paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/0102010 | To establish the first relation we compute, using REF , for all MATH: MATH . Next we recall the formula for MATH from REF , formula for MATH from REF , Lemma (REF d), and that MATH: MATH . The second identity follows from the first by REF , since the symmetry of MATH and the anti-(co)multiplicative properties of the antipode imply : MATH . |
math/0102010 | NAME follows from REF . We have established all the axioms of a weak NAME algebra except REF , which we show next. At a point below, we let MATH, at another MATH, and use REF as well as REF . Let MATH be the element from REF implementing the inner automorphism MATH, then for all MATH, MATH . |
math/0102010 | From REF it follows that MATH satisfies the measuring axiom. From REF it follows that MATH. The action of MATH on MATH is a left module action of an algebra by the NAME relations and MATH for MATH. Recall that MATH. Since MATH, it is clear that MATH for every MATH. We compute for every MATH: MATH whence REF follows. Thus the action of MATH on MATH coincides with the standard left action of a weak NAME algebra MATH on its dual MATH as in REF . Since the invariant subalgebra MATH is MATH, it follows that MATH. |
math/0102010 | We subsequently use REF , Lemma (REF d) and its opposite (obtained by applying MATH), REF , and REF in the next computation: for every MATH, MATH . Next note that MATH for all MATH, which follows from an application of MATH to REF . Then let MATH and compute: MATH . |
math/0102010 | That MATH is a linear isomorphism follows from REF . That MATH is a homomorphism follows almost directly from REF and the conjugation formula in REF : MATH since for all MATH: MATH. |
math/0102010 | First we check that MATH given MATH. Let MATH, MATH, denote the coaction dual to the action MATH above. Then MATH. It follows from REF that MATH restricted to MATH is the comultiplication: MATH . Since MATH is shown above to be the invariant subalgebra of this action of MATH on MATH, it is also precisely the coinvariant subalgebra of MATH. We then compute using REF : MATH whence MATH. Since MATH, we compute that MATH measures MATH: MATH . We note also that MATH and that MATH by the homomorphism and anti-homomorphism properties of MATH and MATH. Finally, MATH since both MATH and MATH belong to MATH, while MATH. |
math/0102010 | That MATH is a linear isomorphism follows from REF . That MATH is a homomorphism follows from the conjugation formula in REF : MATH since for all MATH: MATH. |
math/0102010 | If MATH, then for every MATH: MATH using the definition of a module algebra over a weak NAME algebra. We similarly compute for each MATH: MATH . From the bijectivity of MATH and MATH, it follows that MATH, so that MATH, whence MATH. |
math/0102010 | The proof in CITE that MATH, MATH and MATH is valid here as it only makes use of the MATH-algebra MATH, the subalgebra of MATH-generated by MATH, and an obvious involution on it. Note that the theorem is true for MATH (where MATH). Assume inductively that the proposition holds for MATH and less. We use the induction hypothesis in the second step below, and the NAME identities for sets MATH in the fifth step: MATH where the last step is by CITE. Let MATH denote the shift map of MATH induced by MATH. It follows from the induction hypothesis that MATH is the NAME idempotent for the composite expectation MATH . Let MATH and MATH. For the computation below, we note that MATH and by CITE: MATH . We compute: MATH . |
math/0102010 | Let MATH be dual bases of MATH. It follows that MATH via MATH. By the symmetry condition on MATH, MATH restricted to MATH is a trace with values in MATH. Then MATH is the symmetric separability element and MATH gives a MATH-linear projection of MATH onto MATH coinciding with MATH, since MATH is an NAME MATH-algebra CITE. Let MATH and MATH in MATH: these are dual bases of MATH by the Basic Construction Theorem. But we see that MATH. Next we compute that there are dual bases MATH for MATH. By the construction of the last paragraph, it follows that MATH has dual bases in MATH, whence MATH has depth MATH. We let MATH and MATH, both in MATH. It suffices to compute for MATH: MATH . Similarly we compute MATH by using the equivalent expressions MATH and MATH. |
math/0102010 | By the results of CITE, the center of MATH is trivial and MATH via MATH for MATH. But by the hypothesis MATH has non-degenerate trace MATH with dual bases MATH. It follows that MATH defined by MATH, where MATH, has dual bases in MATH. The conclusion now follows readily from the proposition. |
math/0102011 | Let MATH be a family witnessing that MATH is perfectly meager. Let MATH consist of those ordinals of cofinality MATH that for every MATH, MATH. The usual argument involving NAME theorem shows that MATH has the required property. |
math/0102011 | Let MATH be a meager set, and let MATH. Fix a descending sequence of open sets MATH such that each MATH is dense in MATH and MATH. By induction build a sequence MATH such that MATH, and for every MATH, CASE: MATH, CASE: MATH. Suppose that MATH is given. For every MATH find MATH such that MATH. Let MATH. REF has the required property. Suppose that MATH is a possible witness that MATH is perfectly meager, and let MATH. Find MATH such that MATH and let MATH be such that MATH. Clearly, MATH . In particular, MATH . |
math/0102011 | The first part is a special case of a more general fact. For MATH let MATH be a maximal antichain below MATH such that MATH . Use REF to find MATH such that for every MATH, MATH is finite. Let MATH. Without loss of generality we can assume that MATH. It follows that MATH is clopen in MATH for every MATH. Define MATH as MATH if for every MATH there exists MATH such that MATH and MATH. It is easy to see that MATH is a continuous function that has the required properties. To show the second part we need to build MATH in such a way that for every MATH, there is MATH such that MATH. |
math/0102011 | Consider the following two cases. CASE: MATH. Let MATH and MATH be such that MATH. Clearly MATH is constant with value MATH. CASE: MATH. Build by induction a sequence of REF such that MATH and for every MATH, CASE: MATH, CASE: sets MATH are pairwise disjoint and have diameter MATH . Suppose that MATH is given. Note that MATH is uncountable for every MATH. For MATH choose pairwise different reals MATH. It is not important now but will be relevant in the sequel, that we can choose these reals ``effectively" from a fixed countable subset of MATH. Let MATH be such that sequences MATH are also pairwise different. For every MATH let MATH be such that for every MATH, MATH. If MATH is as in the second part of REF then we can find MATH in MATH. Define MATH . Observe that MATH has the required property. |
math/0102011 | MATH. Fix MATH and for MATH let MATH be a maximal antichain below MATH such that CASE: MATH is compact. CASE: MATH. Fix a sequence MATH such that for MATH, CASE: MATH, CASE: MATH, CASE: MATH. By induction build a sequence MATH such that for MATH, CASE: MATH is compact, CASE: MATH, CASE: MATH. Let MATH. As in the proof of REF we show that there exists a continuous function MATH (encode elements of MATH as reals) such that MATH . Consider MATH. Clearly, MATH is compact in MATH. Remaining requirements are met as well. CASE: MATH is limit. Given MATH fix sequences MATH and MATH such that CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH. By induction build a sequence MATH such that for MATH, CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH is compact in MATH. Let MATH. Note that MATH is as required. |
math/0102011 | As before, without loss of generality we can assume that MATH is countable. Induction on MATH. CASE: MATH. Suppose that MATH and MATH is meager, and let MATH and MATH be given. Let MATH . Using the fact that MATH is homeomorphic to MATH via homeomorphism respecting vertical sections, and by NAME theorem, we conclude that MATH is a meager set in MATH. Recall the following classical lemma: Suppose that MATH is a NAME set. CASE: Assume MATH is meager for all MATH. Then there exists a sequence of NAME sets MATH such that CASE: MATH is a closed nowhere dense set for all MATH, CASE: MATH. By the inductive hypothesis we can find MATH such that MATH. By REF for every MATH there exists MATH such that MATH. Moreover, by REF, the mapping MATH is can be chosen to be NAME, and subsequently, by shrinking MATH, continuous. Let MATH be defined such that MATH and MATH. It is clear that MATH has the required properties. CASE: MATH is limit. Fix sequences MATH and MATH such that CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH. By induction build a sequence MATH such that for MATH, CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH, where MATH. As before REF is possible by NAME theorem. Let MATH. It is clear that MATH. |
math/0102011 | We have three cases: CASE: There exists MATH such that MATH. Without loss of generality we can assume that for some MATH. It follows that MATH is constant. CASE: There exists MATH such that MATH. By shrinking MATH we can assume that there exists a continuous function MATH such that MATH. In particular, for MATH, MATH. If MATH was minimal then, using the argument below, we can also assume that MATH is one-to-one. Suppose that MATH, MATH, and MATH. Without loss of generality we can assume that for every MATH, MATH determines the value of MATH (up to finitely many values). Suppose that MATH is a function such that MATH for MATH. Let MATH be the condition defined as MATH . Let MATH be the finite set of all mappings MATH satisfying the requirements. Suppose that MATH, MATH and MATH . There exists MATH such that the sets MATH are pairwise disjoint. Induction on MATH and MATH. If MATH this is essentially REF . Let MATH be an enumeration of MATH. For MATH choose pairwise different reals MATH. Note that this choice can be made canonically from, for example, the countable dense let of leftmost branches of subtrees of MATH. Let MATH be such that sequences MATH are also pairwise different. Define conditions MATH, MATH such that for every MATH, CASE: MATH, CASE: MATH, CASE: MATH, CASE: MATH. Let MATH and MATH. Suppose that MATH and let MATH. By the part already proved, for each MATH find a condition MATH such that the sets MATH are pairwise disjoint. Note that we can do it in such a way that the mapping MATH is continuous (As before we first choose MATH in a NAME way, and then shrink MATH to make this mapping continuous). That defines a MATH-name for an element of MATH, which we call MATH. Next, let MATH and apply the inductive hypothesis to find MATH such that MATH are pairwise disjoint. Let MATH be defined as MATH and MATH. It is clear that MATH is as required. CASE: MATH. Let MATH be an increasing sequence of finite sets such that MATH. By induction build a sequence of REF such that MATH and for every MATH, CASE: MATH, CASE: sets MATH are pairwise disjoint. Let MATH. Suppose that MATH and MATH are two distinct points in MATH. Let MATH be the first ordinal such that MATH. Let MATH be so large that MATH and there are two distinct MATH such that MATH and MATH. Since MATH, it follows that MATH. |
math/0102013 | Define MATH by MATH . If we use a different representative MATH to represent MATH, then MATH and it is easily verified that MATH. Hence, MATH . This shows that MATH is well defined. It has the obvious inverse map MATH . |
math/0102013 | Writing MATH instead of MATH, it follows from the proposition that MATH . |
math/0102013 | A point in the fiber of MATH above MATH is of the form MATH. For MATH, since MATH, MATH . Thus, under the isomorphism MATH the torus MATH acts on MATH by MATH. |
math/0102013 | By REF the restriction of the line bundle MATH to the fixed point MATH gives rise to a commutative Cartesian diagram of MATH-equivariant maps MATH . Taking the homotopy quotients results in the Cartesian diagram MATH . But MATH is precisely the line bundle MATH over MATH associated to the character MATH of MATH. |
math/0102013 | Let MATH be the constant map. Then MATH, the identity map on MATH. The elements MATH in MATH are really MATH. Hence, MATH . In REF , take MATH to be MATH. Then MATH . Since MATH and MATH are both algebra homomorphisms, for any polynomial MATH in MATH variables, MATH . |
math/0102013 | For MATH, let MATH denote left multiplication by MATH. Then MATH is a diffeomorphism and its differential MATH is an isomorphism. Now define MATH by MATH . To show that MATH is well-defined, pick another representative of MATH, say MATH for some MATH. Then MATH since right multiplication MATH on MATH is the identity map. Since MATH is clearly a surjective bundle map of two vector bundles of the same rank, it is an isomorphism. |
