paper stringlengths 9 16 | proof stringlengths 0 131k |
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math/0004054 | We argue as follows: assume MATH; we recall estimate REF: MATH where MATH is given by REF and MATH satisfies REF. REF implies MATH, and thus REF simplifies as MATH . REF of MATH implies that MATH . Moreover, relation REF leads to MATH . This relation implies immediately that MATH and therefore, thanks to REF MATH . If ... |
math/0004054 | We go back to the original scales and time MATH; then MATH . Therefore, in coordinates MATH (see REF ), we have the relations MATH . We have also the estimate MATH . If MATH, REF is a consequence of REF. Otherwise, we observe that MATH belongs to MATH for all large enough MATH; therefore, we are able to use the equival... |
math/0004056 | The spinor group MATH is completely defined in terms of the algebra MATH: MATH where MATH is a special NAME - NAME group, and MATH . Let MATH is a representation of the algebra MATH in a vector space MATH. The representation MATH induces via REF a representation of the group MATH, and also via REF a representation MATH... |
math/0004056 | REF immediately follows from REF . Further, follows to CITE we see that for some point MATH and a vector field MATH the NAME formula with respect to a decomposition MATH gives MATH or MATH . Let MATH be a local orthonormal tangent frame of the submanifold MATH at the point MATH and let MATH be a local orthonormal frame... |
math/0004056 | In the case of the Lorentzian manifold MATH we have the following immersions MATH and MATH. At this point in accordance with REF on the surfaces MATH and MATH there exist the spinor fields MATH and MATH, respectively. Let us find a NAME operator on the surface MATH. First of all, in accordance with REF and the formulae... |
math/0004061 | Let MATH be an integral basis of MATH; complete it with integral vectors MATH to a basis of MATH. In a neighborhood of MATH the vectors MATH generate MATH independent, commuting, locally Hamiltonian vector fields MATH. The closed forms MATH define a family of MATH functions MATH. The commuting relations of such functio... |
math/0004061 | From REF it is obvious that the MATH belong to MATH. Let MATH, for MATH, be the list of real infinitesimal weights at MATH. Assume that the MATH are a set of MATH-generators of MATH. An element of MATH stabilizes all points of MATH if and only if MATH for all MATH (MATH means MATH) if and only if MATH for all MATH, if ... |
math/0004061 | The notion of local maximum and local minimum makes perfect sense for closed REF-forms. A proof of this fact would require a long digression hence we omit it. A partial result (for rational REF-forms) is used in CITE. A full proof can be found in CITE. |
math/0004061 | Let MATH be a point fixed by the action. The real infinitesimal weights of the isotropy representation at MATH, MATH, must MATH-generate MATH and are exactly in number of MATH. By a MATH change of coordinates in MATH we can assume that MATH, where MATH is an integral basis for MATH and the MATH indicates the dual eleme... |
math/0004061 | Let MATH be a basis of the NAME algebra MATH. This basis is associated to MATH REF-forms MATH; to each of them is associated a finite dimensional MATH-vector space MATH. For any MATH-tuple of real numbers MATH such that the set MATH of MATH-subvector spaces of MATH is in direct sum, the form associated to the vector MA... |
math/0004061 | Assume that the generic Hamiltonian has rationality degree MATH. Then we can choose a basis MATH of the free part of the first homology group of MATH so that MATH . This follows from the Theorem of Elementary Divisors and REF . If the generic Hamiltonian was less than MATH-rational then the matrix MATH would have less ... |
math/0004061 | The thesis is equivalent to the fact that, if a quasi-periodic function generates a MATH-action, then MATH. The negation of this last statement can be proven to be equivalent to the following situation: a torus MATH, with MATH, acts on a closed symplectic manifold. This action is such that MATH for some MATH maximally ... |
math/0004061 | In REF we proved that the smallest covering from which it is possible to define the momentum map as a function is that in which the pull back of a generic Hamiltonian is exact. Such covering is regular, with group of deck transformations which is free Abelian of rank equal to the rationality degree of the generic Hamil... |
