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math/0102035 | We set MATH . Then MATH . If MATH, we have MATH, and so MATH implying MATH, a contradiction. This proves REF. Now, since MATH and MATH and MATH, REF implies MATH that is, REF holds. |
math/0102035 | Setting MATH, we have MATH. Then REF implies MATH . So by REF we have MATH . On the other hand, we have MATH. Again REF implies MATH since MATH. By REF , we have MATH . By REF, we get the assertion. |
math/0102035 | We argue by induction. If MATH, the result follows from REF with MATH. Now suppose the result always holds for MATH. Set MATH . Then, by REF , MATH . On the other hand, we have MATH, so by the inductive assumption, MATH . By REF, we get the assertion. |
math/0102035 | Let MATH be a developing map of MATH with the monodromy representation MATH on MATH. MATH can be lifted to a MATH representation MATH so that the two following properties hold: CASE: Let MATH REF and MATH REF be deck transformations on MATH corresponding to loops about MATH and MATH, respectively. Then it holds that MATH . CASE: The eigenvalues of the matrix MATH (respectively, MATH) are MATH (respectively, MATH). This is proven in CITE for MATH, and the same argument will work for general MATH and MATH. We include an outline of the argument here: one chooses a solution MATH to REF (with MATH and MATH), then MATH has a monodromy representation MATH, where MATH about loops MATH. Then MATH, and we simply choose the lift MATH so that MATH. The first property is then clear. To show the second property, we note that when MATH and MATH are all given the value MATH, then MATH is identically MATH and so MATH is constant and all MATH. Hence the eigenvalues MATH (respectively, MATH) of MATH (respectively, MATH) are MATH (respectively, MATH) in this case. Then, as MATH and MATH are deformed back to their original values, the matrices MATH (respectively, MATH) change analytically and so the sign of the eigenvalues cannot change, showing the second property. We have MATH and since MATH is an integer, we have MATH . Assume MATH is an odd number. Then, by the assumption, MATH is an even integer, and by REF above we have MATH, so by REF , MATH implying the corollary when MATH is odd. Now suppose that MATH is even. We have MATH, because MATH is odd. Hence MATH, and since MATH, REF implies MATH implying the corollary when MATH is even. |
math/0102035 | Suppose that MATH. Then by REF, MATH proving the theorem when MATH. Next, suppose that MATH. In this case, MATH. Hence again by REF, we have REF, and the theorem follows. Thus we may assume MATH and MATH at all ends. Then by REF we have all MATH. So, by REF, the corresponding pseudometric MATH has divisor MATH where MATH at each umbilic point MATH (MATH). Then by REF , MATH and so REF implies the theorem. |
math/0102035 | We may set MATH. Since the MATH, the developing map MATH of MATH is well-defined on MATH, and so MATH is meromorphic on MATH. As MATH has finite total curvature, MATH extends to MATH as a holomorphic mapping. In particular, MATH. |
math/0102035 | Since MATH has only two non-integral conical singularities, it is reducible, and REF shows that the map MATH is written in the form MATH where MATH and MATH are relatively prime polynomials with MATH and MATH. Note that MATH can have a multiple root only at a conical singularity of MATH, hence only at MATH. Thus MATH for all roots MATH of MATH. Since the change MATH preserves MATH, we may assume that MATH. By a direct calculation, we have MATH . Note that MATH. Let MATH. If MATH, then MATH, and since MATH is not a singularity of MATH, we have MATH, and hence MATH. If MATH, then MATH and MATH, so MATH. Hence the only root of the polynomial MATH is MATH: MATH . We claim that MATH. If MATH, then MATH (or MATH) having order MATH at MATH means that MATH, by REF and the above form of MATH. Suppose MATH. Then we have MATH, where MATH is a polynomial in MATH with MATH and MATH. Furthermore, MATH, where MATH is a polynomial with MATH, since MATH. So MATH. Then by REF, we have MATH. Suppose that MATH. Since MATH, the top term of MATH must vanish. Thus we have MATH, contradicting that MATH. So MATH is constant. Similarly, if MATH, then MATH. Hence MATH, and MATH is as in REF. MATH and MATH follow from REF. |
math/0102035 | MATH is reducible only if the image of the representation MATH is simultaneously diagonalizable, so we may choose a developing map MATH such that MATH . Thus we have MATH . Differentiating this gives MATH which implies that MATH is single-valued on MATH. On the other hand, by REF, there is a complex coordinate MATH around each end MATH such that MATH for some MATH (MATH). Let MATH be the deck transformation of MATH corresponding to a loop surrounding MATH. Then MATH . Hence MATH when MATH and MATH. By REF, MATH in REF is diagonal, so MATH . Hence MATH has poles of order MATH at MATH, and thus MATH where MATH is meromorphic. Integrating this gives the assertion. |
math/0102037 | Suppose that MATH. It implies that the NAME expansion of MATH is given by MATH because the residue of MATH must be real. Moreover, it follows from REF that MATH . Therefore we have MATH . Hence we can choose an orthonormal basis MATH of MATH so that MATH for some real constants MATH, MATH. With respect to this basis, we have MATH . Then using the polar coordinate MATH, we have MATH where MATH is a base point. Here, we have dropped the constant terms in MATH by a suitable parallel translation. By REF implies that the surface MATH is asymptotic to the catenoid (respectively, the plane) for the sufficiently small MATH if MATH (respectively, if MATH). Conversely, suppose that MATH. It implies that MATH REF and MATH . It is obvious that the end is asymptotic to neither a catenoid-type end nor a planar end. From now on, we shall prove that an end is embedded if it is asymptotic to a catenoid-type end or a planar end. Assume that the end is not embedded. Then there exist two sequences MATH, MATH convergent to MATH such that MATH for all MATH. Then by REF, there exists a positive constant MATH such that MATH where MATH and MATH (MATH). With these estimates, we have MATH and then, MATH holds. However the left hand side of REF diverges to MATH as MATH. This is a contradiction. |
math/0102040 | Let MATH, and suppose for a given MATH that there is a MATH which satisfies REF and such that REF is satisfied. Given that MATH, and given that MATH, there is a nonsingular MATH such that MATH. Hence, MATH. Upon showing that MATH is nonsingular, REF will then follow from REF. If MATH is singular, then there are nonzero vectors MATH such that MATH, and such that MATH. Let MATH, MATH denote solutions of REF with MATH, MATH. Then, MATH . Using REF, and recalling that MATH is defined in REF, we obtain MATH . Thus, by REF , MATH. This contradicts the assumption that MATH. Conversely, if MATH for a given MATH, then for MATH let MATH. One observes that REF is satisfied and that MATH. Moreover, MATH. If for this choice of MATH, REF is not yet satisfied, one introduces MATH and observes that MATH, MATH, and that MATH satisfies all requirements of REF. |
math/0102040 | Let MATH, MATH denote solutions of REF with MATH, MATH. Then MATH as in REF. This implies MATH with MATH defined in REF. Moreover, by the definition of MATH given in REF, one obtains MATH . By REF , one infers that MATH. To prove REF, let MATH, where MATH is defined in REF. Then, by REF, MATH which implies MATH and hence MATH . Thus one concludes MATH from which REF follows immediately by REF . |
math/0102040 | Let MATH and MATH be defined in REF with MATH and MATH respectively. Then, MATH . By the rank REF , MATH for nonsingular MATH. Thus, by REF, and by the uniqueness of solution of REF, there is a nonsingular MATH for which MATH . By REF, MATH and hence, by REF we see that MATH from which REF immediately follows. |
