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math/0010155 | First note that MATH is a bounded operator for MATH which is given by the integral MATH if MATH . This gives an estimate MATH and shows that the integral in REF converges as a NAME integral. It is clear that we only need establish the formula if MATH (see REF ) for some MATH . To do this we compute MATH . Now using the Dominated Convergence Theorem we obtain REF . |
math/0010155 | Necessity follows immediately from REF for the functions MATH . Conversely, by REF , if MATH where MATH we obtain by REF MATH independent of MATH . This implies that MATH . |
math/0010155 | As before we consider MATH so that MATH . It suffices to show MATH . Referring to REF with some fixed MATH and MATH for MATH with MATH, we obtain the estimate for MATH: MATH where MATH . Suppose MATH is the MATH-boundedness constant of MATH . Then MATH . Hence by REF we have MATH . |
math/0010155 | We define MATH and note that MATH this follows easily from the integral representation REF . Our conditions and REF ensure that MATH . It is only necessary to check that this implies MATH and of course MATH . But this follows directly from REF and the remarks thereafter. |
math/0010155 | Let us assume that MATH is NAME with constant MATH where MATH, and that MATH . We will show that MATH admits a MATH-calculus. We use REF . Fix some MATH . We can assume that there exists MATH so that MATH and so that MATH admits a MATH-calculus for some MATH . Now suppose MATH and MATH . Then for any MATH and MATH we have MATH . By the resolvent equation, MATH . Since MATH has a MATH-calculus we can define MATH and note that MATH where MATH is independent of MATH . Thus, by the NAME of MATH . It follows that MATH and this gives MATH . Combined with a similar estimate for MATH we obtain the result by using REF . |
math/0010155 | It is shown in CITE that MATH and in REF, that MATH . Hence we can apply REF . |
math/0010155 | CASE: Assume that MATH admits a MATH-calculus. Suppose MATH . Suppose MATH and let MATH . We then can argue by REF that MATH independent of MATH . Hence by REF the family MATH is NAME with constant bounded independent of MATH . Now by REF it follows that if MATH then MATH is NAME. Indeed for MATH and MATH for MATH we have MATH . It follows that MATH . Now clearly MATH and so REF follows from REF is very similar and we omit it. CASE: Here we use REF . Suppose MATH admits a MATH-calculus and suppose MATH . We show that the sequence MATH is NAME with constant independent of MATH . To do this we note that if MATH . Let MATH . Let MATH . We observe that MATH independent of MATH by REF . Applying REF yields that MATH is NAME with constant independent of MATH . But this implies that MATH is also NAME with constant independent of MATH and hence (taking limits) so is MATH . A similar argument for MATH and an application of REF shows that MATH . Hence MATH . The proof is finished as in REF . |
math/0010155 | Adapt the proof of REF using the fact that the set MATH is NAME, again by REF . |
math/0010155 | We need only observe that if MATH then the family MATH is NAME and then apply REF . |
math/0010155 | Choose MATH with MATH and MATH . The function MATH is in MATH and the set MATH for MATH is an NAME family. Applying REF we have MATH . We can see this implies REF either by applying REF or by the following simple direct argument based on our construction of the functional calculus. Defining MATH as in REF we note that MATH and MATH are bounded operators since MATH for any MATH . Now if MATH we have MATH . Thus MATH where MATH . Letting MATH yields the result. Now assume MATH has property MATH. For MATH and MATH consider the functions MATH . Note that MATH . Since MATH is bounded uniformly for MATH and MATH and also MATH this collection of operators in NAME. Now, REF yields that the set MATH is NAME for MATH . |
math/0010155 | Since MATH has (UMD), we have that MATH is MATH-sectorial and MATH (see for example, CITE). It is easy to see that MATH so the result is a consequence of REF . |
math/0010155 | This is just a version of a classical result of CITE on uniqueness of unconditional bases in MATH . Let MATH . Then for some constant MATH independent of MATH we have if MATH for MATH by REF . Now since MATH has cotype REF, there exists a constant MATH independent of MATH so that MATH . Now suppose MATH and MATH and choose by the NAME theorem MATH with MATH and MATH . Consider the map MATH defined by MATH . By REF MATH is bounded and MATH . Hence for some constant MATH we have MATH . Now MATH . If we integrate for MATH we obtain the right-half of REF (compare REF ). The left-half follows from the equation: MATH . |
math/0010155 | We use the notation of REF . In particular notice that REF implies the existence of an isomorphic embedding MATH defined by MATH . Fix MATH . Then MATH maps MATH isomorphicly onto MATH (with the graph norm). Let MATH; then MATH is an infinite-dimensional reflexive subspace of MATH. Then (since MATH contains no copy of MATH) the set MATH is equi-integrable. We show that this implies that the (bounded) operator MATH satisfies a lower bound on MATH . Indeed if not there exists a sequence MATH in MATH so that MATH and MATH . But, by the resolvent equation, MATH for all MATH . Now by REF and equi-integrability, we have MATH which gives a contradiction. Now applying the same argument to MATH gives a lower bound on MATH on MATH . Thus MATH has a lower bound on MATH . Since MATH this implies the result. |
math/0010155 | Let us assume that MATH is a NAME of cotype REF, the other case is similar. By REF if MATH we can write MATH and this leads immediately to MATH which implies the boundedness of MATH . |
math/0010155 | Let us prove this for MATH a NAME of cotype REF as the other case is dual. Suppose MATH and MATH . If MATH then for MATH by REF MATH . Now suppose MATH . We will estimate the norm of MATH where MATH . If we let MATH then we have an estimate MATH where MATH and so MATH . Now if MATH with MATH we have MATH . Hence the left-hand side is estimated by MATH where MATH is the NAME maximal function. Hence MATH . However REF implies that we have an estimate MATH . Substituting in we have MATH and since MATH is bounded on MATH this establishes a uniform bound on the operators MATH . Letting MATH yields the result. |
math/0010156 | If the result is false we can clearly inductively construct an infinite normalized block basic sequence MATH so that there is no constant MATH so that for all finitely nonzero sequences MATH we have: MATH . It therefore suffices to show that REF holds for every normalized block basic sequence MATH . We can clearly then suppose MATH. We next use REF and CITE combined with CITE (see also CITE p. REF) that, since MATH has nontrivial type for every MATH there exists MATH so that any subspace MATH of MATH with dimension MATH has a subspace MATH of dimension MATH which is REF-complemented in MATH and MATH-isomorphic to MATH . Assume REF is false. Then we can inductively find a sequence MATH and an increasing sequence MATH with MATH so that MATH for MATH, MATH and either MATH or MATH . In order to create new NAME decompositions of MATH, we will need the following elementary lemma, that we state without a proof: Let MATH be a NAME decomposition of a NAME space MATH. Assume that each MATH has a finite NAME decomposition MATH with a uniform bound on the decomposition constant. Then MATH is also a NAME decomposition of MATH . We denote the induced decomposition by MATH . Now by REF which has dimension at least MATH contains a subspace MATH which is MATH-Hilbertian and MATH-complemented in MATH . Let MATH be the complement of MATH in MATH by the projection of norm REF. At the same time MATH is REF-complemented (by the NAME theorem) in MATH for MATH and let MATH be its associated complement. We thus have a new NAME decomposition: MATH . If we write MATH then we have a NAME decomposition MATH . Next select a normalized basis MATH of MATH which is MATH-equivalent to the canonical basis of MATH . It is easy to see that we can obtain a new NAME decomposition by interlacing the MATH with the MATH that is : MATH . Now again using REF we can form two further decompositions: MATH and MATH . Now we can apply REF . If we use decomposition REF we note that MATH and so for a suitable MATH and all MATH . However, using decomposition REF there is also a constant MATH so that MATH . This leads to an estimate: MATH . If we use decomposition REF instead we obtain an estimate: MATH . Combining gives us REF and completes the proof. |
