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math-ph/0008012 | By the previous corollary MATH . |
math-ph/0008012 | It is sufficient to prove this proposition for a standard elementary domain of class MATH. Fix MATH and choose a sequence MATH, MATH for all MATH. Since MATH is a NAME space, one may assume without loss of generality that the sequence MATH weakly converges in MATH to some function MATH. Using REF for almost all MATH in... |
math-ph/0008012 | Choose a sequence MATH, MATH for all MATH. Let MATH and MATH. Then MATH , MATH, MATH, MATH. Because the embedding operator MATH is compact we can choose a subsequence MATH of the sequence MATH which converges in MATH to a function MATH. Because the second embedding operator MATH is also compact we can choose a subseque... |
math-ph/0008012 | Let MATH be an elementary domain of class MATH. Then there exists an elementary domain MATH of class MATH and a quasiisometrical homeomorphism MATH. By the previous theorem operators MATH and MATH are bounded. By REF the embedding operator MATH is compact. The embedding operator MATH is the composition MATH. Therefore ... |
math-ph/0008012 | Let MATH be an elementary domain of class MATH. Then there exists an elementary domain MATH of class MATH and a REF-quasiisometrical homeomorphism MATH such that MATH. By REF operators MATH and MATH are bounded. By REF the embedding operator MATH is compact. The embedding operator MATH is equal to the composition MATH.... |
math-ph/0008012 | Necessity: REF is clearly necessary: if MATH: MATH is compact, and MATH, MATH, then MATH: MATH is compact. To prove REF, assume the contrary: there exists MATH, MATH, and MATH such that MATH for all MATH. Since MATH, one concludes from previous inequality that MATH as MATH and MATH in MATH, MATH stands for weak converg... |
math-ph/0008013 | The heart of REF is the relation REF, so let us begin with a derivation of this equation. Fix MATH and MATH. Recall that the NAME function, MATH is the unique square summable solution to the equation MATH . A natural guess is that the solution factors: MATH . With this ansatz, REF yields for MATH and MATH: MATH and MAT... |
math-ph/0008022 | From the selfadjointness of the operator MATH it follows that the matrix MATH is invertible for all MATH (see CITE). The relation REF implies that MATH . Similarly, for the operator MATH we have MATH . Taking the complex conjugate and multiplying by MATH from the left we obtain MATH . We multiply this relation by MATH ... |
math-ph/0008022 | REF is an easy consequence of the singular values decomposition (see for example, CITE). Thus we have MATH. REF now follows from the fact that the geometric and algebraic multiplicities of an eigenvalue MATH are unequal iff MATH is non-trivial. |
math-ph/0008022 | Without loss of generality we may assume that MATH. By REF MATH . Therefore by well-known inequalities for the singular values of compact operators (see for example, CITE) we have MATH where MATH denotes the maximal singular value of a compact operator MATH, that is, the maximal eigenvalue of the self-adjoint non-negat... |
math-ph/0008022 | Let MATH, MATH be a (not necessarily orthogonal) basis in MATH. For all MATH we have MATH . Multiplying these equations by MATH from the left we obtain MATH . Thus MATH . By REF it follows from REF that MATH . Hence MATH and therefore by REF MATH. Let MATH, MATH be some basis in MATH. We have MATH for all MATH. Multipl... |
math-ph/0008022 | Since MATH is unitary by the definitions of MATH and MATH it suffices to prove REF . By the definition of MATH we have that MATH for any MATH and any MATH. Thus MATH for all MATH. Conversely, by REF for any MATH there is a unique MATH which satisfies the equation MATH. This proves REF . REF is proved similarly. |
math-ph/0008022 | We recall that for MATH the MATH matrix MATH is not of maximal rank REF iff there is a vector MATH such that MATH. For MATH the matrix MATH is not of maximal rank iff there is a vector MATH such that MATH. Let us suppose that the matrix MATH is not MATH-compatible with MATH. Then by REF it follows that all off-diagonal... |
math-ph/0008022 | Let us suppose that MATH. Then the MATH matrix MATH is not of maximal rank iff there is a non-zero vector MATH such that MATH. From the unitarity of MATH it follows that MATH and therefore MATH . Thus MATH and MATH. From this it follows that the MATH matrix MATH is not of maximal rank, and thus the MATH matrix MATH is ... |
