely contained in MATH or which has one less intersections with MATH than MATH does. In the cases of REF , we have the same regions among the nine regions of MATH. In other cases, MATH has an intersection with MATH. (Primeness) Assume that MATH is not prime. Then we have a simple closed curve MATH which intersects MATH in just two points belonging to different arcs of MATH (here we call such a curve a separating curve). If MATH is in a tangle area, then it is easy to see that MATH is not prime as well. However, MATH, since no pair of nine regions share two different arcs outside of tangle areas. We may assume that MATH has minimal intersection with MATH. Take a look at one of the outermost intersections of MATH and MATH. Then the intersection is one of the figures in REF . In REF , we can eliminate the intersection, which is a contradiction. In the cases of REF , we can obtain another separating curve MATH which is entirely contained in MATH or which has one less intersections with MATH than MATH does. In the cases of REF , we have the same regions among the nine regions of MATH. In the case of REF , MATH has an intersection with MATH and here we may assume that the other intersection is outside of MATH from the minimality. Then regions MATH and MATH must share arcs inside and outside of MATH, and thus regions MATH and MATH must share an arc inside of MATH, which contradicts our assumption. CASE: If MATH is REF, then we obtain an alternating diagram of MATH by reducing move I. Thus, MATH is non-splittable, which is a contradiction. Assume that MATH is REF. Then we can find a part in the diagram which we can apply reducing move II to REF . We have four possibilities; MATH, MATH, MATH, or MATH. In the first case, regions MATH and MATH must share a crossing so that MATH contains the part in REF . Since MATH and MATH, we have regions MATH and MATH, and then MATH might be the same as MATH or MATH (see REF ). However, the regions which can share a crossing with region MATH are MATH, MATH, or MATH. If MATH, then we obtain a non-prime diagram. Therefore, this case does not occure. In the second case, regions MATH and MATH must share a crossing. Thus, we can decide the inside of MATH more precisely and then we can see that MATH is non-reduced REF . In the third case, regions MATH and MATH must share a crossing REF . The regions which can share a crossing with MATH are MATH, MATH, or MATH. We also have three pssibilities that MATH, MATH, or MATH. Thus, we obtain that MATH or MATH. In the former case, we have a non-prime diagram. In the latter case, we can decide the inside of MATH more precisely and then we can see that MATH is non-reduced REF . We can prove the fourth case similarly. |
math/9905172 | CASE: Take a block of type MATH and put names MATH, MATH, MATH, MATH, and MATH to it as shown in REF . Assume that there is a pair of vertices coming from saddles in a same bubble. From the almost alternating property, we have that MATH, MATH, or MATH. The first case contradicts the mimimality of REF and the last two cases contradict the primeness. CASE: Take two blocks of type MATH. Put names MATH, MATH, and MATH to one of them and MATH, MATH, and MATH to the other following REF . Then, we have that MATH and MATH or that MATH and MATH. The former case contradicts the minimality of REF and the latter case contradicts the primeness unless the claim holds. |
math/9905172 | CASE: It contradicts the mimimality of REF . CASE: Assume that the boundary curves of MATH and the face MATH which is MATH-adjacent to it have been replaced on the diagram. Put names MATH, MATH, and MATH to the face of degree MATH as shown in REF . We may assume that MATH is below the dealternator. Here note that MATH is on the boundary cycle of MATH. Therefore we have that MATH is surrounded by boundary curve MATH or that MATH and MATH. Similarly, we have that MATH is surrounded by boundary curve MATH or that MATH and MATH. It is easy to see that none of four cases can be held considering the length of the boundary curve of MATH. CASE: Put names MATH, MATH, MATH as shown in REF . From symmetricity, we may assume that the face, say MATH, with MATH as its boundary cycle is MATH-adjacent to MATH at MATH or MATH. Assume that the boundary curves of MATH and MATH have been replaced on the diagram. In the first case, we have that MATH, MATH, MATH, or MATH considering replacing the boundary cycle MATH on the diagram. The first three cases contradicts the mimimality of MATH (REF or V) and the last case contradicts the primeness. In the second case, we have that MATH, MATH, or MATH considering replacing boundary cycle MATH on the diagram. The first and the third cases contradict the mimimality of REF and the second case contradicts the primeness. |
