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cs/9909019 | Soundness is direct from REF . For completeness, suppose that MATH is not a theorem of Section . Since Section is complete for the class of all hypercubes, there exists a hypercube MATH, where MATH is a singleton, an interpretation MATH on MATH, and a world MATH such that MATH. We show that it is possible to construct a homogeneous broadcast environment MATH whose decomposition into a union of Cartesian products contains MATH as one of its components. Indeed MATH will be the component consisting of all the traces of length one, that is, the component characterizing the initial state of knowledge of the agents. We define the environment MATH as follows. For each agent MATH, we take the both the set of external actions MATH and the set of internal actions MATH to be the set MATH. Thus, the set of actions of each agent is also a singleton, viz MATH. The components of the environment are as follows: CASE: The set of states MATH where MATH. Thus, the set MATH of instantaneous private states of agent MATH is exactly the set of local states MATH of agent MATH in MATH. CASE: All states are initial, that is, MATH. CASE: Since the set actions MATH of agent MATH, the environment, is a singleton, the protocol of the environment is the unique function MATH. CASE: The transition function MATH is defined by MATH for (the unique) joint action MATH and state MATH. (Thus, similarly, the local transition functions MATH satisfy MATH for (the unique) joint external action MATH, (the unique) internal action MATH and private state MATH.) CASE: The definition of the observation function MATH is determined by the fact that MATH is a broadcast environment, that is, MATH for each MATH. CASE: The valuation MATH is defined by MATH. This is a homogeneous broadcast environment by construction. It is now straightforward to establish that for every joint perfect recall protocol MATH, the connected component of MATH consisting of all traces of length one is isomorphic to MATH. (We remark that our choice of action sets and transition function above are not actually relevant to this conclusion.) |
cs/9909019 | Similar to the proof of REF . Note that the construction of this proof uses only the initial component of the frame. The valuation of the environment may be chosen to operate as required on this initial component. |
math-ph/9909002 | Induction on MATH gives existence and uniqueness of MATH: For MATH, REF is simply MATH. Once MATH has been determined for all MATH with MATH, REF is solved in the form MATH . The right hand side of REF solves REF. |
math-ph/9909002 | When REF is inserted to replace MATH, MATH takes the form of a polymer partition function, with the nonempty subsets of MATH as polymers and disjointness as the compatibility relation. REF is the standard polymer formula for the logarithm of the partition function CITE. REF follows by noting that for all MATH, the connectedness condition in the function MATH implies that after differentiation, some factors MATH remain, so that evaluating at zero picks out the term MATH from the sum (see REF). |
math-ph/9909002 | See REF. |
math-ph/9909002 | Let MATH. By REF and the binomial theorem, MATH . |
math-ph/9909002 | Expand and permute. |
math-ph/9909002 | The statement about minors is trivial; REF is NAME 's inequality (see for example, REF). If MATH and MATH, then MATH is also a NAME matrix, with NAME constant MATH. |
math-ph/9909002 | All eigenvalues of MATH are nonnegative, so there is a real matrix MATH, with MATH such that MATH. If MATH is the MATH row vector of MATH, this means MATH, thus in particular MATH. The NAME inequality implies MATH, so REF holds. |
math-ph/9909002 | By REF , MATH is a NAME matrix with NAME constant MATH. Let MATH be its NAME representation. By REF, MATH . As in the proof of REF , the NAME bound implies the statement. |
math-ph/9909002 | It is obvious that the diagonal elements remain unchanged and that the matrix remains symmetric. By permuting the rows and columns of MATH with the same permutation, which amounts to a change of basis and therefore does not change positivity properties, we can assume that MATH for some MATH, and thus get, with MATH, MATH . The blockdiagonal matrix inherits positivity from MATH. Thus MATH is a convex combination of two positive matrices, hence positive. |
math-ph/9909002 | If we write the monomials as MATH with the antisymmetrization operator MATH becomes the field - independent term of the special case of MATH where MATH consists only of the term MATH, with coefficient function MATH . Integrating over the MATH and MATH variables only removes the delta functions; in particular MATH . We now consider the contribution MATH of one tree MATH in the sum in REF to the absolute value of the connected correlation MATH. The only differences to REF are that CASE: there is no sum over MATH and MATH. CASE: because we now consider the field - independent part (MATH), all fields are integrated over; this picks out the term MATH and MATH in REF . The second condition implies that MATH vanishes unless MATH, which we assume from now on. Then MATH . Here we denoted thoses integration variables on which MATH factors depend by MATH, the others by MATH. By definition, the lines in the tree can only connect distinct MATH and MATH. By REF, MATH . By REF, the factor MATH combines with the other powers of MATH to MATH. By REF, the pseudodistances appearing in the sum are all of the form MATH, so MATH . We can now bound the integral by MATH using REF and then sum over all trees. Again, the only dependence on the tree left is in the incidence numbers. As in the proof of REF , we get factors MATH or MATH, depending on how we do the bounds. Thus REF hold. |
math-ph/9909002 | Induction on MATH, with REF, and MATH for all MATH, as the inductive hypotheses. The statement for MATH is REF, with MATH. MATH . In the remainder term, the sum over MATH includes a sum over MATH. Let MATH, and MATH. Then MATH by REF . Now apply REF to MATH. The second summand in REF gives the new remainder term MATH. The first summand in REF is MATH . Because MATH, MATH, so MATH. Thus by REF, MATH . By the inductive REF , MATH, hence does not depend on MATH, so its exponential can be taken out of the integral. |
math-ph/9909003 | Note that, since MATH is arbitrary, it suffices to establish the asserted continuity for MATH. Consider first the case where MATH is such that MATH and, to simplify notation, set MATH. Let MATH be a decreasing sequence in MATH which converges to MATH. Then MATH is an increasing family of wedges and REF implies that the sequence MATH converges strongly to the modular conjugation MATH of MATH, where MATH . In view of the specific geometric action of products of the modular conjugations on the wedge algebras, given above, and the fact that MATH, one has for any fixed MATH provided MATH is sufficiently large. As MATH is weakly closed, one can proceed from this inclusion to MATH and thence to MATH, so that MATH . But this implies MATH and hence MATH. Next, let MATH be an increasing sequence in MATH converging to MATH. Note that MATH, so that, in view of the NAME duality of the net MATH, one has MATH. Moreover, MATH is an increasing family of wedges. The same argument presented in the first paragraph therefore yields the strong convergence of MATH to MATH. Finally, let MATH be an arbitrary sequence in MATH converging to MATH. Since any such sequence contains monotone subsequences, for which the strong convergence of the corresponding modular conjugations has already been established, and since the respective limits coincide, the continuity of the operators MATH at MATH follows for the special choice of MATH. For arbitrary MATH, pick a MATH such that MATH and MATH. Then MATH and MATH are continuous in MATH. According to relation REF one has MATH so the strong continuity of MATH follows from the continuity properties of the antiunitary involutions appearing on the right - hand side of this equality. |
math-ph/9909003 | Let MATH. As MATH is product of two such antiunitary involutions, it is unitary. Moreover, one has MATH and similarly MATH. So one obtains, for MATH, MATH where in the second equality relation REF has been used. Consequently, one has MATH . Similarly, one finds MATH . From these relations one sees, in particular, that for MATH and MATH, MATH . Since MATH is thus a homomorphism on the subgroup of the rationals and is continuous on MATH according to the preceding lemma, it is a continuous homomorphism on MATH. |
math-ph/9909003 | The fact that the rotations in the time - zero hyperplane are induced by unitary operators in MATH will be employed. If MATH is a rotation by MATH about REF-axis, say, one obtains, first of all, from REF the equalities MATH since MATH. Next, one notes that the unitary operators MATH induce the same time translation MATH on the net. So the differences MATH map, by their adjoint action, each wedge algebra MATH, MATH, onto itself and leave the vector MATH fixed. These differences therefore commute with all modular conjugations MATH, MATH, CITE and are thus contained in the center of MATH. In particular, MATH for some central element MATH. Finally, one has MATH since MATH. Putting these three facts together, one arrives at the following relations in MATH . Thus MATH, and in a similar way one proves that MATH. |
math-ph/9909003 | Consider, for example, the operator MATH. It leaves MATH invariant and satisfies MATH since MATH for all translations MATH along REF - axis. But this implies that MATH commutes with the modular conjugations MATH and hence with MATH, MATH. Thus MATH commutes in particular with MATH and since MATH according to REF , it also commutes with MATH. By the same argument one can establish the commutativity of the remaining unitaries. |
math-ph/9909013 | Since MATH is infinite, there is a properly infinite projection MATH CITE. Since MATH is properly infinite, we may apply the halving lemma CITE repeatedly to obtain a countably infinite family MATH of mutually orthogonal projections such that MATH for all MATH and MATH. (Halve MATH as MATH; then halve MATH as MATH, and so on. Now replace MATH by MATH; compare CITE.) Let MATH. For each MATH, let MATH denote the partial isometry with initial space MATH and final space MATH. By the same reasoning, there is a countable family MATH of mutually orthogonal projections in MATH and partial isometries MATH with MATH and MATH. For each MATH, let MATH and let MATH . Define MATH and MATH as in the discussion preceding this proposition, let MATH be the corresponding NAME operator, and let the unit vector MATH be the MATH eigenvector for MATH. Now, let MATH be any unit vector in MATH. Since MATH, we have MATH in the strong-operator topology. Similarly, MATH in the strong-operator topology. Therefore if we let MATH we have MATH . Note that the inner product MATH, and thus MATH is a unit vector for all MATH. Since MATH, it suffices to observe that each MATH is NAME correlated for MATH. Recall that MATH, and thus MATH. A simple calculation then reveals that MATH . |
math-ph/9909013 | From CITE, there is a normal state MATH of MATH such that MATH. But since all normal states are in the (norm) closed convex hull of vector states CITE, and since MATH is norm continuous and convex, there is a vector MATH such that MATH. By the continuity of MATH (on MATH), there is an open neighborhood MATH of MATH in MATH such that MATH for all MATH. Since MATH is cyclic for MATH, there is a MATH such that MATH. Thus, MATH which entails that MATH is a nonseparable state for MATH. This, by the preceding lemma, entails that MATH is nonseparable. |
math-ph/9909013 | Let MATH be precompact open subsets of MATH such that MATH, and such that MATH for some neighborhood MATH of the origin. In an irreducible vacuum representation MATH, local algebras are of infinite type CITE, and since MATH, the NAME property holds for MATH CITE. If MATH is any representation in the local quasiequivalence class of MATH, these properties hold for MATH as well. Thus, we may apply REF to conclude that the set of vectors inducing NAME correlated states for MATH is dense in MATH. Finally, note that any state NAME correlated for MATH is NAME correlated for MATH. |
math-ph/9909013 | The regular diamonds (in the sense of CITE) form a basis for the topology on MATH. Thus, we may choose regular diamonds MATH such that MATH and MATH. The nonfiniteness of the local algebras MATH is established in CITE, and the split property for these algebras is established in CITE. Since the split property entails the NAME property, it follows from REF that the set of vectors inducing NAME correlated states for MATH [and thereby NAME correlated for MATH] is dense in MATH. |
