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math/0010041 | Let MATH be the MATH-algebra generated by symbols MATH, MATH, MATH, and MATH subject to the relations MATH . Giving each MATH degree -REF and giving MATH degree REF makes MATH into a graded MATH-algebra. In fact, the natural quotient map MATH preserves this grading. Let MATH be the kernel of MATH. Then MATH is generated by homogeneous elements. Suppose that MATH is a homogeneous generator of MATH of degree MATH. If MATH, then MATH is a homogeneous element of MATH of degree REF. If MATH, then MATH is likewise a homogeneous element of MATH of degree MATH. Since neither MATH nor MATH are zero divisors in MATH, neither MATH nor MATH are zero. Hence every homogeneous generator of MATH is a factor of some degree MATH element of MATH. Now let us restrict our attention to degree MATH. The map MATH takes the ring MATH of degree REF elements of MATH to the ring MATH with kernel MATH. Using REF , we can write every element MATH as a polynomial in MATH, MATH, and MATH. Since MATH is commutative, the commutators of any two elements of MATH must be in MATH. Hence, MATH for any MATH and MATH. The only remaining identity in MATH comes from the fact that MATH and MATH are multiplicative inverses. Since MATH for any MATH, we have MATH . Hence, MATH . Let us recapitulate: In MATH, we have only the standard relations MATH the commutator relations MATH and the special relation MATH . Since all relations in MATH come from elements in MATH, and all generators of MATH are factors of elements in MATH, the only other possible relations in MATH are MATH and MATH. Inspecting these last two expressions on MATH shows that they do indeed hold. |
math/0010041 | Let MATH denote the MATH-algebra generated by MATH and the set MATH. Given MATH and MATH in terms of monomials consisting of MATH, let MATH be such that MATH. We show that MATH. We can think of MATH as MATH, where MATH . By REF , for each MATH, we can find a 'MATH-integral' MATH of MATH in MATH. Since MATH we have MATH. Thus, we can find a MATH such that MATH for all MATH. That is, MATH which is contained in MATH. Hence the theorem. |
math/0010041 | Let the algebra generated over MATH by the set mentioned in the statement of the lemma be MATH. Clearly, members of the set MATH map MATH to itself. Note that MATH . Thus, elements of MATH (degree REF endomorphisms in MATH) map MATH to itself. If MATH, then MATH and MATH and MATH. If MATH, then MATH, where MATH. Therefore MATH is generated by MATH. Thus, MATH. Similarly, if MATH, then MATH where MATH. Again MATH. Hence the lemma. |
math/0010041 | First we note that MATH as MATH-algebras. So, it suffices to show that if MATH then MATH for some finite number of commutators. Suppose MATH is such that MATH for some MATH. Then for any MATH, we have MATH. Thus, the MATH-commutators MATH. Now suppose that MATH. Then, MATH. Induction and the fact that MATH completes the lemma. |
math/0010041 | We show by induction that MATH is surjective for MATH. When MATH, note that MATH is generated over MATH by MATH. The map MATH is clearly surjective. Let MATH. Consider MATH. Since MATH is surjective, there exists a MATH such that MATH. Consider MATH. Since MATH, we have MATH for MATH by REF . Consider MATH. Again by surjectivity of MATH, there exists a MATH such that MATH. Since MATH, we have MATH. Thus, MATH. |
math/0010042 | In this proof, we let MATH and MATH. The MATH-bimodule MATH of first order MATH-differential operators is generated as a MATH-module by homogeneous MATH such that MATH for any MATH (see REF). Since MATH is generated by the two elements MATH and MATH, it is enough to consider those homogeneous MATH such that MATH and MATH are in MATH. We need to show that MATH is in the MATH span of the MATH's and MATH's. Since MATH is in MATH, we can replace MATH with MATH and so assume that MATH. We consider the following cases separately: CASE: If MATH is a first order MATH-differential operator such that MATH then MATH implies that MATH. Call such pairs MATH type CREF. Since MATH, we can conclude MATH where MATH. Hence MATH. Similarly, if MATH is a first order MATH-differential operator such that MATH then MATH implies that MATH. Call such pairs MATH type CREF. Here again we can conclude MATH where MATH. Thus we have MATH. CASE: Let MATH be homogeneous such that MATH for MATH. As MATH for MATH, we have MATH for any MATH and MATH. Hence, we can assume without loss of generality that MATH. This implies that MATH where MATH. It follows that MATH. Now we can use the fact that MATH to obtain MATH . Let MATH . We have, MATH for all nonnegative integers MATH and MATH. This further implies that MATH for all MATH. If MATH is not empty for MATH, then we can fix a MATH such that MATH for some MATH. Note that if MATH and MATH, then MATH. This implies that either MATH or MATH for MATH which contradicts our assumption. Hence, MATH where MATH (that is, type CREF). Such a MATH is sum of first order differential operators covered in the first case. The case where MATH is a homogeneous homomorphism such that MATH for MATH, is dealt very similarly to the case just discussed. CASE: Let MATH be homogeneous such that MATH for MATH. Since MATH is homogeneous, we have MATH and MATH for some integers MATH and MATH. This implies that MATH is a factor of MATH. Whenever MATH, we can rewrite MATH as MATH for some MATH (because MATH). Since MATH we have MATH . Hence, we can assume that MATH satisfies MATH where MATH and MATH. The relation MATH implies that MATH . Since MATH, we obtain MATH for all MATH; whereas, MATH for all MATH. This implies that for all MATH, MATH . Hence the pairs MATH are of type MATH or MATH for all MATH. Either conclusion takes us back to the second case. This proves the theorem. |
math/0010042 | For any MATH we have, MATH for some integer MATH. Thus, MATH for any graded homogenous endomorphism MATH. |
math/0010042 | Since the given collection of operators are graded, it is enough to check for linear independence of homogeneous operators. This reduces to check that the set MATH is linearly independent over MATH which is clear. |
math/0010042 | By the corollary above there are MATH, MATH, MATH, MATH which are multiples of MATH and such that MATH . Put MATH and MATH. The operators MATH and MATH commute, so we have MATH for each MATH. We will prove the lemma by descending induction on MATH. To start, MATH, so MATH is in MATH. Hence MATH where MATH. However, MATH, so for each MATH, MATH, and MATH, either MATH or MATH commutes with MATH (by REF ). Hence, each MATH is in MATH, and thus MATH is in MATH. Now suppose that MATH is in MATH. Then so is MATH by our assumptions on the MATH. Then we have MATH for some MATH. By CITE, we have MATH in MATH such that MATH . Put MATH . Then MATH, and we have MATH . As MATH, MATH, and MATH all commute with MATH, MATH. This implies that MATH. By applying the base case to MATH, we conclude MATH. |
math/0010042 | Let MATH be a homogeneous MATH-differential operator of order MATH. Then MATH where each MATH has the property that for some MATH, MATH is in MATH for any MATH. Then MATH will be in the algebra generated by MATH, MATH, MATH, MATH, and MATH if each MATH is so. Thus we will assume that there is a MATH such that for any MATH, MATH and proceed by induction on MATH. The base case is immediate, so we will suppose already that all MATH-differential operators of order MATH or less can be expressed with the generators in the set above. In particular, MATH is in the span of MATH and MATH. Since MATH for some MATH which depends on MATH and MATH, we have MATH is in the span of MATH and MATH. Since MATH, and MATH, we can assume without loss of generality that MATH for some MATH and MATH. By REF, there are MATH such that MATH. Put MATH. Then MATH. Hence MATH. It follows that MATH is in the ring generated by MATH and MATH. |
math/0010042 | Let MATH be an ideal of MATH. Every element MATH of MATH can be written as MATH where MATH and MATH for all MATH. Since MATH is in MATH for every integer MATH, we conclude there is some MATH such that each MATH is in MATH. Similarly, we can assume that each MATH is in MATH. When we consider the commutators of MATH with MATH and MATH, we conclude MATH contains MATH. |
math/0010042 | Note that the elements of MATH and those of MATH commute with each other and that the algebras MATH and MATH generate MATH. Hence, we have a surjective map from MATH to MATH. This map is injective by the proposition above. Hence the theorem. |
math/0010042 | For each MATH, let MATH denote the set of all homogeneous operators in MATH of degree MATH. Then MATH where MATH and MATH are respectively the ring of operators in MATH and MATH of degree REF. By REF, MATH and MATH are localised polynomial rings. Hence, MATH is a localised polynomial ring and in particular a domain. The operator MATH is not a left zero-divisor in MATH because the ring MATH is a domain. Similarly, MATH is not a right zero-divisor. Suppose that MATH is a right zero-divisor in MATH and MATH. Then for any MATH, MATH where MATH. It follows that for any MATH, MATH, MATH, MATH in MATH there is some integer MATH such that MATH . Since MATH is not a left zero-divisor, this cannot be REF. Hence, by REF , MATH is not in MATH. A similar argument shows that MATH cannot be a zero-divisor on the left. Finally, suppose MATH. It is sufficient to consider the case when MATH are homogeneous of degrees MATH respectively, in the sense that MATH, and similarly for MATH. By suitably multiplying by (or factoring of) powers of MATH and MATH, we can assume that MATH and MATH are in MATH, which we know is a domain. Hence the theorem. |
math/0010042 | Let MATH be a MATH-dimensional irreducible module of MATH, for MATH. Without loss of generality, we can assume that a MATH-basis of MATH is MATH where MATH is the highest weight vector, and that the action of MATH is given by the map MATH. By using the map MATH, we get the required MATH action on MATH. |
math/0010043 | Since a tight tree set is undirected there must exist a cut MATH in MATH that contains MATH. By REF every sequence MATH of cuts in MATH containing MATH must be finite. The last cut in such a sequence of maximal length must point at MATH. Suppose that there are two cuts MATH and MATH in MATH, such that MATH, MATH and MATH. By REF of a tree set we distinguish between four cases. MATH, MATH and MATH would immediately imply a contradiction. If MATH then there must exist some MATH such that MATH since MATH. If MATH then MATH does not point at MATH; if MATH does not lie in MATH then MATH does not point at MATH. |
math/0010043 | By REF MATH is equivalent to MATH. |
math/0010043 | If these MATH pre-images are non-empty there exists a MATH for which MATH. The statement now is a consequence of MATH. |
math/0010043 | We can find an authomorphism MATH such that MATH and MATH. |
math/0010043 | Let MATH be a cut in a structure tree set MATH of the structure tree MATH. By REF there is a natural number MATH such that for every edge MATH there exist no more than MATH cuts in MATH which contain MATH but do not contain MATH. This constant is the same for all cuts in MATH. By this argument and by MATH we obtain MATH for every vertex MATH. Thus, by REF , MATH cannot be a quasi-isometry if the region of any vertex in MATH has infinite diameter. Now we assume that there exists a cut MATH in MATH such that MATH and MATH both have finite diameter in MATH. By REF this implies that all regions of vertices in MATH have finite diameter in MATH. To prove the theorem we now have to show that MATH is a quasi-isometry. We now construct a function MATH which will turn out to be a quasi-inverse of MATH. If a vertex MATH in MATH is not contained in MATH then let MATH be an arbitrary vertex in MATH which is adjacent to MATH. Otherwise we set MATH. If MATH is an arbitrary element of MATH then we have MATH. We will now prove that MATH and MATH are quasi-isometries that are quasi-inverse to each other by checking the four axioms in REF . CASE: If there is no cut MATH in MATH which separates two adjacent vertices MATH and MATH then MATH. Otherwise there must be an automorphism MATH such that MATH is in the edge-boundary MATH. By REF and since MATH is finite the distance MATH can have only finitely many values for adjacent vertices MATH and MATH. Thus the set MATH has a maximal element MATH. For any two vertices MATH and MATH in MATH there is a path MATH of length MATH and thus we obtain MATH . CASE: MATH . Let MATH and MATH be two vertices in MATH with MATH and let MATH be the vertex in MATH which is adjacent to MATH and MATH. Then MATH . By REF the latter sum does not depend on the choice of the vertices. Thus it is a constant which we denote by MATH. For two vertices MATH and MATH in MATH and a path MATH of length MATH such that MATH for MATH we finally have MATH . CASE: MATH . In this case the proof of REF works analogously by using the inequality MATH for any adjacent vertices MATH and MATH in MATH. REF A vertex MATH in MATH and the vertex MATH always lie in the same MATH-pre-image of some vertex in MATH. By defining MATH for any MATH we obtain MATH for all vertices MATH in MATH. CASE: By MATH we obtain MATH for all vertices MATH of the structure tree MATH. |
