paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/9906111 | Consider a point MATH, in other words a natural transformation MATH for MATH. By putting together the maps MATH we get a function MATH. Next, for any MATH we have projections MATH and we can use the resulting maps to identify MATH with MATH and MATH with MATH and MATH with MATH. There are evident maps MATH and MATH, and using the naturality of MATH with respect to these maps we find that MATH is a semiring homomorphism. Similarly, we have maps MATH in MATH and the naturality of MATH with respect to these maps implies that MATH commutes with MATH-operations. It is clear that MATH, so MATH. We define a map MATH by MATH. Because MATH we see that MATH is injective. Now suppose we start with a point MATH. Let MATH be the restriction of MATH, and put MATH . We need to check that this gives a natural transformation. As MATH and MATH, it suffices to check naturality for maps MATH in MATH, or equivalently (by REF ) for homomorphisms MATH. We need to show that the left hand square in the following diagram commutes: MATH . The right hand square commutes and the two right hand horizontal maps are injective so it suffices to show that the two composite maps MATH are the same. We call these two maps MATH and MATH. Let MATH be the set of all functions from MATH to MATH, thought of as a ring with pointwise operations. It is formal to check that MATH and MATH, so MATH is a homomorphism of semirings from MATH to MATH. It can thus be extended to a ring map MATH, and the same applies to MATH. It is well-known that MATH so it suffices to check that MATH on MATH for all MATH. This reduces us to the case where MATH say. It is also well-known that MATH, so it suffices to check that MATH, which is true because MATH is a homomorphism of MATH-rings. This shows that MATH, and clearly MATH. Thus MATH is surjective and hence an isomorphism. |
math/9906111 | Write MATH as a disjoint union of orbits. |
math/9906111 | Suppose MATH, and define MATH by MATH. By hypothesis, this has the same character as MATH, so there exists an intertwining operator MATH such that MATH for all MATH. As MATH is an irreducible representation of MATH we see that MATH and thus MATH is unique up to multiplication by a scalar matrix. We can thus define a map MATH by MATH; this is a homomorphism making the following diagram commute. MATH . Put MATH and MATH. As MATH is irreducible we know that MATH divides MATH. Put MATH, and note that MATH is surjective and MATH. Let MATH act on MATH by MATH and on MATH by MATH. We claim that there is a MATH-equivariant map MATH such that MATH and MATH on MATH. Clearly MATH as MATH-sets and MATH is MATH-equivariant, so it suffices to define MATH on MATH. Fix MATH, and choose MATH as before. After multiplying by a suitable scalar, we may assume that MATH so MATH. By REF , it will suffice to show that MATH. Suppose that MATH stabilises MATH, so MATH, so MATH. By the definition of MATH we have MATH, or in other words MATH, so MATH stabilises MATH, as required. Now define MATH by MATH. Clearly MATH, and the kernel of MATH is the group MATH of MATH'th roots of unity, so we can regard MATH as a map MATH. As MATH is equivariant, it is easy to check that MATH for MATH, so we get an induced map MATH. One also sees directly that for MATH we have MATH so MATH is a MATH-cocycle. On the other hand MATH divides MATH and thus is coprime to MATH, so we have MATH. We can thus choose a function MATH such that MATH for all MATH. By putting MATH we see that MATH and thus MATH for MATH. We define MATH; this clearly gives a homomorphism MATH with MATH, as required. |
math/9906111 | For each irreducible representation MATH of MATH, let MATH denote the sum of the inequivalent MATH-conjugates of MATH. Any MATH-invariant character is a direct sum of copies of the characters of the representations MATH, so it suffices to show that MATH extends to a representation of MATH. Let MATH be the stabiliser of MATH, so MATH. REF implies that MATH can be extended to a representation MATH of MATH, and one sees from the NAME formula that MATH, so MATH is the required extension of MATH. |
math/9906113 | CASE: By induction on MATH. CASE: Similarly. |
math/9906113 | By induction on MATH. |
math/9906119 | The proof in CITE for MATH extends with minor modifications to the general case. |
math/9906119 | Computed using NAME. |
math/9906119 | It follows from REF that the map REF identifies MATH with the curve class MATH in MATH, so the NAME are integrals over MATH. Localization. Consider the standard action of MATH on MATH. Since the integrands are polynomials in NAME classes of equivariant vector bundles with respect to the induced MATH-action on MATH, they may be evalued using NAME 's residue formula (see CITE). As there are only finitely many fixed points and curves in MATH, the formulae involved are similar to the ones found in CITE. Details are left to the reader. The two-point integrals with MATH were evaluated in this manner. Reconstruction. The number MATH is however more easily obtained from MATH numbers using the following NAME: MATH . In fact, a further analysis shows that the MATH-point MATH numbers appearing in REF are in turn determined by the MATH-point MATH numbers above. |
math/9906119 | From REF it follows that MATH. Using REF it is straightforward to check that MATH is the differential operator governing MATH (see for instance CITE). The rest follows from REF . |
math/9906124 | See CITE. See also CITE. |
math/9906124 | This is REF. |
math/9906124 | Sufficiency is due to CITE and necessity to CITE. |
math/9906124 | See CITE. See also CITE. |
math/9906124 | MATH is well defined: Suppose MATH, where MATH. Then MATH is one to one: If MATH, then for some MATH one has MATH . Thus MATH, and MATH. Let MATH. Then MATH, and MATH for some MATH since this subgroup is normal in MATH. Thus MATH. MATH is onto: Let MATH. Since the MATH are onto we have MATH and MATH for some MATH, MATH. Thus MATH . |
math/9906124 | Suppose MATH. Let MATH. Then MATH for some MATH, MATH, MATH. So MATH and MATH. Thus MATH . Similarly MATH implies that MATH . Hence MATH. Now suppose that MATH. From the proof that MATH is onto in REF we may assume that MATH for some MATH. Let MATH. Then MATH for some MATH. So MATH and MATH. Hence MATH and MATH. So MATH. Hence MATH. |
math/9906124 | First assume that every splitting homomorphism MATH with MATH and MATH has an essential factorization. Let MATH be a closed, orientable REF-manifold with MATH finite. We may assume that MATH is irreducible. Let MATH be a NAME splitting of MATH of minimal genus MATH. Let MATH be the associated splitting homomorphism. By REF MATH, and so if MATH were greater than two, then MATH would have an essential factorization, and hence MATH would be reducible. Since MATH is irreducible this would yield a NAME splitting of lower genus, contradicting the choice of MATH. Thus MATH, and we are done. Now assume that the NAME Conjecture holds in the finite fundamental group case. Let MATH be a splitting homomorphism with MATH and genus MATH. Realize MATH by a NAME splitting MATH. Then by REF MATH is finite. Suppose MATH does not have an essential factorization. Then MATH is irreducible. By the NAME Conjecture MATH is a NAME fibered space. Since MATH is finite MATH has a NAME fibration over a REF-sphere with at most three exceptional fibers. (See CITE or CITE. Note that it may have a NAME fibration over a projective plane with one exceptional fiber, but then it also has a NAME fibration of the given type.) If there were fewer than three exceptional fibers, then MATH would be MATH or a lens space. But by results of CITE and of CITE the irreducible NAME splittings of these spaces have, respectively, genus zero and one, contradicting our choice of MATH. Thus there are three exceptional fibers MATH of multiplicities MATH, MATH, MATH, MATH. Moreover, up to ordering, MATH must be one of MATH, MATH, MATH, MATH, or MATH. We now recall two constructions for NAME splittings of closed, orientable NAME fibered spaces over orientable base surfaces. For simplicity we restrict to the special case at hand. See CITE and CITE for the general case and a more detailed description. First choose two of the three exceptional fibers. Join their image points in the base REF-sphere by an arc which misses the image point of the other exceptional fiber. Lift this arc to an arc in MATH joining the two chosen exceptional fibers. A regular neighborhood MATH of the resulting graph is a cube with two handles. It turns out that the closure of its complement is also a cube with handles, and so MATH is a genus two NAME splitting of MATH. It is called a vertical NAME splitting. We remark that in the general case all vertical NAME splittings have the same genus MATH. Next choose one exceptional fiber MATH, and let MATH be a regular neighborhood of it. The closure MATH of the complement of MATH is bundle over MATH with fiber a surface MATH CITE, CITE. Moreover, MATH is a branched covering space of the base surface of the NAME fibration; the branch points are the images of the exceptional fibers and have branching indices equal to the indices of the exceptional fibers. Suppose MATH is connected and has intersection number MATH with a meridian of the solid torus MATH. Let MATH be a regular neighborhood of MATH in MATH. Then MATH is a cube with handles whose genus is twice the genus of MATH. It turns out that the closure of the complement of MATH in MATH is also a cube with handles, and thus MATH is a NAME splitting of MATH. It is called a horizontal NAME splitting at MATH. Denote its genus by MATH. Note that if either of the two conditions on MATH is violated, then by REF does not have a horizontal NAME splitting at MATH. Let MATH be the least common multiple of MATH and MATH, where MATH and MATH are the other two exceptional fibers. CITE have shown that every irreducible NAME splitting of a closed, orientable NAME fibered space over an orientable base surface is either vertical or horizontal. Since MATH and MATH our splitting MATH must be horizontal. CITE has determined precisely which vertical and horizontal NAME splittings are irreducible. In particular a horizontal splitting is irreducible if and only if either MATH or MATH. In our case the first condition is impossible, and the second condition holds only for MATH, where MATH. But in this case MATH is NAME fibered over a disk with two exceptional fibers of index two and so must be a twisted MATH-bundle over a NAME bottle; it follows that the fiber MATH is an annulus, and so MATH does not have a horizontal NAME splitting at MATH. Thus MATH must be reducible, and so MATH must have an essential factorization. |
math/9906124 | Suppose the condition on the MATH holds. Let MATH be a closed, orientable, irreducible REF-manifold and MATH a NAME splitting of minimal genus MATH. If MATH, then the NAME Conjecture holds for MATH. So assume MATH. If MATH had an essential factorization, then MATH would be reducible, and, since MATH is irreducible MATH would have a NAME splitting of lower genus, contradicting the choice of MATH. So MATH does not have an essential factorization, and we must be in REF . In REF MATH has a MATH subgroup. By NAME 's version of the torus theorem CITE either MATH contains an incompressible torus, and we are done, or MATH contains a normal MATH subgroup. In the latter case the proof of the NAME fibered space conjecture by CITE and by CITE gives that MATH is a NAME fibered space, and again we are done. In REF MATH is isomorphic to a discrete, non-trivial, torsion-free subgroup of MATH. Since the kernel of the projection to MATH is a finite group this subgroup projects isomorphically to a discrete subgroup MATH of MATH. A subgroup of MATH is discrete and torsion free if and only if its natural action on hyperbolic REF-space MATH is free (no non-trivial element has a fixed point) and discontinuous (each compact set meets only finitely many of its translates) CITE. Thus the quotient space MATH is a hyperbolic REF-manifold CITE. Since MATH is closed and aspherical MATH, and so MATH is closed and orientable. By the topological rigidity of hyperbolic REF-manifolds, due to CITE we have that MATH and MATH are homeomorphic, and we are done. Now suppose that the NAME Conjecture is true. Let MATH be a splitting homomorphism of genus MATH. Assume that MATH has no essential factorization. Let MATH be a NAME splitting which realizes MATH. Then MATH is irreducible, and hence MATH is irreducible. By REF we may assume that MATH is infinite. Since MATH is irreducible, MATH is torsion-free CITE. If MATH is hyperbolic, then MATH for some subgroup MATH of MATH acting freely and discontinuously on MATH. Thus MATH is discrete and torsion free. By a result of NAME MATH lifts isomorphically to a subgroup of MATH. (See CITE.) If MATH contains an incompressible torus, then clearly MATH contains a MATH subgroup. If MATH is NAME fibered, then MATH infinite implies that MATH has a covering space which is homeomorphic to a MATH bundle over a closed, orientable surface MATH of positive genus CITE, and so again MATH contains a MATH subgroup. (See also CITE.) |
math/9906124 | If MATH is covered by a bundle with fiber a surface MATH, then the image of MATH in MATH is a good subgroup. If MATH contains a MATH subgroup, then a summand is a good subgroup. The converse follows from CITE. For convenience we give the relevant portion of the proof of that result. Let MATH be a good subgroup and MATH the covering of MATH corresponding to MATH. Since the group of covering translations is isomorphic to MATH CITE there is a covering translation of infinite order. Let MATH be the quotient of MATH by this covering translation. MATH has a normal subgroup which is isomorphic to MATH and has infinite cyclic quotient. The NAME compact core CITE MATH of MATH is a compact submanifold of MATH with MATH isomorphic to MATH. Since MATH is irreducible CITE we may assume that MATH is irreducible. It then follows from the NAME fibration theorem CITE that MATH is a surface bundle over MATH. If MATH, then we are done. If MATH, then MATH consists of tori which are incompressible in MATH and so MATH has a MATH subgroup. |
