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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 the domain of REF of continuous function MATH we get MATH . Integrating this inequality over MATH we obtain MATH . Therefore all conditions of REF hold and the embedding operator MATH is compact. |
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 subsequence MATH of the sequence MATH which converges in MATH to a function MATH. It is evident that MATH almost everywhere in MATH and the function MATH which is defined as MATH on MATH and MATH on MATH belongs to MATH. Hence MATH . Therefore MATH for MATH . |
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 the embedding operator MATH is compact. Because any domain MATH of class MATH is a finite union of elementary domains of class MATH the result follows from REF . |
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. Therefore the embedding operator MATH is compact. Because any domain MATH of class MATH is a finite union of elementary domains of class MATH, the result follows from REF . |
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 convergence. Since MATH: MATH is compact, it follows that MATH . Since MATH it follows that MATH and MATH. This is a contradiction: by REF the inequality MATH holds. Necessity of REF is established. Sufficiency: if MATH is compact then MATH implies that a subsequence MATH (denoted again MATH) converges in MATH, that is MATH as MATH. REF implies MATH. Fix an arbitrary small MATH. Note that MATH. Choose MATH and fix it. Then choose MATH so large that MATH. Then MATH. This implies convergence of MATH in MATH. The sufficiency is proved. |
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 MATH . It is now an easy exercise to solve these equations using the NAME functions for MATH and MATH: REF gives MATH as a function of MATH, MATH while REF shows that MATH is a multiple of the NAME function for MATH, MATH with MATH an arbitrary factor which drops out of the resulting solution: MATH . Setting MATH and MATH in this expression yields REF with MATH . Because MATH is finite, MATH is a rational function with finitely many simple real poles (which occur at the zeros of MATH). Hence, the partial fraction expansion (alternatively, the represenation theory of NAME functions) shows that MATH is of the form displayed in REF : MATH with MATH and MATH. (The coefficient of MATH is one, since MATH as MATH.) We now turn to the verification of the relation between the spectral measures expressed in REF. First consider the situation when MATH is finite. For a self-adjoint operator MATH on a finite dimensional vector space there is a useful formula for the spectral measure, MATH, associated to a vector MATH: MATH where MATH is the NAME ``function." (This formula offers a simple route to the ``spectral averaging" principle discussed in ref. CITE; its derivation is an instructive exercise which we leave to the reader.) Coupled with REF easily yields: MATH . When MATH is infinite we must turn to a more abstract derivation of REF . Writing each side of REF as a spectral integral we find that MATH . Expanding the right side of this equality with partial fractions yields MATH . This equation can be viewed as a special case of: MATH and indeed REF implies REF, for all MATH, since by the NAME Theorem the set of finite sums of the form MATH is dense in MATH. As mentioned previously, REF is the statement claimed in REF. Finally, the spectral inclusion MATH follows since MATH which may be verified using REF. If MATH is a cyclic family for MATH then REF shows further that MATH. In case MATH is a cyclic vector for MATH, this family is easily seen to be cyclic for MATH. This completes the proof of the proposition. |
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 from the right and make use of the unitarity of the scattering matrix in the form MATH thus obtaining MATH . Equivalently, this relation can be written in a form analogous to REF, MATH . In CITE we proved that REF has a solution for all MATH. If MATH is not an eigenvalue of the operator MATH then it has a unique solution. If MATH is an eigenvalue of MATH this solution is non-unique, but still determines the scattering matrix uniquely. Therefore from comparison of REF the relation REF follows. If MATH is real by the remark preceding the theorem we can choose the matrices MATH and MATH to be real. From this and from REF the second claim of the theorem follows. |
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-negative operator MATH. This gives MATH. From MATH it follows that MATH is a maximal eigenvalue of MATH and MATH. By the min-max principle (see for example, CITE) any MATH, MATH maximizing MATH satisfies MATH. Moreover, if MATH with some MATH, then MATH and MATH. Therefore since MATH we obtain MATH and MATH. The relations REF imply that MATH. |
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. Multiplying these equations by MATH we obtain MATH and thus MATH . Again by REF it follows from REF that MATH . Thus MATH and therefore by REF MATH. So we have proved that MATH. The inclusion REF and the equality REF imply that MATH . The inclusion REF and the equality REF imply that MATH . The proof of the relations MATH is similar and will therefore be omitted. We turn to the proof of REF - REF . By the unitarity of MATH from REF it follows that MATH. Since MATH for any linear operator MATH we obtain REF . By the unitarity of MATH from REF it follows that MATH which proves REF . As already noted the vectors MATH and MATH also satisfy MATH . A final application of REF yields MATH which by the unitarity of MATH and MATH implies REF which completes the proof of the lemma. |
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 blocks of MATH and MATH are not of maximal rank. |
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 also not of maximal rank. Now let us suppose that MATH. Then the matrix MATH is not of maximal rank iff there is a nontrivial vector MATH such that MATH. Again from the unitarity we have MATH and therefore MATH . Thus, the MATH matrix MATH is not of maximal rank. |
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 REF it follows that MATH. |
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 left. Next we multiply the resulting expression by MATH thus obtaining MATH . By REF we have that MATH is invertible and thus MATH . From the third relation in REF we obtain MATH . The fourth relation in REF gives MATH . The second relation in REF determines MATH. It remains to prove that MATH is unitary. By the unitarity of the matrix MATH we have MATH . Therefore MATH and thus by REF we obtain MATH . The relations MATH, MATH, and MATH can be proved similarly. The left inverse is constructed similarly and by means of the obvious relation MATH is easily shown to be equal to the right inverse. |
