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gr-qc/0105066 | If MATH, MATH or MATH is zero the lemma is trivial. Assume they are non-zero. On the left hand side the last two deltas combine to give MATH . By summing over all possible positions of the dummy indices MATH we get MATH where MATH depending on whether the index configuration is an even or an odd permutation. It is now clear that we get only the two types of terms that are on the right hand side of the theorem; the last type when both dummy indices are on MATH and the other type when at least one of them is on a MATH. What remains is to confirm the coefficients, which essentially means counting how many of each kind occurs and what sign they have. When both dummy indices are on MATH they can be moved to the first position. That means that they have moved a total of MATH positions and changed sign equally many times. That gives a factor MATH. There are a total of MATH such terms. When both dummy indices are on the same MATH they have moved equally many positions; thus such terms are added. There are MATH such terms and we also get a factor MATH since MATH. When both dummy indices are on different MATH one of them can be absorbed thereby distorting the order of the other indices. Once that order is restored we have overall an odd permutation; thus such terms are subtracted. There is a total of MATH such terms. The same situation occurs when one of the dummy indices is on a MATH and the other one is on MATH. There are MATH such terms with the upper index on MATH and MATH with the lower index on MATH. Taking all this together gives the identity in the lemma. |
gr-qc/0105066 | The basic idea is to repeatedly take traces of REF getting a sequence of equations; in the last of these no MATH remains. The result follows from substituting equations at the end of this sequence into earlier ones. However, care must be taken so that no unwanted cancelling of terms occur which forces this process to stop prematurely. To make the proof easier to overview we define the following notation. Let MATH and if MATH and MATH are less then the actual number of indices on MATH then we contract over the remaining ones. Thus REF can be written as MATH and MATH itself as MATH and the trace of MATH as MATH. We observe that the coefficients in REF are non-zero if MATH is non-zero and MATH and MATH are non-zero respectively. First assume that MATH is even, then the coefficients in REF are non-negative. Start with REF and multiply with MATH and use REF . Then we get MATH where we have omitted the coefficients. Doing that once more gives us MATH. Repeating this process gives a sequence of equations as illustrated in REF where each row corresponds to one equation. The last equation is MATH (assuming MATH). Putting this into the second last equation gives MATH by using that MATH. Feeding the new information into earlier equations gives the desired conclusion MATH. If MATH is odd then there is a minus sign in REF which means that there is a risk of cancellation in the above process. However, checking how the signs propagate gives us REF where the sign at a node is the sign of the term and the sign at the edge is the sign of the coefficient in the identity in REF . It is now clear that there will be no cancellations since each term originates as the difference between two terms with different sign. |
gr-qc/0105066 | MATH: Follows directly from REF MATH: The case MATH is trivial. Assume MATH. Contracting MATH once and using REF gives MATH . The theorem now follows from REF applied to the tensor MATH . |
gr-qc/0105066 | If MATH or MATH is zero the lemma is trivial. Assume they are non-zero. Absorbing the last delta on the left hand side gives MATH . By summing over all possible positions of the lower dummy index MATH we get MATH where MATH depending on whether the index configuration is an even or an odd permutaion. When the MATH is on a delta it can be absorbed giving terms of the same type as the first term on the right hand side of REF. There are MATH such terms and on each of them the indices have been moved a total of MATH steps. When the MATH is on MATH it can be moved to the first position of the lower indices giving terms of the same type as the second term on the right hand side of REF. There are MATH such terms and on each of them the index has been moved a total of MATH steps. |
hep-th/0105072 | We can check the well-definedness of the equivalence relation using REF . Because MATH freely acts on MATH, MATH is indeed a line bundle. The function MATH is MATH-valued, so the standard Hermitian structure on MATH defines that on MATH. |
hep-th/0105072 | If we take the other section MATH, then there exists a unique MATH such that MATH. By REF we have MATH . Hence the CS action is well-defined. Obviously it is of unit norm. |
hep-th/0105072 | If an orientation preserving diffeomorphism MATH is given, then we have the orientation preserving diffeomorphism MATH. This diffeomorphism induces the orientation preserving diffeomorphism MATH. This proves REF . REF is obvious from the definition of MATH. |
hep-th/0105072 | We can easily check the formula MATH . Using REF we can prove this lemma. |
hep-th/0105072 | These properties are the consequences of the properties of the disk MATH. REF are proved directly using REF is proved by the definition of the disk MATH. We can prove REF by writing explicitly the contraction of the line bundles MATH. |
hep-th/0105072 | For REF , we construct the functor by MATH. For REF , we construct the functor by MATH. By the definition of the product of MATH-gerbes these functors give the isomorphism of MATH-gerbes. For REF , the functor MATH is given by MATH. For REF , the functor MATH is given by MATH. Because MATH acts on MATH simply transitively, these functors are isomorphisms of MATH-gerbes. |
hep-th/0105072 | Since MATH is the center of MATH, we can easily show this lemma. |
hep-th/0105072 | The map MATH defined by MATH gives the isomorphism of the liftings. We can also construct the isomorphism MATH easily. These isomorphisms give rise to a MATH-action on MATH. For liftings MATH the set of all morphisms from MATH to MATH give rise to a MATH-torsor which we denote by MATH. There is a unique isomorphism MATH. Using this MATH-torsor we can prove that the MATH-action is simply transitive. |
hep-th/0105072 | By the definition of the product we have MATH. So MATH are expressed as an element of MATH and the map MATH is well-defined. The map is a homomorphism because MATH is the center. We can easily verify that the homomorphism is an isomorphism of groups which makes the following diagram commutative. MATH . |
hep-th/0105072 | We can directly show this proposition using REF . |
hep-th/0105072 | We can show this lemma in the proof of REF . |
hep-th/0105072 | For REF , let MATH and MATH be objects in MATH and MATH respectively. We define the functor MATH by assigning the formal object MATH to the product of liftings MATH. By the definition of MATH-action on MATH the functor is well-defined. The quasi-inverse functor of this functor is constructed as follows. Let MATH be a lifting of MATH for MATH. Take a trivialization MATH. This induces the trivialization MATH. We put MATH for MATH. The trivialization MATH induces the natural isomorphism of liftings MATH. The lifting MATH is unique up to the unique isomorphisms defined below. If MATH is the other trivialization, there exists a unique element MATH such that MATH for all MATH. By multiplying MATH we have the unique isomorphism MATH. The unique isomorphism commute with the natural isomorphism from MATH to the product of liftings. Then the assignment MATH gives the quasi-inverse functor. For REF , let MATH be a lifting of MATH for the trivial central extension MATH. Since the quotient space MATH is a MATH-torsor, we have a functor MATH. If a MATH-torsor MATH is given, then MATH is clearly a lifting of MATH for the trivial central extension. This gives the quasi-inverse functor. For REF , by the help of REF and the isomorphism of REF , a lifting MATH for MATH defines a functor MATH. The assignment MATH gives the isomorphism of MATH-gerbes MATH. |
