paper stringlengths 9 16 | proof stringlengths 0 131k |
|---|---|
math/9907175 | The first and second identities were essentially established in REF together with REF , where the components are viewed as operators acting on MATH or MATH. Note that MATH. We observe from definition that the adjoint MATH-action of MATH and MATH with respect to the bilinear form MATH are MATH and MATH respectively. The third and fourth identities are obtained by applying the adjoint action MATH to the first two identities. |
math/9907175 | It is a routine computation to see that: MATH where the MATH-analog of the power series MATH is defined in REF . In particular, we have MATH . Then the theorem is proved by observing that MATH. |
math/9907175 | Using the usual method of MATH-vertex operator calculus CITE and REF we see that the vertex operators MATH satisfy relations REF . Observe further that the above vertex operators at MATH have the same form as those in the basic representations of the quantum affine algebras (see CITE). Thus the relations REF are also verified. For each fixed MATH or MATH we have shown that the operators MATH give a level one representation of the quantum toroidal algebra MATH (see also CITE). |
math/9907179 | The fundamental group MATH is normally generated by a boundary circle of a normal disk to MATH. Since MATH, we may assume that this circle lies on a copy MATH in the boundary MATH. We claim that there are generators MATH, MATH, for MATH which bound vanishing cycles (disks of self-intersection MATH) in MATH. (Note that here we are identifying MATH with MATH.) This claim can be seen to be true inside a MATH nucleus, that is, in a regular neighborhood of the union of MATH and a cusp fiber. NAME calculus diagram for the nucleus is given in REF below. The section MATH is the union of the disk spanned by the circle labelled MATH and the core of REF-handle which is attached to it. The torus MATH is obtained as follows. The circle labelled `REF' bounds a disk MATH which is punctured in two points by one of the dotted circles. Remove a pair of disks from MATH at these intersection points, and connect the boundaries of these disks with an annulus which surrounds the path MATH in the diagram. The torus MATH is the union of the twice-punctured MATH and this annulus. We can see that the loop MATH of the diagram lies on MATH, and it is easy to deform MATH to also lie on MATH. (When this is done, MATH and MATH will intersect in a point.) Thus MATH is generated by the classes represented by the loops MATH and MATH. The vanishing cycles are the cores of the (MATH)-framed REF-handles which are attached to MATH and MATH. This proves the claim. This means that MATH is the zero map; hence MATH is also the zero map. However, MATH is normally generated by the image of this map; so MATH is simply connected. Thus MATH . Because MATH is normally generated by the meridian MATH, MATH is normally generated by MATH. An inductive application of NAME 's theorem shows that MATH is normally generated by MATH. Thus MATH is normally generated by MATH, and so MATH. |
math/9907179 | A rim torus on MATH has the form MATH, for some loop MATH on MATH. Recall that there is a fiber bundle MATH with fiber MATH. Let MATH. We see that MATH bounds REF-chain MATH in MATH. |
math/9907179 | Let MATH denote a nucleus in MATH which contains the fiber MATH and section MATH from the construction of MATH. We see that MATH. The homology MATH, where the negative definite MATH forms are generated by the classes of embedded spheres of self-intersection MATH, and the two hyperbolic pairs MATH are each generated by a torus MATH of self-intersection REF, and a sphere MATH of self-intersection -REF which meet transversely in a single point. The homology MATH is generated by the image of MATH together with MATH and the classes of rim tori. Next consider MATH (MATH copies) where MATH is the torus fiber of the fibration of MATH over the torus. Each MATH has the homology of MATH. Each fiber sum in the construction of MATH increases the first betti number MATH by MATH - the base of the fibration has genus increased by MATH - and MATH is increased by the addition of two hyperbolic pairs as follows: Choose a standard basis MATH for MATH. For example, MATH is represented by a loop as shown in REF . Each of the curves MATH bounds a punctured torus MATH in MATH. In REF , MATH is isotopic to the pictured curve MATH, and the punctured torus is composed of the twice-punctured disk which MATH bounds, together with a REF-handle which pipes around the knot. Let MATH be a push-off of MATH in MATH. Then the linking number of MATH and MATH is MATH. (Here we are using the fact that MATH is a left-hand trefoil knot.) This means that the self-intersection number of MATH (as a surface in MATH, say), keeping its boundary in MATH, is MATH. Thus in MATH one produces genus REF surfaces MATH, MATH, of self-intersection MATH which are formed from pairs of these tori. Let MATH, MATH be the rim tori corresponding to MATH, MATH (reversed on purpose). Then in MATH two hyperbolic pairs are generated by the pairs MATH. Each further fiber sum adds two such hyperbolic pairs to MATH. It follows that MATH, is generated by the MATH, MATH, and the section class MATH. Using our observations above, if MATH is a basic class of MATH we can write MATH where MATH and MATH he adjunction inequality (see for example, CITE) states that if MATH is a basic class and MATH is an embedded surface of genus MATH and self-intersection MATH then MATH . In particular, this implies that if the self-intersection of MATH is MATH, then any basic class MATH must be orthogonal to it: MATH. Since MATH is a torus of self-interesection MATH, it follows that MATH, and, since MATH is also represented by an embedded torus of self-intersection MATH, MATH. The same argument applies to show that MATH for each MATH. Thus MATH. Apply the adjunction inequality to the genus MATH surface MATH and the genus REF surface representing MATH to obtain: MATH . Because MATH is a basic class, MATH, hence MATH . Since MATH, and MATH, it is easy to check that MATH. Furthermore, MATH lies in the negative definite space MATH; so if MATH then MATH . This contradiction implies that MATH; so MATH. Any of the (MATH)-spheres generating the MATH's is orthogonal to MATH; hence orthogonal to each basic class of MATH. It now follows from CITE that MATH has simple type; MATH. Thus we have MATH . It now follows from REF that MATH and MATH, as claimed. Finally, we apply REF, NAME, and NAME to calculate MATH. Since MATH, CITE applies to give MATH where the the sum is taken for all distinct classes MATH for MATH a rim torus. Thus the first equality follows from REF which shows that each MATH in MATH. Now MATH is a symplectic manifold with MATH; so CITE implies that MATH. Since we are assuming that MATH, CITE implies that MATH, and this completes the proof. |
math/9907179 | The hypothesis implies that the only nontrivial NAME invariant of MATH has value MATH; so CITE implies that MATH has no symplectic structure. Since MATH contains an embedded sphere of self-intersection MATH, neither does MATH with its opposite orientation admit a symplectic structure. |
math/9907180 | Simply carry out the above construction with a MATH of order MATH. Then MATH, but MATH. |
math/9907180 | It follows from REF that when MATH, any MATH-sphere component of the fixed point set of a cyclic group action represents a nontrivial element of MATH. A singular MATH-plane near a fixed point is part of a singular MATH, and if the group action were to reverse orientation locally, it would send MATH to MATH. For the second claim, we need to rule out the possibility that MATH contains REF-dimensional components. Suppose for a contradiction that MATH is a circle component of MATH. We may assume MATH. By local linearity, MATH acts preserving orientation on the linking sphere MATH to MATH, so MATH is polyhedral. Moreover, the stabilizer of each point on MATH is cyclic. If MATH itself is cyclic, it has a global fixed point on MATH, and hence MATH is REF-dimensional near MATH. Otherwise, for any non-cyclic subgroup MATH of MATH, MATH is also a component of MATH. Since MATH is a subgroup of each of MATH, MATH, MATH, and MATH, for MATH even, it suffices to rule out dihedral groups. But in a MATH action on MATH, an involution reverses the orientation on the order MATH axis of rotation, which in turn reverses the orientation on the fixed sphere of the group element of order MATH. |
math/9907180 | It follows from results of CITE that MATH is tautly embedded in MATH. Thus we have an isomorphism MATH, as MATH varies over invariant neighborhoods of MATH. Now consider the projection MATH. If we quotient by the MATH-action, we obtain a map MATH such that for any MATH, MATH. In particular, when restricted to the complement of the singular set, MATH becomes a fibration with contractible fiber. A trivial application of the relative spectral sequence of this fibration then shows that MATH for all MATH, whenever MATH is an invariant open neighborhood of MATH. By excision again, the latter group is isomorphic to MATH. But since MATH is four-dimensional, MATH is trivial for MATH. The lemma follows from the long exact cohomology sequence of the pair MATH and tautness. |
math/9907180 | The group operation in K will be written additively. Write MATH, where MATH, and let MATH be generators of the factors in this decomposition. Let MATH. Observe that MATH must be invariant under MATH, so MATH is trivial. If MATH, then MATH has exponent MATH, and the proof is complete. So assume MATH. In this case, MATH. Write MATH, where MATH. Then MATH. On the other hand, MATH, so MATH, and MATH, so we have MATH, where MATH and MATH. Now, MATH. But MATH has order MATH, so MATH. Hence MATH. Since MATH, we have MATH for some MATH. Then by the binomial theorem, MATH . But MATH divides all the terms with MATH, and the MATH term is MATH. So MATH, and then MATH. Hence MATH. Now MATH, so MATH. Since MATH, MATH, so MATH. The same argument applies to any other element of order MATH, so MATH is trivial on all of MATH and hence cannot have order MATH. |
