paper stringlengths 9 16 | proof stringlengths 0 131k |
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math-ph/0012012 | By REF A direct calculation gives then MATH . The coordinates of the commutator vanish for the following reasons: CASE: If MATH then it vanishes due to antisymmetry in MATH, MATH; CASE: If MATH, MATH, MATH are all different then the NAME MATH-s vanish and MATH commutes with MATH, thus everything vanishes; CASE: If MATH then MATH vanishes and MATH and the resulting term cancels with MATH; CASE: If MATH then use antisymmetry in MATH to transform it into the previous case. |
math-ph/0012012 | Due to the transitive action of the symmetries on the hyperboloid, it is sufficient to calculate the formulae in one point only which may conveniently be chosen to be MATH . Then a simple calculation gives MATH . |
math-ph/0012012 | Note first that the symmetry generators MATH preserve the hyperboloid MATH in MATH-dimensional NAME space. The NAME space NAME operator MATH commutes with the symmetry generators MATH. From the previous corollary, MATH can be understood to be a restriction of MATH to a subset of functions behaving radially as MATH. |
math-ph/0012018 | By REF , and since MATH, for large MATH REF is contractive in MATH and has a unique solution there. It thus has a unique solution in MATH, by REF . Since by the same argument REF is contractive in MATH and since MATH, the unique MATH solution of REF is in MATH as well. The rest is straightforward. |
math-ph/0012018 | We write MATH and get for MATH an equation of the form MATH where MATH, and MATH. The existence of a solution for small MATH is an immediate consequence of continuity and the fact that MATH changes sign in an interval of size MATH. If MATH we note that the equation MATH is contractive for small MATH and thus has a unique root. If instead MATH we have, if MATH are two roots, then for some MATH independent of MATH, MATH whence the conclusion. MATH . |
math-ph/0012018 | CASE: Since MATH we get MATH, implying that REF We have MATH . If MATH as MATH, then the proof is as in REF . If on the contrary, for some large constant MATH we have MATH then by REF we have MATH so that MATH and MATH. But then MATH . |
math-ph/0012018 | CASE: Note first that, taking MATH and writing MATH as a NAME transform, compare REF MATH we have by our assumptions that MATH where we denoted MATH and MATH is its NAME transform. Now define MATH . We have, by the formula for the inverse NAME transform MATH where by construction we have MATH, MATH is analytic in MATH and MATH. We write MATH . We first need to estimate MATH the transformation is well defined, since the function is just MATH. We need to write MATH which defines the function MATH: MATH . Since MATH we have MATH A similar inequality holds for MATH . Indeed, we have MATH . It is easy to check that for MATH and small enough MATH this equation is contractive in the norm MATH. But now, for constants independent of MATH, REF We now use REF to write MATH and get MATH and thus, proceeding as in the proof of REF we get for some MATH. To evaluate MATH for large MATH we resort again to MATH as defined in REF which satisfies REF . This time we note that the equation is contractive in the norm MATH when MATH is small enough. MATH . |
math-ph/0012018 | Consider first the function MATH we see that (compare REF ) MATH and thus, for some positive constants MATH, MATH and thus, since MATH we have MATH . Note now that the function MATH vanishes for MATH. Note furthermore that MATH . It is easy to check that this integral equation is contractive in the norm MATH for small enough MATH; the proof of the proposition is complete. |
math-ph/0012018 | We have, by REF MATH . The estimate of the last term is done as in REF . MATH . |
math-ph/0012018 | We have MATH or, letting MATH, MATH, MATH, MATH . Consider a square centered at MATH with side MATH. For both REF for MATH considered in REF, note that in our assumptions and by the choice of the square we have MATH (on all sides of the square). In REF on the boundary of the rectangle we have by construction of the rectangle, MATH. Also by construction, on the sides of the rectangle we have MATH. Still by assumption, MATH and the ratio in REF is MATH. In REF , we have MATH . Thus, on the boundary of the square, the variation of the argument of the functions MATH and that of MATH differ by at most MATH and thus have to agree exactly (being integer multiples of MATH); thus MATH has exactly one root in the square. The same argument shows that MATH has no root in any other region in its analyticity domain except in the square constructed in the beginning of the proof. MATH . |
math-ph/0012018 | First we note that by taking the NAME transform of REF and solving for the resolvent of MATH we get that MATH with MATH and MATH analytic in MATH by our REF , and the assumed analyticity of MATH, MATH. Hence the existence and uniqueness of the pole of MATH follows from REF , with MATH. |
math-ph/0012020 | For MATH and MATH set MATH. This is defined as a strong integral since MATH acts strongly continuously. For MATH let MATH be the left-ideal defined by the set MATH . If MATH is a state on MATH with MATH then MATH defines a particle representation. If MATH denotes the spectral family of the corresponding representation of the translation group one has MATH where MATH. Conversely we know that the set of all states, fulfilling MATH for some MATH, is norm-dense in the set of all normal states (folium) of the universal particle representation. Since MATH and the translations fulfill REF, we have MATH . Hence we get MATH. This implies MATH maps a norm-dense set of normal states of the universal particle representation onto itself. This implies the invariance of the set of particle representations. The vacuum states are characterized by the annihilation of MATH. This ideal is MATH invariant, implying the invariance of the vacuum representations. For details on the universal particle representation see CITE. The left-ideals MATH were introduced in CITE. |
math-ph/0012020 | Let MATH be a locally normal state on MATH with respect to a vacuum representation MATH. Since this representation is MATH-invariant there exists an anti-unitary operator MATH on MATH with MATH . Since MATH is a vacuum representation it has the NAME property. Consequently every normal state on MATH is a vector state. In particular local normality implies that for every MATH there exists a vector MATH with MATH for all MATH. Let now MATH be symmetric with respect to the origin. Then MATH implies MATH and then obtain MATH . Hence MATH is a normal functional on MATH. This implies the first statement. For the second statement notice that in a particle representation the translations are unitarily implemented and we know the commutation relations REF of MATH with the translations. Since every double cone can be translated in such a way that its center is at the origin we find that MATH is normal on MATH if and only if it is normal on MATH. Hence it is sufficient to look at the complement of symmetric double cones. Since the algebras if these sets are invariant under MATH the method of the locally normal case can be applied. Consequently the set of particle representations is MATH-invariant. |
math-ph/0012020 | Let MATH be a MATH-KMS-state, then for MATH the expression MATH can be analytically continued into MATH. At the lower boundary this function has the value MATH. Since the algebra is translation invariant and since the translations act strongly continuously, there is a norm-dense sub-algebra MATH such that for MATH the expression MATH is entire analytic in MATH. From the relation MATH we see that MATH is invariant under MATH. Moreover, the antilinearity of MATH implies for complex MATH the equation MATH . Inserting for MATH the operators MATH and MATH into the expression for MATH one obtains for complex MATH . Since MATH commutes with the time translations it follows that also MATH is translation invariant. Hence we obtain MATH . From MATH we see that MATH fulfills the MATH-KMS-condition for elements MATH. Since MATH is norm-dense in MATH, the MATH-KMS-conditions holds for arbitrary elements. This implies that MATH is again a MATH-KMS-state. Hence the set of MATH-KMS-representations is MATH invariant. |
math-ph/0012020 | Let a neighborhood of the identity MATH be a subset of MATH such that MATH implies MATH. Since MATH is a neighborhood of the identity there exists MATH, such that the generators of MATH are linearly independent. Choosing MATH such that MATH for MATH then the statements of the lemma are fulfilled. |
math-ph/0012020 | In the variable MATH the above function can be analytically extended in the into the strip MATH. provided we keep MATH real and in their proper domain. An analytic continuation in all MATH variables is obtained with help of the NAME - NAME theorem (see for example, CITE) The domain into which the function can be continued has still to be determined. It is clear from the construction that the real function is the boundary value of the analytic continuation. Next we want to determine the domain of holomorphy of this function. This calculation will be done by mapping the strip MATH bi - holomorphically onto itself in such a way that the interval MATH is mapped onto MATH and the rest of the boundary onto MATH. This is achieved by the transformation MATH . With these new variables the domain of holomorphy becomes MATH . If the elements MATH are properly chosen then an interior point of the MATH variables corresponds to an interior point in the MATH variables. In the MATH - variable the domain REF is convex and hence simply connected. Since the transformation REF is bi - holomorphic, it follows that also the image in the MATH - variables is simply connected. Hence there are no monodromy problems in these variables. Note that the symbol MATH denotes a representation of the NAME group and therefore, the first expression of REF can be written as MATH . The arguments inside of MATH are defined for all elements of the complex NAME group. MATH applied to this product is defined if it belongs to the domain REF (transformed into the variable MATH), eventually multiplied from the left with an element of the real NAME group. In particular we can look at the product MATH for MATH sufficiently close to MATH in REF and obtain MATH . This implies MATH. Choosing MATH in such a way that it contains with MATH also its inverse, then the statement of the proposition is obtained by choosing MATH. |
math-ph/0012020 | From the above discussion we know that the statement is true for MATH in a sufficiently small neighborhood of the identity in MATH. But this implies by varying the wedges and the neighborhoods that it is true for all MATH. |
math-ph/0012020 | From REF we know MATH is independent of MATH as long as MATH belongs to the set MATH (see REF ). Hence we obtain MATH with MATH . Next observe that a translation in the characteristic two-plane of MATH commutes with MATH and is mapped onto its negative by MATH. This implies MATH . Using the fact that MATH is symmetric and that MATH holds we get the result of the theorem. |
math-ph/0012020 | First a remark: If MATH and MATH has an analytic continuation into the strip MATH then the same holds for MATH because MATH has an analytic continuation and the relation MATH holds. (For a detailed proof see CITE.) Since all groups in REF leave the wedge MATH invariant it follows that they commute. If MATH then MATH acts locally on MATH that is, it maps MATH into MATH and is independent of MATH, as long as MATH. By REF we can replace MATH by MATH. This implies that MATH acts locally on MATH and is independent of MATH, as long as MATH. Replacing MATH by its adjoint MATH and observing that the NAME conjugation MATH acts locally on MATH and is independent of MATH in the same range as before, we find, that MATH has the good properties, that is, it maps MATH into MATH and is independent of MATH for MATH. We define MATH . Notice that MATH commutes with MATH. Hence we obtain MATH. This implies MATH. By assumption the fully analytic elements applied to the vacuum are a core for MATH and hence also for MATH and MATH. (Remember that MATH.) Denoting the closure by the same symbol, MATH has a polar decomposition MATH . By the uniqueness of the decomposition and the positivity of MATH it follows that the relation MATH holds. Since MATH maps MATH onto MATH the same must hold for MATH. Since by analytic continuation these expression are independent of MATH we get that MATH is the NAME and NAME is independent of MATH and maps MATH onto itself. |