math/9906124 | Suppose the extended conjecture is true. If MATH, then by the proof of REF we reduce to the situation in which either MATH has a MATH subgroup or MATH is hyperbolic; in the latter case we apply the Virtual Bundle Conjecture and so conclude in both cases that MATH has a good subgroup. If MATH, then a similar argument shows that either MATH has a good subgroup or MATH is a NAME fibered space with MATH finite, in which case MATH. Now assume that the conditions on the MATH hold. Let MATH be a closed, orientable, irreducible REF-manifold. By arguments similar to those in the proof of REF we reduce to the case that MATH has a good subgroup and does not have a MATH subgroup. Then by REF MATH is finitely covered by a surface bundle. Note that this is the case if MATH is assumed to be hyperbolic CITE, CITE, and so the Virtual Bundle Conjecture holds. For the general situation note that by the fibered case of NAME 's hyperbolization theorem CITE, CITE the surface bundle is hyperbolic. We conclude from the following result that the NAME Conjecture holds for MATH. Let MATH be a closed, orientable REF-manifold which has a finite sheeted covering space MATH which is hyperbolic. Then MATH is hyperbolic. This is well known. It follows immediately from the topological rigidity of hyperbolic REF-manifolds CITE and REF that MATH is homotopy equivalent to a hyperbolic REF-manifold. See CITE or the next section for a proof. This concludes the proof of REF . |
math/9906124 | We may assume that the covering is regular CITE. As pointed out by NAME, the covering translation in MATH corresponding to an element of MATH is by NAME rigidity homotopic to a unique isometry. The lifts of these isometries to the universal cover MATH give a subgroup MATH of MATH. The quotient MATH is then the desired hyperbolic REF-manifold. To fill out this sketch one needs to verify that MATH and that MATH acts freely and discontinuously on MATH. It will be convenient to first establish the following result. Let MATH be a closed hyperbolic REF-manifold and MATH a homotopy such that MATH for all MATH. Fix MATH. Let MATH for all MATH. Then the class MATH of MATH in MATH is trivial. Let MATH be represented by the loop MATH. Then the map MATH given by MATH shows that MATH. Hence MATH, which is trivial for a closed hyperbolic REF-manifold CITE. Returning to the proof of REF , we have covering maps MATH. Choose a basepoint MATH, and let MATH and MATH. Given MATH, let MATH and MATH be path classes lifting MATH with MATH and MATH, respectively. There is a covering translation MATH of MATH such that MATH and a lifting MATH of MATH such that MATH. By NAME rigidity there is a unique isometry MATH of MATH which is homotopic to MATH. Let MATH be a homotopy from MATH to MATH. It lifts to a homotopy MATH of MATH to an isometry MATH of MATH. We claim that MATH is independent of the choice of homotopy MATH. Let MATH be another homotopy from MATH to MATH. Define MATH by MATH for MATH and MATH for MATH. This homotopy satisfies the conditions of REF , and so the loop MATH is homotopically trivial. This implies that the paths MATH and MATH are path homotopic, and so their liftings MATH and MATH have the same endpoint MATH. Thus MATH. Thus the function MATH given by MATH is well defined. Note that if MATH, then MATH, and MATH. Let MATH and MATH. We next show that MATH is a homomorphism. Suppose MATH and MATH. Let MATH be the image of MATH under MATH. Then MATH, and so MATH. Thus MATH, and MATH is a homotopy from this map to the isometry MATH. By the uniqueness of MATH we have that MATH. By choosing the homotopy MATH from MATH to MATH to be MATH, we see that MATH, and so MATH. We now show that MATH is one to one. Suppose MATH. If MATH, then MATH, and so MATH is trivial. So assume that this is not the case. Then MATH is the identity of MATH, and MATH is not. Since the covering is finite sheeted MATH is the identity for some MATH. Define MATH by MATH for MATH. Then MATH. By REF the class MATH of the loop MATH is trivial in MATH. Let MATH be the path class of the path MATH joining MATH and MATH. Then MATH is non-trivial, but MATH is trivial. Thus MATH has torsion, contradicting the fact that MATH is aspherical CITE. MATH is discrete because it contains the finite index discrete subgroup MATH CITE. Thus it acts discontinuously on MATH. It acts freely because otherwise it would have torsion, contradicting the asphericity of MATH. Hence MATH is a REF-manifold with MATH. It follows from asphericity that MATH and MATH are homotopy equivalent, and so MATH is closed and orientable; hence MATH. |
math/9906128 | Let MATH be fixed, and let MATH satisfy MATH . By MATH, there exist MATH, such that MATH . If MATH is an admissible set then MATH, so this inclusion holds for all MATH . Hence MATH . Applying consecutively REF , the above inclusion and condition MATH, we obtain condition MATH. |
math/9906128 | Condition MATH is trivially satisfied, because the uniform base MATH of MATH consists of only one element MATH. Now we observe that MATH is continuous as a linear map on a finite-dimensional space, so MATH of REF holds for MATH the class of all continuous maps. |
math/9906128 | First, MATH is normal, so MATH is NAME. Next, for any MATH, let MATH . Then the set MATH determines the equivalence relation MATH . To ensure that MATH has a uniformity, we need to check that for any MATH any MATH there are MATH such that MATH which means that the equivalence relation MATH and the uniformity of MATH are weakly compatible (see CITE), so REF holds. We will show that MATH . Consider MATH . Suppose MATH. Assume for simplicity that MATH. Choose MATH such that MATH and MATH for all MATH . To get MATH as above, let MATH . Then REF above is satisfied. Now, let MATH (it means that we obtain MATH and MATH by ``shifting" coordinates of MATH and MATH steps to the right). Then REF above are also satisfied. Therefore we have MATH . If MATH and MATH then MATH for all MATH and some MATH, and MATH . Then for any MATH we have MATH . Therefore we have MATH and we conclude that MATH. |
math/9906128 | By REF MATH is NAME and NAME is compact. Hence MATH is closed in MATH . Then, there exists an open neighborhood MATH of MATH and a continuous function MATH that extends MATH and MATH is uniformly continuous. Now if we extend MATH on the whole MATH by putting MATH for MATH then MATH is uniformly u.s.c. (since MATH is open). It is routine to check that MATH satisfies conditions MATH - MATH. |
math/9906128 | Let MATH . Then condition MATH reads as follows: there exists MATH such that MATH . Let MATH. By definition of MATH, MATH has a subcover MATH with MATH. But MATH is l.s.c., so MATH consists of open sets. Moreover, if MATH then MATH for some MATH. Hence MATH belongs to MATH, so MATH is an open cover of MATH. By REF , we have MATH. Therefore, by definition of MATH, MATH has a locally finite open refinement MATH . Since MATH, we can assume that MATH (MATH for some MATH). From the definitions of MATH it follows that for all MATH where MATH for some MATH (here we assign an index to MATH and MATH according to this inclusion). Then let MATH . From the fact that MATH is normal and NAME 's Lemma it follows that there exists a partition of unity subordinate to MATH, that is, there are continuous functions MATH, satisfying MATH . Now let MATH where MATH, are the vertices of MATH . Then MATH is a continuous function. Since MATH is locally finite, for each MATH there are a neighborhood MATH of MATH and a finite set MATH such that MATH is nonzero only for MATH. Therefore we have MATH, so REF is satisfied. Moreover, if REF holds, then MATH is finite and so REF holds-MATH . Take MATH. Let MATH . If MATH is such that MATH then MATH, or MATH. Hence MATH. By definition of MATH, this implies that MATH where MATH, are the unit vectors of MATH. Next, consider MATH . By definition of MATH we have MATH for all MATH, or MATH . From REF , it follows that MATH . Since MATH is admissible, MATH. Hence by REF , we have MATH and, therefore, MATH is well defined. Thus, from REF we have REF . To finish the proof, we notice that REF implies that MATH. Therefore we have MATH . |
math/9906128 | Let MATH. Without loss of generality we can assume that MATH . Then according to REF (with MATH), for any nonempty convex-valued l.s.c. map MATH for any MATH, there is a continuous MATH with MATH . We inductively construct a sequence of continuous functions MATH, such that MATH . By REF , there is MATH so that REF holds for MATH . Assume that we have constructed MATH satisfying these conditions. Then let MATH . Then MATH is l.s.c. By REF for MATH, we have MATH, so MATH is nonempty, and it is convex because the base in convex. Therefore by REF , there is a continuous map MATH with MATH . Then there is a MATH such that MATH . By REF , we have MATH so REF implies REF . From REF , it also follows that MATH. Therefore we have MATH and now, from REF , and REF , it follows that MATH . Hence REF holds. Thus, we have constructed a sequence MATH satisfying the required conditions. Now REF implies that this is a NAME sequence. Therefore MATH converges to a map MATH . And from REF , it follows that MATH for all MATH . To finish the proof we observe that MATH is the uniform limit of MATH with respect to the uniformity of MATH. Therefore MATH is continuous, since MATH is finer than MATH. |
math/9906128 | In REF , we let MATH and notice that MATH, so REF holds-MATH . |
math/9906128 | In REF we let MATH be discrete and MATH. Then MATH is l.s.c., so REF holds. But since MATH, REF also holds. Finally, we notice that an almost selection with respect to the discrete topology is in fact a selection. |
math/9906128 | By REF with (MATH), for any MATH there exist MATH and a continuous function MATH with MATH finite such that MATH for all MATH . Since the convexity is global, we can define a map MATH by MATH . Since MATH, the multifunction MATH has a fixed point, that is, there exists a MATH. Therefore, there exist such MATH and MATH that MATH and MATH . Hence MATH . |
math/9906128 | Suppose MATH . By REF , for any MATH, there is a MATH . Now we need to show that MATH for some MATH. By REF , the net MATH has a convergent subnet. Therefore, we can simply assume that MATH. Then for any MATH, there exists a MATH such that MATH for all MATH, or MATH . By REF , MATH for all MATH. Therefore MATH . Since MATH and the multifunction MATH is u.s.c., then for any MATH, there exists a MATH such that MATH . Now using REF , we obtain MATH . Hence MATH, and the proof is complete. |
math/9906128 | In REF we let MATH . Then REF hold because the convexity is regular: MATH. Also, since MATH is compact, we have MATH and MATH, so REF holds. |
math/9906128 | Notice that if MATH is a discrete uniform space, then a multifunction MATH has open fibers if and only if it is l.s.c. Therefore REF of this theorem are exactly REF follows from the fact that MATH is u.s.c. with respect to the discrete topology. Also, since MATH is compact and MATH is discrete, we have MATH and MATH, so REF holds. |
math/9906138 | The theorem follows from the equality MATH . Hence the value of MATH on a DD(MATH)-singular NAME having MATH pairs of double points descends to the sum of values of MATH on singular links each of which has MATH double points. |
math/9906138 | The proof of REF is best illustrated by an example shown in REF , where we show a singular link on the left hand side and its corresponding double dating diagram on the right hand side. For the proof of REF we show how to build a singular link for a given DD diagram. Let D be a DD diagram with some full lines and pairs of chords. We will call the chords belonging to a certain pair ``pals". In REF we see a double dating diagram on the left hand side. Use ``finger moves" along the chords in order to get from REF to REF . Following this procedure to all pairs of pals chords will give a singular link as desired. One may think of the finger move as axon connecting two neurons by a synapse. Clearly this link is not unique and it can be obtained in any linking class. However now we show that any two singular links respecting the same DD diagram and within the same class differ by a finite sequence of double crossing flips. This is what is claimed in REF. We say that a link (singular or not), is in a neural position if it is embedded in MATH so that each component is an unknotted circle remote from all other components, knotted and linked to other components by a network of axons and synapses. Every link is isotopic to a link in a neural position. To see this one can take a disk neighborhood of every crossing of a given link diagram. The crossing can be regular or singular. Making the disk thicker one gets a MATH-ball neighborhood of the crossing. It is possible to move this ball-neighborhood around in MATH-space thus creating an axon of the neuro-link and the crossing neighborhood becomes the synapse. A self intersection point belonging to a singular link will give a singular synapse . A non singular synapse, is either positive or negative according to its contribution (MATH for positive, MATH for negative) to the linking matrix. See REF for examples of a link and a knot in neural positions, and REF for the distinguishing between the signs of non-singular synapse and also for an example of a singular synapse. Given two links, MATH, respecting the same DD diagram, D and the same linking matrix, we first bring both of them to neural positions. Suppose MATH are the MATH and MATH components of MATH respectively. The numbers MATH and MATH of both links are given by the diagram MATH. First, we will make sure that modulo double crossing change, both MATH and MATH have the same number of synapse involving each pair. Since the links respect the same diagram we do not have to worry about singular synapses. The number of them is given from the diagram. suppose that the pair MATH has more non-singular synapses than MATH. Since the links MATH have the same linking matrix, and the fact that each synapse adds MATH to the linking matrix we have that, MATH . It follows that we can find a set of synapses involving MATH representing this difference which can be divided into pairs, each of them consists of one positive and one negative synapse. Every such pair gives us a double crossing flip that allows us to cancel those synapses and hence decreases the number of synapses between MATH by MATH. We repeat this argument for every pair of components of MATH. This way we change MATH into a link, MATH, every two components of which has the same number of synapses in the corresponding pair of components in MATH. In order to finish the proof we still have to show that we can find a sequence of double flips making MATH . The same as MATH. Since an axon is made of two strings going in two opposite directions it is transparent to other axons and circles in the sense that we can flip it through axons and circles using only double crossing flips. Therefore the combinatorics of the setting which is defined entirely by the diagram determines the topology. MATH . The links MATH and MATH have the same combinatorial setting and the axons of MATH can be moved freely, (up to double crossing flips), in space onto the axons of MATH so they become isotopic links. This completes the proof. |