math/0010156 | Using REF (compare CITE) we can block the given (FDD) to produce an (FDD) MATH so that MATH and MATH are both (UFDD)'s. Let us denote, as in REF , the dual (FDD) of MATH by MATH . Now it follows applying REF to both MATH and MATH (which also has (MRP)) that there exists a constant MATH so that if MATH and MATH are two finitely nonzero sequences MATH for MATH . Hence MATH . Now for given MATH we may find MATH with MATH and MATH . Let MATH (where MATH is the projection associated with the FDD MATH). Then MATH where MATH . Hence if MATH is finitely nonzero, we have MATH . Thus MATH so that we obtain the lower estimate: MATH . This completes the proof. |
math/0010156 | Let MATH be the given (FDD) of MATH . We will show first that there is a blocking MATH of MATH which satisfies an upper MATH-estimate that is, if there is a constant MATH so that if MATH is block basic with respect to MATH and finitely non-zero then MATH . Once this is done, the proof can be completed easily. Indeed if MATH is the dual decomposition to MATH for MATH then we can apply the fact that MATH also has (MRP) (MATH is reflexive) to block MATH to obtain a decomposition which also has an upper MATH-estimate. Thus we can assume MATH and MATH both have an upper REF-estimate and then repeat the argument used in REF to deduce that MATH . Since MATH necessarily has type MATH, we can apply REF and assume MATH obeys REF . We now introduce a particular type of tree in the space MATH . Let MATH for MATH be the sub-algebra of the NAME sets of MATH generated by the dyadic intervals MATH for MATH . Let MATH denote the conditional expectation operator MATH . We will say that a tree MATH is a martingale difference tree or (MDT) if CASE: each MATH is MATH measurable, CASE: if MATH then MATH CASE: there exists MATH so that if MATH then MATH . In such a tree the partial sums along any branch form a dyadic martingale which is eventually constant. We will prove the following Lemma: There is a constant MATH so that if MATH is a weakly null (MDT), there is a full subtree MATH so that for any branch MATH we have: MATH . For each MATH we define integers MATH and MATH. If MATH we set MATH to be the greatest MATH so that MATH and MATH to be the least MATH so that MATH . If MATH we set MATH and MATH; if MATH where MATH we set MATH to be the last member of MATH and MATH . Since MATH is weakly null we have MATH for every MATH . It is then easy to pick a full subtree MATH so that MATH whenever MATH . Now let MATH . Then MATH . For any branch MATH of MATH, we have that MATH is a block basic sequence with respect to MATH for every MATH . Hence MATH . Integrating again we have MATH . From this we get MATH . Estimating the last term by the NAME inequality and using the fact that MATH is (UMD) we get the Lemma. Now we introduce a functional MATH on MATH by defining MATH to be the infimum of all MATH so that for every weakly null (MDT) MATH with MATH we have a full subtree MATH so that for any branch MATH . Note that since MATH we have an estimate MATH. By considering the null tree we have MATH . It is clear that MATH is continuous and MATH-homogeneous. Most importantly we observe that MATH is convex; the proof of this is quite elementary and we omit it. It follows that we can define an equivalent norm by MATH and MATH for MATH . Next we prove that if MATH and MATH is a weakly null sequence then MATH . We first note that we can suppose MATH and MATH all exist. Now suppose MATH . Then we can find weakly null (MDT)'s MATH with MATH so that for every full subtree MATH we have a branch MATH on which: MATH . In fact by easy induction we can pick a full subtree so that REF holds for every branch. Hence we suppose the original tree satisfies REF for every branch. Similarly we may find weakly null (MDT)'s MATH with MATH and for every branch MATH . We next consider the (MDT) defined by MATH, MATH and if MATH then MATH . Now for every branch of the (MDT) MATH with initial element MATH we have MATH . However, from the definition of MATH it follows that there exists MATH so that if MATH we can find a branch MATH whose initial element is MATH and such that MATH . Combining gives the equation for MATH, MATH . This proves REF . But note that if MATH is weakly null we have MATH and so we deduce: MATH . Using this equation it is now easy to block the NAME decomposition MATH to produce a NAME decomposition MATH with the property that for any MATH if MATH and MATH then MATH where MATH are chosen to be decreasing and so that MATH . Next suppose MATH is any finitely non-zero block basic sequence with respect to MATH . By an easy induction we obtain for MATH: MATH . Hence MATH . This establishes REF and as shown earlier this suffices to complete the proof. |
math/0010156 | It follows clearly from the work of CITE, that if MATH then MATH is a regularity pair. So let now MATH be a regularity pair. Since MATH does not have (MRP) CITE, we have that MATH. Then, solving our NAME problem with MATH, we obtain that MATH. Thus we can limit ourselves to the case MATH and MATH. Then by the closed graph Theorem, for any MATH so that MATH is the infinitesimal generator of a bounded analytic semigroup on MATH, there is a constant MATH such that for any MATH: MATH . Using the inclusion MATH for MATH, we can now state the following analogue of REF : Let MATH be a NAME decomposition of MATH. Assume that MATH is a regularity pair. Then there is a constant MATH so that whenever MATH are such that MATH then MATH . Then our first step will be to show that the NAME system satisfies some lower-REF estimates in MATH in the following sense: If there exists MATH such that MATH is a regularity pair, and if MATH or MATH then there is a constant MATH such that for any normalized block basic sequence MATH of MATH and for any MATH in MATH: MATH . We first observe that if MATH it follows from the work of CITE on MATH-stable random variables that there is a sequence MATH in MATH which is equivalent to the canonical basis of MATH in any MATH for MATH. Thus MATH is weakly null in MATH, and by a gliding hump argument, we may assume that MATH is actually a block basic sequence with respect to the NAME basis. If MATH then the NAME functions already form a block basic sequence in every MATH for MATH . Now assume the Lemma is false. We pick a normalized block basic sequence MATH of MATH and MATH in MATH so that MATH . Then pick MATH such that MATH is a block basic sequence of MATH. By induction, we pick a normalized block basic sequence MATH of MATH, MATH in MATH and MATH so that MATH is a block basic sequence of MATH and MATH . So we can find MATH and MATH two sequences of finite intervals of MATH such that MATH is a partition of MATH and for all MATH, MATH and MATH. Then set MATH . Then MATH is an unconditional NAME decomposition of MATH. Each MATH can be decomposed into MATH, where MATH, MATH and the corresponding projections are uniformly bounded. So, by REF , MATH is a NAME decomposition of MATH. We can now make use of REF . If we decompose MATH in MATH, we obtain that there is a constant MATH such that for all MATH: MATH . Since MATH is an unconditional basic sequence in MATH, there is a constant MATH so that for all MATH: MATH which is in contradiction with our construction. We now conclude the proof of REF . The NAME basis of MATH has a block basic sequence equivalent to the standard basis of MATH . Hence REF shows that MATH whenever MATH or MATH . Thus MATH . |
math/0010157 | Recall the integral representation: MATH where MATH is the following contour on the complex plane MATH Therefore MATH . Notice that because MATH for MATH the first MATH . NAME coefficients of MATH do not change if one takes MATH as lower limit in summation. Now after making a change of variables MATH, .,MATH, MATH and noticing that MATH in appropriate distributional sense one arrives at REF. We leave details to the interested reader. Similar formulas can be found in CITE. Alternatively one can notice that both MATH and MATH satisfy MATH-st order differential equation MATH. It follows from expansions MATH and MATH as MATH that we have two MATH -tuples of linearly independent solutions, which therefore must be related by a linear transformation. |