math/9905172 | CASE: We say a subgraph of MATH a subblock of type MATH, MATH, MATH, and MATH (respectively, MATH, MATH, MATH, and MATH) if its boundary curve constracts a diagram MATH, MATH, MATH, and MATH (respectively, the mirror image MATH, MATH, MATH, and MATH) of REF , respectively. It is easy to see that it contradicts the mimimality of REF if graph MATH has MATH (respectively, MATH) and MATH, MATH and MATH (respectively, MATH, MATH and MATH). Here note that any block of a type of MATH consists of one of MATH, MATH and MATH and one of MATH, MATH and MATH (for instance, a block of type MATH consists of a subblock of type MATH and a subblock of type MATH). In addition, any block of MATH (respectively, MATH) contains a subblock of type MATH (respectively, MATH). Therefore, any block of MATH and any block of MATH cannot coexist in graph MATH. From the primeness, the type of a block whose boundary curve can exist inside of the boundary curve of MATH is only MATH among MATH, MATH, and MATH, and then their boundary curves pass the same bubbles. Next, assume that there is the boundary curve MATH of a face of degree MATH inside of a boundary curve MATH of a subblock of type MATH. Then we have that MATH and MATH, MATH and MATH, MATH, or MATH. The last three cases contradicts the primeness. Therefore consider the first case. If the face is of a subblock of type MATH, then it contradicts primeness. If the face is of a subblock of type MATH, then their boundary curves pass the same MATH bubbles from the primeness. If the face is of a subblock of type MATH, then we cannot connect bubbles MATH and MATH with an arc for MATH (see REF ). Now assume that there is the boundary curve MATH of a face of degree MATH inside of the boundary curve MATH of a subblock of type MATH. Then we have that MATH, MATH, MATH, or MATH. The last two cases contradicts the primeness. Consider the first case. Then, we also obtain that MATH from the primeness. If the face is of a subblock of type MATH or MATH, then it contradicts the primeness. If the face is of a subblock of type MATH, then their boundary curves pass the same MATH bubbles also from the primeness. Next, consider the second case. Note that we are now considering the coexistence of blocks of MATH. Therefore, we have a subblock of type MATH, MATH, or MATH. Then it contradicts the minimality of REF . Now we need to show that a block of type MATH and a block of type MATH (or MATH and MATH) do not coexist in graph MATH. If graph MATH has MATH, then the boundary curve of the top and bottom face of any block of type MATH must pass the same MATH bubbles as the boundary curve of the top and bottom face of the block of type MATH, respectively. However, MATH has two top faces whose boundary curves do not pass the same bubbles. It is a contradiction. CASE: If any of MATH and any of MATH coexist in graph MATH, then we have MATH and MATH on the diagram, which contradicts the reducedness. It is easy to see the last part following the previous case. |
math/9905172 | CASE: Assume that we have a face of degree MATH with boundary cycle MATH. Considering the length of the cycle, we have that MATH and MATH or MATH. The former case contradicts the reducedness and the latter case contradicts the mimimality of REF . CASE: Assume that we have a face of degree MATH which is MATH- and MATH-adjacent to a block of type MATH and its boundary cycle is MATH (it can be similarly shown the case that the face is MATH- and MATH-adjacent to blocks of type MATH). Considering the length of the cycle and the almost alternating property, we have that MATH, MATH, and MATH. Then, it contradicts the reducedness. CASE: We show only the case that we have a face of degree MATH which is MATH-adjacent to a block of type MATH and MATH-, and MATH-adjacent to another block of type MATH. And let MATH be the boundary cycle of the face. Considering the length of the cycle and the almost alternating property, we have that the face is MATH-adjacent to the block at MATH or MATH. The former case contradicts the reducedness. In the latter case, the face must be MATH-adjacent to the block at MATH, and then it contradicts the reducedness, again. |
