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math/0106118 | It is sufficient to prove the claim for MATH. We prove our claim by induction on MATH. We note that there is a section MATH of MATH such that MATH. Hence MATH, MATH. Let MATH be a wall and MATH belongs to MATH. Let MATH be MATH-divisors which are very close to MATH and MATH. We consider NAME filtration REF , where MATH. Since MATH, we get MATH. Assume that MATH and MATH for different MATH and MATH. Since MATH, MATH in MATH. Then MATH, which is a contradiction. Therefore MATH and MATH and MATH, MATH. Since MATH or MATH is supported on some fibers, we see that MATH . Since MATH, by the same argument as in REF , we get MATH where MATH satisfies that MATH. By induction hypothesis, MATH . Also we know that MATH does not depend on MATH. Therefore we get our claim. |
math/0106118 | Let MATH be an element of MATH such that MATH. We set MATH. We consider NAME transform induced by a universal family MATH on MATH. Then MATH, MATH, MATH. Replacing MATH by MATH, MATH, we may assume that MATH and MATH. By REF , MATH is isomorphic to MATH. By REF , MATH. By using REF again, we see that MATH is isomorphic to NAME scheme of points. |
math/0106118 | Let MATH be a simple pure dimension REF sheaf of MATH. It is sufficient to show that MATH is an isomorphism. We set MATH and MATH. Since MATH is supported on fibers, MATH. By the simpleness of MATH, MATH. We note that MATH. Hence MATH is injective and MATH is isomorphic. Since MATH is a line bundle on MATH, MATH is isomorphic. Therefore MATH must be isomorphic. |
math/0106118 | If MATH, MATH is primitive, then MATH is primitive. Since MATH is a unimodular lattice, there is a divisor MATH such that MATH. Hence MATH, MATH. Therefore MATH. |
math/0106118 | Let MATH be a quotient such that MATH is of pure dimension MATH. Let MATH be the induced generalized parabolic structure. Then MATH . Hence MATH. |
math/0106118 | We set MATH . Since MATH is a bounded set, for a sufficiently large MATH which depends on MATH, CASE: MATH, MATH, MATH and CASE: MATH . Then MATH. Since MATH is semi-stable, in the same way as in CITE, we see that there is an integer MATH such that for MATH and a generalized parabolic subsheaf MATH of MATH, MATH and the equality holds, if and only if MATH. Hence if the equality holds, then MATH is semi-stable and we may assume that REF holds for MATH. In particular, MATH. We note that MATH . By REF, if the inequality in REF is strict, then we get MATH . If the equality holds in REF, then MATH, and hence REF implies that L.H.S. of REF is MATH. Therefore our claim holds. |
math/0106118 | We choose a MATH so that MATH. We shall prove that MATH is proper. Let MATH be a discrete valuation ring and MATH the quotient field of MATH. We set MATH and MATH. Let MATH be a morphism such that MATH is extended to a morphism MATH. Since MATH is a closed subscheme of MATH, there is a morphism MATH, that is, there is a flat family of a sequence of quotients MATH . Let MATH and MATH be the quotient bundles of MATH and MATH corresponding to the morphism MATH. We set MATH, MATH and MATH. MATH is injective. Indeed, we set MATH. Then MATH. By REF, MATH . Therefore MATH. There is a rational number MATH which depends on MATH and MATH such that MATH is of type MATH. Proof of the claim: Let MATH be a quotient of MATH. Let MATH be the kernel of MATH. We note that MATH is injective. We set MATH. Then MATH. By REF, MATH . Let MATH be a subsheaf of MATH generated by MATH. Then MATH belongs to MATH. Let MATH be a positive number such that MATH. Since MATH is a bounded set, for a sufficiently large MATH which depends on MATH and MATH, we have MATH, MATH and MATH . Therefore MATH where MATH. Since MATH, we get MATH . Since MATH, MATH . There is a rational number MATH and an integer MATH which depend on MATH, MATH and MATH such that MATH for MATH. By CITE, there is a purely MATH-dimensional sheaf MATH of NAME polynomial MATH and a map MATH whose kernel is a coherent sheaf of dimension MATH. Let MATH be a quotient such that MATH is semi-stable. We set MATH and MATH. Since MATH and MATH, MATH . Since MATH is semi-stable, CITE implies that MATH where MATH is a constant which only depends on MATH. Since MATH, we get MATH and MATH, which means that MATH is of type MATH. Replacing MATH, we may assume that for all type MATH sheaves MATH of NAME polynomial MATH, MATH is generated by global sections and MATH, MATH. In particular MATH. Assume that MATH is not injective and let MATH be the kernel. Then we get a contradiction from REF and MATH. Thus MATH is injective, and hence it is isomorphic. Since MATH is generated by global sections, MATH must be surjective, which implies that it is isomorphic. Therefore MATH is of pure dimension MATH, of type MATH and MATH is an isomorphism. Thus we complete the proof of REF . We assume that MATH. Then MATH is surjective. Thus MATH is surjective and define a morphism MATH. Since MATH as elements of MATH, we get MATH. Assume that there is a generalized parabolic quotient MATH which destabilizes semi-stability. Since MATH satisfies MATH, we get that MATH, MATH and MATH . Since MATH are rational numbers, for a sufficiently small MATH, we get MATH which is a contradiction. Therefore MATH is generalized parabolic semi-stable. Thus we get a lifting of MATH and conclude that MATH is proper. |
math/0106122 | Any integral cohomology class on a surface can be represented by the intersection index with a two-sided curve. The intersection of MATH with such a curve is zero by the definition of MATH. |
math/0106122 | If MATH is orientable, there is nothing to prove. Otherwise, let MATH be any mod REF cohomology class. The obstruction to lifting it to an integral cohomology class lives in MATH. Consider the commutative diagram MATH where the vertical arrows are restrictions to MATH and horizontal arrows reductions modulo MATH. It follows that the map MATH is trivial. Hence, the restriction of MATH to MATH lifts to an integral cohomology class on MATH and has zero pairing with MATH by the previous lemma. Thus, MATH by NAME duality over MATH. |
math/0106122 | Note that MATH and therefore MATH by the NAME formula. Hence, MATH, which proves the lemma. |
math/0106122 | The real wedge product MATH is a non-vanishing section of the real determinant bundle MATH restricted to MATH. Note first that it can be extended to a global non-vanishing section of the complexification MATH. Indeed, the obstruction is given by the relative characteristic number MATH which is zero because the complement to the characteristic circle MATH is orientable. It remains to note that, for the totally real surface MATH, the map MATH defined by replacing the real wedge product on MATH by the complex one on MATH is an isomorphism. |
math/0106122 | To any characteristic two-dimensional submanifold MATH there is associated a quadratic function MATH on the kernel of the inclusion homomorphism MATH called the NAME - NAME - NAME form of MATH (see CITE, CITE, and REF). The value of this function on the characteristic homology class MATH (assuming that MATH) satisfies the congruence MATH where MATH is the signature of the four-manifold MATH. This formula is due to NAME (at least in the case when MATH, see CITE, MATH REF). It may be obtained from the generalised NAME - NAME - NAME congruence (see REF , or CITE, NAME REF) by reducing it modulo REF and plugging in an elementary algebraic property of the NAME invariant (see CITEEF ). Applying REF to the characteristic submanifolds MATH and MATH, respectively, and using the orientability of MATH, we obtain that MATH and MATH . (The second congruence follows also from NAME 's lemma, see REF.) Thus, MATH whence MATH because MATH by the definition of the NAME - NAME - NAME form. |
math/0106122 | Note first that the pushoff of MATH inside MATH is the MATH-normal pushoff of MATH by REF of MATH. Hence, MATH by the definition of the NAME index. Thus, we need to show that MATH . The obstruction MATH can be computed as follows. Let MATH be a generic tangent vector field on MATH such that MATH on MATH. (Note that MATH is transverse to MATH.) Then MATH fails to give a non-vanishing normal extension of MATH at the points MATH such that MATH. These points are, firstly, the zeroes of MATH and, secondly, the complex points of MATH. Neglecting the signs involved, we get the modulo MATH formula MATH where MATH is the NAME characteristic of MATH and MATH is the number of complex points on MATH. Similarly, let us consider the modulo MATH obstruction to extending the section MATH (from REF ) to a non-vanishing section of MATH. On the one hand, it is equal to MATH because MATH is the transverse zero set of such an extension MATH. On the other hand, observe that MATH . As we have already seen in the proof of REF , the obstruction to extending MATH from MATH to a non-vanishing section of MATH is the sum of two obstructions. Firstly, MATH can be non-trivial. Secondly, the map MATH degenerates at the complex points of MATH. Altogether, we see that MATH . Combining REF gives the congruence MATH . The right hand side is equal to MATH for any surface MATH with a single boundary component, and the lemma follows. |