math/9907180 | Recall REF (See CITE): If MATH is a NAME MATH-subgroup of a finite group MATH such that MATH, then MATH has a normal complement MATH. Assume MATH is not a MATH-group. Then every NAME subgroup is proper, and hence abelian. If each NAME subgroup is normal, then MATH must be abelian. So suppose MATH is a NAME MATH-subgroup which is not normal in MATH. Then MATH must be abelian, so MATH, and NAME 's theorem gives us a normal complement MATH such that MATH. Observe that for any other MATH, a NAME MATH-subgroup will be contained in MATH and hence be normal in MATH. By minimality, MATH must be a MATH-group. Also by minimality, MATH must be cyclic, say of order MATH, and the semidirect product automorphism MATH must have order MATH. One more application of minimality shows that MATH must be trivial on any proper submodule of MATH, and then the lemma applies. |
math/9907180 | The subgroup MATH is cyclic, and by the NAME theorem on induced characters, each irreducible complex representation takes the form MATH, where MATH is a complex representation of MATH. In particular, each faithful one has dimension MATH. Now, if MATH, then it has a representation on MATH which splits as a sum of irreducibles. Since MATH is not a nontrivial direct product, at least one must be faithful, so MATH. And if MATH, MATH would necessarily have a three-dimensional real representation. But the finite subgroups of MATH are well-known, and MATH is not among them. |
math/9907180 | The sequence has nonzero rows MATH and MATH only, so it suffices to show that the differentials MATH, MATH, and MATH vanish. But MATH and MATH vanish in odd dimensions, and MATH vanishes in even dimensions. The conclusion follows. |
math/9907180 | Let MATH. By REF , MATH . And since the sequence for MATH also collapses, we have MATH . By comparing the MATH-ranks, we find that MATH. Since the action has a fixed point, MATH, so we must have MATH and MATH. On the other hand, MATH. By the NAME theorem, MATH. This is possible only if MATH or MATH and MATH - in other words, when MATH. When MATH, as we have assumed, there are no actions with a fixed point. |
math/9907180 | A nonabelian compact NAME group MATH has a nontrivial NAME group MATH, where MATH is a maximal torus. MATH is itself a compact NAME group whose identity component MATH is (of course) abelian. A structure theorem for such groups (see CITE or CITE) states that there is a finite subgroup MATH of MATH such that MATH. In particular, there is some MATH of finite order MATH, say, which normalizes MATH but does not centralize it. Since MATH acts non-trivially on MATH, there is a closed one-parameter subgroup MATH such that MATH. Choose MATH of prime order MATH which is not fixed by MATH. Then MATH and its conjugates under powers of MATH generate a subgroup MATH which is normalized by MATH. Thus MATH is a finite, nonabelian subgroup of MATH. |
math/9907180 | If an action by a nonabelian group exists, then by REF , an action by a minimal nonabelian finite group MATH exists. But we have seen that for every such MATH which acts, MATH. The existence of fixed points for abelian groups follows easily from the NAME Fixed Point theorem and the result of CITE used in REF above. |
math/9907180 | By REF , MATH must be abelian of rank at most two, and have a fixed point. But if MATH has rank two, it cannot act freely on the linking sphere to the fixed point, and so cannot act pseudofreely. |
math/9907180 | As before, we suppose MATH is minimal nonabelian. REF implies that the action must be fixed-point-free. We briefly consider the possibilities: CASE: MATH, with MATH. The arguments of REF rule out these actions as before, since MATH must have period REF or REF. CASE: MATH, with MATH, or MATH. As before, we show easily in the first case that MATH must equal REF. Choose generators so that MATH. Then MATH consists of two spheres on which MATH acts by an order two rotation. Denote the fixed points of this rotation by MATH, and the rotation angles of the MATH action around MATH by MATH and MATH. According to the MATH-signature theorem, MATH. But at each point, one of MATH or MATH equals MATH, a contradiction. The same argument applies to MATH. CASE: The dihedral group MATH. The MATH-signature theorem implies that some component MATH must have MATH. But MATH, and since the action is homologically trivial, MATH. So MATH. Together, these elements generate MATH, so MATH, a contradiction. CASE: MATH has rank two. Let MATH, and MATH be as in REF. MATH contains four spheres; since they cannot all be left individually invariant by MATH, two opposing spheres are left invariant and two are interchanged by some MATH. Thus MATH fixes four points (two on each invariant sphere), while MATH fixes two spheres. A contradiction follows just as in REF above. |
math/9907182 | The NAME transform MATH respects the characteristic cycle: we have MATH, using the natural identification MATH (this is given as REF). Since the multiplicity of the zero section MATH in MATH is the NAME characteristic of the stalk cohomology of MATH at a point in the open stratum, it will be enough to show that for generic inclusions MATH, MATH of a point MATH into MATH and MATH, respectively, the restrictions MATH and MATH are isomorphic. We can consider MATH as an element of MATH. Let MATH be given by MATH. Then MATH is naturally isomorphic to following the top row in the following diagram: MATH . Here MATH, MATH are the specialization functors (using the natural identifications MATH and MATH), the functors marked MATH are the appropriate NAME transforms, and MATH. The left quadrilateral and the triangle naturally commute by REF , and the right hand square commutes by REF . Thus MATH. It is easy now to see that MATH and MATH both leave the dimensions of the stalk cohomology at a generic point unchanged (for MATH, use the fact that MATH is conical). |
math/9907182 | Just note that MATH. |
math/9907182 | Since we are free to choose the point MATH, we choose it as follows. The dual cone MATH to MATH is defined to be MATH; MATH and MATH are cones over dual projective varieties in MATH and MATH. By REF is a divisor and so it is not contained in any MATH for any stratum MATH. Let MATH be a smooth point of MATH. Then MATH is a line through the origin contained in MATH; let MATH be any nonzero point in MATH. In microlocal language, the point MATH represents a codimension one point of MATH: it is a smooth point of both MATH and MATH, they intersect transversely there, and it lies in no other component of MATH. Let MATH be the linear function with MATH. Our choice of MATH implies that MATH has singularities only on MATH and that if we choose a normal slice MATH to MATH at MATH, the singularity of MATH is NAME at MATH. Let MATH and embed MATH into MATH by the map MATH. Put MATH; it is a conical perverse sheaf supported on MATH. The local system on MATH can be computed using stratified NAME theory: MATH where MATH. But the local system MATH is easy to compute by NAME theory; its monodromy is multiplication by MATH, where MATH. Thus the monodromy of MATH is nontrivial. An elementary application of the theory of perverse sheaves on a complex line (see CITE and the example following the proof of REF) implies that a perverse sheaf with nontrivial monodromy around the origin must have a nonzero vanishing cycle at MATH; thus MATH. Then by REF , MATH, so we are done. |
math/9907182 | This is equivalent to showing that the dual cone MATH is a divisor. Using the inner product on MATH given by matrix coordinates, MATH is naturally identified again with MATH, with the action of MATH given by MATH . Certainly MATH is MATH-stable, and it is easily checked that MATH lies in MATH, where again MATH is any nonzero nilpotent matrix. Its stabilizer is MATH, a two-dimensional group, so MATH. |
math/9907182 | Let MATH, where MATH if MATH, and MATH if MATH. Checking that this gives the required action is an exercise in local NAME coordinates, similar to arguments in CITE and CITE. |
math/9907182 | Acting on MATH by the loop MATH, we obtain the loop MATH; that is, the point in MATH stays fixed and the covector goes around by a conical action. We can then slide the loop MATH in MATH to get a loop of the form MATH for some MATH. Acting now by MATH now produces a trivial loop. |
math/9907182 | Since MATH is a MATH-equivariant perverse sheaf, the local systems MATH must also be MATH-equivariant. |
math/9907191 | REF are standard results on Generic Geometry and can be found for instance in CITE. We will prove REF . Suppose that MATH is a parametrization of MATH by arc length, so that MATH and MATH. Since MATH is a orthonormal basis of MATH, we have that MATH . In particular, provided that MATH, we have that MATH, where MATH . Thus, MATH if and only if MATH if and only if MATH is a critical point of MATH. On the other hand, let MATH be a parametrization of MATH by arc length, with MATH. In this case, MATH for all MATH and thus MATH. This gives that MATH is also equal to the normal vector to MATH in MATH. In particular, MATH is a critical point of MATH if and only if the two curves are tangent. Suppose now that MATH is a critical point of MATH. We have that MATH where MATH is the normal curvature of MATH. This gives that MATH . But we can do the same with MATH, getting MATH . This implies that MATH . Since MATH, we conclude that MATH is a non-degenerate critical point of MATH if and only if MATH. The fact that the cases MATH and MATH correspond to an island or a bridge respectively can be deduced from the way we have chosen the orientation on MATH. |
