paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0105133 | CASE: For any MATH we have that: MATH . We deduce that MATH . Replacing MATH by MATH and denoting MATH, we obtain: MATH . Notice now that MATH . To justify the last inequality, notice that if MATH is a monomial from MATH, with MATH, then we have: MATH . For the rest of the proof, ``degree" and ``degree zero term" will refer to the ``variables" MATH. Decompose MATH as MATH where MATH denotes the sum of all terms of degree MATH, MATH. Recall that the degree zero term of MATH is MATH with MATH. The degree zero term of MATH is MATH. From the vanishing of the latter we obtain that MATH for all MATH, hence MATH. Also the sum of the terms of degree REF in MATH is REF, hence we have MATH . Exactly as before, this implies MATH. Since MATH, we can continue the algorithm until we obtain MATH, hence MATH . Now we let MATH vary and deduce that MATH . CASE: When computing MATH, only the following two relations will be used: MATH and MATH . Hence we have to show that: MATH . This follows immediately from REF and the fact that MATH . |
math/0105133 | Take MATH with MATH and MATH minimal. From MATH it follows that MATH and then MATH . On the other hand, we have that MATH hence we must have MATH for all MATH. From the fact that MATH, it follows that MATH and MATH for any MATH and finally that MATH . The assertion stated in the lemma follows from the minimality of MATH. |
math/0105137 | We first construct a functor from quasi-coherent sheaves over MATH to MATH-comodules. Suppose that MATH is a quasi-coherent sheaf over MATH. Then MATH is in particular a quasi-coherent sheaf over MATH, so corresponds to a MATH-module MATH. Then if MATH is a point of MATH defined over MATH, with MATH and MATH, MATH . Let us denote by MATH the isomorphism of sheaves MATH. Then, MATH defines an isomorphism MATH of MATH-modules. Taking MATH to be the identity map MATH of MATH, we define MATH to be the composite MATH . We must show that MATH is counital and coassociative. Note first that MATH, thought of as a morphism in the groupoid MATH, is the identity morphism of the object MATH, and so in particular is idempotent. The cocycle condition implies that MATH is also idempotent, and since it is an isomorphism, it follows that MATH is the identity of MATH. Now, MATH defines a map from the point MATH to the point MATH of MATH. Since MATH is a map of sheaves over MATH, we conclude that MATH is the identity map. From this it follows easily that MATH is counital. To see that MATH is coassociative, let MATH denote the map that takes MATH to MATH. Let MATH denote the map that takes MATH to MATH. Then we have MATH and so MATH makes sense. A calculation shows that MATH, the diagonal map. If MATH is an arbitrary point of MATH with MATH and MATH, there is a map from MATH to MATH. Since MATH is a map of sheaves, we find that MATH is the composite MATH . This description allows us to compute MATH and MATH, and so also their composite. We find that MATH takes MATH to MATH. Similarly MATH takes MATH to MATH. The cocycle condition forces these to be equal, and so MATH is coassociative. We have now constructed a comodule MATH associated to any quasi-coherent sheaf MATH over MATH. We leave to the reader the striaghtforward check that this is functorial. Our next goal is to construct a functor from MATH-comodules to quasi-coherent sheaves over MATH. Suppose MATH is a MATH-comodule with structure map MATH. Then, in particular, MATH is a MATH-module, so there is an associated quasi-coherent sheaf MATH over MATH, defined by MATH, where MATH is a ring homomorphism. Given a point MATH of MATH with MATH and MATH, we have MATH . We define MATH by letting MATH be the composite MATH . We leave to the reader the check that MATH is a map of sheaves. It remains to show that MATH satisfies the cocycle condition and is an isomorphism. We begin with the cocycle condition. Suppose that MATH are ring homomorphisms with MATH, MATH, and MATH. Consider the following commutative diagram, in which all tensor products that occur are taken over MATH, and MATH. MATH . The outer clockwise composite in this diagram is MATH, and the outer counterclockwise composite is MATH, using the description of MATH given above. Thus MATH satisfies the cocycle condition. We must still show that MATH is an isomorphism for all MATH. Since MATH satisfies the cocycle condition and MATH is itself an isomorphism, it suffices to show that MATH is an isomorphism, where MATH is the identity morphism of MATH. That is, MATH is the composite MATH . But one can check, using the fact that MATH is counital, that MATH is the identity of MATH. This completes the proof that MATH is a quasi-coherent sheaf over MATH. We leave to the reader the check that it is functorial in MATH. We also leave to the reader the check that these constructions define inverse equivalences of categories. |
math/0105137 | Suppose MATH is a map of sheaves on MATH such that MATH. This means that MATH for all points MATH of MATH. We must show that MATH for all points MATH of MATH. We first show that MATH for all MATH in the essential image of MATH. Indeed, suppose MATH is a morphism from MATH to MATH. Then, since MATH commutes with the structure map MATH, we get the commutative diagram below. MATH . It follows that MATH. Now suppose MATH is a general point of MATH. Since MATH is in the sheaf-theoretic essential image of MATH, we can choose a covering MATH such that MATH is in the essential image of MATH for all MATH. Thus MATH for all MATH. We then have a commutative diagram MATH . The horizontal arrows are monomorphisms, since MATH and MATH are sheaves in MATH, so MATH as well. |
math/0105137 | Suppose we have a map MATH. This means we have maps MATH for all points MATH of MATH. We need to construct maps MATH for all points MATH of MATH such that MATH. Suppose first that MATH is in the essential image of MATH, so that there is a morphism MATH from MATH to MATH for some point MATH of MATH. If MATH were to exist, then we would have the commutative diagram below, MATH so we define MATH. We claim that this definition of MATH is independent of the choice of MATH. Indeed, suppose MATH is a morphism from MATH to MATH. Then MATH is a morphism from MATH to MATH, and so, since MATH is full, there is a morphism MATH from MATH to MATH such that MATH. Since MATH is a map of sheaves, MATH. On the other hand, by the cocycle condition we have MATH. Combining these two equations gives MATH so MATH is independent of the choice of MATH. In particular, if MATH, we can take MATH to be the identity map of MATH. The cocycle condition implies that MATH and MATH are identity maps, and so MATH. We now show that MATH commutes with the structure maps of MATH and MATH on the essential image of MATH. Suppose that MATH is a map of points of MATH, and that MATH is in the essential image of MATH. Choose a morphism MATH from MATH to MATH for some point MATH of MATH. Let MATH, so that MATH is a morphism from MATH to MATH, where MATH. Since MATH is a map of sheaves, we get the commutative square below. MATH . We would like to know that the square below is commutative. MATH . We claim that is an isomorphism from the top square to the bottom square, and so the bottom square must be commutative. Indeed, in the upper left corner this isomorphism is MATH, in the upper right corner it is MATH, in the lower left corner it is MATH, and in the lower right corner it is MATH. All the required diagrams commute to make this a map of squares. This uses the fact that MATH and MATH are maps of sheaves and the well-definedness of MATH. We now check that MATH commutes with MATH, on the essential image of MATH. Suppose we have a morphism MATH in MATH, and that MATH is in the essential image of MATH. Let MATH be a morphism from MATH to MATH for some point MATH of MATH. Consider the following diagram. MATH . By definition of MATH, the left-hand square is commutative. The cocycle condition implies that MATH, so the definition of MATH also implies that the outside square commutes. Since the horizontal maps are isomorphisms, the right-hand square must also be commutative. We now extend the definition of MATH to an arbitrary point MATH of MATH. The sheaf-theoretic essential image of MATH is all of MATH, we can choose a cover MATH of MATH in the topology MATH such that MATH is in the essential image of MATH for all MATH. Let MATH denote the image of MATH in MATH. We then have a commutative diagram MATH where the right-hand horizontal maps are the difference of the two restriction maps. Thus each row expresses its left-hand entry as a kernel. The diagram commutes since MATH is a map of sheaves on the essential image of MATH. Thus, there is a unique map MATH making the diagram commute. We now check that MATH is independent of the choice of cover. It suffices to show that MATH is unchanged if we replace the cover MATH by a refinement MATH, since any two covers have a common refinement. If we denote the map coming from the refinement by MATH, then we would have to have MATH, since some of the MATH form a cover of MATH and MATH is a map of sheaves on the essential image of MATH. Then the sheaf condition forces MATH as well. In particular, if MATH is already in the essential image of MATH, then we can take the identity cover to find that the new definition of MATH is an extension of our old definition. We now show that MATH is a map of sheaves over MATH. Suppose we have a map MATH of points of MATH. Choose a cover MATH of MATH such that MATH is in the essential image of MATH for all MATH. Then there is an induced cover MATH of MATH. The map MATH induces corresponding maps MATH, where MATH. Since MATH is a map of sheaves on the essential image of MATH, we have the commutative diagram below. MATH . The sheaf condition and the definition of MATH then show that the diagram MATH is commutative, and so MATH is a map of sheaves over MATH. The proof that MATH commutes with MATH, and so is a map of sheaves over MATH, is similar. |
math/0105137 | Suppose that MATH is a sheaf over MATH. We must construct a sheaf MATH over MATH and an isomorphism MATH of sheaves. We first construct MATH for MATH in the essential image of MATH, and show that it has the desired properties there. For every point MATH in the essential image of MATH, choose a point MATH of MATH and a morphism MATH from MATH to MATH. Note that this only requires choosing over a set, since MATH is a small category. Define MATH. We now construct the restriction of the structure map MATH to the essential image of MATH. Suppose that we have a map MATH between points of MATH, where MATH is in the essential image of MATH. Let MATH, so that MATH is a morphism from MATH to MATH, where MATH. Then MATH is a morphism from MATH to MATH. Since MATH is full and faithful, there is a unique morphism MATH of MATH from MATH to MATH such that MATH, We then define MATH to be the composite MATH . We must check the functoriality conditions for MATH (restricted to the essential image of MATH). First of all, if MATH is the identity map, then MATH will be the identity morphism of MATH. Since MATH is faithful, it follows that MATH is the identity morphism of MATH. The cocycle condition forces MATH to be the identity map, and so MATH is the identity as required. If MATH is another map of points of MATH, a diagram chase involving the cocycle condition for MATH and the fact that MATH is a map of sheaves shows that MATH is the composition MATH. We now show that MATH is a sheaf on the essential image of MATH. Indeed, suppose MATH is a point in the essential image of MATH, and MATH is a cover of MATH in MATH. We must check that MATH is an equalizer diagram. We have an equalizer diagram MATH since MATH is a sheaf. We construct an isomorphism from the bottom diagram to the top, from which it follows that the top is also an equalizer diagram. The morphism MATH induces a morphism MATH. We also have the morphism MATH. The composition MATH for a unique MATH, since MATH is full and faithful. Then MATH defines the desired isomorphism MATH. One constructs the isomorphism MATH in the same manner, using the morphisms MATH and MATH. The proof that the diagram below MATH is commutative is a computation using the fact that MATH is a map of sheaves, the cocycle condition, and the fact that MATH is faithful. We now construct the restriction of the map MATH to the essential image of MATH. Suppose MATH is a morphism from MATH to MATH, where MATH is in the essential image of MATH. Then MATH is a morphism from MATH to MATH. Since MATH is full and faithful, there is a unique morphism MATH from MATH to MATH such that MATH. Hence we can define MATH. We leave to the reader the diagram chase showing that MATH is a map of sheaves. We now construct the desired isomorphism of sheaves MATH. (Since MATH is determined by the restriction of MATH to the image of MATH, we can do this even though we have not completed the definition of MATH). Suppose MATH is a point of MATH. Then MATH is a morphism from MATH to MATH. Since MATH is full and faithful, there is a unique morphism MATH from MATH to MATH such that MATH. We define MATH . Obviously MATH is an isomorphism, but we must check it is compatible with the structure maps. We leave these checks to the reader; both are diagram chases. We have now defined a sheaf MATH on the essential image of MATH, and to complete the proof we need only extend it to a sheaf on all of