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math/0012263 | If MATH is a coherent sheaf then MATH is in MATH by the above REF . If MATH is in MATH then by REF MATH . If MATH then the latter is zero for MATH and hence MATH has zero cohomology except in the component of degree MATH. But then MATH and MATH are isomorphic in MATH. |
math/0012263 | This is straightforward. |
math/0012263 | Condition i. implies by REF that MATH has a filtration MATH with the cokernels MATH in the filtration of the form MATH which is a complex (displayed with components of degree MATH and MATH) MATH . Conditions i,ii., and iii. now imply all but the last statement in a. There is an exact sequence MATH giving an exact sequence MATH . Now by REF , MATH is quasi-isomorphic to MATH and hence its sheafification MATH is zero. Since MATH is a bounded complex of cofree MATH modules of finite corank, and MATH is exact, it follows easily that MATH. Thus MATH the latter according to REF . Hence the last statement in a. follows. CASE: Let MATH. Then clearly i. and ii. in a. holds. Also MATH is isomorphic to MATH by REF . Hence the conditions on MATH gives that MATH for MATH say. Taking a projective resolution of MATH we get a complex MATH in MATH with MATH. Also there are natural maps MATH which are homotopy equivalences. By the uniqueness REF we must have a homotopy equivalence MATH . |
math/0012263 | Let MATH and consider it as a complex in MATH. By REF there is a quasi-isomorphism MATH . The middle complex is (displayed with components of degree MATH and MATH) MATH and so we get the proposition by sheafifying this complex and letting MATH be MATH. |
math/0012263 | We get a functor MATH . We shall prove that this functor is fully faithful and that every object in MATH is isomorphic to MATH for some MATH in MATH. This will establish that MATH is an equivalence of categories, by CITE. Let MATH and MATH be in MATH. The exact sequence MATH gives a triangle in MATH and an exact sequence MATH which gives an isomorphism MATH . We also get from the triangle in MATH an exact sequence MATH . Now since MATH is homotopy equivalent to a minimal complex, MATH will be homotopy equivalent to a bounded above complex. Since MATH is acyclic, REF gives an isomorphism MATH . Together with REF this gives that MATH is a fully faithful functor. Also MATH since MATH gives an equivalence of categories. Hence MATH is fully faithful. Now given MATH in MATH. We want to show that MATH for some MATH in MATH. Let MATH. Then MATH is a bounded below minimal complex. Also MATH . So MATH is nonzero only if MATH. This gives MATH. We now claim that MATH can be completed to a minimal complex MATH in MATH with MATH. This follows by looking at MATH. We see that CASE: MATH is non-zero for only a finite number of MATH's. CASE: MATH is a finite length module. CASE: MATH non-zero implies MATH. Hence MATH can be completed to a minimal complex MATH in MATH with MATH. But then MATH. This shows that MATH is an equivalence of categories. By REF we now see that if MATH is in MATH then MATH . By the main REF we see that MATH is a quasi-inverse to MATH. |
math/0012263 | Let MATH. By REF and what is stated just before this proposition, it is clear that MATH for MATH if MATH and that MATH for MATH if MATH. Thus MATH exists and its sheafification is quasi-isomorphic to MATH by REF . Now we have the short exact sequence MATH . We see by REF , that if MATH is non-zero then MATH for some MATH. This implies MATH. Furthermore we see by REF that MATH is non-zero only if MATH for some MATH which implies MATH. Thus we get that MATH is zero for MATH not in the interval MATH. Since MATH is minimal this gives MATH for MATH not in the interval MATH. Let MATH. Then MATH and by REF a. we get that the kernels MATH in the filtration of MATH are isomorphic to MATH. But also MATH. Thus in the cofiltration of MATH the kernels MATH for MATH are equal to MATH. Hence we get MATH for MATH since the functor REF in Subsection REF is an isomorphism of categories. |
math/0012263 | This follows since MATH is MATH and thus by REF , MATH is isomorphic to MATH. |
math/0012263 | The component MATH is MATH. Suppose that MATH is non-zero for some MATH and MATH and let MATH be minimal such. Since MATH is a minimal complex, MATH is in the kernel of the differential MATH. But again since MATH is minimal this is not in the image of MATH whose domain is MATH. Since MATH is acyclic this must mean that MATH for all MATH and MATH and this proves ii. Thus MATH for MATH so the complex MATH is acyclic for MATH. Thus MATH is surjective. But this is just the map MATH and this proves i. |
math/0012263 | If MATH is a coherent sheaf, this is just the definition of MATH. Let MATH be a triangle in MATH. The functor MATH takes triangles to triangles. Taking cohomology we then get a long exact sequence of cohomology. Thus if the lemma holds for MATH and MATH it holds for MATH. Since the coherent sheaves generate the triangulated category MATH, we get the lemma. |
math/0012263 | By REF , MATH is isomorphic to MATH in MATH. Since MATH is the complex (displayed with components of degree MATH and MATH) MATH we get that MATH . |
math/0012263 | If MATH is a coherent sheaf and MATH then the equality clearly holds. Since MATH is generated as a triangulated category by coherent sheaves, and the right hand side in the equality above is invariant for homotopic complexes, the theorem follows. |
math/0012263 | Suppose first that MATH so MATH is a hyperplane in MATH. Let MATH be a locally free resolution of MATH, and let MATH. Then there are exact sequences MATH . Since MATH is a locally free sheaf on MATH and no component of MATH (in reduced form) is contained in MATH, the element MATH is not contained in any associated prime of MATH. In the diagram MATH the right vertical map is therefore injective. Thus the cokernels MATH is an exact sequence. This gives that MATH is a locally free resolution of MATH and hence MATH is isomorphic to a coherent sheaf. By cutting down with hyperplanes, this proves the if part of the statement in generality. To prove the converse we may assume by first cutting down with a linear subspace intersecting MATH properly, that a component of MATH (in reduced form) is contained in MATH. Let MATH be in MATH. Then MATH is not left exact. Applying this again to the exact sequence MATH where MATH is the kernel, and proceeding, we see that the sequence MATH is not left exact. Hence MATH is not a resolution of MATH. |
math/0012263 | We may assume MATH is a hyperplane inclusion so MATH is a one-dimensional subspace of MATH. Let MATH be in MATH and let MATH be in MATH be such that MATH has bounded cohomology. (Note that this is always true if MATH for some MATH in MATH.) Then to give a morphism MATH corresponds to a morphism MATH which gives a morphism MATH in MATH. Such a map can be represented by a diagram of morphisms of complexes MATH where MATH is a quasi-isomorphism. By the proof of REF we may assume that MATH is a bounded complex of vector bundles. But then we clearly get a diagram MATH where MATH is a quasi-isomorphism. This gives us a morphism MATH in MATH and thus a morphism in MATH . By REF in Subsection REF there is an isomorphism of functors MATH . Thus we get an isomorphism MATH . From this isomorphism and REF we get, using the adjointness of MATH and MATH, a morphism in MATH . Taking MATH this gives our natural transformation of functors. It remains to prove that the natural transformation is an isomorphism. There is a homotopy equivalence MATH . By the constructions above MATH is an isomorphism in MATH and hence MATH is an isomorphism in MATH. By REF , MATH is a homotopy equivalence, and hence MATH is a homotopy equivalence. |
math/0012263 | Immediate. |
math/0012263 | By REF there is a "twisted" quasi-isomorphism MATH and by REF in Subsection REF there is an isomorphism of complexes of MATH-modules MATH . By the previous REF there is a homotopy equivalence of complexes of MATH-modules MATH . Now for any MATH in MATH there is a quasi-isomorphism of complexes of MATH-modules MATH . Then by letting MATH be MATH we get a quasi-isomorphism MATH and this proves the proposition. |
math/0012263 | First note that MATH and MATH both denote the fiber of MATH at the point MATH. Let MATH be a locally free resolution. Then MATH may be identified with the complex MATH. Since this is sheaves over a point MATH, MATH is isomorphic to MATH for any MATH and the latter may be identified with MATH . Since MATH by REF , we get the theorem. |
math/0012263 | The projective dimension of MATH is the largest integer MATH such that MATH is non-zero, by CITE. Hence c. follows, and a. and b. are clear. |
math/0012263 | Immediate. |
math/0012263 | Let MATH be the degeneracy locus of MATH. By REF , MATH is not in MATH for any point MATH in MATH. Hence MATH is empty, and MATH and MATH. The converse part is also clear. |
math/0012263 | This follows by REF. |
math/0012263 | Note that if MATH is a coherent sheaf on MATH, then the local projective dimension MATH . Now if MATH for some MATH, we may assume that the codimension of the support of MATH is equal to MATH for some MATH. (Else we may replace MATH with a larger integer.) By REF there is an open subset MATH of MATH with the codimension of MATH greater or equal to MATH such that the projective dimension of all localizations MATH is less or equal to MATH for all integers MATH and points MATH in MATH. Let MATH and MATH. Also let MATH be minimal such that MATH. (Note that such a MATH exists by our assumptions on MATH and MATH for MATH.) We have an exact sequence MATH . Since MATH for MATH we get MATH. By the NAME Theorem CITE, the local projective dimension of MATH is MATH. Hence MATH is also MATH. By the exact sequence MATH we get MATH. By the exact sequence REF for MATH we get that either MATH or MATH. In the latter case we may continue and eventually get MATH for some MATH. This is not possible however since MATH. Thus we have proved the lemma. |
math/0012263 | REF says, as in REF b., that when MATH then MATH for all points MATH. Thus when MATH, clearly MATH is empty and so MATH. For MATH a point in MATH the rank of MATH must then be constant for all MATH. Since MATH we get that MATH is a sheaf of rank MATH. |
