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math/0107073 | Given MATH with MATH, the elements of the fiber MATH are all of the form MATH. They are mapped to MATH which describes a REF-plane in MATH. |
math/0107073 | We have shown now that the bundle MATH is diffeomorphic to the NAME fibration. The bundle MATH is the principal circle bundle associated to the complex line bundle MATH. So we must have MATH isomorphic to the line bundle associated to the NAME fibration, which is the line bundle MATH. Therefore MATH is isomorphic to MATH, as a complex vector bundle on a real manifold, and its NAME number is MATH . By REF , we find that MATH is biholomorphic to MATH via a biholomorphism taking MATH to MATH. The rest follows from REF . |
math/0107073 | Consider the line bundle over MATH soldered by the expression MATH. This is the line bundle whose fiber over a point MATH consists of MATH where the MATH is the complex determinant for the complex structure MATH on MATH determined by that point of MATH. Then REF-form MATH is a connection REF-form for that line bundle, and its curvature REF-form is MATH . Therefore the line bundle MATH has nonnegative NAME classes. Moreover if it has vanishing first NAME class, then we must have MATH, so must have MATH. From the structure equations we see that all of the invariants of MATH vanish, so by the NAME - NAME theorem and NAME 's theorem, we can easily see that MATH is the NAME - NAME equations. Now we have only to ascertain the relation between the NAME classes of MATH and those of MATH. We leave it to the reader to show that MATH as complex line bundles (evident from the soldering representations). |
math/0107073 | First we apply the great circle fibration REF to identify the fiber MATH with MATH diffeomorphically, and identify MATH with MATH diffeomorphically. This determines the topology of MATH and MATH completely, and allows us to apply REF to see that via biholomorphism, we can identify MATH with MATH and MATH with MATH. This determines the NAME class of the canonical bundle, allowing us to employ REF . |
math/0107074 | CASE: Fix an element MATH of unit norm, and pick a MATH such that MATH. If MATH are nets of complete contractions such that MATH in the point norm topology, define MATH, MATH by MATH for MATH. These maps are complete contractions, and MATH in the point norm topology, showing that MATH has the CCFP. To show that MATH has the CCFP, just define MATH and MATH. CASE: This is a simple exercise in the composition of maps. CASE: If MATH has the CBAP then there exists a net MATH of finite rank maps converging in the point norm topology to MATH and satisfying MATH. Given MATH and MATH, we can select MATH and MATH such that MATH and the result follows, since MATH and all such maps are finite rank. CASE: It was shown in CITE that nuclearity of MATH is equivalent to having an approximate point norm factorization of MATH through matrix algebras by complete contractions. This in turn is clearly equivalent to MATH having the CCFP, so the result follows from the transitivity of REF , taking MATH to be MATH. CASE: Suppose that MATH has the general slice map property. Let MATH be a MATH-algebra with a closed subspace MATH and consider an element MATH, all of whose right slices lie in MATH. If MATH, then the composition of MATH with MATH is a right slice map on MATH. By REF, so all right slices of MATH lie in MATH. Since MATH has the general slice map property, MATH, so MATH for all MATH. Take a limit over MATH to show that MATH, proving that MATH has the general slice map property. CASE: This is a special case of REF . REF We prove only REF since the argument also applies to REF . Let MATH be any MATH-algebra, and suppose that MATH has the strong OAP. Given MATH, we may choose MATH so that MATH . Then we may choose, by hypothesis, a finite rank map MATH such that MATH . Applying MATH to REF , and using REF and the triangle inequality, gives MATH . This shows that MATH has the strong OAP. CASE: Let MATH be a norm closed ideal in a MATH-algebra MATH with quotient map MATH, and suppose that MATH is exact. To show that MATH is exact, we need only prove that the kernel of MATH is contained in MATH, CITE. Consider an element MATH, and observe that MATH, which is MATH by exactness of MATH. Apply MATH and take a limit over MATH to obtain MATH. This shows that MATH is exact. |
math/0107074 | Consider MATH and MATH. We fix a positive number MATH to be chosen later. For each MATH, MATH, let MATH denote the element of MATH whose value at MATH is MATH. Then define MATH by MATH . Then MATH . Each map MATH is a complete contraction, so MATH may be viewed as a vector integral of such maps, giving MATH. For each MATH, let MATH be the associated normal vector functional on MATH. Then the left slice map MATH is well defined on MATH, which we identify with MATH, and MATH. Let MATH be the restriction of MATH to MATH. Then MATH. We first show that the range of MATH is contained in MATH. In the following calculation, our assumptions on continuity of MATH and MATH will automatically imply that the integrands are integrable. If MATH, then MATH . It follows from REF that MATH and this last integral is an element of MATH. If we let MATH denote MATH, then we have shown that MATH, and MATH maps MATH into MATH, since the span of the elements MATH, MATH, MATH, is norm dense in MATH. For the given elements MATH we now wish to choose MATH and MATH so that MATH . We restrict attention to MATH, MATH, and MATH, MATH, so that we already have MATH, MATH. For each MATH, the map MATH is continuous on MATH, so we may choose a symmetric neighborhood MATH of MATH such that MATH . Now choose a non-negative MATH, MATH, whose support is contained in MATH, and let MATH denote the left translate MATH of MATH. The map MATH is continuous on MATH, by REF, so there is a neighborhood MATH of MATH, contained in MATH, within which MATH. In particular, the NAME inequality shows that MATH . Finally we choose MATH, MATH, and having support in MATH. We are now ready to show that, with these choices, MATH is close to MATH for MATH. Let MATH be arbitrary vectors of unit norm. Then MATH . We now estimate these integrals separately. Since MATH the first may be rewritten as MATH . Here we have used REF and NAME 's theorem, which is permissible because the integrand is a continuous function of compact support on MATH. For the second integral, we change the order of integration, and observe that the MATH-variable can be restricted to MATH (because of the MATH term). This yields MATH again using REF . Returning to REF , the two estimates REF lead to MATH . The unit vectors were arbitrary in REF , so MATH . The proof is completed by choosing MATH so small that MATH . Thus MATH has the CCFP. |
math/0107074 | Apply REF to the pair MATH, which has the CCFP by REF . |
math/0107074 | Since MATH is abelian, the NAME duality theorem, CITE, states that there is a dual action MATH of MATH on MATH, and MATH is isomorphic to MATH. Two applications of REF give MATH the last equality being a special case of REF. |
math/0107074 | It is clearly sufficient to consider a finite number of elements MATH, MATH, where MATH and MATH, MATH, MATH. For MATH, define MATH by MATH . Since MATH is a homomorphism, it is clear that MATH. For each MATH, we define a map MATH, MATH, by slicing in the second copy of MATH by the vector functional MATH. A priori, MATH maps into MATH, but it will become apparent from the subsequent calculations that the range of MATH lies in MATH. From REF , MATH . Let MATH be defined by MATH . Then we may integrate first with respect to MATH in REF to conclude that MATH . This establishes that each MATH maps into MATH. It also follows from REF that MATH so to show that MATH is approximately the identity on the elements MATH, it suffices to choose MATH and MATH so that the right-hand side of REF is small in norm. A simple estimate gives MATH since MATH, so given MATH, it suffices to find MATH, MATH, MATH, MATH, such that MATH on the combined supports of the MATH's. We now use the hypothesis that MATH is amenable. Let MATH, a compact subset of MATH, be the union of the supports of the MATH's. By REF, there is a unit vector MATH which we may take to be in MATH such that MATH satisfies MATH . Here MATH is defined by MATH. Let MATH be the support of MATH, and select MATH, MATH, and MATH on MATH. For these choices, it follows from REF that MATH . An immediate consequence of REF is MATH completing the proof. |