math/0010043 | Since MATH is a subset of MATH we need only prove that MATH is finite if MATH has finite diameter. By the definition of a structure tree set the stabilizer MATH of the set MATH has at most two orbits MATH and MATH on MATH. For every cut MATH in MATH the set MATH has a maximal element. The same holds for the orbit MATH. The larger of these two maxima is the maximal distance between a vertex in MATH and a vertex in MATH. Since MATH has finite diameter the same must hold for MATH. |
math/0010043 | Let MATH be the centre of MATH. In MATH there is a sequence MATH of vertices such that MATH. Since only finitely many vertices of the sequence lie in MATH and for all other elements of the sequence there exist components in MATH which contain them, MATH must be a star ball, too. |
math/0010043 | A graph that ramifies uniformly has infinite diameter. If MATH consisted only of components with finite diameter then MATH would be a star ball. If there were only connected components with finite diameter in MATH then MATH would be a star ball since MATH has infinite diameter. |
math/0010043 | For some MATH in MATH let MATH be a component of infinite diameter in MATH. By induction we now choose a sequence MATH of components having infinite diameter such that MATH . The existence of such a sequence is a consequence of REF . Let MATH be an arbitrary vertex in the inner vertex-boundary MATH. Since MATH is connected we can find a path from MATH to some vertex MATH in MATH that does not leave MATH. Again by induction we obtain a sequence of paths whose union is a metric ray. |
math/0010043 | The set of automorphisms MATH acts transitively on both bipartite blocks of MATH. Every ball MATH in MATH with radius at least REF is a covering ball of MATH. Since MATH is a quasi-isometry the pre-image MATH, by REF , has a finite diameter. Every ball in MATH containing MATH is a covering ball. |
math/0010043 | We assume that there is a star ball MATH. Let MATH be a radius such that MATH is a covering ball. In MATH there is a component MATH containing a vertex MATH whose distance to MATH is greater then MATH. Let MATH denote a vertex of the MATH-orbit such that MATH is an element of the covering ball MATH. MATH is again a star ball which is now contained in MATH. Since MATH is connected it is completely contained in one of the components of MATH. All other components of MATH are contained in MATH. Thus MATH cannot be a star ball. |
math/0010043 | By connecting the MATH-images of adjacent vertices of a ray MATH of infinite diameter in MATH by geodesic paths whose lengths are at most the constant MATH in REF of quasi-isometry, we obtain a path MATH in MATH. By REF its diameter is infinite. Again by REF all MATH-balls in MATH, as a subgraph of MATH, contain at most finitely many MATH-images of vertices in MATH. Since we have constructed MATH only with paths of length at most MATH, the subgraph MATH of MATH must be locally finite. By REF , MATH must contain some ray MATH. The end of MATH has to be the MATH-image of MATH. By REF there is no infinite sequence of cuts in a structure cut set with nonempty intersection. Thus an end containing a vertex in MATH cannot be mapped by MATH onto an end of MATH. |
math/0010043 | We have to prove that a ray MATH which is not equivalent to any vertex has infinite diameter. Let MATH be a cut containing MATH. Every vertex MATH in MATH can be separated from MATH by a cut MATH. The intersection MATH is again an edge-cut containing MATH. By induction we obtain a strictly decreasing sequence MATH with empty intersection such that MATH lies in all the cuts MATH and the inner vertex-boundaries of these cuts are pairwise disjoint. The distance of a vertex in MATH to any vertex in MATH is greater than MATH. Since MATH lies in all cuts MATH it must have an infinite diameter. |
math/0010043 | The implications REF are trivial. REF implies REF . We will now prove REF . CASE: If MATH is quasi-isometric to MATH then, by REF , there cannot exist a mixed end in MATH. Thick proper ends cannot be mapped onto MATH under MATH since thick ends cannot be described by a sequence of MATH-cuts. By REF they also cannot be mapped onto MATH and therefore MATH. Another consequence of REF is MATH. The only remaining ends with rays of infinite diameter are the thin proper ends. By REF MATH is bijective on MATH. Thus we have MATH. Now it is also clear that MATH equals MATH. By REF MATH is almost transitive. CASE: Assuming that MATH we want to prove that a graph with uniform ramification is quasi-isometric under the vertex structure mapping MATH to a structure tree MATH. We suppose that there exists a vertex MATH in MATH having region MATH of infinite diameter. The stabilizer MATH of MATH maps the neighbours of MATH in MATH onto themselves. The set MATH of the corresponding automorphisms in MATH has at most two orbits MATH and MATH on MATH where MATH is the function defined in REF. For some cut MATH we now choose a finite and connected subgraph MATH of MATH with MATH so that MATH connects all pairs of vertices in MATH by paths of minimal length that do not leave MATH. Note that these paths are not necessarily geodesic. We define MATH . MATH has finite diameter, because the distance of every vertex in MATH to MATH is at most MATH. For every MATH we now replace the restriction of MATH onto MATH by an automorphic image of MATH. With the possibly existing orbit MATH we proceed analogously. Thereby we obtain a connected subgraph MATH of MATH. We set MATH . Let MATH and MATH be two vertices in MATH and MATH a MATH-geodesic path connecting them. All parts of maximal length in MATH that are completely contained in one of the orbits MATH and MATH can be replaced by a path in MATH of the same length. Thus we have MATH for all pairs of vertices MATH and MATH in MATH. The automorphism group MATH generates no more than two orbits on MATH. In order to prove the existence of a ray with infinite diameter, we now proceed analogously to the proof of REF . Assuming that there is no ray with infinite diameter in MATH, by REF , there must exist some star ball MATH. By REF , MATH can be chosen so that it contains an element of both MATH and MATH. There is no radius MATH such that all sets in MATH which are contained in components of MATH are subsets of MATH, because then, by REF , MATH would also be a star ball in MATH, which would be a contradiction to the uniform ramification of MATH. Let MATH be a set of vertices which is contained in a component MATH in MATH such that its distance to MATH is at least MATH. Since MATH contains an element of both orbits MATH and MATH there must exist an automorphism MATH such that MATH is contained in MATH. Now MATH is completely contained in MATH. MATH is connected and therefore part of a component in MATH. But this is impossible, because then MATH would have one component of infinite diameter, all other components would be contained in MATH and MATH would not be a star ball in MATH. Thus, by REF , there exists a ray MATH with infinite diameter in MATH. Since MATH for all vertices MATH and MATH in MATH, the ray MATH also has infinite diameter in MATH. The ray MATH has finite intersection with every cut in MATH, because the intersection of these cuts with MATH is finite. Hence the MATH-image of the end of MATH must be MATH. This is a contradiction to the condition MATH. We have now proved that, assuming MATH, there is no vertex in MATH whose region has infinite diameter. So, by REF , the graph MATH is quasi-isometric to its structure tree MATH. |
math/0010043 | If MATH acts almost transitively on MATH, then also MATH must act almost transitively on MATH. Thus MATH must be an end in MATH. First we prove that all MATH-pre-images of vertices in MATH have finite diameter. If there is a constant MATH such that any two vertices in MATH with MATH-distance at least MATH have a different MATH-image, then MATH for all vertices MATH. This is equivalent to the condition that for all pairs of vertices in MATH with MATH-distance at least MATH there exists a cut in MATH which separates them. Let MATH be a cut containing MATH and let MATH be a covering ball of MATH with respect to MATH. We define MATH . MATH has a finite diameter. Let MATH be a vertex in MATH with MATH. The ball MATH contains a vertex MATH of the MATH-orbit with respect to MATH. As MATH is also a subset of MATH, we have MATH . We define MATH and choose two arbitrary vertices MATH and MATH with distance larger then MATH. Since MATH is the whole set of vertices MATH, there is a cut MATH containing MATH for which MATH. If MATH is an element of MATH there is nothing more to prove because then MATH is the desired cut which separates MATH and MATH. So we suppose that MATH is an element of MATH. Let MATH be a cut that contains MATH and for which MATH. The edge-boundaries MATH and MATH are disjoint and MATH is a subset of MATH. Thus MATH must be a subset the component of MATH which contains MATH. Thus MATH and MATH. To prove the theorem we have to show that the region of any vertex MATH has finite diameter. By REF we just have to deal with the case MATH. Let MATH be the ray which starts at MATH and lies in MATH. We furthermore define MATH . The MATH-pre-image of any vertex MATH in MATH is non-empty and has a finite diameter. Thus it is contained in some covering ball MATH with respect to MATH. Let MATH be the set of all MATH-pre-images of a vertices in MATH which have a non-empty intersection with MATH. The diameter of MATH is finite. By REF this also holds for MATH. Let MATH be the smallest index such that no vertex in MATH lies in MATH. We define MATH . For every MATH there exists an automorphism MATH with MATH. Since such an automorphism MATH must fix MATH and MATH to a vertex in MATH which has a distance to MATH that is at most MATH, it causes a translation on MATH of maximal length MATH and therefore it must map MATH onto a vertex in MATH. Hence MATH . This implies MATH . Since MATH we finally have MATH . |
math/0010044 | Let MATH be any strongly fillable contact REF-manifold, MATH one of its strong fillings and MATH a Legendrian knot in MATH . It is not hard to find a neighborhood MATH of MATH in MATH and MATH of MATH in MATH and a diffeomorphism MATH such that MATH along MATH, and MATH . One may then use standard arguments (see REF) to extend MATH to a symplectomorphism MATH, where MATH is a neighborhood of MATH and MATH is a neighborhood of MATH. Finally note that the contact planes MATH and MATH agree on MATH and are close together near MATH. By a small perturbation of MATH near MATH (small enough to keep MATH contact so we may use NAME 's Theorem), we may therefore assume MATH near MATH . Hence, if MATH is the expanding vector field for MATH, then MATH will be the desired vector field for MATH. |