math/0106122 | By REF , the meridian MATH is nullhomologous in MATH (modulo REF). The NAME - NAME sequence for MATH shows that there exists a homology class MATH such that CASE: MATH, CASE: MATH, where MATH and MATH are inclusion maps. It follows from the second property that MATH . Therefore, REF shows that MATH differs from the homology class of the MATH-normal pushoff MATH (see REF ). The pre-image MATH consists of exactly two homology classes, represented by the curves MATH and MATH. The first curve gives the MATH-normal pushoff of the meridian by REF . Thus, MATH and therefore MATH bounds in MATH. Now consider the fibre MATH of the projection MATH. The NAME twist MATH maps this fibre to the curve MATH. It follows that the disc MATH bounded by the fibre in MATH and the chain bounded by MATH in MATH are glued together into a REF-cycle in MATH. The intersection index of this cycle with MATH equals MATH, and therefore MATH is homologically non-trivial in MATH. |
math/0106122 | Consider the following long exact sequences in cohomology with compact support: MATH and MATH . The first map in the first sequence is trivial because MATH is nullhomologous in MATH. Hence, MATH . On the other hand, the first map in the second sequence is onto because MATH in MATH, and therefore MATH . Since MATH and MATH are diffeomorphic, we conclude that MATH and the result follows because MATH for any four-manifold MATH by NAME duality. |
math/0106126 | It follows at once that MATH. For MATH, MATH . On the other hand, MATH . Note that the terms MATH match the terms MATH and the remaining terms in the sums for MATH and MATH also match. Thus, MATH . |
math/0106126 | This follow from REF and maps REF . |
math/0106126 | This follows immediately, since both MATH and MATH are induced by antisymmetrization maps, and both MATH and MATH are induced by projections. |
math/0106126 | There is a surjective MATH-module homomorphism MATH given on homogeneous elements by MATH . The result now follows from the commutativity of the diagram MATH . |
math/0106126 | The summand of MATH isomorphic to MATH can be represented via chains which are MATH-linear combinations of terms: MATH where MATH is the cyclic shift given by the cycle MATH CITE. It is enough to compute MATH . Thus, any element in MATH can be represented as a chain in MATH. |
math/0106126 | From REF MATH . From REF the summand of MATH isomorphic to MATH can be computed from the complex MATH, and similarly for MATH. Consider the MATH-linear homomorphism MATH . We have the following commutative diagram with exact rows: MATH where the dimensions in the top row are inherited from the chain complex for NAME homology. By REF-lemma, MATH. Let MATH denote an element in the mapping cone MATH of the form MATH where MATH is the cyclic shift in MATH and MATH is the cyclic shift in MATH. The chain map MATH in REF may be defined on the respective mapping cones, and MATH where the latter is in fact an element of MATH. Also, at the level of mapping cones, we have MATH . The theorem follows, since the isomorphism MATH is realized by sending the class of MATH to the class of MATH. |
math/0106126 | From CITE, the map MATH is zero. |
math/0106128 | For REF , write MATH, and differentiate MATH with respect to MATH to obtain MATH (so MATH and MATH are homogeneous coordinates for MATH). We note that the clinant may be thought of as the ``complex slope" of MATH, since the equation for the tangent line to MATH at MATH may be written MATH. Next, taking MATH to be an arclength parameter along MATH, differentiation of MATH with respect to MATH gives MATH, and solving for MATH gives REF follows similarly by straightforward computation, giving REF as well. For simplicity, we have specialized the formula MATH, MATH by assuming the second of the two curves in first order contact has been conformally mapped to a straight line (see CITE, footnote REF on the `natural measure of the horn angle' MATH between two curves in first order contact). Conformal invariance of MATH may be derived by (double) application of the ``chain rule" for Schwarzian derivative of MATH: MATH . We recall that MATH iff MATH is a NAME transformation, so this well-known formula generalizes NAME invariance of MATH. |
math/0106128 | All claims may be regarded as symmetric space formalism - see the appendix. We verify REF here, as multiplication of NAME functions is fundamental. Thus, let MATH and MATH. Then MATH so MATH is the NAME function of the curve MATH - in particular, NAME functions are closed under the above multiplication. Now, MATH, so MATH, the curve whose NAME function is MATH. |
math/0106128 | REF are formal consequences of the representation REF of NAME functions as transvections. To prove REF , suppose MATH and MATH. Then MATH, so the function MATH is involutive with fixed point MATH. Differentiation of MATH at MATH implies MATH. If MATH then MATH. On the other hand, if MATH note that MATH defines a local diffeomorphism REF satisfying MATH. Thus, MATH near MATH, hence, MATH. |
math/0106128 | First suppose REF holds, with initial NAME functions MATH, MATH. We make the inductive assumption that MATH is a NAME function whenever MATH . Then MATH implies MATH and hence MATH. Thus, MATH is also a NAME function, and REF holds. Conversely, assume REF . Suppose, for some MATH, REF holds whenever MATH. (Note that the identity is obvious when either MATH or MATH is zero - in particular, when MATH). Let MATH be integers satisfying MATH. Then MATH so REF for MATH follows by induction. |
math/0106128 | By REF , we have MATH . Thus, MATH, so MATH lies on MATH. (In terms of symmetric space formalism, the appearance of the factor MATH may be traced to the special form of the NAME immersion on cosets of transvections.) Since MATH, a regular parametrization of MATH is thus obtained, and REF follows from REF . The last statement follows at once, and gives direct geometric meaning to the function MATH as the initial rate of displacement of MATH from MATH. (Similar interpretation of the extension MATH is obtained below.) |
math/0106128 | We begin with a more direct derivation of REF . Assume REF holds. Then differentiation of the identity MATH with respect to MATH and MATH yields two equations: MATH, and MATH. We thus obtain the equation MATH. Therefore, MATH holds, hence, REF . In view of the previous proposition, REF holds as well. We remark that MATH also satisfies an ODE in MATH, for fixed MATH, obtained by combining REF : MATH . Note that MATH is tangent to the foliation MATH and REF follows from MATH. (Here it is convenient to identify a MATH -complex vector MATH with its real part MATH; but one could as well denote tangent and normal vectors to MATH by MATH and MATH, respectively, and write the equation for equipotentials of MATH as MATH.) For the last statement, we use the fact that a holomorphic differential MATH is closed, so can be written locally as MATH for some complex potential MATH (and the dual vector to MATH is MATH). REF is the derivative of MATH along MATH. |
math/0106128 | First consider a local metric determined as above by coordinate MATH. Then MATH, while MATH. Now if MATH is replaced by a conformally related metric, say MATH, then the MATH-gradient, MATH satisfies MATH. |
math/0106128 | Since MATH lies on the curve MATH, for all MATH, we may define a function of time MATH, which REF allows us to write as MATH. Setting MATH in MATH, we see that MATH satisfies the ODE: MATH, which is equivalent to REF . Next, when MATH, we have MATH since MATH is a point of tangency. Therefore, evaluating MATH at MATH, gives MATH. The MATH-derivative of REF at MATH is thus MATH. Similarly, MATH, and the MATH-derivative of REF at MATH gives MATH. Since MATH, we find that MATH and MATH, and arrive at MATH. The expression involving MATH is readily verified; a more comprehensive discussion of the significance of the residue MATH is given in CITE. |
math/0106128 | Care must be taken here to distinguish between conjugates of numbers and functions! Consider MATH with MATH. Differentiating MATH, we obtain MATH. Letting MATH, we obtain MATH. Now set MATH, and note that MATH, for MATH; thus, MATH along MATH. Observe that MATH is an arbitrary analytic function along MATH, so the first description of MATH follows. Using the same representation of MATH and MATH in terms of MATH, and using the fact that MATH, we have MATH . For the normal variation, MATH, we use the positively oriented orthonormal frame MATH, MATH along MATH. Note that the normal variation corresponds to an arbitrary analytic function along MATH. |