math-ph/0008022 | Assume the converse, that is, let there be MATH, MATH such that MATH or, equivalently, MATH . We multiply this equation by MATH from the left and use the unitarity of MATH which in particular implies MATH . This yields MATH . Again by unitarity we have MATH. Recall that MATH by the hypothesis of the lemma. Thus, from R... |
math-ph/0008022 | By REF both matrices MATH and MATH have maximal rank. We will discuss only the second of the relations REF. In block notation this relation has the form MATH . By REF the matrix MATH must be compatible with MATH such that MATH and MATH are both well defined. We multiply the first of the relations REF by MATH from the l... |
math-ph/0008022 | We split the proof into several steps. CASE: First we suppose that MATH is compatible with MATH and prove that the composition REF holds. Let MATH, MATH for any MATH be the solution of the NAME equation with the operator MATH, MATH (see REF) at energy MATH. Let MATH be MATH matrices MATH such that MATH . Observe that M... |
math-ph/0008022 | Obviously the coefficients MATH in REF satisfy the relation MATH . From the unitarity of the scattering matrix it follows that MATH or, equivalently, MATH . This relation and REF complete the proof of the lemma. |
math-ph/0008022 | From the well-known determinant formula for block matrices (see for example . CITE) MATH which follows from the decomposition MATH it follows that MATH . By REF we have MATH and thus MATH. |
math-ph/0008022 | The external part of any solution to the NAME equation with the operator MATH satisfying the conditions of the theorem is a linear combination of the columns of the matrix-valued function MATH where MATH is given by REF. Thus, the columns of REF have to satisfy REF, that is, MATH . The solution of this equation can be ... |
math-ph/0008022 | Suppose that MATH. Then by REF we get MATH, which is a contradiction. Thus, MATH and therefore by REF MATH. The representation REF follows from REF. |
math-ph/0008024 | The proof is based on the relation MATH where MATH is the Lagrangian REF on MATH and MATH is the Lagrangian REF on MATH CITE. |
math-ph/0008024 | The proof is based on the fact that such a Hamiltonian form MATH defines the global section MATH of the fibred manifold REF , and MATH. Then the constrained NAME equations can be written as MATH . They are obviously weaker than the NAME REF restricted to MATH. |
math-ph/0008024 | By virtue of the equality REF , the projection REF yields a surjection of MATH onto MATH. Given a section MATH of the fibred manifold REF , we have the morphism MATH. In accordance with REF , this is a surjection such that MATH . Hence, MATH is a bundle isomorphism over MATH which is independent of the choice of a glob... |
math-ph/0008024 | Let us consider the NAME REF , written as the equality MATH for a section MATH of the NAME bundle MATH. The Hamiltonian map MATH reads MATH . Due to the projections MATH, REF , the NAME REF break in two parts MATH . Let MATH be an arbitrary section of MATH, for example, a solution of the NAME - NAME equations. There ex... |
math-ph/0008029 | We will establish the relation MATH which is equivalent to REF. Using the chart, we identify MATH with MATH, and we identify the components MATH of MATH and MATH of MATH with respect to a local trivialization of MATH near MATH with elements of MATH where MATH is the chart range. (Note that it is no restriction to assum... |
math-ph/0008029 | Since there is essentially no deviation from the proof of this statement given for the scalar case in the works CITE, we shall be content with giving only a sketch of the proof. Let MATH be another NAME with causal normal neighbourhood MATH. We assume first that MATH. Let MATH have compact closure and set MATH. Then ch... |
math-ph/0008029 | REF is easy to see: Since MATH is an invertible matrix, one can use REF (for the case of a bundle morphism) to reduce the proof of the statement to the scalar case, where the claimed property is well-known (compare CITE). To prove REF , note first that again by REF it is sufficient to show MATH for each past pointing, ... |