math/0010157 | It follows from MATH . |
math/0010157 | It follows from REF that MATH . Therefore one has MATH for some MATH. On the other hand, MATH . Therefore MATH . The coordinates MATH were chosen so that MATH is linear in MATH. Therefore in these coordinates MATH. |
math/0010157 | One has for MATH from any sector MATH, MATH and MATH: MATH where sets MATH,MATH are frames of relative cycles in MATH, MATH correspondingly. Expansion for the integrals MATH, MATH at MATH can be evaluated using the steepest-descent method. Recall that given a non-degenerate critical point MATH of MATH and a metric one has two relative cycles MATH, MATH from MATH and MATH respectively which are formed by gradient lines MATH of MATH such that MATH as MATH and MATH as MATH respectively. Then one has the following asymptotic expansions MATH . Recall that MATH has MATH non-degenerate critical points MATH, MATH: MATH and that the relative cycles MATH, (respectively, MATH ) form basis in MATH, (respectively, MATH). It follows that MATH . |
math/0010157 | REF follows from MATH. Notice that for any MATH: MATH . It follows that MATH is symmetric. NAME of MATH follows from the analogous property of MATH. Notice that MATH is obviously MATH-linear and therefore one has induced pairing MATH with values in MATH on elements of MATH. It follows from REF that MATH . By definition of the coordinates MATH one has MATH as an element of MATH. |
math/0010157 | Notice that MATH is locally constant with respect to the NAME connection MATH. Therefore MATH . In the coordinates MATH one has MATH . The pairing MATH satisfies MATH . |
math/0010157 | It follows from MATH. The arguments are similar to the arguments from the proof of REF . |
math/0010157 | It follows from MATH and MATH respectively. |
math/0010157 | One has MATH, MATH. Therefore MATH . Differentiating this equality with respect to MATH and using REF we see that MATH. |
math/0010157 | REF are easy to check. In order to check REF let us notice that the subspace MATH is generated by elements MATH and that MATH . To check REF let us notice that since MATH is dual to locally constant basis MATH and MATH are single-valued sections therefore for any MATH the pairing MATH does not depend on MATH: MATH . Differentiating this equality with respect to MATH one gets MATH . Therefore MATH for MATH. |
math/0010157 | One has MATH . Therefore MATH and MATH . This is an exact equality by REF . |
math/0010157 | One has MATH we see that MATH . Therefore MATH and MATH . |
math/0010157 | MATH with MATH. |
math/0010157 | The form of the term which contains MATH follows from REF follows from REF . Therefore expansion REF can be rewritten as expansion in powers of MATH. To prove that this expansion contains only integral powers of MATH let us introduce an additional parameter. Namely, let us consider family MATH where MATH and MATH is defined by REF, MATH, MATH. Identifying MATH with MATH via multiplication of every coordinate by MATH we see that MATH as deformations of MATH. It follows from REF that MATH we see that MATH and that MATH . Therefore MATH corresponds to MATH, MATH, in the moduli space of deformations of MATH. One can now repeat the above story starting with MATH, MATH and consider element MATH representing periods of the variation of semi-infinite NAME structures associated with moduli space of deformations of MATH. The normalization condition is given by MATH and the parameters MATH are specified via MATH . Notice that this is correctly defined since MATH and therefore MATH is invariant under monodromy around MATH. It follows from REF that MATH . Therefore expansion of MATH can contain only integral powers of MATH-which gives REF . |
math/0010157 | It follows from REF from CITE (This remark is a part of a joint work with NAME) It is an interesting open question whether a homological mirror symmetry conjecture can be formulated in this setting in order to explain the new mirror phenomena. A natural analog of derived category of coherent sheaves associated with MATH does not seem to catch all the necessary information since it is most likely that it has no non-trivial objects. A resolution of this problem might be in introducing a further generalization of MATH-categories where objects exist only in some virtual sense. On the other hand, homological mirror symmetry conjecture in opposite direction which identifies MATH with a version of derived NAME category can be formulated. A sketch of the appropriate definition of the derived NAME category associated with MATH can be found in CITE (some technical details of the construction were checked in recent preprints math. SG MATH by NAME). Whether there exists some numerical mirror symmetry related with the opposite homological mirror symmetry conjecture (that is, MATH versus a NAME category) is an interesting open question deserving further study. |
math/0010161 | We proceed by induction on MATH. For MATH the transformation is true by NAME 's general transformation REF. So, suppose the identity is already shown for MATH. Then, MATH . Now we expand the last MATH-shifted factorial by the terminating MATH-binomial theorem (compare CITE), MATH which is just the MATH case of REF. That is, we apply MATH and obtain for the expression in REF MATH . Note that MATH is bounded by MATH, thus the interchange of summations in REF is justified (provided MATH). Now we can apply the inductive hypothesis to the inner sum which gives us MATH . In this expression, we again interchange summations, and obtain MATH . We simplify the inner sums, according to REF, MATH and eventually deduce the proposition. |
math/0010161 | We proceed by induction on MATH. For MATH the transformation is true by NAME 's transformation REF for well-poised series. So, suppose the identity is already shown for MATH. Then, MATH . Now we expand the last quotient of MATH-shifted factorials by the terminating MATH-Pfaff - NAME summation REF and obtain for the expression in REF MATH . Note that MATH is bounded by MATH, thus the interchange of summations in REF is justified (provided MATH). Now we can apply the inductive hypothesis to the inner sum which gives us MATH . In this expression, we again interchange summations, and obtain MATH . We simplify the last inner sum, according to REF, MATH and eventually deduce the proposition. |
math/0010162 | We apply NAME 's argument successively to the parameters MATH using REF. The multiple series identity in REF is analytic in each of the parameters MATH in a domain around the origin. Now, the identity is true for MATH and MATH, by the MATH-binomial theorem in REF (see below for the details). This holds for all MATH. Since MATH is an interior point in the domain of analycity of MATH, by analytic continuation, we obtain an identity for MATH. By iterating this argument for MATH, we establish REF for general MATH. The details are displayed as follows. Setting MATH, for MATH, the left side of REF becomes MATH . We shift the summation indices in REF by MATH, for MATH and obtain MATH . Next, we apply the MATH, MATH, case of CITE, specifically MATH and the MATH, MATH, MATH, case of the multidimensional summation theorem in REF to simplify the expression obtained in REF to MATH . Now, this can easily be further transformed into MATH which is exactly the MATH, MATH, case of the right side of REF. |
math/0010162 | We have, for MATH, MATH by NAME 's MATH transformation in REF. Now we apply REF to the MATH's on the left and on the right side of this transformation. Specifically, we rewrite the MATH on left side of REF by the MATH, MATH, and MATH case of REF . The MATH on the right side of REF is rewritten by the MATH, MATH, MATH, and MATH case of REF . Finally, we divide both sides of the resulting equation by MATH and simplify to obtain REF. |