math/0106128 | Using MATH, REF follows at once from MATH. The main content of REF, therefore, is the interpretation of MATH as a connection. The latter is included in REF which will be treated in REF; however, it is useful to include a direct computational argument for REF here. Given our use of the embedding MATH, the main point we need to verify is that MATH is in fact tangent to MATH. We introduce a notational shorthand suggested by the lemma: for a given curve MATH, an analytic function MATH can be turned into a tangent vector MATH to MATH at MATH, by the operator MATH. (The formula MATH is valid only along MATH, though MATH does indeed represent an element of MATH, by analytic continuation). Now consider a two-parameter family of non-singular analytic curves MATH and the corresponding two-parameter variation of NAME functions satisfying MATH. We compute successive derivatives of the latter, suppressing all arguments: MATH, MATH, MATH, MATH, MATH, MATH (and similar formulas for MATH and MATH). Rearranging, and applying the vector operator to MATH, MATH, etc., one gets: MATH . These formulas are substituted into MATH (obtained by setting MATH, MATH in REF ), and coefficients of the second order terms MATH are collected. After a key cancellation and subsequent division by MATH, these coefficients turn out to be real: MATH . For future reference, we re-express this: MATH . In particular, after precomposing with MATH, the right-hand-side has the form of a tangent vector to MATH at MATH. Finally, replacing MATH with general vectorfields MATH on MATH, one obtains MATH, a bilinear operation satisfying the required further properties for an affine connection: MATH and MATH, for MATH. |
math/0106128 | Setting MATH in REF , we obtain the geodesic equation: MATH . One easily verifies that MATH, and the corollary follows. To relate the corollary more directly to REF , we note that MATH. That the geodesic equation may be written in the form MATH is not surprising, given that MATH is an evolving unparametrized curve, so the above equation is missing ``half" of the information required to govern MATH. Up to initial parametrization, this information is provided by the normal motion requirement in the second statement of the corollary. |
math/0106128 | The curve MATH parametrized by MATH, MATH, has tangent vector MATH. Setting MATH, it follows that the degree of the unit tangent to MATH is the winding number of MATH about the origin. Applying the argument principle to the curve MATH and the function MATH, we find that MATH . We remark that the curve MATH may be replaced by an arbitrary simple closed curve, as in the argument principle, but such a generalization is of no importance to us presently. |
math/0106128 | We prove the second of the two equivalent statements in REF by induction. For a given positive integer MATH, assume the equation holds for all MATH, whenever MATH (the case MATH being known). Using REF and (both versions of) the definition of MATH, we obtain MATH, so the result holds for MATH. One argues similarly for MATH. Thus, REF follows by induction, and MATH is a homomorphism. The first equality in REF now follows easily from the homomorphism property, and MATH gives MATH (which is the definition of powers given in CITE). To prove REF, set MATH, and MATH. Note MATH, MATH, and MATH. Thus, we make the identification MATH, that is, MATH, the first equation in REF - note that in MATH, the MATH power of MATH is MATH. Setting MATH gives the second equation. The third equation follows by a similar argument using, instead, MATH, MATH, and MATH. |
math/0106131 | It will suffice to assume that MATH . Let MATH be defined as MATH and MATH by MATH . Arguing as in CITE or CITE one sees that MATH is completely positive and hence can be extended to a completely positive map on all of MATH which we still denote by MATH . Using the fact that MATH fixes MATH and again arguing as in CITE, we see that there exists MATH such that MATH . Clearly, MATH and MATH must be completely positive and MATH extends MATH . Since MATH for all MATH in MATH by NAME 's rigidity result MATH for all MATH in MATH . Thus, MATH fixes the MATH-subalgebra, MATH and so by CITE CITE (see also CITE) MATH must be a bimodule map over this algebra. Thus, CITE MATH and we have that MATH . Finally, MATH . The proof for right MATH-module maps is similar. For the bimodule case let MATH and deduce that MATH is a MATH-bimodule map. If MATH contains an invertible element e, then MATH and so MATH is unique. |
math/0106131 | There exists MATH in MATH with MATH for all MATH in MATH . Since MATH and hence, MATH for all MATH in MATH . |
math/0106131 | By the NAME extension theorem for completely bounded MATH-bimodule maps CITE (see also CITE, CITE), it follows that MATH is MATH-injective, MATH-injective and MATH-injective. Thus, REF implies REF. It will suffice to prove that REF implies REF, the other implications are similar. If MATH is MATH-injective then the identity map from MATH to MATH extends to a completely bounded left MATH-module projection from MATH to MATH . Letting MATH play the role of MATH in the rigidity theorem yields the result. |
math/0106131 | We prove the equivalence of REF,b and REF, the remaining arguments are similar. We have that REF implies REF by NAME 's NAME extension theorem for module maps. Clearly, REF implies REF. We now prove that REF implies REF. Since MATH is a MATH-injective module, the identity map on MATH extends to a completely bounded left MATH-module map from MATH into MATH. But by the Rigidity Theorem, this extended map must be the identity map on MATH and hence MATH. Thus, MATH is injective. |
math/0106131 | Since MATH and MATH are MATH-injective, there exist completely bounded left MATH-module maps MATH and MATH which fix MATH . By the rigidity of MATH is the identity on MATH and hence MATH is the identity restricted to MATH range MATH . This makes MATH a MATH-injective module and hence MATH and MATH . |
math/0106131 | Since MATH is MATH-injective the identity map on MATH extends to a completely bounded MATH-bimodule map from MATH to MATH . Composing with the projection onto MATH gives a completely bounded left MATH-module map from MATH to MATH which is the identity on MATH . By rigidity REF MATH and hence MATH is injective. |
math/0106131 | Since MATH is MATH-injective, the identity map from MATH to MATH has a completely bounded left MATH-module extension to MATH . This map is clearly a projection. Hence MATH is injective by REF . |
math/0106131 | By REF such a MATH exists, it remains to show that MATH is unique. To this end consider, MATH which is clearly a right MATH-submodule of MATH . It will suffice to show that MATH . Let MATH be a contractive approximate identity for MATH . For MATH we have, MATH with the last equality using the fact that MATH is essential. The same calculation for matrices shows that the quotient map MATH is a complete isometry on MATH and a right MATH-module map. Now since MATH is injective, there exists a completely contractive right MATH-module map MATH. Hence by rigidity MATH for all MATH in MATH and it follows that MATH . |
math/0106131 | By REF , there exist unique elements MATH in MATH such that MATH for all MATH in MATH . But MATH and so MATH for all MATH . Applying REF we conclude that MATH for all MATH and so MATH . |
math/0106131 | For each MATH in MATH there exists a unique MATH in MATH implementing MATH . By this uniqueness the map MATH must be a MATH-monomorphism on MATH . Furthermore, let MATH be essential ideals, and let MATH in MATH be implement by MATH . If MATH and MATH on MATH then, since MATH is essential, we must have MATH . This shows that the inclusions of MATH into MATH are coherent and allows us to extend these MATH-monomorphism to the direct limit, MATH . Now assume that MATH is any MATH-monomorphism with MATH for all MATH in MATH . Then for MATH in MATH and MATH from which it follows that MATH is the unique element implementing MATH . Finally, since MATH is exactly the image of MATH we have the last claim. |
math/0106131 | Since MATH is necessarily an essential ideal of MATH and MATH we have MATH . By REF we have a MATH-monomorphism MATH of MATH into MATH . Hence, MATH as MATH-algebras. Since MATH is MATH-injective, by REF , we have MATH and the result follows. |
math/0106131 | By CITE MATH is a commutative MATH-algebra. However, commutative MATH-algebras are injective by CITE since bounded linear maps between C*-algebras are positive whenever their norm equals their evaluation at the identity of the C*-algebra. Consequently, the MATH-monomorphism of MATH into MATH must be onto. |
math/0106131 | Since MATH is MATH-isomorphically embedded into MATH extending the canonical MATH-monomorphism of MATH into MATH by REF , the MATH-algebra MATH serves as an injective extension of the MATH-algebra MATH, compare CITE. However, the identity map on MATH admits a unique extension to a completely positive map of MATH into itself with the same completely bounded norm one since MATH by construction and MATH is the injective envelope of MATH. So MATH has to be the injective envelope of MATH, too. |