hep-th/0105072 | By using REF the assignment MATH defines the functor above. We construct a quasi-inverse functor of this functor as follows. Let MATH be a lifting of MATH for the central extension MATH. By the help of REF we can write MATH, where MATH is a lifting of MATH for the central extension MATH. Since MATH, we obtain the lifting MATH of MATH for MATH. Similarly we obtain the lifting MATH of MATH for MATH. Then the assignment MATH gives the quasi-inverse functor. |
hep-th/0105072 | The relation MATH is indeed an equivalence relation because of REF . Clearly the right action is well-defined and is simply transitive. The map MATH is also well-defined since MATH. We can easily check that it commutes with the actions of the groups. |
hep-th/0105072 | For REF , MATH denotes the diffeomorphism covered by MATH. Let MATH be a lifting of the MATH-bundle MATH. We define a MATH-action on MATH via the isomorphism MATH induced from REF . If we define a map MATH by MATH, then MATH is a lifting of the MATH-bundle MATH. This assignment of liftings gives rise to the isomorphism REF . If a bundle map MATH is given, then the map MATH defined by MATH induces the map MATH. By the functoriality of the integration over manifolds the induced map gives the isomorphism of the liftings of MATH . For REF , we have the isomorphism MATH of the central extensions of MATH from REF . By the help of REF we have the isomorphism of REF . Since the integration over manifolds satisfies MATH, we can easily construct the isomorphism REF . For REF , we prove the case of MATH. It is clear that MATH. By using REF we get the isomorphism MATH. We can easily see that MATH. Applying REF we have the isomorphism REF . We also have the isomorphism REF using the additivity of the integration over manifolds. For REF , we describe the image under the isomorphism MATH first. Let MATH be a lifting of MATH-bundle MATH. By choosing a trivialization of MATH we write MATH, where MATH and MATH are liftings of MATH and MATH respectively. Using REF we identify MATH with a lifting on MATH for the central extension MATH. Under the isomorphism MATH we obtain the corresponding MATH-torsor MATH. Therefore we have MATH. This construction seems to depend on the choice of the trivialization. Recall the proof of REF . If we take the other trivialization, then we have the unique isomorphism MATH. This induces the unique isomorphism MATH. Hence the image of MATH is uniquely determined up to the unique isomorphism. Now we put MATH. The trivialization is induced by choosing a section MATH. We put MATH. Though MATH in general, we can choose a section such that MATH by using the action of MATH. We fix such a section MATH and put MATH. Under this choice we can put MATH and MATH, where the equivariant maps are defined by MATH respectively. We can easily see that the inverse lifting MATH is naturally isomorphic to MATH. Hence we have natural isomorphisms MATH and MATH. If we take the other section MATH, then there exists a unique element MATH such that MATH. The isomorphism MATH is defined by MATH. Note that MATH and MATH give the corresponding elements MATH and MATH. Using REF and the property of the integration, we have MATH. Hence the unique isomorphism MATH is defined by MATH. Therefore the isomorphism of liftings MATH defined by MATH gives rise to the isomorphism REF . |
hep-th/0105259 | Notice that the linear part of MATH is MATH and also, that the above expansion preserves the MATH-degree. We choose a linear basis in MATH with the following ordering MATH namely, in blocks of increasing MATH-degree and, within each block, non-decreasing MATH-degree. The above remarks then show that the matrix MATH, defined by MATH where MATH is also ordered as in REF , is upper triangular, with units along the diagonal and hence invertible. |
hep-th/0105259 | We have MATH . In the above inner product, only the MATH-linear terms in MATH contribute, since the MATH's vanish on products and the unit function. One particular way of evaluating the MATH-fold coproduct is to apply MATH always on the rightmost tensor factor. It is then clear that, in this case, we may instead apply MATH, since MATH and MATH only differ by terms containing products of the MATH's or units (this is only true if MATH is applied in the rightmost factor). Notice now that the MATH-product of the MATH's is dual to MATH . Repeated application of this equation and use of the definition of MATH, REF , completes the proof. |
hep-th/0105259 | Consider the algebra MATH of formal power series MATH, MATH, with the usual product. Define a basis MATH of MATH, the dual of MATH, via MATH that is, MATH reads off the coefficient of MATH in MATH and MATH. For MATH we have MATH which implies the coproduct MATH in MATH. Endowing MATH with a commutative product, we arrive at the isomorphism MATH, as NAME algebras, with MATH. Define a new basis MATH in MATH by MATH and MATH. Then MATH implying the coproduct MATH. The MATH's, under the above isomorphism, correspond to the MATH in MATH. Solving the equation MATH for MATH, one arrives at REF . |
hep-th/0105259 | We apply the inverse NAME lemma to MATH. For a given MATH-degree MATH, only MATH of MATH-degree up to MATH enter in the formulas - we denote them collectively by MATH (for example, MATH denotes the standard expression of MATH in terms of the MATH while MATH denotes the same expression with every MATH multiplied by MATH). Consider the family of diffeomorphisms MATH, MATH. Then MATH is the identity map while MATH is the zero map. The corresponding velocity field is MATH where MATH. We have MATH . However, MATH and, taking into account the closure of MATH, we find MATH . This is the inverse NAME lemma. We concentrate now on the action of MATH on MATH. We have MATH . In this latter (implied) sum, all terms in the coproduct of MATH appear except the first one, MATH, which is annihilated by MATH. Notice now that MATH. Since MATH is linear in its second factor we conclude that MATH . Substituting back into REF and putting MATH we find MATH which, upon performing the integration over MATH, gives MATH. The remarks preceding the theorem complete the proof. |
hep-th/0105259 | The various powers of the coproduct of MATH can be computed by iteration of the second of REF , MATH . This shows that in MATH, the MATH-linear term can only be produced by the MATH-nested commutator MATH . The latter, however, has no MATH component, since MATH can be written as a MATH-nested commutator at most. It is also clear, for the same reason, that there are no terms of higher MATH-degree in the MATH's, as those would correspond to even longer nested commutators. MATH then must have at least one unit tensor factor in each of its terms. On the other hand, the MATH-linear term in MATH is not zero, because, by definition, the corresponding MATH-nested commutator has a MATH component. |
hep-th/0105259 | Any linear combination of the MATH is primitive, while (sums of) products of them are not, due to REF . Therefore, the MATH constitute a linear basis in the vector space of primitive elements of MATH. To the given MATH, REF associates a primitive element MATH, with MATH as its linear part. The unique linear combination of the MATH (and, hence, of the MATH) with this linear part is MATH. |
math-ph/0105001 | See CITE and CITE. |
math-ph/0105001 | The claim follows from a combination of REF with REF below. |
math-ph/0105001 | Let MATH be a continuous function. Define MATH by MATH. Then also MATH is a continuous function. Therefore MATH where the fourth equality follows from MATH . Since REF holds for any continuous function MATH, the statement of the lemma follows from the NAME representation theorem. |