math/9906138 | The proof of this theorem is basically identical to that of REF. The only difference occurs when we reduce the number of axons in MATH so that it would be the same as for MATH. as already mentioned there, we cancel a pair of axons by the use of double crossing flips. In general these crossings do not have to be close. If the two axons are adjacent then those crossings are close. If they are not and we have two axons separated by a third one, as below, then we slide one of them as indicated and make the double crossing flip encircled which is a wedge type flip, so that they become adjacent. MATH . |
math/9906138 | The invariants of type MATH are the constants. There is a unique NAME invariant of type MATH for knots, which is MATH (the second coefficient of the NAME invariant). Therefore for a link with MATH component, we have MATH each one for each component of the link. The extra invariant in MATH is the NAME MATH. If MATH then we also have MATH such invariants. The list of finite type invariants given above, consists of the invariant of type MATH which come from NAME invariant of links and tangles. This list gives lower bounds on the dimensions claimed in the theorem. The completion of the proof is done by calculating the dimensions of the space of double dating diagrams up to degree MATH. This calculation is the content of the next theorem. |
math/9906138 | The proof of the theorem is done by using REF-T, REF-T and REF-T which give sequences of equalities between double dating diagrams. Those equalities are given in REF. |
math/9906138 | In order to prove the theorem we introduce another space MATH having MATH as a subspace. Further we will define a map MATH from MATH into MATH which extends MATH and we will show that MATH is mapped into the subspace of strutless diagrams. The objects of MATH are diagrams having trivalent vertices some of which external where a chord meets a circle component of the diagram, and some are internal where three chords meet. Some of the internal vertices are marked by MATH. The chords that meet at an internal marked vertex are required that at least two of them are attached to the same circle-component of the diagram. those two chords are also signed by MATH signs. In REF there is an example of a diagram in MATH. Now define MATH by MATH and if a diagram contains trivalent vertices non of them is a marked vertex, then it is mapped to zero. Notice that the following holds. MATH . Suppose now that there is a marked vertex for which there are other strands touching the circle where the two end points are, on the arc between those two points. Then after applying MATH as much as necessary, the following equality. MATH . Hence the only thing to worry about is an object as in the right hand side of the previous equation. The claim is finally proven by the two following equalities. MATH . On the other hand, the fact that MATH is onto MATH follows from the following equality. MATH . Using STU we have that every strutless diagram is a sum of diagrams in which every trivalent vertex is adjacent to a circle as in the figure, and each such a diagram is in the image of MATH . |
math/9906147 | Without any loss of generality one can suppose that the curve is reduced. Therefore one can choose the infinite line in the complex projective plane MATH so that it intersects the curve at MATH different points. Let the affine part of the curve MATH be given by the equation MATH (where MATH is a polynomial of degree MATH). The surface MATH is nonsingular and the negative inertia index of the intersection form on its second homology group MATH just coincides with the right hand side of the inequality in the statement this follows, for example, from the result of NAME on the intersection form for quasihomogeneous singularities CITE, applied to the (isolated) singularity MATH, where MATH is the homogeneous part of degree MATH of the polynomial MATH. Now the statement follows from the facts that the intersection form of the MATH - singularity of MATH variables is negative defined and, if the considered curve has a MATH - singularity, then the vanishing homology group of this singularity is embedded into the homology group MATH (see, for example, CITE). |
math/9906147 | Let MATH, MATH, and let an (affine plane) curve MATH be given by the equation MATH where MATH. The degree MATH of the curve MATH is equal to MATH. Let MATH (MATH and MATH are local coordinates on the plane near the origin). Then MATH . Since MATH terms of higher degree, one has MATH terms of higher degree. Substituting this expression into REF one gets MATH where MATH and the sum contains only members with powers MATH which lie over the segment MATH, that is, those for which MATH. This proves the statement. |
math/9906153 | Let MATH, where MATH. For MATH define MATH. It is clear that MATH. Define MATH if MATH and MATH otherwise. Define MATH to be the sum of all letters MATH such that MATH. Then form the following system of equations: MATH . There are MATH right linear equations in MATH unknowns satisfying the conditions of REF . Therefore they have a unique solution. |
math/9906153 | The semigroup defined is the set of equivalence classes of MATH with respect to the second two relations (that is, the NAME extension rules MATH and MATH) with a zero adjoined and multiplication of any two classes of MATH defined to be zero. |
math/9906153 | Define an automaton MATH where MATH, MATH and MATH is defined as follows: MATH . It is clear from the definitions that the extended state transition MATH is such that MATH if and only if MATH. Hence MATH. Therefore MATH is regular over MATH. |
math/9906153 | Recall the (`target') functions MATH and MATH. We use the following definition to restrict sets to those elements whose `target' is MATH. MATH . Then define MATH be the set of irreducible forms of the terms MATH with respect to MATH. For each object MATH we define an incomplete non-deterministic automaton MATH with input alphabet MATH, and language MATH. This automaton rejects only the irreducible elements of MATH, that is, it accepts all terms that do not represent elements of MATH, terms that do not have `target' MATH and terms that are reducible by MATH. We will use the following notation: These are the set of all left hand side of rules, the set of all prefixes of left hand sides of rules and the set of all proper prefixes of left hand sides of rules respectively. Now define MATH where Let MATH so that MATH and MATH. Define the transition MATH by: MATH . The extended state transition function MATH is such that the intersection of MATH with MATH is non-empty if and only if MATH is an element of MATH which is not an element of MATH or is reducible. Thus for each object MATH, and automaton MATH can be constructed, where MATH. The results quoted in REF allow us to make MATH deterministic REF and take its complement. The language MATH recognised by the resulting automaton MATH is MATH, that is, MATH. Hence (by REF ) MATH is regular. Since MATH is a complete rewrite system on MATH there exists a unique irreducible term in each class of MATH with respect to MATH. Therefore the set MATH is bijective with MATH. The automaton MATH gives rise to a system of right linear language equations REF with a unique solution, which is a regular expression for the language MATH accepted by the automaton. The regular expression can be obtained by applying NAME 's Theorem REF to solve the language equations. Given that each set MATH is bijective with a regular language MATH, the action is described as follows: let MATH and MATH for MATH, then MATH. |
math/9906159 | We use additive notation for the group operation in MATH. Let MATH be a prime factorization. Then MATH, and MATH restricts to an automorphism of each of the factors. If MATH is odd, MATH is cyclic, so each order two automorphism of MATH sends a generator to its inverse. However, MATH, so if MATH, MATH has three automorphisms of order two: MATH, MATH, and MATH. Thus MATH is encoded by a pair MATH, where MATH, MATH, and MATH. For MATH, we simply have MATH. With this notation, we write MATH. Then MATH can be viewed as a product MATH where MATH is the (odd) ``MATH-eigenspace" for MATH, and MATH is the (odd) ``MATH-eigenspace". Since MATH fixes MATH, MATH for each odd MATH with MATH, so MATH. On the other hand, any element of MATH is a multiple of REF, so MATH may be normalized to be trivial mod MATH, as well. We may therefore assume that MATH, and we have a subextension MATH. These are classified by MATH, where MATH defines the module structure on MATH. CITE computes that if MATH, then MATH, so every extension is split. Thus if MATH, the main extension MATH is a semidirect product. Henceforth we assume MATH. We now have MATH, so MATH, as well. The split extensions are again semidirect products. But for each MATH, there is one non-split extension. Since MATH is central in MATH, the sequence can be written MATH so we may pretend for the moment that MATH, and focus attention on MATH. The structure of MATH depends on the sign of MATH. If MATH, then MATH is central, and MATH. If MATH, then MATH. By appending the MATH factor, we recover MATH in each case. |
math/9906159 | Recall that MATH, MATH, and MATH. A group MATH is said to be complete if MATH is trivial and every automorphism of MATH is inner. For MATH, MATH is complete (see CITE). Alternating groups are never complete: for MATH, MATH, and conjugation by the transposition MATH is an automorphism of MATH which is not inner in MATH. Claim: MATH. We know MATH, since any conjugation by an element of MATH leaves MATH invariant, and since MATH is trivial. MATH has the presentation MATH. Any automorphism must send MATH to an element of order REF, and MATH to an element of order REF. Since MATH contains REF elements of order REF elements of order REF, we see that MATH. But MATH. Claim: MATH. This follows by an argument similar to the preceding one: the presentation MATH gives an upper bound MATH. But MATH, a group of order REF. Thus MATH, with the outer automorphism realized by conjugation by a transposition in MATH or MATH. MATH is trivial, since MATH is complete. Now, each of MATH, MATH, and MATH has trivial center, so in each case, MATH. Thus if an abstract kernel admits a realization, it will be unique. Extensions do exist in all cases: the two extensions of MATH have MATH and MATH; the extensions of MATH have MATH and MATH; and the unique extension of MATH has MATH. All are split extensions. |
math/9906159 | We use additive notation for the group operation in MATH, but multiplicative notation in the (possibly nonabelian) group MATH. Let MATH generate MATH. MATH is generated by MATH, together with MATH and MATH, where MATH and MATH. We assume MATH and MATH are normalized so that MATH and MATH lie in the NAME REF-subgroup of MATH, and CASE: MATH is either REF, or a generator of this subgroup. CASE: If MATH, then MATH is either REF, or a generator. CASE: If MATH, then either MATH, or it has order two. With this in mind, the problem easily reduces to the case MATH, since the restriction of MATH to the subgroup MATH of MATH is known. We assume henceforth that MATH is a power of REF. If MATH, the sequence splits, so we assume MATH. Once MATH is known, MATH is determined by the automorphism MATH of MATH. In the proof of REF , we observed that MATH is described by a pair MATH, where MATH, and also that if MATH, then MATH splits. So we may assume MATH and consider the various cases for MATH. Note that MATH for some MATH, and that MATH. Hence MATH is determined mod MATH by MATH. If MATH, then the argument cited in REF shows that MATH splits if MATH. So we assume MATH, depending on MATH. If MATH, then MATH. If MATH, then MATH, and MATH, so MATH splits. If MATH, we have several cases: CASE: If MATH and MATH, then MATH. MATH contains three elements of order REF, but all are contained in MATH, so the sequence MATH does not split. CASE: If MATH and MATH, then MATH. CASE: If MATH and MATH, then MATH, and MATH, with MATH. Although MATH contains involutions, the extension MATH does not split if MATH. CASE: If MATH and MATH, then MATH, so the sequence splits. |
math/9906159 | We use the commutative diagram MATH where the vertical maps are induced by the coefficient reduction. The right-hand map is an isomorphism for all the subgroups except for MATH, where MATH. The fact that MATH and MATH are squares of one-dimensional classes makes calculation easy for every column except MATH. In that case, since the generator of MATH is indecomposable, the map MATH is zero. It follows immediately that each of MATH and MATH is either MATH or MATH. Determining which actually occurs requires a closer look at the NAME spectral sequence. Details are in CITE; our calculation later on actually only requires that each be an even multiple of MATH. |