math-ph/9909018 | For the NAME domain MATH of MATH, we observe that MATH . But the centroid of MATH is REF. Thus MATH . The lemma follows by decomposing MATH into the MATH affine copies of MATH, one for each site, and adding the inequalities obtained from REF . |
math-ph/9909022 | The MATH-mapping is a unitary mapping conserving spectral invariants, that is, spectrum and spectral multiplicities, compare CITE. Hence all the elements of the orbit MATH are density matrices with the same spectra and multiplicities. The spectral resolution of any density matrix MATH of the same spectral invariants as MATH in REF has the form: MATH with equal dimensions of MATH and MATH. Then there is a unitary operator MATH mapping all the MATH's onto the corresponding MATH's for all MATH. For example, one can choose orthonormal bases MATH, respectively, MATH in MATH containing subbases, for all MATH, of MATH, and MATH, respectively, to order them in accordance with orderings of MATH's, that is, so that MATH and define MATH by the formula MATH . Then MATH, what proves the lemma. |
math-ph/9909022 | We shall prove that the NAME subspace MATH has a topological complement CITE in MATH , that is, MATH the topological direct sum with a NAME subspace MATH of MATH. Let MATH be expressed in the form REF. We shall use also the projection MATH corresponding to the eigenvalue MATH = REF, hence always MATH. Let MATH be a projection of MATH onto MATH defined by the strongly convergent series. One has MATH hence the projection MATH is continuous, what implies CITE the complementability of MATH. The NAME group MATH can be modeled (as a manifold) by its NAME algebra MATH via the inverse of the exponential mapping CITE, and the subgroup MATH is modeled via the same mapping by the complementable subspace MATH. This gives the result (CITE, CITE). |
math-ph/9909022 | MATH is a NAME of MATH , and MATH is a linear isometry (hence homeomorphism) of MATH (MATH ) (with the norm MATH ) onto MATH, what follows directly from definitions, compare Notes REF. This gives the first assertion. The second one follows because of complementability of the space MATH, MATH, and the inverse mapping of the mapping MATH, if restricted to an open neighbourhood of the zero point of MATH, can be chosen as a chart of the manifold MATH . CASE: MATH : If MATH, then REF shows that also MATH for any MATH, since MATH is an ideal in MATH . The application of REF to MATH shows that MATH. Also, MATH leaves MATH REF MATH pointwise invariant. Hence MATH. CASE: MATH : It follows now that for MATH the set MATH is a subset of MATH . Since MATH is closed in the norm-topology of MATH and on the subset MATH the trace-topology determined by MATH is finer than the topology of MATH determined by MATH , MATH, it follows that MATH REF is closed also in trace-topology, that is, REF . CASE: MATH : Let MATH. Let MATH be an infinite orthonormal set in MATH such that MATH, compare REF. Let us define MATH by the formula (in the NAME notation, CITE) MATH . We have MATH for any bounded real sequence MATH, but for some choices of MATH (for example . MATH) one has MATH. This proves that MATH. Let us make now a technical remark providing an alternative proof of the last statement, as well as a device to further work: Let us choose in REF MATH, where MATH for MATH, but still MATH. Such a choice of strictly positive divergent sequence MATH, for any given MATH, is always possible. Then MATH. Let us now calculate MATH-according to REF : MATH . Due to divergence of MATH, it is clear that the result MATH diverges for MATH, that is, we can obtain in this way at the best an unbounded operator. This shows that our MATH, although it is still in MATH . This is another proof of the inequality MATH, because MATH is MATH - continuous. MATH CASE: Since the sequence MATH converges strongly to MATH, it is seen that MATH (MATH ) (considered as a subspace of MATH) is strongly dense in MATH. This proves REF . CASE: MATH : Let us choose MATH. The preceding considerations also show that MATH REF is not closed in MATH if MATH; namely, according to REF , and REF , one can choose MATH such that the sequence MATH converges to zero. This means that the sequence MATH converges to MATH. CASE: MATH : The restriction of the projection MATH is continuous also in the trace-norm topology, what follows from continuity of MATH in that topology: For positive operators MATH, all MATH, hence MATH and the continuity of MATH follows. The equivalence of the norms MATH and MATH on MATH REF in the case of MATH-can be shown as follows: Let MATH, and from REF and the definition of the norm MATH we obtain MATH where the sum is taken over a finite index set. The opposite inequality is obtained by the known property of the trace-norm: MATH since MATH, and MATH = REF. This fact, and the derived implications of finite dimensionality of MATH give the validity of REF . It is clear that REF cannot be true if REF were not valid. CASE: If MATH, then the NAME MATH REF is a NAME space, compare REF , hence MATH REF is reflexive. |
math-ph/9909022 | Any density matrix MATH is approximated in MATH by finite dimensional ones, what is seen, for example, from its spectral resolution: MATH . |
math-ph/9909022 | The first three properties are immediate consequences of REF , compare also REF. The validity of REF follows immediately from REF and from REF for such functions MATH which have form of polynomials in the specific type of functions MATH, REF. For general MATH one can prove REF directly as follows: Let us first express MATH according to REF, MATH where the second derivatives in any point MATH are symmetric bilinear MATH -continuous functions on MATH . Hence, the linear mapping MATH can be (and is here) considered as an element of MATH . We need to calculate MATH . With a help of the notation REF and of the above derived formula for MATH we obtain MATH . From the symmetry of second derivatives, and from validity of NAME identity for commutators of operators in MATH , we obtain the result. |
math-ph/9909022 | REF shows that the kernel of the operator MATH contains the kernel of MATH , and also is bounded. Hence, it is defined as bounded linear functional on MATH, that is, as an element of MATH. |
math-ph/9909022 | Unique solvability of REF on each interval MATH on which the function MATH is uniformly bounded follows from general theory of differential equations in NAME spaces, compare CITE. NAME and REF can be proved, for example, by the method of the proof of CITE using the NAME expansion, since MATH is norm-continuous. Finally, REF can be verified by differentiation and by the uniqueness of the local flow MATH of the vector field MATH . |
math-ph/9909022 | The result follows from the relation REF showing that MATH can be realized by the MATH-action, and MATH consists of the MATH-orbits MATH , MATH. |
math-ph/9909022 | The equivalence of the norms MATH , and MATH , as well as the completeness of MATH REF was proved in REF . To prove equivalence of norms MATH , and MATH , let us write MATH, with MATH, as before. Let c= REF := MATH. Then MATH, where the degeneracy of MATH equals MATH for MATH, and the number MATH of mutually different eigenvalues MATH of MATH is finite. This proves equivalence of the norm MATH with MATH , hence also their equivalence with MATH , since always MATH. We have further MATH. On the other hand, since MATH, one has MATH . These inequalities together with the previously proved equivalences show also the desired equivalence of MATH . This proves also nondegeneracy of MATH; its analytic dependence on the point MATH of the orbit MATH can be proved from its dependence on elements of the group MATH acting on MATH . The explicit form of MATH shows, after inserting into it MATH, and MATH, that it can be expressed by our NAME bracket REF: we obtain REF, according to REF. The closedness MATH follows from the proved NAME identity for the NAME brackets REF . The mapping MATH is an isomorphism, what is a consequence of the proved equivalence of topologies on MATH (MATH ), of the surjective property of the mapping MATH, as well as of the reflexivity of the NAME space (MATH (MATH ); MATH ). This proves that MATH is strongly nondegenerate. |
math-ph/9909022 | REF is proved in CITE. It implies, that MATH, where the products with MATH are considered as the corresponding (unique) bounded extensions in MATH . From these facts we see, that, for the considered MATH , the expectation MATH exists, compare REF, what in turn implies MATH, that is, REF . With MATH as in REF , MATH and MATH are defined on the domain MATH, and the range of MATH is in MATH; hence, both products are densely defined finite - range operators, the first one in MATH . The last REF is valid due the fact that the set of analytic vectors of MATH is invariant also with respect to the action of the operator MATH. |
math-ph/9909022 | CASE: From REF , and the definition in REF , as well as from the corresponding definitions of MATH, with the help of spectral decompositions of MATH, the first assertion of REF follows immediately. It is sufficient to prove the density for MATH. Density of the set MATH in MATH will be proved from its density in MATH in MATH - topology, because MATH is dense in MATH in this topology. But it suffices to prove arbitrary close approximatebility of one - dimensional projections by such projections from MATH , that is, by MATH. Since MATH is linear and dense in MATH , unit vectors in MATH are dense in unit sphere of MATH (by triangle inequality). Then, for two unit vectors MATH, we can use: MATH where the second equation is proved by calculation of eigenvalues of MATH is selfadjoint with trace zero, and range two - dimensional, hence its two eigenvalues are opposite reals MATH; then, by calculating MATH one obtains the desired equation. This easily leads to a proof of density of MATH in MATH . The density of MATH in MATH follows then by a use of convexity of both sets. CASE: This is a consequence of REF , as well as of our constructions in REF, see esp. CASE: For any MATH, it is MATH. Due to REF we know, that the linear mapping MATH is continuous and can be uniquely extended to the whole MATH (MATH), the range of the extended mapping being the whole MATH . This leads eventually to validity of the statement. |
math-ph/9909022 | There is a projection MATH such that MATH (MATH might be chosen to be the range projection of MATH). Then MATH, hence MATH . The expression consists of a sum of elements of MATH with ranges contained in the NAME subspace determined by the orthogonal projection MATH, where we used the spectral projections MATH of MATH . Hence MATH. Due to unitarity of the transformation of MATH , we have also MATH. This proves the assertion. |
math-ph/9909022 | The proof is contained in the text following REF . |
math-ph/9909022 | Due to the continuity of MATH, and because MATH is a NAME space, MATH is a closed (hence NAME) subgroup of MATH. This implies that MATH can be considered as a bijective mapping of the analytic manifold MATH onto the orbit MATH . This mapping is analytic, and its differential (that is, the tangent map) maps the tangent space MATH onto a finite - dimensional subspace of MATH , which is complementable. This fact together with the MATH - invariance of MATH implies, CITE, that MATH is an embedded submanifold of MATH . The second, and the third assertions are implied by the considerations developed in the Subsection REF, since the vector - fields MATH := MATH REF generate the flows MATH which were used to formation of the orbit MATH . The existence of a dense subset of MATH of analytic elements lying in MATH with respect to the norm - topology of MATH in MATH implies the fourth assertion. Differentiation of these flows demonstrates also validity of the last statement. |
math-ph/9909022 | By a use of the identity MATH as well as of the relation MATH compare REF, and REF , we obtain MATH . After a subsequent application of REF with MATH, the preceding calculation gives the result. |
math-ph/9909022 | The second equation in REF is just the first equation of REF. The unrestricted NAME bracket on MATH occurring on the left side of REF is equal, according to REF, to the derivative of f along to the vector field REF at each point MATH. This implies, that the derivative MATH in any point MATH of an arbitrary function only depends on its restriction MATH to MATH . One has MATH on MATH , and the last derivative is expressed by the NAME bracket REF on MATH . This proves the first equation. |