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 MATH and MATH are linearly independent functions and therefore the scattering matrices may uniquely be recovered from MATH. The columns of MATH define the external part of solutions of the NAME equation for MATH at energy MATH. We are looking for a square matrix MATH such that its MATH columns defines the external parts of a solution to the NAME equation for MATH. Here the indices are assumed to be arranged such that the first indices are those of MATH followed by the indices of MATH. The aim is to obtain MATH from MATH, MATH, and the lengths MATH of the new internal lines MATH. By the above observation this will determine the scattering matrix MATH. The strategy will be to find new solutions of the NAME equations for MATH with incoming plane waves REF in the channels MATH which agree suitably in the channels MATH and MATH. With the conventions made above we write MATH where the superscripts denote the sizes of the blocks. For arbitrary MATH matrices MATH and MATH, respectively, consider the MATH and MATH matrices MATH . Here MATH stands for the MATH unit matrix and MATH stands for the MATH zero matrix. Obviously, the columns of MATH are the external parts of linear combinations of the columns of MATH, and thus define the external parts of solutions of the NAME equations for the operators MATH, MATH. Note that MATH has an incoming plane wave in any of the channels MATH and MATH has no incoming plane wave in all channels MATH. Now we make the coordinate transformations MATH on the lines MATH (MATH). The reason for this is as follows. Under the gluing process MATH (see REF ) the two half-lines corresponding to MATH and MATH are replaced by the interval MATH, giving the new lines in MATH. This may be realized by identifying a point MATH on the half-line corresponding to MATH and with coordinate MATH REF with the point MATH on the half-line corresponding to MATH with coordinate MATH. In particular MATH on the MATH-line corresponds to MATH on the MATH-line and vice versa. Applying this transformation to MATH we obtain in this new coordinate system MATH . We now require that MATH and MATH agree on the lines labeled by MATH. This will fix MATH and MATH. Indeed, we then obtain MATH . By the linear independence of the functions MATH and MATH it follows that MATH and thus MATH . Since for any invertible MATH and MATH the identities MATH and MATH hold, we have MATH . Since MATH is assumed to be compatible with MATH the inverses in REF are well defined. Similar to REF and according to the ordering convention made above we write the scattering matrix MATH for the graph MATH in the block form MATH where the superscripts denote again the sizes of the blocks. Since MATH has an incoming plane wave in any of the first MATH channels MATH, REF allow one to determine MATH and MATH: MATH . To determine the remaining blocks of the scattering matrix MATH instead of REF we consider the MATH and MATH matrices MATH with arbitrary MATH matrices MATH and MATH. Again MATH are the external parts of some solutions of the NAME equations with the operators MATH, MATH. Now MATH has no incoming plane waves in the channels MATH, but MATH has an incoming plane wave in any of the channels MATH. Repeating the arguments used above we obtain the following matching conditions for MATH and MATH: MATH and thus MATH . Since MATH is compatible with MATH the inverses are again well defined. From this and from REF it follows that MATH . By the definition of the generalized star product REF we obtain REF. CASE: Now suppose that MATH. We prove that the composition REF remains valid. Also MATH and the multipicity of MATH equals MATH . The assumption MATH implies that MATH and MATH have nontrivial kernels. This implies that the homogeneous form of REF , MATH and MATH respectively, have nontrivial solutions. It is easy to prove that the inhomogeneous REF still have solutions in this case. Consider for instance REF , which is equivalent to MATH . By the NAME alternative this equation has a non-trivial solution iff MATH for any MATH satisfying MATH . By REF with MATH and MATH we have MATH . From the unitarity of MATH, which states in particular that MATH and from REF it follows that MATH . Since MATH for any operator MATH we obtain REF is discussed similarly. From REF it follows that the NAME equation with the operator MATH for given value of the spectral parameter MATH has (nonunique) solutions which have the form MATH and MATH where MATH and MATH (MATH and MATH) solve REF (REF, respectively). Note that MATH and MATH. On the lines in the set MATH the quantity MATH coincides with MATH and MATH and MATH. We will now prove that MATH . Thus, the functions REF are zero on all external lines of the graph MATH and their support has nontrivial overlap with the interval lines MATH. From REF it follows that MATH. By REF we have MATH . By unitarity it follows that MATH and thus MATH. The second relation in REF is proved similarly. Now we note that from REF it follows that MATH . The columns of REF correspond to linear independent eigenfunctions of MATH for the eigenvalue MATH. There are precisely MATH such eigenfunctions and the supports of all them have nontrivial overlap with the internal lines MATH. CASE: Let MATH and let MATH where the linear subspace MATH is defined by REF. This means that there are precisely MATH eigenfunctions of MATH which disappear if we cut the internal lines MATH. We will prove that MATH and that MATH which in turn implies that MATH . From the existence of the above mentioned eigenfunctions it follows that these eigenfunctions can be constructed by means of superposition and matching of the solutions REF of the NAME equation for the operators MATH and MATH at energy MATH. For any vectors MATH the functions MATH define the external parts of solutions of the NAME equations for the operators MATH, MATH. Since the eigenfunctions are supported on internal lines of the graph MATH REF the vectors MATH and MATH must satisfy MATH such that MATH vanishes in any of the channels MATH and MATH vanishes in all channels MATH. Making the coordinate transformation MATH on the lines MATH and requiring that MATH and MATH agree on the lines labeled by MATH, we obtain MATH or equivalently MATH . Linear independent solutions of REF correspond to linear independent eigenfunctions of MATH and vise versa. Thus REF implies REF. This completes the proof of the theorem. |
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 wiritten in the form MATH where MATH stands for the NAME pseudoinverse. |
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 global section MATH. Combining REF gives MATH that leads to the desired equivalence. |
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 exists a connection REF such that the relation REF holds, namely, MATH where MATH is a connection on MATH which has MATH as an integral section. It is easily seen that, in this case, the Hamiltonian map REF satisfies the relation REF for MATH. Hence, the Hamiltonian forms REF constitute a complete set. |
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 assume that MATH trivialises on the chart domain since only the behaviour of MATH in any arbitrarily small neighbourhood of MATH is relevant here.) With the indicated identifications provided by the chart, the required relation REF reads MATH and MATH in components of the local trivialization, where we have introduced MATH . Now let MATH for all MATH. This means that there is a function MATH with MATH and an open conic neighbourhood MATH of MATH (in MATH) so that, for all MATH, MATH holds for all MATH with suitable MATH. Now choose some MATH with MATH and MATH. Then the MATH are analytic functions of MATH, hence bounded on compact sets. This implies that it suffices to show that there are an open conic neighbourhood MATH of MATH and some MATH so that for all MATH holds for all MATH with suitable MATH, in order to conclude that MATH for all MATH. To prove that REF holds, we observe that MATH . Furthermore, since REF holds and since the cone MATH is scale-invariant, we see that MATH for all MATH. Thus, if MATH, we obtain from REF , for all MATH, MATH . Observing that MATH for MATH and using also that MATH holds for all MATH, MATH, the desired bound REF is implied by the last estimate. |