math/0010162 | We have, for MATH, MATH by NAME 's MATH transformation in REF. Now we apply REF to the MATH's on the left and on the right side of this transformation. Specifically, we rewrite the MATH on left side of REF by the MATH, MATH, and MATH case of REF . The MATH on the right side of REF is rewritten by the MATH, MATH, MATH, and MATH case of REF . Finally, we divide both sides of the resulting equation by REF and simplify to obtain REF. |
math/0010162 | We have, for MATH, MATH by NAME 's MATH transformation in REF. Now we apply REF to the MATH's on the left and on the right side of this transformation. Specifically, we rewrite the MATH on left side of REF by the MATH, MATH, and MATH case of REF . The MATH on the right side of REF is rewritten by the MATH, MATH, MATH, and MATH case of REF . Finally, we divide both sides of the resulting equation by MATH and simplify to obtain REF. |
math/0010162 | We have, for MATH, MATH by NAME 's MATH transformation in REF. Now we apply REF to the MATH's on the left and on the right side of this transformation. Specifically, we rewrite the MATH on left side of REF by the MATH, MATH, MATH, and MATH case of REF . The MATH on the right side of REF is rewritten by the MATH, MATH, MATH, MATH, and MATH case of REF . Finally, we divide both sides of the resulting equation by MATH and simplify to obtain REF. |
math/0010162 | We have, for MATH, MATH by NAME 's MATH transformation in REF. Now we apply REF to the MATH's on the left and on the right side of this transformation. Specifically, we rewrite the MATH on left side of REF by the MATH, MATH, and MATH case of REF . The MATH on the right side of REF is rewritten by the MATH, MATH, MATH, and MATH case of REF . Finally, we divide both sides of the resulting equation by REF and simplify to obtain REF. |
math/0010162 | We have, for MATH, MATH by NAME 's MATH transformation in REF. Now we apply REF to the MATH's on the left and on the right side of this transformation. Specifically, we rewrite the MATH on left side of REF by the MATH, MATH, MATH, and MATH case of REF . The MATH on the right side of REF is rewritten by the MATH, MATH, MATH, MATH, and MATH case of REF . Finally, we divide both sides of the resulting equation by REF and simplify to obtain REF. |
math/0010162 | We have, for MATH, MATH by the MATH summation in REF. Now we apply REF to the MATH of this summation. Specifically, we rewrite the MATH in REF by the MATH, MATH, and MATH case of REF . Finally, we divide both sides of the resulting equation by REF and simplify to obtain REF. For an alternative proof, set MATH and MATH in REF . In this case the multilateral series on the right side of REF reduces to MATH the last evaluation by the MATH case of REF . |
math/0010162 | We have, for MATH, MATH by the MATH summation in REF. Now we apply REF to the MATH of this summation. Specifically, we rewrite the MATH in REF by the MATH, MATH, and MATH case of REF . Finally, we divide both sides of the resulting equation by MATH and simplify to obtain REF. For an alternative proof, set MATH, MATH, MATH, and MATH in REF . In this case the multilateral series on the right side of REF is terminated from below and from above and reduces just to one term, REF. In the resulting equation, we replace MATH by MATH and MATH by MATH. |
math/0010162 | We have, for MATH, MATH by the MATH summation in REF. Now we apply REF to the MATH of this summation. Specifically, we rewrite the MATH in REF by the MATH, MATH, and MATH case of REF . Finally, we divide both sides of the resulting equation by REF and simplify to obtain REF. For an alternative proof, set MATH and MATH in REF . In this case the multilateral series on the right side of REF reduces to MATH the last evaluation by REF . |
math/0010162 | We utilize the MATH summation in REF and apply REF to the MATH in that summation. Specifically, we rewrite the MATH in REF by the MATH, MATH, MATH, and MATH case of REF . Finally, we divide both sides of the resulting equation by MATH and simplify to obtain REF. For an alternative proof, set MATH, and MATH, MATH, in REF . In this case the multilateral series on the right side of REF is terminated from below and from above and reduces just to one term, REF. In the resulting summation, replace MATH by MATH. |
math/0010162 | Write MATH and MATH. We have, for MATH, MATH by the MATH summation in REF. Now we apply REF to the MATH of this summation. Specifically, we rewrite the MATH in REF by the MATH, MATH, MATH, and MATH case of REF . Finally, we divide both sides of the resulting equation by MATH and simplify to obtain REF. For an alternative proof, set MATH, MATH, and MATH in REF . In this case the multilateral series on the right side of REF is terminated from below and from above and reduces just to one term, REF. In the resulting summation, replace MATH by MATH, MATH, and MATH by MATH. |
math/0010162 | Write MATH. We have, for MATH, MATH by the MATH summation in REF. Now we apply REF to the MATH of this summation. Specifically, we rewrite the MATH in REF by the MATH, MATH, and MATH case of REF . Finally, we divide both sides of the resulting equation by MATH and simplify to obtain REF. For an alternative proof, set MATH, and MATH, MATH, in REF . In this case the multilateral series on the right side of REF is terminated from below and from above and reduces just to one term, REF. In the resulting summation, replace MATH by MATH, and MATH by MATH, for MATH. |
math/0010167 | The first assertion follows from the monotonicity of the line-closure operator. The second is a consequence of the fact that MATH for any subset MATH of MATH. |
math/0010167 | Suppose MATH is not line-closed. Then there exists a line-closed set MATH which is not closed. Let MATH, and choose a linear order on MATH such that MATH precedes MATH. Now, let MATH be the lexicographically first ordered basis for the flat MATH which is contained in MATH. Then, by the choice of ordering, MATH, by REF . We claim MATH. Suppose not. Then, for some MATH, MATH is less than MATH. Since MATH is line-closed, MATH. Also, by the exchange axiom in MATH, MATH is a basis for MATH, and is lexicographically smaller than MATH. This contradicts the choice of MATH. Thus MATH, so MATH. Conversely, if MATH is line-closed, then MATH by REF . |
math/0010167 | The ideal MATH is generated by elements MATH where MATH is dependent. Since MATH has no multiple points, MATH is a circuit. Then MATH, each being equal to MATH. This shows that MATH is homogeneous in the grading above. Thus MATH, and the result follows. |
math/0010167 | With REF in hand the proof is identical to the argument in the proof of REF . It is enough to prove the result for nbb sets of a fixed size MATH. Then we induct on MATH. Suppose MATH . By REF we may assume that MATH for a fixed element MATH and all MATH in the sum. Setting MATH, we have MATH for all MATH, by definition of MATH. Write MATH. Then we have MATH . Applying the derivation MATH we obtain MATH . Using again the definition of MATH, we have that MATH for MATH. Then, applying REF once more, we have MATH . Since MATH implies MATH, we conclude MATH for all MATH by the inductive hypothesis. |
math/0010167 | If MATH is not line-closed, then by REF there is a linear ordering of MATH such that the cardinality of MATH is strictly greater than that of MATH. Then, by REF , we have MATH, so MATH is not quadratic. |
math/0010167 | Such a basis MATH would form a line-closed set by hypothesis, but it cannot be closed in MATH since MATH. |
math/0010167 | We show that MATH is quadratic by verifying directly that MATH for all MATH. Let MATH be a circuit, MATH, and suppose MATH. Then MATH . Thus MATH lies in MATH. Then MATH, and it follows from the NAME rule that MATH lies in MATH. |
math/0010167 | The crucial points are REF that MATH is generated by boundaries of circuits of size at most MATH, and REF that MATH-closure agrees with matroid closure on sets of size at most MATH. Using these observations the proof of REF is easy to adapt to the more general setting. |
math/0010167 | In our setting the atom ordering MATH is the natural linear ordering on MATH. In this context a set MATH is bounded below if and only if there exists MATH with MATH. Suppose MATH is nbb and MATH. Let MATH. Then MATH, so MATH. Since MATH is nbb we conclude MATH, so MATH is not bounded below. Conversely, if MATH is not nbb then for some MATH, MATH is bounded below by MATH and thus MATH is not NBB. |