math-ph/9909022 | The mapping MATH is affine, and MATH is convex, since MATH REF is convex and MATH is affine. Hence Ran REF is convex, and conv-MATH. One can see from the definitions that conv-MATH, and that MATH is norm - dense in MATH. The MATH - invariance follows from REF, and from the MATH - invariance of MATH . Let MATH. The closedness of Ran REF can be proved by construction of a projection - valued measure on MATH representing the commutative group (linear space) MATH , respectively, the commutative algebra of ``classical observables" MATH (generated by the functions MATH), CITE. The support of this measure is identical with Ran(MATH ), hence Ran REF is closed. |
math-ph/9909022 | The flow MATH leaves the MATH - orbits invariant, since it is a NAME flow and the MATH - orbits are symplectic leaves of the NAME ( - NAME) structure on MATH, CITE. Hence MATH for all MATH, and MATH. The vectors MATH form a right - invariant vector field on MATH for each MATH, and MATH are values of MATH - dependent vector fields (for any MATH) on G. Their infinite differentiability follows from the properties of MATH. The existence and uniqueness of the solution MATH of REF fulfilling REF are then consequences of the theory of ordinary differential equations on manifolds, compare CITE. Let MATH. The derivative of MATH at MATH equals, according to REF, to MATH, what can be rewritten in the form of NAME bracket for MATH: MATH . This, together with REF, proves REF. The generator MATH generates, on the other hand, a NAME flow MATH on MATH . Since MATH, REF is proved by REF, and REF. |
math-ph/9909022 | The algebraic properties of MATH , and also the MATH - invariance of MATH are consequences of REF, and of the cocycle identities REF. Relation REF is a consequence of REF, and of the relation REF. The MATH - continuity, that is .that for all MATH, MATH the functions MATH are continuous, and MATH, follows from REF, the continuity properties of MATH , MATH, and MATH, as well as from the NAME dominated convergence theorem. |
math-ph/9909022 | Recall that (compare REF ) MATH and the pull - back has trivial kernel in MATH. Since Ran REF consisting of MATH - orbits is dense in MATH , a continuous function MATH identically vanishes on each orbit lying in MATH , hence vanishes on MATH , iff there vanishes MATH. It follows that the association MATH is a bijection. It is linear in MATH, and REF , and REF show the conservation of the NAME brackets, hence the validity of REF. It remains to prove, that the ``deformed" flows MATH are generated by MATH. Let MATH. Then MATH where we define MATH, the dual mapping of the inner differentiation of the NAME algebra, MATH. This proves the proposition. |
math-ph/9909022 | Due to the simplicity of MATH , the abelian subalgebra MATH of MATH coincides with its center MATH. The center MATH is invariant with respect to any MATH-automorphism of MATH , hence the restriction of MATH to MATH is also an automorphism. The NAME - NAME theory of commutative MATH-algebras (compare CITE, and also REF ) implies that the MATH - automorphisms of MATH are in a bijective correspondence with homeomorphisms MATH-of MATH onto itself. CASE: Let MATH-Aut-MATH. Then the corresponding homeomorphism MATH is defined by: MATH . Let an arbitrary MATH be considered as a constant function - an element MATH, where MATH. Then the value MATH of MATH will be denoted by MATH . The pointwise character of algebraic operations in MATH-implies that in this way defined MATH-is a mapping to MATH - morphisms of MATH into itself. We shall show that MATH is a nonzero morphism (hence an isomorphism, due to simplicity of MATH ) for any MATH. A general element MATH of MATH is uniformly approximated by elements of the form MATH hence also by the elements of the form MATH, since MATH is a MATH-automorphism of MATH . For the elements of the form REF one has MATH . For a zero morphism MATH it would be MATH-for all MATH, what cannot happen, since both MATH , and MATH are unital. It follows that MATH-Aut-MATH. REF implies then REF due to continuity of all the MATH, as well as of MATH: MATH . The continuity of MATH follows from the continuity of each function MATH. CASE: Let us now have given any homeomorphism MATH of the NAME compact MATH onto itself, as well as an arbitrary strongly continuous family MATH-Aut-MATH. Let us define the mappings MATH, and MATH as follows: MATH . The continuity and the morphism properties of the given MATH,and MATH show that both the mappings MATH, and MATH introduced in REF are MATH - automorphisms of MATH . Hence REF determines an automorphism MATH as the composition of these two automorphisms: MATH-Aut-MATH. |
math-ph/9909022 | For any MATH, and any MATH, it is MATH iff MATH, hence according to REF, MATH is injective. For any MATH and MATH, we have MATH (since MATH), and MATH, hence MATH is bijective. Let MATH. For MATH the identity MATH implies the bicontinuity of MATH. For any MATH, and for MATH, we have MATH and we can see, compare CITE, that the mapping MATH is a complex analytic function. |
math-ph/9909022 | The mapping MATH defined in REF for MATH has the form MATH, where MATH is represented (compare REF ) by a bounded operator. If a vector MATH corresponds to the curve MATH then the corresponding operator is MATH . By a use of REF , one obtains MATH. Inserting these expressions to REF, we obtain the relation MATH what is identical with the result of the corresponding insertions from REF into REF. |
math-ph/9909022 | Conservation of d means conservation of the ``transition probabilities" MATH, MATH; this means also conservation of the metric tensor MATH. According to the NAME theorem there is unitary or antiunitary bijection MATH, as stated in the proposition. But the symplectic form is invariant with respect to unitary transformation, as was shown in REF . The last part of the proposition is a consequence of the fact that antiunitary mappings u change the value of the scalar product in MATH to its complex conjugate: MATH. For more details compare CITE. |
math-ph/9909022 | The differential MATH can be represented, according to considerations in Subsection REF, on its domain by the bounded operator MATH and its vanishing implies commutativity of MATH with MATH, that is, invariance of the one - dimensional subspace x with respect to the action of MATH. For proof of the converse, the arguments go in the reversed order. |
math-ph/9909022 | CASE: The proof of integrability trivially follows from REF , since the integral curves of the Hamiltonian vector field corresponding to MATH leave all MATH invariant. CASE: The second assertion follows from REF , and from its proof. A more intuitive argument is seen from the Momentum mapping MATH restricted to any MATH with a help of REF: In view of REF, the MATH - th component of MATH can be expressed as: MATH what proves bijection of MATH onto MATH. CASE: The (densely defined) vector fields MATH corresponding to the basis MATH of MATH form a basis of MATH for any MATH-for all MATH. These vector fields are proportional to the Hamiltonian vector fields corresponding to MATH for the selfadjoint generators MATH of the representation MATH. The vector field MATH is identical zero. According to REF, and REF, one has MATH respectively, from REF one has MATH . The NAME - NAME symplectic form on a MATH orbits through MATH has the form REF: MATH . From the CCR REF one has MATH and for the remaining indices MATH. For the NAME - NAME form we have MATH what corresponds to REF in accordance with REF ; this proves the symplectomorphism property of MATH . The commutation relations REF show nondegeneracy of the restricted form MATH :=MATH-for all relevant MATH . It remains to prove that MATH maps all MATH onto a unique orbit. The basis MATH determines global coordinates on the dual MATH,MATH being the coordinates of MATH in the dual basis. It is clear from REF that on any MATH there is a point MATH such that MATH. The coordinates of other points on those orbits are then MATH, and the remaining coordinate MATH, REF, hence it is constant on the orbit and of equal value on all orbits, that is, on all the images MATH. This proves the last statement. |
math-ph/9909022 | The analytic domain MATH is a common invariant domain for all operators from MATH. According to NAME 's theorem (compare CITE) it suffices to prove essential selfadjointness of the operator MATH, what is sum of squares of a basis of MATH. We choose here the basis consisting of the generators MATH of MATH, and of all their symmetrized products MATH. Then MATH where we used notation from REF, and the CCR REF. From the known properties of the Hamiltonians MATH of independent linear oscillators, we conclude (with a help of, for example, CITE on operators on tensor products of NAME spaces) that MATH is essentially selfadjoint. The NAME 's theorem states now integrability of MATH onto a unitary representation. NAME and uniqueness now easily follow. |
math-ph/9909022 | MATH is MATH - invariant, MATH, hence also MATH for all MATH. The function MATH-from REF leads to the identity MATH what is a consequence of REF , and REF . Hence we have MATH. Differentiation of REF with a help of REF , and of the group - representation property of MATH gives: MATH what is the relation REF with MATH from REF . Insertion of REF into REF gives REF , what proves that the function MATH from REF solves REF . Let MATH be some global solution of REF with MATH, and fulfilling MATH. Then it satisfies REF , what follows from the differentiation of MATH with a help of REF , and REF . Consequently, this MATH-satisfies also REF . Conversely, let MATH be some global solution of REF . Again by differentiation of MATH, one obtains, as above, the identity (compare also Notation REF) MATH with MATH. This means that each global solution of REF fulfills also REF . |
math-ph/9909022 | The NAME bracket MATH is a norm - continuous bilinear form of the variables MATH, hence (with the above mentioned identification) the linear functionals: MATH, are norm continuous on MATH , representing some vectors MATH. Let MATH be an arbitrary element. Then the (bounded linear) function MATH is smooth, MATH, and its differential MATH (in any point MATH) is identified with MATH itself. Hence, the mapping MATH is onto. Since the functions MATH are smooth (in the sense of the underlying norm - topology), all the functions MATH are also smooth. This, due the NAME property of derivatives, implies smoothness of MATH . |
math/9909002 | Let MATH be the ball in MATH with centre MATH and radius MATH. Take a smooth function MATH with MATH . Such a function may be obtained by smoothing the function MATH where MATH for MATH, MATH for MATH and MATH for MATH. The form MATH has compact support, so MATH. We want to show that as MATH these forms converge in MATH to MATH. Consider MATH . Since MATH, and MATH, then MATH and hence MATH . As MATH vanishes outside MATH and is identically MATH on MATH, we have MATH . But MATH and since MATH, the right hand side tends to zero as MATH, thus so does the left hand side. From REF we see that MATH converges in MATH to MATH. Now MATH vanishes on MATH and outside MATH, and on the annulus in between we have the estimates MATH and MATH. Thus MATH . This again converges to zero as MATH since MATH. We thus have convergence of both terms on the right hand side of REF and consequently MATH converges in MATH to MATH. Hence MATH lies in the closure of MATH and its MATH-cohomology class vanishes. |
math/9909002 | Since MATH is covariant constant, MATH is constant. Thus from REF , the linear growth of MATH implies that the map MATH defined by MATH is zero. By NAME theory this means that if MATH is a MATH harmonic MATH-form, then the MATH harmonic MATH-form MATH vanishes for MATH. But by linear algebra, the map MATH is an isomorphism, hence the only non-zero harmonic forms occur when MATH. |
math/9909002 | The group MATH acts unitarily on the NAME space of MATH harmonic forms. This may possibly be infinite-dimensional. Nevertheless, one knows that for a unitary representation the space of analytic vectors - the ones for which MATH is analytic - is dense CITE. If MATH is a MATH harmonic form which lies in this subspace, then it has a well-defined MATH . NAME derivative MATH which is also a harmonic form. As in REF , we write MATH . We now proceed as before: MATH converges in MATH to MATH, and from the linear growth of MATH, MATH converges in MATH to zero. Thus from REF , MATH converges to MATH, whose MATH-cohomology class is therefore zero. Hence MATH acts trivially on a dense subspace of MATH harmonic forms and by continuity is trivial on the whole space. |