math/9906159 | The main claim that needs to be established is the exactness of the following sequence: MATH . Since a rotation in MATH is homotopic to the identity, it is clear that any MATH is in the kernel of MATH. For the reverse inclusion, suppose MATH. Since MATH acts trivially on homology, it has a fixed point MATH. A fairly standard calculation shows that the only sections of MATH with sectional curvature MATH are those tangent to the two factors. Since MATH is an isometry, it preserves the splitting of MATH, so it acts on MATH by a pair of rotations MATH and MATH. Thus MATH agrees with an element of MATH on a REF-dimensional subspace of MATH. Since MATH, MATH. This finishes the proof of exactness. Now, the quotient MATH is easily seen to be generated by MATH, the isometry which acts by the antipodal map in each factor, and MATH, which switches coordinates. MATH is central in MATH, so MATH extends MATH with a direct product. However, MATH is not central, so the product is only semidirect. |
math/9906159 | Since MATH acts trivially on homology, we must have MATH . Let MATH. Since each of MATH and MATH preserves orientation on MATH, each has a fixed point on MATH. By the assumption of pseudofreeness, each fixed set must be REF-dimensional unless MATH. Claim: Let MATH and MATH denote the two obvious projections from MATH. Then MATH . Observe that for any MATH, there is a unique MATH so that MATH. For if MATH and MATH are both in MATH and MATH, then MATH. But MATH, so it has two fixed points on MATH. But then MATH, contradicting pseudofreeness. Now we can define MATH by MATH . MATH is clearly a homomorphism, and by symmetry, an isomorphism. This proves the claim. To simplify notation, write MATH and MATH. MATH acts pseudofreely on MATH, so it must be a polyhedral group. Moreover, the projection MATH must be injective, since if MATH, then MATH, and MATH also. Similar considerations apply to MATH. Now, it is well known (see CITE) that any two isomorphic finite subgroups of MATH are conjugate. That is, there is a change of coordinates with respect to which we actually have MATH. Without loss of generality, assume we have applied it. (In doing so, we have fixed, once and for all, a particular identification between the two factors of MATH. For later reference, this also defines a particular choice of coordinate switch MATH. Abstractly, MATH is only well-defined mod MATH.) Then for any MATH, we have MATH. |
math/9906159 | We begin with some observations about the geometry of MATH. Suppose MATH and MATH are nontrivial elements of MATH. Each has a well defined axis of rotation - we denote them respectively by MATH and MATH. Observe that MATH for some MATH if and only if MATH leaves MATH invariant. If MATH fixes MATH, then MATH, and MATH. If MATH inverts MATH, then MATH, and MATH is an order REF rotation. In this case, MATH. REF tells us that any MATH with MATH has the form MATH, where MATH. Also, an element MATH acting homologically trivially and pseudofreely has the form MATH, where neither coordinate is MATH. It follows from the previous observations that if MATH and MATH for some MATH, then MATH and MATH are both nontrivial order REF rotations, and MATH fixes a torus. So if MATH acts pseudofreely, any MATH with MATH and MATH must be central in MATH. Similar considerations show that a MATH of order greater than REF leaves invariant a unique axis in each factor, so at most one cyclic subgroup of MATH can be normalized by MATH, and in fact, the generator of such a group commutes with MATH. Thus for each MATH, one of the following holds: CASE: MATH has order REF, and MATH is trivial. CASE: MATH has order greater than REF, and MATH normalizes (at most) a single maximal cyclic subgroup of MATH, which it centralizes. Now consider the possibilities for MATH: CASE: MATH. The only groups MATH satisfying REF are MATH and MATH. If MATH is odd, these two groups are isomorphic, and the extension is realized by choosing MATH, where MATH has order REF and shares an axis with the generator of MATH. If MATH is even, this construction realizes the case MATH. MATH is realized if both MATH and MATH are order MATH rotations. Note that if MATH is odd, the latter construction still yields a MATH action, but if MATH is odd, then MATH fixes a torus. CASE: MATH. If some MATH exists satisfying REF , then MATH has order REF but fails to satisfy REF . If no MATH satisfies REF , then MATH, and either MATH, MATH, MATH, or MATH has both MATH and MATH nontrivial, and thus fixes a torus. CASE: MATH, MATH, or MATH. By REF , the sequence MATH splits, so we may pick MATH of order REF. By REF , MATH is trivial. Now let MATH be an involution. Then MATH also has order REF, but MATH is nontrivial, contradicting REF . |
math/9906159 | As a special case of NAME 's theorem CITE, an involution MATH on MATH with MATH on MATH has a fixed point set MATH with MATH. If the sequence MATH splits, then MATH contains such an involution. The second statement follows from the observation that for MATH, Fix-MATH . Fix-MATH. |
math/9906159 | This follows immediately from REF . |
math/9906159 | We know from REF that if MATH splits, then MATH cannot act pseudofreely. It follows from REF that MATH must be of type REF or REF (and that MATH is even). Recall from the beginning of REF that the structure of MATH is determined by the data MATH and MATH. By REF , a hypothetical MATH has the form MATH, where MATH, and MATH. Our construction of a pseudofree MATH-action uses two ingredients: CASE: A choice of a particular semi-diagonal MATH-action, which makes it possible to realize MATH as MATH. CASE: An appropriate choice of MATH which ensures that MATH. As in the proof of REF , conjugation by MATH must either fix or invert the axis of rotation of MATH in each factor, so MATH is either MATH, or simply MATH. For simplicity, we will assume that MATH and MATH share the axis of MATH, so MATH. The construction can also be carried out with minor modifications if MATH. A semi-diagonal embedding of MATH into MATH takes the form MATH for some MATH. Observe that MATH, and also that MATH. Since MATH, choosing MATH allows us to realize MATH as MATH: MATH . Next observe that MATH. Thus if we choose MATH and MATH so that MATH, then MATH (recall that MATH fixes MATH). It follows from the second part of REF that the extended action is still pseudofree. |
math/9906159 | We know it is necessary that the sequence not split. To prove the converse, we assume it does not split and we construct a pseudofree MATH-action. Let the data MATH and MATH be given. As in the previous proof, MATH, and we will choose MATH and MATH appropriately. In this case, choose any MATH and MATH so that MATH, and each MATH leaves MATH invariant. (For example, simply take MATH.) Let MATH, and let MATH embed MATH in MATH via MATH. We show that the subgroup of MATH generated by MATH and MATH is isomorphic to MATH, and then appeal to REF to see that the resulting action of MATH is pseudofree. For the first claim, it suffices to verify that MATH and that MATH. We have MATH. On the other hand, MATH, because MATH fixes MATH. Note that as a consequence, MATH. For MATH, MATH. But MATH. |
math/9906159 | Let MATH and MATH, and suppose an extension MATH exists, with MATH acting pseudofreely. Let MATH, and MATH. By REF , we know that MATH must be cyclic of even order, with MATH or MATH, and that the sequence MATH must not split. REF describes those groups MATH which satisfy these criteria. If MATH, the geometry of MATH puts one more restriction on MATH. Let MATH, where MATH. Then MATH, where MATH. If both MATH and MATH are nontrivial, then MATH fixes a torus. If both are trivial, then the product of MATH with the unique element of order two in MATH fixes a torus. So exactly one of them is nontrivial. Suppose it is MATH. The element MATH has the form MATH, so MATH. Thus MATH and MATH can not commute in the linear case. REF then shows that MATH necessarily has one of the forms given in the statement of the theorem. We must now construct pseudofree actions of these groups. For simplicity, we assume all rotations used in the constructions have the same axes. CASE: MATH. We start with an action of MATH as provided by REF . Extend MATH (the semidiagonalizing automorphism of MATH) to an automorphism of MATH. Let MATH, where MATH is a generator of the NAME REF-subgroup of MATH. Then MATH acts pseudofreely. We need only verify that MATH to conclude that the subgroup of MATH generated by MATH and MATH is isomorphic to MATH: MATH . CASE: MATH with MATH. Again, start with an action of MATH as in REF . Let MATH, where MATH is an order two rotation. Then MATH, so MATH. And we have MATH, chosen so that MATH generates REF-subgroup of MATH. To show that MATH, we observe that MATH as required. NAME of the extended actions follows from REF , as usual. |
math/9906159 | We have already seen that REF is necessary and sufficient in the homologically trivial case, and that REF is necessary and sufficient in the MATH case. REF is necessary by REF (NAME 's theorem). On the other hand, this condition implies that MATH does not split, and REF show this to be sufficient in the MATH case. Finally, REF enumerates those groups MATH with MATH which satisfy the first three conditions, and REF shows that the last condition is necessary and sufficient to finish the classification. |
math/9906159 | It is shown in CITE that, without the pseudofree assumption, such an isotropy group MATH must be abelian of rank MATH or MATH. But a rank MATH group cannot act freely on the linking sphere to MATH, and thus cannot act pseudofreely on MATH. |
math/9906159 | Since MATH for some MATH, MATH. Observe that for any MATH, if MATH, then MATH. In other words, MATH acts on Fix-MATH. Two points of Fix-MATH are in the same MATH-orbit if and only if they are in the same MATH-orbit, for if MATH, MATH, and MATH, then MATH. But each MATH-orbit of the action on Fix-MATH has cardinality MATH. Since they all have the same size, the number of orbits must divide MATH. |
math/9906159 | Since the groups occur in pairs, MATH is even. Let MATH, and rearrange the list so that MATH, MATH, and so forth. Since MATH, it follows that MATH. As in the MATH case, the possible MATH are MATH, and MATH. In each case, these numbers represent the sizes of maximal cyclic subgroups of MATH, and each conjugacy class of maximal cyclic subgroups occurs in the list. In CITE, NAME describes the groups that can correspond to this data. Proofs are omitted for some of his assertions, and some details are different in our case and his, so we repeat the argument here. MATH clearly corresponds to a cyclic group of order MATH. For the remaining cases, we make the following observations: If a maximal cyclic subgroup MATH of MATH is also normal, it has index MATH in MATH. For MATH operates freely on Fix-MATH, a set of four points. But since orbits come in pairs, Fix-MATH must meet REF or REF MATH-orbits, so MATH or REF. By a similar argument, any maximal cyclic subgroup intersecting the center of MATH non-trivially has index MATH. MATH corresponds to a group of order MATH. It has a cyclic, index REF subgroup MATH of order MATH, which must be normal. NAME assumes that his groups operate on a space with MATH, and uses this fact to prove that each MATH must have order REF. In our case, we use the observation above. Now assume MATH is fixed. Since each MATH has order REF, MATH, so MATH. It follows that MATH is dihedral. In the MATH case, MATH. G is nonabelian. There are three nonabelian groups of order REF, and two of them contain elements of order REF. The third is MATH. In the MATH case, MATH has order REF and trivial center. With this in mind, a look at a table of groups (for example, CITE) easily shows that MATH. Finally, in the MATH case, MATH. NAME theory shows that G contains five copies of MATH, intersecting trivially, REF copies of MATH, and six copies of MATH. If any proper subgroup of MATH were normal, it would contain every conjugate of each element. Counting arguments show this to be impossible, so MATH is simple. Thus MATH. |
math/9906159 | Say MATH is nice if for each singular point MATH, MATH meets REF or REF MATH-orbits. If MATH is nice, then its isotropy groups are repeated in pairs. We assume inductively that every proper subgroup of MATH is nice, but that MATH is not. Then by REF , there is some MATH so that MATH; that is, MATH acts transitively on MATH. By REF , MATH is cyclic - say MATH, and Fix-MATH. Let MATH, MATH, and MATH. By minimality, MATH is generated by MATH, and MATH, and since MATH, MATH is normal in MATH. By minimality again, MATH for some prime MATH. Now, MATH acts freely on MATH, so MATH. Thus MATH is an extension of MATH by MATH or MATH. The remainder of the proof is an analysis of these extensions. Most can be ruled out by elementary group theory considerations. The two more difficult cases use arguments essentially due to NAME in CITE. In the following cases, consideration of the possible automorphism actions of MATH on MATH shows that some element of MATH must be central, and then that MATH is contained in a cyclic subgroup of order MATH, contradicting minimality. CASE: MATH, for MATH. CASE: Any non-split extension of MATH by MATH. CASE: MATH, for MATH. The case MATH must actually be abelian, so MATH is a direct product. MATH contains two cyclic subgroups of order REF. They intersect non-trivially, and therefore have the same fixed set. It follows that MATH must act semifreely. Two cases remain: MATH, and MATH, where MATH. Suppose MATH admits a non-nice action. MATH has seven cyclic subgroups. Since MATH does not act freely on MATH, their fixed-point sets are disjoint, and each has MATH stabilizer. Since MATH is abelian, it acts on each fixed-point set, so each constitutes an orbit. The action has NAME data MATH. Let MATH be MATH minus a small invariant neighborhood of the singular set, and let MATH. MATH is a compact REF-manifold with seven MATH boundary components MATH. The cohomology long exact sequence for the pair MATH (with MATH-coefficients) shows that MATH has rank REF. The covering MATH is classified by a map MATH. This inducesa map MATH, which factors through MATH. Since it factors, MATH. On the other hand, each MATH maps to a different subgroup of MATH under the natural inclusion, and hence each MATH corresponds to a different nontrivial element of MATH. Since MATH, each nontrivial element of MATH restricts non-trivially to MATH for some MATH. By the NAME theorem and the cohomology structure of MATH, each of these has a nonzero cube which maps to the top class of MATH. Thus MATH, a contradiction. A somewhat similar argument covers the remaining case. Let MATH, with MATH. The semidirect product automorphism must have order REF; otherwise MATH contains a cyclic subgroup MATH. Thus MATH, where MATH. MATH has MATH different subgroups of order REF, all of which are conjugate, so if an action exists, it has NAME data MATH. Define MATH and MATH as before. Then MATH has boundary consisting of five lens spaces MATH, and MATH, with associated inclusions MATH. Once again, the covering MATH is classified by a map MATH, with induced maps MATH. However, the cohomology calculation is just a bit subtler this time. For any coefficient module MATH, the transfer map gives an isomorphism MATH. With MATH coefficients, the ring MATH is generated by elements MATH and MATH, where MATH is the image of MATH under the NAME map. MATH therefore inherits a MATH-module structure from the action of MATH on MATH given by MATH. Thus MATH, and MATH, so the action of MATH on MATH is given by multiplication by MATH. This has the unfortunate consequence that MATH. To compensate for the MATH-action on MATH, we replace MATH with a twisted coefficient module MATH, where MATH acts by MATH. Note that MATH as MATH-modules, since the restriction of the MATH-action to its subgroup MATH is trivial. With this twisting, we observe: CASE: Restriction gives an isomorphism MATH. CASE: For MATH, the maps MATH are trivial. Now, since the coboundary map MATH is NAME dual to the augmentation MATH, we see that MATH consists of all MATH in MATH such that MATH. In particular, if MATH, then MATH, as well. But by the observations above, there are elements MATH which restrict trivially to each MATH, but non-trivially to MATH. This rules out the MATH-actions in question. |