math/0012263 | Let MATH correspond to MATH in MATH. The condition a. gives that the ranks of the maps MATH are constant for all MATH and all MATH. Thus the kernels and cokernels of these maps are bundles. We may therefore assume we have a complex MATH . Now we use condition b. We may apply REF and get that MATH for MATH. Thus letting MATH we have the theorem. |
math/0012263 | We may assume that MATH in the proof above is MATH . Thus MATH. Furthermore there is an exact sequence MATH where MATH is torsion free and MATH is the torsion part of MATH. Let MATH be the codimension of the support MATH. Since MATH degenerates in codimension MATH by REF we must have MATH. Also the projective dimension of MATH is MATH for MATH in MATH by the NAME theorem CITE. Then the projective dimension of MATH for MATH in MATH. But this is impossible since the locus of MATH in MATH such that the projective dimension of MATH is MATH has codimension MATH according to REF . Hence MATH and MATH. |
math/0012263 | The latter follows from the former since MATH is generated by such coherent sheaves MATH and for these MATH is coherent. Consider the functors MATH . By CITE if MATH is injective in MATH, then MATH is injective. Also since MATH is right adjoint to MATH and the latter is exact, if MATH is injective in MATH then MATH is injective. Let MATH be an injective resolution in MATH. Then MATH is an injective resolution in MATH and MATH is an injective resolution in MATH. We now claim that MATH is a resolution in MATH. To see this, note that MATH is a resolution for MATH. Hence it follows that MATH must be a resolution (see CITE). Since MATH is coherent on MATH, we are done. |
math/0012263 | Consider the diagram MATH . We first claim that there is a natural isomorphism of functors MATH . Let MATH be an injective resolution in MATH. By the first part in the proof of REF , MATH is an injective resolution in MATH and MATH is an injective resolution in MATH. Since the natural map MATH is a quasi-isomorphism, MATH is a homotopy equivalence and so MATH is an isomorphism in MATH. Now MATH is MATH and since the latter is a complex of flasque sheaves, there is a quasi-isomorphism MATH . Composing MATH with MATH shows REF . The theorem now follows from the diagram MATH where the second square gives a natural isomorphism of functors MATH . |
math/0012263 | For a coherent sheaf MATH on MATH denote MATH for short as MATH. Giving an isomorphism MATH is equivalent to giving an isomorphism (where MATH is the homomorphism given by MATH) MATH of MATH-modules (with action on the first factors). Now the map MATH induces a map of MATH-modules (with MATH-action on the right module via MATH) MATH . Composing with MATH this gives a map MATH which is a MATH-module map where the MATH-module structure on MATH is given via MATH. We next need to show that this map MATH makes MATH into a MATH-comodule. That is, that the following diagram commutes. MATH . This may be verified by translating the cocycle REF to statements for MATH-modules. CASE: The automorphism MATH of MATH is induced from the automorphism MATH which is the composition MATH . Suppose now MATH is a MATH. Then the isomorphism MATH is a map of MATH-modules, where the MATH-module structure on the right is given via the map MATH in REF . The composition, which we denote by MATH, MATH now gives an isomorphism of MATH-modules (the action on the first module is given by acting naturally on the first factor). We note that this map is determined by MATH . NAME REF we get an isomorphism MATH . We next need to show that this morphism fulfills the cocycle REF . For this it will be enough to show that the following map MATH of MATH-modules MATH coincides with the map MATH of MATH-modules which is the composition MATH . But the map MATH sends MATH and the map MATH sends MATH . Since MATH is a MATH-comodule REF are equal. |
math/0012263 | Given MATH a MATH map, we get by the functoriality of REF a commutative diagram of MATH-modules MATH NAME this and using that MATH we get a morphism in MATH. Conversely, given a morphism MATH in MATH. This gives a commutative diagram of quasi-coherent sheaves on MATH. MATH . Observe that MATH is the sheafification of MATH and MATH. Also MATH is the sheafification of MATH and MATH . Now we are going to apply the adjunction REF noting that it is valid on MATH for any MATH-algebra MATH. Thus we get a diagram MATH . Combined with the commutative diagram of MATH-modules (where the MATH-module structure on the lower row is given via MATH) MATH this gives a MATH map MATH. |
math/0012263 | First MATH is quasi-coherent by CITE. Let MATH. Since MATH is the sheafification of the MATH-module MATH then MATH is the sheafification of the MATH-module MATH where MATH acts by multiplication on the second factor. By REF b. we thus need to show that this module is naturally a MATH. Let MATH be the antipode, that is, the algebra homomorphism corresponding to MATH given by MATH. We claim that there is a map MATH making the first module into a MATH-comodule. The map MATH is given by MATH . To check that the map is well defined, one verifies (using that MATH is a NAME algebra) MATH . It is also easily seen that MATH is a MATH-module map (with MATH acting via MATH on the second module and then on the first and third factor). Finally it is easily seen that MATH is a MATH-comodule map. Hence by REF , MATH, which is the sheafification of MATH is a MATH-equivariant quasi-coherent sheaf. |
math/0012263 | Let MATH be in MATH and let MATH be in MATH. Given a morphism MATH, we get a MATH-equivariant morphism MATH . Now given any morphism of schemes MATH, the functor MATH is left adjoint to the functor MATH. Hence the isomorphism MATH corresponds to a morphism MATH which is MATH-equivariant (as can be checked). Composing with REF gives us the MATH-equivariant morphism MATH. Conversely given a MATH-equivariant morphism MATH this corresponds to a morphism MATH. Taking the fiber at MATH in MATH gives a morphism MATH. Now starting out from MATH, we get a map MATH and again a map MATH which is easily seen to be MATH. Conversely, starting with a MATH-equivariant map MATH we get MATH and then again by our construction a map MATH. We must show that MATH. So let MATH. Then MATH is a MATH-equivariant map such that the associated MATH is zero. We must show that then MATH. So let MATH and MATH. Then MATH with MATH acting by multiplication on the second factor. The map MATH corresponds to a map of MATH-s MATH such that composing with the counit MATH gives the zero map MATH. Since MATH is a MATH-comodule map there is a commutative diagram MATH . So let MATH be in the image of MATH. Then MATH . The composition of the maps MATH and MATH is zero since this composition is just MATH tensor the map MATH which is zero. Hence REF maps to zero by MATH . Since MATH we get that this is MATH . Composing with the map MATH we get that MATH . NAME the map MATH with MATH we get a map MATH which maps REF to MATH where we have used that the map MATH maps MATH to MATH. But MATH and MATH and so REF is MATH . This gives MATH. Hence the adjunction is proven. |
math/0012263 | This is just the general fact that a right adjoint to an exact functor between abelian categories takes injectives to injectives. |
math/0012263 | Since MATH by the projection formula, it will be enough to show that the right derived functors of MATH vanish. The functor MATH is the composition of the functors MATH and MATH. Note first that MATH takes injectives to objects acyclic for MATH. This is because if MATH is an injective quasi-coherent sheaf on MATH, then MATH is flasque CITE and hence MATH is also flasque and hence acyclic for MATH. Also MATH is quasi-coherent since MATH is NAME. The composition of MATH and MATH therefore gives a NAME spectral sequence for the sheaf MATH. MATH . Now since MATH is an affine morphism, MATH for MATH. Thus MATH . We show that the latter is zero for MATH and this will finish the proof. But MATH and MATH is an isomorphism. Thus it is enough to show that MATH for MATH. But the same argument as above also gives an isomorphism MATH . By the projection formula MATH. Since MATH is a NAME scheme, a direct sum of injectives is injective, and so REF vanish for MATH. |
math/0012263 | Let MATH be an injective resolution of MATH in MATH such that MATH is acyclic for the functor MATH. Such a resolution exists by REF . Then MATH considered as an object in MATH is isomorphic to MATH. By REF we then get that MATH is acycylic, which proves the proposition. |
math/0012263 | This is analogous to the proof of REF . |
math/0012263 | CASE: It is clear that the complex is exact in degrees MATH. It is also easily seen that it is exact in degree MATH, since there are no relations of MATH of linear forms in MATH. So consider degree MATH. It is straight forward to check that the image of MATH has dimension MATH and so the sequence is exact in degree MATH. Now taking the graded dual of the complex and using the identification MATH we see that the complex becomes isomorphic to the original complex twisted with MATH. That the complex is exact in degrees MATH follows then from this fact. CASE: This is clear from the considerations above. |
math/0012263 | MATH corresponds to a line MATH in MATH. The matrix of MATH is obtained from the matrix of MATH by letting the entries map to quadratic forms in MATH. Let MATH be the image of MATH in MATH. Clearly in degrees MATH the cohomology of REF vanishes. Consider the complex in degree MATH. We must prove that the differential MATH has no linear relations in MATH. By a suitable permutation of MATH we may assume that a relation between the MATH involves only MATH for MATH. We may find such a permutation so that the new matrix MATH obtained from MATH also may be obtained from MATH by permuting its rows and columns. Now for each MATH it is then easily checked that there is no linear relation of MATH. Hence the complex is exact in degree MATH. So consider the complex in degree MATH. We shall show that MATH is surjective in degree MATH. First a piece of notation. If MATH and MATH is the element in MATH not in MATH, denote MATH by MATH. Assume that MATH is independent. The images of MATH by MATH are the following : MATH . Now if MATH is dependent, then MATH, and we easily see that the image of MATH is the whole of MATH. If MATH is independent, then the image of MATH is the transpose of MATH, which is nonzero. This also implies that the image of MATH is the whole of MATH. Now by taking the graded dual of the complex MATH and using the isomorphism MATH, we get a complex isomorphic to MATH. This gives that MATH is injective in degree MATH and MATH is exact in degrees MATH. In degree MATH we see that the cohomology has dimension MATH. |