math-ph/0105001 | We first approximate our process by finite-volume pure-jump processes and use NAME 's formula to obtain the densities of these processes with respect to the independent spin-flip process. Indeed, denote by MATH the path-space measure of the finite-volume approximation with generator REF and by MATH the path-space measure of the independent spin-flip process in MATH, that is, the process with generator MATH . We have for MATH such that MATH, MATH . Since MATH is the path-space measure of the independent spin-flip process, the transformed measure MATH equals MATH. Abbreviate MATH . Then we obtain MATH where MATH can be computed from NAME 's formula (see CITE p. REF) and for MATH large enough reads MATH where MATH is the number of spin flips at site MATH up to time MATH along the trajectory MATH. We thus obtain the representation of REF by observing that MATH does not depend on MATH for MATH large enough and using the convergence of MATH to MATH as MATH. Indeed, the only point to check is that MATH so that the conditional expectation in REF is well-defined. However, this is a consequence of the following two observations: CASE: MATH is uniformly bounded because MATH. CASE: For MATH we have the bound MATH where, as before, MATH and MATH are the maximum and minimum rates, MATH is the total number of spin flips in the region MATH up to time MATH along the trajectory MATH. Since the rates are bounded from above, the expectation of the Right-hand side of REF over MATH is finite uniformly in MATH. |
math-ph/0105001 | This is an immediate consequence of REF . See also CITE. |
math-ph/0105001 | Let us first compute MATH in the case of the finite-volume reversible NAME chain with generator REF . For the sake of notational simplicity, we omit the indices MATH referring to the finite volume, and abbreviate MATH: MATH where MATH is defined in REF . |
math-ph/0105001 | See CITE. REF is expected to be true without the requirement of monotonicity but this has not been proved. |
math-ph/0105001 | During the proof we abbreviate MATH. We prove that the limit MATH converges uniformly in MATH for MATH small enough when MATH. The MATH depends on both MATH and MATH. Let us write MATH to denote the range of MATH (see REF). I: MATH, MATH. To warm up, we first deal with unbiased independent spin-flip dynamics. For this dynamics the distribution of MATH under MATH coincides with the distribution of MATH under MATH. Therefore we can write MATH where MATH is a continuous function of MATH, the sum runs over MATH while MATH . The notation MATH, MATH, is defined by MATH with MATH the configuration obtained from MATH by flipping all the spins in MATH. Suppose first that MATH. Then MATH . For MATH we can decompose MATH into disjoint nearest-neighbor connected subsets MATH and thus rewrite REF as follows: MATH with MATH and MATH. Note that MATH for all MATH that do not contain MATH. Next, since MATH with MATH we have the estimate MATH . A similar estimate holds for MATH. Since MATH as MATH, it follows that for MATH small enough we can expand the logarithm of both the numerator and the denominator in REF in a uniformly convergent cluster expansion: MATH . By the estimate REF we have, for MATH small enough, MATH and hence we obtain uniform convergence of the limit in REF . The case MATH is treated in the same way. We only have to redefine the MATH's as the MATH-connected decomposition of MATH. Note that MATH depends on MATH and converges to zero when MATH. II: MATH, MATH. Next we prove that the limit REF converges uniformly if both interactions MATH are finite range. For the sake of notational simplicity we first restrict ourselves to the case MATH. We abbreviate MATH. The idea is that we go back to the independent spin-flip dynamics via NAME 's formula. After that we can again set up a cluster expansion, which includes additional factors in the weights due to the dynamics. The first step is to rewrite REF in terms of the independent spin-flip dynamics: MATH . For a given realization MATH of the independent spin-flip process, we say that a site MATH is MATH-active if the spin at that site has flipped at least once. The set of all MATH-active sites is denoted by MATH. Let MATH denote the trajectory that stays fixed at MATH over the time interval MATH. For MATH, define MATH and put MATH . Also define MATH where the trajectory MATH is defined as MATH . With this notation we can rewrite the right-hand side of REF as MATH where MATH is a continuous function of MATH. In order to obtain the uniform convergence of REF , it suffices now to prove the uniform convergence of the expression between brackets in REF . As in REF, we decompose the set of MATH-active sites into disjoint nearest-neighbor connected sets MATH and rewrite, using the product character of MATH, MATH . The cluster weights are now given by MATH and an analogous expression for MATH after we replace MATH by MATH. The factor MATH arises from the probability MATH . Having arrived at this point, we can proceed as in the case of the independent spin-flip dynamics. Namely, we estimate the weights MATH and prove that MATH with MATH as MATH. To obtain this estimate, note that MATH . Then apply to REF NAME, the bounds MATH on the flip rates, and the estimate MATH to obtain MATH . This clearly implies REF . The case MATH is straightforward after redefining the MATH's. |
math-ph/0105001 | The joint distribution of MATH is NAME with Hamiltonian (recall REF ) MATH . For fixed MATH, the last term is constant in MATH and can therefore be forgotten. Since MATH differs from MATH only in the single-site interaction, MATH satisfies REF if and only if MATH satisfies REF . Hence MATH for any MATH, and we conclude from REF that MATH is NAME. |
math-ph/0105001 | The proof uses REF . CASE: For small MATH the dynamical field MATH is large and, for given MATH, forces MATH in the direction of MATH. Rewrite the joint Hamiltonian in REF as MATH . For MATH small enough, MATH has the unique ground state MATH and so, for MATH large enough, MATH satisfies REF (see REF , p. REF). Therefore, for MATH such that MATH, MATH has a unique NAME measure for any MATH. Hence, MATH is NAME by REF . For large MATH the dynamical field MATH is small and cannot cancel the effect of the external field MATH. Rewrite the joint Hamiltonian as MATH . For MATH large enough (independently of MATH), MATH has the unique ground state MATH. Hence, for MATH large enough, MATH has a unique NAME measure by REF (again, see REF , p. REF). Hence, MATH is NAME by REF . This fact is a consequence of the results in REF, for the single-site NAME transformation. Since the joint Hamiltonian in REF is ferromagnetic, it suffices to show that there is a special configuration MATH such that MATH. We choose MATH to be the alternating configuration. For MATH large enough, MATH has two ground states, and by an application of NAME theory (see REF), it follows that, for MATH large enough, MATH. Therefore MATH is a bad configuration for MATH, implying that MATH is not NAME by REF . In this case we rewrite the Hamiltonian in REF as MATH . For ``intermediate" MATH we have that MATH and MATH are of the same order. As explained in REF, we can find a bad configuration MATH such that the term MATH in the Hamiltonian ``compensates" the effect of the homogeneous-field term MATH and for which MATH has two ground states which are predominatly plus and minus. Since the proof of existence of MATH requires analysis of the random field NAME model, we have to restrict to the case MATH (unlike the previous case MATH is not constructed, but chosen from a measure one set). Then for MATH large enough, by a NAME argument (see REF, Theorem B REF) MATH, implying that MATH is not NAME by REF . |
math-ph/0105001 | The last term in REF being irrelevant, we can drop it and study the Hamiltonian MATH . This Hamiltonian is of the same form as REF , but with MATH becoming MATH-dependent. We have MATH and MATH with MATH so that, in the regime where MATH, MATH, MATH, we find that the effect of MATH dominates. Hence we can find a special configuration that compensates the effect of the field MATH and for which the Hamiltonian REF has two ground states, implying that MATH. Similarly, when MATH we can find MATH intermediate such that MATH is ``compensated" by MATH. |
math-ph/0105001 | For fixed MATH, the Hamiltonian MATH of REF corresponds to an interaction MATH. By REF , this interaction satisfies MATH . Therefore, for MATH small enough, REF is satisfied for the interaction MATH for all MATH and all MATH. Hence MATH, and we conclude from REF that MATH. |