math-ph/0012020 | During the proof of the last theorem we had constructed the operator MATH. By the uniqueness of the polar decomposition we see that the positive operator MATH and the NAME MATH act locally and are independent of MATH. This implies that MATH acts locally and maps MATH onto MATH. Writing MATH then since MATH is independent of MATH we conclude that also MATH is independent of MATH. This implies MATH for all MATH. We know that MATH maps MATH onto itself for every MATH. Choosing MATH and varying MATH we find that MATH maps MATH onto itself. Since MATH commutes with the translations it is a local gauge. Hence also MATH acts locally for every MATH. Therefore MATH must define a representation of the NAME group. Since this group representation is Abelian it must be trivial. This implies that MATH is the minimal representation. |
math/0012001 | Since folding reduces size and annuli come in pairs (a subdivision annulus followed by a folding annulus), the number of annuli is bounded above by MATH. A simple application of NAME characteristics shows that MATH has no more than MATH edges because the valence of each vertex is at least three. Subdividing and folding, however, may create additional edges as well as vertices of valence one or two, so we need to understand the effect of subdividing and folding on the number of edges. To this end, we introduce the notion of partial folds, that is, folds where both participating edges have to be subdivided, and full folds, that is, folds where at least one of the participating edges is not subdivided CITE. Clearly, a subdivision followed by a full fold does not increase the number of edges, whereas a subdivision followed by a partial fold increases the number of edges by one. A partial fold reduces the number of possible folds by one because the map resulting from it is an immersion around the new vertex created by the fold, so the only folds that are possible after a partial fold are those that were available before. Similarly, a full fold does not increase the number of possible folds. This means that the number of partial folds that occurs in our construction is bounded by the number of folds that the map MATH admits. Since MATH is tight, the number of folds at one vertex MATH is bounded by MATH. Summing up, we see that the number of possible folds is bounded by MATH . Hence, the number of edges after a fold is bounded by MATH. Subdivisions increase the number of edges, so the number of edges at any point in our construction is bounded by MATH . The number of tetrahedra belonging to one annulus is bounded above by MATH (four tetrahedra per edge), which gives us a theoretical upper bound of MATH on the number of tetrahedra in our triangulation of MATH. |
math/0012001 | Suppose, to obtain a contradiction, that MATH is not isotopic to MATH. By NAME 's theorem we may isotope MATH so that the induced foliation has only saddle singularities. Let MATH be the infinite cyclic cover of MATH coming from MATH. By assumption we may lift MATH to MATH. Note that projection onto the second factor MATH gives a NAME function when restricted to MATH. However this NAME function must have a maximum. The induced foliation of MATH has a singularity at this maximum, but this singularity cannot possibly be a saddle singularity. This is a contradiction. |
math/0012004 | The idea of the proof is that MATH is a ``maximal metric" in the sense of CITE, subject to the constraint that MATH agrees locally with the given metric on MATH. Given MATH and MATH, define MATH to be the infimum of the path lengths of all MATH-coarse paths joining MATH to MATH which have word length MATH. Note that the sequence MATH is nonincreasing and has limit MATH. Fix MATH for the moment. Since MATH is proper, the infimum defining MATH is acheived by some MATH-coarse path MATH with a minimal word length MATH, MATH. Note that the path MATH cannot have a subpath MATH such that each of MATH, MATH is MATH because then MATH which would produce a MATH-coarse path MATH of path length MATH whose word length is smaller than MATH, a contradiction. It follows that at least MATH of the distances MATH are MATH (where MATH denotes the greatest integer function). We therefore have MATH . Now the sequence MATH is evidently nondecreasing; moreover, MATH if and only if MATH. If MATH is not bounded above it follows that MATH diverges to MATH, a contradiction. Therefore MATH is eventually constant, proving that MATH is eventually constant and equal to MATH. This shows that MATH is a coarse geodesic metric. Now we compare MATH to MATH. Obviously MATH . For the other direction, we have seen that MATH is realized by some MATH-coarse path of least word length MATH, and the argument shows that MATH . |
math/0012004 | If we substitute ``finitely generated" for ``compactly generated", and ``finite index" for ``cocompact", then this is a standard result, and the proof goes through unchanged, with one caveat. To show that two finite generating sets MATH determine quasi-isometric word metrics one must prove that MATH and MATH for some integers MATH. We must prove the same when MATH are compact generating sets. First we reduce to the case of compact generating sets containing a neighborhood of the identity MATH. Supposing that MATH is any compact generating set, it follows that MATH, and so by the NAME category theorem some MATH contains an open ball MATH. Also, some MATH contains MATH. Therefore, MATH contains a neighborhood of MATH, and we can replace MATH by MATH. Letting MATH be two compact generating sets each containing a neighborhood of MATH, for each MATH there exists MATH such that MATH contains a neighborhood of MATH, and by compactness of MATH it follows that MATH for some MATH; similarly MATH. |
math/0012004 | By a result of CITE and of CITE, we know that there exists a left invariant metric MATH on MATH yielding the topology on MATH. Let MATH be a compact generating set for MATH; by enlarging MATH we may assume that MATH contains the MATH-ball of some radius MATH about MATH, and that MATH. Let MATH. Applying REF , the MATH-coarse geodesic metric MATH agrees with MATH on each MATH-ball of radius MATH, and MATH is quasi-isometric to the compactly generated word metric MATH. Moreover, MATH is clearly left invariant. |
math/0012004 | Obvious. |
math/0012004 | Suppose that MATH for some uniform cobounded subgroup MATH of MATH. Choose an induced cobounded quasi-action MATH; as remarked at the end of REF, MATH is unique up to bounded distance in the sup norm. Now we apply the main result of CITE: If MATH is a bounded valence, bushy tree and MATH is a quasi-action of a group MATH on MATH, then there exists an action MATH of MATH on a bounded valence, bushy tree MATH, and there exists a quasiconjugacy MATH from MATH to MATH. We obtain a quasiconjugacy MATH from the quasi-action MATH to an injective cobounded action MATH. Restricting to MATH gives an injective action MATH which is quasiconjugate via MATH to the canonical action of MATH on MATH. Since MATH is cobounded on MATH it follows that MATH is cobounded, that is, cocompact, on MATH. Now we need a lemma from CITE: Given a bounded valence, bushy tree MATH, a sequence MATH converges in MATH if and only if MATH satisfies the following property: CASE: There is a number MATH so that for any MATH there is a MATH so that the set MATH has diameter at most MATH. A convergent sequence MATH clearly satisfies coarse convergence. Since coarse convergence is clearly invariant under quasiconjugacy, the image sequence MATH also satisfies coarse convergence. Applying REF it follows that MATH converges in MATH, proving that MATH is continuous. Also, MATH is proper, for suppose MATH is compact. Choose a sequence MATH. Passing to a subsequence, MATH converges to some MATH. It follows that MATH satisfies coarse convergence in MATH, and again by quasiconjugacy invariance it follows that MATH satisfies coarse convergence in MATH. Applying REF it follows that MATH converges in MATH to some MATH. By continuity of MATH we have MATH and so MATH, proving that MATH is compact and so MATH is proper. Having proved that MATH is continuous, proper, and cocompact, applying REF it follows that MATH is surjective, which implies that MATH, proving that MATH is maximal in MATH. |
math/0012004 | Assuming REF is true suppose that we have an embedding MATH as in REF . Let MATH be a compact fundamental domain for MATH and consider the compact generating set MATH for MATH. The left-invariant word metric on MATH determined by the generating set MATH is quasi-isometric to the tree MATH. Specifically, the map MATH, taking a vertex MATH to any isometry MATH such that MATH, is a quasi-isometry from MATH to MATH. We have a quasi-isometry MATH, and applying REF the injection MATH is a quasi-isometry. The left action of MATH on itself is clearly a cobounded quasi-action, and quasiconjugating via MATH we obtain a cobounded quasi-action of MATH on MATH. Applying REF produces a quasiconjugacy MATH from the MATH quasi-action on MATH to a cobounded action MATH for some bounded valence, bushy tree MATH. Repeating the argument above using REF , the homomorphism MATH is continuous, proper, and cocompact. Properness implies that the kernel is compact, but the group MATH having no compact normal subgroup, it follows that MATH is an embedding. Letting MATH be a coarse inverse of MATH, this shows that MATH, and the latter is clearly a uniform, cobounded subgroup of MATH. Applying REF it follows that MATH, which implies that the composition of injections MATH is an isomorphism, and so MATH is surjective. |
math/0012004 | We may assume, by subdividing MATH if necessary, that MATH and MATH act without edge inversions, and so the maps MATH, MATH are cellular. Let MATH denote stabilizer subgroups of MATH respectively. Consider a vertex MATH of MATH, the image vertices MATH, MATH, and MATH; we must show that MATH is evenly covered by MATH. Let MATH, choose MATH, and so the left hand side of the even covering equation for MATH is MATH. Let MATH, and let MATH; choosing MATH, the right hand side of the even covering equation for MATH equals the sum of MATH. But MATH is the disjoint union of MATH. |
math/0012004 | Suppose that some edge MATH of MATH is subdivided by inserting at least two distinct vertices in MATH, and so MATH has an edge MATH contained in the interior of MATH, with endpoints MATH. Consider the edge MATH of MATH, whose ends are located at vertices MATH, MATH. From the property ``folding of subdivision vertices" it follows that MATH and MATH both have valence REF and total index REF; this implies furthermore that MATH is the unique edge of MATH. |
math/0012004 | Consider a composition of covering maps MATH. Let MATH be the subdivision needed for MATH, and let MATH be the subdivision needed for MATH. Pulling back the subdivision points of MATH defines a further subdivision MATH of MATH. The map MATH from MATH to MATH now satisfies cellularity and subdivision normalization, and even covering is easily checked. |
math/0012004 | The second part is simply a special case of REF . We leave the proof of the first part to the reader, except to verify that if MATH is a nonisomorphic covering map then MATH. To see why this is true, let MATH be the subdivision of MATH, and note that there must two cells MATH of MATH, either both edges or both vertices, such that MATH. Choosing cells MATH of MATH lying over MATH respectively, MATH contains an isometry of MATH taking MATH to MATH. |
math/0012004 | Consider the covering map MATH. Apply REF to obtain a covering map factorization MATH of MATH. Consider any covering map MATH, and so the composition MATH is a universal covering map. We have MATH, which implies in turn that MATH factors as a product of covering maps MATH. |
math/0012004 | Let MATH be the endpoints of MATH incident to the ends MATH of index MATH respectively. Let MATH be a lift of MATH. Let MATH be the incident lifts of MATH, with ends MATH lifting MATH and incident to MATH, and ends MATH lifting MATH and incident to vertices MATH lifting MATH, respectively. Since MATH is thornless and MATH it follows that there is an end MATH. Let MATH. For each MATH, let MATH, be the lifts of MATH located at MATH, and let MATH. The subgroup MATH of MATH stabilizing MATH clearly acts as the symmetric group MATH on the set MATH. The subgroup MATH acts on MATH preserving the decomposition MATH, and so MATH acts on MATH as the semidirect product MATH . Now consider the vertex MATH and its lift MATH. Note that the end MATH has index MATH. Also, MATH, and letting MATH and MATH, it follows that MATH acts on MATH as the symmetric group MATH. On the other hand, clearly MATH, and the image of MATH is the subgroup of MATH that preserves the decomposition MATH. If MATH it follows that MATH is not surjective, whereas when MATH it follows that MATH is surjective. |