math/0102013 | By REF the tangent bundle of MATH is the homogeneous vector bundle associated to the adjoint representation of MATH on MATH. In the notation of REF, MATH . By REF , at a fixed point MATH the normal bundle MATH is MATH . It follows that the equivariant NAME class of MATH is MATH . Each simple reflection MATH in MATH carries exactly one positive root to a negative root. Hence, MATH . Since MATH is the product of MATH reflections, MATH . |
math/0102013 | The coset MATH is a fixed point MATH . Since any two maximal tori in MATH are conjugate by an element of MATH, there is an element MATH such that MATH . Therefore, MATH, and MATH. Thus any fixed point can be represented as MATH for some MATH. Conversely, if MATH, then MATH is a fixed point of the action of MATH on MATH. It follows that there is a surjective map MATH and hence a bijection MATH . |
math/0102013 | Write MATH. The commutative diagram of MATH-equivariant maps MATH induces a commutative diagram in equivariant cohomology MATH . Since MATH is an injection, the restriction MATH is given by the same formula as the restriction MATH. The theorem then follows from REF . |
math/0102013 | Let MATH and MATH denote the equivalence classes of MATH in MATH and MATH respectively. Define MATH by MATH . If MATH is any element of MATH, then MATH. Hence, MATH for some MATH and MATH which shows that MATH is surjective. A surjective bundle map between two vector bundles of the same rank is an isomorphism. |
math/0102013 | MATH . |
math/0102013 | Let MATH be the projection map. Then MATH . |
math/0102013 | The normal bundle MATH at the point MATH is the tangent space MATH. By REF , the equivariant NAME class of MATH is MATH . |
math/0102013 | The commutative diagram MATH induces by the push-pull formula a commutative diagram in cohomology MATH . In degree REF, the restriction MATH is an isomorphism. Hence, if MATH has degree MATH in MATH, then by the commutative REF MATH . |
math/0102013 | Since MATH is an equivariant extension of MATH, the cohomology class MATH is an equivariant extension of MATH. Combining REF , the formula for the ordinary characteristic numbers of MATH follows. |
math/0102013 | In REF , take MATH to be MATH. Then MATH . By REF, MATH . By REF, MATH . |
math/0102014 | Let MATH be a such product and MATH the ( nonabelian ) subalgebras which define it. As any nilpotent NAME algebra has at least two generators, and the product by generators combines precisely these vectors, the product MATH has at least four generators. Now the adjoined vectors are central, and as MATH for MATH, the assertion follows. Now, if MATH were filiform, we would have MATH . |
math/0102014 | The lowest dimension for which CNLA exist is seven CITE . Thus, as these algebras have at least two generators and a one dimensional center, the minimal dimension for a product MATH is REF. |
math/0102014 | Let MATH be arbitrary elements of MATH. Then MATH . From the brackets of MATH it follows that MATH . This shows that MATH is a derivation of MATH and that MATH. From the third identity it follows moreover that MATH. Reasoning with elements MATH we obtain that MATH and MATH let MATH and MATH. Then MATH and MATH and reordering as before MATH which shows that MATH for MATH. If MATH and MATH and as MATH and MATH, we have MATH . This proves that MATH. Similarly we obtain MATH, taking MATH generators of MATH and MATH, respectively, we have MATH . This shows that MATH and MATH, from which the last assertion follows. |
math/0102014 | Let MATH be a generator of MATH satisfying the first condition. From the preceding proof we have MATH for MATH MATH be the generator such that MATH with MATH. Then MATH. If MATH are arbitrary generators of MATH, then by the definition of MATH it follows that MATH if and only if MATH and MATH. This shows that MATH thus MATH and MATH second assertion follows in a similar way. |
math/0102014 | As for MATH we have MATH and MATH, the result follows from the fact that MATH. |
math/0102014 | Let MATH and MATH. By the definition of MATH, MATH. Then MATH . It follows that MATH . From the relations MATH to MATH and the form of MATH in MATH it is easy to see that the last identity simplifies to MATH . Thus, if MATH, then MATH. |
math/0102014 | Let MATH be the smallest integers such that MATH. Without loss of generality suppose that MATH. Then for any MATH we have MATH. Thus any derivation of MATH is nilpotent. |
math/0102014 | Let MATH be such that MATH. Let MATH be any nonsplit characteristically nilpotent MATH-algebra with MATH generators-MATH . Consider the MATH-power MATH, where MATH . Then we have MATH . |
math/0102018 | Since MATH is the strong limit of MATH, one has MATH where MATH . Thus MATH satisfies REF and MATH (which is equivalent to REF ) since MATH does (see [REF ]). |
math/0102018 | Let MATH, MATH both satisfy REF . Then MATH . By REF and the definition of MATH one has Range-MATH and so MATH . Kernel-MATH. But REF implies the injectivity of MATH (see [REF ]). Therefore MATH. The proof is then concluded observing that MATH-independence follows by REF . |
math/0102018 | For brevity we define MATH and MATH . Then one has MATH . Since MATH is a bounded operator satisfying REF , and MATH is self-adjoint, by REF and [REF ], MATH has a bounded inverse for any MATH. Therefore MATH and MATH . Let us now prove the reverse inclusion. Given MATH, MATH we define MATH by MATH . Thus one has MATH with MATH . In conclusion MATH . Since, by REF , MATH we have MATH . Thus MATH coincides with the operator constructed in [REF ]; therefore, by REF , this operator is self-adjoint, has resolvent given by MATH and is equal to MATH on the kernel of MATH. |
math/0102018 | Given MATH, the sequence MATH is a NAME one in MATH and so it converges to some MATH. Then, by [REF ], there exists MATH such that MATH. Conversely let MATH; since MATH has a dense range there exists a (unique in MATH) sequence MATH such that MATH converges in MATH to MATH. |
math/0102018 | MATH-independence is an immediate consequence of REF . By the definition of MATH and by REF there follows MATH and the proof is concluded by the relation MATH which can be obtained proceeding as in [REF ]. |
math/0102021 | Given MATH, we compute MATH which proves that MATH is a MATH-equivariant map. It is straightforward to check that that MATH is an isometry. |
math/0102021 | Let us assume that MATH and MATH, and then choose MATH. Let MATH and MATH be as in REF . Then by taking even smaller MATH we will get MATH . Therefore MATH. |
math/0102021 | The first relation is obvious because the operator MATH is unitary and moves supports by the action of MATH. To prove the second estimate note that it is obvious if MATH. If MATH, then we get MATH and we can use the previous Lemma. |
math/0102021 | Clearly MATH, where MATH is spanned by MATH with fixed MATH. But we have MATH. Indeed, the eigenfunctions MATH are linearly independent and real analytic, so the corresponding MATH are also linearly independent because MATH near MATH. It follows that MATH. For any fixed MATH denote by MATH the orthonormal system which is obtained from the system MATH by the NAME orthogonalization process. Then MATH is an orthonormal basis in MATH, and MATH is an orthonormal basis in MATH. Note that MATH and it is supported near MATH. Denote by MATH and MATH the orthogonal projections on MATH and MATH respectively, MATH and MATH their NAME kernels. Then MATH . In particular, MATH . Similarly we find MATH and MATH . It follows that MATH for all MATH, therefore MATH which proves the lemma. |
math/0102021 | Note first that if the estimate REF holds with some MATH, it holds also for all smaller values of MATH. Now we should argue as in the proof of REF to conclude that MATH and MATH . Applying the variational principle REF we conclude that the estimate REF holds with MATH instead of MATH. It remains to notice that MATH takes all values from MATH when MATH. |
math/0102022 | We will prove that, under REF - REF , the fact that MATH is universally tight on MATH implies that MATH is tight on MATH. The proof, as will be clear, easily adapts to any finite cover of MATH, hence showing that the universal cover is tight, provided MATH is residually finite (which is the case, since MATH is NAME - see REF for a discussion of residual finiteness). Assume, on the contrary, that MATH is not tight and that MATH is an overtwisted disk in MATH. Then, after a possible contact isotopy, we can assume that MATH intersects MATH transversally along Legendrian curves and arcs and that MATH (see REF). Note that closed curves in MATH are homotopically trivial on MATH since MATH is incompressible. We would like to argue that, starting with innermost closed curves, we can eliminate them by pushing MATH across MATH (we do not push MATH across MATH to avoid introducing self-intersections into MATH). Since MATH is assumed to be irreducible, REF-sphere formed by two disks (one on MATH and one on MATH) bounding an innermost curve of intersection MATH on MATH bounds a ball across which MATH can be isotoped. We will include the sketch of the following fact (see REF), as we will need to refer to some steps in the proof of it later. We can push MATH across MATH to eliminate MATH in a finite number of steps, each of which is a bypass along an arc of the circle MATH. Consider the subdisk MATH of MATH bounded by MATH. Since MATH is a homotopically trivial Legendrian curve on MATH, MATH must be negative. After possible perturbation rel boundary, MATH is convex with Legendrian boundary satisfying MATH. MATH consists of properly embedded arcs with endpoints on MATH and no closed components, since the interior of MATH lies in the tight manifold MATH. We push MATH to engulf a bypass along MATH corresponding to a MATH-parallel dividing curve on MATH. We can continue until there is only one arc left. It is not hard then to see that the last bit of the isotopy is a contact isotopy. In a similar manner we can eliminate outermost arcs of intersection, by engulfing the half-disk of MATH bounded by such an arc in a ball, and then pushing MATH across that ball as in the previous lemma. Therefore, starting with innermost circles MATH and outermost arcs, we can push MATH across MATH to reduce MATH in steps, at the expense of modifying the dividing curve configuration on MATH. Changes in dividing curve configurations can be reduced to a sequence of bypass attachment moves which must now be analyzed. Without the assumption that the dividing set on the cutting surface consists of MATH-parallel arcs the following lemma would not be true. Let MATH be a convex surface with Legendrian boundary in a contact manifold MATH, such that MATH is MATH-parallel and MATH tight. Then any convex surface MATH obtained from MATH by a sequence of bypasses will have MATH obtained from MATH by possibly adding pairs of parallel nontrivial curves (up to isotopy rel boundary). This follows by examing the possible bypasses. One possibility is that the bypass is trivial, that is it is contained in a disk and produces no change in the dividing curves. Some bypasses will produce an overtwisted disk, and hence cannot exist inside a tight manifold. Some bypasses introduce a pair of parallel curves, and finally a bypass may change one pair of parallel curves into another pair or remove a pair of parallel curves. We need one more simple lemma: Let MATH be a convex surface with Legendrian boundary in a contact manifold MATH, such that MATH is MATH-parallel. If a convex surface MATH is obtained from MATH by a bypass such that MATH is isotopic to MATH, then MATH and MATH are contact isotopic, and in particular MATH is tight if and only if MATH is. Left to the reader. Consider a single bypass move on MATH with MATH-parallel. It is either trivial or increases MATH. If MATH is increased, the arc of attachment MATH starts on an arc MATH of MATH and comes back to MATH, thereby generating a non-trivial element of MATH. There exists a large enough finite cover MATH which ``expands" MATH to MATH, such that a lift of MATH becomes a trivial arc of attachment, that is, it connects two different components of MATH. The existence of the large finite cover follows from the facts that REF an incompressible surface MATH has MATH which injects into MATH and REF a NAME REF-manifold has residually finite MATH. Therefore, by passing through a finite succession of covers, we construct a cover MATH, together with the preimage MATH of MATH and a lift MATH of the overtwisted disk MATH, in which all the bypasses needed to isotop MATH across MATH are trivial. The surface MATH with MATH, obtained from MATH via trivial bypass attachments, has the same dividing set as MATH and is contact isotopic to MATH by REF . It follows that MATH is tight. This contradicts the existence of an overtwisted disk MATH and finishes the proof of REF . |