math/0012263 | That MATH is a rank MATH bundle follows from REF in conjunction with REF . By REF b. and REF we find that the NAME polynomial MATH and by standard computations this gives MATH and MATH. |
math/0012264 | Consider the pairing MATH . Denoting the element in REF by MATH, we show that MATH is zero. MATH . Now note that for an element MATH in MATH we have MATH . Since MATH in MATH, we get the lemma. |
math/0012264 | It is straight forward to check that MATH . Let us now prove that MATH. We find MATH and we conclude by REF . |
math/0012264 | The first statement follows from REF . Calculating MATH explicitly this also follows from REF by noticing that MATH and that MATH . |
math/0012264 | Since MATH is MATH and MATH is MATH this is just the standard MATH adjunctions. Explicitly both complexes have MATH'th term equal to MATH and differential MATH given as follows. Let MATH be in the above product and MATH in MATH. Then MATH . |
math/0012264 | Both sides are the cycles of degree MATH in the complexes above. |
math/0012264 | We must show that MATH and MATH take homotopic morphisms to homotopic morphisms. But MATH is nullhomotopic if and only if the inclusion MATH is split and this condition is preserved by the additive functors MATH and MATH. Since MATH and MATH preserve cones we are done. |
math/0012264 | Given a dg-module MATH in MATH. Denote the piece of cohomological degree MATH and MATH-degree MATH by MATH. Then MATH and for MATH in MATH . Now let MATH . With the cohomological grading of MATH given by MATH and the MATH-grading of MATH given by MATH, then MATH becomes an object of MATH. Similarly if MATH is in MATH, then with exactly the same change of grading, MATH becomes and object of MATH. Via these isomorphisms of categories it is furthermore easy to check that the functors MATH and MATH correspond. |
math/0012264 | This is clear. |
math/0012264 | The complex MATH has a filtration MATH. We claim that for MATH then MATH for MATH and zero otherwise. This follows by induction from the exact sequence MATH by noting that the right term is a homogeneous part of the NAME complex for MATH and MATH tensored with MATH (over MATH) MATH with differential MATH . Since now MATH is MATH and MATH is exact in the category of vector spaces, we get the lemma. |
math/0012264 | CASE: The complex MATH is MATH which by the adjunction of REF is MATH. Since MATH is a complex of free MATH-modules of finite rank, this is equal to MATH . Here MATH is a free resolution of MATH since it is isomorphic as a complex of vector spaces to MATH which is MATH and this is quasi-isomorphic to MATH. CASE: MATH is a resolution of MATH of finite rank free modules. Therefore MATH . But MATH is the complex MATH which is MATH. By analogy of REF MATH . |
math/0012264 | We start with the first one. For a complex MATH let MATH be the truncation MATH and MATH be the truncation MATH. We have a short exact sequence MATH REF If MATH is a bounded complex it follows by induction, using the above sequence, and the fact that MATH and MATH are exact on short exact sequences, that MATH is a quasi-isomorphism. CASE: Suppose now that MATH is bounded above so MATH where the MATH are bounded. By REF we see that MATH commutes with such colimits, that is, MATH . Since MATH is a left adjoint it also commutes with colimits so MATH is a quasi-isomorphism since MATH is exact on the category of vector spaces. Before proceeding note that for a module MATH over MATH, MATH is exact in cohomological degrees MATH by the proof of REF . CASE: Suppose now that MATH is bounded below, say MATH. By the remark just above we see that MATH is exact in cohomological degrees MATH. The functor MATH commutes with MATH since it is a right adjoint and the MATH also commutes with MATH since the MATH are finite dimensional. So MATH is a quasi-isomorphism in cohomological degrees MATH and so MATH is exact in this range. Since MATH is MATH the same also holds true for MATH. CASE: Now let MATH be an arbitrary complex. From the diagram MATH we get that MATH is a quasi-isomorphism in cohomological degrees MATH. Since MATH can be chosen arbitrary we are done. We now prove the second part. The complex MATH is the complex MATH . By the same argument as in REF the map MATH is a quasi-isomorphism. So if MATH we have MATH and so MATH is also a quasi-isomorphism. By induction using truncations, we get that MATH is a quasi-isomorphism for bounded MATH. If MATH is bounded above then MATH. Since MATH is MATH and MATH is seen to commute with this colimit (MATH is bounded above), we get that MATH is a quasi-isomorphism. If MATH is any differential graded MATH-module, then MATH and since MATH commutes with this inverse limit and MATH also does so, we get that MATH is a quasi-isomorphism. |
math/0012264 | Consider the composition MATH . To show that this factors through MATH let MATH be in MATH. Then MATH is in MATH and we need to show in addition that MATH is in MATH. But this follows from the triangle REF , since both MATH and MATH are in MATH and hence also MATH must be in MATH. Hence we get the functor MATH and similarly MATH. To show that there is an adjoint equivalence of categories we show, see REF that the functors are adjoint and REF that the canonical morphism MATH and MATH are isomorphisms. CASE: The elements of MATH are equivalence classes of diagrams in MATH. MATH where MATH is in the multiplicative system determined by MATH. Such a diagram will, via the adjunction between MATH and MATH correspond to a diagram MATH where MATH is in the multiplicative system determined by MATH. Equivalently REF give equivalent REF . Hence we get a map MATH . Correspondingly we get a map MATH in the other direction by letting a diagram MATH with MATH in the multiplicative system determined by MATH map to a diagram MATH . That MATH is the identity follows by REF of multiplicative systems, that is, we can complete REF to a diagram MATH . Similarly MATH is the identity. CASE: Consider MATH in MATH. We get a triangle MATH where MATH is in MATH. We then get a triangle in MATH . From the adjunction between MATH and MATH there is also a natural transformation MATH and the composition here is an isomorphism. Since the first map is in the multiplicative system determined by MATH, the second map MATH is also. Hence MATH is in MATH. Therefore MATH is in MATH and so MATH is an isomorphism in MATH. Similarly MATH is an isomorphism in MATH. |
math/0012264 | REF applies so we get an adjoint equivalence MATH . Hence MATH is an equivalence. Now the natural transformation MATH given by MATH become an isomorphism on objects in MATH. Hence the functor MATH when composed with the equivalence MATH becomes an equivalence. Thus MATH becomes an equivalence. |
math/0012264 | If MATH is acyclic the identity map in MATH is in the image of MATH and hence MATH is nullhomotopic. Conversely, if MATH is nullhomotopic, let MATH be a homotopy. Then a cycle MATH in MATH is the image of MATH in MATH as an easy calculation shows. |
math/0012264 | There is a sequence MATH . Now we can easily verify from REF that MATH is a direct product of homogeneous parts of the NAME complex for MATH and MATH taking the product over all MATH greater or equal to maximum of MATH and MATH. Hence MATH is acyclic in all cohomological degrees MATH. Since MATH is MATH we get MATH exact in cohomological degrees MATH. |
math/0012264 | Clearly i. implies ii. and iii. We next show that ii. implies i. Let MATH be MATH such that MATH is an acyclic complex MATH . Then MATH is the total direct product complex of a double complex with terms MATH. The vertical differentials are here given by MATH so the columns are bounded above and exact. Then by CITE the total direct product complex is acyclic and we conclude by REF . We now show that iii. implies ii. Assume MATH is not acyclic with non-zero cohomology in maximal degree MATH which we assume to be MATH. By the sequence MATH we get MATH acyclic. Hence MATH is nullhomotopic and MATH and MATH have isomorphic cohomology. Replacing MATH with MATH, we may assume that MATH has non-zero cohomology in degree MATH, that MATH is zero for MATH, and we want to show that MATH is not acyclic. Now MATH is not surjective. But then it is easily checked explicitly by REF that MATH has non-zero cohomology in degree MATH. |
math/0012264 | ii. follows from i. To prove i., the triangle MATH in MATH gives a triangle MATH . Since MATH is equal to MATH which is MATH it will be enough to show that MATH is acyclic. Replace MATH with MATH now assumed to be in MATH. We will show that MATH is acyclic, so assume the contrary, that it has non-zero cohomology in degree MATH which we assume to be MATH. Consider MATH . Since now MATH has non-zero cohomology, by REF , MATH is not acyclic and so has non-zero cohomology in some degree MATH. Since MATH is acyclic, MATH gets non-zero cohomology in some degree MATH. But this contradicts REF . |
math/0012264 | When MATH is zero for MATH, then MATH. Hence MATH . Since all MATH are nullhomotopic, MATH will be acylic and hence MATH is nullhomotopic. |
math/0012264 | This is analog to REF . In order to prove the analog of REF one should use that MATH is MATH and that MATH commutes with inverse limits. |
math/0012264 | CASE: Apply REF to the cone of MATH. CASE: Let MATH be MATH. There is an injective map MATH. Let MATH be the cokernel. Then MATH is a bounded above MATH-module with MATH zero. But then MATH must be zero. |