math/0010167 | Assume without loss that the natural order on MATH is a linear extension of MATH. Then the proof of REF goes through without change. |
math/0010167 | Suppose MATH is a bounded below set. Let MATH with MATH for all MATH. Then MATH precedes MATH in any linear extension of MATH. Since MATH, it follows that MATH contains a broken circuit. |
math/0010167 | Suppose MATH is a quotient of a matroid MATH with the same points and lines as MATH. Since the closure of MATH in MATH contains the line-closure of MATH in MATH, which agrees with the line-closure of MATH in MATH, we conclude that MATH is a basis for MATH. Thus MATH. It follows that MATH. |
math/0010167 | Thus is an immediate consequence of REF . |
math/0010167 | This is a consequence of REF, which asserts that MATH is determined by its essential flats, along with their ranks. A flat of MATH is essential if it is a truncation of a matroid of higher rank. Truncation is in particular a quotient map, and preserves points and lines, so long as the image has rank at least two. Thus, if MATH is locally taut, the only essential flats of MATH are the nontrivial lines. So MATH is determined by the number of points and the list of nontrivial lines. |
math/0010172 | Since MATH and MATH, as a consequence of REF the following identities hold: MATH as a consequence of REF. The cyclicity of MATH implies MATH. We recall now that in REF before the products of the MATH's there is a push-forward; thus, in order to compute MATH, we will have to apply the generalized NAME REF . The boundary of the MATH-simplex can be written as the union of other MATH-simplices (the faces of the simplex), corresponding to the collapsing of successive points, plus two other faces, where the first point tends to MATH or the last tends to MATH. The faces of the first type give MATH, because they yield terms containing MATH, which vanish for dimensional reasons. The remaining two faces give MATH again, for MATH odd, the cyclicity of MATH implies that these terms cancel each other. This also works for MATH even, in case MATH is odd. On the other hand, when both MATH and MATH are even, these two terms have the same sign, and therefore they do not cancel each other. |
math/0010172 | We prove the identity for homogeneous functionals; the general case follows by linearity. We begin by computing the functional derivatives of MATH and MATH: MATH . Next, we note that the integral selects the part of the integrand whose form degree in MATH is equal to MATH, and that the super test form MATH is the sum of the usual test forms (with some signs to be considered). We write MATH as MATH where by MATH we denote the degree MATH part of MATH; that is, MATH, MATH and so on. The signs MATH are the same as in REF of MATH; namely, MATH. Similarly we introduce signs MATH as in MATH. We can then write: MATH where the subscript denotes the restriction to the term of the indicated form degree. We recall that MATH; then we obtain for example, for the MATH-th term in the last expression of the above identity (recalling the definition of the total degree of the functional derivative of MATH with respect to MATH): MATH . By confronting the two expressions in REF , and doing similar computations in the other cases, we obtain for MATH: MATH . We cast then all these expressions in the definition of the super BV antibracket MATH . Then we use the above expressions, and, after rewriting MATH as MATH, we compute the products MATH, separately for the case MATH even and MATH odd. In order for the superbracket to coincide with the ordinary bracket, these products must be MATH for MATH odd. It can be readily computed that the choice of signs made in REF and in REF is consistent with the above rules; therefore, the proof then follows. |
math/0010172 | The proof of this identity is similar to the proof of REF ; in fact, we have to compute the functional derivatives of MATH and MATH with respect to MATH and MATH, and express them via the functional derivatives with respect to the usual fields of the theory. We therefore recall the formulae for the functional derivatives, and we apply them to MATH, obtaining: MATH . Then the following holds, if we go from the dot product to the ordinary product: MATH . By confronting the terms in REF , and operating similarly for the other cases, we obtain the following identities for MATH: MATH . We can finally cast all these expressions in the explicit formula for the super BV antibracket, and, by recalling the explicit values of the chosen signs MATH and MATH, we can finally obtain the desired identity (recall the form degree selection rule imposed by the pushforwards). |
math/0010172 | By definition, a flat observable MATH satisfies separately MATH . This implies that MATH for all MATH. The second equation above together with the NAME identity implies that MATH is a coboundary operator. Since MATH and MATH square to zero and MATH, we obtain MATH . The second claim then follows since MATH squares to zero. For MATH invariant, we also have MATH. So by NAME we obtain MATH. |
math/0010172 | We begin by computing the left and right partial derivatives with respect to MATH and MATH; for example, the left partial derivative of MATH with respect to MATH is given by MATH . It follows that: MATH . If we now insert the above functional derivatives in the formula for the BV antibracket, we obtain MATH by the invariance of MATH (MATH is a superconnection). The first term vanishes by NAME 's Theorem, and the second by the super NAME identity MATH . So the claim follows. |
math/0010172 | The above formulae follow from REF. Let us begin with MATH: MATH . By definition however MATH provided MATH is a test form of total degree MATH. Since MATH has total degree MATH, we cannot apply the above formula directly. We use then the following trick. Let MATH be a scalar of total degree MATH. Then MATH . Thus, MATH where we have used REF. Similarly, we have MATH by REF. |
math/0010172 | One first builds a global angular form MATH on MATH with the correct behavior under the antipodal map on the fibers: one may construct it as in REF using the NAME connection for a given Riemannian metric, which also allows to identify MATH with the unit sphere bundle MATH. Next one extends MATH to the complement of the zero section of MATH and multiplies it by a function MATH that is identically one in a neighborhood MATH of the zero section and identically zero outside a second neighborhood MATH. One then defines MATH as the extension by zero of MATH. Since MATH, MATH is the extension by zero of MATH. The last form may be extended to the zero section of MATH; hence, the extension by zero of MATH can be seen as a covariantly closed element of MATH. The general NAME theorem implies MATH. So REF implies that there is a form MATH such that MATH, with MATH the projection MATH. Also observe that one may choose MATH to satisfy MATH. Finally, define MATH. An easy check shows that it satisfies all properties above (with MATH determined by the restriction of MATH to the diagonal). |
math/0010172 | First of all, we write down the left partial derivatives of MATH: MATH . With the help of REF and by the definition of the super BV antibracket, we get MATH . By the invariance of MATH it follows MATH by NAME 's theorem. So we have proved REF follow from the definitions of the super BV antibracket and of the super BV Laplacian MATH and from the fact that MATH depends only on MATH. |
math/0010172 | By above reasonings, we can consider MATH as a variation of the (flat) connection MATH. The cyclicity of the trace allows to replace the exterior derivative by the covariant derivative MATH. MATH has the same form as MATH of REF, where we have set MATH, and we have replaced MATH by MATH and the wedge product by the dot product. Accordingly to the sign rules for the dot product and repeating almost verbatim the arguments used in the proof of REF , we get MATH . Recalling REF and the NAME rule, it is then not difficult to verify that MATH which yields the desired identity. |
math/0010172 | From the definition of the super BV antibracket, we get MATH . REF follow respectively from the definition of the BV Laplacian and of the super BV antibracket, and from the fact that the functionals MATH do not depend on MATH. |
