paper stringlengths 9 16 | proof stringlengths 0 131k |
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hep-th/0003114 | Let MATH be the algebra of functions depending on MATH only. Any function from MATH is a pullback of some function on MATH to the entire extended phase space MATH REF. Let also MATH be a BRST observable. Then one can check that there exist functions MATH and MATH such that MATH . As a matter of REF can not depend on MA... |
hep-th/0003114 | REF evidently holds in the lowest order with respect to degree REF provided respective classical BRST charge MATH satisfies MATH. At the classical level the higher order terms in expansion of MATH with respect to MATH do not depend on the momenta MATH. Thus these terms belong to MATH. It is useful to assume that the sa... |
hep-th/0003114 | Consider an expansion of MATH in the homogeneous components MATH . The boundary REF implies MATH. REF obviously holds in the first degree. In the higher degrees we have MATH where MATH is given by MATH and MATH are terms of the expansion REF of MATH with respect to degree. Similarly to the proof of REF the necessary an... |
hep-th/0003239 | First of all the volume of MATH is infinite. It can be seen by calculating: MATH where we have used again the parameterization of the Euclidean NAME manifold given in the previous section. This implies that there are no MATH harmonic MATH- or equivalently MATH-forms. Now, as MATH is NAME and complete, REF implies that ... |
hep-th/0003243 | Since the set of vectors MATH is a core fore MATH and MATH is closed, there is a sequence MATH such that MATH and MATH, strongly. Thus if MATH, one also has MATH and MATH. So the operator MATH, given by MATH is well defined. Its adjoint MATH also has the dense set of vectors MATH in its domain and MATH . This shows tha... |
hep-th/0003243 | CASE: If MATH is affiliated with the wedge algebra MATH, say, the operators MATH are affiliated with MATH by covariance. Now for MATH varying in some open, bounded region MATH, there is a wedge MATH, hence the operators MATH, MATH, contain the common dense subspace MATH in their domains. Thus for MATH . Since MATH, the... |
hep-th/0003243 | Let MATH be any vector in the domain of temperateness of MATH with compact spectral support. Then MATH where MATH is defined as a strong integral. Now by the NAME - NAME property of MATH, there is for any MATH a MATH and a test function MATH whose NAME transform has support in any given neighbourhood of MATH such that ... |
hep-th/0003243 | The argument is very similar to the proof of the preceding lemma and it therefore suffices to indicate the main steps. In REF one has, in view of the fact that MATH, MATH, and the asymptotic relation REF , MATH where, in the last step, MATH have been replaced by their asymptotically dominant parts. Since MATH, the regi... |
hep-th/0003243 | Let MATH be any local operator which is localized in MATH. Since MATH and MATH are solutions of the NAME - NAME equation of mass MATH, the commutator function MATH can be represented in the form MATH . Here the functions MATH are given by MATH where MATH is the momentum space wave function of the single particle vector... |
hep-th/0003243 | The proof is based on induction in MATH. For MATH, one has MATH because of the support properties of MATH in momentum space. Assuming that the statement holds for MATH, let MATH be a test function whose NAME transform has support about points on the mass shell such that MATH. It then follows from relation REF that MATH... |
math-ph/0003001 | Obviously, MATH and MATH. To bound MATH we return to REF and note the bound (which holds in MATH) MATH . The second inequality holds because the convolution of two symmetric decreasing functions is symmetric decreasing. Next we turn to MATH, which is a bit more complicated. We split the integration region in REF into M... |
math-ph/0003001 | Thanks to REF we only need to establish the contraction property. We first note that for positive real numbers MATH where MATH is some point on the line between MATH and MATH. Thus we get MATH where MATH as in REF . In the last line we have simply noted that the integrals are smaller than the corresponding integral in ... |
math-ph/0003001 | Take MATH and MATH so that MATH maps MATH into itself. Then, if MATH, the contraction condition will be satisfied. |