math/0102040 | Let MATH, and for a given MATH suppose that there is a MATH satisfying REF such that REF holds. The matrices MATH, MATH, are invertible by REF, and by REF it follows that MATH . By REF, one then concludes that MATH and hence that MATH whenever REF holds. Upon showing that MATH is nonsingular, REF will follow from REF. If MATH is singular, then there is a nonzero vector MATH such that MATH. By the nonsingularity of MATH, MATH, MATH, and as a result, REF yields MATH and hence, a contradiction given REF (compare REF). Conversely, if MATH for a given MATH, then for MATH, MATH, MATH, defined by REF, are nonsingular. Indeed, if either MATH or MATH are singular, then there is a MATH, MATH, such that MATH, a contradiction. Next, let MATH and let MATH. Then, for these MATH, MATH, REF holds. REF now implies that MATH for MATH and MATH. For this choice, MATH does not satisfy REF. However, one can normalize MATH as described in the proof of REF . |
math/0102040 | Let MATH and MATH denote bases for MATH and MATH, respectively. By REF, one has MATH for MATH. We now assume that MATH. One observes that MATH and MATH are linearly independent in MATH, where MATH. Consequently, there is some MATH and some MATH such that MATH . By NAME 's identity, MATH is constant with respect to MATH. On the other hand, an application of NAME 's inequality shows that the left-hand side of REF is in MATH. By REF one obtains a contradiction and hence concludes that MATH . The analogous argument then also yields MATH and hence the limit point property of NAME systems with MATH in REF. |
math/0102040 | Let MATH, and let MATH, MATH with MATH. By REF, MATH . By REF, this yields MATH . The integral expression in REF is strictly positive by REF . This yields the equivalence of MATH, and hence of MATH, with the condition given in REF. The equivalence of REF follows from the observation that MATH . The analogous characterization of MATH now follows from REF . |
math/0102040 | Let MATH and suppose that MATH for MATH and MATH, MATH. Then, MATH . By REF, an immediate contradiction results if MATH. However, if MATH, then either MATH or MATH, MATH. In either case, a contradiction results since MATH by REF and MATH satisfies the first-order system REF. Hence, MATH is well-defined on MATH. For MATH and MATH, REF implies that MATH . This is equivalent to MATH on MATH. Given the nonsingularity of MATH on MATH, REF implies REF, with nonnegativity holding at MATH. Next, let MATH, and suppose that MATH, defined by REF, is well-defined on MATH, and satisfies REF. Then, on MATH, REF and consequently REF follow, which implies that REF holds, and hence that MATH. The analogous characterization of MATH follows from REF . |
math/0102040 | In this proof, we consider only the case corresponding to MATH, that is, MATH in REF. The other case follows in a similar manner. For this reason, we let MATH in the remaining discussion. We also let MATH be the greatest value such that REF holds for MATH and show that REF must hold for some MATH with MATH, thus proving MATH . The solution of REF, MATH, presumed to be defined on MATH, can be continued onto some MATH with MATH; MATH then satisfies MATH for MATH and for some MATH. For brevity, let MATH denote REF with MATH, and let MATH denote REF in the following. Integrating REF, one obtains MATH . We note that MATH where, MATH . Thus, for MATH, REF implies that as MATH and together with REF that MATH . (Here MATH denotes a norm on MATH.) Hence, by REF - REF, one infers that for MATH and as MATH, MATH . Next, one notes that MATH where MATH . By REF, for MATH, MATH and hence by REF - REF, MATH . Of course, by REF as MATH, MATH . Thus, by REF, one concludes for MATH and as MATH, that MATH . NAME 's inequality applied to REF together with a consideration of the effect of the variable change REF, as illustrated in REF, yields MATH uniformly for MATH. Thus by REF, the contradiction results that for all MATH, MATH satisfies REF. |
math/0102040 | We note that REF is equivalent to REF with MATH. By the variable change REF is also equivalent to REF with MATH. Next, we recall the connection between the NAME REF , and REF by means of the NAME transformation REF. Solution matrices of REF which statisfy REF at MATH thus correspond to solution matrices, MATH, of REF for which MATH. Moreover, solutions of REF for which for which MATH are obtainable from solutions of REF, with MATH, by means of REF with MATH. Thus, by utilizing this connection between explicit exponential solutions of REF with MATH and solutions of the NAME REF , and by performing on the resulting solution of REF the conformal mapping REF followed by the variable transformation REF, one obtains the following solution for REF, MATH for MATH, MATH, and MATH. By hypothesis, MATH. Thus the exponential term in REF will result in MATH unless MATH thus implying REF. |
math/0102040 | We note that the system REF is independent of the reference point MATH. Next, we recall that MATH, defined in REF is determined by an apriori choice of the sequence MATH, subject only to MATH being in MATH (compare REF). Moreover, we note that MATH, defined as a limit in REF, described explicity in REF , and which gives solutions of REF satisfying REF for MATH, is also independent of the reference point MATH. Thus, had we chosen a different point of reference, MATH, at the start, the asymptotic analysis begun in REF and continued through REF, would remain the same after the variable change in REF, except for the integral expression in REF in which MATH would be replaced by MATH. However, given the local integrability assumption on MATH in REF , one concludes that this integral expression is uniformly continuous with respect to MATH whenever MATH is confined to a compact subset of MATH. Thus REF, and consequently REF, are uniform with respect to MATH and with respect to MATH whenever both are confined to compact subsets of MATH. Consequently, REF holds for elements MATH, that this asymptotic expansion is uniform with respect to MATH for MATH in MATH, and that it is uniform with respect to MATH when MATH is confined to compact subsets of MATH. |
math/0102040 | In the following let MATH, and MATH. The existence of an expansion of the type REF is shown as follows. First one considers a matrix NAME integral equation of the type MATH where MATH and MATH abbreviates the MATH . NAME 's kernel MATH . Clearly, MATH solves the NAME system REF. In addition, it satisfies MATH. Thus, up to normalization, MATH represents the NAME solution associated with MATH on the half-line MATH. Next, introducing MATH one rewrites REF in the form MATH where MATH . Thus, one infers, MATH . Introducing MATH where MATH REF results in MATH . This yields MATH for some MATH, MATH, depending on MATH. Integrating by parts in REF, repeatedly applying REF to MATH for all MATH then results in the existence of an asymptotic expansion for MATH of the type MATH . Inserting the expansions for MATH and MATH into REF (using a geometric series expansion for MATH) then yields the existence of an expansion of the type REF for MATH. The actual expansion coefficients and the associated recursion relation REF then follow upon inserting expansion REF into the NAME REF . The stated uniformity assertions concerning the asymptotic expansion REF then follow from iterating the system of NAME integral integral REF . |
math/0102040 | This is obvious from REF. |
math/0102040 | Define MATH, MATH, by MATH then MATH due to the uniform nature of the asymptotic expansion REF for MATH varying in compact intervals. Next, introduce MATH then MATH . Using MATH a standard NAME iteration argument in REF then yields MATH and hence REF. |
math/0102040 | Define for MATH, MATH, MATH and for MATH and a.e. MATH, MATH . By REF , MATH and hence by REF , MATH where MATH and MATH are fundamental solution matrices of MATH respectively, with MATH . By REF , MATH as MATH, MATH. Thus, as MATH, MATH, MATH for some constant MATH by REF, and REF. |