math/0106134 | According to the NAME theorem, on fractional integration, (see CITE, for example), we have MATH where MATH, provided MATH. We let MATH denote the operator norm of this map on scalar-valued functions. Now, given MATH and MATH as in the Lemma, we define MATH by MATH. We apply the NAME theorem with exponents MATH and MATH to conclude that MATH . Taking the MATH-th root gives the first inequality in the Lemma. The operator norm for MATH was computed by CITE. The behavior near MATH asserted in the Lemma follows from NAME 's result, the boundedness for MATH and the NAME interpolation theorem. |
math/0106134 | We consider the integral with respect to MATH in MATH and apply the NAME inequality using exponents MATH and MATH to obtain MATH . Here, the function MATH is defined by MATH . One may make the change of variables MATH in the integral with respect to MATH to obtain this representation of MATH. The function MATH is defined by MATH . Now, we can rewrite the integral with respect to MATH in MATH as MATH which defines MATH. Inserting the definitions of MATH and MATH into MATH gives the inequality relating MATH and MATH. We now establish the estimates for MATH and MATH. The estimate MATH follows from REF with MATH and MATH. The hypothesis MATH follows from REF . To estimate MATH, observe that MATH from REF and thus we may apply REF with MATH and MATH to obtain the first inequality below. Next, we use NAME 's inequality and REF with MATH and MATH to obtain the second and third inequalities: MATH . These are the estimates of the Lemma. |
math/0106134 | It suffices to prove the Lemma when all of the functions are positive. We first claim that if MATH or MATH, we have MATH . We use the sequence of exponents defined in REF . By REF applied MATH times, we obtain MATH where the function MATH satisfies MATH and the function MATH satisfies MATH . We write out the integral defining MATH and obtain MATH . We have MATH, MATH and MATH where the exponents MATH, MATH and MATH satisfy MATH . Thus NAME 's inequality and the estimates for MATH and MATH imply that MATH . Since MATH as MATH, REF tells us that MATH. Thus, if MATH, then we obtain REF with MATH. To obtain the result with MATH, we simply need to change variables to replace MATH by MATH. Now that we have the estimate REF , we apply a simple interpolation argument to obtain the Lemma. We define an operator MATH by MATH . According to the estimate REF , MATH maps MATH and MATH. Hence, by the NAME interpolation theorem, MATH is bounded on MATH, with the same bound. This implies the Lemma. |
math/0106134 | We begin by indicating how we construct the solutions MATH when the potential MATH is nice, MATH, say, and MATH. We can write MATH as the infinite sum MATH . We consider one entry in the matrix MATH with MATH. Note that for MATH even, the off-diagonal part MATH is zero. Also, note that the operator MATH involves multiplying by exponentials of modulus REF, and thus the entries of MATH have the same MATH-norm as the entries of MATH. We apply REF with MATH and MATH to obtain that MATH . The last inequality uses the elementary inequality that MATH. A similar argument holds for the remaining entries. Thus the infinite sum in REF will converge in MATH if MATH. We substitute the sum for MATH into REF . Since MATH and MATH are off-diagonal, only the even terms of the series for MATH are needed in the expression for MATH. Also, note that the operator MATH when acting on diagonal matrices. Using these observations and that MATH is off-diagonal, we can write MATH . The term with MATH in this series is essentially the NAME transform. If we consider one entry, we obtain MATH and we have a similar expression for MATH. Thus, MATH by the standard NAME identity. The higher order terms require a bit more work. In order to simplify the notation, we consider the upper-right entry in MATH . We take a sufficiently regular (scalar-valued) test function MATH and write out the expression for MATH to obtain MATH . Note that if each of the functions MATH and MATH are in MATH, say, then the integral on the right of the previous expression converges absolutely. Thus, we may use the NAME theorem to carry out the integration with respect to MATH first and obtain the expression MATH . Here, MATH is the NAME transform of MATH and is defined by MATH. Thus we have that MATH . Now, REF implies for each MATH, there is a constant MATH so that MATH and hence the series MATH converges in MATH if MATH. We now turn to the proof that the map MATH is NAME continuous. Let MATH and MATH be two matrix potentials. We write MATH, then we have MATH where the difference occurs in the MATH-th spot and MATH if MATH is even or MATH if MATH is odd. Thus, REF implies that if MATH and MATH, then for each MATH, there is a constant MATH so that MATH . Since we have MATH, we have the NAME continuity also. The inverse map is handled similarly so we only give a sketch of the argument. The map which takes MATH to MATH is given by MATH where the map MATH depends on MATH and was defined in REF . We have that MATH may be represented as MATH where MATH is the NAME transform MATH acting on matrix valued functions. As before, we can write MATH as the series, MATH . Substituting REF into REF gives a series representation for MATH, MATH where the MATH-th term is given by MATH . Arguing as before, we can show that for each MATH, there is constant MATH so that MATH and also obtain that MATH depends continuously on MATH. The statements that MATH and MATH follow from continuity and the corresponding results for nice potentials as proved by Sung CITE. However, one must replace REF. The uniqueness of MATH needed to carry Sung's arguments holds for MATH, thanks to the estimate for MATH in REF (see also REF ). Also, the extension of the non-linear NAME identity to functions in MATH is immediate from the result for smooth functions and the continuity of the map MATH. |
math/0106135 | Since the spaces MATH and MATH have the same exterior algebra parts of their cohomologies, obviously MATH and from REF it follows that MATH . Thus, the spectral sequence of the fibration collapses. Now the NAME theorem implies that the restriction to the fiber is surjective in real cohomology. |
math/0106135 | Define MATH to be the sum of all the monomials in the MATH which are homogeneous of degree MATH. This is clearly a symmetric polynomial. The relations REF that we have to prove amount to MATH . We shall prove more, namely that MATH holds for all MATH. First we prove that MATH for all MATH. We know MATH because all the symmetric functions in MATH vanish. Thus the vanishing of MATH implies the vanishing of MATH. We now prove REF by induction on MATH. The case MATH is what we just proved. Suppose we have proved the statement up to MATH. Then consider MATH . As soon as MATH both the left hand side and the first summand on the right hand side vanish by the induction hypothesis. Therefore the second summand on the right also vanishes, which is what is to be proved in the inductive step. Having proved the multiplicative relations, it remains to prove the statement about the vector space basis of the cohomology. This can be proved by induction on the degree. A vector space basis for MATH is given by MATH. Suppose we have proved the statement up to degree MATH. Now MATH is linearly generated by all homogeneous monomials of degree MATH in the MATH. However, by induction there are linear relations expressing MATH as a linear combination of monomials containing MATH at most linearly, expressing MATH as a linear combination of monomials containing MATH at most in the second power, and so on up to MATH. We also have a new relation in this degree, namely MATH. This allows us to replace MATH by a linear combination of monomials containing only smaller powers of MATH. We now have eliminated all monomials not listed in REF. The remaining ones must be linearly independent because their number in each degree is seen to equal the respective NAME number by inspection of the NAME polynomial. |
math/0106135 | To prove the relations REF we can proceed as in the proof of REF . To prove the relations REF we will proceed by backward induction on MATH. For MATH the left hand side is MATH, which obviously vanishes. Now suppose we have proved the statement down to MATH. Consider the relation MATH which was proved already. Multiplying it by MATH and splitting the resulting sum into two sums corresponding to the cases MATH and MATH we get MATH . The first sum vanishes by the induction hypothesis. Therefore the second sum vanishes, which is what we need to prove in the inductive step. To prove the statement about the vector space basis of the cohomology we can use the same argument as in the proof of REF . |