math-ph/0003001 | Define MATH to be the subset of MATH on which MATH for all MATH. If we can show that MATH also leaves MATH invariant, then we have shown the wanted REF on the solution in MATH (by the uniqueness of the solution on MATH) except for the possibility that equality also can occur. This is so, since we can apply the fixed po... |
math-ph/0003015 | The bundle MATH over NAME spacetime MATH is the trivial bundle MATH. Thus we can simplify our notation by identifying the spinor spaces over all points. Also, the covariant derivative coincides with the partial derivative of the individual components, and the curvature scalar vanishes. Therefore the functions MATH and ... |
math-ph/0003015 | A first (rather involved) proof for the NAME equation for spinor half densities was given by CITE. Here we give a simple proof that holds for the NAME equation on MATH. Once we haven chosen coordinate frames for the tangent and cotangent bundle, the operator MATH has principal and subprincipal symbols MATH . We choose ... |
math-ph/0003020 | Choose MATH such that MATH. Then MATH and MATH annihilates MATH or vanishes identically. To prove independence, it is enough to notice that REF determining MATH, projected on eigenspaces of MATH reads MATH . Thus, MATH with MATH are linear in NAME variable of scaling degree MATH with successive terms involving variable... |
math-ph/0003020 | The order of extension does not depend on the basis chosen. In the basis of REF MATH. Then MATH as a straightforward consequence of REF. |
math-ph/0003020 | Let us take densities of Hamiltonians MATH, MATH. Their scaling degree are correspondingly MATH. These densities are gauge invariant, as follows from REF . Due to corollary to REF they are independent. Thus they may be taken as a part of coordinates on MATH. So MATH. In particular this implies that the multiplicity of ... |
math-ph/0003020 | Brute force checking of data presented in REF. We lack more elegant proof so far. |
math-ph/0003020 | As follows from scaling weight grading of NAME structures MATH . Since MATH . Recall that we have assumed ascending ordering of MATH, and so matrix MATH is lower triangular with respect to anti diagonal MATH. This proves that MATH equals to product of those constants. Since, by the choice of coordinates, we know that M... |
math-ph/0003020 | By the implicit function theorem one may resolve the system of polynomial equations MATH locally with respect to variables MATH. Consider the Hamiltonian flows in involution generated by MATH: MATH . Then one has MATH . Restricting on MATH and using the non degeneracy of MATH the result follows. |
math-ph/0003020 | Consider a ring of polynomial functions in MATH. It is obviously generated by elements of MATH. Using the adjoint action of nilpotent group with NAME algebra being MATH one may always reduce MATH to canonical form MATH (see REF ). Indeed, projecting REF on MATH eigenspaces we get MATH where MATH and MATH. This allows t... |
math-ph/0003020 | Let MATH corresponds to group element fixing MATH, that is, MATH . Projecting on MATH eigenspaces we gain the system of REF . Since MATH, we have MATH . The rest of REF for MATH determines MATH. Given positive MATH there are MATH scalar equations. Due to injectivity of map MATH for negative MATH we may unambiguously so... |
math-ph/0003020 | The action of NAME group MATH are inner in MATH and in MATH. For any root MATH, such that MATH, one has a reflection MATH in the hyperplane perpendicular to the root acting on MATH as follows: MATH . These reflections act canonically on NAME subalgebra MATH. Since all elements of MATH can be expressed as a product of t... |
math-ph/0003020 | Consider the set MATH. Since MATH we have MATH. The set MATH is stable under the homomorphism MATH because MATH and MATH are stable. So MATH and, hence, MATH maps MATH in MATH. The surjectivity is obvious. |
math-ph/0003020 | By REF MATH generate the ring of gauge invariant polynomials in MATH. Restricting them on MATH and restricting adjoint group to MATH the proof follows. |
math-ph/0003020 | Indeed, let us choose another set of algebraically independent polynomials MATH . They are known to be of the same degrees. Using them we define new coordinates MATH: MATH . Starting from functions MATH of the lowest degree and proceeding up we conclude that MATH may be chosen to coincide with MATH. |
math-ph/0003020 | Indeed, NAME MATH, being invariant with respect to adjoint action of MATH, are not left invariant by action of group MATH. Let MATH. Then MATH . This implies that MATH results MATH invariant and hence depends on MATH only. |