math/0102040 | Define MATH and apply REF with MATH, MATH. Then (in obvious notation) MATH as MATH, MATH, and hence the asymptotic expansion REF for MATH in REF coincides with that of MATH. |
math/0102040 | To see that uniformity holds for this expansion, first recall the role of REF in providing uniformity in the asymptotic REF which then leads to REF holding uniformly with respect to MATH varying within compact subsets of MATH and with respect to MATH for MATH in MATH. This in turn leads to a similar uniformity holding for REF which is the key to REF holding with respect to MATH varying within compact subsets of MATH and with respect to MATH for MATH in MATH. |
math/0102040 | Since MATH and MATH are fixed throughout this proof, we will temporarily suppress these variables whenever possible to simplify notations. Introducing MATH the NAME integral equation for MATH (compare REF), MATH can be rewritten in terms of that of MATH in the form MATH where we abbreviated MATH . Using the elementary algebraic facts MATH for any MATH, iterating REF yields MATH . Applying the NAME lemma to REF then proves REF assuming MATH, only. Assuming also REF one can compute the next term in the asymptotic expansion REF and then obtains REF using REF and the finite-interval variant of REF, whenever MATH is a right NAME point of MATH and MATH is a left NAME point of MATH. Exactly the same arguments apply to MATH. Introducing MATH the NAME integral equation for MATH, MATH can be rewritten in terms of that of MATH in the form MATH . Iterating REF, taking into account REF, yields MATH . Applying the NAME lemma to REF the proves REF assuming MATH, only. Assuming also REF one can compute the next term in the asymptotic expansion REF and then obtains REF using REF and the finite-interval variant of REF, whenever MATH is a right NAME point of MATH and MATH is a left NAME point of MATH. Finally, we turn to MATH. Introducing MATH the NAME integral equation for MATH, MATH can be rewritten in terms of that of MATH in the form MATH . Iterating REF, taking into account REF, yields MATH . Next, we take into account the different normalizations of MATH and MATH. Using MATH (compare , REF and MATH), one readily verifies the relationship MATH . Thus, applying the NAME lemma to REF then proves REF (in agreement with REF), assuming MATH, only. Assuming also REF one can compute the next term in the asymptotic expansion REF and then obtains REF using REF, whenever MATH is a right NAME point of MATH. |
math/0102040 | Using REF one estimates MATH since by REF MATH . Moreover, since MATH by REF, one infers REF from REF and MATH as MATH, MATH, MATH (compare REF). |
math/0102040 | Since REF follows from REF , it suffices to focus on the proof of REF. Moreover, applying REF , we may without loss of generality assume for the rest of the proof that MATH . In the following, we will adapt the principal ingredients of a recent proof of the local NAME uniqueness theorem for scalar NAME operators (that is, for MATH) by CITE, to the current NAME situation. First we recall that by REF holds along the rays MATH, MATH for all MATH satisfying REF. To simplify notations in the following we will again suppress MATH and MATH whenever possible and hence abbreviate, MATH, MATH, and MATH by MATH, MATH, and MATH, respectively. Next, denoting in obvious notation by MATH the solutions associated with MATH, MATH, which are defined in REF, we introduce MATH . Using the asymptotic expansions REF - REF for MATH, MATH, and MATH, and the analogous ones for MATH, MATH, and MATH, one verifies for each fixed MATH, MATH assuming MATH for all MATH, MATH, only. Next, using the fact that for each fixed MATH, MATH by REF, one concludes MATH using REF, and MATH. Combining REF , one infers MATH along the rays MATH, MATH. Thus, REF implies MATH along the rays MATH, MATH. The analogous estimate REF holds along the complex conjugate rays MATH, MATH, in the lower complex half-plane MATH. To simplify notations we denote the open sector generated by MATH and its complex conjugate MATH by MATH, the open sector generated by the MATH and its complex conjugate MATH by MATH, the remaining sector in MATH is denoted by MATH, and its complex conjugate sector in MATH is denoted by MATH. Thus, one obtains a partition of MATH into MATH where each sector MATH, MATH, has opening angle strictly less than MATH. Since (each matrix element of) the expression under the norm in REF is entire and of order less or equal to one, one can apply the NAME principle (compare , for example, CITE) to each sector MATH, MATH, and obtains that each matrix element under the norm in REF is uniformly bounded in each sector and hence on all of MATH. By NAME 's theorem, these matrix elements are all equal to certain constants. By the right-hand side of REF, these constants all vanish. Thus, we proved MATH and hence MATH . Differentiating MATH, MATH, with respect to MATH yields MATH . Multiplying REF by MATH and using REF, and similarly, multiplying REF by MATH and using REF then yields MATH . In exactly the same way one derives MATH using REF - REF. Thus, REF implies MATH for a.e. MATH. Thus far we only used MATH for all MATH, MATH and REF. In the special case MATH, each of REF - REF allows for the completion of the proof of REF. Indeed, using the fact that MATH and taking for instance REF, one infers for a.e. MATH, that MATH . Since all zeros (and poles) of the left-hand side of REF have even multiplicity, while all zeros (and poles) of the right-hand side of REF are simple, one concludes, assuming only that MATH for all MATH, MATH, that MATH . Thus for the case MATH, we see by REF, and REF, and REF, for a.e. MATH, that MATH . Now, REF, and REF show, for a.e. MATH, that MATH . By REF we see that MATH . Thus, by REF, and REF, MATH . Together, REF imply REF in the special case MATH. Unfortunately, the case MATH appears to be quite a bit more involved. To deal with this case we first note that taking determinants in REF yields MATH for a.e. MATH. Next, we intend to prove that MATH . Given the fact that MATH, MATH, is self-adjoint, showing REF is equivalent to showing that MATH and MATH are unitarily equivalent for a.e. MATH. Arguing by contradiction, we assume that at least one pair of eigenvalues of MATH and MATH differs. Thus, fixing MATH, let MATH be an eigenvalue of MATH but not of MATH. Then REF implies, for all MATH satisfying REF, that MATH . Next, for MATH and MATH define MATH . Then, MATH is strictly positive definite, MATH . Indeed, suppose MATH for some MATH, then MATH implies MATH and hence MATH . Thus, MATH by REF, and hence MATH proves REF. Introducing MATH and MATH defined by MATH one verifies MATH (by REF) and MATH (by REF). Thus, MATH satisfies REF. Next, since MATH as a special case of REF, one derives MATH using REF. This contradiction to REF proves REF. Hence for MATH and for a.e. MATH by REF implies that for a.e. MATH, the family of NAME operators MATH in MATH, defined by MATH with MATH, have identical spectra for all boundary data MATH satisfying REF. Hence, assuming MATH, MATH, one can apply REF and obtains REF. |
math/0102040 | REF is proved by combining REF, and REF, and REF then follows by combining REF, and REF, taking into account the asymptotic expansions MATH along any ray with MATH in the case of NAME operators (compare REF). |
math/0102040 | By REF, MATH . Next, suppose that MATH is a left and right NAME point of MATH. By REF one obtains MATH and hence MATH . Since a.e. MATH is a NAME point of MATH, one concludes REF. |