math/9909002 | The three MATH-forms MATH define by exterior multiplication three commuting operators MATH on the algebra of differential forms. There are three adjoints MATH, and it is a matter of linear algebra to show that these satisfy the following commutation relations: MATH and similar ones by cyclic permutation. Here MATH is the standard basis of the NAME algebra MATH satisfying MATH etc. Its action is induced from the action of the unit quaternions on the exterior algebra. (As noted by CITE, MATH and MATH generate an action of the NAME algebra MATH). From REF , each MATH harmonic form MATH lies in the middle dimension MATH. But MATH and MATH commute with the Laplacian, and map to MATH and MATH forms respectively, so we must have for each MATH . But then REF implies that MATH. For a complex structure MATH corresponding to MATH, the eigenspaces of MATH are the forms of type MATH with eigenvalue MATH, so if MATH, then MATH. The condition MATH is the statement that the form is primitive, so we see that MATH is of type MATH and primitive with respect to all complex structures. In Riemannian terms this implies that MATH is anti-self-dual if MATH is odd and self-dual if MATH is even. |
math/9909002 | Let MATH be a Killing vector field on MATH generated by the action of MATH. Then MATH is tangential to MATH. Let MATH be the corresponding vector field on the quotient MATH, then the metric on MATH is the induced inner product on the horizontal space in MATH: the orthogonal complement of the tangent space of the orbit of MATH. Thus at MATH . The distance MATH between points MATH is the length in MATH of the horizontal lift from MATH to MATH of a geodesic in MATH, and this is greater than or equal to the straight line distance between MATH and MATH, thus MATH . But the vector field MATH is defined by a group of affine transformations of a flat space and so is of the form MATH which has linear growth, so from REF , MATH and using REF , MATH . |
math/9909006 | Without any loss of generality we assume that MATH and MATH . For MATH, MATH holds for any MATH, MATH . Therefore if MATH, then MATH and MATH . Substituting MATH for MATH into REF , we obtain that MATH . The assertion now follows. |
math/9909008 | This follows from REF (because the NAME map MATH is an isomorphism, provided by the smoothness of MATH). Actually, since MATH is smooth, and the intersection homology group is independent on the stratification REF , all the intersection homology groups are the same and equal to the usual homology. |
math/9909008 | CASE: Let MATH and MATH, where MATH. From the properties of the intersection see, for example, REF , page REF, one obtains that MATH . Then the sum of the following two contributions in MATH is zero: MATH . REF follows from the definition of MATH and MATH, including the sign MATH when MATH. |
math/9909008 | The result follows from the definition of the intersection MATH. |
math/9909008 | Use the definition of MATH and MATH and II. REF |
math/9909008 | In a neighbourhood of an intersection point MATH has a product structure MATH where MATH is a ball (MATH a half ball) in MATH, and MATH is a real REF-disc transversal to MATH. We denote the boundary of MATH by MATH. Let MATH-small open disc of radius MATH and origin MATH; that is, MATH. Then we have the diffeomorphism MATH. Since MATH, where MATH, we get: MATH . |
math/9909008 | The contribution MATH is clear. Next we want to determine the coefficients of the simplices which lie in MATH. It is enough to work modulo MATH, where the union is over all MATH, since it intersects MATH in codimension REF and its inverse image intersects MATH in codimension REF. Consider the composition: MATH . Again, the union MATH is over all MATH. If MATH, MATH, and MATH, then the first map is the boundary operator MATH, and the second is the excision isomorphism MATH. The composed map MATH and the map MATH are connected by the following diagram, which commutes up to sign: MATH . The commutativity of the diagram (up to a sign), basically comes from the fact that for a manifold with boundary, the NAME duality identifies the boundary operator in homology with the restriction map (to the boundary) in cohomology (see for example, CITE, page REF). The above sign is universal, depends only on the orientation conventions. Therefore, it can be determined using MATH-transversal chains, as in REF . |
math/9909008 | Let MATH be the natural map. Recall that the MATH term in NAME 's cohomological mixed NAME complex associated with the space MATH is the ``NAME resolution" MATH: MATH . This is considered with its ``bête" filtration MATH. Consider now the dual double complex MATH of MATH. Then MATH and MATH are quasi-isomorphic. This follows from the discussion II. REF. Therefore, their spectral sequence (for MATH) are isomorphic. This gives MATH and MATH. Finally notice that by a result of NAME, the weight spectral sequence (over MATH) of a mixed NAME complex degenerates at rank two, which provides MATH. |
math/9909008 | The proof is similar as in the case of REF. In this case, MATH is MATH . Actually this (and the whole cohomologically mixed NAME complex of MATH) can be constructed as a mixed cone of the complexes of MATH, respectively of MATH. This is compatible with the construction of MATH. |
math/9909008 | The main result here is the degeneration of the spectral sequence at level REF, which follows via duality MATH from the NCD case. Fix a triangulation MATH and let MATH be its first barycentric subdivision. Then the NAME duality map (compare II. REF) can be organized in the following morphism of double complexes. Let MATH, which form a double complex with MATH and MATH, similarly as in III. REF. Set the double complex MATH with dual morphisms MATH and MATH. Let MATH with boundary morphisms MATH and MATH (similarly as MATH defined above). Then MATH satisfies MATH (compare II. REF) and MATH (compare II. REF). Now, MATH is quasi-isomorphic to MATH (that is, their spectral sequences are the same for MATH), and the later is dual to MATH. Using II. REF, MATH induces an isomorphism at the level of the MATH term, hence it is an quasi-isomorphism, and it induces isomorphism at the level of any MATH (MATH). On the other hand, MATH for MATH. Hence the result follows. |
math/9909008 | The image of any cycle MATH, with MATH, via the composite map MATH has MATH-dimensional support. Actually, this support is the image of MATH by the natural map MATH. Hence it supports no cycle in MATH. |
math/9909008 | For the first part notice that MATH that is, MATH (compare the identification MATH above). The second part follows from the above REF (and its proof). |
math/9909008 | For MATH and MATH use II. REF and II. REF and the corresponding definitions. MATH follows from the results of the previous subsections and from the construction of the mixed cone. |
math/9909016 | Note first that MATH and MATH exist because MATH and MATH are simply connected and infinity is an apparent singularity of the system. Let MATH, MATH, and let MATH, MATH. From REF , it follows that MATH is also analytic and invertible at MATH. Hence MATH and MATH are analytic and invertible on all of MATH and MATH, respectively. It follows from REF that in a neighborhood of the singularity MATH, the entries of the functions MATH and MATH behave as MATH and the entries of the functions MATH and MATH behave as MATH, where MATH is taken from the eigenvalues of MATH and MATH refers to their multiplicity. By assumption we have MATH. This implies that MATH, MATH, MATH and MATH. The functions MATH and MATH can be extended analytically onto the boundary MATH. The values of the functions defined by MATH and MATH with MATH coincide with the boundary values of the above functions MATH and MATH, which are defined on MATH and MATH. We introduce the function MATH, MATH, and observe that this representation is a factorization of MATH in the space MATH. On the other hand, we have MATH, and because MATH and MATH are solutions of the system, it follows that MATH is a piecewise constant function on MATH with jumps only at MATH. Now let us express the monodromy of the system in terms of MATH. We start from the solution MATH on MATH and continue analytically by going once around the point MATH (see the construction of the deck transformations MATH). When we first cross MATH, the proper analytic continuation is the function MATH on MATH. After the second crossing of MATH, we obtain the function MATH on MATH. Hence, up to simultaneous similarity, we have MATH. On the other hand, by REF , the monodromy is given by MATH. Hence MATH with some MATH. This implies that MATH, hence MATH for each MATH. Because both MATH and MATH are piecewise constant, there is a MATH such that MATH. Hence MATH, which implies that MATH is a factorization in the space MATH. By REF , this is even a MATH-factorization. |
math/9909016 | From REF it follows that there exists a system with singularities only at MATH whose monodromy is given by MATH. Let MATH be a solution of this system, that is, MATH is analytic and has the given monodromy representation. We are going to modify MATH step by step in order to obtain an analytic matrix function MATH which also satisfies REF . Let us consider MATH near the point MATH. Because the monodromy is given as above, the solution changes by a matrix similar to MATH after going once around MATH in positive direction. Hence, restricted to MATH, the function MATH can be written in the form MATH, where MATH is a certain analytic function. We apply REF with MATH and MATH. We obtain that MATH, MATH, where MATH is analytic and invertible on MATH and MATH is analytic and invertible on MATH. Hence MATH for MATH. We define the function MATH for MATH. This function satisfies REF , and also REF at the point MATH. The next steps consist in doing the same for MATH. The essential point is that, at the MATH-step, we define the function MATH, where MATH is analytic and invertible on MATH. This modification does not produce additional singularities (except at infinity), it does not change the monodromy, and it changes the local behavior only at the MATH (in the desired way), but not at the other singularities. So we obtain a function MATH which satisfies REF - REF . Finally, we consider the point infinity. Restricting MATH to MATH we can write MATH, where MATH is analytic and invertible on MATH. We apply REF with MATH and factor MATH, MATH, where MATH is analytic and invertible on MATH, MATH is the diagonal matrix, and MATH is analytic and invertible on MATH. It follows that MATH. We define MATH, MATH. Because MATH is analytic and invertible on MATH, the function MATH satisfies also the required REF - REF . Since MATH on MATH, this function also satisfies REF after an appropriate permutation of the entries of MATH. Finally, we define the corresponding matrix function MATH by MATH. |
math/9909016 | Because the systems have the same singularities and the same monodromy, there exist an analytic function MATH and a MATH such that MATH. We are first going to show that MATH is analytic and invertible on all of MATH. Let us restrict the solutions MATH and MATH onto a neighborhood MATH of the point MATH. From REF it follows that there are MATH such that MATH where MATH and MATH are analytic and invertible at MATH. Hence MATH where MATH. By going once around the point MATH in positive direction it follows that MATH where MATH. Combining both equations it follows that MATH, that is, MATH and MATH commute with each other. Now we use the fact that the matrix MATH is non-resonant. Because MATH, this implies that MATH can be written as a polynomial in the matrix MATH. Hence also MATH and MATH commute. Using this in connection with REF we obtain MATH . Hence MATH, which implies that MATH is analytic and invertible at MATH. So far we have shown that MATH and MATH are entire analytic functions. We analyze their behavior at infinity by restricting the solutions MATH and MATH onto MATH. From REF , it follows that MATH where MATH and MATH and MATH are analytic and invertible on a neighborhood of infinity. Here we assume that MATH and MATH are of the form MATH where MATH, MATH, MATH, MATH. Here MATH stands for the identity matrix of size MATH. Now the above equations give MATH . This means that MATH where MATH is analytic and invertible in a neighborhood of infinity. Now we introduce block partitions of the matrices MATH and MATH: MATH where MATH is of size MATH and MATH is of size MATH. Using the structure of MATH and MATH, it follows that MATH and MATH as MATH. Because MATH and MATH are entire analytic functions, we obtain that MATH and MATH for MATH, and that MATH and MATH are matrix polynomials of degree at most MATH if MATH. In particular, the diagonal blocks are constant matrices. The block triangular structure of MATH and MATH implies that MATH and MATH. Hence MATH and MATH are invertible square matrices, that is, MATH. Hence the indices corresponding to both systems are the same. Thereby, we have also proved that MATH is of the desired form. |