math/0107074 | We have already shown that MATH. For the converse, let MATH be the dual coaction. By REF , the pair MATH has the CCFP, so by REF MATH . But MATH is isomorphic to MATH, CITE, so REF becomes MATH by REF, proving REF . |
math/0107074 | By CITE, there is a dual action MATH of MATH on MATH such that MATH is isomorphic to MATH. Using REF , with MATH in place of MATH, we see that MATH proving REF . |
math/0107074 | Since the pair MATH has the CCFP for any NAME space MATH REF , these properties all transfer from MATH to MATH. The verification that these properties all transfer from MATH to MATH is essentially routine, based on the nuclearity of MATH. To give the flavor, we will prove this for REF , and leave the others to the reader. Suppose that MATH has the general slice map property and fix a MATH-algebra MATH with a closed subspace MATH. Let MATH be an element whose right slices lie in MATH. If MATH and MATH, then slicing by MATH is the same as slicing by MATH and then by MATH. Thus the element MATH obtained from MATH by slicing by MATH has the property that all right slices are in MATH. Since MATH has the general slice map property, by nuclearity, it follows that MATH. Since MATH was arbitrary, the general slice map property for MATH implies that MATH as required. We have already shown in REF that the pairs MATH and MATH have the CCFP, so we conclude from REF and the preceding remarks that these properties transfer between MATH and MATH in both directions. Applying this to the pair MATH and MATH, we see that they also transfer in both directions between MATH and MATH. |
math/0107075 | For any MATH, MATH and so MATH . To estimate MATH, we may replace MATH, MATH, in REF by unitaries MATH, and we may further assume that MATH has the form MATH for some unitary MATH. Thus it suffices to estimate MATH as MATH ranges over the unitary group of MATH. Write MATH, MATH. Then MATH, and MATH. Thus REF becomes MATH . The result follows by taking the supremum over all unitaries MATH in REF . |
math/0107075 | Let MATH be any unitary, and regard MATH and MATH as projections in MATH. Both are positive operators, so the inequality MATH follows by applying states to this difference. Then MATH, since MATH by taking MATH in REF . Now choose a projection MATH, MATH, and choose a partial isometry MATH such that MATH . The element MATH is a unitary in MATH satisfying MATH or, equivalently, MATH. Then MATH, which forces MATH, since these operators commute. Thus MATH . This choice of unitary shows that the inequality in REF fails for each MATH, and it follows that MATH. |
math/0107075 | Each MATH is a finite linear combination of group elements, so it suffices to prove, for fixed MATH, that the equation MATH is satisfied by all but a finite number of MATH. The modular properties of MATH show that REF always holds when either MATH or MATH is in MATH, so we may assume that both elements are not. In this case the right hand side of REF is REF, and so we only need establish that MATH for only finitely many MATH. If MATH and MATH are two such elements, then MATH. The hypotheses then imply that MATH, showing that REF fails for at most one MATH. |
math/0107075 | If we can prove REF for MATH, then the MATH - norm continuity of conditional expectations and the NAME density theorem will show that it holds generally. Thus we assume that MATH. Then, by REF , we may choose MATH so that MATH . For this choice of MATH, MATH where the last equality follows from REF , since MATH. Now write MATH . Since MATH, we may apply MATH to conclude that MATH and MATH commute. Thus MATH . We now estimate the last term in REF . By REF , MATH . Using REF becomes MATH . Replacing MATH and MATH from REF gives REF . If MATH is a unitary, REF follows immediately from REF by replacing MATH with REF. |
math/0107075 | CASE: For MATH, MATH, we have MATH . Taking the supremum in REF over MATH implies that MATH . Applying REF with MATH gives MATH which proves REF . CASE: Assume now that MATH is abelian. The estimate in REF shows that MATH is a strongly singular masa in MATH (although, a priori, it was not clear that MATH was maximal). Thus, MATH. We now estimate MATH. Let MATH and MATH be orthogonal projections in MATH which are equivalent in MATH. We may choose a nilpotent partial isometry MATH such that MATH and MATH. Then MATH . By REF , MATH since MATH. It follows that MATH, and since MATH is always true, equality is immediate. |
math/0107075 | Any generator satisfies the hypotheses of REF . |
math/0107075 | One direction is clear. Conversely, suppose that the hypotheses on the normalizers are fulfilled, but suppose that there is a MATH such that, for some MATH and MATH, MATH. Then MATH and so MATH. Hence MATH for some MATH, since one of MATH, MATH is in MATH. Since MATH is prime, so too is MATH, forcing MATH. Thus MATH normalizes MATH, a contradiction which proves the result. |
math/0107075 | By CITE, (see also REF), a prime element MATH in a non - elementary hyperbolic group MATH satisfies MATH for all MATH. REF them shows that the hypotheses of REF are satisfied, and the result follows. |
math/0107075 | The subgroup MATH of MATH satisfies the hypotheses of REF . |
math/0107075 | NAME, CITE, and NAME, CITE, have both given examples of countable amenable discrete I.C.C. groups containing abelian subgroups which satisfy the hypotheses of REF , and the result is then immediate. For the reader's convenience, we briefly describe NAME 's example. Let MATH be an infinite field that is the countable union of finite subfields (the algebraic closure of a finite field has this property). Let MATH be the group of affine transformations of the linear space of dimension REF over MATH, and let MATH be the abelian subgroup of homotheties about REF. The calculations of CITE show that the hypotheses of REF are satisfied. |
math/0107075 | For each MATH define a bounded bilinear map MATH by MATH for MATH. We consider first the case where MATH and MATH are group elements in MATH. If either one is in MATH then the module properties of MATH imply that MATH for all MATH. Now suppose that MATH. By hypothesis, there exists MATH such that MATH for MATH, so both terms on the right hand side of REF are REF, showing that MATH for all MATH. It then follows that, for MATH, MATH for MATH sufficiently large. The estimate MATH for MATH, MATH is immediate from REF , so if MATH, MATH, MATH, and MATH, we obtain MATH . Thus, for each MATH, MATH since MATH for MATH sufficiently large. Let MATH in REF to see that MATH for MATH and MATH. REF also gives the estimate MATH . We then repeat the previous argument, this time in the second variable, to obtain-MATH for all MATH. This completes the proof. |
math/0107075 | From the third section, MATH satisfies the hypotheses of REF , and the result follows. |
math/0107075 | In proving REF , it clearly suffices to assume that MATH, MATH. Now fix MATH. By the asymptotic homomorphism hypothesis we may choose a unitary MATH such that MATH . Then, using REF , MATH . Since MATH commute, MATH and so REF follows by substituting REF into REF and letting MATH. |