math/0102040 | We start with the unitary transformation MATH in MATH defined by MATH which maps MATH to MATH, where MATH . Next, we introduce the unitary operator MATH in MATH defined by MATH where the unitary MATH matrices MATH are solutions of the first-order system REF. Since by REF, MATH, the solutions of REF are well-defined and MATH, MATH. One computes MATH where MATH and MATH . Finally, defining MATH, one arrives at REF - REF. |
math/0102040 | Define MATH . Using the notation employed in the proof of REF one verifies that MATH . |
math/0102040 | The fact that MATH for all MATH, that a.e. MATH is a NAME point of MATH, and the trace REF , imply REF. Together with REF this yields REF. |
math/0102040 | This is clear from REF, which imply MATH . |
math/0102052 | Let MATH, then the finiteness of the isotropy group of MATH in MATH is equivalent to: MATH. As there are only finitely many isotropy groups for a torus action on an algebraic variety, the finiteness of all isotropy groups for the MATH-action on MATH is equivalent to: MATH. Since MATH this amounts to: MATH for all MATH. Now MATH has a MATH-invariant rational structure, defined by the lattice of differentials at MATH of one-parameter subgroups of MATH; the rational subspaces are exactly the NAME algebras of subtori. Moreover, any rational subspace MATH intersecting trivially all subspaces MATH is contained in a rational complement to all these subspaces. This proves equivalence of REF , and the assertion on existence of subtori MATH and their dimension. For any such subtorus MATH, all orbits in the affine variety MATH are closed, and the isotropy groups are finite abelian groups. This implies the latter assertion. |
math/0102052 | We use equivariant cohomology, see for example, CITE. Consider the action of MATH on MATH, then the equivariant cohomology ring MATH is clearly isomorphic to MATH. Since MATH is a free module of rank MATH over MATH, the NAME spectral sequence (see CITE III. REF) yields an isomorphism MATH that is, MATH . This is a commutative, positively graded algebra, finite and free of rank MATH over its subring MATH. The latter is a NAME ring of dimension MATH. Thus, the ring MATH is NAME of dimension MATH as well, with NAME series MATH . Since the subtorus MATH of MATH acts on MATH with finite isotropy groups, we have MATH . This is a finitely generated module over MATH. But MATH is a torus of dimension MATH, so that MATH is a polynomial ring in MATH variables. Since MATH is NAME of dimension MATH and finite over MATH, it is a free module over that ring, by the NAME formula (see CITE). By the NAME spectral sequence again, it follows that the canonical map MATH is an isomorphism. Therefore, we have MATH . But MATH; thus, we obtain MATH. Moreover, MATH is the quotient of MATH by a regular sequence consisting of MATH homogeneous elements of degree MATH. Therefore, the usual NAME polynomial of MATH equals MATH . It remains to compare cohomology of MATH with its NAME group. For this, we use equivariant intersection theory, see CITE and also CITE. The equivariant NAME group with complex coefficients (graded by codimension) MATH is again isomorphic to MATH, by REF . Moreover, for any scheme MATH with an action of MATH, the natural map MATH is an isomorphism (to see this, one reduces to the case where the quotient MATH exists and is a principal MATH-bundle, and one argues as in CITE, p. REF). As a consequence, the map MATH is an isomorphism; it follows that the cycle map MATH defined in CITE, is an isomorphism as well. Finally, MATH by REF . |
math/0102052 | The former assertion is a (well-known) consequence of NAME 's formula for the invariant ring MATH: MATH . For the latter assertion, recall that MATH is a graded polynomial ring with homogeneous generators of degrees MATH, where MATH. Thus, the degree of MATH is MATH. Moreover, denoting MATH the finite group MATH, we have an exact sequence MATH . Thus, MATH acts on MATH with invariant subring MATH. Since MATH is a graded polynomial ring, it contains a graded MATH-stable subspace MATH such that the map MATH is an isomorphism. It follows that MATH decomposes as a direct sum of homogeneous components MATH ; the increasing sequence of their degrees (with multiplicities given by the dimensions of the MATH) is the same as MATH. Now MATH is a sum of rational functions of the same degree, equal to MATH. |
math/0102052 | In the case where MATH is a unique orbit, we have MATH, whence MATH. So the assertion follows from REF . In the general case, choose a closed orbit MATH in MATH, of codimension MATH, with complement MATH. The inclusion map MATH defines a NAME morphism MATH of degree MATH. By CITE, this map and the restriction map MATH fit into a short exact sequence MATH . It follows that MATH . Since MATH, our assertion follows by induction. |
math/0102052 | We argue by induction on MATH. If MATH, then MATH so that both MATH and MATH are trivial. In the general case, we use the notation of the proof of REF . The (surjective) restriction map MATH sends MATH (respectively, MATH) onto MATH (respectively, MATH). Let MATH. Since MATH by the induction assumption, we may assume that MATH for some MATH. Now we have in MATH: MATH by the projection formula. Moreover, MATH is the equivariant cohomology class of MATH in MATH. Since MATH is a transversal intersection of MATH boundary divisors, say MATH, we have MATH, and MATH as well. |
math/0102052 | Since the algebra MATH is positively graded, it suffices to prove that the quotient MATH is a finitely generated MATH-module. By REF , MATH is a quotient of MATH for some positive integer MATH. Consider the finite filtration of MATH by the powers of the image of MATH, and notice that all the subquotients MATH are finite modules over MATH. Since the latter is isomorphic to MATH, the assertion follows. |
math/0102052 | The argument is similar to that of the classical NAME normalization theorem, see CITE; we present it for completeness. We argue by induction on MATH, the case where MATH being trivial. In the general case, we may assume that MATH are algebraically dependent, and we choose a polynomial relation MATH . We may assume that this relation is homogeneous and involves MATH. Let MATH be the degrees of MATH. Define MATH by MATH where MATH are in MATH. Then MATH . Regarding the right-hand side as a polynomial in MATH, the coefficient of the leading term equals MATH. Since MATH is infinite and by our assumptions on MATH, we may choose MATH so that this coefficient is non-zero. Then MATH is integral over the subring MATH of MATH generated by MATH and MATH. We conclude by the induction assumption for MATH. |
math/0102055 | (Here MATH denotes linear equivalence.) In CITE, it is shown that MATH (there called MATH) is the intersection of MATH with a quartic. Therefore MATH. Applying the same result at other points in the sequence REF gives MATH for all MATH. Since MATH and MATH are NAME, this equation then implies that MATH is NAME for all MATH. MATH and MATH generate the nef cone of MATH; again, we can translate this along the sequence REF to give the last statement of the theorem. |
math/0102055 | In CITE it is proved that MATH. The multiplication table above shows that MATH, MATH, MATH, and MATH are linearly independent, so they form a basis. All are fixed by MATH, so MATH acts trivially. The intersections of MATH with these basis elements do not depend on the singular fibre used, so the numerical class of MATH does not. |
math/0102055 | Any such automorphism fixes MATH, and so comes from a projective automorphism of MATH. These are exactly MATH, and MATH does not act trivially. |