math/0004061 | The NAME algebras of the two groups are the same and the fundamental vector fields are the same. |
math/0004061 | Assume the form MATH has a local maximum at the point MATH. We claim that MATH must then have a local maximum at MATH. In fact the form MATH is orthogonal to vectors of the form MATH, for any MATH (in other words MATH for any MATH); hence the Hessian of the form MATH, at the point MATH, has same signature and same rank... |
math/0004061 | In CITE was proven that the set MATH is a convex polytope MATH, and that MATH. We just have to show that MATH . We plan to use the fact that MATH is a MATH-manifold, and that the momentum map MATH, restricted to MATH, has values in MATH. Assume REF is false, let MATH be a point in MATH and MATH such that the line MATH ... |
math/0004061 | Let MATH be a small perturbation of the given form MATH (MATH is a small closed REF-form). The theorem states that REF-form MATH is exact if and only if REF-form MATH is exact. There is one obvious implication in this statement: non exact Hamiltonians remain non exact Hamiltonians. If MATH is not exact then it must be ... |
math/0004062 | CASE: Since MATH is a morphism, we prove it for the generators. Let MATH, MATH. Then MATH, and we have MATH and analogously for MATH. CASE: As the previous item, we can prove it for generators MATH, MATH. We have MATH . CASE: Simply MATH . We remark that these equalities are easily understood with drawings. CASE: By in... |
math/0004062 | By induction, for MATH being MATH . If the equality holds for MATH, we have MATH . |
math/0004062 | Let MATH. We consider MATH, where MATH is the adjoint in MATH. It is easy to see by induction that this element is the action of the left hand side of REF on MATH. Now, the right hand side acts on MATH as a scalar, namely MATH . The assertion follows at once. |
math/0004062 | Let MATH, and let us define inductively MATH. Taking on MATH the bi-degree given by MATH, MATH, we have MATH. Furthermore, it is proved in REF that MATH and MATH. It is immediate to see by induction (it follows also from REF) that MATH, and MATH, where MATH. Thus, MATH if MATH. We shall prove that the set MATH is linea... |
math/0004062 | Let MATH be the space of the representation MATH. For MATH we denote also by MATH the element MATH. Let MATH be a basis of MATH. We have MATH, and hence MATH. We have an inclusion MATH (in fact, MATH is free as a right MATH-module by CITE) and hence MATH is finite dimensional. Now, MATH is of diagonal GT with matrix MA... |
math/0004062 | Since MATH has a finite number of simple modules up to isomorphism, it is enough to see that each simple module MATH can appear no more than MATH times (MATH depending on MATH) in a module MATH for it to be of finite NAME rank. Let then MATH be the space affording MATH, MATH and MATH such that MATH. Suppose MATH (MATH)... |
math/0004062 | Straightforward (see CITE). |
math/0004062 | We have to prove that the group-likes associated with the basis MATH of MATH generate a finite group. These group-likes are nothing but MATH, MATH. Let MATH, let MATH be the least common multiple of the MATH's and let MATH be the group of MATH-roots of unity. Let MATH and MATH the symmetric group on the (finite) set MA... |
math/0004070 | We may assume that MATH, since otherwise MATH . But then actually MATH, since on MATH we have MATH, so that on this set MATH, which is integrable. Assume first that MATH. Fix MATH, and let MATH . Notice that MATH since MATH implies MATH. Thus for a very large MATH, we can break up MATH into convenient strings of terms ... |
math/0004070 | It is enough to show that MATH . For then, letting MATH, applying this to MATH gives MATH so that MATH and hence MATH . Consider first MATH and its associated MATH, denoted by MATH. For any invariant function MATH such that MATH, for example MATH, we have MATH, so the Theorem gives MATH . Thus MATH is integrable (and, ... |
math/0004079 | Without loss of generality, we may assume that MATH is a smooth point of the connection. Let MATH be a point of multiplicity MATH, and let MATH be a local coordinate at MATH. Note the effect of twisting (that is, replacing MATH by MATH) is to replace the local connection matrix MATH at MATH with respect to a basis MATH... |