math/0106135 | The real cohomology of MATH was calculated by CITE; we use the presentation in CITE. There are two linearly independent generators MATH and MATH, which satisfy the relations MATH and MATH . On the other hand, MATH generates the top-dimensional cohomology MATH. Suppose that MATH admits a formal Riemannian metric. By an obvious abuse of notation, we denote by MATH and MATH the harmonic representatives of the above cohomology classes. Then the above relations for MATH and MATH hold at the level of differential forms. In particular MATH is a volume form on MATH. On the other hand, it follows from REF that both kernel distributions MATH have rank at least MATH. Therefore, we can locally choose linearly independent vector fields MATH and MATH. It follows from REF that MATH. But this implies MATH, contradicting the fact that MATH is a volume form. |
math/0106135 | Note that the first relation in REF, for MATH, gives MATH. Using this, the second relation, for MATH, implies MATH. Assume that, under the assumptions of the lemma, MATH is a volume form on some open subset MATH. Then MATH, but MATH is not zero on MATH. Thus MATH is of constant rank MATH in MATH. By the same argument, MATH is of constant rank equal to MATH or MATH in MATH. The kernel distributions MATH are of ranks MATH and MATH or MATH. As MATH and MATH are closed, the NAME theorem implies that the kernel distributions are integrable. We proceed by induction on MATH. For MATH REF gives the following relation between MATH and MATH: MATH . From the above discussion, the ranks of MATH and MATH are MATH. Thus locally there are linearly independent vectors MATH and MATH. From REF it follows that MATH, which implies MATH. This implies that MATH can not be a volume form anywhere, and therefore vanishes identically. Assume that the statement holds for MATH. Let us consider a manifold MATH of dimension MATH and let MATH be forms on MATH satisfying REF, such that MATH is a volume form on some open subset MATH. Since MATH is integrable, it defines a foliation. Let MATH be a leaf of this foliation. Then MATH is of dimension MATH, and the forms MATH restricted to MATH satisfy relations REF and MATH is a volume form on MATH. This contradicts the induction hypothesis. |
math/0106135 | Assume that the flag manifold MATH is geometrically formal, that is, there is a metric for which all products of harmonic forms are harmonic. If for the classes MATH we choose their harmonic representatives (denoted by the same letters), geometric formality implies that the relations REF hold at the level of differential forms. The dimension of MATH is MATH, and from REF we see that MATH is a volume form on MATH. This gives a contradiction with the above lemma. |
math/0106135 | We proceed as in the proof of REF . The first relation in REF, for MATH, gives MATH. Using this, the second relation, for MATH, implies MATH. If we assume that MATH is a volume form on some open subset MATH, then in MATH we conclude that MATH is of constant rank MATH and MATH is of constant rank equal to MATH, MATH or MATH. So, the kernel distributions MATH are of ranks MATH and MATH, MATH or MATH, and are integrable. We proceed by induction on MATH. For MATH REF gives the following relation between MATH and MATH: MATH . From the above discussion , the rank of MATH is MATH, thus locally there is a non-zero vector field MATH. From REF it follows that MATH, which implies MATH. This implies that MATH can not be a volume form. Assume that the statement holds for MATH. Let us consider a manifold MATH of dimension MATH and let MATH be forms on MATH satisfying REF, such that MATH is a volume form on MATH. Since MATH is integrable, it defines a foliation. Let MATH be a leaf of this foliation. Then MATH is of dimension MATH, and the forms MATH restricted to MATH satisfy relations REF and MATH is a volume form on MATH. This contradicts the induction hypothesis. |
math/0106135 | Assume that MATH or MATH is geometrically formal. If for the classes MATH we choose their harmonic representatives, geometric formality implies that the relations REF hold at the level of differential forms. The dimension of MATH is MATH, and from REF we see that MATH is a volume form on MATH. This contradicts the above lemma. |
math/0106135 | We will proceed by backward induction on MATH. To prove this for MATH note that MATH on MATH and the assumption that MATH is a volume form on MATH implies that MATH has constant rank equal to MATH, so, being a leaf of its kernel foliation, MATH has dimension MATH and MATH is a volume form on MATH. Assume that the lemma has been proved for all MATH. Since MATH is a volume form on MATH we conclude that MATH, for all MATH. As MATH, denote by MATH a distribution complementary to MATH in MATH. Relation REF implies that the form MATH vanishes identically on MATH. If we evaluate this form on MATH vectors from MATH, MATH vectors from MATH on which MATH does not vanish, MATH vectors form MATH on which MATH does not vanish, and so on, and finally MATH vectors from MATH on which MATH does not vanish, we conclude that MATH vanishes identically on MATH. Since MATH, for MATH, the choice of such a vectors is always possible. Since MATH is a volume form on MATH it follows that MATH restricted to MATH has constant rank equal to MATH. Thus, MATH, so MATH is a volume form on MATH. |
math/0106135 | Assume that under the conditions given in the proposition, MATH is a volume form on MATH. Then the above lemma implies that we have the following situation: a manifold MATH of dimension MATH and MATH closed MATH-forms satisfying relations REF such that MATH restricted to MATH has constant rank equal to MATH and MATH is a volume form on MATH. Note that the forms MATH satisfy the relations REF on MATH. We prove by induction on MATH that this situation gives a contradiction. For MATH we have that MATH is a volume on MATH and REF implies that on MATH we have following relations: MATH . As in the proofs of REF this gives a contradiction. Let us assume that the statement holds for all MATH. Consider a manifold MATH of dimension MATH and assume we have MATH-forms MATH satisfying above conditions. Then, obviously, we have on MATH the same situation with MATH two-forms MATH giving the contradiction. |
math/0106135 | Assume that MATH, MATH is geometrically formal. If for the classes MATH we choose their harmonic representatives (denoted by the same letters), geometric formality implies that the relations REF hold at the level of differential forms. The dimension of MATH is MATH, and from REF we see that MATH is a volume form on MATH. This contradicts REF . |
math/0106135 | By REF the cohomology algebra of MATH is the tensor product of a polynomial algebra MATH, which is the cohomology algebra of MATH, and an exterior algebra MATH, which is the cohomology algebra of MATH. The inclusions MATH induce a fibration MATH with fiber MATH. As the base and the total space are generalised symmetric spaces, REF implies that all cohomology classes on MATH are restrictions of classes on MATH. We shall use the basis for MATH used in the proof of REF . Assume that MATH is geometrically formal and identify all the elements of the cohomology algebra with their harmonic representatives. Then the harmonic MATH-forms MATH and MATH on MATH satisfy MATH, but MATH, and therefore have kernels of rank MATH. As the codimension of the fiber MATH in MATH is MATH, it follows that the restrictions of MATH and MATH to MATH have kernels of rank at least MATH everywhere. Thus at every point of a fiber we can find linearly independent local vector fields MATH and MATH contained in the kernels of MATH and MATH respectively. As the restrictions of MATH and MATH to MATH satisfy relation REF, we conclude MATH as in the proof of REF . This shows that the restriction of MATH to MATH vanishes identically. This contradicts the fact that restriction to MATH is surjective in cohomology. |
math/0106135 | To prove the first statement, let us consider the fibration MATH with fiber MATH. The base is a symmetric space, so REF shows that the restriction to the fiber is surjective in cohomology. REF implies that MATH . Assume that MATH is geometrically formal. For the cohomology classes MATH we take their harmonic representatives with respect to a formal metric. Then the relations REF hold for the harmonic forms, as forms. If we restrict these forms to the fiber MATH, REF implies that the form MATH vanishes. This contradicts the fact that the restriction is surjective in cohomology. For the second case, we consider the fibration MATH with fiber MATH, where, as above, restriction to the fiber is surjective in cohomology. Again REF implies MATH and, as in the first case, the assumption of geometric formality for MATH contradicts the fact that the restriction to the fiber is surjective in cohomology. For the third case, we have the fibration MATH with fiber MATH, and we can proceed as above. |
math/0106136 | MATH . Note that MATH . Conversely suppose that MATH . From REF we conclude that MATH . MATH . Choose an element MATH . By hypothesis we know that MATH and so MATH . From MATH we know that MATH and so MATH . MATH . We know that MATH . It is clear that MATH for all MATH . From MATH we conclude that MATH . By hypothesis we know that MATH so MATH . |