math/0102055 | CASE: The set of vertices given is the orbit of MATH under the group generated by MATH and MATH. We described one of the faces (that given by MATH) above; taking this information around by the action of the group gives the whole tiling. This tells us everything about the cone within the convex hull of the tiling, and on the other hand the boundary of the tiling is just the ray generated by MATH, which certainly is nef, so MATH is as claimed. REF follows from the description of the cone in REF - the translations MATH are enough to give transitivity of the action on faces and on vertices. CASE: Any automorphism MATH of MATH must preserve the pencil of Abelian surfaces, since it must, from the shape of the cone, preserve the class MATH. It must preserve the individual surfaces in that pencil, since the general member of the pencil is not isomorphic to any other member. Since the general surface in the pencil has no automorphisms other than those all Abelian surfaces have - translation and reflection about a point - MATH must act in one of these two ways. If the former, it commutes with other translations, in particular MATH and MATH, and so acts on the tiling as a translation. If the latter, MATH is a translation. This shows that MATH-automorphisms acting trivially on MATH is generated by translations and MATH. But by REF , the automorphisms that act trivially on MATH are exactly MATH. |
math/0102055 | The cubic form must be preserved by all automorphisms, and MATH and MATH generate the ring of polynomial forms preserved by the automorphism group, so MATH must be a linear combination of MATH and MATH. Suppose MATH. Then we can work out the values of MATH and MATH on MATH and MATH, and MATH give MATH, MATH. |
math/0102055 | We are now considering the lower triangular faces of the cuboctohedron of REF . Taking any one of the edges of the flopping face, the description of the movable fan inside the cuboctohedron centred on that edge tells us that half the boundary of the pyramid coincides with part of the boundary of MATH; these four descriptions put together give the whole of the boundary of the pyramid, so the nef cone must be the pyramid described. The automorphism group must fix the only square face of the pyramid, so MATH consists exactly of those elements of MATH that fix the flopping face. |
math/0102055 | We are now looking at the left-hand-side square face of the cuboctohedron in REF . This immediately gives five vertices: MATH, MATH, MATH, MATH, and MATH. The cone also fits into a side square face of the equivalent cuboctohedron centred at MATH, since MATH is also a resolution of MATH; there is a symmetry that interchanges MATH and MATH, which must also swap MATH and MATH, and this determines its effect completely; it takes MATH to MATH. Now, looking from the point of view of the vertex MATH, the pyramidal cone of MATH, as discussed above, fits into the top square face of the cuboctohedron, so MATH, one flop away, fits into an upper triangular face. Similarly, the pyramid that sits in the top face of MATH's cuboctohedron is in a lower triangular face from the point of view of MATH, and MATH, two flops away, must again be in an upper triangular face. Symmetry shows that the same holds for MATH and MATH. This means that we must have a symmetry of the given slice of MATH any of these to any other, which forces the cone to be the shape described. |
math/0102055 | REF , and REF construct the three types of cone, and the construction describes what is on the other side of any face of any of these cones - there are square faces where tilings meet pyramids, triangular faces where pyramids meet lozengoids and quadrilateral faces where lozengoids meet other lozengoids, twisted by a quarter-turn. (This is hard to visualize, since these faces are not square on the picture REF , but of course these are affine pictures and we have chosen different affine slice for the different models.) Every minimal model must be a finite sequence of flops away from MATH, so we have described all minimal models. The nef cone conjecture is clear, and a fundamental domain is easily seen from the descriptions in the three cases. For the last statement, the two models of tiling type are the resolutions of MATH and MATH, those of pyramid type are one flop from each of these (all such flops being isomorphic because MATH acts transitively on the faces of MATH); lozengoid-type models come from flopping two adjacent faces of MATH, which can be chosen in two ways because of the lack of automorphisms that act as order-MATH rotations on the nef cone of MATH. |
math/0102055 | Recall from REF that MATH is the group of birational automorphisms that act trivially on MATH: therefore MATH is the group of all symmetries of the movable fan arising from birational automorphisms. The construction above identifies the group of all symmetries of the fan with a subgroup of finite index in the group of symmetries of the uniform tiling MATH given by MATH, which is larger than the NAME group MATH because the graph has a non-trivial automorphism. But this graph automorphism exchanges the two MATH-orbits of square tiling cells in MATH, and the construction above identifies these two orbits with the class of marked models isomorphic to MATH and the class of marked models isomorphic to MATH, so birational automorphisms do not swap them over. Therefore MATH is isomorphic to a subgroup of MATH. In fact, the subgroup in question is of index MATH in MATH - a factor of MATH comes from the fact that the creased face diagonals must be preserved (alternatively, because the two classes of vertices corresponding to MATH and MATH must be preserved) and a further factor of MATH because there are no automorphisms of MATH that induce rotations of order MATH or reflections. |
math/0102055 | These two points MATH and MATH can be moved by the automorphisms of MATH to get infinitely many points on the boundary of MATH, accumulating at MATH. REF are preserved by the automorphisms of MATH, so there are points accumulating at MATH on two different quadric cones tangent at MATH. These two sets of points constrain the second derivative, if it were defined, to take two different values, so it cannot be defined. |
math/0102055 | First consider the hyperbolic uniform tiling constructed in REF by distorting the movable fan of MATH. The standard fundamental domain for the action of the reflection group on this tiling is a tetrahedron with a vertex at the centre of one cell in each orbit - the centres of two cubes and two square-tiled paraboloids (where the `centre' of a paraboloid is the unique point it contains that is in the boundary of hyperbolic MATH-space). The thick-line triangles on REF show how eight such fundamental domains can be combined into a single fundamental domain for the action of MATH. Its vertices are seven in number: CASE: MATH, the centre of the paraboloid drawn (in this view, the point at infinity vertically downwards), CASE: MATH, MATH, MATH, and MATH, the centres of the square-tiled paraboloids meeting the one drawn at the points MATH, MATH, MATH, and MATH, CASE: MATH and MATH, the centres of the cubes meeting the paraboloid drawn at MATH and MATH. The facets of this fundamental domain are the triangles MATH, MATH, MATH, and MATH, and the quadrilaterals MATH, MATH, and MATH. Now consider the movable fan of MATH. For each of the points MATH, the centre of a square-tiled paraboloid or a cube, let MATH be a generator of the central ray of the corresponding square-tiled paraboloid or lozengoid in the movable fan, all chosen in the same affine slice MATH. It is clear from symmetry that the three quadrilaterals MATH, MATH, and MATH are planar, and the triangles MATH, MATH, MATH, and MATH are certainly planar, so there is a (possibly non-convex) polyhedron with these seven faces. The cone on this polyhedron will be a rational polyhedral fundamental domain for the action of MATH on MATH as long as it is convex. Examination of the shape of the polyhedron shows that it will be convex as long as the three quadrilaterals are convex and the line MATH passes through the interior of MATH. We will therefore check these facts. Choosing the right copy of the fundamental domain, we have, for some positive scalars MATH, the following expressions (the expression for MATH is the sum of the vertices of the lozengoid described in REF ; MATH was described in REF, and the remainder follow by translating these two by appropriate symmetries of MATH) MATH . Now the convexity of the quadrilaterals follows from MATH while the condition on MATH follows from MATH . All these equalities can be checked from REF . |