math/9909016 | Assume REF - REF holds. Then also REF of this proposition are fulfilled. As to REF , remark that MATH implies MATH because MATH is analytic and invertible at MATH. At infinity we have MATH. Hence, similarly, MATH where MATH. Since MATH is analytic and invertible, MATH as MATH. This implies REF . In order to prove the converse, we remark first that REF of this proposition imply REF with the exception of the apparentness of the singularity at infinity, which will follow from REF . REF of this proposition in connection with REF immediately implies REF . Note that the mere similarity of the residue to MATH does not cause problems because MATH. Finally, let us consider the singularity at infinity. Because of the asymptotics of MATH at infinity, there exists an analytic solution MATH of the system MATH near infinity. Now define MATH, MATH. Then, MATH . Hence MATH is a solution of our original system considered on the domain MATH. However, the relation of such a ``restricted" solution to any solution of the system is given by MATH with some MATH. This is exactly REF . Obviously, this also shows that infinity is an apparent singularity. |
math/9909016 | In regard to the previous proposition, we have to show that the characterization given by REF is equivalent to the property that MATH is analytic on MATH, has simple poles at MATH and behaves at infinity as described in REF . The fact that MATH is analytic on MATH and has simple poles at MATH is equivalent to writing the entries of MATH as MATH, where MATH is an entire analytic function. The characterization of MATH as a certain polynomial is now equivalent to REF . |
math/9909016 | Let MATH be projections such that the image of MATH is the subspace MATH. Then MATH, and we obtain MATH . Choose any point MATH and consider the solution MATH for which MATH. Hence MATH. Because the matrix function MATH is a solution of the above first order system, it follows that MATH for all MATH. This means that MATH is an invariant subspace of MATH. Hence MATH is also an invariant subspace of the inverse MATH. Let MATH be the monodromy representation of MATH and MATH. Then MATH. Hence MATH is an invariant subspace of MATH. |
math/9909016 | Suppose that this condition is not fulfilled for some MATH, that is, MATH. Because the indices are ordered decreasingly, it follows that MATH for each MATH and each MATH. REF implies that for those MATH and MATH, the entries MATH of the matrix MATH vanish identically. This means that the matrix MATH is of block triangular form, that is, it possesses the non-trivial invariant subspace MATH. By the previous lemma, the monodromy representation of some solution has also this invariant subspace. Hence the monodromy of the system is reducible, which contradicts the assumption. |
math/9909016 | Because MATH the residue MATH of MATH at MATH has nonzero entries on the diagonal below the main diagonal. All entries below this diagonal are zero. By considering the powers of MATH, it follows that MATH. By REF we have MATH, which implies MATH. Because MATH and MATH is non-resonant, we obtain MATH by considering the NAME normal forms. |
math/9909016 | CASE: Suppose we are given a MATH-th order linear differential REF with the above properties. Let MATH be MATH linear independent solutions. We introduce MATH . Then MATH is analytic on MATH, and it is well known that the Wronskian MATH does not vanish on all of MATH. In particular, MATH is a solution of the MATH system MATH, where MATH and MATH are the coefficients of the scalar differential equation. Obviously, the system for MATH has the same singularities MATH (and possibly an apparent singularity at infinity) and the same monodromy as the scalar differential equation. Below we will replace the system MATH by a modified system MATH which will have the desired properties. However, in order to analyze the behavior at the singular points MATH, we will first consider differently modified systems MATH near these singularities. For MATH, we introduce MATH and write MATH for each MATH. Because the singularities of the scalar equation are assumed to be Fuchsian, the functions MATH are analytic at MATH. For MATH, we define MATH. Then MATH is a solution of the system MATH with MATH. From this we obtain that MATH``analytic term at MATH" with MATH . A straightforward calculation shows that the characteristic equation MATH for the matrix MATH coincides with the indical REF for the local exponents of the scalar equation with respect to the singularity MATH. Hence the eigenvalues of MATH are just MATH. For MATH we make a modification. We define MATH, which is a solution of MATH. Again MATH``analytic term at MATH", but now MATH contains an additional term MATH. Consequently, MATH coincides with the indical equation for the singularity MATH. Because the assumption on the local exponents is also different, we obtain nevertheless that the eigenvalues of MATH are MATH. In fact, the converse is also true. The scalar differential equation is Fuchsian at MATH if and only if the system MATH obtained in the above way is Fuchsian at MATH. Next we are going to analyze the singularity at infinity. We make a change of variables MATH and introduce MATH by MATH for MATH. These new functions are solutions of a modified scalar linear differential equation. The point MATH corresponds to the point MATH. By assumption they are not singular points. Hence MATH is analytic in a neighborhood of MATH and MATH. We remark that the identity MATH holds, which can be verified by a direct calculation. Here MATH is a certain invertible upper triangular MATH matrix. This matrix does not depend on any parameters and can (in principle) be evaluated explicitly. Now we are prepared to define the system which will satisfy the conditions of REF . Let MATH and define MATH. Because the function MATH is analytic and invertible on MATH, it follows from the afore-mentioned properties of the system MATH that the new system MATH has singularities only at MATH (and possibly an apparent singularity at infinity) and has the same monodromy as the scalar differential equation. Next we analyze the local behavior of the new system at infinity. Combining REF , it follows that MATH for MATH, where MATH is analytic and invertible on MATH. Hence the behavior at infinity is as required for systems of standard form with indices MATH. (Recall that MATH is assumed.) In order to analyze the behavior at the singularity MATH, we make the connection with the systems MATH. First write MATH, where MATH can be shown to be analytic and invertible on all of MATH. In fact, one uses that MATH is an upper triangular matrix and that MATH is a diagonal matrix function with particular entries. Now, recalling how MATH and MATH were obtained from MATH, it follows that MATH where MATH is a certain matrix function analytic and invertible on MATH. Hence the local properties of MATH and MATH are essentially the same. In particular, MATH has a simple pole at MATH as so has MATH, and the residue of MATH at MATH is similar to MATH. Using the characterizations given in REF , it follows that MATH is a system of standard form with ``local" data MATH and the properties mentioned above. In particular, we obtain that MATH is of the form REF . The assertion that MATH is a consequence of the special structure of MATH given in REF and the fact that MATH is an upper triangular matrix function. Observe that MATH . Finally, we need to show that the ``local" data is given also by MATH, that is, MATH. In fact, REF implies that MATH. Because by REF, we obtain MATH. Again by assumption, the eigenvalues of MATH are MATH, hence they are the same as those of MATH. Now we can conclude MATH. CASE: The essence of the above construction was the passage from the matrix function MATH to the matrix function MATH by means of the transformation REF with MATH. The fact that the entries below the first row of MATH are of a special form is reason for the connection of MATH to the scalar equation. Now consider the singularities MATH and the indices MATH as fixed and the functions MATH as arbitrary. It follows from the above argumentation that the matrices MATH obtained by REF are necessarily of the form REF . Moreover, a straightforward computation (using the special structure of MATH and MATH) shows that the entries below the first row of MATH are also of a special form, that is, they only depend on MATH and MATH but not on MATH. Moreover, one obtains that MATH. In this sense, these entries of MATH have to be considered as ``fixed". In principle, they can be computed. We want to make the converse transformation with MATH, that is, to pass from MATH to MATH by means of MATH . Obviously, the matrix MATH will be only of the desired form REF if the entries below the first row of MATH are chosen ``properly". The crucial point is that if these entries of MATH are given indeed in the appropriate way, then the matrix MATH will be of the form REF with certain MATH. This can be shown again by a straightforward computation. Hence we cannot start with just any matrix MATH of the form REF . Fortunately, the following result is true. Any system MATH with MATH given by REF and MATH is equivalent to some system with another matrix MATH which is of the same form and whose entries below the first row are prescribed arbitrarily. In order to prove this statement we need only recall REF . We have to prove that there exists a matrix MATH of the form REF and the properties stated there such that a transformation with MATH takes the matrix MATH to the matrix MATH. Note that the entries of MATH are polynomials of a certain degree. We can split this transformation into several steps. First, let MATH be a diagonal matrix with suitable diagonal entries such that MATH is taken into a matrix with entries MATH. At the MATH-step, MATH, we then choose the matrix MATH to have ones on the main diagonal, suitable polynomials on the MATH-th diagonal above the main diagonal and zero entries everywhere else. In fact, a straightforward computation (analyzing MATH) shows that these polynomials can be chosen in such a way that the entries of the MATH-th diagonal above the main diagonal of MATH (except the entry in the first row) can be modified in any desired way. At the MATH-th step, all entries of MATH below this diagonal remain unaltered. Hence after the MATH-th step we arrive at a matrix MATH of the same form and with desired entries below the first row. After this, we may assume that the matrix MATH of the system described in REF is of such a form that the transformation REF gives a matrix MATH which is of the form REF with certain MATH. Hence this new system gives raise to a MATH-th order linear differential equation. We have to show that this equation has the properties stated in REF . In fact, the argumentation given above can be reversed. Using the properties of MATH (and employing the systems MATH at an intermediate step) it follows that MATH are analytic on MATH and that MATH has at most a MATH-th order pole at MATH. Hence the scalar equation is a Fuchsian equation and it follows also from above that the local exponents are equal to the eigenvalues of the residue of MATH, which is a matrix similar to MATH. Obviously, the monodromy is the same as for the given system. The assertion that the scalar equation has no singularity at infinity is only slightly more difficult. We know that the solution MATH of the system can be written as REF . Introducing a modified scalar equation by change of variables as above, we see that REF holds with MATH being the solution for the system with MATH, or, equivalently, the matrix REF corresponding to the solution of the scalar equation. Combining this with REF and the fact that MATH, it is possible to conclude that MATH is analytic and invertible near MATH. Hence MATH is no singular point for the modified scalar equation, and neither is MATH a singular point for the desired scalar equation. |
math/9909016 | The MATH-tuples MATH and MATH are equivalent if and only if there exists a matrix MATH such that MATH. If all matrices MATH are scalar matrices, then the assertion is trivial. Otherwise, if at least one MATH is not scalar, one rewrites MATH, and now a straightforward computation shows that MATH is an upper triangular matrix. Using this, the relation MATH is easily seen to be equivalent to REF . |