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 choose any open, relatively compact neighbourhood MATH, in MATH, of MATH. Denote the set MATH by MATH, and denote the set MATH by MATH. Then MATH, endowed with the appropriate restriction of MATH as spacetime metric, is a globally hyperbolic sub-spacetime of MATH. Thus there is a foliation MATH of MATH in NAME. Each MATH possesses a causal normal neighbourhood MATH in MATH, so MATH forms an open covering of MATH. Now consider two causal normal neighbourhoods MATH and MATH of MATH and MATH (in MATH) such that their closures are contained in MATH and MATH, respectively. Then let MATH . We write MATH for the open interior of MATH; it is also a globally hyperbolic sub-spacetime. Now MATH is a compact subset of MATH and hence there is a finite subfamily MATH of MATH covering MATH. It is not very difficult to see that one may choose such a family with the following properties: CASE: MATH holds for the corresponding NAME of which the MATH are causal normal neighbourhoods; CASE: For all MATH there is some MATH with MATH; CASE: MATH covers MATH and MATH covers MATH (by enlarging MATH, MATH, MATH and MATH if necessary). Now by assumption, MATH is of NAME form on MATH, and thus certainly MATH is of NAME form when restricted to MATH (that is, MATH restricted to MATH). By construction, MATH covers the part MATH of MATH. On the other hand, MATH is a globally hyperbolic sub-spacetime of MATH, so there is a NAME form MATH on MATH. Therefore, on MATH we have MATH mod MATH. Now MATH contains a NAME MATH for MATH (owing to the properties of causal normal neighbourhoods). And hence, since MATH is a NAME form on MATH and thus a bisolution of the wave-operator mod MATH, as likewise is MATH by assumption, this implies that MATH mod MATH on MATH, as follows by a straightforward generalization of REF. But this entails that MATH mod MATH on all of MATH, and thus MATH is of NAME form MATH. From here onwards, one iterates the just given argument to show inductively that if MATH is of NAME form on MATH, then it is also of NAME form on MATH, and hence MATH is of NAME form on all MATH, MATH ; compare the argument of ``NAME in small steps" in CITE. Finally, since MATH covers the part MATH of MATH, one concludes in a like manner that MATH is also of NAME form on MATH. And since the relatively compact set MATH entering in the definition of MATH was arbitrary, this shows that MATH is of NAME form on all of MATH. This establishes the statement of the Theorem for the case that MATH, but it is obvious that an analogous proof establishes the statement also in the case MATH. Now let MATH be an arbitrary NAME. Then one can choose a NAME MATH. One concludes first that MATH is of NAME form on any causal normal neighbourhood MATH of MATH, and then that MATH is of NAME form on MATH. |
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, lightlike MATH in NAME space, where MATH . Assume the wavefront set of MATH were empty at the base-point MATH. Then MATH is MATH near MATH and we find that MATH with MATH and MATH. But MATH is scale invariant, that is, MATH for all MATH, MATH, so that we are forced to conclude that MATH for all MATH. This entails MATH for each MATH and all MATH, which implies MATH for all MATH, where MATH is the Euclidean Laplacian in MATH. But this is clearly a contradiction since MATH is an elliptic differential operator and thus preserves the wavefront set of the distribution MATH and this is non-empty at coinciding base points as remarked above. Therefore, there are elements MATH. However, every such element must be of the form MATH since this is so for MATH and MATH results from MATH by application of a derivative operator. So, there is some MATH in MATH. Now we use that MATH is invariant under spatial coordinate rotations with respect to MATH (that is, rotations in the MATH-hyperplane) together with REF to conclude that each MATH, is contained in MATH. |
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-integral kernel, for all MATH. Therefore one obtains as in the proof of REF that MATH. Denoting by MATH the restriction of MATH to MATH, it follows that MATH whenever MATH is the two-point function of a NAME state on MATH. (For the CCR case this is immediate as in this case MATH for some NAME form MATH on MATH. For the CAR case this follows since then MATH for some NAME form MATH on MATH, and application of differential operators cannot increase the wavefront set.) For any quasifree state MATH fulfilling the CCR or CAR it holds that MATH is a bisolution for the wave-operator (it would be sufficient for the subsequent arguments that MATH be a bisolution mod MATH). This means that one can apply the PST, REF , in order to show that REF already implies MATH owing to the fact that MATH is a neighbourhood of a NAME MATH: Let MATH be an element of MATH. Then the first part of REF shows that MATH are both lightlike. As any inextendible lightlike geodesic intersects MATH, we can - provided that both MATH and MATH are non-zero - use the second part of REF to conclude that MATH, where MATH is the (unique) element of MATH with MATH. But then, because of REF, MATH with MATH future pointing. Thus we conclude that MATH. Now we will show that the PST in combination with the propagation of NAME form entails that MATH and MATH are absent from MATH. We will give an indirect proof and thus assume that MATH contains an element of the form MATH. Then there will be a NAME MATH passing through MATH; this NAME possesses a causal normal neighbourhood MATH. By the propagation of NAME form, MATH, being a bisolution (mod MATH would suffice) for the wave-operator, will be of NAME form on MATH. Moreover, MATH, and so there is some point MATH, MATH, with MATH. Since MATH is of NAME form on MATH, it follows (see above) that MATH, and thus MATH can only be contained in MATH if MATH and MATH. By the PST, this contradicts the assumption that MATH. Thus elements of the form MATH are absent from MATH, and by an analogous argument, also pairs of covectors of the form MATH aren't contained in MATH. Thus we have established the inclusion REF. Now we have to establish the reverse inclusion MATH . In order to prove this we use REF together with REF , showing that for any NAME state MATH on the NAME one has MATH . The same result can be derived for any NAME state MATH on the NAME. According to the PST, this implies that MATH for all MATH with MATH, MATH. Since this holds for all MATH in the causal normal neighbourhood MATH of a NAME surface, relation REF now follows. Thus we have proved MATH. CASE: Now let MATH be a state on the CCR or CAR algebra associated to some (wave or NAME) operator with the property that REF holds. Let MATH be a causal normal neighbourhood of any given NAME. According to REF , there is, in the NAME, a NAME form MATH on MATH fulfilling REF, and in the NAME, there is a NAME form MATH on MATH obeying REF. According to REF of the proof, it holds in either case, MATH so that one obtains MATH . Now we have in the NAME MATH. Introducing the flip morphism MATH and some bundle morphism MATH covering MATH, as well as MATH, this implies MATH . But because of the anti-symmetry of the set MATH, its intersection with its image under MATH, MATH, is empty, so one finds MATH . The same reasoning applies to the NAME, where MATH. Thus MATH is shown to be of NAME form on MATH in both cases. This shows that MATH is a NAME state. |