math-ph/0003020 | Since MATH are MATH invariant polynomials and so is Killing metrics, we may make use of NAME theorem MATH stating that MATH is a polynomial in MATH invariant polynomials MATH. Now using definition of coordinates MATH REF and changing variables to MATH we obtain MATH . By another NAME theorem MATH are invariant with res... |
math-ph/0003020 | Consider MATH. Rewrite REF as follows MATH . Since functions MATH are not altered we have the same pairing and due to REF MATH action preserves constant non degenerate submatrix of the matrix MATH. It is only possible if MATH acts linearly. |
math-ph/0003020 | Fix a MATH orbit MATH in MATH. It suffices to show that MATH permutes solutions of REF . Notice, that MATH invariant polynomials MATH, corresponding to abelian constituents of MATH are linear and may be chosen to be corresponding NAME variables as follows from REF. Then, due to REF MATH groups acts on them linearly in ... |
math-ph/0003020 | Since MATH is a subgroup, there exist two transformations product of which is identity. Since they correspond to the same matrix MATH, the proof follows. |
math-ph/0003020 | NAME group MATH preserves metrics MATH, and is generated by reflections REF corresponding to simple roots of zero MATH grade. Group MATH, restricted to MATH, also preserves the metrics and is generated by reflections associated with simple roots of MATH grade REF, due to REF . We conclude that MATH is a finite group ge... |
math-ph/0003020 | Due to REF, MATH is the momentum for the hierarchies in question, so their dynamical equations admit constant solutions, because of REF . Now we address the non degeneracy statement. MATH . Restricting on MATH and using REF and MATH we conclude MATH . Thus, due to assumption of REF , it suffices to prove the non degene... |
math-ph/0003020 | Due to REF dispersionless limit of the second bracket remains linear in MATH. The result follows. |
math-ph/0003020 | MATH are NAME of finite dimensional NAME bracket REF and thus due to REF the result follows. |
math-ph/0003020 | Let us choose coordinates MATH as follows. Take the first MATH coordinates to be MATH and choose the rest to be canonical for finite dimensional bracket. The advantage of this choice is the simplicity of constraint REF MATH . Due to REF we have MATH . Both set of coordinates MATH and MATH are local coordinates on MATH,... |
math-ph/0003020 | Indeed, due to REF MATH admits a restriction of MATH and preserves the latter. Hence, MATH were MATH invariant in MATH and so they remain restricted on MATH. |
math-ph/0003020 | Consider the following polynomial in MATH . When all MATH but MATH vanish it simplifies to MATH. At this point MATH and thus MATH is non degenerate. Indeed, let MATH be eigenvector of some representative, which always exists, of MATH in MATH with eigenvalue MATH. Then, due to homogeneity of MATH and from their MATH inv... |
math-ph/0003020 | Let MATH be the metrics of the second NAME structure and let MATH - of the first. Introduce function MATH, and introduce the following vector fields MATH . Notice, that with this choice MATH and MATH and MATH as follows from considerations above, and where we have assumed MATH to be chosen anti diagonal with all nonzer... |
math-ph/0003020 | Since obtained pencil of Hamiltonian structures satisfies NAME identity we have a flat pencil of metrics. It is quasihomogeneous as was shown in REF . Thus, following ref. CITE, it is enough to show that the degree of quasihomogeneity MATH. As was said in the proof of REF MATH. Due to quasihomogeneity of flat pencil we... |
math-ph/0003026 | Let MATH: MATH. This leads to MATH and, therefore, MATH must have the form MATH. Consequently, MATH and there exists MATH such that MATH and MATH. Therefore, in the symplectic decomposition MATH we obtain MATH . Moreover, MATH, so MATH and then MATH with MATH. Similarly, MATH with MATH. Since both MATH and MATH are eff... |
math-ph/0003026 | For all MATH and MATH, MATH. Consequently, MATH, MATH. Recall that MATH is into. This leads to the relation MATH which holds for all MATH. Note that MATH since MATH is effective. Therefore, MATH, for all MATH and MATH holds for all MATH. Applying REF , we can deduce that there exists MATH such that MATH. Let then MATH ... |
math-ph/0003026 | MATH is non-degenerate on any subspace MATH of MATH such that MATH. Therefore, MATH must be even. Moreover, MATH is not zero, so MATH. Since MATH, MATH. |