MATH. For each point MATH of MATH, choose a cover MATH such that MATH is in the essential image of MATH for all MATH, making sure to choose the identity cover when MATH is already in the essential image of MATH. Once again, we can do this since MATH is a small category. We then define MATH as we must if we are going to get a sheaf, as the equalizer of the two maps of MATH-modules MATH . This definition of MATH will of course depend on the choice of cover MATH. Suppose MATH is some other cover such that MATH is in the essential image of MATH for all MATH. We claim that there is a canonical equalizer diagram MATH . To see this, let MATH denote the pullback of the two arrows MATH . We claim that there is a canonical isomorphism MATH. It suffices to check this when MATH is a refinement of MATH, since any two covers have a common refinement. In this case, there is a diagram MATH where the first map is induced by first mapping to MATH, and then using the structure maps of MATH restricted to the essential image of MATH to map further to MATH. It suffices to prove that this diagram is an equalizer. It is easy to check that MATH maps into the equalizer. If MATH maps to MATH in each MATH, then, using the fact that MATH restricted to the essential image of MATH is a sheaf, we find that MATH maps to MATH in each MATH. By definition of MATH, then, MATH. Similarly, suppose MATH is in the equalizer. Again using the fact that MATH restricted to the essential image of MATH is a sheaf, we construct an element MATH. The images of MATH and MATH in MATH coincide, since they coincide after restriction to the induced cover. Thus we get an element MATH restricting to the MATH. It follows that MATH restricts to the MATH as well, and so MATH is the desired equalizer. Now we can construct the structure maps of MATH. Suppose MATH is a map of points of MATH. The cover MATH of MATH induces a cover MATH of MATH, and the restriction MATH of MATH is in the essential image of MATH for all MATH, since MATH is so. Thus we get a map from MATH to MATH and so an induced map MATH on the equalizers. After composing this with the canonical isomorphism MATH, we get the desired structure map MATH. Since we chose the identity cover when MATH was already in the essential image of MATH, this extends the definition we have already given in that case. We leave it to the reader to check the functoriality of MATH. We now show that MATH is a sheaf. Suppose MATH is a point of MATH and MATH is a cover of MATH. Let MATH be the given cover of MATH, so that each MATH is in the essential image of MATH. Then MATH is a cover of MATH, and each MATH is the essential image of MATH since each MATH is. Similarly, MATH is a cover of MATH. Thus we get the commutative diagram below. MATH . The subscripts MATH, MATH, and MATH all refer to the MATH, and the subscripts MATH and MATH all refer to the MATH. So, for example, MATH is the image of MATH in MATH. The right-hand horizontal arrows are all the differences of the two restriction maps. This means that the second and third rows express their left-hand entries as kernels, since MATH restricted to the essential image of MATH is a sheaf. Similarly, the bottom vertical arrows are also differences of the two restriction maps. It follows that each column expresses its top entry as a kernel, since the definition of MATH does not depend on which cover we choose, up to isomorphism. A diagram chase then shows that the top row expresses MATH as a kernel, which means that MATH is a sheaf. We now construct the isomorphism MATH. Suppose MATH is a morphism in MATH. Let MATH be the given cover of MATH, so that each MATH is in the essential image of MATH. It follows that MATH is also in the essential image of MATH for all MATH. Let MATH denote the image of MATH in MATH, and similarly let MATH denote the image of MATH in MATH. Then we have a commutative diagram MATH . Here the right-hand horizontal arrows are differences of restriction maps, as usual. The top row is an equalizer by definition, and we have proved that the bottom row is also an equalizer diagram. Hence there is a unique map MATH, necessarily an isomorphism, making the diagram commute. The facts that MATH satisfies the cocycle condition and is a map of sheaves are the usual sheaf-theoretic diagram chases, and we leave them to the reader. |
math/0105137 | Let MATH. Since the product is finite, it suffices to show that MATH is an equalizer for all MATH-modules MATH. Since MATH is faithfully flat, it suffices to show that MATH is an equalizer for all MATH. But, before tensoring with MATH, this sequence is just the beginning of the bar resolution of MATH as a MATH-algebra; since the bar resolution is contractible, this diagram remains an equalizer after tensoring with MATH. |
math/0105137 | Suppose MATH is a point of MATH, and MATH is a cover in the flat topology. We must show that the diagram MATH is an equalizer diagram. But, since MATH is quasi-coherent, MATH is isomorphic to the diagram MATH which is an equalizer diagram by REF . |
math/0105137 | Suppose MATH is an arbitrary MATH-algebra. Then MATH is an equalizer diagram since MATH is pure, but MATH . Thus MATH is also an equalizer diagram, being a direct sum of equalizer diagrams. Since MATH is faithfully flat over MATH, it follows that MATH is an equalizer diagram. |
math/0105137 | Since MATH is an equivalence of categories, and quasi-coherent sheaves are a full subcategory of sheaves in the flat topology by REF , we find immediately that MATH is full and faithful. It remains to show that it is essentially surjective. Suppose MATH is a quasi-coherent sheaf over MATH. Because MATH is an equivalence of categories, there is a sheaf MATH in the flat topology, over MATH, such that MATH. We will show that MATH is in fact quasi-coherent, so that MATH is essentially surjective on quasi-coherent sheaves. To do so, we must show that, if MATH is a map of points of MATH, then the adjoint MATH of the structure map of MATH is an isomorphism. First suppose that MATH is in the essential image of MATH. Then there is a MATH and a map MATH. Let MATH, so that MATH. Then we have the commutative diagram below. MATH . The top square of this diagram commutes because MATH as sheaves, and the bottom square commutes because MATH is a map of sheaves. The vertical maps are isomorphisms, and the top horizontal map is an isomorphism since MATH is quasi-coherent. Hence the bottom horizontal map is an isomorphism as well. In fact, if MATH is in the essential image of MATH and MATH is a cover of MATH in the flat topology, we claim that the equalizer diagram MATH is pure. Indeed, suppose MATH is a MATH-algebra, so we have MATH. Then MATH is a cover of MATH in the flat topology. It follows from what we have just done (and the fact that covers in the flat topology are finite), that the diagram MATH is isomorphic to MATH, and so is still an equalizer diagram. Now suppose MATH is an arbitrary point of MATH. Since the sheaf-theoretic essential image of MATH is all of MATH, we can choose a cover MATH such that each MATH is in the essential image of MATH. There is an induced cover MATH of MATH, and maps MATH, so each MATH is also in the essential image of MATH. We then get the commutative diagram below, which is a map from the diagram MATH to MATH. MATH . Here the map MATH is the difference between the two restriction maps, so the bottom row expresses MATH as a kernel. We have already seen that the maps MATH and MATH are isomorphisms, so if we knew that MATH were an equalizer diagram, we would be able to conclude that MATH is an isomorphism, and therefore that MATH is quasi-coherent. In particular, if MATH is flat over MATH, we conclude that the diagram MATH is isomorphic to the equalizer diagram MATH. In case MATH is in the essential image of MATH, we have proved that MATH is pure. In particular, MATH is a pure equalizer diagram for all MATH. Since MATH is faithfully flat over MATH, it follows from REF that the equalizer diagram MATH is pure. Thus, for any MATH, MATH is an equalizer diagram, and so MATH is quasi-coherent. |
math/0105137 | Given MATH, MATH if and only if MATH and MATH have the same domain and codomain when thought of as morphisms of MATH and MATH. The proposition follows. |
math/0105137 | The map MATH is full if and only if every morphism MATH is equal to MATH for some morphism MATH of MATH. NAME another way, MATH is full if and only if every ring homomorphism MATH can be extended through MATH to a ring homomorphism MATH. This is equivalent to MATH being a split monomorphism. |
math/0105137 | Any map MATH of rings that is both a split monomorphism and a ring epimorphism is an isomorphism. Indeed, MATH is monic since MATH is a ring epimorphism and epic since MATH is a split monomorphism, so is an isomorphism for all MATH. |
math/0105137 | We first determine when MATH is in the essential image of MATH. For this to happen we need an object MATH and a morphism MATH from MATH to MATH. A morphism MATH from MATH to anywhere is equivalent to the composite MATH which we also denote, by abuse of notation, by MATH. The codomain of MATH is the composite MATH. Altogether then, MATH is in the essential image of MATH if and only if there is a map MATH such that MATH. Now, suppose the sheaf-theoretic essential image of MATH is all of MATH. Then there must be a cover MATH such that the image of the identity map of MATH, namely MATH, is in the essential image of MATH for all MATH. By the preceding paragraph, this is true if and only if there exist maps MATH such that MATH. Let MATH be the product of the MATH and let MATH be the product of the MATH. Then MATH is the product of the MATH, which displays MATH as a faithfully flat extension of MATH since MATH is a cover of MATH. Conversely, suppose there is a ring map MATH such that MATH exhibits MATH as a faithfully flat extension of MATH. Suppose MATH is an arbitrary point of MATH. Then MATH is a cover of MATH. One can easily check that the image of MATH in MATH is the composite MATH . Since MATH, the image of MATH is in the essential image of MATH, and so MATH is in the sheaf-theoretic essential image of MATH. |
math/0105137 | Let MATH. Let MATH denote the ring homomorphism defined by MATH, and let MATH denote the ring homomorphism defined by MATH. Then MATH and the composite MATH induce two formal group laws MATH and MATH over MATH, both MATH-typical and of strict height MATH. Furthermore, MATH. A result of NAME, as modified by CITE, then implies that there is a faithfully flat graded ring extension MATH and a strict isomorphism from MATH to MATH. This strict isomorphism is represented by a graded ring homomorphism MATH. Let MATH be the composite MATH. Since the domain of MATH is MATH, MATH. This means that there is a well-defined map MATH . Furthermore, MATH represents the codomain of MATH, so is MATH. We know already that MATH is a faithfully flat ring extension, and we claim that MATH is also a faithfully flat ring extension. Indeed, since MATH is a graded field, MATH is a free MATH-module, and so MATH makes MATH into a free MATH-module. REF complete the proof. |
math/0105137 | By REF, MATH is the direct limit of comodules MATH, where MATH is finitely presented and MATH-nilpotent. Since we are assuming either that MATH is finitely presented or that MATH, in either case we may as well take MATH. Then REF reduces us to the case MATH. In this case, one can check using the cobar resolution (as in CITE) that we have canonical isomorphisms MATH and MATH . Now REF implies that MATH . This completes the proof. |
math/0105138 | We assume that MATH, as the complex form of the NAME series is more convenient. MATH is equipped with its standard positive definite Hermitian inner product MATH where MATH and MATH. This agrees with the usual inner product on MATH. The norm of MATH is given by MATH, and MATH. The facts about NAME series required for the proof are as follows. If MATH is locally square integrable then it has a NAME expansion MATH (the convergence is in MATH and the series may not converge pointwise). The MATH norm of MATH is given by MATH . If MATH is absolutely continuous and MATH is locally square integrable then MATH has the NAME expansion MATH and therefore MATH as the contribution to the middle sum from the term MATH is zero. Let MATH be the NAME expansion of MATH, where MATH. Then MATH . Therefore, using REF, we have MATH . Also, by REF, MATH . Subtracting REF from REF, we set MATH . REF (below) implies that MATH with equality if and only if MATH is a multiple of MATH, or MATH for all MATH. The latter occurs if and only if MATH where MATH and MATH. |
math/0105138 | If MATH, for some integer MATH, then equality holds in REF. If MATH is not an integer multiple of MATH, we set MATH. Then MATH, and the addition formula for sine yields MATH . Since MATH, we then have MATH by induction, which completes the proof. |