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 see that MATH is indeed a well defined distribution and independent of MATH. The limit of MATH for MATH is not locally integrable, so before we can apply REF to argue as above, we will have to rewrite REF, using integration by parts. To this end, introduce coordinates MATH on MATH, where MATH and MATH stands for some coordinatization of MATH. In these coordinates MATH . In case MATH is even, we carry out a MATH-fold partial integration with respect to MATH and arrive at MATH where MATH is some constant. Now, the limit of the integrand is locally integrable and turns out to be independent of MATH, too. Hence REF can be applied to show the desired result. In case MATH is odd, we carry out a MATH-fold partial integration with respect to MATH and arrive at MATH where again MATH is a suitable constant. Since the limit of the integrand is not locally integrable in MATH, we are not ready to apply REF yet. But we have already written the integrand in a suggestive form as to the next partial integration. This turns MATH into a logarithm and the MATH to MATH, at worst, thus rendering the integrand locally integrable in the limit MATH. We also get a boundary term at the integration boundary MATH, which is integrable in the limit MATH as well. Inspection of these limits shows that they are indeed independent of MATH, whence MATH is well defined and independent of MATH also in this case. As a consequence of the independence of MATH, we can write the distributions in the form given in the statement of the Lemma: Using the trivial time-function MATH and the abbreviation MATH, we get MATH . By exchanging the integration and the limit, we finally get the desired result. |
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 dealt with in exactly the same manner. We begin by considering MATH in case MATH is odd: By using the results of REF, we see that MATH where MATH is some constant depending on MATH and MATH . Now we change the integration variables from MATH to MATH, which we denote by MATH. We get MATH where MATH. To compute the limit MATH, we have to investigate the behaviour of MATH. We use REF and the fact that MATH where we have set MATH, MATH (see CITE for details), to conclude MATH with MATH given by REF. Using this, and MATH we get the desired result. The case MATH even is treated exactly the same way, with the exception that there is a term MATH in MATH which seems to blow up for MATH. Using partial integration, it is easy to see, though, that the term in MATH resulting from this term in MATH vanishes for any MATH. Therefore, the rest of the argument goes through unchanged. The argument for MATH runs along the same lines. |
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 these relations MATH . Thus, using the fact that MATH, we have MATH while MATH and MATH which completes the proof. |
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 MATH as acting formally like MATH. In any case, we arrive at MATH . Now use the NAME formula for sl REF mentioned above, and since the MATH and MATH commute with MATH, and MATH, the stated result follows. |
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 the result for MATH in REF . |
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 commutator with MATH, the first relation in REF yields MATH. And from the middle relation in REF , MATH yields MATH. |
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 relation MATH . Using the radiation condition we obtain that this integral tends to zero when MATH. Thus, MATH . |
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 the radiation REF we have MATH . The last term above vanishes when MATH and is negative when MATH. In both REF is proved. |
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 to the operator MATH . Consequently the integrals MATH taken over the sphere MATH (see the proof of REF ) decrease when MATH . Since MATH commute with MATH we obtain that the integral MATH taken over the sphere MATH also decreases when MATH . In the case when MATH is a zero divisor the radiation REF becomes MATH . Since the behavior of MATH in REF is of the type MATH when MATH and since the expression MATH contains the multiplication from the right-hand side by MATH, and in this case MATH, it can be seen that the integral REF taken over the sphere MATH decreases at infinity also. |
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 in the set MATH. Let MATH be the open subset of MATH consisting of curves MATH such that MATH is disjoint from some open neighbourhood of MATH. We show that the boundary MATH of MATH is empty. Thus MATH is open and closed and therefore coincides with MATH. Suppose the boundary MATH of MATH were non-empty and let MATH be a curve-segment in MATH. This implies that there is a point MATH which is met by MATH (see also REF ). Since MATH there is a curve-segment MATH which is sufficiently close to MATH, so that we can choose an open neighbourhood MATH of MATH and a local diffeomorphism MATH with the properties listed in REF , in particular such that MATH. We may even choose MATH and MATH such that MATH has a continuous extension MATH to MATH with MATH. Then there exists an open ball MATH with radius MATH such that MATH and MATH with MATH. We take a point MATH and a neighbourhood MATH of MATH such that MATH is a diffeomorphism onto an open neighbourhood of MATH. Because MATH has the properties listed in REF (in particular the second property), the smooth hypersurface MATH is timelike and hence non-characteristic. Therefore there is unique continuation for solutions across this surface. The solution MATH vanishes on MATH and by unique continuation the point MATH cannot be in MATH which is a contradiction. Hence, MATH is empty. Since the envelope MATH is the smallest set containing MATH with the properties that have now been shown for the complement of MATH, it follows that MATH vanishes on MATH. |
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. By REF vanishes in MATH and by REF it vanishes in MATH. Hence, the theorem is proved. |
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 in MATH, and therefore an isomorphism MATH. |
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 seen that MATH is defined by MATH equations, and therefore every irreducible component of MATH has dimension at least MATH. We deduce that MATH is pure dimensional and locally a complete intersection. If MATH, this implies that MATH, so that MATH is pure dimensional. The above decomposition of MATH shows that MATH is irreducible. |
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 consequence of REF . |
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 and MATH, for all MATH. It follows from the definition and from REF , that the pair MATH is canonical if and only if MATH for all MATH. Since MATH is locally a complete intersection, hence NAME, a result of CITE (see also REF ) says that MATH is canonical if and only if the pair MATH is canonical near MATH. Since MATH is smooth, this means precisely that MATH is canonical. On the other hand, since MATH is locally a product of MATH and an affine space, it follows that MATH is canonical if and only if MATH is canonical, which completes the proof of the theorem. |