math/0102022 | Observe that MATH can be factored (after possibly isotoping relative to the boundary) into MATH, where MATH is convex, MATH is rotative on MATH, non-rotative on MATH and the dividing set induced on the annulus MATH consists of one MATH-parallel arc and nonseparating arcs (see REF ). The factorization can be chosen so that the MATH-parallel arc straddles one of the marked dividing curves adjacent to the unmarked curve across which the bypass of the layer MATH is added. Note that MATH. Finally, consider MATH. It is isotopic to a product neighborhood MATH of MATH through an isotopy which fixes MATH. This is because a non-rotative contact structure is determined by the dividing set on a horizontal annulus (see REF ). |
math/0102022 | For each MATH, take MATH to be MATH filled with solid tori MATH along each MATH except for MATH. This means MATH. Using the relative homology sequence and NAME duality for the manifold MATH and its boundary MATH, we find a class MATH which is nonzero under the boundary map MATH. Represent it by a surface MATH in MATH. NAME on MATH let MATH and consider MATH for MATH (assume transversality). If MATH we can pair off the intersections and get MATH. Now, take a suitable linear combination of the MATH to obtain MATH for which MATH is nonzero when restricted to each MATH. Finally pick a norm-minimizing representative of MATH. The well-grooming is simply asking that MATH be parallel curves oriented in the same direction. (If not already well-groomed, we can pair off oppositely oriented parallel curves without altering the norm.) |
math/0102022 | Observe that MATH. Let MATH, MATH be parallel copies of MATH on MATH which do not intersect MATH, and let MATH be the annulus bounded by MATH and MATH. Then take the annulus MATH to be a union of MATH and copies of MATH and MATH. Also let MATH be an annulus parallel to MATH with MATH, MATH as boundary. We may use the Legendrian realization principle to realize copies of MATH and MATH on MATH (and MATH) as Legendrian divides. It is at this point that the assumption that every connected component of MATH has at least two boundary components is needed to conclude that the nonisolating condition in the Legendrian realization principle is met. Now, after rounding, MATH and MATH are disjoint except along their common boundary MATH. Let MATH be the solid torus region bounded by MATH and MATH. The dividing curves on the meridian of MATH will be as in REF . With this explicit description, it is immediate that the attachment of MATH is equivalent to a folding operation. Finally, repeat this procedure with all annuli MATH, MATH, in succession. |
math/0102022 | Assume without loss of generality that MATH has collared Legendrian boundary. Let MATH be the connected component of MATH in whose direction twisting was added. Now let MATH be a properly embedded arc on MATH satisfying the following: CASE: MATH is essential (that is, not MATH-compressible). CASE: The endpoints MATH, MATH lie on the Legendrian divides of the collared Legendrian boundary of MATH. CASE: At least one of MATH, MATH (both if possible) lies on MATH. CASE: MATH has minimal geometric intersection MATH amongst arcs on MATH isotopic to MATH with endpoints on the same component of MATH as MATH. CASE: MATH is Legendrian. (This is made possible by the Legendrian realization principle.) Since MATH consists of MATH-parallel arcs and closed curves parallel to MATH, any choice of MATH will have no intersections with the MATH-parallel dividing arcs and minimally intersect the closed dividing curves parallel to the boundary, that is, MATH times if only one of MATH, MATH lie on MATH, and MATH times if they both do. Define MATH to be the set of Legendrian arcs MATH satisfying the following: CASE: There exists a properly embedded convex surface MATH with collared Legendrian boundary. CASE: MATH is efficient with respect to MATH. CASE: The endpoints of MATH lie on Legendrian divides of the collared Legendrian boundary of MATH. CASE: There exists an isotopy (not necessarily a contact isotopy) which sends MATH to MATH and MATH to MATH, where MATH is isotoped to MATH along MATH. We are now ready to define our invariant which distinguishes the MATH. For any MATH we define the twisting number MATH to be MATH, where MATH is the convex surface containing MATH in the definition of MATH above. Define the maximum twisting for MATH to be MATH . We claim that MATH. Assume on the contrary that there exists a Legendrian arc MATH with MATH, sitting on the convex surface MATH. There exists an isotopy MATH, MATH, such that MATH and MATH. Since MATH and MATH are efficient with respect to MATH and are isotopic on MATH through efficient curves, it is possible to ``slide" MATH along MATH without altering the isotopy class of MATH, and we may assume that MATH and MATH, MATH are fixed. As usual, pass to a large finite cover MATH of MATH to unwrap MATH to MATH. For a large enough finite cover MATH, there exists a lift/extension of MATH, MATH, to MATH, where MATH is the identity and the support of the isotopy MATH is compact and is contained inside an embedded REF-ball. The key property we use here is that the isotopy MATH, MATH, can be lifted to a large finite cover MATH so that the trace of the lift MATH is contained in a REF-ball MATH. This follows from the fact that MATH is NAME, which implies that MATH has universal cover MATH and MATH is residually finite. Let MATH, MATH, be a time-dependent vector field along MATH which induces the isotopy MATH. Damp MATH out away from MATH and extend it to all of MATH. Then MATH, where the MATH is the support of a vector field (that is, MATH such that MATH). Let MATH, MATH, be the flow corresponding to MATH. It is supported inside MATH from our construction. By assumption, MATH is strictly smaller than MATH (or MATH). If we can prove that, for each component MATH of MATH which is on MATH, there exist at least MATH closed curves in MATH parallel to MATH, we have a contradiction. In fact it is clearly enough to obtain the contradiction for any large enough cover. Take the cover MATH of MATH given by the previous claim, and use the technique of isotopy discretization, which first appeared in CITE, and is a method which works extremely well in the context of convex surfaces and state transitions (due to bypasses). We split MATH into small time intervals MATH, MATH, so that all the MATH are are all ``sufficiently close" if MATH, in the following sense: there exists an embedding MATH such that MATH and MATH is a submersion for all MATH. (Here MATH is the first projection.) This implies (by NAME 's convex movie approach in CITE and the equivalence of NAME 's `retrogradient switch' with the notion of a bypass) that to get from MATH to MATH we move via a sequence of bypass moves. We may need to take an even larger cover of MATH to expand MATH (without changing names) and extend MATH so that we can ``localize" all the bypasses, that is, all the allowable bypasses are either trivial or reduce the number of dividing curves. The bypasses which potentially give us trouble are the bypasses which increase the number of dividing curves. Such a bypass occurs only when the bypass ``wraps around" a closed curve which is homotopically nontrivial - by unwrapping in this direction in a large enough cover, we may avoid a dividing curve increase. More precisely, since MATH is incompressible in MATH, MATH injects into MATH, and we may unwrap MATH in the direction of any homotopically nontrivial curve in a finite cover of MATH using residual finiteness. We can show that there are no bypasses reducing the number of dividing curves as in the proof of REF . Therefore, we keep attaching trivial bypasses, and eventually find a contact isotopy of MATH with MATH, which contradicts the assumption that MATH. |
math/0102022 | Recall that, according to the classification of tight contact structures on toric annuli (see CITE), the universal tightness of MATH simply means that it is obtained by taking the standard rotative contact structure MATH on MATH with coordinates MATH, truncating for a certain interval MATH, making the boundary convex with MATH dividing curves each, and finally folding to attach a non-rotative layer. Without loss of generality, we may take both MATH and MATH to satisfy the MATH-Parallel Assumption with MATH-parallel arcs along MATH. We will prove that MATH is tight by contradiction. The proof of tightness of finite covers of MATH is identical, since any finite cover of MATH also satisfies conditions of REF . This implies the universal tightness of MATH, since NAME REF-manifolds have residually finite fundamental group, so any overtwisted disk that would exist in the universal cover could be REF projected into a finite cover. Assume MATH is overtwisted with an overtwisted disk MATH. As before, we look at a large finite cover MATH that sufficiently expands MATH. Let MATH be connected components of MATH and MATH be a lift of MATH to MATH. In MATH we may extricate MATH from MATH essentially without any penalty, as we shall see. As in the proof of REF , we push MATH across MATH in stages. We can eliminate components of MATH by pushing across balls, and this is equivalent to a sequence of bypass moves. If we expand MATH sufficiently, that is, we take a large enough cover, the only bypass operations are either REF trivial or REF reduce MATH. In particular, there are no dividing curve increases or changes in slope. Let MATH be a fixed MATH. Denote by MATH the connected component of MATH where MATH. Suppose the bypass attached onto MATH from the MATH side decreases MATH. Let MATH be the new torus we obtain after the bypass attachment. Then MATH is tight by the Attach = Dig Principle REF . Repeated application of the Attach = Dig Principle proves REF . |
math/0102022 | We define the invariant which will allow us to distinguish the MATH up to isotopy. First we choose a suitable isotopy class of closed curves. If MATH separates, then take the next splitting surfaces MATH of MATH, MATH. Since MATH cannot be a disk with boundary on MATH, we may take arcs MATH with endpoints on MATH which are not MATH-compressible on MATH as well as in MATH. Glue MATH and MATH to obtain MATH. If MATH is nonseparating and there is an (incompressible) splitting surface MATH which spans between MATH and MATH, then take MATH to be some arc on MATH with endpoints on MATH and MATH and glue the endpoints to obtain MATH. If there is no single surface MATH, then take MATH to have boundary component REF on MATH and MATH to have boundary components on MATH, find MATH, MATH, as before, and glue to get MATH. Let MATH be the set of Legendrian curves which are isotopic to MATH (but not necessarily Legendrian isotopic). Assume without loss of generality that we have two surfaces MATH and MATH. Recall MATH, MATH, consist of efficient parallel oriented curves on MATH which have nonzero geometric intersection with MATH. We construct a surface MATH by performing a band sum at each (geometric) intersection of MATH and MATH. Whether we do a positive band sum or a negative band sum depends on the following. In order to define the band sum, temporarily assume MATH, MATH, MATH are all linear on MATH, with slopes MATH, MATH, MATH, respectively. (That is, for the time being, forget the fact that MATH and MATH are Legendrian, and treat them just as curves on MATH.) Thicken MATH to MATH so that MATH and MATH. Each band intersects MATH, MATH, in a linear arc of slope MATH, which interpolates between MATH and MATH and is never MATH. In the case MATH, take suitable multiples of MATH and MATH (still call them MATH and MATH) so MATH, and call this MATH. We now define the framing for MATH to be one which comes from MATH. Here we are assuming without loss of generality that MATH, MATH, have endpoints on the same band. A natural Legendrian representative of MATH will have subarcs MATH, MATH, where the endpoints of MATH lie on half-elliptic points on MATH, which are also on MATH. This means that the actual Legendrian curves MATH and MATH intersect in a tangency at the endpoints of MATH. As before, define the maximal twisting MATH to be the maximum twisting number of Legendrian curves in MATH. The following proposition proves REF . |