math/0012004 | We break the algorithm into two subroutines. CASE: Given an edge-indexed graph MATH which is its own minimal subcover, and given any blowup and proper subcover MATH, produce a blowup and proper subcover MATH so that MATH has no index REF edges, and so that MATH. Refer to the commutative diagram below. The fact that MATH follows because if not then MATH contradicting that MATH has no proper subcovers. We may assume that MATH does have at least one index REF edge; note that each of them is of type REF, because MATH for a type REF - MATH edge MATH of MATH is a union of type REF - MATH edges of MATH, but each index REF edge of MATH has type REF. Let MATH collapse a maximal forest MATH of index REF edges. Let MATH be a minimal subcover, and so MATH has no index REF edge. Consider MATH, a subgraph of MATH in MATH, and so MATH is itself a type REF forest; in fact, MATH induces an isomorphism between each component of MATH and a component of MATH. The type REF forest collapse MATH can be factored as a composition of type REF forest collapses MATH where MATH; we remark that no vertex of MATH has valence and total index in MATH both equal to REF, and so the same is true of vertices of MATH in MATH. The covering map MATH induces a covering map MATH so that MATH. Note that properness of MATH implies properness of MATH: if some component of MATH has more than one component in its preimage under MATH then this produces a point of MATH which has more than one component in its preimage under MATH; otherwise, some cell MATH of MATH which is disjoint from MATH has more than one preimage under MATH, and it follows that MATH has more than one preimage under MATH. We thus obtain a blowup and proper subcover MATH so that MATH has no index REF edges. MATH . This completes the description of Subroutine REF. CASE: Given successive blowup and subcover operations MATH and MATH so that neither MATH nor MATH has any index REF edge, produce another blowup and subcover MATH. Moreover, MATH. First we consider the case where MATH, in which case we do a pushout as follows: MATH . As a subset of MATH, MATH is the set of ordered pairs MATH such that MATH, and the maps MATH, MATH are projections. We denote the vertices and edges of MATH as ``tensor products". Each vertex of MATH is of the form MATH where MATH, MATH, and MATH. There are two types of edges in MATH. First, letting MATH, for each vertex MATH such that MATH there is an edge MATH. Second, for each edge MATH and MATH such that MATH, there is an edge MATH. NAME indexing on MATH is defined as follows: each edge of the form MATH has both ends of index REF; and the indexing of every other edge MATH is obtained by pullback under the projection MATH. We first check that MATH is a covering map, and the only thing to verify is that each end MATH is evenly covered. Noting that MATH is the disjoint union of the edges MATH, it follows that each end of MATH is evenly covered. If MATH is not an end of MATH, let MATH for MATH, and consider a vertex MATH. Note that MATH. Note also that MATH induces an index preserving bijection between the set MATH and the set MATH. Since MATH is evenly covered, it follows that MATH is evenly covered. Next we check that MATH is a blowup of MATH. Setting MATH to be the set of all edges of MATH of the form MATH, note that these edges are pairwise disjoint and so form an index REF forest, and MATH is the map which collapses each edge MATH to the point MATH. The only important issue to resolve is whether MATH has a vertex of valence REF and total index REF. Suppose there is such a vertex MATH. Since MATH is a blowup of MATH it follows that MATH has no vertex of valence REF and total index REF, which implies that one of the two edges of MATH incident to MATH is an edge MATH of MATH, and so MATH and MATH is an endpoint of MATH. The two ends of MATH incident to MATH are mapped distinctly to MATH, and since we've already proved that MATH is a covering map it follows that MATH has valence REF and total index REF in MATH. But MATH is a blowup of MATH, which implies that MATH has no vertices of valence REF and total index REF, a contradiction. Finally, we must check that the composition MATH is a blowup of MATH. Clearly MATH is an index REF forest in MATH, and the composition MATH is just collapsing of MATH. And we have already checked above that MATH has no vertex of valence REF and total index REF. Note also that MATH. We therefore have the required blowup and subcover MATH when MATH is single edge. More generally we can proceed inductively, doing successive pushouts, to obtain the blowup and subcover for general MATH. This completes the description of Subroutine REF. To prove REF , consider an edge-indexed graph MATH without index REF edges which is its own minimal subcover. If MATH has no blowup and proper subcover, the algorithm is finished. Otherwise, choose a blowup and proper subcover MATH, and immediately apply Subroutine REF to obtain one for which MATH has no index REF edges and MATH. Note that MATH is its own minimal subcover. Now proceed inductively as follows: assume we have a sequence of blowup and proper subcover operations MATH for which MATH has no index REF edges and is its own minimal subcover, and such that the cardinalities MATH are strictly increasing. If MATH has no blowup and proper subcover then the algorithm is finished. Otherwise, choose a blowup and proper subcover MATH, and immediately apply Subroutine REF to obtain one for which MATH has no index REF edges and MATH. Now apply Subroutine REF to obtain a blowup and proper subcover MATH, and note that MATH. As the cardinalities of the blowup trees MATH are increasing, the blowup and subcovers MATH are all distinct. But MATH has only finitely many blowup and subcovers, and so the algorithm must stop. |
math/0012004 | Let MATH, MATH, MATH be as in the statement of the proposition. We must produce an isometry MATH such that MATH, or in other words MATH is MATH-equivariant meaning that for any MATH we have MATH. Keeping in mind the results of CITE mentioned above, our main difficulties are to understand index REF ends of MATH. Note that since MATH and MATH are their own minimal subcovers, we have MATH, MATH. We collect some facts about the tree MATH. Notation: the subgroups of MATH stabilizing a vertex MATH and an edge MATH are denoted MATH, MATH respectively. We may assume that MATH (and similarly MATH) acts without edge inversions, and so for each vertex MATH and each incident edge MATH we have MATH; if an edge is inverted by some isometry, simply subdivide the edge at its midpoint. For each vertex MATH and each edge MATH of MATH with endpoint MATH and end MATH incident to MATH, we have MATH. The group MATH acts on the edges incident to MATH, and it follows that MATH equals the cardinality of the MATH-orbit of MATH. An edge MATH of MATH with endpoints MATH is called an index REF edge if MATH has index REF in at least one of MATH or MATH, or equivalently if the image of MATH in MATH has an index REF end. Given two edges MATH of MATH, the following are equivalent: CASE: One of MATH is a subgroup of the other. CASE: MATH. CASE: Letting MATH be the unique embedded edge path in MATH from MATH to MATH, and letting MATH, for each MATH the set MATH forms a single orbit (of cardinality REF) for the action of MATH on set MATH of edges incident to MATH. Obviously REF implies REF . To show that REF implies REF it suffices to observe that if MATH are both incident to a vertex MATH and if MATH forms an orbit of the action of MATH then MATH. To prove that REF implies REF , suppose that MATH and let MATH be the edge path as in REF . It follows that MATH for MATH. By induction on MATH we easily reduce to REF assuming that MATH are incident to MATH and MATH, we must show that MATH is a MATH orbit of the action of MATH on MATH. If this is not true, then there exists an edge MATH incident to MATH such that MATH are in the same MATH orbit (this uses the fact that all orbits of the MATH action on MATH have cardinality MATH). Let MATH be the closures of the components of MATH containing MATH, respectively. Any element of MATH taking MATH to MATH restricts to an isomorphism MATH. Let MATH be the isomorphism whose restriction to MATH is the identity, so that MATH and MATH. Then we have MATH, contradicting that MATH, and therefore showing that MATH do form a MATH orbit. The condition MATH is obviously an equivalence relation on edges, called stabilizer equivalence. REF in the lemma shows that there are three types of stabilizer equivalence classes. First is a singleton, a class consisting of a single edge MATH; this occurs when MATH has no index REF ends. Second is a doubleton, a pair of edges MATH sharing an endpoint; this occurs when MATH has exactly one end of index REF. Third is a line in MATH, which occurs when the image of the line is an edge of MATH both of whose ends have index REF. REF shows that if MATH are inequivalent edges then neither of MATH is contained in the other. We shall now define the map MATH. To define MATH on MATH, note that the map MATH is a bijection between MATH and the maximal compact subgroups of MATH. Pick a representative vertex MATH of each orbit of MATH; the subgroup MATH of MATH is compact and must therefore fix some vertex of MATH, and we define MATH be any such vertex. Extend MATH to a MATH-equivariant map MATH. Now extend to a MATH-equivariant map MATH by mapping each edge of MATH to a constant speed geodesic in MATH. First we show that MATH is surjective. Since MATH is connected, nonsurjectivity of MATH would imply that there is a vertex MATH of MATH so that some component MATH of MATH is disjoint from MATH. Since MATH is thornless, MATH is unbounded. But this contradicts cocompactness of MATH in MATH. Next we show that MATH is injective on MATH. If not, consider MATH such that MATH. If MATH is the closure of the subgroup of MATH generated by MATH, then using the fact that MATH and MATH are maximal compact subgroups, it follows that MATH is a closed, noncompact subgroup of MATH. Also, MATH is a closed subgroup of MATH, by properness of MATH. But MATH is contained in MATH which is compact, and so MATH is compact, contradicting properness of MATH. Next we show that for any vertex MATH, if MATH then MATH. Arguing by contradiction, suppose MATH. We may assume MATH for some edge MATH of MATH. Arguing as above using properness of MATH, since MATH is compact it follows that the closure of the subgroup of MATH generated by MATH is compact, but MATH is a maximal compact subgroup and so MATH. Let MATH be the unique embedded edge path in MATH which starts at MATH and ends with MATH. Since MATH stabilizes MATH it follows that MATH stabilizes each edge in this edge path, and by applying REF it follows that the edges MATH are all stabilizer equivalent. Obviously MATH is not identified with any point in the interior of MATH, and so MATH. Let MATH be the stabilizer equivalence class of MATH, and so either MATH and MATH is a doubleton, or MATH is a line in MATH; in either case we derive a contradiction. CASE: Let MATH be the subgroup of MATH stabilizing MATH. The restriction of MATH to MATH is a standard MATH reflection on an arc, and the map MATH is MATH equivariant. The image MATH is a subtree expressed as a union of two arcs MATH sharing at least one endpoint MATH. By MATH-equivariance it follows that for MATH there is a subsegment MATH of MATH incident to MATH such that MATH, and the sets MATH, MATH are disjoint from each other and from MATH. Moreover, MATH if and only if MATH. But this contradicts that MATH identifies MATH with an interior point of MATH. CASE: Say MATH. Let MATH, MATH. The restriction to MATH of the stabilizer of MATH is a standard MATH action on the line MATH. The fixed points of the reflections in MATH are precisely the vertices; let MATH be the reflection fixing MATH. The even vertices MATH form one orbit under MATH, and the odd vertices MATH form another orbit. The image MATH is the subtree of MATH spanned by MATH, the group MATH acts on MATH, and the map MATH is MATH equivariant. The action of MATH on MATH is evidently proper and cobounded, and since MATH is a tree it follows that there is a MATH-invariant line MATH on which the MATH action is standard. Let MATH be the fixed point of MATH. Note that for each MATH, one of the following two properties holds, and these properties are equivariant with respect to MATH: either MATH; or MATH and MATH is the closest point to MATH on MATH. In either case, it is evident from this description that the arc MATH contains none of the MATH's except for MATH and MATH, and therefore contains none of the MATH's except for MATH and MATH. But this contradicts that MATH lies on MATH. The argument in the last paragraph gives a little more information: it shows that for any stabilizer equivalence class which is a line MATH, any two nonadjacent edges of MATH have disjoint images under MATH. We have shown that MATH does not identify any vertex of MATH with any other point of MATH. Next we show, for any two nonadjacent edges MATH of MATH, that MATH. Suppose not: there exists MATH, MATH such that MATH. Letting