math/9906159 | Let MATH act homologically trivially, so that MATH is exact. Claim: MATH acts freely on MATH. Let MATH, so that MATH generates MATH. If, for some MATH, MATH lies in the same orbit as MATH, then for some MATH, MATH. But MATH has NAME number MATH, so if it has a fixed point, its fixed point set must be REF-dimensional. For the same reason, MATH acts freely on the set of singular points in MATH, and therefore it identifies the paired orbits of REF . Choose a small MATH-invariant open neighborhood MATH of the singular set of the MATH-action, and let MATH. Let MATH. It follows from the claim that MATH is a manifold with boundary consisting of two or three lens spaces MATH. (These MATH's are exactly those which appear in pairs in the NAME data.) Since MATH is a cover of MATH, we can use the NAME spectral sequence REF to compute MATH. Note that CASE: Since MATH is finite, MATH for any free MATH-module MATH. CASE: In general, MATH, the submodule of MATH fixed by MATH. In our case, MATH acts by MATH on MATH, so MATH. CASE: No nonzero differentials enter or leave MATH. Thus MATH. Similar arguments show that, for each component MATH of MATH, MATH, and that the restriction MATH is given by the corresponding map on subgroups. By NAME duality in Y, we have an exact sequence MATH . This becomes MATH . And since the restriction and inclusion maps factor through MATH, we also have a sequence MATH . This sequence is not exact in general. However, from the previous sequence, it follows that MATH injects into MATH, and that MATH injects into a quotient of MATH. The remainder of the proof consists of cohomology calculations showing that this is impossible unless MATH is abelian and MATH is cyclic. CASE: MATH is cyclic. Since MATH maps onto MATH, MATH. Thus if MATH is nonabelian, the kernel of MATH is nontrivial. In the exact sequence MATH the image of MATH lies in the diagonal subgroup of MATH. If MATH is nonabelian, the kernel of the inclusion map does not. CASE: MATH, MATH, or MATH. To every polyhedral group MATH, there corresponds a binary polyhedral group MATH such that MATH is exact. The NAME spectral sequence (MATH, in this case) relates the cohomologies of the groups in this sequence. It shows, in particular, that MATH injects into MATH. But since each MATH acts freely on MATH, NAME duality shows that MATH. Thus MATH. By computing the sizes of MATH in each case, we find that MATH, MATH, and MATH. On the other hand, MATH. Finally, MATH, MATH, and MATH. In each case, MATH is too large to inject into a quotient of MATH. CASE: MATH. Recall that MATH and MATH . Since MATH, and MATH, MATH. Except in the cases MATH and MATH, it is immediate that MATH cannot inject into any quotient of MATH. We consider the remaining possibilities: Suppose MATH. Using REF , we find that the only nonabelian extension MATH has MATH, with MATH. Thus we have an exact sequence MATH where the three MATH subgroups are generated by MATH, MATH, and MATH. Since the restrictions and inclusions factor through MATH, REF shows that MATH. This contradicts exactness. MATH also has two abelian extensions: MATH, and MATH. In each case, the restriction and inclusion maps can be explicitly calculated, and the sequence is seen not to be exact. MATH is the only candidate for MATH which is abelian, but not cyclic. Finally, suppose MATH. Then we have MATH, MATH, and MATH. Consulting REF again, we see that MATH has matrix MATH relative to the bases discussed there. And MATH has matrix MATH, as is easily checked. Now, if MATH is not onto, then the element counts which applied for most MATH also apply for MATH. But if MATH is onto, then MATH should inject into MATH, and hence we should have MATH. However, the element MATH is in the kernel, but not in the image. |
math/9906159 | Given the necessity of REF , and REF , it suffices to rule out the possibility that MATH, where MATH. In the case of the linear models REF , the argument divided into two parts: If MATH, then MATH fixes a torus, contradicting pseudofreeness. And if MATH or MATH, then the fact that MATH exchanges factors of MATH means that MATH and MATH cannot commute, so no action exists, even with a two-dimensional singular set. Assume, then, that MATH and MATH commute. Now, MATH must act freely, so MATH is a manifold which inherits an action by MATH. As motivation, we note that the linear models for MATH are distinguished by the intersection forms (with MATH coefficients) of the quotient spaces: In MATH, the diagonal MATH maps to an embedded MATH with self-intersection MATH, so MATH. On the other hand, generators of second homology for MATH are given by the images of MATH, (where MATH is some point fixed by MATH), and MATH, (where MATH is any point in the first factor). Each of these has trivial self-intersection, so MATH. To complete the proof, we will show (just as in the linear models) that if MATH acts on MATH, then MATH. NAME 's theorem then guarantees a two-dimensional singular set. Let MATH and MATH denote the standard generators for MATH, and consider the cohomology spectral sequence of the covering: MATH. It follows from the multiplicative structure of the spectral sequence that the behavior of the entire MATH page is determined by MATH and MATH. At least one must be nonzero, since the sequence converges to the cohomology of a four-manifold. And since MATH acts on the quotient, the differentials must respect the induced action of MATH on MATH. Hence MATH, and MATH is generated by MATH. It is easy to check that MATH, so MATH is a permanent cocycle. This spectral sequence is identified by a homotopy equivalence with that of the fibration MATH (compare CITE), and under this identification, the cocycles which live to MATH are those in the image of MATH. Thus there is a class MATH which lifts to MATH. We claim that MATH. To see this from the homological point of view, let MATH denote the singular chain complex of MATH. Then MATH, and MATH. The covering projection induces MATH via MATH. Moreover, because MATH is a free MATH-module, there is a natural isomorphism MATH, so that MATH. Every class in MATH is integral. Thus we can choose a cochain MATH so that MATH in MATH and such that MATH is a cocycle representing MATH in MATH. (Note that MATH itself need not be an integral cocycle. For related discussion, see CITE.) Then MATH, and MATH represents an integral lift of MATH - say MATH. So MATH. But MATH is also an equivariant cochain and hence must be of the form MATH, where MATH. So MATH, and MATH represents MATH. Thus MATH, so MATH. A heuristic geometrical argument is more direct: the NAME dual of MATH can be represented by a surface MATH, and MATH will represent an integral lift of MATH, so MATH. The intersection points will be paired up by the MATH-action, so MATH, and MATH. |
math/9906160 | Suppose the sequence of constituent lengths starts with MATH where MATH is even. We will prove that this is followed by another short constituent. We have to prove MATH . Since MATH is even, we can consider the integer MATH . We will obtain REF from the expansion of the following expression with the generalized NAME identity REF: MATH . Note that the total weight of the commutators in REF is that of the left-hand side of REF: MATH . We now look at the resulting terms in REF. Since MATH is the first centralizer, the term MATH will vanish most of the time, except when MATH occurs in a weight that marks the end of a constituent. Keeping in mind the information we already have on the previous constituents, the sum reduces to MATH . NAME 's Theorem yields MATH by REF MATH . Therefore REF yields MATH as MATH. This is precisely REF. |
math/9906160 | Let the intermediate constituent have length MATH. Suppose we have already proved that this intermediate constituent is followed by MATH short ones ending in MATH, with MATH, so that there is a segment MATH in the sequence of constituent lengths of MATH. We will show that the next constituent is also short and ends in MATH. Let MATH. According to REF , take a homogeneous element MATH at the beginning of the MATH-th constituent before the intermediate one. What has to be proved is MATH . We assume this is not the case, and obtain a contradiction. We expand MATH here MATH because of our assumption on the initial segment of the sequence of two-step centralizers. Note that this will give us terms of weight MATH higher than the left-hand term of REF. We get, using REF and the fact that MATH, MATH . Here we have used several times NAME 's Theorem and its consequences. Now if REF does not hold we obtain a contradiction. In fact REF can fail in two ways. It might be that MATH is centralized by MATH. This means that the MATH-th constituent after the intermediate one is not short, so by REF it has length at least MATH as MATH. However, we would have just obtained a constituent of length MATH, a contradiction. As an alternative, it could be that MATH for some MATH, with MATH. We would then have a short constituent not ending in MATH; but we have shown that this constituent would be followed by a constituent of length MATH, and this is a final contradiction. |
math/9906160 | Deflating MATH times, according to REF , we may assume MATH. Write MATH, where MATH is a power of MATH (possibly MATH), and MATH. By REF , we know that the first intermediate constituent is followed by at least MATH short ones ending in MATH. Let MATH . If MATH, we have MATH. Therefore we obtain MATH . We will prove MATH so that the algebra is finite-dimensional, since constituents have length greater than MATH. This contradiction will give that MATH, so that MATH as claimed. We consider MATH . Here MATH is odd, because of REF . Note that MATH so that MATH has the same weight as the commutators in REF. We expand, using REF and the information about the constituents, MATH . Now MATH as MATH for MATH. As to the second summation, we have MATH . The total coefficient of MATH is thus MATH as claimed. |
math/9906160 | By REF and induction, we may assume that we have the subsequence MATH in the sequence of two-step centralizers of MATH, for some MATH. Let MATH, so that MATH, and let MATH be a homogeneous element at the end of the MATH-th two-step centralizer before the long one, so that MATH. We have to prove MATH . We have MATH, from our assumption on the initial segment of the sequence of two-step centralizers of MATH. We compute with REF MATH . |
math/9906161 | Suppose first that MATH and MATH, MATH are as above. By shrinking MATH if necessary, we can make sure that MATH implies MATH for every MATH . By shrinking MATH further if necessary, we can arrange that the exponential map of MATH at MATH identifies a neighborhood MATH of the origin in MATH and MATH. Moreover, the elements MATH for which MATH, are identified with MATH, since these MATH are totally geodesic. This proves that MATH is locally linearizable. Conversely, if MATH is locally linearizable, then the choice of an inner product on MATH induces a metric on MATH, hence on MATH via the diffeomorphism MATH, and as linear subspaces of MATH are totally geodesic with respect to this metric on MATH, the same holds for MATH over MATH. |
math/9906161 | We claim that for any MATH REF-tuple MATH, MATH, is a normal collection in the sense of NAME. Indeed, this is clear if MATH is disjoint from MATH; otherwise MATH by our assumption. So assume that MATH. By the normality of MATH, near any point MATH in MATH there are local coordinates MATH, MATH, on MATH such that MATH and MATH, MATH. Similarly, by the normality of MATH there are local coordinates MATH, MATH near MATH on MATH such that MATH . Thus, MATH . Thus, the differentials of the coordinates MATH, MATH, and MATH, MATH, span the conormal bundle of MATH. The same holds for the differentials of MATH, MATH, MATH, MATH. It follows that the differentials of MATH, MATH, MATH, MATH, MATH, MATH, MATH, MATH are independent at MATH in a coordinate neighborhood of MATH, so these functions give local coordinates on MATH near MATH in terms of which MATH, MATH, MATH and MATH have common product decomposition: MATH, MATH and MATH given by REF, and MATH by MATH . This proves that MATH is indeed a normal collection. Hence, by CITE, it lifts to a normal collection of p-submanifolds on MATH. Writing MATH for the blow-down map, and MATH for the lift of MATH, etc., we see in particular that MATH is a normal collection whose intersection is MATH if MATH, and is empty otherwise. Putting together these facts we see that we have proved the lemma. |