math-ph/0003026 | Let MATH be a vector satisfying MATH and let MATH be the symplectic decomposition of MATH in MATH. If MATH with MATH, we can write MATH, MATH. Then, if we assume that MATH for such a MATH, it is straightforward to obtain MATH and therefore, MATH. However, since MATH, MATH is non-degenerate on MATH and then MATH. It is ... |
math-ph/0003026 | We can assume that MATH is a polynomial map of degree less than MATH. Denote MATH. We assume first that there exists MATH such that MATH. Since MATH, MATH is an isomorphism and then MATH is a local diffeomorphism. Let MATH be defined by MATH . Denote MATH. MATH is a homogeneous polynom of degree MATH. It is easy to che... |
math-ph/0003026 | MATH is surjective. Moreover, MATH . Therefore, MATH is a local diffeomorphism. |
math-ph/0003026 | Since MATH, MATH satisfies: MATH . Therefore, MATH and MATH . After observing that for any form MATH the relation MATH must hold (MATH), we get the result. |
math-ph/0003026 | Suppose first that MATH. Then there exists MATH with MATH such that MATH. So, one has MATH . Therefore, MATH with MATH, MATH, MATH, MATH and MATH. Conversely, if there exist MATH and MATH such that MATH with MATH and MATH, then MATH. It means that MATH and, therefore, MATH with MATH, MATH. |
math-ph/0003027 | In coordinates, any vector field MATH of MATH is of the type MATH, where MATH are functions. The time form writes as MATH. Hence, MATH. Therefore, MATH if and only if MATH and MATH. But the first condition on MATH means that the vector field MATH is projectable and, together with the second condition, that MATH is even... |
math-ph/0003027 | If MATH where MATH, then, MATH. MATH . |
math-ph/0003027 | In coordinates, we have MATH . Since MATH is projectable, this yields MATH. This does not depend on the coefficients MATH. The result turns out to be independent of the chart. MATH . |
math-ph/0003027 | MATH is a (scaled) fibred morphism over MATH. Thus, REF yields the equality MATH where the components of MATH were given by MATH. MATH . |
math-ph/0003027 | The coordinate expression shows that because of the projectabilty of MATH the vertical restriction does not contain coefficients of the type MATH. The result turns out to be independent of the chart. MATH . |
math-ph/0003027 | The proof of MATH follows easily in virtue of the NAME rule and the fact that MATH is a symmetry of the contact maps. The proof of MATH can be obtained easily by considering the expressions of REF which are polynomial in the coordinates MATH or MATH. MATH . |
math-ph/0003027 | REF yields, MATH. The proof of MATH can be seen directly because of the fact that MATH is a symmetry of MATH. For the proof of MATH we consider the coordinate expressions MATH and MATH, where we have set MATH to be the coefficient of MATH in REF and MATH the coefficient of MATH in REF . It can be easily seen that, if t... |
math-ph/0003027 | REF yields MATH. In analogy to the proof of REF this yields the equivalence between the condition MATH and the two conditions MATH, MATH. Clearly, MATH is equivalent to MATH. Thus, REF yield the result. MATH . |
math-ph/0003027 | This follows directly from NAME 's formula using MATH and the closure of MATH. MATH . |
math-ph/0003027 | Let MATH be a second order connection such that MATH. This implies that there exists a local function MATH such that MATH. Using MATH, we get MATH. Let us set MATH; MATH is valued into MATH, and has coordinate expression MATH, with MATH. We have to show that a local function MATH only exists if MATH. Calculating in coo... |
math-ph/0003027 | MATH is locally equivalent to the closure of MATH. Hence, there is a local function MATH such that MATH. Therefore, MATH. MATH . |
math-ph/0003027 | MATH. This is equivalent to the equation MATH. It follows directly from REF that MATH is a conserved quantity. MATH . |
math-ph/0003027 | Both directions can be proved in analogy to the proof of the equivalence MATH. MATH . |
math-ph/0003027 | By recalling that MATH we obtain that MATH. Hence, by observing that two vector fields MATH are equal if and only if MATH and MATH, we obtain the result. MATH . |
math-ph/0003027 | The equalities MATH and MATH. yield the result. MATH . |
math-ph/0003027 | The first expression follows simply from the contact splitting of MATH and REF . The observer dependent expression of MATH follows simply from the splitting of MATH through the observer. The coordinate expression MATH with respect to a basis MATH yields the second expression. MATH . |