math/0102057 | Take MATH and MATH in the NAME theorem. Since MATH and MATH, we know that there exist functions MATH for MATH such that for every partial function MATH from MATH to MATH whose domain is finite, there is some MATH such that MATH. Let MATH be an enumeration with infinitely many repetitions of each MATH for MATH. For each MATH, define MATH by MATH. The family MATH is as required: for if MATH are distinct and MATH are given, then the set MATH is a finite function, so there is some MATH such that MATH and it is now easy to see that MATH is infinite. |
math/0102057 | NAME: see REF. |
math/0102057 | Let MATH be the set of prime numbers, and let MATH be a family of almost disjoint (infinite) subsets of MATH: MATH. By REF , there exists MATH such that MATH, and if MATH are distinct and MATH, then MATH is infinite. Define functions MATH and MATH in MATH as follows. Let MATH where MATH, and let MATH where MATH (=REF if MATH). Note that MATH. For MATH, let MATH be the subgroup of MATH generated by MATH. We show that the family MATH satisfies the conclusions of REF : MATH is pure in MATH. Proof of REF : Suppose that MATH for some MATH, and MATH. Say MATH, with MATH. Without loss of generality (adding more elements from MATH to the Right-hand side if necessary), MATH for some MATH, and MATH. Relabelling (if necessary), we may assume that MATH, and because MATH if MATH, we may write MATH . Fix MATH. Since MATH are distinct MATH, letting MATH (NAME delta), we know that the set MATH is infinite. For large enough MATH in this set (for example, MATH is zero if and only if MATH. So for infinitely many MATH, for MATH, and MATH. Unfix MATH. For each MATH, for infinitely many MATH. Since MATH, we must have MATH for some MATH in MATH, and therefore MATH is torsion-free). Hence MATH is pure in MATH, which establishes REF : MATH has cardinality MATH, so REF holds. Proof of REF : If MATH, then for some MATH. Pick MATH with MATH and MATH; so the set MATH and MATH is infinite, and if MATH is bigger than max-MATH, then MATH, since MATH is non-zero and divisible by MATH but by no prime in MATH, and MATH is non-zero and divisible by MATH but by no prime in MATH. It follows that MATH has cardinality MATH. After this observation, a second's reflection on the element types of MATH and MATH (for MATH) should convince the reader that the groups are neither isomorphic nor free. REF : REF holds: if MATH, then MATH and MATH are almost disjoint. Proof of REF : Suppose (towards a contradiction) that for some MATH, for some non-free abelian group MATH, there exist isomorphisms MATH range-MATH. Since MATH is MATH-free, MATH must have cardinality MATH. Let MATH be a MATH-filtration of MATH. Without loss of generality, we may assume that each MATH is pure in MATH, so that MATH is torsion-free. Let MATH for MATH and MATH. Note that MATH is a MATH-filtration of MATH, since it is increasing and continuous with union MATH, and each MATH is countable. For large enough MATH, the set MATH defined by MATH is a club of MATH well-known, or see REF. Note that if MATH, then MATH maps MATH into MATH. Since MATH is not free, it follows by REF that MATH is not MATH-free-MATH is stationary. By NAME 's Criterion, for each MATH has a non-free (torsion-free) subgroup MATH of finite rank MATH such that every subgroup of MATH of rank less than MATH is free. Let MATH be a pure subgroup of MATH of rank MATH. Then MATH is free with basis MATH say. So MATH is a torsion-free rank-REF group which is not free, and hence there is a non-zero element MATH which is divisible in MATH by infinitely many natural numbers. Call this set of natural numbers MATH. For MATH, for large enough MATH, and MATH is an element of the subgroup of MATH generated by MATH for all MATH. Taking large enough MATH, we may assume that min-MATH. Since MATH, we can show the following claims: MATH: The set MATH does not contain infinitely many powers of one prime. MATH: The set MATH. Now MATH is true because non-zero sums of elements in MATH are divisible by at most finitely many powers of any given prime by the definition of the elements MATH. Note that MATH, where the characteristics are taken relative to MATH, MATH and MATH respectively. Hence MATH holds. By MATH, since MATH is infinite, the set MATH is infinite. Also, the same characteristic inequality implies that MATH. So MATH is true. Hence, MATH which is finite (since the family MATH is almost disjoint). This is a contradiction, and so REF follows, completing the proof of REF . |
math/0102057 | By NAME 's characterisation of slender groups (see REF for example), MATH must contain a copy of the NAME group. |
math/0102058 | Toward contradiction let MATH witness Pr-MATH so MATH is of cardinality MATH. Let MATH list MATH; choose an increasing sequence MATH such that MATH and MATH. We choose MATH by induction on MATH such that: CASE: MATH CASE: MATH CASE: MATH CASE: MATH if MATH and Dom-MATH is unbounded in Dom-MATH, then Dom-MATH. For clause REF note that if Dom-MATH is unbounded in Dom-MATH then there is MATH unbounded in Dom-MATH of order-type MATH and such MATH determines MATH in MATH uniquely Now choose MATH such that MATH maps MATH into MATH when MATH and be increasing MATH contradict the choice of MATH. |
math/0102062 | The second statement follows from the first one with all intervals MATH with a fixed MATH. For the first statement, it suffices to show that MATH . Moreover, since each MATH is an additive process with freely independent increments, the expression MATH is multi-linear in its arguments, and all mixed cumulants are equal to MATH, it suffices to show that MATH with MATH. First suppose that MATH, one of the intervals in a uniform subdivision, with MATH. Then using the same properties as above, MATH where in the last equality we have used that fact that by stationarity of MATH, MATH does not depend on MATH. Therefore MATH . By stationarity, it follows that REF holds for MATH and consequently for any rational MATH. The result follows for general MATH by the continuity assumption on MATH. |
math/0102062 | Denote MATH where MATH. Then MATH where MATH. Then as in CITE, MATH . At least one MATH is centered, so MATH and MATH (since MATH). Thus MATH . Therefore MATH and so MATH, which converges to MATH as MATH. |
math/0102062 | By the free independence assumption on MATH, the joint distribution of the MATH-tuple MATH is, for each MATH, entirely determined by the distribution of MATH and the joint distribution of MATH. These distributions, and hence the conclusion of the lemma, are not changed if we assume in addition that the family MATH is freely independent from MATH. MATH, where MATH denotes the vector MATH with each term multiplied by MATH on both sides. Since MATH and MATH are freely independent, if MATH is centered then so is MATH. Then the previous lemma implies the result. |
math/0102062 | Let MATH. MATH where MATH is a vector of length MATH such that MATH . For MATH, at least one of the MATH is centered. So REF applies and the limit, as MATH goes to MATH, of the appropriate term is MATH. On the other hand, it follows from CITE that for any MATH the limit of MATH is MATH. We conclude that, denoting MATH, MATH where MATH is well-defined for the free NAME stochastic measure by CITE. On the other hand, MATH where the sum is taken over all partitions MATH of MATH which contain the class MATH and such that for all j, the restriction of MATH to MATH is MATH. The only noncrossing partition satisfying these requirements is MATH, so MATH . |