MATH be the closure of the subgroup of MATH generated by MATH, it follows that MATH stabilizes the point MATH. Using properness of MATH it follows that MATH is compact. This implies that MATH stabilizes some vertex MATH of MATH, and so MATH. Let MATH be the shortest edge path from MATH to MATH, and similarly for MATH. It follows that MATH, MATH and MATH, MATH. Applying REF it follows that the stabilizer equivalence classes MATH and MATH of MATH and MATH contain MATH and MATH, respectively, and so MATH and MATH are both incident to the vertex MATH. Also, we already know that MATH are not stabilizer equivalent to each other because disjoint edges of a stabilizer equivalence class have disjoint images, and so MATH. Using the description above of a stabilizer equivalence class and its image under MATH, and using the fact that MATH and MATH are disjoint from MATH, we may reduce to the case that MATH; and since MATH are not adjacent at least one of MATH is MATH. Consider the case where one of MATH equals MATH, say MATH (the other case, where MATH, is similar and is left to the reader). Then there must be MATH such that MATH in MATH. Let MATH and choose MATH which interchanges MATH with MATH, and so MATH interchanges MATH with MATH. By equivariance under MATH, the edge MATH contains a point MATH such that MATH, and by the argument just given it follows that the stabilizer equivalence classes MATH of MATH are distinct but are adjacent to a common vertex. But this is impossible, because MATH are separated from each other by the edges MATH of the line MATH, and the lines MATH are distinct stabilizer equivalence classes in MATH. Next we show that each edge MATH of MATH contains a point denoted MATH such that MATH for any MATH; we may choose the points MATH equivariantly with respect to MATH. The point MATH is called the midpoint of MATH and the closures of the two components of MATH are called the halves of MATH. To see why MATH exists, let MATH be the endpoints of MATH. Let MATH be the longest subsegment of MATH which is identified via MATH with a subsegment of another edge incident to MATH, and similarly for MATH. It follows that MATH for otherwise there would be edges MATH incident to MATH such that MATH, contradiction. We can then take MATH to be any point of MATH. Now we may give a global description of the map MATH. Given a vertex MATH, define MATH to be the union, over all edges MATH incident to MATH, of the half of MATH containing MATH. Note that MATH is invariant under MATH. The restriction of MATH to MATH is a MATH-equivariant family of partial NAME folds CITE: the half-edges forming MATH are subdivided and then folded, no two half-edges being entirely folded together. These are the only identifications made by MATH. Note that for any half-edge MATH incident to MATH, if MATH is partially folded with any other half-edge then all half-edges in the MATH orbit of MATH are partially folded together to form a single path in MATH; these paths, one for each partially folded orbit of MATH-half-edges, may then undergo further partial foldings among each other. The description of the map MATH shows that the edge-indexed graph MATH is a blowup of MATH: for any MATH, the partial folds performed on MATH are represented downstairs in MATH by a blowup of the vertex MATH of MATH, and doing this for each vertex of MATH we obtain a blowup MATH. The induced map MATH is a covering map, by REF . By hypothesis, MATH has no blowup and proper subcover, and so MATH is isomorphic to MATH. Also by hypothesis, MATH has no index REF edges, and so the blowup is trivial and the map MATH is an isometry. This completes the proof of REF . |
math/0012004 | Let MATH be a virtually free group of finite rank MATH. There is a finite graph of finite groups MATH whose fundamental group is isomorphic to MATH CITE, and so MATH acts properly discontinuously and cocompactly on the NAME tree MATH of MATH. The kernel of the action MATH is finite. By REF , the group MATH is contained in a maximal uniform subgroup of MATH, and that subgroup is equal to MATH for some maximally symmetric tree MATH and some quasi-isometry MATH. The action MATH is properly discontinuous and cocompact, and the graph of groups MATH has fundamental group isomorphic to MATH and NAME tree MATH. |
math/0012004 | One way to proceed with the proof is to construct sufficiently many examples, that is, to describe some special scheme for constructing sufficiently many edge-indexings of MATH with the desired properties. Instead we shall give a general scheme for enumerating all of the appropriate edge indexings of MATH, in effect enumerating the maximally symmetric trees MATH with MATH isomorphic to MATH. From the description, the infinitude of appropriate edge-indexings of MATH will follow. The enumeration scheme is carried out completely for one example MATH, after the conclusion of the proof. The reader may want to refer to this example while perusing the proof. Instead of indexing the ends of MATH with actual numerical values, index them with variables MATH. An actual edge-indexing of MATH without index REF ends corresponds to an assignment of MATH. We shall construct MATH, a finite union of affine subspaces of MATH defined over MATH, such that an edge-indexing MATH is in MATH if and only if there exists a blowup and proper subcover for the edge indexing MATH. Then, when MATH is not a finite tree, we shall construct a certain homogeneous subvariety MATH of degree MATH defined over MATH, not contained in any degree REF subvariety except for all of MATH, such that an edge-indexing MATH is in MATH if and only if MATH is a unimodular. The conclusions of the theorem will quickly follow from the form of MATH. Let MATH denote all possible blowups of MATH, where each end of MATH is indexed with one of the variables MATH or with the integer MATH, as follows. As usual MATH denotes a subtree of MATH consisting of edges of index REF, the cellular map MATH collapses each component of MATH to a point and is otherwise one-to-one, and under this collapse there is a bijection between MATH and MATH. This bijection is used to index MATH with the variable edge indices MATH. We assume that one of these blowups, say MATH, is actually the identity map on MATH, and so MATH. Given MATH, let MATH, MATH denote all the proper subcovers of MATH, where MATH is a graph with a variable MATH indexing each end MATH. The even covering equations for MATH form a system MATH of first degree equations in the variables MATH. We examine the even covering system MATH more carefully. To each end MATH located at a vertex MATH, and to each vertex MATH, there corresponds an even covering equation whose right hand side is MATH and whose left hand side is a sum of those variables MATH labelling ends of MATH plus an integer equal to the cardinality of MATH; let MATH denote this equation. Let MATH be the system of REF, MATH; and let MATH be the system of REF, MATH. If the system MATH is inconsistent then we may discard MATH as a candidate blowup and proper subcover. We may characterize inconsistency of MATH as follows. Note that for MATH the subsystems MATH and MATH have disjoint variable sets, and so the system MATH is consistent if and only if each of the subsystems MATH is consistent. The subsystem MATH is inconsistent if and only if there exist MATH such that each of the two sets MATH, MATH lies entirely in MATH but these two sets have different cardinalities. Assume now that the system MATH is consistent. In each subsystem MATH, since the variable MATH occurs alone on the right hand side of each equation in MATH, we may eliminate MATH; after this elimination, if there are any equations of the form (constant)MATH(constant) we may eliminate them as well. This produces a new system of REF in the variables MATH. The system MATH is the union of the systems MATH for MATH; we call MATH the system of reduced even covering equations for MATH. Let MATH denote the solution set of MATH, an affine subspace of MATH defined over MATH. We claim that if MATH is vacuous, meaning that MATH, then MATH is a graph isomorphism and so is not a proper subcover. To prove the claim, assume that MATH is vacuous. Since the variables in distinct subsystems MATH are distinct, it follows that each MATH is vacuous. Let MATH be the vertex incident to MATH. Vacuity of MATH implies one of two possibilities. In the first possibility, MATH; in this case each equation MATH has the form (constant)MATH, and so elimination removes all these equations. In the second possibility, MATH is a single vertex MATH; in this case MATH consists of the single equation MATH, with right hand side MATH, and elimination of MATH removes this equation. In all other cases, MATH is nonvacuous. Partition the set MATH into two sets: MATH consists of all vertices incident to some variably indexed end; and MATH consist of all the rest, namely those vertices incident only to ends of index REF, the interior vertices of MATH. It follows from the previous paragraph that MATH and that MATH is one-to-one on MATH. Since MATH has neither loops nor bigons it follows further that MATH is one-to-one on the union of all edges with both ends in the set MATH. Consider now the restriction of MATH to MATH; we have already seen that MATH is one-to-one on MATH and the remaining vertices of MATH are mapped disjointly from MATH. But this implies that MATH is one-to-one on vertices and edges of MATH. Thus we see that MATH is an isomorphism, proving the claim. To summarize, we have shown that the edge-indexings of MATH which do have a blowup and proper subcover are those which lie on a finite union MATH of codimension MATH affine subspaces of MATH defined over MATH. It follows that there are infinitely many edge indexings with integer values MATH that do not lie in MATH, and so infinitely many edge-indexings with no index REF edge and no blowup and proper subcover. If MATH is a finite tree then this finishes the proof, because every edge-indexing of MATH is unimodular. Suppose now that MATH is not a finite tree, but instead is a graph of rank MATH. Let MATH be simple, closed, oriented edge paths whose corresponding REF-cycles give a basis for MATH. Let MATH be the number of edges in MATH. Applying the canonical cocycle MATH to MATH we obtain an equation of the form MATH where all MATH of the variables are distinct. Setting this equal to MATH and clearing the denominator we thus obtain the following homogeneous equation of degree MATH: MATH . Let MATH be the simultaneous solution variety of this system of equations for MATH; the points on MATH with integer coordinates MATH are precisely the unimodular edge-indexings of MATH. Note that rational points are dense in MATH. Clearly MATH has codimension MATH and so there are infinitely many edge-indexings in the complement of MATH, that is, infinitely many nonunimodular edge-indexings which have no index REF edge and no blowup and proper subcover. For the unimodular case, note that the homogeneous variety MATH is not of degree REF, indeed MATH is not contained in any linear subspace of MATH except for MATH itself. To see why we rewrite the defining equations for MATH as follows. Choosing a maximal tree MATH of MATH we may push all variables in MATH to the right hand side and all variables not in MATH to the left hand side, and then reorder the variables, obtaining a set of defining equations for MATH of the form MATH where each MATH is a quotient of homogeneous monomials of equal degree MATH with no variable occurring more than once in MATH. From this it is obvious that MATH is not contained in any proper linear subspace of MATH. Decompose MATH as MATH where MATH is the union of those MATH which are homogeneous and MATH is the union of those which are not homogeneous. As we have just seen, MATH, and it follows that the rational rays lying in MATH are dense in MATH. For each such ray MATH the intersection MATH is finite, and so assuming that MATH points into the positive orthant of MATH it follows that the number of integer points on MATH with coordinates MATH is infinite. |
math/0012007 | The proof follows from MATH. |