math/0102022 | It is clear that MATH or MATH, for these bounds are achieved by MATH on MATH. We need to show that MATH or MATH. We will assume that there exist two surfaces MATH and MATH - the other case is similar. Suppose MATH is a Legendrian curve in MATH with MATH. There exists an isotopy MATH, MATH, for which MATH and MATH. We pass to a large cover of MATH to remove extra intersections of MATH with MATH. In order to reduce from the case where MATH is a closed manifold with incompressible torus MATH to the manifold-with-torus-boundary case, we construct an auxiliary contact manifold MATH which has the following properties: CASE: MATH is a large finite cover of MATH (not necessarily NAME) and MATH the pullback of MATH. CASE: MATH, MATH, lift to closed curves MATH. In other words, MATH maps REF down to MATH under the covering projection MATH. CASE: MATH has been expanded sufficiently on MATH so that the intersection of MATH with any component MATH of MATH is (either empty or) contained inside a ``small" disk MATH which does not intersect most of the dividing curves of MATH. CASE: If we denote MATH and MATH, then MATH and MATH, where MATH and MATH are measured with respect to preimages MATH and MATH of MATH and MATH. The construction of MATH will be done in the next section. In the meantime we continue with the proof of REF , assuming MATH has been constructed. MATH gives rise to a time-dependent vector field MATH along MATH; extend MATH and hence MATH to all of MATH by damping out outside a small neighborhood of MATH. Then the support MATH of MATH is contained in MATH. For each MATH, MATH remains constant outside a disk MATH which we may take to be convex with Legendrian boundary. If we discretize this isotopy into small time intervals MATH, then each step MATH to MATH corresponds to attaching a bypass which either is trivial or decreases the number of dividing curves while keeping the slope unchanged. Using the ``attach = dig" principle, we find that the contact manifold MATH is obtained from MATH by attaching dividing-curve-decreasing bypasses or ``unfolding". Observe that MATH intersects each MATH at most once, say at MATH. For each MATH, we may assume MATH and the isotopic copies MATH of MATH, MATH, which contain MATH are convex with efficient Legendrian boundary (that is, efficient with respect to the torus containing the boundary component). In order to compare the twisting number of MATH cut open along MATH and that of MATH cut open along MATH, we need to modify the cut-open MATH slightly by attaching unfolding layers onto MATH and extending this cut-open arc slightly on the attached toric annulus without modifying the twisting number. Once we do this, we are comparing (the modified) MATH with MATH, both on MATH, which reduces the problem to the torus boundary case. |
math/0102022 | REF and its proof imply REF , once we look at the mapping class group MATH of the closed toroidal manifold MATH. Our argument is a little different from that of CITE in that we do not bound the torsion, and instead use facts about the mapping class group to pass from ``infinitely many isotopy classes" to ``infinitely many isomorphism classes". Let MATH be the incompressible tori as in the Torus Decomposition Theorem and MATH the connected components of MATH. Apply NAME 's sutured manifold decomposition to each MATH, obtain the corresponding universally tight contact structure on MATH, and add appropriate MATH along each MATH. This is MATH. Note that any element of MATH must fix the isotopy class of MATH, but may permute the MATH (as well as MATH). Since there are only finitely many such permutations, we may assume that the elements of MATH we consider fix each connected component MATH (as well as MATH). The hyperbolic components MATH present no problem, since MATH is finite modulo NAME twists on the boundary, according to REF . This means that, for any compact surface MATH with boundary on MATH, there are finitely many isotopy classes of images of MATH under MATH, up to isotoping MATH along MATH. Let MATH and MATH be adjacent hyperbolic components with common boundary MATH, and MATH, MATH be norm-minimizing surfaces in MATH, MATH with boundary on MATH. Both MATH and MATH are not disks or annuli (by REF ). Then take curves MATH on MATH and MATH on MATH to form a closed curve MATH as in the proof of REF . Since MATH, MATH are not disks or annuli, we may take MATH, MATH to be nontrivial curves which begin and end on the same boundary component of MATH, MATH, respectively. Now consider MATH as well as MATH, for all MATH. Since there are finitely many isotopy classes MATH modulo NAME twists along tori, and NAME twisting does not change MATH, there exists a finite number MATH. In a manner similar to REF , we can prove that MATH. (Note that our current situation is not quite the same as the situation in REF , but the proof goes through, if we first extricate an isotopic copy of MATH from components MATH with MATH. This extrication can be done as before without penalty, by passing to a large finite cover.) If we inductively choose the next MATH so that each MATH, the new MATH cannot be isomorphic to the ones previously chosen. On the other hand, the NAME fibered components MATH have larger mapping class groups and require a little more care. Let MATH be the NAME fibration map which projects to the base MATH. In case MATH is toroidal - see the paragraph after REF for exceptions - the torus MATH is vertical, and there exists another torus MATH which intersects MATH persistently. This case is already treated in CITE, where the torsion for MATH and MATH are shown to be finite. The two remaining cases are: MATH is MATH with at most three singular points or punctures, or MATH is MATH with at most two singular points or punctures. In the MATH case, if there are two punctures and no singular points, then MATH is MATH and the minimality of the Torus Decomposition of MATH implies that MATH must be a torus bundle over MATH, in which case the theorem is already proved by CITE. If there is one puncture and one or no singular points, then the torus is compressible. Therefore, for MATH, we assume that there is a total of three singular points or punctures (with at least one puncture). For MATH, if there is one puncture and no singular points, we have a twisted MATH-bundle over a NAME bottle which is separated by the torus boundary. We will treat this case separately. Let MATH be a norm-minimizing, oriented, surface on MATH that is not homologous to a vertical annulus. It follows that MATH may be made horizontal, and by the NAME formula, MATH cannot be an annulus or a disk. A diffeomorphism MATH of MATH must take MATH to another horizontal surface MATH, since MATH is not an annulus and hence cannot be vertical REF . NAME formula calculation reveals that MATH and MATH have the same number of sheets. Moreover, MATH is isotopic to a diffeomorphism which preserves the NAME fibration. Now, if we construct MATH as before from adjacent MATH and MATH (MATH not necessarily NAME fibered), then MATH remains invariant under various choices of MATH up to diffeomorphism. This is because a diffeomorphism MATH of MATH which does not permute the boundary components will induce a map MATH which sends MATH, MATH, on each boundary component MATH. The rest of the argument goes through as in the hyperbolic case. Finally assume MATH is a twisted MATH-bundle over a NAME bottle and MATH. In this case we may distinguish the universally tight contact structures up to isomorphism by passing to a double cover MATH of MATH which is obtained by gluing two copies of MATH onto the double cover MATH of MATH. (We may need to take larger covers to unwind all the twisted MATH-bundle components over NAME bottles.) The contact structures on MATH we construct lift to nonisomorphic universally tight contact structures on MATH, and therefore it follows that there are infinitely many universally tight contact structures on MATH up to isomorphism. |
math/0102022 | Consider the universal cover MATH. We claim that a lift MATH of MATH has endpoints on different MATH-components of MATH. (Call them MATH, MATH.) Assume otherwise; namely the universal cover MATH (connected component) of MATH intersects a particular MATH along at least two coherently oriented lines. This follows from the well-grooming of MATH - all the components of MATH were oriented in the same direction. Now, MATH must separate MATH into two half-planes. However, the coherent orientation of MATH contradicts the fact that MATH separates. This implies that MATH. To conclude the proof, we simply note that MATH injects into MATH. |
math/0102024 | Let MATH be in MATH. NAME showed that MATH can be represented by a fibration over MATH if and only if the kernel of the map MATH is finitely generated CITE. So as MATH does not fiber, the kernel of the MATH is not finitely generated. As MATH has finite index in MATH, it follows that the kernel of the restricted map MATH is also not finitely generated. So MATH cannot represent a fibration. |
math/0102024 | First, let's make some basic observations. Throughout, all (co)homology will have coefficients in MATH. Suppose MATH is a regular finite cover of REF. Let MATH be the covering group. The homomorphism MATH is injective, and MATH is exactly the MATH-invariant cohomology. Now let's prove the theorem. Suppose MATH is a common regular cover of MATH and MATH. Let MATH be the covering maps, and MATH the covering groups. By NAME rigidity, we can assume that the covering groups MATH act via isometries of some fixed hyperbolic metric on MATH. As the isometry group of MATH is finite, so is the group MATH. From now on, assume that MATH has only one cusp. We will show there is a non-zero MATH-invariant class MATH in MATH. This gives a contradiction for the following reason. Every non-zero class in MATH can be represented by a fibration, while by REF no non-zero class in MATH can be represented by a fibration. But because MATH is MATH-invariant, MATH is in MATH for both MATH, which is impossible. Let MATH be a NAME surface for MATH which is a fiber, and let MATH be the lift to MATH. The surface MATH represents a non-trivial class in MATH. Moreover, since MATH in MATH is nontrivial, so is MATH is in MATH. Look at the the class in MATH which is MATH . Consider the restricted coverings MATH. Now the covering group MATH acts freely on the torus MATH. Hence MATH induces a rational isomorphism on MATH, and MATH acts identically on MATH. Therefore MATH acts identically on MATH. Thus MATH . So MATH is non-zero. If MATH is the dual class in MATH, then MATH is the non-zero MATH-invariant class we sought. |
math/0102024 | As the MATH are non-arithmetic, they cover a common orientable commensurator orbifold MATH CITE. Let MATH be the (orbifold) covering maps. The inclusion of MATH into MATH induces a map MATH via restriction of representations. Because the MATH are generic, MATH has a flexible cusp, and the variety MATH is also a complex curve. In fact, MATH is a birational isomorphism, though we will not need this CITE. The main step is: The map MATH is onto. The map MATH is a non-constant map of irreducible affine algebraic curves over MATH. Let MATH denote the smooth projective model of MATH. The curve MATH is the normalization of MATH compactified by adding an ideal point for each end of MATH CITE. Similarly, let MATH be the smooth projective model of MATH. The map MATH induces a regular map of the same name between MATH and MATH (this map is just a branched covering of closed NAME surfaces). Let MATH be a point in MATH which corresponds to a character - that is, not an ideal point. As the map from MATH to MATH is surjective, choose MATH in MATH with MATH. We need to show that MATH is not an ideal point. Suppose that MATH is an ideal point. By REF there is some MATH in MATH for which MATH. That is, there is some element of MATH which acts by a hyperbolic isometry on the simplicial tree associated to the ideal point MATH. Now for any MATH, MATH also acts by a hyperbolic isometry on the tree and so MATH. As MATH is of finite index in MATH, we can choose MATH so that MATH is in MATH. But then MATH, contradicting that MATH is the character of a representation. So MATH is not an ideal point and hence MATH is onto. Now to finish the proof of the theorem, let MATH. By the lemma, there is some character MATH in MATH with MATH. Let MATH be a representation with character MATH. Then the restrictions of MATH to the subgroups MATH give a pair of compatible representations with the required properties. |