math-ph/0008029 | CASE: Let the element MATH of MATH be defined by MATH where MATH is a MATH-regularizing function. Using the arguments of REF of the proof of REF in combination with REF , it is straightforward to deduce MATH. Thus, if MATH denotes a NAME form on MATH, then by the very definition of NAME form MATH is given by a MATH-int... |
math-ph/0008029 | As first step, we will prove independence of MATH by generalizing an argument given in CITE to arbitrary dimensions: For MATH note that MATH converges for MATH to a locally integrable function which does not depend on MATH anymore. As we can use REF to conclude that we may interchange integration and limit in REF, we s... |
math-ph/0008029 | Since for each MATH the support of MATH is for MATH shrinking to MATH, it suffices to prove the statement for the case that MATH. We will demonstrate the statement only for simple tensors MATH with MATH since this results in slightly simpler notation, but it will be obvious from the argument that general MATH can be de... |
math-ph/0008035 | Let MATH and MATH. Then, by the NAME identity, MATH using the property that the MATH operators are central. |
math-ph/0008035 | Write MATH in the form MATH. Fix MATH and let MATH. Since MATH commutes with MATH, we have MATH. |
math-ph/0008035 | The converse is clear by construction and our previous observations. What must be checked is that given a representation of MATH, setting MATH yields an sl REF algebra that commutes with MATH. From REF , we have MATH and similar relations for MATH and MATH show that MATH commutes with MATH. Now, using REF , we note the... |
math-ph/0008035 | Use REF in the relation MATH . |
math-ph/0008035 | In the formulation of REF first set MATH. Then use the special realization as in REF with MATH. As in REF MATH which combines to yield MATH up to the factor MATH. Now observe that MATH where on MATH, MATH acts simply as multiplication by MATH. Combining with the above observations yields the result. |
math-ph/0008035 | From REF , MATH . Now we have the generating function for the inner products MATH depending only on the pair products MATH, MATH. Hence orthogonality. Expanding the left-hand side of the equation yields the squared norms. |
math-ph/0008035 | First substitute MATH for MATH in REF . And observe that MATH . Now use the special realization, REF , taking MATH in REF , with MATH . After substituting accordingly and simplifying, one finds the stated result. |
math-ph/0008035 | First, use the fact that MATH commutes with MATH together with the NAME formula for the NAME to yield MATH . Now write, using the standard form, REF , MATH, with MATH. Since MATH commutes with MATH, we have MATH . We can reconstitute MATH as MATH since MATH commutes with MATH, and use REF , or, equivalently, think of M... |
math-ph/0008038 | Since the MATH is abelian, conjugating it by a fixed element of the group will yield an abelian algebra. Calculating the adjoint group action MATH (with a use of commutation relations) yields the indicated operators. For MATH, the result is skew-symmetric, thus requiring the factor of MATH for those MATH. |
math-ph/0008038 | First compute MATH . When this is expressed in factored form (compare REF ), taking inner products with MATH eliminates all factors except for MATH. In general, this is MATH with MATH a function of MATH's and MATH's. In the matrix realization above, applying MATH to MATH shows that MATH. The rest follows from REF using... |
math-ph/0008038 | To start, note that MATH entails MATH. Now we must solve for the MATH's in REF . First, MATH . And with MATH, we square and re-sum on the left-hand side to yield MATH from which MATH. Expressing MATH in terms of MATH's in the above expressions for the MATH yields the result. |
math-ph/0008038 | REF provides us the NAME function for our representation of so(MATH,REF) MATH . Differentiating, we obtain MATH . Combining these, we find the system of partial differential equations MATH (implied summation over MATH in the middle term) from which we can read off the result stated for MATH. Since MATH, taking the comm... |
math-ph/0008042 | Let MATH be a sphere with center at the origin and radius MATH sufficiently large so that MATH is contained in the ball MATH with boundary MATH REF . According to REF in each point MATH of the domain MATH we have the equality MATH . We now consider the limit of this equality when MATH. We have the following asymptotic ... |