math/9907191 | The formula for MATH is a direct consequence of REF . We will see the first part of the statement. Given MATH, MATH is a NAME function if and only if MATH is a regular value of the NAME map MATH (just note that the points MATH where MATH are the critical points of the NAME map). Therefore, the NAME Theorem implies that for a generic MATH, MATH is a NAME function. The other two conditions are also an easy consequence of this theorem. |
math/9907191 | Let MATH and let MATH be the domain with boundary in MATH given by MATH, where MATH are small geodesic balls centered at MATH which do not intersect MATH. It follows from REF that MATH . On the other hand, MATH, which concludes the proof. |
math/9907198 | If MATH is a retract of MATH, then the tower MATH is a retract of MATH, so if MATH is null after mapping in each MATH, then so is MATH. (Given MATH, then consider MATH . Then the composite MATH factors through MATH.) The condition MATH involves numbers MATH and MATH, and we write MATH if we want to specify the numbers. Given a cofibration sequence MATH in which MATH and MATH satisfy conditions MATH and MATH, respectively, we show that MATH satisfies MATH. Consider the following commutative diagram, in which the rows are cofibrations: MATH . We fix MATH and assume that MATH, so that we have MATH . Given any map MATH, then since MATH is null, the composite MATH factors through MATH, giving MATH . Since MATH is null, though, then the composite MATH is null. |
math/9907198 | As above, we write MATH for the map MATH and MATH for the map MATH, so that MATH is the image of MATH . CASE: Assume that if MATH, then MATH. In the case MATH, if MATH, then MATH; so we see that MATH. By the long exact sequence REF, we conclude that MATH when MATH. Reindexing, we find that MATH when MATH; that is, REF holds. The case MATH is similar; in this case, the long exact sequence implies that MATH. CASE: Assume that MATH. If MATH whenever MATH, then MATH when MATH. So by the exact sequence REF, we see that MATH is an isomorphism under the same condition. This map is induced by MATH, so we conclude that when MATH, we have MATH . But since MATH is MATH - complete, then MATH, so MATH. Reindexing gives MATH when MATH; that is, MATH implies MATH. If MATH, then a similar argument shows that MATH. REF are similar. |
math/9907198 | This follows immediately from REF . More precisely, to show that REF is generic, one applies REF to the set MATH. For REF , one applies it to the set MATH. |
math/9907200 | By NAME 's classical theorem we know MATH . Using MATH and from the definition of MATH it follows that MATH . Now by the remarks after REF , MATH and since the canonical divisor of a genus MATH curve has degree MATH it follows that MATH . Moreover in the top dimension MATH commutes with evaluation and recalling MATH the result follows. |
math/9907200 | That the line bundles are dual amounts to an identification of the positive and negative harmonic forms of MATH with the positive and negative eigenspaces for the Hermitian form MATH; then compare the formulae defining MATH in REF . But by definition the MATH are defined to be eigenspaces for an operator whose index is signature and the MATH comprising MATH are the definite subspaces for the Hermitian form arising from the symplectic signature pairing. That the trivialisations agree is a consequence of REF . For if the two trivialisations differed then their difference would define a map from the set of components of the boundary MATH to MATH depending only on the particular monodromies associated to these components. Moreover since we know that there are identities REF for all fibrations MATH we always know that MATH . Since the values of the trivialisations on loops depend only and naturally on the monodromies, this difference map defines MATH which (taking the above relation for MATH a sphere with three discs removed) is a homomorphism. But by REF we know the mapping class groups admit no such non-trivial homomorphisms. |
math/9907200 | The proof of the proposition is reasonably straightforward and versions due to CITE and NAME - CITE have now appeared (an independent proof was given in CITE). The idea is simple; on choosing a metric all the smooth fibres are hyperelliptic NAME surfaces and admit natural double branched covers over spheres. These can be patched together smoothly except near separating singular fibres. The point is that the map fibrewise is given by sections of the canonical bundle on the fibre, and the nodal points in reducible fibres are base points of the canonical system on a stable curve; for smooth or stable irreducible curves the canonical linear system has no base points and the branch locus varies smoothly. Near reducible fibres we can assume the complex structure is integrable and graft in a local holomorphic model. This argument (and those of NAME and NAME - NAME) gives the base of the double cover the structure of a sphere bundle over the sphere but does not identify it beyond that. The signature formulae allow one to do precisely this. The ruled surface MATH for suitable rank two MATH and any line bundle MATH. Since MATH and MATH, to find the diffeomorphism type of the base from a monodromy equation we need only understand the parity of the first NAME class of any suitable bundle MATH above. Specifically, in the complex case the map MATH is the map defined by the sheaf MATH and we know MATH and more generally the sphere fibres of the ruled surface are the projectivisations of spaces of holomorphic sections of the canonical bundles down the fibres of MATH. Thus for any fibration MATH we can take MATH. From REF we know MATH . It follows that the parity of MATH and the parity of MATH coincide, and that is precisely what we require. |
math/9907200 | Suppose such a MATH exists; we work by contradiction. We mentioned before that another approach to signature uses NAME 's non-additivity; an easy case of this approach CITE shows that if a fibration has MATH separating vanishing cycles and no others, then its signature is given by MATH. Comparing to REF this forces MATH. We will prove that this quantity is positive. We have MATH for any NAME fibration with MATH singular fibres; in our case, since all the vanishing cycles are homologically trivial, MATH and hence MATH. Now the number of disjoint exceptional MATH - spheres in MATH is bounded by MATH since each contributes a new homology class. The fibre of the fibration has self-intersection zero; fibre-summing MATH with itself by the identity if necessary, we can see that if any fibration with monodromy group contained in the NAME group exists, then one exists for which a section also has self-intersection even and in particular is not exceptional. Thus the number of MATH - spheres in MATH may be assumed to be bounded above by MATH. We invoke REF: for any minimal symplectic REF - manifold MATH which is not irrational ruled, MATH. In particular, blowing down all MATH - spheres in MATH, we see that MATH and hence MATH provided MATH is not irrational ruled. But we also know MATH . Rearranging this gives MATH which is the required contradiction. That leaves only the case of MATH irrational ruled, but simple topological conditions show that such a manifold can have no NAME fibration. For any NAME fibration of MATH has fibres of genus MATH by considering MATH, and then looking at MATH we see that in fact all the vanishing cycles must be nullhomotopic and hence MATH was a trivial product. |
math/9907215 | of REF . We fix MATH throughout the proof of REF and write MATH for MATH. We also fix a finitely generated MATH - module, MATH. The finiteness of the global dimension of MATH means that MATH admits a projective resolution of finite length (at most equal to MATH, by definition.) Because MATH is a local ring, the only projective MATH - modules are the free modules. Thus MATH admits a resolution of finite length, MATH say, by free modules. Fix such a resolution: MATH . It follows that the groups MATH can be computed as the homology of a complex of finite length of the form MATH immediately implying that each MATH is finitely generated over MATH and that MATH . On the other hand, since MATH is a flat MATH - module, the alternating sum of MATH - ranks adds to zero along exact sequences. Thus it follows from REF that MATH completing the proof of REF . MATH . |
math/9907215 | This is immediate from the fact that MATH is free over MATH, of rank MATH, and so a free resolution of MATH as a MATH - module also gives a free resolution of MATH as a MATH - module, with alternating sum of MATH - ranks equal to MATH times the alternating sum of MATH - ranks. MATH . |
math/9907215 | This follows by almost exactly the same arguments used above to prove REF . Since MATH is a local ring and has finite global dimension, we may again take a finite, free resolution of MATH, this time as a MATH - module, MATH . The only extra point needed is that since MATH is not a projective MATH - module, we must first calculate MATH. Consider the sequence MATH . Taking MATH - homology, and recalling MATH for MATH, we obtain MATH and MATH vanishes for MATH. Since MATH is both annihilated by MATH and contained in MATH, it must be zero also, while MATH is simply MATH. Thus MATH equals MATH and similarly MATH equals MATH. It follows that we can calculate the MATH by working in the category of finitely generated MATH - modules and using the projective resolution REF . The remainder of the proof of REF now follows exactly as before for REF , in particular MATH is again flat as a MATH - module and so the MATH - rank of MATH is equal to MATH, from which the proposition is clear. MATH . |