math/0102024 | As MATH is a knot complement in a MATH-homology sphere, the cohomology group MATH vanishes and every representation into MATH lifts to MATH, so we're free to think about MATH-representations instead. Consider a reducible representation MATH. Since MATH is generated by MATH, we see that the primary eigenvalue of any MATH is a power of the primary eigenvalue of MATH. Thus, it follows from REF that MATH has a non-integral reducible representation if and only if MATH has a root which is not an algebraic unit. Suppose that MATH is not monic. Then MATH has a non-integral root provided that MATH is not an integer multiple of a monic integer polynomial. As we're in the MATH-homology sphere case, MATH, and this can't happen. So MATH has a non-integral root, and thus a non-abelian reducible representation which is non-integral. Now suppose that MATH is monic. Then all the roots of MATH are algebraic integers. Let MATH be a root of MATH. Because MATH is symmetric, MATH is also a root of MATH and so is integral. Thus all the roots of MATH are algebraic units. So all the non-abelian reducible representations of MATH are integral, completing the proof of the theorem. |
math/0102024 | Let MATH be a lift of a given non-abelian reducible MATH representation. As MATH fibers over MATH, the universal abelian cover of MATH is of the form MATH where MATH is a compact surface (here MATH is some finite abelian cover of a fiber in the fibration of MATH). As MATH is the commutator subgroup of MATH, the representation MATH takes MATH to a finitely generated abelian subgroup MATH consisting of translations. The subgroup MATH is non-trivial as MATH is non-abelian. For each MATH we need to show that if MATH then the homothety MATH is an algebraic integer. The action of MATH by conjugation on the normal subgroup MATH takes an element MATH to MATH. So MATH is a group automorphism of the lattice MATH. Thought of as an element of MATH, the map MATH satisfies its characteristic polynomial MATH, which is a monic polynomial with integer coefficients. Let MATH be a non-identity element of MATH. If we act on MATH via MATH we get that MATH. Thus MATH and MATH is an algebraic integer. So MATH is integral. |
math/0102024 | Suppose that MATH is commensurable to another knot complement MATH in a MATH-homology sphere. Call the common finite cover MATH. We will show that MATH has a non-abelian reducible representation which is non-integral, and so cannot fiber. Let MATH in MATH be the character of a non-integral reducible representation. As MATH is generic, by REF , there are representations MATH which agree on MATH where the character of MATH is equal to MATH. In particular, MATH is reducible and non-integral (MATH may be abelian, because we don't get to pick MATH, just MATH). Also, the character of MATH is in MATH. We will show The representation MATH of MATH is reducible. Assuming the claim, let's prove that MATH is not fibered. Pick MATH in MATH such that MATH has non-integral trace. Then for any MATH, the matrix MATH also has non-integral trace as its eigenvalues are powers of those of MATH. Since MATH is of finite index in MATH, choose a MATH such that MATH is in MATH. But then MATH is in MATH as well, and so MATH has non-integral trace. Thus MATH is non-integral. Since the character of MATH is in MATH, by REF there is a non-abelian reducible representation MATH which has the same character as MATH. As MATH has the same character as MATH, it is non-integral. By REF , MATH does not fiber over MATH. This completes the proof of the theorem modulo the claim. Now let's go back and prove REF . Let MATH and MATH. Now MATH restricted to MATH is the same as MATH restricted to MATH, and MATH is reducible. Thus MATH is reducible on MATH. The subgroup MATH is of finite index in MATH, so we can replace it by a finite index normal subgroup of MATH. Let MATH and MATH, two subgroups of MATH. Note that MATH is not the trivial subgroup because MATH is non-trivial, in fact non-integral, on any finite index subgroup of MATH. Now suppose that MATH is irreducible, that is, the fixed point set of MATH acting on MATH is empty. As MATH is reducible, MATH is either REF or REF points. As MATH is normal in MATH, the set MATH is MATH-invariant. So if MATH consisted of a single point, MATH would be reducible as well. So MATH is REF points. Look at the homomorphism MATH where MATH is thought of as the symmetric group on MATH. The homomorphism MATH is non-trivial as MATH is irreducible. Any MATH leaves invariant the geodesic MATH joining the two points MATH. If MATH then MATH acts on MATH by an orientation reversing isometry, and MATH has order MATH. Note that MATH is meta-abelian, as the kernel of MATH is abelian because it consists of isometries which fix the pair of points MATH. To finish the proof of the claim, we look at MATH. We claim that MATH is finite. Let MATH in MATH be a meridian, that is, NAME filling in along MATH yields a MATH-homology sphere. Let MATH in MATH be a longitude, that is, a generator of the kernel MATH. If MATH were the complement of a knot in a MATH-homology sphere, MATH would be a basis of MATH. In general, MATH generate a finite index subgroup of MATH. As MATH generates MATH, we must have MATH and MATH has order two. We claim that since MATH is a knot complement in a MATH-homology sphere, if MATH is the kernel of the unique surjection MATH then MATH. Consider a NAME surface MATH for MATH. The surface MATH has MATH equal to MATH in MATH. We can explicitly construct the cover MATH corresponding to MATH by gluing together two copies of MATH cut along MATH. Thus we see that MATH lifts to MATH. This shows that the boundary of MATH, namely MATH, is MATH in MATH. Thus MATH is in MATH. Therefore, as MATH is an abelian group of isometries fixing MATH, we have MATH. So the subgroup of MATH generated by the images of MATH is finite, in fact has order REF. Thus MATH itself is finite. Now we'll argue that MATH is infinite, yielding a contradiction. Look at MATH and in particular at MATH. Let MATH be a meridian in MATH. As MATH is non-integral, it is easy to see from the homomorphism MATH that the MATH in MATH with non-integral trace are exactly those MATH which are non-zero in MATH. Therefore, MATH has non-integral trace. In particular, MATH has infinite order, and hence MATH is infinite. As MATH shares a finite index subgroup with MATH, MATH shares a finite index subgroup with MATH. Thus MATH is infinite. But we've already shown that MATH is finite. This is a contradiction, and so MATH must be reducible. This proves REF and thus the theorem. |
math/0102025 | If the two elements, MATH and MATH commute then the fixed point set of one is invariant under the other, so both have a fixed point or neither does. If both do it is a common fixed point. If neither has a fixed point then nothing in the group they generate has a fixed point since the element with a fixed point would also commute with MATH and MATH . Conversely, if MATH and MATH generate a group acting freely, then they commute by NAME 's theorem. And if MATH and MATH have a common fixed point MATH then they generate a group acting freely on the intervals MATH and MATH . By NAME 's theorem again we see they commute. |
math/0102025 | The points MATH and MATH are two distinct fixed points of MATH. It follows that MATH . |
math/0102025 | Replacing MATH by MATH if necessary we may suppose that MATH is positive. By conjugating MATH we may assume that MATH is the fixed point of MATH and that MATH for MATH and that for MATH we either have MATH or MATH . Suppose first that for all MATH we have MATH . We will show that for large MATH . Note that the graphs of MATH and MATH can intersect in at most one point by REF . Hence if MATH then the origin is the unique point of intersection of these graphs. Since MATH it follows that for MATH sufficienly large MATH and hence MATH . So we have the desired result if MATH . If MATH then MATH and since MATH we may choose MATH such that MATH for all MATH . Since MATH there is a MATH with MATH. Note that the graphs of MATH and MATH can intersect in at most one point by REF , so the fact that MATH implies that MATH for all MATH . Thus MATH . Similarly if MATH there is a MATH such that MATH and hence for any MATH there is a MATH which is the unique point of intersection of the graphs of MATH and of MATH. Since MATH it follows that MATH for all MATH . So again MATH . Consider now the case that MATH for MATH . If MATH then the argument above remains valid, so we may assume the graph of MATH intersects the MATH and MATH axes in distinct points. Letting MATH go to infinity, the graphs of MATH limit on the union of positive MATH-axis and the negative MATH-axis. Also letting MATH go to negative infinity these graphs limit on the union of positive MATH-axis and the negative MATH-axis. From this it is easy to see that there exists a MATH such that the graph of MATH intersects the graph of MATH in two points near the two points where the graph of MATH intersects the axes. But this is impossible by REF . |
math/0102025 | Recall that MATH and MATH are commensurate if and only if MATH and MATH are commensurate. Hence replacing MATH and MATH by their inverses if necessary we can assume both are positive. By REF there is a MATH such that MATH . So MATH and we have MATH . The other inequality is similar. |
math/0102025 | Without loss of generality we may assume both MATH and MATH are positive. Suppose MATH and note that MATH implies immediately that MATH. But also MATH since otherwise there would be a MATH and a MATH with either MATH or MATH. This would imply MATH or MATH contadicting the fact that MATH . We have shown that MATH . Since MATH and MATH are commensurate there is a MATH such that MATH . From this it is immediate that MATH . Since MATH and MATH were arbitary positive elements we conclude that MATH . To prove the second assertion of the lemma, suppose MATH is commensurate with MATH and MATH . We may assume both MATH and MATH are positive. Then there exists MATH such that MATH. It follows that MATH for any MATH and hence MATH . |
math/0102025 | First we note that if MATH acts freely then the result is immediate from NAME 's theorem and the definitions. Hence we may assume that MATH contains a non-trivial positive element MATH with a fixed point. Clearly we may assume there is a non-trivial element MATH . Since MATH if and only if MATH we need only show MATH is closed under multiplication in order to show it is a subgroup. If MATH we must show MATH . We may, without loss of generality assume both are positive and that MATH. It follows that MATH and hence that MATH for all MATH . So for all MATH we have MATH and consequently MATH . So MATH . We note that REF implies that any non-trivial element of MATH is fixed point free, so MATH acts freely and by NAME 's theorem it is abelian. The fact that MATH is normal follows from the observation that MATH and MATH implies MATH . But as MATH and MATH both have fixed points they are commensurate by REF and hence MATH by REF . |
math/0102025 | If MATH then MATH is MATH-invariant and MATH is well defined. If MATH then MATH has no fixed points by REF . If MATH then for each MATH is bounded as a function of MATH. This implies MATH has infinitely many fixed points and hence must be the trivial element. Suppose now that MATH. Replacing MATH by MATH if necessary, we may assume MATH. We will show that the assumption that MATH has no fixed points leads to a contradiction. Let MATH and assume without loss of generality that MATH for all MATH, so MATH . Then MATH . Hence for every MATH is finite. But this is not possible, as follows. Since MATH is non-abelian, there is an element MATH in the kernel of MATH. The element MATH has the property that MATH and MATH preserves the measure MATH. Thus MATH which imples that for every MATH that MATH is either MATH or infinite, a contradiction. Thus we conclude that MATH has a fixed point. Since MATH it is not the identity, so it has exactly one fixed point. |
math/0102025 | Choose MATH so MATH. We will prove by induction that MATH for any MATH . This is clear for MATH . Assume inductively that MATH for any MATH . We know by REF and the induction hypothesis that MATH . Also by the mean value theorem there is a point MATH such that MATH . Therefore by REF for every MATH . Using the mean value theorem again and the fact that MATH, we see MATH . This completes the induction step, so MATH for all MATH . The fact that MATH then follows from the fact that MATH . |