math/9906161 | Since any total order compatible with inclusion can be obtained from any other one by repeatedly interchanging the order of adjacent elements, but keeping the order compatible with inclusion, it suffices to show that MATH are naturally isomorphic if both of these total orders respect inclusion. Now, either MATH, in which case the statement is clearly true, or MATH for some MATH. Since inclusion is respected, we must have MATH. But upon the blow up of their intersection, any two closed p-submanifolds with normal intersection lift to be disjoint. Hence, on MATH the lifts MATH and MATH are disjoint, and thus they can be blown up in either order. This proves the lemma. |
math/9906161 | It suffices to show that MATH lifts to be an element of MATH where MATH is minimal with respect to inclusion and MATH . Taking into account that for any MATH, MATH is a normal collection of p-submanifolds of MATH, this claim follows from CITE, or it can be checked directly by using projective coordinates on MATH. |
math/9906161 | By the first part of the statement, for any MATH, MATH, the commutator MATH is in MATH. |
math/9906161 | It is convenient to use the explicit description of MATH in terms of localization and quantization REF. Thus, we may assume that MATH. Note that the pull-back of MATH to MATH is a polyhomogeneous symbol of order MATH which we denote by MATH. Then the kernel of MATH is MATH where MATH is the kernel of MATH. But by the fundamental theorem of calculus MATH and MATH is a polyhomogeneous symbol of order MATH. Taking into account the rapid decay of MATH in MATH we immediately conclude that MATH vanishing with all derivatives at MATH, so, returning to the global setting, MATH. |
math/9906161 | This result essentially reduces to the fact that MATH is an algebra. Indeed, write MATH, and note that MATH where MATH denotes the composite of MATH with the multiplication operator MATH by MATH. Since the latter is in MATH, hence in MATH, we conclude that MATH. Thus, we only have to analyze MATH. Again, we can reduce the discussion to a local one. But writing MATH as the left quantization of a symbol MATH as in REF, MATH satisfying REF with MATH replaced by MATH, and writing the oscillatory integral explicitly as a convergent integral, we see that MATH for MATH, MATH. Changing the variables: MATH . This is a convergent integral with MATH dependence only in MATH. Since MATH we conclude that MATH. Hence, returning to the global setting, MATH as claimed. |
math/9906161 | We prove this statement by finding MATH explicitly in terms of local coordinates. To simplify the notation we assume that MATH. We identify MATH with MATH locally so that MATH is given by MATH, MATH. In the interior of MATH we can use the same coordinates as at the front face of MATH, that is, the ones given in REF. So suppose that MATH is supported in the region of validity of these coordinates. Then MATH with MATH as in REF. Here the integral is understood as a distributional pairing in general, but it actually converges if MATH. We now consider the coordinates REF on the both factors, that is, we take MATH corresponding to MATH, and MATH corresponding to MATH. Expressing MATH in terms of MATH and MATH (using MATH) gives MATH where we wrote MATH and MATH, MATH denote the components of MATH and MATH respectively. Thus, REF yields MATH . Evaluating at MATH gives MATH . Since MATH is the inverse NAME transform in the MATH variable of the symbol MATH whose left quantization is MATH, and since the MATH integral above can be understood as the NAME transform in MATH evaluated at the origin, we deduce that MATH . Thus, the indicial operator MATH where MATH is the zero covector above MATH is given by MATH that is, by the left quantization in MATH of MATH. Similar results hold for MATH in general, namely MATH . Though the local coordinates are only valid in the interior of MATH, hence not at MATH, the continuity of MATH up to MATH shows that REF also holds with MATH. The explicit expression, REF shows, in particular, that MATH is indeed independent of the extension MATH of MATH that we chose, and also of the choice of MATH with MATH prescribed at MATH. Moreover, also from REF, for each MATH, MATH, MATH here we wrote MATH for MATH for simplicity. In fact, REF shows the more precise statement which encodes the smooth dependence of MATH on MATH, namely that MATH . In the Euclidean setting the many-body space MATH can be identified with MATH, and we can write MATH and correspondingly MATH as we have claimed. |
math/9906161 | Choose MATH with principal symbol MATH and indicial operators MATH. Note that these match up under the projections MATH since those of MATH match up. Thus, they indeed specify the restriction of some MATH to the boundary of MATH. Quantizing MATH gives an operator MATH with the required indicial operators and principal symbol. Hence, MATH has vanishing principal symbol and indicial operators, so it is in MATH. Summing the NAME series MATH asymptotically to some MATH and letting MATH gives the required left parametrix. A right parametrix can be constructed similarly, and then the usual argument shows that they can be taken to be the same. |
math/9906161 | Since MATH is bounded, we can replace MATH by a function MATH such that MATH on the spectrum of MATH. Now MATH is invertible for MATH, so MATH for MATH. Again, MATH can be analyzed uniformly up to the real axis, and then the NAME integral representation of MATH now proves the proposition. |
math/9906161 | Suppose that MATH and MATH. With MATH as above, let MATH, so MATH as MATH. Then MATH and MATH. With MATH as above, let MATH. Then MATH and MATH . But MATH by assumption, so by REF MATH . Hence, there exist MATH (in place of MATH), MATH, etc., as in REF , MATH, and the indicial operator of MATH at MATH is just the composite of those of MATH and MATH, hence invertible, showing that MATH. |
math/9906161 | As MATH in MATH, we have for all MATH that MATH . Thus, with MATH, the previous proposition implies that for any MATH, MATH is differentiable from both the left and the right at MATH, and both of these derivatives are equal to MATH. (We remark that this is proved directly in the Appendix as a first step to the proof of the proposition.) Thus, MATH is MATH and it satisfies the ODE MATH. But, given say MATH, this ODE has a unique solution which is MATH. The last statement follows by writing down the solution of the ODE explicitly, which, if MATH for some MATH, takes the form MATH, MATH, for an appropriate constant MATH. |
math/9906161 | The previous corollary and the above remarks show that for all MATH, MATH. Let MATH. Thus, MATH vanishes at MATH for all MATH. Since the base variables MATH and MATH are MATH-invariant, we conclude that MATH vanishes identically, hence MATH is constant, and similarly for MATH, proving that MATH for all MATH. The last statement of the proposition follows since MATH vanishes at MATH. |
math/9906161 | Let MATH . REF implies REF immediately. Let MATH be a MATH-invariant function. Let MATH here we slightly abuse the notation and write MATH. Then MATH is MATH-invariant, so REF applies and gives MATH. Since MATH, the chain rule and a short calculation of MATH gives REF . The first equation in REF follows since along MATH, MATH. As MATH, the second equation follows as well. Since MATH along MATH as MATH and MATH is decreasing, we deduce the last statement. |
math/9906161 | Our strategy consists of constructing a MATH-invariant function MATH with MATH in a neighborhood of MATH. Thus, by REF , MATH for MATH, MATH sufficiently small, so MATH is increasing there. This will allow us to draw the desired conclusion for the correct choice of MATH. We remark that this MATH will reappear in the proof of the propagation estimate in REF . Moreover, it is essentially the same as the corresponding function in the three-body propagation estimate CITE, though we will use slightly different methods to estimate MATH. In fact, first we find a MATH-invariant function MATH such that MATH will be appropriately small near MATH. So introduce coordinates centered at MATH as after REF . Then the metric function takes the form MATH with MATH and, due to the assumption that MATH is totally geodesic, MATH . We write MATH for the restriction of the tangential part of the metric function to MATH, so MATH . Now, the NAME vector field of MATH is given by MATH with MATH. Hence, if MATH then MATH . Now, MATH, hence MATH, is small near MATH, so to model MATH we introduce the vector field MATH locally (near MATH) on MATH. Thus, we have MATH which is small if MATH is small. We define MATH as follows. First, MATH, and MATH since MATH, so near MATH, MATH, that is, MATH is transversal to the hypersurface MATH. Thus, near MATH in MATH we can solve the NAME problem MATH . Since MATH and MATH vanish at MATH, the same holds on the bicharacteristic of MATH through MATH, but MATH and the Hessian is still positive in directions transversal to the bicharacteristics as these hold at MATH. Moreover, by CITE, MATH . Let MATH so MATH. At MATH we have MATH, so MATH when MATH, and then MATH implies that this inequality holds everywhere. Therefore, MATH . Now, MATH . Thus, using REF, for some MATH we have MATH . Next, note that MATH so by REF, MATH . For MATH let MATH . Thus, MATH . We next estimate MATH. First, as MATH, MATH is positive definite in a small neighborhood of MATH and MATH there. On the other hand, MATH so MATH . Moving MATH to the right hand side and completing the square gives MATH so MATH . We can finally estimate MATH, using REF as well: MATH . Note that MATH, so near MATH, MATH is small. So now suppose that MATH and MATH . Then MATH so MATH, MATH. Hence, under the additional assumption MATH that is, that MATH sufficiently close to MATH, we have MATH . Since MATH, we have MATH in a neighborhood of MATH. Now choose MATH sufficiently small, so that MATH. Note that this requirement is independent of MATH. We thus conclude that for MATH, MATH satisfying REF, we have MATH . Now, using the result of REF , let MATH be the unique points such that MATH and for all MATH-invariant MATH . Choosing a sufficiently small open interval MATH around MATH, MATH, hence MATH, automatically satisfies REF for MATH, while REF holds automatically as MATH. Thus, applying REF with MATH in place of MATH, we see that, with MATH we have MATH . As MATH is continuous and MATH, this shows that MATH is increasing on MATH. To see this, first note that MATH on MATH, for otherwise MATH is not empty, MATH is closed, so taking MATH and MATH. Thus, for MATH, MATH is differentiable from either side at MATH and the derivatives are both positive, so MATH is increasing on MATH, hence MATH contradicting MATH. Thus, MATH on MATH, so MATH is increasing here, so MATH for MATH. Taking into account the definition of MATH we immediately deduce that MATH . Since MATH is arbitrary, we conclude that MATH for MATH, so MATH for such MATH. Similarly, MATH for such MATH, so by the construction of MATH, MATH is the integral curve of MATH through MATH (for MATH, MATH). Of course, a similar argument (with a change of sign in MATH in REF) works for MATH, so we conclude that MATH and MATH is an integral curve of MATH. As MATH preserves MATH (being essentially its rescaled NAME vector field), MATH, MATH, so MATH, and hence at MATH, MATH and MATH agree and MATH is a bicharacteristic of MATH as claimed. |
math/9906161 | Let MATH be as above and introduce local coordinates centered at MATH. We may assume that MATH is given by MATH for a suitable splitting MATH. Thus, MATH is of the form MATH, and as MATH, MATH. By REF , taking into account that MATH is MATH-invariant, MATH . Since MATH, there exist MATH, MATH, such that MATH for MATH, while for any MATH there exists MATH such that MATH for MATH. In particular, for any MATH there exists MATH such that for MATH we have MATH. By choosing MATH sufficiently small we can thus make sure that MATH for MATH if MATH. Hence, MATH can be regarded as a curve in MATH, MATH taken with respect to MATH, if we let MATH. Of course, MATH is a generalized broken bicharacteristic, broken at MATH (since it has no points above MATH). Thus, by REF , MATH extends to a generalized broken bicharacteristic, broken at MATH, defined on MATH; by continuity of MATH this must coincide with MATH, so MATH is a generalized broken bicharacteristic, broken at MATH, as claimed. |
math/9906161 | If MATH then MATH for MATH near MATH by REF , hence near MATH, MATH is a (MATH-projected) bicharacteristic of MATH (as MATH vanishes at MATH). If MATH then REF applies and proves the result. If MATH, then with MATH as in REF , MATH is a generalized broken bicharacteristic, broken at MATH, with MATH (prime taken with respect to MATH). Thus, REF applies again and proves the result. |
math/9906161 | Let MATH . Thus, MATH, so MATH. Since the spectrum of MATH is a subset of MATH and MATH, we have MATH where MATH and MATH if MATH is in the spectrum of MATH. By REF , MATH . Let MATH be identically MATH near MATH and vanish near MATH. Then MATH . Now let MATH be identically MATH near MATH and vanish near MATH. Let MATH . Multiplying REF from both sides by MATH then proves REF. |
math/9906161 | We apply a parameter dependent version of the previous lemma to the indicial operators to conclude that for each MATH there exists MATH with MATH . It follows from the NAME integral formula construction of the square root in the calculus and the explicit formulae REF that the indicial operators MATH match up so that there exists MATH with indicial operators MATH. Here note that the set where MATH does not vanish has compact closure, hence MATH is bounded below on it by a positive constant. Thus, we can take the same smooth function MATH in REF for the square root for every MATH and MATH. By REF, MATH with MATH. Since MATH, rearranging this proves the proposition. |