math/0105138 | This method is an adaptation of NAME 's original approach CITE. In that paper, he solves a discrete version of the problem, showing that the average squared distance between the vertices of a MATH-gon of constant side length is maximized by the regular MATH-gon. He then obtains the main result by approximation. We go directly to the continuum case, which turns out to be simpler. To simplify notation, let MATH. Let MATH . We claim that MATH is MATH with MATH and initial conditions MATH . These formulas are clear when MATH is MATH and hold in the general case by approximating by MATH functions. The explicit formula for MATH makes it clear that MATH is MATH. Next we derive a differential inequality for MATH, using an elementary geometric fact (which appears in a slightly different form in NAME 's paper as REF ): For any tetrahedron MATH, MATH, MATH, MATH in MATH, MATH with equality if and only if MATH and MATH are parallel as vectors. Denote the vectors MATH, MATH, MATH, MATH by MATH, MATH, MATH, MATH. Then MATH, and MATH . Equality holds if and only if MATH for some MATH, which is equivalent to MATH and MATH being parallel as vectors. For any MATH, MATH and MATH, we can apply REF to the tetrahedron MATH, MATH, MATH, MATH to derive the equation MATH . Holding MATH fixed and integrating with respect to MATH, MATH by the NAME inequality. Therefore MATH. For any fixed MATH, this can be rewritten MATH . When MATH is not a multiple of MATH, MATH and MATH is well-defined for small MATH. Further, MATH has a local minimum at MATH, and so the second derivative of MATH is non-negative at zero. Using REF, this tells us that MATH . Meanwhile, MATH satisfies the differential equation MATH . We are trying to show that MATH and that if equality holds for any MATH, then MATH. Let MATH . In these terms, we want to show that MATH and that if MATH for some MATH then MATH. Using REF, MATH . By two applications of NAME 's rule, we compute MATH. Thus MATH, as well. So if MATH is ever positive, it will have a positive local maximum at some point MATH. At that point, MATH which is a contradiction. So MATH is non-positive on MATH. Further, if MATH is zero at any point in MATH, the strong maximum principle CITE implies that MATH vanishes on the entire interval. Thus MATH with equality at any point of MATH if and only if MATH. Last, we show that if MATH then MATH is an ellipse. By our work above, if MATH, then for each fixed MATH, MATH maximizes MATH subject to the constraint that MATH is held constant. The NAME multiplier equation for this variational problem is MATH where MATH is a constant depending on MATH. When MATH we can use the fact that MATH has period MATH and this becomes MATH . Differentiating twice with respect to MATH, and using both the periodicity and the equation, MATH . So MATH satisfies the equation MATH and has period MATH. This implies that MATH for some MATH, and MATH with MATH and MATH in MATH. But MATH, for otherwise MATH, a contradiction. Taking two antiderivatives, MATH with MATH in MATH. Periodicity implies that MATH, completing the proof. |
math/0105138 | The shortest distance between MATH and MATH along the curve is MATH. Thus, the squared chord length MATH is in MATH, except when MATH. Being undefined at this point does not affect the existence of the integrals. Using NAME 's inequality for concave functions REF , that MATH is increasing, and that MATH for almost all MATH, we have MATH . If equality holds in REF, then the above string of inequalities implies that equality holds between the two middle terms, that is, equality holds in REF. Thus, since MATH, we may apply REF to conclude that MATH must be as in REF. Since MATH has unit speed, it follows that MATH is a unit vector for all MATH, which forces the vectors MATH and MATH to be orthonormal, and so implies that MATH is the unit circle. Conversely, if MATH is the unit circle, then MATH for all MATH and therefore equality holds in REF. |
math/0105138 | In both cases equality is clear for the unit circle. By the mean value property of integrals and REF , MATH establishing REF. Further, equality in REF implies equality in REF, which, by REF , happens if and only if MATH is the unit circle. The proof of REF follows easily from REF: MATH and again equality implies in particular that MATH, which, by REF, happens if and only if MATH is the unit circle. |
math/0105138 | Making the substitution MATH, MATH, changing the order of integration, and using the fact that MATH for any MATH, we have MATH . For each MATH, if we let MATH, then MATH and MATH is increasing and concave on MATH. By REF , MATH with equality if and only if MATH is the unit circle. Integrating this from MATH to MATH tells us that MATH is greater than or equal to the corresponding value for the unit circle, with equality if and only if REF holds for almost all MATH. But if equality holds for any MATH, then MATH is the unit circle. |
math/0105138 | If we let MATH then using REF, we see that MATH is the renormalization energy based on MATH. We must show that MATH is convex and decreasing in MATH for MATH for all MATH. It suffices to check the signs of the first and second partial derivatives of MATH with respect to MATH on MATH. When MATH, MATH, and MATH, MATH and MATH . Since MATH can be arbitrarily close to MATH if the curve is nearly straight, examining this equation shows that the condition MATH is required to enforce the convexity of MATH. So for every MATH, MATH is decreasing and convex on MATH. Further, a direct calculation shows that MATH when MATH. Thus MATH satisfies the hypotheses of REF . Computing the energy of the round circle by changing the variable MATH and noting that the resulting integrand is symmetric about MATH, we have MATH with equality if and only if MATH is the unit circle. |
math/0105138 | Sallee's stretching theorem CITE (see also CITE) says that for any closed unit-speed space curve MATH of length MATH, there exists a corresponding closed, convex, unit-speed plane curve MATH of length MATH such that for every MATH, MATH in MATH, MATH with equality for all MATH and MATH iff MATH is convex and planar. Since the integrand defining MATH is an increasing function of chord length for MATH, this implies that every maximizer of MATH must be convex and planar. Let MATH denote the space of closed, convex, planar, unit-speed curves of length MATH which pass through the origin, with the MATH norm. It now suffices to show that a maximizer of MATH exists in MATH. NAME 's selection principle CITE implies that this space of parametrized curves is compact in the MATH norm. It easy to see that MATH is MATH-continuous for MATH in MATH (in fact, it is jointly continuous in MATH and MATH on the product MATH), completing the proof. |
math/0105138 | We know from the proof of REF that MATH is a MATH-continuous functional on the space MATH. If we choose any MATH, and choose a compact interval MATH containing MATH, then MATH contains a neighborhood of MATH. We now show MATH is compact. Take any sequence MATH. Since MATH is compact, we may assume that the MATH converge to some MATH. Since MATH is MATH-compact (see the proof of REF ), we may also assume that the MATH converge to some MATH. It remains to show that MATH is a maximizer for MATH. If not, there exists some MATH with MATH. But then MATH since MATH is continuous in MATH. On the other hand, since the MATH are maximizers for the MATH, we have MATH for each MATH, and so MATH . |
math/0105142 | According to REF an operator MATH is a weak solution to REF if REF holds. Meanwhile, REF implies MATH . Thus, by REF the operator MATH is a weak solution to the dual NAME REF . The converse statement is proven in a similar way. |
math/0105142 | CASE: Assume that the NAME REF has a strong solution MATH, that is, REF hold. Let MATH be a finite interval such that MATH and MATH. Applying to both sides of REF the spectral projection MATH, a short computation yields MATH for any MATH. Since MATH, by REF one concludes that MATH belongs to the resolvent set of the operator MATH. Hence, REF implies MATH . Next, let MATH be a finite interval and MATH a finite system of mutually disjoint intervals such that MATH. For those MATH such that MATH pick a point MATH. Using REF one obtains MATH . The left hand side of REF can be computed explicitly: MATH . The first term on the right-hand side of REF is the integral sum for the NAME integral REF. More precisely, since MATH is analytic in a complex neighborhood of MATH, by REF one infers MATH . The last term on the right hand side of REF vanishes MATH . This can be seen as follows. For any MATH we have the estimate MATH . Here we have used the estimate MATH . Passing to the limit MATH in REF, by REF - REF one concludes that for any finite interval MATH . Since REF implies REF , which, in particular, proves the uniqueness of a strong solution to the NAME REF . In order to prove the converse statement of REF , assume that the NAME integral on the right-hand side part of REF converges as MATH and MATH in the strong operator topology. Denote the resulting integral by MATH. Then, REF holds for any finite MATH and MATH. This implies that for any MATH we have MATH . Hence, MATH and MATH converges to MATH as MATH and MATH. Therefore, MATH and, hence, MATH . Then REF can be rewritten in the form MATH . Combining REF proves that MATH is a strong solution to the NAME REF . CASE: Assume that the dual NAME REF has a strong solution MATH. As in the proof of REF , choose a finite interval MATH such that MATH. Since MATH, we have MATH for any MATH by the definition of a strong solution. Take a point MATH. It follows from REF that MATH. As in the proof of REF , it is easy to check the validity of the representation MATH which holds MATH . Next, let MATH be a finite interval and MATH a finite system of mutually disjoint intervals such that MATH. For those MATH such that MATH pick a point MATH. Using REF one then finds that MATH . The equality REF implies MATH . Thus, passing in REF to the limit as MATH one infers that MATH . Since for any MATH one concludes that the integral on the right-hand side part of REF converges as MATH and MATH in the strong operator topology and REF holds, which gives a unique strong solution to the dual NAME REF . In order to prove the converse statement of REF , assume that there exists the strong operator limit MATH . For any finite MATH and MATH such that MATH we have MATH . By REF any point MATH belongs to the resolvent set of the operator MATH and, hence, to the one of MATH. Picking such a MATH, MATH, the operator REF can be split into two parts MATH where MATH . Using the functional calculus for the self-adjoint operator MATH one obtains MATH . Thus, for MATH one concludes that MATH . That is, MATH since MATH by REF. It follows from REF that MATH for any MATH and, thus, MATH . Applying MATH to the both sides of the resulting equality REF one infers that MATH is a strong solution to the dual NAME REF which completes the proof. |
math/0105142 | By REF, and REF the operator NAME integrals in REF can be understood in the operator norm topology. Thus, MATH given REF and MATH given by REF are unique strong solutions to the NAME REF by REF . Therefore, the operators MATH and MATH are unique weak solutions and MATH by REF . In order to prove REF it suffices to note that under REF for any MATH and for any MATH, MATH, due to REF the following estimate holds MATH . Thus, MATH which proves that MATH and, hence, the inclusion REF is proven. It remains to prove the inclusion REF. Given MATH, we represent MATH for some finite MATH, MATH, in the form REF where MATH and MATH are just the same ones as in REF , respectively. Under REF , by REF one concludes that the operator NAME integral in REF converges as MATH and MATH in the uniform operator topology to some operator MATH. Then, from REF one learns that for any MATH and, thus, MATH which proves REF. The proof is complete. |
math/0105142 | CASE: The operator MATH is a weak solution to REF if and only if MATH is a weak solution to the equation MATH where MATH . Applying REF completes the proof of REF . CASE: The operator MATH is a strong solution to REF if and only if MATH is a strong solution to the equation MATH where MATH . Applying REF completes the proof of REF . CASE: The operator MATH is a strong solution to REF if and only if MATH is a strong solution to the equation MATH where MATH . Applying REF completes the proof of REF . The proof is complete. |
math/0105142 | The proof is based on an application of NAME 's Fixed Point Theorem. CASE: Let MATH be a continuous function on MATH except at zero such that MATH . Introducing the function MATH by REF any fixed point of the map MATH given by MATH where the improper NAME integral is understood in the weak sense, is a weak solution to the NAME REF . Taking into account that MATH from REF one concludes that MATH and MATH . Clearly, MATH maps the ball MATH into itself whenever MATH and MATH is a strict contraction of the ball MATH whenever MATH . Since the extremal problem for the NAME transform, which is to find the infimum of MATH over all functions MATH such that MATH for MATH, has the solution (compare REF ) MATH one concludes that MATH maps the ball MATH into itself whenever MATH and MATH is a strict contraction of MATH whenever MATH . Solving REF one concludes that if the radius MATH of the ball MATH is within the bounds MATH then MATH is a strictly contractive mapping of the ball MATH into itself. Applying NAME 's Fixed Point Theorem proves REF . CASE: Given MATH, under REF we have the identity MATH which implies the estimate MATH whenever MATH . Since REF holds and the operator MATH has a finite MATH-norm, the mapping MATH where the integral is understood in the strong sense, is well defined on the domain MATH . Since for MATH one clearly has the estimate MATH any fixed point of the map MATH is a strong solution to the NAME REF by REF . Using REF we have the following two estimates MATH and MATH . Clearly, by REF MATH maps the ball MATH into itself whenever MATH and by REF MATH is a strict contraction on MATH whenever MATH . Solving REF simultaneously, one concludes that if the radius of the ball MATH is within the bounds MATH then MATH is a strictly contracting mapping of the ball MATH into itself. Applying NAME 's Fixed Point Theorem we infer that REF has a unique solution in any ball MATH whenever MATH satisfies REF. Therefore, the fixed point does not depend upon the radii satisfying REF and hence it belongs to the smallest of these balls. This observation proves the estimate MATH . Finally, using REF , for the fixed point MATH one obtains the estimate MATH . Then REF yields MATH which completes the proof. |