math/0004079 | We show first that MATH is independent of the local bases. We work locally around a point MATH which is a singular point of the connection with multiplicity MATH. We assume first that MATH for a local coordinate, and that the connection matrix is MATH with MATH and MATH. (This includes both the admissible and pseudolog... |
math/0004079 | We have MATH . The formulas in the lemma follow easily from this. |
math/0004079 | The connection MATH is vertical. Vanishing for curvature terms involving MATH for MATH implies MATH . The tactic is to eliminate first MATH from MATH, then MATH etc. One easily verifies matrix relations MATH . In particular, MATH . Write MATH . This yields MATH . Applying again now the relations REF yields MATH . Assum... |
math/0004079 | Straightforward. |
math/0004079 | The first two traces are equal by the diagram. For the third, note that since MATH, it follows that one has an exact sequence compatible with the endomorphism multiplication by MATH . Since MATH has no constant term in MATH, it acts nilpotently on the right. |
math/0004079 | Write MATH. Note that neither side of the identity changes if we replace MATH by MATH so we may assume MATH. Also, by linearity, we may assume MATH. The matrix for the action of MATH on MATH, the entries of which are themselves matrices, is MATH . The matrix for MATH is MATH. We write MATH for the naive trace, that is,... |
math/0004079 | By REF , we may replace MATH with MATH. By REF MATH is the sum of the MATH on MATH. By REF this is the same as MATH. Note the factor MATH on the right is because as a matrix MATH acts on MATH while the trace one wants is the action on MATH. |
math/0004079 | Define absolute forms MATH, MATH and MATH: MATH . The local term at MATH is MATH . Applying trace to MATH yields MATH . (Note here that MATH is not closed as an absolute form!) Also, modulo MATH, we have MATH because the residue of an exact form vanishes. The local term thus becomes MATH . We have MATH . Thus MATH . Ex... |
math/0004082 | That the sequence MATH is nonincreasing, invariant under reordering, and nonnegative when MATH follows from the fact that it is a scaled limit of MATH (see below); it suffices therefore to show that MATH with probability REF. We first observe that MATH . Since decreasing MATH cannot decrease MATH, we conclude that it s... |
math/0004082 | This follows immediately from the corresponding fact for MATH. |
math/0004082 | By REF above, MATH . Then MATH where the last step is valid since MATH and MATH. The theorem then follows by taking the limit MATH. |
math/0004082 | Let MATH. Then MATH, so we can extend MATH by adjoining MATH with multiplicity MATH of distribution MATH; denote the resulting random multiset by MATH. But then by REF , we can change the total ordering MATH so that MATH becomes maximal instead of minimal. We then find that MATH induces an increasing subsequence of MAT... |
math/0004082 | An increasing subsequence in MATH can pass through a point on a strict row or column at most once; thus MATH is unchanged if we remove any excess multiplicity in those rows and columns. Let MATH be the resulting multiset, then MATH . It will thus suffice to prove that MATH. But the moment generating function of MATH is... |
math/0004083 | We will prove REF by constructing a sign-reversing involution on the set of summands appearing on the right-hand side of REF , for which the condition MATH is violated. The argument will be first presented for the special case MATH, and then extended to a general MATH. The following general construction will be useful ... |
math/0004083 | Consider the right-hand side of REF . First we note that the loop-erased walks MATH are pairwise vertex-disjoint, and therefore MATH must be the identity permutation, under the assumptions of REF . It remains to show that the restrictions on the MATH may be relaxed as indicated. Assume that the walks MATH,, MATH satisf... |
math/0004083 | This theorem can be proved by a direct argument similar to the one used in the proof of REF . To save an effort, we will instead use a simple observation that will reduce REF to REF . Let us define a new network MATH by splitting every boundary vertex MATH into a source MATH and a sink MATH, converting all outgoing edg... |