math-ph/0105001 | CASE: This a consequence of REF . CASE: This is proved in exactly the same way as the corresponding point in REF . CASE: Here we cannot rely on monotonicity as was the case in REF . It is therefore not sufficient to show that for the fully alternating configuration MATH, the Hamiltonian MATH exhibits a phase transition. We have to show the following slightly stronger fact: if MATH is any NAME measure corresponding to the interaction MATH, then MATH . This proof of this fact relies on NAME theory for the Hamiltonian MATH. The first step is to prove that the all-plus-configuration is the unique ground state of this Hamiltonian. Since the NAME Hamiltonian satisfies the NAME condition, we conclude from REF that the set of ground states of MATH is a subset of MATH. If we drop the term MATH (that is, if MATH), then the remaining Hamiltonian has as the unique ground state the all-plus-configuration and satisfies the NAME condition. Therefore, for MATH small enough, we conclude from REF that MATH has the all-plus-configuration as the only possible ground state. From REF it is easy to verify that the all-plus-configuration is actually a ground state for MATH small enough. In order to conclude that for MATH large enough, the unique phase of MATH is a weak perturbation of the all plus configuration (uniformly in MATH), we can rely on the theory developed in CITE, or CITE which allows exponentially decaying perturbations of a finite range interaction satisfying the NAME condition (see for example, REF). Similarly, MATH has a unique phase which is a weak perturbation of the all minus configuration. This is sufficient to conclude that no version of the conditional probabilities is continuous at MATH, see the discussion CITE p. CASE: We can use the same argument as developed in REF, introducing a random perturbation of the alternating configuration to ``compensate the uniform magnetic field" (since this requires analysis of the random field NAME model, we have the restriction MATH). The only complication is the extra term in the Hamiltonian arising from MATH. This requires NAME theory for the interaction MATH, where MATH is a random modification of the fully alternating configuration obtained by flipping the spins in the alternating configuration with probability MATH for a flip from MATH to MATH. Since the couplings between MATH and MATH are not finite range, we cannot apply directly REF for the random Hamiltonian MATH. However, as the interaction decays exponentially fast and NAME analyses do not distinguish between finite range and exponentially decaying interactions, similar arguments as those developed in CITE still work in our case and yield the analogue of REF. However we have not written out the details. |
math-ph/0105003 | This result is similar to the classification of all three-line locus configurations on the plane (REF from CITE), and can be proven in a similar way. To begin with, notice that REF are equivalent to the vanishing along MATH of some first odd normal derivatives of the potential MATH. More explicitly, they can be reformulated as follows: for all MATH . Applying this for our particular case, we arrive at the following conditions: MATH plus similar conditions for MATH and MATH. Suppose that MATH, then from the first two relations in REF we deduce immediately that MATH, or MATH, and hence MATH. As a result, we see that if at least two of the multiplicities MATH are greater than MATH, then they all must be the same and MATH, which up to sign flips gives us the NAME case MATH. Another possibility is MATH with MATH. As we know already, in this case MATH which provides REF . The two remaining locus conditions are: MATH which gives MATH and leads to the families MATH and MATH. Finally, for the remaining case MATH we have three locus conditions MATH . It is easy to see that there are only two possibilities: either MATH or MATH. The first one gives us the NAME case while the second one leads to the family MATH. |
math-ph/0105003 | As we mentioned already, all we have to do is to check the locus conditions for every two-dimensional subsystem in MATH. So, let us take any two-dimensional plane MATH in MATH and consider the set MATH assuming that it is nonempty. Then there are three possibilities: CASE: MATH consists of one vector; CASE: MATH consists of two perpendicular vectors; CASE: MATH consists of three vectors MATH for some MATH. In the first two cases MATH is a locus system for trivial reasons. So, essentially, the locus conditions for the system MATH reduce to the requirement that for all MATH the subsystem MATH as in REF must be one of those listed in REF . Let us consider first the case MATH. We have four different subsystems as in REF. The main point is that they cannot all be of the type REF. Indeed, in that case we would have that MATH which would imply that MATH. Thus, either all of these subsystems are of the NAME type REF, or at least one of them is of type REF - REF. In the latter case, if we suppose that MATH and MATH then for MATH we have only two possibilities: either MATH, or MATH. Therefore, all possible locus configurations for MATH are listed in the theorem. The general MATH case follows in a similar manner. |
math-ph/0105003 | Let us identify MATH with its dual using the standard Euclidean structure. Then the bilinear form MATH associated to the system REF according to REF looks as follows: MATH . The associated matrix which we will denote by the same symbol MATH has the form MATH where MATH and MATH. A straightforward check shows that its inverse has the form MATH where MATH and MATH for all MATH. To verify MATH-conditions, we should deal with two-dimensional planes MATH containing at least two of the vectors MATH. Altogether we have the following MATH different types of such planes: CASE: MATH; CASE: MATH; CASE: MATH; CASE: MATH. Let us consider the first case. Let us fix a basis in MATH as MATH and MATH. Then the corresponding plane MATH is spanned by MATH and MATH. Using the explicit formula for MATH one easily finds that (up to a nonessential factor) MATH (with MATH appearing in the MATH-th component) and similarly MATH . Now we should check that the restrictions of the forms MATH and MATH onto MATH are proportional. Here MATH is given by the formula MATH . After some calculations one finds that MATH . On the other hand, evaluating MATH we obtain that MATH . This demonstrates that MATH and MATH are proportional and gives MATH-condition for REF . In REF we take MATH, MATH and have only to check that the Euclidean product MATH is zero. The latter becomes obvious after calculating MATH which is proportional to the vector MATH. Two other cases are completely analogous. As a result, we conclude that the system REF is a MATH-system for any values of the parameters MATH and the corresponding function REF is a solution of the generalised WDVV equation. |
math-ph/0105010 | Since the cohomology class of MATH is killed by MATH, MATH is a coboundary. In other words, there is a MATH such that MATH, where MATH is the coboundary operator. Since MATH, one can choose MATH such that MATH. Let MATH. Then MATH is in the same cohomology class as MATH, and MATH. In terms of MATH and MATH, this means that MATH, or MATH. |
math-ph/0105010 | Since MATH is a finitely generated free Abelian group, the short exact sequence MATH leads to the short exact sequence MATH and MATH, with MATH. Since MATH is uniquely divisible, its NAME cohomology groups vanish, so the long exact sequence of NAME cohomology gives MATH . |
math-ph/0105010 | According to REF , the natural map from MATH to MATH is an isomorphism, where MATH as in Notation REF. The finiteness hypotheses on MATH and MATH imply that MATH is a finite group. It follows that the pairing in the theorem takes values in MATH and that MATH. Roughly speaking, the finiteness hypotheses imply that one can replace MATH with MATH throughout. According to CITE, there is a duality pairing between MATH and MATH that identifies each with the dual of the other (in the sense of Notation REF). Up to a sign, this pairing agrees with the one in the statement of the theorem by CITE and CITE. In particular, this shows that MATH is a finite group, so MATH. Thus the duality of the theorem is just a restatement of the duality between MATH and MATH. |
math-ph/0105010 | This is simply a restatement of the injectivity of the map MATH. |