math/0102025 | In REF we showed the function MATH defined by MATH is a surjective semi-conjugacy. It suffices to prove that MATH is one-to-one. Note that if MATH is the MATH-quasi-invariant measure provided by REF , then MATH is NAME measure on MATH, the unique quasi-invariant measure for the standard action of MATH on MATH . The function MATH is monotonic so MATH consists of either a single point or a closed interval. If it is an interval MATH then MATH. The endpoints MATH are in the support of MATH but the interior MATH is disjoint from the support of MATH . We note that any interval MATH must be wandering. That is, we must have MATH for all non-trivial MATH . This is because if MATH the fact that MATH is a semi-conjugacy implies MATH so MATH would have two fixed points MATH and MATH. Thus it suffices to show that the action of MATH has no wandering intervals of the form MATH. Of course to do this we need only do it for a subgroup of MATH. In the proof of REF we showed that the function MATH defined by MATH, where MATH and MATH is an injective homomorphism. In particular if MATH then MATH . Since we are assuming MATH is not abelian, MATH is not trivial and neither is its kernel. Hence we will focus on a subgroup MATH generated by two non-trivial elements, MATH with MATH and MATH with MATH and MATH. Replacing MATH by MATH if necessary, we may assume MATH for all MATH. Consider the quotient space of MATH by the action of MATH. There is a MATH diffeomorphism from this quotient to MATH. We lift this diffeomorphism to a MATH diffeomorphism of MATH and use it to conjugate our MATH action to a MATH action on MATH for which MATH . We know MATH for some MATH but composing MATH with an inner automorphism of MATH (that is, an affine change of co-ordinates) we may assume MATH and MATH for some MATH . We first consider the case that MATH is irrational. Let MATH. Since MATH is in the kernel of MATH, a normal abelian subgroup, we have that MATH commutes with MATH. We also have that MATH. Hence MATH is the lift of a MATH diffeomorphism of the circle MATH with irrational rotation number. By NAME 's theorem this circle diffeomorphism has a dense orbit from which it follows that if MATH is the group generated by MATH and MATH then the set MATH is dense in MATH for any MATH. Hence MATH has no wandering intervals at all. Thus we may assume MATH is rational. Note that MATH implies MATH or MATH . Differentiating we see that for all MATH we have MATH and MATH. In particular MATH and the derivative of MATH are periodic and hence uniformly bounded. We now want to show by contradiction that MATH has no wandering intervals of the form MATH. Suppose MATH is a wandering interval. Then for every non-trivial MATH because otherwise MATH and the endpoints of this MATH invariant interval would be two fixed points for MATH . We can conclude, in particular, that the elements MATH are all distinct, where MATH denotes MATH minus the greateset integer in MATH. This is because otherwise MATH would equal MATH for some MATH . It follows that the intervals MATH are all pairwise disjoint. Since MATH is a semiconjugcay and is MATH-equivariant, MATH for all MATH. As MATH is monotonic and MATH for all MATH, we have MATH. In particular MATH so since MATH and MATH we have MATH for all MATH . Recall that MATH denotes the length of MATH and observe that MATH for some integer MATH, so MATH . Hence MATH as the intervals MATH are pairwise disjoint and all in MATH . Let MATH be the value provided by REF and choose an interval MATH which contains MATH in its interior and has length MATH . Define MATH and note that it is a non-trivial closed interval with MATH in its interior. We may shrink MATH slightly so that MATH of each endpoint is a single point (because MATH is a non-trivial interval for at most countably many values of MATH . ) We redefine MATH to be MATH of this shrunken MATH and note that MATH . It follows that MATH for all MATH . We observe that from REF that MATH . But MATH and for MATH sufficiently large MATH which implies there is an interval MATH with MATH . This means that MATH . Hence MATH for MATH sufficiently large. This clearly contradicts REF above. We conclude that the semi-conjugacy MATH is one-to-one and hence a homeomorphism. |
math/0102029 | By NAME 's Flexibility Theorem, it suffices to find a characteristic foliation MATH on MATH with (an isotopic copy of) MATH which is represented by Legendrian curves and arcs. We remark here that these Legendrian curves and arcs constructed will always pass through singular points of MATH. Consider a component MATH of MATH - let us assume MATH, so all the elliptic singular points are sources. Denote MATH, where MATH consists of closed curves MATH which intersect MATH, and MATH consists of closed curves MATH. This means that for MATH, either MATH or MATH, where MATH, MATH, are subarcs of MATH, MATH, MATH, are subarcs of MATH, and the endpoint of MATH is the initial point of MATH. Since MATH is nonisolating, MATH is nonempty. What the MATH provide are `escape routes' for the flows whose sources are MATH or the singular set of MATH - in other words, the flow would be exiting along MATH. Construct MATH so that REF the subarcs of MATH coming from MATH are now Legendrian, with a single positive half-hyperbolic point in the interior of the arc, REF the curves of MATH contained in MATH are Legendrian curves, with one positive half-elliptic point and one positive half-hyperbolic point. If MATH intersects MATH, then we give a neighborhood MATH a characteristic foliation as in REF . After filling in this collar, we may assume that MATH is transverse to and flows out of MATH. If MATH is empty, then we introduce a positive elliptic singular point on the interior of MATH, and let MATH be a small closed loop around the singular point, transverse to the flow. At any rate, we may assume the flow enters through MATH and exits through MATH - by filling in appropriate positive hyperbolic points we may extend MATH to all of MATH. |
math/0102029 | The argument is almost identical to the MATH case. Using the notation from above, let MATH, MATH, be the three intersections of MATH and MATH be components of MATH containing MATH. As above, if MATH and MATH, then we have a decrease REF or a positive NAME twist REF . Suppose MATH. The chief difference between MATH and higher genus MATH is that the arc MATH from MATH to MATH does not always bound a half-disk, together with an arc MATH - if it does, then we have REF or REF . We have two remaining cases: MATH, which we call the `mystery move' REF , and MATH, which gives either REF or REF . See REF for these possibilities. |
math/0102029 | The key ingredient is the Legendrian realization principle. We apply it to MATH, fixing MATH and MATH, so that we may assume that there exists a convex disk MATH with collared Legendrian boundary, MATH, MATH, and MATH. All the operations which follow will now take place inside a MATH-invariant neighborhood MATH, where MATH. Let MATH be two points on the same component of MATH. On MATH, take a Legendrian arc MATH with endpoints on MATH which extends MATH as in the left-hand diagram of REF . On MATH, take another Legendrian arc MATH depicted in the right-hand diagram of REF , with endpoints MATH. Note that both MATH and MATH may be slightly modified using NAME 's Flexibility Theorem or the Legendrian realization principle to realize MATH, MATH as Legendrian arcs. Now, form the closed Legendrian curve MATH. Let MATH be a convex disk with Legendrian boundary MATH. Since MATH, there are two possibilities for bypasses along MATH, namely the trivial bypass and the disallowed bypass. The attachment must be trivial since MATH is tight. |
math/0102029 | As in the Existence Lemma, isolate MATH inside a convex disk MATH with collared Legendrian boundary and MATH, after possible perturbation of MATH, fixing MATH and MATH. Here, we modify the characteristic foliation on MATH rel MATH, so that it matches that of the local model in the Existence Lemma. Now, a thickened neighborhood MATH of MATH is tight since a model exists. Moreover, it is MATH-invariant due to the uniqueness of the tight contact structure on REF-ball with fixed boundary REF . Finally, we conclude that MATH is MATH-invariant. |
math/0102029 | Perturb and push MATH off the Legendrian skeleton MATH of MATH and make MATH transverse to MATH. The set MATH consists of the singular points and trajectories connecting between singular points of the same sign, assuming that the flow on MATH has been perturbed into a NAME one without closed orbits - MATH is a deformation retract of MATH. We can then push MATH into the dividing set as follows: Along the trajectory from MATH to MATH containg MATH, the contact structure MATH has the form MATH, on MATH with coordinates MATH, where MATH is the neighborhood in MATH of the trajectory MATH containing MATH, and MATH corresponds to the dividing set of MATH. Then consider the front projection onto MATH using the MATH, MATH coordinates. Near MATH, we may assume MATH is a straight line MATH through MATH with negative slope MATH. Modify MATH fixing endpoints, to MATH with MATH everywhere except for MATH, where the smooth curve has vertical slope. Since we may assume that the MATH occur on distinct trajectories of MATH, we are done after perturbing MATH again to make it transverse to MATH. |
math/0102029 | First note that a closed curve MATH of MATH on MATH cannot be homotopically nontrivial - this is because MATH bounds a subdisk of MATH, contradicting the incompressibility of MATH. Next, assume MATH and MATH bounds a disk MATH in MATH. If MATH is the innermost intersection MATH on MATH, then MATH can be capped off to reduce the number of components of MATH. If there exists an arc of MATH on MATH, then MATH could not have been empty. Therefore, the conditions of the Legendrian realization principle are satisfied, and MATH can be made Legendrian while keeping the endpoints of arcs fixed. |
math/0102029 | In the proof we will alternate between viewing MATH as being parallel to MATH near MATH (called MATH) and perpendicular to MATH (called MATH). We may alternate between the two viewpoints by making modifications to MATH near MATH. See REF . Since the endpoints of MATH on MATH and MATH overlap, this enables us to compare MATH and MATH. If MATH, then MATH is already contact isotopic to MATH. This follows from the fact that the tight contact structure on MATH is unique REF , coupled with the observation that the MATH-invariant contact structure on MATH is tight. Otherwise, let MATH be a MATH-parallel dividing curve on MATH, and MATH be the `corresponding' bypass half-disk which contains MATH. Now, take a parallel copy MATH of MATH as well as a parallel copy MATH of MATH. Attach MATH onto MATH. Then the new MATH agrees with MATH along MATH and MATH. If we modify MATH and MATH, then we can induct on MATH. |
math/0102029 | Let MATH, MATH, be a REF-parameter family of embeddings, where MATH maps MATH identically onto MATH, MATH, and MATH is independent of MATH along MATH. Break MATH into small intervals MATH with MATH, and MATH, so that for each interval MATH, MATH can be sandwiched inside a MATH with boundary MATH and MATH. Write MATH. The strategy is to compare MATH to MATH, and to show that we can pass between then via allowable state transitions, and then compare MATH to MATH. This approach has the advantage that every time we are comparing nonintersecting copies of MATH. Assume therefore that MATH and MATH are nonintersecting (except along their boundary), and that they bound a layer MATH. To show that there exists a sequence of bypass moves from MATH to MATH, we use an important idea due to CITE, the convex movie. According to NAME, we may assume (after some perturbations) that there exist MATH, MATH, with MATH, such that: CASE: MATH is convex if MATH does not equal any MATH. CASE: MATH is not convex because there is a retrogradient saddle-saddle connection connecting from a negative hyperbolic singularity to a positive hyperbolic singularity. CASE: The saddle-saddle connection serves as a switch as we move from MATH to MATH (MATH small). Moreover, this switching is equivalent to a bypass move. |
math/0102029 | This is because the slopes of the new meridional disks in MATH are MATH, and we can patch MATH copies of the meridional disk onto MATH to make it into an overtwisted disk. We are essentially unlinking MATH from MATH after surgery. |