math/9905172 | Let MATH be the boundary cycle of MATH. Note that the boundary curve must pass the dealternator and curve MATH. Thus, MATH must be MATH-adjacent to a block of type MATH at MATH from the almost alternating property. However then, it contradicts the minimality of REF . We can similarly show that there does not exist MATH. Let MATH be the boundary cycle of MATH and let MATH (respectively, MATH) be inside of MATH (respectively, MATH). Since the length of curve MATH is MATH and it must pass MATH and MATH, we have that MATH, MATH or that MATH, MATH. However if MATH or MATH, then it contradicts the primeness. And if MATH or MATH, then it contradicts the minimality of REF . In the case of MATH, let MATH be its boundary cycle. Following the previous argument, we have that MATH, MATH and MATH, MATH. From the length of the cycle, MATH must be MATH-adjacent to a block of type MATH at MATH. Then, we have that MATH or MATH. The first case contradicts the minimality of REF and the second case contradicts the primeness. We can similarly show that there does not exist MATH. Let MATH be the boundary cycle of MATH. It is MATH-adjacent to MATH at MATH or MATH from the almost alternating property. The former case contradicts the reducedness. In the latter case, we have that MATH and MATH from the reducedness and the minimality of REF . Therefore, MATH and the boundary curve of MATH of MATH is in one and in the other of the two regions of MATH, respectively. Therefore, MATH cannot be MATH-adjacent to MATH. We can similarly show that there does not exist MATH. |
math/9905172 | We show the proof only for MATH and MATH. Let MATH be the boundary cycle of MATH. Note that the boundary curve must pass the dealternator and curve MATH. Thus, MATH must be MATH-adjacent to a block of type MATH at MATH or MATH from the almost alternating property. Then we have the former case and that MATH from the minimality of REF , which completes the proof. Moreover, we have that MATH from the primeness and then we obtain the diagram shown in REF . Let MATH be the boundary cycle of MATH. Since the length of the boundary curve is MATH and it must pass the boundary curves of the top and bottom faces of the block of type MATH of MATH, we have that MATH, MATH, or MATH. If MATH, then we have that MATH, MATH or that MATH, MATH. The first and fourth cases contradict the mimimality of REF and the second case contradicts the primeness. Therefore we have that MATH and then, we similarly obtain that MATH. However, then it contradicts the primeness, since MATH is adjacent to a face of degree MATH at MATH or at MATH. This completes the proof. Moreover, if MATH, then MATH or MATH. The former case contradicts the mimimality of REF . In the latter case, MATH must be adjacent to a face of degree MATH at MATH from the minimality of REF . Then, we have that MATH, MATH, that MATH, MATH, or that MATH, MATH. All the five cases but the third contradict the primeness or the minimality of REF . Therefore, we obtain the diagram shown in REF . In the case that MATH, we obtain a mirror image of the diagram. |
math/9905172 | Take a look at the diagram of MATH with MATH and MATH REF . From the mimimality of REF , MATH must be adjacent to a face of degree MATH at MATH. Then, it is easy to see that there does not exist MATH, since the length of the boundary curve of MATH is MATH and it must pass MATH, MATH, and MATH. We can similarly show that there does not exist MATH or MATH. |
math/9905172 | Assume that MATH contains MATH. Then, any arc passing MATH and MATH must passes MATH. Thus, MATH does not contain MATH or MATH. Since MATH passes MATH (respectively, MATH) seeing it at the opposite (respectively, same) side as MATH, it must be surrounded by (respectively, it must surround) the cycle containing MATH. It is a contradiction. Thus MATH does not contain MATH. Assume that MATH contains MATH and MATH. Then the diagram contains MATH. It contradicts the mimimality of REF , since MATH contains MATH, MATH, or MATH. We can similarly show other cases. |
