paper stringlengths 9 16 | proof stringlengths 0 131k |
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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... |
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 h... |
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 h... |
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/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 ch... |
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/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 le... |
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... |
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 ... |
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/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 gr... |
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 ... |
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 ... |
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... |
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 . Di... |
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 de... |
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 MAT... |
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. Tha... |
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 exp... |
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. Si... |
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, ... |
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, ... |
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, ... |
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/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, ... |
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/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 ... |
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, M... |
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 ... |
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 RE... |
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 alternati... |
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 M... |
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 ... |
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... |
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 n... |
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,... |
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... |
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 o... |
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 derivativ... |
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 z... |
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 inva... |
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, MAT... |
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... |
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 L... |
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... |
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 expressio... |
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 comm... |
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 normalizati... |
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 orientatio... |
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 ... |
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 im... |
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 isomor... |
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 ... |
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 mai... |
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 hyper... |
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 c... |
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/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 ... |
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 generalit... |
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 b... |
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 si... |
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 lowe... |
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 perpend... |
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