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 Conjecture REF can be reduced to the following: if MATH is a normal, l.c.i. variety and MATH is a normal, NAME divisor on MATH which is log canonical, then MATH is log canonical around MATH. Applying Conjecture REF to MATH, we get MATH, for all MATH. We may assume that MATH is defined by one equation, so that MATH is defined by MATH equations. Therefore, if MATH is an irreducible component of MATH, with MATH, then MATH. Since by REF we have MATH, we deduce that MATH. We conclude that there is an open neighbourhood MATH of MATH such that MATH. By REF , it is possible to pick MATH (depending on an embedded resolution of MATH) such that if MATH, then MATH is log-canonical, which finishes the proof of the proposition. |
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 regular system of parameters MATH in MATH, we get an étale ring homomorphism MATH. Let MATH be an equation for MATH at MATH. In general MATH. However, if MATH is such that MATH, then MATH. Since MATH induces an isomorphism of the associated graded rings, we can find MATH as before such that MATH. Therefore, in order to prove REF , we may assume that MATH. Since MATH is étale, by replacing MATH and MATH with suitable open neighbourhoods of MATH and MATH, respectively, we can apply REF to reduce to the case when MATH, MATH and MATH is defined by a polynomial MATH. We use the equations for MATH described in the first section. With the notation MATH for every MATH, we have MATH, where MATH is defined in MATH by the polynomials MATH, with MATH. We prove that MATH by computing the dimension of a deformation of this set. Note that if we put MATH for every MATH and MATH, then each polynomial MATH is homogeneous of degree MATH. Consider the family MATH over MATH defined in MATH by the polynomials MATH, with MATH. The fiber of MATH over every MATH is isomorphic to MATH, while the fiber over MATH is the corresponding scheme obtained by replacing MATH with its homogeneous component of degree MATH. Since all the rings are graded (for the grading we defined above), the semicontinuity theorem for the dimension of the fibers of a morphism (see REF ) shows that in order to prove that MATH, we may assume that MATH is homogeneous of degree MATH. Consider now on MATH the reverse lexicographic order where we order the variables such that MATH if MATH or if MATH and MATH (see, for example, REF). Let MATH be the initial term of MATH is this order. It is then easy to see that MATH, for MATH. Therefore the initial ideal of the ideal defining MATH has dimension MATH. Since the dimension of an ideal is equal with the dimension of its initial ideal, we deduce that MATH, which concludes the proof of the lemma. |
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 definition, we get MATH, where MATH . Every monomial which appears in the MATH term of MATH, has degree bounded above by MATH and below by MATH, where MATH for all MATH (recall the convention that MATH). Moreover, we always have precisely one monomial of degree MATH, namely MATH, whose coefficient is MATH, the number of irreducible components of maximal dimension of MATH. Similarly, every monomial which appears in the MATH term of MATH has degree bounded above by MATH and below by MATH, where MATH for all MATH. We always have exactly one monomial of degree MATH, namely MATH, whose coefficient is MATH. A first consequence of this and REF is that MATH is, indeed, integrable. Using condition MATH, it is an easy computation to show that we have MATH, for every MATH. Moreover, REF gives MATH, for every MATH, and REF implies that the inequality is strict for infinitely many MATH. We deduce that MATH, for every MATH and equality holds if and only if MATH and MATH (therefore, the inequality is strict for infinitely many MATH). We conclude from the above inequalities first that in MATH, the term MATH appears precisely once for every MATH, and has coefficient MATH. Similarly, in MATH, the term MATH appears at most once. It appears if and only if MATH and MATH, and in this case it has coefficient MATH. We use the change of variable formula to compute the integral of MATH, as MATH . In this form, MATH can be explicitely computed, since MATH has normal crossings. For every subset MATH, let MATH. With this notation, we have MATH . We just sketch the proof of this formula, as it is similar to that of REF, or that of REF. Since MATH is additive, we may work locally on MATH. In order to compute the part MATH in MATH which corresponds to arcs over MATH, we may assume that there is a regular system of parameters MATH on MATH, such that MATH is defined by MATH, for all MATH. We have MATH . For every MATH, if MATH, then MATH where MATH is locally trivial over MATH, with fiber MATH, and our formula for MATH follows. Every monomial in the term of MATH corresponding to MATH, has degree bounded above by MATH, and below by MATH, where MATH and MATH. Note that if MATH, then MATH. We introduce one more piece of notation: for MATH, let MATH. For every MATH, we have MATH. We see that for MATH, we have MATH . Moreover, REF implies that if MATH, then MATH and that MATH. In particular, this implies that the only monomial of the form MATH which can appear in the term corresponding to MATH and MATH is for MATH. To prove the implication MATH in REF , suppose that MATH, for all MATH and assume that for some MATH, we have MATH. The above inequalities show that MATH does not appear in the sum MATH, for every MATH. As we have seen, this imples that MATH. In particular, we have MATH. Continuing in this way, we get MATH, for every MATH, a contradiction with REF . Therefore we must have MATH, for all MATH, so MATH. Suppose next that MATH, for all MATH. The above argument shows that MATH, for every MATH. In particular, the coefficient of MATH in MATH is MATH, for every MATH. From the above inequalities, we see that for every MATH, the term MATH appears in MATH if and only if MATH and in this case it has coefficient MATH, since MATH is irreducible. Therefore MATH, for every MATH. This proves the implication MATH in REF . We next turn to the implication MATH. Suppose that for some MATH, with MATH, for all MATH, we have MATH, and that for some MATH, MATH. We pick the function MATH such that in addition to MATH it satisfies the following requirement. For every MATH, consider the set MATH of all the pairs MATH, such that MATH. This is clearly a finite set. We require that for every MATH and every MATH, for some MATH with MATH, we have MATH or equivalently, we have MATH. On the other hand, if MATH, for some MATH, then by MATH we have MATH . Note that the top degree monomials which appear in different terms of the sums MATH (for possibly different MATH) don't cancel each other, because they have positive coefficients. Let MATH be the highest degree of a monomial which appears in a term corresponding to some MATH. By the previous remark, the corresponding monomial does not cancel with a monomial in a term corresponding to MATH. Since MATH, our hypothesis implies that MATH, while from MATH we deduce MATH. Moreover, we deduce from MATH and MATH that the monomial of degree MATH does not cancel with monomials in terms corresponding to MATH, if MATH. This shows that in MATH there is indeed a monomial of degree MATH, where MATH, a contradiction. Suppose next that for some MATH, with MATH, for all MATH, we have MATH and MATH, and that for some MATH, MATH. The above argument shows that MATH, for every MATH. Note that since MATH, we have in the expression of MATH the monomial MATH, with coefficient at least MATH, once from MATH (with MATH), and once from MATH (with MATH). This gives a contradiction. |