math/0010044 | The map MATH has order four with two fixed points and two points of order two (which are interchanged under MATH). One may thus conclude that the torus bundle is a NAME fibered space over MATH with NAME invariants MATH . To determine the sign of invariants, let MATH be a small disk about one of the fixed points with MATH, and consider MATH. If MATH, then a regular fiber in the NAME fibered structure will be given by MATH. One may pick a product structure on MATH so that a regular fiber will be a MATH-curve on MATH . From this we see that the gluing matrix MATH associated to this singular fiber is MATH . Thus two of the NAME invariants are MATH . Similarly, one may check that the third invariant is MATH . |
math/0010044 | The proof can be found in CITE, but we reproduce it here for completeness. We first show the existence of a double cover MATH for which MATH is overtwisted. If MATH, then take two copies MATH and MATH of MATH and map MATH to MATH by MATH and MATH to MATH by MATH. In other words, MATH is the double cover of MATH given by MATH, where MATH. We may assume MATH, MATH, and MATH, MATH. If the relative NAME class on MATH is MATH, then MATH, and MATH, which is not a possible relative NAME class for a tight contact structure with the given boundary slopes. (See CITE for a discussion of the relative NAME class.) Therefore MATH is overtwisted. We leave it as an exercise for the reader to explicitly find an overtwisted disk in MATH. We now prove the tightness of MATH. To construct MATH on MATH, we started with a tight contact structure MATH. If there is an overtwisted disk MATH, then necessarily MATH. Below we show how to inductively choose a different torus MATH isotopic to MATH that does not intersect MATH, and for which MATH is tight. This would contradict the initial assumption of the existence of an overtwisted disk. Without loss of generality we may assume that MATH and MATH is a disjoint collection of arcs and circles. We first show how to eliminate an outermost arc of intersection. Let MATH be an outermost arc of MATH on MATH. The arc MATH then separates off a half-disk MATH from MATH that does not intersect MATH except along MATH. If we cut open MATH to obtain MATH, then MATH is attached to either MATH or MATH . We assume the latter (the argument in the other case is similar). Let MATH be a small closed neighborhood of MATH in MATH . The neighborhood MATH is a toric annulus which, by altering the product structure, we may call MATH . By taking MATH to be small enough, we can ensure that MATH has the same (isotopy class of) intersections with MATH as MATH, except for the absence of MATH . Moreover, we may assume that MATH is convex. (For the moment we assume that MATH. Below we show how to handle a convex torus with more than two dividing curves.) Now glue MATH to MATH using the map MATH to obtain a toric annulus which we denote MATH . Note that MATH can be reconstructed by gluing the two boundary components of MATH, and MATH is a convex torus that has one fewer arc of intersection with MATH than MATH. Finally, we prove the contact structure MATH is tight. By the classification of tight contact structures on toric annuli CITE, any convex torus MATH in MATH has slope MATH . Let MATH. If MATH, then we effectively did not modify the contact structure above. MATH is a toric annulus with convex boundary having boundary slopes MATH and MATH that was obtained by gluing MATH a toric annulus with (universally) tight contact structure and boundary slopes MATH and REF, and MATH a toric annulus with (universally) tight contact structure and boundary slopes REF and MATH . Using the classification of tight contact structures on toric annuli, we claim there is a tight contact structure on MATH with boundary slopes MATH and MATH that may be split along the convex torus MATH with MATH, so that the pieces are contactomorphic to the contact structures induced on MATH and MATH. This may easily be seen using REF which discusses the gluing of tight contact structures on toric annuli. We have therefore eliminated one outermost arc of intersection from MATH with a convex torus in MATH that cuts MATH into a toric annulus with a tight contact structure. It is not hard to continue this procedure to eliminate other outermost arcs of intersection. We now eliminate circle components of MATH . If MATH is an innermost circle of MATH on MATH, then MATH bounds disks MATH and MATH, and MATH bounds a ball MATH. We may now apply the same argument as above to a small neighborhood MATH of MATH to eliminate MATH (and possibly other curves) from MATH. At each step of this reduction process, there exists a convex torus MATH which splits MATH into a toric annulus with tight contact structure. It is easy to see that by eliminating outermost arcs and innermost circles of MATH, we can eventually make MATH, thus proving MATH is tight. Recall we must still discuss what happens if the new splitting torus MATH constructed above has more than two dividing curves. In this case, REF implies there is a contact structure on a neighborhood MATH of MATH which is MATH-invariant in the direction given by MATH and contains a convex MATH disjoint from MATH, with MATH and MATH. If we use MATH as the new splitting torus, then MATH will be tight (this is just the argument above). But then MATH will also be tight since there exists a contact embedding MATH . |
math/0010044 | To prove MATH is universally tight, we will first show that it can be made transverse to the MATH-fibers of the NAME fibration. We then show that this implies that MATH is covered by a tight contact structure on REF-torus. Since all of the tight contact structures on REF-torus are universally tight (see CITE), MATH must therefore be universally tight. REF . (Normalization of contact structure MATH.) Let MATH be a Legendrian curve isotopic to a regular fiber with MATH which we take to be maximal among Legendrian curves isotopic to a regular fiber. Let MATH, MATH, be Legendrian curves (simultaneously) isotopic to the singular fibers MATH with MATH, and let MATH be a standard convex neighborhood of MATH; assume also that MATH is maximal among Legendrian curves isotopic to MATH with negative twisting number. After making the Legendrian ruling curves on MATH vertical (that is, parallel to the regular MATH-fibers), take a convex annulus MATH with Legendrian boundary for which one component of MATH is a ruling curve on MATH and the other component is a ruling curve on MATH. If not all dividing curves on MATH connect between MATH and MATH, then the Imbalance Principle (see CITE) gives rise to a bypass along a ruling curve for, say, MATH. Provided MATH, the Twist Number Lemma (see CITE) implies the existence of a Legendrian curve isotopic to MATH with larger twisting number. Therefore, we conclude that either MATH and MATH has no MATH-parallel dividing curves, or MATH. Assume MATH and MATH has no MATH-parallel dividing curves. If MATH is a convex neighborhood of MATH, then MATH will have a piecewise smooth convex torus boundary. Rounding the corners in the standard way (see CITE), MATH will be a convex torus with boundary slope MATH measured using the identification MATH. Now MATH is a solid torus with convex boundary with slope MATH, measured using MATH, which is equivalent to slope MATH measured using MATH. This implies that the contact structure on MATH is the unique contact structure on the standard neighborhood of the Legendrian knot MATH with MATH. Now assume MATH and MATH has MATH-parallel dividing curves. Then we use the corresponding bypasses to thicken MATH, MATH to MATH, MATH so that the boundary slopes (measured on MATH) are MATH or MATH and the dividing curves on MATH between MATH and MATH do not have MATH-parallel curves. The former case gives an overtwisted contact structure and the latter yields a Legendrian curve isotopic to a regular fiber with MATH. Summarizing, MATH has been normalized so that MATH and MATH are standard neighborhoods of Legendrian curves with twisting number MATH, MATH has no MATH-parallel dividing curves, and MATH is a standard neighborhood of a Legendrian curve with twisting number MATH. REF . (Making MATH transverse to the fibers.) Initially MATH has Legendrian rulings by vertical curves. We perturb MATH slightly so that the characteristic foliation becomes nonsingular NAME, and MATH and MATH become transverse to MATH. Since MATH, MATH, and MATH are all convex in standard form, it is possible to perturb MATH along the Legendrian divides as in CITE to accomplish this. Now, it is a question of isotoping MATH so that MATH is transverse to the fibers on each MATH. Let us consider MATH, for example, and use the identification MATH to measure slope. The regular fibers of the NAME fibration have slope MATH, and the nonsingular NAME characteristic foliation has dividing curves of slope MATH. Since MATH, it is clearly possible to extend MATH so that the contact structure is transverse to the NAME fibers. Moreover, this extension is contact isotopic to MATH rel MATH. In this way, we isotop MATH so that MATH is transverse to the MATH-fibers of MATH. REF . (Pulling back to MATH.) Since the monodromy of MATH as a torus bundle has order four, there exists a REF-fold cover MATH with the property that MATH are lifts of fibers of MATH. The pullback MATH is therefore transverse to the fibers MATH of MATH. REF . (Universal tightness.) A tight contact structure on MATH is universally tight by the classification results of CITE and CITE. Therefore, in order to show MATH is universally tight, it suffices to show that MATH is tight. We prove the well-known fact that a contact structure MATH on MATH which is transverse to the fibers must be tight. (The proof extends easily to any circle bundle.) Let MATH, MATH be the projections of MATH to MATH and MATH respectively. If MATH, MATH are area forms on MATH and MATH, then MATH is a symplectic form on MATH that restricts to a symplectic form on MATH. The contact structure MATH is symplectically fillable and therefore tight. |
math/0010044 | Let MATH be a Legendrian knot isotopic to a regular fiber with MATH as in REF . Let MATH, MATH, be disjoint solid tori isotopic to tubular neighborhoods of MATH, for which MATH contains a contact-isotopic copy of MATH. By perturbing MATH we may assume MATH is convex with vertical (that is, parallel to the regular fibers) dividing curves, and, furthermore, we may assume that MATH, after possibly taking a smaller solid torus. In order to increase the twisting number of a Legendrian curve, we need to find a bypass. We will find a bypass along, say, MATH by patching together meridional disks of MATH and MATH to obtain a punctured torus MATH and showing the existence of a MATH-parallel dividing curve on MATH. REF . (Normalizing MATH on the complement.) Let us first normalize the tight contact structure on MATH. The contact structure on MATH is contactomorphic to a MATH-invariant tight contact structure on MATH with convex boundary, MATH, MATH, and slopes MATH, (that is, the tight contact structure induced on MATH, thought of as a neighborhood of a convex torus in standard form) and a standard (open) neighborhood of a vertical (that is, isotopic to MATH) Legendrian curve with REF twisting removed. This lemma is proved in CITE (compare CITE), but we sketch the basic idea. Since MATH is tight on MATH, no dividing curve on MATH can be MATH-parallel. This leaves us with two possibilities (depending on signs) for the dividing curves on MATH, modulo spiraling. Cutting MATH open along MATH, we obtain a tight contact structure on a genus two handlebody with a fixed set of dividing curves. In addition the dividing curves are such that one can use techniques in CITE to show there is a unique tight contact structure on the handlebody. From this we can conclude that there is a unique tight contact structure on MATH with the given dividing curve data. Now, since the contact structure described in the lemma also has this dividing curve data, our contact structure must be contactomorphic to it. REF . (Patching meridional disks.) If we measure slopes of MATH, MATH, using the identification MATH, then the slopes are MATH, MATH and MATH respectively. After making the ruling curves on MATH meridional, a convex meridional disk MATH for MATH will have, respectively, MATH, MATH, MATH, and also REF dividing curves. We would like to patch copies of the meridional disks together to create a convex surface and moreover relate information about the dividing curves on this