math-ph/0105010 | Let MATH denote the subgroup of MATH generated by MATH. We claim that MATH is trivial in MATH. By REF , it suffices to show that MATH for all MATH. According to REF below, MATH, where MATH. Therefore the hypothesis MATH justifies the claim. Since MATH is trivial in MATH, there is some MATH for which MATH. Then MATH represents the same class in MATH, and MATH. |
math-ph/0105010 | It is easier to work with homology of MATH than the cohomology of MATH, so consider MATH. The only vector in MATH fixed by a non-trivial rotation is the zero vector. According to REF, this shows that MATH. The result now follows from REF . |
math-ph/0105010 | By REF , it suffices to compute MATH. By REF , MATH. Then REF implies that MATH. According to REF , below, MATH is a unit unless MATH is a prime power, in which case its norm is MATH. Thus MATH, and MATH, unless MATH. Suppose now that MATH. Then MATH is a vector space over MATH, the field with MATH elements. If MATH, this justifies the last claim in the statement. Now assume that MATH is odd. Decompose MATH into eigenspaces for MATH: MATH. On MATH, MATH, so MATH (since multiplication by MATH is an isomorphism on a MATH-vector space when MATH is odd), and MATH by REF. On the other hand, MATH, so MATH as well. Therefore, MATH. |
math-ph/0105010 | Let MATH denote the cyclotomic polynomial of order MATH. That is, MATH is the monic, irreducible polynomial whose roots are the primitive MATH-th roots of unity. Since these are exactly the conjugates of MATH over MATH, it follows that MATH . Since the roots of MATH are all the MATH-th roots of unity, MATH. Dividing by MATH and setting MATH leads to MATH. The lemma now follows by induction on MATH. |
math-ph/0105010 | The case MATH is considered in REF , so assume that MATH. REF shows that the cohomology group vanishes if MATH is not a power of REF, so assume now that MATH. By REF , MATH is dual to MATH, so it suffices to show that this homology group has the stated form. By REF and the exact sequence REF, MATH. Since MATH, it follows from REF that MATH. Note that, although MATH is antilinear, MATH is linear as a map on MATH, so Notation REF applies. An easy calculation shows that each MATH . NAME block contributes a one-dimensional space to MATH and that each MATH . NAME block contributes nothing. |
math-ph/0105010 | This follows from associativity of cup and cap products, as described above, and from the compatibility MATH between the duality pairing and the cup product. |
math-ph/0105033 | Denote by MATH the right hand side of REF . Then MATH sends every vector in MATH to zero. Now the invariance of MATH implies that MATH, so by the NAME Lemma MATH is proportional to the identity on MATH. Since MATH, MATH restricted to MATH is the identity and MATH. Accordingly, MATH. |
math-ph/0105033 | Let MATH be the adjoint representation of MATH on MATH and MATH as defined in REF. Then MATH transforms as MATH where MATH is acting on MATH and MATH, that is, MATH is invariant under the action of MATH on MATH. This implies for MATH that MATH and finally for the first component of the NAME character MATH for all MATH. So MATH is an invariant element of MATH. Reducing this latter space as a MATH - module shows that MATH appears only once, as MATH. Consequently, the subspace of invariant two-forms is one dimensional and, since MATH is invariant, MATH can be written as claimed. |
math-ph/0105033 | With MATH as above the left hand side is by REF MATH . Now MATH are the components of the equivariant multiliear map MATH from MATH to MATH which is constant. This implies the assertion. |
math-ph/0105034 | It follows from REF that MATH can be written as a product of blocks of type MATH. The strategy proposed by NAME is to study the evolution of the initial vector MATH under these elementary blocks, that is, to enlarge the orbit in question. This enlarged orbit will be denoted by MATH, where MATH is the vector obtained after applying the first MATH blocks of type MATH to MATH. It was shown in CITE that if MATH for some MATH, then MATH for every MATH and MATH diverges super-exponentially in every coordinate. This readily gives MATH. Of course, MATH is obvious. To show that MATH, we need some preparation. Let MATH, MATH, be chosen such that MATH and define MATH. Since MATH preserve the invariant MATH, we have MATH for every MATH. It is easily seen (compare CITE) that, if MATH has at least two coordinates whose modulus is greater that one, where MATH denotes the norm MATH. Thus every MATH obeying REF has to be of one of the following forms, MATH . If a MATH satisfies REF, we say it is of type I, II, III, or IV if it satisfies the respective condition in REF. Let us now describe a scenario sufficient for entry in MATH (compare CITE for similar reasoning). Suppose MATH obeys REF and MATH where MATH. We consider the case MATH, the other case is similar. If MATH is of type I, II, or IV, then MATH is in MATH. This follows by simple direct calculations considering the cases MATH and MATH (compare REF as well). If MATH is of type III, then we proceed as follows (compare CITE). The iterate faces MATH for some MATH. If MATH obeys REF, then it must be one of the types I - IV. It is easy to see that MATH fixes the second component of MATH. Thus, MATH must be of type III as well. Using this, it is straightforward to see that MATH belongs to MATH. Thus, we infer that indeed the scenario CASE: MATH and MATH forces entry in MATH. Our next step is to investigate how growing of MATH (or MATH) can force REF . We only consider MATH. Consider the situation MATH . Thus we have at least one situation of the form REF between the two iterates and among these, we consider the one with largest MATH for definiteness. To make sure that MATH obeys REF it suffices to assume that MATH has norm sufficiently large. More precisely, let MATH be given with MATH whenever MATH. Then REF together with MATH forces that MATH obeys REF. To make sure that MATH obeys REF, note that MATH if MATH and MATH if MATH respectively. Thus, as MATH is an isometry, MATH obeys REF if MATH and MATH have norm larger than MATH. Let us summarize these considerations as follows: REF together with sufficient largeness of MATH and MATH force REF and thus entry in the ecape set. A completely analoguous arguments applies after interchanging MATH and MATH. To conclude the proof, we show now that if REF fails, then REF fails with explicit upper bound on the orbit. This will prove both MATH and the last statement of the Proposition. Thus, assume that REF fails for some energy MATH and write MATH for MATH and MATH for MATH. By definition of MATH, MATH and the invariance of MATH, we have MATH (A similar argument is used in the proof of REF .) Now, we have for every MATH where REF occurs with MATH large enough for REF to hold the following situation: If MATH is large, then we have a scenario sufficient for entry in MATH as explained above, which is a contradiction to failure of REF . If MATH is large (but MATH is not), then we consider the extended orbit between MATH and MATH and we are again in a scenario sufficient for entry in MATH because MATH can be chosen such that the role of the intermediate vector MATH is played by MATH whose norm is large by assumption, again a contradiction. This shows that if REF fails, we get an explicit upper bound for MATH in all situations where we have REF. In particular, REF fails and we have an upper bound on the MATH - orbit in this case. Note that the upper bound we get is an explicit function of MATH and MATH and hence is continuous in MATH. |
math-ph/0105034 | Recall from REF that MATH is equal to MATH for every MATH. It follows from a standard strong approximation procedure that MATH . In fact, the set MATH is just MATH where MATH is a MATH-periodic operator whose potential values over one period are given by MATH. Clearly, any operator MATH is a strong limit of such operators, so we can apply REF. REF to get REF. Using REF , we get the following chain of inclusions: MATH . |