math/0003002 | Here is a well-known explicit description of the group MATH of points of order MATH on MATH. Let us denote by MATH the MATH-divisor MATH on MATH if MATH is odd and the MATH-divisor MATH if MATH is even. In both cases MATH is an effective divisor of degree MATH. Namely, let MATH be a subset of even cardinality. Then (CI... |
math/0003002 | Let us put MATH . It follows from REF that the MATH-module MATH is very simple. Now the result follows readily from REF . |
math/0003002 | Clearly, MATH . This implies easily that MATH where MATH. Since MATH is a proper subset of MATH, we have MATH . Recall that there exists MATH such that MATH lies in MATH and the exact multiplicative order of MATH is MATH. This implies that MATH. Since MATH, we conclude that MATH. Therefore MATH. |
math/0003002 | Let us put MATH. We have MATH . Clearly, for each MATH one may find a MATH matrix with determinant MATH and trace MATH. This implies that MATH satisfies the conditions of REF . The construction described in REF allows us to construct a MATH-dimensional MATH-representation MATH of MATH for each subset MATH of of MATH. I... |
math/0003002 | Let us put MATH. We know that MATH satisfies the conditions of REF . The construction described in REF allows us to construct a MATH-dimensional MATH-representation MATH of MATH for each subset MATH of of MATH. It is known (CITE, pp. REF) that MATH's exhaust the list of all absolutely irreducible MATH-representations o... |
math/0003002 | We may assume that MATH. Clearly, MATH is a faithful MATH-module and MATH . CASE: MATH is a semisimple MATH-module. Indeed, let MATH be a simple MATH-submodule. Then MATH is a non-zero MATH-stable subspace in MATH and therefore must coincide with MATH. On the other hand, each MATH is also a MATH-submodule in MATH, beca... |
math/0003002 | Since MATH is isotypic, there exist a simple MATH-module MATH, a positive integer MATH and an isomorphism MATH of MATH-modules. Let us put MATH . The isomorphism MATH gives rise to the isomorphism of MATH-vector spaces MATH . We have MATH . Clearly, MATH is isomorphic to the matrix algebra MATH of size MATH over MATH. ... |
math/0003002 | We have MATH. Clearly, it suffices to check that the MATH-module MATH is very simple. First, notice that MATH acts doubly transitively on MATH. Indeed, each subgroup of MATH (except MATH itself) has index MATH (REF , p. CASE: This implies that MATH acts transitively on MATH. If the stabilizer MATH of a point MATH has i... |
math/0003002 | We have MATH. First, notice that MATH acts doubly transitively on MATH. Indeed, the classification of subgroups of NAME groups REF , p. REF implies that each subgroup of MATH (except MATH itself) has index MATH. This implies that MATH acts transitively on MATH. If the stabilizer MATH of a point MATH has index MATH then... |
math/0003002 | REF follow from REF respectively applied to MATH. |
math/0003002 | Assume that MATH. Since MATH contains a subgroup isomorphic to MATH CITE, p. REF, it suffices to check the case of MATH, in light of REF . The group MATH has two conjugacy classes of maximal subgroups of index MATH and all other subgroups in MATH have index greater than MATH CITE, p. CASE: Therefore all subgroups in MA... |
math/0003008 | The first formula is REF and the second one is established in the proof of CITE; compare also CITE. |
math/0003008 | We will apply the Lemma with the roles of MATH and MATH interchanged. Since MATH holds for the counit MATH (for example, CITE), this requires replacing MATH by MATH. First assume that MATH is the centrally primitive idempotent of MATH corresponding to some irreducible MATH-module MATH. Then, by the Lemma, MATH. Thus, M... |
math/0003010 | Using only the facts that MATH and MATH define a measure (that is, not necessarily a probability measure), the proofs of REF will establish the equations: MATH . To get a recurrence relation for the MATH's, one simply plugs the second equation into the first. Similarly one obtains a recurrence relation for the MATH's. ... |
math/0003010 | The arguments are completely analogous to those for MATH in CITE REF , using the cycle index of the finite symplectic groups CITE. |
math/0003010 | The most important observation (see CITE for a readable proof) is that the fixed space of an element MATH of MATH has isometry type MATH precisely when the partition corresponding to the polynomial MATH in the rational canonical form of MATH satisfies MATH and MATH. In other words, the partition has MATH parts and MATH... |