math/9907215 | We consider the short exact sequences MATH obtained from the filtration MATH for MATH sufficiently large. The modules MATH are finitely generated MATH - modules for all MATH and so, by REF MATH . In particular, these numbers are all defined. Since MATH vanishes, the NAME characteristic MATH is finite by REF . It then follows inductively from the long exact sequence in cohomology that MATH is finite for all MATH, equal to the product of MATH and MATH. Multiplying these together, we obtain REF. MATH . |
math/9907215 | We start with the observation that if MATH and MATH are MATH - torsion then the lemma follows from REF , since all NAME characteristics are defined and alternating products of NAME characteristics along exact sequences multiply together to give REF. In general we have MATH, which, together with this remark gives the inequality. However, exactness can fail on the right. Let MATH denote the image of MATH in MATH. It follows from the first remark that MATH. We show that if MATH is MATH - torsion, then MATH and thus MATH. The surjection MATH induces a surjection MATH. By assumption, MATH is MATH - torsion, and thus MATH is also MATH - torsion. By REF it is also MATH - torsion for any choice of uniform, open, normal subgroup of MATH. Let MATH be any such subgroup. Then MATH is MATH - torsion and MATH - torsion free. Since MATH is in the centre of MATH, it follows that for every element MATH of MATH there exists an element, MATH, of MATH which is not contained in MATH and which annihilates MATH. In particular, this is true for all elements of MATH. Let MATH be a uniform, pro-MATH group. Suppose MATH is a MATH - torsion MATH - module, generated by a single element, MATH, and such that there exists an element MATH of MATH, not contained in MATH, for which MATH. Then MATH. Since MATH is in the centre of MATH we may consider the modules MATH, equal to MATH, separately for each MATH and thus, without loss of generality, we may assume MATH is annihilated by MATH, that is, MATH is a MATH - module generated by a single element, MATH. Furthermore, this generator MATH has a non trivial annihilator in MATH given by the image of MATH under the canonical surjection of MATH onto MATH. But then it is immediate this means MATH has rank zero as a MATH - module, as required. MATH . By the definition of MATH - ranks, for any finitely generated MATH - module, MATH, the MATH-invariant, MATH, calculated as a MATH - module vanishes if and only if it vanishes when considering MATH as a MATH - module. The module MATH is finitely generated over MATH, as MATH is NAME and NAME is finitely generated over MATH. Thus the proposition follows by induction on the number of generators of MATH. MATH . |
math/9907215 | We begin by considering a finitely generated MATH - module, MATH, which is actually MATH - torsion. Since MATH is MATH - torsion, it follows from REF that MATH, MATH and MATH are all finite, where MATH is the NAME characteristic in the category of MATH - modules as defined by REF . Upon consideration of the exact sequence : MATH and since the alternating product of NAME characteristics along an exact sequence multiplies together to make REF, this shows that MATH . However, both MATH and MATH actually have exponent MATH and so, by REF , this shows that they have equal rank as MATH - modules. Thus REF holds for finitely generated MATH - torsion modules. In particular, it holds for MATH. Because MATH is MATH - torsion free, the sequence MATH is exact, and so MATH . Since the MATH - rank of MATH equals that of MATH, it only remains to show that if MATH is a MATH - torsion free MATH - module, then the MATH - rank of MATH equals the MATH - rank of N. But this follows immediately from REF and the long exact sequence in MATH-homology of MATH . Indeed, the multiplication by MATH map on N induces multiplication by MATH on the homology groups, MATH. Since on a finite Abelian group the kernel and cokernel of multiplication by MATH have the same order, we easily see that MATH . By REF , the term on the right equals the MATH - rank of MATH, while by REF the term on the left equals the MATH - rank of MATH. MATH . |
math/9907215 | From the remark just prior to the statement of REF , we know MATH contains a pro-MATH, normal subgroup of finite index, Thus the finiteness of the MATH - rank of MATH follows immediately from the case of a pro-MATH, MATH-adic, NAME group and the NAME spectral sequence for group cohomology. MATH . |
math/9907215 | This is an immediate consequence of NAME 's Lemma together with REF . Recall that in general if MATH is a closed subgroup of MATH and MATH is a discrete MATH - module then MATH and MATH are naturally isomorphic (see for example CITE, I. REF) Similarly for the homology of compact MATH - modules. MATH . |
math/9907215 | The MATH-torsion submodule MATH is a finitely generated MATH - submodule, and so all of the sub-quotients, MATH, have rank REF over MATH. MATH . |
math/9907215 | We have the following fundamental diagram, with vertical maps given by restriction maps and exact rows as a consequence of our assumption concerning Conjecture REF: MATH . This diagram is analysed in detail in CITE, where the following is proved (REF .) CASE: The kernel and cokernel of MATH are both finite. CASE: The cokernel of MATH is finite and the kernel of MATH has finite MATH - corank, at most equal to MATH. CASE: This is an equality if MATH contains MATH. Since MATH has finite MATH - rank, by assumption, it follows from the snake lemma that MATH has finite MATH - rank, given by MATH with equality if MATH contains MATH. Since we are assuming Conjecture REF for MATH, the top row of REF extends to a long exact sequence in cohomology: MATH . Since the map MATH is surjective and MATH is finite, it follows from REF and the snake lemma that the cokernel of MATH is finite. We assume MATH. Then it follows from our assumption of the vanishing of MATH that the groups MATH are finite for all MATH. We exploit the fact that it is known that MATH for MATH at least MATH. Then it follows from the NAME spectral sequence that we have the exact sequence MATH for MATH at least MATH. We have assumed that MATH vanishes, whilst for MATH the groups MATH vanish since the MATH cohomological dimension of MATH equals MATH under our assumption that MATH is not equal to MATH. The groups MATH are easily seen to be finite for all MATH. See, for example, CITE. MATH . Assume that MATH. Then for all MATH one has MATH . It is proven in CITE that REF is true with MATH replaced by MATH (REF for MATH and REF for MATH.) Since MATH the NAME spectral sequence gives rise to short exact sequences MATH . Since MATH is MATH - primary and has the discrete topology, and since MATH is isomorphic to MATH, the vanishing of MATH for MATH implied by REF ensures that MATH vanishes for all MATH. MATH . Thus we see from the exact sequence REF and the finiteness of MATH that for all MATH the MATH are finite. The duality between homology and cohomology described in REF extends to this situation, even though MATH is not finitely generated as a MATH - module. Indeed, taking NAME duals passes between the categories of compact MATH - modules and discrete MATH - modules, without any finite generation assumption. This duality again sends compact, projective modules to discrete, injective modules, and we still have MATH for any compact MATH - module, without assuming MATH to be finitely generated. In fact, since finitely generated MATH - modules are pseudo-compact as MATH - modules, this is explained in greater detail in CITE. Thus we also still have a canonical isomorphism MATH . It follows that MATH is finite for all MATH, and so the homological MATH - rank of MATH is just the MATH - rank of MATH, given by REF . This completes the proof of REF MATH . |
math/9907215 | This follows closely the ideas of NAME in the cyclotomic case given in detail for elliptic curves in the appendix to CITE. We proceed similarly to the proof of REF above, with a close analysis of a diagram, this time REF . The snake lemma gives MATH . The first thing to remark is that the NAME dual of the kernels and cokernels of MATH and MATH (and hence of MATH by the snake lemma) are all finitely generated MATH - modules, annihilated by a finite power of MATH. Note that although REF is given in terms of group homology, because of REF it can also be rephrased in terms of cohomology as for REF . For discrete, cofinitely generated MATH - modules, we will use also the notation MATH to denote MATH where the choice between cohomology and homology is dictated by whether the module is discrete or compact, respectively. Since either choice gives zero for finite modules when MATH is pro-MATH, there will be no ambiguity. Thus if MATH is a finitely generated MATH - module then MATH also. Then the multiplicativity of MATH along exact sequences gives MATH where all NAME characteristics are defined. Since MATH and MATH are both annihilated by a finite power of MATH, by REF the number we require for the Theorem is then the logarithm to base MATH of this. For the first term, recall that we know MATH vanishes for MATH, by REF and the assumption that MATH. Thus by the long exact sequence in cohomology of which the central vertical arrow is a part, we have MATH. Since isogenies are surjective MATH, surjects onto MATH, and so MATH. The full NAME spectral sequence: MATH is bounded since MATH, MATH and MATH all have finite cohomological dimension at MATH. It follows from this that MATH . Precise