math/0004084 | Since the map MATH is differentiable for all MATH, there exists a vector field MATH on MATH satisfying REF. We need to check that MATH is MATH-vertical and right invariant. We have that MATH which proves that MATH is MATH-vertical, by definition. Also, MATH which proves that MATH is also right invariant. |
math/0004084 | For manifolds without corners, this is a result from CITE. For manifolds with corners the proof is the same. |
math/0004084 | If MATH is integrable, then REF is REF is Proposition REF , and REF is REF. The proof of REF in the general case are the same. |
math/0004084 | If MATH is a differentiable groupoid (not just a local one), this is REF. For local groupoids the proof is the same. |
math/0004084 | Because each MATH is a differential groupoid, REF follows from REF on each MATH. From this we obtain the desired relation everywhere on MATH. REF follows from the relation MATH valid on each MATH, and hence everywhere on MATH. |
math/0004084 | We begin by writing MATH. Then, using REF , we can find sections MATH of MATH, MATH, MATH, depending smoothly on MATH, such that MATH . To complete REF , we now use a local groupoid MATH integrating MATH. The existence of such a MATH is ensured by CITE. By replacing MATH with an open neighborhood of MATH, if necessary,... |
math/0004084 | Suppose first that the family MATH is a differentiable atlas. We begin by showing that the groupoid structure on MATH, induced from the groupoids MATH, is compatible with the differentiable structure defined by MATH. That is, we need to check that the structural morphisms are differentiable. We now check this. First, t... |
math/0004084 | Consider the curve MATH where MATH and MATH is chosen such that MATH. By assumption, MATH is a closed curve on MATH, and hence, by the assumption that MATH is MATH - simply connected, we can continuously deform this curve to the constant curve MATH, within MATH, through closed curves based at MATH. More precisely, we c... |
math/0004084 | Let MATH be a local groupoid integrating MATH. Choose MATH and a neighborhood MATH of MATH small enough such that MATH is a diffeomorphism onto an open neighborhood MATH of MATH in MATH. Let MATH be a compact neighborhood of MATH. Because MATH is a compact subset of the open set MATH and MATH is continuous, MATH, we ca... |
math/0004084 | By REF , we see that we it is enough to show that the family MATH, defined using the maps MATH of REF , is an atlas. Let MATH and MATH be two such maps, defined on MATH and, respectively, on MATH, such that their images intersect. Let MATH be an element of this intersection. As in the proof of REF , we can arrange that... |
math/0004084 | Recall that MATH, the path groupoid of the foliation MATH, consists of fixed end point homotopy classes of paths MATH that are fully contained in a single leaf, with respect to homotopies within that leaf. As explained above, the morphism MATH defines a leafwise flat connection on MATH that preserves the NAME bracket. ... |
math/0004084 | We may assume that MATH is connected. Since semisimple NAME algebras are rigid, all fibers of MATH will be isomorphic NAME algebras. Fix one of these algebras and denote it by MATH. Also, let MATH be the group of automorphisms of MATH. Then, if we define MATH (the fibers are the sets of NAME algebra isomorphisms MATH),... |
math/0004084 | Using REF or REF , we see that each of the algebroids MATH, obtained by restricting MATH to the stratum MATH, is integrable. By REF , the NAME algebroid MATH is then integrable. |
math/0004084 | The set of points satisfying REF at a face MATH is closed under the NAME bracket, by definition of a NAME flag. Since MATH is also closed under the NAME bracket, it follows that MATH is a NAME algebra. Fix a point MATH in the interior of a face MATH of codimension MATH with defining functions MATH. Let MATH be the corr... |
math/0004084 | Let MATH and MATH be the foliation and, respectively, the boundary NAME data defining MATH. Observe that the sections of MATH are vector fields that are tangent to all faces of MATH, and hence each open face of MATH is MATH-invariant. This means that the stratification of MATH by open faces is a MATH-invariant stratifi... |