math/9905172 | From REF , it is sufficient to show that we have an arc MATH, MATH, or MATH on the diagram under the assumptionthat there exists MATH, MATH, MATH, or MATH. Let MATH be MATH-adjacent to MATH. Then, its boundary curve MATH must pass MATH or MATH. In the former case, we have an arc MATH or MATH from the mimimality of REF . In the latter case, we have that MATH and MATH from the mimimality of REF and the primeness. Moreover, it can be adjacent to at most one MATH at MATH from the primeness. Thus, we do not have MATH in this case. If we have MATH or MATH, then the boundary curve of MATH must pass MATH and MATH, and thus we have MATH. If we have MATH, then the boundary curve of another MATH must pass MATH, MATH, and MATH from the primeness. Then, we have MATH or MATH. |
math/9905172 | CASE: It is easy to see that MATH is MATH, MATH, MATH, or MATH. If MATH, MATH, MATH, MATH, then MATH. From REF , we do not have MATH. Let MATH be the boundary cycle of MATH. Then, the face is adjacent to MATH at MATH. From the almost alternating property, it must be MATH-adjacent to MATH at MATH. Then, we have that MATH or MATH. The former case contradicts the reducedness and the latter case contradicts the minimality of REF . Therefore, if MATH, then MATH is MATH or MATH. In the first case, let MATH be the boundary cycle of MATH and let MATH be MATH-adjacent to MATH at MATH. From the almost alternating property, we have that MATH or MATH. Since the length of curve MATH is MATH, we have that MATH and MATH. Then we have that MATH, MATH or that MATH, MATH. The first and the fourth cases contradict the reducedness. In the second and the third case, we obtain the diagram of type MATH and type MATH, respectively. For the case of MATH, let MATH be the boundary cycle of MATH and let MATH be adjacent to MATH at MATH. Then, we do not have that MATH or MATH, since the length of curve MATH is MATH and it must pass arc MATH and bubble MATH. If MATH, then we have that MATH considering the length of the curve. In the former case, we have that MATH, MATH, MATH, or MATH. The second and the third cases contradict the mimimality of MATH from REF . In the first and the fourth case, we obtain MATH of type MATH and MATH, respectively. MATH . Note that MATH is MATH of type MATH, and MATH and MATH are type MATH. Now it is easy to see that those statements hold. CASE: We can see this from the primeness and the reducedness of MATH. |
math/9905172 | CASE: Let MATH be the boundary cycle of the face of degree MATH. If MATH, then it contradicts the minimality of REF . If MATH, then there exists a face which has two black vertices on its boundary, which contradicts REF Let MATH be the boundary cycle of the face of degree MATH. Similarly to the previous case, we can show the cases of that MATH and MATH. If MATH, then it also contradicts the minimality of REF . Therefore, we have that MATH. Now let MATH and MATH be the boundary curve of the face of degree MATH and MATH in the statement, respectively. Then, note that each bounday curve is surrounded by that of the face of degree MATH which is MATH-adjacent to a block of type MATH. If MATH, MATH, or MATH, then it contradicts the minimality of MATH (REF or VII). If MATH, then it contradicts the primeness. |
math/9905172 | The block of type I is MATH, MATH, MATH, MATH, MATH, MATH, or MATH. We show only the cases of MATH, MATH, MATH. In the case of MATH, let MATH be the boundary curve of MATH. We can show the cases that MATH and MATH similarly to REF . Assume that MATH. Considering the face of degree MATH of MATH, we can see that it contradicts the minimality (VI or VII) or primeness of MATH in both cases that two faces of MATH are adjacent to a face of degree MATH at MATH and MATH. Next assume that MATH. Then the two faces of MATH are adjacent to a face of degree MATH at MATH or MATH. The latter case contradicts the minimality of REF . In the former case, let MATH be the boundary cycle of the face of degree MATH. Then, we have that MATH, MATH, MATH, or MATH. The first two cases contradict the primeness and the last case contradicts the minimality of REF . In the third case, we obtain diagram MATH. In the case of MATH, considering the face of degree MATH whose boundary curve passes the leftside of the dealternator on the diagram and following the previous case, we have that MATH and the two faces of degree MATH