math/0105142 | Under REF we have the inequalities MATH . The NAME function MATH is a strictly increasing continuous function on MATH and MATH not withstanding the possibility for the one-sided limits MATH and MATH to turn into MATH and MATH, respectively. Therefore, the equation MATH has a unique root MATH, the function MATH is an element of MATH, and, hence, the NAME REF has a weak/strong solution, since the existence REF is satisfied. In order to prove that the constant MATH in the upper bound REF is sharp, it suffices to show that for any MATH there exists a function MATH such that MATH and the NAME REF has no solutions MATH. Let MATH be a positive continuous function on MATH such that MATH . Given MATH, introduce the functions MATH and MATH . One infers MATH and MATH . Hence for any MATH, one can find a MATH such that the following inequality holds MATH . Introducing MATH one obviously concludes that MATH . Meanwhile, REF implies the estimate MATH . Therefore, the NAME function MATH given by MATH does not vanish on MATH (note that MATH as MATH) and hence REF is violated for all MATH. Since MATH is a continuous function and it does not vanish on MATH, REF is violated for all MATH. Hence, the NAME REF has no weak/strong solutions in this case since the existence REF is violated. |
math/0105142 | By the definition of an admissible operator, the difference of the resolvents of MATH and MATH is a trace class operator and the following representation holds for some MATH such that MATH which implies REF, since MATH. |
math/0105142 | REF implies that REF MATH and REF the operators MATH and MATH are resolvent comparable. Thus, MATH is an admissible operator. By REF the operator MATH is a trace class perturbation of MATH and hence MATH and MATH are resolvent comparable. Therefore, REF holds by the definition of the spectral shift function for a pair of resolvent comparable admissible operators, which proves REF . CASE: We start with the representation MATH where MATH makes sense by REF. By REF MATH which proves that MATH by REF. Therefore, the NAME determinant of the operator MATH is well defined and MATH . Since REF holds, the operator MATH is well defined on the whole NAME space MATH as a closed operator being the product of two closed operators. Hence, MATH is bounded by the Closed Graph Theorem. In particular, the following representation holds MATH . Using REF, and the fact that MATH one proves MATH . Further, using REF of MATH one computes MATH which by REF proves REF. Moreover, REF yield MATH which completes the proof of REF . CASE: One infers MATH . Hence MATH by REF and the fact that MATH since MATH, which proves that the perturbation determinant MATH is well defined. Moreover, MATH . By NAME 's REF we have MATH and hence REF holds by REF. |
math/0105142 | If MATH given by REF is a strong solution of REF, this means that MATH and MATH . Under REF , and REF we have the inclusions MATH . Moreover, the NAME REF splits into a pair of the equations MATH . Rewriting these equations in the form MATH one immediately observes that REF combined with REF is equivalent to invariance of the subspaces MATH, MATH for the operator MATH. In turn, REF implies the invariance of the subspace MATH for MATH, which proves the lemma. |
math/0105142 | CASE: By REF MATH and, thus, the spectrum of MATH is a subset of the imaginary axis. This means that zero does not belong to the spectrum of MATH and, hence, MATH has a bounded inverse. CASE: Since by REF MATH has a bounded inverse, REF is equivalent to the representation MATH which, in turn, taking into account REF, is equivalent to the NAME REF . Then, applying REF , the validity of REF - REF is equivalent to the fact that the graph subspaces MATH, MATH are reducing subspaces. CASE: Taking into account REF, by inspection one gets MATH . Since MATH and MATH, the validity of REF - REF is an immediate consequence of REF - REF. |
math/0105142 | By REF the normal operator MATH has a bounded inverse. Due to REF the spectrum of MATH is purely point. Thus, by REF MATH where MATH is the zero trace commutator class introduced by REF . By REF one concludes MATH . Therefore, since by REF the operator MATH is a relatively trace class perturbation of MATH, one concludes that the operators MATH and MATH are resolvent comparable. By REF MATH and MATH are also resolvent comparable, and, therefore, by REF the operator MATH is admissible with the self-adjoint representative MATH. Thus, the stability property holds MATH by the definition of the spectral shift function for resolvent comparable admissible operators. Next, let MATH be the polar decomposition of MATH. By REF the operator MATH is diagonal with respect to the decomposition REF. Using representation REF one infers that MATH is a trace class operator, since MATH is the NAME - NAME operator by the hypothesis. Therefore, MATH where MATH, and, hence, MATH by REF. The operator MATH is similar to the self-adjoint operator MATH: MATH . Using REF, one concludes that MATH and MATH are resolvent comparable. Therefore, taking into account that MATH one infers that MATH is a self-adjoint representative of the admissible operator MATH and, hence, MATH by REF . By REF the operator MATH is diagonal with respect to decomposition REF MATH where MATH, MATH are self-adjoint operators in the NAME spaces MATH introduced by REF. Since MATH is a block-diagonal operator, by additivity of the spectral shift function associated with a pair of self-adjoint operators with respect to direct sum decompositions (which follows from the definition of the spectral shift function by the trace REF ) one obtains that MATH . By REF the operator MATH is diagonal with respect to the decomposition REF MATH where MATH, MATH, MATH are operators similar to self-adjoint operators MATH given by REF: MATH . Here MATH, MATH, MATH are the entries in the matrix representation for the operator MATH . By REF MATH is a NAME operator, which proves that MATH . Therefore, the operators MATH, MATH, MATH are admissible with the self-adjoint representatives MATH. Since MATH and MATH are resolvent comparable, so MATH and MATH, MATH, MATH are. Hence, we have the following representation by REF MATH . Combining REF, and REF proves REF. |
math/0105142 | Assume, for definiteness, that the operator MATH has a finite norm with respect to the spectral measure of the diagonal entry MATH of MATH and the inequality holds MATH . Recall that by REF and hence MATH. By REF the NAME equation MATH has a unique strong solution MATH . Therefore, the dual NAME equation MATH has a unique strong solution MATH by REF , and, moreover, MATH. Introducing the notations MATH and MATH, REF can be rewritten in the form MATH . Therefore, the NAME REF has a strong solution of the form REF. Applying REF one proves that the subspaces MATH, MATH are reducing subspaces for MATH, which proves the first assertion of the theorem under REF . In the case where MATH has a finite norm with respect to the spectral measure of the diagonal entry-MATH and the inequality MATH holds, the proof can be performed in an analogous way. Applying REF proves estimate REF which, in turn, proves that MATH . The last assertion of the theorem is a corollary of REF . The proof is complete. |
math/0105142 | The proof is analogous to that of REF . The only difference is that now we refer to REF , since for bounded MATH, MATH. the concepts of the weak, strong and operator solutions of the NAME REF coincide. |
math/0105142 | Any spectral subspace for MATH is its reducing subspace. Thus, by REF the subspaces REF are reducing graph subspaces for MATH. Then REF implies that the angular operators MATH and MATH from the right-hand side parts of REF are strong solutions to REF , respectively. A proof of REF can be found in CITE. |
math/0105142 | Under REF or REF this assertion is an immediate consequence of REF or REF respectively, and REF . Therefore, assume REF . Since the operator MATH is bounded, and the interval MATH does not contain points of the spectrum of MATH for all MATH, by a result by CITE (see also REF ) the spectral projection MATH continuously depends on MATH in the uniform operator topology. By REF the projection MATH admits matrix representation with respect to the direct sum of the NAME spaces MATH where MATH, MATH. In particular, the continuity of the family MATH implies the continuity of the families of operators MATH and MATH in the uniform operator topology of the spaces MATH and MATH, respectively. Since the family MATH is continuous, the family MATH is also continuous. Multiplying the operator MATH by MATH from the left proves the continuity of the angular operators MATH as a function of MATH in the uniform operator topology. Recalling now that MATH proves the continuity of the family MATH, MATH, MATH as a function of the parameter MATH in the uniform operator topology. The proof is complete. |
math/0105142 | We recall that the strong solvability of the NAME REF under constraint REF is equivalent to the strong solvability of the following pair of REF . Therefore, the assumption MATH for some MATH implies MATH. Hence, the right-hand side of REF is an element of the space MATH. Under REF one concludes that MATH by REF (in particular by estimate REF), and, thus, MATH, since REF holds. Further, assume that REF holds for some MATH. By REF the operator MATH, MATH is similar to a self-adjoint operator MATH. That is, the representation holds MATH for some MATH such that MATH (see REF). Therefore, REF can be rewritten in the form MATH and, hence, the operator MATH is a strong solution to the NAME equation MATH . By REF one infers MATH . Meanwhile, the assumption MATH for some MATH implies MATH and, hence, MATH, MATH. Applying REF once more, one deduces that MATH. Hence, MATH, MATH. Finally, by REF one concludes that MATH. The proof is complete. |
math/0105142 | The existence of the one-parameter family of the spectral shift functions MATH, MATH that is continuous in the topology of the weighted space MATH is proven in CITE. Next, since MATH belongs to the spectral gap of MATH for any MATH, the spectral shift function MATH is a constant MATH a.e. on the interval MATH. Integrating the difference MATH over MATH with the weight MATH yields the estimate MATH which proves that MATH is a continuous integer-valued function of MATH. Since MATH, it follows that MATH for all MATH. |
math/0105142 | Under the assumptions of the theorem the existence of a strong solution MATH of the NAME REF is guaranteed by REF or REF . Since, by hypothesis, MATH, one infers MATH by REF . Thus, REF holds. Therefore, MATH is a trace class operator, and hence REF holds. REF holds by hypothesis and, therefore, MATH, MATH are admissible operators, MATH and MATH, MATH are resolvent comparable and the decomposition REF takes place by REF . Introducing the family MATH, MATH, by REF one infers the existence of the operators MATH that continuously depend on MATH in the topology of the space MATH and are such such that MATH, MATH has reducing graph subspaces MATH . Therefore, by REF the NAME equation MATH has a strong solution MATH which reads with respect to the decomposition REF as MATH and MATH, MATH. Hence, each entry MATH, MATH, in REF is a strong solution of the NAME equation MATH . Since MATH is continuous in the norm operator topology, the right-hand side of REF depends continuously on MATH in the topology of the space MATH. Therefore, by REF (estimate REF) the path MATH, MATH, is continuous in the topology of the space MATH, and, thus, the family MATH is continuous in the topology of the space MATH. Clearly, the map MATH is continuous in the topology of the space MATH, MATH . Taking into account that the family MATH is continuous in the topology of MATH, MATH, and introducing the self-adjoint representatives of the admissible operators MATH, MATH, MATH, MATH one concludes that the map MATH is also continuous in the topology of MATH, MATH . Let MATH . Obviously MATH . Our claim is that MATH, MATH belongs to the resolvent set of MATH, MATH for all MATH, that is, MATH . Under REF is a consequence of REF . Assume, therefore, either REF or REF . Under REF , applying REF one obtains the following uniform bounds MATH . Thus, one concludes that MATH . Under REF the operator MATH, MATH, MATH, MATH, is a strict contraction by REF . Therefore, MATH and MATH . By REF the operators MATH and MATH, MATH, MATH, are similar to each other, which proves REF under REF or/and REF. Applying REF one proves that there is a family of spectral shift functions MATH, MATH continuous in the topology of the weighted space MATH such that MATH for any interval MATH, MATH. By REF the operators MATH and MATH, MATH, MATH, are resolvent comparable and, hence, by REF one has the representation MATH since MATH are self-adjoint representatives of the admissible operators MATH, MATH, MATH. It follows that the spectral shift functions MATH associated with the pairs MATH can be chosen in such a way that for any interval MATH, MATH which, in particular, implies REF , since REF holds. |