math-ph/0105034 | Assume there exists MATH such that MATH. Pick some MATH for which the limit in REF exists. By NAME 's theorem CITE there exists a solution MATH of MATH such that MATH decays exponentially at MATH at the rate MATH, where MATH. Now, it follows from REF and the fact that the words MATH occur infinitely often in MATH CITE that the words MATH, MATH, occur infinitely often in MATH and hence in MATH. Since MATH there is a constant MATH such that, for every MATH, we have MATH . Pick MATH such that, for every MATH and every MATH, the solution MATH obeys MATH . Choose MATH such that MATH. Look for an occurrence of MATH in MATH, that is, MATH, such that MATH. It follows from the NAME theorem that MATH which in turn implies by REF MATH contradicting REF . |
math-ph/0105034 | This is well known CITE. |
math-ph/0105034 | As explained above, REF and compactness of MATH provide uniform bounds on MATH for MATH, MATH, that is, uniform bounds on MATH for every odd MATH and every MATH. By MATH this gives uniform bounds on MATH and MATH for every MATH and we can then use the invariant MATH to derive a uniform bound for MATH, MATH and every MATH. |
math-ph/0105034 | Let MATH be a quasi-Sturmian sequence and let MATH. Since MATH is necessarily recurrent, its restriction MATH to MATH has the form REF for some Sturmian MATH, some aperiodic substitution MATH, and some finite word MATH; see the appendix and REF in particular for details. Let MATH. It was shown in CITE that MATH starts with infinitely many squares conjugate to MATH or MATH. Thus the lemma follows immediately from REF. |
math-ph/0105034 | Using REF , we see that REF applies to MATH for every MATH and every MATH. Thus for every MATH, the operator MATH has empty point spectrum. |
math-ph/0105034 | Using REF this can be shown in the same way as in the Sturmian case. Details can be found in CITE (compare CITE as well). |
math-ph/0105034 | The proof is similar to the proof of the corresponding result in the Sturmian case (compare CITE). However, it requires some additional effort as the quasi-Sturmian systems are not reflection invariant. To treat this case as well we will need the results (and the notation) of the discussion at the end of REF. Recall that MATH. Mimicking the argument of CITE, which only relies on trace map bounds and recursions, one can easily infer that there exists MATH and MATH with MATH for every prefix MATH of MATH and every MATH. Now, every prefix of MATH can be written as MATH with a prefix MATH of MATH and a word MATH of length MATH. Combining these estimates, we see that there exist MATH and MATH with MATH . By, REF , we have MATH. Thus, replacing MATH by MATH but keeping the same set of energies, we can again apply the above reasoning and infer MATH . Now, the proof can be finished as follows. By results of CITE, every factor MATH of MATH can be written as MATH, where MATH is a suffix of a suitable MATH and MATH is a prefix of MATH. This implies that every factor MATH of a sequence in MATH can be written as MATH with MATH a suffix of MATH and MATH a prefix of MATH for a suitable MATH. Assume without loss of generality MATH. (The case MATH can be treated directly.) By REF , MATH with MATH and MATH. We will assume that MATH. The other case is similar (and, in fact, even simpler). Then MATH can be written as MATH, where MATH is a suffix of MATH. Now, we can estimate MATH . We will provide suitable bounds for all these factors. The factor MATH is just a constant. The factor MATH can be estimated by REF as MATH is a prefix MATH. We find MATH. It remains to estimate the factor MATH. As MATH is a suffix of MATH, the word MATH is a prefix of MATH. Since MATH is a palindrome, we have MATH. This means that MATH is a prefix of MATH. By REF , we can thus estimate MATH. By general principles, however, we have MATH (compare CITE) and we see that MATH can be estimated by MATH. Combining these estimates we arrive at the desired statement. |
math-ph/0105034 | The claim follows from REF along with NAME REF. |
math-ph/0105038 | It is clear that MATH preserves the complex polarisation of MATH. To verify that MATH preserves the symplectic structure, it is enough to verify that REF is indeed a Hamiltonian. Without the boundary term, this calculation is in NAME - NAME REF . To verify that MATH is a Hamiltonian for the vector field MATH on MATH. Let MATH represent a variation in MATH. Then we have MATH and so MATH . Integrating by parts in the first term, using MATH, we obtain MATH as required. Next we compare the NAME bracket of MATH and MATH, for two elements MATH of MATH, with MATH. The NAME bracket is equal to the variation of MATH, where we have written MATH for MATH. From above, this is simply MATH using the definition of curvature, MATH and writing MATH in the boundary term. |
math-ph/0105038 | From REF, MATH . In order to integrate by parts, note MATH . But from REF and the fact that MATH is trace-free, the first term on the right hand side is equal to MATH, one of the terms in the Hamiltonian. On the left hand side use REF to write MATH . Hence we find MATH . |
math-ph/0105038 | This is not completely trivial as vector fields in MATH are not in general tangent to the boundary. We get around this by representing MATH as a quotient of complexified vector fields that are tangent to the boundary whose MATH-part is holomorphic, by complexified vector fields, tangent to the boundary, whose MATH-part is zero. The formula can be extended to complex vector fields (that is, MATH with MATH independent of MATH, except on the boundary where MATH) by requiring complex linearity. We can now restrict to the NAME algebra of complex vector fields such that MATH is holomorphic on MATH. This latter subalgebra acts holomorphically as it is defined by the condition MATH, and so our formula simplifies to the boundary integral REF analogous to the case for complex gauge transformations. Thus, naive geometric quantization does quantize the action of this algebra of complex vector fields tangent to MATH with holomorphic MATH part. To obtain an action of MATH, observe that the subalgebra of complex vector fields of the form MATH, with MATH, is a NAME algebra ideal, acts trivially and the quotient of the algebra of complex vector fields tangent to MATH and with MATH holomorphic by this subalgebra is MATH. We have therefore obtained the desired holomorphic action of MATH on the determinant line bundle. |
math-ph/0105042 | Taking REF into account, we have to show that REF, if it is well defined, dominates the inner square. Assuming that the limit in question exists, it is easy to show that MATH majorizes the inner square MATH of MATH. Namely, using REF we can express MATH as MATH using REF of REF , and the fact that MATH which follows from it. Now, in every term in the sum above we have the elementary estimate for complex numbers MATH. If the first term MATH in REF has a limit at all, then it tends to MATH for a certain MATH, showing MATH in the limit MATH. It remains to show that all the limits involved in REF exist. In order to show finiteness of the first term it suffices to show that the decomposition REF of MATH converges in MATH for MATH, since it then tends to MATH and is thus necessarily finite as we have just seen. For the sum defining the decomposition, we have by REF MATH taking REF into account, showing that claim. By REF of the inner product, the MATH-th term in the sum in REF becomes MATH . For the first term we find, using the NAME - NAME estimate, by REF, and of course the assumption of REF : MATH . Further using REF, the second term is bounded by MATH . Finally the third term satisfies MATH . All three terms decay faster than exponentially in MATH, making the overall sum in REF convergent in the limit MATH. |
math-ph/0105042 | To show REF we estimate MATH . In the last step we used that we have MATH by REF . REF follows from REF. By REF it suffices to show that MATH entails MATH to show REF. For that, by the second to last inequality mentioned previously, it suffices to approximate MATH in the MATH-norm with elements of MATH. Such an approximation can be easily constructed, for example, as MATH for MATH, with the cut-off functions MATH of REF . That MATH is surjective is now clear, since MATH is the identity on MATH. For MATH and MATH, the scalar product reduces to MATH since MATH, and because MATH and MATH have decompositions with vanishing MATH and MATH, respectively. Continuity of MATH then implies REF . REF follows from REF and the fact that the sum in REF converges to an element of MATH. |