math/0106136 | As every circuit with MATH or more elements has a chord the result is clear by induction on the number of elements of the circuits. |
math/0106136 | MATH is a consequence of MATH . MATH . Suppose that MATH has a chord MATH and let MATH be two circuits such that MATH and MATH . From the equivalence MATH we know that MATH . It follows from MATH that MATH . MATH . We will prove it by contradiction. Let MATH be a circuit without a chord and suppose that MATH . From the definitions we know that MATH . For this to be possible, there must exist a circuit MATH such that MATH . Set MATH . As MATH is binary we know that MATH is also a circuit. We conclude that MATH and MATH and so MATH is a chord of MATH a contradiction. |
math/0106136 | MATH is a special case of REF . MATH . From the definitions we know that for all MATH . We conclude from REF that every circuit with MATH elements has a chord, that is, MATH is MATH-chordal. |
math/0106136 | Remark that REF imply REF . So the result is a consequence of REF . |
math/0106137 | Let MATH and MATH be sets of NAME generators of MATH, according to REF . Applying the NAME algorithm we get the following presentation for MATH: MATH . For every MATH and MATH, let MATH and MATH be the longitudes associated to the components of MATH corresponding to the meridians MATH and MATH respectively (as usual we consider longitudes which are homologically trivial in the complement of the relative component). Then we have the relations: MATH and MATH . Introducing MATH and MATH, we obtain the following new presentation for MATH: MATH . Therefore, the fundamental group of MATH admits the presentation: MATH . Since MATH and MATH, there exist certain integers MATH, MATH, MATH and MATH such that MATH and MATH. For MATH and MATH we define MATH . Since MATH and MATH (respectively, MATH and MATH) commute, we have MATH . Using these relations we can eliminate all the generators of the previous presentation of MATH, replacing them with the set MATH. The first four types of relations of the above presentation disappear and the statement is obtained. |
math/0106137 | From REF we see that the link MATH admits a strongly invertible involution MATH whose axis (pictured with dashed line) intersects each component of the link in two points. Thus, in virtue of the Montesinos theorem CITE, the manifold MATH can be obtained as the MATH-fold covering of MATH, branched over some link. Applying the Montesinos algorithm, we get the link depicted in REF . Obviously, this branching set is equivalent to the link presented in REF . |
math/0106137 | Both the link MATH and the surgery coefficients defining the manifold MATH (and so, also the manifold) are invariant with respect to an obvious rotation symmetry MATH of order MATH. Denote by MATH the cyclic group of order MATH generated by this rotation. Observe that the fixed-point set of the action of MATH on MATH is a trivial knot disjoint from MATH. Therefore, we have an action of MATH on MATH, with a knot MATH as fixed-point set. The underlying space of the quotient orbifold MATH is precisely the manifold MATH, which can be obtained by NAME surgery on MATH, with coefficients MATH and MATH, MATH, along the MATH-component link MATH depicted in REF . The components of MATH are unlinked, unknotted, and form a trivial link with MATH components. Therefore the underlying space of the quotient orbifold is homeomorphic to the connected sum of MATH lens spaces MATH (see CITE). Moreover, it is obvious from the action of MATH that the singular set MATH of the quotient orbifold is a knot which does not depend on MATH. |
math/0106137 | From REF we see that MATH admits an invertible involution MATH whose axis intersects each component in two points and the rotation symmetry MATH of order MATH which was discussed in REF . These symmetries induce symmetries (also denoted by MATH and MATH) of the generalized periodic NAME manifold MATH, such that MATH. As mentioned above, MATH induces the symmetry (also denoted by MATH) of the orbifold MATH (whose singular set is given by REF ), and the covering MATH is given by REF . The covering MATH is given by REF . As we see from REF , MATH induces the strongly invertible involution (also denoted by MATH) of the link MATH. Using the Montesinos algorithm we see that MATH (note that the part of the singular set of MATH having index MATH can be obtained as a connected sum of MATH two-bridge links corresponding to the rational tangles MATH). |
math/0106137 | From REF , MATH is the MATH-fold cyclic covering of MATH, branched over a knot MATH. REF - REF shows how to deform MATH to a NAME 's normal form of a two-bridge knot with NAME parameters MATH by ambient isotopy (from REF to REF ) and surgery calculus CITE (from REF to REF ). |
math/0106137 | From REF , the group MATH is generated by the MATH elements MATH and has relations of two types: MATH where MATH and MATH, and all the indices are taken mod MATH and MATH respectively. Denote MATH and MATH for MATH and MATH. Then we have MATH relations of the two following types: MATH and MATH . Therefore, the defining relations for the group are: MATH and MATH for MATH. Denoting MATH, MATH, we will eliminate all other generators in the following order: MATH according to the above formulae. At the end of this process we will get MATH relations arising from MATH. That completes the proof. |
math/0106139 | We claim that the span of MATH is two - dimensional. Assume for some pair of integers MATH as integer homology classes. Hence we can find a MATH - chain MATH in MATH such that MATH forms a MATH - cycle in MATH. We test it against MATH and MATH which yields the same number since these represent the same homology class. By the assumption, MATH and therefore MATH . Thus MATH, verifying the statement. Hence, MATH, which excludes that MATH was the NAME bottle. |
math/0106139 | By a result of NAME in CITE there exists a symplectic isotopy of MATH which maps the MATH-holomorphic spheres MATH to projective lines. Hence the image of MATH under this map, which we denote by abuse of notation by MATH again, is a Lagrangian in MATH in the complement of MATH complex lines, without loss of generality, MATH. The lines correspond to MATH and MATH, respectively. Hence the intersection numbers with the symplectic disks show that the induced homomorphism MATH is an isomorphism. But MATH symplectomorphically. The standard NAME torus MATH is thereby mapped to the zero section of the cotangent bundle MATH. Moreover, the Lagrangian embedding MATH induces an isomorphism MATH. By a result of CITE it follows that MATH is smoothly isotopic to MATH in MATH and hence the original MATH is smoothly isotopic to the NAME torus. |
math/0106139 | The first statement is obvious since there are no contractible closed geodesics for a flat structure on MATH. Let MATH be a two - punctured holomorphic curve with asymptotics MATH which intersects MATH. Since geodesics parallel to MATH and MATH foliate MATH there is a MATH with MATH parallel to them which intersects MATH in MATH. Unless MATH coincides with MATH (in which case we are done) they intersect in a discrete set of points with a finite positive (algebraic) intersection index in each of them see CITE. Pick one of these points then it will persist if we change MATH slightly. Thus we may assume that MATH. There is an involution on MATH which preserves all geodesics parallel to MATH but reverses their orientations. That gives rise to a canonical involution on MATH which preserves MATH. Thus we obtain a curve MATH with asymptotics MATH. Since MATH we obtain a MATH - cycle in MATH via MATH which non-trivially intersects MATH. This is impossible, since the latter can be moved away from the zero section through shifting it by a nowhere vanishing MATH - form and MATH is, of course, generated by MATH. For the third statement we perform an explicit deformation of MATH which intersects MATH once transversally. There is a one form MATH which vanishes on the geodesics parallel to MATH and intersects with MATH in one geodesic MATH perpendicular to MATH. Thus MATH and MATH intersect in MATH, and this intersection can be made easily transversal. To prove the last assertion notice first of all that the two simple odd closed geodesics of the flat NAME bottle are not homotopic. Hence, if a two - punctured holomorphic curve MATH is asymptotic to an odd geodesic MATH at one puncture it has to be asymptotic to MATH at the other. Therefore, MATH forms a MATH - cycle in MATH which intersects MATH homological trivially. Thus, due to the previous fact the intersection index of MATH and MATH with MATH - coefficients is non-trivial. From the second assertion we conclude that MATH. |