math/0102062 | Such a product decomposition is valid for any subdivision MATH. |
math/0102062 | The first statement is based on a purely combinatorial observation that MATH and the fact that MATH for MATH; see REF. The second statement is based on a purely combinatorial observation that MATH and the same fact. |
math/0102062 | Let MATH be a subdivision of MATH. Let MATH be another such subdivision, and let MATH be their common refinement. Temporarily denote by MATH the index MATH such that MATH. Denote MATH, and similarly for MATH. MATH . The above expression MATH is a sum with positive coefficients. Hence so is MATH. Therefore its expectation is a sum over a collection of indices, with weights given by products of MATH, all of which are independent of the distribution of MATH, of free cumulants of MATH of order MATH. Each of those free cumulants is bounded in norm by MATH, where MATH. Since for the free NAME process all such cumulants are equal to MATH, the result is at most MATH times the corresponding quantity for the free NAME process, for which we denote MATH by MATH. That is, MATH and so MATH which implies in particular that MATH . By assumption, the net MATH converges in norm, and MATH . Therefore the net MATH is a NAME net, and so converges. |
math/0102062 | Let MATH be a free NAME stochastic measure. By REF , MATH is well-defined for MATH with a single outer class. Since for such MATH and MATH, MATH, MATH also contains only one outer class, by REF we conclude that for such MATH, MATH is well-defined as well. By REF , MATH is then well-defined for an arbitrary MATH, and applying REF again implies that MATH is well-defined for an arbitrary MATH as well. Finally, by REF the same is true for an arbitrary consistent MATH-tuple MATH of free stochastic measures. |
math/0102062 | The free increments property and stationarity follow immediately from the corresponding properties of MATH. For a general MATH-tuple MATH of free stochastic measures that has these two properties, by stationarity the continuity property is equivalent to the continuity of the function MATH for all MATH. By NAME inversion, this is equivalent to the continuity of MATH for all MATH. By additivity and the free increments property, this is equivalent to the continuity of this function, for all MATH, at MATH, and so to the same property for MATH. Thus finally, for the MATH-tuple in the hypothesis, it suffices to prove that MATH as MATH. Note that we do not need to put in a multi-index MATH since MATH is already an arbitrary collection of subsets of MATH. Denote by MATH the partition MATH with interval classes MATH and let MATH. Clearly MATH is a consistent MATH-tuple. Then MATH by REF , and so goes to MATH as MATH. |
math/0102062 | Let MATH be a subdivision of MATH. For each MATH, let MATH be a subdivision of MATH, and MATH be the subdivision of MATH obtained by combining MATH. Then as MATH, also MATH. Therefore MATH and so MATH is also the limit of the left-hand-side if in addition MATH. Here MATH . On the other hand, MATH and MATH . Therefore the difference MATH is the limit, as MATH and then as MATH, of MATH . This expression is a sum with positive coefficients. Also, for the free NAME process, MATH . By the same estimates as in REF , the result follows. |
math/0102062 | Let MATH be an inner class of MATH, and let MATH be the restriction of MATH to MATH. Then it suffices to prove that MATH . Denote MATH. MATH where MATH . Denote MATH. Let MATH be the support of MATH identified as a sub-partition of MATH, and let MATH. Then by REF , MATH . Fix MATH. Let MATH, MATH, where MATH and MATH. Denote MATH the class of MATH corresponding to MATH. Since MATH is an inner class of MATH, the condition MATH implies that MATH and MATH are classes of MATH for MATH. Let MATH map such a MATH to the partition in MATH obtained by removing from MATH the classes MATH for MATH and MATH for MATH. It is easy to see that MATH is a bijection onto MATH, and that MATH. Therefore MATH since the first sum equals to MATH. |
math/0102062 | The statement of the theorem holds for MATH. From now on, assume MATH. The proof will proceed by induction on MATH. The statement of the theorem is vacuous for MATH; assume that it holds for all tuples of less than MATH elements. By REF , MATH . Therefore MATH . Then by REF , MATH . In its turn, MATH . Since MATH, MATH has at most MATH classes, so the induction hypothesis applies to MATH. Thus MATH . Define the map MATH by MATH. Note that the outer classes of MATH are in one-to-one correspondence with the outer classes of MATH, and each inner class of MATH corresponds to a unique inner class of either MATH or MATH. It is easy to see that MATH is in fact a bijection onto MATH. Combining REF with REF , we see that MATH with MATH. Therefore MATH . On the other hand, for all MATH, MATH. Note that the NAME inversion formula for MATH involves only MATH. Therefore, applying this formula, MATH . |
math/0102063 | For REF , by stationarity, it suffices to prove that MATH. Since the result is to be proven for all MATH, it suffices to do so for all even integer MATH. Moreover, the statement is not about the stochastic measure but about its distributions MATH. Namely, what needs to be shown is that MATH. Indeed, MATH . Thus MATH as MATH. Under the positivity assumption on the free cumulants, MATH . This is MATH iff the radius of convergence of the power series defining MATH is infinite, in other words if MATH is analytic in the complex plane. But the only functions that are entire, map MATH into MATH and satisfy MATH are of the form MATH, and these are the MATH-transforms of non-scaled, non-centered semicircular distributions. |
math/0102063 | CASE: |
math/0102063 | CASE: |
math/0102063 | Decompose MATH, with MATH. Let MATH, MATH. Take MATH, MATH. Then MATH . Both sides are linear in MATH, so the equality holds for arbitrary simple adapted biprocesses. |
math/0102063 | The formula holds for simple adapted biprocesses. Denote MATH, MATH. MATH, so MATH is continuous on MATH, and the left-hand-side of REF is continuous. If MATH, both sides of the equation are MATH. If MATH, MATH. Finally, since MATH is normal, the hypotheses imply that MATH. |
math/0102063 | It suffices to show that for MATH, MATH, MATH . Moreover, we may assume that MATH and either MATH or MATH. In the first case it is easy to see that both sides of REF are equal to MATH. In the second case, the statement to be proven is MATH . Now MATH where the limit is taken in the operator norm, and MATH REF is as in the definition of MATH. The sum is equal to MATH . The operator norm limit of the third term is MATH by REF. Now consider the first term. It can be written as MATH, with MATH . For MATH, by the proof of REF MATH uniformly in MATH. Therefore MATH converges to MATH in the MATH-norm. This implies the convergence of integrals. The argument for the second term is similar. |
math/0102063 | It suffices to prove the formula for a monomial MATH. Note that in this case the only non-zero terms on the right-hand-side of REF are for MATH, MATH, so the sum has finitely many terms. We proceed by induction on MATH. Denote by MATH the coefficients in the expansion MATH . Using REF with MATH, we know that MATH and we need to show that MATH satisfy these equations. Indeed, MATH . The second term of this sum is equal to MATH . Putting these three terms together we obtain REF . |
math/0102063 | This is a slight generalization of REF ; the proof proceeds as in REF . |
math/0102063 | MATH is a simple biprocess. |