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 MATH-valued points. A MATH-point of MATH is given by an algebra homomorphism MATH of the form MATH, where MATH corresponds to MATH. The condition that MATH is an algebra homomorphism is equivalent with saying that for every MATH, with MATH, the morphisms MATH mapping MATH to MATH are MATH-valued points of MATH. If MATH is a MATH-valued point of MATH, then MATH if and only if the morphism MATH, given by MATH factors through MATH. This condition is satisfied in particular if MATH is a MATH-valued point of MATH. The proof of the second assertion is similar, giving for the projection MATH, an isomorphism MATH. |
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 MATH is irreducible for some MATH, then MATH is locally complete intersection, and then use REF . But for every MATH, REF implies that if MATH, then MATH, and therefore MATH is not irreducible. We conclude that MATH at every singular point MATH, in particular that MATH is locally complete intersection. |
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 consisting of ring homomorphisms MATH with MATH (we make the convention MATH). We obviously have MATH. Set MATH and assume that MATH is irreducible. If MATH, for every MATH, then MATH can be chosen arbitrarily with order MATH, so that MATH. Since MATH, we get MATH. Next suppose that MATH and that MATH. We will show that MATH is irreducible. Consider first the case when MATH, for all MATH. As above, in this case MATH can be chosen arbitrarily with order MATH. Therefore we get MATH . On the other hand, if MATH, and if MATH, then by hypothesis there is MATH, such that MATH, and MATH. For every choice of MATH with order MATH, MATH, we have MATH, where the image of MATH in MATH can take finitely many values. Moreover, MATH is uniquely determined by the image of MATH in MATH. Therefore we have MATH . It follows that MATH, so MATH is irreducible. Since this is true for all MATH with MATH, it follows by REF that MATH is irreducible for all 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 MATH of MATH. The inequality MATH for an arbitrary point MATH gives MATH . This implies that MATH, and therefore MATH, hence MATH. The inequality MATH for MATH gives now MATH. |
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, rational, polyhedral cone of maximal dimension in MATH, for some lattice MATH of rank MATH. Let MATH, where MATH is the dual lattice. For every face MATH, we have a corresponding orbit MATH of dimension MATH and a distinguished point MATH defined by the semigroup morphism MATH, MATH, if MATH and MATH, otherwise. By REF , it is enough to show that if MATH, then MATH . We use the inductive description of MATH, due to NAME, for the case when MATH is locally complete intersection (see CITE). There are MATH and MATH, with MATH, such that MATH can be obtained as follows: take MATH, MATH and for every MATH, MATH, there is MATH, such that MATH where MATH. We show by induction on MATH that if MATH is a face of MATH, then there is a nonsingular fan MATH refining MATH, with rays MATH such that: CASE: MATH span a cone of MATH. CASE: For every MATH, MATH span a cone of MATH. CASE: MATH has at least MATH elements. The assertion is trivial when MATH. For MATH, let MATH be the refinement corresponding to MATH and MATH. MATH consists of the cones spanned by MATH and MATH, where MATH is a cone in MATH. Notice that MATH, with equality if and only if MATH or MATH. If we take MATH to be MATH or MATH (we pick the one in MATH, if possible), and MATH to be the other one, then this refinement satisfies the requirements for MATH. In order to prove MATH, notice that MATH consists of semigroup homomorphisms MATH, such that the composition with the projection onto MATH is MATH. Let MATH and consider the refinemet MATH constructed above. Let MATH be integers such that MATH, if MATH, and MATH, if MATH. Take MATH to be the locally closed subset of MATH consisting of morphisms MATH, with MATH. We clearly have MATH. Fix MATH. Since MATH is a nonsingular fan, by REF and MATH, there is a relation MATH . This shows that for MATH, the image of MATH in MATH is uniquely determined by the values of MATH on MATH, MATH. Therefore we get MATH since MATH, and there is at least one MATH, with MATH (as MATH, we have MATH, and going downward, we get MATH). |
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 the fibers of MATH, it is also surjective. Hence the corresponding map of jet schemes MATH is surjective. Given two points MATH in the same fiber of MATH, let MATH be a point in the (non-empty) fiber MATH. Then MATH. Hence MATH acts transitively along the fibers of the map MATH. Since MATH is normal, this implies that the ring of MATH - invariant functions on MATH equals MATH. Because MATH is a smooth open subset of MATH, we obtain that MATH is dense in MATH, and so any MATH - invariant function on MATH is determined by its restriction to MATH. This proves the proposition in the case of the finite jet schemes. The same argument works in the case the infinite jet scheme as well. |
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 with finite-dimensional homogeneous components, we obtain that any graded lifting of a MATH - basis of MATH to MATH is a MATH - basis of MATH. Conversely, the image of any MATH - basis of MATH in MATH is a MATH - basis of MATH. Now choose any graded basis MATH of MATH over MATH. Then the image MATH of MATH in MATH is a MATH - basis of MATH. According to REF , the image of MATH in MATH can be extended to a MATH - basis of MATH. Hence the image of MATH in MATH (we denote it also by MATH) can be extended to a MATH - basis MATH of MATH. Thus, we obtain a directed system MATH, of sets of graded elements of MATH and embeddings MATH. Let MATH be the union of all of the sets MATH. Note that by REF , MATH is the inductive limit of the directed system MATH. We claim that MATH is a basis of MATH over MATH. Indeed, consider the multiplication map MATH. This map is surjective: take any homogeneous element MATH of MATH of degree MATH. Then it already belongs to MATH. Hence by construction MATH lies in the image of MATH. The map MATH is also injective. Indeed, suppose there is an element MATH in the kernel of MATH. Without loss of generality we can assume that MATH is homogeneous of degree MATH. But then MATH belongs to the kernel of the map MATH, and so MATH by construction. This completes the proof. |
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 invariant function on MATH (respectively, MATH) is a constant. |