patched-together surface to the dividing curves on the meridional disks. We view the MATH (minus MATH) from REF as the region between MATH and MATH (minus MATH). Write MATH, MATH, as before. Assume MATH and MATH. Since MATH is MATH-invariant, we have (for example) a REF-parameter family of positive regions MATH. We may then isotop MATH, MATH, away from MATH (that is, on MATH) to arrange the slopes of the Legendrian ruling curves so that the meridional disk MATH in MATH has Legendrian boundary. Now, take one copy of MATH and two copies MATH, MATH of MATH, and arrange them so that MATH and MATH, where MATH is a union of Legendrian arcs on MATH with endpoints on opposite ends of MATH. Let MATH, which is a torus with an open disk removed. See REF . After smoothing the corners using the `elliptic monodromy lemma' or the `pivot lemma' of CITE, MATH will have smooth Legendrian boundary. Since MATH, we shall think of MATH as having its boundary on MATH . REF . (Combinatorics of MATH, MATH and MATH.) Since MATH has two dividing curves, it either has two positive regions and one negative region, or one positive and two negative regions. We assume the former - the argument for the latter is identical. The rotation number MATH, MATH, satisfies the formula MATH in CITE. Therefore, MATH can attain values MATH. CASE: Assume that at least one of MATH or MATH (say MATH, after possible relabeling) satisfies MATH. We first show that MATH, after possibly isotoping rel MATH, will have a positive MATH-parallel region. If MATH or MATH, there is no problem. If MATH, the dividing curves on MATH may be either of the two types shown in REF . If we have a configuration shown on the right-hand side of REF , then we may isotop MATH rel MATH so that the dividing curves on MATH are as shown on the left-hand side of REF . This follows from the classification of tight contact structures on solid tori in CITE or CITE. If MATH, then we take the dividing curves on MATH to be as shown on the left-hand side of REF . The dividing curves on MATH will then be as shown in REF . Note we have a MATH-parallel component and hence a bypass along MATH . The cases MATH are similar, by observing that any positive MATH-parallel component of MATH must necessarily be connected to a positive MATH-parallel component on one of the copies of MATH, yielding a MATH-parallel component on MATH. The slope of MATH on MATH is MATH and the slope of MATH on MATH is MATH . This implies that the slope of MATH on MATH is MATH . We therefore have a bypass on MATH attached along a ruling curve of slope MATH (as measured from MATH). Using this bypass, we may thicken MATH to MATH with standard convex boundary having boundary slope MATH . Thus, when measured from the product structure MATH on MATH, the slope is MATH showing that MATH is the standard neighborhood of a Legendrian curve MATH isotopic to MATH with twist number REF. CASE: We are left with the case where MATH. Now the dividing curves on the punctured torus MATH constructed from MATH and two copies of MATH will be as in REF . Capping MATH off with MATH, we obtain a closed contractible dividing curve on the torus MATH which contradicts tightness. This completes the proof of REF . |
math/0010044 | From REF we have a tight contact structure MATH on MATH . Now assume that MATH is weakly symplectically semi-fillable. From REF we have a Legendrian knot MATH in MATH that is isotopic to, say, MATH with twist number REF. We may assume that the neighborhood MATH was chosen so that MATH . As discussed above, if we perform Legendrian surgery on MATH, we remove a small neighborhood of MATH (take this neighborhood to lie in MATH) and re-glue it by MATH . We easily see this has the same effect as changing the MATH to MATH . Thus, after Legendrian surgery we obtain a weakly symplectically semi-fillable contact structure on MATH, the NAME fibered space over MATH with NAME invariants MATH . This contradicts NAME 's Theorem REF , thus proving MATH is not weakly symplectically semi-fillable. |
math/0010046 | Since MATH is abelian, every homomorphism MATH factors through MATH. By REF, we have MATH. Furthermore, every homomorphism MATH factors through MATH. Hence: MATH . By NAME 's enumeration principle REF, we have: MATH . Clearly, MATH, where MATH. The NAME function of a subgroup MATH of MATH can be computed from NAME 's REF , as follows. Let MATH be the NAME subgroup of MATH. Then MATH, and so the associated partition is MATH, where MATH and MATH. If a subgroup MATH of MATH does not contain MATH, then MATH. If MATH contains MATH, then the partition MATH is between MATH and MATH, that is, MATH or MATH. Order the set MATH as MATH. Clearly, MATH, and so MATH. The number of subgroups MATH such that MATH and MATH equals MATH, where MATH. Set MATH and MATH. A simple calculation now shows: MATH . This, together with REF , and REF, yields REF. |
math/0010046 | By NAME 's Lemma (compare CITE), MATH is isomorphic to MATH, the first homology of the chain complex REF, tensored over MATH with the module MATH. Under the identification MATH, the boundary map MATH is the integral MATH matrix MATH. Noting that MATH finishes the proof. |
math/0010046 | Let MATH be a finite presentation. Since MATH, we have MATH. As in the proof of NAME 's REF , MATH is the first homology of the chain complex MATH obtained by tensoring REF with MATH (viewed as a MATH-module via the representation MATH). In other words, MATH. Let MATH. In view of REF, the chain complex REF decomposes as: MATH . The twisted Jacobian matrices MATH are of size MATH, and have entries in MATH. By REF , MATH. Hence, MATH. Now notice that MATH has rank equal to MATH. Hence, MATH. Therefore, MATH . Since MATH, we get REF . |
math/0010046 | Assume MATH. Let MATH be the cyclic group generated by MATH. Then there is an automorphism MATH such that MATH. The linear extension MATH is an isomorphism, taking MATH to MATH. Consequently, the MATH-modules presented by these two matrices are isomorphic. Hence, MATH, and so the contributions of MATH and MATH to the sum REF are equal. |
math/0010046 | CASE: The first inequality was already noted in REF. To prove the second inequality, pick a finite presentation of MATH with MATH generators, so that the Jacobian matrix MATH has MATH columns. For each representation MATH, the matrix MATH also has MATH columns, since MATH is one-dimensional. Hence, MATH, for all MATH. CASE: Each MATH is of the form MATH, for some integer MATH dividing MATH. Hence, MATH, for all MATH. |
math/0010046 | By REF , we have MATH . Since MATH is sufficiently large, MATH. Let MATH be the order of a non-trivial representation MATH. It is readily seen that MATH. Thus, the assignment MATH establishes a bijection between MATH and the set of non-unit divisors of MATH. Moreover, MATH, where MATH. Now consider the representation MATH. Clearly, the order of MATH is MATH. Let MATH be the dual homomorphism. We then have MATH. Furthermore, by REF, we have MATH . The conclusion follows at once. |
math/0010046 | CASE: The cyclotomic polynomial MATH factors over the field MATH into MATH distinct, monic, irreducible polynomials of degree MATH. If MATH is any one of those factors, then the field MATH is isomorphic to MATH. Let MATH be the automorphism of MATH induced by multiplication by MATH in MATH. Clearly, MATH has order MATH. Note the following: If we view MATH as the MATH-vector space with basis MATH, then MATH may be identified with the companion matrix of MATH, and so MATH, the characteristic polynomial of MATH. Alternatively, if we view MATH as the additive group of the field MATH, where MATH is a primitive MATH-th root of unity, then MATH. CASE: Notice that MATH is isomorphic to MATH, for any MATH: The mapping MATH, MATH, where MATH in the multiplicative group MATH, provides such an isomorphism. Now let MATH be an arbitrary matrix of order MATH in MATH. The characteristic polynomial of MATH must be one of the MATH irreducible factors of MATH. All such factors are the characteristic polynomials of some power of MATH. Thus, MATH, for some MATH. An exercise in linear algebra shows that there is a matrix MATH such that MATH. Hence, MATH. CASE: Let MATH. Every element in the semi-direct product MATH has unique normal form MATH, for some MATH and MATH. Write MATH. Straightforward computations show that MATH leaves the subgroup MATH invariant, and that the restriction MATH satisfies MATH. The number of solutions MATH of this equation is MATH. The count of automorphisms of MATH then follows from the following claim: There are precisely MATH values MATH for which MATH. To prove the claim, notice that MATH (both polynomials factor completely over MATH into linear factors, and the factorizations coincide). But, over MATH, the polynomial MATH has MATH distinct irreducible factors, all with multiplicity one, and so MATH has MATH irreducible factors, each appearing exactly MATH times. |
math/0010046 | The proof is by induction on the length of the word MATH. If MATH, the equality holds trivially. Suppose MATH, where MATH. Put MATH. We then have: MATH. Rewriting this last word in normal form (in the semidirect product MATH), we obtain the following equality (in the additive group MATH): MATH . Taking NAME derivatives of MATH, and applying MATH, gives: MATH . Now apply MATH, evaluate at MATH, and sum over MATH: MATH . |
math/0010046 | Let MATH be a presentation for MATH. Let MATH be a representation, given by MATH. We want to lift it to a representation MATH. Such a representation is given by MATH, where MATH. In view of REF , we must solve the following system of equations over MATH: MATH where MATH is the NAME matrix of MATH. This system has MATH solutions. Starting now with a non-trivial representation MATH, all such solutions give rise to surjective representations MATH, except MATH of them, which give rise to abelian representations. |
math/0010046 | Clearly, MATH, and the first identity follows from REF . NAME 's REF gives MATH. Recall that the subgroup lattice of the symmetric group MATH is MATH (see REF ). Thus, NAME 's REF gives MATH. Using REF , we get: MATH . The second identity now follows from REF . |
math/0010046 | The only group of order MATH is MATH; the only groups of order MATH are MATH and MATH; the only groups of order MATH are MATH (if MATH), and MATH and MATH (if MATH). The formulas follow from REF . |
math/0010046 | The proof is based on the computations displayed in REF . The first column lists the rigid isotopy classes (mirror pairs identified) of MATH-arrangements MATH (with MATH, or MATH and MATH horizontal), according to the classification by NAME, NAME\ĭ, and CITE. The next two columns list the NAME invariants MATH and MATH for the corresponding MATH-arrangement groups, MATH. These invariants are computed from the varieties MATH and MATH, using REF , as in REF . |
math/0010048 | REF will be proven by using the compactness properties of MATH and by comparing MATH with a suitable family MATH-converging to MATH. Let MATH be such that MATH. Upon extracting a subsequence, we can suppose that this liminf is actually a limit, so that in particular MATH. We will modify MATH so as to obtain a comparison sequence MATH in MATH. We first modify MATH to obtain piecewise-affine functions MATH. On MATH the function MATH is defined as MATH . Let MATH . For all MATH we define MATH on the interval MATH as the polynomial of degree MATH satisfying MATH . For such MATH the constant second derivative is given on MATH by MATH . On the remaining part of MATH we simply set MATH . Note that MATH but MATH. Moreover, MATH . If MATH then MATH . Finally, to complete the description of MATH, note that on the two intervals MATH and MATH is affine. Let MATH be defined on MATH, where MATH . By construction, we immediately obtain that MATH . In fact, recalling REF , we have MATH . It is easily checked that the MATH-limit of MATH is MATH. Note that MATH tends to MATH in MATH. Since MATH, by the coerciveness properties of MATH we obtain that MATH is precompact in MATH and each limit of a converging subsequence belongs to MATH (see CITE). Moreover, if MATH then also MATH, and by the MATH-convergence of MATH to MATH we obtain MATH and REF is proved. To prove REF , by a density argument it suffices to show it for MATH piecewise MATH and with bounded second derivative. Then, upon a slight piecewise-affine change of variable tending uniformly to the identity as MATH, we can reason as if for all MATH we have MATH. We can then take MATH on MATH (except at MATH, where we set MATH); that is, we choose as MATH in REF the piecewise-constant interpolation of MATH. We have: REF if MATH or MATH then REF if MATH then REF in all other cases, MATH with all the rests tending to MATH uniformly as MATH. Note that we have, in REF while in REF MATH as MATH. Taking into account REF - REF and the remark above, we immediately obtain MATH so that REF is proved. The final statement of the theorem follows easily from the well-known property of convergence of minima of MATH-limits. In fact, from REF and the lower semicontinuity of MATH with respect to the MATH convergence, we obtain MATH. On the other hand, if MATH is a solution of MATH, by REF we can find MATH as in REF so that MATH . Finally, the pre-compactness property of minimizing sequences follows from REF . |