math/0102063 | Denote MATH. Since MATH is simple, MATH for MATH large enough, so it suffices to prove the Proposition for finite MATH. That is, we need to show that the distribution of MATH is the unique probability measure MATH determined by MATH. For any finite MATH, MATH is well-defined by the results in subsection REF gives MATH . In terms of generating functions REF corresponds to the differential equation MATH where MATH are the usual partial derivatives, MATH is the NAME transform of the distribution of MATH, and MATH is the MATH-transform of MATH. See more on this in subsection REF. But MATH is a solution to REF , with the same initial conditions. Indeed, that equation can be obtained by the same method as the original one in CITE. Namely, start with the equation MATH where MATH. Differentiating it with respect to MATH, we obtain MATH . But MATH. Substituting MATH for MATH in REF , we obtain REF . A posteriori, MATH is the distribution of MATH and therefore a positive measure. Since MATH is bounded, its free cumulants determine a unique probability distribution. |
math/0102063 | MATH is a MATH-th root of a homogeneous polynomial of degree MATH, with positive coefficients, in various MATH-norms of MATH, with no constant term. Therefore it is homogeneous and MATH only at MATH. Now we prove the triangle inequality. First take two functions MATH in MATH. Since all the free cumulants MATH are non-negative, MATH, and so MATH. Now let MATH be simple biprocesses, MATH, MATH. Then MATH . By approximation, MATH is a norm on MATH. For MATH, the triangle inequality and homogeneity follow by the limiting procedure, and for a non-negative function MATH, MATH iff MATH. |
math/0102063 | We do this by induction. For MATH, by REF MATH . By REF , MATH and so MATH where we have used the induction hypothesis as well as the equality for positive scalar functions. |
math/0102063 | Apply REF to MATH for MATH odd, MATH for MATH even. We get MATH where in the second equality we have used REF . |
math/0102063 | For MATH the distribution of MATH, MATH. Since all the free cumulants are positive, MATH . Therefore MATH. In particular MATH. |
math/0102063 | First assume MATH. MATH so by hypothesis the series defining MATH and MATH converge absolutely. Also, MATH . Finally, MATH implies that MATH . So the sums on the right-hand-side of REF converge absolutely as well. The formula holds for finitely many MATH, and since both sides of that equation are continuous, the formula can be extended to MATH. |
math/0102063 | Let MATH, taken a priory as a formal power series. Assume that MATH. MATH . For MATH, MATH . Therefore the sum on the right-hand-side of REF converges absolutely. By continuity, the formula holds also for MATH. |
math/0102069 | Note that MATH. The conclusion follows from the injective operad-morphism (see CITE) MATH which is an isomorphism in this case (since MATH has finitely generated components). |
math/0102069 | This is just the morphism of MATH-operads induced by the chain-map MATH . |
math/0102069 | Identify MATH with the operad MATH, by REF. Now form the composite MATH where CASE: MATH is the canonical coproduct defined in CITE, CASE: the morphism MATH is defined in REF, CASE: MATH, CASE: MATH is the structure map of the m-structure of MATH. |
math/0102069 | Recall that MATH where MATH is the basepoint and MATH is the basepoint of MATH. This means that the basepoint of MATH is MATH and MATH. |
math/0102069 | This follows from REF and the definition of the m-structure of the (geometric) suspension. Let MATH be an m-coalgebra over MATH with structure map MATH and let the structure map of the unit interval MATH be MATH . Then the suspension has a structure map that is the composite MATH . Where CASE: MATH is the operad morphism defined in CITE CASE: MATH is induced by the quotient map MATH . Now observe that the left column is MATH and that the right is the structure map of MATH. |
math/0102069 | Just desuspend the left square of REF (and interchange its columns) to get MATH and splice in the commutative square representing a morphism MATH (in MATH) MATH . The ``outer rim" of the result is precisely REF. |
math/0102069 | This follows by an inductive application of REF. |
math/0102069 | We begin by proving the if-part: If there exists a morphism MATH then, by REF, MATH and MATH in MATH. Since MATH, the morphism MATH induces a morhism MATH in MATH. One the other hand, suppose there exists a morphism MATH in MATH, of level MATH (see REF). Then there exists a sequence MATH where each arrow, MATH , points to the left or right and the left-pointing arrows induce homology isomorphisms. Suppose MATH, where MATH. Then MATH. We convert REF into a sequence MATH which defines a morphism MATH in MATH. The conclusion follows by REF. |
math/0102069 | This follows from REF, which implies that MATH is equivalent to MATH in MATH and CITE. |
math/0102072 | The full proof of the mapping property of MATH is tedious and we do not give it here. To see how to proceed, let us prove REF in detail. First note that the expression on the Right-hand side of REF is well defined and independent of the ambiguities in the decomposition of MATH. To prove the equality, let MATH be as above. Then, writing again MATH the NAME formula yields MATH . The NAME derivative of the metric can now be calculated using the decomposition of the metric and NAME 's formula: MATH where we have used REF , especially that MATH. The calculation of the contribution of the term MATH in the decomposition of MATH is a little bit more involved: First, write MATH . We can assume that MATH is (a sum of elements) of the form MATH. Note that for elements in MATH of this type we have MATH . Now MATH and the first and the third summand here are MATH leaving us with MATH . Summing the two results, the equation is proved. The proof of REF uses the same method. For MATH, MATH and MATH one can check that MATH and direct calculation, using the decomposition of MATH as above, shows that the Right-hand side is of order MATH. REF follow easily. |
math/0102072 | To prove REF we will use REF . Thus, for MATH we have MATH . To prove REF , we just have to observe that the MATH-term in REF is MATH if MATH are in MATH. This should be clear from the proof of REF is just a reformulation of REF . To prove REF , let MATH. Then first MATH . The different terms in this expression are MATH . Now the first summand in REF and the last two summands in REF give MATH and we get the same contribution fom the first summand in REF and the last two summands in REF . Putting everything together gives the result. Note that we have made extensive use of the fact that MATH and MATH are tangent to the boundary, since we needed every summand to be regular in MATH! |
math/0102072 | REF is clear by construction, REF follow from REF . Consider REF : Taking two extensions MATH of MATH it is clear that MATH. Thus MATH by REF . |
math/0102072 | REF is straightforward. To prove REF , we first show that MATH is well defined by showing that it is well defined for the different terms in the decomposition of MATH. To do this note first that MATH that is, the expression MATH is of order MATH. Thus MATH . To see that the expression MATH only depends on MATH, note that the extension MATH pairs with any element MATH to MATH and there is a natural map MATH. But the pairing of MATH with any element in MATH clearly only depends on MATH. The corresponding statement for MATH is clear. To prove the equality in REF note that it is trivially true for vertical MATH. On the other hand MATH is MATH by definition of the extension. This also proves REF . |
math/0102072 | We use the method developed in REF. For simplicity, assume that MATH. Also write MATH. Then MATH . The different parts of the NAME derivative of the metric can be calculated as in REF : MATH . This proves the claim for MATH. It is easy to see that it also holds for forms of the type MATH. REF are just reformulations of REF . |
math/0102072 | This is an application of the NAME formula using REF . We only give an example of how the calculation differs from the one in REF : MATH and so on. |
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