math/0004084 | This follows from REF . |
math/0004087 | Let MATH, and let us consider the linear operator MATH associated to it defined by MATH . Let us suppose without loss of generality that MATH. Since MATH is completely continuous, we get that MATH . |
math/0004087 | If both MATH and MATH are scattered, then MATH is a NAME space and therefore MATH has the DPP. Now, let us suppose that one of them, say MATH, is not scattered. Since MATH is infinite, MATH is not NAME, and therefore there exist two sequences MATH and MATH such that MATH is weakly null and MATH for every MATH. Also, si... |
math/0004087 | If MATH is scattered, then MATH is a NAME space for every MATH, and therefore MATH has the DPP. Now, if MATH is not scattered, we can consider the trilinear form MATH defined by MATH that is, the symmetrized respect to the two first variables of the trilinear form used in REF . Now we can apply analogous reasonings as ... |
math/0004088 | This follows immediately from CITE, where it is stated for the irreducible unitary representation with parameter MATH of the NAME group. |
math/0004088 | It is sufficient to observe that REF implies that the map MATH defines a continuous linear functional on MATH for MATH. |
math/0004088 | For MATH, we have MATH and therefore MATH . |
math/0004088 | For real MATH this is the infinitesimal version of the previous proposition, just differentiate MATH with respect to MATH and set MATH. For complex MATH it follows by linearity. |
math/0004088 | We will show the result for MATH, the general case can be shown similarly (see also the proof of REF ). Let MATH be a NAME function. Let MATH. Then we have MATH . Similarly, we get MATH and together these two inequalities imply MATH for MATH. |
math/0004088 | Let MATH be any sequence such that MATH and MATH for some MATH in the weak topology. To show that MATH is closable, we have to show that this implies MATH. Let us evaluate MATH between two exponential vectors MATH, MATH, MATH, then we get MATH and therefore MATH, as desired. |
math/0004088 | For MATH and MATH such that also MATH, we get MATH where we used REF . The first equality of the proposition now follows, since both MATH and MATH are derivations. The second equality follows immediately. |
math/0004088 | This formula is a consequence of the fact that MATH for all MATH, we get MATH . |
math/0004088 | This is obvious, since MATH is a derivation. |
math/0004088 | We have to show that for any sequence MATH in MATH with MATH and MATH, we get MATH. Let MATH. Set MATH, MATH, MATH, and MATH, then we have MATH and MATH. Thus we get MATH for all MATH, where MATH . But this implies MATH, since MATH is dense in MATH. |
math/0004088 | It is not difficult to check this directly on the NAME operators MATH, MATH. We get MATH and MATH . By linearity and continuity it therefore extends to all of MATH. |
math/0004088 | These properties can be deduced easily from the definition of the gradient and the properties of the derivation operator MATH and the inner product MATH. |
math/0004088 | We assume that such an operator MATH exists and show that this leads to a contradiction. Let MATH be the operator defined by MATH, it is easy to see that MATH and that MATH is given by MATH . Therefore, if MATH existed, we would have MATH which is clearly impossible. |
math/0004088 | From REF we get MATH . But since MATH and MATH commute with MATH, we can pull them out of the anti-commutator, and we get MATH . |
math/0004088 | Let MATH. Recalling the definition of MATH we get the following alternative expression for MATH, MATH . Evaluating this between two exponential vectors, we obtain MATH . |
math/0004088 | Let MATH be a sequence such that MATH and MATH in the weak topology. Then we get MATH for all MATH, and thus MATH. |
math/0004088 | CASE: Let MATH. We set MATH then we have MATH, and therefore MATH . On the other hand we have MATH, and MATH . Taking the difference of these two expressions, we get MATH . CASE: A straightforward computation gives MATH where we used that MATH defines a bounded operator, since MATH. REF can be shown similarly. |
math/0004088 | It is sufficient to check this for functionals of the form MATH, MATH. We get MATH . |