math/0010172 | By definition of the super BV antibracket, we can write MATH . REF are consequences of the fact that the MATH-s do not depend on MATH and of the definitions of the super BV antibracket and of the super BV Laplacian. |
math/0010172 | We begin by computing the exterior derivative of one of the factors of the above sum. With the help of the generalized NAME Theorem we obtain MATH . We consider the first term on the right-hand side of REF ; the NAME rule for the dot product implies MATH . We recall that MATH by REF . We compute the following expression MATH . For the second term on the right-hand side of the above equation, we obtain, repeating (almost) the same arguments used in the proof of REF , MATH; for the first term, we obtain analogously MATH . Summing up all these contributions with the right signs and using repeatedly REF , we obtain, for the first term on the right-hand side of REF , the result MATH . By the invariance of MATH, and since MATH and MATH have odd total degree, we get MATH . We now consider the second term on the right hand side of REF . We recall the orientation choices for the MATH-simplex made in REF; with these in mind we obtain (once again with the same arguments of the proof of REF ) MATH . Since the trace is cyclic in the arguments and MATH has even total degree, the first two terms in the above expression cancel each other. In summary, we have obtained MATH . Recalling formulae REF , we now apply MATH to MATH. We obtain by repeating (almost) the same arguments as in the proof of REF MATH with the same unifying notation of REF . After repeated application of REF , we get the following result MATH so the claim follows. |
math/0010172 | Since MATH does not depend on MATH, we have MATH . We compute the right super functional derivative of MATH with respect to MATH getting MATH . The left super functional derivative with respect to MATH of MATH reads MATH . So it follows by MATH that the claim is true. |
math/0010172 | When differentiating a form given as in REF or REF, we apply the following rules: CASE: MATH is odd with respect to exterior derivative; CASE: MATH behaves ``as if" it were covariantly closed. To justify the second rule, we first notice that, given any MATH matrix MATH, integration by parts shows that MATH . (With commuting variables we would have the same relation with a minus sign on the right-hand side) As a consequence, MATH with MATH because MATH takes values in MATH. Therefore, MATH where the new exterior derivative MATH is defined by MATH. Introducing the covariant derivative MATH we get from REF that MATH, that is, REF . In particular, we have MATH since on MATH-variables MATH, and MATH by the NAME identity. Therefore, MATH with MATH and MATH defined in REF. Now some simple manipulations and the use of REF show that MATH . Similarly, we get MATH where we have used the constraint MATH and the ensuing identity MATH . |
math/0010172 | The forms MATH and MATH are related by the formula MATH . The first property a global angular form has to satisfy is MATH. By the surjectivity of MATH and by REF , it suffices to show that MATH. Since MATH selects the MATH-independent part in MATH this property is satisfied if and only if we set the correct normalization: MATH where MATH is the volume of the unit MATH-sphere; that is, MATH . Next we use REF to get MATH . Now recall that the differential of a global angular form must be basic on MATH (in particular it has to be the pullback with respect to MATH of a representative of the NAME class). By REF , together with the surjectivity of MATH, it is sufficient to show the identity MATH, where MATH is a representative of the NAME class. All the MATH with MATH contain a form on MATH, so they cannot be MATH-basic (that is, MATH-independent). Therefore, we must choose the coefficients MATH so that the terms in square brackets vanish. This yields a recursion rule that, once the initial condition is fixed by REF, has the unique solution REF. Now observe that the last term MATH vanishes when MATH is odd. Therefore, MATH is closed in this case, and this is enough to prove that it is a global angular form. If MATH is even, however, MATH with MATH denoting the Pfaffian, and the recursion fixes MATH . As a consequence, in this case we get MATH . Since the right-hand side is minus (a representative of the pullback to MATH of) the NAME class, the lemma is proved. |
math/0010172 | We shall apply NAME Theorem to the push-forward with respect to the maps MATH; we note that the MATH-simplex MATH has a boundary, and that this boundary can be written as MATH where each MATH. With our choice of orientation of the simplices - see after REF - the first face of the boundary comes with opposite orientation, while the second has the right one, the third has opposite orientation again, and so forth: MATH . We apply the covariant derivative with respect to MATH to the MATH-th term of the series, and we obtain: MATH where MATH denotes the projection onto the first factor, while MATH is the canonical injection of the boundary of the simplex into the simplex itself. We have used implicitly the identity MATH which follows from REF. We now consider the two terms on the right hand side of REF separately, and we begin with the second term, which we call ``the MATH-th boundary term" from now on. Since MATH, we can write MATH and MATH is the canonical injection of the MATH-th face of the boundary. Considering the orientations of the faces, we obtain for the MATH-th boundary term the following expression: MATH . We begin with the first face MATH; it is not difficult to prove the following identities MATH similarly, one shows for MATH . We have denoted by MATH, respectively, MATH, the injection of MATH into MATH given by MATH, respectively, MATH. for MATH, it holds MATH . It follows therefore (MATH by its very definition) MATH . We consider now the first term under the action of the push-forward with respect to MATH: MATH . Similarly, we obtain for MATH and for MATH we obtain MATH . Finally, we obtain the following expression for the MATH-th boundary term of REF : MATH . We then consider the first term of REF , and by the NAME rule we obtain MATH here we have used MATH which is a consequence of REF . Summing up all the two contributions to REF with the correct signs, we obtain for the left hand side of REF MATH . We begin by summing up all the terms which contain before them MATH, and we obtain MATH; similarly, by summing up all the terms which have MATH on the right, we obtain MATH. By recalling the definition of the curvature of the connection MATH, the sum of the remaining terms will give us MATH . For MATH, we shall now write the projection MATH as the composition of three projections, that is, MATH, where the projections are defined as follows: MATH . We notice for MATH the useful identity MATH, and for MATH holds MATH, for MATH. We then use the following identity (which follows from MATH and REF ): MATH note the appearance of signs in the above identity: this is due to the fact that the three projections above do reverse the product orientation of the fiber of the trivial bundle over MATH given by the projection MATH. It is finally useful to introduce the commutative diagram MATH this diagram allows us to apply REF to MATH, in order to get the pullback with respect to MATH before the pushforward with respect to MATH. We can then apply the first identity of REF , when we integrate with respect to MATH. We shall use once again such a commutative diagram, after integration with respect to MATH, along with REF , in order to obtain the desired identity, accordingly to the notation introduced in REF . |
math/0010175 | A necessary and sufficient condition for a MATH-quasigroup to be MATH-reducible is that the function MATH from REF satisfies the system of partial differential REF . Write REF for any two fixed different values MATH and MATH: MATH . For the MATH-quasigroup defined by REF takes the form MATH . First we assume that MATH . Then MATH . Since the left-hand side does not depend on MATH, then MATH. But MATH implies that MATH, that is, MATH. This along with MATH leads to MATH. These equalities are possible if and only if MATH . However, in this REF does not define a local MATH-quasigroup since it is not solvable with respect to the variables MATH. Leaving this case aside, we find from REF that MATH . We can see again that both sides of this equations are constant: MATH . Integration gives MATH . Since MATH and MATH are arbitrary numbers from MATH, this proves that MATH . Thus MATH . |