math/0107075 | Let MATH be an arbitrary unitary. To show strong singularity, we will apply REF with MATH and MATH. Then MATH . Thus MATH is strongly singular. We now estimate MATH. Let MATH be a nilpotent partial isometry such that MATH and MATH are orthogonal projections in MATH. From REF , MATH, so the choices of MATH and MATH in REF lead to MATH . This proves that MATH, and the reverse inequality always holds. |
math/0107075 | This is an immediate consequence of MATH for any pair of orthogonal vectors in a NAME space. |
math/0107075 | Let MATH with MATH. Write MATH and let MATH. Note that MATH and MATH. Then MATH . Here we have used the module properties of conditional expectations and that MATH and MATH commute. The last expression in REF is no greater than MATH since MATH. The estimates REF combine to yield MATH letting MATH vary over the unit ball of MATH. Since MATH (see CITE), the last inequality gives MATH . The result follows by letting MATH vary over the unit ball of MATH in REF . |
math/0107075 | Let MATH be a nilpotent partial isometry such that MATH and MATH are orthogonal projections in MATH, and define a unitary MATH by MATH . Then MATH and MATH . Thus MATH . By the orthogonality of MATH and MATH, and the modularity of MATH, MATH . The two terms on the right hand side of REF are equal because the map MATH implements an isometry from MATH to MATH in the norms MATH and MATH. Thus REF becomes MATH . From REF , MATH, and so the definition of MATH gives MATH . Thus MATH . Then REF give MATH . In the NAME algebra MATH, consider the masas MATH and MATH. We may apply REF to obtain MATH . Since MATH and MATH have orthogonal ranges, MATH using REF . Note that these expectations are defined on MATH, but viewing them on MATH by first applying MATH does not change REF . We now combine REF to obtain MATH which shows that MATH. |
math/0107076 | Suppose that there is a unitary MATH for which MATH when MATH and MATH are arbitrary elements of MATH. Then REF will hold for MATH, and the MATH-norm continuity of MATH will extend its validity to all MATH. Thus we may restrict attention to MATH. Let MATH be the unitary MATH, MATH, which generates MATH, and let MATH be the corresponding unitary generator of MATH. Since MATH is an isometry for the NAME space norms, we have MATH . From this it follows that MATH because the right hand side of REF is the conjugate of the MATH . NAME coefficient of MATH. Fix MATH, let MATH, and choose MATH, by REF , so that MATH . From REF , we may now choose MATH so that MATH . Since MATH is an orthonormal basis, we may write MATH, which gives MATH . We split this sum at MATH and estimate each part separately. By REF , MATH provided that MATH. Since MATH we may use the NAME inequality to estimate MATH by the choice of MATH (see REF ). Then REF allow us to obtain the inequality MATH from REF . This proves REF , and the proof is complete. |
math/0107076 | CASE: Clear. CASE: These are the result of counting the terms in REF and noting that the cardinalities of MATH and MATH in REF are respectively MATH and MATH. REF show that MATH and so these differences are independent of MATH. Since MATH, we conclude that MATH . Subtraction of REF from REF gives MATH using REF . Since MATH, a simple induction argument based on REF , shows that MATH . It follows from REF that MATH . There are MATH words in MATH, so counting these words according to whether they lie in a sum of the form MATH, MATH or MATH leads to MATH which, after cancellation, is MATH . Substituting from REF gives MATH . Now REF follows from REF if we define MATH to be MATH. |
math/0107076 | First suppose that these are one point sets. Then each MATH is either MATH or MATH, so the difference is estimated by MATH, using REF . For the general case, REF shows that this difference can be realized as a sum of MATH differences for one point sets. This gives the estimate MATH . Let MATH be the largest possible right hand side in REF , which is MATH. |
math/0107076 | We first define MATH and MATH, for MATH, MATH, and then verify that they have the required properties. Let MATH . For MATH and MATH, let MATH . Note that the pairs MATH and MATH are distinct, since otherwise cancellations would occur in MATH or MATH. Thus the cardinalities of the sets in REF are respectively MATH and MATH, and they depend only on MATH and MATH. If the reduced word MATH results from one of these cancellations, then the first letter of MATH must not be MATH (if MATH, then this constraint disappears). The original reduced word in MATH which canceled to this was MATH which requires the first letter of MATH to be different from MATH (this constraint disappears for MATH). Thus the first letter of MATH must lie in MATH. Conversely, such a MATH allows exactly MATH cancellations on the left in REF . A similar analysis on the right shows that a word of the form REF is both reduced and allows the correct number of cancellations precisely when MATH starts with an element of MATH and ends with an element of MATH. This proves the result. |
math/0107076 | Let MATH and MATH in MATH be fixed words of lengths MATH and MATH respectively. We require that MATH since otherwise REF , for which we are aiming, is trivial. Let MATH be an arbitrary word of length MATH, and suppose throughout that MATH. If a word MATH has length MATH then it is orthogonal to MATH for MATH. From this and REF , it follows that MATH . Thus, with the notation of REF , MATH . By REF MATH for MATH. Thus there exist constants MATH, uniformly bounded by MATH, such that MATH . Now MATH is the number of terms in MATH, so MATH . Thus MATH so MATH for MATH. This last inequality and REF then give the estimate MATH . Let MATH be the constant on the right hand side of REF . If we sum REF over all words MATH of length MATH (of which there are MATH), we obtain MATH . Since MATH and MATH, REF implies that MATH . If we let MATH, then the terms of the series in REF are bounded by MATH, and the series is clearly summable. The result now follows from REF . |
math/0107076 | Apply REF and CITE. |
math/0107076 | The elements MATH generate a copy of MATH inside MATH, and we write MATH for the resulting subfactor of MATH. As before, we let MATH be the sum of all words of length MATH in MATH, and we note that MATH is an orthonormal basis for MATH. The result will follow from REF if we can show that MATH for all MATH. If MATH, MATH, then MATH using modularity of MATH. If MATH, then MATH, so all terms in REF vanish. If MATH, then the argument of REF shows the validity of REF . A similar analysis holds if we begin by supposing that MATH. We may thus assume that MATH. In this case MATH, so REF becomes MATH . The elements MATH and MATH may be written as products MATH where MATH, MATH, and MATH, MATH. Let MATH where MATH is the length of MATH in terms of MATH. We will show that the cardinality MATH is at most MATH. Consider a word MATH of length MATH and let MATH be written in the free product MATH with each MATH a non - zero power of some MATH, and MATH for all MATH. Note that MATH. We observe that the product MATH is in MATH if and only if one of the following conditions is satisfied: REF there exists MATH, MATH, such that MATH for MATH, and MATH for MATH (with obvious modifications if MATH or MATH), and the element MATH is in MATH; REF there exists MATH, MATH, such that MATH for MATH, and MATH for MATH (with obvious modifications if MATH or MATH), and the element MATH is in MATH. In REF , MATH and MATH are in the same MATH since further cancellation must occur, while in REF the same conclusion applies to the elements MATH, MATH, and MATH. In REF , there are at most MATH values of MATH, and thus at most the same number of possibilities for MATH. In REF , MATH takes up to MATH values, and the remaining term MATH is a non - zero power MATH for some MATH and some MATH with MATH. This gives an upper bound of MATH possibilities. The two cases combine to give an upper estimate (which could undoubtedly be lowered by a more detailed analysis) of MATH for MATH. It now follows that MATH since MATH for each MATH. Thus the terms in REF are dominated by MATH, so convergence is guaranteed by REF . This proves that MATH is an asymptotic homomorphism, and the other statements now follow from CITE. |