math/0012264 | There are short exact sequences of complexes MATH where MATH is the truncation of MATH and MATH is the kernel complex MATH which is quasi-isomorphic to MATH. We get a sequence MATH . Now assume by induction that there is a homotopy equivalence MATH with MATH having zero differential. Since MATH is a surjective map, MATH must be surjective because of the form of MATH. Let MATH be the kernel of the composite surjection MATH . Then MATH is zero and so by REF , MATH is in MATH. Also the map MATH is a quasi-isomorphism after applying MATH. Hence it is a homotopy equivalence. Composing its inverse with MATH we get a pushout diagram MATH . Applying MATH we see that MATH also has zero differential and the middle vertical map is a quasi-isomorphism after applying MATH. We now get diagrams MATH . Taking inverse limits we get a map MATH which becomes a quasi-isomorphism when applying MATH. This proves the theorem. |
math/0012264 | REF . clearly holds. REF . follows by the exact sequence MATH which induces a triangle. Now MATH is a product like REF and we would like to have a product over MATH. However using REF ii., it is easily shown that MATH is actually isomorphic to a product like REF over MATH. For REF. suppose given a diagram MATH in MATH where MATH is in the multiplicative system defined by the null system. We may assume MATH has the form REF and MATH has the form REF for MATH. It will be sufficient to prove that if MATH is in MATH then MATH is an isomorphism. Now suppose MATH is an isomorphism where MATH as in REF . Then the cone MATH is in MATH and by REF , MATH is acyclic. Hence MATH is a quasi-isomorphism when applied to MATH. Then it is also a quasi-isomorphism when applied to MATH . But then MATH applied to its cone is acyclic and its cone is therefore nullhomotopic by REF . Thus REF becomes an isomorphism an therefore also REF . |
math/0012264 | It is clear that MATH takes MATH to MATH and MATH to MATH. |
math/0012267 | To prove the lemma, we take advantage of the fact that the expression on the left-hand side of REF is the solution of an ordinary differential equation. By the semi-group property, we may consider separately the following regimes: For MATH, we distinguish the cases MATH, MATH, MATH, MATH, MATH, MATH and for MATH, we deal separately with MATH, MATH, MATH, MATH, MATH, MATH. On each of these time intervals the claimed behaviour follows easily by elementary calculus, see also the similar result CITE. |
math/0012267 | Let MATH and MATH be constants to be chosen later, and denote by MATH the first exit time of MATH from the strip MATH. Set MATH. Then, for MATH, we get from REF that MATH for some constants MATH. The relations REF yield the existence of constants MATH such that MATH and MATH for MATH. From REF we obtain MATH . For any given MATH, we can choose MATH large enough for the term in brackets to be larger than MATH. Then, by REF , there exists a constant MATH such that MATH for all MATH. Therefore, if MATH, then MATH follows. The lower bound can be obtained in exactly the same way. |
math/0012267 | Again, we will only show how to obtain an upper bound, since the corresponding lower bound can be established in exactly the same way. First we fix a constant MATH. We denote by MATH the first exit time of MATH from the strip MATH. For MATH, we have MATH with constants MATH. Choosing MATH large enough, we get MATH which implies MATH by partial integration. Here MATH satisfies MATH. We want to estimate the contribution of the middle term on the right-hand side of REF. Assume first that MATH and consider MATH. By convexity, MATH . Now, MATH . Since MATH as MATH, we also have MATH provided MATH is large enough. This shows the existence of constants MATH and MATH such that MATH . For MATH, MATH is immediate, and REF shows that REF also holds for MATH. Note that in the case MATH, REF holds trivially. Since MATH is a direct consequence of REF and our choice of MATH, MATH follows, and, therefore, the upper bound REF holds for all MATH. |
math/0012267 | In order to show REF, we rescale space and time in the following way: MATH . Let MATH. Then MATH, and MATH satisfies the differential equation MATH where MATH . REF is a perturbation of order MATH of the NAME equation MATH . Using NAME 's inequality, one easily shows that on a MATH-time scale of order MATH, the solution of REF differs by MATH from the solution of REF, which is MATH . This function is bounded away from zero, and remains of order one for MATH of order one, which shows that MATH on MATH. Since MATH and MATH, MATH and MATH necessarily cross at some time MATH. |
math/0012267 | The proof is similar to the one of REF . |
math/0012267 | By NAME 's formula we get MATH . Consider first the case MATH for MATH large enough. REF implies that MATH for a constant MATH. On the other hand, REF shows that the second term on the right-hand side of REF is bounded in absolute value by MATH for a constant MATH. Thus if MATH we obtain that MATH. We consider next the case MATH. For MATH, the above argument can be repeated. The non-trivial case occurs for MATH. By rescaling variables as in REF , we obtain that MATH . We have to show that MATH which is equivalent to MATH for MATH of order one. The lower bound is trivial as MATH and MATH. In order to show the upper bound, first note that for MATH, we have MATH which implies MATH. Therefore, it is sufficient to consider MATH. Taking into account REF , we find that showing the upper bound amounts to showing that MATH . Since MATH is proportional to MATH, choosing a priori a large enough MATH also makes MATH large. Thus it is in fact sufficient to verify that MATH for all MATH. Optimizing the left-hand side with respect to MATH and MATH shows that we may assume MATH and that REF holds. |
math/0012267 | Let MATH, with some MATH, be a partition of MATH. In CITE, we show that the probability REF is bounded above by MATH . Now we choose the partition by requiring that MATH . Since MATH, we have MATH, and thus MATH . If MATH is such that MATH, then by REF, there is a constant MATH such that MATH and hence by REF (choosing the same MATH for brevity of notation) MATH . For all other MATH, we have MATH . In both cases, we find MATH which leads to the result, using the definition of MATH. |
math/0012267 | The proof is based on the fact that the variable MATH satisfies the relation MATH . Consider first the case MATH. Let MATH be a constant to be chosen later, and set MATH. We define the first exit time MATH . Pick any MATH and MATH. Then we have MATH for all MATH. From REF, we obtain the existence of a constant MATH such that MATH for these MATH. Hence, by REF we get the estimate MATH and thus, by REF , MATH where we use again the same MATH for brevity of notation. Using REF once again, we arrive at the bound MATH . Now we choose MATH which implies MATH for all MATH, by the definition of MATH. Hence MATH for almost all MATH. Since we have MATH whenever MATH, we conclude that MATH, and thus MATH for almost all MATH, which implies that MATH for MATH and these MATH. This completes the proof of REF. The proof of REF is almost the same. In the case MATH, we take MATH. The estimate REF has to be replaced by MATH and thus we get, instead of REF, the bound MATH . The remainder of the proof is similar. |
math/0012267 | REF implies that for all MATH, MATH . Let MATH. For MATH, the variable MATH satisfies MATH . Applying NAME 's inequality, we obtain MATH where MATH. This proves the result for MATH. Now if MATH is negative, the result is trivially satisfied, and if MATH becomes positive again, the above argument can be repeated. Note that MATH is immediate. This proves the first assertion, and the second assertion can be proved directly, without use of MATH. |
math/0012267 | M. enumREF CASE: We define a partition MATH of the interval MATH by requiring MATH . Note that similar arguments as in the proof of REF yield MATH . Now let MATH and MATH as usual. Define MATH for MATH, and MATH . Then MATH . CASE: In order to estimate MATH, we introduce the stochastic process MATH defined by MATH where MATH is the Brownian motion MATH. Note that MATH is the solution of the SDE REF with initial condition MATH at time MATH. We define the stopping times MATH describing the time when MATH either reaches the MATH-axis or the upper boundary MATH. Now, REF implies that if MATH, then MATH for MATH. This shows that MATH for MATH, and MATH . Each of these terms depends only on MATH, and can be easily estimated. Let MATH denote the variance of MATH. Then by symmetry (as in REF), we have MATH . The second term on the right-hand side of REF or REF, respectively, can be estimated using the symmetry (in distribution) of REF under the map MATH: MATH . In order to estimate the third term on the right-hand side of REF, we will use the fact that for MATH . REF thus yield the existence of a constant MATH such that MATH . This allows us to estimate (for MATH) MATH . CASE: The estimates REF, inserted in REF, imply that MATH with MATH . By REF, for each MATH, there exists a MATH such that MATH . Together with REF, this implies that MATH for all MATH, and thus the result follows from REF with MATH. |