math-ph/0105042 | That the vectors MATH exist and are unique in MATH follows from NAME 's representation theorem applied to the bounded linear functionals MATH on the NAME space MATH. We have to show that they are in MATH. Choose a sequence MATH in MATH that approximates MATH, that is, MATH for MATH. Using the decomposition REF for the MATH we calculate (adopting NAME 's summation convention for repeated upper and lower indices) MATH . The last term on the right-hand side vanishes for all MATH due to REF. The third term is zero since MATH. We use the NAME - NAME estimate for the scalar product MATH and the fact that MATH to estimate the second term as follows: MATH with some constant MATH independent of MATH. In fact, since MATH, and using REF we see that the sum is finite for all MATH. Since the sequence MATH is convergent in the norm MATH and by REF of this norm, the sum must actually converge and therefore admits a global bound MATH as above. In conclusion, since MATH is MATH-convergent to MATH, that is, MATH for MATH, we must have MATH by necessity, and thus already MATH is MATH-convergent to MATH. This shows the claim. |
math-ph/0105042 | REF is clear from the defining property of MATH, except for the last equality that says MATH. This will soon turn out to be true. Let MATH be a sequence converging to MATH in MATH, which exists by REF . Then with REF, and since MATH for all MATH we have MATH . Now MATH by REF, so that the left-hand side tends to MATH for MATH (here we assume that the denominators are nonzero which can be achieved by choosing MATH suitably). Since the denominators stay bounded, we must necessarily have MATH for MATH showing REF, and also MATH showing REF since MATH converges to MATH with respect to MATH and MATH majorizes the inner square. Incidentally, this also shows MATH and this proves the last equality in REF, since the norm of MATH and that of the linear functional MATH coincide by NAME 's theorem. To show REF, we consider again the decomposition REF of a vector MATH which yields MATH . In this expression we find MATH, since by REF MATH converges strongly to MATH in MATH, and due to REF . By arguments similar to that in the proof of REF , the sum stays bounded independently of MATH, and since every single term in it converges to MATH by REF, the sum also tends to MATH. This leaves us with the second term which converges to MATH by REF. This shows REF. |
math-ph/0105042 | First, we must show that the decomposition is indeed possible because MATH. To this end, note that the topology MATH is stronger than the restriction of MATH to MATH. In fact, if a sequence in MATH converges in the norm MATH then it converges in the MATH-norm by REF, and by the action of the indefinite product on MATH it is easy to see that this suffices to ensure convergence in the norm MATH. Now, taking REF into account, we have to show that the MATH-orthogonal decomposition MATH holds. First, observe that the vectors MATH form a MATH-complete orthonormal system in MATH. Now, for MATH, MATH we have MATH by the definition of MATH and REF. This shows that a sequence MATH in MATH converges to a limit MATH if and only if MATH and independently the MATH-orthogonal projection of MATH onto the closed subspace MATH of MATH tends to zero. Denote by MATH the orthogonal complement of MATH in MATH with respect to MATH. By the above-given argument, the subset MATH of MATH is dense in MATH with respect to the topology MATH. This shows that the proposed decomposition is indeed MATH-orthogonal. In conclusion, a MATH-Cauchy sequence in MATH can be identified with a pair MATH with a MATH and MATH for some MATH. This shows MATH. |
math-ph/0105042 | The strategy of the proof will be as follows: The metric operator exists by REF , and we have seen in REF that we can write down its action in the decomposition MATH. We can then explicitly demonstrate that the operator MATH on MATH acts as stated. This special form of MATH immediately implies that it is a bounded, completely invertible operator on MATH. Thus by REF , MATH is a NAME space and since MATH is also bounded, its NAME space structure MATH is maximal by REF . Now, by definition of the MATH we have MATH for all MATH, showing MATH. On the other hand, by REF we have MATH, showing MATH. It remains to consider the restriction of MATH to MATH. Take MATH, MATH (see the proof of REF ) and note that MATH for those vectors. Since these vectors are dense in MATH, it follows that the restriction of MATH to MATH is the identity. This shows the claim. |
math-ph/0105042 | Under the given conditions, the function MATH can be constructed using the well-known facts about the NAME - NAME spaces, for which we refer to CITE. The space MATH contains the space MATH which consists of functions of compact support and is nontrivial for MATH. Furthermore, for MATH we have MATH. Thus there exists a MATH and a nonzero function MATH with MATH in MATH , such that MATH for some MATH. Then MATH is an element of MATH for MATH, see CITE. It has MATH-norm MATH and support in MATH. Since convolution with MATH-functions does not change the regularity, the function MATH is an element of MATH and therefore a fortiori of MATH with all the desired properties. |
math-ph/0105045 | First, one shows that MATH see CITE. Set MATH for MATH. (Note again that REF has a factor MATH.) The potential analysis implies MATH for some constant MATH, see CITE. Clearly MATH is a harmonic function. Hence REF imply that MATH for some constant MATH. That is, MATH has the following representation formula MATH . The above results and REF imply MATH . Furthermore, we can show that MATH . See, for example, CITE. |
math-ph/0105045 | Set MATH . We can check that MATH is a holomorphic function as follows: MATH . In the first equality we have used the symmetry of the matrix MATH. Using REF , we have the following expansion of MATH near infinity MATH . Hence, MATH is a constant (zero, in fact) and MATH . |
math-ph/0105045 | Define MATH by REF . From the assumption, we have MATH . A direct computation shows that MATH where MATH means that a matrix MATH with entries satisfying MATH as MATH. Here, we have used the condition that MATH for any MATH. Similarly, MATH . Hence, MATH . Now it is easy to compute the holonomy of MATH. |
math-ph/0105045 | It follows from REF . |
math-ph/0105046 | The proofs of REF are contained in REF because REF imply REF . REF is a consequence of considerations in CITE, see also REF, and of a straightforward generalization to non-zero vector potentials. REF follows from REF . |
math-ph/0105046 | For REF see CITE, which generalizes CITE to the case of continuously differentiable vector potentials MATH. Note that the assumption of a vanishing divergence, MATH, in CITE is not needed in the argument. REF is a straightforward generalization of CITE to continuously differentiable MATH, see also CITE. |
math-ph/0105046 | See CITE. |
math-ph/0105046 | See CITE. |
math-ph/0105046 | Since we follow exactly the strategy of the proof of CITE, we only remark that the two main steps in this proof remain valid in the presence of a vector potential MATH. The first step, used in REF, concerns the lowering of the eigenvalues of the operator MATH by so-called NAME bracketing in case MATH and by the (subsequent) insertion of interfaces in MATH with the requirement of NAME boundary conditions. For MATH, supplied with REF , the validity of these two techniques is established in REF. The second step is an application of a spectral-averaging estimate of CITE, which is re-phrased as REF below. Since there the operator MATH is only required to be self-adjoint and does not enter the right-hand side of REF , it makes no difference if MATH is taken as MATH (as is done in CITE) or as MATH for each MATH. |
math-ph/0105046 | Since the assumption MATH implies the operator inequalities MATH, the lemma is proven as REF for any positive bounded function MATH with compact support. It extends to positive bounded functions with arbitrary supports by a monotone-convergence argument. |