math/9909016 | CASE: Because MATH is of triangular form we obtain from REF that there exists a solution MATH which is also of triangular form. So let us write MATH as MATH . For MATH, we have MATH where the last equality defines the numbers MATH. They satisfy MATH. Because MATH is similar to the residue of MATH at MATH, the numbers MATH and MATH are the eigenvalues of MATH. Once we have shown that MATH, it follows that MATH and MATH. In fact, from REF we obtain that the solutions of MATH are given by MATH up to a redundant constant. Regardless of the precise expression for MATH, it follows that the monodromy representation for MATH is given by REF with MATH and with certain MATH. Hence MATH is equivalent to REF and the MATH and MATH are indeed properly ordered. CASE: We consider all MATH-tuples of the form REF with fixed MATH, and we fix also the numbers MATH (that is, the eigenvalues of MATH). Observing that MATH and MATH, we introduce the functions MATH and MATH by REF . In what follows we consider all systems of the form REF with those MATH and MATH and with MATH where MATH runs through all polynomials in MATH. These systems are indeed of standard form because MATH and because MATH are assumed to be non-resonant. From the first part of this proof we already know that the monodromy representation of the corresponding solutions are of the form REF with certain vectors MATH. It remains to show that if MATH ranges through all of MATH, then the corresponding vectors MATH take values in all equivalence classes (defined by the equivalence relation REF ). We first examine the question when two systems with MATH and MATH given by MATH with MATH are equivalent. By definition this is the case if and only if there exists a MATH such that MATH, where MATH is of the form MATH with MATH and MATH if MATH, and MATH if MATH. If MATH, then MATH implies that MATH has to be an upper triangular matrix (except for the case where MATH is given with MATH and MATH, which can easily be dealt with separately). Using this information about MATH, it follows that the above systems are equivalent if and only if MATH for some MATH and MATH. This, in turn, can be rephrased by saying that MATH for some MATH, where MATH is the image of the linear mapping MATH. We conclude that MATH. If MATH (hence MATH), then MATH is even the zero mapping. Now decompose MATH as a direct sum, and remark that MATH and even MATH if MATH. Finally, we arrive at the following statement: if MATH, then the above systems are equivalent if and only if MATH for some MATH. Now choose any basis MATH in MATH where MATH. Assume that a solution of the system REF with MATH is given by some MATH, MATH. The corresponding monodromy representation is supposed to be given by REF with some vector MATH. Now consider an arbitrary polynomial MATH and write MATH. A straightforward computation shows that a solution of REF with MATH is given with MATH (notice that there are certainly further solutions). Moreover, the monodromy representation of this solution is given by REF with MATH. In this way, we have defined a linear mapping MATH from MATH into the set of all admissible vectors MATH, which characterize the reducible monodromy. Now we invoke the fact that there is a one-to-one correspondence between equivalence classes of systems of standard form and equivalence classes of given data (see REF ). This fact specialized to the mapping MATH means the following: for MATH, the vector MATH is equivalent to the vector MATH in the sense of REF if and only if MATH for some MATH. This allows us easily to conclude that the kernel of MATH is trivial. Hence, in case MATH, we have MATH, and we are done because the set of all admissible vectors MATH is an (MATH)-dimensional subspace of MATH. On the other hand, MATH means that MATH for some MATH. Hence (by the non-resonance assumption) the vector MATH, occurring in REF , is not the zero vector. Because MATH, the assertion is proved if we have shown that MATH. Indeed, if MATH, then the equivalence of MATH to the zero vector implies that MATH. Hence MATH, which is a contradiction. |
math/9909016 | REF is a conclusion of the implication MATH . The second condition of REF is just relation REF . Now assume that MATH. Then REF implies that the matrix MATH of the corresponding system is the form REF . Using the implication MATH , it follows that MATH is of the form REF with MATH, contradicting the assumption. |
math/9909016 | Again, REF are an immediate consequence of REF . In order to prove REF we argue as follows. The indices for the given data are MATH and MATH if and only if this data is associated to some system of standard form for which MATH can be written as MATH with MATH, MATH, MATH and MATH (see REF ). If MATH, then we can apply REF and conclude that the monodromy is reducible and MATH, which contradicts our assumption on the data. Hence MATH. Now we can use REF , which shows the desired equivalence. |
math/9909016 | These triples are equivalent if and only if there exists an invertible matrix MATH such that MATH. It follows that the matrix MATH maps the invariant subspaces of MATH into the invariant subspaces of MATH. Because of REF , MATH and MATH are the only non-trivial invariant subspaces of both triples. Hence MATH and MATH are also invariant subspaces of MATH, and this implies that MATH is of triangular form, MATH . Now the equality MATH gives the following two conditions, MATH and a third condition, which will be stated later. We first claim that MATH. Indeed, REF can be rewritten as MATH. If MATH, then REF applied to the triple of MATH matrices obtained from MATH by removing the third rows and columns implies that MATH is equivalent to a triple of the form REF with MATH. This contradicts REF . Similarly, we obtain MATH. A simple thought shows that we may assume MATH. Hence REF simplify to MATH and the third condition can be expressed as MATH . REF lead to four distinct cases. If MATH for some MATH, then MATH, and if MATH for some MATH, then MATH. Otherwise, MATH and MATH, respectively, can be arbitrary. Combining these statements with REF proves the assertion. |
math/9909016 | CASE: Because MATH is of triangular form, there exists a solution which is also of triangular form (see REF ). Hence we can write MATH as MATH . For MATH, we have MATH where the numbers MATH are defined by the last equality and satisfy MATH. Because MATH is similar to the residue of MATH at MATH, the numbers MATH are the eigenvalues of MATH. Once it is shown that they are properly ordered, it follows that MATH for MATH. The diagonal entries of the solution MATH satisfy MATH. Therefore they are given by MATH up to a redundant constant. Regardless of the precise expressions for the off-diagonal entries of MATH it follows that the corresponding monodromy representation is given by REF with MATH and with certain MATH. Hence MATH is equivalent to REF and the above relation between MATH and MATH implies that the eigenvalues of MATH are properly numbered. CASE: Suppose we are given MATH, which is equivalent to a triple of the form REF , and assume that the numbers MATH and MATH are defined appropriately. We introduce MATH by REF and consider systems REF with those MATH and yet unspecified MATH. We have to show that there exist MATH (being the ratio MATH of polynomials with properties as described above) such that the monodromy of this system is given by MATH. We first consider some trivial cases. Assume that MATH is equivalent to REF with MATH. The proof of REF tells us that there exists a MATH such that the system MATH has monodromy given by MATH . Moreover, there exists a MATH such that the system MATH has monodromy given by MATH . In this connection we remark that REF are indeed ``admissible" triples in the sense that the product of the occurring matrices is equal to MATH. Moreover, MATH and MATH where the polynomials MATH and MATH have the proper degree. Now we consider the system REF with those MATH and MATH and with MATH. This system has a solution of triangular form as given in REF with MATH. The monodromy representation with respect to this solution is of the form REF with desired values MATH and MATH. However, we may have different values MATH and MATH (instead of the desired values MATH and MATH). Analyzing the relation between the monodromy of REF and the monodromy of the ``subsystems" REF , it follows that REF is equivalent to REF with MATH replaced by MATH and that REF is equivalent to REF with MATH replaced by MATH. We apply REF and obtain that MATH with certain MATH and MATH. A simple thought shows that a representation REF with MATH and having above MATH and MATH as entries is equivalent to the same representation but with above MATH and MATH. In fact, the equivalence is established by a certain upper triangular matrix having MATH-entry equal to zero. Hence, the monodromy of the constructed system REF is given by MATH, and this concludes this case. The other trivial case, where MATH is equivalent to a triple of the form REF with MATH can be treated in the same way. Here we consider the system REF with MATH and MATH and the appropriate ``subsystems". Now assume that MATH does not fall into these trivial classes. In other words, MATH satisfies REF concerning the reducibility type. Arguing similar as above, we find an appropriate function MATH such that the system MATH has monodromy given by MATH and we also find a function MATH such that the system REF has the monodromy REF . With those functions MATH and MATH and an arbitrary function MATH we consider the system REF . This system has a solution with certain functions MATH and MATH. The monodromy representation of this solution is given by REF with the desired values MATH, but possibly different values MATH, MATH and MATH. Analyzing the relation to the ``subsystems", for which we know the monodromy, it follows again from REF that MATH with certain MATH and MATH. By means of the construction of a suitable triangular matrix (as above), we obtain that MATH is equivalent to a triple REF having off-diagonal entries MATH, MATH and certain MATH. In other word, the monodromy of the system we have constructed so far is ``almost" as desired, only the values MATH might not yet be the desired ones MATH. In fact, in REF and also in the ``generic" subcase of REF , which has been singled out in REF , we are already done because (as has been remarked in the paragraphs following this proposition) there exists only a single equivalence class of such triples. Hence the monodromy of the constructed system is equivalent to MATH. In the remaining cases we must still modify the system, and we proceed as follows. Again the proof of REF tells us that there exists a suitable function MATH such that the system MATH has monodromy given by MATH . In this connection we remark that both MATH and MATH satisfy a linear relation REF with the same right hand side. The differences satisfy the homogeneous linear relation, which corresponds to REF . Hence REF is an admissible triple, that is, the product of the matrices is equal to one. The monodromy representation of the solution in REF is also given by REF except that the upper right entries are replaced by certain values MATH. The relation between these and the former values is (see REF ) MATH with certain MATH and MATH. Now we consider a new system REF where MATH has the same entries as the old one except that MATH is replaced by MATH where we define MATH. It is easy to see that the solution of this new system has the same entries as the old solution except that MATH is replaced by a certain MATH. Moreover, the monodromy representation with respect to this solution is of the form REF with off-diagonal entries MATH, MATH (as in the old one) and certain MATH. What remains to show is that this monodromy representation is equivalent to MATH. Because we already pointed out that MATH is equivalent to REF with off-diagonal entries MATH, MATH and MATH, we are done as soon as we have shown that MATH and MATH are related in the way described in REF . In order to show this, we have to consider the defining relations for the MATH-entries of the solutions and of the monodromy representations explicitly. As to the old system, we have MATH where MATH, MATH, are the deck transformations. The corresponding relations for the new system are MATH . We also have to state the conditions on the MATH-entries in REF , MATH . Combining REF and using MATH, we obtain that MATH . Comparing this with REF we may conclude without loss of generality that MATH. (The solution of the differential equation is not unique but we may take any of them.) Using this identity and combining REF with REF , we arrive at MATH . Comparing this with REF we conclude that MATH where the last equality follows from REF . Now we conclude from REF that the triple REF with MATH, MATH and MATH, which is the monodromy representation for the new system, is equivalent to the triple REF with MATH, MATH and MATH, which in turn is equivalent to MATH. Hence the monodromy of the new system is given by MATH. |