math/0102029 | We apply a variant of the Gluing Theorem. Instead of cutting along MATH disks, we will cut along MATH disks MATH, MATH, where MATH are arcs on MATH drawn in dotted lines in REF . Orient MATH, where the normal orientation is indicated by the arrows in REF . After cutting, MATH is a solid torus, and its boundary will have two copies each of MATH, which we denote MATH and MATH, depending on whether the orientation induced from MATH agrees or disagrees with the boundary orientation of MATH. In order to prove tightness, we will need to use our knowledge about tight contact structures on solid tori, in exchange for simplifying the combinatorics somewhat. (Refer to CITE for a discussion of tight contact structures on solid tori.) Initial configuration MATH: For each of the MATH, the initial configuration of dividing curves is as in REF . (The portions with MATH are not pictured here.) This is due to the MATH-invariance of the tight contact structure on MATH. If we cut MATH along MATH, MATH, MATH and rounded the edges, the resulting solid torus will have slope MATH, which is isomorphic to MATH. A bypass which is attached to MATH `from the outside' is attached from the direction of the oriented normal of MATH, and a bypass which is attached `from the inside' is attached from the opposite direction. For each MATH, there are three possible dividing curve configurations, given in REF as MATH, MATH, and MATH - they are denoted in suggestive group-theoretic notation to indicate that each outer bypass attachment is an action by MATH, and three bypasses in a row gives back the original configuration. We then write the initial configuration MATH as MATH (the MATH-th component is for MATH). Configuration changes from MATH: CASE: There can be no (nontrivial) bypasses attached to any MATH from the inside - any such bypasses would create a homotopically trivial curve. This is an easy check once we attach a bypass from the inside and round the edges in REF . Therefore, there is no MATH. CASE: There can exist bypasses from the outside. Attaching a bypass to the outside of REF along MATH does not change the dividing set of the portion pictured in REF , after rounding. MATH on MATH is given in REF . Peeling off the bypass on the `outside' (= bypass `on the inside' of REF ) adds an extra twist to the dividing curves of MATH, so the slopes change from MATH to MATH. This proves that MATH, MATH, MATH are the only allowed state transitions from MATH. Next assume we have a configuration MATH where the entries MATH are either MATH or MATH, and the set of indices MATH with MATH is a subset of either MATH or MATH. In other words, assume MATH. Configuration changes from MATH: We claim that if MATH, then any allowable state transition will remain in MATH. We are trying to attach bypasses onto MATH from the inside or from the outside. CASE: If MATH, then it is possible to find a bypass on the inside, and use it to perform a state change MATH. This is precisely the inverse process of MATH, where the bypass is attached from the outside. CASE: If MATH, it is not possible to attach a second bypass from the outside. Peeling off a second bypass from MATH corresponds to reducing the boundary slope from MATH (MATH is the number of MATH's in MATH) to MATH. However, the sign of the bypass is opposite that of the first bypass, and it is not possible to find two bypasses of opposite signs on a basic MATH layer from slope MATH to slope MATH. Recall from CITE that a basic slice MATH with boundary slopes MATH and MATH has relative NAME class MATH (for a suitable section MATH on the boundary). If MATH admits a splitting MATH with MATH, and MATH has relative NAME class MATH, then the relative NAME class for MATH is dictated to be MATH. A second bypass from the outside violates this consistency condition. (Another way to obtain a contradiction is to observe that stacking a MATH layer atop MATH, with MATH, MATH, gives an overtwisted disk on MATH.) REF Assume MATH. If MATH, then at least two MATH, and a bypass on the inside gives rise to an overtwisted disk. If MATH, then the configuration in the neighborhood of MATH, after smoothing MATH and MATH, is equivalent to the configuration in REF . There cannot exist a bypass from the inside for this case, because there will be two bypasses of opposite signs on a basic MATH as in REF . CASE: If MATH, then it is possible to find a bypass on the outside, but only if MATH or MATH, and MATH or MATH. This follows from the same reason as REF . All of the cut-open solid tori are then tight. |
math/0102029 | Consider the following convex decomposition MATH, where MATH, and MATH. Properly speaking, MATH is an annulus since the first cutting has already taken place in the decomposition. We prove the theorem by tracing back along the convex decomposition. Note that we will often make modifications using the Flexibility Theorem or the Legendrian realization principle without explicitly stating that they will be used. Initial Configuration MATH. Initially, MATH will consist of MATH parallel dividing curves with holonomy MATH and MATH will consist of MATH parallel curves with holonomy MATH. Inductive Assumption. Suppose we have inductively (via state transitions) at a configuration MATH where: CASE: MATH consists of MATH, MATH, parallel dividing curves with holonomy MATH. CASE: MATH is obtained by `folding' inside a MATH-invariant neighborhood MATH of MATH. The folding operation is described in detail in REF, and is used to increase the number of parallel dividing curves. CASE: MATH consists of MATH parallel arcs which go across from one side of the annulus to the other, together with MATH one-sided (MATH both endpoints on one side of the annulus) arcs. Half of the one-sided arcs have endpoints on one side of the annulus, and half have endpoints on the other. Moreover, MATH, when glued up along MATH, will become MATH parallel curves with holonomy MATH. CASE: The tight contact structure on MATH, after rounding, is MATH. Let MATH be the set of such configurations. State changes along MATH. We claim there can be no state transitions. To do this, we examine all possible locations on MATH where a non-trivial bypass may be attached. Let MATH be an arc of attachment for the candidate bypass, MATH, MATH, the points of intersection with MATH, and MATH be the dividing curve containing MATH. We enumerate all the possible cases and eliminate them in turn, by examining MATH. The following key observation helps reduce the potentially infinite number of possibilities to a finite number: NAME observation. Any MATH-parallel arc on MATH which does not intersect MATH may be pushed into the MATH portion. See REF . The various cases are enumerated in REF . The dark solid lines are the dividing curves and the light solid lines are the candidate MATH. The candidate bypasses may be attached to the front or the back. Note that all of the extraneous dividing curves have already been pushed across. All the cases fail because the candidate bypass (attached from the interior of MATH) gives rise to either REF a homotopically trivial dividing curve or REF a convex torus on the interior of MATH with zero boundary slope, which is a contradiction, since the boundary slope of MATH is MATH and the meridional slope is MATH. We will treat a few cases, and leave the rest to the reader. CASE: See REF . Here, regardless of which side we attach a bypass along MATH, there will exist a homotopically trivial disk after attachment. In the figure only the relevent portions of MATH are shown, and the edges have already been rounded. The two thin lines are the arc of attachment MATH, and the hypothetical bypasses are attached from the front onto either of the two thin lines. CASE: See REF . Here we have depicted the case MATH. The left and right sides of the large rectangle are identified to give MATH. The MATH to the right immediately gives a homotopically trivial dividing curve after attachment. The MATH to the left yields a dividing set with slope zero. This gives us a contradiction. CASE: See REF . This is similar to REF . The MATH to the right gives a homotopically trivial curve, and the MATH to the left yields zero slope. CASE: In this case we may use the key observation to reduce to the case when each of the MATH has exactly MATH dividing curves (all the one-sided components can be pushed across). The attachment of a bypass along MATH is tantamount to modifying the boundary slope of MATH from MATH to MATH. Therefore we have proved that there are no state transitions which modify MATH. State changes along MATH. We show that any state transition still leaves us in MATH. Assume MATH. Here MATH is parallel to and disjoint from MATH in most cases except when MATH. (See REF below for the exception.) Let MATH be the MATH-bundle bounded by MATH and MATH. We have three cases. CASE: MATH. Let MATH be a vertical Legendrian curve on MATH with MATH, and let MATH be a vertical Legendrian curve on MATH with MATH. Let MATH be an annulus with MATH and MATH. Then, using the Imbalance Principle in CITE, we find a bypass along MATH. Attaching the bypass is equivalent to finding a MATH-invariant layer MATH with MATH, where MATH. We find that MATH is therefore a MATH-invariant tight (one-sided) neighborhood of MATH. Now, since MATH itself is contained in a MATH-invariant neighborhood of MATH, we conclude the same for MATH. This proves REF of the inductive assumption. REF are immediate from the invariance of MATH with fixed MATH. CASE: MATH. Assume MATH or MATH. Take a vertical Legendrian MATH on MATH with MATH and a vertical Legendrian MATH on MATH with MATH. Take an annulus MATH with MATH and MATH. MATH can be extended to an annulus MATH isotopic to MATH. However, since no state transitions can occur on MATH, we may assume that MATH. Moreover, we may assume that MATH is obtained from MATH by attaching a bypass which lies on MATH. Now, take all the one-sided dividing arcs on MATH which end on a fixed boundary component of MATH. Attach all the bypasses corresponding to these one-sided dividing arcs to get an isotopic copy of MATH. Do the same with the other boundary component. Then we get another isotopic copy of MATH, and MATH is sandwiched inbetween. This proves the conditions of the inductive assumption. CASE: We have one more case left, if MATH (MATH). Without loss of generality, assume that MATH. Then the slope of MATH is MATH. A non-trivial bypass attachment will give MATH with slope MATH, MATH. If we consider the tight contact structure on MATH, then, by the classification of tight contact structures on MATH (compare CITE), we may take MATH to be large positive. Now consider vertical NAME MATH on MATH with MATH and MATH on MATH with MATH. Take an annulus MATH with MATH and MATH. Then there will exist bypasses along MATH which allow us to find MATH with slope MATH. Now, cutting along MATH (and using the fact that MATH does not depend on the cut), we see that the bypass attachment corresponding to MATH will yield a slope MATH solid torus inside the slope MATH solid torus MATH, a contradiction. A similar but easier argument shows that there cannot be a dividing curve decrease if MATH and MATH. |
math/0102034 | Fix a matrix MATH, such that MATH, and define MATH, for MATH. Then MATH, so MATH is a group homomorphism. Since MATH, for all MATH, there is no continuous function MATH, such that MATH and MATH. Hence, MATH does not have a continuous extension to MATH. |
math/0102034 | Let MATH. The inclusion MATH (virtually) extends to a continuous homomorphism MATH. Since MATH, and MATH is NAME dense in MATH, it is reasonable to expect that MATH. (Actually, this need not quite be true, but it is close to correct.) Then MATH. |
math/0102034 | Let CASE: MATH, CASE: MATH be the subspace of the vector space MATH spanned by MATH, and CASE: MATH be the natural projection onto the first factor. We have MATH. Note that: CASE: MATH is connected (because MATH is connected and MATH is continuous); CASE: MATH is an additive subgroup of MATH (because MATH is an additive subgroup, and MATH is an additive homomorphism); and CASE: MATH contains MATH (because MATH contains MATH). Since MATH the desired conclusion follows. We have MATH. Because MATH is discrete, we know that MATH is continuous, so the MATH-image of any compact subset of MATH is compact. This implies that MATH, the restriction of MATH to MATH, is a proper map. (That is, the inverse image of every compact set is compact.) It is a fact that MATH therefore, MATH differs from MATH by only a compact amount. Since MATH is proper (and MATH is a homomorphism), this implies that MATH is proper. Therefore MATH is compact. Since MATH is a homomorphism, we conclude that MATH is a compact subgroup of MATH. However, MATH so we conclude that MATH is trivial, as desired. Completion of the proof. From REF , and the fact that MATH is a closed subgroup of MATH, we see that MATH is the graph of a well-defined continuous homomorphism MATH. Also, because MATH, we know that MATH extends MATH. |