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 . Hence MATH is a non increasing function on MATH and so for every MATH . CASE: Consider the function MATH defined by MATH . We are going to show that MATH. We have MATH . If we denote MATH, we get that MATH . We have MATH whenever MATH, since MATH is concave and MATH. It is clear that for every MATH, MATH and MATH belong to the interval MATH, therefore MATH on MATH. Hence we conclude that MATH for all MATH, and so MATH, for all 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 (see CITE or CITE) then we get that MATH where MATH . Notice that SREF and SREF can be deduced from the identity MATH applied to MATH and MATH respectively. Now we are going to study each summand separately: SREF: It is easy to check that MATH SREF: It can be shown that MATH SREF: MATH. Notice that since MATH we get that for every MATH and every MATH . On the other hand MATH . Therefore we obtain that for each MATH, MATH for all 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 function we have MATH . Consider the function MATH defined for MATH and MATH. MATH . Then MATH . Since MATH, we achieve MATH . Hence if MATH, MATH for all MATH and if MATH for all MATH and thus, the result holds. |
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 we have MATH since the function MATH is non increasing for MATH. Therefore the result follows. |
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 , for every MATH we get that MATH, therefore it is enough to prove REF for MATH. Furthermore, since MATH is log-convex on MATH, the function MATH is a non decreasing funtion on MATH, so MATH for all MATH. On the other hand, since MATH, we can use REF , MATH and MATH, and we obtain that MATH . Thus, it sufficies to show that MATH for all integer MATH and this is 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 prove that MATH . |
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 neighborhood basis of MATH consisting of closed sets. This proves REF . That REF holds is obvious (just take MATH). Conversely, let MATH satisfy REF . Then there exists an ideal MATH of MATH such that MATH and MATH. Let MATH be a positive element of the NAME ideal MATH of MATH. Then MATH is continuous on MATH because MATH. Since MATH vanishes off of MATH, it is continuous on all of MATH. Thus MATH, whence MATH, and MATH. |
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 point. It follows that MATH is an ideal of MATH with continuous trace and MATH. Finally, if MATH is a neighborhood basis of MATH consisting of closed sets then MATH is a neighborhood basis of MATH with the same properties. Thus MATH by REF . We have shown that MATH and hence MATH are MATH-invariant. The final assertion follows from REF . |
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 the form MATH for some closed ideal MATH of MATH. We can shrink MATH a bit if need be, and assume that there are open neighborhoods MATH of MATH in MATH and MATH of MATH in MATH such that MATH is homeomorphic to MATH. Suppose that MATH and MATH are distinct points in MATH. Note that each orbit is closed in MATH because MATH is liminal CITE. Thus, for each MATH, the points MATH and MATH are distinct in MATH. Since MATH is NAME and NAME is continuous, we can separate MATH and MATH by MATH-invariant open sets and it follows that MATH is NAME. Thus, MATH is a MATH-invariant neighborhood of MATH which is NAME because MATH is NAME and the stability subgroups vary continuously CITE. |
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 net lifts to a net MATH in MATH with MATH. Now MATH and MATH, so that MATH for some MATH and MATH, that is, MATH. Now MATH is a net in MATH. Since MATH is MATH-wandering MATH is relatively compact. By passing to a subnet and relabeling, we may assume that for some MATH the net MATH converges in MATH. Since MATH and MATH are fixed, MATH also converges. Since MATH and MATH, we conclude that MATH converges. |
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 relabel so that there are MATH such that MATH in MATH. Then MATH and MATH is closed. |
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 MATH, and MATH has continuous trace. The action of MATH on MATH is MATH-proper by CITE. Note that MATH, and let MATH be a neighborhood of MATH which is compact in MATH. Then MATH is MATH-wandering relative to MATH, and since MATH is MATH-invariant MATH is also MATH-wandering relative to MATH. Conversely, assume each point in MATH has a MATH-wandering neighborhood. Then REF implies that the orbits are closed, and MATH is postliminal CITE (even liminal CITE). In particular, each MATH is of the form MATH for some MATH and MATH. Let MATH be a MATH-wandering open neighborhood of MATH. By REF the action of MATH on MATH is MATH-proper. Since the stability subgroups vary continuously it follows from CITE that MATH is an ideal of MATH which has continuous trace. Thus MATH is a Fell point of MATH, whence it is also a Fell point of MATH. |
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 follows that the map MATH from the set of MATH-invariant open subsets of MATH to the set of ideals of MATH is a bijection. By REF , MATH where MATH is an open MATH-invariant subset of MATH. Let MATH be as in REF. Suppose that MATH. Since MATH has continuous trace, it is certainly postliminal, and MATH for MATH and MATH. Furthermore, CITE implies that the stabilizer map MATH is continuous on MATH and that the action of MATH on MATH is MATH-proper. Let MATH be a compact neighborhood of MATH in MATH. Then MATH is MATH-wandering relative to MATH, and since MATH is MATH-invariant, MATH is also MATH-wandering relative to MATH. Let MATH be the quotient map. We claim there is a closed neighborhood MATH of MATH in MATH such that MATH. To prove the claim, we identify MATH with MATH. Then REF implies MATH has a closed neighborhood MATH. The map MATH is continuous by CITE, and factors through MATH by CITE. Thus we get a continuous map MATH. Let MATH. Then MATH is a closed neighborhood of MATH. To prove the claim, it remains to see that MATH. But if MATH, then there is a MATH such that MATH. In particular, MATH. Since MATH is open and MATH-invariant, it follows that MATH. (We have MATH for MATH.) Thus MATH and MATH have the same closures in MATH. But MATH is liminal and each orbit must be closed in MATH CITE. Thus MATH as claimed. With MATH as above, set MATH. Note that MATH is compact and MATH-wandering and MATH is closed. Finally, MATH is NAME because MATH, and MATH is NAME since MATH has continuous trace CITE. This implies that MATH. Therefore MATH, and MATH. To prove the reverse implication notice that MATH is a Fell algebra by REF . In particular, it is postliminal, and every irreducible representation of MATH is of the form MATH for MATH and MATH. We will show that MATH by verifying REF . Since MATH is a Fell algebra MATH has a NAME open neighborhood MATH, where MATH is a closed ideal of MATH CITE. Note that MATH is a Fell algebra with NAME spectrum. Hence MATH has continuous trace. This establishes REF . Let MATH be a compact MATH-wandering neighborhood of MATH as in REF. We identify MATH with MATH. Note that MATH is a closed neighborhood of MATH (first consider the complement and recall that the quotient map is open). That MATH is NAME follows from CITE because MATH is NAME and the stability subgroups vary continuously on MATH by REF . Let MATH be a neighborhood basis of MATH in MATH consisting of compact sets. Since a compact subset of a NAME space is closed, MATH is a neighborhood basis of MATH in MATH consisting of closed sets. This establishes REF . Since MATH was an arbitrary irreducible representation of MATH, we must have MATH. Therefore MATH and we're done. |