math-ph/0008042 | Consider the expression MATH . Consequently, MATH . Due to REF we have (using the notation of REF ) MATH . The first integral on the right-hand side is some constant MATH. In order to simplify the second we use MATH and observe that MATH . Then MATH we obtain that MATH . Substituting this expression in REF and using th... |
math-ph/0008042 | First we suppose that MATH. Multiplying REF by MATH from the right-hand side we obtain MATH adding and subtracting REF we obtain MATH . Which can be writen as follows MATH . Thus, REF is equivalent to REF , from which it can be seen that MATH fulfills the radiation REF and MATH fulfills its conjugate which corresponds ... |
math-ph/0008043 | A distributional solution on a globally hyperbolic manifold MATH vanishing in a neighbourhood of a NAME surface MATH vanishes in MATH. This is a general property of hyperbolic wave equations (see for example, CITE). We will show, that if a solution vanishes in a neighbourhood MATH of MATH for some MATH then it vanishes... |
math-ph/0008043 | Let MATH be the scalar product on MATH inducing the state MATH. By REF it is sufficient to show that MATH is MATH-dense in MATH. We show that a MATH-continuous linear form MATH on MATH vanishing on MATH vanishes on the set MATH. Note that MATH is a real-valued distribution in MATH and a solution to the NAME equation. B... |
math/0008002 | The assertion follows by adjointness from the fact that MATH is also formally étale. |
math/0008002 | If we look at MATH-valued points, then MATH is a pseudotorsor over MATH. Indeed, suppose that MATH, MATH corresponds to a MATH-valued point of MATH. Any other such morphism is of the form MATH, where MATH is a MATH-valued point of MATH. If MATH, then for a fixed closed point in MATH, we get induced MATH-valued points i... |
math/0008002 | It is enough to notice that since MATH is an irreducible component of MATH, we have MATH and therefore MATH. |
math/0008002 | We have a decomposition MATH and in general MATH is an irreducible component of MATH of dimension MATH. Therefore the ``only if" part of both assertions is obvious and holds without the l.c.i. hypothesis. Suppose now that MATH. Working locally, we may assume that MATH and that MATH is defined by MATH equations. We have... |
math/0008002 | By REF , MATH is l.c.i., hence NAME. Since MATH is smooth, MATH is generically reduced, and we conclude by NAME 's theorem (see REF ). |
math/0008002 | Again, we may assume that MATH is defined by MATH equations. It follows from the equations of MATH that we have MATH, such that MATH is defined by MATH equations. The first assertion follows from this once we notice that by REF , for every irreducible component MATH of MATH, we have MATH. The last statement is a conseq... |
math/0008002 | Since MATH is in particular NAME, by NAME 's Criterion (see REF ) it is enough to show that MATH. If MATH is irreducible, by REF , we may assume that MATH. But if MATH, since for every MATH we have MATH, it follows that MATH, contradicting REF . |
math/0008002 | Notice first that we have MATH. Indeed, in order to compute the coefficient of MATH, we may restrict to an open subset whose intersection with MATH is nonempty and smooth, in which case the formula is well-known. Consider the divisor MATH on MATH, defined by MATH. We can write MATH. But we have MATH . Therefore we have... |
math/0008002 | As in the proof of REF , we have MATH log canonical if and only if MATH is log canonical and on the other hand MATH for all MATH if and only if the pair MATH is log canonical. It follows from REF that if MATH is log canonical, then MATH is log canonical. |
math/0008002 | Notice that by REF and by the definition of canonical singularities, either condition implies that MATH is normal, so that REF applies. Moreover, by REF , if MATH, then MATH is pure dimensional, so that an application of REF completes the proof. |
math/0008002 | Again, REF shows that MATH is pure dimensional if and only if MATH and we apply REF . |