math/0004088 | Let MATH and set MATH then we have MATH for all NAME functions MATH. Therefore MATH for all MATH, since MATH and MATH for all MATH, and thus MATH . But this implies that the density of MATH is contained in the NAME spaces MATH for all MATH. |
math/0004088 | Since MATH, we have MATH for all polynomials MATH. We therefore get MATH . The hypotheses of the proposition assure MATH and therefore allow us to get the estimate MATH for all polynomials MATH. But this implies that MATH admits a bounded density. |
math/0004088 | To prove this, we show that the NAME integrals satisfy the same formula for the matrix elements between exponential vectors. Let MATH be such that its creation integral in the sense of NAME and NAME is defined with a domain containing the exponential vectors. Then we get MATH . For the annihilation integral we deduce t... |
math/0004089 | REF follow from the greedy algorithm (see REF ). By REF of the exchange capacity, we have MATH. Since MATH, it follows from REF that MATH. Thus we obtain MATH. |
math/0004089 | If MATH, for each MATH the first MATH vertices in MATH must belong to MATH. Then it follows from REF that MATH. Since MATH and MATH, this implies MATH. |
math/0004089 | For any MATH we have MATH and MATH, and hence MATH. |
math/0004089 | When the MATH-scaling phase finishes with MATH, we have MATH for every MATH, which implies MATH as well as MATH. Similarly, when the MATH-scaling phase finishes with MATH, we have MATH for every MATH, which implies MATH as well as MATH. When the MATH-scaling phase ends with MATH due to REF , then MATH and MATH. By the ... |
math/0004089 | By REF , the output MATH of the algorithm satisfies MATH. For any MATH, the weak duality in REF asserts MATH. Thus we have MATH, which implies by the integrality of MATH that MATH minimizes MATH. |
math/0004089 | The algorithm starts with MATH and ends with MATH, so the algorithm consists of MATH scaling phases. Each scaling phase finds MATH-augmenting paths. To find an augmenting path, we perform at most MATH pushes per extreme base. A saturating push requires MATH time while a nonsaturating one MATH time. Here, note that ther... |
math/0004089 | Let MATH be any minimizer of MATH. There exists a vector MATH with MATH such that MATH. Note that MATH for each MATH. By REF , there exists a subset MATH such that MATH. Then we have MATH. This implies MATH due to the assumption, and hence MATH. |
math/0004089 | Each time we call the procedure MATH, the algorithm adds a new arc to MATH or deletes a set of vertices. This can happen at most MATH times. Thus the overall running time of the algorithm is MATH, which is strongly polynomial. |
math/0004091 | MATH is a distance function. Indeed, MATH is symmetric and MATH if and only if MATH. We have MATH and MATH . Hence, MATH. Let MATH be a MATH - dispersed subspace of MATH. Since MATH is surjective, for any MATH, there is MATH such that MATH. Let MATH. By REF , MATH for all MATH. Thus MATH establishes an isometry between... |
math/0004091 | Let MATH. We define MATH by MATH for MATH. (Since MATH,MATH.) We have, by the triangle inequality, MATH . On the other hand, MATH for MATH. Therefore, MATH for all MATH. |
math/0004098 | Immediate from REF , and REF . |
math/0004098 | Most of the details of the proof are contained in CITE and CITE, so we only sketch points not already covered there. The essential step (for the present applications) is REF , which shows that MATH intertwines the isometry MATH-with the restriction of the unitary operator MATH to the resolution subspace MATH. We have: ... |
math/0004098 | The correspondence is MATH with MATH and in the reverse direction, MATH does the job, as can be checked by direct substitution. |
math/0004098 | It is clear that the operator MATH in REF is a projection if the family MATH consists of mutually commuting projections. We now prove the converse by induction starting with two given projections MATH such that MATH is given to be a projection. Then the commutator MATH satisfies MATH. Using that MATH we conclude that M... |
math/0004098 | We refer the reader to CITE. |
math/0004098 | The corollary is applied in REF below, so we postpone its proof to REF. The argument is in REF , and it is based on the ordered factorizations REF - REF . |
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