math/0010051 | Only the ``if"-part remains to be shown. So let MATH be a leftinvariant subspace. Let MATH be an admissible vector for the regular representation, provided by REF . If MATH denotes the projection into MATH, then MATH is an admissible vector for MATH, since MATH, for all MATH. |
math/0010051 | Suppose MATH is an admissible vector. Then MATH, and the adjoint of MATH is given by MATH, where MATH, which defines a MATH function (here we use unimodularity). Hence, for every MATH, MATH, and the right-hand side is a continuous function. Hence for every MATH-function MATH there exists a continuous function which coincides with MATH almost everywhere. But then MATH is discrete. |
math/0010051 | For MATH, consider the measurable field of operators MATH, defined by MATH. Then, by the boundedness condition, MATH. Moreover MATH exists and defines a field of trace class operators with integrable trace class norm. Hence we may apply the inversion formula and obtain for the inverse NAME transform MATH of MATH: MATH . |
math/0010051 | The operator MATH commutes with left translation and hence has a decomposition MATH with a field MATH of essentially uniformly bounded operators. Of course the aim is to show that MATH-almost everywhere. For this purpose let MATH be a sequence with dense span in MATH. Then, since the NAME transform also intertwines the representations of the convolution algebra MATH arising from the left action of MATH, we have for all MATH that MATH for all MATH belonging to a common conull subset MATH. Moreover, possibly after passing to a suitable conull subset of MATH, we can assume that all MATH and MATH are bounded, and that the span of MATH is dense in MATH, for every MATH. But then by the continuity of the operators it follows that MATH for all MATH. |
math/0010051 | For the sufficiency we note that MATH, and we let MATH be the inverse NAME transform of that. Then REF shows that MATH, which means that MATH is the identity operator on MATH, and MATH is admissible. Now let MATH be an admissible vector for MATH, and define MATH. Then we have MATH and MATH (note that MATH is a bounded operator on all of MATH). Applying REF first to MATH and then to MATH yields MATH-almost everywhere. Hence MATH . |
math/0010052 | We construct MATH locally; patching together the local constructions in order to obtain a globally defined almost-complex structure still satisfying the same type of bounds is an easy task left to the reader (recall that the space of MATH-compatible almost-complex structures is pointwise contractible). Let MATH be a local tangent vector field of unit MATH-length and with MATH (observe that, because of the rescaling process, MATH is almost flat for large MATH). We define MATH, and observe that MATH has unit MATH-length (MATH is MATH-unitary and hence MATH-unitary) and MATH. Next, we proceed inductively, assuming that we have defined local vector fields MATH with the following properties for all MATH : MATH have unit MATH-length; MATH; MATH if MATH; MATH; MATH; MATH; MATH and MATH. We choose MATH to be a MATH-unit vector field which is MATH-orthogonal to MATH. The bound on MATH implies that we can choose MATH in such a way that MATH. Next, we define MATH . By construction, MATH for all MATH. Moreover, MATH. Since MATH, and because MATH is a component of MATH, we have MATH. Similarly, we have MATH; the first term differs from MATH by a MATH-term and is therefore bounded by MATH, while the bound on MATH implies that the second term is also bounded by MATH. Therefore we have MATH. Finally, using the lower bound on MATH we obtain that MATH. Finally, it is trivial that MATH ; therefore we can proceed with the induction process. We now define the almost-complex structure MATH by the identities MATH and MATH. By construction, MATH is compatible with MATH, and the corresponding Riemannian metric MATH admits MATH as an orthonormal frame. The required bounds on MATH immediately follow from the available estimates. |
math/0010052 | The argument is very similar to that used by CITE, except that one needs to be slightly more careful in showing that the various bounds hold uniformly in MATH. Fix a point MATH : then we can find a neighborhood MATH of MATH and a local coordinate map MATH, such that MATH contains a ball of fixed uniform MATH-radius around MATH, and such that the expressions of MATH and MATH in these local coordinates satisfy uniform bounds independently of MATH (these uniformity properties follow from the compactness of MATH). A linear transformation can be used to ensure that the differential of MATH at the origin is MATH-linear with respect to MATH. Next, we rescale the coordinates by MATH to obtain a new coordinate map MATH, in which MATH coincides with the standard almost-complex structure at the origin and has derivatives bounded by MATH, while the expression of MATH is bounded between fixed constants and has derivatives bounded by MATH. Next, we observe that the bound on MATH and the lower bound on MATH imply that the expression of MATH at the origin of the coordinate chart is bounded from above and below by uniform constants. Therefore, after composing MATH with a suitable element of MATH, we can assume without affecting the bounds on MATH and MATH that MATH coincides with the standard NAME form MATH of MATH at the origin. Define over MATH the symplectic form MATH. By construction, MATH. Observe that, in the chosen coordinates, the NAME connection of MATH differs from the trivial connection by MATH ; therefore, the bounds on MATH imply that the derivatives of MATH are also bounded by MATH, and that MATH. In particular, decreasing the size of MATH by at most a fixed factor if necessary, we obtain that the closed MATH-forms MATH over MATH are all symplectic, and we can apply NAME 's argument to construct in a controlled way a symplectomorphism between a subset of MATH and a subset of MATH. More precisely, it follows immediately from NAME 's lemma that we can choose a MATH-form MATH such that MATH, and such that MATH, MATH, MATH and MATH. Next, we define vector fields MATH by the identity MATH; clearly MATH and its derivatives satisfy the same bounds as MATH. Integrating the flow of the vector fields MATH we obtain diffeomorphisms MATH, and it is a classical fact that the map MATH is a local symplectomorphism between MATH and MATH and therefore defines NAME coordinates. Because MATH over a ball of fixed MATH-radius around MATH, the vector fields MATH satisfy a uniform bound of the type MATH for some constant MATH, so that MATH, and therefore MATH is well-defined over a ball of fixed MATH-radius around MATH. Moreover, the bounds MATH, obtained by integrating the bounds on MATH, and MATH, obtained from the bounds on the expression on MATH in the local coordinates, imply that MATH. Similarly, the bounds MATH and MATH for all MATH imply that MATH. This completes the proof of REF . |
math/0010052 | The argument is a direct adaptation of the proof of REF in NAME 's paper CITE. Pick a value of MATH and a point MATH. We work in the approximately MATH-holomorphic NAME coordinates given by REF , and use a trivialization of MATH in which the connection MATH-form becomes MATH. Then, we define a local section of MATH by MATH and observe that MATH is holomorphic with respect to the standard complex structure of MATH. Multiplying MATH by a cut-off function which equals MATH over the ball of radius MATH around the origin, we obtain a globally defined section of MATH; because of the estimates on the NAME coordinates one easily checks that the families of sections constructed in this way are asymptotically holomorphic and have uniform Gaussian decay properties CITE. |
math/0010052 | Let MATH be the stratification of MATH in which the only stratum is the zero section of MATH (these stratifications are obviously asymptotically holomorphic). By REF , starting from any asymptotically holomorphic sections of MATH (for example, the zero sections) we can obtain for large MATH asymptotically holomorphic sections of MATH which are uniformly transverse to MATH, that is, uniformly transverse to MATH. It is then a simple observation that the zero sets of these sections are, for large MATH, smooth approximately MATH-holomorphic (and therefore symplectic) submanifolds of MATH CITE. Finally, the uniqueness of the construction up to isotopy is a direct consequence of the one-parameter result REF. |
math/0010052 | Let MATH, and let MATH be the stratification of MATH consisting of strata MATH, MATH, defined as follows : viewing the points of MATH as elements of MATH with coefficients in MATH, each MATH is the set of all elements in MATH whose rank is equal to MATH. By REF , the stratifications MATH are asymptotically holomorphic. Applying REF to these stratifications and starting from the zero sections, we obtain asymptotically holomorphic sections of MATH which are uniformly transverse to MATH. The determinantal locus MATH is precisely the set of points where the graph of MATH intersects the stratum MATH. The result of uniqueness up to isotopy is obtained by applying REF . |
math/0010052 | By construction the NAME stratifications of MATH satisfy the assumptions of REF , as in every fiber of MATH they can be identified with the same holomorphic quasi-stratification of MATH. As a consequence, they are asymptotically holomorphic, and the existence of asymptotically holomorphic sections of MATH with the desired transversality properties is an immediate consequence of REF . The properties of MATH follow immediately from the uniform transversality to the stratum MATH of vanishing sections, while the properties of MATH are direct consequences of the uniform transversality to the NAME strata (recall that each MATH is smooth and is a union of strata). Finally, the uniqueness result is obtained by applying REF . |
math/0010055 | First we note the well-known facts that MATH . Recalling REF , we set MATH . This is a quadratic nonlinearity of the form REF. Thus, if MATH is null for each MATH, then the result follows by induction. In fact, if MATH, then MATH is the MATH-fold commutator MATH. A simple calculation shows that MATH . Thus, these commutators are null if MATH is null. We can express the angular momentum operators as MATH, MATH, where MATH is the tensor with value MATH, MATH if MATH is an even, respectively odd, permutation of REF, and with value REF otherwise. Using this, we find that the MATH component of MATH is MATH with MATH . To see that this commutator is also null, write MATH . Then MATH with MATH. Now the null condition says that MATH for MATH. But since MATH is tangent to MATH at MATH, we have MATH for MATH. This implies that MATH is null. |
math/0010055 | Spatial derivatives have the decomposition MATH . So if we introduce the two operators MATH and the null vectors MATH we obtain MATH . On the other hand, if we write MATH REF can be transformed into MATH . Thus, we have MATH . Now, we may assume that MATH, for otherwise the estimates are trivial. But then it follows that MATH and MATH are bounded by MATH, and as a consequence we have MATH . Using REF, we have MATH . The first term in REF vanishes since MATH obeys the null condition, and by REF the remaining terms in REF have the estimate REF. The proof of REF is similar. |