math/9909016 | NAME data is exactly the data which is not covered by REF . As was already remarked in the paragraph after REF , the data which is covered by this theorem is exactly the data for which MATH is reducible with the condition MATH or is diagonalizable without another condition. |
math/9909016 | The triples MATH and MATH are equivalent if and only if there exists a matrix MATH such that MATH. We write MATH in the same block form as the matrices in the triple REF , MATH and inspect the MATH-block entry in the resulting equality MATH. We obtain MATH. Because of REF , it follows that MATH. Otherwise, MATH is an invariant subspace for all MATH with the eigenvalue MATH. Hence MATH is also of block triangular form. Now we look at the MATH-block entry of MATH, and it follows that MATH. We claim that MATH is a scalar matrix. Assume the contrary. Then MATH is either similar to a diagonal matrix with two different diagonal entries or it is similar to a NAME block. We may assume without loss of generality that MATH is actually equal to such a diagonal matrix or to a NAME block. (Otherwise one applies an appropriate similarity transformation simultaneously to MATH and MATH, which will equally result in a contradiction.) By simple computations it follows in the first case that all MATH are of diagonal form. In the second case, it follows that all MATH are either NAME blocks or scalar matrices. Both contradict the assumption on the triple MATH. With this information about the entries of MATH, it is now easy to conclude that MATH is equivalent to REF . |
math/9909016 | CASE: Because MATH is of block-triangular form, there exists a solution which is also of block triangular form. Hence let us write MATH as MATH . We first consider the scalar subsystem MATH. We know that MATH. If MATH denote the residues of MATH at MATH, then the monodromy is given by MATH, where MATH. Moreover, it follows that MATH. Now we consider the MATH subsystem MATH, where MATH is given by the lower right block in REF . We claim that this system is not equivalent to a system of the form REF . Indeed, if it were equivalent, then a simple thought shows that the system MATH is equivalent to a system of the form REF , which contradicts the assumption that it is a system of MATH-block form. Hence we are in a position to apply REF . We obtain that the data of the system MATH, MATH and MATH, is quasi-block data. Note that MATH and that MATH is similar to the residue of MATH at MATH. Hence, in particular, MATH. Combining the information about the two subsystems, we obtain immediately that the monodromy data MATH of the original system is equivalent to a triple REF with certain MATH. The eigenvalues of the matrix MATH, which is similar to the residue of MATH at MATH, are given by the number MATH and the eigenvalues of MATH because they are the residues of MATH and MATH, respectively. Hence, since the data of the MATH subsystem is quasi-block data, the data of the original system is MATH-quasi-block data. Moreover, from the values for MATH and MATH given above, it follows that MATH and MATH. CASE: Assume that MATH is a triple of the form REF such that (on defining MATH and MATH as well as MATH and MATH by REF ) the appropriate conditions for MATH-quasi-block data are satisfied. In particular, MATH and MATH is quasi-block data, and MATH, MATH. We have to show that there exists a system of MATH-block form which has monodromy MATH, the indices MATH, and the residues of MATH at MATH are similar to MATH. We first define MATH noting that MATH. Furthermore, because MATH and MATH is quasi-block data, REF shows that there exists a MATH system having this data with singularities MATH such that this system is not equivalent to a system of the form REF . Denote this system by MATH, and thus define the matrix function MATH. Let MATH and MATH be the indices of this MATH system. We claim that MATH and MATH. Indeed, because the MATH system is not of triangular form, we have MATH, and by assumption we also have MATH. Moreover, MATH, where the last relation follows from REF applied in the case of the MATH system. These relations for the integers prove the stated equality. Now we consider systems MATH given by REF with those MATH and MATH and yet unspecified MATH, the entries of which have, of course, to be the ratio of appropriate polynomials. We note that these systems are of the form REF . They are also systems of standard form with indices MATH, because, in particular, MATH is assumed. The monodromy representation of a suitable solution of these systems is given by REF with certain MATH. The residue MATH of MATH at MATH has the eigenvalues given by MATH and the eigenvalues of MATH. Hence, as the eigenvalues of MATH are also given in this way, it follows that MATH is similar to MATH. (In case of coinciding eigenvalues, one has also to employ that MATH and MATH.) What remains to show is, firstly, that the systems constructed in this way are indeed of MATH-block form, that is, that they are not equivalent to systems REF . Secondly, one has to show that if MATH runs through all admissible functions, then corresponding vectors MATH appearing in the monodromy representation REF run through all possibilities modulo the equivalence described in REF . Let us turn to the first problem. Assume the contrary, namely that our system with MATH is equivalent to a system with MATH being of the form REF . REF says that both systems have the same indices and that the equivalence is established by a matrix function MATH, which has a particular structure. We also know that the MATH subsystem MATH of MATH is not equivalent to a triangular system, and we are going to show a contradiction to this assertion. If MATH, then the MATH- and MATH-entries of MATH vanish. Using MATH, it follows that MATH has MATH-entry equal to zero, which is a contradiction. If MATH, then the MATH is of the same block structure as MATH. Now using MATH, it is easy to see that the lower right MATH submatrix in MATH establishes an equivalence between the subsystem MATH and a triangular MATH system, namely the one which is given by the lower right MATH matrix in MATH. In the remaining case, MATH, the function MATH is a constant matrix MATH. As before, let MATH be the residues of MATH at MATH, and let MATH be those of MATH. The equivalence between MATH and MATH can now be restated as MATH. We claim that each one-dimensional invariant subspace of MATH is contained in some two-dimensional invariant subspace of this triple. Indeed, this is obvious if this triple has at least two one-dimensional invariant subspaces. If it has exactly one, then it is necessarily the space MATH, which is contained in MATH (because of the triangular form of MATH). The above relation between MATH and MATH establishes a one-to-one correspondence between invariant subspaces of the previous triple and MATH. Hence each one-dimensional invariant subspace of the latter triple is contained in some two-dimensional invariant subspace. A simple thought shows that the MATH matrices obtained from MATH by removing the first rows and columns are reducible. But those matrices are the residues of the system MATH, and thus we obtain a contradiction. Finally, we turn to the problem of showing that if MATH runs through all admissible vector functions, then the vector MATH runs through all possibilities modulo the equivalence stated in REF (see also REF ). Before we have to show that the assumptions of REF are satisfied. We know that MATH and MATH is quasi-block data. Hence MATH is not diagonalizable. If this triple possesses one non-trivial invariant subspace and all MATH have only a single eigenvalue, then so have MATH. It follows that MATH, which is a contradiction (see REF in the definition of quasi-block data). As to REF , assume that REF is not satisfied. Then it is easy to see (using the nonresonance of MATH) that MATH. Because MATH and MATH by REF of quasi-block data, it follows that MATH. On the other hand, MATH and MATH. We obtain that MATH and MATH. Hence MATH, which conflicts with MATH. So we have proved that the assumptions of REF are satisfied. Next we examine the question when two system with MATH and MATH given by REF with MATH and MATH being introduced above, but with possibly different functions MATH and MATH, respectively, are equivalent. Assume that MATH where MATH and MATH. This is the case (see REF ) if there exists a function MATH of a particular form such that MATH. We first claim that the MATH- and MATH-entries of MATH, MATH and MATH, are equal to zero. This is obvious in the case MATH. In case MATH, we have MATH and the upper left MATH matrix of MATH is constant. In order to show that MATH, we write MATH, and inspect the MATH-entries. It is equal to zero for MATH and also for MATH. The MATH-entry of MATH equals MATH times the MATH-entry of MATH. We conclude that MATH because otherwise the MATH-entry of MATH is zero, which contradicts the assumption that MATH is not of triangular form. In the case, MATH, we have MATH with a constant matrix MATH. This can be rephrased in terms of the residues MATH and MATH of MATH and MATH, respectively, namely MATH. The triples MATH and MATH possess exactly one one-dimensional invariant subspace MATH. (If there were another, then we could argue similar to above and conclude that the MATH matrices obtained from these triples by removing the first rows and columns are reducible.) The matrix MATH must map this invariant subspace into itself. Hence it is also an invariant subspace of MATH, which implies that MATH. So we have shown that MATH with the same block structure as MATH and MATH and with entries having certain properties. The next claim is that MATH is a constant scalar matrix. From MATH, it follows that MATH, and we know that the MATH system with MATH is not equivalent to a triangular system. If the indices MATH coincide, then MATH is a constant matrix, and the above can be rephrased in terms of the residues MATH of MATH. Namely, MATH. If MATH is not a scalar matrix, then we can argue similar as in the second paragraph of the proof of REF and conclude that the triple MATH is reducible, which contradicts the assumption on MATH. In case MATH, MATH is of triangular form with constant values on the diagonal. Now a straightforward analysis of MATH (using that MATH is not triangular) implies that MATH is a scalar constant matrix. Having this information about MATH, we can now reformulate the condition MATH as follows. The systems MATH and MATH are equivalent if and only if MATH for some MATH and some MATH where MATH and MATH. Referring to REF this can be expressed as MATH . Hence the two systems with MATH and MATH are equivalent if and only if there exists MATH such that MATH where MATH is the image of the linear mapping defined by MATH . Now we decompose MATH as a direct sum. We arrive at the following statement: If MATH and MATH, then the corresponding systems are equivalent if and only if MATH for some MATH. Obviously, we have MATH. However, we claim that MATH in the case where the MATH matrix appearing in REF has rank MATH, that is, if MATH is equivalent to a triple REF . Indeed, this is shown as soon as one has shown that the kernel of MATH is non-trivial. For this we consider the MATH system MATH with MATH. It is easy to see that this system is also of standard form with indices MATH and MATH, and with data MATH and MATH. The solution of this system is given by MATH. From MATH it follows immediately that MATH by multiplying on both sides with MATH. We may assume that the monodromy representation of MATH is given by MATH . Now we define MATH to be the last row in MATH. Obviously, this vector function satisfies MATH. What remains to show is that MATH and MATH are polynomials of appropriate degree. From the above monodromy representation for MATH, it follows that MATH is a single valued function. This function is analytic on MATH. Because of the non-resonance of MATH and because MATH is an eigenvalue of MATH, it follows that MATH is an eigenvalue of MATH. Hence the matrices MATH have an eigenvalue equal to zero and are also non-resonant. But these matrices describe the local