math/0107077 | Let MATH be a diagonal, where the series is norm convergent and MATH and MATH are strongly independent. Since MATH is a norm convergent series, we may choose MATH so that MATH and we set MATH. From the NAME series we know that MATH. Now define two constants MATH and MATH by MATH . Since, for each MATH, the series MATH are norm convergent, we may apply, by CITE, an element MATH to REF to obtain MATH . From the strong independence of MATH, we may choose linear functionals MATH, MATH, such that MATH where MATH denotes the canonical orthonormal basis for MATH. It now follows that MATH for MATH and for all MATH. Using REF , we have that MATH for MATH, and for all MATH. Since MATH, we may choose MATH sufficiently large that MATH holds for MATH, and for all MATH. The inequality MATH follows from REF , and so the relation MATH is a consequence of multiplying the expression in REF on the left by MATH and summing over MATH. Now multiply REF on the left by MATH and use REF to obtain MATH . Define a finite dimensional subspace of MATH by MATH . REF implies that the NAME space quotient map from MATH to MATH has norm at most MATH, which can only happen when MATH. We conclude that MATH is finite dimensional. |
math/0107077 | For any element in the algebraic tensor product MATH, we have that its projective tensor norm is at least as large as its NAME norm. Thus, the identity map on MATH extends to a contractive map from the projective tensor product to the NAME tensor product. It is easily checked that if MATH is a diagonal in the projective tensor product, then its image under this map is a diagonal in the NAME tensor product, and the result follows from REF . A similar argument applies to the operator space projective tensor product. |
math/0107077 | Since MATH can be represented as the algebra of left multiplication operators on itself, we may assume that MATH is a subalgebra of MATH for some MATH. Now suppose that MATH is an invariant orthogonal projection for MATH, that is, MATH for all MATH in MATH. Then it is easily seen that MATH is a MATH-bimodule, and that the equation MATH defines a derivation of MATH into MATH. By hypothesis, there exists MATH such that MATH . Combining these last two equations, we see that MATH commutes with MATH. Since MATH, the element MATH is invertible in MATH with inverse MATH. The equations MATH and MATH show that MATH commutes with MATH, and thus reduces this algebra. By inductively choosing such projections MATH and conjugating by the corresponding invertible elements, we may assume that the representation MATH is a finite direct sum of representations, MATH, where the image MATH is a subalgebra of MATH that has no non-trivial invariant projections. Thus, MATH is a transitive subalgebra, and hence MATH by NAME 's theorem, CITE. Using the simplicity of each matrix algebra, it is now easy to argue that MATH is isomorphic to a finite direct sum of matrix algebras. To see this, note that if MATH, then MATH is either MATH or MATH and argue by induction on MATH. |
math/0107079 | Consider REF. Note that the jump matrix has determinant MATH. Thus MATH, and hence MATH is an entire function. Also MATH as MATH. Thus by NAME 's theorem, we have MATH, and in particular, MATH is analytic in MATH, and is continuous in MATH. Set MATH and set MATH and MATH . Set MATH . Note that by REF, MATH . Now the jump matrix for MATH is a constant matrix : MATH . Equations : By differentiating REF, MATH satisfies the same jump condition. Hence MATH has no jump cross MATH. Also from REF, MATH has a double pole at MATH. Therefore using REF, we have MATH for some constant matrices MATH which depend on MATH. On the other hand, MATH again has no jump cross MATH, and it is now entire. Thus using REF, we have MATH for some constant matrix MATH. Now we take `cross differentiation' of REF in two different ways : MATH and MATH . Thus we obtain an equation MATH . By plugging in the formulas of MATH and MATH and comparing the coefficients in MATH, we obtain the relations for MATH : MATH . Constant matrices MATH : Now we express MATH in terms of MATH. We plug in REF into REF. This determines the constant matrices MATH. Using REF for REF, MATH . Using REF for REF, MATH . Using REF for REF, MATH . Using REF for REF, MATH . Symmetry : Note that the jump matrix MATH for MATH satisfies MATH . Thus it is standard to show that MATH satisfies the symmetry condition MATH . By taking MATH, and considering terms of order MATH and MATH, we obtain the symmetry relations MATH . Since MATH and MATH, we can set MATH and set MATH . Equations from REF : Now we re-write REF in terms MATH and MATH. First, if we use REF, use REF yields MATH . Second, if we use REF yields the same REF . Finally, REF is trivial if we use REF. Equations from REF : We also have the condition that the two different REF for MATH are equal. Similarly REF for MATH are equal. Also there is an additional REF . From REF, we have MATH . Using REF, this yields MATH . If we use REF , and the relation MATH, this equation becomes MATH . From REF, we have MATH which yields MATH . Thus using REF yields MATH . On the other hand, REF implies MATH which yields MATH . Thus using REF yields MATH . |
math/0107082 | Observe that MATH so that integration by parts yields MATH . REF follows by repeating this procedure. |
math/0107082 | The case MATH is simply the known result MATH . For MATH, the function MATH satisfies MATH, for MATH. Then REF , with MATH, yields MATH so that REF follows from REF . |
math/0107082 | Same as the proof for Example MATH with MATH . |
math/0107082 | Use the identity REF . |
math/0107082 | The NAME zeta function satisfies MATH so the result follows from REF with MATH, since in that case the integral on the right-hand side equals the one on the left-hand side, except for the prefactor MATH. |
math/0107082 | Divide REF by MATH and then sum over MATH to produce MATH . The result follows by interchanging the order of summation. |
math/0107082 | Use the identity REF and MATH in REF to produce MATH . The generating function REF is now employed to see that the sum from MATH to infinity is independent of MATH, so it is absorbed into the implicit constant of integration. |
math/0107082 | Differentiate both sides of the identity MATH with respect to MATH at MATH. |
math/0107082 | Take the limit MATH in REF and use the fact that MATH. |
math/0107082 | This is a direct consequence of MATH valid for MATH. |
math/0107082 | We have MATH . |
math/0107082 | Differentiate MATH at MATH and use the identity MATH . |
math/0107082 | REF follows directly from REF at MATH. |
math/0107082 | Perform the derivative and use MATH, in addition to REF and the identity REF . |
math/0107082 | Use REF and the analogous result for the NAME polynomials, MATH for MATH. |
math/0107082 | For MATH, REF is a direct consequence of REF and the well-known property of the NAME polynomials, MATH . The case MATH follows directly from REF since MATH . |
math/0107082 | In REF of MATH substitute, depending on the parity of MATH, the NAME expansions REF or REF for the functions MATH and REF or REF for the NAME polynomials. |
math/0107082 | In REF use REF and the property MATH satisfied by the NAME polynomials. |
math/0107082 | Set MATH and evaluate REF at MATH. Then use REF . |
math/0107082 | The basic identity MATH appears in CITE:REF:REF. The identity of the lemma is obtained by replacing MATH by MATH and MATH by MATH in REF . |