are adjacent to a face of degree MATH at MATH. Now let MATH be the boundary curve of the other face of degree MATH. Since it is also MATH-adjacent to a block of type MATH, we have that MATH or MATH from the almost alternating property. The former case contradicts the primeness. In the latter case, we obtain diagram MATH. In the case of MATH, considering the face of degree MATH and following the proof of REF , we have that MATH. Now let MATH be the boundary cycle of the face of degree MATH. Since the face is also MATH-adjacent to a block of type MATH, we have that MATH or MATH considering the almost alternating property. The former case contradicts the mimimality of REF . In the latter case, we obtain diagram MATH. |
math/9905172 | From REF , it is sufficient to show only the last three cases. If there exists a block of type MATH, then it contradicts the primeness, the reducedness, or the minimality of MATH (REF or VII). Let MATH be the boundary cycle of MATH. NAME the length of the curve, we can assume that MATH, MATH, MATH, or MATH. Then, it contradicts the primeness or the minimality of MATH (REF or VII). Take a look at a block of type MATH and let MATH be its boundary cycle. From the minimality of MATH (REF or VII), we obtain that MATH and thus the curve surrounds the boundary curve of the face MATH which is adjacent to two faces of degree MATH with MATH. Therefore, face MATH cannot be MATH-adjacent to a block of type MATH. |
math/9905172 | The diagram obtained from any block of MATH and MATH contains MATH and MATH, respectively. It contradicts the minimality of MATH (REF or VII), since we have MATH as well. The diagram obtained from MATH (respectively, MATH) contains an arc connecting MATH and MATH (respectively, MATH and MATH). However, since MATH (respectively, MATH) is in one of the two regions of MATH and MATH is in the other, no arc can connect them. |
math/9905172 | We omit the proof for the first three and the last cases. Take the boundary cycle MATH of MATH. From the almost alternating property and the length of the curve, MATH is MATH-adjacent to MATH at MATH or MATH. Since the first case contradicts the minimality of REF , we may consider the second case. Then we have that MATH from the minimality of REF . Then, we can see that the boundary curve passes the leftside of the dealternator. Therefore, we do not have MATH. Considering the diagram, it is also easy to see that we do not have MATH, MATH or MATH. |
math/9905178 | For an infinitesimal deformation it is enough to consider MATH, MATH, MATH where MATH. First we show that the triple MATH defines an infinitesimal deformation if and only if MATH is a cocycle. As the first row and the first column in MATH are simply NAME complexes, a standard algebra deformation theory argument shows that the associativity of MATH and MATH modulo MATH is equivalent to the conditions MATH. In view of this fact we need to show that MATH satisfies REF if and only if MATH and MATH. Expanding REF for MATH in powers of MATH one easily finds that the MATH-order terms are simply REF for MATH. Therefore only terms of order MATH require further study. The MATH-order term in the first of REF is MATH . This is precisely the statement that MATH. Evaluating this condition at MATH one easily finds that MATH, that is, MATH, where MATH is the unit in the infinitesimal deformation MATH. Thus the second of REF holds. Similarly one shows that REF hold for MATH modulo MATH if and only if MATH. Therefore the necessary and sufficient condition for MATH to be an infinitesimal deformation of MATH is that MATH be a REF-cocycle in MATH as required. Let MATH and MATH be two infinitesimal deformations of an algebra factorisation MATH given by the cocycles MATH and MATH respectively. We need to show that these two deformations are equivalent to each other modulo MATH if and only if the corresponding cocycles differ by a coboundary. In view of the NAME theory, MATH and MATH are the algebra isomorphisms modulo MATH if and only if MATH and MATH. Thus it remains to be shown that MATH is an algebra isomorphism modulo MATH if and only if MATH . Suppose that MATH is an isomorphism of algebras, and let MATH be the MATH-order term in the expansion of MATH, that is, MATH . Since MATH is an algebra map MATH . Note that the product on the left hand side of REF is in MATH while on the right hand side is in MATH. One can use REF to find the MATH-order term on the left hand side of REF MATH . All the products are in MATH now. On the other hand the MATH-order term on the right hand side is MATH . Thus if MATH is an algebra map we have MATH as required. To prove the converse one needs to repeat the same computations in reversed order. |