math/0012009 | The proof has a few steps. Setting MATH, the parametrix MATH for MATH constructed in CITE satisfies MATH where MATH and MATH are residual, and is related to MATH by the formula MATH . The desired uniform estimate for MATH then follows from the standard uniform MATH estimate MATH and from appropriate uniform estimates for MATH, MATH and MATH in weighted spaces which we now derive. The main work involves demonstrating a uniform estimate in weighted spaces for the resolvent of the Laplacian in hyperbolic space, and so we turn to this first. Uniform boundedness of MATH is equivalent to the uniform boundedness of the conjugated operator MATH on MATH. The NAME kernel of this conjugated operator is MATH where MATH is the NAME kernel of the resolvent of the Laplacian in MATH. Recall from CITE that there is an explicit formula for MATH: MATH . Here we use that the resolvent is a point-pair invariant, so MATH only involves the Riemannian (hyperbolic) distance between MATH and MATH, and equivalently, may be written in terms of the elementary point-pair invariant MATH defined by MATH . There are two regions in MATH where MATH behaves slightly differently as MATH. These are one near the diagonal, say MATH, and one away from the diagonal, say MATH. We use a partition of unity to divide the NAME kernel into two parts. It suffices to prove MATH boundedness of each piece separately. In the former region, MATH, the factor MATH is bounded, independent of MATH, hence the uniform MATH boundedness of MATH, and thus REF can be proved just as the MATH boundedness of MATH, directly from properties of the NAME kernel. More explicitly, the result in this region may be deduced either by invoking the standard extension of the symbol calculus with spectral parameter, or else simply by direct calculation. On the other hand, in MATH, MATH decays exponentially as MATH, so the additional factor MATH, which takes the form MATH, where MATH is some fixed point in the interior of MATH, does not make much difference: the NAME lemma shows the desired bound REF (and in fact better bounds for this piece!). To proceed, we recall that MATH is constructed in stages, and as a sum of two terms, MATH. Here MATH is in the small calculus, and is a slight modification of the operator MATH obtained from the use of the symbol calculus to solve away the conormal singularity along MATH. By standard methods, such a MATH may be constructed so as to depend holomorphically on MATH, and to have uniformly bounded norm on MATH. It also acts on MATH for each MATH, with norm depending on MATH, but not MATH. Next, MATH is obtained after solving away the first MATH terms of the NAME series expansion of the error term MATH, MATH, where MATH is sufficiently large (greater than MATH). This approximate solution of MATH is found as follows: first solve the normal equation MATH which is the restriction of the equation to the front face MATH. But MATH is naturally identified with the Laplacian on MATH, and so we may apply the result of the discussion above to choose MATH with this property, which satisfies the estimate REF , and such that MATH vanishes to first order along MATH. Thus MATH . Applying the first MATH terms of the NAME series for MATH, MATH on the right on both sides of this equation yields MATH where MATH . Clearly MATH with MATH independent of MATH. We also obtain a remainder term MATH from applying MATH on the right of MATH, and this satisfies uniform bounds (from MATH to MATH) of the same form. These facts taken all together finish the proof. |
math/0012009 | This is simply a limiting case of the argument of the previous section, but which has additional simplifying features. Thus, we simply consider a smooth contour MATH that runs along the real axis on the interval MATH, but avoids the poles of MATH as well as MATH minus the poles of MATH, see REF . Such a contour is simply the limit of the contours MATH that we have considered before. Note, in particular, that with MATH as in the previous section, MATH everywhere, and it is strictly negative when MATH. Since the critical point of MATH lies in MATH, the choice of the contour outside this interval is irrelevant (except that it should be far from the real axis near infinity, as before, to ensure convergence). In particular, MATH, for all MATH, except if MATH, when MATH (but MATH); of course a similar statement holds at MATH with MATH replaced by MATH. Thus, the residues of the poles of MATH and MATH also provide lower order contributions than the stationary phase term in most regions in MATH, except that the poles of MATH give the same order as the stationary phase term at MATH, and similarly for MATH. The standard stationary phase lemma now gives the desired results, much as in the previous section. To deal with the points MATH and MATH, that is, the endpoints of MATH, one introduces MATH, respectively, MATH, just as in the previous section, and notes that in MATH, the phase with respect to MATH is never stationary, while with respect to MATH it is only stationary at MATH, hence the left end point gives a rapidly decreasing contribution in this region, while the right end point gives asymptotically non-trivial terms only at MATH, exactly as expected. |
math/0012010 | Let MATH be defined by REF. We are going to prove the formula under the assumption that MATH. Fix MATH, MATH and MATH, and assume MATH. First we are going to assume that MATH is contained in a product of the form MATH contained in a relatively compact open subset MATH. In this case, define MATH . Clearly, MATH is MATH on the square MATH and MATH if MATH or MATH. By NAME formula, MATH where we have set MATH and MATH. This formula translates into MATH . We integrate this formula with respect to MATH to obtain MATH . For each of these integrals, we can use the bounded convergence theorem to take the limit as MATH, provided that we choose MATH, where MATH denotes the least integer MATH for which REF holds on MATH and to obtain an estimate of the form MATH where MATH. Now we pass to the case of general MATH by covering with finitely many sets of the form considered above and using a partition of unity. The details are easy and left to the reader. |
math/0012010 | By the NAME rule, MATH is a sum of products of derivatives of MATH and MATH. It is clear that such a sum fulfills REF. To see that it also fulfills REF, note that by REF every derivative of MATH vanishes to infinite order on MATH. To see that MATH we use NAME development to write MATH (uniformly on compact subsets of MATH). Now choose MATH and substitute into REF for MATH with MATH. The claim follows now by taking the limit and using REF . |
math/0012010 | For the moment, fix MATH; for simplicity, assume that MATH, so that MATH. We shall write MATH. Then MATH - in fact, MATH, MATH. We let MATH be the real Laplacian in the MATH variables MATH, that is, MATH . We then have that MATH. Recall that we write MATH. By REF , we see that MATH. We apply the integral REF from REF for MATH, and some MATH, which implies that MATH . We now replace MATH by MATH in all three integrals above. Then we integrate by parts and estimate, where we choose MATH (see REF for the definition of this number) with MATH. Since all the estimates are easy, we do not write them out; the reader can easily check them. |
math/0012010 | Let MATH. Choose a function MATH which is equal to MATH in some open neighbourhood of MATH. By REF , since MATH and MATH, we have that MATH for all MATH. Hence, MATH is smooth (see for Example CITE), and so MATH is smooth in some neighbourhood of MATH, since MATH there. Since MATH was arbitrary, the claim follows. |
math/0012010 | Let us write MATH where MATH are the underlying real coordinates in MATH, as usual identified by MATH. Let us also choose a neighbourhood MATH of MATH with the property that MATH. We extend MATH in the first MATH variables almost holomorphically; that is, we have a function MATH with the property that MATH and, if we introduce complex coordinates MATH, MATH, MATH, then MATH . Also, MATH is still polynomial in MATH. We introduce new coordinates MATH by MATH and write MATH. MATH is smooth in the first MATH complex variables in some neighbourhood of the origin, and polynomial in MATH. We will now compute the real Jacobian of MATH with respect to MATH at MATH. At MATH, MATH and MATH, so that we have MATH by assumption. Hence, by the implicit function theorem, there exists a smooth function MATH defined in some neighbourhood of MATH, valued in MATH, such that MATH solves the equation MATH uniquely. Here we have already taken into account that MATH depends holomorphically on MATH, a fact that the reader will easily check. Since MATH, this implies that if MATH, then MATH. We let MATH and claim that MATH satisfies REF. In fact, computation shows that MATH, where the right hand side is evaluated at MATH. This formula shows that each MATH is a sum of products each of which contains a factor which is a derivative of MATH with respect to MATH or MATH. By the definition of MATH, we have that MATH . By REF every derivative of those vanishes if MATH, which is in turn the case if MATH and MATH. But this is clearly fulfilled if MATH. The proof is now finished by applying the NAME rule, the chain rule and the observations made above. |
math/0012010 | Let MATH denote the MATH-th standard basis vector in MATH, MATH. If MATH, then clearly MATH. For MATH consider the vectors MATH, MATH. For MATH small enough, these are linearly independent. We now consider the linear change of coordinates given by MATH, MATH, MATH. By REF it is enough to show that there exist positive constants MATH and MATH such that MATH. The existence of MATH is clear. But if MATH, then MATH. An appropriate choice for MATH finishes the argument. |
math/0012010 | By all the choices above, MATH satisfies the smoothness assumptions. Let us first check that every derivative of MATH is of slow growth. Since MATH is holomorphic in MATH and continuous on its closure, the NAME estimates imply that we have an estimate of the form MATH for each MATH, where MATH denotes MATH. By the chain rule, MATH is a sum of products of derivatives of MATH (which are bounded) and a derivative of MATH with respect to MATH, evaluated at MATH, of order at most MATH. Hence, by REF we conclude that there exists a positive constant MATH such that MATH . We now have to estimate the derivatives of MATH for MATH. But MATH. Hence, if we take an arbitrary derivative of MATH, we get a sum of products of derivatives of components of MATH and a derivative of MATH with respect to MATH each of which contains a term of the form MATH. By REF we conclude that for each compact set MATH and each MATH there exists a positive constant MATH with MATH. This proves REF . |
math/0012010 | By the chain rule, we have smooth functions MATH for MATH, MATH, defined in a neighbourhood of MATH in MATH, polynomial in the last MATH variables, such that MATH and MATH. By REF we can choose MATH and MATH such that if we set MATH, then MATH is invertible. Hence, we can apply REF ; let us call the solution MATH. Then MATH satisfies REF, and we shrink MATH and MATH and choose MATH in such a way that MATH is well defined and continuous in a neighbourhood of MATH. It is easily checked that MATH is a function in MATH as a consequence of REF and the fact that each MATH. First note that this implies MATH, and by REF , MATH for each MATH. Now, each derivative MATH of MATH is a sum of products of derivatives of MATH (which are uniformly bounded on MATH) and derivatives of MATH, MATH, and MATH, all of which fulfill the analog of REF on MATH. So MATH fulfills the analog of REF on MATH. Next, we compute the derivative of MATH with respect to MATH. We have that MATH . Applying any derivative MATH, we see that the first sum gives rise to products of derivatives of MATH and derivatives of MATH, MATH, and MATH. Now the derivatives of MATH fulfill REF. Since on MATH, MATH, we conclude that MATH. But by REF, any derivative of MATH is MATH, so that derivatives of MATH evaluated at MATH are MATH. All the other terms in the product are MATH for some MATH, so that the terms coming from the first sum are actually MATH. For the second and third sum, a similar argument using that MATH and MATH are in MATH implies that all the terms arising from them are MATH. All in all, we conclude that MATH, which finishes the proof. |
math/0012017 | Suppose not. Then for some point MATH the orbit MATH is tangent to the contact distribution. Therefore the tangent space MATH is isotropic in the symplectic vector space MATH where MATH. We now argue that this forces the action of MATH not to be effective. More precisely we argue that the slice representation of the connected component of identity MATH of the isotropy group of the point MATH is not effective. The group MATH acts on MATH preserving the symplectic form MATH and preserving MATH. Since MATH is isotropic, MATH as a symplectic representation of MATH. Here MATH denotes the symplectic perpendicular to MATH in MATH. Note that since MATH is a torus, the action of MATH on MATH is trivial. Hence it is trivial on MATH. Observe next that the dimension of the symplectic vector space MATH is MATH. On the other hand, since MATH is a compact connected Abelian group acting symplecticly on MATH, its image in the group of symplectic linear transformations MATH lies in a maximal torus MATH of a maximal compact subgroup of MATH. The dimension of MATH is MATH. Therefore the representation of MATH on MATH is not faithful. Since the fiber at MATH of the normal bundle of MATH in MATH is MATH, the slice representation of MATH is not faithful. Consequently the action of MATH in not effective in a neighborhood of an orbit MATH. Contradiction. |