math/0012267 | M. enumREF CASE: Let MATH and define a partition MATH of MATH by MATH . We would like to control the probability of not reaching the MATH-axis during the time interval MATH. Let MATH . Then the probability on the left-hand side of REF is MATH . If we manage to estimate each MATH by a constant less than MATH (say, MATH), then the probability will be exponentially small in MATH. In the sequel, we shall estimate MATH uniformly in MATH, and bound MATH by MATH, since the last interval of the partition may be too small to get a good bound. So let MATH from now on. CASE: We consider first the case MATH. We define the process MATH as the solution of the linearized SDE MATH where MATH is the Brownian motion MATH. Let MATH denote the variance of MATH. Then MATH . We can now apply REF in the particular case MATH to show that if MATH for MATH, then MATH in the same interval. We thus obtain MATH . The probability on the right-hand side satisfies MATH yielding MATH . By making MATH small enough, we can guarantee that this bound is smaller than some imposed constant of order MATH, say MATH. This shows that the length of MATH has been chosen large enough that the probability of reaching the MATH-axis during this time interval is appreciable. CASE: We examine now the case MATH. We introduce a time MATH, defined by MATH . Our strategy will be to show that MATH is likely to cross MATH before time MATH, which will allow us to use the previous result. REF implies the existence of a constant MATH such that MATH . Let MATH be the solution of the linear SDE MATH where MATH is again the Brownian motion MATH. The variance of MATH is MATH . REF shows that if MATH on the interval MATH, then MATH on that interval. If we introduce the stopping time MATH then we have MATH . The second term on the right-hand side can be bounded, as in REF, by MATH . Using REF, the first term on the right-hand side of REF can be estimated in the following way: MATH . Using REF , it is easy to show that the expression MATH is uniformly bounded by a constant independent of MATH and MATH. The sum of REF and of the last term in REF provides an upper bound for MATH. CASE: Using the fact that for MATH and MATH, one has MATH and MATH, we arrive at the bound MATH where the constant MATH can be chosen independent of MATH because MATH by assumption. Thus if we choose MATH and MATH, we obtain that MATH for MATH. This yields MATH and the result follows from our choice of MATH. |
math/0012267 | The proof follows along the lines of the one of CITE, the main difference being the quadratic behaviour MATH of MATH in our case as opposed to the linear one in CITE. We start by defining a partition MATH of MATH, given by MATH . On each interval MATH, we consider a Gaussian approximation MATH of MATH, defined by MATH where MATH. If MATH for all MATH, then by REF, there is a constant MATH such that MATH for all MATH, provided the condition MATH holds for all MATH. Now, MATH where MATH is a constant satisfying MATH for all MATH. This shows that there exists a constant MATH such that REF is satisfied whenever MATH . This condition is equivalent to REF. Assume MATH for the moment. Then, MATH where MATH denotes the variance of MATH. By partial integration, we find MATH . Now, the NAME property yields MATH and the bound REF follows by a straightforward calculation. |
math/0012267 | Let MATH. We first observe that by NAME 's formula, MATH . This shows that MATH . Thus, with the rescaling MATH we obtain that MATH obeys a perturbation of order MATH of the NAME equation MATH . One easily shows that the solution satisfies MATH for MATH of order REF, and this property carries over to the perturbed equation with the help of NAME 's inequality. Finally, since MATH and MATH, these curves necessarily cross at a time MATH. |
math/0012267 | The proof is a direct consequence of REF and the properties of MATH, and thus much simpler than the proof of REF . |
math/0012267 | M. enumREF CASE: Let MATH be the deterministic solution tracking the saddle at MATH and set MATH. Then MATH where REF imply MATH and MATH . Let MATH and define a partition MATH of MATH by MATH . Let MATH . Then we have, as in REF, MATH . The result will thus be proved if we manage to choose MATH in such a way that MATH is bounded away from MATH for MATH. CASE: We will estimate the MATH in a similar way as in REF , but we shall distinguish three cases instead of two. These cases correspond to MATH reaching the levels MATH, MATH and MATH. We introduce a subdivision MATH defined by MATH and stopping times MATH . Then we can write, similarly as in REF, MATH . The first term can be bounded by comparing with the solution of the SDE REF linearized around MATH, with the help of REF . As in REF, we obtain the upper bound MATH where MATH is a constant such that MATH for all MATH. Now if MATH, we also have MATH . Comparing with the solution of the SDE REF linearized around MATH, the first term can be bounded, as in REF, by MATH . This estimate shows that a path starting on MATH at time MATH has an appreciable probability to reach the saddle before time MATH. Note, however, that we cannot obtain directly a similar estimate for the probability to reach MATH as well, which is why we restart the process in MATH. CASE: In order to estimate the second summand in REF, let MATH be the process starting in MATH at time MATH which satisfies the linear equation MATH with MATH. The variance MATH of MATH satisfies, as in REF, MATH . Thus we obtain, using REF , MATH . Here the introduction of the stopping time MATH turns out to play a crucial role. The above probability is indeed close to MATH when MATH is larger than a constant times MATH, which shows that once a path has reached the saddle, it also has about fifty percent chance to reach the level MATH in a time of order MATH. CASE: From REF and the fact that MATH, we obtain the existence of a constant MATH such that MATH . Since MATH, the properties of MATH, MATH, MATH and MATH imply the existence of another constant MATH such that MATH . Now choosing MATH and MATH, we get MATH for MATH and thus MATH and the result follows from our choice of MATH. |
math/0012267 | Let MATH be defined by MATH . By NAME 's inequality, it is easy to see, as in REF , that if MATH for all MATH, then MATH for those MATH. We thus have MATH . Note, however, that for MATH, MATH so that the second term in REF is equal to zero. The first term equals MATH by NAME 's submartingale inequality. |
nlin/0012034 | One obtains a solution of the meromorphic NAME REF by making an ansatz of the form REF and observing that the residue REF with REF are equivalent to REF , that is, by reversing our steps. The solvability of the linear system for the coefficients implied by REF is guaranteed by the distinctness of the MATH and the conditions MATH CITE. Under an ansatz of the form REF , the relation REF is equivalent to REF . Note that the same solution of the nonlinear NAME equation is obtained from the matrix MATH for both cases MATH by REF . Uniqueness follows from NAME 's theorem. |
nlin/0012034 | The existence part of the proof of this proposition follows from the corresponding existence result for the meromorphic NAME REF whose solution MATH yields a solution MATH of the holomorphic problem by REF , and REF . The uniqueness part of the proof follows from the continuity of the boundary values and NAME 's theorem: the ratio MATH of any two solutions is analytic in MATH and continuous in MATH. Therefore this ratio is entire and from the normalization condition we learn that MATH. |
nlin/0012034 | One finds a solution of the phase-conjugated problem by solving the holomorphic NAME REF for MATH and then obtains MATH from REF . The analyticity and boundary behavior follow from the analogous properties of MATH and MATH. The jump conditions for MATH are verified using REF and the boundary conditions satisfied by MATH, taking into account the discrepancy in boundary values of MATH along the contour and the symmetry of MATH. Finally the normalization condition follows from the corresponding property of MATH and the decay of MATH. Therefore, MATH so defined solves the phase-conjugated NAME problem. Uniqueness of solutions for the NAME REF follows as before from NAME 's theorem using the continuity of the boundary values and the normalization at infinity. Clearly the whole procedure can be reversed, and the unique solution of the holomorphic NAME REF can be obtained from the solution MATH of the phase-conjugated NAME REF by the same formula, REF . |
nlin/0012034 | Uniqueness follows from the continuity of boundary values and the normalization condition via NAME 's theorem. The symmetry of the solution then follows from the corresponding symmetry of the jump relations and uniqueness. |
nlin/0012034 | Uniqueness and symmetry are proved as for MATH. The equivalence is established by the explicit triangular change of variables via REF ; the transformation is clearly invertible, and it is a direct calculation to show that any solution of the NAME REF leads via REF to a solution of the NAME REF . |
nlin/0012034 | Observe that by direct calculation, the matrix MATH is analytic for all MATH except on the positive real axis, where it satisfies the jump relation MATH. Since both MATH and MATH have determinant one and have smooth boundary values except at MATH, it follows that the quotient MATH is analytic in MATH. But by construction, the matrix MATH obtained in REF is MATH, and consequently MATH is MATH at the origin since MATH is an analytic mapping. Therefore, the quotient is bounded at MATH and hence analytic throughout the interior of MATH. |
nlin/0012034 | We first introduce an auxiliary NAME problem. Let MATH be the oriented contour illustrated in REF . For MATH, we define a jump matrix MATH as follows. For MATH, set MATH for MATH, set MATH and, for MATH, set MATH . Consider the following problem. Find a matrix MATH satisfying: CASE: Analyticity: MATH is analytic for MATH and takes continuous boundary values on MATH including self-intersection points. CASE: Boundary behavior: MATH takes continuous boundary values from each connected component of MATH, with continuity holding also at corner points corresponding to self-intersections of MATH. CASE: NAME conditions: The boundary values taken on MATH satisfy MATH with the jump matrix MATH defined by REF , and REF . CASE: Normalization: MATH is normalized at infinity: MATH uniformly with respect to direction. Observe that the jump matrix MATH has the following properties: CASE: MATH has determinant one for all MATH. CASE: MATH is smooth on each open arc, and in particular is NAME. CASE: At each point MATH of self-intersection of MATH, let the intersecting arcs be enumerated in counterclockwise order (beginning with any arc) as MATH where MATH is even. The limits MATH exist and satisfy MATH . CASE: MATH as MATH. In fact, the decay is exponentially fast in MATH. The first two conditions are obvious. Checking the third condition is a direct computation that we omit, and the fourth condition follows from the fact that the MATH growth of the conjugating factors MATH and MATH is controlled easily by the exponential decay of MATH for MATH, and of MATH for MATH. It then follows from REF proved in the appendix that there will exist a unique solution MATH of the NAME REF that additionally satisfies: CASE: MATH is uniformly bounded and satisfies MATH as MATH for all MATH, and, CASE: the boundary values MATH taken on each component of MATH are NAME continuous for all exponents strictly less than REF, if and only if the corresponding homogeneous NAME problem has only the trivial solution, that is, the NAME alternative applies. If there exists a solution MATH, then define MATH by MATH . The function so defined has a holomorphic extension through the circle MATH, since it takes boundary values there from both sides that are continuous and equal. It is easy to check that it solves the NAME REF . Uniqueness follows from the analogous property of MATH. So, we must now show that such a matrix MATH exists by proving that all solutions of the homogeneous problem are trivial. Let us define this homogeneous problem. Let MATH be given. Find a matrix MATH satisfying CASE: Analyticity: MATH is analytic for MATH. CASE: Boundary behavior: MATH takes boundary values from each connected component of its domain of analyticity that are NAME continuous with exponent MATH, including at self-intersection (corner) points. CASE: NAME conditions: The boundary values MATH that MATH assumes on any smooth oriented component of MATH satisfy MATH . CASE: Homogeneous normalization: The matrix function MATH vanishes for large MATH, satisfying the precise estimate MATH holding for some MATH and all sufficiently large MATH. Thus, a solution of the homogeneous NAME REF , is similar to MATH, but vanishes for large MATH. The identity matrix in the normalization condition for MATH is replaced with the zero matrix. Note that, according to the discussion following the statement of REF in the appendix, it suffices to find a MATH such that all nontrivial solutions of the homogeneous NAME REF with exponents MATH can be ruled out. Unfortunately, the jump matrix MATH lacks the symmetry needed to apply the general theory described in the appendix, so we must construct a specific argument. We will suppose that the NAME exponent satisfies MATH. Let MATH be a corresponding solution of the homogeneous NAME REF . First, set MATH . This matrix is analytic in each sector for MATH, and on the real axis and the rays MATH satisies the same jump conditions as MATH. As MATH, we have for some MATH the estimate MATH . Next, set MATH . Since the matrices multiplying MATH above are uniformly bounded, MATH retains the decay properties of MATH. Also, MATH is analytic for MATH. On the real axis, oriented from right to left, there is the jump condition MATH . Now, the matrix function MATH is also analytic for MATH, and since MATH for large MATH, we can apply NAME 's theorem for all MATH to deduce that MATH . Since for MATH real, MATH, REF becomes, using the relations REF MATH . Adding this equation to its conjugate-transpose, and looking at the MATH entry of the resulting matrix equation, one finds MATH where MATH is the MATH-th column of MATH, and consequently MATH for MATH. From REF , it then follows immediately that MATH for MATH. The jump relations REF then relate the boundary values of the remaining, possibly nonzero, entries of MATH by MATH . So, defining scalar functions MATH for MATH by MATH we see that both functions are analytic for MATH, both are MATH for large MATH, and both take continuous boundary values on MATH, where they satisfy MATH with the ray considered oriented from infinity to the origin. We now show that necessarily MATH. Given MATH satisfying the above properties, define a scalar function MATH that is analytic in the extended plane MATH by setting MATH . We are using the notation MATH for the class representative of MATH with MATH. From the jump relation for MATH it follows that MATH is analytic for MATH and MATH. In the extended plane where MATH, we have MATH; moreover, from the mere algebraic decay of MATH for large MATH, we find that MATH decays exponentially for large MATH in these regions, and in particular on the boundaries MATH and MATH. Finally, define an analytic function of MATH for MATH by MATH . From the exponential decay of MATH for MATH and MATH, it follows that MATH for all MATH. Also MATH is uniformly bounded for all MATH. Now, we recall NAME 's theorem CITE: Suppose that MATH is a complex-valued function defined and continuous for MATH and analytic for MATH. Suppose that MATH for MATH and MATH for all MATH, where MATH. Then MATH is identically zero. Applying this result of complex analysis for MATH and MATH, we deduce that MATH. This in turn implies that MATH, and in conjunction with our eariler results that MATH. Consequently we find that MATH. Therefore all solutions of the homogeneous NAME REF with NAME exponents MATH are trivial, and the required function MATH exists by the NAME alternative (compare REF ). Because MATH has NAME continuous boundary values, the matrix MATH defined by REF is a solution of the NAME REF taking uniformly continuous boundary values on MATH. Finally, we notice that since the jump matrix MATH is analytic on each ray of MATH and decays exponentially to the identity as MATH, all of the order MATH moments vanish, and it follows from REF that MATH is uniformly order MATH for large MATH. Using REF , we see that this in turn implies the decay estimate REF , which completes the proof of REF . |
nlin/0012034 | This is an elementary consequence of the fact that for all MATH, MATH satisfies the exact same jump relations as MATH. |
nlin/0012034 | Recall the exact representation REF of the function MATH related to the outer solution MATH obtained in REF by a change of variables. It follows from the construction in REF that MATH blows up like MATH with a leading coefficient that is uniformly bounded as MATH tends to zero. Now for MATH small, the NAME expansion for the analytic map MATH gives MATH, where MATH is a constant that is bounded as MATH goes to zero. Approximating MATH on the right-hand side of REF by such a formula, one sees that the holomorphic prefactor MATH must be of the form MATH with MATH being a matrix analytic in MATH that is uniformly bounded as MATH tends to zero. Now, since MATH is bounded only by MATH for large MATH, we get a uniform estimate for all MATH of the form MATH as well. These bounds, along with the definition of MATH yield the desired estimate. |
nlin/0012034 | By definition, we have for all MATH, MATH . Now, as MATH tends to zero, MATH for all MATH; in particular MATH for all MATH. Therefore, directly from the large MATH asymptotic properties of the matrix MATH in the estimate REF , we have for MATH, MATH . From the proof of REF , the conjugating factors are each uniformly bounded for MATH by MATH. Using this fact in REF yields the desired bound. |
nlin/0012034 | Rather than repeating similar arguments to those used in the proof of REF , we simply use the matrix MATH whose existence is guaranteed by that same lemma to construct a solution MATH of the NAME REF . For MATH, set MATH . It is a direct matter to check that the jump relations and normalization condition for MATH, along with the smoothness and decay of the boundary values given in REF imply that the matrix so-defined is a solution of the NAME REF with the desired properties. Uniqueness follows from the uniform boundedness for finite MATH, continuity of the boundary values, and NAME 's theorem. |
nlin/0012034 | We begin by introducing an auxiliary NAME problem. Let MATH be the contour illustrated in REF for both choices of the parity MATH. We define a jump matrix for MATH as follows. MATH . Find a matrix MATH satisfying CASE: Analyticity: MATH is analytic for MATH. CASE: Boundary behavior: MATH assumes continuous boundary values from each connected component of MATH that are continuous, including corner points corresponding to self-intersections. CASE: NAME condition: On the oriented contour MATH minus the origin, the boundary values satisfy MATH . CASE: Normalization: MATH is normalized at infinity: MATH . The NAME REF differs from the NAME REF of interest only in the orientation of the contour and the introduction of some contour components supporting identity jump matrices. Therefore, the solutions of these two problems are in one-to-one correspondence: MATH. The re-orientation of the contour and introduction of the real axis and the circle are simply to rewrite the problem in precisely the form to which the general results from the appendix can be applied. We now proceed to apply the NAME theory developed in the appendix. First observe that on smooth component of MATH, and for each MATH, the jump matrix MATH is NAME continuous with exponent MATH. Indeed, in the interior of each component, the jump matrix is analytic, and it is easy to see that the only obstruction to arbitrary smoothness is in the limiting behavior at the origin. Here, the term that determines the smoothness is the factor MATH in MATH. But at MATH, this term is in all NAME classes with exponents MATH strictly less than one. Next, note that on each ray of MATH, the jump matrix decays to the identity matrix as MATH at least as fast as MATH. Indeed, from the asymptotic formulae REF , we see that for MATH, the decay to the identity is MATH. On the two rays on either side of MATH in the same quadrant, the decay of the jump matrix to the identity is exponential for large MATH since MATH is real and negative for MATH. For MATH, one sees from REF that the jump matrix decays exponentially to the identity like MATH. The symmetry that determines the jump matrix in the lower half-plane in terms of that in the upper half-plane ensures that similar decay properties hold in the lower half-plane, and of course on the real axis the jump matrix is exactly the identity. Next, we observe that the jump matrices are consistent at the origin, in the following sense. If we number the rays in counter-clockwise order starting with the positive real axis as MATH, and define MATH, then the cyclic relation MATH holds for all values of the parameters MATH, MATH, MATH, MATH, and MATH. It is easy to see that the same property holds at each intersection point MATH of the circle with a ray (that is, MATH), since by definition, MATH . Finally, we observe that for all MATH with MATH, the relation MATH implies that MATH . This is