math/0107082 | In view of relationship REF , we set MATH in REF and take the limit MATH. For a general value of MATH there will be three kinds of terms to consider in the sum on the right-hand side of REF : CASE: For MATH no singularities will arise and we simply have MATH . CASE: For MATH we use the results MATH and MATH where MATH is the usual digamma function, to obtain MATH . CASE: For MATH, say MATH with MATH, we use MATH and MATH to obtain MATH where we have used REF of the balanced egapolygamma function and REF . When all the terms are added up, we find, in view of REF , that the coefficients of the MATH singularity and the term proportional to the harmonic number MATH reduce to a MATH-independent constant, which can be dropped. This proves the theorem. |
math/0107082 | Use REF and proceed along the same lines as the proof above. |
math/0107082 | Differentiate REF at MATH and use the result MATH to obtain the following result for the moments of the function MATH: MATH . This result is equivalent to the statement of the theorem in view of REF of the balanced negapolygammas and result REF for the moments of the NAME polynomials. |
math/0107082 | Apply REF to MATH. REF correspond to MATH and MATH respectively. |
math/0107082 | Use the reflection formula for MATH, results REF and REF to produce REF . The term that corresponds to the NAME polynomials disappears in view of MATH . |
math/0107082 | Divide REF by MATH and sum over MATH. |
math/0107082 | Set MATH in REF and use the identity REF and the hypothesis MATH to get rid of the last term in REF in the limit MATH. |
math/0107082 | Direct evaluation of the right-hand side of REF gives, in view of REF , MATH . Clearly only the terms with MATH odd, say MATH, survive in the sum. REF then follows directly from REF . |
math/0107082 | Use REF . |
math/0107082 | Use the representation REF of the balanced negapolygammas to obtain the desired result, by direct differentiation of the formula MATH valid for real MATH, given in CITE. |
math/0107083 | Since the differential system for MATH, MATH, and MATH is NAME in any of these cases, one can choose initial values MATH of MATH, MATH, and MATH arbitrarily at the specified point and have a (unique) solution to the system, globally defined on a simply connected neighborhood. Choosing MATH in a bounded region in the quarter-space defined by MATH and MATH will then yield a congruence class of immersions realizing MATH on a (possibly smaller) fixed neighborhood of the given point in MATH. |
math/0107083 | Choose a principal curve from the first family and use its intersections with the principal curves from the second family as initial points to construct via integration a function MATH on MATH so that MATH. Now set MATH, so that MATH. Since MATH (the first equation of REF), it follows that MATH . Thus, one can find (by quadrature) a function MATH on MATH so that MATH, that is, so that MATH. The function MATH is a principal coordinate and its fibers are the principal curves of the second family. In particular, the mapping MATH embeds MATH as a domain in MATH that lies inside MATH. Note that if MATH for some functions MATH and MATH, then MATH for some function MATH of one variable with positive derivative and MATH. Conversely, for any function MATH with MATH, the functions MATH satisfy the conditions MATH and MATH. Now, since MATH (by the third equation of REF), it follows that MATH so there exists a function MATH defined on MATH so that MATH. Next, by construction, MATH is a multiple of MATH and, moreover, since MATH (the second equation of REF), it follows that the MATH-form MATH is closed. Since this MATH-form is closed and a multiple of MATH, it must be of the form MATH for some function MATH defined on MATH. Thus, MATH . Now, consider any coordinate system MATH in which MATH for some functions MATH and MATH on the range of MATH. As has been noted already, there is a MATH so that MATH and MATH. Using this to compare the two formulae for MATH yields MATH . Thus, choosing MATH so that MATH satisfies MATH (which can be done by a sequence of two quadratures) yields a coordinate change MATH for which MATH. This yields the normal form of the lemma. Note that if MATH and MATH, then MATH, so that MATH is of the form MATH for some constants MATH and MATH, as claimed in the lemma. Finally, that the MATH-forms and function defined in REF do satisfy REF can be safely left to the reader. |
math/0107083 | All but the last statement has been verified already. To prove the last statement, it suffices to note that the spherical center of the geodesic circle tangent to MATH with geodesic curvature MATH is given by MATH . Computation now shows that MATH where MATH and MATH and, moreover, that MATH. It follows that MATH moves on the great circle perpendicular to the fixed vector MATH. |
math/0107083 | Write MATH and note that these components satisfy the equations MATH . This is the content of the proposition. |
math/0107085 | Consider the function MATH . It has the property that MATH . We also note that MATH. To see this is independent of MATH note for any MATH . But MATH preserves the measure MATH and the orientation of MATH so MATH . Hence MATH . Since MATH is the lift of a probability measure on MATH we know that MATH for any MATH . So if MATH we see that MATH . To see that MATH we note MATH . |
math/0107085 | To see that MATH is a homomorphism we suppose MATH and consider MATH as defined above. Then MATH . Since MATH we know that MATH is also a homomorphism. There is a natural homomorphism MATH which assigns to MATH its projection on MATH. If MATH is the natural projection then for any MATH . Hence MATH is a homomorphism. |
math/0107085 | Since MATH is abelian it is amenable and there is a NAME probability measure MATH on MATH invariant under MATH. Let MATH be the lift of this measure to MATH . Then REF implies that MATH and MATH are homomorphisms. As above let MATH be the natural projection. The kernel of MATH is MATH, where MATH, that is, the lifts of the identity. If MATH then MATH is in the kernel of MATH so MATH for some MATH . But since MATH is a homomorphism MATH which implies MATH . Hence MATH is abelian. |
math/0107085 | Let MATH be the lift of the measure MATH to MATH . Pick lifts MATH and MATH of MATH and MATH respectively which satisfy MATH . Then for any MATH we have MATH . Suppose that MATH. It then follows that MATH so for a sufficiently small MATH and MATH . Applying MATH to MATH we see that MATH. Hence MATH which is not possible if MATH and hence MATH is in the support of MATH . We have shown that MATH implies MATH . The inequality MATH is similar. Projecting back to MATH we conclude that MATH on MATH . |
math/0107085 | Let MATH be an element of MATH. Then the group generated by MATH and MATH is abelian and hence amenable so there it has an invariant NAME probability measure. But NAME measure is the unique NAME probability measure invariant by the irrational rotation MATH. Since MATH preserves orientation and NAME measure, it is a rotation. |
math/0107085 | The proof is by contradiction. Assume MATH is a component of MATH and MATH . Since MATH and MATH commute MATH so MATH implies MATH . Let MATH be an element of MATH and without loss of generality assume MATH . Define MATH . Then MATH and MATH are fixed under both MATH and MATH and MATH has no fixed points in MATH . Then NAME 's Lemma REF implies MATH for all MATH, contradicting the hypothesis that MATH is a component of MATH. The observation that MATH follows from the fact that MATH implies that either MATH is a component of MATH or MATH is the endpoint of an interval which is a component of MATH so, in either case, MATH . |