math/0106139 | There is a compact set MATH such that MATH for all MATH sufficiently large. Since MATH converge in MATH, the bundle MATH does. On the other hand MATH is trivialized over MATH containing all MATH for MATH. By the choice of MATH we may define NAME numbers for MATH, which are equal to the NAME number of MATH, MATH by that remark. The second statement is a corollary of the first. We find that MATH . Here MATH denotes the number of connected components of the first level MATH. The second line is due to REF and the fact, that MATH, the third line uses that the projection to MATH, MATH provides a MATH - chain in MATH whose boundary is MATH therefore providing the cancellation of the NAME indices in the formula. Finally the last inequality uses the fact that there are no contractible geodesics in MATH. Therefore the NAME characteristic of each connected component of MATH is non - positive. Equality holds therefore if all components of MATH consist of annuli. But notice that there the only holomorphic cylinders in MATH (punctured holomorphic spheres with one positive and one negative puncture), due to the simple fact that closed geodesic (NAME orbits) which are homotopic have the same length (action). Therefore, in this case MATH consist of MATH - punctured holomorphic spheres in MATH. The last statement follows from the general principle we will use in the proof of the main result. Fix a non - singular point MATH on the top level, and a complex line MATH. For any MATH there is a (unique) MATH - holomorphic sphere MATH through MATH and tangent to MATH representing the generator of MATH. For a subsequence of MATH of the compactness statement, we have convergence of the MATH in that sense, to a broken holomorphic sphere MATH passing through MATH. Moreover, either the image is tangent to MATH or it has a cusp singularity at MATH. In both cases the images of the two punctured holomorphic spheres do not coincide. Hence there is a neighborhood MATH of MATH such that MATH and they intersect at MATH with a finite algebraic intersection index (see CITE), which is equal to the product MATH of the multiplicities of the curves at MATH. For MATH large enough the sum of intersection indices of points in MATH is equal to that same number. Since each further point of MATH would contribute a positive number to the total sum of intersection indices and the algebraic intersection index of MATH is MATH we conclude that MATH . |
math/0106141 | From REF , for any immersion MATH, there exists a unique frame MATH such that MATH where MATH. It is sufficient to construct an adapted polynomial Killing field MATH, MATH such that MATH . There exists a regular algebraic description of these polynomial Killing fields through a formal Killing field (see CITE,CITE), which is in our case a MATH-valued formal power series solution MATH of MATH, where MATH . Since for all MATH, MATH is again a formal Killing field in MATH. The diagonal terms MATH solve the elliptic equation MATH . If MATH is a doubly-periodic solution of REF: MATH then only the finite vector fields MATH are linearly independent. All MATH are also doubly-periodic. REF determines an elliptic linear operator MATH on the torus MATH: MATH . By the linear elliptic theory, the compactness of the torus implies that the spectrum of this operator is discrete and all the eigenspaces are finite dimensional. All vector fields MATH belong to the kernel of MATH. This observation proves the lemma. Thus it follows from the proof of REF that such a polynomial Killing field MATH exists. Now it is sufficient to prove the existence of a formal Killing field MATH, MATH of MATH. Following the standard procedure in CITE we get a formal vector field MATH with values in MATH satisfying MATH. Since MATH, MATH also satisfies the same equation. Setting MATH we obtain a formal Killing field of MATH as required, which completes the proof of REF . |
math/0106147 | If MATH is smooth, set MATH, where MATH is a large positive integer. Then we have MATH. By the NAME Index Theorem it follows that MATH hence MATH, since by our assumptions MATH. So equality holds in the Index Theorem and therefore MATH and MATH are proportional: MATH for some rational number MATH. Since MATH and MATH, we conclude that MATH. That ends the proof in the smooth case. If MATH is singular, let MATH be a desingularisation of MATH and let MATH be a component of MATH which maps surjectively onto MATH. Note that the fiber of MATH need no longer be connected and consider the NAME factorisation MATH . It follows immediately from the construction that MATH, that MATH has degree REF on MATH and on the general fiber of MATH. The argumentation above therefore yields that MATH is trivial on MATH. The claim follows. |
math/0106147 | Fix a very ample line bundle MATH and set MATH. Let MATH be a general member cut out by MATH elements of MATH. The dimension of MATH will thus be MATH, and since MATH, the restriction MATH is big (and nef). Consequence: MATH, since otherwise MATH would be covered by curves MATH which are contained in general fibers of MATH, so that MATH, contradicting the bigness of MATH. In particular, we have MATH and our claim is shown. |
math/0106147 | Since MATH is nef, it carries a singular metric with positive curvature current MATH in particular MATH . Since MATH and since every MATH-form MATH on MATH is closed, we conclude that MATH since MATH for a fiber MATH of MATH. Hence MATH. Let MATH be the maximal open subset of MATH over which MATH is a submersion and let MATH. Then by CITE MATH where MATH are fibers of MATH and MATH. Hence MATH, an analytic set of dimension REF. Hence by a classical theorem (see for example, CITE), MATH where MATH are irreducible components of MATH, that is, of fibers, of dimension REF and where MATH . Since MATH, NAME 's lemma shows that MATH is a multiple of fibers, whence MATH for a suitable MATH-divisor MATH on MATH. Here is an algebraic proof, which however does not easily extend to higher dimensions as the previous (see REFEF). Fix an ample line bundle MATH on MATH and choose a positive rational number MATH such that MATH . Let MATH . Then MATH and MATH . Introducing MATH and MATH, the ample line bundles MATH and MATH fulfill the equation MATH hence MATH and MATH are numerically proportional, and so do MATH and MATH. Consequently MATH as before. |
math/0106148 | The generating function MATH can be written as follows: MATH . For the proof, we have to show that it is possible to change the order of the summation. So it is sufficient to prove that for any MATH converges absolutely. It can be proved by virtue of the idea of CITE. Put MATH for MATH and MATH. Making use of the inequality MATH we have MATH . Let MATH be in a compact set which dose not involve positive integers. Then there exists a positive constant MATH such that MATH . Hence MATH . |
math/0106148 | We use the partial-fractions expansion: MATH CASE: We set MATH by MATH . Then using REF we have MATH . CASE: Using REF, it can be proved in the same manner as REF. CASE: We set MATH and MATH by MATH . Then MATH . We divide the range of sum of the third term into two parts as MATH . The later sum is equal to zero because of MATH. Thus we have MATH . CASE: Repeating shift of MATH, we have MATH . CASE: Similarly as in the previous cases, MATH . |
math/0106148 | CASE: Applying REF succesively MATH CASE: Use REF MATH times, and REF . Remaining relations and the relations of MATH's can be proved quite similarly. |
math/0106151 | In obtaining MATH from MATH in NAME, we do not allow deletion of cuts that were active at the solution MATH of REF . Using MATH and MATH to denote the active rows in MATH and MATH, we have that MATH is also the solution of the following linear program (in which the inactive cuts are not present): MATH . The subproblem to be solved for MATH differs from REF in two ways. First, additional rows may be added to MATH and MATH, consisting of function values and subgradients obtained at MATH and also inactive cuts carried over from the previous REF . Second, the trust-region radius MATH may be smaller than MATH. Hence, the feasible region of the problem to be solved for MATH is a subset of the feasible region for REF , so the optimal objective value cannot be smaller. |
math/0106151 | The first equality follows immediately from REF , while the second REF follows immediately from REF . We now prove REF . Consider the following subproblem in the scalar MATH: MATH . Denoting the solution of this problem by MATH, we have by comparison with REF that MATH . If MATH is feasible in REF , we have from REF that MATH . Therefore, when MATH is feasible for REF , we have from REF that MATH so that REF holds in this case. When MATH is infeasible for REF , consider setting MATH (which is certainly feasible for REF ). We have from REF , the definition of MATH, REF that MATH underestimates MATH, and convexity of MATH that MATH . Therefore, using REF , we have MATH verifying REF in this case as well. |