not a general fact, but a consequence of the special structure of the jump matrices for this NAME problem. Also, since the jump matrix is the identity on the real axis, MATH is strictly positive definite for real MATH. These facts allow us to apply REF proved in the appendix to deduce the existence of a matrix MATH that is analytic in MATH, that for each MATH takes on boundary values on MATH that are uniformly bounded and NAME continuous with exponent MATH, that for each MATH satisfies MATH as MATH, and whose boundary values satisfy the jump relation MATH. Note that the meaning of the subscripts MATH and MATH in regard to the boundary values of MATH refer to the contour MATH oriented as shown in REF . Since the boundary values are uniformly continuous and since MATH for all real MATH and for all MATH on the circle of radius MATH (except at the self-intersection points where the jump matrix is not defined), the matrix function MATH is in fact analytic at these points. The function defined by MATH is easily seen to be the unique solution of the NAME REF , since it is analytic in MATH, and since it has NAME continuous and uniformly bounded boundary values that satisfy MATH. Note that here the subscripts MATH and MATH of the boundary values refer to the orientation of MATH as shown in REF . It remains only to obtain the decay estimate REF . For this, we return to the auxiliary problem for MATH, and we compute the moments (compare REF in the appendix) of the jump matrix MATH. First, observe that as MATH tends to infinity along any ray of MATH except those satisfying MATH, MATH . Now when MATH we can use REF to obtain MATH . On the other hand, when MATH, REF implies that MATH . Therefore in both cases, the higher-order compatibility REF holds, and since the jump matrix on each ray is uniformly analytic in a sufficiently thin parallel strip surrounding the ray, REF gives a uniform decay rate for MATH of MATH. This implies the decay estimate REF and completes the proof. |
nlin/0012034 | It follows from REF that the factor MATH is uniformly bounded for all MATH. Next, we study the behavior of MATH in MATH. Along with its inverse, this matrix is bounded for MATH. To deform into MATH, we need only consider the behavior of the exponential factors MATH, since MATH is uniformly bounded. Now, since MATH is smooth and tangent to the ray MATH, it will be the case for sufficiently small MATH that MATH uniformly for MATH because the radius of MATH is MATH, and MATH for MATH. Therefore, uniformly for MATH we have MATH. Similar arguments hold for MATH, and the proof is complete since we always have MATH. |
nlin/0012034 | From the definition of MATH, we have MATH for MATH. The continuity of the boundary values then implies that MATH is analytic for MATH in MATH. This latter statement assumes that MATH does not agree identically with its tangent line, in which case the regions MATH would be empty and we would have MATH. The remaining contours of nonanalyticity for MATH have all been taken to be straight line segments within the fixed disk neighborhood MATH, and therefore agree with the corresponding contours for MATH. The statement carries over to MATH because for sufficiently small MATH the regions MATH where MATH differs from MATH will only meet these contours at the origin. This proves that the domains of analyticity for MATH and MATH agree within MATH. The estimate REF follows from our asymptotic analysis of the jump matrix MATH. On the straight line segments MATH and their images in the lower half-plane, we have MATH, and therefore the simple NAME approximations REF give MATH which is MATH for all MATH. On the two straight line segments MATH, we again find that MATH, but we have an additional contribution from the asymptotic approximation of MATH to take into account. The errors that dominate are determined by the choice of MATH since for MATH and MATH, we have (using REF , and REF ) MATH with the first error term coming from REF and the second error term coming from the NAME approximations REF . Similarly, for MATH and MATH, we have MATH . That both of these estimates are uniformly small in MATH now follows from the boundedness and asymptotic properties of MATH. By symmetry of the definition of the jump matrices, both of these estimates also hold on MATH. Finally, we note that to verify the estimate REF for MATH requires the same sort of analysis as above, with the additional observation that the jump matrix for MATH, denoted by MATH above, is uniformly bounded in MATH according to the arguments in the proof of REF . |
nlin/0012034 | The factors MATH, MATH, and MATH are clearly all uniformly bounded in MATH, essentially since the solution MATH of the outer model NAME REF obtained in REF has bounded boundary values at MATH on MATH. To analyze MATH, we observe that the factor MATH is uniformly bounded for all MATH, and in particular for MATH. The remaining factor in the definition of MATH is controlled as described above, essentially because of the small size of the neighborhood MATH. The fact that MATH follows from the analogous properties of the individual factors, and that the determinant is unchanged by conjugation by MATH. |
nlin/0012034 | Using the fact that MATH is analytic in MATH, we find MATH . The estimate REF then follows from REF , and the uniform boundedness in MATH of MATH and its inverse. |
nlin/0012034 | From the definition of the local approximation, we have for MATH, MATH for some MATH where we have used the uniform boundedness of MATH and its inverse. Now, for all MATH that are not on the boundary of MATH, we have MATH, and then from the decay REF of MATH established in REF the estimate REF follows from REF . For the other values of MATH where MATH, we have to take into account an additional factor. But for MATH on the boundary of MATH, it follows from the asymptotic behavior REF of MATH and the geometry of the region MATH that this factor is uniformly MATH, and again REF yields the required estimate REF . |
nlin/0012034 | We begin with the smaller oriented contour MATH and consider its smooth arc components one type at a time. For all MATH, let MATH denote the jump matrix for MATH relative to the orientation of MATH rather than that of MATH. On each arc of MATH, we have either MATH or MATH. First, consider the clockwise-oriented boundaries of the circular disks MATH, MATH, and MATH. On these boundaries, the jump matrix is simply MATH since MATH is analytic at these boundaries and thus the corresponding jump matrix is the identity. The plus boundary value refers here to the exterior of the disk and therefore by REF , we obtain MATH with similar estimates holding for the clockwise-oriented boundaries of MATH. Likewise, by REF , MATH . Next, we consider the parts of MATH in the interiors of the various circular disks. First consider one of the fixed disks MATH. Here we obtain MATH where we have used REF (or their counterparts for odd numbered endpoints REF ). Now, since each disk is bounded away from the origin as MATH, it follows from REF that MATH . Consequently, we have MATH with same estimate holding in MATH. Next, we examine the interior of the shrinking disk MATH. For MATH, we find, applying first REF and then REF that MATH . It remains to examine the components of MATH that lie strictly outside the closures of all disks. We call this exterior component MATH. In all of MATH we have a uniform bound on MATH, and in particular on the boundary values, that is independent of MATH. Therefore, we have MATH . For MATH in a band of MATH, we have from REF that MATH . This quantity is MATH for fixed MATH, but becomes larger near the outer boundary of the shrinking disk MATH. From REF , we see that the error achieved at the cost of the shrinking radius of MATH is the estimate MATH . For MATH in a gap of MATH, we have from REF that MATH . The function MATH evaluates to a real constant in each gap. Using the definition of MATH, we find that MATH where the scalar MATH is defined by MATH . Now MATH is uniformly bounded, and by REF , we find that for MATH we have MATH. The term MATH is, however, exponentially small with its maximum size MATH being attained on the outside boundary of MATH. Consequently, we have the estimate MATH for some MATH. Similarly, on the boundaries of the ``lenses" in MATH, the matrix MATH is analytic, so we have MATH . Here, MATH is the explicit ``lens transformation" matrix which differs from the identity matrix only inside the lenses, where it is defined by REF . These factors are exponentially small perturbations of the identity matrix away from the endpoints and the origin. It is again the proximity of the origin at the outside boundary of MATH that dominates this error; ultimately we obtain an estimate of the form MATH for some MATH. We now combine the estimates REF , and REF . The overall bound is optimized by taking MATH, which determines the radius of the neighborhood MATH as MATH. With this choice, all bounds except for the exponentially small contributions REF are MATH. Finally, we note that the required estimate for the jump matrix MATH for MATH in the completed contour MATH follows from the corresponding result for MATH along with the facts that MATH for all MATH and MATH for all MATH. This completes the proof. |
nlin/0012034 | The analyticity, boundary behavior, and normalization properties follow directly from REF . The jump condition is equivalent to REF of the matrix MATH. Uniqueness of the solution follows from the continuity of the boundary values and NAME 's theorem. |