math/9905178 | The first part of the theorem can be proven in the following way (standard in the deformation theory of algebras, which also asserts that MATH and MATH are NAME cocycles). Let MATH . The proof hinges on two observations. Firstly, one easily finds that MATH . Secondly one should notice that MATH where MATH is obtained by replacing MATH, MATH and MATH in definition of MATH with MATH, MATH and MATH. Expanding MATH, with MATH expressed entirely in terms of the tilded structure maps, one discovers that the term-by-term cancellations yield MATH. Thus MATH is a cocycle as asserted. (This expansion is a straightforward procedure, one only has to remember to take the inclusions of NAME cocycles into MATH properly into account.) It follows from the NAME theory that MATH and MATH are deformations of MATH and MATH respectively modulo MATH if and only if MATH and MATH are coboundaries in the NAME cohomology, that is, there exist MATH and MATH such that MATH and MATH. Thus only the conditions arising from REF require further study. Gathering all the terms of order MATH in REF one easily finds that MATH is a deformation modulo MATH if and only if MATH . All this means that the necessary and sufficient condition for MATH to be a deformation modulo MATH is that MATH that is, MATH is a coboundary, as required. |
math/9905184 | Since MATH defines a system of standard coordinates on MATH, it is birational. The equivariance and the statement on the image are straightforward. |
math/9905184 | By REF , MATH is NAME open in MATH. By straightforward calculations, MATH is a normal form for MATH and hence, by REF , a normal form for MATH. |
math/9905184 | The mapping MATH is birational as the composition of MATH and the map REF . The other statements are straightforward. |
math/9905184 | The proof is straightforward. |
math/9905184 | It follows from the general position condition that MATH in the notation of REF . Let MATH be arbitrary elements. Since MATH is MATH-homogeneous, there exists MATH such that MATH. Then there exists MATH with MATH and MATH. By the construction, MATH and the proof is finished. |
math/9905184 | For MATH generic, MATH. Hence there exists MATH with MATH and MATH. Clearly MATH extends to an isomorphism from the action by MATH. Without loss of generality, MATH. Similarly there exists an isomorphism MATH such that MATH and MATH on MATH for all MATH. Thus we may assume that MATH. By REF , we may also assume that MATH. Again, for MATH generic, MATH, where MATH. Therefore there exists an isomorphism MATH such that MATH and MATH. This proves the lemma. |
math/9905184 | The inverse of MATH is given by MATH where MATH is birational by REF and MATH . The equivariance is straightforward. |
math/9905184 | In the above notation the inverse MATH can be calculated as follows: MATH for MATH, MATH for MATH, MATH and finally MATH . The equivariance is straightforward. |
math/9905184 | The required birational isomorphism is given by MATH where MATH is birational by REF and MATH is birational by REF . |
math/9905184 | Set MATH as above. We write the spaces MATH as in REF in the form of a MATH matrix with MATH blocks as follow: MATH where MATH denotes the identity matrix of the size MATH. Let MATH be subspace of all matrices of the form REF . Clearly MATH fulfil the assumptions of REF . Then it follows from REF , that MATH is a normal form for MATH, where MATH acts on MATH as the diagonal subgroup of MATH, that is, each MATH block is conjugated by the same matrix. By the definition of a normal form, it is sufficient to prove that the invariant field of MATH is generated by the restrictions of MATH's. Let MATH be given by REF . By the obvious calculations, MATH and hence MATH . By REF, the polynomial invariants of MATH-tuples MATH with respect to the diagonal MATH-conjugations are generated by the monomials of the form REF with MATH. Since all points of MATH are semi-stable (see CITE), the categorical quotient MATH exists and is given by these monomials. Then the rational invariants on MATH are pullbacks of rational functions on MATH and the proof is finished. |