math/0010055 | This result is essentially REF. |
math/0010055 | Recall that the weighted norm involves derivatives in the form MATH . In the case when MATH, the result was given in REF. Otherwise, if MATH, then the result is an immediate consequence of REF. |
math/0010055 | By REF , we must estimate terms of the form MATH but since MATH, we will consider MATH with MATH, and MATH. Let MATH. We separate two cases: either MATH or MATH. In the first case, REF is estimated as follows using REF: MATH . Otherwise, we use REF to estimate REF by: MATH . |
math/0010055 | Let MATH, MATH. Since MATH, we have MATH. Thus, by REF , we find using our assumption MATH . Thus, if MATH is small enough, the bound REF results. Again since MATH, we have MATH. From REF we now have MATH . If we apply REF and our assumption, then MATH from which REF follows. |
math/0010056 | Take MATH, MATH, and MATH. By REF , MATH . This proves the first part of the corollary, and the second is immediate. |
math/0010056 | The point MATH belongs to MATH. Since this point is nonconstant, it cannot be a torsion point. |
math/0010056 | The first statement is immediate from REF . The second statement follows without difficulty by applying REF first to the extension MATH, and then to the extension MATH. |
math/0010056 | This follows directly from REF (with MATH and MATH), REF , and REF . |
math/0010056 | This follows directly from REF , and REF . |
math/0010056 | Both sides have the same divisor, and evaluate to MATH at MATH. |
math/0010056 | By REF , MATH. As in the proof of REF , MATH and the identity of the proposition follows. |
math/0010056 | Suppose first that we are in REF . Let MATH be the linear fractional transformation which (after identifying the roots of MATH with the nonzero elements of MATH) agrees with the given automorphism of MATH on the roots of MATH. It follows from the NAME of the automorphism that MATH. If MATH, then (since MATH) we must have MATH, and then the set of roots of MATH must be of the form MATH for some nonzero MATH. But this contradicts the fact that MATH, so MATH cannot be a linear polynomial. Hence in this case the theorem follows from REF . Now suppose we are in REF . Let MATH be the quotient of MATH by the given rational subgroup. Then MATH is an elliptic curve defined over MATH by a NAME model MATH, and there is an isogeny MATH of degree MATH, also defined over MATH. Let MATH where MATH is the MATH-coordinate of the isogeny MATH (as in REF ) and MATH is the the linear fractional transformation which maps the roots of MATH to the roots of MATH in the same way as the dual isogeny MATH maps MATH to MATH. Since MATH is defined over MATH, MATH. If MATH, then after replacing MATH by MATH we may assume that MATH. Then multiplication by MATH sends the roots of MATH to the roots of MATH, so MATH is the twist of MATH by MATH. Let MATH be an isomorphism over MATH. Then MATH and MATH. This is impossible since we assumed that MATH, so MATH cannot be a linear polynomial. Now the theorem follows in this case from REF . |
math/0010056 | That these points belong to MATH can be checked directly. Since they are nonconstant, both points have infinite order. The automorphism of MATH which sends MATH to MATH fixes the first point and sends the second point to its inverse, so they are independent in MATH. Since MATH, REF show that the rank cannot be greater than two. |
math/0010056 | As with REF , the simplest proof is a direct calculation. |
math/0010056 | Let MATH, MATH, and MATH, and apply REF with MATH and MATH. |
math/0010056 | Take MATH to be the permutation of MATH which switches MATH and MATH, and MATH to be the permutation which switches MATH and MATH. Then the linear fractional transformations MATH act on MATH as MATH and MATH do, respectively. One computes in REF that MATH and MATH where MATH . If MATH, then MATH and MATH are MATH-linearly independent. Setting MATH, and using that MATH, one obtains MATH . These formulas give us a rational point on the curve of genus zero defined by MATH. Using this point one computes that MATH where MATH and then MATH. Hence if MATH is as in the statement of the theorem, then MATH and the theorem follows from REF . The MATH points of infinite order are computed by taking points with MATH-coordinates MATH, MATH, and MATH, and expressing MATH in terms of MATH. |
math/0010056 | Take MATH to be the permutation of MATH which switches MATH and MATH, MATH to be the permutation which switches MATH and MATH, and MATH to be the cyclic permutation MATH. Let MATH be the corresponding linear fractional transformation. Then in REF we have MATH where MATH . Now suppose MATH with MATH. Then MATH and MATH are MATH-linearly independent, and setting MATH we find MATH . These formulas give us a rational point on the curve MATH. If MATH with MATH, then MATH and MATH are MATH-linearly independent, and setting MATH we find MATH . These formulas give us a rational point on the curve MATH. The theorem now follows from REF . |
math/0010056 | We may write MATH as MATH where MATH has MATH rational roots. If MATH denotes the rational cyclic subgroup of order MATH, then MATH contains a rational point, and we may choose our model so that this point is MATH. Denote the other roots of MATH by MATH and MATH. If MATH is a generator of MATH and MATH is its MATH-coordinate, then MATH and a computation gives MATH. Hence MATH is a square, and we write MATH with MATH. Thus MATH is given by MATH. The quotient of MATH by the group generated by MATH is MATH . The isogeny from MATH to MATH is MATH where MATH . The linear fractional transformation MATH sends the roots of MATH to the roots of MATH. Set MATH. Let MATH be the permutation of MATH which switches MATH and MATH, and let MATH be the corresponding linear fractional transformation. One computes in REF that MATH and MATH where MATH . Setting MATH we find MATH . These formulas give us a rational point on the curve defined by MATH. Using this point one computes that MATH where MATH . We can solve for MATH in terms of MATH (see CITE). The theorem then follows from REF . |
math/0010056 | The simplest proof is a direct computation. To construct this example one takes MATH to be MATH and proceeds exactly as in the proofs of REF , with MATH, MATH, which gives MATH . The curve defined by MATH has a rational point MATH, and using this one computes that MATH where MATH. REF with this input leads to the data above. |
math/0010056 | Without loss of generality we may assume that MATH, since if not, MATH by REF and there is nothing to prove. Let MATH, a homogeneous polynomial of degree MATH. REF are immediate from REF, respectively, applied to MATH. Suppose now that the Parity Conjecture holds, the irreducible factors of MATH all have degree at most MATH, and MATH is such that MATH and MATH. Choose a closed interval MATH with rational endpoints which contains MATH but does not contain any roots of MATH, and let MATH be a linear fractional transformation which maps MATH onto MATH and (for simplicity) such that MATH. Replace MATH by the polynomial MATH of degree at most MATH. Then we still have that MATH, and our construction guarantees that this new polynomial MATH also satisfies: CASE: the constant term of MATH and the coefficient of MATH are both nonzero, CASE: the irreducible factors of MATH have degree at most MATH, CASE: MATH and MATH, CASE: MATH is positive if MATH. Further, multiply MATH by the square of an integer to clear denominators of the coefficients. If MATH is an elliptic curve over MATH, write MATH for its conductor. If further MATH and MATH is relatively prime to the conductor of the character MATH associated to the quadratic extension MATH, then MATH. Applying this with MATH and MATH for MATH and MATH positive integers congruent to MATH modulo an integer MATH sufficiently divisible by the prime divisors of MATH, and using REF above, gives that MATH . Let MATH . By REF, for MATH, MATH . (Note that as stated, REF does not include the restriction MATH in our definition of MATH. However, the proof in CITE does restrict to positive MATH.) Theorem C of CITE implies that MATH for all but finitely many MATH. However, by REF, if MATH then MATH so the Parity Conjecture tells us that MATH. Hence MATH for all but finitely many MATH, and so REF follows from the NAME bound REF. |
math/0010056 | If MATH has a real root then MATH contains both positive and negative values (MATH has no multiple roots because it was assumed to be squarefree). Thus by a result of NAME REF we have MATH . Now the corollary follows immediately from REF . |
math/0010056 | This is immediate from REF . |
math/0010056 | Suppose first that we are in REF . Translating the given rational root of MATH we may assume that MATH with MATH. Since MATH has MATH real roots we also have MATH. In particular, MATH. Let MATH be as in REF . Then MATH is divisible by MATH. We compute that MATH, but MATH is positive for large MATH, so MATH, and hence MATH, has real roots. Hence the Corollary in this case follows from REF . Similarly, suppose we are in REF . Then as discussed before REF , MATH has a model MATH with MATH, MATH. The discriminant of this model is MATH. Since all the MATH-torsion on MATH is defined over MATH, we have MATH. Let MATH be as in REF . Then MATH is positive for large MATH, but MATH. Hence MATH has real roots, so the Corollary in this case follows from REF . |
math/0010056 | This is immediate from REF , and REF (the last to handle the excluded value MATH in REF ). |
math/0010056 | REF follows directly from REF . The polynomial MATH of REF has degree MATH, and hence it has a real root, so REF follows from REF . |
math/0010056 | If MATH and MATH, let MATH denote the squarefree part of MATH, that is, the unique squarefree integer MATH such that MATH for some integer MATH. For every squarefree integer MATH let MATH denote the hyperelliptic curve MATH of genus MATH. The map MATH defines an injection MATH (where MATH denotes the hyperelliptic involution on MATH). Thus by Conjecture REF the order of the set on the left is bounded by MATH. Let MATH . There is a constant MATH such that if MATH then MATH. It follows that MATH for MATH. But it is standard to show that MATH for MATH, and the proposition follows. |
math/0010066 | Suppose the ring has length REF. Then the component is a rooted tree with a loop at the root. The loop has no effect on the permutation generated. Ignore it, and suppress the corresponding transposition. This leaves us with a permutation of the form MATH whose associated graph is a tree MATH with vertex set, MATH say, of cardinality MATH, and with edge set MATH . We wish to prove by induction on MATH that REF is a MATH-cycle. For MATH this is true. For MATH larger, let MATH be any edge of MATH with MATH a leaf vertex. Deleting MATH and MATH leaves a tree on the vertex set MATH of cardinality MATH, so, by the induction hypothesis, the permutation MATH is a cycle. Write MATH . Then MATH is conjugate to MATH . However, a conjugate of a cycle is a cycle. This completes the induction. Next, we treat the case of a ring of length REF or greater. Let MATH . Choose an edge MATH of the ring. Deleting it changes our graph component into a tree, so MATH is of the form REF . Then, as we just proved, MATH is a MATH-cycle. However, a product of a MATH-cycle and a transposition of two of its members will equal a product of two disjoint cycles whose lengths sum to MATH, thus proving the lemma. In fact, if we write MATH will be conjugate to MATH a product of two disjoint cycles whose lengths sum to MATH. But conjugation does not change cycle structure, so MATH also has this form. |
math/0010066 | Any representation of MATH as an exchange shuffle will determine a partition of the set MATH into one- and two-element blocks, the one-element blocks being those cycles represented as trees, and the two-element blocks those pairs of cycles represented as unicycles. Think of such a partition as an involution, by letting an involution MATH partition MATH into its orbits under MATH. The product of MATH over the fixed points MATH of MATH multiplied by the product of MATH over the two-element orbits MATH of MATH will then equal the number of representations of MATH yielding this MATH, and summing this over all involutions MATH will yield MATH. |