details of how NAME Characteristics behave in spectral sequences, and in particular how to obtain this formula, are given in CITE. But since MATH is just MATH and in particular is finite, as remarked in REF it has NAME characteristic equal to one, and thus MATH . Locally, for any prime MATH of MATH in MATH we define MATH where the direct limits are taken with respect to the restriction homomorphisms. Recall that MATH is the image of MATH under the reduction map at MATH. As explained in the local calculations in REF, it follows from the results in CITE that MATH . Let MATH be MATH if MATH, whilst if MATH then MATH denotes MATH. Then the groups MATH vanish (also explained in REF) and as described above for the global map MATH, we have the simple descriptions MATH and MATH. For each prime MATH in MATH we fix a choice of prime MATH lying above, which will also be denoted by MATH. It follows from NAME 's Lemma that MATH where MATH is as above and MATH is the decomposition group of MATH in MATH, that is the NAME group of the extension MATH. Thus MATH and similarly for MATH. Then, as we saw when considering MATH, the NAME spectral sequence MATH gives MATH . Since MATH is a MATH - group, it is known, due to REF , that the NAME Characteristic corresponding to primes in MATH which do not lie above MATH equals one. Then substituting REF into REF and taking the logarithm to base MATH gives MATH . We recall here NAME 's formulae for calculating such NAME characteristics. If MATH is a finite MATH - module such that all primes MATH of MATH dividing the order of MATH are contained in MATH, then (see REF ) MATH . If MATH is a non-Archimedean prime of MATH and MATH is a finite MATH - module, then (see REF ) MATH . Substituting these into REF gives precisely REF in the Theorem. MATH . |
math/9907215 | This follows immediately from REF . Since we are now assuming the MATH are MATH - torsion (and thus also MATH - torsion for all finite extensions, MATH of MATH) we know from REF that MATH . Replacing MATH by MATH and MATH by MATH in REF , and substituting the formula thus resulting from REF into REF , gives the Corollary. MATH . |
quant-ph/9907019 | Let MATH be the maximum number such that there is a family of (distinct) subsets MATH with the desired properties. Let MATH. For each MATH we count the number of subsets MATH with cardinality MATH but MATH. This number is MATH . Defining MATH we could add another set to our family of subsets if MATH. Therefore: MATH . |
quant-ph/9907019 | By definition of MATH there is a MATH Q code MATH with MATH if only MATH is large enough. Using MATH as ground set, REF provides us with subsets MATH of cardinality MATH with pairwise intersections smaller than MATH. Here we have for the number MATH of those sets: MATH . We construct a simultaneous Q-ID code MATH by taking as MATH the uniform distribution on MATH and as MATH the sum of the corresponding MATH-s: MATH . It's now straight forwards to calculate that the errors are small: MATH and for MATH: MATH . |
quant-ph/9907019 | Let MATH be a MATH-achievable resolution rate for MATH, and let MATH. For a fixed process MATH there is a minimum MATH such that for some process MATH: MATH . We have to prove that MATH. Suppose the opposite and let MATH be a sequence of processes such that MATH is strictly monotonically increasing (hence divergent). Define a new process MATH by MATH . Consider the minimum MATH for which MATH. Since for MATH we have MATH there exist probability distributions MATH for which MATH and MATH. By definition of MATH there are such MATH for MATH, too. This contradicts MATH being chosen as minimum number. |
quant-ph/9907019 | By the previous remark, MATH is a uniform MATH-achievable resolution rate for MATH. So, for MATH there is some MATH and a sequence of processes MATH such that for all MATH and MATH: MATH . For fixed MATH let's assume that MATH for some MATH. This leads to MATH . Since this is a contradiction it follows that MATH for all MATH. Now, as there are not more than MATH probability distributions on MATH of type MATH, and as for all MATH and MATH the resolution of MATH is smaller than MATH . It follows MATH (for all MATH), and MATH. |
quant-ph/9907019 | For each block length MATH let MATH be a simultaneous MATH Q-ID code of maximum size MATH and MATH with MATH the common refinement of the MATH (compare REF ). Consider a sequence MATH of processes with MATH where MATH is arbitrary for MATH. We have for all MATH: MATH . So, REF implies MATH. |
quant-ph/9907019 | By REF we have MATH. In the following section we will see (compare REF ) that MATH. Analogously we obtain MATH (compare the previous remark and REF). |
quant-ph/9907019 | Let's assume that this is false and that there is a pair of processes MATH such that for some MATH for infinitely many integers MATH. With this assumption we will be able to construct for those integers - if only they are large enough - a MATH Q code, with MATH some integer fulfilling MATH . As the first inequality contradicts to the definition of MATH the lemma will be proved. Given MATH let for every MATH . We choose the codewords MATH successively by the random selection method with probability MATH . The decoding operator of the codeword MATH is defined to be the operator MATH where MATH . The success probability is MATH . For the expected value of the last summands MATH holds, where MATH . Here the inequality holds because MATH . So, we get for the expected success probability MATH . For some MATH and random variable MATH clearly MATH holds. Therefore: MATH . Hence MATH - if only MATH is large enough - , and there is certainly one codeword MATH with the desired success probability. |
quant-ph/9907019 | Let MATH. We show that by the random selection method there is a process MATH such that MATH . This works as follows: For fixed MATH let MATH. Each MATH-tuple MATH of codewords gives rise to the MATH-type probability distribution MATH . We will show that MATH interpreting the MATH as independent outcomes of a random experiment with underlying probability distribution MATH. This directly implies our claim. Recall that MATH (compare REF ), and by the previous lemma it is enough to show that for every MATH the following expression goes to MATH as MATH tends to infinity: MATH . Since all MATH summands are equal this is just MATH . Here MATH, and the last inequality holds because MATH . The first summand of REF is easy to handle: MATH . For the last inequality (which holds for large MATH) recall that MATH. Now, since for every MATH we can apply NAME 's Inequality: MATH . And we get the following upper bound for the second summand of REF : MATH . |
quant-ph/9907035 | There is a universal quantum NAME machine MATH in the standard enumeration MATH such that for every quantum NAME machine MATH in the enumeration there is a self-delimiting program MATH (the index of MATH) and MATH for all MATH. Setting MATH proves the theorem. |
quant-ph/9907035 | Let MATH be such that MATH . Denote the program MATH that minimizes the righthand side by MATH and the program MATH that minimizes the expression in the statement of the theorem by MATH. By running MATH on all binary strings (candidate programs) simultaneously dovetailed-fashion one can enumerate all objects that are directly computable given MATH in order of their halting programs. Assume that MATH is also given a MATH length program MATH to compute MATH - that is, enumerate the basis vectors in MATH. This way MATH computes MATH, the program MATH computes MATH. Now since the vectors of MATH are mutually orthogonal MATH . Since MATH is one of the basis vectors we have MATH is the length of a prefix code (the NAME code) to compute MATH from MATH and MATH. Denoting this code by MATH we have that the concatenation MATH is a program to compute MATH: parse it into MATH and MATH using the self-delimiting property of MATH and MATH. Use MATH to compute MATH and use MATH to compute MATH, determine the probabilities MATH for all basis vectors MATH in MATH. Determine the NAME code words for all the basis vectors from these probabilities. Since MATH is the code word for MATH we can now decode MATH. Therefore, MATH which was what we had to prove. |
quant-ph/9907035 | For every state MATH in MATH-dimensional NAME space with basis vectors MATH we have MATH. Hence there is a MATH such that MATH. Let MATH be a MATH-bit program to construct a basis state MATH given MATH. Then MATH. Then MATH. |