behavior of MATH (and of its inverse) near the point MATH. A straightforward computation (using the fact that MATH is single valued) shows that MATH is actually analytic at MATH. Hence we have shown that MATH is entire analytic. That MATH and MATH are polynomials of degree not greater than MATH and MATH, respectively, can be obtained by employing the behavior of MATH at infinity, which is described by the integers MATH and MATH. In summary, we can conclude that MATH lies in the kernel of MATH. Similar as in the proof of REF , we can define a linear mapping MATH from the space MATH into the set of all ``admissible" vectors MATH (see also REF ). This mapping is characterized by the following property. For a system of the form REF with above defined MATH and MATH and with MATH given by REF , where MATH, the corresponding monodromy is given by REF , where MATH are given by MATH. We have a one-to-one correspondence between equivalence classes of systems of standard form and equivalence classes of data (see REF ). In our situation we can apply REF (see also REF ) and what has been stated above about the equivalence of the systems under consideration. We obtain the following statement. For MATH, the vectors MATH and MATH are equivalent in the sense of REF (or, REF ) if and only if there exists a MATH such that MATH. The kernel of MATH is trivial. Indeed, MATH implies immediately MATH. Hence MATH or even MATH depending on whether the MATH matrix appearing in REF has rank MATH or MATH. So we are done as soon as we have shown that the sum of MATH and the following linear subspace of MATH, MATH exhausts the whole four-dimensional subspace of MATH consisting of all admissible vectors MATH (see again REF ). As we have estimates for the dimensions, this follows from the statement that the intersection of REF with MATH contains only the zero vector. Indeed, if MATH is contained in the above set, then MATH is equivalent to MATH, whence again follows that MATH. Hence MATH. |
math/9909016 | First of all, the reader can easily verify that the classification into REF - REF is complete. Some of these cases do overlap (namely, REF with MATH is stated in REF with MATH and MATH or MATH is stated multiply in REF , and REF with MATH and MATH is given in both REF ), but in these cases, the description of the partial indices is the same. We remark in particular that the cases given in REF - REF are exactly the cases which are not contained in REF - REF . In REF we have completely characterized the data which is associated to systems of the form REF . We obtain that given data is associated to such a system if and only if MATH is equivalent to a certain triple REF such that for the corresponding integers MATH (which are defined by resorting to the properly numbered eigenvalues of MATH) the inequality MATH holds. This inequality follows from the statement that MATH for MATH, and from the ordering MATH. We note that in some cases, there exist essentially different ways to establish an equivalence between MATH and a triangular triple REF , which in turn results into different definitions of MATH. REF is true if and only if such a relationship exists. Examining the different reducibility types for a triple REF under the condition MATH we arrive at the following cases. First of all, only the reducibility types REF - REF are covered. REF is covered exactly for MATH as the enumeration of these integers is unique. REF is covered exactly for MATH as MATH is defined uniquely and MATH without loss of generality. Similarly, REF is covered exactly for MATH as MATH is unique and MATH without loss of generality. REF is obviously covered completely. So we arrive at REF . However, there are some other possibilities leading to REF . In REF , the integers MATH are uniquely defined, but an equivalence with a triple REF can be established in three different ways. The first way is if the triple has entries MATH. This gives MATH, MATH, MATH, and hence the first subcase in REF . The second possibility is if the triple has entries MATH. Here we obtain MATH, MATH, MATH, which is the second subcase in REF . Finally, we may have MATH, which leads to MATH, MATH, MATH, and the third subcase in REF . We have now identified all the data associated to system REF in terms of the classification established at the beginning of this section. Next we do the same for the systems of MATH-block form (see REF ). The data of such systems has been described in REF . With the numbers MATH and MATH defined by REF we obtain that the indices are MATH, MATH, and MATH. The latter follows from the fact that MATH and MATH. The ordering MATH can be rephrased by MATH, which is equivalent to MATH. We arrive at conclusion that the data of such systems is precisely the data which is of MATH-quasi-block form and satisfies MATH. Now we have to examine what does this mean in terms of the classification given at the beginning of this section. In the triple REF , there appears a triple MATH which along with MATH has to be quasi-block data. We have to distinguish REF stated in the definition of quasi-block data. In REF , the triple REF possesses exactly one one-dimensional invariant subspace MATH and no two-dimensional invariant subspace containing it. Hence the data MATH and MATH is precisely the data of reducibility type REF or REF with the condition MATH, where MATH and MATH are now defined by REF . In fact, these numbers are the same as the former ones defined in REF . So we arrive at REF . In REF , we may assume MATH to be of triangular form. We have MATH and MATH with the integers MATH and MATH defined in terms of the eigenvalues of MATH. Hence the data is given by triples REF with MATH and MATH. These triples are of reducibility type REF or REF . We do not obtain REF or REF because of the non-diagonalizability of MATH. In case of reducibility type REF , we have MATH, MATH and MATH, which gives the condition MATH and MATH. In REF we have the same identification, but in addition we may also have (interchanging MATH and MATH) MATH, MATH and MATH. This would give the condition MATH and MATH, which, however, conflicts with the assumption MATH. So we arrive at REF . In REF , the only possibility not contradicting the non-diagonalizability of MATH is if the triple MATH is equivalent to a triple REF with MATH. This results in the identification MATH, MATH and MATH. Hence we obtain the conditions MATH and MATH, and this gives REF . So we have also settled the cases of data corresponding to systems of MATH-block form. The analysis of the data corresponding to systems of MATH-block form REF is completely analogous and will therefore be omitted. Here we arrive at REF instead of REF . Finally, we have to examine the indices for the data in the remaining cases. We know that this data is associated to certain systems which are neither of MATH-block form, MATH-block form nor of triangular form. It follows that MATH because otherwise (by REF ) the MATH- and MATH-entries of the matrix MATH of the system vanish. Similarly, it follows that MATH. From the fact that MATH and MATH, we immediately obtain REF . In order to prove REF , we have to examine when the indices are such that MATH. Again by REF , this is exactly the case if the system is of the form MATH where MATH, MATH, MATH and MATH. If MATH or MATH, then MATH is of block-triangular or even triangular form, which has been excluded. Hence MATH. (In fact, if MATH, then MATH is not equivalent to a system of triangular or block-triangular form.) Applying REF , we obtain the desired REF . |
math/9909019 | By CITE and definitions. |
math/9909019 | We define a function MATH by MATH . Then evidently MATH. |
math/9909019 | Let MATH and MATH where MATH is defined in REF . On the other hand by REF . Hence by the isomorphism MATH and REF this corollary holds. |
math/9909019 | If MATH then the proposition is trivial. We proceed by induction. Let MATH, and we want to calculate the cardinality of the set MATH. Assume that MATH. We have MATH and the sets in above relation are disjoint. Hence by induction we obtain MATH. |
math/9909019 | Let MATH and assume that MATH, which means that MATH contains a subsequence MATH. So by REF contains a subsequence MATH such that MATH is order-isomorphic to MATH, a contradiction. Hence MATH. On the other hand, let MATH contain MATH, which means that there exist MATH such that MATH is order-isomorphic to MATH. Let MATH be the maximal element in MATH. If MATH then the subsequence MATH of MATH is order-isomorphic to some permutation in the set MATH, so MATH. Hence we can assume that MATH. Let MATH be the maximal element in MATH; by the same reason we see that the only nontrivial case is MATH an so on. So MATH is just a permutation of the numbers MATH. Now, since MATH, there exists MATH such that the subsequence MATH, MATH of MATH is order-isomorphic to some permutation in the set MATH. Hence in any case MATH contains some permutation MATH with MATH, which means that if MATH contains MATH then MATH. |
math/9909019 | By REF, and by REF we obtain MATH. Hence, again by definition, we obtain MATH. |
math/9909019 | By definitions, REF and induction. |
math/9909019 | Assume first that MATH contains MATH. By definitions we have MATH. On the other hand MATH, and by REF we obtain MATH, which means that MATH, so again by REF we obtain MATH. Now let MATH, then MATH, which means that if MATH then MATH. By taking MATH, we get that MATH must contain MATH. |
math/9909019 | By REF and CITE. |
math/9909019 | By REF and definitions. |
math/9909019 | MATH . Let MATH, and let us consider the possible value of MATH: CASE: MATH. Evidently there exist MATH, MATH, hence MATH either contains MATH which is order- isomorphic to MATH, or contains MATH which is order-MATH-isomorphic to MATH, a contradiction. CASE: MATH. Similarly to case MATH we have MATH or MATH. If MATH then MATH if and only if MATH. If MATH then MATH, since otherwise MATH contains MATH, hence MATH if and only if MATH. CASE: MATH. Evidently MATH if and only if MATH. Since the above cases are disjoint we obtain MATH. MATH . Let MATH, and let us consider the possible value of MATH: CASE: MATH. Evidently there exist MATH, MATH. Since MATH avoids MATH we get MATH, hence MATH contains MATH which is order-MATH isomorphic to MATH, a contradiction. CASE: MATH. Let MATH, MATH. Since MATH avoids MATH we get MATH. If MATH or MATH then MATH contains MATH, a contradiction. So MATH. Hence MATH if and only if MATH. CASE: MATH. If MATH then MATH contains MATH, so we have MATH. Hence MATH if and only if MATH. CASE: MATH. Evidently MATH if and only if MATH. Since the above cases are disjoint we obtain MATH. MATH . Let MATH, and let us consider the possible value of MATH: CASE: MATH. Since MATH avoids MATH we get, similarly to case MATH, that MATH contains MATH, which is order-MATH-isomorphic to MATH, a contradiction. CASE: MATH. If MATH then MATH contains MATH, and if MATH then MATH contains MATH, so we have that MATH. If MATH then MATH contains MATH, so MATH. Hence MATH. That is, MATH if and only if MATH. CASE: MATH. Similarly to case MATH we have MATH, so MATH. Hence MATH if and only if MATH. CASE: MATH. Evidently MATH if and only if MATH. Since the above cases are disjoint we obtain MATH. |
math/9909019 | MATH . Let MATH and let us consider the possible value of MATH: CASE: MATH. Impossible, similarly to case MATH in REF . CASE: MATH. Similarly to case MATH in REF we have that MATH or MATH. If MATH then we get MATH since otherwise MATH contains MATH. Besides, MATH for all MATH since otherwise MATH contains MATH. So there is only one such permutation, namely MATH. If MATH then MATH if and only if MATH. CASE: MATH. Evidently MATH if and only if MATH. Since the above cases are disjoint we obtain MATH. By the transformation MATH we get MATH. Besides MATH and MATH, which means that MATH, hence MATH for all MATH. It is easy to see that for MATH the same formula holds. MATH . Let MATH, and let us consider the possible value of MATH: CASE: MATH. If MATH then MATH contains MATH. Let MATH, if MATH then MATH contains MATH otherwise MATH contains MATH. Hence MATH. If MATH then MATH contains MATH, so we have that MATH. If MATH then MATH contains MATH, otherwise MATH contains MATH, a contradiction. CASE: MATH. Since MATH avoids MATH, we have only one permutation MATH. CASE: MATH. Since MATH avoids MATH, we have that MATH. Hence MATH if and only if MATH. CASE: MATH. Evidently MATH if and only if MATH. Since the above cases are disjoint we obtain MATH; besides MATH and MATH. Similarly to the first part of the proof we obtain MATH for all MATH. |
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