math/0010066 | We have MATH . For MATH, an involution on MATH can either fix the first element, leaving MATH ways of moving the remaining elements, or swap it with a member of MATH, leaving MATH ways of moving the remaining elements. Hence MATH . If we set MATH then induction on MATH proves that, for all MATH, MATH . From the above, MATH, so MATH will be strictly increasing for MATH. REF require a calculation, which we omit. |
math/0010066 | See NAME and NAME. |
math/0010066 | We have just seen that MATH. Since MATH, the first inequality is true for MATH. We have the following form of NAME 's formula CITE: MATH . For MATH, this yields the estimate MATH, which implies the result. For the second inequality, let MATH be the maximum of MATH over all permutations MATH of MATH. We will induce on MATH to prove that MATH for all MATH. For MATH to be nonzero, MATH must be a product of two disjoint cycles, MATH, say. Without loss of generality, let MATH. Represent MATH in the form REF , and let the first transposition be MATH, MATH. The remaining transpositions must represent MATH, and they will still do so if we precede them by MATH. If MATH, MATH is a MATH-cycle, so the representation is as a tree, and the tree has a loop at REF. We thus get at most MATH choices for the remaining transpositions; by REF , this is no more than MATH. If MATH, MATH . Our representation of MATH has a loop at REF, so it must contain a tree component rooted at REF representing MATH; for the representation of MATH to be unicyclic, it must also contain a unicyclic component representing MATH, so by REF , the number of ways of filling in the remaining transpositions is then no more than MATH. Then MATH and since for all possible MATH, MATH, MATH . MATH for MATH can be checked via direct computer enumeration (or using some of our results below), and using the bound REF recursively then yields MATH for MATH; so let MATH. Then by induction, MATH . Substitution into the generating function for MATH yields MATH, and estimation with NAME 's formula as above yields MATH for MATH, that is, MATH. This completes the proof. |
math/0010066 | Let MATH be an arbitrary permutation of MATH, and let MATH be its decomposition into disjoint cycles. By REF , MATH . Plugging in the bounds from REF above, MATH that is, MATH . From REF , it follows that for MATH, MATH is maximized at MATH, so MATH unless MATH, that is, unless MATH is the identity permutation. However the multiplicity of the identity permutation is MATH, by CITE, or directly from REF . This completes the proof. |
math/0010066 | First, let the component be a tree. We will prove the induced permutation is a cycle, but now without using conjugation. As before, let us suppose we have a permutation of the form REF : MATH . We wish to prove by induction on MATH that REF is a MATH-cycle. We will need a stronger induction hypothesis. Let MATH, , MATH be thought of as an increasing sequence of times of day, and let a postman travel on the set MATH. If he is at MATH at time MATH, he will move to MATH, or, if he is at MATH at time MATH, he will move to MATH. Otherwise, he will stay fixed. Our induction hypothesis is that, regardless of where and at what time of day the postman starts, he will return to the same place at the same time of day MATH days later, and that his intervening journey will, for any time of day MATH and MATH, find him at MATH at time MATH exactly once. For MATH this is clear, since the postman simply stays put for one day. Otherwise, let MATH be any edge of MATH. Deleting MATH will split MATH into subtrees; let the subtree where the postman starts his journey be MATH, with MATH vertices, say, and let the other subtree be MATH, with MATH vertices. By the induction hypothesis on MATH, the postman will travel until he reaches an endpoint of MATH at time of day MATH, when he will travel to the other endpoint of MATH in MATH. Then, by the induction hypothesis on MATH, he will find himself at the endpoint of MATH at time of day MATH exactly MATH days later, when he will travel back to MATH. His journey there will then go on as if he had never left, so he will arrive back at his starting point and starting time of day after having spent MATH days in MATH and MATH days in MATH, for a total of MATH days. In fact, his cyclical journey in MATH simply consists of his journeys in MATH and MATH spliced together, so it's clear that the induction hypothesis is still satisfied. If we let the postman be at MATH at the start of the day (before any time MATH), then he will be at MATH at the end of the day (after all times MATH.) We may say that he stops for the night there. At any rate, he is still at MATH at the start of the next day; he is at MATH at the start of the day after that; and so on. Our induction hypothesis then tells us that MATH, MATH, , MATH are distinct, so MATH is a MATH-cycle, as desired. If the component is unicyclic, suppose at first that there are no edges other than those in the ring, and draw the ring in the plane so that it is oriented counterclockwise. If the postman spends the night at an upper vertex, he will leave via the incoming edge, in a clockwise direction, and continue to travel clockwise, passing through all the lower vertices he meets on the way, before spending the next night at the next upper vertex. Therefore, the upper vertices form a cycle in the induced permutation. Similarly, if he spends a night at a lower vertex, the postman will spend his next night at the next lower vertex in a counterclockwise direction, so the lower vertices also form a cycle. We illustrate this in REF . In the general case, the unicyclic component will contain trees attached to the ring by edges. However, we already know that each such tree will, on its own, generate a cyclical journey. When attached to the ring, these journeys will be spliced into the journeys around the cycles of upper and lower vertices, as explained above in the proof that the induced permutation on a tree is a cycle. This completes the proof. |
math/0010066 | First we comment on what is to be proved. Start with the left-hand side above. Suppose we are given a representation of MATH as a unicyclic exchange shuffle, and suppose that in the associated directed graph MATH is the upper and MATH the lower cycle. Break up the ring into maximal segments of consecutive vertices that are all upper or all lower. Say there are MATH such segments. Beginning with an arbitrary upper segment, write the vertices of the ring as MATH . Of course, there are MATH ways of doing this, depending on which upper segment we begin with. We construct a forest of MATH trees by, for each MATH, , MATH, doing the following: CASE: Start with the digraph representing MATH . CASE: Follow the postman as he traverses the part MATH of the ring, and certain trees hanging off the ring (see REF ). CASE: Use the dummy vertex REF for any ring vertices the postman traverses that are not in MATH . In this way, we get upper trees MATH for MATH. By a similar process, to be described below, we get lower trees MATH for MATH. We get corresponding decompositions MATH and MATH . The details of these decompositions are illustrated in REF . These MATH trees and MATH sequences satisfy the properties implied by the right-hand side above: CASE: The tree MATH represents MATH . The tree MATH represents MATH . CASE: The largest neighbor of MATH in the tree MATH is greater than the first element of MATH. The smallest neighbor of MATH in the tree MATH is less than the last element of MATH. We let these indices wrap around modulo MATH, so that MATH. The content of REF is that this decomposition exists, and, conversely, that any choice of positive integer MATH and choice of MATH subsequences and MATH tree representations thereof satisfying REF above give rise to a unicyclic representation of MATH where MATH is the upper and MATH the lower cycle. Next we proceed to the proof. Consider a representation of MATH as a unicycle, where MATH is the upper cycle. Take the vertices of the ring to be, in order, MATH, MATH, , MATH, where MATH is upper and MATH lower. We will let these indices wrap around modulo MATH, so MATH, MATH, and so on. Start with a maximal consecutive subsequence MATH of upper vertices among the MATH's, and let MATH be the following maximal consecutive subsequence of lower vertices. Suppose that our postman has just started out for the day from MATH which is an upper vertex, by maximality. What happens to him in the remainder of his journey? Since MATH is upper, MATH, so he will eventually be leaving along the edge labeled MATH; but first, he may traverse subtrees of the unicyclic component attached to the ring at MATH. Call a collection of subtrees attached to the ring by edges labeled MATH, where MATH, a MATH. We can now say that our postman will traverse the MATH incident on MATH. Following this, he arrives at MATH. This is a lower vertex, so MATH. He will therefore traverse the incident MATH and continue to MATH without stopping for the night. If this vertex is a lower vertex, he will traverse MATH and continue to MATH, and so on. Eventually, after traversing MATH, he will come to MATH. This is an upper vertex, so MATH, meaning that he will traverse the incident MATH, and then stop for the night at MATH. The next day, he will traverse the MATH before traveling to MATH. Here he will traverse MATH, stop at MATH, and traverse MATH before passing to MATH, and he will continue in this way until reaching MATH and stopping for the night. Since we are now in the same position we started out in (although with different values of MATH, MATH, and MATH), we end our description here. This travelogue generates a sequence which forms a portion of the cycle MATH. Equating our MATH's with the sequence of vertices they generate when traversed, we may write it as MATH . Generate a tree MATH by the following procedure: take the vertices MATH, MATH, , MATH, and the edges connecting them; add to this all the MATH's above rooted on these vertices. Add an edge from MATH to REF, and take all the trees MATH, MATH, , MATH appearing above and reroot them at REF. Walking the postman through MATH tells us immediately that it is a representation of MATH. Observe that MATH, and that MATH. All this is illustrated in REF . Let us interchange `lower' and `upper' above, so that our starting subsequence is a maximal consecutive subsequence MATH of lower vertices, MATH, , MATH are upper, and MATH is lower. Supposing that our postman is just about to stop for the night at MATH, he will travel along the sequence MATH before stopping for the night at MATH, where we will leave him. As before, we may view MATH as being part of MATH. Generate a tree MATH by taking the vertices MATH, MATH, , MATH, and the edges connecting them; add to this all the MATH's above rooted on these vertices. Add an edge from MATH to REF, and take all the trees MATH, , MATH, MATH appearing above and reroot them at REF. Walking the postman through MATH tells us that it is a representation of MATH. In this case, MATH, and MATH. This is illustrated in REF . Split the vertices occurring in the ring into blocks of upper and lower vertices, so that MATH for some sequences MATH, , MATH of upper and MATH, , MATH of lower vertices. We can then apply the above transformations to generate MATH, , MATH, MATH, , MATH and associated tree representations MATH of the MATH's and MATH of the MATH's. Looking at the postman's journey tells us that MATH and MATH. As above, let the first vertex of MATH be MATH. This must be upper, so MATH; but MATH is the last vertex of MATH, which is MATH (letting indices wrap around modulo MATH), so MATH. Similarly, MATH. In this way, we get a split of MATH and MATH into subsequences, and tree representations thereof, of exactly the kind counted on the right-hand side of REF . In fact, for each unicyclic representation whose ring contains MATH blocks of upper vertices