quant-ph/9907035 | The uncomputability follows a fortiori from the classical case. The semicomputability follows because we have established an upper bound on the quantum NAME complexity, and we can simply enumerate all halting classical programs up to that length by running their computations dovetailed fashion. The idea is as follows: Let the target state be MATH of MATH qubits. Then, MATH. (The unconditional case MATH is similar with MATH replaced by MATH.) We want to identify a program MATH such that MATH minimizes MATH among all candidate programs. To identify it in the limit, for some fixed MATH satisfying REF below for given MATH, repeat the computation of every halting program MATH with MATH at least MATH times and perform the assumed projection and measurement. For every halting program MATH in the dovetailing process we estimate the probability MATH from the fraction MATH: the fraction of MATH positive outcomes out of MATH measurements. The probability that the estimate MATH is off from the real value MATH by more than a MATH is given by NAME 's bound: for MATH, MATH . This means that the probability that the deviation MATH exceeds MATH vanishes exponentially with growing MATH. Every candidate program MATH satisfies REF with its own MATH or MATH. There are MATH candidate programs MATH and hence also MATH outcomes MATH with halting computations. We use this estimate to upper bound the probability of error MATH. For given MATH, the probability that some halting candidate program MATH satisfies MATH is at most MATH with MATH . The probability that no halting program does so is at least MATH. That is, with probability at least MATH we have MATH for every halting program MATH. It is convenient to restrict attention to the case that all MATH's are large. Without loss of generality, if MATH then consider MATH instead of MATH. Then, MATH . The approximation algorithm is as follows: CASE: Set the required degree of approximation MATH and the number of trials MATH to achieve the required probability of error MATH. CASE: Dovetail the running of all candidate programs until the next halting program is enumerated. Repeat the computation of the new halting program MATH times REF : If there is more than one program MATH that achieves the current minimum then choose the program with the smaller length (and hence least number of successfull observations). If MATH is the selected program with MATH successes out of MATH trials then set the current approximation of MATH to MATH . This exceeds the proper value of the approximation based on the real MATH instead of MATH by at most REF bit for all MATH. CASE: Goto REF . |
quant-ph/9907035 | Every orthonormal basis of MATH has MATH basis vectors and there are at most MATH programs of length less than MATH. Hence there are at most MATH programs available to approximate the basis vectors. We construct an orthonormal basis satisfying the lemma: The set of directly computed pure quantum states MATH span a MATH-dimensional subspace MATH with MATH in the MATH-dimensional NAME space MATH such that MATH. Here MATH is a MATH-dimensional subspace of MATH such that every vector in it is perpendicular to every vector in MATH. We can write every element MATH as MATH where the MATH's form an orthonormal basis of MATH and the MATH's form an orthonormal basis of MATH so that the MATH's and MATH's form an orthonormal basis MATH for MATH. For every directly computable state MATH and basis vector MATH we have MATH implying that MATH and therefore MATH (MATH). This proves the lemma. |
quant-ph/9907035 | Use the notation of the proof of REF . Assume to the contrary that there are MATH basis vectors MATH with MATH. Then at least two of them, say MATH and MATH and some pure quantum state MATH directly computed from a MATH-length program satisfy MATH . (MATH). This means that MATH since not both MATH and MATH can be equal to MATH. Hence for every directly computed pure quantum state of complexity MATH there is at most one basis state of the same complexity (in fact only if that basis state is identical with the directly computed state.) Now eliminate all directly computed pure quantum states MATH of complexity MATH together with the basis states MATH that stand in relation REF . We are now left with MATH basis states that stand in relation of REF with the remaining at most MATH remaining directly computable pure quantum states of complexity MATH. Repeating the same argument we end up with MATH basis vector that stand in relation of REF with REF directly computable pure quantum states of complexity MATH which is impossible. |
quant-ph/9907035 | By REF we there is a program MATH to compute MATH with MATH and a program MATH to compute MATH from MATH with MATH up to additional constants. Use MATH to construct two copies of MATH and MATH to construct MATH from one of the copies of MATH. The separation between these concatenated binary programs is taken care of by the self-delimiting property of the subprograms. The additional constant term takes care of the couple of MATH-bit programs that are required. |
quant-ph/9907035 | MATH by the proof of REF . Then, the lemma follows by REF . |
quant-ph/9907035 | Give the reference universal machine MATH as input where MATH is the index of the identity quantum NAME machine that transports the attached pure quantum state MATH to the output. |
quant-ph/9907035 | Transfer the conditional MATH to the input using a MATH-bit program. |
quant-ph/9907081 | All assertions but the last one are obvious. So, for MATH strictly positive (MATH is no eigenvalue) and MATH it holds that MATH with equality (for example ) for MATH. (This is NAME Inequality!) Therefore: MATH and this immediately implies that for MATH . : MATH . |
quant-ph/9907081 | Let MATH and MATH, and define operators MATH on MATH. Given now MATH let MATH large enough such that: MATH . MATH . |
quant-ph/9907081 | MATH . Therefore: MATH . |
quant-ph/9907081 | We remark that MATH and MATH are NAME spaces with inner product MATH. Define a linear map MATH by MATH for all MATH. MATH is a contraction, in fact: MATH . For MATH define MATH by MATH for all MATH. (It is positive because MATH.) It holds MATH (MATH), and MATH: MATH . So, MATH, and MATH . |
quant-ph/9907081 | Let MATH, MATH, MATH and MATH, where without loss of generality MATH. As MATH and MATH . REF implies that: MATH . Consequently, MATH, and by the limit MATH: MATH . |
quant-ph/9907081 | This is only a question of notation: MATH . To reduce the claim to REF notice that MATH. |
quant-ph/9907081 | Because of MATH this is just a consequence of MATH . |
solv-int/9907002 | We use the notation introduced above. Let MATH be a solution to the linear problem MATH with a constant vector MATH. Since MATH the zero curvature condition stays valid and the system is solvable. The additional term MATH will give rise to an additional rotation around MATH in the dIHM model. The importance of this possibility will be clarified in the next section. Moreover define MATH . Note that this implies that MATH . In other words: MATH with MATH. We will show, that the MATH solve the dIHM model (if MATH). To do so we use MATH as a gauge field: MATH . If one writes MATH one gets MATH and one can conclude that MATH . This clearly coincides with MATH up to the irrelevant normalization factor MATH. On the other hand one gets for the gauge transform of MATH . But with above substitution for MATH one gets MATH and since MATH we get MATH . Remember that MATH and MATH. Using REF and the fact that MATH and MATH anti-commute we conclude MATH . Combining this and REF one obtains for the gauge transform of MATH . Since the first term is a multiple of the identity and independent of MATH it cancels in the zero curvature condition and therefore can be dropped. This gives the desired result if MATH. |
solv-int/9907002 | This is already covered by the proof of REF . |
solv-int/9907002 | The method is more or less the same as in the singly discrete case although this time we start from the other side: Start with a solution MATH of the ddIHM model. Choose MATH such that MATH . This is always possible since the first equation leaves a gauge freedom of rotating around MATH. Moreover define MATH and normalize MATH in such a way that MATH takes the form MATH REF ensure that MATH and thus can be written MATH for some complex MATH. Equipped with this we can gauge a normalized version of MATH with MATH and get MATH if we write MATH as before. On the other hand we get for an - again renormalized - MATH . But the zero curvature condition MATH yields that MATH must be lower and MATH upper triangular. Thus MATH has the MATH-dependency as required in REF . |
solv-int/9907008 | The ``if" portion is easy enough: MATH since the polynomials are assumed linearly independent. Now, suppose REF holds. It will turn out to be convenient to restate the equation in terms of exterior products. Define a vector-valued function MATH by MATH . Then we can write MATH where MATH stands for the standard duality between MATH-forms and MATH-forms. Consider this as a function of MATH as the other variables range over the support of MATH; we have: MATH . Now, since MATH has at least MATH elements in its support, these functions span a MATH-dimensional space (this follows, for instance, from NAME 's interpolation formula). On the other hand, the functions must clearly be linear combinations of the MATH. Since there only MATH functions MATH, it follows that we can write the MATH as linear combinations of the functions MATH, MATH. But this is precisely what we wanted to prove. |