math/0008002 | REF implies that Conjectures REF are equivalent, and we have seen in the previous section that Conjecture REF implies Conjecture REF. It is therefore enough to prove that if Conjecture REF is true for all normal, l.c.i. varieties, then so is Conjecture REF. Using a trick due to NAME (see REF ), the assertion in Conject... |
math/0008002 | The second assertion follows from the first one, since working locally we may assume that MATH, for some divisor MATH and if MATH, then MATH. To prove the first statement, note that the case MATH is trivial, and therefore we may assume that MATH. It is enough to show that for every MATH we have MATH . If we pick a regu... |
math/0008002 | We have already seen that MATH is a cylinder for every integer MATH. Moreover, MATH and REF gives MATH . |
math/0008002 | We fix a function MATH, such that for every MATH, MATH where MATH is a constant with MATH, for all MATH. We extend it by defining MATH. For the proof of the implication MATH we will put later an extra condition. It follows from REF that if MATH, then MATH for MATH, and MATH. Computing the integral of MATH from the defi... |
math/0008002 | We prove the first assertion. Let MATH be the canonical projection. Using REF , it is enough to show that there is an isomorphism MATH, which maps MATH into MATH. Here MATH denotes the image of MATH by the canonical section of MATH and MATH is the tangent cone to MATH at MATH. We give the isomorphism at the level of MA... |
math/0008002 | Indeed, if MATH is a singular point, then MATH, and by REF , it follows that for every MATH, MATH, and therefore MATH gives an irreducible component of MATH. |
math/0008002 | One of the characterizations of rational double points is that they are locally complete intersection rational singularities (see CITE). Therefore MATH follows from REF and the fact that NAME singularities are rational if and only if they are canonical (see REF ). In order to prove MATH, it is enough to show that if MA... |
math/0008002 | The hypothesis says that MATH is smooth; in particular, MATH is integral. By REF , MATH is irreducible for all MATH if and only if MATH for all MATH, and by REF , it is enough to check this for infinitely many MATH. For every MATH and integers MATH, with MATH for all MATH, let MATH be the locally closed subset of MATH ... |
math/0008002 | We show first that if MATH, then MATH . To see this, we may work locally and assume that MATH is defined by MATH, where MATH. MATH is defined by MATH, and if MATH, then the Jacobian MATH is MATH for some MATH matrix MATH. Therefore MATH and MATH follows. Suppose now that MATH. Consider an open connected neighbourhood M... |
math/0008002 | We use notation and results from CITE. Since all the semigroups we use are saturated, we make no distinction between the semigroup and the cone it generates. In general, for two varieties MATH and MATH, we have MATH. Using this, we reduce immediately to the case when MATH is affine, where MATH is a strongly convex, rat... |
math/0008002 | Let MATH be the smooth open subset of MATH consisting of regular elements. It is known that the morphism MATH is smooth and surjective (see CITE). Therefore the morphism MATH is also smooth and surjective. Consider the map MATH defined by the formula MATH. The map MATH is smooth, and since MATH acts transitively along ... |
math/0008002 | Let MATH be the open dense MATH - orbit of regular elements in MATH. By REF , MATH is irreducible. Hence MATH is dense in MATH. Since MATH is smooth, MATH is surjective. Therefore the map MATH is dominant. |
math/0008002 | Since MATH is a complete intersection, it is NAME. Therefore MATH is a flat module over MATH. Since MATH is MATH - graded with finite-dimensional homogeneous components, flatness implies that MATH is free over MATH (see, for example, REF). This proves REF . Since MATH is free over MATH, and both rings are MATH - graded... |
math/0008002 | The jet scheme MATH (respectively, MATH) of the open dense MATH - orbit MATH of MATH is an orbit of the group MATH (respectively, MATH) in MATH (respectively, MATH). Therefore any invariant function on it is a constant. But according to the proof of REF , it is a dense subvariety in MATH (respectively, MATH). Hence any... |
math/0008007 | We are only going to prove REF because REF can be shown in a similar way. CASE: It is enough to prove that for every MATH . Let us compute MATH. MATH where MATH. Now using that MATH for MATH, (see for instance REF) and considering MATH we obtain that MATH where MATH . Since MATH is concave on MATH and MATH, we get MATH... |