math/9905185 | MATH . It is clear that a partial homeomorphism of the above form MATH is in MATH and, since MATH is full, this is still true for a partial homeomorphism locally of that form. On the other hand, the inverse of MATH, which is MATH, is of the same form. The product of MATH and MATH is also of the same form: if MATH, it is of the form MATH, where MATH is an open set on which MATH is injective and if MATH, it is of the form MATH, where MATH is an open set on which MATH is injective. This implies that the set of partial homeomorphisms locally of the above form is a full pseudogroup, and therefore is MATH. MATH . We may assume that MATH, MATH and MATH . Then MATH . |
math/9905185 | MATH . By assumption, the relation MATH holds on a nonempty open set. The essential freeness implies that MATH. MATH . We have seen in the previous proposition that the product of MATH and MATH is of the form MATH or MATH and that the inverse of MATH is MATH. This shows that MATH is a homomorphism. By construction, MATH is locally constant. |
math/9905185 | If MATH is essentially free, we may define the map MATH from MATH to MATH by MATH. This map is an inverse for MATH. Conversely, suppose that MATH is an isomorphism. Let MATH and MATH be a nonempty open set such that MATH. Then for MATH, the germ MATH is a unit. This implies that MATH. |
math/9905185 | MATH . We will check measurewise amenability ( according to CITEEF, it is equivalent to topological amenability for étale groupoids). We first show that MATH is amenable. It is the increasing union of MATH. According to REF, it suffices to show that MATH is amenable. This is a NAME equivalence relation with countable equivalence classes. There is a countable family of open sets MATH which covers MATH and such that, for each MATH, the restrictions MATH are one-to-one. Therefore, the equivalence relation MATH is countably separated, its quotient space MATH is analytic . This implies that MATH is a proper NAME groupoid (CITEEF), hence it is amenable. We can now apply a result on the amenability of an extension (CITE; REFEF) to conclude that MATH is amenable. Indeed, we may write MATH as the disjoint union of the invariant NAME subsets MATH and MATH, where MATH is the intersection of the domains of the MATH's and MATH is its complement. The homomorphism MATH is strongly surjective in the sense given there on MATH. On the other hand, the reduction of MATH to MATH is a proper principal groupoid having as quotient space the complement MATH of the domain MATH. MATH . This is a well-known property of the MATH-algebra of an amenable groupoid (see for example, CITE; REFEF). MATH . This is also a well-known property of the MATH-algebra of an amenable groupoid (see for example, CITE; REFEF). |
math/9905185 | This graph is the space MATH endowed with the natural left and right actions of MATH. One can check directly the axioms of CITE. |
math/9905185 | This is clear since, keeping the notation of the above discussion, MATH is in the closure of MATH iff MATH. |
math/9905185 | Suppose that MATH is dense. If the intersection MATH, where MATH are finite subsets of MATH, contains only a finite union MATH, then the complement of MATH in MATH, which is open, must be empty. On the other hand, suppose that MATH holds. We shall show that MATH, which by REF is equivalent to the density of MATH. If MATH is not a cluster point of the map MATH introduced in the discussion following REF, there exist finite subsets MATH of MATH such that MATH and the conditions MATH are satisfied for only a finite number of MATH's. This means that MATH contains only finitely many MATH's; by MATH, it is contained in MATH. By definition, MATH does not belong to MATH. |