math/9906161 | We define MATH if MATH, otherwise we define MATH as in the previous proposition. The only additional ingredient is the analysis of MATH near MATH with MATH. To do this analysis, we follow the construction of MATH in detail. So let MATH and let MATH . Thus, MATH. Let MATH . By our assumption, there exists MATH such that the norm of MATH in MATH is bounded by MATH. Now choose MATH such that MATH on MATH. Then MATH, so MATH . By our assumptions, the seminorms of MATH in MATH, MATH, remain uniformly bounded as MATH, so the NAME integral representation of MATH, via an almost analytic extension, shows that MATH remains uniformly bounded. Thus, MATH is continuous as a function on MATH with values in MATH. A similar argument also holds for the derivatives of MATH. Let MATH be identically MATH near MATH and vanish near MATH, and let MATH . Again, the MATH match up so there exists MATH with these indicial operators. We can also make sure that the lower order terms also vanish where MATH does, that is, that MATH. Then the indicial operators of MATH and MATH are the same, so MATH with MATH, proving the proposition. |
math/9906161 | First, MATH follows from REF. Moreover, the infinite order vanishing of MATH at MATH implies that for every MATH, MATH and MATH, MATH . Thus, NAME 's rule shows that for MATH acting in MATH, MATH acting in MATH, MATH and MATH . But this means precisely that MATH and it vanishes to infinite order at the boundary in the last factor. |
math/9906161 | This can be proved directly from the definition of the indicial operators, that is, by computing MATH where MATH and MATH, similarly to CITE. Since this is equal to MATH, and MATH, we can assume that MATH, the calculation being very similar in the general case. To compute the commutator, it suffices to commute both MATH and MATH for every MATH modulo terms that vanish with their first derivatives in MATH. A straight-forward calculation can be performed just as in REF, where only the REFth order terms were kept. That shows with our coordinates that MATH . Here MATH denotes the operator with kernel given by MATH applied to that of MATH. Since in our notation the kernel of MATH is MATH with the integral being convergent, rewriting this with the coordinates on the compactification MATH, REF, so that MATH takes the form MATH proves that all terms of REF satisfy the stated estimate, completing the proof. Another approach to compute MATH-indicial operators is to use that near MATH, MATH can be regarded as a (non-classical!) pseudo-differential operator in the free variables MATH with values in bounded operators on MATH (in fact, with values in MATH). More precisely, MATH. This allows us to use the scattering calculus for the computation of the commutators to give the stated result. |
math/9906161 | Note that the estimate REF is trivial if MATH (with MATH, MATH arbitrary) since then both sides vanish as MATH denoting the subsystem Hamiltonian as in REF, and MATH by the assumption on the absence of bound states of all subsystem Hamiltonians (including MATH with MATH). We prove REF by induction on MATH. First, REF is certainly satisfied for MATH. In fact, as MATH, we can use the commutator formula in the scattering calculus, REF, to find MATH. Since MATH vanishes at the free face, MATH, it does not contribute to MATH, so we indeed have, by REF, MATH away from MATH under the assumption that MATH . So suppose now that REF has been proved for all MATH with MATH, MATH. This implies that all indicial operators of MATH, MATH satisfy an inequality like REF. In fact, the indicial operators are of the form MATH with MATH, MATH. Such a MATH is of the form MATH where MATH is given by MATH, MATH, so MATH give coordinates along MATH. Note that as MATH is compact, so is MATH and as MATH is independent of MATH at MATH, MATH, so we can apply the inductive hypothesis. Taking into account that the estimate REF is trivial at MATH for MATH with MATH, we see that for all MATH with MATH, we have MATH . Since MATH is MATH-invariant on MATH, it is independent of MATH for each fixed MATH, and if it vanishes at MATH, then so does MATH by REF. Thus, by REF , MATH where the seminorms of MATH are bounded by those of MATH and by MATH. By REF the former are bounded by the latter, so MATH satisfies the estimate MATH with MATH independent of MATH and MATH. We now use our hypothesis on the absence of bound states. So suppose that MATH, MATH near MATH, MATH near MATH. By assumption, MATH is not an eigenvalue of the subsystem Hamiltonian, MATH. Thus, MATH strongly as MATH. Since MATH is compact, and the inclusion map MATH is compact, for MATH with sufficiently small support we have MATH for all MATH. Thus, MATH . Multiplying by MATH from both left and right we finally conclude that MATH . Relabelling MATH and MATH as MATH and MATH (thereby putting stronger restrictions on MATH) provides the inductive step and completes the proof of REF. |
math/9906161 | Let MATH be MATH-invariant, MATH, satisfy estimates REF, and such that MATH and MATH. (Here MATH can be regarded as a function on MATH.) Let MATH identically MATH near MATH, and let MATH be such that MATH and MATH. For example, MATH can be constructed as in REF . The indicial operators of MATH are MATH since MATH. Thus, by REF and as MATH, we have MATH . Thus, taking into account REF and the remark following REF gives MATH with MATH, MATH, MATH so the second statement of REF holds. Moreover, writing MATH, and expanding the left hand side of REF, every term but the one given in REF has operator wave front set disjoint from MATH. Letting MATH be the sum of these terms proves the corollary. |
math/9906161 | The main step in the proof is the construction of an operator which has a microlocally positive commutator with MATH near MATH. In fact, we construct the symbol of this operator. This symbol will not be a scattering symbol, that is, it will not be in MATH, only due to its behavior as MATH corresponding to its MATH-invariance. This will be accommodated by composing its quantization with a cutoff in the spectrum of MATH, MATH, MATH supported near MATH, as discussed in REF . This approach simply extends the one taken in the three-body scattering proof of CITE, though the actual construction is different due to the more complicated geometry. Employing an iterative argument as usual, we may assume that MATH and we need to show that MATH. First we define a distance function to MATH. Thus, we let MATH denoting the Euclidean norm. Then MATH vanishes quadratically at MATH, so MATH. In particular, MATH . Next, we use the variable MATH to measure propagation. Let MATH . Since the MATH component of MATH at MATH is MATH, we see that MATH . In addition, MATH so we conclude that MATH . For MATH, MATH, with other restrictions to be imposed later on, let MATH so MATH is a MATH-invariant function. Let MATH be equal to MATH on MATH and MATH for MATH. Thus, MATH. Let MATH be MATH on MATH, MATH on MATH, with MATH satisfying MATH. Furthermore, for MATH large, to be determined, let MATH . Thus, on MATH we have MATH and MATH. Since MATH, the first of these inequalities implies that MATH, so on MATH . Hence, MATH . We now proceed to estimate MATH. First, by REF, MATH . So let MATH . Under the additional assumptions MATH we have MATH, so we conclude that MATH, hence MATH . This at once gives a positivity estimate for MATH near MATH. Namely, MATH . Thus, MATH with MATH . Hence, with MATH we have MATH . Moreover, MATH since MATH on MATH, so MATH . On the other hand, MATH is supported where MATH so, for MATH sufficiently small, in the region which we know is disjoint from MATH. Moreover, on MATH, MATH so, for MATH sufficiently small, we deduce from the inductive hypothesis that MATH (hence MATH) is disjoint from MATH. In addition, by choosing MATH sufficiently small, we can assume that the support of MATH, MATH and MATH are all disjoint from MATH. Moreover, with MATH denoting a partial derivative with respect to one of MATH, MATH . As MATH is outside the support of the second term, and as MATH vanishes at MATH, we conclude that for any multiindex MATH, MATH . More generally, at any MATH with MATH, defined by MATH, MATH, as above, MATH is independent of MATH at MATH so outside MATH . In fact, outside MATH, but in the set where MATH is positive, MATH so the uniform bounds of REF also follow. Let MATH be identically MATH near MATH and supported sufficiently close to MATH so that the product decomposition of MATH near MATH is valid on MATH. We also define MATH . Thus, MATH is a MATH-invariant function satisfying REF. Let MATH be the operator given by REF with MATH in place of MATH, so in particular its indicial operators are MATH. Note that REF holds with MATH. So suppose that MATH and MATH. Choose MATH so large that MATH. By REF , we deduce the following statement. For any MATH compact with MATH there exists MATH, MATH, MATH with MATH such that if MATH is supported in MATH then MATH . Let MATH so MATH for MATH and it is uniformly bounded in MATH. The last statement follows from MATH being uniformly bounded as a REFth order symbol, that is, from MATH uniformly (MATH independent of MATH). We also define MATH . Then, with MATH identically MATH near MATH, MATH where MATH is uniformly bounded in MATH. Note that MATH arises by commuting MATH, powers of MATH and MATH through other operators, but as the indicial operators of MATH and MATH are a multiple of the identity, MATH, MATH and MATH commute with these operators to top order, and in case of MATH, the commutator is uniformly bounded as an operator of one lower order. Then, multiplying REF by MATH from the left and right and rearranging the terms we obtain the following estimate of bounded self-adjoint operators on MATH: MATH where MATH and MATH is uniformly bounded in MATH as MATH. Now, MATH is uniformly bounded in MATH, hence as a bounded operator on MATH. Thus, if MATH is chosen sufficiently large, then MATH for all MATH, so MATH . Adding this to REF shows that MATH . The point of the commutator calculation is that in MATH the pairing makes sense for MATH since MATH. Now apply REF to MATH and pair it with MATH in MATH. Then for MATH . Letting MATH now keeps the right hand side of REF bounded. In fact, MATH remains bounded in MATH as MATH. Similarly, by REF, MATH remains bounded in MATH as MATH if we chose MATH so large that MATH. Also, MATH is bounded in MATH, so MATH stays bounded by REF as well. These estimates show that MATH is uniformly bounded in MATH. Since MATH strongly on MATH, we conclude that MATH. By REF this implies that for every MATH, MATH . This is exactly the iterative step we wanted to prove. In the next step we decrease MATH slightly to ensure that MATH is disjoint from MATH. |
math/9906161 | We again employ an iterative argument, so we assume that MATH and we need to show that MATH. We first construct a MATH function MATH of MATH, MATH, MATH, MATH and MATH which measures the distance of bicharacteristics of MATH in MATH from MATH. Thus, MATH will be small along these bicharacteristics. We will take MATH of the form MATH where MATH only depends on MATH, MATH, MATH and MATH. Note that MATH is the squared distance of the integral curves of MATH, which are just straight lines, from MATH, so near MATH the second term in MATH gives the fourth power of this distance. Pushing forward MATH by the map MATH at some point MATH, we obtain the vector MATH . Since we are interested in what happens near MATH, where MATH, we are led to consider the constant coefficient vector field MATH in the variables MATH, so MATH . Note that the MATH component of MATH is nonzero. Let MATH so MATH gives a curve through MATH with tangent vector MATH. Now we define MATH by MATH so MATH vanishes exactly quadratically along MATH and is positive elsewhere, and MATH . Note that by the triangle inequality MATH for sufficiently large MATH. Since for MATH we have MATH, we see that MATH that is, when MATH, MATH, MATH, MATH, MATH, so MATH . Hence, MATH . Similar conclusions hold for MATH and MATH, so MATH . Next, we calculate MATH. Since the function we are differentiating vanishes quadratically at MATH, the same follows for its derivatives with respect to any vector field tangent to MATH. Since the MATH component of MATH (and of MATH) is of the form MATH with MATH vanishing at MATH, MATH (that is, at MATH), we conclude that MATH . On the other hand, MATH . But, as in REF, MATH . Thus, MATH . Our results thus far imply that MATH . Now let MATH, MATH, with other restrictions to be imposed on these later, and let MATH . We use MATH to measure propagation along the bicharacteristics; MATH would also work. We again let MATH be equal to MATH on MATH and MATH for MATH and we let MATH be MATH on MATH, MATH on MATH, with MATH satisfying MATH. Furthermore, for MATH large, to be determined, MATH, let MATH . We usually simply write MATH in place of MATH. We only use MATH to slightly shrink the support of MATH in our inductive proof (that is, as MATH is increasing), instead of adjusting MATH as in the proof of REF . Thus, on MATH we have MATH and MATH. Since MATH, the first of these inequalities implies that MATH, so MATH . Hence, MATH, which together with MATH gives MATH since the MATH component of MATH in non-zero. Since we also have MATH we conclude that MATH. Thus, under the additional assumption MATH we deduce that MATH, so MATH . Hence, for MATH, MATH appropriately chosen and for MATH, MATH satisfying MATH, we have MATH . Again, this directly gives a positivity estimate for MATH near MATH. Now MATH . Hence, with MATH we have MATH . In addition, similarly to REF, we see that MATH . Moreover, with MATH denoting a partial derivative with respect to one of MATH, MATH . Thus, REF hold, and hence the uniform bounds of REF also follow. Now MATH is supported where MATH so near the backward direction along bicharacteristics through MATH, in the region which we know is disjoint from MATH. In addition, by choosing MATH sufficiently small, we can assume that the support of MATH, MATH and MATH are all disjoint from MATH. From this point we can simply follow the proof of REF . Thus, we conclude that for every MATH, MATH . This is exactly the iterative step we wanted to prove. In the next step we decrease MATH slightly to ensure that MATH is disjoint from MATH. |