interleaved with MATH blocks of lower vertices, we get MATH such splits and representations, since we had MATH choices for MATH. This gives a map from the set of things counted on the left-hand side of REF to the set of things counted on the right-hand side of REF . To find an inverse map, we need to be able to splice a number of tree representations together into a unicycle. Suppose that we start with a tree MATH which represents MATH. Our first task is to construct the corresponding MATH, say. We may let MATH be the largest vertex with an edge to REF. Set MATH to be the largest vertex less than MATH with an edge to MATH, set MATH to the largest vertex less than MATH with an edge to MATH, and so on, until we reach a vertex MATH such that no MATH has an edge to MATH. The sequence of vertices constructed in this way obviously has MATH. The tree MATH will consist of the path of edges from MATH to REF, together with a MATH rooted at MATH, a MATH and a MATH rooted at MATH, a MATH and a MATH rooted at MATH, and so on, until we reach REF, which will have a MATH rooted at it. We therefore have MATH and MATH. Similarly, if we start with a tree MATH representing MATH, we need to construct MATH. Let MATH be the smallest vertex with an edge to REF, MATH the smallest vertex greater than MATH with an edge to MATH, MATH the smallest vertex greater than MATH with an edge to MATH, and so on, until we reach a vertex MATH such that no MATH has an edge to MATH. In this case, MATH, MATH, and MATH. Now that we have constructed the MATH's and MATH's, we can concatenate them into a ring by placing an edge from the last vertex of each MATH to the first vertex of MATH, and the last vertex of MATH to the first of MATH. We delete the edges from the last vertices of the MATH's and MATH's to REF, and leave the trees incident upon the vertices of the MATH's and MATH's in place. Let MATH be the first vertex of MATH. The inequality MATH then tells us that MATH exceeds the last vertex of MATH, that is, MATH. Similarly, if MATH is the first vertex of MATH, the inequality MATH tells us that MATH. This tells us that the vertices we wish to be upper are indeed upper, and the ones we wish to be lower, lower. The MATH of MATH rooted at REF must be unrooted and rerooted on the vertices of MATH and the first vertex of MATH, so as to represent MATH in the form REF ; similarly, the MATH of MATH rooted at REF must be rerooted on the vertices of MATH and the first vertex of MATH, so that MATH is represented in the form REF . From the inequalities we have established, we see that this can be done in a unique way. This construction of a unicyclic representation of MATH shows us that we have a map from the set of things counted by the right-hand side of REF to the set of things counted by its left-hand side. This map is inverse to our previous map, so both maps are bijections, and the left- and right-hand sides of REF are equal. This completes the proof. |
math/0010066 | If every member of MATH is less than every member of MATH, then MATH must always vanish, yielding MATH. On the other hand, if MATH is a tree with root REF representing MATH, then the condition MATH is vacuous, so MATH . Similarly, MATH and substituting into REF yields the desired result. Of course, the first equation is obvious and does not require REF . |
math/0010066 | Represent MATH as a product of transpositions, MATH . We will exhibit a sequence MATH with MATH . Let MATH be the permutation of MATH that moves the first MATH terms past the last MATH. Then by REF and similarly MATH . Using MATH, REF becomes MATH . Rearrange terms to get: MATH . Now we use repeatedly the conjugation rule MATH to obtain MATH where MATH . Now rearrange again to get MATH . This shows us that REF induces a map from the representations of MATH to those of MATH. Interchanging the roles of MATH and MATH, and of MATH and MATH, in REF will give us a map from representations of MATH to those of MATH, and it will be inverse to this map. This shows that the pairing MATH yields a bijection of representations of MATH and MATH. |
math/0010066 | Without loss of generality, let MATH end in its smallest element, which we call MATH. We claim that MATH and MATH where in these sums MATH, but not MATH, MATH, or the MATH's, may be empty. The lemma will follow from these claims. To see where MATH is maximized, use REF . If MATH is decreasing, then MATH, MATH, , MATH, and MATH will also be decreasing, so by REF , each term on the right-hand side of REF will be maximized. If MATH, then the term MATH will appear on the right-hand side of REF . By REF , this will be maximum only when MATH is a cyclic permutation of a decreasing subsequence, so REF is maximized only when claimed. MATH and MATH must be handled together by induction on the length of MATH. For each possible length of MATH, if MATH and MATH are maximized when MATH is decreasing, MATH obviously will be as well, and, if MATH, this will be the only place the maximum occurs, since it is the only place where MATH is maximized. For MATH, use REF . The induction hypothesis and REF will then show that the right-hand side of REF is maximized when MATH is decreasing. We now prove our claims. Proof of REF . To prove REF , look at its left-hand side. It counts MATH-tuples MATH of trees rooted at REF, representing MATH, , MATH, respectively, where MATH and MATH, , MATH do not contain the transposition MATH. The MATH's must contain a unique transposition MATH, where MATH; if this transposition is in MATH, removing it from MATH will leave us with a tree MATH rooted at MATH and a tree MATH rooted at REF which jointly represent MATH. Then MATH and REF must appear in separate cycles in MATH, so MATH appears after MATH in MATH. Recall that the sequence MATH ends with the element MATH. Let MATH be the initial segment of MATH ending in MATH, and let MATH be the remainder of MATH, so MATH. Since MATH appears after MATH in MATH, it follows that MATH is a subsequence of MATH. Delete MATH from MATH to give MATH. Now MATH so MATH represents MATH and MATH represents MATH. In fact, MATH, , MATH, MATH, MATH, , MATH represent MATH, , MATH, MATH, MATH, , MATH, respectively, so MATH is a representation of the form counted by MATH. We have already remarked that MATH is a tree rooted at MATH representing MATH. This gives a map from the set counted by the left-hand side of REF to the set counted by the right-hand side of REF . This map is easily seen to be invertible, proving REF . Proof of REF . The left-hand side of REF counts the same thing as the left-hand side of REF , except that the transposition MATH is required instead of forbidden. Also, we assume that MATH contains MATH; this gives the factor of MATH. Let MATH, where MATH ends with MATH. If the transposition MATH is deleted from MATH, we will be left with a tree MATH rooted at MATH, representing MATH, and a tree MATH rooted at REF, representing MATH. However, MATH, so the MATH-tuple of trees MATH is of the form counted by the right-hand side of REF . This gives us a map from the set counted by the left-hand side of REF to the set counted by the right-hand side of REF . This map is invertible, so REF holds. |
math/0010066 | Let MATH and MATH be sequences of distinct positive integers with disjoint ranges, and let MATH. We wish to show that, for fixed MATH and MATH, MATH is maximized iff we can write MATH . We will induce on MATH. Without loss of generality, let REF be the smallest element in the domain of MATH. Exchange MATH and MATH, if necessary, to put REF in MATH, so that we can set MATH. In REF , then, we must have MATH, so MATH. Let MATH be a unicyclic representation of MATH, with ring MATH. MATH will contain some transposition MATH, MATH. If REF is not in MATH, then deleting the edge from REF to MATH from MATH will leave a graph with a tree component MATH rooted at REF and a unicyclic component MATH. This graph will represent MATH. Considering the postman for a moment, we see that in MATH, after stopping for the night at MATH, he must proceed immediately to REF and traverse MATH. Hence MATH and REF are in the same cycle of MATH, and MATH for some MATH. Evidently, MATH must represent MATH, and MATH must represent MATH. Conversely, given a tree rooted at REF representing MATH, and a unicycle representing MATH, we can insert an edge from REF to MATH to give a unicyclic representation of MATH. Hence MATH where MATH is the portion of MATH coming from graphs with REF not in MATH. By REF , the MATH term will be maximized just when MATH, which will certainly happen in REF , and in this case, by the induction hypothesis, the MATH term will be maximized as well. Hence MATH is never any bigger than it is in REF . Let MATH count the remaining unicyclic representations, that is, those where REF is in MATH. Let there be edges in MATH from MATH to REF and from REF to MATH. Since MATH and MATH, REF is lower and MATH is upper, so MATH must be the upper cycle and MATH the lower. Also, using the notation of the proof of REF , the last element of some MATH is REF; that is, MATH for some MATH. Conversely, if MATH for some MATH, REF will certainly appear in MATH. Therefore, we can think of MATH as the right-hand side of REF , if we apply the following constraint: in the sets of tree representations that this right-hand side counts, delete all tree representations not containing a tree MATH with MATH. In REF , the inequalities between MATH, MATH, MATH, and MATH in REF are vacuous, so we may remove them, and think of MATH as the right-hand side of REF , subject to our constraint. In all other cases, relaxing these inequalities does not decrease the right-hand side of REF , so it will do to prove that the right-hand side of REF , subject to our constraint, is never bigger than in REF . Our constraint is equivalent to demanding that a transposition MATH occur somewhere in the tree representations of the MATH's, so the right-hand side of REF with our constraint applied is MATH . It follows from REF that REF is maximized in REF . We still need to show exactly when the maximum occurs. For all MATH, MATH, and MATH is positive by CITE. Define MATH to be the number of tree representations of MATH with REF as the root and a transposition MATH present, for MATH the smallest element of MATH. Then, for any MATH, let MATH be MATH cyclically rotated so that its smallest element, MATH, is at the end. In this case, MATH. As in the proof of REF , if a tree representation of MATH rooted at REF contains the transposition MATH, we can delete it to get a tree representation of MATH rooted at MATH. This transformation is reversible, so MATH and by CITE, this is positive. Hence MATH and MATH are always positive. Also, by REF , we know that the maximum of MATH and MATH can only occur when MATH is a cyclic permutation of a decreasing sequence. Looking at the MATH term in REF , then, we see that for the maximum to occur, we must be able to write MATH . If MATH, we are done. Otherwise, write MATH, and look at the MATH term in REF . By the induction hypothesis, if we are at the maximum, REF must be satisfied for MATH and MATH. Therefore, either MATH, and we are done, or MATH. If MATH, look at the representations of MATH with REF in MATH. Each MATH must be exceeded by a MATH; therefore, REF is the only possible value for any MATH, so MATH in REF and MATH. Ignoring the other conditions on MATH, MATH, MATH, and MATH that occur in REF , we see that MATH . Let MATH denote a shifted version of MATH with every element greater than every element of MATH. Then, as we know, MATH and since we know that MATH, it will do to show that MATH . Now, MATH is a sum of MATH over cyclic permutations MATH of MATH. Since MATH, MATH, so it will do to show that MATH; but we know this from above. This shows that the case MATH cannot occur, completing the proof. |
math/0010066 | Let MATH be an arbitrary permutation of MATH, and let MATH be its decomposition into disjoint cycles, where MATH is a MATH-cycle, MATH is a MATH-cycle, , and MATH is a MATH-cycle. By REF , MATH . Using the bounds in REF then shows that MATH is no bigger than REF and that the equality condition is as claimed. |
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