math/0102034 | We are given a homomorphism MATH. Assume MATH is a lattice in MATH. Let CASE: MATH, CASE: MATH, CASE: MATH, and CASE: MATH be the natural projection onto the first factor. We use the proof of REF , so there are only two issues to address. First, we need to show that MATH has a syndetic hull MATH in MATH. Second, because MATH may not be MATH-connected, we do not have REF , the analogue of REF. Recall that NAME closures are virtually connected. (This is stated formally in REF below.) Hence, MATH has only finitely many components, so, by passing to a finite-index subgroup of MATH, we may assume that MATH, so MATH. By assumption, MATH contains a maximal compact subgroup MATH of MATH, and, by definition, MATH contains a maximal compact subgroup MATH of MATH. Therefore, the projection of MATH to each factor of MATH contains a maximal compact subgroup of that factor. However, MATH is diagonally embedded in MATH, so it probably does not contain the product MATH, which is a maximal compact subgroup of MATH. Thus, REF probably does not apply directly. However, MATH is contained in MATH, so the rather technical REF below, which can be proved in almost exactly the same way as REF , does apply. So we conclude that some finite-index subgroup of MATH has a syndetic hull MATH in MATH, as desired. (Note that, because MATH contains a finite-index subgroup of MATH, the homomorphism MATH virtually extends MATH.) REF asserts that we may take the syndetic hull MATH to be simply connected; thus, MATH has no nontrivial compact subgroups. Hence, the subgroup MATH also has no compact subgroups. REF was used only to obtain this conclusion, so we have no need for REF. The general case. From REF , we know that MATH has a syndetic hull MATH. So MATH is a lattice in MATH, and, by assumption, MATH. Therefore, REF implies that MATH virtually extends to a continuous homomorphism MATH. Now, because MATH is connected, and MATH, one can show that MATH. So it is not hard to extend MATH to a continuous homomorphism MATH. |
math/0102034 | Let MATH be a maximal compact subgroup of MATH that is contained in MATH. Assume MATH. Because MATH normalizes MATH (see REF), we know that MATH normalizes MATH, so MATH is a subgroup of MATH. Since MATH contains the maximal compact subgroup MATH of MATH, we see that MATH is simply connected (see REF). Hence MATH is simply connected, so MATH is connected (see REF). The general case. Because MATH contains the maximal compact subgroup MATH of MATH, REF implies that MATH is connected, as desired. |
math/0102034 | To simplify the notation, let us assume MATH contains a maximal compact subgroup of MATH, ignoring the adjoint representation. (Without this simplification, the proof would use REF , with MATH, instead of using REF , as we do here.) Under this assumption, REF implies that if MATH is any NAME closed subgroup of MATH that contains MATH, then MATH is connected. We apply this fact with: CASE: MATH (see REF); so MATH is connected. CASE: MATH (see REF); so MATH is connected. CASE: MATH (see REF); so MATH is connected. |
math/0102034 | Let us first prove uniqueness: suppose MATH and MATH are syndetic hulls of MATH. We have MATH for MATH (see REF); so MATH. Therefore MATH and MATH normalize each other (see REF), so MATH is a subgroup of MATH. It is simply connected (see REF), and MATH is compact (because MATH), so REF implies that MATH; thus MATH. Similarly, MATH. Therefore MATH, so the syndetic hull, if it exists, is unique. We now prove existence. We may assume, by induction on MATH (see REF), that MATH has a unique syndetic hull MATH in MATH (note that MATH is unipotent, so it has no nontrivial compact subgroups). Assume MATH is abelian. Because MATH is a normal subgroup of MATH, we may consider the quotient group MATH: let MATH be a maximal compact subgroup of MATH. By definition, MATH is compact. Also, MATH is simply connected (see REF), so MATH is connected (see REF). Therefore MATH is a syndetic hull of MATH. Assume MATH. Because MATH is connected (see REF), we know, from REF , that MATH has a syndetic hull MATH in MATH. Then MATH is a syndetic hull of MATH in MATH. Assume MATH is abelian. We have MATH, and MATH is connected (see REF), so, from REF , we know that MATH has a syndetic hull MATH in MATH. Then MATH is also a syndetic hull of MATH in MATH. Assume MATH is a normal subgroup of MATH. Let MATH be the natural homomorphism. Then, because MATH is compact, we see that MATH is closed, so MATH is a closed subgroup of MATH. Also, because MATH, we have MATH so MATH is abelian. Thus, from REF , we know that MATH has a syndetic hull MATH in MATH. Then MATH is a syndetic hull of MATH in MATH. The general case. The uniqueness of the syndetic hull MATH implies that MATH normalizes MATH; that is, MATH. Now MATH is connected (see REF), so, from REF , we know that MATH has a syndetic hull MATH in MATH; then MATH is also a syndetic hull of MATH in MATH. |
math/0102034 | Write MATH and MATH, where MATH is some discrete, normal subgroup of the center of the universal cover MATH of MATH. If MATH is any syndetic hull of MATH, then MATH is a syndetic hull of MATH. |
math/0102034 | By replacing MATH with MATH, we may assume MATH. Also, by replacing MATH with a subgroup, we may assume MATH is finite. From the structure theory of solvable NAME closed subgroups CITE, we have MATH and MATH, where MATH is the subgroup generated by the elements of MATH, all of whose eigenvalues are real and positive; then, because MATH contains the maximal compact subgroup MATH of MATH, REF implies MATH is connected. This is a finite-index subgroup of MATH, because MATH, and MATH is virtually connected. Therefore, MATH is virtually connected. |
math/0102035 | In CITE it is shown that a complete conformal MATH immersion of MATH with regular ends is a finite cover of a catenoid cousin or an immersion determined by MATH where MATH is a nonzero integer and MATH, MATH and MATH are complex numbers, which satisfy MATH for a positive integer MATH and MATH. (The proof in CITE contains typographical errors: The exponents MATH and MATH in REF should be reversed. If MATH, then the last paragraph of REF is correct. If MATH, then one must consider a possibility that is included in REF in that proof, and the result follows.) Changing MATH to MATH if necessary, we may assume MATH. Choose MATH so that MATH. Doing the MATH transformation MATH and replacing MATH with MATH produces the same surface, and one has MATH . Thus MATH and MATH are as desired. To study the symmetry group of the surface, we consider the transformations MATH of the plane. Then the NAME differential and secondary NAME map change as MATH where MATH . Hence MATH and MATH represent isometries of the surface. One can then check that there are no other isometries of the surface, that is, that there are no other anti-conformal bijections MATH of MATH so that MATH and MATH for some MATH. Thus the symmetry group is MATH. To see that the warped catenoid cousins have the explicit representation described in the theorem, one needs only to verify that MATH satisfies REF. |
math/0102035 | By REF, MATH . Since MATH by REF, we have MATH . Thus the only possibilities are MATH . CASE: By REF, we have MATH. Thus the end MATH is regular, and MATH is meromorphic on MATH. By REF, MATH. If MATH, the end has non-vanishing flux, and the surface does not exist, by REF. If MATH or MATH, by REF there is at most one umbilic point. Since any branch point of MATH is at an end or an umbilic point, REF is contradicted. Hence a surface of this type does not exist. CASE: Here the surface is simply connected, so it has a canonical isometrically corresponding minimal surface in MATH with the same total absolute curvature. We conclude the surface is a horosphere or an NAME cousin. CASE: Here, by REF, we have MATH. On the other hand, by REF, we have MATH. Thus MATH is either MATH or MATH, and we now consider these two cases separately: CASE: If MATH, then there are no umbilic points, by REF. If MATH, then the ends are regular, and REF implies the surface is a MATH-fold cover of an embedded catenoid cousin with MATH, or a warped catenoid cousin with MATH. Now assume that MATH . Then we have MATH by REF. By REF, we cannot have just one MATH, so also MATH. Then MATH is single-valued on MATH. Since MATH and MATH are both single-valued on MATH, the lift MATH is also (see REF), and so the dual immersion MATH is also single-valued on MATH. Since MATH, MATH is a MATH immersion with dual total absolute curvature MATH and of type MATH. Such a MATH cannot exist by REF, so such a MATH does not exist. CASE: If MATH, then the surface has only one umbilic point MATH with MATH, by REF, and we can set MATH, MATH, MATH, and MATH. By REF, MATH. Then by REF at least one of MATH and MATH is not an integer. Hence both are not integers, by REF. Then REF implies we may assume MATH and MATH. by REF, the metric MATH is the pull-back of the NAME metric on MATH by the map MATH . On the other hand, the NAME differential is of the form MATH . Thus MATH can be written in the form MATH . Consider the equation (which is introduced in CITE as REF ) MATH . We expand the coefficients MATH and MATH as MATH . Then the origin MATH is a regular singularity of REF . Let MATH and MATH be the solutions of the corresponding indicial equation MATH with MATH. If the surface exists, then REF implies that MATH must be a positive integer and the log-term coefficient of the solutions of REF must vanish. When MATH, the log-term coefficient vanishes if and only if MATH where MATH and MATH are given recursively by MATH as in REF of REF. Here we have MATH and so the log-term coefficient never vanishes at the end MATH, because MATH. Thus this type of surface does not exist. CASE: This is the only remaining case. But this type of surface does not exist, by the following REF . |
math/0102035 | We suppose MATH, and will arrive at a contradiction. Without loss of generality, we may set MATH and MATH, MATH, and MATH. CASE: Since MATH and MATH, REF implies MATH . Since MATH for all MATH, REF implies MATH for all MATH. Hence MATH by REF. Then REF implies MATH for all MATH, and as REF imply MATH, we have MATH and so the ends are regular. On the other hand, since MATH, REF imply MATH . Then by REF, we have MATH and furthermore at least two of the MATH are less than MATH. We may arrange the ends so that MATH . Moreover, by REF (note that the MATH there equal MATH), the metric MATH is reducible (as defined in REF of the present paper). Then, by REF and the relation REF in the appendices here, the secondary NAME map MATH can be expressed in the form MATH where MATH, MATH are relatively prime polynomials without zeros at MATH and MATH, and MATH . Note that the order of MATH at MATH is MATH and is also MATH. If MATH, MATH holds. Thus either MATH or MATH is an integer, but this contradicts REF, so MATH: MATH . Thus, by REF, we have MATH . Hence either MATH holds because of REF. To get more specific information about MATH and MATH, we now consider MATH: CASE: Since MATH is holomorphic on MATH with two zeroes (by REF), REF implies MATH with the MATH as in REF, as pointed out in CITE. Note that MATH because MATH and MATH. Let MATH and MATH be the two roots of MATH . In the case of a double root, we write MATH. Using REF, MATH has only the four following possibilities: MATH where MATH is a non-negative integer and the points MATH are mutually distinct. In the first case REF, the order of MATH at infinity REF is given by MATH . So MATH or MATH. Hence MATH and the order of MATH at MATH is MATH in the first case. In the other three REF , the orders of MATH at infinity are MATH respectively. These orders must equal either MATH or MATH, so none of these three cases can occur. We conclude that MATH is of the form MATH . Since the order of MATH at MATH is MATH, REF holds. CASE: Now we determine the polynomials MATH, MATH in REF . Differentiating REF, we have MATH where MATH . Since MATH and MATH are relatively prime, MATH does not divide MATH when MATH. But REF imply that MATH divides MATH, so MATH is constant, and we may assume MATH. Here, as seen in the previous step, REF holds, and then, MATH. Thus we have MATH . CASE: By REF we have MATH . REF also has roots MATH and MATH, so MATH . By REF and the first equation of REF, we have MATH. Substituting the first equation of REF into the second, we have MATH . Since MATH, REF implies MATH, contradicting REF and completing the proof. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.