math/0105143 | Let MATH be an isolated NAME stable set of a vector field MATH. We first observe that there is an open set sequence MATH such that MATH, MATH, for all MATH, and MATH. To see this, let MATH be the ball of radius MATH around MATH, that is, MATH. By the NAME stability of MATH, for each MATH, there is an open neighborhood MATH of MATH such that MATH for MATH. Setting MATH we are done. Now, let MATH be an isolating block for MATH and MATH be such that MATH. Then, MATH for every MATH. So, MATH since MATH is an isolating block of MATH. On the other hand, as MATH is forward invariant, we have that MATH for MATH. Then, MATH and so MATH . Applying REF we obtain MATH. But MATH and MATH is invariant. Then, MATH for every MATH and so MATH. This proves MATH. Recalling that MATH is a neighborhood of MATH, we see that MATH is an attracting set since it is compact. This proves the lemma. |
math/0105143 | Clearly MATH. By assumption we have MATH (disjoint union). Suppose that MATH for some MATH. Then, MATH since MATH is connected and the above union is disjoint. As MATH is NAME stable and MATH is transitive, we conclude that MATH. We conclude that MATH. Next suppose that MATH for some MATH. Then, MATH by connectedness once more. We conclude that MATH. To finish we note that MATH must intersect either MATH or MATH (for some MATH) since MATH and MATH. This proves the claim. |
math/0105143 | Note that MATH is a normally hyperbolic manifold of MATH (see CITE). It follows that there is a neighborhood MATH of MATH where the normally expanding manifold MATH has a continuation MATH, for every MATH. Moreover, for every MATH which is MATH close to MATH there is a MATH diffeomorphism MATH. MATH varies continuously with MATH and MATH is the identity map. Using MATH we define MATH by MATH . The map MATH is continuous in MATH (see the remark in REF) and MATH. As MATH is residual in MATH we have that MATH for some open-dense set sequence MATH of MATH. Define MATH and MATH for all MATH. Clearly both MATH and MATH are open sets, and MATH. It is also clear that every MATH has a unique repeller close to MATH. We claim that MATH is dense in MATH for every MATH. Indeed, fix MATH and MATH. Thus, MATH. The denseness of MATH in MATH allows us to choose a sequence MATH such that MATH, as MATH. Define MATH by MATH . It follows that MATH, as MATH. Then, for every MATH large there is an extension MATH of MATH to MATH so that the sequence MATH converges to MATH (see, for instance, CITE or the proof of REF). Indeed, there is a sequence MATH such that MATH, as MATH, MATH, and MATH for all MATH large. In particular, MATH. The uniqueness of the continuation of MATH for MATH close to MATH then implies that MATH, MATH, and MATH. This proves that MATH. The claim follows since MATH, as MATH. By the claim we have that MATH is residual in MATH. Observe that if MATH, then MATH, for all MATH, yielding MATH. We conclude that MATH has infinitely many sinks. As MATH and MATH are conjugated we conclude that MATH has infinitely many sinks too. But MATH is normally expanding, so, the infinitely many sinks of MATH correspond to infinitely many saddle-type hyperbolic periodic orbits of MATH whose homoclinic classes are trivial. We conclude that every MATH has infinitely many homoclinic classes and the lemma follows. |
math/0105143 | Consider MATH as two solid tori MATH glued along their boundary tori. Let MATH be as in REF . Note that MATH has four fixed points: two saddles, one sink, and one source, denoted by MATH in REF . The orbits in MATH exit MATH and go to MATH except for the ones in the middle circle. We obtain the required diffeomorphism MATH by modifying MATH inside MATH. First, we observe that the configuration inside the solid torus MATH is as in REF . In particular, MATH is contracted inside it in the way described in REF . Two fixed points MATH are indicated in that figure. We deform MATH, the image of MATH by MATH, to obtain MATH as in REF . The modification produces a horseshoe in the meridian disks of MATH as indicated; observe that MATH and MATH are fixed by MATH. The modification also produces two fixed points MATH of MATH described in REF . Note that MATH (respectively, MATH) and MATH (respectively, MATH) are homoclinically related. We also require the resulting map MATH to be dissipative (Jacobian MATH) in MATH. Next, we modify MATH to obtain MATH by performing a rotation supported on the middle solid torus MATH indicated in REF . This rotation is done to create a heterodimensional cycle involving MATH (see CITE). We still have that the map MATH is dissipative in MATH. Finally, we modify MATH in order to obtain MATH such that the expanding eigenvalues of MATH (respectively, contracting eigenvalues of MATH) become complex conjugated. This can be done without destroying the heterodimensional cycle involving MATH. Unfolding the cycle we create a blender (``mélangeur") containing MATH (see CITE). The existence of the blender guarantees that MATH and MATH are persistently connected. This procedure is similar to the one described in CITE. Define MATH. By REF it follows that there is a neighborhood MATH of MATH and MATH residual in MATH such that every MATH has infinitely many sinks. We observe that every MATH has a unique repeller: the continuation of the source MATH of MATH indicated in REF (note that MATH is also a source of MATH). This is because any other repeller of MATH close to MATH must belong to the dissipative region MATH but this is impossible by the dissipative condition. This proves the result. |
math/0105145 | The equality REF is due to REF . The ones REF follow from the fact MATH. |
math/0105145 | The first REF . We evaluate the coefficient for MATH in the LHS of REF as follows: MATH where we used REF to get the last line. Thus, REF is proved. The second REF . By a simple calculation, we have MATH where MATH, and the sum is taken over all the subsets MATH of MATH. Therefore, the LHS of REF is written as MATH . By a similar residue calculation as above, the coefficient for MATH of REF is evaluated as (MATH and MATH) MATH . Thus, REF is proved. |
math/0105145 | First, we remark that the infinite product REF exists, because its MATH-th projection image reduces to the finite product MATH due to REF . Let us show that the family MATH in REF is a solution of the MATH-system REF . With the substitution of REF , the MATH-th projection image of the first term in the LHS of REF is MATH . In the second equality above, we exchanged the order of the products. It is allowed because the double product is a finite one (compare REF ). The second term in the LHS of REF can be calculated in a similar way as follows: MATH . From REF , we conclude that REF reduces to REF . Furthermore, by REF , we have MATH . Then, substituting REF , we obtain REF . Therefore, MATH is a canonical solution of REF . Next, we show the uniqueness. Suppose that MATH is a canonical solution of REF . We define MATH as MATH . Then, by the inversion REF , we have MATH . Also, by REF , MATH . With REF , the same calculation as REF shows that MATH is the (unique) solution of REF . Therefore, by REF , MATH is unique. |
math/0105145 | By REF , it is enough to show that MATH . By REF - REF further reduces to the following equality: MATH where MATH is the MATH-truncation of MATH. |
math/0105145 | Since the map MATH is continuous, it preserves the infinite product. Therefore, the specialization of the canonical solution of REF gives a canonical solution of REF . Let us show the uniqueness. By repeating the same proof for REF , the uniqueness is reduced to the one for the standard case MATH. Let us write REF for MATH as REF MATH . These equations uniquely determine MATH mod MATH. Since MATH is arbitrary, MATH is unique. |
math/0105145 | Let MATH be a solution of REF . The same calculation as REF shows that REF is equivalent to the following equality for each MATH (compare REF ): MATH . Clearly, REF follows from REF . Conversely, assume REF . By REF , we have MATH . Because of REF , the both hand sides of REF are REF mod MATH. Thus, we have MATH (MATH). Therefore, REF holds. |
math/0105145 | By comparing REF , it is enough to prove the equality MATH . Due to REF , for given MATH and MATH, there is some number MATH such that MATH holds for any MATH. Then, for MATH, we have MATH . Therefore, the LHS of REF is evaluated as MATH . |
math/0105145 | The proof is the same as the one for REF. |
math/0105145 | It is obtained from the relations among MATH and MATH for MATH CITE: MATH where MATH and MATH for the upper label in REF . |
math/0105145 | We take the standard orthonormal basis MATH of MATH. Let MATH be the element such that MATH. Then, MATH . |
math/0105145 | Because of REF , it is enough to prove the equality MATH . We remark that MATH where MATH are arbitrary constants depending on MATH, MATH, MATH. Set MATH. Then, REF is obtained as MATH where we used REF , and REF . |
math/0105145 | By REF , we have MATH . Using REF , and REF , we obtain MATH . |
math/0105145 | We recall the following four facts: CASE: By REF , we have MATH . CASE: By REF and the proof therein, the canonical solution MATH of REF and the solution MATH of the corresponding standard MATH-system are related as MATH . CASE: By REF , and REF , the series MATH in REF satisfies MATH where MATH is the one in REF . Note that MATH depends on MATH. CASE: By the proof of REF , it holds that MATH . Combining REF - REF , we immediately have MATH mod MATH. Thus, MATH holds. Therefore, taking the limit MATH of REF with the help of REF , we obtain MATH . The equality REF is obtained from REF in the same way as the proof of REF . |
math/0105151 | We use some results of CITE. Our arguments are completely parallel to the arguments in the second proof of REF. In REF, Pfaffian rings are introduced and investigated. It is shown there that for a suitable term order (order on monomials), the ideal MATH of leading monomials of MATH is generated by square-free monomials. Thus MATH may be viewed as a NAME ring of a certain simplicial complex MATH. The faces of this simplicial complex MATH are described in REF. Namely, translated into a less formal language, the faces are sets MATH of integer lattice points in the (upper) triangular region MATH, such that MATH . An example of such a point set MATH, with MATH and MATH, is the set MATH . A geometric realization of this set is contained in REF , the elements of the set MATH being indicated by bold dots, the small dots indicating the triangular region of which MATH is a subset. As usual, let MATH be the number of MATH-dimensional faces of MATH, that is, the number of such sets MATH of cardinality MATH. REF says that the dimensions of the homogeneous components of the Pfaffian ring MATH can be expressed in terms of the face numbers MATH, namely there holds MATH . Now consider such a set MATH. For convenience we apply to it the mapping MATH (that is, the reflection in the main diagonal followed by a shift by MATH in the negative MATH-direction). Thus we obtain a point set, MATH say, in the (lower) triangular region MATH. See REF for the result when this mapping is applied to our example point set in REF . Next we apply to MATH a variant of NAME 's ``light and shadow procedure" (see CITE). This variant defines, for each such point set MATH (and, thus, for each point set MATH in the upper triangular region), a family MATH of MATH nonintersecting lattice paths, MATH running from MATH to MATH, MATH, in the following way. First we ignore everything which is in the left-bottom corner of the triangular region to the left/below of the line MATH, and everything which is in the right-top corner of the triangular region to the right/above of the line MATH. In our running example, these two lines are indicated as dotted lines in REF . Next we suppose that there is a light source being located in the top-left corner. The shadow of a point MATH is defined to be the set of points MATH (MATH denoting the set of real numbers) with MATH and MATH. We consider the (top-left) border of the union of the shadows of all the points of the set MATH that are located inside the strip between the two lines MATH and MATH. We also include the shadows of the points MATH and MATH. This border is a lattice path, MATH say, from MATH to MATH. Now we remove all the points of the set that lie on this path. Then the light and shadow procedure is repeated with the remaining points. In the second run we also include the shadows of MATH and MATH, etc. We stop after a total of MATH iterations. Thus we obtain exactly MATH lattice paths, the MATH-th path, MATH say, running from MATH to MATH. Clearly, by construction, the paths have the property that they are nonintersecting and that they never pass above the diagonal MATH. In addition, a moment's thought shows that REF guarantees that after these MATH iterations all the points of the set MATH are exhausted. REF displays the lattice paths which in our example are obtained by this procedure. On the other hand, if we are given a family MATH of MATH nonintersecting lattice paths, MATH running from MATH to MATH, MATH, with a total number of exactly MATH NAME, how many point sets MATH of cardinality MATH in the (upper) triangular region MATH satisfying REF are there which, after the transformation MATH and subsequent light and shadow as described above, generate the given family of nonintersecting lattice paths? Clearly, every NAME of a path of the family must be occupied by a point of MATH. Aside from that, any point on any of the MATH paths, any point in the bottom-left corner cut off by MATH, and any point in the top-right corner cut off by MATH may or may not be in MATH. Hence, if we denote by MATH the total number of points in the union of the MATH paths and these two corner regions, then there are exactly MATH sets MATH of cardinality MATH that reduce to MATH under light and shadow. As an easy computation shows, we have MATH. Hence, if MATH denotes the number of all families MATH of MATH nonintersecting lattice paths, MATH running from MATH to MATH, MATH, with a total number of exactly MATH NAME, we see that the NAME series equals MATH and if we sum the inner sum by means of the NAME - NAME summation (see for example, CITE), then we obtain MATH . In the sum over MATH, the terms for MATH vanish, so that we may sum over MATH. Application of the binomial theorem then yields MATH . This is exactly REF. |