math/0010175 | In fact, the equation of a hypersurface MATH is MATH where MATH is a constant. In a Cartesian coordinate system of MATH, the normal vector MATH at an arbitrary point of the hypersurface MATH has the coordinates MATH . Thus if two of these coordinates are equal, this implies MATH. As we saw in the proof of REF , this implies that MATH, and the local MATH-quasigroup is reducible. The converse statement is trivial: it follows from REF if one calculate the coordinates of a normal vector. |
math/0010182 | First, we can write MATH with MATH. Then MATH . Thus by REF , and REF , the assertion follows immediately. |
math/0010182 | This is observed in CITE. First we consider the surjective homomorphism MATH. By the assumption, we have an isomorphism MATH. If MATH is not trivial, we get an obvious contradiction to the NAME property of the free groups (see REF , CITE). Now the assertion follows from Five Lemma: MATH . |
math/0010182 | In fact, the degeneration family gives a surjective homomorphism MATH which is an isomorphism on the first homology. Now take another degeneration families MATH such that MATH is a generic torus curve with REF MATH and MATH. Such a family always exists. We get a surjective homomorphism MATH which also induces an isomorphism on the first homology group. Note that MATH by CITE. Now we apply REF to conclude that the composition MATH is an isomorphism. This implies both of MATH are isomorphisms. |
math/0010182 | Using a degeneration family MATH from the generic torus curves as in the proof of REF , we have always a surjective homomorphism MATH. Thus the assertion follows from REF . |
math/0010182 | It is NAME who has first observed that MATH with MATH. His method is to use the existence of a certain finite covering CITE. For the proof, we use the pencil MATH. First the discriminant polynomial of MATH is given by MATH . Observe that MATH has three real roots for MATH, where two of them make the multiple root MATH for MATH. Thus we have REF (respectively REF) singular pencil lines for MATH (respectively, for MATH): MATH. In MATH, the node disappears and the singular line MATH splits into two simple tangent lines MATH with MATH. Hereafter we consider the case MATH. The line MATH is passing the node MATH, MATH is a bitangent line and MATH is a simple tangent line. The graphs of MATH and MATH provide us the necessary informations. We take REF generators MATH on the pencil line MATH, MATH as in REF . Put MATH and MATH. The big circle relation is MATH. The monodromy relation at MATH is given by MATH . At MATH, we get a simple tangent relation: MATH . The line MATH is a bi-tangent line and the relation is: MATH . Thus we can eliminate MATH from generators. At MATH, we get the commuting relation: MATH . Putting MATH, the monodromy relations at MATH is given by MATH . Now we can rewrite REF using REF as the braid relations: MATH . The second relation of REF reduces to MATH. Thus we can eliminate MATH and we take MATH as generators. They satisfy REF . The relation MATH reduces to MATH and we can see easily that other relations follow from REF . Thus we have proved MATH . The NAME polynomial of MATH is equal to MATH. This can be shown by the exact same computation as in CITE or a direct NAME calculus CITE from the above relations. This completes the proof. |
math/0010184 | Consider a non-free action of a finite group MATH on MATH by orientation preserving diffeomorphisms. Let MATH be the quotient orbifold. If MATH is irreducible then the equivariant NAME lemma implies that any REF-suborbifold with infinite fundamental group has a compression disc. Hence MATH is small and we apply the main theorem. Suppose that MATH is reducible. Since MATH does not contain a bad MATH-suborbifold, there is a prime decomposition along a family of spherical REF-suborbifolds, see REF. These lift to a family of REF in MATH. Consider an innermost REF-sphere; it bounds a ball MATH. The quotient MATH of MATH by its stabilizer MATH in MATH has one boundary component which is a spherical REF-orbifold. We close it by attaching a discal REF-orbifold. The resulting closed REF-orbifold MATH is a prime factor of MATH. The orbifold MATH is irreducible, and hence spherical. The action of MATH on MATH is standard and preserves the sphere MATH. Thus the action is a suspension and MATH is discal. This contradicts the minimality of the prime decomposition. |
math/0010184 | Let MATH be a compact orientable connected irreducible topologically atoroidal MATH-orbifold. By CITE, CITE there exists in MATH a (possibly empty) maximal collection MATH of disjoint embedded pairwise non-parallel essential turnovers. Since MATH is irreducible and topologically atoroidal, any turnover in MATH is hyperbolic (that is, has negative NAME characteristic). When MATH is empty, the orbifold theorem reduces either to the main theorem or to REF according to whether MATH is small or NAME. When MATH is not empty, we first cut open the orbifold MATH along the turnovers of the family MATH. By maximality of the family MATH, the closure of each component of MATH is a compact orientable irreducible topologically atoroidal MATH-orbifold that does not contain any essential embedded turnover. Let MATH be one of these connected components. By the previous case MATH is either hyperbolic, Euclidean or NAME fibred. Since, by construction, MATH contains at least one hyperbolic turnover MATH, MATH must be hyperbolic. Moreover any such hyperbolic turnover MATH in MATH is a Fuchsian MATH-suborbifold, because there is a unique conjugacy class of faithful representations of the fundamental group of a turnover in MATH. We assume first that all the connected components of MATH have MATH-dimensional convex cores. In this case the totally geodesic hyperbolic turnovers are the boundary components of the convex cores. Hence the hyperbolic structures on the components of MATH can be glued together along the hyperbolic turnovers of the family MATH to give a hyperbolic structure on the MATH-orbifold MATH. If the convex core of MATH is REF-dimensional, then MATH is either a product MATH, where MATH is a hyperbolic turnover, or a quotient of MATH by an involution. When MATH, then the MATH-orbifold MATH is NAME fibred, because the mapping class group of a turnover is finite. When MATH is the quotient of MATH, then it has only one boundary component and it is glued either to another quotient of MATH or to a component with MATH-dimensional convex core. When we glue two quotients of MATH by an involution, we obtain a NAME fibred orbifold. Finally, gluing MATH to a hyperbolic orbifold with totally geodesic boundary is equivalent to quotient this boundary by an isometric involution. |
math/0010184 | The assertion is clear in the smooth case and we therefore assume that MATH has cone points. Due to NAME, there can be at most three cone points. If MATH has only one cone point MATH, then MATH is simply connected and hence can be developed (isometrically immersed) into MATH. A circle of small radius centered at MATH cannot close up under the developing map and we obtain a contradiction. Thus MATH must have two or three cone points. If MATH has two cone points, we connect them by a minimizing segment MATH. By cutting MATH open along MATH we obtain a spherical surface which is topologically a disc and whose boundary consists of two edges of equal length. It can be developed into MATH as well, and it follows that the surface is a spherical bigon, that is, the metric suspension of an arc. We obtain the second alternative of our assertion. If MATH has three cone points, we connect any two of them by a minimizing geodesic segment. The segments don't intersect and they divide MATH into two spherical triangles. The triangles are isometric because they have the same side lengths, and we obtain the third alternative. |
math/0010184 | This is a direct implication of the lower curvature bound MATH because the circumference of geodesic triangles has length MATH. |