math/9906161 | First note that there is nothing to prove if MATH, so from now on we assume that MATH. The proof is very similar to the previous one and the positive commutator construction is exactly the same as in three-body scattering CITE, based on the MATH-invariant function MATH used here in the proof of REF . Thus, we take local coordinates centered at MATH as above, that is, of the form MATH, and let MATH be defined by REF, so in particular MATH is MATH-invariant. In the proof of REF we showed that there exists MATH such that for any MATH and any MATH imply that MATH satisfies REF, so MATH . We define MATH as in REF. Then REF, hence REF also hold. Since MATH can be taken arbitrarily small, we can choose it and MATH so that MATH is a small neighborhood of MATH; in particular, MATH is disjoint from MATH for each MATH. We can then apply the compactness argument of REF to prove REF for the operators MATH, MATH, etc., defined in that proof, and conclude that MATH. |
math/9906161 | The proof is very similar to the previous ones and now the positive commutator construction follows that of CITE in three-body scattering. Thus, we take local coordinates as above, that is, of the form MATH with MATH defined by linear equations in MATH. Then we construct MATH (defined near MATH) to measure the squared distance from integral curves of MATH this is achieved by solving a NAME problem as in CITE and in REF here. (Indeed, an approximate construction, like that of MATH in the normal case discussed above, would also work). Then we extend MATH to a function on MATH (using the coordinates MATH near MATH), let MATH and define MATH as in REF. The difference in the powers of MATH and MATH in this definition of MATH in the (general) tangential setting and that in the normal case (given in REF) arises since in the normal setting MATH approximates the fourth power of the distance from the generalized bicharacteristics while here it approximates the squared distance. The estimates on MATH are just as in CITE, see also the proof of REF here in the similar totally geodesic setting (the estimates are simply better but no different in nature under the totally geodesic assumption since now we do not have REF), giving a slightly better result than in the totally geodesic normal case: it is MATH, not MATH, that has to be bounded below by an appropriate positive constant. The difference arises as the model integral curves in the tangential setting are closer to the actual ones than in the normal setting. Thus, one obtains REF here as well. The functional analysis part, under the assumption that there are no bound states, is exactly as in the normal case. |
math/9906161 | We only need to prove that for every MATH, if MATH and MATH then there exists a generalized broken bicharacteristic MATH, MATH, with MATH and such that MATH for MATH. In fact, if this statement holds for all MATH with MATH, let MATH and put the natural partial order on MATH, so MATH if the domains satisfy MATH and MATH. Then MATH is not empty and every non-empty totally ordered subset of MATH has an upper bound, so an application of NAME 's lemma gives a maximal generalized broken bicharacteristic of MATH in the intersection of MATH with MATH which passes through MATH. A similar maximal statement holds if we replace MATH by MATH. Indeed, it suffices to show that for any MATH, if MATH then MATH for the existence of a generalized broken bicharacteristic on MATH can be demonstrated similarly by replacing the forward propagation estimates by backward ones, and, directly from REF , piecing together the two generalized broken bicharacteristics gives one defined on MATH. Note that if every element of MATH is totally geodesic and MATH then, due to REF. Indeed, to prove this statement for MATH, we only need that MATH be totally geodesic for MATH with MATH (since the result is local); in particular, it always holds at MATH. We proceed to prove that REF implies REF by induction on MATH. As remarked above, this is certainly true for MATH. We only prove the implication here under the assumption that all elements of MATH are totally geodesic, the general case repeats the argument of NAME. Thus, we already know the implication if MATH. So suppose that REF has been proved for all MATH with MATH and that MATH satisfies REF. As noted in the first paragraph, we thus know that the intersection of MATH with MATH is a union of maximally extended generalized broken bicharacteristics of MATH. We use the notation of the proof of REF below. Let MATH be a neighborhood of MATH in MATH which is given by equations of the form MATH, MATH, MATH, MATH, MATH, such that MATH on MATH and MATH. Such a neighborhood exists since MATH and MATH for every MATH. Also let MATH be a subset of MATH defined by replacing MATH by a smaller MATH. By REF , there is a sequence of points MATH such that MATH, MATH as MATH, and MATH for all MATH, so we may assume that MATH for all MATH. By the inductive hypothesis, there exist generalized broken bicharcteristics MATH with MATH and such that if for all MATH, MATH, we have MATH, then MATH for MATH. But MATH is increasing on generalized broken bicharacteristics in MATH since MATH there, so we conclude that MATH for MATH, hence MATH, so MATH for all MATH. By REF , applied with MATH, there is a subsequence of MATH converging uniformly to a generalized broken bicharacteristic MATH. In particular, MATH and MATH for all MATH, providing the inductive step. Note that this argument for normal points MATH does not use that the elements of MATH are totally geodesic; it works equally well in the general case. Thus, for non-totally geodesic MATH now we only need to consider MATH, and, as mentioned above, this can be done by the NAME argument. |
math/9906161 | This result is a weak form of the limiting absorption principle and can be proved by a NAME estimate. In the Euclidean setting, it is a combination of the NAME estimate, proved by CITE, and its microlocalized version obtained by CITE. In the geometric setting, the NAME estimate describes the commutator of MATH with a self-adjoint first order differential operator MATH such that MATH (this is of course a restriction only at MATH). Namely, it says that for MATH supported sufficiently close to MATH, we have MATH where MATH, hence compact on MATH. It was proved in the geometric three-body setting (with an appropriate adjustment to allow bound states of subsystems) in CITE, following the Euclidean argument of CITE. The proof given there goes through essentially unchanged for more than three bodies. Under our REF , the symbolic commutator calculation in the scattering calculus, MATH, and a slight modification of REF , prove the NAME estimate. The argument of CITE then proves the existence of the limits MATH in MATH, MATH, and MATH shows that the same holds in MATH for every MATH and for every MATH. To show the flavor of the arguments, we prove here a version of the estimate of CITE. Such arguments as this can be combined to prove the limiting absorption principle without a direct use of the NAME estimate as was done in the geometric two-body type setting by CITE and in the geometric three-body setting in CITE. Here, however, we concentrate on proving the wave front set result. The major difference between the propagation estimates of the previous section and the ones near MATH is that MATH is radial at MATH: it has the form MATH. Thus, we need to use a weight MATH to obtain a positive symbol estimate. So for MATH, let MATH where MATH is identically MATH near MATH and is supported in a bigger neighborhood of MATH (it is simply a cutoff near MATH), MATH vanishes on MATH, identically MATH on MATH, MATH, MATH, and MATH vanishes with all derivatives at every MATH with MATH. Then for sufficiently small MATH, MATH implies MATH . Thus, both MATH and MATH are MATH-invariant. Let MATH be a quantization of MATH as in REF , except that now MATH is not supported in a single coordinate chart, so we need to define MATH as the sum of localized operators (of course, this is not necessary in the actual Euclidean setting). Thus, roughly speaking, MATH is the product of a quantization of MATH and MATH, MATH. The fact that MATH does not cause any trouble, and the argument of REF shows that for MATH supported sufficiently close to MATH we have MATH where MATH . Let MATH . Since MATH we conclude that MATH . Since MATH, the second term on the left hand side can be dropped. Since MATH in MATH for MATH, we conclude that for MATH the right hand side stays bounded as MATH. Thus, MATH is uniformly bounded in MATH, and as MATH in MATH, we conclude that MATH. But then REF shows that for any MATH with MATH, we have MATH for every MATH. This proves that the incoming radial set, MATH, is disjoint from MATH, MATH. Iterating the argument, as in the proof of REF , gives that MATH. Since MATH is closed, the same conclusion holds for a neighborhood of MATH. Finally, as all generalized broken bicharacteristics of MATH tend to MATH as MATH and MATH, the propagation of singularities theorem, REF , implies that MATH. The existence of the asymptotic expansions is a local question, so at MATH we can work in the scattering calculus to prove it, see CITE for details of the proof. |
math/9906161 | As mentioned above, the first part follows from the self-adjointness of MATH, so that for MATH, MATH, MATH, we have MATH; recall that the distributional pairing is the real pairing, not the complex (that is, MATH) one. The wave front statement of REF and the assumption on MATH show the existence of the limit MATH in MATH and that in addition MATH for every MATH. The positive commutator argument of REF then applies and shows that MATH. In the Euclidean setting these results follow from a microlocalized version of the NAME estimate due to CITE; see CITE for a detailed argument. Finally, since MATH is closed, a neighborhood of MATH in MATH is disjoint from MATH. Since all generalized broken bicharacteristics approach MATH as MATH by REF , the last part follows from MATH and REF . It can be also proved by modifying the argument of REF along the lines of our proof of REF . Namely, we consider the family MATH, MATH, and note that for MATH, MATH, so MATH. Let MATH, etc., be defined as MATH with MATH where MATH is given by REF (that is, we do not need to use the approximating factor MATH). Then MATH . Note that the pairings make sense since now MATH is disjoint from MATH, MATH. Thus, MATH . Since MATH, the second term can be dropped from the left hand side. Thus, knowing that MATH in MATH as MATH, and assuming that MATH has already been proved and REF is satisfied by MATH, we conclude that MATH. The iteration of this argument of REF and the similar arguments for tangential propagation allow us to conclude the forward propagation estimates which can then be turned into maximal statements as we did in REF . This argument also shows the influence of the sign of MATH: if MATH, REF cannot be used to derive results on MATH. Instead, the signs are then correct in the backward estimate, just as expected. |
math/9906161 | Just as in the proof of REF , the positive commutator estimate of REF (but now applied with a regularizing factor in MATH) shows that MATH, and then REF shows that MATH . We remark that although we need a regularizing factor here which requires some changes in the proof, for example, see the argument of the paragraph below, the regularizing factor (whether MATH or another one) commutes with MATH, so the additional arguments for dealing with it are essentially the same as the two-body ones. Thus, the regularization part of the proof of MATH essentially follows CITE. We proceed to show that MATH . We give the details below since regularity arguments for distributions which are large at infinity seem to appear less often in the literature than the `finer ones'; in particular, CITE assumes MATH with MATH. We essentially follow the proof of CITE below. So suppose that REF has been shown for some MATH; we now show it with MATH replaced by MATH. This time we consider MATH where MATH is identically MATH near MATH and is supported in a bigger neighborhood of MATH (it is simply a cutoff near MATH), MATH identically MATH on MATH, vanishes on MATH, MATH, and MATH vanishes with all derivatives at every MATH with MATH. Then MATH . The first key point now is that on MATH we have MATH, so MATH. Let MATH as in REF again shows that for MATH supported sufficiently close to MATH we have MATH where MATH . Let MATH so MATH analogous statements also hold for MATH and MATH. Thus, MATH where MATH is uniformly bounded in MATH and MATH, MATH on MATH, so MATH is uniformly bounded in MATH. Thus, we need to estimate the commutator MATH, and now we do not have a large MATH as in the proof of REF to help us deal with it. The other key point is thus that we have MATH and MATH identically MATH on MATH, vanishes on MATH, MATH. Let MATH be the quantization of MATH multiplied by MATH as in REF , and define MATH similarly but with MATH replaced by MATH. Thus, as MATH is uniformly bounded in the symbol class MATH, MATH with MATH and MATH uniformly bounded in MATH, MATH for MATH, MATH for MATH, and MATH uniformly bounded in MATH. Moreover, MATH uniformly due to the disjoint operator wave front sets. Thus, MATH with MATH uniformly bounded in MATH, so MATH, MATH uniformly bounded in MATH. Combining REF, we see that for MATH we have MATH . We deduce as at the end of the proof of REF that MATH for every MATH and for every MATH, so REF holds. In particular, MATH for every MATH and for every MATH. In the Euclidean setting we can now simply refer to NAME 's uniqueness theorem CITE to conclude that MATH. Here we give some details to indicate how this conclusion can be reached in general. The crucial step is improving the estimate past the critical regularity MATH. In the Euclidean setting this was done by CITE and his argument was adapted to the geometric setting in CITE. We thus conclude that MATH for MATH. This is the point where MATH, and not just MATH is used. Now we can apply a commutator estimate like that of REF but near MATH. Thus, we conclude that MATH, so MATH. REF and CITE on the absence of bound states with positive energy adapted to the geometric setting, as discussed in CITE, concludes that MATH. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.