math/0012017 | Consider the symplectization MATH of MATH. As usual MATH denotes the coordinate on MATH. The contact action of MATH on MATH extends trivially to a Hamiltonian action on the symplectization. The corresponding moment map MATH is given by MATH . The symplectic manifold MATH is a symplectization of MATH. The manifold MATH is a connected symplectic manifold with a Hamiltonian action of MATH, the map MATH is a corresponding moment map for the action of MATH. Moreover, it has the following two properties: CASE: MATH is contained in the convex open subset MATH of MATH; CASE: MATH is proper. Therefore REF applies. We conclude that the fibers of MATH are connected and that the image MATH is convex. Next, since the action of the torus MATH on MATH is effective, it is free on a dense open subset of MATH. This is a consequence of the principal orbit type theorem and the fact that MATH is abelian. Consequently the action of MATH on MATH is free on a dense open subset. Hence the image MATH has non-empty interior. Also, since MATH is compact and MATH is abelian, the number of subgroups of MATH that occur as isotropy groups of points of MATH is finite. Therefore not only does [LMTW, REF ] imply that MATH is the intersection a locally polyhedral subset of MATH with the open half-space MATH, but that in fact MATH is a polyhedral cone. |
math/0012017 | It is clear that MATH is a compact path-connected topological space and that the moment map MATH descends to a continuous map MATH . Moreover, by REF , for any open half-space MATH and any component MATH of MATH, the map MATH is a topological embedding which is a homeomorphism on an open dense set. This gives us a way to define a length structure on MATH: We define the class MATH to be the set of all curves MATH such that MATH is a rectifiable curve in the unit sphere MATH. For MATH we set MATH where MATH is the length functional on the rectifiable curves in the sphere defined by the standard metric. Let MATH be the corresponding metric on MATH. Then, since for any half-space MATH and any component MATH of MATH the set MATH is convex in the sphere MATH, the map MATH is an isometric embedding. Thus MATH is locally an isometric embedding. |
math/0012017 | The image of MATH lies in a connected component of MATH. |
math/0012017 | The idea of the proof is to show that there is an open half-space MATH containing MATH such that MATH and MATH lie in the same connected component of MATH. For then by REF MATH. Given a path MATH in MATH we write MATH for the path MATH in MATH. Since by assumption the geodesics MATH and MATH trace out two distance great circles in MATH, MATH. On the other hand we clearly have MATH, MATH, MATH and MATH. Since MATH, there is an open half-space MATH containing the points MATH, MATH and MATH. By REF , MATH and MATH lie in the same connected component of MATH as MATH. By REF there a geodesic MATH in MATH connecting MATH to MATH such that MATH traces out a short geodesic connecting MATH to MATH. Choose an open half-space MATH containing the points MATH, MATH and MATH. Note that by construction MATH is connected to MATH by MATH, MATH is connected to MATH by a piece of MATH and MATH is connected to MATH by a piece of MATH. By REF MATH and MATH lie in the same connected component of MATH. By REF we have MATH. Choose a half-space MATH containing MATH, MATH and MATH. Since MATH, since MATH is connected to MATH by a piece of MATH and since MATH is connected to MATH by a piece of MATH, MATH and MATH lie in the same connected component of MATH. By REF there a geodesic MATH in MATH connecting MATH to MATH such that MATH traces out a short geodesic connecting MATH to MATH. Finally choose a half-space MATH containing MATH, MATH and MATH. Arguing as above we see that MATH and MATH lie in the same connected component of MATH. Hence, by REF , MATH. |
math/0012017 | Suppose MATH are two points with MATH. We want to show that MATH. Suppose not. Then the distance MATH between MATH and MATH is positive. Let MATH be a short geodesic connecting MATH and MATH, so that MATH and MATH. Then MATH is a geodesic in the unit sphere MATH starting and ending at MATH. Therefore MATH multiply covers a great circle in MATH (and so MATH is an integer multiple of MATH). Suppose that we can construct a geodesic MATH connecting MATH to MATH so that MATH covers a great circle distinct from the one covered by MATH. Then by REF MATH contradicting the choice of MATH as a short geodesic. Now we construct MATH with the required properties. Pick an open half-space MATH containing MATH. Let MATH denote the connected component of MATH containing MATH. By REF the set MATH is convex with nonempty interior. Pick a point MATH in MATH so that MATH is not in the image of the geodesic MATH. By REF there is a geodesic MATH connecting MATH to MATH with the image of MATH lying entirely in MATH. Let MATH be a short geodesic connecting MATH to MATH. If the image of MATH lies entirely in a half-space containing MATH and MATH then by REF we have MATH. Otherwise MATH traces out a long geodesic connecting MATH to MATH. If MATH passes through MATH then the piece of MATH starting at MATH and ending at MATH is the desired geodesic MATH. If MATH does not pass through MATH, concatenate MATH with MATH. The concatenation MATH is the desired geodesic. |
math/0012017 | Suppose MATH are two points in the image of MATH. Then either MATH and MATH lie in some open half-space MATH or MATH. In the former case, by REF , MATH is connected. Hence, by REF , MATH is convex and consequently MATH is convex. In the latter case we argue as follows. The sets MATH, MATH consists of single points; denote these points by MATH. Connect MATH and MATH by a short geodesic MATH. Then the image of MATH contains an arc of a great circle in MATH passing through MATH and MATH (in fact it follows from the proof of REF that the image of MATH is exactly such an arc). |
math/0012017 | By REF for any open half-space MATH of MATH there exist vectors MATH in the integral lattice MATH (MATH depends on MATH) such that MATH . Moreover, we may and will assume that the set of MATH's is minimal. Thus no MATH is strictly positive on MATH. Since the moment cone is a cone on a compact set, there exist finitely many half-spaces MATH such that MATH contains MATH. For each such half-space MATH, let MATH be the minimal set of integral vectors so that MATH . We claim that MATH . As a first step we argue that for any MATH we have MATH . By choice of MATH there exists a point MATH such that MATH (since MATH, MATH). Suppose there exists a point MATH with MATH. Since MATH is convex, MATH for all MATH. On the other hand MATH for all MATH. Since MATH is open there is MATH so that MATH for all MATH. Therefore for all MATH we have MATH a contradiction. We conclude that MATH . Next we argue that the reverse inclusion holds as well: MATH. By construction for each MATH . Since MATH covers the image cone MATH, we have MATH . Therefore MATH . Finally, since MATH is closed and convex, its intersection with the unit sphere MATH is closed and connected. On the other hand MATH . It follows from REF that the set MATH is a disjoint union of two closed sets. Therefore the set MATH is empty. We conclude that MATH . |
math/0012018 | Recall that the dualizing sheaf of a smooth elliptic curve is the structure sheaf. Using NAME duality as presented in CITE and using CITE we obtain the statement of the Lemma with the additional assumption that MATH is locally free. Let now MATH be an arbitrary coherent sheaf. Because MATH is a smooth projective curve, there exists an exact sequence MATH with locally free sheaves MATH and MATH. By applying the covariant functors MATH and MATH we obtain the following diagram with exact rows: MATH . The commutativity of the left square ensures the existence of the right vertical arrow, which is an isomorphism by the five lemma. By construction, this isomorphism is functorial with respect to MATH. To get functoriality with respect to MATH we have to show commutativity of the diagram MATH arising from a morphism of coherent shaves MATH and resolutions MATH and MATH as above. A priori the vertical isomorphism in REF might depend on the resolutions chosen. But once we have shown commutativity for the identity MATH (but arbitrary resolutions) we know that the isomorphism MATH does not depend on the resolution of MATH. In particular, if MATH is locally free, this is already known. To show commutativity of REF one usually lifts MATH to a morphism of complexes: MATH but this will not be possible in general. We consider first the situation where MATH is a torsion sheaf, MATH and MATH for a line bundle MATH on MATH. Observe that MATH since MATH is a torsion sheaf. If the degree of MATH is large enough we have MATH and obtain a lift for any morphism MATH. This shows that we obtain the same NAME duality isomorphism, if we twist the resolution of a torsion sheaf by a line bundle of sufficiently high degree. If now MATH is an arbitrary torsion sheaf we can apply the same argument to show that we can lift any morphism MATH to a morphism from MATH to a twisted resolution MATH where MATH is of sufficiently high degree. This shows commutativity of REF in case MATH is torsion and so also the independence on the resolution of the NAME duality isomorphism if MATH is a torsion sheaf. If MATH is locally free, any torsion direct summand of MATH will be mapped to zero in MATH. So we can assume MATH to be locally free in this case. Then the known functoriality of the NAME duality isomorphism implies the commutativity of REF . Using the additivity of the functors involved, we obtain the required commutativity in general. |
math/0012018 | Since MATH and MATH it is enough to prove the claim in case MATH and MATH separately. But in the case MATH the statements are obvious. For the rest of the proof we assume MATH. If MATH the first statement is the usual adjointness of MATH and MATH for sheaves of vector spaces. To obtain the second statement in this case, we have to use that MATH is a local homeomorphism. If MATH we have MATH where we used the canonical isomorphisms MATH and MATH. Suppose MATH is another object in MATH with MATH and MATH. Let a morphism MATH be given and consider the induced diagram: MATH . To get functoriality we have to show commutativity of this diagram. This involves the definition of composition in MATH. Assume the given morphism is represented simply by MATH where MATH. The left vertical arrow sends the morphism MATH with MATH to MATH whose component corresponding to MATH is MATH. The sum is performed over equivalence classes of certain maps MATH sending a boundary point to MATH and MATH is obtained from MATH by composition with the identification MATH. Our isomorphism sends this to the morphism on the right hand side whose component in MATH corresponding to MATH is given by the above sum. This coincides with the sum we would write down for the composition on the right hand side, because MATH defines a bijection between the images of the lifts of the holomorphic maps MATH to the universal covers MATH of MATH and MATH respectively. These images are triangles and if we denote their Euclidean area by MATH and MATH we obtain from the definition of MATH the equation MATH ant this shows that both sums have the same weights. To get functoriality in general we have to consider situations involving both cases considered above. In all such situations the commutativity of the corresponding diagram follows immediately from the definition of composition in MATH. A similar consideration gives the second statement. |
math/0012018 | If we apply REF to MATH we obtain a functorial isomorphism: MATH . On the other hand, by applying REF to the maps MATH we obtain: MATH . To get the claim we need to see that the functors MATH and MATH are isomorphic, but this follows easily from the definitions because MATH and MATH are local homeomorphisms. |
math/0012020 | Any point MATH has a neighbourhood in MATH outside of which MATH can be extended linearly to give a diffeomorphism of MATH. Choose an finite collection of such neighbourhoods MATH which cover MATH and, for each MATH, let MATH denote a diffeomorphism which is linear outside a compact region and satisfies MATH for all MATH. Thus MATH is an open cover of MATH. Choosing any partition of unity MATH subordinate to this cover we clearly have MATH on MATH. REF now give MATH for all MATH. Symmetry completes the result. |
math/0012020 | For each MATH choose MATH with MATH. Then, for each MATH, set MATH (here we are considering MATH to be a function on MATH which is independent MATH and extending MATH by REF outside MATH). For MATH and MATH . REF (with the obvious minor modification) and REF now give MATH and MATH . Our assumption on the MATH's implies MATH. Using the partition of unity on MATH given by MATH, the remainder of the result can be completed with an argument similar to that above. |