solv-int/9907008 | Again the ``if" case is straightforward. In the other direction, we can clearly divide each MATH by MATH, and thus may assume Without loss of generality that MATH on MATH. We first consider the case MATH, for which MATH . As MATH varies over MATH, this spans a REF-dimensional function space; it follows that as MATH varies, MATH spans a REF-dimensional space (the dimension must be either REF or REF; REF clearly leads to a contradiction). In other words, there must be a linear dependence between the coefficients of MATH. By replacing the MATH with an orthogonal linear combination, we find that this dependence is Without loss of generality of the form MATH for some constant MATH. Now, if MATH were REF, then we would have MATH . Now, let MATH be an open interval in MATH. If either MATH or MATH were identically REF on MATH, our determinant would be identically REF on MATH (contradiction); it follows that we may choose MATH so that both MATH and MATH are nonzero. Then we can divide both sides of REF by MATH and integrate; we find that MATH on MATH. But this again makes the determinant REF. We conclude that the linear dependence satisfied by MATH must take the form MATH with MATH. In particular, we find that the MATH-form MATH orthogonal to MATH is not itself in the span of MATH. In particular, any MATH-form can be written as a linear combination of MATH and some of the MATH. Taking the inner product with MATH, we conclude that for MATH, we have MATH for some polynomial MATH of degree at most REF. Now, since MATH, we find: MATH for all MATH. Similarly, since MATH , we have MATH . Now, since MATH is linearly independent of MATH, we can solve these two equations for MATH and MATH as rational multiples of MATH, subsitute into the equation MATH, then solve for MATH. We find: MATH where MATH . We observe that each numerator has degree at most REF, as does the polynomial MATH. In particular, if we exclude any given MATH, we can express the squares of the other MATH as rational functions with common denominator of degree at most REF. It follows that the functions MATH have at most REF poles between them, and thus that we can write MATH where MATH, MATH, MATH, and MATH are polynomials of degree at most REF and MATH is a polynomial of degree at most REF. We now need to show that, in fact, each MATH is a polynomial of degree at most REF. By the usual factorization, we find: MATH valid on MATH. Without loss of generality, we may assume that the constant of proportionality is REF, and that MATH. Dividing both sides by MATH and taking the limit as MATH, we find: MATH . Applying a suitable linear transformation to the polynomials MATH, we have, without loss of generality, MATH . We can then solve for MATH by taking MATH above; we find MATH . At this point, we can compare coefficients on both sides of REF, obtaining a number of polynomial equations relating the coefficients MATH, MATH, MATH. The resulting ideal can be verified (using magma, for instance) to contain the polynomials MATH, MATH, and MATH; passing to the radical, we can then solve for MATH, MATH, MATH. Substituting in, we find that MATH is now a square, and that each MATH is a multiple of MATH. In other words, each MATH is a polynomial of degree at most REF, and we are done with the case MATH. It remains only to show that we can reduce the cases MATH to cases of lower dimension. Choose a particular element MATH. By replacing the MATH with appropriate linear transformations, we may assume MATH for some nonzero constant MATH. In particular, we find that MATH . By induction, it follows that for MATH, there exist polynomials MATH of degree at most MATH such that MATH on MATH. Undoing our linear transformations, we find that for every polynomial MATH of degree at most MATH vanishing to second order at MATH, we can write MATH as a linear combination of the MATH. But this was independent of our choice of MATH. In particular, taking MATH to be any other element of MATH, we have MATH and MATH . It follows that for any polynomial MATH of degree at most MATH, MATH is a linear combination of the MATH. By dimensionality, it follows that each MATH is itself of the form MATH, and we are done. |
solv-int/9907008 | For each MATH, we have MATH . Multiplying over MATH, we are done. |
solv-int/9907008 | When MATH, the LFT reverses the order of integration, thus justifying the extra factor of MATH introduced for MATH and MATH. For MATH, there is a more subtle difficulty, namely that the relative order of the eigenvalues is significant, and can change. If we simply reverse the order, this is not a problem (the total effect is MATH, thus cancelling out the sign of MATH). So we can restrict to the case MATH. The effect of the LFT is then to cyclically shift the ordering, taking the eigenvalues with MATH and making them smallest. If there are MATH such eigenvalues, the sign of the NAME matrix is changed by MATH; thus if MATH is odd, there is no problem. On the other hand, if MATH is odd, we have a problem unless the eigenvalues are restricted to only one side of MATH, or equivalently that MATH has constant sign over the support of MATH. Since MATH we may take this sign to be positive. |
solv-int/9907008 | Consider what happens when we exchange MATH and MATH in a term of either equation. If MATH or MATH, then MATH since MATH is alternating. Otherwise, we see that every factor MATH with MATH is taken to another such factor, except for the factor MATH or MATH, whichever is present. So each term in our sum is taken to the negative of a term from our sum; it follows that the sum is alternating under parity-preserving permutations. It follows that it must be a multiple of MATH . By degree considerations, it remains only to verify the constant, which we can do by considering the coefficient of largest degree in MATH, and applying induction. |
solv-int/9907008 | We need to integrate this over the variables MATH, and thus need to evaluate the determinant MATH where we take MATH to be the left endpoint of MATH, and MATH to be the right endpoint of MATH. In particular, we need to determine when there exists a function MATH with MATH . As in REF, we may use row operations to transform this to: MATH where we define MATH . We cannot quite apply REF , however, since the last column of our determinant is constant. However, we clearly have MATH, so we can eliminate that column, obtaining MATH . This, then, satisfies the hypotheses of REF ; there thus exist linear polynomials MATH such that MATH where we have set MATH . Differentiating twice and using the definition of MATH, we find: MATH . We can thus solve this for MATH; we find that MATH has the form MATH with MATH, MATH, and MATH. We observe that these conditions are, naturally, preserved by linear fractional transformations. In particular, by applying a suitable linear fractional transformation, we may insist that MATH be strictly cubic, and that both endpoints of MATH be finite (possibly equal). (The result may very well no longer be a matrix ensemble, but as we noted above, this does not affect any algebraic conclusions.) Now, consider how MATH and MATH must behave at MATH and MATH . Differentiating REF once and taking a limit MATH we find, since each MATH, MATH . But this is just MATH. If MATH, then we must have MATH. Then MATH . The only way this can happen is if MATH after all. Similarly, we have MATH. Suppose first that MATH. Then up to LFT, we may insist that MATH and MATH, and thus MATH. We thus have two possibilities. The first is that MATH has an additional zero, neither REF nor REF. In this case, integrating MATH and taking into account the constraints on MATH, we obtain MATH . Now, for MATH not to diverge at MATH, we must have MATH, and similarly MATH. But then MATH; it follows that MATH must be nonzero on MATH; in particular, MATH. The other possibility is that MATH has a double root, Without loss of generality at REF. Upon integrating MATH, we obtain MATH and find MATH, MATH. The possibilities for MATH then follow by LFT. The other case we must consider is MATH, and thus MATH, MATH. If MATH had a simple root in MATH, it would have two (Without loss of generality REF), and thus MATH would have the form MATH with MATH. But then the integral for MATH would diverge. Similarly, if MATH had a double root at REF, MATH would have the form MATH, which would diverge on one side of MATH. Thus either MATH has a pair of complex roots, or MATH. In the first case, a linear transformation takes the roots to MATH, and thus MATH . For MATH to be well-defined, we must have MATH. The other possiblity gives MATH; thus a linear transformation gives MATH . |
solv-int/9907008 | As for MATH, the issue is when MATH takes the form of an orthogonal ensemble. Applying an LFT as necessary, we may assume that MATH. Now differentiate with respect to MATH, divide by MATH, and take a limit as MATH. On the one hand, this operation takes orthogonal ensembles to orthogonal ensembles. On the other hand, we can then expand along the first column, finding that MATH must take the form of an orthogonal ensemble. By induction, we find that MATH must satisfy the constraints valid for MATH. Upon undoing the LFT, we obtain the desired ``only if" result. It remains to show that each of the above weight functions actually do work. We need only consider the following possibilities: MATH (on MATH, MATH, MATH, and MATH respectively) since the others are all images of these under LFTs. For MATH, observe that MATH this is true for MATH, and both sides have the same derivative. In particular, for each MATH, we have a polynomial MATH of degree MATH and a constant MATH with MATH . In particular, this must be true for MATH, and thus MATH as required. We thus find that we obtain a unitary ensemble with weight function proportional to MATH. Similarly, for MATH, we have MATH so MATH. For MATH, we find MATH . This allows us to solve for each MATH except MATH; we obtain MATH. Finally, for MATH, we have: MATH and MATH. |
solv-int/9907008 | The only tricky aspect of this case is that the determinant we must analyze is no longer of the form to which REF applies; to be precise, we need MATH to have orthogonal ensemble form. But this determinant is clearly equal to the determinant of the block matrix MATH . Adding the first column to the other columns, we can then apply REF , and argue as above. |