math/0105151 | In view of REF , we only have to solve the problem of enumeration, with respect to NAME, of nonintersecting lattice paths that are bounded by a diagonal line. This has been previously accomplished in CITE and in CITE. To be precise, to show that the generating function MATH in the numerator on the right-hand side of REF can be expressed by the determinant on the right-hand side of REF, one sets MATH, MATH, MATH, MATH in REF, then multiplies the resulting expression by MATH, and sums over all MATH. To show that it can be expressed by the numerator on the right-hand side of REF, respectively of REF, we prepend MATH horizontal steps and append MATH vertical steps to MATH. Then, out of a family of nonintersecting paths as in the statement of REF , we obtain a family MATH of nonintersecting lattice paths, where MATH runs from MATH to MATH and does not pass above MATH, MATH. See REF for the corresponding path family which is obtained out of the one in REF . Clearly, the number of the latter families is exactly the same as the number of the former, because the prepended and appended portions are ``forced," that is, if MATH are nonintersecting, then they must contain these prepended and appended portions. Now one can either again apply REF, this time with MATH, MATH, MATH, MATH, multiply the resulting expression by MATH, sum over all MATH, and thus obtain the numerator in REF, or apply REF in conjunction with REF , in CITE with MATH, and thus obtain the numerator in REF. |
math/0105151 | It is well-known that, if the NAME series of a finitely generated graded MATH-algebra MATH is written in the form MATH, where MATH is a polynomial with rational coefficients such that MATH, then the dimension of MATH equals MATH and the multiplicity of MATH equals MATH (see for example, CITE). (Equivalently, the multiplicity is the sum of the components of the MATH-vector, the latter being, by definition, the vector of coefficients of MATH.) Using NAME - NAME summation (see for example, CITE) again, the numerator on the right-hand side of REF specialized at MATH is MATH . This determinant can be evaluated by using for example, REF , in CITE, with MATH replaced by MATH, MATH, MATH, and MATH. The result is MATH . This expression can be transformed into the one given in REF. |
math/0105153 | Since MATH is compact there exists a constant MATH such that MATH for all MATH. For any MATH it follows that the unbounded operator MATH in MATH with dense domain MATH is positive definite and so in particular injective. Moreover, it is selfadjoint and therefore also surjective with real spectrum. When viewed as a bounded operator from MATH to MATH the open mapping theorem guarantees existence of a bounded inverse. Together with a standard compact NAME embedding we obtain that the resolvent operator is compact: MATH . Compactness implies discrete spectrum, say MATH, with finite multiplicities and possible accumulation point at MATH. We observe that MATH eigenvalue of MATH if and only if MATH is an eigenvalue of MATH. We already know that MATH and conclude MATH for MATH. Clearly, MATH . Hence the eigenvalues MATH of MATH tend to MATH for MATH which proves the NAME index theorem. |
math/0105153 | To show equality of MATH and MATH we choose the following smooth variation of MATH in order to linearize REF as described in subsection REF. With this choice MATH and MATH is a zero of the linear REF , which however might not close up at time MATH. In other words the linearized flow along a flow line (zero of the nonlinear equations) provides a zero of the linearized equations. With respect to the unitary frame MATH these are given by MATH and this proves the claim. |
math/0105153 | CASE: Consider the transition maps MATH and MATH, then MATH where in the second equality we used REF as well as the loop REF . Since MATH is a constant loop the corresponding term vanishes. It remains to show MATH: Since MATH we can write MATH and by REF the following is true: MATH . Now in case MATH we have MATH and are done. If MATH a calculation shows that MATH REF MATH where we used REF as well as the loop REF in the second equality. The third one follows from REF . |
math/0105153 | For each MATH there is the path of symplectic matrices MATH defined in REF . Together these give rise to a REF-parameter family of Lagrangian subspaces MATH and we consider the NAME index MATH of the loop of NAME MATH along the boundary of the parameter domain MATH. We may break down the loop MATH into the four subpaths MATH indicated in REF . MATH is contractible and so its NAME index is zero. Together with the catenation property of MATH this leads to MATH . Since MATH, MATH and so MATH=REF. Moreover, as indicated in REF the NAME index of MATH is precisely the NAME index of the corresponding symplectic path. We get a minus sign in the case of MATH, because of its reversed orientation MATH . It remains to show MATH . In order to prove REF we need to derive two identities. Recall from REF that MATH is determined by MATH . Now for fixed MATH the path MATH, MATH, leads to another path of symmetric matrices MATH . Note that MATH, since MATH. Using these equations we obtain MATH . Integration over MATH from MATH to MATH, together with MATH, leads to the first identity we are looking for, namely MATH . Use skew-symmetry of MATH and MATH as well as MATH and MATH to obtain the second identity MATH where in the last equality we used the boundary conditions for MATH and the skewsymmetric family of matrices MATH. We apply both identities to derive one more crucial result. Let MATH and use the symmetry of the projection operator MATH to obtain MATH where we used REF as well as orthogonality of MATH in the fifth equality, the identity REF in the last but one equality and MATH in the last one. We are ready to prove REF . The first two equalities are by definition of the spectral flow and the crossing operator MATH where the third equality is REF . Here it is important that the sums are over the same set of MATH. This follows from the equivalence of REF MATH . Equality four has been derived in REF , equality five is by definition of MATH and the last one is by definition of MATH. This completes the proof of REF . |
math/0105153 | Let MATH be a nondegenerate critical point, that is, MATH, and denote by MATH the negative eigenvalues of MATH counted with multiplicities. Fix a real number MATH. It will be convenient in REF to assume in addition MATH. Let MATH be a smooth cut-off function which equals MATH near MATH and MATH near MATH and is strictly decreasing elsewhere. For MATH define MATH by MATH. Let MATH be the path of symmetric matrices defined in REF and modify MATH, if necessary, such that its value at each MATH remains the same but MATH. This technical condition ensures regularity of crossings. For MATH consider two families of selfadjoint operators in MATH with dense domain MATH and MATH where MATH and MATH. The reason to use two families instead of only one is that regularity can be checked easily that way. Both families fit into the framework of REF: Using the boundary conditions for MATH and MATH a short calculation shows that the second family consists for each MATH of a positive definite operator and so the injectivity REF is satisfied. The family is also regular, because there are no crossings at all. To check for the first family we observe that MATH is positive definite and MATH is precisely the model operator REF for MATH. In view of MATH this confirms the injectivity REF . Our choice of cut-off function now guarantees regularity of the family. More precisely, assume MATH is a crossing and let MATH, then the first condition implies MATH . Now MATH implies MATH. Differentiating REF with respect to MATH leads at MATH to MATH and in view of the regularity of the matrix MATH we get MATH. Now we use the second condition which says that MATH is also in the kernel of the second order differential operator MATH. Since its boundary conditions are zero, it follows MATH. This proves nondegeneracy of the crossing operator MATH at each crossing MATH. We are in position to apply REF to both families. Let us start with the first one and observe that its spectral flow is given by MATH. We obtain MATH where according to MATH in REF for each MATH the path MATH is determined by MATH and MATH where MATH. Observe that MATH is precisely the matrix MATH in REF and so MATH equals MATH from REF . In other words MATH. Since MATH contains the MATH-dependent matrices MATH we make use of the second family MATH in order to further simplify the problem of calculating MATH. As before we get for each MATH a path MATH which is determined by MATH where MATH . The spectral flow of a family of positive definite operators is zero and so REF gives MATH where we used MATH in the last step. Hence it remains to show that the NAME index of the path MATH equals MATH. In order to do so we would like to treat the two cases MATH and MATH separately. CASE: MATH . We conclude MATH where we used REF in the first equality, REF in the third and the fact that there are no further crossings in the second equality. It remains to show that there are indeed no crossings MATH. This is equivalent to MATH not being in the spectrum of MATH for any MATH. Because the matrix MATH is constant in MATH, we can integrate the corresponding differential REF for MATH and obtain with MATH . We study the characteristic polynomial of MATH and obtain its eigenvalues (of multiplicity MATH each) MATH . Finally the sum (difference) of the hyperbolic cosine and sine is MATH if and only if MATH, which shows that there are no further crossings. Let us remark that, because MATH is symplectic, it follows MATH and this is reflected in the key identity for hyperbolic functions MATH. CASE: MATH . We obtain MATH where we used REF in the first equality, REF in the third and the fact that there are no further crossings in the second equality: To see this let us rearrange coordinates MATH of MATH in the form MATH such that MATH gets identified with MATH and the path of symmetric matrices MATH from REF with MATH . The symplectic path generated by the second term in the direct sum does not meet the NAME cycle for any MATH as was shown in REF . Let us denote the first term in the direct sum by MATH. It remains to investigate the path in MATH determined by MATH and show that there are no crossings with the NAME cycle MATH in MATH for any MATH. A numerical plot of MATH confirming this in case MATH is shown in REF . To give a proof we introduce the notation MATH and observe that MATH . Now MATH is a solution of REF if and only if MATH is a solution of MATH where we used the identity MATH. The solution MATH is known from case MATH above. We are done once we have shown that the following function is strictly positive for MATH . Note that MATH and MATH. The function is clearly positive for MATH if MATH. This is true since MATH . We calculate the first derivative MATH which is positive on MATH since both coefficients are, due to REF . Moreover, we get MATH and MATH so that MATH is a local minimum and it follows that MATH is strictly increasing on MATH. This finishes the proof of case MATH and of the index REF . |
math/0105153 | Since MATH is closed the quotient MATH inherits the structure of a NAME space from MATH and the projection MATH is continous. By REF is finite dimensional and so is its subspace MATH. Hence the latter space is closed and so is its preimage MATH under the continous map MATH, which is MATH. Finally, the complement of MATH in MATH is contained in the finite dimensional space MATH. CASE: It is well known, compare REF, that any finite dimensional subspace of a NAME space admits a topological complement. Let MATH be a topological complement of MATH. Moreover, since MATH is closed with finite dimensional complement, we can write MATH. NAME of MATH implies that MATH and so we may choose a basis MATH of MATH, where MATH is a set of linearly independent elements of MATH. Now define the linear operator MATH where MATH is determined uniquely by MATH and the coefficients MATH by MATH. It is not hard to check that a topological complement of MATH is given by MATH. Actually, since existence of a right inverse of a bounded linear operator is equivalent to surjectivity and existence of a topological complement of the kernel, we observe that MATH is indeed a right inverse of MATH. Now assume MATH, then MATH which proves that MATH. Define MATH then MATH and this set is closed: MATH is closed and so is its preimage under the continuous map MATH. Finally we obtain MATH where the first isomorphism is induced by MATH, namely MATH and the subsequent identity is due to MATH and given by MATH . |