math/0107085 | First consider the case that MATH . The assertion that MATH preserves components of MATH is then trivial. Also, MATH is topologically conjugate to an irrational rotation which by REF implies that MATH is topologically conjugate to a rotation and hence MATH or MATH . In either case we have the desired result. Thus we may assume both MATH and MATH have periodic points. Since for any circle homeomorphism all periodic points must have the same period, we can let MATH be the least common multiple of the periods and observe that MATH and MATH . Since MATH and MATH commute and both MATH and MATH have fixed points, if MATH then MATH will exist and be a common fixed point for MATH and MATH. If we split the circle at MATH we obtain two commuting diffeomorphisms, MATH and MATH of an interval to which we may apply REF and obtain the desired result. |
math/0107085 | We first observe the result is easy if MATH has no periodic points. In this case MATH is conjugate in MATH to an irrational rotation by NAME 's Theorem, and the centralizer of an irrational rotation is abelian by REF . So the centralizer of MATH in MATH is abelian. But then MATH restricted to an abelian subgroup of MATH is a homomorphism by REF . Thus we may assume MATH has a periodic point, say of period MATH. Then MATH has a fixed point. Let MATH be a lift of MATH to MATH which has a fixed point. Let MATH denote the group of all lifts to MATH of all elements of MATH. Note that every element of this group commutes with MATH because by REF any lifts to MATH of commuting homeomorphisms of MATH commute. Let MATH and observe that MATH for all MATH, so MATH acts on the unbounded closed set MATH and MATH acts on MATH which is the image of MATH under the covering projection to MATH . We will show that if MATH then MATH . Indeed applying REF to MATH and map MATH of MATH which MATH covers, we observe that MATH which implies MATH if MATH has has a fixed point. It follows that if MATH is the stabilizer of MATH then MATH acts freely on MATH and hence is abelian by NAME 's theorem. Also MATH acts on MATH and hence there is a measure MATH supported on MATH and invariant under MATH. Clearly this measure is also invariant under the action of MATH on MATH. The measure MATH lifts to a MATH-invariant measure MATH . It follows from REF that the translation number MATH and the rotation number MATH are homomorphisms. |
math/0107085 | By REF we may conclude that for each MATH we have MATH . Call this set of fixed points MATH. Clearly MATH is the set of global fixed points of MATH. Fix a value of MATH and consider MATH . Restricting to any component MATH of the complement of MATH we consider the possiblity that there is a MATH with a fixed point in MATH. NAME 's Lemma REF , applied to the closure of the open inteval MATH, tells us that such a MATH is the identity on MATH. Thus the restriction of MATH to MATH is free and hence abelian by NAME 's Theorem. But MATH was an arbitrary component of the complement of MATH and obviously elements of MATH commute on their common fixed set MATH. So we conclude that MATH is abelian. We have hence shown that if MATH and MATH are joined by a path of length two in the commutativity graph, they are joined by a path of length one. A straightforward induction shows that any two generators are joined by a path of length one, that is, any two commute. |
math/0107085 | By REF we may conclude that for each MATH we have MATH . But since each MATH has a fixed point MATH . Hence there is a common fixed point for all of the generators. Splitting MATH at a common fixed point we see MATH is isomorphic to a subgroup of MATH satisfying the hypothesis of REF . It follows that MATH is abelian. |
math/0107085 | Suppose MATH and suppose MATH . We will show that this implies that MATH . Let MATH . Observe that MATH implies MATH. Hence we may assume without loss of generality that MATH is in the subinterval of MATH with endpoints MATH and MATH (otherwise replace MATH and MATH by their inverses). In particular MATH . Note that MATH . This implies MATH and since MATH we can conclude that MATH . Since MATH was arbitrary we see MATH . |
math/0107085 | The relation MATH implies MATH, so MATH and MATH are conjugate. Therefore REF implies MATH is conjugate to MATH and MATH is conjugate to MATH and MATH. This proves the result. |
math/0107085 | Let MATH denote the surface of genus MATH with MATH punctures. Let MATH denote the geometric intersection number of the (isotopy classes of) closed curves MATH and MATH on MATH. Let MATH with MATH be NAME twists about essential, simple closed curves MATH with the following properties: CASE: MATH for MATH. CASE: MATH for each MATH. CASE: MATH for MATH. It is known (see, for example . CITE) that there exist such MATH which generate MATH; we choose such elements. These are also nontrivial in MATH. Since NAME twists about simple closed curves with intersection number zero commute, and since NAME twists MATH about essential, simple closed curves with intersection number one satisy MATH (see, for example, REF), REF follow. We note that the proofs of these relations do not depend on the location of the puncture, as long as it is chosen off the curves MATH. To prove REF, consider the surface of genus MATH and with MATH boundary components obtained by cutting MATH along MATH and MATH. The loop MATH becomes a pair of arcs connecting MATH pairs of boundary components, and MATH becomes an arc connecting MATH boundary components. The genus, boundary components, and combinatorics of arcs is the exact same when cutting MATH along MATH and MATH. Hence by the classification of surfaces it follows that there is a homeomorphism between resulting surfaces, inducing a homeomorphism MATH with MATH and MATH. The homotopy class of MATH gives the required element MATH. To prove REF , let MATH be an essential, separating, simple closed curve on MATH such that one of the components MATH of MATH has genus one and contains MATH and MATH. Then there is a homeomorphism MATH taking any element of MATH to any other element, and which is the identity on MATH. Then the isotopy class of MATH, as an element of MATH, is the required conjugate, since it lies in the subgroup of MATH of diffeomorphisms supported on MATH, which is isomorphic to MATH and equals MATH. |
math/0107085 | We may choose a set of generators MATH for MATH with the properties listed in REF . If MATH has global fixed points other than the endpoints of MATH we wish to consider the restriction of the action to the closure of a component of the complement of these global fixed points. If all of these restricted actions are abelian then the original action was abelian. Hence it suffices to prove that these restrictions are abelian. Thus we may consider an action of MATH on a closed interval MATH with no interior global fixed points. None of the generators above can act trivially on MATH since the fact that they are all conjugate would mean they all act trivially. We first consider the case that one generator has a nontrivial interval of fixed points in MATH and show this leads to a contradiction. Since they are all conjugate, each of the generators has a nontrivial interval of fixed points. Choose an interval MATH of fixed points which is maximal among all intervals of fixed points for all of the MATH. That is, MATH is a nontrivial interval of fixed points for one of the MATH, which we assume (without loss of generality) is MATH, and there is no MATH which properly contains MATH and which is pointwise fixed by some MATH. Suppose MATH. At least one of MATH and MATH is not an endpoint of MATH since MATH is not the identity. Suppose it is MATH which is in MATH . Then MATH, so by REF we know that MATH is fixed by all of the generators except possibly MATH, since MATH is the only generator with which MATH does not commute. The point MATH cannot be fixed by MATH since otherwise it would be an interior point of MATH fixed by all the generators, but there are no global fixed points in MATH other than the endpoints. The identity MATH together with REF then tells us that MATH . Now assume without loss of generality that MATH (otherwise replace all generators by their inverses). Define MATH . Since MATH we have MATH and hence MATH . But MATH is a fixed point of MATH (as well as the other generators which commute with MATH) and MATH commutes with MATH . Hence MATH and MATH are fixed by MATH (since MATH preserves MATH). We can now apply NAME 's Lemma REF to conclude that MATH is the identity on MATH which contradicts the fact that MATH is a maximal interval of fixed points among all the MATH. We have thus contradicted the supposition that one of the generators has a nontrivial interval of fixed points in MATH, so we may assume that each of the generators, when restricted to MATH, has fixed point set with empty interior. That is, each generator is fully supported on MATH . We now note that given any two of the generators above, there is another generator MATH with which they commute. Thus we may conclude from REF that the action of MATH on MATH abelian. |