math/0012020 | The first part of this result follows easily from REF applied to the atlas MATH for MATH and any partition of unity MATH which is subordinate to the covering MATH of MATH (n.b. MATH and MATH for each MATH in this case). Choose MATH with MATH and set MATH. The second part of the result follows from the first part by taking MATH. |
math/0012020 | REF for MATH means we can find MATH such that MATH for all MATH and MATH. With the notation of REF it follows that MATH for any MATH and MATH. Combining this with REF and the fact that MATH is finite, we then get MATH for all MATH and MATH. This completes the proof. |
math/0012020 | The first isomorphism is an immediate consequence of REF whilst the second then follows from REF. Now define a smooth function MATH on MATH by MATH. It follows that MATH is a smooth function on MATH with MATH for MATH and MATH for MATH. Hence MATH, so multiplication by MATH defines a continuous map MATH by REF . Combining this observation with REF , the identity MATH and the second part of the present result, we now have MATH for all MATH (n.b. MATH is non-negative whilst MATH is bounded away from REF). A similar argument for MATH now completes the result. |
math/0012020 | Clearly REF implies the norm MATH is invariant under translations with respect to the first variable of MATH. With the help of REF it follows that the local inclusion MATH leads to an estimate MATH for any MATH and MATH, where MATH is independent of MATH. On the other hand MATH on MATH. Combined with REF we now get MATH for any MATH. Clearly a similar argument can be used for MATH. |
math/0012020 | Let MATH. Therefore MATH. Now REF gives MATH . However REF (with MATH) combine to give MATH . Together with the definition of local inclusion and REF we now have MATH the last inequality following from a further application of REF . |
math/0012020 | The continuous inclusion MATH given by REF simply means that we can find MATH such that MATH for all MATH. Now let MATH with MATH. Also let MATH and choose MATH with MATH. Therefore MATH for any multi-index MATH whilst MATH. Hence MATH for all MATH. The result now follows from the fact that if MATH then we can approximate MATH arbitrarily closely (in the MATH norm) by MATH for some MATH. |
math/0012020 | Without loss of generality we may assume that the semi-norms are defined so that MATH for all MATH with MATH. Therefore REF can be rewritten as MATH for all MATH. Suppose an estimate of this form is not valid. It follows that we can find a sequence MATH in MATH such that MATH and MATH as MATH. Now the continuity of the map MATH gives us MATH and a constant MATH such that MATH for all MATH. It follows that the set MATH is bounded in MATH (by the maximum of MATH and MATH). The Uniform Boundedness Theorem (see REF for example) then implies MATH must also be bounded. The result now follows by contradiction. |
math/0012020 | Using the inclusion MATH and the dual pairing MATH we can define a continuous bilinear map MATH by MATH. By REF we can thus find MATH and a constant MATH such that MATH for all MATH and MATH. Now suppose MATH and choose MATH with MATH. Thus MATH and MATH so MATH for all MATH, where we have used the continuous inclusion MATH in the last inequality. Therefore MATH for all MATH and MATH. Now let MATH with MATH. Choose MATH so that MATH. Now the NAME multiplier MATH defines an isomorphism on MATH whose inverse is simply the constant coefficient differential operator MATH of order MATH (recall that MATH is even). Therefore MATH for all MATH and MATH, where the last inequality follows from REF , the fact that MATH defines an isomorphism MATH and the inequality MATH. Since MATH is dense in MATH, REF implies that for any MATH we have MATH (the dual space of MATH) with a corresponding norm estimate. Since we have a continuous inclusion MATH whilst MATH was arbitrary, we finally arrive at a local inclusion MATH. Let MATH be a partition of unity of MATH where MATH and, for each MATH, MATH. Using REF we thus have MATH for all MATH, where MATH is independent of MATH. On the other hand it is clear that MATH (where for MATH we can take MATH). The fact that we have a continuous inclusion MATH now follows from the existence of a local inclusion. |
math/0012020 | Consider the notation of REF and let MATH and MATH. Now MATH is a first order differential operator on MATH whose coefficients are contained in MATH (in fact the coefficients do not depend upon the first variable and are compactly supported with respect to the rest). Thus we have MATH for any MATH. On the other hand MATH so MATH for any MATH, MATH and MATH. Since we clearly have MATH, the above results combine to give MATH for all MATH. Let MATH. Using REF we get MATH for all MATH. On the other hand we can write MATH where MATH is a first order differential operator on MATH whose coefficients are independent of MATH. By REF we thus have MATH for all MATH. On the other hand, using REF and the fact that multiplication by an element of MATH defines a continuous map on MATH (see REF ), we get MATH for all MATH. Clearly the last two estimates complete the result. |
math/0012020 | Induction clearly reduces the proof to the case MATH. Now let MATH and MATH. Since MATH and multiplication by such functions defines continuous maps on MATH and MATH (see REF ), REF gives us MATH . Combining this with REF we thus have MATH and MATH for all MATH. The result follows. |
math/0012020 | For any MATH it is easy to check that MATH whilst the isometric inclusion MATH leads to an isometric inclusion MATH. Since MATH is complete it thus remains to show that MATH is dense in MATH. In turn, using the isomorphism MATH (see REF) and the definition of the norm on MATH, it is clear that the first part of the result is completed by the following. Choose MATH with MATH and set MATH. Now REF gives us MATH so, given any MATH, we can find MATH with MATH. On the other hand MATH so REF implies MATH where MATH is independent of MATH. Setting MATH we thus get MATH by REF , where MATH is again independent of MATH. This completes the claim. The second part of the result follows from REF and the fact that MATH. Arguing as above it is clear that MATH is a closed subspace of MATH which contains MATH. Now let MATH. By REF MATH and MATH so, given MATH, we can find MATH and MATH with MATH. Thus MATH. On the other hand REF gives MATH whilst from REF we have MATH . Combining REF we get MATH . The fact that MATH is independent of MATH completes the result. |
math/0012020 | REF reduce our task to proving the result under the assumptions that MATH and MATH. Let MATH, MATH, MATH, MATH and MATH be as given in the introduction to REF. Now the pull back of the density MATH on MATH is a density on MATH so we can write MATH for some positive function MATH defined on MATH. In particular MATH for any measurable function MATH which is supported on MATH. Now MATH is independent of MATH whilst MATH is compact. Therefore MATH is bounded and bounded away from REF on MATH. Hence we have MATH for any measurable function MATH on MATH. From the definition of MATH (see REF) and the finiteness of MATH it follows that MATH where we have used the identity MATH (recall that MATH is a partition of unity). From REF we have MATH so REF now gives MATH for all measurable functions MATH on MATH. Finally REF gives MATH for all measurable functions MATH on MATH, where the last line follows from the existence of constants MATH such that MATH for all MATH. |
math/0012020 | Once again REF reduce our task to proving the result in the case MATH and MATH. In MATH is a continuous function on MATH then REF give MATH where we have used the finiteness of MATH and REF in the second last and last steps respectively. From REF we now get MATH for all continuous functions MATH on MATH, whilst REF finally gives MATH for all continuous functions MATH on MATH. |
math/0012020 | Since MATH we have MATH for all MATH. It follows that MATH for all MATH. If MATH is any continuous function on MATH we thus have MATH for all MATH. Hence MATH . On the other hand MATH whenever MATH. Therefore MATH completing the result. |
math/0012020 | For any MATH . REF gives us MATH and MATH for all MATH. Now suppose MATH with MATH. Thus MATH and MATH both belong to the ball of radius MATH centred at the origin. Combining this observation with REF and the fact that MATH we then get MATH . On the other hand, if MATH with MATH then MATH so MATH by REF. Combining the above estimates we thus get MATH for all MATH with MATH. Suppose MATH is a continuous function and MATH with MATH. Using REF we thus have MATH . The result now follows from the definition of MATH. |
math/0012020 | REF reduce our task to proving the result in the case MATH and MATH. Consider the notation introduced in the first paragraph of REF and let MATH be as given in REF. Now, for each MATH, define open sets MATH by MATH and MATH. Thus the map MATH is a diffeomorphism. Also define a function MATH by MATH for MATH and MATH for MATH. It is easy to see that MATH is smooth and independent of MATH for sufficiently large MATH; it follows from REF that MATH. Finally set MATH, so MATH. Since MATH is a partition of unity on MATH we have MATH. Together with REF and the fact that MATH is finite we then get MATH for all continuous functions MATH on MATH. On the other hand, the definition of MATH means we have MATH for all MATH. The following claims thus complete the result. This is a straightforward consequence of REF and the fact that MATH for MATH (n.b. MATH is bounded). If we write MATH in the form MATH with MATH and MATH then MATH (where we are considering MATH to be the unit sphere in MATH). Thus, for all MATH with MATH, MATH where the inequalities MATH and MATH have been used in the second last and last lines respectively. For each MATH set MATH . Estimate REF implies there exists MATH such that MATH for all MATH. Further use of REF together with REF then gives MATH for all continuous functions MATH with MATH. |
math/0012020 | Let MATH and consider the notation of REF . Now, for MATH, the pull-back under MATH of the density MATH is a smooth density on MATH. Thus we can write MATH where MATH (in fact MATH is independent of MATH and compactly supported in the remaining variables). It follows that we have MATH for all MATH and MATH. Using REF (to deal with MATH), REF , we therefore have MATH for all MATH and MATH. Now let MATH. By REF . Hence we can find MATH such that MATH n.b. MATH may depend on MATH but MATH does not (we can define MATH to be the number of elements in MATH). Using REF we can now choose MATH such that MATH . Define MATH by MATH and set MATH. Using REF (n.b. MATH), it follows that MATH . On the other hand the definition of MATH gives us MATH . Combining REF, we then get MATH . This completes the result. |
math/0012020 | Let MATH and choose MATH. Now MATH, MATH and MATH so, for any MATH, MATH while MATH by REF . Together with REF we then get MATH for any MATH. By the second part of REF we have MATH . Consider the following cases. Using REF choose MATH with MATH and MATH. Set MATH so MATH, MATH and MATH by REF , where MATH is independent of MATH and MATH. Using REF choose MATH with MATH and MATH. Set MATH so MATH, MATH and MATH by REF , where MATH is independent of MATH and MATH. By combining REF cases it follows that we can find MATH with MATH . This completes the result. |
math/0012020 | We can prove multiplication defines a continuous bilinear map MATH by an argument identical to that given for REF . The first two parts of the result now follow from REF respectively. Now MATH and MATH so MATH and MATH. However MATH, so REF and the continuity of multiplication as a map MATH and as a map MATH give MATH for any MATH and MATH. |
math/0012020 | Initially suppose MATH. Let MATH be a bounded sequence. By REF it follows that MATH is a bounded sequence in MATH. By local compactness we can thus find a subsequence MATH which is convergent in MATH. REF then implies this subsequence must also be convergent in MATH. It follows that multiplication by MATH defines a compact map MATH. Now let MATH be an arbitrary element of MATH. Let MATH and, using REF , choose MATH with MATH. REF then implies multiplication by MATH defines a map in MATH with norm at most MATH, where MATH is independent of MATH. The result now follows from the fact that the set of compact maps in MATH is closed (see REF for example). |
math/0012020 | The equivalence of REF follows from REF . Let MATH. Using REF and the NAME rule we have MATH where the constants are independent of MATH. With the help of REF and the fact that MATH we now get MATH . Suppose REF is not satisfied. Thus we can find some multi-index MATH with MATH and a sequence of points MATH with MATH such that MATH . Choose a sequence MATH with MATH as MATH and MATH. Therefore MATH on neighbourhood of MATH so REF gives MATH . The fact that REF is not satisfied now follows from REF. |