math/0008007 | First of all note that for every MATH, REF holds for MATH and MATH, simply using REF and because MATH and MATH . Now we only have to prove that for every MATH the function MATH defined by MATH is concave. If we compute its derivate, we obtain MATH . Next we use that there exists a function MATH such that for every MATH... |
math/0008007 | CASE: Let MATH. By using NAME 's formula it is easy to see that MATH . Since MATH, we have MATH for all MATH, (note that this result can be extended to MATH strictly smaller than MATH). Therefore MATH is a non increasing function in MATH and so it is MATH. CASE: Let MATH. If we use again NAME 's expression of the Gamma... |
math/0008007 | We apply NAME formula and so, we only need to achieve MATH and MATH . REF is deduced from the fact that the function MATH is convex for MATH. In particular since MATH and MATH we deduce MATH for all MATH and so the inequality is true for MATH (consider MATH). In order to show REF we use the corresponding expansion and ... |
math/0008007 | Let MATH be a hyperplane in MATH. A well known result (see CITE) ensures that MATH where MATH (the isotropy constant) is MATH (see CITE). Hence it is enough to prove that MATH for all MATH and all MATH. Notice that this follows from REF . |
math/0008007 | NAME to the NAME 's result quoted in the introduction, we only have to consider the case MATH. NAME 's inequality implies that MATH (in fact MATH is the ellipsoid of maximal volume contained in MATH). Hence it is enough to show that MATH for all MATH and for all MATH, that is, MATH (see for instance CITE). By using REF... |
math/0008007 | Following the same methods than REF , we only have to prove MATH for MATH. The case MATH is REF and MATH can be checked directly. |
math/0008007 | We use the results from CITE MATH . By REF MATH is non increasing with MATH and this implies the result, since MATH . Note that this value belongs to MATH. Indeed MATH and MATH . Finally we show that the result is sharp. It is easy to check that MATH and if we consider REF-dimensional subspace MATH then it is easy to p... |
math/0008013 | Let MATH. There exists a positive element MATH such that MATH. It follows that the set MATH is a closed neighborhood of MATH and MATH. Let MATH be a compact neighborhood basis of MATH in MATH. Notice that MATH is NAME since MATH is. Thus MATH is closed in MATH, and therefore in MATH as well. It follows that MATH is a n... |
math/0008013 | We use REF to show that MATH whenever MATH and MATH. If MATH then there exists an ideal MATH of MATH with continuous trace such that MATH. Note that MATH, where MATH. Since MATH has continuous trace each element MATH of MATH is a Fell point and MATH is NAME. Thus MATH is also NAME, and each point MATH in MATH is a Fell... |
math/0008013 | Since MATH is postliminal, we can identify MATH and MATH. We can view MATH as the appropriate quotient of MATH, and then the map MATH is an open surjection onto MATH CITE. In particular, (the class of) MATH is a typical element of MATH. Since MATH is a Fell algebra, MATH has an open NAME neighborhood CITE which is of t... |
math/0008013 | Let MATH be a compact set in MATH and choose MATH such that MATH. It suffices to show that for each MATH and MATH, MATH is relatively compact in MATH. Let MATH be a net in the set described in REF . It will suffice to find a convergent subnet. Since MATH is open, we can pass to a subnet, relabel, and assume that this n... |
math/0008013 | Suppose that MATH. Let MATH be a MATH-wandering neighborhood of MATH. Then there are MATH such that MATH and MATH for all MATH. We may replace MATH by MATH for some MATH, and assume that MATH. Then MATH . Since the right-hand side of REF has relatively compact image in MATH and MATH is open, we can pass to a subnet and... |
math/0008013 | The free case is treated in CITE. Now suppose that MATH is abelian, that the stability groups vary continuously and that MATH is a Fell algebra. Fix MATH and let MATH. By REF , MATH has an open NAME MATH-invariant neighborhood MATH, where MATH is an ideal of MATH. Thus MATH for some MATH-invariant open subset MATH of M... |