math/0105153 | We have to show that for every MATH it holds MATH . Assume the right hand side was true and choose any MATH. By surjectivity of MATH we find MATH such that MATH. Then the pair MATH is indeed element of MATH and is mapped to MATH under MATH. To prove the other direction assume the left hand side was true, choose any MATH and apply the surjectivity assumption MATH of REF to conclude the existence of a pair MATH being mapped to MATH by MATH; in other words MATH . Given MATH, surjectivity of MATH implies existence of MATH such that MATH, that is, MATH. Now define MATH and observe that MATH. |
math/0105153 | We prove that REF holds. Let MATH. Recall that MATH is the perturbed NAME MATH and MATH by the NAME index theorem. Moreover, since MATH is selfadjoint MATH . This shows that MATH is NAME of index MATH for any MATH. Together with the assumption MATH it follows that the condition in REF on the differentiability of MATH is satisfied: MATH. It remains to verify assumption MATH, that is, we have to show that MATH and MATH together imply MATH. The first condition says that MATH. So it satisfies a linear second order ODE with coefficients of class MATH and therefore MATH. Now assume by contradiction that there is MATH such that MATH. In five steps we are going to construct MATH such that MATH . As our construction will be local, we may choose geodesic normal coordinates MATH around MATH. Let MATH denote the injectivity radius of MATH. The piece of the loop MATH which lies inside the coordinate patch determines the curve MATH via MATH so that MATH. CASE: Because MATH is continuous, we may choose a constant MATH sufficiently small such that MATH . CASE: Because MATH is continuous and MATH, we may choose a constant MATH sufficiently small such that MATH . CASE: Set MATH and choose a cut-off function MATH such that MATH . CASE: Choose a cut-off function MATH such that MATH . CASE: We are ready to define MATH . Putting everything together we get MATH . The third equality follows from the definition of MATH REF , and the fourth one from REF as well as a straight forward calculation. In the fifth equality we used that for MATH . REF implies MATH and therefore, by REF , MATH and MATH. REF gives the final strict inequality. REF dense: Since MATH and MATH is a residual subset of MATH by MATH, the set MATH is too. By NAME 's category theorem it is therefore dense. open: We fix some regular MATH and construct an open neighbourhood MATH of MATH in MATH such that MATH. Recall that MATH is a NAME map of class MATH. Since MATH is open in MATH, it follows that the restriction MATH is NAME of class MATH too. Now rewrite MATH as MATH and observe that this set is compact (compare our remarks in the introduction on NAME 's uniform MATH-bound); in other words MATH is a proper map. As will be discussed later surjectivity is an open condition and so there exists an open neighbourhood MATH of MATH in MATH such that MATH is surjective for all MATH. Now use continouity of MATH and compactness of MATH to conclude the existence of an open neighbourhood MATH of MATH in MATH such that MATH. It follows MATH. It remains to show that surjectivity of MATH is an open condition (in MATH): Note that for MATH near MATH the operators MATH and MATH (strictly speaking their representatives with respect to a trivialization of MATH at MATH) differ by a bounded operator, whose norm can be made arbitrarily small by choosing MATH sufficiently close to MATH. In view of the subsequent lemma and the selfadjointness of the linear operators we get MATH . |
math/0105153 | Let MATH be a topological complement of MATH. It suffices to show MATH . The restriction MATH is a bounded bijection and therefore admits a bounded inverse MATH by the open mapping theorem. Let MATH, that is, MATH, then MATH . Choose MATH, then it follows MATH. |
math/0105153 | ad MATH dense: Given any MATH we have to construct a sequence MATH such that MATH . The idea will be to approximate MATH by regular MATH's in the MATH-topology and then approximate MATH by smooth regular elements MATH in the MATH-topology. Finally we make use of the observation that in order to control the metric MATH we essentially have to control only finitely many MATH-norms in its series, because the strong weights MATH take care of all the other ones. CASE: Because MATH is dense in MATH for any integer MATH, we can find MATH with MATH. For MATH let us define MATH. CASE: Because MATH is open in MATH for any integer MATH, we can choose MATH sufficiently small such that MATH, the open MATH-ball around MATH, is contained in MATH. CASE: Because MATH is compact, MATH is dense in MATH for MATH, compare CITE. Hence we can find MATH for MATH. For MATH we define MATH. CASE: Now pick MATH and choose MATH sufficiently large such that MATH. Choose MATH and observe that by REF for MATH . Note that this implies MATH for any MATH. We get for any MATH . MATH open: Pick MATH and set MATH. Exploiting openess of MATH in MATH we are able to choose a constant MATH such that for any MATH of class MATH with MATH it follows MATH. Now define MATH . Let MATH of class MATH be such that MATH. Therefore each term in the series on the left hand side has to be strictly smaller then MATH, in particular the second one MATH . But this is equivalent to MATH and therefore MATH. Finally MATH implies MATH. CASE: Since MATH is open and dense in MATH it is residual. The identity MATH implies the claim, because any countable intersection of residual sets is again residual and therefore dense. |
math/0105157 | For the proof we will use the notation introduced in REF . Let us consider a horizontal curve MATH. The elements MATH freely generate a subgroup MATH, which is normal since MATH. Note that MATH are meridians about (all) the (irreducible) components of MATH. Since MATH is normal, it is generated by all the meridians about all irreducible components of MATH. The group MATH is, hence, identified with the fundamental group of the generic fiber. Therefore, the basis MATH is geometric. Note that MATH is the boundary of a sufficiently big disk on the fiber. One obtains a natural exact sequence: MATH . Given any two curves MATH as in the statement, let us take the elements MATH and MATH described in REF . We recall that the elements MATH form a geometric basis of MATH and MATH is a geometric basis of MATH. Let us choose a big disk MATH of radius MATH, such that the values of the non-transversal vertical lines to MATH are in the interior of MATH. Note that we can construct a polydisk MATH well adapted to MATH and two polydisks MATH and MATH well adapted to MATH such that MATH . Let us define MATH . From the discussion above, MATH is a basis for the free group MATH. Moreover, let us suppose that the generic fiber MATH (of MATH) is very close to a non-transversal vertical line MATH (if no such line exists there is nothing to prove). Since the boundary of a big disk in MATH is sent to the boundary of a big disk in MATH, we easily deduce that MATH is homotopic to the boundary of a big disk in a generic fiber for MATH. This argument proves that MATH is a geometric basis of MATH. Let us define MATH . By the naturality of the exact sequence REF we deduce that MATH, , MATH is a basis of the free group MATH. Since the extension of MATH to MATH preserves the line at infinity and since the inverse of MATH is a meridian about the point at infinity in MATH, we deduce that it is also the case for MATH. Then, MATH is a pseudogeometric basis of MATH. From the exact sequence REF and the system of generators MATH, we obtain a representation of MATH in MATH which provides an element of MATH. We have proven that this element represents the braid monodromy of MATH. Because of the way the generators are chosen, one can apply the same argument to MATH, obtaining an element MATH representing the braid monodromy of MATH. Since the two families of generators are related by MATH, we deduce that MATH . |
math/0105157 | This is a straightforward consequence of REF . |
math/0105157 | It is enough to apply the final method described in strategy REF using the representation MATH . The orbits of both braid monodromies have size MATH and are disjoint as shown by the GAPREF program in the Appendix. |
math/0105157 | Let MATH and MATH in the statement of REF correspond to MATH and MATH respectively. By REF there is no homeomorphism MATH preserving orientations on MATH, MATH and MATH. Since both curves have real equations, they are invariant by complex conjugation which preserves orientations on MATH but exchanges orientations on the curves. Hence there is no homeomorphism MATH preserving the orientation of MATH but reversing it on MATH and MATH. Since algebraic braids always turn in the same direction - that is, they are positive - it is not possible to have a homeomorphism MATH preserving the orientation of MATH and only some, but not all, of the components of the curves. |
math/0105163 | This result is very close to REF, and for the sake of completeness we present the proof, which proceeds by induction. Introduce the set MATH . Then MATH by definition, and MATH since MATH consists of a single vector MATH. In order to pass from MATH to MATH, we argue as follows. Let MATH denote the set of all vectors in MATH whose last entry is MATH. Consider a mapping MATH defined as follows: for MATH we have MATH and for MATH we have MATH. It is clear that the mapping MATH is well - defined. Moreover, it is a surjection and every element of MATH has exactly two preimages - one in MATH and one in its complement. Applying the definition of the operator MATH we prove the inductive step - MATH is the contribution of vectors from MATH and MATH is the contribution of vectors belonging to MATH. This completes the proof of the first part of the theorem. For obtaining the closed REF , let us rewrite the recurrent relations REF in the following way: MATH; MATH. Proceeding by induction and recalling that MATH we get REF . Finally, a simple proof by induction using REF gives REF . |
math/0105163 | Using REF , by repeated partial integrations, we get that the kernel of MATH equals MATH . Specializing to the diagonal, we get the statement of the lemma. |
math/0105163 | The result follows by an application of REF to REF , since, as MATH, MATH has an asymptotic expansion in the integer powers of MATH. |
math/0105163 | When MATH, it is true that MATH and MATH are MATH-functions of MATH with values in MATH, with NAME expansions at MATH, MATH . If we substitute REF in the sum in the right-hand side of REF we get that it is equal to MATH where MATH . Now, using REF we see that MATH where MATH in MATH. We have therefore proved that MATH . |
math/0105165 | By the NAME expansion of the exponential one obtains MATH with MATH. Using the control REF on the conditional brackets of the martingale it is easy to obtain by induction on the integrand and the NAME property that MATH . Combining this with REF and using the fact that MATH with MATH one obtains that MATH with MATH. Developing the product in MATH one obtains by integration, induction and straightforward combinatorial computation that MATH . Which leads to REF by REF . |
math/0105165 | Put MATH and write for MATH, MATH . It will be shown here that MATH where MATH the increasing sequence defined by MATH and MATH and converging to MATH the smallest positive solution of MATH. REF is then obtained for MATH and using MATH and MATH. Write MATH (note that MATH) and consider for MATH the function MATH . By NAME expansion, for MATH, MATH and since MATH with a normal convergence of the series, the order of the limits can be changed, which leads to MATH . It follows that MATH . Now, for MATH; the constant MATH does exist and MATH (thus MATH). Thus from REF expansion, the series MATH converges towards f for MATH and in that interval MATH . From which one deduces that MATH . On the other hand if one considers the sequence MATH then it is an exercise to show that MATH is increasing and will converge towards MATH, which leads to REF. |
math/0105165 | The estimate on the heat kernel MATH will follow from the chain rule and decomposing it the probability of moving away from MATH to "a well chosen set containing MATH in the time MATH" and its complement. More precisely, writing MATH and MATH, using REF one obtains that for MATH, MATH and MATH, MATH . Let's choose MATH and MATH (we will use the notation MATH). For MATH (which basically says that the heat kernel is far from its diagonal behavior) one has MATH and MATH. Using the NAME type estimate REF one controls the second term in REF MATH . By REF one controls the first term in REF: for MATH with MATH, MATH, and MATH one has MATH . Combining REF and using the value of MATH and MATH given above one easily obtains the estimate REF under REF . |