math/0003010 | The idea for all of the proofs is the same; hence we prove part two as follows: MATH as desired. Note that the meat of the lemma is the second equality, which follows from the formulas for MATH and MATH. The third equality is simply a relabelling of subscripts. |
math/0003010 | The MATH probability of choosing a partition with MATH for all MATH is MATH . Since MATH is equal to MATH by definition, it it is enough to prove two claims: first that for every choice of MATH, MATH is equal to the asserted transition rule probability for moving from MATH to MATH, and second that the transition rule p... |
math/0003014 | Integrating by parts, we obtain MATH . In view of REF , we have MATH . Now the lemma is proved by passing to the limit as MATH. |
math/0003014 | The identity REF imply that MATH . Therefore, in view of REF , MATH . Taking into account REF and the second REF , we obtain MATH . REF implies that MATH and MATH satisfy the conditions of REF and that MATH. Obviously, REF follows from REF . |
math/0003014 | In view of REF we have MATH . This estimates, REF imply that the functions MATH, MATH and MATH are uniformly bounded on MATH. Since MATH whenever MATH, integrating by parts with respect to MATH we obtain MATH . Now REF follows from REF . |
math/0003014 | REF are obviously fulfilled; REF follow from the fact that MATH. Finally, REF holds true because the MATH-th derivative of the extended function MATH coincides with a linear combination of a MATH-function and two MATH-functions. |
math/0003014 | Let MATH be the multiplication operator and MATH be the NAME multiplier generated by the characteristic function of the interval MATH. Then the NAME norm of the operator MATH acting in MATH is equal to MATH. Therefore MATH which implies that MATH . |
math/0003014 | If MATH and MATH is the norm in MATH then MATH . One can easily see that MATH . Therefore REF implies the required estimate. |
math/0003014 | According to REF , the NAME transform of MATH coincides on the interval MATH with the NAME transform of MATH . Since MATH is even, this implies that MATH for all MATH. Estimating MATH in the integrals on the right hand sides, we arrive at REF - REF . |
math/0003014 | If we define MATH as in REF with MATH (see REF ) then REF follow from REF with MATH, REF . |
math/0003014 | Since MATH, REF with MATH, MATH, MATH and MATH implies MATH . Let MATH be defined as in REF with MATH. Then REF follows from REF . Since MATH, REF imply that MATH . Estimating MATH with MATH and applying REF , we obtain REF . |
math/0003014 | REF are proved by straightforward integration of REF . REF imply that MATH . Now, applying the first REF with MATH, we arrive at REF . |
math/0003032 | A rank one algebraic factor has to have fibres of positive dimension. Hence the pre - image of the origin under the factor map is a union of finitely many rational tori of positive dimension and by REF MATH cannot be irreducible. |
math/0003032 | By diagonalizing MATH over MATH and taking the real form of it, one immediately sees that the centralizer of MATH in MATH acts transitively on vectors with nonzero projections on all eigenspaces and thus has a single open and dense orbit. Since the centralizer over MATH is the closure of the centralizer over MATH, the ... |
math/0003032 | Choose MATH such that MATH. Let MATH be cyclic vectors for MATH and MATH, respectively. Now consider the integer vector MATH and find MATH commuting with MATH such that MATH. We have MATH. The conjugacy MATH maps bijectively the MATH - span of the MATH - orbit of MATH to MATH - span of the MATH - orbit of MATH. By cycl... |
math/0003032 | If MATH for MATH, then MATH has MATH as an eigenvalue, and hence has a rational subspace consisting of all invariant vectors. This subspace must be invariant under MATH which contradicts its irreducibility. |
math/0003032 | MATH is a ring because MATH is a ring. As we pointed out above images of integer matrices are algebraic integers and images of matrices with determinant MATH are algebraic units. Hence MATH. Finally, for every polynomial MATH with integer coefficients, MATH is an integer matrix, hence MATH. |
math/0003032 | By REF , MATH is isomorphic to the group of units in the order MATH, the statement follows from the NAME Unit Theorem (CITE, Ch. REF). |
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