math-ph/0105046 | Let MATH be bounded and open. Then REF together with CITE implies that MATH for MATH-almost all MATH. Therefore, by the non-randomness of the density-of-states measure MATH and NAME 's lemma we have MATH . Here we used REF and the assumption that the constants involved there do not depend on MATH. Now the NAME theorem yields the claimed absolute continuity of MATH. |
math-ph/0105046 | We proceed as in the proof of CITE and define MATH for MATH and MATH. The NAME inequality CITE yields MATH . We then evaluate the trace on the right-hand side in an orthonormal eigenbasis of MATH. Using NAME 's theorem, the probabilistic expectation of the quantum-mechanical expectation of MATH with respect to a normalized eigenfunction of MATH is estimated by MATH, which is smaller than the second factor on the right-hand side of REF since MATH. The proof is completed by noting that the l.h.s. of REF converges for MATH to the trace on the l.h.s. of REF by monotone convergence of forms CITE, similar to the proof of CITE. |
math-ph/0105046 | For each MATH, the choice MATH and MATH yields a MATH-decomposition of MATH in the sense of REF . It remains to verify the three assumptions of REF is guaranteed by REF . REF is fulfilled with MATH. To verify REF , we make use of REF and observe that MATH. |
math-ph/0105046 | The proof is analogous to that of REF . To verify the assumptions of REF we note that REF is guaranteed by REF . REF is fulfilled with MATH if MATH. As for REF , we make use of REF and explicitly compute the involved expectation if MATH. |
math-ph/0105046 | The key input is the fact that every Gaussian random potential MATH admits a MATH-decomposition in the sense of REF . More precisely, MATH is a standard Gaussian random variable with NAME density MATH. This random variable and the Gaussian random field MATH, where MATH is defined in REF , are stochastically independent. For details see the proof of CITE. To obtain the specific form MATH, which is independent of the magnetic field, we used REF . |
math-ph/0105046 | Since MATH and MATH is dense in MATH, the form MATH is densely defined. Its symmetry and positivity are obvious from the definition. To prove that MATH is also closed we have to show that the space MATH is complete with respect to the (metric induced by the form-) norm MATH . To this end, we proceed along the lines of REF and let MATH be a sequence in MATH which is NAME with respect to the norm REF . By completeness of MATH, there exist functions MATH, MATH, MATH, such that MATH and MATH strongly in MATH as MATH. Since MATH in the sense of distributions on MATH as MATH, we have MATH and hence MATH. The existence and uniqueness of MATH follow now from the one-to-one correspondence between densely defined, symmetric, bounded below, closed forms and self-adjoint, bounded below operators, see CITE. |
math-ph/0105046 | The domain MATH of MATH is dense in MATH, because both MATH and MATH contain MATH. Hence MATH is well defined as a form sum of MATH and MATH. Moreover, MATH is symmetric, positive and closed, because it is the sum of two of such forms. Since MATH, the negative part MATH of MATH is also a form perturbation of MATH. The proof of the lemma is then completed by the CITE. |
math-ph/0105046 | The proofs of REF for the free case carry over to the case MATH and MATH. In particular, the inclusion relations between the various form domains for MATH and MATH hold analogously for the form domains in the case MATH and MATH. |
math-ph/0105046 | NAME 's theorem and the NAME inequality show that MATH. Therefore, the induced multiplication operator on its maximal domain MATH is densely defined and self-adjoint. Moreover, since MATH implies MATH, we are allowed to use the product and chain rule for distributional derivatives CITE which yield MATH. |
math-ph/0105046 | For MATH see CITE. The proof for MATH consists of three steps. In the first step, we assume MATH to be bounded from below. In this case MATH is a form sum of MATH operators each of which is bounded from below, recall REF . Hence we may employ the strong NAME product formula generalized to form sums of several operators CITE and write MATH . Gauge equivalence REF now shows that MATH for all MATH and all MATH. By the distributional inequality MATH, valid for all MATH CITE, the operator MATH obeys a NAME criterion CITE and hence is the generator of a positivity-preserving semigroup. It follows that MATH for all MATH and all MATH. This together with REF implies REF (with MATH) for scalar potentials MATH which are bounded from below. In the second step, we prove that if MATH is a form perturbation of MATH then it is also one of MATH with form bound not exceeding the one for MATH (see CITE or CITE for the case MATH). This follows from REF below with MATH and MATH together with the fact that the form bound of MATH relative to MATH can be expressed as MATH see CITE. Here MATH denotes the (uniform) norm of bounded operators on MATH. In the third step, we extend the validity of REF (with MATH) to scalar potentials MATH with MATH and MATH being a form perturbation of MATH. To this end, we approximate MATH by MATH defined through MATH, MATH , MATH. Monotone convergence for forms CITE yields the convergence of MATH to MATH in the strong resolvent sense as MATH. It follows that MATH for all MATH. Since REF (with MATH) holds for each MATH by the first step, the proof is complete. |
math/0105001 | Let MATH be a formal deformation of MATH, and let MATH be a deformation of MATH with respect to MATH. Then MATH and MATH are isomorphic as MATH-modules. In fact, for MATH, MATH full idempotent, and MATH given by REF , an explicit isomorphism is (see CITE) MATH . In general, any fixed isomorphism MATH induces a formal deformation MATH of MATH, and different choices of REF lead to isomorphic deformations (not necessarily equivalent). Note that an isomorphism REF also defines a left module structure MATH on MATH over MATH, and we can choose REF so that MATH is a deformation of the module structure of MATH, MATH (this is the case for REF ). A simple computation shows that any two isomorphisms REF yielding deformations of MATH define equivalent deformations of MATH. Moreover, it follows from the uniqueness part of REF that this equivalence class does not depend on the choice of MATH or element in MATH. Hence this procedure defines a canonical map MATH which is a bijection CITE. It follows from the construction of MATH and REF that formal deformations related by MATH are NAME equivalent. |
math/0105001 | Since MATH and MATH, we can write MATH, where MATH . A simple computation shows that MATH and MATH . The equations MATH and MATH imply that MATH . It follows from REF that MATH, for all MATH. Note that if MATH and MATH, then MATH, and hence MATH . But MATH. Thus MATH by REF , and the result follows. |
math/0105001 | The bracket MATH satisfies REF . As a result, MATH and MATH. It is easy to check that the following NAME rule holds for MATH in MATH: MATH . Combining these identities, we get MATH for MATH. |
math/0105001 | If MATH, then MATH and MATH are NAME equivalent by REF . Conversely, if MATH and MATH are NAME equivalent, then there exists a full idempotent MATH so that MATH. By REF , MATH with MATH full. We know that (see REF ) MATH is isomorphic to MATH as a MATH-module, and since it is also isomorphic to MATH, we must have MATH. As in REF , MATH is a MATH-equivalence bimodule satisfying MATH, for all MATH and MATH, and MATH is isomorphic to the deformations in the class MATH. The result then follows from REF . |
math/0105001 | Let MATH be a formal deformation of MATH. Then MATH itself, regarded as a right module over MATH, provides a deformation of MATH. Since MATH, it follows that MATH. |
math/0105001 | Let MATH, MATH, and MATH be formal deformations of MATH so that MATH and MATH. Let MATH be a deformation of MATH corresponding to MATH, and MATH be a deformation corresponding to MATH . We know that MATH and MATH. As discussed in REF, MATH is a MATH-equivalence bimodule, so MATH. Since MATH is a f.g.p.m. over MATH, it follows (see REF ) that it is of the form MATH, where MATH as a MATH-module. But MATH . Hence MATH is a deformation of MATH, and the conclusion follows. |
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