math/0008013 | Again, the free case is dealt with in CITE. In any event, the largest Fell ideal of MATH is MATH where MATH. Since MATH is invariant under the dual action, it follows that MATH for some open MATH-invariant subset MATH of MATH. Now apply REF . |
math/0008013 | Suppose that MATH for MATH and MATH. We want to show that MATH converges to MATH. Since this happens if and only if every subnet converges to MATH, we can pass to a subnet, relabel and assume that MATH. Since MATH is NAME, MATH for some MATH. Thus by assumption, MATH converges to MATH. |
math/0008013 | Our proof is modeled on the proof of CITE. Here we'll give the proof for MATH abelian and remark that the free case follows from the same sort of argument together with the following observation. If the action is free, then MATH-regularity implies that MATH is homeomorphic to the MATH-ization MATH of MATH CITE. It foll... |
math/0008013 | Let MATH, and MATH such that MATH. If MATH then MATH and MATH can be separated by disjoint relative open subsets of MATH because MATH is NAME. Since MATH is open these relative open sets are open. Now suppose that MATH. Fix a positive element MATH of MATH such that MATH and let MATH be the (continuous) map MATH. Note t... |
math/0008013 | If MATH is the largest liminal ideal then MATH. If MATH is abelian then MATH is invariant under the dual action, and we have MATH for some open MATH-invariant subset MATH of MATH. This is trivial in the free case. Let MATH be as in REF. Note that every MATH has a neighborhood MATH (namely MATH) such that MATH is closed... |
math/0008013 | If MATH is abelian, the largest postliminal ideal of MATH is invariant under the dual action, so equals MATH for some MATH-invariant open subset MATH of MATH. Let MATH be as in REF. Every MATH has an open MATH-invariant neighborhood MATH (namely MATH) such that MATH is MATH by CITE. Thus MATH. Let MATH and MATH an open... |
math/0008014 | Assume that there is a Legendrian embedding into MATH which admits no NAME chord of length not bigger than MATH. Consider a Lagrangian embedding as constructed in the previous section. It will be displaced from itself by MATH. Fix any almost complex structure MATH as in the theorem. Due to monotonicity of MATH there ar... |
math/0008015 | The general theory of Schwarzian derivatives shows CITE that for a linearly independent pair MATH, MATH of solutions of REF, the function MATH satisfies REF. Conversely, any function MATH satisfying MATH is obtained in this way. If MATH and MATH, then there is a fundamental system of solutions of REF in a neighborhood ... |
math/0008015 | The hyperbolic NAME map of the dual surface MATH is equal to the secondary NAME map MATH of MATH. Thus REF are equivalent to the condition that MATH is single valued at MATH, by REF. On the other hand, as seen in the proof of REF is also equivalent to the condition that MATH is single valued at MATH. |
math/0008015 | We write MATH where MATH and MATH are nonzero and holomorphic at MATH, for some integers MATH and MATH. If MATH, so MATH and MATH for some MATH, then the difference of the solutions of the indicial equations is MATH in all three cases, hence the three statements are clearly equivalent. If MATH, then the indicial equati... |
math/0008015 | Since MATH, we need to consider only the cases MATH and MATH. If MATH, then the hyperbolic NAME map is constant, so REF implies MATH. Thus MATH is a totally umbilic MATH immersion, so both MATH and MATH are horospheres. So we consider the remaining case MATH. Then MATH is meromorphic of degree MATH on MATH, which impli... |
math/0008015 | In this case, the hyperbolic NAME map MATH is of degree MATH. Without loss of generality, we may assume that MATH is an end of the surface. Moreover, by REF we may assume that MATH is a pole of MATH. As MATH is a branch point of MATH (since MATH), MATH has a pole of order MATH at MATH. Up to a constant multiple, the fu... |
math/0008015 | For given MATH, there is a meromorphic function MATH on MATH so that MATH since the right-hand side of REF has no residue. One can check that MATH when MATH. Hence, by REF , there exists a MATH-reducible MATH immersion MATH with MATH and MATH as in REF and secondary NAME map MATH satisfying REF. Conversely, let MATH be... |
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