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 that MATH. Now MATH and MATH are disjoint open neighborhoods of MATH and MATH, respectively. Thus MATH and since MATH is open we have MATH. |
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 in MATH for every MATH by CITE, so MATH. Let MATH. Let MATH be an open neighborhood of MATH such that MATH is closed in MATH for each MATH. Not every orbit in MATH can be closed in MATH because MATH is the largest liminal ideal. Suppose that MATH is not closed in MATH. Then there exists MATH and MATH such that MATH. Since MATH, MATH has a neighborhood MATH such that MATH is closed in MATH for all MATH. But we can assume that MATH for some MATH and then MATH must be closed in MATH. Thus MATH, and this is a contradiction. Hence MATH and we are done. |
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 neighborhood of MATH such that MATH is MATH. Note that MATH cannot be MATH by the maximality of MATH. Choose distinct points MATH and MATH in MATH such that every open neighborhood MATH of MATH contains MATH and every open neighborhood MATH of MATH contains MATH. If MATH and MATH then MATH is an open neighborhood of MATH which does not contain MATH, which is a contradiction. If MATH and MATH both belong to MATH or if MATH and MATH both belong to MATH then we get an immediate contradiction because MATH and MATH are open and MATH. Hence MATH. |
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 are no MATH - holomorphic spheres in MATH. Then an argument given by NAME in CITE produces a non - constant MATH - holomorphic disk in MATH with boundary on MATH and (symplectic) area smaller than or equal to MATH. But due to the injectivity of MATH there is another disk completely lying in MATH. Since MATH is monotone the symplectic areas of both disks agree. The latter is an integer multiple of MATH. NAME holomorphic disks have positive symplectic area, hence MATH. |
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 of MATH of the form MATH where MATH and MATH are holomorphic and nonzero at MATH. Then MATH satisfies MATH where MATH is the covering transformation which corresponds to a small loop around MATH, implying MATH. So for REF , we have MATH for all MATH, and therefore also for MATH, which implies that MATH is a meromorphic function on MATH. By REF , there exists a conformal MATH immersion MATH on MATH with the secondary NAME map MATH for all MATH. If MATH, then MATH coincides with MATH by REF, so we have that the MATH-parameter family MATH are complete conformal MATH immersions with hyperbolic NAME map MATH and NAME differential MATH. We remark here that if MATH and MATH, then the monodromy matrix MATH defined by MATH is not diagonalizable and is not even in MATH. So any MATH immersion on MATH (with MATH and MATH) cannot be well-defined on MATH when some MATH. Next we consider REF , that is MATH. There exists a fundamental system of solutions of REF of the form MATH where MATH and MATH are holomorphic and nonzero at MATH. When MATH is the covering transformation induced from a small loop about MATH, MATH satisfies MATH . In particular, MATH. On the other hand, in the proof of REF, we have seen that MATH for MATH. Hence MATH and are diagonal matrices for all MATH, and we are in the MATH-reducible case. Note that this remains true when MATH is replaced by MATH where MATH is the set of positive reals. So we have a one-parameter family of complete conformal MATH immersions with hyperbolic NAME map MATH and NAME differential MATH and secondary NAME maps MATH for MATH. (MATH and MATH for MATH will not produce equivalent surfaces, as MATH.) Furthermore, REF implies there is only a one-parameter family of MATH immersions with data MATH. |
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 equation in the first case (respectively, second case, third case) is MATH . Hence the difference of the roots is MATH (respectively, MATH, MATH), and so all three statements hold. |
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 implies MATH. Hence we may choose MATH, and by REF, we may assume MATH. Since MATH has no branch points, REF implies there are no umbilic points, and REF implies MATH at each end MATH. By REF and the fact that MATH, we have MATH . By REF, we have MATH, so MATH or MATH. CASE: In this case, REF implies MATH. We may put the end at MATH, and then MATH has a single pole of order MATH at MATH and no zeroes. Thus MATH for some MATH. A MATH surface in MATH with secondary NAME map MATH and NAME differential MATH is called an NAME cousin CITE. So a surface with data MATH is the dual of an NAME cousin CITE. (Recall that dualizing switches the two NAME maps, and changes the NAME differential only by a sign.) CASE: In this case, REF becomes MATH. Then MATH (MATH), by REF. Hence the immersion MATH is a MATH surface of genus MATH whose two ends must both be regular CITE, and this type of surface is classified in CITE. In particular, MATH is in the case MATH of REF. So the surface is either a catenoid cousin CITE or a warped catenoid cousin with embedded ends (the case MATH in REF). |
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 function MATH is uniquely characterized as a degree MATH meromorphic function on MATH with a pole of order MATH at the origin CITE. Thus we have MATH, and we can normalize MATH, by REF. Suppose MATH generates MATH. Then the branch points of MATH are MATH, MATH, MATH and MATH modulo MATH, which are the ends and umbilic points. We assume MATH and MATH are the ends. (If MATH is an end, for example, we may change the generator MATH to MATH.) Thus the umbilic points are MATH and MATH. Next we find the NAME differential MATH, using the following fact: Let MATH and MATH be points in MATH such that MATH, MATH, and MATH. Then MATH is a meromorphic function on MATH such that MATH respectively, MATH are the set of zeroes (respectively, poles), that is, the divisor of MATH is MATH. Conversely, any elliptic function on MATH with the same divisor is of this form. The meromorphic function MATH should have poles of order MATH at MATH, MATH (ends) and zeroes of order MATH at MATH, MATH (umbilic points). Thus MATH can be written as in REF. |
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 a MATH-reducible MATH immersion of type MATH with MATH. Then MATH and MATH are as in REF. Let MATH (respectively, MATH) be the difference of the roots of the indicial equation of REF at MATH (respectively, MATH) for such MATH and MATH. Then we have MATH and MATH. Since MATH is MATH-reducible, MATH and MATH are positive integers (so also MATH). We may assume MATH. (If not, we can exchange the two ends MATH and MATH, by changing MATH and MATH to MATH and MATH. Using REF, we see that REF is unchanged.) Suppose that MATH, then MATH is not branched at both MATH and MATH. Noting that the branching orders of MATH and MATH are equal at any finite point of the surface (this follows from REF ), we see that MATH has branch points of order MATH at MATH and MATH and no other branch points. So MATH has degree MATH and MATH for some MATH and so MATH, which is impossible. Thus MATH, and it follows that MATH is a positive integer. By setting MATH, we have MATH . Thus MATH and MATH are as in REF with MATH. |
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