math/0010184 | CASE: Suppose that MATH and that MATH is a cone point MATH. Any geodesic triangle MATH has angle MATH at MATH. We denote MATH. Since MATH, hinge comparison implies that MATH. CASE: In the case of equality it follows that the cone angle at MATH equals MATH and that one of the points MATH and MATH, say MATH, has distance MATH from MATH. If MATH were not a cone point, then it would lie on the segment connecting the two cone points MATH and only MATH would have distance MATH from MATH, contradicting MATH. Hence MATH must be a cone point, and it follows that MATH lies on the segment joining MATH and the cone point MATH. |
math/0010184 | MATH is the double of a spherical triangle MATH with two angles MATH and third angle MATH. Since the angle sum of a spherical triangle is MATH, all angles of MATH are MATH. Such triangles can NAME converge to a point, but not to a segment. Hence the NAME closure of the space of turnovers as in the lemma is compact and contains as only additional space the point. It follows that the diameter assumes a maximum MATH on this space of turnovers. By REF , we have MATH. |
math/0010184 | The turnover MATH is the double of a spherical triangle MATH with acute angles MATH and a lower diameter bound. Since the angle sum of spherical triangles is MATH, we also have the positive lower bound MATH for the angles of MATH. Such triangles have a lower bound on their inradius, whence the claim. |
math/0010184 | It follows from the classification of links, compare REF , that MATH contains a smooth standard disc with radius bounded below in terms of MATH, and hence the ball MATH contains an embedded smooth standard ball with a lower bound on its radius in terms of MATH and MATH. We may therefore assume without loss of generality that MATH is a smooth point. CASE: We have a lower bound MATH because MATH contains a smooth standard ball of radius MATH. Let MATH denote the subset of initial directions of minimizing geodesic segments with length MATH. The lower bound for the volume of the annulus MATH implies a lower bound MATH. CASE: By triangle comparison, there exists for MATH a number MATH such that: Any geodesic loop of length MATH based in MATH has angle MATH with all directions in MATH. The same holds for the angles of MATH with segments of length MATH starting in MATH and perpendicular to the singular locus MATH. Thus, if MATH, then minimizing segments from MATH to closest cut points must have angles MATH with all directions in MATH, compare our discussion of the cut locus in REF. We use this observation to obtain lower bounds for the injectivity radius. For MATH there exists MATH such that: Let MATH be a spherical cone surface homeomorphic to REF-sphere and with cone angles MATH. Let MATH be a subset with MATH. Then MATH if MATH is a turnover. If MATH has REF or REF cone points, then there exists a point MATH such that MATH. In the case that MATH has two cone points, MATH can be chosen as a cone point. When MATH is a turnover, the description in REF of segments of maximal length MATH implies: Points in MATH with radius (NAME distance from MATH) close to MATH must be close to one of the three minimizing segments connecting cone points, that is, lie in a region of small area. Hence MATH contains points with radius MATH for sufficiently small MATH depending on MATH. If MATH has REF or REF cone points then it is isometric to the unit sphere MATH or the metric suspension of a circle with length MATH, compare the classification in REF . In both cases the assertion is easily verified. We choose MATH with MATH as in REF , and accordingly MATH. CASE: For a singular vertex MATH . REF implies that MATH. CASE: Assume that MATH is a singular point with MATH at distance MATH from all singular vertices, and choose the singular direction MATH according to REF . By the assumption on the injectivity radius, there exists a geodesic loop MATH of length MATH based at MATH or a segment MATH of length MATH meeting MATH orthogonally at at a point MATH. Either of them has angles MATH with the directions in MATH and therefore angles MATH with the direction MATH. In the case of a loop, consider the geodesic variation of MATH moving its base point with unit speed in the direction MATH. Since both ends of the loop have angle MATH with MATH, the first variation formula implies that the length of MATH decreases at a rate MATH. Similarly, in the case of a segment, we obtain that MATH decreases at a rate MATH. It follows that MATH. CASE: Suppose now that MATH is a smooth point with MATH at distance MATH from MATH. We choose the direction MATH according to REF . As in REF , we see that MATH decays in the direction MATH with rate MATH. It follows that MATH. Conclusion. The assertion holds for MATH and MATH. |
math/0010184 | CASE: The convexity of the NAME polyhedron MATH, compare REF, implies that MATH is convex. CASE: Suppose that MATH is a cone point and MATH is a point in MATH with MATH. We consider first the case when the cone angle at MATH is MATH. If MATH is the metric suspension of a circle then there exists a loop of length MATH based at MATH and surrounding MATH. It follows that MATH is contained in the convex hull of MATH and hence MATH. If MATH is a spherical turnover, we cut MATH open along MATH and obtain a convex spherical polygon with MATH as cone point. Inside the polygon we find a loop as before. We are left with the case that the cone angle at MATH equals MATH. Let MATH be the singular ray in direction MATH. Observe that, if MATH is a point on MATH different from its initial point MATH, MATH is a segment perpendicular to MATH and MATH is a (small) ball around MATH, then the convex hull of MATH in MATH contains MATH. Now the ray MATH is contained in MATH. Since MATH is convex and has non-empty interior, arbitrarily close to every point of MATH we find interior points of MATH. Our observation therefore implies that MATH and MATH. |
math/0010184 | Let MATH be a ray starting in MATH. In the case of a loop MATH, the assertion follows by applying angle comparison to the isosceles geodesic triangle with MATH as one of its side and twice the segment MATH as the other two sides, and by letting MATH. Comparison is applied to the angles adjacent to the non-minimizing side MATH. In the second case, the argument is similar. We consider instead the geodesic triangle with sides MATH, MATH and a minimizing segment MATH as third side, and use that every direction at the singular point MATH has angle MATH with MATH. |
math/0010184 | If MATH, we find a thick smooth point on MATH. If MATH, there are no singular vertices and the lower bounds on MATH and the cone angles imply thickness as well. |
math/0010184 | According to REF each thin submanifold contributes a definite quantum to the volume of MATH. Thus MATH can have only finitely many components. Finiteness of volume implies moreover that thin submanifolds are compact or cusps with compact cross sections. Consider a (globally minimizing) ray MATH. There is a uniform lower bound on the volume of balls with radius MATH and centers outside MATH, where MATH is the constant in REF . Hence, by volume reasons, MATH enters MATH after finite time. A thin submanifold containing a ray is noncompact and must therefore be a cusp. We conclude that the complement of all cusp components of MATH is compact, because otherwise it would contain a ray which would end up in yet another cusp, a contradiction. |
math/0010184 | We have MATH. The immersion MATH into model space given by the developing map must be an isometry onto a round ball. Therefore also MATH. |
math/0010184 | Since MATH, there exists a point MATH with MATH. Let MATH be a point in MATH closest to MATH. Either MATH is the midpoint of a geodesic loop MATH of length MATH based at MATH, or MATH belongs to a singular edge with cone angle MATH and there is a (unique) minimizing geodesic segment MATH of length MATH which is perpendicular to the singular locus at MATH, compare our discussion of the cut locus in REF. In both cases, we have a geodesic triangle MATH with MATH. By our assumption on the diameter of MATH holds moreover MATH. Triangle comparison yields a positive lower bound MATH for MATH. |
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