math/0012020 | By REF it suffices to show MATH iff MATH as MATH. Let MATH and MATH. Thus we can find MATH with MATH. Now suppose MATH is sufficiently large so that MATH. Since MATH on the latter set we have MATH. REF and the continuity of multiplication as a bilinear map MATH now imply MATH for some MATH which is independent of MATH. It follows that MATH as MATH. On the other hand, suppose MATH as MATH. Let MATH and choose MATH so that MATH. Now REF gives us a constant MATH such that MATH for all MATH. Since MATH with MATH we can find MATH with MATH and MATH. It follows that MATH so REF gives MATH. However MATH so MATH, completing the result. |
math/0012020 | The first part of the result follows from the fact that the NAME transform defines an isomorphism MATH. Now suppose MATH and choose any multi-index MATH and MATH. Then REF gives MATH for all MATH. The assumption that MATH immediately implies MATH forms a bounded subset of MATH as MATH varies over MATH. The second part of the result now follows from REF and the continuity of the dual pairing of MATH and MATH. |
math/0012020 | The NAME kernel of the operator MATH is MATH where MATH is the NAME transform of MATH. By standard properties of the NAME transform of symbols (see REF for example) MATH is smooth and rapidly decaying away from REF, along with all its derivatives. On the other hand MATH and MATH have disjoint supports, at least one of which is compact. It follows that MATH. REF now completes the result. |
math/0012020 | Given MATH define a new symbol MATH as the symbol of the pseudo-differential operator MATH. Standard results on the calculus of pseudo-differential operators (see REF ) imply that the assignment MATH defines a continuous map MATH. On the other hand MATH for any MATH. The result now follows from REF on the admissible space MATH. |
math/0012020 | The only non trivial Conditions are REF (n.b. REF is covered by REF ). The former is established in REF whilst the latter uses the fact that conjugation of a pseudo-differential operator by a diffeomorphism which is linear outside a compact region gives another pseudo-differential operator of the same order (see REF for example). This is an easy consequence of the identity MATH. Suppose MATH and let MATH be a multi-index with MATH. Thus MATH whilst MATH defines a symbol in MATH so MATH defines a continuous map on MATH (by REF for the admissible space MATH). Hence MATH and we have a norm estimate of the form MATH . The existence of a continuous inclusion MATH now follows from the definition of MATH. On the other hand we can write MATH for some constants MATH. Now suppose MATH. Therefore MATH with MATH (see REF ) whilst MATH so MATH defines a continuous map on MATH. Hence MATH and MATH . With the help of REF if follows that MATH and MATH. This completes the proof of REF . The fact that a continuous inclusion MATH induces a continuous inclusion MATH is trivial. REF now gives the converse. Let MATH and choose MATH with MATH and MATH on a neighbourhood of MATH. Setting MATH it follows that MATH and MATH. Now let MATH. Therefore MATH or, equivalently, MATH. Using the local inclusion MATH it follows that MATH and MATH where MATH may depend on MATH (through MATH) but not on MATH. On the other hand REF implies MATH defines a continuous map MATH. REF for the model spaces MATH and MATH then shows that MATH defines a continuous map MATH. Therefore MATH and MATH where MATH may depend on MATH (through MATH and MATH) but not on MATH. Since MATH, REF combine to establish the existence of a local inclusion MATH. REF now gives the converse. REF for MATH follows from the same Condition for MATH and the fact that pseudo-differential operators preserve the set MATH (which is dense in MATH). For the dual space we can use REF and the fact that the map which sends a pseudo-differential operator MATH to its adjoint induces an (anti-linear) isomorphism on MATH (see REF ). Clearly MATH whilst MATH is an isomorphism which preserves the set MATH. The fact that MATH is dense in MATH now implies the same is true for MATH, with the first identity following immediately. The second identity can be obtained from REF and the expression MATH which is valid for all MATH and MATH. |
math/0012020 | For each MATH set MATH. Since MATH for any MATH, REF gives MATH for any MATH. On the other hand, a straightforward check shows MATH in MATH for any MATH. REF for the admissible space MATH then implies MATH in MATH for any MATH; by definition this means MATH in MATH. |
math/0012020 | Since multiplication by MATH defines an isomorphism MATH for any MATH, we may prove the claim assuming MATH. Consider the notation of REF and define MATH for each MATH. Now REF implies MATH so REF gives us a sequence MATH with MATH in MATH for any MATH. REF and the fact that MATH then give MATH in MATH for any MATH. Setting MATH for any MATH, REF implies MATH whilst MATH in MATH for any MATH. This completes the claim. Let MATH. By REF we have MATH and MATH so REF and the above Claim (coupled with REF) give us sequences MATH and MATH with MATH in MATH and MATH in MATH for any MATH. For any MATH define further functions by MATH and MATH. Another application of REF now shows MATH and MATH in MATH for any MATH. By REF we can find MATH so that we have a continuous inclusion MATH for all MATH. It follows that MATH. Therefore MATH. Now let MATH. A straightforward application of REF shows MATH in MATH for any MATH. REF and the convergence MATH in MATH then implies MATH in MATH. |
math/0012020 | For the density of MATH in MATH see REF. Now let MATH and, for each MATH, set MATH. Thus MATH has compact support (contained in MATH) whilst the symbol MATH is contained in MATH so REF gives MATH. Therefore MATH. A straightforward check shows MATH in MATH so REF gives MATH in MATH. Since the operator (of multiplication by) MATH defines a continuous map on MATH we now get MATH in MATH. |
math/0012020 | Let MATH and choose a sequence MATH such that MATH in MATH (which is possible by REF since MATH). It follows that the sequence of symbols MATH converges to MATH in MATH and so the pseudo-differential operator MATH converges to MATH in MATH (by REF for the admissible space MATH). However the map MATH given by multiplication by MATH can be written as MATH where MATH acts as an isomorphism MATH. Using the fact that the set of compact operators is closed in MATH it therefore suffices to show that the operator MATH defines a compact map MATH for any MATH. Now MATH so REF implies MATH defines a continuous map MATH. On the other hand REF for the admissible space MATH gives us continuous inclusions MATH and MATH. By composing these maps and using REF it follows that MATH defines a compact map MATH. |
math/0012020 | By REF we know that multiplication defines a continuous map MATH for all sufficiently large MATH. Furthermore MATH so MATH by REF . The result now follows directly from an application of REF . |
math/0012020 | Without loss of generality we may assume MATH on a neighbourhood of MATH (if this were not the case we could simply replace MATH with MATH where MATH is chosen so that MATH on MATH). It follows that we can find MATH with MATH and MATH. For each MATH let MATH denote the MATH-th entry of MATH and MATH its (full) symbol. Also let MATH denote the MATH matrix with entries MATH and MATH the determinant of this matrix; in particular MATH where MATH. Now, by the definition of uniform ellipticity, we can find MATH such that MATH. Let MATH denote the MATH-th cofactor of the matrix MATH. Hence MATH. Also set MATH, define MATH to be the pseudo-differential operator with symbol MATH and let MATH denote the MATH system of pseudo-differential operators with entries MATH. If MATH the differential operator MATH is contained in MATH and has symbol MATH. Therefore the MATH-th entry of MATH is contained in MATH and, modulo an element of MATH, has symbol MATH . Hence MATH where MATH is a MATH matrix of pseudo-differential operators whose MATH-th entry is contained in MATH. In particular MATH and MATH define continuous maps MATH . Using the assumption of uniform ellipticity on MATH we can choose MATH so that MATH is bounded away from REF uniformly for MATH. For each MATH set MATH and let MATH denote the MATH matrix of multiplication operators whose MATH-th entry is given by MATH . Now, as matrices, MATH (where MATH is simply the diagonal matrix with a REF in the MATH-th position if MATH and REF otherwise). Since MATH while MATH is independent of MATH (n.b. if MATH then MATH is a zeroth order differential operator) we have MATH as differential operators. Coupled with the relation MATH and the fact that MATH is a differential operator we then get MATH . On the other hand, REF and the fact that MATH clearly imply MATH defines a continuous map MATH . Using REF, the relations MATH and the fact that MATH is a differential operator, we get MATH . We also observe that MATH defines a continuous map MATH . The first part of the result now follows if we apply REF to MATH and use REF , the continuous inclusion MATH and the mapping properties given by REF. On the other hand we can combine norm estimates for the various continuous maps to get MATH for all such MATH. |
math/0012020 | We can obtain the result with MATH by applying REF on coordinate charts and patching the conclusion together with the help of observations similar to REF . Induction, REF then complete the result for general MATH. |
math/0012020 | Suppose MATH with MATH. Thus we can choose a chart MATH of MATH with MATH. Let MATH be the corresponding chart of MATH and choose further functions MATH with MATH. Also define MATH by MATH (extended by REF) for MATH. Using the transformation properties of the principal symbol of a differential operator under diffeomorphisms (see REF for example) it is straightforward to check that the operator MATH is uniformly elliptic on the open set MATH. REF then gives us the following; if MATH for some MATH then MATH and we have an estimate of the form MATH . Putting MATH and using the fact that MATH is a differential operator we get MATH . Combining these observations with REF we now get the following; if MATH for some MATH then MATH and we have an estimate of the form MATH the second inequality coming from REF and the fact that MATH. The result now follows by allowing MATH to vary of the elements of a suitable partition of unity on MATH. |
math/0012020 | Induction clearly reduces the result to the case MATH. Now suppose we have MATH and MATH for some MATH. Therefore MATH whilst MATH and MATH is a differential operator so MATH . By REF we know that multiplication by MATH defines a continuous map MATH with norm bounded independently of MATH. On the other hand REF implies MATH maps MATH continuously with operator norm bounded independently of MATH. Combining these observations with REF and our original hypothesis we thus get MATH with a norm estimate of the form MATH where MATH is independent of MATH. REF (applied to MATH) now implies MATH and MATH where MATH is independent of MATH. A further application of REF clearly completes the result. |
math/0012020 | REF is known to hold for MATH; see REF for example. Duality and induction now reduce our task to proving that if REF holds for MATH, MATH, then it also holds for MATH where MATH. Denote the maps MATH and MATH by MATH and MATH respectively, and assume MATH is an isomorphism. Since MATH, MATH must also be injective. Now let MATH and set MATH. Therefore MATH whilst MATH. REF then implies MATH, thereby establishing the surjectivity of MATH. The Open Mapping Theorem now shows that MATH is an isomorphism. |
math/0012020 | By REF it suffices to prove that the operator REF is contained in MATH. Since we obviously have MATH the following claim completes the result. Clearly it suffices to prove the claim assuming MATH is given by REF. Then MATH . Now, by REF , MATH and MATH for some MATH. On the other hand MATH. Therefore each term in REF is contained in MATH. |
math/0012020 | Since we know that MATH defines a continuous map MATH (see REF ) we can apply a characterisation of pseudo-differential operators given in CITE. Let MATH and suppose, for MATH, MATH is either the operator MATH or (multiplication by) MATH for some MATH. Let MATH be the number of MATH which are of zero order. Choose MATH with MATH and define MATH by MATH (extended by REF) for MATH. Now the fact that MATH are differential operators implies MATH . For MATH set MATH. Therefore MATH whilst REF gives MATH . Combining this with REF it follows that the MATH-th entry of MATH defines a continuous map MATH for any MATH. REF now completes the result. |
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