math/0107085 | Closed case. We first consider the group MATH, and choose generators MATH for MATH with the properties listed in REF . All of these elements are conjugate by REF , so they all have the same rotation number. We wish to consider a subgroup MATH of type MATH . We let MATH and for MATH we set MATH and MATH. For MATH we let MATH . Since MATH commutes with any element of MATH we have the relations CASE: MATH for all MATH CASE: MATH if MATH, CASE: MATH if MATH, CASE: MATH and MATH whenever MATH. We now define MATH to be the subgroup of MATH generated by MATH. The fact that MATH commutes with each of MATH and has the opposite rotation number implies that the rotation number of every element of MATH is MATH . Thus each of these elements has a fixed point. We note that given any two of these generators of MATH there is another generator with which they commute. If all of these generators have fixed point sets with empty interior then we may conclude from REF that any two of them have equal fixed point sets, that is, that MATH for all MATH . So in this case we have found a common fixed point for all generators of MATH . If we split MATH at this point we get an action of MATH on an interval MATH which must be abelian by REF . It follows that all of the original generators MATH commute with each other except possibly MATH may not commute with MATH and MATH, and MATH and MATH may not commute. Let MATH be the putative action, and consider the element MATH guaranteed by REF . We have that MATH and MATH and MATH . Hence MATH is abelian, hence trivial since (see CITE) abelian quotients of MATH are trivial. Thus we are left with the case that one generator of MATH has a nontrivial interval of fixed points in MATH . Since they are all conjugate, each of the generators of MATH has has a nontrivial interval of fixed points. Also we may assume no generator fixes every point of MATH since if one did the fact that they are all conjugate would imply they all act trivially. Choose a maximal interval of fixed points MATH for any of the subset of generators MATH . That is, MATH is a nontrivial interval of fixed points for one of the MATH, which we assume (without loss of generality) is MATH, and there is no MATH which properly contains MATH and which is pointwise fixed by some MATH. Suppose MATH and MATH are the endpoints of MATH. Then MATH so by REF , MATH and MATH are fixed points for all of the generators except MATH, since MATH is the only one with which MATH does not commute. If the point MATH we have found a common fixed point for all the generators of MATH and we can split MATH at this point obtaining an action of MATH on an interval which implies MATH is abelian by REF . So suppose MATH. Recall that the diffeomorphism MATH has rotation number MATH, so it fixes some point. Thus the identity MATH, together with the fact that MATH cannot fix MATH, implies by REF that MATH and MATH . Define MATH . We will denote by MATH the interval in MATH with endpoints MATH and MATH which contains MATH. Note that MATH properly contains MATH and that MATH . Since MATH and MATH commutes with MATH we conclude MATH and MATH are fixed by MATH. We can now apply NAME 's Lemma REF to the interval MATH with diffeomorphisms MATH and MATH to conclude that MATH is the identity on MATH. But MATH is a proper subinterval of MATH which contradicts the fact that MATH was a maximal interval of fixed points for any one of the MATH. Thus in this case too we have arrived at a contradiction. Punctured case. We now consider the group MATH, and choose elements MATH as above. The argument above shows that the subgroup MATH of MATH generated by these elements acts trivially on MATH. We claim that the normal closure of MATH in MATH is all of MATH, from which it follows that MATH acts trivially, finishing the proof. To prove the claim, let MATH denote the closed surface of genus MATH and recall the exact sequence (see, for example, CITE): MATH where MATH is generated by finitely many ``pushing the point" homeomorphisms MATH around generating loops MATH in MATH with basepoint the puncture. Each generating loop MATH has intersection number one with exactly one of the loops, say MATH, corresponding to one of the twist generators of MATH. Conjugating MATH by MATH gives a twist about a loop MATH which together with MATH bounds an annulus containing the puncture. The twist about MATH composed with a negative twist about MATH gives the isotopy class of MATH. In this way we see that the normal closure of MATH in MATH contains MATH together with each of the generators of the kernel of the above exact sequence, proving the claim. |
math/0107085 | Let MATH be the generators of MATH and define automorphisms MATH and MATH by MATH . Then MATH with MATH generate the index two subgroup of MATH given by those automorphisms which induce on the abelianization MATH of MATH an automorphism of determinant one (see, for example, CITE). A straightforward but tedious computation shows that relations REF - REF are satisified. |
math/0107085 | We fix MATH and show MATH has a fixed point. Since MATH there is a MATH with MATH . Let MATH denote the centralizer of MATH, so MATH . Also, since MATH there is MATH which is distinct from MATH so MATH. By REF we know that the rotation number MATH is a homomorphism so MATH . This implies that MATH has a fixed point. |
math/0107085 | Let MATH be the index two subgroup of MATH from REF . Let MATH be a homomorphism MATH to MATH or MATH and let MATH and MATH where MATH and MATH are the generators from REF . Then MATH are generators for MATH . We will show in fact that the subgroup MATH generated by MATH is trivial, as is the subgroup MATH generated by MATH . The arguments are identical so we consider only the set MATH . Since MATH, we have by REF that the commutavity graph for the generators MATH of MATH is connected. Hence by REF , if these generators are fully supported then MATH is abelian. But the relations MATH then imply that MATH is trivial. Hence we may assume at least one of these generators is not fully supported. Let MATH be an interval which is a maximal component of fixed point sets for any of the MATH. More precisely, we choose MATH so that there is MATH with MATH a component of MATH and so that there is no MATH such that MATH properly contains MATH. We wish to show that MATH is fixed pointwise by each MATH. In case MATH we note that at least the endpoints of MATH are fixed by MATH because MATH and MATH commute and the endpoints of MATH are in MATH . So REF or REF implies these endpoints are fixed by MATH . For the general case we first show MATH . In fact there is no MATH with the property that MATH because if there were then the interval MATH defined to be the smallest interval containing MATH is a MATH-invariant interval with no interior fixed points for MATH. (In case MATH we need the fact that MATH