math/0105165 | For MATH let MATH be the functions introduced in REF. Write MATH the filtration associated to Brownian motion appearing in the SDE solved by MATH. MATH is a MATH-continuous local martingale vanishing at MATH such that (NAME calculus) MATH with MATH. Since MATH a.s. by NAME, NAME representation theorem MATH is a MATH-Brownian motion with MATH and MATH . The idea of the proof is then to show that probability of MATH to move away from MATH behaves like the probability of a BM of variance MATH to move away, to achieve this it will be sufficient to show that MATH becomes negligible in front of MATH using REF to control MATH. More precisely we will use the following lemma. For MATH one has MATH . Let MATH, from the representation theorem MATH with MATH. It follows that for MATH . It follows from REF that for MATH, MATH. From which one deduces MATH . Combining REF one obtains that MATH . Which leads to REF under the last condition in REF. Now let us show that For MATH one has MATH where MATH and MATH depend only on MATH and MATH. Write MATH . Since MATH one obtains from NAME formula that MATH . Thus MATH satisfies the conditions of REF with MATH, MATH and MATH, which leads to REF by observing that MATH is upper bounded by a constant depending only on MATH and MATH. It follows from REF that under the additional conditions, MATH (where MATH depends only on MATH and MATH) one has MATH . Choosing MATH and MATH for MATH and MATH REF are satisfied and MATH and it follows from REF that MATH . Which proves REF . |
math/0105165 | Write MATH, the expectation with respect to the law of probability associated to the generator MATH. By REF one obtains that MATH . Bounding, MATH by MATH and for MATH, MATH by MATH one obtains that for MATH . Writing MATH corresponds to the maximum number of periods of the scale MATH included in the segment MATH: MATH; one obtains MATH . Using MATH, REF one obtains REF. |
math/0105165 | The proof follows trivially from the following equation MATH . |
math/0105165 | The proof of REF is based on the following functional mixing estimate (obtained from REF ):for MATH and MATH one has MATH . Then by the explicit REF and a straightforward induction on MATH one obtains that (using REF) MATH . Which leads to REF by REF. |
math/0105165 | The limit REF is a direct consequence of the following theorem that is an application of the theory of level-REF large deviations (we refer to CITE for a sufficient reminder). Let MATH (NAME continuous with exponent MATH). Let MATH, MATH. Then MATH . We have written MATH is the pressure associated to the scaling shift induced by MATH on the torus: For MATH one can see the torus as a shift space equipped with the transformation MATH where for each MATH, MATH is a vector in MATH and for each MATH is the expression of MATH in base MATH (MATH). Give MATH with the discrete topology and MATH with the product topology. Write MATH the probability measure on MATH affecting identical weight MATH to each element of MATH and write MATH the associated product measure on MATH. With respect to the probability space MATH the coordinate representation process MATH is a sequence of i.i.d. random variables distributed by MATH. When MATH is seen as an element of the torus MATH then the probability measure induced by MATH on the torus is the NAME measure. Define the empirical measure MATH associated to the process MATH by MATH where MATH is the periodic point in MATH obtained by repeating MATH periodically. For each MATH, MATH is an element of the space MATH of measures on MATH and invariant by the shift MATH. Then by REF, MATH, the MATH distribution on MATH of the empirical process MATH has a large deviation property with speed MATH and entropy function MATH. We remind that for MATH, MATH where MATH denotes the marginal distribution of MATH associated to MATH and MATH is the relative entropy of MATH with respect to MATH: MATH. Choosing MATH, NAME continuous with exponent MATH, one deduces from the large deviation property of MATH and NAME 's theorem that MATH where MATH is the pressure of MATH. We remind that MATH where MATH is the space of measures on MATH invariant by the shift MATH. Since MATH is NAME continuous MATH . And one obtains REF from REF . |
math/0105165 | The proof of REF is based on analytical inequalities that allow to control the stability of the homogenization process on the smaller scales under the perturbation of larger ones. More precisely we will first work on an abstract decomposition of MATH given by REF into effective scales MATH perturbation scales MATH and drift scales MATH: MATH with MATH, MATH, MATH and MATH shall correspond to fluctuating scales. Write MATH the solution of the cell problem associated to MATH (MATH, MATH) and MATH. Since MATH is harmonic with respect to MATH one obtains by NAME formula that MATH from which one obtains that MATH . Write MATH the solution of the cell problem associated to MATH and MATH. We will show that MATH with MATH and MATH . REF is a direct consequence of the explicit formula MATH. REF follows from the explicit formula MATH noticing that the period of MATH and MATH are MATH and MATH and REF . The long time behavior of MATH is a perturbation of MATH as shown in the following REF MATH then for MATH and MATH . For the proof of REF by scaling one can assume that MATH and MATH. Write for MATH . Using REF to separates the scales in REF, it is an easy exercise to obtain that if MATH then CASE: for MATH one has MATH CASE: for MATH one has MATH . Observing that MATH one deduces REF by applying NAME formula. Combining REF and choosing MATH, MATH, MATH (MATH, MATH) and MATH as defined in REF one obtains that for MATH . Which leads to REF by the choice REF for MATH. |
math/0105165 | The proof of REF is based on a control of the NAME transform of MATH, more precisely it is well known that for MATH, and MATH one has MATH. Observing that MATH and using MATH one deduces by the NAME inequality that MATH . If MATH is a positive bounded random variable, MATH and MATH it is easy to show by integrating by part over MATH and using MATH that MATH . Applying this inequality to REF with MATH, MATH and MATH and observing by NAME formula that MATH one obtains MATH . Now observe that MATH satisfies the conditions of REF with MATH, and MATH. It follows that for MATH one has MATH . Assuming MATH and choosing MATH the condition on MATH in REF is satisfied under the right inequality in REF and one obtains MATH . From this the result REF follows easily by assuming the left inequality in REF (that basically says that the influence of the drift scales MATH is small in front of the influence of the fluctuating scales) and REF . |
math/0105172 | Let MATH denote a monic polynomial (not necessarily irreducible) over MATH. The left hand side is equal to MATH . The coefficient of MATH REF vanishes because given MATH the distribution of MATH for a random monic degree MATH polynomial is uniform over all MATH possible values. |
math/0105172 | To prove REF we recall some work of CITE on the cycle index of MATH (a survey of applications of cycle indices of the finite classical groups can be found in CITE). As the textbook CITE explains in the section on rational canonical forms, the conjugacy classes of MATH correspond to the following combinatorial data: to each monic non-constant irreducible polynomial MATH over MATH, associate a partition (perhaps the trivial partition) MATH of a non-negative integer MATH. We write MATH if MATH is a partition of the integer MATH. The only restrictions necessary for this data to represent a conjugacy class are CASE: MATH CASE: MATH. The size of the conjugacy class corresponding to the data MATH is equal to MATH where MATH is a function of MATH and MATH which depends on MATH only through the degree of MATH. Define MATH to be one if MATH is the partition corresponding to MATH in the rational canonical form of MATH and to be zero otherwise. It follows that MATH . We now use the fact (derived in CITE) that MATH . Other elementary derivations can be found in CITE. (There are similar factorizations for all irreducible polynomials for all finite classical groups). Consequently MATH . REF now follows from REF follows from REF using NAME 's identity MATH . |
math/0105172 | The paper CITE proves (see the remark on page REF) by very elementary means (using only the NAME decomposition of MATH with respect to a parabolic subgroup) that MATH . Clearly MATH where the sum is over all multiplicative characters of MATH. From REF and the equation MATH for non-trivial MATH it follows that MATH . Note that the second term on the right-hand side arises from MATH trivial. |
math/0105172 | The left hand side is equal to MATH . The terms corresponding to MATH all vanish. The degree REF term is equal to MATH. The degree REF term is equal to MATH which simplifies to MATH. |
math/0105172 | Letting MATH denote a monic polynomial with coefficients in MATH, the left hand side is equal to MATH . Observe that for MATH, the expression MATH vanishes because (from the explicit description of invariant P) MATH is equidistributed over all elements of MATH given the value of MATH. The computations for MATH and MATH are straightforward. The second assertion follows from the first assertion by replacing MATH by MATH and using the fact that all irreducible polynomials invariant under MATH have odd degree. |
math/0105172 | Arguing as in CITE and using the fact that NAME sums are multiplicative on polynomials, it follows that MATH is equal to MATH . This product can be broken down into terms according to whether MATH is even or odd and the result follows from REF . |
math/0105173 | These are essentially proved in CITE. So our proof is sketchy. CASE: Since MATH equals MATH plus the sum of lower monomials, the first assertion follows by induction on MATH. The second assertion follows from the argument in CITE, where we use the standard module MATH instead of simple modules. CASE: Let MATH be a monomial appearing in MATH, which is not MATH. It is not l - dominant by the assumption. By REF , MATH appears in MATH for some monomial MATH appearing in MATH. In particular, we have MATH. Repeating the argument for MATH, we have MATH. The coefficient of MATH in MATH is equal to the sum of coefficients of MATH in MATH for all possible MATH's. (MATH is fixed.) Again by induction on MATH, we can determine the coefficient inductively. CASE: By REF , the transition matrix between MATH and the dual base of MATH above is uppertriangular with diagonal entries MATH. CASE: By REF , we may assume that MATH is not a root of unity. Consider the case MATH, MATH for MATH for some MATH. By REF implies that the MATH for MATH does not contains l - dominant terms other than MATH. (See REF below for a geometric proof.) In particular, MATH is uniquely determined by above REF in this case. We use REF to ``calculate" MATH for arbitrary MATH as follows. We order inverses of roots (counted with multiplicities) of MATH REF as MATH, MATH, , so that MATH for MATH if MATH. This is possible since MATH is not a root of unity. For each MATH, we define a NAME polynomial MATH by MATH if MATH is a root of MATH. Therefore we have MATH. By our choice, we have MATH by REF . Each MATH is uniquely determined by the above discussion. Therefore MATH is also uniquely determined. CASE: It is enough to check the case MATH. In this case, MATH is the standard module with NAME polynomial MATH. The assertion follows from the axioms. CASE: This also follows from the axioms. By REF , we may assume MATH is a root of unity. By REF , we may assume MATH is an l - fundamental representation. In this case, the assertion follows from REF , since MATH is a polynomial in MATH. |
math/0105173 | Since MATH is injective, it is enough to show that MATH. In fact, it is easy to prove this equality directly from the geometric defintion in REF. However, we prove it only from Axioms. By REF , we may assume MATH is not a root of unity. We order inverses of roots (counted with multiplicities) of MATH REF as in the proof of REF. Then we have MATH by REF . The product can be taken in any order, since MATH is commutative. Each MATH is either the inverse of a root of MATH or MATH. We divide MATH's into two sets accordingly. Then the products of MATH over groups are equal to MATH and MATH again by REF . Therefore we get the assertion. |
math/0105173 | For simplicity, we assume that MATH is not a root of unity. The proof for the case when MATH is a root of unity can be given by a straightforward modification. Let us show MATH for MATH, MATH. By induction and REF we may assume that MATH is of form MATH where MATH is a monomial with MATH, MATH for MATH, and that MATH is as REF. By a direct calculation, we get MATH where MATH. If MATH, then the right hand side is zero, so the assertion is obvious. If MATH, then we have MATH . Therefore the above expression is equal to MATH. Next we show the closedness of the image under the involution. By REF and the above assertion, we may assume MATH as above. We further assume MATH does not contain MATH, MATH. Then we get MATH . This is contained in MATH. Now we can define MATH and MATH on MATH so that MATH where we have assumed that MATH is not a root of unity for the second equality. By the above discussion together with REF, the right hand sides are contained in MATH, and therefore in the image of MATH by REF. Since MATH is injective by REF, MATH, MATH are well-defined. |
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