math/0106151 | We prove the result by showing that the value MATH has the desired property, where MATH is from REF and MATH is from REF . Suppose for contradiction that there are indices MATH and MATH such that MATH . Since the trust region can be reduced by at most a factor of MATH by Procedure Reduce-MATH, there must be an earlier trust region radius MATH (with MATH) such that MATH and MATH in REF , that is, MATH . By applying REF , and using REF , we have MATH where the last equality follows from MATH and REF . By combining REF with REF , we have that MATH . By using standard properties of subgradients, we have MATH . By combining this expression with REF , and using REF again, we obtain that MATH . However, since MATH, we have from REF that MATH, giving a contradiction. |
math/0106151 | Suppose for contradiction that the none of the minor iterations following MATH satisfies either REF or REF ; that is, MATH . It follows from this bound, together with REF and Procedure NAME, that none of the cuts generated at minor iterations MATH is deleted. We assume in the remainder of the proof that MATH and MATH are generic minor iteration indices that satisfy MATH . Because the function and subgradients from minor iterations MATH, MATH are retained throughout the major iteration MATH, we have MATH . By definition of the subgradient, we have MATH . Therefore, from REF , it follows that MATH so that MATH . Since MATH, we have from REF that MATH. Therefore, from REF and the fact that MATH, we have that MATH. It follows from REF that MATH . Since MATH is rejected by the test REF , we have from REF that the following inequalities hold: MATH . By rearranging this expression, we obtain MATH . Consider now all points MATH satisfying MATH . Using this bound together with REF , we obtain MATH . By combining this bound with REF , we find that the following bound is satisfied for all MATH in the neighborhood REF : MATH . It follows from this bound, in conjunction with REF , that MATH (the solution of the trust-region problem with model function MATH) cannot lie in the neighborhood REF . Therefore, we have MATH . But since MATH for all MATH, it is impossible for an infinite sequence MATH to satisfy REF . We conclude that REF must hold for some MATH, as claimed. |
math/0106151 | Suppose for the moment that the inner iteration sequence is infinite, that is, the test REF always fails. By applying REF recursively, with any constant MATH satisfying the properties stated in REF , we can identify a sequence of indices MATH such that MATH . When MATH, we have from REF that MATH so the right-hand side of REF is strictly positive. Hence for MATH sufficiently large, we have that MATH . But this inequality contradicts REF , proving REF . For the case of MATH, there are two possibilities. If the inner iteration sequence terminates finitely at some MATH, we have MATH and indeed that MATH . Because of REF , we have that MATH for all MATH in a neighborhood of MATH, implying that MATH. Therefore, termination under these circumstances yields a guarantee that MATH. When the algorithm does not terminate, it follows from REF that MATH. By applying REF , we verify our REF of monotonic convergence. |
math/0106151 | If the claim does not hold, there are two possibilities. The first is that the sequence of major iterations terminates finitely at some MATH. However, REF ensures, however, that the minor iteration sequence will terminate at some new major iteration MATH under these circumstances, so we can rule out this possibility. The second possibility is that the sequence MATH is infinite but that there is some MATH and an infinite subsequence of indices MATH such that MATH . Since the sequence MATH is infinite, decreasing, and bounded below, it converges to some value MATH. Moreover, since the entire sequence MATH is monotone decreasing, it follows that MATH and therefore MATH . Hence, by boundedness of the subgradients (see REF ), we can identify a constant MATH such that MATH . It follows from REF that MATH . For each major iteration index MATH, let MATH be the minor iteration index that passes the acceptance test REF . By combining REF with REF , we have that MATH . Since MATH, we deduce that MATH . By REF , we have MATH which contradicts REF . We conclude that the second possibility (an infinite sequence MATH not converging to MATH) cannot occur either, so the proof is complete. |
math/0106151 | We show first that the algorithm cannot ``get stuck" at a particular MATH, generating an infinite sequence of minor iterations at MATH without eventually satisfying either REF or the acceptance test REF . We see from the reasoning in the proof of REF together with the monotonicity property of REF that an infinite sequence of minor iterations must satisfy that MATH . Since the right-hand side of REF is bounded below by MATH, the test REF must be satisfied for some MATH. Therefore, the minor iteration sequence cannot be infinite. Now consider the other possibility of an infinite sequence of major iterations MATH. Since we have MATH for all MATH and MATH, and since the acceptance test REF is satisfied at all MATH, we have MATH . But this relation is inconsistent with the fact that MATH is bounded below (by MATH), so this possibility can also be ruled out, and the proof is complete. |
math/0106151 | Consider first what happens in ATR before the first function evaluation is complete. Up to this point, all the iterates MATH in the basket are generated in the basket-filling part and therefore satisfy MATH, with MATH. When the first evaluation is completed (by MATH, say), it trivially passes the test to be accepted as the new incumbent. Hence, the first noninfinite incumbent value becomes MATH, and by definition we have MATH. Since all later incumbents must have objective values smaller than this first MATH, they all must lie in the level set MATH, proving our first statement. All points MATH generated within actoncompletedtask lie within a distance MATH either of MATH or of one of the later incumbents MATH. Since all the incumbents, including MATH, lie in MATH, we conclude that the second claim in the theorem is also true. |
math/0106151 | We select any MATH and prove that MATH. Since MATH and MATH have the same parent incumbent (MATH, say), no new incumbent has been accepted between the generation of these two iterates, so the wholesale cut deletion that may occur with the adoption of a new incumbent cannot have occurred. There may, however, have been a call to Model-Update-MATH. The first ``else if" clause in NAME would have ensured that cuts active at the solution of trsub-MATH were still present in the model when we solved trsub-MATH to obtain MATH. Moreover, since no new incumbent was accepted, MATH cannot have been increased, and we have MATH. We now use the same argument as in the proof of REF to deduce that MATH. |
math/0106152 | The maps MATH induce an inverse. |
math/0106152 | Apply the small object argument CITE. |
math/0106152 | First suppose that MATH is a MATH-fibration. The generating trivial cofibrations are contained in MATH, so MATH is also a fibration. Now we must show that MATH is also a MATH-equivalence. Consider test diagrams of the form MATH where MATH. Let MATH be a constant map with image MATH, and let MATH be the fiber of MATH over MATH. Since lifts exist in the test diagram, MATH for MATH. Using the long exact sequence of homotopy groups of a fibration, it follows that MATH is a MATH-equivalence. Now suppose that MATH is a fibration and MATH-equivalence. It follows from the long exact sequence of homotopy groups that MATH for MATH, where MATH is any fiber of MATH. There are lifts in the test diagrams shown above for MATH because the obstructions to such lifts belong to MATH. Hence MATH is a MATH-fibration. |
math/0106152 | Since trivial fibrations are MATH-fibrations, MATH-cofibrations are also cofibrations. Suppose that MATH is a relative MATH-cell complex CITE. Then MATH is a weak equivalence since MATH is obtained from MATH by gluing along maps of the form MATH. Note that MATH is a MATH-equivalence, so MATH is also a MATH-equivalence. Arbitrary MATH-cofibrations are retracts of relative MATH-cell complexes CITE, so all MATH-cofibrations are MATH-equivalences. Conversely, suppose that MATH is a cofibration and MATH-equivalence. We show that MATH is a MATH-cofibration by demonstrating that it has the left lifting property with respect to all maps that are both fibrations and MATH-equivalences. By REF , this means that MATH has the left lifting property with respect to all MATH-fibrations. Factor MATH as MATH where MATH is a trivial cofibration and MATH is a fibration. Note that MATH is also a MATH-equivalence. Let MATH be a map that is both a fibration and MATH-equivalence, and consider a diagram MATH . The dashed arrow exists because MATH is a trivial cofibration and MATH is a fibration. This gives the diagram MATH . There is no obstruction to lifting over MATH since MATH is surjective. Obstructions to finding a lift over the higher relative skeletons of MATH belong to MATH, where MATH is some fiber of MATH. These obstructions lie in the image of MATH, where MATH is a fiber of MATH. For every MATH, either MATH or MATH is zero. Hence there are no obstructions to lifting. |
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