solv-int/9907008 | Consider the determinants associated to MATH and MATH. Up to cyclic shift, only one column differs between the two determinants, thus allowing us to express their sum as a determinant. When MATH is even, the `special' column takes the form MATH here MATH. Taking appropriate linear combinations, we obtain the determinant MATH . When MATH is odd, the special column takes the form MATH this leads (up to sign) to the MATH block determinant MATH . We first analyze the case MATH odd. In this case, the usual theory tells us that there exist polynomials MATH and MATH of degree at most MATH with MATH for all MATH, with MATH as above. Now, evaluating this at an endpoint of MATH, we find that the polynomials MATH must have a common root (possibly MATH). In particular, it follows that MATH must satisfy the conditions of REF . On the other hand, we find that MATH for each MATH; in particular, MATH must be a rational function. We therefore have the following possibilities to consider: MATH on MATH and MATH respectively. In the first case, we find that each MATH is a polynomial of degree MATH. In particular, MATH is a polynomial of degree MATH, contradicting the bound on MATH. In the second case, we observe that MATH for polynomials MATH of degree MATH. In particular, we find that MATH, implying, since MATH, that MATH . Since MATH, we have MATH, the only integral solution of which is MATH. In this case, the relevant degree bounds all hold, and thus the determinant is indeed of the correct form, giving MATH as required. For MATH even, we must have polynomials MATH of degree at most MATH with MATH where we write MATH for MATH. We can rewrite this as: MATH using the fact that MATH. For MATH, we conclude that MATH must satisfy the conditions of REF . Now, if the endpoints of MATH are different, then we find, since MATH at both endpoints, that each MATH at both endpoints. But this causes the polynomials to be linearly dependent, a contradiction. On the other hand, in the other cases, we know that MATH and MATH. In both cases, we obtain from the identity for MATH a differential equation for MATH. For MATH, no polynomial solution to the equation exists. For MATH, we can find an explicit power series solution to the equation, and find that a polynomial solution exists only when MATH is half-integral, when the solution has degree MATH. As above, this leaves only one possibility for MATH, namely MATH, as required. It remains to consider MATH. Here we can twice differentiate the equation MATH (with MATH and MATH linear) to deduce that MATH with MATH, MATH, and MATH. So up to LFT, MATH must have one of the forms MATH . (note that if we exchange MATH and MATH in the first case, we replace MATH with MATH, justifying our restriction on MATH.) In the first case, if MATH, the following: MATH must be a linear polynomial for suitable constants MATH and MATH respectively proportional to MATH and MATH (and thus MATH). We deduce therefore that MATH. But then we readily determine that only the empty interval satisfies the requirements. For MATH and MATH, there are not even appropriate choices for MATH and MATH (since both MATH and MATH are transcendental functions). Finally, for the third choice, we readily verify that decimation indeed works as required. |
solv-int/9907008 | Consider, for instance, MATH. Once we integrate along the largest, REFrd largest, etc. variables and do some simplification, the resulting matrix has columns MATH and MATH, with the last column given by MATH. In particular, we note that aside from the last, constant, column, the columns come in pairs, one the derivative of the other. Thus the determinant is essentially of the form considered in REF . In particular, it has a symplectic ensemble form if and only if the determinant MATH (of which our determinant is a derivative) has an orthogonal ensemble form. But by the proof of REF , this precisely what we needed to show. (The statement about the resulting weight functions is straightforward.) Similar arguments apply for the remaining equivalences. |
solv-int/9907008 | We will consider only the deductions from REF , as the other cases are similar. Let MATH be a single interval which includes an endpoint of the support of MATH and MATH. From the first statement in REF we see that the event of a sequence of eigenvalues from MATH not being contained in MATH occurs in three ways relative to the ensemble MATH: REF the eigenvalues from MATH and those from MATH are not contained in MATH; or REF one eigenvalue from MATH is contained in MATH and no eigenvalue from MATH is contained in MATH (note that the one eigenvalue must be either the largest (smallest) eigenvalue when MATH contains the right (left) hand end point); or REF one eigenvalue from MATH is contained in MATH and no eigenvalue from MATH is contained in MATH. This gives the first equation in REF . From the second statement in REF we see that the event of a sequence of eigenvalues from MATH containing MATH eigenvalues in MATH can occur in two ways relative to MATH: REF there are MATH eigenvalues from MATH in MATH; or REF there are MATH eigenvalues from MATH in MATH of which MATH are integrated out in forming even(MATH). This implies the second equation in REF . |
solv-int/9907008 | From the definitions of MATH and MATH we see that REF is proportional to MATH . The biorthogonal property allows the integrations required by REF to be computed to give REF . |
solv-int/9907008 | This is a simple consequence of the property CITE MATH . |
solv-int/9907008 | Because the polynomials MATH are monic we can add multiples of columns in REF to obtain MATH . Application of REF and the formula qdetMATH-MATHqdet-MATH gives REF with MATH as specified and formulas for MATH and MATH which are easily seen to be expressible in terms of MATH as stated. |
solv-int/9907009 | REF immediately yield REF are clear from REF yield REF, and REF prove REF. MATH and REF then yield REF - REF follows from REF. Finally, by REF, MATH is clearly meromorphic on MATH away from the poles MATH of MATH . By REF one concludes MATH and hence MATH is meromorphic on MATH . |
solv-int/9907009 | By REF - REF one infers for MATH . In connection with the MATH term in REF we used MATH . REF , and the dominated convergence theorem (for discrete measures). Together with MATH REF yields MATH and hence REF comparing REF. |
solv-int/9907009 | It suffices to combine REF, and REF. |
solv-int/9907009 | REF is a direct consequence of REF, and REF immediately implies REF. |
solv-int/9907009 | We only need to prove REF - REF. Starting from MATH which is equivalent to REF due to the recursion REF, one computes from REF, MATH . Thus, MATH and hence MATH where MATH is independent of MATH (but may depend on MATH and MATH). Next, combining REF and the NAME REF , one observes that MATH has an asymptotic expansion of the type MATH uniformly with respect to MATH as long as MATH varies in compact subsets of MATH . Inserting REF into REF and integrating with respect to MATH then contradicts the asymptotic expansion REF for MATH unless MATH . This proves REF is then obvious from REF. Using REF, one obtains MATH and hence REF. |
solv-int/9907009 | REF follow upon use of MATH, REF - REF, and REF. Since the numerator in REF is entire in MATH, and MATH again by REF, one concludes REF. By inspection, MATH satisfies REF (and hence REF) with MATH replaced by MATH and MATH by MATH. Consequently, MATH being a MATH potential implies that MATH is one as well. |
solv-int/9907009 | Our starting point will be REF and a careful case distinction taking into account whether or not MATH is a branch point, and distinguishing the cases MATH, MATH, and MATH. CASE: MATH and MATH: One computes from REF, MATH and upon comparison with MATH, MATH one concludes that no cancellations can occur in REF, proving the first statement in REF. CASE: MATH and MATH: Combining REF, and REF one computes MATH using MATH. From MATH one concludes again that no cancellations can occur in REF. Thus, the second statement in REF holds. The remainder of the proof requires a more refined argument, the basis of which will be derived next. Writing MATH a comparison of the powers MATH and MATH in REF yields MATH and MATH . Inserting REF into REF, a little algebra proves the basic identity MATH . CASE: MATH and MATH: Then REF yields MATH with MATH since MATH. In this case there is a cancellation in REF. For instance, choosing MATH one computes from REF, MATH since MATH . It remains to show that MATH does not vanish identically in MATH. Arguing by contradiction we assume MATH . Differentiating REF with respect to MATH and inserting the ensuing expression for MATH and the one for MATH from REF into REF then results in the contradiction MATH . Moreover, since MATH one concludes that precisely one factor of MATH cancels in REF. Hence, the third relation in REF holds. The case MATH in treated analogously. CASE: MATH, MATH, and MATH: Taking into account that MATH (using REF and MATH) is independent of MATH, REF yield MATH since MATH . Thus we infer again that precisely one factor of MATH cancels in REF. Hence, the fourth relation in REF is proved. CASE: MATH, MATH, and MATH (compare REF): One calculates as in REF, MATH since MATH . Next we show that MATH does not vanish identically in MATH. Arguing again by contradiction we suppose that MATH . Thus MATH for some constant MATH. Insertion of REF and its MATH-derivative into REF then again yields the contradiction MATH . Hence the last relation in REF holds in this case. CASE: MATH, MATH (compare REF): As in REF one obtains MATH since MATH . The remainder of the proof of REF is now just a special case of REF (with MATH). |
solv-int/9907012 | Using REF we obtain MATH which, by comparing coefficients in front of the vectors MATH, MATH, leads to REF . |
solv-int/9907012 | Let us rewrite REF in a backward form MATH which gives MATH where MATH is the MATH-th component of the vector MATH. Comparing REF we infer that the co-vectors MATH satisfy the backward NAME equations MATH where MATH and therefore (see REF ) also the forward NAME equations. Finally, since MATH, then the corresponding co-dimension MATH subspaces MATH and MATH of hyperplane lattices are co-parallel in the sense of REF . |
solv-int/9907012 | Assume a decomposition of MATH in the basis MATH, MATH where MATH . Using REF we obtain that MATH which, together with REF , concludes the proof. |
Subsets and Splits
No community queries yet
The top public SQL queries from the community will appear here once available.