has a fixed point, which follows from REF , in order to know this interval exists.) But since MATH there is some MATH which commutes with both MATH and MATH, and it leaves both MATH and MATH invariant. Applying NAME 's Lemma REF to MATH and MATH we conclude that MATH, which contradicts the maximality of MATH. Hence we have shown that MATH for all MATH . Now by REF the relation MATH holds so we may apply REF to conclude that MATH is fixed pointwise by MATH . We next consider the generator MATH . If MATH is an endpoint of MATH then since MATH at least one of MATH and MATH must be in MATH. Hence the relations MATH and MATH imply that MATH . This holds for the other endpoint of MATH as well, so MATH is invariant under MATH for all MATH . The same argument shows MATH is invariant under MATH for all MATH . But since MATH is the identity on MATH we conclude from MATH that MATH is the identity on MATH. Similarly MATH is the identity on MATH. Next the relation MATH together with REF implies that MATH is the identity on MATH. A similar argument gives the same result for MATH. Finally, the relation MATH implies that MATH is the identity on MATH . Thus we have shown that any subgroup of MATH or MATH which is a homomorphic image of MATH, the index two subgroup of MATH has an interval of global fixed points. In the case of MATH we can split at a global fixed point to get a subgroup of MATH which is a homomorphic image of MATH. In the MATH case we can restrict the action to a subinterval on which the action has no global fixed point. But our result then says that for the restricted action there is an interval of global fixed points, which is a contradiction. We conclude that the subgroup of MATH generated by MATH is trivial and the same argument applies to the subgroup MATH generated by MATH . So MATH is trivial. Since MATH has order two any homomorphism MATH from MATH to MATH or MATH has an image whose order is at most two. The fact that MATH has no elements of finite order implies that in this case MATH is trivial. Since there is a natural homomorphism from MATH onto the group MATH, the statement of REF for MATH holds. |
math/0107085 | We have MATH and MATH . So MATH . |
math/0107085 | If we put no restrictions on the integers MATH and, in particular allow them to be MATH, then it is trivially true that any element can be put in the form above. If for any MATH and MATH then the fact that MATH allows us to substitute and obtain another expression in the form above, representing the same element of MATH, but with fewer occurrences of the terms MATH and MATH. Similarly if MATH and MATH then MATH can be replaced with MATH further reducing the occurrences of the terms MATH and MATH. These substitutions can be repeated at most a finite number of times after which we have the desired form. |
math/0107085 | By construction MATH is surjective; we now prove injectivity. To this end, consider an arbitrary nontrivial element written in the form of REF . After conjugating we may assume that MATH (replacing MATH by MATH). So we must show that (MATH applied to) the element MATH acts nontrivially on MATH . If MATH, that is, if MATH, then MATH clearly acts nontrivially if MATH, so we may assume MATH . Let MATH be an element of the interior of MATH . We will prove by induction on MATH that MATH . This clearly implies MATH . Let MATH . Then assuming MATH, we have MATH . But MATH implies MATH . So REF implies MATH, and then REF implies MATH . Thus MATH implies MATH . One shows similarly if MATH then MATH . Now as induction hypothesis assume that MATH and that either MATH . We wish to establish the induction hypothesis for MATH . There are four cases corresponding to the values of MATH for each of MATH and MATH . If MATH and MATH then MATH so MATH and REF implies that MATH. Hence by REF , MATH as desired. On the other hand if MATH and MATH then MATH. Hence MATH implies MATH with MATH. Consequently MATH by REF . So REF implies MATH . Thus we have verified the induction hypothesis for MATH in the case MATH . If MATH and MATH then MATH so MATH and REF implies that MATH. Hence by REF MATH as desired. Finally for the case MATH and MATH we have MATH. Hence MATH implies MATH with MATH. Consequently MATH by REF . So REF implies MATH . Thus we have verified the induction hypothesis for MATH in the case MATH also. |
math/0107085 | Consider the group of all lifts to MATH of elements of MATH acting on MATH. Choose lifts MATH and MATH of MATH and MATH respectively and let MATH denote the covering translation. We have MATH for some integer MATH . Each lift of an element of MATH has a well defined translation number in MATH and these are topological conjugacy invariants. Hence MATH . Solving we conclude MATH . Since MATH has a rational rotation number MATH has a point whose period is a divisor of MATH . |
math/0107085 | Assume, without loss of generality, that MATH and MATH and MATH . Then MATH . But MATH . For MATH sufficiently large this implies MATH and for MATH sufficiently negative that MATH . The intermediate value theorem implies MATH has a fixed point. |
math/0107085 | If MATH then MATH so MATH is a periodic point for MATH . But the only periodic points of MATH are fixed. We conclude that MATH . Let MATH be any component of the complement of MATH in MATH . Since MATH we have that MATH . Let MATH be the centralizer of MATH. Note that MATH and MATH are both in MATH . Let MATH be any element of MATH. Then if MATH has a fixed point in MATH, since it commutes with MATH we know by NAME 's Lemma REF that MATH . In other words MATH acts freely (though perhaps not effectively) on MATH . Hence the restrictions of any elements of MATH to MATH commute by NAME 's Theorem. In particular the restrictions of MATH and MATH to MATH commute. But MATH was an arbitrary component of the complement of MATH. Since the restrictions of MATH and MATH to MATH are both the identity we conclude that MATH and MATH commute on all of MATH. |
math/0107085 | If there is such a subgroup, it has generators MATH and MATH satisfying MATH. REF asserts that MATH and MATH commute. But the commutator MATH is MATH and this element is nontrivial by REF . This contradicts the existence of such a subgroup. |
math/0107085 | Suppose there is such a subgroup, and we have generators MATH satisfying MATH. By REF , MATH has a periodic point. Let MATH be the closed set of periodic points of MATH (all of which have the same period, say MATH). If MATH then MATH which shows MATH . The omega limit set MATH of a point MATH under MATH is equal to a subset of the set of periodic points of MATH if MATH has periodic points, and is independent of MATH if MATH has no periodic points. Since MATH is MATH we know by NAME 's theorem that MATH is either all of MATH or is a subset of MATH (see CITE). In the case at hand MATH cannot be all of MATH since MATH implies MATH and this would imply MATH . Hence there exists MATH . Then MATH where MATH is the period of points in MATH . The fact that MATH has period MATH for MATH implies that its rotation number under MATH well defined as an element of MATH is MATH for some MATH which is relatively prime to MATH. Using the fact that MATH, we conclude that MATH . Consequently MATH and since MATH and MATH are relatively prime MATH . The fact that MATH and MATH for some MATH tells us that MATH because if it were then MATH would be a multiple of MATH. Similarly MATH . Let MATH be the group generated by MATH and MATH. Now MATH has a global fixed point MATH, so we can split MATH at MATH to obtain a MATH action of MATH on MATH. Note that MATH. We can apply REF to conclude that MATH and MATH commute. Thus MATH and MATH commute. But their commutator MATH . Since MATH and MATH this element is nontrivial by REF . This contradicts the existence of such a subgroup. |
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