nlin/0012034 | For MATH sufficiently small there is a unique MATH such that MATH. This value satisfies MATH as MATH. Let MATH. We divide each contour into two pieces as follows. Let MATH denote the contour parametrized by MATH for MATH and MATH denote the contour parametrized by MATH for MATH. Then, let MATH denote the part of MATH with MATH and MATH the part of MATH with MATH. Note that MATH tends to MATH as MATH tends to zero, so for small MATH we are nearly dividing the circle in half along the tangent line to MATH at the origin. These subdivisions of the contours are illustrated in REF . Let MATH denote the characteristic function of a contour segment MATH; then for MATH and MATH, the NAME operator MATH is decomposed as: MATH . Likewise, for MATH and MATH, we have MATH . It is therefore sufficient to prove the uniform boundedness of each operator on the right-hand side of each expression. Below, we will give details only for the terms involving MATH, since the reader will see that the bounds involving MATH are obtained in exactly the same way. We now introduce arc length parametrizations of MATH, MATH and MATH, and at the same time use the invariance of the NAME integral under translations and rotations to bring the intersection point MATH to the origin and to make MATH tangent to the positive real axis there. Therefore, using tildes to denote the translation and rotation (see REF ), for MATH we have the parametrization: MATH where MATH is an angle converging to MATH as MATH. For MATH we have MATH and for MATH we have MATH where MATH converges to MATH as MATH. Note that at the origin, MATH is tangent to a ray making an angle of MATH with the positive real axis, and MATH is tangent to a ray making an angle of MATH with the positive real axis. These angles converge to MATH and MATH respectively as MATH. The reader should note that the scales have not been changed by these transformations. The circle MATH still has radius MATH, and the subsequent estimates will be uniform for MATH sufficiently small. For MATH and MATH, we define a kernel MATH by writing MATH where the map defined by MATH is a unitary isomorphism between MATH and MATH. The reader will observe that the left-hand side of REF is the third term on the right-hand side of REF . Therefore we have written this term as a sum of an explicit NAME integral in the new coordinate system, plus an error term involving the kernel MATH. Then, for MATH and MATH, we have an expression in terms of the same kernel of the ``reciprocal" NAME integral: MATH where the map MATH is a unitary isomorphism between MATH and MATH. The left-hand side of REF is just the second term on the right-hand side of REF written in the new coordinates, which has similarly been split into a NAME integral and a remainder term involving MATH. Similarly, we define a kernel MATH by writing for MATH and MATH from which we then obtain for the reciprocal NAME integral, for MATH and MATH, MATH where MATH denotes the unitary isomorphism from MATH to MATH defined by MATH. The decomposition REF is a representation of the first term on the right-hand side of REF . Likewise, REF is a representation of the first term on the right-hand side of REF . Having represented each term involving MATH on the right-hand sides of REF as a sum of a NAME integral operator in the new coordinate system and a remainder type integral operator, we now need to prove that these are in fact all bounded operators. The NAME operators are handled by an argument of CITE that uses the theory of NAME transforms in MATH spaces on straight rays. Their methods show that regardless of the value of MATH, MATH as long as MATH and MATH are both nonzero. Since these angles converge to MATH and MATH respectively as MATH, this will be the case for all sufficiently small MATH. Thus, the NAME integral operators appearing as the first terms on the right-hand sides of REF , and REF are bounded, with bounds that are independent of MATH. We now turn to the estimation of the ``remainder" operators with kernels MATH and MATH. In this connection, we first note that these kernels, which are explicitly written as MATH are bounded functions on their respective domains of definition, and MATH . Therefore, for all MATH, MATH and MATH while for MATH, MATH and for MATH, MATH . These norm estimates are finite for all MATH, and we must control their dependence on MATH as MATH. We now claim that MATH . These limits are finite because the integrands are bounded near MATH (in particular they are both less than MATH for all MATH), decay like MATH for large MATH, and are uniformly bounded elsewhere. To prove the claim, first rescale in both integrals by setting MATH and MATH. This modifies the integrand through the Jacobian by multiplication by MATH. For the MATH integral, the region of integration tends to the fixed rectangle MATH, while for the MATH integral, the region of integration tends to the semi-infinite strip MATH. Using the fact that the curve MATH is twice differentiable, one sees that the integrands MATH and MATH converge pointwise as MATH to the integrands on the right-hand side of REF . The convergence is in fact uniform for the MATH integral, and therefore this part of the claim follows immediately. For the MATH integral the claim follows from a dominated convergence argument. Since the limits REF exist, the operators having kernels MATH, MATH, MATH and finally MATH are bounded in the appropriate MATH spaces, uniformly as MATH. Combining these estimates with the NAME estimates of the NAME kernels and the results of a parallel analysis involving the circular arc MATH completes the proof of the lemma. |
nlin/0012034 | We first note that, modulo self-intersection points, the contour MATH can be written as a union of a MATH-independent part MATH, and several arcs making up the shrinking circle MATH. Let MATH, and decompose it into a sum MATH where the support of MATH is contained in MATH and that of MATH is contained in MATH. Then, for almost every MATH, MATH and for almost every MATH, MATH . Integrating to compute the norm, we first estimate MATH where MATH . First, we estimate the ``diagonal" terms MATH and MATH. Because the contour MATH is independent of MATH, there is some MATH-independent constant MATH such that MATH . Now for the integral MATH the contour depends on MATH, but in a simple way that can be scaled out. Thus, rescaling, MATH . With the contour rescaled to a radius independent of MATH, we see that there exists a MATH-independent constant MATH such that MATH . Next, we turn to the estimation of the ``cross terms" MATH and MATH. For this purpose, we again decompose the MATH-independent contour MATH into two MATH-independent parts by cutting it at the boundary of the fixed but small disk MATH (this disk is illustrated in REF ). Thus, let MATH (respectively MATH) denote the part of MATH inside (respectively outside) MATH. With this decomposition, we have by NAME, MATH where MATH is the MATH-independent total arc length of MATH. The supremum in the last line is bounded as MATH tends to zero because distance between MATH and MATH increases as MATH decreases. Therefore, there is a MATH-independent constant MATH such that MATH for all sufficiently small MATH. We momentarily delay the estimation of the term involving MATH. Applying the same decomposition to MATH and using NAME, we find with the same constant MATH . Finally, to estimate the remaining terms in MATH and MATH, we appeal to REF . Note that MATH is a union of eight smooth curve segments (in fact all but two of the segments are exactly straight ray segments) meeting at the origin. Therefore the lemma applies to the interaction of each curve segment with the circle MATH of radius MATH. Summing the MATH-independent estimates guaranteed by REF finally gives constants MATH and MATH such that MATH and MATH . Assembling the estimates of the diagonal terms and the cross terms finally completes the proof. |
nlin/0012034 | For MATH, set MATH and MATH. Then from REF , we observe that the conditions of REF in the appendix are met. This guarantees the existence of a matrix MATH satisfying the global error NAME REF in the MATH sense. According to REF , this solution MATH satisfies the error estimate REF , which by virtue of the bound on MATH afforded by REF and the fact that the total length of the contour MATH remains uniformly bounded as MATH tends to zero implies REF via the equivalence MATH. |
nlin/0012034 | Start with the exact representation of MATH: MATH which implies for the MATH entry MATH . Multiplying by MATH and passing to the limit MATH for fixed MATH using the fact that for large MATH, MATH and the fact that MATH yields MATH . The theorem is thus established upon using the estimate stated in REF . |
nlin/0012034 | The uniqueness and the symmetry REF are proved in exactly the same way. Consider the related boundary value problem of seeking a function MATH that is analytic in MATH and satisfies REF , and takes NAME continuous boundary values on both sides of MATH with exponent MATH that satisfy MATH for MATH or MATH in a band of MATH and MATH for MATH or MATH in a gap of MATH. Recall the function MATH, first used in REF, that satisfies MATH is analytic except at the bands of MATH (where the branch cuts are placed), and satisfies MATH as MATH. With this choice, the boundary value MATH relative to the oriented contour MATH is the positive value MATH . The function defined from any solution of this boundary value problem by the formula MATH is again analytic in MATH with at worst inverse square-root singularities at the endpoints of the bands and gaps on MATH. The boundary values on MATH, continuous except possibly at the isolated band and gap endpoints, satisfy MATH. The function MATH decays like MATH as MATH. The relatively mild nature of the singularities, together with the agreement of the boundary values where they are continuous implies that MATH is actually entire, and then by NAME 's theorem the decay condition implies that MATH. It then follows that MATH, so that all solutions are trivial. To use this result to prove uniqueness for MATH, one considers MATH to be the difference of two solutions and finds that this difference satisfies the above problem and is therefore zero. Similarly, to prove the symmetry REF , one considers the function MATH and again finds that MATH. Now, the function defined for MATH by MATH is by construction NAME continuous on MATH with exponent MATH and satisfies MATH. It follows that MATH. The proof is complete upon observing that by continuity of boundary values and decay at infinity, MATH necessarily agrees with the NAME integral of the function MATH. That is, the relation MATH is a consequence of NAME 's theorem. It follows that the function MATH for MATH leads to a solution of REF as required. |
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