math/9905185 | (Compare REF) For MATH, the condition MATH is satisfied by construction. The conditions MATH are a restatement of above properties MATH. For MATH, we give the sketch of the proof and refer to CITE, CITE and CITE for details. Let MATH be such a family of partial isometries. A direct computation (see REF) shows that the MATH's (together with REF) generate an inverse semigroup MATH of partial isometries. Let MATH be the spectrum of the (commutative) MATH-algebra MATH generated by the idempotents of MATH and let MATH be the NAME isomorphism. For each MATH, let MATH [respectively, MATH] be the support of MATH [respectively, MATH]. The isomorphism MATH from MATH onto MATH induces a homeomorphism MATH and we define MATH on the (disjoint) union MATH by MATH if MATH. Then MATH is a NAME partition for MATH with matrix MATH . According to REF, MATH is isomorphic to MATH. Let MATH be the associated family of bisections of MATH. It satisfies the same relations MATH as the family MATH. Let MATH be the inverse semigroup of bisections of MATH generated by the MATH's (and MATH). Given two finite paths MATH and MATH and an idempotent MATH of MATH, we define MATH . One checks that every element of MATH can be written under the form MATH and that this form is unique when we require MATH and the lengths of MATH to be minimal. The same is true for MATH. Therefore the map MATH is a representation MATH of the inverse semigroup MATH onto MATH. This representation extends to a representation of MATH. Indeed, every MATH can be written as a finite sum MATH, where MATH and MATH is a compact open bisection. One can check that MATH is well defined and that MATH is a representation of the MATH-algebra MATH continuous for the inductive limit topology and therefore extends to MATH. |
math/9905185 | We have seen indeed the equivalence of the conditions MATH and MATH. |
math/9905185 | It suffices to show that the representation MATH of MATH defined by MATH is faithful. Since its restriction MATH to MATH is faithful and the groupoid MATH is nuclear and essentially principal, we may apply REF to conclude. |
math/9905185 | The SGDS MATH and MATH are essentially free and we may apply REF. Every nonempty open set MATH contains a cylinder set, that is, a set MATH of all terminal paths starting with the finite path MATH. Let MATH be an infinite path MATH. There exist a finite path MATH. Then MATH belongs to MATH, hence to MATH, and MATH. Let MATH be a finite terminal path distinct from MATH. There exists an integer MATH such that MATH where MATH is nonempty. We choose MATH and a finite path MATH. Then MATH belongs to MATH and MATH. |
math/9905185 | We apply REF to the SGDS MATH. First observe that if the loop MATH has an outgoing edge, then MATH is a proper subset of MATH. We have seen that every nonempty open set MATH contains a cylinder set MATH. By assumption, there exists a loop MATH and a finite path MATH. Let MATH. Then MATH and MATH is a proper subset of MATH. |
math/9905187 | Follows from the NAME theorem. |
math/9905187 | Given MATH, then since MATH is a homomorphisms, MATH. Thus MATH, so MATH. Hence the NAME bracket is defined. Given MATH such that MATH then MATH. Hence MATH, so MATH, and the NAME bracket is well defined. From REF we know there exist MATH such that MATH, and hence the NAME bracket is non trivial. The derivative property follows from expanding MATH. The NAME identity follows from the NAME identity for commutators. Since all NAME structures my be written in terms of a bivector, the NAME structure can be extended to MATH. |
math/9905187 | Choose a sequence of self-conjugate elements in MATH which is a basis of MATH as a MATH-dimensional vector space. For example MATH now remove any elements from the sequence that are permutation of previous elements, or are in the span of preceding elements and the ideal generated by the quotients elements MATH. This gives a sequence of MATH. Each element in MATH can be uniquely written as a sum MATH. Set MATH. To construct a unitary ordering set MATH. MATH is far from unique. For example we can always set a new ordering as MATH. |
math/9905187 | Trivial. |
math/9905187 | Let MATH be the subalgebra of MATH generated by star products of MATH. Let MATH be generated by MATH. Let the function MATH be the algebraic homomorphism satisfying MATH . Now we show there is a map MATH such that MATH. Let MATH be any map such that MATH, this can be constructed similar to proof of REF . For MATH let MATH and MATH. Thus MATH and MATH. Let MATH. We can say this converges in MATH and it is easy to show that MATH. Now let MATH and let MATH where MATH is any map such that MATH. The ordering is given by MATH. This defines the map MATH. Since MATH is a bijective homomorphism, then MATH is unique. |
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